[{"Name":"Infinite Interval, Poisson Formula","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Heat Equation and Poisson\u0026#39;s Formula","Duration":"5m 58s","ChapterTopicVideoID":29212,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.035","Text":"New topic, the heat equation from physics and the Poisson formula for solving it."},{"Start":"00:08.035 ","End":"00:11.230","Text":"This is the partial differential equation,"},{"Start":"00:11.230 ","End":"00:18.620","Text":"du by dt equals a^2 second derivative of u with respect to x."},{"Start":"00:18.620 ","End":"00:22.180","Text":"This is called the one-dimensional heat equation."},{"Start":"00:22.180 ","End":"00:23.890","Text":"There is a dimension of time,"},{"Start":"00:23.890 ","End":"00:28.015","Text":"but there\u0027s only one space dimension, x."},{"Start":"00:28.015 ","End":"00:34.045","Text":"In physics, it could represent how heat disperses"},{"Start":"00:34.045 ","End":"00:40.495","Text":"in a metal bar or a rod or something we can look at as one-dimensional."},{"Start":"00:40.495 ","End":"00:43.920","Text":"There are higher-dimensional heat equations,"},{"Start":"00:43.920 ","End":"00:46.070","Text":"there is a 3D equation,"},{"Start":"00:46.070 ","End":"00:49.730","Text":"which is just simply an extension of this instead of x,"},{"Start":"00:49.730 ","End":"00:52.270","Text":"we have x, y, and z."},{"Start":"00:52.270 ","End":"00:58.709","Text":"This describes how heat disperses in space or in a body that\u0027s heated."},{"Start":"00:59.420 ","End":"01:04.910","Text":"Here\u0027s an example of a setup from real life for physics."},{"Start":"01:04.910 ","End":"01:10.090","Text":"You have a rod considered one-dimensional,"},{"Start":"01:10.090 ","End":"01:15.215","Text":"and it\u0027s heated up here and here with 2 candles."},{"Start":"01:15.215 ","End":"01:18.650","Text":"You can expect that it will get hotter here and here,"},{"Start":"01:18.650 ","End":"01:23.860","Text":"and then it will flow from these 2 points inwards and towards the end."},{"Start":"01:23.860 ","End":"01:27.470","Text":"It could represent temperature."},{"Start":"01:27.470 ","End":"01:33.125","Text":"We could mark it as darker blue means hotter and white is cooler."},{"Start":"01:33.125 ","End":"01:42.005","Text":"But we\u0027ll actually use a graph x will be the displacement along the rod,"},{"Start":"01:42.005 ","End":"01:45.260","Text":"so we mark one end as 0,"},{"Start":"01:45.260 ","End":"01:48.680","Text":"and then just measure it in certain units,"},{"Start":"01:48.680 ","End":"01:55.250","Text":"and the shade of blue will be replaced by vertical axis, the temperature,"},{"Start":"01:55.250 ","End":"01:58.490","Text":"like here where it\u0027s hottest, this is highest,"},{"Start":"01:58.490 ","End":"02:02.915","Text":"it\u0027s cooler here, down here, hotter here."},{"Start":"02:02.915 ","End":"02:08.700","Text":"We\u0027ll use this form and we\u0027re going to look for a solution u(x,"},{"Start":"02:08.700 ","End":"02:12.760","Text":"t), which satisfies the PDE."},{"Start":"02:12.760 ","End":"02:19.910","Text":"Now, this equation came from physicists but I\u0027d like to show you why it\u0027s plausible,"},{"Start":"02:19.910 ","End":"02:22.580","Text":"how to make a bit of sense of it."},{"Start":"02:22.580 ","End":"02:25.940","Text":"In this situation here where it\u0027s hot,"},{"Start":"02:25.940 ","End":"02:29.765","Text":"you\u0027d expect the heat to flow to the sides,"},{"Start":"02:29.765 ","End":"02:33.860","Text":"the left and to the right, and similarly here."},{"Start":"02:33.860 ","End":"02:36.380","Text":"Whereas here where it\u0027s cold,"},{"Start":"02:36.380 ","End":"02:38.675","Text":"you\u0027d expect it to attract the heat,"},{"Start":"02:38.675 ","End":"02:40.280","Text":"this will get hotter."},{"Start":"02:40.280 ","End":"02:43.055","Text":"Now, if you look at the graph,"},{"Start":"02:43.055 ","End":"02:49.590","Text":"we see that here it\u0027s convex or concave down."},{"Start":"02:50.450 ","End":"02:58.795","Text":"The temperature in this interval would start off by getting lower,"},{"Start":"02:58.795 ","End":"03:04.670","Text":"so the derivative of u with respect to t would be negative,"},{"Start":"03:04.670 ","End":"03:09.240","Text":"so u_xx is negative and u_t is negative,"},{"Start":"03:09.240 ","End":"03:11.495","Text":"and the more convex,"},{"Start":"03:11.495 ","End":"03:14.180","Text":"the more the temperature will fall,"},{"Start":"03:14.180 ","End":"03:21.905","Text":"so it makes sense that u_t is proportional to u_xx with some positive constant,"},{"Start":"03:21.905 ","End":"03:24.080","Text":"which we call a^2."},{"Start":"03:24.080 ","End":"03:26.617","Text":"Here, where it\u0027s cool,"},{"Start":"03:26.617 ","End":"03:30.960","Text":"u_xx is positive because it\u0027s concave up."},{"Start":"03:30.960 ","End":"03:36.845","Text":"Just convex, u_xx is positive and the temperature increases here."},{"Start":"03:36.845 ","End":"03:39.830","Text":"Again, they go in the same direction,"},{"Start":"03:39.830 ","End":"03:41.885","Text":"makes sense that they\u0027re proportional,"},{"Start":"03:41.885 ","End":"03:43.925","Text":"and here it\u0027s just like here."},{"Start":"03:43.925 ","End":"03:51.515","Text":"Now, just the PDE alone is not enough information for solving the heat equation,"},{"Start":"03:51.515 ","End":"03:56.720","Text":"which maps u as a function of x and t. In other words,"},{"Start":"03:56.720 ","End":"04:00.125","Text":"it\u0027s like a function u(x) that changes over time."},{"Start":"04:00.125 ","End":"04:02.585","Text":"We need more information."},{"Start":"04:02.585 ","End":"04:04.070","Text":"We need, for example,"},{"Start":"04:04.070 ","End":"04:06.035","Text":"the position and length of the rod."},{"Start":"04:06.035 ","End":"04:12.505","Text":"Let\u0027s say it\u0027s a finite rod of length L and the left endpoint is at 0."},{"Start":"04:12.505 ","End":"04:18.515","Text":"But we also very importantly need the starting temperature distribution."},{"Start":"04:18.515 ","End":"04:20.330","Text":"Like an initial condition,"},{"Start":"04:20.330 ","End":"04:25.730","Text":"we need to know the temperature along the rod at time 0."},{"Start":"04:25.730 ","End":"04:29.720","Text":"The first case we\u0027ll consider is an infinite rod,"},{"Start":"04:29.720 ","End":"04:34.310","Text":"which is presumed to lie on the x-axis,"},{"Start":"04:34.310 ","End":"04:37.475","Text":"so the rod represents the x-axis."},{"Start":"04:37.475 ","End":"04:40.850","Text":"Again, we consider it to be just one-dimensional,"},{"Start":"04:40.850 ","End":"04:45.020","Text":"which will mean that the temperature at any cross-section is the same."},{"Start":"04:45.020 ","End":"04:50.285","Text":"It doesn\u0027t vary with the width only with the length of x."},{"Start":"04:50.285 ","End":"04:54.515","Text":"There\u0027s a closed formula due to Poisson."},{"Start":"04:54.515 ","End":"04:57.155","Text":"The solution to the PDE,"},{"Start":"04:57.155 ","End":"05:02.480","Text":"this is our heat equation on the interval for x,"},{"Start":"05:02.480 ","End":"05:04.490","Text":"which is all the real line,"},{"Start":"05:04.490 ","End":"05:08.720","Text":"and with the initial condition like we mentioned,"},{"Start":"05:08.720 ","End":"05:11.810","Text":"u(x) when t is 0,"},{"Start":"05:11.810 ","End":"05:17.495","Text":"so you get the temperature function along the rod at time 0."},{"Start":"05:17.495 ","End":"05:27.820","Text":"The answer is that u(x) and t in general after time t is equal to this integral."},{"Start":"05:27.820 ","End":"05:29.760","Text":"There\u0027s a y here,"},{"Start":"05:29.760 ","End":"05:30.810","Text":"but it\u0027s dy,"},{"Start":"05:30.810 ","End":"05:33.570","Text":"so the y will disappear,"},{"Start":"05:33.570 ","End":"05:41.250","Text":"and this will be some function of x and t. I should say u(x) naught also could be"},{"Start":"05:41.250 ","End":"05:45.590","Text":"any well-behaved function of x. I will prove"},{"Start":"05:45.590 ","End":"05:50.675","Text":"this formula in the next clip and later after that,"},{"Start":"05:50.675 ","End":"05:55.415","Text":"there will be solved examples and that\u0027s where you\u0027ll see how we use this."},{"Start":"05:55.415 ","End":"05:59.220","Text":"That concludes this clip."}],"ID":30796},{"Watched":false,"Name":"Proof of Poisson Formula Proof (part 1)","Duration":"6m 19s","ChapterTopicVideoID":29209,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.640","Text":"In this clip, we\u0027ll prove the Poisson formula but to do that,"},{"Start":"00:05.640 ","End":"00:08.610","Text":"we\u0027ll use the Fourier transform."},{"Start":"00:08.610 ","End":"00:11.730","Text":"Now, if you haven\u0027t studied the Fourier transform,"},{"Start":"00:11.730 ","End":"00:14.834","Text":"you might want to skip this clip or if you\u0027re brave,"},{"Start":"00:14.834 ","End":"00:18.780","Text":"you could make do with just the review and in any case,"},{"Start":"00:18.780 ","End":"00:21.270","Text":"if you have studied Fourier transform,"},{"Start":"00:21.270 ","End":"00:24.900","Text":"I recommend going through this because it also reviews"},{"Start":"00:24.900 ","End":"00:29.880","Text":"the technique of how to solve differential equations using the Fourier transform,"},{"Start":"00:29.880 ","End":"00:34.349","Text":"which up to now you might have seen in the case of ordinary differential equations."},{"Start":"00:34.349 ","End":"00:36.480","Text":"Any way, let\u0027s get started."},{"Start":"00:36.480 ","End":"00:41.585","Text":"We\u0027re going to define the Fourier transform on a space known as this"},{"Start":"00:41.585 ","End":"00:46.805","Text":"or this piecewise continuous functions which are absolutely integrable."},{"Start":"00:46.805 ","End":"00:49.330","Text":"I won\u0027t go over what these terms mean,"},{"Start":"00:49.330 ","End":"00:50.480","Text":"you should know that."},{"Start":"00:50.480 ","End":"00:54.800","Text":"Let\u0027s not just assume that all the functions that we encounter have these properties."},{"Start":"00:54.800 ","End":"00:59.150","Text":"Now, the Fourier transform maps a function to a function,"},{"Start":"00:59.150 ","End":"01:04.220","Text":"and it maps a function f to another function."},{"Start":"01:04.220 ","End":"01:06.515","Text":"We\u0027ll use the notation f hat,"},{"Start":"01:06.515 ","End":"01:09.295","Text":"f caret, f circumflex,"},{"Start":"01:09.295 ","End":"01:11.030","Text":"whatever you want to call it."},{"Start":"01:11.030 ","End":"01:16.320","Text":"Where this f caret is a function of Omega."},{"Start":"01:16.320 ","End":"01:18.825","Text":"F is a function of x,"},{"Start":"01:18.825 ","End":"01:20.780","Text":"f caret is a function of Omega."},{"Start":"01:20.780 ","End":"01:27.268","Text":"Often in physics, x represents time and omega represents frequency,"},{"Start":"01:27.268 ","End":"01:35.450","Text":"and use that notation sometimes that f(x) is transformed by f to f hat."},{"Start":"01:35.450 ","End":"01:40.880","Text":"It turns out that there is an inverse transform that we can get back"},{"Start":"01:40.880 ","End":"01:46.620","Text":"from f hat to f using this formula which is similar to this,"},{"Start":"01:46.620 ","End":"01:52.160","Text":"except there is no 1 over 2Pi and instead of the minus here it\u0027s a plus."},{"Start":"01:52.160 ","End":"01:56.320","Text":"So it\u0027s like a two-way transformation mapping."},{"Start":"01:56.320 ","End":"02:01.790","Text":"This is not quite precise when we map back,"},{"Start":"02:01.790 ","End":"02:04.580","Text":"we don\u0027t always get back to f(x),"},{"Start":"02:04.580 ","End":"02:11.480","Text":"but to the average of the right limit and left limit of f(x),"},{"Start":"02:11.480 ","End":"02:13.670","Text":"the average, this plus this over 2,"},{"Start":"02:13.670 ","End":"02:15.650","Text":"But if f is continuous at x,"},{"Start":"02:15.650 ","End":"02:18.110","Text":"then this is just f(x)."},{"Start":"02:18.110 ","End":"02:21.470","Text":"Now some examples and mainly properties,"},{"Start":"02:21.470 ","End":"02:24.335","Text":"at least the ones we\u0027re going to need for the proof."},{"Start":"02:24.335 ","End":"02:28.550","Text":"First of all, it\u0027s linear that takes sums to"},{"Start":"02:28.550 ","End":"02:32.975","Text":"sums and a constant can come out of the transformation,"},{"Start":"02:32.975 ","End":"02:42.725","Text":"that\u0027s linearity, is also a thing that convolutions go to products except for the 2Pi."},{"Start":"02:42.725 ","End":"02:47.040","Text":"In case you forgotten or don\u0027t know what convolution of functions is,"},{"Start":"02:47.040 ","End":"02:49.339","Text":"here\u0027s the definition,"},{"Start":"02:49.339 ","End":"02:51.115","Text":"I won\u0027t dwell on it."},{"Start":"02:51.115 ","End":"02:55.400","Text":"There\u0027s another property that if we"},{"Start":"02:55.400 ","End":"02:59.465","Text":"take the Fourier transform of the nth derivative of a function,"},{"Start":"02:59.465 ","End":"03:06.895","Text":"it goes to the transform multiplied by i Omega to the power of n. Now the property,"},{"Start":"03:06.895 ","End":"03:14.990","Text":"sometimes called similarity is that if you transform f of a constant times x, ax,"},{"Start":"03:14.990 ","End":"03:19.955","Text":"which means that the graph is compressed by a factor of a,"},{"Start":"03:19.955 ","End":"03:24.680","Text":"then it goes to the function which is expanded by a factor of"},{"Start":"03:24.680 ","End":"03:29.975","Text":"a and also this extra constant that a can\u0027t be 0 here, of course."},{"Start":"03:29.975 ","End":"03:33.050","Text":"Now this property is important for us for the proof."},{"Start":"03:33.050 ","End":"03:35.825","Text":"We\u0027re going to take a specific example that we need."},{"Start":"03:35.825 ","End":"03:39.230","Text":"It turns out that the Fourier transform of e to"},{"Start":"03:39.230 ","End":"03:44.090","Text":"the minus x squared is the following function of Omega."},{"Start":"03:44.090 ","End":"03:51.150","Text":"Now let\u0027s apply this formula with a equaling 1 over"},{"Start":"03:51.150 ","End":"04:00.128","Text":"2a square root of t. That gives us 1 over a is just 2a square root of t,"},{"Start":"04:00.128 ","End":"04:05.780","Text":"and also here, 1 over a means that the 2a square root of t comes to the numerator"},{"Start":"04:05.780 ","End":"04:11.668","Text":"so that the Omega over a here is Omega times this,"},{"Start":"04:11.668 ","End":"04:16.890","Text":"and this simplifies to 2 squared is 4 cancels with the"},{"Start":"04:16.890 ","End":"04:22.610","Text":"4 a squared and Omega squared t. Now using the linearity property,"},{"Start":"04:22.610 ","End":"04:25.760","Text":"let\u0027s take this constant and move it to"},{"Start":"04:25.760 ","End":"04:30.860","Text":"the other side and put it inside the transform and we"},{"Start":"04:30.860 ","End":"04:35.300","Text":"get 1 over a square root of Pi over t times"},{"Start":"04:35.300 ","End":"04:40.355","Text":"this is equal to just the e to the minus a squared Omega squared t and also,"},{"Start":"04:40.355 ","End":"04:47.045","Text":"we\u0027ve opened up the square here so we get x squared over 4a squared t. Now,"},{"Start":"04:47.045 ","End":"04:51.620","Text":"another thing is that if we have a function of 2 variables,"},{"Start":"04:51.620 ","End":"04:54.230","Text":"as we often do in partial differential equations,"},{"Start":"04:54.230 ","End":"04:56.197","Text":"say u is u(x,"},{"Start":"04:56.197 ","End":"05:06.210","Text":"t), then we define the Fourier transform to act just on the first parameter in the 2."},{"Start":"05:06.210 ","End":"05:08.595","Text":"It\u0027s as if t was like a constant,"},{"Start":"05:08.595 ","End":"05:11.890","Text":"and for each t we define the Fourier transform"},{"Start":"05:11.890 ","End":"05:17.300","Text":"of u(x) and t just by treating x as the variable."},{"Start":"05:17.300 ","End":"05:22.205","Text":"So t just sits there and we just use the same formula"},{"Start":"05:22.205 ","End":"05:27.875","Text":"as for f(x) going to f hat of Omega."},{"Start":"05:27.875 ","End":"05:31.055","Text":"That\u0027s guaranteed, when we put the hat and take the Fourier transform,"},{"Start":"05:31.055 ","End":"05:33.950","Text":"it\u0027s the first variable that it acts on."},{"Start":"05:33.950 ","End":"05:36.830","Text":"The second one just comes along for the ride."},{"Start":"05:36.830 ","End":"05:40.370","Text":"Then we have the following rule that if we"},{"Start":"05:40.370 ","End":"05:45.395","Text":"differentiate this function u with respect to t,"},{"Start":"05:45.395 ","End":"05:47.075","Text":"the second variable,"},{"Start":"05:47.075 ","End":"05:49.513","Text":"and apply the Fourier transform,"},{"Start":"05:49.513 ","End":"05:52.370","Text":"it\u0027s like first applying the Fourier transform and"},{"Start":"05:52.370 ","End":"05:55.340","Text":"then taking the derivative with respect to"},{"Start":"05:55.340 ","End":"06:01.685","Text":"t. The order of Fourier transform and derivative with respect to t can be reversed."},{"Start":"06:01.685 ","End":"06:06.130","Text":"If you don\u0027t like this notation with this funny d,"},{"Start":"06:06.130 ","End":"06:09.985","Text":"well, we could write it this way."},{"Start":"06:09.985 ","End":"06:13.670","Text":"That\u0027s it for this review of the Fourier transform,"},{"Start":"06:13.670 ","End":"06:20.040","Text":"let\u0027s take a break and then get into the proof of the Poisson formula."}],"ID":30797},{"Watched":false,"Name":"Proof of Poisson Formula Proof (part 2)","Duration":"6m 56s","ChapterTopicVideoID":29211,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:03.330","Text":"Okay. We\u0027re continuing after the break,"},{"Start":"00:03.330 ","End":"00:06.450","Text":"the proof of the Poisson formula."},{"Start":"00:06.450 ","End":"00:09.360","Text":"We just reviewed the Fourier transform."},{"Start":"00:09.360 ","End":"00:14.500","Text":"Here\u0027s the heat equation with its initial condition."},{"Start":"00:14.500 ","End":"00:18.150","Text":"The Poisson formula says that u(x,"},{"Start":"00:18.150 ","End":"00:20.265","Text":"t) is given by this."},{"Start":"00:20.265 ","End":"00:22.305","Text":"That\u0027s what we have to prove."},{"Start":"00:22.305 ","End":"00:25.560","Text":"The idea of the proof is to take"},{"Start":"00:25.560 ","End":"00:31.380","Text":"the Fourier transform of the equation and then solve the new equation that we"},{"Start":"00:31.380 ","End":"00:36.030","Text":"get to find the Fourier transform of u and hopefully"},{"Start":"00:36.030 ","End":"00:40.875","Text":"the equation becomes much simpler after the Fourier transform."},{"Start":"00:40.875 ","End":"00:44.015","Text":"When we found the transform of u,"},{"Start":"00:44.015 ","End":"00:47.525","Text":"we just have to take the inverse transform to get back to u."},{"Start":"00:47.525 ","End":"00:49.115","Text":"That\u0027s the plan."},{"Start":"00:49.115 ","End":"00:53.865","Text":"F(u, t) is f(a^2 u_xx),"},{"Start":"00:53.865 ","End":"00:56.115","Text":"f being the Fourier transform."},{"Start":"00:56.115 ","End":"01:01.070","Text":"Using linearity, we can take a^2 in front of the f. Now,"},{"Start":"01:01.070 ","End":"01:03.245","Text":"we\u0027re going to use 2 formulas."},{"Start":"01:03.245 ","End":"01:09.530","Text":"1 is the formula for the nth derivative of a function which is given like"},{"Start":"01:09.530 ","End":"01:15.845","Text":"so and we\u0027re going to use that for the derivative with respect to x."},{"Start":"01:15.845 ","End":"01:20.240","Text":"This formula, we\u0027re going to use for the derivative with respect to"},{"Start":"01:20.240 ","End":"01:27.485","Text":"t. Fourier transform of du by dt is the Fourier transform of,"},{"Start":"01:27.485 ","End":"01:30.890","Text":"well, like this, du by dt with this notation,"},{"Start":"01:30.890 ","End":"01:36.215","Text":"which is the derivative with respect to t of the transform of u."},{"Start":"01:36.215 ","End":"01:40.640","Text":"That\u0027s the left-hand side and now here\u0027s the a^2 from here."},{"Start":"01:40.640 ","End":"01:46.015","Text":"Then u_xx, it\u0027s the 2nd derivative of u with respect to x. X is the main variable"},{"Start":"01:46.015 ","End":"01:52.790","Text":"and that\u0027s where we can use this formula with n=2 and instead of this f,"},{"Start":"01:52.790 ","End":"01:55.205","Text":"we have u(x, t)."},{"Start":"01:55.205 ","End":"01:58.135","Text":"We get i Omega^2,"},{"Start":"01:58.135 ","End":"02:01.320","Text":"the transform of u(omega t)."},{"Start":"02:01.320 ","End":"02:03.690","Text":"Now, i^2 is minus 1,"},{"Start":"02:03.690 ","End":"02:08.570","Text":"so this simplifies to minus a^2 Omega^2 u hat."},{"Start":"02:08.570 ","End":"02:10.040","Text":"Now, let\u0027s see what we have here."},{"Start":"02:10.040 ","End":"02:14.900","Text":"We have an ordinary differential equation of the first order in"},{"Start":"02:14.900 ","End":"02:21.505","Text":"u hat considered as a function of t. We\u0027ll treat Omega like a parameter."},{"Start":"02:21.505 ","End":"02:23.465","Text":"We have the derivative,"},{"Start":"02:23.465 ","End":"02:30.275","Text":"first derivative of u hat is some constant times u hat."},{"Start":"02:30.275 ","End":"02:34.925","Text":"If we think of this like function y(t),"},{"Start":"02:34.925 ","End":"02:38.450","Text":"we have y\u0027(t) is Alpha y(t)."},{"Start":"02:38.450 ","End":"02:43.885","Text":"That\u0027s the general pattern where Alpha is all this stuff here."},{"Start":"02:43.885 ","End":"02:46.085","Text":"We know the solution to this."},{"Start":"02:46.085 ","End":"02:50.270","Text":"It\u0027s y equals any constant times e to the Alpha"},{"Start":"02:50.270 ","End":"02:55.820","Text":"t. This is much easier than solving this original equation,"},{"Start":"02:55.820 ","End":"02:59.504","Text":"which is a 2nd order and with partial derivatives,"},{"Start":"02:59.504 ","End":"03:02.510","Text":"here we have simple 1st order ODE."},{"Start":"03:02.510 ","End":"03:06.905","Text":"In our case, the solution comes out to be as follows,"},{"Start":"03:06.905 ","End":"03:13.865","Text":"where Alpha is this minus a^2 Omega^2 and then the question is what is a?"},{"Start":"03:13.865 ","End":"03:16.025","Text":"If we let t=0,"},{"Start":"03:16.025 ","End":"03:19.850","Text":"we get that u hat(Omega 0) is a."},{"Start":"03:19.850 ","End":"03:22.290","Text":"Returning to u,"},{"Start":"03:22.290 ","End":"03:28.220","Text":"we have u(x naught) equals f(x) and we can apply the Fourier transform to"},{"Start":"03:28.220 ","End":"03:35.225","Text":"both sides and get that u hat(Omega 0) is f hat(Omega)."},{"Start":"03:35.225 ","End":"03:37.820","Text":"Comparing this and this,"},{"Start":"03:37.820 ","End":"03:41.600","Text":"we have 2 versions of u hat(Omega 0),"},{"Start":"03:41.600 ","End":"03:44.645","Text":"1\u0027s a and 1\u0027s f hat(Omega)."},{"Start":"03:44.645 ","End":"03:47.300","Text":"So a equals f hat(Omega)."},{"Start":"03:47.300 ","End":"03:53.900","Text":"Now we can put a in here and get our expression for what u hat is."},{"Start":"03:53.900 ","End":"03:55.745","Text":"It comes out to be a,"},{"Start":"03:55.745 ","End":"03:58.790","Text":"which is f hat(Omega) times this."},{"Start":"03:58.790 ","End":"04:01.640","Text":"Now, this exponent is something we did in"},{"Start":"04:01.640 ","End":"04:06.050","Text":"the previous clip with examples of Fourier transform."},{"Start":"04:06.050 ","End":"04:14.180","Text":"We got that e^ minus a^2 Omega^2 t is the Fourier transform of this function of x."},{"Start":"04:14.180 ","End":"04:20.840","Text":"What we get is that u hat(Omega t) equals f hat(Omega) times this,"},{"Start":"04:20.840 ","End":"04:24.905","Text":"replaced by this, this and this are the same."},{"Start":"04:24.905 ","End":"04:30.835","Text":"Now, here we have the product of 2 transformed functions."},{"Start":"04:30.835 ","End":"04:34.925","Text":"There\u0027s a rule that convolution goes to a product,"},{"Start":"04:34.925 ","End":"04:37.220","Text":"but as a matter of 2 Pi,"},{"Start":"04:37.220 ","End":"04:39.163","Text":"so we need to adjust this a bit,"},{"Start":"04:39.163 ","End":"04:40.980","Text":"what we can say is,"},{"Start":"04:40.980 ","End":"04:46.285","Text":"1st of all, on the left-hand side is the Fourier transform of u."},{"Start":"04:46.285 ","End":"04:51.860","Text":"This f hat(Omega) is the Fourier transform of f(x)."},{"Start":"04:51.860 ","End":"04:56.180","Text":"We can put 2 pi and 1 over 2 pi here."},{"Start":"04:56.180 ","End":"05:00.500","Text":"We need the 2 pi for the formula on the convolution."},{"Start":"05:00.500 ","End":"05:06.545","Text":"Anyway, here\u0027s a reminder of what the formula is for the Fourier transform of"},{"Start":"05:06.545 ","End":"05:12.125","Text":"a convolution 2 pi times the transforms of the 2 functions."},{"Start":"05:12.125 ","End":"05:14.725","Text":"We get 1 over 2 pi."},{"Start":"05:14.725 ","End":"05:21.410","Text":"Then this, this, and this is Fourier transform of the convolution of these 2 functions."},{"Start":"05:21.410 ","End":"05:25.670","Text":"F(x), star, this function of x."},{"Start":"05:25.670 ","End":"05:29.390","Text":"We can put the 1 over 2 pi inside the transform,"},{"Start":"05:29.390 ","End":"05:34.625","Text":"so we\u0027ve got f(1 over 2 pi) this, star, this."},{"Start":"05:34.625 ","End":"05:38.120","Text":"Now, we can take the inverse Fourier transform of"},{"Start":"05:38.120 ","End":"05:41.960","Text":"both sides and just drop the f here and here."},{"Start":"05:41.960 ","End":"05:43.850","Text":"It\u0027s not entirely precise."},{"Start":"05:43.850 ","End":"05:48.000","Text":"It\u0027s true provided these functions are continuous."},{"Start":"05:48.000 ","End":"05:51.170","Text":"Well, this will turn out to be continuous and that will justify it."},{"Start":"05:51.170 ","End":"05:54.710","Text":"Drop the Fourier transforms and now we"},{"Start":"05:54.710 ","End":"05:58.100","Text":"have 1 over 2 pi times this convolution and we still"},{"Start":"05:58.100 ","End":"06:00.920","Text":"have to compute this convolution is"},{"Start":"06:00.920 ","End":"06:04.943","Text":"a reminder of definition of convolution of 2 functions."},{"Start":"06:04.943 ","End":"06:06.565","Text":"G will be this 1."},{"Start":"06:06.565 ","End":"06:08.810","Text":"What we get is 1 over 2 pi,"},{"Start":"06:08.810 ","End":"06:12.245","Text":"the integral from minus infinity to infinity, f(y)."},{"Start":"06:12.245 ","End":"06:13.820","Text":"Here instead of x,"},{"Start":"06:13.820 ","End":"06:16.295","Text":"put x minus y."},{"Start":"06:16.295 ","End":"06:18.020","Text":"Now we\u0027ll do a couple of things."},{"Start":"06:18.020 ","End":"06:20.090","Text":"First of all, we\u0027ll take this"},{"Start":"06:20.090 ","End":"06:26.080","Text":"constant in front of the integral and combine it with this constant."},{"Start":"06:26.080 ","End":"06:30.245","Text":"Secondly, remember what f(y) is."},{"Start":"06:30.245 ","End":"06:36.870","Text":"F is just u of wherever this is, comma, 0."},{"Start":"06:36.870 ","End":"06:40.235","Text":"We get this as u(y) comma, 0."},{"Start":"06:40.235 ","End":"06:47.690","Text":"Just compute it. This times this gives us this and this is the Poisson formula."},{"Start":"06:47.690 ","End":"06:52.145","Text":"If you go back to the beginning and see what we wanted to get, it\u0027s exactly this."},{"Start":"06:52.145 ","End":"06:56.550","Text":"That completes this proof and we are done."}],"ID":30798},{"Watched":false,"Name":"The Error Function","Duration":"2m 19s","ChapterTopicVideoID":29213,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.300","Text":"This clip is about the error function."},{"Start":"00:03.300 ","End":"00:08.205","Text":"It\u0027s a mathematical function with notation erf"},{"Start":"00:08.205 ","End":"00:15.269","Text":"and not necessarily related to partial differential equations or the heat equation."},{"Start":"00:15.269 ","End":"00:18.585","Text":"Although it will be useful for the heat equation."},{"Start":"00:18.585 ","End":"00:24.260","Text":"Anyway, it\u0027s defined as follows as an integral."},{"Start":"00:24.260 ","End":"00:26.030","Text":"Here\u0027s its graph,"},{"Start":"00:26.030 ","End":"00:27.665","Text":"what it looks like."},{"Start":"00:27.665 ","End":"00:32.000","Text":"It goes to minus 1 at minus infinity at 0, at 0,"},{"Start":"00:32.000 ","End":"00:36.995","Text":"and then it tends to plus 1 at infinity."},{"Start":"00:36.995 ","End":"00:44.410","Text":"Another way of describing the error function is using this integral."},{"Start":"00:44.410 ","End":"00:48.690","Text":"If we remove the 2,"},{"Start":"00:48.690 ","End":"00:53.645","Text":"and then we could say that it\u0027s this integral,"},{"Start":"00:53.645 ","End":"00:56.595","Text":"but the 2 makes it double."},{"Start":"00:56.595 ","End":"01:00.500","Text":"If we take the 1 over root Pi to the minus t squared,"},{"Start":"01:00.500 ","End":"01:03.450","Text":"it\u0027s twice the integral from 0-x,"},{"Start":"01:03.450 ","End":"01:06.710","Text":"so it\u0027s like this symmetrical bit."},{"Start":"01:06.710 ","End":"01:13.310","Text":"In fact, this function with the 2 here is the derivative of this."},{"Start":"01:13.310 ","End":"01:18.305","Text":"Now there\u0027s no direct formula, computational formula,"},{"Start":"01:18.305 ","End":"01:21.260","Text":"an expression with an integral but it\u0027s not something that\u0027s"},{"Start":"01:21.260 ","End":"01:25.340","Text":"computable and to get values,"},{"Start":"01:25.340 ","End":"01:28.580","Text":"we use a calculator or computer."},{"Start":"01:28.580 ","End":"01:35.255","Text":"You can also find tables of its values from the days before there were computers."},{"Start":"01:35.255 ","End":"01:37.890","Text":"Here are some of its properties."},{"Start":"01:37.890 ","End":"01:44.000","Text":"It\u0027s an odd function and shape looks like an odd function."},{"Start":"01:44.000 ","End":"01:46.205","Text":"At 0 it\u0027s 0."},{"Start":"01:46.205 ","End":"01:53.890","Text":"As we mentioned, the limit at infinity is 1 and at minus infinity it\u0027s minus 1."},{"Start":"01:53.890 ","End":"02:00.380","Text":"The derivative of the error function is a bell-shaped function."},{"Start":"02:00.380 ","End":"02:03.610","Text":"It\u0027s basically this but doubled."},{"Start":"02:03.610 ","End":"02:08.210","Text":"That\u0027s basically it for this introduction to the error function."},{"Start":"02:08.210 ","End":"02:16.340","Text":"I remarked that you\u0027ll see will be used in the exercises for the heat equation PDE."},{"Start":"02:16.340 ","End":"02:19.500","Text":"Okay. That\u0027s it."}],"ID":30799},{"Watched":false,"Name":"Exercise 1","Duration":"9m 7s","ChapterTopicVideoID":29214,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.465","Text":"In this exercise, we have an IVP."},{"Start":"00:03.465 ","End":"00:06.690","Text":"This is the PDE, which is the heat equation,"},{"Start":"00:06.690 ","End":"00:08.640","Text":"this is the domain,"},{"Start":"00:08.640 ","End":"00:11.595","Text":"and this is the initial value,"},{"Start":"00:11.595 ","End":"00:14.130","Text":"which is split definition."},{"Start":"00:14.130 ","End":"00:19.605","Text":"It\u0027s equal to constant T1 on the positive x axis,"},{"Start":"00:19.605 ","End":"00:24.810","Text":"and it\u0027s equal to T2 on the negative X axis."},{"Start":"00:24.810 ","End":"00:28.440","Text":"That\u0027s the starting condition and we\u0027ll see what happens"},{"Start":"00:28.440 ","End":"00:32.355","Text":"over time to the temperature along the rod."},{"Start":"00:32.355 ","End":"00:34.980","Text":"In fact, I\u0027m going to give you a spoiler."},{"Start":"00:34.980 ","End":"00:37.800","Text":"In time as T goes to infinity,"},{"Start":"00:37.800 ","End":"00:40.049","Text":"the rug turns to a uniform temperature,"},{"Start":"00:40.049 ","End":"00:42.990","Text":"which is the average of these 2."},{"Start":"00:44.390 ","End":"00:50.825","Text":"Our task is to solve the IVP to find u(x) and u(t),"},{"Start":"00:50.825 ","End":"00:55.145","Text":"and then to compute the limit as T goes to infinity."},{"Start":"00:55.145 ","End":"00:57.050","Text":"Sometimes as in this case,"},{"Start":"00:57.050 ","End":"00:59.150","Text":"the error function crops up in"},{"Start":"00:59.150 ","End":"01:04.700","Text":"these heat value problems so here\u0027s a reminder of what the error function is."},{"Start":"01:04.700 ","End":"01:07.190","Text":"Since this is on an infinite interval,"},{"Start":"01:07.190 ","End":"01:09.695","Text":"we can use the Poisson formula,"},{"Start":"01:09.695 ","End":"01:14.505","Text":"which gives us U in terms of the initial condition,"},{"Start":"01:14.505 ","End":"01:18.240","Text":"U at time 0 and this is the formula."},{"Start":"01:18.240 ","End":"01:20.370","Text":"We\u0027ve proven it even,"},{"Start":"01:20.370 ","End":"01:22.320","Text":"and we have u(y),0."},{"Start":"01:22.320 ","End":"01:25.185","Text":"Well, we have u(x),0."},{"Start":"01:25.185 ","End":"01:30.555","Text":"Just replace X by Y, first copy it."},{"Start":"01:30.555 ","End":"01:35.930","Text":"Now note that because U is split to positive and negative,"},{"Start":"01:35.930 ","End":"01:41.420","Text":"what we\u0027ll do is we\u0027ll split this integral into 2 parts,"},{"Start":"01:41.420 ","End":"01:46.280","Text":"from minus infinity to 0 and from 0 to infinity right away."},{"Start":"01:46.280 ","End":"01:52.310","Text":"We can say it\u0027s the integral from minus infinity to 0 and from 0 to infinity."},{"Start":"01:52.310 ","End":"01:56.295","Text":"Now on this part u(y) 0 is T1,"},{"Start":"01:56.295 ","End":"01:59.910","Text":"and then this part u(y) 0 is T2."},{"Start":"01:59.910 ","End":"02:02.560","Text":"Now the other way around,"},{"Start":"02:02.660 ","End":"02:08.160","Text":"T2 here and T1 here."},{"Start":"02:08.160 ","End":"02:13.539","Text":"Also, lightly rearrange this to make it something squared,"},{"Start":"02:13.539 ","End":"02:18.470","Text":"and now let\u0027s make a substitution for what\u0027s in the brackets here."},{"Start":"02:18.470 ","End":"02:26.315","Text":"Letter T is taken so let\u0027s use letter S. S will be Y minus X over"},{"Start":"02:26.315 ","End":"02:31.710","Text":"2A root T and then DS in terms of"},{"Start":"02:31.710 ","End":"02:38.425","Text":"DY is 1 over 2A square root of T, dY."},{"Start":"02:38.425 ","End":"02:42.485","Text":"We also have to substitute the limits of integration."},{"Start":"02:42.485 ","End":"02:46.760","Text":"There are 3 of them, minus infinity is 0 and infinity."},{"Start":"02:46.760 ","End":"02:55.920","Text":"When y is 0, then S is minus X over 2A root T when Y is"},{"Start":"02:55.920 ","End":"03:00.050","Text":"infinity so we get infinity minus X over"},{"Start":"03:00.050 ","End":"03:04.985","Text":"a positive number is infinity and similarly when Y is minus infinity,"},{"Start":"03:04.985 ","End":"03:11.360","Text":"S is minus infinity so after these substitutions this goes from"},{"Start":"03:11.360 ","End":"03:17.900","Text":"minus infinity to this minus X over 2A root T. Similarly,"},{"Start":"03:17.900 ","End":"03:21.630","Text":"this is here and here infinity."},{"Start":"03:21.740 ","End":"03:28.865","Text":"This bit here becomes S so we have minus S squared here and minus S squared here."},{"Start":"03:28.865 ","End":"03:33.050","Text":"Now look, I\u0027ve split this root pi of T as"},{"Start":"03:33.050 ","End":"03:37.654","Text":"root pi root T. Now I\u0027ve colored the parts that will go together."},{"Start":"03:37.654 ","End":"03:46.215","Text":"Look at the 2A root T and the denominator and DY altogether that will give us DS."},{"Start":"03:46.215 ","End":"03:51.015","Text":"The green color will give us DS and here also."},{"Start":"03:51.015 ","End":"03:54.660","Text":"We\u0027re left with just 1 over root pi,"},{"Start":"03:54.660 ","End":"03:59.240","Text":"and then here DS and here Ds and now, once again,"},{"Start":"03:59.240 ","End":"04:03.950","Text":"we can split integrals up into different ranges."},{"Start":"04:03.950 ","End":"04:08.095","Text":"We would like to have minus infinity to 0 here."},{"Start":"04:08.095 ","End":"04:11.330","Text":"We can write from minus infinity to 0,"},{"Start":"04:11.330 ","End":"04:17.935","Text":"but then we have to subtract the bit from minus X over 2A root T to 0."},{"Start":"04:17.935 ","End":"04:20.989","Text":"The other one, this is already negative,"},{"Start":"04:20.989 ","End":"04:28.565","Text":"so we just break it up normally from here to 0 and from 0 to infinity to give us this."},{"Start":"04:28.565 ","End":"04:32.453","Text":"They don\u0027t want to look at it and see that it\u0027s okay,"},{"Start":"04:32.453 ","End":"04:35.870","Text":"and here\u0027s this same line again."},{"Start":"04:35.870 ","End":"04:40.460","Text":"I want to go back and talk about the integral from"},{"Start":"04:40.460 ","End":"04:44.675","Text":"minus infinity to infinity of each of the minus X squared DS."},{"Start":"04:44.675 ","End":"04:48.410","Text":"This is the famous integral and it\u0027s known that"},{"Start":"04:48.410 ","End":"04:53.160","Text":"this definite integral is equal to square root of Pi."},{"Start":"04:53.160 ","End":"04:55.325","Text":"There are several ways of showing it,"},{"Start":"04:55.325 ","End":"05:01.630","Text":"we\u0027re going to just accept this and there is a little illustration here."},{"Start":"05:01.630 ","End":"05:09.135","Text":"This blue line is E to the minus X squared."},{"Start":"05:09.135 ","End":"05:13.490","Text":"We have X here, we have S as a matter and"},{"Start":"05:13.490 ","End":"05:20.230","Text":"the red color is the area under the curve all the way from minus infinity to infinity,"},{"Start":"05:20.230 ","End":"05:23.904","Text":"although it gets very close to 0 very quickly."},{"Start":"05:23.904 ","End":"05:27.760","Text":"Also note something else that this function,"},{"Start":"05:27.760 ","End":"05:30.685","Text":"the E to the minus X squared,"},{"Start":"05:30.685 ","End":"05:32.679","Text":"is an even function."},{"Start":"05:32.679 ","End":"05:36.475","Text":"It\u0027s symmetrical around the y-axis."},{"Start":"05:36.475 ","End":"05:40.229","Text":"Because of the symmetry each of the parts,"},{"Start":"05:40.229 ","End":"05:43.990","Text":"the bit on the left and the bit on the right are half the hole,"},{"Start":"05:43.990 ","End":"05:49.560","Text":"so we can say that the integral from 0 to infinity of each of"},{"Start":"05:49.560 ","End":"05:55.610","Text":"the minus S squared DS is root pi over 2 and also from minus infinity to 0,"},{"Start":"05:55.610 ","End":"06:00.380","Text":"each of these 2 halves is root pi over 2 and we can use that"},{"Start":"06:00.380 ","End":"06:05.780","Text":"here because we have here from minus infinity to 0 and here from 0 to infinity."},{"Start":"06:05.780 ","End":"06:10.855","Text":"This bit, and this bit will be root pi over 2,"},{"Start":"06:10.855 ","End":"06:14.255","Text":"just substituting that in here."},{"Start":"06:14.255 ","End":"06:15.575","Text":"Now let\u0027s simplify a bit."},{"Start":"06:15.575 ","End":"06:20.015","Text":"Collect like terms, root pi over 2 we have here and here"},{"Start":"06:20.015 ","End":"06:25.990","Text":"so it\u0027s T2 plus T1 times root pi over 2."},{"Start":"06:25.990 ","End":"06:28.665","Text":"The other part,"},{"Start":"06:28.665 ","End":"06:31.485","Text":"this is the same integral as this,"},{"Start":"06:31.485 ","End":"06:36.645","Text":"so we have it here minus T2 times."},{"Start":"06:36.645 ","End":"06:39.035","Text":"Here we have it plus T1 times."},{"Start":"06:39.035 ","End":"06:45.035","Text":"T1 minus T2 times this integral and all this times 1 over root pi."},{"Start":"06:45.035 ","End":"06:49.040","Text":"We are getting close to using the error function."},{"Start":"06:49.040 ","End":"06:55.760","Text":"Let\u0027s modify this a bit by multiplying and dividing by root pi over 2."},{"Start":"06:55.760 ","End":"07:01.060","Text":"We\u0027ll put a root Pi over 2 here and 2 over root pi here."},{"Start":"07:01.060 ","End":"07:06.170","Text":"This is very much looking like error function and in fact,"},{"Start":"07:06.170 ","End":"07:08.870","Text":"if you go back to the definition of error function,"},{"Start":"07:08.870 ","End":"07:14.655","Text":"it\u0027s just like this except that we have this instead of X."},{"Start":"07:14.655 ","End":"07:18.500","Text":"This is the error function of what\u0027s written here,"},{"Start":"07:18.500 ","End":"07:23.990","Text":"minus X over 2 root and I\u0027ve colored the root pi here,"},{"Start":"07:23.990 ","End":"07:27.370","Text":"here and here because they\u0027re all going to cancel."},{"Start":"07:27.370 ","End":"07:34.625","Text":"We\u0027ll get the u(x, t) is T1 plus T2 over 2 plus."},{"Start":"07:34.625 ","End":"07:38.870","Text":"Now we can get rid of the minus here and"},{"Start":"07:38.870 ","End":"07:43.175","Text":"here because the error function is an odd function."},{"Start":"07:43.175 ","End":"07:44.450","Text":"This minus is gone,"},{"Start":"07:44.450 ","End":"07:45.715","Text":"it becomes a plus."},{"Start":"07:45.715 ","End":"07:53.465","Text":"We have T1 minus T2 over 2 and then the error function of this without the minus."},{"Start":"07:53.465 ","End":"07:56.930","Text":"That\u0027s the answer to the first question but remember we had"},{"Start":"07:56.930 ","End":"08:00.470","Text":"another question about the limit of u(x,"},{"Start":"08:00.470 ","End":"08:03.285","Text":"t ) as t goes to infinity."},{"Start":"08:03.285 ","End":"08:08.780","Text":"Just copy this expression and just concentrate on the T part which is here."},{"Start":"08:08.780 ","End":"08:10.685","Text":"This is the constant,"},{"Start":"08:10.685 ","End":"08:12.890","Text":"doesn\u0027t change in the limit."},{"Start":"08:12.890 ","End":"08:15.920","Text":"This is a constant also so here,"},{"Start":"08:15.920 ","End":"08:20.833","Text":"all we have to do is take the limit of the error function,"},{"Start":"08:20.833 ","End":"08:24.260","Text":"but the error function is continuous so we can put the limit"},{"Start":"08:24.260 ","End":"08:31.175","Text":"inside and get error function of the limit as T goes to infinity of this."},{"Start":"08:31.175 ","End":"08:32.900","Text":"T goes to infinity,"},{"Start":"08:32.900 ","End":"08:39.110","Text":"this goes to 0 and we know the error function of 0 is 0."},{"Start":"08:39.110 ","End":"08:46.600","Text":"This whole second part disappears and we\u0027re left with just T1 plus T2 over 2."},{"Start":"08:46.600 ","End":"08:48.080","Text":"Like I said in the beginning,"},{"Start":"08:48.080 ","End":"08:56.870","Text":"if the rod is initially equal to T2 for negative X and T1 for positive X at infinity,"},{"Start":"08:56.870 ","End":"09:02.555","Text":"the whole rod is at a temperature of T1 plus T2 over 2,"},{"Start":"09:02.555 ","End":"09:05.075","Text":"at asymptotically it is."},{"Start":"09:05.075 ","End":"09:07.950","Text":"That concludes this clip."