Classifying Linear 2nd Order PDEs
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- Tutorial - Classifying Second Order Linear PDEs
- Differentiation Formulas
- Example 1 (hyperbolic)
- Example 2 (parabolic)
- Example 3 (elliptic)
- Exercise 1
- Exercise 2ab
- Exercise 2c
- Exercise 3
- Exercise 4a
- Exercise 4b
- Exercise 5
- Exercise 6
- Exercise 7abc
- Exercise 7d
- Exercise 8ab
- Exercise 8cd
- Exercise 9
- Exercise 10

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[{"Name":"Classifying Linear 2nd Order PDEs","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Classifying Second Order Linear PDEs","Duration":"4m 14s","ChapterTopicVideoID":29120,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29120.jpeg","UploadDate":"2022-06-06T07:38:40.8370000","DurationForVideoObject":"PT4M14S","Description":null,"MetaTitle":"Tutorial - Classifying Second Order Linear PDEs: Video + Workbook | Proprep","MetaDescription":"Classifying Linear 2nd Order PDEs - Classifying Linear 2nd Order PDEs. 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/classifying-linear-2nd-order-pdes/classifying-linear-2nd-order-pdes/vid30690","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.610","Text":"Now we come to a new topic,"},{"Start":"00:02.610 ","End":"00:05.789","Text":"second-order partial differential equations."},{"Start":"00:05.789 ","End":"00:09.645","Text":"We\u0027ll start with linear ones and how to classify them."},{"Start":"00:09.645 ","End":"00:13.875","Text":"First, some examples are 3 famous examples."},{"Start":"00:13.875 ","End":"00:17.210","Text":"Thus the wave equation, which is this,"},{"Start":"00:17.210 ","End":"00:18.845","Text":"the heat equation,"},{"Start":"00:18.845 ","End":"00:21.635","Text":"and the Laplace equation."},{"Start":"00:21.635 ","End":"00:24.695","Text":"Here, u is a function of t and x,"},{"Start":"00:24.695 ","End":"00:28.085","Text":"t and x, and here u as a function of x and y."},{"Start":"00:28.085 ","End":"00:36.545","Text":"The general form of a second-order linear partial differential equation is the following."},{"Start":"00:36.545 ","End":"00:39.680","Text":"Where these coefficients are not constants."},{"Start":"00:39.680 ","End":"00:41.980","Text":"They are functions of x and y."},{"Start":"00:41.980 ","End":"00:43.310","Text":"They could be constants,"},{"Start":"00:43.310 ","End":"00:45.350","Text":"but in general, they\u0027re functions of x and y."},{"Start":"00:45.350 ","End":"00:53.570","Text":"The discriminant of this is defined as a_12^2 minus a_11a_22."},{"Start":"00:53.570 ","End":"00:56.185","Text":"The notation for it is Delta."},{"Start":"00:56.185 ","End":"01:00.350","Text":"Similar to the discriminant in quadratic equations."},{"Start":"01:00.350 ","End":"01:06.725","Text":"Also note that here there\u0027s a 2 so that a_12 is half the coefficient of u(x, y)."},{"Start":"01:06.725 ","End":"01:10.460","Text":"The first step in solving such a PDE is to solve"},{"Start":"01:10.460 ","End":"01:16.040","Text":"an ordinary differential equation called the characteristic ODE."},{"Start":"01:16.040 ","End":"01:20.930","Text":"It\u0027s the derivative of y with respect to x is the following expression,"},{"Start":"01:20.930 ","End":"01:22.400","Text":"which if you simplify,"},{"Start":"01:22.400 ","End":"01:28.610","Text":"it comes out to be a_12 plus or minus the square root of Delta over a_11."},{"Start":"01:28.610 ","End":"01:33.305","Text":"According to the sign of Delta, the discriminant,"},{"Start":"01:33.305 ","End":"01:38.365","Text":"we classify the equation to be the hyperbolic with Delta is positive."},{"Start":"01:38.365 ","End":"01:40.740","Text":"Then we get 2 real solutions."},{"Start":"01:40.740 ","End":"01:42.195","Text":"Could be parabolic."},{"Start":"01:42.195 ","End":"01:44.670","Text":"But Delta is 0, and we just get"},{"Start":"01:44.670 ","End":"01:51.205","Text":"1 real solution because this is 0 or elliptic when Delta is negative."},{"Start":"01:51.205 ","End":"01:56.120","Text":"Then there are 2 complex conjugate solutions to the ODE."},{"Start":"01:56.120 ","End":"02:03.250","Text":"The solution to this differential equation gives us characteristic curves."},{"Start":"02:03.250 ","End":"02:06.280","Text":"Call them Phi of x, y equals c_1."},{"Start":"02:06.280 ","End":"02:09.425","Text":"If you put everything except the constant onto the left-hand side."},{"Start":"02:09.425 ","End":"02:12.590","Text":"The other one we\u0027ll call Psi of x, y equals c_2."},{"Start":"02:12.590 ","End":"02:17.330","Text":"The case of parabolic there is only 1 because the plus or minus 0."},{"Start":"02:17.330 ","End":"02:19.175","Text":"In the elliptic case,"},{"Start":"02:19.175 ","End":"02:21.850","Text":"Phi and Psi are just complex conjugates."},{"Start":"02:21.850 ","End":"02:26.360","Text":"Phi and Phi bar and these are the 2 characteristic curves,"},{"Start":"02:26.360 ","End":"02:29.465","Text":"families of curves really because according to c_1 and c_2,"},{"Start":"02:29.465 ","End":"02:31.385","Text":"we get a family of curves."},{"Start":"02:31.385 ","End":"02:34.360","Text":"These equations with Phi and Psi,"},{"Start":"02:34.360 ","End":"02:38.185","Text":"are what enabled us to make a change of variables."},{"Start":"02:38.185 ","End":"02:42.320","Text":"The plan is to get the original equation into what is called canonical form,"},{"Start":"02:42.320 ","End":"02:43.430","Text":"which is much simpler."},{"Start":"02:43.430 ","End":"02:47.735","Text":"The way to do it is with change of variables as follows."},{"Start":"02:47.735 ","End":"02:52.445","Text":"In this case, we just let Psi equal Phi of x,"},{"Start":"02:52.445 ","End":"02:55.835","Text":"y, and Eta equals Psi of x, y."},{"Start":"02:55.835 ","End":"02:59.510","Text":"In this case, Psi is Phi and Eta,"},{"Start":"02:59.510 ","End":"03:01.645","Text":"you can choose either x or y."},{"Start":"03:01.645 ","End":"03:04.540","Text":"Maybe it\u0027s simpler to solve it with one of the choices,"},{"Start":"03:04.540 ","End":"03:05.965","Text":"but it doesn\u0027t really matter."},{"Start":"03:05.965 ","End":"03:14.050","Text":"In this case, Psi is the real part of Phi and Eta is the imaginary part of Phi."},{"Start":"03:14.050 ","End":"03:17.515","Text":"When we do this, we will get a canonical form,"},{"Start":"03:17.515 ","End":"03:21.685","Text":"which in the hyperbolic case will look as follows."},{"Start":"03:21.685 ","End":"03:28.955","Text":"The canonical form is a PDE in w as a function of Psi and Eta."},{"Start":"03:28.955 ","End":"03:33.715","Text":"It\u0027s much simpler than the original PDE for the hyperbolic."},{"Start":"03:33.715 ","End":"03:35.560","Text":"This is the form for the parabolic."},{"Start":"03:35.560 ","End":"03:37.600","Text":"This is the form for the elliptic."},{"Start":"03:37.600 ","End":"03:38.770","Text":"This is the form,"},{"Start":"03:38.770 ","End":"03:46.605","Text":"I should say that dot dot dot means that it\u0027s terms of order less than 2,"},{"Start":"03:46.605 ","End":"03:48.390","Text":"like w Psi,"},{"Start":"03:48.390 ","End":"03:52.369","Text":"w Eta and w itself."},{"Start":"03:52.369 ","End":"03:55.910","Text":"Now all of these won\u0027t make much sense until you\u0027ve seen the examples."},{"Start":"03:55.910 ","End":"03:58.910","Text":"I suggest, don\u0027t worry about whether you"},{"Start":"03:58.910 ","End":"04:02.225","Text":"understood this on the examples will explain it all."},{"Start":"04:02.225 ","End":"04:04.310","Text":"First, we\u0027ll have one each of hyperbolic,"},{"Start":"04:04.310 ","End":"04:05.540","Text":"parabolic, and elliptic."},{"Start":"04:05.540 ","End":"04:07.940","Text":"Then there\u0027ll be some other exercises."},{"Start":"04:07.940 ","End":"04:12.140","Text":"You can return to this later and then it will make sense."},{"Start":"04:12.140 ","End":"04:15.150","Text":"Well, that\u0027s it for this clip."}],"ID":30690},{"Watched":false,"Name":"Differentiation Formulas","Duration":"2m 42s","ChapterTopicVideoID":29121,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"In this clip, we\u0027ll develop some formulas,"},{"Start":"00:03.210 ","End":"00:10.035","Text":"which we will need in pretty much all of the following clips for solving PDEs,"},{"Start":"00:10.035 ","End":"00:14.505","Text":"which are linear and of the second order."},{"Start":"00:14.505 ","End":"00:18.810","Text":"In the introduction, we talked about a substitution where"},{"Start":"00:18.810 ","End":"00:24.915","Text":"we have Psi as a function of x and y and Eta."},{"Start":"00:24.915 ","End":"00:29.490","Text":"We re-write u so that its function w of Psi and Eta."},{"Start":"00:29.490 ","End":"00:32.160","Text":"We will need some formulas."},{"Start":"00:32.160 ","End":"00:35.295","Text":"We want to express the derivatives of u,"},{"Start":"00:35.295 ","End":"00:38.280","Text":"at least the ones of order 0, 1, and 2,"},{"Start":"00:38.280 ","End":"00:48.690","Text":"in terms of the derivatives of w. Let\u0027s develop these."},{"Start":"00:48.690 ","End":"00:52.212","Text":"Start with u(x, y) equals w(Psi(x,"},{"Start":"00:52.212 ","End":"00:55.050","Text":"y) and Eta(x,"},{"Start":"00:55.050 ","End":"00:58.250","Text":"y) Ultimately, it\u0027s a function of x and y,"},{"Start":"00:58.250 ","End":"01:01.050","Text":"but through Psi and Eta."},{"Start":"01:01.490 ","End":"01:06.155","Text":"If we differentiate with respect to x and use the chain rule,"},{"Start":"01:06.155 ","End":"01:09.875","Text":"this will be w with respect to Psi and Psi with respect to x."},{"Start":"01:09.875 ","End":"01:11.150","Text":"The same thing with Eta,"},{"Start":"01:11.150 ","End":"01:12.410","Text":"w with respect to Eta,"},{"Start":"01:12.410 ","End":"01:14.360","Text":"Eta with respect to x."},{"Start":"01:14.360 ","End":"01:17.315","Text":"That\u0027s our first formula."},{"Start":"01:17.315 ","End":"01:21.755","Text":"If we differentiate with respect to x, again,"},{"Start":"01:21.755 ","End":"01:26.585","Text":"you get the derivative of this with respect to x,"},{"Start":"01:26.585 ","End":"01:28.250","Text":"which is all of this."},{"Start":"01:28.250 ","End":"01:30.695","Text":"Psi with respect to x,"},{"Start":"01:30.695 ","End":"01:34.775","Text":"plus w with respect to Psi, Psi xx."},{"Start":"01:34.775 ","End":"01:42.660","Text":"Similarly for the other one with respect to where we go via Eta."},{"Start":"01:42.740 ","End":"01:46.740","Text":"We collect terms and arrange it."},{"Start":"01:46.740 ","End":"01:48.300","Text":"We get u_xx is this."},{"Start":"01:48.300 ","End":"01:50.490","Text":"That\u0027s our second formula."},{"Start":"01:50.490 ","End":"01:56.790","Text":"If we take this one and differentiate with respect to y,"},{"Start":"01:56.790 ","End":"01:59.285","Text":"I\u0027ll spare you all the details, we get this."},{"Start":"01:59.285 ","End":"02:03.020","Text":"Collect terms and simplifying,"},{"Start":"02:03.020 ","End":"02:10.880","Text":"we get u respect to x and y. I just copy this formula again here, you want to continue."},{"Start":"02:10.880 ","End":"02:13.960","Text":"There\u0027s 2 missing. There\u0027s u_y and u_yy."},{"Start":"02:13.960 ","End":"02:18.165","Text":"The u_y is the w_Psi,"},{"Start":"02:18.165 ","End":"02:22.135","Text":"the Psi_y and similarly with Eta."},{"Start":"02:22.135 ","End":"02:26.030","Text":"Then differentiate again with respect to y."},{"Start":"02:26.030 ","End":"02:31.645","Text":"You can study this on your own and simplify and we get u_yy."},{"Start":"02:31.645 ","End":"02:35.450","Text":"The useful ones are the ones colored in blue,"},{"Start":"02:35.450 ","End":"02:37.550","Text":"1, 2, 3,"},{"Start":"02:37.550 ","End":"02:40.125","Text":"4, 5 formulas."},{"Start":"02:40.125 ","End":"02:43.100","Text":"That\u0027s it for this clip."}],"ID":30691},{"Watched":false,"Name":"Example 1 (hyperbolic)","Duration":"3m 20s","ChapterTopicVideoID":29122,"CourseChapterTopicPlaylistID":294424,"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":"Here\u0027s an example of a linear partial differential equation of"},{"Start":"00:05.190 ","End":"00:11.505","Text":"the second-order and we\u0027re going to bring it to canonical form but first,"},{"Start":"00:11.505 ","End":"00:15.210","Text":"to classify it this coefficient is a_11,"},{"Start":"00:15.210 ","End":"00:17.790","Text":"this is twice a_12,"},{"Start":"00:17.790 ","End":"00:19.875","Text":"so a_12 is 5,"},{"Start":"00:19.875 ","End":"00:21.900","Text":"and this is a_22."},{"Start":"00:21.900 ","End":"00:27.750","Text":"Then we compute Delta which is (a12)^2- a_11a_22 and that"},{"Start":"00:27.750 ","End":"00:33.780","Text":"gives us 25 minus 16 which is 9 and it\u0027s positive."},{"Start":"00:33.780 ","End":"00:36.405","Text":"Positive means it\u0027s hyperbolic."},{"Start":"00:36.405 ","End":"00:40.160","Text":"I\u0027ll copy the row from the table for the hyperbolic."},{"Start":"00:40.160 ","End":"00:43.700","Text":"When the discriminant is positive it\u0027s hyperbolic."},{"Start":"00:43.700 ","End":"00:47.285","Text":"Next, we\u0027re going to find the characteristic curves and we do this with"},{"Start":"00:47.285 ","End":"00:49.790","Text":"the differential equation dy by"},{"Start":"00:49.790 ","End":"00:54.640","Text":"dx equals a_12 plus or minus square root of delta over a_11."},{"Start":"00:54.640 ","End":"00:56.730","Text":"In our case Delta is 9,"},{"Start":"00:56.730 ","End":"00:58.545","Text":"square root of Delta is 3,"},{"Start":"00:58.545 ","End":"01:03.690","Text":"5 plus or minus 3 over 2 gives us 2 solutions, 4 and 1."},{"Start":"01:03.690 ","End":"01:12.450","Text":"So if it\u0027s 4 then we get y\u0027 minus 4 is 0 and if it\u0027s 1 we get y\u0027 minus 1 is 0."},{"Start":"01:12.450 ","End":"01:14.780","Text":"In the first case, we integrate it,"},{"Start":"01:14.780 ","End":"01:18.455","Text":"we get y minus 4x is a constant."},{"Start":"01:18.455 ","End":"01:22.505","Text":"Here y minus x is a different constant."},{"Start":"01:22.505 ","End":"01:27.900","Text":"Then we make the substitution according to this and"},{"Start":"01:27.900 ","End":"01:33.650","Text":"we get that Psi is y minus 4x and Eta is y minus x."},{"Start":"01:33.650 ","End":"01:37.655","Text":"We\u0027ll need the partial derivatives of these up to order 2."},{"Start":"01:37.655 ","End":"01:42.614","Text":"So if we differentiate with respect to x we get 0 minus 4,"},{"Start":"01:42.614 ","End":"01:45.845","Text":"with respect to y we get just the 1, and so on."},{"Start":"01:45.845 ","End":"01:49.880","Text":"These are the partial derivatives of order 1 and 2."},{"Start":"01:49.880 ","End":"01:56.075","Text":"The idea is to find w of Psi and Eta and later on substitute Psi and Eta."},{"Start":"01:56.075 ","End":"01:59.840","Text":"We\u0027ll just get up to the canonical form which is the P, D,"},{"Start":"01:59.840 ","End":"02:01.995","Text":"and W. Remember,"},{"Start":"02:01.995 ","End":"02:04.265","Text":"not remember, just go back and look it up."},{"Start":"02:04.265 ","End":"02:10.235","Text":"We had 5 formulas for partial derivatives of u in terms of those in w,"},{"Start":"02:10.235 ","End":"02:12.440","Text":"those 2 more, ux, and uy,"},{"Start":"02:12.440 ","End":"02:16.565","Text":"we don\u0027t need those because our equation just had these 3 quantities."},{"Start":"02:16.565 ","End":"02:18.590","Text":"These are the formulas."},{"Start":"02:18.590 ","End":"02:23.495","Text":"If we look at these values and substitute them here,"},{"Start":"02:23.495 ","End":"02:29.330","Text":"all the partial derivatives of Psi and Eta with respect to x and y and so on."},{"Start":"02:29.330 ","End":"02:31.955","Text":"So it\u0027s all technical,"},{"Start":"02:31.955 ","End":"02:36.995","Text":"make these substitutions and then collect and we have here u_xx,"},{"Start":"02:36.995 ","End":"02:44.570","Text":"for example here minus 4^2 =16 minus 1 times minus 4 times 2 is 8, and so on."},{"Start":"02:44.570 ","End":"02:47.390","Text":"So we get these 3 equations."},{"Start":"02:47.390 ","End":"02:54.465","Text":"Now we substitute those into our PDE and what we get is the following,"},{"Start":"02:54.465 ","End":"02:58.475","Text":"just substituting this instead of u_xx and so on."},{"Start":"02:58.475 ","End":"03:03.480","Text":"Then collect together like terms and all that\u0027s left is"},{"Start":"03:03.480 ","End":"03:10.100","Text":"w Psi Eta is 0 which is our canonical form and it does satisfy what\u0027s in the table."},{"Start":"03:10.100 ","End":"03:11.375","Text":"Let\u0027s go back and see."},{"Start":"03:11.375 ","End":"03:14.497","Text":"Yeah, w Psi Eta plus terms of lower-order,"},{"Start":"03:14.497 ","End":"03:17.270","Text":"well, we don\u0027t have any here, equals 0."},{"Start":"03:17.270 ","End":"03:21.000","Text":"So we\u0027re okay and that concludes this clip."}],"ID":30692},{"Watched":false,"Name":"Example 2 (parabolic)","Duration":"3m 3s","ChapterTopicVideoID":29123,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.295","Text":"Now a second example of a linear second-order PDE."},{"Start":"00:05.295 ","End":"00:09.045","Text":"We\u0027re going to classify it and bring it to canonical form."},{"Start":"00:09.045 ","End":"00:12.075","Text":"We\u0027ve identified the coefficients,"},{"Start":"00:12.075 ","End":"00:13.980","Text":"is only a_11,"},{"Start":"00:13.980 ","End":"00:16.230","Text":"a_12, and a_22."},{"Start":"00:16.230 ","End":"00:19.095","Text":"The lower orders are not present."},{"Start":"00:19.095 ","End":"00:24.330","Text":"Just remember to divide this one by 2 because this is 2a_12."},{"Start":"00:24.330 ","End":"00:29.145","Text":"Then we compute the discriminant Delta by this formula."