[{"Name":"Exam-Level Summarizing Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"6m 48s","ChapterTopicVideoID":29310,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29310.jpeg","UploadDate":"2022-06-06T10:48:44.3370000","DurationForVideoObject":"PT6M48S","Description":null,"MetaTitle":"Exercise 1 - Exam-Level Summarizing Exercises: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Exam-Level Summarizing Exercises practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/partial-differential-equations/exam_level-summarizing-exercises/exam_level-summarizing-exercises/vid30866","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.170","Text":"In this exercise, we\u0027re given a boundary value problem for u(x, y) as follows."},{"Start":"00:07.170 ","End":"00:15.120","Text":"This is a quasi-linear equation of the first order defined on the first quadrant,"},{"Start":"00:15.120 ","End":"00:17.745","Text":"where x and y are both positive."},{"Start":"00:17.745 ","End":"00:27.785","Text":"We\u0027re given boundary values along the positive x-axis and the positive y-axis."},{"Start":"00:27.785 ","End":"00:34.310","Text":"We use the method of characteristics just to remind you the notation quasi-linear"},{"Start":"00:34.310 ","End":"00:40.740","Text":"equation is Au_x plus Bu_y equals D. In our case,"},{"Start":"00:40.740 ","End":"00:42.330","Text":"A is the constant 1,"},{"Start":"00:42.330 ","End":"00:43.905","Text":"B is the constant 1,"},{"Start":"00:43.905 ","End":"00:46.870","Text":"and C is the function u."},{"Start":"00:46.880 ","End":"00:50.505","Text":"We solve it with the usual 4 steps."},{"Start":"00:50.505 ","End":"00:53.330","Text":"Step 1, we parameterize the initial curve."},{"Start":"00:53.330 ","End":"00:58.825","Text":"Now, the initial curve is the positive x-axis and the positive y-axis."},{"Start":"00:58.825 ","End":"01:02.285","Text":"We\u0027ll initialize it as follows."},{"Start":"01:02.285 ","End":"01:06.140","Text":"We\u0027ll take the line for Tau and bend it"},{"Start":"01:06.140 ","End":"01:11.008","Text":"90 degrees so that the negative Tau falls along the y-axis,"},{"Start":"01:11.008 ","End":"01:14.845","Text":"and the positive Tau falls along the x axis."},{"Start":"01:14.845 ","End":"01:17.135","Text":"This curve is called Gamma,"},{"Start":"01:17.135 ","End":"01:20.415","Text":"and this is where t=0."},{"Start":"01:20.415 ","End":"01:23.430","Text":"We can write it as follows,"},{"Start":"01:23.430 ","End":"01:26.420","Text":"to be precise, Gamma is not this,"},{"Start":"01:26.420 ","End":"01:29.165","Text":"but it\u0027s the curve in space above this,"},{"Start":"01:29.165 ","End":"01:32.975","Text":"where we take the third axis as u."},{"Start":"01:32.975 ","End":"01:36.395","Text":"We have Gamma is x(0, Tau),"},{"Start":"01:36.395 ","End":"01:37.993","Text":"y(0, Tau),"},{"Start":"01:37.993 ","End":"01:39.680","Text":"and u(0, Tau)."},{"Start":"01:39.680 ","End":"01:44.203","Text":"It\u0027s a curve in 3D and its projection is this, where x(0,"},{"Start":"01:44.203 ","End":"01:49.220","Text":"Tau) is just Tau along the positive x-axis."},{"Start":"01:49.220 ","End":"01:50.510","Text":"When Tau is 1,"},{"Start":"01:50.510 ","End":"01:53.015","Text":"x is 1, Tau is 2, x is 2."},{"Start":"01:53.015 ","End":"01:57.530","Text":"Along the y-axis, x=0,"},{"Start":"01:57.530 ","End":"01:59.210","Text":"so that\u0027s this part."},{"Start":"01:59.210 ","End":"02:04.940","Text":"As for y, y=0 along this axis,"},{"Start":"02:04.940 ","End":"02:06.510","Text":"where Tau is positive."},{"Start":"02:06.510 ","End":"02:08.255","Text":"But when Tau is negative,"},{"Start":"02:08.255 ","End":"02:10.055","Text":"y is minus Tau,"},{"Start":"02:10.055 ","End":"02:11.530","Text":"when Tau is minus 3,"},{"Start":"02:11.530 ","End":"02:12.790","Text":"y is plus 3."},{"Start":"02:12.790 ","End":"02:16.485","Text":"That\u0027s this part, and u, we would give them,"},{"Start":"02:16.485 ","End":"02:20.390","Text":"we go back and look,=0 along"},{"Start":"02:20.390 ","End":"02:26.905","Text":"the y-axis and 0 along the x-axis except for the part between 0 and 1,"},{"Start":"02:26.905 ","End":"02:28.760","Text":"where it\u0027s equal to 1."},{"Start":"02:28.760 ","End":"02:30.994","Text":"We can say that u(0,"},{"Start":"02:30.994 ","End":"02:37.190","Text":"Tau) is 1 and it falls exactly when Tau is between 0 and 1,"},{"Start":"02:37.190 ","End":"02:39.440","Text":"and everything else is 0."},{"Start":"02:39.440 ","End":"02:42.715","Text":"It\u0027s when Tau is bigger than 1 or less than 0."},{"Start":"02:42.715 ","End":"02:44.810","Text":"Now, Step 2 where we have to find"},{"Start":"02:44.810 ","End":"02:49.570","Text":"the characteristic curves by solving the characteristic ODEs,"},{"Start":"02:49.570 ","End":"02:51.425","Text":"dx by dt equals A,"},{"Start":"02:51.425 ","End":"02:53.150","Text":"dy by dt equals B,"},{"Start":"02:53.150 ","End":"02:54.800","Text":"there\u0027s actually a third one,"},{"Start":"02:54.800 ","End":"02:56.450","Text":"du by dt equals C,"},{"Start":"02:56.450 ","End":"02:59.000","Text":"but we\u0027ll only use that in the following step."},{"Start":"02:59.000 ","End":"03:04.140","Text":"Here, we get that dx by dt is A,"},{"Start":"03:04.140 ","End":"03:05.520","Text":"which is 1,"},{"Start":"03:05.520 ","End":"03:08.545","Text":"dy by dt is B, which is 1."},{"Start":"03:08.545 ","End":"03:12.990","Text":"We can integrate with respect to t. For x,"},{"Start":"03:12.990 ","End":"03:17.495","Text":"we get that x=t and not a constant,"},{"Start":"03:17.495 ","End":"03:19.745","Text":"but an arbitrary function of Tau."},{"Start":"03:19.745 ","End":"03:25.705","Text":"Similarly, y is t plus the different arbitrary function of Tau."},{"Start":"03:25.705 ","End":"03:29.040","Text":"It\u0027s a constant as far as t goes, if you like."},{"Start":"03:29.040 ","End":"03:34.077","Text":"Then we can say that if we put t=0, we get that x(0,"},{"Start":"03:34.077 ","End":"03:35.820","Text":"Tau) is c1(t),"},{"Start":"03:35.820 ","End":"03:39.045","Text":"y(0, Tau) is c2(t)."},{"Start":"03:39.045 ","End":"03:43.500","Text":"Now, x(0, Tau) is also equal to this."},{"Start":"03:43.500 ","End":"03:48.120","Text":"We get that c1(t) is equal to this,"},{"Start":"03:48.120 ","End":"03:52.685","Text":"and c2(t) is equal to this."},{"Start":"03:52.685 ","End":"03:56.610","Text":"Yeah, x is t plus c1,"},{"Start":"03:56.610 ","End":"03:59.520","Text":"so x is t plus this,"},{"Start":"03:59.520 ","End":"04:01.530","Text":"and y is t plus c2,"},{"Start":"04:01.530 ","End":"04:04.490","Text":"so y is t plus this."},{"Start":"04:04.490 ","End":"04:09.455","Text":"That gives us x and y in terms of t and Tau."},{"Start":"04:09.455 ","End":"04:11.840","Text":"These are called the characteristic curves."},{"Start":"04:11.840 ","End":"04:14.545","Text":"Let\u0027s get on to Step 3."},{"Start":"04:14.545 ","End":"04:18.010","Text":"In Step 3, we\u0027re going to find u in terms of t and"},{"Start":"04:18.010 ","End":"04:22.195","Text":"Tau from the third characteristic equation, ut=c."},{"Start":"04:22.195 ","End":"04:25.610","Text":"In our case, c=u,"},{"Start":"04:25.610 ","End":"04:31.035","Text":"so du by dt equals u, that\u0027s from this."},{"Start":"04:31.035 ","End":"04:38.670","Text":"We had the initial condition that u of 0 and Tau is 1 between 0 and 1,"},{"Start":"04:38.670 ","End":"04:40.799","Text":"and 0 elsewhere,"},{"Start":"04:40.799 ","End":"04:42.685","Text":"du by dt equals u,"},{"Start":"04:42.685 ","End":"04:45.970","Text":"we get u is some constant e^t,"},{"Start":"04:45.970 ","End":"04:47.290","Text":"but not a constant,"},{"Start":"04:47.290 ","End":"04:49.515","Text":"an arbitrary function of Tau."},{"Start":"04:49.515 ","End":"04:54.175","Text":"C3 of Tau is what we get if we put t=0 here,"},{"Start":"04:54.175 ","End":"04:57.580","Text":"because e^t is 1 when t is 0."},{"Start":"04:57.580 ","End":"05:00.910","Text":"We get u of 0 and Tau, which is this,"},{"Start":"05:00.910 ","End":"05:03.825","Text":"so c3 of Tau is this,"},{"Start":"05:03.825 ","End":"05:09.065","Text":"which means that u(t, Tau) is c3 of Tau times e^t."},{"Start":"05:09.065 ","End":"05:13.460","Text":"We can just stick the e^t on top of this 1 here,"},{"Start":"05:13.460 ","End":"05:15.580","Text":"so this is u(t, Tau)."},{"Start":"05:15.580 ","End":"05:17.370","Text":"Now, we come to the final step."},{"Start":"05:17.370 ","End":"05:20.870","Text":"Step 4, where we have to get back to u(x,"},{"Start":"05:20.870 ","End":"05:23.105","Text":"y) from u of t and Tau."},{"Start":"05:23.105 ","End":"05:26.540","Text":"We have x and y in terms of t and Tau,"},{"Start":"05:26.540 ","End":"05:29.290","Text":"but we don\u0027t have t and Tau in terms of x, y."},{"Start":"05:29.290 ","End":"05:32.985","Text":"Note that if Tau is bigger than 0,"},{"Start":"05:32.985 ","End":"05:35.345","Text":"then we can compute x minus y,"},{"Start":"05:35.345 ","End":"05:40.975","Text":"because x will be t plus Tau and y will be just t,"},{"Start":"05:40.975 ","End":"05:44.970","Text":"so x minus y will be just Tau."},{"Start":"05:44.970 ","End":"05:47.910","Text":"On the other hand, if Tau is less than 0,"},{"Start":"05:47.910 ","End":"05:54.375","Text":"then x minus y is t minus Tau,"},{"Start":"05:54.375 ","End":"05:56.670","Text":"which also comes out to be Tau."},{"Start":"05:56.670 ","End":"06:01.070","Text":"In any case, Tau is equal to x minus y,"},{"Start":"06:01.070 ","End":"06:05.854","Text":"what we could do is put x minus y of Tau here, not just yet,"},{"Start":"06:05.854 ","End":"06:11.090","Text":"first of all, we have to figure out what t is when Tau is between 0 and 1."},{"Start":"06:11.090 ","End":"06:13.835","Text":"For convenience, copy this down here."},{"Start":"06:13.835 ","End":"06:22.880","Text":"Now, notice that when Tau is bigger than 0 from here, then y=t."},{"Start":"06:22.880 ","End":"06:25.690","Text":"In this part, Tau is bigger than 0,"},{"Start":"06:25.690 ","End":"06:28.100","Text":"so we can replace t by y."},{"Start":"06:28.100 ","End":"06:30.395","Text":"Now, we can replace both things."},{"Start":"06:30.395 ","End":"06:32.360","Text":"Tau is x minus y,"},{"Start":"06:32.360 ","End":"06:33.875","Text":"t is y,"},{"Start":"06:33.875 ","End":"06:39.470","Text":"so what we get is that u of x and y is e to the y,"},{"Start":"06:39.470 ","End":"06:45.540","Text":"0 less than x minus y less than 1 and 0, else or otherwise."},{"Start":"06:45.540 ","End":"06:49.420","Text":"That\u0027s the answer, and we are done."}],"ID":30866},{"Watched":false,"Name":"Exercise 2","Duration":"5m 43s","ChapterTopicVideoID":29311,"CourseChapterTopicPlaylistID":294426,"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 have the following problem which"},{"Start":"00:04.170 ","End":"00:08.910","Text":"is a wave equation on a semi-infinite interval,"},{"Start":"00:08.910 ","End":"00:13.455","Text":"and it has a Neumann boundary condition."},{"Start":"00:13.455 ","End":"00:16.800","Text":"These are the initial conditions: U is f,"},{"Start":"00:16.800 ","End":"00:18.570","Text":"and U by the t is g,"},{"Start":"00:18.570 ","End":"00:21.975","Text":"where g is defined piecewise."},{"Start":"00:21.975 ","End":"00:24.765","Text":"When we have a Neumann boundary condition,"},{"Start":"00:24.765 ","End":"00:31.440","Text":"we extend these functions to even functions."},{"Start":"00:31.440 ","End":"00:35.710","Text":"If it was Dirichlet, then we\u0027d extend to odd functions."},{"Start":"00:37.310 ","End":"00:40.125","Text":"I\u0027ll show you a diagram."},{"Start":"00:40.125 ","End":"00:43.620","Text":"G is the part in blue here."},{"Start":"00:43.620 ","End":"00:47.420","Text":"We take a mirror image to make it an even function,"},{"Start":"00:47.420 ","End":"00:49.370","Text":"so extend it like so."},{"Start":"00:49.370 ","End":"00:52.565","Text":"So what we have is that between minus 1 and 1,"},{"Start":"00:52.565 ","End":"00:56.515","Text":"it\u0027s equal to 1, and 0 outside of that."},{"Start":"00:56.515 ","End":"00:59.990","Text":"F, which is 0, just extends to 0."},{"Start":"00:59.990 ","End":"01:02.645","Text":"We\u0027re going to use d\u0027Alembert\u0027s formula,"},{"Start":"01:02.645 ","End":"01:04.310","Text":"which is the following."},{"Start":"01:04.310 ","End":"01:06.770","Text":"We\u0027re only going to apply this to the case where x"},{"Start":"01:06.770 ","End":"01:10.265","Text":"is bigger than 0 so that u tilde is equal to u."},{"Start":"01:10.265 ","End":"01:14.510","Text":"Note that this part is 0 because f tilde is 0."},{"Start":"01:14.510 ","End":"01:21.860","Text":"Also, a equals 1 in this formula because u_tt is a^2u_xx,"},{"Start":"01:21.860 ","End":"01:24.700","Text":"but here, a is 1."},{"Start":"01:24.700 ","End":"01:27.420","Text":"So a is 1. This part is 0."},{"Start":"01:27.420 ","End":"01:30.510","Text":"What we get is the following,"},{"Start":"01:30.510 ","End":"01:33.725","Text":"and we need to handle this case by case."},{"Start":"01:33.725 ","End":"01:35.810","Text":"The first case is,"},{"Start":"01:35.810 ","End":"01:37.800","Text":"when this interval x minus t,"},{"Start":"01:37.800 ","End":"01:42.368","Text":"x plus t is completely to the right of the part where this is 1,"},{"Start":"01:42.368 ","End":"01:45.500","Text":"this comes out to be where x minus t is"},{"Start":"01:45.500 ","End":"01:49.244","Text":"bigger or equal to 1 or x is bigger or equal to t plus 1."},{"Start":"01:49.244 ","End":"01:54.560","Text":"Then U is just the integral of 0, so it\u0027s 0."},{"Start":"01:54.560 ","End":"01:56.935","Text":"Next, case 2,"},{"Start":"01:56.935 ","End":"02:01.320","Text":"which is when x minus t is on one side of 1,"},{"Start":"02:01.320 ","End":"02:03.720","Text":"and x plus t is on the other side,"},{"Start":"02:03.720 ","End":"02:05.570","Text":"we can express this as minus 1,"},{"Start":"02:05.570 ","End":"02:07.010","Text":"less than or equal to x minus t,"},{"Start":"02:07.010 ","End":"02:08.405","Text":"less than or equal to 1,"},{"Start":"02:08.405 ","End":"02:10.296","Text":"less than or equal to x plus t,"},{"Start":"02:10.296 ","End":"02:13.100","Text":"and that comes out to be that"},{"Start":"02:13.100 ","End":"02:17.660","Text":"the maximum of 1 minus t and t minus 1 is less than or equal to x."},{"Start":"02:17.660 ","End":"02:22.010","Text":"We get this from this inequality and from this inequality,"},{"Start":"02:22.010 ","End":"02:24.980","Text":"and this, we get from this inequality."},{"Start":"02:24.980 ","End":"02:26.490","Text":"X minus t less than or equal to 1,"},{"Start":"02:26.490 ","End":"02:28.475","Text":"so x less than or equal to t plus 1."},{"Start":"02:28.475 ","End":"02:33.490","Text":"If we want, we could rewrite this as absolute value of t minus 1."},{"Start":"02:33.490 ","End":"02:40.115","Text":"In this case, the integral is just the integral of 1 from x minus t to 1."},{"Start":"02:40.115 ","End":"02:44.015","Text":"That comes out to be one-half times the length of this interval,"},{"Start":"02:44.015 ","End":"02:49.283","Text":"which is 1 minus x plus t. Here\u0027s our third case,"},{"Start":"02:49.283 ","End":"02:55.710","Text":"where the interval falls inside this interval from minus 1 to 1."},{"Start":"02:55.710 ","End":"02:57.590","Text":"This is bigger or equal to minus 1,"},{"Start":"02:57.590 ","End":"02:59.465","Text":"and this is less than or equal to 1,"},{"Start":"02:59.465 ","End":"03:02.930","Text":"and it comes out to be t minus 1, less than or equal to x,"},{"Start":"03:02.930 ","End":"03:06.530","Text":"less than or equal to 1 minus t. In this case,"},{"Start":"03:06.530 ","End":"03:11.015","Text":"the integral is just 1 everywhere on this interval."},{"Start":"03:11.015 ","End":"03:14.870","Text":"So it comes out to be the length of the interval over 2,"},{"Start":"03:14.870 ","End":"03:16.700","Text":"which is 2t over 2,"},{"Start":"03:16.700 ","End":"03:20.658","Text":"which is t. Case 4,"},{"Start":"03:20.658 ","End":"03:22.679","Text":"let me show the picture."},{"Start":"03:22.679 ","End":"03:26.630","Text":"That\u0027s when this right endpoint is between minus 1 and 1,"},{"Start":"03:26.630 ","End":"03:30.170","Text":"and the left endpoint is less than minus 1."},{"Start":"03:30.170 ","End":"03:31.475","Text":"You could write it this way,"},{"Start":"03:31.475 ","End":"03:33.260","Text":"which reduces to this."},{"Start":"03:33.260 ","End":"03:37.820","Text":"Now, the minimum of 1 minus t and t minus 1 has to be less than or equal to 0."},{"Start":"03:37.820 ","End":"03:39.170","Text":"The minimum of something,"},{"Start":"03:39.170 ","End":"03:42.470","Text":"and it\u0027s negative, is always less than or equal to 0."},{"Start":"03:42.470 ","End":"03:45.090","Text":"It\u0027s also bigger or equal to 0,"},{"Start":"03:45.090 ","End":"03:47.385","Text":"so x has to be 0,"},{"Start":"03:47.385 ","End":"03:50.460","Text":"and if x is 0, then t equals 1."},{"Start":"03:50.460 ","End":"03:53.346","Text":"We only have 1 solution."},{"Start":"03:53.346 ","End":"03:57.920","Text":"It turns out that this interval is exactly then from minus 1 to 1,"},{"Start":"03:57.920 ","End":"03:59.840","Text":"which is covered by this case in"},{"Start":"03:59.840 ","End":"04:04.155","Text":"the extreme when x minus t is here and x plus t is here,"},{"Start":"04:04.155 ","End":"04:07.665","Text":"so we can just ignore this case."},{"Start":"04:07.665 ","End":"04:09.605","Text":"There\u0027s 1 more case,"},{"Start":"04:09.605 ","End":"04:16.575","Text":"and that is when this interval completely contains this interval,"},{"Start":"04:16.575 ","End":"04:21.260","Text":"and we can write that as follows: x minus t is less than or equal to 1,"},{"Start":"04:21.260 ","End":"04:22.820","Text":"x plus t is bigger or equal to 1,"},{"Start":"04:22.820 ","End":"04:26.855","Text":"which means that x is between 0 and t minus 1,"},{"Start":"04:26.855 ","End":"04:30.065","Text":"and t has to be bigger than 1 in this case."},{"Start":"04:30.065 ","End":"04:34.190","Text":"The integral comes out to be the integral of 1 from here to here,"},{"Start":"04:34.190 ","End":"04:35.765","Text":"which is just 1."},{"Start":"04:35.765 ","End":"04:37.640","Text":"These are all the 5 cases,"},{"Start":"04:37.640 ","End":"04:40.595","Text":"except one of the cases is dropped out."},{"Start":"04:40.595 ","End":"04:43.527","Text":"So if we summarize,"},{"Start":"04:43.527 ","End":"04:47.720","Text":"then here are the 4 cases that we have,"},{"Start":"04:47.720 ","End":"04:50.225","Text":"and we\u0027ll leave the solution like this."},{"Start":"04:50.225 ","End":"04:52.625","Text":"I don\u0027t think it can be simplified."},{"Start":"04:52.625 ","End":"04:56.150","Text":"Let\u0027s finish with an animation of what this function looks"},{"Start":"04:56.150 ","End":"05:01.800","Text":"like for various values of t starting from 0 and increasing."}],"ID":30867},{"Watched":false,"Name":"Exercise 3","Duration":"8m 23s","ChapterTopicVideoID":29312,"CourseChapterTopicPlaylistID":294426,"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":"In this exercise, we\u0027re given that the function u is"},{"Start":"00:04.500 ","End":"00:09.030","Text":"a solution of this initial boundary and value problem."},{"Start":"00:09.030 ","End":"00:14.250","Text":"This is the PDE is defined on an interval x"},{"Start":"00:14.250 ","End":"00:19.575","Text":"from 0 to 1 and t presumably time from 0 onwards."},{"Start":"00:19.575 ","End":"00:21.930","Text":"This is the initial condition."},{"Start":"00:21.930 ","End":"00:25.800","Text":"It\u0027s what happens when time equals 0 for both u"},{"Start":"00:25.800 ","End":"00:30.270","Text":"and u_t and at the end points of x, when x is 0,"},{"Start":"00:30.270 ","End":"00:34.350","Text":"x equals 1, we get boundary conditions of the Neumann type because"},{"Start":"00:34.350 ","End":"00:39.165","Text":"the instead of u of we have du by dx at these 2 end points."},{"Start":"00:39.165 ","End":"00:47.030","Text":"That we\u0027re also given that the limit as t goes to infinity of the solution u of x,"},{"Start":"00:47.030 ","End":"00:51.580","Text":"t exists and is finite and we call it U(x)."},{"Start":"00:51.580 ","End":"00:58.230","Text":"Our task is to find the constant A that here and this function U(x)."},{"Start":"00:58.230 ","End":"01:05.030","Text":"This is like the asymptotic solution for the equation when t is very, very large."},{"Start":"01:05.030 ","End":"01:10.475","Text":"We\u0027ll start by taking the limit as t goes to infinity of this PDE."},{"Start":"01:10.475 ","End":"01:14.495","Text":"Take the limit of everything as t goes to infinity."},{"Start":"01:14.495 ","End":"01:23.145","Text":"Now, assuming it\u0027s okay to interchange the differentiation with the limit of u,"},{"Start":"01:23.145 ","End":"01:26.430","Text":"and then differentiate it twice with respect to t,"},{"Start":"01:26.430 ","End":"01:32.130","Text":"and similarly here respect to t and here twice with respect to x,"},{"Start":"01:32.130 ","End":"01:36.920","Text":"and this doesn\u0027t depend on t. What we get,"},{"Start":"01:36.920 ","End":"01:40.430","Text":"this is given to us to be U(x),"},{"Start":"01:40.430 ","End":"01:44.675","Text":"and this is the same thing that\u0027s what this u is,"},{"Start":"01:44.675 ","End":"01:47.570","Text":"and each of these is going to be 0 because"},{"Start":"01:47.570 ","End":"01:51.140","Text":"the function of x differentiated with respect to t. Here"},{"Start":"01:51.140 ","End":"01:58.145","Text":"we have second derivative of u with respect to x and here the limit of Ax is just Ax."},{"Start":"01:58.145 ","End":"02:00.875","Text":"Like we said, this is 0, this is 0."},{"Start":"02:00.875 ","End":"02:02.750","Text":"This is just the second derivative."},{"Start":"02:02.750 ","End":"02:06.760","Text":"We can write it as u double prime of x plus Ax."},{"Start":"02:06.760 ","End":"02:09.220","Text":"We get the differential equation,"},{"Start":"02:09.220 ","End":"02:12.460","Text":"u double prime of x is minus Ax."},{"Start":"02:12.460 ","End":"02:19.605","Text":"Integrate once, and we get u prime is minus A over 2x^2 plus a constant B."},{"Start":"02:19.605 ","End":"02:24.415","Text":"We could integrate again but it\u0027s just turns out not to be useful,"},{"Start":"02:24.415 ","End":"02:26.185","Text":"at least not at this stage."},{"Start":"02:26.185 ","End":"02:32.815","Text":"Let\u0027s take the limit as t goes to infinity of the boundary conditions."},{"Start":"02:32.815 ","End":"02:36.120","Text":"These will give us the A and B that we need."},{"Start":"02:36.120 ","End":"02:40.465","Text":"So these are the boundary conditions."},{"Start":"02:40.465 ","End":"02:44.560","Text":"Now, take the limit as t goes to infinity of each of them,"},{"Start":"02:44.560 ","End":"02:46.705","Text":"welcome them in parallel."},{"Start":"02:46.705 ","End":"02:48.795","Text":"This is 2, so this is 2."},{"Start":"02:48.795 ","End":"02:50.980","Text":"This is 1, this is 1."},{"Start":"02:50.980 ","End":"02:58.955","Text":"Now, we can reverse the order of the limit and the derivative here, and also here."},{"Start":"02:58.955 ","End":"03:07.450","Text":"Now, this limit is the derivative with respect to x(U) at the point u prime of 0."},{"Start":"03:07.450 ","End":"03:11.250","Text":"This one is u prime of 1U."},{"Start":"03:11.250 ","End":"03:13.545","Text":"Let\u0027s copy this over here."},{"Start":"03:13.545 ","End":"03:17.915","Text":"Now we can substitute these 2 values to get A and B."},{"Start":"03:17.915 ","End":"03:21.380","Text":"If we substitute 0 here,"},{"Start":"03:21.380 ","End":"03:23.715","Text":"then we get just B,"},{"Start":"03:23.715 ","End":"03:25.635","Text":"so B equals 2."},{"Start":"03:25.635 ","End":"03:27.345","Text":"If we substitute 1,"},{"Start":"03:27.345 ","End":"03:30.540","Text":"we get minus A over 2 plus B."},{"Start":"03:30.540 ","End":"03:32.135","Text":"From these 2 together,"},{"Start":"03:32.135 ","End":"03:37.435","Text":"it\u0027s easy to get that A equals 2 and B equals 2."},{"Start":"03:37.435 ","End":"03:41.925","Text":"But the question we were asked was find the value of A."},{"Start":"03:41.925 ","End":"03:45.200","Text":"Now substitute A and B in this equation,"},{"Start":"03:45.200 ","End":"03:49.760","Text":"and we get that u prime is minus x^2 plus 2."},{"Start":"03:49.760 ","End":"03:52.985","Text":"If we integrate minus x^2 plus 2,"},{"Start":"03:52.985 ","End":"03:58.860","Text":"we get U(x) equals minus 1/3x^3 plus 2x plus a constant"},{"Start":"03:58.860 ","End":"04:04.670","Text":"C. Now we need to do some work to find this C. We\u0027ll define a function M(t)."},{"Start":"04:04.670 ","End":"04:05.990","Text":"M stands for moment,"},{"Start":"04:05.990 ","End":"04:07.505","Text":"but I don\u0027t want to get into it."},{"Start":"04:07.505 ","End":"04:08.990","Text":"Just define it."},{"Start":"04:08.990 ","End":"04:12.565","Text":"M(t) equals the integral from 0 to 1,"},{"Start":"04:12.565 ","End":"04:18.945","Text":"0 and 1 being the endpoints of the interval of U(x, t)dx."},{"Start":"04:18.945 ","End":"04:20.900","Text":"M prime of t,"},{"Start":"04:20.900 ","End":"04:27.900","Text":"we can differentiate under the integral will be the integral of u_t(x) and t dx."},{"Start":"04:27.900 ","End":"04:29.070","Text":"M double prime,"},{"Start":"04:29.070 ","End":"04:31.970","Text":"second derivative will be the integral of u_tt."},{"Start":"04:31.970 ","End":"04:35.465","Text":"I recall the original PDE,"},{"Start":"04:35.465 ","End":"04:41.340","Text":"which is u_tt plus u_t equals u_xx plus Ax."},{"Start":"04:41.480 ","End":"04:47.785","Text":"This plus this is the integral from 0 to 1 of u_tt plus u_t."},{"Start":"04:47.785 ","End":"04:50.450","Text":"Now, as it is not very useful because this"},{"Start":"04:50.450 ","End":"04:53.075","Text":"is derivative with respect to t and the integral dx."},{"Start":"04:53.075 ","End":"04:56.655","Text":"But if we replace this by the right hand side,"},{"Start":"04:56.655 ","End":"04:58.090","Text":"u_xx plus 2x,"},{"Start":"04:58.090 ","End":"04:59.795","Text":"then we can get somewhere."},{"Start":"04:59.795 ","End":"05:05.885","Text":"This is equal to the integral of this is du by dx integral of this is x^2."},{"Start":"05:05.885 ","End":"05:08.915","Text":"We want to take this between 0 and 1,"},{"Start":"05:08.915 ","End":"05:13.670","Text":"and so we get u_x(1, t) minus u_x(0,"},{"Start":"05:13.670 ","End":"05:19.880","Text":"t) and x^2 from 0 to 1 is 1^2 minus 0 squared is 1."},{"Start":"05:19.880 ","End":"05:26.600","Text":"Recall the boundary conditions at 0 and at 1, given these,"},{"Start":"05:26.600 ","End":"05:29.450","Text":"we can replace those numbers here and get,"},{"Start":"05:29.450 ","End":"05:33.195","Text":"this is 1, this is 2."},{"Start":"05:33.195 ","End":"05:34.815","Text":"We get 1 minus 2 plus 1,"},{"Start":"05:34.815 ","End":"05:36.420","Text":"and this is equal to 0,"},{"Start":"05:36.420 ","End":"05:41.110","Text":"so we\u0027ve got that M double prime plus M prime equals 0."},{"Start":"05:41.110 ","End":"05:44.840","Text":"Simple differential equation. Several ways to do it,"},{"Start":"05:44.840 ","End":"05:47.555","Text":"let\u0027s use the integrating factor."},{"Start":"05:47.555 ","End":"05:51.400","Text":"We\u0027ll multiply both sides by e to the power of 1t."},{"Start":"05:51.400 ","End":"05:55.230","Text":"It\u0027s like if there was plus A here,"},{"Start":"05:55.230 ","End":"05:57.645","Text":"then it would be e to the At."},{"Start":"05:57.645 ","End":"05:59.270","Text":"Integrating factor."},{"Start":"05:59.270 ","End":"06:04.505","Text":"Now, the left-hand side is the derivative of e to the tM prime."},{"Start":"06:04.505 ","End":"06:12.850","Text":"Now, integrate both sides and we get e^tM prime of t is some constant, call it c_1."},{"Start":"06:12.850 ","End":"06:16.650","Text":"Take it to the t to the other side and we have M prime is"},{"Start":"06:16.650 ","End":"06:20.030","Text":"equal to c_1e to the minus t. Now"},{"Start":"06:20.030 ","End":"06:23.270","Text":"integrate this again and we\u0027ve got M(t) is"},{"Start":"06:23.270 ","End":"06:27.985","Text":"minus c_1e to the minus t plus another constant c_2."},{"Start":"06:27.985 ","End":"06:34.535","Text":"Now, we need to find c_1 and c_2 and we\u0027ll do this by means of the initial conditions."},{"Start":"06:34.535 ","End":"06:36.695","Text":"We have that for all x, u(x,"},{"Start":"06:36.695 ","End":"06:41.255","Text":"0) is 0 and d by dt of 0 is 0."},{"Start":"06:41.255 ","End":"06:44.405","Text":"M(0) is the integral of u(x,"},{"Start":"06:44.405 ","End":"06:47.310","Text":"0)dx, and for each x this is 0,"},{"Start":"06:47.310 ","End":"06:56.300","Text":"so the integral is 0 and M prime of 0 is the integral of du by dt and this is also 0."},{"Start":"06:56.300 ","End":"07:03.785","Text":"This gives us by looking here and putting t equals 0 minus c_1 plus c_2,"},{"Start":"07:03.785 ","End":"07:07.160","Text":"so that\u0027s equal to 0, and using this,"},{"Start":"07:07.160 ","End":"07:13.475","Text":"when t equals 0, we get that c_1 equals 0 because this equals 0."},{"Start":"07:13.475 ","End":"07:15.560","Text":"From these 2 together,"},{"Start":"07:15.560 ","End":"07:19.528","Text":"we get that c_1 and c_2 are both 0."},{"Start":"07:19.528 ","End":"07:23.940","Text":"In that case, M(t) is this,"},{"Start":"07:23.940 ","End":"07:25.760","Text":"but c_1 and c_2 is 0,"},{"Start":"07:25.760 ","End":"07:27.680","Text":"so it\u0027s identically 0."},{"Start":"07:27.680 ","End":"07:29.704","Text":"M(t) is the 0 function."},{"Start":"07:29.704 ","End":"07:32.000","Text":"Recall the definition of M(t),"},{"Start":"07:32.000 ","End":"07:37.795","Text":"integral of 0 to 1 of u(x t)dx, so that\u0027s 0."},{"Start":"07:37.795 ","End":"07:41.750","Text":"We take the limit as t goes to infinity,"},{"Start":"07:41.750 ","End":"07:43.340","Text":"that\u0027s equal to 0,"},{"Start":"07:43.340 ","End":"07:45.940","Text":"and putting the limit inside,"},{"Start":"07:45.940 ","End":"07:50.595","Text":"we get that the integral of U(x)dx is 0."},{"Start":"07:50.595 ","End":"07:54.195","Text":"Recall that we found U(x) to be this,"},{"Start":"07:54.195 ","End":"07:58.010","Text":"we can substitute that into the integral that\u0027s going to give us"},{"Start":"07:58.010 ","End":"08:05.060","Text":"C. We get the integral of this is 0 and this integral L straightforward,"},{"Start":"08:05.060 ","End":"08:07.085","Text":"I\u0027ll spare you the details,"},{"Start":"08:07.085 ","End":"08:11.210","Text":"comes out that C is minus 11/12."},{"Start":"08:11.210 ","End":"08:13.220","Text":"Put that in here,"},{"Start":"08:13.220 ","End":"08:16.520","Text":"and we get the solution that U(x) is minus"},{"Start":"08:16.520 ","End":"08:20.854","Text":"1/3x^2 plus 2x minus 11/12 and that is the answer,"},{"Start":"08:20.854 ","End":"08:23.730","Text":"and that concludes this exercise."}],"ID":30868},{"Watched":false,"Name":"Exercise 4","Duration":"7m 30s","ChapterTopicVideoID":29313,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.530","Text":"In this exercise, we\u0027re given the following boundary value problem."},{"Start":"00:04.530 ","End":"00:06.675","Text":"It\u0027s a Poisson equation."},{"Start":"00:06.675 ","End":"00:08.520","Text":"It\u0027s like a Laplace equation,"},{"Start":"00:08.520 ","End":"00:10.275","Text":"but it\u0027s not homogeneous."},{"Start":"00:10.275 ","End":"00:18.570","Text":"It\u0027s in an annulus radii between 1 and 2 and it has Dirichlet boundary conditions,"},{"Start":"00:18.570 ","End":"00:21.870","Text":"and we\u0027re working in polar coordinates."},{"Start":"00:21.870 ","End":"00:27.465","Text":"Usual trick, the non-homogeneous is to split it up into 2 problems."},{"Start":"00:27.465 ","End":"00:34.270","Text":"One with a homogeneous PDE and the other with homogeneous boundary conditions."},{"Start":"00:34.270 ","End":"00:39.890","Text":"We\u0027ll let u equal v plus w as follows."},{"Start":"00:39.890 ","End":"00:41.510","Text":"Now, you might say,"},{"Start":"00:41.510 ","End":"00:44.660","Text":"why is w not a function of r and Theta?"},{"Start":"00:44.660 ","End":"00:48.110","Text":"Well, because of the way this is setup and we just"},{"Start":"00:48.110 ","End":"00:51.635","Text":"have functions of r on the right and no Theta,"},{"Start":"00:51.635 ","End":"00:56.570","Text":"usually we can try to get it to be a function just of r,"},{"Start":"00:56.570 ","End":"00:58.760","Text":"and then it\u0027s easier to solve,"},{"Start":"00:58.760 ","End":"01:01.475","Text":"v is a Laplace equation,"},{"Start":"01:01.475 ","End":"01:04.760","Text":"and w is a Poisson equation,"},{"Start":"01:04.760 ","End":"01:07.220","Text":"but with homogeneous boundary conditions."},{"Start":"01:07.220 ","End":"01:10.725","Text":"Let\u0027s solve for w first,"},{"Start":"01:10.725 ","End":"01:13.880","Text":"and recall that the Laplacian of a function in"},{"Start":"01:13.880 ","End":"01:17.780","Text":"polar coordinates is given by this formula."},{"Start":"01:17.780 ","End":"01:22.015","Text":"This is what we get when we convert this,"},{"Start":"01:22.015 ","End":"01:25.185","Text":"the Laplacian using this formula,"},{"Start":"01:25.185 ","End":"01:27.920","Text":"and here the 2 boundary conditions,"},{"Start":"01:27.920 ","End":"01:32.300","Text":"because w is a function of r,"},{"Start":"01:32.300 ","End":"01:37.820","Text":"then w with respect to Theta is 0,"},{"Start":"01:37.820 ","End":"01:41.395","Text":"and w with respect to r is w\u0027,"},{"Start":"01:41.395 ","End":"01:43.935","Text":"and this is w\u0027\u0027."},{"Start":"01:43.935 ","End":"01:49.770","Text":"What we get, we replace that here is w\u0027\u0027 plus 1 over rw\u0027"},{"Start":"01:49.770 ","End":"01:56.900","Text":"equals r. We can multiply both sides by r, get the following."},{"Start":"01:56.900 ","End":"02:00.800","Text":"On the left-hand side is the derivative of a product,"},{"Start":"02:00.800 ","End":"02:03.275","Text":"the derivative of rw\u0027."},{"Start":"02:03.275 ","End":"02:05.315","Text":"Check that we get this."},{"Start":"02:05.315 ","End":"02:11.120","Text":"Now, we can integrate both sides and get that rw\u0027 is the integral of this,"},{"Start":"02:11.120 ","End":"02:14.420","Text":"which is r^3 over 3 plus a constant a,"},{"Start":"02:14.420 ","End":"02:17.335","Text":"divide both sides by r,"},{"Start":"02:17.335 ","End":"02:21.180","Text":"and we get this and integrate again,"},{"Start":"02:21.180 ","End":"02:26.580","Text":"and we get r^3 over 9 plus a natural log of r plus b,"},{"Start":"02:26.580 ","End":"02:30.695","Text":"and we have 2 constants we need to figure out."},{"Start":"02:30.695 ","End":"02:32.430","Text":"But we have 2 conditions,"},{"Start":"02:32.430 ","End":"02:35.705","Text":"w of 1 and w of 2 are both 0."},{"Start":"02:35.705 ","End":"02:38.440","Text":"From these, we can deduce a and b."},{"Start":"02:38.440 ","End":"02:42.540","Text":"If we substitute r=1,"},{"Start":"02:42.540 ","End":"02:52.440","Text":"then we get 1 over 9 plus a natural log of 1 which is 0 plus b equals 0."},{"Start":"02:52.440 ","End":"02:55.560","Text":"If we substitute r equals 2,"},{"Start":"02:55.560 ","End":"03:03.315","Text":"we get 8 over 9 plus a natural log of 2 plus b equals 0, these 2 equations."},{"Start":"03:03.315 ","End":"03:06.920","Text":"From the first one we get b is minus 1/9."},{"Start":"03:06.920 ","End":"03:08.210","Text":"Put that in here,"},{"Start":"03:08.210 ","End":"03:11.450","Text":"we get a natural log of 2 plus 7/9 is 0."},{"Start":"03:11.450 ","End":"03:17.420","Text":"Extracting a, we get a is minus 7/9 natural log 2."},{"Start":"03:17.420 ","End":"03:19.435","Text":"We have a and b."},{"Start":"03:19.435 ","End":"03:26.880","Text":"We can substitute them in here and we now have w of r. Next,"},{"Start":"03:26.880 ","End":"03:28.840","Text":"we have to compute v,"},{"Start":"03:28.840 ","End":"03:30.380","Text":"copying from above,"},{"Start":"03:30.380 ","End":"03:32.675","Text":"this is what we had."},{"Start":"03:32.675 ","End":"03:35.990","Text":"This is the Laplace equation on an annulus,"},{"Start":"03:35.990 ","End":"03:38.510","Text":"and that\u0027s the standard formula for that,"},{"Start":"03:38.510 ","End":"03:40.415","Text":"which is the following."},{"Start":"03:40.415 ","End":"03:46.115","Text":"Now, we\u0027re going to substitute r equals 1 and r equals 2 respectively."},{"Start":"03:46.115 ","End":"03:47.675","Text":"Because we have this."},{"Start":"03:47.675 ","End":"03:49.785","Text":"When r equals 1,"},{"Start":"03:49.785 ","End":"03:53.750","Text":"we get 1 plus sine Theta equals a naught."},{"Start":"03:53.750 ","End":"03:55.910","Text":"Natural log of 1 is 0,"},{"Start":"03:55.910 ","End":"03:57.695","Text":"and everywhere we see r here,"},{"Start":"03:57.695 ","End":"04:00.725","Text":"replace it by 1, so disappears."},{"Start":"04:00.725 ","End":"04:02.570","Text":"If we compare coefficients,"},{"Start":"04:02.570 ","End":"04:04.810","Text":"we can deduce a lot of things."},{"Start":"04:04.810 ","End":"04:07.875","Text":"We\u0027ve got that a naught equals 1,"},{"Start":"04:07.875 ","End":"04:09.870","Text":"that\u0027s the free coefficient,"},{"Start":"04:09.870 ","End":"04:14.810","Text":"and then the coefficient of sine Theta is 1 here,"},{"Start":"04:14.810 ","End":"04:18.580","Text":"and here it\u0027s b_1 plus b_ minus 1."},{"Start":"04:18.580 ","End":"04:22.110","Text":"Then for all the others,"},{"Start":"04:22.110 ","End":"04:25.860","Text":"for all the cosine n Theta and sine n Theta,"},{"Start":"04:25.860 ","End":"04:27.150","Text":"we have 0,"},{"Start":"04:27.150 ","End":"04:34.310","Text":"so a_n plus a_ minus n is 0 and b_n plus b_ minus n is 0."},{"Start":"04:34.310 ","End":"04:37.300","Text":"But this one is only true from n bigger or equal to 2,"},{"Start":"04:37.300 ","End":"04:41.660","Text":"because n equals 1 is taking care of here."},{"Start":"04:41.680 ","End":"04:46.195","Text":"Now, when we substitute r equals 2, we get this,"},{"Start":"04:46.195 ","End":"04:49.080","Text":"and again, comparing coefficients,"},{"Start":"04:49.080 ","End":"04:50.820","Text":"we get this set of equalities."},{"Start":"04:50.820 ","End":"04:53.675","Text":"Now, we want to solve these to get all the a naught,"},{"Start":"04:53.675 ","End":"04:55.310","Text":"b naught, a_n,"},{"Start":"04:55.310 ","End":"04:57.080","Text":"a_ minus n, etc."},{"Start":"04:57.080 ","End":"05:02.115","Text":"Let\u0027s divide into cases according to n. If n equals 0,"},{"Start":"05:02.115 ","End":"05:04.785","Text":"we get a naught equals 1,"},{"Start":"05:04.785 ","End":"05:10.380","Text":"and from here, a naught plus b naught natural log 2 is 1,"},{"Start":"05:10.380 ","End":"05:15.695","Text":"and from these we get that a naught is 1 and b naught is 0."},{"Start":"05:15.695 ","End":"05:19.295","Text":"Next, we\u0027ll take the case where n equals 1."},{"Start":"05:19.295 ","End":"05:25.170","Text":"From here, we get that 2a_1 plus 1/2a minus 1 is 2."},{"Start":"05:25.170 ","End":"05:30.509","Text":"We can also take n equals 1 here and get a_1 plus a_ minus 1 is 0."},{"Start":"05:30.509 ","End":"05:34.320","Text":"That\u0027s for a\u0027s and for b\u0027s."},{"Start":"05:34.320 ","End":"05:38.655","Text":"When n is 1, we get from here b_1 plus p_ minus 1 is 1,"},{"Start":"05:38.655 ","End":"05:46.335","Text":"and from this, we get 2b_1 plus 1/2b minus 1 is 0."},{"Start":"05:46.335 ","End":"05:48.330","Text":"If we solve these,"},{"Start":"05:48.330 ","End":"05:49.890","Text":"these 4 unknowns, a_1,"},{"Start":"05:49.890 ","End":"05:51.955","Text":"a_ minus 1, b_1, b_ minus 1,"},{"Start":"05:51.955 ","End":"05:56.690","Text":"I\u0027ll leave it to you to check that these are the values of a_1,"},{"Start":"05:56.690 ","End":"05:59.255","Text":"a_ minus 1, b_1, b_ minus 1."},{"Start":"05:59.255 ","End":"06:02.990","Text":"The final case will be n bigger or equal to 2,"},{"Start":"06:02.990 ","End":"06:04.490","Text":"and if n is bigger or equal to 2,"},{"Start":"06:04.490 ","End":"06:06.745","Text":"it\u0027s also bigger or equal to 1."},{"Start":"06:06.745 ","End":"06:12.065","Text":"We can take this together with this for a,"},{"Start":"06:12.065 ","End":"06:16.105","Text":"and for b, we can take this and this."},{"Start":"06:16.105 ","End":"06:19.370","Text":"If we solve these again,"},{"Start":"06:19.370 ","End":"06:20.960","Text":"there\u0027s 4 unknowns,"},{"Start":"06:20.960 ","End":"06:24.740","Text":"then we get all of them are 0."},{"Start":"06:24.740 ","End":"06:33.150","Text":"Actually, the only ones that are not 0 are a naught and a_1,"},{"Start":"06:33.150 ","End":"06:34.425","Text":"a_ minus 1,"},{"Start":"06:34.425 ","End":"06:37.405","Text":"and b_1, b_ minus 1."},{"Start":"06:37.405 ","End":"06:44.300","Text":"Yes, rewriting that, now we can substitute these values in the formula for v(r,"},{"Start":"06:44.300 ","End":"06:46.685","Text":"Theta) If we do that,"},{"Start":"06:46.685 ","End":"06:49.340","Text":"we get here, we get 1."},{"Start":"06:49.340 ","End":"06:52.455","Text":"Here, we don\u0027t get anything."},{"Start":"06:52.455 ","End":"06:56.000","Text":"Anyway, I\u0027ll just give you the answer and you can check."},{"Start":"06:56.000 ","End":"07:01.010","Text":"For example, a_1 is 4/3."},{"Start":"07:01.010 ","End":"07:05.410","Text":"We have 4/3 r^1 cosine 1 Theta."},{"Start":"07:05.410 ","End":"07:11.655","Text":"We also have for a_ minus n minus 4/3 r to the minus 1,"},{"Start":"07:11.655 ","End":"07:14.070","Text":"also cosine 1 Theta."},{"Start":"07:14.070 ","End":"07:16.500","Text":"This is what we have for v,"},{"Start":"07:16.500 ","End":"07:20.325","Text":"and we already found w to be this."},{"Start":"07:20.325 ","End":"07:24.230","Text":"Since u is v plus w,"},{"Start":"07:24.230 ","End":"07:27.260","Text":"just add these 2 and this is what we get,"},{"Start":"07:27.260 ","End":"07:30.630","Text":"and that concludes this exercise."}],"ID":30869},{"Watched":false,"Name":"Exercise 5","Duration":"6m 58s","ChapterTopicVideoID":29314,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.280","Text":"In this exercise, we\u0027re going to solve the following equation,"},{"Start":"00:03.280 ","End":"00:08.130","Text":"it\u0027s a partial differential equation which is quasi linear."},{"Start":"00:08.130 ","End":"00:10.635","Text":"This is the equation part,"},{"Start":"00:10.635 ","End":"00:13.740","Text":"this is the initial condition part,"},{"Start":"00:13.740 ","End":"00:20.625","Text":"and we\u0027re only taking the part where y is bigger or equal to e^-x."},{"Start":"00:20.625 ","End":"00:23.100","Text":"This is the curve y equals e^-x,"},{"Start":"00:23.100 ","End":"00:25.050","Text":"so it\u0027s the shaded part."},{"Start":"00:25.050 ","End":"00:27.000","Text":"We\u0027re not given little gamma,"},{"Start":"00:27.000 ","End":"00:30.975","Text":"but clearly gamma is set of all x, y."},{"Start":"00:30.975 ","End":"00:34.245","Text":"It\u0027s this curve, is y is e^-x."},{"Start":"00:34.245 ","End":"00:37.290","Text":"We\u0027re given the value of u on gamma,"},{"Start":"00:37.290 ","End":"00:39.330","Text":"we want to identify A, B,"},{"Start":"00:39.330 ","End":"00:41.940","Text":"and C. In our case,"},{"Start":"00:41.940 ","End":"00:46.625","Text":"A is the constant 1 from here coefficient."},{"Start":"00:46.625 ","End":"00:49.615","Text":"Also we have 1 here that\u0027s B,"},{"Start":"00:49.615 ","End":"00:53.500","Text":"and C I\u0027ll just bring you over to the other side,"},{"Start":"00:53.500 ","End":"00:56.705","Text":"so we have this minus u."},{"Start":"00:56.705 ","End":"01:00.770","Text":"We\u0027re going to parameterize the surface using 2 parameters;"},{"Start":"01:00.770 ","End":"01:05.570","Text":"t and tau, such that the initial curve big gamma,"},{"Start":"01:05.570 ","End":"01:07.100","Text":"this is big gamma."},{"Start":"01:07.100 ","End":"01:10.795","Text":"We want it so that when t is 0,"},{"Start":"01:10.795 ","End":"01:13.620","Text":"we get gamma, this varies with tau,"},{"Start":"01:13.620 ","End":"01:17.070","Text":"x equals tau, that\u0027s the simplest."},{"Start":"01:17.070 ","End":"01:21.635","Text":"Then y is equal to e to the minus tau,"},{"Start":"01:21.635 ","End":"01:23.120","Text":"you get that from here,"},{"Start":"01:23.120 ","End":"01:27.960","Text":"and then u have 0 tau from here is equal"},{"Start":"01:27.960 ","End":"01:33.140","Text":"2 because x is tau e to the tau squared plus e to the minus tau."},{"Start":"01:33.140 ","End":"01:41.670","Text":"The next step is to find the characteristic curves by solving dx by dt equals A,"},{"Start":"01:41.670 ","End":"01:44.655","Text":"and dy by dt is equal to B."},{"Start":"01:44.655 ","End":"01:46.980","Text":"We also have du by dt equals C,"},{"Start":"01:46.980 ","End":"01:49.260","Text":"but we don\u0027t need that just yet."},{"Start":"01:49.260 ","End":"01:53.670","Text":"This gives us that dx by dt equals 1,"},{"Start":"01:53.670 ","End":"01:57.090","Text":"that\u0027s A, and dy by dt equals 1, that\u0027s B."},{"Start":"01:57.090 ","End":"01:59.315","Text":"Then integrating with respect to t,"},{"Start":"01:59.315 ","End":"02:04.900","Text":"we get that x equals t plus not a constant but a function of tau,"},{"Start":"02:04.900 ","End":"02:08.000","Text":"and y is t plus another function of tau,"},{"Start":"02:08.000 ","End":"02:09.740","Text":"call them C1, and C2."},{"Start":"02:09.740 ","End":"02:12.020","Text":"If we put t equals 0,"},{"Start":"02:12.020 ","End":"02:18.200","Text":"then we get that x of 0 tau is C1 of tau, and similarly with y."},{"Start":"02:18.200 ","End":"02:22.205","Text":"Now, remember that gamma is equal to,"},{"Start":"02:22.205 ","End":"02:24.320","Text":"well, I just copied it from here."},{"Start":"02:24.320 ","End":"02:32.260","Text":"Now that we have this, we can see comparing x of 0 tau that we get C1 of tau is tau,"},{"Start":"02:32.260 ","End":"02:37.410","Text":"and C2 of tau is e to the minus tau."},{"Start":"02:37.410 ","End":"02:41.075","Text":"We have this, and now we can substitute this in this,"},{"Start":"02:41.075 ","End":"02:43.670","Text":"here and here, we found x,"},{"Start":"02:43.670 ","End":"02:47.855","Text":"and y, and now we need to find u."},{"Start":"02:47.855 ","End":"02:51.065","Text":"We get that from the third 1,"},{"Start":"02:51.065 ","End":"02:56.780","Text":"I said that u with respect to t is equal to C. In our case,"},{"Start":"02:56.780 ","End":"02:58.505","Text":"C is equal to this,"},{"Start":"02:58.505 ","End":"03:05.355","Text":"so what we get is u is 2x plus 1 e^x^2-u."},{"Start":"03:05.355 ","End":"03:09.000","Text":"We also have initial condition from here."},{"Start":"03:09.000 ","End":"03:12.990","Text":"What we get bring u over to the other side is u by dt"},{"Start":"03:12.990 ","End":"03:16.620","Text":"plus u equals this which is a function of x,"},{"Start":"03:16.620 ","End":"03:19.125","Text":"but we want in terms of t, and tau,"},{"Start":"03:19.125 ","End":"03:24.375","Text":"so we just replace x by t plus tau,"},{"Start":"03:24.375 ","End":"03:27.285","Text":"and then we have this expression."},{"Start":"03:27.285 ","End":"03:29.930","Text":"Think of tau like a constant,"},{"Start":"03:29.930 ","End":"03:35.390","Text":"and we have an ordinary differential equation u as a function of t. What we\u0027ll do is,"},{"Start":"03:35.390 ","End":"03:38.345","Text":"we\u0027ll multiply this by an integrating factor,"},{"Start":"03:38.345 ","End":"03:41.170","Text":"which will be e^t,"},{"Start":"03:41.170 ","End":"03:43.955","Text":"in case you\u0027ve forgotten integrating factors,"},{"Start":"03:43.955 ","End":"03:46.190","Text":"I\u0027ll do the calculation for you."