}],"ID":30800},{"Watched":false,"Name":"Exercise 2","Duration":"5m 16s","ChapterTopicVideoID":29215,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.850","Text":"In this exercise, we\u0027re going to solve"},{"Start":"00:02.850 ","End":"00:07.050","Text":"the following initial value problem with the heat equation,"},{"Start":"00:07.050 ","End":"00:12.315","Text":"which is this and it\u0027s on the whole real line and t"},{"Start":"00:12.315 ","End":"00:20.820","Text":"non-negative and we have the initial conditions at time 0, u(x,0) is e^x."},{"Start":"00:20.820 ","End":"00:23.415","Text":"We\u0027re going to solve it with Poisson\u0027s formula,"},{"Start":"00:23.415 ","End":"00:29.595","Text":"which gives us the answer u(x,t) in terms of the initial condition."},{"Start":"00:29.595 ","End":"00:32.325","Text":"Just copying this first of all."},{"Start":"00:32.325 ","End":"00:35.355","Text":"Now, we know what u(y,0) is."},{"Start":"00:35.355 ","End":"00:39.090","Text":"It\u0027s just like u(x,0) but with x replaced by y"},{"Start":"00:39.090 ","End":"00:43.310","Text":"so that would be e^y and everything else is the same."},{"Start":"00:43.310 ","End":"00:49.440","Text":"Now, expand y minus x^2 is y^2 minus 2xy plus x^2."},{"Start":"00:49.440 ","End":"00:53.405","Text":"We\u0027re going to take out something that doesn\u0027t involve y."},{"Start":"00:53.405 ","End":"00:59.970","Text":"The last bit, x^2 over 4a^2t has no y in it and"},{"Start":"00:59.970 ","End":"01:03.140","Text":"so we can pull it out front and"},{"Start":"01:03.140 ","End":"01:07.235","Text":"also addition in the exponent is multiplication, of course."},{"Start":"01:07.235 ","End":"01:11.030","Text":"Next, we can put this over a common denominator,"},{"Start":"01:11.030 ","End":"01:18.515","Text":"put the y into the numerator so it becomes because there\u0027s a minus here, minus 4a^2ty."},{"Start":"01:18.515 ","End":"01:22.085","Text":"This last bit is from the y."},{"Start":"01:22.085 ","End":"01:27.635","Text":"Now let\u0027s simplify this numerator bit without the minus to take this part,"},{"Start":"01:27.635 ","End":"01:33.065","Text":"we can take 2y out of the last couple of terms and we get this."},{"Start":"01:33.065 ","End":"01:41.870","Text":"Now we\u0027re going to complete the square y^2 minus 2y times this plus this thing squared,"},{"Start":"01:41.870 ","End":"01:47.615","Text":"that will be a whole square and then I\u0027m just going to subtract this to even it out."},{"Start":"01:47.615 ","End":"01:57.475","Text":"Now, this becomes the square of y minus x plus 2a^2t minus this and then"},{"Start":"01:57.475 ","End":"02:02.250","Text":"put this back here in this integral so now we have"},{"Start":"02:02.250 ","End":"02:08.060","Text":"this and note that the 2nd term in the numerator here also is without y."},{"Start":"02:08.060 ","End":"02:12.505","Text":"Once again, we can use the trick of pulling out of the integral."},{"Start":"02:12.505 ","End":"02:18.230","Text":"This time, what we get the minus with the minus here gives us a plus so we get"},{"Start":"02:18.230 ","End":"02:24.425","Text":"x plus 2a^2t^2 over 4a^2t and then e^minus,"},{"Start":"02:24.425 ","End":"02:29.390","Text":"just this part, but slight change instead of this squared over this,"},{"Start":"02:29.390 ","End":"02:32.690","Text":"we can write this over the square root of the denominator,"},{"Start":"02:32.690 ","End":"02:36.130","Text":"which is 2a square root of t all squared."},{"Start":"02:36.130 ","End":"02:39.825","Text":"Now let\u0027s combine these to exponents."},{"Start":"02:39.825 ","End":"02:44.855","Text":"Minus this plus this so if we put it over a common denominator,"},{"Start":"02:44.855 ","End":"02:53.389","Text":"we get minus x^2 plus x^2 cancels plus twice this times this plus this squared"},{"Start":"02:53.389 ","End":"03:03.245","Text":"over 4a^2t and it simplifies 4a^2t goes into this x times and into this a^2t times."},{"Start":"03:03.245 ","End":"03:13.085","Text":"What we have now is e to the power of all this is x plus a^2t and the same integral."},{"Start":"03:13.085 ","End":"03:17.225","Text":"I will solve this integral with the help of a substitution."},{"Start":"03:17.225 ","End":"03:23.040","Text":"We\u0027ll take this and call it z or z in America."},{"Start":"03:23.040 ","End":"03:28.875","Text":"Z is y minus this over this and so dz in terms of dy"},{"Start":"03:28.875 ","End":"03:36.155","Text":"is 1 over 2a square root of t dy also the limits of integration."},{"Start":"03:36.155 ","End":"03:38.390","Text":"When y goes to infinity,"},{"Start":"03:38.390 ","End":"03:40.115","Text":"that goes to infinity,"},{"Start":"03:40.115 ","End":"03:41.210","Text":"y goes to minus infinity,"},{"Start":"03:41.210 ","End":"03:43.460","Text":"that goes to minus infinity."},{"Start":"03:43.460 ","End":"03:46.705","Text":"Also, dy in terms of dz,"},{"Start":"03:46.705 ","End":"03:52.360","Text":"is just bringing this to the other side would be 2a square root of t dz."},{"Start":"03:52.360 ","End":"03:54.530","Text":"We end up with this,"},{"Start":"03:54.530 ","End":"03:56.750","Text":"bringing that constant out of the integral,"},{"Start":"03:56.750 ","End":"04:00.680","Text":"the 2a square root of t and this is the familiar integral."},{"Start":"04:00.680 ","End":"04:05.575","Text":"We know that this is equal to the square root of Pi."},{"Start":"04:05.575 ","End":"04:07.055","Text":"After we cancel everything,"},{"Start":"04:07.055 ","End":"04:09.260","Text":"look 2a square root to t square root of Pi,"},{"Start":"04:09.260 ","End":"04:11.690","Text":"so here 2a square root of Pi,"},{"Start":"04:11.690 ","End":"04:16.170","Text":"square root of t. Everything cancels except this."},{"Start":"04:16.170 ","End":"04:22.460","Text":"The answer is e^x plus a^2t and we\u0027re done except that for a change,"},{"Start":"04:22.460 ","End":"04:26.650","Text":"let\u0027s check if this is the correct solution."},{"Start":"04:26.650 ","End":"04:31.460","Text":"Remember this is the original equation and initial condition."},{"Start":"04:31.460 ","End":"04:34.280","Text":"Let\u0027s see if these conditions are met."},{"Start":"04:34.280 ","End":"04:38.960","Text":"Now, du/dx is just the same thing"},{"Start":"04:38.960 ","End":"04:41.870","Text":"because the inner derivative is 1 and the 2nd derivative"},{"Start":"04:41.870 ","End":"04:49.195","Text":"is also the same thing but the derivative with respect to t is this multiplied by a^2."},{"Start":"04:49.195 ","End":"04:53.925","Text":"Now notice that u_t is exactly a^2 times"},{"Start":"04:53.925 ","End":"04:59.590","Text":"u_xx so we\u0027re okay with that condition and the other condition,"},{"Start":"04:59.590 ","End":"05:05.580","Text":"u(x) naught is e^x plus a^2 times 0,"},{"Start":"05:05.580 ","End":"05:11.230","Text":"which is just e^x and that\u0027s what we wanted it to be."},{"Start":"05:11.230 ","End":"05:13.575","Text":"That\u0027s the 2nd condition."},{"Start":"05:13.575 ","End":"05:16.980","Text":"We\u0027ve checked the solution and now we\u0027re really done."}],"ID":30801},{"Watched":false,"Name":"Exercise 3","Duration":"4m 35s","ChapterTopicVideoID":29216,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.565","Text":"In this exercise, we\u0027re going to solve the following initial value problem."},{"Start":"00:05.565 ","End":"00:15.285","Text":"This is a heat equation on the whole real line and this is the initial condition at t=0,"},{"Start":"00:15.285 ","End":"00:18.180","Text":"and here\u0027s a picture of what it looks like."},{"Start":"00:18.180 ","End":"00:21.660","Text":"Just a reminder of what the error function is,"},{"Start":"00:21.660 ","End":"00:25.005","Text":"which we may find of use, and let\u0027s start."},{"Start":"00:25.005 ","End":"00:27.780","Text":"Reminder of the Poisson formula."},{"Start":"00:27.780 ","End":"00:30.240","Text":"Because it\u0027s on the infinite interval,"},{"Start":"00:30.240 ","End":"00:36.960","Text":"we can use this formula which gives us what u(x,t) is in terms of the initial condition,"},{"Start":"00:36.960 ","End":"00:39.030","Text":"just have to figure out this integral."},{"Start":"00:39.030 ","End":"00:40.870","Text":"Well, u(y,"},{"Start":"00:40.870 ","End":"00:49.080","Text":"0) is the same as u(x,0) just x replaced by y and a=1."},{"Start":"00:49.080 ","End":"00:51.165","Text":"u_t is a^2 u_xx,"},{"Start":"00:51.165 ","End":"00:52.380","Text":"so a^2 is 1,"},{"Start":"00:52.380 ","End":"00:53.740","Text":"so a is 1."},{"Start":"00:53.740 ","End":"00:56.015","Text":"Now we can simplify this."},{"Start":"00:56.015 ","End":"01:00.380","Text":"First of all, plugging in a=1, we get this."},{"Start":"01:00.380 ","End":"01:05.060","Text":"Then because of this formula,"},{"Start":"01:05.060 ","End":"01:09.560","Text":"the part where it\u0027s 0 we don\u0027t need and we\u0027re just going to take the integral"},{"Start":"01:09.560 ","End":"01:14.840","Text":"from 0-1 and subtract the integral from 1 to infinity,"},{"Start":"01:14.840 ","End":"01:18.110","Text":"so we get this."},{"Start":"01:18.110 ","End":"01:22.310","Text":"I forgot to say we\u0027re going to write as something squared"},{"Start":"01:22.310 ","End":"01:29.295","Text":"like so and then we get the integral from 0-1 corresponding to this,"},{"Start":"01:29.295 ","End":"01:32.390","Text":"and minus the integral from 1 to infinity,"},{"Start":"01:32.390 ","End":"01:34.340","Text":"the minus because it\u0027s plus minus"},{"Start":"01:34.340 ","End":"01:43.535","Text":"1 times and these integrals will do with a substitution will let z equal what\u0027s in here,"},{"Start":"01:43.535 ","End":"01:46.160","Text":"y minus x over 2 square root of t,"},{"Start":"01:46.160 ","End":"01:50.095","Text":"dz is 1 over 2 root t dy."},{"Start":"01:50.095 ","End":"01:52.670","Text":"The limits of integration,"},{"Start":"01:52.670 ","End":"01:55.700","Text":"we have 0, 1, and infinity."},{"Start":"01:55.700 ","End":"02:02.210","Text":"I\u0027ll just check that 0 goes to minus x over 2 root t. 1 goes"},{"Start":"02:02.210 ","End":"02:08.129","Text":"to 1 minus x over 2 root t and when y goes to infinity,"},{"Start":"02:08.129 ","End":"02:10.590","Text":"z goes to infinity."},{"Start":"02:10.590 ","End":"02:13.295","Text":"After substituting all these,"},{"Start":"02:13.295 ","End":"02:14.915","Text":"this is what we get."},{"Start":"02:14.915 ","End":"02:21.035","Text":"Note that dy is 2 root t times dz and the 2 root t"},{"Start":"02:21.035 ","End":"02:24.110","Text":"can be pulled out in front of the integrals"},{"Start":"02:24.110 ","End":"02:27.450","Text":"in both cases and so we can take it out in brackets."},{"Start":"02:27.450 ","End":"02:31.565","Text":"We\u0027re going to be using the error function to express these."},{"Start":"02:31.565 ","End":"02:34.010","Text":"To make it closer to the error function,"},{"Start":"02:34.010 ","End":"02:38.330","Text":"we need a 2 over root Pi in front of each of these integrals."},{"Start":"02:38.330 ","End":"02:41.360","Text":"We\u0027ll put it out here as root Pi over 2."},{"Start":"02:41.360 ","End":"02:44.209","Text":"Now, just merging all these constants,"},{"Start":"02:44.209 ","End":"02:47.245","Text":"we get 1/2 times this."},{"Start":"02:47.245 ","End":"02:52.010","Text":"Here we are. Let\u0027s do each of the pieces separately."},{"Start":"02:52.010 ","End":"02:56.330","Text":"Now the first part here we can split it up into 2 integrals,"},{"Start":"02:56.330 ","End":"03:01.205","Text":"from this to 0 and from 0 to this, like so."},{"Start":"03:01.205 ","End":"03:07.115","Text":"Now, this is the error function of 1 minus x over 2 root t. But what about this?"},{"Start":"03:07.115 ","End":"03:09.125","Text":"Well, by the symmetry,"},{"Start":"03:09.125 ","End":"03:16.130","Text":"from minus something to 0 is the same area as from 0 to the plus something."},{"Start":"03:16.130 ","End":"03:21.480","Text":"This becomes the error function of plus x over 2 root t,"},{"Start":"03:21.480 ","End":"03:26.335","Text":"and this like we said is the error function of 1 minus x over 2 root t,"},{"Start":"03:26.335 ","End":"03:28.160","Text":"and the other one similarly,"},{"Start":"03:28.160 ","End":"03:31.295","Text":"we break up into 2 integrals."},{"Start":"03:31.295 ","End":"03:37.070","Text":"We\u0027ll take it from 0 to infinity and then subtract the part from"},{"Start":"03:37.070 ","End":"03:44.590","Text":"0-1 minus x over 2 root t. This is the error function of infinity or the limit."},{"Start":"03:44.590 ","End":"03:47.530","Text":"We know that that\u0027s equal to 1."},{"Start":"03:47.530 ","End":"03:51.290","Text":"Alternately, you could break it up and say that this integral we know is"},{"Start":"03:51.290 ","End":"03:55.580","Text":"root Pi over 2 and 2 over root Pi times root Pi over 2 is 1."},{"Start":"03:55.580 ","End":"03:56.960","Text":"Either way, it comes out 1,"},{"Start":"03:56.960 ","End":"04:01.085","Text":"and this error function of 1 minus x over 2 root t,"},{"Start":"04:01.085 ","End":"04:02.930","Text":"just like over here."},{"Start":"04:02.930 ","End":"04:04.430","Text":"Now we piece these together."},{"Start":"04:04.430 ","End":"04:07.115","Text":"We want this minus this,"},{"Start":"04:07.115 ","End":"04:08.450","Text":"and also the 1/2."},{"Start":"04:08.450 ","End":"04:15.020","Text":"We have this and then plus this and again minus, minus this,"},{"Start":"04:15.020 ","End":"04:24.155","Text":"so twice this and then minus the one from here and so just rewriting it,"},{"Start":"04:24.155 ","End":"04:30.920","Text":"we\u0027ve got 1/2 error function of this plus error function of this minus 1/2."},{"Start":"04:30.920 ","End":"04:35.640","Text":"That\u0027s the answer and we are done."}],"ID":30802},{"Watched":false,"Name":"Exercise 4","Duration":"5m 44s","ChapterTopicVideoID":29217,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.045","Text":"In this exercise, we have an initial value problem involving the heat equation"},{"Start":"00:06.045 ","End":"00:12.254","Text":"here on an infinite interval and we\u0027re given the initial condition at time 0,"},{"Start":"00:12.254 ","End":"00:15.180","Text":"it\u0027s e to the minus x^2."},{"Start":"00:15.180 ","End":"00:20.285","Text":"We\u0027re going to use the Poisson formula, which is this."},{"Start":"00:20.285 ","End":"00:27.285","Text":"This computes u(x,t) depending on the initial condition u(y,0),"},{"Start":"00:27.285 ","End":"00:31.604","Text":"which is just u(x,0) with x replaced by y."},{"Start":"00:31.604 ","End":"00:36.060","Text":"In our case a which is here and here is equal to 1,"},{"Start":"00:36.060 ","End":"00:39.015","Text":"because this would normally be a squared u_xx,"},{"Start":"00:39.015 ","End":"00:42.135","Text":"but a squared is 1, so a is 1."},{"Start":"00:42.135 ","End":"00:45.840","Text":"This simplifies a bit to this."},{"Start":"00:45.840 ","End":"00:48.950","Text":"Rest of this is just calculations."},{"Start":"00:48.950 ","End":"00:52.175","Text":"We\u0027ll start by combining the exponents here."},{"Start":"00:52.175 ","End":"00:57.640","Text":"We have e to the minus y squared plus y minus x^2 over 4t."},{"Start":"00:57.640 ","End":"01:01.715","Text":"Now, let\u0027s work on what\u0027s in the exponent here."},{"Start":"01:01.715 ","End":"01:03.185","Text":"This is here."},{"Start":"01:03.185 ","End":"01:10.010","Text":"Now, expand the brackets here and make a common denominator of 4t."},{"Start":"01:10.010 ","End":"01:13.165","Text":"This is 4ty^2,"},{"Start":"01:13.165 ","End":"01:17.009","Text":"and here y^2 minus 2xy plus x^2."},{"Start":"01:17.009 ","End":"01:20.790","Text":"Take the y^2 out of the brackets from these 2 terms,"},{"Start":"01:20.790 ","End":"01:23.940","Text":"we have 4t plus 1y^2."},{"Start":"01:23.940 ","End":"01:29.450","Text":"The idea is to try and get this to be a difference of squares."},{"Start":"01:29.450 ","End":"01:34.879","Text":"We have this formula that a^2 minus 2ab plus b^2 is a minus b squared."},{"Start":"01:34.879 ","End":"01:37.400","Text":"This will be a squared,"},{"Start":"01:37.400 ","End":"01:43.980","Text":"so a is y times square root of 4t plus 1,"},{"Start":"01:43.980 ","End":"01:45.960","Text":"and this is squared."},{"Start":"01:45.960 ","End":"01:50.050","Text":"Now, we need minus 2ab."},{"Start":"01:50.960 ","End":"01:53.895","Text":"Here\u0027s the minus 2,"},{"Start":"01:53.895 ","End":"01:56.400","Text":"and the a is this."},{"Start":"01:56.400 ","End":"02:02.575","Text":"What are we missing here to get this to come out to be minus 2xy?"},{"Start":"02:02.575 ","End":"02:05.190","Text":"We have minus 2 and we have y,"},{"Start":"02:05.190 ","End":"02:06.900","Text":"so we need an x here,"},{"Start":"02:06.900 ","End":"02:10.070","Text":"and we also have to divide by this square root,"},{"Start":"02:10.070 ","End":"02:13.310","Text":"so x over the square root and then we\u0027re okay."},{"Start":"02:13.310 ","End":"02:16.405","Text":"This cancels with this and we have minus 2xy."},{"Start":"02:16.405 ","End":"02:20.520","Text":"Then we need plus b^2,"},{"Start":"02:20.520 ","End":"02:23.775","Text":"b is this part here,"},{"Start":"02:23.775 ","End":"02:26.145","Text":"so the a plus b^2."},{"Start":"02:26.145 ","End":"02:32.055","Text":"Now, we have to compensate by subtracting this b^2,"},{"Start":"02:32.055 ","End":"02:35.100","Text":"and we still have an x^2 at the end."},{"Start":"02:35.100 ","End":"02:38.830","Text":"All this is over 4t."},{"Start":"02:38.830 ","End":"02:43.400","Text":"Now, applying this basic algebra rule,"},{"Start":"02:43.400 ","End":"02:51.515","Text":"we get a minus b squared and that covers this part and the last part,"},{"Start":"02:51.515 ","End":"02:55.310","Text":"just color it because we\u0027re going to work on this bit next."},{"Start":"02:55.310 ","End":"02:58.880","Text":"Yeah, and over 4t. This part,"},{"Start":"02:58.880 ","End":"03:03.560","Text":"just rearrange the order x squared minus x squared over 4t plus 1."},{"Start":"03:03.560 ","End":"03:08.530","Text":"Pull the x squared out and we have 1 minus 1 over 4t plus 1."},{"Start":"03:08.530 ","End":"03:14.280","Text":"Now this becomes 4t plus 1 minus 1 over 4t plus 1."},{"Start":"03:14.280 ","End":"03:20.205","Text":"So it\u0027s 4t over 4t plus 1 here."},{"Start":"03:20.205 ","End":"03:23.190","Text":"Yeah, and I forgot the over 4t."},{"Start":"03:23.190 ","End":"03:30.085","Text":"Now, this 4t will cancel with this 4t and we\u0027ll have x squared times 1 over 4t plus 1."},{"Start":"03:30.085 ","End":"03:32.584","Text":"Now let\u0027s remember what we\u0027re doing."},{"Start":"03:32.584 ","End":"03:36.380","Text":"We have this expression and we\u0027re just"},{"Start":"03:36.380 ","End":"03:40.850","Text":"simplifying what\u0027s in this brackets and the exponent."},{"Start":"03:40.850 ","End":"03:42.630","Text":"There\u0027s a minus still here,"},{"Start":"03:42.630 ","End":"03:45.570","Text":"so what we need is all this,"},{"Start":"03:45.570 ","End":"03:50.445","Text":"but with this part replaced by this."},{"Start":"03:50.445 ","End":"03:53.660","Text":"This doesn\u0027t involve y,"},{"Start":"03:53.660 ","End":"03:56.105","Text":"so we can bring that in the front."},{"Start":"03:56.105 ","End":"04:02.885","Text":"This is here, e to the minus x squared over 4t plus 1."},{"Start":"04:02.885 ","End":"04:08.280","Text":"This is here, there\u0027s a minus."},{"Start":"04:08.420 ","End":"04:11.255","Text":"Now we\u0027ll make a substitution."},{"Start":"04:11.255 ","End":"04:15.835","Text":"We\u0027ll let z equal what\u0027s in the brackets here,"},{"Start":"04:15.835 ","End":"04:19.005","Text":"y square root of 4t plus 1"},{"Start":"04:19.005 ","End":"04:23.595","Text":"minus x over square root of 4t plus 1 over 2 root t, that\u0027s the z."},{"Start":"04:23.595 ","End":"04:30.585","Text":"Then dz would just be square root of 4t plus 1 over 2 root t dy."},{"Start":"04:30.585 ","End":"04:36.295","Text":"dy equals dz times the reciprocal of this."},{"Start":"04:36.295 ","End":"04:40.505","Text":"Now, limits of integration when y goes to infinity,"},{"Start":"04:40.505 ","End":"04:45.280","Text":"z goes to infinity because this is a positive number."},{"Start":"04:45.280 ","End":"04:47.060","Text":"So it won\u0027t go to minus infinity,"},{"Start":"04:47.060 ","End":"04:48.560","Text":"it will go to plus infinity."},{"Start":"04:48.560 ","End":"04:51.410","Text":"Similarly, when y goes to minus infinity,"},{"Start":"04:51.410 ","End":"04:53.980","Text":"z goes to minus infinity."},{"Start":"04:53.980 ","End":"04:59.200","Text":"This integral becomes e to the minus z^2."},{"Start":"04:59.200 ","End":"05:03.320","Text":"The reciprocal of this can come out in front of the integral."},{"Start":"05:03.320 ","End":"05:06.200","Text":"That\u0027s 2 root t over root 4t plus 1."},{"Start":"05:06.200 ","End":"05:10.090","Text":"This integral we know is square root of Pi,"},{"Start":"05:10.090 ","End":"05:12.500","Text":"and let\u0027s just color some things."},{"Start":"05:12.500 ","End":"05:13.850","Text":"Color this 2 root t,"},{"Start":"05:13.850 ","End":"05:20.080","Text":"color 2 root Pit because these pieces together will all cancel."},{"Start":"05:20.080 ","End":"05:23.610","Text":"2 root t root Pi and here 2 root Pit,"},{"Start":"05:23.610 ","End":"05:27.035","Text":"same thing, so these just disappear and become 1."},{"Start":"05:27.035 ","End":"05:31.830","Text":"And so we get 1 over square root of 4t plus"},{"Start":"05:31.830 ","End":"05:37.165","Text":"1 times e to the minus x squared over 4t plus 1."},{"Start":"05:37.165 ","End":"05:40.380","Text":"That\u0027s the answer for u of xt,"},{"Start":"05:40.380 ","End":"05:44.410","Text":"and that concludes this exercise."}],"ID":30803},{"Watched":false,"Name":"Exercise 5a","Duration":"2m 14s","ChapterTopicVideoID":29218,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:04.185","Text":"In this exercise, we have 2 parts,"},{"Start":"00:04.185 ","End":"00:06.660","Text":"but each of them shares a common bit."},{"Start":"00:06.660 ","End":"00:13.380","Text":"The PDE is the heat equation on the whole real line,"},{"Start":"00:13.380 ","End":"00:15.570","Text":"but there\u0027s 2 different initial conditions."},{"Start":"00:15.570 ","End":"00:18.165","Text":"In part a, this is our initial condition."},{"Start":"00:18.165 ","End":"00:21.070","Text":"In part b, well, when we get to it, we\u0027ll see."},{"Start":"00:21.070 ","End":"00:23.465","Text":"Let\u0027s start with part a."},{"Start":"00:23.465 ","End":"00:26.045","Text":"We\u0027ll use Poisson\u0027s formula,"},{"Start":"00:26.045 ","End":"00:29.045","Text":"which is this but in our case,"},{"Start":"00:29.045 ","End":"00:30.620","Text":"a equals 1."},{"Start":"00:30.620 ","End":"00:32.045","Text":"That\u0027s from here."},{"Start":"00:32.045 ","End":"00:33.425","Text":"It\u0027s normally a^2,"},{"Start":"00:33.425 ","End":"00:36.410","Text":"u_xx, a^2 is 1 so a is 1,"},{"Start":"00:36.410 ","End":"00:39.830","Text":"and u(y, 0) is the same as u( x,"},{"Start":"00:39.830 ","End":"00:41.975","Text":"0,) x replaced by y."},{"Start":"00:41.975 ","End":"00:44.870","Text":"It\u0027s also 2. Substituting these in here,"},{"Start":"00:44.870 ","End":"00:48.920","Text":"it simplifies somewhat and we get this integral."},{"Start":"00:48.920 ","End":"00:52.950","Text":"2 cancels with the 2 and the exponent,"},{"Start":"00:52.950 ","End":"00:56.120","Text":"we can write it as minus something squared,"},{"Start":"00:56.120 ","End":"01:00.445","Text":"y minus x over 2 root t^2."},{"Start":"01:00.445 ","End":"01:03.245","Text":"Now we\u0027ll do a substitution as usual."},{"Start":"01:03.245 ","End":"01:11.380","Text":"We\u0027ll let z equal y minus x over 2 root 2 and then dz will be 1 over 2 root 2 dy."},{"Start":"01:11.380 ","End":"01:13.430","Text":"Also the limits of integration,"},{"Start":"01:13.430 ","End":"01:14.900","Text":"we\u0027ve seen this before,"},{"Start":"01:14.900 ","End":"01:18.200","Text":"z is also from minus infinity to infinity."},{"Start":"01:18.200 ","End":"01:19.760","Text":"If you let y go to infinity,"},{"Start":"01:19.760 ","End":"01:23.175","Text":"then z goes to infinity because this is positive."},{"Start":"01:23.175 ","End":"01:27.700","Text":"Note that dy is 2 root t times dz,"},{"Start":"01:27.700 ","End":"01:31.270","Text":"which we can also bring out in front of the integral,"},{"Start":"01:31.270 ","End":"01:32.655","Text":"so we get this,"},{"Start":"01:32.655 ","End":"01:34.880","Text":"the 2 root 2, as I said comes out."},{"Start":"01:34.880 ","End":"01:41.150","Text":"This part is z and this is the integral we get and we know what this integral is;"},{"Start":"01:41.150 ","End":"01:43.930","Text":"that\u0027s the square root of Pi."},{"Start":"01:43.930 ","End":"01:46.130","Text":"Altogether we get this,"},{"Start":"01:46.130 ","End":"01:48.710","Text":"everything cancels except for the 2,"},{"Start":"01:48.710 ","End":"01:51.500","Text":"so the answer is that u(x,"},{"Start":"01:51.500 ","End":"01:54.470","Text":"t) is the constant function 2."},{"Start":"01:54.470 ","End":"01:57.575","Text":"I\u0027ll just add this makes sense because"},{"Start":"01:57.575 ","End":"02:03.495","Text":"the initial condition is that u is constant 2 along the x-axis."},{"Start":"02:03.495 ","End":"02:05.300","Text":"As time goes by,"},{"Start":"02:05.300 ","End":"02:06.800","Text":"if the temperature is constant,"},{"Start":"02:06.800 ","End":"02:12.200","Text":"nothing\u0027s going to change and it\u0027s going to remain constant as t gets large as you want."},{"Start":"02:12.200 ","End":"02:14.730","Text":"That concludes part a."}],"ID":30804},{"Watched":false,"Name":"Exercise 5b","Duration":"4m 10s","ChapterTopicVideoID":29205,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.715","Text":"Continuing with the same exercise we\u0027ve just finished part a in the previous clip,"},{"Start":"00:05.715 ","End":"00:07.830","Text":"and this was the answer."},{"Start":"00:07.830 ","End":"00:13.350","Text":"Now we\u0027re doing part b where we\u0027ve changed the initial condition to this one."},{"Start":"00:13.350 ","End":"00:15.660","Text":"U(x_naught) equals x^2,"},{"Start":"00:15.660 ","End":"00:17.055","Text":"and we\u0027re given a hint."},{"Start":"00:17.055 ","End":"00:20.265","Text":"Now if we try doing part b with Poisson\u0027s formula,"},{"Start":"00:20.265 ","End":"00:25.605","Text":"the integral turns out to be rather complicated so we\u0027re going to use the hint,"},{"Start":"00:25.605 ","End":"00:32.010","Text":"which is to define a new function q to be the derivative of u with respect to t. Anyway,"},{"Start":"00:32.010 ","End":"00:37.040","Text":"if q=ut, differentiate both sides with respect to t,"},{"Start":"00:37.040 ","End":"00:38.570","Text":"and this is what we get."},{"Start":"00:38.570 ","End":"00:41.465","Text":"Replace u_t by u_xx,"},{"Start":"00:41.465 ","End":"00:43.295","Text":"which is equal 2."},{"Start":"00:43.295 ","End":"00:47.214","Text":"Now we can change the order of the differentiation so instead of"},{"Start":"00:47.214 ","End":"00:50.930","Text":"xx t we could do it as first t, then x,"},{"Start":"00:50.930 ","End":"00:55.085","Text":"then x because there\u0027s a theorem due to Clairaut or Schwarz"},{"Start":"00:55.085 ","End":"01:00.650","Text":"that you can rearrange the order of the partial derivatives."},{"Start":"01:00.650 ","End":"01:03.515","Text":"Since u_t is q,"},{"Start":"01:03.515 ","End":"01:07.940","Text":"what we get is q_t=q_xx."},{"Start":"01:07.940 ","End":"01:09.710","Text":"That\u0027s also the heat equation."},{"Start":"01:09.710 ","End":"01:11.735","Text":"Q satisfies it also."},{"Start":"01:11.735 ","End":"01:17.615","Text":"What we need to solve it is also the initial condition q(x_naught )."},{"Start":"01:17.615 ","End":"01:21.995","Text":"Q is defined to be u_t so q(x_naught) is u_t(x_naught)."},{"Start":"01:21.995 ","End":"01:23.795","Text":"We want to figure out what this is,"},{"Start":"01:23.795 ","End":"01:25.400","Text":"but we don\u0027t have u,"},{"Start":"01:25.400 ","End":"01:31.340","Text":"so we can differentiate u to get u_t(x_t) and then substitute 0,"},{"Start":"01:31.340 ","End":"01:32.960","Text":"go about it another way."},{"Start":"01:32.960 ","End":"01:35.760","Text":"U(x, 0) is x^2."},{"Start":"01:35.760 ","End":"01:36.945","Text":"We\u0027re given that."},{"Start":"01:36.945 ","End":"01:40.030","Text":"What you shouldn\u0027t do in case you attempted to,"},{"Start":"01:40.030 ","End":"01:41.630","Text":"is to think that,"},{"Start":"01:41.630 ","End":"01:47.450","Text":"this is a function just of x and if we differentiate with respect to t we\u0027ll get 0."},{"Start":"01:47.450 ","End":"01:51.980","Text":"Well no, you can\u0027t substitute and then differentiate."},{"Start":"01:51.980 ","End":"01:55.653","Text":"You have to first differentiate and then substitute,"},{"Start":"01:55.653 ","End":"01:57.950","Text":"so this doesn\u0027t work."},{"Start":"01:57.950 ","End":"02:02.825","Text":"Now recall that u_t is u_xx heat equation."},{"Start":"02:02.825 ","End":"02:07.720","Text":"What we can do is we can compute u_xx from u,"},{"Start":"02:07.720 ","End":"02:12.845","Text":"even though we only know u on the x-axis on x, 0."},{"Start":"02:12.845 ","End":"02:18.410","Text":"What we said no to here applies to differentiating with respect to t. But we"},{"Start":"02:18.410 ","End":"02:24.435","Text":"can differentiate with respect to x because we have u(x_naught) on all of the x-axis,"},{"Start":"02:24.435 ","End":"02:26.750","Text":"and the partial derivative with respect to x just"},{"Start":"02:26.750 ","End":"02:30.590","Text":"depends on values to the left and right of a point."},{"Start":"02:30.590 ","End":"02:33.230","Text":"From u(x_naught) equals x^2,"},{"Start":"02:33.230 ","End":"02:34.910","Text":"differentiate with respect to x."},{"Start":"02:34.910 ","End":"02:36.110","Text":"We get 2x,"},{"Start":"02:36.110 ","End":"02:39.970","Text":"then differentiate again with respect to x and we get 2."},{"Start":"02:39.970 ","End":"02:43.410","Text":"Now we have u_xx,"},{"Start":"02:43.410 ","End":"02:48.750","Text":"and since u_t=u_xx we also have u_t=2."},{"Start":"02:48.750 ","End":"02:56.495","Text":"Now we have the heat equation and an initial condition because u_t=q."},{"Start":"02:56.495 ","End":"02:58.040","Text":"These are the two things that matter."},{"Start":"02:58.040 ","End":"03:05.010","Text":"We have the heat equation satisfied by q and also the initial condition."},{"Start":"03:05.010 ","End":"03:08.600","Text":"This is exactly like the problem in part a,"},{"Start":"03:08.600 ","End":"03:11.210","Text":"except with q instead of u."},{"Start":"03:11.210 ","End":"03:15.800","Text":"We can use the result of part a to say that q (x,"},{"Start":"03:15.800 ","End":"03:17.180","Text":"t) is equal to 2,"},{"Start":"03:17.180 ","End":"03:20.475","Text":"just like u was in part a."},{"Start":"03:20.475 ","End":"03:23.550","Text":"Now, q is u_t,"},{"Start":"03:23.550 ","End":"03:26.010","Text":"so u_t(x, t) is equal to 2,"},{"Start":"03:26.010 ","End":"03:30.485","Text":"and now we can integrate this with respect to t and get that u(x,"},{"Start":"03:30.485 ","End":"03:34.100","Text":"t) is 2 t plus constant,"},{"Start":"03:34.100 ","End":"03:37.040","Text":"but not exactly a constant and arbitrary function of x,"},{"Start":"03:37.040 ","End":"03:39.335","Text":"which is a constant as far as t goes."},{"Start":"03:39.335 ","End":"03:41.285","Text":"Now how do we find c(x)?"},{"Start":"03:41.285 ","End":"03:45.735","Text":"We can substitute t=0 and use this,"},{"Start":"03:45.735 ","End":"03:48.810","Text":"that u(x_naught) is x^2,"},{"Start":"03:48.810 ","End":"03:53.990","Text":"so u(x_naught) is x^2 means that if we put t here,"},{"Start":"03:53.990 ","End":"03:57.560","Text":"we also get twice 0 plus c(x), so we get c(x)."},{"Start":"03:57.560 ","End":"03:59.270","Text":"C(x) is x^2,"},{"Start":"03:59.270 ","End":"04:06.360","Text":"and now put c(x) here and we have that u(x,t ) is 2t plus x^2,"},{"Start":"04:06.360 ","End":"04:10.690","Text":"and that\u0027s the solution to part b, and we\u0027re done."}],"ID":30805},{"Watched":false,"Name":"Exercise 6a","Duration":"6m 32s","ChapterTopicVideoID":29206,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:05.550","Text":"In this exercise, we\u0027re given the following initial value problem,"},{"Start":"00:05.550 ","End":"00:12.210","Text":"which is not a heat equation because there\u0027s this extra minus u."},{"Start":"00:12.210 ","End":"00:14.040","Text":"Anyway, we have to find u(x,"},{"Start":"00:14.040 ","End":"00:19.620","Text":"t) and there\u0027s a hint or suggestion to define v(x,"},{"Start":"00:19.620 ","End":"00:21.900","Text":"t) according to this rule."},{"Start":"00:21.900 ","End":"00:25.980","Text":"Then in part b, we\u0027ll compute the integral that\u0027s written here"},{"Start":"00:25.980 ","End":"00:30.900","Text":"based on the result of part a. I will start with part a."},{"Start":"00:30.900 ","End":"00:34.080","Text":"We\u0027re going to use this suggestion,"},{"Start":"00:34.080 ","End":"00:36.750","Text":"and if we get Alpha and Beta,"},{"Start":"00:36.750 ","End":"00:41.255","Text":"this should reduce to a regular heat equation."},{"Start":"00:41.255 ","End":"00:43.070","Text":"In other words, like this,"},{"Start":"00:43.070 ","End":"00:45.830","Text":"but with v and without this extra term."},{"Start":"00:45.830 ","End":"00:52.255","Text":"The hint was originally given as u in terms of v. Later we\u0027ll use v in terms of u."},{"Start":"00:52.255 ","End":"00:57.150","Text":"We want some partial derivatives in order to substitute here du"},{"Start":"00:57.150 ","End":"01:02.450","Text":"by dt is using the product rule for differentiation."},{"Start":"01:02.450 ","End":"01:07.040","Text":"This times the derivative of this plus the derivative of this,"},{"Start":"01:07.040 ","End":"01:09.050","Text":"which is just Beta,"},{"Start":"01:09.050 ","End":"01:14.480","Text":"because we\u0027re differentiating with respect to t times v. Similarly,"},{"Start":"01:14.480 ","End":"01:17.375","Text":"du by dx, we get this."},{"Start":"01:17.375 ","End":"01:21.005","Text":"Then we need also second derivative with respect to"},{"Start":"01:21.005 ","End":"01:27.139","Text":"x. I\u0027ll let you follow this pause and see that this is okay."},{"Start":"01:27.139 ","End":"01:33.680","Text":"Note that we get twice this term and so we can just write a 2 in front of it."},{"Start":"01:33.680 ","End":"01:40.100","Text":"What\u0027s more all the terms here contain this e^Alpha x plus Beta t,"},{"Start":"01:40.100 ","End":"01:43.070","Text":"so we can take them out the brackets,"},{"Start":"01:43.070 ","End":"01:46.130","Text":"like here and also here,"},{"Start":"01:46.130 ","End":"01:49.085","Text":"we can take this out the brackets."},{"Start":"01:49.085 ","End":"01:52.040","Text":"Now we\u0027re going to substitute in, where is it?"},{"Start":"01:52.040 ","End":"01:57.705","Text":"This equation, the PDE."},{"Start":"01:57.705 ","End":"01:59.820","Text":"We have ut here,"},{"Start":"01:59.820 ","End":"02:01.185","Text":"we have u here,"},{"Start":"02:01.185 ","End":"02:03.375","Text":"we have u_xx here,"},{"Start":"02:03.375 ","End":"02:09.980","Text":"and this is what we get because we can divide everything by e^Alpha x plus Beta t,"},{"Start":"02:09.980 ","End":"02:12.845","Text":"and then we get the following."},{"Start":"02:12.845 ","End":"02:21.890","Text":"This is equal to this minus v. These 3 terms which are just v is collecting like terms,"},{"Start":"02:21.890 ","End":"02:26.730","Text":"there\u0027s only these 3 that are like and rearranging,"},{"Start":"02:26.730 ","End":"02:28.790","Text":"so we just leave v_t on one side,"},{"Start":"02:28.790 ","End":"02:30.305","Text":"everything else on the right,"},{"Start":"02:30.305 ","End":"02:34.850","Text":"we have v_xx here 2 Alpha v_x here."},{"Start":"02:34.850 ","End":"02:39.290","Text":"Here we have Alpha squared minus 1 and then minus Beta"},{"Start":"02:39.290 ","End":"02:45.320","Text":"v. What we\u0027d really like is for this to be 0 and this to be 0."},{"Start":"02:45.320 ","End":"02:48.830","Text":"Because then we\u0027ll have just v_t equals v_xx,"},{"Start":"02:48.830 ","End":"02:50.705","Text":"which is the heat equation."},{"Start":"02:50.705 ","End":"02:56.585","Text":"2 unknowns is easy 2 Alpha 0, so Alpha is 0."},{"Start":"02:56.585 ","End":"02:58.700","Text":"If Alpha is 0, minus 1,"},{"Start":"02:58.700 ","End":"03:00.080","Text":"minus Beta is 0,"},{"Start":"03:00.080 ","End":"03:02.065","Text":"so Beta is minus 1."},{"Start":"03:02.065 ","End":"03:04.290","Text":"Which means that u(x,"},{"Start":"03:04.290 ","End":"03:07.550","Text":"t) is e^Alpha x plus Beta t,"},{"Start":"03:07.550 ","End":"03:17.285","Text":"which is just e^t times v. If we want it the other way round v is equal to e^tu,"},{"Start":"03:17.285 ","End":"03:21.575","Text":"and so v satisfies the following IVP,"},{"Start":"03:21.575 ","End":"03:25.595","Text":"v_t equals v_xx and v(x, 0)."},{"Start":"03:25.595 ","End":"03:30.325","Text":"Just using this e^t is e^0."},{"Start":"03:30.325 ","End":"03:34.400","Text":"Using this formula, we have set of 0,1,"},{"Start":"03:34.400 ","End":"03:37.445","Text":"it\u0027s still 0 and 1 because e^0 is 1."},{"Start":"03:37.445 ","End":"03:39.260","Text":"Writing it out in full,"},{"Start":"03:39.260 ","End":"03:41.330","Text":"this is the initial value problem."},{"Start":"03:41.330 ","End":"03:43.045","Text":"We have for v,"},{"Start":"03:43.045 ","End":"03:46.685","Text":"and for this, we can use the Poisson formula."},{"Start":"03:46.685 ","End":"03:48.845","Text":"In our case, a is 1,"},{"Start":"03:48.845 ","End":"03:53.480","Text":"and the fact that v is this means that we can just take the"},{"Start":"03:53.480 ","End":"03:59.565","Text":"integral from 0 to infinity of 1 times the rest of it."},{"Start":"03:59.565 ","End":"04:03.525","Text":"In other words just this part,"},{"Start":"04:03.525 ","End":"04:07.430","Text":"1 from 0 to infinity and is 0 from minus infinity to 0,"},{"Start":"04:07.430 ","End":"04:09.065","Text":"so we don\u0027t need that part."},{"Start":"04:09.065 ","End":"04:14.480","Text":"We can rewrite this as something squared, like so."},{"Start":"04:14.480 ","End":"04:16.490","Text":"Now we want to do a substitution."},{"Start":"04:16.490 ","End":"04:19.130","Text":"Let z be this,"},{"Start":"04:19.130 ","End":"04:23.555","Text":"so dz is 1 over 2 root t, dy."},{"Start":"04:23.555 ","End":"04:29.100","Text":"Dy would be 2 root t dz if we substitute."},{"Start":"04:29.330 ","End":"04:32.790","Text":"We can take the 2 root t in front of the integral,"},{"Start":"04:32.790 ","End":"04:36.510","Text":"so this is the integral we now have."},{"Start":"04:36.510 ","End":"04:40.595","Text":"We also have to substitute the limits of integration."},{"Start":"04:40.595 ","End":"04:42.260","Text":"As y goes to infinity,"},{"Start":"04:42.260 ","End":"04:45.250","Text":"z goes to infinity but when y is 0,"},{"Start":"04:45.250 ","End":"04:50.985","Text":"z is minus x over 2 root t. That\u0027s this here."},{"Start":"04:50.985 ","End":"04:52.970","Text":"I\u0027m just combining these,"},{"Start":"04:52.970 ","End":"04:57.415","Text":"we\u0027re left with 1 over the square root of Pi times the integral."},{"Start":"04:57.415 ","End":"05:00.620","Text":"Now we\u0027ll break the integral up into 2 parts from"},{"Start":"05:00.620 ","End":"05:05.045","Text":"this to 0 and from 0 to infinity, like so."},{"Start":"05:05.045 ","End":"05:10.925","Text":"We can write this 1 over root Pi as a 1/2 times 2 over root pi."},{"Start":"05:10.925 ","End":"05:15.125","Text":"The reason for this is we want to use the error function."},{"Start":"05:15.125 ","End":"05:22.295","Text":"Because the error function is 2 over root Pi times the integral from 0 to x."},{"Start":"05:22.295 ","End":"05:24.710","Text":"There\u0027s something else I need to explain."},{"Start":"05:24.710 ","End":"05:29.900","Text":"This isn\u0027t from 0 to x over 2 root t. But if you look at the picture,"},{"Start":"05:29.900 ","End":"05:33.440","Text":"there\u0027s a symmetry that the integral from minus"},{"Start":"05:33.440 ","End":"05:37.505","Text":"something to 0 is the same as from 0 to plus the something."},{"Start":"05:37.505 ","End":"05:42.560","Text":"You could do it in 2 steps by putting a minus in front and switching"},{"Start":"05:42.560 ","End":"05:47.975","Text":"the order of these 2 and then using the fact that the error function is odd anyway,"},{"Start":"05:47.975 ","End":"05:50.050","Text":"just get to this right away."},{"Start":"05:50.050 ","End":"05:52.670","Text":"Here, of course we\u0027re using the fact that this integral is"},{"Start":"05:52.670 ","End":"05:55.460","Text":"well known and it\u0027s root Pi over 2."},{"Start":"05:55.460 ","End":"05:58.880","Text":"If it was minus infinity to infinity, it would be root Pi."},{"Start":"05:58.880 ","End":"06:01.130","Text":"That\u0027s only 1/2 of the root Pi."},{"Start":"06:01.130 ","End":"06:03.440","Text":"Simplifying this, it comes out to a 1/2,"},{"Start":"06:03.440 ","End":"06:05.795","Text":"bring the 1/2 out front of the brackets,"},{"Start":"06:05.795 ","End":"06:08.950","Text":"got error function of this plus 1."},{"Start":"06:08.950 ","End":"06:11.265","Text":"This is v(x, t)."},{"Start":"06:11.265 ","End":"06:14.310","Text":"Remember that u(x, t) is e to the minus t, v(x,"},{"Start":"06:14.310 ","End":"06:17.880","Text":"t) and so we get that u(x,"},{"Start":"06:17.880 ","End":"06:23.015","Text":"t) is like this or like this I mean,"},{"Start":"06:23.015 ","End":"06:26.375","Text":"but with an e to the minus t here."},{"Start":"06:26.375 ","End":"06:28.415","Text":"That\u0027s the answer."},{"Start":"06:28.415 ","End":"06:32.940","Text":"To part a at least and part b in the following clip."}],"ID":30806},{"Watched":false,"Name":"Exercise 6b","Duration":"2m 35s","ChapterTopicVideoID":29207,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.000","Text":"This is a continuation clip."},{"Start":"00:03.000 ","End":"00:05.595","Text":"We just did part a,"},{"Start":"00:05.595 ","End":"00:08.385","Text":"and this was the result we got."},{"Start":"00:08.385 ","End":"00:13.575","Text":"Now we\u0027re going to do part b where we have to compute the following integral."},{"Start":"00:13.575 ","End":"00:17.999","Text":"The first thing we\u0027ll do is find out what is du by dx."},{"Start":"00:17.999 ","End":"00:19.470","Text":"Since we have u already,"},{"Start":"00:19.470 ","End":"00:21.510","Text":"we have to differentiate with respect to x."},{"Start":"00:21.510 ","End":"00:24.615","Text":"This is a constant as far as x goes."},{"Start":"00:24.615 ","End":"00:26.805","Text":"That stays."},{"Start":"00:26.805 ","End":"00:33.495","Text":"Then we get the derivative of the error function times inner derivative,"},{"Start":"00:33.495 ","End":"00:36.030","Text":"derivative of 1 is 0."},{"Start":"00:36.030 ","End":"00:37.830","Text":"This is what we have."},{"Start":"00:37.830 ","End":"00:41.225","Text":"We know what the derivative of the error function is."