[{"Name":"Infinite Interval","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Infinite Interval","Duration":"8m 35s","ChapterTopicVideoID":29168,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.520","Text":"New topic, The Wave Equation,"},{"Start":"00:02.520 ","End":"00:05.655","Text":"but specifically on an infinite interval."},{"Start":"00:05.655 ","End":"00:08.040","Text":"This is what the equation is."},{"Start":"00:08.040 ","End":"00:09.990","Text":"In general, It\u0027s a hyperbolic linear,"},{"Start":"00:09.990 ","End":"00:15.420","Text":"second-order partial differential equation with initial conditions."},{"Start":"00:15.420 ","End":"00:20.145","Text":"You can think of t is time and x is space one-dimensional."},{"Start":"00:20.145 ","End":"00:23.325","Text":"You have a guitar string or piano wire or whatever,"},{"Start":"00:23.325 ","End":"00:29.430","Text":"to infinite in length and you\u0027re given at time 0 at each point."},{"Start":"00:29.430 ","End":"00:33.000","Text":"We\u0027re also given d u by d t when t is 0."},{"Start":"00:33.000 ","End":"00:40.520","Text":"That\u0027s the reason that we specifically take the case where x is on the whole real line,"},{"Start":"00:40.520 ","End":"00:43.760","Text":"we get a simple formula, as you\u0027ll see."},{"Start":"00:43.760 ","End":"00:47.845","Text":"A is just some parameter is A Here."},{"Start":"00:47.845 ","End":"00:51.950","Text":"D\u0027Alembert gave a closed formula"},{"Start":"00:51.950 ","End":"00:56.060","Text":"that just gives you the answer if you just substitute, I\u0027ll show you what it is."},{"Start":"00:56.060 ","End":"00:58.550","Text":"The solution is the following."},{"Start":"00:58.550 ","End":"01:00.215","Text":"U(x) t equals."},{"Start":"01:00.215 ","End":"01:02.990","Text":"It looks a bit messy but it\u0027s not too bad at all,"},{"Start":"01:02.990 ","End":"01:04.625","Text":"we\u0027ll see it in the example."},{"Start":"01:04.625 ","End":"01:06.530","Text":"Basically just take f,"},{"Start":"01:06.530 ","End":"01:09.980","Text":"the function from here and a, of course,"},{"Start":"01:09.980 ","End":"01:12.155","Text":"is just the a from here,"},{"Start":"01:12.155 ","End":"01:16.995","Text":"G is the function from here and big F is this."},{"Start":"01:16.995 ","End":"01:21.860","Text":"By the way, function f(x) t happens to be missing,"},{"Start":"01:21.860 ","End":"01:28.980","Text":"meaning it\u0027s 0 then this is the homogeneous version of the wave equation."},{"Start":"01:28.980 ","End":"01:32.315","Text":"We\u0027re going to actually prove, not in this clip,"},{"Start":"01:32.315 ","End":"01:35.925","Text":"that the formula works at least for the homogeneous case,"},{"Start":"01:35.925 ","End":"01:40.235","Text":"then this will also drop out if f is 0."},{"Start":"01:40.235 ","End":"01:43.234","Text":"Let\u0027s do our first example."},{"Start":"01:43.234 ","End":"01:45.875","Text":"We have the following problem."},{"Start":"01:45.875 ","End":"01:51.905","Text":"U_tt is u _xx plus xt u(x,0) is x squared and u_t( x,"},{"Start":"01:51.905 ","End":"01:53.330","Text":"0 ) is x."},{"Start":"01:53.330 ","End":"01:59.030","Text":"Again run this interval where time is non-negative."},{"Start":"01:59.030 ","End":"02:00.210","Text":"We\u0027re starting from now,"},{"Start":"02:00.210 ","End":"02:05.770","Text":"and going into the future and x everything on the real line."},{"Start":"02:05.770 ","End":"02:09.455","Text":"If we compare our equation to the general equation,"},{"Start":"02:09.455 ","End":"02:11.150","Text":"we can say whatever thing is."},{"Start":"02:11.150 ","End":"02:14.905","Text":"We can say that a is one because we have a squared,"},{"Start":"02:14.905 ","End":"02:16.995","Text":"u_xx and we have 1U_xx."},{"Start":"02:16.995 ","End":"02:21.675","Text":"We see that f( x) is x squared, that\u0027s this."},{"Start":"02:21.675 ","End":"02:24.150","Text":"G (x) is x,"},{"Start":"02:24.150 ","End":"02:28.245","Text":"and big F(xt) is xt."},{"Start":"02:28.245 ","End":"02:31.625","Text":"Here again, is D\u0027Alembert formula."},{"Start":"02:31.625 ","End":"02:34.670","Text":"In our case, if we substitute a, f,"},{"Start":"02:34.670 ","End":"02:36.170","Text":"g and big F,"},{"Start":"02:36.170 ","End":"02:38.885","Text":"what we get is the following."},{"Start":"02:38.885 ","End":"02:44.525","Text":"F(x plus at) is (x plus t)squared, and so on."},{"Start":"02:44.525 ","End":"02:50.390","Text":"Here, g(s) is equal to s. This is what we get."},{"Start":"02:50.390 ","End":"02:52.100","Text":"Again, a is one,"},{"Start":"02:52.100 ","End":"02:54.595","Text":"so x plus at is x plus t,"},{"Start":"02:54.595 ","End":"03:01.290","Text":"F(st) is S times T. Now it\u0027s just calculations,"},{"Start":"03:01.290 ","End":"03:06.495","Text":"computations, so x plus t squared plus x minus t squared."},{"Start":"03:06.495 ","End":"03:12.780","Text":"The 2xt cancels with minus 2xt and we\u0027re left with 2x squared plus 2t squared."},{"Start":"03:12.780 ","End":"03:17.505","Text":"Here the integral of s is 1/2s squared."},{"Start":"03:17.505 ","End":"03:20.930","Text":"The 1/2 with the 1/2 makes a 1/4 and s squared"},{"Start":"03:20.930 ","End":"03:24.850","Text":"evaluated here and here and subtracted gives us this."},{"Start":"03:24.850 ","End":"03:31.550","Text":"In the last one, let\u0027s start by first doing the integral ds and we can bring"},{"Start":"03:31.550 ","End":"03:34.790","Text":"Tau in front of the integral because it doesn\u0027t depend on"},{"Start":"03:34.790 ","End":"03:39.240","Text":"s. Now the integral of s is half 1/2s squared,"},{"Start":"03:39.240 ","End":"03:40.845","Text":"just like we said here."},{"Start":"03:40.845 ","End":"03:43.740","Text":"The 1/2 with the 1/2 makes a 1/4 and then we"},{"Start":"03:43.740 ","End":"03:48.455","Text":"have s squared evaluated here and here and subtracted."},{"Start":"03:48.455 ","End":"03:51.954","Text":"We get this one squared minus this squared."},{"Start":"03:51.954 ","End":"03:55.155","Text":"Like I said, the 1/2 goes with the 1/2 to make a 1/4."},{"Start":"03:55.155 ","End":"03:56.620","Text":"We have the Tau,"},{"Start":"03:56.620 ","End":"04:01.100","Text":"this upper limit squared minus this lower limit squared."},{"Start":"04:01.100 ","End":"04:03.740","Text":"This over 2 is x squared plus t squared."},{"Start":"04:03.740 ","End":"04:07.220","Text":"Now here we get x squared plus 2xt plus t squared,"},{"Start":"04:07.220 ","End":"04:10.400","Text":"and here x squared minus 2xt plus t squared."},{"Start":"04:10.400 ","End":"04:12.680","Text":"When you subtract, we get 2 minus,"},{"Start":"04:12.680 ","End":"04:15.595","Text":"minus 2 is 4xt."},{"Start":"04:15.595 ","End":"04:20.030","Text":"Similarly, x plus instead of t,"},{"Start":"04:20.030 ","End":"04:23.750","Text":"we have something else squared an x minus that same thing squared."},{"Start":"04:23.750 ","End":"04:26.120","Text":"Just like here we got 4xt,"},{"Start":"04:26.120 ","End":"04:29.165","Text":"he will get 4xt minus."},{"Start":"04:29.165 ","End":"04:31.100","Text":"If you\u0027re not sure about that."},{"Start":"04:31.100 ","End":"04:34.190","Text":"You can just expand it all and collect, and you\u0027ll see."},{"Start":"04:34.190 ","End":"04:42.285","Text":"We have x squared plus t squared plus xt plus the 4 with the 4 will cancel."},{"Start":"04:42.285 ","End":"04:51.085","Text":"We\u0027ll get plus the integral of xt minus Tau times Tau d tau. Here we are."},{"Start":"04:51.085 ","End":"04:54.560","Text":"The 4 with the 4 cancels, like I said,"},{"Start":"04:54.560 ","End":"05:00.925","Text":"the x and come out in front of the integral it doesn\u0027t depend on Tau."},{"Start":"05:00.925 ","End":"05:03.545","Text":"We just have to integrate this."},{"Start":"05:03.545 ","End":"05:10.570","Text":"This we can write as t times the integral of Tau minus the integral of Tau squared."},{"Start":"05:10.570 ","End":"05:14.600","Text":"This is Tau squared over 2 between"},{"Start":"05:14.600 ","End":"05:19.625","Text":"these limits makes it t squared over 2 times t is t cubed over 2."},{"Start":"05:19.625 ","End":"05:22.495","Text":"Here we\u0027ve got t cubed over 3,"},{"Start":"05:22.495 ","End":"05:26.535","Text":"and plug in 0-t we got t cubed over 3."},{"Start":"05:26.535 ","End":"05:31.360","Text":"The only thing to simplify as to say that a 1/2 minus a 1/3 is a 1/6."},{"Start":"05:31.360 ","End":"05:33.375","Text":"This is the answer,"},{"Start":"05:33.375 ","End":"05:35.460","Text":"and what we\u0027ll do now,"},{"Start":"05:35.460 ","End":"05:37.280","Text":"we\u0027ll check this answer."},{"Start":"05:37.280 ","End":"05:40.250","Text":"In most exercises, you don\u0027t bother to check the answer,"},{"Start":"05:40.250 ","End":"05:42.980","Text":"but in the tutorial let\u0027s do the full thing."},{"Start":"05:42.980 ","End":"05:45.829","Text":"We\u0027ll check if this,"},{"Start":"05:45.829 ","End":"05:50.525","Text":"this u in terms of x and t satisfies this equation,"},{"Start":"05:50.525 ","End":"05:52.430","Text":"and the initial conditions."},{"Start":"05:52.430 ","End":"05:54.565","Text":"We need partial derivatives,"},{"Start":"05:54.565 ","End":"06:00.665","Text":"we\u0027ll need U_tt and U_xx and we will also need ut for this."},{"Start":"06:00.665 ","End":"06:02.435","Text":"Let\u0027s compute all those."},{"Start":"06:02.435 ","End":"06:07.790","Text":"Start of with u and then differentiate with respect to t. This is 0,"},{"Start":"06:07.790 ","End":"06:10.370","Text":"this is 2t, this is just x, here,"},{"Start":"06:10.370 ","End":"06:14.630","Text":"3 times 1/6 is a 1/2 and it\u0027s just t squared."},{"Start":"06:14.630 ","End":"06:18.210","Text":"Differentiate again with respect to t. Well,"},{"Start":"06:18.210 ","End":"06:24.730","Text":"we just get 2 plus xt and you differentiate u with respect to x twice."},{"Start":"06:24.730 ","End":"06:27.475","Text":"The second derivative of this is 2,"},{"Start":"06:27.475 ","End":"06:29.739","Text":"second derivative of this is nothing."},{"Start":"06:29.739 ","End":"06:31.410","Text":"First derivative of this is t,"},{"Start":"06:31.410 ","End":"06:33.890","Text":"so the second derivative is 0."},{"Start":"06:33.890 ","End":"06:41.860","Text":"Here also the derivative with respect to x is 6t cubed and then derivative again is 0."},{"Start":"06:41.860 ","End":"06:44.155","Text":"We\u0027re only left with the 2 from here."},{"Start":"06:44.155 ","End":"06:45.460","Text":"Now that we have all these,"},{"Start":"06:45.460 ","End":"06:48.335","Text":"we can substitute here."},{"Start":"06:48.335 ","End":"06:52.875","Text":"The first one to check is u( x a, 0)."},{"Start":"06:52.875 ","End":"06:55.570","Text":"We put t equals 0 here."},{"Start":"06:55.570 ","End":"06:57.490","Text":"Everything is 0 here, here,"},{"Start":"06:57.490 ","End":"07:01.220","Text":"and here, and we\u0027re just left with x squared."},{"Start":"07:01.250 ","End":"07:07.790","Text":"What we have to get this when t is 0 is equal to this so we\u0027re okay there."},{"Start":"07:07.790 ","End":"07:09.905","Text":"Next one is this,"},{"Start":"07:09.905 ","End":"07:13.690","Text":"du by dt at (x, 0)."},{"Start":"07:13.690 ","End":"07:16.475","Text":"We substitute t equals 0 in this,"},{"Start":"07:16.475 ","End":"07:18.860","Text":"that\u0027s 0, that\u0027s 0."},{"Start":"07:18.860 ","End":"07:21.740","Text":"We just have the x, which is what we wanted,"},{"Start":"07:21.740 ","End":"07:23.959","Text":"so that\u0027s good also."},{"Start":"07:23.959 ","End":"07:29.900","Text":"Now we\u0027re going to check this equality, the actual PDE."},{"Start":"07:29.900 ","End":"07:34.625","Text":"The question is, does u_tt equal u_xx plus xt?"},{"Start":"07:34.625 ","End":"07:37.430","Text":"Well, let\u0027s see. Let\u0027s substitute all the things,"},{"Start":"07:37.430 ","End":"07:47.010","Text":"u_tt is 2 plus xt and u_xx is 2 and xt is xt."},{"Start":"07:47.010 ","End":"07:50.000","Text":"Sure this is equal too, so that\u0027s okay."},{"Start":"07:50.000 ","End":"07:53.690","Text":"Also, we\u0027ve done a full example including checking."},{"Start":"07:53.690 ","End":"07:57.450","Text":"You\u0027ve learned the formula of D\u0027Alembert and that"},{"Start":"07:57.450 ","End":"07:58.920","Text":"concludes this introductory"},{"Start":"07:58.920 ","End":"08:17.890","Text":"clip."}],"ID":30709},{"Watched":false,"Name":"d\u0026#39;Alembert\u0026#39;s Formula","Duration":"7m 26s","ChapterTopicVideoID":29169,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.690","Text":"In this clip, we\u0027ll prove d\u0027Alembert\u0027s formula,"},{"Start":"00:03.690 ","End":"00:05.370","Text":"but not in the general case,"},{"Start":"00:05.370 ","End":"00:07.455","Text":"just in the homogeneous case."},{"Start":"00:07.455 ","End":"00:12.990","Text":"To remind you, this is the heat equation on an infinite interval that\u0027s important."},{"Start":"00:12.990 ","End":"00:16.095","Text":"We gave d\u0027Alembert\u0027s formula as follows."},{"Start":"00:16.095 ","End":"00:19.890","Text":"Now the homogeneous case means that there is no plus"},{"Start":"00:19.890 ","End":"00:25.605","Text":"f(x,t) and we also drop the last term in d\u0027Alembert\u0027s formula."},{"Start":"00:25.605 ","End":"00:30.045","Text":"We have to prove that this is the solution in this case."},{"Start":"00:30.045 ","End":"00:35.625","Text":"What we\u0027ll do is we\u0027ll start with a change of variables and if you check the Jacobian,"},{"Start":"00:35.625 ","End":"00:38.280","Text":"it\u0027s non-0. We\u0027re okay."},{"Start":"00:38.280 ","End":"00:40.560","Text":"We\u0027ll call the new function w,"},{"Start":"00:40.560 ","End":"00:48.600","Text":"so that u(x,t) is the same as w of Psi and Eta if you plug in Psi and Eta to be these 2."},{"Start":"00:48.600 ","End":"00:55.750","Text":"Now we need the partial derivatives u_x by the chain rule for function of 2 variables."},{"Start":"00:55.750 ","End":"00:57.840","Text":"It\u0027s w with respect to Psi,"},{"Start":"00:57.840 ","End":"00:59.430","Text":"Psi with respect to x,"},{"Start":"00:59.430 ","End":"01:02.835","Text":"and we\u0027ll also get from w to x via Eta."},{"Start":"01:02.835 ","End":"01:06.720","Text":"These 2 have to be equal each time."},{"Start":"01:06.720 ","End":"01:13.150","Text":"We know that Psi x is 1 and Eta x is also 1,"},{"Start":"01:13.150 ","End":"01:16.515","Text":"so we end up with just w Psi plus w Eta."},{"Start":"01:16.515 ","End":"01:21.140","Text":"With respect to t, it\u0027s very similar except that"},{"Start":"01:21.140 ","End":"01:28.055","Text":"the derivative Psi with respect to t is a and Eta with respect to t is minus a."},{"Start":"01:28.055 ","End":"01:29.600","Text":"This is what we get."},{"Start":"01:29.600 ","End":"01:36.605","Text":"We need u_xx to differentiate this with respect to x. I wrote it out here."},{"Start":"01:36.605 ","End":"01:38.705","Text":"Unlike before, this is 1,"},{"Start":"01:38.705 ","End":"01:40.535","Text":"this is 1, this is 1,"},{"Start":"01:40.535 ","End":"01:42.690","Text":"this is 1,"},{"Start":"01:42.690 ","End":"01:44.330","Text":"and these 2 are equal."},{"Start":"01:44.330 ","End":"01:48.065","Text":"The mixed partial derivative of order 2,"},{"Start":"01:48.065 ","End":"01:51.155","Text":"Psi Eta or Eta Psi, same thing."},{"Start":"01:51.155 ","End":"01:57.120","Text":"We end up with this plus twice this plus one of these,"},{"Start":"01:57.120 ","End":"01:59.530","Text":"and we still need u_tt."},{"Start":"01:59.530 ","End":"02:05.300","Text":"From this, we get a times the derivative of w Psi with respect to"},{"Start":"02:05.300 ","End":"02:12.925","Text":"t minus a derivative of w Eta with respect to t. This is what we get."},{"Start":"02:12.925 ","End":"02:21.795","Text":"Once again, we can substitute that Psi t is a and Eta t is minus a."},{"Start":"02:21.795 ","End":"02:26.555","Text":"Calculations, we end up with a^2 times this,"},{"Start":"02:26.555 ","End":"02:28.460","Text":"which is very similar to this,"},{"Start":"02:28.460 ","End":"02:30.793","Text":"but here there\u0027s a plus and here there\u0027s a minus,"},{"Start":"02:30.793 ","End":"02:32.645","Text":"and the a^2 outside."},{"Start":"02:32.645 ","End":"02:37.475","Text":"This is our PDE. Let\u0027s substitute and we get that"},{"Start":"02:37.475 ","End":"02:44.535","Text":"u_tt from here is equal to a^2 u_xx."},{"Start":"02:44.535 ","End":"02:48.560","Text":"This bit from here."},{"Start":"02:48.560 ","End":"02:53.300","Text":"Notice everything\u0027s the same except here there\u0027s a minus and here there\u0027s a plus."},{"Start":"02:53.300 ","End":"02:58.185","Text":"We subtract both sides and divide by 4 or 4a^2,"},{"Start":"02:58.185 ","End":"03:01.534","Text":"we get that w Psi Eta is 0."},{"Start":"03:01.534 ","End":"03:04.435","Text":"If we integrate with respect to Eta,"},{"Start":"03:04.435 ","End":"03:10.325","Text":"we get that w Psi is a constant with respect to Eta,"},{"Start":"03:10.325 ","End":"03:13.250","Text":"but it\u0027s an arbitrary function of Psi."},{"Start":"03:13.250 ","End":"03:15.800","Text":"Now let\u0027s integrate again with respect to Psi,"},{"Start":"03:15.800 ","End":"03:20.640","Text":"the indefinite integral or primitive of F,"},{"Start":"03:20.640 ","End":"03:27.800","Text":"we\u0027ll call that Phi and you have to add another constant which is not really a constant,"},{"Start":"03:27.800 ","End":"03:32.449","Text":"it\u0027s a function of Eta and call that one Psi, Greek letters,"},{"Start":"03:32.449 ","End":"03:33.920","Text":"Phi and Psi,"},{"Start":"03:33.920 ","End":"03:35.705","Text":"the Greek of this letter,"},{"Start":"03:35.705 ","End":"03:39.350","Text":"which is like PS, like in psychology."},{"Start":"03:39.350 ","End":"03:44.210","Text":"Now, remember that w Psi of Eta is the same as"},{"Start":"03:44.210 ","End":"03:50.170","Text":"u(x,t) and Psi is x plus at and Eta is x minus at."},{"Start":"03:50.170 ","End":"03:56.885","Text":"Now, we have what u is in terms of the functions Phi and Psi, which are arbitrary."},{"Start":"03:56.885 ","End":"04:00.740","Text":"But what we haven\u0027t used yet is the initial conditions,"},{"Start":"04:00.740 ","End":"04:05.840","Text":"and the 2 initial conditions will help us specify Phi and Psi."},{"Start":"04:05.840 ","End":"04:08.640","Text":"We\u0027ll need ut for the initial conditions,"},{"Start":"04:08.640 ","End":"04:12.620","Text":"so let\u0027s differentiate this with respect to t. From here we get"},{"Start":"04:12.620 ","End":"04:17.195","Text":"Phi\u0027 and then the inner derivative with respect to t, which is a."},{"Start":"04:17.195 ","End":"04:22.370","Text":"From here, Psi\u0027 and we get a minus a here."},{"Start":"04:22.370 ","End":"04:26.330","Text":"We have these 2 formulas. Here they are."},{"Start":"04:26.330 ","End":"04:30.050","Text":"Now we\u0027ll substitute in 1 of the initial conditions,"},{"Start":"04:30.050 ","End":"04:33.170","Text":"this one, and what we get,"},{"Start":"04:33.170 ","End":"04:36.499","Text":"if t=0 in this first equation,"},{"Start":"04:36.499 ","End":"04:40.465","Text":"then we get u(x,0)=Phi(x),"},{"Start":"04:40.465 ","End":"04:43.465","Text":"because t is 0 plus Psi(x)."},{"Start":"04:43.465 ","End":"04:46.175","Text":"On the other hand, u(x,0) is f(x),"},{"Start":"04:46.175 ","End":"04:49.195","Text":"so we get that this plus this equals this."},{"Start":"04:49.195 ","End":"04:55.670","Text":"Now we have the other initial condition that ut(x,0) is g(x),"},{"Start":"04:55.670 ","End":"05:01.030","Text":"so put in t equals 0 here and we\u0027ll get what\u0027s here,"},{"Start":"05:01.030 ","End":"05:02.560","Text":"but without the at,"},{"Start":"05:02.560 ","End":"05:06.245","Text":"because t is 0, equals g(x)."},{"Start":"05:06.245 ","End":"05:11.630","Text":"These 2 formulas will allow us to compute Phi and Psi."},{"Start":"05:11.630 ","End":"05:14.135","Text":"Let\u0027s integrate this one."},{"Start":"05:14.135 ","End":"05:19.010","Text":"First, divide by a and integrate it and we get Phi(x)"},{"Start":"05:19.010 ","End":"05:23.960","Text":"minus Psi(x) and we want to integrate 1 over a of this,"},{"Start":"05:23.960 ","End":"05:32.160","Text":"so it\u0027s 1 over a, the integral from 0 to x g(s) ds and constant of integration."},{"Start":"05:32.160 ","End":"05:38.390","Text":"This other second equation in this pair is taken from here."},{"Start":"05:38.390 ","End":"05:43.775","Text":"Now, if we add then the left-hand side will give us twice Phi(x)."},{"Start":"05:43.775 ","End":"05:45.500","Text":"If we add and divide by 2,"},{"Start":"05:45.500 ","End":"05:52.580","Text":"we get Phi(x) and if we subtract and divide by 2, we\u0027ll get Psi(x)."},{"Start":"05:52.580 ","End":"05:55.860","Text":"Phi(x) is this plus this over 2,"},{"Start":"05:55.860 ","End":"06:03.975","Text":"so it\u0027s half this plus half this and then Psi(x) is this minus this over 2."},{"Start":"06:03.975 ","End":"06:08.310","Text":"Very similar. This should be a minus, not a plus."},{"Start":"06:08.920 ","End":"06:15.420","Text":"Remember that u(x,t) is Phi(x) plus at plus Psi(x) minus at."},{"Start":"06:15.420 ","End":"06:20.180","Text":"Now we can substitute because we know what Phi(x) and Psi(x) are."},{"Start":"06:20.180 ","End":"06:23.250","Text":"We get, for this bit,"},{"Start":"06:23.250 ","End":"06:27.450","Text":"we get f(x) plus at over 2 plus 1 over 2a,"},{"Start":"06:27.450 ","End":"06:29.640","Text":"the integral from 0 instead of x,"},{"Start":"06:29.640 ","End":"06:31.230","Text":"we have x plus at."},{"Start":"06:31.230 ","End":"06:33.785","Text":"Now, I didn\u0027t write the plus c over 2,"},{"Start":"06:33.785 ","End":"06:36.280","Text":"because in a moment we\u0027ll have minus c over 2 and"},{"Start":"06:36.280 ","End":"06:39.200","Text":"these 2 will cancel and we\u0027re adding this plus this."},{"Start":"06:39.200 ","End":"06:44.750","Text":"The second one is just x minus at instead of x plus at."},{"Start":"06:44.750 ","End":"06:47.350","Text":"Like I said, the constants cancel."},{"Start":"06:47.350 ","End":"06:49.585","Text":"Now we can organize this."},{"Start":"06:49.585 ","End":"06:57.905","Text":"This and this together will give us this and this integral minus this integral,"},{"Start":"06:57.905 ","End":"07:03.564","Text":"this difference will give us the integral from x minus at to x plus at."},{"Start":"07:03.564 ","End":"07:07.663","Text":"Here we have from 0 to x plus at,"},{"Start":"07:07.663 ","End":"07:09.665","Text":"and here because it\u0027s a minus,"},{"Start":"07:09.665 ","End":"07:12.470","Text":"it\u0027s from x minus at to 0,"},{"Start":"07:12.470 ","End":"07:14.120","Text":"flip this upside down."},{"Start":"07:14.120 ","End":"07:20.780","Text":"Then we can cut out the middleman 0 and just go straight from x minus at to x plus at."},{"Start":"07:20.780 ","End":"07:27.090","Text":"This is exactly the expression we were trying to get and so that concludes this proof."}],"ID":30710},{"Watched":false,"Name":"Exercise 1","Duration":"7m 30s","ChapterTopicVideoID":29170,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.195","Text":"In this exercise, we have a heat equation and initial conditions,"},{"Start":"00:06.195 ","End":"00:10.125","Text":"and it\u0027s defined on the whole real line for x,"},{"Start":"00:10.125 ","End":"00:13.499","Text":"and as usual, time is only positive."},{"Start":"00:13.499 ","End":"00:18.930","Text":"This is a homogeneous equation because there\u0027s no plus f(x),"},{"Start":"00:18.930 ","End":"00:22.470","Text":"and t term goes here."},{"Start":"00:22.470 ","End":"00:25.635","Text":"If it\u0027s 0, then it\u0027s homogeneous."},{"Start":"00:25.635 ","End":"00:28.470","Text":"In our case a=1,"},{"Start":"00:28.470 ","End":"00:30.705","Text":"which is from here."},{"Start":"00:30.705 ","End":"00:33.270","Text":"u_tt is a squared u_xx,"},{"Start":"00:33.270 ","End":"00:38.400","Text":"so a is 1. f(x) is the function here,"},{"Start":"00:38.400 ","End":"00:44.400","Text":"u(x) naught, and g(x) is this one, u_t(x) naught."},{"Start":"00:44.400 ","End":"00:46.295","Text":"Here\u0027s a reminder."},{"Start":"00:46.295 ","End":"00:50.990","Text":"Just copy pasted it of the original form of the equation."},{"Start":"00:50.990 ","End":"00:54.695","Text":"What\u0027s in faint here is the bit for nonhomogeneous,"},{"Start":"00:54.695 ","End":"00:56.270","Text":"in our case this is zero."},{"Start":"00:56.270 ","End":"00:59.465","Text":"When we do the solution with d\u0027Alembert formula,"},{"Start":"00:59.465 ","End":"01:06.875","Text":"we\u0027ll also ignore the bit with f 0. u(x,t) is equal to just this."},{"Start":"01:06.875 ","End":"01:08.300","Text":"Let\u0027s break it down."},{"Start":"01:08.300 ","End":"01:09.800","Text":"From the formula,"},{"Start":"01:09.800 ","End":"01:13.015","Text":"it\u0027s f(x) plus at and so on."},{"Start":"01:13.015 ","End":"01:15.765","Text":"But f is the 0 function,"},{"Start":"01:15.765 ","End":"01:18.470","Text":"so all this comes out 0 and the last bit,"},{"Start":"01:18.470 ","End":"01:22.120","Text":"like we said, that\u0027s faded out is 0."},{"Start":"01:22.120 ","End":"01:23.565","Text":"We\u0027re left with this bit."},{"Start":"01:23.565 ","End":"01:25.230","Text":"Now, a is 1,"},{"Start":"01:25.230 ","End":"01:30.290","Text":"so x minus t to x plus t. Here a is one, so it\u0027s a half,"},{"Start":"01:30.290 ","End":"01:38.355","Text":"so half the integral from x minus c to x plus t. g(s) is absolute value of sds."},{"Start":"01:38.355 ","End":"01:40.560","Text":"This is what u(x, t) is."},{"Start":"01:40.560 ","End":"01:44.435","Text":"Now, the small technical problem, it\u0027s not difficult,"},{"Start":"01:44.435 ","End":"01:48.565","Text":"but because absolute value of s is defined piecewise,"},{"Start":"01:48.565 ","End":"01:54.650","Text":"it\u0027s something for s less than 0 and something for s bigger than 0."},{"Start":"01:54.650 ","End":"02:01.580","Text":"But we don\u0027t know in this x minus t to x plus t where the 0 is,"},{"Start":"02:01.580 ","End":"02:03.770","Text":"then we would break it up into two."},{"Start":"02:03.770 ","End":"02:05.900","Text":"But if 0 was not inside it,"},{"Start":"02:05.900 ","End":"02:07.640","Text":"that would be something else."},{"Start":"02:07.640 ","End":"02:11.810","Text":"In short, I\u0027ll draw some cases in the picture,"},{"Start":"02:11.810 ","End":"02:13.705","Text":"and that should explain it."},{"Start":"02:13.705 ","End":"02:19.715","Text":"The interval x minus t to plus t could be completely to the right of the 0."},{"Start":"02:19.715 ","End":"02:22.820","Text":"The 0 is not in here, and in this case,"},{"Start":"02:22.820 ","End":"02:29.615","Text":"absolute value of s o is equal to s. The other possibility would be that the 0"},{"Start":"02:29.615 ","End":"02:32.420","Text":"is inside the interval x minus t to x"},{"Start":"02:32.420 ","End":"02:36.470","Text":"plus t. Then we\u0027ll have to split the integral up into 2 pieces,"},{"Start":"02:36.470 ","End":"02:37.700","Text":"from here to here,"},{"Start":"02:37.700 ","End":"02:41.600","Text":"where absolute value of s is minus s. From here to here,"},{"Start":"02:41.600 ","End":"02:48.020","Text":"our absolute value of s is plus s. Then the third case is when it\u0027s like this,"},{"Start":"02:48.020 ","End":"02:52.520","Text":"then absolute value of s is minus s. I will start with Case 1."},{"Start":"02:52.520 ","End":"02:54.290","Text":"This will be Case 1."},{"Start":"02:54.290 ","End":"02:58.745","Text":"This is where x minus t is bigger than 0,"},{"Start":"02:58.745 ","End":"03:02.270","Text":"then x plus t will surely be bigger than 0."},{"Start":"03:02.270 ","End":"03:04.745","Text":"What we get in this case, like I said,"},{"Start":"03:04.745 ","End":"03:11.370","Text":"the absolute value of s is just s. The integral of s is a half s^2."},{"Start":"03:11.370 ","End":"03:14.460","Text":"Bring the half in front so it\u0027s a quarter s^2,"},{"Start":"03:14.460 ","End":"03:18.760","Text":"but the s^2 has to be evaluated in these limits."},{"Start":"03:18.760 ","End":"03:22.110","Text":"It\u0027s this squared minus this squared,"},{"Start":"03:22.110 ","End":"03:26.400","Text":"and that gives us x plus t^2 minus x minus t^2."},{"Start":"03:26.400 ","End":"03:28.770","Text":"I explained about the quarter already."},{"Start":"03:28.770 ","End":"03:30.350","Text":"You don\u0027t have to simplify this,"},{"Start":"03:30.350 ","End":"03:32.005","Text":"but I\u0027d like to."},{"Start":"03:32.005 ","End":"03:36.218","Text":"This is x^2 plus 2 x t plus t squared,"},{"Start":"03:36.218 ","End":"03:40.400","Text":"and this part is x squared minus 2xt plus t squared."},{"Start":"03:40.400 ","End":"03:43.040","Text":"You subtract them, you\u0027ll get 2x t minus,"},{"Start":"03:43.040 ","End":"03:45.335","Text":"minus 2xt, which is 4xt."},{"Start":"03:45.335 ","End":"03:49.555","Text":"4xt times a quarter is xt."},{"Start":"03:49.555 ","End":"03:51.330","Text":"That\u0027s the Case 1."},{"Start":"03:51.330 ","End":"03:53.475","Text":"Case 1, u(x, t) is xt."},{"Start":"03:53.475 ","End":"03:56.699","Text":"Let\u0027s go on to next case."},{"Start":"03:56.699 ","End":"04:01.835","Text":"Case 2 is where x minus t is negative,"},{"Start":"04:01.835 ","End":"04:04.865","Text":"but x plus t is positive."},{"Start":"04:04.865 ","End":"04:08.345","Text":"I\u0027ll say now already what Case 3 will be,"},{"Start":"04:08.345 ","End":"04:15.075","Text":"that will be this one where x plus t is negative."},{"Start":"04:15.075 ","End":"04:18.380","Text":"We\u0027ll do Case 2, then we\u0027ll do Case 3."},{"Start":"04:18.380 ","End":"04:25.720","Text":"Yes. Also, these three inequalities can be simplified a bit to say x minus t bigger than"},{"Start":"04:25.720 ","End":"04:34.490","Text":"0 tells us that x is bigger than t. Well,"},{"Start":"04:34.490 ","End":"04:39.790","Text":"the first bit tells us that x is less than t,"},{"Start":"04:39.790 ","End":"04:42.175","Text":"if you bring this over to the other side,"},{"Start":"04:42.175 ","End":"04:45.940","Text":"and this bit tells us that x is bigger than minus t,"},{"Start":"04:45.940 ","End":"04:49.000","Text":"less than t and bigger than minus t. That\u0027s what it is,"},{"Start":"04:49.000 ","End":"04:51.730","Text":"x is between minus t and t. Here,"},{"Start":"04:51.730 ","End":"04:53.560","Text":"x bigger than t, like we said."},{"Start":"04:53.560 ","End":"04:57.745","Text":"The remaining case will obviously be x less than minus t,"},{"Start":"04:57.745 ","End":"05:02.229","Text":"which you can get just by bringing the t over to the other side."},{"Start":"05:02.229 ","End":"05:05.110","Text":"I think we\u0027ll just highlight these,"},{"Start":"05:05.110 ","End":"05:07.700","Text":"I\u0027ll need them afterwards."},{"Start":"05:07.700 ","End":"05:11.010","Text":"Case 2, and this is Case 3."},{"Start":"05:11.010 ","End":"05:13.335","Text":"In Case 2,"},{"Start":"05:13.335 ","End":"05:15.910","Text":"like I said, we break the integral up into two."},{"Start":"05:15.910 ","End":"05:18.910","Text":"We need the absolute value of s from here to here,"},{"Start":"05:18.910 ","End":"05:20.410","Text":"and also from here to here."},{"Start":"05:20.410 ","End":"05:24.265","Text":"But in the first bit absolute value of s is minus s, and then the second bit,"},{"Start":"05:24.265 ","End":"05:28.975","Text":"absolute value of x is equal to s. This integral,"},{"Start":"05:28.975 ","End":"05:36.430","Text":"we can take the minus away and reverse the order of the top and bottom limits."},{"Start":"05:36.430 ","End":"05:38.530","Text":"Now in each case we have the integral of s,"},{"Start":"05:38.530 ","End":"05:40.300","Text":"which is a half s^2."},{"Start":"05:40.300 ","End":"05:43.280","Text":"In each case the 1/2 will go with the 1/2 to make a 1/4."},{"Start":"05:43.280 ","End":"05:49.125","Text":"For the first one we have a 1/4 times x minus t squared,"},{"Start":"05:49.125 ","End":"05:51.645","Text":"the minus 0^2 we don\u0027t have to bother with."},{"Start":"05:51.645 ","End":"05:53.580","Text":"There\u0027s the x minus t squared,"},{"Start":"05:53.580 ","End":"05:54.770","Text":"and for the other one,"},{"Start":"05:54.770 ","End":"05:58.010","Text":"we get x plus t squared minus 0^2."},{"Start":"05:58.010 ","End":"06:02.785","Text":"We can simplify this plus this is just twice"},{"Start":"06:02.785 ","End":"06:08.550","Text":"x^2 plus twice t^2 because the minus 2xt cancels with plus 2xt."},{"Start":"06:08.550 ","End":"06:10.935","Text":"The 2/4 gives a 1/2,"},{"Start":"06:10.935 ","End":"06:13.850","Text":"so this is what we have simplified for this."},{"Start":"06:13.850 ","End":"06:16.865","Text":"We\u0027ve done Case 1 and Case 2,"},{"Start":"06:16.865 ","End":"06:18.850","Text":"and now Case 3."},{"Start":"06:18.850 ","End":"06:22.250","Text":"In this case, everything is negative."},{"Start":"06:22.250 ","End":"06:27.045","Text":"u(x, t) is the integral of a 1/2,"},{"Start":"06:27.045 ","End":"06:36.740","Text":"integral of minus s from x minus t to x plus t. Just bring the minus out in front."},{"Start":"06:36.740 ","End":"06:38.966","Text":"We\u0027ve done this integral before,"},{"Start":"06:38.966 ","End":"06:41.195","Text":"let me just go back a bit."},{"Start":"06:41.195 ","End":"06:50.335","Text":"The integral of sds from x minus t to plus t comes out to be xt."},{"Start":"06:50.335 ","End":"06:54.590","Text":"The only difference is that here we have an extra minus in front,"},{"Start":"06:54.590 ","End":"06:58.680","Text":"so the answer will be minus xt."},{"Start":"06:59.120 ","End":"07:02.340","Text":"Summarizing it as follows."},{"Start":"07:02.340 ","End":"07:03.780","Text":"When x was bigger than t,"},{"Start":"07:03.780 ","End":"07:05.965","Text":"which was our first case, we got xt."},{"Start":"07:05.965 ","End":"07:09.980","Text":"The second case was when x is between minus t and t,"},{"Start":"07:09.980 ","End":"07:11.975","Text":"and we got 1/2 x^2 plus t^2."},{"Start":"07:11.975 ","End":"07:16.800","Text":"Our third case was x less than minus t,"},{"Start":"07:16.800 ","End":"07:18.460","Text":"and then we got the answer that u(x,"},{"Start":"07:18.460 ","End":"07:21.040","Text":"t) is minus xt."},{"Start":"07:21.500 ","End":"07:28.690","Text":"This is the answer written piecewise, and we are done."}],"ID":30711},{"Watched":false,"Name":"Exercise 2","Duration":"5m 9s","ChapterTopicVideoID":29171,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.735","Text":"In this exercise, we have a wave equation with initial conditions."},{"Start":"00:06.735 ","End":"00:13.800","Text":"In other words, we have u and u_t at t=0 and it\u0027s defined on"},{"Start":"00:13.800 ","End":"00:22.530","Text":"the whole real line for x and time is positive using the notation from the tutorial a=1."},{"Start":"00:22.530 ","End":"00:27.705","Text":"That comes from here, u_t_t is a^2u_x_x, and f(x)=0,"},{"Start":"00:27.705 ","End":"00:32.730","Text":"this is f and this is g(x), so it\u0027s this,"},{"Start":"00:32.730 ","End":"00:38.490","Text":"and we know that the solution according to d\u0027Alembert formula is the following."},{"Start":"00:38.490 ","End":"00:41.280","Text":"This is in the homogeneous case,"},{"Start":"00:41.280 ","End":"00:43.820","Text":"let\u0027s just start computing it."},{"Start":"00:43.820 ","End":"00:46.175","Text":"We know that f is 0,"},{"Start":"00:46.175 ","End":"00:51.655","Text":"so that all this first part disappears and a is 1,"},{"Start":"00:51.655 ","End":"00:53.340","Text":"so we have 1 over 2,"},{"Start":"00:53.340 ","End":"00:58.260","Text":"the integral from x minus t to x plus t of g,"},{"Start":"00:58.260 ","End":"00:59.715","Text":"and g is from here,"},{"Start":"00:59.715 ","End":"01:01.850","Text":"so this is g of s, ds."},{"Start":"01:01.850 ","End":"01:03.410","Text":"We just have to evaluate this,"},{"Start":"01:03.410 ","End":"01:05.510","Text":"and that will be our answer."},{"Start":"01:05.510 ","End":"01:08.045","Text":"The only tricky part is"},{"Start":"01:08.045 ","End":"01:14.210","Text":"the absolute value because absolute value of s depends on where s is."},{"Start":"01:14.210 ","End":"01:17.000","Text":"Now it\u0027s going to be in this interval from x minus t"},{"Start":"01:17.000 ","End":"01:19.700","Text":"to x plus t but it depends where 0 is,"},{"Start":"01:19.700 ","End":"01:21.545","Text":"because when s crosses 0,"},{"Start":"01:21.545 ","End":"01:23.690","Text":"the sine of this thing changes,"},{"Start":"01:23.690 ","End":"01:27.620","Text":"so really, there are three cases."},{"Start":"01:27.620 ","End":"01:32.510","Text":"The 0 could be totally outside x minus t, x plus t,"},{"Start":"01:32.510 ","End":"01:33.800","Text":"it could be to the right,"},{"Start":"01:33.800 ","End":"01:35.315","Text":"could be to the left,"},{"Start":"01:35.315 ","End":"01:38.960","Text":"or 0 could be right inside this interval."},{"Start":"01:38.960 ","End":"01:42.920","Text":"In this case, the absolute value of s is equal to s everywhere."},{"Start":"01:42.920 ","End":"01:45.950","Text":"In this case, absolute value of s is minus s,"},{"Start":"01:45.950 ","End":"01:48.710","Text":"and in this case we\u0027ll have to split the integral up into two,"},{"Start":"01:48.710 ","End":"01:51.230","Text":"because from here to here it will be minus s,"},{"Start":"01:51.230 ","End":"01:56.625","Text":"and from here to here it will be plus s. Let\u0027s take the three cases,"},{"Start":"01:56.625 ","End":"01:58.890","Text":"the case 1, case 2, case 3."},{"Start":"01:58.890 ","End":"02:03.005","Text":"In case 1, x minus t is bigger than 0."},{"Start":"02:03.005 ","End":"02:05.045","Text":"You could write it as x bigger than t,"},{"Start":"02:05.045 ","End":"02:09.210","Text":"and in that case, the absolute value of s is equal to s,"},{"Start":"02:09.210 ","End":"02:14.680","Text":"so we have this integral and the integral of s over 1 plus x^2,"},{"Start":"02:14.680 ","End":"02:18.665","Text":"if you put a 2 here and make it a 4 here,"},{"Start":"02:18.665 ","End":"02:22.540","Text":"then 2s is the derivative of 1 plus s^2,"},{"Start":"02:22.540 ","End":"02:29.845","Text":"so we have the natural log of the denominator between this and this,"},{"Start":"02:29.845 ","End":"02:32.735","Text":"so what we get is one-quarter."},{"Start":"02:32.735 ","End":"02:36.305","Text":"Now if we put in x plus t instead of s, we get this,"},{"Start":"02:36.305 ","End":"02:40.490","Text":"put x minus t instead of s and we get this and subtract them,"},{"Start":"02:40.490 ","End":"02:44.124","Text":"so that\u0027s the case 1 solution."},{"Start":"02:44.124 ","End":"02:46.490","Text":"Now case 2,"},{"Start":"02:46.490 ","End":"02:50.390","Text":"where x minus t is less than 0,"},{"Start":"02:50.390 ","End":"02:53.510","Text":"but x plus t is greater than 0,"},{"Start":"02:53.510 ","End":"02:59.150","Text":"and we can also write this as follows if you think about it,"},{"Start":"02:59.150 ","End":"03:08.945","Text":"and so u(x,t) is the integral from x minus t to 0 of minus s,"},{"Start":"03:08.945 ","End":"03:12.410","Text":"that\u0027s what absolute value of s is and everything else like"},{"Start":"03:12.410 ","End":"03:16.280","Text":"so and here the same thing except with plus s in the numerator,"},{"Start":"03:16.280 ","End":"03:21.590","Text":"and it\u0027s from 0 to x plus t. Get rid of the minus here by reversing"},{"Start":"03:21.590 ","End":"03:25.230","Text":"the upper and lower limits of integration and that\u0027s all I did"},{"Start":"03:25.230 ","End":"03:29.105","Text":"from here to here and then we can say that this,"},{"Start":"03:29.105 ","End":"03:33.380","Text":"it\u0027s pretty much like in this case we have a quarter of"},{"Start":"03:33.380 ","End":"03:38.795","Text":"natural log of 1 plus s^2 between this and this."},{"Start":"03:38.795 ","End":"03:41.075","Text":"The only difference is,"},{"Start":"03:41.075 ","End":"03:44.210","Text":"that when s is 0, it\u0027s just like here,"},{"Start":"03:44.210 ","End":"03:47.090","Text":"if you put s equals 0, natural log of 1 plus"},{"Start":"03:47.090 ","End":"03:50.525","Text":"0 is natural log of one is 0 so we don\u0027t get the other part."},{"Start":"03:50.525 ","End":"03:53.015","Text":"The second term here is what gives us this,"},{"Start":"03:53.015 ","End":"03:54.500","Text":"and it\u0027s very similar."},{"Start":"03:54.500 ","End":"03:57.860","Text":"I\u0027m going to just notice this is x minus t, yeah,"},{"Start":"03:57.860 ","End":"03:59.930","Text":"I\u0027ve got these two the wrong way round,"},{"Start":"03:59.930 ","End":"04:03.980","Text":"so just to reverse the minus and the plus here, sorry."},{"Start":"04:03.980 ","End":"04:06.665","Text":"Okay, now let\u0027s move on to the third case,"},{"Start":"04:06.665 ","End":"04:11.090","Text":"where x plus t is less than 0,"},{"Start":"04:11.090 ","End":"04:16.535","Text":"which we can also write as x less than minus t. In this case,"},{"Start":"04:16.535 ","End":"04:18.740","Text":"the absolute value of s is minus s,"},{"Start":"04:18.740 ","End":"04:20.315","Text":"so this is the integral we get."},{"Start":"04:20.315 ","End":"04:23.330","Text":"We can swap upper and lower limits,"},{"Start":"04:23.330 ","End":"04:27.485","Text":"and get rid of the minus here, so as before,"},{"Start":"04:27.485 ","End":"04:31.250","Text":"this is a half natural log of 1 plus x squared,"},{"Start":"04:31.250 ","End":"04:37.775","Text":"the half with the half collects and we get one-quarter natural log from here to here,"},{"Start":"04:37.775 ","End":"04:40.550","Text":"and then just substitute the upper and lower limits."},{"Start":"04:40.550 ","End":"04:45.710","Text":"Put x equals x minus t here and s equals x plus t and subtract."},{"Start":"04:45.710 ","End":"04:48.245","Text":"Now we have three cases."},{"Start":"04:48.245 ","End":"04:55.775","Text":"We have the case where x is bigger than t and then we get this for u(x,t)."},{"Start":"04:55.775 ","End":"04:57.560","Text":"In this case, we get this,"},{"Start":"04:57.560 ","End":"04:58.783","Text":"in this case we get this,"},{"Start":"04:58.783 ","End":"05:01.085","Text":"and we just want to collect it all together,"},{"Start":"05:01.085 ","End":"05:02.420","Text":"this is what we get,"},{"Start":"05:02.420 ","End":"05:04.325","Text":"the three different cases well collected,"},{"Start":"05:04.325 ","End":"05:09.990","Text":"just copied and that\u0027s the answer, and we\u0027re done."}],"ID":30712},{"Watched":false,"Name":"Exercise 3","Duration":"4m 50s","ChapterTopicVideoID":29172,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.375","Text":"In this exercise, we have a wave equation on an infinite interval,"},{"Start":"00:06.375 ","End":"00:09.030","Text":"and these are the initial conditions."},{"Start":"00:09.030 ","End":"00:10.800","Text":"What happens when t equals 0?"},{"Start":"00:10.800 ","End":"00:13.500","Text":"We\u0027re given u and d u by d t,"},{"Start":"00:13.500 ","End":"00:15.780","Text":"and we will solve it with D\u0027Alembert\u0027s formula."},{"Start":"00:15.780 ","End":"00:18.000","Text":"First, let\u0027s identify some things."},{"Start":"00:18.000 ","End":"00:19.620","Text":"A equals 1,"},{"Start":"00:19.620 ","End":"00:23.640","Text":"I\u0027ll show you what the general form is."},{"Start":"00:23.640 ","End":"00:25.560","Text":"Here in our case,"},{"Start":"00:25.560 ","End":"00:27.015","Text":"a is 1,"},{"Start":"00:27.015 ","End":"00:30.030","Text":"then f(x) is x,"},{"Start":"00:30.030 ","End":"00:32.430","Text":"this is g(x),"},{"Start":"00:32.430 ","End":"00:37.605","Text":"and big F is x^2 plus t^2 from here."},{"Start":"00:37.605 ","End":"00:44.130","Text":"This is the solution in general using D\u0027Alembert\u0027s formula,"},{"Start":"00:44.130 ","End":"00:49.175","Text":"and let\u0027s substitute what happens in our case when a=1, etc."},{"Start":"00:49.175 ","End":"00:52.590","Text":"We get f(x) is x,"},{"Start":"00:52.590 ","End":"00:56.845","Text":"so here\u0027s x plus t plus x minus t over 2."},{"Start":"00:56.845 ","End":"00:58.760","Text":"Here we have the integral."},{"Start":"00:58.760 ","End":"01:01.250","Text":"Again, a is 1 here, here, here,"},{"Start":"01:01.250 ","End":"01:04.640","Text":"g(s) is s,"},{"Start":"01:04.640 ","End":"01:06.820","Text":"and for f(s)Tau,"},{"Start":"01:06.820 ","End":"01:10.715","Text":"this will be s^2 plus Tau squared."},{"Start":"01:10.715 ","End":"01:13.370","Text":"Now, x plus t plus x minus t over 2,"},{"Start":"01:13.370 ","End":"01:16.295","Text":"this comes down to just x."},{"Start":"01:16.295 ","End":"01:19.775","Text":"Here we have the integral of s is s squared over 2."},{"Start":"01:19.775 ","End":"01:24.255","Text":"We have 1/4s squared from x minus t to x plus t,"},{"Start":"01:24.255 ","End":"01:27.305","Text":"and here we\u0027ll do the ds integral first."},{"Start":"01:27.305 ","End":"01:32.165","Text":"This comes out to be s^3 over 3 plus Tau squared s,"},{"Start":"01:32.165 ","End":"01:33.785","Text":"between here and here."},{"Start":"01:33.785 ","End":"01:37.170","Text":"This is still x,"},{"Start":"01:37.170 ","End":"01:41.420","Text":"this is 1/4x plus t squared minus x minus t squared."},{"Start":"01:41.420 ","End":"01:43.250","Text":"The last one is the integral."},{"Start":"01:43.250 ","End":"01:49.460","Text":"We have to substitute instead of s x plus t minus Tau and then x minus t minus Tau."},{"Start":"01:49.460 ","End":"01:54.105","Text":"For the plus, we get this for s^3 over 3,"},{"Start":"01:54.105 ","End":"01:55.640","Text":"and here Tau squared s,"},{"Start":"01:55.640 ","End":"01:57.440","Text":"and here minus, minus,"},{"Start":"01:57.440 ","End":"01:59.720","Text":"and there\u0027s a minus here and here."},{"Start":"01:59.720 ","End":"02:02.405","Text":"Now for computing this minus this,"},{"Start":"02:02.405 ","End":"02:07.610","Text":"we can use the formula for a plus b cubed minus a minus b cubed."},{"Start":"02:07.610 ","End":"02:11.120","Text":"Well, we\u0027ll develop it from the formula for a plus b cubed is this,"},{"Start":"02:11.120 ","End":"02:12.925","Text":"a minus b cubed is this."},{"Start":"02:12.925 ","End":"02:14.675","Text":"This is what we get."},{"Start":"02:14.675 ","End":"02:18.035","Text":"Now we\u0027ll apply that for these two terms,"},{"Start":"02:18.035 ","End":"02:20.380","Text":"and we get 6."},{"Start":"02:20.380 ","End":"02:22.985","Text":"This is the a squared b part,"},{"Start":"02:22.985 ","End":"02:26.195","Text":"x^2 times t minus Tau,"},{"Start":"02:26.195 ","End":"02:30.495","Text":"and then 2t minus Tau cubed, that\u0027s the b."},{"Start":"02:30.495 ","End":"02:33.260","Text":"Then we combine these two."},{"Start":"02:33.260 ","End":"02:36.127","Text":"You just look at the part without the Tau squared,"},{"Start":"02:36.127 ","End":"02:40.385","Text":"we have x plus something and x minus something."},{"Start":"02:40.385 ","End":"02:46.775","Text":"You subtract them, we get twice t minus Tau with the Tau squared."},{"Start":"02:46.775 ","End":"02:48.730","Text":"This is what we have."},{"Start":"02:48.730 ","End":"02:53.490","Text":"Then here we have 6/3 divided by 2,"},{"Start":"02:53.490 ","End":"02:56.690","Text":"is just 1, x^2 t minus Tau."},{"Start":"02:56.690 ","End":"02:59.600","Text":"Here we have 2/3 times 2,"},{"Start":"02:59.600 ","End":"03:02.465","Text":"is 1/3, t minus Tau cubed,"},{"Start":"03:02.465 ","End":"03:08.910","Text":"and here 2 over 2 is 1 Tau squared t minus Tau squared Tau,"},{"Start":"03:08.910 ","End":"03:10.125","Text":"which is Tau cubed."},{"Start":"03:10.125 ","End":"03:17.930","Text":"Now the integral of t minus Tau is 1/2t minus Tau squared,"},{"Start":"03:17.930 ","End":"03:22.075","Text":"but divided by the minus 1, the inner derivative."},{"Start":"03:22.075 ","End":"03:23.930","Text":"This is what we get."},{"Start":"03:23.930 ","End":"03:29.600","Text":"For this, we get t minus Tau to the fourth, over 4."},{"Start":"03:29.600 ","End":"03:31.580","Text":"The 4 with the 3 is 12,"},{"Start":"03:31.580 ","End":"03:33.665","Text":"and we also have a minus."},{"Start":"03:33.665 ","End":"03:41.240","Text":"Then here we just get Tau cubed over 3 and there\u0027s still a t there and here,"},{"Start":"03:41.240 ","End":"03:43.550","Text":"Tau to the fourth over 4."},{"Start":"03:43.550 ","End":"03:46.130","Text":"This is what we have, and we just have to substitute"},{"Start":"03:46.130 ","End":"03:49.705","Text":"the upper and lower limits of integration and subtract."},{"Start":"03:49.705 ","End":"03:52.820","Text":"Now, these upper and lower limits of the Tau."},{"Start":"03:52.820 ","End":"03:55.190","Text":"Should have written Tau equals 0 to Tau equals"},{"Start":"03:55.190 ","End":"03:58.780","Text":"t. Let\u0027s first of all substitute Tau equals t,"},{"Start":"03:58.780 ","End":"04:03.319","Text":"so this becomes 0 and this becomes 0."},{"Start":"04:03.319 ","End":"04:06.560","Text":"Here we get 1/3(t) to the fourth,"},{"Start":"04:06.560 ","End":"04:09.095","Text":"and here minus 1/4t to the fourth."},{"Start":"04:09.095 ","End":"04:12.655","Text":"If we substitute Tau equals 0,"},{"Start":"04:12.655 ","End":"04:15.180","Text":"then these two become 0,"},{"Start":"04:15.180 ","End":"04:17.030","Text":"and what we have here, well,"},{"Start":"04:17.030 ","End":"04:20.150","Text":"these both become plus because we\u0027re subtracting, it\u0027s the lower limit."},{"Start":"04:20.150 ","End":"04:22.850","Text":"So plus and when Tau equals 0,"},{"Start":"04:22.850 ","End":"04:25.195","Text":"we get x squared, t squared,"},{"Start":"04:25.195 ","End":"04:31.620","Text":"and here we get t to the fourth plus a 12th."},{"Start":"04:31.620 ","End":"04:33.705","Text":"That\u0027s what we have."},{"Start":"04:33.705 ","End":"04:36.755","Text":"There\u0027s just one step just to simplify a bit."},{"Start":"04:36.755 ","End":"04:38.900","Text":"Notice that we have t to the fourth here,"},{"Start":"04:38.900 ","End":"04:39.940","Text":"here, and here,"},{"Start":"04:39.940 ","End":"04:41.300","Text":"and a simple exercise,"},{"Start":"04:41.300 ","End":"04:45.860","Text":"infractions gives us that we have 1/6 of t to the fourth,"},{"Start":"04:45.860 ","End":"04:47.925","Text":"and so this is our answer,"},{"Start":"04:47.925 ","End":"04:50.980","Text":"and that concludes this exercise."}],"ID":30713},{"Watched":false,"Name":"Exercise 4","Duration":"1m 25s","ChapterTopicVideoID":29173,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.225","Text":"In this exercise, we\u0027re going to solve the homogeneous wave equation defined on"},{"Start":"00:06.225 ","End":"00:13.475","Text":"the whole real line and for time non-negative and as usual,"},{"Start":"00:13.475 ","End":"00:18.860","Text":"this is called f and this is called g. We have the formula,"},{"Start":"00:18.860 ","End":"00:22.460","Text":"this is d\u0027Alembert formula for the solution and in"},{"Start":"00:22.460 ","End":"00:26.545","Text":"our case we just substitute f is e^minus."},{"Start":"00:26.545 ","End":"00:32.480","Text":"We have this, then we have this integral where g(s) is just s,"},{"Start":"00:32.480 ","End":"00:34.970","Text":"because g(x) is x."},{"Start":"00:34.970 ","End":"00:39.335","Text":"Then we can take e^minus x out of this,"},{"Start":"00:39.335 ","End":"00:47.190","Text":"and here the integral of s is 1/2 s^2 so put the 1/2 together with the 1 over 2a."},{"Start":"00:47.190 ","End":"00:52.440","Text":"Then we get x plus at^2 minus x minus at^2."},{"Start":"00:52.440 ","End":"00:54.480","Text":"If we simplify it,"},{"Start":"00:54.480 ","End":"00:58.460","Text":"this is cosine hyperbolic of at"},{"Start":"00:58.460 ","End":"01:02.855","Text":"using this formula for the hyperbolic cosine in terms of the exponential."},{"Start":"01:02.855 ","End":"01:10.260","Text":"Here, we get 2 at x minus 2 at x with a minus."},{"Start":"01:10.260 ","End":"01:16.740","Text":"So we get 4 axt and then 1 over 4a times this will give us"},{"Start":"01:16.740 ","End":"01:25.420","Text":"just xt and the solution is as follows and that concludes this exercise."}],"ID":30714},{"Watched":false,"Name":"Exercise 5","Duration":"1m 6s","ChapterTopicVideoID":29174,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"In this exercise, we\u0027re going to solve the following problem."},{"Start":"00:03.420 ","End":"00:07.590","Text":"This is a wave equation on an infinite interval,"},{"Start":"00:07.590 ","End":"00:10.695","Text":"and these are the initial conditions."},{"Start":"00:10.695 ","End":"00:13.545","Text":"We\u0027re going to use d\u0027Alembert\u0027s formula."},{"Start":"00:13.545 ","End":"00:20.340","Text":"In this case, f(x) is sine(x) and g(x) is cosine(x)."},{"Start":"00:20.340 ","End":"00:22.035","Text":"This is the formula,"},{"Start":"00:22.035 ","End":"00:24.210","Text":"so we\u0027re going to substitute,"},{"Start":"00:24.210 ","End":"00:26.310","Text":"by the way, a is not given,"},{"Start":"00:26.310 ","End":"00:29.805","Text":"it\u0027s for general parameter a bigger than 0."},{"Start":"00:29.805 ","End":"00:32.790","Text":"In our case, f is sine,"},{"Start":"00:32.790 ","End":"00:34.515","Text":"so this is what we get here,"},{"Start":"00:34.515 ","End":"00:36.210","Text":"and g is cosine."},{"Start":"00:36.210 ","End":"00:39.270","Text":"The integral of cosine is sine,"},{"Start":"00:39.270 ","End":"00:42.645","Text":"so we have sine of this minus sign of this."},{"Start":"00:42.645 ","End":"00:45.920","Text":"Then let\u0027s put this over a common denominator,"},{"Start":"00:45.920 ","End":"00:49.960","Text":"which will be 2_ a and here also over 2_ a."},{"Start":"00:49.960 ","End":"00:54.335","Text":"Then the first term in this numerator and in this numerator"},{"Start":"00:54.335 ","End":"00:58.745","Text":"can combine to be a plus 1 sine(x plus at),"},{"Start":"00:58.745 ","End":"01:02.200","Text":"and here a minus 1 sine(x minus at)."},{"Start":"01:02.200 ","End":"01:03.890","Text":"Anyway, this is the answer,"},{"Start":"01:03.890 ","End":"01:07.050","Text":"and that concludes this exercise."}],"ID":30715},{"Watched":false,"Name":"Exercise 6","Duration":"3m 19s","ChapterTopicVideoID":29175,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.190","Text":"In this exercise, we\u0027re given that u(x,"},{"Start":"00:04.190 ","End":"00:07.440","Text":"t) is a solution to the following problem."},{"Start":"00:07.440 ","End":"00:12.030","Text":"This is a wave equation on an infinite interval,"},{"Start":"00:12.030 ","End":"00:19.074","Text":"and it has the initial condition for u(x,0) and u_t of x naught,"},{"Start":"00:19.074 ","End":"00:22.335","Text":"x naught is a piecewise-defined function."},{"Start":"00:22.335 ","End":"00:30.990","Text":"We have to compute the integral from minus infinity to 4 of u(x, 1) dx."},{"Start":"00:30.990 ","End":"00:33.605","Text":"Let\u0027s identify some of the parts."},{"Start":"00:33.605 ","End":"00:38.670","Text":"A is 1 because u_tt is a^2 u_xx."},{"Start":"00:38.670 ","End":"00:41.520","Text":"f(x) is u of x naught,"},{"Start":"00:41.520 ","End":"00:46.890","Text":"so it\u0027s this, and g(x) is this, which is 0."},{"Start":"00:46.890 ","End":"00:50.405","Text":"We\u0027re going to use d\u0027Alembert formula, which is this."},{"Start":"00:50.405 ","End":"00:52.160","Text":"But since g is 0,"},{"Start":"00:52.160 ","End":"00:55.670","Text":"this part disappears and we just have this."},{"Start":"00:55.670 ","End":"01:02.060","Text":"Just rewrite it as follows with the half of this and half of this and then a is 1,"},{"Start":"01:02.060 ","End":"01:06.410","Text":"f(x plus t) according to this is 0 or 4,"},{"Start":"01:06.410 ","End":"01:09.440","Text":"depending on whether x plus t is less than 2 or bigger than 2."},{"Start":"01:09.440 ","End":"01:14.220","Text":"Similarly for the other part with x minus t,"},{"Start":"01:14.220 ","End":"01:16.655","Text":"and we can rewrite this,"},{"Start":"01:16.655 ","End":"01:18.560","Text":"bring the t over to the other side,"},{"Start":"01:18.560 ","End":"01:21.245","Text":"so x less than 2 minus t or bigger than,"},{"Start":"01:21.245 ","End":"01:25.625","Text":"and here x less than 2 plus t or bigger than."},{"Start":"01:25.625 ","End":"01:27.755","Text":"A diagram will help."},{"Start":"01:27.755 ","End":"01:29.930","Text":"Only three possibilities really,"},{"Start":"01:29.930 ","End":"01:34.490","Text":"even though you would think there\u0027s 2 possibilities here times 2 possibilities here is 4."},{"Start":"01:34.490 ","End":"01:39.415","Text":"It\u0027s really only 3 because 2 minus t is less than 2 plus t,"},{"Start":"01:39.415 ","End":"01:43.040","Text":"so it can\u0027t be less than this and bigger than this."},{"Start":"01:43.040 ","End":"01:46.355","Text":"Three possibilities, either it\u0027s less than 2 minus t"},{"Start":"01:46.355 ","End":"01:50.270","Text":"or between 2 minus t and 2 plus t or if it\u0027s here,"},{"Start":"01:50.270 ","End":"01:54.335","Text":"bigger than 2 plus t. If you compute it in each case,"},{"Start":"01:54.335 ","End":"01:55.950","Text":"you get the following."},{"Start":"01:55.950 ","End":"01:59.700","Text":"If x is less than 2 minus t and it\u0027s also"},{"Start":"01:59.700 ","End":"02:05.045","Text":"less than 2 plus t and we get 0 here and 0 here, so 0 here."},{"Start":"02:05.045 ","End":"02:10.290","Text":"In this case, we get a 4 here and a 0 here."},{"Start":"02:10.290 ","End":"02:13.886","Text":"A half 4 plus a half times 0 is 2,"},{"Start":"02:13.886 ","End":"02:15.740","Text":"and then the last case,"},{"Start":"02:15.740 ","End":"02:18.680","Text":"x is bigger than this and x is bigger than this and we"},{"Start":"02:18.680 ","End":"02:21.845","Text":"get a half of 4 plus 4, so it\u0027s 4."},{"Start":"02:21.845 ","End":"02:24.289","Text":"These are the three possibilities."},{"Start":"02:24.289 ","End":"02:26.650","Text":"Let\u0027s continue with the exercise."},{"Start":"02:26.650 ","End":"02:30.650","Text":"u (x, 1) will be just letting t equals 1."},{"Start":"02:30.650 ","End":"02:32.780","Text":"This is 1, this is 1 and 3,"},{"Start":"02:32.780 ","End":"02:35.930","Text":"and this is 3, so we get the following cases."},{"Start":"02:35.930 ","End":"02:39.755","Text":"The integral from minus infinity to 4,"},{"Start":"02:39.755 ","End":"02:41.755","Text":"we can split it up into 3,"},{"Start":"02:41.755 ","End":"02:46.120","Text":"because 1 and 3 are the points where it changes definition."},{"Start":"02:46.120 ","End":"02:51.955","Text":"So from minus infinity to 1 plus from 1-3 plus from 3-4."},{"Start":"02:51.955 ","End":"02:56.925","Text":"Here it\u0027s 0, here this is 2,"},{"Start":"02:56.925 ","End":"02:58.335","Text":"and here it\u0027s 4."