[{"Name":"Review - Type I Line Integral [and more]","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Review - Type I Line Integrals","Duration":"5m 13s","ChapterTopicVideoID":29241,"CourseChapterTopicPlaylistID":294435,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:05.175","Text":"In this clip, we\u0027ll give a quick review of type 1 line integrals."},{"Start":"00:05.175 ","End":"00:08.115","Text":"Here\u0027s what such a line integral looks like."},{"Start":"00:08.115 ","End":"00:14.940","Text":"I forgot to say that this is in 2D because you could also have 3D or higher dimensions."},{"Start":"00:14.940 ","End":"00:22.470","Text":"The function f is a function of 2 real variables and takes scalar values,"},{"Start":"00:22.470 ","End":"00:25.200","Text":"other words, just real values."},{"Start":"00:25.200 ","End":"00:29.930","Text":"ds is defined as the square root of dx squared plus dy squared."},{"Start":"00:29.930 ","End":"00:32.390","Text":"It\u0027s a differential of curve length."},{"Start":"00:32.390 ","End":"00:34.465","Text":"We\u0027ll see this in a moment."},{"Start":"00:34.465 ","End":"00:37.865","Text":"Gamma is a curve,"},{"Start":"00:37.865 ","End":"00:40.399","Text":"it\u0027s directed means it has a direction,"},{"Start":"00:40.399 ","End":"00:42.230","Text":"if you go from a to b,"},{"Start":"00:42.230 ","End":"00:44.660","Text":"it\u0027s not the same as going from b back to a."},{"Start":"00:44.660 ","End":"00:47.120","Text":"There\u0027s a narrow on the curve, smooth,"},{"Start":"00:47.120 ","End":"00:50.165","Text":"meaning it has a derivative."},{"Start":"00:50.165 ","End":"00:52.490","Text":"Parameterizable, we can express x,"},{"Start":"00:52.490 ","End":"00:55.565","Text":"y as functions of t. We\u0027ll define that in a moment."},{"Start":"00:55.565 ","End":"00:59.525","Text":"Here\u0027s an example of what a line integral type 1 looks like."},{"Start":"00:59.525 ","End":"01:03.530","Text":"This is it. We have to always say what gamma is."},{"Start":"01:03.530 ","End":"01:07.520","Text":"Gamma is the unit circle x^2 plus y^2 equals 1."},{"Start":"01:07.520 ","End":"01:10.310","Text":"Often we describe gamma simply as the set of"},{"Start":"01:10.310 ","End":"01:14.420","Text":"points and we ignore the fact that there\u0027s a parameterization."},{"Start":"01:14.420 ","End":"01:18.125","Text":"That\u0027s why I say parameterizable that it could be parameterized."},{"Start":"01:18.125 ","End":"01:21.430","Text":"Now there\u0027s also a direction because I said directed."},{"Start":"01:21.430 ","End":"01:26.345","Text":"We\u0027ll say we go around counterclockwise and this is in fact a default."},{"Start":"01:26.345 ","End":"01:27.739","Text":"If you don\u0027t say anything,"},{"Start":"01:27.739 ","End":"01:31.055","Text":"it\u0027s counterclockwise, if it\u0027s a closed curve."},{"Start":"01:31.055 ","End":"01:33.820","Text":"Now the definition of what this thing equals."},{"Start":"01:33.820 ","End":"01:37.760","Text":"What we do is we find a parametrization of gamma."},{"Start":"01:37.760 ","End":"01:47.730","Text":"We express gamma as a function of t. T goes from a to b. Gamma (t) =(x (t), y (t))."},{"Start":"01:47.730 ","End":"01:55.865","Text":"Then we define the integral to be the integral from a to b of f (x (t), y (t))."},{"Start":"01:55.865 ","End":"02:00.740","Text":"I won\u0027t read this out, but this is essentially what ds is."},{"Start":"02:00.740 ","End":"02:03.980","Text":"I told you before, the ds is square root of dx^2 plus dy^2."},{"Start":"02:03.980 ","End":"02:05.435","Text":"It\u0027s easier to remember it this way."},{"Start":"02:05.435 ","End":"02:07.310","Text":"In fact it\u0027s equal to it."},{"Start":"02:07.310 ","End":"02:12.420","Text":"Because if x is x (t),"},{"Start":"02:12.420 ","End":"02:15.025","Text":"then dx is x\u0027 (t) dt,"},{"Start":"02:15.025 ","End":"02:18.065","Text":"and dy is y\u0027 (t) dt."},{"Start":"02:18.065 ","End":"02:20.300","Text":"As you substitute that here,"},{"Start":"02:20.300 ","End":"02:29.120","Text":"the dt comes out of the square root and we get the square root of (x\u0027)^2 plus (y\u0027)^2."},{"Start":"02:29.120 ","End":"02:33.800","Text":"It all works out. But this is the definition used in practice."},{"Start":"02:33.800 ","End":"02:36.950","Text":"I have to say something though, because you could argue that what if"},{"Start":"02:36.950 ","End":"02:40.085","Text":"we chose a different parametrization we\u0027d get a different answer?"},{"Start":"02:40.085 ","End":"02:44.435","Text":"Well no, it turns out that it doesn\u0027t depend on the parameterization."},{"Start":"02:44.435 ","End":"02:46.250","Text":"Not a hundred percent precise,"},{"Start":"02:46.250 ","End":"02:49.160","Text":"but good enough to say that this provided that"},{"Start":"02:49.160 ","End":"02:52.985","Text":"the curve is traversed only once from a to b,"},{"Start":"02:52.985 ","End":"02:55.010","Text":"or if it\u0027s a circular,"},{"Start":"02:55.010 ","End":"02:58.400","Text":"like in our case, just go around the circle once."},{"Start":"02:58.400 ","End":"03:01.040","Text":"You go around twice, then you\u0027d get a different answer."},{"Start":"03:01.040 ","End":"03:03.680","Text":"Now let\u0027s do an example exercise where we\u0027ll take"},{"Start":"03:03.680 ","End":"03:08.150","Text":"exactly this line integral here and compute what it\u0027s equal to."},{"Start":"03:08.150 ","End":"03:13.720","Text":"Exercise: Compute the integral on Gamma (x^2 + y) ds,"},{"Start":"03:13.720 ","End":"03:19.355","Text":"where Gamma is the unit circle and we assume it\u0027s traversed counterclockwise."},{"Start":"03:19.355 ","End":"03:22.480","Text":"We start with a parameterization of the curve Gamma."},{"Start":"03:22.480 ","End":"03:26.810","Text":"The usual parameterization for the unit circle, at least the simplest 1,"},{"Start":"03:26.810 ","End":"03:33.550","Text":"is to take Gamma (t) to be cosine t sine t. When t goes from 0 to 2Pi,"},{"Start":"03:33.550 ","End":"03:35.735","Text":"t is like the angle in radians,"},{"Start":"03:35.735 ","End":"03:38.885","Text":"then cosine t sine t just goes around the unit circle."},{"Start":"03:38.885 ","End":"03:40.880","Text":"It starts at the point 1,"},{"Start":"03:40.880 ","End":"03:43.580","Text":"0 and ends at the point 1,0."},{"Start":"03:43.580 ","End":"03:45.590","Text":"Now we use this formula,"},{"Start":"03:45.590 ","End":"03:50.210","Text":"so we substitute and what we get is x^2 is cosine^2 t,"},{"Start":"03:50.210 ","End":"03:51.785","Text":"y is sine t,"},{"Start":"03:51.785 ","End":"03:53.825","Text":"ds is the square root."},{"Start":"03:53.825 ","End":"03:58.550","Text":"X\u0027 of t is the derivative of this,"},{"Start":"03:58.550 ","End":"04:01.460","Text":"which is minus sine (t)^2."},{"Start":"04:01.460 ","End":"04:05.975","Text":"Then the derivative of this which is cosine (t)^2 dt."},{"Start":"04:05.975 ","End":"04:09.550","Text":"This from 0 to 2Pi, which is from here."},{"Start":"04:09.550 ","End":"04:11.420","Text":"This is just a regular integral."},{"Start":"04:11.420 ","End":"04:13.355","Text":"Now it\u0027s not a line integral or anything,"},{"Start":"04:13.355 ","End":"04:15.635","Text":"just an ordinary integral."},{"Start":"04:15.635 ","End":"04:18.650","Text":"This comes out to be sine squared plus cosine squared,"},{"Start":"04:18.650 ","End":"04:20.330","Text":"which is 1 square root of 1 is 1."},{"Start":"04:20.330 ","End":"04:22.325","Text":"This thing just drops off."},{"Start":"04:22.325 ","End":"04:24.679","Text":"Now we can use trigonometric identities."},{"Start":"04:24.679 ","End":"04:29.435","Text":"Cosine^2 t is 1 plus cosine 2t over 2."},{"Start":"04:29.435 ","End":"04:32.119","Text":"We can break the integral up into 2 parts."},{"Start":"04:32.119 ","End":"04:35.030","Text":"Now, we can also break this into"},{"Start":"04:35.030 ","End":"04:38.375","Text":"2 parts and take the half in front of each of the integrals."},{"Start":"04:38.375 ","End":"04:40.265","Text":"Now we have 3 pieces."},{"Start":"04:40.265 ","End":"04:42.019","Text":"Let\u0027s actually do the integration."},{"Start":"04:42.019 ","End":"04:44.035","Text":"The integral of 1 is t,"},{"Start":"04:44.035 ","End":"04:48.420","Text":"the integral of cosine 2t is a quarter sine 2t."},{"Start":"04:48.420 ","End":"04:51.260","Text":"It\u0027s a half, but with the half it gives a quarter."},{"Start":"04:51.260 ","End":"04:56.140","Text":"Sorry. The integral of sine is minus cosine."},{"Start":"04:56.140 ","End":"04:59.900","Text":"All these integrals are evaluated from 0 to 2Pi."},{"Start":"04:59.900 ","End":"05:03.695","Text":"What we get, I\u0027ll leave you to check the details of the following."},{"Start":"05:03.695 ","End":"05:06.350","Text":"The answer is Pi."},{"Start":"05:06.350 ","End":"05:10.655","Text":"This line integral type 1 is equal to Pi."},{"Start":"05:10.655 ","End":"05:14.339","Text":"That\u0027s all for this quick refresher."}],"ID":30814},{"Watched":false,"Name":"Review - The 2D Divergence Theorem","Duration":"8m 44s","ChapterTopicVideoID":29239,"CourseChapterTopicPlaylistID":294435,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:04.770","Text":"In this clip, we\u0027ll review the 2D divergence theorem,"},{"Start":"00:04.770 ","End":"00:08.130","Text":"also known as Gauss\u0027s theorem."},{"Start":"00:08.130 ","End":"00:11.760","Text":"We\u0027re going to build on the previous clip where we did"},{"Start":"00:11.760 ","End":"00:15.435","Text":"a review of what a Type 1 line integral is."},{"Start":"00:15.435 ","End":"00:19.125","Text":"The 2D divergence theorem will be used in the following clip,"},{"Start":"00:19.125 ","End":"00:21.750","Text":"we\u0027ll need it for a certain proof."},{"Start":"00:21.750 ","End":"00:26.190","Text":"The setup is that we have a bounded region called D,"},{"Start":"00:26.190 ","End":"00:32.445","Text":"and its boundary is a simple closed piecewise smooth curve,"},{"Start":"00:32.445 ","End":"00:36.345","Text":"which is also called Gamma."},{"Start":"00:36.345 ","End":"00:39.095","Text":"As a subtle difference here,"},{"Start":"00:39.095 ","End":"00:41.345","Text":"boundary of D is just a set of points."},{"Start":"00:41.345 ","End":"00:46.190","Text":"Gamma is a parameterizable curve and it has an orientation,"},{"Start":"00:46.190 ","End":"00:48.230","Text":"but we often mix the notation."},{"Start":"00:48.230 ","End":"00:52.310","Text":"If you\u0027re wondering why piecewise smooth, just think of,"},{"Start":"00:52.310 ","End":"00:53.900","Text":"for example, a rectangle,"},{"Start":"00:53.900 ","End":"00:55.610","Text":"it has corners, so it\u0027s not smooth,"},{"Start":"00:55.610 ","End":"00:57.470","Text":"it\u0027s just piecewise smooth."},{"Start":"00:57.470 ","End":"01:01.295","Text":"Part of the setup is that we\u0027re given a vector field,"},{"Start":"01:01.295 ","End":"01:04.320","Text":"F, whose components are P and Q,"},{"Start":"01:04.320 ","End":"01:06.665","Text":"and it\u0027s continuously differentiable,"},{"Start":"01:06.665 ","End":"01:09.230","Text":"not just in D but in a neighborhood of D,"},{"Start":"01:09.230 ","End":"01:14.000","Text":"meaning a bit more than D. With a hat on it is"},{"Start":"01:14.000 ","End":"01:18.970","Text":"the outward facing unit normal vector to Gamma."},{"Start":"01:18.970 ","End":"01:23.270","Text":"A hat usually indicates unit vector and n is"},{"Start":"01:23.270 ","End":"01:27.890","Text":"a function of x and y or the point on the boundary."},{"Start":"01:27.890 ","End":"01:29.300","Text":"All this was the setup."},{"Start":"01:29.300 ","End":"01:32.000","Text":"The theorem claims the following."},{"Start":"01:32.000 ","End":"01:34.010","Text":"The integral along Gamma,"},{"Start":"01:34.010 ","End":"01:36.290","Text":"sometimes written as boundary of D,"},{"Start":"01:36.290 ","End":"01:39.935","Text":"and then it\u0027s understood that it\u0027s counterclockwise of"},{"Start":"01:39.935 ","End":"01:45.155","Text":"the dot-product of the vector field with the outer normal."},{"Start":"01:45.155 ","End":"01:51.410","Text":"The integral of that over the boundary is equal to the integral on all"},{"Start":"01:51.410 ","End":"01:57.555","Text":"of the region of the dot-product Del operator,"},{"Start":"01:57.555 ","End":"01:58.725","Text":"called a Nabla,"},{"Start":"01:58.725 ","End":"02:01.035","Text":"dot f, and dA,"},{"Start":"02:01.035 ","End":"02:03.060","Text":"which is dx, dy."},{"Start":"02:03.060 ","End":"02:07.280","Text":"Just to remind you what ds and da are."},{"Start":"02:07.280 ","End":"02:12.790","Text":"Here\u0027s a definition of the operator grad Del Nabla."},{"Start":"02:12.790 ","End":"02:21.005","Text":"What this comes down to is actually dP by dx plus dQ by dy, anyway."},{"Start":"02:21.005 ","End":"02:29.480","Text":"This is also called the divergence of the vector field F. That\u0027s it for the theory part."},{"Start":"02:29.480 ","End":"02:30.980","Text":"Let\u0027s just do an example."},{"Start":"02:30.980 ","End":"02:34.550","Text":"We will take an actual case and evaluate"},{"Start":"02:34.550 ","End":"02:38.780","Text":"left-hand side and right-hand side and show that they really are equal."},{"Start":"02:38.780 ","End":"02:45.895","Text":"As an example, we\u0027ll take our vector field F to be x^3, y^3."},{"Start":"02:45.895 ","End":"02:49.880","Text":"We\u0027ll take Gamma to be the unit circle."},{"Start":"02:49.880 ","End":"02:53.630","Text":"Doesn\u0027t say anything, so it\u0027s counterclockwise."},{"Start":"02:53.630 ","End":"03:00.695","Text":"Now, note that the unit circle is the boundary of the unit disk,"},{"Start":"03:00.695 ","End":"03:04.190","Text":"which in Cartesian coordinates is this,"},{"Start":"03:04.190 ","End":"03:07.490","Text":"but in polar coordinates it\u0027s r less than 1."},{"Start":"03:07.490 ","End":"03:09.410","Text":"Now the question is,"},{"Start":"03:09.410 ","End":"03:12.725","Text":"is the divergence theorem satisfied in this case?"},{"Start":"03:12.725 ","End":"03:15.020","Text":"In other words, does this equation hold?"},{"Start":"03:15.020 ","End":"03:18.665","Text":"We\u0027ll evaluate the right-hand side, and see what we get."},{"Start":"03:18.665 ","End":"03:22.680","Text":"Then we\u0027ll evaluate the left-hand side and see if we get the same thing."},{"Start":"03:22.680 ","End":"03:26.631","Text":"The right-hand side is Del dot,"},{"Start":"03:26.631 ","End":"03:29.470","Text":"x^3, y^3 from here, dA."},{"Start":"03:29.470 ","End":"03:32.060","Text":"This is d by dx, d by dy,"},{"Start":"03:32.060 ","End":"03:35.420","Text":"so we get the partial derivative with respect to x"},{"Start":"03:35.420 ","End":"03:40.735","Text":"plus the partial derivative with respect to y of the second coordinate."},{"Start":"03:40.735 ","End":"03:45.800","Text":"Now we\u0027ll make a substitution because it\u0027s easier in polar coordinates."},{"Start":"03:45.800 ","End":"03:47.555","Text":"After all it is a disk."},{"Start":"03:47.555 ","End":"03:50.255","Text":"So the usual x= r cos Theta,"},{"Start":"03:50.255 ","End":"03:51.643","Text":"y= r sin Theta,"},{"Start":"03:51.643 ","End":"03:54.475","Text":"and dxdy, which is dA,"},{"Start":"03:54.475 ","End":"03:56.580","Text":"is rdr d Theta."},{"Start":"03:56.580 ","End":"03:59.940","Text":"Going straight away say that dA is rdr d Theta, it\u0027s not."},{"Start":"03:59.940 ","End":"04:05.025","Text":"Of course, x^2 plus y^2 is r^2."},{"Start":"04:05.025 ","End":"04:09.640","Text":"What we get is 3r^2 rdr d Theta."},{"Start":"04:09.640 ","End":"04:16.715","Text":"The unit disk in polar coordinates can be described as r goes from 0 to 1,"},{"Start":"04:16.715 ","End":"04:19.520","Text":"and Theta goes from 0 to 2Pi."},{"Start":"04:19.520 ","End":"04:23.570","Text":"It\u0027s like a rectangular region when you\u0027re talking polar."},{"Start":"04:23.570 ","End":"04:28.565","Text":"We can actually completely separate r from Theta and get the product of 2 integrals."},{"Start":"04:28.565 ","End":"04:31.625","Text":"We can get r^3dr from 0 to 1."},{"Start":"04:31.625 ","End":"04:33.260","Text":"All this has nothing to do with Theta,"},{"Start":"04:33.260 ","End":"04:36.005","Text":"so this whole integral can be pulled out in front,"},{"Start":"04:36.005 ","End":"04:40.195","Text":"and then we\u0027re left with just the integral of 1d Theta from 0 to Pi."},{"Start":"04:40.195 ","End":"04:46.440","Text":"This is equal to integral of r^3 from 0 to 1 is r^4 over 4,"},{"Start":"04:46.440 ","End":"04:47.940","Text":"but substitute 1 and 0,"},{"Start":"04:47.940 ","End":"04:49.185","Text":"we\u0027re left with 1/4."},{"Start":"04:49.185 ","End":"04:51.375","Text":"This integral is 2Pi."},{"Start":"04:51.375 ","End":"04:54.645","Text":"Altogether we have 3Pi over 2."},{"Start":"04:54.645 ","End":"04:59.235","Text":"That was the right hand side for this."},{"Start":"04:59.235 ","End":"05:04.460","Text":"Now we\u0027ll evaluate the left hand side and hopefully we\u0027ll also get 3Pi over 2."},{"Start":"05:04.460 ","End":"05:07.025","Text":"This is what the left hand side says."},{"Start":"05:07.025 ","End":"05:10.805","Text":"Now, we need to know what the unit normal vector is,"},{"Start":"05:10.805 ","End":"05:12.590","Text":"the outward facing 1."},{"Start":"05:12.590 ","End":"05:14.675","Text":"Here\u0027s a picture disk,"},{"Start":"05:14.675 ","End":"05:19.025","Text":"the circle and the outward facing vector."},{"Start":"05:19.025 ","End":"05:26.145","Text":"In general, a normal vector to a curve of the following form."},{"Start":"05:26.145 ","End":"05:27.890","Text":"F(x, y) is a constant,"},{"Start":"05:27.890 ","End":"05:30.260","Text":"is given by grad f,"},{"Start":"05:30.260 ","End":"05:35.235","Text":"where grad is this Del or Nabla."},{"Start":"05:35.235 ","End":"05:37.800","Text":"Not dot f, but just nabla f,"},{"Start":"05:37.800 ","End":"05:41.710","Text":"which means df by dx, df by dy."},{"Start":"05:42.080 ","End":"05:44.370","Text":"In our case, f(x,"},{"Start":"05:44.370 ","End":"05:46.440","Text":"y) is x^2 plus y^2."},{"Start":"05:46.440 ","End":"05:53.130","Text":"The gradient is (2x, 2y)."},{"Start":"05:53.130 ","End":"05:54.670","Text":"We need a unit vector,"},{"Start":"05:54.670 ","End":"05:58.224","Text":"so we take the vector and divide it by its norm."},{"Start":"05:58.224 ","End":"06:01.345","Text":"The 2 cancels and we get this."},{"Start":"06:01.345 ","End":"06:03.070","Text":"Or we could use this expression,"},{"Start":"06:03.070 ","End":"06:08.815","Text":"but we can simplify it because we know that on Gamma x^2 plus y^2=1."},{"Start":"06:08.815 ","End":"06:12.159","Text":"It\u0027s easy to see that this is an outward facing integral by substituting,"},{"Start":"06:12.159 ","End":"06:14.755","Text":"for example, (1, 0)."},{"Start":"06:14.755 ","End":"06:21.010","Text":"We can just write that n= (x, y) the vector."},{"Start":"06:21.010 ","End":"06:24.880","Text":"Now back here, F.n,"},{"Start":"06:24.880 ","End":"06:33.000","Text":"this is F and this is n. Doing the dot product gives us x^4 plus y^4."},{"Start":"06:33.000 ","End":"06:35.900","Text":"Now this is a Type 1 line integral,"},{"Start":"06:35.900 ","End":"06:38.315","Text":"which we know how to do."},{"Start":"06:38.315 ","End":"06:43.850","Text":"The usual parameterization for this circle is this cos t,"},{"Start":"06:43.850 ","End":"06:46.890","Text":"sin t from 0 to 2Pi."},{"Start":"06:47.290 ","End":"06:56.610","Text":"What we get is x^4 is cos to the 4 t. Y^4 is sin to the 4 t from 0 to Pi,"},{"Start":"06:56.610 ","End":"07:01.430","Text":"and ds is the square root of"},{"Start":"07:01.430 ","End":"07:07.850","Text":"the derivative of cos t squared plus the derivative of sin t squared."