[{"Name":"The Energy Integrals and Uniqueness of Solutions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"5m 55s","ChapterTopicVideoID":29234,"CourseChapterTopicPlaylistID":294425,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29234.jpeg","UploadDate":"2022-06-06T10:04:34.3200000","DurationForVideoObject":"PT5M55S","Description":null,"MetaTitle":"Exercise 1 - The Energy Integrals and Uniqueness of Solutions: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on The Energy Integrals and Uniqueness of Solutions practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/partial-differential-equations/energy-integrals-and-uniqueness-of-solutions/the-energy-integrals-and-uniqueness-of-solutions/vid30809","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"In this exercise, we\u0027ll use an energy"},{"Start":"00:03.210 ","End":"00:06.735","Text":"integral to show the uniqueness of the following problem."},{"Start":"00:06.735 ","End":"00:09.090","Text":"Now the concept energy integral,"},{"Start":"00:09.090 ","End":"00:11.339","Text":"you\u0027ll get to learn through the examples;"},{"Start":"00:11.339 ","End":"00:13.085","Text":"I won\u0027t give a formal definition."},{"Start":"00:13.085 ","End":"00:15.870","Text":"After you\u0027ve seen 2 or 3 of these exercises,"},{"Start":"00:15.870 ","End":"00:17.370","Text":"you\u0027ll get the hang of it."},{"Start":"00:17.370 ","End":"00:19.890","Text":"We\u0027re given a hint to define E(t),"},{"Start":"00:19.890 ","End":"00:22.965","Text":"E for energy, as the following."},{"Start":"00:22.965 ","End":"00:25.560","Text":"This will be different in each exercise."},{"Start":"00:25.560 ","End":"00:27.585","Text":"There\u0027s an art to finding these,"},{"Start":"00:27.585 ","End":"00:29.340","Text":"a bit of trial and error and some method,"},{"Start":"00:29.340 ","End":"00:32.925","Text":"but here you\u0027ll always be given it in the exercise."},{"Start":"00:32.925 ","End":"00:36.000","Text":"General method is the same in all of these exercises,"},{"Start":"00:36.000 ","End":"00:40.250","Text":"we show that if there are solutions u and v,"},{"Start":"00:40.250 ","End":"00:43.850","Text":"then necessarily u is the same as v,"},{"Start":"00:43.850 ","End":"00:45.755","Text":"though there is only 1 solution."},{"Start":"00:45.755 ","End":"00:50.630","Text":"The way we do this is we define w=u minus v,"},{"Start":"00:50.630 ","End":"00:52.775","Text":"and this is the w hint to that here."},{"Start":"00:52.775 ","End":"00:56.210","Text":"We prove that w is identically 0."},{"Start":"00:56.210 ","End":"00:57.950","Text":"If w is identically 0,"},{"Start":"00:57.950 ","End":"01:03.515","Text":"then u must equal v. Now consider the PDE part;"},{"Start":"01:03.515 ","End":"01:08.285","Text":"u satisfies this and v satisfies this,"},{"Start":"01:08.285 ","End":"01:16.580","Text":"w to remind you is u minus v. This minus this equals this minus this,"},{"Start":"01:16.580 ","End":"01:18.159","Text":"and the F cancels."},{"Start":"01:18.159 ","End":"01:22.040","Text":"We get this. By linearity of differentiation."},{"Start":"01:22.040 ","End":"01:26.900","Text":"This is equal to second derivative respect to t(w)."},{"Start":"01:26.900 ","End":"01:29.580","Text":"Similarly here w x,"},{"Start":"01:29.580 ","End":"01:31.850","Text":"x and that gives us the PDE for"},{"Start":"01:31.850 ","End":"01:35.390","Text":"w. Next we\u0027re going to find the initial and boundary conditions."},{"Start":"01:35.390 ","End":"01:39.350","Text":"These are the initial and boundary conditions for u."},{"Start":"01:39.350 ","End":"01:43.010","Text":"Similarly, just replace u by v and we get"},{"Start":"01:43.010 ","End":"01:50.300","Text":"the initial and boundary conditions for v. If we subtract this minus this,"},{"Start":"01:50.300 ","End":"01:52.154","Text":"we get that u minus v(x,"},{"Start":"01:52.154 ","End":"01:56.815","Text":"naught) is f minus f, which is 0."},{"Start":"01:56.815 ","End":"01:59.535","Text":"If we subtract this minus this,"},{"Start":"01:59.535 ","End":"02:08.620","Text":"we get this g minus g is 0 and subtract this minus this and this minus this."},{"Start":"02:08.620 ","End":"02:11.160","Text":"Just copied again the PDE."},{"Start":"02:11.160 ","End":"02:17.050","Text":"Now we have the complete problem for w PDE initial values, boundary values."},{"Start":"02:17.050 ","End":"02:21.200","Text":"This is a much simpler problem than the original problem."},{"Start":"02:21.200 ","End":"02:24.755","Text":"To remind you, we\u0027re going to show that w is necessarily 0."},{"Start":"02:24.755 ","End":"02:30.005","Text":"The energy integral, just from the hint we were given is the following."},{"Start":"02:30.005 ","End":"02:31.430","Text":"We don\u0027t really need the half."},{"Start":"02:31.430 ","End":"02:35.045","Text":"There are reasons why the half is there and what this has to do with energy."},{"Start":"02:35.045 ","End":"02:36.305","Text":"Don\u0027t worry about that."},{"Start":"02:36.305 ","End":"02:39.200","Text":"Now, E of 0 is the following,"},{"Start":"02:39.200 ","End":"02:41.795","Text":"just replacing t by 0."},{"Start":"02:41.795 ","End":"02:44.480","Text":"Now I claim that both of these are 0,"},{"Start":"02:44.480 ","End":"02:47.055","Text":"wt (x, 0) is 0."},{"Start":"02:47.055 ","End":"02:49.035","Text":"That\u0027s straight from here."},{"Start":"02:49.035 ","End":"02:53.680","Text":"From this, we can differentiate it and get that wx(x,"},{"Start":"02:53.680 ","End":"02:56.665","Text":"0) is 0, this is 0, this is 0."