},{"Start":"03:46.190 ","End":"03:54.455","Text":"We\u0027re looking for mu of t such that if we multiply it by u by dt plus u,"},{"Start":"03:54.455 ","End":"04:01.020","Text":"we\u0027ll get the derivative with respect to t of mu of t times u,"},{"Start":"04:01.020 ","End":"04:03.410","Text":"that\u0027s how the integrating factor works."},{"Start":"04:03.410 ","End":"04:04.760","Text":"You know what?"},{"Start":"04:04.760 ","End":"04:06.590","Text":"You\u0027re supposed to know this stuff,"},{"Start":"04:06.590 ","End":"04:10.340","Text":"I\u0027ll just leave the computation\u0027s up here,"},{"Start":"04:10.340 ","End":"04:16.070","Text":"and in the end we get that mu of t is e^t."},{"Start":"04:16.070 ","End":"04:20.870","Text":"Indeed, this left hand side is the derivative with respect to"},{"Start":"04:20.870 ","End":"04:26.025","Text":"t of mu of t times u. Differentiate this,"},{"Start":"04:26.025 ","End":"04:35.915","Text":"and we get e^t times u plus e^t times u\u0027 or u with respect to t. On the right hand side,"},{"Start":"04:35.915 ","End":"04:40.180","Text":"we can combine the e^t by adding the t over here."},{"Start":"04:40.180 ","End":"04:44.390","Text":"Now we have a derivative with respect to t equals a function of t,"},{"Start":"04:44.390 ","End":"04:47.565","Text":"so we just need the integral of this."},{"Start":"04:47.565 ","End":"04:51.870","Text":"The integral of this plus C of tau,"},{"Start":"04:51.870 ","End":"04:56.510","Text":"this is easy because what we have here,"},{"Start":"04:56.510 ","End":"05:00.005","Text":"is exactly the derivative of what we have here."},{"Start":"05:00.005 ","End":"05:05.010","Text":"This is just e^t plus tau squared plus t,"},{"Start":"05:05.010 ","End":"05:08.905","Text":"now just bring the e to the t over to the other side,"},{"Start":"05:08.905 ","End":"05:12.610","Text":"and it cancels with this t here,"},{"Start":"05:12.610 ","End":"05:15.065","Text":"and then we have e to the minus t here."},{"Start":"05:15.065 ","End":"05:17.940","Text":"That gives us u in terms of t,"},{"Start":"05:17.940 ","End":"05:21.575","Text":"and tau, and we have the initial condition here."},{"Start":"05:21.575 ","End":"05:26.734","Text":"Now, if we substitute t equals 0 here,"},{"Start":"05:26.734 ","End":"05:28.530","Text":"we should get this function here."},{"Start":"05:28.530 ","End":"05:30.725","Text":"If we put t equals 0,"},{"Start":"05:30.725 ","End":"05:35.975","Text":"we get e to the tau squared plus C of tau,"},{"Start":"05:35.975 ","End":"05:37.655","Text":"this becomes 1,"},{"Start":"05:37.655 ","End":"05:40.630","Text":"and this is going to equal what\u0027s written here,"},{"Start":"05:40.630 ","End":"05:43.890","Text":"so e to the tau squared cancels,"},{"Start":"05:43.890 ","End":"05:47.180","Text":"we have C of tau equals e to the minus tau."},{"Start":"05:47.180 ","End":"05:52.925","Text":"If we substitute C of tau here to be e to the minus tau,"},{"Start":"05:52.925 ","End":"05:56.610","Text":"what we get is this part is the same,"},{"Start":"05:56.610 ","End":"05:57.870","Text":"and e to the minus tau,"},{"Start":"05:57.870 ","End":"06:02.330","Text":"combined with e to the minus t to give e to the minus t plus tau."},{"Start":"06:02.330 ","End":"06:05.705","Text":"That gives us u in terms of t, and tau."},{"Start":"06:05.705 ","End":"06:09.260","Text":"The last step is to get back from t tau to x,"},{"Start":"06:09.260 ","End":"06:11.760","Text":"y, so that\u0027s Step 4."},{"Start":"06:11.760 ","End":"06:13.200","Text":"Now, we had x,"},{"Start":"06:13.200 ","End":"06:15.300","Text":"and y in terms of t, and tau."},{"Start":"06:15.300 ","End":"06:17.025","Text":"We want to reverse this,"},{"Start":"06:17.025 ","End":"06:18.995","Text":"let\u0027s check if we can do that."},{"Start":"06:18.995 ","End":"06:21.950","Text":"The determinant of the Jacobian, well,"},{"Start":"06:21.950 ","End":"06:26.430","Text":"it comes out to be non 0, so we\u0027re okay."},{"Start":"06:26.430 ","End":"06:30.365","Text":"Well, actually we don\u0027t need to do that because we\u0027re fortunate,"},{"Start":"06:30.365 ","End":"06:33.710","Text":"that this function doesn\u0027t depend on t,"},{"Start":"06:33.710 ","End":"06:36.785","Text":"and tau separately, it\u0027s all t plus tau."},{"Start":"06:36.785 ","End":"06:39.485","Text":"We have that t plus tau equals x,"},{"Start":"06:39.485 ","End":"06:43.500","Text":"so we can save the computation of getting t,"},{"Start":"06:43.500 ","End":"06:44.955","Text":"and tau in terms of x, and y,"},{"Start":"06:44.955 ","End":"06:47.700","Text":"and just say that t plus tau is x."},{"Start":"06:47.700 ","End":"06:49.305","Text":"So plug that in,"},{"Start":"06:49.305 ","End":"06:50.540","Text":"we get that u of x,"},{"Start":"06:50.540 ","End":"06:55.165","Text":"and y is e to the x squared plus e^-x,"},{"Start":"06:55.165 ","End":"06:58.690","Text":"and that\u0027s the answer, and we\u0027re done."}],"ID":30870},{"Watched":false,"Name":"Exercise 6","Duration":"5m 18s","ChapterTopicVideoID":29315,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.880","Text":"In this exercise, we\u0027re given that u(x,"},{"Start":"00:02.880 ","End":"00:05.880","Text":"t) is a solution to the following problem,"},{"Start":"00:05.880 ","End":"00:08.640","Text":"which looks a bit like a heat equation,"},{"Start":"00:08.640 ","End":"00:11.205","Text":"but it\u0027s got some extra stuff."},{"Start":"00:11.205 ","End":"00:14.385","Text":"In fact, we\u0027ll reduce it to a heat equation."},{"Start":"00:14.385 ","End":"00:17.385","Text":"That\u0027s 3 parts, the PDE,"},{"Start":"00:17.385 ","End":"00:21.990","Text":"the initial condition, and couple of boundary conditions."},{"Start":"00:21.990 ","End":"00:25.740","Text":"Our task is to prove that u at the 0.5,"},{"Start":"00:25.740 ","End":"00:29.970","Text":"1 is less than 1 over 4 square root of e,"},{"Start":"00:29.970 ","End":"00:33.630","Text":"and we\u0027re given a hint to define h(x,"},{"Start":"00:33.630 ","End":"00:38.490","Text":"t) as u(x,t) times e^Delta x for a suitable constant Delta."},{"Start":"00:38.490 ","End":"00:42.270","Text":"From this equation, we can get u in terms of h by bringing"},{"Start":"00:42.270 ","End":"00:47.435","Text":"e^Delta x to the other side and we have u equals he^minus Delta."},{"Start":"00:47.435 ","End":"00:49.855","Text":"Now we want the derivatives of u."},{"Start":"00:49.855 ","End":"00:51.420","Text":"We need u_t,"},{"Start":"00:51.420 ","End":"00:54.300","Text":"u_xx, and u_x."},{"Start":"00:54.300 ","End":"00:57.285","Text":"Let\u0027s go for a u_t first of all."},{"Start":"00:57.285 ","End":"01:02.775","Text":"This is just dh by dt and this doesn\u0027t involve t like a constant."},{"Start":"01:02.775 ","End":"01:05.195","Text":"We have h_t^minus Delta x."},{"Start":"01:05.195 ","End":"01:08.120","Text":"Now with respect to x, we use the product rule,"},{"Start":"01:08.120 ","End":"01:10.835","Text":"derivative of h times this,"},{"Start":"01:10.835 ","End":"01:17.995","Text":"and then minus Delta times e^minus Delta x times this part."},{"Start":"01:17.995 ","End":"01:22.910","Text":"Then we can take the second derivative with respect to x."},{"Start":"01:22.910 ","End":"01:27.455","Text":"We can do it in 1 step if we use the formula"},{"Start":"01:27.455 ","End":"01:31.715","Text":"for second derivative of a product and we get this,"},{"Start":"01:31.715 ","End":"01:34.400","Text":"or you could just do it 1 step at a time."},{"Start":"01:34.400 ","End":"01:37.205","Text":"I\u0027ll leave you to check the calculation."},{"Start":"01:37.205 ","End":"01:39.500","Text":"Anyway, we have what we need."},{"Start":"01:39.500 ","End":"01:41.885","Text":"We have u, we have u_t,"},{"Start":"01:41.885 ","End":"01:43.865","Text":"we have u_x,"},{"Start":"01:43.865 ","End":"01:45.875","Text":"and we have u_xx."},{"Start":"01:45.875 ","End":"01:51.570","Text":"We can substitute these in our PDE, which is this."},{"Start":"01:52.000 ","End":"01:54.200","Text":"This is just algebra,"},{"Start":"01:54.200 ","End":"01:56.280","Text":"I will leave you to check the calculations."},{"Start":"01:56.280 ","End":"02:00.260","Text":"But note that every term has e^minus Delta x in it,"},{"Start":"02:00.260 ","End":"02:03.559","Text":"so we can throw that out and it becomes simpler."},{"Start":"02:03.559 ","End":"02:06.490","Text":"Now, open the brackets and collect like terms,"},{"Start":"02:06.490 ","End":"02:10.730","Text":"and we have h_t is h_xx plus this times h_x plus"},{"Start":"02:10.730 ","End":"02:15.200","Text":"this times h. Now we would really like this and this to be 0."},{"Start":"02:15.200 ","End":"02:20.500","Text":"In fact, we are lucky that we can do that because this is twice 2 minus Delta."},{"Start":"02:20.500 ","End":"02:22.620","Text":"This is Delta minus 2^2,"},{"Start":"02:22.620 ","End":"02:24.700","Text":"so if Delta equals 2,"},{"Start":"02:24.700 ","End":"02:26.615","Text":"and we\u0027ve got rid of both of these,"},{"Start":"02:26.615 ","End":"02:27.950","Text":"but we happen to be lucky here."},{"Start":"02:27.950 ","End":"02:29.690","Text":"If we didn\u0027t have a common factor,"},{"Start":"02:29.690 ","End":"02:31.550","Text":"we couldn\u0027t proceed this way."},{"Start":"02:31.550 ","End":"02:33.210","Text":"Delta equals 2,"},{"Start":"02:33.210 ","End":"02:41.070","Text":"and that gives us that h_t is h_xx because this disappears and we have that h(x,"},{"Start":"02:41.070 ","End":"02:47.410","Text":"t) which is ue^Delta, is actually ue^2x."},{"Start":"02:47.410 ","End":"02:55.280","Text":"The initial and boundary conditions for u were as follows but if h is ue^2x,"},{"Start":"02:55.280 ","End":"02:59.045","Text":"we can multiply everything here by e^2x and get"},{"Start":"02:59.045 ","End":"03:04.160","Text":"the initial and boundary conditions for h. These 2 together with this,"},{"Start":"03:04.160 ","End":"03:05.780","Text":"which is a heat equation,"},{"Start":"03:05.780 ","End":"03:09.680","Text":"will give us a problem in h as"},{"Start":"03:09.680 ","End":"03:12.920","Text":"follows and we\u0027re going to use"},{"Start":"03:12.920 ","End":"03:17.569","Text":"the maximum principle for the heat equation on a finite interval."},{"Start":"03:17.569 ","End":"03:19.400","Text":"Here\u0027s a picture."},{"Start":"03:19.400 ","End":"03:23.240","Text":"We have to show that h(0.5,1) is less than something."},{"Start":"03:23.240 ","End":"03:27.658","Text":"We\u0027ll take a rectangle that completely contains this,"},{"Start":"03:27.658 ","End":"03:32.615","Text":"and let\u0027s take it from 0-1 here and from 0-2 here."},{"Start":"03:32.615 ","End":"03:35.650","Text":"That\u0027s our rectangle."},{"Start":"03:35.650 ","End":"03:38.130","Text":"The parabolic boundary,"},{"Start":"03:38.130 ","End":"03:41.300","Text":"what we call, is the boundary of this,"},{"Start":"03:41.300 ","End":"03:44.245","Text":"this, and this without the top."},{"Start":"03:44.245 ","End":"03:47.405","Text":"It consists of 3 line segments."},{"Start":"03:47.405 ","End":"03:50.225","Text":"The first one is the bottom,"},{"Start":"03:50.225 ","End":"03:52.370","Text":"this 1 is the left,"},{"Start":"03:52.370 ","End":"03:54.985","Text":"and this 1 is the right."},{"Start":"03:54.985 ","End":"03:57.455","Text":"By the maximum principle,"},{"Start":"03:57.455 ","End":"04:02.540","Text":"the value of h at an interior point is going to"},{"Start":"04:02.540 ","End":"04:08.150","Text":"be less than the maximum value of h along the parabolic boundary,"},{"Start":"04:08.150 ","End":"04:12.320","Text":"and it\u0027s strictly less than provided h is not a constant function."},{"Start":"04:12.320 ","End":"04:17.620","Text":"It\u0027s obviously not constant because it\u0027s not constant along this lower edge."},{"Start":"04:17.620 ","End":"04:21.350","Text":"Now, the maximum on the parabolic boundary"},{"Start":"04:21.350 ","End":"04:25.250","Text":"is the same as the maximum on the lower edge because here it\u0027s 0,"},{"Start":"04:25.250 ","End":"04:28.384","Text":"here it\u0027s 0, here it\u0027s bigger or equal to 0."},{"Start":"04:28.384 ","End":"04:33.080","Text":"In fact, here\u0027s the graph of what it looks like on the lower edge."},{"Start":"04:33.080 ","End":"04:37.980","Text":"The maximum occurs when x is equal to"},{"Start":"04:37.980 ","End":"04:44.460","Text":"0.5 and the height here is 0.5 times 1 minus 0.5 times root e,"},{"Start":"04:44.460 ","End":"04:46.485","Text":"which is root e over 4,"},{"Start":"04:46.485 ","End":"04:49.510","Text":"or 1/4 root e. Now,"},{"Start":"04:49.510 ","End":"04:51.860","Text":"we don\u0027t want h, we want u."},{"Start":"04:51.860 ","End":"04:54.605","Text":"Remember that h is ue^2x."},{"Start":"04:54.605 ","End":"04:57.950","Text":"If x equals 0.5,"},{"Start":"04:57.950 ","End":"05:01.430","Text":"then e^2x is just e,"},{"Start":"05:01.430 ","End":"05:06.025","Text":"so we can replace h by u if we put e^1 here."},{"Start":"05:06.025 ","End":"05:09.260","Text":"Now, divide both sides by e,"},{"Start":"05:09.260 ","End":"05:10.910","Text":"and we get u over 0.5,"},{"Start":"05:10.910 ","End":"05:13.700","Text":"1 less than 1 over 4 root e,"},{"Start":"05:13.700 ","End":"05:18.060","Text":"and that\u0027s what we had to show and so we are done."}],"ID":30871},{"Watched":false,"Name":"Exercise 7","Duration":"6m 40s","ChapterTopicVideoID":29316,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.645","Text":"In this exercise we\u0027re given this initial and boundary value problem,"},{"Start":"00:05.645 ","End":"00:08.745","Text":"and it involves a function h here,"},{"Start":"00:08.745 ","End":"00:12.930","Text":"and our task is to find a condition on h such that"},{"Start":"00:12.930 ","End":"00:18.420","Text":"the solution u has a finite limit as t goes to infinty,"},{"Start":"00:18.420 ","End":"00:23.340","Text":"and we\u0027re given a hint to define v equals u"},{"Start":"00:23.340 ","End":"00:29.315","Text":"times this exponential for our suitable constants Alpha and Beta."},{"Start":"00:29.315 ","End":"00:32.970","Text":"The idea is that v will satisfy the heat equation,"},{"Start":"00:32.970 ","End":"00:35.350","Text":"u doesn\u0027t because of this 10u."},{"Start":"00:35.350 ","End":"00:39.370","Text":"Hopefully v will get rid of the extra bit."},{"Start":"00:39.370 ","End":"00:43.170","Text":"U in terms of v is v,"},{"Start":"00:43.170 ","End":"00:47.323","Text":"e^Alpha x plus Beta t,"},{"Start":"00:47.323 ","End":"00:52.350","Text":"and then we want the derivatives u_t and u_xx."},{"Start":"00:52.350 ","End":"00:55.880","Text":"u_t just differentiate with respect to t and use"},{"Start":"00:55.880 ","End":"01:01.195","Text":"the product rule is v_t and from here we get the extra Beta,"},{"Start":"01:01.195 ","End":"01:03.165","Text":"and then U_xx,"},{"Start":"01:03.165 ","End":"01:06.620","Text":"well we can use the formula for the second derivative of"},{"Start":"01:06.620 ","End":"01:10.640","Text":"a product and get the second derivative of v,"},{"Start":"01:10.640 ","End":"01:13.415","Text":"and then the first derivative of v,"},{"Start":"01:13.415 ","End":"01:20.200","Text":"and the first derivative of this with respect to x which is Alpha and there is 2 here."},{"Start":"01:20.200 ","End":"01:28.110","Text":"Then second derivative of this is Alpha^2 e to the same thing and the v just stays,"},{"Start":"01:28.110 ","End":"01:30.075","Text":"that\u0027s the f part."},{"Start":"01:30.075 ","End":"01:32.280","Text":"Now we can substitute u, u_t,"},{"Start":"01:32.280 ","End":"01:37.883","Text":"t and u_xx in the PDE,"},{"Start":"01:37.883 ","End":"01:40.985","Text":"and what we get is the following."},{"Start":"01:40.985 ","End":"01:42.680","Text":"You can check that."},{"Start":"01:42.680 ","End":"01:49.040","Text":"Note that every term has E^Alpha x plus Beta t in it and this is non-zero,"},{"Start":"01:49.040 ","End":"01:54.590","Text":"so we can just cancel everything by that factor and we get the following."},{"Start":"01:54.590 ","End":"01:59.060","Text":"Now put the Beta v on the right-hand side and collect like terms."},{"Start":"01:59.060 ","End":"02:05.934","Text":"We have v_xx, we have V_x and we have v. Now we want this to be 0 and this to be 0,"},{"Start":"02:05.934 ","End":"02:10.760","Text":"and if that\u0027s the case then festival from here Alpha is 0."},{"Start":"02:10.760 ","End":"02:13.040","Text":"If Alpha is 0 10 minus Beta is 0,"},{"Start":"02:13.040 ","End":"02:14.840","Text":"so beta equals 10."},{"Start":"02:14.840 ","End":"02:17.620","Text":"Now we\u0027ve got Alpha and Beta,"},{"Start":"02:17.620 ","End":"02:23.940","Text":"and we also have that v_t equals v_xx because all this part is"},{"Start":"02:23.940 ","End":"02:30.305","Text":"0 and v which is ue to the minus this exponent."},{"Start":"02:30.305 ","End":"02:33.765","Text":"This exponent is just minus 10t."},{"Start":"02:33.765 ","End":"02:40.140","Text":"We have the initial condition that when t is 0 u is equal to h(x),"},{"Start":"02:40.140 ","End":"02:48.980","Text":"so if we multiply by e^minus 10t v(x, 0) is he^0."},{"Start":"02:48.980 ","End":"02:51.110","Text":"That\u0027s the initial condition for v,"},{"Start":"02:51.110 ","End":"02:53.120","Text":"now we also need a boundary condition."},{"Start":"02:53.120 ","End":"02:59.240","Text":"We get the boundary condition on v at 0 and at 1 is also 0."},{"Start":"02:59.240 ","End":"03:02.000","Text":"It\u0027s 0 times something which is 0."},{"Start":"03:02.000 ","End":"03:04.880","Text":"Now we have everything about v that we need."},{"Start":"03:04.880 ","End":"03:07.120","Text":"We have the PDE,"},{"Start":"03:07.120 ","End":"03:13.460","Text":"the initial condition, and we have the boundary condition."},{"Start":"03:13.460 ","End":"03:18.050","Text":"This is the problem for v to the heat equation on"},{"Start":"03:18.050 ","End":"03:22.250","Text":"a finite interval with Dirichlet boundary conditions."},{"Start":"03:22.250 ","End":"03:24.355","Text":"There\u0027s a formula for that,"},{"Start":"03:24.355 ","End":"03:25.670","Text":"it goes like this."},{"Start":"03:25.670 ","End":"03:27.245","Text":"It\u0027s expressed in terms of u,"},{"Start":"03:27.245 ","End":"03:29.630","Text":"but we can just change u to v. If we have"},{"Start":"03:29.630 ","End":"03:33.920","Text":"the heat equation on a finite interval and we\u0027re given"},{"Start":"03:33.920 ","End":"03:38.120","Text":"the initial condition is f and the Dirichlet boundary conditions are"},{"Start":"03:38.120 ","End":"03:42.740","Text":"0 then u is given as the following series."},{"Start":"03:42.740 ","End":"03:47.435","Text":"In that case just change u to v. This is what we have,"},{"Start":"03:47.435 ","End":"03:50.590","Text":"but in our case L is 1 and a is 1."},{"Start":"03:50.590 ","End":"03:57.090","Text":"So v simplifies to this and then if we let t=0,"},{"Start":"03:57.090 ","End":"04:01.755","Text":"we get v(x, 0), which is h(x) is the following."},{"Start":"04:01.755 ","End":"04:07.655","Text":"When t is 0, this exponent disappears because e^0 is 1."},{"Start":"04:07.655 ","End":"04:10.670","Text":"Now we have a sine series for"},{"Start":"04:10.670 ","End":"04:15.350","Text":"h. Now there\u0027s a formula for the coefficients of a sine series."},{"Start":"04:15.350 ","End":"04:19.145","Text":"If f is given as the following sine series,"},{"Start":"04:19.145 ","End":"04:22.845","Text":"then the coefficient b_n is given by this formula."},{"Start":"04:22.845 ","End":"04:32.310","Text":"In our case f is h and b_n is AN and l equals L here, here and here."},{"Start":"04:32.310 ","End":"04:41.690","Text":"We get that A_N using this formula is twice the integral from 0-1 h(x) sine of nPixdx."},{"Start":"04:41.690 ","End":"04:47.959","Text":"Now we can put A_N into here and we get the following,"},{"Start":"04:47.959 ","End":"04:55.245","Text":"but this is v and we want u. U is ve^10t,"},{"Start":"04:55.245 ","End":"05:00.075","Text":"so we just put an extra 10 here."},{"Start":"05:00.075 ","End":"05:10.175","Text":"We get 10 minus n squared Pi squared times t. This is the coefficient of sine(nPix)."},{"Start":"05:10.175 ","End":"05:18.110","Text":"Now, the limit of this part is either 0 or infinity depending if n is 1 or bigger than 1."},{"Start":"05:18.110 ","End":"05:23.168","Text":"If n is 1 here we have 10 minus Pi squared,"},{"Start":"05:23.168 ","End":"05:24.880","Text":"Pi squared is less than 10."},{"Start":"05:24.880 ","End":"05:29.660","Text":"Square root of 10 is 3.16 and Pi is 3.14."},{"Start":"05:29.660 ","End":"05:36.985","Text":"This is positive so E to the power something positive times t goes to infinity."},{"Start":"05:36.985 ","End":"05:40.240","Text":"If n is bigger than 1 then n"},{"Start":"05:40.240 ","End":"05:43.550","Text":"squared Pi squared is already bigger than 10 so this is negative,"},{"Start":"05:43.550 ","End":"05:45.890","Text":"so this goes to 0."},{"Start":"05:45.890 ","End":"05:48.830","Text":"Now this integral doesn\u0027t depend on t for."},{"Start":"05:48.830 ","End":"05:50.600","Text":"As far as t goes, it\u0027s a constant."},{"Start":"05:50.600 ","End":"05:58.550","Text":"This times each of the 10 minus n squared Pi squared t also goes to 0 if n is 2, 3, 4,"},{"Start":"05:58.550 ","End":"06:02.690","Text":"etc and because this goes to infinity when"},{"Start":"06:02.690 ","End":"06:07.190","Text":"t goes to infinity the only way this is possible is if"},{"Start":"06:07.190 ","End":"06:16.650","Text":"this integral is=0 because then we have 0 times E^10 minus 2Pi^2 t which is 0 for all t,"},{"Start":"06:16.650 ","End":"06:19.260","Text":"and so the limit is 0."},{"Start":"06:19.260 ","End":"06:27.705","Text":"The condition that we need is that this for n=1 is 0."},{"Start":"06:27.705 ","End":"06:30.840","Text":"When n is 1 it\u0027s just h(x) sine Pix,"},{"Start":"06:30.840 ","End":"06:32.720","Text":"the integral from 0-1."},{"Start":"06:32.720 ","End":"06:34.970","Text":"That\u0027s what we had to find,"},{"Start":"06:34.970 ","End":"06:36.680","Text":"a condition on h,"},{"Start":"06:36.680 ","End":"06:41.520","Text":"and here it is, and that concludes this exercise."}],"ID":30872},{"Watched":false,"Name":"Exercise 8","Duration":"4m 22s","ChapterTopicVideoID":29317,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"In this exercise, we are given this boundary value problem to solve."},{"Start":"00:05.340 ","End":"00:07.260","Text":"This looks like a plastic equation,"},{"Start":"00:07.260 ","End":"00:12.165","Text":"but it\u0027s not homogeneous so it\u0027s actually a passonic equation on the unit disk."},{"Start":"00:12.165 ","End":"00:15.435","Text":"This is the boundary value on the unit circle."},{"Start":"00:15.435 ","End":"00:18.705","Text":"It\u0027s easier to solve in polar coordinates."},{"Start":"00:18.705 ","End":"00:22.035","Text":"X^2 plus y^2 is r^2."},{"Start":"00:22.035 ","End":"00:26.235","Text":"1 plus x is 1 plus r cosine Theta,"},{"Start":"00:26.235 ","End":"00:29.280","Text":"but r=1 on the unit circle,"},{"Start":"00:29.280 ","End":"00:31.800","Text":"so it\u0027s 1 plus cosine Theta."},{"Start":"00:31.800 ","End":"00:36.770","Text":"Now this is not a homogeneous equation and the boundary condition is not homogeneous."},{"Start":"00:36.770 ","End":"00:41.990","Text":"What we can do is split it up into the sum of two functions,"},{"Start":"00:41.990 ","End":"00:47.015","Text":"one which is homogeneous and one which has a homogeneous boundary condition."},{"Start":"00:47.015 ","End":"00:50.345","Text":"We\u0027ll let u=v plus w,"},{"Start":"00:50.345 ","End":"00:55.055","Text":"where v gets the boundary condition,"},{"Start":"00:55.055 ","End":"00:58.730","Text":"and w gets the non-homogeneous PDE."},{"Start":"00:58.730 ","End":"01:03.440","Text":"When I look for w as a function of r and not Theta because everything"},{"Start":"01:03.440 ","End":"01:08.255","Text":"here is just dependent on r. So we can just throw Theta out."},{"Start":"01:08.255 ","End":"01:14.728","Text":"The Laplacian in polar coordinates as the following and equals r^2,"},{"Start":"01:14.728 ","End":"01:19.080","Text":"and just copy this because w just depends on r,"},{"Start":"01:19.080 ","End":"01:21.570","Text":"and not on Theta w Theta,"},{"Start":"01:21.570 ","End":"01:22.995","Text":"Theta is 0,"},{"Start":"01:22.995 ","End":"01:25.685","Text":"and here we just have the derivatives of"},{"Start":"01:25.685 ","End":"01:29.465","Text":"our second derivative here and first derivative here."},{"Start":"01:29.465 ","End":"01:32.405","Text":"Multiplying by r, we get this."},{"Start":"01:32.405 ","End":"01:36.455","Text":"The left-hand side is the derivative of a product."},{"Start":"01:36.455 ","End":"01:40.085","Text":"Check it that it\u0027s the derivative of rw\u0027."},{"Start":"01:40.085 ","End":"01:42.035","Text":"Using the product rule, we get this."},{"Start":"01:42.035 ","End":"01:46.010","Text":"Now, take the integral of both sides and we get that"},{"Start":"01:46.010 ","End":"01:51.020","Text":"rw\u0027 is r^4/4 plus constant of integration."},{"Start":"01:51.020 ","End":"01:53.360","Text":"And dividing everything by r,"},{"Start":"01:53.360 ","End":"01:58.010","Text":"we get that w\u0027 is r^3/4 plus a/r."},{"Start":"01:58.010 ","End":"02:02.945","Text":"Now, w\u0027 prime has to be differentiable at the origin."},{"Start":"02:02.945 ","End":"02:05.555","Text":"A has to equal 0,"},{"Start":"02:05.555 ","End":"02:07.265","Text":"even just for continuity,"},{"Start":"02:07.265 ","End":"02:08.600","Text":"a has to be 0."},{"Start":"02:08.600 ","End":"02:13.355","Text":"That means that w\u0027(r) is just the r^3/4."},{"Start":"02:13.355 ","End":"02:20.775","Text":"Then another integration gives us that w(r) is r^4/16 plus another constant, call it b."},{"Start":"02:20.775 ","End":"02:24.995","Text":"We have the boundary condition that w(1) is 0,"},{"Start":"02:24.995 ","End":"02:29.065","Text":"which gives us that b is minus 1/16."},{"Start":"02:29.065 ","End":"02:34.970","Text":"That will give us w(r) is out on the r^4/16 of 16 minus 1/16,"},{"Start":"02:34.970 ","End":"02:41.280","Text":"and his is w. Now we have to find v. Here again is the problem for"},{"Start":"02:41.280 ","End":"02:43.200","Text":"v. This time it\u0027s"},{"Start":"02:43.200 ","End":"02:48.440","Text":"a homogeneous Laplace equation on the unit disk with boundary condition."},{"Start":"02:48.440 ","End":"02:53.315","Text":"Now the general solution for the plastic equation on the disk is the following,"},{"Start":"02:53.315 ","End":"02:56.105","Text":"where the disk is of radius Rho, but in our case,"},{"Start":"02:56.105 ","End":"03:00.230","Text":"Rho=1 so this becomes the following."},{"Start":"03:00.230 ","End":"03:02.530","Text":"If we substitute r=1,"},{"Start":"03:02.530 ","End":"03:05.935","Text":"on the one hand, it\u0027s 1 plus cosine Theta."},{"Start":"03:05.935 ","End":"03:10.050","Text":"On the other hand, it\u0027s this replacing r by 1,"},{"Start":"03:10.050 ","End":"03:12.690","Text":"which means that the r^n drops out."},{"Start":"03:12.690 ","End":"03:15.260","Text":"Now we can compare coefficients."},{"Start":"03:15.260 ","End":"03:17.900","Text":"There are only two non-zero coefficients."},{"Start":"03:17.900 ","End":"03:22.900","Text":"This would relate to 0 and this relates to 1."},{"Start":"03:22.900 ","End":"03:25.730","Text":"Because we\u0027ve written it as a Naught /2,"},{"Start":"03:25.730 ","End":"03:29.360","Text":"we get the that a_Naught=2 and a_1=1."},{"Start":"03:29.360 ","End":"03:31.175","Text":"This is 1 cosine Theta,"},{"Start":"03:31.175 ","End":"03:33.454","Text":"and everything else is 0."},{"Start":"03:33.454 ","End":"03:36.365","Text":"Now back to this equation."},{"Start":"03:36.365 ","End":"03:40.040","Text":"A_Naught is 2 and a_1 is 1,"},{"Start":"03:40.040 ","End":"03:48.300","Text":"and that gives us that v(r) Theta is 1 plus r^1 times 1 cosine 1 Theta,"},{"Start":"03:48.300 ","End":"03:50.435","Text":"just 1 plus r cosine Theta."},{"Start":"03:50.435 ","End":"03:55.080","Text":"Now we have v and we also have w from above,"},{"Start":"03:55.080 ","End":"03:59.340","Text":"which was r^4 over 16 minus 1/16."},{"Start":"03:59.340 ","End":"04:02.595","Text":"Together, this is u just simplifying a bit,"},{"Start":"04:02.595 ","End":"04:05.960","Text":"1 minus 1/16 is 15 over 16,"},{"Start":"04:05.960 ","End":"04:08.689","Text":"and that\u0027s the solution in polar,"},{"Start":"04:08.689 ","End":"04:13.910","Text":"we just have to convert to Cartesian r cosine Theta is x,"},{"Start":"04:13.910 ","End":"04:18.170","Text":"r^4=r^2,2, which is x^2 plus y^2,2,"},{"Start":"04:18.170 ","End":"04:23.090","Text":"and this is the answer in Cartesian, and we\u0027re done."}],"ID":30873},{"Watched":false,"Name":"Exercise 9","Duration":"3m 36s","ChapterTopicVideoID":29318,"CourseChapterTopicPlaylistID":294426,"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":"In this exercise, we\u0027re given the following initial value problem."},{"Start":"00:04.500 ","End":"00:09.900","Text":"It\u0027s a wave equation on the infinite interval with initial values,"},{"Start":"00:09.900 ","End":"00:12.510","Text":"we have to compute u of x, 1."},{"Start":"00:12.510 ","End":"00:15.660","Text":"We want the function u just when t equals 1,"},{"Start":"00:15.660 ","End":"00:17.520","Text":"not the general solution."},{"Start":"00:17.520 ","End":"00:21.790","Text":"We\u0027re going to use d\u0027Alembert\u0027s formula for the solution of"},{"Start":"00:21.790 ","End":"00:26.635","Text":"the wave equation on an infinite interval and this is the formula."},{"Start":"00:26.635 ","End":"00:29.635","Text":"In our case a equals 1."},{"Start":"00:29.635 ","End":"00:31.085","Text":"That\u0027s because in general,"},{"Start":"00:31.085 ","End":"00:36.080","Text":"we have u_ tt equals a squared u_xx and here a squared is 1,"},{"Start":"00:36.080 ","End":"00:40.220","Text":"so a is 1 and f of x is just the function"},{"Start":"00:40.220 ","End":"00:46.970","Text":"1 and t equals 1 from the statement of the problem here, t equals 1."},{"Start":"00:46.970 ","End":"00:50.580","Text":"What we get is f is 1,"},{"Start":"00:50.580 ","End":"00:53.280","Text":"so it\u0027s 1 plus 1 over 2 is 1,"},{"Start":"00:53.280 ","End":"00:55.245","Text":"a is 1, so this is a half,"},{"Start":"00:55.245 ","End":"01:01.200","Text":"and g of s is the following function it\u0027s just g of x,"},{"Start":"01:01.200 ","End":"01:10.760","Text":"but x replaced by s. Here it\u0027s equal to 1 minus x squared and outside it\u0027s equal to 0,"},{"Start":"01:10.760 ","End":"01:12.665","Text":"that\u0027s g of x from here."},{"Start":"01:12.665 ","End":"01:20.675","Text":"What we need to do is split up into different cases depending where x is on this axis."},{"Start":"01:20.675 ","End":"01:25.100","Text":"For example, if x is such that"},{"Start":"01:25.100 ","End":"01:30.725","Text":"the whole interval from x minus 1 to x plus 1 is to the right of 1, that\u0027s one thing."},{"Start":"01:30.725 ","End":"01:34.970","Text":"This comes out to be the case where x minus 1 is bigger or equal to"},{"Start":"01:34.970 ","End":"01:39.455","Text":"1 so x is bigger or equal to 2 and in this case,"},{"Start":"01:39.455 ","End":"01:43.655","Text":"the integral is just the integral of 0,"},{"Start":"01:43.655 ","End":"01:48.105","Text":"I mean, g of s is 0 on this whole interval."},{"Start":"01:48.105 ","End":"01:50.520","Text":"What we get is that u have x,"},{"Start":"01:50.520 ","End":"01:51.985","Text":"1 is equal to 1."},{"Start":"01:51.985 ","End":"01:53.465","Text":"That\u0027s the first case."},{"Start":"01:53.465 ","End":"01:58.730","Text":"Now, the second case where this overlaps like so,"},{"Start":"01:58.730 ","End":"02:03.360","Text":"it just means that x minus 1 is between minus"},{"Start":"02:03.360 ","End":"02:08.300","Text":"1 and 1 and that means that x is between 0 and 2."},{"Start":"02:08.300 ","End":"02:12.500","Text":"In this case, we just take the integral of this function between x"},{"Start":"02:12.500 ","End":"02:17.775","Text":"minus 1 and 1 so this is the integral we get."},{"Start":"02:17.775 ","End":"02:21.200","Text":"The integral of 1 minus x squared is x minus x cubed"},{"Start":"02:21.200 ","End":"02:24.770","Text":"over 3 between these limits and you know what?"},{"Start":"02:24.770 ","End":"02:26.690","Text":"It\u0027s just technical."},{"Start":"02:26.690 ","End":"02:29.165","Text":"I\u0027ll leave you with the computations."},{"Start":"02:29.165 ","End":"02:32.390","Text":"The answer comes out to be this."},{"Start":"02:32.390 ","End":"02:34.145","Text":"That\u0027s the second case."},{"Start":"02:34.145 ","End":"02:41.790","Text":"Now onto the third case where it similar to this but it overlaps on the left so that x"},{"Start":"02:41.790 ","End":"02:44.970","Text":"plus 1 falls in the gap between minus 1 and"},{"Start":"02:44.970 ","End":"02:49.800","Text":"1 and that means that x is between minus 2 and 0."},{"Start":"02:49.800 ","End":"02:56.105","Text":"In this case, we get the integral from minus 1 to x plus 1 of 1 minus x squared ds."},{"Start":"02:56.105 ","End":"02:59.520","Text":"Very similar computation to this the answer comes"},{"Start":"02:59.520 ","End":"03:03.170","Text":"out to be the same except with a minus sign here as"},{"Start":"03:03.170 ","End":"03:06.185","Text":"the third case and now the fourth and last case"},{"Start":"03:06.185 ","End":"03:10.520","Text":"is when the interval is completely to the left of minus 1,"},{"Start":"03:10.520 ","End":"03:14.330","Text":"so the integrant is 0 so what we"},{"Start":"03:14.330 ","End":"03:19.460","Text":"get is the following and just like in the very first case,"},{"Start":"03:19.460 ","End":"03:22.090","Text":"this comes out to be equal to 1."},{"Start":"03:22.090 ","End":"03:27.905","Text":"Now, all we have to do is summarize the 4 cases and this is what we get,"},{"Start":"03:27.905 ","End":"03:29.360","Text":"just copying all the cases."},{"Start":"03:29.360 ","End":"03:33.260","Text":"I think I combined 1 plus 2 over 3 to be 5 over 3,"},{"Start":"03:33.260 ","End":"03:37.320","Text":"and then that misses the answer and we\u0027re done."}],"ID":30874},{"Watched":false,"Name":"Exercise 10","Duration":"2m 36s","ChapterTopicVideoID":29319,"CourseChapterTopicPlaylistID":294426,"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 have a boundary value problem consisting of"},{"Start":"00:04.605 ","End":"00:09.330","Text":"a Laplace equation with Dirichlet boundary conditions."},{"Start":"00:09.330 ","End":"00:15.435","Text":"This is on the unit disk and we just have to find u at the origin."},{"Start":"00:15.435 ","End":"00:19.455","Text":"We don\u0027t have to actually solve for u in general."},{"Start":"00:19.455 ","End":"00:22.725","Text":"What we\u0027re going to do is use the mean value property,"},{"Start":"00:22.725 ","End":"00:26.400","Text":"which says that the value of u at a point is the average of"},{"Start":"00:26.400 ","End":"00:30.570","Text":"the values of u on a circle centered at that point."},{"Start":"00:30.570 ","End":"00:36.425","Text":"That helps us because we have the values of u on the circle of radius 1,"},{"Start":"00:36.425 ","End":"00:42.020","Text":"and we\u0027ll use that to get the average and hence the value of u at the origin."},{"Start":"00:42.020 ","End":"00:47.165","Text":"Now, another thing is that it doesn\u0027t have to be necessarily from 0 to 2Pi."},{"Start":"00:47.165 ","End":"00:52.425","Text":"We just have to take the integral on the full circle counterclockwise,"},{"Start":"00:52.425 ","End":"00:56.750","Text":"so we could take from any a to a plus 2Pi."},{"Start":"00:56.750 ","End":"00:59.060","Text":"In our case, x naught,"},{"Start":"00:59.060 ","End":"01:02.000","Text":"y naught is the 0,0."},{"Start":"01:02.000 ","End":"01:04.285","Text":"The radius Rho is 1."},{"Start":"01:04.285 ","End":"01:08.735","Text":"We\u0027ll take a to be equal to minus Pi,"},{"Start":"01:08.735 ","End":"01:12.655","Text":"so that we get the integral from minus Pi to Pi."},{"Start":"01:12.655 ","End":"01:15.440","Text":"This will help us because this is a symmetric interval"},{"Start":"01:15.440 ","End":"01:18.065","Text":"and we\u0027ll show that this is an odd function."},{"Start":"01:18.065 ","End":"01:21.725","Text":"U of cosine Theta sine Theta is"},{"Start":"01:21.725 ","End":"01:25.925","Text":"cosine hyperbolic cosine Theta sine hyperbolic sine Theta."},{"Start":"01:25.925 ","End":"01:28.100","Text":"We still have the formula here."},{"Start":"01:28.100 ","End":"01:31.685","Text":"The cosine is an even function,"},{"Start":"01:31.685 ","End":"01:34.820","Text":"then the function of an even function is even."},{"Start":"01:34.820 ","End":"01:38.410","Text":"Sine and sine hyperbolic are both odd,"},{"Start":"01:38.410 ","End":"01:41.210","Text":"and when you concatenate odd functions,"},{"Start":"01:41.210 ","End":"01:43.175","Text":"it\u0027s still get an odd function."},{"Start":"01:43.175 ","End":"01:48.850","Text":"Altogether this is an odd function because even times odd is odd."},{"Start":"01:48.850 ","End":"01:55.120","Text":"We have the integral of an odd function on a symmetric interval and the answer is 0."},{"Start":"01:55.120 ","End":"01:59.960","Text":"Now, I\u0027ll just explain more detail why an odd is odd,"},{"Start":"01:59.960 ","End":"02:02.720","Text":"and why this is even."},{"Start":"02:02.720 ","End":"02:06.470","Text":"Any function of an even function is an even function."},{"Start":"02:06.470 ","End":"02:13.145","Text":"In this case, any function of cosine of minus Theta is the same as f of cosine of Theta,"},{"Start":"02:13.145 ","End":"02:14.350","Text":"because cosine is even,"},{"Start":"02:14.350 ","End":"02:16.475","Text":"so the whole thing comes out to be even."},{"Start":"02:16.475 ","End":"02:18.275","Text":"Now in the case of odd,"},{"Start":"02:18.275 ","End":"02:21.619","Text":"if we have odd function of an odd function,"},{"Start":"02:21.619 ","End":"02:23.780","Text":"we put minus Theta instead of Theta,"},{"Start":"02:23.780 ","End":"02:26.105","Text":"the minus comes out in front of the sign,"},{"Start":"02:26.105 ","End":"02:28.700","Text":"and then it comes in front of the sine hyperbolic."},{"Start":"02:28.700 ","End":"02:32.240","Text":"Altogether, this minus comes out in front."},{"Start":"02:32.240 ","End":"02:33.740","Text":"That explains the even,"},{"Start":"02:33.740 ","End":"02:37.720","Text":"odd, and that concludes this exercise."}],"ID":30875},{"Watched":false,"Name":"Exercise 11","Duration":"2m 51s","ChapterTopicVideoID":29320,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.120","Text":"In this exercise, u satisfies the following boundary value problem."},{"Start":"00:06.120 ","End":"00:14.115","Text":"This is in an annulus and satisfies Laplace\u0027s equation and there are boundary conditions."},{"Start":"00:14.115 ","End":"00:21.030","Text":"We have to prove that if x and y are in the annulus,"},{"Start":"00:21.030 ","End":"00:23.160","Text":"then u of x,"},{"Start":"00:23.160 ","End":"00:27.480","Text":"y is between 0 and 2,022."},{"Start":"00:27.480 ","End":"00:32.205","Text":"We\u0027ll do this using the minimum and maximum principles for a harmonic function."},{"Start":"00:32.205 ","End":"00:36.495","Text":"After all, u is harmonic because it satisfies Laplace\u0027s equation."},{"Start":"00:36.495 ","End":"00:40.020","Text":"Now, it\u0027s easier to work in polar coordinates."},{"Start":"00:40.020 ","End":"00:42.250","Text":"In polar coordinates,"},{"Start":"00:42.250 ","End":"00:47.915","Text":"we have that the annulus is represented by r between 1 and 2"},{"Start":"00:47.915 ","End":"00:54.335","Text":"and u on the inner circle is natural log of 2 plus,"},{"Start":"00:54.335 ","End":"00:58.745","Text":"x is 1 cosine Theta which is just cosine Theta,"},{"Start":"00:58.745 ","End":"01:02.790","Text":"and here x is r cosine Theta,"},{"Start":"01:02.790 ","End":"01:06.694","Text":"which is 2 cosine Theta that\u0027s on the larger circle."},{"Start":"01:06.694 ","End":"01:11.780","Text":"Now we want to find the maximum and the minimum of u on the boundary,"},{"Start":"01:11.780 ","End":"01:14.255","Text":"which is made up of 2 separate circles."},{"Start":"01:14.255 ","End":"01:17.600","Text":"First of all, on the inner circle,"},{"Start":"01:17.600 ","End":"01:21.205","Text":"cosine Theta is between minus 1 and 1."},{"Start":"01:21.205 ","End":"01:26.490","Text":"The biggest that 2 plus cosine Theta could be is 2 plus 1,"},{"Start":"01:26.490 ","End":"01:27.790","Text":"which is 3,"},{"Start":"01:27.790 ","End":"01:30.980","Text":"because the natural log is an increasing function."},{"Start":"01:30.980 ","End":"01:36.275","Text":"The natural log on this maximum is natural log of 3."},{"Start":"01:36.275 ","End":"01:41.120","Text":"The minimum on this inner circle is when cosine Theta is least,"},{"Start":"01:41.120 ","End":"01:42.860","Text":"which is minus 1."},{"Start":"01:42.860 ","End":"01:45.874","Text":"We have natural log of 1 which is 0."},{"Start":"01:45.874 ","End":"01:50.285","Text":"Now let\u0027s find the maximum and minimum on the outer circle of radius 2."},{"Start":"01:50.285 ","End":"01:54.650","Text":"Similarly, the maximum is when cosine Theta is 1,"},{"Start":"01:54.650 ","End":"01:58.055","Text":"which gives us 2,022."},{"Start":"01:58.055 ","End":"02:01.870","Text":"The minimum is when cosine Theta is minus 1,"},{"Start":"02:01.870 ","End":"02:05.370","Text":"which gives us that 2 cosine Theta is minus 2,"},{"Start":"02:05.370 ","End":"02:11.875","Text":"minus 2 minus 2 is minus 4 natural log of e to the 2,022 minus 4."},{"Start":"02:11.875 ","End":"02:14.375","Text":"Now if we want the overall maximum,"},{"Start":"02:14.375 ","End":"02:16.490","Text":"take the biggest of these 2,"},{"Start":"02:16.490 ","End":"02:19.340","Text":"and for the minimum, the least of these 2."},{"Start":"02:19.340 ","End":"02:26.780","Text":"We get that the maximum on the annulus on the boundary is the maximum of these 2,"},{"Start":"02:26.780 ","End":"02:29.255","Text":"which is 2,022,"},{"Start":"02:29.255 ","End":"02:32.680","Text":"and the minimum of these 2 which is 0."},{"Start":"02:32.680 ","End":"02:37.335","Text":"So u is between this and this,"},{"Start":"02:37.335 ","End":"02:40.040","Text":"but because u is not a constant function,"},{"Start":"02:40.040 ","End":"02:49.110","Text":"strict inequalities apply and it\u0027s actually between 0 and 2,022 strictly."},{"Start":"02:49.110 ","End":"02:51.820","Text":"Okay, that\u0027s it."}],"ID":30876},{"Watched":false,"Name":"Exercise 12","Duration":"6m 22s","ChapterTopicVideoID":29321,"CourseChapterTopicPlaylistID":294426,"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 want to find the general solution of"},{"Start":"00:04.170 ","End":"00:08.835","Text":"the following differential equation on the first quadrant."},{"Start":"00:08.835 ","End":"00:11.040","Text":"It\u0027s the general solution."},{"Start":"00:11.040 ","End":"00:14.400","Text":"No initial or boundary conditions are given."},{"Start":"00:14.400 ","End":"00:16.685","Text":"We\u0027ll start with the classification."},{"Start":"00:16.685 ","End":"00:20.270","Text":"These are the second-order terms and we define a_1,"},{"Start":"00:20.270 ","End":"00:23.614","Text":"a_12, and a_22 accordingly."},{"Start":"00:23.614 ","End":"00:25.445","Text":"A_11 is x squared,"},{"Start":"00:25.445 ","End":"00:27.470","Text":"a_12 is just x,"},{"Start":"00:27.470 ","End":"00:29.035","Text":"y without the 2,"},{"Start":"00:29.035 ","End":"00:31.805","Text":"and a_22 is y squared."},{"Start":"00:31.805 ","End":"00:38.720","Text":"Now, the discriminant Delta is a_12 squared minus a_11 a_22,"},{"Start":"00:38.720 ","End":"00:41.975","Text":"and that is equal to 0,"},{"Start":"00:41.975 ","End":"00:45.485","Text":"which means that we have a parabolic equation."},{"Start":"00:45.485 ","End":"00:49.310","Text":"Next, the procedure is that we solve the differential equation dy by"},{"Start":"00:49.310 ","End":"00:54.695","Text":"dx equals a_12 plus or minus square root of Delta/a_ 11."},{"Start":"00:54.695 ","End":"00:58.135","Text":"This is equal 2xy."},{"Start":"00:58.135 ","End":"01:02.987","Text":"Since Delta is 0, the plus or minus doesn\u0027t make any difference here."},{"Start":"01:02.987 ","End":"01:05.485","Text":"Over 2x^2 which is y/x."},{"Start":"01:05.485 ","End":"01:08.455","Text":"There\u0027s only one solution as opposed to two."},{"Start":"01:08.455 ","End":"01:11.155","Text":"Let\u0027s continue with the one that we do have,"},{"Start":"01:11.155 ","End":"01:14.440","Text":"which is dy/dx equals y/x,"},{"Start":"01:14.440 ","End":"01:19.390","Text":"so we can separate the variables and get dy/y equals dx/x,"},{"Start":"01:19.390 ","End":"01:23.740","Text":"which gives us natural log of y is natural log of x plus a."},{"Start":"01:23.740 ","End":"01:26.140","Text":"Then taking the exponential of both sides,"},{"Start":"01:26.140 ","End":"01:29.620","Text":"we get that y is e^ax,"},{"Start":"01:29.620 ","End":"01:33.255","Text":"and e^a, we can call it a constant c_1."},{"Start":"01:33.255 ","End":"01:36.145","Text":"Actually, c_1 has to be positive,"},{"Start":"01:36.145 ","End":"01:41.545","Text":"which is true anyway, because y and x are positive in the first quadrant."},{"Start":"01:41.545 ","End":"01:45.045","Text":"This gives us y/x equals c_1."},{"Start":"01:45.045 ","End":"01:47.338","Text":"Let\u0027s refer to our table,"},{"Start":"01:47.338 ","End":"01:52.340","Text":"and we are in tight parabolic and we just have the one solution."},{"Start":"01:52.340 ","End":"01:54.695","Text":"This is Phi(x, y)."},{"Start":"01:54.695 ","End":"01:59.630","Text":"The way we do a change of variables as we let Psi equals Phi (x,"},{"Start":"01:59.630 ","End":"02:06.480","Text":"y) and choose either Eta equals x or Eta equals y. Phi(x,"},{"Start":"02:06.480 ","End":"02:08.910","Text":"y) is this y/x, and Eta (x,"},{"Start":"02:08.910 ","End":"02:12.305","Text":"y), let\u0027s go with Eta equals x."},{"Start":"02:12.305 ","End":"02:16.160","Text":"One thing we want to do is verify that the Jacobian is non-zero,"},{"Start":"02:16.160 ","End":"02:20.710","Text":"and really this is a change of variables transformation."},{"Start":"02:20.710 ","End":"02:23.870","Text":"The Jacobian, which is"},{"Start":"02:23.870 ","End":"02:29.485","Text":"the determinant of the first-order partial derivatives is, let\u0027s see."},{"Start":"02:29.485 ","End":"02:33.860","Text":"Derivative of this with respect to x and with respect to y,"},{"Start":"02:33.860 ","End":"02:37.805","Text":"the derivative of this with respect to x and with respect to y."},{"Start":"02:37.805 ","End":"02:47.260","Text":"It comes out to be 0 minus 1/ x^2 and that\u0027s not 0 because x is positive."},{"Start":"02:47.260 ","End":"02:53.345","Text":"The plan is now to solve the transformed equation."},{"Start":"02:53.345 ","End":"02:55.250","Text":"Instead of u (x, y),"},{"Start":"02:55.250 ","End":"02:57.650","Text":"we\u0027ll call that w(Psi,"},{"Start":"02:57.650 ","End":"03:01.665","Text":"Eta) in the domain of Psi Eta."},{"Start":"03:01.665 ","End":"03:07.020","Text":"At the end because of this has a nonzero Jacobian determinant,"},{"Start":"03:07.020 ","End":"03:10.051","Text":"we can reverse at least in theory,"},{"Start":"03:10.051 ","End":"03:14.810","Text":"this and get x in terms of Psi and Eta and y in terms of Psi and Eta."},{"Start":"03:14.810 ","End":"03:18.035","Text":"Then substitute that into the w that we find."},{"Start":"03:18.035 ","End":"03:22.095","Text":"Continuing, this is the original equation."},{"Start":"03:22.095 ","End":"03:24.770","Text":"We were up to the stage where we said that u (x,"},{"Start":"03:24.770 ","End":"03:27.425","Text":"y) is w(Psi, Eta)."},{"Start":"03:27.425 ","End":"03:32.120","Text":"We computed some of the first-order partial derivatives,"},{"Start":"03:32.120 ","End":"03:36.125","Text":"but we also want the second-order partial derivatives."},{"Start":"03:36.125 ","End":"03:38.334","Text":"I\u0027ll let you check these."