},{"Start":"00:41.225 ","End":"00:43.040","Text":"When we learned the error function,"},{"Start":"00:43.040 ","End":"00:46.295","Text":"there were some properties and this is 1 of the properties."},{"Start":"00:46.295 ","End":"00:49.475","Text":"Now we can substitute instead of x,"},{"Start":"00:49.475 ","End":"00:51.335","Text":"x over 2 root t,"},{"Start":"00:51.335 ","End":"00:53.510","Text":"and we get this."},{"Start":"00:53.510 ","End":"00:55.400","Text":"It all simplifies."},{"Start":"00:55.400 ","End":"00:57.980","Text":"We take all the constants,"},{"Start":"00:57.980 ","End":"01:00.025","Text":"we get the following,"},{"Start":"01:00.025 ","End":"01:03.810","Text":"1 over 2 root Pi,"},{"Start":"01:03.810 ","End":"01:06.750","Text":"root t, this 2 with this 2 cancels."},{"Start":"01:06.750 ","End":"01:14.795","Text":"We also have an e^minus t. The integral of this thing squared is the following."},{"Start":"01:14.795 ","End":"01:20.020","Text":"E^minus 2t and 4Pi t with squaring this."},{"Start":"01:20.020 ","End":"01:25.745","Text":"Then here we just have to double the exponent in order to square the whole thing here."},{"Start":"01:25.745 ","End":"01:32.690","Text":"Now we can take this part in front of the integral and we can put the 2 inside."},{"Start":"01:32.690 ","End":"01:34.850","Text":"We get, well, if you think about it,"},{"Start":"01:34.850 ","End":"01:37.310","Text":"it\u0027s 2 over 2^2, which is 1/2."},{"Start":"01:37.310 ","End":"01:41.034","Text":"If you take the square root of that is 1 over root 2 and we can combine."},{"Start":"01:41.034 ","End":"01:42.080","Text":"Anyway, think about this."},{"Start":"01:42.080 ","End":"01:47.794","Text":"This is what we can get to simplify it as minus something squared."},{"Start":"01:47.794 ","End":"01:50.345","Text":"Now we\u0027re going to substitute this,"},{"Start":"01:50.345 ","End":"01:51.950","Text":"we\u0027ll call z,"},{"Start":"01:51.950 ","End":"01:55.700","Text":"and dz is 1 over root 2t dx."},{"Start":"01:55.700 ","End":"02:01.440","Text":"Also, dx is root 2t times dz."},{"Start":"02:01.510 ","End":"02:09.755","Text":"We get the root 2t here and we take it in front of the integral also e^minus z^2."},{"Start":"02:09.755 ","End":"02:11.780","Text":"This is a well-known integral."},{"Start":"02:11.780 ","End":"02:14.665","Text":"We know that this is the square root of Pi."},{"Start":"02:14.665 ","End":"02:18.035","Text":"This is our answer. We just simplify it a bit."},{"Start":"02:18.035 ","End":"02:24.370","Text":"We have 1 over 2^2 and we have 2^1/2, altogether 2^-1/1/2."},{"Start":"02:24.370 ","End":"02:28.015","Text":"Then the root Pi over Pi is 1 over root Pi."},{"Start":"02:28.015 ","End":"02:30.770","Text":"The root t over t is 1 over root t. Anyway,"},{"Start":"02:30.770 ","End":"02:33.335","Text":"this is the simplified result."},{"Start":"02:33.335 ","End":"02:36.240","Text":"That concludes this clip."}],"ID":30807},{"Watched":false,"Name":"Exercise 7","Duration":"3m 47s","ChapterTopicVideoID":29208,"CourseChapterTopicPlaylistID":294429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.940","Text":"In this exercise, we have following initial value problem involving"},{"Start":"00:05.940 ","End":"00:14.130","Text":"the heat equation and we\u0027re given a hint to define w equals du by dt,"},{"Start":"00:14.130 ","End":"00:17.730","Text":"and we\u0027ll see why we need this in a moment."},{"Start":"00:17.730 ","End":"00:20.325","Text":"If we use Poisson\u0027s formula,"},{"Start":"00:20.325 ","End":"00:24.495","Text":"we get the following integral for u(x,t),"},{"Start":"00:24.495 ","End":"00:28.485","Text":"and this is not an easy integral to do,"},{"Start":"00:28.485 ","End":"00:30.255","Text":"so we\u0027re going to use the trick,"},{"Start":"00:30.255 ","End":"00:31.695","Text":"which is the hint."},{"Start":"00:31.695 ","End":"00:37.095","Text":"We\u0027re going to define w equals du by dt and we\u0027ll see how this helps."},{"Start":"00:37.095 ","End":"00:39.270","Text":"If we differentiate with respect to t,"},{"Start":"00:39.270 ","End":"00:42.075","Text":"we get du by dt with respect to t,"},{"Start":"00:42.075 ","End":"00:46.785","Text":"but u_t is equal to u_xx,"},{"Start":"00:46.785 ","End":"00:49.114","Text":"so we get this."},{"Start":"00:49.114 ","End":"00:54.305","Text":"Now we can switch the order of the partial derivatives using"},{"Start":"00:54.305 ","End":"00:57.500","Text":"Clairaut\u0027s theorem or Schwarz\u0027s theorem and"},{"Start":"00:57.500 ","End":"01:01.640","Text":"get du by dt second derivative with respect to x."},{"Start":"01:01.640 ","End":"01:03.625","Text":"But this is w,"},{"Start":"01:03.625 ","End":"01:06.305","Text":"so w _t is w_xx."},{"Start":"01:06.305 ","End":"01:09.050","Text":"It means that w satisfies the heat equation."},{"Start":"01:09.050 ","End":"01:14.545","Text":"But we still don\u0027t have an initial condition for w in order to solve it."},{"Start":"01:14.545 ","End":"01:16.410","Text":"w is u_t,"},{"Start":"01:16.410 ","End":"01:18.780","Text":"you need to figure out u_t."},{"Start":"01:18.780 ","End":"01:22.390","Text":"We have, u(x,0) equals cosine x,"},{"Start":"01:22.390 ","End":"01:25.400","Text":"but what we can\u0027t do is just say,"},{"Start":"01:25.400 ","End":"01:28.550","Text":"oh, the derivative of this with respect to t is 0."},{"Start":"01:28.550 ","End":"01:32.090","Text":"You can\u0027t do that. You can\u0027t first substitute then differentiate."},{"Start":"01:32.090 ","End":"01:33.455","Text":"This is not right,"},{"Start":"01:33.455 ","End":"01:38.400","Text":"but we can say that u_t is u_xx."},{"Start":"01:38.720 ","End":"01:41.090","Text":"Therefore from this equation,"},{"Start":"01:41.090 ","End":"01:42.470","Text":"we get this equation."},{"Start":"01:42.470 ","End":"01:45.470","Text":"Now it\u0027s okay to differentiate with respect to x because we\u0027re"},{"Start":"01:45.470 ","End":"01:49.585","Text":"given u along the x-axis here."},{"Start":"01:49.585 ","End":"01:53.795","Text":"First derivative is sine x, from cosine x."},{"Start":"01:53.795 ","End":"01:57.710","Text":"Second derivative is minus cosine x."},{"Start":"01:57.710 ","End":"01:59.930","Text":"Sorry, I forgot a minus here."},{"Start":"01:59.930 ","End":"02:04.490","Text":"Now we have that w(x,0) is equal to this,"},{"Start":"02:04.490 ","End":"02:05.945","Text":"which is equal to this,"},{"Start":"02:05.945 ","End":"02:09.455","Text":"so w(x,0) is minus cosine x,"},{"Start":"02:09.455 ","End":"02:12.830","Text":"and we have w_t equals w_xx,"},{"Start":"02:12.830 ","End":"02:14.945","Text":"the heat equation from here."},{"Start":"02:14.945 ","End":"02:21.050","Text":"Now we have the full initial value problem and we can use Poisson\u0027s formula."},{"Start":"02:21.050 ","End":"02:25.505","Text":"Now emphasize the minus here because if we go back and look,"},{"Start":"02:25.505 ","End":"02:29.925","Text":"this is our Poisson formula for you."},{"Start":"02:29.925 ","End":"02:33.560","Text":"It appears that we\u0027re just going in circles because"},{"Start":"02:33.560 ","End":"02:36.500","Text":"we haven\u0027t got anything special for w. I mean,"},{"Start":"02:36.500 ","End":"02:39.020","Text":"it\u0027s just minus what we had for u."},{"Start":"02:39.020 ","End":"02:41.240","Text":"But here\u0027s the trick."},{"Start":"02:41.240 ","End":"02:43.130","Text":"Ignore this integral."},{"Start":"02:43.130 ","End":"02:50.735","Text":"What\u0027s important is that we have that w(x,t) equals minus u(x,t)."},{"Start":"02:50.735 ","End":"02:55.635","Text":"But w(x,t) is du by dt of x,t."},{"Start":"02:55.635 ","End":"02:58.310","Text":"Now we have this equation,"},{"Start":"02:58.310 ","End":"03:01.115","Text":"which is an ordinary differential equation."},{"Start":"03:01.115 ","End":"03:04.204","Text":"If we think of x like a constant,"},{"Start":"03:04.204 ","End":"03:08.015","Text":"think of it like y\u0027 equals minus y."},{"Start":"03:08.015 ","End":"03:13.695","Text":"The solution is u(x,t) is some constant times e to the minus t,"},{"Start":"03:13.695 ","End":"03:18.110","Text":"but not a constant really or arbitrary function of just x."},{"Start":"03:18.110 ","End":"03:20.375","Text":"If we substitute t equals 0,"},{"Start":"03:20.375 ","End":"03:23.110","Text":"we get u(x,0) equals,"},{"Start":"03:23.110 ","End":"03:25.910","Text":"on the 1 hand, we were given that it\u0027s equal to cosine x."},{"Start":"03:25.910 ","End":"03:30.065","Text":"On the other hand, you have u(x,0) equals just C(x)."},{"Start":"03:30.065 ","End":"03:32.570","Text":"0 is t, so that makes this 1."},{"Start":"03:32.570 ","End":"03:35.810","Text":"We get that C(x) is cosine x,"},{"Start":"03:35.810 ","End":"03:43.100","Text":"and then we can just put that here and get u(x,t) equals e to the minus t cosine x."},{"Start":"03:43.100 ","End":"03:45.380","Text":"That is the answer."},{"Start":"03:45.380 ","End":"03:48.090","Text":"That concludes this clip."}],"ID":30808}],"Thumbnail":null,"ID":294429},{"Name":"Finite Interval Separation of Variables","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Heat Equation on a Finite Interval (part 1)","Duration":"4m 40s","ChapterTopicVideoID":29224,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.910","Text":"This is a brief introduction to the heat equation on a finite interval."},{"Start":"00:05.910 ","End":"00:09.360","Text":"Previously, we learned it on the infinite interval."},{"Start":"00:09.360 ","End":"00:15.390","Text":"The technique used in the finite interval case it\u0027s called separation of variables."},{"Start":"00:15.390 ","End":"00:18.735","Text":"But we have to have homogeneous boundary conditions."},{"Start":"00:18.735 ","End":"00:20.744","Text":"We have an unknown function,"},{"Start":"00:20.744 ","End":"00:24.375","Text":"u(x, t) which satisfies the following."},{"Start":"00:24.375 ","End":"00:26.940","Text":"The heat equation is as follows."},{"Start":"00:26.940 ","End":"00:29.865","Text":"du by dt is a^2,"},{"Start":"00:29.865 ","End":"00:33.030","Text":"second derivative of du by dx,"},{"Start":"00:33.030 ","End":"00:39.530","Text":"a is positive plus function of x and t. This is on a domain."},{"Start":"00:39.530 ","End":"00:44.484","Text":"Well, it\u0027s basically the semi-infinite strip shown in the picture."},{"Start":"00:44.484 ","End":"00:49.730","Text":"x is between 0 and L and t is from 0 up to infinity."},{"Start":"00:49.730 ","End":"00:55.310","Text":"The homogeneous heat equation is when this function f is 0."},{"Start":"00:55.310 ","End":"00:58.315","Text":"We just have u_t =a^2 u_xx."},{"Start":"00:58.315 ","End":"01:01.235","Text":"There\u0027s an initial condition and boundary conditions."},{"Start":"01:01.235 ","End":"01:05.825","Text":"The initial condition is that we\u0027re given the value of u on"},{"Start":"01:05.825 ","End":"01:11.240","Text":"this part of the x-axis between 0 and L. I should have said here x is between 0 and L,"},{"Start":"01:11.240 ","End":"01:14.930","Text":"and the boundary conditions where we\u0027re given the function"},{"Start":"01:14.930 ","End":"01:19.635","Text":"along this line and this line, semi-infinite lines."},{"Start":"01:19.635 ","End":"01:21.545","Text":"Well, there\u0027s different cases."},{"Start":"01:21.545 ","End":"01:24.890","Text":"Sometimes we\u0027re given the value of the function itself"},{"Start":"01:24.890 ","End":"01:29.375","Text":"u along here it\u0027s a(t) and along here it\u0027s b(t)."},{"Start":"01:29.375 ","End":"01:35.854","Text":"But sometimes we\u0027re given the value of the derivative of u with respect to x."},{"Start":"01:35.854 ","End":"01:39.140","Text":"This type of condition is called the von Neumann condition,"},{"Start":"01:39.140 ","End":"01:41.330","Text":"and this is of type Dirichlet."},{"Start":"01:41.330 ","End":"01:44.245","Text":"This is where the endpoints are fixed."},{"Start":"01:44.245 ","End":"01:50.165","Text":"The other condition, the von Neumann gives us that the derivative here,"},{"Start":"01:50.165 ","End":"01:52.460","Text":"and the derivative here are known."},{"Start":"01:52.460 ","End":"01:56.385","Text":"In these pictures, a(t) and b(t) are 0."},{"Start":"01:56.385 ","End":"01:59.790","Text":"Here this point is on the x-axis."},{"Start":"01:59.790 ","End":"02:06.050","Text":"In this case, this tangent is horizontal and this tangent is horizontal."},{"Start":"02:06.050 ","End":"02:08.480","Text":"We\u0027ll come to this in a moment."},{"Start":"02:08.480 ","End":"02:11.450","Text":"Let me just say that you could also have mixed endpoints."},{"Start":"02:11.450 ","End":"02:17.300","Text":"You could have this condition on the left and this condition on the right."},{"Start":"02:17.300 ","End":"02:19.010","Text":"In this section we\u0027re going to assume"},{"Start":"02:19.010 ","End":"02:25.390","Text":"homogeneous boundary conditions that a(t) and b(t) are 0."},{"Start":"02:25.390 ","End":"02:28.399","Text":"This or this, or it could be a mixture."},{"Start":"02:28.399 ","End":"02:30.725","Text":"The technique we\u0027re going to use,"},{"Start":"02:30.725 ","End":"02:34.220","Text":"like I said is called separation of variables."},{"Start":"02:34.220 ","End":"02:38.255","Text":"It\u0027s similar to what you learned in the wave equation."},{"Start":"02:38.255 ","End":"02:40.940","Text":"We also had separation of variables."},{"Start":"02:40.940 ","End":"02:47.750","Text":"I\u0027m going to assume you have some knowledge of Sturm-Liouville equations."},{"Start":"02:47.750 ","End":"02:50.430","Text":"If you don\u0027t know about Sturm-Liouville problems,"},{"Start":"02:50.430 ","End":"02:53.900","Text":"you could understand 95 percent of this."},{"Start":"02:53.900 ","End":"02:58.945","Text":"You might want to continue anyway and then catch up later with Sturm-Liouville."},{"Start":"02:58.945 ","End":"03:02.165","Text":"We\u0027re going to solve in 2 main stages."},{"Start":"03:02.165 ","End":"03:04.685","Text":"This is just an outline when the examples,"},{"Start":"03:04.685 ","End":"03:06.080","Text":"you\u0027ll see the details."},{"Start":"03:06.080 ","End":"03:09.260","Text":"The first stage is we find independent solutions,"},{"Start":"03:09.260 ","End":"03:12.730","Text":"I mean linearly independent to the heat equation."},{"Start":"03:12.730 ","End":"03:15.740","Text":"The solutions are of the form some function of"},{"Start":"03:15.740 ","End":"03:19.400","Text":"x times some function of t. I say linear independent,"},{"Start":"03:19.400 ","End":"03:21.409","Text":"I mean you have a function,"},{"Start":"03:21.409 ","End":"03:24.390","Text":"you don\u0027t take 3 times that function as well,"},{"Start":"03:24.390 ","End":"03:25.550","Text":"or if you have 2 of them,"},{"Start":"03:25.550 ","End":"03:27.680","Text":"the sum of them is also a solution,"},{"Start":"03:27.680 ","End":"03:29.420","Text":"but we don\u0027t take that."},{"Start":"03:29.420 ","End":"03:37.055","Text":"In practice, it turns out there\u0027s a countable number or a sequence of such solutions."},{"Start":"03:37.055 ","End":"03:39.485","Text":"We\u0027re going to label them u_n,"},{"Start":"03:39.485 ","End":"03:43.955","Text":"where n goes usually 1,2,3 up to infinity."},{"Start":"03:43.955 ","End":"03:45.830","Text":"Sometimes from 0 to infinity."},{"Start":"03:45.830 ","End":"03:47.165","Text":"It could be variations,"},{"Start":"03:47.165 ","End":"03:49.975","Text":"but usually there\u0027s an infinite number."},{"Start":"03:49.975 ","End":"03:55.190","Text":"The general solution to the original equation will be"},{"Start":"03:55.190 ","End":"04:01.220","Text":"a linear combination or an infinite linear combination of these as follows."},{"Start":"04:01.220 ","End":"04:04.700","Text":"The second stage of the solution is to find out what"},{"Start":"04:04.700 ","End":"04:08.775","Text":"these constants A_n are, these coefficients."},{"Start":"04:08.775 ","End":"04:13.055","Text":"These will be found out in terms of the initial condition,"},{"Start":"04:13.055 ","End":"04:15.260","Text":"which is where is it?"},{"Start":"04:15.260 ","End":"04:19.160","Text":"Using this we\u0027ll find the coefficients A_n."},{"Start":"04:19.160 ","End":"04:22.130","Text":"I forgot to add. In practice,"},{"Start":"04:22.130 ","End":"04:29.330","Text":"these equations and conditions are presented altogether like so, not spread out."},{"Start":"04:29.330 ","End":"04:32.180","Text":"This is what a typical problem would look like."},{"Start":"04:32.180 ","End":"04:34.520","Text":"Sometimes it\u0027s called an IBVP,"},{"Start":"04:34.520 ","End":"04:39.300","Text":"initial boundary and value problem like so."}],"ID":30786},{"Watched":false,"Name":"The Heat Equation on a Finite Interval (part 2)","Duration":"7m 13s","ChapterTopicVideoID":29227,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.835","Text":"In this clip, we\u0027ll discuss some of the technique of separation of variables."},{"Start":"00:05.835 ","End":"00:09.210","Text":"We mentioned in the introduction that it has 2 stages,"},{"Start":"00:09.210 ","End":"00:14.655","Text":"we call them Stage 1 and Stage 2 and in this clip we\u0027ll do the complete Stage 1."},{"Start":"00:14.655 ","End":"00:20.700","Text":"Stage 2 will leave for the solved exercises following this clip."},{"Start":"00:20.700 ","End":"00:25.680","Text":"As an example, we\u0027ll take homogeneous boundary conditions of type Dirichlet,"},{"Start":"00:25.680 ","End":"00:29.430","Text":"that\u0027s here, it\u0027s u no ux here and here."},{"Start":"00:29.430 ","End":"00:32.925","Text":"This is the regular heat equation, it\u0027s homogeneous."},{"Start":"00:32.925 ","End":"00:40.145","Text":"I\u0027ve grayed out the initial condition because we don\u0027t need that for Stage 1."},{"Start":"00:40.145 ","End":"00:42.860","Text":"It\u0027s very important for Stage 2."},{"Start":"00:42.860 ","End":"00:45.805","Text":"In fact, Stage 2 relies mostly on this."},{"Start":"00:45.805 ","End":"00:48.380","Text":"Remember in Stage 1 in the beginning,"},{"Start":"00:48.380 ","End":"00:52.879","Text":"we want to find a solution with variables separated."},{"Start":"00:52.879 ","End":"00:55.085","Text":"We\u0027re looking for a solution u,"},{"Start":"00:55.085 ","End":"01:01.100","Text":"that\u0027s of the type some function of x times some function of t. So to do this,"},{"Start":"01:01.100 ","End":"01:06.665","Text":"we want to substitute in here but we first have to find derivatives u_t and u_xx."},{"Start":"01:06.665 ","End":"01:09.769","Text":"The derivative of this product are fairly straightforward."},{"Start":"01:09.769 ","End":"01:11.330","Text":"If we differentiate with respect to x,"},{"Start":"01:11.330 ","End":"01:15.380","Text":"we just differentiate X and the derivative with"},{"Start":"01:15.380 ","End":"01:19.840","Text":"respect to t is just an ordinary derivative of t. We"},{"Start":"01:19.840 ","End":"01:24.650","Text":"get u_t is a derivative here and this the"},{"Start":"01:24.650 ","End":"01:32.365","Text":"same and u_xx means the t part is the same and this we differentiate twice."},{"Start":"01:32.365 ","End":"01:36.874","Text":"Now we substitute this in the PDE, which is this,"},{"Start":"01:36.874 ","End":"01:42.990","Text":"we get, this is u_t a^2, u_xx."},{"Start":"01:43.270 ","End":"01:47.945","Text":"Rearranging this, bringing stuff from 1 side to the other,"},{"Start":"01:47.945 ","End":"01:52.525","Text":"we can separate the variables and get this."},{"Start":"01:52.525 ","End":"01:59.660","Text":"We get X\" over x equals 1 over a^2 T\u0027 over t. Now this part is a function of x."},{"Start":"01:59.660 ","End":"02:04.294","Text":"This part is a function of t and when this happens,"},{"Start":"02:04.294 ","End":"02:08.030","Text":"must both be equal to the same constant,"},{"Start":"02:08.030 ","End":"02:10.475","Text":"and we\u0027ll call that minus Lambda."},{"Start":"02:10.475 ","End":"02:14.965","Text":"Now, let\u0027s forget about t for a moment and we\u0027ll return to it later."},{"Start":"02:14.965 ","End":"02:18.080","Text":"We get X\" over X is minus Lambda,"},{"Start":"02:18.080 ","End":"02:23.630","Text":"and then rearrange, we get X\" equals minus Lambda X."},{"Start":"02:23.630 ","End":"02:28.580","Text":"Then bring it over X\" plus Lambda X equals 0."},{"Start":"02:28.580 ","End":"02:32.630","Text":"This is a second-order ODE in X."},{"Start":"02:32.630 ","End":"02:35.330","Text":"We need some boundary conditions."},{"Start":"02:35.330 ","End":"02:39.590","Text":"We\u0027ll take them from boundary conditions for u."},{"Start":"02:39.590 ","End":"02:46.940","Text":"Given u(0,t) equals u(L,t) equals 0."},{"Start":"02:46.940 ","End":"02:50.195","Text":"Now we know what in general u (x, t) is."},{"Start":"02:50.195 ","End":"02:52.489","Text":"Let\u0027s see if we can scroll back."},{"Start":"02:52.489 ","End":"02:55.790","Text":"Yeah, this is what it\u0027s equal to,"},{"Start":"02:55.790 ","End":"02:58.160","Text":"which is just X (x), T (t)."},{"Start":"02:58.160 ","End":"03:01.880","Text":"Going back here, this is X(0),"},{"Start":"03:01.880 ","End":"03:08.960","Text":"T(t) equals X(L), T(t) equals 0."},{"Start":"03:08.960 ","End":"03:12.350","Text":"Here we can divide everything by T(t)."},{"Start":"03:12.350 ","End":"03:14.135","Text":"We have 2 things."},{"Start":"03:14.135 ","End":"03:21.140","Text":"We have the ordinary differential equation second-order with x."},{"Start":"03:21.140 ","End":"03:26.510","Text":"We can call these also either initial or boundary conditions."},{"Start":"03:26.510 ","End":"03:31.565","Text":"Together, these form a Sturm–Liouville problem,"},{"Start":"03:31.565 ","End":"03:33.805","Text":"which should look familiar."},{"Start":"03:33.805 ","End":"03:38.230","Text":"This problem will look a bit more familiar if we use the letter y."},{"Start":"03:38.230 ","End":"03:45.430","Text":"So y\" plus Lambda y equals 0 and y(0) equals y(l) equals 0."},{"Start":"03:45.430 ","End":"03:49.315","Text":"We know from Sturm–Liouville problems that"},{"Start":"03:49.315 ","End":"03:54.022","Text":"these are the eigenvalues where n goes from 1,"},{"Start":"03:54.022 ","End":"03:56.125","Text":"2, 3 to infinity."},{"Start":"03:56.125 ","End":"04:03.160","Text":"For each n, we have this eigenvalue and for each n we have a corresponding eigenfunction,"},{"Start":"04:03.160 ","End":"04:08.705","Text":"which is sine and Pi x over L. This becomes in our case,"},{"Start":"04:08.705 ","End":"04:13.730","Text":"the following eigenvalues and x_n the same as what I just erased,"},{"Start":"04:13.730 ","End":"04:15.920","Text":"same as the y_n,"},{"Start":"04:15.920 ","End":"04:19.204","Text":"n goes along the natural numbers."},{"Start":"04:19.204 ","End":"04:22.940","Text":"Now that we have the x_n let\u0027s return to the T_ns the"},{"Start":"04:22.940 ","End":"04:27.590","Text":"equation for T(t) was the following."},{"Start":"04:27.590 ","End":"04:31.400","Text":"This time we ignore the x part and just focus on the t"},{"Start":"04:31.400 ","End":"04:38.735","Text":"and get an equation like this for each n. The Lambda ns are the same for x as for t,"},{"Start":"04:38.735 ","End":"04:42.730","Text":"we get Lambda n equals Pi n over L^2."},{"Start":"04:42.730 ","End":"04:52.580","Text":"That gives us, with a bit of rearranging that T_n\u0027 equals T_n(t) times minus Lambda,"},{"Start":"04:52.580 ","End":"04:57.305","Text":"which you bring it over to this side and replace Lambda by nPi over L^2."},{"Start":"04:57.305 ","End":"05:02.675","Text":"This is of the type y\u0027 plus Ay equals 0."},{"Start":"05:02.675 ","End":"05:06.440","Text":"When our case a is nPi over L^2."},{"Start":"05:06.440 ","End":"05:13.520","Text":"We know that the solution is y equals ce^minus At,"},{"Start":"05:13.520 ","End":"05:19.010","Text":"where A is from here and C is just an arbitrary constant."},{"Start":"05:19.010 ","End":"05:22.520","Text":"But in our case we\u0027ll take C=1. It\u0027s the easiest."},{"Start":"05:22.520 ","End":"05:24.680","Text":"We\u0027re looking for linearly independent solutions."},{"Start":"05:24.680 ","End":"05:27.995","Text":"A multiple of a solution really is the same solution."},{"Start":"05:27.995 ","End":"05:29.960","Text":"We can leave it as c_n for now,"},{"Start":"05:29.960 ","End":"05:32.780","Text":"but afterwards we\u0027ll just take c_n=1."},{"Start":"05:32.780 ","End":"05:40.560","Text":"Summarizing so far, we got that X_n(x) is sin(Pinx over L)."},{"Start":"05:40.560 ","End":"05:51.005","Text":"We just found that T_n(t) is e^minus nPi A over L^2 times t. At this point we took C_n=1."},{"Start":"05:51.005 ","End":"05:53.360","Text":"We have X_n and we have T_n."},{"Start":"05:53.360 ","End":"05:57.960","Text":"If we multiply a pair of these, we get u_n."},{"Start":"05:58.460 ","End":"06:03.620","Text":"This times this is u_n and at the end we take"},{"Start":"06:03.620 ","End":"06:08.705","Text":"the sum of a_n u_n to be the most general solution."},{"Start":"06:08.705 ","End":"06:11.705","Text":"If we just take a pair this times this,"},{"Start":"06:11.705 ","End":"06:16.435","Text":"we could call this u_n(t) is a solution and because of the homogeneity,"},{"Start":"06:16.435 ","End":"06:18.935","Text":"a linear combination of the solution,"},{"Start":"06:18.935 ","End":"06:21.980","Text":"even infinite linear combination is also a solution."},{"Start":"06:21.980 ","End":"06:26.905","Text":"All that remains is to find the coefficients A_n."},{"Start":"06:26.905 ","End":"06:29.310","Text":"This will do in Stage 2,"},{"Start":"06:29.310 ","End":"06:36.260","Text":"but Stage 2 will do in the examples in the exercises and not here."},{"Start":"06:36.260 ","End":"06:41.330","Text":"To emphasize this general solution applies for"},{"Start":"06:41.330 ","End":"06:44.420","Text":"the Dirichlet boundary conditions"},{"Start":"06:44.420 ","End":"06:47.555","Text":"when both the left one on the right one or of the type Dirichlet,"},{"Start":"06:47.555 ","End":"06:50.270","Text":"not the most general."},{"Start":"06:50.270 ","End":"06:55.070","Text":"It\u0027s a general answer for Stage 1, for Dirichlet Dirichlet."},{"Start":"06:55.070 ","End":"06:57.230","Text":"I\u0027ll just mention again Stage 2,"},{"Start":"06:57.230 ","End":"06:59.750","Text":"we find the coefficients A_n from the initial condition,"},{"Start":"06:59.750 ","End":"07:01.955","Text":"which is u(x_naught) equals f(x)."},{"Start":"07:01.955 ","End":"07:03.200","Text":"But to see this,"},{"Start":"07:03.200 ","End":"07:08.975","Text":"there are solved exercises or examples following this clip,"},{"Start":"07:08.975 ","End":"07:11.225","Text":"and that\u0027s where we\u0027ll learn how to do that."},{"Start":"07:11.225 ","End":"07:13.980","Text":"Meanwhile here we\u0027re done."}],"ID":30787},{"Watched":false,"Name":"Exercise 1","Duration":"5m ","ChapterTopicVideoID":29228,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.195","Text":"In this exercise, we\u0027re given the following problem with the heat equation,"},{"Start":"00:06.195 ","End":"00:10.905","Text":"homogeneous and homogeneous boundary conditions."},{"Start":"00:10.905 ","End":"00:14.370","Text":"We can use the technique of separation of variables."},{"Start":"00:14.370 ","End":"00:18.090","Text":"In principle, we could start from the following formula"},{"Start":"00:18.090 ","End":"00:21.975","Text":"which we derived in the introduction clip,"},{"Start":"00:21.975 ","End":"00:26.385","Text":"which works whenever we have both Dirichlet conditions,"},{"Start":"00:26.385 ","End":"00:27.690","Text":"but we need the practice,"},{"Start":"00:27.690 ","End":"00:29.400","Text":"we\u0027ll start from scratch."},{"Start":"00:29.400 ","End":"00:31.440","Text":"These are the stages for solving,"},{"Start":"00:31.440 ","End":"00:34.305","Text":"just repeating them from the tutorial."},{"Start":"00:34.305 ","End":"00:36.690","Text":"Let\u0027s start with the first stage of course,"},{"Start":"00:36.690 ","End":"00:41.705","Text":"and we\u0027re going to find solutions which are separated variables,"},{"Start":"00:41.705 ","End":"00:47.615","Text":"some function of x times some function of t. Now we want it to fit the PDE."},{"Start":"00:47.615 ","End":"00:56.840","Text":"We need u_t and that\u0027s gotten by differentiating T. We also need second derivative u_xx."},{"Start":"00:56.840 ","End":"00:59.780","Text":"We just differentiate this twice,"},{"Start":"00:59.780 ","End":"01:05.690","Text":"substitute into the equation u_t=u_xx and we get the following."},{"Start":"01:05.690 ","End":"01:07.475","Text":"Rearranging it a bit."},{"Start":"01:07.475 ","End":"01:09.635","Text":"We get this over this,"},{"Start":"01:09.635 ","End":"01:11.910","Text":"equals this over this."},{"Start":"01:11.910 ","End":"01:16.250","Text":"Whenever a function of x is identically equal to a function of t,"},{"Start":"01:16.250 ","End":"01:18.140","Text":"has to be a constant function,"},{"Start":"01:18.140 ","End":"01:21.320","Text":"which for convenience will call minus Lambda."},{"Start":"01:21.320 ","End":"01:23.689","Text":"Now first we\u0027ll just concentrate on the X part,"},{"Start":"01:23.689 ","End":"01:30.130","Text":"later we\u0027ll return to T. We get X\u0027\u0027 equal minus Lambda X."},{"Start":"01:30.130 ","End":"01:33.725","Text":"Then rearranging, we get the following."},{"Start":"01:33.725 ","End":"01:35.780","Text":"Now we can substitute"},{"Start":"01:35.780 ","End":"01:42.260","Text":"the boundary conditions and this gives us the following. Won\u0027t read it out."},{"Start":"01:42.260 ","End":"01:45.500","Text":"We can cancel by T(t)."},{"Start":"01:45.500 ","End":"01:49.160","Text":"This is good unless T(t) is identically 0,"},{"Start":"01:49.160 ","End":"01:50.360","Text":"which is not interesting."},{"Start":"01:50.360 ","End":"01:54.560","Text":"We don\u0027t care about 0 solutions, so cancel that."},{"Start":"01:54.560 ","End":"01:59.615","Text":"Then this together with this,"},{"Start":"01:59.615 ","End":"02:03.635","Text":"gives us a Sturm–Liouville problem as follows."},{"Start":"02:03.635 ","End":"02:09.410","Text":"We know that the solution to this Sturm–Liouville problem is the following."},{"Start":"02:09.410 ","End":"02:11.735","Text":"We have it in terms of y."},{"Start":"02:11.735 ","End":"02:20.630","Text":"Here we have big X and we have Pi instead of the general L. We get the following,"},{"Start":"02:20.630 ","End":"02:23.540","Text":"that the eigenvalues are,"},{"Start":"02:23.540 ","End":"02:25.760","Text":"well, if L=Pi,"},{"Start":"02:25.760 ","End":"02:28.370","Text":"this comes out to n^2,"},{"Start":"02:28.370 ","End":"02:30.920","Text":"and here also if L=Pi,"},{"Start":"02:30.920 ","End":"02:33.305","Text":"we just get nx here."},{"Start":"02:33.305 ","End":"02:36.655","Text":"These are the eigenvalues and eigenfunctions."},{"Start":"02:36.655 ","End":"02:39.600","Text":"What we need now are the corresponding T_n\u0027s."},{"Start":"02:39.600 ","End":"02:41.495","Text":"The Lambda n are the same."},{"Start":"02:41.495 ","End":"02:45.035","Text":"The equation for big T is the following."},{"Start":"02:45.035 ","End":"02:47.855","Text":"In general, for each n,"},{"Start":"02:47.855 ","End":"02:53.884","Text":"we have a T_n or responding to the Lambda n. This is minus n^2."},{"Start":"02:53.884 ","End":"02:56.690","Text":"Rearranging that we have the following,"},{"Start":"02:56.690 ","End":"02:57.890","Text":"and this is similar to"},{"Start":"02:57.890 ","End":"03:03.810","Text":"the first-order differential equation, linear constant coefficients."},{"Start":"03:03.810 ","End":"03:05.720","Text":"We know how to solve this."},{"Start":"03:05.720 ","End":"03:12.410","Text":"The solution to this is that y is some constant times e to the minus at,"},{"Start":"03:12.410 ","End":"03:15.485","Text":"where a in our case is n^2."},{"Start":"03:15.485 ","End":"03:17.060","Text":"We don\u0027t need the constant."},{"Start":"03:17.060 ","End":"03:19.100","Text":"You could take it to be equal to 1"},{"Start":"03:19.100 ","End":"03:22.729","Text":"because we are only looking for linearly independent solutions."},{"Start":"03:22.729 ","End":"03:26.120","Text":"If we take a different constant is still the same solution really."},{"Start":"03:26.120 ","End":"03:28.220","Text":"That\u0027s T_n. To repeat,"},{"Start":"03:28.220 ","End":"03:31.090","Text":"X_n is sin (nx)."},{"Start":"03:31.090 ","End":"03:35.115","Text":"We let u_n be the product of the two."},{"Start":"03:35.115 ","End":"03:37.850","Text":"Each of these is actually a solution,"},{"Start":"03:37.850 ","End":"03:42.365","Text":"but the general solution is an infinite linear combination of these,"},{"Start":"03:42.365 ","End":"03:47.670","Text":"u, which will be the sum of some constant times this,"},{"Start":"03:47.670 ","End":"03:49.580","Text":"and that\u0027s stage one."},{"Start":"03:49.580 ","End":"03:52.745","Text":"Stage two, we need to find these A_n."},{"Start":"03:52.745 ","End":"03:56.765","Text":"To do that, we\u0027re going to use the initial condition, which is,"},{"Start":"03:56.765 ","End":"04:02.320","Text":"if you go back and look, it\u0027s u(x) naught equals sin(x)."},{"Start":"04:02.320 ","End":"04:07.410","Text":"We have that u(x) naught is the sum of A_n sin(nx),"},{"Start":"04:07.410 ","End":"04:09.810","Text":"because when t is equal to 0,"},{"Start":"04:09.810 ","End":"04:13.670","Text":"e^-0 is 1 and we just get that."},{"Start":"04:13.670 ","End":"04:16.925","Text":"Now we can compare coefficients of the sine series,"},{"Start":"04:16.925 ","End":"04:19.340","Text":"because this is equal to this."},{"Start":"04:19.340 ","End":"04:23.630","Text":"The only non-zero term here is when n=1,"},{"Start":"04:23.630 ","End":"04:25.910","Text":"and then we get that A_n is 1."},{"Start":"04:25.910 ","End":"04:29.525","Text":"We have A_1=1 and all the other A_n\u0027s are 0."},{"Start":"04:29.525 ","End":"04:32.800","Text":"Now back to this equation, we can substitute."},{"Start":"04:32.800 ","End":"04:35.930","Text":"We only have to take the term where n=1."},{"Start":"04:35.930 ","End":"04:37.390","Text":"Then we get that u(x,"},{"Start":"04:37.390 ","End":"04:41.090","Text":"t) is A_1 e to the minus 1^2 t sin(1x)."},{"Start":"04:41.090 ","End":"04:43.880","Text":"A_1 is 1,"},{"Start":"04:43.880 ","End":"04:46.190","Text":"so the solution is u(x,"},{"Start":"04:46.190 ","End":"04:49.235","Text":"t) is e to the minus t sin(x)."},{"Start":"04:49.235 ","End":"04:51.874","Text":"I recommend checking this solution,"},{"Start":"04:51.874 ","End":"04:56.315","Text":"checking that it satisfies the PDE and the initial condition."},{"Start":"04:56.315 ","End":"04:57.500","Text":"It\u0027s easy enough to do,"},{"Start":"04:57.500 ","End":"04:58.768","Text":"but I won\u0027t do it here."},{"Start":"04:58.768 ","End":"05:01.260","Text":"Anyway, that concludes this clip."}],"ID":30788},{"Watched":false,"Name":"Exercise 2","Duration":"5m 45s","ChapterTopicVideoID":29229,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.610","Text":"In this exercise we\u0027re going to solve the following initial boundary value problem."},{"Start":"00:05.610 ","End":"00:12.552","Text":"It\u0027s a heat equation homogeneous with homogeneous boundary conditions,"},{"Start":"00:12.552 ","End":"00:19.830","Text":"only this time we have a Von Neumann condition at each of the endpoints 0 and Pi."},{"Start":"00:19.830 ","End":"00:21.510","Text":"Since it\u0027s homogeneous,"},{"Start":"00:21.510 ","End":"00:23.610","Text":"both the equation and the boundary conditions,"},{"Start":"00:23.610 ","End":"00:27.315","Text":"we can use the separation of variables method."},{"Start":"00:27.315 ","End":"00:28.830","Text":"Here are the 2 stages."},{"Start":"00:28.830 ","End":"00:30.975","Text":"I won\u0027t read them out, they should be familiar."},{"Start":"00:30.975 ","End":"00:35.115","Text":"Start with Stage 1 where we look for a solution of the form"},{"Start":"00:35.115 ","End":"00:40.800","Text":"something of x times something of t. This is the separation of the variables."},{"Start":"00:40.800 ","End":"00:44.900","Text":"We differentiate twice with respect to x to get this."},{"Start":"00:44.900 ","End":"00:49.010","Text":"I also left the first derivative in because we\u0027ll need it later."},{"Start":"00:49.010 ","End":"00:51.470","Text":"We also need the derivative with respect to"},{"Start":"00:51.470 ","End":"00:55.220","Text":"t which has gotten by differentiating the T part."},{"Start":"00:55.220 ","End":"01:00.485","Text":"Now we can substitute in the PDE and get the following."},{"Start":"01:00.485 ","End":"01:03.185","Text":"Rearranging, we get this,"},{"Start":"01:03.185 ","End":"01:05.105","Text":"this is the function of just x."},{"Start":"01:05.105 ","End":"01:08.510","Text":"This is the function of just t. It"},{"Start":"01:08.510 ","End":"01:12.395","Text":"must each be a constant and we\u0027ll call that constant minus Lambda,"},{"Start":"01:12.395 ","End":"01:17.030","Text":"so rearranging we can get this equation."},{"Start":"01:17.030 ","End":"01:20.225","Text":"It\u0027s a second-order ODE."},{"Start":"01:20.225 ","End":"01:23.175","Text":"We want some boundary values."},{"Start":"01:23.175 ","End":"01:26.270","Text":"Let\u0027s start with these u_x(0,"},{"Start":"01:26.270 ","End":"01:29.060","Text":"t)= u_x(Pi, t)= 0,"},{"Start":"01:29.060 ","End":"01:32.440","Text":"and we have u_x here."},{"Start":"01:32.440 ","End":"01:39.020","Text":"We get the following double equation and the T part cancels."},{"Start":"01:39.020 ","End":"01:43.475","Text":"Now, if we look at this together with this,"},{"Start":"01:43.475 ","End":"01:52.235","Text":"we have an ODE with end point conditions which are a Sturm-Liouville problem as follows."},{"Start":"01:52.235 ","End":"01:58.420","Text":"This time, the initial conditions or boundary conditions are with x prime."},{"Start":"01:58.420 ","End":"02:01.925","Text":"It\u0027s a bit different than the previous exercise."},{"Start":"02:01.925 ","End":"02:06.260","Text":"We know from the study of Sturm-Liouville equation is that"},{"Start":"02:06.260 ","End":"02:11.270","Text":"the solution to this problem which is what this essentially is,"},{"Start":"02:11.270 ","End":"02:12.929","Text":"is the following,"},{"Start":"02:12.929 ","End":"02:17.915","Text":"these are the eigenvalues and these are the eigenfunctions,"},{"Start":"02:17.915 ","End":"02:22.280","Text":"we\u0027re just adapting to our case where L equals Pi,"},{"Start":"02:22.280 ","End":"02:26.535","Text":"we get the eigenvalues are n^2 because Pi/L is 1."},{"Start":"02:26.535 ","End":"02:33.330","Text":"So here Pi/L is 1 so the eigenfunctions are cosine (nx)."},{"Start":"02:33.330 ","End":"02:34.880","Text":"In the case of the cosine,"},{"Start":"02:34.880 ","End":"02:37.795","Text":"the n starts from 0."},{"Start":"02:37.795 ","End":"02:39.780","Text":"These are the x\u0027s,"},{"Start":"02:39.780 ","End":"02:42.305","Text":"now let\u0027s turn to the t\u0027s."},{"Start":"02:42.305 ","End":"02:51.375","Text":"Lambda n are the same and we have the equation T\u0027 (t)/T (t) is minus Lambda."},{"Start":"02:51.375 ","End":"02:56.945","Text":"For each n we have this equation and we know that the Lambda n,"},{"Start":"02:56.945 ","End":"03:00.960","Text":"they are equal to n^2 so here we have T_"},{"Start":"03:00.960 ","End":"03:06.810","Text":"n\u0027/T_n= minus n^2 which gives us the following ODE."},{"Start":"03:06.810 ","End":"03:09.395","Text":"This is of the following form."},{"Start":"03:09.395 ","End":"03:16.310","Text":"Simple first-order ODE linear with constant coefficient homogeneous."},{"Start":"03:16.310 ","End":"03:18.425","Text":"I know the solution is this."},{"Start":"03:18.425 ","End":"03:23.040","Text":"In our case, T_n (t) is e to the minus n^2t."},{"Start":"03:23.040 ","End":"03:25.050","Text":"We don\u0027t need this constant,"},{"Start":"03:25.050 ","End":"03:27.725","Text":"I mean, we needed to take it wherever we want."},{"Start":"03:27.725 ","End":"03:34.485","Text":"Say 1, anything non-0 and the x_n just recall from here are as follows."},{"Start":"03:34.485 ","End":"03:36.885","Text":"We have the T_n and we have the X_n,"},{"Start":"03:36.885 ","End":"03:41.025","Text":"so now we can let u_n be the product of these 2,"},{"Start":"03:41.025 ","End":"03:43.215","Text":"n= 0, 1, 2, etc."},{"Start":"03:43.215 ","End":"03:46.850","Text":"We want a linear combination of these or"},{"Start":"03:46.850 ","End":"03:50.690","Text":"an infinite linear combination to give us u(x, t)."},{"Start":"03:50.690 ","End":"03:58.850","Text":"It\u0027s the sum of A_n U_n but it\u0027s customary to write the constant not as A_0,"},{"Start":"03:58.850 ","End":"04:02.090","Text":"but A_0/2, it doesn\u0027t really matter how you write it."},{"Start":"04:02.090 ","End":"04:04.248","Text":"That concludes Stage 1."},{"Start":"04:04.248 ","End":"04:06.415","Text":"In Stage 2,"},{"Start":"04:06.415 ","End":"04:11.855","Text":"we\u0027re going to find the coefficients A_n using the initial condition."},{"Start":"04:11.855 ","End":"04:14.770","Text":"The initial condition is the following,"},{"Start":"04:14.770 ","End":"04:16.345","Text":"go back and look."},{"Start":"04:16.345 ","End":"04:23.170","Text":"On the other hand, u (x,0) can be gotten from here by letting t=0."},{"Start":"04:23.170 ","End":"04:27.140","Text":"When t is 0 this exponent is 0,"},{"Start":"04:27.140 ","End":"04:32.960","Text":"so e^0 is 1 so we\u0027re just left with the following."},{"Start":"04:32.960 ","End":"04:38.705","Text":"Now we have that this equals this and we can compare coefficients of a cosine series."},{"Start":"04:38.705 ","End":"04:41.630","Text":"But on the right we don\u0027t have a cosine series,"},{"Start":"04:41.630 ","End":"04:43.160","Text":"so we need to adapt it."},{"Start":"04:43.160 ","End":"04:45.200","Text":"We use trigonometric identities,"},{"Start":"04:45.200 ","End":"04:47.135","Text":"the following one in particular."},{"Start":"04:47.135 ","End":"04:48.860","Text":"If we do that,"},{"Start":"04:48.860 ","End":"04:53.405","Text":"then we get the following, simplifies to this."},{"Start":"04:53.405 ","End":"04:58.840","Text":"Now we can see that the constant Part A_0 over 2=2."},{"Start":"04:58.840 ","End":"05:00.990","Text":"From here when n is 2,"},{"Start":"05:00.990 ","End":"05:07.725","Text":"we\u0027re going to get that A_n is minus 3/2 and all the others will be 0."},{"Start":"05:07.725 ","End":"05:12.930","Text":"That\u0027s right, A_0 /2 is 2."},{"Start":"05:12.930 ","End":"05:15.575","Text":"A_2 is minus 3/2 and the other A_n\u0027s are 0,"},{"Start":"05:15.575 ","End":"05:17.510","Text":"return to this equation."},{"Start":"05:17.510 ","End":"05:20.045","Text":"Now we can substitute the A_n\u0027s."},{"Start":"05:20.045 ","End":"05:25.390","Text":"We get A_0/2 but we only have 1 term from this series,"},{"Start":"05:25.390 ","End":"05:29.150","Text":"that\u0027s the term where n=2 so we get the following."},{"Start":"05:29.150 ","End":"05:37.670","Text":"Now, A_0/2 is 2 and A_2 is 3/2,"},{"Start":"05:37.670 ","End":"05:39.845","Text":"2^2 is 4, so we get this,"},{"Start":"05:39.845 ","End":"05:46.050","Text":"and this is the answer for what u(xt) is and that concludes this exercise."