},{"Start":"02:56.665 ","End":"03:02.845","Text":"We have 0^2 plus 0^2 is equal to 0 and the integral of 0 is 0."},{"Start":"03:02.845 ","End":"03:09.520","Text":"Summarizing, we have an energy integral defined like this and E(0) is 0."},{"Start":"03:09.520 ","End":"03:15.415","Text":"Next thing we will explore is differentiating the energy integral e(t)."},{"Start":"03:15.415 ","End":"03:18.010","Text":"Well, we can only differentiate with respect to t,"},{"Start":"03:18.010 ","End":"03:20.590","Text":"there is only 1 variable is equal to,"},{"Start":"03:20.590 ","End":"03:24.760","Text":"and we\u0027re assuming it\u0027s okay to differentiate under the integral sign."},{"Start":"03:24.760 ","End":"03:32.245","Text":"The derivative of wt^2 is twice wt times the derivative of wt, which is this."},{"Start":"03:32.245 ","End":"03:39.980","Text":"Here, twice wx times wx t. We can change the order of the differentiation here,"},{"Start":"03:39.980 ","End":"03:44.840","Text":"and we can also replace wtt with wxx."},{"Start":"03:44.840 ","End":"03:47.065","Text":"We get the following."},{"Start":"03:47.065 ","End":"03:51.890","Text":"We\u0027re going to use integration by parts on the second term."},{"Start":"03:51.890 ","End":"03:54.950","Text":"The first term is nothing much to be done because this"},{"Start":"03:54.950 ","End":"03:58.640","Text":"is wt and the integrals with respect to x."},{"Start":"03:58.640 ","End":"04:02.735","Text":"Copy the first term and break this up into 2 according to this."},{"Start":"04:02.735 ","End":"04:06.380","Text":"We won\u0027t go into the minor details of integration by parts."},{"Start":"04:06.380 ","End":"04:11.480","Text":"But just notice that v\u0027 is wxx,"},{"Start":"04:11.480 ","End":"04:16.495","Text":"and u is the integral of this with respect to x, which is wt."},{"Start":"04:16.495 ","End":"04:17.975","Text":"After we\u0027ve done this,"},{"Start":"04:17.975 ","End":"04:20.150","Text":"we see that this term and this term are the"},{"Start":"04:20.150 ","End":"04:23.359","Text":"same with a plus and with a minus or they cancel."},{"Start":"04:23.359 ","End":"04:25.220","Text":"We\u0027re just left with this."},{"Start":"04:25.220 ","End":"04:27.365","Text":"This comes out to be 0."},{"Start":"04:27.365 ","End":"04:30.410","Text":"Since you look at wx at 0,"},{"Start":"04:30.410 ","End":"04:33.680","Text":"it\u0027s equal to 0 and at l is equal to 0."},{"Start":"04:33.680 ","End":"04:37.435","Text":"Here we have 0 minus 0, which is 0."},{"Start":"04:37.435 ","End":"04:43.005","Text":"E(0) is 0, and E\u0027 is 0."},{"Start":"04:43.005 ","End":"04:47.180","Text":"This differential equation has only 1 solution. E(t) is 0."},{"Start":"04:47.180 ","End":"04:52.880","Text":"From here we get that E(t) is a constant and at 0 it\u0027s 0, the constant is 0."},{"Start":"04:52.880 ","End":"04:55.790","Text":"E(t) is identically 0."},{"Start":"04:55.790 ","End":"04:58.280","Text":"E(t) we defined it this way,"},{"Start":"04:58.280 ","End":"05:00.095","Text":"so this is equal to 0."},{"Start":"05:00.095 ","End":"05:03.500","Text":"Now we have the integral of a non-negative integrant."},{"Start":"05:03.500 ","End":"05:06.440","Text":"Obviously this is non-negative and it\u0027s continuous."},{"Start":"05:06.440 ","End":"05:09.410","Text":"If we have the integral of a continuous non-negative function"},{"Start":"05:09.410 ","End":"05:12.380","Text":"and it\u0027s 0 must be that this function,"},{"Start":"05:12.380 ","End":"05:15.260","Text":"the integrant, is identically 0."},{"Start":"05:15.260 ","End":"05:19.600","Text":"Now simple algebra, if a^2 plus b^2 is 0,"},{"Start":"05:19.600 ","End":"05:22.575","Text":"both a and b have to be 0."},{"Start":"05:22.575 ","End":"05:27.545","Text":"From here we get that wt is 0 and wx is 0."},{"Start":"05:27.545 ","End":"05:32.840","Text":"If we have a function whose partial derivative with respect to t and x are both 0,"},{"Start":"05:32.840 ","End":"05:35.465","Text":"then the function has to be a constant function."},{"Start":"05:35.465 ","End":"05:39.035","Text":"But we know that w of x 0 is 0."},{"Start":"05:39.035 ","End":"05:41.599","Text":"If it\u0027s a constant sometimes it\u0027s 0,"},{"Start":"05:41.599 ","End":"05:43.715","Text":"that constant has to be 0."},{"Start":"05:43.715 ","End":"05:47.834","Text":"We have that w(xt) is identically 0, but w,"},{"Start":"05:47.834 ","End":"05:50.145","Text":"recall is u minus v,"},{"Start":"05:50.145 ","End":"05:56.400","Text":"u is identically equal to v. That concludes this exercise."}],"ID":30809},{"Watched":false,"Name":"Exercise 2","Duration":"6m 11s","ChapterTopicVideoID":29235,"CourseChapterTopicPlaylistID":294425,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"In this exercise, we\u0027re going to use an energy integral to"},{"Start":"00:04.080 ","End":"00:07.935","Text":"show the uniqueness of the solution to the following problem."},{"Start":"00:07.935 ","End":"00:11.730","Text":"This problem is not a wave equation."},{"Start":"00:11.730 ","End":"00:14.490","Text":"It\u0027s similar to a non-homogeneous wave equation."},{"Start":"00:14.490 ","End":"00:16.890","Text":"This minus u spoils it."},{"Start":"00:16.890 ","End":"00:22.910","Text":"It\u0027s the PDE with initial conditions and boundary conditions at"},{"Start":"00:22.910 ","End":"00:25.370","Text":"either end points 0 and L. It\u0027s on"},{"Start":"00:25.370 ","End":"00:30.020","Text":"a finite interval and we\u0027re given a hint as to what the energy integral is."