},{"Start":"00:29.145 ","End":"00:31.305","Text":"It comes out to be 0,"},{"Start":"00:31.305 ","End":"00:34.270","Text":"which means that we have a parabolic type."},{"Start":"00:34.270 ","End":"00:37.865","Text":"Here\u0027s the row from the table."},{"Start":"00:37.865 ","End":"00:41.390","Text":"Characteristic ODE is this one."},{"Start":"00:41.390 ","End":"00:43.160","Text":"This is the formula for it."},{"Start":"00:43.160 ","End":"00:45.665","Text":"Get dy over dx equals 1."},{"Start":"00:45.665 ","End":"00:48.430","Text":"There\u0027s only one solution here."},{"Start":"00:48.430 ","End":"00:51.290","Text":"This is y prime minus 1 is 0."},{"Start":"00:51.290 ","End":"00:55.475","Text":"Integrate both sides and we have y minus x is a constant."},{"Start":"00:55.475 ","End":"00:58.820","Text":"This y minus x is Phi of x,"},{"Start":"00:58.820 ","End":"01:02.300","Text":"y and then we make the substitution as follows."},{"Start":"01:02.300 ","End":"01:05.630","Text":"Xi is Phi of x,y and Eta we can choose to be x or y."},{"Start":"01:05.630 ","End":"01:08.390","Text":"We\u0027ll take Eta equals x."},{"Start":"01:08.390 ","End":"01:12.190","Text":"We have this and here are all the partial derivatives of"},{"Start":"01:12.190 ","End":"01:18.155","Text":"Xi and Eta derivative with respect to x,y and then the 3 second-order derivatives,"},{"Start":"01:18.155 ","End":"01:20.755","Text":"we\u0027ll need those in a moment."},{"Start":"01:20.755 ","End":"01:24.995","Text":"Next we\u0027re going to figure out the canonical PDE,"},{"Start":"01:24.995 ","End":"01:27.485","Text":"which is the equation in w,"},{"Start":"01:27.485 ","End":"01:31.220","Text":"and assume that we\u0027ve converted u to w,"},{"Start":"01:31.220 ","End":"01:35.040","Text":"w is in Xi and Eta, these."},{"Start":"01:35.040 ","End":"01:38.750","Text":"We\u0027ll see what differential equation W satisfies,"},{"Start":"01:38.750 ","End":"01:42.800","Text":"we\u0027ll use the formulas that we derived in an earlier clip,"},{"Start":"01:42.800 ","End":"01:45.920","Text":"we\u0027ll need u_xx, u_xy, and u_yy."},{"Start":"01:45.920 ","End":"01:48.770","Text":"Now here we have the partial derivatives of"},{"Start":"01:48.770 ","End":"01:52.625","Text":"Xi and Eta with respect to x and y up to second order."},{"Start":"01:52.625 ","End":"01:55.960","Text":"We just substitute those here wherever we see them."},{"Start":"01:55.960 ","End":"02:01.010","Text":"Then we evaluate and we get the following."},{"Start":"02:01.010 ","End":"02:04.445","Text":"Here minus 1 squared is 1,"},{"Start":"02:04.445 ","End":"02:08.920","Text":"twice 1 times minus 1 is minus 2, and so on."},{"Start":"02:08.920 ","End":"02:12.470","Text":"These are the 3 second-order partial derivatives."},{"Start":"02:12.470 ","End":"02:16.640","Text":"Then remember the original PDE that we had, which is this."},{"Start":"02:16.640 ","End":"02:19.120","Text":"We substitute u_xx is this,"},{"Start":"02:19.120 ","End":"02:22.460","Text":"u_xy is this, u_yy is this."},{"Start":"02:22.460 ","End":"02:30.320","Text":"Then we get the following and just collect terms and it all boils down to the following,"},{"Start":"02:30.320 ","End":"02:34.130","Text":"w second derivative with respect to Eta is 0,"},{"Start":"02:34.130 ","End":"02:36.425","Text":"and that\u0027s our canonical equation."},{"Start":"02:36.425 ","End":"02:38.870","Text":"Later, if we were to solve this,"},{"Start":"02:38.870 ","End":"02:41.690","Text":"we could then get a formula for the solution in"},{"Start":"02:41.690 ","End":"02:44.870","Text":"terms of Xi and Eta and then we could go back"},{"Start":"02:44.870 ","End":"02:51.785","Text":"to these and substitute Xi and Eta and then we\u0027d get the result in terms of x and y."},{"Start":"02:51.785 ","End":"02:53.600","Text":"Anyway, we\u0027re not going to do that here."},{"Start":"02:53.600 ","End":"02:57.980","Text":"Let\u0027s just go and check in the table that we had the correct canonical form,"},{"Start":"02:57.980 ","End":"02:59.210","Text":"w Eta, Eta,"},{"Start":"02:59.210 ","End":"03:01.250","Text":"and we didn\u0027t have any extras."},{"Start":"03:01.250 ","End":"03:03.930","Text":"That concludes this clip."}],"ID":30693},{"Watched":false,"Name":"Example 3 (elliptic)","Duration":"2m 49s","ChapterTopicVideoID":29124,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.460","Text":"Here\u0027s our third example."},{"Start":"00:02.460 ","End":"00:06.105","Text":"This is a second-order linear PDE."},{"Start":"00:06.105 ","End":"00:09.555","Text":"We\u0027re going to classify it and then bring it to canonical form."},{"Start":"00:09.555 ","End":"00:12.240","Text":"First we identify the coefficients,"},{"Start":"00:12.240 ","End":"00:13.995","Text":"a_11 is 1,"},{"Start":"00:13.995 ","End":"00:15.750","Text":"a_12 is 0,"},{"Start":"00:15.750 ","End":"00:17.115","Text":"it\u0027s not present here,"},{"Start":"00:17.115 ","End":"00:20.055","Text":"and a_22 is x^2."},{"Start":"00:20.055 ","End":"00:21.930","Text":"Up to now, we\u0027ve always had constants."},{"Start":"00:21.930 ","End":"00:26.205","Text":"There\u0027s no reason it can\u0027t be a function of x and y in this case it\u0027s x^2."},{"Start":"00:26.205 ","End":"00:29.265","Text":"Then we compute Delta by this formula."},{"Start":"00:29.265 ","End":"00:32.040","Text":"Delta comes out to be minus x^2."},{"Start":"00:32.040 ","End":"00:34.200","Text":"It\u0027s elliptic everywhere."},{"Start":"00:34.200 ","End":"00:37.440","Text":"Here\u0027s the row from the table that we had."},{"Start":"00:37.440 ","End":"00:40.290","Text":"Elliptic when discriminant is negative."},{"Start":"00:40.290 ","End":"00:43.745","Text":"We\u0027re going to compute the characteristic ODE"},{"Start":"00:43.745 ","End":"00:47.030","Text":"and the solutions will be the characteristic curves."},{"Start":"00:47.030 ","End":"00:49.730","Text":"This time because Delta\u0027s negative,"},{"Start":"00:49.730 ","End":"00:52.465","Text":"we need to go into complex numbers."},{"Start":"00:52.465 ","End":"00:57.005","Text":"Here we get the solution as plus or minus ix."},{"Start":"00:57.005 ","End":"01:03.035","Text":"There\u0027s 2 equations, y\u0027 plus ix equals 0 and y\u0027 minus ix equals 0."},{"Start":"01:03.035 ","End":"01:08.120","Text":"Integrating, we get y plus ix^2 over 2 is"},{"Start":"01:08.120 ","End":"01:13.505","Text":"1 constant and y minus ix^2 over 2, it\u0027s another constant."},{"Start":"01:13.505 ","End":"01:17.705","Text":"This is Phi and this is the conjugate of Phi,"},{"Start":"01:17.705 ","End":"01:19.670","Text":"of course x and y are real."},{"Start":"01:19.670 ","End":"01:22.040","Text":"Anyway, in this case, looking here,"},{"Start":"01:22.040 ","End":"01:26.695","Text":"we get our psi and eta from these formulas."},{"Start":"01:26.695 ","End":"01:30.890","Text":"We get that psi equals just y."},{"Start":"01:30.890 ","End":"01:35.765","Text":"That\u0027s the real part and the imaginary part is the x^2 over 2, that\u0027s eta."},{"Start":"01:35.765 ","End":"01:39.845","Text":"We\u0027ll need the partial derivatives of these."},{"Start":"01:39.845 ","End":"01:41.360","Text":"Here they all are."},{"Start":"01:41.360 ","End":"01:44.465","Text":"The ones that are grayed out are the ones that we don\u0027t actually use."},{"Start":"01:44.465 ","End":"01:48.695","Text":"What we\u0027ll need, u_xx and u_yy."},{"Start":"01:48.695 ","End":"01:52.685","Text":"From here we can fill out some of the values."},{"Start":"01:52.685 ","End":"01:56.420","Text":"The ones that are interesting are the ones that are not 0,"},{"Start":"01:56.420 ","End":"01:59.180","Text":"that\u0027s this one and these two."},{"Start":"01:59.180 ","End":"02:05.505","Text":"What that gives us is u_xx is x^2 w eta,"},{"Start":"02:05.505 ","End":"02:09.495","Text":"eta plus w eta, that\u0027s this."},{"Start":"02:09.495 ","End":"02:15.680","Text":"Then from here we get that u_yy is just 1w psi, psi."},{"Start":"02:15.680 ","End":"02:18.905","Text":"This was our original PDE."},{"Start":"02:18.905 ","End":"02:22.520","Text":"Now substitute u_xx and u_yy here."},{"Start":"02:22.520 ","End":"02:28.625","Text":"We get this divide by x^2 and slightly rearrange and we have this."},{"Start":"02:28.625 ","End":"02:31.020","Text":"Now this involves wx psi."},{"Start":"02:31.020 ","End":"02:38.535","Text":"We don\u0027t need 1 over x^2, but remember that eta is equals to x^2 over 2 from here."},{"Start":"02:38.535 ","End":"02:47.040","Text":"1 over x^2 is eta over 2 we can put that here and that gives our answer as the following."},{"Start":"02:47.040 ","End":"02:50.110","Text":"That concludes this clip."}],"ID":30694},{"Watched":false,"Name":"Exercise 1","Duration":"8m 13s","ChapterTopicVideoID":29125,"CourseChapterTopicPlaylistID":294424,"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":"In this exercise, we have a second-order linear PDE,"},{"Start":"00:04.965 ","End":"00:08.310","Text":"this one defined on this domain."},{"Start":"00:08.310 ","End":"00:11.040","Text":"We have to classify it parabolic,"},{"Start":"00:11.040 ","End":"00:15.390","Text":"hyperbolic, elliptic, and bring it to canonical form."},{"Start":"00:15.390 ","End":"00:17.106","Text":"We won\u0027t actually solve it,"},{"Start":"00:17.106 ","End":"00:18.380","Text":"in future exercises,"},{"Start":"00:18.380 ","End":"00:21.195","Text":"we\u0027ll get up to the stage of actually solving."},{"Start":"00:21.195 ","End":"00:25.365","Text":"Meanwhile, we classify and bring to canonical form."},{"Start":"00:25.365 ","End":"00:29.610","Text":"The first step is to write down the coefficients a_11, a_12,"},{"Start":"00:29.610 ","End":"00:32.780","Text":"a_22 from here, here and here,"},{"Start":"00:32.780 ","End":"00:37.200","Text":"but remember that this coefficient is twice a_12."},{"Start":"00:37.200 ","End":"00:40.920","Text":"Instead of 4xy, we only have 2xy."},{"Start":"00:40.920 ","End":"00:45.165","Text":"A discriminant Delta is defined like this,"},{"Start":"00:45.165 ","End":"00:46.620","Text":"and in our case,"},{"Start":"00:46.620 ","End":"00:51.630","Text":"it\u0027s 2xy^2 minus this, times this,"},{"Start":"00:51.630 ","End":"00:56.540","Text":"and this comes out to be 0 because it\u0027s 4x^2 y^2,"},{"Start":"00:56.540 ","End":"01:03.635","Text":"and here also 4x^2 y^2 and Delta equals 0 is parabolic, I\u0027ll remind you."},{"Start":"01:03.635 ","End":"01:06.815","Text":"Here\u0027s the table we had in the tutorial,"},{"Start":"01:06.815 ","End":"01:10.355","Text":"so this row parabolic for Delta equals 0."},{"Start":"01:10.355 ","End":"01:15.620","Text":"Continuing, we have the characteristic ODE,"},{"Start":"01:15.620 ","End":"01:21.050","Text":"which is this dy by dx equals a_12 plus or minus the square root of Delta,"},{"Start":"01:21.050 ","End":"01:24.880","Text":"where Delta is this over a_11."},{"Start":"01:24.880 ","End":"01:29.770","Text":"This comes out to be 2xy over 2x^2,"},{"Start":"01:29.770 ","End":"01:32.255","Text":"which is y over x."},{"Start":"01:32.255 ","End":"01:35.065","Text":"This is a separable equation,"},{"Start":"01:35.065 ","End":"01:38.790","Text":"dy over y equals dx over x,"},{"Start":"01:38.790 ","End":"01:41.600","Text":"and then we can integrate and get"},{"Start":"01:41.600 ","End":"01:45.200","Text":"natural log of y equals natural log of x plus a constant."},{"Start":"01:45.200 ","End":"01:46.890","Text":"Bring this to the other side,"},{"Start":"01:46.890 ","End":"01:48.960","Text":"and write the log of the difference, sorry,"},{"Start":"01:48.960 ","End":"01:51.900","Text":"the difference of the logs is the log of the quotient,"},{"Start":"01:51.900 ","End":"01:56.210","Text":"so we get that this equals c. Now,"},{"Start":"01:56.210 ","End":"01:58.640","Text":"take the anti-natural log,"},{"Start":"01:58.640 ","End":"02:01.100","Text":"which is the exponential of both sides,"},{"Start":"02:01.100 ","End":"02:05.010","Text":"and we get that y over x is e to the power of c,"},{"Start":"02:05.010 ","End":"02:07.095","Text":"which we\u0027ll call c_1."},{"Start":"02:07.095 ","End":"02:09.035","Text":"If you go back to that table,"},{"Start":"02:09.035 ","End":"02:14.645","Text":"we see that the change of variables that we make is psi equals this,"},{"Start":"02:14.645 ","End":"02:19.925","Text":"and eta we can choose to be either x or y, we\u0027ll take y."},{"Start":"02:19.925 ","End":"02:25.070","Text":"Let\u0027s check that really this is a transformation that\u0027s invertible."},{"Start":"02:25.070 ","End":"02:29.735","Text":"We didn\u0027t do this in the previous exercises or examples, but we\u0027ll do it now."},{"Start":"02:29.735 ","End":"02:35.990","Text":"The Jacobian of the transformation is the following, it\u0027s the determinant."},{"Start":"02:35.990 ","End":"02:41.965","Text":"Now that I think about, it\u0027s possible that the Jacobian is the whole matrix,"},{"Start":"02:41.965 ","End":"02:45.065","Text":"so I better put bars around this also."},{"Start":"02:45.065 ","End":"02:48.980","Text":"We don\u0027t have these partial derivatives of the first-order,"},{"Start":"02:48.980 ","End":"02:50.750","Text":"so let\u0027s compute them."},{"Start":"02:50.750 ","End":"02:52.670","Text":"Well, I\u0027ll just write them out."},{"Start":"02:52.670 ","End":"02:55.415","Text":"It\u0027s easy to check we have 4 of them,"},{"Start":"02:55.415 ","End":"02:58.820","Text":"each one of these 2 with respect to x and with respect to y."},{"Start":"02:58.820 ","End":"03:00.589","Text":"Now we can compute the determinant,"},{"Start":"03:00.589 ","End":"03:03.970","Text":"the product of these 2 minus the product of these 2,"},{"Start":"03:03.970 ","End":"03:08.060","Text":"and that comes out to be minus y over x^2,"},{"Start":"03:08.060 ","End":"03:12.840","Text":"which is not 0 because y and x are non-zero bigger than 0 in fact."},{"Start":"03:12.840 ","End":"03:17.340","Text":"Next we\u0027ll consider the function w of psi and eta,"},{"Start":"03:17.340 ","End":"03:20.150","Text":"which you get if you switched from x,"},{"Start":"03:20.150 ","End":"03:22.250","Text":"y to psi eta."},{"Start":"03:22.250 ","End":"03:24.530","Text":"Let me see if I can explain this a bit better."},{"Start":"03:24.530 ","End":"03:26.390","Text":"Because the passage from x,"},{"Start":"03:26.390 ","End":"03:32.405","Text":"y to psi eta is reversible or invertible because of the Jacobian, in principle,"},{"Start":"03:32.405 ","End":"03:36.365","Text":"we could write the inverse of this transformation as x"},{"Start":"03:36.365 ","End":"03:40.805","Text":"as some function of psi and eta and y is some function of psi and eta,"},{"Start":"03:40.805 ","End":"03:43.865","Text":"then we can substitute those here and here,"},{"Start":"03:43.865 ","End":"03:46.670","Text":"and we get a different function of psi and eta,"},{"Start":"03:46.670 ","End":"03:52.055","Text":"that\u0027s our w. You don\u0027t have to actually compute it just in principle, it\u0027s there."},{"Start":"03:52.055 ","End":"03:55.900","Text":"We\u0027ll need the original PDE,"},{"Start":"03:55.900 ","End":"03:58.580","Text":"and we\u0027ll also need"},{"Start":"03:58.580 ","End":"04:03.990","Text":"the second-order partial derivatives of psi and eta with respect to x and y,"},{"Start":"04:03.990 ","End":"04:06.630","Text":"we already computed the first-order."},{"Start":"04:06.920 ","End":"04:09.810","Text":"Again, I\u0027ll just write them out,"},{"Start":"04:09.810 ","End":"04:12.950","Text":"it\u0027s easy enough to check by differentiating,"},{"Start":"04:12.950 ","End":"04:14.611","Text":"I\u0027ll just do one example."},{"Start":"04:14.611 ","End":"04:19.195","Text":"Psi, second derivative with respect to x is the derivative of this with respect to x,"},{"Start":"04:19.195 ","End":"04:21.785","Text":"so this is minus y,"},{"Start":"04:21.785 ","End":"04:23.585","Text":"x to the minus 2,"},{"Start":"04:23.585 ","End":"04:28.745","Text":"so you multiply by minus 2 and then make it x to the minus 3,"},{"Start":"04:28.745 ","End":"04:32.030","Text":"minus 2 times minus is plus 2,"},{"Start":"04:32.030 ","End":"04:34.175","Text":"so it\u0027s 2y over x^3."},{"Start":"04:34.175 ","End":"04:40.490","Text":"We\u0027ll also need the partial derivatives of u with respect to x, y,"},{"Start":"04:40.490 ","End":"04:44.360","Text":"and so on in terms of the partial derivatives of w with respect"},{"Start":"04:44.360 ","End":"04:48.210","Text":"to psi and eta and second-order as well,"},{"Start":"04:48.210 ","End":"04:51.950","Text":"and this is what we had in the tutorial."},{"Start":"04:51.950 ","End":"04:55.565","Text":"There are others, but these are the 4 we need from here,"},{"Start":"04:55.565 ","End":"04:58.400","Text":"here, here, and here."},{"Start":"04:58.400 ","End":"05:01.550","Text":"From these partial derivatives,"},{"Start":"05:01.550 ","End":"05:04.175","Text":"we can just substitute them, for example,"},{"Start":"05:04.175 ","End":"05:09.536","Text":"psi with respect to x is minus y over x^2,"},{"Start":"05:09.536 ","End":"05:13.300","Text":"eta with respect to x is 0."},{"Start":"05:13.300 ","End":"05:16.870","Text":"Here all the others,"},{"Start":"05:17.480 ","End":"05:22.700","Text":"ux comes out to be minus y over x^2 W psi."},{"Start":"05:22.700 ","End":"05:27.800","Text":"Uxx is get something from here and here,"},{"Start":"05:27.800 ","End":"05:31.320","Text":"we get minus y over x^2^2."},{"Start":"05:31.320 ","End":"05:33.965","Text":"It\u0027s y squared over x to the fourth W psi,"},{"Start":"05:33.965 ","End":"05:37.685","Text":"psi plus 2y over x^3 W psi."},{"Start":"05:37.685 ","End":"05:39.169","Text":"I won\u0027t continue,"},{"Start":"05:39.169 ","End":"05:43.005","Text":"I\u0027ll just write them out and leave you to check."