},{"Start":"07:07.850 ","End":"07:09.890","Text":"But this comes out to be 1,"},{"Start":"07:09.890 ","End":"07:11.980","Text":"so that drops out."},{"Start":"07:11.980 ","End":"07:16.365","Text":"We get the integral of cos to the 4 t plus sin to the 4 t dt."},{"Start":"07:16.365 ","End":"07:18.470","Text":"Now this is just a regular integral."},{"Start":"07:18.470 ","End":"07:20.810","Text":"We use some trigonometrical identities,"},{"Start":"07:20.810 ","End":"07:25.910","Text":"we write cos to the 4 of cos squared squared and similarly, sin squared squared."},{"Start":"07:25.910 ","End":"07:29.615","Text":"Then cos squared t is 1 plus cos 2t over 2."},{"Start":"07:29.615 ","End":"07:34.915","Text":"Sin squared t is 1 minus cos 2t over 2 squared squared."},{"Start":"07:34.915 ","End":"07:36.450","Text":"Just basic algebra,"},{"Start":"07:36.450 ","End":"07:40.865","Text":"take the 2^2 in the denominator out at a 1/4 in front."},{"Start":"07:40.865 ","End":"07:43.535","Text":"Square these quantities."},{"Start":"07:43.535 ","End":"07:45.080","Text":"Let\u0027s see what cancels here."},{"Start":"07:45.080 ","End":"07:47.980","Text":"The middle term cancels here and here."},{"Start":"07:47.980 ","End":"07:50.730","Text":"There\u0027ll be 2 of these and 2 of these,"},{"Start":"07:50.730 ","End":"07:57.620","Text":"so take the 2 out front and we have 1 plus cos squared t. Again,"},{"Start":"07:57.620 ","End":"08:00.965","Text":"we\u0027ll use the formula for cos squared."},{"Start":"08:00.965 ","End":"08:04.780","Text":"This time we\u0027ll get 1 plus cos 4t over 2."},{"Start":"08:04.780 ","End":"08:08.200","Text":"Bring the 2 out and write this as 2 plus 1 plus cos 4t,"},{"Start":"08:08.780 ","End":"08:14.160","Text":"which is 3 plus cos 4t dt."},{"Start":"08:14.160 ","End":"08:17.660","Text":"Now, the integral of cos 4t will"},{"Start":"08:17.660 ","End":"08:21.800","Text":"come out to be 0 because it\u0027s going to be sin 4t over 4,"},{"Start":"08:21.800 ","End":"08:25.700","Text":"but sine at multiples of Pi is 0."},{"Start":"08:25.700 ","End":"08:30.130","Text":"This is 0 and the integral of this is 3 times 2Pi,"},{"Start":"08:30.130 ","End":"08:34.935","Text":"so we get this expression which is 3Pi over 2."},{"Start":"08:34.935 ","End":"08:37.940","Text":"That\u0027s what we got before for the right-hand side,"},{"Start":"08:37.940 ","End":"08:40.175","Text":"and this is the left-hand side and they\u0027re equal."},{"Start":"08:40.175 ","End":"08:44.700","Text":"That concludes this exercise and this clip."}],"ID":30815},{"Watched":false,"Name":"A Necessary Condition for Solvability of the Neumann Problem","Duration":"5m 2s","ChapterTopicVideoID":29240,"CourseChapterTopicPlaylistID":294435,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.134","Text":"In this clip, we\u0027ll learn about a necessary condition"},{"Start":"00:03.134 ","End":"00:06.150","Text":"for the solution of the Neumann problem,"},{"Start":"00:06.150 ","End":"00:08.100","Text":"which we\u0027ll state in a moment."},{"Start":"00:08.100 ","End":"00:12.180","Text":"We\u0027ll start with reminding you what the Laplace Operator is."},{"Start":"00:12.180 ","End":"00:16.650","Text":"If we apply the Laplace Operator to a function of 2 variables,"},{"Start":"00:16.650 ","End":"00:18.675","Text":"could be more variables,"},{"Start":"00:18.675 ","End":"00:20.010","Text":"there\u0027s a higher-order version."},{"Start":"00:20.010 ","End":"00:21.510","Text":"This is the 2D version."},{"Start":"00:21.510 ","End":"00:26.775","Text":"The Laplacian Delta of u is u_xx plus u_yy,"},{"Start":"00:26.775 ","End":"00:31.290","Text":"the sum of the 2 non-mixed partial derivatives."},{"Start":"00:31.290 ","End":"00:38.630","Text":"Happens that Delta is Del squared or formal product of Del dot with Del."},{"Start":"00:38.630 ","End":"00:42.125","Text":"Don\u0027t worry about that Neumann problem."},{"Start":"00:42.125 ","End":"00:48.350","Text":"We apply it to the Poisson PDE and there\u0027s a thing called Neumann boundary condition."},{"Start":"00:48.350 ","End":"00:51.930","Text":"The Poisson PDE is basically"},{"Start":"00:51.930 ","End":"00:56.780","Text":"the non-homogeneous Laplace equation on the region"},{"Start":"00:56.780 ","End":"01:03.050","Text":"D. This is sometimes written in full as u_xx plus u_yy."},{"Start":"01:03.050 ","End":"01:05.660","Text":"Note that it\u0027s different from the wave equation."},{"Start":"01:05.660 ","End":"01:09.915","Text":"In the wave equation we had u_xx equals u_yy."},{"Start":"01:09.915 ","End":"01:13.950","Text":"Well, we have new u_tt equals u_xx, but same thing."},{"Start":"01:13.950 ","End":"01:18.125","Text":"Here, they\u0027re on the same side of the equals."},{"Start":"01:18.125 ","End":"01:23.600","Text":"The Neumann boundary condition in this problem is the following."},{"Start":"01:23.600 ","End":"01:28.550","Text":"The directional derivative of u in the direction of the"},{"Start":"01:28.550 ","End":"01:33.860","Text":"outward normal is a given function g. This is notation."},{"Start":"01:33.860 ","End":"01:36.905","Text":"There are many notations for directional derivative."},{"Start":"01:36.905 ","End":"01:38.165","Text":"This is one of them."},{"Start":"01:38.165 ","End":"01:42.635","Text":"One formula for the directional derivative is to take"},{"Start":"01:42.635 ","End":"01:49.345","Text":"grad of the function and dot product it with the vector whose direction you want."},{"Start":"01:49.345 ","End":"01:54.545","Text":"By the way, there is also something called a Dirichlet boundary condition,"},{"Start":"01:54.545 ","End":"01:57.135","Text":"where instead of du by dn,"},{"Start":"01:57.135 ","End":"02:01.235","Text":"the directional derivative we have the function u itself."},{"Start":"02:01.235 ","End":"02:03.945","Text":"You\u0027ll see this in the exercises."},{"Start":"02:03.945 ","End":"02:07.700","Text":"There\u0027s also a Laplace partial differential equation"},{"Start":"02:07.700 ","End":"02:09.710","Text":"with the Neumann boundary conditions."},{"Start":"02:09.710 ","End":"02:14.675","Text":"This is like this just the homogeneous version instead of f,"},{"Start":"02:14.675 ","End":"02:17.780","Text":"we put 0 and then it has a new name,"},{"Start":"02:17.780 ","End":"02:21.815","Text":"the Laplace equation with Neumann boundary condition."},{"Start":"02:21.815 ","End":"02:28.135","Text":"It\u0027s the same boundary condition here and here, non-homogeneous, homogeneous."},{"Start":"02:28.135 ","End":"02:30.000","Text":"Now, the theorem."},{"Start":"02:30.000 ","End":"02:33.470","Text":"The theorem says that if u is a solution to"},{"Start":"02:33.470 ","End":"02:37.760","Text":"the Poisson PDE with Neumann boundary conditions,"},{"Start":"02:37.760 ","End":"02:39.230","Text":"i.e, this problem,"},{"Start":"02:39.230 ","End":"02:43.385","Text":"then necessarily the following equality holds,"},{"Start":"02:43.385 ","End":"02:48.290","Text":"the integral of the function g on the boundary of D,"},{"Start":"02:48.290 ","End":"02:56.820","Text":"is equal to the double integral on D of the function f. That\u0027s this here."},{"Start":"02:56.820 ","End":"03:00.080","Text":"We can add a special case of Part 2 of this theorem."},{"Start":"03:00.080 ","End":"03:03.635","Text":"If f of x is equal to 0,"},{"Start":"03:03.635 ","End":"03:06.905","Text":"then we have the Laplace equation."},{"Start":"03:06.905 ","End":"03:09.080","Text":"The condition becomes,"},{"Start":"03:09.080 ","End":"03:13.550","Text":"this integral is equal to 0 because if f is 0,"},{"Start":"03:13.550 ","End":"03:15.200","Text":"this integral is 0,"},{"Start":"03:15.200 ","End":"03:19.385","Text":"so really this is just a special case of this."},{"Start":"03:19.385 ","End":"03:25.255","Text":"But this is named after Poisson and this is named after Laplace."},{"Start":"03:25.255 ","End":"03:29.285","Text":"Now the proof. Really only have to prove the Poisson case."},{"Start":"03:29.285 ","End":"03:32.420","Text":"Because like I said, if you just let f equals 0,"},{"Start":"03:32.420 ","End":"03:35.900","Text":"we automatically get the Laplace case from the Poisson."},{"Start":"03:35.900 ","End":"03:44.360","Text":"Here\u0027s a reminder of Gauss\u0027s theorem and in our case we\u0027ll take big F as grad view."},{"Start":"03:44.360 ","End":"03:47.330","Text":"I forgot to remind you earlier what dS and dA,"},{"Start":"03:47.330 ","End":"03:48.470","Text":"okay, it\u0027s written here."},{"Start":"03:48.470 ","End":"03:52.580","Text":"Applying this to the vector field F,"},{"Start":"03:52.580 ","End":"03:55.265","Text":"which is the gradient of u,"},{"Start":"03:55.265 ","End":"03:58.790","Text":"we get that F here."},{"Start":"03:58.790 ","End":"04:01.550","Text":"We put grad u,"},{"Start":"04:01.550 ","End":"04:03.500","Text":"and instead of F here,"},{"Start":"04:03.500 ","End":"04:06.058","Text":"we put grad u,"},{"Start":"04:06.058 ","End":"04:11.840","Text":"and then grad u dot unit vector n is the directional derivative of"},{"Start":"04:11.840 ","End":"04:18.995","Text":"u in the direction of n. Grad or Del dot Del is Del squared,"},{"Start":"04:18.995 ","End":"04:21.905","Text":"which is Delta, the Laplacian."},{"Start":"04:21.905 ","End":"04:24.020","Text":"We have the Laplacian of u_dA."},{"Start":"04:24.020 ","End":"04:27.560","Text":"Now, u satisfies the Poisson PDE,"},{"Start":"04:27.560 ","End":"04:30.155","Text":"so we know what the Laplacian of u is."},{"Start":"04:30.155 ","End":"04:32.940","Text":"Where is it? Oh, up here."},{"Start":"04:32.940 ","End":"04:36.360","Text":"It\u0027s equal to f of x, y."},{"Start":"04:36.360 ","End":"04:38.689","Text":"We were also given,"},{"Start":"04:38.689 ","End":"04:40.129","Text":"if you go back again,"},{"Start":"04:40.129 ","End":"04:43.700","Text":"that directional derivative of u in the direction of n is g of x,"},{"Start":"04:43.700 ","End":"04:45.175","Text":"y. Where is that?"},{"Start":"04:45.175 ","End":"04:52.745","Text":"That\u0027s this one. If used both of these,"},{"Start":"04:52.745 ","End":"04:54.965","Text":"Del u is f of x y,"},{"Start":"04:54.965 ","End":"04:58.025","Text":"du by dn g of x, y."},{"Start":"04:58.025 ","End":"05:00.095","Text":"That\u0027s what we have to show."},{"Start":"05:00.095 ","End":"05:02.880","Text":"That concludes this clip."}],"ID":30816}],"Thumbnail":null,"ID":294435},{"Name":"The Laplace Equation in a Disk","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - The Laplace Equation in a Disk","Duration":"1m 18s","ChapterTopicVideoID":29284,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.180","Text":"In this clip, we\u0027ll learn the general solution of the Laplace equation in a disk."},{"Start":"00:06.180 ","End":"00:14.415","Text":"Laplace equation in a disk of radius Rho centered at the origin is the following."},{"Start":"00:14.415 ","End":"00:18.620","Text":"The Laplacian of u is equal to 0,"},{"Start":"00:18.620 ","End":"00:25.515","Text":"and this is just the domain which is the interior of the disc of radius Rho."},{"Start":"00:25.515 ","End":"00:29.370","Text":"The general solution is best expressed in polar coordinates."},{"Start":"00:29.370 ","End":"00:34.710","Text":"It\u0027s a Fourier series in r and Theta as follows."},{"Start":"00:34.710 ","End":"00:37.685","Text":"Now the coefficients a_n and b_n,"},{"Start":"00:37.685 ","End":"00:41.365","Text":"are computed from the boundary conditions."},{"Start":"00:41.365 ","End":"00:44.450","Text":"There are 2 main kinds of boundary conditions."},{"Start":"00:44.450 ","End":"00:49.340","Text":"There are Dirichlet boundary conditions where we\u0027re given the value of"},{"Start":"00:49.340 ","End":"00:54.200","Text":"u as some function g on the circle itself,"},{"Start":"00:54.200 ","End":"00:56.605","Text":"which is the boundary of the disk,"},{"Start":"00:56.605 ","End":"01:00.545","Text":"and the Neumann boundary conditions are where we\u0027re given"},{"Start":"01:00.545 ","End":"01:04.145","Text":"the directional derivative in the direction of the"},{"Start":"01:04.145 ","End":"01:08.795","Text":"outward normal on the boundary of the disk."},{"Start":"01:08.795 ","End":"01:13.355","Text":"Anyway, the main thing is that this is the general formula,"},{"Start":"01:13.355 ","End":"01:19.050","Text":"and we\u0027ll see examples in the exercises that follow this clip."}],"ID":30817},{"Watched":false,"Name":"Exercise 1","Duration":"3m 16s","ChapterTopicVideoID":29285,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"In this exercise, we\u0027re given the following problem."},{"Start":"00:03.480 ","End":"00:10.890","Text":"We have a Laplace equation here and we have a Dirichlet boundary condition here."},{"Start":"00:10.890 ","End":"00:15.105","Text":"We have to find the solution in polar coordinates."},{"Start":"00:15.105 ","End":"00:16.740","Text":"We\u0027re given a hint,"},{"Start":"00:16.740 ","End":"00:19.785","Text":"or we can use the following trigonometric identity."},{"Start":"00:19.785 ","End":"00:24.165","Text":"Note that this domain is the unit disk,"},{"Start":"00:24.165 ","End":"00:29.970","Text":"the interior, and this is the circle that is the boundary of the unit disk."},{"Start":"00:29.970 ","End":"00:35.595","Text":"We start by writing the general formula for the solution of the Laplace equation."},{"Start":"00:35.595 ","End":"00:38.895","Text":"That will automatically satisfy this."},{"Start":"00:38.895 ","End":"00:44.255","Text":"Now we just have to find the a_n and b_n to get the boundary condition."},{"Start":"00:44.255 ","End":"00:48.465","Text":"Note that Rho=1 in our case,"},{"Start":"00:48.465 ","End":"00:54.565","Text":"so r/Rho is just r and this is the general form."},{"Start":"00:54.565 ","End":"00:59.720","Text":"The boundary, which is the unit circle x^2 plus y^2 = 1."},{"Start":"00:59.720 ","End":"01:01.610","Text":"In polar coordinates, it\u0027s much simpler,"},{"Start":"01:01.610 ","End":"01:03.580","Text":"it\u0027s just r=1,"},{"Start":"01:03.580 ","End":"01:05.975","Text":"and Theta can be whatever it wants."},{"Start":"01:05.975 ","End":"01:14.395","Text":"The function 1 plus y^3 from here in polar coordinates is 1 plus r sine (Theta) ^3."},{"Start":"01:14.395 ","End":"01:18.650","Text":"That gives us that on the boundary"},{"Start":"01:18.650 ","End":"01:26.205","Text":"u(1)and Theta is whatever is 1 plus sine cubed (Theta),"},{"Start":"01:26.205 ","End":"01:27.540","Text":"because r is 1."},{"Start":"01:27.540 ","End":"01:28.740","Text":"On the other hand,"},{"Start":"01:28.740 ","End":"01:31.705","Text":"you have 1 Theta can be gotten by"},{"Start":"01:31.705 ","End":"01:38.985","Text":"substituting r=1 in this equation, the general solution."},{"Start":"01:38.985 ","End":"01:41.985","Text":"This r^n drops out,"},{"Start":"01:41.985 ","End":"01:43.360","Text":"and we have this."},{"Start":"01:43.360 ","End":"01:46.975","Text":"Now we can compare these 2 equal things and get"},{"Start":"01:46.975 ","End":"01:52.160","Text":"that this series is equal to 1 plus sine cubed Theta."},{"Start":"01:52.160 ","End":"01:54.160","Text":"We\u0027d like to compare coefficients,"},{"Start":"01:54.160 ","End":"01:56.500","Text":"but the right hand side isn\u0027t in the form of"},{"Start":"01:56.500 ","End":"02:00.035","Text":"the sum of cosine (n) Theta or sine (n)Theta."},{"Start":"02:00.035 ","End":"02:02.560","Text":"That\u0027s where the hint comes in,"},{"Start":"02:02.560 ","End":"02:10.410","Text":"because this sine cubed Theta was given to be 3/4 sine Theta minus 1/4 sine 3 Theta."},{"Start":"02:10.410 ","End":"02:14.080","Text":"Now, this is in the form that it is here."},{"Start":"02:14.080 ","End":"02:16.024","Text":"We can compare coefficients."},{"Start":"02:16.024 ","End":"02:18.505","Text":"a_0 over 2 is 1,"},{"Start":"02:18.505 ","End":"02:23.720","Text":"which gives us a_0 is 2 and the rest of it is signs."},{"Start":"02:23.720 ","End":"02:25.470","Text":"When n is 1,"},{"Start":"02:25.470 ","End":"02:28.620","Text":"we get the b_1 coefficient,"},{"Start":"02:28.620 ","End":"02:32.205","Text":"and that\u0027s 3/4 and when n is 3,"},{"Start":"02:32.205 ","End":"02:36.030","Text":"we get that b_3 is minus a 1/4,"},{"Start":"02:36.030 ","End":"02:41.205","Text":"and all the other a_n and b_n except for these are 0."},{"Start":"02:41.205 ","End":"02:48.020","Text":"Now we return to the general solution and we just substitute all the a_ns."},{"Start":"02:48.020 ","End":"02:54.075","Text":"Most of them are 0, but these 3 are not so what we get is a_0 over 2,"},{"Start":"02:54.075 ","End":"03:01.890","Text":"which is 1, and here the 3/4 goes with the n=1 sine n Theta."},{"Start":"03:01.890 ","End":"03:04.230","Text":"It\u0027s just sine Theta, and when n is 3,"},{"Start":"03:04.230 ","End":"03:12.345","Text":"we get sine 3 Theta and r^3 and we get the minus a 1/4 from here."},{"Start":"03:12.345 ","End":"03:14.360","Text":"That\u0027s the solution."},{"Start":"03:14.360 ","End":"03:17.040","Text":"That concludes the exercise."}],"ID":30818},{"Watched":false,"Name":"Exercise 2","Duration":"3m 54s","ChapterTopicVideoID":29286,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.405","Text":"In this exercise, we have the following problem to solve."},{"Start":"00:04.405 ","End":"00:07.150","Text":"We have here Laplace equation."},{"Start":"00:07.150 ","End":"00:11.210","Text":"This is the open disk of Radius 2."},{"Start":"00:11.210 ","End":"00:14.410","Text":"We\u0027re given the boundary condition,"},{"Start":"00:14.410 ","End":"00:18.880","Text":"which is a Dirichlet condition on the circle of Radius 2,"},{"Start":"00:18.880 ","End":"00:20.410","Text":"which is the boundary of the disk."},{"Start":"00:20.410 ","End":"00:22.195","Text":"Okay, let\u0027s start."},{"Start":"00:22.195 ","End":"00:28.100","Text":"This is the general solution of the Laplace equation on a disk of radius Rho."},{"Start":"00:28.100 ","End":"00:34.005","Text":"In our case, Rho is equal to 2 square root of 4."},{"Start":"00:34.005 ","End":"00:42.710","Text":"The general solution for a disk of radius 2 centered at the origin is the following."},{"Start":"00:42.710 ","End":"00:46.270","Text":"The boundary, the circle radius 2 in"},{"Start":"00:46.270 ","End":"00:51.145","Text":"polar coordinates is just r=2 no restriction on Theta."},{"Start":"00:51.145 ","End":"00:56.290","Text":"The boundary condition, this x^2 plus y to the fourth in"},{"Start":"00:56.290 ","End":"01:04.510","Text":"polar coordinates is r cosine Theta squared plus r sine Theta to the fourth."},{"Start":"01:04.510 ","End":"01:09.355","Text":"That means that u of 2 Theta, means r equals 2,"},{"Start":"01:09.355 ","End":"01:12.045","Text":"is place r by 2 here,"},{"Start":"01:12.045 ","End":"01:16.195","Text":"4 cosine squared Theta plus 16 sine to the fourth Theta."},{"Start":"01:16.195 ","End":"01:22.580","Text":"On the other hand, we can get u of 2 Theta by replacing r equals 2 here,"},{"Start":"01:22.580 ","End":"01:24.735","Text":"the r/2 is 1,"},{"Start":"01:24.735 ","End":"01:29.880","Text":"that cancels out, so we\u0027re left with u of 2 Theta is the following."},{"Start":"01:29.880 ","End":"01:32.820","Text":"Now we have 2 expressions for the same thing."},{"Start":"01:32.820 ","End":"01:34.980","Text":"We can equate between them."},{"Start":"01:34.980 ","End":"01:39.335","Text":"We get that this series equals this."},{"Start":"01:39.335 ","End":"01:42.245","Text":"What we\u0027re going to do is comparison of coefficients."},{"Start":"01:42.245 ","End":"01:44.810","Text":"The thing is, the right-hand side isn\u0027t in"},{"Start":"01:44.810 ","End":"01:49.570","Text":"the form of the sum of cosine n Theta and sine n Theta."