},{"Start":"03:38.334 ","End":"03:40.727","Text":"In the tutorial,"},{"Start":"03:40.727 ","End":"03:44.450","Text":"we had the following formulas for the second-order derivatives."},{"Start":"03:44.450 ","End":"03:46.820","Text":"We also had them for the first order,"},{"Start":"03:46.820 ","End":"03:50.600","Text":"but we just need the second-order ones here."},{"Start":"03:50.600 ","End":"03:57.440","Text":"Now we can substitute the derivatives of Psi and Eta with respect to x and y from here."},{"Start":"03:57.440 ","End":"04:01.075","Text":"Well, I\u0027ll just give them to you,"},{"Start":"04:01.075 ","End":"04:04.420","Text":"and you can verify those later."},{"Start":"04:04.420 ","End":"04:07.354","Text":"Now we can simplify these."},{"Start":"04:07.354 ","End":"04:09.960","Text":"We need the three of these."},{"Start":"04:10.310 ","End":"04:15.050","Text":"Again, I\u0027ll leave you to check that this is correct."},{"Start":"04:15.050 ","End":"04:16.430","Text":"It\u0027s just like the first term."},{"Start":"04:16.430 ","End":"04:22.765","Text":"We have w_psi, psi and minus y/x^2^2 is y^2/x^4,"},{"Start":"04:22.765 ","End":"04:25.710","Text":"and so on throughout and some stuff cancels."},{"Start":"04:25.710 ","End":"04:28.820","Text":"Then mark the ones which is 0 so there\u0027s no point in"},{"Start":"04:28.820 ","End":"04:33.770","Text":"computing the ones next to a 0 because the product is only going to be 0."},{"Start":"04:33.770 ","End":"04:38.300","Text":"Now back to the original equation,"},{"Start":"04:38.300 ","End":"04:41.335","Text":"we can now substitute u_xx from here,"},{"Start":"04:41.335 ","End":"04:43.405","Text":"and u_xy from here,"},{"Start":"04:43.405 ","End":"04:45.670","Text":"and u _yy from here."},{"Start":"04:45.670 ","End":"04:48.200","Text":"This is what we get."},{"Start":"04:48.200 ","End":"04:51.435","Text":"Now we want to expand this,"},{"Start":"04:51.435 ","End":"04:57.815","Text":"and it turns out that everything cancels except the w_Eta Eta term."},{"Start":"04:57.815 ","End":"05:02.645","Text":"For example, if you want to compute the coefficient of W_Psi Psi,"},{"Start":"05:02.645 ","End":"05:07.920","Text":"from here we get y^2/ x^2."},{"Start":"05:07.920 ","End":"05:14.190","Text":"From here, we get minus 2y^2 /x^2 squared."},{"Start":"05:14.190 ","End":"05:17.655","Text":"Here, we get plus y^2/ x^2."},{"Start":"05:17.655 ","End":"05:20.000","Text":"It\u0027s 1 minus 2 plus 1 of these,"},{"Start":"05:20.000 ","End":"05:22.895","Text":"which is 0 of these, and so on."},{"Start":"05:22.895 ","End":"05:25.355","Text":"Everything cancels except for this."},{"Start":"05:25.355 ","End":"05:32.024","Text":"Now we divide both sides by x^2 and we have W_Eta Eta equals 4."},{"Start":"05:32.024 ","End":"05:33.050","Text":"This is easy to solve."},{"Start":"05:33.050 ","End":"05:36.680","Text":"First step, we can integrate with respect to Eta and get"},{"Start":"05:36.680 ","End":"05:41.035","Text":"W_Eta is 4 Eta plus like a constant,"},{"Start":"05:41.035 ","End":"05:43.800","Text":"but an arbitrary function of xy."},{"Start":"05:43.800 ","End":"05:51.335","Text":"Integrate again, and we get W equals this integral with respect to Eta is 2Eta^2."},{"Start":"05:51.335 ","End":"05:57.065","Text":"Then here, this times Eta plus another arbitrary function of Psi."},{"Start":"05:57.065 ","End":"06:04.590","Text":"At this point, we can substitute Psi and Eta and then we get that u (x,"},{"Start":"06:04.590 ","End":"06:05.892","Text":"y) is 2,"},{"Start":"06:05.892 ","End":"06:08.219","Text":"Eta is x squared,"},{"Start":"06:08.219 ","End":"06:10.515","Text":"f of Psi is f(y/x),"},{"Start":"06:10.515 ","End":"06:15.470","Text":"Eta is x, and g(Psi) is g(y/x)."},{"Start":"06:15.470 ","End":"06:17.180","Text":"That\u0027s the answer."},{"Start":"06:17.180 ","End":"06:22.650","Text":"That\u0027s the most general solution of the PDE. We are done."}],"ID":30877},{"Watched":false,"Name":"Exercise 13","Duration":"6m 35s","ChapterTopicVideoID":29322,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.900","Text":"In this exercise, we\u0027re given this IBVP initial boundary value problem."},{"Start":"00:06.900 ","End":"00:09.930","Text":"We have to show that the solution to this is unique,"},{"Start":"00:09.930 ","End":"00:13.125","Text":"and we\u0027ll do this by means of an energy integral,"},{"Start":"00:13.125 ","End":"00:16.830","Text":"which is given in the hint. Let\u0027s start."},{"Start":"00:16.830 ","End":"00:19.335","Text":"The idea is the usual."},{"Start":"00:19.335 ","End":"00:20.760","Text":"When we show uniqueness,"},{"Start":"00:20.760 ","End":"00:24.870","Text":"we show that if u and v are solutions of the problem that"},{"Start":"00:24.870 ","End":"00:29.190","Text":"necessarily u is identically equal to v. There aren\u0027t 2."},{"Start":"00:29.190 ","End":"00:31.545","Text":"It appears there\u0027s really only 1."},{"Start":"00:31.545 ","End":"00:35.965","Text":"The usual method is to define a 1/3 w,"},{"Start":"00:35.965 ","End":"00:37.710","Text":"which is u minus v,"},{"Start":"00:37.710 ","End":"00:41.760","Text":"and to prove that w is identically 0."},{"Start":"00:41.760 ","End":"00:51.275","Text":"Let us satisfy this PDE and v satisfies the same thing just with v replacing u."},{"Start":"00:51.275 ","End":"00:56.470","Text":"Then w is u minus v. Let\u0027s see what equation w satisfies."},{"Start":"00:56.470 ","End":"00:59.160","Text":"If we subtract the 2 equations,"},{"Start":"00:59.160 ","End":"01:03.410","Text":"then big F of x,t cancels and we get this plus"},{"Start":"01:03.410 ","End":"01:08.800","Text":"this minus this minus this equals this minus this."},{"Start":"01:08.800 ","End":"01:10.910","Text":"If we collect together,"},{"Start":"01:10.910 ","End":"01:17.585","Text":"we get an equation in w because u_tt minus v_tt is w_tt."},{"Start":"01:17.585 ","End":"01:20.795","Text":"This minus this is Beta times wt,"},{"Start":"01:20.795 ","End":"01:24.660","Text":"and this minus this is w_xx."},{"Start":"01:24.660 ","End":"01:29.990","Text":"That\u0027s the PDE for w. Now we want to convert the initial and boundary conditions."},{"Start":"01:29.990 ","End":"01:34.595","Text":"For f of x, we have these conditions and for v of x,"},{"Start":"01:34.595 ","End":"01:35.960","Text":"we have these conditions."},{"Start":"01:35.960 ","End":"01:37.555","Text":"I mean, it\u0027s the same."},{"Start":"01:37.555 ","End":"01:41.660","Text":"If we subtract these equations pairwise,"},{"Start":"01:41.660 ","End":"01:45.568","Text":"we\u0027ll get the boundary and initial conditions for w,"},{"Start":"01:45.568 ","End":"01:50.930","Text":"and we can add the PDE so we\u0027ll get the problem for w as follows."},{"Start":"01:50.930 ","End":"01:56.490","Text":"Copy the PDE and then u of x naught minus v of x naught is 0,"},{"Start":"01:56.490 ","End":"01:58.365","Text":"that\u0027s w of x naught."},{"Start":"01:58.365 ","End":"02:03.695","Text":"All of these for initial and boundary conditions are 0."},{"Start":"02:03.695 ","End":"02:08.825","Text":"Now we\u0027ll recall the energy integral that we were given in the hint."},{"Start":"02:08.825 ","End":"02:13.180","Text":"Note that if we let t equal 0,"},{"Start":"02:13.180 ","End":"02:20.885","Text":"then w_t of x naught is equal to naught because we have it here."},{"Start":"02:20.885 ","End":"02:26.635","Text":"But what about the derivative with respect to x at x,0?"},{"Start":"02:26.635 ","End":"02:29.435","Text":"Well, we\u0027re not given that x equals to 0."},{"Start":"02:29.435 ","End":"02:33.875","Text":"But from this, if we differentiate with respect to x,"},{"Start":"02:33.875 ","End":"02:37.115","Text":"we\u0027ll get that w_x is 0."},{"Start":"02:37.115 ","End":"02:40.260","Text":"Like we said, w_t is 0,"},{"Start":"02:40.260 ","End":"02:41.790","Text":"and w of x,0 is 0,"},{"Start":"02:41.790 ","End":"02:44.180","Text":"which gives us w_x of x,0 is 0,"},{"Start":"02:44.180 ","End":"02:47.530","Text":"which means that the whole integrant is 0."},{"Start":"02:47.530 ","End":"02:52.230","Text":"E of 0 is equal to 0."},{"Start":"02:52.230 ","End":"02:55.230","Text":"We just showed that E of 0 is 0."},{"Start":"02:55.230 ","End":"02:59.330","Text":"I should have mentioned this earlier that E of t"},{"Start":"02:59.330 ","End":"03:04.685","Text":"is bigger or equal to 0 because the integrant is bigger or equal to 0."},{"Start":"03:04.685 ","End":"03:06.635","Text":"We\u0027ll need this later."},{"Start":"03:06.635 ","End":"03:08.520","Text":"E prime of t,"},{"Start":"03:08.520 ","End":"03:14.045","Text":"we can get by differentiating under the integral sign and what we get,"},{"Start":"03:14.045 ","End":"03:17.580","Text":"well, should be a 2 here and a 2 here,"},{"Start":"03:17.580 ","End":"03:19.290","Text":"which canceled with the 1/2."},{"Start":"03:19.290 ","End":"03:21.530","Text":"But the derivative of something squared is twice"},{"Start":"03:21.530 ","End":"03:23.570","Text":"that something times the anti-derivative."},{"Start":"03:23.570 ","End":"03:28.170","Text":"Here also twice w_x times w_xt."},{"Start":"03:28.190 ","End":"03:31.995","Text":"We\u0027re going to change the order of differentiation."},{"Start":"03:31.995 ","End":"03:34.890","Text":"We could write w_tx instead."},{"Start":"03:34.890 ","End":"03:37.950","Text":"The PDE that w satisfies is this."},{"Start":"03:37.950 ","End":"03:47.265","Text":"We can replace the w_tt here by w_xx minus Beta w_t, like so."},{"Start":"03:47.265 ","End":"03:49.130","Text":"The other 1/2,"},{"Start":"03:49.130 ","End":"03:50.915","Text":"which we write as a separate integral,"},{"Start":"03:50.915 ","End":"03:54.440","Text":"we can do it using integration by parts."},{"Start":"03:54.440 ","End":"03:56.425","Text":"This is the formula."},{"Start":"03:56.425 ","End":"04:01.355","Text":"Here we\u0027ll take this is v and this is u prime."},{"Start":"04:01.355 ","End":"04:04.520","Text":"We can split this integral up into 2 also."},{"Start":"04:04.520 ","End":"04:09.200","Text":"This, using the integration by parts formula is,"},{"Start":"04:09.200 ","End":"04:10.565","Text":"well, I wrote it in the other order,"},{"Start":"04:10.565 ","End":"04:14.570","Text":"uv is w_t, w_x,"},{"Start":"04:14.570 ","End":"04:20.170","Text":"and uv prime is w_t, w_xx."},{"Start":"04:20.170 ","End":"04:23.900","Text":"Now, this cancels with this and what we\u0027re left"},{"Start":"04:23.900 ","End":"04:28.220","Text":"with is this plus this path. This is also 0."},{"Start":"04:28.220 ","End":"04:34.550","Text":"The reason this is 0 is this part is what we get when we substitute x equals l,"},{"Start":"04:34.550 ","End":"04:36.980","Text":"and here we substitute x equals 0."},{"Start":"04:36.980 ","End":"04:38.610","Text":"Now the w_x of l,"},{"Start":"04:38.610 ","End":"04:40.560","Text":"t is equal to 0,"},{"Start":"04:40.560 ","End":"04:42.015","Text":"1 of the boundary conditions."},{"Start":"04:42.015 ","End":"04:43.865","Text":"This is the other boundary conditions."},{"Start":"04:43.865 ","End":"04:46.250","Text":"The boundary conditions, this is 0, this is 0."},{"Start":"04:46.250 ","End":"04:51.550","Text":"All we\u0027re left with is minus Beta integral of something squared,"},{"Start":"04:51.550 ","End":"04:54.195","Text":"and Beta was given to be positive."},{"Start":"04:54.195 ","End":"04:56.460","Text":"This is greater or equal to 0."},{"Start":"04:56.460 ","End":"05:00.305","Text":"Altogether we have something less than or equal to 0."},{"Start":"05:00.305 ","End":"05:08.150","Text":"Now, because the derivative of E is less than or equal to 0 and E of 0 is 0,"},{"Start":"05:08.150 ","End":"05:13.565","Text":"we start at 0 and it\u0027s decreasing or non-increasing,"},{"Start":"05:13.565 ","End":"05:16.310","Text":"and so it has to stay less than or equal to 0."},{"Start":"05:16.310 ","End":"05:22.200","Text":"On the other hand, we showed just above that E of t is bigger or equal to 0."},{"Start":"05:22.200 ","End":"05:25.685","Text":"If it\u0027s bigger or equal to 0 and less than or equal to 0,"},{"Start":"05:25.685 ","End":"05:27.664","Text":"then it is 0."},{"Start":"05:27.664 ","End":"05:31.355","Text":"We\u0027ve shown that E of t is identically 0."},{"Start":"05:31.355 ","End":"05:40.130","Text":"If it\u0027s identically 0 and if the integrant is bigger or equal to 0 and continuous,"},{"Start":"05:40.130 ","End":"05:46.255","Text":"the only way we\u0027re going to get 0 here is if this is always 0."},{"Start":"05:46.255 ","End":"05:53.550","Text":"The integrant has to equal 0 for all x and t and in algebra,"},{"Start":"05:53.550 ","End":"05:56.880","Text":"if a sum of 2 squares is 0,"},{"Start":"05:56.880 ","End":"05:59.190","Text":"then each of them has to be 0."},{"Start":"05:59.190 ","End":"06:03.545","Text":"We have that w_t is 0 and w_x is 0."},{"Start":"06:03.545 ","End":"06:07.939","Text":"If we have a function whose both of whose partial derivatives is 0,"},{"Start":"06:07.939 ","End":"06:10.985","Text":"then the function is a constant."},{"Start":"06:10.985 ","End":"06:18.470","Text":"If w of x, t is a constant and at some point it\u0027s equal to 0,"},{"Start":"06:18.470 ","End":"06:21.275","Text":"then that constant has to be 0 everywhere."},{"Start":"06:21.275 ","End":"06:23.150","Text":"Otherwise, it wouldn\u0027t be a constant."},{"Start":"06:23.150 ","End":"06:24.710","Text":"We get that w of x,"},{"Start":"06:24.710 ","End":"06:27.595","Text":"t is identically equal to 0."},{"Start":"06:27.595 ","End":"06:29.628","Text":"W was u minus v,"},{"Start":"06:29.628 ","End":"06:32.210","Text":"so this means that u is identically equal to v,"},{"Start":"06:32.210 ","End":"06:33.500","Text":"and that\u0027s what we had to show,"},{"Start":"06:33.500 ","End":"06:35.610","Text":"and so we are done."}],"ID":30878},{"Watched":false,"Name":"Exercise 14","Duration":"4m 31s","ChapterTopicVideoID":29323,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.915","Text":"In this exercise, we\u0027re going to solve the following problem,"},{"Start":"00:03.915 ","End":"00:09.435","Text":"which has a heat equation and initial condition and boundary conditions."},{"Start":"00:09.435 ","End":"00:13.200","Text":"Can use a technique of separation of variables."},{"Start":"00:13.200 ","End":"00:16.770","Text":"What we have is a homogeneous heat equation on"},{"Start":"00:16.770 ","End":"00:20.955","Text":"a finite interval with Neumann boundary conditions."},{"Start":"00:20.955 ","End":"00:23.550","Text":"Now we\u0027re not going to solve it from scratch because we already"},{"Start":"00:23.550 ","End":"00:26.475","Text":"developed some formulas. Here they are."},{"Start":"00:26.475 ","End":"00:33.390","Text":"What we want is the case where both boundary conditions are of the Neumann type."},{"Start":"00:33.390 ","End":"00:36.730","Text":"We are going to use this formula."},{"Start":"00:36.920 ","End":"00:40.365","Text":"Just change capital A to little a."},{"Start":"00:40.365 ","End":"00:42.660","Text":"This is the formula we had."},{"Start":"00:42.660 ","End":"00:45.720","Text":"If we let t=0,"},{"Start":"00:45.720 ","End":"00:49.320","Text":"then we have the u(x, 0)=x."},{"Start":"00:49.320 ","End":"00:52.770","Text":"On the other hand, when t=0,"},{"Start":"00:52.770 ","End":"00:57.450","Text":"this exponent e^minus 0 just drops out."},{"Start":"00:57.450 ","End":"00:58.970","Text":"This is what we have."},{"Start":"00:58.970 ","End":"01:01.555","Text":"We have a cosine series for x."},{"Start":"01:01.555 ","End":"01:08.000","Text":"Now, there is a formula for a Fourier cosine series that gives us the coefficients a_n."},{"Start":"01:08.000 ","End":"01:09.965","Text":"This is the formula."},{"Start":"01:09.965 ","End":"01:12.220","Text":"We just have to compute this."},{"Start":"01:12.220 ","End":"01:14.450","Text":"We\u0027ll use integration by parts."},{"Start":"01:14.450 ","End":"01:16.180","Text":"We\u0027ll let this be g,"},{"Start":"01:16.180 ","End":"01:18.105","Text":"and this part be f\u0027."},{"Start":"01:18.105 ","End":"01:20.870","Text":"This is the formula I\u0027m talking about."},{"Start":"01:20.870 ","End":"01:25.110","Text":"What we get is this is g,"},{"Start":"01:25.110 ","End":"01:26.295","Text":"this is f,"},{"Start":"01:26.295 ","End":"01:30.120","Text":"this is also f and g\u0027 is 1."},{"Start":"01:30.120 ","End":"01:31.670","Text":"It doesn\u0027t appear here."},{"Start":"01:31.670 ","End":"01:35.600","Text":"Now, this is only going to work when n is not 0."},{"Start":"01:35.600 ","End":"01:38.600","Text":"We will have to return to the case n=0,"},{"Start":"01:38.600 ","End":"01:40.355","Text":"and we\u0027re done with this case."},{"Start":"01:40.355 ","End":"01:48.125","Text":"Now, this part is just 0 because when x is 0, well, this is 0."},{"Start":"01:48.125 ","End":"01:51.080","Text":"But in any event, when x is 0, this is also 0."},{"Start":"01:51.080 ","End":"01:56.080","Text":"When x is L, then we have n Pi and sine of n Pi is 0."},{"Start":"01:56.080 ","End":"01:59.630","Text":"We just get the integral of the second part."},{"Start":"01:59.630 ","End":"02:02.480","Text":"We can take the n Pi/L in front,"},{"Start":"02:02.480 ","End":"02:05.090","Text":"where it becomes L Pi."},{"Start":"02:05.090 ","End":"02:08.585","Text":"L cancels with the L and we have 2 Pi."},{"Start":"02:08.585 ","End":"02:15.695","Text":"The integral of minus sine of n Pi/L, x dx."},{"Start":"02:15.695 ","End":"02:20.000","Text":"The integral of minus sine is cosine."},{"Start":"02:20.000 ","End":"02:23.545","Text":"But we also have to divide by the anti-derivative again."},{"Start":"02:23.545 ","End":"02:29.760","Text":"We have to evaluate this from 0 to L. When x is L,"},{"Start":"02:29.760 ","End":"02:32.100","Text":"we have cosine n Pi."},{"Start":"02:32.100 ","End":"02:36.455","Text":"When x is 0, we have cosine 0, which is 1."},{"Start":"02:36.455 ","End":"02:38.465","Text":"This is what we get."},{"Start":"02:38.465 ","End":"02:41.505","Text":"Cosine n Pi minus 1."},{"Start":"02:41.505 ","End":"02:45.485","Text":"The n Pi/L combines with this."},{"Start":"02:45.485 ","End":"02:49.090","Text":"We invert this, we get 2L/(n Pi)^2."},{"Start":"02:49.090 ","End":"02:54.585","Text":"We know that cosine n Pi is minus 1^n."},{"Start":"02:54.585 ","End":"03:00.535","Text":"We just showed that a_n is 2L/(n Pi)^2 minus 1^n minus 1."},{"Start":"03:00.535 ","End":"03:03.830","Text":"But this is only so when n is not 0."},{"Start":"03:03.830 ","End":"03:08.225","Text":"We return to this formula with n=0 this time."},{"Start":"03:08.225 ","End":"03:12.185","Text":"We get a_0 is this integral because when n is 0,"},{"Start":"03:12.185 ","End":"03:14.865","Text":"cosine of 0 is 1, just disappears."},{"Start":"03:14.865 ","End":"03:16.925","Text":"This integral is straightforward."},{"Start":"03:16.925 ","End":"03:19.220","Text":"Integral of x is x^2/2,"},{"Start":"03:19.220 ","End":"03:25.290","Text":"so it\u0027s L^2/2 times 2/L is equal to L. Now,"},{"Start":"03:25.290 ","End":"03:27.375","Text":"we have a_n and a_0."},{"Start":"03:27.375 ","End":"03:32.120","Text":"What we have to do now is substitute them in this formula."},{"Start":"03:32.120 ","End":"03:39.615","Text":"I notice L, so this is this L and also a_n is this."},{"Start":"03:39.615 ","End":"03:42.065","Text":"We just put that here."},{"Start":"03:42.065 ","End":"03:45.560","Text":"Now, notice that minus 1^n minus 1,"},{"Start":"03:45.560 ","End":"03:47.480","Text":"we\u0027ve seen this many times,"},{"Start":"03:47.480 ","End":"03:49.950","Text":"is either 2 or 0,"},{"Start":"03:49.950 ","End":"03:53.240","Text":"depending on whether n is odd or even."},{"Start":"03:53.240 ","End":"03:55.550","Text":"When n is even, we get the 0 case."},{"Start":"03:55.550 ","End":"04:00.660","Text":"We just have to take the odd n\u0027s and odd n\u0027s like 1,"},{"Start":"04:00.660 ","End":"04:01.860","Text":"3, 5, etc."},{"Start":"04:01.860 ","End":"04:06.300","Text":"we can get as n=2k minus 1 and k=1,"},{"Start":"04:06.300 ","End":"04:08.535","Text":"2, 3, etc."},{"Start":"04:08.535 ","End":"04:11.790","Text":"We take k from 1 to infinity."},{"Start":"04:11.790 ","End":"04:15.165","Text":"Instead of n, we have 2k minus 1."},{"Start":"04:15.165 ","End":"04:19.334","Text":"Also here and also here."},{"Start":"04:19.334 ","End":"04:21.465","Text":"This becomes 2,"},{"Start":"04:21.465 ","End":"04:24.165","Text":"combines with the 2 to become 4."},{"Start":"04:24.165 ","End":"04:26.430","Text":"This is the answer."},{"Start":"04:26.430 ","End":"04:29.100","Text":"I\u0027ll just highlight it,"},{"Start":"04:29.100 ","End":"04:32.260","Text":"and we are done."}],"ID":30879},{"Watched":false,"Name":"Exercise 15","Duration":"5m 23s","ChapterTopicVideoID":29324,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.310","Text":"In this exercise, we\u0027re going to find"},{"Start":"00:02.310 ","End":"00:06.840","Text":"the canonical form of the partial differential equation,"},{"Start":"00:06.840 ","End":"00:13.020","Text":"2u_xx plus 2yu_yy plus u_y equals 0,"},{"Start":"00:13.020 ","End":"00:15.945","Text":"and this is on the upper half-plane."},{"Start":"00:15.945 ","End":"00:19.320","Text":"First of all, we want to classify it to see if it\u0027s hyperbolic,"},{"Start":"00:19.320 ","End":"00:21.105","Text":"parabolic, or elliptic."},{"Start":"00:21.105 ","End":"00:26.475","Text":"We label the coefficients as a_11, a_12, a_22."},{"Start":"00:26.475 ","End":"00:29.955","Text":"Notice that the middle coefficient is 2a_12."},{"Start":"00:29.955 ","End":"00:31.440","Text":"In this case, it doesn\u0027t matter,"},{"Start":"00:31.440 ","End":"00:33.585","Text":"a_12 will come out to be 0."},{"Start":"00:33.585 ","End":"00:36.885","Text":"Anyway, the discriminant is the following,"},{"Start":"00:36.885 ","End":"00:38.235","Text":"and so in our case,"},{"Start":"00:38.235 ","End":"00:41.880","Text":"it\u0027s 0^2 minus 2 2y,"},{"Start":"00:41.880 ","End":"00:45.340","Text":"which comes out to be minus 4y."},{"Start":"00:45.340 ","End":"00:46.700","Text":"Since y is positive,"},{"Start":"00:46.700 ","End":"00:48.628","Text":"minus 4y is negative,"},{"Start":"00:48.628 ","End":"00:51.635","Text":"and so it\u0027s always elliptic in the domain."},{"Start":"00:51.635 ","End":"00:57.080","Text":"The next step is to solve the differential equation dy by dx equals a_12"},{"Start":"00:57.080 ","End":"01:02.370","Text":"plus or minus root Delta over a_11."},{"Start":"01:02.370 ","End":"01:05.640","Text":"In our case, Delta is minus 4y,"},{"Start":"01:05.640 ","End":"01:09.800","Text":"it\u0027s negative so we\u0027re going to have to work with imaginary or complex numbers,"},{"Start":"01:09.800 ","End":"01:15.065","Text":"and we get plus or minus square root of y times i."},{"Start":"01:15.065 ","End":"01:18.590","Text":"The 4 comes out here as 2 and it cancels with the 2."},{"Start":"01:18.590 ","End":"01:22.670","Text":"If we divide both sides by root y and bring the dx over,"},{"Start":"01:22.670 ","End":"01:24.748","Text":"we\u0027ve separated the variables,"},{"Start":"01:24.748 ","End":"01:28.925","Text":"and we have dy over root y equals plus or minus i dx."},{"Start":"01:28.925 ","End":"01:31.475","Text":"Then integrate both sides."},{"Start":"01:31.475 ","End":"01:34.550","Text":"The integral of 1 over root y is twice root y,"},{"Start":"01:34.550 ","End":"01:38.030","Text":"integral of this is this plus the constant."},{"Start":"01:38.030 ","End":"01:39.920","Text":"Let\u0027s rewrite it like so."},{"Start":"01:39.920 ","End":"01:44.179","Text":"Then we could say that c_1 is with a plus and c_2 is with a minus."},{"Start":"01:44.179 ","End":"01:48.035","Text":"Now let\u0027s refer to the table from the tutorial,"},{"Start":"01:48.035 ","End":"01:51.990","Text":"and we look up elliptic where Delta is less than 0."},{"Start":"01:51.990 ","End":"01:54.945","Text":"If we have Phi(x, y)=c_1,"},{"Start":"01:54.945 ","End":"01:58.268","Text":"like 2 root y plus ix is c_1,"},{"Start":"01:58.268 ","End":"02:00.300","Text":"and with a minus ix is c_2."},{"Start":"02:00.300 ","End":"02:08.730","Text":"What we have to do is take Psi to be the real part and Eta to be the imaginary part."},{"Start":"02:08.730 ","End":"02:13.700","Text":"Let\u0027s take Psi equals 2 root y and Eta could be,"},{"Start":"02:13.700 ","End":"02:15.795","Text":"doesn\u0027t matter if we take the plus or the minus,"},{"Start":"02:15.795 ","End":"02:18.530","Text":"let\u0027s take the plus x. I need to make sure this"},{"Start":"02:18.530 ","End":"02:22.265","Text":"really is a transformation that\u0027s invertible."},{"Start":"02:22.265 ","End":"02:25.153","Text":"We look at the Jacobian matrix,"},{"Start":"02:25.153 ","End":"02:27.370","Text":"and take its determinant."},{"Start":"02:27.370 ","End":"02:31.290","Text":"Psi x is 0,"},{"Start":"02:31.290 ","End":"02:36.290","Text":"Psi y is the derivative of 2 root y is 1 over root y,"},{"Start":"02:36.290 ","End":"02:37.978","Text":"derivative of x is 1,"},{"Start":"02:37.978 ","End":"02:40.640","Text":"derivative of x with respect to y is 0."},{"Start":"02:40.640 ","End":"02:46.984","Text":"This minus 1 over root y is not 0."},{"Start":"02:46.984 ","End":"02:49.280","Text":"So we\u0027re okay with the transformation."},{"Start":"02:49.280 ","End":"02:51.860","Text":"This means that in principle, this is invertible."},{"Start":"02:51.860 ","End":"02:53.720","Text":"In this case just for interest\u0027s sake."},{"Start":"02:53.720 ","End":"02:55.940","Text":"It\u0027s not only in principle invertible,"},{"Start":"02:55.940 ","End":"02:58.340","Text":"but it\u0027s practically easy to do that."},{"Start":"02:58.340 ","End":"02:59.660","Text":"X is equal to Eta,"},{"Start":"02:59.660 ","End":"03:05.020","Text":"from here divided by 2 square both sides and we have that y is Psi over Eta squared."},{"Start":"03:05.020 ","End":"03:08.985","Text":"We don\u0027t need that, we just want to know that it is invertible."},{"Start":"03:08.985 ","End":"03:12.120","Text":"We could get back from a function of Psi,"},{"Start":"03:12.120 ","End":"03:14.010","Text":"Eta to a function of xy."},{"Start":"03:14.010 ","End":"03:18.530","Text":"Because what we\u0027re going to do is let u(xy) be w of Psi Eta."},{"Start":"03:18.530 ","End":"03:22.460","Text":"That\u0027s what happens to the function u when we do the transformation."},{"Start":"03:22.460 ","End":"03:25.790","Text":"Now, we need the partial derivatives of Psi and Eta with"},{"Start":"03:25.790 ","End":"03:28.835","Text":"respect to x and y. Here they are."},{"Start":"03:28.835 ","End":"03:33.878","Text":"Just copy them from the determinant here, from the Jacobian."},{"Start":"03:33.878 ","End":"03:37.685","Text":"We\u0027ll also need the second-order partial derivatives."},{"Start":"03:37.685 ","End":"03:40.025","Text":"Here are all of them."},{"Start":"03:40.025 ","End":"03:42.120","Text":"Most of them are 0,"},{"Start":"03:42.120 ","End":"03:45.220","Text":"the second-order ones, except for this 1."},{"Start":"03:45.220 ","End":"03:48.335","Text":"Here again, is the original equation."},{"Start":"03:48.335 ","End":"03:57.725","Text":"Now, we need these partial derivatives to compute u_xx in terms of w Psi and w Eta, etc."},{"Start":"03:57.725 ","End":"04:01.640","Text":"We don\u0027t have to start from scratch because in the tutorial,"},{"Start":"04:01.640 ","End":"04:04.400","Text":"we had 5 of them figured out u_x,"},{"Start":"04:04.400 ","End":"04:07.010","Text":"u_i, u_xx, u_xy, and u_yy."},{"Start":"04:07.010 ","End":"04:10.980","Text":"We only need 3 out of the 5, and here they are."},{"Start":"04:11.300 ","End":"04:17.225","Text":"I already put the partial derivatives correspond too from this table."},{"Start":"04:17.225 ","End":"04:20.075","Text":"Psi y is 1 over root y,"},{"Start":"04:20.075 ","End":"04:24.020","Text":"wherever we had 1 of them as 0 we didn\u0027t need to compute the other 1."},{"Start":"04:24.020 ","End":"04:25.730","Text":"If 1 of them is 0,"},{"Start":"04:25.730 ","End":"04:27.515","Text":"there\u0027s no need to compute,"},{"Start":"04:27.515 ","End":"04:30.770","Text":"here I don\u0027t need Eta x for example."},{"Start":"04:30.770 ","End":"04:33.485","Text":"Having said that, I don\u0027t need this."},{"Start":"04:33.485 ","End":"04:36.305","Text":"Now, just collecting stuff together,"},{"Start":"04:36.305 ","End":"04:38.420","Text":"I\u0027ll let you check the calculations."},{"Start":"04:38.420 ","End":"04:40.010","Text":"Some stuff cancels."},{"Start":"04:40.010 ","End":"04:42.245","Text":"This is what it reduces to."},{"Start":"04:42.245 ","End":"04:48.390","Text":"Now we can substitute these in the original equation."},{"Start":"04:48.390 ","End":"04:51.785","Text":"What we get is the following."},{"Start":"04:51.785 ","End":"04:54.035","Text":"If we expand the brackets,"},{"Start":"04:54.035 ","End":"04:56.060","Text":"this is what we get."},{"Start":"04:56.060 ","End":"04:58.880","Text":"When I say y to the minus a half,"},{"Start":"04:58.880 ","End":"05:02.720","Text":"that\u0027s the same as 1 over root y together with w Psi."},{"Start":"05:02.720 ","End":"05:06.470","Text":"This cancels with this and all we\u0027re left with is this."},{"Start":"05:06.470 ","End":"05:10.370","Text":"Of course, we can divide by 2 so we get the following,"},{"Start":"05:10.370 ","End":"05:13.310","Text":"which is our canonical equation."},{"Start":"05:13.310 ","End":"05:16.385","Text":"Let\u0027s just check in the table."},{"Start":"05:16.385 ","End":"05:21.305","Text":"Here we are to see if we\u0027ve got the right canonical form and yes, we do."},{"Start":"05:21.305 ","End":"05:24.330","Text":"That concludes this clip."}],"ID":30880},{"Watched":false,"Name":"Exercise 16","Duration":"7m 54s","ChapterTopicVideoID":29325,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.090","Text":"In this exercise, we\u0027re given the following initial boundary and value problem to solve."},{"Start":"00:06.090 ","End":"00:08.940","Text":"Let\u0027s see what kind it is."},{"Start":"00:08.940 ","End":"00:12.420","Text":"Well, this is a wave equation."},{"Start":"00:12.420 ","End":"00:15.825","Text":"It\u0027s nonhomogeneous because it\u0027s got this extra term here,"},{"Start":"00:15.825 ","End":"00:18.870","Text":"and it\u0027s on a semi-infinite interval."},{"Start":"00:18.870 ","End":"00:24.915","Text":"U and u_t at time 0 are the initial conditions."},{"Start":"00:24.915 ","End":"00:26.940","Text":"We only have 1 boundary condition,"},{"Start":"00:26.940 ","End":"00:29.790","Text":"not 2 because we\u0027re on the semi-infinite interval."},{"Start":"00:29.790 ","End":"00:34.635","Text":"The thing is that we\u0027d really like the boundary condition to be 0."},{"Start":"00:34.635 ","End":"00:38.385","Text":"Then we can extend the problem to either an"},{"Start":"00:38.385 ","End":"00:42.120","Text":"odd or an even function on the whole interval."},{"Start":"00:42.120 ","End":"00:45.215","Text":"There is a way we can get this to be 0."},{"Start":"00:45.215 ","End":"00:50.210","Text":"The standard trick, we can replace u by v plus"},{"Start":"00:50.210 ","End":"00:57.020","Text":"w. W doesn\u0027t have to be a function of x and t. It can be just a function of t. It works."},{"Start":"00:57.020 ","End":"01:03.000","Text":"We want to find w(t) such that v(0,t) is 0,"},{"Start":"01:03.000 ","End":"01:07.505","Text":"that v will have a homogeneous boundary condition."},{"Start":"01:07.505 ","End":"01:09.830","Text":"We will find what w is,"},{"Start":"01:09.830 ","End":"01:17.510","Text":"and v is to substitute x=0 here and bring v to the left and u to the right."},{"Start":"01:17.510 ","End":"01:21.875","Text":"We get v(0,t)=u(0,t) minus w(t)."},{"Start":"01:21.875 ","End":"01:25.745","Text":"We want to have that v(0, t) is 0."},{"Start":"01:25.745 ","End":"01:28.370","Text":"Let this be 0 and u(0,"},{"Start":"01:28.370 ","End":"01:32.565","Text":"t) is given to be this. This is what we get."},{"Start":"01:32.565 ","End":"01:37.680","Text":"Then we can get w(t) just by moving w(t) to the left."},{"Start":"01:37.680 ","End":"01:41.435","Text":"We have that it\u0027s equal to e to the minus t minus 1."},{"Start":"01:41.435 ","End":"01:43.145","Text":"Now that we have w,"},{"Start":"01:43.145 ","End":"01:45.440","Text":"we can put w here."},{"Start":"01:45.440 ","End":"01:50.270","Text":"We have that u(x,t) is v(x,t) plus e to the minus t minus 1."},{"Start":"01:50.270 ","End":"01:53.610","Text":"We want to find the problem for v,"},{"Start":"01:53.610 ","End":"01:55.560","Text":"that is the problem for u."},{"Start":"01:55.560 ","End":"01:58.035","Text":"Now we have the pd for u,"},{"Start":"01:58.035 ","End":"02:04.635","Text":"but we need it for v. We\u0027ll need u_tt and u_xx in terms of v_tt and v_xx."},{"Start":"02:04.635 ","End":"02:12.139","Text":"U_tt from here is v_tt plus w double prime of t,"},{"Start":"02:12.139 ","End":"02:16.835","Text":"which is e to the minus t. Here\u0027s the computation."},{"Start":"02:16.835 ","End":"02:20.450","Text":"W(t) differentiate once, differentiate a second time."},{"Start":"02:20.450 ","End":"02:24.925","Text":"That\u0027s e to the minus t. With respect to x,"},{"Start":"02:24.925 ","End":"02:29.090","Text":"even the first derivative of this with respect to x is 0."},{"Start":"02:29.090 ","End":"02:31.355","Text":"U_xx is just v_xx."},{"Start":"02:31.355 ","End":"02:34.430","Text":"Substituting into the PDE for u,"},{"Start":"02:34.430 ","End":"02:42.375","Text":"we get a PDE for v. We get from here u_tt is v_tt plus e to the minus t from there."},{"Start":"02:42.375 ","End":"02:46.515","Text":"From here we have that u_xx is v_xx."},{"Start":"02:46.515 ","End":"02:48.200","Text":"This copied from here."},{"Start":"02:48.200 ","End":"02:50.845","Text":"Now the e to the minus t cancels."},{"Start":"02:50.845 ","End":"02:56.240","Text":"The PDE for v is the simple wave equation, the homogeneous."},{"Start":"02:56.240 ","End":"03:01.385","Text":"Let\u0027s see, we have a PDE for v. We have an initial condition."},{"Start":"03:01.385 ","End":"03:03.780","Text":"We still don\u0027t have boundary conditions."},{"Start":"03:03.780 ","End":"03:06.050","Text":"Let\u0027s just summarize so far."},{"Start":"03:06.050 ","End":"03:08.255","Text":"This we know, this we know,"},{"Start":"03:08.255 ","End":"03:10.865","Text":"and it\u0027s homogeneous like we wanted."},{"Start":"03:10.865 ","End":"03:13.130","Text":"We need to still figure out these,"},{"Start":"03:13.130 ","End":"03:15.240","Text":"the boundary conditions for v. But,"},{"Start":"03:15.240 ","End":"03:20.165","Text":"let\u0027s use the boundary conditions for u to get those for v. V is u."},{"Start":"03:20.165 ","End":"03:21.950","Text":"Well, we have it written."},{"Start":"03:21.950 ","End":"03:24.200","Text":"Where is it? Here,"},{"Start":"03:24.200 ","End":"03:25.430","Text":"but we switch u and v,"},{"Start":"03:25.430 ","End":"03:27.785","Text":"so this will become a minus."},{"Start":"03:27.785 ","End":"03:32.660","Text":"Then substitute t=0."},{"Start":"03:32.660 ","End":"03:34.070","Text":"This is what we get."},{"Start":"03:34.070 ","End":"03:36.490","Text":"U(x,0) is 1,"},{"Start":"03:36.490 ","End":"03:39.770","Text":"and this one cancels with the minus 1 and we still have plus 1,"},{"Start":"03:39.770 ","End":"03:42.545","Text":"so this is 1. What about v_t?"},{"Start":"03:42.545 ","End":"03:46.820","Text":"Well, differentiating gives us v_t in general."},{"Start":"03:46.820 ","End":"03:52.600","Text":"Then substituting t=0. We have u_t(x,0)."},{"Start":"03:52.600 ","End":"03:55.890","Text":"If you go back and see it\u0027s 2 sine x minus 1."},{"Start":"03:55.890 ","End":"03:58.760","Text":"Then together with the e to the minus 0,"},{"Start":"03:58.760 ","End":"04:01.310","Text":"which is 1, it\u0027s 2 sine x."},{"Start":"04:01.310 ","End":"04:05.175","Text":"We have that this is, let\u0027s just write it."},{"Start":"04:05.175 ","End":"04:07.430","Text":"We found that this is one,"},{"Start":"04:07.430 ","End":"04:08.980","Text":"we\u0027ll call that f(x),"},{"Start":"04:08.980 ","End":"04:11.220","Text":"and this is 2 sine x,"},{"Start":"04:11.220 ","End":"04:14.760","Text":"we\u0027ll call that g(x). Here we are again."},{"Start":"04:14.760 ","End":"04:19.220","Text":"This boundary condition is of type Dirichlet,"},{"Start":"04:19.220 ","End":"04:23.000","Text":"which means that we\u0027re going to look for an odd extension."},{"Start":"04:23.000 ","End":"04:26.535","Text":"If it was Neumann, we\u0027d use an even extension."},{"Start":"04:26.535 ","End":"04:28.820","Text":"How do we find the odd extension?"},{"Start":"04:28.820 ","End":"04:35.240","Text":"We look for odd extensions of f and g to the whole real line."},{"Start":"04:35.240 ","End":"04:38.660","Text":"F tilde we call the extension,"},{"Start":"04:38.660 ","End":"04:41.420","Text":"is just being one on the positives."},{"Start":"04:41.420 ","End":"04:45.005","Text":"It\u0027s 1 or minus 1 depending on positive or negative."},{"Start":"04:45.005 ","End":"04:49.265","Text":"But for g, 2 sine x already is an odd function."},{"Start":"04:49.265 ","End":"04:51.410","Text":"We can just keep it as is."},{"Start":"04:51.410 ","End":"04:55.040","Text":"We don\u0027t really need v tilde extension of v because"},{"Start":"04:55.040 ","End":"04:58.810","Text":"it\u0027s going to be the same as v for positive x."},{"Start":"04:58.810 ","End":"05:00.920","Text":"We can use d\u0027Alembert\u0027s formula,"},{"Start":"05:00.920 ","End":"05:04.895","Text":"just make sure we only use it for positive x."},{"Start":"05:04.895 ","End":"05:07.445","Text":"This is the usual formula."},{"Start":"05:07.445 ","End":"05:09.590","Text":"Let\u0027s try and develop it."},{"Start":"05:09.590 ","End":"05:13.255","Text":"Simplify it. We\u0027ll just work on the G part first."},{"Start":"05:13.255 ","End":"05:18.849","Text":"G tilde of s is 2 sine s. I mean just g tilde of"},{"Start":"05:18.849 ","End":"05:24.100","Text":"x is 2 sine x replace x by s. The 1/2 cancels with the 2."},{"Start":"05:24.100 ","End":"05:27.710","Text":"The integral of sine is minus cosine."},{"Start":"05:27.710 ","End":"05:31.345","Text":"We want to evaluate this between these 2 limits."},{"Start":"05:31.345 ","End":"05:34.363","Text":"What we get, this one will come out in plus,"},{"Start":"05:34.363 ","End":"05:36.850","Text":"and this one will come out in minus because of the minus here."},{"Start":"05:36.850 ","End":"05:40.610","Text":"We get cosine x minus t minus cosine x plus t."},{"Start":"05:40.610 ","End":"05:45.155","Text":"There is a trigonometric identity, this one."},{"Start":"05:45.155 ","End":"05:49.085","Text":"We can convert this to 2 sine x,"},{"Start":"05:49.085 ","End":"05:56.875","Text":"sine t. This is v. Let\u0027s go from v to u. U is v plus w,"},{"Start":"05:56.875 ","End":"05:59.695","Text":"and w is e to the t minus 1."},{"Start":"05:59.695 ","End":"06:03.835","Text":"Now we have u. Finally, you will get to work on this expression."},{"Start":"06:03.835 ","End":"06:06.130","Text":"Now, remember the definition of f tilde."},{"Start":"06:06.130 ","End":"06:10.795","Text":"It\u0027s 1 when x is positive and minus 1 when x is negative,"},{"Start":"06:10.795 ","End":"06:14.710","Text":"we need to split up into cases to see where x plus t could"},{"Start":"06:14.710 ","End":"06:19.315","Text":"be positive or negative and where x minus t is positive or negative."},{"Start":"06:19.315 ","End":"06:22.450","Text":"There\u0027s really only 2 cases or 3,"},{"Start":"06:22.450 ","End":"06:23.905","Text":"but 1 drops out."},{"Start":"06:23.905 ","End":"06:26.785","Text":"One case is when this is positive,"},{"Start":"06:26.785 ","End":"06:30.075","Text":"the x plus t and x minus t is positive."},{"Start":"06:30.075 ","End":"06:34.100","Text":"Of course, we\u0027re still working under the restriction that x is positive."},{"Start":"06:34.100 ","End":"06:40.190","Text":"Altogether, this simplifies to just x bigger than t. Then all of this is satisfied."},{"Start":"06:40.190 ","End":"06:44.700","Text":"Then f(x) plus t will be 1,"},{"Start":"06:44.700 ","End":"06:46.620","Text":"and x minus t is also positive."},{"Start":"06:46.620 ","End":"06:48.900","Text":"F tilde of x minus t is 1,"},{"Start":"06:48.900 ","End":"06:50.985","Text":"1 plus 1 over 2 is 1."},{"Start":"06:50.985 ","End":"06:55.070","Text":"In Case 2, move a bit to the left,"},{"Start":"06:55.070 ","End":"06:56.705","Text":"this will stay positive,"},{"Start":"06:56.705 ","End":"06:58.565","Text":"but this will become negative."},{"Start":"06:58.565 ","End":"07:00.995","Text":"However, we still want to keep x positive."},{"Start":"07:00.995 ","End":"07:04.580","Text":"This boils down to x less than t,"},{"Start":"07:04.580 ","End":"07:06.470","Text":"but x is still bigger than 0."},{"Start":"07:06.470 ","End":"07:10.340","Text":"In this case, we have that this is positive,"},{"Start":"07:10.340 ","End":"07:12.320","Text":"and this is negative."},{"Start":"07:12.320 ","End":"07:15.050","Text":"This becomes 1,"},{"Start":"07:15.050 ","End":"07:18.920","Text":"and this is minus 1. This becomes 0."},{"Start":"07:18.920 ","End":"07:24.140","Text":"Actually are no other cases because the other cases involve x being negative."},{"Start":"07:24.140 ","End":"07:26.525","Text":"We\u0027ll leave it at this."},{"Start":"07:26.525 ","End":"07:31.745","Text":"What we get is u(x,t) is if x is bigger than t,"},{"Start":"07:31.745 ","End":"07:37.220","Text":"it\u0027s equal to this expression here with this replaced by 1."},{"Start":"07:37.220 ","End":"07:39.785","Text":"The 1 cancels with the minus 1."},{"Start":"07:39.785 ","End":"07:41.870","Text":"It leaves us just with this."},{"Start":"07:41.870 ","End":"07:43.475","Text":"In the other case,"},{"Start":"07:43.475 ","End":"07:47.180","Text":"where this is 0,"},{"Start":"07:47.180 ","End":"07:49.310","Text":"the minus 1 remains."},{"Start":"07:49.310 ","End":"07:52.400","Text":"We have this. Finally,"},{"Start":"07:52.400 ","End":"07:55.200","Text":"that\u0027s the answer and we\u0027re done."}],"ID":30881},{"Watched":false,"Name":"Exercise 17","Duration":"4m 47s","ChapterTopicVideoID":29326,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.695","Text":"In this exercise, we\u0027re given the following problem to solve."},{"Start":"00:04.695 ","End":"00:08.835","Text":"This is a wave equation on a finite interval."},{"Start":"00:08.835 ","End":"00:11.504","Text":"These are initial conditions,"},{"Start":"00:11.504 ","End":"00:16.965","Text":"and these are boundary conditions which are homogeneous."},{"Start":"00:16.965 ","End":"00:24.235","Text":"Our task is to compute the limit as t goes to infinity of this integral."},{"Start":"00:24.235 ","End":"00:27.500","Text":"We\u0027re going to do this without actually solving for u."},{"Start":"00:27.500 ","End":"00:31.805","Text":"This integral dx leaves us just with a function of t,"},{"Start":"00:31.805 ","End":"00:34.105","Text":"and we\u0027ll call that E(t)."},{"Start":"00:34.105 ","End":"00:35.490","Text":"E stands for energy,"},{"Start":"00:35.490 ","End":"00:37.320","Text":"but there\u0027s no worry about that."},{"Start":"00:37.320 ","End":"00:39.311","Text":"E(t) is the following,"},{"Start":"00:39.311 ","End":"00:40.940","Text":"let\u0027s start working on it."},{"Start":"00:40.940 ","End":"00:43.385","Text":"First thing we\u0027ll do is differentiate,"},{"Start":"00:43.385 ","End":"00:47.750","Text":"and let\u0027s assume that it\u0027s okay to differentiate under the integral sign."},{"Start":"00:47.750 ","End":"00:53.160","Text":"What we get is twice u_t times the derivative of u_t,"},{"Start":"00:53.160 ","End":"00:58.080","Text":"which is u_tt, plus twice u_x times"},{"Start":"00:58.080 ","End":"01:04.060","Text":"the derivative of ux with respect to t, so it\u0027s u_xt."},{"Start":"01:04.190 ","End":"01:06.675","Text":"The arrows here,"},{"Start":"01:06.675 ","End":"01:11.335","Text":"because I\u0027m going to switch the order of integration in the next line,"},{"Start":"01:11.335 ","End":"01:17.325","Text":"the thing we\u0027re going to do is swap this u_tt for u_xx,"},{"Start":"01:17.325 ","End":"01:20.500","Text":"that is from the original pde,"},{"Start":"01:20.500 ","End":"01:23.480","Text":"the wave equation, as it is,"},{"Start":"01:23.480 ","End":"01:28.925","Text":"it would be useless because we\u0027re having an integral dx and the derivative tt,"},{"Start":"01:28.925 ","End":"01:32.845","Text":"so wouldn\u0027t help, but this will make it useful."},{"Start":"01:32.845 ","End":"01:38.445","Text":"As I said, the xt becomes tx split up into 2 integrals."},{"Start":"01:38.445 ","End":"01:41.280","Text":"u_tt is replaced by u_xx,"},{"Start":"01:41.280 ","End":"01:45.885","Text":"and for convenience, I put the 2 out of the way as a 1/2 here."},{"Start":"01:45.885 ","End":"01:48.950","Text":"We\u0027re going to use integration by parts on this."},{"Start":"01:48.950 ","End":"01:51.325","Text":"This is the formula,"},{"Start":"01:51.325 ","End":"01:53.115","Text":"and in our case,"},{"Start":"01:53.115 ","End":"01:55.830","Text":"this will be v and this will be u\u0027,"},{"Start":"01:55.830 ","End":"01:58.685","Text":"so after we\u0027ve applied the formula,"},{"Start":"01:58.685 ","End":"02:02.080","Text":"we get this as is,"},{"Start":"02:02.080 ","End":"02:05.330","Text":"and here u times v,"},{"Start":"02:05.330 ","End":"02:07.400","Text":"it\u0027s confusing with a letters here."},{"Start":"02:07.400 ","End":"02:10.010","Text":"I should have used f and g here."},{"Start":"02:10.010 ","End":"02:14.435","Text":"But we want the integral of this,"},{"Start":"02:14.435 ","End":"02:16.175","Text":"and that is u_t,"},{"Start":"02:16.175 ","End":"02:18.350","Text":"which I wrote over here,"},{"Start":"02:18.350 ","End":"02:21.840","Text":"and the u_x in the second place here."},{"Start":"02:21.840 ","End":"02:25.740","Text":"Here we have the same u_t here,"},{"Start":"02:25.740 ","End":"02:29.785","Text":"and the derivative of this, which is u_xx."},{"Start":"02:29.785 ","End":"02:32.595","Text":"Then this cancels with this,"},{"Start":"02:32.595 ","End":"02:34.715","Text":"so we\u0027re just left with the middle term."},{"Start":"02:34.715 ","End":"02:36.500","Text":"u is the function u(x,"},{"Start":"02:36.500 ","End":"02:39.635","Text":"t), and if we put x = 1,"},{"Start":"02:39.635 ","End":"02:45.360","Text":"then we get u_t(1, t ) or u_x(1, t )."},{"Start":"02:45.360 ","End":"02:48.420","Text":"When we put 0, it\u0027s x which is 0,"},{"Start":"02:48.420 ","End":"02:52.500","Text":"so we have u_t(0, t ) and u_x(0, t )."},{"Start":"02:52.500 ","End":"02:58.960","Text":"Now this is 0 and this is 0 because these are our boundary conditions."},{"Start":"02:58.960 ","End":"03:03.435","Text":"Let\u0027s see, go back. Here they are both 0."},{"Start":"03:03.435 ","End":"03:05.125","Text":"I forgot to say earlier,"},{"Start":"03:05.125 ","End":"03:10.990","Text":"our goal is to find the limit as t goes to infinity of E(t)."},{"Start":"03:10.990 ","End":"03:15.085","Text":"Now 0 minus 0 is 0."},{"Start":"03:15.085 ","End":"03:22.680","Text":"Half E\u0027(t) is 0 so E\u0027(t) is 0 so E(t) is a constant,"},{"Start":"03:22.680 ","End":"03:26.140","Text":"and the constant will be E at any particular point."},{"Start":"03:26.140 ","End":"03:32.131","Text":"But E(0), we can compute a different way so E(t) is constantly equal to E(0),"},{"Start":"03:32.131 ","End":"03:36.785","Text":"so it remains to compute this because"},{"Start":"03:36.785 ","End":"03:41.860","Text":"the limit as t goes to infinity of a constant is just that constant,"},{"Start":"03:41.860 ","End":"03:44.605","Text":"so the limit will also be E(0)."},{"Start":"03:44.605 ","End":"03:47.875","Text":"Now, E(0) by the definition is this."