},{"Start":"02:58.335 ","End":"03:01.670","Text":"The integrals come out to be the integral of 0 is 0."},{"Start":"03:01.670 ","End":"03:04.180","Text":"The integral of 2 from 1-3,"},{"Start":"03:04.180 ","End":"03:05.780","Text":"3 minus 1 is 2,"},{"Start":"03:05.780 ","End":"03:07.525","Text":"2 times 2 is 4."},{"Start":"03:07.525 ","End":"03:09.540","Text":"Here 4 minus 3 is 1,"},{"Start":"03:09.540 ","End":"03:11.550","Text":"1 times 4 is 4."},{"Start":"03:11.550 ","End":"03:15.600","Text":"We get 0 plus 4 plus 4 is 8."},{"Start":"03:15.600 ","End":"03:19.300","Text":"That\u0027s the answer and we are done."}],"ID":30716},{"Watched":false,"Name":"Exercise 7a","Duration":"4m 50s","ChapterTopicVideoID":29176,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.170","Text":"In this exercise, we\u0027re given that the function u of x"},{"Start":"00:04.170 ","End":"00:08.175","Text":"and t is a solution to the following problem."},{"Start":"00:08.175 ","End":"00:11.970","Text":"This is a wave equation on"},{"Start":"00:11.970 ","End":"00:18.480","Text":"an infinite interval with initial condition for u and du by dt."},{"Start":"00:18.480 ","End":"00:24.810","Text":"Given all this, we have to find the solution u of x and t and secondly,"},{"Start":"00:24.810 ","End":"00:32.985","Text":"to compute the integral from 0 to 10 of du by dx of x and 5dx."},{"Start":"00:32.985 ","End":"00:35.985","Text":"Note that a equals 1,"},{"Start":"00:35.985 ","End":"00:38.265","Text":"f of x equals 0."},{"Start":"00:38.265 ","End":"00:45.260","Text":"That\u0027s the 0 here and g of x is e to the minus absolute value of x from here,"},{"Start":"00:45.260 ","End":"00:49.190","Text":"the a equals 1 is because u_tt equals a^2 u_xx,"},{"Start":"00:49.190 ","End":"00:51.190","Text":"so that makes a equal 1."},{"Start":"00:51.190 ","End":"00:56.930","Text":"Then we know that the d\u0027Alembert formula is this so the solution for u(x,"},{"Start":"00:56.930 ","End":"00:59.675","Text":"t) is given by the closed form here."},{"Start":"00:59.675 ","End":"01:03.470","Text":"We just have to substitute the values we have."},{"Start":"01:03.470 ","End":"01:07.340","Text":"Well, this part is going to be 0 because F is 0."},{"Start":"01:07.340 ","End":"01:13.880","Text":"We just get this where g(s) is e to the minus absolute value of s,"},{"Start":"01:13.880 ","End":"01:18.395","Text":"and the limits are x plus t and x minus t because a equals 1."},{"Start":"01:18.395 ","End":"01:24.185","Text":"Now, the absolute value of s is piecewise defined,"},{"Start":"01:24.185 ","End":"01:27.530","Text":"so we\u0027re going to have to divide into cases."},{"Start":"01:27.530 ","End":"01:32.360","Text":"We have actually 3 cases for this interval from x minus t to x"},{"Start":"01:32.360 ","End":"01:37.310","Text":"plus t. It could be that the whole interval is to the left of 0."},{"Start":"01:37.310 ","End":"01:41.839","Text":"That\u0027s 1 case. It could be that 0 is in the middle of the interval,"},{"Start":"01:41.839 ","End":"01:45.950","Text":"or it could be that the interval is to the right of the 0,"},{"Start":"01:45.950 ","End":"01:48.800","Text":"and each case will have a different formula for e to"},{"Start":"01:48.800 ","End":"01:52.575","Text":"the minus absolute value of s to the individual parts."},{"Start":"01:52.575 ","End":"01:54.630","Text":"Well, I\u0027ll explain what I mean."},{"Start":"01:54.630 ","End":"01:56.385","Text":"In the first case,"},{"Start":"01:56.385 ","End":"01:59.310","Text":"x plus t is less than 0,"},{"Start":"01:59.310 ","End":"02:05.210","Text":"which means that x is less than minus t. In this case,"},{"Start":"02:05.210 ","End":"02:10.310","Text":"everything here is negative s in this interval is negative."},{"Start":"02:10.310 ","End":"02:15.500","Text":"E to the minus absolute value of s will be e to the minus minus s,"},{"Start":"02:15.500 ","End":"02:17.570","Text":"because absolute value of s is minus s,"},{"Start":"02:17.570 ","End":"02:19.730","Text":"it will get just e to the s,"},{"Start":"02:19.730 ","End":"02:22.445","Text":"is the straightforward integral to compute,"},{"Start":"02:22.445 ","End":"02:25.480","Text":"because the integral of e to the s is just e to the s itself,"},{"Start":"02:25.480 ","End":"02:29.780","Text":"so we just have to substitute x plus t and x minus t and"},{"Start":"02:29.780 ","End":"02:35.520","Text":"subtract the lower from the upper and the half here is the 2 in the denominator."},{"Start":"02:35.520 ","End":"02:38.480","Text":"That\u0027s Case 1 and you know what,"},{"Start":"02:38.480 ","End":"02:39.665","Text":"we\u0027ll jump to Case 3,"},{"Start":"02:39.665 ","End":"02:41.905","Text":"because these are the simple cases."},{"Start":"02:41.905 ","End":"02:45.500","Text":"Case 2 is where we have to divide the 2 pieces."},{"Start":"02:45.500 ","End":"02:52.460","Text":"In this case it\u0027s very similar except e to the minus s this time and then we\u0027ll get,"},{"Start":"02:52.460 ","End":"02:56.540","Text":"considering that the integral is minus e to the minus s,"},{"Start":"02:56.540 ","End":"02:59.870","Text":"what we\u0027ll get is like here."},{"Start":"02:59.870 ","End":"03:01.220","Text":"Well, not quite like,"},{"Start":"03:01.220 ","End":"03:06.200","Text":"but we\u0027ll have e to the minus this takeaway e to the minus that."},{"Start":"03:06.200 ","End":"03:10.475","Text":"But we also need an extra minus because of the minus here."},{"Start":"03:10.475 ","End":"03:13.910","Text":"We\u0027ll put this minus here down here and we can remove"},{"Start":"03:13.910 ","End":"03:19.400","Text":"the brackets and also switch the positions of these 2 and get rid of the minus,"},{"Start":"03:19.400 ","End":"03:22.415","Text":"so we end up with this for Case 3."},{"Start":"03:22.415 ","End":"03:24.710","Text":"What about Case 2?"},{"Start":"03:24.710 ","End":"03:28.910","Text":"x minus t is negative and x plus t is positive,"},{"Start":"03:28.910 ","End":"03:33.200","Text":"which comes down to minus t less than x less than t,"},{"Start":"03:33.200 ","End":"03:36.470","Text":"x is between minus t and t and in this case,"},{"Start":"03:36.470 ","End":"03:38.450","Text":"we break up the integral into 2 parts,"},{"Start":"03:38.450 ","End":"03:43.199","Text":"from x minus t to 0 and from 0 to x plus t. Here,"},{"Start":"03:43.199 ","End":"03:46.265","Text":"the function is e to the s, like here."},{"Start":"03:46.265 ","End":"03:50.225","Text":"In this part, the function is e to the minus s,"},{"Start":"03:50.225 ","End":"03:51.965","Text":"just like here,"},{"Start":"03:51.965 ","End":"03:54.785","Text":"so we get 2 separate parts."},{"Start":"03:54.785 ","End":"03:59.715","Text":"We get half of e to the s. The integral is e to the s,"},{"Start":"03:59.715 ","End":"04:03.170","Text":"plug in 0 and x minus t and here we get"},{"Start":"04:03.170 ","End":"04:05.630","Text":"the extra minus because the integral of e to"},{"Start":"04:05.630 ","End":"04:08.780","Text":"the minus s is minus e to the minus s. Again,"},{"Start":"04:08.780 ","End":"04:11.550","Text":"substitute the parts e to the 0 is 1,"},{"Start":"04:11.550 ","End":"04:17.345","Text":"so this becomes a plus 1 when it fits over the minus 2 and here also,"},{"Start":"04:17.345 ","End":"04:23.020","Text":"this becomes minus and also we open the brackets and this is also a minus."},{"Start":"04:23.020 ","End":"04:24.980","Text":"This is Case 1,"},{"Start":"04:24.980 ","End":"04:27.860","Text":"this is Case 2,"},{"Start":"04:27.860 ","End":"04:32.180","Text":"this is Case 3 and these are the 3 conditions in Case 1,"},{"Start":"04:32.180 ","End":"04:33.710","Text":"Case 2, Case 3."},{"Start":"04:33.710 ","End":"04:36.410","Text":"If we summarize it all, this is what we get,"},{"Start":"04:36.410 ","End":"04:38.960","Text":"it\u0027s just taking the data we had,"},{"Start":"04:38.960 ","End":"04:42.575","Text":"the results we had and putting them in a table."},{"Start":"04:42.575 ","End":"04:48.365","Text":"Yeah, piecewise definition of u and x and t and that concludes Part A."},{"Start":"04:48.365 ","End":"04:51.060","Text":"We\u0027ll continue in the following clip."}],"ID":30717},{"Watched":false,"Name":"Exercise 7b","Duration":"3m 26s","ChapterTopicVideoID":29177,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.965","Text":"We just finished part a and now we\u0027ll do part b."},{"Start":"00:04.965 ","End":"00:08.175","Text":"This was the solution to part a."},{"Start":"00:08.175 ","End":"00:10.110","Text":"Now on to part b,"},{"Start":"00:10.110 ","End":"00:13.830","Text":"we need to compute du by dx."},{"Start":"00:13.830 ","End":"00:19.140","Text":"First, we can do the substitution because this is"},{"Start":"00:19.140 ","End":"00:21.480","Text":"the value of t. It doesn\u0027t matter if you"},{"Start":"00:21.480 ","End":"00:24.210","Text":"substitute first and then differentiate or vice versa."},{"Start":"00:24.210 ","End":"00:29.235","Text":"We can differentiate this after we\u0027ve substituted t= 5."},{"Start":"00:29.235 ","End":"00:32.175","Text":"What we get is just a straightforward."},{"Start":"00:32.175 ","End":"00:37.460","Text":"For example, here, we get the thing itself because e^x plus"},{"Start":"00:37.460 ","End":"00:43.655","Text":"5 derivative of itself and so does e^x minus 5 and a derivative is 1."},{"Start":"00:43.655 ","End":"00:46.160","Text":"Here, the 2 drops off."},{"Start":"00:46.160 ","End":"00:48.035","Text":"This stays as is,"},{"Start":"00:48.035 ","End":"00:50.420","Text":"but here we get a minus."},{"Start":"00:50.420 ","End":"00:54.455","Text":"It becomes plus, and here they both get minuses."},{"Start":"00:54.455 ","End":"00:56.540","Text":"This is minus and this is plus."},{"Start":"00:56.540 ","End":"00:58.640","Text":"Now, the integral from 0-10,"},{"Start":"00:58.640 ","End":"01:01.105","Text":"we can break up into 2 parts,"},{"Start":"01:01.105 ","End":"01:05.620","Text":"from 0-5 and from 5-10."},{"Start":"01:05.620 ","End":"01:08.930","Text":"Note that this part won\u0027t appear well,"},{"Start":"01:08.930 ","End":"01:12.320","Text":"this part really, because it\u0027s less than minus 5."},{"Start":"01:12.320 ","End":"01:13.910","Text":"So from 0-5,"},{"Start":"01:13.910 ","End":"01:16.040","Text":"we get from here, and from 5-10,"},{"Start":"01:16.040 ","End":"01:17.725","Text":"we get from here."},{"Start":"01:17.725 ","End":"01:21.515","Text":"This is the expression we have to compute now."},{"Start":"01:21.515 ","End":"01:24.920","Text":"Like I said, this will take from here,"},{"Start":"01:24.920 ","End":"01:27.605","Text":"and this will take from here."},{"Start":"01:27.605 ","End":"01:31.195","Text":"What we get is the following."},{"Start":"01:31.195 ","End":"01:40.045","Text":"Then we can simplify by taking minus e^minus 5 over 2 outside of the integral."},{"Start":"01:40.045 ","End":"01:45.125","Text":"What we\u0027re left with is the minus disappears and the minus 5 disappears."},{"Start":"01:45.125 ","End":"01:49.345","Text":"We\u0027re left with just e^x and here, e^minus x."},{"Start":"01:49.345 ","End":"01:52.415","Text":"Here, the common part is,"},{"Start":"01:52.415 ","End":"01:55.520","Text":"well, everything except the e^minus x."},{"Start":"01:55.520 ","End":"01:59.840","Text":"We get from here, e^minus 5,"},{"Start":"01:59.840 ","End":"02:05.340","Text":"and from here minus e^plus 5 over 2."},{"Start":"02:05.340 ","End":"02:07.935","Text":"Now we\u0027ll do the integrations."},{"Start":"02:07.935 ","End":"02:12.180","Text":"This one becomes e^x plus e^minus x,"},{"Start":"02:12.180 ","End":"02:13.965","Text":"we get a minus from here."},{"Start":"02:13.965 ","End":"02:17.730","Text":"This integral is minus e^minus x."},{"Start":"02:17.730 ","End":"02:22.175","Text":"The minus will serve to just change the position of these two."},{"Start":"02:22.175 ","End":"02:24.250","Text":"This is what we have now."},{"Start":"02:24.250 ","End":"02:29.875","Text":"Now, we need to substitute the limits of integration here from 0-5, here from 5-10."},{"Start":"02:29.875 ","End":"02:35.210","Text":"What we get here is e^5 plus e^minus 5,"},{"Start":"02:35.210 ","End":"02:36.380","Text":"and then the 0,"},{"Start":"02:36.380 ","End":"02:38.135","Text":"but with a minus sign."},{"Start":"02:38.135 ","End":"02:44.870","Text":"E^0 with a minus and e^minus 0 with the minus with both 1."},{"Start":"02:44.870 ","End":"02:47.900","Text":"Here, e^minus x."},{"Start":"02:47.900 ","End":"02:53.010","Text":"It\u0027s e^minus 10, subtract e^minus 5."},{"Start":"02:53.110 ","End":"03:00.290","Text":"Now, let\u0027s multiply everything by minus e^minus 5."},{"Start":"03:00.290 ","End":"03:02.830","Text":"This is routine."},{"Start":"03:02.830 ","End":"03:05.025","Text":"This is what we get."},{"Start":"03:05.025 ","End":"03:07.545","Text":"Then collecting like terms,"},{"Start":"03:07.545 ","End":"03:11.415","Text":"minus 1 minus 1 is minus 2."},{"Start":"03:11.415 ","End":"03:15.150","Text":"Minus e^minus 10 plus e^minus 10 cancels."},{"Start":"03:15.150 ","End":"03:17.879","Text":"We have 1, 2, 3 of these."},{"Start":"03:17.879 ","End":"03:22.710","Text":"Finally, minus e^minus 15, all over 2."},{"Start":"03:22.710 ","End":"03:26.650","Text":"That\u0027s the answer to part b, and we\u0027re done."}],"ID":30718},{"Watched":false,"Name":"Exercise 8","Duration":"3m 11s","ChapterTopicVideoID":29178,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.720","Text":"In this exercise, we\u0027re given that the function"},{"Start":"00:03.720 ","End":"00:07.755","Text":"u(x) and t is a solution of the following problem."},{"Start":"00:07.755 ","End":"00:09.390","Text":"This is a pd,"},{"Start":"00:09.390 ","End":"00:13.320","Text":"it\u0027s a wave equation on an infinite interval with"},{"Start":"00:13.320 ","End":"00:18.675","Text":"initial conditions for u(x) naught and u_t(x_naught)."},{"Start":"00:18.675 ","End":"00:21.525","Text":"We have to find the function u,"},{"Start":"00:21.525 ","End":"00:28.020","Text":"and to compute the integral from 0-2 of u(x,1)dx."},{"Start":"00:28.020 ","End":"00:31.695","Text":"Let\u0027s start by noting a is 1,"},{"Start":"00:31.695 ","End":"00:33.440","Text":"f is this function,"},{"Start":"00:33.440 ","End":"00:35.705","Text":"and g is 0."},{"Start":"00:35.705 ","End":"00:40.220","Text":"We use d\u0027Alembert\u0027s formula, which is this."},{"Start":"00:40.220 ","End":"00:41.990","Text":"But because g is 0,"},{"Start":"00:41.990 ","End":"00:44.405","Text":"the second part drops off."},{"Start":"00:44.405 ","End":"00:45.950","Text":"We\u0027re left with this."},{"Start":"00:45.950 ","End":"00:50.300","Text":"Now a is 1, so this is equal to the following,"},{"Start":"00:50.300 ","End":"00:56.240","Text":"also taking the 1/2 out front of each 1 separately and u(x) and 1,"},{"Start":"00:56.240 ","End":"00:59.195","Text":"will get by putting t=1 here and here."},{"Start":"00:59.195 ","End":"01:01.130","Text":"Now f(x) plus 1,"},{"Start":"01:01.130 ","End":"01:02.945","Text":"is equal to this,"},{"Start":"01:02.945 ","End":"01:07.165","Text":"just replacing x by x plus 1 here and here."},{"Start":"01:07.165 ","End":"01:10.220","Text":"Set of absolute value of x plus 1 less than 1,"},{"Start":"01:10.220 ","End":"01:14.330","Text":"we can write it as x between minus 2 and 0."},{"Start":"01:14.330 ","End":"01:17.630","Text":"F(x) minus 1 similarly,"},{"Start":"01:17.630 ","End":"01:22.580","Text":"and this reduces to x between 0 and 2."},{"Start":"01:22.580 ","End":"01:27.170","Text":"Now, we can write what u is in terms of 3 cases;"},{"Start":"01:27.170 ","End":"01:29.000","Text":"x between minus 2 and 0,"},{"Start":"01:29.000 ","End":"01:32.845","Text":"x between 0 and 2 and otherwise else."},{"Start":"01:32.845 ","End":"01:35.165","Text":"This is what we get the 1/2,"},{"Start":"01:35.165 ","End":"01:36.830","Text":"which is here and here."},{"Start":"01:36.830 ","End":"01:39.995","Text":"Between minus 2 and 0,"},{"Start":"01:39.995 ","End":"01:44.380","Text":"we get this from here and the 0 from here."},{"Start":"01:44.380 ","End":"01:46.950","Text":"Between 0 and 2,"},{"Start":"01:46.950 ","End":"01:51.450","Text":"we get this from here and 0 from here, so this."},{"Start":"01:51.450 ","End":"01:54.080","Text":"When both of these are false,"},{"Start":"01:54.080 ","End":"01:56.705","Text":"then it\u0027s 0 plus 0 is 0."},{"Start":"01:56.705 ","End":"01:58.865","Text":"That\u0027s the answer to part a."},{"Start":"01:58.865 ","End":"02:01.060","Text":"Let\u0027s move on to part b."},{"Start":"02:01.060 ","End":"02:03.925","Text":"This is the answer to part a,"},{"Start":"02:03.925 ","End":"02:06.025","Text":"and we want to compute this integral."},{"Start":"02:06.025 ","End":"02:09.130","Text":"Note that the integral is from 0-2."},{"Start":"02:09.130 ","End":"02:12.295","Text":"This is irrelevant, and this is irrelevant."},{"Start":"02:12.295 ","End":"02:16.925","Text":"We\u0027re just going to take our value of u from here."},{"Start":"02:16.925 ","End":"02:22.685","Text":"What we get is the integral of 1 minus x minus 1^2,"},{"Start":"02:22.685 ","End":"02:24.519","Text":"which you can split up into 2 parts,"},{"Start":"02:24.519 ","End":"02:28.810","Text":"the one separately and the x minus 1^2 separately with a minus here."},{"Start":"02:28.810 ","End":"02:30.865","Text":"The integral of 1 is x."},{"Start":"02:30.865 ","End":"02:35.000","Text":"The integral of x minus 1^2 is a third x minus 1^3,"},{"Start":"02:35.000 ","End":"02:39.055","Text":"and each of these we have to evaluate between 0 and 2."},{"Start":"02:39.055 ","End":"02:44.730","Text":"Here we get 1/2 of 2 minus a 1/2(0) is 1."},{"Start":"02:44.730 ","End":"02:49.650","Text":"Here we get minus a 6th of 2 minus 1^3,"},{"Start":"02:49.650 ","End":"02:52.350","Text":"take away 0 minus 1^3."},{"Start":"02:52.350 ","End":"02:56.700","Text":"This is 1^3 is 1,"},{"Start":"02:56.700 ","End":"03:03.450","Text":"and this is minus 1^3 is minus 1.1 minus minus 1 is 2."},{"Start":"03:03.450 ","End":"03:05.090","Text":"2 over 6 is a 1/3."},{"Start":"03:05.090 ","End":"03:08.465","Text":"1 minus a 1/3 is 2/3."},{"Start":"03:08.465 ","End":"03:11.970","Text":"That\u0027s the answer, and we\u0027re done."}],"ID":30719},{"Watched":false,"Name":"Exercise 9","Duration":"6m 58s","ChapterTopicVideoID":29179,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.795","Text":"In this exercise, we\u0027re going to solve the following problem,"},{"Start":"00:03.795 ","End":"00:10.500","Text":"which is a wave equation on an infinite interval with initial conditions."},{"Start":"00:10.500 ","End":"00:17.235","Text":"This is not homogeneous because of this term here."},{"Start":"00:17.235 ","End":"00:21.960","Text":"This is what we call big F of x and"},{"Start":"00:21.960 ","End":"00:27.965","Text":"t. We\u0027re given a hint useful formula that we might want to use."},{"Start":"00:27.965 ","End":"00:30.670","Text":"You\u0027ll see that we use it twice."},{"Start":"00:30.670 ","End":"00:36.555","Text":"We\u0027ll start by identifying the parts like in the tutorial."},{"Start":"00:36.555 ","End":"00:40.365","Text":"a is 1 because u_tt is a^2,"},{"Start":"00:40.365 ","End":"00:42.480","Text":"u_xx, a^2 is 1,"},{"Start":"00:42.480 ","End":"00:45.285","Text":"then a is 1 because a is positive."},{"Start":"00:45.285 ","End":"00:47.790","Text":"Then f(x) is what\u0027s here,"},{"Start":"00:47.790 ","End":"00:53.830","Text":"that\u0027s 0. g(x) is cosine x. F(x,"},{"Start":"00:53.990 ","End":"00:56.190","Text":"t) is what\u0027s here,"},{"Start":"00:56.190 ","End":"00:59.175","Text":"that\u0027s the t cosine x."},{"Start":"00:59.175 ","End":"01:03.805","Text":"Now we have d\u0027Alembert\u0027s\u0027 formula."},{"Start":"01:03.805 ","End":"01:07.565","Text":"It\u0027s a French name and we don\u0027t pronounce the t,"},{"Start":"01:07.565 ","End":"01:11.600","Text":"d\u0027Alembert, never mind."},{"Start":"01:11.600 ","End":"01:22.380","Text":"In our case, we get that all this part is 0 because f is 0."},{"Start":"01:22.380 ","End":"01:31.125","Text":"Then here, we substitute g(s) equals cosine s, a=1."},{"Start":"01:31.125 ","End":"01:34.155","Text":"I forgot to replace the a."},{"Start":"01:34.155 ","End":"01:36.750","Text":"The a just drops off actually."},{"Start":"01:36.750 ","End":"01:44.040","Text":"For this part, a is 1, so take 2."},{"Start":"01:44.040 ","End":"01:47.985","Text":"For the second integral, we have here,"},{"Start":"01:47.985 ","End":"01:52.150","Text":"when a is 1, x plus t minus Tau here,"},{"Start":"01:52.150 ","End":"01:54.170","Text":"x minus t minus Tau."},{"Start":"01:54.170 ","End":"02:00.325","Text":"The a here is 1 and f is what\u0027s here."},{"Start":"02:00.325 ","End":"02:07.710","Text":"We get, when we replace x and t with s and Tau,"},{"Start":"02:07.710 ","End":"02:13.145","Text":"we get Tau cosine s. Now,"},{"Start":"02:13.145 ","End":"02:17.360","Text":"the integral of cosine is sine."},{"Start":"02:17.360 ","End":"02:22.610","Text":"Here we have sine s with the limits x minus t,"},{"Start":"02:22.610 ","End":"02:25.980","Text":"x plus t. Here,"},{"Start":"02:25.980 ","End":"02:30.125","Text":"we\u0027ll do the integration with respect to s first."},{"Start":"02:30.125 ","End":"02:33.980","Text":"t is like a constant and the integral of cosine x is"},{"Start":"02:33.980 ","End":"02:39.450","Text":"sine s. These are the limits for the sine."},{"Start":"02:41.960 ","End":"02:45.030","Text":"Then we get from here,"},{"Start":"02:45.030 ","End":"02:50.615","Text":"a 1/2 sine of x plus t minus sine of x minus t plus a 1/2,"},{"Start":"02:50.615 ","End":"02:53.435","Text":"the integral from 0-t."},{"Start":"02:53.435 ","End":"02:58.470","Text":"The Tau stays here."},{"Start":"02:58.470 ","End":"03:02.420","Text":"Then this sine is evaluated between these two"},{"Start":"03:02.420 ","End":"03:08.360","Text":"subtracted get sine of the upper minus sine of the lower here."},{"Start":"03:08.360 ","End":"03:15.880","Text":"Now, twice, we have sine of something minus sine of something."},{"Start":"03:15.880 ","End":"03:21.380","Text":"We have sine of a plus b minus sine of a minus b. I mean"},{"Start":"03:21.380 ","End":"03:26.930","Text":"if it\u0027s the template and here also sine of a plus b minus sine of a minus b."},{"Start":"03:26.930 ","End":"03:33.475","Text":"That\u0027s because we\u0027re going to use the formula that we had above."},{"Start":"03:33.475 ","End":"03:36.340","Text":"I\u0027ll give it again here."},{"Start":"03:36.650 ","End":"03:42.035","Text":"Different a and b here and here just to show you how to use the formula."},{"Start":"03:42.035 ","End":"03:44.875","Text":"This is a reminder of what it is."},{"Start":"03:44.875 ","End":"03:47.374","Text":"As I said, we\u0027ll use it twice."},{"Start":"03:47.374 ","End":"03:51.380","Text":"Here we get cosine a sine b,"},{"Start":"03:51.380 ","End":"03:56.720","Text":"which is cosine x sine t. In this part,"},{"Start":"03:56.720 ","End":"03:58.915","Text":"this is a, and this is b."},{"Start":"03:58.915 ","End":"04:02.085","Text":"We have sine a plus b minus sine a minus b."},{"Start":"04:02.085 ","End":"04:05.945","Text":"We get cosine of a,"},{"Start":"04:05.945 ","End":"04:08.700","Text":"which is cosine x, sine of b,"},{"Start":"04:08.700 ","End":"04:11.040","Text":"which is sine of t minus Tau."},{"Start":"04:11.040 ","End":"04:17.435","Text":"The Tau just stays the 1/2 gets swallowed into the formula here,"},{"Start":"04:17.435 ","End":"04:19.795","Text":"and the 1/2 here also disappears."},{"Start":"04:19.795 ","End":"04:26.050","Text":"We can bring the cosine x in front of the integral because this is d Tau."},{"Start":"04:26.050 ","End":"04:30.995","Text":"This integral can be done using integration by parts."},{"Start":"04:30.995 ","End":"04:32.945","Text":"This is the formula,"},{"Start":"04:32.945 ","End":"04:34.535","Text":"1 of its variants."},{"Start":"04:34.535 ","End":"04:39.445","Text":"In this case, we\u0027ll take u equals Tau,"},{"Start":"04:39.445 ","End":"04:44.220","Text":"and v\u0027 is sine of t minus Tau."},{"Start":"04:44.220 ","End":"04:47.130","Text":"We should have replaced the x here with Tau,"},{"Start":"04:47.130 ","End":"04:48.885","Text":"but you get the idea."},{"Start":"04:48.885 ","End":"04:51.600","Text":"What we get is here u,"},{"Start":"04:51.600 ","End":"04:55.110","Text":"v, this is u, this is v\u0027,"},{"Start":"04:55.110 ","End":"04:58.310","Text":"so v is the same thing,"},{"Start":"04:58.310 ","End":"05:03.500","Text":"but with cosine, actually the integral of sine is minus cosine."},{"Start":"05:03.500 ","End":"05:09.245","Text":"But we also get a minus 1 from the inner derivative and the minus is canceled."},{"Start":"05:09.245 ","End":"05:12.260","Text":"This is okay as is."},{"Start":"05:12.260 ","End":"05:14.840","Text":"We\u0027ll substitute Tau equals 0,"},{"Start":"05:14.840 ","End":"05:16.925","Text":"Tau equals t, and subtract."},{"Start":"05:16.925 ","End":"05:24.440","Text":"Here, u\u0027v is 1 times the cosine of t minus Tau."},{"Start":"05:24.440 ","End":"05:25.490","Text":"It should be minus,"},{"Start":"05:25.490 ","End":"05:28.355","Text":"and this is also be a minus from the minus Tau,"},{"Start":"05:28.355 ","End":"05:30.785","Text":"but the 2 minuses cancel again."},{"Start":"05:30.785 ","End":"05:38.510","Text":"Continuing, what we get here is that when Tau equals t,"},{"Start":"05:38.510 ","End":"05:44.480","Text":"and here we get t times cosine of 0,"},{"Start":"05:44.480 ","End":"05:46.895","Text":"which is just t,"},{"Start":"05:46.895 ","End":"05:48.425","Text":"That\u0027s this t here."},{"Start":"05:48.425 ","End":"05:52.830","Text":"When Tau is 0, then the whole thing is 0."},{"Start":"05:52.830 ","End":"05:55.130","Text":"That\u0027s the t minus 0."},{"Start":"05:55.130 ","End":"05:56.585","Text":"Now we have this,"},{"Start":"05:56.585 ","End":"05:59.090","Text":"the integral of cosine is sine,"},{"Start":"05:59.090 ","End":"06:04.975","Text":"but there\u0027s also a minus 1 from the inner derivative which cancels with this minus,"},{"Start":"06:04.975 ","End":"06:09.165","Text":"so it makes it plus sine of t minus Tau."},{"Start":"06:09.165 ","End":"06:13.700","Text":"Also, we have to substitute the upper and lower limits."},{"Start":"06:13.700 ","End":"06:18.650","Text":"Tau equals 0, Tau equals t. When Tau equals t,"},{"Start":"06:18.650 ","End":"06:23.110","Text":"then sine 0, which is 0."},{"Start":"06:23.110 ","End":"06:25.445","Text":"When Tau equals 0,"},{"Start":"06:25.445 ","End":"06:27.545","Text":"we get sine t,"},{"Start":"06:27.545 ","End":"06:31.250","Text":"but we subtract it because it\u0027s the lower limit of integration."},{"Start":"06:31.250 ","End":"06:33.410","Text":"It\u0027s getting simpler, now,"},{"Start":"06:33.410 ","End":"06:39.560","Text":"cosine x with t is t cosine x and cosine x with minus"},{"Start":"06:39.560 ","End":"06:47.345","Text":"sine t is minus cosine x sine t. Note that this is exactly the same as this,"},{"Start":"06:47.345 ","End":"06:48.620","Text":"but this is with a minus,"},{"Start":"06:48.620 ","End":"06:50.660","Text":"so this cancels with this."},{"Start":"06:50.660 ","End":"06:54.085","Text":"All we\u0027re left with is t cosine x."},{"Start":"06:54.085 ","End":"06:56.205","Text":"That\u0027s the answer."},{"Start":"06:56.205 ","End":"06:59.500","Text":"That completes this exercise."}],"ID":30720},{"Watched":false,"Name":"Exercise 10","Duration":"4m 48s","ChapterTopicVideoID":29180,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"In this exercise, we have a wave equation"},{"Start":"00:03.060 ","End":"00:09.600","Text":"non-homogeneous on an infinite interval and these are the initial conditions."},{"Start":"00:09.600 ","End":"00:13.500","Text":"I\u0027ll start by identifying the usual part,"},{"Start":"00:13.500 ","End":"00:17.910","Text":"usual notation a is 1 that\u0027s here, u_tt is a^2,"},{"Start":"00:17.910 ","End":"00:22.425","Text":"u_xx of a is 1, f is this,"},{"Start":"00:22.425 ","End":"00:24.480","Text":"g is this,"},{"Start":"00:24.480 ","End":"00:28.395","Text":"and big f(x) and t is this part here."},{"Start":"00:28.395 ","End":"00:30.780","Text":"Minus 2e^minus x sin(t)."},{"Start":"00:30.780 ","End":"00:33.510","Text":"We use the d\u0027Alembert formula,"},{"Start":"00:33.510 ","End":"00:35.340","Text":"which in general is this."},{"Start":"00:35.340 ","End":"00:39.180","Text":"In our case, it comes out to be the following."},{"Start":"00:39.180 ","End":"00:40.815","Text":"That if a equals 1,"},{"Start":"00:40.815 ","End":"00:44.745","Text":"here and here, g(s) equals 1,"},{"Start":"00:44.745 ","End":"00:48.400","Text":"and f(s) and Tau is just like this,"},{"Start":"00:48.400 ","End":"00:52.960","Text":"but with x replaced by s and t replaced by Tau."},{"Start":"00:52.960 ","End":"00:58.855","Text":"That\u0027s here. Simplify this using the following trigonometric identity."},{"Start":"00:58.855 ","End":"01:06.760","Text":"Here, a is x plus t and b is x minus t. So a plus b/2 will"},{"Start":"01:06.760 ","End":"01:14.795","Text":"be x and a minus b/2 will be t. That gives us sin(x) cos(t)."},{"Start":"01:14.795 ","End":"01:20.095","Text":"Now here the integral of 1 ds is just s and we get this minus this,"},{"Start":"01:20.095 ","End":"01:26.090","Text":"which is 2t/2 is t. The t is this integral."},{"Start":"01:26.090 ","End":"01:28.495","Text":"Now we have this double integral."},{"Start":"01:28.495 ","End":"01:32.494","Text":"Let\u0027s start by doing the inner integral."},{"Start":"01:32.494 ","End":"01:37.475","Text":"We can also take sin Tau in front of this integral because the integral is ds."},{"Start":"01:37.475 ","End":"01:42.680","Text":"We have e^minus s and the 2 cancels with the half,"},{"Start":"01:42.680 ","End":"01:45.635","Text":"and we take the minus out in front."},{"Start":"01:45.635 ","End":"01:49.175","Text":"Now the integral of e^minus s is minus"},{"Start":"01:49.175 ","End":"01:55.850","Text":"e^minus s. The minus will cancel with this minus gives us a plus here."},{"Start":"01:55.850 ","End":"01:59.480","Text":"It\u0027s e to the power of minus what\u0027s this is,"},{"Start":"01:59.480 ","End":"02:02.950","Text":"takeaway e to the power of what this is."},{"Start":"02:02.950 ","End":"02:06.395","Text":"Break this up into 2 integrals."},{"Start":"02:06.395 ","End":"02:08.585","Text":"Integral of this minus the integral of this."},{"Start":"02:08.585 ","End":"02:12.455","Text":"But we can also take in front of the integral what doesn\u0027t depend on Tau."},{"Start":"02:12.455 ","End":"02:18.680","Text":"Here we have e^minus x minus t. Here we have e^minus x plus t,"},{"Start":"02:18.680 ","End":"02:21.005","Text":"so we get the following,"},{"Start":"02:21.005 ","End":"02:25.730","Text":"this minus this, and these are tricky integrals."},{"Start":"02:25.730 ","End":"02:29.135","Text":"Anyway, I won\u0027t get into the technical details."},{"Start":"02:29.135 ","End":"02:30.755","Text":"I\u0027ll just give you the answers."},{"Start":"02:30.755 ","End":"02:34.745","Text":"This integral, the indefinite integral of this is this,"},{"Start":"02:34.745 ","End":"02:37.925","Text":"and the indefinite integral of this is this."},{"Start":"02:37.925 ","End":"02:43.040","Text":"Let\u0027s just take that on trust or look it up in Table of Integrals or do it yourself."},{"Start":"02:43.040 ","End":"02:46.370","Text":"Of course, in our case, we have to replace all the x\u0027s here by"},{"Start":"02:46.370 ","End":"02:50.880","Text":"Tau because we have the integral with respect to Tau."},{"Start":"02:50.990 ","End":"02:55.970","Text":"What we get just using these formulas,"},{"Start":"02:55.970 ","End":"02:57.994","Text":"this gives us this,"},{"Start":"02:57.994 ","End":"03:00.650","Text":"and we need to evaluate between 0 and"},{"Start":"03:00.650 ","End":"03:05.180","Text":"t. The other one using this formula gives us this, again,"},{"Start":"03:05.180 ","End":"03:11.295","Text":"evaluated between 0 and t. What we get in the first one,"},{"Start":"03:11.295 ","End":"03:13.275","Text":"if Tau equals t,"},{"Start":"03:13.275 ","End":"03:17.760","Text":"then we get sin(t) minus cos(t)."},{"Start":"03:17.760 ","End":"03:20.010","Text":"If you let Tau equals 0,"},{"Start":"03:20.010 ","End":"03:25.750","Text":"we get 0 minus 1."},{"Start":"03:25.750 ","End":"03:27.830","Text":"But this is subtracted because it\u0027s the lower limit,"},{"Start":"03:27.830 ","End":"03:29.620","Text":"so it becomes plus 1."},{"Start":"03:29.620 ","End":"03:36.200","Text":"Here similarly we get T and then you substitute 0 and we get 0 plus 1,"},{"Start":"03:36.200 ","End":"03:37.640","Text":"but it\u0027s minus 1."},{"Start":"03:37.640 ","End":"03:42.140","Text":"Next, we multiply e^minus t by this brackets."},{"Start":"03:42.140 ","End":"03:45.995","Text":"e^minus t times e^t is 1 that just disappears."},{"Start":"03:45.995 ","End":"03:49.685","Text":"But here we get plus e^minus t,"},{"Start":"03:49.685 ","End":"03:52.520","Text":"and a similar thing with the last term."},{"Start":"03:52.520 ","End":"03:56.390","Text":"Now, here and here we have a half e^minus x."},{"Start":"03:56.390 ","End":"04:00.845","Text":"Let\u0027s just add what\u0027s in this brackets to what\u0027s in this brackets."},{"Start":"04:00.845 ","End":"04:04.145","Text":"Sin(t) plus sin(t) is 2 sin(t)."},{"Start":"04:04.145 ","End":"04:12.005","Text":"Cos(t) cancels, and the e^minus t minus e^t stays."},{"Start":"04:12.005 ","End":"04:17.615","Text":"What we get 2 sin(t) plus e^minus t minus e^t."},{"Start":"04:17.615 ","End":"04:19.760","Text":"We can simplify it a bit more."},{"Start":"04:19.760 ","End":"04:24.680","Text":"The 2 with the half cancels and we get e^minus x sin(t)."},{"Start":"04:24.680 ","End":"04:29.050","Text":"Here, this minus this over 2."},{"Start":"04:29.050 ","End":"04:32.530","Text":"We can use the formula for sine hyperbolic."},{"Start":"04:32.530 ","End":"04:34.940","Text":"We have sine hyperbolic here,"},{"Start":"04:34.940 ","End":"04:36.895","Text":"except that it\u0027s in the reverse order."},{"Start":"04:36.895 ","End":"04:44.237","Text":"It\u0027s minus e^minus x times hyperbolic t. This is about as simple as it gets."},{"Start":"04:44.237 ","End":"04:46.757","Text":"This is the final answer."},{"Start":"04:46.757 ","End":"04:48.960","Text":"We are done."}],"ID":30721},{"Watched":false,"Name":"Exercise 11a","Duration":"3m 39s","ChapterTopicVideoID":29181,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.370","Text":"This exercise has two parts,"},{"Start":"00:02.370 ","End":"00:05.220","Text":"so read each one and then solve it."},{"Start":"00:05.220 ","End":"00:09.840","Text":"Part A is to show that the general solution to the wave equation,"},{"Start":"00:09.840 ","End":"00:11.775","Text":"utt equals uxx,"},{"Start":"00:11.775 ","End":"00:16.620","Text":"is u of x t equals some function of x plus t plus another"},{"Start":"00:16.620 ","End":"00:22.080","Text":"function of x minus t. We\u0027ll start working on part a."},{"Start":"00:22.080 ","End":"00:26.865","Text":"This equation can be written in the form uxx minus utt equals 0,"},{"Start":"00:26.865 ","End":"00:33.645","Text":"and then we can use the technique of a previous section where we first classify it."},{"Start":"00:33.645 ","End":"00:37.590","Text":"We take A11, A12, A22."},{"Start":"00:37.590 ","End":"00:43.725","Text":"There is no A12 because that\u0027s the coefficient of uxt, and that\u0027s 0."},{"Start":"00:43.725 ","End":"00:50.339","Text":"Then we get the characteristic ordinary differential equation using this formula,"},{"Start":"00:50.339 ","End":"00:55.100","Text":"and substituting everything, we get 0 plus or minus the square"},{"Start":"00:55.100 ","End":"01:00.645","Text":"root of 0^2-1 times minus 1, so it\u0027s 1."},{"Start":"01:00.645 ","End":"01:02.250","Text":"Now we get two solutions,"},{"Start":"01:02.250 ","End":"01:03.930","Text":"1 and minus 1."},{"Start":"01:03.930 ","End":"01:07.775","Text":"That gives us two characteristic equations."},{"Start":"01:07.775 ","End":"01:10.910","Text":"Rewriting dx by dt as x prime."},{"Start":"01:10.910 ","End":"01:14.600","Text":"This gives us x minus t is a constant,"},{"Start":"01:14.600 ","End":"01:17.320","Text":"and this gives us x plus t is a constant."},{"Start":"01:17.320 ","End":"01:22.760","Text":"Then we substitute psi equals this one and eta equals this one."},{"Start":"01:22.760 ","End":"01:25.280","Text":"We will need the partial derivatives,"},{"Start":"01:25.280 ","End":"01:27.820","Text":"the first order derivatives of these,"},{"Start":"01:27.820 ","End":"01:29.780","Text":"and from these we can just note that"},{"Start":"01:29.780 ","End":"01:33.980","Text":"the determinant of the Jacobian is non-zero, so we\u0027re okay."},{"Start":"01:33.980 ","End":"01:38.750","Text":"The second-order derivatives are all 0 because these are all constants."},{"Start":"01:38.750 ","End":"01:45.680","Text":"Now substitute and get from u to w. What we have again from a previous section,"},{"Start":"01:45.680 ","End":"01:48.140","Text":"we get these two formulas for"},{"Start":"01:48.140 ","End":"01:54.034","Text":"the second-order ugt and uxx in terms of partial derivatives of w,"},{"Start":"01:54.034 ","End":"01:55.235","Text":"this is what we get."},{"Start":"01:55.235 ","End":"01:57.875","Text":"Then we know what psi t,"},{"Start":"01:57.875 ","End":"02:01.505","Text":"eta t and all those are, we\u0027ve got them here."},{"Start":"02:01.505 ","End":"02:04.250","Text":"Using these four, we have the following."},{"Start":"02:04.250 ","End":"02:08.840","Text":"Here also these two are 0, the second-order ones."},{"Start":"02:08.840 ","End":"02:10.790","Text":"Here we have either 1 or minus 1."},{"Start":"02:10.790 ","End":"02:16.790","Text":"Anyway, we have the equation uxx minus utt equals 0, and substituting."},{"Start":"02:16.790 ","End":"02:20.075","Text":"Here we have W psi psi,"},{"Start":"02:20.075 ","End":"02:21.935","Text":"minus 1 squared is 1."},{"Start":"02:21.935 ","End":"02:28.320","Text":"Then we have 2 times W psi eta plus 1 times W eta eta,"},{"Start":"02:28.320 ","End":"02:31.460","Text":"that\u0027s this and the other one comes out to be this."},{"Start":"02:31.460 ","End":"02:35.305","Text":"If we subtract everything cancels except the W psi eta,"},{"Start":"02:35.305 ","End":"02:37.530","Text":"2 minus minus 2 is 4."},{"Start":"02:37.530 ","End":"02:39.995","Text":"That gives us the W psi eta,"},{"Start":"02:39.995 ","End":"02:42.140","Text":"the mixed second-order derivative is 0."},{"Start":"02:42.140 ","End":"02:44.210","Text":"We\u0027ll integrate once with respect to Eta,"},{"Start":"02:44.210 ","End":"02:45.935","Text":"and then with respect to psi,"},{"Start":"02:45.935 ","End":"02:53.660","Text":"integrate this with respect to eta and we get W psi as a constant as far as eta goes,"},{"Start":"02:53.660 ","End":"02:56.225","Text":"but a function of psi actually."},{"Start":"02:56.225 ","End":"03:00.095","Text":"Integrate with respect to psi and we get, well,"},{"Start":"03:00.095 ","End":"03:05.090","Text":"let\u0027s say that the indefinite integral of F is little"},{"Start":"03:05.090 ","End":"03:10.595","Text":"f and we get f of psi and then an arbitrary function of eta."},{"Start":"03:10.595 ","End":"03:12.650","Text":"Now we\u0027ve returned from w to u,"},{"Start":"03:12.650 ","End":"03:14.210","Text":"remembering that u of xy is"},{"Start":"03:14.210 ","End":"03:19.900","Text":"W psi and eta and replace psi equals x minus d eta equals x plus t."},{"Start":"03:19.900 ","End":"03:24.030","Text":"What we get is u of xy is f of x minus t"},{"Start":"03:24.030 ","End":"03:28.040","Text":"plus g of x plus t. That\u0027s not quite what was requested."},{"Start":"03:28.040 ","End":"03:30.455","Text":"You got the minus and the plus the wrong way round."},{"Start":"03:30.455 ","End":"03:36.545","Text":"Well, never mind, just switched the labeling f and g and get it the way we wanted."},{"Start":"03:36.545 ","End":"03:40.080","Text":"Okay. That concludes this exercise."}],"ID":30722},{"Watched":false,"Name":"Exercise 11b","Duration":"2m 25s","ChapterTopicVideoID":29163,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.000","Text":"Now we come to part b of the exercise."},{"Start":"00:03.000 ","End":"00:09.625","Text":"In part a, we showed that the solution is the following."},{"Start":"00:09.625 ","End":"00:14.048","Text":"Now let\u0027s take the first of the two conditions, this one,"},{"Start":"00:14.048 ","End":"00:16.740","Text":"and replacing u with its definition,"},{"Start":"00:16.740 ","End":"00:21.064","Text":"we get that a(t) is u(t, t)."},{"Start":"00:21.064 ","End":"00:22.760","Text":"Letting x equal t here,"},{"Start":"00:22.760 ","End":"00:26.805","Text":"we get f(2t) plus g(0)."},{"Start":"00:26.805 ","End":"00:28.995","Text":"If we let t equals 0,"},{"Start":"00:28.995 ","End":"00:32.360","Text":"here we get f(0) plus g(0)."},{"Start":"00:32.360 ","End":"00:34.185","Text":"Here we get a(0),"},{"Start":"00:34.185 ","End":"00:36.705","Text":"which were given to be 0."},{"Start":"00:36.705 ","End":"00:38.550","Text":"We\u0027ll need this later."},{"Start":"00:38.550 ","End":"00:42.390","Text":"Now, if we replace in this line,"},{"Start":"00:42.390 ","End":"00:44.505","Text":"s equals to t,"},{"Start":"00:44.505 ","End":"00:48.300","Text":"then we get that f(s) is a(t),"},{"Start":"00:48.300 ","End":"00:51.700","Text":"which is s over 2 minus g(0)."},{"Start":"00:51.700 ","End":"00:54.425","Text":"Next, we take this equation here,"},{"Start":"00:54.425 ","End":"00:56.470","Text":"the other thing that we were given,"},{"Start":"00:56.470 ","End":"00:59.125","Text":"u(minus t, t) is b(t)."},{"Start":"00:59.125 ","End":"01:01.390","Text":"We know that u(minus t,"},{"Start":"01:01.390 ","End":"01:09.890","Text":"t) is this when you replace x with minus t. We get f(0) plus g(minus 2t)."},{"Start":"01:10.790 ","End":"01:14.795","Text":"If we replace s equals minus 2t,"},{"Start":"01:14.795 ","End":"01:17.365","Text":"then we get the following."},{"Start":"01:17.365 ","End":"01:22.405","Text":"We get g(s) is b(t) is minus s over 2."},{"Start":"01:22.405 ","End":"01:25.445","Text":"Then the f(0) is to the other side."},{"Start":"01:25.445 ","End":"01:30.475","Text":"We have g(s) and we have f(s)."},{"Start":"01:30.475 ","End":"01:35.680","Text":"Now we can substitute these in the solution u(x, t)."},{"Start":"01:35.680 ","End":"01:37.960","Text":"What we get is u(x,"},{"Start":"01:37.960 ","End":"01:41.410","Text":"t) equals f(x) plus t,"},{"Start":"01:41.410 ","End":"01:44.350","Text":"which comes out to be using this formula,"},{"Start":"01:44.350 ","End":"01:49.165","Text":"a(x) plus t over 2 minus g(0) and g(x minus t),"},{"Start":"01:49.165 ","End":"01:53.530","Text":"using this comes out to b(minus x minus t over 2) minus f(0)."},{"Start":"01:53.530 ","End":"01:58.105","Text":"Collecting terms, getting the right order of things,"},{"Start":"01:58.105 ","End":"02:00.295","Text":"we get this plus this,"},{"Start":"02:00.295 ","End":"02:02.605","Text":"minus this, plus this."},{"Start":"02:02.605 ","End":"02:06.814","Text":"We know that f(0) plus g(0) is 0."},{"Start":"02:06.814 ","End":"02:12.215","Text":"Also, we can get rid of minus by switching the order here, t minus x."},{"Start":"02:12.215 ","End":"02:14.270","Text":"Just make it look nice,"},{"Start":"02:14.270 ","End":"02:16.370","Text":"we\u0027ll change it here also."},{"Start":"02:16.370 ","End":"02:17.780","Text":"We get that u(x,"},{"Start":"02:17.780 ","End":"02:21.620","Text":"t) is a(t plus x over 2) plus b(t minus x over 2),"},{"Start":"02:21.620 ","End":"02:25.890","Text":"and that\u0027s the answer to part b, and we\u0027re done."}],"ID":30723},{"Watched":false,"Name":"Exercise 12","Duration":"5m 39s","ChapterTopicVideoID":29164,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.470","Text":"In this exercise, we\u0027re given the following problem to solve."},{"Start":"00:04.470 ","End":"00:06.450","Text":"It\u0027s a wave equation,"},{"Start":"00:06.450 ","End":"00:13.245","Text":"but not the usual one because we have these two elements that don\u0027t belong."},{"Start":"00:13.245 ","End":"00:19.920","Text":"The usual wave equation for which we can use the d\u0027Alembert\u0027s formula is this."},{"Start":"00:19.920 ","End":"00:24.809","Text":"But we are given a hint to make the following substitution,"},{"Start":"00:24.809 ","End":"00:28.115","Text":"and we\u0027ll find what Alpha and Beta are later."},{"Start":"00:28.115 ","End":"00:30.960","Text":"The idea is that this substitution will bring us to"},{"Start":"00:30.960 ","End":"00:34.395","Text":"a standard wave equation that we\u0027re familiar with."},{"Start":"00:34.395 ","End":"00:37.350","Text":"Given this, we can write u in terms of v by"},{"Start":"00:37.350 ","End":"00:41.745","Text":"bringing the exponential to the other side and we get it with a plus."},{"Start":"00:41.745 ","End":"00:43.610","Text":"Then we can differentiate."},{"Start":"00:43.610 ","End":"00:46.025","Text":"We need to find u_t,"},{"Start":"00:46.025 ","End":"00:47.780","Text":"u _t _t,"},{"Start":"00:47.780 ","End":"00:49.315","Text":"and u _x _x."},{"Start":"00:49.315 ","End":"00:52.640","Text":"U _t, using derivative of a product,"},{"Start":"00:52.640 ","End":"00:55.610","Text":"derivative with respect to t we get this,"},{"Start":"00:55.610 ","End":"00:58.715","Text":"then differentiate again with respect to t,"},{"Start":"00:58.715 ","End":"01:03.110","Text":"this gives us these two terms and this gives us this and this."},{"Start":"01:03.110 ","End":"01:05.950","Text":"But the two middle ones are the same."},{"Start":"01:05.950 ","End":"01:11.585","Text":"We get this after we take out e to the Alpha x plus Beta t outside the brackets."},{"Start":"01:11.585 ","End":"01:13.895","Text":"Now we still need u_x_x."},{"Start":"01:13.895 ","End":"01:17.000","Text":"It\u0027s just going to be the same as u_t_t except"},{"Start":"01:17.000 ","End":"01:20.735","Text":"that instead of t we have x and instead of Beta, we have Alpha."},{"Start":"01:20.735 ","End":"01:22.375","Text":"We get this one."},{"Start":"01:22.375 ","End":"01:27.155","Text":"Then we go back to this equation which we had up here."},{"Start":"01:27.155 ","End":"01:28.760","Text":"Now we can replace u_t _t,"},{"Start":"01:28.760 ","End":"01:32.230","Text":"u_x _x, u_t and u itself."},{"Start":"01:32.230 ","End":"01:39.855","Text":"What we get is u_t_t from here equals u_x_x from here,"},{"Start":"01:39.855 ","End":"01:44.760","Text":"minus 2_u _t from here,"},{"Start":"01:44.760 ","End":"01:47.460","Text":"and then minus u,"},{"Start":"01:47.460 ","End":"01:50.910","Text":"which is this v_e to the, yeah."},{"Start":"01:50.910 ","End":"01:55.940","Text":"You can divide everything by this e to the power of 2_p is everywhere,"},{"Start":"01:55.940 ","End":"01:58.865","Text":"in every term so just cancels out."},{"Start":"01:58.865 ","End":"02:00.755","Text":"It\u0027s non-zero, of course."},{"Start":"02:00.755 ","End":"02:03.680","Text":"Then let\u0027s collect together like terms,"},{"Start":"02:03.680 ","End":"02:06.455","Text":"which I\u0027ve color-coded to make it easier."},{"Start":"02:06.455 ","End":"02:11.000","Text":"What we get is v_t_t and everything else on the other side."},{"Start":"02:11.000 ","End":"02:14.580","Text":"V_x we have just here, two Alpha."},{"Start":"02:14.580 ","End":"02:18.600","Text":"V_t we have here and here, that\u0027s this,"},{"Start":"02:18.600 ","End":"02:21.230","Text":"and then just v we have here,"},{"Start":"02:21.230 ","End":"02:23.630","Text":"here, here, and here."},{"Start":"02:23.630 ","End":"02:24.935","Text":"That gives us this."},{"Start":"02:24.935 ","End":"02:27.160","Text":"Now we\u0027re free to choose Alpha and Beta."},{"Start":"02:27.160 ","End":"02:30.995","Text":"What we\u0027ll do is make this 0 and this 0"},{"Start":"02:30.995 ","End":"02:35.795","Text":"by choosing Alpha equals 0 and Beta equals minus 1."},{"Start":"02:35.795 ","End":"02:40.220","Text":"Luckily, this one also turns out to be 0."},{"Start":"02:40.220 ","End":"02:42.950","Text":"If you substitute Alpha equals 0,"},{"Start":"02:42.950 ","End":"02:45.995","Text":"Beta equals 1 here, we get 0."},{"Start":"02:45.995 ","End":"02:50.650","Text":"Everything here is 0 except for the v_t_t equals v_x_x,"},{"Start":"02:50.650 ","End":"02:54.330","Text":"and we still need the initial conditions."},{"Start":"02:54.330 ","End":"02:59.600","Text":"Remember that v equals u times e to the minus this."},{"Start":"02:59.600 ","End":"03:02.390","Text":"Then we can replace Alpha equals 0,"},{"Start":"03:02.390 ","End":"03:03.815","Text":"Beta equals minus 1."},{"Start":"03:03.815 ","End":"03:09.200","Text":"This just becomes e to the t. V is u_e to"},{"Start":"03:09.200 ","End":"03:14.645","Text":"the t. Now we have an initial condition on u(x) naught equal sin(x)."},{"Start":"03:14.645 ","End":"03:16.985","Text":"Put t equals 0 here,"},{"Start":"03:16.985 ","End":"03:22.610","Text":"and we get v (x) naught equals sin(x) e to the 0."},{"Start":"03:22.610 ","End":"03:25.620","Text":"Yeah, just sin(x)."},{"Start":"03:25.620 ","End":"03:28.340","Text":"Now differentiate with respect to t,"},{"Start":"03:28.340 ","End":"03:30.470","Text":"and we get routine."},{"Start":"03:30.470 ","End":"03:35.705","Text":"The following product rule and differentiate with respect to t,"},{"Start":"03:35.705 ","End":"03:39.020","Text":"this derivative times this,"},{"Start":"03:39.020 ","End":"03:42.390","Text":"plus this times this derivative."},{"Start":"03:42.500 ","End":"03:48.620","Text":"Now we\u0027ll use the other initial condition on u_t (x) naught is naught."},{"Start":"03:48.620 ","End":"03:51.695","Text":"Again, put t equals 0 here,"},{"Start":"03:51.695 ","End":"03:55.600","Text":"and we get v (x) 0 equals sin(x)."},{"Start":"03:55.600 ","End":"04:04.055","Text":"Notice that v (x) naught is sin(x) and v_t (x) naught equals sin(x)."},{"Start":"04:04.055 ","End":"04:06.815","Text":"And also, we need this."},{"Start":"04:06.815 ","End":"04:09.635","Text":"Now if we take all these together,"},{"Start":"04:09.635 ","End":"04:12.530","Text":"we have the following problem involving v,"},{"Start":"04:12.530 ","End":"04:16.535","Text":"v_ t_t equals v_x _x with initial conditions these."},{"Start":"04:16.535 ","End":"04:19.355","Text":"Now, this is the standard form of the heat equation."},{"Start":"04:19.355 ","End":"04:24.750","Text":"We can apply d\u0027Alembert\u0027s formula to get the following."},{"Start":"04:26.090 ","End":"04:32.685","Text":"Well, we have a rule for sin(a) plus b minus sin(a) minus b,"},{"Start":"04:32.685 ","End":"04:35.700","Text":"and let a equals x and b equals t,"},{"Start":"04:35.700 ","End":"04:38.775","Text":"so this gives us sin(x) cosine t,"},{"Start":"04:38.775 ","End":"04:41.915","Text":"and here the integral of sine is minus cosine,"},{"Start":"04:41.915 ","End":"04:49.140","Text":"which we have to take from x minus t to x plus t. This becomes cosine of x plus t"},{"Start":"04:49.140 ","End":"04:52.610","Text":"minus cosine of x minus t. We\u0027ll use"},{"Start":"04:52.610 ","End":"04:57.305","Text":"another trigonometric identity for cos(a) minus cos(b)."},{"Start":"04:57.305 ","End":"05:00.665","Text":"This time this will be a, this is b."},{"Start":"05:00.665 ","End":"05:03.912","Text":"A plus b^2 is going to be x,"},{"Start":"05:03.912 ","End":"05:10.490","Text":"and a minus b^2 is going to be t. We get the following."},{"Start":"05:10.490 ","End":"05:13.970","Text":"The minus also cancels with the minus."},{"Start":"05:13.970 ","End":"05:17.720","Text":"We get plus 2 sin(x), sin(t)."},{"Start":"05:17.720 ","End":"05:23.555","Text":"Now we can take sin(x) out of the brackets and we have cos(t) plus 2 sin(t)."},{"Start":"05:23.555 ","End":"05:24.990","Text":"This is the solution for v,"},{"Start":"05:24.990 ","End":"05:32.975","Text":"but we want to get back to u. V is u_e to the t. We get that u is v_e to the minus t,"},{"Start":"05:32.975 ","End":"05:39.480","Text":"which is this e to the minus t. This is the answer. We\u0027re done."}],"ID":30724},{"Watched":false,"Name":"Exercise 13","Duration":"9m 20s","ChapterTopicVideoID":29165,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.690","Text":"In this exercise, we have to solve the following problem,"},{"Start":"00:03.690 ","End":"00:08.760","Text":"which is a wave equation but it\u0027s not our standard wave equation."},{"Start":"00:08.760 ","End":"00:14.955","Text":"What we\u0027re used to looks like this and there are several differences."},{"Start":"00:14.955 ","End":"00:18.720","Text":"For example, here, t is bigger or equal to 0,"},{"Start":"00:18.720 ","End":"00:23.280","Text":"whereas here t has conditional definitions not the same."},{"Start":"00:23.280 ","End":"00:29.850","Text":"Also, u(x, 0) is one function for all x and here, u(x,"},{"Start":"00:29.850 ","End":"00:34.890","Text":"0) is one thing if x is bigger or equal to 0 and something,"},{"Start":"00:34.890 ","End":"00:38.070","Text":"well isn\u0027t even given as u(x, 0)."},{"Start":"00:38.070 ","End":"00:41.520","Text":"We have u of x minus 3/2x so for various reasons,"},{"Start":"00:41.520 ","End":"00:47.660","Text":"it\u0027s not our standard form and so we can\u0027t use the d\u0027Alembert formula."},{"Start":"00:47.660 ","End":"00:52.970","Text":"However, we are given a hint that the general solution of such an equation,"},{"Start":"00:52.970 ","End":"00:55.955","Text":"regardless of initial and boundary conditions,"},{"Start":"00:55.955 ","End":"00:58.310","Text":"is of this form, u(x,"},{"Start":"00:58.310 ","End":"01:01.880","Text":"t) is f of x minus at plus g of x plus at."},{"Start":"01:01.880 ","End":"01:06.920","Text":"Well, in our case, a=1 and so this is the general solution to"},{"Start":"01:06.920 ","End":"01:13.970","Text":"the equation and what we have left is to find f and g. We\u0027ll do it in stages or steps."},{"Start":"01:13.970 ","End":"01:18.695","Text":"We take first the case where x is bigger or equal to 0."},{"Start":"01:18.695 ","End":"01:23.175","Text":"That\u0027s this and we have u(x,"},{"Start":"01:23.175 ","End":"01:28.640","Text":"0) equals sine x when x is bigger or equal to 0 and in this case,"},{"Start":"01:28.640 ","End":"01:30.785","Text":"if we let t equals 0 here,"},{"Start":"01:30.785 ","End":"01:35.545","Text":"we get u(x, 0) equals f(x) plus g(x)."},{"Start":"01:35.545 ","End":"01:37.170","Text":"On the other hand, u(x,"},{"Start":"01:37.170 ","End":"01:39.540","Text":"0) is sine x so equating the two we get,"},{"Start":"01:39.540 ","End":"01:41.620","Text":"sine x is f(x) plus g(x)."},{"Start":"01:41.620 ","End":"01:44.650","Text":"This is still for x bigger or equal to 0."},{"Start":"01:44.650 ","End":"01:49.610","Text":"Now du by dt by differentiating this,"},{"Start":"01:49.610 ","End":"01:51.440","Text":"we get the following."},{"Start":"01:51.440 ","End":"01:54.560","Text":"It\u0027s just putting a prime on each of f and"},{"Start":"01:54.560 ","End":"01:57.905","Text":"g and also there\u0027s an internal derivative which is here minus 1"},{"Start":"01:57.905 ","End":"02:05.810","Text":"and here just 1 so we have that minus here and we\u0027re given that du by dt(x,"},{"Start":"02:05.810 ","End":"02:13.740","Text":"0) is sine x and so if we put t equals 0 here,"},{"Start":"02:13.740 ","End":"02:19.620","Text":"we get that sine x is minus f\u0027(x) because"},{"Start":"02:19.620 ","End":"02:25.970","Text":"t=0 plus g\u0027(x) and then if we integrate this with respect to x,"},{"Start":"02:25.970 ","End":"02:32.270","Text":"we get minus cosine x is minus f(x) plus g(x) plus constant of integration."},{"Start":"02:32.270 ","End":"02:34.805","Text":"Copying this equation here,"},{"Start":"02:34.805 ","End":"02:37.330","Text":"we have the following two."},{"Start":"02:37.330 ","End":"02:39.250","Text":"Adding these two,"},{"Start":"02:39.250 ","End":"02:41.570","Text":"we get sine x minus cosine x."},{"Start":"02:41.570 ","End":"02:43.835","Text":"The f(x)s cancel out."},{"Start":"02:43.835 ","End":"02:49.785","Text":"We get 2g(x) plus c. If we subtract this,"},{"Start":"02:49.785 ","End":"02:50.910","Text":"take away this,"},{"Start":"02:50.910 ","End":"02:52.880","Text":"we get sine x minus minus cosine x,"},{"Start":"02:52.880 ","End":"02:55.310","Text":"which is sine of x plus cosine x."},{"Start":"02:55.310 ","End":"02:56.850","Text":"The gs cancel."},{"Start":"02:56.850 ","End":"03:01.415","Text":"f minus minus f is 2f and it\u0027s minus c. Now,"},{"Start":"03:01.415 ","End":"03:04.445","Text":"from here we can extract g and from here we can extract"},{"Start":"03:04.445 ","End":"03:07.500","Text":"f. Bring the c over to the other side,"},{"Start":"03:07.500 ","End":"03:10.200","Text":"divide by 2 and here also bring this c to the other side,"},{"Start":"03:10.200 ","End":"03:12.185","Text":"divide by 2, switch sides."},{"Start":"03:12.185 ","End":"03:18.325","Text":"This is what we get for f and g on the condition that x is bigger or equal to 0."},{"Start":"03:18.325 ","End":"03:23.630","Text":"Let\u0027s see if we can get f and g for x less than 0."},{"Start":"03:23.630 ","End":"03:26.825","Text":"Well, let\u0027s try at least and see what happen."},{"Start":"03:26.825 ","End":"03:30.260","Text":"In Stage 2 where x is negative,"},{"Start":"03:30.260 ","End":"03:32.405","Text":"go back and see."},{"Start":"03:32.405 ","End":"03:39.830","Text":"We have this, that u of x minus 3/2 x is 0 when x is negative so we have this,"},{"Start":"03:39.830 ","End":"03:46.745","Text":"and if we substitute these in this equation where t is minus 3/2x,"},{"Start":"03:46.745 ","End":"03:52.145","Text":"we get f of x minus t. This minus this is 5/2x."},{"Start":"03:52.145 ","End":"03:58.035","Text":"This plus this is minus 1/2x and that is equal to 0,"},{"Start":"03:58.035 ","End":"04:00.675","Text":"at least for x negative."},{"Start":"04:00.675 ","End":"04:02.970","Text":"Then we can make a substitution."},{"Start":"04:02.970 ","End":"04:05.655","Text":"If we let s equals minus 1/2x,"},{"Start":"04:05.655 ","End":"04:08.130","Text":"this is minus 5s."},{"Start":"04:08.130 ","End":"04:14.735","Text":"Also notice that x negative means that s is positive and vice versa."},{"Start":"04:14.735 ","End":"04:18.785","Text":"We get f of minus 5s,"},{"Start":"04:18.785 ","End":"04:26.400","Text":"which is this, plus g(s) or bring the g(s) to the other side so that equals minus g(s)."},{"Start":"04:26.400 ","End":"04:28.520","Text":"Since s is positive,"},{"Start":"04:28.520 ","End":"04:31.880","Text":"we can use the formula for g(s) from here."