},{"Start":"00:30.020 ","End":"00:33.612","Text":"This will vary from problem to problem."},{"Start":"00:33.612 ","End":"00:37.925","Text":"You don\u0027t expect that at this stage to be able to produce such an integral,"},{"Start":"00:37.925 ","End":"00:40.415","Text":"just how to use it once you\u0027ve given it."},{"Start":"00:40.415 ","End":"00:44.000","Text":"The uniqueness works by assuming, on the contrary,"},{"Start":"00:44.000 ","End":"00:48.185","Text":"that there are 2 solutions of the problem and then showing"},{"Start":"00:48.185 ","End":"00:50.120","Text":"that in such a case u is identically"},{"Start":"00:50.120 ","End":"00:52.930","Text":"equal to v so there really aren\u0027t 2, there\u0027s only 1."},{"Start":"00:52.930 ","End":"00:59.115","Text":"We do this by defining w equals u minus v and then proving that w is identically 0."},{"Start":"00:59.115 ","End":"01:00.480","Text":"If w is identically 0,"},{"Start":"01:00.480 ","End":"01:03.115","Text":"then u is identically equal to v. Now,"},{"Start":"01:03.115 ","End":"01:07.025","Text":"this is the PDE that u satisfies and just replacing u with v,"},{"Start":"01:07.025 ","End":"01:09.680","Text":"same equation v satisfies."},{"Start":"01:09.680 ","End":"01:12.140","Text":"If we subtract equations,"},{"Start":"01:12.140 ","End":"01:18.470","Text":"we get u_tt minus v_tt is u_xx minus v_xx minus u"},{"Start":"01:18.470 ","End":"01:21.710","Text":"minus v. This part cancels the F."},{"Start":"01:21.710 ","End":"01:26.555","Text":"The derivative of u minus v is the derivative of w, differentiation is linear."},{"Start":"01:26.555 ","End":"01:30.740","Text":"This is w_tt equals w_xx minus"},{"Start":"01:30.740 ","End":"01:36.500","Text":"w. That\u0027s the PD for w. We need the initial and boundary conditions."},{"Start":"01:36.500 ","End":"01:40.700","Text":"We start with the initial and boundary conditions"},{"Start":"01:40.700 ","End":"01:44.525","Text":"for u and then just copy that and replace"},{"Start":"01:44.525 ","End":"01:47.970","Text":"u with v to get the corresponding conditions for v."},{"Start":"01:47.970 ","End":"01:53.150","Text":"We can subtract pairwise if we subtract u of x naught minus v of x_naught."},{"Start":"01:53.150 ","End":"02:02.090","Text":"We\u0027ll get w of x_naught and that will be equal to 0 because it\u0027s f minus f so that\u0027s 0."},{"Start":"02:02.090 ","End":"02:07.220","Text":"Similarly, w_t of x_naught will be g minus g, which is 0."},{"Start":"02:07.220 ","End":"02:12.860","Text":"Everything is 0 for w and this again is the PDE for w. PDE"},{"Start":"02:12.860 ","End":"02:18.815","Text":"initial conditions and boundary conditions for w. We\u0027ll turn to the energy integral."},{"Start":"02:18.815 ","End":"02:20.375","Text":"We\u0027re given this in the hint,"},{"Start":"02:20.375 ","End":"02:21.440","Text":"you can actually drop the 0.5,"},{"Start":"02:21.440 ","End":"02:22.970","Text":"it doesn\u0027t really matter."},{"Start":"02:22.970 ","End":"02:25.515","Text":"Now let\u0027s see what E(0) is."},{"Start":"02:25.515 ","End":"02:28.880","Text":"Just put 0 instead of t everywhere."},{"Start":"02:28.880 ","End":"02:33.540","Text":"I claim that all these quantities are 0."},{"Start":"02:33.640 ","End":"02:37.400","Text":"W_t(x_ naught) is naught because of this,"},{"Start":"02:37.400 ","End":"02:47.100","Text":"w_x, that\u0027s not quite clear,"},{"Start":"02:47.100 ","End":"02:50.580","Text":"but w(x_ naught) from here is 0."},{"Start":"02:50.580 ","End":"02:57.840","Text":"You can differentiate partially with respect to x and get that w_x(x_ naught) is naught."},{"Start":"02:57.840 ","End":"03:00.642","Text":"We\u0027ve shown that this is 0,"},{"Start":"03:00.642 ","End":"03:02.655","Text":"0^2, this is 0, this is 0."},{"Start":"03:02.655 ","End":"03:10.355","Text":"Altogether E(naught) is just 0^2 plus 0^2 plus 0^2 integral, and so it\u0027s 0."},{"Start":"03:10.355 ","End":"03:12.570","Text":"Summarizing what we know about E,"},{"Start":"03:12.570 ","End":"03:14.450","Text":"E(t) is the following."},{"Start":"03:14.450 ","End":"03:18.440","Text":"E(0)=0. We continue to explore E. Next thing to do is to"},{"Start":"03:18.440 ","End":"03:22.250","Text":"try to find the derivative of E. E is only a function of t,"},{"Start":"03:22.250 ","End":"03:25.820","Text":"so derivative with respect to t and we"},{"Start":"03:25.820 ","End":"03:29.420","Text":"assume that we can differentiate under the integral sign."},{"Start":"03:29.420 ","End":"03:30.980","Text":"There are certain conditions that need to be met."},{"Start":"03:30.980 ","End":"03:34.384","Text":"Just ignore those technical details and assume everything\u0027s okay."},{"Start":"03:34.384 ","End":"03:40.355","Text":"Derivative of this is twice w_t times the derivative of w_t, which is w_tt."},{"Start":"03:40.355 ","End":"03:43.460","Text":"Similarly here 2w_x, w_xt,"},{"Start":"03:43.460 ","End":"03:46.095","Text":"and here 2w, w_t."},{"Start":"03:46.095 ","End":"03:49.030","Text":"Certain things we can do here the 2 will"},{"Start":"03:49.030 ","End":"03:52.555","Text":"cancel with the 2\u0027s everywhere here, here, and here."