},{"Start":"01:49.570 ","End":"01:55.160","Text":"We\u0027ll use some trigonometric identities to transform it into such a form."},{"Start":"01:55.160 ","End":"01:57.725","Text":"We\u0027ll just work on the right-hand side."},{"Start":"01:57.725 ","End":"02:02.200","Text":"4 cosine squared Theta plus 16 sine to the fourth Theta is 4,"},{"Start":"02:02.200 ","End":"02:04.110","Text":"1 plus cosine 2 Theta over 2,"},{"Start":"02:04.110 ","End":"02:07.830","Text":"that\u0027s cosine squared Theta and 1 minus cosine 2 Theta over"},{"Start":"02:07.830 ","End":"02:12.585","Text":"2 is sine squared Theta sine squared, squared."},{"Start":"02:12.585 ","End":"02:14.280","Text":"There\u0027s a squared here."},{"Start":"02:14.280 ","End":"02:16.950","Text":"Now, 2 into 4 goes 2,"},{"Start":"02:16.950 ","End":"02:18.750","Text":"so it\u0027s twice the numerator."},{"Start":"02:18.750 ","End":"02:21.960","Text":"Here 2 squared into 16 goes 4."},{"Start":"02:21.960 ","End":"02:24.365","Text":"It\u0027s 4 times the numerator,"},{"Start":"02:24.365 ","End":"02:29.280","Text":"which is this minus this squared comes out to be this."},{"Start":"02:29.750 ","End":"02:36.380","Text":"Skip 2 step cosine squared 2 Theta is using the same formula,"},{"Start":"02:36.380 ","End":"02:41.815","Text":"1 plus cosine twice 2 Theta over 2."},{"Start":"02:41.815 ","End":"02:44.595","Text":"This simplifies to the following."},{"Start":"02:44.595 ","End":"02:46.349","Text":"Now we have constants,"},{"Start":"02:46.349 ","End":"02:48.290","Text":"cosines, and sines."},{"Start":"02:48.290 ","End":"02:50.645","Text":"There aren\u0027t any, but it is in this form."},{"Start":"02:50.645 ","End":"02:54.695","Text":"Now we can compare coefficients with this and this."},{"Start":"02:54.695 ","End":"02:58.045","Text":"We see a naught over 2 is 8,"},{"Start":"02:58.045 ","End":"03:01.040","Text":"that gives us that a naught is 16."},{"Start":"03:01.040 ","End":"03:05.490","Text":"Also, we have the cosine 2 Theta coefficient."},{"Start":"03:05.490 ","End":"03:07.755","Text":"Here, it\u0027s a_2,"},{"Start":"03:07.755 ","End":"03:09.720","Text":"here it\u0027s minus 6."},{"Start":"03:09.720 ","End":"03:12.450","Text":"The cosine 4 Theta on the right,"},{"Start":"03:12.450 ","End":"03:15.375","Text":"it\u0027s 2, on the left it\u0027s a_4."},{"Start":"03:15.375 ","End":"03:21.230","Text":"These are the 3 non-zero coefficients we get from the rest of them is 0."},{"Start":"03:21.230 ","End":"03:26.155","Text":"Now, the general formula on the disk of radius 2 is this."},{"Start":"03:26.155 ","End":"03:32.990","Text":"We have to do is substitute the 3 values that we know and the rest to 0."},{"Start":"03:32.990 ","End":"03:35.120","Text":"We get the following."},{"Start":"03:35.120 ","End":"03:37.760","Text":"We get the a naught over 2 is 8,"},{"Start":"03:37.760 ","End":"03:42.120","Text":"we get when n is 2,"},{"Start":"03:42.120 ","End":"03:43.860","Text":"we get a_2,"},{"Start":"03:43.860 ","End":"03:46.870","Text":"which is minus 6,"},{"Start":"03:46.870 ","End":"03:49.620","Text":"r over 2 squared cosine 2 Theta."},{"Start":"03:49.620 ","End":"03:52.545","Text":"Similarly for the last 1 with n equals 4."},{"Start":"03:52.545 ","End":"03:55.480","Text":"That concludes this exercise."}],"ID":30819},{"Watched":false,"Name":"Exercise 3","Duration":"2m 47s","ChapterTopicVideoID":29287,"CourseChapterTopicPlaylistID":294436,"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 have a Laplace equation with"},{"Start":"00:06.075 ","End":"00:10.695","Text":"a Dirichlet boundary condition on the disk of"},{"Start":"00:10.695 ","End":"00:15.700","Text":"radius 4 and we have to find the solution in polar coordinates,"},{"Start":"00:15.700 ","End":"00:18.690","Text":"and we\u0027re given a hint with the trigonometric identity,"},{"Start":"00:18.690 ","End":"00:20.250","Text":"which we might want to use."},{"Start":"00:20.250 ","End":"00:24.360","Text":"The general formula in polar coordinates on a disk of"},{"Start":"00:24.360 ","End":"00:30.015","Text":"radius Rho is the following function of r and Theta."},{"Start":"00:30.015 ","End":"00:33.120","Text":"In our case, Rho is 4 square root of"},{"Start":"00:33.120 ","End":"00:38.795","Text":"16 and so we can replace Rho by 4 and get this formula."},{"Start":"00:38.795 ","End":"00:44.420","Text":"The boundary x^2 plus y^2 equals 16 in polar coordinates is just r equals"},{"Start":"00:44.420 ","End":"00:51.980","Text":"4 and the function which is the value on the boundary is x^2 y,"},{"Start":"00:51.980 ","End":"00:53.705","Text":"but in polar coordinates,"},{"Start":"00:53.705 ","End":"00:58.705","Text":"that comes out to be r^3 cosine^2 Theta sine Theta."},{"Start":"00:58.705 ","End":"01:03.640","Text":"Now, this is what u(x,y) equals on the boundary."},{"Start":"01:03.640 ","End":"01:06.180","Text":"On the boundary r is 4."},{"Start":"01:06.180 ","End":"01:08.070","Text":"We get that u(4,"},{"Start":"01:08.070 ","End":"01:11.640","Text":"Theta) is r^3 is 4^3 is"},{"Start":"01:11.640 ","End":"01:17.990","Text":"64 cosine^2 Theta sine Theta and using this hint that we were given,"},{"Start":"01:17.990 ","End":"01:22.710","Text":"we can replace cosine^2 Theta sine Theta by a quarter sine Theta plus sine 3 Theta."},{"Start":"01:22.710 ","End":"01:25.910","Text":"The quarter and the 64 canceled give 16."},{"Start":"01:25.910 ","End":"01:27.425","Text":"This is what we have."},{"Start":"01:27.425 ","End":"01:31.354","Text":"On the other hand, u(4 Theta) is this formula."},{"Start":"01:31.354 ","End":"01:33.650","Text":"Just replace r by 4,"},{"Start":"01:33.650 ","End":"01:35.705","Text":"so 4 over 4 is 1."},{"Start":"01:35.705 ","End":"01:37.450","Text":"We don\u0027t need that."},{"Start":"01:37.450 ","End":"01:40.740","Text":"Now, we have 2 expressions for u(4 and Theta),"},{"Start":"01:40.740 ","End":"01:45.290","Text":"this and this and that means we can compare them and get that"},{"Start":"01:45.290 ","End":"01:51.635","Text":"this series is equal to 16 sine Theta plus 16 sine 3 Theta."},{"Start":"01:51.635 ","End":"01:54.185","Text":"Now we can compare coefficients."},{"Start":"01:54.185 ","End":"01:57.200","Text":"There\u0027s only 2 coefficients here in the both sines,"},{"Start":"01:57.200 ","End":"01:58.790","Text":"so we\u0027ll get 2 of the bn,"},{"Start":"01:58.790 ","End":"02:01.765","Text":"in fact we\u0027ll get b1 and b3."},{"Start":"02:01.765 ","End":"02:04.830","Text":"B1 is 16 from here,"},{"Start":"02:04.830 ","End":"02:11.075","Text":"b3 is 16 from here and all the other an\u0027s and bn\u0027s are 0."},{"Start":"02:11.075 ","End":"02:14.630","Text":"Now we return to the formula we had above. Where is it?"},{"Start":"02:14.630 ","End":"02:16.220","Text":"Yeah, this one."},{"Start":"02:16.220 ","End":"02:20.765","Text":"All we have to do now is replace the coefficients with the ones we found here."},{"Start":"02:20.765 ","End":"02:23.150","Text":"Most of them is 0 except for 2 of them,"},{"Start":"02:23.150 ","End":"02:24.875","Text":"so we\u0027ll only have 2 terms,"},{"Start":"02:24.875 ","End":"02:28.745","Text":"we\u0027ll have the term with b1 and with b3."},{"Start":"02:28.745 ","End":"02:33.360","Text":"What we get b1 is 16,"},{"Start":"02:33.360 ","End":"02:41.475","Text":"so we get 16 sine Theta and here we get 16 sine 3 Theta and there\u0027s also the r over 4^n,"},{"Start":"02:41.475 ","End":"02:44.415","Text":"r over 4^1, r over 4^3."},{"Start":"02:44.415 ","End":"02:47.590","Text":"Now this is the answer and we\u0027re done."}],"ID":30820},{"Watched":false,"Name":"Exercise 4","Duration":"5m 12s","ChapterTopicVideoID":29288,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"02:28.130 ","End":"02:33.540","Text":"2Pi."},{"Start":"02:33.540 ","End":"02:39.275","Text":"Since f of Theta is 0 between minus Pi and 0,"},{"Start":"02:39.275 ","End":"02:45.575","Text":"all these integrals will be from 0 to Pi instead of minus Pi to Pi."},{"Start":"02:45.575 ","End":"02:51.875","Text":"What\u0027s more? We can replace f of Theta by 1 on the interval from 0 to Pi."},{"Start":"02:51.875 ","End":"02:54.580","Text":"What we get, let\u0027s start with a_n,"},{"Start":"02:54.580 ","End":"03:01.290","Text":"is the integral from 0 to Pi cosine n Theta d Theta, and 1 over Pi."},{"Start":"03:01.290 ","End":"03:04.415","Text":"This is equal to the following;"},{"Start":"03:04.415 ","End":"03:06.470","Text":"integral of cosine is sine,"},{"Start":"03:06.470 ","End":"03:08.698","Text":"we have to divide by the anti-derivative."},{"Start":"03:08.698 ","End":"03:11.960","Text":"Now sine of 0 and sine of n Pi are both 0,"},{"Start":"03:11.960 ","End":"03:15.410","Text":"so this integral is 0 minus 0 is 0."},{"Start":"03:15.410 ","End":"03:20.000","Text":"A naught, which we have to do separately because we can\u0027t divide by 0,"},{"Start":"03:20.000 ","End":"03:24.560","Text":"we get the integral of 1d Theta from 0 to Pi,"},{"Start":"03:24.560 ","End":"03:28.530","Text":"that\u0027s equal to Pi divided by Pi, so it\u0027s 1."},{"Start":"03:28.530 ","End":"03:31.020","Text":"Bn is the integral of sine nTheta,"},{"Start":"03:31.020 ","End":"03:36.095","Text":"dTheta and this comes out to be the integral of sine is minus co-sine."},{"Start":"03:36.095 ","End":"03:39.335","Text":"Change the order of the 0 and the Pi."},{"Start":"03:39.335 ","End":"03:44.885","Text":"Substitute 0 and then subtract what we get when we substitute Pi."},{"Start":"03:44.885 ","End":"03:54.780","Text":"What we get is 0 we get 1 and for Pi we get minus cosine n Pi,"},{"Start":"03:54.780 ","End":"03:58.350","Text":"but cosine nPi is minus 1 to the n,"},{"Start":"03:58.350 ","End":"03:59.735","Text":"we\u0027ve seen this so many times."},{"Start":"03:59.735 ","End":"04:06.335","Text":"Note that if n is even this comes out to be 0 and if n is odd,"},{"Start":"04:06.335 ","End":"04:08.885","Text":"this numerator comes out to be 2."},{"Start":"04:08.885 ","End":"04:12.725","Text":"We can write this by separating odds and evens as"},{"Start":"04:12.725 ","End":"04:16.745","Text":"n equals 2k minus 1 for odd and n equals 2k for evens,"},{"Start":"04:16.745 ","End":"04:24.585","Text":"so it\u0027s equal to 2 over Pi and then 1 over n is 1 over 2k minus 1 and 0 for the evens."},{"Start":"04:24.585 ","End":"04:27.360","Text":"Now back to u(r, Theta)."},{"Start":"04:27.360 ","End":"04:33.245","Text":"This was the general formula on a disk of radius 3,"},{"Start":"04:33.245 ","End":"04:36.590","Text":"and now we have the ans and the bns,"},{"Start":"04:36.590 ","End":"04:38.365","Text":"so we can substitute them."},{"Start":"04:38.365 ","End":"04:43.475","Text":"Remember, the a n\u0027s are all 0 except for a naught, which is 1."},{"Start":"04:43.475 ","End":"04:47.900","Text":"What we get is for a naught equals 1 we get 1.5,"},{"Start":"04:47.900 ","End":"04:49.730","Text":"we don\u0027t get anymore cosine."},{"Start":"04:49.730 ","End":"04:52.060","Text":"We get the signs from bn,"},{"Start":"04:52.060 ","End":"04:56.265","Text":"but only when n is 2k minus 1."},{"Start":"04:56.265 ","End":"05:01.125","Text":"We let k go from 1 to infinity when n be 2k minus 1,"},{"Start":"05:01.125 ","End":"05:03.900","Text":"and we get sine of nTheta over n,"},{"Start":"05:03.900 ","End":"05:07.185","Text":"replacing n by 2k minus 1."},{"Start":"05:07.185 ","End":"05:08.930","Text":"Now this is the answer,"},{"Start":"05:08.930 ","End":"05:12.120","Text":"not the end of this exercise."}],"ID":30821},{"Watched":false,"Name":"Exercise 5","Duration":"4m 33s","ChapterTopicVideoID":29289,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.525","Text":"In this exercise, we have a boundary value problem."},{"Start":"00:03.525 ","End":"00:06.525","Text":"This is the PDE, the Laplace equation."},{"Start":"00:06.525 ","End":"00:08.550","Text":"This is the boundary value,"},{"Start":"00:08.550 ","End":"00:11.250","Text":"the equations on the interior of the disc,"},{"Start":"00:11.250 ","End":"00:13.500","Text":"and the boundary values on the boundary of the disc,"},{"Start":"00:13.500 ","End":"00:14.700","Text":"which is the circle."},{"Start":"00:14.700 ","End":"00:18.030","Text":"Note that the boundary condition is defined piecewise."},{"Start":"00:18.030 ","End":"00:22.110","Text":"On the top half of the unit circle,"},{"Start":"00:22.110 ","End":"00:23.490","Text":"it\u0027s equal to 1."},{"Start":"00:23.490 ","End":"00:27.060","Text":"On the bottom half, it\u0027s equal to minus 1."},{"Start":"00:27.060 ","End":"00:29.715","Text":"U is not continuous on the boundary."},{"Start":"00:29.715 ","End":"00:31.470","Text":"Of course, in the interior it will be,"},{"Start":"00:31.470 ","End":"00:33.825","Text":"not only continuous but differentiable."},{"Start":"00:33.825 ","End":"00:38.390","Text":"By the way, the solution will come out to be 0 at these 2 points,"},{"Start":"00:38.390 ","End":"00:41.090","Text":"the average between the 1 and the minus 1."},{"Start":"00:41.090 ","End":"00:44.510","Text":"We start with the general formula for the solution to"},{"Start":"00:44.510 ","End":"00:49.730","Text":"the Laplace equation in a disc centered at the origin of radius Rho."},{"Start":"00:49.730 ","End":"00:51.515","Text":"In our case, Rho is 1,"},{"Start":"00:51.515 ","End":"00:55.145","Text":"so this simplifies a bit to the following."},{"Start":"00:55.145 ","End":"00:57.920","Text":"We want the solution in polar coordinates."},{"Start":"00:57.920 ","End":"01:01.760","Text":"That\u0027s how the Laplace equation on the disc works."},{"Start":"01:01.760 ","End":"01:04.500","Text":"The boundary x^2 plus y^2 is 1,"},{"Start":"01:04.500 ","End":"01:07.510","Text":"and polar coordinates, is r equals 1."},{"Start":"01:07.510 ","End":"01:12.650","Text":"That means that you in polar coordinates on the boundary,"},{"Start":"01:12.650 ","End":"01:15.875","Text":"is you have 1 Theta equals 1,"},{"Start":"01:15.875 ","End":"01:18.245","Text":"Theta between 0 and Pi,"},{"Start":"01:18.245 ","End":"01:20.030","Text":"and minus 1 on the bottom half,"},{"Start":"01:20.030 ","End":"01:23.345","Text":"which we express as Theta between minus Pi and 0,"},{"Start":"01:23.345 ","End":"01:27.365","Text":"the whole circle is taken to be as Theta from minus Pi to Pi."},{"Start":"01:27.365 ","End":"01:29.225","Text":"Sometimes we take 0 to 2Pi."},{"Start":"01:29.225 ","End":"01:31.325","Text":"Here we take minus Pi to Pi."},{"Start":"01:31.325 ","End":"01:36.465","Text":"Now we can also express u(1) Theta using this formula,"},{"Start":"01:36.465 ","End":"01:39.285","Text":"and replacing r equals 1."},{"Start":"01:39.285 ","End":"01:41.640","Text":"So we get the following,"},{"Start":"01:41.640 ","End":"01:46.280","Text":"and since these 2 are equal to the common thing, you have 1 Theta,"},{"Start":"01:46.280 ","End":"01:53.525","Text":"we can compare between them and say that this series is equal to this,"},{"Start":"01:53.525 ","End":"01:55.460","Text":"which we\u0027ll call f(Theta)."},{"Start":"01:55.460 ","End":"01:56.570","Text":"It\u0027s a function of Theta,"},{"Start":"01:56.570 ","End":"01:58.700","Text":"where Theta goes from minus Pi to Pi,"},{"Start":"01:58.700 ","End":"02:01.220","Text":"and it\u0027s minus 1 or 1."},{"Start":"02:01.220 ","End":"02:08.060","Text":"Now the left-hand side is just a Fourier series for a function defined on minus Pi to Pi."},{"Start":"02:08.060 ","End":"02:10.940","Text":"We can find the coefficients using"},{"Start":"02:10.940 ","End":"02:14.920","Text":"the formula for the coefficients of the Fourier series."},{"Start":"02:14.920 ","End":"02:18.180","Text":"The formula is that a_n is this integral,"},{"Start":"02:18.180 ","End":"02:20.225","Text":"and b_n is this integral."},{"Start":"02:20.225 ","End":"02:24.100","Text":"In our case, what we get when we substitute f of Theta,"},{"Start":"02:24.100 ","End":"02:27.980","Text":"is we get the integral from 0 to Pi of"},{"Start":"02:27.980 ","End":"02:32.630","Text":"cosine n Theta d Theta minus the integral from minus Pi to 0."},{"Start":"02:32.630 ","End":"02:35.035","Text":"Here we are again, just rewrote them."},{"Start":"02:35.035 ","End":"02:38.590","Text":"Now, note that f of Theta is an odd function."},{"Start":"02:38.590 ","End":"02:39.880","Text":"It\u0027s 1 or minus 1,"},{"Start":"02:39.880 ","End":"02:42.220","Text":"depending on whether Theta is positive or negative,"},{"Start":"02:42.220 ","End":"02:44.830","Text":"and cosine is an even function."},{"Start":"02:44.830 ","End":"02:48.505","Text":"An odd function times an even function is an odd function."},{"Start":"02:48.505 ","End":"02:51.385","Text":"The integral on a symmetric interval will be 0."},{"Start":"02:51.385 ","End":"02:55.070","Text":"In this case, we have odd times odd is even,"},{"Start":"02:55.070 ","End":"02:57.370","Text":"and for an even function on a symmetric interval,"},{"Start":"02:57.370 ","End":"03:01.700","Text":"what we can do is take twice the integral from 0 to Pi."},{"Start":"03:01.700 ","End":"03:04.425","Text":"What we get is 2/Pi,"},{"Start":"03:04.425 ","End":"03:09.845","Text":"the integral from 0 to Pi and f(Theta) is then 1."},{"Start":"03:09.845 ","End":"03:11.920","Text":"So we need the integral of sine(n),"},{"Start":"03:11.920 ","End":"03:13.330","Text":"Theta d Theta,"},{"Start":"03:13.330 ","End":"03:14.960","Text":"and this is equal to,"},{"Start":"03:14.960 ","End":"03:17.670","Text":"integral of sine is minus cosine,"},{"Start":"03:17.670 ","End":"03:20.320","Text":"and then we substitute,"},{"Start":"03:20.320 ","End":"03:24.870","Text":"we get minus minus cosine 0 Theta, which is 1,"},{"Start":"03:24.870 ","End":"03:28.515","Text":"you get minus cosine of n Pi,"},{"Start":"03:28.515 ","End":"03:31.905","Text":"which is minus 1^n."},{"Start":"03:31.905 ","End":"03:34.410","Text":"If n is odd, we get 1 minus,"},{"Start":"03:34.410 ","End":"03:36.750","Text":"minus 1 is 4,"},{"Start":"03:36.750 ","End":"03:39.745","Text":"and if n is even, we get 0."},{"Start":"03:39.745 ","End":"03:41.740","Text":"To express the odd n\u0027s,"},{"Start":"03:41.740 ","End":"03:46.540","Text":"we just take n equals 2k minus 1 and k equals 1, 2, 3, etc."},{"Start":"03:46.540 ","End":"03:48.970","Text":"So replacing n by 2k minus 1,"},{"Start":"03:48.970 ","End":"03:50.470","Text":"this is what we get."},{"Start":"03:50.470 ","End":"03:54.460","Text":"Now, returning to the formula for u of our Theta,"},{"Start":"03:54.460 ","End":"03:57.665","Text":"we now know what all the coefficients are."},{"Start":"03:57.665 ","End":"04:00.255","Text":"The an\u0027s are all 0,"},{"Start":"04:00.255 ","End":"04:05.925","Text":"so all we have is bn, r^n sine(n) Theta."},{"Start":"04:05.925 ","End":"04:08.730","Text":"Even then we don\u0027t get them for all n,"},{"Start":"04:08.730 ","End":"04:13.305","Text":"these terms, we only get for n equals 2k minus 1."},{"Start":"04:13.305 ","End":"04:16.140","Text":"So we get r^2k minus 1,"},{"Start":"04:16.140 ","End":"04:19.005","Text":"sine of 2k minus 1 Theta,"},{"Start":"04:19.005 ","End":"04:21.360","Text":"over 2k minus 1,"},{"Start":"04:21.360 ","End":"04:24.795","Text":"and the r^n becomes r^2k minus 1."},{"Start":"04:24.795 ","End":"04:29.140","Text":"We also get 4 over Pi."},{"Start":"04:30.260 ","End":"04:34.300","Text":"This is the answer, and we are done."}],"ID":30822},{"Watched":false,"Name":"Exercise 6a","Duration":"3m 5s","ChapterTopicVideoID":29290,"CourseChapterTopicPlaylistID":294436,"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\u0027re given"},{"Start":"00:02.340 ","End":"00:07.710","Text":"the following Neumann problem with the Laplace equation on the unit disk."