},{"Start":"05:43.005 ","End":"05:47.240","Text":"Now that we have the partial derivatives of u with respect to x,"},{"Start":"05:47.240 ","End":"05:52.025","Text":"y in terms of the partial derivative of w with in terms of psi and eta,"},{"Start":"05:52.025 ","End":"05:56.135","Text":"we can now return to the original PDE,"},{"Start":"05:56.135 ","End":"05:59.810","Text":"make the substitutions in 4 places here, here, here,"},{"Start":"05:59.810 ","End":"06:02.260","Text":"and here from these 4,"},{"Start":"06:02.260 ","End":"06:06.400","Text":"what we get is the following."},{"Start":"06:06.400 ","End":"06:08.970","Text":"For example, Uxx,"},{"Start":"06:08.970 ","End":"06:15.145","Text":"we look it up here and that\u0027s what we put in this brackets after the 2x^2."},{"Start":"06:15.145 ","End":"06:17.000","Text":"Similarly for the others,"},{"Start":"06:17.000 ","End":"06:18.695","Text":"let\u0027s collect like terms."},{"Start":"06:18.695 ","End":"06:20.090","Text":"Let\u0027s take for example,"},{"Start":"06:20.090 ","End":"06:23.600","Text":"W psi, psi and see how many we have."},{"Start":"06:23.600 ","End":"06:30.885","Text":"From here we have multiply this by this 2 y^2 over x^2."},{"Start":"06:30.885 ","End":"06:33.380","Text":"From here, if you multiply out,"},{"Start":"06:33.380 ","End":"06:37.355","Text":"we also get y^2 over x^2 this time minus 4."},{"Start":"06:37.355 ","End":"06:43.970","Text":"We have 2 minus 4 and here we have another 2y^2 over x^2,"},{"Start":"06:43.970 ","End":"06:45.680","Text":"so 2 minus 4 plus 2 is 0."},{"Start":"06:45.680 ","End":"06:48.245","Text":"We don\u0027t have any W psi, psi."},{"Start":"06:48.245 ","End":"06:50.615","Text":"Also, it\u0027s tedious,"},{"Start":"06:50.615 ","End":"06:54.124","Text":"I\u0027ll tell you the answer for this stage."},{"Start":"06:54.124 ","End":"07:04.145","Text":"2y^2 W eta, eta plus 4y^2 over x^2 w psi plus 2y e to the y over x,"},{"Start":"07:04.145 ","End":"07:06.605","Text":"and divide the whole thing by 2."},{"Start":"07:06.605 ","End":"07:08.700","Text":"It\u0027s just 1 here,"},{"Start":"07:08.700 ","End":"07:10.565","Text":"2 here, and 1 here."},{"Start":"07:10.565 ","End":"07:14.195","Text":"Now the partial derivative of w with respect to eta and"},{"Start":"07:14.195 ","End":"07:19.055","Text":"psi and we have y and x in the coefficients."},{"Start":"07:19.055 ","End":"07:23.557","Text":"Remember that psi is y over x and eta is y,"},{"Start":"07:23.557 ","End":"07:32.495","Text":"y^2 is eta squared from here and y over x is psi,"},{"Start":"07:32.495 ","End":"07:38.920","Text":"so 2 psi squared w psi and then y is eta again,"},{"Start":"07:38.920 ","End":"07:42.995","Text":"e to the y over x is psi equals 0."},{"Start":"07:42.995 ","End":"07:47.605","Text":"Final step is just to divide everything by eta squared."},{"Start":"07:47.605 ","End":"07:50.280","Text":"This is the answer."},{"Start":"07:50.280 ","End":"07:55.745","Text":"Let\u0027s just check in the table of canonical forms that we\u0027ve got the right thing."},{"Start":"07:55.745 ","End":"08:01.073","Text":"Note that the second-order term is this."},{"Start":"08:01.073 ","End":"08:04.520","Text":"The parabolic, it\u0027s w eta,"},{"Start":"08:04.520 ","End":"08:07.250","Text":"eta plus lower-order terms,"},{"Start":"08:07.250 ","End":"08:09.875","Text":"terms of order less than 2."},{"Start":"08:09.875 ","End":"08:14.460","Text":"With that, we conclude this exercise."}],"ID":30695},{"Watched":false,"Name":"Exercise 2ab","Duration":"5m 57s","ChapterTopicVideoID":29126,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.320","Text":"In this exercise, we have a second-order linear"},{"Start":"00:04.320 ","End":"00:08.655","Text":"PDE and there are three parts to the question."},{"Start":"00:08.655 ","End":"00:10.860","Text":"Show that the equation is hyperbolic,"},{"Start":"00:10.860 ","End":"00:13.140","Text":"find its canonical form,"},{"Start":"00:13.140 ","End":"00:16.875","Text":"and then find the general solution to the equation."},{"Start":"00:16.875 ","End":"00:19.275","Text":"Well, in this clip we\u0027ll do A and B."},{"Start":"00:19.275 ","End":"00:20.730","Text":"A is very short,"},{"Start":"00:20.730 ","End":"00:23.535","Text":"and C will be in the next clip."},{"Start":"00:23.535 ","End":"00:32.055","Text":"Let\u0027s start by identifying the coefficients according to the notation we use previously."},{"Start":"00:32.055 ","End":"00:36.215","Text":"A11 is the coefficient of u_xx, so that\u0027s 2."},{"Start":"00:36.215 ","End":"00:37.850","Text":"A12, well,"},{"Start":"00:37.850 ","End":"00:39.530","Text":"it belongs to u_xy,"},{"Start":"00:39.530 ","End":"00:42.070","Text":"which is absent, so it\u0027s 0."},{"Start":"00:42.070 ","End":"00:44.865","Text":"A22 goes with u_yy,"},{"Start":"00:44.865 ","End":"00:46.280","Text":"and it\u0027s all of this."},{"Start":"00:46.280 ","End":"00:48.635","Text":"Then we compute the discriminant Delta,"},{"Start":"00:48.635 ","End":"00:51.380","Text":"which is given by the following formula."},{"Start":"00:51.380 ","End":"00:55.040","Text":"This comes out to be the following."},{"Start":"00:55.040 ","End":"00:58.835","Text":"Which is, well, we see it\u0027s positive."},{"Start":"00:58.835 ","End":"01:03.350","Text":"That means that it\u0027s hyperbolic and that answers part A,"},{"Start":"01:03.350 ","End":"01:05.225","Text":"but we will continue."},{"Start":"01:05.225 ","End":"01:10.910","Text":"Next thing we do is to solve the characteristic ODE,"},{"Start":"01:10.910 ","End":"01:16.340","Text":"which is dy by dx is A12 plus or minus Delta,"},{"Start":"01:16.340 ","End":"01:24.035","Text":"where Delta is from here over A11 and that equals A12 is 0,"},{"Start":"01:24.035 ","End":"01:28.120","Text":"Delta was this from here."},{"Start":"01:28.120 ","End":"01:29.905","Text":"If you take the square root,"},{"Start":"01:29.905 ","End":"01:34.290","Text":"4 becomes 2 the squared disappears and the 2 with the 2 cancels"},{"Start":"01:34.290 ","End":"01:39.315","Text":"so we\u0027re just left with plus or minus 1 plus y^2."},{"Start":"01:39.315 ","End":"01:43.845","Text":"Then the equation is separable if we rearrange it,"},{"Start":"01:43.845 ","End":"01:49.965","Text":"we can say dy over 1 plus y^2 equals plus or minus dx."},{"Start":"01:49.965 ","End":"01:54.510","Text":"Integral of this is arc tangent and the integral"},{"Start":"01:54.510 ","End":"01:58.910","Text":"of dx is just x so we have these 2 solutions,"},{"Start":"01:58.910 ","End":"02:04.000","Text":"really has to be a different constant for each plus x or minus x."},{"Start":"02:04.000 ","End":"02:07.015","Text":"Anyway, now that we have these 2,"},{"Start":"02:07.015 ","End":"02:09.770","Text":"these are the functions we called Phi 1 and Phi"},{"Start":"02:09.770 ","End":"02:14.615","Text":"2 and these are what give us arc Psi and Eta."},{"Start":"02:14.615 ","End":"02:18.815","Text":"Psi is this, an eta is this."},{"Start":"02:18.815 ","End":"02:24.920","Text":"This will be for the change of variables to get our equation into the canonical form,"},{"Start":"02:24.920 ","End":"02:31.970","Text":"so u(x, y) becomes w of Psi and Eta if we do the change of variables."},{"Start":"02:31.970 ","End":"02:36.500","Text":"Just to remind you of what the original equation was and now we\u0027re working"},{"Start":"02:36.500 ","End":"02:41.540","Text":"towards converting u in x and y to w in Psi and Eta."},{"Start":"02:41.540 ","End":"02:48.320","Text":"Note that we have u_xx here and u_yy and uy."},{"Start":"02:48.320 ","End":"02:51.725","Text":"The other ones are missing like ux and u_xy."},{"Start":"02:51.725 ","End":"02:56.425","Text":"We need these 3 partial derivatives"},{"Start":"02:56.425 ","End":"03:01.205","Text":"and we get them from our table that we computed in the previous clip."},{"Start":"03:01.205 ","End":"03:04.790","Text":"These are the following and we need"},{"Start":"03:04.790 ","End":"03:10.495","Text":"all these partial derivatives of psi and eta with respect to x and y."},{"Start":"03:10.495 ","End":"03:12.860","Text":"From these two formulas,"},{"Start":"03:12.860 ","End":"03:17.975","Text":"we just differentiate as needed and we can compute the ones that we need."},{"Start":"03:17.975 ","End":"03:21.500","Text":"This is 1 because the derivative with respect"},{"Start":"03:21.500 ","End":"03:25.580","Text":"to x is 1 and if you differentiate again with respect to x, you get 0."},{"Start":"03:25.580 ","End":"03:31.094","Text":"Derivative with respect to y is 1 over 1 plus y^2."},{"Start":"03:31.094 ","End":"03:32.700","Text":"That\u0027s from the arctangent."},{"Start":"03:32.700 ","End":"03:36.995","Text":"The derivative with respect to y again gives us this."},{"Start":"03:36.995 ","End":"03:39.455","Text":"Similarly with Eta, very similar,"},{"Start":"03:39.455 ","End":"03:43.630","Text":"except that there\u0027s a minus here and otherwise the same."},{"Start":"03:43.630 ","End":"03:48.514","Text":"We substitute all these here where needed,"},{"Start":"03:48.514 ","End":"03:53.025","Text":"and we get the following three equations."},{"Start":"03:53.025 ","End":"03:57.690","Text":"Like here, Psi x is 1, 1^2 is 1."},{"Start":"03:57.690 ","End":"04:00.950","Text":"This is just W Psi, Psi."},{"Start":"04:00.950 ","End":"04:03.700","Text":"Here we have Eta x,"},{"Start":"04:03.700 ","End":"04:06.540","Text":"Psi x, which is the product of these 2,"},{"Start":"04:06.540 ","End":"04:08.010","Text":"which is minus 1."},{"Start":"04:08.010 ","End":"04:10.050","Text":"So this becomes minus 2,"},{"Start":"04:10.050 ","End":"04:12.930","Text":"and so on with all the others."},{"Start":"04:12.930 ","End":"04:20.030","Text":"Now we want to substitute these three into our original PDE, which is this."},{"Start":"04:20.030 ","End":"04:23.255","Text":"Substitute this from here, this from here,"},{"Start":"04:23.255 ","End":"04:24.965","Text":"and this from here,"},{"Start":"04:24.965 ","End":"04:26.780","Text":"or rather this from here."},{"Start":"04:26.780 ","End":"04:29.150","Text":"I got the last two the wrong way around, never mind."},{"Start":"04:29.150 ","End":"04:33.825","Text":"What we get is the following, which looks messy."},{"Start":"04:33.825 ","End":"04:38.810","Text":"The u_yy is all of this,"},{"Start":"04:38.810 ","End":"04:41.149","Text":"but it\u0027s not as bad as it looks."},{"Start":"04:41.149 ","End":"04:43.435","Text":"Let\u0027s see if we can expand."},{"Start":"04:43.435 ","End":"04:48.855","Text":"The first row, will give us just this,"},{"Start":"04:48.855 ","End":"04:51.005","Text":"2 minus 4 plus 2."},{"Start":"04:51.005 ","End":"04:57.680","Text":"In this row, the 1 plus y^2 squared cancels with the 1 over 1 plus y^2 squared."},{"Start":"04:57.680 ","End":"05:02.705","Text":"If all this cancels we just have minus 2 times these,"},{"Start":"05:02.705 ","End":"05:05.195","Text":"what are in front of the brackets."},{"Start":"05:05.195 ","End":"05:12.120","Text":"We get minus 2w Psi Psi minus 4 this plus 4 of this,"},{"Start":"05:12.120 ","End":"05:14.370","Text":"it\u0027s minus 2 with minus 2."},{"Start":"05:14.370 ","End":"05:16.905","Text":"Here we have a minus 2 and here,"},{"Start":"05:16.905 ","End":"05:20.010","Text":"again, minus 2 is minus 2 is plus 4."},{"Start":"05:20.010 ","End":"05:21.630","Text":"For the last row again,"},{"Start":"05:21.630 ","End":"05:24.444","Text":"the 1 plus y^2 cancels with the 1 plus y^2,"},{"Start":"05:24.444 ","End":"05:31.040","Text":"so we have minus 4y times w psi and w Eta, which is this."},{"Start":"05:31.040 ","End":"05:36.785","Text":"Now, everything pretty much cancels out in pairs except for the ones that I\u0027ve colored."},{"Start":"05:36.785 ","End":"05:38.450","Text":"This cancels with this,"},{"Start":"05:38.450 ","End":"05:40.925","Text":"this cancels with this,"},{"Start":"05:40.925 ","End":"05:43.730","Text":"this, with this, and this with this."},{"Start":"05:43.730 ","End":"05:50.105","Text":"All we\u0027re left with is minus 8w Psi eta is 0 divide by 8."},{"Start":"05:50.105 ","End":"05:53.755","Text":"This is our canonical form."},{"Start":"05:53.755 ","End":"05:57.820","Text":"We\u0027ll do part C in the next clip."}],"ID":30696},{"Watched":false,"Name":"Exercise 2c","Duration":"1m 40s","ChapterTopicVideoID":29108,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.330","Text":"We\u0027re continuing with the same exercise we had before."},{"Start":"00:03.330 ","End":"00:08.288","Text":"We just completed a and b and now we\u0027re going to do part c. In part b,"},{"Start":"00:08.288 ","End":"00:11.766","Text":"we found that the canonical equation was this."},{"Start":"00:11.766 ","End":"00:16.629","Text":"Let\u0027s solve this one and afterwards we\u0027ll get back from Psi Eta to x, y."},{"Start":"00:16.629 ","End":"00:19.998","Text":"Integrating both sides with respect to Eta,"},{"Start":"00:19.998 ","End":"00:21.649","Text":"here we get just w Psi,"},{"Start":"00:21.649 ","End":"00:25.075","Text":"and here we get an arbitrary function of Psi."},{"Start":"00:25.075 ","End":"00:27.694","Text":"We\u0027ll do the same thing again,"},{"Start":"00:27.694 ","End":"00:32.520","Text":"integrate with respect to Psi and we get the integral of this,"},{"Start":"00:32.520 ","End":"00:35.297","Text":"meaning the indefinite integral,"},{"Start":"00:35.297 ","End":"00:37.419","Text":"we\u0027ll call it F of Psi,"},{"Start":"00:37.419 ","End":"00:40.342","Text":"plus some arbitrary function of Eta,"},{"Start":"00:40.342 ","End":"00:45.796","Text":"and we\u0027ll call that G. This is our general solution with Psi and Eta."},{"Start":"00:45.796 ","End":"00:47.509","Text":"Remember that u of x,"},{"Start":"00:47.509 ","End":"00:49.243","Text":"y was w of Psi and Eta."},{"Start":"00:49.243 ","End":"00:50.994","Text":"We just found w here."},{"Start":"00:50.994 ","End":"00:54.138","Text":"This is equal to F of Psi plus G of Eta,"},{"Start":"00:54.138 ","End":"00:55.737","Text":"which is Eta of x and y."},{"Start":"00:55.737 ","End":"01:00.518","Text":"In part b, we already found the functions Psi and Eta,"},{"Start":"01:00.518 ","End":"01:08.126","Text":"or at least the change of variables Psi in terms of y and x and Eta in terms of y and x."},{"Start":"01:08.126 ","End":"01:12.793","Text":"We just have to substitute these here and here,"},{"Start":"01:12.793 ","End":"01:14.691","Text":"and we get that u(x,"},{"Start":"01:14.691 ","End":"01:19.292","Text":"y) is F of arc tangent of y plus x plus G of"},{"Start":"01:19.292 ","End":"01:26.312","Text":"arc tangent y minus x. F and G are any differentiable functions."},{"Start":"01:26.312 ","End":"01:33.260","Text":"That\u0027s the general solution to the original PDE,"},{"Start":"01:33.260 ","End":"01:41.340","Text":"and that concludes this exercise and this clip."}],"ID":30697},{"Watched":false,"Name":"Exercise 3","Duration":"4m 55s","ChapterTopicVideoID":29109,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"In this exercise,"},{"Start":"00:01.740 ","End":"00:06.225","Text":"we have a second order linear PDE, this one,"},{"Start":"00:06.225 ","End":"00:11.910","Text":"and we have to show that the equation is hyperbolic to find its canonical form,"},{"Start":"00:11.910 ","End":"00:15.820","Text":"and then to find the general solution to the equation."},{"Start":"00:15.920 ","End":"00:21.255","Text":"Let\u0027s start. First of all, let\u0027s identify the coefficients."},{"Start":"00:21.255 ","End":"00:22.680","Text":"Like in the tutorial,"},{"Start":"00:22.680 ","End":"00:25.380","Text":"this one is a_11, this one,"},{"Start":"00:25.380 ","End":"00:27.960","Text":"be careful, it\u0027s 2a_12,"},{"Start":"00:27.960 ","End":"00:30.495","Text":"this one is a_22."},{"Start":"00:30.495 ","End":"00:35.580","Text":"That just writing it out, a_12 is 2."},{"Start":"00:35.580 ","End":"00:40.275","Text":"Then we compute the discriminant Delta using this formula."},{"Start":"00:40.275 ","End":"00:44.310","Text":"What it gives us in our case is 16,"},{"Start":"00:44.310 ","End":"00:45.645","Text":"which is positive,"},{"Start":"00:45.645 ","End":"00:48.540","Text":"which makes it a hyperbolic equation."},{"Start":"00:48.540 ","End":"00:50.415","Text":"That was part a."},{"Start":"00:50.415 ","End":"00:52.920","Text":"Let\u0027s continue with part b."},{"Start":"00:52.920 ","End":"00:59.430","Text":"This is the characteristic ODE where Delta is as above."},{"Start":"00:59.430 ","End":"01:04.905","Text":"What we get is dy by dx equals 2 plus or 1 minus square root of 16 over 2."},{"Start":"01:04.905 ","End":"01:07.095","Text":"Square root of 16 is 4."},{"Start":"01:07.095 ","End":"01:09.030","Text":"We can also divide by 2,"},{"Start":"01:09.030 ","End":"01:12.615","Text":"so what we get is 1 plus or minus 2."},{"Start":"01:12.615 ","End":"01:19.140","Text":"Then we integrate and we\u0027ve got that y is 1 plus or minus 2 times x plus"},{"Start":"01:19.140 ","End":"01:27.000","Text":"c. We can write this as c equals y minus 1 plus or minus 2x."},{"Start":"01:27.000 ","End":"01:31.650","Text":"Strictly speaking, we should have taken each case separately in a different constants."},{"Start":"01:31.650 ","End":"01:37.695","Text":"Really, c_1 is y minus 3x and c_2 is y plus x."},{"Start":"01:37.695 ","End":"01:41.140","Text":"But practice it doesn\u0027t much matter."},{"Start":"01:41.140 ","End":"01:43.980","Text":"We can make a change of variables."},{"Start":"01:43.980 ","End":"01:48.015","Text":"We\u0027re going to let Psi equals y minus 3x from here,"},{"Start":"01:48.015 ","End":"01:51.690","Text":"and Eta equals y plus x."