},{"Start":"03:47.875 ","End":"03:50.290","Text":"We had the definition with t here,"},{"Start":"03:50.290 ","End":"03:52.105","Text":"replace t by 0,"},{"Start":"03:52.105 ","End":"03:55.280","Text":"and this is the integral we have to compute"},{"Start":"03:56.160 ","End":"04:02.500","Text":"u_t(x,0) was given as an initial condition to be x^2(1- x)."},{"Start":"04:02.500 ","End":"04:05.380","Text":"U_x(x, 0) wasn\u0027t given,"},{"Start":"04:05.380 ","End":"04:07.420","Text":"but u(x, 0) is 0,"},{"Start":"04:07.420 ","End":"04:11.980","Text":"which means that u is 0 along the x-axis when t is 0,"},{"Start":"04:11.980 ","End":"04:15.905","Text":"which means that the derivative along the x-axis is also 0."},{"Start":"04:15.905 ","End":"04:18.225","Text":"We have u_t and u_x now,"},{"Start":"04:18.225 ","End":"04:20.525","Text":"and we can substitute them here."},{"Start":"04:20.525 ","End":"04:23.060","Text":"Here we substitute x^2 (1-x),"},{"Start":"04:23.060 ","End":"04:24.410","Text":"here we substitute 0,"},{"Start":"04:24.410 ","End":"04:26.525","Text":"and these are squared and added."},{"Start":"04:26.525 ","End":"04:30.705","Text":"Expand this, this is 1 minus 2x plus x squared."},{"Start":"04:30.705 ","End":"04:33.390","Text":"Multiplying by x squared squared,"},{"Start":"04:33.390 ","End":"04:37.540","Text":"which is x^4, we get x^4 - 2x^5 + x^6."},{"Start":"04:37.760 ","End":"04:41.805","Text":"Then it\u0027s just a simple computation."},{"Start":"04:41.805 ","End":"04:47.790","Text":"The answer comes out to be 1/105, and we\u0027re done."}],"ID":30882},{"Watched":false,"Name":"Exercise 18","Duration":"9m 24s","ChapterTopicVideoID":29327,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.125","Text":"In this exercise, we\u0027re given that the function u(x,"},{"Start":"00:04.125 ","End":"00:07.949","Text":"t) is a solution of the following initial value problem."},{"Start":"00:07.949 ","End":"00:11.880","Text":"This is a heat equation on the infinite interval."},{"Start":"00:11.880 ","End":"00:14.219","Text":"This is the initial condition."},{"Start":"00:14.219 ","End":"00:17.100","Text":"Our task is to compute this limit,"},{"Start":"00:17.100 ","End":"00:19.740","Text":"but without actually solving for u,"},{"Start":"00:19.740 ","End":"00:21.675","Text":"which in any event is difficult."},{"Start":"00:21.675 ","End":"00:23.460","Text":"If we did try to solve for u,"},{"Start":"00:23.460 ","End":"00:25.500","Text":"would use Poisson\u0027s formula."},{"Start":"00:25.500 ","End":"00:27.660","Text":"This is the formula I mean."},{"Start":"00:27.660 ","End":"00:32.190","Text":"We\u0027d get an integral that\u0027s really too hard to solve in practice."},{"Start":"00:32.190 ","End":"00:34.560","Text":"We did learn a technique for finding"},{"Start":"00:34.560 ","End":"00:40.350","Text":"the asymptotic solution under the condition that the absolute value of h,"},{"Start":"00:40.350 ","End":"00:42.300","Text":"which is this function here,"},{"Start":"00:42.300 ","End":"00:44.450","Text":"is less than or equal to Beta e to"},{"Start":"00:44.450 ","End":"00:49.880","Text":"the minus Alpha absolute value of x for some constants Alpha and Beta."},{"Start":"00:49.880 ","End":"00:53.765","Text":"Now in fact, this condition does not hold in our problem."},{"Start":"00:53.765 ","End":"00:56.540","Text":"Because if the condition held,"},{"Start":"00:56.540 ","End":"01:01.190","Text":"then the limit as x goes to plus or minus infinity of h is 0,"},{"Start":"01:01.190 ","End":"01:03.155","Text":"but that\u0027s not the case here."},{"Start":"01:03.155 ","End":"01:05.210","Text":"When x goes to infinity,"},{"Start":"01:05.210 ","End":"01:06.875","Text":"the limit is 2."},{"Start":"01:06.875 ","End":"01:09.965","Text":"Dominating factors are e^x."},{"Start":"01:09.965 ","End":"01:12.650","Text":"When x goes to minus infinity,"},{"Start":"01:12.650 ","End":"01:14.750","Text":"then the limit is 1,"},{"Start":"01:14.750 ","End":"01:18.290","Text":"because e to the minus x dominates."},{"Start":"01:18.290 ","End":"01:20.750","Text":"Then the case at least one of these is not 0."},{"Start":"01:20.750 ","End":"01:22.565","Text":"In fact, both of them are not 0."},{"Start":"01:22.565 ","End":"01:24.680","Text":"This condition does not hold."},{"Start":"01:24.680 ","End":"01:33.110","Text":"However, we\u0027re going to try and fix that by defining a function G as follows,"},{"Start":"01:33.110 ","End":"01:35.885","Text":"and letting f=h minus g,"},{"Start":"01:35.885 ","End":"01:42.700","Text":"where g(x) is 2 when x is positive and 1 when x is negative,"},{"Start":"01:42.890 ","End":"01:46.140","Text":"then f(x) is equal to,"},{"Start":"01:46.140 ","End":"01:47.480","Text":"if we spell it out,"},{"Start":"01:47.480 ","End":"01:54.335","Text":"h is this minus g is 2 when x is positive and this minus 1 when x is negative."},{"Start":"01:54.335 ","End":"01:57.185","Text":"If we put under a common denominator,"},{"Start":"01:57.185 ","End":"01:58.520","Text":"we get this and this."},{"Start":"01:58.520 ","End":"02:01.165","Text":"I\u0027ll leave you to check the computations."},{"Start":"02:01.165 ","End":"02:06.905","Text":"We can pull e to the minus x in front here and e^x in front here."},{"Start":"02:06.905 ","End":"02:12.305","Text":"This condition does hold for f because for all x,"},{"Start":"02:12.305 ","End":"02:14.210","Text":"f(x), we can write it."},{"Start":"02:14.210 ","End":"02:16.190","Text":"If x is positive,"},{"Start":"02:16.190 ","End":"02:18.540","Text":"then x is absolute value of x."},{"Start":"02:18.540 ","End":"02:19.805","Text":"If x is negative,"},{"Start":"02:19.805 ","End":"02:21.760","Text":"minus x is absolute value of x,"},{"Start":"02:21.760 ","End":"02:23.660","Text":"so e to the absolute value of x."},{"Start":"02:23.660 ","End":"02:27.070","Text":"Minus 1 here we can write as plus or minus."},{"Start":"02:27.070 ","End":"02:30.080","Text":"Here also, when x is negative,"},{"Start":"02:30.080 ","End":"02:33.865","Text":"this is e to the twice absolute value of x and the same here."},{"Start":"02:33.865 ","End":"02:36.360","Text":"F(x) to the plus or minus this,"},{"Start":"02:36.360 ","End":"02:38.510","Text":"but when we take absolute value of f(x),"},{"Start":"02:38.510 ","End":"02:40.505","Text":"it\u0027s equal exactly to this."},{"Start":"02:40.505 ","End":"02:43.825","Text":"Then 1 over this is less than or equal to 1."},{"Start":"02:43.825 ","End":"02:47.495","Text":"This is less than or equal e to the minus absolute value of x."},{"Start":"02:47.495 ","End":"02:52.280","Text":"It is in the form Beta e to the minus Alpha absolute value of x."},{"Start":"02:52.280 ","End":"02:55.175","Text":"We take Beta is 1 and Alpha is 1."},{"Start":"02:55.175 ","End":"03:05.240","Text":"Now, h is equal to f plus g. That allows us to split our problem up into 2, u= v+w,"},{"Start":"03:05.240 ","End":"03:08.765","Text":"where v will have the same equation as u,"},{"Start":"03:08.765 ","End":"03:12.745","Text":"but with g(x) here instead of h. For w,"},{"Start":"03:12.745 ","End":"03:17.555","Text":"we\u0027ll have f (x) instead of h. Both of these are on the same domain."},{"Start":"03:17.555 ","End":"03:22.115","Text":"The infinite interval and time is positive."},{"Start":"03:22.115 ","End":"03:26.930","Text":"Limit we have to calculate is equal to the following."},{"Start":"03:26.930 ","End":"03:30.810","Text":"Just split u up into v+w."},{"Start":"03:30.810 ","End":"03:34.070","Text":"Next, we\u0027ll do some side computations to figure out v,"},{"Start":"03:34.070 ","End":"03:37.825","Text":"w, vx, and wx."},{"Start":"03:37.825 ","End":"03:40.033","Text":"Let\u0027s start with v,"},{"Start":"03:40.033 ","End":"03:43.250","Text":"and later we\u0027ll do w. This is the problem for"},{"Start":"03:43.250 ","End":"03:46.850","Text":"v and the initial condition is fairly simple,"},{"Start":"03:46.850 ","End":"03:50.280","Text":"so we can use Poisson\u0027s formula."},{"Start":"03:50.280 ","End":"03:51.950","Text":"This is the formula,"},{"Start":"03:51.950 ","End":"03:54.255","Text":"I mean, and in our case,"},{"Start":"03:54.255 ","End":"03:56.150","Text":"well, a is 1."},{"Start":"03:56.150 ","End":"04:01.705","Text":"We get this, also use s instead of y and v here instead of u."},{"Start":"04:01.705 ","End":"04:03.820","Text":"Split this integral up into 2,"},{"Start":"04:03.820 ","End":"04:06.935","Text":"from minus infinity to 0 and from 0 to infinity."},{"Start":"04:06.935 ","End":"04:12.985","Text":"From minus infinity to 0, v(s,0)=1."},{"Start":"04:12.985 ","End":"04:15.920","Text":"From 0 to infinity,"},{"Start":"04:15.920 ","End":"04:19.610","Text":"it\u0027s equal to 2 when we can bring the 1 on the 2 in front."},{"Start":"04:19.610 ","End":"04:21.515","Text":"This is what we get."},{"Start":"04:21.515 ","End":"04:25.445","Text":"The other thing is we can write this as something squared t is positive."},{"Start":"04:25.445 ","End":"04:26.990","Text":"Now do a substitution."},{"Start":"04:26.990 ","End":"04:31.340","Text":"We\u0027ll let z be x-s/2 root t,"},{"Start":"04:31.340 ","End":"04:35.150","Text":"if you\u0027re in America, make that zee instead of zed."},{"Start":"04:35.150 ","End":"04:41.305","Text":"Dz is minus 1 over 2 root t ds."},{"Start":"04:41.305 ","End":"04:44.165","Text":"We also need ds,"},{"Start":"04:44.165 ","End":"04:48.325","Text":"which is going to be minus twice root tdz."},{"Start":"04:48.325 ","End":"04:50.885","Text":"Also, we have to substitute the limits,"},{"Start":"04:50.885 ","End":"04:54.025","Text":"we\u0027ll need minus infinity 0 and infinity."},{"Start":"04:54.025 ","End":"04:55.650","Text":"When s goes to infinity,"},{"Start":"04:55.650 ","End":"04:59.900","Text":"z goes to minus infinity and vice versa, when s is 0,"},{"Start":"04:59.900 ","End":"05:06.020","Text":"that is x over 2 root t. After substituting all that,"},{"Start":"05:06.020 ","End":"05:07.815","Text":"this is what we get."},{"Start":"05:07.815 ","End":"05:10.490","Text":"We can get rid of the minus here and here if we"},{"Start":"05:10.490 ","End":"05:13.525","Text":"switch the order of the upper and lower limits."},{"Start":"05:13.525 ","End":"05:15.125","Text":"This is what we have."},{"Start":"05:15.125 ","End":"05:19.820","Text":"Next, we\u0027ll adjust this a bit to get 2 over root Pi in front of each integral."},{"Start":"05:19.820 ","End":"05:22.535","Text":"That\u0027s because we\u0027re going to use the error function,"},{"Start":"05:22.535 ","End":"05:27.065","Text":"which is defined as 2 over root Pi times this integral."},{"Start":"05:27.065 ","End":"05:30.950","Text":"If we have the integral put it from 0 to x from a to b,"},{"Start":"05:30.950 ","End":"05:35.075","Text":"we take the error function of b minus the error function of a."},{"Start":"05:35.075 ","End":"05:38.990","Text":"What we have here is error function of infinity,"},{"Start":"05:38.990 ","End":"05:45.820","Text":"meaning the error function of x as x tends to infinity minus the error function of x over"},{"Start":"05:45.820 ","End":"05:49.055","Text":"2 root t from here plus error function"},{"Start":"05:49.055 ","End":"05:53.830","Text":"here of x over 2 root t minus error function of minus infinity."},{"Start":"05:53.830 ","End":"05:57.330","Text":"This is equal to 1,"},{"Start":"05:57.330 ","End":"06:00.960","Text":"and this is equal to minus 1."},{"Start":"06:00.960 ","End":"06:05.715","Text":"Here we have a half plus 1 which is 3/2."},{"Start":"06:05.715 ","End":"06:12.390","Text":"Then minus a half plus 1 which is half error function of x over 2 root t. Replace x by 0,"},{"Start":"06:12.390 ","End":"06:15.000","Text":"which is what we want and we have that v (0,"},{"Start":"06:15.000 ","End":"06:17.710","Text":"t) is just 3/2,"},{"Start":"06:17.710 ","End":"06:21.905","Text":"because the error function of 0 is 0."},{"Start":"06:21.905 ","End":"06:25.010","Text":"We also need v(x), (0, t)."},{"Start":"06:25.010 ","End":"06:27.290","Text":"Differentiating this with respect to x,"},{"Start":"06:27.290 ","End":"06:28.855","Text":"we get the following."},{"Start":"06:28.855 ","End":"06:33.335","Text":"Using the formula that the derivative of error function is this."},{"Start":"06:33.335 ","End":"06:36.560","Text":"Then the 2 with the 2 cancels."},{"Start":"06:36.560 ","End":"06:41.910","Text":"This 2 root t comes here and the root t with root Pi combine,"},{"Start":"06:41.910 ","End":"06:43.425","Text":"and we have this."},{"Start":"06:43.425 ","End":"06:45.970","Text":"Also square this again."},{"Start":"06:45.970 ","End":"06:49.325","Text":"I don\u0027t remember why I squared that, but never mind."},{"Start":"06:49.325 ","End":"06:51.845","Text":"Plug in x=0,"},{"Start":"06:51.845 ","End":"06:54.260","Text":"we get v x(0,"},{"Start":"06:54.260 ","End":"06:57.960","Text":"t)= 1/2 times 1 over root Pi t,"},{"Start":"06:57.960 ","End":"07:00.720","Text":"then e^0, which is 1,"},{"Start":"07:00.720 ","End":"07:01.940","Text":"so this drops off."},{"Start":"07:01.940 ","End":"07:07.670","Text":"Next, you want to do some computations for w. This is the PDE"},{"Start":"07:07.670 ","End":"07:14.030","Text":"and the initial condition for w. We showed earlier that f,"},{"Start":"07:14.030 ","End":"07:17.945","Text":"this function does satisfy this condition."},{"Start":"07:17.945 ","End":"07:24.530","Text":"In that case, we can expand w as a series as follows."},{"Start":"07:24.530 ","End":"07:27.780","Text":"The actual coefficients here don\u0027t much matter."},{"Start":"07:27.780 ","End":"07:35.850","Text":"If you\u0027re interested Mn(x) from n=0 on-wards is equal to this."},{"Start":"07:35.850 ","End":"07:38.805","Text":"If we let x=0,"},{"Start":"07:38.805 ","End":"07:40.935","Text":"then all these are of order."},{"Start":"07:40.935 ","End":"07:44.085","Text":"Big O, 1/t to the 3/2,"},{"Start":"07:44.085 ","End":"07:46.725","Text":"because all the other t\u0027s only bigger than 3/2."},{"Start":"07:46.725 ","End":"07:52.490","Text":"We get this, which is good enough because we\u0027re going to let t go to infinity anyway."},{"Start":"07:52.490 ","End":"07:55.010","Text":"Then differentiating with respect to x,"},{"Start":"07:55.010 ","End":"07:56.420","Text":"derivative of this is 0,"},{"Start":"07:56.420 ","End":"07:57.965","Text":"it doesn\u0027t contain x."},{"Start":"07:57.965 ","End":"08:02.645","Text":"Then we have minus derivative of M1 for x is 0."},{"Start":"08:02.645 ","End":"08:04.145","Text":"I forgot the minus here."},{"Start":"08:04.145 ","End":"08:12.310","Text":"Okay? Then plus M2\u0027 (0) over 64 root Pi1 over t to the 5 over 2, etc."},{"Start":"08:12.310 ","End":"08:18.830","Text":"Just replace like next one will be driven 3(x) and 3\u0027(0)."},{"Start":"08:18.830 ","End":"08:23.650","Text":"Okay? You can write 1/2 to the 3/2 as 1/ t root t. All"},{"Start":"08:23.650 ","End":"08:28.300","Text":"these are big O of one over t to the 5/2."},{"Start":"08:28.300 ","End":"08:31.060","Text":"That gives us w and wx as 0,"},{"Start":"08:31.060 ","End":"08:33.460","Text":"t. Now let\u0027s summarize."},{"Start":"08:33.460 ","End":"08:36.880","Text":"These are all these four results in a table."},{"Start":"08:36.880 ","End":"08:41.680","Text":"Now let\u0027s get back to computing this limit will replace Vx,"},{"Start":"08:41.680 ","End":"08:45.730","Text":"Wx, V and W from the table here."},{"Start":"08:45.730 ","End":"08:48.870","Text":"This is what we get, color-coded it."},{"Start":"08:48.870 ","End":"08:51.490","Text":"It help you to see what\u0027s, what."},{"Start":"08:51.490 ","End":"08:56.890","Text":"Let\u0027s multiply this part of the numerator by root t. What we get,"},{"Start":"08:56.890 ","End":"08:59.080","Text":"this root t disappears,"},{"Start":"08:59.080 ","End":"09:06.645","Text":"this root t disappears and Big O of 1/t to the 5/2 becomes o of 1 over t*2."},{"Start":"09:06.645 ","End":"09:10.730","Text":"Notice that when t goes to infinity,"},{"Start":"09:10.730 ","End":"09:14.080","Text":"these go to 0 and so do these."},{"Start":"09:14.080 ","End":"09:17.080","Text":"All we\u0027re left with is the following."},{"Start":"09:17.080 ","End":"09:21.340","Text":"That simplifies to 1/3 root Pi,"},{"Start":"09:21.340 ","End":"09:24.680","Text":"and that\u0027s the answer and we\u0027re done."}],"ID":30883},{"Watched":false,"Name":"Exercise 19","Duration":"5m 7s","ChapterTopicVideoID":29328,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.925","Text":"In this exercise, we want to find the general solution of this equation,"},{"Start":"00:05.925 ","End":"00:08.760","Text":"which is the second-order linear equation."},{"Start":"00:08.760 ","End":"00:11.220","Text":"We start by classifying it."},{"Start":"00:11.220 ","End":"00:14.025","Text":"We\u0027ll label the coefficients as usual,"},{"Start":"00:14.025 ","End":"00:15.390","Text":"this is a_11,"},{"Start":"00:15.390 ","End":"00:17.490","Text":"this is twice a_12."},{"Start":"00:17.490 ","End":"00:19.140","Text":"Be careful with the 2 here."},{"Start":"00:19.140 ","End":"00:25.065","Text":"A_12 is just minus sine x plus a_22 u_yy."},{"Start":"00:25.065 ","End":"00:27.510","Text":"We don\u0027t label the rest of it,"},{"Start":"00:27.510 ","End":"00:29.355","Text":"just the second-order terms."},{"Start":"00:29.355 ","End":"00:34.740","Text":"Delta the discriminant is given by this formula and in our case,"},{"Start":"00:34.740 ","End":"00:36.480","Text":"this is what it is."},{"Start":"00:36.480 ","End":"00:41.275","Text":"It comes out to be sine squared x plus cosine squared x, which is 1."},{"Start":"00:41.275 ","End":"00:43.825","Text":"The point is that it\u0027s positive everywhere,"},{"Start":"00:43.825 ","End":"00:46.790","Text":"so this equation is everywhere hyperbolic."},{"Start":"00:46.790 ","End":"00:50.765","Text":"Usual procedure is to solve this differential equation."},{"Start":"00:50.765 ","End":"00:55.340","Text":"It\u0027s actually a pair of equations replacing a_12 and a_11,"},{"Start":"00:55.340 ","End":"00:58.175","Text":"et cetera, from what we have here."},{"Start":"00:58.175 ","End":"01:01.795","Text":"We get minus sine x plus or minus square root of 1 over 1."},{"Start":"01:01.795 ","End":"01:04.920","Text":"It\u0027s minus sine x plus or minus 1."},{"Start":"01:04.920 ","End":"01:07.025","Text":"Solve both equations together,"},{"Start":"01:07.025 ","End":"01:12.005","Text":"y is cosine x plus or minus x plus a constant."},{"Start":"01:12.005 ","End":"01:14.250","Text":"But really it should be 2 different constants,"},{"Start":"01:14.250 ","End":"01:15.890","Text":"1 for the plus and 1 for the minus,"},{"Start":"01:15.890 ","End":"01:17.315","Text":"but we get lazy here."},{"Start":"01:17.315 ","End":"01:22.760","Text":"Then c is y minus cosine x plus or minus x."},{"Start":"01:22.760 ","End":"01:26.060","Text":"Referring to the table from the tutorial,"},{"Start":"01:26.060 ","End":"01:29.240","Text":"we see we\u0027re in a hyperbolic case when we"},{"Start":"01:29.240 ","End":"01:34.535","Text":"have a function equals c_1 and the function equals c_2 as we do here."},{"Start":"01:34.535 ","End":"01:40.360","Text":"Then the substitution we make is psi equals 1 of the functions and eta is the other."},{"Start":"01:40.360 ","End":"01:44.240","Text":"We\u0027ll take psi to be with the plus,"},{"Start":"01:44.240 ","End":"01:47.650","Text":"and eta will be with the minus here."},{"Start":"01:47.650 ","End":"01:53.015","Text":"Let\u0027s just note that we expect the canonical form to be w psi eta,"},{"Start":"01:53.015 ","End":"01:56.840","Text":"possibly with lower-order terms equals 0."},{"Start":"01:56.840 ","End":"02:02.020","Text":"A quick check to see that this transformation is invertible."},{"Start":"02:02.020 ","End":"02:06.890","Text":"We take the determinant of the Jacobian matrix,"},{"Start":"02:06.890 ","End":"02:08.810","Text":"and that comes out to be 2,"},{"Start":"02:08.810 ","End":"02:11.585","Text":"which is not 0 so we\u0027re okay."},{"Start":"02:11.585 ","End":"02:16.925","Text":"Now we\u0027re going to transform u in x and y to"},{"Start":"02:16.925 ","End":"02:22.100","Text":"the function w and psi and eta we\u0027ll need the partial derivatives,"},{"Start":"02:22.100 ","End":"02:25.145","Text":"the first-order 1 we just copy from here."},{"Start":"02:25.145 ","End":"02:26.990","Text":"The second-order 1."},{"Start":"02:26.990 ","End":"02:29.930","Text":"Just a straightforward computation derivative of sine x is"},{"Start":"02:29.930 ","End":"02:34.205","Text":"cosine x and everything else gives us 0 here, here, here, here."},{"Start":"02:34.205 ","End":"02:38.900","Text":"Now, we copy the formulas from the tutorial page."},{"Start":"02:38.900 ","End":"02:40.550","Text":"They were 5 of them altogether,"},{"Start":"02:40.550 ","End":"02:42.560","Text":"but we don\u0027t need du by dx."},{"Start":"02:42.560 ","End":"02:45.709","Text":"Next, we\u0027re going to substitute all the partial derivatives"},{"Start":"02:45.709 ","End":"02:50.095","Text":"of psi and eta from this table here."},{"Start":"02:50.095 ","End":"02:53.090","Text":"I\u0027ll just write them in."},{"Start":"02:53.090 ","End":"02:56.360","Text":"You can just check, we\u0027re just copying the appropriate ones."},{"Start":"02:56.360 ","End":"02:59.360","Text":"Then after we make all these substitutions,"},{"Start":"02:59.360 ","End":"03:01.639","Text":"what we get is the following."},{"Start":"03:01.639 ","End":"03:04.145","Text":"For uy, u_xx u_xy and u_yy,"},{"Start":"03:04.145 ","End":"03:07.445","Text":"I\u0027ll leave you to check that I substituted correctly."},{"Start":"03:07.445 ","End":"03:09.605","Text":"After a bit of simplification,"},{"Start":"03:09.605 ","End":"03:11.000","Text":"this is what we get."},{"Start":"03:11.000 ","End":"03:15.740","Text":"We\u0027re going to substitute these in our original PDE."},{"Start":"03:15.740 ","End":"03:18.275","Text":"What we get, it\u0027s a bit long."},{"Start":"03:18.275 ","End":"03:22.460","Text":"Is the following, I\u0027ve color-coded it so we can collect like"},{"Start":"03:22.460 ","End":"03:28.745","Text":"terms like W psi psi will be here and here and here,"},{"Start":"03:28.745 ","End":"03:34.130","Text":"this coefficient and then w psi-eta we have here and here and here."},{"Start":"03:34.130 ","End":"03:35.825","Text":"We collect them all here."},{"Start":"03:35.825 ","End":"03:37.730","Text":"This is just routine."},{"Start":"03:37.730 ","End":"03:39.440","Text":"I\u0027ll leave you to check it."},{"Start":"03:39.440 ","End":"03:43.985","Text":"All of them cannot be 0 except for the W psi eta."},{"Start":"03:43.985 ","End":"03:47.845","Text":"For example here, this cancels with this,"},{"Start":"03:47.845 ","End":"03:54.335","Text":"and then we have 1 minus 1 sine squared minus cosine squared, which is 0."},{"Start":"03:54.335 ","End":"03:58.670","Text":"Here also, the 2 sine x cancel with the 2 sine x,"},{"Start":"03:58.670 ","End":"04:01.970","Text":"we have sine squared minus 2 sine squared is minus sine"},{"Start":"04:01.970 ","End":"04:05.570","Text":"squared minus cosine squared plus 1 again it\u0027s 0."},{"Start":"04:05.570 ","End":"04:08.615","Text":"Here is obviously 0 here it\u0027s obviously 0."},{"Start":"04:08.615 ","End":"04:10.270","Text":"Just left with this."},{"Start":"04:10.270 ","End":"04:12.668","Text":"We have minus 4 sine squared,"},{"Start":"04:12.668 ","End":"04:15.405","Text":"and minus 4 cosine squared."},{"Start":"04:15.405 ","End":"04:21.795","Text":"Cosine squared plus sine squared is 1 so we have minus 4 W psi eta is 0."},{"Start":"04:21.795 ","End":"04:26.435","Text":"W psi eta is 0 because this is a function of psi and eta,"},{"Start":"04:26.435 ","End":"04:28.340","Text":"we can solve this differential equation,"},{"Start":"04:28.340 ","End":"04:30.365","Text":"integrate with respect to eta,"},{"Start":"04:30.365 ","End":"04:32.945","Text":"and we get W psi is"},{"Start":"04:32.945 ","End":"04:37.715","Text":"an arbitrary function of psi because it\u0027s a constant as far as eta goes,"},{"Start":"04:37.715 ","End":"04:40.190","Text":"integrating with respect to psi,"},{"Start":"04:40.190 ","End":"04:45.530","Text":"we get big F of psi is any primitive of little f. Don\u0027t worry about the constant."},{"Start":"04:45.530 ","End":"04:46.970","Text":"The constant can go here or here,"},{"Start":"04:46.970 ","End":"04:50.390","Text":"it\u0027s swallowed up in one of these plus an arbitrary function of eta,"},{"Start":"04:50.390 ","End":"04:55.730","Text":"call it big G. Now we return to u of x y from"},{"Start":"04:55.730 ","End":"05:01.385","Text":"w of psi and eta by remembering that psi is this, and eta is this."},{"Start":"05:01.385 ","End":"05:04.895","Text":"Substituting them here, we get the following,"},{"Start":"05:04.895 ","End":"05:07.650","Text":"and that\u0027s the answer."}],"ID":30884},{"Watched":false,"Name":"Exercise 20","Duration":"9m 50s","ChapterTopicVideoID":29329,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.225","Text":"In this exercise, we\u0027re given the following problem."},{"Start":"00:04.225 ","End":"00:06.820","Text":"This looks like a heat equation,"},{"Start":"00:06.820 ","End":"00:09.310","Text":"but it\u0027s not because it\u0027s got this extra term."},{"Start":"00:09.310 ","End":"00:14.005","Text":"This is the initial condition at time t=0,"},{"Start":"00:14.005 ","End":"00:20.855","Text":"and these are the boundary conditions for the ends of the interval 0,1."},{"Start":"00:20.855 ","End":"00:24.325","Text":"What we have to do is to compute the following limit"},{"Start":"00:24.325 ","End":"00:28.465","Text":"without actually computing you explicitly."},{"Start":"00:28.465 ","End":"00:31.330","Text":"The first step is to try and reduce this to"},{"Start":"00:31.330 ","End":"00:35.860","Text":"a heat equation using one of the standard tricks or transformations,"},{"Start":"00:35.860 ","End":"00:41.530","Text":"letting u equal v times an exponent e to the Alpha x plus Beta"},{"Start":"00:41.530 ","End":"00:44.660","Text":"T. Then later we assign Alpha and"},{"Start":"00:44.660 ","End":"00:48.030","Text":"Beta according to whatever is most convenient, or whatever works."},{"Start":"00:48.030 ","End":"00:51.080","Text":"Of course, this is invertible because we can get v back"},{"Start":"00:51.080 ","End":"00:54.740","Text":"from u by taking e to the power of the negative."},{"Start":"00:54.740 ","End":"00:57.065","Text":"I will need u_t, u_x,"},{"Start":"00:57.065 ","End":"01:03.365","Text":"and u_xx because we want to substitute here in the PDE,"},{"Start":"01:03.365 ","End":"01:08.705","Text":"let\u0027s see, du by dt is the derivative of this product rule."},{"Start":"01:08.705 ","End":"01:10.940","Text":"Well, we\u0027ve done this before and it\u0027s also straightforward."},{"Start":"01:10.940 ","End":"01:15.050","Text":"I\u0027ll just give you the answers here, you can check them."},{"Start":"01:15.050 ","End":"01:19.190","Text":"Now we can substitute these in this PDE."},{"Start":"01:19.190 ","End":"01:21.345","Text":"This is u_t,"},{"Start":"01:21.345 ","End":"01:24.930","Text":"next 2 terms are du by dx,"},{"Start":"01:24.930 ","End":"01:30.360","Text":"and the last term is du by dx by dx,"},{"Start":"01:30.360 ","End":"01:32.945","Text":"u_xx I\u0027d like to call it for short."},{"Start":"01:32.945 ","End":"01:37.340","Text":"Now we can divide all the terms containing e to the Alpha x with Beta t,"},{"Start":"01:37.340 ","End":"01:39.140","Text":"so we can throw those out."},{"Start":"01:39.140 ","End":"01:41.720","Text":"Then we can collect like terms."},{"Start":"01:41.720 ","End":"01:43.550","Text":"First step, I just mark them."},{"Start":"01:43.550 ","End":"01:47.120","Text":"All the ones with v is here and here,"},{"Start":"01:47.120 ","End":"01:50.105","Text":"and here is 2 terms with v_x,"},{"Start":"01:50.105 ","End":"01:52.405","Text":"and then everything is on its own."},{"Start":"01:52.405 ","End":"01:57.800","Text":"If we collect these together and just keeping v_t on the left,"},{"Start":"01:57.800 ","End":"02:01.430","Text":"we get v_xx and then 2 Alpha plus 1,"},{"Start":"02:01.430 ","End":"02:03.665","Text":"which is minus 1 on the other side."},{"Start":"02:03.665 ","End":"02:07.864","Text":"Then here Alpha squared and then minus"},{"Start":"02:07.864 ","End":"02:13.670","Text":"Alpha minus Beta for v. We want this to be the heat equation,"},{"Start":"02:13.670 ","End":"02:15.125","Text":"just v_t equals v_xx,"},{"Start":"02:15.125 ","End":"02:17.960","Text":"and we can get that if we find"},{"Start":"02:17.960 ","End":"02:22.040","Text":"Alpha and Beta such that this coefficient is 0 and this is 0."},{"Start":"02:22.040 ","End":"02:24.530","Text":"If we do that, make them both 0,"},{"Start":"02:24.530 ","End":"02:31.580","Text":"then easy to see that the answer is Alpha equals 1/2 and Beta equals minus 1/4."},{"Start":"02:31.580 ","End":"02:33.950","Text":"You can substitute and check."},{"Start":"02:33.950 ","End":"02:36.785","Text":"This gives us that v_t equals v_xx,"},{"Start":"02:36.785 ","End":"02:38.120","Text":"the heat equation,"},{"Start":"02:38.120 ","End":"02:42.710","Text":"and also because we had v in terms of u as"},{"Start":"02:42.710 ","End":"02:47.510","Text":"multiplied by e to the minus Alpha x minus Beta t. Now that we have Alpha and Beta,"},{"Start":"02:47.510 ","End":"02:50.120","Text":"we can get that v equals u times e to"},{"Start":"02:50.120 ","End":"02:56.495","Text":"the minus 1/2 x plus 1/4t using the Alpha and Beta here."},{"Start":"02:56.495 ","End":"02:58.970","Text":"Of course, if we need it the other way round,"},{"Start":"02:58.970 ","End":"03:01.325","Text":"we just bring the exponent to the other side,"},{"Start":"03:01.325 ","End":"03:03.109","Text":"and change the sign."},{"Start":"03:03.109 ","End":"03:05.120","Text":"We have the PDE for v,"},{"Start":"03:05.120 ","End":"03:08.435","Text":"we still need the boundary and initial conditions."},{"Start":"03:08.435 ","End":"03:11.375","Text":"The initial condition for you is this."},{"Start":"03:11.375 ","End":"03:16.785","Text":"The initial condition for v using this is"},{"Start":"03:16.785 ","End":"03:23.625","Text":"the same e^1/2x multiplied by this e^-1/2x plus 1/4t but t is 0,"},{"Start":"03:23.625 ","End":"03:25.935","Text":"so it\u0027s just e^1/2 x,"},{"Start":"03:25.935 ","End":"03:28.575","Text":"e^-1/2x, and this cancels and it\u0027s just 1."},{"Start":"03:28.575 ","End":"03:32.375","Text":"Next to the boundary conditions when x is 0 and x is 1."},{"Start":"03:32.375 ","End":"03:34.265","Text":"For u they\u0027re both 0."},{"Start":"03:34.265 ","End":"03:37.790","Text":"For v, we just take the 0 and multiply it."},{"Start":"03:37.790 ","End":"03:41.105","Text":"Well, it doesn\u0027t matter, we\u0027re multiplying 0, so it\u0027s 0."},{"Start":"03:41.105 ","End":"03:43.085","Text":"We have the problem for v now."},{"Start":"03:43.085 ","End":"03:45.865","Text":"This is the PDE,"},{"Start":"03:45.865 ","End":"03:52.850","Text":"and we have the initial condition v of x naught equals 1,"},{"Start":"03:52.850 ","End":"03:59.615","Text":"and we have the boundary conditions that these are both equal to 0."},{"Start":"03:59.615 ","End":"04:04.190","Text":"Collecting these together, this is the problem that we have for v,"},{"Start":"04:04.190 ","End":"04:10.630","Text":"a heat equation on a finite interval with initial and boundary conditions."},{"Start":"04:10.630 ","End":"04:14.660","Text":"What we use here is the Poisson formula,"},{"Start":"04:14.660 ","End":"04:16.880","Text":"which in general is this but in our case,"},{"Start":"04:16.880 ","End":"04:18.515","Text":"we know that L is 1,"},{"Start":"04:18.515 ","End":"04:20.240","Text":"and that a is 1,"},{"Start":"04:20.240 ","End":"04:23.375","Text":"so what we get is this."},{"Start":"04:23.375 ","End":"04:26.075","Text":"If we let t=0,"},{"Start":"04:26.075 ","End":"04:28.625","Text":"then on the one hand, we know that v(x,"},{"Start":"04:28.625 ","End":"04:31.760","Text":"0) is 1, and on the other hand,"},{"Start":"04:31.760 ","End":"04:37.145","Text":"we can substitute t equals 0 here so this whole exponent disappears."},{"Start":"04:37.145 ","End":"04:38.825","Text":"We get the following,"},{"Start":"04:38.825 ","End":"04:43.045","Text":"and we have 1 expressed as a sine series,"},{"Start":"04:43.045 ","End":"04:47.735","Text":"and we know the formula for coefficients of the Fourier sine series."},{"Start":"04:47.735 ","End":"04:49.670","Text":"In our case, it\u0027s this,"},{"Start":"04:49.670 ","End":"04:51.830","Text":"but in general, we start with this,"},{"Start":"04:51.830 ","End":"04:54.605","Text":"and then L is equal to 1,"},{"Start":"04:54.605 ","End":"04:58.435","Text":"and the function here is just the function 1."},{"Start":"04:58.435 ","End":"05:02.210","Text":"We have a formula for the coefficients b_n,"},{"Start":"05:02.210 ","End":"05:05.750","Text":"but won\u0027t compute them now. We compute them later."},{"Start":"05:05.750 ","End":"05:07.400","Text":"We may not need them,"},{"Start":"05:07.400 ","End":"05:09.545","Text":"I\u0027ll tell you what will happen."},{"Start":"05:09.545 ","End":"05:11.510","Text":"We\u0027ll need b1 and not the others,"},{"Start":"05:11.510 ","End":"05:14.525","Text":"but okay, let\u0027s just keep this formula for later."},{"Start":"05:14.525 ","End":"05:22.445","Text":"Let\u0027s return to this formula for v and we\u0027ll just convert it to u"},{"Start":"05:22.445 ","End":"05:31.100","Text":"by multiplying by e^1/2x minus 1/4 t. What we get is this adjusted."},{"Start":"05:31.100 ","End":"05:40.300","Text":"We put e^1/2x in front and inside the exponent we can put another minus 1/4."},{"Start":"05:40.300 ","End":"05:46.615","Text":"The original question related to u_t so let\u0027s differentiate this term by term,"},{"Start":"05:46.615 ","End":"05:50.670","Text":"and we get in front of each term, this exponent."},{"Start":"05:50.670 ","End":"05:53.090","Text":"Now, we\u0027ll break this sum up into 2."},{"Start":"05:53.090 ","End":"05:56.195","Text":"We\u0027ll take the first term for n equals 1,"},{"Start":"05:56.195 ","End":"06:01.180","Text":"then we\u0027ll take the tail of the series from 2 to infinity."},{"Start":"06:01.180 ","End":"06:03.090","Text":"When n is 1,"},{"Start":"06:03.090 ","End":"06:06.540","Text":"we get up to here this."},{"Start":"06:06.540 ","End":"06:09.020","Text":"I\u0027m going to start marking the t in red because we\u0027re"},{"Start":"06:09.020 ","End":"06:11.840","Text":"following what happens when t goes to infinity,"},{"Start":"06:11.840 ","End":"06:14.045","Text":"so t is the most important here."},{"Start":"06:14.045 ","End":"06:18.695","Text":"Then from 2 to infinity of the same thing."},{"Start":"06:18.695 ","End":"06:23.974","Text":"Notice that as n increases from 2 onwards,"},{"Start":"06:23.974 ","End":"06:30.895","Text":"this inside the brackets increases and is bigger than a 1/4 plus 4 pi squared."},{"Start":"06:30.895 ","End":"06:34.950","Text":"This minus is decreasing and the exponent is decreasing,"},{"Start":"06:34.950 ","End":"06:39.730","Text":"and so it\u0027s less than or equal to what we would get if we just put 2 here."},{"Start":"06:39.730 ","End":"06:48.210","Text":"What we can say is that this is O of e to the minus 1/4 plus 4 pi squared,"},{"Start":"06:48.210 ","End":"06:49.980","Text":"put 4 instead of the n squared."},{"Start":"06:49.980 ","End":"06:53.480","Text":"All the rest of this is not related to"},{"Start":"06:53.480 ","End":"06:57.980","Text":"t because we are assuming that when we take the limit as t goes to infinity,"},{"Start":"06:57.980 ","End":"07:01.535","Text":"that we can put the limit inside the summation."},{"Start":"07:01.535 ","End":"07:04.180","Text":"Let\u0027s assume that that\u0027s correct."},{"Start":"07:04.180 ","End":"07:07.100","Text":"Now that we have an expression for u_t,"},{"Start":"07:07.100 ","End":"07:08.915","Text":"we want to compute this."},{"Start":"07:08.915 ","End":"07:11.570","Text":"This was the given expression that we had to"},{"Start":"07:11.570 ","End":"07:14.900","Text":"figure out limit as t goes to infinity of all this."},{"Start":"07:14.900 ","End":"07:18.360","Text":"U_t is this,"},{"Start":"07:18.360 ","End":"07:22.320","Text":"and here we have u_t(x,"},{"Start":"07:22.320 ","End":"07:31.380","Text":"t), and this is equal to this together with the minus."},{"Start":"07:31.380 ","End":"07:37.340","Text":"I also copied again what the expression was that we had to evaluate."},{"Start":"07:37.340 ","End":"07:43.490","Text":"Now so far, we\u0027ve just got the first term up to the plus O of."},{"Start":"07:43.490 ","End":"07:46.905","Text":"We still need that part and that\u0027s here."},{"Start":"07:46.905 ","End":"07:51.454","Text":"Now if we look at the 2 expressions colored in green,"},{"Start":"07:51.454 ","End":"07:53.690","Text":"we multiply them together,"},{"Start":"07:53.690 ","End":"08:01.820","Text":"then this becomes limit of e to the minus 3 Pi squared t, which is 0."},{"Start":"08:01.820 ","End":"08:06.109","Text":"The limit as t goes to infinity of this is 0."},{"Start":"08:06.109 ","End":"08:09.080","Text":"As for this, most of the stuff cancels."},{"Start":"08:09.080 ","End":"08:14.030","Text":"Look, e to the 1/4 plus Pi squared t cancels with e to the minus 1/4 plus Pi"},{"Start":"08:14.030 ","End":"08:20.545","Text":"squared t and e^1/2x with e^1/2x cancels."},{"Start":"08:20.545 ","End":"08:26.330","Text":"All that we\u0027re left with is the minus sign and the 1/4 plus Pi squared,"},{"Start":"08:26.330 ","End":"08:31.800","Text":"and b_1 an integral of sine Pi X dx."},{"Start":"08:31.800 ","End":"08:35.465","Text":"Here we are again. Now we\u0027re going to compute this integral."},{"Start":"08:35.465 ","End":"08:40.205","Text":"Integral of minus sine is cosine,"},{"Start":"08:40.205 ","End":"08:44.170","Text":"cosine Pi X over Pi from 0 to 1."},{"Start":"08:44.170 ","End":"08:46.455","Text":"When x is 1,"},{"Start":"08:46.455 ","End":"08:48.825","Text":"we get minus 1."},{"Start":"08:48.825 ","End":"08:51.180","Text":"When x is 0, we get 1."},{"Start":"08:51.180 ","End":"08:53.880","Text":"We get minus 1 minus 1,"},{"Start":"08:53.880 ","End":"08:57.170","Text":"which is minus 2 over Pi,"},{"Start":"08:57.170 ","End":"08:59.210","Text":"a 1/4 plus Pi squared b_1."},{"Start":"08:59.210 ","End":"09:07.595","Text":"But we\u0027ll need for later again that this integral is equal to 2 over Pi."},{"Start":"09:07.595 ","End":"09:09.990","Text":"What we\u0027re still missing is b_1."},{"Start":"09:09.990 ","End":"09:13.400","Text":"Like I mentioned earlier we have a general formula for b_n,"},{"Start":"09:13.400 ","End":"09:14.870","Text":"and we didn\u0027t need to compute it then,"},{"Start":"09:14.870 ","End":"09:18.040","Text":"it turns out we only need b_1 from this."},{"Start":"09:18.040 ","End":"09:20.505","Text":"So b_1 is twice this integral,"},{"Start":"09:20.505 ","End":"09:24.840","Text":"but we have this integral from here it\u0027s 2 over Pi."},{"Start":"09:24.840 ","End":"09:28.170","Text":"We get twice 2 over Pi which is 4 over Pi."},{"Start":"09:28.170 ","End":"09:31.770","Text":"If we put b_1 here as 4 over Pi,"},{"Start":"09:31.770 ","End":"09:38.580","Text":"then we get that this limit is minus 2 over Pi 1/4 plus Pi squared 4 over Pi."},{"Start":"09:38.580 ","End":"09:42.420","Text":"Just combine the 2 over Pi with the 4 over Pi,"},{"Start":"09:42.420 ","End":"09:45.605","Text":"and what we get is minus 8 over Pi squared,"},{"Start":"09:45.605 ","End":"09:47.500","Text":"a quarter plus Pi squared."},{"Start":"09:47.500 ","End":"09:48.745","Text":"That\u0027s the answer,"},{"Start":"09:48.745 ","End":"09:51.050","Text":"and we are done."}],"ID":30885},{"Watched":false,"Name":"Exercise 21","Duration":"5m 32s","ChapterTopicVideoID":29330,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.275","Text":"In this exercise, we have the following equation."},{"Start":"00:04.275 ","End":"00:08.385","Text":"It\u0027s a quasi linear equation of the first-order,"},{"Start":"00:08.385 ","End":"00:14.850","Text":"and we have this condition and that corresponds to the shaded area"},{"Start":"00:14.850 ","End":"00:22.035","Text":"in the picture are also given this initial condition on x,"},{"Start":"00:22.035 ","End":"00:25.095","Text":"x^2, where x is less than or equal to 0,"},{"Start":"00:25.095 ","End":"00:29.775","Text":"which corresponds to this curve here, half a Parabola."},{"Start":"00:29.775 ","End":"00:31.815","Text":"The picture is not important."},{"Start":"00:31.815 ","End":"00:35.145","Text":"It\u0027s quasi-linear, which in general is like this"},{"Start":"00:35.145 ","End":"00:39.240","Text":"with A and B and C functions of x, y, u."},{"Start":"00:39.240 ","End":"00:42.330","Text":"Here, A is the constant 1, B is the constant,"},{"Start":"00:42.330 ","End":"00:45.495","Text":"2 and 3 is the function u,"},{"Start":"00:45.495 ","End":"00:50.345","Text":"and the first step is to find a parameterization of the initial curve,"},{"Start":"00:50.345 ","End":"00:52.275","Text":"which is big Gamma."},{"Start":"00:52.275 ","End":"00:59.540","Text":"Little Gamma is the 2D curve and big Gamma is the 3D curve where we"},{"Start":"00:59.540 ","End":"01:07.285","Text":"assign u to be u of x and y and the initial curve means at time 0 in 2D,"},{"Start":"01:07.285 ","End":"01:09.865","Text":"this can be parametrized as tau,"},{"Start":"01:09.865 ","End":"01:13.820","Text":"tau^2, but we\u0027re given the value of u on this curve,"},{"Start":"01:13.820 ","End":"01:17.615","Text":"which is this so we\u0027ve got third parameter u,"},{"Start":"01:17.615 ","End":"01:20.495","Text":"which is t plus sine of tau^3."},{"Start":"01:20.495 ","End":"01:24.475","Text":"Now the second step is to find the characteristic curves."},{"Start":"01:24.475 ","End":"01:33.185","Text":"In general, it\u0027s X_t equals A and d y by d t equals B and there\u0027s a third 1,"},{"Start":"01:33.185 ","End":"01:34.780","Text":"d u by d t equals c,"},{"Start":"01:34.780 ","End":"01:36.400","Text":"but we use that in the next step,"},{"Start":"01:36.400 ","End":"01:38.785","Text":"integrating with respect to t,"},{"Start":"01:38.785 ","End":"01:42.780","Text":"we get that x is 1 t plus"},{"Start":"01:42.780 ","End":"01:48.705","Text":"some function C_1 of tau and y is 2 t plus another function of tau."},{"Start":"01:48.705 ","End":"01:51.775","Text":"Now if we let t equals 0,"},{"Start":"01:51.775 ","End":"01:57.550","Text":"we get that x of 0 tau is C_1 of t and y of"},{"Start":"01:57.550 ","End":"02:04.450","Text":"0 tau equals C_2 of t but we also have x of 0 tau,"},{"Start":"02:04.450 ","End":"02:08.900","Text":"y of 0 tau from these 2 and so combining,"},{"Start":"02:08.900 ","End":"02:12.380","Text":"we get that x of t tau,"},{"Start":"02:12.380 ","End":"02:15.160","Text":"which is t plus C_1 of tau,"},{"Start":"02:15.160 ","End":"02:24.405","Text":"is equal to t plus tau and the other 1 get 2 t from here and tau^2 from here is this."},{"Start":"02:24.405 ","End":"02:26.175","Text":"Now step 3."},{"Start":"02:26.175 ","End":"02:27.890","Text":"The third 1 of these,"},{"Start":"02:27.890 ","End":"02:32.255","Text":"we have u t is equal to big C,"},{"Start":"02:32.255 ","End":"02:37.545","Text":"which is u so d u by d t equals u"},{"Start":"02:37.545 ","End":"02:43.615","Text":"and we also have time 0 that u of 0 tau is tau plus sine t^3."},{"Start":"02:43.615 ","End":"02:46.625","Text":"This is a differential equation in u,"},{"Start":"02:46.625 ","End":"02:48.305","Text":"and the solution to this,"},{"Start":"02:48.305 ","End":"02:54.455","Text":"is u equals e to the t times a constant c e^t."},{"Start":"02:54.455 ","End":"02:56.510","Text":"If we let t equals 0,"},{"Start":"02:56.510 ","End":"03:04.145","Text":"we get that u of 0 tau is C. This C is u of 0 tau so we get u of t tau"},{"Start":"03:04.145 ","End":"03:12.225","Text":"is e^t here and C which is u of 0 tau is tau plus sine of t^3."},{"Start":"03:12.225 ","End":"03:15.340","Text":"Now we have u in terms of t and tau."},{"Start":"03:15.340 ","End":"03:19.495","Text":"Now it\u0027s time to get back to x and y"},{"Start":"03:19.495 ","End":"03:25.040","Text":"so we have that x and y in terms of t and tau are as follows."},{"Start":"03:25.040 ","End":"03:29.570","Text":"We would like to reverse this to get t and tau in terms of x and y."},{"Start":"03:29.570 ","End":"03:33.665","Text":"Let\u0027s first of all make sure that it\u0027s possible."},{"Start":"03:33.665 ","End":"03:38.630","Text":"We\u0027ll do this by checking the Jacobian so the 4 partial derivatives,"},{"Start":"03:38.630 ","End":"03:45.135","Text":"x_t x_tau, y_t y_tau R 1,1,2 and 2 tau."},{"Start":"03:45.135 ","End":"03:48.360","Text":"The product is 2 tau minus 2,"},{"Start":"03:48.360 ","End":"03:53.710","Text":"which gives tau equals 1 but there is no solution in tau less than or equal to 0,"},{"Start":"03:53.710 ","End":"03:56.570","Text":"so it\u0027s always non-zero in our domain."},{"Start":"03:56.570 ","End":"03:58.630","Text":"Now, I can say I want to solve this,"},{"Start":"03:58.630 ","End":"04:00.305","Text":"and now that we know it\u0027s possible,"},{"Start":"04:00.305 ","End":"04:03.020","Text":"if we take twice this minus this,"},{"Start":"04:03.020 ","End":"04:10.555","Text":"we get rid of the t and we\u0027ve got y minus 2 x is equal to tau^2 minus 2 tau."},{"Start":"04:10.555 ","End":"04:14.345","Text":"Then we can get a quadratic equation in tau,"},{"Start":"04:14.345 ","End":"04:17.090","Text":"which has 2 solutions."},{"Start":"04:17.090 ","End":"04:20.855","Text":"Tau equals, well the following with a plus or minus,"},{"Start":"04:20.855 ","End":"04:25.580","Text":"and we can divide top and bottom by 2 inside the square root sign,"},{"Start":"04:25.580 ","End":"04:27.220","Text":"we have to divide by 4,"},{"Start":"04:27.220 ","End":"04:30.365","Text":"so we have the following for tau."},{"Start":"04:30.365 ","End":"04:33.260","Text":"Yeah, we don\u0027t have to take the plus,"},{"Start":"04:33.260 ","End":"04:39.365","Text":"just the minus because we know that tau is less than or equal to 0."},{"Start":"04:39.365 ","End":"04:43.265","Text":"Well, strictly speaking, in order for tau to be negative,"},{"Start":"04:43.265 ","End":"04:46.640","Text":"what\u0027s under the square root needs to be bigger than 1 so,"},{"Start":"04:46.640 ","End":"04:50.350","Text":"we need to restrict the domain to y bigger than 2 x."},{"Start":"04:50.350 ","End":"04:52.505","Text":"Anyway, if not worry about that."},{"Start":"04:52.505 ","End":"04:57.589","Text":"Summarizing so far, we have tau is this,"},{"Start":"04:57.589 ","End":"05:01.505","Text":"and we still have to get t in terms of x and y,"},{"Start":"05:01.505 ","End":"05:07.565","Text":"so 1 thing we could do would be to subtract x minus tau,"},{"Start":"05:07.565 ","End":"05:09.115","Text":"that would give us t,"},{"Start":"05:09.115 ","End":"05:11.045","Text":"so t is x minus tau,"},{"Start":"05:11.045 ","End":"05:17.810","Text":"and then tau from here is equal to this and we already found u in terms of t and tau,"},{"Start":"05:17.810 ","End":"05:24.005","Text":"which is this, so you just have to substitute t and tau from here and we\u0027ll get this."},{"Start":"05:24.005 ","End":"05:28.160","Text":"It looks a bit messy, but that\u0027s it and that\u0027s the answers."},{"Start":"05:28.160 ","End":"05:29.885","Text":"Just highlight it,"},{"Start":"05:29.885 ","End":"05:32.700","Text":"and we are done."}],"ID":30886},{"Watched":false,"Name":"Exercise 22","Duration":"5m 6s","ChapterTopicVideoID":29331,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.415","Text":"In this exercise, we have the following initial and boundary value problem."},{"Start":"00:05.415 ","End":"00:09.180","Text":"It\u0027s a non-homogeneous heat equation on"},{"Start":"00:09.180 ","End":"00:15.455","Text":"a finite interval with Neumann boundary conditions and this is the initial condition."