},{"Start":"04:31.880 ","End":"04:33.260","Text":"Here it\u0027s with x,"},{"Start":"04:33.260 ","End":"04:39.565","Text":"we can replace x by s because s is positive and we would get from here that"},{"Start":"04:39.565 ","End":"04:43.530","Text":"the left-hand side is still f(minus 5s) but the right-hand side"},{"Start":"04:43.530 ","End":"04:47.460","Text":"is g(s) with the sign reversed so just make this minus,"},{"Start":"04:47.460 ","End":"04:52.790","Text":"plus, plus and this is what we get with an s instead of the x."},{"Start":"04:52.790 ","End":"04:54.800","Text":"Now we can make another substitution."},{"Start":"04:54.800 ","End":"04:58.200","Text":"Let x equal minus 5s."},{"Start":"04:58.200 ","End":"05:02.030","Text":"So x is now back to being negative."},{"Start":"05:02.030 ","End":"05:05.370","Text":"Just substitute in this equation."},{"Start":"05:05.370 ","End":"05:10.410","Text":"Minus 5s is x and s is minus x over 5,"},{"Start":"05:10.410 ","End":"05:12.240","Text":"just the reverse of this."},{"Start":"05:12.240 ","End":"05:14.960","Text":"Or just substituting in here."},{"Start":"05:14.960 ","End":"05:18.470","Text":"This is what we get and small simplification."},{"Start":"05:18.470 ","End":"05:20.180","Text":"We know that sine is an odd function,"},{"Start":"05:20.180 ","End":"05:22.265","Text":"so this minus and this minus cancel."},{"Start":"05:22.265 ","End":"05:24.050","Text":"Also, cosine is an even function,"},{"Start":"05:24.050 ","End":"05:31.305","Text":"so this minus drops off so this is the equation we get for f(x) when x is negative."},{"Start":"05:31.305 ","End":"05:38.850","Text":"We have 3 out of the 4 and it turns out we don\u0027t need g(x) for x negative, as you\u0027ll see."},{"Start":"05:38.850 ","End":"05:44.735","Text":"Now, let\u0027s move to the next step where we summarize what we have and reach a result."},{"Start":"05:44.735 ","End":"05:51.979","Text":"We had this formula for u in terms of f and g and then we had 3 equations,"},{"Start":"05:51.979 ","End":"05:56.820","Text":"g(x) what it equals and f(x) what it equals for x positive,"},{"Start":"05:56.820 ","End":"06:01.710","Text":"and we also have f(x) for x negative and we want to find what u(x,"},{"Start":"06:01.710 ","End":"06:04.365","Text":"t) equals in all cases."},{"Start":"06:04.365 ","End":"06:09.890","Text":"We\u0027ll divide into two cases according to whether x minus t is positive or negative."},{"Start":"06:09.890 ","End":"06:11.915","Text":"First, the positive case,"},{"Start":"06:11.915 ","End":"06:16.530","Text":"which also means that x is bigger or equal to t and in any event,"},{"Start":"06:16.530 ","End":"06:18.600","Text":"t is bigger or equal to 0."},{"Start":"06:18.600 ","End":"06:21.075","Text":"We get that u(x,"},{"Start":"06:21.075 ","End":"06:24.780","Text":"t) is equal to f of x minus t,"},{"Start":"06:24.780 ","End":"06:29.810","Text":"but x minus t is positive so we copy it from here,"},{"Start":"06:29.810 ","End":"06:33.425","Text":"just replacing x by x minus t. Now,"},{"Start":"06:33.425 ","End":"06:39.530","Text":"c/2 cancels with minus c/2 and we can slightly rearrange this by taking 1/2 out"},{"Start":"06:39.530 ","End":"06:42.500","Text":"of the brackets here and here and what we"},{"Start":"06:42.500 ","End":"06:46.055","Text":"end up with is that we have the formula for u(x,"},{"Start":"06:46.055 ","End":"06:53.690","Text":"t) equals this but on condition that t is between 0 and x."},{"Start":"06:53.690 ","End":"06:56.255","Text":"Let\u0027s go now to Case 2,"},{"Start":"06:56.255 ","End":"07:00.365","Text":"where x minus t is negative."},{"Start":"07:00.365 ","End":"07:05.330","Text":"In this case, x is less than t. To go back to the very beginning of the problem,"},{"Start":"07:05.330 ","End":"07:07.640","Text":"we were given the restriction on t,"},{"Start":"07:07.640 ","End":"07:09.860","Text":"that t is bigger or equal to."},{"Start":"07:09.860 ","End":"07:14.550","Text":"This of x is negative and this of x is positive."},{"Start":"07:14.550 ","End":"07:16.790","Text":"In any event, t comes out to be bigger or equal to"},{"Start":"07:16.790 ","End":"07:20.089","Text":"0 because when x is negative, this is positive."},{"Start":"07:20.089 ","End":"07:25.430","Text":"From this inequality, we get that minus 2/3t is less than or equal"},{"Start":"07:25.430 ","End":"07:31.490","Text":"to x just by bringing the 3 over 2 to the other side, the minus."},{"Start":"07:31.490 ","End":"07:34.100","Text":"If we add t to both sides,"},{"Start":"07:34.100 ","End":"07:38.720","Text":"here we get a 1/3t is less than or equal to x plus"},{"Start":"07:38.720 ","End":"07:43.700","Text":"t so x plus t is bigger or equal to 1/3t and t is positive,"},{"Start":"07:43.700 ","End":"07:45.565","Text":"so this is positive."},{"Start":"07:45.565 ","End":"07:51.950","Text":"In the other case also x is positive and t is non-negative,"},{"Start":"07:51.950 ","End":"07:55.280","Text":"so x plus t is positive so in all of Case 2,"},{"Start":"07:55.280 ","End":"08:00.630","Text":"we have that x plus t is positive and that means that, let\u0027s just go back,"},{"Start":"08:00.630 ","End":"08:03.695","Text":"that we can compute g of x plus t,"},{"Start":"08:03.695 ","End":"08:09.170","Text":"even though we only have the case of g for positive argument but this is all we get."},{"Start":"08:09.170 ","End":"08:12.980","Text":"We have u(x, t) as f of x minus t plus g of x plus"},{"Start":"08:12.980 ","End":"08:16.535","Text":"t. x minus t is negative but from here,"},{"Start":"08:16.535 ","End":"08:18.635","Text":"x plus t is positive."},{"Start":"08:18.635 ","End":"08:21.010","Text":"We have all those formulas."},{"Start":"08:21.010 ","End":"08:25.335","Text":"We get from the x minus t negative."},{"Start":"08:25.335 ","End":"08:27.120","Text":"For f, we get this,"},{"Start":"08:27.120 ","End":"08:29.070","Text":"and for x plus t positive,"},{"Start":"08:29.070 ","End":"08:32.670","Text":"we got this for g. Just go back and see."},{"Start":"08:32.670 ","End":"08:37.490","Text":"These two cancel and we can just throw out the brackets and put them all together"},{"Start":"08:37.490 ","End":"08:43.195","Text":"and this holds for minus 2/3t is less than or equal to x,"},{"Start":"08:43.195 ","End":"08:47.260","Text":"and also x is less than t. Now,"},{"Start":"08:47.260 ","End":"08:51.890","Text":"we have this formula and this formula and we just have to"},{"Start":"08:51.890 ","End":"08:56.640","Text":"put them together in one formula and we have u(x,"},{"Start":"08:56.640 ","End":"09:00.525","Text":"t ) piecewise but in only 2 pieces."},{"Start":"09:00.525 ","End":"09:04.515","Text":"When x is in this range,"},{"Start":"09:04.515 ","End":"09:11.525","Text":"we get this expression and when t is between 0 and x,"},{"Start":"09:11.525 ","End":"09:13.835","Text":"we get the other piece,"},{"Start":"09:13.835 ","End":"09:16.730","Text":"and this is our u(x, t),"},{"Start":"09:16.730 ","End":"09:21.120","Text":"and that\u0027s what we wanted to find, and so we\u0027re done."}],"ID":30725},{"Watched":false,"Name":"Exercise 14","Duration":"4m 9s","ChapterTopicVideoID":29166,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.950","Text":"In this exercise, we have a wave equation on an infinite interval,"},{"Start":"00:04.950 ","End":"00:07.650","Text":"similar to the ones we\u0027ve done before."},{"Start":"00:07.650 ","End":"00:12.045","Text":"The setup is such that we can apply d\u0027Alembert\u0027s formula."},{"Start":"00:12.045 ","End":"00:14.295","Text":"Let\u0027s identify the parts."},{"Start":"00:14.295 ","End":"00:16.230","Text":"We have a, we have f(x),"},{"Start":"00:16.230 ","End":"00:17.835","Text":"we have g(x),"},{"Start":"00:17.835 ","End":"00:22.380","Text":"and we don\u0027t have the f(x,t) or we could say it\u0027s"},{"Start":"00:22.380 ","End":"00:28.545","Text":"0. d\u0027Alembert\u0027s formula gives us the following."},{"Start":"00:28.545 ","End":"00:32.130","Text":"In our case, f(x) is 0,"},{"Start":"00:32.130 ","End":"00:34.799","Text":"so this part comes out to be 0."},{"Start":"00:34.799 ","End":"00:37.700","Text":"G(s) is the absolute value of s,"},{"Start":"00:37.700 ","End":"00:44.940","Text":"so what we get is just this integral of absolute value of s on this interval."},{"Start":"00:44.940 ","End":"00:50.480","Text":"To remind you that absolute value of s is a piecewise defined function,"},{"Start":"00:50.480 ","End":"00:56.210","Text":"depends on whether the argument s is bigger or equal to 0 or less than or equal to 0."},{"Start":"00:56.210 ","End":"00:59.120","Text":"What it means is that we have to divide up into 3 cases."},{"Start":"00:59.120 ","End":"01:05.495","Text":"It really all depends whether 0 is in this interval or above it, or below it."},{"Start":"01:05.495 ","End":"01:07.010","Text":"Case a will be this,"},{"Start":"01:07.010 ","End":"01:09.035","Text":"and then we\u0027ll get a case b,"},{"Start":"01:09.035 ","End":"01:11.930","Text":"and case c. As you can see,"},{"Start":"01:11.930 ","End":"01:15.680","Text":"all depends on how 0 falls relative to this interval."},{"Start":"01:15.680 ","End":"01:19.610","Text":"In case a, which is this one,"},{"Start":"01:19.610 ","End":"01:25.580","Text":"then the absolute value of s is just equal to s. Because everything is positive here."},{"Start":"01:25.580 ","End":"01:27.395","Text":"This is the integral we get."},{"Start":"01:27.395 ","End":"01:30.050","Text":"The integral of s is 1/2s^2,"},{"Start":"01:30.050 ","End":"01:36.010","Text":"substituting the upper and lower limits and also combining the 1/2 with the 1/2 Alpha."},{"Start":"01:36.010 ","End":"01:40.955","Text":"This part minus this part comes out to be"},{"Start":"01:40.955 ","End":"01:47.350","Text":"2Alpha xt minus minus 2Alpha xt, so 4Alpha xt."},{"Start":"01:47.350 ","End":"01:49.290","Text":"When you divide by the 4Alpha,"},{"Start":"01:49.290 ","End":"01:51.465","Text":"we\u0027re just left with xt."},{"Start":"01:51.465 ","End":"01:55.025","Text":"That\u0027s case a. Let\u0027s do the last one."},{"Start":"01:55.025 ","End":"01:56.825","Text":"Let\u0027s do the easy ones first."},{"Start":"01:56.825 ","End":"02:06.630","Text":"Case c is here where the absolute value of s is going to be minus s. If it\u0027s minus s,"},{"Start":"02:07.040 ","End":"02:11.570","Text":"well, similar to here except that it\u0027s minus and instead of putting a minus,"},{"Start":"02:11.570 ","End":"02:15.170","Text":"we can just reverse the order of the terms."},{"Start":"02:15.170 ","End":"02:19.615","Text":"We\u0027ve got this minus this and it comes out to be minus xt."},{"Start":"02:19.615 ","End":"02:21.230","Text":"Now the middle case,"},{"Start":"02:21.230 ","End":"02:27.424","Text":"which is just slightly harder when we have these 2 inequalities,"},{"Start":"02:27.424 ","End":"02:30.020","Text":"which comes down to this inequality,"},{"Start":"02:30.020 ","End":"02:33.770","Text":"we get the integral in 2 parts."},{"Start":"02:33.770 ","End":"02:36.800","Text":"We have to take the part from here to here"},{"Start":"02:36.800 ","End":"02:42.095","Text":"separately because they\u0027re absolute value of s is minus s. From here to here,"},{"Start":"02:42.095 ","End":"02:47.570","Text":"absolute value of s is plus s. Here\u0027s what we end up with."},{"Start":"02:47.570 ","End":"02:50.940","Text":"Minus here and plus here."},{"Start":"02:53.810 ","End":"02:57.650","Text":"Well, they\u0027re both equal to 1/2s^2."},{"Start":"02:57.650 ","End":"03:03.545","Text":"In this case, we get 0^2 minus x minus Alpha t^2."},{"Start":"03:03.545 ","End":"03:08.125","Text":"Here, x plus Alpha t^2 minus 0^2."},{"Start":"03:08.125 ","End":"03:13.605","Text":"The 1/2 goes with the 1/2 Alpha to become 1/4Alpha in both cases."},{"Start":"03:13.605 ","End":"03:15.225","Text":"If we do this,"},{"Start":"03:15.225 ","End":"03:20.795","Text":"we get the minus with minus cancels and we just get x minus Alpha t^2."},{"Start":"03:20.795 ","End":"03:26.440","Text":"Here we get plus x plus Alpha t^2 and everything is over 4Alpha."},{"Start":"03:26.440 ","End":"03:31.984","Text":"This is equal to x^2 plus Alpha^2t^2."},{"Start":"03:31.984 ","End":"03:37.135","Text":"The middle part minus 2Alpha xt cancels out with plus 2Alpha xt."},{"Start":"03:37.135 ","End":"03:39.360","Text":"We just get the last term squared,"},{"Start":"03:39.360 ","End":"03:42.085","Text":"Alpha^2t^2 over 4Alpha,"},{"Start":"03:42.085 ","End":"03:46.040","Text":"which slightly simplifies it and divide top and bottom by 2 to this."},{"Start":"03:46.040 ","End":"03:47.774","Text":"Now we have 3 cases,"},{"Start":"03:47.774 ","End":"03:51.840","Text":"all we have to do is collect them together and summarize."},{"Start":"03:51.980 ","End":"04:01.100","Text":"The first case was this one where x is bigger than Alpha t, where we got xt."},{"Start":"04:01.100 ","End":"04:06.085","Text":"Then case c was this one and case b is this one."},{"Start":"04:06.085 ","End":"04:10.000","Text":"That\u0027s the answer and we\u0027re done."}],"ID":30726},{"Watched":false,"Name":"Exercise 15","Duration":"2m 33s","ChapterTopicVideoID":29167,"CourseChapterTopicPlaylistID":294442,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.580","Text":"In this exercise we have"},{"Start":"00:02.580 ","End":"00:07.170","Text":"a wave equation which we\u0027re going to solve with d\u0027Alembert\u0027s formula."},{"Start":"00:07.170 ","End":"00:10.380","Text":"This is the equation on an infinite interval,"},{"Start":"00:10.380 ","End":"00:12.810","Text":"but t is positive."},{"Start":"00:12.810 ","End":"00:17.986","Text":"We\u0027re given the initial conditions; u(x,0), t(x,0)."},{"Start":"00:17.986 ","End":"00:21.300","Text":"This is just a picture of u(x,0)."},{"Start":"00:21.300 ","End":"00:23.280","Text":"Let\u0027s identify the parts."},{"Start":"00:23.280 ","End":"00:25.470","Text":"We have a=1,"},{"Start":"00:25.470 ","End":"00:28.410","Text":"this is f(x),"},{"Start":"00:28.410 ","End":"00:34.130","Text":"g(x) is 0, and the formula is the following."},{"Start":"00:34.130 ","End":"00:38.645","Text":"The g(s) part is 0 because g is 0,"},{"Start":"00:38.645 ","End":"00:40.655","Text":"so we just have to compute this."},{"Start":"00:40.655 ","End":"00:42.710","Text":"Our f is defined piece-wise,"},{"Start":"00:42.710 ","End":"00:45.130","Text":"so it\u0027s a little bit messy."},{"Start":"00:45.130 ","End":"00:50.149","Text":"First of all plug a=1 and get rid of that,"},{"Start":"00:50.149 ","End":"00:51.710","Text":"so we have the following."},{"Start":"00:51.710 ","End":"00:54.860","Text":"Now we\u0027ll compute each of these separately."},{"Start":"00:54.860 ","End":"01:02.885","Text":"It start with the f(x) plus t. Just substituting x plus t instead of x here we get this,"},{"Start":"01:02.885 ","End":"01:07.120","Text":"and let\u0027s just rewrite the condition."},{"Start":"01:07.120 ","End":"01:11.150","Text":"This would be that x plus t is between 1 and minus 1,"},{"Start":"01:11.150 ","End":"01:14.990","Text":"and then subtract t and we get x between minus 1 minus t,"},{"Start":"01:14.990 ","End":"01:19.100","Text":"and 1 minus t. Similarly for f(x)"},{"Start":"01:19.100 ","End":"01:24.905","Text":"minus t again just substitute x minus t in place of x here,"},{"Start":"01:24.905 ","End":"01:26.795","Text":"and this is what we get."},{"Start":"01:26.795 ","End":"01:33.320","Text":"Again, rewriting the condition on x more explicitly,"},{"Start":"01:33.320 ","End":"01:36.155","Text":"x is between this and this."},{"Start":"01:36.155 ","End":"01:38.270","Text":"Now that we have these 2 pieces,"},{"Start":"01:38.270 ","End":"01:44.195","Text":"we can just add them and divide by 2 or divide each one by 2 separately and then add,"},{"Start":"01:44.195 ","End":"01:47.540","Text":"and what we get is the following answer."},{"Start":"01:47.540 ","End":"01:50.010","Text":"Get this over 2;"},{"Start":"01:50.010 ","End":"01:52.140","Text":"the 0 over 2 is just 0,"},{"Start":"01:52.140 ","End":"01:54.165","Text":"and similarly here and here."},{"Start":"01:54.165 ","End":"01:58.570","Text":"This is the answer even though it\u0027s not very intuitive."},{"Start":"01:58.570 ","End":"02:02.420","Text":"Perhaps a little clip will illustrate it."},{"Start":"02:02.420 ","End":"02:05.875","Text":"What this function looks like as a function of x,"},{"Start":"02:05.875 ","End":"02:08.470","Text":"and then vary t."},{"Start":"02:29.980 ","End":"02:33.450","Text":"That concludes this clip."}],"ID":30727}],"Thumbnail":null,"ID":294442},{"Name":"Semi-Infinite Interval","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Infinite Interval","Duration":"6m 33s","ChapterTopicVideoID":29156,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.195","Text":"Continuing with the wave equation,"},{"Start":"00:03.195 ","End":"00:06.194","Text":"we\u0027ve learned some techniques for an infinite interval."},{"Start":"00:06.194 ","End":"00:09.284","Text":"Now let\u0027s turn to a semi-infinite interval."},{"Start":"00:09.284 ","End":"00:14.355","Text":"It\u0027s a hyperbolic linear second-order PDE wave equation"},{"Start":"00:14.355 ","End":"00:17.655","Text":"with initial boundary conditions as follows."},{"Start":"00:17.655 ","End":"00:20.355","Text":"This is the PDE."},{"Start":"00:20.355 ","End":"00:22.920","Text":"These are the initial conditions."},{"Start":"00:22.920 ","End":"00:26.580","Text":"The interval is from 0 to infinity."},{"Start":"00:26.580 ","End":"00:30.945","Text":"I suppose it could be from minus infinity to 0 and it would work similarly."},{"Start":"00:30.945 ","End":"00:35.265","Text":"We\u0027ll just take it as the right semi-infinite interval."},{"Start":"00:35.265 ","End":"00:38.430","Text":"The boundary conditions is 2 choices,"},{"Start":"00:38.430 ","End":"00:41.150","Text":"it\u0027s either the Dirichlet boundary condition,"},{"Start":"00:41.150 ","End":"00:44.615","Text":"which is u(0,t)=0,"},{"Start":"00:44.615 ","End":"00:51.050","Text":"or the Neumann boundary condition du by dx (0,t)=0."},{"Start":"00:51.050 ","End":"00:58.300","Text":"These are the conditions on the positive y-axis or in this case t-axis."},{"Start":"00:58.300 ","End":"01:03.895","Text":"As usual, a is a parameter bigger than 0,"},{"Start":"01:03.895 ","End":"01:08.720","Text":"and IC means initial condition and BC is boundary condition."},{"Start":"01:08.720 ","End":"01:13.220","Text":"Often they get mixed up it\u0027s not that big of a distinction."},{"Start":"01:13.220 ","End":"01:16.700","Text":"Let\u0027s first deal with the Dirichlet boundary condition and"},{"Start":"01:16.700 ","End":"01:20.120","Text":"then we\u0027ll move on to the Neumann boundary condition."},{"Start":"01:20.120 ","End":"01:25.130","Text":"What we\u0027ll do is we\u0027ll extend the problem to the whole interval"},{"Start":"01:25.130 ","End":"01:30.865","Text":"minus infinity to infinity by extending our functions to odd functions."},{"Start":"01:30.865 ","End":"01:34.275","Text":"We\u0027ll call the extension of f, f tilde,"},{"Start":"01:34.275 ","End":"01:44.250","Text":"and this will be the same as f when x is positive and -f(-x) when x is negative."},{"Start":"01:44.250 ","End":"01:45.765","Text":"0 we\u0027ll let it be 0."},{"Start":"01:45.765 ","End":"01:51.830","Text":"Similarly with g, we extend it to g tilde and F,"},{"Start":"01:51.830 ","End":"01:53.765","Text":"we extend it to F tilde,"},{"Start":"01:53.765 ","End":"01:55.760","Text":"but it\u0027s going to be odd in x."},{"Start":"01:55.760 ","End":"01:57.860","Text":"Similar definition."},{"Start":"01:57.860 ","End":"02:02.105","Text":"Now we get the infinite interval problem in"},{"Start":"02:02.105 ","End":"02:08.945","Text":"u tilde looks the same as our original problem except with the tilde on every function."},{"Start":"02:08.945 ","End":"02:11.075","Text":"Now we\u0027re in familiar territory,"},{"Start":"02:11.075 ","End":"02:16.340","Text":"and we can use the d\u0027Alembert formula to find the solution u tilde."},{"Start":"02:16.340 ","End":"02:17.825","Text":"When we\u0027ve done that,"},{"Start":"02:17.825 ","End":"02:22.610","Text":"after we found u tilde we then just restrict it to x bigger or equal to 0,"},{"Start":"02:22.610 ","End":"02:25.550","Text":"the same function but restrict the domain."},{"Start":"02:25.550 ","End":"02:33.005","Text":"U tilde satisfies the PDE and the initial conditions for all x,"},{"Start":"02:33.005 ","End":"02:37.850","Text":"so u(x,t), the restriction satisfies the PDE,"},{"Start":"02:37.850 ","End":"02:40.880","Text":"and the initial conditions for x bigger or equal to 0."},{"Start":"02:40.880 ","End":"02:45.319","Text":"As 1 thing that still remains is the boundary condition."},{"Start":"02:45.319 ","End":"02:51.110","Text":"We have to show that u satisfies the Dirichlet boundary condition."},{"Start":"02:51.110 ","End":"02:57.080","Text":"To see that, we won\u0027t prove it but it\u0027s not hard to prove that u tilde is"},{"Start":"02:57.080 ","End":"02:59.360","Text":"an odd function in x because"},{"Start":"02:59.360 ","End":"03:03.410","Text":"all the other functions are odd turns out that u tilde is odd."},{"Start":"03:03.410 ","End":"03:10.670","Text":"In other words, ũ(-x,t) is -ũ(x,t) and since it\u0027s odd,"},{"Start":"03:10.670 ","End":"03:17.765","Text":"then at 0 it\u0027s equal to 0. u(x,t) also satisfies the Dirichlet boundary conditions."},{"Start":"03:17.765 ","End":"03:19.910","Text":"It satisfies all the conditions."},{"Start":"03:19.910 ","End":"03:21.800","Text":"The PDE, the initial conditions,"},{"Start":"03:21.800 ","End":"03:23.599","Text":"and the boundary conditions."},{"Start":"03:23.599 ","End":"03:31.040","Text":"Now the d\u0027Alembert formula gives us the solution for u tilde as follows."},{"Start":"03:31.040 ","End":"03:35.435","Text":"Therefore, we can just restrict x to being bigger or equal to 0,"},{"Start":"03:35.435 ","End":"03:37.970","Text":"and we get that u(x,t) is"},{"Start":"03:37.970 ","End":"03:43.145","Text":"the same expression just that the functions here have tildes on them."},{"Start":"03:43.145 ","End":"03:47.540","Text":"Just because x is bigger or equal to 0 doesn\u0027t mean that,"},{"Start":"03:47.540 ","End":"03:50.180","Text":"say x minus at will be bigger than 0."},{"Start":"03:50.180 ","End":"03:54.140","Text":"In fact, it could be negative so we need the tildes here we can\u0027t just drop them."},{"Start":"03:54.140 ","End":"03:57.905","Text":"Now to the second case Neumann\u0027s boundary condition,"},{"Start":"03:57.905 ","End":"04:02.060","Text":"where du by dx(0,t) is 0."},{"Start":"04:02.060 ","End":"04:07.685","Text":"This time instead of taking an odd extension we\u0027ll take an even extension."},{"Start":"04:07.685 ","End":"04:16.505","Text":"In other words we define f tilde and g tilde and F tilde as follows where the value at x,"},{"Start":"04:16.505 ","End":"04:19.190","Text":"when x is non-negative is as is,"},{"Start":"04:19.190 ","End":"04:20.854","Text":"and when x is negative,"},{"Start":"04:20.854 ","End":"04:25.800","Text":"we take the value of minus x which is bigger or equal to 0."},{"Start":"04:26.330 ","End":"04:31.189","Text":"Then we consider the infinite interval problem in u tilde."},{"Start":"04:31.189 ","End":"04:34.003","Text":"Just taking the original problem,"},{"Start":"04:34.003 ","End":"04:40.220","Text":"and replacing u with u tilde and leaving out the boundary condition meanwhile."},{"Start":"04:40.220 ","End":"04:44.510","Text":"This is our familiar problem with an infinite interval and so we"},{"Start":"04:44.510 ","End":"04:49.760","Text":"can use the d\u0027Alembert\u0027s formula and find a solution ũ(x,t)."},{"Start":"04:49.760 ","End":"04:52.160","Text":"When we have u tilde,"},{"Start":"04:52.160 ","End":"04:56.895","Text":"we can restrict it to x bigger or equal to 0 and call that u(x,t)."},{"Start":"04:56.895 ","End":"05:04.145","Text":"u(x,t) satisfies the PDE and the initial condition for x bigger or equal to 0,"},{"Start":"05:04.145 ","End":"05:07.010","Text":"because u tilde satisfies the condition and u is the"},{"Start":"05:07.010 ","End":"05:10.264","Text":"same as u tilde when x is bigger or equal to 0."},{"Start":"05:10.264 ","End":"05:14.195","Text":"Now, u tilde is an even function in x."},{"Start":"05:14.195 ","End":"05:15.620","Text":"That\u0027s not hard to show,"},{"Start":"05:15.620 ","End":"05:19.295","Text":"but we won\u0027t do it here because we make everything else even,"},{"Start":"05:19.295 ","End":"05:22.085","Text":"then the solution will also be even."},{"Start":"05:22.085 ","End":"05:26.900","Text":"Now, the derivative of an even function is an odd function."},{"Start":"05:26.900 ","End":"05:28.595","Text":"I found the proof,"},{"Start":"05:28.595 ","End":"05:31.655","Text":"try copy pasted here and you can study this if you"},{"Start":"05:31.655 ","End":"05:35.485","Text":"want to prove that this really is an odd function."},{"Start":"05:35.485 ","End":"05:39.125","Text":"Since u tilde x is an odd function,"},{"Start":"05:39.125 ","End":"05:42.610","Text":"then at 0 it\u0027s equal to 0,"},{"Start":"05:42.610 ","End":"05:47.450","Text":"and so u(x,t) satisfies the Neumann boundary condition as well."},{"Start":"05:47.450 ","End":"05:49.655","Text":"This is the Neumann boundary condition."},{"Start":"05:49.655 ","End":"05:55.370","Text":"Now the d\u0027Alembert formula gives us for u tilde the following."},{"Start":"05:55.370 ","End":"05:58.010","Text":"If x is bigger or equal to 0,"},{"Start":"05:58.010 ","End":"06:00.755","Text":"it gives us the formula for u(x,t)."},{"Start":"06:00.755 ","End":"06:05.660","Text":"It\u0027s important to keep the tildes here because even though x is bigger or equal to 0,"},{"Start":"06:05.660 ","End":"06:07.530","Text":"x minus at, for example,"},{"Start":"06:07.530 ","End":"06:09.120","Text":"could be less than 0,"},{"Start":"06:09.120 ","End":"06:10.865","Text":"so we need the tilde."},{"Start":"06:10.865 ","End":"06:17.120","Text":"Now in summary, if we have a Dirichlet boundary condition,"},{"Start":"06:17.120 ","End":"06:19.550","Text":"then we use an odd extension,"},{"Start":"06:19.550 ","End":"06:21.425","Text":"and a Neumann boundary condition,"},{"Start":"06:21.425 ","End":"06:24.794","Text":"we\u0027ll use an even extension."},{"Start":"06:24.794 ","End":"06:29.712","Text":"The Dirichlet problem or the extension Neumann even extension,"},{"Start":"06:29.712 ","End":"06:31.235","Text":"and that\u0027s basically it."},{"Start":"06:31.235 ","End":"06:33.750","Text":"That\u0027s the conclusion of this clip."}],"ID":30728},{"Watched":false,"Name":"Exercise 1","Duration":"4m 44s","ChapterTopicVideoID":29157,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.495","Text":"In this exercise, we\u0027re given the following problem,"},{"Start":"00:03.495 ","End":"00:09.300","Text":"which is a wave equation on a semi-infinite interval."},{"Start":"00:09.300 ","End":"00:12.690","Text":"These are the initial conditions and"},{"Start":"00:12.690 ","End":"00:17.445","Text":"the boundary condition is the Dirichlet boundary condition."},{"Start":"00:17.445 ","End":"00:23.265","Text":"Our task is to find u(x, 1/2)."},{"Start":"00:23.265 ","End":"00:24.600","Text":"Now as you may remember,"},{"Start":"00:24.600 ","End":"00:29.565","Text":"the Dirichlet boundary condition is one where we make an odd extension,"},{"Start":"00:29.565 ","End":"00:34.965","Text":"or odd completion of all the functions to minus infinity, infinity."},{"Start":"00:34.965 ","End":"00:37.455","Text":"Well, f(x) is 0,"},{"Start":"00:37.455 ","End":"00:40.110","Text":"it\u0027s extension is also 0,"},{"Start":"00:40.110 ","End":"00:44.760","Text":"but g(x) does need to be extended or"},{"Start":"00:44.760 ","End":"00:50.460","Text":"completed and it will become 1 when x is between 0 and 1,"},{"Start":"00:50.460 ","End":"00:56.230","Text":"and minus 1 when x is between minus 1 and 0 and 0 otherwise."},{"Start":"00:57.800 ","End":"01:00.230","Text":"Here\u0027s a picture."},{"Start":"01:00.230 ","End":"01:03.595","Text":"This was the original g,"},{"Start":"01:03.595 ","End":"01:05.900","Text":"1 here and 0 here."},{"Start":"01:05.900 ","End":"01:09.005","Text":"When we extend it to an odd function,"},{"Start":"01:09.005 ","End":"01:15.740","Text":"we make it minus 1 here and 0 here, rotated 180 degrees."},{"Start":"01:15.740 ","End":"01:17.614","Text":"Now the solution u(x,"},{"Start":"01:17.614 ","End":"01:19.265","Text":"t) using the formula,"},{"Start":"01:19.265 ","End":"01:22.190","Text":"the d\u0027Alembert formula, we get u tilde,"},{"Start":"01:22.190 ","End":"01:26.555","Text":"but then we restrict it to you when x is just non-negative."},{"Start":"01:26.555 ","End":"01:28.970","Text":"This is what we get."},{"Start":"01:28.970 ","End":"01:32.090","Text":"F is 0, so this part is 0."},{"Start":"01:32.090 ","End":"01:35.480","Text":"We just get this part also, a is 1."},{"Start":"01:35.480 ","End":"01:41.225","Text":"We\u0027d get 1/2 the integral from x minus t to x plus t of g tilde."},{"Start":"01:41.225 ","End":"01:47.550","Text":"Then u(x, 1/2) just means replace t by a half,"},{"Start":"01:47.550 ","End":"01:48.900","Text":"so x plus a half,"},{"Start":"01:48.900 ","End":"01:54.190","Text":"x minus a half and because g tilde is piece wise defined,"},{"Start":"01:54.190 ","End":"01:59.440","Text":"we\u0027ll have to divide up into cases depending on where x is."},{"Start":"01:59.440 ","End":"02:02.665","Text":"For example, x could be such that"},{"Start":"02:02.665 ","End":"02:05.800","Text":"all this interval from x minus a half to x plus the half is in"},{"Start":"02:05.800 ","End":"02:12.490","Text":"this bit or it could be from somewhere here to somewhere here or it could be like so."},{"Start":"02:12.490 ","End":"02:15.900","Text":"Then we\u0027ll just draw a diagram,"},{"Start":"02:15.900 ","End":"02:22.090","Text":"here it is, and we\u0027ll just divide it up according to the picture."},{"Start":"02:22.090 ","End":"02:32.005","Text":"In case 1, this is completely to the right of 1 so x minus a half is bigger than 1,"},{"Start":"02:32.005 ","End":"02:33.520","Text":"which you can write in other words,"},{"Start":"02:33.520 ","End":"02:35.350","Text":"as x bigger than one-and-a-half."},{"Start":"02:35.350 ","End":"02:38.350","Text":"In this case, u(x,"},{"Start":"02:38.350 ","End":"02:45.970","Text":"1/2) is the integral of g. G Is 0 over here."},{"Start":"02:45.970 ","End":"02:50.105","Text":"The answer comes out to be 0 that\u0027s an easy case."},{"Start":"02:50.105 ","End":"02:52.110","Text":"In case 2,"},{"Start":"02:52.110 ","End":"02:57.120","Text":"it overlaps between this part and this part,"},{"Start":"02:57.120 ","End":"02:59.890","Text":"so it\u0027s 1 here and 0 here."},{"Start":"02:59.890 ","End":"03:06.860","Text":"This comes out to be when x minus a half is between 0 and 1,"},{"Start":"03:06.860 ","End":"03:11.415","Text":"which means that x is between a half and one-and-a-half."},{"Start":"03:11.415 ","End":"03:19.030","Text":"In this case, we get the integral of 1 on this part from x minus a half to 1,"},{"Start":"03:19.030 ","End":"03:22.170","Text":"and from 1 to x plus a half it\u0027s 0,"},{"Start":"03:22.170 ","End":"03:23.895","Text":"so this part is 0."},{"Start":"03:23.895 ","End":"03:26.360","Text":"Here we just take the length of the interval,"},{"Start":"03:26.360 ","End":"03:29.645","Text":"which is 1 minus x plus a half,"},{"Start":"03:29.645 ","End":"03:32.005","Text":"and multiply it by a half."},{"Start":"03:32.005 ","End":"03:37.969","Text":"It comes out to be a half times one-and-a-half minus x."},{"Start":"03:37.969 ","End":"03:41.600","Text":"The third case in the picture would be like this."},{"Start":"03:41.600 ","End":"03:46.430","Text":"In this part it\u0027s equal to 1 and this part it\u0027s equal to minus 1."},{"Start":"03:46.430 ","End":"03:54.320","Text":"The part here would be from x minus a half to 0 and here from 0 to x plus a half,"},{"Start":"03:54.320 ","End":"03:57.560","Text":"so minus 1 here and 1 here."},{"Start":"03:57.560 ","End":"04:00.245","Text":"What we get is for this part,"},{"Start":"04:00.245 ","End":"04:02.824","Text":"minus 1 times the length of the interval,"},{"Start":"04:02.824 ","End":"04:05.930","Text":"which is 0 minus x minus a half,"},{"Start":"04:05.930 ","End":"04:07.835","Text":"which is minus x plus a half."},{"Start":"04:07.835 ","End":"04:11.270","Text":"Here we have one times the length of the interval,"},{"Start":"04:11.270 ","End":"04:15.785","Text":"which is just x plus a half minus 0 which is x plus a half."},{"Start":"04:15.785 ","End":"04:19.940","Text":"Doing the computations, the answer comes out to be x."},{"Start":"04:19.940 ","End":"04:24.155","Text":"Now if we combine this case with the previous two cases,"},{"Start":"04:24.155 ","End":"04:25.716","Text":"we get the following,"},{"Start":"04:25.716 ","End":"04:29.315","Text":"when x is bigger than one and a half it\u0027s 0."},{"Start":"04:29.315 ","End":"04:35.750","Text":"When x is between a half and one-and-a-half it\u0027s equal to a half times 3 over 2 minus x."},{"Start":"04:35.750 ","End":"04:41.175","Text":"Where it\u0027s between 0 and a half it\u0027s equal to x,"},{"Start":"04:41.175 ","End":"04:44.470","Text":"and that\u0027s the answer and we\u0027re done."}],"ID":30729},{"Watched":false,"Name":"Exercise 2","Duration":"2m 37s","ChapterTopicVideoID":29158,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:04.830","Text":"In this exercise, we have the following problem which is"},{"Start":"00:04.830 ","End":"00:08.820","Text":"a wave equation on a semi-infinite interval,"},{"Start":"00:08.820 ","End":"00:12.300","Text":"and these are the initial conditions."},{"Start":"00:12.300 ","End":"00:15.390","Text":"We have a boundary condition which if you"},{"Start":"00:15.390 ","End":"00:19.080","Text":"think about it is the Dirichlet boundary condition."},{"Start":"00:19.080 ","End":"00:23.370","Text":"The other one, the Neumann, has du/dx."},{"Start":"00:23.370 ","End":"00:27.075","Text":"If you remember when we have a Dirichlet boundary condition,"},{"Start":"00:27.075 ","End":"00:33.945","Text":"then we extend our function to an odd function on minus infinity to infinity."},{"Start":"00:33.945 ","End":"00:38.610","Text":"Now in this case both these functions are already odd functions,"},{"Start":"00:38.610 ","End":"00:41.660","Text":"so the extensions are just the same."},{"Start":"00:41.660 ","End":"00:43.910","Text":"F tilde is x, g tilde,"},{"Start":"00:43.910 ","End":"00:48.380","Text":"sine x, but on the whole real line."},{"Start":"00:48.380 ","End":"00:52.385","Text":"If you tilde desire extended solution,"},{"Start":"00:52.385 ","End":"00:57.605","Text":"then this is what it\u0027s equal to using the d\u0027Alembert formula,"},{"Start":"00:57.605 ","End":"00:59.495","Text":"and a is 1,"},{"Start":"00:59.495 ","End":"01:01.835","Text":"and also here and here."},{"Start":"01:01.835 ","End":"01:04.180","Text":"We get the following,"},{"Start":"01:04.180 ","End":"01:06.420","Text":"f(x) is x,"},{"Start":"01:06.420 ","End":"01:08.265","Text":"so here is just x plus t,"},{"Start":"01:08.265 ","End":"01:11.085","Text":"x minus t. Here,"},{"Start":"01:11.085 ","End":"01:14.940","Text":"g(s) is sine s. Then what we get,"},{"Start":"01:14.940 ","End":"01:19.005","Text":"this comes out to be just x, and this,"},{"Start":"01:19.005 ","End":"01:23.040","Text":"the integral of sine s is minus cosine s,"},{"Start":"01:23.040 ","End":"01:29.810","Text":"and we take it between x minus t and x plus t. Instead of the minus here,"},{"Start":"01:29.810 ","End":"01:38.660","Text":"we can invert the order and we just get cosine of this one minus cosine of this one."},{"Start":"01:38.660 ","End":"01:41.210","Text":"There\u0027s a trigonometric formula."},{"Start":"01:41.210 ","End":"01:43.420","Text":"Identity, the following,"},{"Start":"01:43.420 ","End":"01:45.335","Text":"and if we apply this,"},{"Start":"01:45.335 ","End":"01:50.360","Text":"where a is x minus t and b is x plus t,"},{"Start":"01:50.360 ","End":"01:57.015","Text":"then a plus b/2 is this plus this over 2, which is x."},{"Start":"01:57.015 ","End":"01:59.595","Text":"This minus this,"},{"Start":"01:59.595 ","End":"02:02.010","Text":"is minus 2t,"},{"Start":"02:02.010 ","End":"02:09.410","Text":"over 2 is minus t. There would be a minus t here,"},{"Start":"02:09.410 ","End":"02:12.380","Text":"except that there\u0027s a minus here."},{"Start":"02:12.380 ","End":"02:14.840","Text":"The 2, of course, cancels with the 1/2."},{"Start":"02:14.840 ","End":"02:16.310","Text":"Anyway, but think about it."},{"Start":"02:16.310 ","End":"02:19.270","Text":"Applying this formula, this is what we get."},{"Start":"02:19.270 ","End":"02:27.785","Text":"What we have to do now is restrict u tilde to non-negative x and just call it u,"},{"Start":"02:27.785 ","End":"02:32.450","Text":"so we get that u(x,t) is x plus sine x,"},{"Start":"02:32.450 ","End":"02:37.600","Text":"sine t. That concludes this exercise."}],"ID":30730},{"Watched":false,"Name":"Exercise 3","Duration":"6m 37s","ChapterTopicVideoID":29159,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.800","Text":"In this exercise, we have to solve this wave equation."},{"Start":"00:04.800 ","End":"00:08.205","Text":"This is the PDE part,"},{"Start":"00:08.205 ","End":"00:12.180","Text":"and it\u0027s on a semi-infinite interval."},{"Start":"00:12.180 ","End":"00:16.200","Text":"These are the initial conditions."},{"Start":"00:16.200 ","End":"00:18.390","Text":"What happens when t is 0?"},{"Start":"00:18.390 ","End":"00:21.765","Text":"We\u0027re given u and du by dt,"},{"Start":"00:21.765 ","End":"00:24.720","Text":"and because it\u0027s a semi-infinite interval,"},{"Start":"00:24.720 ","End":"00:29.790","Text":"we also need a boundary condition of what happens when x is 0."},{"Start":"00:29.790 ","End":"00:35.880","Text":"This is like the positive t-axis or y-axis and we\u0027re given this."},{"Start":"00:35.880 ","End":"00:40.645","Text":"There\u0027s 2 variations which we can have in this kind of problem."},{"Start":"00:40.645 ","End":"00:45.740","Text":"We either given u of 0t or ux of 0t,"},{"Start":"00:45.740 ","End":"00:47.155","Text":"u of 0t,"},{"Start":"00:47.155 ","End":"00:51.410","Text":"it\u0027s the Dirichlet boundary condition and in this case,"},{"Start":"00:51.410 ","End":"00:54.370","Text":"we use the odd extension."},{"Start":"00:54.370 ","End":"00:59.210","Text":"If we use an odd extension and get an odd solution,"},{"Start":"00:59.210 ","End":"01:07.100","Text":"the odd solution to just this part without the boundary condition since it\u0027s odd,"},{"Start":"01:07.100 ","End":"01:11.135","Text":"it will automatically satisfy that u of 0,"},{"Start":"01:11.135 ","End":"01:13.220","Text":"and whatever is 0."},{"Start":"01:13.220 ","End":"01:15.705","Text":"An odd function at 0 is 0."},{"Start":"01:15.705 ","End":"01:21.455","Text":"We\u0027re going to do the odd completion or extension to the whole real line."},{"Start":"01:21.455 ","End":"01:26.165","Text":"As usual, we denote with a tilde the completion."},{"Start":"01:26.165 ","End":"01:34.475","Text":"Now, g tilde is also sine x just on a larger domain because g is already odd."},{"Start":"01:34.475 ","End":"01:37.490","Text":"I mean, if you put minus x,"},{"Start":"01:37.490 ","End":"01:40.400","Text":"that of x you get the minus in front."},{"Start":"01:40.400 ","End":"01:44.780","Text":"But x^2 is not an odd function."},{"Start":"01:44.780 ","End":"01:48.860","Text":"We have to use this formula for getting the odd extension."},{"Start":"01:48.860 ","End":"01:57.095","Text":"It becomes x^2 when x is positive but minus x^2 when x is negative,"},{"Start":"01:57.095 ","End":"02:01.235","Text":"it\u0027s actually minus of minus x^2."},{"Start":"02:01.235 ","End":"02:03.955","Text":"Minus x^2 is x^2,"},{"Start":"02:03.955 ","End":"02:08.855","Text":"and there\u0027s a picture of this odd completion of x^2,"},{"Start":"02:08.855 ","End":"02:12.305","Text":"is x^2 here and minus x^2 here."},{"Start":"02:12.305 ","End":"02:20.330","Text":"Then we apply this formula of Dalenberg and we get this expression for u tilde,"},{"Start":"02:20.330 ","End":"02:23.330","Text":"which is the odd extension of u,"},{"Start":"02:23.330 ","End":"02:25.480","Text":"and let\u0027s evaluate it."},{"Start":"02:25.480 ","End":"02:27.720","Text":"Note that a is 1,"},{"Start":"02:27.720 ","End":"02:29.890","Text":"so here and here,"},{"Start":"02:29.890 ","End":"02:35.770","Text":"and here, and here we can replace a with 1 and get this."},{"Start":"02:35.770 ","End":"02:44.020","Text":"Also, replace g tilde of s with sine s. Really only care what happens when"},{"Start":"02:44.020 ","End":"02:47.770","Text":"x is bigger than 0 here because we\u0027re going to replace"},{"Start":"02:47.770 ","End":"02:52.615","Text":"u tilde by u for when x is positive."},{"Start":"02:52.615 ","End":"02:56.095","Text":"Let\u0027s look at each of these 2 pieces."},{"Start":"02:56.095 ","End":"03:01.525","Text":"F tilde of x plus t, using this formula,"},{"Start":"03:01.525 ","End":"03:05.290","Text":"is either x plus t^2 or minus x plus t^2,"},{"Start":"03:05.290 ","End":"03:07.915","Text":"depending on which of these 2 holds."},{"Start":"03:07.915 ","End":"03:12.815","Text":"But if we\u0027re going to only consider x bigger than 0,"},{"Start":"03:12.815 ","End":"03:17.390","Text":"then x plus t is even bigger than x,"},{"Start":"03:17.390 ","End":"03:18.830","Text":"so certainly bigger than 0."},{"Start":"03:18.830 ","End":"03:20.750","Text":"This we can rule out,"},{"Start":"03:20.750 ","End":"03:29.745","Text":"so f tilde of x plus t is always x plus t^2 if x is bigger than 0."},{"Start":"03:29.745 ","End":"03:33.075","Text":"Now f tilde of x minus t,"},{"Start":"03:33.075 ","End":"03:35.220","Text":"again using this formula,"},{"Start":"03:35.220 ","End":"03:38.470","Text":"it\u0027s either x minus t^2 or minus x minus t^2,"},{"Start":"03:38.470 ","End":"03:41.900","Text":"depending on whether x minus t is positive or negative."},{"Start":"03:41.900 ","End":"03:45.450","Text":"Which means depending on whether x is bigger than t,"},{"Start":"03:45.450 ","End":"03:50.375","Text":"or less than t. But we\u0027re considering x bigger than 0."},{"Start":"03:50.375 ","End":"03:54.840","Text":"The 2 cases we get are like this x"},{"Start":"03:54.840 ","End":"03:59.330","Text":"minus t^2 when x is bigger than t. But then the other case,"},{"Start":"03:59.330 ","End":"04:01.040","Text":"instead of just x less than t,"},{"Start":"04:01.040 ","End":"04:05.675","Text":"we\u0027re going to restrict it to being also positive because we\u0027re only,"},{"Start":"04:05.675 ","End":"04:09.420","Text":"like I said, interested in positive x."},{"Start":"04:09.670 ","End":"04:17.070","Text":"Now that we\u0027ve figured out f tilde of x plus t and of x minus t,"},{"Start":"04:17.070 ","End":"04:20.143","Text":"we can substitute them in here,"},{"Start":"04:20.143 ","End":"04:29.005","Text":"and what we\u0027ll get will be a conditional x bigger than t or x between 0 and t. In 1 case,"},{"Start":"04:29.005 ","End":"04:33.830","Text":"we take this and the other case we take this."},{"Start":"04:33.830 ","End":"04:35.855","Text":"But for x plus t,"},{"Start":"04:35.855 ","End":"04:37.685","Text":"we always take this."},{"Start":"04:37.685 ","End":"04:42.200","Text":"This x plus t^2 appears twice and then the condition is we either take x"},{"Start":"04:42.200 ","End":"04:47.615","Text":"minus t^2 or minus x minus t^2 depending on where x is."},{"Start":"04:47.615 ","End":"04:54.745","Text":"Also the integral of sine s is minus cosine s. Continuing,"},{"Start":"04:54.745 ","End":"05:03.260","Text":"bit of algebra shows that this is equal to x^2 plus t^2 and this is equal to 2xt."},{"Start":"05:03.260 ","End":"05:08.190","Text":"It\u0027s like 2xt minus minus 2xt over 2,"},{"Start":"05:08.190 ","End":"05:10.810","Text":"which is 4xt over 2 is 2xt."},{"Start":"05:10.810 ","End":"05:14.520","Text":"Here it\u0027s also the xt part cancels."},{"Start":"05:14.520 ","End":"05:19.065","Text":"We just get 2x^2 plus 2t^2 divided by 2 is this."},{"Start":"05:19.065 ","End":"05:25.475","Text":"For the cosine, we just substitute upper and lower limits and subtract."},{"Start":"05:25.475 ","End":"05:27.095","Text":"Just instead of u tilde,"},{"Start":"05:27.095 ","End":"05:33.985","Text":"we can just write u because u equals u tilde when x is positive."},{"Start":"05:33.985 ","End":"05:36.950","Text":"This is the expression we have so far,"},{"Start":"05:36.950 ","End":"05:39.020","Text":"we need to work on it more."},{"Start":"05:39.020 ","End":"05:44.105","Text":"What we can do next is use a trigonometric identity to simplify this."},{"Start":"05:44.105 ","End":"05:48.800","Text":"There\u0027s a formula for cosine a minus cosine b,"},{"Start":"05:48.800 ","End":"05:57.830","Text":"will take a to be x plus t and b to be x minus t. Then a plus b over 2 will be"},{"Start":"05:57.830 ","End":"06:02.100","Text":"this plus this over 2 will be x and a minus b over"},{"Start":"06:02.100 ","End":"06:06.740","Text":"2 will come out t. What we get is this,"},{"Start":"06:06.740 ","End":"06:11.120","Text":"and this part becomes sine x, sine t. Yes,"},{"Start":"06:11.120 ","End":"06:16.810","Text":"the minus 1/2 cancels with the minus 2, so that disappears."},{"Start":"06:16.810 ","End":"06:19.433","Text":"This is what we have,"},{"Start":"06:19.433 ","End":"06:21.440","Text":"and it\u0027s the answer,"},{"Start":"06:21.440 ","End":"06:28.715","Text":"except that we can maybe put the sine x sine t inside the piecewise-defined function,"},{"Start":"06:28.715 ","End":"06:31.500","Text":"and write it like this."},{"Start":"06:32.780 ","End":"06:35.055","Text":"Yeah, that\u0027s the answer."},{"Start":"06:35.055 ","End":"06:37.780","Text":"That concludes this clip."}],"ID":30731},{"Watched":false,"Name":"Exercise 4","Duration":"6m 47s","ChapterTopicVideoID":29160,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.080","Text":"In this exercise, we have a wave equation and it\u0027s on a semi-infinite interval."},{"Start":"00:07.080 ","End":"00:11.730","Text":"We need the extra boundary condition."},{"Start":"00:11.730 ","End":"00:17.046","Text":"This kind, where it\u0027s du/dx as opposed to just plain u,"},{"Start":"00:17.046 ","End":"00:23.115","Text":"this is what makes it a Neumann boundary condition rather than a Dirichlet condition."},{"Start":"00:23.115 ","End":"00:26.144","Text":"In the case of a Neumann boundary condition,"},{"Start":"00:26.144 ","End":"00:32.610","Text":"we make an even completion even as an even function to minus infinity,"},{"Start":"00:32.610 ","End":"00:36.495","Text":"infinity for both f and g, and u."},{"Start":"00:36.495 ","End":"00:43.580","Text":"The completion of f to an even function is just itself because x^2 is already even."},{"Start":"00:43.580 ","End":"00:46.130","Text":"But for g(x),"},{"Start":"00:46.130 ","End":"00:50.420","Text":"we use the definition of the even extension completion."},{"Start":"00:50.420 ","End":"00:57.050","Text":"What we get is sine x for positive x and sine of minus x,"},{"Start":"00:57.050 ","End":"01:02.000","Text":"which is actually the same as minus sine x for x negative."},{"Start":"01:02.000 ","End":"01:05.255","Text":"You might want to see a picture of g tilde."},{"Start":"01:05.255 ","End":"01:07.040","Text":"Not important, just for interest sake."},{"Start":"01:07.040 ","End":"01:08.630","Text":"It\u0027s just like sine here."},{"Start":"01:08.630 ","End":"01:11.075","Text":"Then we take an even extension,"},{"Start":"01:11.075 ","End":"01:15.845","Text":"like mirroring it in the y-axis or the t-axis."},{"Start":"01:15.845 ","End":"01:18.640","Text":"The extension of u will be u tilde."},{"Start":"01:18.640 ","End":"01:21.155","Text":"By the d\u0027Alembert formula,"},{"Start":"01:21.155 ","End":"01:25.580","Text":"we get that u tilde is equal to the following expression."},{"Start":"01:25.580 ","End":"01:28.955","Text":"We know that a is 1 in our case."},{"Start":"01:28.955 ","End":"01:30.890","Text":"We get the following,"},{"Start":"01:30.890 ","End":"01:33.135","Text":"just replacing a by 1."},{"Start":"01:33.135 ","End":"01:37.380","Text":"Now f tilde of x is x^2 for every x."},{"Start":"01:37.380 ","End":"01:45.595","Text":"In this case, what we get for this expression is x plus t^2 plus x minus t^2 over 2,"},{"Start":"01:45.595 ","End":"01:47.150","Text":"which when you simplify,"},{"Start":"01:47.150 ","End":"01:50.465","Text":"it comes down to x^2 plus t^2."},{"Start":"01:50.465 ","End":"01:56.200","Text":"So we have that this expression is equal to this."},{"Start":"01:56.200 ","End":"02:00.150","Text":"Next, we\u0027ll compute what this expression is."},{"Start":"02:00.150 ","End":"02:03.590","Text":"Then we can add them together and get u tilde."},{"Start":"02:03.590 ","End":"02:10.465","Text":"U tilde is x^2 plus t^2 from here plus this."},{"Start":"02:10.465 ","End":"02:14.165","Text":"If we take x to be non-negative,"},{"Start":"02:14.165 ","End":"02:19.340","Text":"then we can replace u tilde by u and we get the following."},{"Start":"02:19.340 ","End":"02:23.540","Text":"It\u0027s the same thing provided that x is positive or non-negative."},{"Start":"02:23.540 ","End":"02:27.530","Text":"It remains to calculate what this integral is."},{"Start":"02:27.530 ","End":"02:29.990","Text":"Because g tilde is piecewise-defined,"},{"Start":"02:29.990 ","End":"02:32.450","Text":"we\u0027re going to have to divide into cases."},{"Start":"02:32.450 ","End":"02:33.980","Text":"A picture will help."},{"Start":"02:33.980 ","End":"02:35.930","Text":"This x minus tx plus t,"},{"Start":"02:35.930 ","End":"02:39.469","Text":"it all depends where 0 is in relation to this interval."},{"Start":"02:39.469 ","End":"02:41.690","Text":"0 could be to the left of the interval,"},{"Start":"02:41.690 ","End":"02:43.310","Text":"it could be inside of the interval,"},{"Start":"02:43.310 ","End":"02:45.151","Text":"it could be to the right of the interval."},{"Start":"02:45.151 ","End":"02:47.705","Text":"We\u0027ll actually rule out the last case, as you\u0027ll see."},{"Start":"02:47.705 ","End":"02:51.365","Text":"Anyway, let\u0027s take this call it Case 1,"},{"Start":"02:51.365 ","End":"02:54.815","Text":"where x minus t is bigger than 0,"},{"Start":"02:54.815 ","End":"02:58.670","Text":"which is the same as x bigger than t. Often doesn\u0027t matter."},{"Start":"02:58.670 ","End":"03:00.835","Text":"Bigger than, bigger or equal to,"},{"Start":"03:00.835 ","End":"03:04.205","Text":"value of function at a single point doesn\u0027t really matter."},{"Start":"03:04.205 ","End":"03:08.110","Text":"Not strict about strong or weak inequalities."},{"Start":"03:08.110 ","End":"03:12.910","Text":"In this Case 1 where x is bigger or equal to t,"},{"Start":"03:12.910 ","End":"03:15.640","Text":"we get that this whole interval,"},{"Start":"03:15.640 ","End":"03:19.015","Text":"when s is in it, is going to be positive."},{"Start":"03:19.015 ","End":"03:21.080","Text":"S is in here."},{"Start":"03:21.080 ","End":"03:27.745","Text":"The value of g tilde is just sine s. The integral of sine is minus cosine."},{"Start":"03:27.745 ","End":"03:29.785","Text":"We need to evaluate this."},{"Start":"03:29.785 ","End":"03:35.860","Text":"We get cosine of x plus t minus cosine of x minus t. Now,"},{"Start":"03:35.860 ","End":"03:38.740","Text":"we\u0027re going to use a trigonometric identity."},{"Start":"03:38.740 ","End":"03:43.165","Text":"Cosine a minus cosine b is the following."},{"Start":"03:43.165 ","End":"03:47.020","Text":"This will be a, this will be b."},{"Start":"03:47.020 ","End":"03:50.040","Text":"A plus b over 2 will be x,"},{"Start":"03:50.040 ","End":"03:57.515","Text":"a minus b over 2 will be t. What we get here is just sine x sine t. Yeah,"},{"Start":"03:57.515 ","End":"04:01.165","Text":"also the minus 2 cancels with the minus 1/2."},{"Start":"04:01.165 ","End":"04:03.375","Text":"That was Case 1."},{"Start":"04:03.375 ","End":"04:11.720","Text":"Case 2 is this one where x minus t is to the left of 0 and x plus t is to the right of 0."},{"Start":"04:11.720 ","End":"04:16.700","Text":"But we\u0027re only interested in the case where x is non-negative."},{"Start":"04:16.700 ","End":"04:20.270","Text":"This can\u0027t be too much to the left because x,"},{"Start":"04:20.270 ","End":"04:23.495","Text":"the middle of the interval, still has to be bigger than 0."},{"Start":"04:23.495 ","End":"04:25.880","Text":"It\u0027s x minus t negative,"},{"Start":"04:25.880 ","End":"04:28.835","Text":"but x has to be positive."},{"Start":"04:28.835 ","End":"04:35.330","Text":"That translates to x between 0 and t. In this case,"},{"Start":"04:35.330 ","End":"04:40.734","Text":"what we get is on this part from here to here,"},{"Start":"04:40.734 ","End":"04:45.080","Text":"it\u0027s minus sine s. That\u0027s what g tilde is."},{"Start":"04:45.080 ","End":"04:46.385","Text":"But on the other side,"},{"Start":"04:46.385 ","End":"04:50.015","Text":"it\u0027s just regular sine s. If we evaluate this,"},{"Start":"04:50.015 ","End":"04:54.050","Text":"in each case, the integral of sine is minus cosine."},{"Start":"04:54.050 ","End":"04:56.210","Text":"Minus sine becomes plus cosine,"},{"Start":"04:56.210 ","End":"04:58.550","Text":"and sine becomes minus cosine."},{"Start":"04:58.550 ","End":"05:02.225","Text":"These are the limits of integration here and here."},{"Start":"05:02.225 ","End":"05:04.100","Text":"Substitute them."},{"Start":"05:04.100 ","End":"05:11.615","Text":"Cosine 0 minus cosine of x minus t. Here I changed the minus to a plus,"},{"Start":"05:11.615 ","End":"05:15.110","Text":"did the subtraction in the reverse order."},{"Start":"05:15.110 ","End":"05:18.310","Text":"Cosine 0 is 1, and here also,"},{"Start":"05:18.310 ","End":"05:24.350","Text":"so this is a 1/2 plus a 1/2 is 1 minus 1/2 of this,"},{"Start":"05:24.350 ","End":"05:25.835","Text":"minus a 1/2 of this,"},{"Start":"05:25.835 ","End":"05:28.930","Text":"which is minus a 1/2 of this plus this."},{"Start":"05:28.930 ","End":"05:33.820","Text":"And now we\u0027re going to use another trigonometric formula for cosine a plus"},{"Start":"05:33.820 ","End":"05:38.950","Text":"cosine b as opposed to cosine a minus cosine b."},{"Start":"05:38.950 ","End":"05:43.055","Text":"Quite different here we had sines and here we\u0027re going to have cosines,."},{"Start":"05:43.055 ","End":"05:44.920","Text":"If we substitute those,"},{"Start":"05:44.920 ","End":"05:49.750","Text":"we get 1 minus cosine x cosine t,"},{"Start":"05:49.750 ","End":"05:51.460","Text":"and that\u0027s Case 2."},{"Start":"05:51.460 ","End":"05:54.474","Text":"Case 3, we can rule out."},{"Start":"05:54.474 ","End":"05:57.080","Text":"Let\u0027s go back to the picture."},{"Start":"05:57.080 ","End":"05:59.780","Text":"Yeah, this is Case 3."},{"Start":"05:59.780 ","End":"06:03.460","Text":"If this is the way it is an x which is the middle of the interval is"},{"Start":"06:03.460 ","End":"06:07.870","Text":"negative and we\u0027re only collecting results for x positive."},{"Start":"06:07.870 ","End":"06:09.295","Text":"That\u0027s ruled out."},{"Start":"06:09.295 ","End":"06:11.200","Text":"We only have the 2 cases."},{"Start":"06:11.200 ","End":"06:13.175","Text":"If we put them together,"},{"Start":"06:13.175 ","End":"06:15.100","Text":"when x is bigger than t,"},{"Start":"06:15.100 ","End":"06:19.614","Text":"we get sine x sine t. When x is less than t, but still positive,"},{"Start":"06:19.614 ","End":"06:24.900","Text":"1 minus cosine x cosine t. That\u0027s the integral of g tilde."},{"Start":"06:24.900 ","End":"06:30.640","Text":"I remember that u was equal to this for positive x,"},{"Start":"06:30.640 ","End":"06:35.910","Text":"of course, and this translates to this or this."},{"Start":"06:35.910 ","End":"06:40.095","Text":"Just combining, adding x^2 plus t^2 here and here,"},{"Start":"06:40.095 ","End":"06:48.700","Text":"this is the result we have for u(x) and t. That concludes this exercise."}],"ID":30732},{"Watched":false,"Name":"Exercise 5","Duration":"10m 6s","ChapterTopicVideoID":29161,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.265","Text":"In this exercise, we have the following initial boundary value problem."},{"Start":"00:05.265 ","End":"00:11.325","Text":"You can see from this that it\u0027s a wave equation and this is the semi-infinite interval."},{"Start":"00:11.325 ","End":"00:13.425","Text":"These are the initial values,"},{"Start":"00:13.425 ","End":"00:17.505","Text":"u and u by dt at x, 0."},{"Start":"00:17.505 ","End":"00:25.015","Text":"This is a boundary value when x=0 along the t-axis."},{"Start":"00:25.015 ","End":"00:30.470","Text":"This is the Dirichlet boundary condition as opposed to the Neumann."},{"Start":"00:30.470 ","End":"00:33.095","Text":"We know how to stop the Dirichlet boundary condition."},{"Start":"00:33.095 ","End":"00:37.460","Text":"What we do is we make u and f and g odd functions."},{"Start":"00:37.460 ","End":"00:44.030","Text":"We take extensions to them completions onto the whole interval,"},{"Start":"00:44.030 ","End":"00:46.060","Text":"or the real line."},{"Start":"00:46.060 ","End":"00:49.230","Text":"Then we solve this and at the end,"},{"Start":"00:49.230 ","End":"00:54.120","Text":"we just restrict u to x bigger or equal to 0."},{"Start":"00:54.120 ","End":"00:59.360","Text":"This is taken care of automatically once we satisfy these conditions."},{"Start":"00:59.360 ","End":"01:01.070","Text":"The odd extension of f,"},{"Start":"01:01.070 ","End":"01:03.785","Text":"call it f tilde is the following,"},{"Start":"01:03.785 ","End":"01:09.650","Text":"which gives us either 2x^2 minus x or minus 2x^2 minus x."},{"Start":"01:09.650 ","End":"01:16.425","Text":"If you replace here x by minus x and then take a minus to the whole thing,"},{"Start":"01:16.425 ","End":"01:17.970","Text":"this is what we get."},{"Start":"01:17.970 ","End":"01:20.735","Text":"As for g, same thing,"},{"Start":"01:20.735 ","End":"01:25.040","Text":"except that in this case g(0) is not 0,"},{"Start":"01:25.040 ","End":"01:27.275","Text":"so we just force it to be 0."},{"Start":"01:27.275 ","End":"01:31.070","Text":"That won\u0027t make any difference because when we take an integral,"},{"Start":"01:31.070 ","End":"01:33.090","Text":"1 point makes no difference."},{"Start":"01:33.090 ","End":"01:37.380","Text":"This comes out to be 4x plus 1 or 4x minus 1."},{"Start":"01:37.380 ","End":"01:43.175","Text":"Here, we get this by replacing x by minus x so we get minus 4x plus 1."