},{"Start":"00:07.710 ","End":"00:12.090","Text":"We have to find out for which value of c does"},{"Start":"00:12.090 ","End":"00:14.130","Text":"the boundary value problems satisfy"},{"Start":"00:14.130 ","End":"00:17.490","Text":"the necessary conditions for solvability that we learned."},{"Start":"00:17.490 ","End":"00:19.775","Text":"Part B, we\u0027ll read when we come to it."},{"Start":"00:19.775 ","End":"00:21.325","Text":"The problem is Neumann,"},{"Start":"00:21.325 ","End":"00:23.505","Text":"because we\u0027re not given u itself,"},{"Start":"00:23.505 ","End":"00:27.405","Text":"but the directional derivative in the outer normal direction."},{"Start":"00:27.405 ","End":"00:31.305","Text":"Now, our domain D is the open unit disc,"},{"Start":"00:31.305 ","End":"00:33.140","Text":"and the boundary,"},{"Start":"00:33.140 ","End":"00:35.000","Text":"which we call the path gamma,"},{"Start":"00:35.000 ","End":"00:37.240","Text":"is the unit circle."},{"Start":"00:37.240 ","End":"00:41.450","Text":"To remind you, the necessary condition we\u0027re talking about is the following."},{"Start":"00:41.450 ","End":"00:48.545","Text":"The line integral on the closed path gamma of du by dn is 0."},{"Start":"00:48.545 ","End":"00:51.560","Text":"Let\u0027s start by parameterizing this path,"},{"Start":"00:51.560 ","End":"00:55.655","Text":"which is the unit circle in a counterclockwise direction,"},{"Start":"00:55.655 ","End":"00:57.335","Text":"we can take Theta,"},{"Start":"00:57.335 ","End":"01:02.945","Text":"the angle as the argument and the general point will be cosine Theta sine Theta."},{"Start":"01:02.945 ","End":"01:08.990","Text":"But Theta is usually from 0-2 Pi or from minus Pi to Pi."},{"Start":"01:08.990 ","End":"01:12.995","Text":"But here we\u0027ll take it from minus Pi over 2 to 3 Pi over 2."},{"Start":"01:12.995 ","End":"01:15.190","Text":"It\u0027s okay as long as the length is 2 Pi."},{"Start":"01:15.190 ","End":"01:17.750","Text":"This will be more convenient when we come to"},{"Start":"01:17.750 ","End":"01:21.215","Text":"defining when x is bigger than 0 and x is less than 0."},{"Start":"01:21.215 ","End":"01:23.150","Text":"To compute this integral,"},{"Start":"01:23.150 ","End":"01:27.165","Text":"replace x, y by cosine Theta, sine Theta,"},{"Start":"01:27.165 ","End":"01:33.650","Text":"and ds is the square root of dx squared plus dy squared,"},{"Start":"01:33.650 ","End":"01:37.010","Text":"which comes out to be minus sine Theta."},{"Start":"01:37.010 ","End":"01:41.225","Text":"That\u0027s the derivative of cosine squared plus cosine Theta,"},{"Start":"01:41.225 ","End":"01:45.010","Text":"the derivative of sine Theta also squared d Theta."},{"Start":"01:45.010 ","End":"01:48.460","Text":"This expression under the square root is equal to 1,"},{"Start":"01:48.460 ","End":"01:52.100","Text":"so this simplifies to the following."},{"Start":"01:52.100 ","End":"01:54.850","Text":"Forgot to uncover the picture."},{"Start":"01:54.850 ","End":"02:01.250","Text":"Here\u0027s a reminder of what du by dn is equal to on the boundary."},{"Start":"02:01.250 ","End":"02:04.445","Text":"It\u0027s equal to either Y or C,"},{"Start":"02:04.445 ","End":"02:08.255","Text":"depending on whether x is positive or x is negative."},{"Start":"02:08.255 ","End":"02:12.950","Text":"Now x is positive on the right half of the circle,"},{"Start":"02:12.950 ","End":"02:18.250","Text":"which means that Theta goes from minus Pi over 2 up to Pi over 2."},{"Start":"02:18.250 ","End":"02:21.760","Text":"Now we can split this integral up into 2 integrals,"},{"Start":"02:21.760 ","End":"02:24.540","Text":"apart from minus Pi over 2 to Pi over 2,"},{"Start":"02:24.540 ","End":"02:28.555","Text":"plus the part from Pi over 2 to 3 Pi over 2."},{"Start":"02:28.555 ","End":"02:31.820","Text":"This part is equal to y,"},{"Start":"02:31.820 ","End":"02:33.845","Text":"which is sine Theta,"},{"Start":"02:33.845 ","End":"02:38.180","Text":"and this is equal to C. This integral is equal"},{"Start":"02:38.180 ","End":"02:42.590","Text":"to 0 because this is an odd function on a symmetric interval."},{"Start":"02:42.590 ","End":"02:44.555","Text":"Here we have a constant,"},{"Start":"02:44.555 ","End":"02:48.995","Text":"so is the constant times the length of the interval, which is Pi."},{"Start":"02:48.995 ","End":"02:52.305","Text":"We get Pi c. Altogether,"},{"Start":"02:52.305 ","End":"02:58.190","Text":"the integral is equal to Pi c. In order for the integral to be 0,"},{"Start":"02:58.190 ","End":"02:59.900","Text":"Pi c has to be 0,"},{"Start":"02:59.900 ","End":"03:05.670","Text":"which means that c has to be 0 and that\u0027s the answer to Part a."}],"ID":30823},{"Watched":false,"Name":"Exercise 6b","Duration":"8m 32s","ChapterTopicVideoID":29277,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.130","Text":"Now we come to part b,"},{"Start":"00:02.130 ","End":"00:04.095","Text":"we just completed part a,"},{"Start":"00:04.095 ","End":"00:06.570","Text":"which was to find out the value of c,"},{"Start":"00:06.570 ","End":"00:09.660","Text":"and it turns out that c=0."},{"Start":"00:09.660 ","End":"00:14.040","Text":"Let\u0027s just copy this time with c=0,"},{"Start":"00:14.040 ","End":"00:17.250","Text":"and our task is to solve this problem."},{"Start":"00:17.250 ","End":"00:21.885","Text":"We\u0027ll use the general formula for the solution to a Laplace equation"},{"Start":"00:21.885 ","End":"00:26.745","Text":"in a disk of radius Rho here, Rho=1."},{"Start":"00:26.745 ","End":"00:30.015","Text":"Now we\u0027re given this boundary condition,"},{"Start":"00:30.015 ","End":"00:32.070","Text":"in the case of anointment condition,"},{"Start":"00:32.070 ","End":"00:33.870","Text":"turns out it\u0027s not quite enough,"},{"Start":"00:33.870 ","End":"00:36.330","Text":"we can only find u up to a constant,"},{"Start":"00:36.330 ","End":"00:42.225","Text":"so given an extra condition that u at the origin is 0."},{"Start":"00:42.225 ","End":"00:46.065","Text":"Now if we substitute Rho=1 here,"},{"Start":"00:46.065 ","End":"00:51.240","Text":"then this comes out to be just r^n."},{"Start":"00:51.240 ","End":"00:57.980","Text":"How do we compute the u by dn, the directional derivative?"},{"Start":"00:57.980 ","End":"01:02.930","Text":"Note that the direction of the outward normal to the circle is"},{"Start":"01:02.930 ","End":"01:08.110","Text":"exactly the direction of the increase of r, the radius."},{"Start":"01:08.110 ","End":"01:10.840","Text":"Du by dn is du by dr,"},{"Start":"01:10.840 ","End":"01:14.630","Text":"which we also write as u sub r. What we"},{"Start":"01:14.630 ","End":"01:20.030","Text":"get if we differentiate this with respect to r is the following."},{"Start":"01:20.030 ","End":"01:23.390","Text":"That du by dn is this."},{"Start":"01:23.390 ","End":"01:28.565","Text":"The boundary condition that we\u0027re given relates to x^2 plus y^2 equal 1,"},{"Start":"01:28.565 ","End":"01:31.460","Text":"which is the same as r equal 1."},{"Start":"01:31.460 ","End":"01:33.455","Text":"This is what we were given."},{"Start":"01:33.455 ","End":"01:35.704","Text":"Let\u0027s convert this to polar,"},{"Start":"01:35.704 ","End":"01:38.075","Text":"that this is r =1,"},{"Start":"01:38.075 ","End":"01:41.390","Text":"so we can say it\u0027s u_r of 1 Theta,"},{"Start":"01:41.390 ","End":"01:43.105","Text":"and there\u0027s 2 cases,"},{"Start":"01:43.105 ","End":"01:44.600","Text":"x is bigger than 0,"},{"Start":"01:44.600 ","End":"01:46.790","Text":"meaning cosine Theta bigger than 0,"},{"Start":"01:46.790 ","End":"01:50.770","Text":"cosine Theta less than 0 and y is sine Theta."},{"Start":"01:50.770 ","End":"01:55.469","Text":"We can replace u_r of 1 Theta by this,"},{"Start":"01:55.469 ","End":"01:58.155","Text":"replacing r by 1,"},{"Start":"01:58.155 ","End":"02:00.930","Text":"and so we get this,"},{"Start":"02:00.930 ","End":"02:04.065","Text":"just the r^n minus 1 drops out."},{"Start":"02:04.065 ","End":"02:08.410","Text":"Instead of saying cosine Theta bigger than 0 or less than 0,"},{"Start":"02:08.410 ","End":"02:10.960","Text":"we can make it a condition about Theta."},{"Start":"02:10.960 ","End":"02:14.755","Text":"This is the right half of the circle,"},{"Start":"02:14.755 ","End":"02:17.845","Text":"which is from minus 90-90 degrees,"},{"Start":"02:17.845 ","End":"02:19.420","Text":"and the other part,"},{"Start":"02:19.420 ","End":"02:24.455","Text":"the left side, is from 90-270 degrees."},{"Start":"02:24.455 ","End":"02:29.390","Text":"Like in part a, instead of using the integral from 0-2Pi,"},{"Start":"02:29.390 ","End":"02:32.710","Text":"we can use any interval of length 2Pi, and in this case,"},{"Start":"02:32.710 ","End":"02:36.990","Text":"it\u0027s convenient take from minus Pi/2 to 3Pi/2."},{"Start":"02:36.990 ","End":"02:41.750","Text":"Now the left-hand side is a Fourier series on this interval."},{"Start":"02:41.750 ","End":"02:46.430","Text":"We have the usual formula for the coefficients a_n and b_n."},{"Start":"02:46.430 ","End":"02:51.035","Text":"In this case, the coefficient of cosine n Theta is not a_n, it\u0027s na_n,"},{"Start":"02:51.035 ","End":"02:57.230","Text":"and it\u0027s equal to the integral of the function times cosine n Theta,"},{"Start":"02:57.230 ","End":"03:03.545","Text":"the function is sine Theta from minus Pi/2-3Pi/2."},{"Start":"03:03.545 ","End":"03:07.770","Text":"Well, it\u0027s 0 from Pi/2-3Pi/2,"},{"Start":"03:07.770 ","End":"03:14.895","Text":"so we just have to take the sine Theta from minus Pi/2-Pi/2 and ignore the upper part."},{"Start":"03:14.895 ","End":"03:18.260","Text":"Because this is an odd function times an even function,"},{"Start":"03:18.260 ","End":"03:21.910","Text":"which is an odd function on a symmetric interval now,"},{"Start":"03:21.910 ","End":"03:24.705","Text":"then we can say it\u0027s equal to 0."},{"Start":"03:24.705 ","End":"03:26.670","Text":"If n is not 0,"},{"Start":"03:26.670 ","End":"03:28.320","Text":"if it\u0027s 1, 2, 3, etc,"},{"Start":"03:28.320 ","End":"03:33.190","Text":"we can divide both sides by n and get that a_n equal 0."},{"Start":"03:33.190 ","End":"03:35.545","Text":"That\u0027s only from 1 on-wards."},{"Start":"03:35.545 ","End":"03:36.980","Text":"We don\u0027t know what a 0 is,"},{"Start":"03:36.980 ","End":"03:38.255","Text":"at least not from this."},{"Start":"03:38.255 ","End":"03:41.990","Text":"Now the formula for the coefficient of sine"},{"Start":"03:41.990 ","End":"03:46.965","Text":"n Theta which is nb_n is equal to similar thing,"},{"Start":"03:46.965 ","End":"03:48.770","Text":"but this time instead of cosine n Theta,"},{"Start":"03:48.770 ","End":"03:50.450","Text":"we have sine n Theta,"},{"Start":"03:50.450 ","End":"03:53.215","Text":"and this time we have an even function."},{"Start":"03:53.215 ","End":"03:55.290","Text":"We can\u0027t say it\u0027s 0,"},{"Start":"03:55.290 ","End":"04:00.200","Text":"what we can do is say that it\u0027s double the integral from 0-Pi/2."},{"Start":"04:00.200 ","End":"04:04.900","Text":"We can use the trigonometric identity for sine Alpha, sine Beta,"},{"Start":"04:04.900 ","End":"04:06.620","Text":"and if we do that here,"},{"Start":"04:06.620 ","End":"04:07.925","Text":"this is what we get,"},{"Start":"04:07.925 ","End":"04:10.070","Text":"then the 2 cancels,"},{"Start":"04:10.070 ","End":"04:12.320","Text":"and this is what we have now."},{"Start":"04:12.320 ","End":"04:14.990","Text":"The integral of cosine is sine,"},{"Start":"04:14.990 ","End":"04:18.125","Text":"but we also have to divide by the inner derivative."},{"Start":"04:18.125 ","End":"04:20.090","Text":"This won\u0027t work if n equals 1,"},{"Start":"04:20.090 ","End":"04:22.775","Text":"so we\u0027d restrict this to n bigger than 1,"},{"Start":"04:22.775 ","End":"04:25.820","Text":"and we\u0027ll do the case n equals 1 separately afterwards."},{"Start":"04:25.820 ","End":"04:28.820","Text":"If we substitute when Theta is 0,"},{"Start":"04:28.820 ","End":"04:30.860","Text":"everything is 0 here and here,"},{"Start":"04:30.860 ","End":"04:33.665","Text":"so we just substitute Theta equals Pi/2,"},{"Start":"04:33.665 ","End":"04:35.905","Text":"so it\u0027s 1 minus nPi/2,"},{"Start":"04:35.905 ","End":"04:40.210","Text":"which is Pi/2 minus Pi/2n sine of that,"},{"Start":"04:40.210 ","End":"04:42.740","Text":"and here, same thing with plus."},{"Start":"04:42.740 ","End":"04:44.885","Text":"Now the trigonometric identity,"},{"Start":"04:44.885 ","End":"04:48.530","Text":"the sine of Pi/2 minus Theta is cosine Theta,"},{"Start":"04:48.530 ","End":"04:50.425","Text":"but it also works with a plus,"},{"Start":"04:50.425 ","End":"04:55.205","Text":"so we can replace this sine by cosine here and here."},{"Start":"04:55.205 ","End":"04:59.715","Text":"Now we can take the cosine Pi/2n outside the brackets,"},{"Start":"04:59.715 ","End":"05:09.120","Text":"and we have 1/1 minus n minus 1/1 plus n. This comes out to be 1 plus n minus 1 minus n,"},{"Start":"05:09.120 ","End":"05:12.900","Text":"which is 2n over the product which is 1 minus n^2,"},{"Start":"05:12.900 ","End":"05:16.955","Text":"and the n here is because we had nb_n,"},{"Start":"05:16.955 ","End":"05:19.085","Text":"now we have just b_n,"},{"Start":"05:19.085 ","End":"05:28.330","Text":"so we divide by n. Now the cosine of n times Pi/2 is periodic every 4 it repeats itself."},{"Start":"05:28.330 ","End":"05:30.750","Text":"Like cosine of 0 is 1,"},{"Start":"05:30.750 ","End":"05:32.745","Text":"then for Pi/2 is 0,"},{"Start":"05:32.745 ","End":"05:34.550","Text":"then minus 1, 0, then 1,"},{"Start":"05:34.550 ","End":"05:36.465","Text":"0 minus 1, 0."},{"Start":"05:36.465 ","End":"05:44.775","Text":"The pattern is that the odd multiples of Pi/2 are 0 and the even multiples of Pi/2,"},{"Start":"05:44.775 ","End":"05:47.895","Text":"are either 1 or minus 1 alternating."},{"Start":"05:47.895 ","End":"05:54.230","Text":"We can write that using odds and evens when n equals 2k or 2k minus 1,"},{"Start":"05:54.230 ","End":"05:59.195","Text":"and the alternating bit we can get by minus 1^k,"},{"Start":"05:59.195 ","End":"06:02.345","Text":"other than that, we replace n by 2k here."},{"Start":"06:02.345 ","End":"06:07.700","Text":"Also remember that we had the condition that n is bigger than 1, and that\u0027s this."},{"Start":"06:07.700 ","End":"06:09.950","Text":"Now back here,"},{"Start":"06:09.950 ","End":"06:12.560","Text":"this time with n equals 1,"},{"Start":"06:12.560 ","End":"06:15.130","Text":"we want to find out what b_1 is."},{"Start":"06:15.130 ","End":"06:18.180","Text":"The first term here in the integral,"},{"Start":"06:18.180 ","End":"06:19.950","Text":"1 minus n is 0,"},{"Start":"06:19.950 ","End":"06:22.125","Text":"cosine of 0 is 1,"},{"Start":"06:22.125 ","End":"06:23.640","Text":"and here if n is 1,"},{"Start":"06:23.640 ","End":"06:24.660","Text":"1 plus 1 is 2,"},{"Start":"06:24.660 ","End":"06:27.240","Text":"so we have cosine of 2 Theta."},{"Start":"06:27.240 ","End":"06:29.240","Text":"Integral of 1 is Theta,"},{"Start":"06:29.240 ","End":"06:33.575","Text":"the integral of cosine 2 Theta is sine 2 Theta/2."},{"Start":"06:33.575 ","End":"06:36.890","Text":"Substitute the limits of integration when Theta is 0,"},{"Start":"06:36.890 ","End":"06:40.435","Text":"both of these is 0 when Theta is Pi/2,"},{"Start":"06:40.435 ","End":"06:44.355","Text":"this is 0, so we\u0027re just left with the Pi/2 here,"},{"Start":"06:44.355 ","End":"06:46.590","Text":"1/Pi times Pi/2, which is a half,"},{"Start":"06:46.590 ","End":"06:48.360","Text":"so b_1 is a half."},{"Start":"06:48.360 ","End":"06:51.305","Text":"Now summarizing what we have so far,"},{"Start":"06:51.305 ","End":"06:55.430","Text":"a_n is 0 when n is bigger than 0,"},{"Start":"06:55.430 ","End":"06:57.830","Text":"note that we still don\u0027t know what a_0 is,"},{"Start":"06:57.830 ","End":"07:00.780","Text":"but we know what b_1 is, from here."},{"Start":"07:00.780 ","End":"07:04.050","Text":"We know what b_n is when n is bigger than 1,"},{"Start":"07:04.050 ","End":"07:05.595","Text":"either 0 or this,"},{"Start":"07:05.595 ","End":"07:07.785","Text":"according to the cases."},{"Start":"07:07.785 ","End":"07:11.030","Text":"Next, let\u0027s find out what a_0 is."},{"Start":"07:11.030 ","End":"07:14.540","Text":"We can get this by returning to this formula."},{"Start":"07:14.540 ","End":"07:17.420","Text":"We have all the b_n\u0027s from here,"},{"Start":"07:17.420 ","End":"07:19.715","Text":"and we have all the a_n\u0027s,"},{"Start":"07:19.715 ","End":"07:21.655","Text":"all the a_n\u0027s are 0."},{"Start":"07:21.655 ","End":"07:24.455","Text":"What we\u0027re left with sine 2/2,"},{"Start":"07:24.455 ","End":"07:29.150","Text":"then the case n equals 1 is separately for b_1, it\u0027s half."},{"Start":"07:29.150 ","End":"07:32.225","Text":"We can substitute that in the moment and the rest of it,"},{"Start":"07:32.225 ","End":"07:35.390","Text":"the sum n goes from 2 to infinity,"},{"Start":"07:35.390 ","End":"07:37.070","Text":"but all the odd ones or zeros,"},{"Start":"07:37.070 ","End":"07:38.810","Text":"so it\u0027s just 2, 4, 6, 8,"},{"Start":"07:38.810 ","End":"07:43.560","Text":"etc, which is 2k when k goes from 1 to infinity,"},{"Start":"07:43.560 ","End":"07:49.755","Text":"so just replace the n here with 2k and the n here with 2k."},{"Start":"07:49.755 ","End":"07:53.323","Text":"Now if we substitute r Theta as 0,"},{"Start":"07:53.323 ","End":"07:55.025","Text":"0, which were given,"},{"Start":"07:55.025 ","End":"07:57.410","Text":"actually, we don\u0027t even need the Theta,"},{"Start":"07:57.410 ","End":"08:00.195","Text":"we just need to know what happens when r equal 0,"},{"Start":"08:00.195 ","End":"08:02.590","Text":"because the origin is r equal 0,"},{"Start":"08:02.590 ","End":"08:06.165","Text":"so r here is 0 and out of the 2k is 0,"},{"Start":"08:06.165 ","End":"08:09.270","Text":"so we\u0027re just left with a_0/2,"},{"Start":"08:09.270 ","End":"08:12.510","Text":"and if this is 0, then a_0 is 0."},{"Start":"08:12.510 ","End":"08:14.285","Text":"Now we can rewrite this."},{"Start":"08:14.285 ","End":"08:16.340","Text":"The a naught is 0,"},{"Start":"08:16.340 ","End":"08:21.315","Text":"so this part drops off and the b_1 is a half,"},{"Start":"08:21.315 ","End":"08:24.645","Text":"and the b_2k is this."},{"Start":"08:24.645 ","End":"08:26.630","Text":"This in fact is the final answer."},{"Start":"08:26.630 ","End":"08:28.100","Text":"Let\u0027s just decorate it,"},{"Start":"08:28.100 ","End":"08:29.675","Text":"put it in a red box,"},{"Start":"08:29.675 ","End":"08:33.120","Text":"and declare that we are done."}],"ID":30824},{"Watched":false,"Name":"Exercise 7a","Duration":"3m 20s","ChapterTopicVideoID":29278,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.028","Text":"In this exercise,"},{"Start":"00:02.028 ","End":"00:04.965","Text":"we have a Neumann problem with"},{"Start":"00:04.965 ","End":"00:11.580","Text":"a Laplace equation on a disk and there\u0027s a parameter Alpha here."},{"Start":"00:11.580 ","End":"00:15.300","Text":"We have to find out which value of Alpha will give us"},{"Start":"00:15.300 ","End":"00:20.730","Text":"a solvable equation using what we learned about the necessary condition for solvability."},{"Start":"00:20.730 ","End":"00:27.255","Text":"In Part b, we\u0027ll use the value of Alpha we found in Part a and solve the BVP."},{"Start":"00:27.255 ","End":"00:33.735","Text":"The domain of this problem is an open disk centered at the origin with radius 4,"},{"Start":"00:33.735 ","End":"00:38.005","Text":"and the boundary is the circle of radius 4."