}],"ID":30789},{"Watched":false,"Name":"Exercise 3","Duration":"8m 6s","ChapterTopicVideoID":29230,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.319","Text":"In this exercise, we\u0027re going to solve the following problem."},{"Start":"00:04.319 ","End":"00:08.085","Text":"This is a heat equation, homogeneous."},{"Start":"00:08.085 ","End":"00:10.335","Text":"This is the initial condition,"},{"Start":"00:10.335 ","End":"00:14.205","Text":"and these are homogeneous boundary conditions"},{"Start":"00:14.205 ","End":"00:19.125","Text":"of the Von Neumann type because of the derivative with respect to x."},{"Start":"00:19.125 ","End":"00:22.379","Text":"Here\u0027s a reminder of the 2 stages for solving."},{"Start":"00:22.379 ","End":"00:24.779","Text":"I won\u0027t read it out. They should be familiar."},{"Start":"00:24.779 ","End":"00:27.015","Text":"We\u0027ll start with stage 1."},{"Start":"00:27.015 ","End":"00:30.558","Text":"We\u0027re looking for a solution which has the variables separated,"},{"Start":"00:30.558 ","End":"00:34.605","Text":"some function of x times some function of t. We\u0027ll need"},{"Start":"00:34.605 ","End":"00:39.255","Text":"the second derivative with respect to x, which is this,"},{"Start":"00:39.255 ","End":"00:42.110","Text":"and also the first derivative we\u0027ll need later,"},{"Start":"00:42.110 ","End":"00:48.305","Text":"and the derivative with respect to t. Now we\u0027ll substitute it into the PDE,"},{"Start":"00:48.305 ","End":"00:49.575","Text":"which is this,"},{"Start":"00:49.575 ","End":"00:51.155","Text":"and we get the following."},{"Start":"00:51.155 ","End":"00:53.465","Text":"This is 16 times this."},{"Start":"00:53.465 ","End":"01:01.550","Text":"Here we can really separate the variables X double prime over X equals T prime over T,"},{"Start":"01:01.550 ","End":"01:05.383","Text":"and the 16 goes in the denominator on this side."},{"Start":"01:05.383 ","End":"01:07.587","Text":"Because these are equal,"},{"Start":"01:07.587 ","End":"01:10.175","Text":"function of x equals to function of t must be a constant."},{"Start":"01:10.175 ","End":"01:12.110","Text":"We\u0027ll call it minus Lambda."},{"Start":"01:12.110 ","End":"01:14.660","Text":"Just concentrating on the X first,"},{"Start":"01:14.660 ","End":"01:17.855","Text":"we get X double prime is minus Lambda X."},{"Start":"01:17.855 ","End":"01:19.310","Text":"Bring this to the other side,"},{"Start":"01:19.310 ","End":"01:20.825","Text":"and this is what we have."},{"Start":"01:20.825 ","End":"01:28.820","Text":"We have the boundary conditions for du by dx and we have du by dx here."},{"Start":"01:28.820 ","End":"01:31.400","Text":"Just write it again here."},{"Start":"01:31.400 ","End":"01:36.695","Text":"This is X prime times T, where x is 0."},{"Start":"01:36.695 ","End":"01:41.995","Text":"Similarly here, where x is 3 and the T of t cancels,"},{"Start":"01:41.995 ","End":"01:44.450","Text":"we get a Sturm-Liouville problem."},{"Start":"01:44.450 ","End":"01:49.505","Text":"This, together with this give us this."},{"Start":"01:49.505 ","End":"01:54.119","Text":"Now assuming you\u0027re familiar with Sturm-Liouville problems,"},{"Start":"01:54.119 ","End":"01:55.825","Text":"if we take this problem,"},{"Start":"01:55.825 ","End":"01:57.680","Text":"which is just like this,"},{"Start":"01:57.680 ","End":"02:03.185","Text":"except the y instead of X and L instead of 3,"},{"Start":"02:03.185 ","End":"02:08.070","Text":"then we know that the eigenvalues are like so,"},{"Start":"02:08.070 ","End":"02:12.120","Text":"and the eigenfunctions are these,"},{"Start":"02:12.120 ","End":"02:15.060","Text":"and n goes from 0."},{"Start":"02:15.060 ","End":"02:17.860","Text":"We substitute L equals 3,"},{"Start":"02:17.860 ","End":"02:22.745","Text":"we get the following eigenvalues and eigenfunctions."},{"Start":"02:22.745 ","End":"02:26.450","Text":"Now we turn to the functions T. Got the X\u0027s,"},{"Start":"02:26.450 ","End":"02:30.660","Text":"now we need the T. The Lambda n are the same,"},{"Start":"02:30.660 ","End":"02:35.235","Text":"so we now have that 1 over 16,"},{"Start":"02:35.235 ","End":"02:38.345","Text":"T prime over T is minus Lambda,"},{"Start":"02:38.345 ","End":"02:41.088","Text":"which is really many equations in 1."},{"Start":"02:41.088 ","End":"02:43.623","Text":"For each n, we have this,"},{"Start":"02:43.623 ","End":"02:49.880","Text":"and we know that Lambda n is nPi over 3 squared."},{"Start":"02:49.880 ","End":"02:55.370","Text":"Rearranging again, putting the 4^2 in here to make it 4."},{"Start":"02:55.370 ","End":"02:57.328","Text":"This is what we have,"},{"Start":"02:57.328 ","End":"02:59.075","Text":"and this problem,"},{"Start":"02:59.075 ","End":"03:03.927","Text":"first order linear homogeneous ODE with constant coefficients,"},{"Start":"03:03.927 ","End":"03:05.675","Text":"we know how to do this."},{"Start":"03:05.675 ","End":"03:08.660","Text":"The solution is as follows."},{"Start":"03:08.660 ","End":"03:11.630","Text":"In our case, we don\u0027t need the c. I mean,"},{"Start":"03:11.630 ","End":"03:14.570","Text":"we can take it to be anything non-zero like 1,"},{"Start":"03:14.570 ","End":"03:18.810","Text":"and we get that T_n of t is 1,"},{"Start":"03:18.810 ","End":"03:20.355","Text":"e to the minus,"},{"Start":"03:20.355 ","End":"03:29.415","Text":"and A is this times t is copying the X_n\u0027s from here."},{"Start":"03:29.415 ","End":"03:32.336","Text":"We now have the T_n and X_n."},{"Start":"03:32.336 ","End":"03:34.845","Text":"u_n is T_n times X_n."},{"Start":"03:34.845 ","End":"03:37.485","Text":"We have u_n is this,"},{"Start":"03:37.485 ","End":"03:41.090","Text":"and each of the u_n\u0027s is a solution."},{"Start":"03:41.090 ","End":"03:44.030","Text":"But the general solution is a linear combination"},{"Start":"03:44.030 ","End":"03:47.815","Text":"of these possibly infinite linear combination."},{"Start":"03:47.815 ","End":"03:49.925","Text":"We can write this,"},{"Start":"03:49.925 ","End":"03:52.880","Text":"although the sum not quite accurate."},{"Start":"03:52.880 ","End":"03:56.360","Text":"The A naught, we like to write as A naught over 2."},{"Start":"03:56.360 ","End":"03:59.030","Text":"You don\u0027t have to, but it\u0027s customary."},{"Start":"03:59.030 ","End":"04:03.005","Text":"You\u0027ll see why later when we find a cosine series."},{"Start":"04:03.005 ","End":"04:06.125","Text":"For convenience, A naught over 2 plus the sum."},{"Start":"04:06.125 ","End":"04:10.253","Text":"This time it\u0027s from 1 to infinity of a_n,"},{"Start":"04:10.253 ","End":"04:12.242","Text":"whatever is written here."},{"Start":"04:12.242 ","End":"04:14.340","Text":"We\u0027ll need u of x_0."},{"Start":"04:14.340 ","End":"04:18.800","Text":"If we plug in 0, t is 0,"},{"Start":"04:18.800 ","End":"04:21.130","Text":"this exponent is 0,"},{"Start":"04:21.130 ","End":"04:24.790","Text":"so this e^0 is 1,"},{"Start":"04:24.790 ","End":"04:28.085","Text":"drops out and we\u0027re left with just this."},{"Start":"04:28.085 ","End":"04:30.740","Text":"That was stage 1. Now we come to stage 2 where we\u0027re going to"},{"Start":"04:30.740 ","End":"04:33.665","Text":"find the A_ n from the initial condition."},{"Start":"04:33.665 ","End":"04:35.785","Text":"Initial condition was this."},{"Start":"04:35.785 ","End":"04:40.310","Text":"On the other hand, we just showed that u of x naught is the following series,"},{"Start":"04:40.310 ","End":"04:46.745","Text":"so we can compare coefficients for cosine series except that this isn\u0027t a cosine series,"},{"Start":"04:46.745 ","End":"04:50.600","Text":"so we\u0027ll have to expand x as a cosine series,"},{"Start":"04:50.600 ","End":"04:53.720","Text":"and this is the formula."},{"Start":"04:53.720 ","End":"04:55.445","Text":"If we have a function,"},{"Start":"04:55.445 ","End":"04:59.035","Text":"we can write it as a cosine series like so,"},{"Start":"04:59.035 ","End":"05:03.635","Text":"and the coefficients are given by this integral."},{"Start":"05:03.635 ","End":"05:06.205","Text":"If we didn\u0027t write A naught over 2,"},{"Start":"05:06.205 ","End":"05:09.735","Text":"this would be just 1 over L."},{"Start":"05:09.735 ","End":"05:14.085","Text":"Uniformity that\u0027s why we take A naught over 2 instead of just A naught here."},{"Start":"05:14.085 ","End":"05:18.462","Text":"In our case, f(x) equals x, L=3,"},{"Start":"05:18.462 ","End":"05:25.635","Text":"so x is the following series where the A_n are given by this formula,"},{"Start":"05:25.635 ","End":"05:28.965","Text":"and that\u0027s pretty much the same as this."},{"Start":"05:28.965 ","End":"05:31.875","Text":"So a_n, little a,"},{"Start":"05:31.875 ","End":"05:34.005","Text":"is the same as A_ n, big A,"},{"Start":"05:34.005 ","End":"05:39.890","Text":"and we can use this formula for A_n this integral."},{"Start":"05:39.890 ","End":"05:42.350","Text":"We\u0027ll use integration by parts."},{"Start":"05:42.350 ","End":"05:45.770","Text":"This is boring I\u0027ll leave you to check the details."},{"Start":"05:45.770 ","End":"05:47.170","Text":"We get this,"},{"Start":"05:47.170 ","End":"05:50.130","Text":"and then this part is 0,"},{"Start":"05:50.130 ","End":"05:51.870","Text":"it\u0027s 0 minus 0."},{"Start":"05:51.870 ","End":"05:57.180","Text":"When x=3, then we get sine of nPi, which is 0."},{"Start":"05:57.180 ","End":"05:59.820","Text":"When x=0, well, we have a 0 here."},{"Start":"05:59.820 ","End":"06:01.890","Text":"It\u0027s just the second part,"},{"Start":"06:01.890 ","End":"06:05.350","Text":"the integral of sine is minus cosine,"},{"Start":"06:05.350 ","End":"06:11.330","Text":"so the integral of minus sine is cosine and we need to divide by another nPi over 3,"},{"Start":"06:11.330 ","End":"06:14.980","Text":"so that makes this squared and the 2/3 is from here."},{"Start":"06:14.980 ","End":"06:17.970","Text":"This comes out to be, let\u0027s see,"},{"Start":"06:17.970 ","End":"06:23.600","Text":"when x=3, we have cosine nPi here."},{"Start":"06:23.600 ","End":"06:28.160","Text":"When x=0, we have cosine 0, which is 1."},{"Start":"06:28.160 ","End":"06:32.665","Text":"This 3^2 can come out here as 3^2."},{"Start":"06:32.665 ","End":"06:35.970","Text":"2/3 times 3^2 is 6."},{"Start":"06:35.970 ","End":"06:39.060","Text":"Cosine nPi is minus 1^n."},{"Start":"06:39.060 ","End":"06:42.719","Text":"nPi squared we can write it as n squared Pi squared."},{"Start":"06:42.719 ","End":"06:49.680","Text":"Now, cosine nPi is minus 1^n plus 1 when n is even,"},{"Start":"06:49.680 ","End":"06:51.215","Text":"so we only need the odd n\u0027s,"},{"Start":"06:51.215 ","End":"06:55.970","Text":"which gives us minus 1 minus 1 is minus 2 times 6 is minus"},{"Start":"06:55.970 ","End":"07:01.190","Text":"12 and odd numbers are covered by 2k minus 1,"},{"Start":"07:01.190 ","End":"07:05.220","Text":"where k goes from 1,2,3 to infinity,"},{"Start":"07:05.220 ","End":"07:06.905","Text":"so we can write it this way."},{"Start":"07:06.905 ","End":"07:13.445","Text":"I forgot to say this formula only works when n is not 0 because we can\u0027t divide by 0,"},{"Start":"07:13.445 ","End":"07:18.275","Text":"so we have to compute A_0 separately from the formula."},{"Start":"07:18.275 ","End":"07:20.540","Text":"If you go back and look at the formula when n=0,"},{"Start":"07:20.540 ","End":"07:22.210","Text":"this is what we get."},{"Start":"07:22.210 ","End":"07:26.355","Text":"It an easy computation we get A_0 naught is 3."},{"Start":"07:26.355 ","End":"07:31.525","Text":"Summarizing, A_0 is 3 and A_n is given by the following."},{"Start":"07:31.525 ","End":"07:38.135","Text":"Now we return to this formula which we had for u(x,t) in terms of the coefficients A_n,"},{"Start":"07:38.135 ","End":"07:43.510","Text":"we have to do is substitute now A_0 over 2 is 3 over 2,"},{"Start":"07:43.510 ","End":"07:50.705","Text":"and we replaced the indexing with n by indexing with k. We let n=2k minus 1,"},{"Start":"07:50.705 ","End":"07:54.155","Text":"just the odd numbers because the even ones are 0."},{"Start":"07:54.155 ","End":"07:57.330","Text":"Wherever we see n here,"},{"Start":"07:57.330 ","End":"08:00.750","Text":"we put 2k minus 1,"},{"Start":"08:00.750 ","End":"08:02.805","Text":"n is 2k minus 1,"},{"Start":"08:02.805 ","End":"08:05.270","Text":"and this is the answer."},{"Start":"08:05.270 ","End":"08:07.740","Text":"That concludes this exercise."}],"ID":30790},{"Watched":false,"Name":"Exercise 4","Duration":"6m 40s","ChapterTopicVideoID":29232,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.584","Text":"In this exercise, we\u0027re going to solve the following problem."},{"Start":"00:03.584 ","End":"00:09.120","Text":"We have a heat equation on the finite interval from 0 to 2."},{"Start":"00:09.120 ","End":"00:12.285","Text":"We have an initial condition at t=0,"},{"Start":"00:12.285 ","End":"00:17.385","Text":"which is this and we have 2 boundary conditions,"},{"Start":"00:17.385 ","End":"00:19.875","Text":"both of the Von Neumann type."},{"Start":"00:19.875 ","End":"00:23.340","Text":"Note that both the PDE and the boundary conditions"},{"Start":"00:23.340 ","End":"00:26.760","Text":"are homogeneous here because of the 0 here,"},{"Start":"00:26.760 ","End":"00:32.579","Text":"because there\u0027s no extra plus f(x) and t. Given that these are both homogeneous,"},{"Start":"00:32.579 ","End":"00:35.985","Text":"we can use the separation of variables technique."},{"Start":"00:35.985 ","End":"00:39.755","Text":"Remind you what they are not going to read it out and we\u0027ll start with stage 1"},{"Start":"00:39.755 ","End":"00:43.295","Text":"where we look for a solution of the type."},{"Start":"00:43.295 ","End":"00:45.860","Text":"Some function of x times some function of t,"},{"Start":"00:45.860 ","End":"00:48.715","Text":"that\u0027s where the variables are separated."},{"Start":"00:48.715 ","End":"00:51.530","Text":"We need to substitute in the PDE,"},{"Start":"00:51.530 ","End":"00:54.290","Text":"we need u_xx, we also have u_x."},{"Start":"00:54.290 ","End":"00:59.150","Text":"We just have to differentiate the X part the T(t) stays like a constant."},{"Start":"00:59.150 ","End":"01:04.120","Text":"We also need u_t, where we just differentiate the T part."},{"Start":"01:04.120 ","End":"01:09.295","Text":"Now, the PDE is u_t=9u_xx."},{"Start":"01:09.295 ","End":"01:16.450","Text":"Now we can substitute both this and this in here and this is what we get."},{"Start":"01:16.450 ","End":"01:19.805","Text":"Rearranging, we get the following,"},{"Start":"01:19.805 ","End":"01:27.350","Text":"where we\u0027ve actually separated the variables X\u0027\u0027 over X =9 over here,"},{"Start":"01:27.350 ","End":"01:28.940","Text":"the t over here switch, anyway,"},{"Start":"01:28.940 ","End":"01:32.990","Text":"we get 1 over 9t\u0027 over t. When you have a function"},{"Start":"01:32.990 ","End":"01:37.850","Text":"of x equal to function of t must be a constant."},{"Start":"01:37.850 ","End":"01:41.120","Text":"Next, let\u0027s just forget about the T for a moment."},{"Start":"01:41.120 ","End":"01:42.410","Text":"We\u0027ll return to it later."},{"Start":"01:42.410 ","End":"01:43.955","Text":"We\u0027ll just look at the xs."},{"Start":"01:43.955 ","End":"01:48.639","Text":"We get this, which rearranges to this,"},{"Start":"01:48.639 ","End":"01:52.875","Text":"a second-order ODE in x."},{"Start":"01:52.875 ","End":"01:58.550","Text":"Then we also have the boundary conditions and we know what u_x is."},{"Start":"01:58.550 ","End":"02:01.390","Text":"It\u0027s still here."},{"Start":"02:01.390 ","End":"02:07.260","Text":"This gives us the following but T(t) cancels"},{"Start":"02:07.260 ","End":"02:12.830","Text":"and if we take this ODE together with this condition,"},{"Start":"02:12.830 ","End":"02:16.223","Text":"we get a Sturm-Liouville problem as follows"},{"Start":"02:16.223 ","End":"02:20.270","Text":"on the assumption that you\u0027re familiar with Sturm-Liouville problems."},{"Start":"02:20.270 ","End":"02:25.445","Text":"This one is of this type and we know the solution for this problem."},{"Start":"02:25.445 ","End":"02:28.625","Text":"Solution is given by eigenvalues and eigenfunctions."},{"Start":"02:28.625 ","End":"02:30.560","Text":"These are the eigenvalues."},{"Start":"02:30.560 ","End":"02:34.536","Text":"There\u0027s a whole sequence of them and goes from 0,"},{"Start":"02:34.536 ","End":"02:36.395","Text":"1, 2, 3 up to infinity."},{"Start":"02:36.395 ","End":"02:41.390","Text":"The eigenfunctions are these and our cases just like this,"},{"Start":"02:41.390 ","End":"02:48.270","Text":"but with X instead of Y and we have 2 instead of L. In our case,"},{"Start":"02:48.270 ","End":"02:54.960","Text":"these are the eigenvalues and X_n is cosine nPi over 2x."},{"Start":"02:54.960 ","End":"02:56.580","Text":"Now that we have the X_n,"},{"Start":"02:56.580 ","End":"02:59.180","Text":"let\u0027s go and look for the corresponding T_n."},{"Start":"02:59.180 ","End":"03:02.180","Text":"For each n, there\u0027ll be an X_n and a corresponding T_n,"},{"Start":"03:02.180 ","End":"03:07.345","Text":"and they\u0027re connected by means of the Lambda_n, the same eigenvalue."},{"Start":"03:07.345 ","End":"03:10.354","Text":"The general form was this."},{"Start":"03:10.354 ","End":"03:12.935","Text":"This time we\u0027re ignoring the X."},{"Start":"03:12.935 ","End":"03:16.745","Text":"For each n, we have 1 over 9,"},{"Start":"03:16.745 ","End":"03:20.750","Text":"t\u0027 over t=minus Lambda_n,"},{"Start":"03:20.750 ","End":"03:23.750","Text":"which is minus nPi over 2^2."},{"Start":"03:23.750 ","End":"03:29.530","Text":"We can bring this 3^2 over to the other side and put it inside as a 3,"},{"Start":"03:29.530 ","End":"03:32.195","Text":"like so and then switch sides."},{"Start":"03:32.195 ","End":"03:36.845","Text":"This is what we have for T. It\u0027s also an ordinary differential equation,"},{"Start":"03:36.845 ","End":"03:39.215","Text":"but of the first order."},{"Start":"03:39.215 ","End":"03:45.895","Text":"This type of equation is like y\u0027 plus constant times y=0."},{"Start":"03:45.895 ","End":"03:53.180","Text":"Familiar ODE solution is that y equals some constant e^minus At,"},{"Start":"03:53.180 ","End":"03:55.670","Text":"where A is our constant here."},{"Start":"03:55.670 ","End":"03:58.505","Text":"Now we can take any non-zero constant,"},{"Start":"03:58.505 ","End":"04:00.070","Text":"take it to be 1."},{"Start":"04:00.070 ","End":"04:07.310","Text":"We have T_n is e^minus At and A is this thing squared."},{"Start":"04:07.310 ","End":"04:10.250","Text":"This is our solution for T_n and this is also for"},{"Start":"04:10.250 ","End":"04:15.465","Text":"each n. As the T_n we had X_n here as we write that."},{"Start":"04:15.465 ","End":"04:17.400","Text":"Now that we have T_n and X_n,"},{"Start":"04:17.400 ","End":"04:18.810","Text":"we can write u_n,"},{"Start":"04:18.810 ","End":"04:23.475","Text":"which is the product of T_n times X_n."},{"Start":"04:23.475 ","End":"04:28.370","Text":"We have the u_n\u0027s and the general solution is going to"},{"Start":"04:28.370 ","End":"04:33.110","Text":"be a linear combination of these or an infinite linear combination."},{"Start":"04:33.110 ","End":"04:36.230","Text":"We get the sum of A_n, u_n."},{"Start":"04:36.230 ","End":"04:41.855","Text":"But customary to write the constant term not as a 0, but as a 0 over 2."},{"Start":"04:41.855 ","End":"04:45.185","Text":"This is the end of stage 1."},{"Start":"04:45.185 ","End":"04:49.520","Text":"All we need now is to find the constants A_0,"},{"Start":"04:49.520 ","End":"04:52.775","Text":"A_n, and we\u0027ll do that in stage 2."},{"Start":"04:52.775 ","End":"04:56.839","Text":"We\u0027re going to find the A_n from the initial condition,"},{"Start":"04:56.839 ","End":"04:59.690","Text":"which I\u0027ll remind you, is as follows."},{"Start":"04:59.690 ","End":"05:05.905","Text":"Now we can also get u(x,0) by substituting t=0 here."},{"Start":"05:05.905 ","End":"05:09.750","Text":"We\u0027ll put this t=0."},{"Start":"05:09.750 ","End":"05:17.740","Text":"Then the whole exponent is 0 and e^0 is,1 so we get this expression for u(x,0)."},{"Start":"05:17.740 ","End":"05:20.990","Text":"Now we have 2 expressions for u(x,0) and what we\u0027re going to do"},{"Start":"05:20.990 ","End":"05:24.260","Text":"is compare coefficients for the cosine series."},{"Start":"05:24.260 ","End":"05:27.229","Text":"Because cosine series are unique."},{"Start":"05:27.229 ","End":"05:31.325","Text":"Writing these side-by-side, we can see that"},{"Start":"05:31.325 ","End":"05:35.870","Text":"the only thing we need to compare is we have A_0 over 2,"},{"Start":"05:35.870 ","End":"05:37.295","Text":"we compare it with 6."},{"Start":"05:37.295 ","End":"05:40.260","Text":"Also here, only when n is 3,"},{"Start":"05:40.260 ","End":"05:43.370","Text":"do we get something from the right so n is 0 and n is 3 are"},{"Start":"05:43.370 ","End":"05:47.240","Text":"the only ones we get and all the others will be 0. Let\u0027s write that."},{"Start":"05:47.240 ","End":"05:50.485","Text":"A_0 over 2 is 6,"},{"Start":"05:50.485 ","End":"05:53.520","Text":"A_3 will be 4,"},{"Start":"05:53.520 ","End":"05:56.130","Text":"and all the other A_n\u0027s will be 0."},{"Start":"05:56.130 ","End":"05:58.740","Text":"In other words, for n not equal to 0 or 3."},{"Start":"05:58.740 ","End":"06:01.010","Text":"Now recalling this expansion,"},{"Start":"06:01.010 ","End":"06:07.085","Text":"all we have to do is to substitute n=3 here,"},{"Start":"06:07.085 ","End":"06:10.084","Text":"here, and here."},{"Start":"06:10.084 ","End":"06:13.630","Text":"The A_naught term is already taken care of here."},{"Start":"06:13.630 ","End":"06:16.310","Text":"We just have to take care of the 0 and the 3,"},{"Start":"06:16.310 ","End":"06:18.170","Text":"the rest are 0."},{"Start":"06:18.170 ","End":"06:21.880","Text":"This comes out to be here, A_3,"},{"Start":"06:21.880 ","End":"06:26.295","Text":"here cosine 3Pi over 2 here, 3 times 3,"},{"Start":"06:26.295 ","End":"06:29.040","Text":"comes out to be 9Pi over 2^2,"},{"Start":"06:29.040 ","End":"06:35.010","Text":"or 81Pi^2 over 4 minus t. A_naught over 2 is 6,"},{"Start":"06:35.010 ","End":"06:37.880","Text":"and this is the answer."},{"Start":"06:37.880 ","End":"06:40.680","Text":"That concludes this exercise."}],"ID":30791},{"Watched":false,"Name":"Exercise 5","Duration":"8m 40s","ChapterTopicVideoID":29219,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.670","Text":"In this exercise, we have an initial boundary value problem consisting of PDE,"},{"Start":"00:05.670 ","End":"00:07.305","Text":"which is the heat equation."},{"Start":"00:07.305 ","End":"00:09.540","Text":"On the finite interval,"},{"Start":"00:09.540 ","End":"00:14.025","Text":"we have an initial condition of what happens at t=0,"},{"Start":"00:14.025 ","End":"00:17.190","Text":"and we have 2 boundary conditions,"},{"Start":"00:17.190 ","End":"00:20.250","Text":"what happens at the endpoints through time."},{"Start":"00:20.250 ","End":"00:24.050","Text":"These are both homogeneous and this is also"},{"Start":"00:24.050 ","End":"00:28.645","Text":"considered homogeneous because it doesn\u0027t have the extra plus f(x, t)."},{"Start":"00:28.645 ","End":"00:32.000","Text":"In the case of the homogeneous here and here,"},{"Start":"00:32.000 ","End":"00:35.570","Text":"we can use the technique of separation of variables,"},{"Start":"00:35.570 ","End":"00:38.885","Text":"just written it down here for reference, I won\u0027t read it out."},{"Start":"00:38.885 ","End":"00:41.630","Text":"Let\u0027s get started straightaway with stage 1,"},{"Start":"00:41.630 ","End":"00:46.010","Text":"which is to find a solution which has variables separated,"},{"Start":"00:46.010 ","End":"00:47.360","Text":"meaning in this form,"},{"Start":"00:47.360 ","End":"00:49.430","Text":"calculate sum of the partial derivative."},{"Start":"00:49.430 ","End":"00:51.860","Text":"This is routine. We need u_xx and we need"},{"Start":"00:51.860 ","End":"00:56.335","Text":"u_t because we\u0027re going to substitute in the PDE."},{"Start":"00:56.335 ","End":"00:59.075","Text":"When we put this and this in here,"},{"Start":"00:59.075 ","End":"01:00.930","Text":"this is what we get."},{"Start":"01:00.930 ","End":"01:04.820","Text":"Then we rearrange a bit and we get a function of"},{"Start":"01:04.820 ","End":"01:08.780","Text":"X equals a function of T. This is separated."},{"Start":"01:08.780 ","End":"01:12.725","Text":"When this happens, they each have to equal a constant,"},{"Start":"01:12.725 ","End":"01:15.935","Text":"the same constant, which we\u0027ll call minus Lambda."},{"Start":"01:15.935 ","End":"01:17.990","Text":"Let\u0027s just leave out the T for now,"},{"Start":"01:17.990 ","End":"01:19.520","Text":"we\u0027ll return to t later."},{"Start":"01:19.520 ","End":"01:21.155","Text":"Just looking at the X,"},{"Start":"01:21.155 ","End":"01:24.770","Text":"rearrange to get this and then put this on the other side,"},{"Start":"01:24.770 ","End":"01:27.110","Text":"and we wanted to have this with a plus,"},{"Start":"01:27.110 ","End":"01:29.330","Text":"which is why we chose minus Lambda here."},{"Start":"01:29.330 ","End":"01:37.915","Text":"We have also the boundary conditions for u and this is u_x copied from up here."},{"Start":"01:37.915 ","End":"01:40.260","Text":"Substituting in this formula,"},{"Start":"01:40.260 ","End":"01:42.870","Text":"here we have X\u0027(0,"},{"Start":"01:42.870 ","End":"01:48.060","Text":"t(t)) and here X\u0027(2t(t)), t(t) cancels,"},{"Start":"01:48.060 ","End":"01:49.770","Text":"so we get X\u0027(0) because X\u0027(2)=0,"},{"Start":"01:49.770 ","End":"01:54.800","Text":"and then this equation or double equation"},{"Start":"01:54.800 ","End":"02:00.470","Text":"together with this will give us a Sturm Liouville problem as follows."},{"Start":"02:00.470 ","End":"02:04.700","Text":"Now, it\u0027s assumed you know how to solve Sturm Liouville problems, basic ones."},{"Start":"02:04.700 ","End":"02:07.670","Text":"This one is of this type,"},{"Start":"02:07.670 ","End":"02:11.840","Text":"just that we have y here and have a general l, that of the 2,"},{"Start":"02:11.840 ","End":"02:14.405","Text":"and we know that the solution to this problem,"},{"Start":"02:14.405 ","End":"02:19.594","Text":"solution in the sense of eigenvalues and eigenfunctions is the following."},{"Start":"02:19.594 ","End":"02:21.900","Text":"For each n from 0,"},{"Start":"02:21.900 ","End":"02:23.280","Text":"1, 2, 3, etc."},{"Start":"02:23.280 ","End":"02:27.600","Text":"We have the eigenvalues Lambda nPi"},{"Start":"02:27.600 ","End":"02:33.035","Text":"over L^2 and the eigenfunctions y_n are cosine nPi over L x,"},{"Start":"02:33.035 ","End":"02:38.435","Text":"which in our case putting L=2 and replacing y by big X."},{"Start":"02:38.435 ","End":"02:43.595","Text":"These are the eigenvalues and these are the eigenfunctions X_n."},{"Start":"02:43.595 ","End":"02:44.900","Text":"Now that we\u0027ve found the X_n,"},{"Start":"02:44.900 ","End":"02:46.925","Text":"we want to go back and find the T_n."},{"Start":"02:46.925 ","End":"02:50.345","Text":"They have the same lambda n, X_n and T_n,"},{"Start":"02:50.345 ","End":"02:53.340","Text":"so we\u0027ll return to this equation,"},{"Start":"02:53.340 ","End":"02:55.650","Text":"this time we concentrated on the T part,"},{"Start":"02:55.650 ","End":"02:58.815","Text":"we can write 9 as 3^2."},{"Start":"02:58.815 ","End":"03:06.240","Text":"This is an equation for each n. We know that Lambda n is minus nPi over 2^2."},{"Start":"03:06.240 ","End":"03:12.185","Text":"We can put the 3 in here and bring this to the other side and we get this equation."},{"Start":"03:12.185 ","End":"03:17.419","Text":"This is the first order ODE of the following type."},{"Start":"03:17.419 ","End":"03:19.895","Text":"We know that the solution is this,"},{"Start":"03:19.895 ","End":"03:21.864","Text":"so in our case,"},{"Start":"03:21.864 ","End":"03:24.525","Text":"it comes to be this."},{"Start":"03:24.525 ","End":"03:27.265","Text":"We can take any c we want, except 0."},{"Start":"03:27.265 ","End":"03:29.780","Text":"We\u0027ll take c=1 to make it easiest."},{"Start":"03:29.780 ","End":"03:32.360","Text":"Because these T_n are linearly independent."},{"Start":"03:32.360 ","End":"03:35.105","Text":"If we take one that\u0027s just a constant times that,"},{"Start":"03:35.105 ","End":"03:36.865","Text":"it\u0027s the same one really."},{"Start":"03:36.865 ","End":"03:39.060","Text":"Just to repeat what X_n is,"},{"Start":"03:39.060 ","End":"03:42.810","Text":"so we now have T_n and X_n."},{"Start":"03:42.810 ","End":"03:46.755","Text":"We\u0027re going to let u_n be the product of T_n times X_n."},{"Start":"03:46.755 ","End":"03:48.750","Text":"U_n equals the following,"},{"Start":"03:48.750 ","End":"03:52.415","Text":"and this is actually a solution for each n,"},{"Start":"03:52.415 ","End":"03:58.790","Text":"but the general solution is an infinite linear combination of these."},{"Start":"03:58.790 ","End":"04:01.240","Text":"What we have is in general, u(x,"},{"Start":"04:01.240 ","End":"04:06.815","Text":"t) is the sum of A_n times u_n,"},{"Start":"04:06.815 ","End":"04:08.210","Text":"which is like this,"},{"Start":"04:08.210 ","End":"04:10.430","Text":"except that in the case of n=0,"},{"Start":"04:10.430 ","End":"04:12.005","Text":"we don\u0027t write A_0."},{"Start":"04:12.005 ","End":"04:14.540","Text":"It\u0027s customary to write A_0 over 2,"},{"Start":"04:14.540 ","End":"04:16.310","Text":"but you could just write A_0."},{"Start":"04:16.310 ","End":"04:18.800","Text":"It would work out right in the end."},{"Start":"04:18.800 ","End":"04:20.550","Text":"That was stage 1."},{"Start":"04:20.550 ","End":"04:25.280","Text":"Now we come to stage 2 where we have to find the actual values of"},{"Start":"04:25.280 ","End":"04:30.964","Text":"the coefficients A_n and we use the initial condition, which is this."},{"Start":"04:30.964 ","End":"04:33.230","Text":"You know what? Let me go back a moment."},{"Start":"04:33.230 ","End":"04:35.165","Text":"I forgot to say,"},{"Start":"04:35.165 ","End":"04:38.530","Text":"we can substitute t=0 here,"},{"Start":"04:38.530 ","End":"04:40.530","Text":"then we get the following,"},{"Start":"04:40.530 ","End":"04:43.365","Text":"get A_0 over 2 plus this sum,"},{"Start":"04:43.365 ","End":"04:46.085","Text":"all that changes is if t is 0,"},{"Start":"04:46.085 ","End":"04:49.985","Text":"this whole exponent is 0 and e(0) is 1,"},{"Start":"04:49.985 ","End":"04:54.800","Text":"so this bit drops out and we just have A_n times the cosine."},{"Start":"04:54.800 ","End":"04:57.350","Text":"Back here. We have u(x_0) is this,"},{"Start":"04:57.350 ","End":"04:59.915","Text":"on the other hand, u(x_0) is x."},{"Start":"04:59.915 ","End":"05:03.725","Text":"That means that this is"},{"Start":"05:03.725 ","End":"05:10.545","Text":"the cosine series expansion of the function x on the interval 0, 2."},{"Start":"05:10.545 ","End":"05:13.395","Text":"Now, we know about cosine series,"},{"Start":"05:13.395 ","End":"05:14.960","Text":"but I\u0027ll remind you,"},{"Start":"05:14.960 ","End":"05:18.365","Text":"if we have a function f(x) on the interval 0, L,"},{"Start":"05:18.365 ","End":"05:22.895","Text":"it\u0027s cosine series is of the following form,"},{"Start":"05:22.895 ","End":"05:28.990","Text":"and the coefficients A_n given by this integral."},{"Start":"05:28.990 ","End":"05:32.370","Text":"Now, in our case, f(x) is x,"},{"Start":"05:32.370 ","End":"05:34.680","Text":"L is 2,"},{"Start":"05:34.680 ","End":"05:39.440","Text":"and a_n is the same as A_n."},{"Start":"05:39.440 ","End":"05:43.850","Text":"I mean, this expression is the same as this, just little a,"},{"Start":"05:43.850 ","End":"05:51.085","Text":"big A and 2 instead of L. We can compute A_n using this same integral."},{"Start":"05:51.085 ","End":"05:55.790","Text":"L is 2, so put 2 here and here and here,"},{"Start":"05:55.790 ","End":"05:58.895","Text":"and now we have this integral to compute."},{"Start":"05:58.895 ","End":"06:01.965","Text":"We\u0027ll do this integral by parts,"},{"Start":"06:01.965 ","End":"06:04.625","Text":"and this is what it comes out to be."},{"Start":"06:04.625 ","End":"06:07.535","Text":"I\u0027ll spare you the boring details."},{"Start":"06:07.535 ","End":"06:09.080","Text":"Check this afterwards."},{"Start":"06:09.080 ","End":"06:12.170","Text":"Just note that this only works if n"},{"Start":"06:12.170 ","End":"06:15.260","Text":"is not 0 because we can\u0027t have a 0 in the denominator."},{"Start":"06:15.260 ","End":"06:18.145","Text":"We\u0027ll return later to do the case n=0."},{"Start":"06:18.145 ","End":"06:20.735","Text":"This is equal to 0."},{"Start":"06:20.735 ","End":"06:23.450","Text":"If you substitute 0, we certainly get 0."},{"Start":"06:23.450 ","End":"06:25.790","Text":"If you substitute 2 for x,"},{"Start":"06:25.790 ","End":"06:28.775","Text":"we get nPi, sine nPi is also 0."},{"Start":"06:28.775 ","End":"06:31.085","Text":"We just have to concentrate on this integral."},{"Start":"06:31.085 ","End":"06:33.800","Text":"The integral of minus sine is cosine,"},{"Start":"06:33.800 ","End":"06:37.820","Text":"and we need to divide again by nPi over 2,"},{"Start":"06:37.820 ","End":"06:40.060","Text":"so it\u0027s nPi over 2^2."},{"Start":"06:40.060 ","End":"06:42.225","Text":"Put the 2^2 over here,"},{"Start":"06:42.225 ","End":"06:44.850","Text":"leave nPi^2 and the denominator."},{"Start":"06:44.850 ","End":"06:46.200","Text":"If x is 2,"},{"Start":"06:46.200 ","End":"06:48.645","Text":"we get cosine nPi here."},{"Start":"06:48.645 ","End":"06:54.900","Text":"If x is 0, we get cosine of 0 is 1, but subtracted."},{"Start":"06:54.900 ","End":"06:57.630","Text":"We have cosine nPi minus 1 here,"},{"Start":"06:57.630 ","End":"07:02.995","Text":"2^2 is 4, cosine nPi is minus 1^n."},{"Start":"07:02.995 ","End":"07:05.570","Text":"Note that when n is even,"},{"Start":"07:05.570 ","End":"07:07.775","Text":"this gives us 1 minus 1 is 0,"},{"Start":"07:07.775 ","End":"07:14.320","Text":"so we only get something non-zero for odd n. In other words, n=2k minus 1."},{"Start":"07:14.320 ","End":"07:18.570","Text":"This is equal to minus 1^n is minus 1,"},{"Start":"07:18.570 ","End":"07:19.740","Text":"minus 1 is minus 2,"},{"Start":"07:19.740 ","End":"07:21.645","Text":"times 4 is minus 8."},{"Start":"07:21.645 ","End":"07:25.530","Text":"N is 2k minus 1, so this."},{"Start":"07:25.530 ","End":"07:27.250","Text":"For the even n,"},{"Start":"07:27.250 ","End":"07:32.955","Text":"we get 0 but we still haven\u0027t done the case where n=0."},{"Start":"07:32.955 ","End":"07:34.955","Text":"Let\u0027s take a look what the integral was."},{"Start":"07:34.955 ","End":"07:36.395","Text":"This is the integral."},{"Start":"07:36.395 ","End":"07:42.655","Text":"Substitute n=0 here, this becomes 1."},{"Start":"07:42.655 ","End":"07:45.180","Text":"This part is 1."},{"Start":"07:45.180 ","End":"07:48.240","Text":"We just get x dx."},{"Start":"07:48.240 ","End":"07:52.680","Text":"The integral of x dx is x^2 over 2 is between 0 and 2,"},{"Start":"07:52.680 ","End":"07:54.600","Text":"comes out to be 2."},{"Start":"07:54.600 ","End":"07:58.260","Text":"This is A_n and this is A_0."},{"Start":"07:58.260 ","End":"08:00.885","Text":"Let\u0027s just repeat that."},{"Start":"08:00.885 ","End":"08:02.760","Text":"That\u0027s what we have now."},{"Start":"08:02.760 ","End":"08:07.245","Text":"A_0 2 and A_n is the following."},{"Start":"08:07.245 ","End":"08:11.975","Text":"Now, we just have to substitute these in the series, which is this,"},{"Start":"08:11.975 ","End":"08:15.220","Text":"A_0 over 2 is 2 over 2 is 1,"},{"Start":"08:15.220 ","End":"08:19.170","Text":"and here we change the indexing."},{"Start":"08:19.170 ","End":"08:21.870","Text":"Instead of n, we only want the odd ones,"},{"Start":"08:21.870 ","End":"08:26.430","Text":"so we take 2k minus 1 where k goes from 1 to infinity."},{"Start":"08:26.430 ","End":"08:30.140","Text":"Then replace n here and here,"},{"Start":"08:30.140 ","End":"08:35.910","Text":"and here by 2k minus 1 and here and here."},{"Start":"08:35.910 ","End":"08:40.900","Text":"This is the answer and that concludes the exercise."}],"ID":30792},{"Watched":false,"Name":"Summary of Cases and Formulas","Duration":"4m 9s","ChapterTopicVideoID":29226,"CourseChapterTopicPlaylistID":294430,"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 clip, we\u0027ll learn about a shortcut for how to"},{"Start":"00:05.190 ","End":"00:10.425","Text":"solve an IBVP with the heat equation,"},{"Start":"00:10.425 ","End":"00:17.040","Text":"which is homogeneous and homogeneous boundary conditions by not repeating"},{"Start":"00:17.040 ","End":"00:20.160","Text":"the same work over and over but storing some results in"},{"Start":"00:20.160 ","End":"00:24.045","Text":"a table that will be this table of different cases."},{"Start":"00:24.045 ","End":"00:29.580","Text":"Basically, we will be able to skip stage 1 almost and go right into stage 2."},{"Start":"00:29.580 ","End":"00:32.715","Text":"Let me give some examples from past exercises."},{"Start":"00:32.715 ","End":"00:36.285","Text":"In 1 of the exercises at the end of stage 1,"},{"Start":"00:36.285 ","End":"00:42.890","Text":"we got the following general form of the solution and then in stage 2,"},{"Start":"00:42.890 ","End":"00:47.165","Text":"we use the initial condition to find the A_n."},{"Start":"00:47.165 ","End":"00:49.325","Text":"The initial conditions happened to be this."},{"Start":"00:49.325 ","End":"00:52.580","Text":"This problem had Dirichlet boundary conditions."},{"Start":"00:52.580 ","End":"00:55.550","Text":"Let\u0027s go back and see."},{"Start":"00:55.550 ","End":"00:58.250","Text":"This is a Dirichlet condition."},{"Start":"00:58.250 ","End":"01:00.125","Text":"This is a Dirichlet condition."},{"Start":"01:00.125 ","End":"01:01.520","Text":"Turns out that in general,"},{"Start":"01:01.520 ","End":"01:05.120","Text":"if we have this heat equation and if"},{"Start":"01:05.120 ","End":"01:10.205","Text":"the boundary conditions are Dirichlet then the form of the solution,"},{"Start":"01:10.205 ","End":"01:13.909","Text":"we can say right away is this."},{"Start":"01:13.909 ","End":"01:20.600","Text":"We have an infinite sum of coefficient times a sine of"},{"Start":"01:20.600 ","End":"01:24.230","Text":"something x and e to the power of minus"},{"Start":"01:24.230 ","End":"01:28.495","Text":"something squared t. There\u0027s an n in each of these."},{"Start":"01:28.495 ","End":"01:30.920","Text":"Looking again at the example,"},{"Start":"01:30.920 ","End":"01:33.095","Text":"we see that this is of that form,"},{"Start":"01:33.095 ","End":"01:37.345","Text":"except that I possibly wrote the sine before the exponent."},{"Start":"01:37.345 ","End":"01:43.400","Text":"In general, if we have a problem of Dirichlet-type boundary conditions,"},{"Start":"01:43.400 ","End":"01:47.885","Text":"then we can straight away jump to this and just substitute"},{"Start":"01:47.885 ","End":"01:53.170","Text":"L and a from the given and then we straight away get the general form,"},{"Start":"01:53.170 ","End":"01:58.160","Text":"and then all we have left is to find the A_n which is stage 2."},{"Start":"01:58.160 ","End":"02:00.560","Text":"As I recall, there was a clip after"},{"Start":"02:00.560 ","End":"02:03.335","Text":"the introductory clip on the technique of separation of"},{"Start":"02:03.335 ","End":"02:08.974","Text":"variables and we prove this formula for this case."},{"Start":"02:08.974 ","End":"02:15.185","Text":"Let\u0027s look at another example where we have both Von Neumann conditions."},{"Start":"02:15.185 ","End":"02:17.510","Text":"In this case, is something very similar,"},{"Start":"02:17.510 ","End":"02:20.150","Text":"the difference is that instead of a sine there\u0027s"},{"Start":"02:20.150 ","End":"02:24.080","Text":"a cosine and there\u0027s also a constant part."},{"Start":"02:24.080 ","End":"02:27.620","Text":"Let\u0027s look at 1 of the examples we had for this."},{"Start":"02:27.620 ","End":"02:31.460","Text":"In this exercise at the end of the stage 1,"},{"Start":"02:31.460 ","End":"02:36.750","Text":"we got u(x,t) equals all this."},{"Start":"02:36.970 ","End":"02:40.340","Text":"This is exactly of the form,"},{"Start":"02:40.340 ","End":"02:44.175","Text":"you check it, that we have here."},{"Start":"02:44.175 ","End":"02:47.310","Text":"Our problem, let\u0027s go back to it."},{"Start":"02:47.310 ","End":"02:57.415","Text":"If we go back, we\u0027ll see that this was 1 of those with Neumann conditions at both ends."},{"Start":"02:57.415 ","End":"02:59.060","Text":"In those two examples,"},{"Start":"02:59.060 ","End":"03:02.315","Text":"we could have saved a lot of work if we\u0027d had these tables,"},{"Start":"03:02.315 ","End":"03:05.465","Text":"then we can jump straight away to stage 2."},{"Start":"03:05.465 ","End":"03:06.890","Text":"Just take these formulas,"},{"Start":"03:06.890 ","End":"03:09.455","Text":"substitute L and a."},{"Start":"03:09.455 ","End":"03:14.300","Text":"We have the general form and then we use the initial condition for the problem,"},{"Start":"03:14.300 ","End":"03:16.170","Text":"this one in stage 2."},{"Start":"03:16.170 ","End":"03:20.330","Text":"For reference, I\u0027ve added 2 more entries to the table."},{"Start":"03:20.330 ","End":"03:21.885","Text":"They are mixed kind,"},{"Start":"03:21.885 ","End":"03:27.165","Text":"Dirichlet here and Neumann here and there\u0027s the other way round."},{"Start":"03:27.165 ","End":"03:30.760","Text":"Each of these also has a formula."},{"Start":"03:30.760 ","End":"03:32.920","Text":"They\u0027re pretty much the same."},{"Start":"03:32.920 ","End":"03:35.165","Text":"Here there\u0027s a sine here that the cosine."},{"Start":"03:35.165 ","End":"03:39.490","Text":"In these cases, n runs over the odd numbers."},{"Start":"03:39.490 ","End":"03:44.580","Text":"We have an index k and if n is 2k-1,"},{"Start":"03:44.580 ","End":"03:46.420","Text":"it\u0027s like 1,"},{"Start":"03:46.420 ","End":"03:47.850","Text":"3, 5, 7, etc."},{"Start":"03:47.850 ","End":"03:50.860","Text":"here. I don\u0027t know if we have an example of 1 of these,"},{"Start":"03:50.860 ","End":"03:55.894","Text":"but for completeness, I\u0027ve written all 4 of these formulas."},{"Start":"03:55.894 ","End":"03:59.530","Text":"Okay, after this as another couple of exercises and we\u0027ll use"},{"Start":"03:59.530 ","End":"04:05.590","Text":"this table to show you how much shorter it is because basically,"},{"Start":"04:05.590 ","End":"04:07.180","Text":"we skipped stage 1."},{"Start":"04:07.180 ","End":"04:10.430","Text":"Okay, that\u0027s it for this clip."}],"ID":30793},{"Watched":false,"Name":"Exercise 6","Duration":"4m 34s","ChapterTopicVideoID":29220,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.399","Text":"In this exercise, we have the following initial and boundary value problem."