},{"Start":"01:43.175 ","End":"01:45.100","Text":"Then take minus of the whole thing,"},{"Start":"01:45.100 ","End":"01:46.940","Text":"so it\u0027s plus 4x minus 1,"},{"Start":"01:46.940 ","End":"01:49.045","Text":"the 0 it\u0027s 0."},{"Start":"01:49.045 ","End":"01:51.560","Text":"There\u0027s a picture of both of them."},{"Start":"01:51.560 ","End":"01:55.010","Text":"This one is f and this one,"},{"Start":"01:55.010 ","End":"01:58.070","Text":"the broken 1 is g tilde rather."},{"Start":"01:58.070 ","End":"02:00.679","Text":"Using the d\u0027Alembert formula,"},{"Start":"02:00.679 ","End":"02:02.750","Text":"this is the solution for u tilde,"},{"Start":"02:02.750 ","End":"02:07.610","Text":"which is the completion of u to an odd function,"},{"Start":"02:07.610 ","End":"02:11.250","Text":"adding x, and a is 1 in our case,"},{"Start":"02:11.250 ","End":"02:14.290","Text":"so it simplifies to this."},{"Start":"02:14.290 ","End":"02:23.480","Text":"If we replace x by x plus t and then by x minus t here and here,"},{"Start":"02:23.480 ","End":"02:29.270","Text":"what we get is for x plus t bigger or equal to 0,"},{"Start":"02:29.270 ","End":"02:31.820","Text":"we get this expression with x plus t instead"},{"Start":"02:31.820 ","End":"02:34.294","Text":"of x and in the other case we get this expression"},{"Start":"02:34.294 ","End":"02:39.800","Text":"again with x replaced by x plus t. For the other 1,"},{"Start":"02:39.800 ","End":"02:43.940","Text":"we also need f(x) minus t. Well,"},{"Start":"02:43.940 ","End":"02:47.300","Text":"similar thing, practically the same."},{"Start":"02:47.300 ","End":"02:51.740","Text":"We can rewrite these inequalities as follows."},{"Start":"02:51.740 ","End":"02:53.915","Text":"Just a bit more convenient."},{"Start":"02:53.915 ","End":"02:55.640","Text":"There are 3 cases."},{"Start":"02:55.640 ","End":"03:00.710","Text":"It looks like there\u0027s 4 cases because 2 possibilities here with 2 possibilities here,"},{"Start":"03:00.710 ","End":"03:02.480","Text":"but not all of them can be."},{"Start":"03:02.480 ","End":"03:07.670","Text":"If you put the points minus t and t on the number line,"},{"Start":"03:07.670 ","End":"03:09.830","Text":"then x can only be in 1 of 3 places."},{"Start":"03:09.830 ","End":"03:12.680","Text":"It\u0027s either less than the left 1 or between"},{"Start":"03:12.680 ","End":"03:17.660","Text":"the 2 values or to the right of the rightmost 1, those 3 cases."},{"Start":"03:17.660 ","End":"03:22.445","Text":"But we\u0027re only going to be considering x bigger or equal to 0 because at the end,"},{"Start":"03:22.445 ","End":"03:26.270","Text":"we restrict everything to non-negative x."},{"Start":"03:26.270 ","End":"03:27.410","Text":"There\u0027s only 2 cases."},{"Start":"03:27.410 ","End":"03:30.920","Text":"This one drops out and we have this one and this one."},{"Start":"03:30.920 ","End":"03:35.885","Text":"This one\u0027s also reduces to just 0 less than or equal to x less than t,"},{"Start":"03:35.885 ","End":"03:38.455","Text":"because x is bigger or equal to 0."},{"Start":"03:38.455 ","End":"03:40.470","Text":"We\u0027ll call this one case 1,"},{"Start":"03:40.470 ","End":"03:42.990","Text":"and then we\u0027ll do this one as case 2."},{"Start":"03:42.990 ","End":"03:45.240","Text":"Let\u0027s start with case 1,"},{"Start":"03:45.240 ","End":"03:49.655","Text":"x bigger or equal to t. Remember some of the formulas we got"},{"Start":"03:49.655 ","End":"03:56.385","Text":"earlier for f tilde of x plus t of x minus t and also g tilde of x."},{"Start":"03:56.385 ","End":"04:00.115","Text":"What we can say is f tilde of x plus t,"},{"Start":"04:00.115 ","End":"04:03.350","Text":"we take from the top row here,"},{"Start":"04:03.350 ","End":"04:07.130","Text":"because if x is bigger or equal to t,"},{"Start":"04:07.130 ","End":"04:10.940","Text":"it\u0027s also bigger or equal to minus t. This is the 1 we need,"},{"Start":"04:10.940 ","End":"04:15.500","Text":"and f(x) minus t will also take from the"},{"Start":"04:15.500 ","End":"04:20.330","Text":"top 1 because x is bigger or equal to t so this is what we get."},{"Start":"04:20.330 ","End":"04:21.755","Text":"Here just to remind you,"},{"Start":"04:21.755 ","End":"04:28.015","Text":"we\u0027re going to need this integral so that s is between x minus t and x plus t,"},{"Start":"04:28.015 ","End":"04:32.510","Text":"s between x minus t and x plus t. This is bigger or equal to"},{"Start":"04:32.510 ","End":"04:36.560","Text":"0 because x is bigger or equal to t. In all this interval,"},{"Start":"04:36.560 ","End":"04:38.345","Text":"s is bigger or equal to 0,"},{"Start":"04:38.345 ","End":"04:40.175","Text":"so we take from the top row,"},{"Start":"04:40.175 ","End":"04:44.560","Text":"and then g(s) will be 4s plus 1,"},{"Start":"04:44.560 ","End":"04:45.845","Text":"g tilde, I mean."},{"Start":"04:45.845 ","End":"04:50.870","Text":"Now we\u0027re all ready to substitute into here because we have f tilde of x plus t,"},{"Start":"04:50.870 ","End":"04:54.115","Text":"f tilde of x minus t, and g tilde."},{"Start":"04:54.115 ","End":"04:59.971","Text":"What we get is this plus this on the numerator here."},{"Start":"04:59.971 ","End":"05:02.000","Text":"Here, instead of g tilde of s,"},{"Start":"05:02.000 ","End":"05:03.725","Text":"we have 4s plus 1."},{"Start":"05:03.725 ","End":"05:06.245","Text":"You have to calculate this."},{"Start":"05:06.245 ","End":"05:10.030","Text":"This divided by 2 is this,"},{"Start":"05:10.030 ","End":"05:13.050","Text":"and this divided by 2 is this."},{"Start":"05:13.050 ","End":"05:17.600","Text":"Then we have minus x plus t minus x minus t,"},{"Start":"05:17.600 ","End":"05:20.270","Text":"which is minus 2x/2,"},{"Start":"05:20.270 ","End":"05:22.198","Text":"so minus x,"},{"Start":"05:22.198 ","End":"05:25.175","Text":"and here the integral is 2s^2 plus"},{"Start":"05:25.175 ","End":"05:30.800","Text":"s. Then we just substitute the upper and lower limits and subtract,"},{"Start":"05:30.800 ","End":"05:34.760","Text":"substitute x plus t in place of s and we get the first 2 terms,"},{"Start":"05:34.760 ","End":"05:37.970","Text":"x minus t and we get the last 2 terms."},{"Start":"05:37.970 ","End":"05:40.810","Text":"Just need some simplifying to do."},{"Start":"05:40.810 ","End":"05:44.415","Text":"This over 2 is just x plus t^2."},{"Start":"05:44.415 ","End":"05:48.110","Text":"This one over 2 is minus x minus t^2."},{"Start":"05:48.110 ","End":"05:53.280","Text":"These 2 combine to give 2t,"},{"Start":"05:53.280 ","End":"06:00.350","Text":"which divided by 2 gives us t. This combines with this to get twice x plus t^2."},{"Start":"06:00.350 ","End":"06:02.810","Text":"But this with this cancels,"},{"Start":"06:02.810 ","End":"06:08.765","Text":"so what we\u0027re left with after that is minus x plus t. Now we have u tilde of x,"},{"Start":"06:08.765 ","End":"06:11.950","Text":"t in case 1,"},{"Start":"06:11.950 ","End":"06:13.430","Text":"which to remind you,"},{"Start":"06:13.430 ","End":"06:15.495","Text":"is the case x bigger or equal to t. Now,"},{"Start":"06:15.495 ","End":"06:17.810","Text":"let\u0027s go on to case 2."},{"Start":"06:17.810 ","End":"06:19.550","Text":"Now case 2,"},{"Start":"06:19.550 ","End":"06:21.725","Text":"x is between 0 and t,"},{"Start":"06:21.725 ","End":"06:25.575","Text":"and a reminder of some of the results we had."},{"Start":"06:25.575 ","End":"06:28.580","Text":"We need f tilde of x plus t,"},{"Start":"06:28.580 ","End":"06:31.265","Text":"and we\u0027ll take this from the top row here,"},{"Start":"06:31.265 ","End":"06:33.170","Text":"because x is bigger or equal to 0,"},{"Start":"06:33.170 ","End":"06:35.738","Text":"so bigger or equal to minus t,"},{"Start":"06:35.738 ","End":"06:37.745","Text":"and f of x minus t,"},{"Start":"06:37.745 ","End":"06:39.389","Text":"f tilde, I mean,"},{"Start":"06:39.389 ","End":"06:46.288","Text":"is from the bottom row because x is less than t. This is the 1 we need."},{"Start":"06:46.288 ","End":"06:52.600","Text":"For g(s), there are 2 cases."},{"Start":"06:52.600 ","End":"06:57.044","Text":"Look, if x is between 0 and t,"},{"Start":"06:57.044 ","End":"07:03.000","Text":"then x minus t is less than 0 from this one,"},{"Start":"07:03.000 ","End":"07:07.690","Text":"and from here, x is bigger or equal to 0,"},{"Start":"07:07.690 ","End":"07:11.635","Text":"so certainly x plus t is bigger or equal to 0."},{"Start":"07:11.635 ","End":"07:14.480","Text":"0 is somewhere between x minus t and"},{"Start":"07:14.480 ","End":"07:19.940","Text":"x plus t. The integral we have to compute this one can be broken up into 2 pieces,"},{"Start":"07:19.940 ","End":"07:29.310","Text":"from x minus t to 0 and from 0 to x plus t. Then this part will take less than 0,"},{"Start":"07:29.310 ","End":"07:31.545","Text":"so it will be 4s minus 1,"},{"Start":"07:31.545 ","End":"07:34.517","Text":"and in this part it will be 4s plus 1."},{"Start":"07:34.517 ","End":"07:39.110","Text":"This is the expression from d\u0027Alembert formula."},{"Start":"07:39.110 ","End":"07:41.585","Text":"Now, this is the integral part,"},{"Start":"07:41.585 ","End":"07:45.680","Text":"like I said here it\u0027s 4s minus 1 and here it\u0027s 4s plus 1."},{"Start":"07:45.680 ","End":"07:49.000","Text":"This we take from, where is it?"},{"Start":"07:49.000 ","End":"07:53.150","Text":"You\u0027ve got f tilde of x plus t here,"},{"Start":"07:53.150 ","End":"07:56.245","Text":"plus f tilde of x minus t,"},{"Start":"07:56.245 ","End":"07:58.095","Text":"which is this here,"},{"Start":"07:58.095 ","End":"08:00.720","Text":"over 2 plus the integrals."},{"Start":"08:00.720 ","End":"08:02.775","Text":"This is equal to,"},{"Start":"08:02.775 ","End":"08:04.470","Text":"I\u0027m just combining these two,"},{"Start":"08:04.470 ","End":"08:11.640","Text":"we get minus 2x and the integral of 4s minus 1 is 2x^2 minus s,"},{"Start":"08:11.640 ","End":"08:15.680","Text":"and this integral is 2s^2 plus s. In each of them we have a half and we"},{"Start":"08:15.680 ","End":"08:21.090","Text":"have limits of integration here and here."},{"Start":"08:22.310 ","End":"08:25.230","Text":"This one simplifies 2."},{"Start":"08:25.230 ","End":"08:30.676","Text":"This comes out to x plus t^2 minus x minus t^2 and then minus x,"},{"Start":"08:30.676 ","End":"08:32.780","Text":"then this one, we substitute 0,"},{"Start":"08:32.780 ","End":"08:36.755","Text":"you get 0, so it\u0027s just minus the lower limit."},{"Start":"08:36.755 ","End":"08:42.320","Text":"We can put the minus in front and replace s by x minus t here."},{"Start":"08:42.320 ","End":"08:48.318","Text":"Here the 0 gives nothing so we just substitute the x plus t instead of s, get this."},{"Start":"08:48.318 ","End":"08:52.050","Text":"A half of twice this is just this,"},{"Start":"08:52.050 ","End":"08:55.715","Text":"and also here a half of this is this."},{"Start":"08:55.715 ","End":"08:59.390","Text":"Then we get a half of minus,"},{"Start":"08:59.390 ","End":"09:02.405","Text":"minus is plus x minus t,"},{"Start":"09:02.405 ","End":"09:07.260","Text":"and here plus x plus t over 2."},{"Start":"09:09.050 ","End":"09:14.270","Text":"This with this gives twice whatever it is and minus,"},{"Start":"09:14.270 ","End":"09:16.925","Text":"minus the same thing is minus twice."},{"Start":"09:16.925 ","End":"09:22.830","Text":"Then we have minus x from here and this part just gives us 2x/2 which is x,"},{"Start":"09:22.830 ","End":"09:25.310","Text":"so the x cancels with the x."},{"Start":"09:25.310 ","End":"09:31.850","Text":"The x^2 plus t^2 parts cancel and we just get from here 4xt in the middle term,"},{"Start":"09:31.850 ","End":"09:36.290","Text":"minus minus 4xt is 8xt."},{"Start":"09:36.290 ","End":"09:38.405","Text":"This is case 2,"},{"Start":"09:38.405 ","End":"09:42.260","Text":"which is the case where x is between 0 and t. Now,"},{"Start":"09:42.260 ","End":"09:43.310","Text":"we have both cases,"},{"Start":"09:43.310 ","End":"09:46.370","Text":"we\u0027re just going to combine them and recall"},{"Start":"09:46.370 ","End":"09:49.520","Text":"that x is bigger or equal to 0 in all our computation,"},{"Start":"09:49.520 ","End":"09:55.440","Text":"so we can replace u tilde by u because that\u0027s what it is,"},{"Start":"09:55.440 ","End":"09:58.885","Text":"u is the u tilde restricted to x bigger or equal to 0."},{"Start":"09:58.885 ","End":"10:00.190","Text":"From the first case,"},{"Start":"10:00.190 ","End":"10:03.985","Text":"we get this and in the second case, we get this."},{"Start":"10:03.985 ","End":"10:07.070","Text":"This is the solution."}],"ID":30733},{"Watched":false,"Name":"Exercise 6","Duration":"5m 5s","ChapterTopicVideoID":29162,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.580","Text":"In this exercise, we have 2 initial boundary value problems,"},{"Start":"00:05.580 ","End":"00:14.415","Text":"1 with u and 1 with v. Each of them is a wave equation on the semi-infinite interval."},{"Start":"00:14.415 ","End":"00:24.060","Text":"Our task is to compute the limit as t goes to infinity of u(1,t) over v(1,t)."},{"Start":"00:24.060 ","End":"00:29.400","Text":"Note that we don\u0027t have to compute the solutions u(x,t) and v(x,t),"},{"Start":"00:29.400 ","End":"00:34.455","Text":"we only have to compute u(1,t) and v(1,t)."},{"Start":"00:34.455 ","End":"00:35.970","Text":"It\u0027s sufficient to do this."},{"Start":"00:35.970 ","End":"00:38.290","Text":"That\u0027s all we\u0027ll need in this limit."},{"Start":"00:38.290 ","End":"00:42.470","Text":"For u, we have the following equations."},{"Start":"00:42.470 ","End":"00:45.830","Text":"Let\u0027s call this 1 f and this 1 is g. This is"},{"Start":"00:45.830 ","End":"00:52.370","Text":"a Dirichlet problem because of the form u(0,t) rather than du by dx."},{"Start":"00:52.370 ","End":"00:56.720","Text":"We can see already that the other 1 with v will be the other kind,"},{"Start":"00:56.720 ","End":"00:58.970","Text":"the Neumann boundary condition."},{"Start":"00:58.970 ","End":"01:03.410","Text":"Anyway, because of the Dirichlet boundary condition,"},{"Start":"01:03.410 ","End":"01:06.710","Text":"we need odd completions for f,"},{"Start":"01:06.710 ","End":"01:08.150","Text":"g, and u."},{"Start":"01:08.150 ","End":"01:11.120","Text":"When we complete them, we put a tilde over them."},{"Start":"01:11.120 ","End":"01:14.075","Text":"Then by the d\u0027Alembert formula,"},{"Start":"01:14.075 ","End":"01:16.115","Text":"we\u0027ll get the u tilde."},{"Start":"01:16.115 ","End":"01:17.825","Text":"Well, is the following."},{"Start":"01:17.825 ","End":"01:21.750","Text":"Now, a equals 1 in our case, because usually,"},{"Start":"01:21.750 ","End":"01:24.060","Text":"it\u0027s u_tt equals a squared u_xx,"},{"Start":"01:24.060 ","End":"01:26.020","Text":"so here a is 1."},{"Start":"01:26.020 ","End":"01:29.255","Text":"If we do that, comes out simpler also,"},{"Start":"01:29.255 ","End":"01:31.775","Text":"because g is 0."},{"Start":"01:31.775 ","End":"01:35.995","Text":"Well, the extended function g tilde is also 0,"},{"Start":"01:35.995 ","End":"01:37.340","Text":"so just have this."},{"Start":"01:37.340 ","End":"01:42.610","Text":"Now, f tilde is also x cubed because it\u0027s already an odd function,"},{"Start":"01:42.610 ","End":"01:51.055","Text":"so f tilde is x cubed on all the real line and the extension or completion of 0 is 0."},{"Start":"01:51.055 ","End":"01:55.350","Text":"This drops off and we just have this."},{"Start":"01:55.350 ","End":"01:58.485","Text":"Since f(x) is x cubed,"},{"Start":"01:58.485 ","End":"02:05.775","Text":"we have that f of 1 plus t is 1 plus t cubed, the 1 here."},{"Start":"02:05.775 ","End":"02:07.560","Text":"Put x equals 1 in here,"},{"Start":"02:07.560 ","End":"02:10.860","Text":"1 minus t cubed over 2."},{"Start":"02:10.860 ","End":"02:13.240","Text":"Now let\u0027s tackle the other 1,"},{"Start":"02:13.240 ","End":"02:19.280","Text":"v. This is the set of equations for v. This time,"},{"Start":"02:19.280 ","End":"02:21.820","Text":"it\u0027s a Neumann boundary condition."},{"Start":"02:21.820 ","End":"02:23.930","Text":"Instead of the odd completion,"},{"Start":"02:23.930 ","End":"02:26.000","Text":"we need an even completion."},{"Start":"02:26.000 ","End":"02:31.790","Text":"Then again, we\u0027ll apply the d\u0027Alembert formula and get the following."},{"Start":"02:31.790 ","End":"02:35.915","Text":"Note that f tilde, in this case,"},{"Start":"02:35.915 ","End":"02:40.400","Text":"is piece wise defined because uneven completion"},{"Start":"02:40.400 ","End":"02:44.950","Text":"means we replace x by minus x. I mean it\u0027s not an even function,"},{"Start":"02:44.950 ","End":"02:46.600","Text":"it\u0027s an odd function."},{"Start":"02:46.600 ","End":"02:49.630","Text":"When x is positive, we\u0027re all right, when x is negative,"},{"Start":"02:49.630 ","End":"02:52.435","Text":"we need to replace x by minus x."},{"Start":"02:52.435 ","End":"02:57.040","Text":"This is actually minus x cubed and g is still 0."},{"Start":"02:57.040 ","End":"03:02.360","Text":"In fact, the odd and even completions of the 0 function is the 0 function."},{"Start":"03:02.360 ","End":"03:04.215","Text":"Let\u0027s continue."},{"Start":"03:04.215 ","End":"03:06.285","Text":"A equals 1."},{"Start":"03:06.285 ","End":"03:08.369","Text":"This simplifies to this."},{"Start":"03:08.369 ","End":"03:12.390","Text":"Also g is 0 and g tilde is 0."},{"Start":"03:12.390 ","End":"03:16.468","Text":"Replace x by 1 and we get v(1,t),"},{"Start":"03:16.468 ","End":"03:19.630","Text":"and we know what f tilde is,"},{"Start":"03:19.630 ","End":"03:22.780","Text":"it\u0027s the following to remind you."},{"Start":"03:22.780 ","End":"03:25.685","Text":"What we get in our case,"},{"Start":"03:25.685 ","End":"03:33.455","Text":"this part is going to always be 1 plus t cubed because 1 plus t is positive."},{"Start":"03:33.455 ","End":"03:35.060","Text":"But the other part,"},{"Start":"03:35.060 ","End":"03:41.915","Text":"f tilde of 1 minus t depends on if 1 minus t is less than 0 or bigger than 0."},{"Start":"03:41.915 ","End":"03:45.373","Text":"If it\u0027s bigger than 0, we just get 1 minus t cubed,"},{"Start":"03:45.373 ","End":"03:50.110","Text":"if it\u0027s less than 0, we have minus 1 minus t cubed and the minus here."},{"Start":"03:50.110 ","End":"03:53.670","Text":"This simplifies to t bigger than 1,"},{"Start":"03:53.670 ","End":"03:57.465","Text":"this simplifies to t less than 1."},{"Start":"03:57.465 ","End":"04:01.670","Text":"Now, we\u0027re going to be taking the limit as t goes to infinity."},{"Start":"04:01.670 ","End":"04:05.060","Text":"We only need the case where t is bigger than 1,"},{"Start":"04:05.060 ","End":"04:11.360","Text":"so we get that v tilde(1,t) is the following. That\u0027s v tilde."},{"Start":"04:11.360 ","End":"04:15.725","Text":"Now recall, we already computed u tilde as this."},{"Start":"04:15.725 ","End":"04:18.004","Text":"It\u0027s now ready to do the limit."},{"Start":"04:18.004 ","End":"04:23.585","Text":"But note that u tilde is the same as u and v tilde is the same as v,"},{"Start":"04:23.585 ","End":"04:26.000","Text":"which is true when x is bigger than 0,"},{"Start":"04:26.000 ","End":"04:27.680","Text":"x is in this case is 1."},{"Start":"04:27.680 ","End":"04:31.055","Text":"In both cases, x is bigger than 0."},{"Start":"04:31.055 ","End":"04:39.220","Text":"The limit for u over v is the same as u tilde over v tilde and we get this over this."},{"Start":"04:39.220 ","End":"04:41.990","Text":"Half doesn\u0027t really matter, it cancels out."},{"Start":"04:41.990 ","End":"04:47.000","Text":"This is what we get using the formula for a plus b cubed,"},{"Start":"04:47.000 ","End":"04:52.525","Text":"we get this, which simplifies to this."},{"Start":"04:52.525 ","End":"04:55.280","Text":"Then we have a polynomial over polynomial,"},{"Start":"04:55.280 ","End":"04:59.285","Text":"but the degree in the numerator is less than a degree in the denominator."},{"Start":"04:59.285 ","End":"05:02.345","Text":"The limit is 0."},{"Start":"05:02.345 ","End":"05:05.850","Text":"That concludes this exercise."}],"ID":30734},{"Watched":false,"Name":"Exercise 7","Duration":"4m 7s","ChapterTopicVideoID":29152,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.790","Text":"In this exercise, we have a wave equation on a semi-infinite interval,"},{"Start":"00:05.790 ","End":"00:10.740","Text":"but the boundary condition is not one of our usual ones."},{"Start":"00:10.740 ","End":"00:12.240","Text":"We want u(0,"},{"Start":"00:12.240 ","End":"00:16.890","Text":"t) to be 0, or du by dx(0, t) to be 0."},{"Start":"00:16.890 ","End":"00:19.980","Text":"But here we have 1/6^3,"},{"Start":"00:19.980 ","End":"00:21.525","Text":"not the same as 0,"},{"Start":"00:21.525 ","End":"00:26.535","Text":"so what we\u0027re going to do is use a little trick to make the boundary conditions 0."},{"Start":"00:26.535 ","End":"00:31.280","Text":"We define v to be u(x t) plus w(t),"},{"Start":"00:31.280 ","End":"00:37.424","Text":"and we find w(t) such that the boundary condition for v is 0."},{"Start":"00:37.424 ","End":"00:39.200","Text":"Now here\u0027s how we do that,"},{"Start":"00:39.200 ","End":"00:41.720","Text":"v(0, t) is u(0,"},{"Start":"00:41.720 ","End":"00:43.505","Text":"t) plus w(t),"},{"Start":"00:43.505 ","End":"00:47.735","Text":"this is just replacing x by 0 here and here."},{"Start":"00:47.735 ","End":"00:50.960","Text":"Now v(0, t) is 0,"},{"Start":"00:50.960 ","End":"00:52.895","Text":"that\u0027s what we want, u(0,"},{"Start":"00:52.895 ","End":"00:57.095","Text":"t) is a 6t^3 plus w(t)."},{"Start":"00:57.095 ","End":"01:00.900","Text":"That gives us that w(t) is minus 6t^3."},{"Start":"01:00.900 ","End":"01:03.210","Text":"Now, put that here,"},{"Start":"01:03.210 ","End":"01:07.095","Text":"and we get that v(x,t) is u(x,"},{"Start":"01:07.095 ","End":"01:09.090","Text":"t) minus a 6t^3."},{"Start":"01:09.090 ","End":"01:15.185","Text":"You can write this in short as u equals v plus 6t^3."},{"Start":"01:15.185 ","End":"01:24.020","Text":"I should say our goal now is to convert this initial boundary value problem to 1 in v,"},{"Start":"01:24.020 ","End":"01:26.895","Text":"u is v plus 1/6t^3,"},{"Start":"01:26.895 ","End":"01:31.130","Text":"will need the derivatives u_tt is"},{"Start":"01:31.130 ","End":"01:37.530","Text":"v_tt plus the second derivative of this with respect to t. Differentiate it,"},{"Start":"01:37.530 ","End":"01:39.050","Text":"once we get to 1/2 t squared,"},{"Start":"01:39.050 ","End":"01:42.320","Text":"differentiate it again, we get t, that\u0027s that."},{"Start":"01:42.320 ","End":"01:47.630","Text":"With respect to x, this has a 0 derivative with respect to x,"},{"Start":"01:47.630 ","End":"01:49.910","Text":"and if you differentiate again still 0,"},{"Start":"01:49.910 ","End":"01:52.970","Text":"so u_xx equals v_xx."},{"Start":"01:52.970 ","End":"01:57.945","Text":"Now we want to substitute into this PDE in u,"},{"Start":"01:57.945 ","End":"02:02.700","Text":"and we\u0027ll get that u_tt is v_tt plus t,"},{"Start":"02:02.700 ","End":"02:06.285","Text":"and here u_xx is v_xx plus t,"},{"Start":"02:06.285 ","End":"02:08.130","Text":"and then the t cancels."},{"Start":"02:08.130 ","End":"02:10.275","Text":"That\u0027s just luck."},{"Start":"02:10.275 ","End":"02:14.315","Text":"We actually get a homogeneous wave equation."},{"Start":"02:14.315 ","End":"02:18.065","Text":"Now, v(0, t) is 0 and v(x,"},{"Start":"02:18.065 ","End":"02:20.735","Text":"0) is u(x, 0)."},{"Start":"02:20.735 ","End":"02:22.805","Text":"Let\u0027s go back and check, u(x,"},{"Start":"02:22.805 ","End":"02:29.870","Text":"0) is x minus 1/6t^3, the t is 0,"},{"Start":"02:29.870 ","End":"02:31.895","Text":"so this just disappears,"},{"Start":"02:31.895 ","End":"02:38.449","Text":"and v_t(x_0) is u_t(x_0), which is 0,"},{"Start":"02:38.449 ","End":"02:43.299","Text":"minus the derivative of"},{"Start":"02:44.030 ","End":"02:49.585","Text":"this 1/2t^2 when t is 0, so it\u0027s just 0."},{"Start":"02:49.585 ","End":"02:52.720","Text":"Call this one f(x), call this one g(x)."},{"Start":"02:52.720 ","End":"02:56.095","Text":"Now this is the Dirichlet boundary condition,"},{"Start":"02:56.095 ","End":"03:00.370","Text":"so we need to work with odd extensions."},{"Start":"03:00.370 ","End":"03:02.875","Text":"f and g are already odd,"},{"Start":"03:02.875 ","End":"03:04.810","Text":"I mean, x is an odd function,"},{"Start":"03:04.810 ","End":"03:05.920","Text":"and so is 0,"},{"Start":"03:05.920 ","End":"03:09.190","Text":"so the tilde of both of these is the same,"},{"Start":"03:09.190 ","End":"03:12.025","Text":"just extended to the whole real line."},{"Start":"03:12.025 ","End":"03:16.900","Text":"Now let\u0027s apply the d\u0027Alembert formula to v tilde,"},{"Start":"03:16.900 ","End":"03:19.060","Text":"and we get the following."},{"Start":"03:19.060 ","End":"03:20.950","Text":"I should say that a is 1,"},{"Start":"03:20.950 ","End":"03:22.630","Text":"that\u0027s why we don\u0027t have the a t,"},{"Start":"03:22.630 ","End":"03:25.645","Text":"and the a here, and an a here, and an a here."},{"Start":"03:25.645 ","End":"03:30.185","Text":"A is 1. f(x) plus t,"},{"Start":"03:30.185 ","End":"03:33.335","Text":"or f tilde(x plus t) is just x plus t,"},{"Start":"03:33.335 ","End":"03:35.255","Text":"because f(x) is x."},{"Start":"03:35.255 ","End":"03:37.055","Text":"We have this plus this,"},{"Start":"03:37.055 ","End":"03:38.490","Text":"and g tilde is 0,"},{"Start":"03:38.490 ","End":"03:39.815","Text":"so this whole bit,"},{"Start":"03:39.815 ","End":"03:41.600","Text":"the integral drops out,"},{"Start":"03:41.600 ","End":"03:45.790","Text":"and this is 2x/2, which is x."},{"Start":"03:45.790 ","End":"03:47.460","Text":"That\u0027s v tilde,"},{"Start":"03:47.460 ","End":"03:52.290","Text":"and v is just the same as v tilde only restricted to x bigger or equal to 0,"},{"Start":"03:52.290 ","End":"03:53.855","Text":"so we found v,"},{"Start":"03:53.855 ","End":"03:56.165","Text":"but we still need to find u."},{"Start":"03:56.165 ","End":"04:00.350","Text":"Well, u is v plus the 1/6t^3, so u(x,"},{"Start":"04:00.350 ","End":"04:02.960","Text":"t) is x plus 1/6t^3,"},{"Start":"04:02.960 ","End":"04:05.060","Text":"and that\u0027s the answer."},{"Start":"04:05.060 ","End":"04:08.160","Text":"That concludes this exercise."}],"ID":30735},{"Watched":false,"Name":"Exercise 8","Duration":"6m 34s","ChapterTopicVideoID":29153,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.190","Text":"In this exercise, we are given that u(x,t) is a solution"},{"Start":"00:05.190 ","End":"00:10.485","Text":"to the following wave equation with initial and boundary conditions."},{"Start":"00:10.485 ","End":"00:12.960","Text":"It\u0027s our semi-infinite interval,"},{"Start":"00:12.960 ","End":"00:15.695","Text":"and our task is to compute the following limit."},{"Start":"00:15.695 ","End":"00:21.660","Text":"Limit as x goes to infinity of each of the x times u(x,2 ), when t equals 2,"},{"Start":"00:21.660 ","End":"00:23.370","Text":"we get a function of x,"},{"Start":"00:23.370 ","End":"00:27.825","Text":"and we\u0027re given a hint to define v in terms of u as follows."},{"Start":"00:27.825 ","End":"00:29.700","Text":"An Alpha and Beta will be constants,"},{"Start":"00:29.700 ","End":"00:32.745","Text":"which we\u0027ll determine later."},{"Start":"00:32.745 ","End":"00:38.880","Text":"This is not the standard wave equation we had that we applied d\u0027Alembert\u0027s formula to,"},{"Start":"00:38.880 ","End":"00:40.755","Text":"we have these 2 extra bits."},{"Start":"00:40.755 ","End":"00:48.535","Text":"The standard form is when there\u0027s just a function of x and t besides the u_xx,"},{"Start":"00:48.535 ","End":"00:50.205","Text":"can\u0027t have you also."},{"Start":"00:50.205 ","End":"00:51.620","Text":"We need a trick,"},{"Start":"00:51.620 ","End":"00:54.080","Text":"and the trick is just the hint that is given,"},{"Start":"00:54.080 ","End":"00:56.660","Text":"if we bring this exponent to the other side,"},{"Start":"00:56.660 ","End":"00:59.060","Text":"we get u in terms of v as follows,"},{"Start":"00:59.060 ","End":"01:02.345","Text":"and then we\u0027re going to compute the partial derivatives,"},{"Start":"01:02.345 ","End":"01:06.155","Text":"du by dt is using the product rule."},{"Start":"01:06.155 ","End":"01:08.221","Text":"Well, we\u0027ve done this before, It\u0027s as follows."},{"Start":"01:08.221 ","End":"01:11.410","Text":"Then we differentiate again with respect to t,"},{"Start":"01:11.410 ","End":"01:12.895","Text":"this is what we get."},{"Start":"01:12.895 ","End":"01:16.400","Text":"As usual, there\u0027s a middle term, it appears twice."},{"Start":"01:16.400 ","End":"01:22.990","Text":"We can rewrite it as twice this and we can take the exponent outside the brackets."},{"Start":"01:22.990 ","End":"01:25.790","Text":"Now similarly for x, well,"},{"Start":"01:25.790 ","End":"01:29.555","Text":"I won\u0027t go into the details, ux and u_xx are similarly obtained."},{"Start":"01:29.555 ","End":"01:32.885","Text":"Now we recall our PDE,"},{"Start":"01:32.885 ","End":"01:34.355","Text":"this one here,"},{"Start":"01:34.355 ","End":"01:37.305","Text":"and we\u0027ll substitute u_tt,"},{"Start":"01:37.305 ","End":"01:40.485","Text":"u_xx, ux from here,"},{"Start":"01:40.485 ","End":"01:42.735","Text":"and u from here."},{"Start":"01:42.735 ","End":"01:45.275","Text":"What we get is the following."},{"Start":"01:45.275 ","End":"01:47.330","Text":"We can take all these exponents,"},{"Start":"01:47.330 ","End":"01:51.245","Text":"e to the Alpha Beta t out the brackets and then it will cancel."},{"Start":"01:51.245 ","End":"01:58.640","Text":"We can also then open brackets here and color-code the different types vt,"},{"Start":"01:58.640 ","End":"02:01.280","Text":"v, vx, and so on."},{"Start":"02:01.280 ","End":"02:04.225","Text":"Now we can collect like terms together,"},{"Start":"02:04.225 ","End":"02:06.375","Text":"and this is what we get."},{"Start":"02:06.375 ","End":"02:11.895","Text":"This will be the simplest if we get vx and vt to be 0,"},{"Start":"02:11.895 ","End":"02:17.910","Text":"and we\u0027ll get this by letting Alpha equal minus 1 and Beta equal 0."},{"Start":"02:17.910 ","End":"02:22.040","Text":"Then by luck, this thing also comes out to be 0."},{"Start":"02:22.040 ","End":"02:24.385","Text":"Just substitute and check."},{"Start":"02:24.385 ","End":"02:29.295","Text":"What we\u0027re left with here is just v_tt equals v_xx,"},{"Start":"02:29.295 ","End":"02:31.740","Text":"and also substituting alpha equals minus 1,"},{"Start":"02:31.740 ","End":"02:37.715","Text":"and Beta equal 0 in the original formula for v in terms of u."},{"Start":"02:37.715 ","End":"02:42.775","Text":"Then we get that v equals u(x,t) e^x."},{"Start":"02:42.775 ","End":"02:49.485","Text":"Let\u0027s remember these, this one and this equals this,"},{"Start":"02:49.485 ","End":"02:51.325","Text":"we\u0027ll need these later."},{"Start":"02:51.325 ","End":"02:54.425","Text":"Continuing on a clean page."},{"Start":"02:54.425 ","End":"02:58.340","Text":"Now, we had these equations for you,"},{"Start":"02:58.340 ","End":"03:01.640","Text":"the 2 initial conditions and the boundary condition,"},{"Start":"03:01.640 ","End":"03:06.660","Text":"and we can get from these corresponding equations for v just by"},{"Start":"03:06.660 ","End":"03:11.985","Text":"multiplying u by e^x is1."},{"Start":"03:11.985 ","End":"03:14.910","Text":"This multiplied by e^x is this,"},{"Start":"03:14.910 ","End":"03:17.430","Text":"0 times anything is 0."},{"Start":"03:17.430 ","End":"03:24.270","Text":"Now what we get is the problem in v, the PDE."},{"Start":"03:24.270 ","End":"03:26.310","Text":"Remember we had that here,"},{"Start":"03:26.310 ","End":"03:30.360","Text":"and these 3 copy from here."},{"Start":"03:30.360 ","End":"03:35.025","Text":"Let\u0027s denote this as f(x) and this as g(x),"},{"Start":"03:35.025 ","End":"03:37.459","Text":"and we have a Dirichlet boundary condition,"},{"Start":"03:37.459 ","End":"03:38.840","Text":"and we will solve this problem."},{"Start":"03:38.840 ","End":"03:41.360","Text":"Let\u0027s just remember what we were asked to do."},{"Start":"03:41.360 ","End":"03:43.475","Text":"We\u0027re asked to compute this limit."},{"Start":"03:43.475 ","End":"03:48.050","Text":"Let\u0027s see if we can convert this to a condition in terms of v. Well,"},{"Start":"03:48.050 ","End":"03:56.110","Text":"v(x,t) is u(x,t) e^x and v(x,2) is u(x,2) times e^x."},{"Start":"03:56.110 ","End":"04:00.155","Text":"The limit as x goes to infinity of this"},{"Start":"04:00.155 ","End":"04:05.065","Text":"is simply the limit as x goes to infinity of v(x,2)."},{"Start":"04:05.065 ","End":"04:10.130","Text":"Let\u0027s compute v. We\u0027ll solve this set of equations for v."},{"Start":"04:10.130 ","End":"04:12.515","Text":"Now we need an odd extension of the problem"},{"Start":"04:12.515 ","End":"04:15.739","Text":"because we\u0027re working with Dirichlet boundary condition."},{"Start":"04:15.739 ","End":"04:19.535","Text":"Get v_tt equals v_xx also for the tilde"},{"Start":"04:19.535 ","End":"04:25.165","Text":"and f tilde and g tilde need to be extended to odd functions."},{"Start":"04:25.165 ","End":"04:27.230","Text":"If it\u0027s 1 when x is positive,"},{"Start":"04:27.230 ","End":"04:28.970","Text":"it\u0027s minus 1 when x is negative,"},{"Start":"04:28.970 ","End":"04:31.520","Text":"if e to the minus x when x is positive,"},{"Start":"04:31.520 ","End":"04:35.135","Text":"then the minus becomes plus here and we get a minus here."},{"Start":"04:35.135 ","End":"04:37.805","Text":"Strictly speaking, we haven\u0027t defined it at 0."},{"Start":"04:37.805 ","End":"04:41.390","Text":"Well, you could say f tilde of 0 is 0 and u tilde of 0 is 0."},{"Start":"04:41.390 ","End":"04:43.410","Text":"But the function at 1 point,"},{"Start":"04:43.410 ","End":"04:44.990","Text":"not going to make any difference."},{"Start":"04:44.990 ","End":"04:49.451","Text":"In case you\u0027re wondering why we didn\u0027t bring this over,"},{"Start":"04:49.451 ","End":"04:50.885","Text":"well, this we get for free."},{"Start":"04:50.885 ","End":"04:52.970","Text":"Once we find a solution for the odd extension,"},{"Start":"04:52.970 ","End":"04:54.905","Text":"we get this for free."},{"Start":"04:54.905 ","End":"05:01.055","Text":"D\u0027Alembert\u0027s formula gives us that v tilde is as follows,"},{"Start":"05:01.055 ","End":"05:03.740","Text":"v tilde(x,2) is the same thing,"},{"Start":"05:03.740 ","End":"05:06.620","Text":"just replacing t with 2."},{"Start":"05:06.620 ","End":"05:09.050","Text":"Now we\u0027re concerned with what happens as x goes to infinity."},{"Start":"05:09.050 ","End":"05:11.765","Text":"So we can assume x is bigger than 2,"},{"Start":"05:11.765 ","End":"05:17.820","Text":"which means that both x plus 2 and x minus 2 are positive for here and here,"},{"Start":"05:17.820 ","End":"05:20.370","Text":"and also s is positive for here because it\u0027s"},{"Start":"05:20.370 ","End":"05:24.525","Text":"between x plus 2 and x minus plus 2 also positive."},{"Start":"05:24.525 ","End":"05:30.525","Text":"F tilde and f tilde(x,-2) are both"},{"Start":"05:30.525 ","End":"05:36.665","Text":"1 here and here and g tilde here is e to the minus s from here."},{"Start":"05:36.665 ","End":"05:41.449","Text":"Well, instead of x we have s. This becomes 1 plus 1 over"},{"Start":"05:41.449 ","End":"05:47.100","Text":"2 plus 1/2 the integral of e to the minus s ds."},{"Start":"05:47.100 ","End":"05:49.610","Text":"1 plus 1/2 is 1. The integral of e to the minus s is"},{"Start":"05:49.610 ","End":"05:53.615","Text":"e to the minus s over minus 1 between this limit and this limit,"},{"Start":"05:53.615 ","End":"05:55.970","Text":"and then well,"},{"Start":"05:55.970 ","End":"05:58.880","Text":"we can substitute the lower 1 minus the upper 1."},{"Start":"05:58.880 ","End":"06:00.140","Text":"Get rid of the minus 1,"},{"Start":"06:00.140 ","End":"06:02.410","Text":"just the reverse order of subtraction."},{"Start":"06:02.410 ","End":"06:07.610","Text":"Then the limit as x goes to infinity of v(x,2) which is"},{"Start":"06:07.610 ","End":"06:14.495","Text":"e^x u(x,2) is equal to the limit of the following expression as x goes to infinity,"},{"Start":"06:14.495 ","End":"06:16.480","Text":"1 is a constant,"},{"Start":"06:16.480 ","End":"06:19.115","Text":"e to the minus x goes to 0,"},{"Start":"06:19.115 ","End":"06:21.185","Text":"doesn\u0027t matter if you put a minus 2,"},{"Start":"06:21.185 ","End":"06:22.850","Text":"x minus 2 also goes to infinity,"},{"Start":"06:22.850 ","End":"06:24.365","Text":"x plus 2 goes to infinity."},{"Start":"06:24.365 ","End":"06:26.570","Text":"E to the minus infinity is 0."},{"Start":"06:26.570 ","End":"06:28.925","Text":"This comes out to be just 1."},{"Start":"06:28.925 ","End":"06:31.225","Text":"This is the answer we\u0027re looking for,"},{"Start":"06:31.225 ","End":"06:34.840","Text":"and that concludes this exercise."}],"ID":30736},{"Watched":false,"Name":"Exercise 9","Duration":"8m 27s","ChapterTopicVideoID":29154,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.450","Text":"In this exercise, we\u0027re given the wave equation u_tt equals u_xx."},{"Start":"00:06.450 ","End":"00:12.230","Text":"It\u0027s on the semi-infinite interval and it has initial conditions,"},{"Start":"00:12.230 ","End":"00:16.430","Text":"u(x, 0) and u_t(x, 0)."},{"Start":"00:16.430 ","End":"00:20.100","Text":"We have a boundary condition which is of"},{"Start":"00:20.100 ","End":"00:26.760","Text":"the Dirichlet type because it\u0027s u and not du by dx but it\u0027s not homogeneous."},{"Start":"00:26.760 ","End":"00:29.730","Text":"We\u0027ve learned what to do if it\u0027s a o here."},{"Start":"00:29.730 ","End":"00:31.350","Text":"In our case it\u0027s not o,"},{"Start":"00:31.350 ","End":"00:41.935","Text":"it\u0027s t. There\u0027s a standard technique or trick that we can perform in this case."},{"Start":"00:41.935 ","End":"00:46.110","Text":"What we\u0027ll do is add"},{"Start":"00:46.110 ","End":"00:52.960","Text":"a correction function w(t) to u in the hopes that v,"},{"Start":"00:52.960 ","End":"00:59.730","Text":"which is the sum, will satisfy the homogeneous Dirichlet boundary condition."},{"Start":"01:00.230 ","End":"01:05.450","Text":"Let\u0027s see what we get if we substitute x equals 0."},{"Start":"01:05.450 ","End":"01:07.700","Text":"We get v(0,"},{"Start":"01:07.700 ","End":"01:11.225","Text":"t) equals u (0, t) plus w(t)."},{"Start":"01:11.225 ","End":"01:15.095","Text":"Now we know what u(0, t) is,"},{"Start":"01:15.095 ","End":"01:17.360","Text":"it\u0027s t and v(0, t),"},{"Start":"01:17.360 ","End":"01:20.945","Text":"we know what we want it to be, which is 0."},{"Start":"01:20.945 ","End":"01:25.655","Text":"So we have the following and that gives us immediately that w(t)"},{"Start":"01:25.655 ","End":"01:32.250","Text":"is minus t. If you put w(t) here minus t,"},{"Start":"01:32.250 ","End":"01:35.055","Text":"then we get that v(x,"},{"Start":"01:35.055 ","End":"01:37.230","Text":"t) equals u(x,"},{"Start":"01:37.230 ","End":"01:43.860","Text":"t) minus t. Note that v_tt is the same as"},{"Start":"01:43.860 ","End":"01:52.165","Text":"u_tt because the second derivative of minus t with respect to t is 0."},{"Start":"01:52.165 ","End":"01:56.060","Text":"As a matter of fact, the second derivative with respect to x"},{"Start":"01:56.060 ","End":"02:00.125","Text":"is 0 already the first derivative is 0 with respect to x."},{"Start":"02:00.125 ","End":"02:03.175","Text":"So we have this equals this also."},{"Start":"02:03.175 ","End":"02:05.400","Text":"Let\u0027s see what we get."},{"Start":"02:05.400 ","End":"02:12.635","Text":"We start with these equations for u and let\u0027s convert them to equations for"},{"Start":"02:12.635 ","End":"02:21.180","Text":"v. Like we said because v_tt equals u_tt and v_xx equals u_xx,"},{"Start":"02:21.180 ","End":"02:27.405","Text":"this will give us the differential equation v_tt equals v_xx."},{"Start":"02:27.405 ","End":"02:32.820","Text":"Also, we already dealt with this and we have that v(0,"},{"Start":"02:32.820 ","End":"02:38.060","Text":"t) satisfies the Dirichlet boundary conditions are homogeneous."},{"Start":"02:38.060 ","End":"02:39.995","Text":"Now what we\u0027re missing is v(x,"},{"Start":"02:39.995 ","End":"02:44.840","Text":"0) and v_t(x, 0)."},{"Start":"02:44.840 ","End":"02:46.220","Text":"We know that v(x, t) is u(x,"},{"Start":"02:46.220 ","End":"02:50.150","Text":"t) minus t. All we have to do is subtract t and we"},{"Start":"02:50.150 ","End":"02:54.740","Text":"get that v(x) nought equals sine^2 x minus t,"},{"Start":"02:54.740 ","End":"02:56.119","Text":"which is 0 in this case."},{"Start":"02:56.119 ","End":"02:59.035","Text":"So this is just sine^2 x."},{"Start":"02:59.035 ","End":"03:02.820","Text":"We need also v_t(x) naught."},{"Start":"03:02.820 ","End":"03:06.780","Text":"Let\u0027s first of all see what v_t (x)_t in general is,"},{"Start":"03:06.780 ","End":"03:11.390","Text":"it\u0027s equal to this derivative with"},{"Start":"03:11.390 ","End":"03:16.760","Text":"respect to t minus derivative of this with respect to t, which is 1."},{"Start":"03:16.760 ","End":"03:23.420","Text":"What we get is this minus 1 sine x plus 1 minus 1."},{"Start":"03:23.420 ","End":"03:29.840","Text":"This 1 is just sine^2 x and this 1 is just sine x."},{"Start":"03:29.840 ","End":"03:36.590","Text":"This homogeneous Dirichlet boundary condition means that we\u0027ll be using"},{"Start":"03:36.590 ","End":"03:43.550","Text":"odd extensions to the whole line and those are denoted with a tilde on top."},{"Start":"03:43.550 ","End":"03:49.805","Text":"So v tilde by Dallin bears formula is the following."},{"Start":"03:49.805 ","End":"03:53.570","Text":"We haven\u0027t said what our f tilde and g tilde."},{"Start":"03:53.570 ","End":"03:56.255","Text":"First you have to say what our f and g. Well,"},{"Start":"03:56.255 ","End":"03:59.660","Text":"this is f and this is g. In other words,"},{"Start":"03:59.660 ","End":"04:06.340","Text":"f(x) is sine^2 x and g(x) is sine x and"},{"Start":"04:06.340 ","End":"04:12.950","Text":"the odd completion of g is itself because sine x is already odd."},{"Start":"04:12.950 ","End":"04:19.100","Text":"Sine^2 isn\u0027t so we have to take a piecewise definition for positive x,"},{"Start":"04:19.100 ","End":"04:22.085","Text":"the same thing, sine^2 x for negative x,"},{"Start":"04:22.085 ","End":"04:26.285","Text":"we replace x by minus x and then put a minus in front."},{"Start":"04:26.285 ","End":"04:31.490","Text":"This simplifies to just minus sine^2 x for negative x."},{"Start":"04:31.490 ","End":"04:36.815","Text":"We need to figure out what is f tilde (x) plus t and f tilde (x) minus t,"},{"Start":"04:36.815 ","End":"04:40.625","Text":"because this will depend on the values (x) and t,"},{"Start":"04:40.625 ","End":"04:47.150","Text":"f tilde (x) plus t will either be sine^2 (x) plus t or minus sine^2 (x) plus t,"},{"Start":"04:47.150 ","End":"04:51.545","Text":"depending on whether x is bigger than minus t. It should say"},{"Start":"04:51.545 ","End":"04:56.345","Text":"x plus t is bigger than 0 as the same as x bigger than minus t and here,"},{"Start":"04:56.345 ","End":"04:58.325","Text":"x plus t less than 0,"},{"Start":"04:58.325 ","End":"05:00.985","Text":"same thing as x less than minus t."},{"Start":"05:00.985 ","End":"05:04.850","Text":"Similarly for f tilde (x) minus is equal to this or this,"},{"Start":"05:04.850 ","End":"05:08.840","Text":"according to whether x minus t is bigger than 0 or less than 0,"},{"Start":"05:08.840 ","End":"05:11.640","Text":"which means x is bigger or less than t. Now we\u0027re going"},{"Start":"05:11.640 ","End":"05:14.585","Text":"to be restricting x to be positive."},{"Start":"05:14.585 ","End":"05:17.420","Text":"Not everything\u0027s possible."},{"Start":"05:17.420 ","End":"05:20.990","Text":"Here we have no problem if x is bigger than 0,"},{"Start":"05:20.990 ","End":"05:25.505","Text":"it\u0027s certainly bigger than minus t. If x"},{"Start":"05:25.505 ","End":"05:32.450","Text":"is positive then necessarily going to land on this case but here,"},{"Start":"05:32.450 ","End":"05:38.690","Text":"the possibilities that either x is bigger than t or if x is less than t,"},{"Start":"05:38.690 ","End":"05:40.610","Text":"then we write it as x less than t,"},{"Start":"05:40.610 ","End":"05:42.895","Text":"but x bigger than 0."},{"Start":"05:42.895 ","End":"05:45.135","Text":"These are the 2 cases."},{"Start":"05:45.135 ","End":"05:46.570","Text":"In both of these cases,"},{"Start":"05:46.570 ","End":"05:48.040","Text":"or in each of these cases,"},{"Start":"05:48.040 ","End":"05:53.290","Text":"x is going to be bigger than minus t. We read off the top line."},{"Start":"05:53.290 ","End":"05:54.910","Text":"Here it\u0027s definitely this,"},{"Start":"05:54.910 ","End":"05:57.385","Text":"but here it could be this or this."},{"Start":"05:57.385 ","End":"06:00.235","Text":"In the case where x is bigger than t,"},{"Start":"06:00.235 ","End":"06:05.005","Text":"we got that f tilde (x) plus t is sine^2 x plus"},{"Start":"06:05.005 ","End":"06:11.125","Text":"t and f tilde (x) minus t is the sine^2 (x) minus t. Well,"},{"Start":"06:11.125 ","End":"06:13.180","Text":"this is going to be always and this is either plus"},{"Start":"06:13.180 ","End":"06:15.745","Text":"or minus and we saw that in this case it\u0027s plus."},{"Start":"06:15.745 ","End":"06:19.695","Text":"G(s) is always sine s. This is what we have."},{"Start":"06:19.695 ","End":"06:23.050","Text":"This part it\u0027s already simplified as much as we can."},{"Start":"06:23.050 ","End":"06:25.535","Text":"We just have to work on the integral now."},{"Start":"06:25.535 ","End":"06:28.580","Text":"This integral is minus cosine x,"},{"Start":"06:28.580 ","End":"06:32.435","Text":"it\u0027s the integral of sine and these are the limits of integration."},{"Start":"06:32.435 ","End":"06:34.640","Text":"We substitute those to get rid of"},{"Start":"06:34.640 ","End":"06:38.450","Text":"the minus and the switch these 2 around and that\u0027s okay."},{"Start":"06:38.450 ","End":"06:41.840","Text":"So cosine (x) minus t minus cosine (x) plus"},{"Start":"06:41.840 ","End":"06:46.070","Text":"t. There\u0027s a formula for difference of cosines."},{"Start":"06:46.070 ","End":"06:51.005","Text":"I won\u0027t go into detail but if you apply that to here,"},{"Start":"06:51.005 ","End":"06:54.830","Text":"then we get minus twice, sine."},{"Start":"06:54.830 ","End":"07:00.574","Text":"This plus this over 2 is x and this minus this over 2"},{"Start":"07:00.574 ","End":"07:07.370","Text":"is minus t. The minus will cancel with the minus and the half with 2."},{"Start":"07:07.370 ","End":"07:09.830","Text":"That gives us the following."},{"Start":"07:09.830 ","End":"07:12.950","Text":"That brings us to the other case where x is"},{"Start":"07:12.950 ","End":"07:16.760","Text":"between 0 and t. These are all cases where x is positive."},{"Start":"07:16.760 ","End":"07:18.905","Text":"We saw that in that case,"},{"Start":"07:18.905 ","End":"07:25.040","Text":"f (x) plus t is still sine^2 (x) plus t but f (x) minus t comes out to be minus"},{"Start":"07:25.040 ","End":"07:28.490","Text":"sine^2 x minus t. Because we got that"},{"Start":"07:28.490 ","End":"07:32.225","Text":"x minus t was negative and so we needed the minus here."},{"Start":"07:32.225 ","End":"07:34.820","Text":"Everything else is the same."},{"Start":"07:34.820 ","End":"07:38.600","Text":"Just like before, because everything is the same"},{"Start":"07:38.600 ","End":"07:42.165","Text":"except for the sign here, sign, S-I-G-N."},{"Start":"07:42.165 ","End":"07:44.290","Text":"This is a minus."},{"Start":"07:44.330 ","End":"07:47.240","Text":"This boils down to sine x,"},{"Start":"07:47.240 ","End":"07:49.205","Text":"sine t, just like here."},{"Start":"07:49.205 ","End":"07:50.960","Text":"The only difference is here we have a plus,"},{"Start":"07:50.960 ","End":"07:52.565","Text":"here we have a minus."},{"Start":"07:52.565 ","End":"07:55.670","Text":"Combining those 2, we get v(x,"},{"Start":"07:55.670 ","End":"07:58.730","Text":"t) is in the first case it was this and"},{"Start":"07:58.730 ","End":"08:02.735","Text":"the second case it was this and the only difference is the sign here,"},{"Start":"08:02.735 ","End":"08:08.945","Text":"plus or minus but this is not the answer because we wanted u."},{"Start":"08:08.945 ","End":"08:10.640","Text":"But u(x, t) is v(x,"},{"Start":"08:10.640 ","End":"08:14.990","Text":"t) plus t. We just copy this here,"},{"Start":"08:14.990 ","End":"08:17.735","Text":"but add a plus t here and here."},{"Start":"08:17.735 ","End":"08:23.630","Text":"We drop the tilde because we\u0027re taking the case where x is positive."},{"Start":"08:23.630 ","End":"08:27.660","Text":"That\u0027s the answer for this exercise and we\u0027re done."}],"ID":30737},{"Watched":false,"Name":"Exercise 10","Duration":"6m 15s","ChapterTopicVideoID":29155,"CourseChapterTopicPlaylistID":294443,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.990","Text":"In this exercise, we have a wave equation,"},{"Start":"00:03.990 ","End":"00:07.710","Text":"which is this, and we have initial conditions,"},{"Start":"00:07.710 ","End":"00:09.150","Text":"these 2, and we have"},{"Start":"00:09.150 ","End":"00:14.340","Text":"a boundary condition which is needed because this is on a semi-infinite interval,"},{"Start":"00:14.340 ","End":"00:17.265","Text":"and we need a boundary where x is 0."},{"Start":"00:17.265 ","End":"00:21.750","Text":"This is a Dirichlet boundary condition which is homogeneous,"},{"Start":"00:21.750 ","End":"00:23.404","Text":"meaning it\u0027s 0 here."},{"Start":"00:23.404 ","End":"00:25.110","Text":"We know how to do this."},{"Start":"00:25.110 ","End":"00:31.890","Text":"We need to find an odd completion for all the functions f, g and u."},{"Start":"00:31.890 ","End":"00:33.960","Text":"For f, we get the following."},{"Start":"00:33.960 ","End":"00:39.285","Text":"It\u0027s x^2 when x is positive and minus of minus x^2,"},{"Start":"00:39.285 ","End":"00:42.810","Text":"which is just minus x^2 when x is negative."},{"Start":"00:42.810 ","End":"00:46.670","Text":"Similarly, for g sine(x)^2 from here,"},{"Start":"00:46.670 ","End":"00:48.110","Text":"when x is positive,"},{"Start":"00:48.110 ","End":"00:49.475","Text":"when x is negative,"},{"Start":"00:49.475 ","End":"00:51.695","Text":"we put a minus here and a minus here,"},{"Start":"00:51.695 ","End":"00:57.086","Text":"and it comes out minus sine^2x,"},{"Start":"00:57.086 ","End":"01:02.180","Text":"then we can apply d\u0027Alembert\u0027s formula and get what u Tilde is the odd completion of u,"},{"Start":"01:02.180 ","End":"01:03.945","Text":"which is the following."},{"Start":"01:03.945 ","End":"01:07.730","Text":"We really care only to find a solution where x"},{"Start":"01:07.730 ","End":"01:11.690","Text":"is positive because we\u0027re going t restrict u Tilde to positive x,"},{"Start":"01:11.690 ","End":"01:13.015","Text":"and then get u."},{"Start":"01:13.015 ","End":"01:15.060","Text":"When x is positive,"},{"Start":"01:15.060 ","End":"01:17.565","Text":"f Tilde of x plus Alpha t,"},{"Start":"01:17.565 ","End":"01:22.380","Text":"this term is x plus Alpha t^2 because if x is positive,"},{"Start":"01:22.380 ","End":"01:29.430","Text":"so is x plus Alpha t and f of x minus Alpha t depends,"},{"Start":"01:29.430 ","End":"01:33.965","Text":"x minus Alpha t could be positive or negative."},{"Start":"01:33.965 ","End":"01:38.360","Text":"Positive when x is bigger than Alpha t and negative when x is less"},{"Start":"01:38.360 ","End":"01:42.635","Text":"than Alpha t. That\u0027s this term and this term."},{"Start":"01:42.635 ","End":"01:44.690","Text":"What about g Tilde of s?"},{"Start":"01:44.690 ","End":"01:46.910","Text":"Well here, there are several cases,"},{"Start":"01:46.910 ","End":"01:49.945","Text":"and this is best explained with a diagram."},{"Start":"01:49.945 ","End":"01:52.080","Text":"The interval x minus Alpha t,"},{"Start":"01:52.080 ","End":"01:55.940","Text":"x plus Alpha t could be completely to the right of 0,"},{"Start":"01:55.940 ","End":"01:59.690","Text":"or it could be from before to after 0,"},{"Start":"01:59.690 ","End":"02:02.179","Text":"and it could be all to the left of 0."},{"Start":"02:02.179 ","End":"02:07.880","Text":"In this case, we have that x minus Alpha t is bigger or equal to 0,"},{"Start":"02:07.880 ","End":"02:10.415","Text":"so x bigger or equal to Alpha t,"},{"Start":"02:10.415 ","End":"02:14.150","Text":"and here x is also bigger than 0 automatically."},{"Start":"02:14.150 ","End":"02:20.600","Text":"In this case, what we get x minus Alpha t negative x plus Alpha t"},{"Start":"02:20.600 ","End":"02:23.450","Text":"positive gets us that x is between minus Alpha t and"},{"Start":"02:23.450 ","End":"02:28.205","Text":"Alpha t. But if we want it to be also bigger than 0,"},{"Start":"02:28.205 ","End":"02:34.530","Text":"then x just is between 0 and Alpha t, just this part."},{"Start":"02:34.570 ","End":"02:39.820","Text":"Similarly, for this case,"},{"Start":"02:39.820 ","End":"02:42.495","Text":"x less than minus Alpha t,"},{"Start":"02:42.495 ","End":"02:45.365","Text":"this will give us the empty set because we only care about"},{"Start":"02:45.365 ","End":"02:48.260","Text":"x positive and there is no positive x in this."},{"Start":"02:48.260 ","End":"02:50.540","Text":"Really, only these 2 cases count;"},{"Start":"02:50.540 ","End":"02:53.720","Text":"this possibility, or this possibility."},{"Start":"02:53.720 ","End":"02:55.490","Text":"In the first case,"},{"Start":"02:55.490 ","End":"02:57.820","Text":"x bigger or equal to Alpha t,"},{"Start":"02:57.820 ","End":"03:00.485","Text":"we have that s is positive."},{"Start":"03:00.485 ","End":"03:01.753","Text":"Go and look at the picture,"},{"Start":"03:01.753 ","End":"03:04.610","Text":"and you see that the whole interval is positive,"},{"Start":"03:04.610 ","End":"03:06.095","Text":"so s is positive,"},{"Start":"03:06.095 ","End":"03:08.930","Text":"so g Tilde of s is sine^2(s),"},{"Start":"03:08.930 ","End":"03:14.235","Text":"and this is the integral we have to compute using the trig identity for sine^2."},{"Start":"03:14.235 ","End":"03:17.330","Text":"This is 1 minus cosine twice the angle over 2."},{"Start":"03:17.330 ","End":"03:22.940","Text":"The 2 will put in with the 2 Alpha and this gives us 2 separate integrals."},{"Start":"03:22.940 ","End":"03:32.070","Text":"Integral of 1 is s over the 4 Alpha and the integral of cosine 2s is sine(2s)/2,"},{"Start":"03:32.070 ","End":"03:36.095","Text":"so it\u0027s over 8 Alpha between these 2 limits."},{"Start":"03:36.095 ","End":"03:40.415","Text":"We put in the upper and lower limits in s,"},{"Start":"03:40.415 ","End":"03:43.490","Text":"we get x plus Alpha t minus x minus Alpha t,"},{"Start":"03:43.490 ","End":"03:48.925","Text":"which is 2 Alpha t and 2 Alpha t/4 Alpha is t/2."},{"Start":"03:48.925 ","End":"03:56.290","Text":"For this we have sine of twice this minus sine of twice this."},{"Start":"03:56.290 ","End":"03:58.640","Text":"I won\u0027t do all the calculations."},{"Start":"03:58.640 ","End":"04:03.605","Text":"We have to use some trig identities for sine a minus sine b,"},{"Start":"04:03.605 ","End":"04:07.849","Text":"and this gives us something in terms of cosine and sine."},{"Start":"04:07.849 ","End":"04:10.535","Text":"Refer to trigonometric formulas."},{"Start":"04:10.535 ","End":"04:13.235","Text":"This is the answer we get for Case 1,"},{"Start":"04:13.235 ","End":"04:17.390","Text":"and that will give us that u(x,t) is the following."},{"Start":"04:17.390 ","End":"04:19.250","Text":"This is f(x) plus Alpha t,"},{"Start":"04:19.250 ","End":"04:25.895","Text":"which we computed as this f(x) minus Alpha t. This is just taken from here."},{"Start":"04:25.895 ","End":"04:27.595","Text":"If we open the brackets here,"},{"Start":"04:27.595 ","End":"04:31.413","Text":"the 2 Alpha t cancels with minus 2 Alpha t,"},{"Start":"04:31.413 ","End":"04:34.720","Text":"and we\u0027re left with 2x^2 plus 2Alpha^2 t^2 and the 2"},{"Start":"04:34.720 ","End":"04:38.510","Text":"disappears because there\u0027s a 2 here plus this bit,"},{"Start":"04:38.510 ","End":"04:40.610","Text":"and that\u0027s Case 1."},{"Start":"04:40.610 ","End":"04:42.575","Text":"Now let\u0027s do Case 2,"},{"Start":"04:42.575 ","End":"04:46.550","Text":"where x is between 0 and Alpha t. I forgot to say,"},{"Start":"04:46.550 ","End":"04:51.710","Text":"here we can drop the Tilde because we have x bigger than 0 in our computation."},{"Start":"04:51.710 ","End":"04:56.345","Text":"When x is bigger than 0, u Tilde is u. K is 2."},{"Start":"04:56.345 ","End":"05:00.530","Text":"Again, we need to figure out what the integral of g Tilde of s is."},{"Start":"05:00.530 ","End":"05:03.215","Text":"In this case, we break it up into 2 pieces;"},{"Start":"05:03.215 ","End":"05:06.685","Text":"the integral from x minus Alpha t to 0,"},{"Start":"05:06.685 ","End":"05:11.625","Text":"plus the integral from 0 to x plus Alpha t. Here,"},{"Start":"05:11.625 ","End":"05:13.575","Text":"s is negative,"},{"Start":"05:13.575 ","End":"05:17.920","Text":"so g Tilde of s is minus sine^2(s),"},{"Start":"05:17.920 ","End":"05:19.130","Text":"whereas here it\u0027s positive,"},{"Start":"05:19.130 ","End":"05:21.505","Text":"so it\u0027s plus sine^2(s)."},{"Start":"05:21.505 ","End":"05:25.025","Text":"Again, as per the computations,"},{"Start":"05:25.025 ","End":"05:27.575","Text":"using trigonometric formulas,"},{"Start":"05:27.575 ","End":"05:29.150","Text":"we get in the end,"},{"Start":"05:29.150 ","End":"05:34.150","Text":"x over 2 Alpha minus sine(2x) cosine 2 Alpha t over 4 Alpha."},{"Start":"05:34.150 ","End":"05:36.585","Text":"This is for when x is positive."},{"Start":"05:36.585 ","End":"05:39.720","Text":"This is the same as u of x and t,"},{"Start":"05:39.720 ","End":"05:42.795","Text":"f(x) plus Alpha t came out to be x plus Alpha t^2,"},{"Start":"05:42.795 ","End":"05:47.010","Text":"but f of x minus Alpha t came out to be minus x minus Alpha t^2."},{"Start":"05:47.010 ","End":"05:49.100","Text":"This part is just from here."},{"Start":"05:49.100 ","End":"05:52.095","Text":"Then simplifying here,"},{"Start":"05:52.095 ","End":"05:56.310","Text":"the x^2 and the x^2 cancel and the Alpha^2 t^2 cancels,"},{"Start":"05:56.310 ","End":"06:05.730","Text":"and we\u0027re just left with 4 Alpha xt over 2 is only 2 Alpha xt plus this minus this."