},{"Start":"01:51.690 ","End":"01:53.370","Text":"Or if you just looked from here,"},{"Start":"01:53.370 ","End":"01:54.840","Text":"y minus, if you take the plus,"},{"Start":"01:54.840 ","End":"01:56.610","Text":"it\u0027s 3x, and if you take the minus,"},{"Start":"01:56.610 ","End":"01:57.780","Text":"it\u0027s y minus minus x,"},{"Start":"01:57.780 ","End":"01:59.070","Text":"which is y plus x."},{"Start":"01:59.070 ","End":"02:03.290","Text":"We want to make this change of variables and get u(x,"},{"Start":"02:03.290 ","End":"02:07.440","Text":"y) in terms of indifferent function w(Psi, Eta)."},{"Start":"02:07.440 ","End":"02:08.730","Text":"Just to remember,"},{"Start":"02:08.730 ","End":"02:13.400","Text":"this is the equation that we\u0027re solving and we want to convert u(x,"},{"Start":"02:13.400 ","End":"02:16.170","Text":"y) to w(Psi, Eta)."},{"Start":"02:16.170 ","End":"02:19.905","Text":"We\u0027ll need u_xx, u_xy and u_yy."},{"Start":"02:19.905 ","End":"02:23.490","Text":"From the formulas in the tutorial,"},{"Start":"02:23.490 ","End":"02:25.440","Text":"we need these 3 equations,"},{"Start":"02:25.440 ","End":"02:27.405","Text":"just copied them from there."},{"Start":"02:27.405 ","End":"02:31.050","Text":"We need all of these partial derivatives of Psi and Eta with respect to"},{"Start":"02:31.050 ","End":"02:34.845","Text":"x and y like this, this, and this."},{"Start":"02:34.845 ","End":"02:38.160","Text":"We can get those from the formula for Psi and Eta."},{"Start":"02:38.160 ","End":"02:44.850","Text":"We just differentiate psi with respect to x minus 3 with respect to y, it\u0027s 1."},{"Start":"02:44.850 ","End":"02:48.450","Text":"Second derivatives are all 0 because these are constants."},{"Start":"02:48.450 ","End":"02:51.090","Text":"With Eta, we get with respect to x,"},{"Start":"02:51.090 ","End":"02:53.250","Text":"It\u0027s 1, with respect to y it\u0027s 1."},{"Start":"02:53.250 ","End":"02:56.340","Text":"All the second order ones are also 0."},{"Start":"02:56.340 ","End":"02:59.070","Text":"We just have to plug these in here."},{"Start":"02:59.070 ","End":"03:07.305","Text":"Let\u0027s see, u_xx is w(Psi Psi) is times this squared minus 3 squared is 9."},{"Start":"03:07.305 ","End":"03:10.065","Text":"We get 9w_Psi Psi."},{"Start":"03:10.065 ","End":"03:13.440","Text":"Here, we get Eta_x, Psi_x,"},{"Start":"03:13.440 ","End":"03:16.110","Text":"which is this times,"},{"Start":"03:16.110 ","End":"03:17.985","Text":"this times the 2,"},{"Start":"03:17.985 ","End":"03:20.865","Text":"gives us minus 6, and so on."},{"Start":"03:20.865 ","End":"03:23.880","Text":"Just substitutions into all we\u0027re left with is"},{"Start":"03:23.880 ","End":"03:27.525","Text":"the partial derivatives of w without the Psi and Eta."},{"Start":"03:27.525 ","End":"03:29.115","Text":"That we have these 3,"},{"Start":"03:29.115 ","End":"03:32.100","Text":"we\u0027re going to substitute them in the original."},{"Start":"03:32.100 ","End":"03:33.465","Text":"Write it again,"},{"Start":"03:33.465 ","End":"03:35.685","Text":"this one in the PDE."},{"Start":"03:35.685 ","End":"03:37.440","Text":"Put u_xx here,"},{"Start":"03:37.440 ","End":"03:39.180","Text":"what we get is"},{"Start":"03:39.180 ","End":"03:45.240","Text":"twice all of this plus 4 times this minus"},{"Start":"03:45.240 ","End":"03:49.130","Text":"6 times this equals a note that the right-hand side,"},{"Start":"03:49.130 ","End":"03:56.655","Text":"y minus 3x is Psi and 2x plus 2y will be 2 Eta."},{"Start":"03:56.655 ","End":"03:59.867","Text":"We want everything in Psi and Eta,"},{"Start":"03:59.867 ","End":"04:02.405","Text":"so that\u0027s why we do these conversions."},{"Start":"04:02.405 ","End":"04:06.680","Text":"Now let\u0027s just collect together like terms w_Psi Psi."},{"Start":"04:06.680 ","End":"04:11.580","Text":"We have 2 times 9 is 18 minus 12 minus 6."},{"Start":"04:11.580 ","End":"04:13.335","Text":"I wrote it all down,"},{"Start":"04:13.335 ","End":"04:17.190","Text":"and 18 minus 12 minus 6 is 0,"},{"Start":"04:17.190 ","End":"04:19.440","Text":"and 2 plus 4 minus 6 is 0."},{"Start":"04:19.440 ","End":"04:21.360","Text":"We\u0027re just left with this one."},{"Start":"04:21.360 ","End":"04:30.290","Text":"Minus 12, minus 12 minus 8 is minus 32 w_Psi Eta equals whatever it was."},{"Start":"04:30.290 ","End":"04:36.050","Text":"Divide by minus 32 and we have the following canonical form."},{"Start":"04:36.050 ","End":"04:38.345","Text":"Minus 4 over minus 32 is an 8th,"},{"Start":"04:38.345 ","End":"04:41.510","Text":"and 2 over minus 32 is minus the 16th."},{"Start":"04:41.510 ","End":"04:46.669","Text":"Technically speaking, I should have put everything on the left-hand side equals 0,"},{"Start":"04:46.669 ","End":"04:48.650","Text":"like this, but it doesn\u0027t matter,"},{"Start":"04:48.650 ","End":"04:50.060","Text":"you can leave it like this."},{"Start":"04:50.060 ","End":"04:51.600","Text":"That concludes part b."},{"Start":"04:51.600 ","End":"04:53.645","Text":"Let\u0027s take a break before part c,"},{"Start":"04:53.645 ","End":"04:55.830","Text":"which is being in the next clip."}],"ID":30698},{"Watched":false,"Name":"Exercise 4a","Duration":"1m 43s","ChapterTopicVideoID":29110,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.670","Text":"We\u0027re continuing this exercise,"},{"Start":"00:02.670 ","End":"00:04.815","Text":"we just finished a and b, now,"},{"Start":"00:04.815 ","End":"00:07.470","Text":"we\u0027re up to part c. In part b,"},{"Start":"00:07.470 ","End":"00:10.900","Text":"we found the canonical form was this,"},{"Start":"00:12.080 ","End":"00:17.160","Text":"and we can integrate with respect to Eta to get W Psi"},{"Start":"00:17.160 ","End":"00:21.620","Text":"is 1/8 Psi^2 Eta is a constant as far as Eta goes,"},{"Start":"00:21.620 ","End":"00:23.275","Text":"it\u0027s multiplied by Eta."},{"Start":"00:23.275 ","End":"00:27.015","Text":"The integral of minus sin is cos,"},{"Start":"00:27.015 ","End":"00:31.200","Text":"but we also have to divide by the inner derivatives,"},{"Start":"00:31.200 ","End":"00:35.040","Text":"so the integral is cos( 2 Eta) divided by 2,"},{"Start":"00:35.040 ","End":"00:37.410","Text":"the 2 with the 16 gives 32,"},{"Start":"00:37.410 ","End":"00:41.715","Text":"and we also have to add an arbitrary function of Psi."},{"Start":"00:41.715 ","End":"00:46.200","Text":"We need to do another integration this time with respect to Psi."},{"Start":"00:46.200 ","End":"00:48.950","Text":"From here we get Psi^3 over 3,"},{"Start":"00:48.950 ","End":"00:51.625","Text":"which means that it\u0027s 24 here,"},{"Start":"00:51.625 ","End":"00:55.740","Text":"and this is just a constant now,"},{"Start":"00:55.740 ","End":"00:58.605","Text":"as far as Psi goes, is multiplied by Psi,"},{"Start":"00:58.605 ","End":"01:02.745","Text":"that\u0027s called the indefinite integral of F,"},{"Start":"01:02.745 ","End":"01:05.720","Text":"and then we need a arbitrary function of Eta,"},{"Start":"01:05.720 ","End":"01:11.310","Text":"we\u0027ll call it big G. We need to go back from w(Psi,"},{"Start":"01:11.310 ","End":"01:13.410","Text":"Eta) to u in x and y."},{"Start":"01:13.410 ","End":"01:15.480","Text":"Remember from part b,"},{"Start":"01:15.480 ","End":"01:19.620","Text":"we had the Psi equals y minus 3x and Eta is y plus x."},{"Start":"01:19.620 ","End":"01:23.745","Text":"Just substitute Psi and Eta everywhere here,"},{"Start":"01:23.745 ","End":"01:25.446","Text":"we get u(x,"},{"Start":"01:25.446 ","End":"01:29.068","Text":"y) is 1 over 24, y minus 3x^3,"},{"Start":"01:29.068 ","End":"01:32.197","Text":"Eta as y plus x, and so on."},{"Start":"01:32.197 ","End":"01:40.112","Text":"2 Eta is 2x plus 2y here Psi is y minus 3x, just replace everything."},{"Start":"01:40.112 ","End":"01:41.325","Text":"This is the answer,"},{"Start":"01:41.325 ","End":"01:44.320","Text":"and that concludes this exercise."}],"ID":30699},{"Watched":false,"Name":"Exercise 4b","Duration":"5m 57s","ChapterTopicVideoID":29111,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.215","Text":"Here we have a 2nd order linear PDE,"},{"Start":"00:04.215 ","End":"00:07.695","Text":"and we have to show that it\u0027s parabolic."},{"Start":"00:07.695 ","End":"00:11.250","Text":"We have to find its canonical form and then we have to find"},{"Start":"00:11.250 ","End":"00:15.330","Text":"the general solution to the equation. Let\u0027s start with a."},{"Start":"00:15.330 ","End":"00:17.100","Text":"Just like in the tutorial,"},{"Start":"00:17.100 ","End":"00:19.800","Text":"we identify these coefficients,"},{"Start":"00:19.800 ","End":"00:23.145","Text":"a_11, a_12, and a_22."},{"Start":"00:23.145 ","End":"00:27.945","Text":"a_12 is only half of this coefficient."},{"Start":"00:27.945 ","End":"00:29.430","Text":"This is 1,"},{"Start":"00:29.430 ","End":"00:30.750","Text":"this is 3,"},{"Start":"00:30.750 ","End":"00:32.310","Text":"and this is 9."},{"Start":"00:32.310 ","End":"00:34.695","Text":"Then we\u0027re going to compute the discriminant,"},{"Start":"00:34.695 ","End":"00:38.120","Text":"which is given by this formula and in our case,"},{"Start":"00:38.120 ","End":"00:41.675","Text":"it comes out to be 3^2 minus 1 times 9,"},{"Start":"00:41.675 ","End":"00:46.010","Text":"which is 0, and 0 means parabolic."},{"Start":"00:46.010 ","End":"00:48.109","Text":"Onto part b. Next,"},{"Start":"00:48.109 ","End":"00:52.790","Text":"we solve the characteristic ODE which is this,"},{"Start":"00:52.790 ","End":"00:58.830","Text":"we computed it and so we get 3 plus or minus 0 over 1,"},{"Start":"00:58.830 ","End":"01:00.510","Text":"that\u0027s dy by dx."},{"Start":"01:00.510 ","End":"01:05.164","Text":"This is only 1 possibility because plus or minus zeros just nothing,"},{"Start":"01:05.164 ","End":"01:07.315","Text":"so 3 over 1 is 3."},{"Start":"01:07.315 ","End":"01:10.150","Text":"Then we get dy equals 3dx,"},{"Start":"01:10.150 ","End":"01:14.060","Text":"integrate and we have y equals 3x plus a constant"},{"Start":"01:14.060 ","End":"01:19.090","Text":"and we can write this as c equals y minus 3x."},{"Start":"01:19.090 ","End":"01:21.700","Text":"Whenever we only have one equation"},{"Start":"01:21.700 ","End":"01:24.800","Text":"as opposed to c1 equals something and c2 equals something,"},{"Start":"01:24.800 ","End":"01:31.355","Text":"then the way we make a substitution of variables is to let Psi equal this,"},{"Start":"01:31.355 ","End":"01:34.430","Text":"and Eta we can choose either x or y."},{"Start":"01:34.430 ","End":"01:36.920","Text":"Let\u0027s go with Eta equals x."},{"Start":"01:36.920 ","End":"01:41.450","Text":"Let\u0027s also compute the Jacobian or its determinant"},{"Start":"01:41.450 ","End":"01:46.830","Text":"to see if this transformation is 1 to 1 and invertible."},{"Start":"01:46.830 ","End":"01:50.340","Text":"The Psi_x is minus 3,"},{"Start":"01:50.340 ","End":"01:52.605","Text":"Psi_y is 1,"},{"Start":"01:52.605 ","End":"01:54.120","Text":"Eta_x is 1,"},{"Start":"01:54.120 ","End":"02:00.140","Text":"Eta_y is 0 and then we can compute the value of the Jacobian,"},{"Start":"02:00.140 ","End":"02:01.740","Text":"the determinant of it,"},{"Start":"02:01.740 ","End":"02:04.980","Text":"and straightforward, minus 3 times 0,"},{"Start":"02:04.980 ","End":"02:07.035","Text":"minus 1 times 1 minus 1."},{"Start":"02:07.035 ","End":"02:10.670","Text":"It\u0027s not 0 anywhere and so we\u0027re safe to look"},{"Start":"02:10.670 ","End":"02:14.540","Text":"for the change of variables from xy to Psi,"},{"Start":"02:14.540 ","End":"02:20.285","Text":"Eta will go from function u to function w. Continuing,"},{"Start":"02:20.285 ","End":"02:23.915","Text":"just a reminder of what the original PDE is."},{"Start":"02:23.915 ","End":"02:26.710","Text":"Now we have here u_xx,"},{"Start":"02:26.710 ","End":"02:30.600","Text":"we have u_xy and we have u_yy."},{"Start":"02:30.600 ","End":"02:33.950","Text":"In the tutorial, we had formulas for all of"},{"Start":"02:33.950 ","End":"02:37.520","Text":"these partial derivatives in terms of"},{"Start":"02:37.520 ","End":"02:42.220","Text":"the partial derivative of w with respect to Psi and Eta."},{"Start":"02:42.220 ","End":"02:45.810","Text":"We need all of these, Psi_x,"},{"Start":"02:45.810 ","End":"02:50.610","Text":"Psi_xx, Psi_xy, and so on."},{"Start":"02:50.610 ","End":"02:55.440","Text":"We already did the first derivative,"},{"Start":"02:55.440 ","End":"02:57.050","Text":"now we just need to continue to get"},{"Start":"02:57.050 ","End":"03:01.140","Text":"the 3 second-order derivatives because these are constants,"},{"Start":"03:01.140 ","End":"03:04.650","Text":"all of these 3 are 0 for both Psi and for Eta,"},{"Start":"03:04.650 ","End":"03:07.925","Text":"that will make work easier when we substitute here,"},{"Start":"03:07.925 ","End":"03:10.580","Text":"what we\u0027ll get is the following."},{"Start":"03:10.580 ","End":"03:12.140","Text":"Let\u0027s do some of the coefficients,"},{"Start":"03:12.140 ","End":"03:15.020","Text":"for example, W_Psi Psi,"},{"Start":"03:15.020 ","End":"03:20.420","Text":"so we need Psi_x^2 minus 3^2, that\u0027s 9."},{"Start":"03:20.420 ","End":"03:27.305","Text":"Let\u0027s take this one and say we have Eta_y Psi_x,"},{"Start":"03:27.305 ","End":"03:33.485","Text":"which is this times this is 0, Psi_y, Eta_x."},{"Start":"03:33.485 ","End":"03:39.110","Text":"That\u0027s 1 and that gives us the 1 here. The rest of these are 0."},{"Start":"03:39.110 ","End":"03:42.515","Text":"For example, Psi_xy is 0,"},{"Start":"03:42.515 ","End":"03:46.585","Text":"Eta_x Eta_y, 1 times 0 is 0."},{"Start":"03:46.585 ","End":"03:51.080","Text":"You can check all of these and we get the following."},{"Start":"03:51.080 ","End":"03:57.560","Text":"We have to substitute these in our original PDE which is this."},{"Start":"03:57.560 ","End":"04:01.050","Text":"If we substitute this for u_xx,"},{"Start":"04:01.050 ","End":"04:02.820","Text":"this for u_xy,"},{"Start":"04:02.820 ","End":"04:04.470","Text":"this for u_yy,"},{"Start":"04:04.470 ","End":"04:08.540","Text":"what we get is the following,"},{"Start":"04:08.540 ","End":"04:13.640","Text":"a straightforward substitution and just going to collect like terms."},{"Start":"04:13.640 ","End":"04:16.220","Text":"Let\u0027s collect the terms for W_Psi Psi,"},{"Start":"04:16.220 ","End":"04:22.610","Text":"we get 9 minus 6 times 3 plus 9. This comes out to 0."},{"Start":"04:22.610 ","End":"04:28.550","Text":"Similarly, we get minus 6 plus 6 for W_Psi Eta."},{"Start":"04:28.550 ","End":"04:31.865","Text":"We also need to substitute the right-hand side."},{"Start":"04:31.865 ","End":"04:34.505","Text":"X is Eta,"},{"Start":"04:34.505 ","End":"04:38.630","Text":"you can see it here and y minus 3x is Psi."},{"Start":"04:38.630 ","End":"04:40.190","Text":"On the right-hand side,"},{"Start":"04:40.190 ","End":"04:42.930","Text":"we get e^Eta sin(Psi)."},{"Start":"04:42.930 ","End":"04:47.585","Text":"Collecting like terms, we just have W_Eta Eta equals"},{"Start":"04:47.585 ","End":"04:53.480","Text":"e^Eta sin(Psi) and this is the canonical form,"},{"Start":"04:53.480 ","End":"04:55.520","Text":"and that\u0027s part b."},{"Start":"04:55.520 ","End":"04:59.900","Text":"Let\u0027s get on with c. We start with this, the canonical form."},{"Start":"04:59.900 ","End":"05:03.890","Text":"We want to solve this and afterwards we\u0027re going to go back to x and y."},{"Start":"05:03.890 ","End":"05:06.935","Text":"Here we need to integrate twice with respect to Eta."},{"Start":"05:06.935 ","End":"05:09.200","Text":"Integrate once we have W_Eta,"},{"Start":"05:09.200 ","End":"05:13.640","Text":"this is a constant sin(Psi) the integral of e^Eta is e^Eta,"},{"Start":"05:13.640 ","End":"05:17.090","Text":"but we get an arbitrary function of Psi."},{"Start":"05:17.090 ","End":"05:25.040","Text":"Then when we integrate again to get that w is again the integral of e^Eta is e^ Eta."},{"Start":"05:25.040 ","End":"05:26.270","Text":"This one is a constant,"},{"Start":"05:26.270 ","End":"05:31.295","Text":"so it multiplies by Eta and we get another arbitrary function of Psi."},{"Start":"05:31.295 ","End":"05:34.000","Text":"We\u0027re ready to compute what u is."},{"Start":"05:34.000 ","End":"05:39.710","Text":"It\u0027s w of Psi and Eta but we\u0027re going to substitute what Psi and Eta are,"},{"Start":"05:39.710 ","End":"05:40.910","Text":"and we\u0027re going to put them in here,"},{"Start":"05:40.910 ","End":"05:46.510","Text":"so e^Eta is e^x sin(y minus 3x)."},{"Start":"05:46.510 ","End":"05:49.365","Text":"Let\u0027s put everywhere, Psi is this,"},{"Start":"05:49.365 ","End":"05:52.700","Text":"Eta is this and this is the expression we get for u(x,"},{"Start":"05:52.700 ","End":"05:57.870","Text":"y) and that concludes this exercise."}],"ID":30700},{"Watched":false,"Name":"Exercise 5","Duration":"6m 35s","ChapterTopicVideoID":29112,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.985","Text":"In this exercise, we have the following linear 2nd-order PDE."},{"Start":"00:05.985 ","End":"00:09.780","Text":"There are 3 parts to this exercise."},{"Start":"00:09.780 ","End":"00:15.795","Text":"First, to show that the equation is hyperbolic as opposed to say parabolic or elliptic."},{"Start":"00:15.795 ","End":"00:18.930","Text":"Secondly, we have to find its canonical form and"},{"Start":"00:18.930 ","End":"00:22.710","Text":"then to find the general solution to the equation."},{"Start":"00:22.710 ","End":"00:24.915","Text":"Let\u0027s start with part A."},{"Start":"00:24.915 ","End":"00:27.410","Text":"From this equation, we identify"},{"Start":"00:27.410 ","End":"00:32.008","Text":"the coefficients according to the notation from the tutorial,"},{"Start":"00:32.008 ","End":"00:33.355","Text":"a_11 is 1,"},{"Start":"00:33.355 ","End":"00:38.550","Text":"a_12 is also 1 because this is 2a_12,"},{"Start":"00:38.550 ","End":"00:41.365","Text":"and a_22 is minus 3."