},{"Start":"00:38.005 ","End":"00:43.550","Text":"We take the boundary as a closed path Gamma, which circles counterclockwise."},{"Start":"00:43.550 ","End":"00:47.630","Text":"A necessary condition I\u0027ll remind you is that the"},{"Start":"00:47.630 ","End":"00:53.100","Text":"integral of du by dn along this path is 0."},{"Start":"00:53.100 ","End":"01:00.125","Text":"This is the directional derivative in the direction of the outward unit normal."},{"Start":"01:00.125 ","End":"01:05.350","Text":"Let\u0027s see if we can figure out what this integral is in terms of Alpha,"},{"Start":"01:05.350 ","End":"01:08.495","Text":"and then we\u0027ll find what Alpha will make it 0."},{"Start":"01:08.495 ","End":"01:11.795","Text":"We can parameterize the circle in terms of the angle"},{"Start":"01:11.795 ","End":"01:17.410","Text":"Theta and get x=4 cosine Theta, y=4 sine Theta."},{"Start":"01:17.410 ","End":"01:20.910","Text":"We\u0027ll take Theta from 0-2 Pi."},{"Start":"01:20.910 ","End":"01:25.280","Text":"Now, when we substitute in this equation in the left-hand side,"},{"Start":"01:25.280 ","End":"01:31.860","Text":"we get the integral is equal to x is 4 cosine Theta,"},{"Start":"01:31.860 ","End":"01:33.870","Text":"y is 4 sine Theta,"},{"Start":"01:33.870 ","End":"01:39.403","Text":"and ds is the square root of dx^2 plus dy^2"},{"Start":"01:39.403 ","End":"01:47.520","Text":"which is the square root of dx by d Theta^2 plus dy by d Theta^2 d Theta."},{"Start":"01:47.520 ","End":"01:50.365","Text":"We\u0027ve done this before, this thing."},{"Start":"01:50.365 ","End":"01:54.355","Text":"Anyway, this comes down to just 4 d Theta."},{"Start":"01:54.355 ","End":"01:57.200","Text":"Applying the function du by dn which we\u0027re"},{"Start":"01:57.200 ","End":"02:00.410","Text":"given on the boundary to be Alpha x^2 minus 1,"},{"Start":"02:00.410 ","End":"02:01.484","Text":"so we get Alpha."},{"Start":"02:01.484 ","End":"02:06.330","Text":"This is x^2 minus 1 and here 4 d Theta."},{"Start":"02:06.330 ","End":"02:09.142","Text":"This comes out to be 4."},{"Start":"02:09.142 ","End":"02:10.745","Text":"From here, put it in front."},{"Start":"02:10.745 ","End":"02:20.446","Text":"This squared is 16 cosine^2 Theta times Alpha minus 1 d Theta,"},{"Start":"02:20.446 ","End":"02:25.175","Text":"and we can split it into 2 integrals with the minus here."},{"Start":"02:25.175 ","End":"02:29.701","Text":"4 times 16 is 64 times Alpha that comes out in front,"},{"Start":"02:29.701 ","End":"02:37.655","Text":"and then we have cosine^2 Theta d Theta and here minus 4 times the integral of 1 d Theta."},{"Start":"02:37.655 ","End":"02:39.725","Text":"This is a well-known integral,"},{"Start":"02:39.725 ","End":"02:42.070","Text":"it comes out to be Pi."},{"Start":"02:42.070 ","End":"02:45.465","Text":"Another way you can do it and so how I think of it."},{"Start":"02:45.465 ","End":"02:49.485","Text":"Cosine^2 goes between 0 and 1 and the average is 1/2."},{"Start":"02:49.485 ","End":"02:52.204","Text":"Take 1/2 times the length of the interval 2Pi;"},{"Start":"02:52.204 ","End":"02:55.950","Text":"1/2 times 2Pi is Pi, it\u0027s a rule of thumb."},{"Start":"02:55.950 ","End":"03:02.535","Text":"Then integral of 1 on an interval is 1 times the length of the interval, 1 times 2Pi."},{"Start":"03:02.535 ","End":"03:09.170","Text":"We have this and we want this to be 0 when it\u0027s equal to 0."},{"Start":"03:09.170 ","End":"03:14.700","Text":"Isolate Alpha comes out to be 8Pi over 64Pi,"},{"Start":"03:14.700 ","End":"03:17.545","Text":"and that equals 1/8."},{"Start":"03:17.545 ","End":"03:20.830","Text":"That\u0027s the answer to Part a."}],"ID":30825},{"Watched":false,"Name":"Exercise 7b","Duration":"4m 3s","ChapterTopicVideoID":29279,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.825","Text":"Now we come to part B of this problem."},{"Start":"00:03.825 ","End":"00:09.990","Text":"In part A we found that the value of Alpha was equal to 1/8,"},{"Start":"00:09.990 ","End":"00:12.270","Text":"and using this value of Alpha,"},{"Start":"00:12.270 ","End":"00:15.360","Text":"we want to solve the boundary value problem."},{"Start":"00:15.360 ","End":"00:18.495","Text":"Just put Alpha equals 1/8 in here,"},{"Start":"00:18.495 ","End":"00:21.000","Text":"and this is the problem we now have."},{"Start":"00:21.000 ","End":"00:27.150","Text":"We\u0027ll use the general solution for the Laplace equation in a disk centered at the origin."},{"Start":"00:27.150 ","End":"00:28.890","Text":"If it has radius Rho,"},{"Start":"00:28.890 ","End":"00:31.289","Text":"this is the general formula."},{"Start":"00:31.289 ","End":"00:38.905","Text":"In our case, Rho is equal to 4 and so we get the following just replacing Rho by 4."},{"Start":"00:38.905 ","End":"00:45.750","Text":"Now we\u0027re given the value of the directional derivative du by dn on the boundary,"},{"Start":"00:45.750 ","End":"00:53.705","Text":"but this is the same as the partial derivative of u with respect to r. Here\u0027s a reminder"},{"Start":"00:53.705 ","End":"00:56.870","Text":"why the direction of the outer"},{"Start":"00:56.870 ","End":"01:02.360","Text":"normal in this case is just the direction in which the radius increases,"},{"Start":"01:02.360 ","End":"01:06.450","Text":"so du by dn is du by dr. We"},{"Start":"01:06.450 ","End":"01:11.180","Text":"can compute u_r just by differentiating this with respect to r,"},{"Start":"01:11.180 ","End":"01:13.190","Text":"and all we get is,"},{"Start":"01:13.190 ","End":"01:18.065","Text":"instead of r to the n we get n r to the n minus 1 here."},{"Start":"01:18.065 ","End":"01:22.600","Text":"We can apply the boundary condition we were given,"},{"Start":"01:22.600 ","End":"01:25.270","Text":"which was for x^2 plus y^2 equals 16,"},{"Start":"01:25.270 ","End":"01:28.805","Text":"which is the same as the circle r equals 4,"},{"Start":"01:28.805 ","End":"01:33.385","Text":"so one way to do it would be just to use the condition that was given."},{"Start":"01:33.385 ","End":"01:36.880","Text":"It was given to be x^2 over 8 minus 1,"},{"Start":"01:36.880 ","End":"01:41.480","Text":"which turns out to be cosine 2 Theta."},{"Start":"01:41.480 ","End":"01:45.340","Text":"The other way of computing this is just by substituting"},{"Start":"01:45.340 ","End":"01:49.745","Text":"r equals 4 here and what we get then,"},{"Start":"01:49.745 ","End":"01:52.810","Text":"4 to the n minus 1 over 4 to the n is a quarter,"},{"Start":"01:52.810 ","End":"01:55.705","Text":"so n over 4 and everything else is the same."},{"Start":"01:55.705 ","End":"01:59.142","Text":"Now that we have two different expressions for u_r(4 Theta),"},{"Start":"01:59.142 ","End":"02:03.050","Text":"we can compare them, so cosine 2 Theta equals this."},{"Start":"02:03.050 ","End":"02:10.270","Text":"Now we can compare coefficients because we have a Fourier series on the right."},{"Start":"02:10.270 ","End":"02:18.695","Text":"We get that all the b_n\u0027s are 0 because there are no signs at all here."},{"Start":"02:18.695 ","End":"02:20.090","Text":"That means from 1,"},{"Start":"02:20.090 ","End":"02:21.795","Text":"2, 3, etc."},{"Start":"02:21.795 ","End":"02:27.595","Text":"The a_n\u0027s are all 0 except for when n equals 2."},{"Start":"02:27.595 ","End":"02:30.780","Text":"For n equals 1, 3, 4,"},{"Start":"02:30.780 ","End":"02:34.560","Text":"5, etc., a_n is 0."},{"Start":"02:34.560 ","End":"02:37.740","Text":"Now, to figure out what a_2 is,"},{"Start":"02:37.740 ","End":"02:41.855","Text":"compare 1 to the coefficient of cosine 2 Theta here,"},{"Start":"02:41.855 ","End":"02:43.805","Text":"which has gotten when n equals 2,"},{"Start":"02:43.805 ","End":"02:46.100","Text":"we get a_2 here and 2/4 here."},{"Start":"02:46.100 ","End":"02:51.575","Text":"For some reason, I wrote it as a quarter times 2 instead of 2 over 4,"},{"Start":"02:51.575 ","End":"02:53.045","Text":"anyway same thing,"},{"Start":"02:53.045 ","End":"02:56.300","Text":"a_2 comes out to be equal to 2."},{"Start":"02:56.300 ","End":"02:58.955","Text":"Now back to this formula,"},{"Start":"02:58.955 ","End":"03:02.285","Text":"we have all the coefficients except for a_0,"},{"Start":"03:02.285 ","End":"03:05.034","Text":"so we substitute all of them,"},{"Start":"03:05.034 ","End":"03:11.190","Text":"there\u0027s only one thing that\u0027s not 0 and that\u0027s the a_2 coefficient,"},{"Start":"03:11.190 ","End":"03:13.290","Text":"that goes with cosine 2 Theta,"},{"Start":"03:13.290 ","End":"03:20.220","Text":"and here with r^2 over 4^2 which comes out to be 1/2 a_0"},{"Start":"03:20.220 ","End":"03:27.765","Text":"plus r^2 over 8 because a_2 is equal to 2 and 2/16 is an 8."},{"Start":"03:27.765 ","End":"03:29.570","Text":"But we don\u0027t know a_0."},{"Start":"03:29.570 ","End":"03:32.165","Text":"In fact, there\u0027s no way to figure it out."},{"Start":"03:32.165 ","End":"03:38.150","Text":"We need more information if we want to get a_0 so it\u0027s unique up to a constant,"},{"Start":"03:38.150 ","End":"03:40.640","Text":"but really solution is not unique."},{"Start":"03:40.640 ","End":"03:44.855","Text":"You can even call this letter C just to emphasize it."},{"Start":"03:44.855 ","End":"03:46.340","Text":"For any constant C,"},{"Start":"03:46.340 ","End":"03:47.900","Text":"this is a solution."},{"Start":"03:47.900 ","End":"03:54.200","Text":"If like in the previous problem we were given the value of u at some point,"},{"Start":"03:54.200 ","End":"03:58.100","Text":"like u(0,0) then we could substitute and get what C is."},{"Start":"03:58.100 ","End":"04:01.288","Text":"But as it is, it just has to remain like so."},{"Start":"04:01.288 ","End":"04:03.330","Text":"We are done."}],"ID":30826},{"Watched":false,"Name":"Exercise 8a","Duration":"2m 29s","ChapterTopicVideoID":29280,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.840","Text":"In this exercise, we\u0027re given the following Neumann problem."},{"Start":"00:03.840 ","End":"00:06.150","Text":"On the disk of radius R,"},{"Start":"00:06.150 ","End":"00:10.185","Text":"the Laplacian is equal to 0."},{"Start":"00:10.185 ","End":"00:14.895","Text":"On the boundary where little r equals big R,"},{"Start":"00:14.895 ","End":"00:20.550","Text":"the directional derivative in the normal direction is given as follows."},{"Start":"00:20.550 ","End":"00:27.390","Text":"In part a, we have to find a condition on the parameters a,"},{"Start":"00:27.390 ","End":"00:29.730","Text":"b, and c, so that the problem"},{"Start":"00:29.730 ","End":"00:33.615","Text":"satisfies what we call the necessary condition for solvability."},{"Start":"00:33.615 ","End":"00:36.600","Text":"It happens to also be a sufficient condition."},{"Start":"00:36.600 ","End":"00:38.360","Text":"Then part b, well,"},{"Start":"00:38.360 ","End":"00:40.610","Text":"we\u0027ll read it and we come to it."},{"Start":"00:40.610 ","End":"00:46.760","Text":"Domain is the disk of radius R and"},{"Start":"00:46.760 ","End":"00:51.980","Text":"the boundary is the circle of radius r. To remind you,"},{"Start":"00:51.980 ","End":"00:55.400","Text":"the necessary condition we talked about is the following,"},{"Start":"00:55.400 ","End":"01:04.414","Text":"that the integral along the boundary of this directional derivative is 0."},{"Start":"01:04.414 ","End":"01:10.055","Text":"We could work all in polar coordinates at 1 way to do it."},{"Start":"01:10.055 ","End":"01:13.505","Text":"In this case, if we use Theta as a parameter,"},{"Start":"01:13.505 ","End":"01:21.775","Text":"then the circle of radius R could be described as r of Theta is the constant R,"},{"Start":"01:21.775 ","End":"01:24.810","Text":"Theta goes from 0-2 Pi."},{"Start":"01:24.810 ","End":"01:30.280","Text":"In polar coordinates, ds is given as follows."},{"Start":"01:30.280 ","End":"01:37.280","Text":"What we get for this integral is this which we\u0027re given as"},{"Start":"01:37.280 ","End":"01:45.480","Text":"A sine Theta plus B cosine Theta plus C. Then ds comes out to be using this formula,"},{"Start":"01:45.480 ","End":"01:46.800","Text":"R of Theta is R,"},{"Start":"01:46.800 ","End":"01:48.750","Text":"so the derivative is 0."},{"Start":"01:48.750 ","End":"01:51.415","Text":"All together this comes out to be just R,"},{"Start":"01:51.415 ","End":"01:55.010","Text":"which we can pull in front of the integral sign."},{"Start":"01:55.010 ","End":"02:00.740","Text":"We can split this integral up into 3 separate integrals as follows."},{"Start":"02:00.740 ","End":"02:07.210","Text":"At the integral of sine from 0-2 Pi is 0 and so is the integral of cosine."},{"Start":"02:07.210 ","End":"02:12.410","Text":"The integral of 1 is just the length of the interval which is 2 Pi."},{"Start":"02:12.410 ","End":"02:13.940","Text":"This is what we get."},{"Start":"02:13.940 ","End":"02:17.683","Text":"Now, the condition that we want is that this equals 0."},{"Start":"02:17.683 ","End":"02:22.575","Text":"This equals 0 gives us that 2Pi RC is 0,"},{"Start":"02:22.575 ","End":"02:29.290","Text":"which means that C=0 and that\u0027s the solution for Part a."}],"ID":30827},{"Watched":false,"Name":"Exercise 8b","Duration":"3m 40s","ChapterTopicVideoID":29281,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.795","Text":"Now we come to Part B of this problem."},{"Start":"00:03.795 ","End":"00:07.905","Text":"We have to solve the boundary value problem assuming that the condition of"},{"Start":"00:07.905 ","End":"00:12.135","Text":"a holds and we want the answer in Cartesian coordinates."},{"Start":"00:12.135 ","End":"00:19.785","Text":"In Part A, we showed that the condition is that C equal 0."},{"Start":"00:19.785 ","End":"00:22.755","Text":"Let\u0027s just rewrite this with C equals 0."},{"Start":"00:22.755 ","End":"00:24.540","Text":"This is what we have,"},{"Start":"00:24.540 ","End":"00:30.585","Text":"and the general solution for the Laplace equation in the disk is the following."},{"Start":"00:30.585 ","End":"00:34.485","Text":"In our case, Rho is just big R,"},{"Start":"00:34.485 ","End":"00:37.530","Text":"so replacing it, we get the following."},{"Start":"00:37.530 ","End":"00:40.400","Text":"Note that du by dn,"},{"Start":"00:40.400 ","End":"00:44.630","Text":"the directional derivative in the direction of the normal in our case,"},{"Start":"00:44.630 ","End":"00:49.160","Text":"is just the derivative with respect to the radius."},{"Start":"00:49.160 ","End":"00:53.930","Text":"This is because the normal facing outwards on a circle"},{"Start":"00:53.930 ","End":"00:59.290","Text":"falls exactly on a radius in the direction of the increasing r,"},{"Start":"00:59.290 ","End":"01:01.700","Text":"and du by dr, we get from this by"},{"Start":"01:01.700 ","End":"01:05.570","Text":"differentiating with respect to r. R only appears here,"},{"Start":"01:05.570 ","End":"01:10.000","Text":"so the derivative of r to the n is nr to the n minus 1."},{"Start":"01:10.000 ","End":"01:12.060","Text":"If we let little r,"},{"Start":"01:12.060 ","End":"01:14.480","Text":"be big R as in here,"},{"Start":"01:14.480 ","End":"01:19.280","Text":"then what we get is r to the n minus 1 over r to"},{"Start":"01:19.280 ","End":"01:24.050","Text":"the n is simply 1 over R. On the other hand,"},{"Start":"01:24.050 ","End":"01:28.145","Text":"we have this expression ur of r Theta from here,"},{"Start":"01:28.145 ","End":"01:33.305","Text":"because we said that du by dr is du by dn,"},{"Start":"01:33.305 ","End":"01:36.990","Text":"so this is equal to that."},{"Start":"01:36.990 ","End":"01:40.350","Text":"We have u_r of R Theta equal to 2 different things,"},{"Start":"01:40.350 ","End":"01:43.115","Text":"they have to be equal amongst themselves."},{"Start":"01:43.115 ","End":"01:45.590","Text":"We have the following."},{"Start":"01:45.590 ","End":"01:51.470","Text":"What we have here on the right is a Fourier series with cosine and Theta,"},{"Start":"01:51.470 ","End":"01:55.240","Text":"and we can use the method of comparing coefficients."},{"Start":"01:55.240 ","End":"01:58.125","Text":"What we get, let\u0027s see."},{"Start":"01:58.125 ","End":"02:01.235","Text":"For A, we can look at sine Theta,"},{"Start":"02:01.235 ","End":"02:03.710","Text":"which is this part when n is 1,"},{"Start":"02:03.710 ","End":"02:08.125","Text":"that\u0027s b_1 times 1 times 1 over R,"},{"Start":"02:08.125 ","End":"02:15.240","Text":"and the B we can get from 1 times 1 over R times a_1."},{"Start":"02:15.240 ","End":"02:17.455","Text":"When n is bigger than 1,"},{"Start":"02:17.455 ","End":"02:20.210","Text":"we get 0 here,"},{"Start":"02:20.210 ","End":"02:26.175","Text":"which is either n times 1 over R times a_n or b_n."},{"Start":"02:26.175 ","End":"02:32.925","Text":"These are all zeros. What we see is that a_1 is RB,"},{"Start":"02:32.925 ","End":"02:41.185","Text":"b_1 is RA and a_n and b_n are all 0 for n bigger than 1."},{"Start":"02:41.185 ","End":"02:44.915","Text":"Now we can substitute these values in here."},{"Start":"02:44.915 ","End":"02:50.405","Text":"We only get 2 non-zero coefficients in this sum,"},{"Start":"02:50.405 ","End":"02:52.620","Text":"when n is 1,"},{"Start":"02:53.060 ","End":"02:55.590","Text":"r over r,"},{"Start":"02:55.590 ","End":"02:59.085","Text":"and then a_n is RB,"},{"Start":"02:59.085 ","End":"03:04.740","Text":"a_1 is BR and b_n we get RA or AR,"},{"Start":"03:04.740 ","End":"03:10.725","Text":"r over r. The R cancels with the R,"},{"Start":"03:10.725 ","End":"03:13.125","Text":"the R cancels with the R here,"},{"Start":"03:13.125 ","End":"03:15.260","Text":"and this is what we have left."},{"Start":"03:15.260 ","End":"03:18.185","Text":"Now we want the answer in Cartesian."},{"Start":"03:18.185 ","End":"03:20.795","Text":"R cosine Theta is x,"},{"Start":"03:20.795 ","End":"03:23.935","Text":"our sine Theta is y,"},{"Start":"03:23.935 ","End":"03:26.130","Text":"and the a naught over 2,"},{"Start":"03:26.130 ","End":"03:28.609","Text":"we could leave it or we could just call it another constant"},{"Start":"03:28.609 ","End":"03:31.190","Text":"c. But idea to use the letter c,"},{"Start":"03:31.190 ","End":"03:33.140","Text":"we had the letter c already in the beginning."},{"Start":"03:33.140 ","End":"03:38.345","Text":"Let\u0027s call it k. Now this is the general solution up to a constant k,"},{"Start":"03:38.345 ","End":"03:40.770","Text":"and we are done."}],"ID":30828},{"Watched":false,"Name":"Exercise 9","Duration":"5m 27s","ChapterTopicVideoID":29282,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.890","Text":"In this exercise, we\u0027re given the following boundary value problem to solve."},{"Start":"00:04.890 ","End":"00:06.750","Text":"This is the Laplacian,"},{"Start":"00:06.750 ","End":"00:09.495","Text":"but it\u0027s non-homogeneous,"},{"Start":"00:09.495 ","End":"00:11.339","Text":"so it\u0027s not a Laplace equation,"},{"Start":"00:11.339 ","End":"00:13.425","Text":"it\u0027s a Poisson equation."},{"Start":"00:13.425 ","End":"00:17.370","Text":"It\u0027s on the disk of radius 3."},{"Start":"00:17.370 ","End":"00:22.335","Text":"We\u0027re given the boundary condition for u,"},{"Start":"00:22.335 ","End":"00:24.490","Text":"which is as follows."},{"Start":"00:24.490 ","End":"00:29.865","Text":"We\u0027re going to use our usual technique of splitting the problem into two."},{"Start":"00:29.865 ","End":"00:35.835","Text":"One of the problems will have a non-homogeneous PDE,"},{"Start":"00:35.835 ","End":"00:38.730","Text":"but it will have homogeneous boundary conditions,"},{"Start":"00:38.730 ","End":"00:41.095","Text":"and the other one will be the other way around,"},{"Start":"00:41.095 ","End":"00:49.025","Text":"it will have a homogeneous PDE and a non-homogeneous boundary condition."