},{"Start":"06:05.730 ","End":"06:07.260","Text":"We\u0027ve done Case 1 and Case 2."},{"Start":"06:07.260 ","End":"06:09.395","Text":"Now we just have to combine them here."},{"Start":"06:09.395 ","End":"06:10.760","Text":"This was Case 1,"},{"Start":"06:10.760 ","End":"06:12.275","Text":"this is Case 2,"},{"Start":"06:12.275 ","End":"06:15.510","Text":"and this is the answer, and we\u0027re done."}],"ID":30738}],"Thumbnail":null,"ID":294443},{"Name":"Finite Interval, Homogeneous Equation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Finite Interval, Homogeneous Equation","Duration":"2m 56s","ChapterTopicVideoID":29138,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"Continuing with the wave equation,"},{"Start":"00:03.510 ","End":"00:07.815","Text":"we\u0027ve done an infinite interval and a semi-infinite interval,"},{"Start":"00:07.815 ","End":"00:10.470","Text":"and now we\u0027re going to do"},{"Start":"00:10.470 ","End":"00:12.960","Text":"the finite interval using"},{"Start":"00:12.960 ","End":"00:17.085","Text":"a specific technique called the method of separation of variables."},{"Start":"00:17.085 ","End":"00:19.065","Text":"Can\u0027t see what the problem is,"},{"Start":"00:19.065 ","End":"00:24.075","Text":"the wave equation u_tt equals a^2u_xx,"},{"Start":"00:24.075 ","End":"00:26.727","Text":"and this time finite interval,"},{"Start":"00:26.727 ","End":"00:29.775","Text":"I\u0027ll take it from 0 to L,"},{"Start":"00:29.775 ","End":"00:33.320","Text":"but it really could be any finite interval from A to B."},{"Start":"00:33.320 ","End":"00:37.639","Text":"You can just adjust it to make a substitution or something."},{"Start":"00:37.639 ","End":"00:42.830","Text":"The initial conditions will be similar to the infinite and semi-infinite case."},{"Start":"00:42.830 ","End":"00:47.165","Text":"We\u0027re going to be given u(x_naught) and u_t(x_naught)."},{"Start":"00:47.165 ","End":"00:48.740","Text":"In other words, at time 0,"},{"Start":"00:48.740 ","End":"00:50.765","Text":"when x is in our interval,"},{"Start":"00:50.765 ","End":"00:52.230","Text":"we\u0027ve got 2 functions,"},{"Start":"00:52.230 ","End":"00:54.255","Text":"f(x) and g(x),"},{"Start":"00:54.255 ","End":"00:56.400","Text":"and the boundary conditions,"},{"Start":"00:56.400 ","End":"00:59.745","Text":"2 of them, because we have 2 end points,"},{"Start":"00:59.745 ","End":"01:08.180","Text":"and there\u0027s what we call Dirichlet boundary conditions will be given u(0,t) and u(L,t),"},{"Start":"01:08.180 ","End":"01:10.315","Text":"in this case both 0."},{"Start":"01:10.315 ","End":"01:16.245","Text":"What this means is that these 2 end points at 0 and L are bound to 0."},{"Start":"01:16.245 ","End":"01:18.560","Text":"They\u0027re fixed and as time goes by,"},{"Start":"01:18.560 ","End":"01:20.480","Text":"the wave may go up and down,"},{"Start":"01:20.480 ","End":"01:23.525","Text":"but these 2 points remain fixed."},{"Start":"01:23.525 ","End":"01:26.990","Text":"There\u0027s also the von Neumann boundary condition."},{"Start":"01:26.990 ","End":"01:32.645","Text":"That\u0027s where we\u0027re given the partial derivatives at the end points."},{"Start":"01:32.645 ","End":"01:38.270","Text":"In this case, the endpoints are free to move as illustrated here,"},{"Start":"01:38.270 ","End":"01:41.030","Text":"but the angle will be the same."},{"Start":"01:41.030 ","End":"01:46.195","Text":"It\u0027ll be horizontal when it comes out here and it\u0027ll be horizontal here when it goes in."},{"Start":"01:46.195 ","End":"01:49.265","Text":"But other than that, it can move up and down,"},{"Start":"01:49.265 ","End":"01:53.195","Text":"0 being a horizontal gradient."},{"Start":"01:53.195 ","End":"01:55.025","Text":"We can also mix and match."},{"Start":"01:55.025 ","End":"01:57.739","Text":"I mean, you could take for the left end point"},{"Start":"01:57.739 ","End":"02:01.865","Text":"Dirichlet and for the right von Neumann or vice versa."},{"Start":"02:01.865 ","End":"02:04.550","Text":"Actually, there\u0027s 4 possibilities here."},{"Start":"02:04.550 ","End":"02:05.869","Text":"We\u0027ve got this possibility,"},{"Start":"02:05.869 ","End":"02:06.957","Text":"this one, this one,"},{"Start":"02:06.957 ","End":"02:11.010","Text":"and this one, is just 1 times 1 combinatorically."},{"Start":"02:11.010 ","End":"02:15.560","Text":"Like I said, we\u0027ll learn to do these using separation of variables,"},{"Start":"02:15.560 ","End":"02:19.745","Text":"which you might have seen before and the stages for solving."},{"Start":"02:19.745 ","End":"02:22.595","Text":"Well, you\u0027ll see them in the exercises,"},{"Start":"02:22.595 ","End":"02:24.270","Text":"but I\u0027ll just give you an outline,"},{"Start":"02:24.270 ","End":"02:28.655","Text":"and though this can\u0027t really be understood until you actually get into an example."},{"Start":"02:28.655 ","End":"02:34.595","Text":"We first find solutions of the form a function of x times a function of t,"},{"Start":"02:34.595 ","End":"02:38.465","Text":"and after we found a sequence of these,"},{"Start":"02:38.465 ","End":"02:41.450","Text":"then we take a linear combination,"},{"Start":"02:41.450 ","End":"02:43.685","Text":"finite or infinite of these,"},{"Start":"02:43.685 ","End":"02:46.175","Text":"and that gives us a more general solution."},{"Start":"02:46.175 ","End":"02:50.570","Text":"We find these coefficients using the initial and boundary conditions."},{"Start":"02:50.570 ","End":"02:53.255","Text":"But like I said, you\u0027ll see this in the exercises."},{"Start":"02:53.255 ","End":"02:56.400","Text":"That\u0027s all for this introductory clip."}],"ID":30739},{"Watched":false,"Name":"Exercise 1","Duration":"6m 15s","ChapterTopicVideoID":29139,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.610","Text":"In this exercise, we\u0027re going to solve"},{"Start":"00:02.610 ","End":"00:07.560","Text":"the wave equation using the technique of separation of variables."},{"Start":"00:07.560 ","End":"00:10.050","Text":"This is the wave equation,"},{"Start":"00:10.050 ","End":"00:18.600","Text":"and in this case it\u0027s defined from 0-1 for x and time from 0 onwards to infinity."},{"Start":"00:18.600 ","End":"00:24.990","Text":"We have a couple of initial conditions and some boundary conditions."},{"Start":"00:24.990 ","End":"00:28.203","Text":"Now I\u0027ll remind you the stages for solving,"},{"Start":"00:28.203 ","End":"00:29.820","Text":"and let\u0027s get started."},{"Start":"00:29.820 ","End":"00:33.110","Text":"Looking for a solution u of this form,"},{"Start":"00:33.110 ","End":"00:35.960","Text":"this is the separation of variables as in"},{"Start":"00:35.960 ","End":"00:40.925","Text":"the name x separately and t separately and the product of the 2."},{"Start":"00:40.925 ","End":"00:45.050","Text":"We want to apply this wave equation, u_tt=u_xx."},{"Start":"00:45.050 ","End":"00:46.900","Text":"Les see u_xx,"},{"Start":"00:46.900 ","End":"00:50.135","Text":"you just have to differentiate the x part twice,"},{"Start":"00:50.135 ","End":"00:53.765","Text":"u_tt just differentiate the t part twice."},{"Start":"00:53.765 ","End":"00:57.965","Text":"Now we\u0027re going to make these 2 equal so that\u0027s this"},{"Start":"00:57.965 ","End":"01:03.920","Text":"and bring the x to the other side and the t to the other side."},{"Start":"01:03.920 ","End":"01:10.625","Text":"This is what we have and if a function of x=a function of t identically,"},{"Start":"01:10.625 ","End":"01:13.520","Text":"then it has to equal a constant function."},{"Start":"01:13.520 ","End":"01:16.700","Text":"We\u0027ll call the function minus Lambda,"},{"Start":"01:16.700 ","End":"01:18.390","Text":"the minus for convenience,"},{"Start":"01:18.390 ","End":"01:20.550","Text":"ignore the t part for a moment."},{"Start":"01:20.550 ","End":"01:22.840","Text":"We have x\u0027\u0027 over x is minus Lambda,"},{"Start":"01:22.840 ","End":"01:25.535","Text":"so x\u0027\u0027 is minus Lambda x."},{"Start":"01:25.535 ","End":"01:26.780","Text":"Bring this over to the other side,"},{"Start":"01:26.780 ","End":"01:29.585","Text":"x\u0027\u0027 plus Lambda x is 0."},{"Start":"01:29.585 ","End":"01:31.595","Text":"That\u0027s a differential equation,"},{"Start":"01:31.595 ","End":"01:34.025","Text":"ordinary just in x."},{"Start":"01:34.025 ","End":"01:40.025","Text":"Let\u0027s get an initial condition or 2 from this boundary condition."},{"Start":"01:40.025 ","End":"01:44.450","Text":"Just replace u with x times t and plug"},{"Start":"01:44.450 ","End":"01:50.345","Text":"in 0 and 1 for x and we get this and to this equals 0."},{"Start":"01:50.345 ","End":"01:52.615","Text":"The T(t) cancels,"},{"Start":"01:52.615 ","End":"01:56.525","Text":"so we get x(0) equals x(1)=0 and now"},{"Start":"01:56.525 ","End":"02:00.620","Text":"this together with this gives us an initial value problem,"},{"Start":"02:00.620 ","End":"02:03.980","Text":"second order differential equation in x."},{"Start":"02:03.980 ","End":"02:07.075","Text":"This is a Sturm Louisville boundary problem,"},{"Start":"02:07.075 ","End":"02:09.470","Text":"and we\u0027ve done exercises like this."},{"Start":"02:09.470 ","End":"02:11.435","Text":"We did something very similar."},{"Start":"02:11.435 ","End":"02:13.475","Text":"This one, in fact,"},{"Start":"02:13.475 ","End":"02:17.180","Text":"just about the same except that we don\u0027t have L,"},{"Start":"02:17.180 ","End":"02:20.315","Text":"it\u0027s just 1 and the variable,"},{"Start":"02:20.315 ","End":"02:22.175","Text":"the dependent one is y,"},{"Start":"02:22.175 ","End":"02:23.645","Text":"whereas here it\u0027s big X."},{"Start":"02:23.645 ","End":"02:26.360","Text":"The same thing. These are the eigenvalues,"},{"Start":"02:26.360 ","End":"02:28.934","Text":"these are the eigenfunctions,"},{"Start":"02:28.934 ","End":"02:31.250","Text":"and if you plug in L=1,"},{"Start":"02:31.250 ","End":"02:35.630","Text":"then the eigenvalues don\u0027t need the L here, it\u0027s 1,"},{"Start":"02:35.630 ","End":"02:42.020","Text":"so it\u0027s just nPi^2 and the L here also disappears so sin(nPi x),"},{"Start":"02:42.020 ","End":"02:45.170","Text":"and this is for n=1, 2, 3, etc."},{"Start":"02:45.170 ","End":"02:48.860","Text":"Now let\u0027s work on the other function T(t)."},{"Start":"02:48.860 ","End":"02:51.230","Text":"We had this, forget about the x now,"},{"Start":"02:51.230 ","End":"02:55.685","Text":"T\u0027\u0027 over t is minus Lambda n,"},{"Start":"02:55.685 ","End":"03:00.280","Text":"which is nPi^2 minus multiplied by t,"},{"Start":"03:00.280 ","End":"03:03.785","Text":"bring it over to the other side and we have this equation,"},{"Start":"03:03.785 ","End":"03:05.420","Text":"I called this Omega,"},{"Start":"03:05.420 ","End":"03:09.510","Text":"because we know how to solve this equation,"},{"Start":"03:09.510 ","End":"03:12.095","Text":"y\u0027\u0027 plus Omega^2 y is 0."},{"Start":"03:12.095 ","End":"03:15.950","Text":"It\u0027s some constant times cosine Omega x,"},{"Start":"03:15.950 ","End":"03:18.935","Text":"another constant times sine Omega x."},{"Start":"03:18.935 ","End":"03:20.705","Text":"Bring that back here."},{"Start":"03:20.705 ","End":"03:24.560","Text":"Let\u0027s call c_1 b_n and call c_2 a_n."},{"Start":"03:24.560 ","End":"03:32.960","Text":"This is the general solution to this for a given n and we\u0027ll use the initial condition."},{"Start":"03:32.960 ","End":"03:41.710","Text":"This one, u_t x_0 is 0 so this becomes X(x) T\u0027(0) and if this is 0,"},{"Start":"03:41.710 ","End":"03:48.170","Text":"then this is a constant and X(x) is a function unless it\u0027s identically 0,"},{"Start":"03:48.170 ","End":"03:50.675","Text":"T\u0027(0) has to be 0."},{"Start":"03:50.675 ","End":"03:53.540","Text":"Let\u0027s require that it\u0027s not going to be interesting if x"},{"Start":"03:53.540 ","End":"03:56.335","Text":"is identically 0 instead of 0 function."},{"Start":"03:56.335 ","End":"03:58.590","Text":"If T\u0027(0) is 0,"},{"Start":"03:58.590 ","End":"04:00.105","Text":"this is true for T_n,"},{"Start":"04:00.105 ","End":"04:03.870","Text":"for all the n so T_n\u0027(t),"},{"Start":"04:03.870 ","End":"04:06.005","Text":"in general, is this,"},{"Start":"04:06.005 ","End":"04:08.345","Text":"which we get by differentiating this."},{"Start":"04:08.345 ","End":"04:10.745","Text":"Now if we want T_n\u0027(0),"},{"Start":"04:10.745 ","End":"04:13.805","Text":"but n is 0, this is 0,"},{"Start":"04:13.805 ","End":"04:19.625","Text":"and the cos(0) is 1 so we just get nPi a_n,"},{"Start":"04:19.625 ","End":"04:21.800","Text":"so nPi a_n is 0,"},{"Start":"04:21.800 ","End":"04:25.555","Text":"then divide by nPi and we get that a_n is 0."},{"Start":"04:25.555 ","End":"04:29.310","Text":"If a_n is 0, T_n(t) is just this."},{"Start":"04:29.310 ","End":"04:33.225","Text":"So T_n(t) is b_n cos(nPi t)."},{"Start":"04:33.225 ","End":"04:39.960","Text":"Remember that X_n is sin(nPi x) so multiplying these 2,"},{"Start":"04:39.960 ","End":"04:42.430","Text":"and taking a linear combination,"},{"Start":"04:42.430 ","End":"04:46.646","Text":"we get u(x) t is the sum of this times this sin(nPi"},{"Start":"04:46.646 ","End":"04:52.000","Text":"x) times cos(nPi t) times a constant b_n."},{"Start":"04:52.000 ","End":"04:55.090","Text":"It remains to find the b_n and we\u0027ll use"},{"Start":"04:55.090 ","End":"05:00.550","Text":"the initial condition for u(x,0) is this function."},{"Start":"05:00.550 ","End":"05:06.490","Text":"We\u0027re given that u(x_naught) is equal to this in the initial condition."},{"Start":"05:06.490 ","End":"05:08.175","Text":"On the other hand,"},{"Start":"05:08.175 ","End":"05:13.115","Text":"we have that u(x,t) is equal to this and we can substitute t=0,"},{"Start":"05:13.115 ","End":"05:16.045","Text":"which will make this equal 1."},{"Start":"05:16.045 ","End":"05:20.010","Text":"We just get the sum b_n sin(nPi x), like so."},{"Start":"05:20.010 ","End":"05:22.910","Text":"Now, this is equal to this and this equals this so we can compare"},{"Start":"05:22.910 ","End":"05:26.210","Text":"these 2 and get this equation."},{"Start":"05:26.210 ","End":"05:27.710","Text":"Now, here\u0027s an infinite sum."},{"Start":"05:27.710 ","End":"05:29.405","Text":"Here, there are only 2 terms."},{"Start":"05:29.405 ","End":"05:33.860","Text":"Terms from here with n=1 and 5,"},{"Start":"05:33.860 ","End":"05:35.900","Text":"we can compare coefficients."},{"Start":"05:35.900 ","End":"05:37.730","Text":"Most of the b_n will be 0,"},{"Start":"05:37.730 ","End":"05:40.265","Text":"except when n=1 or 5,"},{"Start":"05:40.265 ","End":"05:41.900","Text":"b_1 will be a half,"},{"Start":"05:41.900 ","End":"05:43.820","Text":"b_5 minus 7,"},{"Start":"05:43.820 ","End":"05:46.010","Text":"and all the rest are 0."},{"Start":"05:46.010 ","End":"05:47.770","Text":"Now that we have the b_n,"},{"Start":"05:47.770 ","End":"05:52.070","Text":"we can plug back into this equation for u(x,t),"},{"Start":"05:52.070 ","End":"05:54.335","Text":"and we\u0027ll get just 2 terms."},{"Start":"05:54.335 ","End":"05:55.745","Text":"Again, when n is 1,"},{"Start":"05:55.745 ","End":"05:57.875","Text":"and then it\u0027s 5."},{"Start":"05:57.875 ","End":"05:59.950","Text":"When n is 1,"},{"Start":"05:59.950 ","End":"06:10.920","Text":"we get the b_1 times sin(1Pi x) cos(1Pi t) and here minus 7 sin(5Pi x) cos(5Pi t)."},{"Start":"06:10.920 ","End":"06:15.730","Text":"This is the answer, and we\u0027re done."}],"ID":30740},{"Watched":false,"Name":"Exercise 2","Duration":"9m 17s","ChapterTopicVideoID":29140,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.640","Text":"In this exercise, we\u0027re going to solve"},{"Start":"00:02.640 ","End":"00:07.515","Text":"the wave equation using the technique of separation of variables."},{"Start":"00:07.515 ","End":"00:13.725","Text":"Here are the stages for solving using separation of variables."},{"Start":"00:13.725 ","End":"00:15.285","Text":"In step 1,"},{"Start":"00:15.285 ","End":"00:19.815","Text":"we look for all solutions of the differential equation,"},{"Start":"00:19.815 ","End":"00:25.110","Text":"which has the form sum function of x times sum function of t. Usually,"},{"Start":"00:25.110 ","End":"00:27.315","Text":"there\u0027s a whole sequence of those."},{"Start":"00:27.315 ","End":"00:32.820","Text":"Then we write the general solution as a linear combination of these solutions."},{"Start":"00:32.820 ","End":"00:34.890","Text":"Let\u0027s say each of these is u_n,"},{"Start":"00:34.890 ","End":"00:38.085","Text":"then we have the sum A_n,"},{"Start":"00:38.085 ","End":"00:40.920","Text":"u_n and the A_ns are unknown constants."},{"Start":"00:40.920 ","End":"00:45.590","Text":"Then we use the initial land or boundary conditions to find these A_n."},{"Start":"00:45.590 ","End":"00:47.525","Text":"So let\u0027s start."},{"Start":"00:47.525 ","End":"00:54.260","Text":"u(x,t) will be X(x)T(t) We\u0027ll look and see if we can find solution of this form."},{"Start":"00:54.260 ","End":"00:58.405","Text":"What we have is u_tt=u_xx."},{"Start":"00:58.405 ","End":"01:02.015","Text":"So compute both u_xx and u_tt."},{"Start":"01:02.015 ","End":"01:05.165","Text":"You differentiate with respect to X^2,"},{"Start":"01:05.165 ","End":"01:07.445","Text":"you just get X\u0027\u0027 here."},{"Start":"01:07.445 ","End":"01:14.670","Text":"If you differentiate with respect to T^2 this T becomes T.\u0027\u0027 Now we equate this."},{"Start":"01:14.670 ","End":"01:19.140","Text":"We have this and rearrange and put the"},{"Start":"01:19.140 ","End":"01:24.545","Text":"X\u0027\u0027 on the denominator and the T\u0027\u0027 on the denominator."},{"Start":"01:24.545 ","End":"01:28.730","Text":"So we have a function of X identically equal to a function of"},{"Start":"01:28.730 ","End":"01:32.945","Text":"T. So it must be a constant function."},{"Start":"01:32.945 ","End":"01:38.285","Text":"Let\u0027s call it minus Lambda minus for convenience, you\u0027ll see."},{"Start":"01:38.285 ","End":"01:40.760","Text":"Ignore the T part for a moment."},{"Start":"01:40.760 ","End":"01:43.610","Text":"We have the X\u0027\u0027 over X equals minus Lambda."},{"Start":"01:43.610 ","End":"01:46.580","Text":"So X\u0027\u0027 is minus lambda X."},{"Start":"01:46.580 ","End":"01:49.805","Text":"To bring this to the other side and we get this."},{"Start":"01:49.805 ","End":"01:54.200","Text":"That\u0027s why I put a minus in front of the Lambda because I wanted a plus here."},{"Start":"01:54.200 ","End":"01:58.670","Text":"This is an ordinary differential equation on the second order but we don\u0027t have"},{"Start":"01:58.670 ","End":"02:03.815","Text":"any initial conditions or boundary conditions so we\u0027ll use this."},{"Start":"02:03.815 ","End":"02:08.570","Text":"We spell this out in terms of T(t)X(x)."},{"Start":"02:08.570 ","End":"02:16.910","Text":"We get X(0)T(t) equals X(1)T(t) is 0 divide by T(t)."},{"Start":"02:16.910 ","End":"02:20.240","Text":"We get X(0) equals X(1) is 0."},{"Start":"02:20.240 ","End":"02:23.240","Text":"We also have this here."},{"Start":"02:23.240 ","End":"02:25.505","Text":"So this is the differential equation part."},{"Start":"02:25.505 ","End":"02:27.110","Text":"This is the initial condition."},{"Start":"02:27.110 ","End":"02:32.015","Text":"What we have here should look familiar to Sturm–Liouville problem."},{"Start":"02:32.015 ","End":"02:33.830","Text":"In previous exercise,"},{"Start":"02:33.830 ","End":"02:36.815","Text":"we solved the problem of the following form,"},{"Start":"02:36.815 ","End":"02:39.155","Text":"which looks pretty much like this,"},{"Start":"02:39.155 ","End":"02:43.760","Text":"except that we have L in place of 1,"},{"Start":"02:43.760 ","End":"02:48.420","Text":"also 0 less than or equal to X less than or equal to 1 is what applies here."},{"Start":"02:48.420 ","End":"02:54.245","Text":"For this, we found eigenvalues and eigenfunctions as follows."},{"Start":"02:54.245 ","End":"02:58.130","Text":"For our case, just plug in L equals 1,"},{"Start":"02:58.130 ","End":"03:01.790","Text":"so here it\u0027s (nπ)^2,"},{"Start":"03:01.790 ","End":"03:07.150","Text":"and here sin(nπx) because that is 1,"},{"Start":"03:07.150 ","End":"03:10.535","Text":"so write those eigenvalues,"},{"Start":"03:10.535 ","End":"03:13.484","Text":"eigenfunction and n runs from 1,"},{"Start":"03:13.484 ","End":"03:15.590","Text":"2, 3, etc., to infinity."},{"Start":"03:15.590 ","End":"03:20.295","Text":"Now, let\u0027s focus on T we focused on X earlier."},{"Start":"03:20.295 ","End":"03:24.690","Text":"We have the T\u0027\u0027 over T equals minus Lambda^2."},{"Start":"03:24.690 ","End":"03:27.770","Text":"Now we have Lambda ns,"},{"Start":"03:27.770 ","End":"03:29.045","Text":"a sequence of them,"},{"Start":"03:29.045 ","End":"03:34.300","Text":"and minus Lambda_n^2 would be minus (nπ)^2."},{"Start":"03:34.300 ","End":"03:36.980","Text":"Can bring this to the other side."},{"Start":"03:36.980 ","End":"03:38.030","Text":"Forget about Lambda."},{"Start":"03:38.030 ","End":"03:39.440","Text":"Now just bring this over,"},{"Start":"03:39.440 ","End":"03:45.905","Text":"so we have T\u0027\u0027 plus (nπ)^2T is 0."},{"Start":"03:45.905 ","End":"03:48.995","Text":"If we let this be omega,"},{"Start":"03:48.995 ","End":"03:56.180","Text":"then it\u0027s a familiar second-order differential equation like in the Sturm–Liouville."},{"Start":"03:56.180 ","End":"04:02.830","Text":"The solution for this we saw was the combination of cosine Omega X, sine Omega X."},{"Start":"04:02.830 ","End":"04:04.590","Text":"Let\u0027s do that here."},{"Start":"04:04.590 ","End":"04:12.320","Text":"Only we don\u0027t have X this time we have T. We\u0027ll call the constants here b_n and a_n."},{"Start":"04:12.320 ","End":"04:17.060","Text":"So now we\u0027re going to use the initial condition that at time 0,"},{"Start":"04:17.060 ","End":"04:21.665","Text":"the partial derivative with respect to T is 0,"},{"Start":"04:21.665 ","End":"04:24.945","Text":"u_t is X times T\u0027."},{"Start":"04:24.945 ","End":"04:30.065","Text":"You just differentiate with respect to T and plug-in X(0) here this x here with 0."},{"Start":"04:30.065 ","End":"04:31.385","Text":"This is a constant,"},{"Start":"04:31.385 ","End":"04:32.645","Text":"this is a function."},{"Start":"04:32.645 ","End":"04:34.460","Text":"Function times a constant is 0,"},{"Start":"04:34.460 ","End":"04:36.454","Text":"the constant is 0."},{"Start":"04:36.454 ","End":"04:41.340","Text":"There is another possibility that X(x) is identically 0,"},{"Start":"04:41.340 ","End":"04:43.050","Text":"then it\u0027s not interesting."},{"Start":"04:43.050 ","End":"04:45.770","Text":"We get a trivial function,"},{"Start":"04:45.770 ","End":"04:47.990","Text":"just 0 even after we multiply it."},{"Start":"04:47.990 ","End":"04:51.485","Text":"So we\u0027ll go with T\u0027(0) equals 0."},{"Start":"04:51.485 ","End":"04:54.590","Text":"From this, if we differentiate it,"},{"Start":"04:54.590 ","End":"04:59.070","Text":"we have the following and if we plug in 0,"},{"Start":"04:59.070 ","End":"05:02.900","Text":"then get T_n\u0027(0) is 0,"},{"Start":"05:02.900 ","End":"05:07.885","Text":"but it\u0027s also equal to this with T equals 0."},{"Start":"05:07.885 ","End":"05:13.485","Text":"The sine of this is 0 because T is 0,"},{"Start":"05:13.485 ","End":"05:19.800","Text":"so this is 0, and cosine(nπt) T is 0 if cosine(0) is1."},{"Start":"05:19.800 ","End":"05:22.020","Text":"We just have the nπa_n,"},{"Start":"05:22.020 ","End":"05:27.095","Text":"so this is 0, nπ(0) 0 so a_n is 0."},{"Start":"05:27.095 ","End":"05:31.025","Text":"If you plug that in here that the a_n is 0,"},{"Start":"05:31.025 ","End":"05:33.783","Text":"then we get that T_n is just b_n"},{"Start":"05:33.783 ","End":"05:38.585","Text":"cosine(nπt) and a_n with the sine, the a_n just have to be 0."},{"Start":"05:38.585 ","End":"05:41.250","Text":"So this is T_n."},{"Start":"05:41.290 ","End":"05:45.415","Text":"Still here, X_n is this."},{"Start":"05:45.415 ","End":"05:48.330","Text":"Now that we have T_n and X_n,"},{"Start":"05:48.330 ","End":"05:53.750","Text":"we can have U_n by multiplying this by this."},{"Start":"05:53.750 ","End":"05:55.340","Text":"I\u0027ll skip this stage."},{"Start":"05:55.340 ","End":"05:57.080","Text":"U_n is this product,"},{"Start":"05:57.080 ","End":"06:01.930","Text":"and then we take a linear combination and get the general u."},{"Start":"06:01.930 ","End":"06:04.290","Text":"The next thing is to find the b_n."},{"Start":"06:04.290 ","End":"06:07.550","Text":"There\u0027s a condition we haven\u0027t used yet,"},{"Start":"06:07.550 ","End":"06:11.690","Text":"which is this one but for that,"},{"Start":"06:11.690 ","End":"06:15.880","Text":"Let\u0027s just see what is u(x,0) from this series."},{"Start":"06:15.880 ","End":"06:18.360","Text":"The cosine(0) is 1,"},{"Start":"06:18.360 ","End":"06:21.060","Text":"so the cosine drops off so we just have this."},{"Start":"06:21.060 ","End":"06:23.650","Text":"We now have two representations of u(x,0)."},{"Start":"06:23.650 ","End":"06:25.550","Text":"On the one hand, it\u0027s this series,"},{"Start":"06:25.550 ","End":"06:27.535","Text":"on the other hand, it\u0027s equal to 1."},{"Start":"06:27.535 ","End":"06:32.885","Text":"We get this series equals 1 and this is essentially"},{"Start":"06:32.885 ","End":"06:38.500","Text":"a Fourier sine series for the function which is constantly 1."},{"Start":"06:38.500 ","End":"06:40.935","Text":"Yes, basically what I said."},{"Start":"06:40.935 ","End":"06:45.950","Text":"We can find the coefficients b_n using formula,"},{"Start":"06:45.950 ","End":"06:51.020","Text":"I\u0027ll remind you for sine series of a function f on an interval 0,"},{"Start":"06:51.020 ","End":"06:54.550","Text":"L is given as follows."},{"Start":"06:54.550 ","End":"06:59.240","Text":"That works out in our case because when L is 1,"},{"Start":"06:59.240 ","End":"07:01.640","Text":"sine(nπx), which is what we have."},{"Start":"07:01.640 ","End":"07:05.300","Text":"This gives us a formula for finding the coefficients b_n."},{"Start":"07:05.300 ","End":"07:06.560","Text":"These are the ones we\u0027re looking for."},{"Start":"07:06.560 ","End":"07:11.990","Text":"So just have to figure out an integral for each n. We\u0027ll use"},{"Start":"07:11.990 ","End":"07:19.110","Text":"this formula with L equals 1 and f(x) equals 1."},{"Start":"07:19.110 ","End":"07:20.565","Text":"Let\u0027s see what we get."},{"Start":"07:20.565 ","End":"07:22.950","Text":"Just brought this over."},{"Start":"07:22.950 ","End":"07:30.735","Text":"Now b_n is equal to 2 over 1 which is 2. f(x) is 1."},{"Start":"07:30.735 ","End":"07:32.210","Text":"L here is 1,"},{"Start":"07:32.210 ","End":"07:33.935","Text":"so it\u0027s just n Pi x."},{"Start":"07:33.935 ","End":"07:37.920","Text":"So we have that b_n is this integral at c,"},{"Start":"07:37.920 ","End":"07:46.320","Text":"the integral of sine(πx) is minus cosine(nπx) divided by nπ,"},{"Start":"07:46.320 ","End":"07:49.330","Text":"and the 2 thrown it in here."},{"Start":"07:49.330 ","End":"07:53.060","Text":"I have to evaluate this between 0 and 1."},{"Start":"07:53.060 ","End":"07:58.695","Text":"If you plug in 1, it\u0027s just cosine(nπ)."},{"Start":"07:58.695 ","End":"08:01.455","Text":"Plug in 0, it\u0027s 1."},{"Start":"08:01.455 ","End":"08:10.945","Text":"So we have 1 minus cosine(nπx) and we know that this is minus 1 to b_n."},{"Start":"08:10.945 ","End":"08:18.364","Text":"We separate the odd from the even in terms of n. If n is even,"},{"Start":"08:18.364 ","End":"08:21.320","Text":"we get 1 minus 1 is 0 but if n is odd,"},{"Start":"08:21.320 ","End":"08:25.210","Text":"we get 1 plus 1 is 2."},{"Start":"08:25.210 ","End":"08:29.160","Text":"So n is odd means n is 2k minus 1,"},{"Start":"08:29.160 ","End":"08:33.960","Text":"so this is 2 times minus 2 is minus 4 and"},{"Start":"08:33.960 ","End":"08:39.570","Text":"replace n with 2k minus 1 and otherwise 0 when n is 2k."},{"Start":"08:39.570 ","End":"08:43.190","Text":"We have a formula for the coefficients b_n."},{"Start":"08:43.190 ","End":"08:46.325","Text":"Now returning to this,"},{"Start":"08:46.325 ","End":"08:50.240","Text":"what we had to do here was substitute the b_n."},{"Start":"08:50.240 ","End":"08:51.470","Text":"So now that we\u0027ve found them,"},{"Start":"08:51.470 ","End":"08:54.350","Text":"plug that in here and instead of b_n,"},{"Start":"08:54.350 ","End":"08:58.090","Text":"we have exactly this minus 4 over here,"},{"Start":"08:58.090 ","End":"09:03.905","Text":"but not n 2k minus 1 because we\u0027re only taking it over the odd numbers."},{"Start":"09:03.905 ","End":"09:06.200","Text":"So k goes from 1 to infinity."},{"Start":"09:06.200 ","End":"09:12.610","Text":"2k minus 1 runs over the odd numbers and also 2k minus 1 here in place of this n,"},{"Start":"09:12.610 ","End":"09:18.090","Text":"and 2k minus 1 in place of this n. That\u0027s the end of this exercise."}],"ID":30741},{"Watched":false,"Name":"Exercise 3","Duration":"9m 5s","ChapterTopicVideoID":29141,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.975","Text":"In this exercise, we\u0027re going to solve the wave equation"},{"Start":"00:03.975 ","End":"00:07.860","Text":"using the technique of separation of variables."},{"Start":"00:07.860 ","End":"00:11.010","Text":"This is the differential equation,"},{"Start":"00:11.010 ","End":"00:12.765","Text":"this is the domain,"},{"Start":"00:12.765 ","End":"00:15.390","Text":"and these are boundary conditions."},{"Start":"00:15.390 ","End":"00:21.015","Text":"We\u0027ve done this in previous exercises."},{"Start":"00:21.015 ","End":"00:24.075","Text":"You know what, I\u0027ll just leave it up here and I won\u0027t read it out."},{"Start":"00:24.075 ","End":"00:28.410","Text":"I\u0027ll just emphasize that the separation of variables means trying to find"},{"Start":"00:28.410 ","End":"00:31.110","Text":"a solution of the form some function of"},{"Start":"00:31.110 ","End":"00:35.055","Text":"x times some function of t. Let\u0027s start with that."},{"Start":"00:35.055 ","End":"00:37.650","Text":"Now, u(x, t),"},{"Start":"00:37.650 ","End":"00:40.605","Text":"just copying is X(x)T(t)."},{"Start":"00:40.605 ","End":"00:45.840","Text":"Then we wanted to meet this condition, u_tt equals 4u_xx."},{"Start":"00:45.840 ","End":"00:52.805","Text":"u_xx is gotten simply by differentiating X twice,"},{"Start":"00:52.805 ","End":"00:55.565","Text":"because T(t) is a constant as far as X goes."},{"Start":"00:55.565 ","End":"01:02.050","Text":"Similarly, u_tt is differentiating the function T twice."},{"Start":"01:02.050 ","End":"01:04.865","Text":"Now, setting these equal,"},{"Start":"01:04.865 ","End":"01:08.820","Text":"we\u0027ll let this be 4 times this."},{"Start":"01:08.820 ","End":"01:15.540","Text":"X T\u0027\u0027 is 4X\u0027\u0027T."},{"Start":"01:15.540 ","End":"01:18.695","Text":"Rearranging a bit, we get this."},{"Start":"01:18.695 ","End":"01:22.880","Text":"Now we have a function of X which is identically equal to a function of T,"},{"Start":"01:22.880 ","End":"01:27.605","Text":"so the only way that can happen is if each of these is the same constant."},{"Start":"01:27.605 ","End":"01:30.395","Text":"We\u0027ll call it minus Lambda,"},{"Start":"01:30.395 ","End":"01:33.210","Text":"for convenience, you\u0027ll see."},{"Start":"01:34.520 ","End":"01:36.770","Text":"Forget about the T part."},{"Start":"01:36.770 ","End":"01:38.825","Text":"Meanwhile, just look at this equals this."},{"Start":"01:38.825 ","End":"01:43.700","Text":"We can multiply by X(x) and then bring it to the other side,"},{"Start":"01:43.700 ","End":"01:45.065","Text":"and we have this,"},{"Start":"01:45.065 ","End":"01:47.585","Text":"which is an ordinary differential equation."},{"Start":"01:47.585 ","End":"01:50.860","Text":"It will be nice to have some boundary conditions."},{"Start":"01:50.860 ","End":"01:53.720","Text":"Well, we can get them from this condition."},{"Start":"01:53.720 ","End":"01:58.415","Text":"This gives us in terms of X and T the following."},{"Start":"01:58.415 ","End":"02:01.910","Text":"We can divide by T(t),"},{"Start":"02:01.910 ","End":"02:08.540","Text":"and that leaves us with this equation and these boundary conditions."},{"Start":"02:08.540 ","End":"02:10.460","Text":"This should be familiar."},{"Start":"02:10.460 ","End":"02:14.335","Text":"This is a Sturm-Liouville boundary value problem,"},{"Start":"02:14.335 ","End":"02:17.550","Text":"almost exactly except that when we studied it,"},{"Start":"02:17.550 ","End":"02:20.685","Text":"we had a y and not X."},{"Start":"02:20.685 ","End":"02:22.470","Text":"There was a general L,"},{"Start":"02:22.470 ","End":"02:24.445","Text":"which in our case is 1."},{"Start":"02:24.445 ","End":"02:27.290","Text":"Anyway, the solutions were as follows."},{"Start":"02:27.290 ","End":"02:31.230","Text":"The eigenvalues were nPi/L squared,"},{"Start":"02:31.230 ","End":"02:32.835","Text":"n equals 1, 2, 3, etc."},{"Start":"02:32.835 ","End":"02:37.680","Text":"The corresponding eigenfunctions were sine nPi/L x."},{"Start":"02:37.680 ","End":"02:40.245","Text":"If we put l equals 1 here,"},{"Start":"02:40.245 ","End":"02:42.555","Text":"this is just nPi,"},{"Start":"02:42.555 ","End":"02:46.690","Text":"and this is just nPi times x."},{"Start":"02:47.180 ","End":"02:50.960","Text":"These are the eigenvalues in our case,"},{"Start":"02:50.960 ","End":"02:55.355","Text":"and these are the eigenfunctions written as X_n."},{"Start":"02:55.355 ","End":"02:57.350","Text":"This is for n equals 1, 2,"},{"Start":"02:57.350 ","End":"02:59.615","Text":"3, etc, to infinity."},{"Start":"02:59.615 ","End":"03:04.895","Text":"Now that we\u0027ve found the X_n we\u0027re going to find the corresponding T_n."},{"Start":"03:04.895 ","End":"03:07.860","Text":"We\u0027ll return to this equation,"},{"Start":"03:08.470 ","End":"03:10.730","Text":"T\u0027\u0027/4T is minus Lambda."},{"Start":"03:10.730 ","End":"03:13.760","Text":"Now we know that Lambda has to be of this form."},{"Start":"03:13.760 ","End":"03:16.620","Text":"So from this equals this,"},{"Start":"03:16.620 ","End":"03:18.350","Text":"just rearranging it a bit,"},{"Start":"03:18.350 ","End":"03:20.765","Text":"we get to this equation."},{"Start":"03:20.765 ","End":"03:28.925","Text":"I labeled 2nPi as Omega because we know how to solve this equation like this,"},{"Start":"03:28.925 ","End":"03:32.690","Text":"just with y instead of T. We know that the solution to"},{"Start":"03:32.690 ","End":"03:39.230","Text":"this is the linear combination of cosine Omega x and sine Omega x."},{"Start":"03:39.230 ","End":"03:42.680","Text":"We\u0027ll call those constants here b_n and a_n,"},{"Start":"03:42.680 ","End":"03:46.610","Text":"they depend on n. Now we\u0027re going to use the initial condition."},{"Start":"03:46.610 ","End":"03:51.525","Text":"From this, we get X times T\u0027 is 0."},{"Start":"03:51.525 ","End":"03:53.550","Text":"The product is 0,"},{"Start":"03:53.550 ","End":"03:55.935","Text":"so this is 0 or this is 0."},{"Start":"03:55.935 ","End":"03:57.570","Text":"But this is a function."},{"Start":"03:57.570 ","End":"04:01.035","Text":"It\u0027s for all x in the domain."},{"Start":"04:01.035 ","End":"04:07.090","Text":"We have either some constant is 0 or that some function is identically 0."},{"Start":"04:07.090 ","End":"04:10.340","Text":"We\u0027ll throw this out, it\u0027s not an interesting case of X is 0,"},{"Start":"04:10.340 ","End":"04:12.080","Text":"then T times X is 0."},{"Start":"04:12.080 ","End":"04:13.610","Text":"Trivial case not interesting,"},{"Start":"04:13.610 ","End":"04:18.395","Text":"so we\u0027ll take this condition where T\u0027(0) is 0."},{"Start":"04:18.395 ","End":"04:25.575","Text":"To compute T\u0027, we could differentiate this and substitute 0."},{"Start":"04:25.575 ","End":"04:33.425","Text":"T_n\u0027, derivative of this cosine gives us minus sine and the anti-derivative 2nPi."},{"Start":"04:33.425 ","End":"04:35.990","Text":"Similarly, sine gives us cosine."},{"Start":"04:35.990 ","End":"04:37.755","Text":"Now plug in 0,"},{"Start":"04:37.755 ","End":"04:41.410","Text":"so T_n\u0027 of 0 is 0."},{"Start":"04:41.720 ","End":"04:44.310","Text":"This is 0,"},{"Start":"04:44.310 ","End":"04:46.110","Text":"sine of 0 is 0,"},{"Start":"04:46.110 ","End":"04:47.925","Text":"so all this is 0."},{"Start":"04:47.925 ","End":"04:49.950","Text":"Cosine of 0 is 1,"},{"Start":"04:49.950 ","End":"04:51.935","Text":"so we\u0027re just left with this bit."},{"Start":"04:51.935 ","End":"04:54.010","Text":"This bit is equal to 0."},{"Start":"04:54.010 ","End":"04:56.010","Text":"Since 2nPi is not 0,"},{"Start":"04:56.010 ","End":"04:58.020","Text":"we get that a_n is 0."},{"Start":"04:58.020 ","End":"05:01.140","Text":"Now substitute a_n equals 0 here,"},{"Start":"05:01.140 ","End":"05:04.245","Text":"so we\u0027re just left with the b_n part."},{"Start":"05:04.245 ","End":"05:08.040","Text":"T_n(t) is b_n cosine 2nPit."},{"Start":"05:08.040 ","End":"05:12.915","Text":"That\u0027s the T_n. We already have the X_n, a sine nPix."},{"Start":"05:12.915 ","End":"05:15.660","Text":"u_n is X_n times T_n,"},{"Start":"05:15.660 ","End":"05:17.970","Text":"which is this times this."},{"Start":"05:17.970 ","End":"05:23.585","Text":"Then we take infinite linear combination,"},{"Start":"05:23.585 ","End":"05:26.015","Text":"the sum from 1 to infinity of this,"},{"Start":"05:26.015 ","End":"05:29.165","Text":"and that\u0027s going to be the form of our solution,"},{"Start":"05:29.165 ","End":"05:34.590","Text":"and all we have to do still is to find the coefficients b_n."},{"Start":"05:34.590 ","End":"05:38.180","Text":"We\u0027ll use the initial condition we haven\u0027t used yet."},{"Start":"05:38.180 ","End":"05:42.890","Text":"u(x, naught) is minus 2x minus 1. u(x,"},{"Start":"05:42.890 ","End":"05:46.450","Text":"naught) means that t is 0,"},{"Start":"05:46.450 ","End":"05:49.715","Text":"so we go here, let t equals 0."},{"Start":"05:49.715 ","End":"05:53.000","Text":"Here\u0027s t. If t is 0,"},{"Start":"05:53.000 ","End":"05:57.440","Text":"then all of these 2nPit is 0 and cosine of it is 1,"},{"Start":"05:57.440 ","End":"06:03.700","Text":"so we basically could just drop this cosine off and get this, b_n sine nPix."},{"Start":"06:03.700 ","End":"06:05.985","Text":"This is going to equal minus 2x minus 1,"},{"Start":"06:05.985 ","End":"06:08.610","Text":"because that\u0027s what u(x, naught) is equal to."},{"Start":"06:08.610 ","End":"06:10.770","Text":"These two are equal,"},{"Start":"06:10.770 ","End":"06:14.234","Text":"and so this equals this."},{"Start":"06:14.234 ","End":"06:19.730","Text":"If we look at it this way, this looks exactly like a Fourier sine series."},{"Start":"06:19.730 ","End":"06:24.995","Text":"It is a sine series for f(x) equals minus 2x minus 1."},{"Start":"06:24.995 ","End":"06:27.380","Text":"I\u0027ll show you why this is so."},{"Start":"06:27.380 ","End":"06:31.790","Text":"We have the formula for a sine series of a function f on the interval 0,"},{"Start":"06:31.790 ","End":"06:34.615","Text":"L as the following formula."},{"Start":"06:34.615 ","End":"06:38.250","Text":"First of all let\u0027s see if the argument of the sine is correct."},{"Start":"06:38.250 ","End":"06:39.645","Text":"If L is 1,"},{"Start":"06:39.645 ","End":"06:41.490","Text":"which it is in our case, this is L,"},{"Start":"06:41.490 ","End":"06:45.390","Text":"this is 1, then nPi/L is just nPi,"},{"Start":"06:45.390 ","End":"06:46.875","Text":"so this is nPix."},{"Start":"06:46.875 ","End":"06:51.885","Text":"It has the right form, b_n sine nPix."},{"Start":"06:51.885 ","End":"06:58.370","Text":"That means that we can get these coefficients b_n that we need from this formula,"},{"Start":"06:58.370 ","End":"07:03.940","Text":"2/L integral from 0 to L f(x) sine nPi/L x dx."},{"Start":"07:03.940 ","End":"07:06.945","Text":"In this formula, L equals 1,"},{"Start":"07:06.945 ","End":"07:10.980","Text":"and f(x) is minus 2x minus 1."},{"Start":"07:10.980 ","End":"07:16.960","Text":"b_n is this integral just from here with these substitutions."},{"Start":"07:16.960 ","End":"07:21.350","Text":"We\u0027ll do this integral using the formula for integration by parts."},{"Start":"07:21.350 ","End":"07:22.640","Text":"I\u0027ve done it so many times,"},{"Start":"07:22.640 ","End":"07:24.710","Text":"I won\u0027t really go into detail about that."},{"Start":"07:24.710 ","End":"07:26.060","Text":"This is the one that we\u0027re going to"},{"Start":"07:26.060 ","End":"07:29.335","Text":"differentiate and this is the one that we\u0027re going to integrate,"},{"Start":"07:29.335 ","End":"07:33.305","Text":"and we get this times the integral of this,"},{"Start":"07:33.305 ","End":"07:41.150","Text":"which is this evaluated from 0-1 minus the integral of this derivative,"},{"Start":"07:41.150 ","End":"07:44.120","Text":"which is minus 2 times this again."},{"Start":"07:44.120 ","End":"07:46.025","Text":"There\u0027s a 2 in the beginning."},{"Start":"07:46.025 ","End":"07:49.520","Text":"Here we can simplify it a bit by taking the minus"},{"Start":"07:49.520 ","End":"07:53.605","Text":"here with the minuses here and getting it to be 2x plus 1."},{"Start":"07:53.605 ","End":"07:57.755","Text":"Here, again, the minus with the minus will make it a plus."},{"Start":"07:57.755 ","End":"08:00.430","Text":"We can bring the 2 out in front."},{"Start":"08:00.430 ","End":"08:02.880","Text":"Here when x is 1,"},{"Start":"08:02.880 ","End":"08:04.410","Text":"this is twice,"},{"Start":"08:04.410 ","End":"08:08.355","Text":"1 plus 1 is 3 cosine 1nPi."},{"Start":"08:08.355 ","End":"08:12.615","Text":"Then when it\u0027s 0, this is 1."},{"Start":"08:12.615 ","End":"08:16.680","Text":"3 cosine nPi minus 1 and the denominator nPi."},{"Start":"08:16.680 ","End":"08:19.670","Text":"Here the integral of cosine is sine,"},{"Start":"08:19.670 ","End":"08:21.710","Text":"but we have to divide by another nPi,"},{"Start":"08:21.710 ","End":"08:25.535","Text":"which makes it squared here and substitute 0,1."},{"Start":"08:25.535 ","End":"08:29.450","Text":"This part comes out 0 because when you substitute x is 0 or 1,"},{"Start":"08:29.450 ","End":"08:30.605","Text":"we get 0,"},{"Start":"08:30.605 ","End":"08:32.270","Text":"so we\u0027re just left with this."},{"Start":"08:32.270 ","End":"08:38.970","Text":"Cosine of nPi is minus 1^n,"},{"Start":"08:38.970 ","End":"08:40.530","Text":"we know this well."},{"Start":"08:40.530 ","End":"08:44.930","Text":"We get the b_n is this. We\u0027re almost done."},{"Start":"08:44.930 ","End":"08:47.795","Text":"We have b_n and we have the formula for u(x,"},{"Start":"08:47.795 ","End":"08:52.090","Text":"t) in terms of b_n and these functions."},{"Start":"08:52.090 ","End":"08:57.025","Text":"We just have to substitute for b_n, what\u0027s here."},{"Start":"08:57.025 ","End":"08:58.720","Text":"This is basically it,"},{"Start":"08:58.720 ","End":"09:01.270","Text":"just replace b_n by what we wrote here."},{"Start":"09:01.270 ","End":"09:03.040","Text":"This is the answer,"},{"Start":"09:03.040 ","End":"09:06.020","Text":"and that concludes this exercise."}],"ID":30742},{"Watched":false,"Name":"Exercise 4","Duration":"8m 38s","ChapterTopicVideoID":29142,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.970","Text":"In this exercise, we have a wave equation on a finite interval,"},{"Start":"00:05.970 ","End":"00:10.770","Text":"which we\u0027re going to solve using the technique of separation of variables."},{"Start":"00:10.770 ","End":"00:17.490","Text":"We\u0027re given here the initial conditions and the boundary conditions are mixed."},{"Start":"00:17.490 ","End":"00:22.605","Text":"On the left, we haven\u0027t Neumann condition with du by dx."},{"Start":"00:22.605 ","End":"00:26.655","Text":"On the right we have a Dirichlet condition with just plain u,"},{"Start":"00:26.655 ","End":"00:30.735","Text":"and here\u0027s a reminder of the stages for solving."},{"Start":"00:30.735 ","End":"00:33.485","Text":"I won\u0027t read it out as before,"},{"Start":"00:33.485 ","End":"00:35.805","Text":"you start with Step 1."},{"Start":"00:35.805 ","End":"00:43.565","Text":"We want to find x and t such that u(xt) is product of X(x) times T(t)."},{"Start":"00:43.565 ","End":"00:46.730","Text":"This is the second derivative with respect to x."},{"Start":"00:46.730 ","End":"00:49.055","Text":"Just differentiate this twice."},{"Start":"00:49.055 ","End":"00:51.810","Text":"With respect to t, we differentiate this twice,"},{"Start":"00:51.810 ","End":"00:57.725","Text":"and now we have our equation that this is equal to a^2 times this,"},{"Start":"00:57.725 ","End":"01:00.035","Text":"so this is what we get."},{"Start":"01:00.035 ","End":"01:04.835","Text":"We can just rearrange and get it in the following form."},{"Start":"01:04.835 ","End":"01:08.240","Text":"Because a function of x is equal to the function of t,"},{"Start":"01:08.240 ","End":"01:12.010","Text":"it must be a constant and we\u0027ll call this constant minus Lambda."},{"Start":"01:12.010 ","End":"01:14.335","Text":"If we just look at the x part,"},{"Start":"01:14.335 ","End":"01:18.560","Text":"we get x\u0027\u0027 equals minus Lambda x,"},{"Start":"01:18.560 ","End":"01:20.490","Text":"and bring it over to the other side."},{"Start":"01:20.490 ","End":"01:25.720","Text":"This differential equation we have we the boundary condition."},{"Start":"01:25.720 ","End":"01:29.135","Text":"Applying this to x times t,"},{"Start":"01:29.135 ","End":"01:30.545","Text":"which is what u is,"},{"Start":"01:30.545 ","End":"01:40.280","Text":"we get du by dx is the derivative of x0 times T(t) equals x at L times T(t)."},{"Start":"01:40.280 ","End":"01:44.375","Text":"This is going to be true for all t so we can cancel this,"},{"Start":"01:44.375 ","End":"01:46.800","Text":"T(t) is not identically 0."},{"Start":"01:46.800 ","End":"01:47.990","Text":"Well, it could be,"},{"Start":"01:47.990 ","End":"01:49.960","Text":"but that\u0027s not an interesting case."},{"Start":"01:49.960 ","End":"01:54.200","Text":"We get that x\u0027\u0027 plus lambda x equals 0,"},{"Start":"01:54.200 ","End":"01:56.315","Text":"which is our differential equation."},{"Start":"01:56.315 ","End":"01:59.420","Text":"This will be our initial condition or"},{"Start":"01:59.420 ","End":"02:02.975","Text":"boundary condition for the ordinary differential equation."},{"Start":"02:02.975 ","End":"02:08.420","Text":"We\u0027ve done very similar exercises in the chapter on Sturm Louisville."},{"Start":"02:08.420 ","End":"02:10.310","Text":"We\u0027ve solved maybe not exactly this,"},{"Start":"02:10.310 ","End":"02:15.020","Text":"but something basically like this and the solution is that"},{"Start":"02:15.020 ","End":"02:21.920","Text":"the eigenvalues are like so and the eigenfunctions are like so."},{"Start":"02:21.920 ","End":"02:27.950","Text":"In our case, we\u0027ll just call it big X rather than phi and this"},{"Start":"02:27.950 ","End":"02:34.430","Text":"is what we have lambda n and x_n and n is 1,2,3, etc."},{"Start":"02:34.430 ","End":"02:38.360","Text":"We found eigenvalues for x and"},{"Start":"02:38.360 ","End":"02:44.015","Text":"eigenvectors now let\u0027s turn to t. Forget about the x we have this equation in t,"},{"Start":"02:44.015 ","End":"02:47.765","Text":"but the Lambdas will be shared the same eigenvalues."},{"Start":"02:47.765 ","End":"02:52.745","Text":"Lambda will be 1 of these where n equals 1,2,3, etc."},{"Start":"02:52.745 ","End":"02:56.015","Text":"Rearrange this, we get the following."},{"Start":"02:56.015 ","End":"03:00.215","Text":"This is a straightforward second-order differential equation."},{"Start":"03:00.215 ","End":"03:05.150","Text":"This is the general form of the equation and we know that the solution to"},{"Start":"03:05.150 ","End":"03:11.120","Text":"such an equation is the combination of cosine omega x plus sine omega x."},{"Start":"03:11.120 ","End":"03:14.240","Text":"In this case, omega is what\u0027s in the brackets."},{"Start":"03:14.240 ","End":"03:15.760","Text":"This is omega."},{"Start":"03:15.760 ","End":"03:22.870","Text":"Now, the c_1 and c_2 will depend on n so let\u0027s call them b_n and a_n."},{"Start":"03:22.870 ","End":"03:27.250","Text":"For each n, we have this pair of constants and then we have"},{"Start":"03:27.250 ","End":"03:32.500","Text":"a cosine of omega t. Well, here it\u0027s an x,"},{"Start":"03:32.500 ","End":"03:36.580","Text":"and here we have t as the independent variable plus a_n"},{"Start":"03:36.580 ","End":"03:41.290","Text":"sine of omega t. Now we have X_n and T_n,"},{"Start":"03:41.290 ","End":"03:43.360","Text":"and we want a linear combination of them."},{"Start":"03:43.360 ","End":"03:46.180","Text":"We don\u0027t need the constant a_n we mentioned because"},{"Start":"03:46.180 ","End":"03:49.015","Text":"we already have constants here and here,"},{"Start":"03:49.015 ","End":"03:53.920","Text":"we can dispense with it and we know that n goes from 1 to infinity and"},{"Start":"03:53.920 ","End":"03:59.195","Text":"these are our X_n and T_n so if we substitute them,"},{"Start":"03:59.195 ","End":"04:08.010","Text":"this is X_n and this whole thing is T_n is actually complete Stage 1 of our instructions."},{"Start":"04:08.010 ","End":"04:12.230","Text":"Now we have Stage 2 where we have to find what these constants are,"},{"Start":"04:12.230 ","End":"04:16.275","Text":"the b_n and the a_n using the initial conditions,"},{"Start":"04:16.275 ","End":"04:23.945","Text":"I will take this one first u(x) notice cosine pi over 2 L times x and we do that,"},{"Start":"04:23.945 ","End":"04:28.365","Text":"we get substituting t equals 0 here and here,"},{"Start":"04:28.365 ","End":"04:34.685","Text":"this comes out to be 0 because we have sine of 0."},{"Start":"04:34.685 ","End":"04:37.040","Text":"Here we have cosine of 0,"},{"Start":"04:37.040 ","End":"04:41.345","Text":"which is 1 so all that\u0027s left of this square bracket is the b_n,"},{"Start":"04:41.345 ","End":"04:45.800","Text":"which we\u0027ll put in front of the cosine so we get that cosine of"},{"Start":"04:45.800 ","End":"04:50.975","Text":"pi over 2Lx is the sum 1 to infinity of this."},{"Start":"04:50.975 ","End":"04:55.265","Text":"Now we\u0027ll do something called comparing coefficients."},{"Start":"04:55.265 ","End":"04:58.070","Text":"If we expand the right-hand side,"},{"Start":"04:58.070 ","End":"05:03.390","Text":"we have cosine of 1 pi over 2 L and 3 pi over 2L,"},{"Start":"05:03.390 ","End":"05:07.940","Text":"5 Pi over 2L times x cosine of each of these."},{"Start":"05:07.940 ","End":"05:10.535","Text":"Anyway, this has to equal,"},{"Start":"05:10.535 ","End":"05:14.980","Text":"this because his is a sine series,"},{"Start":"05:14.980 ","End":"05:17.045","Text":"like a Fourier series,"},{"Start":"05:17.045 ","End":"05:26.120","Text":"and it\u0027s unique so b_1 has to equal 1 and the b_2 and b_3 and so on are going to equal 0."},{"Start":"05:26.120 ","End":"05:28.290","Text":"So 1 equals b_1,"},{"Start":"05:28.290 ","End":"05:30.675","Text":"0 is b_2, b_3, b_4, etc."},{"Start":"05:30.675 ","End":"05:33.030","Text":"Now that we have the b_n\u0027s,"},{"Start":"05:33.030 ","End":"05:36.035","Text":"we want to find the a_n\u0027s and for that,"},{"Start":"05:36.035 ","End":"05:40.790","Text":"we\u0027ll use the other initial condition, which was this."},{"Start":"05:40.790 ","End":"05:44.330","Text":"We don\u0027t have u_t would need to form"},{"Start":"05:44.330 ","End":"05:49.295","Text":"a differentiation of u with respect to t, this is u."},{"Start":"05:49.295 ","End":"05:51.845","Text":"This part just depends on x,"},{"Start":"05:51.845 ","End":"05:56.180","Text":"like a constant with respect to t derivative of cosine is minus"},{"Start":"05:56.180 ","End":"06:02.600","Text":"sine and then the antiderivative is this part here, written here."},{"Start":"06:02.600 ","End":"06:08.000","Text":"The other ones similar sine gives us cosine and this part here,"},{"Start":"06:08.000 ","End":"06:10.969","Text":"except for the t, comes out in front."},{"Start":"06:10.969 ","End":"06:15.485","Text":"If we let t equals 0 to get u_t (x) and 0,"},{"Start":"06:15.485 ","End":"06:19.100","Text":"we have this place that over here."},{"Start":"06:19.100 ","End":"06:22.370","Text":"Now, here, if we let t equals 0,"},{"Start":"06:22.370 ","End":"06:26.570","Text":"the sine part comes out 0 and the cosine of this is"},{"Start":"06:26.570 ","End":"06:31.730","Text":"1 so we just have this that we can put in frontier,"},{"Start":"06:31.730 ","End":"06:38.365","Text":"well split off the an first and then the pi 2 n minus 1 over 2L a."},{"Start":"06:38.365 ","End":"06:41.340","Text":"On the other hand, we have this."},{"Start":"06:41.340 ","End":"06:46.325","Text":"We have an equation that this equals this sum and again,"},{"Start":"06:46.325 ","End":"06:49.855","Text":"we\u0027ll do the comparison of coefficients."},{"Start":"06:49.855 ","End":"06:52.792","Text":"We expand the right-hand side,"},{"Start":"06:52.792 ","End":"06:55.520","Text":"we get when n is 1,"},{"Start":"06:55.520 ","End":"06:56.716","Text":"we have a 1 here,"},{"Start":"06:56.716 ","End":"06:58.010","Text":"when n is 2,"},{"Start":"06:58.010 ","End":"07:00.020","Text":"we have a 3, when n is 3,"},{"Start":"07:00.020 ","End":"07:01.595","Text":"we get a 5."},{"Start":"07:01.595 ","End":"07:07.160","Text":"I\u0027ve color-coded it so you can see what we\u0027re comparing so we get that 1 is equal"},{"Start":"07:07.160 ","End":"07:13.750","Text":"to this coefficient and also 1 equals this."},{"Start":"07:13.750 ","End":"07:17.125","Text":"Just writing it out,"},{"Start":"07:17.125 ","End":"07:18.950","Text":"we get the following."},{"Start":"07:18.950 ","End":"07:22.000","Text":"This gives us an a_2, is the following."},{"Start":"07:22.000 ","End":"07:27.850","Text":"Just bring it over to the other side and inverted and a_3 is this and the other a_n\u0027s"},{"Start":"07:27.850 ","End":"07:34.540","Text":"are 0 a_n times this is 0 and then divide by this,"},{"Start":"07:34.540 ","End":"07:36.415","Text":"we get that a_n is 0."},{"Start":"07:36.415 ","End":"07:39.670","Text":"These are the a_n\u0027s and the b_n\u0027s,"},{"Start":"07:39.670 ","End":"07:44.490","Text":"either 1 or 0 depending on whether n is 1 or not and"},{"Start":"07:44.490 ","End":"07:50.355","Text":"remember our formula for u only now we have all the a_n\u0027s and b_n\u0027s."},{"Start":"07:50.355 ","End":"07:58.665","Text":"It\u0027s only going to be 3 non-zero terms when n is not 1 or 2 or 3,"},{"Start":"07:58.665 ","End":"08:05.775","Text":"then all the a_n\u0027s and b_n\u0027s will be 0 so we only have 3 terms when n is 1,"},{"Start":"08:05.775 ","End":"08:10.145","Text":"then what we get is b_1, which is here."},{"Start":"08:10.145 ","End":"08:19.315","Text":"This is equal to pi over 2L x cosine and we just get this with n equals 1,"},{"Start":"08:19.315 ","End":"08:24.700","Text":"which is cosine of pi over 2L a_t."},{"Start":"08:24.700 ","End":"08:26.540","Text":"Then when n is 2 or 3,"},{"Start":"08:26.540 ","End":"08:31.670","Text":"we get this cosine with this sine with the appropriate n"},{"Start":"08:31.670 ","End":"08:38.550","Text":"straightforward and this is our final answer so we are done."}],"ID":30743},{"Watched":false,"Name":"Exercise 5","Duration":"7m 14s","ChapterTopicVideoID":29143,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.975","Text":"In this exercise, we have a wave equation which will solve using separation of variables."},{"Start":"00:06.975 ","End":"00:10.110","Text":"It\u0027s on a finite interval and"},{"Start":"00:10.110 ","End":"00:15.435","Text":"the boundary conditions are mixed and these are the initial conditions."},{"Start":"00:15.435 ","End":"00:21.120","Text":"Now, this is pretty much similar to previous exercises that we\u0027ve done."},{"Start":"00:21.120 ","End":"00:25.335","Text":"These are the 2 stages and you know what?"},{"Start":"00:25.335 ","End":"00:29.895","Text":"I\u0027ll just do this more quickly because it\u0027s very similar to previous exercises."},{"Start":"00:29.895 ","End":"00:37.500","Text":"This part is common to this exercise up to this point here, it\u0027s all the same."},{"Start":"00:37.500 ","End":"00:40.740","Text":"Go check previous few exercises and"},{"Start":"00:40.740 ","End":"00:44.880","Text":"the difference when we put the boundary conditions in."},{"Start":"00:44.880 ","End":"00:51.590","Text":"In this case, we get u of 0 and t is X of 0 T of t,"},{"Start":"00:51.590 ","End":"00:54.800","Text":"and du by dx because it\u0027s separated,"},{"Start":"00:54.800 ","End":"00:56.990","Text":"we just differentiate the x part,"},{"Start":"00:56.990 ","End":"01:04.805","Text":"x prime of L, T of t. Cancel the T of t and we get x Naught equals x prime of L equals 0."},{"Start":"01:04.805 ","End":"01:09.410","Text":"Together with the differential equation, this one,"},{"Start":"01:09.410 ","End":"01:12.575","Text":"we have an ordinary differential equation with"},{"Start":"01:12.575 ","End":"01:18.185","Text":"boundary conditions and this should be familiar in the Sturm Liouville\u0027s exercises,"},{"Start":"01:18.185 ","End":"01:19.550","Text":"we had the following problem,"},{"Start":"01:19.550 ","End":"01:24.800","Text":"which is practically the same and if you go back and look for it,"},{"Start":"01:24.800 ","End":"01:26.750","Text":"I\u0027m not sure what exercise number it is."},{"Start":"01:26.750 ","End":"01:32.075","Text":"We get the solution that the eigenvalues are as follows."},{"Start":"01:32.075 ","End":"01:36.200","Text":"Pi times an odd number over 2L^2,"},{"Start":"01:36.200 ","End":"01:38.240","Text":"where n is 1, 2, 3, etc."},{"Start":"01:38.240 ","End":"01:41.110","Text":"This is 1, 3, 5, 7, etc."},{"Start":"01:41.110 ","End":"01:43.625","Text":"The eigenfunctions are as follows."},{"Start":"01:43.625 ","End":"01:47.975","Text":"Writing this again with X instead of Phi,"},{"Start":"01:47.975 ","End":"01:50.030","Text":"these are our eigenvalues,"},{"Start":"01:50.030 ","End":"01:53.050","Text":"these are our eigenfunctions."},{"Start":"01:53.050 ","End":"01:57.785","Text":"These are the eigenfunctions for x, what about T?"},{"Start":"01:57.785 ","End":"02:00.830","Text":"We go back to this equation here,"},{"Start":"02:00.830 ","End":"02:03.200","Text":"but this time focus on the T part."},{"Start":"02:03.200 ","End":"02:06.250","Text":"The eigenvalues will be shared."},{"Start":"02:06.250 ","End":"02:08.370","Text":"Here\u0027s the equation."