},{"Start":"00:41.365 ","End":"00:43.400","Text":"Now that we have these,"},{"Start":"00:43.400 ","End":"00:45.890","Text":"we can figure out the discriminant,"},{"Start":"00:45.890 ","End":"00:48.080","Text":"which is given by this formula."},{"Start":"00:48.080 ","End":"00:54.035","Text":"This discriminant comes out to be 1^2 minus 1 times minus 3,"},{"Start":"00:54.035 ","End":"00:55.865","Text":"which is 4,"},{"Start":"00:55.865 ","End":"00:59.990","Text":"which is positive and positive means hyperbolic,"},{"Start":"00:59.990 ","End":"01:02.030","Text":"0 is parabolic,"},{"Start":"01:02.030 ","End":"01:04.670","Text":"and negative is elliptic."},{"Start":"01:04.670 ","End":"01:06.545","Text":"Now on to part B,"},{"Start":"01:06.545 ","End":"01:11.150","Text":"the characteristic ordinary differential equation is the following,"},{"Start":"01:11.150 ","End":"01:15.950","Text":"where Delta, we\u0027ve computed it, it\u0027s general this."},{"Start":"01:15.950 ","End":"01:21.395","Text":"In our case what we get is 1 plus or minus root 4/1,"},{"Start":"01:21.395 ","End":"01:28.400","Text":"so it\u0027s 1 plus or minus 2 and we can solve this just by integrating both sides."},{"Start":"01:28.400 ","End":"01:38.835","Text":"We get y is this times x plus c. C is y minus 1 plus or minus 2x,"},{"Start":"01:38.835 ","End":"01:41.360","Text":"really should break it up into 2."},{"Start":"01:41.360 ","End":"01:42.980","Text":"If we have a plus here,"},{"Start":"01:42.980 ","End":"01:44.930","Text":"then it\u0027s y minus 3x,"},{"Start":"01:44.930 ","End":"01:48.050","Text":"otherwise y plus x and strictly speaking,"},{"Start":"01:48.050 ","End":"01:50.060","Text":"they should be different constants here,"},{"Start":"01:50.060 ","End":"01:52.640","Text":"different family of characteristic curves."},{"Start":"01:52.640 ","End":"01:59.040","Text":"But it\u0027s okay because what we\u0027re going do is take this to be Ksi and this to be Eta,"},{"Start":"01:59.040 ","End":"02:01.955","Text":"and that\u0027s going to be our change of variables."},{"Start":"02:01.955 ","End":"02:06.710","Text":"This change of variables will transform our PDE into a canonical 1."},{"Start":"02:06.710 ","End":"02:09.905","Text":"Probably best to check that the Jacobian has"},{"Start":"02:09.905 ","End":"02:16.385","Text":"a non-zero determinant that guarantees that the proper transformation is invertible."},{"Start":"02:16.385 ","End":"02:20.420","Text":"In this case, Ksi x is minus 3,"},{"Start":"02:20.420 ","End":"02:22.440","Text":"all the other partial derivatives,"},{"Start":"02:22.440 ","End":"02:26.455","Text":"1 and it comes out 0."},{"Start":"02:26.455 ","End":"02:30.200","Text":"If we convert from u of x,"},{"Start":"02:30.200 ","End":"02:31.550","Text":"y to another function,"},{"Start":"02:31.550 ","End":"02:32.900","Text":"w of Ksi and Eta,"},{"Start":"02:32.900 ","End":"02:35.170","Text":"that will be invertible."},{"Start":"02:35.170 ","End":"02:38.570","Text":"Continuing, you want to figure out what is w,"},{"Start":"02:38.570 ","End":"02:41.690","Text":"and this is just a reminder of what the original PDE is,"},{"Start":"02:41.690 ","End":"02:44.690","Text":"will need these partial derivatives,"},{"Start":"02:44.690 ","End":"02:46.865","Text":"u_xx, u_xy and u_yy."},{"Start":"02:46.865 ","End":"02:49.805","Text":"These are the 3 we need because these are the 3 that appear."},{"Start":"02:49.805 ","End":"02:54.380","Text":"We\u0027re going to need all these derivatives of Ksi and Eta with"},{"Start":"02:54.380 ","End":"02:59.195","Text":"respect to x and y. Ksi is this, Eta is this."},{"Start":"02:59.195 ","End":"03:03.215","Text":"We already computed these 4 in the Jacobian."},{"Start":"03:03.215 ","End":"03:05.210","Text":"Then we have the second-order ones,"},{"Start":"03:05.210 ","End":"03:08.830","Text":"and they\u0027re all 0s because these are all constants."},{"Start":"03:08.830 ","End":"03:13.490","Text":"The ones here that I colored in gray are the ones that are 0,"},{"Start":"03:13.490 ","End":"03:18.545","Text":"just makes it easier in the computation just ignore the ones which have a gray in them,"},{"Start":"03:18.545 ","End":"03:23.285","Text":"substitute all these here that will help us do the calculation."},{"Start":"03:23.285 ","End":"03:28.140","Text":"Here we get minus 3^2 is 9w Ksi,"},{"Start":"03:28.140 ","End":"03:34.790","Text":"Ksi, and 1 times minus 3 times 2 is minus 6, 1^2 is 1."},{"Start":"03:34.790 ","End":"03:37.640","Text":"That gives us a 9 minus 6 1."},{"Start":"03:37.640 ","End":"03:40.775","Text":"Similarly with the other 6 coefficients,"},{"Start":"03:40.775 ","End":"03:42.590","Text":"just routine computations,"},{"Start":"03:42.590 ","End":"03:44.570","Text":"so we get these 3."},{"Start":"03:44.570 ","End":"03:50.030","Text":"Now we substitute these in the partial differential equation that we have,"},{"Start":"03:50.030 ","End":"03:53.225","Text":"and what we\u0027ll get is the following."},{"Start":"03:53.225 ","End":"03:56.915","Text":"This plus twice this minus 3 times this."},{"Start":"03:56.915 ","End":"03:59.930","Text":"Now we just have to collect like terms w Ksi,"},{"Start":"03:59.930 ","End":"04:05.560","Text":"Ksi and then w Ksi Eta and then w Eta Eta."},{"Start":"04:05.560 ","End":"04:07.385","Text":"These are the coefficients,"},{"Start":"04:07.385 ","End":"04:10.915","Text":"and 9 minus 6 minus 3 is 0,"},{"Start":"04:10.915 ","End":"04:13.350","Text":"and 1 plus 2 minus 3 is 0."},{"Start":"04:13.350 ","End":"04:14.850","Text":"It\u0027s only the middle one,"},{"Start":"04:14.850 ","End":"04:17.340","Text":"minus 6 minus 4 minus 6,"},{"Start":"04:17.340 ","End":"04:21.005","Text":"comes out to be minus 16,"},{"Start":"04:21.005 ","End":"04:28.085","Text":"just divide by the minus 16 and this is our canonical PDE, very simple."},{"Start":"04:28.085 ","End":"04:31.370","Text":"Partial derivative of w with respect to Ksi and Eta,"},{"Start":"04:31.370 ","End":"04:34.520","Text":"the mixed second-order derivative is 0."},{"Start":"04:34.520 ","End":"04:36.470","Text":"That was part B. Now part C,"},{"Start":"04:36.470 ","End":"04:38.075","Text":"we\u0027re going to actually solve it."},{"Start":"04:38.075 ","End":"04:39.616","Text":"We have this,"},{"Start":"04:39.616 ","End":"04:41.255","Text":"then we integrate,"},{"Start":"04:41.255 ","End":"04:43.400","Text":"let\u0027s say first with respect to Eta,"},{"Start":"04:43.400 ","End":"04:48.710","Text":"we get w Ksi is some function of Ksi,"},{"Start":"04:48.710 ","End":"04:51.065","Text":"like a constant with respect to Eta,"},{"Start":"04:51.065 ","End":"04:53.345","Text":"integrate with respect to Ksi."},{"Start":"04:53.345 ","End":"05:00.710","Text":"The integral of this, we\u0027ll call it big F and we get another arbitrary function of Eta."},{"Start":"05:00.710 ","End":"05:03.770","Text":"We have F of Ksi plus G of Eta."},{"Start":"05:03.770 ","End":"05:05.735","Text":"Now that we have w,"},{"Start":"05:05.735 ","End":"05:09.740","Text":"we want to go back to u(x, y)."},{"Start":"05:09.740 ","End":"05:15.740","Text":"What we do is substitute Ksi equals y minus 3x Eta is y plus x."},{"Start":"05:15.740 ","End":"05:19.640","Text":"Substitute them then we\u0027ve got u(x,y) is F of y minus"},{"Start":"05:19.640 ","End":"05:24.287","Text":"3x plus G of y plus x and that\u0027s the answer."},{"Start":"05:24.287 ","End":"05:25.580","Text":"You can stop here,"},{"Start":"05:25.580 ","End":"05:26.705","Text":"but I\u0027m going to do an extra,"},{"Start":"05:26.705 ","End":"05:29.210","Text":"I\u0027m going to verify that this is the solution."},{"Start":"05:29.210 ","End":"05:31.930","Text":"Let\u0027s see if it satisfies our PDE."},{"Start":"05:31.930 ","End":"05:34.080","Text":"You have optional check."},{"Start":"05:34.080 ","End":"05:39.570","Text":"The equation is u_xx plus 2u_xy minus 3u_yy, we\u0027ll substitute this."},{"Start":"05:39.570 ","End":"05:44.285","Text":"U_xx is going to be F double-prime."},{"Start":"05:44.285 ","End":"05:46.460","Text":"When we differentiate twice with respect to x,"},{"Start":"05:46.460 ","End":"05:51.000","Text":"we get minus 3 appearing twice as a factor, minus 3^2."},{"Start":"05:51.000 ","End":"05:53.150","Text":"Here the inner derivative is 1,"},{"Start":"05:53.150 ","End":"05:55.835","Text":"so no change, just second derivative here."},{"Start":"05:55.835 ","End":"05:59.525","Text":"Similarly, the mixed derivative, x, and y,"},{"Start":"05:59.525 ","End":"06:01.880","Text":"we just get the minus 3 once,"},{"Start":"06:01.880 ","End":"06:05.195","Text":"once with respect to x, once with respect to y."},{"Start":"06:05.195 ","End":"06:08.315","Text":"Then the last one we just have this as is."},{"Start":"06:08.315 ","End":"06:10.980","Text":"Now again, just collect together"},{"Start":"06:10.980 ","End":"06:16.800","Text":"the F prime prime terms and that\u0027s 9 minus 6 minus 3 at 0."},{"Start":"06:16.800 ","End":"06:22.900","Text":"From here we get 1 plus 2 minus 3."},{"Start":"06:22.900 ","End":"06:26.075","Text":"For the G double prime, that\u0027s also 0."},{"Start":"06:26.075 ","End":"06:29.480","Text":"Altogether it\u0027s 0, which is what we wanted."},{"Start":"06:29.480 ","End":"06:32.150","Text":"We\u0027ve done part A, B, and C,"},{"Start":"06:32.150 ","End":"06:36.390","Text":"and we\u0027ve even checked the solution, and that\u0027s it."}],"ID":30701},{"Watched":false,"Name":"Exercise 6","Duration":"5m 47s","ChapterTopicVideoID":29113,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.330","Text":"In this exercise, we have a linear of 2nd order partial differential equation here."},{"Start":"00:06.330 ","End":"00:10.215","Text":"We have to show that this is of the parabolic type,"},{"Start":"00:10.215 ","End":"00:13.020","Text":"we have to find its canonical form,"},{"Start":"00:13.020 ","End":"00:16.845","Text":"and then we have to find a general solution to the equation."},{"Start":"00:16.845 ","End":"00:18.900","Text":"We start with part a,"},{"Start":"00:18.900 ","End":"00:25.050","Text":"where we identify coefficients a_11, a_12, a_22."},{"Start":"00:25.050 ","End":"00:31.080","Text":"They\u0027re all 1 as the a_12 is half of this coefficient."},{"Start":"00:31.080 ","End":"00:37.550","Text":"Then we compute the discriminant Delta using this formula and it comes out to be 0,"},{"Start":"00:37.550 ","End":"00:40.280","Text":"which means that we have a parabolic equation."},{"Start":"00:40.280 ","End":"00:45.085","Text":"Bigger than 0 is hyperbolic and less than 0 is elliptic."},{"Start":"00:45.085 ","End":"00:52.190","Text":"We go on to part b and we start with the characteristic ODE which is this,"},{"Start":"00:52.190 ","End":"00:55.280","Text":"dy by dx is, well, that\u0027s written."},{"Start":"00:55.280 ","End":"00:57.320","Text":"We know that Delta is 0,"},{"Start":"00:57.320 ","End":"01:00.070","Text":"so we\u0027re only going to get 1 solution here,"},{"Start":"01:00.070 ","End":"01:03.630","Text":"and dy by dx is equal to 1,"},{"Start":"01:03.630 ","End":"01:06.650","Text":"so integrating it y equals x plus c,"},{"Start":"01:06.650 ","End":"01:10.580","Text":"put the c on 1 side and we have c equals y minus x."},{"Start":"01:10.580 ","End":"01:14.190","Text":"When we only have 1 solution and for the change"},{"Start":"01:14.190 ","End":"01:17.765","Text":"of variables to take Psi equals this and for Eta,"},{"Start":"01:17.765 ","End":"01:20.105","Text":"we can choose x or y."},{"Start":"01:20.105 ","End":"01:27.050","Text":"We\u0027ll go with x. Sometimes we skip the check but really should check the Jacobian"},{"Start":"01:27.050 ","End":"01:30.584","Text":"to make sure that it has a non-zero determinant"},{"Start":"01:30.584 ","End":"01:34.220","Text":"and then this is a proper reversible mapping."},{"Start":"01:34.220 ","End":"01:37.580","Text":"We can get back in the end from Psi Eta to x,"},{"Start":"01:37.580 ","End":"01:40.039","Text":"y, it comes out to be non-zero,"},{"Start":"01:40.039 ","End":"01:41.300","Text":"so we\u0027re fine there."},{"Start":"01:41.300 ","End":"01:44.930","Text":"That means that we can start looking for w(Psi,"},{"Start":"01:44.930 ","End":"01:48.035","Text":"Eta), which is equivalent to u(x, y)."},{"Start":"01:48.035 ","End":"01:52.240","Text":"This will be in canonical form which will be easier to solve and"},{"Start":"01:52.240 ","End":"01:56.795","Text":"after we\u0027ve solved this we go back to x and y, that\u0027s the idea."},{"Start":"01:56.795 ","End":"02:02.205","Text":"Our original PDE is this and we\u0027ll need to go from u_xx,"},{"Start":"02:02.205 ","End":"02:09.155","Text":"u_xy, and so on to the derivatives of w with respect to Psi and Eta."},{"Start":"02:09.155 ","End":"02:12.425","Text":"I\u0027m going to use the formulas we had in the tutorial."},{"Start":"02:12.425 ","End":"02:14.030","Text":"Usually we just needed these 3,"},{"Start":"02:14.030 ","End":"02:16.790","Text":"but this time we\u0027ll need all 5 of these because we have u_xx,"},{"Start":"02:16.790 ","End":"02:19.195","Text":"u_xy, u_yy, u_x and u_y."},{"Start":"02:19.195 ","End":"02:22.070","Text":"These are the formulas just copied them."},{"Start":"02:22.070 ","End":"02:24.680","Text":"Here they are again."},{"Start":"02:24.680 ","End":"02:28.935","Text":"What we\u0027ll need is Psi_x, Psi_y,"},{"Start":"02:28.935 ","End":"02:37.335","Text":"Psi_xx, all these we\u0027re going to evaluate from Psi in terms of y and x."},{"Start":"02:37.335 ","End":"02:41.615","Text":"We already computed 4 of the partial derivatives,"},{"Start":"02:41.615 ","End":"02:44.450","Text":"we need the 2nd order ones."},{"Start":"02:44.450 ","End":"02:48.110","Text":"They\u0027re all zero because we have constants here."},{"Start":"02:48.110 ","End":"02:50.956","Text":"We have now the complete list of 1st and 2nd"},{"Start":"02:50.956 ","End":"02:54.785","Text":"order partial derivatives, the ones that are 0,"},{"Start":"02:54.785 ","End":"02:58.210","Text":"I shaded in gray to make it easier to compute,"},{"Start":"02:58.210 ","End":"03:02.900","Text":"so we don\u0027t have to take any term which has a gray part."},{"Start":"03:02.900 ","End":"03:11.280","Text":"Psi_x is minus 1, Eta_x is 1."},{"Start":"03:11.280 ","End":"03:16.590","Text":"I did it for each of these rows and gray for 0s."},{"Start":"03:16.590 ","End":"03:20.190","Text":"From this all we can get,"},{"Start":"03:20.190 ","End":"03:24.045","Text":"u_x is w_Psi times minus 1,"},{"Start":"03:24.045 ","End":"03:26.835","Text":"that\u0027s minus w_Psi plus w_Eta."},{"Start":"03:26.835 ","End":"03:28.320","Text":"This one is 0,"},{"Start":"03:28.320 ","End":"03:30.150","Text":"so we just have 1w_Psi."},{"Start":"03:30.150 ","End":"03:33.600","Text":"We get minus 1^2 is 1."},{"Start":"03:33.600 ","End":"03:41.950","Text":"Here, 2 times 1 times minus 1 is minus 2 and here we have 1^2. You get the idea."},{"Start":"03:41.950 ","End":"03:43.774","Text":"That we have all these,"},{"Start":"03:43.774 ","End":"03:51.105","Text":"we can substitute them in our PDE and what we get is the following,"},{"Start":"03:51.105 ","End":"03:58.725","Text":"like u_x, we copy this piece here then ux_y this, so on."},{"Start":"03:58.725 ","End":"04:06.935","Text":"Then collect like terms and everything cancels except for Eta plus w_Eta."},{"Start":"04:06.935 ","End":"04:13.125","Text":"For example, w_Psi Psi minus twice plus once is 0."},{"Start":"04:13.125 ","End":"04:14.760","Text":"For w_Psi Eta,"},{"Start":"04:14.760 ","End":"04:16.155","Text":"we have minus 2,"},{"Start":"04:16.155 ","End":"04:18.230","Text":"and from here plus 2 also cancels."},{"Start":"04:18.230 ","End":"04:19.790","Text":"Everything cancels except this."},{"Start":"04:19.790 ","End":"04:23.000","Text":"This is the canonical PDE."},{"Start":"04:23.000 ","End":"04:26.210","Text":"We\u0027ll continue to part c. We start with"},{"Start":"04:26.210 ","End":"04:33.050","Text":"this equation and we use the techniques of ordinary differential equations."},{"Start":"04:33.050 ","End":"04:37.725","Text":"We can use an integrating factor and the integrating factor will be e^Eta."},{"Start":"04:37.725 ","End":"04:40.285","Text":"If we multiply everything by e^Eta,"},{"Start":"04:40.285 ","End":"04:42.725","Text":"this whole thing will be the derivative of something."},{"Start":"04:42.725 ","End":"04:46.100","Text":"It\u0027ll be the derivative of e^Eta w_Eta."},{"Start":"04:46.100 ","End":"04:53.745","Text":"Integrate this and we get that e^Eta w_Eta is some function of Psi."},{"Start":"04:53.745 ","End":"05:00.370","Text":"We want to solve this bring the e^Eta to the other side and it\u0027s e^minus Eta."},{"Start":"05:00.380 ","End":"05:03.665","Text":"F(Psi) is a constant as far as eta is concerned,"},{"Start":"05:03.665 ","End":"05:05.765","Text":"we just integrate e^minus Eta,"},{"Start":"05:05.765 ","End":"05:08.450","Text":"which is minus e^minus Eta."},{"Start":"05:08.450 ","End":"05:14.855","Text":"This constant, which is f(Psi) stays and then we need another constant,"},{"Start":"05:14.855 ","End":"05:17.530","Text":"meaning another function of Psi."},{"Start":"05:17.530 ","End":"05:22.250","Text":"This is the solution for w in terms of Eta and Psi."},{"Start":"05:22.250 ","End":"05:23.960","Text":"But we want u(x,"},{"Start":"05:23.960 ","End":"05:25.955","Text":"y) in terms of x and y,"},{"Start":"05:25.955 ","End":"05:28.445","Text":"not terms of Psi and Eta."},{"Start":"05:28.445 ","End":"05:33.680","Text":"We remember that Psi equals y minus x and Eta equals x."