},{"Start":"03:52.555 ","End":"03:55.930","Text":"We can exchange the order of the x and the t for"},{"Start":"03:55.930 ","End":"03:59.710","Text":"the partial differentiation and we can replace"},{"Start":"03:59.710 ","End":"04:07.485","Text":"w_tt from the differential equation from the PDE as w_xx minus w. We\u0027ve done all that,"},{"Start":"04:07.485 ","End":"04:11.015","Text":"we get this and that cancels."},{"Start":"04:11.015 ","End":"04:15.595","Text":"For example, here we have minus w,"},{"Start":"04:15.595 ","End":"04:17.820","Text":"w_t, and here we have plus w,"},{"Start":"04:17.820 ","End":"04:19.525","Text":"w_t so that cancels."},{"Start":"04:19.525 ","End":"04:22.945","Text":"Now collecting together what we have left,"},{"Start":"04:22.945 ","End":"04:26.215","Text":"we still have w_t with w_xx,"},{"Start":"04:26.215 ","End":"04:30.210","Text":"and we have here w_x with w_tx."},{"Start":"04:30.210 ","End":"04:32.360","Text":"This, as you can guess,"},{"Start":"04:32.360 ","End":"04:34.760","Text":"we\u0027re going to do with integration by parts."},{"Start":"04:34.760 ","End":"04:36.665","Text":"I\u0027ve already marked v\u0027 and u\u0027."},{"Start":"04:36.665 ","End":"04:40.445","Text":"This is the formula and what we get, well,"},{"Start":"04:40.445 ","End":"04:49.870","Text":"I\u0027ll just emphasize the important part that u is the integral of u\u0027 so that will be w_t."},{"Start":"04:49.870 ","End":"04:55.945","Text":"This is with respect to x and v\u0027 is w_xx."},{"Start":"04:55.945 ","End":"05:01.100","Text":"This first term cancels with the last term and we\u0027re left with this middle bit."},{"Start":"05:01.100 ","End":"05:10.390","Text":"But this comes out to be 0 because w_x is 0 when x is 0 or x is L. So we get 0 minus 0."},{"Start":"05:10.390 ","End":"05:12.720","Text":"This all reduces to 0."},{"Start":"05:12.720 ","End":"05:17.164","Text":"We have that E\u0027(t) is 0 and E(0) is 0."},{"Start":"05:17.164 ","End":"05:18.260","Text":"From these 2 things,"},{"Start":"05:18.260 ","End":"05:20.315","Text":"from this differential equation,"},{"Start":"05:20.315 ","End":"05:24.580","Text":"we can integrate this and say that E(t) is a constant,"},{"Start":"05:24.580 ","End":"05:26.370","Text":"but E(0) is 0,"},{"Start":"05:26.370 ","End":"05:28.515","Text":"so the constant has to be 0."},{"Start":"05:28.515 ","End":"05:33.780","Text":"The solution to this is that E(t) is identically 0."},{"Start":"05:33.780 ","End":"05:36.620","Text":"Now, E(t) was defined as follows,"},{"Start":"05:36.620 ","End":"05:38.839","Text":"so this is 0,"},{"Start":"05:38.839 ","End":"05:44.080","Text":"the integrand is non-negative and continuous."},{"Start":"05:44.080 ","End":"05:48.095","Text":"If the integral of a non-negative continuous function is 0,"},{"Start":"05:48.095 ","End":"05:50.345","Text":"then the integrand is 0."},{"Start":"05:50.345 ","End":"05:53.810","Text":"We get that this sum is 0."},{"Start":"05:53.810 ","End":"05:55.880","Text":"Now by basic algebra,"},{"Start":"05:55.880 ","End":"05:59.310","Text":"if like a^2 plus b^2 plus c^2 is 0,"},{"Start":"05:59.310 ","End":"06:01.670","Text":"then a, b, and c are all 0."},{"Start":"06:01.670 ","End":"06:03.615","Text":"In particular, c is 0,"},{"Start":"06:03.615 ","End":"06:05.760","Text":"so w(x,t) is 0,"},{"Start":"06:05.760 ","End":"06:12.150","Text":"but w is just u minus v so u is identically equal to v and that concludes this exercise."}],"ID":30810},{"Watched":false,"Name":"Exercise 3","Duration":"4m 51s","ChapterTopicVideoID":29236,"CourseChapterTopicPlaylistID":294425,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.170","Text":"In this exercise, we\u0027ll use an energy integral to show that"},{"Start":"00:04.170 ","End":"00:08.055","Text":"the solution to this problem is unique."},{"Start":"00:08.055 ","End":"00:11.670","Text":"It\u0027s a nonhomogeneous heat equation and we\u0027re"},{"Start":"00:11.670 ","End":"00:15.435","Text":"given a hint to use the following energy integral."},{"Start":"00:15.435 ","End":"00:18.390","Text":"You probably noticed that in each of these exercises,"},{"Start":"00:18.390 ","End":"00:23.290","Text":"the energy integral is a bit different and as usual we\u0027ll show"},{"Start":"00:23.290 ","End":"00:29.405","Text":"the uniqueness by showing that if u and v are 2 solutions to the problem,"},{"Start":"00:29.405 ","End":"00:34.235","Text":"then u and v are identical and we show that u and v are identical"},{"Start":"00:34.235 ","End":"00:40.630","Text":"by defining w to be their difference and proving that w is identically 0."},{"Start":"00:40.630 ","End":"00:43.910","Text":"Let\u0027s see what PDE w satisfies."},{"Start":"00:43.910 ","End":"00:45.785","Text":"U satisfies this one,"},{"Start":"00:45.785 ","End":"00:47.600","Text":"and V satisfies this one."},{"Start":"00:47.600 ","End":"00:52.469","Text":"The same thing except with the letter u replaced by the letter v. We subtract"},{"Start":"00:52.469 ","End":"00:57.725","Text":"these 2 equations and let me subtract the f cancels so we have this."},{"Start":"00:57.725 ","End":"00:59.735","Text":"This is exactly wt,"},{"Start":"00:59.735 ","End":"01:04.340","Text":"this is wxx because differentiation is linear."},{"Start":"01:04.340 ","End":"01:07.640","Text":"Whether with respect to t or twice with respect to x,"},{"Start":"01:07.640 ","End":"01:10.445","Text":"the difference of the derivatives is the derivative of the difference."},{"Start":"01:10.445 ","End":"01:12.815","Text":"Now that we have the PDE,"},{"Start":"01:12.815 ","End":"01:17.210","Text":"we want the initial conditions and boundary conditions also for u."