},{"Start":"02:08.370 ","End":"02:15.929","Text":"Now Lambda is 1 of the Lambda n\u0027s so it\u0027s 1 of these where n goes 1, 2, 3, etc."},{"Start":"02:15.929 ","End":"02:18.300","Text":"This gives us the equation,"},{"Start":"02:18.300 ","End":"02:19.640","Text":"just bring this to the other side,"},{"Start":"02:19.640 ","End":"02:23.060","Text":"multiply by T of t and the a"},{"Start":"02:23.060 ","End":"02:27.485","Text":"squared goes into here a squared so this is the equation we have,"},{"Start":"02:27.485 ","End":"02:30.710","Text":"and this is of the form y double prime plus Omega"},{"Start":"02:30.710 ","End":"02:34.234","Text":"squared y equals 0 that\u0027s a familiar equation."},{"Start":"02:34.234 ","End":"02:38.110","Text":"Omega is what\u0027s inside the brackets here."},{"Start":"02:38.110 ","End":"02:42.255","Text":"Using the solution, which is this in our case,"},{"Start":"02:42.255 ","End":"02:45.600","Text":"we get instead of c_1 we\u0027ll call it b_n and c_2,"},{"Start":"02:45.600 ","End":"02:49.410","Text":"we\u0027ll call it a_n and this of course depends on n. There\u0027s an n here,"},{"Start":"02:49.410 ","End":"02:53.390","Text":"for each n, it\u0027s a different solution so these are the constants."},{"Start":"02:53.390 ","End":"02:57.200","Text":"Normally, we look for a general solution of the form the sum of"},{"Start":"02:57.200 ","End":"03:01.220","Text":"X_n T_n but in the case of T_n we left the constants in,"},{"Start":"03:01.220 ","End":"03:03.005","Text":"so we don\u0027t need another constant."},{"Start":"03:03.005 ","End":"03:04.730","Text":"These already are constants."},{"Start":"03:04.730 ","End":"03:07.145","Text":"Now just to remind you, these are the X_n,"},{"Start":"03:07.145 ","End":"03:11.763","Text":"here are the T_n and we want the sum of X_n and T_n."},{"Start":"03:11.763 ","End":"03:14.135","Text":"This comes out to be the following."},{"Start":"03:14.135 ","End":"03:16.320","Text":"This part is the X_n,"},{"Start":"03:16.320 ","End":"03:20.115","Text":"this part is the T_n and n goes from 1 to infinity."},{"Start":"03:20.115 ","End":"03:22.665","Text":"That\u0027s the end of Stage 1."},{"Start":"03:22.665 ","End":"03:25.910","Text":"Stage 2, we\u0027re going to find what these constants are."},{"Start":"03:25.910 ","End":"03:31.865","Text":"That\u0027s missing, finding U is finding all the b_n and the a_n and for this,"},{"Start":"03:31.865 ","End":"03:34.850","Text":"we\u0027re going to use the initial conditions."},{"Start":"03:34.850 ","End":"03:38.075","Text":"The first initial condition is this one we also have another."},{"Start":"03:38.075 ","End":"03:40.090","Text":"Let\u0027s start with this one."},{"Start":"03:40.090 ","End":"03:45.420","Text":"Sine of 5 Pi over 2L x is u of x Naught,"},{"Start":"03:45.420 ","End":"03:48.035","Text":"just put t equals 0 here."},{"Start":"03:48.035 ","End":"03:50.330","Text":"The sine becomes 0,"},{"Start":"03:50.330 ","End":"03:52.685","Text":"we just have the cosine."},{"Start":"03:52.685 ","End":"03:55.515","Text":"Cosine is 1, sine is 0,"},{"Start":"03:55.515 ","End":"04:01.160","Text":"so all that\u0027s left from these square brackets is the b_n so put the b_n in front of"},{"Start":"04:01.160 ","End":"04:04.430","Text":"the sign here and this is what we get and if we"},{"Start":"04:04.430 ","End":"04:07.790","Text":"expand this so there have Sigma notation using a dot,"},{"Start":"04:07.790 ","End":"04:10.510","Text":"dot, dot, this is what we get."},{"Start":"04:10.510 ","End":"04:13.535","Text":"Then we\u0027re going to compare coefficients."},{"Start":"04:13.535 ","End":"04:16.790","Text":"The only thing that\u0027s in common is this and"},{"Start":"04:16.790 ","End":"04:20.300","Text":"this so b_3 has to equal the coefficient here,"},{"Start":"04:20.300 ","End":"04:24.475","Text":"which is 1, and all the others will be 0."},{"Start":"04:24.475 ","End":"04:29.180","Text":"Here we found all the b_n are all 0 except for b_3, which is 1."},{"Start":"04:29.180 ","End":"04:33.410","Text":"Now we\u0027ll use the other initial condition to find the a_n."},{"Start":"04:33.410 ","End":"04:36.904","Text":"To find the a_n, we\u0027ll use the other initial condition,"},{"Start":"04:36.904 ","End":"04:40.460","Text":"which is this one and we don\u0027t have u_t,"},{"Start":"04:40.460 ","End":"04:46.655","Text":"we have u differentiate this with respect to t and we get the following."},{"Start":"04:46.655 ","End":"04:49.820","Text":"This part is a constant as far as t goes."},{"Start":"04:49.820 ","End":"04:55.925","Text":"The derivative of the cosine is minus sine together with the anti-derivative,"},{"Start":"04:55.925 ","End":"04:57.685","Text":"which is this,"},{"Start":"04:57.685 ","End":"05:02.585","Text":"and derivative of sine is cosine again with an inner derivative."},{"Start":"05:02.585 ","End":"05:05.075","Text":"If we let t equal 0,"},{"Start":"05:05.075 ","End":"05:09.785","Text":"then this part is 0 because we have a sine of 0,"},{"Start":"05:09.785 ","End":"05:11.795","Text":"so we just have the second part,"},{"Start":"05:11.795 ","End":"05:16.860","Text":"which is a_n times this constant times 1,"},{"Start":"05:16.860 ","End":"05:22.700","Text":"because cosine 0 is 1 times this sine of something x."},{"Start":"05:22.700 ","End":"05:26.570","Text":"On the other hand, we know that u_t of"},{"Start":"05:26.570 ","End":"05:30.440","Text":"x Naught is sine of Pi over 2L x so instead of this,"},{"Start":"05:30.440 ","End":"05:33.665","Text":"we put this, the right-hand side is the same."},{"Start":"05:33.665 ","End":"05:37.175","Text":"Now we\u0027ll use the method of comparing coefficients."},{"Start":"05:37.175 ","End":"05:41.810","Text":"Expand this without the Sigma and we noticed that the only thing in"},{"Start":"05:41.810 ","End":"05:48.485","Text":"common is this sign here with this sign here so from this we can get a_1."},{"Start":"05:48.485 ","End":"05:50.570","Text":"All the others will be 0."},{"Start":"05:50.570 ","End":"05:54.030","Text":"We get that 1 equals a_1 times this,"},{"Start":"05:54.030 ","End":"05:58.565","Text":"which gives us that a_1 is this,"},{"Start":"05:58.565 ","End":"06:02.245","Text":"and the others are 0 when n is not equal to 1."},{"Start":"06:02.245 ","End":"06:04.820","Text":"We have all the a_n another reminder,"},{"Start":"06:04.820 ","End":"06:06.245","Text":"we have all the b_n,"},{"Start":"06:06.245 ","End":"06:09.455","Text":"which is as follows b_3 is 1 and all the others are 0."},{"Start":"06:09.455 ","End":"06:13.340","Text":"Now that we have the a_n\u0027s and b_n\u0027s we can substitute them"},{"Start":"06:13.340 ","End":"06:18.740","Text":"into this and note that we mostly get 0s."},{"Start":"06:18.740 ","End":"06:23.435","Text":"Only when n equals 1 or 3 do we get something non-zero."},{"Start":"06:23.435 ","End":"06:29.130","Text":"When n is 1 then we get that a_n is 1,"},{"Start":"06:29.130 ","End":"06:35.400","Text":"but b_n is 0 so we get sine of Pi twice 1 minus"},{"Start":"06:35.400 ","End":"06:41.730","Text":"1 is 1 to Pi over 2L a_t and here also 2n minus 1 is 1,"},{"Start":"06:41.730 ","End":"06:50.640","Text":"so Pi over 2L times x and the coefficient a_n or a_1 is 2L over Pi a."},{"Start":"06:50.640 ","End":"06:53.115","Text":"Similarly, when n equals 3,"},{"Start":"06:53.115 ","End":"06:55.260","Text":"then a_n is 0,"},{"Start":"06:55.260 ","End":"06:59.490","Text":"b_n is 1 so we get 1 and 2n"},{"Start":"06:59.490 ","End":"07:04.395","Text":"minus 1 is twice 3 minus 1 is 5 so we get a 5 Pi over 2L x here,"},{"Start":"07:04.395 ","End":"07:07.605","Text":"and 5 Pi over 2L at from here."},{"Start":"07:07.605 ","End":"07:14.620","Text":"This is what we get and that\u0027s actually the final answer so we are done."}],"ID":30744},{"Watched":false,"Name":"Summary of Cases","Duration":"1m 53s","ChapterTopicVideoID":29137,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:04.230","Text":"This clip, you can learn to save a lot of work in"},{"Start":"00:04.230 ","End":"00:08.550","Text":"these exercises with homogeneous wave equation on finite interval."},{"Start":"00:08.550 ","End":"00:13.560","Text":"There are 4 main cases that we\u0027ve studied depending on the boundary conditions."},{"Start":"00:13.560 ","End":"00:15.480","Text":"The basic setup is this."},{"Start":"00:15.480 ","End":"00:18.435","Text":"This is the wave equation on the finite interval,"},{"Start":"00:18.435 ","End":"00:21.000","Text":"and these are the initial conditions and"},{"Start":"00:21.000 ","End":"00:23.700","Text":"the boundary conditions as the left and the right one,"},{"Start":"00:23.700 ","End":"00:27.000","Text":"each of them could be Dirichlet of von Neumann,"},{"Start":"00:27.000 ","End":"00:30.980","Text":"and in each of the problems with the similar type,"},{"Start":"00:30.980 ","End":"00:33.380","Text":"we always reach this same point."},{"Start":"00:33.380 ","End":"00:35.420","Text":"We might as well put it in a table."},{"Start":"00:35.420 ","End":"00:39.565","Text":"The first case is when the left and right are both Dirichlet conditions,"},{"Start":"00:39.565 ","End":"00:45.095","Text":"and then the general form of the solution is that u(x,t) equals the following."},{"Start":"00:45.095 ","End":"00:52.160","Text":"Then we find the a_n and b_n using the initial conditions."},{"Start":"00:52.160 ","End":"00:55.145","Text":"The second case, the one they\u0027re both for Neumann,"},{"Start":"00:55.145 ","End":"00:57.950","Text":"and this is the general form of the solution."},{"Start":"00:57.950 ","End":"01:02.000","Text":"Again, we\u0027ll find b_n and a_n using the initial conditions."},{"Start":"01:02.000 ","End":"01:07.915","Text":"The third case is the von Neumann at 0 and a Dirichlet at L,"},{"Start":"01:07.915 ","End":"01:10.975","Text":"and this is the general form of the solution."},{"Start":"01:10.975 ","End":"01:15.390","Text":"We have to find a_n and b_n from the initial conditions."},{"Start":"01:15.390 ","End":"01:19.305","Text":"The last one is like this but switched."},{"Start":"01:19.305 ","End":"01:24.155","Text":"All that difference is instead of a cosine here we have a sine here."},{"Start":"01:24.155 ","End":"01:30.665","Text":"In the remaining exercises in this chapter we\u0027ll use this table to save work."},{"Start":"01:30.665 ","End":"01:33.640","Text":"I\u0027d like to make a small correction."},{"Start":"01:33.640 ","End":"01:40.035","Text":"It\u0027s customary to have the a_n with the cosine and the b_n with the sine."},{"Start":"01:40.035 ","End":"01:44.790","Text":"I just switch them all around and so on."},{"Start":"01:44.790 ","End":"01:48.740","Text":"Just to keep it in line with what\u0027s customary."},{"Start":"01:48.740 ","End":"01:54.060","Text":"Use this table, not the previous one. That concludes this clip."}],"ID":30745},{"Watched":false,"Name":"Exercise 6","Duration":"2m 46s","ChapterTopicVideoID":29144,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.360","Text":"In this exercise, we\u0027re given a wave equation and the finite interval from 0-1."},{"Start":"00:06.360 ","End":"00:08.490","Text":"These are the initial conditions,"},{"Start":"00:08.490 ","End":"00:12.315","Text":"and the boundary conditions are both of the Neumann type."},{"Start":"00:12.315 ","End":"00:14.070","Text":"We look at the table."},{"Start":"00:14.070 ","End":"00:15.870","Text":"This is the entry we want,"},{"Start":"00:15.870 ","End":"00:18.195","Text":"Neumann on both sides."},{"Start":"00:18.195 ","End":"00:21.690","Text":"This will be the formula we use."},{"Start":"00:21.690 ","End":"00:25.485","Text":"Well, u(x, t) equals this."},{"Start":"00:25.485 ","End":"00:28.680","Text":"Back here, just copied that formula,"},{"Start":"00:28.680 ","End":"00:34.275","Text":"you have to figure out the a_n and the b_n using the initial conditions."},{"Start":"00:34.275 ","End":"00:35.880","Text":"First of all, let\u0027s substitute,"},{"Start":"00:35.880 ","End":"00:39.630","Text":"in our case, L=1 and a=1."},{"Start":"00:39.630 ","End":"00:41.690","Text":"This is what we get."},{"Start":"00:41.690 ","End":"00:44.870","Text":"Let\u0027s go for this initial condition first,"},{"Start":"00:44.870 ","End":"00:47.015","Text":"so we need du by dt."},{"Start":"00:47.015 ","End":"00:49.430","Text":"That\u0027s a straightforward differentiation."},{"Start":"00:49.430 ","End":"00:51.140","Text":"This disappears, this becomes b_0."},{"Start":"00:51.140 ","End":"00:53.735","Text":"Derivative of cosine is minus sine,"},{"Start":"00:53.735 ","End":"00:56.915","Text":"derivative of sine is cosine, and so on."},{"Start":"00:56.915 ","End":"00:59.675","Text":"Our initial condition is this."},{"Start":"00:59.675 ","End":"01:04.010","Text":"Substitute t=0 here."},{"Start":"01:04.010 ","End":"01:05.765","Text":"The sine is 0,"},{"Start":"01:05.765 ","End":"01:07.610","Text":"the cosine is 1."},{"Start":"01:07.610 ","End":"01:11.705","Text":"We just get the n Pi b_n from this square bracket,"},{"Start":"01:11.705 ","End":"01:15.035","Text":"and the n Pi b_n we put in front of the cosine here."},{"Start":"01:15.035 ","End":"01:17.030","Text":"This is what we have now."},{"Start":"01:17.030 ","End":"01:22.040","Text":"Then we\u0027ll compare coefficients so that b_n has to"},{"Start":"01:22.040 ","End":"01:27.755","Text":"be 1 when n is 0 and all the other b_n\u0027s are 0."},{"Start":"01:27.755 ","End":"01:29.600","Text":"If we substitute these in here,"},{"Start":"01:29.600 ","End":"01:34.335","Text":"the b_0 is 1 and the others are 0,"},{"Start":"01:34.335 ","End":"01:37.110","Text":"so we just get the a_n part."},{"Start":"01:37.110 ","End":"01:40.370","Text":"This is what we have now."},{"Start":"01:40.370 ","End":"01:42.620","Text":"Next we have to find the a_n,"},{"Start":"01:42.620 ","End":"01:46.450","Text":"so we use the other initial condition, which is this."},{"Start":"01:46.450 ","End":"01:48.600","Text":"Again, we\u0027ll compare coefficients,"},{"Start":"01:48.600 ","End":"01:54.360","Text":"we have 1 plus cosine Pi x plus cosine 3 Pi x equals this when t is 0."},{"Start":"01:54.360 ","End":"01:59.520","Text":"When t is 0 means this cosine is 1 that drops off and this t is 0,"},{"Start":"01:59.520 ","End":"02:03.465","Text":"so this falls and this is what we\u0027re left with."},{"Start":"02:03.465 ","End":"02:06.095","Text":"Now we\u0027re going to compare coefficients again."},{"Start":"02:06.095 ","End":"02:08.749","Text":"A_n, if n is 0,"},{"Start":"02:08.749 ","End":"02:13.450","Text":"is 2, because something over 2=1."},{"Start":"02:13.450 ","End":"02:17.040","Text":"Then for n=1 and 3,"},{"Start":"02:17.040 ","End":"02:21.170","Text":"we get that a_n is 1, and all the others are 0."},{"Start":"02:21.170 ","End":"02:23.720","Text":"Now returning to this equation we had,"},{"Start":"02:23.720 ","End":"02:27.625","Text":"we have the a_ns now so we can just substitute them."},{"Start":"02:27.625 ","End":"02:29.490","Text":"A_0 is 2,"},{"Start":"02:29.490 ","End":"02:30.555","Text":"so this is 1."},{"Start":"02:30.555 ","End":"02:35.100","Text":"A_n here is equal to 1 when n is 1 or 3,"},{"Start":"02:35.100 ","End":"02:39.760","Text":"so we get cosine n Pi x cosine n Pi t for n=1 or 3,"},{"Start":"02:39.760 ","End":"02:41.185","Text":"which is this and this."},{"Start":"02:41.185 ","End":"02:43.120","Text":"Well, the others are 0."},{"Start":"02:43.120 ","End":"02:47.330","Text":"This is the solution. We\u0027re done."}],"ID":30746},{"Watched":false,"Name":"Exercise 7","Duration":"2m 58s","ChapterTopicVideoID":29145,"CourseChapterTopicPlaylistID":294444,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.822","Text":"In this exercise we\u0027re going to solve the wave equation"},{"Start":"00:03.822 ","End":"00:07.965","Text":"which is on a finite interval from 0-1."},{"Start":"00:07.965 ","End":"00:16.020","Text":"These are the initial conditions and this is the boundary conditions on both sides."},{"Start":"00:16.020 ","End":"00:18.164","Text":"It\u0027s the von Neumann condition."},{"Start":"00:18.164 ","End":"00:20.220","Text":"Look this up in the table."},{"Start":"00:20.220 ","End":"00:22.110","Text":"This is the entry we want,"},{"Start":"00:22.110 ","End":"00:24.420","Text":"Neumann in both sides."},{"Start":"00:24.420 ","End":"00:27.915","Text":"This will be the formula we use."},{"Start":"00:27.915 ","End":"00:31.485","Text":"Well, u(x,t) equals this."},{"Start":"00:31.485 ","End":"00:33.315","Text":"So, yeah, write that here."},{"Start":"00:33.315 ","End":"00:37.650","Text":"We know that L=1, and a=1."},{"Start":"00:37.650 ","End":"00:41.935","Text":"A^2 here is 1, so a=1 and L=1."},{"Start":"00:41.935 ","End":"00:44.435","Text":"Simplifies to this."},{"Start":"00:44.435 ","End":"00:49.000","Text":"We\u0027re going to use the initial conditions to figure out the a_n and the b_n."},{"Start":"00:49.000 ","End":"00:52.245","Text":"Let\u0027s go for this condition first."},{"Start":"00:52.245 ","End":"00:55.035","Text":"U(x,0) equals cosine Pi x."},{"Start":"00:55.035 ","End":"01:01.245","Text":"So cosine Pi x is what\u0027s here with t=0."},{"Start":"01:01.245 ","End":"01:06.425","Text":"So the sin(0) is 0 and cosine"},{"Start":"01:06.425 ","End":"01:12.505","Text":"0 is 1 So we just get the a_n and the cosine nPi x."},{"Start":"01:12.505 ","End":"01:21.860","Text":"Comparing coefficients, we get that a_n=1 because we have a cosine 1Pi x here,"},{"Start":"01:21.860 ","End":"01:24.485","Text":"and all the others are 0."},{"Start":"01:24.485 ","End":"01:27.980","Text":"Next we\u0027ll work on the other initial condition, this one."},{"Start":"01:27.980 ","End":"01:32.330","Text":"But for that we\u0027ll need the derivative of u with respect to t. Here again is"},{"Start":"01:32.330 ","End":"01:37.895","Text":"u and derivative with respect to t is the followings pretty routine."},{"Start":"01:37.895 ","End":"01:40.175","Text":"Now our initial condition is this."},{"Start":"01:40.175 ","End":"01:47.940","Text":"So substitute t=0 and we get that cos(2Pi x) is this with t=0."},{"Start":"01:47.940 ","End":"01:51.115","Text":"The sine is 0 and the cosine is 1."},{"Start":"01:51.115 ","End":"01:54.065","Text":"So we just get nPi b_n from this,"},{"Start":"01:54.065 ","End":"01:55.820","Text":"put that in front of the cosine."},{"Start":"01:55.820 ","End":"01:57.110","Text":"This is what we have."},{"Start":"01:57.110 ","End":"02:03.345","Text":"Again, we\u0027re going to compare coefficients so the only relevant one here is when n=2,"},{"Start":"02:03.345 ","End":"02:07.500","Text":"we get that 2Pi b_n=1."},{"Start":"02:07.500 ","End":"02:11.145","Text":"If 2Pi b_n=1, then b_n is 1 over 2Pi."},{"Start":"02:11.145 ","End":"02:15.255","Text":"That\u0027s for n=2 and the others are 0."},{"Start":"02:15.255 ","End":"02:22.340","Text":"Just to remember, a_n was equal to 1 or 0 depending on whether n=1 or not."},{"Start":"02:22.340 ","End":"02:26.353","Text":"Now returning to the formula for u(x,t) in terms of the a_n and b_n,"},{"Start":"02:26.353 ","End":"02:29.330","Text":"we can now substitute them all because we have them all."},{"Start":"02:29.330 ","End":"02:34.765","Text":"We only get terms when n=2 or 1 and when n=1,"},{"Start":"02:34.765 ","End":"02:36.760","Text":"we get that a1=1,"},{"Start":"02:36.760 ","End":"02:39.370","Text":"but b1=0 and n=1 here."},{"Start":"02:39.370 ","End":"02:46.000","Text":"So it\u0027s cosine Pi x cosine Pi t. When n=2,"},{"Start":"02:46.000 ","End":"02:49.705","Text":"the b_n=1 over 2Pi and the a_n=0."},{"Start":"02:49.705 ","End":"02:59.230","Text":"So we get the 1 over 2Pi cosine 2Pi x sine 2Pi t. That\u0027s the answer, and we\u0027re done."}],"ID":30747}],"Thumbnail":null,"ID":294444},{"Name":"Finite Interval, Nonhomogeneous Equation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1 (part 1)","Duration":"7m 38s","ChapterTopicVideoID":29146,"CourseChapterTopicPlaylistID":294445,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.580","Text":"In this exercise, we\u0027re going to solve"},{"Start":"00:02.580 ","End":"00:08.325","Text":"the following nonhomogeneous wave equation on a finite interval."},{"Start":"00:08.325 ","End":"00:11.565","Text":"It\u0027s nonhomogeneous in two respects."},{"Start":"00:11.565 ","End":"00:17.220","Text":"First of all, the PDE itself has this element in it,"},{"Start":"00:17.220 ","End":"00:23.459","Text":"and secondly, one of the boundary conditions also has a nonhomogeneous entry."},{"Start":"00:23.459 ","End":"00:24.750","Text":"This one is homogeneous,"},{"Start":"00:24.750 ","End":"00:26.160","Text":"but the other 1 isn\u0027t,"},{"Start":"00:26.160 ","End":"00:29.415","Text":"and there\u0027ll be two different tricks to deal with these."},{"Start":"00:29.415 ","End":"00:33.480","Text":"Well, for this, it\u0027s a trick that we\u0027ve used before as written"},{"Start":"00:33.480 ","End":"00:38.865","Text":"here for this will be a new technique and prepare for a lengthy clip."},{"Start":"00:38.865 ","End":"00:42.015","Text":"There are 3 stages in solving this."},{"Start":"00:42.015 ","End":"00:47.125","Text":"Although the middle stage is going to be the bulky one."},{"Start":"00:47.125 ","End":"00:53.915","Text":"The first stage is to use a trick to get rid of the sine 3t here, and make it 0."},{"Start":"00:53.915 ","End":"00:58.610","Text":"We\u0027ll convert from u to v by adding a correction function."},{"Start":"00:58.610 ","End":"00:59.945","Text":"This is a standard technique,"},{"Start":"00:59.945 ","End":"01:03.575","Text":"but here we\u0027re given a hint what to try for a correction function,"},{"Start":"01:03.575 ","End":"01:06.890","Text":"to try A(t) plus B(t)x,"},{"Start":"01:06.890 ","End":"01:10.790","Text":"and then we\u0027ll get homogeneous boundary conditions."},{"Start":"01:10.790 ","End":"01:15.350","Text":"Then we will solve the equation and get v,"},{"Start":"01:15.350 ","End":"01:18.800","Text":"and that\u0027s where we\u0027ll learn a new technique because it still won\u0027t"},{"Start":"01:18.800 ","End":"01:22.660","Text":"be homogeneous as far as the PDE itself."},{"Start":"01:22.660 ","End":"01:27.425","Text":"After we\u0027ve got you, the simple thing is just to get from v to u"},{"Start":"01:27.425 ","End":"01:32.000","Text":"by adding this w. During the course of finding w,"},{"Start":"01:32.000 ","End":"01:35.300","Text":"of course we\u0027ll find the functions A(t) and B(t)."},{"Start":"01:35.300 ","End":"01:37.955","Text":"Let\u0027s get started with stage 1,"},{"Start":"01:37.955 ","End":"01:41.060","Text":"finding the correction function w will use"},{"Start":"01:41.060 ","End":"01:44.895","Text":"the hint that we were given and take w of this form,"},{"Start":"01:44.895 ","End":"01:47.880","Text":"and let\u0027s see how we can find A and B."},{"Start":"01:47.880 ","End":"01:50.943","Text":"One of the boundary conditions on u was this,"},{"Start":"01:50.943 ","End":"01:58.969","Text":"let\u0027s convert it to something on v. If we replace x=0 here,"},{"Start":"01:58.969 ","End":"02:02.165","Text":"then v(0)t is 0."},{"Start":"02:02.165 ","End":"02:04.745","Text":"I mean that\u0027s what we\u0027re looking for,"},{"Start":"02:04.745 ","End":"02:06.320","Text":"we\u0027re looking for v to be homogeneous,"},{"Start":"02:06.320 ","End":"02:09.825","Text":"so we set it 0 A(f)t, don\u0027t know."},{"Start":"02:09.825 ","End":"02:12.240","Text":"But if x is 0,"},{"Start":"02:12.240 ","End":"02:16.500","Text":"then the regardless of what b is we get here is 0."},{"Start":"02:16.500 ","End":"02:20.845","Text":"We\u0027ve got that A(t) equals 0."},{"Start":"02:20.845 ","End":"02:25.325","Text":"Now the other boundary condition on u was this one."},{"Start":"02:25.325 ","End":"02:30.560","Text":"The boundary condition on the right at L. Sine of 3 t is U of"},{"Start":"02:30.560 ","End":"02:35.945","Text":"L and t just substitute x=L here."},{"Start":"02:35.945 ","End":"02:39.800","Text":"V(l)t, we want it to equal 0,"},{"Start":"02:39.800 ","End":"02:41.915","Text":"that\u0027s what we\u0027re looking for."},{"Start":"02:41.915 ","End":"02:47.295","Text":"A(t) we already found in here that it\u0027s 0,"},{"Start":"02:47.295 ","End":"02:53.150","Text":"and here just leave it as B(t) times L. Then we can extract B(t)"},{"Start":"02:53.150 ","End":"02:59.135","Text":"just by dividing both sides by L and putting B(t) on the left,"},{"Start":"02:59.135 ","End":"03:00.980","Text":"and this is what we get."},{"Start":"03:00.980 ","End":"03:03.965","Text":"We now have A(t)and B(t),"},{"Start":"03:03.965 ","End":"03:08.494","Text":"and then we\u0027ve got w as a plus b x,"},{"Start":"03:08.494 ","End":"03:14.940","Text":"which means that w is 1 over L sine 3 of tx."},{"Start":"03:14.940 ","End":"03:17.695","Text":"Now that we have w,"},{"Start":"03:17.695 ","End":"03:21.725","Text":"remember that u is equal to v plus w,"},{"Start":"03:21.725 ","End":"03:28.105","Text":"and so we have u equals v plus 1 over l sine of 3 tx,"},{"Start":"03:28.105 ","End":"03:31.990","Text":"and that concludes the first stage of 3."},{"Start":"03:31.990 ","End":"03:39.680","Text":"Now we come to stage 2 where we\u0027re going to solve the PDE for V. In stage 1,"},{"Start":"03:39.680 ","End":"03:43.970","Text":"we got that u equals v plus 1 over L sine 3tx."},{"Start":"03:43.970 ","End":"03:47.945","Text":"We want to substitute this in the PDE."},{"Start":"03:47.945 ","End":"03:56.400","Text":"The PDE for you is this so we need utt and uxx in terms of v, utt,"},{"Start":"03:56.400 ","End":"04:00.260","Text":"is obtained by differentiating this twice with"},{"Start":"04:00.260 ","End":"04:03.635","Text":"respect to t. After we differentiate, once,"},{"Start":"04:03.635 ","End":"04:07.160","Text":"we get co-sine and the 3 comes out,"},{"Start":"04:07.160 ","End":"04:10.385","Text":"and when we differentiate the cosine and we get minus sine,"},{"Start":"04:10.385 ","End":"04:12.005","Text":"and again a 3 comes out."},{"Start":"04:12.005 ","End":"04:15.980","Text":"We get the minus 9 in addition to what we have here."},{"Start":"04:15.980 ","End":"04:19.295","Text":"When we differentiate twice with respect to x,"},{"Start":"04:19.295 ","End":"04:23.195","Text":"we just get vxx because all this is a constant,"},{"Start":"04:23.195 ","End":"04:24.680","Text":"derivative of x is 1,"},{"Start":"04:24.680 ","End":"04:26.540","Text":"second derivative is 0."},{"Start":"04:26.540 ","End":"04:28.690","Text":"So we just have vxx,"},{"Start":"04:28.690 ","End":"04:32.855","Text":"and then plug these 2 into here and here."},{"Start":"04:32.855 ","End":"04:38.315","Text":"What we get is that this is vtt minus, well,"},{"Start":"04:38.315 ","End":"04:43.195","Text":"this equals a^2 uxx is just vxx,"},{"Start":"04:43.195 ","End":"04:44.960","Text":"and the rest of it is the same."},{"Start":"04:44.960 ","End":"04:49.805","Text":"Notice that we have 2 of the same term on the left and on the right."},{"Start":"04:49.805 ","End":"04:55.485","Text":"This bit here is the same as this here."},{"Start":"04:55.485 ","End":"05:00.730","Text":"That the x is here and the x is here. It\u0027s the same thing."},{"Start":"05:00.730 ","End":"05:03.480","Text":"This simplifies to this,"},{"Start":"05:03.480 ","End":"05:05.810","Text":"and that\u0027s our PDE for v,"},{"Start":"05:05.810 ","End":"05:08.795","Text":"which is a bit simpler than the one for u."},{"Start":"05:08.795 ","End":"05:10.760","Text":"Anyway, that\u0027s the PDE."},{"Start":"05:10.760 ","End":"05:13.415","Text":"What about initial conditions for v?"},{"Start":"05:13.415 ","End":"05:15.800","Text":"Initial when t equals 0?"},{"Start":"05:15.800 ","End":"05:18.890","Text":"Well, we have the initial condition for you."},{"Start":"05:18.890 ","End":"05:20.780","Text":"We have two of them in fact,"},{"Start":"05:20.780 ","End":"05:25.895","Text":"let\u0027s see if we can convert them to conditions in v. If u of x naught is naught,"},{"Start":"05:25.895 ","End":"05:30.215","Text":"then because v(x)t equals this,"},{"Start":"05:30.215 ","End":"05:38.149","Text":"then we get that v(0)x naught is naught because when t is 0,"},{"Start":"05:38.149 ","End":"05:41.680","Text":"sine of 3t is also 0."},{"Start":"05:41.680 ","End":"05:44.220","Text":"We\u0027re just left with u of x0,"},{"Start":"05:44.220 ","End":"05:45.705","Text":"which is 0,"},{"Start":"05:45.705 ","End":"05:49.095","Text":"and the other one what we get."},{"Start":"05:49.095 ","End":"05:52.855","Text":"First, we need vt dv by dt,"},{"Start":"05:52.855 ","End":"05:57.890","Text":"which we get by doing a partial derivative of this with respect to t."},{"Start":"05:57.890 ","End":"06:03.065","Text":"Here we get ut and here differentiate with respect to t of sine."},{"Start":"06:03.065 ","End":"06:08.735","Text":"We have cosine, but an extra 3 comes out in a derivative."},{"Start":"06:08.735 ","End":"06:12.290","Text":"Now we can put t=0 here."},{"Start":"06:12.290 ","End":"06:18.239","Text":"We get v of x 0 is if we put t=0 here,"},{"Start":"06:18.239 ","End":"06:21.030","Text":"we get 3x over l,"},{"Start":"06:21.030 ","End":"06:23.480","Text":"and here also cosine of 0 is 1,"},{"Start":"06:23.480 ","End":"06:28.220","Text":"so that\u0027s also 3x over L. It comes out to be 0."},{"Start":"06:28.220 ","End":"06:31.565","Text":"I\u0027ve written it in faint if you want to check the work again."},{"Start":"06:31.565 ","End":"06:36.455","Text":"Now let\u0027s collect together the information on v. We have the whole thing."},{"Start":"06:36.455 ","End":"06:41.495","Text":"We have vtt from here, the PDE part."},{"Start":"06:41.495 ","End":"06:45.710","Text":"We have the initial conditions from here and here,"},{"Start":"06:45.710 ","End":"06:47.585","Text":"and the boundary conditions,"},{"Start":"06:47.585 ","End":"06:49.295","Text":"we already tailored them."},{"Start":"06:49.295 ","End":"06:51.320","Text":"We\u0027d get 0 and 0"},{"Start":"06:51.320 ","End":"06:57.145","Text":"homogeneous both boundary conditions that both of dirichlet type, of course."},{"Start":"06:57.145 ","End":"07:03.440","Text":"We have a nonhomogeneous equation in v,"},{"Start":"07:03.440 ","End":"07:07.195","Text":"but the boundary conditions are homogeneous."},{"Start":"07:07.195 ","End":"07:11.150","Text":"Now here\u0027s the technique"},{"Start":"07:11.150 ","End":"07:14.960","Text":"that we\u0027re going to use for such equations when we have homogeneous boundary conditions,"},{"Start":"07:14.960 ","End":"07:19.155","Text":"but non-homogeneous part in the PDE."},{"Start":"07:19.155 ","End":"07:22.265","Text":"What we do is make it homogeneous."},{"Start":"07:22.265 ","End":"07:27.740","Text":"Just throw out the non-homogeneous part and start by"},{"Start":"07:27.740 ","End":"07:34.055","Text":"trying to solve this using separation of variables on the homogeneous PDE."},{"Start":"07:34.055 ","End":"07:37.770","Text":"You know what? Let\u0027s take a break and continue after."}],"ID":30748},{"Watched":false,"Name":"Exercise 1 (part 2)","Duration":"13m 9s","ChapterTopicVideoID":29147,"CourseChapterTopicPlaylistID":294445,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.180","Text":"We\u0027re continuing after the break and we got to"},{"Start":"00:03.180 ","End":"00:10.605","Text":"the homogeneous equation in v. This will help us to solve the non-homogeneous."},{"Start":"00:10.605 ","End":"00:14.295","Text":"We\u0027ll use the method of separation of variables."},{"Start":"00:14.295 ","End":"00:18.270","Text":"We\u0027ll write v as X(x)T(t),"},{"Start":"00:18.270 ","End":"00:23.220","Text":"and then the equation here becomes the following."},{"Start":"00:23.220 ","End":"00:24.840","Text":"This is all familiar."},{"Start":"00:24.840 ","End":"00:27.585","Text":"Rearranging, we get this,"},{"Start":"00:27.585 ","End":"00:31.110","Text":"ignore the path with t. We just care about x at the moment,"},{"Start":"00:31.110 ","End":"00:37.304","Text":"so we get that X\u0027\u0027 plus Lambda X equals 0."},{"Start":"00:37.304 ","End":"00:44.330","Text":"The first boundary condition gives us that X(0)T(t) is 0."},{"Start":"00:44.330 ","End":"00:47.500","Text":"This is for all t,"},{"Start":"00:47.500 ","End":"00:51.212","Text":"so must mean that X(0) is 0."},{"Start":"00:51.212 ","End":"00:57.890","Text":"The other boundary condition gives us that X(L) is 0."},{"Start":"00:57.890 ","End":"01:01.595","Text":"If we take this and this together with this,"},{"Start":"01:01.595 ","End":"01:05.390","Text":"then we have a second-order ODE with"},{"Start":"01:05.390 ","End":"01:10.360","Text":"these boundary conditions and we recognize this as a Sturm Liouville type of problem."},{"Start":"01:10.360 ","End":"01:13.180","Text":"We\u0027ve seen this problem many times,"},{"Start":"01:13.180 ","End":"01:19.190","Text":"and the solution comes out to be X_n is sine n Pi x over L,"},{"Start":"01:19.190 ","End":"01:26.855","Text":"so that v is infinite combination of X_n(x)T_n(t)."},{"Start":"01:26.855 ","End":"01:28.520","Text":"We found X_n,"},{"Start":"01:28.520 ","End":"01:30.845","Text":"but we haven\u0027t yet found T_n."},{"Start":"01:30.845 ","End":"01:35.105","Text":"Now, continuing with the technique that I\u0027m showing you,"},{"Start":"01:35.105 ","End":"01:38.224","Text":"our problem is not the homogeneous problem."},{"Start":"01:38.224 ","End":"01:41.755","Text":"We want to solve the non-homogeneous,"},{"Start":"01:41.755 ","End":"01:42.930","Text":"which is this,"},{"Start":"01:42.930 ","End":"01:45.345","Text":"with the non-homogeneous part."},{"Start":"01:45.345 ","End":"01:50.435","Text":"What we\u0027re going to do is use the same linear combination,"},{"Start":"01:50.435 ","End":"01:52.880","Text":"but not to solve the homogeneous 1,"},{"Start":"01:52.880 ","End":"01:55.180","Text":"to solve the non-homogeneous."},{"Start":"01:55.180 ","End":"01:58.988","Text":"Then the T_n will be different than if we solve the homogeneous"},{"Start":"01:58.988 ","End":"02:02.960","Text":"but this is where this technique comes in."},{"Start":"02:02.960 ","End":"02:04.880","Text":"Up to this point, it\u0027s the same,"},{"Start":"02:04.880 ","End":"02:09.616","Text":"but now we\u0027re going to solve it for the non-homogeneous."},{"Start":"02:09.616 ","End":"02:15.620","Text":"We\u0027re looking for solution of the following form to the non-homogeneous, which is this."},{"Start":"02:15.620 ","End":"02:21.650","Text":"Note that the initial conditions are these for v,"},{"Start":"02:21.650 ","End":"02:26.950","Text":"and we\u0027ll use these initial conditions to help us find the functions T_n."},{"Start":"02:26.950 ","End":"02:29.330","Text":"First 1 is this one,"},{"Start":"02:29.330 ","End":"02:32.135","Text":"and if we substitute that here,"},{"Start":"02:32.135 ","End":"02:34.725","Text":"we get that 0 equals."},{"Start":"02:34.725 ","End":"02:37.590","Text":"We put t=0 here,"},{"Start":"02:37.590 ","End":"02:41.535","Text":"we get T_n(0) instead of T_n(t)."},{"Start":"02:41.535 ","End":"02:48.210","Text":"We can compare coefficients because this is a Fourier series,"},{"Start":"02:48.210 ","End":"02:50.809","Text":"and the coefficients are unique."},{"Start":"02:50.809 ","End":"02:55.220","Text":"For each sine n Pi x over L on the left, the coefficient is 0."},{"Start":"02:55.220 ","End":"02:56.990","Text":"On the right, it\u0027s T_n(0)."},{"Start":"02:56.990 ","End":"03:02.296","Text":"T_n(0) equals 0 for all n. For the other initial condition,"},{"Start":"03:02.296 ","End":"03:04.850","Text":"we\u0027ll need dv by dt,"},{"Start":"03:04.850 ","End":"03:10.610","Text":"so differentiating this, we get the x part doesn\u0027t change."},{"Start":"03:10.610 ","End":"03:14.075","Text":"But instead of T_n(t), we have T_n\u0027(t)."},{"Start":"03:14.075 ","End":"03:17.075","Text":"Then if we put in t=0,"},{"Start":"03:17.075 ","End":"03:20.490","Text":"then we get 0 equals."},{"Start":"03:20.490 ","End":"03:24.920","Text":"Again comparing coefficients, we get that"},{"Start":"03:24.920 ","End":"03:31.660","Text":"T_n\u0027(0) equals 0 for each n. Let\u0027s do a summary of what we have so far."},{"Start":"03:31.660 ","End":"03:34.010","Text":"Here\u0027s the summary of the essentials so far."},{"Start":"03:34.010 ","End":"03:42.365","Text":"We just found that T_n(0) is 0 and T_n\u0027(0) is 0 for all n. This is our equation,"},{"Start":"03:42.365 ","End":"03:50.645","Text":"the non-homogeneous equation in v. We\u0027re looking for a solution of this form,"},{"Start":"03:50.645 ","End":"03:54.630","Text":"the sum of the following."},{"Start":"03:54.630 ","End":"03:57.290","Text":"Just like we were going to do for the homogeneous,"},{"Start":"03:57.290 ","End":"04:00.065","Text":"we\u0027re going to do it for the non-homogeneous."},{"Start":"04:00.065 ","End":"04:04.025","Text":"Let\u0027s substitute v in the PDE,"},{"Start":"04:04.025 ","End":"04:06.665","Text":"make sure it satisfies that."},{"Start":"04:06.665 ","End":"04:10.440","Text":"V_tt from here is this,"},{"Start":"04:10.440 ","End":"04:15.110","Text":"and we just put a double prime in front of the T_n(t)."},{"Start":"04:15.110 ","End":"04:20.625","Text":"Then a^2 first. V_xx,"},{"Start":"04:20.625 ","End":"04:25.400","Text":"we differentiate this part twice with respect to x."},{"Start":"04:25.400 ","End":"04:27.185","Text":"When we differentiate sine,"},{"Start":"04:27.185 ","End":"04:29.045","Text":"we get cosine,"},{"Start":"04:29.045 ","End":"04:32.720","Text":"but we also get an extra inner derivative."},{"Start":"04:32.720 ","End":"04:34.580","Text":"When we differentiate the cosine,"},{"Start":"04:34.580 ","End":"04:36.245","Text":"we get minus sine,"},{"Start":"04:36.245 ","End":"04:41.570","Text":"so we have minus sine and we have this thing squared because we differentiated twice."},{"Start":"04:41.570 ","End":"04:45.005","Text":"Lastly, this piece just put it on the end here."},{"Start":"04:45.005 ","End":"04:51.200","Text":"Now here just take the a^2 and put the a here instead."},{"Start":"04:51.200 ","End":"04:57.080","Text":"Now we\u0027re going to compare coefficients of the sine n Pi x over L,"},{"Start":"04:57.080 ","End":"05:00.610","Text":"but just the case where n=3."},{"Start":"05:00.610 ","End":"05:02.355","Text":"When n is 3,"},{"Start":"05:02.355 ","End":"05:09.530","Text":"then the coefficient of sine 3 Pi x over L is just this bit with n=3."},{"Start":"05:09.530 ","End":"05:14.560","Text":"This side we have T_3\u0027\u0027(t), and here,"},{"Start":"05:14.560 ","End":"05:18.770","Text":"when n=3, we get the sine 3 Pi x over L,"},{"Start":"05:18.770 ","End":"05:20.240","Text":"we want just the coefficient,"},{"Start":"05:20.240 ","End":"05:21.815","Text":"so it\u0027s this bit and this bit."},{"Start":"05:21.815 ","End":"05:32.135","Text":"This bit is 3 Pi a over L squared minus and this part is T_3(t)."},{"Start":"05:32.135 ","End":"05:36.020","Text":"Then we have the last bit which is already in the form sine 3 Pi x over L,"},{"Start":"05:36.020 ","End":"05:37.835","Text":"so we need this part."},{"Start":"05:37.835 ","End":"05:44.045","Text":"These are the coefficients of sine 3 Pi x over L compared on the left and on the right."},{"Start":"05:44.045 ","End":"05:46.460","Text":"Now, bring this part to the left-hand side,"},{"Start":"05:46.460 ","End":"05:48.605","Text":"and this is the equation we have."},{"Start":"05:48.605 ","End":"05:55.195","Text":"This is a non-homogeneous 2nd-order linear equation in T_3."},{"Start":"05:55.195 ","End":"05:57.565","Text":"When we have a non-homogeneous,"},{"Start":"05:57.565 ","End":"06:00.470","Text":"we find the general solution to the homogeneous"},{"Start":"06:00.470 ","End":"06:03.905","Text":"plus a particular solution to the non-homogeneous."},{"Start":"06:03.905 ","End":"06:05.825","Text":"For the homogeneous case,"},{"Start":"06:05.825 ","End":"06:08.915","Text":"if we replace T_3 by y,"},{"Start":"06:08.915 ","End":"06:15.175","Text":"then we have y plus Omega squared y equals 0,"},{"Start":"06:15.175 ","End":"06:18.530","Text":"where Omega is this 3 Pi a over L,"},{"Start":"06:18.530 ","End":"06:23.431","Text":"and the solution to this is the following solution to the homogeneous."},{"Start":"06:23.431 ","End":"06:27.985","Text":"The solution in general will be the solution to the homogeneous,"},{"Start":"06:27.985 ","End":"06:35.480","Text":"which is this just replacing Omega with 3 Pi L over L plus a particular solution."},{"Start":"06:35.480 ","End":"06:40.520","Text":"Now we still have to find a particular solution y_p, and from experience,"},{"Start":"06:40.520 ","End":"06:42.875","Text":"you learn what to try because this is"},{"Start":"06:42.875 ","End":"06:46.640","Text":"a linear 1st-order expression in t. We\u0027re going to try"},{"Start":"06:46.640 ","End":"06:50.290","Text":"a general linear 1st-order At plus B."},{"Start":"06:50.290 ","End":"06:54.740","Text":"We\u0027ll try this and see if we can get coefficients A and B to make it work."},{"Start":"06:54.740 ","End":"06:56.270","Text":"Back to this equation,"},{"Start":"06:56.270 ","End":"06:59.735","Text":"replace T_3 by At plus B."},{"Start":"06:59.735 ","End":"07:05.735","Text":"Second derivative of this with respect to t is 0 plus this bit,"},{"Start":"07:05.735 ","End":"07:08.210","Text":"again, T_3 is At plus B,"},{"Start":"07:08.210 ","End":"07:09.620","Text":"and here t plus 1."},{"Start":"07:09.620 ","End":"07:13.115","Text":"Now we\u0027ll compare coefficients."},{"Start":"07:13.115 ","End":"07:17.120","Text":"They\u0027re both linear expressions in t. We\u0027ll"},{"Start":"07:17.120 ","End":"07:21.665","Text":"compare the coefficients of 1 and of t. In the case of the t,"},{"Start":"07:21.665 ","End":"07:24.170","Text":"on the right we have 1, on the left,"},{"Start":"07:24.170 ","End":"07:27.250","Text":"we have this expression times A."},{"Start":"07:27.250 ","End":"07:29.835","Text":"This times A is 1,"},{"Start":"07:29.835 ","End":"07:32.475","Text":"then A is 1 over this,"},{"Start":"07:32.475 ","End":"07:35.055","Text":"which is L over 3 Pi a squared."},{"Start":"07:35.055 ","End":"07:37.505","Text":"Similarly for the free coefficient."},{"Start":"07:37.505 ","End":"07:42.330","Text":"We get this times B equals 1,"},{"Start":"07:42.330 ","End":"07:44.625","Text":"and so B is also this."},{"Start":"07:44.625 ","End":"07:47.660","Text":"A and B come out the same and come out to be this."},{"Start":"07:47.660 ","End":"07:50.300","Text":"If A equals B equals this, then y_p,"},{"Start":"07:50.300 ","End":"07:57.550","Text":"which is At plus B will equal L over 3 Pi a squared times t plus 1."},{"Start":"07:57.550 ","End":"08:07.205","Text":"Now back here, T_3 is the solution to the homogeneous plus a particular y."},{"Start":"08:07.205 ","End":"08:13.740","Text":"We\u0027re not there yet, what we still have to do is find c_1 and c_2,"},{"Start":"08:13.740 ","End":"08:15.675","Text":"that\u0027s what we\u0027re missing."},{"Start":"08:15.675 ","End":"08:21.920","Text":"Recall that we found that T_n(0) is 0 and T_n\u0027(0) is 0 for all n."},{"Start":"08:21.920 ","End":"08:30.005","Text":"If we take this and substitute instead of t 0,"},{"Start":"08:30.005 ","End":"08:32.810","Text":"then we get on the left-hand side 0,"},{"Start":"08:32.810 ","End":"08:35.240","Text":"on the right, if t is 0,"},{"Start":"08:35.240 ","End":"08:37.940","Text":"this cosine is 1."},{"Start":"08:37.940 ","End":"08:40.475","Text":"1 times c_1 is c_1."},{"Start":"08:40.475 ","End":"08:42.920","Text":"But this term, the sine is 0,"},{"Start":"08:42.920 ","End":"08:44.720","Text":"so this term disappears."},{"Start":"08:44.720 ","End":"08:46.445","Text":"Here if t is 0,"},{"Start":"08:46.445 ","End":"08:49.190","Text":"we just get this constant here."},{"Start":"08:49.190 ","End":"08:53.030","Text":"We have c_1 as minus this."},{"Start":"08:53.030 ","End":"08:54.830","Text":"Now we need to use the other 1,"},{"Start":"08:54.830 ","End":"08:57.560","Text":"so we need T_3\u0027."},{"Start":"08:57.560 ","End":"08:59.090","Text":"We\u0027re just differentiating this,"},{"Start":"08:59.090 ","End":"09:02.185","Text":"the cosine is minus sine and an anti-derivative,"},{"Start":"09:02.185 ","End":"09:05.125","Text":"and here derivative of sine is cosine,"},{"Start":"09:05.125 ","End":"09:07.070","Text":"and the derivative of this, well,"},{"Start":"09:07.070 ","End":"09:09.530","Text":"derivative of t plus 1 is just 1."},{"Start":"09:09.530 ","End":"09:14.450","Text":"Now let t =0 and use this here, so on the left,"},{"Start":"09:14.450 ","End":"09:16.070","Text":"we have 0 and on the right,"},{"Start":"09:16.070 ","End":"09:18.720","Text":"we have the sine part is 0,"},{"Start":"09:18.720 ","End":"09:20.055","Text":"so this drops off."},{"Start":"09:20.055 ","End":"09:22.350","Text":"The cosine of this is 1,"},{"Start":"09:22.350 ","End":"09:23.520","Text":"so we have c_2,"},{"Start":"09:23.520 ","End":"09:27.830","Text":"3 Pi a over L. This is just itself,"},{"Start":"09:27.830 ","End":"09:29.975","Text":"so that gives us what c_2 is."},{"Start":"09:29.975 ","End":"09:32.140","Text":"Now we have c_1 and c_2,"},{"Start":"09:32.140 ","End":"09:33.830","Text":"and so back to this,"},{"Start":"09:33.830 ","End":"09:39.220","Text":"we can now substitute c_1 and c_2 from here and here,"},{"Start":"09:39.220 ","End":"09:41.820","Text":"and what we get is the following."},{"Start":"09:41.820 ","End":"09:45.590","Text":"Now, let\u0027s go back to an equation we already had."},{"Start":"09:45.590 ","End":"09:50.930","Text":"We did comparison of coefficients for n=3 on this equation."},{"Start":"09:50.930 ","End":"09:53.860","Text":"This was v_tt,"},{"Start":"09:53.860 ","End":"09:56.265","Text":"this was a^2 v_x x,"},{"Start":"09:56.265 ","End":"09:58.625","Text":"and this is the non-homogeneous bit."},{"Start":"09:58.625 ","End":"10:05.230","Text":"We want to still compare coefficients when n is not 3 because we only got T_3 so far."},{"Start":"10:05.230 ","End":"10:06.790","Text":"We need all the T_ns."},{"Start":"10:06.790 ","End":"10:07.935","Text":"If n is not 3,"},{"Start":"10:07.935 ","End":"10:10.595","Text":"then we don\u0027t need this."},{"Start":"10:10.595 ","End":"10:17.415","Text":"We just need the coefficients of sine n Pi x over L from here."},{"Start":"10:17.415 ","End":"10:23.880","Text":"For this n, we get T_n(t) and we get minus n Pi over L squared."},{"Start":"10:23.880 ","End":"10:29.720","Text":"On the left, we just have T_n\u0027\u0027 as the coefficients of this."},{"Start":"10:29.720 ","End":"10:35.710","Text":"This gives us a differential equation in T_n."},{"Start":"10:35.710 ","End":"10:39.830","Text":"Just bring it over to the other side we have T_n is like y."},{"Start":"10:39.830 ","End":"10:43.915","Text":"y\u0027\u0027 plus Omega squared y equals 0,"},{"Start":"10:43.915 ","End":"10:46.900","Text":"and these are boundary conditions for y."},{"Start":"10:46.900 ","End":"10:50.870","Text":"Think of T_n as y. This is basically what we get where Omega is"},{"Start":"10:50.870 ","End":"10:55.880","Text":"n Pi a over L. We know the solution to this is the following."},{"Start":"10:55.880 ","End":"11:00.005","Text":"It\u0027s the sum of c_1 cosine plus c_2 sine,"},{"Start":"11:00.005 ","End":"11:03.590","Text":"which in our case is the c_1 cosine Omega t,"},{"Start":"11:03.590 ","End":"11:06.065","Text":"c_2 sine Omega t,"},{"Start":"11:06.065 ","End":"11:08.375","Text":"that\u0027s T_n, which is y,"},{"Start":"11:08.375 ","End":"11:14.060","Text":"and then T_n prime is this straightforward differentiation with respect to t,"},{"Start":"11:14.060 ","End":"11:20.115","Text":"and we have that T_n(0) is 0 for all n. If we use this,"},{"Start":"11:20.115 ","End":"11:22.020","Text":"we get c_1 times 1,"},{"Start":"11:22.020 ","End":"11:24.705","Text":"plus c_2 times 0 is 0,"},{"Start":"11:24.705 ","End":"11:27.240","Text":"and so c_1 is 0."},{"Start":"11:27.240 ","End":"11:30.815","Text":"If we use this one and let t=0,"},{"Start":"11:30.815 ","End":"11:37.485","Text":"then we get that 0 equals minus c_1 times 0"},{"Start":"11:37.485 ","End":"11:44.270","Text":"plus c_2 times n Pi over L. We already know that c_1 is 0 from here,"},{"Start":"11:44.270 ","End":"11:47.284","Text":"so we\u0027re just left with this equals 0,"},{"Start":"11:47.284 ","End":"11:53.670","Text":"which gives us that c_2 is 0 and so T_n(t),"},{"Start":"11:53.670 ","End":"11:58.635","Text":"which is c_1 cosine plus c_2 sine if both c_1 and c_2 are 0,"},{"Start":"11:58.635 ","End":"12:05.345","Text":"then that gives that T_n is identically 0 when n is not 3,"},{"Start":"12:05.345 ","End":"12:08.660","Text":"and T_3(t), where n=3,"},{"Start":"12:08.660 ","End":"12:11.060","Text":"we already found is equal to this."},{"Start":"12:11.060 ","End":"12:14.015","Text":"Now, v is the following sum."},{"Start":"12:14.015 ","End":"12:15.555","Text":"We want to replace T_n."},{"Start":"12:15.555 ","End":"12:17.020","Text":"We found all the T_ns."},{"Start":"12:17.020 ","End":"12:18.860","Text":"Instead of getting an infinite sum,"},{"Start":"12:18.860 ","End":"12:24.200","Text":"we\u0027ll just get this when n=3 because all the other T_ns are 0."},{"Start":"12:24.200 ","End":"12:25.930","Text":"When n is 3,"},{"Start":"12:25.930 ","End":"12:29.360","Text":"we get sine of 3 Pi x over L,"},{"Start":"12:29.360 ","End":"12:33.235","Text":"T_3 of t, and we already have T_3(t)."},{"Start":"12:33.235 ","End":"12:36.230","Text":"From here, let\u0027s just slightly rearrange the orders."},{"Start":"12:36.230 ","End":"12:39.635","Text":"I put the plus term first, this one here,"},{"Start":"12:39.635 ","End":"12:43.370","Text":"and then the 2 minus 1 after that slight rearrangement."},{"Start":"12:43.370 ","End":"12:45.045","Text":"We have v(xt)."},{"Start":"12:45.045 ","End":"12:48.320","Text":"We\u0027re almost finished because our task was to find u,"},{"Start":"12:48.320 ","End":"12:50.510","Text":"not v. But,"},{"Start":"12:50.510 ","End":"12:53.315","Text":"we know that u is v plus w,"},{"Start":"12:53.315 ","End":"12:58.050","Text":"and we already found that w was the following."},{"Start":"12:58.050 ","End":"13:01.845","Text":"This is w(xt)."},{"Start":"13:01.845 ","End":"13:03.570","Text":"Okay, this is the answer."},{"Start":"13:03.570 ","End":"13:10.080","Text":"Let\u0027s just highlight it and declare that we are done."}],"ID":30749},{"Watched":false,"Name":"Exercise 2","Duration":"8m 49s","ChapterTopicVideoID":29148,"CourseChapterTopicPlaylistID":294445,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.415","Text":"In this exercise, we\u0027re going to solve the following nonhomogeneous wave equation."},{"Start":"00:05.415 ","End":"00:10.140","Text":"It\u0027s on a finite interval and it\u0027s non-homogeneous because we have"},{"Start":"00:10.140 ","End":"00:17.880","Text":"this extra term here but note that the boundary conditions are homogeneous."},{"Start":"00:17.880 ","End":"00:20.595","Text":"We have a 0 here and a 0 here."},{"Start":"00:20.595 ","End":"00:22.680","Text":"If it wasn\u0027t so we\u0027d have to do"},{"Start":"00:22.680 ","End":"00:27.555","Text":"some extra work finding a correction function but here we don\u0027t."},{"Start":"00:27.555 ","End":"00:31.530","Text":"These are the main stages for solving the problem."},{"Start":"00:31.530 ","End":"00:33.090","Text":"I won\u0027t read it out,"},{"Start":"00:33.090 ","End":"00:37.080","Text":"but I suggest you return after the exercise back here again,"},{"Start":"00:37.080 ","End":"00:40.860","Text":"it will make more sense after we\u0027ve actually solved it."},{"Start":"00:40.860 ","End":"00:43.420","Text":"Let\u0027s begin with Stage 1."},{"Start":"00:43.420 ","End":"00:48.195","Text":"We\u0027re going to separate the variables on the homogeneous PDF."},{"Start":"00:48.195 ","End":"00:50.780","Text":"We\u0027ve omitted this extra term."},{"Start":"00:50.780 ","End":"00:55.345","Text":"This is the PDF together with the boundary conditions."},{"Start":"00:55.345 ","End":"00:57.770","Text":"It\u0027s the same thing we always do."},{"Start":"00:57.770 ","End":"01:01.205","Text":"Let u be a function of x times a function of t,"},{"Start":"01:01.205 ","End":"01:04.400","Text":"differentiate with respect to T twice and here with"},{"Start":"01:04.400 ","End":"01:07.915","Text":"respect to X twice and we get the following."},{"Start":"01:07.915 ","End":"01:10.700","Text":"A function of T equals a function of x,"},{"Start":"01:10.700 ","End":"01:11.960","Text":"so it has to be a constant,"},{"Start":"01:11.960 ","End":"01:14.785","Text":"and we\u0027ll call that constant minus Lambda."},{"Start":"01:14.785 ","End":"01:18.785","Text":"Then we get the following second-order differential equation,"},{"Start":"01:18.785 ","End":"01:27.740","Text":"ordinary in X and from the boundary conditions we get X(0) times T(t) is X(L)."},{"Start":"01:27.740 ","End":"01:32.995","Text":"T(t) is 0, but the T(t) cancels so we have X of 0 equals X(L),"},{"Start":"01:32.995 ","End":"01:35.960","Text":"and we recognize this as a Sturm Louisville problem."},{"Start":"01:35.960 ","End":"01:43.260","Text":"The solution is following sequence of X_ns where n goes from 1, 2, 3, etc,"},{"Start":"01:43.260 ","End":"01:47.990","Text":"to infinity and we\u0027ll look for a solution in the form of a linear combination"},{"Start":"01:47.990 ","End":"01:53.710","Text":"of X_n times T_n but we know that X_n is the following."},{"Start":"01:53.710 ","End":"01:55.700","Text":"What we\u0027re going to do next is use"},{"Start":"01:55.700 ","End":"02:00.380","Text":"the same format to find a solution to the nonhomogeneous."},{"Start":"02:00.380 ","End":"02:02.225","Text":"This was normally for the homogeneous,"},{"Start":"02:02.225 ","End":"02:05.135","Text":"but it will also work for the nonhomogeneous."},{"Start":"02:05.135 ","End":"02:07.370","Text":"Of course we\u0027ll get different T_ns."},{"Start":"02:07.370 ","End":"02:09.005","Text":"Here we are again,"},{"Start":"02:09.005 ","End":"02:10.700","Text":"this is a form of U(x,t)."},{"Start":"02:10.700 ","End":"02:13.885","Text":"We\u0027re going to substitute in the nonhomogeneous."},{"Start":"02:13.885 ","End":"02:15.680","Text":"What I wrote here is misleading."},{"Start":"02:15.680 ","End":"02:18.725","Text":"It\u0027s a solution for both the homogeneous or non-homogeneous,"},{"Start":"02:18.725 ","End":"02:20.465","Text":"whichever way you want to go."},{"Start":"02:20.465 ","End":"02:23.090","Text":"We want the nonhomogeneous, which is this."},{"Start":"02:23.090 ","End":"02:25.340","Text":"Now we\u0027ll substitute u into here."},{"Start":"02:25.340 ","End":"02:27.505","Text":"We need u_tt and u_xx."},{"Start":"02:27.505 ","End":"02:29.505","Text":"Well, that\u0027s easy enough."},{"Start":"02:29.505 ","End":"02:33.560","Text":"U_tt, this part is a constant as far as T goes,"},{"Start":"02:33.560 ","End":"02:37.655","Text":"so it\u0027s just second derivative of T_n."},{"Start":"02:37.655 ","End":"02:40.345","Text":"Of course, the sum stays the sum."},{"Start":"02:40.345 ","End":"02:44.780","Text":"For u_xx, we need to differentiate this twice with respect to x,"},{"Start":"02:44.780 ","End":"02:46.670","Text":"derivative of sine is cosine,"},{"Start":"02:46.670 ","End":"02:52.700","Text":"derivative of cosine is minus sine but we also get nPi over L of the factor twice."},{"Start":"02:52.700 ","End":"02:54.124","Text":"It\u0027s squared."},{"Start":"02:54.124 ","End":"02:56.405","Text":"There\u0027s the minus sign."},{"Start":"02:56.405 ","End":"03:00.020","Text":"This is just as is,"},{"Start":"03:00.020 ","End":"03:06.185","Text":"but I put a 1 here just so we can see that it\u0027s like nPix with n equals 1."},{"Start":"03:06.185 ","End":"03:11.870","Text":"Now we\u0027ll compare coefficients of sine nPix over L on both sides."},{"Start":"03:11.870 ","End":"03:13.550","Text":"We distinguish two cases."},{"Start":"03:13.550 ","End":"03:16.340","Text":"If n is 1, this term is included,"},{"Start":"03:16.340 ","End":"03:17.750","Text":"but if n isn\u0027t 1,"},{"Start":"03:17.750 ","End":"03:19.690","Text":"then this is irrelevant."},{"Start":"03:19.690 ","End":"03:22.290","Text":"If n is not equal to 1,"},{"Start":"03:22.290 ","End":"03:26.480","Text":"then we get that the term of sine nPix over L is T_n"},{"Start":"03:26.480 ","End":"03:30.660","Text":"double-prime and on here, sine of nPix."},{"Start":"03:30.660 ","End":"03:32.540","Text":"We have this together with this,"},{"Start":"03:32.540 ","End":"03:36.515","Text":"it\u0027s minus npi over L squared,"},{"Start":"03:36.515 ","End":"03:43.470","Text":"sorry, with an a inside here also because the a squared goes in here times T_n(t)."},{"Start":"03:43.470 ","End":"03:45.015","Text":"This is for n not equal to 1."},{"Start":"03:45.015 ","End":"03:47.125","Text":"What happens when n equals 1?"},{"Start":"03:47.125 ","End":"03:49.190","Text":"We get an extra term here."},{"Start":"03:49.190 ","End":"03:51.965","Text":"This has a coefficient 1."},{"Start":"03:51.965 ","End":"03:56.315","Text":"When n is 1, we get the same thing, but plus 1."},{"Start":"03:56.315 ","End":"04:00.430","Text":"Here we can put explicitly that n is 1, here and here."},{"Start":"04:00.430 ","End":"04:02.300","Text":"I forgot to say, here,"},{"Start":"04:02.300 ","End":"04:05.660","Text":"just put this term over on the left and make it equal 0."},{"Start":"04:05.660 ","End":"04:10.430","Text":"Similarly here, let\u0027s bring the minus onto the left so it\u0027s equal to 1."},{"Start":"04:10.430 ","End":"04:18.800","Text":"Notice that we have second-order differential equations in T_n and in T_1."},{"Start":"04:18.800 ","End":"04:20.920","Text":"That was Stage 2."},{"Start":"04:20.920 ","End":"04:25.880","Text":"Stage 3, we\u0027re going to use the initial conditions for you to get initial conditions for"},{"Start":"04:25.880 ","End":"04:31.790","Text":"the ordinary differential equations in T_n(t) back to this equation."},{"Start":"04:31.790 ","End":"04:34.520","Text":"If we differentiate it with respect to t,"},{"Start":"04:34.520 ","End":"04:36.950","Text":"we just get a prime here."},{"Start":"04:36.950 ","End":"04:41.875","Text":"Now, the initial condition for u(x) naught is this,"},{"Start":"04:41.875 ","End":"04:43.800","Text":"from which we get this."},{"Start":"04:43.800 ","End":"04:45.785","Text":"Then by comparing coefficients,"},{"Start":"04:45.785 ","End":"04:49.820","Text":"we get that all the T_n of 0 are 0."},{"Start":"04:49.820 ","End":"04:56.025","Text":"Our other initial condition is u_t(x)naught equals naught,"},{"Start":"04:56.025 ","End":"04:58.440","Text":"from which we get this."},{"Start":"04:58.440 ","End":"05:00.605","Text":"Again, by comparing coefficients,"},{"Start":"05:00.605 ","End":"05:06.170","Text":"we get that T_n\u0027(0) equals 0 for all n. Now let\u0027s use these,"},{"Start":"05:06.170 ","End":"05:14.195","Text":"this and this in the results of Part 2 festival in the case where n is not equal to 1."