},{"Start":"00:05.399 ","End":"00:07.860","Text":"This is a Heat Equation,"},{"Start":"00:07.860 ","End":"00:11.565","Text":"it\u0027s homogeneous and on a finite interval,"},{"Start":"00:11.565 ","End":"00:18.360","Text":"the initial condition and the boundary conditions which are also homogeneous."},{"Start":"00:18.360 ","End":"00:24.090","Text":"This time we\u0027ll use the shortcut and we\u0027ll basically skip Stage 1."},{"Start":"00:24.090 ","End":"00:28.410","Text":"Here is the table of boundary conditions and the form of"},{"Start":"00:28.410 ","End":"00:33.320","Text":"the solution that we made in a previous clip."},{"Start":"00:33.320 ","End":"00:41.240","Text":"In our case, we have a Dirichlet condition at 0 and at L. This is the formula,"},{"Start":"00:41.240 ","End":"00:43.655","Text":"the first one of the 4."},{"Start":"00:43.655 ","End":"00:49.625","Text":"We bring this over to our page and here it is."},{"Start":"00:49.625 ","End":"00:53.090","Text":"Now we have some of the parameters."},{"Start":"00:53.090 ","End":"00:55.985","Text":"We have that L is Pi,"},{"Start":"00:55.985 ","End":"00:59.435","Text":"and we have that a is 1,"},{"Start":"00:59.435 ","End":"01:04.080","Text":"because this is Ut=a^2Uxx."},{"Start":"01:04.080 ","End":"01:07.460","Text":"If we substitute a=1, L=Pi here,"},{"Start":"01:07.460 ","End":"01:13.610","Text":"simplifies as Pi over L is 1 here and here,"},{"Start":"01:13.610 ","End":"01:14.900","Text":"and so a is 1,"},{"Start":"01:14.900 ","End":"01:19.610","Text":"all we\u0027re left with here is minus n^2 times t. Now we want"},{"Start":"01:19.610 ","End":"01:24.919","Text":"to get these constants by means of the initial condition, which is this."},{"Start":"01:24.919 ","End":"01:26.810","Text":"We want u (x) and 0."},{"Start":"01:26.810 ","End":"01:29.360","Text":"If you put t equals 0 here,"},{"Start":"01:29.360 ","End":"01:33.905","Text":"e to the minus n^2 t is e to the 0, which is 1."},{"Start":"01:33.905 ","End":"01:37.295","Text":"All we\u0027re left with is this."},{"Start":"01:37.295 ","End":"01:40.790","Text":"U(x, 0) is this series."},{"Start":"01:40.790 ","End":"01:44.945","Text":"But u(x,0) is also equal to x."},{"Start":"01:44.945 ","End":"01:47.405","Text":"This is equal to x,"},{"Start":"01:47.405 ","End":"01:51.055","Text":"and that means that we have the sine series for x."},{"Start":"01:51.055 ","End":"01:54.950","Text":"There\u0027s a formula for the coefficients in a sine series,"},{"Start":"01:54.950 ","End":"02:01.040","Text":"and the formula is that the coefficients b_n are given in"},{"Start":"02:01.040 ","End":"02:07.490","Text":"general as follows in terms of L and the function f(x)."},{"Start":"02:07.490 ","End":"02:11.200","Text":"But in our case, we know that f(x) is x and L is Pi."},{"Start":"02:11.200 ","End":"02:13.810","Text":"This simplifies to the following,"},{"Start":"02:13.810 ","End":"02:17.365","Text":"f(x) is x and L is Pi."},{"Start":"02:17.365 ","End":"02:21.930","Text":"Similarly here, well, Pi over L is 1, it\u0027s just n_x."},{"Start":"02:21.930 ","End":"02:26.810","Text":"Now we have a formula for the b_n and we can do this integral by parts."},{"Start":"02:26.810 ","End":"02:30.875","Text":"This is the product x times the sine."},{"Start":"02:30.875 ","End":"02:32.960","Text":"This will be,"},{"Start":"02:32.960 ","End":"02:34.430","Text":"what we call u,"},{"Start":"02:34.430 ","End":"02:36.590","Text":"and this is v\u0027."},{"Start":"02:36.590 ","End":"02:40.460","Text":"This is u times v. V is the integral of this,"},{"Start":"02:40.460 ","End":"02:43.295","Text":"which is minus cosine."},{"Start":"02:43.295 ","End":"02:47.780","Text":"Minus integral of u\u0027v,"},{"Start":"02:47.780 ","End":"02:49.730","Text":"so derivative of the x is 1,"},{"Start":"02:49.730 ","End":"02:54.485","Text":"and this from here as the integral from 0 to Pi."},{"Start":"02:54.485 ","End":"03:02.080","Text":"For convenience, we can make this minus into a plus and change the limits of integration."},{"Start":"03:02.080 ","End":"03:06.125","Text":"We have a 0 up here and the Pi down here."},{"Start":"03:06.125 ","End":"03:12.580","Text":"Here, we can also cancel minus with minus so that becomes a plus."},{"Start":"03:12.580 ","End":"03:15.495","Text":"Now we can do this integration."},{"Start":"03:15.495 ","End":"03:18.230","Text":"Integral of cosine is sine,"},{"Start":"03:18.230 ","End":"03:21.470","Text":"but we need to divide by another n, making it n^2."},{"Start":"03:21.470 ","End":"03:31.100","Text":"Now this is equal to 0 because sine of nPi is 0 and sine of n times 0 is also 0,"},{"Start":"03:31.100 ","End":"03:33.170","Text":"so we just need for here,"},{"Start":"03:33.170 ","End":"03:35.360","Text":"if we put x equals 0,"},{"Start":"03:35.360 ","End":"03:36.610","Text":"this thing becomes 0,"},{"Start":"03:36.610 ","End":"03:41.395","Text":"so we just have to subtract the case x equals Pi."},{"Start":"03:41.395 ","End":"03:43.399","Text":"Minus the subtraction."},{"Start":"03:43.399 ","End":"03:47.525","Text":"Then x is Pi and cosine nPi over n,"},{"Start":"03:47.525 ","End":"03:51.050","Text":"and cosine of nPi is minus 1 to"},{"Start":"03:51.050 ","End":"03:56.855","Text":"the n. It\u0027s just minus 2 because the Pi with the Pi cancels,"},{"Start":"03:56.855 ","End":"04:03.965","Text":"minus 1 to the n over n. This is what b_n is equal to."},{"Start":"04:03.965 ","End":"04:11.260","Text":"Now we just have to put the coefficients b_n into our series."},{"Start":"04:11.260 ","End":"04:16.575","Text":"This one, we have u(x) t in terms of b_n."},{"Start":"04:16.575 ","End":"04:21.905","Text":"This is a summary, we have b_n and we have u(x t), substitute this."},{"Start":"04:21.905 ","End":"04:25.850","Text":"The minus 2 comes out in front and instead of the b_n,"},{"Start":"04:25.850 ","End":"04:31.740","Text":"we have minus 1 to the n over n. This is the answer,"},{"Start":"04:31.740 ","End":"04:35.020","Text":"and that concludes this exercise."}],"ID":30794},{"Watched":false,"Name":"Exercise 7","Duration":"9m 39s","ChapterTopicVideoID":29222,"CourseChapterTopicPlaylistID":294430,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.825","Text":"In this exercise, we have an initial and boundary value problem."},{"Start":"00:04.825 ","End":"00:08.065","Text":"This is the PDE, which is the heat equation,"},{"Start":"00:08.065 ","End":"00:11.335","Text":"but it\u0027s not homogeneous because of this extra bit."},{"Start":"00:11.335 ","End":"00:14.260","Text":"The boundary conditions are homogeneous,"},{"Start":"00:14.260 ","End":"00:16.090","Text":"but this is not."},{"Start":"00:16.090 ","End":"00:19.975","Text":"So we can\u0027t use separation of variables technique."},{"Start":"00:19.975 ","End":"00:24.370","Text":"What we can do is find the correction function w,"},{"Start":"00:24.370 ","End":"00:26.935","Text":"which is just a function of x,"},{"Start":"00:26.935 ","End":"00:31.945","Text":"such that u can be written as v plus w,"},{"Start":"00:31.945 ","End":"00:40.480","Text":"and we will get a homogeneous heat equation in v. U=v plus w. If we need it,"},{"Start":"00:40.480 ","End":"00:45.660","Text":"v equals u minus w. Differentiate with respect to"},{"Start":"00:45.660 ","End":"00:51.820","Text":"x and we get ux=vx plus w prime."},{"Start":"00:51.820 ","End":"00:58.540","Text":"The second derivative, u x x is v x x plus w double prime of x."},{"Start":"00:58.540 ","End":"01:02.710","Text":"They also need u t. U t=V T"},{"Start":"01:02.710 ","End":"01:08.510","Text":"because this function of x has a 0 derivative with respect to t,"},{"Start":"01:08.510 ","End":"01:11.150","Text":"you have to satisfy the PDE."},{"Start":"01:11.150 ","End":"01:17.585","Text":"This, now we can replace u by v. U t is v t. From here,"},{"Start":"01:17.585 ","End":"01:27.280","Text":"Uxx is V_x_x plus W double-prime so we get this plus this term which is just copied."},{"Start":"01:27.410 ","End":"01:31.465","Text":"Now we\u0027re looking for a homogeneous equation in v,"},{"Start":"01:31.465 ","End":"01:34.570","Text":"like this, v t equals 4Vxx."},{"Start":"01:34.570 ","End":"01:39.370","Text":"If we can make this last bit 0, we\u0027ll have it."},{"Start":"01:39.370 ","End":"01:48.415","Text":"Let\u0027s try 4w double-prime of x plus 1/2 sine of 2x=0."},{"Start":"01:48.415 ","End":"01:53.995","Text":"Now let\u0027s bring this term to the right-hand side and then divide by 4."},{"Start":"01:53.995 ","End":"02:00.725","Text":"We get that w double-prime is minus an 1/8 sine 2x."},{"Start":"02:00.725 ","End":"02:03.310","Text":"Integrating this with respect to x,"},{"Start":"02:03.310 ","End":"02:09.310","Text":"we get this because the integral of minus sine is cosine."},{"Start":"02:09.310 ","End":"02:14.150","Text":"We\u0027re not adding any constants because we\u0027re just looking for a solution w,"},{"Start":"02:14.150 ","End":"02:15.995","Text":"we\u0027re not looking for a general solution."},{"Start":"02:15.995 ","End":"02:17.960","Text":"Any solution w will do."},{"Start":"02:17.960 ","End":"02:25.640","Text":"Integrating again, we get that w(x) is 1 over 32 sine of 2x."},{"Start":"02:25.640 ","End":"02:28.370","Text":"Now because we found a solution w,"},{"Start":"02:28.370 ","End":"02:30.425","Text":"if we take w to be this,"},{"Start":"02:30.425 ","End":"02:37.550","Text":"then we will have that the equation for v is a heat equation which is homogeneous."},{"Start":"02:37.550 ","End":"02:39.860","Text":"All this stuff vanishes."},{"Start":"02:39.860 ","End":"02:46.085","Text":"Continuing, u=v plus w so that v of x t"},{"Start":"02:46.085 ","End":"02:53.295","Text":"is u(x t) minus w( x) which is minus 1 over 32 sine 2x."},{"Start":"02:53.295 ","End":"03:00.995","Text":"We saw that v_t=4v_x_x if we define w the way we did."},{"Start":"03:00.995 ","End":"03:03.020","Text":"Now let\u0027s review and see where we are."},{"Start":"03:03.020 ","End":"03:09.160","Text":"The original problem was this the PDE domain,"},{"Start":"03:09.160 ","End":"03:12.260","Text":"initial condition and boundary conditions."},{"Start":"03:12.260 ","End":"03:19.396","Text":"Then we defined v(x t ) to be u(x t) minus 1 over 32 sine 2x,"},{"Start":"03:19.396 ","End":"03:24.995","Text":"and then v satisfied the homogeneous PDE,"},{"Start":"03:24.995 ","End":"03:29.060","Text":"like this, but homogeneous vt=4v_x_x."},{"Start":"03:29.060 ","End":"03:33.710","Text":"What we still need is the whole problem for v. In other words,"},{"Start":"03:33.710 ","End":"03:38.000","Text":"we need initial condition and boundary conditions also."},{"Start":"03:38.000 ","End":"03:39.980","Text":"First, the boundary conditions."},{"Start":"03:39.980 ","End":"03:42.963","Text":"Note that just repeating this from here,"},{"Start":"03:42.963 ","End":"03:46.265","Text":"if we let x=0,"},{"Start":"03:46.265 ","End":"03:49.835","Text":"then this term comes out to be 0."},{"Start":"03:49.835 ","End":"03:53.650","Text":"If we let x=Pi,"},{"Start":"03:53.650 ","End":"03:55.960","Text":"and also because sine 2Pi is 0."},{"Start":"03:55.960 ","End":"03:58.305","Text":"Also, this comes out to be 0."},{"Start":"03:58.305 ","End":"04:01.635","Text":"Then because u is 0,"},{"Start":"04:01.635 ","End":"04:06.319","Text":"you have t is 0 and you have Pi of t are 0,"},{"Start":"04:06.319 ","End":"04:13.835","Text":"then v of 0 t and v of Pi t will also be 0 because the difference is 0."},{"Start":"04:13.835 ","End":"04:18.975","Text":"Both these are 0,1 over 32 sine twice 0 is 0."},{"Start":"04:18.975 ","End":"04:21.170","Text":"Also if I put Pi here,"},{"Start":"04:21.170 ","End":"04:26.170","Text":"should have said 0 or Pi from this 0 and this Pi,"},{"Start":"04:26.170 ","End":"04:31.160","Text":"we have boundary conditions for v. What about initial conditions?"},{"Start":"04:31.160 ","End":"04:32.735","Text":"We need v of x naught."},{"Start":"04:32.735 ","End":"04:37.045","Text":"Well, it\u0027s=u(x ) naught minus w(x), which is this."},{"Start":"04:37.045 ","End":"04:42.105","Text":"We were given u(x) naught is equal to sine of x1 minus cosine of 3x,"},{"Start":"04:42.105 ","End":"04:44.180","Text":"and now minus this bit."},{"Start":"04:44.180 ","End":"04:46.505","Text":"So this is v(x) naught."},{"Start":"04:46.505 ","End":"04:47.990","Text":"Perhaps we could simplify it."},{"Start":"04:47.990 ","End":"04:49.745","Text":"I just leave it as it is."},{"Start":"04:49.745 ","End":"04:51.605","Text":"That\u0027s the initial condition."},{"Start":"04:51.605 ","End":"04:56.600","Text":"Now putting everything together we have, here\u0027s the PDE,"},{"Start":"04:56.600 ","End":"04:59.480","Text":"here\u0027s the initial condition that we just got,"},{"Start":"04:59.480 ","End":"05:02.870","Text":"and here are the boundary conditions from up here."},{"Start":"05:02.870 ","End":"05:08.110","Text":"Now we\u0027re going to use our table of formulas for the 4 cases."},{"Start":"05:08.110 ","End":"05:12.125","Text":"Just note that here we have homogeneous,"},{"Start":"05:12.125 ","End":"05:14.930","Text":"both the PDE and the boundary conditions,"},{"Start":"05:14.930 ","End":"05:18.520","Text":"and these are of the Dirichlet type."},{"Start":"05:18.520 ","End":"05:22.180","Text":"This is what our boundary conditions look like."},{"Start":"05:22.180 ","End":"05:26.335","Text":"That means the form of the solution is this."},{"Start":"05:26.335 ","End":"05:29.480","Text":"We know what a and LR,"},{"Start":"05:29.480 ","End":"05:32.450","Text":"we just have to find the coefficients an."},{"Start":"05:32.450 ","End":"05:37.990","Text":"Back here, a is 2 and L is Pi."},{"Start":"05:37.990 ","End":"05:40.235","Text":"That simplifies this formula,"},{"Start":"05:40.235 ","End":"05:42.395","Text":"leaves us with just this."},{"Start":"05:42.395 ","End":"05:49.615","Text":"Pi over L is 1 and here also and a is 2."},{"Start":"05:49.615 ","End":"05:56.630","Text":"That gives us 2n^2 for n^2 minus t. This is what we get."},{"Start":"05:56.630 ","End":"05:58.595","Text":"You just have to find the b_n."},{"Start":"05:58.595 ","End":"06:03.590","Text":"If we put t=0, we\u0027ll get v(x,0)."},{"Start":"06:03.590 ","End":"06:06.110","Text":"V(x,0) is also equal to this,"},{"Start":"06:06.110 ","End":"06:09.212","Text":"except without this e to the minus 4n^2t,"},{"Start":"06:09.212 ","End":"06:10.685","Text":"because when t=0,"},{"Start":"06:10.685 ","End":"06:12.950","Text":"this factor is 1."},{"Start":"06:12.950 ","End":"06:20.315","Text":"What we get is v(x,0) on the 1 hand is equal to this sum."},{"Start":"06:20.315 ","End":"06:22.685","Text":"That\u0027s this, without this factor."},{"Start":"06:22.685 ","End":"06:28.200","Text":"On the other hand, it\u0027s equal to copying from the right-hand side of this."},{"Start":"06:28.200 ","End":"06:31.880","Text":"What we\u0027re saying is that this series is"},{"Start":"06:31.880 ","End":"06:39.454","Text":"the sine series expansion of this function on the interval from 0 to Pi."},{"Start":"06:39.454 ","End":"06:42.230","Text":"Let\u0027s expand the brackets a bit."},{"Start":"06:42.230 ","End":"06:47.075","Text":"Here we can expand the sine x times this minus this."},{"Start":"06:47.075 ","End":"06:50.180","Text":"We can expand this part here,"},{"Start":"06:50.180 ","End":"06:59.160","Text":"the sine x cosine 3x using the formula for sine times cosine, which is this."},{"Start":"06:59.160 ","End":"07:04.430","Text":"What we get is the half well here and here."},{"Start":"07:04.430 ","End":"07:11.880","Text":"Sine of Alpha plus Beta is sine(x) plus 3x is sine 4x."},{"Start":"07:12.080 ","End":"07:15.770","Text":"I put it second and the minus goes with it."},{"Start":"07:15.770 ","End":"07:23.450","Text":"Then we want minus 1/2 sine of x minus 3x minus 1/2 sine minus 2x."},{"Start":"07:23.450 ","End":"07:26.720","Text":"We can make them both plus because sine is an odd function,"},{"Start":"07:26.720 ","End":"07:29.215","Text":"so it plus 1/2 sine 2x."},{"Start":"07:29.215 ","End":"07:31.040","Text":"There\u0027s the last term here."},{"Start":"07:31.040 ","End":"07:35.420","Text":"Now we can compare coefficients and get all the b_n."},{"Start":"07:35.420 ","End":"07:40.010","Text":"We see that B1 is 1 from the sine 1x."},{"Start":"07:40.010 ","End":"07:46.920","Text":"We also have that b4 would be minus 1/2."},{"Start":"07:46.920 ","End":"07:50.475","Text":"As for b2, both these belongs."},{"Start":"07:50.475 ","End":"07:53.715","Text":"B2 would be1/2 minus 1 over 32,"},{"Start":"07:53.715 ","End":"07:56.145","Text":"which is 15 over 32,"},{"Start":"07:56.145 ","End":"07:58.260","Text":"write those down, b1 is 1,"},{"Start":"07:58.260 ","End":"08:00.750","Text":"b2 is 15 over 32 b4 is minus 1/2,"},{"Start":"08:00.750 ","End":"08:03.145","Text":"and the other b_ns are 0."},{"Start":"08:03.145 ","End":"08:06.890","Text":"Those are where n is not equal to 1, 2 or 4."},{"Start":"08:06.890 ","End":"08:13.745","Text":"Now, recall this formula for v in terms of the coefficients b_n."},{"Start":"08:13.745 ","End":"08:15.905","Text":"We\u0027ve now found the b_n,"},{"Start":"08:15.905 ","End":"08:17.350","Text":"and this is what they are."},{"Start":"08:17.350 ","End":"08:21.275","Text":"All we have to do is substitute those here and we get the answer."},{"Start":"08:21.275 ","End":"08:26.080","Text":"We v(x, t ) is equal to when n=1,"},{"Start":"08:26.080 ","End":"08:30.075","Text":"then this exponent is minus 4t."},{"Start":"08:30.075 ","End":"08:32.895","Text":"Here we have b1,"},{"Start":"08:32.895 ","End":"08:35.070","Text":"which is 1, and here n is 1."},{"Start":"08:35.070 ","End":"08:36.615","Text":"We get this."},{"Start":"08:36.615 ","End":"08:39.675","Text":"Then when n=2,"},{"Start":"08:39.675 ","End":"08:42.320","Text":"4x 2^2 is 16,"},{"Start":"08:42.320 ","End":"08:47.690","Text":"so it\u0027s e to the minus 16 t. It\u0027s sine of 2x because n is 2 and we"},{"Start":"08:47.690 ","End":"08:53.550","Text":"have this 15 over 32 and the last, 1 minus 1/2."},{"Start":"08:53.550 ","End":"08:55.695","Text":"Then when n is 4,"},{"Start":"08:55.695 ","End":"08:59.320","Text":"4x4^2 is 64, and we have sine 4x."},{"Start":"08:59.320 ","End":"09:01.580","Text":"This is what we have. We\u0027re almost there."},{"Start":"09:01.580 ","End":"09:05.035","Text":"We have v(x, t) but we want u(x, t)."},{"Start":"09:05.035 ","End":"09:10.205","Text":"To go back and see we defined u(x, t)=v(x,"},{"Start":"09:10.205 ","End":"09:18.090","Text":"t) plus 1 over 32 sine 2x or rather we define v as u minus this, but same thing."},{"Start":"09:18.090 ","End":"09:23.615","Text":"All we have to do is add 1 over 32 sine 2x to this expression,"},{"Start":"09:23.615 ","End":"09:27.385","Text":"and we can combine it in with this sine 2x."},{"Start":"09:27.385 ","End":"09:31.365","Text":"We can even put the 1 over the 32."},{"Start":"09:31.365 ","End":"09:34.700","Text":"This is the expression we have now for u(x,"},{"Start":"09:34.700 ","End":"09:39.870","Text":"t) and that is the solution and we are done."}],"ID":30795}],"Thumbnail":null,"ID":294430},{"Name":"Duhamel\u0027s Principle","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Duhamel`s Principle","Duration":"3m 42s","ChapterTopicVideoID":29199,"CourseChapterTopicPlaylistID":294431,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29199.jpeg","UploadDate":"2022-06-06T09:17:13.6030000","DurationForVideoObject":"PT3M42S","Description":null,"MetaTitle":"Tutorial - Duhamel`s Principle: Video + Workbook | Proprep","MetaDescription":"The Heat Equation - Duhamel\u0027s Principle. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/partial-differential-equations/the-heat-equation/duhamel%27s-principle/vid30779","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.150","Text":"In this clip, we\u0027ll learn about Duhamel\u0027s principle in the heat equation."},{"Start":"00:06.150 ","End":"00:11.085","Text":"We already did the Duhamel\u0027s principle in the wave equation."},{"Start":"00:11.085 ","End":"00:15.615","Text":"It\u0027s a principle that applies to various types of equation."},{"Start":"00:15.615 ","End":"00:17.160","Text":"Let\u0027s get started."},{"Start":"00:17.160 ","End":"00:20.745","Text":"We have a non-homogeneous heat equation."},{"Start":"00:20.745 ","End":"00:24.810","Text":"Non-homogeneous because of the f(x,t) here,"},{"Start":"00:24.810 ","End":"00:27.780","Text":"which may not be 0, but we want"},{"Start":"00:27.780 ","End":"00:32.009","Text":"the initial condition and the boundary conditions to both be 0."},{"Start":"00:32.009 ","End":"00:35.910","Text":"Under these conditions, we can give the solution u(x,"},{"Start":"00:35.910 ","End":"00:40.650","Text":"t) as the integral of function of 3 variables, p(x,"},{"Start":"00:40.650 ","End":"00:46.099","Text":"t, Tau) where p is the solution to an easier problem."},{"Start":"00:46.099 ","End":"00:52.520","Text":"I.e., the following. This is a heat equation which is homogeneous."},{"Start":"00:52.520 ","End":"00:58.500","Text":"The initial condition is equal to exactly this non-homogeneous part."},{"Start":"00:58.500 ","End":"01:02.990","Text":"It moves from the PDE to the initial condition."},{"Start":"01:02.990 ","End":"01:07.415","Text":"Here also, the boundary values are 0."},{"Start":"01:07.415 ","End":"01:12.830","Text":"If you\u0027re wondering about this extra third variable as opposed to 2 variables,"},{"Start":"01:12.830 ","End":"01:15.140","Text":"what you could do is just fixed Tau."},{"Start":"01:15.140 ","End":"01:16.670","Text":"For any given Tau,"},{"Start":"01:16.670 ","End":"01:19.880","Text":"we have this problem in 2 variables,"},{"Start":"01:19.880 ","End":"01:23.150","Text":"x and t with fixed Tau."},{"Start":"01:23.150 ","End":"01:25.540","Text":"Then we solve this,"},{"Start":"01:25.540 ","End":"01:26.870","Text":"and after we\u0027ve solved that,"},{"Start":"01:26.870 ","End":"01:28.655","Text":"we can let Tau vary again."},{"Start":"01:28.655 ","End":"01:31.775","Text":"It\u0027s not really a problem that is a third variable."},{"Start":"01:31.775 ","End":"01:39.485","Text":"What we do is we solve this IBVP using our techniques for homogeneous heat equation,"},{"Start":"01:39.485 ","End":"01:41.225","Text":"and after we found P,"},{"Start":"01:41.225 ","End":"01:46.030","Text":"we just figure out the integral and then we get u."},{"Start":"01:46.030 ","End":"01:50.570","Text":"Here are a few words on what Duhamel means in this context,"},{"Start":"01:50.570 ","End":"01:52.280","Text":"it\u0027s named after someone anyway."},{"Start":"01:52.280 ","End":"01:53.935","Text":"You could read this if you want."},{"Start":"01:53.935 ","End":"01:57.410","Text":"We\u0027re going to prove this technique shortly."},{"Start":"01:57.410 ","End":"01:58.550","Text":"They might be wondering,"},{"Start":"01:58.550 ","End":"02:03.425","Text":"what happens if our boundary conditions are not homogeneous?"},{"Start":"02:03.425 ","End":"02:08.284","Text":"Where are they? They\u0027re not equal to 0,"},{"Start":"02:08.284 ","End":"02:11.195","Text":"then we use a correcting function."},{"Start":"02:11.195 ","End":"02:12.830","Text":"We\u0027ve discussed this previously."},{"Start":"02:12.830 ","End":"02:14.000","Text":"I won\u0027t go into details,"},{"Start":"02:14.000 ","End":"02:19.085","Text":"but you can try correcting function to make the boundary conditions homogeneous."},{"Start":"02:19.085 ","End":"02:24.020","Text":"Another thing I should point out is how do we get those boundary conditions for p?"},{"Start":"02:24.020 ","End":"02:27.470","Text":"Well, we just copy them from u,"},{"Start":"02:27.470 ","End":"02:30.920","Text":"except that we add a third variable, Tau."},{"Start":"02:30.920 ","End":"02:35.210","Text":"Now, this equation doesn\u0027t have to be Dirichlet type."},{"Start":"02:35.210 ","End":"02:37.460","Text":"It could be von Neumann type or mixed."},{"Start":"02:37.460 ","End":"02:40.810","Text":"It could be a du/dx here or here."},{"Start":"02:40.810 ","End":"02:48.170","Text":"In that case, we also transfer that partial derivative with respect to x to p from u."},{"Start":"02:48.170 ","End":"02:50.870","Text":"If we had this boundary condition for u,"},{"Start":"02:50.870 ","End":"02:54.035","Text":"this would be the boundary condition for p. Similarly, if this is for u,"},{"Start":"02:54.035 ","End":"03:00.815","Text":"this is for p. They might also be wondering what happens if this initial value is not 0."},{"Start":"03:00.815 ","End":"03:03.529","Text":"Well, there\u0027s a workaround for that also."},{"Start":"03:03.529 ","End":"03:07.200","Text":"If you have a problem with this initial values,"},{"Start":"03:07.200 ","End":"03:12.605","Text":"say, g(x), then you can split our problem into 2 sub-problems."},{"Start":"03:12.605 ","End":"03:15.695","Text":"We let u equal v plus w,"},{"Start":"03:15.695 ","End":"03:21.200","Text":"where v has a 0 initial value and"},{"Start":"03:21.200 ","End":"03:26.840","Text":"w is homogeneous but with the same initial value as u."},{"Start":"03:26.840 ","End":"03:30.815","Text":"Each of these is easier to solve than this,"},{"Start":"03:30.815 ","End":"03:35.155","Text":"so, we make 1 problem into 2 easier problems."},{"Start":"03:35.155 ","End":"03:38.270","Text":"Let\u0027s take a break here and after that we\u0027ll prove"},{"Start":"03:38.270 ","End":"03:42.480","Text":"the technique above using Duhamel\u0027s principle."}],"ID":30779},{"Watched":false,"Name":"Proof of Duhamel`s Principle","Duration":"4m 36s","ChapterTopicVideoID":29198,"CourseChapterTopicPlaylistID":294431,"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":"We\u0027re continuing after the break,"},{"Start":"00:02.310 ","End":"00:05.205","Text":"we presented Duhamel\u0027s principle in the heat equation."},{"Start":"00:05.205 ","End":"00:06.885","Text":"Now we\u0027re going to prove it."},{"Start":"00:06.885 ","End":"00:09.630","Text":"First, let\u0027s review what it said."},{"Start":"00:09.630 ","End":"00:12.360","Text":"We start with an IBVP as follows,"},{"Start":"00:12.360 ","End":"00:15.090","Text":"with a non-homogeneous heat equation."},{"Start":"00:15.090 ","End":"00:21.390","Text":"We solve an easier problem to find the function p which will present in a moment."},{"Start":"00:21.390 ","End":"00:23.565","Text":"Once we have p,"},{"Start":"00:23.565 ","End":"00:26.415","Text":"u is given by this integral."},{"Start":"00:26.415 ","End":"00:31.560","Text":"Now p satisfies an easier problem which is homogeneous."},{"Start":"00:31.560 ","End":"00:34.995","Text":"This is the problem that p satisfies,"},{"Start":"00:34.995 ","End":"00:37.785","Text":"we\u0027ll break the proof up into 4 parts."},{"Start":"00:37.785 ","End":"00:43.830","Text":"Really what we have to do is to show that this u satisfies 4 things."},{"Start":"00:43.830 ","End":"00:51.315","Text":"satisfies 2 boundary conditions and the PDE."},{"Start":"00:51.315 ","End":"00:53.175","Text":"Let\u0027s write that down."},{"Start":"00:53.175 ","End":"00:56.640","Text":"We have these 4 things to prove."},{"Start":"00:56.640 ","End":"01:02.940","Text":"The first one says that u (x,0) is the following integral."},{"Start":"01:02.940 ","End":"01:06.160","Text":"Because it\u0027s the integral from 0-0,"},{"Start":"01:06.160 ","End":"01:09.265","Text":"certainly it\u0027s 0, so that\u0027s okay."},{"Start":"01:09.265 ","End":"01:16.270","Text":"The 2nd and third conditions are very similar here with an x = 0."},{"Start":"01:16.270 ","End":"01:18.645","Text":"Here with an x = L,"},{"Start":"01:18.645 ","End":"01:21.655","Text":"we have to show that these 2 integrals are both 0."},{"Start":"01:21.655 ","End":"01:25.775","Text":"But they\u0027re actually just the boundary conditions. Not exactly."},{"Start":"01:25.775 ","End":"01:30.980","Text":"This says that this integrant is 0 and this integrant is 0."},{"Start":"01:30.980 ","End":"01:34.535","Text":"The integral of both of them is 0."},{"Start":"01:34.535 ","End":"01:38.015","Text":"We still have to show that the PDE is satisfied."},{"Start":"01:38.015 ","End":"01:41.135","Text":"We\u0027ll need ut and Uxx."},{"Start":"01:41.135 ","End":"01:47.810","Text":"U is this ut We can get by differentiating under the integral sign."},{"Start":"01:47.810 ","End":"01:51.200","Text":"I copy, pasted this from the Wikipedia."},{"Start":"01:51.200 ","End":"01:56.900","Text":"There is a Leibnitz integral rule that says basically that you can differentiate under"},{"Start":"01:56.900 ","End":"01:59.210","Text":"the integral sign as long as you\u0027re not"},{"Start":"01:59.210 ","End":"02:03.140","Text":"differentiating with respect to the variable of the integral,"},{"Start":"02:03.140 ","End":"02:05.120","Text":"so not d by dt,"},{"Start":"02:05.120 ","End":"02:07.480","Text":"but we can do d by dx."},{"Start":"02:07.480 ","End":"02:10.130","Text":"This is the formula that is used,"},{"Start":"02:10.130 ","End":"02:14.450","Text":"we\u0027ll apply that here because this integral as d tau,"},{"Start":"02:14.450 ","End":"02:18.110","Text":"we can differentiate with respect to either t or x, in this case,"},{"Start":"02:18.110 ","End":"02:22.700","Text":"t. If we apply this formula to our case,"},{"Start":"02:22.700 ","End":"02:25.310","Text":"the x here presents the t here."},{"Start":"02:25.310 ","End":"02:30.805","Text":"In our case, b(x) is t and a(x) is 0."},{"Start":"02:30.805 ","End":"02:37.910","Text":"We need the derivative of t here which is one derivative of 0, which is 0."},{"Start":"02:37.910 ","End":"02:41.585","Text":"We put t and 0 in place of tau,"},{"Start":"02:41.585 ","End":"02:44.030","Text":"the variable that\u0027s being integrated."},{"Start":"02:44.030 ","End":"02:48.650","Text":"The last part, we keep the 0 on the t and we"},{"Start":"02:48.650 ","End":"02:54.805","Text":"differentiate this with respect to the t. That\u0027s the variable here."},{"Start":"02:54.805 ","End":"02:57.800","Text":"Should we write this? I don\u0027t need the one on the term with"},{"Start":"02:57.800 ","End":"03:01.225","Text":"the 0 drops off so we have this."},{"Start":"03:01.225 ","End":"03:03.775","Text":"We also need Uxx."},{"Start":"03:03.775 ","End":"03:07.490","Text":"Again, we can use this formula, in this case,"},{"Start":"03:07.490 ","End":"03:11.870","Text":"a(x) and b(x) are constants as far as x goes."},{"Start":"03:11.870 ","End":"03:15.440","Text":"Both these derivatives would be 0."},{"Start":"03:15.440 ","End":"03:17.900","Text":"We only get the last term,"},{"Start":"03:17.900 ","End":"03:20.720","Text":"which is a derivative with respect to x here."},{"Start":"03:20.720 ","End":"03:21.830","Text":"We do this twice."},{"Start":"03:21.830 ","End":"03:25.655","Text":"So we get 2nd derivative with respect to x."},{"Start":"03:25.655 ","End":"03:28.385","Text":"Now that we have ut and uxx,"},{"Start":"03:28.385 ","End":"03:35.495","Text":"we can substitute these in the heat equation and see if we really get equality."},{"Start":"03:35.495 ","End":"03:44.855","Text":"Ut is this a^2 times Uxx is this plus f(x, t)."},{"Start":"03:44.855 ","End":"03:51.290","Text":"Let\u0027s see if we can show that these are equal with a bit of algebra we can rearrange."},{"Start":"03:51.290 ","End":"03:55.475","Text":"We can also put the a^2 inside the integral."},{"Start":"03:55.475 ","End":"04:00.470","Text":"Inside the integral we get pt minus a^2 pxx."},{"Start":"04:00.470 ","End":"04:06.365","Text":"The integral is 0 because p satisfies the heat equation."},{"Start":"04:06.365 ","End":"04:09.340","Text":"This was part of the conditions on P,"},{"Start":"04:09.340 ","End":"04:13.685","Text":"Pt a^2 pxx. This is 0."},{"Start":"04:13.685 ","End":"04:16.940","Text":"Also, the right-hand side is 0,"},{"Start":"04:16.940 ","End":"04:18.365","Text":"because if you look back,"},{"Start":"04:18.365 ","End":"04:25.820","Text":"the initial condition is that p(x,t,t) = to f(x,t) of this equals this,"},{"Start":"04:25.820 ","End":"04:29.660","Text":"the difference is 0 so 0 equals 0."},{"Start":"04:29.660 ","End":"04:32.360","Text":"We\u0027ve checked that this is true,"},{"Start":"04:32.360 ","End":"04:37.470","Text":"that was the 4th thing we had to check and that concludes the proof."}],"ID":30780},{"Watched":false,"Name":"Exercise 1","Duration":"6m 59s","ChapterTopicVideoID":29200,"CourseChapterTopicPlaylistID":294431,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.325","Text":"In this exercise, we\u0027re going to solve the following problem using Duhamel\u0027s principle."},{"Start":"00:05.325 ","End":"00:10.470","Text":"Here we have a heat equation which is non-homogeneous due to this term"},{"Start":"00:10.470 ","End":"00:16.275","Text":"and both the initial condition and the boundary conditions are homogeneous."},{"Start":"00:16.275 ","End":"00:19.380","Text":"According to Duhamel\u0027s principle, u(x,"},{"Start":"00:19.380 ","End":"00:25.055","Text":"t) is the integral of a function p(x, t, Tau),"},{"Start":"00:25.055 ","End":"00:31.385","Text":"which is a solution to an easier homogeneous problem as follows,"},{"Start":"00:31.385 ","End":"00:35.000","Text":"where the non-homogeneous term from"},{"Start":"00:35.000 ","End":"00:41.060","Text":"the original problem becomes the initial value of this new problem."},{"Start":"00:41.060 ","End":"00:44.690","Text":"We\u0027re going to solve for P in two different ways."},{"Start":"00:44.690 ","End":"00:45.950","Text":"In the first method,"},{"Start":"00:45.950 ","End":"00:51.650","Text":"it\u0027s a common technique to convert the function p to"},{"Start":"00:51.650 ","End":"00:58.820","Text":"a function P by substituting T=t minus Tau,"},{"Start":"00:58.820 ","End":"01:05.130","Text":"we consider Tau is fixed so really P as a function of 2 variables,"},{"Start":"01:05.130 ","End":"01:12.430","Text":"and we can get P. This is basically the definition which is the reverse of this,"},{"Start":"01:12.430 ","End":"01:19.250","Text":"the second variable minus the third variable is what we have here."},{"Start":"01:19.250 ","End":"01:22.370","Text":"Like here, this second variable is t,"},{"Start":"01:22.370 ","End":"01:26.210","Text":"which is t minus Tau, this minus this."},{"Start":"01:26.210 ","End":"01:28.385","Text":"If we do this substitution,"},{"Start":"01:28.385 ","End":"01:32.465","Text":"this problem modifies to this one."},{"Start":"01:32.465 ","End":"01:37.520","Text":"The derivative with respect to t is the derivative with respect to"},{"Start":"01:37.520 ","End":"01:42.500","Text":"t because they differ by a constant P(x,"},{"Start":"01:42.500 ","End":"01:48.180","Text":"0), 0 is t minus Tau so t=Tau."},{"Start":"01:48.180 ","End":"01:49.815","Text":"We\u0027re looking here,"},{"Start":"01:49.815 ","End":"01:51.960","Text":"when t=Tau, we have p(x, tau)."},{"Start":"01:51.960 ","End":"01:55.380","Text":"Tau is this, so that\u0027s this."},{"Start":"01:55.380 ","End":"01:59.475","Text":"Here, x is equal to 0 or L,"},{"Start":"01:59.475 ","End":"02:02.030","Text":"like here, x is 0 or L,"},{"Start":"02:02.030 ","End":"02:04.385","Text":"or these are equal to 0."},{"Start":"02:04.385 ","End":"02:09.065","Text":"We can solve this problem using the separation of variables method,"},{"Start":"02:09.065 ","End":"02:12.140","Text":"but we can jump straight to the formula because we have"},{"Start":"02:12.140 ","End":"02:17.610","Text":"a Dirichlet type boundary conditions,"},{"Start":"02:17.610 ","End":"02:23.370","Text":"we just have to use this formula and then we\u0027ll copy this back to our page."},{"Start":"02:23.370 ","End":"02:25.880","Text":"Here we are."},{"Start":"02:25.880 ","End":"02:30.560","Text":"Just I use the letter b_n instead of A_n and we have"},{"Start":"02:30.560 ","End":"02:36.395","Text":"T instead of t and we\u0027ll determine the b_n using this initial condition."},{"Start":"02:36.395 ","End":"02:43.575","Text":"Let t=0, 0 just means that this exponent disappears,"},{"Start":"02:43.575 ","End":"02:47.190","Text":"it\u0027s e^0 which is 1."},{"Start":"02:47.190 ","End":"02:49.140","Text":"On the other hand,"},{"Start":"02:49.140 ","End":"02:52.830","Text":"p(x_naught) is P(x,"},{"Start":"02:52.830 ","End":"02:56.850","Text":"Tau, Tau), which is this."},{"Start":"02:56.850 ","End":"03:00.800","Text":"We can compare coefficients of the sine series,"},{"Start":"03:00.800 ","End":"03:05.630","Text":"this is like a term from here with n=2,"},{"Start":"03:05.630 ","End":"03:08.725","Text":"b_2 is 6,"},{"Start":"03:08.725 ","End":"03:12.275","Text":"and all the other b_n\u0027s are 0."},{"Start":"03:12.275 ","End":"03:15.155","Text":"Substituting these back in this formula,"},{"Start":"03:15.155 ","End":"03:16.895","Text":"instead of the infinite series,"},{"Start":"03:16.895 ","End":"03:20.140","Text":"we just get a single term where n=2,"},{"Start":"03:20.140 ","End":"03:23.520","Text":"which is b_2, which is 6,"},{"Start":"03:23.520 ","End":"03:26.370","Text":"sin(2Pi over L x),"},{"Start":"03:26.370 ","End":"03:29.770","Text":"and here again n replaced by 2."},{"Start":"03:29.770 ","End":"03:31.370","Text":"Now that we have P,"},{"Start":"03:31.370 ","End":"03:34.265","Text":"we can return to p using this formula."},{"Start":"03:34.265 ","End":"03:38.865","Text":"Just replace T here by t minus Tau."},{"Start":"03:38.865 ","End":"03:40.550","Text":"Using the law of exponents,"},{"Start":"03:40.550 ","End":"03:43.950","Text":"we can break this up into the product of"},{"Start":"03:43.950 ","End":"03:49.725","Text":"2 exponents and that\u0027s our first way of finding p(x, t, Tau)."},{"Start":"03:49.725 ","End":"03:51.255","Text":"Now let\u0027s find p(x, t,"},{"Start":"03:51.255 ","End":"03:53.720","Text":"Tau) using the other method."},{"Start":"03:53.720 ","End":"04:01.355","Text":"We can straight away write it as a sine series using that same formula we used above."},{"Start":"04:01.355 ","End":"04:07.070","Text":"Again, we\u0027re thinking of Tau like a constant so b_n could actually be a function of Tau."},{"Start":"04:07.070 ","End":"04:09.500","Text":"Now if we let t=Tau,"},{"Start":"04:09.500 ","End":"04:12.170","Text":"we get here p(x,"},{"Start":"04:12.170 ","End":"04:18.125","Text":"Tau, Tau), which by the initial condition for p is this."},{"Start":"04:18.125 ","End":"04:22.340","Text":"Again, we can compare coefficients for the sine series,"},{"Start":"04:22.340 ","End":"04:24.095","Text":"I\u0027ve emphasized the n here."},{"Start":"04:24.095 ","End":"04:29.850","Text":"You only get something when n is 2 so that 6 is"},{"Start":"04:29.850 ","End":"04:37.245","Text":"equal to b_2 times e^minus(2Pi over L)^2 Tau."},{"Start":"04:37.245 ","End":"04:43.180","Text":"Just comparing the signs so this part and this part are part of the coefficient."},{"Start":"04:43.370 ","End":"04:45.770","Text":"As for the other ends,"},{"Start":"04:45.770 ","End":"04:51.005","Text":"0 is b_n e^minus this."},{"Start":"04:51.005 ","End":"04:55.565","Text":"That\u0027s when n is not equal to 2, so b_2,"},{"Start":"04:55.565 ","End":"04:57.740","Text":"just bringing the exponent to the other side,"},{"Start":"04:57.740 ","End":"04:59.060","Text":"so make the minus a plus,"},{"Start":"04:59.060 ","End":"05:02.635","Text":"b_2 is this, and b_n is 0."},{"Start":"05:02.635 ","End":"05:06.480","Text":"Substituting these coefficients into the formula,"},{"Start":"05:06.480 ","End":"05:13.715","Text":"this one, we only have a term when n=2 so this is what we get."},{"Start":"05:13.715 ","End":"05:15.770","Text":"This is where n=2,"},{"Start":"05:15.770 ","End":"05:22.335","Text":"here and here and here, that\u0027s b_2."},{"Start":"05:22.335 ","End":"05:24.960","Text":"Now we can unfix Tau,"},{"Start":"05:24.960 ","End":"05:26.625","Text":"consider it a variable again."},{"Start":"05:26.625 ","End":"05:27.860","Text":"Nothing that really happens here,"},{"Start":"05:27.860 ","End":"05:30.320","Text":"it\u0027s just a psychological viewpoint."},{"Start":"05:30.320 ","End":"05:34.345","Text":"Yeah, so now we can look at this as true for a variable Tau."},{"Start":"05:34.345 ","End":"05:39.425","Text":"Note that this is the same answer that we got the other way here."},{"Start":"05:39.425 ","End":"05:43.475","Text":"It\u0027s good to know that we get the same thing in 2 different methods."},{"Start":"05:43.475 ","End":"05:48.790","Text":"All we have to do now is compute u using this formula."},{"Start":"05:48.790 ","End":"05:51.780","Text":"We\u0027re going to just compute this integral and then we\u0027ll be done."},{"Start":"05:51.780 ","End":"05:53.325","Text":"Substituting for p,"},{"Start":"05:53.325 ","End":"05:59.420","Text":"it\u0027s this but we took out of the integral everything that doesn\u0027t have Tau in it,"},{"Start":"05:59.420 ","End":"06:03.170","Text":"the part with t and the part with x that don\u0027t have Tau."},{"Start":"06:03.170 ","End":"06:09.845","Text":"Now, this integral is the same thing divided by the coefficient"},{"Start":"06:09.845 ","End":"06:17.030","Text":"of Tau so just same thing but inverted (L over 2Pi a)^2."},{"Start":"06:17.030 ","End":"06:25.980","Text":"Then we have to evaluate this between 0 and t. This becomes, when Tau=t,"},{"Start":"06:25.980 ","End":"06:31.620","Text":"it\u0027s just this to the power of t and when tau is 0, this is equal to 1,"},{"Start":"06:31.620 ","End":"06:38.855","Text":"and all the rest of it the same except that put this constant up here by the 6."},{"Start":"06:38.855 ","End":"06:46.070","Text":"Notice that this is the same as this almost except for the minus sign so we can"},{"Start":"06:46.070 ","End":"06:54.675","Text":"multiply this by this brackets and what we get is 1 minus e^plus this thing."},{"Start":"06:54.675 ","End":"06:59.410","Text":"That\u0027s it. That\u0027s the answer and we are done."}],"ID":30781},{"Watched":false,"Name":"Exercise 2","Duration":"12m 31s","ChapterTopicVideoID":29201,"CourseChapterTopicPlaylistID":294431,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.320","Text":"In this exercise, we have an initial value problem,"},{"Start":"00:04.320 ","End":"00:07.500","Text":"which we\u0027re going to solve using Duhamel\u0027s Principle."},{"Start":"00:07.500 ","End":"00:12.885","Text":"This is a heat equation non-homogeneous because of this term."},{"Start":"00:12.885 ","End":"00:19.500","Text":"It\u0027s on the whole real interval and that\u0027s why we don\u0027t need boundary conditions."},{"Start":"00:19.500 ","End":"00:25.305","Text":"We only need an initial condition and that\u0027s why I called it an IVP and not an IBVP."},{"Start":"00:25.305 ","End":"00:27.075","Text":"Using Duhamel\u0027s Principle,"},{"Start":"00:27.075 ","End":"00:34.760","Text":"what we do is we express u or we compute u as the following integral in terms of p,"},{"Start":"00:34.760 ","End":"00:41.060","Text":"where p is a solution to a simpler IVP as follows."},{"Start":"00:41.060 ","End":"00:44.630","Text":"This is a homogeneous heat equation."},{"Start":"00:44.630 ","End":"00:51.485","Text":"The non-homogeneous term from the PDE moves to the initial condition."},{"Start":"00:51.485 ","End":"00:55.595","Text":"The initial condition is not your usual where something is 0,"},{"Start":"00:55.595 ","End":"01:01.220","Text":"it\u0027s putting the 2nd and 3rd variables to be equal, t equals Tau."