},{"Start":"00:49.025 ","End":"00:52.825","Text":"We\u0027ll split u up into v plus w,"},{"Start":"00:52.825 ","End":"01:00.760","Text":"where v has the non-homogeneous PDE and w has the non-homogeneous boundary condition."},{"Start":"01:00.760 ","End":"01:03.840","Text":"Let\u0027s work on v first."},{"Start":"01:03.840 ","End":"01:06.495","Text":"Note that there\u0027s only r,"},{"Start":"01:06.495 ","End":"01:09.195","Text":"there\u0027s no Theta on the right-hand sides."},{"Start":"01:09.195 ","End":"01:13.230","Text":"One obvious thing to try is to find the radial solution,"},{"Start":"01:13.230 ","End":"01:14.810","Text":"v as a function of r,"},{"Start":"01:14.810 ","End":"01:16.580","Text":"not depending on Theta."},{"Start":"01:16.580 ","End":"01:20.125","Text":"Now, the Laplacian in polar coordinates is this,"},{"Start":"01:20.125 ","End":"01:22.660","Text":"and v now just depends on r,"},{"Start":"01:22.660 ","End":"01:25.325","Text":"so we can drop the Theta parameter."},{"Start":"01:25.325 ","End":"01:29.150","Text":"This part will be 0 because v just depends on r,"},{"Start":"01:29.150 ","End":"01:31.955","Text":"so dv by dTheta is 0."},{"Start":"01:31.955 ","End":"01:34.550","Text":"What we get is the following,"},{"Start":"01:34.550 ","End":"01:37.435","Text":"just rewriting v_ rr as v\u0027\u0027,"},{"Start":"01:37.435 ","End":"01:42.335","Text":"and v_ r as v\u0027 because it only depends on"},{"Start":"01:42.335 ","End":"01:50.985","Text":"r. This is a second-order differential equation with initial condition."},{"Start":"01:50.985 ","End":"01:53.720","Text":"Multiply this equation by r,"},{"Start":"01:53.720 ","End":"01:55.504","Text":"and we get the following."},{"Start":"01:55.504 ","End":"02:01.580","Text":"We can write this left-hand side as the derivative of a product."},{"Start":"02:01.580 ","End":"02:06.240","Text":"If you look at it, it\u0027s the derivative of rv\u0027,"},{"Start":"02:06.260 ","End":"02:09.680","Text":"because it\u0027s r times the derivative of this,"},{"Start":"02:09.680 ","End":"02:12.575","Text":"v\u0027\u0027, and derivative of r,"},{"Start":"02:12.575 ","End":"02:15.595","Text":"which is just 1 times v\u0027."},{"Start":"02:15.595 ","End":"02:18.050","Text":"Now that we have it as a derivative,"},{"Start":"02:18.050 ","End":"02:23.250","Text":"we can integrate both sides with respect to r and get that rv\u0027"},{"Start":"02:23.250 ","End":"02:28.935","Text":"is r^3 over 3 root 3 plus a constant of integration a,"},{"Start":"02:28.935 ","End":"02:31.200","Text":"divide by r,"},{"Start":"02:31.200 ","End":"02:33.570","Text":"and we get this."},{"Start":"02:33.570 ","End":"02:39.180","Text":"We have to take a equals 0 because v\u0027 is differentiable in the disk,"},{"Start":"02:39.180 ","End":"02:42.990","Text":"and in particular at r=0."},{"Start":"02:42.990 ","End":"02:47.025","Text":"That means that v(r) is the integral of this,"},{"Start":"02:47.025 ","End":"02:50.360","Text":"so it\u0027s r^3 and divide it by another 3,"},{"Start":"02:50.360 ","End":"02:53.785","Text":"and we get another constant of integration."},{"Start":"02:53.785 ","End":"02:59.910","Text":"But our boundary condition is that v(3) equals 0,"},{"Start":"02:59.910 ","End":"03:01.980","Text":"so substitute 3 equate to 0,"},{"Start":"03:01.980 ","End":"03:04.375","Text":"so c equals minus this,"},{"Start":"03:04.375 ","End":"03:07.315","Text":"which comes out to be minus square root of 3."},{"Start":"03:07.315 ","End":"03:08.815","Text":"Plug that here,"},{"Start":"03:08.815 ","End":"03:12.215","Text":"and we have v(r) is this."},{"Start":"03:12.215 ","End":"03:20.150","Text":"That\u0027s v, now let\u0027s move on to w. We have a Laplace equation on a disk of radius 3,"},{"Start":"03:20.150 ","End":"03:25.385","Text":"so the general solution is given by the usual formula and in our case,"},{"Start":"03:25.385 ","End":"03:26.930","Text":"Rho equals 3,"},{"Start":"03:26.930 ","End":"03:30.285","Text":"so just replace Rho with 3."},{"Start":"03:30.285 ","End":"03:33.495","Text":"If we substitute r equals 3 here,"},{"Start":"03:33.495 ","End":"03:34.650","Text":"on the one hand,"},{"Start":"03:34.650 ","End":"03:37.220","Text":"w(3, Theta) is this."},{"Start":"03:37.220 ","End":"03:41.105","Text":"On the other hand, we can just substitute r equals 3 here."},{"Start":"03:41.105 ","End":"03:47.775","Text":"We get that 4 cosine(2Theta) minus 8 sine(4Theta) is equal to,"},{"Start":"03:47.775 ","End":"03:50.340","Text":"r over 3 is 1,"},{"Start":"03:50.340 ","End":"03:54.575","Text":"so we just get this sum here with this bit missing."},{"Start":"03:54.575 ","End":"03:57.740","Text":"Now we use our method of comparing coefficients because we"},{"Start":"03:57.740 ","End":"04:00.890","Text":"have a Fourier series of sines and cosines."},{"Start":"04:00.890 ","End":"04:03.845","Text":"There are only two non-0 coefficients;"},{"Start":"04:03.845 ","End":"04:08.395","Text":"the coefficient of cosine(2Theta) and the coefficient of sine(4Theta)."},{"Start":"04:08.395 ","End":"04:11.940","Text":"This will give us a_ 2 and b_ 4."},{"Start":"04:11.940 ","End":"04:17.715","Text":"We get that a_ 2 is 4,"},{"Start":"04:17.715 ","End":"04:20.955","Text":"and b_ 4 is minus 8,"},{"Start":"04:20.955 ","End":"04:24.055","Text":"and all the rest of the coefficients are 0."},{"Start":"04:24.055 ","End":"04:28.445","Text":"If we substitute these coefficients back in this formula,"},{"Start":"04:28.445 ","End":"04:31.315","Text":"we\u0027ll just get two terms."},{"Start":"04:31.315 ","End":"04:38.810","Text":"The only non-0 coefficients are a_ 2 and b_ 4."},{"Start":"04:39.470 ","End":"04:43.125","Text":"For n equals 2 here,"},{"Start":"04:43.125 ","End":"04:44.670","Text":"we\u0027ll get a_ 2,"},{"Start":"04:44.670 ","End":"04:46.500","Text":"r over 3^2,"},{"Start":"04:46.500 ","End":"04:50.010","Text":"and the 4 from here, cosine(2Theta)."},{"Start":"04:50.010 ","End":"04:51.975","Text":"In the case of the b_ 4,"},{"Start":"04:51.975 ","End":"04:55.110","Text":"we\u0027ll get sine(4Theta),"},{"Start":"04:55.110 ","End":"04:57.252","Text":"and the minus 8,"},{"Start":"04:57.252 ","End":"05:00.675","Text":"and r over 3^4,"},{"Start":"05:00.675 ","End":"05:03.460","Text":"so that\u0027s w. Now,"},{"Start":"05:03.460 ","End":"05:05.170","Text":"we also have v,"},{"Start":"05:05.170 ","End":"05:07.630","Text":"we just did it above."},{"Start":"05:07.630 ","End":"05:11.365","Text":"Adding w plus v to get u,"},{"Start":"05:11.365 ","End":"05:14.480","Text":"we get the answer as follows."},{"Start":"05:14.480 ","End":"05:16.890","Text":"This part here is w,"},{"Start":"05:16.890 ","End":"05:18.540","Text":"and this is v,"},{"Start":"05:18.540 ","End":"05:24.570","Text":"let\u0027s see above, r^3 over 9 root 3 minus root 3,"},{"Start":"05:24.570 ","End":"05:28.120","Text":"that\u0027s it. We are done."}],"ID":30829},{"Watched":false,"Name":"Exercise 10","Duration":"4m 9s","ChapterTopicVideoID":29283,"CourseChapterTopicPlaylistID":294436,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.000","Text":"In this exercise, we have the following boundary value problem."},{"Start":"00:04.000 ","End":"00:06.460","Text":"We\u0027re working in polar coordinates."},{"Start":"00:06.460 ","End":"00:09.505","Text":"This is a Poisson equation,"},{"Start":"00:09.505 ","End":"00:14.065","Text":"like a non-homogeneous Laplace equation on the unit dis."},{"Start":"00:14.065 ","End":"00:18.350","Text":"This is the boundary value on the unit circle."},{"Start":"00:18.350 ","End":"00:23.860","Text":"The usual technique is to split this up into 2 problems."},{"Start":"00:23.860 ","End":"00:27.205","Text":"One that gets the non-homogeneous PDE"},{"Start":"00:27.205 ","End":"00:30.775","Text":"and the other gets the non-homogeneous boundary condition."},{"Start":"00:30.775 ","End":"00:32.215","Text":"This is what I mean."},{"Start":"00:32.215 ","End":"00:38.875","Text":"Let u be v plus w. V will have a homogeneous PDE,"},{"Start":"00:38.875 ","End":"00:41.910","Text":"but a non-homogeneous boundary condition and the other way round"},{"Start":"00:41.910 ","End":"00:45.270","Text":"with w. This is not homogeneous and this is."},{"Start":"00:45.270 ","End":"00:47.325","Text":"Let\u0027s work on w first."},{"Start":"00:47.325 ","End":"00:52.310","Text":"Note that there is no Theta in the right hand side."},{"Start":"00:52.310 ","End":"00:54.830","Text":"Everything just depends on r. The way to go"},{"Start":"00:54.830 ","End":"00:57.560","Text":"about it is to try and find a radial solution."},{"Start":"00:57.560 ","End":"01:02.375","Text":"In other words, w as a function just of r. If we do that,"},{"Start":"01:02.375 ","End":"01:07.385","Text":"the Laplacian in polar coordinates is the following."},{"Start":"01:07.385 ","End":"01:11.925","Text":"That is equal to r^4th and w(1) equals 0."},{"Start":"01:11.925 ","End":"01:18.065","Text":"We can write this as w\u0027\u0027 wr(w\u0027)."},{"Start":"01:18.065 ","End":"01:19.895","Text":"This part is 0,"},{"Start":"01:19.895 ","End":"01:22.010","Text":"because if w is a function of r,"},{"Start":"01:22.010 ","End":"01:25.130","Text":"then the derivatives with respect to Theta r 0."},{"Start":"01:25.130 ","End":"01:31.520","Text":"Now we can take this and multiply both sides by r. I get the following."},{"Start":"01:31.520 ","End":"01:35.870","Text":"The left-hand side is the derivative of a product."},{"Start":"01:35.870 ","End":"01:39.725","Text":"Check this. If you use the product rule on this, you get this."},{"Start":"01:39.725 ","End":"01:44.425","Text":"Now we can integrate both sides with respect to r and get w\u0027 equals"},{"Start":"01:44.425 ","End":"01:49.130","Text":"r^6 over 6 plus constant of integration."},{"Start":"01:49.130 ","End":"01:52.445","Text":"If we substitute r=0,"},{"Start":"01:52.445 ","End":"01:55.145","Text":"then we get a=0."},{"Start":"01:55.145 ","End":"01:57.695","Text":"Now dividing by r,"},{"Start":"01:57.695 ","End":"02:01.955","Text":"we get w\u0027 is r^5 over 6."},{"Start":"02:01.955 ","End":"02:11.330","Text":"Then integrating, we get that w is r^6 over 36 plus another constant of integration."},{"Start":"02:11.330 ","End":"02:15.185","Text":"But we\u0027re given that w(1) is 0."},{"Start":"02:15.185 ","End":"02:21.195","Text":"Substituting that, we get that c is minus 1 over 36."},{"Start":"02:21.195 ","End":"02:22.445","Text":"Putting that in here,"},{"Start":"02:22.445 ","End":"02:24.169","Text":"putting it over a common denominator,"},{"Start":"02:24.169 ","End":"02:26.090","Text":"36, we get the following,"},{"Start":"02:26.090 ","End":"02:28.960","Text":"1 over 36 r^6 minus 1."},{"Start":"02:28.960 ","End":"02:36.665","Text":"That\u0027s w. Now we want to find v. This was what the equation for v was."},{"Start":"02:36.665 ","End":"02:43.940","Text":"We\u0027ll solve this because it\u0027s a Laplace equation on the disk using the standard formula,"},{"Start":"02:43.940 ","End":"02:45.990","Text":"which is the following."},{"Start":"02:45.990 ","End":"02:50.730","Text":"In our case, Rho is equal to 1."},{"Start":"02:50.730 ","End":"02:52.810","Text":"Putting Rho equals 1 here,"},{"Start":"02:52.810 ","End":"02:54.445","Text":"it reduces to the following."},{"Start":"02:54.445 ","End":"02:59.900","Text":"That if our overall we have just r. Now if we substitute r=1,"},{"Start":"02:59.900 ","End":"03:03.085","Text":"on the one hand, we\u0027ll get cosine Theta."},{"Start":"03:03.085 ","End":"03:06.815","Text":"On the other hand we get what is here with r= 1."},{"Start":"03:06.815 ","End":"03:11.410","Text":"That cosine Theta equals a naught over 2 plus the sum,"},{"Start":"03:11.410 ","End":"03:14.380","Text":"same as here, except that without the r^n,"},{"Start":"03:14.380 ","End":"03:16.870","Text":"because it\u0027s 1^n, which is 1."},{"Start":"03:16.870 ","End":"03:20.770","Text":"Now this is a Fourier series of sines and cosines,"},{"Start":"03:20.770 ","End":"03:22.525","Text":"so we can compare coefficients."},{"Start":"03:22.525 ","End":"03:24.760","Text":"There is only one term on the left,"},{"Start":"03:24.760 ","End":"03:29.180","Text":"cosine Theta and that will give us a_1."},{"Start":"03:29.180 ","End":"03:34.695","Text":"We\u0027ll get that a_1 is equal to 1 and everything else is 0."},{"Start":"03:34.695 ","End":"03:36.800","Text":"Copying this equation here,"},{"Start":"03:36.800 ","End":"03:40.655","Text":"now substitute a_1 equals 1 and everything else is 0."},{"Start":"03:40.655 ","End":"03:47.385","Text":"What we get is that v(r)Theta is r^1 times a_1,"},{"Start":"03:47.385 ","End":"03:51.165","Text":"which is 1 cosine 1 Theta."},{"Start":"03:51.165 ","End":"03:56.325","Text":"Now we found v(r) Theta and we already had w(r)."},{"Start":"03:56.325 ","End":"03:58.695","Text":"We can combine them."},{"Start":"03:58.695 ","End":"04:03.950","Text":"Remember that u is w plus v and what we get just"},{"Start":"04:03.950 ","End":"04:10.080","Text":"adding the two together is this and this is our answer. We\u0027re done."}],"ID":30830}],"Thumbnail":null,"ID":294436},{"Name":"The Laplace Equation in an Annulus","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Laplace Annulus intro","Duration":"6m 50s","ChapterTopicVideoID":29264,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.805","Text":"We\u0027ve just done the section on the Laplace equation in a disk."},{"Start":"00:05.805 ","End":"00:09.825","Text":"Now we come to the Laplace equation in an annulus."},{"Start":"00:09.825 ","End":"00:14.550","Text":"An annulus is the area bounded between 2 concentric circles."},{"Start":"00:14.550 ","End":"00:17.820","Text":"Here we\u0027ll take the circles centered at the origin."},{"Start":"00:17.820 ","End":"00:20.550","Text":"There\u0027s an inner radius and an outer radius,"},{"Start":"00:20.550 ","End":"00:22.710","Text":"and we\u0027ll call them a and b."},{"Start":"00:22.710 ","End":"00:27.682","Text":"We mostly work in polar coordinates for this section."},{"Start":"00:27.682 ","End":"00:33.150","Text":"The Laplace equation in such an annulus in polar coordinates is given"},{"Start":"00:33.150 ","End":"00:39.315","Text":"as the Laplacian of u is equal to 0 in the annulus."},{"Start":"00:39.315 ","End":"00:45.242","Text":"This is the formula for the Laplacian of u in polar coordinates."},{"Start":"00:45.242 ","End":"00:46.789","Text":"In case, you\u0027re wondering,"},{"Start":"00:46.789 ","End":"00:50.615","Text":"or you have forgotten what the Cartesian coordinate version is,"},{"Start":"00:50.615 ","End":"00:55.775","Text":"this is what it is and we can derive this from this,"},{"Start":"00:55.775 ","End":"00:58.805","Text":"but it\u0027s just technical and we\u0027ll just"},{"Start":"00:58.805 ","End":"01:03.205","Text":"accept that this is the Laplacian in polar coordinates."},{"Start":"01:03.205 ","End":"01:08.210","Text":"We can rewrite this a little bit if we differentiate"},{"Start":"01:08.210 ","End":"01:12.500","Text":"the product here using the product rule and then multiply by 1 over"},{"Start":"01:12.500 ","End":"01:17.960","Text":"r. Then we could write it in the simpler notation with"},{"Start":"01:17.960 ","End":"01:23.795","Text":"the subscript u_rr plus 1 over r u_r plus 1 over r^2 u_Theta Theta."},{"Start":"01:23.795 ","End":"01:29.330","Text":"That\u0027s the Laplacian in polar coordinates just a little variation on this."},{"Start":"01:29.330 ","End":"01:33.560","Text":"Now the general solution to the Laplace equation in"},{"Start":"01:33.560 ","End":"01:37.955","Text":"the annulus is given by the following formula."},{"Start":"01:37.955 ","End":"01:43.010","Text":"Note that there are coefficients a_n and a_minus"},{"Start":"01:43.010 ","End":"01:49.175","Text":"n. We also have r^minus n besides r^n,"},{"Start":"01:49.175 ","End":"01:50.795","Text":"which we had for a disk."},{"Start":"01:50.795 ","End":"01:54.155","Text":"Basically this goes for the interior of a disk,"},{"Start":"01:54.155 ","End":"01:56.950","Text":"and this goes for the exterior of the disk."},{"Start":"01:56.950 ","End":"02:01.055","Text":"In an annulus were both inside and outside of a disk."},{"Start":"02:01.055 ","End":"02:03.410","Text":"There\u0027s also an natural log here."},{"Start":"02:03.410 ","End":"02:06.590","Text":"This is the formula and the coefficients."},{"Start":"02:06.590 ","End":"02:09.480","Text":"The a_naught b naught a_n, a_minus n, b_n,"},{"Start":"02:09.480 ","End":"02:14.165","Text":"b_minus n are computed from the boundary conditions."},{"Start":"02:14.165 ","End":"02:17.420","Text":"There\u0027s different boundary conditions possible."},{"Start":"02:17.420 ","End":"02:20.090","Text":"Common one is the Dirichlet boundary condition"},{"Start":"02:20.090 ","End":"02:21.950","Text":"where we are given the value of the solution"},{"Start":"02:21.950 ","End":"02:27.380","Text":"u on the inner circle and on the outer circle as 2 functions,"},{"Start":"02:27.380 ","End":"02:29.615","Text":"f of Theta and g of Theta."},{"Start":"02:29.615 ","End":"02:36.170","Text":"The Neumann boundary conditions are where we\u0027re given the derivative with respect to r,"},{"Start":"02:36.170 ","End":"02:39.225","Text":"partial derivative, that of the function itself."},{"Start":"02:39.225 ","End":"02:41.810","Text":"This turns out to be the same as"},{"Start":"02:41.810 ","End":"02:46.850","Text":"the directional derivative in the outward direction on the boundary,"},{"Start":"02:46.850 ","End":"02:47.975","Text":"which is the circle."},{"Start":"02:47.975 ","End":"02:53.960","Text":"Because d u by d r is the radial derivative in the direction of"},{"Start":"02:53.960 ","End":"02:57.440","Text":"the increasing radius and the increasing radius from around"},{"Start":"02:57.440 ","End":"03:03.199","Text":"the boundary of a circle is just the outward normal."},{"Start":"03:03.199 ","End":"03:08.910","Text":"You could also have mixed boundary conditions on one side it could be Dirichlet,"},{"Start":"03:08.910 ","End":"03:11.900","Text":"in the other side it could be Neumann."},{"Start":"03:11.900 ","End":"03:16.220","Text":"In fact, it could even be that some linear combination"},{"Start":"03:16.220 ","End":"03:21.970","Text":"of u and d u by d r equals a given function."},{"Start":"03:21.970 ","End":"03:24.850","Text":"Now it\u0027s time for an example."},{"Start":"03:24.850 ","End":"03:28.505","Text":"Here\u0027s our example boundary value problem."},{"Start":"03:28.505 ","End":"03:32.480","Text":"We are going to solve the following in polar coordinates."},{"Start":"03:32.480 ","End":"03:36.785","Text":"This is the PDE, the Laplace equation,"},{"Start":"03:36.785 ","End":"03:42.980","Text":"and the domain is the annulus between radius 1 and radius 2."},{"Start":"03:42.980 ","End":"03:46.940","Text":"We\u0027re given that on the inner circle,"},{"Start":"03:46.940 ","End":"03:55.265","Text":"that u=1 constant and on the outer circle, u=0."},{"Start":"03:55.265 ","End":"03:58.325","Text":"Here\u0027s the formula that we were given."},{"Start":"03:58.325 ","End":"04:02.330","Text":"Could have written it as u(r) Theta here and here,"},{"Start":"04:02.330 ","End":"04:05.825","Text":"because in any event here r is 1, here r is 2."},{"Start":"04:05.825 ","End":"04:10.010","Text":"When r is 1 yet you have 1 Theta equals 1."},{"Start":"04:10.010 ","End":"04:13.880","Text":"On the other hand, you have 1 Theta from this formula is equal"},{"Start":"04:13.880 ","End":"04:19.190","Text":"to a naught r if it\u0027s 1 natural log is 0,"},{"Start":"04:19.190 ","End":"04:24.560","Text":"so this term drops out and r^n is just 1."},{"Start":"04:24.560 ","End":"04:26.300","Text":"We get the following."},{"Start":"04:26.300 ","End":"04:29.210","Text":"We compare coefficients of the series."