},{"Start":"05:14.195 ","End":"05:16.475","Text":"This was the ODE we had,"},{"Start":"05:16.475 ","End":"05:19.145","Text":"and these are now our initial conditions."},{"Start":"05:19.145 ","End":"05:20.945","Text":"In the other case,"},{"Start":"05:20.945 ","End":"05:23.130","Text":"where n equals 1."},{"Start":"05:23.130 ","End":"05:27.795","Text":"We get the same thing except that there\u0027s a 1 here and n is 1."},{"Start":"05:27.795 ","End":"05:30.800","Text":"Now, in this case we can guess a solution."},{"Start":"05:30.800 ","End":"05:37.265","Text":"Everything is 0, so just guess that T_n(t) is 0 and it works."},{"Start":"05:37.265 ","End":"05:39.170","Text":"It\u0027s a pretty common thing. Everything is 0."},{"Start":"05:39.170 ","End":"05:40.910","Text":"You try the 0 solution."},{"Start":"05:40.910 ","End":"05:43.850","Text":"Now for n equals 1, we need to work a little bit more."},{"Start":"05:43.850 ","End":"05:45.769","Text":"This should look familiar."},{"Start":"05:45.769 ","End":"05:51.080","Text":"We can simplify it a bit if we use y instead of T_1 so we have a function y(t)."},{"Start":"05:51.080 ","End":"05:53.720","Text":"Also, let\u0027s call this Omega."},{"Start":"05:53.720 ","End":"05:55.640","Text":"We have the following ODE,"},{"Start":"05:55.640 ","End":"05:58.960","Text":"y\u0027\u0027 plus Omega squared y equals 1."},{"Start":"05:58.960 ","End":"06:02.060","Text":"The initial conditions, or boundary conditions,"},{"Start":"06:02.060 ","End":"06:05.795","Text":"y(0) equals y\u0027(0) equals 0."},{"Start":"06:05.795 ","End":"06:09.880","Text":"This is a non-homogeneous equation because of the 1 here."},{"Start":"06:09.880 ","End":"06:13.870","Text":"When we solve it, we know that the solution will be the general solution of the"},{"Start":"06:13.870 ","End":"06:18.295","Text":"homogeneous plus a particular solution for the homogeneous."},{"Start":"06:18.295 ","End":"06:19.840","Text":"We know what the solution is."},{"Start":"06:19.840 ","End":"06:22.660","Text":"It\u0027s familiar at c_1 cosine Omega t plus"},{"Start":"06:22.660 ","End":"06:27.190","Text":"c2 sine Omega t. A guess for a particular solution."},{"Start":"06:27.190 ","End":"06:28.900","Text":"Since this is a constant,"},{"Start":"06:28.900 ","End":"06:34.720","Text":"let\u0027s guess another constant and see if we can find what a is."},{"Start":"06:34.720 ","End":"06:44.695","Text":"If we substitute in the equation with y particular than we get that y_p\u0027\u0027 is 0,"},{"Start":"06:44.695 ","End":"06:46.480","Text":"equals Omega squared y_p,"},{"Start":"06:46.480 ","End":"06:48.175","Text":"which is Omega squared A."},{"Start":"06:48.175 ","End":"06:53.455","Text":"If this is 1, then we get that A is equal to 1 over Omega squared."},{"Start":"06:53.455 ","End":"06:59.855","Text":"Our solution is the general solution of the homogenic plus 1 over Omega squared."},{"Start":"06:59.855 ","End":"07:01.536","Text":"The particular solution."},{"Start":"07:01.536 ","End":"07:04.185","Text":"We still need to find c_1 and c_2."},{"Start":"07:04.185 ","End":"07:06.140","Text":"Well, we have 2 equations for that."},{"Start":"07:06.140 ","End":"07:09.350","Text":"We have y(0) equals y\u0027(0) equals 0."},{"Start":"07:09.350 ","End":"07:11.930","Text":"These will help us to find the two constants."},{"Start":"07:11.930 ","End":"07:14.540","Text":"From y(0) equals 0."},{"Start":"07:14.540 ","End":"07:18.080","Text":"We get that c_1,"},{"Start":"07:18.080 ","End":"07:19.430","Text":"the cosine 0 is 1,"},{"Start":"07:19.430 ","End":"07:20.990","Text":"but sine 0 is 0,"},{"Start":"07:20.990 ","End":"07:25.520","Text":"so it\u0027s c_1 from here plus 1 over Omega squared equals 0,"},{"Start":"07:25.520 ","End":"07:29.450","Text":"giving us that c_1 is minus 1 over Omega squared."},{"Start":"07:29.450 ","End":"07:31.145","Text":"From the other one,"},{"Start":"07:31.145 ","End":"07:33.290","Text":"y prime of 0 equals 0."},{"Start":"07:33.290 ","End":"07:36.440","Text":"Here we get Omega c_2 equals 0,"},{"Start":"07:36.440 ","End":"07:38.945","Text":"because if we differentiate this,"},{"Start":"07:38.945 ","End":"07:43.730","Text":"we get something sine Omega t and the sine is 0."},{"Start":"07:43.730 ","End":"07:47.700","Text":"Here we get Omega c_2 cosine 0,"},{"Start":"07:47.700 ","End":"07:49.575","Text":"which is Omega c_2."},{"Start":"07:49.575 ","End":"07:52.350","Text":"This derivative is just 0."},{"Start":"07:52.350 ","End":"07:54.780","Text":"We get Omega C_2 is 0,"},{"Start":"07:54.780 ","End":"07:57.195","Text":"which means that c_2 is 0."},{"Start":"07:57.195 ","End":"08:00.135","Text":"Here we\u0027ve found c_1 and c_2."},{"Start":"08:00.135 ","End":"08:06.780","Text":"We get c_1 cosine Omega t is minus 1 over Omega squared cosine Omega t. The c_2 is 0."},{"Start":"08:06.780 ","End":"08:10.250","Text":"We don\u0027t need that plus one over omega squared."},{"Start":"08:10.250 ","End":"08:17.880","Text":"Now, let\u0027s go back to T_1(t) and replace Omega by Pi a over L,"},{"Start":"08:17.880 ","End":"08:20.870","Text":"1 over Omega is L over Pi a."},{"Start":"08:20.870 ","End":"08:27.125","Text":"Now recall that we found that T_n(t) is identically 0 for all n except for one."},{"Start":"08:27.125 ","End":"08:29.165","Text":"When we take this sum,"},{"Start":"08:29.165 ","End":"08:31.700","Text":"all the terms except 1 are 0."},{"Start":"08:31.700 ","End":"08:36.040","Text":"We just take the term where n equals 1 and we get sine of"},{"Start":"08:36.040 ","End":"08:40.990","Text":"1Pix over L T_1(t) but we know what T_1 is,"},{"Start":"08:40.990 ","End":"08:42.770","Text":"we just found it here."},{"Start":"08:42.770 ","End":"08:46.935","Text":"The final solution is the following."},{"Start":"08:46.935 ","End":"08:50.140","Text":"That concludes this exercise."}],"ID":30750},{"Watched":false,"Name":"Exercise 3","Duration":"11m 28s","ChapterTopicVideoID":29149,"CourseChapterTopicPlaylistID":294445,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.860","Text":"In this exercise, we have a non-homogeneous wave equation."},{"Start":"00:04.860 ","End":"00:11.070","Text":"It\u0027s non-homogeneous because we have this non-homogeneous term."},{"Start":"00:11.070 ","End":"00:13.605","Text":"It\u0027s on a finite interval."},{"Start":"00:13.605 ","End":"00:18.180","Text":"Note that the boundary conditions are homogeneous."},{"Start":"00:18.180 ","End":"00:21.165","Text":"Here we have a 0 and here we have a 0."},{"Start":"00:21.165 ","End":"00:23.655","Text":"This is not of concern."},{"Start":"00:23.655 ","End":"00:26.175","Text":"The initial conditions, I mean,"},{"Start":"00:26.175 ","End":"00:28.950","Text":"homogeneous or not, doesn\u0027t make a big difference."},{"Start":"00:28.950 ","End":"00:31.365","Text":"Here are the stages for solving."},{"Start":"00:31.365 ","End":"00:33.845","Text":"You\u0027ve had them before. I won\u0027t read them out."},{"Start":"00:33.845 ","End":"00:36.590","Text":"Just do each one as we come to it."},{"Start":"00:36.590 ","End":"00:38.105","Text":"In the first step,"},{"Start":"00:38.105 ","End":"00:42.994","Text":"we do a separation of variables on the homogeneous problem,"},{"Start":"00:42.994 ","End":"00:46.220","Text":"which means that we\u0027ve eliminated, where is it?"},{"Start":"00:46.220 ","End":"00:50.755","Text":"Yeah, just remove this and then it\u0027s the homogeneous problem."},{"Start":"00:50.755 ","End":"00:53.915","Text":"Here are the boundary conditions."},{"Start":"00:53.915 ","End":"00:57.860","Text":"First, usual separation of variables means we look for a solution"},{"Start":"00:57.860 ","End":"01:01.550","Text":"of the form or X(x) time T(t)."},{"Start":"01:01.550 ","End":"01:06.110","Text":"The PDE says that the second derivative with"},{"Start":"01:06.110 ","End":"01:11.555","Text":"respect to t is a^2 times second derivative with respect to x."},{"Start":"01:11.555 ","End":"01:17.660","Text":"Then rearranging it and ignoring the T part will just be concerned with"},{"Start":"01:17.660 ","End":"01:24.060","Text":"the x part at the moment we get x\u0027\u0027 plus lambda x=0."},{"Start":"01:24.060 ","End":"01:28.340","Text":"Now using the boundary conditions,"},{"Start":"01:28.340 ","End":"01:31.200","Text":"we get that x(0)=X(L)=0."},{"Start":"01:32.680 ","End":"01:38.060","Text":"We recognize this to be a storm Leo Ville problem and we know that"},{"Start":"01:38.060 ","End":"01:43.620","Text":"the solution is an infinite number of them."},{"Start":"01:43.620 ","End":"01:46.755","Text":"X_n is sin(n) pi x/L,"},{"Start":"01:46.755 ","End":"01:48.030","Text":"where n is 1, 2,"},{"Start":"01:48.030 ","End":"01:50.030","Text":"3, etc to infinity."},{"Start":"01:50.030 ","End":"01:58.555","Text":"We\u0027re looking for a solution which is linear combination infinite of these X_n, T_n."},{"Start":"01:58.555 ","End":"02:02.400","Text":"In our case we already have X_n so this is the form."},{"Start":"02:02.400 ","End":"02:05.645","Text":"This is the form for the homogeneous problem."},{"Start":"02:05.645 ","End":"02:08.540","Text":"But actually this will also serve as"},{"Start":"02:08.540 ","End":"02:12.730","Text":"a solution for the non-homogeneous problem, the same form."},{"Start":"02:12.730 ","End":"02:15.015","Text":"Now we\u0027re in Step 2."},{"Start":"02:15.015 ","End":"02:18.800","Text":"We take this general form for the"},{"Start":"02:18.800 ","End":"02:24.830","Text":"homogeneous and use the same form in the non-homogeneous, which is this."},{"Start":"02:24.830 ","End":"02:28.875","Text":"Let\u0027s substitute this in here and see what we get."},{"Start":"02:28.875 ","End":"02:35.570","Text":"This just second derivative of the T_n(t) so that\u0027s the double prime here."},{"Start":"02:35.570 ","End":"02:39.870","Text":"Because the X part is a constant as far as T goes =a^2."},{"Start":"02:40.130 ","End":"02:46.465","Text":"Then we need the second derivative with respect to x."},{"Start":"02:46.465 ","End":"02:50.915","Text":"From sine, we get cosine and then minus sine."},{"Start":"02:50.915 ","End":"02:53.870","Text":"But each time we get an n pi/L,"},{"Start":"02:53.870 ","End":"03:02.725","Text":"so it\u0027s minus the sign than the n pi/L^2 and sine of n pi x/L T_n(t),"},{"Start":"03:02.725 ","End":"03:07.825","Text":"that\u0027s U_xx plus the non-homogeneous bit."},{"Start":"03:07.825 ","End":"03:11.270","Text":"Now we\u0027ll do a comparison of coefficients,"},{"Start":"03:11.270 ","End":"03:13.910","Text":"but we need to split up into cases."},{"Start":"03:13.910 ","End":"03:16.940","Text":"Case where n is 2 and the case where n is not 2,"},{"Start":"03:16.940 ","End":"03:20.170","Text":"because this term is relevant only when n is 2."},{"Start":"03:20.170 ","End":"03:22.220","Text":"If n is not equal to 2,"},{"Start":"03:22.220 ","End":"03:24.200","Text":"we ignore that last term."},{"Start":"03:24.200 ","End":"03:28.565","Text":"Put the a inside the bracket here,"},{"Start":"03:28.565 ","End":"03:32.210","Text":"and we have T_n\u0027\u0027 from here."},{"Start":"03:32.210 ","End":"03:37.570","Text":"That\u0027s the coefficient of sine n pi x/L and here we need this bit with this bit."},{"Start":"03:37.570 ","End":"03:45.000","Text":"n pi a/L^ t_n(t)."},{"Start":"03:45.000 ","End":"03:47.700","Text":"Bring this to the left-hand side."},{"Start":"03:47.700 ","End":"03:53.120","Text":"This is an ordinary differential equation in t_n second-order,"},{"Start":"03:53.120 ","End":"03:55.925","Text":"but only for n not equal to 2."},{"Start":"03:55.925 ","End":"03:58.145","Text":"Now when n=2,"},{"Start":"03:58.145 ","End":"04:01.745","Text":"then we have to include this also."},{"Start":"04:01.745 ","End":"04:04.775","Text":"We have to add the plus t over here,"},{"Start":"04:04.775 ","End":"04:12.003","Text":"similar to this except that we have a plus t and wherever it says n, we put 2."},{"Start":"04:12.003 ","End":"04:15.920","Text":"If you bring this to the left-hand side we have"},{"Start":"04:15.920 ","End":"04:21.485","Text":"the following second-order ordinary differential equation in T_2."},{"Start":"04:21.485 ","End":"04:25.580","Text":"For these two equations will need initial conditions in"},{"Start":"04:25.580 ","End":"04:30.850","Text":"order to solve them and that will bring us to Step 3,"},{"Start":"04:30.850 ","End":"04:34.670","Text":"where we\u0027ll get these initial conditions for T_n and"},{"Start":"04:34.670 ","End":"04:38.355","Text":"T_2 from the initial conditions for u."},{"Start":"04:38.355 ","End":"04:43.040","Text":"We\u0027ll need u_t partial derivative of u with respect to t also,"},{"Start":"04:43.040 ","End":"04:46.310","Text":"because one of the initial conditions is in u_t,"},{"Start":"04:46.310 ","End":"04:50.750","Text":"the first initial condition is this one."},{"Start":"04:50.750 ","End":"04:57.215","Text":"We use this with t=0 and this is what we get."},{"Start":"04:57.215 ","End":"05:01.460","Text":"Comparing coefficients here of n pi x/L,"},{"Start":"05:01.460 ","End":"05:04.850","Text":"We have two cases where n is 1,"},{"Start":"05:04.850 ","End":"05:08.920","Text":"like on the right is like a one here and then not equal to 1."},{"Start":"05:08.920 ","End":"05:13.010","Text":"If n=1, then the coefficient"},{"Start":"05:13.010 ","End":"05:19.150","Text":"of n=1 will be the t_n(0), so t_n(0)=1."},{"Start":"05:19.150 ","End":"05:21.820","Text":"But if n is not equal to 1,"},{"Start":"05:21.820 ","End":"05:25.205","Text":"then the coefficient has to be 0."},{"Start":"05:25.205 ","End":"05:31.955","Text":"The other initial condition involves u_t and so we\u0027ll substitute t=0 here."},{"Start":"05:31.955 ","End":"05:35.675","Text":"We get this and again we compare coefficients,"},{"Start":"05:35.675 ","End":"05:40.565","Text":"but this time there is no separate cases on the right at 0."},{"Start":"05:40.565 ","End":"05:50.330","Text":"All the T_n\u0027(0) must be 0 for all n. Now we have to distinguish three cases."},{"Start":"05:50.330 ","End":"05:53.645","Text":"The case where n is not one or two."},{"Start":"05:53.645 ","End":"05:56.195","Text":"The case where n=1,"},{"Start":"05:56.195 ","End":"06:00.490","Text":"as we see here, it\u0027s different because of the one here."},{"Start":"06:00.490 ","End":"06:05.150","Text":"Also n=2 is different from the boundary condition."},{"Start":"06:05.150 ","End":"06:08.270","Text":"The differential equation satisfied by T_2"},{"Start":"06:08.270 ","End":"06:12.115","Text":"is different than the equation satisfied by the other T_n."},{"Start":"06:12.115 ","End":"06:13.890","Text":"Yeah, three cases."},{"Start":"06:13.890 ","End":"06:16.050","Text":"If n is not 1 or 2,"},{"Start":"06:16.050 ","End":"06:21.860","Text":"then this is the differential equation and these are the boundary conditions."},{"Start":"06:21.860 ","End":"06:24.530","Text":"We can immediately see because everything is 0,"},{"Start":"06:24.530 ","End":"06:29.045","Text":"the solution is T_n(t) is 0."},{"Start":"06:29.045 ","End":"06:33.650","Text":"If n=1, the same differential equation."},{"Start":"06:33.650 ","End":"06:40.960","Text":"But the initial condition is different because here we have that T_1(0) is1."},{"Start":"06:40.960 ","End":"06:43.220","Text":"T_1\u0027(0) is still 0,"},{"Start":"06:43.220 ","End":"06:46.640","Text":"and this one is different than what we had here."},{"Start":"06:46.640 ","End":"06:47.900","Text":"Let\u0027s leave it for a moment."},{"Start":"06:47.900 ","End":"06:49.685","Text":"We\u0027ll solve it soon."},{"Start":"06:49.685 ","End":"06:52.205","Text":"Just want to get the three cases written."},{"Start":"06:52.205 ","End":"06:54.290","Text":"The case where n=2."},{"Start":"06:54.290 ","End":"06:57.050","Text":"If you go back, we had a different differential equation."},{"Start":"06:57.050 ","End":"07:02.509","Text":"We had the T here and the initial conditions are 0."},{"Start":"07:02.509 ","End":"07:04.070","Text":"But in each of these two,"},{"Start":"07:04.070 ","End":"07:06.800","Text":"It\u0027s not the simple zero solution."},{"Start":"07:06.800 ","End":"07:09.035","Text":"Let\u0027s take the case where n=1."},{"Start":"07:09.035 ","End":"07:10.490","Text":"Just copied this."},{"Start":"07:10.490 ","End":"07:13.535","Text":"This looks like the familiar equation."},{"Start":"07:13.535 ","End":"07:15.725","Text":"Y\u0027\u0027 plus omega^2,"},{"Start":"07:15.725 ","End":"07:20.060","Text":"y=0, where we know that the solution is of this form."},{"Start":"07:20.060 ","End":"07:26.015","Text":"It\u0027s some linear combination of cosine omega t and sine omega t. In our case,"},{"Start":"07:26.015 ","End":"07:28.235","Text":"omega is what\u0027s in here."},{"Start":"07:28.235 ","End":"07:32.255","Text":"Yeah, we don\u0027t need the n here because n is 1, sorry."},{"Start":"07:32.255 ","End":"07:38.470","Text":"We know that T_1(t) is of the form c_1 cosine Omega,"},{"Start":"07:38.470 ","End":"07:43.565","Text":"which is Pi a/L(t) plus c_2 sine of."},{"Start":"07:43.565 ","End":"07:45.700","Text":"To save time."},{"Start":"07:45.700 ","End":"07:48.745","Text":"I\u0027ll give you the answer for c_1 and c_2."},{"Start":"07:48.745 ","End":"07:52.309","Text":"I mean, all you have to do is compute"},{"Start":"07:52.309 ","End":"07:58.545","Text":"T_1\u0027 and substitute t=0 in this one and the derivative."},{"Start":"07:58.545 ","End":"08:05.200","Text":"For example, if you put t=0 in this one, the sine disappears."},{"Start":"08:05.200 ","End":"08:07.165","Text":"Cosine(0) is 1,"},{"Start":"08:07.165 ","End":"08:10.670","Text":"so we just get c_1=1."},{"Start":"08:10.670 ","End":"08:14.435","Text":"For the other one, we need to substitute it in,"},{"Start":"08:14.435 ","End":"08:17.930","Text":"well here minus sign and here cosine."},{"Start":"08:17.930 ","End":"08:20.585","Text":"Okay, leave you to do it. This is what we get."},{"Start":"08:20.585 ","End":"08:24.305","Text":"Putting c_1 and c_2 in here and here,"},{"Start":"08:24.305 ","End":"08:29.430","Text":"we get that T_1(t) is just cosine pi a/ L(t)."},{"Start":"08:29.430 ","End":"08:32.170","Text":"Now the case where n=2,"},{"Start":"08:32.170 ","End":"08:37.025","Text":"this was the differential equation not homogeneous and"},{"Start":"08:37.025 ","End":"08:41.965","Text":"these are initial conditions because it\u0027s non-homogeneous,"},{"Start":"08:41.965 ","End":"08:46.910","Text":"will need to find the homogeneous solution plus a particular solution."},{"Start":"08:46.910 ","End":"08:51.050","Text":"T_2 will be, let\u0027s call it T_2(h) for the"},{"Start":"08:51.050 ","End":"08:57.755","Text":"homogeneous and T_2(p) for the particular solution."},{"Start":"08:57.755 ","End":"09:00.620","Text":"Now, we know that the homogeneous,"},{"Start":"09:00.620 ","End":"09:06.395","Text":"just like above is cosine and sine."},{"Start":"09:06.395 ","End":"09:08.390","Text":"Only this time instead of a 1,"},{"Start":"09:08.390 ","End":"09:12.784","Text":"we have a 2 for n. For a particular solution"},{"Start":"09:12.784 ","End":"09:17.660","Text":"we\u0027ll guess some constant times t. You get experience with this."},{"Start":"09:17.660 ","End":"09:20.510","Text":"If this is t, well,"},{"Start":"09:20.510 ","End":"09:22.790","Text":"you could have tried a t plus b."},{"Start":"09:22.790 ","End":"09:25.039","Text":"Maybe you want to try something of first-order,"},{"Start":"09:25.039 ","End":"09:27.135","Text":"but you\u0027d get that b is 0, but yeah,"},{"Start":"09:27.135 ","End":"09:29.540","Text":"a t plus B would be another good guess."},{"Start":"09:29.540 ","End":"09:36.275","Text":"Substituting this in the non-homogeneous differential equation, we get this."},{"Start":"09:36.275 ","End":"09:39.965","Text":"The second derivative is 0."},{"Start":"09:39.965 ","End":"09:45.210","Text":"This squared as is T_2(p) is a t on"},{"Start":"09:45.210 ","End":"09:51.620","Text":"the right t. This gives us that a comparing coefficients of t,"},{"Start":"09:51.620 ","End":"09:55.655","Text":"we get that a is equal to 1 over this,"},{"Start":"09:55.655 ","End":"09:58.145","Text":"which is l over 2 Pi a^2."},{"Start":"09:58.145 ","End":"10:02.000","Text":"We know that the general form of T_2 with"},{"Start":"10:02.000 ","End":"10:07.420","Text":"constants c_1 and c_2 is the following homogeneous plus particular."},{"Start":"10:07.420 ","End":"10:10.645","Text":"We need to find c_1 and c_2."},{"Start":"10:10.645 ","End":"10:13.160","Text":"But we have two equations to help us."},{"Start":"10:13.160 ","End":"10:15.605","Text":"That would be these two."},{"Start":"10:15.605 ","End":"10:22.724","Text":"After the calculations we get c_1 is 0,"},{"Start":"10:22.724 ","End":"10:25.620","Text":"c_2 is minus L/2 Pi a6^3."},{"Start":"10:25.620 ","End":"10:29.595","Text":"Now putting c_1 and c_2 here and here,"},{"Start":"10:29.595 ","End":"10:39.240","Text":"we get that T_2 is the following and recall that T_1 is this."},{"Start":"10:39.240 ","End":"10:45.280","Text":"Also T_n is 0 for n not equal to 1 and 2."},{"Start":"10:45.280 ","End":"10:48.155","Text":"These are the only non-zero Ts."},{"Start":"10:48.155 ","End":"10:53.540","Text":"U(x,t ) instead of being the sum from one to infinity,"},{"Start":"10:53.540 ","End":"11:00.445","Text":"we can just take from 1 to 2 because we only want the two elements for n=1 and for n=2."},{"Start":"11:00.445 ","End":"11:05.625","Text":"That will be sine of 1 Pi x/L"},{"Start":"11:05.625 ","End":"11:12.400","Text":"times T_1(1) plus sine 2pi x/L T_2(t),"},{"Start":"11:12.400 ","End":"11:16.490","Text":"which is this just rearranged the order,"},{"Start":"11:16.490 ","End":"11:21.980","Text":"took this part first and the minus part second."},{"Start":"11:21.980 ","End":"11:29.010","Text":"That is actually the answer and so that concludes this exercise."}],"ID":30751},{"Watched":false,"Name":"Exercise 4","Duration":"8m 45s","ChapterTopicVideoID":29150,"CourseChapterTopicPlaylistID":294445,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.765","Text":"In this exercise, we have another nonhomogeneous wave equation on a finite interval."},{"Start":"00:06.765 ","End":"00:10.230","Text":"This is the non-homogeneous part,"},{"Start":"00:10.230 ","End":"00:15.880","Text":"but at least we have homogeneous boundary conditions."},{"Start":"00:16.970 ","End":"00:23.915","Text":"Here, for reference are the stages for solving the usual ones who\u0027ve had these before."},{"Start":"00:23.915 ","End":"00:32.550","Text":"Let\u0027s just start with Stage 1 which is separation of variables on the homogeneous PDF."},{"Start":"00:32.750 ","End":"00:38.325","Text":"Here\u0027s the homogeneous, it\u0027s without that extra non-homogeneous bit."},{"Start":"00:38.325 ","End":"00:41.750","Text":"The separation of variables means we\u0027re going to look for solution"},{"Start":"00:41.750 ","End":"00:47.090","Text":"of this form and when we do that,"},{"Start":"00:47.090 ","End":"00:48.560","Text":"plug it into the equation,"},{"Start":"00:48.560 ","End":"00:49.640","Text":"this is what we get."},{"Start":"00:49.640 ","End":"00:51.335","Text":"This is familiar."},{"Start":"00:51.335 ","End":"00:53.450","Text":"Rearrange it."},{"Start":"00:53.450 ","End":"00:59.960","Text":"Forget about the T part and we have that X\u0027\u0027 over X is minus Lambda."},{"Start":"00:59.960 ","End":"01:05.385","Text":"This gives us from the equation,"},{"Start":"01:05.385 ","End":"01:12.980","Text":"X\u0027\u0027 plus Lambda X equals 0 and from the boundary conditions,"},{"Start":"01:12.980 ","End":"01:18.710","Text":"we get that X\u0027 of 0 is 0 and X\u0027 of L is 0."},{"Start":"01:18.710 ","End":"01:21.860","Text":"This is a Sturm Louisville problem."},{"Start":"01:21.860 ","End":"01:24.110","Text":"We\u0027ve done this kind of thing before."},{"Start":"01:24.110 ","End":"01:31.320","Text":"The solutions turn out to be of the whole sequence of them, X_n,"},{"Start":"01:31.320 ","End":"01:39.990","Text":"which is cosine n Pi x over L. N starts from 0 this time and when n is 0,"},{"Start":"01:39.990 ","End":"01:42.970","Text":"it\u0027s not a cosine it\u0027s just 1."},{"Start":"01:43.130 ","End":"01:45.950","Text":"Then if we were doing the homogeneous,"},{"Start":"01:45.950 ","End":"01:49.300","Text":"we would look for a solution of this form"},{"Start":"01:49.300 ","End":"01:55.475","Text":"and we already have the Xn\u0027s and we would then look for the T_n."},{"Start":"01:55.475 ","End":"02:01.890","Text":"The thing is that this same form is also good for the nonhomogeneous."},{"Start":"02:02.660 ","End":"02:11.450","Text":"This is the non-homogeneous so let\u0027s just substitute this in here and see what we get."},{"Start":"02:11.450 ","End":"02:14.300","Text":"This part is u_tt."},{"Start":"02:14.300 ","End":"02:21.195","Text":"We just differentiate the part with T twice here and then we have"},{"Start":"02:21.195 ","End":"02:29.290","Text":"a^2 second derivative with respect to x. Cosine gives us minus sign,"},{"Start":"02:29.290 ","End":"02:32.150","Text":"which gives us minus cosine, so minus the cosine."},{"Start":"02:32.780 ","End":"02:37.405","Text":"We also get this antiderivative n Pi over L twice"},{"Start":"02:37.405 ","End":"02:42.830","Text":"squared plus this just copied from here."},{"Start":"02:43.740 ","End":"02:51.275","Text":"Now we\u0027re going to compare coefficients of n Pi x over L. Well, there\u0027s 2 cases."},{"Start":"02:51.275 ","End":"02:53.190","Text":"If n is 1,"},{"Start":"02:53.190 ","End":"02:55.200","Text":"then we include this term if not,"},{"Start":"02:55.200 ","End":"02:57.435","Text":"then we throw this term out."},{"Start":"02:57.435 ","End":"03:00.195","Text":"If n is not equal to 1,"},{"Start":"03:00.195 ","End":"03:04.870","Text":"ignore this and the coefficient of cosine Pi X over"},{"Start":"03:04.870 ","End":"03:11.345","Text":"L here is T_n double prime of t and here,"},{"Start":"03:11.345 ","End":"03:14.820","Text":"put the a^2 here, just a here,"},{"Start":"03:14.820 ","End":"03:21.350","Text":"so N Pi a over L squared with a minus and also the T_n of"},{"Start":"03:21.350 ","End":"03:25.250","Text":"t part which gives us"},{"Start":"03:25.250 ","End":"03:31.070","Text":"the following ordinary differential equation of the second order in T_n."},{"Start":"03:31.070 ","End":"03:34.835","Text":"The other case of n is equal to 1."},{"Start":"03:34.835 ","End":"03:37.760","Text":"We get, well, same thing as here,"},{"Start":"03:37.760 ","End":"03:41.640","Text":"except that we get another plus 1 at the end."},{"Start":"03:41.680 ","End":"03:44.255","Text":"If we rearrange that,"},{"Start":"03:44.255 ","End":"03:50.235","Text":"we get the other ordinary differential equation for T_1."},{"Start":"03:50.235 ","End":"03:56.290","Text":"All the ends except 1 from here and n equals 1 from here."},{"Start":"03:56.460 ","End":"03:58.810","Text":"We have a differential equations,"},{"Start":"03:58.810 ","End":"04:01.780","Text":"but we don\u0027t have boundary conditions or initial conditions."},{"Start":"04:01.780 ","End":"04:05.575","Text":"We need 2 conditions for a second-order equation."},{"Start":"04:05.575 ","End":"04:13.080","Text":"We\u0027ll get these conditions from the conditions for u."},{"Start":"04:13.080 ","End":"04:18.570","Text":"This is u again and we\u0027ll need du by dt,"},{"Start":"04:18.570 ","End":"04:21.335","Text":"so that\u0027s for the T_n prime."},{"Start":"04:21.335 ","End":"04:26.390","Text":"This is the first of the initial conditions."},{"Start":"04:27.230 ","End":"04:37.875","Text":"If we substitute t equals 0 here that will give us this sum with T_n of 0"},{"Start":"04:37.875 ","End":"04:48.590","Text":"is equal to cosine Pi X over L and that will give us by comparing coefficients of n Pi x"},{"Start":"04:48.590 ","End":"04:58.560","Text":"over L that T_n"},{"Start":"04:58.560 ","End":"05:03.690","Text":"of 0 will equal 1 if n=1,"},{"Start":"05:03.690 ","End":"05:08.430","Text":"then we have cosine 1 Pi x over L and otherwise 0."},{"Start":"05:08.430 ","End":"05:12.935","Text":"The other initial condition"},{"Start":"05:12.935 ","End":"05:20.480","Text":"u_t(x,0) is 0 so we substitute in this equation t=0,"},{"Start":"05:20.480 ","End":"05:22.660","Text":"and we get this."},{"Start":"05:22.660 ","End":"05:24.950","Text":"We don\u0027t have to divide into cases."},{"Start":"05:24.950 ","End":"05:31.460","Text":"It\u0027s clear that all the coefficients of the cosine n Pi x over L is 0."},{"Start":"05:31.460 ","End":"05:38.820","Text":"In other words, T_n prime is 0 for all n without the exception that we had here."},{"Start":"05:39.110 ","End":"05:42.285","Text":"If n is not equal to 1,"},{"Start":"05:42.285 ","End":"05:50.905","Text":"then this is the equation that we had and these are the initial conditions."},{"Start":"05:50.905 ","End":"05:57.415","Text":"From here it\u0027s clear that the only solution is the zero solution because everything is 0."},{"Start":"05:57.415 ","End":"06:01.020","Text":"If n is equal to 1,"},{"Start":"06:01.020 ","End":"06:03.245","Text":"then we have a 1 here."},{"Start":"06:03.245 ","End":"06:06.500","Text":"It\u0027s not homogeneous so we\u0027ll have to use the technique"},{"Start":"06:06.500 ","End":"06:10.975","Text":"of solution to the homogeneous plus a particular solution."},{"Start":"06:10.975 ","End":"06:14.065","Text":"Let\u0027s find T_1 now."},{"Start":"06:14.065 ","End":"06:18.115","Text":"We have the differential equation and initial conditions."},{"Start":"06:18.115 ","End":"06:21.880","Text":"This is non-homogeneous so we\u0027re going to break it"},{"Start":"06:21.880 ","End":"06:26.455","Text":"up into the homogeneous part and a particular solution."},{"Start":"06:26.455 ","End":"06:29.800","Text":"The homogeneous part will be,"},{"Start":"06:29.800 ","End":"06:32.425","Text":"as we know, C_1 cosine,"},{"Start":"06:32.425 ","End":"06:40.425","Text":"this t plus C_2 sine this times T. Particular solution we\u0027ll make a guess."},{"Start":"06:40.425 ","End":"06:41.774","Text":"It\u0027s a constant here,"},{"Start":"06:41.774 ","End":"06:47.680","Text":"let\u0027s guess that the particular solution will also be a constant or substitute into here,"},{"Start":"06:47.680 ","End":"06:54.570","Text":"this a T_1 double prime plus this squared T_1 equals,"},{"Start":"06:54.570 ","End":"06:57.400","Text":"oops, it\u0027s a 1 here."},{"Start":"06:57.400 ","End":"07:02.300","Text":"The second derivative of 0 so we get that this thing squared times a is 1,"},{"Start":"07:02.300 ","End":"07:06.520","Text":"so a is 1 over the squared, just inverted."},{"Start":"07:06.520 ","End":"07:10.870","Text":"Here\u0027s the T particular,"},{"Start":"07:10.870 ","End":"07:14.075","Text":"and here\u0027s the T homogeneous."},{"Start":"07:14.075 ","End":"07:19.730","Text":"We need to find C_1 and C_2 but for that we have 2 equations."},{"Start":"07:19.730 ","End":"07:26.275","Text":"We have these 2 that we haven\u0027t used yet and these will help us to find C_1 and C_2."},{"Start":"07:26.275 ","End":"07:31.200","Text":"For example, if we let t=0,"},{"Start":"07:31.200 ","End":"07:34.325","Text":"then this whole thing is 0 because it\u0027s a sine,"},{"Start":"07:34.325 ","End":"07:44.670","Text":"the cosine is 1 so we get C_1 plus L over Pi a squared equals 0 so C_1 equals this."},{"Start":"07:44.670 ","End":"07:50.090","Text":"If we differentiate and let t=0,"},{"Start":"07:50.090 ","End":"07:51.680","Text":"we\u0027ll get the other condition,"},{"Start":"07:51.680 ","End":"07:53.975","Text":"which is that C_2 is 0."},{"Start":"07:53.975 ","End":"07:55.850","Text":"I\u0027ll spare you the details."},{"Start":"07:55.850 ","End":"07:58.220","Text":"Now that we have C_1 and C_2,"},{"Start":"07:58.220 ","End":"08:07.040","Text":"we can put them here and here and this is what we get for T_1 and all the other Ts,"},{"Start":"08:07.040 ","End":"08:09.790","Text":"all the other T_n are 0."},{"Start":"08:09.790 ","End":"08:13.455","Text":"Now we can substitute in our sum,"},{"Start":"08:13.455 ","End":"08:19.770","Text":"but everything 0 except when n=1 and so we get"},{"Start":"08:19.770 ","End":"08:29.010","Text":"just cosine of 1 Pi x over L T_1 of t and we know what T_1 of t is, it\u0027s this."},{"Start":"08:29.010 ","End":"08:37.310","Text":"What we get is cosine Pi x over L times this whole thing and this is the answer."},{"Start":"08:37.310 ","End":"08:40.355","Text":"Well, together with u(x,t) equals,"},{"Start":"08:40.355 ","End":"08:45.600","Text":"and that concludes this exercise."}],"ID":30752},{"Watched":false,"Name":"Exercise 5","Duration":"9m 43s","ChapterTopicVideoID":29151,"CourseChapterTopicPlaylistID":294445,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.155","Text":"Here we have a non-homogeneous wave equation to solve."},{"Start":"00:04.155 ","End":"00:12.195","Text":"It\u0027s non-homogeneous because of the extra plus e^t here finite interval."},{"Start":"00:12.195 ","End":"00:18.970","Text":"The good thing is that the boundary conditions are homogeneous here and here."},{"Start":"00:19.340 ","End":"00:22.470","Text":"Here are the stages for solving."},{"Start":"00:22.470 ","End":"00:23.760","Text":"I won\u0027t read them out."},{"Start":"00:23.760 ","End":"00:25.230","Text":"You can pause and read them."},{"Start":"00:25.230 ","End":"00:28.335","Text":"Let\u0027s just get into stage 1 already."},{"Start":"00:28.335 ","End":"00:32.160","Text":"We use separation of variables on the homogeneous PDF."},{"Start":"00:32.160 ","End":"00:39.780","Text":"Homogeneous means we drop the non-homogeneous bit so we just have the a^2u_xx."},{"Start":"00:39.780 ","End":"00:43.700","Text":"These are the boundary conditions."},{"Start":"00:43.700 ","End":"00:46.190","Text":"Here\u0027s the separation of variables."},{"Start":"00:46.190 ","End":"00:48.335","Text":"We look for you of this form."},{"Start":"00:48.335 ","End":"00:52.520","Text":"Then we substitute into this equation and this is what we get."},{"Start":"00:52.520 ","End":"00:55.730","Text":"Then rearrange and we get this."},{"Start":"00:55.730 ","End":"00:59.545","Text":"Forget about the T part go for the X part."},{"Start":"00:59.545 ","End":"01:01.515","Text":"If we rearrange this,"},{"Start":"01:01.515 ","End":"01:05.419","Text":"we get X\u0027\u0027 plus Lambda x is 0."},{"Start":"01:05.419 ","End":"01:10.820","Text":"The boundary conditions here they are again, are as follows,"},{"Start":"01:10.820 ","End":"01:17.180","Text":"so that we can cancel the T(t) and we get X\u0027(0) equals X\u0027(L)."},{"Start":"01:17.180 ","End":"01:22.345","Text":"This is a Sturm Louisville ordinary differential equation in X."},{"Start":"01:22.345 ","End":"01:23.810","Text":"We know how to solve these,"},{"Start":"01:23.810 ","End":"01:26.285","Text":"so I\u0027ll just give you the solution."},{"Start":"01:26.285 ","End":"01:33.425","Text":"These are the eigenfunctions as a whole sequence of them and goes from 0 to infinity."},{"Start":"01:33.425 ","End":"01:35.090","Text":"That\u0027s usual when there\u0027s a cosine."},{"Start":"01:35.090 ","End":"01:37.670","Text":"Remember to start from 0 because when n is 0,"},{"Start":"01:37.670 ","End":"01:39.490","Text":"this is just the constant 1."},{"Start":"01:39.490 ","End":"01:43.010","Text":"Now we know that u(x,t) and t is going to be"},{"Start":"01:43.010 ","End":"01:48.740","Text":"an infinite linear combination of these X_n times T_n."},{"Start":"01:48.740 ","End":"01:54.144","Text":"It turns out that this general form is also good for the non-homogeneous."},{"Start":"01:54.144 ","End":"01:56.580","Text":"We\u0027ll solve for T_n and we\u0027ll get different T_n."},{"Start":"01:56.580 ","End":"01:58.400","Text":"If we solve for the homogeneous,"},{"Start":"01:58.400 ","End":"01:59.810","Text":"but for the non-homogeneous,"},{"Start":"01:59.810 ","End":"02:01.915","Text":"this is also the form."},{"Start":"02:01.915 ","End":"02:05.670","Text":"In our case, this is what it is because we know the X_n."},{"Start":"02:05.670 ","End":"02:08.945","Text":"We still need to find the T_ns."},{"Start":"02:08.945 ","End":"02:10.915","Text":"Here we are again."},{"Start":"02:10.915 ","End":"02:15.110","Text":"Here again is the non-homogeneous PDE."},{"Start":"02:15.110 ","End":"02:20.000","Text":"We want this to work for this u_tt,"},{"Start":"02:20.000 ","End":"02:25.520","Text":"just differentiate twice with respect to t. The cosine is a constant,"},{"Start":"02:25.520 ","End":"02:30.470","Text":"just contains x. T_n(t) differentiated twice comes out T_n\u0027\u0027(t)."},{"Start":"02:30.470 ","End":"02:34.820","Text":"That\u0027s the u_tt. Now u_xx is this."},{"Start":"02:34.820 ","End":"02:37.609","Text":"We just differentiate term wise."},{"Start":"02:37.609 ","End":"02:42.341","Text":"Second derivative of the cosine is minus cosine."},{"Start":"02:42.341 ","End":"02:46.930","Text":"The anti-derivative is n Pi over L. But there\u0027s 2 differentiations,"},{"Start":"02:46.930 ","End":"02:53.105","Text":"so we get nPi over L squared as the minus cosine and is the T_n."},{"Start":"02:53.105 ","End":"02:55.100","Text":"We also have this term,"},{"Start":"02:55.100 ","End":"03:00.560","Text":"second derivative with respect to t is just itself e^t."},{"Start":"03:00.560 ","End":"03:05.600","Text":"But for consistency, that\u0027s also write it as cosine nPi x over L,"},{"Start":"03:05.600 ","End":"03:07.970","Text":"where n is 0. This thing is 1."},{"Start":"03:07.970 ","End":"03:12.055","Text":"But now we have consistently same terms."},{"Start":"03:12.055 ","End":"03:18.485","Text":"This is so we can compare coefficients of the nPi x over L is going to be 2n."},{"Start":"03:18.485 ","End":"03:21.070","Text":"N is 0 and n is not 0."},{"Start":"03:21.070 ","End":"03:26.190","Text":"When n is not 0, we don\u0027t need to look at this term and we get"},{"Start":"03:26.190 ","End":"03:32.130","Text":"that T_n\u0027\u0027(t) is the coefficient part."},{"Start":"03:32.130 ","End":"03:39.370","Text":"Here, the coefficient of cosine nPi x over L is this minus this squared T_n."},{"Start":"03:39.370 ","End":"03:45.110","Text":"This gives us the following second-order ODE in T_n."},{"Start":"03:45.110 ","End":"03:48.845","Text":"When n is 0, we also take this term."},{"Start":"03:48.845 ","End":"03:53.960","Text":"We have here just T_n\u0027\u0027(0),"},{"Start":"03:53.960 ","End":"03:57.210","Text":"but it\u0027s n is 0, so it\u0027s T_0\u0027\u0027."},{"Start":"03:57.650 ","End":"04:02.160","Text":"This when n is 0, is just 0."},{"Start":"04:02.160 ","End":"04:06.570","Text":"All we\u0027re left with is this, the e^t."},{"Start":"04:06.570 ","End":"04:11.810","Text":"We get T_0\u0027\u0027 is e^t second-order differential equation n_t(0)."},{"Start":"04:11.810 ","End":"04:14.795","Text":"We have these 2 differential equations,"},{"Start":"04:14.795 ","End":"04:18.865","Text":"but we don\u0027t have boundary or initial conditions on them."},{"Start":"04:18.865 ","End":"04:23.510","Text":"We\u0027ll get those from the initial conditions on u(x,t)."},{"Start":"04:23.510 ","End":"04:27.425","Text":"Here again is u(x,t) the general form and will need its"},{"Start":"04:27.425 ","End":"04:31.340","Text":"derivative with respect to t. We have 2 initial conditions."},{"Start":"04:31.340 ","End":"04:33.845","Text":"The first one is u (x naught),"},{"Start":"04:33.845 ","End":"04:37.070","Text":"and the other one is going to be u_t( x naught)."},{"Start":"04:37.070 ","End":"04:38.765","Text":"Deal with this one."},{"Start":"04:38.765 ","End":"04:41.824","Text":"Just substitute t equals 0."},{"Start":"04:41.824 ","End":"04:45.260","Text":"Here, what we get is the same thing as this,"},{"Start":"04:45.260 ","End":"04:50.040","Text":"except the 0 instead of the t. This is going to equal cosine Pi x"},{"Start":"04:50.040 ","End":"04:55.520","Text":"over L. If we compare coefficients of cosine nPi x over L,"},{"Start":"04:55.520 ","End":"04:59.135","Text":"the only thing we get here is where n equals 1."},{"Start":"04:59.135 ","End":"05:01.265","Text":"You can imagine a 1 written here."},{"Start":"05:01.265 ","End":"05:02.815","Text":"When n is 1,"},{"Start":"05:02.815 ","End":"05:05.985","Text":"we get that T_n(0) equals 1."},{"Start":"05:05.985 ","End":"05:07.975","Text":"For all the others,"},{"Start":"05:07.975 ","End":"05:10.414","Text":"T_n(0) is = 0."},{"Start":"05:10.414 ","End":"05:13.220","Text":"Now the other initial condition for this,"},{"Start":"05:13.220 ","End":"05:16.695","Text":"we put t equals 0 here."},{"Start":"05:16.695 ","End":"05:21.260","Text":"We get the sum of cosine T_n\u0027(0),"},{"Start":"05:21.260 ","End":"05:25.725","Text":"and this is = 0, which is 0."},{"Start":"05:25.725 ","End":"05:27.980","Text":"All the T_n\u0027s is 0."},{"Start":"05:27.980 ","End":"05:29.735","Text":"When we compare coefficients,"},{"Start":"05:29.735 ","End":"05:33.860","Text":"collecting this and this and just rearranging a bit,"},{"Start":"05:33.860 ","End":"05:37.860","Text":"we get that when n is not equal to 1,"},{"Start":"05:37.860 ","End":"05:42.545","Text":"then T_n(0) is 0 and T_n\u0027(0) is 0."},{"Start":"05:42.545 ","End":"05:45.080","Text":"When n = 1,"},{"Start":"05:45.080 ","End":"05:46.850","Text":"T_n(0) is 1,"},{"Start":"05:46.850 ","End":"05:50.490","Text":"and T_n\u0027(0) is 0 instead of n,"},{"Start":"05:50.490 ","End":"05:52.095","Text":"we write 1 because n is 1."},{"Start":"05:52.095 ","End":"05:54.320","Text":"We actually have 3 cases."},{"Start":"05:54.320 ","End":"05:58.595","Text":"The initial conditions depend on whether n is 1 or not,"},{"Start":"05:58.595 ","End":"06:02.600","Text":"but the ODE depends on whether n is 0 or not."},{"Start":"06:02.600 ","End":"06:04.430","Text":"Really the 3 cases,"},{"Start":"06:04.430 ","End":"06:07.730","Text":"case where n is not equal to 0 or 1, the case where n equals 0."},{"Start":"06:07.730 ","End":"06:09.650","Text":"In the case where n equals 1."},{"Start":"06:09.650 ","End":"06:15.620","Text":"For this case, this was the ODE we had for n not equal to 0,"},{"Start":"06:15.620 ","End":"06:19.835","Text":"and the initial condition, because n is also not equal to 1 is this."},{"Start":"06:19.835 ","End":"06:23.330","Text":"Now everything is a 0 here and here."},{"Start":"06:23.330 ","End":"06:29.485","Text":"We can guess the solution that T_n(t) is identically 0."},{"Start":"06:29.485 ","End":"06:33.425","Text":"Just to repeat n is not equal to 0 or 1 for this."},{"Start":"06:33.425 ","End":"06:37.495","Text":"Now we\u0027ll take the case where n equals 1."},{"Start":"06:37.495 ","End":"06:40.680","Text":"In this case, n is still not 0."},{"Start":"06:40.680 ","End":"06:43.730","Text":"This is the differential equation,"},{"Start":"06:43.730 ","End":"06:47.150","Text":"but this is the initial condition, we\u0027ll solve it in a moment."},{"Start":"06:47.150 ","End":"06:50.555","Text":"Let\u0027s just get the last case written where n is 0."},{"Start":"06:50.555 ","End":"06:52.970","Text":"This was our ODE."},{"Start":"06:52.970 ","End":"06:56.315","Text":"This is the initial condition,"},{"Start":"06:56.315 ","End":"06:58.130","Text":"same like in here,"},{"Start":"06:58.130 ","End":"07:00.125","Text":"where n is not equal to 1."},{"Start":"07:00.125 ","End":"07:04.100","Text":"We have this case all we still need to solve this case and this case."},{"Start":"07:04.100 ","End":"07:08.590","Text":"Continuing. The n=1 case was this."},{"Start":"07:08.590 ","End":"07:10.985","Text":"We know how to solve this equation."},{"Start":"07:10.985 ","End":"07:12.530","Text":"Just think of this as Omega."},{"Start":"07:12.530 ","End":"07:13.985","Text":"We have Omega^2."},{"Start":"07:13.985 ","End":"07:16.955","Text":"Y\u0027\u0027 plus Omega^2 Y is 0."},{"Start":"07:16.955 ","End":"07:18.365","Text":"We know how to solve that."},{"Start":"07:18.365 ","End":"07:27.735","Text":"We have a combination of cosine Omega t and sine Omega t. If you substitute t = 0,"},{"Start":"07:27.735 ","End":"07:31.395","Text":"this cosine is 1 and we get c_1 and this is 0."},{"Start":"07:31.395 ","End":"07:37.035","Text":"We get T_1 of 0 is 1, c_1 is 1."},{"Start":"07:37.035 ","End":"07:42.335","Text":"Take the derivative of this and substitute t = 0."},{"Start":"07:42.335 ","End":"07:45.625","Text":"Well, you\u0027ll see that we get that c_2 is 0."},{"Start":"07:45.625 ","End":"07:48.390","Text":"Well, c_2 times this constant is 0,"},{"Start":"07:48.390 ","End":"07:49.965","Text":"so c_2 is 0."},{"Start":"07:49.965 ","End":"07:52.575","Text":"We have c_1 and c_2."},{"Start":"07:52.575 ","End":"07:55.295","Text":"If we take the linear combination,"},{"Start":"07:55.295 ","End":"07:57.020","Text":"c_1 is 1, c_2 is 0."},{"Start":"07:57.020 ","End":"07:58.700","Text":"We just have this bit."},{"Start":"07:58.700 ","End":"08:01.700","Text":"I felt like writing the t on the numerator here."},{"Start":"08:01.700 ","End":"08:05.330","Text":"The last case, where n=0."},{"Start":"08:05.330 ","End":"08:08.270","Text":"This is a non-homogeneous second-order equation,"},{"Start":"08:08.270 ","End":"08:14.420","Text":"but it\u0027s very easy because there\u0027s no T naught\u0027\u0027 or T not is just the second derivative."},{"Start":"08:14.420 ","End":"08:18.465","Text":"We have to do is integrate twice and we get e^t,"},{"Start":"08:18.465 ","End":"08:20.510","Text":"integrates to e^t each time,"},{"Start":"08:20.510 ","End":"08:22.205","Text":"but we get an extra constant."},{"Start":"08:22.205 ","End":"08:29.140","Text":"We get e^t plus A and then e^t plus At plus B. I should have written down here."},{"Start":"08:29.140 ","End":"08:31.705","Text":"T_0\u0027 is e^t plus A."},{"Start":"08:31.705 ","End":"08:39.290","Text":"Now we can substitute the initial conditions because T naught of 0 is 0."},{"Start":"08:39.290 ","End":"08:42.185","Text":"We get that B is equal to minus 1."},{"Start":"08:42.185 ","End":"08:43.985","Text":"This is 0, this is 1."},{"Start":"08:43.985 ","End":"08:46.340","Text":"To make it 0, this has to be minus 1."},{"Start":"08:46.340 ","End":"08:52.840","Text":"T\u0027(0) is 0, which means that e^t,"},{"Start":"08:52.840 ","End":"08:55.020","Text":"which is 1 plus A is 0."},{"Start":"08:55.020 ","End":"08:56.625","Text":"A is minus 1."},{"Start":"08:56.625 ","End":"08:59.120","Text":"Putting A and B here,"},{"Start":"08:59.120 ","End":"09:04.995","Text":"what we get is T naught of t is e^t minus 1, t minus 1."},{"Start":"09:04.995 ","End":"09:09.570","Text":"That\u0027s right. The other cases T_1(t) is this."},{"Start":"09:09.570 ","End":"09:15.000","Text":"We had that T_n(t) for the other ends is just 0."},{"Start":"09:15.000 ","End":"09:20.885","Text":"These are the 3 cases now returned to this equation where we have the infinite sum."},{"Start":"09:20.885 ","End":"09:25.820","Text":"But only when n is 0 or 1 do we get something non-zero."},{"Start":"09:25.820 ","End":"09:31.115","Text":"We get the term for n equals 0 and the term for n equals 1."},{"Start":"09:31.115 ","End":"09:36.065","Text":"We know what T_0 and T_1 are they are here and here."},{"Start":"09:36.065 ","End":"09:39.050","Text":"Put them in, and this is what we get,"},{"Start":"09:39.050 ","End":"09:41.075","Text":"and that\u0027s the answer."},{"Start":"09:41.075 ","End":"09:43.290","Text":"We are done."}],"ID":30753}],"Thumbnail":null,"ID":294445},{"Name":"Domain of Dependence","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Domain of Dependence","Duration":"7m 58s","ChapterTopicVideoID":29136,"CourseChapterTopicPlaylistID":294446,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29136.jpeg","UploadDate":"2022-06-06T08:27:12.3370000","DurationForVideoObject":"PT7M58S","Description":null,"MetaTitle":"Domain of Dependence: Video + Workbook | Proprep","MetaDescription":"The Wave Equation - Domain of Dependence. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/partial-differential-equations/the-wave-equation/domain-of-dependence/vid30754","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.900","Text":"In this clip, we\u0027ll talk about a new concept,"},{"Start":"00:03.900 ","End":"00:05.925","Text":"the domain of dependence,"},{"Start":"00:05.925 ","End":"00:10.260","Text":"also known as the characteristic triangle."},{"Start":"00:10.260 ","End":"00:14.670","Text":"What it will do for us it will allow us to use"},{"Start":"00:14.670 ","End":"00:19.662","Text":"the D\u0027Alembert formula sometimes in cases where we normally couldn\u0027t,"},{"Start":"00:19.662 ","End":"00:24.510","Text":"like when we have a finite interval and not an infinite interval."},{"Start":"00:24.510 ","End":"00:28.935","Text":"If we\u0027re lucky, we can still use D\u0027Alembert\u0027s formula."},{"Start":"00:28.935 ","End":"00:32.730","Text":"At the start, we\u0027re given the nonhomogeneous wave equation on"},{"Start":"00:32.730 ","End":"00:37.154","Text":"an infinite interval as follows."},{"Start":"00:37.154 ","End":"00:39.325","Text":"We have 3 functions,"},{"Start":"00:39.325 ","End":"00:44.690","Text":"the initial conditions for u and du by dt."},{"Start":"00:44.690 ","End":"00:46.564","Text":"These are at time 0."},{"Start":"00:46.564 ","End":"00:48.800","Text":"We don\u0027t need boundary conditions in"},{"Start":"00:48.800 ","End":"00:53.110","Text":"this case because there is no boundary to the interval."},{"Start":"00:53.110 ","End":"00:57.560","Text":"D\u0027Alembert\u0027s formula for finding u(x,t) at"},{"Start":"00:57.560 ","End":"01:03.710","Text":"any point where t is bigger or equal to 0 is given by this formula."},{"Start":"01:03.710 ","End":"01:06.170","Text":"I want to modify it a little bit."},{"Start":"01:06.170 ","End":"01:08.750","Text":"Just change variable names."},{"Start":"01:08.750 ","End":"01:14.290","Text":"Let\u0027s suppose that we just want to find the solution u at a particular point,"},{"Start":"01:14.290 ","End":"01:17.080","Text":"x naught, t naught, and not in general."},{"Start":"01:17.080 ","End":"01:19.314","Text":"Then we can replace x naught,"},{"Start":"01:19.314 ","End":"01:27.150","Text":"t naught by x and t. Then we also have to change s and Tau,"},{"Start":"01:27.150 ","End":"01:32.990","Text":"s will be changed to x and Tau will be t. There\u0027s something else."},{"Start":"01:32.990 ","End":"01:40.985","Text":"This double integral is really an integral over a 2-dimensional domain,"},{"Start":"01:40.985 ","End":"01:47.045","Text":"call it Delta of x naught and t naught because it depends on x naught and t naught."},{"Start":"01:47.045 ","End":"01:51.230","Text":"This integral will depend on x naught and t naught."},{"Start":"01:51.230 ","End":"01:56.215","Text":"The t and the x will disappear after we\u0027ve done the integral."},{"Start":"01:56.215 ","End":"01:59.233","Text":"If we look at this triangle,"},{"Start":"01:59.233 ","End":"02:00.975","Text":"and it\u0027s marked here,"},{"Start":"02:00.975 ","End":"02:04.610","Text":"we can see that this is the same as this double integral."},{"Start":"02:04.610 ","End":"02:09.440","Text":"What we\u0027re going to do is add layers going from"},{"Start":"02:09.440 ","End":"02:14.520","Text":"the bottom to the top and from the left slope to the right slope."},{"Start":"02:14.520 ","End":"02:16.805","Text":"Here\u0027s a typical slice."},{"Start":"02:16.805 ","End":"02:23.485","Text":"The left part will be x naught minus a times t minus t naught."},{"Start":"02:23.485 ","End":"02:31.240","Text":"This height part will be t naught minus t, because this is t, this is t naught."},{"Start":"02:31.240 ","End":"02:33.110","Text":"You could look at similar triangles."},{"Start":"02:33.110 ","End":"02:40.160","Text":"The height here of this triangle is t naught,"},{"Start":"02:40.160 ","End":"02:43.590","Text":"and the width of this part is x naught,"},{"Start":"02:43.590 ","End":"02:48.130","Text":"and from here to here is a times t naught."},{"Start":"02:48.130 ","End":"02:54.875","Text":"Actually, the run over the rise is a."},{"Start":"02:54.875 ","End":"03:00.785","Text":"Here also this over this or the other way round,"},{"Start":"03:00.785 ","End":"03:02.540","Text":"this will be a times this."},{"Start":"03:02.540 ","End":"03:05.300","Text":"That\u0027s where we get the minus a from."},{"Start":"03:05.300 ","End":"03:06.770","Text":"Anyway, think about it."},{"Start":"03:06.770 ","End":"03:09.080","Text":"Similarly on the other side,"},{"Start":"03:09.080 ","End":"03:13.600","Text":"except that we have a plus here instead of the minus here."},{"Start":"03:13.600 ","End":"03:21.425","Text":"That shows us that this iterated integral from 0 to t and from x naught"},{"Start":"03:21.425 ","End":"03:29.390","Text":"minus at naught minus t to x naught plus at naught minus t. As we go in layers upwards,"},{"Start":"03:29.390 ","End":"03:31.250","Text":"we get the whole triangle."},{"Start":"03:31.250 ","End":"03:33.274","Text":"Now what about the other parts?"},{"Start":"03:33.274 ","End":"03:36.679","Text":"I claim everything just depends on stuff going on in the triangle."},{"Start":"03:36.679 ","End":"03:40.760","Text":"For example, this single integral is"},{"Start":"03:40.760 ","End":"03:45.840","Text":"from x naught minus at naught to x naught plus at naught,"},{"Start":"03:45.840 ","End":"03:51.280","Text":"which is from here to here."},{"Start":"03:51.280 ","End":"03:55.550","Text":"This first term is just the average of 2 quantities,"},{"Start":"03:55.550 ","End":"04:02.006","Text":"the value of f at this point,"},{"Start":"04:02.006 ","End":"04:07.335","Text":"it\u0027s this plus this divided by 2."},{"Start":"04:07.335 ","End":"04:10.649","Text":"It\u0027s average of f at these 2 points."},{"Start":"04:10.649 ","End":"04:16.090","Text":"Altogether, to compute what\u0027s going on at x naught, t naught,"},{"Start":"04:16.090 ","End":"04:20.300","Text":"we used values of the different functions f, g,"},{"Start":"04:20.300 ","End":"04:26.540","Text":"and F, only from this triangle including the sides."},{"Start":"04:26.540 ","End":"04:32.765","Text":"Again, we have this point takes into account this point,"},{"Start":"04:32.765 ","End":"04:38.136","Text":"this edge, and this whole triangle,"},{"Start":"04:38.136 ","End":"04:40.220","Text":"so this, this, and this."},{"Start":"04:40.220 ","End":"04:43.805","Text":"What it means is that the value of x naught,"},{"Start":"04:43.805 ","End":"04:47.300","Text":"t naught only depends on stuff going on in this triangle,"},{"Start":"04:47.300 ","End":"04:49.360","Text":"and not anything outside."},{"Start":"04:49.360 ","End":"04:53.630","Text":"If for example, we were given not the infinite interval,"},{"Start":"04:53.630 ","End":"04:58.715","Text":"but only from 1 point here to another point here, closed interval,"},{"Start":"04:58.715 ","End":"05:02.990","Text":"it wouldn\u0027t matter as long as we had all the values we needed from"},{"Start":"05:02.990 ","End":"05:08.710","Text":"this triangle called the domain of dependence,"},{"Start":"05:08.710 ","End":"05:11.030","Text":"or the characteristic triangle."},{"Start":"05:11.030 ","End":"05:15.930","Text":"Because it turns out that these lines are characteristic lines."},{"Start":"05:15.930 ","End":"05:18.590","Text":"Going back to what we said earlier,"},{"Start":"05:18.590 ","End":"05:23.315","Text":"suppose we only want to find the value of u at a single point,"},{"Start":"05:23.315 ","End":"05:27.230","Text":"and suppose that our problem was given not on an infinite interval,"},{"Start":"05:27.230 ","End":"05:34.490","Text":"but only on a finite interval from 0-L. We also would normally need"},{"Start":"05:34.490 ","End":"05:43.415","Text":"these boundary conditions when x equals naught and x equals L. If we\u0027re lucky,"},{"Start":"05:43.415 ","End":"05:46.865","Text":"then we can find u at x naught,"},{"Start":"05:46.865 ","End":"05:50.540","Text":"t naught using d\u0027Alembert\u0027s formula."},{"Start":"05:50.540 ","End":"05:53.455","Text":"Let\u0027s see what this condition is."},{"Start":"05:53.455 ","End":"05:56.550","Text":"The condition is that the point x naught,"},{"Start":"05:56.550 ","End":"06:02.325","Text":"t naught should be inside this triangle,"},{"Start":"06:02.325 ","End":"06:06.583","Text":"defined by the x axis;"},{"Start":"06:06.583 ","End":"06:10.470","Text":"the line x minus at equals naught and the line x"},{"Start":"06:10.470 ","End":"06:15.435","Text":"plus at equals L. The value of u at x naught,"},{"Start":"06:15.435 ","End":"06:18.965","Text":"t naught depends on the values of the function f,"},{"Start":"06:18.965 ","End":"06:22.925","Text":"g, and F in this triangle here."},{"Start":"06:22.925 ","End":"06:26.500","Text":"This triangle is inside this triangle here."},{"Start":"06:26.500 ","End":"06:32.000","Text":"This larger triangle is all in the region where t is"},{"Start":"06:32.000 ","End":"06:39.860","Text":"non-negative and x is between 0 and L. That is the condition."},{"Start":"06:39.860 ","End":"06:43.388","Text":"Let\u0027s just write it in a more practical form,"},{"Start":"06:43.388 ","End":"06:46.665","Text":"1 way of writing it is that x naught,"},{"Start":"06:46.665 ","End":"06:52.789","Text":"t naught should be in the domain of dependence of this point here."},{"Start":"06:52.789 ","End":"06:58.025","Text":"This happens to be L/2, L/2a."},{"Start":"06:58.025 ","End":"07:00.365","Text":"That doesn\u0027t matter so much."},{"Start":"07:00.365 ","End":"07:06.200","Text":"The better way of writing this condition is the following double inequality"},{"Start":"07:06.200 ","End":"07:12.280","Text":"that x naught should be between at naught and L minus at naught."},{"Start":"07:12.280 ","End":"07:15.840","Text":"This is the condition for which x naught,"},{"Start":"07:15.840 ","End":"07:21.270","Text":"t naught is in the domain of dependence of this point here."},{"Start":"07:21.270 ","End":"07:27.335","Text":"Then we can use the d\u0027Alembert\u0027s formula to figure out u(x naught, t naught)."},{"Start":"07:27.335 ","End":"07:29.120","Text":"By the way, in this case,"},{"Start":"07:29.120 ","End":"07:31.745","Text":"we don\u0027t need the functions A(t)"},{"Start":"07:31.745 ","End":"07:36.395","Text":"and B(t) for the boundaries because we don\u0027t have the boundaries."},{"Start":"07:36.395 ","End":"07:40.010","Text":"This boundary here and the vertical line through here are"},{"Start":"07:40.010 ","End":"07:44.265","Text":"outside this triangle and so they\u0027re not needed."},{"Start":"07:44.265 ","End":"07:46.895","Text":"This is the introduction tutorial."},{"Start":"07:46.895 ","End":"07:51.170","Text":"After this, there\u0027ll be a few solved exercises,"},{"Start":"07:51.170 ","End":"07:55.945","Text":"and that\u0027s where you really understand how to do it."