},{"Start":"01:01.220 ","End":"01:09.680","Text":"A common technique is to let T equal t minus Tau and then we get a different function,"},{"Start":"01:09.680 ","End":"01:12.530","Text":"call it big P of just 2 variables,"},{"Start":"01:12.530 ","End":"01:15.665","Text":"x and t. You could define it this way."},{"Start":"01:15.665 ","End":"01:17.990","Text":"You put t plus Tau here and Tau here."},{"Start":"01:17.990 ","End":"01:20.900","Text":"Tau is fixed for the time being."},{"Start":"01:20.900 ","End":"01:27.110","Text":"We just fixed Tau and at the end when we\u0027ve solved it that we can let Tau vary again."},{"Start":"01:27.110 ","End":"01:29.300","Text":"For this given but arbitrary Tau,"},{"Start":"01:29.300 ","End":"01:30.470","Text":"this is how we define it."},{"Start":"01:30.470 ","End":"01:33.320","Text":"If you do it the other way around p(x, t,"},{"Start":"01:33.320 ","End":"01:36.920","Text":"Tau) is big P(x) and t minus Tau,"},{"Start":"01:36.920 ","End":"01:38.530","Text":"or capital T here."},{"Start":"01:38.530 ","End":"01:40.285","Text":"This minus this,"},{"Start":"01:40.285 ","End":"01:46.762","Text":"is big T. We want to convert this problem to a problem in big P and"},{"Start":"01:46.762 ","End":"01:53.180","Text":"big T. Let\u0027s note that the derivative of capital P with respect to capital T,"},{"Start":"01:53.180 ","End":"02:02.060","Text":"is little p derived with respect to little t. Here\u0027s an explanation of that,"},{"Start":"02:02.060 ","End":"02:03.470","Text":"I won\u0027t dwell on it."},{"Start":"02:03.470 ","End":"02:07.260","Text":"What it means is that dp by dt,"},{"Start":"02:07.260 ","End":"02:12.090","Text":"small letters is DP by DT, capital letters."},{"Start":"02:12.090 ","End":"02:15.665","Text":"If we differentiate by x once or twice,"},{"Start":"02:15.665 ","End":"02:19.865","Text":"it\u0027s going to be the same as derivative of little p with respect to x."},{"Start":"02:19.865 ","End":"02:23.150","Text":"The domain for big T is T bigger or equal to"},{"Start":"02:23.150 ","End":"02:26.630","Text":"0 because little t is bigger or equal to Tau."},{"Start":"02:26.630 ","End":"02:28.280","Text":"This is t minus Tau,"},{"Start":"02:28.280 ","End":"02:30.545","Text":"so bigger than 0."},{"Start":"02:30.545 ","End":"02:34.775","Text":"Now we\u0027ve converted this problem to this problem."},{"Start":"02:34.775 ","End":"02:40.985","Text":"Now we have a homogeneous heat equation on the whole real line."},{"Start":"02:40.985 ","End":"02:45.060","Text":"In principle, we could use Poisson\u0027s formula and then"},{"Start":"02:45.060 ","End":"02:49.100","Text":"we would get the following integral u(y,"},{"Start":"02:49.100 ","End":"02:53.880","Text":"0) well, u is like p and we\u0027d get to the x^2 Tau,"},{"Start":"02:53.880 ","End":"02:57.155","Text":"we get y^2 Tau because it\u0027s in terms of y."},{"Start":"02:57.155 ","End":"02:59.450","Text":"But this is a difficult integral."},{"Start":"02:59.450 ","End":"03:02.525","Text":"There is a trick we can do to make this simpler."},{"Start":"03:02.525 ","End":"03:08.740","Text":"Not really a trick it\u0027s just that it\u0027s known and we\u0027ll show this that if a function u(x,"},{"Start":"03:08.740 ","End":"03:13.250","Text":"t) not here, it\u0027s in general satisfies the heat equation."},{"Start":"03:13.250 ","End":"03:18.640","Text":"Then du by dt and du by dx also satisfy the heat equation."},{"Start":"03:18.640 ","End":"03:22.355","Text":"Not only that but also the higher-order derivatives."},{"Start":"03:22.355 ","End":"03:25.970","Text":"For example, the 3 derivatives of order 2,"},{"Start":"03:25.970 ","End":"03:30.620","Text":"but all the partial derivatives satisfy the heat equation."},{"Start":"03:30.620 ","End":"03:36.515","Text":"For example, u_xx would satisfy the following equation."},{"Start":"03:36.515 ","End":"03:39.440","Text":"I single this one out because we\u0027re going to use this one."},{"Start":"03:39.440 ","End":"03:41.615","Text":"The asterisk here is a note."},{"Start":"03:41.615 ","End":"03:43.895","Text":"I\u0027m going to prove this at the end of the clip."},{"Start":"03:43.895 ","End":"03:46.684","Text":"Adapting this to our case."},{"Start":"03:46.684 ","End":"03:48.305","Text":"What we get, well,"},{"Start":"03:48.305 ","End":"03:51.635","Text":"u is replaced by capital P. In our case,"},{"Start":"03:51.635 ","End":"03:59.240","Text":"t is big T. This becomes P_xx T equals a^2 (P_xx)_xx."},{"Start":"03:59.240 ","End":"04:00.770","Text":"Now we make a substitution."},{"Start":"04:00.770 ","End":"04:08.430","Text":"We\u0027ll let function v(x,t) be p_xx and then we get from here,"},{"Start":"04:10.780 ","End":"04:17.105","Text":"v_T equals a^2(v_xx), the heat equation for v. What about the initial condition?"},{"Start":"04:17.105 ","End":"04:20.660","Text":"Well, we have this initial condition for P,"},{"Start":"04:20.660 ","End":"04:23.150","Text":"that p(x, 0) equals x^2 Tau."},{"Start":"04:23.150 ","End":"04:28.085","Text":"Normally, you can\u0027t differentiate after you\u0027ve substituted,"},{"Start":"04:28.085 ","End":"04:31.070","Text":"you have to first differentiate and then substitute."},{"Start":"04:31.070 ","End":"04:32.105","Text":"But in this case,"},{"Start":"04:32.105 ","End":"04:39.375","Text":"we\u0027re differentiating with respect to x so we\u0027re only taking values along the x-axis."},{"Start":"04:39.375 ","End":"04:41.660","Text":"It doesn\u0027t matter we can just differentiate"},{"Start":"04:41.660 ","End":"04:44.610","Text":"this with respect to x and just keep the 0 here."},{"Start":"04:44.610 ","End":"04:48.210","Text":"The derivative of this with respect to x is 2x Tau."},{"Start":"04:48.210 ","End":"04:51.370","Text":"Second derivative is just 2Tau."},{"Start":"04:51.370 ","End":"04:56.680","Text":"That\u0027s a lot simpler for working with an x^2 Tau."},{"Start":"04:56.680 ","End":"05:03.245","Text":"This time we get initial value problem with a simpler initial condition,"},{"Start":"05:03.245 ","End":"05:06.950","Text":"the problem for v. We have here"},{"Start":"05:06.950 ","End":"05:14.285","Text":"the heat equation on the infinite interval and the initial condition 2Tau."},{"Start":"05:14.285 ","End":"05:16.700","Text":"Now with this simpler initial condition,"},{"Start":"05:16.700 ","End":"05:20.420","Text":"we can try again to use Poisson\u0027s formula."},{"Start":"05:20.420 ","End":"05:22.990","Text":"This time we get 2Tau here,"},{"Start":"05:22.990 ","End":"05:26.844","Text":"which is like a constant as far as y is concerned,"},{"Start":"05:26.844 ","End":"05:29.100","Text":"v(y, 0) is really v(x,"},{"Start":"05:29.100 ","End":"05:31.825","Text":"0) with x replaced by y."},{"Start":"05:31.825 ","End":"05:35.080","Text":"You\u0027ll see that this is an easy integral to compute."},{"Start":"05:35.080 ","End":"05:41.560","Text":"First thing we can do is to write this as minus something squared."},{"Start":"05:41.560 ","End":"05:49.000","Text":"It\u0027s y minus x over 2a root T. Also the 2Tau can be brought in front of the integral."},{"Start":"05:49.000 ","End":"05:56.220","Text":"Here, the obvious thing is to make a substitution that this will be, call it z."},{"Start":"05:56.220 ","End":"05:59.400","Text":"Z is y minus x over 2a root t,"},{"Start":"05:59.400 ","End":"06:01.560","Text":"z if you\u0027re in America,"},{"Start":"06:01.560 ","End":"06:07.545","Text":"and dz is 1 over 2a root T dy."},{"Start":"06:07.545 ","End":"06:11.275","Text":"Note that dy is 2a root T dz."},{"Start":"06:11.275 ","End":"06:13.415","Text":"That\u0027s what we\u0027ll use."},{"Start":"06:13.415 ","End":"06:15.200","Text":"What we get here is,"},{"Start":"06:15.200 ","End":"06:23.000","Text":"here\u0027s the 2a root T. This is e to the minus z^2 and the rest of it is the same."},{"Start":"06:23.000 ","End":"06:25.700","Text":"Now this is a famous integral."},{"Start":"06:25.700 ","End":"06:29.020","Text":"We know that this integral is root Pi."},{"Start":"06:29.020 ","End":"06:30.920","Text":"Replace this by root Pi,"},{"Start":"06:30.920 ","End":"06:36.050","Text":"and now we can start canceling and everything cancels except the 2Tau."},{"Start":"06:36.050 ","End":"06:42.265","Text":"We have v(x,t) is 2Tau and we want to get back to big P,"},{"Start":"06:42.265 ","End":"06:45.045","Text":"and from there to little p from there to u."},{"Start":"06:45.045 ","End":"06:49.910","Text":"V is the second derivative of p with respect to x."},{"Start":"06:49.910 ","End":"06:51.815","Text":"We integrate twice."},{"Start":"06:51.815 ","End":"06:55.085","Text":"This first of all is equal to 2Tau."},{"Start":"06:55.085 ","End":"07:03.140","Text":"Now integrating once, we get 2Tau times x plus a constant in x,"},{"Start":"07:03.140 ","End":"07:11.090","Text":"which is really an arbitrary function of T. Then integrating again 2xTau becomes x^2Tau."},{"Start":"07:11.090 ","End":"07:17.180","Text":"This is A(T) times x and another constant which is an arbitrary function of"},{"Start":"07:17.180 ","End":"07:23.990","Text":"T. We know that P(x,0) is x^2Tau."},{"Start":"07:23.990 ","End":"07:25.895","Text":"Go back and check."},{"Start":"07:25.895 ","End":"07:29.505","Text":"Now we can also get P(x,0) from here,"},{"Start":"07:29.505 ","End":"07:33.600","Text":"substituting t equals 0."},{"Start":"07:33.600 ","End":"07:38.590","Text":"Then we get P(x,0) is x^2Tau plus A(0)x plus B(0)."},{"Start":"07:38.590 ","End":"07:44.135","Text":"Comparing these, it follows that this is 0."},{"Start":"07:44.135 ","End":"07:46.850","Text":"This is not a simple equality for some x."},{"Start":"07:46.850 ","End":"07:50.690","Text":"It\u0027s true for all x. This is really an identity."},{"Start":"07:50.690 ","End":"07:55.475","Text":"If we have Ax plus B equals 0 as a function,"},{"Start":"07:55.475 ","End":"08:00.065","Text":"we know that the coefficients have to be 0, both of them."},{"Start":"08:00.065 ","End":"08:04.610","Text":"That\u0027s good, but we still have to find A (t) in general and B(t)."},{"Start":"08:04.610 ","End":"08:08.180","Text":"We\u0027ll use the heat equation that P satisfies,"},{"Start":"08:08.180 ","End":"08:13.624","Text":"which is this and then we can differentiate this."},{"Start":"08:13.624 ","End":"08:15.615","Text":"Well, the whole thing really."},{"Start":"08:15.615 ","End":"08:18.320","Text":"A derivative with respect to t, well,"},{"Start":"08:18.320 ","End":"08:25.594","Text":"this becomes 0 when we have A(t,x) plus B\u0027(t) and with respect to x twice."},{"Start":"08:25.594 ","End":"08:28.475","Text":"The first time we differentiate with respect to x,"},{"Start":"08:28.475 ","End":"08:32.645","Text":"this disappears, this just becomes A(t) and here we have 2xTau."},{"Start":"08:32.645 ","End":"08:38.000","Text":"The second time we get 2Tau and this disappears."},{"Start":"08:38.000 ","End":"08:40.190","Text":"This is what we have now."},{"Start":"08:40.190 ","End":"08:42.470","Text":"This is true for all x and you get"},{"Start":"08:42.470 ","End":"08:46.405","Text":"another inequalities and identities so we can compare coefficients."},{"Start":"08:46.405 ","End":"08:52.040","Text":"The coefficient of x is 0 because there\u0027s no coefficient of x here."},{"Start":"08:52.040 ","End":"08:56.000","Text":"The free coefficient is B\u0027(t) and here it\u0027s this,"},{"Start":"08:56.000 ","End":"08:59.270","Text":"we just rewrite it slightly as 2a^2 Tau."},{"Start":"08:59.270 ","End":"09:01.355","Text":"If we integrate this,"},{"Start":"09:01.355 ","End":"09:04.880","Text":"we get that A(t) is some constant,"},{"Start":"09:04.880 ","End":"09:07.505","Text":"but that constant has to be 0 from here."},{"Start":"09:07.505 ","End":"09:15.275","Text":"Similarly here, P(t) is 2a^2Tau times big T plus a constant."},{"Start":"09:15.275 ","End":"09:20.660","Text":"But that constant also has to be 0 to make it work."},{"Start":"09:20.660 ","End":"09:28.205","Text":"We found A(t) is 0 and B(t) is 2a^2 Tau t. Copying this down here,"},{"Start":"09:28.205 ","End":"09:32.740","Text":"we get that P(x,t) is x^2Tau,"},{"Start":"09:32.740 ","End":"09:36.520","Text":"A(t) is 0, so that disappears."},{"Start":"09:36.520 ","End":"09:40.805","Text":"Then B(t) is 2a^2t Tau, which is this."},{"Start":"09:40.805 ","End":"09:43.525","Text":"This is our expression for P(x,t)."},{"Start":"09:43.525 ","End":"09:50.590","Text":"Then we can return to p by replacing T by T minus Tau."},{"Start":"09:50.590 ","End":"09:53.160","Text":"Good, so now we\u0027ve got little p and there\u0027s"},{"Start":"09:53.160 ","End":"09:57.060","Text":"one more stage to get back to we have to get from little p"},{"Start":"09:57.060 ","End":"10:04.675","Text":"to u and we do that by means of this integral that\u0027s part of the Duhamel method."},{"Start":"10:04.675 ","End":"10:07.005","Text":"Let\u0027s compute this integral."},{"Start":"10:07.005 ","End":"10:10.520","Text":"Just replace p by what\u0027s written here."},{"Start":"10:10.520 ","End":"10:17.465","Text":"Also expanded the brackets so it\u0027s 2a^2Tau t minus 2a^2Tau^2."},{"Start":"10:17.465 ","End":"10:21.125","Text":"The integral of Tau is Tau^2 over 2."},{"Start":"10:21.125 ","End":"10:24.260","Text":"Similarly here, Tau gives us Tau^2 over 2."},{"Start":"10:24.260 ","End":"10:27.435","Text":"The 2 just cancels with this 2."},{"Start":"10:27.435 ","End":"10:31.195","Text":"The integral of Tau^2 is Tau^3 over 3."},{"Start":"10:31.195 ","End":"10:37.650","Text":"That gives us this to evaluate between 0 and t. When we put in Tau equals 0,"},{"Start":"10:37.650 ","End":"10:38.760","Text":"this is all 0."},{"Start":"10:38.760 ","End":"10:41.515","Text":"We just have to put Tau equals t here,"},{"Start":"10:41.515 ","End":"10:45.590","Text":"basically just copy this but put t instead of Tau.,"},{"Start":"10:45.590 ","End":"10:48.575","Text":"and here we get t^3."},{"Start":"10:48.575 ","End":"10:50.300","Text":"We slightly simplify."},{"Start":"10:50.300 ","End":"10:54.290","Text":"This minus this 1 minus 2/3 is a 1/3."},{"Start":"10:54.290 ","End":"11:00.640","Text":"We get u(x,t) is a 1/2x^2 t^2 plus a 1/3 a^2 t^3."},{"Start":"11:00.640 ","End":"11:02.745","Text":"That is the answer."},{"Start":"11:02.745 ","End":"11:05.690","Text":"But don\u0027t go yet I still owe you the matter"},{"Start":"11:05.690 ","End":"11:08.810","Text":"of proving that if you satisfy the heat equation,"},{"Start":"11:08.810 ","End":"11:11.030","Text":"so we do its partial derivatives."},{"Start":"11:11.030 ","End":"11:16.300","Text":"Let\u0027s start with the heat equation that u satisfies and"},{"Start":"11:16.300 ","End":"11:22.295","Text":"we\u0027ll show that the first-order partial derivatives satisfy the heat equation."},{"Start":"11:22.295 ","End":"11:25.625","Text":"First of all, let\u0027s do u_x."},{"Start":"11:25.625 ","End":"11:29.615","Text":"(u_x)t just swap the order of the derivatives,"},{"Start":"11:29.615 ","End":"11:34.405","Text":"get (u_t)x. U_t is a^2 u_xx."},{"Start":"11:34.405 ","End":"11:38.235","Text":"Then we can put 1 of these xs outside."},{"Start":"11:38.235 ","End":"11:40.790","Text":"We get this and that\u0027s what we wanted to show,"},{"Start":"11:40.790 ","End":"11:45.905","Text":"so u_x satisfies the heat equation and u_t"},{"Start":"11:45.905 ","End":"11:51.260","Text":"with respect to t is a^2 u_xx with respect to t. Again,"},{"Start":"11:51.260 ","End":"11:57.170","Text":"we can switch the order of the partial derivatives and get (u_t) xx,"},{"Start":"11:57.170 ","End":"11:59.540","Text":"so u_t also satisfies."},{"Start":"11:59.540 ","End":"12:01.520","Text":"Once we have this,"},{"Start":"12:01.520 ","End":"12:05.690","Text":"we can get by induction all the partial derivatives."},{"Start":"12:05.690 ","End":"12:12.290","Text":"The partial derivatives can be reached first from u_x or first from u_t and"},{"Start":"12:12.290 ","End":"12:19.265","Text":"then a chain of differentiations by x or by t. I think that\u0027s fairly clear."},{"Start":"12:19.265 ","End":"12:22.290","Text":"That concludes this clip."}],"ID":30782},{"Watched":false,"Name":"Exercise 3","Duration":"7m 41s","ChapterTopicVideoID":29203,"CourseChapterTopicPlaylistID":294431,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.340","Text":"In this exercise, we have"},{"Start":"00:02.340 ","End":"00:07.350","Text":"an initial value problem and we\u0027re going to solve it using Duhamel\u0027s principle."},{"Start":"00:07.350 ","End":"00:12.630","Text":"What we have is a heat equation but non-homogeneous,"},{"Start":"00:12.630 ","End":"00:16.350","Text":"and it\u0027s defined on the whole real line,"},{"Start":"00:16.350 ","End":"00:18.600","Text":"so we don\u0027t need boundary conditions,"},{"Start":"00:18.600 ","End":"00:20.460","Text":"just an initial condition,"},{"Start":"00:20.460 ","End":"00:25.455","Text":"and this also is homogeneous u(x,0)=0."},{"Start":"00:25.455 ","End":"00:29.400","Text":"This allows us to apply the Duhamel\u0027s principle, which says,"},{"Start":"00:29.400 ","End":"00:34.020","Text":"that we can find u as this integral of a function p(x,"},{"Start":"00:34.020 ","End":"00:41.240","Text":"t, Tau) and this is a solution to an easier IVP,"},{"Start":"00:41.240 ","End":"00:42.755","Text":"which is the following."},{"Start":"00:42.755 ","End":"00:46.070","Text":"We have a homogeneous heat equation and"},{"Start":"00:46.070 ","End":"00:50.720","Text":"the non-homogeneous part moves to the initial value here."},{"Start":"00:50.720 ","End":"00:55.790","Text":"Initial value, but not when something equals 0, but when t=Tau."},{"Start":"00:55.790 ","End":"00:58.880","Text":"Now we\u0027re going to define another function,"},{"Start":"00:58.880 ","End":"01:02.085","Text":"P. Let fixed Tau."},{"Start":"01:02.085 ","End":"01:04.110","Text":"For this given Tau,"},{"Start":"01:04.110 ","End":"01:08.910","Text":"we can define a function p(x,"},{"Start":"01:08.910 ","End":"01:10.740","Text":"T) as p(x,"},{"Start":"01:10.740 ","End":"01:12.510","Text":"t plus Tau,"},{"Start":"01:12.510 ","End":"01:14.740","Text":"Tau) for this particular Tau."},{"Start":"01:14.740 ","End":"01:18.650","Text":"If you look at it the other way around p(x, t,"},{"Start":"01:18.650 ","End":"01:24.290","Text":"Tau) is p(x, t minus Tau)."},{"Start":"01:24.290 ","End":"01:29.235","Text":"T is t minus Tau."},{"Start":"01:29.235 ","End":"01:31.545","Text":"So really p(x, t,"},{"Start":"01:31.545 ","End":"01:35.095","Text":"Tau) is p(x, T)."},{"Start":"01:35.095 ","End":"01:39.485","Text":"We want to convert this problem to a problem in P,"},{"Start":"01:39.485 ","End":"01:44.370","Text":"so we\u0027ll need P_t and P_xx."},{"Start":"01:44.560 ","End":"01:52.405","Text":"We\u0027ve done this before, so I\u0027ll just say that this P_t is this P_t,"},{"Start":"01:52.405 ","End":"01:55.875","Text":"and this P_xx is this P_xx,"},{"Start":"01:55.875 ","End":"02:01.084","Text":"so this IVP becomes the following."},{"Start":"02:01.084 ","End":"02:03.635","Text":"The initial condition, this Tau,"},{"Start":"02:03.635 ","End":"02:07.445","Text":"Tau means that t minus Tau is 0,"},{"Start":"02:07.445 ","End":"02:10.870","Text":"so this is really T=0."},{"Start":"02:10.870 ","End":"02:17.205","Text":"The domain T equal to Tau means that T minus Tau is bigger or equal to 0,"},{"Start":"02:17.205 ","End":"02:20.405","Text":"so T is bigger than or equal to 0."},{"Start":"02:20.405 ","End":"02:23.210","Text":"The obvious thing to do here is to use"},{"Start":"02:23.210 ","End":"02:30.520","Text":"the Poisson formula because we have homogeneous heat equation on an infinite interval."},{"Start":"02:30.520 ","End":"02:34.530","Text":"So what we would do is try this,"},{"Start":"02:34.530 ","End":"02:43.624","Text":"and from experience when this initial value has x in it gets often a difficult integral."},{"Start":"02:43.624 ","End":"02:45.710","Text":"Here that x becomes y,"},{"Start":"02:45.710 ","End":"02:49.135","Text":"so the y together with the exponent,"},{"Start":"02:49.135 ","End":"02:51.379","Text":"it\u0027s a bit difficult to integrate,"},{"Start":"02:51.379 ","End":"02:56.510","Text":"so we can use a standard trick to try and get rid of the x here."},{"Start":"02:56.510 ","End":"03:00.950","Text":"We know that the derivative of a function that"},{"Start":"03:00.950 ","End":"03:04.865","Text":"satisfies the heat equation also satisfies the heat equation,"},{"Start":"03:04.865 ","End":"03:07.880","Text":"the partial derivative with respect to x or T,"},{"Start":"03:07.880 ","End":"03:10.145","Text":"or even higher-order derivatives,"},{"Start":"03:10.145 ","End":"03:12.380","Text":"just prove it again, it\u0027s so short."},{"Start":"03:12.380 ","End":"03:15.080","Text":"We take P_x derivative with respect to"},{"Start":"03:15.080 ","End":"03:19.250","Text":"T using the formula around the mixed second-order derivatives,"},{"Start":"03:19.250 ","End":"03:20.905","Text":"we can swap the order,"},{"Start":"03:20.905 ","End":"03:25.415","Text":"and then the heat equation on this means that this is P_xx."},{"Start":"03:25.415 ","End":"03:28.550","Text":"We can just again rearrange,"},{"Start":"03:28.550 ","End":"03:30.770","Text":"instead of twice with respect to x,"},{"Start":"03:30.770 ","End":"03:34.220","Text":"and once more we can take first once and then another twice."},{"Start":"03:34.220 ","End":"03:41.865","Text":"So this P_x with respect to T is a^2 P_x with respect to xx."},{"Start":"03:41.865 ","End":"03:44.040","Text":"If we let v=P_x,"},{"Start":"03:44.040 ","End":"03:49.605","Text":"then what that gives us is that v_T equals a^2 v_xx,"},{"Start":"03:49.605 ","End":"03:53.345","Text":"which is the heat equation for v. The initial condition"},{"Start":"03:53.345 ","End":"03:58.090","Text":"p(x,0) is 2x sine t. So we can differentiate"},{"Start":"03:58.090 ","End":"04:08.080","Text":"this and get that v(x,0) is 2 sine Tau without the x."},{"Start":"04:08.080 ","End":"04:13.119","Text":"So we get an IVP for the function v. This is the PDE,"},{"Start":"04:13.119 ","End":"04:15.485","Text":"this is the initial condition,"},{"Start":"04:15.485 ","End":"04:20.990","Text":"and this is a lot easier this time when we apply Poisson\u0027s formula."},{"Start":"04:20.990 ","End":"04:23.905","Text":"2 sine Tau is like a constant as far as y goes,"},{"Start":"04:23.905 ","End":"04:30.120","Text":"so we can pull it out and get the following integral."},{"Start":"04:33.040 ","End":"04:36.775","Text":"First we write this as something squared,"},{"Start":"04:36.775 ","End":"04:39.420","Text":"and then we do our substitution,"},{"Start":"04:39.420 ","End":"04:42.705","Text":"which is z equals what\u0027s in here,"},{"Start":"04:42.705 ","End":"04:45.675","Text":"and dz is the following."},{"Start":"04:45.675 ","End":"04:48.990","Text":"So dy is 2a root T dz,"},{"Start":"04:48.990 ","End":"04:51.950","Text":"and we can make the substitution."},{"Start":"04:51.950 ","End":"04:54.035","Text":"This is now minus z^2,"},{"Start":"04:54.035 ","End":"04:57.365","Text":"and this is the 2a root T from here."},{"Start":"04:57.365 ","End":"04:59.360","Text":"This is the famous integral."},{"Start":"04:59.360 ","End":"05:02.689","Text":"We know its value is the square root of Pi."},{"Start":"05:02.689 ","End":"05:04.670","Text":"Now we can cancel out."},{"Start":"05:04.670 ","End":"05:10.525","Text":"All this stuff cancels with all this stuff and we\u0027re just left with 2 sine Tau."},{"Start":"05:10.525 ","End":"05:13.420","Text":"This is v(x, T),"},{"Start":"05:13.420 ","End":"05:17.415","Text":"so it\u0027s the P_x(x, T)."},{"Start":"05:17.415 ","End":"05:22.340","Text":"2 sine Tau, so we want to integrate with respect to x and we get"},{"Start":"05:22.340 ","End":"05:30.575","Text":"2x sine Tau plus some function of T. If we substitute T=0,"},{"Start":"05:30.575 ","End":"05:32.750","Text":"then we get the following."},{"Start":"05:32.750 ","End":"05:36.740","Text":"On the other hand, we had an initial condition for P. If you go back and see"},{"Start":"05:36.740 ","End":"05:41.030","Text":"that P(x,0) is 2x sine Tau."},{"Start":"05:41.030 ","End":"05:44.600","Text":"So if this equals this for all x,"},{"Start":"05:44.600 ","End":"05:48.555","Text":"then a(0) has to be 0."},{"Start":"05:48.555 ","End":"05:52.114","Text":"That\u0027s good, but we want a of T altogether."},{"Start":"05:52.114 ","End":"05:56.580","Text":"From the PDE, the heat equation,"},{"Start":"05:56.620 ","End":"05:59.540","Text":"A\u0027(T) is dp/dt,"},{"Start":"05:59.540 ","End":"06:02.015","Text":"and if we look here,"},{"Start":"06:02.015 ","End":"06:03.470","Text":"differentiating with respect to T,"},{"Start":"06:03.470 ","End":"06:05.390","Text":"this is just 0, is like a constant."},{"Start":"06:05.390 ","End":"06:07.235","Text":"This gives us A\u0027(T)."},{"Start":"06:07.235 ","End":"06:10.640","Text":"If we differentiate with respect to x twice,"},{"Start":"06:10.640 ","End":"06:15.230","Text":"this is 0, and this first derivative is 2 sine Tau,"},{"Start":"06:15.230 ","End":"06:17.179","Text":"second derivative is 0."},{"Start":"06:17.179 ","End":"06:18.410","Text":"So this is what we get,"},{"Start":"06:18.410 ","End":"06:22.295","Text":"which gives us that A\u0027(T) is 0."},{"Start":"06:22.295 ","End":"06:25.835","Text":"Now if A(0) is 0 and A\u0027(T) is 0,"},{"Start":"06:25.835 ","End":"06:29.569","Text":"follows that A(T) is 0."},{"Start":"06:29.569 ","End":"06:32.630","Text":"From this, we get that A(T) is a constant and that"},{"Start":"06:32.630 ","End":"06:36.080","Text":"constant has to be 0 because it has to work here also."},{"Start":"06:36.080 ","End":"06:38.780","Text":"If we put this A(T) here,"},{"Start":"06:38.780 ","End":"06:40.420","Text":"it gives us that p(x,"},{"Start":"06:40.420 ","End":"06:43.455","Text":"T) is 2x sine Tau."},{"Start":"06:43.455 ","End":"06:49.365","Text":"So that\u0027s P. Now we want to work our way back to p and then to u,"},{"Start":"06:49.365 ","End":"06:51.420","Text":"if that\u0027s a p, p(x,"},{"Start":"06:51.420 ","End":"06:55.110","Text":"t, Tau) is 2x sine Tau."},{"Start":"06:55.110 ","End":"06:56.930","Text":"There\u0027s no T here."},{"Start":"06:56.930 ","End":"06:58.850","Text":"It doesn\u0027t participate."},{"Start":"06:58.850 ","End":"07:00.445","Text":"Tau is like a constant,"},{"Start":"07:00.445 ","End":"07:02.270","Text":"so it\u0027s just the same p(x,"},{"Start":"07:02.270 ","End":"07:05.380","Text":"t, Tau) is also 2x sine Tau."},{"Start":"07:05.380 ","End":"07:09.860","Text":"What we have to do now is find u using this integral."},{"Start":"07:09.860 ","End":"07:19.995","Text":"It\u0027s the integral of p dTau from 0 to T. The integral of sine Tau is minus cosine Tau,"},{"Start":"07:19.995 ","End":"07:23.480","Text":"and then substitute these limits of integration."},{"Start":"07:23.480 ","End":"07:27.115","Text":"Just forget about the minus and subtract the other way."},{"Start":"07:27.115 ","End":"07:30.555","Text":"When Tau is 0, we get 1."},{"Start":"07:30.555 ","End":"07:32.190","Text":"When Tau is t,"},{"Start":"07:32.190 ","End":"07:35.280","Text":"we get cosine t minus,"},{"Start":"07:35.280 ","End":"07:36.570","Text":"there\u0027s 2x here,"},{"Start":"07:36.570 ","End":"07:38.700","Text":"and this is the answer,"},{"Start":"07:38.700 ","End":"07:42.370","Text":"and that completes this solution."}],"ID":30783},{"Watched":false,"Name":"Exercise 4","Duration":"9m 58s","ChapterTopicVideoID":29204,"CourseChapterTopicPlaylistID":294431,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.434","Text":"In this exercise, we\u0027re going to solve the following IBVP using Duhamel\u0027s principle."},{"Start":"00:06.434 ","End":"00:13.380","Text":"What we have here is a heat equation nonhomogeneous on a finite interval,"},{"Start":"00:13.380 ","End":"00:16.740","Text":"and we have homogeneous initial condition,"},{"Start":"00:16.740 ","End":"00:19.575","Text":"and homogeneous boundary conditions."},{"Start":"00:19.575 ","End":"00:22.830","Text":"The Duhamel principle says that we can convert"},{"Start":"00:22.830 ","End":"00:26.895","Text":"this problem to hopefully, a simpler problem."},{"Start":"00:26.895 ","End":"00:28.455","Text":"Is two parts to it,"},{"Start":"00:28.455 ","End":"00:33.600","Text":"there\u0027s a problem for p and then a solution u,"},{"Start":"00:33.600 ","End":"00:35.340","Text":"which is this integral."},{"Start":"00:35.340 ","End":"00:39.960","Text":"The problem for p is also a heat equation,"},{"Start":"00:39.960 ","End":"00:41.770","Text":"but homogeneous this time."},{"Start":"00:41.770 ","End":"00:46.940","Text":"In fact, the non-homogeneous part comes to the initial condition here."},{"Start":"00:46.940 ","End":"00:51.370","Text":"The boundary conditions become boundary conditions on p,"},{"Start":"00:51.370 ","End":"00:53.980","Text":"you just have the extra variable tau,"},{"Start":"00:53.980 ","End":"00:57.590","Text":"and since these are Von Neumann conditions,"},{"Start":"00:57.590 ","End":"01:00.275","Text":"these also a Von Neumann conditions."},{"Start":"01:00.275 ","End":"01:02.555","Text":"D by dx here and here,"},{"Start":"01:02.555 ","End":"01:07.085","Text":"and the domain that if t bigger or equal to 0,"},{"Start":"01:07.085 ","End":"01:09.020","Text":"we get t bigger or equal to tau,"},{"Start":"01:09.020 ","End":"01:11.450","Text":"so that the border is t=tau,"},{"Start":"01:11.450 ","End":"01:15.260","Text":"and initial condition is on this border."},{"Start":"01:15.260 ","End":"01:19.385","Text":"Now, a common strategy for this type of problem is to say,"},{"Start":"01:19.385 ","End":"01:21.605","Text":"let tau be fixed,"},{"Start":"01:21.605 ","End":"01:24.565","Text":"arbitrary but fixed for the moment."},{"Start":"01:24.565 ","End":"01:28.390","Text":"We\u0027ll substitute T=t minus tau,"},{"Start":"01:28.390 ","End":"01:35.980","Text":"and then we get a function P(x,t) such that P(x,t) is little p(x,t) plus tau,"},{"Start":"01:35.980 ","End":"01:44.060","Text":"tau, and the other way round p(x,tau) is P(x,t-tau)."},{"Start":"01:45.030 ","End":"01:55.480","Text":"This minus this is T. Now because T and t just differ by plus or minus a constant,"},{"Start":"01:55.480 ","End":"01:59.110","Text":"then the derivatives are basically the same,"},{"Start":"01:59.110 ","End":"02:00.790","Text":"and we\u0027ve checked this before,"},{"Start":"02:00.790 ","End":"02:03.730","Text":"that dp by dt,"},{"Start":"02:03.730 ","End":"02:05.770","Text":"the capital letters is p,"},{"Start":"02:05.770 ","End":"02:10.735","Text":"t. Similarly for x and x,x."},{"Start":"02:10.735 ","End":"02:18.360","Text":"What we get, this converts to the following problem in P. Well, it\u0027s basically the same,"},{"Start":"02:18.360 ","End":"02:22.400","Text":"and just note that t bigger or equal to tau,"},{"Start":"02:22.400 ","End":"02:27.185","Text":"convert to T big or equal to 0 because this is t minus tau."},{"Start":"02:27.185 ","End":"02:31.580","Text":"Now this kind of problem can be done by separation of variables."},{"Start":"02:31.580 ","End":"02:35.025","Text":"It\u0027s homogeneous finite interval,"},{"Start":"02:35.025 ","End":"02:38.465","Text":"and the boundary conditions are also homogeneous."},{"Start":"02:38.465 ","End":"02:42.245","Text":"In fact we even made a table for different possibilities."},{"Start":"02:42.245 ","End":"02:44.060","Text":"You might recall this."},{"Start":"02:44.060 ","End":"02:45.905","Text":"Anyway, in our case,"},{"Start":"02:45.905 ","End":"02:52.125","Text":"we have Von Neumann conditions for both the left and right end points,"},{"Start":"02:52.125 ","End":"02:55.040","Text":"and the general form of the solution is this,"},{"Start":"02:55.040 ","End":"02:58.220","Text":"where we just have to find the coefficients."},{"Start":"02:58.220 ","End":"03:01.865","Text":"Getting back, this is what we get."},{"Start":"03:01.865 ","End":"03:04.430","Text":"We know that a is 1 and L is 1."},{"Start":"03:04.430 ","End":"03:06.470","Text":"In our case, here is the a,"},{"Start":"03:06.470 ","End":"03:08.285","Text":"a^2 should be here,"},{"Start":"03:08.285 ","End":"03:11.020","Text":"and the L is here."},{"Start":"03:11.020 ","End":"03:13.610","Text":"It simplifies a bit to this."},{"Start":"03:13.610 ","End":"03:17.480","Text":"Now if we substitute t=0 here,"},{"Start":"03:17.480 ","End":"03:20.585","Text":"what we get is P(x,0) is the following."},{"Start":"03:20.585 ","End":"03:26.680","Text":"Basically, what happens is that this exponent drops out because it\u0027s e^0, which is 1."},{"Start":"03:26.680 ","End":"03:28.585","Text":"On the other hand,"},{"Start":"03:28.585 ","End":"03:31.480","Text":"P(x,0) we can get from here."},{"Start":"03:31.480 ","End":"03:34.915","Text":"Now we have two versions of P(x,0)."},{"Start":"03:34.915 ","End":"03:38.150","Text":"That means that we can use comparison of coefficients."},{"Start":"03:38.150 ","End":"03:40.070","Text":"They\u0027re both cosine series."},{"Start":"03:40.070 ","End":"03:43.535","Text":"This just has a single term where n=1."},{"Start":"03:43.535 ","End":"03:46.935","Text":"What we get is, when n=1,"},{"Start":"03:46.935 ","End":"03:50.315","Text":"here we have e^tau cosine 1Pix,"},{"Start":"03:50.315 ","End":"03:54.660","Text":"and here we have a1 sine 1Pix."},{"Start":"03:54.660 ","End":"03:58.185","Text":"So a_1 is e^tau,"},{"Start":"03:58.185 ","End":"04:01.135","Text":"and all the others are 0."},{"Start":"04:01.135 ","End":"04:09.740","Text":"Just summarizing, so far we have that P(x,t) is the following cosine series,"},{"Start":"04:09.740 ","End":"04:14.945","Text":"and we discovered that only a_1 is non-zero, it\u0027s e^tau."},{"Start":"04:14.945 ","End":"04:21.785","Text":"All the others are 0 and so we just have to take the n=1 term from here,"},{"Start":"04:21.785 ","End":"04:25.310","Text":"and we get that P(x,t) is just a_1,"},{"Start":"04:25.310 ","End":"04:26.980","Text":"times e to the,"},{"Start":"04:26.980 ","End":"04:31.595","Text":"just put n=1 here, and n=1 here."},{"Start":"04:31.595 ","End":"04:33.995","Text":"A_1 is e^tau,"},{"Start":"04:33.995 ","End":"04:36.400","Text":"cosine 1Pix is cosine Pix."},{"Start":"04:36.400 ","End":"04:38.790","Text":"This is what we have for P(x,t),"},{"Start":"04:38.790 ","End":"04:44.870","Text":"and we can return to p by letting T equal t minus tau."},{"Start":"04:44.870 ","End":"04:48.670","Text":"We get that P(x,t) tau is the following,"},{"Start":"04:48.670 ","End":"04:52.510","Text":"and the last step is to go back from p to u,"},{"Start":"04:52.510 ","End":"04:58.070","Text":"and we do that by means of this integral that\u0027s part of the Duhamel\u0027s principle."},{"Start":"04:58.070 ","End":"05:01.375","Text":"This is the integral we have to compute."},{"Start":"05:01.375 ","End":"05:04.490","Text":"We can take the stuff without tau in front of the"},{"Start":"05:04.490 ","End":"05:07.820","Text":"integral like the cosine Pix can go here,"},{"Start":"05:07.820 ","End":"05:14.255","Text":"and we can also take the e to the minus Pi^2t in front of the integral."},{"Start":"05:14.255 ","End":"05:17.615","Text":"What we\u0027re left with is e^tau,"},{"Start":"05:17.615 ","End":"05:18.800","Text":"and e to the minus,"},{"Start":"05:18.800 ","End":"05:21.290","Text":"minus Pi^2 tau,"},{"Start":"05:21.290 ","End":"05:25.880","Text":"which means e^1 plus Pi^2 tau, just algebra."},{"Start":"05:25.880 ","End":"05:31.670","Text":"Now, this integral is an exponent e to the power of a constant times tau."},{"Start":"05:31.670 ","End":"05:36.365","Text":"It\u0027s just 1 over that constant times the same thing."},{"Start":"05:36.365 ","End":"05:40.760","Text":"Now we have to substitute 0 and t and subtract."},{"Start":"05:40.760 ","End":"05:45.185","Text":"What we get is we put in tau=t."},{"Start":"05:45.185 ","End":"05:47.075","Text":"We get this."},{"Start":"05:47.075 ","End":"05:49.010","Text":"You put tau=0,"},{"Start":"05:49.010 ","End":"05:51.335","Text":"we get 1 on this part,"},{"Start":"05:51.335 ","End":"05:52.700","Text":"the same for both."},{"Start":"05:52.700 ","End":"05:54.215","Text":"We can simplify a bit,"},{"Start":"05:54.215 ","End":"05:59.590","Text":"take this e to the minus Pi^2 t and multiply it here and here."},{"Start":"05:59.590 ","End":"06:04.265","Text":"E to the minus Pi^2 t cancels out with the Pi^2 here just gives"},{"Start":"06:04.265 ","End":"06:10.295","Text":"e^t and also minus e to the minus Pi^2t."},{"Start":"06:10.295 ","End":"06:14.555","Text":"This will put on the denominator and we still have a cosine of Pix,"},{"Start":"06:14.555 ","End":"06:17.420","Text":"and that\u0027s the answer."},{"Start":"06:17.420 ","End":"06:21.215","Text":"That concludes this exercise but wait, don\u0027t go yet."},{"Start":"06:21.215 ","End":"06:22.400","Text":"Let\u0027s for a change,"},{"Start":"06:22.400 ","End":"06:26.420","Text":"do a verification of the solution if you\u0027re interested."},{"Start":"06:26.420 ","End":"06:31.710","Text":"Here\u0027s our solution, here\u0027s the IBVP."},{"Start":"06:31.710 ","End":"06:34.710","Text":"Let\u0027s see if this satisfies this."},{"Start":"06:34.710 ","End":"06:37.210","Text":"We have four equalities to check;"},{"Start":"06:37.210 ","End":"06:39.440","Text":"the PDE, the initial condition,"},{"Start":"06:39.440 ","End":"06:41.630","Text":"and two boundary conditions."},{"Start":"06:41.630 ","End":"06:44.165","Text":"Let\u0027s start with the PDE."},{"Start":"06:44.165 ","End":"06:46.400","Text":"We want to know if this equals this."},{"Start":"06:46.400 ","End":"06:52.340","Text":"We can just check that this minus this is this e^t cosine Pix."},{"Start":"06:52.340 ","End":"06:55.130","Text":"I will start with the left-hand side here,"},{"Start":"06:55.130 ","End":"06:58.295","Text":"and see if we can reach the right-hand side."},{"Start":"06:58.295 ","End":"07:01.415","Text":"Now, we need u_t and u_xx,"},{"Start":"07:01.415 ","End":"07:06.590","Text":"du by dt is just differentiating this bit with"},{"Start":"07:06.590 ","End":"07:12.095","Text":"respect to t. All we get is a minus Pi^2 from here,"},{"Start":"07:12.095 ","End":"07:14.590","Text":"comes here, makes it plus P^2i."},{"Start":"07:14.590 ","End":"07:16.185","Text":"Du by dx,"},{"Start":"07:16.185 ","End":"07:18.740","Text":"just differentiate the cosine Pix,"},{"Start":"07:18.740 ","End":"07:22.385","Text":"so we get minus sine Pix, and there\u0027s a Pi also."},{"Start":"07:22.385 ","End":"07:24.410","Text":"When we differentiate again,"},{"Start":"07:24.410 ","End":"07:27.755","Text":"the sine becomes cosine and we get an additional Pi,"},{"Start":"07:27.755 ","End":"07:29.555","Text":"so it\u0027s minus Pi^2."},{"Start":"07:29.555 ","End":"07:32.825","Text":"You want to check what is this, minus this."},{"Start":"07:32.825 ","End":"07:35.120","Text":"We have this here,"},{"Start":"07:35.120 ","End":"07:37.640","Text":"minus this, which means it\u0027s plus,"},{"Start":"07:37.640 ","End":"07:39.305","Text":"because there already is a minus."},{"Start":"07:39.305 ","End":"07:41.870","Text":"You can take out the common parts."},{"Start":"07:41.870 ","End":"07:44.105","Text":"We have a cosine Pix in both."},{"Start":"07:44.105 ","End":"07:49.500","Text":"We have 1 over 1 plus Pi^2 in both and the rest we can put in brackets."},{"Start":"07:49.500 ","End":"07:53.090","Text":"So we have the numerator is e^t plus P^2,"},{"Start":"07:53.090 ","End":"07:55.165","Text":"e to the minus Pi^2 t,"},{"Start":"07:55.165 ","End":"08:03.125","Text":"plus Pi^2 e^t minus e to the minus Pi^2 t. Now,"},{"Start":"08:03.125 ","End":"08:07.325","Text":"expand these brackets here and we get this,"},{"Start":"08:07.325 ","End":"08:09.430","Text":"and then stuff cancels."},{"Start":"08:09.430 ","End":"08:15.140","Text":"The Pi^2 e to the minus Pi^2 t cancels because here it\u0027s plus and here it\u0027s minus,"},{"Start":"08:15.140 ","End":"08:20.030","Text":"and also what we can do is take e^t outside the brackets."},{"Start":"08:20.030 ","End":"08:25.140","Text":"What remains when we have 1 plus Pi^2 times e^t,"},{"Start":"08:25.140 ","End":"08:29.055","Text":"1 plus Pi^2 cancels with 1 plus Pi^2,"},{"Start":"08:29.055 ","End":"08:34.265","Text":"and we get e^t times cosine Pi x,"},{"Start":"08:34.265 ","End":"08:38.490","Text":"which is exactly what we needed or is it?"},{"Start":"08:38.490 ","End":"08:41.700","Text":"Yeah, that\u0027s what we needed to show."},{"Start":"08:41.700 ","End":"08:44.465","Text":"So check, we\u0027re done with that part."},{"Start":"08:44.465 ","End":"08:47.615","Text":"Next, we\u0027ll check the initial condition,"},{"Start":"08:47.615 ","End":"08:54.980","Text":"when t=0, u(x,0) is just substitute t=0."},{"Start":"08:54.980 ","End":"08:57.350","Text":"E^0 is 1 here, also,"},{"Start":"08:57.350 ","End":"09:00.065","Text":"e^0 is 1,1 minus 1 is 0."},{"Start":"09:00.065 ","End":"09:02.870","Text":"This comes out to be 0, so that\u0027s good."},{"Start":"09:02.870 ","End":"09:07.010","Text":"Now we need the boundary conditions but for"},{"Start":"09:07.010 ","End":"09:11.140","Text":"that we\u0027ll need the derivative with respect to x,"},{"Start":"09:11.140 ","End":"09:15.530","Text":"because we have a Von Neumann condition and we computed this above,"},{"Start":"09:15.530 ","End":"09:17.025","Text":"so just copying it."},{"Start":"09:17.025 ","End":"09:21.735","Text":"We have to show two things that u_x( 0,t) is 0,"},{"Start":"09:21.735 ","End":"09:24.540","Text":"and u_x(1,t) is 0."},{"Start":"09:24.540 ","End":"09:28.930","Text":"Now, substituting x=0 here we get,"},{"Start":"09:28.940 ","End":"09:33.170","Text":"well, we have a sine of Pi times 0 here,"},{"Start":"09:33.170 ","End":"09:34.790","Text":"and that\u0027s sine of 0 is 0."},{"Start":"09:34.790 ","End":"09:36.530","Text":"So this is all 0."},{"Start":"09:36.530 ","End":"09:38.375","Text":"That\u0027s good, that\u0027s what we wanted,"},{"Start":"09:38.375 ","End":"09:43.405","Text":"and as for the other one, we put x=1."},{"Start":"09:43.405 ","End":"09:46.230","Text":"So we get sine of Pi times 1,"},{"Start":"09:46.230 ","End":"09:50.230","Text":"sine of Pi is 0."},{"Start":"09:50.230 ","End":"09:53.070","Text":"Again, we get 0, which is good."},{"Start":"09:53.070 ","End":"09:58.330","Text":"That concludes the checking of the solution, and now we\u0027re really done."}],"ID":30784},{"Watched":false,"Name":"Exercise 5","Duration":"5m 13s","ChapterTopicVideoID":29197,"CourseChapterTopicPlaylistID":294431,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.610","Text":"In this exercise, we\u0027re going to solve"},{"Start":"00:02.610 ","End":"00:07.485","Text":"the following initial boundary value problem using Duhamel\u0027s principle."},{"Start":"00:07.485 ","End":"00:09.588","Text":"This is the problem,"},{"Start":"00:09.588 ","End":"00:13.469","Text":"so heat equation with a non-homogeneous part."},{"Start":"00:13.469 ","End":"00:18.293","Text":"But the initial and boundary conditions are homogeneous,"},{"Start":"00:18.293 ","End":"00:20.280","Text":"and it\u0027s on a finite interval."},{"Start":"00:20.280 ","End":"00:23.715","Text":"We can use the Duhamel\u0027s principle,"},{"Start":"00:23.715 ","End":"00:33.060","Text":"which expresses you as an integral of a function p(xt) and Tau,"},{"Start":"00:33.060 ","End":"00:38.415","Text":"and p is a solution to a hopefully simpler problem, which is this."