},{"Start":"04:29.210 ","End":"04:31.189","Text":"The constant is a_naught,"},{"Start":"04:31.189 ","End":"04:32.750","Text":"and here it\u0027s 1."},{"Start":"04:32.750 ","End":"04:37.285","Text":"Now the coefficient of cosine n Theta."},{"Start":"04:37.285 ","End":"04:38.760","Text":"Here it\u0027s 0,"},{"Start":"04:38.760 ","End":"04:45.840","Text":"here it\u0027s a_n plus a_minus n. This is equal 0 for n bigger or equal to 1."},{"Start":"04:45.840 ","End":"04:49.665","Text":"Similarly with the coefficient of sine n Theta,"},{"Start":"04:49.665 ","End":"04:51.270","Text":"we get the following."},{"Start":"04:51.270 ","End":"04:53.950","Text":"This is also for n bigger or equal 1."},{"Start":"04:53.950 ","End":"05:01.400","Text":"Now, let\u0027s substitute r equal u(2) and Theta that\u0027s equal 0 where given."},{"Start":"05:01.400 ","End":"05:04.189","Text":"On the other hand, from this expression,"},{"Start":"05:04.189 ","End":"05:06.140","Text":"we replace r by 2."},{"Start":"05:06.140 ","End":"05:08.000","Text":"We\u0027ve got the following."},{"Start":"05:08.000 ","End":"05:10.550","Text":"Again, let\u0027s just compare coefficients."},{"Start":"05:10.550 ","End":"05:13.625","Text":"This time, the constant coefficient is this."},{"Start":"05:13.625 ","End":"05:16.490","Text":"On the right and on the left it\u0027s 0."},{"Start":"05:16.490 ","End":"05:20.895","Text":"Coefficient of cosine n Theta gives us this."},{"Start":"05:20.895 ","End":"05:25.770","Text":"The coefficient of sine n Theta gives us this."},{"Start":"05:25.770 ","End":"05:31.125","Text":"Now we have these 3 equations and these 3 equations,"},{"Start":"05:31.125 ","End":"05:33.330","Text":"and we\u0027ve got 6 unknowns."},{"Start":"05:33.330 ","End":"05:38.300","Text":"Well, consider a_n as a single unknown in terms of n. We"},{"Start":"05:38.300 ","End":"05:43.610","Text":"have all these equations and let\u0027s start from this and this,"},{"Start":"05:43.610 ","End":"05:47.810","Text":"we can get that a naught is 1 and then put a naught equals 1."},{"Start":"05:47.810 ","End":"05:52.280","Text":"Here we get b naught is minus 1 over natural log of 2."},{"Start":"05:52.280 ","End":"05:55.010","Text":"Now from this equation and this equation,"},{"Start":"05:55.010 ","End":"05:58.465","Text":"we get that a_n equals a_minus n equals 0."},{"Start":"05:58.465 ","End":"06:01.610","Text":"Because a_minus n is minus a_n."},{"Start":"06:01.610 ","End":"06:04.415","Text":"If you put minus a_n here,"},{"Start":"06:04.415 ","End":"06:11.145","Text":"then we have a_n times 2^n minus 1 over 2^n."},{"Start":"06:11.145 ","End":"06:16.060","Text":"Then even a_n times something non-zero is 0."},{"Start":"06:16.060 ","End":"06:19.925","Text":"A_n has to be 0 and therefore a_minus n is 0."},{"Start":"06:19.925 ","End":"06:23.990","Text":"Similarly with b_n and b_minus n. Now we have"},{"Start":"06:23.990 ","End":"06:30.500","Text":"all the coefficients and we can just plug them back into this equation here."},{"Start":"06:30.500 ","End":"06:33.219","Text":"What we get is the following,"},{"Start":"06:33.219 ","End":"06:35.310","Text":"a naught is 1,"},{"Start":"06:35.310 ","End":"06:40.185","Text":"b naught is minus 1 over natural log of 2,"},{"Start":"06:40.185 ","End":"06:43.230","Text":"and all the others are 0."},{"Start":"06:43.230 ","End":"06:44.790","Text":"We\u0027re just left with this,"},{"Start":"06:44.790 ","End":"06:50.700","Text":"and that\u0027s the answer and that concludes this example and this clip."}],"ID":30831},{"Watched":false,"Name":"Exercise 1","Duration":"4m 23s","ChapterTopicVideoID":29265,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.370","Text":"In this exercise, we\u0027re going to solve the following boundary value problem which is"},{"Start":"00:05.370 ","End":"00:10.605","Text":"given in polar coordinates but we want the solution in Cartesian coordinates."},{"Start":"00:10.605 ","End":"00:17.750","Text":"The Laplacian is 0 in the annulus between the radii 1 and 2."},{"Start":"00:17.750 ","End":"00:19.940","Text":"On the inner circle of radius 1,"},{"Start":"00:19.940 ","End":"00:22.220","Text":"we\u0027re given that u is 0 and on the outer circle of"},{"Start":"00:22.220 ","End":"00:25.780","Text":"radius 2 we\u0027re given that u is 2 cosine Theta."},{"Start":"00:25.780 ","End":"00:31.535","Text":"The formula for the solution on an annulus of the Laplace equation is the following."},{"Start":"00:31.535 ","End":"00:35.090","Text":"Let\u0027s first of all substitute r equals 1 then we\u0027ll substitute r equals 2."},{"Start":"00:35.090 ","End":"00:38.539","Text":"If r equals 1, we get u of 1 Theta equals 0."},{"Start":"00:38.539 ","End":"00:42.200","Text":"On the other hand, if we let r equals 1 here, all these r^n,"},{"Start":"00:42.200 ","End":"00:46.130","Text":"r^minus n r equal to 1 and dropout."},{"Start":"00:46.130 ","End":"00:50.105","Text":"Natural log of 1 is 0 so we get the following."},{"Start":"00:50.105 ","End":"00:52.325","Text":"If we compare coefficients,"},{"Start":"00:52.325 ","End":"00:56.105","Text":"the free coefficient is a naught, so that\u0027s 0."},{"Start":"00:56.105 ","End":"01:03.140","Text":"The coefficient of cosine n Theta is an plus a minus and on the other hand is 0,"},{"Start":"01:03.140 ","End":"01:06.835","Text":"and similarly for the sine and Theta coefficients."},{"Start":"01:06.835 ","End":"01:11.520","Text":"Now let\u0027s substitute the next circle or r equals 2."},{"Start":"01:11.520 ","End":"01:16.255","Text":"We get u of 2 Theta is 2 cosine Theta."},{"Start":"01:16.255 ","End":"01:20.420","Text":"On the other hand, substituting r equals 2 in this formula,"},{"Start":"01:20.420 ","End":"01:26.490","Text":"we get a naught plus v naught natural log of 2 plus the following."},{"Start":"01:26.490 ","End":"01:30.830","Text":"Then comparing coefficients, the free coefficient here is"},{"Start":"01:30.830 ","End":"01:36.215","Text":"a naught plus v naught log 2 and here it\u0027s 0."},{"Start":"01:36.215 ","End":"01:40.970","Text":"The coefficient of cosine Theta,"},{"Start":"01:40.970 ","End":"01:43.235","Text":"which is cosine 1 Theta."},{"Start":"01:43.235 ","End":"01:46.505","Text":"We just take this term when n equals 1,"},{"Start":"01:46.505 ","End":"01:50.945","Text":"so we get a12^1 plus a minus 12 to the minus 1,"},{"Start":"01:50.945 ","End":"01:53.120","Text":"and that\u0027s equal to 2."},{"Start":"01:53.120 ","End":"01:57.845","Text":"Then we have for n bigger or equal to 2,"},{"Start":"01:57.845 ","End":"02:02.330","Text":"both the cosine and the sine of n Theta are going to be 0,"},{"Start":"02:02.330 ","End":"02:07.655","Text":"so we get this is 0 and this is 0."},{"Start":"02:07.655 ","End":"02:12.620","Text":"But this one holds for n bigger or equal to 2 and the b_n holds for n"},{"Start":"02:12.620 ","End":"02:17.240","Text":"bigger or equal to 1 because we don\u0027t have a sine Theta on the left,"},{"Start":"02:17.240 ","End":"02:18.650","Text":"only a cosine Theta."},{"Start":"02:18.650 ","End":"02:24.830","Text":"Now let\u0027s take these equations and these equations together as a system."},{"Start":"02:24.830 ","End":"02:27.545","Text":"This is what we have all of these."},{"Start":"02:27.545 ","End":"02:30.200","Text":"Now let\u0027s separate cases."},{"Start":"02:30.200 ","End":"02:34.270","Text":"When n is 0, we have this and"},{"Start":"02:34.270 ","End":"02:39.605","Text":"this and that easily gives us that a naught equals v naught equals 0."},{"Start":"02:39.605 ","End":"02:41.645","Text":"When n equals 1,"},{"Start":"02:41.645 ","End":"02:47.785","Text":"then we have this one here and this one here also applies when n equals 1."},{"Start":"02:47.785 ","End":"02:50.445","Text":"We\u0027ll leave this one for a moment."},{"Start":"02:50.445 ","End":"02:53.270","Text":"That gives us these equations from which we deduce"},{"Start":"02:53.270 ","End":"02:57.920","Text":"that a1 is 4/3 and a minus 1 is minus 4/3."},{"Start":"02:57.920 ","End":"02:59.500","Text":"I\u0027ll leave the details to you."},{"Start":"02:59.500 ","End":"03:01.825","Text":"When n is bigger or equal to 2,"},{"Start":"03:01.825 ","End":"03:05.210","Text":"then we have this and we also have this,"},{"Start":"03:05.210 ","End":"03:06.875","Text":"leaves the b_n\u0027s for a moment."},{"Start":"03:06.875 ","End":"03:12.110","Text":"Then we get these and deduce that the a_n and a minus n is 0."},{"Start":"03:12.110 ","End":"03:14.780","Text":"Now when n is bigger or equal to 1,"},{"Start":"03:14.780 ","End":"03:18.110","Text":"then we have this as well as this and that"},{"Start":"03:18.110 ","End":"03:21.950","Text":"gives us that the b_n is a 0 and the b minus n is 0."},{"Start":"03:21.950 ","End":"03:26.290","Text":"Notice that the only 1 that are non-zero are these which I\u0027ve colored."},{"Start":"03:26.290 ","End":"03:29.150","Text":"Getting back to this equation,"},{"Start":"03:29.150 ","End":"03:32.510","Text":"we have all the coefficients now we just have to substitute them."},{"Start":"03:32.510 ","End":"03:41.795","Text":"The only ones we get are a_1 and a minus 1 so we only have something with cosine 1 Theta."},{"Start":"03:41.795 ","End":"03:43.220","Text":"From here, when n is 1,"},{"Start":"03:43.220 ","End":"03:46.505","Text":"we get a_1 is 4/3."},{"Start":"03:46.505 ","End":"03:50.595","Text":"Take that out of the brackets and here a minus then is minus 4/3."},{"Start":"03:50.595 ","End":"03:52.485","Text":"Again, take the 4/3 out of the brackets."},{"Start":"03:52.485 ","End":"03:54.955","Text":"Here we have r^1, r^minus1."},{"Start":"03:54.955 ","End":"03:56.900","Text":"In short, this is the answer,"},{"Start":"03:56.900 ","End":"04:01.835","Text":"but it\u0027s in polar coordinates and we were asked for Cartesian coordinates."},{"Start":"04:01.835 ","End":"04:05.854","Text":"We can take out of the brackets and pull the cosine Theta up in front."},{"Start":"04:05.854 ","End":"04:13.790","Text":"Then we have 1 minus 1 over r^2 and r^2 is x^2 plus y^2,"},{"Start":"04:13.790 ","End":"04:18.440","Text":"cosine Theta is x so we end up with the solution in"},{"Start":"04:18.440 ","End":"04:24.060","Text":"Cartesian coordinates as follows and that concludes this exercise."}],"ID":30832},{"Watched":false,"Name":"Exercise 2","Duration":"4m 46s","ChapterTopicVideoID":29266,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.335","Text":"In this exercise, we\u0027re given the following boundary value problem."},{"Start":"00:04.335 ","End":"00:05.880","Text":"We want to solve it,"},{"Start":"00:05.880 ","End":"00:09.105","Text":"but we want to express the answer in polar coordinates."},{"Start":"00:09.105 ","End":"00:12.135","Text":"This thing is actually given in Cartesian coordinates."},{"Start":"00:12.135 ","End":"00:17.220","Text":"This is the Laplace\u0027s equation in the annulus between the radii 1 and 2,"},{"Start":"00:17.220 ","End":"00:21.840","Text":"and we\u0027re given the value of u on"},{"Start":"00:21.840 ","End":"00:29.070","Text":"the inner circle and value of u on the outer circle but in Cartesian form."},{"Start":"00:29.070 ","End":"00:33.000","Text":"The easiest thing to do in this case is to convert"},{"Start":"00:33.000 ","End":"00:34.740","Text":"the problem right away to"},{"Start":"00:34.740 ","End":"00:38.730","Text":"polar coordinates because we have the formula in polar coordinates."},{"Start":"00:38.730 ","End":"00:45.450","Text":"The Laplacian is 0 on the annulus between radius 1 and radius 2."},{"Start":"00:45.450 ","End":"00:47.610","Text":"4 is 2^2 and 1 is 1^2."},{"Start":"00:47.610 ","End":"00:49.950","Text":"This becomes, this."},{"Start":"00:49.950 ","End":"00:52.175","Text":"This part stays the same,"},{"Start":"00:52.175 ","End":"00:55.670","Text":"but the circle of radius 1 becomes r equals 1,"},{"Start":"00:55.670 ","End":"00:58.160","Text":"and here, x^2 plus y^2 is 4,"},{"Start":"00:58.160 ","End":"01:00.190","Text":"is r equals 2,"},{"Start":"01:00.190 ","End":"01:04.275","Text":"2 plus, x is 2 cosine theta."},{"Start":"01:04.275 ","End":"01:10.860","Text":"We end up with a 1/2 cosine theta because we have the 1/4 here and y is 2 sin theta,"},{"Start":"01:10.860 ","End":"01:13.240","Text":"so we get 3/4 sin theta."},{"Start":"01:13.240 ","End":"01:18.830","Text":"The formula we want for the Laplace equation on an annulus is the following."},{"Start":"01:18.830 ","End":"01:22.135","Text":"When r=1, u equals the 16th."},{"Start":"01:22.135 ","End":"01:28.040","Text":"That gives us that this series is equal to 1/16th."},{"Start":"01:28.040 ","End":"01:30.215","Text":"Now let\u0027s compare coefficients."},{"Start":"01:30.215 ","End":"01:32.915","Text":"We get that a naught is 1/16th."},{"Start":"01:32.915 ","End":"01:37.340","Text":"Next, the case where n is bigger or equal to 1,"},{"Start":"01:37.340 ","End":"01:42.110","Text":"and we get that this plus this is 0 and this plus this is 0."},{"Start":"01:42.110 ","End":"01:45.250","Text":"Now we\u0027ll move on to the case where r equals 2."},{"Start":"01:45.250 ","End":"01:49.770","Text":"In this case, we get that 2 plus a 1/2 cos theta plus"},{"Start":"01:49.770 ","End":"01:55.115","Text":"3/4 sin theta is equal to what\u0027s written here,"},{"Start":"01:55.115 ","End":"01:58.040","Text":"but with r replaced by 2."},{"Start":"01:58.040 ","End":"02:00.970","Text":"We get a_n, 2 to the n, etc."},{"Start":"02:00.970 ","End":"02:05.165","Text":"I forgot to say earlier that when r equals 1,"},{"Start":"02:05.165 ","End":"02:10.120","Text":"then r to the n and r to the minus n_r_1 and so they drop out."},{"Start":"02:10.120 ","End":"02:13.035","Text":"Again, we\u0027ll compare coefficients."},{"Start":"02:13.035 ","End":"02:15.940","Text":"The free coefficient is 2,"},{"Start":"02:15.940 ","End":"02:21.320","Text":"but it\u0027s also equal to a naught plus b naught natural log of 2."},{"Start":"02:21.320 ","End":"02:25.020","Text":"Next we have n equals 1,"},{"Start":"02:25.020 ","End":"02:27.630","Text":"where we get the cosine and the sin."},{"Start":"02:27.630 ","End":"02:32.500","Text":"For the cosine, we get this is equal to a 1/2,"},{"Start":"02:32.500 ","End":"02:38.405","Text":"and for the sine, we get this when n equals 1 is equal to 3/4."},{"Start":"02:38.405 ","End":"02:41.165","Text":"Then when n is 2 or more,"},{"Start":"02:41.165 ","End":"02:46.905","Text":"this is 0 and this is 0 and we get this."},{"Start":"02:46.905 ","End":"02:50.630","Text":"Next we\u0027ll put these equations and these equations"},{"Start":"02:50.630 ","End":"02:54.620","Text":"alongside as one set and we get all these equations."},{"Start":"02:54.620 ","End":"02:57.094","Text":"Well, let\u0027s take it case by case."},{"Start":"02:57.094 ","End":"02:59.435","Text":"When n is 0,"},{"Start":"02:59.435 ","End":"03:02.320","Text":"we have this one and this one,"},{"Start":"03:02.320 ","End":"03:04.070","Text":"and if we solve this,"},{"Start":"03:04.070 ","End":"03:05.660","Text":"I\u0027ll just give you the solution,"},{"Start":"03:05.660 ","End":"03:07.655","Text":"a naught is the 16th,"},{"Start":"03:07.655 ","End":"03:12.350","Text":"and b naught is 1 over natural log of 2 to minus the 16th."},{"Start":"03:12.350 ","End":"03:15.820","Text":"This is actually equal to 31/16."},{"Start":"03:15.820 ","End":"03:19.280","Text":"When n equals 1 and just looking at the a_n,"},{"Start":"03:19.280 ","End":"03:22.910","Text":"we have this equation and this equation,"},{"Start":"03:22.910 ","End":"03:26.080","Text":"and that gives us these 2,"},{"Start":"03:26.080 ","End":"03:29.580","Text":"and the solution to those is a_1 equals 1/3,"},{"Start":"03:29.580 ","End":"03:31.575","Text":"a minus 1 is minus a 1/3."},{"Start":"03:31.575 ","End":"03:34.910","Text":"If we consider the b_n\u0027s, where is it?"},{"Start":"03:34.910 ","End":"03:39.035","Text":"Yeah, we have this and we have this."},{"Start":"03:39.035 ","End":"03:41.580","Text":"You get these 2 equations,"},{"Start":"03:41.580 ","End":"03:42.990","Text":"and if we solve those,"},{"Start":"03:42.990 ","End":"03:48.600","Text":"we get b1 and b minus 1,1/2 and minus a 1/2, and the rest."},{"Start":"03:48.600 ","End":"03:51.210","Text":"When n is bigger or equal to 2, this is 0,"},{"Start":"03:51.210 ","End":"03:52.890","Text":"this is 0,"},{"Start":"03:52.890 ","End":"03:57.785","Text":"and the solution comes out to be the day and an a minus n are 0."},{"Start":"03:57.785 ","End":"04:04.400","Text":"Now we have all the coefficients and these are all the non-zero ones."},{"Start":"04:04.400 ","End":"04:07.250","Text":"Similarly when n is bigger or equal to 2,"},{"Start":"04:07.250 ","End":"04:12.995","Text":"the b_n\u0027s also come out to be 0 and b minus n is 0."},{"Start":"04:12.995 ","End":"04:15.200","Text":"Now that we have all the coefficients,"},{"Start":"04:15.200 ","End":"04:20.495","Text":"we just have to substitute them in this expression for u of r and Theta,"},{"Start":"04:20.495 ","End":"04:22.745","Text":"and we get a naught,"},{"Start":"04:22.745 ","End":"04:24.490","Text":"we have b naught,"},{"Start":"04:24.490 ","End":"04:26.130","Text":"which as we said,"},{"Start":"04:26.130 ","End":"04:29.825","Text":"is 1 over natural log of 2 to minus the 16th, which is this."},{"Start":"04:29.825 ","End":"04:34.010","Text":"Then we have a_1 and a minus 1. Yeah, here it is."},{"Start":"04:34.010 ","End":"04:36.605","Text":"I will take the 1/3 in front of the brackets,"},{"Start":"04:36.605 ","End":"04:42.095","Text":"and for the b\u0027s we have a 1/2 and minus a 1/2 and take the half in front of the brackets."},{"Start":"04:42.095 ","End":"04:46.830","Text":"Anyway, we get this and that\u0027s the answer, and we\u0027re done."}],"ID":30833},{"Watched":false,"Name":"Exercise 3a","Duration":"3m 9s","ChapterTopicVideoID":29267,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.169","Text":"In this exercise, we\u0027re given the following boundary value problem."},{"Start":"00:04.169 ","End":"00:07.200","Text":"This is a non-homogeneous Laplace equation,"},{"Start":"00:07.200 ","End":"00:09.585","Text":"also called a Poisson equation."},{"Start":"00:09.585 ","End":"00:11.640","Text":"We have to solve the problem,"},{"Start":"00:11.640 ","End":"00:14.250","Text":"but express the solution in polar coordinates."},{"Start":"00:14.250 ","End":"00:17.715","Text":"As it is, the problem is expressed in Cartesian coordinates."},{"Start":"00:17.715 ","End":"00:21.420","Text":"We\u0027re given a hint to guess a radial solution,"},{"Start":"00:21.420 ","End":"00:25.140","Text":"meaning a function u(r) doesn\u0027t depend on Theta."},{"Start":"00:25.140 ","End":"00:27.450","Text":"We\u0027ll read Part B when we come to it."},{"Start":"00:27.450 ","End":"00:29.560","Text":"Let\u0027s start with Part A."},{"Start":"00:29.560 ","End":"00:33.995","Text":"Let\u0027s first convert the whole problem into polar coordinates."},{"Start":"00:33.995 ","End":"00:37.400","Text":"X^2 plus y^2 is r^2."},{"Start":"00:37.400 ","End":"00:40.292","Text":"This is the circle of radius 1,"},{"Start":"00:40.292 ","End":"00:42.470","Text":"this is the circle of radius 2,"},{"Start":"00:42.470 ","End":"00:44.665","Text":"both centered at the origin."},{"Start":"00:44.665 ","End":"00:49.970","Text":"What we get is the Laplacian of u is r^2,"},{"Start":"00:49.970 ","End":"00:51.380","Text":"r between 1 and 2,"},{"Start":"00:51.380 ","End":"00:53.150","Text":"and we\u0027re given that u is 0,"},{"Start":"00:53.150 ","End":"00:56.750","Text":"both when r equals 1 and r equals 2."},{"Start":"00:56.750 ","End":"01:03.860","Text":"Recall that the Laplacian in polar coordinates is given as follows."