},{"Start":"07:55.945 ","End":"07:58.390","Text":"That concludes this clip."}],"ID":30754},{"Watched":false,"Name":"Exercise 1","Duration":"3m 18s","ChapterTopicVideoID":29131,"CourseChapterTopicPlaylistID":294446,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.840","Text":"In this exercise, we\u0027re given a partial differential equation,"},{"Start":"00:04.840 ","End":"00:06.955","Text":"actually a wave equation."},{"Start":"00:06.955 ","End":"00:13.450","Text":"It\u0027s on a finite interval and our task is not to compute the general solution,"},{"Start":"00:13.450 ","End":"00:18.640","Text":"but just to compute the value of the solution at the given point."},{"Start":"00:18.640 ","End":"00:23.200","Text":"We\u0027ll be using the concept of domain of dependence."},{"Start":"00:23.200 ","End":"00:27.640","Text":"Here is the picture to remind you what we\u0027re talking about."},{"Start":"00:27.640 ","End":"00:33.705","Text":"In our case, a is 1 and L is 2 and the point x_ 0,"},{"Start":"00:33.705 ","End":"00:38.140","Text":"t_ 0 is this point Pi over 2, one quarter."},{"Start":"00:38.140 ","End":"00:43.795","Text":"This line becomes x minus t equals 0 and this line,"},{"Start":"00:43.795 ","End":"00:46.420","Text":"x plus t equals 2."},{"Start":"00:46.420 ","End":"00:50.445","Text":"Our point x_ 0, t_ 0 is this."},{"Start":"00:50.445 ","End":"00:54.380","Text":"By what we learned about domain of dependence,"},{"Start":"00:54.380 ","End":"00:55.770","Text":"if our point x_ 0,"},{"Start":"00:55.770 ","End":"00:59.105","Text":"t_ 0 is inside this triangle, this domain,"},{"Start":"00:59.105 ","End":"01:03.610","Text":"then we can treat the problem as if it was on an infinite interval."},{"Start":"01:03.610 ","End":"01:06.320","Text":"The question is, is this inside or outside?"},{"Start":"01:06.320 ","End":"01:12.425","Text":"The right-hand side of this line is x_ 0 minus t_ 0 is bigger or equal to 0."},{"Start":"01:12.425 ","End":"01:16.235","Text":"Equal to 0 is on the line and to the right of it is bigger than or equal to 0."},{"Start":"01:16.235 ","End":"01:20.890","Text":"Similarly here we want it to be to the left of this line,"},{"Start":"01:20.890 ","End":"01:23.110","Text":"x_ 0 plus t_ 0 less than or equal to 2."},{"Start":"01:23.110 ","End":"01:24.805","Text":"Combining these two,"},{"Start":"01:24.805 ","End":"01:26.710","Text":"we get this inequality,"},{"Start":"01:26.710 ","End":"01:28.365","Text":"x_ 0 is bigger or equal to t_ 0,"},{"Start":"01:28.365 ","End":"01:31.310","Text":"and here it\u0027s less than or equal to 2 minus t_ 0,"},{"Start":"01:31.310 ","End":"01:35.590","Text":"and our point is Pi over 2, one quarter."},{"Start":"01:35.590 ","End":"01:38.900","Text":"To two decimal places Pi over 2 is 1.57,"},{"Start":"01:38.900 ","End":"01:40.870","Text":"a quarter is 0.25,"},{"Start":"01:40.870 ","End":"01:43.310","Text":"2 minus a quarter, 1.75."},{"Start":"01:43.310 ","End":"01:47.540","Text":"It\u0027s obvious that this is between this and this so we\u0027re all right,"},{"Start":"01:47.540 ","End":"01:50.810","Text":"and our point is inside the triangle,"},{"Start":"01:50.810 ","End":"01:53.495","Text":"sometimes called the characteristic triangle."},{"Start":"01:53.495 ","End":"01:55.130","Text":"There\u0027s a shortcut we could take,"},{"Start":"01:55.130 ","End":"01:58.775","Text":"if you remember the formula from the tutorial,"},{"Start":"01:58.775 ","End":"02:01.115","Text":"the condition that the point x_ 0,"},{"Start":"02:01.115 ","End":"02:05.390","Text":"t_ 0 is inside the triangle is the following double inequality."},{"Start":"02:05.390 ","End":"02:07.035","Text":"We could start with this,"},{"Start":"02:07.035 ","End":"02:09.660","Text":"put in a equals 1, L equals 2,"},{"Start":"02:09.660 ","End":"02:12.470","Text":"and then we would get this straight away and"},{"Start":"02:12.470 ","End":"02:15.650","Text":"then we could check if the point is inside or outside."},{"Start":"02:15.650 ","End":"02:19.280","Text":"We have that x_ 0, t_ 0 is in the domain of dependence so that means that we can"},{"Start":"02:19.280 ","End":"02:23.365","Text":"use the d\u0027Alembert\u0027s formula as if it was on an infinite interval."},{"Start":"02:23.365 ","End":"02:25.879","Text":"This is the formula in general."},{"Start":"02:25.879 ","End":"02:29.750","Text":"We have that f(x) equals x and g(x) is cosine(x)."},{"Start":"02:29.750 ","End":"02:33.260","Text":"Go back and look at the initial conditions and a equals 1."},{"Start":"02:33.260 ","End":"02:36.615","Text":"We get the following a is 1,"},{"Start":"02:36.615 ","End":"02:41.480","Text":"0.5 integral from x minus t to x plus t cosine(s)ds."},{"Start":"02:41.480 ","End":"02:45.020","Text":"This is equal to x,"},{"Start":"02:45.020 ","End":"02:49.830","Text":"this whole thing and we can substitute for x,"},{"Start":"02:49.830 ","End":"02:53.025","Text":"t Pi over 2 one quarter."},{"Start":"02:53.025 ","End":"02:54.785","Text":"This is what we get."},{"Start":"02:54.785 ","End":"03:00.200","Text":"Integral of cosine is sine and so we just have to now substitute these values."},{"Start":"03:00.200 ","End":"03:02.045","Text":"Well, we don\u0027t have to actually compute it,"},{"Start":"03:02.045 ","End":"03:06.844","Text":"just put them in like so Pi over 2 plus a quarter,"},{"Start":"03:06.844 ","End":"03:10.145","Text":"Pi over 2 minus a quarter and subtract."},{"Start":"03:10.145 ","End":"03:11.959","Text":"This is the answer."},{"Start":"03:11.959 ","End":"03:19.210","Text":"U(x, t) equals this. End of the clip."}],"ID":30755},{"Watched":false,"Name":"Exercise 2","Duration":"3m 22s","ChapterTopicVideoID":29132,"CourseChapterTopicPlaylistID":294446,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.750","Text":"In this exercise, we\u0027re given the following wave equation."},{"Start":"00:03.750 ","End":"00:06.893","Text":"It\u0027s homogeneous, it\u0027s on a finite interval,"},{"Start":"00:06.893 ","End":"00:08.760","Text":"and these are the initial conditions."},{"Start":"00:08.760 ","End":"00:11.175","Text":"We don\u0027t have to compute the general solution,"},{"Start":"00:11.175 ","End":"00:17.070","Text":"just the value of the solution at the point 1/2, 1/4."},{"Start":"00:17.070 ","End":"00:19.995","Text":"I grayed these out because we don\u0027t actually need them."},{"Start":"00:19.995 ","End":"00:23.670","Text":"The plan is to show that this point, 1/2, 1/4,"},{"Start":"00:23.670 ","End":"00:27.795","Text":"is in the domain of dependence and therefore,"},{"Start":"00:27.795 ","End":"00:30.050","Text":"we can compute u of 1/2,"},{"Start":"00:30.050 ","End":"00:35.840","Text":"1/4 as if it was on an infinite interval by using the d\u0027Alembert formula."},{"Start":"00:35.840 ","End":"00:39.620","Text":"Here\u0027s a picture but we don\u0027t need it to do the calculations."},{"Start":"00:39.620 ","End":"00:43.610","Text":"We want to show that this point is inside this triangle and we actually"},{"Start":"00:43.610 ","End":"00:47.900","Text":"got a formula for it in the tutorial that x_naught,"},{"Start":"00:47.900 ","End":"00:54.185","Text":"t_naught is in the triangle if and only if this inequality is satisfied."},{"Start":"00:54.185 ","End":"00:58.229","Text":"In our case, a=1 and l=1,"},{"Start":"00:58.229 ","End":"01:01.730","Text":"so this double inequality comes out to"},{"Start":"01:01.730 ","End":"01:07.605","Text":"be t_naught less than or equal to x_naught less than or equal to 1 minus t_naught."},{"Start":"01:07.605 ","End":"01:10.020","Text":"x_naught, t_naught is this point here."},{"Start":"01:10.020 ","End":"01:12.135","Text":"Let\u0027s see if it satisfies."},{"Start":"01:12.135 ","End":"01:15.900","Text":"t_naught is 1/4, x_naught is 1/2."},{"Start":"01:15.900 ","End":"01:19.550","Text":"1 minus t_naught is 1 minus 1/4, is 3/4."},{"Start":"01:19.550 ","End":"01:22.010","Text":"Is this double inequality true?"},{"Start":"01:22.010 ","End":"01:24.920","Text":"Clearly, yes, so we\u0027re okay."},{"Start":"01:24.920 ","End":"01:31.280","Text":"We can use the d\u0027Alembert formula to find the value of the solution at that point."},{"Start":"01:31.280 ","End":"01:36.380","Text":"This is the d\u0027Alembert formula in general for the homogeneous case."},{"Start":"01:36.380 ","End":"01:38.870","Text":"There\u0027s also a bit with a F,"},{"Start":"01:38.870 ","End":"01:41.800","Text":"but we don\u0027t have that F in our case."},{"Start":"01:41.800 ","End":"01:44.025","Text":"This is all, we have 2 parts."},{"Start":"01:44.025 ","End":"01:46.995","Text":"F(x) is x(x minus 1)."},{"Start":"01:46.995 ","End":"01:51.525","Text":"F(x) is the condition on u(x_naught),"},{"Start":"01:51.525 ","End":"01:55.950","Text":"and g(x) is the condition on u_t(x_naught )."},{"Start":"01:55.950 ","End":"01:57.840","Text":"One of them is x (x minus 1),"},{"Start":"01:57.840 ","End":"01:59.250","Text":"the other one is 1, and,"},{"Start":"01:59.250 ","End":"02:01.260","Text":"of course, a=1."},{"Start":"02:01.260 ","End":"02:04.410","Text":"This simplifies to,"},{"Start":"02:04.410 ","End":"02:06.615","Text":"every a you put 1 everywhere,"},{"Start":"02:06.615 ","End":"02:08.835","Text":"and for the g(x), we put 1."},{"Start":"02:08.835 ","End":"02:12.320","Text":"Instead of f of this,"},{"Start":"02:12.320 ","End":"02:16.115","Text":"we put this times this minus 1."},{"Start":"02:16.115 ","End":"02:20.330","Text":"We could solve this in general with x and t and then substitute x,"},{"Start":"02:20.330 ","End":"02:22.093","Text":"t is this point,"},{"Start":"02:22.093 ","End":"02:26.645","Text":"but might as well just substitute straight away."},{"Start":"02:26.645 ","End":"02:28.505","Text":"What we have here,"},{"Start":"02:28.505 ","End":"02:30.243","Text":"this time it\u0027s x,"},{"Start":"02:30.243 ","End":"02:32.245","Text":"t not x_naught, t_naught, doesn\u0027t matter."},{"Start":"02:32.245 ","End":"02:36.455","Text":"We\u0027re taking x, t to be the point that we want."},{"Start":"02:36.455 ","End":"02:40.420","Text":"X plus t is 3/4 and x minus t is 1/4,"},{"Start":"02:40.420 ","End":"02:42.500","Text":"those appear a lot."},{"Start":"02:45.020 ","End":"02:47.485","Text":"Here we have 3/4."},{"Start":"02:47.485 ","End":"02:50.110","Text":"3/4 minus 1, 1/4."},{"Start":"02:50.110 ","End":"02:52.145","Text":"1/4 minus 1."},{"Start":"02:52.145 ","End":"02:54.850","Text":"Then 1/2 of this integral."},{"Start":"02:54.850 ","End":"02:58.420","Text":"This integral is just 3/4 minus 1/4,"},{"Start":"02:58.420 ","End":"02:59.715","Text":"which is 1,"},{"Start":"02:59.715 ","End":"03:01.705","Text":"times 1/2, it\u0027s 1/2."},{"Start":"03:01.705 ","End":"03:06.870","Text":"This comes out to be minus 3 over 16."},{"Start":"03:06.870 ","End":"03:09.410","Text":"I\u0027ll show you the computation and we can go through it later."},{"Start":"03:09.410 ","End":"03:11.120","Text":"It\u0027s minus 3 over 16."},{"Start":"03:11.120 ","End":"03:15.380","Text":"So altogether, minus 3 over 16 plus 1/2,"},{"Start":"03:15.380 ","End":"03:18.770","Text":"and that comes out to be 5 over 16."},{"Start":"03:18.770 ","End":"03:22.410","Text":"That\u0027s the answer, and we\u0027re done."}],"ID":30756},{"Watched":false,"Name":"Exercise 3","Duration":"4m 23s","ChapterTopicVideoID":29133,"CourseChapterTopicPlaylistID":294446,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.010","Text":"In this exercise,"},{"Start":"00:02.010 ","End":"00:07.380","Text":"we\u0027re given 2 initial boundary value problems this one,"},{"Start":"00:07.380 ","End":"00:11.325","Text":"and this one both on the domain which is"},{"Start":"00:11.325 ","End":"00:16.500","Text":"a finite interval in x and t as usual is non-negative."},{"Start":"00:16.500 ","End":"00:19.230","Text":"Remind you in physics t usually represents time,"},{"Start":"00:19.230 ","End":"00:21.435","Text":"t equals 0 is the current time."},{"Start":"00:21.435 ","End":"00:23.475","Text":"These 2 are almost the same,"},{"Start":"00:23.475 ","End":"00:29.790","Text":"it\u0027s just that the initial conditions at time 0 are just the other way around."},{"Start":"00:29.790 ","End":"00:32.108","Text":"Other than that it\u0027s all the same,"},{"Start":"00:32.108 ","End":"00:36.930","Text":"and we actually won\u0027t need to use these boundary values."},{"Start":"00:36.930 ","End":"00:40.080","Text":"The question is, which is greater?"},{"Start":"00:40.080 ","End":"00:42.450","Text":"U of 1/2,"},{"Start":"00:42.450 ","End":"00:45.015","Text":"1/2 or v 1/2, 1/2?"},{"Start":"00:45.015 ","End":"00:54.595","Text":"Now the obvious thing to do is to subtract u and v. Define w as u minus v,"},{"Start":"00:54.595 ","End":"00:56.610","Text":"and then the question becomes,"},{"Start":"00:56.610 ","End":"00:58.620","Text":"is w of 1/2,"},{"Start":"00:58.620 ","End":"01:01.310","Text":"1/2 bigger than 0 or less than 0?"},{"Start":"01:01.310 ","End":"01:02.915","Text":"If it\u0027s bigger than 0,"},{"Start":"01:02.915 ","End":"01:06.935","Text":"then the u is bigger than the v, vice versa."},{"Start":"01:06.935 ","End":"01:13.595","Text":"Now let\u0027s see what PDE w satisfies and what initial and boundary conditions."},{"Start":"01:13.595 ","End":"01:17.210","Text":"Well, w_tt is u_tt minus v_tt."},{"Start":"01:17.210 ","End":"01:22.590","Text":"Copy them from here and here and we get this minus this."},{"Start":"01:22.590 ","End":"01:24.720","Text":"This part just cancels,"},{"Start":"01:24.720 ","End":"01:27.510","Text":"so we get u_xx minus v_xx,"},{"Start":"01:27.510 ","End":"01:31.230","Text":"but that\u0027s just w_xx."},{"Start":"01:31.230 ","End":"01:33.375","Text":"Now what about the initial conditions?"},{"Start":"01:33.375 ","End":"01:37.275","Text":"W of x,0 is u of x,0 minus v of x,0,"},{"Start":"01:37.275 ","End":"01:40.318","Text":"which is this minus this,"},{"Start":"01:40.318 ","End":"01:45.030","Text":"and that gives us cosine squared plus sine squared which is 1."},{"Start":"01:45.030 ","End":"01:48.900","Text":"Similarly, w_t of x,0 comes out to"},{"Start":"01:48.900 ","End":"01:55.000","Text":"be minus 1 because we have minus sine squared minus cosine squared."},{"Start":"01:55.000 ","End":"01:58.920","Text":"Now w_x of 0,"},{"Start":"01:58.920 ","End":"02:02.033","Text":"t and w_x of 1,t are both 0,"},{"Start":"02:02.033 ","End":"02:04.590","Text":"but like I said, we won\u0027t be using them."},{"Start":"02:04.590 ","End":"02:07.220","Text":"This is what we have to summarize."},{"Start":"02:07.220 ","End":"02:10.145","Text":"These are the conditions on w, the PDE,"},{"Start":"02:10.145 ","End":"02:12.605","Text":"the domain initial conditions,"},{"Start":"02:12.605 ","End":"02:15.785","Text":"boundary conditions that won\u0027t be used."},{"Start":"02:15.785 ","End":"02:17.930","Text":"This point, x,0,"},{"Start":"02:17.930 ","End":"02:21.680","Text":"y,0 and we have a condition for it to be"},{"Start":"02:21.680 ","End":"02:26.455","Text":"inside the domain of dependence or characteristic triangle,"},{"Start":"02:26.455 ","End":"02:30.855","Text":"then our case we have that a is 1 and L is 1."},{"Start":"02:30.855 ","End":"02:34.985","Text":"The condition of being inside the triangle is this,"},{"Start":"02:34.985 ","End":"02:36.700","Text":"and in our case,"},{"Start":"02:36.700 ","End":"02:38.505","Text":"x,0 is 1/2,"},{"Start":"02:38.505 ","End":"02:39.720","Text":"t,0 is 1/2,"},{"Start":"02:39.720 ","End":"02:42.845","Text":"and 1 minus t,0 is also 1/2,"},{"Start":"02:42.845 ","End":"02:44.870","Text":"and we\u0027re looking at less than or equal to,"},{"Start":"02:44.870 ","End":"02:47.974","Text":"so this holds true or okay."},{"Start":"02:47.974 ","End":"02:49.865","Text":"You want to do it with a picture,"},{"Start":"02:49.865 ","End":"02:52.970","Text":"this line is x plus at equals L,"},{"Start":"02:52.970 ","End":"02:54.830","Text":"but a is 1 and L is 1,"},{"Start":"02:54.830 ","End":"02:59.125","Text":"and this is x minus at equals 0,"},{"Start":"02:59.125 ","End":"03:02.070","Text":"where they meet is 1/2,"},{"Start":"03:02.070 ","End":"03:04.580","Text":"1/2, that\u0027s actually our point."},{"Start":"03:04.580 ","End":"03:07.130","Text":"We are on the corner exactly,"},{"Start":"03:07.130 ","End":"03:11.260","Text":"but it still counts as being inside the domain,"},{"Start":"03:11.260 ","End":"03:14.405","Text":"and that means that we can treat the problem"},{"Start":"03:14.405 ","End":"03:17.983","Text":"as far as this point goes as if it was on an infinite interval,"},{"Start":"03:17.983 ","End":"03:22.850","Text":"and use the d\u0027Alembert formula, which is this."},{"Start":"03:22.850 ","End":"03:26.630","Text":"Sometimes there\u0027s an extra bit involving big F,"},{"Start":"03:26.630 ","End":"03:29.095","Text":"but we didn\u0027t have it in the problem."},{"Start":"03:29.095 ","End":"03:32.250","Text":"We have a homogeneous problem."},{"Start":"03:33.460 ","End":"03:38.930","Text":"Now this will be our f of x and g of x. F of x is 1,"},{"Start":"03:38.930 ","End":"03:40.750","Text":"g of x is minus 1,"},{"Start":"03:40.750 ","End":"03:45.365","Text":"and remember that a is equal to 1 continuing."},{"Start":"03:45.365 ","End":"03:48.690","Text":"This becomes f is 1,"},{"Start":"03:48.830 ","End":"03:55.463","Text":"g of s is minus 1,1 plus 1 over 2 is 1,"},{"Start":"03:55.463 ","End":"03:59.880","Text":"and the integral from 0 to 1,"},{"Start":"03:59.880 ","End":"04:02.565","Text":"this is 1/2 minus 1/2, this 1/2 plus 1/2."},{"Start":"04:02.565 ","End":"04:04.260","Text":"This is just 1,"},{"Start":"04:04.260 ","End":"04:06.315","Text":"so this thing is 1/2."},{"Start":"04:06.315 ","End":"04:10.260","Text":"We\u0027ve got 1 minus 1/2, which is 1/2,"},{"Start":"04:10.260 ","End":"04:13.280","Text":"which is definitely bigger than 0,"},{"Start":"04:13.280 ","End":"04:15.530","Text":"and so that answers our question."},{"Start":"04:15.530 ","End":"04:17.840","Text":"I mean if u minus v is bigger than 0,"},{"Start":"04:17.840 ","End":"04:20.575","Text":"then u is bigger than v at this point,"},{"Start":"04:20.575 ","End":"04:24.400","Text":"and that concludes this exercise."}],"ID":30757},{"Watched":false,"Name":"Exercise 4","Duration":"3m 53s","ChapterTopicVideoID":29135,"CourseChapterTopicPlaylistID":294446,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:04.110","Text":"In this exercise, we\u0027re given the following problem,"},{"Start":"00:04.110 ","End":"00:12.720","Text":"which is the wave equation on a finite interval with initial values and boundary values."},{"Start":"00:12.720 ","End":"00:16.860","Text":"Our task is to compute u(1,1/2),"},{"Start":"00:16.860 ","End":"00:18.150","Text":"not the general solution,"},{"Start":"00:18.150 ","End":"00:23.235","Text":"just a solution at this point and we\u0027re going to use the domain of dependence."},{"Start":"00:23.235 ","End":"00:30.945","Text":"There\u0027s a formula that if our point satisfies this inequality,"},{"Start":"00:30.945 ","End":"00:34.080","Text":"then we can use d\u0027Alembert Formula."},{"Start":"00:34.080 ","End":"00:39.030","Text":"Here, a=1 and L=2,"},{"Start":"00:39.030 ","End":"00:45.985","Text":"this is our inequality and our point is 1,1/2."},{"Start":"00:45.985 ","End":"00:50.460","Text":"If we substitute that t_naught is 1/2,"},{"Start":"00:50.460 ","End":"00:52.650","Text":"x_naught is 1,"},{"Start":"00:52.650 ","End":"00:55.305","Text":"sorry, this should be a 2 here."},{"Start":"00:55.305 ","End":"00:57.860","Text":"We have that 1/2 is less than or equal to 1,"},{"Start":"00:57.860 ","End":"01:01.720","Text":"less than or equal to 1/2 so we\u0027re okay."},{"Start":"01:01.720 ","End":"01:06.845","Text":"That means that we can use the d\u0027Alembert formula as if it was on an infinite interval."},{"Start":"01:06.845 ","End":"01:10.400","Text":"You also have determined this with a picture."},{"Start":"01:10.400 ","End":"01:14.615","Text":"In general, this is what the triangle looks like."},{"Start":"01:14.615 ","End":"01:20.180","Text":"In our case, x plus at=L is x plus t=2,"},{"Start":"01:20.180 ","End":"01:22.925","Text":"and this is x minus t=0."},{"Start":"01:22.925 ","End":"01:26.425","Text":"This point turns out to be the point 1,1."},{"Start":"01:26.425 ","End":"01:34.375","Text":"Our point 1,1/2 is certainly inside and this is d\u0027Alembert formula."},{"Start":"01:34.375 ","End":"01:36.965","Text":"This is function f,"},{"Start":"01:36.965 ","End":"01:38.180","Text":"which is a constant."},{"Start":"01:38.180 ","End":"01:42.500","Text":"This is the function g which is a constant and this bit here is the function"},{"Start":"01:42.500 ","End":"01:48.435","Text":"F as written here and a=1 still."},{"Start":"01:48.435 ","End":"01:52.040","Text":"We can substitute and get, this is 1,"},{"Start":"01:52.040 ","End":"01:57.215","Text":"this is 1, this is 2, and a=1 here."},{"Start":"01:57.215 ","End":"02:03.905","Text":"We could substitute x,t or we could just continue in general and then substitute x,t."},{"Start":"02:03.905 ","End":"02:06.540","Text":"Let\u0027s continue in general,"},{"Start":"02:06.650 ","End":"02:10.995","Text":"what we get is this plus this over 2 is 1,"},{"Start":"02:10.995 ","End":"02:16.215","Text":"2 with the half cancels so we just have ds from x minus t 2x plus t."},{"Start":"02:16.215 ","End":"02:22.595","Text":"Here this integral is just the upper limit minus the lower limit,"},{"Start":"02:22.595 ","End":"02:27.330","Text":"which is twice T minus Tau."},{"Start":"02:28.780 ","End":"02:37.240","Text":"This integral so is the upper minus the lower which is 2t. Here\u0027s the 2t."},{"Start":"02:37.240 ","End":"02:40.055","Text":"Here\u0027s t minus Tau,"},{"Start":"02:40.055 ","End":"02:44.150","Text":"but still times Tau^2 and we still have to do the integral."},{"Start":"02:44.150 ","End":"02:49.730","Text":"This integral multiply I mean open using the distributive law."},{"Start":"02:49.730 ","End":"02:59.670","Text":"We get Tau^2 t minus Tau^3 and again,"},{"Start":"02:59.670 ","End":"03:02.060","Text":"sorry, there should be a 2 here of course,"},{"Start":"03:02.060 ","End":"03:05.945","Text":"because this minus, this is twice t minus Tau."},{"Start":"03:05.945 ","End":"03:08.915","Text":"But then the 2 and the half cancel,"},{"Start":"03:08.915 ","End":"03:11.185","Text":"and this is what we\u0027re left with."},{"Start":"03:11.185 ","End":"03:13.045","Text":"Then you need to do this."},{"Start":"03:13.045 ","End":"03:18.670","Text":"Integral comes out just you can easily check to be this."},{"Start":"03:18.670 ","End":"03:21.585","Text":"Both of these are something t to the fourth,"},{"Start":"03:21.585 ","End":"03:27.250","Text":"1/3 minus 1/4 is 1/12 so this is what we get."},{"Start":"03:27.250 ","End":"03:31.210","Text":"This happens to only depend on t. There\u0027s no x here."},{"Start":"03:31.210 ","End":"03:33.865","Text":"We substitute 1,1/2."},{"Start":"03:33.865 ","End":"03:38.495","Text":"We just have to substitute t=1/2 and this is what we get,"},{"Start":"03:38.495 ","End":"03:40.140","Text":"2 times 1/2 is 1,"},{"Start":"03:40.140 ","End":"03:41.400","Text":"1 plus 1 is 2."},{"Start":"03:41.400 ","End":"03:47.460","Text":"Now here we get 1/12 times 1/16, which is 1/192."},{"Start":"03:47.460 ","End":"03:54.160","Text":"What we get is 2 and 1/192 and that\u0027s the answer, and we\u0027re done."}],"ID":30758},{"Watched":false,"Name":"Exercise 4 (2nd method)","Duration":"2m 27s","ChapterTopicVideoID":29134,"CourseChapterTopicPlaylistID":294446,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.820","Text":"This is the same as the previous exercise,"},{"Start":"00:02.820 ","End":"00:05.325","Text":"we\u0027re just given alternate solution."},{"Start":"00:05.325 ","End":"00:07.410","Text":"Starts out the same."},{"Start":"00:07.410 ","End":"00:15.120","Text":"We check if the 0.15 is in the domain of dependence for this problem."},{"Start":"00:15.120 ","End":"00:21.720","Text":"Just like before, we plug this in the inequality and we find that it\u0027s satisfied."},{"Start":"00:21.720 ","End":"00:24.360","Text":"We could have also done it with a picture,"},{"Start":"00:24.360 ","End":"00:27.854","Text":"and seeing that it\u0027s inside anyway, once it\u0027s inside."},{"Start":"00:27.854 ","End":"00:33.960","Text":"In the previous solution we used the d\u0027Alembert formula."},{"Start":"00:33.960 ","End":"00:41.970","Text":"Here we can use a shortcut if we notice something that in our problem,"},{"Start":"00:41.970 ","End":"00:45.299","Text":"this function, F,"},{"Start":"00:45.299 ","End":"00:47.630","Text":"f, g,"},{"Start":"00:47.630 ","End":"00:50.105","Text":"they all don\u0027t depend on x."},{"Start":"00:50.105 ","End":"00:57.275","Text":"We\u0027ll look for a solution that depends only on t. Say u(x, t) is U(t)."},{"Start":"00:57.275 ","End":"00:59.480","Text":"Then in our domain of dependence,"},{"Start":"00:59.480 ","End":"01:01.880","Text":"we have the equation,"},{"Start":"01:01.880 ","End":"01:09.530","Text":"the wave equation becomes u_ tt is U\u0027\u0027(t) and with respect to x, we get 0."},{"Start":"01:09.530 ","End":"01:12.350","Text":"The derivative of u(t) is just nothing."},{"Start":"01:12.350 ","End":"01:15.460","Text":"It\u0027s a constant as far as x goes plus t squared."},{"Start":"01:15.460 ","End":"01:16.940","Text":"We can integrate twice."},{"Start":"01:16.940 ","End":"01:18.515","Text":"First time we integrate,"},{"Start":"01:18.515 ","End":"01:21.575","Text":"we get this where A is some constant."},{"Start":"01:21.575 ","End":"01:25.220","Text":"The second integration gives us this."},{"Start":"01:25.220 ","End":"01:27.875","Text":"Now we still need to find the 2 constants, A and B."},{"Start":"01:27.875 ","End":"01:32.250","Text":"For that, we can use the initial conditions."},{"Start":"01:32.300 ","End":"01:37.400","Text":"From this condition, we get u(0) is 1,"},{"Start":"01:37.400 ","End":"01:41.120","Text":"which means that we put t is 0 here,"},{"Start":"01:41.120 ","End":"01:44.130","Text":"that B is 1 from u_t(x,"},{"Start":"01:44.130 ","End":"01:45.330","Text":"0) equals 2,"},{"Start":"01:45.330 ","End":"01:50.100","Text":"we get that U\u0027(t) at t=0 is 2"},{"Start":"01:50.100 ","End":"01:56.027","Text":"so A is 2 put A and B in here and we get that u(x,"},{"Start":"01:56.027 ","End":"02:02.645","Text":"t) which is U(t) is 1 plus 2t plus 1/12th t^4."},{"Start":"02:02.645 ","End":"02:06.870","Text":"Now what we have to do is substitute x=1,"},{"Start":"02:06.870 ","End":"02:10.695","Text":"t=1/2, and we get the following."},{"Start":"02:10.695 ","End":"02:14.130","Text":"Just like in the previous version,"},{"Start":"02:14.130 ","End":"02:20.205","Text":"1 plus 1 is 2 and 1/12 times 1/16 is 1/192."},{"Start":"02:20.205 ","End":"02:23.950","Text":"We have 2 and 1/192,"},{"Start":"02:23.950 ","End":"02:27.720","Text":"and that\u0027s the answer and we\u0027re done."}],"ID":30759}],"Thumbnail":null,"ID":294446},{"Name":"Duhamel\u0027s Principle (Wave Eqn)","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Duhamels Principle","Duration":"2m 52s","ChapterTopicVideoID":29130,"CourseChapterTopicPlaylistID":294447,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.560","Text":"In this clip we\u0027ll learn about Duhamel\u0027s principle,"},{"Start":"00:04.560 ","End":"00:10.740","Text":"in particular how to use it to solve certain non-homogeneous wave equations."},{"Start":"00:10.740 ","End":"00:13.070","Text":"This is the general setup,"},{"Start":"00:13.070 ","End":"00:15.225","Text":"this is the wave equation,"},{"Start":"00:15.225 ","End":"00:18.120","Text":"and we have a non-homogeneous part."},{"Start":"00:18.120 ","End":"00:21.335","Text":"These are the 2 initial conditions,"},{"Start":"00:21.335 ","End":"00:26.375","Text":"and they could be non-0 but initially we\u0027ll take them to be 0."},{"Start":"00:26.375 ","End":"00:30.725","Text":"The solution to this problem is given by the following"},{"Start":"00:30.725 ","End":"00:35.330","Text":"integral and p is a function of 3 variables,"},{"Start":"00:35.330 ","End":"00:41.150","Text":"x, t, and Tau which is the solution to the following problem."},{"Start":"00:41.150 ","End":"00:42.940","Text":"The function f(x,"},{"Start":"00:42.940 ","End":"00:50.560","Text":"t) is moved from the PDE to the second initial condition here."},{"Start":"00:50.560 ","End":"00:55.850","Text":"The idea that one can solve a non-homogeneous partial differential equation like"},{"Start":"00:55.850 ","End":"01:01.160","Text":"this non-homogeneous by relating it to a related homogeneous problem,"},{"Start":"01:01.160 ","End":"01:06.031","Text":"this is homogeneous but with non-homogeneous initial condition like this,"},{"Start":"01:06.031 ","End":"01:08.930","Text":"this is known as Duhamel\u0027s principle and it\u0027s named after"},{"Start":"01:08.930 ","End":"01:12.605","Text":"someone called Duhamel, French mathematician."},{"Start":"01:12.605 ","End":"01:16.370","Text":"This problem implies that it\u0027s on the infinite interval,"},{"Start":"01:16.370 ","End":"01:18.095","Text":"because it doesn\u0027t give a limit for x."},{"Start":"01:18.095 ","End":"01:22.850","Text":"But it could be also for a case of a finite interval."},{"Start":"01:22.850 ","End":"01:25.715","Text":"Then we\u0027d have to add boundary conditions."},{"Start":"01:25.715 ","End":"01:28.615","Text":"There could be of type Dirichlet or Von Neumann."},{"Start":"01:28.615 ","End":"01:31.325","Text":"These conditions should be homogeneous,"},{"Start":"01:31.325 ","End":"01:35.899","Text":"meaning equal 0 and if not we\u0027ve learned how to use the correcting function."},{"Start":"01:35.899 ","End":"01:39.800","Text":"We just assume that it\u0027s already been corrected and that these are homogeneous."},{"Start":"01:39.800 ","End":"01:42.940","Text":"Then the boundary conditions for u,"},{"Start":"01:42.940 ","End":"01:48.800","Text":"same as the boundary conditions for p except that there\u0027s an extra variable Tau."},{"Start":"01:48.800 ","End":"01:51.890","Text":"If we have for example u_x(0,"},{"Start":"01:51.890 ","End":"01:55.700","Text":"t) is 0 then the boundary condition on p becomes p_x(0,"},{"Start":"01:55.700 ","End":"01:57.215","Text":"t) with the Tau."},{"Start":"01:57.215 ","End":"02:00.280","Text":"Or for example, the Dirichlet condition u(L,"},{"Start":"02:00.280 ","End":"02:04.640","Text":"t) equals 0, that would be p(Lt) with an extra Tau equals 0."},{"Start":"02:04.640 ","End":"02:08.875","Text":"The conditions are transferred just with an extra comma Tau at the end."},{"Start":"02:08.875 ","End":"02:12.825","Text":"I\u0027ll just mention what we do if these 2 conditions,"},{"Start":"02:12.825 ","End":"02:14.900","Text":"the initial conditions aren\u0027t 0,"},{"Start":"02:14.900 ","End":"02:19.265","Text":"I\u0027ll give you the general idea and one of the solved exercises will do this."},{"Start":"02:19.265 ","End":"02:23.480","Text":"What we do in this case is break it up into 2 sub-problems,"},{"Start":"02:23.480 ","End":"02:30.050","Text":"we split u up into v plus w where v is like this,"},{"Start":"02:30.050 ","End":"02:39.165","Text":"but where these 2 are 0 and w is like this but f is 0."},{"Start":"02:39.165 ","End":"02:42.075","Text":"If we solve these 2 problems,"},{"Start":"02:42.075 ","End":"02:45.290","Text":"then u is just the sum of both solutions."},{"Start":"02:45.290 ","End":"02:48.545","Text":"That\u0027s enough for this introductory tutorial."},{"Start":"02:48.545 ","End":"02:52.560","Text":"You\u0027ll learn more from the solved exercises."}],"ID":30760},{"Watched":false,"Name":"Exercise 1","Duration":"7m 27s","ChapterTopicVideoID":29127,"CourseChapterTopicPlaylistID":294447,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.970","Text":"In this exercise, we\u0027re going to solve the following problem,"},{"Start":"00:03.970 ","End":"00:10.480","Text":"which is a nonhomogeneous wave equation on a finite interval with"},{"Start":"00:10.480 ","End":"00:18.430","Text":"homogeneous initial conditions and homogeneous boundary conditions,"},{"Start":"00:18.430 ","End":"00:21.445","Text":"which are both of the Dirichlet type."},{"Start":"00:21.445 ","End":"00:24.455","Text":"We will be using Duhamel\u0027s principle."},{"Start":"00:24.455 ","End":"00:26.050","Text":"We\u0027ll get that u(x,"},{"Start":"00:26.050 ","End":"00:29.010","Text":"t) is this integral of p,"},{"Start":"00:29.010 ","End":"00:31.325","Text":"and we have to compute p,"},{"Start":"00:31.325 ","End":"00:34.480","Text":"which is the solution to the following problem."},{"Start":"00:34.480 ","End":"00:37.915","Text":"This is the differential equation."},{"Start":"00:37.915 ","End":"00:40.720","Text":"This is one of the initial conditions."},{"Start":"00:40.720 ","End":"00:46.360","Text":"This is the other initial condition from which we take the sin(Pix)."},{"Start":"00:46.360 ","End":"00:48.200","Text":"This here is this here,"},{"Start":"00:48.200 ","End":"00:51.435","Text":"and this is now homogeneous."},{"Start":"00:51.435 ","End":"00:54.965","Text":"Because we have boundary conditions also,"},{"Start":"00:54.965 ","End":"00:57.633","Text":"we have to have boundary conditions here,"},{"Start":"00:57.633 ","End":"00:59.540","Text":"and they\u0027re basically the same,"},{"Start":"00:59.540 ","End":"01:04.445","Text":"except that we have p instead of u and we have a comma Tau."},{"Start":"01:04.445 ","End":"01:07.970","Text":"The way to solve this is to make a substitution."},{"Start":"01:07.970 ","End":"01:11.060","Text":"Just think if Tau is being a parameter is"},{"Start":"01:11.060 ","End":"01:15.800","Text":"fixed so that there really is only 2 variables to"},{"Start":"01:15.800 ","End":"01:23.490","Text":"the function p. Same Tau will keep that T, t minus Tau."},{"Start":"01:23.490 ","End":"01:25.050","Text":"Then we get a function,"},{"Start":"01:25.050 ","End":"01:29.810","Text":"P(x, t) such that this equation holds."},{"Start":"01:29.810 ","End":"01:31.820","Text":"And we have t here,"},{"Start":"01:31.820 ","End":"01:33.590","Text":"we have t minus Tau here."},{"Start":"01:33.590 ","End":"01:38.005","Text":"We can get back from t minus Tau to here by adding Tau."},{"Start":"01:38.005 ","End":"01:40.070","Text":"Tau is just like constant."},{"Start":"01:40.070 ","End":"01:42.485","Text":"P is now function of just 2 variables,"},{"Start":"01:42.485 ","End":"01:46.595","Text":"x and T. We get the following problem."},{"Start":"01:46.595 ","End":"01:50.900","Text":"The derivative with respect to t is the same as the derivative with respect"},{"Start":"01:50.900 ","End":"01:55.270","Text":"to T because the anti-derivative would be 1."},{"Start":"01:55.270 ","End":"02:01.850","Text":"The domain t bigger or equal to Tau is t minus Tau bigger or equal to 0,"},{"Start":"02:01.850 ","End":"02:04.760","Text":"so that\u0027s t bigger or equal to 0."},{"Start":"02:04.760 ","End":"02:11.180","Text":"Here we have p(x) and T is 0 because Tau minus Tau is 0."},{"Start":"02:11.180 ","End":"02:19.200","Text":"Similarly here we get 0 and the same sin(Pix) and this t minus Tau is T. So we have P(0,"},{"Start":"02:19.200 ","End":"02:23.134","Text":"t) equals P(1, T) equals 0."},{"Start":"02:23.134 ","End":"02:26.390","Text":"Now, we learned how to solve this kind of problem."},{"Start":"02:26.390 ","End":"02:30.275","Text":"In general, I\u0027ll show you the table and it will remind you,"},{"Start":"02:30.275 ","End":"02:35.285","Text":"this is the table with different kinds of boundary condition."},{"Start":"02:35.285 ","End":"02:38.975","Text":"I just remind you also what type of problem we\u0027re talking about."},{"Start":"02:38.975 ","End":"02:41.660","Text":"This is the kind of problem that we have,"},{"Start":"02:41.660 ","End":"02:45.320","Text":"the wave equation which this part is homogeneous,"},{"Start":"02:45.320 ","End":"02:47.600","Text":"and the finite interval."},{"Start":"02:47.600 ","End":"02:51.935","Text":"We find ourselves in the Dirichlet type."},{"Start":"02:51.935 ","End":"02:57.365","Text":"So u(x, t) is equal to the following,"},{"Start":"02:57.365 ","End":"02:59.960","Text":"and we just have to find an and bn."},{"Start":"02:59.960 ","End":"03:03.050","Text":"Okay. Back here, this is what we had,"},{"Start":"03:03.050 ","End":"03:07.610","Text":"except that I replaced t with T. Instead of u,"},{"Start":"03:07.610 ","End":"03:10.685","Text":"we have P, the same thing."},{"Start":"03:10.685 ","End":"03:14.945","Text":"In our case, a is 1 and L is 1."},{"Start":"03:14.945 ","End":"03:18.505","Text":"That simplifies this to the following."},{"Start":"03:18.505 ","End":"03:22.285","Text":"Like nPi over L is just nPi,"},{"Start":"03:22.285 ","End":"03:24.965","Text":"and the a also doesn\u0027t appear so it\u0027s nPiT."},{"Start":"03:24.965 ","End":"03:27.575","Text":"If we substitute T=0,"},{"Start":"03:27.575 ","End":"03:29.705","Text":"we get that P(x,"},{"Start":"03:29.705 ","End":"03:32.705","Text":"0) is equal to 0."},{"Start":"03:32.705 ","End":"03:37.235","Text":"So 0 is the following sum where T=0."},{"Start":"03:37.235 ","End":"03:39.680","Text":"Now that makes the sine 0."},{"Start":"03:39.680 ","End":"03:42.005","Text":"So all this part drops off."},{"Start":"03:42.005 ","End":"03:48.765","Text":"We just have the an because the cosine is 1 when T is 0."},{"Start":"03:48.765 ","End":"03:50.355","Text":"This is 0, this is 1."},{"Start":"03:50.355 ","End":"03:53.340","Text":"We just have an and put the an in front of the sine."},{"Start":"03:53.340 ","End":"03:57.260","Text":"By comparison of coefficients when we have a sine or cosine series,"},{"Start":"03:57.260 ","End":"04:00.230","Text":"the coefficients have to agree for each"},{"Start":"04:00.230 ","End":"04:06.435","Text":"sin(nPix) so all the an\u0027s are 0 for n equals 1, 2, 3, etc."},{"Start":"04:06.435 ","End":"04:09.030","Text":"If we put an equals 0 here,"},{"Start":"04:09.030 ","End":"04:10.470","Text":"this part drops out."},{"Start":"04:10.470 ","End":"04:15.630","Text":"This with this gives us the following for P(x, T)."},{"Start":"04:15.630 ","End":"04:17.685","Text":"We still don\u0027t have the bn."},{"Start":"04:17.685 ","End":"04:21.665","Text":"That we\u0027re going to get from the other initial condition that we haven\u0027t used yet,"},{"Start":"04:21.665 ","End":"04:27.410","Text":"we need the derivative of P with respect to T. The sine with x days,"},{"Start":"04:27.410 ","End":"04:34.765","Text":"but the sine with T becomes a cosine and the derivative nPi goes up front."},{"Start":"04:34.765 ","End":"04:37.325","Text":"Now, let T=0."},{"Start":"04:37.325 ","End":"04:39.230","Text":"We know that PT(x,"},{"Start":"04:39.230 ","End":"04:42.840","Text":"0) is sin(Pix) that was given."},{"Start":"04:42.840 ","End":"04:46.725","Text":"Putting T=0 here,"},{"Start":"04:46.725 ","End":"04:50.490","Text":"this cosine becomes 1."},{"Start":"04:50.490 ","End":"04:52.700","Text":"This part just drops off,"},{"Start":"04:52.700 ","End":"04:53.885","Text":"and this is what we have."},{"Start":"04:53.885 ","End":"04:58.145","Text":"Now again, we can compare coefficients of sin(nPix)."},{"Start":"04:58.145 ","End":"05:04.730","Text":"On the left, we only have something for n=1 and all the rest is 0."},{"Start":"05:04.730 ","End":"05:12.840","Text":"So 1=1Pib1 and all the rest 0 for nPibn,"},{"Start":"05:12.840 ","End":"05:15.745","Text":"which means that bn is 0."},{"Start":"05:15.745 ","End":"05:19.190","Text":"Here, means that b1 is 1 over Pi."},{"Start":"05:19.190 ","End":"05:24.160","Text":"In short, bn is 1 over Pi if n is 1 and 0, otherwise."},{"Start":"05:24.160 ","End":"05:26.595","Text":"We can put that into here,"},{"Start":"05:26.595 ","End":"05:30.340","Text":"it\u0027s only going to be one term when n is 1,"},{"Start":"05:35.690 ","End":"05:39.795","Text":"because all the other bn\u0027s are 0."},{"Start":"05:39.795 ","End":"05:41.640","Text":"So we have b1,"},{"Start":"05:41.640 ","End":"05:43.946","Text":"which is 1 over Pi,"},{"Start":"05:43.946 ","End":"05:50.340","Text":"sin(1Pix), sin(1PiT),"},{"Start":"05:50.340 ","End":"05:53.105","Text":"leaving out the 1s this is what we have."},{"Start":"05:53.105 ","End":"06:02.445","Text":"Now we found P and time to go back to p. We just let T=t minus Tau."},{"Start":"06:02.445 ","End":"06:04.245","Text":"That\u0027s what T was."},{"Start":"06:04.245 ","End":"06:07.738","Text":"Up to now, we treated Tau like a parameter or a constant,"},{"Start":"06:07.738 ","End":"06:09.695","Text":"now it\u0027s time to free it up."},{"Start":"06:09.695 ","End":"06:11.855","Text":"So Tau again is a variable."},{"Start":"06:11.855 ","End":"06:15.065","Text":"I don\u0027t need the semicolon which is like to separate it,"},{"Start":"06:15.065 ","End":"06:17.465","Text":"now it\u0027s just a regular variable."},{"Start":"06:17.465 ","End":"06:21.440","Text":"So u(x, t) is the integral of p(x,"},{"Start":"06:21.440 ","End":"06:23.675","Text":"t Tau) d Tau."},{"Start":"06:23.675 ","End":"06:29.675","Text":"Now substitute p equals this and we get the following integral."},{"Start":"06:29.675 ","End":"06:36.635","Text":"It\u0027s d Tau so we can bring the 1 over Pi sin(Pix) in front and get this."},{"Start":"06:36.635 ","End":"06:40.250","Text":"Now we have the integral of sine multiplied out here,"},{"Start":"06:40.250 ","End":"06:43.130","Text":"Pi t minus Pi Tau."},{"Start":"06:43.130 ","End":"06:48.140","Text":"The integral of sine is going to be a minus cosine,"},{"Start":"06:48.140 ","End":"06:52.060","Text":"but the anti-derivative is minus Pi because it\u0027s d Tau."},{"Start":"06:52.060 ","End":"06:54.470","Text":"So here\u0027s the minus cosine,"},{"Start":"06:54.470 ","End":"06:55.760","Text":"here\u0027s the minus Pi."},{"Start":"06:55.760 ","End":"06:57.560","Text":"The minuses will cancel."},{"Start":"06:57.560 ","End":"07:01.295","Text":"The Pi with the Pi will make it 1 over Pi^2."},{"Start":"07:01.295 ","End":"07:04.700","Text":"So we get 1 over Pi^2 sin(Pix)."},{"Start":"07:04.700 ","End":"07:09.245","Text":"This when t equals Tau, or Tau equals t,"},{"Start":"07:09.245 ","End":"07:13.520","Text":"this is 0, cosine 0 is 1,"},{"Start":"07:13.520 ","End":"07:14.765","Text":"so that\u0027s the 1."},{"Start":"07:14.765 ","End":"07:16.730","Text":"When Tau is 0,"},{"Start":"07:16.730 ","End":"07:22.555","Text":"then we get cos(Pit) with a minus because we\u0027re subtracting."},{"Start":"07:22.555 ","End":"07:25.325","Text":"This is the answer for u(x,"},{"Start":"07:25.325 ","End":"07:27.990","Text":"t) and we\u0027re done."}],"ID":30761},{"Watched":false,"Name":"Exercise 2","Duration":"7m 1s","ChapterTopicVideoID":29128,"CourseChapterTopicPlaylistID":294447,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:03.270","Text":"Here we have another one of these problems,"},{"Start":"00:03.270 ","End":"00:09.600","Text":"which is a non-homogeneous wave equation on a finite interval with initial conditions,"},{"Start":"00:09.600 ","End":"00:12.765","Text":"and boundary conditions which are homogeneous,"},{"Start":"00:12.765 ","End":"00:19.290","Text":"that both the Von Neumann type and the PDE is"},{"Start":"00:19.290 ","End":"00:21.990","Text":"the wave equation which is non-homogeneous because of"},{"Start":"00:21.990 ","End":"00:26.954","Text":"this extra term and we\u0027ll use the Duhamel principle."},{"Start":"00:26.954 ","End":"00:31.400","Text":"We get u as the integral of p(x)"},{"Start":"00:31.400 ","End":"00:37.475","Text":"t Tau and P is the solution to the following problem."},{"Start":"00:37.475 ","End":"00:40.280","Text":"P_tt is P_xx."},{"Start":"00:40.280 ","End":"00:43.685","Text":"It\u0027s the same here with u replaced by p,"},{"Start":"00:43.685 ","End":"00:45.245","Text":"and this drops off."},{"Start":"00:45.245 ","End":"00:48.215","Text":"Doesn\u0027t drop off exactly, appears here."},{"Start":"00:48.215 ","End":"00:52.289","Text":"The initial conditions are like this, P(x) Tau."},{"Start":"00:52.289 ","End":"00:58.530","Text":"Tau is 0, and P_t of x Tau,"},{"Start":"00:58.530 ","End":"01:00.945","Text":"Tau is what was here,"},{"Start":"01:00.945 ","End":"01:02.645","Text":"the cosine Pi x."},{"Start":"01:02.645 ","End":"01:06.965","Text":"These boundary conditions become the boundary conditions for"},{"Start":"01:06.965 ","End":"01:14.464","Text":"P. That of u we have P and also we have this extra, Tau, Tau."},{"Start":"01:14.464 ","End":"01:16.895","Text":"Other than that, they\u0027re the same."},{"Start":"01:16.895 ","End":"01:20.465","Text":"This problem is solved with the help of a substitution."},{"Start":"01:20.465 ","End":"01:25.115","Text":"T is T minus Tau and we get a function,"},{"Start":"01:25.115 ","End":"01:27.260","Text":"call it P(x,"},{"Start":"01:27.260 ","End":"01:30.200","Text":"T), such that p(x,"},{"Start":"01:30.200 ","End":"01:35.980","Text":"t ) Tau is P(x, t) minus Tau."},{"Start":"01:35.980 ","End":"01:43.600","Text":"Then we get the following problem for P. We get P_tt equals P_xx,"},{"Start":"01:43.600 ","End":"01:47.870","Text":"because the derivative T is the same as"},{"Start":"01:47.870 ","End":"01:56.165","Text":"the t and p(x) nought is 0 because Tau minus Tau,"},{"Start":"01:56.165 ","End":"02:01.465","Text":"which is T is 0 and here Tau minus Tau is also 0."},{"Start":"02:01.465 ","End":"02:08.225","Text":"T becomes T and the cosine Pi x is here."},{"Start":"02:08.225 ","End":"02:10.970","Text":"What about the boundary conditions?"},{"Start":"02:10.970 ","End":"02:17.340","Text":"Well, t minus Tau is T. Here T minus Tau is"},{"Start":"02:17.340 ","End":"02:26.055","Text":"T. We\u0027re given P_x (0,T) and (1,T) and they\u0027re both 0."},{"Start":"02:26.055 ","End":"02:33.590","Text":"This is the familiar problem and we made a table of the general form of the solution."},{"Start":"02:33.590 ","End":"02:34.790","Text":"Yeah, here it is."},{"Start":"02:34.790 ","End":"02:43.680","Text":"In our case, u of x t equals we have the 2 Von Neumann\u0027s so it\u0027s equal to this,"},{"Start":"02:44.230 ","End":"02:52.337","Text":"which is this just putting T in place of t and p instead of u."},{"Start":"02:52.337 ","End":"02:54.844","Text":"A is 1 and L is 1."},{"Start":"02:54.844 ","End":"02:58.580","Text":"That gives us a little bit simpler form, this."},{"Start":"02:58.580 ","End":"03:01.070","Text":"Now let T equals 0,"},{"Start":"03:01.070 ","End":"03:03.875","Text":"then the sine is 0,"},{"Start":"03:03.875 ","End":"03:05.980","Text":"the cosine is 1."},{"Start":"03:05.980 ","End":"03:07.840","Text":"All we\u0027re left with is a_n,"},{"Start":"03:07.840 ","End":"03:13.010","Text":"which we\u0027ll put in front of the cosine also this is 0 because T is 0."},{"Start":"03:13.010 ","End":"03:20.105","Text":"What we\u0027re left with is a nought over 2 plus the sum of a_n cosine n Pi x."},{"Start":"03:20.105 ","End":"03:25.580","Text":"By comparison of coefficients for a series of cosines,"},{"Start":"03:25.580 ","End":"03:30.230","Text":"we get that all the a_n\u0027s are 0, including a 0."},{"Start":"03:30.230 ","End":"03:32.900","Text":"N equals 0, 1, 2, etc."},{"Start":"03:32.900 ","End":"03:36.725","Text":"If we substitute those in here,"},{"Start":"03:36.725 ","End":"03:41.390","Text":"this drops off and this part drops off."},{"Start":"03:41.390 ","End":"03:46.710","Text":"What we\u0027re left with is this part here, b nought t,"},{"Start":"03:46.710 ","End":"03:51.005","Text":"but also b_n times the cosine of n Pi x"},{"Start":"03:51.005 ","End":"03:55.955","Text":"times sine of n pi T sum n goes from 1 to infinity."},{"Start":"03:55.955 ","End":"03:58.010","Text":"Now we need to find the b_n\u0027s."},{"Start":"03:58.010 ","End":"04:04.040","Text":"We\u0027ll need the derivative of P with respect to T for the other initial condition."},{"Start":"04:04.040 ","End":"04:06.785","Text":"This is a straightforward differentiation."},{"Start":"04:06.785 ","End":"04:08.180","Text":"Nothing special here."},{"Start":"04:08.180 ","End":"04:10.094","Text":"Now let T equal 0."},{"Start":"04:10.094 ","End":"04:12.455","Text":"Our initial condition,"},{"Start":"04:12.455 ","End":"04:18.825","Text":"P_T of x 0 is cosine Pi x that was given."},{"Start":"04:18.825 ","End":"04:27.475","Text":"Here, putting T equals 0 just makes all these cosine n Pi t equal 1."},{"Start":"04:27.475 ","End":"04:29.410","Text":"These drop off."},{"Start":"04:29.410 ","End":"04:34.640","Text":"This is what we have now and we need to compare coefficients."},{"Start":"04:34.640 ","End":"04:42.265","Text":"The only n that will match something here is n equals 1 then we\u0027ll get cosine of 1 Pi x."},{"Start":"04:42.265 ","End":"04:44.445","Text":"For n equals 1,"},{"Start":"04:44.445 ","End":"04:48.780","Text":"we get that 1 Pi b_1 is 1,"},{"Start":"04:48.780 ","End":"04:51.535","Text":"and all the others will be 0."},{"Start":"04:51.535 ","End":"04:53.450","Text":"We can write it like this,"},{"Start":"04:53.450 ","End":"04:59.560","Text":"that b_n is 1 over Pi if n is 1 and 0 otherwise."},{"Start":"04:59.560 ","End":"05:05.915","Text":"If we substitute these b_n\u0027s in this equation here,"},{"Start":"05:05.915 ","End":"05:10.655","Text":"everything drops off except for the case where n equals 1,"},{"Start":"05:10.655 ","End":"05:14.616","Text":"in which case b_n is 1 over Pi."},{"Start":"05:14.616 ","End":"05:21.885","Text":"What we get is 1 over Pi cosine 1 Pi x sine 1 Pi T equals 1. You don\u0027t need the ones."},{"Start":"05:21.885 ","End":"05:29.085","Text":"We found P from P will find p and from p will find u."},{"Start":"05:29.085 ","End":"05:37.110","Text":"Little p is what we get if we take P(x) and t minus Tau."},{"Start":"05:37.850 ","End":"05:44.150","Text":"For the T, we have T minus Tau and at this point we can"},{"Start":"05:44.150 ","End":"05:46.700","Text":"drop the requirement that Tau is fixed like a"},{"Start":"05:46.700 ","End":"05:49.880","Text":"constant so we\u0027ll make it a full-fledged variable again."},{"Start":"05:49.880 ","End":"05:51.800","Text":"Now we have this integral u(x,"},{"Start":"05:51.800 ","End":"05:54.530","Text":"t) equals the integral d Tau."},{"Start":"05:54.530 ","End":"06:00.470","Text":"That\u0027s why we need Tau to vary now and this is equal to p(x,"},{"Start":"06:00.470 ","End":"06:03.435","Text":"t) Tau from here is this."},{"Start":"06:03.435 ","End":"06:07.670","Text":"Bring the 1 over Pi cosine Pi x in front of the integral because it doesn\u0027t"},{"Start":"06:07.670 ","End":"06:12.260","Text":"contain Tau and we have this integral and just also expand the brackets here."},{"Start":"06:12.260 ","End":"06:14.855","Text":"Pi t minus Pi Tau."},{"Start":"06:14.855 ","End":"06:18.850","Text":"The integral of sine is minus cosine."},{"Start":"06:18.850 ","End":"06:20.690","Text":"We also have to divide by"},{"Start":"06:20.690 ","End":"06:26.360","Text":"the anti-derivative minus Pi so we get a minus cosine and we get divided by minus Pi."},{"Start":"06:26.360 ","End":"06:27.980","Text":"The minuses will cancel,"},{"Start":"06:27.980 ","End":"06:31.160","Text":"the Pi will go with this to make Pi squared."},{"Start":"06:31.160 ","End":"06:35.450","Text":"We just have this part to evaluate between Tau equals 0,"},{"Start":"06:35.450 ","End":"06:38.765","Text":"Tau equals t, and then subtract."},{"Start":"06:38.765 ","End":"06:46.655","Text":"What we get is the 1 over Pi squared cosine Pi x and then cosine of Pi t minus Pi Tau."},{"Start":"06:46.655 ","End":"06:50.360","Text":"When Tau is t gives us Pi t minus Pi t is 0."},{"Start":"06:50.360 ","End":"06:52.640","Text":"Cosine is 1."},{"Start":"06:52.640 ","End":"06:59.960","Text":"When Tau is 0, we just get cosine pi t. This is the answer for u (x t),"},{"Start":"06:59.960 ","End":"07:02.160","Text":"and we are done."}],"ID":30762},{"Watched":false,"Name":"Exercise 3","Duration":"4m 46s","ChapterTopicVideoID":29129,"CourseChapterTopicPlaylistID":294447,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.560 ","End":"00:06.855","Text":"In this exercise, we\u0027re going to prove the non-homogeneous case of d\u0027Alembert formula."},{"Start":"00:06.855 ","End":"00:12.134","Text":"We actually proved the homogeneous case the previous section,"},{"Start":"00:12.134 ","End":"00:14.340","Text":"but we didn\u0027t prove the non-homogeneous case."},{"Start":"00:14.340 ","End":"00:19.125","Text":"This will do with the help of Duhamel\u0027s principle,"},{"Start":"00:19.125 ","End":"00:26.055","Text":"together with the solution for the homogeneous case with d\u0027Alembert formula."},{"Start":"00:26.055 ","End":"00:30.510","Text":"We\u0027re going to show that the solution to this problem with"},{"Start":"00:30.510 ","End":"00:35.774","Text":"the F here making it non-homogeneous is the following d\u0027Alembert formula,"},{"Start":"00:35.774 ","End":"00:39.930","Text":"but we can only use d\u0027Alembert formula up to here,"},{"Start":"00:39.930 ","End":"00:42.375","Text":"in the case where f equals 0."},{"Start":"00:42.375 ","End":"00:43.620","Text":"To complete it,"},{"Start":"00:43.620 ","End":"00:47.375","Text":"we\u0027ll use the Duhamel\u0027s principle."},{"Start":"00:47.375 ","End":"00:51.170","Text":"The plan which was mentioned in the introduction,"},{"Start":"00:51.170 ","End":"00:55.400","Text":"is to split this up into two sub-problems."},{"Start":"00:55.400 ","End":"00:57.065","Text":"In the first case,"},{"Start":"00:57.065 ","End":"00:59.810","Text":"we\u0027ll have a problem which is the same as this,"},{"Start":"00:59.810 ","End":"01:02.960","Text":"but with 0 in the initial conditions."},{"Start":"01:02.960 ","End":"01:04.625","Text":"In the second case,"},{"Start":"01:04.625 ","End":"01:06.740","Text":"we\u0027ll leave the initial conditions as they are,"},{"Start":"01:06.740 ","End":"01:11.725","Text":"but we\u0027ll drop the non-homogeneous part of the PDE."},{"Start":"01:11.725 ","End":"01:20.555","Text":"Turns out if we solve these two then v plus w will be a solution to our u."},{"Start":"01:20.555 ","End":"01:23.810","Text":"It\u0027s fairly easy to show that this is the case with just"},{"Start":"01:23.810 ","End":"01:28.070","Text":"using the linearity of differentiation."},{"Start":"01:28.070 ","End":"01:29.885","Text":"Anyway, we can accept that."},{"Start":"01:29.885 ","End":"01:31.715","Text":"So let\u0027s do them."},{"Start":"01:31.715 ","End":"01:34.400","Text":"Let\u0027s start with the v,"},{"Start":"01:34.400 ","End":"01:37.340","Text":"which we\u0027ll do with Duhamel\u0027s principle,"},{"Start":"01:37.340 ","End":"01:42.560","Text":"and write it as follows: P satisfies"},{"Start":"01:42.560 ","End":"01:50.190","Text":"the conditions as in the theory where we take f from here and put it here."},{"Start":"01:50.540 ","End":"01:58.110","Text":"Then we do our usual trick of defining T to equal t minus t"},{"Start":"01:58.110 ","End":"02:05.520","Text":"and fix Tau so that we get p(x,"},{"Start":"02:05.520 ","End":"02:10.730","Text":"t), is really a function of two variables and a fixed parameter Tau,"},{"Start":"02:10.730 ","End":"02:12.800","Text":"which later will vary."},{"Start":"02:12.800 ","End":"02:14.740","Text":"We define P(x,"},{"Start":"02:14.740 ","End":"02:20.630","Text":"t minus Tau) as p(x, t),"},{"Start":"02:20.630 ","End":"02:27.890","Text":"and we get the following problem for P. We\u0027ve seen this thing in earlier exercises,"},{"Start":"02:27.890 ","End":"02:29.760","Text":"so I won\u0027t dwell on it."},{"Start":"02:29.760 ","End":"02:33.820","Text":"We can think of this function as f(x)."},{"Start":"02:33.820 ","End":"02:36.155","Text":"This function of g(x),"},{"Start":"02:36.155 ","End":"02:38.390","Text":"Tau is just a parameter,"},{"Start":"02:38.390 ","End":"02:40.445","Text":"so it really is just a function of x."},{"Start":"02:40.445 ","End":"02:44.300","Text":"So f(x), g(x) means that we can apply"},{"Start":"02:44.300 ","End":"02:49.855","Text":"d\u0027Alembert formula for the homogeneous case as follows."},{"Start":"02:49.855 ","End":"02:59.645","Text":"In our case, f is 0 and g is f with this extra parameter Tau tagging along."},{"Start":"02:59.645 ","End":"03:03.850","Text":"Since f is 0, all this part is 0."},{"Start":"03:03.850 ","End":"03:11.010","Text":"Here we just replace g(s) by F(s, Tau)."},{"Start":"03:11.010 ","End":"03:15.435","Text":"Now we go from P to p,"},{"Start":"03:15.435 ","End":"03:21.285","Text":"by replacing T=t minus Tau."},{"Start":"03:21.285 ","End":"03:24.660","Text":"That\u0027s here and here."},{"Start":"03:24.660 ","End":"03:29.100","Text":"Then v(x, t) is given by the formula, this integral,"},{"Start":"03:29.100 ","End":"03:32.230","Text":"but Tau is now once again a variable,"},{"Start":"03:32.230 ","End":"03:35.230","Text":"because we want to do the integral with Tau."},{"Start":"03:35.230 ","End":"03:37.610","Text":"This comes out to be the integral."},{"Start":"03:37.610 ","End":"03:41.030","Text":"Replace p by this here."},{"Start":"03:41.030 ","End":"03:42.890","Text":"We organize it a bit."},{"Start":"03:42.890 ","End":"03:47.260","Text":"The 1/2a in front and then it becomes a double integral,"},{"Start":"03:47.260 ","End":"03:51.680","Text":"and that\u0027s the solution for v. We still have to find w and"},{"Start":"03:51.680 ","End":"03:56.119","Text":"then get u with v plus w. This is the problem"},{"Start":"03:56.119 ","End":"04:06.095","Text":"for w. We can solve this using the d\u0027Alembert formula for homogeneous wave equation."},{"Start":"04:06.095 ","End":"04:09.470","Text":"Here there\u0027s nothing else besides the a^2u_xx,"},{"Start":"04:09.470 ","End":"04:12.120","Text":"there\u0027s no plus F here."},{"Start":"04:12.120 ","End":"04:15.665","Text":"We\u0027ve got that w is the following,"},{"Start":"04:15.665 ","End":"04:19.235","Text":"applying the shorter d\u0027Alembert formula,"},{"Start":"04:19.235 ","End":"04:24.200","Text":"and you can already see that if we add this to this will get the full d\u0027Alembert formula."},{"Start":"04:24.200 ","End":"04:28.010","Text":"So u is w plus v. I\u0027ll write it the other way around."},{"Start":"04:28.010 ","End":"04:30.095","Text":"I want to take this first and then this,"},{"Start":"04:30.095 ","End":"04:34.525","Text":"and then we get this plus this, plus this."},{"Start":"04:34.525 ","End":"04:37.490","Text":"This is our regular formula for"},{"Start":"04:37.490 ","End":"04:42.770","Text":"the non-homogeneous wave equation, the d\u0027Alembert formula."},{"Start":"04:42.770 ","End":"04:44.510","Text":"That\u0027s what we had to show,"},{"Start":"04:44.510 ","End":"04:46.710","Text":"and we are done."}],"ID":30763}],"Thumbnail":null,"ID":294447}]

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1.1

1