},{"Start":"01:17.210 ","End":"01:19.820","Text":"These are the conditions on u,"},{"Start":"01:19.820 ","End":"01:26.040","Text":"these are the conditions on v and w is the difference so we can subtract."},{"Start":"01:26.040 ","End":"01:28.505","Text":"Subtracting this minus this,"},{"Start":"01:28.505 ","End":"01:30.665","Text":"we get w(x,"},{"Start":"01:30.665 ","End":"01:36.485","Text":"0) is 0, u minus v is w and subtracting this minus this,"},{"Start":"01:36.485 ","End":"01:39.860","Text":"big A cancels and we get that w(0,"},{"Start":"01:39.860 ","End":"01:43.595","Text":"t) is 0 and subtracting this minus this,"},{"Start":"01:43.595 ","End":"01:45.500","Text":"you get that u minus v,"},{"Start":"01:45.500 ","End":"01:48.415","Text":"which is w(L, t )is 0."},{"Start":"01:48.415 ","End":"01:52.670","Text":"This is now the complete problem for w. Notice"},{"Start":"01:52.670 ","End":"01:57.215","Text":"that most everything is 0 and the heat equation is homogeneous."},{"Start":"01:57.215 ","End":"02:01.850","Text":"Now we recall the energy integral that was given to us as a hint."},{"Start":"02:01.850 ","End":"02:04.280","Text":"You want to know 2 things about this,"},{"Start":"02:04.280 ","End":"02:08.120","Text":"E(0) and the derivative E\u0027(t)."},{"Start":"02:08.120 ","End":"02:12.320","Text":"E(0) comes from replacing t by 0."},{"Start":"02:12.320 ","End":"02:18.665","Text":"Now I claim that w(x,0) is just 0 because it\u0027s given here."},{"Start":"02:18.665 ","End":"02:22.175","Text":"E(0) is the integral of 0,"},{"Start":"02:22.175 ","End":"02:26.390","Text":"which is 0 returning to E(t), this is what it is."},{"Start":"02:26.390 ","End":"02:30.529","Text":"Note that it\u0027s bigger or equal to 0 because the integrant"},{"Start":"02:30.529 ","End":"02:35.710","Text":"is bigger or equal to 0 at something ^2 and E\u0027(t),"},{"Start":"02:35.710 ","End":"02:39.170","Text":"we assume we can differentiate under the integral sign so we"},{"Start":"02:39.170 ","End":"02:44.000","Text":"differentiate w^2 with respect to t and we get 2wwt."},{"Start":"02:44.000 ","End":"02:47.330","Text":"Remember that wt is wxx."},{"Start":"02:47.330 ","End":"02:49.265","Text":"That\u0027s the heat equation."},{"Start":"02:49.265 ","End":"02:50.570","Text":"This as it is,"},{"Start":"02:50.570 ","End":"02:51.910","Text":"is not much good."},{"Start":"02:51.910 ","End":"02:55.655","Text":"We have an integral with respect to x and derivative with respect to t"},{"Start":"02:55.655 ","End":"03:00.689","Text":"so we\u0027ll replace wt by wxx and we get this,"},{"Start":"03:00.689 ","End":"03:02.460","Text":"the 1/2 for the 2 cancels."},{"Start":"03:02.460 ","End":"03:05.480","Text":"We\u0027re going to use the integration by parts"},{"Start":"03:05.480 ","End":"03:09.400","Text":"formula on this and to remind you this is what it is."},{"Start":"03:09.400 ","End":"03:13.125","Text":"We\u0027ll take this as g, this as f\u0027."},{"Start":"03:13.125 ","End":"03:15.395","Text":"We\u0027ll need f times g,"},{"Start":"03:15.395 ","End":"03:20.165","Text":"and f is the integral of this with respect to x so it\u0027s just wx,"},{"Start":"03:20.165 ","End":"03:21.635","Text":"just the first derivative."},{"Start":"03:21.635 ","End":"03:24.380","Text":"We also need g\u0027 in the formula."},{"Start":"03:24.380 ","End":"03:28.190","Text":"G is w, so g\u0027 is wx."},{"Start":"03:28.190 ","End":"03:30.890","Text":"Now this first term comes out to be 0,"},{"Start":"03:30.890 ","End":"03:34.185","Text":"because w is 0,"},{"Start":"03:34.185 ","End":"03:37.700","Text":"both when x is 0 and when x is L,"},{"Start":"03:37.700 ","End":"03:40.475","Text":"regardless of what t is."},{"Start":"03:40.475 ","End":"03:46.055","Text":"This is 0-0 and here wx times wx is wx^2."},{"Start":"03:46.055 ","End":"03:51.769","Text":"Looking at this, we see that E\u0027(t )is less than or equal to 0,"},{"Start":"03:51.769 ","End":"03:54.230","Text":"and we also have that E(0) is 0."},{"Start":"03:54.230 ","End":"04:01.985","Text":"Now for function starts off at 0 and it\u0027s decreasing or non-increasing,"},{"Start":"04:01.985 ","End":"04:08.074","Text":"then it\u0027s always less than or equal to 0 so E(t) is less than or equal to 0."},{"Start":"04:08.074 ","End":"04:09.500","Text":"But look, over here,"},{"Start":"04:09.500 ","End":"04:11.600","Text":"we have that E(t) is bigger or equal to"},{"Start":"04:11.600 ","End":"04:15.665","Text":"0 so it\u0027s bigger or equal to 0 and less than or equal to 0,"},{"Start":"04:15.665 ","End":"04:18.090","Text":"then it\u0027s identically equal to 0."},{"Start":"04:18.090 ","End":"04:20.120","Text":"That means that this integral,"},{"Start":"04:20.120 ","End":"04:23.749","Text":"which is what E(t) is, is equal to 0."},{"Start":"04:23.749 ","End":"04:28.130","Text":"The integrant is bigger or equal to 0 and it\u0027s also continuous."},{"Start":"04:28.130 ","End":"04:33.230","Text":"If the integral of a non-negative continuous function is 0,"},{"Start":"04:33.230 ","End":"04:37.085","Text":"then the function itself has to be 0."},{"Start":"04:37.085 ","End":"04:40.130","Text":"The integrant itself has to be 0."},{"Start":"04:40.130 ","End":"04:41.570","Text":"If w squared is 0,"},{"Start":"04:41.570 ","End":"04:44.925","Text":"then w is 0 and if w is 0,"},{"Start":"04:44.925 ","End":"04:46.370","Text":"w is u minus v,"},{"Start":"04:46.370 ","End":"04:51.690","Text":"then u is identically equal to v and that concludes this exercise."