},{"Start":"00:15.455 ","End":"00:20.940","Text":"Our task is to solve it and we\u0027re given a hint to consider the function v,"},{"Start":"00:20.940 ","End":"00:23.550","Text":"which is u minus x."},{"Start":"00:23.550 ","End":"00:27.270","Text":"When I get the problem for v, first of all,"},{"Start":"00:27.270 ","End":"00:31.275","Text":"the PDE starting from this that v is u minus x."},{"Start":"00:31.275 ","End":"00:33.795","Text":"If we differentiate with respect to t,"},{"Start":"00:33.795 ","End":"00:35.800","Text":"the minus x disappears."},{"Start":"00:35.800 ","End":"00:40.700","Text":"Similarly, if we differentiate twice with respect to x, this also disappears."},{"Start":"00:40.700 ","End":"00:43.830","Text":"We get minus x minus 1, then 0."},{"Start":"00:43.830 ","End":"00:49.085","Text":"That means that we actually have the same PDE for v as for u."},{"Start":"00:49.085 ","End":"00:53.675","Text":"Now we want to convert the initial condition and the boundary conditions."},{"Start":"00:53.675 ","End":"00:55.980","Text":"U(x, 0) equals x."},{"Start":"00:55.980 ","End":"00:58.680","Text":"Since v is u minus x,"},{"Start":"00:58.680 ","End":"01:01.485","Text":"then v(x, 0) is 0."},{"Start":"01:01.485 ","End":"01:05.370","Text":"Now the boundary u_x(0,"},{"Start":"01:05.370 ","End":"01:07.875","Text":"t) and u_x(1, t) are equal to 1."},{"Start":"01:07.875 ","End":"01:13.325","Text":"Vx, we just have to subtract that have minus x."},{"Start":"01:13.325 ","End":"01:15.320","Text":"We subtract the derivative with respect to x,"},{"Start":"01:15.320 ","End":"01:16.460","Text":"which is minus 1,"},{"Start":"01:16.460 ","End":"01:19.160","Text":"so the 1 becomes 0."},{"Start":"01:19.160 ","End":"01:21.170","Text":"Now collecting all this,"},{"Start":"01:21.170 ","End":"01:27.661","Text":"this is our problem in v which has homogeneous initial and boundary conditions,"},{"Start":"01:27.661 ","End":"01:30.755","Text":"but the PDE itself is the same as before,"},{"Start":"01:30.755 ","End":"01:33.455","Text":"and it\u0027s non-homogeneous and we have this extra bit."},{"Start":"01:33.455 ","End":"01:36.325","Text":"For the non-homogeneous on a finite interval,"},{"Start":"01:36.325 ","End":"01:41.720","Text":"the technique we learned was to use Duhamel\u0027s principle and v"},{"Start":"01:41.720 ","End":"01:47.315","Text":"is the integral of p. P is actually a function of three variables,"},{"Start":"01:47.315 ","End":"01:51.445","Text":"where p is a solution to the following problem."},{"Start":"01:51.445 ","End":"01:55.340","Text":"Note that the non-homogeneous part of"},{"Start":"01:55.340 ","End":"02:02.225","Text":"the PDE becomes non-homogeneous part of the initial condition."},{"Start":"02:02.225 ","End":"02:05.960","Text":"We really just need two variables and not all three."},{"Start":"02:05.960 ","End":"02:10.685","Text":"What we do is we consider Tau as being some fixed parameter,"},{"Start":"02:10.685 ","End":"02:15.605","Text":"and we substitute T equal t minus Tau."},{"Start":"02:15.605 ","End":"02:18.140","Text":"Then we get a different function, P,"},{"Start":"02:18.140 ","End":"02:21.125","Text":"such that P of x,"},{"Start":"02:21.125 ","End":"02:24.640","Text":"T is the following p( x,"},{"Start":"02:24.640 ","End":"02:25.905","Text":"T plus Tau Tau)."},{"Start":"02:25.905 ","End":"02:28.740","Text":"There\u0027s also reverse formula."},{"Start":"02:28.740 ","End":"02:33.870","Text":"We can get p back from P by adding Tau"},{"Start":"02:33.870 ","End":"02:39.165","Text":"to this to get T and just putting Tau here where Tau is our fixed parameter."},{"Start":"02:39.165 ","End":"02:43.385","Text":"We get the following problem for P as follows."},{"Start":"02:43.385 ","End":"02:49.775","Text":"Still homogeneous, same boundary conditions and the initial condition is cosine Pi x."},{"Start":"02:49.775 ","End":"02:51.650","Text":"We know how to solve this."},{"Start":"02:51.650 ","End":"02:54.875","Text":"We\u0027re going to use Poisson\u0027s formula."},{"Start":"02:54.875 ","End":"02:58.690","Text":"Just remember this, we\u0027ll need it in a moment."},{"Start":"02:58.690 ","End":"03:01.805","Text":"Here\u0027s what we get using the formula."},{"Start":"03:01.805 ","End":"03:04.520","Text":"In our case, a is 1 and L is 1,"},{"Start":"03:04.520 ","End":"03:07.385","Text":"so it simplifies somewhat to this."},{"Start":"03:07.385 ","End":"03:10.825","Text":"Now we\u0027re going to let T equals 0."},{"Start":"03:10.825 ","End":"03:16.595","Text":"Remember I told you to remember that cosine Pi x was p of x and 0."},{"Start":"03:16.595 ","End":"03:20.495","Text":"On the other hand, letting t equals 0 in the series gives us this."},{"Start":"03:20.495 ","End":"03:25.340","Text":"All the e to the power of part drops out when t is 0."},{"Start":"03:25.340 ","End":"03:31.195","Text":"Now we have a Fourier cosine series and we can compare coefficients,"},{"Start":"03:31.195 ","End":"03:35.570","Text":"and we see that the only non-zero coefficient is where n equals 1,"},{"Start":"03:35.570 ","End":"03:39.870","Text":"because here we have 1 cosine 1Pi x."},{"Start":"03:40.180 ","End":"03:46.550","Text":"Then if we put that in here we only get the one term where n equals 1."},{"Start":"03:46.550 ","End":"03:47.990","Text":"We don\u0027t need the series,"},{"Start":"03:47.990 ","End":"03:50.195","Text":"so we just get a_1,"},{"Start":"03:50.195 ","End":"03:51.440","Text":"which is 1,"},{"Start":"03:51.440 ","End":"03:57.265","Text":"e to the minus 1Pi squared times T and cosine 1Pi x."},{"Start":"03:57.265 ","End":"04:01.080","Text":"Then going back from P to p,"},{"Start":"04:01.080 ","End":"04:02.720","Text":"this is what we have."},{"Start":"04:02.720 ","End":"04:04.055","Text":"Well, this is in general,"},{"Start":"04:04.055 ","End":"04:07.325","Text":"and in our case, it\u0027s equal to the following."},{"Start":"04:07.325 ","End":"04:11.095","Text":"Just replace T by t minus Tau."},{"Start":"04:11.095 ","End":"04:16.905","Text":"Now we get back from p to v using this integral."},{"Start":"04:16.905 ","End":"04:24.170","Text":"Substitute p is equal to just copying it from here."},{"Start":"04:24.170 ","End":"04:31.085","Text":"All the stuff without Tau we can bring up front until we just have this integral,"},{"Start":"04:31.085 ","End":"04:36.500","Text":"that\u0027s straightforward, that\u0027s e to the Pi squared Tau over Pi squared."},{"Start":"04:36.500 ","End":"04:42.395","Text":"We want the limits of integration 0 to t. That gives us e to the Pi squared,"},{"Start":"04:42.395 ","End":"04:46.785","Text":"t minus 1 over Pi squared and the rest is the same."},{"Start":"04:46.785 ","End":"04:50.760","Text":"Let\u0027s bring the 1 over Pi squared in front. It looks a bit better."},{"Start":"04:50.760 ","End":"04:54.005","Text":"Now we\u0027re almost there because this is a solution for v(x,"},{"Start":"04:54.005 ","End":"04:56.660","Text":"t) but we want u(x, t)."},{"Start":"04:56.660 ","End":"04:58.685","Text":"Now v is u minus x,"},{"Start":"04:58.685 ","End":"05:01.835","Text":"so u is v plus x."},{"Start":"05:01.835 ","End":"05:04.903","Text":"This is the solution,"},{"Start":"05:04.903 ","End":"05:07.320","Text":"and we are done."}],"ID":30887},{"Watched":false,"Name":"Exercise 23","Duration":"5m 37s","ChapterTopicVideoID":29332,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.090","Text":"In this exercise, we\u0027re given that u of"},{"Start":"00:03.090 ","End":"00:07.680","Text":"r Theta is the solution to the following boundary value problem."},{"Start":"00:07.680 ","End":"00:10.170","Text":"It\u0027s given in polar coordinates."},{"Start":"00:10.170 ","End":"00:15.645","Text":"The PDE is like a non-homogeneous Laplace equation,"},{"Start":"00:15.645 ","End":"00:18.495","Text":"which is actually a Poisson equation."},{"Start":"00:18.495 ","End":"00:24.120","Text":"This is the boundary condition on the unit circle where r equals 1,"},{"Start":"00:24.120 ","End":"00:28.290","Text":"and our task is to find the values of c for which"},{"Start":"00:28.290 ","End":"00:33.240","Text":"the limit as r goes to 0 of u of r Theta over r squared is 1/4."},{"Start":"00:33.240 ","End":"00:38.270","Text":"It could be 1 or more or maybe no solutions. Well, we\u0027ll see."},{"Start":"00:38.270 ","End":"00:45.900","Text":"Now, usual technique for a Poisson equation is to split it into 2,"},{"Start":"00:45.900 ","End":"00:48.980","Text":"a homogeneous one which is a Laplace equation and"},{"Start":"00:48.980 ","End":"00:53.674","Text":"a non-homogeneous one with the homogeneous boundary condition."},{"Start":"00:53.674 ","End":"01:00.650","Text":"So we let u equals v plus w and v satisfies the homogeneous equation,"},{"Start":"01:00.650 ","End":"01:04.430","Text":"non-homogeneous boundary condition and w vice versa,"},{"Start":"01:04.430 ","End":"01:08.405","Text":"non-homogeneous PDE but homogeneous boundary condition."},{"Start":"01:08.405 ","End":"01:10.550","Text":"Let\u0027s go for w first."},{"Start":"01:10.550 ","End":"01:15.050","Text":"It always works that you can find a radial solution if you can find one at all."},{"Start":"01:15.050 ","End":"01:17.645","Text":"We let w equal w(r),"},{"Start":"01:17.645 ","End":"01:19.430","Text":"which doesn\u0027t involve Theta."},{"Start":"01:19.430 ","End":"01:25.200","Text":"There\u0027s a formula for the Laplacian in polar coordinates, and this is it."},{"Start":"01:25.200 ","End":"01:30.375","Text":"This equals 1 and w(1) is 0,"},{"Start":"01:30.375 ","End":"01:32.705","Text":"and because w is a function just of r,"},{"Start":"01:32.705 ","End":"01:37.879","Text":"w Theta Theta will be 0 and w_r is"},{"Start":"01:37.879 ","End":"01:43.640","Text":"just a derivative of w with respect to r. So we can write this equation like so."},{"Start":"01:43.640 ","End":"01:47.570","Text":"Now multiply both sides by r and we have this,"},{"Start":"01:47.570 ","End":"01:53.885","Text":"the left hand side is an exact derivative of a product rw\u0027."},{"Start":"01:53.885 ","End":"01:58.760","Text":"We can take the integral of both sides, the indefinite,"},{"Start":"01:58.760 ","End":"02:05.210","Text":"and we get rw\u0027(r) is r squared over 2 plus constant of integration."},{"Start":"02:05.210 ","End":"02:07.850","Text":"But if we add r equals 0,"},{"Start":"02:07.850 ","End":"02:10.939","Text":"we get 0 equals 0 plus a,"},{"Start":"02:10.939 ","End":"02:13.435","Text":"which forces a to be 0,"},{"Start":"02:13.435 ","End":"02:16.175","Text":"and then dividing by r,"},{"Start":"02:16.175 ","End":"02:19.730","Text":"we get w\u0027(r) is r over 2."},{"Start":"02:19.730 ","End":"02:27.880","Text":"Integrate again, and we have that w(r) is r^2 over 4 plus another constant b."},{"Start":"02:27.880 ","End":"02:30.150","Text":"Now if we let r equals 1,"},{"Start":"02:30.150 ","End":"02:32.985","Text":"here we have a 1/4 plus b, and here,"},{"Start":"02:32.985 ","End":"02:37.350","Text":"w(1) is equal to 0,"},{"Start":"02:37.350 ","End":"02:40.980","Text":"so that 0 equals 1/4 plus b."},{"Start":"02:40.980 ","End":"02:43.155","Text":"So b is minus a quarter,"},{"Start":"02:43.155 ","End":"02:45.420","Text":"and if b is minus 1/4,"},{"Start":"02:45.420 ","End":"02:49.095","Text":"then we have that w(r) is r^2 over 4 minus 1/4,"},{"Start":"02:49.095 ","End":"02:52.125","Text":"which is r^2 minus 1 over 4."},{"Start":"02:52.125 ","End":"02:58.035","Text":"Next we\u0027ll solve for v. The problem was Delta v equals 0,"},{"Start":"02:58.035 ","End":"03:00.905","Text":"v of 1 Theta equals this."},{"Start":"03:00.905 ","End":"03:09.050","Text":"The Poisson formula for the Laplace equation in a disk is this, and in our case,"},{"Start":"03:09.050 ","End":"03:10.280","Text":"Rho is 1,"},{"Start":"03:10.280 ","End":"03:12.980","Text":"so r over Rho is just r,"},{"Start":"03:12.980 ","End":"03:14.675","Text":"and we have this,"},{"Start":"03:14.675 ","End":"03:18.755","Text":"but we have boundary condition for r equals 1."},{"Start":"03:18.755 ","End":"03:20.674","Text":"On the one hand,"},{"Start":"03:20.674 ","End":"03:23.315","Text":"v of 1 Theta equals this."},{"Start":"03:23.315 ","End":"03:24.590","Text":"On the other hand,"},{"Start":"03:24.590 ","End":"03:29.875","Text":"v of 1 Theta is this with r equals 1, so it\u0027s this."},{"Start":"03:29.875 ","End":"03:37.040","Text":"Now we have on the right-hand side the Fourier series with cosines and sines,"},{"Start":"03:37.040 ","End":"03:39.380","Text":"and we can compare coefficients."},{"Start":"03:39.380 ","End":"03:40.760","Text":"On the left there\u0027s only 2."},{"Start":"03:40.760 ","End":"03:44.645","Text":"There\u0027s a free coefficient corresponding to the a naught over 2,"},{"Start":"03:44.645 ","End":"03:50.345","Text":"and there\u0027s the condition for the sign when n is 2,022,"},{"Start":"03:50.345 ","End":"03:53.664","Text":"so we get a naught over 2 is c,"},{"Start":"03:53.664 ","End":"04:00.570","Text":"b 2,022 is equal to 1 because this is 1 sine this."},{"Start":"04:00.570 ","End":"04:03.845","Text":"All the others are 0. That means that if we put"},{"Start":"04:03.845 ","End":"04:07.658","Text":"a naught and b 2022 here and all the rest are 0,"},{"Start":"04:07.658 ","End":"04:14.560","Text":"we get c plus r^2022 sine of 2022 Theta."},{"Start":"04:14.560 ","End":"04:19.350","Text":"That\u0027s v, and we already had w equal to this,"},{"Start":"04:19.350 ","End":"04:26.630","Text":"and u is v plus w so that we have that u is equal to this."},{"Start":"04:26.630 ","End":"04:29.495","Text":"Now, this isn\u0027t what we want to solve for,"},{"Start":"04:29.495 ","End":"04:37.590","Text":"we were asked for what c could possibly be in order that this limit is equal to 1/4."},{"Start":"04:37.590 ","End":"04:44.350","Text":"Slightly rewriting this to put the minus 1/4 together with the c,"},{"Start":"04:44.350 ","End":"04:45.745","Text":"this is what we have."},{"Start":"04:45.745 ","End":"04:50.810","Text":"Now, we can divide both sides by r squared because we want to compute this limit."},{"Start":"04:50.810 ","End":"04:54.030","Text":"We have u of r Theta over r^2 equals this"},{"Start":"04:54.030 ","End":"04:58.250","Text":"times 1 over r^squared subtract 2 from the exponent here,"},{"Start":"04:58.250 ","End":"05:00.665","Text":"and this r^2 becomes 1."},{"Start":"05:00.665 ","End":"05:08.290","Text":"Now, the limit as r goes to 0 is equal to this limit is 0,"},{"Start":"05:08.290 ","End":"05:10.445","Text":"so we have the limit of this,"},{"Start":"05:10.445 ","End":"05:13.385","Text":"the first term plus 1/4."},{"Start":"05:13.385 ","End":"05:17.315","Text":"Now this limit is not going to exist when r goes to 0,"},{"Start":"05:17.315 ","End":"05:21.080","Text":"except if this coefficient is 0,"},{"Start":"05:21.080 ","End":"05:23.765","Text":"meaning that c is equal to 1/4,"},{"Start":"05:23.765 ","End":"05:30.590","Text":"and then what we have is that this limit is equal to 1/4 because if c is 1/4, this is 0."},{"Start":"05:30.590 ","End":"05:33.460","Text":"The limit of 0 over r^2 is 0,"},{"Start":"05:33.460 ","End":"05:35.190","Text":"and that\u0027s as required,"},{"Start":"05:35.190 ","End":"05:38.370","Text":"so this is the answer, and we\u0027re done."}],"ID":30888},{"Watched":false,"Name":"Exercise 24","Duration":"3m 9s","ChapterTopicVideoID":29333,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.470","Text":"In this exercise, we\u0027re given 2 boundary value problems,"},{"Start":"00:04.470 ","End":"00:08.130","Text":"1 for u and 1 for v. We have to"},{"Start":"00:08.130 ","End":"00:13.440","Text":"prove that u at the origin is bigger than v at the origin."},{"Start":"00:13.440 ","End":"00:16.350","Text":"This is given in polar coordinates."},{"Start":"00:16.350 ","End":"00:20.625","Text":"What we\u0027re going to do is use the mean value property."},{"Start":"00:20.625 ","End":"00:24.060","Text":"First, we\u0027ll define w to be the difference of u"},{"Start":"00:24.060 ","End":"00:27.525","Text":"minus v. That\u0027s what we often do when we compare 2 functions."},{"Start":"00:27.525 ","End":"00:30.330","Text":"w satisfies the equation."},{"Start":"00:30.330 ","End":"00:38.570","Text":"Delta w is r^2 minus r^2 is 0. w of our Theta is this minus this,"},{"Start":"00:38.570 ","End":"00:41.180","Text":"this is on the disk of radius 1,"},{"Start":"00:41.180 ","End":"00:44.860","Text":"and this is on the boundary which is the circle of radius 1."},{"Start":"00:44.860 ","End":"00:52.405","Text":"This is the Laplace equation and because w satisfies the Laplace equation, it\u0027s harmonic."},{"Start":"00:52.405 ","End":"00:54.810","Text":"By the mean value property,"},{"Start":"00:54.810 ","End":"00:57.620","Text":"the value of w at the origin is"},{"Start":"00:57.620 ","End":"01:02.015","Text":"the average value of w on a circle centered at the origin."},{"Start":"01:02.015 ","End":"01:05.930","Text":"This circle of radius 1,"},{"Start":"01:05.930 ","End":"01:08.430","Text":"r=1 is centered at the origin."},{"Start":"01:08.430 ","End":"01:14.045","Text":"This is the integral of the values of w along the circle."},{"Start":"01:14.045 ","End":"01:18.784","Text":"Replace w by what it\u0027s equal to from here,"},{"Start":"01:18.784 ","End":"01:24.680","Text":"we can replace the interval of integration instead of 0 to 2 Pi,"},{"Start":"01:24.680 ","End":"01:31.500","Text":"we can take minus Pi to Pi because our functions are all periodic of period 2Pi."},{"Start":"01:31.500 ","End":"01:35.105","Text":"We can take any interval of length 2 Pi here."},{"Start":"01:35.105 ","End":"01:37.580","Text":"The reason I want to do that is to work"},{"Start":"01:37.580 ","End":"01:41.905","Text":"with odd and even functions on a symmetric interval."},{"Start":"01:41.905 ","End":"01:51.860","Text":"In this case, this cosine to the 2022 Theta is an even function because cosine is even,"},{"Start":"01:51.860 ","End":"01:54.985","Text":"doesn\u0027t matter if the exponent is even or not."},{"Start":"01:54.985 ","End":"01:57.390","Text":"In this case here,"},{"Start":"01:57.390 ","End":"01:58.785","Text":"it is important,"},{"Start":"01:58.785 ","End":"02:02.584","Text":"sine is an odd function and if you raise it to an odd power,"},{"Start":"02:02.584 ","End":"02:04.805","Text":"then it\u0027s still odd."},{"Start":"02:04.805 ","End":"02:09.605","Text":"When we take the integral of an even function in a symmetric interval,"},{"Start":"02:09.605 ","End":"02:15.580","Text":"we can double it and take the integral just from 0 to whatever it is."},{"Start":"02:15.580 ","End":"02:21.440","Text":"The integral of an odd function on a symmetric interval is 0."},{"Start":"02:21.440 ","End":"02:24.245","Text":"This part is positive."},{"Start":"02:24.245 ","End":"02:31.535","Text":"The reason is that cosine to an even power or any function to an even power,"},{"Start":"02:31.535 ","End":"02:33.965","Text":"it\u0027s going to be non-negative."},{"Start":"02:33.965 ","End":"02:37.640","Text":"Now how do I get the strictly positive if a function is"},{"Start":"02:37.640 ","End":"02:43.100","Text":"non-negative and it\u0027s not identically 0 and it\u0027s continuous."},{"Start":"02:43.100 ","End":"02:47.000","Text":"I didn\u0027t write that, but it\u0027s important that should be continuous, which it is."},{"Start":"02:47.000 ","End":"02:51.905","Text":"Then in that case we get that the integral is positive."},{"Start":"02:51.905 ","End":"02:56.340","Text":"This is w(00) from here,"},{"Start":"02:56.600 ","End":"03:00.635","Text":"and if w at the origin is positive,"},{"Start":"03:00.635 ","End":"03:06.950","Text":"then u minus v is positive and so u is bigger than v at the origin."},{"Start":"03:06.950 ","End":"03:09.870","Text":"That\u0027s what we had to show, so we\u0027re done."}],"ID":30889},{"Watched":false,"Name":"Exercise 25","Duration":"4m 44s","ChapterTopicVideoID":29334,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.759","Text":"In this exercise, we have the following."},{"Start":"00:03.759 ","End":"00:06.145","Text":"Actually it\u0027s a family of boundary value problems."},{"Start":"00:06.145 ","End":"00:09.385","Text":"For each k, we have a problem like so."},{"Start":"00:09.385 ","End":"00:16.480","Text":"We have to solve it to find u_k and then compute the limit as k goes to infinity of u_k."},{"Start":"00:16.480 ","End":"00:20.770","Text":"The PDE is a non-homogeneous Laplace,"},{"Start":"00:20.770 ","End":"00:23.720","Text":"also called a Poisson equation,"},{"Start":"00:23.720 ","End":"00:28.750","Text":"and usual technique is to split the problem up into 2, 1 time,"},{"Start":"00:28.750 ","End":"00:31.150","Text":"we get a homogeneous PDE and"},{"Start":"00:31.150 ","End":"00:36.305","Text":"a non-homogeneous boundary condition and in the other, it\u0027s vice versa."},{"Start":"00:36.305 ","End":"00:40.540","Text":"U_k will be v plus w. V has"},{"Start":"00:40.540 ","End":"00:47.290","Text":"the homogeneous equation and w has the homogeneous boundary value."},{"Start":"00:47.290 ","End":"00:50.070","Text":"Let\u0027s solve for v first."},{"Start":"00:50.070 ","End":"00:56.515","Text":"I\u0027m just copying the problem for v. Because it\u0027s homogeneous Laplace equation on a disk,"},{"Start":"00:56.515 ","End":"01:00.950","Text":"the standard solution is given by the Poisson formula as follows."},{"Start":"01:00.950 ","End":"01:03.560","Text":"In our case, Rho is equal to 1,"},{"Start":"01:03.560 ","End":"01:10.385","Text":"so we get that v of r Theta is just r^n here instead of r over Rho^n."},{"Start":"01:10.385 ","End":"01:12.480","Text":"If we let r equals 1,"},{"Start":"01:12.480 ","End":"01:14.455","Text":"we get v of 1 Theta,"},{"Start":"01:14.455 ","End":"01:16.295","Text":"which is sine theta."},{"Start":"01:16.295 ","End":"01:19.090","Text":"On the other hand, if we put r equals 1 here,"},{"Start":"01:19.090 ","End":"01:20.850","Text":"we get same thing."},{"Start":"01:20.850 ","End":"01:23.895","Text":"Just the r^n drops out because r is 1."},{"Start":"01:23.895 ","End":"01:31.700","Text":"Now we have sine Theta expressed as a Fourier series and we can compare coefficients."},{"Start":"01:31.700 ","End":"01:34.220","Text":"Well, there\u0027s only 1 non-zero coefficient,"},{"Start":"01:34.220 ","End":"01:36.380","Text":"that\u0027s the coefficient of b_1."},{"Start":"01:36.380 ","End":"01:38.300","Text":"B_1 is 1,"},{"Start":"01:38.300 ","End":"01:41.095","Text":"and all the other coefficients are 0."},{"Start":"01:41.095 ","End":"01:45.635","Text":"If we substitute back in here,"},{"Start":"01:45.635 ","End":"01:49.730","Text":"we only get the coefficient of b_1,"},{"Start":"01:49.730 ","End":"01:53.855","Text":"so it\u0027s b_1 sine 1 Theta and r^1,"},{"Start":"01:53.855 ","End":"01:56.065","Text":"so it\u0027s r sine Theta."},{"Start":"01:56.065 ","End":"01:59.220","Text":"Having found v, let\u0027s turn to w now."},{"Start":"01:59.220 ","End":"02:03.875","Text":"I\u0027m going to look for a radial solution where w just depends on r,"},{"Start":"02:03.875 ","End":"02:07.310","Text":"because the whole problem on the right hand side there is no Theta."},{"Start":"02:07.310 ","End":"02:10.685","Text":"It just depends on r. If it just depends on r,"},{"Start":"02:10.685 ","End":"02:13.570","Text":"then w Theta Theta is 0,"},{"Start":"02:13.570 ","End":"02:17.960","Text":"and the derivative of w with respect to r is just the regular"},{"Start":"02:17.960 ","End":"02:22.775","Text":"derivative because there is only 1 variable r. I forgot to mention here."},{"Start":"02:22.775 ","End":"02:27.620","Text":"This is the formula for the Laplacian of w in polar coordinates."},{"Start":"02:27.620 ","End":"02:31.175","Text":"We can multiply both sides by r,"},{"Start":"02:31.175 ","End":"02:32.970","Text":"get rid of the fraction here,"},{"Start":"02:32.970 ","End":"02:40.460","Text":"and also the left-hand side becomes a complete derivative of the product r times w prime."},{"Start":"02:40.460 ","End":"02:44.645","Text":"Use the product rule to verify that if we differentiate this, we get this."},{"Start":"02:44.645 ","End":"02:49.550","Text":"Now we can integrate both sides with respect to r and get that r"},{"Start":"02:49.550 ","End":"02:54.230","Text":"w prime^1 and divide by the new power,"},{"Start":"02:54.230 ","End":"02:56.030","Text":"that\u0027s the k plus 2 here,"},{"Start":"02:56.030 ","End":"02:58.130","Text":"the constant of integration."},{"Start":"02:58.130 ","End":"03:00.530","Text":"If we let r equals 0,"},{"Start":"03:00.530 ","End":"03:03.535","Text":"then we get that c equals 0,"},{"Start":"03:03.535 ","End":"03:09.010","Text":"and so we can now divide both sides by r because c is gone,"},{"Start":"03:09.010 ","End":"03:14.480","Text":"and we get that w prime is r^k plus 1 over 2^k plus 2,"},{"Start":"03:14.480 ","End":"03:16.990","Text":"Integrate again with respect to r,"},{"Start":"03:16.990 ","End":"03:20.085","Text":"and we get again a constant of integration."},{"Start":"03:20.085 ","End":"03:24.800","Text":"This time, we can use the boundary conditions w(1) equals"},{"Start":"03:24.800 ","End":"03:30.215","Text":"0 to get that a is minus 1 over 2^k,"},{"Start":"03:30.215 ","End":"03:32.480","Text":"k plus 2^2,"},{"Start":"03:32.480 ","End":"03:35.255","Text":"and then we put that here."},{"Start":"03:35.255 ","End":"03:41.215","Text":"We can simplify it to r^k plus 2 minus 1 over 2^k, k plus 2^2."},{"Start":"03:41.215 ","End":"03:43.610","Text":"You can see that if you put r equals 1,"},{"Start":"03:43.610 ","End":"03:46.490","Text":"then we get 0, so that\u0027s fine."},{"Start":"03:46.490 ","End":"03:49.115","Text":"We found v and w,"},{"Start":"03:49.115 ","End":"03:55.530","Text":"and the next step is to recall that u_k is v plus w,"},{"Start":"03:55.530 ","End":"04:03.190","Text":"so we can get that u_k is r sine Theta from here plus r^k plus 2 minus 1 over 2^k,"},{"Start":"04:03.190 ","End":"04:05.960","Text":"k plus 2^2 from here."},{"Start":"04:05.960 ","End":"04:07.790","Text":"That\u0027s not all what we were asked."},{"Start":"04:07.790 ","End":"04:12.425","Text":"We have to find the limit as k goes to infinity of u_k."},{"Start":"04:12.425 ","End":"04:19.920","Text":"Well, r^k plus 2 minus 1 is bounded because this is between 0 and 1,"},{"Start":"04:19.920 ","End":"04:22.310","Text":"so minus 1, it\u0027s bounded between minus 1 and 0."},{"Start":"04:22.310 ","End":"04:25.970","Text":"Any of it\u0027s bounded and the denominator goes to infinity,"},{"Start":"04:25.970 ","End":"04:29.720","Text":"so the limit of this part is 0."},{"Start":"04:29.720 ","End":"04:33.005","Text":"This limit is r sine Theta,"},{"Start":"04:33.005 ","End":"04:34.520","Text":"and that\u0027s the answer."},{"Start":"04:34.520 ","End":"04:38.450","Text":"I\u0027ll just note that if we want to express it in Cartesian coordinates,"},{"Start":"04:38.450 ","End":"04:40.175","Text":"it comes out even neater."},{"Start":"04:40.175 ","End":"04:44.580","Text":"The answer is just y. We\u0027re done."}],"ID":30890},{"Watched":false,"Name":"Exercise 26","Duration":"5m 17s","ChapterTopicVideoID":29335,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.190","Text":"In this exercise, we\u0027re given the following initial boundary and value problem."},{"Start":"00:05.190 ","End":"00:10.455","Text":"We\u0027re going to use an energy integral to show that the solution is unique."},{"Start":"00:10.455 ","End":"00:15.885","Text":"We\u0027re even given the hint as to which energy integral to use like this one,"},{"Start":"00:15.885 ","End":"00:18.285","Text":"will soon see what w is."},{"Start":"00:18.285 ","End":"00:21.990","Text":"Now this is a nonhomogeneous heat equation."},{"Start":"00:21.990 ","End":"00:23.280","Text":"This is the heat equation."},{"Start":"00:23.280 ","End":"00:26.190","Text":"This is a nonhomogeneous part and it\u0027s on"},{"Start":"00:26.190 ","End":"00:29.880","Text":"a finite interval and the boundary conditions are mixed."},{"Start":"00:29.880 ","End":"00:34.710","Text":"This bit is direct way for it\u0027s mixed."},{"Start":"00:34.710 ","End":"00:40.085","Text":"The strategy as usual is to suppose that u and v are both"},{"Start":"00:40.085 ","End":"00:46.210","Text":"solutions to the problem and then showing that u equals v necessarily."},{"Start":"00:46.210 ","End":"00:50.355","Text":"What we do is we define w=u minus v,"},{"Start":"00:50.355 ","End":"00:53.480","Text":"and then we prove that w is 0."},{"Start":"00:53.480 ","End":"00:58.250","Text":"We\u0027re going to work out what the initial boundary value problem is for w,"},{"Start":"00:58.250 ","End":"00:59.540","Text":"what the PDE is,"},{"Start":"00:59.540 ","End":"01:00.695","Text":"what the initial value is,"},{"Start":"01:00.695 ","End":"01:03.635","Text":"what the boundary conditions are?"},{"Start":"01:03.635 ","End":"01:06.170","Text":"We start with the PDE,"},{"Start":"01:06.170 ","End":"01:13.205","Text":"this is the PDE for you and the same thing for v. If we subtract this minus this,"},{"Start":"01:13.205 ","End":"01:16.255","Text":"the function f(x,t ) cancels."},{"Start":"01:16.255 ","End":"01:19.280","Text":"Then since u minus v is w,"},{"Start":"01:19.280 ","End":"01:21.425","Text":"and by linearity of the derivative,"},{"Start":"01:21.425 ","End":"01:27.425","Text":"we get that W_ t= w_ xx, that\u0027s the PDE."},{"Start":"01:27.425 ","End":"01:30.890","Text":"Now what about the initial conditions and boundary conditions?"},{"Start":"01:30.890 ","End":"01:33.590","Text":"Initial condition for u is this for v It\u0027s this,"},{"Start":"01:33.590 ","End":"01:37.865","Text":"subtract them, h cancels and we get w=0."},{"Start":"01:37.865 ","End":"01:40.700","Text":"Similarly for the boundary conditions,"},{"Start":"01:40.700 ","End":"01:44.960","Text":"u_ x minus b_ u= f so is v_ x minus bv and if we"},{"Start":"01:44.960 ","End":"01:49.320","Text":"subtract the f cancels and we get the following."},{"Start":"01:49.320 ","End":"01:57.015","Text":"Then u minus v is w_ x and minus bu plus bv is minus bw."},{"Start":"01:57.015 ","End":"01:59.115","Text":"In short, we get this."},{"Start":"01:59.115 ","End":"02:02.990","Text":"Similarly for the boundary condition at the other end,"},{"Start":"02:02.990 ","End":"02:05.225","Text":"this was 0, this was 1."},{"Start":"02:05.225 ","End":"02:07.280","Text":"Collecting it all together,"},{"Start":"02:07.280 ","End":"02:14.720","Text":"this and this we get the following problem for w. Everything is homogeneous,"},{"Start":"02:14.720 ","End":"02:19.175","Text":"the heat equation as well as the initial and boundary conditions."},{"Start":"02:19.175 ","End":"02:21.875","Text":"This is the energy integral that was given."},{"Start":"02:21.875 ","End":"02:27.420","Text":"Note that E(0) is equal to the following."},{"Start":"02:27.420 ","End":"02:30.450","Text":"Now, w(x,0) is 0."},{"Start":"02:30.450 ","End":"02:31.850","Text":"When we substitute it here,"},{"Start":"02:31.850 ","End":"02:36.424","Text":"we just get the integral of 0^2. The answer is 0."},{"Start":"02:36.424 ","End":"02:39.304","Text":"Another thing to note about the energy integral"},{"Start":"02:39.304 ","End":"02:43.250","Text":"is that since the integrant is bigger or equal to 0,"},{"Start":"02:43.250 ","End":"02:48.200","Text":"that E(t) is bigger or equal to 0 for all"},{"Start":"02:48.200 ","End":"02:52.250","Text":"t. If we differentiate this and assuming"},{"Start":"02:52.250 ","End":"02:56.330","Text":"it\u0027s okay to differentiate under the integral sign, we get w^2."},{"Start":"02:56.330 ","End":"02:58.055","Text":"Well, the 2 cancels with the half."},{"Start":"02:58.055 ","End":"03:00.680","Text":"We get w times the anti-derivative,"},{"Start":"03:00.680 ","End":"03:03.840","Text":"which is w_ t from the PDE,"},{"Start":"03:03.840 ","End":"03:05.220","Text":"which is the heat equation,"},{"Start":"03:05.220 ","End":"03:08.430","Text":"w_ t =w_ xx."},{"Start":"03:08.430 ","End":"03:11.655","Text":"We can replace w_ t with w_ xx."},{"Start":"03:11.655 ","End":"03:15.065","Text":"Now we can use integration by parts on this."},{"Start":"03:15.065 ","End":"03:16.340","Text":"This can be u,"},{"Start":"03:16.340 ","End":"03:19.130","Text":"this can be v prime,"},{"Start":"03:19.130 ","End":"03:27.620","Text":"uv is this and u prime v. Substitute 1 we get this substitute 0, we get this."},{"Start":"03:27.620 ","End":"03:31.025","Text":"We can just rewrite as w_x^2."},{"Start":"03:31.025 ","End":"03:36.005","Text":"Now recall that we have these boundary conditions for"},{"Start":"03:36.005 ","End":"03:40.745","Text":"w. What we can do is replace w_ x (1,"},{"Start":"03:40.745 ","End":"03:44.080","Text":"t ) and w_ x( 0,t)."},{"Start":"03:44.080 ","End":"03:49.080","Text":"Using these just by bringing the bw to the other side,"},{"Start":"03:49.080 ","End":"03:53.055","Text":"w_ x (1,t)is minus bw(1,"},{"Start":"03:53.055 ","End":"03:56.175","Text":"t) and together with the w(1,t) ^2,"},{"Start":"03:56.175 ","End":"03:59.955","Text":"w_ x ( 0,t )is bw (0,"},{"Start":"03:59.955 ","End":"04:05.150","Text":"t) and we get this and this as is."},{"Start":"04:05.150 ","End":"04:09.005","Text":"Now b is positive,"},{"Start":"04:09.005 ","End":"04:14.224","Text":"so that this expression altogether is less than or equal to 0."},{"Start":"04:14.224 ","End":"04:17.690","Text":"We had earlier that E(0)= 0."},{"Start":"04:17.690 ","End":"04:21.870","Text":"Now we have that E prime is less than or equal to 0."},{"Start":"04:21.870 ","End":"04:26.810","Text":"For function starts out at 0 and the derivative is non-positive,"},{"Start":"04:26.810 ","End":"04:29.045","Text":"meaning that it\u0027s decreasing."},{"Start":"04:29.045 ","End":"04:32.255","Text":"Then the function has to stay less than or equal to 0,"},{"Start":"04:32.255 ","End":"04:34.640","Text":"at least for non-negative t,"},{"Start":"04:34.640 ","End":"04:36.435","Text":"which is what we have."},{"Start":"04:36.435 ","End":"04:41.230","Text":"Earlier, we had that E(t) is bigger or equal to 0."},{"Start":"04:41.230 ","End":"04:45.485","Text":"If it\u0027s less than or equal to 0 and bigger or equal to 0,"},{"Start":"04:45.485 ","End":"04:50.060","Text":"then E(t) must identically equal 0."},{"Start":"04:50.060 ","End":"04:52.115","Text":"E(t) is this,"},{"Start":"04:52.115 ","End":"04:54.874","Text":"it follows that w has to be 0."},{"Start":"04:54.874 ","End":"05:00.350","Text":"This is because w^2 is continuous and bigger or equal to 0."},{"Start":"05:00.350 ","End":"05:05.440","Text":"W^2 =0, so w=0,"},{"Start":"05:05.440 ","End":"05:07.680","Text":"then w is u minus v,"},{"Start":"05:07.680 ","End":"05:12.570","Text":"so u minus v is 0, so u=v."},{"Start":"05:12.570 ","End":"05:18.540","Text":"That\u0027s what we needed to show and that shows the uniqueness and we are done."}],"ID":30891},{"Watched":false,"Name":"Exercise 27","Duration":"6m 19s","ChapterTopicVideoID":29336,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.155","Text":"In this exercise, we have the following problem to solve."},{"Start":"00:04.155 ","End":"00:07.410","Text":"Here we have a non-homogeneous heat equation."},{"Start":"00:07.410 ","End":"00:12.555","Text":"This is a non-homogeneous part on the finite interval from 0 to 2."},{"Start":"00:12.555 ","End":"00:15.690","Text":"We have an initial condition at Time 0,"},{"Start":"00:15.690 ","End":"00:18.330","Text":"which is sine Pix."},{"Start":"00:18.330 ","End":"00:23.430","Text":"We have boundary conditions when x is 0 or 2,"},{"Start":"00:23.430 ","End":"00:25.650","Text":"and these are homogeneous."},{"Start":"00:25.650 ","End":"00:28.770","Text":"We\u0027ll split the problem up into 2 pieces."},{"Start":"00:28.770 ","End":"00:30.735","Text":"Let u equal v plus w,"},{"Start":"00:30.735 ","End":"00:35.115","Text":"where each piece gets one of the nonhomogeneous parts."},{"Start":"00:35.115 ","End":"00:37.830","Text":"The problem for v is like the problem for u,"},{"Start":"00:37.830 ","End":"00:42.125","Text":"except that we have a homogeneous heat equation."},{"Start":"00:42.125 ","End":"00:45.455","Text":"The problem for w is like the problem for u,"},{"Start":"00:45.455 ","End":"00:49.370","Text":"except that we have the homogeneous initial condition."},{"Start":"00:49.370 ","End":"00:54.284","Text":"Let\u0027s start with w. We\u0027ll use Duhamel\u0027s principle,"},{"Start":"00:54.284 ","End":"01:01.520","Text":"that will get the non-homogeneous part from the PDE to the initial condition."},{"Start":"01:01.520 ","End":"01:09.065","Text":"The way it works is that w is equal to the integral of p as follows,"},{"Start":"01:09.065 ","End":"01:12.635","Text":"and p is the solution to the problem,"},{"Start":"01:12.635 ","End":"01:17.435","Text":"which is basically copied from here with an extra parameter Tau."},{"Start":"01:17.435 ","End":"01:21.755","Text":"Like I said, this condition moves down to the initial condition."},{"Start":"01:21.755 ","End":"01:26.390","Text":"This problem is in 3 variables with x, t and Tau."},{"Start":"01:26.390 ","End":"01:29.120","Text":"We can get it down to 2 variables."},{"Start":"01:29.120 ","End":"01:38.580","Text":"We just consider a particular Tau is fixed and then we substitute T equals t minus Tau."},{"Start":"01:38.580 ","End":"01:42.435","Text":"Then we get a function P(x,"},{"Start":"01:42.435 ","End":"01:44.150","Text":"T) as follows anyway,"},{"Start":"01:44.150 ","End":"01:46.040","Text":"it\u0027s all in the tutorial."},{"Start":"01:46.040 ","End":"01:52.610","Text":"Then we convert the partial derivatives P with respect to t and so on."},{"Start":"01:52.610 ","End":"01:56.045","Text":"This is just all copied from the tutorial."},{"Start":"01:56.045 ","End":"02:00.380","Text":"We get the following problem for P. Basically,"},{"Start":"02:00.380 ","End":"02:01.580","Text":"looks the same as this,"},{"Start":"02:01.580 ","End":"02:06.260","Text":"this is just 1 variable less still homogeneous heat equation"},{"Start":"02:06.260 ","End":"02:08.750","Text":"with non-homogeneous initial condition."},{"Start":"02:08.750 ","End":"02:12.080","Text":"The solution is of the following form."},{"Start":"02:12.080 ","End":"02:15.440","Text":"We made a table where we had all the different kinds"},{"Start":"02:15.440 ","End":"02:19.130","Text":"of boundary conditions for this and what the form of the solution is."},{"Start":"02:19.130 ","End":"02:24.935","Text":"Here we have the Dirichlet, Dirichlet boundary conditions."},{"Start":"02:24.935 ","End":"02:26.885","Text":"Just change the letters a bit, of course,"},{"Start":"02:26.885 ","End":"02:32.870","Text":"u here is P and change A_n to a_n."},{"Start":"02:32.870 ","End":"02:38.010","Text":"A here is 1 because a squared is 1 and L is 2."},{"Start":"02:38.010 ","End":"02:40.115","Text":"If we substitute those,"},{"Start":"02:40.115 ","End":"02:41.815","Text":"this is what we get."},{"Start":"02:41.815 ","End":"02:45.005","Text":"Then if we let t=0,"},{"Start":"02:45.005 ","End":"02:47.160","Text":"recall that P(x,"},{"Start":"02:47.160 ","End":"02:52.500","Text":"0) is e to the minus Tau sine Pix."},{"Start":"02:52.500 ","End":"02:56.150","Text":"When we put t=0 here,"},{"Start":"02:56.150 ","End":"02:59.075","Text":"the exponent just drops out."},{"Start":"02:59.075 ","End":"03:05.375","Text":"This is what we have and now we have Fourier series and we can compare coefficients."},{"Start":"03:05.375 ","End":"03:11.030","Text":"Well, this is just a single term but when n is 2,"},{"Start":"03:11.030 ","End":"03:15.080","Text":"we get sine Pix here and sine Pix here."},{"Start":"03:15.080 ","End":"03:20.135","Text":"That gives us that e to the minus t is the coefficient a_2."},{"Start":"03:20.135 ","End":"03:22.825","Text":"All the other a_n\u0027s are 0,"},{"Start":"03:22.825 ","End":"03:25.815","Text":"so if we substitute a_2 in an,"},{"Start":"03:25.815 ","End":"03:30.750","Text":"in this we get when n is 2,"},{"Start":"03:30.750 ","End":"03:33.180","Text":"n over 2 is 1, so that\u0027s sine Pix."},{"Start":"03:33.180 ","End":"03:36.090","Text":"Also here, n over 2 is 1,"},{"Start":"03:36.090 ","End":"03:38.265","Text":"so it\u0027s just Pi squared,"},{"Start":"03:38.265 ","End":"03:43.640","Text":"and the a_2 is e to the minus t. We get from P"},{"Start":"03:43.640 ","End":"03:49.365","Text":"to p by replacing T with t minus Tau,"},{"Start":"03:49.365 ","End":"03:52.010","Text":"so that\u0027s what we get."},{"Start":"03:52.010 ","End":"03:54.680","Text":"Want to separate the T\u0027s and the Taus,"},{"Start":"03:54.680 ","End":"03:58.970","Text":"so just bit of algebra and we get it like this."},{"Start":"03:58.970 ","End":"04:02.660","Text":"Then recall that w is this integral."},{"Start":"04:02.660 ","End":"04:04.400","Text":"Now we have p(x, t,"},{"Start":"04:04.400 ","End":"04:06.350","Text":"Tau), so we can put it in."},{"Start":"04:06.350 ","End":"04:11.555","Text":"All this part doesn\u0027t contain Tau and the integral is d Tau,"},{"Start":"04:11.555 ","End":"04:14.720","Text":"so we can put it in front of the integral and just have this,"},{"Start":"04:14.720 ","End":"04:16.535","Text":"which is a much simpler integral."},{"Start":"04:16.535 ","End":"04:18.395","Text":"This we know how to do."},{"Start":"04:18.395 ","End":"04:21.890","Text":"This is just e to the Pi squared minus 1 Tau and we divide by"},{"Start":"04:21.890 ","End":"04:27.260","Text":"the anti-derivative Pi squared minus 1 to take this from 0 to t,"},{"Start":"04:27.260 ","End":"04:31.700","Text":"which gives us when Tau is t,"},{"Start":"04:31.700 ","End":"04:34.040","Text":"we get the same thing just with T here,"},{"Start":"04:34.040 ","End":"04:37.280","Text":"and when Tau is 0, e^0 is 1."},{"Start":"04:37.280 ","End":"04:40.040","Text":"This is what we have now."},{"Start":"04:40.040 ","End":"04:45.270","Text":"Anyway, this is the answer for w( x, t)."},{"Start":"04:45.270 ","End":"04:49.495","Text":"Next, let\u0027s proceed to v(x, t)."},{"Start":"04:49.495 ","End":"04:53.315","Text":"Here again is the problem for v,"},{"Start":"04:53.315 ","End":"04:57.485","Text":"the PDE initial condition, boundary conditions,"},{"Start":"04:57.485 ","End":"05:04.475","Text":"and it\u0027s the same as for w. We have the same general form,"},{"Start":"05:04.475 ","End":"05:08.150","Text":"same method, let t=0,"},{"Start":"05:08.150 ","End":"05:10.520","Text":"we have v(x, 0)."},{"Start":"05:10.520 ","End":"05:13.720","Text":"The initial condition is sine Pix,"},{"Start":"05:13.720 ","End":"05:15.980","Text":"and here when t is 0,"},{"Start":"05:15.980 ","End":"05:19.835","Text":"the exponent drops out because e^0 is 1."},{"Start":"05:19.835 ","End":"05:22.150","Text":"Again, we look and see,"},{"Start":"05:22.150 ","End":"05:24.840","Text":"want to compare coefficients,"},{"Start":"05:24.840 ","End":"05:26.790","Text":"when n is 2,"},{"Start":"05:26.790 ","End":"05:29.393","Text":"we get 2Pi over 2 is just Pi."},{"Start":"05:29.393 ","End":"05:33.585","Text":"So we have sine Pix here and sine Pix here when n is 2,"},{"Start":"05:33.585 ","End":"05:37.730","Text":"so 1 equals a_2 and all the other a_n\u0027s are 0."},{"Start":"05:37.730 ","End":"05:41.930","Text":"Just like before we had only a_2 was non-zero."},{"Start":"05:41.930 ","End":"05:45.650","Text":"Then if we substitute these in this series,"},{"Start":"05:45.650 ","End":"05:50.070","Text":"we only get the one term where n is 2 and in that case,"},{"Start":"05:50.070 ","End":"05:53.580","Text":"a_n is 1 and n over 2 is 1,"},{"Start":"05:53.580 ","End":"05:55.545","Text":"and n over 2 is 1 here."},{"Start":"05:55.545 ","End":"06:00.445","Text":"This is what we have for v. Now let\u0027s bring it all together in a summary."},{"Start":"06:00.445 ","End":"06:04.189","Text":"This is what we found for v and earlier,"},{"Start":"06:04.189 ","End":"06:08.480","Text":"this is what we found for w. All we have to"},{"Start":"06:08.480 ","End":"06:12.950","Text":"do is add them together because u is v plus w,"},{"Start":"06:12.950 ","End":"06:15.830","Text":"so we take this and add it to this,"},{"Start":"06:15.830 ","End":"06:19.380","Text":"and that\u0027s the answer. That\u0027s all."}],"ID":30892},{"Watched":false,"Name":"Exercise 28","Duration":"9m 27s","ChapterTopicVideoID":29291,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.849","Text":"In this exercise we have to solve the following initial and boundary value problem."},{"Start":"00:05.849 ","End":"00:13.230","Text":"What we have here is the non-homogeneous heat equation on a finite interval."},{"Start":"00:13.230 ","End":"00:19.365","Text":"We have an initial condition when t=0 and we have 2 boundary conditions,"},{"Start":"00:19.365 ","End":"00:23.220","Text":"one for x=0 and 1 for x=Pi."},{"Start":"00:23.220 ","End":"00:26.835","Text":"The biggest problem is that the boundary conditions are not"},{"Start":"00:26.835 ","End":"00:31.725","Text":"homogeneous and we\u0027re given a hint to use the correction function,"},{"Start":"00:31.725 ","End":"00:35.345","Text":"and that will help us make these homogeneous."},{"Start":"00:35.345 ","End":"00:39.179","Text":"We\u0027ll define v by u=v plus r;"},{"Start":"00:39.179 ","End":"00:41.315","Text":"this is the correction function,"},{"Start":"00:41.315 ","End":"00:49.955","Text":"then we\u0027ll see what the IBVP is for v. For several parts we need the initial condition;"},{"Start":"00:49.955 ","End":"00:52.835","Text":"the boundary conditions and the PDE."},{"Start":"00:52.835 ","End":"00:56.555","Text":"Let\u0027s start with the boundary conditions here."},{"Start":"00:56.555 ","End":"01:01.874","Text":"The first one is that when x=0, u(0,"},{"Start":"01:01.874 ","End":"01:07.540","Text":"t) is t. On the other hand u(0, t) is v(0,"},{"Start":"01:07.540 ","End":"01:10.035","Text":"t ) plus; let\u0027s see,"},{"Start":"01:10.035 ","End":"01:15.090","Text":"when x is 0 we get v(0, t)."},{"Start":"01:15.090 ","End":"01:22.250","Text":"This disappears and this becomes Pi over Pi as 1 times t is just t. Now,"},{"Start":"01:22.250 ","End":"01:25.670","Text":"if we look at this equation on the left we have t"},{"Start":"01:25.670 ","End":"01:29.925","Text":"and on the right we have t and this is just 0,"},{"Start":"01:29.925 ","End":"01:33.165","Text":"so we\u0027re left with v(0, t)=0."},{"Start":"01:33.165 ","End":"01:38.330","Text":"Very similarly if we take the other end point we get v(Pi,"},{"Start":"01:38.330 ","End":"01:40.160","Text":"t) is 0, and that\u0027s good."},{"Start":"01:40.160 ","End":"01:42.850","Text":"We have homogeneous boundary conditions."},{"Start":"01:42.850 ","End":"01:45.719","Text":"Now let\u0027s do the initial condition."},{"Start":"01:45.719 ","End":"01:49.320","Text":"When t=0 then u(x,"},{"Start":"01:49.320 ","End":"01:52.575","Text":"0) is x over Pi plus sine x."},{"Start":"01:52.575 ","End":"01:57.975","Text":"On the other hand if we substitute t=0 here,"},{"Start":"01:57.975 ","End":"02:00.737","Text":"we get v(x,"},{"Start":"02:00.737 ","End":"02:02.265","Text":"0) and then t is 0."},{"Start":"02:02.265 ","End":"02:03.630","Text":"So this part disappears,"},{"Start":"02:03.630 ","End":"02:05.940","Text":"we just have the x over Pi."},{"Start":"02:05.940 ","End":"02:14.610","Text":"The x over Pi cancels with the x over Pi and we\u0027re left with sin x= v(x,"},{"Start":"02:14.610 ","End":"02:16.335","Text":"0) or vice versa."},{"Start":"02:16.335 ","End":"02:20.370","Text":"So that\u0027s the third thing in the IBVP."},{"Start":"02:20.370 ","End":"02:24.320","Text":"The fourth thing we need is the PDE itself."},{"Start":"02:24.320 ","End":"02:27.710","Text":"This is what u equals in terms of v. Now,"},{"Start":"02:27.710 ","End":"02:31.445","Text":"we\u0027ll need both v_t and v_xx."},{"Start":"02:31.445 ","End":"02:34.870","Text":"Differentiate both sides with respect to t,"},{"Start":"02:34.870 ","End":"02:40.085","Text":"and we get that u_t is v_."