},{"Start":"00:38.415 ","End":"00:44.255","Text":"Note that non-homogeneous part of the PDE becomes"},{"Start":"00:44.255 ","End":"00:50.900","Text":"the initial condition and the boundary conditions are still homogeneous."},{"Start":"00:50.900 ","End":"00:56.000","Text":"The domain changes to t bigger or equal to Tau,"},{"Start":"00:56.000 ","End":"01:01.055","Text":"so that the initial condition really is on a border to a T equals Tau."},{"Start":"01:01.055 ","End":"01:04.420","Text":"There is a standard technique which is often used in this kind of problem,"},{"Start":"01:04.420 ","End":"01:11.220","Text":"and that is to say that Tau is fixed for the moment and later we\u0027ll let it vary again."},{"Start":"01:11.220 ","End":"01:18.890","Text":"Then we\u0027ll let T equal t minus Tau and we get a function p(xt) such that well,"},{"Start":"01:18.890 ","End":"01:22.200","Text":"as follows, we\u0027ve seen this before."},{"Start":"01:22.580 ","End":"01:28.850","Text":"Also, previously we showed that the partial derivatives with respect to T"},{"Start":"01:28.850 ","End":"01:35.150","Text":"and xx are the same for P and p. In short,"},{"Start":"01:35.150 ","End":"01:39.500","Text":"this initial condition stays what looks the"},{"Start":"01:39.500 ","End":"01:44.065","Text":"same except that capital letters here is that the little letters."},{"Start":"01:44.065 ","End":"01:47.210","Text":"Here instead of t, we have Tau,"},{"Start":"01:47.210 ","End":"01:49.910","Text":"because if T is 0,"},{"Start":"01:49.910 ","End":"01:52.430","Text":"that means that t minus Tau is 0."},{"Start":"01:52.430 ","End":"01:54.920","Text":"That means that T equals Tau."},{"Start":"01:54.920 ","End":"02:01.650","Text":"We get Tau instead of T. The boundary conditions stay the same basically."},{"Start":"02:01.650 ","End":"02:05.685","Text":"X is 0 in the case of p,"},{"Start":"02:05.685 ","End":"02:09.390","Text":"it\u0027s just x equals 0 in the case of P and similarly"},{"Start":"02:09.390 ","End":"02:13.415","Text":"for L and L. Now this problem is familiar,"},{"Start":"02:13.415 ","End":"02:16.925","Text":"and we\u0027ve found already a general solution."},{"Start":"02:16.925 ","End":"02:18.830","Text":"We really do have an equation of"},{"Start":"02:18.830 ","End":"02:22.460","Text":"this form and we\u0027re given an initial condition as follows."},{"Start":"02:22.460 ","End":"02:28.115","Text":"In our case we have direct light type for the left endpoint and the right end point."},{"Start":"02:28.115 ","End":"02:30.035","Text":"This is the formula we\u0027ll use,"},{"Start":"02:30.035 ","End":"02:33.475","Text":"instead of AN and we\u0027ll call it bn."},{"Start":"02:33.475 ","End":"02:41.645","Text":"This is what we have just using that formula with T and t and b instead of H,"},{"Start":"02:41.645 ","End":"02:44.515","Text":"just change in notation."},{"Start":"02:44.515 ","End":"02:50.610","Text":"Here we are again. Let\u0027s substitute t=0."},{"Start":"02:50.610 ","End":"02:53.483","Text":"We\u0027ll get that p(x,"},{"Start":"02:53.483 ","End":"02:55.505","Text":"0) equals the following."},{"Start":"02:55.505 ","End":"03:00.439","Text":"Basically what\u0027s happened is this has dropped out because if t is 0,"},{"Start":"03:00.439 ","End":"03:01.955","Text":"then the exponent is 0,"},{"Start":"03:01.955 ","End":"03:03.665","Text":"e to the 0 is 1."},{"Start":"03:03.665 ","End":"03:05.705","Text":"This is what we have."},{"Start":"03:05.705 ","End":"03:09.190","Text":"On the other hand, p(x,"},{"Start":"03:09.190 ","End":"03:13.010","Text":"0) is given by the initial condition as follows."},{"Start":"03:13.010 ","End":"03:19.205","Text":"If we look at this sine series and this is also a sine series with just 1 term."},{"Start":"03:19.205 ","End":"03:22.710","Text":"It\u0027s the term corresponding to n=1."},{"Start":"03:22.710 ","End":"03:29.610","Text":"We can use the comparison of coefficients method and get b1 and all the other b\u0027s,"},{"Start":"03:29.610 ","End":"03:31.035","Text":"which will be 0."},{"Start":"03:31.035 ","End":"03:34.365","Text":"We have b1, n=1."},{"Start":"03:34.365 ","End":"03:35.850","Text":"We have b1,"},{"Start":"03:35.850 ","End":"03:38.970","Text":"and it\u0027s equal to this."},{"Start":"03:38.970 ","End":"03:41.430","Text":"The bn\u0027s are are 0."},{"Start":"03:41.430 ","End":"03:44.205","Text":"Let\u0027s return to this series."},{"Start":"03:44.205 ","End":"03:46.205","Text":"I\u0027ve just copied it."},{"Start":"03:46.205 ","End":"03:50.980","Text":"Now we can substitute b1 and bn."},{"Start":"03:50.980 ","End":"03:55.925","Text":"What we get is just a single term where n=1."},{"Start":"03:55.925 ","End":"03:57.320","Text":"We get b1,"},{"Start":"03:57.320 ","End":"03:58.795","Text":"which is this,"},{"Start":"03:58.795 ","End":"04:04.670","Text":"e to the minus Pi a over L^2 sine Pi x over"},{"Start":"04:04.670 ","End":"04:10.735","Text":"L. We can combine these 2 exponents using rules of exponents."},{"Start":"04:10.735 ","End":"04:14.255","Text":"We get t plus Tau here."},{"Start":"04:14.255 ","End":"04:15.920","Text":"Just think about it."},{"Start":"04:15.920 ","End":"04:18.395","Text":"T plus Tau and then minus these,"},{"Start":"04:18.395 ","End":"04:23.555","Text":"and T plus Tau is t. By the definition,"},{"Start":"04:23.555 ","End":"04:26.285","Text":"we had that T,"},{"Start":"04:26.285 ","End":"04:28.505","Text":"is t minus Tau."},{"Start":"04:28.505 ","End":"04:33.710","Text":"T plus Tau is t. Now that we have P, we\u0027re almost there,"},{"Start":"04:33.710 ","End":"04:38.345","Text":"we just have to take an integral of that to get u. U is the integral of"},{"Start":"04:38.345 ","End":"04:45.665","Text":"pd Tau from 0 to t. This is equal to just copying what P is from here."},{"Start":"04:45.665 ","End":"04:47.600","Text":"We can take in front of the integral,"},{"Start":"04:47.600 ","End":"04:50.194","Text":"anything that doesn\u0027t involve Tau,"},{"Start":"04:50.194 ","End":"04:53.780","Text":"which is this sine and"},{"Start":"04:53.780 ","End":"04:59.260","Text":"this e to the power of none of it involves Tau is basically a constant."},{"Start":"04:59.260 ","End":"05:01.715","Text":"We have the integral from 0 to t d Tau,"},{"Start":"05:01.715 ","End":"05:08.855","Text":"which is just t. I don\u0027t read it out."},{"Start":"05:08.855 ","End":"05:11.170","Text":"This is the answer for u(x,"},{"Start":"05:11.170 ","End":"05:13.890","Text":"t) and we are done."}],"ID":30785}],"Thumbnail":null,"ID":294431},{"Name":"The Maximum and Minimum Principles","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Maximum and Minimum Principles","Duration":"7m 52s","ChapterTopicVideoID":29190,"CourseChapterTopicPlaylistID":294432,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Now we come to a new topic,"},{"Start":"00:02.055 ","End":"00:07.035","Text":"the maximum and minimum principle in the heat equation."},{"Start":"00:07.035 ","End":"00:09.540","Text":"There\u0027s 2 separate principles,"},{"Start":"00:09.540 ","End":"00:12.375","Text":"there\u0027s the maximum principal and a minimum principle."},{"Start":"00:12.375 ","End":"00:14.970","Text":"I will start with the maximum principle."},{"Start":"00:14.970 ","End":"00:17.040","Text":"Let\u0027s take this as a theorem,"},{"Start":"00:17.040 ","End":"00:21.990","Text":"we suppose that u(x,t) is continuous in the following rectangle,"},{"Start":"00:21.990 ","End":"00:25.080","Text":"the x-direction is from 0 to L and"},{"Start":"00:25.080 ","End":"00:30.750","Text":"the T direction is from T_naught to T. Quite commonly, T_naught is 0."},{"Start":"00:30.750 ","End":"00:33.270","Text":"This u(x,t) has to satisfy"},{"Start":"00:33.270 ","End":"00:38.945","Text":"the homogeneous heat equation in the interior of R. The rectangle without the boundary."},{"Start":"00:38.945 ","End":"00:40.715","Text":"Here\u0027s a picture."},{"Start":"00:40.715 ","End":"00:44.660","Text":"There\u0027s a definition of something called a parabolic boundary,"},{"Start":"00:44.660 ","End":"00:46.640","Text":"but don\u0027t worry about that."},{"Start":"00:46.640 ","End":"00:48.425","Text":"I\u0027ll just tell you that in this case,"},{"Start":"00:48.425 ","End":"00:53.930","Text":"it comes out to be the bottom and the 2 sides,"},{"Start":"00:53.930 ","End":"00:57.665","Text":"just the top is missing from the usual boundary,"},{"Start":"00:57.665 ","End":"00:59.780","Text":"made up of 3 parts here."},{"Start":"00:59.780 ","End":"01:03.665","Text":"This part is where x equals T_naught,"},{"Start":"01:03.665 ","End":"01:05.990","Text":"this part is where t is 0,"},{"Start":"01:05.990 ","End":"01:10.646","Text":"and this part is where t is equal to L. We can write it more detail,"},{"Start":"01:10.646 ","End":"01:12.290","Text":"more formally, as follows,"},{"Start":"01:12.290 ","End":"01:15.560","Text":"L is the union of 3 line segments."},{"Start":"01:15.560 ","End":"01:17.075","Text":"This is the bottom,"},{"Start":"01:17.075 ","End":"01:18.940","Text":"this is the left,"},{"Start":"01:18.940 ","End":"01:20.765","Text":"and this is the right."},{"Start":"01:20.765 ","End":"01:24.710","Text":"Maybe I give you 1 more example of a parabolic boundary in a higher dimension."},{"Start":"01:24.710 ","End":"01:26.420","Text":"Suppose we have x, y,"},{"Start":"01:26.420 ","End":"01:28.850","Text":"and t. The base in the x,"},{"Start":"01:28.850 ","End":"01:30.330","Text":"y plane was a circle,"},{"Start":"01:30.330 ","End":"01:31.965","Text":"we\u0027d get a cylinder,"},{"Start":"01:31.965 ","End":"01:36.110","Text":"and the parabolic boundary is the boundary,"},{"Start":"01:36.110 ","End":"01:38.270","Text":"but without the top, without the lid,"},{"Start":"01:38.270 ","End":"01:41.510","Text":"it\u0027s got the base and the sides, but no lid."},{"Start":"01:41.510 ","End":"01:42.860","Text":"It\u0027s just another example,"},{"Start":"01:42.860 ","End":"01:44.405","Text":"let\u0027s not dwell on this term."},{"Start":"01:44.405 ","End":"01:46.565","Text":"Still haven\u0027t stated what the theorem says."},{"Start":"01:46.565 ","End":"01:51.530","Text":"Given all this rectangle R and u(x,t) which"},{"Start":"01:51.530 ","End":"01:56.650","Text":"satisfies the homogeneous heat equation in the interior,"},{"Start":"01:56.650 ","End":"02:04.205","Text":"what it says is that the maximum of u is obtained on the parabolic boundary on L,"},{"Start":"02:04.205 ","End":"02:06.605","Text":"which is made up of these 3 segments."},{"Start":"02:06.605 ","End":"02:09.860","Text":"In other words, the maximum on all of R,"},{"Start":"02:09.860 ","End":"02:18.985","Text":"is the same as the maximum on just L. The maximum doesn\u0027t mean absolute maximum,"},{"Start":"02:18.985 ","End":"02:21.305","Text":"could be a tie, in fact,"},{"Start":"02:21.305 ","End":"02:24.185","Text":"could even be that the function is a constant."},{"Start":"02:24.185 ","End":"02:26.480","Text":"Now, an addition to this,"},{"Start":"02:26.480 ","End":"02:31.145","Text":"that if 1 of the interior points is a maximum,"},{"Start":"02:31.145 ","End":"02:35.285","Text":"then it has to be that the function u is a constant."},{"Start":"02:35.285 ","End":"02:37.640","Text":"If u is not a constant function,"},{"Start":"02:37.640 ","End":"02:41.900","Text":"then the maximum will be somewhere on the parabolic boundary,"},{"Start":"02:41.900 ","End":"02:48.425","Text":"and that will be strictly bigger than all the values inside the rectangle."},{"Start":"02:48.425 ","End":"02:52.505","Text":"Let\u0027s just give more detail what happens in an actual heat equation."},{"Start":"02:52.505 ","End":"02:56.750","Text":"Suppose that we have the following heat equation."},{"Start":"02:56.750 ","End":"03:03.727","Text":"U_t equals a^2 u_xx and x is defined on this interval, t on this interval."},{"Start":"03:03.727 ","End":"03:06.395","Text":"Together x, t is on the rectangle,"},{"Start":"03:06.395 ","End":"03:14.450","Text":"and that we\u0027re given the value of u along the base and along the 2 sides."},{"Start":"03:14.450 ","End":"03:19.445","Text":"Getting back the value of u is g(t) here,"},{"Start":"03:19.445 ","End":"03:22.925","Text":"h(t) here, and f(x) here."},{"Start":"03:22.925 ","End":"03:27.620","Text":"We have these 3 functions for the 3 sides or 2 sides and a base."},{"Start":"03:27.620 ","End":"03:31.010","Text":"Another way of phrasing the maximum principle is that"},{"Start":"03:31.010 ","End":"03:35.315","Text":"the maximum overall points in R of u,"},{"Start":"03:35.315 ","End":"03:40.175","Text":"is the biggest of the following 3 maximums."},{"Start":"03:40.175 ","End":"03:43.220","Text":"The maximum along the base,"},{"Start":"03:43.220 ","End":"03:46.145","Text":"the maximum on the left side,"},{"Start":"03:46.145 ","End":"03:48.560","Text":"and the maximum on the right side."},{"Start":"03:48.560 ","End":"03:50.803","Text":"So you can compute these 3,"},{"Start":"03:50.803 ","End":"03:53.855","Text":"and then take the largest of these 3."},{"Start":"03:53.855 ","End":"03:56.525","Text":"That\u0027s the maximum overall, the rectangle."},{"Start":"03:56.525 ","End":"03:58.400","Text":"That\u0027s it for the maximum principal,"},{"Start":"03:58.400 ","End":"04:00.950","Text":"there will be examples and that will become clearer."},{"Start":"04:00.950 ","End":"04:05.735","Text":"The minimum principle is just the same as the maximum principle,"},{"Start":"04:05.735 ","End":"04:10.340","Text":"but just replaced the word maximum everywhere with minimum,"},{"Start":"04:10.340 ","End":"04:14.510","Text":"and replace the function max with the function min,"},{"Start":"04:14.510 ","End":"04:16.550","Text":"where we see it."},{"Start":"04:16.550 ","End":"04:19.370","Text":"Let\u0027s do an example."},{"Start":"04:19.370 ","End":"04:22.340","Text":"Now, here\u0027s an example where we\u0027ll use"},{"Start":"04:22.340 ","End":"04:25.190","Text":"both the maximum principal and the minimum principle."},{"Start":"04:25.190 ","End":"04:30.335","Text":"We\u0027re given an initial boundary value problem as follows."},{"Start":"04:30.335 ","End":"04:32.690","Text":"We have a heat equation,"},{"Start":"04:32.690 ","End":"04:34.505","Text":"an initial condition,"},{"Start":"04:34.505 ","End":"04:36.710","Text":"and 2 boundary conditions,"},{"Start":"04:36.710 ","End":"04:42.920","Text":"and the domain is on a finite interval for x and t bigger or equal to 0."},{"Start":"04:42.920 ","End":"04:45.770","Text":"This is actually an infinite strip,"},{"Start":"04:45.770 ","End":"04:47.390","Text":"as in the picture."},{"Start":"04:47.390 ","End":"04:50.150","Text":"This includes the boundary,"},{"Start":"04:50.150 ","End":"04:53.975","Text":"but we\u0027ll let D be the interior of that."},{"Start":"04:53.975 ","End":"04:59.300","Text":"In other words, just the part inside without the boundary lines."},{"Start":"04:59.300 ","End":"05:01.775","Text":"We have to show that for every point x,"},{"Start":"05:01.775 ","End":"05:05.157","Text":"t in the interior u of x,"},{"Start":"05:05.157 ","End":"05:08.495","Text":"t is strictly between 0 and 1."},{"Start":"05:08.495 ","End":"05:10.303","Text":"Let\u0027s take an arbitrary x,"},{"Start":"05:10.303 ","End":"05:13.490","Text":"t and we\u0027ll call it x_naught, t_naught."},{"Start":"05:13.490 ","End":"05:15.290","Text":"It\u0027s indeed the interior,"},{"Start":"05:15.290 ","End":"05:19.550","Text":"so we know that x_naught is between 0 and 1 strictly,"},{"Start":"05:19.550 ","End":"05:22.190","Text":"and t_naught is strictly positive."},{"Start":"05:22.190 ","End":"05:27.455","Text":"We want to show what it says here for our x_naught, t_naught."},{"Start":"05:27.455 ","End":"05:31.555","Text":"Choose some T bigger than t_naught,"},{"Start":"05:31.555 ","End":"05:32.885","Text":"like in the picture,"},{"Start":"05:32.885 ","End":"05:35.720","Text":"this dotted line is higher than this point,"},{"Start":"05:35.720 ","End":"05:39.095","Text":"and we\u0027ll let R be the rectangle,"},{"Start":"05:39.095 ","End":"05:43.475","Text":"including the sides this, this, this, this."},{"Start":"05:43.475 ","End":"05:46.625","Text":"This is the same R that we used in the theorem."},{"Start":"05:46.625 ","End":"05:53.840","Text":"Note, that u is not constant in the rectangle R, because for example,"},{"Start":"05:53.840 ","End":"05:56.105","Text":"it\u0027s not constant along the base,"},{"Start":"05:56.105 ","End":"05:59.075","Text":"here it\u0027s equal to sine Pix,"},{"Start":"05:59.075 ","End":"06:02.335","Text":"and the sine function is not a constant."},{"Start":"06:02.335 ","End":"06:05.600","Text":"Here it\u0027s equal to 0 in the middle,"},{"Start":"06:05.600 ","End":"06:07.730","Text":"at half it\u0027s equal to 1,"},{"Start":"06:07.730 ","End":"06:10.670","Text":"and down to 0 again, like so."},{"Start":"06:10.670 ","End":"06:11.930","Text":"It\u0027s not a constant,"},{"Start":"06:11.930 ","End":"06:17.150","Text":"and that tells us that u at this point is strictly less"},{"Start":"06:17.150 ","End":"06:22.765","Text":"than the maximum of u along this parabolic boundary,"},{"Start":"06:22.765 ","End":"06:27.560","Text":"and it\u0027s going to be strictly bigger than"},{"Start":"06:27.560 ","End":"06:33.440","Text":"the minimum of u along these 3 line segments or the parabolic boundary."},{"Start":"06:33.440 ","End":"06:36.600","Text":"To rephrase that u of x_naught, t_naught,"},{"Start":"06:36.600 ","End":"06:38.775","Text":"is bigger than m,"},{"Start":"06:38.775 ","End":"06:40.545","Text":"and less than M,"},{"Start":"06:40.545 ","End":"06:45.710","Text":"little m and big M, where m is the minimum of u along L,"},{"Start":"06:45.710 ","End":"06:46.850","Text":"the parabolic boundary,"},{"Start":"06:46.850 ","End":"06:48.440","Text":"and M is the maximum."},{"Start":"06:48.440 ","End":"06:51.800","Text":"L is this and this and this."},{"Start":"06:51.800 ","End":"06:54.410","Text":"Let\u0027s see what happens on each of these 3."},{"Start":"06:54.410 ","End":"06:56.608","Text":"On the base, well,"},{"Start":"06:56.608 ","End":"06:59.340","Text":"you can look at this picture here,"},{"Start":"06:59.340 ","End":"07:03.550","Text":"the minimum is 0 and the maximum is 1."},{"Start":"07:03.550 ","End":"07:05.885","Text":"On the 2 sides,"},{"Start":"07:05.885 ","End":"07:08.765","Text":"u is just 0 everywhere."},{"Start":"07:08.765 ","End":"07:13.655","Text":"So the very smallest it can be on all these 3 is 0,"},{"Start":"07:13.655 ","End":"07:17.030","Text":"either from here or from these 2 points."},{"Start":"07:17.030 ","End":"07:20.900","Text":"The maximum is 1 at the point 1/2,"},{"Start":"07:20.900 ","End":"07:24.005","Text":"0 that\u0027s this one here."},{"Start":"07:24.005 ","End":"07:27.950","Text":"So that m is 0 and M is 1."},{"Start":"07:27.950 ","End":"07:29.270","Text":"We have that u of x_naught,"},{"Start":"07:29.270 ","End":"07:32.585","Text":"t_naught is strictly between 0 and 1,"},{"Start":"07:32.585 ","End":"07:35.450","Text":"and this is an arbitrary point x_naught,"},{"Start":"07:35.450 ","End":"07:39.716","Text":"t_naught in D. So we can drop the subscripts and say that"},{"Start":"07:39.716 ","End":"07:44.130","Text":"u(x,t) is between 0 and 1 on all of D,"},{"Start":"07:44.130 ","End":"07:52.300","Text":"meaning on all x, t in D. That concludes this example, and this clip."}],"ID":30772},{"Watched":false,"Name":"Exercise 1","Duration":"3m 35s","ChapterTopicVideoID":29191,"CourseChapterTopicPlaylistID":294432,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.775","Text":"In this exercise we\u0027re given two initial boundary value problems here and here,"},{"Start":"00:05.775 ","End":"00:09.180","Text":"both on the domain x from 0-1,"},{"Start":"00:09.180 ","End":"00:14.565","Text":"t positive, and we have to determine which of these two expressions is greater,"},{"Start":"00:14.565 ","End":"00:18.600","Text":"u at the point of half a half or V at the same point."},{"Start":"00:18.600 ","End":"00:23.910","Text":"The way we\u0027ll tackle this is to define a third function w to be u minus"},{"Start":"00:23.910 ","End":"00:30.960","Text":"v. And then the PDE will be homogeneous because this is the same as this."},{"Start":"00:30.960 ","End":"00:33.720","Text":"Let\u0027s just start doing that."},{"Start":"00:33.720 ","End":"00:39.270","Text":"Copying u_t and v_t from here and here, we subtract them."},{"Start":"00:39.270 ","End":"00:46.215","Text":"We get that u_t minus v_t is u_xx minus v_xx."},{"Start":"00:46.215 ","End":"00:50.550","Text":"This gives us that w_t equals w_xx,"},{"Start":"00:50.550 ","End":"00:57.570","Text":"because w is u minus v. The initial value w(x,0) is u(x,0) minus v(x,0)."},{"Start":"00:57.570 ","End":"00:59.570","Text":"This is x^2, this is x."},{"Start":"00:59.570 ","End":"01:01.010","Text":"So we get this."},{"Start":"01:01.010 ","End":"01:05.190","Text":"And then the boundary value w(0,t) would"},{"Start":"01:05.190 ","End":"01:10.070","Text":"be t^2 minus t. And the same thing when x equals 1,"},{"Start":"01:10.070 ","End":"01:11.615","Text":"the other boundary value."},{"Start":"01:11.615 ","End":"01:17.570","Text":"Let\u0027s rewrite this slightly simpler w_t is w_xx,"},{"Start":"01:17.570 ","End":"01:24.255","Text":"w(x,0) is x times x minus 1 and the other two are t(t-1)."},{"Start":"01:24.255 ","End":"01:29.570","Text":"So PDE initial value and two boundary values."},{"Start":"01:29.570 ","End":"01:31.400","Text":"Here\u0027s a picture."},{"Start":"01:31.400 ","End":"01:37.040","Text":"We want to bound this point inside a rectangle so we can use the maximum principle."},{"Start":"01:37.040 ","End":"01:40.670","Text":"So let\u0027s take this t equals 1,"},{"Start":"01:40.670 ","End":"01:42.200","Text":"is a nice round number,"},{"Start":"01:42.200 ","End":"01:43.760","Text":"is bigger than a 0.5."},{"Start":"01:43.760 ","End":"01:47.180","Text":"So we\u0027ll let R be this rectangle where x is"},{"Start":"01:47.180 ","End":"01:51.355","Text":"between 0 and 1 and also t is between 0 and 1."},{"Start":"01:51.355 ","End":"01:54.860","Text":"Then this is the initial condition and these are"},{"Start":"01:54.860 ","End":"01:59.380","Text":"the two boundary conditions on each of these three sides,"},{"Start":"01:59.380 ","End":"02:05.315","Text":"which we\u0027ll denote as L and that\u0027s what we call the parabolic boundary."},{"Start":"02:05.315 ","End":"02:07.700","Text":"It\u0027s the boundary without the top."},{"Start":"02:07.700 ","End":"02:11.390","Text":"We want the maximum on the parabolic boundary,"},{"Start":"02:11.390 ","End":"02:16.160","Text":"which means the maximum on these three segments to take the maximum on each,"},{"Start":"02:16.160 ","End":"02:18.260","Text":"then take the greatest of the three."},{"Start":"02:18.260 ","End":"02:19.999","Text":"These are the three segments."},{"Start":"02:19.999 ","End":"02:25.675","Text":"And here w is equal to x times x minus 1."},{"Start":"02:25.675 ","End":"02:28.760","Text":"Then these two are equal to t(t-1),"},{"Start":"02:28.760 ","End":"02:30.995","Text":"which is essentially the same function."},{"Start":"02:30.995 ","End":"02:33.190","Text":"Let\u0027s see a picture of it."},{"Start":"02:33.190 ","End":"02:37.757","Text":"Here\u0027s a picture of either x(x-1) or t(t-1)."},{"Start":"02:37.757 ","End":"02:40.970","Text":"It\u0027s negative except at the end point where it\u0027s equal to 0."},{"Start":"02:40.970 ","End":"02:43.655","Text":"These are the maximum points."},{"Start":"02:43.655 ","End":"02:50.170","Text":"So the maximum is 0 on all of the parabolic boundary."},{"Start":"02:50.170 ","End":"02:53.445","Text":"The maximum of w on L is 0."},{"Start":"02:53.445 ","End":"02:56.930","Text":"Now, w is not a constant function."},{"Start":"02:56.930 ","End":"02:58.790","Text":"I mean, for example,"},{"Start":"02:58.790 ","End":"03:02.555","Text":"on this side it\u0027s x(x-1) is certainly not a constant."},{"Start":"03:02.555 ","End":"03:08.330","Text":"That means that any point in the interior has a value strictly"},{"Start":"03:08.330 ","End":"03:13.940","Text":"less than the maximum on L. Half a half is in the interior,"},{"Start":"03:13.940 ","End":"03:17.150","Text":"so w of this point is less than 0,"},{"Start":"03:17.150 ","End":"03:18.665","Text":"which is the maximum."},{"Start":"03:18.665 ","End":"03:23.745","Text":"But w is u minus v. So this is less than 0,"},{"Start":"03:23.745 ","End":"03:28.235","Text":"which gives us that u of a half a half is less than v of a half a half."},{"Start":"03:28.235 ","End":"03:30.110","Text":"And this is what we were asked to determine,"},{"Start":"03:30.110 ","End":"03:31.790","Text":"which of these two is greater."},{"Start":"03:31.790 ","End":"03:33.890","Text":"So I guess v is greater."},{"Start":"03:33.890 ","End":"03:36.390","Text":"Okay that concludes this clip."}],"ID":30773},{"Watched":false,"Name":"Exercise 2","Duration":"3m 13s","ChapterTopicVideoID":29192,"CourseChapterTopicPlaylistID":294432,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.575","Text":"In this exercise, we\u0027re given 2 initial boundary and value problems,"},{"Start":"00:04.575 ","End":"00:06.705","Text":"one with u and one with v,"},{"Start":"00:06.705 ","End":"00:10.155","Text":"and we have to decide which is greater,"},{"Start":"00:10.155 ","End":"00:12.420","Text":"u of a 1/2 and 1,"},{"Start":"00:12.420 ","End":"00:14.670","Text":"or v of a 1/2 and 1."},{"Start":"00:14.670 ","End":"00:17.805","Text":"What we\u0027ll do is subtract this minus this."},{"Start":"00:17.805 ","End":"00:20.955","Text":"In other words, we\u0027ll define a new function w,"},{"Start":"00:20.955 ","End":"00:23.505","Text":"which is u minus v,"},{"Start":"00:23.505 ","End":"00:30.840","Text":"and then the PDE will become homogeneous because it\u0027s the same non-homogeneous part."},{"Start":"00:30.840 ","End":"00:35.065","Text":"Copying the 2 PDEs from here and here,"},{"Start":"00:35.065 ","End":"00:37.865","Text":"we see that if we subtract them,"},{"Start":"00:37.865 ","End":"00:43.115","Text":"u_t minus v_t is u_xx minus v_xx, and this cancels."},{"Start":"00:43.115 ","End":"00:47.075","Text":"What we get is that w_t, which is this,"},{"Start":"00:47.075 ","End":"00:50.611","Text":"equals w_xx, w(x,"},{"Start":"00:50.611 ","End":"00:52.850","Text":"0) is this minus this, and w(0,"},{"Start":"00:52.850 ","End":"00:58.640","Text":"t) and w(1,"},{"Start":"00:58.640 ","End":"01:01.220","Text":"t) are equal to this minus this,"},{"Start":"01:01.220 ","End":"01:06.140","Text":"which makes it a plus sine^2 Pi t. Now we"},{"Start":"01:06.140 ","End":"01:08.300","Text":"want to get this point inside"},{"Start":"01:08.300 ","End":"01:11.270","Text":"a rectangle because we\u0027re going to use the minimum principle."},{"Start":"01:11.270 ","End":"01:15.235","Text":"The height of the rectangle has to be strictly bigger than 1."},{"Start":"01:15.235 ","End":"01:17.230","Text":"Let\u0027s make it 2."},{"Start":"01:17.230 ","End":"01:24.915","Text":"Let\u0027s take x between 0 and 1 and t between 0 and 2."},{"Start":"01:24.915 ","End":"01:27.740","Text":"Then our point is smack in the middle of this rectangle,"},{"Start":"01:27.740 ","End":"01:29.365","Text":"certainly in the interior."},{"Start":"01:29.365 ","End":"01:32.300","Text":"We\u0027ll let L be the parabolic boundary,"},{"Start":"01:32.300 ","End":"01:36.215","Text":"which is the boundary excluding the top 3 sides."},{"Start":"01:36.215 ","End":"01:40.370","Text":"We\u0027ve got the value of w on each of these 3 sides."},{"Start":"01:40.370 ","End":"01:43.520","Text":"The minimum will be the minimum on"},{"Start":"01:43.520 ","End":"01:47.125","Text":"each of the 3 sides and then take the least of those 3."},{"Start":"01:47.125 ","End":"01:51.435","Text":"This is just in technical language what L is."},{"Start":"01:51.435 ","End":"01:52.920","Text":"On the bottom,"},{"Start":"01:52.920 ","End":"01:56.400","Text":"value of w is sine Pi x,"},{"Start":"01:56.400 ","End":"01:59.010","Text":"and x is between 0 and 1."},{"Start":"01:59.010 ","End":"02:02.370","Text":"Here it\u0027s sine^2 Pi t,"},{"Start":"02:02.370 ","End":"02:03.900","Text":"t between 0 and 2,"},{"Start":"02:03.900 ","End":"02:05.195","Text":"and the same here."},{"Start":"02:05.195 ","End":"02:09.200","Text":"I claim that this minimum and this minimum and this minimum, they\u0027re all 0."},{"Start":"02:09.200 ","End":"02:12.535","Text":"Let\u0027s draw a picture of the graphs."},{"Start":"02:12.535 ","End":"02:18.095","Text":"Sine Pi x is this red-orange colored graph."},{"Start":"02:18.095 ","End":"02:21.155","Text":"Obviously, the minimum is 0."},{"Start":"02:21.155 ","End":"02:24.145","Text":"That\u0027s the minimum of sine Pi x."},{"Start":"02:24.145 ","End":"02:29.460","Text":"The other one, sine^2 Pi t between 0 and 2, well,"},{"Start":"02:29.460 ","End":"02:32.520","Text":"it\u0027s non-negative, so certainly the minimum is 0,"},{"Start":"02:32.520 ","End":"02:34.805","Text":"and it is 0 here, here, and here."},{"Start":"02:34.805 ","End":"02:42.695","Text":"So the minimum is 0, which means that the minimum on the parabolic boundary, L is 0."},{"Start":"02:42.695 ","End":"02:45.370","Text":"Now, the function is not constant,"},{"Start":"02:45.370 ","End":"02:47.225","Text":"and if it\u0027s not constant,"},{"Start":"02:47.225 ","End":"02:52.445","Text":"then the value at an interior point has to be strictly greater than the minimum."},{"Start":"02:52.445 ","End":"02:53.975","Text":"At a (1/2,1),"},{"Start":"02:53.975 ","End":"02:57.575","Text":"we have that w is bigger than 0."},{"Start":"02:57.575 ","End":"03:00.410","Text":"Since w is u minus v,"},{"Start":"03:00.410 ","End":"03:02.135","Text":"this is bigger than 0,"},{"Start":"03:02.135 ","End":"03:06.245","Text":"which means that u at this point is bigger than v at this point."},{"Start":"03:06.245 ","End":"03:08.360","Text":"We were asked, which is the greater?"},{"Start":"03:08.360 ","End":"03:11.125","Text":"Well, the greater is u of a 1/2 and 1."},{"Start":"03:11.125 ","End":"03:13.990","Text":"That concludes this exercise."}],"ID":30774},{"Watched":false,"Name":"Exercise 3","Duration":"4m 33s","ChapterTopicVideoID":29193,"CourseChapterTopicPlaylistID":294432,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.010","Text":"In this exercise, we\u0027re given"},{"Start":"00:02.010 ","End":"00:07.155","Text":"the following initial boundary value problem on this domain."},{"Start":"00:07.155 ","End":"00:10.095","Text":"Here it is, the PDE,"},{"Start":"00:10.095 ","End":"00:12.600","Text":"initial value, and boundary values."},{"Start":"00:12.600 ","End":"00:17.865","Text":"We have to prove the following inequality about the value of u(1/2,"},{"Start":"00:17.865 ","End":"00:20.745","Text":"1) and we\u0027re given a hint."},{"Start":"00:20.745 ","End":"00:22.995","Text":"Why do we need this hint?"},{"Start":"00:22.995 ","End":"00:26.580","Text":"Because this is not a homogeneous equation."},{"Start":"00:26.580 ","End":"00:30.285","Text":"We have this extra term minus u. Hopefully,"},{"Start":"00:30.285 ","End":"00:33.855","Text":"when we substitute h equals this,"},{"Start":"00:33.855 ","End":"00:39.045","Text":"we\u0027ll get another IBVP with a better PDE,"},{"Start":"00:39.045 ","End":"00:41.160","Text":"which is the heat equation."},{"Start":"00:41.160 ","End":"00:44.730","Text":"We can bring the e^Delta t to the other side,"},{"Start":"00:44.730 ","End":"00:47.430","Text":"so we get u in terms of h. Now,"},{"Start":"00:47.430 ","End":"00:52.350","Text":"we want to compute u_t and u_xx so we can substitute."},{"Start":"00:52.350 ","End":"00:56.185","Text":"Using this product rule, u by dt,"},{"Start":"00:56.185 ","End":"01:00.905","Text":"differentiate this with respect to t. The second time,"},{"Start":"01:00.905 ","End":"01:03.890","Text":"we differentiate this with respect to t,"},{"Start":"01:03.890 ","End":"01:07.025","Text":"which is what gives us the minus Delta here."},{"Start":"01:07.025 ","End":"01:12.385","Text":"U_xx, we just have to differentiate h with respect to x twice."},{"Start":"01:12.385 ","End":"01:15.185","Text":"Now, if we substitute in the PDE,"},{"Start":"01:15.185 ","End":"01:19.520","Text":"which is this, we get the following. Just check."},{"Start":"01:19.520 ","End":"01:21.945","Text":"We put this,"},{"Start":"01:21.945 ","End":"01:23.370","Text":"which is on the left,"},{"Start":"01:23.370 ","End":"01:27.120","Text":"equals this minus u,"},{"Start":"01:27.120 ","End":"01:33.000","Text":"which is he^minus Delta t. Everything has e^minus Delta t in it,"},{"Start":"01:33.000 ","End":"01:37.100","Text":"so we can cancel that factor from all the terms here,"},{"Start":"01:37.100 ","End":"01:38.825","Text":"and we\u0027re left with this."},{"Start":"01:38.825 ","End":"01:42.440","Text":"Note that if Delta equals 1,"},{"Start":"01:42.440 ","End":"01:44.660","Text":"then minus Delta h will be the same as"},{"Start":"01:44.660 ","End":"01:48.504","Text":"minus h and it will cancel and we\u0027ll get the heat equation."},{"Start":"01:48.504 ","End":"01:51.045","Text":"Let Delta equals 1."},{"Start":"01:51.045 ","End":"01:53.055","Text":"Then we have the heat equation."},{"Start":"01:53.055 ","End":"01:55.830","Text":"Also we have that h(x,"},{"Start":"01:55.830 ","End":"01:58.500","Text":"t) which is ue^Delta t,"},{"Start":"01:58.500 ","End":"02:00.465","Text":"is u e^t,"},{"Start":"02:00.465 ","End":"02:01.785","Text":"because Delta is 1."},{"Start":"02:01.785 ","End":"02:09.120","Text":"We want some initial and boundary conditions for h. Put t equals 0 here,"},{"Start":"02:09.120 ","End":"02:14.280","Text":"and we get h(x, 0) equals x(x-1)."},{"Start":"02:14.280 ","End":"02:16.605","Text":"If we let x equals 0,"},{"Start":"02:16.605 ","End":"02:19.250","Text":"then we get h(0,"},{"Start":"02:19.250 ","End":"02:20.910","Text":"t) is h(1,"},{"Start":"02:20.910 ","End":"02:24.830","Text":"t) and that\u0027s equal to the value of u at this point,"},{"Start":"02:24.830 ","End":"02:28.180","Text":"which is 0 times e^t,"},{"Start":"02:28.180 ","End":"02:30.430","Text":"which is just 0."},{"Start":"02:30.430 ","End":"02:34.575","Text":"Also, let\u0027s just copy h_t equals h_xx."},{"Start":"02:34.575 ","End":"02:41.510","Text":"These 3 together give us homogeneous IBVP with the heat equation."},{"Start":"02:41.510 ","End":"02:45.455","Text":"Now, we can use the min-max principle."},{"Start":"02:45.455 ","End":"02:50.545","Text":"We want the 0.5, 1 to be inside the rectangle."},{"Start":"02:50.545 ","End":"02:55.450","Text":"Let\u0027s extend it up to t=2."},{"Start":"02:55.450 ","End":"02:59.345","Text":"Then we have a rectangle with this being"},{"Start":"02:59.345 ","End":"03:03.785","Text":"the parabolic boundary and this being an interior point."},{"Start":"03:03.785 ","End":"03:05.240","Text":"Let\u0027s just write that."},{"Start":"03:05.240 ","End":"03:06.905","Text":"R is the rectangle,"},{"Start":"03:06.905 ","End":"03:09.425","Text":"L is parabolic boundary."},{"Start":"03:09.425 ","End":"03:14.945","Text":"Clearly, h is not constant on this rectangle because,"},{"Start":"03:14.945 ","End":"03:19.550","Text":"for example, along the lower side is equal to x(x-1),"},{"Start":"03:19.550 ","End":"03:21.730","Text":"which is not a constant."},{"Start":"03:21.730 ","End":"03:27.470","Text":"We can use the minimum and maximum principles to say that"},{"Start":"03:27.470 ","End":"03:33.575","Text":"this is between the minimum on L(h) and the maximum on L(h)."},{"Start":"03:33.575 ","End":"03:39.280","Text":"Now, h is exactly 0 on this side and on this side."},{"Start":"03:39.280 ","End":"03:43.355","Text":"Here, it\u0027s equal to x(x-1)."},{"Start":"03:43.355 ","End":"03:46.100","Text":"Here\u0027s a picture of the graph."},{"Start":"03:46.100 ","End":"03:48.170","Text":"The maximum is 0,"},{"Start":"03:48.170 ","End":"03:51.670","Text":"and it doesn\u0027t get any bigger with the 0 here."},{"Start":"03:51.670 ","End":"03:54.800","Text":"The minimum is the value at a half,"},{"Start":"03:54.800 ","End":"03:56.210","Text":"which is a 1/2(1/2-1),"},{"Start":"03:56.210 ","End":"03:57.940","Text":"it\u0027s minus a 1/4."},{"Start":"03:57.940 ","End":"04:01.360","Text":"Let\u0027s write that. Here, it\u0027s minus 1/4."},{"Start":"04:01.360 ","End":"04:04.160","Text":"The minimum, maximum is 0."},{"Start":"04:04.160 ","End":"04:08.030","Text":"Now let\u0027s multiply by e^minus 1,"},{"Start":"04:08.030 ","End":"04:16.010","Text":"recalling that u is h e^minus t. We get minus a 1/4 e^minus 1 is less than h(1/2,"},{"Start":"04:16.010 ","End":"04:21.260","Text":"1) e^minus 1 is less than 0 times e^minus 1."},{"Start":"04:21.260 ","End":"04:25.430","Text":"Now, this is u(1/2, 1),"},{"Start":"04:25.430 ","End":"04:27.620","Text":"so we get the following inequality,"},{"Start":"04:27.620 ","End":"04:30.230","Text":"which is exactly what was asked for."},{"Start":"04:30.230 ","End":"04:34.470","Text":"We\u0027ve proved it, and that concludes this exercise."}],"ID":30775},{"Watched":false,"Name":"Exercise 4","Duration":"7m 7s","ChapterTopicVideoID":29194,"CourseChapterTopicPlaylistID":294432,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.070","Text":"In this exercise, we have"},{"Start":"00:02.070 ","End":"00:08.445","Text":"an initial boundary value problem with a non-homogeneous heat equation here."},{"Start":"00:08.445 ","End":"00:12.945","Text":"We have to prove that u(x,t) is less than 1."},{"Start":"00:12.945 ","End":"00:15.660","Text":"Now there are 2 difficulties here."},{"Start":"00:15.660 ","End":"00:18.659","Text":"One is that this is non-homogeneous,"},{"Start":"00:18.659 ","End":"00:22.095","Text":"and secondly, that the initial value is not homogeneous,"},{"Start":"00:22.095 ","End":"00:24.180","Text":"meaning this is not 0."},{"Start":"00:24.180 ","End":"00:27.480","Text":"What we do in such a case is split it up into"},{"Start":"00:27.480 ","End":"00:31.965","Text":"2 problems so that each has only 1 difficulty."},{"Start":"00:31.965 ","End":"00:34.065","Text":"Let\u0027s put it that way."},{"Start":"00:34.065 ","End":"00:39.225","Text":"We let u equals v plus w, where,"},{"Start":"00:39.225 ","End":"00:44.640","Text":"dv by dt is v_xx minus the non-homogeneous part."},{"Start":"00:44.640 ","End":"00:46.415","Text":"We have this difficulty,"},{"Start":"00:46.415 ","End":"00:49.640","Text":"but the initial value is 0."},{"Start":"00:49.640 ","End":"00:51.980","Text":"For w is the other way round,"},{"Start":"00:51.980 ","End":"00:54.740","Text":"the heat equation is homogeneous,"},{"Start":"00:54.740 ","End":"00:57.005","Text":"but the initial value is not."},{"Start":"00:57.005 ","End":"01:00.340","Text":"It takes the sine Pi x from here."},{"Start":"01:00.340 ","End":"01:06.109","Text":"In all cases the boundary conditions at both end points is 0."},{"Start":"01:06.109 ","End":"01:11.801","Text":"What we\u0027ll do is we\u0027ll use the maximum principle for w because it\u0027s homogeneous,"},{"Start":"01:11.801 ","End":"01:15.020","Text":"and then we\u0027ll use Duhamel\u0027s principle"},{"Start":"01:15.020 ","End":"01:20.030","Text":"on v to get the non-homogeneous part from here down to here."},{"Start":"01:20.030 ","End":"01:24.255","Text":"We\u0027re dealing with w now."},{"Start":"01:24.255 ","End":"01:28.680","Text":"Let x_naught, t_naught be an arbitrary point in the domain."},{"Start":"01:28.680 ","End":"01:35.280","Text":"Meaning that x_naught is between 0 and 1 and t_naught is bigger than 0."},{"Start":"01:35.280 ","End":"01:39.615","Text":"Now, pick any t\u0027 bigger than t_naught."},{"Start":"01:39.615 ","End":"01:47.750","Text":"Then we get a rectangle where x goes from 0-1 and t goes from 0 to t\u0027."},{"Start":"01:47.750 ","End":"01:49.975","Text":"Here\u0027s the picture."},{"Start":"01:49.975 ","End":"01:54.375","Text":"From 0-1 here, from 0 to t\u0027 and here\u0027s x_naught,"},{"Start":"01:54.375 ","End":"01:57.195","Text":"t_naught and this is the rectangle."},{"Start":"01:57.195 ","End":"02:02.305","Text":"We\u0027ll let the parabolic boundary be l to these 3 sides."},{"Start":"02:02.305 ","End":"02:06.080","Text":"Now, w at"},{"Start":"02:06.080 ","End":"02:12.230","Text":"this interior point is less than or equal to the maximum of w on the rectangle,"},{"Start":"02:12.230 ","End":"02:16.220","Text":"which is equal to the maximum on the parabolic boundary."},{"Start":"02:16.220 ","End":"02:18.380","Text":"This maximum is equal to 1."},{"Start":"02:18.380 ","End":"02:23.715","Text":"Here it\u0027s 0, here it\u0027s 0 and here, it\u0027s sine Pi x."},{"Start":"02:23.715 ","End":"02:26.250","Text":"We just take the maximum from here,"},{"Start":"02:26.250 ","End":"02:30.365","Text":"and here\u0027s a picture of the sine Pi x."},{"Start":"02:30.365 ","End":"02:35.300","Text":"The maximum is 1 so 0,1 and 0,"},{"Start":"02:35.300 ","End":"02:37.315","Text":"the biggest is 1."},{"Start":"02:37.315 ","End":"02:40.410","Text":"Now, w isn\u0027t a constant function,"},{"Start":"02:40.410 ","End":"02:46.060","Text":"so we can sharpen this to make the inequality strictly less than 1."},{"Start":"02:46.060 ","End":"02:50.165","Text":"Since x_naught, t_naught was an arbitrary point in the domain,"},{"Start":"02:50.165 ","End":"02:51.620","Text":"we can switch x_naught,"},{"Start":"02:51.620 ","End":"02:53.240","Text":"t_naught to just any x,"},{"Start":"02:53.240 ","End":"02:54.919","Text":"t in the domain."},{"Start":"02:54.919 ","End":"02:57.770","Text":"That deals with w. Next,"},{"Start":"02:57.770 ","End":"03:00.830","Text":"we\u0027ll turn to v and show that v is"},{"Start":"03:00.830 ","End":"03:04.610","Text":"less than or equal to 0 so the sum is less than or equal to 1."},{"Start":"03:04.610 ","End":"03:06.290","Text":"Here is v,"},{"Start":"03:06.290 ","End":"03:08.705","Text":"its equation is problem."},{"Start":"03:08.705 ","End":"03:12.230","Text":"Like I said, we\u0027re going to use the Duhamel principle."},{"Start":"03:12.230 ","End":"03:13.610","Text":"We\u0027ve done this before,"},{"Start":"03:13.610 ","End":"03:15.320","Text":"so I\u0027ll just present it,"},{"Start":"03:15.320 ","End":"03:18.145","Text":"standard presentation of the Duhamel."},{"Start":"03:18.145 ","End":"03:26.599","Text":"We let v be the integral of p. P has the following PDE initial value boundary values."},{"Start":"03:26.599 ","End":"03:30.950","Text":"Note that the non-homogeneous part has gone"},{"Start":"03:30.950 ","End":"03:35.555","Text":"down to the initial condition and we have a homogeneous PDE."},{"Start":"03:35.555 ","End":"03:44.465","Text":"Then there\u0027s a standard technique we use here to make it a function of 2 variables,"},{"Start":"03:44.465 ","End":"03:46.340","Text":"we just, for the moment,"},{"Start":"03:46.340 ","End":"03:48.590","Text":"think of Tau like a constant, we fix it,"},{"Start":"03:48.590 ","End":"03:52.185","Text":"choose any Tau and then define P(x,"},{"Start":"03:52.185 ","End":"03:54.015","Text":"T) equals p(x,"},{"Start":"03:54.015 ","End":"03:55.830","Text":"T plus Tau, Tau)."