}],"ID":30811},{"Watched":false,"Name":"Exercise 4","Duration":"5m 19s","ChapterTopicVideoID":29237,"CourseChapterTopicPlaylistID":294425,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.050","Text":"In this exercise, we\u0027ll be using an energy integral to"},{"Start":"00:04.050 ","End":"00:08.910","Text":"show uniqueness of the solution to the following problem."},{"Start":"00:08.910 ","End":"00:11.475","Text":"This is a heat equation, by the way,"},{"Start":"00:11.475 ","End":"00:17.295","Text":"non-homogeneous and with Von Neumann boundary conditions."},{"Start":"00:17.295 ","End":"00:20.700","Text":"As usual, we\u0027re given the hint as to"},{"Start":"00:20.700 ","End":"00:24.480","Text":"what form the energy integral takes and each exercise is different."},{"Start":"00:24.480 ","End":"00:26.790","Text":"Another remark, when we show uniqueness,"},{"Start":"00:26.790 ","End":"00:28.860","Text":"it doesn\u0027t mean that the solution exists."},{"Start":"00:28.860 ","End":"00:30.735","Text":"There might not be any solution."},{"Start":"00:30.735 ","End":"00:33.570","Text":"It\u0027s just that there can\u0027t be two different ones."},{"Start":"00:33.570 ","End":"00:38.110","Text":"The procedure is the same as in the previous exercises."},{"Start":"00:38.110 ","End":"00:41.705","Text":"We show that if u and v are two solutions,"},{"Start":"00:41.705 ","End":"00:47.070","Text":"then u is identical to v so there really aren\u0027t two, it\u0027s the same one."},{"Start":"00:47.070 ","End":"00:52.505","Text":"We do this by defining w to be the difference of the two and proving a w equals 0,"},{"Start":"00:52.505 ","End":"00:56.750","Text":"therefore, u is identical to v. Let\u0027s, first of all,"},{"Start":"00:56.750 ","End":"00:58.639","Text":"find the PDE for w,"},{"Start":"00:58.639 ","End":"01:01.370","Text":"PDE for u, PDE for v,"},{"Start":"01:01.370 ","End":"01:05.750","Text":"subtract and we get the PDE for w. I want to get"},{"Start":"01:05.750 ","End":"01:11.280","Text":"the boundary conditions and initial conditions for w. These are the ones for u,"},{"Start":"01:11.280 ","End":"01:14.490","Text":"these are the ones for v. Same thing,"},{"Start":"01:14.490 ","End":"01:21.675","Text":"just replace the letter u with a letter v and we subtract this minus this,"},{"Start":"01:21.675 ","End":"01:24.785","Text":"we\u0027ll get w of x naught is 0,"},{"Start":"01:24.785 ","End":"01:28.520","Text":"and this minus this gives us a w_x of 0,"},{"Start":"01:28.520 ","End":"01:30.185","Text":"t is 0,"},{"Start":"01:30.185 ","End":"01:32.105","Text":"similarly for the last one,"},{"Start":"01:32.105 ","End":"01:35.200","Text":"and here again is the PDE."},{"Start":"01:35.200 ","End":"01:40.070","Text":"This is a complete problem for w. In defining the energy integral,"},{"Start":"01:40.070 ","End":"01:43.340","Text":"just like in the hint, as follows,"},{"Start":"01:43.340 ","End":"01:46.730","Text":"we want to find two things about e,"},{"Start":"01:46.730 ","End":"01:52.834","Text":"e of 0 and e prime of t. E of 0 is the following,"},{"Start":"01:52.834 ","End":"01:59.030","Text":"which we get by substituting 0 in place of t. Now,"},{"Start":"01:59.030 ","End":"02:02.315","Text":"e of 0 will turn out to be 0,"},{"Start":"02:02.315 ","End":"02:05.990","Text":"because both of these quantities are 0."},{"Start":"02:05.990 ","End":"02:09.440","Text":"W of x, 0 is 0,"},{"Start":"02:09.440 ","End":"02:18.365","Text":"and therefore the derivative with respect to x along the x axis is also 0."},{"Start":"02:18.365 ","End":"02:20.495","Text":"This is 0, this is 0."},{"Start":"02:20.495 ","End":"02:25.084","Text":"What we get is the integral of 0 squared plus 0 squared,"},{"Start":"02:25.084 ","End":"02:27.710","Text":"which is the integral of 0, which is 0."},{"Start":"02:27.710 ","End":"02:30.410","Text":"Returning to the definition of v of t,"},{"Start":"02:30.410 ","End":"02:35.370","Text":"note that e of t is bigger or equal to 0 for every t. E"},{"Start":"02:35.370 ","End":"02:40.685","Text":"prime of t and let\u0027s assume it\u0027s okay to differentiate under the integral sign,"},{"Start":"02:40.685 ","End":"02:43.340","Text":"w_x squared gives us 2w_x,"},{"Start":"02:43.340 ","End":"02:48.040","Text":"the derivative of w with respect to t and similarly for w twice w,"},{"Start":"02:48.040 ","End":"02:49.270","Text":"w with respect to t. Well,"},{"Start":"02:49.270 ","End":"02:54.020","Text":"the tube is going to cancel with the half and we can switch the order"},{"Start":"02:54.020 ","End":"02:59.520","Text":"of differentiating with respect to x and t. One more thing,"},{"Start":"02:59.520 ","End":"03:04.270","Text":"replace wt by w_xx."},{"Start":"03:04.270 ","End":"03:09.905","Text":"Canceling and separating this into two integrals,"},{"Start":"03:09.905 ","End":"03:11.330","Text":"we get the following,"},{"Start":"03:11.330 ","End":"03:15.620","Text":"and we\u0027re going to do integration by parts twice,"},{"Start":"03:15.620 ","End":"03:18.020","Text":"reuse the letter g and f,"},{"Start":"03:18.020 ","End":"03:22.910","Text":"but the separate problems and each case we\u0027re going to use this formula."},{"Start":"03:22.910 ","End":"03:26.045","Text":"The first integral is this first bracket."},{"Start":"03:26.045 ","End":"03:27.860","Text":"We need fg,"},{"Start":"03:27.860 ","End":"03:32.675","Text":"f is wt and gx is w_xx."