},{"Start":"02:40.085 ","End":"02:41.334","Text":"This part disappears,"},{"Start":"02:41.334 ","End":"02:43.550","Text":"it\u0027s a constant and this;"},{"Start":"02:43.550 ","End":"02:47.245","Text":"something times t so we just have that constant something,"},{"Start":"02:47.245 ","End":"02:50.080","Text":"Pi minus x over Pi."},{"Start":"02:50.080 ","End":"02:54.740","Text":"The other partial derivative we need is u_xx,"},{"Start":"02:54.740 ","End":"02:56.915","Text":"and that\u0027s equal to v_xx."},{"Start":"02:56.915 ","End":"03:00.215","Text":"Now, the second derivative of this with respect to x is 0,"},{"Start":"03:00.215 ","End":"03:01.900","Text":"because this is linear in x."},{"Start":"03:01.900 ","End":"03:06.470","Text":"Similarly, this second derivative with respect to x is 0,"},{"Start":"03:06.470 ","End":"03:08.695","Text":"so we just have this equals this."},{"Start":"03:08.695 ","End":"03:12.800","Text":"Now, the PDE for you is this all we have to do now"},{"Start":"03:12.800 ","End":"03:17.230","Text":"is substitute u_t and u_xx from here and here,"},{"Start":"03:17.230 ","End":"03:20.130","Text":"and what we get is u_t is this,"},{"Start":"03:20.130 ","End":"03:22.296","Text":"u_xx is this,"},{"Start":"03:22.296 ","End":"03:24.450","Text":"plus Pi minus x over Pi."},{"Start":"03:24.450 ","End":"03:30.300","Text":"This Pi minus x over Pi cancels with this Pi minus x over"},{"Start":"03:30.300 ","End":"03:38.190","Text":"Pi and what we\u0027re left with is v_t=v_xx plus 3sin(2x)."},{"Start":"03:38.190 ","End":"03:39.930","Text":"Now let\u0027s collect together."},{"Start":"03:39.930 ","End":"03:44.265","Text":"The PDE, the initial condition,"},{"Start":"03:44.265 ","End":"03:50.540","Text":"and 2 boundary conditions if we put them all together,"},{"Start":"03:50.540 ","End":"03:55.288","Text":"and what we get is the following,"},{"Start":"03:55.288 ","End":"04:00.985","Text":"IBVP for v. We\u0027re going to split this up into 2 like we usually do."},{"Start":"04:00.985 ","End":"04:04.360","Text":"I chose the letters f and g where f has"},{"Start":"04:04.360 ","End":"04:10.150","Text":"the homogeneous heat equation and g has a non-homogeneous,"},{"Start":"04:10.150 ","End":"04:12.804","Text":"but it has the homogeneous initial condition."},{"Start":"04:12.804 ","End":"04:17.740","Text":"The more difficult one or at least more work is g because we\u0027ll use"},{"Start":"04:17.740 ","End":"04:20.770","Text":"Duhamel\u0027s principle and after we apply"},{"Start":"04:20.770 ","End":"04:25.230","Text":"the principle this non-homogeneous part will go into the initial value,"},{"Start":"04:25.230 ","End":"04:29.230","Text":"so it\u0027ll be like this one and then we\u0027ll have to"},{"Start":"04:29.230 ","End":"04:34.150","Text":"solve 2 homogeneous heat equations with non-homogeneous initial value."},{"Start":"04:34.150 ","End":"04:36.030","Text":"Anyway, let\u0027s get started with."},{"Start":"04:36.030 ","End":"04:40.195","Text":"By the Duhamel principle it always starts out the same."},{"Start":"04:40.195 ","End":"04:44.500","Text":"Basically just copy paste from wherever and the only difference is here,"},{"Start":"04:44.500 ","End":"04:47.200","Text":"the initial condition comes from"},{"Start":"04:47.200 ","End":"04:53.090","Text":"the non-homogeneous part of the heat equation here and the end point L,"},{"Start":"04:53.090 ","End":"04:57.690","Text":"and then we reduce the number of variables."},{"Start":"04:57.690 ","End":"05:03.160","Text":"We go from p to P by letting T equals T minus Tau,"},{"Start":"05:03.160 ","End":"05:06.125","Text":"and this is always the same."},{"Start":"05:06.125 ","End":"05:10.425","Text":"The point is we get the following, IBVP."},{"Start":"05:10.425 ","End":"05:15.050","Text":"Essentially what this is just shorten,"},{"Start":"05:15.050 ","End":"05:18.350","Text":"we don\u0027t have this last variable Tau."},{"Start":"05:18.350 ","End":"05:20.060","Text":"We\u0027ve seen this several times before,"},{"Start":"05:20.060 ","End":"05:22.475","Text":"so I didn\u0027t go into all the little details."},{"Start":"05:22.475 ","End":"05:24.365","Text":"Now that we have this,"},{"Start":"05:24.365 ","End":"05:30.515","Text":"we can refer to our table for the heat equation on a finite interval,"},{"Start":"05:30.515 ","End":"05:36.665","Text":"and here we have both Dirichlet boundary conditions."},{"Start":"05:36.665 ","End":"05:40.055","Text":"The general form of the solution is like this,"},{"Start":"05:40.055 ","End":"05:45.070","Text":"so in our case T and we have P instead of u."},{"Start":"05:45.070 ","End":"05:49.940","Text":"Also, we have a=1 and L=Pi."},{"Start":"05:49.940 ","End":"05:51.660","Text":"If we put those here,"},{"Start":"05:51.660 ","End":"05:53.340","Text":"Pi over L is 1,"},{"Start":"05:53.340 ","End":"05:55.335","Text":"so it\u0027s just nx here."},{"Start":"05:55.335 ","End":"05:58.680","Text":"Also Pi over L is 1 and a is 1,"},{"Start":"05:58.680 ","End":"06:01.425","Text":"so we just get each of the minus n^2."},{"Start":"06:01.425 ","End":"06:04.580","Text":"This is expression for P(x,"},{"Start":"06:04.580 ","End":"06:09.890","Text":"T) and what we have to find out now is what these coefficients;"},{"Start":"06:09.890 ","End":"06:11.465","Text":"a_n, what they are."},{"Start":"06:11.465 ","End":"06:15.360","Text":"If we let T=0 here,"},{"Start":"06:15.360 ","End":"06:17.310","Text":"on the one hand P(x,"},{"Start":"06:17.310 ","End":"06:20.955","Text":"0) from here is 3 sin (2x)."},{"Start":"06:20.955 ","End":"06:27.555","Text":"On the other hand if we let T=0 here this e^minus o just disappears,"},{"Start":"06:27.555 ","End":"06:29.600","Text":"and so we have the following."},{"Start":"06:29.600 ","End":"06:34.575","Text":"Now we can compare coefficients for a sine series."},{"Start":"06:34.575 ","End":"06:39.055","Text":"The only non-zero coefficient occurs when n=2,"},{"Start":"06:39.055 ","End":"06:42.895","Text":"and then we get that a_2 is 3,"},{"Start":"06:42.895 ","End":"06:45.960","Text":"and all the other a_ns are 0,"},{"Start":"06:45.960 ","End":"06:48.100","Text":"meaning n is not 2."},{"Start":"06:48.100 ","End":"06:52.145","Text":"Returning to this, we can now substitute the a_ns."},{"Start":"06:52.145 ","End":"06:53.814","Text":"We\u0027ll only get one term;"},{"Start":"06:53.814 ","End":"06:55.610","Text":"the n=2 term,"},{"Start":"06:55.610 ","End":"06:59.910","Text":"and that will come out to be a_2 which is 3,"},{"Start":"06:59.910 ","End":"07:06.070","Text":"e^minus 2^2T which is e^minus 4T and n is 2, so sin 2x."},{"Start":"07:06.070 ","End":"07:13.887","Text":"Now we return from P to p by replacing T with t minus Tau."},{"Start":"07:13.887 ","End":"07:17.190","Text":"If we replace t with t minus Tau this is what we get,"},{"Start":"07:17.190 ","End":"07:20.615","Text":"and then remember that g is this integral."},{"Start":"07:20.615 ","End":"07:22.295","Text":"Just replace p x,"},{"Start":"07:22.295 ","End":"07:23.690","Text":"t) Tau with this,"},{"Start":"07:23.690 ","End":"07:26.105","Text":"and this is the integral we need to compute."},{"Start":"07:26.105 ","End":"07:27.700","Text":"Now it\u0027s d Tau,"},{"Start":"07:27.700 ","End":"07:31.520","Text":"so all the non Tau stuff can be pulled up front and all we"},{"Start":"07:31.520 ","End":"07:36.265","Text":"have is this integral e^4Tau, dTau."},{"Start":"07:36.265 ","End":"07:39.045","Text":"That\u0027s a simple integral,"},{"Start":"07:39.045 ","End":"07:45.103","Text":"e^4Tau over 4 between Tau=0 and Tau=t."},{"Start":"07:45.103 ","End":"07:55.260","Text":"Bring the 4 here under the 3 and we have e^4 Tau minus 1 because that\u0027s e^4 times 0,"},{"Start":"07:55.260 ","End":"08:00.410","Text":"so that\u0027s g(x, t)."},{"Start":"08:00.410 ","End":"08:03.670","Text":"I wrote Tau here instead of t, we fix that. I suppose we could simplify it by combining"},{"Start":"08:03.670 ","End":"08:09.049","Text":"this with this and getting 1 minus e^minus 4t."},{"Start":"08:09.049 ","End":"08:13.030","Text":"I won\u0027t bother, and let\u0027s proceed by solving for f. A reminder,"},{"Start":"08:13.030 ","End":"08:15.580","Text":"this was the problem for f,"},{"Start":"08:15.580 ","End":"08:19.130","Text":"PDE initial conditions, boundary conditions."},{"Start":"08:19.130 ","End":"08:23.830","Text":"The initial condition is non-homogeneous and we\u0027ll solve it the way"},{"Start":"08:23.830 ","End":"08:28.150","Text":"we solved for P. We\u0027ll get the same general series as before,"},{"Start":"08:28.150 ","End":"08:30.085","Text":"just different a_ns,"},{"Start":"08:30.085 ","End":"08:33.290","Text":"letting t=0, f(x,"},{"Start":"08:33.290 ","End":"08:35.160","Text":"0) is sine x,"},{"Start":"08:35.160 ","End":"08:39.040","Text":"and here the exponent drops off."},{"Start":"08:39.040 ","End":"08:43.205","Text":"Now, again we compare coefficients only when n is 1 do"},{"Start":"08:43.205 ","End":"08:47.810","Text":"we get something because then we get that a_1 is 1 and all the others are 0."},{"Start":"08:47.810 ","End":"08:53.645","Text":"If we substitute a_1 and a_n here,"},{"Start":"08:53.645 ","End":"08:55.330","Text":"we only get one term."},{"Start":"08:55.330 ","End":"08:58.024","Text":"When n is 1, here\u0027s e^minus t,"},{"Start":"08:58.024 ","End":"09:01.750","Text":"here\u0027s sin (1x) and a_1 is 1."},{"Start":"09:01.750 ","End":"09:03.880","Text":"We found that f( x,"},{"Start":"09:03.880 ","End":"09:06.045","Text":"t) is e^minus t sin (x)."},{"Start":"09:06.045 ","End":"09:09.000","Text":"G(x, t), this is what it was."},{"Start":"09:09.000 ","End":"09:14.120","Text":"Then v is f plus g, so it\u0027s this."},{"Start":"09:14.120 ","End":"09:20.522","Text":"But remember we have to get back from v to u and u was v plus r;"},{"Start":"09:20.522 ","End":"09:22.535","Text":"so v plus r,"},{"Start":"09:22.535 ","End":"09:24.950","Text":"and this is the answer."},{"Start":"09:24.950 ","End":"09:27.960","Text":"That\u0027s it."}],"ID":30893},{"Watched":false,"Name":"Exercise 29","Duration":"8m 49s","ChapterTopicVideoID":29292,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"In this exercise, we\u0027re asked to solve the following initial and boundary value problem."},{"Start":"00:05.340 ","End":"00:10.980","Text":"What it consists of is a heat equation which is non-homogeneous and we"},{"Start":"00:10.980 ","End":"00:18.150","Text":"have an initial condition and two boundary conditions because it\u0027s on a finite interval."},{"Start":"00:18.150 ","End":"00:23.370","Text":"We have to solve it but we\u0027re given a hint to use the following correction function."},{"Start":"00:23.370 ","End":"00:25.740","Text":"The problem we\u0027re correcting for is"},{"Start":"00:25.740 ","End":"00:29.250","Text":"the non-homogeneous boundary conditions so we\u0027d like"},{"Start":"00:29.250 ","End":"00:33.645","Text":"to get a 0 here if we replace u by v,"},{"Start":"00:33.645 ","End":"00:39.585","Text":"where u is given as v plus t. Let\u0027s see how that works,"},{"Start":"00:39.585 ","End":"00:41.480","Text":"u is equal to v plus t,"},{"Start":"00:41.480 ","End":"00:46.910","Text":"or if you\u0027d like v is u minus t. Let\u0027s see what the IBVP is for v,"},{"Start":"00:46.910 ","End":"00:49.415","Text":"and hopefully it will be homogeneous here."},{"Start":"00:49.415 ","End":"00:51.980","Text":"Let\u0027s take care of the boundary conditions first."},{"Start":"00:51.980 ","End":"00:55.450","Text":"When x equals 0, then we have u(0,"},{"Start":"00:55.450 ","End":"00:59.210","Text":"t) equals t so t is u(0,"},{"Start":"00:59.210 ","End":"01:00.650","Text":"t), which is v(0,"},{"Start":"01:00.650 ","End":"01:04.380","Text":"t) plus t and that gives us v(0,"},{"Start":"01:04.380 ","End":"01:11.010","Text":"t) is 0 because the t is canceled and same for when x equals 1 we get v(1,"},{"Start":"01:11.010 ","End":"01:15.320","Text":"t) equals 0 so we do have homogeneous boundary conditions."},{"Start":"01:15.320 ","End":"01:22.000","Text":"Now let\u0027s go for the initial condition and see what it is for v. When t is 0,"},{"Start":"01:22.000 ","End":"01:25.050","Text":"then we have that u(x,"},{"Start":"01:25.050 ","End":"01:26.600","Text":"0) is sine Pi x,"},{"Start":"01:26.600 ","End":"01:29.045","Text":"but u(x, 0) is v(x,"},{"Start":"01:29.045 ","End":"01:33.655","Text":"0) plus 0, because this is t,"},{"Start":"01:33.655 ","End":"01:37.356","Text":"but t is 0, so this gives us that v(x,"},{"Start":"01:37.356 ","End":"01:39.515","Text":"0) is sine Pi x."},{"Start":"01:39.515 ","End":"01:42.695","Text":"Actually the same initial condition has this."},{"Start":"01:42.695 ","End":"01:46.465","Text":"Now, lastly, we\u0027ll do the PDE part,"},{"Start":"01:46.465 ","End":"01:49.140","Text":"so u(x, t) is v(x,"},{"Start":"01:49.140 ","End":"01:56.910","Text":"t) plus t and we\u0027d like to know what is u_t and what is u_xx."},{"Start":"01:56.910 ","End":"01:59.580","Text":"What I mean is we\u0027d like v_t and v_xx,"},{"Start":"01:59.580 ","End":"02:01.245","Text":"we\u0027ll get them from here."},{"Start":"02:01.245 ","End":"02:05.630","Text":"So u_t is v_t plus 1 because we"},{"Start":"02:05.630 ","End":"02:10.730","Text":"differentiate this with respect to t. If we differentiate with respect to x twice,"},{"Start":"02:10.730 ","End":"02:14.660","Text":"this drops off and we get u_xx is v_xx."},{"Start":"02:14.660 ","End":"02:22.365","Text":"Now we can substitute these in the PDE for u and we\u0027ll get the PDE for v,"},{"Start":"02:22.365 ","End":"02:26.330","Text":"which is this and the 1 cancels,"},{"Start":"02:26.330 ","End":"02:28.340","Text":"and we get the following."},{"Start":"02:28.340 ","End":"02:31.055","Text":"Now let\u0027s collect together the four things,"},{"Start":"02:31.055 ","End":"02:33.665","Text":"this and this, and this,"},{"Start":"02:33.665 ","End":"02:37.190","Text":"and this, and that will give us the problem for v,"},{"Start":"02:37.190 ","End":"02:39.795","Text":"which is like so."},{"Start":"02:39.795 ","End":"02:41.629","Text":"We\u0027ve gotten rid of one difficulty,"},{"Start":"02:41.629 ","End":"02:46.789","Text":"which is the non-homogeneous boundary conditions and now are homogeneous,"},{"Start":"02:46.789 ","End":"02:49.250","Text":"we still have two difficulties."},{"Start":"02:49.250 ","End":"02:55.399","Text":"The non-homogeneous PDE and the non-homogeneous initial value."},{"Start":"02:55.399 ","End":"02:58.280","Text":"What we can do is split it into"},{"Start":"02:58.280 ","End":"03:04.850","Text":"two problems by letting v equals f plus g. Then f just has"},{"Start":"03:04.850 ","End":"03:09.230","Text":"the problem with the initial condition and g as the problem with"},{"Start":"03:09.230 ","End":"03:14.765","Text":"the non-homogeneous PDE so each of them will be easier to solve."},{"Start":"03:14.765 ","End":"03:21.960","Text":"Let\u0027s go first for g. So for g will use Duhamel\u0027s principle,"},{"Start":"03:21.960 ","End":"03:23.490","Text":"refer to the tutorial."},{"Start":"03:23.490 ","End":"03:30.210","Text":"This is standard g is equal to the integral of a function p(x,"},{"Start":"03:30.210 ","End":"03:35.060","Text":"t, Tau), where p is a solution to the following."},{"Start":"03:35.060 ","End":"03:38.225","Text":"Let me just show you what the original problem for g is."},{"Start":"03:38.225 ","End":"03:43.805","Text":"The non-homogeneous part of the PDE becomes"},{"Start":"03:43.805 ","End":"03:52.150","Text":"the non-homogeneous part of the initial value and this is easier to solve than this."},{"Start":"03:52.150 ","End":"03:54.860","Text":"The fact that there\u0027s three variables,"},{"Start":"03:54.860 ","End":"03:56.930","Text":"x, t, and Tau is not convenient."},{"Start":"03:56.930 ","End":"04:00.710","Text":"We can reduce it to two variables by setting big T"},{"Start":"04:00.710 ","End":"04:04.670","Text":"to equal t minus Tau and then we get a function p(x,"},{"Start":"04:04.670 ","End":"04:11.420","Text":"t) and we can get back from P to little p by replacing big T by t minus Tau."},{"Start":"04:11.420 ","End":"04:15.560","Text":"I will do that later when we solve for big P. Now there"},{"Start":"04:15.560 ","End":"04:20.000","Text":"is a general solution for the heat equation,"},{"Start":"04:20.000 ","End":"04:29.285","Text":"which is homogeneous on a finite interval with Dirichlet boundary value conditions."},{"Start":"04:29.285 ","End":"04:33.910","Text":"There\u0027s also one for Neumann boundary conditions and for mixed,"},{"Start":"04:33.910 ","End":"04:40.640","Text":"we made a table such as part of the table where this is the equation,"},{"Start":"04:40.640 ","End":"04:42.080","Text":"that\u0027s what we have."},{"Start":"04:42.080 ","End":"04:44.660","Text":"Then with Dirichlet boundary conditions,"},{"Start":"04:44.660 ","End":"04:49.805","Text":"this is the general solution and so what we get is the following."},{"Start":"04:49.805 ","End":"04:57.710","Text":"We let a equals 1 here and L equals 1 here and here and we get this and all we have to do"},{"Start":"04:57.710 ","End":"05:06.500","Text":"is find the coefficients a_n and we\u0027ll get these using the initial value."},{"Start":"05:06.500 ","End":"05:09.604","Text":"If we put t equals naught,"},{"Start":"05:09.604 ","End":"05:11.345","Text":"then we get P(x,"},{"Start":"05:11.345 ","End":"05:16.255","Text":"0) here, which is sine of 2Pi x."},{"Start":"05:16.255 ","End":"05:19.445","Text":"On the other hand, when t is 0,"},{"Start":"05:19.445 ","End":"05:25.390","Text":"this exponent drops out because e to the 0 is 1 and we get this."},{"Start":"05:25.390 ","End":"05:31.205","Text":"Now we have a Fourier sine series and we can compare coefficients."},{"Start":"05:31.205 ","End":"05:35.965","Text":"The only coefficient that\u0027s non-zero will be a_2,"},{"Start":"05:35.965 ","End":"05:40.910","Text":"and it will equal 1 and all the other a_n\u0027s are 0 when"},{"Start":"05:40.910 ","End":"05:45.785","Text":"n is not 2 and if we substitute these coefficients in this equation,"},{"Start":"05:45.785 ","End":"05:47.540","Text":"then we get that p(x,"},{"Start":"05:47.540 ","End":"05:54.195","Text":"t) is what we have here with n equals 2 so n squared is 4,"},{"Start":"05:54.195 ","End":"06:00.345","Text":"we get e to the minus 4Pi squared t sine and then is 2Pi x."},{"Start":"06:00.345 ","End":"06:01.890","Text":"Now that we have big P,"},{"Start":"06:01.890 ","End":"06:04.415","Text":"we can go back to little p, like we said,"},{"Start":"06:04.415 ","End":"06:11.190","Text":"replace big t by t minus Tau."},{"Start":"06:11.190 ","End":"06:14.320","Text":"If we put here t minus Tau,"},{"Start":"06:14.320 ","End":"06:18.845","Text":"what we get, and they also split it up using the rules of exponents."},{"Start":"06:18.845 ","End":"06:22.550","Text":"We\u0027ll get e to the 4Pi squared Tau and here e to"},{"Start":"06:22.550 ","End":"06:26.690","Text":"the minus 4Pi squared t. Then from little p,"},{"Start":"06:26.690 ","End":"06:30.920","Text":"we go back to g using this integral and this integral,"},{"Start":"06:30.920 ","End":"06:32.945","Text":"if we copy p sit from here,"},{"Start":"06:32.945 ","End":"06:36.990","Text":"gives us the following and so g(x,"},{"Start":"06:36.990 ","End":"06:39.185","Text":"t) is given by this integral."},{"Start":"06:39.185 ","End":"06:40.950","Text":"We have to compute now."},{"Start":"06:40.950 ","End":"06:44.360","Text":"A lot of this integral doesn\u0027t relate to Tau at all."},{"Start":"06:44.360 ","End":"06:46.000","Text":"That\u0027s this part."},{"Start":"06:46.000 ","End":"06:48.650","Text":"So we can pull it in front of the integral."},{"Start":"06:48.650 ","End":"06:51.410","Text":"Now we just have the integral of this, which is easy."},{"Start":"06:51.410 ","End":"06:55.160","Text":"It\u0027s each of the 4Pi square Tau divided by the anti-derivative,"},{"Start":"06:55.160 ","End":"06:58.950","Text":"which is the 4Pi squared between Tau equals 0 and"},{"Start":"06:58.950 ","End":"07:03.120","Text":"Tau equals t. When Tau equals 0, this is 1."},{"Start":"07:03.120 ","End":"07:04.695","Text":"When Tau equals t,"},{"Start":"07:04.695 ","End":"07:09.080","Text":"e to the 4Pi squared t. What we get is the following,"},{"Start":"07:09.080 ","End":"07:12.245","Text":"and we can slightly simplify it if we take the e to the minus"},{"Start":"07:12.245 ","End":"07:16.430","Text":"4Pi squared t and multiply it with the numerator here."},{"Start":"07:16.430 ","End":"07:18.005","Text":"This is what we\u0027ll get,"},{"Start":"07:18.005 ","End":"07:20.735","Text":"and that\u0027s g(x, t)."},{"Start":"07:20.735 ","End":"07:25.055","Text":"Now onto f(x, t)."},{"Start":"07:25.055 ","End":"07:28.535","Text":"This is the IBVP for f,"},{"Start":"07:28.535 ","End":"07:35.540","Text":"which is basically the same as what we had for big P. Essentially it\u0027s the same solution,"},{"Start":"07:35.540 ","End":"07:39.720","Text":"the same series, just different a_n this time."},{"Start":"07:39.720 ","End":"07:41.600","Text":"Putting t equals 0,"},{"Start":"07:41.600 ","End":"07:43.775","Text":"we get on the one hand, f(x,"},{"Start":"07:43.775 ","End":"07:46.490","Text":"0) is sine Pi x."},{"Start":"07:46.490 ","End":"07:49.630","Text":"On the other hand, will this part drops out when"},{"Start":"07:49.630 ","End":"07:53.615","Text":"t equals 0 because the exponent e to the 0 is 1."},{"Start":"07:53.615 ","End":"07:57.050","Text":"Again, we compare coefficients and this time it\u0027s"},{"Start":"07:57.050 ","End":"08:00.695","Text":"the coefficient of n equals 1, which is non-zero."},{"Start":"08:00.695 ","End":"08:04.760","Text":"A_1 equals 1 because this is 1sine 1Pi x."},{"Start":"08:04.760 ","End":"08:12.380","Text":"All the other a_n\u0027s are 0 and if we substitute these in this series,"},{"Start":"08:12.380 ","End":"08:15.215","Text":"then we just get the n equals 1,"},{"Start":"08:15.215 ","End":"08:20.430","Text":"which is 1sine Pi x and e to the minus 1 squared Pi squared t,"},{"Start":"08:20.430 ","End":"08:25.165","Text":"which is e to the minus Pi squared t. I put it in front of the sine."},{"Start":"08:25.165 ","End":"08:29.015","Text":"Now we have here g(x),"},{"Start":"08:29.015 ","End":"08:32.300","Text":"here, f(x), so it\u0027s time to summarize."},{"Start":"08:32.300 ","End":"08:34.190","Text":"From f(x) and g(x),"},{"Start":"08:34.190 ","End":"08:37.670","Text":"we can get v by adding the two."},{"Start":"08:37.670 ","End":"08:43.235","Text":"Then from v we can get u by adding r(x, t),"},{"Start":"08:43.235 ","End":"08:49.950","Text":"which gives us whatever\u0027s here plus t and that\u0027s the answer and we\u0027re done."}],"ID":30894},{"Watched":false,"Name":"Exercise 30","Duration":"5m 6s","ChapterTopicVideoID":29293,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.470","Text":"In this exercise, we\u0027re given the following initial value problem,"},{"Start":"00:04.470 ","End":"00:10.800","Text":"and we have to compute the integral from minus 2 to 2, of u(x3)^2."},{"Start":"00:10.800 ","End":"00:13.215","Text":"Let\u0027s see what the IVP is."},{"Start":"00:13.215 ","End":"00:20.549","Text":"Well, the PDE is a wave equation nonhomogeneous on the infinite interval."},{"Start":"00:20.549 ","End":"00:26.040","Text":"1 initial condition is u(x,0) is cosine x plus a split function,"},{"Start":"00:26.040 ","End":"00:31.879","Text":"and du by dt at the initial time is just equal to 0."},{"Start":"00:31.879 ","End":"00:35.570","Text":"What we\u0027re going to do here is split this problem up"},{"Start":"00:35.570 ","End":"00:39.740","Text":"into 2 problems and each of them is easier than the original problem."},{"Start":"00:39.740 ","End":"00:43.310","Text":"What we\u0027ll do is we\u0027ll let u= v + w,"},{"Start":"00:43.310 ","End":"00:48.440","Text":"where basically v is this problem without the cosines,"},{"Start":"00:48.440 ","End":"00:53.869","Text":"and w will be just the cosines other than the usual elements."},{"Start":"00:53.869 ","End":"00:57.035","Text":"Because of the linearity of the derivative."},{"Start":"00:57.035 ","End":"00:59.120","Text":"If we add these 2 equations,"},{"Start":"00:59.120 ","End":"01:05.345","Text":"we\u0027ll get utt = uxx + cosine x, which is this,"},{"Start":"01:05.345 ","End":"01:07.850","Text":"and similarly for this one,"},{"Start":"01:07.850 ","End":"01:09.830","Text":"and for the last one,"},{"Start":"01:09.830 ","End":"01:15.935","Text":"ut will equal vt+wt, it\u0027s 0."},{"Start":"01:15.935 ","End":"01:19.879","Text":"Okay, let\u0027s start with w. Turns out that w"},{"Start":"01:19.879 ","End":"01:24.770","Text":"actually has an easy solution which you can get by inspection."},{"Start":"01:24.770 ","End":"01:28.400","Text":"Just take w( x t) = cosine x,"},{"Start":"01:28.400 ","End":"01:30.350","Text":"regardless of what t is."},{"Start":"01:30.350 ","End":"01:35.210","Text":"Certainly this condition is met because w (x t) is cosine x."},{"Start":"01:35.210 ","End":"01:43.390","Text":"Also, we differentiate with respect to t. Here we just get 0 so this works."},{"Start":"01:43.390 ","End":"01:49.100","Text":"Here, wxx is the second derivative of cosine x,"},{"Start":"01:49.100 ","End":"01:54.005","Text":"which is minus cosine x and wtt is also 0,"},{"Start":"01:54.005 ","End":"01:57.395","Text":"the same as minus cosine x plus cosine x."},{"Start":"01:57.395 ","End":"02:06.665","Text":"Here it is written out that the PDE works and the first initial condition is satisfied,"},{"Start":"02:06.665 ","End":"02:10.280","Text":"and the other initial condition is satisfied."},{"Start":"02:10.280 ","End":"02:14.795","Text":"That takes care of w. Now what about v?"},{"Start":"02:14.795 ","End":"02:18.645","Text":"For v, we can use D\u0027Alembert\u0027s formula."},{"Start":"02:18.645 ","End":"02:21.680","Text":"As opposed we could have used D\u0027Alembert\u0027s formula for w,"},{"Start":"02:21.680 ","End":"02:23.720","Text":"But then we wouldn\u0027t use this one."},{"Start":"02:23.720 ","End":"02:32.315","Text":"There\u0027s a longer one with plus a double integral of big F of x and t,"},{"Start":"02:32.315 ","End":"02:35.870","Text":"which is the non-homogeneous part."},{"Start":"02:35.870 ","End":"02:38.360","Text":"Anyway, we had a straightforward solution for w,"},{"Start":"02:38.360 ","End":"02:39.860","Text":"so we didn\u0027t need to."},{"Start":"02:39.860 ","End":"02:42.710","Text":"Let\u0027s evaluate this integral."},{"Start":"02:42.710 ","End":"02:45.185","Text":"First of all, what is f and what is g?"},{"Start":"02:45.185 ","End":"02:49.115","Text":"Well, f is the function from here,"},{"Start":"02:49.115 ","End":"02:52.025","Text":"and g is this one which is 0,"},{"Start":"02:52.025 ","End":"02:55.480","Text":"and a is one from the PDE."},{"Start":"02:55.480 ","End":"02:57.270","Text":"This part drops off,"},{"Start":"02:57.270 ","End":"03:01.740","Text":"and this is just x+t and x- t. This is the formula for v( x,"},{"Start":"03:01.740 ","End":"03:05.100","Text":"t ), and for the integral,"},{"Start":"03:05.100 ","End":"03:09.760","Text":"we need v(x,3), which is this."},{"Start":"03:09.800 ","End":"03:13.485","Text":"Okay. We have v(x,3) is this,"},{"Start":"03:13.485 ","End":"03:16.065","Text":"and we know that f(x) is this,"},{"Start":"03:16.065 ","End":"03:21.345","Text":"and we need more precisely f(x)+3 and f(x)-3."},{"Start":"03:21.345 ","End":"03:23.455","Text":"From the formula here,"},{"Start":"03:23.455 ","End":"03:26.620","Text":"just replace x by x +3."},{"Start":"03:26.620 ","End":"03:33.640","Text":"This gives us the condition that it\u0027s 1 if x is between -4 and -2,"},{"Start":"03:33.640 ","End":"03:36.535","Text":"we just subtract 3 from here and here."},{"Start":"03:36.535 ","End":"03:44.165","Text":"Similarly, f (x-3) is either 1 or 0 depending on whether x is between 2 and 4 or not."},{"Start":"03:44.165 ","End":"03:47.950","Text":"If we take the average of these 2,"},{"Start":"03:47.950 ","End":"03:52.420","Text":"the areas where it\u0027s equal to one, they don\u0027t overlap."},{"Start":"03:52.420 ","End":"03:59.010","Text":"It\u0027s going to be a 1/2 when x is either here or here,"},{"Start":"03:59.010 ","End":"04:01.500","Text":"and 0+0 over 2,"},{"Start":"04:01.500 ","End":"04:03.540","Text":"which is 0 otherwise."},{"Start":"04:03.540 ","End":"04:07.560","Text":"Our integral is going to be between -2 and 2,"},{"Start":"04:07.560 ","End":"04:10.145","Text":"so note that v(x,3) is actually"},{"Start":"04:10.145 ","End":"04:15.400","Text":"0 on the interval from -2 to 2 because it isn\u0027t in either of these."},{"Start":"04:15.400 ","End":"04:18.795","Text":"Also recall that we had w(x,"},{"Start":"04:18.795 ","End":"04:20.790","Text":"t) is cosine x,"},{"Start":"04:20.790 ","End":"04:24.045","Text":"so w(x,3) is cosine x."},{"Start":"04:24.045 ","End":"04:26.320","Text":"Now, we can compute u(x,3),"},{"Start":"04:26.320 ","End":"04:28.010","Text":"which we need for the integral,"},{"Start":"04:28.010 ","End":"04:31.495","Text":"it\u0027s just cosine x on this interval."},{"Start":"04:31.495 ","End":"04:37.219","Text":"The integral that we need is the integral of cosine squared x, dx."},{"Start":"04:37.219 ","End":"04:41.465","Text":"Because we have an even function here of x,"},{"Start":"04:41.465 ","End":"04:46.055","Text":"we can take it from 0-2 and double it using"},{"Start":"04:46.055 ","End":"04:51.500","Text":"trigonometry 2 cosine squared is 1 + cosine 2x."},{"Start":"04:51.500 ","End":"04:56.640","Text":"The integral of this is x + 1/2 sine 2x."},{"Start":"04:56.640 ","End":"05:00.090","Text":"You plug in x= 0, both of these is 0."},{"Start":"05:00.090 ","End":"05:01.380","Text":"When we put x=2,"},{"Start":"05:01.380 ","End":"05:04.435","Text":"we get 2 plus 1/2 sine 4,"},{"Start":"05:04.435 ","End":"05:07.680","Text":"and that\u0027s the answer and we\u0027re done."}],"ID":30895},{"Watched":false,"Name":"Exercise 31","Duration":"2m 56s","ChapterTopicVideoID":29294,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.990","Text":"In this exercise, we\u0027re given the following boundary value problem and is it"},{"Start":"00:06.990 ","End":"00:13.455","Text":"possible that this double integral equals 3 Pi over 2 What do we have here?"},{"Start":"00:13.455 ","End":"00:15.375","Text":"This is on the unit disk."},{"Start":"00:15.375 ","End":"00:19.350","Text":"This is the open unit disc and the boundary is the unit circle."},{"Start":"00:19.350 ","End":"00:27.150","Text":"We\u0027re given that the Laplace equation holds inside the unit disc and on the boundary,"},{"Start":"00:27.150 ","End":"00:31.200","Text":"u is equal to x^2 plus xy plus y^2."},{"Start":"00:31.200 ","End":"00:35.000","Text":"The question is, is it possible that the integral of"},{"Start":"00:35.000 ","End":"00:40.145","Text":"the solution on the unit disk is equal to 3 Pi over 2?"},{"Start":"00:40.145 ","End":"00:42.665","Text":"Here we\u0027ll use the maximum principle,"},{"Start":"00:42.665 ","End":"00:48.245","Text":"on the unit disc and that will help us to show that this is actually not possible."},{"Start":"00:48.245 ","End":"00:51.740","Text":"If x, y is inside the disk,"},{"Start":"00:51.740 ","End":"00:54.620","Text":"meaning x^2 plus y^2 is equal less than 1,"},{"Start":"00:54.620 ","End":"00:56.840","Text":"then by the maximum principle,"},{"Start":"00:56.840 ","End":"01:02.285","Text":"u at this point is less than or equal to the maximum on the boundary."},{"Start":"01:02.285 ","End":"01:06.695","Text":"The maximum principle also says that if u is not constant,"},{"Start":"01:06.695 ","End":"01:09.635","Text":"then we can take a strict inequality here,"},{"Start":"01:09.635 ","End":"01:12.915","Text":"that u(x, y) is strictly less than the maximum."},{"Start":"01:12.915 ","End":"01:17.495","Text":"Just copied that here and now we\u0027re going to figure out what is this maximum?"},{"Start":"01:17.495 ","End":"01:22.430","Text":"Well, this is the maximum of x^2 plus xy plus y^2 on the unit circle."},{"Start":"01:22.430 ","End":"01:24.980","Text":"X^2 plus y^2 is 1,"},{"Start":"01:24.980 ","End":"01:26.975","Text":"so this is 1 plus xy."},{"Start":"01:26.975 ","End":"01:32.790","Text":"We can bring the 1 in front of the maximum so we have 1 plus the maximum of x, y."},{"Start":"01:32.790 ","End":"01:34.505","Text":"Now, on the unit circle,"},{"Start":"01:34.505 ","End":"01:38.315","Text":"x equals cosine Theta and y equals sine Theta."},{"Start":"01:38.315 ","End":"01:42.055","Text":"On the circle, Theta goes from 0-2 Pi."},{"Start":"01:42.055 ","End":"01:48.840","Text":"Now, cosine Theta sine Theta is half of sine 2 Theta."},{"Start":"01:48.840 ","End":"01:53.025","Text":"The maximum sine 2 Theta is 1."},{"Start":"01:53.025 ","End":"01:56.280","Text":"We get 1 plus a half of 1,"},{"Start":"01:56.280 ","End":"01:58.470","Text":"and that means that u(x,"},{"Start":"01:58.470 ","End":"02:01.230","Text":"y) is less than this,"},{"Start":"02:01.230 ","End":"02:03.285","Text":"which is 3 over 2."},{"Start":"02:03.285 ","End":"02:06.255","Text":"Now let\u0027s turn to the integral that we had."},{"Start":"02:06.255 ","End":"02:08.475","Text":"The integral of u(x,"},{"Start":"02:08.475 ","End":"02:16.330","Text":"y) on the open unit disc is less than 3 over 2 times the integral of 1."},{"Start":"02:16.330 ","End":"02:21.230","Text":"If one function is strictly less than another on a domain,"},{"Start":"02:21.230 ","End":"02:24.305","Text":"then the integral is going to be strictly less than."},{"Start":"02:24.305 ","End":"02:28.415","Text":"The integral of 1 on a domain is just the area of a domain."},{"Start":"02:28.415 ","End":"02:33.690","Text":"We need the area of the unit disk using the Pi r^2 formula,"},{"Start":"02:33.690 ","End":"02:36.465","Text":"it\u0027s Pi times 1^2, which is Pi."},{"Start":"02:36.465 ","End":"02:40.120","Text":"What we get here is 3 over 2 Pi."},{"Start":"02:40.120 ","End":"02:44.470","Text":"Now, if this integral is less than 3 over 2 Pi,"},{"Start":"02:44.470 ","End":"02:47.950","Text":"then it\u0027s not equal to 3 over 2 Pi."},{"Start":"02:47.950 ","End":"02:50.990","Text":"We\u0027ve just shown that the answer is no,"},{"Start":"02:50.990 ","End":"02:54.585","Text":"that this integral cannot equal 3 over 2 Pi."},{"Start":"02:54.585 ","End":"02:57.490","Text":"That concludes this exercise."}],"ID":30896},{"Watched":false,"Name":"Exercise 32","Duration":"3m 12s","ChapterTopicVideoID":29295,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.130","Text":"In this exercise, we\u0027re given the following initial boundary value problem."},{"Start":"00:05.130 ","End":"00:11.100","Text":"It\u0027s a heat equation on a finite interval with Dirichlet boundary conditions."},{"Start":"00:11.100 ","End":"00:15.480","Text":"Our task is to prove that a solution u at the point 3/2,"},{"Start":"00:15.480 ","End":"00:17.730","Text":"1 is less than 2e."},{"Start":"00:17.730 ","End":"00:21.405","Text":"We\u0027ll do this without actually computing the solution u."},{"Start":"00:21.405 ","End":"00:27.075","Text":"The way we\u0027ll do it is using the maximum principle for the heat equation."},{"Start":"00:27.075 ","End":"00:29.310","Text":"Let\u0027s note that the point 3/2,"},{"Start":"00:29.310 ","End":"00:33.900","Text":"1 is in the interior of the rectangle x from 0-3,"},{"Start":"00:33.900 ","End":"00:35.190","Text":"t from 0-2,"},{"Start":"00:35.190 ","End":"00:37.545","Text":"which smack in the middle in fact."},{"Start":"00:37.545 ","End":"00:41.745","Text":"By the maximum principle for the heat equation,"},{"Start":"00:41.745 ","End":"00:49.010","Text":"the function u attains its maximum on the parabolic boundary of this rectangle."},{"Start":"00:49.010 ","End":"00:52.460","Text":"Let\u0027s see the picture. Here we are."},{"Start":"00:52.460 ","End":"00:59.750","Text":"This is the rectangle and what\u0027s highlighted is the parabolic boundary."},{"Start":"00:59.750 ","End":"01:04.280","Text":"This is our point and these are the 2 boundary conditions,"},{"Start":"01:04.280 ","End":"01:06.725","Text":"and this is the initial condition."},{"Start":"01:06.725 ","End":"01:09.330","Text":"The boundary consists of 3 line segments."},{"Start":"01:09.330 ","End":"01:13.265","Text":"What we\u0027ll do is we\u0027ll take the maximum on each of these,"},{"Start":"01:13.265 ","End":"01:15.350","Text":"it\u0027s easier to compute separately,"},{"Start":"01:15.350 ","End":"01:17.615","Text":"and then take the largest of the 3."},{"Start":"01:17.615 ","End":"01:23.445","Text":"Let\u0027s start with the 1 here which is the left endpoint."},{"Start":"01:23.445 ","End":"01:29.020","Text":"We get that the maximum of u here is the maximum of 3 over root 1 plus t^2."},{"Start":"01:29.020 ","End":"01:35.095","Text":"On this interval, the fraction is maximum when the denominator is minimal,"},{"Start":"01:35.095 ","End":"01:41.795","Text":"and the smallest possible denominator will be when t^2 is 0 which is in our interval."},{"Start":"01:41.795 ","End":"01:45.860","Text":"When t is 0, we get 3 over root 1 is 3."},{"Start":"01:45.860 ","End":"01:49.600","Text":"The next part will be this part here."},{"Start":"01:49.600 ","End":"01:54.530","Text":"The maximum of this upside-down parabola,"},{"Start":"01:54.530 ","End":"01:56.000","Text":"on the interval from 0-3,"},{"Start":"01:56.000 ","End":"01:59.350","Text":"turns out to be 21/4. You know how to do this?"},{"Start":"01:59.350 ","End":"02:01.820","Text":"But, I\u0027ll tell you him quickly in case you\u0027ve forgotten."},{"Start":"02:01.820 ","End":"02:03.335","Text":"We differentiate this,"},{"Start":"02:03.335 ","End":"02:05.030","Text":"you get 3 minus 2x,"},{"Start":"02:05.030 ","End":"02:08.135","Text":"set that to 0, x is 1.5."},{"Start":"02:08.135 ","End":"02:10.610","Text":"Plug in x=1.5,"},{"Start":"02:10.610 ","End":"02:15.945","Text":"and we get 3 plus 4.5 minus 2 and a 1/4,"},{"Start":"02:15.945 ","End":"02:19.635","Text":"which is 5, and a 1/4 which is 21/4."},{"Start":"02:19.635 ","End":"02:22.940","Text":"It\u0027s the maximum because it\u0027s an upside-down parabola, or if you like,"},{"Start":"02:22.940 ","End":"02:27.160","Text":"the second derivative, which is minus 2, is negative."},{"Start":"02:27.160 ","End":"02:30.045","Text":"On the last of the 3 segments,"},{"Start":"02:30.045 ","End":"02:31.680","Text":"the function is constant,"},{"Start":"02:31.680 ","End":"02:32.880","Text":"it\u0027s equal to 3,"},{"Start":"02:32.880 ","End":"02:34.635","Text":"so the maximum is 3."},{"Start":"02:34.635 ","End":"02:36.340","Text":"Now to these 3,"},{"Start":"02:36.340 ","End":"02:40.025","Text":"the biggest is this because it\u0027s bigger than 3,"},{"Start":"02:40.025 ","End":"02:49.925","Text":"and so this is the maximum value of the function on the parabolic boundary."},{"Start":"02:49.925 ","End":"02:52.160","Text":"At the interior point,"},{"Start":"02:52.160 ","End":"02:55.145","Text":"the function is going to be less than this."},{"Start":"02:55.145 ","End":"02:57.800","Text":"Normally you\u0027d say less than or equal to,"},{"Start":"02:57.800 ","End":"03:01.970","Text":"but if it\u0027s not a constant function then it\u0027s strictly less than."},{"Start":"03:01.970 ","End":"03:05.425","Text":"21/4 is less than 2e."},{"Start":"03:05.425 ","End":"03:07.265","Text":"If you\u0027re not sure about that,"},{"Start":"03:07.265 ","End":"03:09.665","Text":"I left you the computations here,"},{"Start":"03:09.665 ","End":"03:11.045","Text":"you can look at this,"},{"Start":"03:11.045 ","End":"03:13.350","Text":"but we are done."}],"ID":30897},{"Watched":false,"Name":"Exercise 33","Duration":"2m 32s","ChapterTopicVideoID":29296,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.350","Text":"In this exercise, we\u0027re given the following problem to solve."},{"Start":"00:04.350 ","End":"00:08.280","Text":"The PDE is a heat equation on a finite interval."},{"Start":"00:08.280 ","End":"00:14.415","Text":"We have an initial condition and a couple of boundary conditions."},{"Start":"00:14.415 ","End":"00:18.645","Text":"We have to prove that u at the point Pi over 2,"},{"Start":"00:18.645 ","End":"00:22.875","Text":"1 is bigger than minus Pi^2 over 4"},{"Start":"00:22.875 ","End":"00:27.615","Text":"and I\u0027m going to do this without actually computing the solution u(x, t)."},{"Start":"00:27.615 ","End":"00:32.280","Text":"The plan is to use the minimum principle for the heat equation,"},{"Start":"00:32.280 ","End":"00:37.440","Text":"not the maximum principle because you have to prove that it\u0027s bigger than something."},{"Start":"00:37.440 ","End":"00:42.655","Text":"You\u0027ll see. This point is in the interior of a rectangle,"},{"Start":"00:42.655 ","End":"00:45.620","Text":"we took one that\u0027s way too big, but it\u0027s okay."},{"Start":"00:45.620 ","End":"00:50.100","Text":"Pi over 2 is between 0 and Pi and 1 is between 0 and 2,"},{"Start":"00:50.100 ","End":"00:55.670","Text":"so we can take this rectangle and the minimum principle says that"},{"Start":"00:55.670 ","End":"01:02.885","Text":"the minimum for the rectangle"},{"Start":"01:02.885 ","End":"01:06.265","Text":"is achieved on the parabolic boundary."},{"Start":"01:06.265 ","End":"01:09.455","Text":"To remind you what the parabolic boundary is,"},{"Start":"01:09.455 ","End":"01:12.320","Text":"it\u0027s the bottom and sides without the top."},{"Start":"01:12.320 ","End":"01:15.640","Text":"The boundary consists of three line segments,"},{"Start":"01:15.640 ","End":"01:16.980","Text":"1, 2,"},{"Start":"01:16.980 ","End":"01:20.690","Text":"3 and we\u0027re going to find the minimum of u on"},{"Start":"01:20.690 ","End":"01:24.665","Text":"each of these three and then take the smallest of the three minima,"},{"Start":"01:24.665 ","End":"01:28.295","Text":"the first one, where t goes from 0-2,"},{"Start":"01:28.295 ","End":"01:31.070","Text":"u( 0, t) and so on."},{"Start":"01:31.070 ","End":"01:38.060","Text":"This one is equal to arctangent of t and the least is when t equals 0,"},{"Start":"01:38.060 ","End":"01:43.130","Text":"the arctangent is an increasing function and when t is 0, it\u0027s 0."},{"Start":"01:43.130 ","End":"01:46.535","Text":"Next one is x, x minus Pi,"},{"Start":"01:46.535 ","End":"01:52.354","Text":"it\u0027s a parabola with zeros at 0 and Pi."},{"Start":"01:52.354 ","End":"01:57.090","Text":"The minimum is at Pi over 2 and at Pi over 2,"},{"Start":"01:57.090 ","End":"02:00.285","Text":"it\u0027s equal to minus Pi^2 over 4."},{"Start":"02:00.285 ","End":"02:02.605","Text":"Then here u is a constant 0,"},{"Start":"02:02.605 ","End":"02:04.685","Text":"so the minimum is also 0."},{"Start":"02:04.685 ","End":"02:06.350","Text":"Now from these three,"},{"Start":"02:06.350 ","End":"02:08.810","Text":"the least is the only negative 1,"},{"Start":"02:08.810 ","End":"02:14.030","Text":"which is this and so the value of u here is going to be"},{"Start":"02:14.030 ","End":"02:19.745","Text":"bigger or equal to the value on the parabolic boundary."},{"Start":"02:19.745 ","End":"02:22.940","Text":"But because the function u is non constant,"},{"Start":"02:22.940 ","End":"02:25.565","Text":"we can take strictly bigger than,"},{"Start":"02:25.565 ","End":"02:28.630","Text":"rather than just bigger or equal to."},{"Start":"02:28.630 ","End":"02:32.820","Text":"That\u0027s what we had to show and we\u0027re done."}],"ID":30898},{"Watched":false,"Name":"Exercise 34","Duration":"3m 46s","ChapterTopicVideoID":29297,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.539","Text":"In this exercise, we\u0027re given the following problem,"},{"Start":"00:03.539 ","End":"00:07.570","Text":"which is a wave equation on a semi-infinite interval,"},{"Start":"00:07.570 ","End":"00:10.410","Text":"and were asked not to solve it,"},{"Start":"00:10.410 ","End":"00:13.110","Text":"but just to compute this integral,"},{"Start":"00:13.110 ","End":"00:18.960","Text":"we have here semi-infinite interval with anointment boundary condition,"},{"Start":"00:18.960 ","End":"00:26.370","Text":"so what we learned to do is to find an even function completion to the whole real line."},{"Start":"00:26.370 ","End":"00:30.380","Text":"Turns out that if we take an extension of"},{"Start":"00:30.380 ","End":"00:35.060","Text":"f and g to the interval from minus infinity to infinity,"},{"Start":"00:35.060 ","End":"00:38.525","Text":"then the solution will be the even extension of view."},{"Start":"00:38.525 ","End":"00:42.895","Text":"Let\u0027s use tilde notation for the even extension."},{"Start":"00:42.895 ","End":"00:46.440","Text":"F tilde will be like f,"},{"Start":"00:46.440 ","End":"00:50.120","Text":"but the domain has changed."},{"Start":"00:50.120 ","End":"00:56.270","Text":"This will be for minus 1 to 1 and this will be everywhere outside minus 1 to 1 or yeah,"},{"Start":"00:56.270 ","End":"00:59.195","Text":"there\u0027s a picture. This is what I mean."},{"Start":"00:59.195 ","End":"01:02.195","Text":"We had initially just this part."},{"Start":"01:02.195 ","End":"01:05.315","Text":"Then we took a mirror image so we get this."},{"Start":"01:05.315 ","End":"01:07.120","Text":"It\u0027s the same formula,"},{"Start":"01:07.120 ","End":"01:11.990","Text":"the 1 minus x^2 because this is even and g tilde is 0,"},{"Start":"01:11.990 ","End":"01:13.910","Text":"just like g was 0."},{"Start":"01:13.910 ","End":"01:16.924","Text":"That\u0027s an even function on the whole interval."},{"Start":"01:16.924 ","End":"01:20.750","Text":"Now, we\u0027re going to assume that x is bigger or equal to 0,"},{"Start":"01:20.750 ","End":"01:26.900","Text":"and then u is the same as u tilde and we can use Poisson\u0027s formula and u tilde,"},{"Start":"01:26.900 ","End":"01:31.415","Text":"but assign it to u and we have the following formula."},{"Start":"01:31.415 ","End":"01:33.710","Text":"A is equal to 1,"},{"Start":"01:33.710 ","End":"01:36.384","Text":"g tilde is 0,"},{"Start":"01:36.384 ","End":"01:43.445","Text":"so what we\u0027re left with is the following and we really only need u of x comma 1."},{"Start":"01:43.445 ","End":"01:50.515","Text":"Let\u0027s t equals 1 and we get this also a reminder as to what f is."},{"Start":"01:50.515 ","End":"01:54.780","Text":"We need f tilde of x plus 1 and f tilde of x minus 1."},{"Start":"01:54.780 ","End":"01:58.745","Text":"Just replace x by x plus 1 here."},{"Start":"01:58.745 ","End":"02:00.275","Text":"This is what we get."},{"Start":"02:00.275 ","End":"02:02.750","Text":"This inequality, we can change it,"},{"Start":"02:02.750 ","End":"02:05.870","Text":"just subtract 1 everywhere and we get,"},{"Start":"02:05.870 ","End":"02:11.695","Text":"this is the formula from minus 2 to 0 and 0 otherwise."},{"Start":"02:11.695 ","End":"02:14.860","Text":"For f tilde of x minus 1,"},{"Start":"02:14.860 ","End":"02:16.385","Text":"we get the following."},{"Start":"02:16.385 ","End":"02:20.030","Text":"Now, when x is between 0 and 2,"},{"Start":"02:20.030 ","End":"02:21.665","Text":"which is what we need,"},{"Start":"02:21.665 ","End":"02:28.370","Text":"between 0 and 2 then here we\u0027ll take this formula and here we take this formula,"},{"Start":"02:28.370 ","End":"02:34.520","Text":"the ones that I\u0027ve colored so that between 0 and 2 u of x,"},{"Start":"02:34.520 ","End":"02:37.510","Text":"1 is equal to the following."},{"Start":"02:37.510 ","End":"02:44.640","Text":"The 0 from here plus this from here over 2,"},{"Start":"02:44.640 ","End":"02:48.065","Text":"I\u0027m talking about this here."},{"Start":"02:48.065 ","End":"02:51.010","Text":"Just summarizing so far we have that u of x,"},{"Start":"02:51.010 ","End":"02:53.740","Text":"1 is the following function,"},{"Start":"02:53.740 ","End":"02:58.375","Text":"where x is between 0 and 2 and we need to calculate this integral."},{"Start":"02:58.375 ","End":"03:00.220","Text":"Let\u0027s just substitute u of x,"},{"Start":"03:00.220 ","End":"03:04.255","Text":"1 in here and then we get the following."