},{"Start":"03:55.830 ","End":"03:58.740","Text":"You could also say that T,"},{"Start":"03:58.740 ","End":"04:01.910","Text":"is t minus Tau, this minus this."},{"Start":"04:01.910 ","End":"04:05.065","Text":"Anyway, P satisfies the following."},{"Start":"04:05.065 ","End":"04:08.420","Text":"We\u0027ve done this before so I won\u0027t go into all the details,"},{"Start":"04:08.420 ","End":"04:13.710","Text":"but basically it\u0027s the same equation just with big letters."},{"Start":"04:14.390 ","End":"04:19.130","Text":"The P(x,0) 0 means that t is 0,"},{"Start":"04:19.130 ","End":"04:22.205","Text":"so it\u0027s Tau, Tau like this."},{"Start":"04:22.205 ","End":"04:28.055","Text":"The value at the end points where x is 0,1 is"},{"Start":"04:28.055 ","End":"04:31.820","Text":"0.Then we\u0027re going to apply the maximum principle to"},{"Start":"04:31.820 ","End":"04:35.735","Text":"P. Choose an arbitrary point x_naught,"},{"Start":"04:35.735 ","End":"04:37.499","Text":"t_naught in the domain,"},{"Start":"04:37.499 ","End":"04:42.360","Text":"and let t\u0027 be bigger than t_naught."},{"Start":"04:42.360 ","End":"04:46.855","Text":"This is very much the same as what we did with w,"},{"Start":"04:46.855 ","End":"04:49.145","Text":"and here\u0027s a picture."},{"Start":"04:49.145 ","End":"04:58.020","Text":"We\u0027re going to let the rectangle R be 0-1 for x and 0 to t\u0027 for T,"},{"Start":"04:58.020 ","End":"05:00.270","Text":"and x_naught, t_naught is here."},{"Start":"05:00.270 ","End":"05:04.830","Text":"The value of P on the boundaries is 0."},{"Start":"05:04.830 ","End":"05:06.585","Text":"On the initial part,"},{"Start":"05:06.585 ","End":"05:09.350","Text":"it\u0027s equal to this minus x,"},{"Start":"05:09.350 ","End":"05:11.645","Text":"e^minus t 1 minus x."},{"Start":"05:11.645 ","End":"05:18.565","Text":"Note that this is always less than or equal to 0 because between 0 and 1,"},{"Start":"05:18.565 ","End":"05:21.280","Text":"x is between 0 and 1."},{"Start":"05:21.280 ","End":"05:23.500","Text":"1 minus x is between 0 and 1,"},{"Start":"05:23.500 ","End":"05:25.650","Text":"and this is positive."},{"Start":"05:25.650 ","End":"05:27.405","Text":"Each of the anything is positive."},{"Start":"05:27.405 ","End":"05:30.740","Text":"Multiplying non negative times positive times"},{"Start":"05:30.740 ","End":"05:35.770","Text":"non-negative is non-negative and with a minus it\u0027s non-positive."},{"Start":"05:35.770 ","End":"05:39.165","Text":"It\u0027s less than or equal to 0."},{"Start":"05:39.165 ","End":"05:47.100","Text":"That means that the maximum is 0 along the parabolic boundary."},{"Start":"05:47.120 ","End":"05:51.245","Text":"P(x_naught, t_naught) is less than or equal to 0."},{"Start":"05:51.245 ","End":"05:55.100","Text":"We could get it sharper to be strictly less than 0,"},{"Start":"05:55.100 ","End":"06:01.270","Text":"but we don\u0027t need that because the inequality for w is sharp inequality."},{"Start":"06:01.270 ","End":"06:04.080","Text":"What we do is now replace x_naught,"},{"Start":"06:04.080 ","End":"06:09.020","Text":"t_naught by the general x,T in the domain."},{"Start":"06:09.020 ","End":"06:16.325","Text":"Now, we get back from P to p. This is also less than or equal to 0."},{"Start":"06:16.325 ","End":"06:22.010","Text":"The domain is different though T big or equal to 0 corresponds to t big or equal to Tau,"},{"Start":"06:22.010 ","End":"06:23.720","Text":"which is our original domain."},{"Start":"06:23.720 ","End":"06:28.330","Text":"Then you want to get back from p to v,"},{"Start":"06:28.330 ","End":"06:32.460","Text":"so unfixed Tau let it be a variable again."},{"Start":"06:32.460 ","End":"06:36.120","Text":"We can take v(x,t) to be this integral."},{"Start":"06:36.120 ","End":"06:40.515","Text":"By the way, from 0 to t,"},{"Start":"06:40.515 ","End":"06:44.570","Text":"Tau is less than or equal to t.,"},{"Start":"06:44.570 ","End":"06:45.905","Text":"like it should be."},{"Start":"06:45.905 ","End":"06:48.500","Text":"Just wanted to remark on that though it\u0027s okay."},{"Start":"06:48.500 ","End":"06:50.720","Text":"V less than or equal to 0."},{"Start":"06:50.720 ","End":"06:54.895","Text":"Recall w strictly less than 1."},{"Start":"06:54.895 ","End":"06:57.855","Text":"U which is v plus w,"},{"Start":"06:57.855 ","End":"07:02.295","Text":"is strictly less than 0 plus 1, which is 1."},{"Start":"07:02.295 ","End":"07:05.385","Text":"This is what we had to show."},{"Start":"07:05.385 ","End":"07:08.800","Text":"That concludes this clip."}],"ID":30776},{"Watched":false,"Name":"Exercise 5","Duration":"6m 39s","ChapterTopicVideoID":29195,"CourseChapterTopicPlaylistID":294432,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.400","Text":"In this exercise, we have"},{"Start":"00:02.400 ","End":"00:07.410","Text":"an initial boundary value problem and it has a non-homogeneous heat equation."},{"Start":"00:07.410 ","End":"00:09.810","Text":"This is a non-homogeneous part."},{"Start":"00:09.810 ","End":"00:17.355","Text":"We have to prove that u(1/2,Pi) is less than 1."},{"Start":"00:17.355 ","End":"00:19.740","Text":"We\u0027ll be using the maximum principle here,"},{"Start":"00:19.740 ","End":"00:26.940","Text":"but we\u0027ll start off with splitting this problem into 2 because not only non-homogeneous,"},{"Start":"00:26.940 ","End":"00:30.600","Text":"but the initial value is also non-homogeneous."},{"Start":"00:30.600 ","End":"00:35.680","Text":"What we do in cases like this is break it up into 2 easier problems."},{"Start":"00:35.680 ","End":"00:38.195","Text":"U is v plus w,"},{"Start":"00:38.195 ","End":"00:43.970","Text":"and then v has all these boundary and initial value is 0,"},{"Start":"00:43.970 ","End":"00:47.035","Text":"but it keeps a non-homogeneous PDE."},{"Start":"00:47.035 ","End":"00:52.250","Text":"Here the opposite, we make the equation homogeneous,"},{"Start":"00:52.250 ","End":"00:56.350","Text":"but we keep all the initial and boundary values."},{"Start":"00:56.350 ","End":"01:01.605","Text":"If you think about it, this works v_t plus w_t."},{"Start":"01:01.605 ","End":"01:04.185","Text":"U_t, is this plus this,"},{"Start":"01:04.185 ","End":"01:09.290","Text":"which is u_xx minus sine Pi x and so on for all the other,"},{"Start":"01:09.290 ","End":"01:11.165","Text":"it\u0027s okay to split like this."},{"Start":"01:11.165 ","End":"01:13.715","Text":"Like I said, we\u0027ll be using the maximum principle."},{"Start":"01:13.715 ","End":"01:15.900","Text":"We\u0027ll use that for w,"},{"Start":"01:15.900 ","End":"01:19.655","Text":"but for v will use Duhamel\u0027s principle to get"},{"Start":"01:19.655 ","End":"01:24.560","Text":"the non-homogeneous part into the initial value and not in the PDE."},{"Start":"01:24.560 ","End":"01:29.565","Text":"Anyway, let\u0027s start with w. Let\u0027s define a rectangle R,"},{"Start":"01:29.565 ","End":"01:32.250","Text":"which is x between 0,"},{"Start":"01:32.250 ","End":"01:36.965","Text":"and 1 will take t from 0 to Pi,"},{"Start":"01:36.965 ","End":"01:39.290","Text":"we might think we need to take it higher than Pi,"},{"Start":"01:39.290 ","End":"01:41.170","Text":"but Pi will suffice."},{"Start":"01:41.170 ","End":"01:43.895","Text":"Let the parabolic boundary of R be l,"},{"Start":"01:43.895 ","End":"01:45.910","Text":"hope the picture will help."},{"Start":"01:45.910 ","End":"01:51.875","Text":"Note that here w is sine Pi t, that\u0027s from here."},{"Start":"01:51.875 ","End":"01:57.815","Text":"From here we get that w is cosine Pi t here and here w is equal to x."},{"Start":"01:57.815 ","End":"02:03.730","Text":"Well, the maximum on this is 1,"},{"Start":"02:03.730 ","End":"02:06.200","Text":"because sine can\u0027t get bigger than 1,"},{"Start":"02:06.200 ","End":"02:08.810","Text":"and it is 1 at t equals a 1/2."},{"Start":"02:08.810 ","End":"02:14.390","Text":"Similarly here, cosine is 1 when t is 0, for example."},{"Start":"02:14.390 ","End":"02:19.735","Text":"Here x is also 1 when it\u0027s here."},{"Start":"02:19.735 ","End":"02:21.770","Text":"Each of these has a maximum of 1,"},{"Start":"02:21.770 ","End":"02:25.525","Text":"so the combined maximum is also 1,"},{"Start":"02:25.525 ","End":"02:30.425","Text":"and It\u0027s just wrote what I said here."},{"Start":"02:30.425 ","End":"02:35.190","Text":"Now, w isn\u0027t a constant function,"},{"Start":"02:35.190 ","End":"02:39.635","Text":"for example, here it\u0027s not constant and here is not constant."},{"Start":"02:39.635 ","End":"02:45.020","Text":"That means that if you take a point that\u0027s not on the parabolic boundary,"},{"Start":"02:45.020 ","End":"02:46.670","Text":"but isn\u0027t the rectangle."},{"Start":"02:46.670 ","End":"02:49.310","Text":"This point counts as that,"},{"Start":"02:49.310 ","End":"02:52.775","Text":"then we get strictly less than the maximum."},{"Start":"02:52.775 ","End":"02:56.480","Text":"We know that w(1/2,Pi) is less than 1."},{"Start":"02:56.480 ","End":"02:59.225","Text":"Now we turn to the other function to v,"},{"Start":"02:59.225 ","End":"03:05.230","Text":"and this is the problem for v. We\u0027ll use Duhamel\u0027s principle here,"},{"Start":"03:05.230 ","End":"03:10.835","Text":"so that means that v is the integral from 0 to t of p(x,T) Tau,"},{"Start":"03:10.835 ","End":"03:14.320","Text":"where p is a solution to the following."},{"Start":"03:14.320 ","End":"03:22.470","Text":"Here is the initial value which is taken from the non-homogeneous part here."},{"Start":"03:22.730 ","End":"03:25.780","Text":"Let\u0027s temporarily fix Tau,"},{"Start":"03:25.780 ","End":"03:26.810","Text":"so it\u0027s not a variable,"},{"Start":"03:26.810 ","End":"03:29.075","Text":"it\u0027s now a constant but arbitrary."},{"Start":"03:29.075 ","End":"03:31.550","Text":"I want to get a function of 2 variables,"},{"Start":"03:31.550 ","End":"03:33.815","Text":"because here we have a function of 3 variables."},{"Start":"03:33.815 ","End":"03:37.045","Text":"We let p of x and t,"},{"Start":"03:37.045 ","End":"03:40.740","Text":"be p(x,T) plus Tau, Tau."},{"Start":"03:40.740 ","End":"03:44.205","Text":"It\u0027s like saying that T,"},{"Start":"03:44.205 ","End":"03:45.730","Text":"t minus Tau,"},{"Start":"03:45.730 ","End":"03:47.989","Text":"the second minus the third variables."},{"Start":"03:47.989 ","End":"03:52.579","Text":"Now, p satisfies the following initial boundary and value problem."},{"Start":"03:52.579 ","End":"03:57.920","Text":"We\u0027ve seen this before, so we just know that it follows the same equation."},{"Start":"03:57.920 ","End":"04:04.270","Text":"The only difference is that t bigger or equal to Tau becomes T bigger or equal to 0."},{"Start":"04:04.270 ","End":"04:06.050","Text":"Instead of Tau, Tau,"},{"Start":"04:06.050 ","End":"04:09.365","Text":"we put 0 because this variable is this minus this."},{"Start":"04:09.365 ","End":"04:12.595","Text":"This is now our problem for P,"},{"Start":"04:12.595 ","End":"04:18.050","Text":"and on this we can use the maximum principle because this is homogeneous to find"},{"Start":"04:18.050 ","End":"04:24.200","Text":"a rectangle R to be x from 0-1 and T from 0 to Pi,"},{"Start":"04:24.200 ","End":"04:27.965","Text":"and that it\u0027s parabolic boundary be l. Let\u0027s see a picture,"},{"Start":"04:27.965 ","End":"04:30.560","Text":"so the red part is the boundary,"},{"Start":"04:30.560 ","End":"04:33.280","Text":"the rectangle is this."},{"Start":"04:33.280 ","End":"04:36.015","Text":"Actually we don\u0027t need this in the picture."},{"Start":"04:36.015 ","End":"04:38.790","Text":"Anyway I\u0027ll just mark the rectangle,"},{"Start":"04:38.790 ","End":"04:41.525","Text":"the rectangle is this."},{"Start":"04:41.525 ","End":"04:43.535","Text":"By the maximum principle,"},{"Start":"04:43.535 ","End":"04:47.240","Text":"the maximum of p on all the rectangle is the same as"},{"Start":"04:47.240 ","End":"04:51.605","Text":"the maximum on the parabolic boundary on the red parts."},{"Start":"04:51.605 ","End":"04:53.585","Text":"But here it\u0027s 0,"},{"Start":"04:53.585 ","End":"04:55.225","Text":"here it\u0027s 0,"},{"Start":"04:55.225 ","End":"04:57.195","Text":"so here the maximum is 0."},{"Start":"04:57.195 ","End":"05:02.105","Text":"Here we just need the maximum of minus sine Pi x from 0-1,"},{"Start":"05:02.105 ","End":"05:03.665","Text":"so here\u0027s a picture."},{"Start":"05:03.665 ","End":"05:05.800","Text":"The maximum is 0,"},{"Start":"05:05.800 ","End":"05:10.560","Text":"so the total maximum is 0."},{"Start":"05:10.560 ","End":"05:13.380","Text":"This is equal to 0,"},{"Start":"05:13.380 ","End":"05:16.020","Text":"and if the maximum is 0,"},{"Start":"05:16.020 ","End":"05:18.510","Text":"then for any particular x,"},{"Start":"05:18.510 ","End":"05:20.445","Text":"T in the rectangle,"},{"Start":"05:20.445 ","End":"05:23.450","Text":"P(x,T) is less than or equal to 0."},{"Start":"05:23.450 ","End":"05:25.415","Text":"Going back from P,"},{"Start":"05:25.415 ","End":"05:28.160","Text":"p, this is what we get."},{"Start":"05:28.160 ","End":"05:30.875","Text":"Just replace T equal to 0."},{"Start":"05:30.875 ","End":"05:33.455","Text":"The condition t bigger than Tau,"},{"Start":"05:33.455 ","End":"05:36.965","Text":"now drop the condition that Tau is fixed,"},{"Start":"05:36.965 ","End":"05:38.285","Text":"let it roam again,"},{"Start":"05:38.285 ","End":"05:43.420","Text":"if we look at the definition of v in terms of the integral of p,"},{"Start":"05:43.420 ","End":"05:46.085","Text":"oops I forgot the P, here it is."},{"Start":"05:46.085 ","End":"05:48.750","Text":"If you\u0027re not sure where I got this from,"},{"Start":"05:50.060 ","End":"05:53.404","Text":"we had this formula."},{"Start":"05:53.404 ","End":"05:54.620","Text":"We don\u0027t need it in general,"},{"Start":"05:54.620 ","End":"05:57.305","Text":"but just need it for the point where x is 1/2,"},{"Start":"05:57.305 ","End":"05:59.255","Text":"and t is Pi."},{"Start":"05:59.255 ","End":"06:04.205","Text":"Now since p is less than or equal to 0 on all the rectangle,"},{"Start":"06:04.205 ","End":"06:07.310","Text":"this integral is going to be less than or equal to 0."},{"Start":"06:07.310 ","End":"06:13.130","Text":"The integral of a non-negative function is non-negative."},{"Start":"06:13.130 ","End":"06:17.330","Text":"Remember we had that w at this point is less than 1."},{"Start":"06:17.330 ","End":"06:19.370","Text":"If v is less than or equal to 0,"},{"Start":"06:19.370 ","End":"06:20.420","Text":"w is less than 1,"},{"Start":"06:20.420 ","End":"06:24.990","Text":"add them, v plus w is u."},{"Start":"06:27.560 ","End":"06:30.120","Text":"This is less than 0,"},{"Start":"06:30.120 ","End":"06:32.205","Text":"this is less than or equal to 1."},{"Start":"06:32.205 ","End":"06:34.159","Text":"See we don\u0027t need the sharp inequality."},{"Start":"06:34.159 ","End":"06:36.290","Text":"This will still be strictly less than 1,"},{"Start":"06:36.290 ","End":"06:39.840","Text":"and that\u0027s what we had to show and we are done."}],"ID":30777},{"Watched":false,"Name":"Exercise 6","Duration":"3m 32s","ChapterTopicVideoID":29196,"CourseChapterTopicPlaylistID":294432,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.925","Text":"This exercise is a proof question."},{"Start":"00:02.925 ","End":"00:06.570","Text":"We\u0027re given the following IBVP but it\u0027s general."},{"Start":"00:06.570 ","End":"00:09.150","Text":"The non-homogeneous heat equation."},{"Start":"00:09.150 ","End":"00:12.495","Text":"There\u0027s initial conditions and boundary conditions."},{"Start":"00:12.495 ","End":"00:17.025","Text":"Our task is to prove that the solution is unique,"},{"Start":"00:17.025 ","End":"00:19.950","Text":"and it turns out that it\u0027s quite easy"},{"Start":"00:19.950 ","End":"00:23.349","Text":"given the maximum principle and the minimum principle."},{"Start":"00:23.349 ","End":"00:24.690","Text":"Let\u0027s start."},{"Start":"00:24.690 ","End":"00:28.800","Text":"The way to go is a proof by contradiction where we assume that there"},{"Start":"00:28.800 ","End":"00:33.075","Text":"are 2 solutions and show that they are equal."},{"Start":"00:33.075 ","End":"00:35.130","Text":"So there really are 2 solutions."},{"Start":"00:35.130 ","End":"00:36.630","Text":"Well, 2 of them are equal,"},{"Start":"00:36.630 ","End":"00:39.510","Text":"so it\u0027s like 1 here, just copied you."},{"Start":"00:39.510 ","End":"00:42.045","Text":"Now let\u0027s say there\u0027s another 1 called v,"},{"Start":"00:42.045 ","End":"00:45.950","Text":"exactly the same, just replace u by v. Now,"},{"Start":"00:45.950 ","End":"00:50.475","Text":"we\u0027ll define w to equal u minus v,"},{"Start":"00:50.475 ","End":"00:53.345","Text":"and we\u0027re going to show that w=0,"},{"Start":"00:53.345 ","End":"00:55.615","Text":"which means that u=v."},{"Start":"00:55.615 ","End":"01:00.350","Text":"We reduce the problem now to showing that w=0."},{"Start":"01:00.350 ","End":"01:03.035","Text":"What problem does w satisfy?"},{"Start":"01:03.035 ","End":"01:06.560","Text":"Let\u0027s first-of-all subtract each equation."},{"Start":"01:06.560 ","End":"01:08.810","Text":"You can subtract equals from equals,"},{"Start":"01:08.810 ","End":"01:12.920","Text":"so I will do this 1 minus this 1 gives us this,"},{"Start":"01:12.920 ","End":"01:16.055","Text":"this minus, this gives us this."},{"Start":"01:16.055 ","End":"01:20.225","Text":"Now notice, that a lot of stuff cancels."},{"Start":"01:20.225 ","End":"01:23.305","Text":"This f cancels with this f,"},{"Start":"01:23.305 ","End":"01:26.415","Text":"and this f cancels with this f,"},{"Start":"01:26.415 ","End":"01:29.460","Text":"and this g with this g and"},{"Start":"01:29.460 ","End":"01:35.060","Text":"this h with this h also recall that differentiation is linear,"},{"Start":"01:35.060 ","End":"01:42.850","Text":"so u_t minus v_t is the derivative of u minus v with respect to t, which is w_t."},{"Start":"01:42.850 ","End":"01:45.795","Text":"All this boils down to this,"},{"Start":"01:45.795 ","End":"01:48.695","Text":"and this is a homogeneous equation in w,"},{"Start":"01:48.695 ","End":"01:54.365","Text":"and all the initial boundary values are 0, they\u0027re also homogeneous."},{"Start":"01:54.365 ","End":"01:56.760","Text":"Remember we want to show w_0."},{"Start":"01:56.760 ","End":"01:58.970","Text":"Let\u0027s take any arbitrary point x_naught,"},{"Start":"01:58.970 ","End":"02:01.205","Text":"t_naught in our domain."},{"Start":"02:01.205 ","End":"02:05.495","Text":"We\u0027re going to show that w of x Naught, t_naught is 0."},{"Start":"02:05.495 ","End":"02:07.025","Text":"That\u0027s all we have to do,"},{"Start":"02:07.025 ","End":"02:11.150","Text":"assuming that this is just any arbitrary point in that domain."},{"Start":"02:11.150 ","End":"02:16.350","Text":"So choose capital T bigger than t_naught,"},{"Start":"02:16.350 ","End":"02:20.820","Text":"and let R be the rectangle from 0-1 in x,"},{"Start":"02:20.820 ","End":"02:23.595","Text":"and from 0 to t in little t,"},{"Start":"02:23.595 ","End":"02:27.320","Text":"and let\u0027s say its parabolic boundary is l. Well,"},{"Start":"02:27.320 ","End":"02:28.865","Text":"a picture could help."},{"Start":"02:28.865 ","End":"02:31.055","Text":"Yeah, here\u0027s a picture."},{"Start":"02:31.055 ","End":"02:36.745","Text":"Note that w is 0 on the parabolic boundary,"},{"Start":"02:36.745 ","End":"02:39.445","Text":"w0 here and here and here."},{"Start":"02:39.445 ","End":"02:44.710","Text":"Because these are just the initial and boundary conditions, everything 0."},{"Start":"02:44.710 ","End":"02:47.110","Text":"If it\u0027s all 0, the maximum on"},{"Start":"02:47.110 ","End":"02:52.030","Text":"this boundary is 0 and so is the minimum thing is constantly 0,"},{"Start":"02:52.030 ","End":"02:54.505","Text":"it\u0027s both minimum and maximum is 0."},{"Start":"02:54.505 ","End":"02:58.375","Text":"Using the maximum principle and the minimum principle,"},{"Start":"02:58.375 ","End":"03:03.160","Text":"we get that the minimum of w of x_t is less than or"},{"Start":"03:03.160 ","End":"03:08.080","Text":"equal to w at any point in the rectangle like x_naught,"},{"Start":"03:08.080 ","End":"03:12.220","Text":"t_naught, and this is less than or equal to the maximum on the parabolic boundary,"},{"Start":"03:12.220 ","End":"03:13.420","Text":"which is also 0."},{"Start":"03:13.420 ","End":"03:15.820","Text":"I have something sandwiched between 0 and 0,"},{"Start":"03:15.820 ","End":"03:17.140","Text":"it has to be 0,"},{"Start":"03:17.140 ","End":"03:19.580","Text":"and that\u0027s all we had to prove."},{"Start":"03:19.580 ","End":"03:21.995","Text":"Everything else just follows in reverse."},{"Start":"03:21.995 ","End":"03:23.960","Text":"It\u0027s true for any x_naught, t_naught,"},{"Start":"03:23.960 ","End":"03:27.140","Text":"so w(x,t) is 0, so w is 0,"},{"Start":"03:27.140 ","End":"03:29.255","Text":"so u minus v is 0,"},{"Start":"03:29.255 ","End":"03:33.180","Text":"and that means that u=v. We\u0027re done."}],"ID":30778}],"Thumbnail":null,"ID":294432},{"Name":"Asymptotic Behaviour on an Infinite Interval","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Asymptotic Behaviour on an Infinite Interval","Duration":"4m 13s","ChapterTopicVideoID":29186,"CourseChapterTopicPlaylistID":294433,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.600","Text":"Continuing with the heat equation we\u0027re now going to study"},{"Start":"00:03.600 ","End":"00:09.510","Text":"the asymptotic behavior of the solution to the heat equation on an infinite interval."},{"Start":"00:09.510 ","End":"00:15.300","Text":"Asymptotic means what happens when t gets very large and it goes to infinity."},{"Start":"00:15.300 ","End":"00:18.480","Text":"Here\u0027s a typical heat equation."},{"Start":"00:18.480 ","End":"00:22.378","Text":"Remember when on an infinite interval we don\u0027t need boundary conditions;"},{"Start":"00:22.378 ","End":"00:24.170","Text":"there\u0027s no boundary to x,"},{"Start":"00:24.170 ","End":"00:26.920","Text":"so we just have an initial condition."},{"Start":"00:26.920 ","End":"00:30.455","Text":"We need to suppose that this function"},{"Start":"00:30.455 ","End":"00:35.300","Text":"h(x) goes to 0 quickly enough when x goes to plus or minus infinity."},{"Start":"00:35.300 ","End":"00:38.540","Text":"This is the requirement that it\u0027s less than"},{"Start":"00:38.540 ","End":"00:44.960","Text":"some exponential function like so where Alpha and Beta are positive numbers."},{"Start":"00:44.960 ","End":"00:48.404","Text":"Like I said our goal is to understand the behavior of the solution,"},{"Start":"00:48.404 ","End":"00:52.355","Text":"meaning u(x, t) when t gets very large."},{"Start":"00:52.355 ","End":"00:56.495","Text":"Now recall the Poisson formula for the solution of such;"},{"Start":"00:56.495 ","End":"01:00.475","Text":"a PDE, and this is what it is."},{"Start":"01:00.475 ","End":"01:06.725","Text":"Now want to express this exponent as a power series just like each of the power of."},{"Start":"01:06.725 ","End":"01:12.375","Text":"Well, we know that each of the z is 1 plus z plus z^2 over 2 factorial and so on,"},{"Start":"01:12.375 ","End":"01:14.550","Text":"z^n over n factorial."},{"Start":"01:14.550 ","End":"01:15.815","Text":"If we do that,"},{"Start":"01:15.815 ","End":"01:18.279","Text":"what we get is the following."},{"Start":"01:18.279 ","End":"01:24.560","Text":"What we\u0027d like to do is to apply this integral term-wise or term by term,"},{"Start":"01:24.560 ","End":"01:27.880","Text":"take the integral of each term separately."},{"Start":"01:27.880 ","End":"01:34.505","Text":"There are conditions that when 1 can do this and it turns out that if h satisfies this,"},{"Start":"01:34.505 ","End":"01:38.150","Text":"then this series converges what we call"},{"Start":"01:38.150 ","End":"01:43.445","Text":"uniformly in which case we can integrate term by term."},{"Start":"01:43.445 ","End":"01:45.680","Text":"We\u0027re not going into the technical details of this,"},{"Start":"01:45.680 ","End":"01:46.970","Text":"we just accept it."},{"Start":"01:46.970 ","End":"01:48.615","Text":"What we get is;"},{"Start":"01:48.615 ","End":"01:50.295","Text":"let\u0027s look at the first term,"},{"Start":"01:50.295 ","End":"01:57.108","Text":"1 over 2a root Pit times the integral of h(y) times 1."},{"Start":"01:57.108 ","End":"02:02.330","Text":"We\u0027ll take the 1 over root t separately as 1 over t^1/2."},{"Start":"02:02.330 ","End":"02:06.030","Text":"I want to separate the t from the x and the next term,"},{"Start":"02:06.030 ","End":"02:09.735","Text":"2a times 4a^2 is 8a^3."},{"Start":"02:09.735 ","End":"02:13.070","Text":"Again, we take the t separately."},{"Start":"02:13.070 ","End":"02:17.525","Text":"We have now t root t which is t^3 over 2,"},{"Start":"02:17.525 ","End":"02:22.370","Text":"and in this term we\u0027ll have t^2 times root."},{"Start":"02:22.370 ","End":"02:25.520","Text":"So t^5 over 2; 2a,"},{"Start":"02:25.520 ","End":"02:29.205","Text":"with (4a^2)^2 is 16a^4."},{"Start":"02:29.205 ","End":"02:32.755","Text":"So times 4 is 64 and a^4 times 8."},{"Start":"02:32.755 ","End":"02:37.730","Text":"This integral here doesn\u0027t have x in it,"},{"Start":"02:37.730 ","End":"02:40.464","Text":"so it\u0027s just a constant M_naught."},{"Start":"02:40.464 ","End":"02:45.750","Text":"This integral we\u0027ll call it M_1(x) because it depends on x,"},{"Start":"02:45.750 ","End":"02:48.285","Text":"and the next one we\u0027ll call it M_2(x),"},{"Start":"02:48.285 ","End":"02:52.095","Text":"and so on, and so we can write this as follows."},{"Start":"02:52.095 ","End":"02:53.790","Text":"Just algebra here."},{"Start":"02:53.790 ","End":"02:59.930","Text":"Now the behavior as t goes to infinity depends on the first non-zero term."},{"Start":"02:59.930 ","End":"03:04.680","Text":"This is something times 1 over t^1/2, 1 over t^3 over 2."},{"Start":"03:04.680 ","End":"03:07.780","Text":"This goes to 0 faster than this."},{"Start":"03:08.540 ","End":"03:15.125","Text":"The asymptotic behavior is roughly like the first non-zero term."},{"Start":"03:15.125 ","End":"03:18.710","Text":"If M_naught is non-zero, then u(x,"},{"Start":"03:18.710 ","End":"03:23.270","Text":"t) asymptotically behaves like this first term."},{"Start":"03:23.270 ","End":"03:28.150","Text":"What we mean if you use the O notation is that u(x,"},{"Start":"03:28.150 ","End":"03:31.235","Text":"t) is O(1 over root t)."},{"Start":"03:31.235 ","End":"03:34.940","Text":"If you really don\u0027t know what the O notation is I copy"},{"Start":"03:34.940 ","End":"03:41.355","Text":"pasted a couple of paragraphs from the Wikipedia so you can take a look at that."},{"Start":"03:41.355 ","End":"03:46.490","Text":"If M_naught is 0 but the next one isn\u0027t 0, then u(x,"},{"Start":"03:46.490 ","End":"03:52.125","Text":"t) behaves like this for a very large t. We can say that u is"},{"Start":"03:52.125 ","End":"03:58.400","Text":"O of 1 over t^3 over 2, and so on."},{"Start":"03:58.400 ","End":"04:03.395","Text":"Usually, in practice it doesn\u0027t happen that both of these are 0 but it could be,"},{"Start":"04:03.395 ","End":"04:05.570","Text":"and then you get 1 over 2^5 over 2,"},{"Start":"04:05.570 ","End":"04:08.205","Text":"etc That\u0027s it,"},{"Start":"04:08.205 ","End":"04:13.500","Text":"and I hope it will be clearer in the following exercises."}],"ID":30767},{"Watched":false,"Name":"Exercise 1","Duration":"2m 23s","ChapterTopicVideoID":29182,"CourseChapterTopicPlaylistID":294433,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.240","Text":"In this exercise we\u0027re given the following initial boundary value problem,"},{"Start":"00:05.240 ","End":"00:09.645","Text":"it\u0027s a heat equation on an infinite interval with initial condition."},{"Start":"00:09.645 ","End":"00:14.310","Text":"I should have written, we\u0027re given that u as a solution to this problem,"},{"Start":"00:14.310 ","End":"00:20.655","Text":"we have to compute the limit as t goes to infinity of t^1/3 u(1),"},{"Start":"00:20.655 ","End":"00:24.255","Text":"t. Let\u0027s call the initial condition h(x),"},{"Start":"00:24.255 ","End":"00:32.150","Text":"and in the tutorial we had the requirement that h(x) is bounded by an exponential,"},{"Start":"00:32.150 ","End":"00:33.810","Text":"so don\u0027t worry about the details,"},{"Start":"00:33.810 ","End":"00:39.640","Text":"it\u0027s possible to do this but let us not get bogged down with the technical stuff."},{"Start":"00:39.640 ","End":"00:42.444","Text":"In that case, we showed that u(x,"},{"Start":"00:42.444 ","End":"00:45.380","Text":"t) can be expressed as follows and"},{"Start":"00:45.380 ","End":"00:51.185","Text":"the asymptotic behavior of u depends on the first of these that\u0027s not 0."},{"Start":"00:51.185 ","End":"00:55.585","Text":"Let\u0027s try the M_ 0 which we know is given by this formula."},{"Start":"00:55.585 ","End":"00:56.930","Text":"M_ 1 is this,"},{"Start":"00:56.930 ","End":"01:01.520","Text":"but I already know that M_ 0 will come out to be non-0 and we\u0027ll show this in a moment."},{"Start":"01:01.520 ","End":"01:03.530","Text":"As such, u(x,"},{"Start":"01:03.530 ","End":"01:11.535","Text":"t) behaves like this constant times 1 over root t so that u is O(1 over root t)."},{"Start":"01:11.535 ","End":"01:13.715","Text":"Now about this integral,"},{"Start":"01:13.715 ","End":"01:16.880","Text":"this is the integral from minus infinity to infinity of h(s)ds,"},{"Start":"01:16.880 ","End":"01:21.290","Text":"which is s^2 e^- absolute value of s. From here,"},{"Start":"01:21.290 ","End":"01:25.070","Text":"just replace x by s and the integrant is strictly"},{"Start":"01:25.070 ","End":"01:29.195","Text":"positive except when s is 0 but in general it\u0027s positive."},{"Start":"01:29.195 ","End":"01:33.110","Text":"If you\u0027re curious as to what this integral actually comes out to be,"},{"Start":"01:33.110 ","End":"01:36.050","Text":"I computed it, it comes out to be 4 but anyway,"},{"Start":"01:36.050 ","End":"01:37.970","Text":"that\u0027s not important, it\u0027s non-0."},{"Start":"01:37.970 ","End":"01:44.330","Text":"Now, this is true for any x in particular for x equals 1 so u(1,"},{"Start":"01:44.330 ","End":"01:50.705","Text":"t) is O of 1 over root t. That means that when t goes to infinity,"},{"Start":"01:50.705 ","End":"01:53.060","Text":"this t^1/3 u(1,"},{"Start":"01:53.060 ","End":"01:58.160","Text":"t) is the limit of t^1/3 something of the order of 1 over"},{"Start":"01:58.160 ","End":"02:03.525","Text":"root t and this comes out to be O(1 over t^1/6)."},{"Start":"02:03.525 ","End":"02:06.120","Text":"If you want the computation,"},{"Start":"02:06.120 ","End":"02:10.830","Text":"t^1/3 over square root of t is t^1/3 minus a 1/2,"},{"Start":"02:10.830 ","End":"02:14.205","Text":"which is t^-1/6, which is this."},{"Start":"02:14.205 ","End":"02:16.830","Text":"When t goes to infinity,"},{"Start":"02:16.830 ","End":"02:21.510","Text":"t^1/6 goes to infinity so this goes to 0,"},{"Start":"02:21.510 ","End":"02:23.590","Text":"and that\u0027s the answer."}],"ID":30768},{"Watched":false,"Name":"Exercise 2","Duration":"2m 2s","ChapterTopicVideoID":29183,"CourseChapterTopicPlaylistID":294433,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.290","Text":"In this exercise, we have an initial boundary value problem,"},{"Start":"00:04.290 ","End":"00:06.000","Text":"which is the heat equation on"},{"Start":"00:06.000 ","End":"00:10.290","Text":"an infinite interval with initial condition. Didn\u0027t write it."},{"Start":"00:10.290 ","End":"00:14.475","Text":"We assume that u is a solution of this IBVP,"},{"Start":"00:14.475 ","End":"00:20.070","Text":"and our task is to compute the limit of root t u(x,"},{"Start":"00:20.070 ","End":"00:22.575","Text":"t) as t goes to infinity."},{"Start":"00:22.575 ","End":"00:26.265","Text":"We call this initial condition h(x), which is this,"},{"Start":"00:26.265 ","End":"00:33.420","Text":"there is a requirement to show that h(x) is bounded by an exponential like this."},{"Start":"00:33.420 ","End":"00:35.490","Text":"It\u0027s possible to do that,"},{"Start":"00:35.490 ","End":"00:36.705","Text":"but we won\u0027t do it."},{"Start":"00:36.705 ","End":"00:44.270","Text":"Basically, the polynomial 2x goes to infinity much slower than e^minus x^2 goes to 0."},{"Start":"00:44.270 ","End":"00:47.410","Text":"This is negligible compared to this."},{"Start":"00:47.410 ","End":"00:50.540","Text":"Given that this is so we know that u(x,"},{"Start":"00:50.540 ","End":"00:54.065","Text":"t) can be written as a series as follows."},{"Start":"00:54.065 ","End":"00:57.228","Text":"M naught is this integral,"},{"Start":"00:57.228 ","End":"00:59.090","Text":"it\u0027s from the tutorial,"},{"Start":"00:59.090 ","End":"01:03.545","Text":"and h(s) is 2se^minus x^2."},{"Start":"01:03.545 ","End":"01:07.860","Text":"This is an odd function on a symmetric interval,"},{"Start":"01:07.860 ","End":"01:10.530","Text":"so the integral is 0,"},{"Start":"01:10.530 ","End":"01:13.470","Text":"u(x, t) therefore starts from M1."},{"Start":"01:13.470 ","End":"01:16.400","Text":"Doesn\u0027t really matter if M1(x) is 0 or not."},{"Start":"01:16.400 ","End":"01:22.225","Text":"In any event, it\u0027s O(1 over t)^3 over 2."},{"Start":"01:22.225 ","End":"01:24.060","Text":"It goes to 0,"},{"Start":"01:24.060 ","End":"01:28.390","Text":"at least as quick as (1 over t)^3 over 2."},{"Start":"01:28.390 ","End":"01:32.460","Text":"Here\u0027s a reminder of what O notation is,"},{"Start":"01:32.460 ","End":"01:34.745","Text":"copied it from the Wikipedia."},{"Start":"01:34.745 ","End":"01:37.910","Text":"The limit as t goes to infinity of root 2u(x,"},{"Start":"01:37.910 ","End":"01:46.175","Text":"t) is the limit of the root t is t^1/2 and something of the order of (1 over t)^3 over 2."},{"Start":"01:46.175 ","End":"01:50.405","Text":"We can write this as O(1 over t),"},{"Start":"01:50.405 ","End":"01:55.940","Text":"t^1/2 over t^3 over 2 is 1 over t. When t goes to infinity,"},{"Start":"01:55.940 ","End":"01:57.965","Text":"this goes to 0,"},{"Start":"01:57.965 ","End":"02:00.365","Text":"so the answer is 0."},{"Start":"02:00.365 ","End":"02:03.450","Text":"That concludes this exercise."}],"ID":30769},{"Watched":false,"Name":"Exercise 3","Duration":"2m 1s","ChapterTopicVideoID":29184,"CourseChapterTopicPlaylistID":294433,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.850","Text":"In this exercise, we\u0027re given that the function"},{"Start":"00:02.850 ","End":"00:08.040","Text":"u(xt) is a solution of the following initial boundary value problem."},{"Start":"00:08.040 ","End":"00:12.509","Text":"This is the PDE on the infinite integral, initial condition."},{"Start":"00:12.509 ","End":"00:17.895","Text":"Our task is to compute the limit as t goes to infinity of root t u(xt)."},{"Start":"00:17.895 ","End":"00:21.750","Text":"As usual we call this function h(x)."},{"Start":"00:21.750 ","End":"00:23.250","Text":"Like in the tutorial,"},{"Start":"00:23.250 ","End":"00:26.880","Text":"we need to show that absolute value of h(x) is less than"},{"Start":"00:26.880 ","End":"00:31.080","Text":"or equal to Beta e to the minus Alpha absolute value of x."},{"Start":"00:31.080 ","End":"00:34.170","Text":"This will work out if we let Alpha equals 1,"},{"Start":"00:34.170 ","End":"00:37.230","Text":"and Beta equals 1 we\u0027ll get exactly this."},{"Start":"00:37.230 ","End":"00:43.400","Text":"This guarantees that we can write u(xt) as a series as follows."},{"Start":"00:43.400 ","End":"00:47.390","Text":"We know that m naught is given by this integral."},{"Start":"00:47.390 ","End":"00:49.355","Text":"Let\u0027s compute it."},{"Start":"00:49.355 ","End":"00:53.120","Text":"This is e to the minus absolute value of s,"},{"Start":"00:53.120 ","End":"00:55.370","Text":"and this is an even function,"},{"Start":"00:55.370 ","End":"00:59.375","Text":"so we can take twice the integral just from 0 to infinity."},{"Start":"00:59.375 ","End":"01:03.470","Text":"The integral of e to the minus s is minus e to the minus s. To"},{"Start":"01:03.470 ","End":"01:07.865","Text":"get the minus and switch the order of 0 and infinity,"},{"Start":"01:07.865 ","End":"01:11.420","Text":"so we get e to the minus 0 which is 1,"},{"Start":"01:11.420 ","End":"01:15.315","Text":"and minus e to the minus infinity which is 0."},{"Start":"01:15.315 ","End":"01:17.205","Text":"This comes out to be 2."},{"Start":"01:17.205 ","End":"01:21.480","Text":"Replace M naught by 2 and we\u0027ve got this."},{"Start":"01:21.480 ","End":"01:23.850","Text":"2/2 is 1."},{"Start":"01:23.850 ","End":"01:26.745","Text":"If we multiply by root t,"},{"Start":"01:26.745 ","End":"01:30.210","Text":"this comes out to be just 1 over a root Pi."},{"Start":"01:30.210 ","End":"01:33.830","Text":"Well, the rest of them, 1 over higher powers of t here,"},{"Start":"01:33.830 ","End":"01:35.270","Text":"we get 1/t,"},{"Start":"01:35.270 ","End":"01:37.785","Text":"1 over t^2, and so on."},{"Start":"01:37.785 ","End":"01:43.070","Text":"We can say that all these are of order 1/t."},{"Start":"01:43.070 ","End":"01:49.070","Text":"In case you\u0027ve forgotten what O is or you need a refresher, copy this view."},{"Start":"01:49.070 ","End":"01:51.865","Text":"Now we want to let t go to infinity."},{"Start":"01:51.865 ","End":"01:54.435","Text":"This is a constant. It stays what it is,"},{"Start":"01:54.435 ","End":"01:56.805","Text":"and this goes to 0."},{"Start":"01:56.805 ","End":"02:00.245","Text":"The answer is 1 over a root Pi,"},{"Start":"02:00.245 ","End":"02:02.490","Text":"and we are done."}],"ID":30770},{"Watched":false,"Name":"Exercise 4","Duration":"3m 10s","ChapterTopicVideoID":29185,"CourseChapterTopicPlaylistID":294433,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.075","Text":"In this exercise, we are given the following initial and boundary value problem."},{"Start":"00:06.075 ","End":"00:11.250","Text":"Homogeneous heat equation on an infinite interval with initial condition."},{"Start":"00:11.250 ","End":"00:14.820","Text":"I should have said, u is assumed to be a solution to this."},{"Start":"00:14.820 ","End":"00:22.020","Text":"We have to compute the limit as t goes to infinity of u of x and t. Now in this exercise,"},{"Start":"00:22.020 ","End":"00:26.415","Text":"h doesn\u0027t satisfy the usual requirement that we need."},{"Start":"00:26.415 ","End":"00:29.955","Text":"It is not smaller or equal to this exponential"},{"Start":"00:29.955 ","End":"00:35.845","Text":"because the limit when x goes to infinity of h(x) is 1."},{"Start":"00:35.845 ","End":"00:42.225","Text":"This goes to 0, but the exponential always goes to 0 when x goes to infinity."},{"Start":"00:42.225 ","End":"00:45.230","Text":"What we\u0027re going to do is split this problem up into 2,"},{"Start":"00:45.230 ","End":"00:47.165","Text":"we\u0027ll take the 1 separately."},{"Start":"00:47.165 ","End":"00:50.525","Text":"What I mean is we\u0027ll let u equal v plus w,"},{"Start":"00:50.525 ","End":"00:55.670","Text":"where v satisfies the equation with initial condition of"},{"Start":"00:55.670 ","End":"01:02.360","Text":"1 and w satisfies the equation with initial condition e to the minus absolute value of x."},{"Start":"01:02.360 ","End":"01:04.730","Text":"If you take v plus w,"},{"Start":"01:04.730 ","End":"01:08.120","Text":"then it will satisfy also the heat equation and"},{"Start":"01:08.120 ","End":"01:12.875","Text":"the initial condition it satisfies will be 1 plus e to the minus absolute value of x."},{"Start":"01:12.875 ","End":"01:14.930","Text":"If we call this g(x),"},{"Start":"01:14.930 ","End":"01:18.275","Text":"this time we\u0027re okay with the initial condition."},{"Start":"01:18.275 ","End":"01:20.300","Text":"It is less than or equal to this,"},{"Start":"01:20.300 ","End":"01:23.885","Text":"if we take Alpha equals 1 and Beta equals 1."},{"Start":"01:23.885 ","End":"01:26.330","Text":"Let\u0027s take care of v first of all."},{"Start":"01:26.330 ","End":"01:33.260","Text":"v is given by Poisson\u0027s formula and v is equal to 1."},{"Start":"01:33.260 ","End":"01:37.135","Text":"What we get is the following integral."},{"Start":"01:37.135 ","End":"01:42.570","Text":"Substitute z equals e^y minus x over"},{"Start":"01:42.570 ","End":"01:47.795","Text":"2a root t. Just what\u0027s inside the brackets here and dz is this."},{"Start":"01:47.795 ","End":"01:57.095","Text":"dy is 2a root t dz and 2a root t cancels and we just have 1 over root Pi,"},{"Start":"01:57.095 ","End":"02:00.185","Text":"the integral of e to the minus z^2 dz."},{"Start":"02:00.185 ","End":"02:02.285","Text":"This is a famous integral,"},{"Start":"02:02.285 ","End":"02:06.010","Text":"and this integral is equal to root Pi."},{"Start":"02:06.010 ","End":"02:09.480","Text":"v(x, t) is equal to the constant 1."},{"Start":"02:09.480 ","End":"02:10.700","Text":"It makes sense."},{"Start":"02:10.700 ","End":"02:13.520","Text":"It\u0027s just equal to the initial condition."},{"Start":"02:13.520 ","End":"02:17.075","Text":"Since it\u0027s a constant, there\u0027s no reason for anything to change."},{"Start":"02:17.075 ","End":"02:19.590","Text":"It just stays equal to 1."},{"Start":"02:19.590 ","End":"02:22.670","Text":"Now, w can be written as a series,"},{"Start":"02:22.670 ","End":"02:25.775","Text":"just like in the tutorial as follows."},{"Start":"02:25.775 ","End":"02:32.805","Text":"The formula for M_0 is the integral of g(s) ds minus infinity to infinity."},{"Start":"02:32.805 ","End":"02:37.745","Text":"It comes out to be strictly positive anyway, non-zero."},{"Start":"02:37.745 ","End":"02:40.310","Text":"If you want to know, it\u0027s actually equal to 2."},{"Start":"02:40.310 ","End":"02:43.580","Text":"It\u0027s twice the integral from 0 to infinity of e to the minus s,"},{"Start":"02:43.580 ","End":"02:46.405","Text":"anyway, point is it\u0027s not 0."},{"Start":"02:46.405 ","End":"02:55.835","Text":"W is of the order of O of 1 over root t. That means that when t goes to infinity,"},{"Start":"02:55.835 ","End":"02:57.935","Text":"w goes to 0."},{"Start":"02:57.935 ","End":"03:03.695","Text":"The limit of u is the limit of v plus w. v is the constant 1,"},{"Start":"03:03.695 ","End":"03:05.555","Text":"w goes to 0,"},{"Start":"03:05.555 ","End":"03:08.460","Text":"so the limit is equal to 1."},{"Start":"03:08.460 ","End":"03:10.989","Text":"That concludes this exercise."}],"ID":30771}],"Thumbnail":null,"ID":294433},{"Name":"Asymptotic Behaviour on a Finite Interval","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"7m 34s","ChapterTopicVideoID":29187,"CourseChapterTopicPlaylistID":294434,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":30764},{"Watched":false,"Name":"Exercise 2","Duration":"5m 24s","ChapterTopicVideoID":29188,"CourseChapterTopicPlaylistID":294434,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":30765},{"Watched":false,"Name":"Exercise 3","Duration":"5m 50s","ChapterTopicVideoID":29189,"CourseChapterTopicPlaylistID":294434,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":30766}],"Thumbnail":null,"ID":294434}]

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1.1

1