},{"Start":"05:33.680 ","End":"05:36.620","Text":"Just substitute those in this equation,"},{"Start":"05:36.620 ","End":"05:39.215","Text":"like here Eta is x,"},{"Start":"05:39.215 ","End":"05:42.185","Text":"here Psi is y minus x, here also."},{"Start":"05:42.185 ","End":"05:48.100","Text":"This is the solution that we get and that concludes this exercise."}],"ID":30702},{"Watched":false,"Name":"Exercise 7abc","Duration":"4m 50s","ChapterTopicVideoID":29114,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.015","Text":"In this exercise, we have a linear second-order partial differential equation,"},{"Start":"00:06.015 ","End":"00:08.655","Text":"a domain y bigger than 0."},{"Start":"00:08.655 ","End":"00:10.065","Text":"There are 4 parts;"},{"Start":"00:10.065 ","End":"00:12.690","Text":"show that the equation is hyperbolic,"},{"Start":"00:12.690 ","End":"00:14.550","Text":"find its canonical form,"},{"Start":"00:14.550 ","End":"00:16.650","Text":"find the general solution,"},{"Start":"00:16.650 ","End":"00:23.055","Text":"and then find a particular solution which satisfies the following initial conditions."},{"Start":"00:23.055 ","End":"00:27.030","Text":"We start by identifying the coefficients a_ 11,"},{"Start":"00:27.030 ","End":"00:28.920","Text":"a_ 12, a_ 22,"},{"Start":"00:28.920 ","End":"00:31.170","Text":"like in the tutorial."},{"Start":"00:31.170 ","End":"00:34.470","Text":"This comes out to be a_ 11 is 1,"},{"Start":"00:34.470 ","End":"00:37.095","Text":"a_ 12 is 0, a_ 22 is minus y^2."},{"Start":"00:37.095 ","End":"00:38.910","Text":"Then we compute the discriminant,"},{"Start":"00:38.910 ","End":"00:45.005","Text":"which is given by this formula and is therefore equal to y^2,"},{"Start":"00:45.005 ","End":"00:49.666","Text":"which is positive and so the equation is hyperbolic."},{"Start":"00:49.666 ","End":"00:51.365","Text":"That\u0027s Part A,"},{"Start":"00:51.365 ","End":"00:53.435","Text":"now let\u0027s get on to Part B."},{"Start":"00:53.435 ","End":"00:56.780","Text":"We start with the characteristic ODE that dy by"},{"Start":"00:56.780 ","End":"01:01.295","Text":"dx is a_ 12 plus or minus root Delta over a_ 11."},{"Start":"01:01.295 ","End":"01:03.395","Text":"Sometimes it\u0027s written in the full form,"},{"Start":"01:03.395 ","End":"01:05.120","Text":"but we already computed Delta,"},{"Start":"01:05.120 ","End":"01:07.226","Text":"so no need for this."},{"Start":"01:07.226 ","End":"01:13.220","Text":"This is dy by dx is 0 plus or minus root of y^2 over 1."},{"Start":"01:13.220 ","End":"01:16.315","Text":"Y prime is plus or minus y,"},{"Start":"01:16.315 ","End":"01:18.200","Text":"and that gives us 2 equations."},{"Start":"01:18.200 ","End":"01:19.790","Text":"Y prime plus y is 0,"},{"Start":"01:19.790 ","End":"01:21.725","Text":"y prime minus y is 0."},{"Start":"01:21.725 ","End":"01:25.430","Text":"This is straightforward ordinary differential equations,"},{"Start":"01:25.430 ","End":"01:31.010","Text":"and we know the solution to this one is y equals some constant e^-x,"},{"Start":"01:31.010 ","End":"01:33.485","Text":"this one is some constant e^-x."},{"Start":"01:33.485 ","End":"01:35.525","Text":"Column c_ 1 and c_ 2."},{"Start":"01:35.525 ","End":"01:42.060","Text":"Then we can isolate the constants and now it\u0027s in the form we want for Xi and Eta."},{"Start":"01:42.060 ","End":"01:44.390","Text":"This is a substitution we\u0027ll make."},{"Start":"01:44.390 ","End":"01:49.040","Text":"This is what we\u0027ll convert our PDE into a canonical one."},{"Start":"01:49.040 ","End":"01:53.165","Text":"If you have time, check that the Jacobian has a non-zero determinant."},{"Start":"01:53.165 ","End":"01:56.090","Text":"It\u0027s optional, you\u0027re supposed to do it."},{"Start":"01:56.090 ","End":"01:59.945","Text":"Y is bigger than 0 so 2y is not 0."},{"Start":"01:59.945 ","End":"02:03.799","Text":"Next, we have to figure out what is the w(Xi,"},{"Start":"02:03.799 ","End":"02:06.840","Text":"Eta) that\u0027s equivalent to our u(x,"},{"Start":"02:06.840 ","End":"02:09.280","Text":"y) after the substitution."},{"Start":"02:09.280 ","End":"02:12.505","Text":"Also let\u0027s remember what the original PDE is."},{"Start":"02:12.505 ","End":"02:16.460","Text":"We\u0027ll need to convert the partial derivatives"},{"Start":"02:16.460 ","End":"02:20.449","Text":"into partial derivatives of w. In the tutorial,"},{"Start":"02:20.449 ","End":"02:26.680","Text":"we had the clip on derivatives and these are the three we\u0027ll need here."},{"Start":"02:26.680 ","End":"02:28.930","Text":"We\u0027ll need u_ y, u_ yy,"},{"Start":"02:28.930 ","End":"02:32.675","Text":"and u_ xx and here\u0027s what they are."},{"Start":"02:32.675 ","End":"02:36.350","Text":"We can compute the partial derivatives of Xi and Eta."},{"Start":"02:36.350 ","End":"02:41.707","Text":"We already had these four for the Jacobian and these are the other four that we need."},{"Start":"02:41.707 ","End":"02:47.540","Text":"Then right down on each one what it\u0027s equal to."},{"Start":"02:47.540 ","End":"02:50.500","Text":"These are the ones which are equal to 0,"},{"Start":"02:50.500 ","End":"02:53.165","Text":"Xi_ yy and Eta_ yy."},{"Start":"02:53.165 ","End":"02:56.815","Text":"After we substitute these,"},{"Start":"02:56.815 ","End":"02:59.750","Text":"we get the following three equations,"},{"Start":"02:59.750 ","End":"03:01.480","Text":"and the original PDE,"},{"Start":"03:01.480 ","End":"03:03.610","Text":"let\u0027s copy it down here."},{"Start":"03:03.610 ","End":"03:06.340","Text":"Substitute these three here,"},{"Start":"03:06.340 ","End":"03:08.800","Text":"and we get the following."},{"Start":"03:08.800 ","End":"03:16.000","Text":"Let\u0027s collect some like terms for each of the partial derivatives of w,"},{"Start":"03:16.000 ","End":"03:24.750","Text":"w_ XiXi is y^2 e^2x minus y^2 e^2x and similarly for the others."},{"Start":"03:24.750 ","End":"03:28.110","Text":"Note that they\u0027re all 0 except for this one."},{"Start":"03:28.110 ","End":"03:30.135","Text":"The only one that\u0027s not 0 is this,"},{"Start":"03:30.135 ","End":"03:33.220","Text":"which is minus 4y^2 w_ XiEta,"},{"Start":"03:33.220 ","End":"03:40.175","Text":"divide both sides by minus 4y^2 and this is our canonical PDE,"},{"Start":"03:40.175 ","End":"03:42.940","Text":"and that\u0027s Part B."},{"Start":"03:42.940 ","End":"03:45.615","Text":"Now, Part C,"},{"Start":"03:45.615 ","End":"03:49.250","Text":"we continue with this canonical PDE,"},{"Start":"03:49.250 ","End":"03:57.170","Text":"integrate with respect to Eta and we get that w_ Xi is an arbitrary function of Xi."},{"Start":"03:57.170 ","End":"04:01.160","Text":"Then we integrate with respect to Xi and we get the w is the indefinite"},{"Start":"04:01.160 ","End":"04:05.989","Text":"integral of f(Xi) plus an arbitrary function of Eta."},{"Start":"04:05.989 ","End":"04:12.140","Text":"Let\u0027s rename these, that the integral of little f be big F,"},{"Start":"04:12.140 ","End":"04:16.735","Text":"and rename little g to big G, so it matches."},{"Start":"04:16.735 ","End":"04:20.985","Text":"We have that w is F(Xi) plus G(Eta)."},{"Start":"04:20.985 ","End":"04:26.000","Text":"Now you want to get back from w to u. U(x,"},{"Start":"04:26.000 ","End":"04:27.380","Text":"y) is w(Psi, Eta)."},{"Start":"04:27.380 ","End":"04:32.075","Text":"We just have to substitute Xi and Eta for what they are in terms of x and y."},{"Start":"04:32.075 ","End":"04:34.430","Text":"If we do that, we get u(x,"},{"Start":"04:34.430 ","End":"04:42.390","Text":"y) is F(ye^x) plus G(ye^-x),"},{"Start":"04:42.390 ","End":"04:44.580","Text":"and that the answer to Part"},{"Start":"04:44.580 ","End":"04:50.280","Text":"C. That concludes this clip and we\u0027ll continue in the following."}],"ID":30703},{"Watched":false,"Name":"Exercise 7d","Duration":"2m 35s","ChapterTopicVideoID":29115,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.460","Text":"Continuing with this exercise,"},{"Start":"00:02.460 ","End":"00:03.930","Text":"we\u0027ve just completed a, b,"},{"Start":"00:03.930 ","End":"00:12.280","Text":"and c. We\u0027re now onto d. The solution from Part C was u(x,"},{"Start":"00:12.280 ","End":"00:15.840","Text":"y) is some function of y e to the x plus another function of"},{"Start":"00:15.840 ","End":"00:20.130","Text":"y e to the minus x will need u_x."},{"Start":"00:20.130 ","End":"00:22.005","Text":"Let\u0027s compute that."},{"Start":"00:22.005 ","End":"00:26.805","Text":"The derivative of this with respect to x is f prime times the inner-derivative,"},{"Start":"00:26.805 ","End":"00:29.594","Text":"which is just y e to the x itself."},{"Start":"00:29.594 ","End":"00:36.855","Text":"Here the derivative with respect to x is y e to the minus x times minus 1."},{"Start":"00:36.855 ","End":"00:38.470","Text":"It\u0027s a minus here."},{"Start":"00:38.470 ","End":"00:40.730","Text":"Let\u0027s take this condition, u(0,"},{"Start":"00:40.730 ","End":"00:47.240","Text":"y) is 2y^2 and what we can do is plug in 0"},{"Start":"00:47.240 ","End":"00:54.015","Text":"here for x so e to the x comes out 1 and e to the minus x is 1."},{"Start":"00:54.015 ","End":"00:59.710","Text":"What we get is that 2y^2 is f(y) plus g(y)."},{"Start":"00:59.710 ","End":"01:02.990","Text":"Now we\u0027ll take the other condition,"},{"Start":"01:02.990 ","End":"01:08.615","Text":"and we already have ux computed here so if we let x = 0,"},{"Start":"01:08.615 ","End":"01:10.280","Text":"e to the x comes out 1,"},{"Start":"01:10.280 ","End":"01:12.895","Text":"e to the minus x comes out 1."},{"Start":"01:12.895 ","End":"01:19.095","Text":"What we get is that 0 is f\u0027(y) times y minus g\u0027(y) times y."},{"Start":"01:19.095 ","End":"01:22.350","Text":"Turn left and right and also divide both sides by y."},{"Start":"01:22.350 ","End":"01:23.885","Text":"This is what we have."},{"Start":"01:23.885 ","End":"01:27.470","Text":"Now, integrating this with respect to y,"},{"Start":"01:27.470 ","End":"01:32.275","Text":"we get that f(y) minus g(y) is a constant."},{"Start":"01:32.275 ","End":"01:37.990","Text":"We also have from here that f(y) plus g(y) is 2y^2."},{"Start":"01:37.990 ","End":"01:40.700","Text":"Now that we have the difference on the sum,"},{"Start":"01:40.700 ","End":"01:43.175","Text":"we can compute each one."},{"Start":"01:43.175 ","End":"01:47.630","Text":"F(y) will be this plus this over 2 and"},{"Start":"01:47.630 ","End":"01:52.070","Text":"g(y) will be this equation minus this equation over 2."},{"Start":"01:52.070 ","End":"01:53.855","Text":"We have f(y) and g(y)."},{"Start":"01:53.855 ","End":"01:59.600","Text":"Y is just any parameter you could switch it for any other variable,"},{"Start":"01:59.600 ","End":"02:02.630","Text":"u(x, y) is the following."},{"Start":"02:02.630 ","End":"02:09.350","Text":"We can just substitute y e to the x in place of y in this equation,"},{"Start":"02:09.350 ","End":"02:13.774","Text":"and y e to the minus x in place of y here,"},{"Start":"02:13.774 ","End":"02:17.230","Text":"and what we get is the following."},{"Start":"02:17.230 ","End":"02:21.644","Text":"The c over 2 cancels, I\u0027m squaring these,"},{"Start":"02:21.644 ","End":"02:23.775","Text":"we get that u(x,"},{"Start":"02:23.775 ","End":"02:28.860","Text":"y) is y^2 e to the 2 x plus y^2 e to the minus 2 x."},{"Start":"02:28.860 ","End":"02:36.100","Text":"That is the answer to Part D and this whole exercise and the clip."}],"ID":30704},{"Watched":false,"Name":"Exercise 8ab","Duration":"8m 53s","ChapterTopicVideoID":29116,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.500","Text":"This exercise is a bit different than the previous one that you\u0027ll see."},{"Start":"00:04.500 ","End":"00:06.735","Text":"We\u0027re given an equation,"},{"Start":"00:06.735 ","End":"00:13.014","Text":"second-order linear PDE in x and y and there are 4 parts."},{"Start":"00:13.014 ","End":"00:16.725","Text":"Part a, we have to show the equation is hyperbolic."},{"Start":"00:16.725 ","End":"00:20.205","Text":"In part b, we have to find its canonical form,"},{"Start":"00:20.205 ","End":"00:28.240","Text":"which I\u0027ll remind you is w_PsiEta plus terms of lower order equals 0."},{"Start":"00:28.240 ","End":"00:31.370","Text":"This is called the first canonical form,"},{"Start":"00:31.370 ","End":"00:33.290","Text":"and just like you\u0027d expect,"},{"Start":"00:33.290 ","End":"00:35.195","Text":"it might also be a second."},{"Start":"00:35.195 ","End":"00:38.675","Text":"Yes, the second canonical form is like this."},{"Start":"00:38.675 ","End":"00:40.820","Text":"Instead of the w_PsiEta,"},{"Start":"00:40.820 ","End":"00:46.840","Text":"we have w_PsiPsi minus w_EtaEta and there are only 2."},{"Start":"00:46.840 ","End":"00:49.565","Text":"This is just for the hyperbolic case,"},{"Start":"00:49.565 ","End":"00:52.925","Text":"not for the elliptic and the parabolic,"},{"Start":"00:52.925 ","End":"00:54.895","Text":"which only have 1 canonical form."},{"Start":"00:54.895 ","End":"00:56.410","Text":"In part c,"},{"Start":"00:56.410 ","End":"00:59.900","Text":"we\u0027re going to convert from the first canonical form to"},{"Start":"00:59.900 ","End":"01:04.310","Text":"the second canonical form by means of a change of variables."},{"Start":"01:04.310 ","End":"01:09.710","Text":"In part B, we do a change of variables to go from xy to PsiEta,"},{"Start":"01:09.710 ","End":"01:15.080","Text":"and the function goes from u to w. Then we\u0027ll do another change of variables."},{"Start":"01:15.080 ","End":"01:20.465","Text":"We\u0027ll go from PsiEta to big X and big T as follows,"},{"Start":"01:20.465 ","End":"01:25.220","Text":"and then we\u0027ll get a function v that will bring it to this form."},{"Start":"01:25.220 ","End":"01:26.570","Text":"Well, that\u0027s what we got to show."},{"Start":"01:26.570 ","End":"01:32.540","Text":"In part d, we\u0027ll use a third substitution in order to simplify this even further."},{"Start":"01:32.540 ","End":"01:34.055","Text":"Now, it looks pretty simple,"},{"Start":"01:34.055 ","End":"01:37.415","Text":"but there could be quite a bit in this ellipsis."},{"Start":"01:37.415 ","End":"01:41.490","Text":"The dot-dot-dot could have something v_t and"},{"Start":"01:41.490 ","End":"01:46.540","Text":"something v_x and something v and then just some function of t and x."},{"Start":"01:46.540 ","End":"01:51.425","Text":"What we\u0027re going to do is get into a simpler form where all that is besides"},{"Start":"01:51.425 ","End":"01:56.030","Text":"the difference of the second derivatives is c times the function,"},{"Start":"01:56.030 ","End":"01:58.265","Text":"in this case zee or zed,"},{"Start":"01:58.265 ","End":"02:04.400","Text":"depending if you\u0027re English or American and this is the substitution that have V(X,"},{"Start":"02:04.400 ","End":"02:06.920","Text":"T) we have z(Z,"},{"Start":"02:06.920 ","End":"02:14.180","Text":"T) times e^Gamma x plus Delta t. Gamma and Delta are constants."},{"Start":"02:14.180 ","End":"02:16.910","Text":"We\u0027ll actually give them numerical values,"},{"Start":"02:16.910 ","End":"02:20.630","Text":"but we leave them as is until we get to a certain point,"},{"Start":"02:20.630 ","End":"02:22.190","Text":"and then we\u0027ll assign the values,"},{"Start":"02:22.190 ","End":"02:25.280","Text":"whatever is convenient to get to this form."},{"Start":"02:25.280 ","End":"02:28.220","Text":"I should say c is real constant."},{"Start":"02:28.220 ","End":"02:29.840","Text":"Anyway, let\u0027s start with a."},{"Start":"02:29.840 ","End":"02:34.120","Text":"As usual, we identify the coefficients a_11, a_12,"},{"Start":"02:34.120 ","End":"02:38.370","Text":"and a_22 from here and a_11 is 1,"},{"Start":"02:38.370 ","End":"02:39.570","Text":"a_12 is 1/2,"},{"Start":"02:39.570 ","End":"02:40.995","Text":"this coefficient is a 1/2,"},{"Start":"02:40.995 ","End":"02:42.690","Text":"a_22 is minus 6,"},{"Start":"02:42.690 ","End":"02:46.920","Text":"then the discriminant usual formula comes out to"},{"Start":"02:46.920 ","End":"02:51.985","Text":"be 1/2^2 minus 1 times minus 6, which is 61/4."},{"Start":"02:51.985 ","End":"02:55.650","Text":"61/4 is 25/4. Anyway,"},{"Start":"02:55.650 ","End":"02:59.075","Text":"it\u0027s positive and therefore hyperbolic."},{"Start":"02:59.075 ","End":"03:04.235","Text":"Now we come to part b and you probably remember the technique."},{"Start":"03:04.235 ","End":"03:07.685","Text":"We solve the ordinary differential equation,"},{"Start":"03:07.685 ","End":"03:15.115","Text":"dy by dx equals a_12 plus or minus the square root of Delta over a_11."},{"Start":"03:15.115 ","End":"03:21.405","Text":"We have all the information 1/2 plus or minus square root of 25/4 over 1,"},{"Start":"03:21.405 ","End":"03:25.110","Text":"so it\u0027s a 1/2 plus or minus 5/2."},{"Start":"03:25.110 ","End":"03:27.975","Text":"We take the plus 6/2 is 3."},{"Start":"03:27.975 ","End":"03:29.325","Text":"We take the minus,"},{"Start":"03:29.325 ","End":"03:32.340","Text":"minus 4/2 is minus 2."},{"Start":"03:32.340 ","End":"03:34.100","Text":"There are two possibilities."},{"Start":"03:34.100 ","End":"03:36.710","Text":"Y\u0027 minus 3 is 0,"},{"Start":"03:36.710 ","End":"03:39.530","Text":"y prime plus 2 equals 0."},{"Start":"03:39.530 ","End":"03:42.695","Text":"We integrate both sides with respect to x."},{"Start":"03:42.695 ","End":"03:47.430","Text":"We get y minus 3x equals the constant and here"},{"Start":"03:47.430 ","End":"03:52.670","Text":"also y plus 2x equals a constant, different constant."},{"Start":"03:52.670 ","End":"03:54.305","Text":"Instead of the constants,"},{"Start":"03:54.305 ","End":"03:56.875","Text":"we put Psi and Eta,"},{"Start":"03:56.875 ","End":"03:59.825","Text":"and this will be a transformation."