},{"Start":"03:04.255 ","End":"03:08.260","Text":"Minus 1/2 and absolute value squared integral,"},{"Start":"03:08.260 ","End":"03:12.490","Text":"the 1/2 cancels with the 1/2 so this is what we have."},{"Start":"03:12.490 ","End":"03:14.360","Text":"You don\u0027t need the minus,"},{"Start":"03:14.360 ","End":"03:16.980","Text":"just raise this to the power of 2,"},{"Start":"03:16.980 ","End":"03:19.714","Text":"so we get x minus 1 to the fourth,"},{"Start":"03:19.714 ","End":"03:21.650","Text":"and 1/2^2, which is a quarter,"},{"Start":"03:21.650 ","End":"03:23.570","Text":"comes in front of the integral."},{"Start":"03:23.570 ","End":"03:28.775","Text":"Now the integral of this raise the power by 1 is 5 divided by 5,"},{"Start":"03:28.775 ","End":"03:32.510","Text":"we get 1/20 x minus 1 to the fifth."},{"Start":"03:32.510 ","End":"03:36.846","Text":"This we evaluate between 0 and 2."},{"Start":"03:36.846 ","End":"03:37.950","Text":"When x is 2,"},{"Start":"03:37.950 ","End":"03:42.285","Text":"we get 1^5 and x is 0 minus 1^5."},{"Start":"03:42.285 ","End":"03:45.255","Text":"The answer comes out to be 1/10,"},{"Start":"03:45.255 ","End":"03:47.500","Text":"and we are done."}],"ID":30899},{"Watched":false,"Name":"Exercise 35","Duration":"3m 39s","ChapterTopicVideoID":29298,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.310","Text":"In this exercise, we\u0027re given the following boundary value problem."},{"Start":"00:05.310 ","End":"00:13.185","Text":"The Laplacian of u is equal to r on the unit disk and u on the boundary,"},{"Start":"00:13.185 ","End":"00:16.440","Text":"which is the unit circle is equal to this."},{"Start":"00:16.440 ","End":"00:20.085","Text":"We have to compute u at the origin."},{"Start":"00:20.085 ","End":"00:23.340","Text":"This of course is given in polar coordinates."},{"Start":"00:23.340 ","End":"00:27.930","Text":"What we\u0027ll do is we\u0027ll split u up into 2 pieces w"},{"Start":"00:27.930 ","End":"00:33.075","Text":"plus v. V will have a homogeneous boundary condition,"},{"Start":"00:33.075 ","End":"00:36.540","Text":"w will have a homogeneous PDE."},{"Start":"00:36.540 ","End":"00:39.855","Text":"In fact, we\u0027ll look for a radial function v,"},{"Start":"00:39.855 ","End":"00:44.870","Text":"which just depends on r. It will work out as you\u0027ll see."},{"Start":"00:44.870 ","End":"00:47.870","Text":"Now let\u0027s go for v first."},{"Start":"00:47.870 ","End":"00:56.425","Text":"The Laplacian of v in polar coordinates is this is equal to r and v(1) equals 0."},{"Start":"00:56.425 ","End":"00:59.625","Text":"Now, if v is a function just of r,"},{"Start":"00:59.625 ","End":"01:06.904","Text":"then this part is 0 and the derivative of v with respect to r is just a plain derivative."},{"Start":"01:06.904 ","End":"01:11.070","Text":"We get v\u0027\u0027(r), that\u0027s 1 over r,"},{"Start":"01:11.070 ","End":"01:15.830","Text":"v\u0027(r) equals r, multiply both sides by r,"},{"Start":"01:15.830 ","End":"01:17.250","Text":"and this is what we get."},{"Start":"01:17.250 ","End":"01:22.445","Text":"The left-hand side now is the derivative of a product."},{"Start":"01:22.445 ","End":"01:24.440","Text":"We\u0027ve seen this before. You can check it out by"},{"Start":"01:24.440 ","End":"01:27.530","Text":"differentiating this using the product rule."},{"Start":"01:27.530 ","End":"01:29.195","Text":"This is equal to r^2."},{"Start":"01:29.195 ","End":"01:32.900","Text":"Now integrate both sides and we\u0027ve got rv\u0027"},{"Start":"01:32.900 ","End":"01:39.260","Text":"equals the integral of this r^3 over 3 plus constant of integration."},{"Start":"01:39.260 ","End":"01:45.065","Text":"Substitute r=0, this shows that a is equal to 0."},{"Start":"01:45.065 ","End":"01:49.370","Text":"In other words, v\u0027 is equal to r^2 over 3."},{"Start":"01:49.370 ","End":"01:53.555","Text":"We just divided both sides by r and then integrate again."},{"Start":"01:53.555 ","End":"01:56.270","Text":"We\u0027ve got v(r) is r^3 over 9,"},{"Start":"01:56.270 ","End":"01:59.335","Text":"plus another constant of integration."},{"Start":"01:59.335 ","End":"02:04.220","Text":"We\u0027ll find c by substituting r=1."},{"Start":"02:04.220 ","End":"02:07.855","Text":"We have that v(1) equals 0."},{"Start":"02:07.855 ","End":"02:13.395","Text":"We\u0027ve got 0 equals 1/9 plus c. C is minus a 1/9."},{"Start":"02:13.395 ","End":"02:17.415","Text":"That means that v(r) is r^3 over 9 minus 1/9,"},{"Start":"02:17.415 ","End":"02:19.800","Text":"which is r^3 minus 1 over 9."},{"Start":"02:19.800 ","End":"02:22.410","Text":"Now we can substitute r=0."},{"Start":"02:22.410 ","End":"02:25.810","Text":"We\u0027ve got v(0) equals minus a 1/9."},{"Start":"02:25.810 ","End":"02:29.390","Text":"Next, we\u0027ll find w(0,0)."},{"Start":"02:29.390 ","End":"02:34.600","Text":"What we\u0027re going to do here is use the mean value property,"},{"Start":"02:34.600 ","End":"02:37.430","Text":"which says that the harmonic function,"},{"Start":"02:37.430 ","End":"02:39.710","Text":"and w is harmonic here,"},{"Start":"02:39.710 ","End":"02:41.420","Text":"has the mean value property,"},{"Start":"02:41.420 ","End":"02:43.160","Text":"which means that, for example,"},{"Start":"02:43.160 ","End":"02:52.565","Text":"w at the origin is the average of w along the unit circle centered at the origin."},{"Start":"02:52.565 ","End":"02:56.320","Text":"The value of w on the circle is this."},{"Start":"02:56.320 ","End":"03:00.620","Text":"We\u0027re going to use the integral from minus Pi to Pi instead of"},{"Start":"03:00.620 ","End":"03:04.880","Text":"from 0-2 Pi because a periodic function with period 2 Pi,"},{"Start":"03:04.880 ","End":"03:08.195","Text":"we can take any interval of length 2 Pi."},{"Start":"03:08.195 ","End":"03:12.600","Text":"The reason we\u0027re doing this is because this is a symmetric interval."},{"Start":"03:12.600 ","End":"03:15.740","Text":"If we have an odd function, which we do,"},{"Start":"03:15.740 ","End":"03:22.820","Text":"then the integral is just equal to 0, u(0,0),"},{"Start":"03:22.820 ","End":"03:27.350","Text":"which is w(0,0) plus v(0) is equal to,"},{"Start":"03:27.350 ","End":"03:29.570","Text":"we have that this is 0 from here,"},{"Start":"03:29.570 ","End":"03:34.425","Text":"and v(0) was minus a 1/9."},{"Start":"03:34.425 ","End":"03:39.760","Text":"The answer is minus 1/9 and that completes this exercise."}],"ID":30900},{"Watched":false,"Name":"Exercise 36","Duration":"5m 24s","ChapterTopicVideoID":29299,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.640","Text":"In this exercise, we\u0027re given"},{"Start":"00:02.640 ","End":"00:06.975","Text":"the following equation with initial condition and we have to solve it."},{"Start":"00:06.975 ","End":"00:11.819","Text":"This happens to be quasi-linear of the first-order, which in general,"},{"Start":"00:11.819 ","End":"00:14.640","Text":"is of this form where the coefficients a,"},{"Start":"00:14.640 ","End":"00:18.180","Text":"b, and c are functions of x, y, and u."},{"Start":"00:18.180 ","End":"00:20.585","Text":"In this case, a is 1,"},{"Start":"00:20.585 ","End":"00:22.555","Text":"b is 1, and c,"},{"Start":"00:22.555 ","End":"00:24.615","Text":"if you bring the u over to the other side,"},{"Start":"00:24.615 ","End":"00:28.185","Text":"is u plus cosine of x minus y."},{"Start":"00:28.185 ","End":"00:30.690","Text":"We\u0027ll follow the steps as in the tutorial."},{"Start":"00:30.690 ","End":"00:36.225","Text":"First, we\u0027ll find a parameterization of the initial curve, Gamma."},{"Start":"00:36.225 ","End":"00:41.265","Text":"We can see that the initial condition is given at the points x,0,"},{"Start":"00:41.265 ","End":"00:42.915","Text":"which is the x-axis."},{"Start":"00:42.915 ","End":"00:44.030","Text":"Along the x-axis,"},{"Start":"00:44.030 ","End":"00:45.815","Text":"the solution is equal to 0."},{"Start":"00:45.815 ","End":"00:53.590","Text":"We take this solution along the x axis to be where t=0 and we still have to vary Tau."},{"Start":"00:53.590 ","End":"00:56.690","Text":"We want the solution in terms of t and Tau."},{"Start":"00:56.690 ","End":"00:59.015","Text":"x is just Tau,"},{"Start":"00:59.015 ","End":"01:03.620","Text":"and then y is 0 and u is 0."},{"Start":"01:03.620 ","End":"01:07.385","Text":"In fact, Gamma is just the x-axis in 3D space."},{"Start":"01:07.385 ","End":"01:10.775","Text":"We know that the solution curve will pass through the x-axis."},{"Start":"01:10.775 ","End":"01:18.605","Text":"Now the second step is to find characteristic curves by solving the characteristic ODEs."},{"Start":"01:18.605 ","End":"01:21.185","Text":"There is a third ODE,"},{"Start":"01:21.185 ","End":"01:23.370","Text":"which is ut=c,"},{"Start":"01:23.370 ","End":"01:25.500","Text":"but we use that in Step 3."},{"Start":"01:25.500 ","End":"01:28.010","Text":"Substituting a and b from here,"},{"Start":"01:28.010 ","End":"01:29.495","Text":"they\u0027re both equal to 1,"},{"Start":"01:29.495 ","End":"01:34.015","Text":"we get dx by dt is 1 and dy by dt is 1."},{"Start":"01:34.015 ","End":"01:37.400","Text":"We solve this, you integrate with respect to t,"},{"Start":"01:37.400 ","End":"01:42.275","Text":"we get that x is t plus an arbitrary function of Tau."},{"Start":"01:42.275 ","End":"01:44.044","Text":"It\u0027s like the constant."},{"Start":"01:44.044 ","End":"01:48.265","Text":"Y is t plus another function of Tau."},{"Start":"01:48.265 ","End":"01:51.840","Text":"That gives us, when we let t=0,"},{"Start":"01:51.840 ","End":"01:54.300","Text":"this is 0, this is 0, we get x of 0,"},{"Start":"01:54.300 ","End":"01:57.570","Text":"Tau is c1, and y of 0 T is c2 Tau."},{"Start":"01:57.570 ","End":"01:59.855","Text":"We know what these two are,"},{"Start":"01:59.855 ","End":"02:02.105","Text":"because just copying from here,"},{"Start":"02:02.105 ","End":"02:04.175","Text":"x of 0 Tau is Tau,"},{"Start":"02:04.175 ","End":"02:06.865","Text":"and y of 0 Tau is 0."},{"Start":"02:06.865 ","End":"02:12.630","Text":"So we have Tau=c1 of Tau and 0=c2 of Tau. It\u0027s right there."},{"Start":"02:12.630 ","End":"02:16.775","Text":"That gives us, by substituting c1 and c2 here and here,"},{"Start":"02:16.775 ","End":"02:24.395","Text":"we get that x is t plus Tau and y is just t. This gives us x and y in terms of t and Tau."},{"Start":"02:24.395 ","End":"02:29.210","Text":"Now, the third step is to take that third characteristic equation, which I mentioned."},{"Start":"02:29.210 ","End":"02:32.360","Text":"That\u0027s du by dt=c,"},{"Start":"02:32.360 ","End":"02:34.310","Text":"and c, if you recall,"},{"Start":"02:34.310 ","End":"02:38.060","Text":"is equal to u plus cosine of x minus y."},{"Start":"02:38.060 ","End":"02:42.740","Text":"That gives us that du by dt=u plus cosine of x minus y."},{"Start":"02:42.740 ","End":"02:47.525","Text":"We have the initial condition that u of 0 Tau is 0."},{"Start":"02:47.525 ","End":"02:53.420","Text":"Subtract u from both sides here and we have u t minus u is cosine of x minus y."},{"Start":"02:53.420 ","End":"02:57.260","Text":"We can substitute x and y in terms of t and Tau from here."},{"Start":"02:57.260 ","End":"02:59.495","Text":"What we get is x is t plus Tau,"},{"Start":"02:59.495 ","End":"03:03.160","Text":"y is t, t plus Tau minus t is just Tau."},{"Start":"03:03.160 ","End":"03:06.170","Text":"We have this first-order differential equation,"},{"Start":"03:06.170 ","End":"03:08.690","Text":"du by dt minus u is cosine Tau."},{"Start":"03:08.690 ","End":"03:10.085","Text":"Many ways to solve this,"},{"Start":"03:10.085 ","End":"03:12.245","Text":"we\u0027ll use the integrating factor."},{"Start":"03:12.245 ","End":"03:15.110","Text":"Integrating factor is e to the minus t,"},{"Start":"03:15.110 ","End":"03:17.625","Text":"the minus 1 from the coefficient here."},{"Start":"03:17.625 ","End":"03:19.800","Text":"Multiply both sides."},{"Start":"03:19.800 ","End":"03:25.700","Text":"Then the left-hand side is exact derivative of e to the minus t times u."},{"Start":"03:25.700 ","End":"03:28.910","Text":"And then we can integrate both sides with respect to t and we get e to"},{"Start":"03:28.910 ","End":"03:35.825","Text":"the minus tu is minus e to the minus t cosine Tau plus,"},{"Start":"03:35.825 ","End":"03:39.320","Text":"like a constant with an arbitrary function of Tau,"},{"Start":"03:39.320 ","End":"03:42.230","Text":"multiply both sides by e to the t,"},{"Start":"03:42.230 ","End":"03:48.650","Text":"and we get that u equals minus cosine Tau plus e to the t, c of Tau."},{"Start":"03:48.650 ","End":"03:53.700","Text":"Just copying that and reminding you of the initial condition."},{"Start":"03:54.130 ","End":"03:58.005","Text":"Letting t=0 in both sides,"},{"Start":"03:58.005 ","End":"04:01.635","Text":"here we get 0 from the initial condition,"},{"Start":"04:01.635 ","End":"04:08.030","Text":"and here, we get minus cosine Tau plus e to the 0c of Tau."},{"Start":"04:08.030 ","End":"04:11.385","Text":"All together, this is 1,"},{"Start":"04:11.385 ","End":"04:15.165","Text":"move to the other side, we get c of Tau is cosine Tau."},{"Start":"04:15.165 ","End":"04:18.480","Text":"Substituting c of Tau here,"},{"Start":"04:18.480 ","End":"04:24.310","Text":"we get that ut Tau is minus cosine Tau plus e to the t cosine Tau."},{"Start":"04:24.520 ","End":"04:28.310","Text":"Simplify this by taking cosine Tau outside"},{"Start":"04:28.310 ","End":"04:32.435","Text":"the bracket and we get e to the t minus 1 cosine Tau."},{"Start":"04:32.435 ","End":"04:35.630","Text":"So we\u0027ve solved you in terms of t and Tau."},{"Start":"04:35.630 ","End":"04:39.430","Text":"The last step is to get back to x and y."},{"Start":"04:39.430 ","End":"04:42.395","Text":"Now, we have x and y in terms of t and Tau."},{"Start":"04:42.395 ","End":"04:44.510","Text":"We want the other way round."},{"Start":"04:44.510 ","End":"04:47.480","Text":"Optionally, you could check the determinant of"},{"Start":"04:47.480 ","End":"04:51.430","Text":"the Jacobian to make sure that it\u0027s not 0,"},{"Start":"04:51.430 ","End":"04:54.560","Text":"but in any event, if we succeed in inverting this,"},{"Start":"04:54.560 ","End":"04:56.615","Text":"we\u0027ll know that\u0027s the case."},{"Start":"04:56.615 ","End":"04:58.715","Text":"From the second equation,"},{"Start":"04:58.715 ","End":"05:01.190","Text":"we get that t=y."},{"Start":"05:01.190 ","End":"05:03.155","Text":"If we put t=y here,"},{"Start":"05:03.155 ","End":"05:05.118","Text":"we get x=y plus Tau."},{"Start":"05:05.118 ","End":"05:06.585","Text":"Tau is x minus y."},{"Start":"05:06.585 ","End":"05:07.700","Text":"We have inverted it."},{"Start":"05:07.700 ","End":"05:11.000","Text":"We\u0027ve got t and Tau in terms of x and y."},{"Start":"05:11.000 ","End":"05:13.675","Text":"Now we can substitute those in here,"},{"Start":"05:13.675 ","End":"05:16.075","Text":"putting t is this, Tau is this."},{"Start":"05:16.075 ","End":"05:21.995","Text":"We get that u of xy equals e to the y minus 1 cosine of x minus y."},{"Start":"05:21.995 ","End":"05:25.110","Text":"That concludes this exercise."}],"ID":30901},{"Watched":false,"Name":"Exercise 37","Duration":"3m 29s","ChapterTopicVideoID":29300,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.105","Text":"In this exercise, there are 3 parts."},{"Start":"00:03.105 ","End":"00:06.735","Text":"In part a, we\u0027re given this equation,"},{"Start":"00:06.735 ","End":"00:09.720","Text":"which is a second-order linear equation,"},{"Start":"00:09.720 ","End":"00:13.590","Text":"and we have to classify it only in this case."},{"Start":"00:13.590 ","End":"00:18.525","Text":"It isn\u0027t always one of the 3 a different regions."},{"Start":"00:18.525 ","End":"00:21.930","Text":"It\u0027s different, so we have to find out where it\u0027s hyperbolic,"},{"Start":"00:21.930 ","End":"00:24.467","Text":"where it\u0027s elliptic, and where it\u0027s parabolic."},{"Start":"00:24.467 ","End":"00:28.580","Text":"In part b, we\u0027ll write the characteristic equations for this PDE,"},{"Start":"00:28.580 ","End":"00:33.740","Text":"and in part c, we\u0027ll sketch the regions where it\u0027s hyperbolic, elliptic,"},{"Start":"00:33.740 ","End":"00:39.320","Text":"and parabolic, and we will also sketch the tangents of these characteristic curves,"},{"Start":"00:39.320 ","End":"00:42.370","Text":"the ones which cross the point 0, 0,"},{"Start":"00:42.370 ","End":"00:45.965","Text":"the origin, so let\u0027s start with part a."},{"Start":"00:45.965 ","End":"00:49.040","Text":"When we classify the only important parts"},{"Start":"00:49.040 ","End":"00:52.010","Text":"are the ones with second-order derivative u_xx,"},{"Start":"00:52.010 ","End":"00:53.770","Text":"u_xy, u_yy,"},{"Start":"00:53.770 ","End":"00:56.270","Text":"and we have a name for these coefficients."},{"Start":"00:56.270 ","End":"00:57.950","Text":"This is a_11,"},{"Start":"00:57.950 ","End":"00:59.960","Text":"this is twice a_12,"},{"Start":"00:59.960 ","End":"01:01.880","Text":"and this is a22,"},{"Start":"01:01.880 ","End":"01:05.615","Text":"the rest are lower-order terms and not used in the classification."},{"Start":"01:05.615 ","End":"01:08.040","Text":"We have therefore that a_11 is 1,"},{"Start":"01:08.040 ","End":"01:09.585","Text":"a_12 is minus 1,"},{"Start":"01:09.585 ","End":"01:11.800","Text":"and a22 is (x plus y)^2."},{"Start":"01:11.800 ","End":"01:15.580","Text":"Now the discriminant Delta is defined as follows."},{"Start":"01:15.580 ","End":"01:21.005","Text":"In our case it comes out to 1^2 minus 1 times x^2 plus y^2."},{"Start":"01:21.005 ","End":"01:24.050","Text":"You have to see where it\u0027s parabolic,"},{"Start":"01:24.050 ","End":"01:29.830","Text":"where it\u0027s hyperbolic, and where it\u0027s elliptic according to the sign of the discriminant."},{"Start":"01:29.830 ","End":"01:32.825","Text":"So equals 0 means this equals 0."},{"Start":"01:32.825 ","End":"01:37.925","Text":"So x^2 plus y^2 is 1 and that\u0027s on the unit circle hyperbolic,"},{"Start":"01:37.925 ","End":"01:40.475","Text":"where the discriminant is positive,"},{"Start":"01:40.475 ","End":"01:45.065","Text":"which means that x^2 plus y^2 less than 1 or inside the circle,"},{"Start":"01:45.065 ","End":"01:48.110","Text":"and it\u0027s elliptic outside the circle."},{"Start":"01:48.110 ","End":"01:52.385","Text":"In part b, we have to write the characteristic equations."},{"Start":"01:52.385 ","End":"01:55.834","Text":"These are given by the following."},{"Start":"01:55.834 ","End":"02:02.645","Text":"Dy/dx is a_12 plus or minus the square root of the discriminant over a_11,"},{"Start":"02:02.645 ","End":"02:03.950","Text":"or if you want it in full,"},{"Start":"02:03.950 ","End":"02:06.650","Text":"this is the more proper way of writing"},{"Start":"02:06.650 ","End":"02:09.635","Text":"it but of course we know that what\u0027s under the square root sign is Delta."},{"Start":"02:09.635 ","End":"02:13.405","Text":"Now for our case, for our partial differential equation,"},{"Start":"02:13.405 ","End":"02:16.710","Text":"a_12 is minus 1,"},{"Start":"02:16.710 ","End":"02:18.150","Text":"a_11 is 1,"},{"Start":"02:18.150 ","End":"02:19.995","Text":"so we can just forget about it,"},{"Start":"02:19.995 ","End":"02:21.935","Text":"and plus or minus square root of Delta."},{"Start":"02:21.935 ","End":"02:23.705","Text":"Well, Delta is this,"},{"Start":"02:23.705 ","End":"02:25.205","Text":"we found this before."},{"Start":"02:25.205 ","End":"02:30.680","Text":"Yeah, here it is, so these are 2 characteristic equations."},{"Start":"02:30.680 ","End":"02:32.060","Text":"If you want, you could write it twice,"},{"Start":"02:32.060 ","End":"02:33.845","Text":"once with the plus 1 for the minus."},{"Start":"02:33.845 ","End":"02:37.925","Text":"Anyway, that\u0027s part b, and in part c,"},{"Start":"02:37.925 ","End":"02:42.695","Text":"we had to sketch the tangent lines that go through the origin."},{"Start":"02:42.695 ","End":"02:44.975","Text":"The origin is 0, 0."},{"Start":"02:44.975 ","End":"02:48.350","Text":"So if you put x=0,"},{"Start":"02:48.350 ","End":"02:51.740","Text":"y=0 in here, in both of the expressions,"},{"Start":"02:51.740 ","End":"02:53.360","Text":"one with a plus 1 for the minus,"},{"Start":"02:53.360 ","End":"03:01.045","Text":"we get minus 1 plus or minus square root of 1 which gives 0 and minus 2,"},{"Start":"03:01.045 ","End":"03:06.478","Text":"and we also have to sketch where it\u0027s hyperbolic, parabolic and elliptic."},{"Start":"03:06.478 ","End":"03:09.411","Text":"Hyperbolic inside the unit circle,"},{"Start":"03:09.411 ","End":"03:11.780","Text":"elliptic outside the unit circle,"},{"Start":"03:11.780 ","End":"03:13.760","Text":"parabolic on the unit circle,"},{"Start":"03:13.760 ","End":"03:15.620","Text":"and here are bits of"},{"Start":"03:15.620 ","End":"03:21.510","Text":"the tangent lines corresponding to slopes 0 and minus 2 through the origin."},{"Start":"03:21.510 ","End":"03:25.680","Text":"This has slope 0, and this slope minus 2."},{"Start":"03:25.680 ","End":"03:29.050","Text":"That concludes this exercise."}],"ID":30902},{"Watched":false,"Name":"Exercise 38","Duration":"10m 3s","ChapterTopicVideoID":29301,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"In this exercise, we\u0027re going to find the general solution"},{"Start":"00:03.150 ","End":"00:06.675","Text":"to the equation, to the problem."},{"Start":"00:06.675 ","End":"00:10.500","Text":"This is the equation and there\u0027s also boundary values,"},{"Start":"00:10.500 ","End":"00:15.510","Text":"using the method of separation of variables after an appropriate transformation."},{"Start":"00:15.510 ","End":"00:19.155","Text":"Note that we\u0027re usually given an initial condition also,"},{"Start":"00:19.155 ","End":"00:21.630","Text":"u(x, naught) equals some function f(x),"},{"Start":"00:21.630 ","End":"00:24.855","Text":"and then we can find an exact solution,"},{"Start":"00:24.855 ","End":"00:26.310","Text":"but if we\u0027re not given this,"},{"Start":"00:26.310 ","End":"00:28.950","Text":"then we have a general solution."},{"Start":"00:28.950 ","End":"00:30.990","Text":"In fact, in Part b,"},{"Start":"00:30.990 ","End":"00:33.270","Text":"we are given a condition u(x,"},{"Start":"00:33.270 ","End":"00:37.530","Text":"naught) and f(x) is sine Pi_x over 2 plus x plus 1,"},{"Start":"00:37.530 ","End":"00:40.100","Text":"and then we\u0027ll have a particular solution."},{"Start":"00:40.100 ","End":"00:43.310","Text":"Let\u0027s see what is meant by appropriate transformation."},{"Start":"00:43.310 ","End":"00:48.620","Text":"We somehow want to make this simpler to get rid of the boundary conditions,"},{"Start":"00:48.620 ","End":"00:50.800","Text":"I mean make them homogeneous."},{"Start":"00:50.800 ","End":"00:53.510","Text":"Note that just by playing around,"},{"Start":"00:53.510 ","End":"00:57.005","Text":"we can get that x plus 1 satisfies"},{"Start":"00:57.005 ","End":"01:03.800","Text":"the PDE because du by dt is 0 and u_xx is 0,"},{"Start":"01:03.800 ","End":"01:06.724","Text":"so this equals this and satisfies the PDE."},{"Start":"01:06.724 ","End":"01:12.930","Text":"Also u(0, t) is 1 because when we put 0 for x, we get 1,"},{"Start":"01:12.930 ","End":"01:15.120","Text":"and also du by dx,"},{"Start":"01:15.120 ","End":"01:16.695","Text":"which is 1,"},{"Start":"01:16.695 ","End":"01:18.070","Text":"is everywhere 1,"},{"Start":"01:18.070 ","End":"01:20.010","Text":"so this is also true."},{"Start":"01:20.010 ","End":"01:22.520","Text":"Now that we have 1 particular solution,"},{"Start":"01:22.520 ","End":"01:25.580","Text":"this will help us find the general solution."},{"Start":"01:25.580 ","End":"01:30.390","Text":"If you\u0027re wondering how more methodically we could get 2x plus 1,"},{"Start":"01:30.390 ","End":"01:32.670","Text":"one way is to replace u(x,"},{"Start":"01:32.670 ","End":"01:34.650","Text":"t) by y(x),"},{"Start":"01:34.650 ","End":"01:37.470","Text":"make it not dependent on t, and then u,"},{"Start":"01:37.470 ","End":"01:39.495","Text":"x is y\u0027,"},{"Start":"01:39.495 ","End":"01:42.860","Text":"and then we have the following, you know what?"},{"Start":"01:42.860 ","End":"01:45.290","Text":"I\u0027ll just leave it here for you to take a look"},{"Start":"01:45.290 ","End":"01:48.245","Text":"at how we might more methodically reach this."},{"Start":"01:48.245 ","End":"01:52.715","Text":"They\u0027re usually trial and error will get you here plus an educated guess."},{"Start":"01:52.715 ","End":"01:54.810","Text":"Now that we have this,"},{"Start":"01:54.810 ","End":"01:56.150","Text":"we\u0027ll let u(x,"},{"Start":"01:56.150 ","End":"01:57.570","Text":"t) be v(x,"},{"Start":"01:57.570 ","End":"01:59.325","Text":"t) plus x plus 1."},{"Start":"01:59.325 ","End":"02:01.520","Text":"We are defining v this way."},{"Start":"02:01.520 ","End":"02:04.325","Text":"If you like, v is u minus this."},{"Start":"02:04.325 ","End":"02:06.455","Text":"Note that u(0,"},{"Start":"02:06.455 ","End":"02:09.950","Text":"t) is v(0, t) plus 1,"},{"Start":"02:09.950 ","End":"02:11.953","Text":"because x is 0,"},{"Start":"02:11.953 ","End":"02:14.840","Text":"and du by dx (0,"},{"Start":"02:14.840 ","End":"02:18.500","Text":"t) is v_x(0, t) plus 1."},{"Start":"02:18.500 ","End":"02:21.065","Text":"The derivative of x plus 1 is 1."},{"Start":"02:21.065 ","End":"02:25.520","Text":"Also, v satisfies the PDE,"},{"Start":"02:25.520 ","End":"02:27.080","Text":"which is the heat equation,"},{"Start":"02:27.080 ","End":"02:32.180","Text":"because v_t is the same as u_t and v_xx is the same as u_xx,"},{"Start":"02:32.180 ","End":"02:34.745","Text":"and so this equals this."},{"Start":"02:34.745 ","End":"02:40.160","Text":"Given all that, we now get a new equation in v,"},{"Start":"02:40.160 ","End":"02:44.540","Text":"a homogeneous heat equation with homogeneous boundary conditions."},{"Start":"02:44.540 ","End":"02:49.445","Text":"Again, no initial condition v(x, 0)."},{"Start":"02:49.445 ","End":"02:54.995","Text":"Now there is a formula for the general solution of this which is mixed."},{"Start":"02:54.995 ","End":"02:57.545","Text":"This is Dirichlet and this Neumann."},{"Start":"02:57.545 ","End":"03:00.020","Text":"There was a table in the tutorial depending"},{"Start":"03:00.020 ","End":"03:02.860","Text":"on the different kinds of boundary conditions,"},{"Start":"03:02.860 ","End":"03:09.275","Text":"and this is the kind that we have Dirichlet and Neumann mixed."},{"Start":"03:09.275 ","End":"03:12.565","Text":"We use this formula here."},{"Start":"03:12.565 ","End":"03:16.770","Text":"In our case, a is 1 and L is 1,"},{"Start":"03:16.770 ","End":"03:19.785","Text":"so this is what we get."},{"Start":"03:19.785 ","End":"03:22.640","Text":"This is the general solution for v,"},{"Start":"03:22.640 ","End":"03:25.130","Text":"but we want the general solution for u,"},{"Start":"03:25.130 ","End":"03:28.700","Text":"so what we have to do is add plus x plus 1,"},{"Start":"03:28.700 ","End":"03:31.145","Text":"because u is v plus x plus 1,"},{"Start":"03:31.145 ","End":"03:34.430","Text":"and this is the general solution for u."},{"Start":"03:34.430 ","End":"03:40.910","Text":"General, meaning that the coefficients A_k are arbitrary."},{"Start":"03:40.910 ","End":"03:43.250","Text":"Well, not completely arbitrary,"},{"Start":"03:43.250 ","End":"03:45.425","Text":"this thing has to converge, but yeah."},{"Start":"03:45.425 ","End":"03:47.270","Text":"Now anyway, I cheated here,"},{"Start":"03:47.270 ","End":"03:48.880","Text":"I used the formula,"},{"Start":"03:48.880 ","End":"03:52.310","Text":"but the question asked for separation of variables,"},{"Start":"03:52.310 ","End":"03:55.025","Text":"so what I\u0027ll do is at the end of the clip,"},{"Start":"03:55.025 ","End":"03:59.645","Text":"I\u0027ll show you how we get this formula using separation of variables."},{"Start":"03:59.645 ","End":"04:03.560","Text":"Meanwhile, let\u0027s move on to Part b."},{"Start":"04:03.560 ","End":"04:06.800","Text":"We found the general solution in Part a,"},{"Start":"04:06.800 ","End":"04:09.950","Text":"and now we\u0027re given an initial condition, u(x,"},{"Start":"04:09.950 ","End":"04:11.540","Text":"naught) equals this,"},{"Start":"04:11.540 ","End":"04:16.160","Text":"so what we can do is substitute t=0 here,"},{"Start":"04:16.160 ","End":"04:19.025","Text":"and then we can equate these 2."},{"Start":"04:19.025 ","End":"04:27.860","Text":"We get that this here is equal to this series with t=0,"},{"Start":"04:27.860 ","End":"04:32.660","Text":"t=0 means this whole exponent disappears and we\u0027re"},{"Start":"04:32.660 ","End":"04:38.290","Text":"just left with the series A_k sine whatever."},{"Start":"04:38.300 ","End":"04:43.935","Text":"Note that the x plus 1 cancels in this equation,"},{"Start":"04:43.935 ","End":"04:46.190","Text":"so now we can compare coefficients."},{"Start":"04:46.190 ","End":"04:49.690","Text":"We have a Fourier sine series here."},{"Start":"04:49.690 ","End":"04:58.905","Text":"Now, Pi_x over 2 is 1 of the 2k minus 1 over 2 Pi_x when k=1."},{"Start":"04:58.905 ","End":"05:03.485","Text":"Because 2k minus 1 is twice 1 minus 1 is 1 over 2, so yeah."},{"Start":"05:03.485 ","End":"05:08.530","Text":"When k=1, we get a non-zero coefficient,"},{"Start":"05:08.530 ","End":"05:10.785","Text":"A_1 is equal to 1."},{"Start":"05:10.785 ","End":"05:14.265","Text":"Here we have 1 sine Pi_x over 2."},{"Start":"05:14.265 ","End":"05:16.695","Text":"The other A_k\u0027s are 0."},{"Start":"05:16.695 ","End":"05:19.385","Text":"We only have 1 non-zero term."},{"Start":"05:19.385 ","End":"05:22.550","Text":"If we go back to the general solution,"},{"Start":"05:22.550 ","End":"05:23.630","Text":"it\u0027s still up here,"},{"Start":"05:23.630 ","End":"05:28.610","Text":"and we substitute the A_k\u0027s that we found from here,"},{"Start":"05:28.610 ","End":"05:32.545","Text":"and we just get the 1 term with k=1,"},{"Start":"05:32.545 ","End":"05:34.440","Text":"and that is that u(x,"},{"Start":"05:34.440 ","End":"05:36.945","Text":"t) is A_1 is 1."},{"Start":"05:36.945 ","End":"05:40.215","Text":"Here we get 2k minus 1 over 2 is just a 1/2,"},{"Start":"05:40.215 ","End":"05:42.330","Text":"so Pi_x over 2."},{"Start":"05:42.330 ","End":"05:44.915","Text":"Here 2k minus 1 over 2 is a 1/2."},{"Start":"05:44.915 ","End":"05:46.410","Text":"We can open this up,"},{"Start":"05:46.410 ","End":"05:49.460","Text":"so it\u0027s 1/2 squared is a 1/4 times Pi^2,"},{"Start":"05:49.460 ","End":"05:52.310","Text":"we get minus Pi^2 t over 4,"},{"Start":"05:52.310 ","End":"05:54.065","Text":"plus x plus 1."},{"Start":"05:54.065 ","End":"05:57.340","Text":"This is the solution to Part b."},{"Start":"05:57.340 ","End":"06:02.045","Text":"Now here\u0027s the optional part of this clip if you want to see how we"},{"Start":"06:02.045 ","End":"06:07.225","Text":"can avoid using that formula in Part a and do it more directly."},{"Start":"06:07.225 ","End":"06:12.025","Text":"Here again is the problem for v,"},{"Start":"06:12.025 ","End":"06:17.245","Text":"a PDE and boundary conditions without initial condition,"},{"Start":"06:17.245 ","End":"06:19.700","Text":"and we\u0027re going to use separation of variables."},{"Start":"06:19.700 ","End":"06:22.760","Text":"Which mean first we look for a solution of the form v(x,"},{"Start":"06:22.760 ","End":"06:25.300","Text":"t) is X(x)T(t),"},{"Start":"06:25.300 ","End":"06:32.255","Text":"and then the partial derivatives v_t is this, v_xx is this."},{"Start":"06:32.255 ","End":"06:34.568","Text":"Here we just differentiate the t part,"},{"Start":"06:34.568 ","End":"06:37.480","Text":"here we differentiate the x part twice,"},{"Start":"06:37.480 ","End":"06:40.970","Text":"and from the equation v_t equals v_xx,"},{"Start":"06:40.970 ","End":"06:45.259","Text":"the heat equation, we get that this equals this."},{"Start":"06:45.259 ","End":"06:49.880","Text":"Then we can separate the variables"},{"Start":"06:49.880 ","End":"06:57.135","Text":"by bringing the x(x) to the denominator and T(t) to the denominator,"},{"Start":"06:57.135 ","End":"07:00.780","Text":"and we get that this quotient equals this quotient,"},{"Start":"07:00.780 ","End":"07:02.990","Text":"and this is a function of x,"},{"Start":"07:02.990 ","End":"07:05.060","Text":"this is a function of t. If they\u0027re equal,"},{"Start":"07:05.060 ","End":"07:08.390","Text":"they must be equal to some constant."},{"Start":"07:08.390 ","End":"07:12.560","Text":"We\u0027ll call that constant minus Lambda for convenience."},{"Start":"07:12.560 ","End":"07:14.630","Text":"Forget about the T for the moment."},{"Start":"07:14.630 ","End":"07:15.920","Text":"We\u0027ll return to it."},{"Start":"07:15.920 ","End":"07:17.495","Text":"Just looking at the x part,"},{"Start":"07:17.495 ","End":"07:21.830","Text":"we get that X\u0027\u0027 is minus Lambda X,"},{"Start":"07:21.830 ","End":"07:25.625","Text":"which gives us the ordinary differential equation,"},{"Start":"07:25.625 ","End":"07:29.465","Text":"X\u0027\u0027 plus Lambda X equals 0."},{"Start":"07:29.465 ","End":"07:38.840","Text":"Take the boundary conditions for v and apply them using that v equals X times T,"},{"Start":"07:38.840 ","End":"07:42.530","Text":"then we get here, X(0)T(t),"},{"Start":"07:42.530 ","End":"07:46.645","Text":"and here the derivative of x at 1 times T(t)."},{"Start":"07:46.645 ","End":"07:48.925","Text":"In any event, T(t) cancels,"},{"Start":"07:48.925 ","End":"07:54.940","Text":"and we get that X(0) equals X\u0027(1) equals 0."},{"Start":"07:54.940 ","End":"08:00.734","Text":"Altogether, this together with the, where is it?"},{"Start":"08:00.734 ","End":"08:06.740","Text":"Here, give us a Sturm–Liouville problem on the interval from 0-1."},{"Start":"08:06.740 ","End":"08:08.630","Text":"There should be a formula for this,"},{"Start":"08:08.630 ","End":"08:14.375","Text":"you need to refer to the notes tutorial or you\u0027re given a formula sheet hopefully,"},{"Start":"08:14.375 ","End":"08:17.810","Text":"and turns out that the eigenfunctions and eigenvalues for"},{"Start":"08:17.810 ","End":"08:22.760","Text":"this problem are eigenvalues Lambda k,"},{"Start":"08:22.760 ","End":"08:26.600","Text":"this where k equal 1,2,3, etc.,"},{"Start":"08:26.600 ","End":"08:35.675","Text":"and eigenfunctions X indexed by k sine 2k minus 1 over 2 Pi_x."},{"Start":"08:35.675 ","End":"08:38.360","Text":"We have X_k,"},{"Start":"08:38.360 ","End":"08:42.170","Text":"we also need to find the corresponding T_k."},{"Start":"08:42.170 ","End":"08:46.220","Text":"They share an eigenvalue in common."},{"Start":"08:46.220 ","End":"08:50.524","Text":"We had T\u0027 over T equals minus Lambda,"},{"Start":"08:50.524 ","End":"08:53.510","Text":"but we\u0027ll take a family of these for each k,"},{"Start":"08:53.510 ","End":"09:00.700","Text":"so we get T_k\u0027 over T_k equals minus Lambda k, which is this."},{"Start":"09:00.700 ","End":"09:03.050","Text":"Just swapping sides and stuff."},{"Start":"09:03.050 ","End":"09:07.550","Text":"This gives us a differential equation, homogeneous,"},{"Start":"09:07.550 ","End":"09:15.439","Text":"T_k prime plus this constant times T_k is 0."},{"Start":"09:15.439 ","End":"09:18.575","Text":"The solution to this is an exponential."},{"Start":"09:18.575 ","End":"09:23.915","Text":"We can take c=1 because T_k is determined up to a constant multiple."},{"Start":"09:23.915 ","End":"09:31.635","Text":"Let\u0027s summarize, X_k is this, T_k is this."},{"Start":"09:31.635 ","End":"09:34.410","Text":"Again, k equal 1,2,3, etc."},{"Start":"09:34.410 ","End":"09:36.635","Text":"We\u0027re looking for a solution now,"},{"Start":"09:36.635 ","End":"09:41.820","Text":"which is an infinite linear combination of the X_k and the T_k."},{"Start":"09:41.820 ","End":"09:49.660","Text":"Like so this infinite series and then just substituting X_k from here and T_k from here,"},{"Start":"09:49.660 ","End":"09:51.395","Text":"this is what we get,"},{"Start":"09:51.395 ","End":"09:54.530","Text":"and this is what we had before for v(x, t),"},{"Start":"09:54.530 ","End":"10:01.625","Text":"so we did it without producing a formula and doing it with separation of variables."},{"Start":"10:01.625 ","End":"10:03.690","Text":"Okay. We are done."}],"ID":30903},{"Watched":false,"Name":"Exercise 39","Duration":"3m 48s","ChapterTopicVideoID":29302,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.110","Text":"In this exercise, we\u0027re given the following problem."},{"Start":"00:04.110 ","End":"00:05.940","Text":"For u(x, y),"},{"Start":"00:05.940 ","End":"00:10.510","Text":"we have to solve it and also compute u(1,1)."},{"Start":"00:11.030 ","End":"00:14.235","Text":"This is the Laplace equation,"},{"Start":"00:14.235 ","End":"00:16.170","Text":"and this is a quarter plane,"},{"Start":"00:16.170 ","End":"00:20.475","Text":"it\u0027s just the first quadrant where x and y are positive."},{"Start":"00:20.475 ","End":"00:27.470","Text":"We\u0027re given a Neumann boundary condition on the y-axis, that u(0,"},{"Start":"00:27.470 ","End":"00:28.905","Text":"y) equals 0,"},{"Start":"00:28.905 ","End":"00:34.625","Text":"and we\u0027re also given an initial condition along the positive x-axis that u(x,"},{"Start":"00:34.625 ","End":"00:36.665","Text":"0) is the following."},{"Start":"00:36.665 ","End":"00:43.750","Text":"In the tutorial, we didn\u0027t cover the case of the Laplace equation on a half-plane."},{"Start":"00:43.750 ","End":"00:49.550","Text":"Here is the formula due to Poisson that deals with that,"},{"Start":"00:49.550 ","End":"00:50.750","Text":"and maybe you\u0027re thinking,"},{"Start":"00:50.750 ","End":"00:52.620","Text":"but we only have a core to play in,"},{"Start":"00:52.620 ","End":"00:54.440","Text":"and what\u0027s the use of a formula for the half-plane?"},{"Start":"00:54.440 ","End":"00:59.975","Text":"Well, you\u0027ll see. We\u0027re also given from some table of integrals,"},{"Start":"00:59.975 ","End":"01:02.615","Text":"the following integral which will need."},{"Start":"01:02.615 ","End":"01:06.425","Text":"What we\u0027re going to do is extend this to the half plane because"},{"Start":"01:06.425 ","End":"01:10.595","Text":"this is a Neumann boundary condition that\u0027s homogeneous."},{"Start":"01:10.595 ","End":"01:13.685","Text":"If we extend to an even function,"},{"Start":"01:13.685 ","End":"01:20.355","Text":"the even function will automatically have derivative 0 when x is 0,"},{"Start":"01:20.355 ","End":"01:24.185","Text":"and this function is already even,"},{"Start":"01:24.185 ","End":"01:29.150","Text":"so we can extend it and we get the following."},{"Start":"01:29.150 ","End":"01:31.640","Text":"The Laplacian of u is 0,"},{"Start":"01:31.640 ","End":"01:34.640","Text":"but this time on the whole half-plane,"},{"Start":"01:34.640 ","End":"01:39.275","Text":"we\u0027ve extended x to be from minus infinity to infinity."},{"Start":"01:39.275 ","End":"01:42.785","Text":"Here it looks like the same formula,"},{"Start":"01:42.785 ","End":"01:47.885","Text":"but this time it\u0027s x from minus infinity to infinity."},{"Start":"01:47.885 ","End":"01:51.980","Text":"Now we can apply that Poisson formula above."},{"Start":"01:51.980 ","End":"01:56.850","Text":"This is what the Poisson formula is only in our case we know what is u(s,"},{"Start":"01:56.850 ","End":"01:59.775","Text":"0), u(x, 0),"},{"Start":"01:59.775 ","End":"02:06.150","Text":"but x replaced by s. This is 1/1 plus s^2,"},{"Start":"02:06.150 ","End":"02:10.490","Text":"also just change the position of these 2."},{"Start":"02:10.490 ","End":"02:16.535","Text":"Now we have this integral here again is that formula copied it down here,"},{"Start":"02:16.535 ","End":"02:24.265","Text":"but we need is A=1, a=0."},{"Start":"02:24.265 ","End":"02:25.800","Text":"When you see x here,"},{"Start":"02:25.800 ","End":"02:31.510","Text":"think of s because we have our integral in terms of s. Also,"},{"Start":"02:31.510 ","End":"02:35.735","Text":"B is going to be y,"},{"Start":"02:35.735 ","End":"02:40.370","Text":"and little b will be x,"},{"Start":"02:40.370 ","End":"02:43.865","Text":"should have really reversed the order s minus x^2,"},{"Start":"02:43.865 ","End":"02:47.380","Text":"like s is x and x is b or any way."},{"Start":"02:47.380 ","End":"02:50.220","Text":"These are the substitutions we make,"},{"Start":"02:50.220 ","End":"02:58.230","Text":"and then the answer comes out to be AB is 1 times y,"},{"Start":"02:58.230 ","End":"03:00.885","Text":"and that cancels with the y here,"},{"Start":"03:00.885 ","End":"03:04.305","Text":"also the pi here with the pi here."},{"Start":"03:04.305 ","End":"03:06.840","Text":"We need A plus B,"},{"Start":"03:06.840 ","End":"03:09.000","Text":"which is 1 plus y,"},{"Start":"03:09.000 ","End":"03:13.230","Text":"and then A plus B^2 is 1 plus y^2,"},{"Start":"03:13.230 ","End":"03:18.795","Text":"and a minus b lowercase is 0 minus xo^2,"},{"Start":"03:18.795 ","End":"03:21.315","Text":"which is just x^2."},{"Start":"03:21.315 ","End":"03:23.710","Text":"This is the answer,"},{"Start":"03:23.710 ","End":"03:28.685","Text":"except that we need to restrict it to the first quadrant,"},{"Start":"03:28.685 ","End":"03:31.340","Text":"so we should really add x bigger than 0,"},{"Start":"03:31.340 ","End":"03:35.000","Text":"y bigger than 0, just to make it compatible with the question."},{"Start":"03:35.000 ","End":"03:38.170","Text":"If we let x=1, y=1,"},{"Start":"03:38.170 ","End":"03:40.980","Text":"then we get 1 plus 1 is 2 here,"},{"Start":"03:40.980 ","End":"03:45.855","Text":"2^2 plus 1^2 is 2/5, or 0.4."},{"Start":"03:45.855 ","End":"03:48.970","Text":"That\u0027s the end of this exercise."}],"ID":30904},{"Watched":false,"Name":"Exercise 40","Duration":"3m 11s","ChapterTopicVideoID":29303,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.490","Text":"In this exercise, we\u0027re going to solve the following initial value problem."},{"Start":"00:05.490 ","End":"00:10.890","Text":"It\u0027s actually a wave equation on the infinite interval,"},{"Start":"00:10.890 ","End":"00:13.620","Text":"but for positive time."},{"Start":"00:13.620 ","End":"00:14.970","Text":"We\u0027re given these two."},{"Start":"00:14.970 ","End":"00:16.650","Text":"They\u0027re both initial conditions,"},{"Start":"00:16.650 ","End":"00:19.920","Text":"u and u_t, when t is 0."},{"Start":"00:19.920 ","End":"00:21.375","Text":"That\u0027s part a."},{"Start":"00:21.375 ","End":"00:23.010","Text":"In part b,"},{"Start":"00:23.010 ","End":"00:28.005","Text":"I\u0027m going to let U(t) be u(1, t)."},{"Start":"00:28.005 ","End":"00:31.905","Text":"We have to show that U(t) has the maximum value,"},{"Start":"00:31.905 ","End":"00:33.555","Text":"call it u-star,"},{"Start":"00:33.555 ","End":"00:36.795","Text":"at some time t-star."},{"Start":"00:36.795 ","End":"00:38.700","Text":"In precise language,"},{"Start":"00:38.700 ","End":"00:46.930","Text":"u-star is the maximum along non-negative t of U(t), which is u(t-star)."},{"Start":"00:47.540 ","End":"00:51.405","Text":"We have to find u-star and t-star."},{"Start":"00:51.405 ","End":"00:53.875","Text":"So let\u0027s start with part a."},{"Start":"00:53.875 ","End":"00:59.165","Text":"The wave equation on an infinite interval suggests d\u0027Alembert\u0027s formula."},{"Start":"00:59.165 ","End":"01:02.660","Text":"This is what it is, but in our case a=1."},{"Start":"01:02.660 ","End":"01:05.030","Text":"So this is what we have."},{"Start":"01:05.030 ","End":"01:13.135","Text":"Just substitute f(x) from here and we have g(x) from here,"},{"Start":"01:13.135 ","End":"01:15.390","Text":"this is what we get."},{"Start":"01:15.390 ","End":"01:21.065","Text":"Now leave the first part alone and just work on the second part, the integral."},{"Start":"01:21.065 ","End":"01:27.630","Text":"This is the exact derivative of 1 over 1 plus s^2."},{"Start":"01:27.630 ","End":"01:30.345","Text":"I mean, if you differentiate 1 over x,"},{"Start":"01:30.345 ","End":"01:32.940","Text":"you get minus 1 over x^2."},{"Start":"01:32.940 ","End":"01:35.855","Text":"Also an anti-derivative of s^2 is 2s."},{"Start":"01:35.855 ","End":"01:38.960","Text":"It\u0027s a definite integral between these limits,"},{"Start":"01:38.960 ","End":"01:41.240","Text":"so we have to substitute those limits,"},{"Start":"01:41.240 ","End":"01:43.970","Text":"and what we get then here,"},{"Start":"01:43.970 ","End":"01:46.040","Text":"substituting x plus t we get this,"},{"Start":"01:46.040 ","End":"01:48.305","Text":"substituting x-t we get this."},{"Start":"01:48.305 ","End":"01:52.115","Text":"There is a minus here because we subtract the lower from the upper."},{"Start":"01:52.115 ","End":"01:54.545","Text":"Whereas here we have a plus."},{"Start":"01:54.545 ","End":"01:57.574","Text":"Essentially these are the same except for the sign."},{"Start":"01:57.574 ","End":"02:03.080","Text":"So this will cancel with this and we\u0027ll get twice this."},{"Start":"02:03.080 ","End":"02:05.095","Text":"The two will cancel with the half."},{"Start":"02:05.095 ","End":"02:09.855","Text":"So what we\u0027re left with is just 1 over 1 plus x plus t^2,"},{"Start":"02:09.855 ","End":"02:12.165","Text":"and that\u0027s the answer for u(x,t)."},{"Start":"02:12.165 ","End":"02:14.055","Text":"That\u0027s part a."},{"Start":"02:14.055 ","End":"02:16.185","Text":"In part b,"},{"Start":"02:16.185 ","End":"02:20.070","Text":"U(t) is u(1, t),"},{"Start":"02:20.070 ","End":"02:23.980","Text":"which means we take u(x,t) and substitute x=1."},{"Start":"02:23.980 ","End":"02:25.940","Text":"So this is what we get."},{"Start":"02:25.940 ","End":"02:31.220","Text":"Note that when t increases in the non-negative range,"},{"Start":"02:31.220 ","End":"02:34.050","Text":"t increases, 1 plus t increases."},{"Start":"02:34.050 ","End":"02:35.530","Text":"So does this square."},{"Start":"02:35.530 ","End":"02:39.830","Text":"1 plus this increases and the reciprocal decreases."},{"Start":"02:39.830 ","End":"02:46.280","Text":"So this is a decreasing function monotonically on t bigger or equal to 0."},{"Start":"02:46.280 ","End":"02:49.865","Text":"The maximum is at the beginning of the interval."},{"Start":"02:49.865 ","End":"02:56.840","Text":"So t-star is 0 and if we substitute that in here,"},{"Start":"02:56.840 ","End":"02:59.600","Text":"we get that u-star is 1/2."},{"Start":"02:59.600 ","End":"03:03.170","Text":"If you want the computation here it is."},{"Start":"03:03.170 ","End":"03:08.220","Text":"Just let t=0 here and you get 1/2."},{"Start":"03:08.220 ","End":"03:11.800","Text":"That\u0027s the end of this exercise."}],"ID":30905},{"Watched":false,"Name":"Exercise 41","Duration":"8m 37s","ChapterTopicVideoID":29304,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.350","Text":"In this exercise, there are 2 parts."},{"Start":"00:03.350 ","End":"00:07.270","Text":"In Part A, we want to find the characteristics Psi and"},{"Start":"00:07.270 ","End":"00:11.485","Text":"Eta in terms of x and y of this equation."},{"Start":"00:11.485 ","End":"00:14.590","Text":"It\u0027s a linear second-order equation in"},{"Start":"00:14.590 ","End":"00:19.615","Text":"2 variables and with constant coefficients, and it\u0027s homogeneous."},{"Start":"00:19.615 ","End":"00:26.155","Text":"Anyway, we want to transform it to canonical form w(Psi and Eta)."},{"Start":"00:26.155 ","End":"00:28.495","Text":"Then in Part B,"},{"Start":"00:28.495 ","End":"00:30.280","Text":"we\u0027ll read it when we come to it."},{"Start":"00:30.280 ","End":"00:31.968","Text":"Let\u0027s start with Part A,"},{"Start":"00:31.968 ","End":"00:34.420","Text":"this is our equation."},{"Start":"00:34.420 ","End":"00:37.630","Text":"The coefficients a_11, a_12,"},{"Start":"00:37.630 ","End":"00:41.830","Text":"a_22 are obtained from these second-order terms."},{"Start":"00:41.830 ","End":"00:47.445","Text":"Just note that uxy it\u0027s coefficient is twice a_12,"},{"Start":"00:47.445 ","End":"00:50.010","Text":"which gives us that a_11 is 1,"},{"Start":"00:50.010 ","End":"00:54.165","Text":"a_12 is 1.5 and a_22 is 2."