},{"Start":"03:32.675 ","End":"03:37.715","Text":"But we can replace wt by w_xx because of the PDE,"},{"Start":"03:37.715 ","End":"03:39.410","Text":"which is a heat equation,"},{"Start":"03:39.410 ","End":"03:42.845","Text":"the second integral goes with the second bracket here."},{"Start":"03:42.845 ","End":"03:47.390","Text":"This time, f is w_x here and here,"},{"Start":"03:47.390 ","End":"03:51.050","Text":"and g prime is w_x."},{"Start":"03:51.050 ","End":"03:58.910","Text":"Now both of these expressions are 0 because of w_x, which is 0,"},{"Start":"03:58.910 ","End":"04:02.090","Text":"both when x equals 0 and when x equals l"},{"Start":"04:02.090 ","End":"04:06.275","Text":"regardless of t. What we\u0027re left with is this and this,"},{"Start":"04:06.275 ","End":"04:11.270","Text":"and each of these integrals are bigger or equal to 0,"},{"Start":"04:11.270 ","End":"04:13.430","Text":"but the whole expression is less than or equal to"},{"Start":"04:13.430 ","End":"04:17.915","Text":"0 and so we\u0027ve just shown that e prime of t is less than or equal to 0."},{"Start":"04:17.915 ","End":"04:20.935","Text":"But before we had e of 0 equals 0,"},{"Start":"04:20.935 ","End":"04:27.515","Text":"these two conditions imply that e of t is less than or equal to 0."},{"Start":"04:27.515 ","End":"04:35.270","Text":"It starts out at 0 and is non-increasing so it has to stay less than or equal to 0."},{"Start":"04:35.270 ","End":"04:36.710","Text":"On the other hand,"},{"Start":"04:36.710 ","End":"04:40.550","Text":"we also showed that e of t is bigger or equal to 0,"},{"Start":"04:40.550 ","End":"04:45.875","Text":"less than or equal to, and bigger or equal to means that it is 0."},{"Start":"04:45.875 ","End":"04:48.635","Text":"If v of t is identically 0,"},{"Start":"04:48.635 ","End":"04:52.355","Text":"then this expression, which is what EFT is,"},{"Start":"04:52.355 ","End":"04:54.605","Text":"is equal to 0."},{"Start":"04:54.605 ","End":"04:58.670","Text":"When you have a non-negative and continuous function"},{"Start":"04:58.670 ","End":"05:01.819","Text":"whose integral is 0, then the function,"},{"Start":"05:01.819 ","End":"05:04.595","Text":"the integrant is 0,"},{"Start":"05:04.595 ","End":"05:06.710","Text":"and then using basic algebra,"},{"Start":"05:06.710 ","End":"05:08.960","Text":"if you have a squared plus b squared equals 0,"},{"Start":"05:08.960 ","End":"05:15.205","Text":"then necessarily b is 0 and so we get that w is identically 0."},{"Start":"05:15.205 ","End":"05:20.730","Text":"That means that u is identically equal to v and that concludes this exercise."}],"ID":30812},{"Watched":false,"Name":"Exercise 5","Duration":"5m 7s","ChapterTopicVideoID":29238,"CourseChapterTopicPlaylistID":294425,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.605","Text":"In this exercise, we\u0027re going to use an energy integral to show"},{"Start":"00:04.605 ","End":"00:08.535","Text":"uniqueness of the solution to the following problem."},{"Start":"00:08.535 ","End":"00:12.345","Text":"It looks like a heat equation, but it\u0027s not."},{"Start":"00:12.345 ","End":"00:15.210","Text":"The energy integral we\u0027re given is this."},{"Start":"00:15.210 ","End":"00:19.650","Text":"It\u0027s different in each exercise and never mind how we got to this,"},{"Start":"00:19.650 ","End":"00:21.315","Text":"it\u0027s just given us a hint."},{"Start":"00:21.315 ","End":"00:26.625","Text":"The solution, it\u0027s the same in all the exercise at the beginning is the same."},{"Start":"00:26.625 ","End":"00:30.555","Text":"We want to show that if 2 solutions u and v exist,"},{"Start":"00:30.555 ","End":"00:32.580","Text":"then they\u0027re actually the same,"},{"Start":"00:32.580 ","End":"00:35.130","Text":"but u is identically equal to v,"},{"Start":"00:35.130 ","End":"00:41.745","Text":"and we do this by defining the difference as w and proving that w is identically 0."},{"Start":"00:41.745 ","End":"00:44.640","Text":"Let\u0027s see what the PDE for w is."},{"Start":"00:44.640 ","End":"00:48.225","Text":"The PDE for u is this, so for v,"},{"Start":"00:48.225 ","End":"00:54.240","Text":"lets replace u with the letter v and subtract the 2 equations."},{"Start":"00:54.240 ","End":"00:59.415","Text":"We get this, u_t minus v_t is w_t,"},{"Start":"00:59.415 ","End":"01:02.810","Text":"the difference of the derivative is the derivative of the difference, and so on,"},{"Start":"01:02.810 ","End":"01:06.080","Text":"and we get this PDE for w. Now we want"},{"Start":"01:06.080 ","End":"01:12.140","Text":"the initial and boundary conditions for w. These are the ones for u and v,"},{"Start":"01:12.140 ","End":"01:14.915","Text":"unlike in previous exercises."},{"Start":"01:14.915 ","End":"01:18.350","Text":"If we subtract say, this minus this,"},{"Start":"01:18.350 ","End":"01:24.170","Text":"we get that u minus v is w(x,0) is f minus f,"},{"Start":"01:24.170 ","End":"01:26.465","Text":"which is 0, and similarly for the other 2,"},{"Start":"01:26.465 ","End":"01:29.180","Text":"and this is the PDE copied from here."},{"Start":"01:29.180 ","End":"01:33.845","Text":"This is the complete problem for w. Now,"},{"Start":"01:33.845 ","End":"01:35.840","Text":"we define the energy integral."},{"Start":"01:35.840 ","End":"01:39.350","Text":"Well, we just copied it from the hint, that\u0027s E(t)."},{"Start":"01:39.350 ","End":"01:41.930","Text":"Now, what is E(0)?"