},{"Start":"03:59.825 ","End":"04:04.910","Text":"Let\u0027s do a quick check that it\u0027s proper transformation that is invertible."},{"Start":"04:04.910 ","End":"04:08.425","Text":"We do that by checking the determinant of the Jacobian."},{"Start":"04:08.425 ","End":"04:09.960","Text":"I\u0027ll let you look at that."},{"Start":"04:09.960 ","End":"04:11.570","Text":"It comes out to be non-zero,"},{"Start":"04:11.570 ","End":"04:14.020","Text":"and that\u0027s the important thing."},{"Start":"04:14.020 ","End":"04:16.460","Text":"When we do the conversion,"},{"Start":"04:16.460 ","End":"04:21.095","Text":"that the new function we get in terms of Psi and Eta we\u0027ll call that w. U(x,"},{"Start":"04:21.095 ","End":"04:27.795","Text":"y) is w( Psi and Eta ) where Psi and Eta are these in relation to y and x."},{"Start":"04:27.795 ","End":"04:32.215","Text":"Let\u0027s remember what the original PDE is. This is it."},{"Start":"04:32.215 ","End":"04:34.075","Text":"We\u0027ll need u_xx,"},{"Start":"04:34.075 ","End":"04:36.400","Text":"u_xy, u_yy,"},{"Start":"04:36.400 ","End":"04:43.605","Text":"and u_x in terms of w and its partial derivatives."},{"Start":"04:43.605 ","End":"04:48.890","Text":"\"U itself is just w.\" 4 things we need,"},{"Start":"04:48.890 ","End":"04:50.540","Text":"like I said, this, this, this,"},{"Start":"04:50.540 ","End":"04:55.790","Text":"and this and I took them from the formulas that"},{"Start":"04:55.790 ","End":"05:02.255","Text":"we developed in an earlier tutorial clip called differentiation formulas."},{"Start":"05:02.255 ","End":"05:03.995","Text":"There were 5 of them,"},{"Start":"05:03.995 ","End":"05:06.220","Text":"but we don\u0027t need u_y."},{"Start":"05:06.220 ","End":"05:08.340","Text":"These are the 4, we need."},{"Start":"05:08.340 ","End":"05:12.830","Text":"All these Psi and Eta derivatives with respect to x and y,"},{"Start":"05:12.830 ","End":"05:18.295","Text":"we can compute because we have Psi and Eta in terms of x and y."},{"Start":"05:18.295 ","End":"05:22.020","Text":"We have these 4 partial derivatives minus 3,"},{"Start":"05:22.020 ","End":"05:23.160","Text":"1, 2, 1."},{"Start":"05:23.160 ","End":"05:25.290","Text":"That\u0027s this minus 3, 1, 2, 1,"},{"Start":"05:25.290 ","End":"05:28.085","Text":"and then we need second-order derivatives."},{"Start":"05:28.085 ","End":"05:30.290","Text":"They are all 0 because all these are"},{"Start":"05:30.290 ","End":"05:34.655","Text":"constants might be an idea to cross out all the ones,"},{"Start":"05:34.655 ","End":"05:40.340","Text":"which are 0 that contain one of these 6 second-order derivatives."},{"Start":"05:40.340 ","End":"05:48.290","Text":"Let\u0027s see. This one will be 0 and this one will be 0 because, Psi_xx and Eta_xx."},{"Start":"05:48.290 ","End":"05:53.000","Text":"Let\u0027s see where else we have a second derivative, yep, here,"},{"Start":"05:53.000 ","End":"05:59.805","Text":"meaning the whole thing and here we have Eta_xy."},{"Start":"05:59.805 ","End":"06:02.735","Text":"This one would be 0,"},{"Start":"06:02.735 ","End":"06:06.720","Text":"and this one will be 0 Eta_yy."},{"Start":"06:07.220 ","End":"06:10.540","Text":"Psi_x is minus 3."},{"Start":"06:10.540 ","End":"06:11.870","Text":"We can write that here."},{"Start":"06:11.870 ","End":"06:16.760","Text":"In fact, I already prepared and I wrote some of"},{"Start":"06:16.760 ","End":"06:20.400","Text":"the values of all of these except for"},{"Start":"06:20.400 ","End":"06:24.935","Text":"the zeros above each one so it makes it easier to compute."},{"Start":"06:24.935 ","End":"06:30.070","Text":"Now, let\u0027s see if we can collect like terms and we have 4 formulas,"},{"Start":"06:30.070 ","End":"06:32.625","Text":"yeah there\u0027s also this formula."},{"Start":"06:32.625 ","End":"06:34.530","Text":"We\u0027ll start with this one then."},{"Start":"06:34.530 ","End":"06:41.505","Text":"So u_x is minus 3w_Psi, plus 2w_Eta."},{"Start":"06:41.505 ","End":"06:47.065","Text":"Next one we have minus 3^2 is 9w_PsiPsi."},{"Start":"06:47.065 ","End":"06:49.820","Text":"I think this is fairly straightforward."},{"Start":"06:49.820 ","End":"06:58.510","Text":"I\u0027ll do one more, 2 times 2 times minus 3 is minus 12w_PsiEta."},{"Start":"06:58.510 ","End":"07:00.420","Text":"That\u0027s this one here."},{"Start":"07:00.420 ","End":"07:01.710","Text":"As I say, it\u0027s a routine."},{"Start":"07:01.710 ","End":"07:03.585","Text":"Just to pause and check."},{"Start":"07:03.585 ","End":"07:06.240","Text":"These are the formulas I now get."},{"Start":"07:06.240 ","End":"07:12.050","Text":"Then we want to substitute these in our original PDE here it is again,"},{"Start":"07:12.050 ","End":"07:14.885","Text":"I just colored things to make it easier."},{"Start":"07:14.885 ","End":"07:21.915","Text":"We have to substitute 4 things from here and that gives us the following,"},{"Start":"07:21.915 ","End":"07:29.820","Text":"u_xx we take from here and then plus u_xy, that\u0027s from here,"},{"Start":"07:29.820 ","End":"07:38.415","Text":"minus 6 times u_yy we take from here and so on."},{"Start":"07:38.415 ","End":"07:42.815","Text":"U itself is w collecting like terms."},{"Start":"07:42.815 ","End":"07:44.240","Text":"Let\u0027s just do one of them."},{"Start":"07:44.240 ","End":"07:46.625","Text":"Let\u0027s say we collect w_PsiPsi."},{"Start":"07:46.625 ","End":"07:50.660","Text":"We have 9 of them minus 3 of them,"},{"Start":"07:50.660 ","End":"07:54.825","Text":"that\u0027s 6 minus 6 of them is 0."},{"Start":"07:54.825 ","End":"07:58.770","Text":"There\u0027s no w_PsiPsi term indeed. I\u0027ll do one more."},{"Start":"07:58.770 ","End":"08:03.270","Text":"I\u0027ll do the W_Psi Eta minus 12 minus 1,"},{"Start":"08:03.270 ","End":"08:11.540","Text":"that\u0027s minus 13 minus 6 times 2 minus 13 minus 12 minus 25."},{"Start":"08:11.540 ","End":"08:12.890","Text":"These are the rest of them."},{"Start":"08:12.890 ","End":"08:15.815","Text":"Divide everything by minus 25,"},{"Start":"08:15.815 ","End":"08:20.265","Text":"and it\u0027s a bit neater to move the other things onto the right side."},{"Start":"08:20.265 ","End":"08:28.580","Text":"We have w_PsiEta is minus 1/25 and then 3 minus 2 and plus 1."},{"Start":"08:28.580 ","End":"08:31.955","Text":"Just reverse the signs here because we\u0027ve got a minus outside,"},{"Start":"08:31.955 ","End":"08:37.580","Text":"25 in the denominator and this is of the form we wanted,"},{"Start":"08:37.580 ","End":"08:41.405","Text":"w_PsiEta plus lower order terms equals 0."},{"Start":"08:41.405 ","End":"08:44.510","Text":"Well, you can put this on the left-hand side equals 0,"},{"Start":"08:44.510 ","End":"08:46.055","Text":"it doesn\u0027t really matter."},{"Start":"08:46.055 ","End":"08:50.255","Text":"That concludes part b and this clip,"},{"Start":"08:50.255 ","End":"08:54.449","Text":"and we\u0027ll continue with part c in the following."}],"ID":30705},{"Watched":false,"Name":"Exercise 8cd","Duration":"9m 34s","ChapterTopicVideoID":29117,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Continuing with this exercise,"},{"Start":"00:01.890 ","End":"00:07.710","Text":"we now come to Part c. We\u0027re going to make a change of variables."},{"Start":"00:07.710 ","End":"00:14.010","Text":"X equals Xi plus Eta over 2 and t equals Xi minus Eta over 2."},{"Start":"00:14.010 ","End":"00:17.940","Text":"We reach an equation in this form which"},{"Start":"00:17.940 ","End":"00:22.995","Text":"is the second canonical form for hyperbolic equations."},{"Start":"00:22.995 ","End":"00:27.810","Text":"This was the canonical form that we got with Xi and Eta,"},{"Start":"00:27.810 ","End":"00:29.580","Text":"this is the first canonical form."},{"Start":"00:29.580 ","End":"00:34.035","Text":"We\u0027re going to make this substitution as instructed."},{"Start":"00:34.035 ","End":"00:40.670","Text":"We\u0027re going to call the new function v. After we\u0027ve substituted Xi and Eta in w,"},{"Start":"00:40.670 ","End":"00:43.550","Text":"we\u0027re going to get v(X, T)."},{"Start":"00:43.550 ","End":"00:48.860","Text":"We are going to need w_Xi Eta in terms of the partial derivatives of v,"},{"Start":"00:48.860 ","End":"00:50.975","Text":"we will need w_Xi,"},{"Start":"00:50.975 ","End":"00:53.390","Text":"we will need w_Eta,"},{"Start":"00:53.390 ","End":"00:54.860","Text":"well, and w itself,"},{"Start":"00:54.860 ","End":"00:58.785","Text":"which will just be v. Using the chain rule,"},{"Start":"00:58.785 ","End":"01:02.190","Text":"w_Xi, or we can get Xi in two ways,"},{"Start":"01:02.190 ","End":"01:12.231","Text":"through X and through T. X Xi is 1/2 and T with respect to Xi is also 1/2,"},{"Start":"01:12.231 ","End":"01:15.000","Text":"so that gives us this."},{"Start":"01:15.000 ","End":"01:17.355","Text":"Then with respect to Eta,"},{"Start":"01:17.355 ","End":"01:23.960","Text":"it\u0027s similar except that T with respect to Eta is minus 1/2. We have these 2."},{"Start":"01:23.960 ","End":"01:26.990","Text":"Now we need a second derivative."},{"Start":"01:26.990 ","End":"01:29.885","Text":"We can either differentiate this with respect to Eta or"},{"Start":"01:29.885 ","End":"01:32.735","Text":"this with respect to Xi, doesn\u0027t matter."},{"Start":"01:32.735 ","End":"01:35.690","Text":"Let\u0027s do this one with respect to Eta."},{"Start":"01:35.690 ","End":"01:39.280","Text":"We have the 1/2 here and we\u0027ll apply the chain rule to v_X."},{"Start":"01:39.280 ","End":"01:43.050","Text":"So v_X with respect to X,"},{"Start":"01:43.050 ","End":"01:46.710","Text":"X with respect to Eta and then v_X X_Eta."},{"Start":"01:46.710 ","End":"01:48.570","Text":"That\u0027s this one."},{"Start":"01:48.570 ","End":"01:50.580","Text":"The second part, again,"},{"Start":"01:50.580 ","End":"01:53.340","Text":"the 1/2 and the v_T with respect to Eta."},{"Start":"01:53.340 ","End":"02:02.690","Text":"We can go either through X and then X_Eta or we can do v_T with T and then T with eta."},{"Start":"02:02.690 ","End":"02:04.880","Text":"Anyways, just the chain rule."},{"Start":"02:04.880 ","End":"02:07.275","Text":"We already did, X_Eta."},{"Start":"02:07.275 ","End":"02:09.255","Text":"We said that that\u0027s equal to 0.5."},{"Start":"02:09.255 ","End":"02:15.060","Text":"This and this are both equal to 1/2 here and here."},{"Start":"02:15.060 ","End":"02:20.430","Text":"T with respect to Eta is minus"},{"Start":"02:20.430 ","End":"02:27.330","Text":"1/2 and here also T with respect to Eta is minus 1/2."},{"Start":"02:27.330 ","End":"02:31.285","Text":"Then we can rearrange, collect like terms."},{"Start":"02:31.285 ","End":"02:40.130","Text":"We get 1/4 v_XX from here and we get minus a quarter v_TT from here."},{"Start":"02:40.130 ","End":"02:45.510","Text":"These two cancel each other out because we have a quarter v_TX"},{"Start":"02:45.510 ","End":"02:51.290","Text":"minus 1/4v_XT and these two mixed second derivatives are equal."},{"Start":"02:51.290 ","End":"02:54.320","Text":"We have the 3 that we\u0027re looking for, this, this,"},{"Start":"02:54.320 ","End":"02:58.160","Text":"and this that we need in here."},{"Start":"02:58.160 ","End":"03:03.275","Text":"This is what just scrolled offscreen this."},{"Start":"03:03.275 ","End":"03:08.210","Text":"We can substitute w_Xi Eta from here,"},{"Start":"03:08.210 ","End":"03:14.380","Text":"w_Xi from here, and w_x goes here."},{"Start":"03:14.380 ","End":"03:17.144","Text":"I say w is v,"},{"Start":"03:17.144 ","End":"03:18.630","Text":"that\u0027s the last one here."},{"Start":"03:18.630 ","End":"03:20.550","Text":"So w_Xi Eta,"},{"Start":"03:20.550 ","End":"03:23.805","Text":"which is this from here,"},{"Start":"03:23.805 ","End":"03:30.630","Text":"equals and then minus 1 over 25 3 times what was here,"},{"Start":"03:30.630 ","End":"03:36.080","Text":"and minus twice w Eta from here."},{"Start":"03:36.080 ","End":"03:37.625","Text":"We get to this,"},{"Start":"03:37.625 ","End":"03:39.785","Text":"just a bit of algebra now,"},{"Start":"03:39.785 ","End":"03:43.795","Text":"multiply both sides by 4."},{"Start":"03:43.795 ","End":"03:46.905","Text":"The 4 will cancel with the 2 everywhere,"},{"Start":"03:46.905 ","End":"03:48.360","Text":"but it\u0027s still 2 remaining,"},{"Start":"03:48.360 ","End":"03:51.710","Text":"so that 2 can be put here above the 25."},{"Start":"03:51.710 ","End":"03:57.720","Text":"Then we have 3v_X plus 3v_T and here minus 2v_X plus"},{"Start":"03:57.720 ","End":"04:04.425","Text":"2v_T and then the 2 also goes on the v. That gives us our answer."},{"Start":"04:04.425 ","End":"04:10.400","Text":"We can just change the order here and put a minus here and collect."},{"Start":"04:10.400 ","End":"04:11.820","Text":"We have v_X,"},{"Start":"04:11.820 ","End":"04:13.800","Text":"3 minus 2 is one of them,"},{"Start":"04:13.800 ","End":"04:15.780","Text":"v_T, we have 3 plus 2,"},{"Start":"04:15.780 ","End":"04:18.000","Text":"so there\u0027s 5 of them, and 2v."},{"Start":"04:18.000 ","End":"04:22.245","Text":"That\u0027s part c. Now we come to the last part,"},{"Start":"04:22.245 ","End":"04:24.240","Text":"d. In part c,"},{"Start":"04:24.240 ","End":"04:26.925","Text":"we got this equation."},{"Start":"04:26.925 ","End":"04:30.555","Text":"Now we\u0027re going to do one more substitution."},{"Start":"04:30.555 ","End":"04:37.960","Text":"We\u0027ll staying with X and T. We\u0027re just going to go from a function v to a function z by"},{"Start":"04:37.960 ","End":"04:45.560","Text":"multiplying by e^Gamma X plus Delta T. Gamma and Delta will choose later."},{"Start":"04:45.560 ","End":"04:46.880","Text":"There\u0027ll be actual numbers,"},{"Start":"04:46.880 ","End":"04:49.720","Text":"but we\u0027ll just leave it in general form for now."},{"Start":"04:49.720 ","End":"04:52.190","Text":"We\u0027re going to convert our equation,"},{"Start":"04:52.190 ","End":"04:55.670","Text":"this one into one of this form,"},{"Start":"04:55.670 ","End":"04:57.370","Text":"which will be simpler."},{"Start":"04:57.370 ","End":"05:01.650","Text":"Here, besides the something times v,"},{"Start":"05:01.650 ","End":"05:03.460","Text":"we have a v_T and v_X."},{"Start":"05:03.460 ","End":"05:08.030","Text":"But here we just have c times the function itself, z."},{"Start":"05:08.030 ","End":"05:10.915","Text":"As a matter of fact,"},{"Start":"05:10.915 ","End":"05:13.130","Text":"just in the US do they say z."},{"Start":"05:13.130 ","End":"05:17.215","Text":"In all other English-speaking countries, Canada,"},{"Start":"05:17.215 ","End":"05:19.975","Text":"the UK, and New Zealand,"},{"Start":"05:19.975 ","End":"05:23.860","Text":"Australia, they say Z."},{"Start":"05:24.350 ","End":"05:30.740","Text":"We\u0027re going from v to z using this substitution."},{"Start":"05:30.740 ","End":"05:35.410","Text":"This is an invertible transformation because to go the other way,"},{"Start":"05:35.410 ","End":"05:38.440","Text":"you would just bring the e to the power of to,"},{"Start":"05:38.440 ","End":"05:41.170","Text":"the other side with the minus."},{"Start":"05:41.170 ","End":"05:45.405","Text":"Anyway, let\u0027s start. We\u0027ll do v_X first."},{"Start":"05:45.405 ","End":"05:47.340","Text":"But there\u0027s 4 things we need,"},{"Start":"05:47.340 ","End":"05:49.070","Text":"v_X, v_T,"},{"Start":"05:49.070 ","End":"05:54.060","Text":"v_XX and v_TT,"},{"Start":"05:54.060 ","End":"05:57.320","Text":"well, and v. Use the product rule."},{"Start":"05:57.320 ","End":"05:58.820","Text":"We have a product here."},{"Start":"05:58.820 ","End":"06:03.930","Text":"If we differentiate this with respect to X, we just get z_X,"},{"Start":"06:04.150 ","End":"06:06.890","Text":"I\u0027m just going to say whichever one I want,"},{"Start":"06:06.890 ","End":"06:09.415","Text":"z_X, e to the,"},{"Start":"06:09.415 ","End":"06:14.435","Text":"plus, this as is times the derivative of this."},{"Start":"06:14.435 ","End":"06:17.135","Text":"Derivative of this with respect to X is e to the,"},{"Start":"06:17.135 ","End":"06:21.080","Text":"whatever is here, times the derivative of this, which is Gamma."},{"Start":"06:21.080 ","End":"06:22.940","Text":"So the derivative of the first bit,"},{"Start":"06:22.940 ","End":"06:25.475","Text":"we get this here,"},{"Start":"06:25.475 ","End":"06:29.870","Text":"and then the second bit we get the Gamma in front."},{"Start":"06:29.870 ","End":"06:33.440","Text":"Almost the same thing with the v_T."},{"Start":"06:33.440 ","End":"06:41.120","Text":"Here, we get the T and Delta here and here\u0027s the second derivative with respect to X."},{"Start":"06:41.120 ","End":"06:45.050","Text":"We can either differentiate this with respect to X"},{"Start":"06:45.050 ","End":"06:50.285","Text":"or you could use the formula for second derivative of a product."},{"Start":"06:50.285 ","End":"06:56.464","Text":"The second derivative of the first times the second gives us this plus twice."},{"Start":"06:56.464 ","End":"06:57.890","Text":"The derivative of this."},{"Start":"06:57.890 ","End":"07:06.920","Text":"The derivative of this gives us the twice and the Gamma and the X here and plus twice."},{"Start":"07:06.920 ","End":"07:09.290","Text":"This times the second derivative of this,"},{"Start":"07:09.290 ","End":"07:15.695","Text":"second derivative of this gives us the Gamma^2 here and similarly with v_TT."},{"Start":"07:15.695 ","End":"07:19.970","Text":"We just have to put all these together in here,"},{"Start":"07:19.970 ","End":"07:21.785","Text":"which I copied again over here."},{"Start":"07:21.785 ","End":"07:30.395","Text":"We need to substitute these 4 quantities and v. So that gives us v_TT copy from here,"},{"Start":"07:30.395 ","End":"07:35.375","Text":"minus and then v_XX from here."