},{"Start":"00:54.165 ","End":"01:03.150","Text":"Then we can compute Delta which comes out to be 0.25 or a quarter."},{"Start":"01:03.150 ","End":"01:06.140","Text":"Then the characteristic equations,"},{"Start":"01:06.140 ","End":"01:09.170","Text":"there\u0027s 2 of them, one through the plus 1 with a minus."},{"Start":"01:09.170 ","End":"01:11.090","Text":"This is the formula."},{"Start":"01:11.090 ","End":"01:12.930","Text":"Well, the a_11 is one,"},{"Start":"01:12.930 ","End":"01:14.280","Text":"so we don\u0027t need it."},{"Start":"01:14.280 ","End":"01:18.455","Text":"A_12 is 1.5 plus or minus square root of a quarter."},{"Start":"01:18.455 ","End":"01:24.890","Text":"This is a 1.5, 1.5 plus or minus 1/2 gives us either 1 or 2."},{"Start":"01:24.890 ","End":"01:27.940","Text":"If we take dy by dx is 2,"},{"Start":"01:27.940 ","End":"01:31.755","Text":"then we get y equals 2x plus a constant."},{"Start":"01:31.755 ","End":"01:33.900","Text":"Dy by dx equals 1,"},{"Start":"01:33.900 ","End":"01:37.890","Text":"y equals x plus a constant, but different constants."},{"Start":"01:37.890 ","End":"01:41.640","Text":"Then if we extract the constants,"},{"Start":"01:41.640 ","End":"01:44.850","Text":"first constant is y minus 2x,"},{"Start":"01:44.850 ","End":"01:46.970","Text":"or you can take 2x minus y,"},{"Start":"01:46.970 ","End":"01:56.220","Text":"any event, we can now get Psi and Eta characteristics as 2x minus y and x minus y."},{"Start":"01:56.220 ","End":"01:59.930","Text":"We use these to get the canonical equation."},{"Start":"01:59.930 ","End":"02:02.720","Text":"We want to convert u(x,"},{"Start":"02:02.720 ","End":"02:06.610","Text":"y) into w(Psi and Eta)."},{"Start":"02:06.610 ","End":"02:14.675","Text":"In the tutorial, we made a table of these partial derivatives with a change of variable."},{"Start":"02:14.675 ","End":"02:16.610","Text":"Now what we\u0027re going to do is substitute"},{"Start":"02:16.610 ","End":"02:24.220","Text":"all the partial derivatives of Psi and Eta with respect to x and y up to second order."},{"Start":"02:24.220 ","End":"02:29.895","Text":"Note that from here we have Psi_x is 2,"},{"Start":"02:29.895 ","End":"02:32.505","Text":"Psi_y is minus 1,"},{"Start":"02:32.505 ","End":"02:33.945","Text":"Eta_x is 1,"},{"Start":"02:33.945 ","End":"02:36.145","Text":"Eta_y is minus 1."},{"Start":"02:36.145 ","End":"02:41.405","Text":"But the second-order derivatives are all 0 because these are constants."},{"Start":"02:41.405 ","End":"02:45.230","Text":"Now we can substitute all these in here."},{"Start":"02:45.230 ","End":"02:50.100","Text":"We\u0027ll put Psi_x is 2 and Eta_x is 1,"},{"Start":"02:50.100 ","End":"02:51.525","Text":"and so on and so on."},{"Start":"02:51.525 ","End":"02:54.830","Text":"What we get is the following."},{"Start":"02:54.830 ","End":"02:58.070","Text":"I just put the numbers on top of them."},{"Start":"02:58.070 ","End":"03:00.300","Text":"The ones that are 0, I marked in red,"},{"Start":"03:00.300 ","End":"03:03.895","Text":"so we can just eliminate them not take them."},{"Start":"03:03.895 ","End":"03:11.170","Text":"Then we can compute this like ux will be 2w_Psi plus w_Eta."},{"Start":"03:12.290 ","End":"03:14.670","Text":"I\u0027ll take another example,"},{"Start":"03:14.670 ","End":"03:20.265","Text":"u_xy will be minus 2 w_Psi, Psi."},{"Start":"03:20.265 ","End":"03:27.015","Text":"Then here minus 2 minus 1 is minus 3 w_Psi Eta."},{"Start":"03:27.015 ","End":"03:30.180","Text":"From here, minus w_Eta,"},{"Start":"03:30.180 ","End":"03:32.250","Text":"Eta, and similarly for the rest."},{"Start":"03:32.250 ","End":"03:36.890","Text":"We have the partial derivatives of u in terms of the partial derivatives of"},{"Start":"03:36.890 ","End":"03:41.990","Text":"w. Now recall that our original PDE is this."},{"Start":"03:41.990 ","End":"03:47.920","Text":"What we have to do is substitute these 5 quantities from here."},{"Start":"03:47.920 ","End":"03:51.834","Text":"What we get just by substituting uxx,"},{"Start":"03:51.834 ","End":"03:54.180","Text":"here it this is 4w_Psi,"},{"Start":"03:54.180 ","End":"03:56.835","Text":"Psi 4w_Psi Eta,"},{"Start":"03:56.835 ","End":"04:01.595","Text":"w_Eta Eta, just copying into here."},{"Start":"04:01.595 ","End":"04:04.429","Text":"Then we want to collect like terms."},{"Start":"04:04.429 ","End":"04:09.310","Text":"We want to collect the w_Psi Psi separately."},{"Start":"04:09.310 ","End":"04:12.493","Text":"Anyway, turns out lots of stuff cancels,"},{"Start":"04:12.493 ","End":"04:14.374","Text":"and the only thing that\u0027s left,"},{"Start":"04:14.374 ","End":"04:22.840","Text":"I\u0027ll leave you to check minus w_Psi Eta minus w_Eta and multiply this by minus 1."},{"Start":"04:22.840 ","End":"04:31.205","Text":"This gives us the canonical form in terms of Psi and Eta that concludes Part A."},{"Start":"04:31.205 ","End":"04:33.710","Text":"Now we come to Part B."},{"Start":"04:33.710 ","End":"04:39.140","Text":"This is the result of Part A we found w, make it canonical."},{"Start":"04:39.140 ","End":"04:43.840","Text":"In Part B, we want to go over from w-v,"},{"Start":"04:43.840 ","End":"04:47.525","Text":"where w and v are related like this."},{"Start":"04:47.525 ","End":"04:51.820","Text":"This is an exponential with unknown Gamma and Delta."},{"Start":"04:51.820 ","End":"04:55.840","Text":"We have to find Gamma and Delta so that the PDE"},{"Start":"04:55.840 ","End":"05:00.275","Text":"for v doesn\u0027t have any first-order derivatives."},{"Start":"05:00.275 ","End":"05:02.855","Text":"Like here we have a first-order derivative,"},{"Start":"05:02.855 ","End":"05:06.880","Text":"w_Eta, but for v will just have second-order terms."},{"Start":"05:06.880 ","End":"05:08.665","Text":"I forgot to add earlier yeah,"},{"Start":"05:08.665 ","End":"05:13.490","Text":"after we\u0027ve done this we should solve the equation for v. From v get back to w,"},{"Start":"05:13.490 ","End":"05:15.920","Text":"and from w get back to u."},{"Start":"05:15.920 ","End":"05:20.850","Text":"Ultimately we will get the solution for u, the original equation."},{"Start":"05:21.000 ","End":"05:28.435","Text":"We\u0027re given that w is v times e^Gamma Psi plus Delta Eta."},{"Start":"05:28.435 ","End":"05:31.205","Text":"We want the partial derivatives,"},{"Start":"05:31.205 ","End":"05:35.505","Text":"these 2, we need w_Eta and W_Psi Eta."},{"Start":"05:35.505 ","End":"05:40.090","Text":"Well, w_Eta we can get from differentiating this"},{"Start":"05:40.090 ","End":"05:45.080","Text":"with respect to Eta using the product rule and a bit of chain rule here."},{"Start":"05:45.080 ","End":"05:46.975","Text":"This is what we get."},{"Start":"05:46.975 ","End":"05:49.480","Text":"We\u0027ll go into the little details."},{"Start":"05:49.480 ","End":"05:52.690","Text":"Then if we differentiate this with respect to Psi,"},{"Start":"05:52.690 ","End":"05:54.340","Text":"we get w_Eta Psi,"},{"Start":"05:54.340 ","End":"05:57.080","Text":"which is the same as w_Psi Eta."},{"Start":"05:57.080 ","End":"05:59.230","Text":"We get this product rule,"},{"Start":"05:59.230 ","End":"06:02.990","Text":"gives these 2 product rule here, gives these 2."},{"Start":"06:02.990 ","End":"06:10.370","Text":"The chain rule gives us the Gamma here and also the Gamma here."},{"Start":"06:10.370 ","End":"06:13.370","Text":"Anyway, put these 2 into this equation,"},{"Start":"06:13.370 ","End":"06:15.935","Text":"meaning add them and get 0."},{"Start":"06:15.935 ","End":"06:17.900","Text":"This is what we get."},{"Start":"06:17.900 ","End":"06:23.120","Text":"But note that every term has this exponent bit."},{"Start":"06:23.120 ","End":"06:25.625","Text":"I\u0027ve shaded them in gray,"},{"Start":"06:25.625 ","End":"06:27.928","Text":"you can just cancel them all out,"},{"Start":"06:27.928 ","End":"06:31.145","Text":"and it comes out much simpler now, just this."},{"Start":"06:31.145 ","End":"06:35.000","Text":"Now we want to collect like terms v_Psi Eta,"},{"Start":"06:35.000 ","End":"06:36.730","Text":"this just the 1."},{"Start":"06:36.730 ","End":"06:39.540","Text":"Delta v_Psi, just this,"},{"Start":"06:39.540 ","End":"06:44.595","Text":"v_Eta, we have Gamma plus 1."},{"Start":"06:44.595 ","End":"06:50.075","Text":"For v, we have Gamma Delta plus Delta."},{"Start":"06:50.075 ","End":"06:53.040","Text":"We said we don\u0027t want any first-order terms,"},{"Start":"06:53.040 ","End":"06:58.880","Text":"so we want this Delta to be 0 and this Gamma plus 1 to be 0."},{"Start":"06:58.880 ","End":"07:00.980","Text":"Note that if delta is 0,"},{"Start":"07:00.980 ","End":"07:02.990","Text":"this also is 0."},{"Start":"07:02.990 ","End":"07:04.790","Text":"That\u0027s just lucky."},{"Start":"07:04.790 ","End":"07:09.710","Text":"We have that Gamma is minus 1, Delta is 0."},{"Start":"07:09.710 ","End":"07:15.790","Text":"Then this equation becomes v_Psi Eta is 0."},{"Start":"07:15.790 ","End":"07:23.580","Text":"It also implies that we have w precisely in terms of v. If Gamma is minus 1 and Delta 0,"},{"Start":"07:23.580 ","End":"07:27.060","Text":"then w is ve^minus Psi."},{"Start":"07:27.060 ","End":"07:30.410","Text":"Now from this, we can solve for v. First of all,"},{"Start":"07:30.410 ","End":"07:32.765","Text":"integrate with respect to Eta,"},{"Start":"07:32.765 ","End":"07:40.100","Text":"so we get v_Psi equals the integral of 0 is not a constant,"},{"Start":"07:40.100 ","End":"07:42.695","Text":"but an arbitrary function of Psi."},{"Start":"07:42.695 ","End":"07:46.760","Text":"Do the same thing again this time integrated with respect to Psi."},{"Start":"07:46.760 ","End":"07:48.665","Text":"We get that v is,"},{"Start":"07:48.665 ","End":"07:52.685","Text":"let\u0027s say that F is a primitive of f. Again,"},{"Start":"07:52.685 ","End":"07:56.240","Text":"not a constant but an arbitrary function of Eta."},{"Start":"07:56.240 ","End":"08:00.800","Text":"This is the general solution for v in terms of Psi and Eta."},{"Start":"08:00.800 ","End":"08:04.520","Text":"Now from v, we\u0027re going to work our way back to w and then to u."},{"Start":"08:04.520 ","End":"08:09.155","Text":"First of all, w is ve^minus Psi from here."},{"Start":"08:09.155 ","End":"08:14.135","Text":"So w equals this thing times e^minus Psi."},{"Start":"08:14.135 ","End":"08:17.780","Text":"Now we\u0027re going to get back from w to u,"},{"Start":"08:17.780 ","End":"08:20.615","Text":"because u(x, y) is w(Psi and Eta)."},{"Start":"08:20.615 ","End":"08:23.090","Text":"We have Psi and Eta in terms of x and y,"},{"Start":"08:23.090 ","End":"08:25.495","Text":"so we just have to substitute them."},{"Start":"08:25.495 ","End":"08:30.210","Text":"Then we get that u is f (2x minus y) plus"},{"Start":"08:30.210 ","End":"08:35.055","Text":"g(x minus y) e to the minus Psi is minus 2x plus y."},{"Start":"08:35.055 ","End":"08:38.110","Text":"That\u0027s the answer, and we\u0027re done."}],"ID":30906},{"Watched":false,"Name":"Exercise 42","Duration":"3m 59s","ChapterTopicVideoID":29305,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.225","Text":"This exercise is difficult,"},{"Start":"00:03.225 ","End":"00:06.330","Text":"not the thing you\u0027d be expected to do on your own."},{"Start":"00:06.330 ","End":"00:10.095","Text":"We\u0027re given the following Poisson PDE."},{"Start":"00:10.095 ","End":"00:13.920","Text":"That\u0027s this with the Neumann boundary condition, which is,"},{"Start":"00:13.920 ","End":"00:17.760","Text":"this is 2 ways of writing the directional derivative and"},{"Start":"00:17.760 ","End":"00:23.130","Text":"a Laplacian is the same as Del squared or nabla squared."},{"Start":"00:23.130 ","End":"00:25.605","Text":"D here is a bounded domain."},{"Start":"00:25.605 ","End":"00:29.340","Text":"Here\u0027s a picture and it has a smooth boundary."},{"Start":"00:29.340 ","End":"00:31.885","Text":"Not sure how to pronounce this."},{"Start":"00:31.885 ","End":"00:35.930","Text":"I actually looked it up in the Wikipedia and has many names."},{"Start":"00:35.930 ","End":"00:37.700","Text":"This is just for interest\u0027s sake."},{"Start":"00:37.700 ","End":"00:43.730","Text":"Anyway, this boundary is also considered as a path which goes counterclockwise."},{"Start":"00:43.730 ","End":"00:49.835","Text":"Our task is to show that the solution to this equation is unique up to a constant."},{"Start":"00:49.835 ","End":"00:53.430","Text":"We\u0027ll need 2 formulas for the solution,"},{"Start":"00:53.430 ","End":"00:59.030","Text":"we\u0027ll need the divergence theorem and one of the vector identities that\u0027s"},{"Start":"00:59.030 ","End":"01:05.754","Text":"a product rule for applying grad to a product of a scalar times a vector function."},{"Start":"01:05.754 ","End":"01:09.709","Text":"The way we\u0027ll go about it is suppose that there are 2 solutions,"},{"Start":"01:09.709 ","End":"01:14.780","Text":"call them u_1 and u_2 and we\u0027ll let v be u_1 minus"},{"Start":"01:14.780 ","End":"01:21.505","Text":"u_2 and the obvious thing to do is to try and show that v is a constant."},{"Start":"01:21.505 ","End":"01:27.125","Text":"The problem that v solves is obtained by subtraction of the right-hand sides,"},{"Start":"01:27.125 ","End":"01:33.320","Text":"we just get that the Laplacian of v is 0 and the boundary condition,"},{"Start":"01:33.320 ","End":"01:36.470","Text":"the directional derivative is also 0."},{"Start":"01:36.470 ","End":"01:42.020","Text":"We\u0027ll apply the divergence theorem and we\u0027ll take the vector function f"},{"Start":"01:42.020 ","End":"01:47.880","Text":"to be scalar v times grad v. Not something intuitive,"},{"Start":"01:47.880 ","End":"01:49.650","Text":"we just pull it out of a hat."},{"Start":"01:49.650 ","End":"01:51.360","Text":"I switch sides here."},{"Start":"01:51.360 ","End":"01:57.935","Text":"Anyway, f is v grad v. Again,"},{"Start":"01:57.935 ","End":"02:04.655","Text":"f is equal to v grad v. Now,"},{"Start":"02:04.655 ","End":"02:06.785","Text":"on the right-hand side,"},{"Start":"02:06.785 ","End":"02:08.600","Text":"we use the formula."},{"Start":"02:08.600 ","End":"02:13.305","Text":"Where is it? This vector identity and we\u0027ll get this."},{"Start":"02:13.305 ","End":"02:16.165","Text":"The scalar function f,"},{"Start":"02:16.165 ","End":"02:23.135","Text":"v and vector F"},{"Start":"02:23.135 ","End":"02:27.080","Text":"will be grad V. Check it,"},{"Start":"02:27.080 ","End":"02:28.820","Text":"and this is what we get."},{"Start":"02:28.820 ","End":"02:30.900","Text":"On the left-hand side,"},{"Start":"02:30.900 ","End":"02:34.480","Text":"we\u0027re using the alternative form of a directional derivative."},{"Start":"02:34.480 ","End":"02:39.050","Text":"On the one hand, grad v.n,"},{"Start":"02:39.050 ","End":"02:44.615","Text":"and that\u0027s this directional derivative in the direction of unit vector"},{"Start":"02:44.615 ","End":"02:50.510","Text":"n. This is equal to 0 because we\u0027re given and grad squared v,"},{"Start":"02:50.510 ","End":"02:52.250","Text":"which is the Laplacian of v,"},{"Start":"02:52.250 ","End":"02:54.200","Text":"is 0 also given,"},{"Start":"02:54.200 ","End":"02:59.930","Text":"and a vector dot-product with itself is the norm of the vector"},{"Start":"02:59.930 ","End":"03:04.520","Text":"squared so this has"},{"Start":"03:04.520 ","End":"03:09.740","Text":"to be 0 because this is 0 and this is 0 so we\u0027re left with this being 0."},{"Start":"03:09.740 ","End":"03:15.605","Text":"If we have a non-negative continuous function on the region and the integral is 0,"},{"Start":"03:15.605 ","End":"03:18.140","Text":"then it has to be 0 everywhere."},{"Start":"03:18.140 ","End":"03:25.310","Text":"The norm of grad v is everywhere 0 and the formula for this would be the"},{"Start":"03:25.310 ","End":"03:29.030","Text":"partial derivative with respect to x^2 plus derivative with respect"},{"Start":"03:29.030 ","End":"03:32.915","Text":"to y^2 and this is going to be 0 over the whole region,"},{"Start":"03:32.915 ","End":"03:37.760","Text":"which means that this is 0 and this is 0."},{"Start":"03:37.760 ","End":"03:41.510","Text":"If both partial derivatives are 0,"},{"Start":"03:41.510 ","End":"03:44.240","Text":"then the function has to be a constant."},{"Start":"03:44.240 ","End":"03:48.760","Text":"If v is a constant and v is u_1 minus u_2,"},{"Start":"03:48.760 ","End":"03:53.750","Text":"then it follows that u_1 is the same as u_2 up"},{"Start":"03:53.750 ","End":"03:59.790","Text":"to an additive constant and that\u0027s what we had to show and we\u0027re done."}],"ID":30907},{"Watched":false,"Name":"Exercise 43","Duration":"1m 35s","ChapterTopicVideoID":29306,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:05.580","Text":"In this exercise, we\u0027re given the following Poisson equation."},{"Start":"00:05.580 ","End":"00:10.080","Text":"Poisson is a non-homogeneous Laplace equation."},{"Start":"00:10.080 ","End":"00:14.429","Text":"We have a homogeneous Neumann boundary condition,"},{"Start":"00:14.429 ","End":"00:17.265","Text":"homogeneous because it\u0027s 0 here."},{"Start":"00:17.265 ","End":"00:22.590","Text":"This is the Neumann boundary condition where we are given the directional derivative."},{"Start":"00:22.590 ","End":"00:26.010","Text":"D here is a bounded domain and"},{"Start":"00:26.010 ","End":"00:31.290","Text":"its boundary is considered as a counterclockwise curve around the region."},{"Start":"00:31.290 ","End":"00:33.980","Text":"We have to show that a necessary condition for"},{"Start":"00:33.980 ","End":"00:38.585","Text":"solvability is that the following integral is 0,"},{"Start":"00:38.585 ","End":"00:40.220","Text":"which means that if there is a solution,"},{"Start":"00:40.220 ","End":"00:43.040","Text":"u, then this is true."},{"Start":"00:43.040 ","End":"00:46.920","Text":"We\u0027re given a hint to use the divergence theorem."},{"Start":"00:46.940 ","End":"00:51.680","Text":"We want to compute this and we want to get to 0."},{"Start":"00:51.680 ","End":"00:53.680","Text":"From the PDE,"},{"Start":"00:53.680 ","End":"00:58.365","Text":"we can get that f is the Laplacian of u,"},{"Start":"00:58.365 ","End":"01:03.005","Text":"while del^2 is the same as del dot product with del,"},{"Start":"01:03.005 ","End":"01:09.365","Text":"and now we apply the divergence theorem with del u being"},{"Start":"01:09.365 ","End":"01:12.080","Text":"F. The other way of writing"},{"Start":"01:12.080 ","End":"01:17.335","Text":"this dot product is to write the directional derivative in this form."},{"Start":"01:17.335 ","End":"01:21.615","Text":"We know that this is 0,"},{"Start":"01:21.615 ","End":"01:23.930","Text":"that\u0027s what\u0027s given here."},{"Start":"01:23.930 ","End":"01:27.230","Text":"We\u0027re given that this directional derivative is 0 everywhere,"},{"Start":"01:27.230 ","End":"01:30.025","Text":"so this integral is 0."},{"Start":"01:30.025 ","End":"01:36.120","Text":"We\u0027ve shown that this integral is 0 and that\u0027s what we had to show and we\u0027re done."}],"ID":30908},{"Watched":false,"Name":"Exercise 44","Duration":"7m 35s","ChapterTopicVideoID":29307,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.840","Text":"In this exercise, which has an asterisk,"},{"Start":"00:03.840 ","End":"00:05.415","Text":"meaning it\u0027s more difficult,"},{"Start":"00:05.415 ","End":"00:10.740","Text":"we\u0027re asked to use separation of variables to find solutions to the Laplace equation,"},{"Start":"00:10.740 ","End":"00:15.555","Text":"Delta u equals 0 in polar coordinates in the unit disk."},{"Start":"00:15.555 ","End":"00:20.670","Text":"Here\u0027s a reminder of what the Laplacian is in polar coordinates."},{"Start":"00:20.670 ","End":"00:24.870","Text":"In part b, we\u0027re given a Neumann problem which is"},{"Start":"00:24.870 ","End":"00:31.020","Text":"this Laplace equation plus a Neumann boundary condition on the unit circle."},{"Start":"00:31.020 ","End":"00:33.840","Text":"Given this, I\u0027m using part a,"},{"Start":"00:33.840 ","End":"00:38.500","Text":"you have to show that the solution can be written in the following format."},{"Start":"00:38.500 ","End":"00:42.980","Text":"Where g is an infinite power series in r with coefficients depending"},{"Start":"00:42.980 ","End":"00:48.440","Text":"on Theta minus Ksi and C is an arbitrary constant."},{"Start":"00:48.440 ","End":"00:51.905","Text":"We also have to show that if the problem is solvable,"},{"Start":"00:51.905 ","End":"00:54.650","Text":"then this integral is 0,"},{"Start":"00:54.650 ","End":"00:57.830","Text":"and this becomes a necessary condition for solvability."},{"Start":"00:57.830 ","End":"01:02.360","Text":"Separation of variables means that we start for looking for solutions of"},{"Start":"01:02.360 ","End":"01:08.030","Text":"the form u of r Theta is some function of r times some function of Theta."},{"Start":"01:08.030 ","End":"01:14.480","Text":"At the end, we\u0027ll take an infinite linear combination of such solutions. You\u0027ll see."},{"Start":"01:14.480 ","End":"01:21.865","Text":"From the Laplacian, we get the following by using this, u_rr,"},{"Start":"01:21.865 ","End":"01:26.090","Text":"we just have to differentiate R. When we see u_Theta Theta,"},{"Start":"01:26.090 ","End":"01:28.535","Text":"we just differentiate with respect to Theta,"},{"Start":"01:28.535 ","End":"01:29.690","Text":"just the Q part."},{"Start":"01:29.690 ","End":"01:31.040","Text":"It\u0027s all separated."},{"Start":"01:31.040 ","End":"01:38.480","Text":"Now let\u0027s multiply this equation by r^2 to get rid of these fractions and divide by RQ."},{"Start":"01:38.480 ","End":"01:40.970","Text":"What we get is the following,"},{"Start":"01:40.970 ","End":"01:43.410","Text":"where we move this term to the right-hand side."},{"Start":"01:43.410 ","End":"01:47.330","Text":"There we\u0027ve separated functions of r on the left,"},{"Start":"01:47.330 ","End":"01:49.490","Text":"functions of Theta on the right,"},{"Start":"01:49.490 ","End":"01:53.810","Text":"which means that this is some constant, call it Lambda."},{"Start":"01:53.810 ","End":"01:57.650","Text":"Now, we can separate this into 2 equations."},{"Start":"01:57.650 ","End":"02:00.200","Text":"If we take this bit equals Lambda,"},{"Start":"02:00.200 ","End":"02:01.610","Text":"we get the following."},{"Start":"02:01.610 ","End":"02:03.740","Text":"If we take this equals Lambda,"},{"Start":"02:03.740 ","End":"02:05.150","Text":"we get the following."},{"Start":"02:05.150 ","End":"02:08.479","Text":"These are ordinary differential equations of the second order."},{"Start":"02:08.479 ","End":"02:11.854","Text":"Note that because Theta is a polar angle,"},{"Start":"02:11.854 ","End":"02:14.705","Text":"Q is a 2Pi periodic function,"},{"Start":"02:14.705 ","End":"02:17.630","Text":"and its derivative will also be 2Pi periodic."},{"Start":"02:17.630 ","End":"02:22.085","Text":"That gives us boundary conditions on this differential equation."},{"Start":"02:22.085 ","End":"02:25.355","Text":"It\u0027s a second-order, so we need 2 conditions and we have them."},{"Start":"02:25.355 ","End":"02:26.840","Text":"Now we can solve this."},{"Start":"02:26.840 ","End":"02:29.645","Text":"This is a Sturm-Liouville problem."},{"Start":"02:29.645 ","End":"02:31.520","Text":"I won\u0027t solve it here."},{"Start":"02:31.520 ","End":"02:35.645","Text":"Assume you know how to solve these or it\u0027s given in the formula sheet."},{"Start":"02:35.645 ","End":"02:42.290","Text":"Then the eigenvalues are Lambda n equals n^2 or n is 0,1,2, etc."},{"Start":"02:42.290 ","End":"02:44.780","Text":"The corresponding eigenfunctions,"},{"Start":"02:44.780 ","End":"02:47.660","Text":"q naught is the function 1."},{"Start":"02:47.660 ","End":"02:51.429","Text":"For each n, there are 2 eigenfunctions,"},{"Start":"02:51.429 ","End":"02:53.400","Text":"Q_n 1 and Q_n 2."},{"Start":"02:53.400 ","End":"02:57.055","Text":"One is cosine n Theta and the other is sine n Theta."},{"Start":"02:57.055 ","End":"02:59.630","Text":"We could quickly mentally check this."},{"Start":"02:59.630 ","End":"03:02.585","Text":"Each of these, if you differentiate it twice,"},{"Start":"03:02.585 ","End":"03:09.860","Text":"you get minus n^2 times the same thing and minus n^2 is minus Lambda."},{"Start":"03:09.860 ","End":"03:12.670","Text":"It will satisfy this equation."},{"Start":"03:12.670 ","End":"03:17.030","Text":"Now the other one, the equation in R is the Cauchy Euler type."},{"Start":"03:17.030 ","End":"03:18.470","Text":"I\u0027ll just present the solutions,"},{"Start":"03:18.470 ","End":"03:22.295","Text":"but I suggest you look up Cauchy-Euler equations,"},{"Start":"03:22.295 ","End":"03:24.905","Text":"for example, in the Wikipedia."},{"Start":"03:24.905 ","End":"03:29.825","Text":"This shows you how to solve them and we happen to fall in a Case 1,"},{"Start":"03:29.825 ","End":"03:34.849","Text":"where m_1 and m_2 are plus or minus n. Anyway, that\u0027s returned."},{"Start":"03:34.849 ","End":"03:37.340","Text":"In our case, if you solve this equation,"},{"Start":"03:37.340 ","End":"03:40.750","Text":"we get that for n=0,"},{"Start":"03:40.750 ","End":"03:47.570","Text":"we have 2 solutions R is the constant 1 or R is natural log of r. For n bigger than 0,"},{"Start":"03:47.570 ","End":"03:50.270","Text":"we get again 2 solutions,"},{"Start":"03:50.270 ","End":"03:53.120","Text":"r to the n and r to the minus n. However,"},{"Start":"03:53.120 ","End":"03:55.145","Text":"we can rule some of these out,"},{"Start":"03:55.145 ","End":"03:59.825","Text":"because our solution has to be bounded at r=0."},{"Start":"03:59.825 ","End":"04:05.485","Text":"We can throw out the minus n and the natural log of r solutions."},{"Start":"04:05.485 ","End":"04:08.825","Text":"The general solution is as follows."},{"Start":"04:08.825 ","End":"04:12.710","Text":"For n=0, we get a combination of 1 and 1,"},{"Start":"04:12.710 ","End":"04:15.575","Text":"so it\u0027s a constant and we call that a_0 over 2."},{"Start":"04:15.575 ","End":"04:17.165","Text":"For n bigger than 0,"},{"Start":"04:17.165 ","End":"04:20.790","Text":"we get a combination of r to the n and either cosine"},{"Start":"04:20.790 ","End":"04:24.630","Text":"n Theta or sine n Theta with appropriate constants,"},{"Start":"04:24.630 ","End":"04:26.060","Text":"so that gives us this."},{"Start":"04:26.060 ","End":"04:28.550","Text":"That\u0027s the solution to Part A."},{"Start":"04:28.550 ","End":"04:32.270","Text":"In Part A, we found the general solution is this."},{"Start":"04:32.270 ","End":"04:37.355","Text":"But in Part B we\u0027re given a boundary condition of Neumann type, which is this."},{"Start":"04:37.355 ","End":"04:41.510","Text":"We\u0027re going to find the coefficients a and b and using the boundary condition,"},{"Start":"04:41.510 ","End":"04:48.200","Text":"let\u0027s differentiate this with respect to r. We get that n comes out in front."},{"Start":"04:48.200 ","End":"04:50.270","Text":"We have n here and n here,"},{"Start":"04:50.270 ","End":"04:55.955","Text":"and r is to the n minus 1 and then substitute r=1."},{"Start":"04:55.955 ","End":"04:58.205","Text":"On the left-hand side, we get g of Theta."},{"Start":"04:58.205 ","End":"05:03.495","Text":"On the right, this drops out because 1 to the n minus 1 is 1."},{"Start":"05:03.495 ","End":"05:08.450","Text":"This gives us the Fourier series for the function g. We can"},{"Start":"05:08.450 ","End":"05:14.000","Text":"use the formulas for the coefficients for n equals 1, 2, 3, etc."},{"Start":"05:14.000 ","End":"05:19.650","Text":"We have, well not a_n but na_n equals this integral."},{"Start":"05:19.650 ","End":"05:22.590","Text":"We get a_n by dividing by n. Similarly,"},{"Start":"05:22.590 ","End":"05:26.900","Text":"nb_n is the integral with a sign n Theta,"},{"Start":"05:26.900 ","End":"05:29.120","Text":"whereas here it was cosine n Theta."},{"Start":"05:29.120 ","End":"05:31.465","Text":"There\u0027s also a formula for a_0,"},{"Start":"05:31.465 ","End":"05:35.960","Text":"and that is that a_0 is the following integral."},{"Start":"05:35.960 ","End":"05:42.455","Text":"But in our case, a_0 is 0 because there is no constant term."},{"Start":"05:42.455 ","End":"05:47.405","Text":"We get the condition that the integral is 0."},{"Start":"05:47.405 ","End":"05:49.550","Text":"We don\u0027t need the 1 over Pi, of course."},{"Start":"05:49.550 ","End":"05:52.475","Text":"This gives us 1 part of part b."},{"Start":"05:52.475 ","End":"05:55.970","Text":"Note that a_0 doesn\u0027t appear in the boundary condition,"},{"Start":"05:55.970 ","End":"05:57.364","Text":"so it\u0027s not constraint."},{"Start":"05:57.364 ","End":"06:00.755","Text":"Now let\u0027s substitute the expressions for a_n and b_n."},{"Start":"06:00.755 ","End":"06:02.960","Text":"Don\u0027t forget to divide by n here."},{"Start":"06:02.960 ","End":"06:05.810","Text":"Then substitute them in the general solution,"},{"Start":"06:05.810 ","End":"06:08.735","Text":"and just collecting and rearranging a little bit,"},{"Start":"06:08.735 ","End":"06:11.000","Text":"we get the following expression."},{"Start":"06:11.000 ","End":"06:16.130","Text":"We also replace Theta by Ksi in the integrals because we already have Theta,"},{"Start":"06:16.130 ","End":"06:17.675","Text":"so we don\u0027t want to mix them."},{"Start":"06:17.675 ","End":"06:21.290","Text":"Now we can simplify this using trigonometry."},{"Start":"06:21.290 ","End":"06:24.380","Text":"We have an identity that cosine of Alpha minus"},{"Start":"06:24.380 ","End":"06:28.025","Text":"Beta is cosine Alpha cosine Beta plus sine Alpha sine Beta."},{"Start":"06:28.025 ","End":"06:34.245","Text":"We have this here except with n Ksi to that have Alpha and then Theta instead of Beta."},{"Start":"06:34.245 ","End":"06:36.695","Text":"What we get is the following."},{"Start":"06:36.695 ","End":"06:41.240","Text":"We just get n Psi minus Theta and the cosine here."},{"Start":"06:41.240 ","End":"06:44.750","Text":"We still get the g of Ksi and everything else is the same."},{"Start":"06:44.750 ","End":"06:50.330","Text":"Now exchanging the order of the summation and the integration,"},{"Start":"06:50.330 ","End":"06:52.125","Text":"we get the following."},{"Start":"06:52.125 ","End":"06:54.169","Text":"We can write this as a constant."},{"Start":"06:54.169 ","End":"06:59.615","Text":"That\u0027s the a_0 over 2 plus the integral of 1 over Pi, g Ksi,"},{"Start":"06:59.615 ","End":"07:04.590","Text":"and this bit is a function of Ksi minus"},{"Start":"07:04.590 ","End":"07:10.680","Text":"Theta and r as a power series in r. This is what we were requested to do."},{"Start":"07:10.680 ","End":"07:12.275","Text":"As to be precise,"},{"Start":"07:12.275 ","End":"07:14.615","Text":"we can write it as follows."},{"Start":"07:14.615 ","End":"07:17.075","Text":"Let\u0027s keep the art to the n separate."},{"Start":"07:17.075 ","End":"07:22.470","Text":"Then g of r Theta is the sumX of 1 over Pi,"},{"Start":"07:22.470 ","End":"07:24.600","Text":"will put this over n,"},{"Start":"07:24.600 ","End":"07:25.950","Text":"the Ksi minus Theta,"},{"Start":"07:25.950 ","End":"07:27.525","Text":"we\u0027ll call it Phi."},{"Start":"07:27.525 ","End":"07:30.615","Text":"We have a function g(r) and Phi."},{"Start":"07:30.615 ","End":"07:35.760","Text":"That completes this problem and we are done."}],"ID":30909},{"Watched":false,"Name":"Exercise 45a","Duration":"3m 9s","ChapterTopicVideoID":29308,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.320","Text":"In this exercise, we\u0027re given that the function u(x,t) is a solution of this PDE."},{"Start":"00:07.320 ","End":"00:09.285","Text":"It\u0027s not a familiar type."},{"Start":"00:09.285 ","End":"00:12.090","Text":"Looks a bit like a heat equation but it\u0027s not."},{"Start":"00:12.090 ","End":"00:18.000","Text":"On the interval from minus infinity to infinity for x and t bigger than 0."},{"Start":"00:18.000 ","End":"00:21.015","Text":"We\u0027re given that for all t,"},{"Start":"00:21.015 ","End":"00:28.545","Text":"the limit of u(x,t) as x goes plus or minus infinity is 0 and similarly for du/dx."},{"Start":"00:28.545 ","End":"00:31.080","Text":"We also define 2 functions,"},{"Start":"00:31.080 ","End":"00:35.010","Text":"Q and S as the following improper integrals."},{"Start":"00:35.010 ","End":"00:39.020","Text":"We have to show that derivative of Q with respect to t is"},{"Start":"00:39.020 ","End":"00:44.935","Text":"0 and the derivative of s with respect to t is less than or equal to 0."},{"Start":"00:44.935 ","End":"00:48.160","Text":"Then part B we\u0027ll read when we come to it."},{"Start":"00:48.160 ","End":"00:50.280","Text":"We have 2 things to show in part a,"},{"Start":"00:50.280 ","End":"00:54.750","Text":"start with this one that dQ by dt is 0,"},{"Start":"00:54.750 ","End":"01:00.265","Text":"so dQ by dt is the derivative of with respect to t of this."},{"Start":"01:00.265 ","End":"01:03.110","Text":"Now, let\u0027s put the derivative"},{"Start":"01:03.110 ","End":"01:06.455","Text":"inside the integral and let\u0027s assume that the conditions are met."},{"Start":"01:06.455 ","End":"01:09.170","Text":"We have du/dt here."},{"Start":"01:09.170 ","End":"01:11.840","Text":"But if we look at the PDE,"},{"Start":"01:11.840 ","End":"01:14.695","Text":"du/dt is equal to this."},{"Start":"01:14.695 ","End":"01:16.095","Text":"We can integrate this,"},{"Start":"01:16.095 ","End":"01:22.620","Text":"the integral of 3u^2u_x is just u^3 and the integral of u_xx is just u_x."},{"Start":"01:22.620 ","End":"01:27.335","Text":"Now we have to evaluate this between minus infinity and infinity."},{"Start":"01:27.335 ","End":"01:34.954","Text":"But we are given that the limits of u and u_x at plus or minus infinity are both 0."},{"Start":"01:34.954 ","End":"01:40.645","Text":"This evaluates to 0 and that\u0027s the first of 2."},{"Start":"01:40.645 ","End":"01:43.380","Text":"Now let\u0027s turn to the other one, ds/dt."},{"Start":"01:43.380 ","End":"01:52.680","Text":"That\u0027s equal to again putting the derivative inside the integral d/dt(1/2u^2)."},{"Start":"01:53.390 ","End":"01:57.095","Text":"If we differentiate this with respect to t,"},{"Start":"01:57.095 ","End":"02:00.355","Text":"1/2u^2 gives u and the anti-derivative is u_t."},{"Start":"02:00.355 ","End":"02:02.565","Text":"We can substitute,"},{"Start":"02:02.565 ","End":"02:06.935","Text":"instead of u_t from the PDE this,"},{"Start":"02:06.935 ","End":"02:12.755","Text":"and then just expand this a bit of Algebra and we have this."},{"Start":"02:12.755 ","End":"02:15.664","Text":"The first term is fairly easy."},{"Start":"02:15.664 ","End":"02:20.540","Text":"The second term we\u0027ll need to use integration by parts and we\u0027ll"},{"Start":"02:20.540 ","End":"02:26.480","Text":"let u be f and u_xx will be g\u0027 in this formula."},{"Start":"02:26.480 ","End":"02:32.450","Text":"The integral of u^3u_x is just u^4 over 4."},{"Start":"02:32.450 ","End":"02:37.805","Text":"With the 3 makes it 3 quarters u^4 between minus infinity and infinity."},{"Start":"02:37.805 ","End":"02:41.250","Text":"Here using this, well I\u0027ll leave you to check."},{"Start":"02:41.250 ","End":"02:42.660","Text":"This is what we get."},{"Start":"02:42.660 ","End":"02:45.260","Text":"Now, from the asymptotic conditions as before,"},{"Start":"02:45.260 ","End":"02:49.745","Text":"we know that u goes to 0 at plus or minus infinity."},{"Start":"02:49.745 ","End":"02:56.720","Text":"Similarly here u and u_x go to 0 when x goes to plus or minus infinity."},{"Start":"02:56.720 ","End":"02:58.565","Text":"We\u0027re just left with this bit."},{"Start":"02:58.565 ","End":"03:02.190","Text":"Now whenever u_x is u_x^2 is non-negative,"},{"Start":"03:02.190 ","End":"03:06.380","Text":"so here we have something that\u0027s less than or equal to 0 and that\u0027s what we had to show."},{"Start":"03:06.380 ","End":"03:09.630","Text":"That concludes part A."}],"ID":30910},{"Watched":false,"Name":"Exercise 45b","Duration":"7m 41s","ChapterTopicVideoID":29309,"CourseChapterTopicPlaylistID":294426,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"Now we come to part B of the exercise."},{"Start":"00:03.045 ","End":"00:05.850","Text":"We\u0027re going to look for solutions to the PDE."},{"Start":"00:05.850 ","End":"00:08.570","Text":"This 1, of the form u(x,"},{"Start":"00:08.570 ","End":"00:11.430","Text":"t) is v(x) plus ct,"},{"Start":"00:11.430 ","End":"00:13.350","Text":"c is a positive constant,"},{"Start":"00:13.350 ","End":"00:17.190","Text":"and v is a function that\u0027s positive everywhere."},{"Start":"00:17.190 ","End":"00:21.960","Text":"Its limit as Ksi goes to minus infinity is 0,"},{"Start":"00:21.960 ","End":"00:27.360","Text":"and similarly, the limit of v\u0027 minus infinity is 0."},{"Start":"00:27.360 ","End":"00:33.780","Text":"The question is, does the solution satisfy the conditions that were in part A?"},{"Start":"00:33.780 ","End":"00:36.135","Text":"And if not, why not?"},{"Start":"00:36.135 ","End":"00:39.235","Text":"We\u0027ll start with u is v(x) plus ct,"},{"Start":"00:39.235 ","End":"00:41.900","Text":"and we want to substitute u in this equation,"},{"Start":"00:41.900 ","End":"00:43.100","Text":"so we\u0027ll also need u_t,"},{"Start":"00:43.100 ","End":"00:45.085","Text":"u_x, and, u_xx."},{"Start":"00:45.085 ","End":"00:52.130","Text":"Well, u by dt is the derivative times the inner derivative with respect to t,"},{"Start":"00:52.130 ","End":"00:56.910","Text":"which is c. U by dx is v\u0027 and the inner derivative is"},{"Start":"00:56.910 ","End":"01:02.335","Text":"1 and u_xx is v\u0027\u0027( x+ct) Now,"},{"Start":"01:02.335 ","End":"01:07.130","Text":"substituting this PDE, and what we get is the following,"},{"Start":"01:07.130 ","End":"01:12.020","Text":"u_t is cv\u0027 and u^2"},{"Start":"01:12.020 ","End":"01:19.840","Text":"is v^2 and u_x is v\u0027 and u_xx is v\u0027\u0027."},{"Start":"01:19.840 ","End":"01:24.335","Text":"This is an autonomous 2nd order ODE."},{"Start":"01:24.335 ","End":"01:29.375","Text":"Autonomous because we don\u0027t see the independent variable anywhere."},{"Start":"01:29.375 ","End":"01:34.625","Text":"V doesn\u0027t depend on Ksi or whatever the independent variable is."},{"Start":"01:34.625 ","End":"01:37.405","Text":"There\u0027s a technique for this."},{"Start":"01:37.405 ","End":"01:39.635","Text":"Even if you haven\u0027t covered this,"},{"Start":"01:39.635 ","End":"01:42.065","Text":"we can just show you right here what we do."},{"Start":"01:42.065 ","End":"01:45.980","Text":"We do what we call a reduction of order substitution."},{"Start":"01:45.980 ","End":"01:53.625","Text":"We let the derivative of v be some function of v. We\u0027ll have to find this function,"},{"Start":"01:53.625 ","End":"01:57.900","Text":"so what we get is the 2nd derivative,"},{"Start":"01:57.900 ","End":"02:00.140","Text":"is the derivative of this with respect to Ksi,"},{"Start":"02:00.140 ","End":"02:05.640","Text":"which is dp by dv times dv by d Ksi chain rule."},{"Start":"02:05.640 ","End":"02:10.340","Text":"But dv by d Ksi is just p. I just changed the order,"},{"Start":"02:10.340 ","End":"02:13.450","Text":"put the p(v) in front dp by dv,"},{"Start":"02:13.450 ","End":"02:20.130","Text":"and what we get is v\u0027 is this p,"},{"Start":"02:20.130 ","End":"02:22.035","Text":"dp by dv,"},{"Start":"02:22.035 ","End":"02:24.120","Text":"and v\u0027 is p,"},{"Start":"02:24.120 ","End":"02:26.285","Text":"so we get this equation."},{"Start":"02:26.285 ","End":"02:28.615","Text":"We can factorize it."},{"Start":"02:28.615 ","End":"02:37.820","Text":"The 2 possibilities, either p is 0 or dp by dv equals c minus 3v^2."},{"Start":"02:37.820 ","End":"02:40.180","Text":"When I rule, this one out,"},{"Start":"02:40.180 ","End":"02:47.105","Text":"you\u0027ll see p=0 gives that v is a constant because p is the derivative of v,"},{"Start":"02:47.105 ","End":"02:50.170","Text":"so v as a function of Ksi is a constant."},{"Start":"02:50.170 ","End":"02:56.015","Text":"This constant has to be 0 because the limit as Ksi goes to minus infinity is 0,"},{"Start":"02:56.015 ","End":"02:57.870","Text":"but it has to be the same constant."},{"Start":"02:57.870 ","End":"03:00.880","Text":"If this constant is 0, here it is 0 also."},{"Start":"03:00.880 ","End":"03:04.090","Text":"But this is not good for us because we were required"},{"Start":"03:04.090 ","End":"03:07.660","Text":"to find v bigger than 0 not equal to 0."},{"Start":"03:07.660 ","End":"03:09.745","Text":"Will go with the other choice,"},{"Start":"03:09.745 ","End":"03:13.580","Text":"dp by dv equals c minus 3v^2."},{"Start":"03:13.580 ","End":"03:18.190","Text":"We can get p from here by integrating both sides with respect to v,"},{"Start":"03:18.190 ","End":"03:19.420","Text":"so we get that p,"},{"Start":"03:19.420 ","End":"03:23.540","Text":"which you recall is dv by d Ksi is equal to some constant of"},{"Start":"03:23.540 ","End":"03:28.255","Text":"integration a plus cv minus v^3."},{"Start":"03:28.255 ","End":"03:34.175","Text":"We\u0027re given that the limit of v and Ksi is 0 minus infinity,"},{"Start":"03:34.175 ","End":"03:36.875","Text":"and the limit of v\u0027,"},{"Start":"03:36.875 ","End":"03:40.010","Text":"which is p, is minus infinity."},{"Start":"03:40.010 ","End":"03:42.560","Text":"Ksi goes to minus infinity,"},{"Start":"03:42.560 ","End":"03:48.815","Text":"this goes to 0, so it means that A is 0."},{"Start":"03:48.815 ","End":"03:52.600","Text":"Dv by d Ksi is cv minus v^3."},{"Start":"03:52.600 ","End":"03:55.370","Text":"We can use separation of variables on this,"},{"Start":"03:55.370 ","End":"04:00.890","Text":"putting the d Ksi to the right and all the vs on the denominator and taking the integral."},{"Start":"04:00.890 ","End":"04:02.480","Text":"We get the following,"},{"Start":"04:02.480 ","End":"04:05.430","Text":"c is given to be positive."},{"Start":"04:05.430 ","End":"04:07.860","Text":"We can let Alpha be the square root of c,"},{"Start":"04:07.860 ","End":"04:10.260","Text":"and then c is equal to Alpha squared."},{"Start":"04:10.260 ","End":"04:13.250","Text":"Then we can break this up using partial fractions."},{"Start":"04:13.250 ","End":"04:17.540","Text":"The denominator decomposes us to v Alpha plus v Alpha minus"},{"Start":"04:17.540 ","End":"04:22.880","Text":"v. Let\u0027s make it v plus Alpha v minus Alpha and put a minus on the numerator."},{"Start":"04:22.880 ","End":"04:25.400","Text":"I\u0027m not going to give the details of this."},{"Start":"04:25.400 ","End":"04:27.875","Text":"What you could do is check,"},{"Start":"04:27.875 ","End":"04:33.120","Text":"put this over a common denominator and we see that we get this in return."},{"Start":"04:33.120 ","End":"04:36.500","Text":"We can verify it, at least the integral of this, well,"},{"Start":"04:36.500 ","End":"04:39.890","Text":"this is twice natural log of v,"},{"Start":"04:39.890 ","End":"04:47.325","Text":"which is natural log of v^2 minus natural log of v plus Alpha,"},{"Start":"04:47.325 ","End":"04:50.130","Text":"absolute value, which makes it go in the denominator,"},{"Start":"04:50.130 ","End":"04:51.620","Text":"and this on the denominator."},{"Start":"04:51.620 ","End":"04:53.390","Text":"Here we have v plus Alpha,"},{"Start":"04:53.390 ","End":"04:55.780","Text":"v minus Alpha, which is this."},{"Start":"04:55.780 ","End":"05:00.260","Text":"This is an indefinite integral of the left-hand side."},{"Start":"05:00.260 ","End":"05:02.600","Text":"On the right-hand side, we get just Ksi,"},{"Start":"05:02.600 ","End":"05:06.980","Text":"but we need to add a constant of integration that into the left-hand side."},{"Start":"05:06.980 ","End":"05:10.070","Text":"This plus Beta equals Ksi."},{"Start":"05:10.070 ","End":"05:16.200","Text":"Now, we\u0027re going to do some algebra to try and extract v from all of this."},{"Start":"05:16.490 ","End":"05:19.755","Text":"Multiplied by 2 Alpha squared,"},{"Start":"05:19.755 ","End":"05:23.780","Text":"then we can invert this fraction and make it minus,"},{"Start":"05:23.780 ","End":"05:28.330","Text":"so we get Beta minus Ksi that x Ksi minus Beta."},{"Start":"05:28.330 ","End":"05:32.370","Text":"This is 1 minus Alpha squared over v^2,"},{"Start":"05:32.370 ","End":"05:36.815","Text":"and the log we can put on the other side of the exponent."},{"Start":"05:36.815 ","End":"05:41.995","Text":"Then we can get rid of the absolute value if you put plus or minus."},{"Start":"05:41.995 ","End":"05:44.190","Text":"Then we can extract v,"},{"Start":"05:44.190 ","End":"05:45.680","Text":"you can put this on this side,"},{"Start":"05:45.680 ","End":"05:50.390","Text":"this on that side, it\u0027s just algebra and we get this."},{"Start":"05:50.390 ","End":"05:52.235","Text":"Now, there\u0027s 2 possibilities here,"},{"Start":"05:52.235 ","End":"05:53.581","Text":"minus or plus,"},{"Start":"05:53.581 ","End":"06:02.395","Text":"but the minus can\u0027t be because we need this to be defined for all Ksi."},{"Start":"06:02.395 ","End":"06:04.500","Text":"When Ksi goes to infinity,"},{"Start":"06:04.500 ","End":"06:07.190","Text":"this is negative, this is going to be sometimes negative,"},{"Start":"06:07.190 ","End":"06:08.675","Text":"and then we have to take the square root,"},{"Start":"06:08.675 ","End":"06:11.960","Text":"so we just take the plus inside here."},{"Start":"06:11.960 ","End":"06:16.130","Text":"Then, recalling that Alpha is the square root of c,"},{"Start":"06:16.130 ","End":"06:17.480","Text":"this is square root of c,"},{"Start":"06:17.480 ","End":"06:21.170","Text":"Alpha squared is c. We can also reverse the order here and"},{"Start":"06:21.170 ","End":"06:25.355","Text":"put a minus and then put the square root over the whole thing."},{"Start":"06:25.355 ","End":"06:27.935","Text":"This is v of Ksi."},{"Start":"06:27.935 ","End":"06:31.215","Text":"We note that when Ksi goes to minus infinity,"},{"Start":"06:31.215 ","End":"06:36.410","Text":"v goes to 0 because this gets large and it\u0027s in the denominator."},{"Start":"06:36.410 ","End":"06:38.900","Text":"A bit of work to differentiate,"},{"Start":"06:38.900 ","End":"06:42.484","Text":"you can see that the derivative also goes to 0."},{"Start":"06:42.484 ","End":"06:48.740","Text":"Now, to get u, we just have to substitute u(x, t) is v(x+ct)."},{"Start":"06:48.740 ","End":"06:50.795","Text":"Instead of Ksi here,"},{"Start":"06:50.795 ","End":"06:53.015","Text":"we put x plus ct,"},{"Start":"06:53.015 ","End":"06:55.745","Text":"and that\u0027s the answer."},{"Start":"06:55.745 ","End":"06:59.270","Text":"But there was another bit we have to show or we\u0027re"},{"Start":"06:59.270 ","End":"07:02.915","Text":"asked if the conditions in part A was satisfied."},{"Start":"07:02.915 ","End":"07:07.530","Text":"I\u0027ll remind you what they are because these conditions,"},{"Start":"07:07.530 ","End":"07:11.350","Text":"is equality is assumed that the integral exists at all."},{"Start":"07:11.350 ","End":"07:17.510","Text":"Integral exists, so is defined or converges only if"},{"Start":"07:17.510 ","End":"07:24.290","Text":"u goes to 0 when x goes to plus or minus infinity."},{"Start":"07:24.290 ","End":"07:26.435","Text":"In our case, well,"},{"Start":"07:26.435 ","End":"07:29.810","Text":"neither of these integrals exists because the limit when x"},{"Start":"07:29.810 ","End":"07:33.265","Text":"goes to infinity of u of here,"},{"Start":"07:33.265 ","End":"07:34.835","Text":"is the square root of c,"},{"Start":"07:34.835 ","End":"07:38.135","Text":"which is not 0, it\u0027s actually positive."},{"Start":"07:38.135 ","End":"07:42.480","Text":"That answers the extra question and now we\u0027re done."}],"ID":30911}],"Thumbnail":null,"ID":294426}]

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1.1

1