},{"Start":"01:41.930 ","End":"01:44.065","Text":"Replace t by 0,"},{"Start":"01:44.065 ","End":"01:46.245","Text":"and this is what we get."},{"Start":"01:46.245 ","End":"01:50.565","Text":"Now, w(x,0) is 0, that\u0027s given."},{"Start":"01:50.565 ","End":"01:52.830","Text":"We can derive from this,"},{"Start":"01:52.830 ","End":"01:57.345","Text":"the w(x,0) is also 0."},{"Start":"01:57.345 ","End":"02:02.435","Text":"E(0) is the integral of 0, which is 0."},{"Start":"02:02.435 ","End":"02:05.870","Text":"The next thing you want to know about E(t) is what\u0027s its derivative?"},{"Start":"02:05.870 ","End":"02:07.925","Text":"What is E\u0027(t)?"},{"Start":"02:07.925 ","End":"02:08.960","Text":"Returning to E(t),"},{"Start":"02:08.960 ","End":"02:10.730","Text":"we want to differentiate this,"},{"Start":"02:10.730 ","End":"02:14.045","Text":"we want d by dt of this expression,"},{"Start":"02:14.045 ","End":"02:18.950","Text":"and let\u0027s just assume that it\u0027s okay to differentiate under the integral sign,"},{"Start":"02:18.950 ","End":"02:22.780","Text":"it is under certain conditions assumed that they are met here."},{"Start":"02:22.780 ","End":"02:25.345","Text":"We differentiate w_x^2,"},{"Start":"02:25.345 ","End":"02:30.830","Text":"we get 2w_x times the derivative of w_x with respect to t is w_xt,"},{"Start":"02:30.830 ","End":"02:41.085","Text":"similarly with w. Here we\u0027re going to replace w_t by what it\u0027s equal to from the PDE."},{"Start":"02:41.085 ","End":"02:47.600","Text":"Doing that, we get the following and we also reverse the order of the mixed derivative."},{"Start":"02:47.600 ","End":"02:51.080","Text":"We break this up into 3 separate integrals."},{"Start":"02:51.080 ","End":"02:52.280","Text":"We have this part,"},{"Start":"02:52.280 ","End":"02:53.720","Text":"then we have to 2w,"},{"Start":"02:53.720 ","End":"03:00.095","Text":"w_xx and we have the minus 2w Beta w, which is this."},{"Start":"03:00.095 ","End":"03:02.900","Text":"We\u0027re going to use integration by parts on this,"},{"Start":"03:02.900 ","End":"03:05.240","Text":"and we\u0027re going to use integration by parts on this,"},{"Start":"03:05.240 ","End":"03:10.065","Text":"that\u0027s a different g and f. This is the formula we\u0027re going to be using,"},{"Start":"03:10.065 ","End":"03:14.460","Text":"and for this one we get f is w_t,"},{"Start":"03:14.460 ","End":"03:16.950","Text":"and g is w_x."},{"Start":"03:16.950 ","End":"03:19.080","Text":"Here we have again,"},{"Start":"03:19.080 ","End":"03:24.120","Text":"f is w_t and g\u0027 is w_xx,"},{"Start":"03:24.120 ","End":"03:27.355","Text":"and similarly with the second integration by part."},{"Start":"03:27.355 ","End":"03:32.510","Text":"Now, this part is 0 and so is this part because w_x is"},{"Start":"03:32.510 ","End":"03:37.770","Text":"0 at the limits just from the boundary conditions,"},{"Start":"03:37.770 ","End":"03:38.910","Text":"and this w_t,"},{"Start":"03:38.910 ","End":"03:41.295","Text":"we replace it by w_xx."},{"Start":"03:41.295 ","End":"03:44.565","Text":"Here we have w_xx^2,"},{"Start":"03:44.565 ","End":"03:47.325","Text":"here we have w_x^2."},{"Start":"03:47.325 ","End":"03:48.840","Text":"Altogether we have minus,"},{"Start":"03:48.840 ","End":"03:50.020","Text":"minus and minus,"},{"Start":"03:50.020 ","End":"03:53.540","Text":"Beta is positive and this is non-negative, non-negative,"},{"Start":"03:53.540 ","End":"03:55.670","Text":"non-negative, so altogether,"},{"Start":"03:55.670 ","End":"03:57.680","Text":"we get minus non-negative,"},{"Start":"03:57.680 ","End":"04:01.210","Text":"which is non-positive or less than or equal to 0."},{"Start":"04:01.210 ","End":"04:07.850","Text":"We have properties of E(t) that its derivative is less than or equal to 0,"},{"Start":"04:07.850 ","End":"04:10.369","Text":"and then 0 it\u0027s equal to 0."},{"Start":"04:10.369 ","End":"04:16.415","Text":"A function which is continuous starts at 0 and is non-increasing,"},{"Start":"04:16.415 ","End":"04:19.205","Text":"remains less than or equal to 0."},{"Start":"04:19.205 ","End":"04:20.900","Text":"On the other hand,"},{"Start":"04:20.900 ","End":"04:24.500","Text":"we know already that E(t) is bigger or equal to 0, we showed that."},{"Start":"04:24.500 ","End":"04:30.350","Text":"From this, we get that E(t) is identically 0."},{"Start":"04:30.350 ","End":"04:33.360","Text":"E(t) is this integral,"},{"Start":"04:33.360 ","End":"04:35.115","Text":"this is 0,"},{"Start":"04:35.115 ","End":"04:39.815","Text":"and the integrand is non-negative and continuous."},{"Start":"04:39.815 ","End":"04:43.595","Text":"The integral of a non-negative continuous function,"},{"Start":"04:43.595 ","End":"04:48.250","Text":"if it\u0027s 0, then the integrand has to be 0."},{"Start":"04:48.250 ","End":"04:51.420","Text":"This expression is 0,"},{"Start":"04:51.420 ","End":"04:53.060","Text":"and from basic algebra,"},{"Start":"04:53.060 ","End":"04:54.890","Text":"if you have a sum of squares is 0,"},{"Start":"04:54.890 ","End":"04:56.720","Text":"then each of the numbers is 0."},{"Start":"04:56.720 ","End":"04:58.985","Text":"In particular here the second one is,"},{"Start":"04:58.985 ","End":"05:01.280","Text":"so w(x,t) is 0,"},{"Start":"05:01.280 ","End":"05:05.060","Text":"which means that u is equal to v or identically equal to,"},{"Start":"05:05.060 ","End":"05:08.070","Text":"and that concludes this exercise."}],"ID":30813}],"Thumbnail":null,"ID":294425}]