},{"Start":"07:35.375 ","End":"07:37.580","Text":"That\u0027s the left-hand side."},{"Start":"07:37.580 ","End":"07:39.245","Text":"Now the right hand side,"},{"Start":"07:39.245 ","End":"07:47.090","Text":"2 over 25 times v_X from here and"},{"Start":"07:47.090 ","End":"07:56.025","Text":"then 2/5 v_T from here and 4 over 25v just from here."},{"Start":"07:56.025 ","End":"07:59.715","Text":"Z times the exponent part."},{"Start":"07:59.715 ","End":"08:02.105","Text":"I should\u0027ve said in each case,"},{"Start":"08:02.105 ","End":"08:05.030","Text":"I\u0027m pulling out the exponent part."},{"Start":"08:05.030 ","End":"08:09.735","Text":"Each of these pieces has something times"},{"Start":"08:09.735 ","End":"08:14.640","Text":"e^Gamma X plus Delta T. Just taking it out wherever we can."},{"Start":"08:14.640 ","End":"08:16.625","Text":"Now we can cancel this."},{"Start":"08:16.625 ","End":"08:22.200","Text":"Just get rid of this and this."},{"Start":"08:23.920 ","End":"08:26.860","Text":"We get this."},{"Start":"08:26.860 ","End":"08:29.205","Text":"Now just rearrange,"},{"Start":"08:29.205 ","End":"08:34.500","Text":"leave the z_TT and z_XX on the left,"},{"Start":"08:34.500 ","End":"08:38.355","Text":"that\u0027s this and this here,"},{"Start":"08:38.355 ","End":"08:40.835","Text":"and put all the other stuff on the right,"},{"Start":"08:40.835 ","End":"08:45.910","Text":"but collected according to the coefficients of z,"},{"Start":"08:45.910 ","End":"08:50.450","Text":"the z by the X and z by the T. I\u0027ll leave you to check"},{"Start":"08:50.450 ","End":"08:55.520","Text":"that it\u0027s just straightforward algebra collecting like terms, distributive law."},{"Start":"08:55.520 ","End":"08:58.860","Text":"But this point we assign Gamma and Delta."},{"Start":"08:58.860 ","End":"09:05.730","Text":"Gamma will take us minus 1 over 25 to make this part 0."},{"Start":"09:05.730 ","End":"09:11.865","Text":"Will take Delta to be 1/5 in order to make this 0."},{"Start":"09:11.865 ","End":"09:15.900","Text":"All we\u0027re left with is z_TT minus z_XX equals"},{"Start":"09:15.900 ","End":"09:20.600","Text":"this thing times z of this is what will cause c. If we do that,"},{"Start":"09:20.600 ","End":"09:29.710","Text":"then all we have left is z_TT minus z_XX equals this constant times z."},{"Start":"09:29.710 ","End":"09:35.410","Text":"That\u0027s part d and that concludes this exercise."}],"ID":30706},{"Watched":false,"Name":"Exercise 9","Duration":"6m 27s","ChapterTopicVideoID":29118,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.940","Text":"In this exercise, we\u0027re given"},{"Start":"00:02.940 ","End":"00:08.175","Text":"the following linear 2nd order partial differential equation,"},{"Start":"00:08.175 ","End":"00:10.890","Text":"a and b are constants."},{"Start":"00:10.890 ","End":"00:15.990","Text":"Note that this is in the 1st canonical form for hyperbolic equations."},{"Start":"00:15.990 ","End":"00:19.740","Text":"In part a, we\u0027re going to convert it to 2nd"},{"Start":"00:19.740 ","End":"00:25.020","Text":"canonical form using the following change of variables,"},{"Start":"00:25.020 ","End":"00:31.425","Text":"and after that, we\u0027re going to simplify this using a substitution as follows,"},{"Start":"00:31.425 ","End":"00:34.335","Text":"where Gamma and Delta will be determined later."},{"Start":"00:34.335 ","End":"00:41.415","Text":"We\u0027ll get it into the form z_xx minus z_tt is some constant times z."},{"Start":"00:41.415 ","End":"00:47.900","Text":"Let\u0027s start with part a, where we\u0027re going to let the new function in x and t"},{"Start":"00:47.900 ","End":"00:54.630","Text":"be called v. We are going to need u_Psi Eta,"},{"Start":"00:54.630 ","End":"01:03.200","Text":"u_Psi and u_Eta in terms of the partial derivatives of v. First of all,"},{"Start":"01:03.200 ","End":"01:08.490","Text":"u_Psi using the chain rule for functions of 2 variables."},{"Start":"01:08.490 ","End":"01:12.650","Text":"We can differentiate v with respect to x and"},{"Start":"01:12.650 ","End":"01:17.850","Text":"then x with respect to Psi or we can go via t, in other words, v_t t_Psi."},{"Start":"01:18.440 ","End":"01:21.600","Text":"V depends on Psi in 2 ways,"},{"Start":"01:21.600 ","End":"01:25.965","Text":"via x and via t. The x_Psi"},{"Start":"01:25.965 ","End":"01:32.145","Text":"is 1/2 and the t_Psi is also 1/2."},{"Start":"01:32.145 ","End":"01:33.780","Text":"We can write it like this."},{"Start":"01:33.780 ","End":"01:39.710","Text":"Then u_Eta, just very similarly except that there\u0027s a minus 1/2 here,"},{"Start":"01:39.710 ","End":"01:41.495","Text":"there is a minus here."},{"Start":"01:41.495 ","End":"01:44.285","Text":"We\u0027re going to differentiate again."},{"Start":"01:44.285 ","End":"01:50.110","Text":"You could either differentiate this with respect to Eta or this with respect to Psi."},{"Start":"01:50.110 ","End":"01:51.716","Text":"I\u0027ll go with this,"},{"Start":"01:51.716 ","End":"01:53.794","Text":"with respect to Eta."},{"Start":"01:53.794 ","End":"01:55.685","Text":"The 1/2 here,"},{"Start":"01:55.685 ","End":"01:59.110","Text":"and then v_x is v_xx,"},{"Start":"01:59.110 ","End":"02:00.875","Text":"x_Eta or v_xt,"},{"Start":"02:00.875 ","End":"02:04.415","Text":"t_Eta and similarly the other one."},{"Start":"02:04.415 ","End":"02:08.250","Text":"This is 1/2 minus 1/2."},{"Start":"02:08.250 ","End":"02:10.795","Text":"This is 1/2. This is minus 1/2."},{"Start":"02:10.795 ","End":"02:12.290","Text":"This is what we get."},{"Start":"02:12.290 ","End":"02:19.760","Text":"Just collecting stuff up and noting that these 2 cancel because v_xt and v_tx are equal."},{"Start":"02:19.760 ","End":"02:23.615","Text":"There\u0027s a theorem on the mixed 2nd order derivatives."},{"Start":"02:23.615 ","End":"02:27.090","Text":"Our original PDE is this."},{"Start":"02:27.090 ","End":"02:30.170","Text":"We can substitute the 3 quantities, this,"},{"Start":"02:30.170 ","End":"02:33.320","Text":"this and this that we computed here, here,"},{"Start":"02:33.320 ","End":"02:37.490","Text":"and here and what we get is the following."},{"Start":"02:37.490 ","End":"02:40.810","Text":"It\u0027s just a straightforward substitution."},{"Start":"02:40.810 ","End":"02:44.230","Text":"Just rearrange it and this is what we get."},{"Start":"02:44.230 ","End":"02:46.615","Text":"Also collecting like terms,"},{"Start":"02:46.615 ","End":"02:51.880","Text":"say v_x has a over 2 plus b over 2,"},{"Start":"02:51.880 ","End":"02:53.813","Text":"which is this,"},{"Start":"02:53.813 ","End":"02:58.240","Text":"and v_t is a times 1/2 minus b times 1/2."},{"Start":"02:58.240 ","End":"03:01.080","Text":"So it\u0027s this, multiply by 4."},{"Start":"03:01.080 ","End":"03:02.910","Text":"This is what we get."},{"Start":"03:02.910 ","End":"03:04.870","Text":"This is good. This is what we wanted."},{"Start":"03:04.870 ","End":"03:06.385","Text":"If we go back and look,"},{"Start":"03:06.385 ","End":"03:09.280","Text":"we had to get it into the form."},{"Start":"03:09.280 ","End":"03:15.520","Text":"The dot-dot-dot is terms of lower order,"},{"Start":"03:15.520 ","End":"03:20.020","Text":"which is what we have except that we put it on the right-hand side."},{"Start":"03:20.020 ","End":"03:23.425","Text":"It doesn\u0027t matter. You can put them on the left also."},{"Start":"03:23.425 ","End":"03:25.670","Text":"That\u0027s part a."},{"Start":"03:25.670 ","End":"03:28.250","Text":"Now continuing to part b."},{"Start":"03:28.250 ","End":"03:30.125","Text":"This is what we had in part a."},{"Start":"03:30.125 ","End":"03:35.855","Text":"We had a hyperbolic equation in the second canonical form."},{"Start":"03:35.855 ","End":"03:38.030","Text":"Now we\u0027re also going to simplify it."},{"Start":"03:38.030 ","End":"03:40.820","Text":"For that, we\u0027re going to make a conversion."},{"Start":"03:40.820 ","End":"03:44.420","Text":"We\u0027re going to define v in terms of z or z in terms of"},{"Start":"03:44.420 ","End":"03:48.185","Text":"v. It\u0027s totally reversible because the e could go over to the other side."},{"Start":"03:48.185 ","End":"03:53.575","Text":"We\u0027re going to make this conversion substitution and we\u0027ll need v_t,"},{"Start":"03:53.575 ","End":"03:56.070","Text":"v_x, v_tt,"},{"Start":"03:56.070 ","End":"03:58.320","Text":"v_xx in terms of z."},{"Start":"03:58.320 ","End":"04:02.540","Text":"First, v_x derivative of a product."},{"Start":"04:02.540 ","End":"04:09.725","Text":"This times the derivative of z with respect to x gives us an extra x here,"},{"Start":"04:09.725 ","End":"04:16.195","Text":"and the e part gives us an extra Gamma that comes out as a derivative."},{"Start":"04:16.195 ","End":"04:17.670","Text":"Similarly, with v_t,"},{"Start":"04:17.670 ","End":"04:19.025","Text":"we get a t here,"},{"Start":"04:19.025 ","End":"04:21.125","Text":"we have a Delta here."},{"Start":"04:21.125 ","End":"04:26.930","Text":"For v_xx, we could just differentiate this with respect to x or we"},{"Start":"04:26.930 ","End":"04:33.535","Text":"could use the 2nd derivative of a product using this formula."},{"Start":"04:33.535 ","End":"04:39.290","Text":"Differentiate this. The first term,"},{"Start":"04:39.290 ","End":"04:45.065","Text":"differentiate it twice gives us z_xx times e, etc."},{"Start":"04:45.065 ","End":"04:48.430","Text":"Middle one, we differentiate this once and this once,"},{"Start":"04:48.430 ","End":"04:52.730","Text":"so we get the z_x and we get the Gamma and the 2 from the formula."},{"Start":"04:52.730 ","End":"04:57.065","Text":"The last part, we differentiate this twice."},{"Start":"04:57.065 ","End":"04:58.820","Text":"We get the Gamma that comes out,"},{"Start":"04:58.820 ","End":"05:05.424","Text":"then a Gamma comes out again and we get the Gamma^2 and similarly for v_tt."},{"Start":"05:05.424 ","End":"05:10.040","Text":"Return to this formula that was here and we\u0027re just"},{"Start":"05:10.040 ","End":"05:15.005","Text":"going to substitute the 4 quantities that are colored here and we get this."},{"Start":"05:15.005 ","End":"05:20.630","Text":"Some algebra, z_xx on the left and minus z_tt on the left."},{"Start":"05:20.630 ","End":"05:23.195","Text":"Everything else gets pushed to the right."},{"Start":"05:23.195 ","End":"05:28.975","Text":"We also separate it by z, z_x and z_t."},{"Start":"05:28.975 ","End":"05:33.782","Text":"We can choose Gamma and Delta because up to now,"},{"Start":"05:33.782 ","End":"05:35.685","Text":"it\u0027s worked for all Gamma and Delta."},{"Start":"05:35.685 ","End":"05:41.570","Text":"We\u0027re going to let this be 0 by letting Gamma equals a plus b."},{"Start":"05:41.570 ","End":"05:46.040","Text":"We can get this to be 0 by letting Delta equals a minus b,"},{"Start":"05:46.040 ","End":"05:47.390","Text":"so this will all be 0,"},{"Start":"05:47.390 ","End":"05:51.485","Text":"this will all be 0 and all this in the square bracket is a constant,"},{"Start":"05:51.485 ","End":"05:55.895","Text":"rearranging it a bit and also substituting Gamma is a plus b."},{"Start":"05:55.895 ","End":"05:59.500","Text":"This is twice a plus b^2 and so on and so on."},{"Start":"05:59.500 ","End":"06:05.960","Text":"Then more simplification, just bit of algebra on what\u0027s in the brackets,"},{"Start":"06:05.960 ","End":"06:09.620","Text":"and we can even go another step because this is a^2 plus 2ab plus b^2."},{"Start":"06:09.620 ","End":"06:11.795","Text":"This is a^2 minus 2ab plus b^2."},{"Start":"06:11.795 ","End":"06:18.270","Text":"So when you subtract, you just get 4ab and 4ab is a constant."},{"Start":"06:18.270 ","End":"06:20.274","Text":"So we have what we wanted,"},{"Start":"06:20.274 ","End":"06:24.540","Text":"z_xx minus z_tt is some constant times z."},{"Start":"06:24.540 ","End":"06:27.730","Text":"That concludes this exercise."}],"ID":30707},{"Watched":false,"Name":"Exercise 10","Duration":"3m 28s","ChapterTopicVideoID":29119,"CourseChapterTopicPlaylistID":294424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.550","Text":"In this exercise, we have the following second-order linear PDE."},{"Start":"00:05.550 ","End":"00:09.660","Text":"You can see that it\u0027s elliptic and a,"},{"Start":"00:09.660 ","End":"00:12.510","Text":"b and c are given constants."},{"Start":"00:12.510 ","End":"00:14.550","Text":"In any particular example,"},{"Start":"00:14.550 ","End":"00:16.884","Text":"you\u0027d actually get numbers like 3,"},{"Start":"00:16.884 ","End":"00:19.665","Text":"7, 4, but we\u0027re going to do this in general."},{"Start":"00:19.665 ","End":"00:22.515","Text":"We\u0027re going to show that a substitution of this form,"},{"Start":"00:22.515 ","End":"00:26.970","Text":"we\u0027ll bring this to an equation that\u0027s also elliptic but simpler."},{"Start":"00:26.970 ","End":"00:31.425","Text":"It\u0027s as if we got rid of the u_x and u_y term."},{"Start":"00:31.425 ","End":"00:34.485","Text":"To the view, we have z,"},{"Start":"00:34.485 ","End":"00:40.005","Text":"z_xx plus z_yy is some constant time z. Gamma and Delta,"},{"Start":"00:40.005 ","End":"00:42.920","Text":"we started out as not knowing what they are and at some point,"},{"Start":"00:42.920 ","End":"00:45.020","Text":"we substitute values for them."},{"Start":"00:45.020 ","End":"00:46.430","Text":"They\u0027ll all be in terms of a, b,"},{"Start":"00:46.430 ","End":"00:49.250","Text":"and c, and that will reduce it to this form."},{"Start":"00:49.250 ","End":"00:53.090","Text":"Also, I\u0027ll note that we\u0027ve used this technique before only with"},{"Start":"00:53.090 ","End":"00:57.605","Text":"the hyperbolic equation in second canonical form,"},{"Start":"00:57.605 ","End":"01:03.290","Text":"which is just like this except that there\u0027s a minus here and we get a minus here."},{"Start":"01:03.290 ","End":"01:07.320","Text":"Let\u0027s start. This is what we have to substitute."},{"Start":"01:07.320 ","End":"01:12.695","Text":"We need the partial derivatives of u in terms of the partial derivatives of z."},{"Start":"01:12.695 ","End":"01:15.410","Text":"We\u0027ve done this before, pretty straightforward."},{"Start":"01:15.410 ","End":"01:17.285","Text":"Derivative of a product,"},{"Start":"01:17.285 ","End":"01:20.795","Text":"derivative of this bit with respect to x."},{"Start":"01:20.795 ","End":"01:24.620","Text":"This as is, and this times this with respect to x,"},{"Start":"01:24.620 ","End":"01:26.060","Text":"brings a Gamma in front."},{"Start":"01:26.060 ","End":"01:27.920","Text":"Suddenly for u_y."},{"Start":"01:27.920 ","End":"01:32.060","Text":"The second derivatives, just by differentiating this again,"},{"Start":"01:32.060 ","End":"01:38.220","Text":"or there\u0027s a formula for second derivative of a product and u_yy, straightforward."},{"Start":"01:38.220 ","End":"01:39.780","Text":"We\u0027ve also done it before."},{"Start":"01:39.780 ","End":"01:41.490","Text":"I\u0027m omitting the details."},{"Start":"01:41.490 ","End":"01:44.450","Text":"Now our original PDE is this,"},{"Start":"01:44.450 ","End":"01:50.985","Text":"and we\u0027re going to substitute all these 5 quantities from here, and u itself."},{"Start":"01:50.985 ","End":"01:54.690","Text":"Here we are. This is u_xx,"},{"Start":"01:54.690 ","End":"01:59.845","Text":"this is u_yy, this is u_x."},{"Start":"01:59.845 ","End":"02:01.625","Text":"Oh yeah, I forgot to say,"},{"Start":"02:01.625 ","End":"02:04.490","Text":"I\u0027m omitting all these e to the Gamma x"},{"Start":"02:04.490 ","End":"02:07.700","Text":"plus Delta y because it\u0027s present in every single term."},{"Start":"02:07.700 ","End":"02:10.165","Text":"It will just cancel in the end."},{"Start":"02:10.165 ","End":"02:15.785","Text":"Continuing, this one is u_y without the e to the power of."},{"Start":"02:15.785 ","End":"02:19.130","Text":"The last one is c times z."},{"Start":"02:19.130 ","End":"02:21.935","Text":"Again, we\u0027ve omitted, this, just cancels."},{"Start":"02:21.935 ","End":"02:25.565","Text":"Now if we open the brackets and just collect like terms,"},{"Start":"02:25.565 ","End":"02:27.384","Text":"this is what we get."},{"Start":"02:27.384 ","End":"02:32.005","Text":"Z_xx plus z_yy."},{"Start":"02:32.005 ","End":"02:37.080","Text":"The coefficient of z_x is 2 Gamma plus a."},{"Start":"02:37.080 ","End":"02:41.305","Text":"It\u0027s the 2 Gamma from here and the a from here, and so on."},{"Start":"02:41.305 ","End":"02:43.930","Text":"Now, why have I covered these in gray?"},{"Start":"02:43.930 ","End":"02:46.340","Text":"Because I\u0027m going to make these 0."},{"Start":"02:46.340 ","End":"02:53.725","Text":"If we choose Gamma to be minus a over 2 and choose Delta to be minus b over 2,"},{"Start":"02:53.725 ","End":"02:56.300","Text":"then these 2 will disappear."},{"Start":"02:56.300 ","End":"03:00.450","Text":"If we substitute Gamma and Delta here,"},{"Start":"03:00.450 ","End":"03:03.285","Text":"what we\u0027ll get is the following,"},{"Start":"03:03.285 ","End":"03:10.080","Text":"and just collect a^2 over 4 minus a^2 over 2 is minus a^2 over 4."},{"Start":"03:10.080 ","End":"03:14.760","Text":"We get c minus a^2 over 4 minus b^2 over 4 here,"},{"Start":"03:14.760 ","End":"03:17.805","Text":"and this is just a constant."},{"Start":"03:17.805 ","End":"03:23.075","Text":"We can say z_xx and z_yy plus some constant times that equals 0."},{"Start":"03:23.075 ","End":"03:25.235","Text":"This is what we had to get to,"},{"Start":"03:25.235 ","End":"03:29.010","Text":"and so that concludes this exercise."}],"ID":30708}],"Thumbnail":null,"ID":294424}]

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