},{"Start":"01:03.860 ","End":"01:07.560","Text":"In our case, Laplacian of u is r^2,"},{"Start":"01:07.560 ","End":"01:11.080","Text":"so we can equate the right-hand side here to r^2."},{"Start":"01:11.080 ","End":"01:17.705","Text":"The hint says we should guess a solution u as a function just of r and not of Theta."},{"Start":"01:17.705 ","End":"01:19.370","Text":"If that\u0027s the case,"},{"Start":"01:19.370 ","End":"01:24.760","Text":"then this becomes 0 because u doesn\u0027t depend on Theta."},{"Start":"01:24.760 ","End":"01:28.280","Text":"What we get is the following."},{"Start":"01:28.280 ","End":"01:32.705","Text":"Multiply both sides by r and we have this."},{"Start":"01:32.705 ","End":"01:36.370","Text":"Now we have the derivative of a function of r is r^3,"},{"Start":"01:36.370 ","End":"01:42.679","Text":"so we can just integrate and say that r times u prime of r is the integral of this,"},{"Start":"01:42.679 ","End":"01:46.780","Text":"r^4 over 4 plus some constant c_1."},{"Start":"01:46.780 ","End":"01:51.385","Text":"Now divide both sides by r, we get this."},{"Start":"01:51.385 ","End":"01:59.220","Text":"Integrate again with respect to r and we have r^4 over 4 times 4 is 16."},{"Start":"01:59.220 ","End":"02:02.060","Text":"Here, c_1 natural log of r,"},{"Start":"02:02.060 ","End":"02:04.375","Text":"and another constant, c_2."},{"Start":"02:04.375 ","End":"02:07.200","Text":"We need to find c_1 and c_2."},{"Start":"02:07.200 ","End":"02:10.115","Text":"For that, we\u0027ll use the boundary conditions."},{"Start":"02:10.115 ","End":"02:13.250","Text":"We know that u(1) is 0,"},{"Start":"02:13.250 ","End":"02:15.650","Text":"meaning when r is 1, u is 0."},{"Start":"02:15.650 ","End":"02:17.254","Text":"If we put that here,"},{"Start":"02:17.254 ","End":"02:19.720","Text":"natural log of 1 is 0."},{"Start":"02:19.720 ","End":"02:25.430","Text":"This term drops out and we have c_2 is minus r^4 over 16."},{"Start":"02:25.430 ","End":"02:27.694","Text":"When r is 1, it\u0027s minus a 16th."},{"Start":"02:27.694 ","End":"02:29.630","Text":"Substitute that here."},{"Start":"02:29.630 ","End":"02:33.095","Text":"Now we have the following expression for u (r)."},{"Start":"02:33.095 ","End":"02:38.390","Text":"Just combine the minus 1 over 16 with the r^4 over 16 common denominator."},{"Start":"02:38.390 ","End":"02:42.094","Text":"Now we have the other condition that u(2) is 0,"},{"Start":"02:42.094 ","End":"02:45.940","Text":"and that gives us 2^4 minus 1 is 15."},{"Start":"02:45.940 ","End":"02:51.600","Text":"Here, c_1 natural log of 2 is 0 so extracting c_1, we get this."},{"Start":"02:51.600 ","End":"02:59.610","Text":"Now, put c_1 into this equation and we have that u(r),"},{"Start":"02:59.610 ","End":"03:02.100","Text":"but strictly speaking, it\u0027s u(r) Theta."},{"Start":"03:02.100 ","End":"03:06.030","Text":"It just doesn\u0027t depend on Theta is the following."},{"Start":"03:06.030 ","End":"03:09.190","Text":"That concludes Part A."}],"ID":30834},{"Watched":false,"Name":"Exercise 3b","Duration":"3m 46s","ChapterTopicVideoID":29268,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.430","Text":"We\u0027re continuing the same exercise."},{"Start":"00:02.430 ","End":"00:06.045","Text":"We\u0027ve just done part a and now we come to part b."},{"Start":"00:06.045 ","End":"00:09.510","Text":"We have to prove that the solution is unique."},{"Start":"00:09.510 ","End":"00:13.575","Text":"We\u0027re given a hint to use Green\u0027s identity."},{"Start":"00:13.575 ","End":"00:14.970","Text":"Sometimes it\u0027s called the first."},{"Start":"00:14.970 ","End":"00:17.460","Text":"I\u0027ve seen places that call it the second identity."},{"Start":"00:17.460 ","End":"00:19.050","Text":"Anyway, this is the 1."},{"Start":"00:19.050 ","End":"00:21.720","Text":"The strategy is the same as usual."},{"Start":"00:21.720 ","End":"00:24.510","Text":"We\u0027ll show that if u_1 and u_2 are"},{"Start":"00:24.510 ","End":"00:29.980","Text":"solutions then necessarily u_1 is identically equal to u_2."},{"Start":"00:29.980 ","End":"00:38.010","Text":"We do this by letting w=u_1 minus u_2 and proving that w is identically 0."},{"Start":"00:38.010 ","End":"00:42.140","Text":"By the way, this itself doesn\u0027t show that there is a solution,"},{"Start":"00:42.140 ","End":"00:45.095","Text":"it just shows that there can\u0027t be more than 1."},{"Start":"00:45.095 ","End":"00:46.470","Text":"However, in part a,"},{"Start":"00:46.470 ","End":"00:49.045","Text":"we found a solution."},{"Start":"00:49.045 ","End":"00:55.685","Text":"A together with b will prove that there is exactly 1 solution and this is it."},{"Start":"00:55.685 ","End":"00:59.465","Text":"Now, u_1 and u_2 both satisfy this equation."},{"Start":"00:59.465 ","End":"01:04.945","Text":"The Laplacian of u_1 and Laplacian of u_2 are both x^2 plus y^2,"},{"Start":"01:04.945 ","End":"01:08.530","Text":"which means that if we let w be u_1 minus u_2,"},{"Start":"01:08.530 ","End":"01:11.480","Text":"the Laplacian of w is the difference,"},{"Start":"01:11.480 ","End":"01:14.615","Text":"this minus this and that is 0."},{"Start":"01:14.615 ","End":"01:17.420","Text":"Also, we have boundary conditions."},{"Start":"01:17.420 ","End":"01:20.215","Text":"We just subtract the boundary conditions"},{"Start":"01:20.215 ","End":"01:25.370","Text":"on the circle of radius 1 and the circle of radius 2."},{"Start":"01:25.370 ","End":"01:32.195","Text":"On each of them both u_1 and u_2 are 0 so u_1 minus u_2 is 0 and w is 0."},{"Start":"01:32.195 ","End":"01:38.540","Text":"Putting this together, this is the boundary value problem we have for w. It satisfies"},{"Start":"01:38.540 ","End":"01:42.215","Text":"the Laplace equation homogeneous in the annulus"},{"Start":"01:42.215 ","End":"01:46.370","Text":"and it\u0027s 0 on the boundary of the annulus."},{"Start":"01:46.370 ","End":"01:52.670","Text":"Now, here again is Green\u0027s identity and we\u0027re going to apply it with u and"},{"Start":"01:52.670 ","End":"01:59.435","Text":"v here both equaling w. What we get is the following."},{"Start":"01:59.435 ","End":"02:04.190","Text":"We\u0027ll take the domain D to be the annulus."},{"Start":"02:04.190 ","End":"02:08.390","Text":"Here is a picture where r=1 and this is where r=2."},{"Start":"02:08.390 ","End":"02:17.500","Text":"This integral becomes 0 because Delta w is 0."},{"Start":"02:17.500 ","End":"02:20.210","Text":"That\u0027s the PDE we have."},{"Start":"02:20.210 ","End":"02:25.550","Text":"Also on the boundary w=0,"},{"Start":"02:25.550 ","End":"02:28.009","Text":"so all we\u0027re left with is this part."},{"Start":"02:28.009 ","End":"02:29.300","Text":"Now, in general,"},{"Start":"02:29.300 ","End":"02:32.540","Text":"a vector dot product with itself is"},{"Start":"02:32.540 ","End":"02:37.460","Text":"the absolute value of the vector or the norm of the vector squared."},{"Start":"02:37.460 ","End":"02:39.530","Text":"What we get is that this integral,"},{"Start":"02:39.530 ","End":"02:43.080","Text":"which is this is equal to 0."},{"Start":"02:43.080 ","End":"02:44.870","Text":"If the integral of"},{"Start":"02:44.870 ","End":"02:52.070","Text":"a non-negative continuous function is 0 then the function itself must be identically 0."},{"Start":"02:52.070 ","End":"02:58.940","Text":"This means that the following is 0 because this is the gradient of w,"},{"Start":"02:58.940 ","End":"03:00.905","Text":"which is the vector w_x,"},{"Start":"03:00.905 ","End":"03:03.050","Text":"w_y and the norm of the vector is"},{"Start":"03:03.050 ","End":"03:06.620","Text":"just the first coordinate squared plus the second coordinate squared."},{"Start":"03:06.620 ","End":"03:08.830","Text":"The sum of the squares is 0."},{"Start":"03:08.830 ","End":"03:15.035","Text":"In algebra, if the sum of squares of 2 real numbers is 0 then each of them is 0."},{"Start":"03:15.035 ","End":"03:18.097","Text":"Now we have that both partial derivatives with w was 0,"},{"Start":"03:18.097 ","End":"03:21.275","Text":"which means that w is a constant."},{"Start":"03:21.275 ","End":"03:24.259","Text":"What we need is the value of w at a point."},{"Start":"03:24.259 ","End":"03:28.040","Text":"We actually know it on all of the boundary is 0,"},{"Start":"03:28.040 ","End":"03:30.275","Text":"so that constant must be 0."},{"Start":"03:30.275 ","End":"03:34.930","Text":"We get that w is identically 0."},{"Start":"03:34.930 ","End":"03:38.895","Text":"That means that u_1 minus u_2 is 0."},{"Start":"03:38.895 ","End":"03:42.850","Text":"That means that u_1 is identically u_2."},{"Start":"03:42.850 ","End":"03:47.280","Text":"That proves the uniqueness and we are done."}],"ID":30835},{"Watched":false,"Name":"Exercise 4","Duration":"7m 32s","ChapterTopicVideoID":29269,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.170","Text":"In this exercise, we have a boundary value problem where"},{"Start":"00:04.170 ","End":"00:08.445","Text":"here we have a Laplace equation which is non-homogeneous,"},{"Start":"00:08.445 ","End":"00:11.010","Text":"also known as Poisson equation."},{"Start":"00:11.010 ","End":"00:15.270","Text":"It\u0027s on the annulus between radii 2 and 3."},{"Start":"00:15.270 ","End":"00:19.800","Text":"We have the value of u on the smaller radius and"},{"Start":"00:19.800 ","End":"00:25.110","Text":"the value of du by dr on the outer radius."},{"Start":"00:25.110 ","End":"00:29.400","Text":"We have to compute the following integral and we\u0027re given a hint to"},{"Start":"00:29.400 ","End":"00:34.170","Text":"use the 2-dimensional Gauss theorem as follows."},{"Start":"00:34.170 ","End":"00:37.950","Text":"We need to choose what is F in this formula."},{"Start":"00:37.950 ","End":"00:41.990","Text":"We\u0027ll choose that to be the gradient of u,"},{"Start":"00:41.990 ","End":"00:43.789","Text":"u is a scalar function,"},{"Start":"00:43.789 ","End":"00:47.360","Text":"but grad u is a vector function,"},{"Start":"00:47.360 ","End":"00:48.760","Text":"which is what we need."},{"Start":"00:48.760 ","End":"00:53.420","Text":"Then domain of course is the annulus where r is between 2 and 3."},{"Start":"00:53.420 ","End":"00:57.395","Text":"Now that will be needing the divergence of the vector field"},{"Start":"00:57.395 ","End":"01:03.455","Text":"F. With a dot is divergence without the dot, it\u0027s gradient."},{"Start":"01:03.455 ","End":"01:06.575","Text":"Gradient takes a scalar to a vector and"},{"Start":"01:06.575 ","End":"01:11.210","Text":"the dot with a vector field brings it to a scalar."},{"Start":"01:11.210 ","End":"01:20.540","Text":"We have the formula that divergence of F is dot Del or grad dot grad of u,"},{"Start":"01:20.540 ","End":"01:23.570","Text":"which is symbolically Del squared of u."},{"Start":"01:23.570 ","End":"01:25.850","Text":"That happens to be the Laplacian of u."},{"Start":"01:25.850 ","End":"01:28.970","Text":"If we do all that and substitute here,"},{"Start":"01:28.970 ","End":"01:32.360","Text":"what we get is Del dot f,"},{"Start":"01:32.360 ","End":"01:35.855","Text":"divergence of F is the Laplacian of u."},{"Start":"01:35.855 ","End":"01:43.699","Text":"Here, the vector field F is just grad u the rest of it dot normal vector ds."},{"Start":"01:43.699 ","End":"01:47.390","Text":"Now the boundary of D is disconnected."},{"Start":"01:47.390 ","End":"01:52.565","Text":"It\u0027s 2 separate circles is the inner circle radius 2 and the outer circle radius 3."},{"Start":"01:52.565 ","End":"01:57.725","Text":"The integral on the boundary breaks up into the sum of 2 integrals,"},{"Start":"01:57.725 ","End":"01:58.880","Text":"1 on the outer circle,"},{"Start":"01:58.880 ","End":"02:00.845","Text":"1 on the inner circle."},{"Start":"02:00.845 ","End":"02:06.500","Text":"Note that the normal vector on the outer circle faces in"},{"Start":"02:06.500 ","End":"02:12.665","Text":"this direction and the normal vector on the inner circle faces in this direction."},{"Start":"02:12.665 ","End":"02:17.465","Text":"This isn\u0027t the same direction as the radial vector,"},{"Start":"02:17.465 ","End":"02:23.365","Text":"where r increases in here is where r decreases opposite directions."},{"Start":"02:23.365 ","End":"02:28.595","Text":"What we can say is that this is equal to the difference."},{"Start":"02:28.595 ","End":"02:30.890","Text":"If we replace the normal vector by"},{"Start":"02:30.890 ","End":"02:35.210","Text":"the radius vector on the outer circle, this equals this,"},{"Start":"02:35.210 ","End":"02:36.695","Text":"but on the inner circle,"},{"Start":"02:36.695 ","End":"02:40.415","Text":"r is minus n,"},{"Start":"02:40.415 ","End":"02:44.035","Text":"r is outward, but n is inward."},{"Start":"02:44.035 ","End":"02:50.270","Text":"Now we note that u dot r its gradient. We\u0027ve seen this before."},{"Start":"02:50.270 ","End":"02:55.070","Text":"This is the directional derivative of u in the direction of r when you take"},{"Start":"02:55.070 ","End":"03:00.469","Text":"grad of a scalar and dot-product it with a vector,"},{"Start":"03:00.469 ","End":"03:05.440","Text":"which is unit and it\u0027s the directional derivative in that direction."},{"Start":"03:05.440 ","End":"03:10.790","Text":"The directional derivative in the direction of the radius unit vector is the same"},{"Start":"03:10.790 ","End":"03:15.875","Text":"as just the derivative of u with respect to r, partial derivative."},{"Start":"03:15.875 ","End":"03:20.765","Text":"In each case, it\u0027s in the direction of increasing r outward on the radius."},{"Start":"03:20.765 ","End":"03:28.474","Text":"Here and here we replace u dot r with du by dr. Now to compute this integral,"},{"Start":"03:28.474 ","End":"03:31.910","Text":"we parameterize the curve r equals 3,"},{"Start":"03:31.910 ","End":"03:35.780","Text":"as usual parameterization of a circle centered at the origin,"},{"Start":"03:35.780 ","End":"03:37.580","Text":"x equals 3 cosine Theta,"},{"Start":"03:37.580 ","End":"03:39.445","Text":"y equals 3 sine Theta."},{"Start":"03:39.445 ","End":"03:42.870","Text":"Theta goes from 0 to 2Pi or from minus pi to Pi,"},{"Start":"03:42.870 ","End":"03:45.285","Text":"any 2 Pi interval, really,"},{"Start":"03:45.285 ","End":"03:51.245","Text":"ds is the square root of x prime Theta squared plus y prime of Theta squared d Theta,"},{"Start":"03:51.245 ","End":"03:57.320","Text":"X prime of Theta is minus 3 sine Theta squared it\u0027s this."},{"Start":"03:57.320 ","End":"04:01.260","Text":"Y prime is 3 cosine Theta squared is this."},{"Start":"04:01.260 ","End":"04:04.890","Text":"This boils down to 3 d Theta."},{"Start":"04:04.890 ","End":"04:07.945","Text":"Similarly for r equals 2,"},{"Start":"04:07.945 ","End":"04:15.079","Text":"same thing except that here we have 2s and here we have a 4 and 4 and then substituting"},{"Start":"04:15.079 ","End":"04:23.260","Text":"ds here and here and also making this integral B in terms of Theta from 0 to 2Pi,"},{"Start":"04:23.260 ","End":"04:25.565","Text":"we get the following."},{"Start":"04:25.565 ","End":"04:30.275","Text":"On the right-hand side replace this by this and this by this."},{"Start":"04:30.275 ","End":"04:37.130","Text":"Now the Laplacian of u is given in the original problem to be r-squared sine cubed"},{"Start":"04:37.130 ","End":"04:44.405","Text":"Theta da in polar coordinates is r dr d Theta and the region D,"},{"Start":"04:44.405 ","End":"04:50.990","Text":"the annulus is the region where r goes from 2 to 3 and Theta goes from 0 to 2Pi."},{"Start":"04:50.990 ","End":"04:56.840","Text":"That gives us replacing the Laplacian of u by this,"},{"Start":"04:56.840 ","End":"05:05.735","Text":"putting it here and replacing the integral on D as the integral from 2 to 3,"},{"Start":"05:05.735 ","End":"05:07.975","Text":"Theta from 0 to 2Pi."},{"Start":"05:07.975 ","End":"05:13.565","Text":"Right-hand side the same except we note that the derivative of u"},{"Start":"05:13.565 ","End":"05:18.995","Text":"by dr on the outer circle is given to be sine squared Theta."},{"Start":"05:18.995 ","End":"05:22.070","Text":"Now this is the integral we have to compute."},{"Start":"05:22.070 ","End":"05:27.205","Text":"Let\u0027s put this on the left-hand side and put this on the right-hand side."},{"Start":"05:27.205 ","End":"05:29.240","Text":"Then a little bit of algebra, for example,"},{"Start":"05:29.240 ","End":"05:33.790","Text":"we can take the 2 in front of the integral and the 3 in front of the integral."},{"Start":"05:33.790 ","End":"05:39.565","Text":"What we get after all that is this here with the 2 in front."},{"Start":"05:39.565 ","End":"05:42.165","Text":"This stays here with the 3 in front."},{"Start":"05:42.165 ","End":"05:47.580","Text":"This can be split up into 2 iterative integrals,"},{"Start":"05:47.580 ","End":"05:52.300","Text":"2 single integrals, because the rs and the Thetas are completely separate."},{"Start":"05:52.300 ","End":"05:56.330","Text":"I\u0027ve color-coded it to see what\u0027s what the r part is in this color"},{"Start":"05:56.330 ","End":"06:00.860","Text":"and that is in front and then the integral respect to Theta."},{"Start":"06:00.860 ","End":"06:04.655","Text":"It just copied that last line here."},{"Start":"06:04.655 ","End":"06:06.395","Text":"We have to compute this integral."},{"Start":"06:06.395 ","End":"06:08.290","Text":"Work on the right-hand side."},{"Start":"06:08.290 ","End":"06:10.100","Text":"Sine squared Theta,"},{"Start":"06:10.100 ","End":"06:12.110","Text":"famous trigonometric formula,"},{"Start":"06:12.110 ","End":"06:15.275","Text":"it\u0027s equal to 1 minus cosine 2 Theta over 2."},{"Start":"06:15.275 ","End":"06:22.370","Text":"Here, rewrite this integral from 0 to 2Pi as the integral from minus Pi to Pi."},{"Start":"06:22.370 ","End":"06:30.754","Text":"It has a period of 2Pi and any 2Pi interval could be put here instead."},{"Start":"06:30.754 ","End":"06:34.445","Text":"The reason we\u0027re doing this is because this is an odd function."},{"Start":"06:34.445 ","End":"06:39.040","Text":"An odd function, it\u0027s symmetric interval has a 0 integral."},{"Start":"06:39.040 ","End":"06:42.600","Text":"Let\u0027s break this 1 up into 2 pieces,"},{"Start":"06:42.600 ","End":"06:48.240","Text":"so 3 over 2 times the 1 and then times minus cosine 2 Theta. This as is."},{"Start":"06:48.240 ","End":"06:53.340","Text":"Now we can start evaluating."},{"Start":"06:53.340 ","End":"06:56.350","Text":"This integral here is 2 Pi."},{"Start":"06:56.350 ","End":"06:58.865","Text":"This integral is 0."},{"Start":"06:58.865 ","End":"07:01.430","Text":"If you integrate this, you get sine 2 Theta over 2."},{"Start":"07:01.430 ","End":"07:05.210","Text":"Anyway, sine 2 Theta is 0 both here and here so this comes"},{"Start":"07:05.210 ","End":"07:09.765","Text":"out to 0. This comes out to be 0."},{"Start":"07:09.765 ","End":"07:11.840","Text":"We don\u0027t need to compute this."},{"Start":"07:11.840 ","End":"07:14.060","Text":"This times this is 0."},{"Start":"07:14.060 ","End":"07:15.725","Text":"All we\u0027re left with is this."},{"Start":"07:15.725 ","End":"07:20.865","Text":"2Pi times 3 over 2 comes out to be 3Pi."},{"Start":"07:20.865 ","End":"07:23.720","Text":"1 more thing to do just divide by this 2 because we"},{"Start":"07:23.720 ","End":"07:26.540","Text":"want just this without the 2 and we get the answer that"},{"Start":"07:26.540 ","End":"07:33.060","Text":"the integral we\u0027re looking for is equal to 3Pi over 2 and we\u0027re done."}],"ID":30836},{"Watched":false,"Name":"Exercise 5","Duration":"3m 24s","ChapterTopicVideoID":29270,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:05.220","Text":"In this exercise, we\u0027re given the following boundary value problem."},{"Start":"00:05.220 ","End":"00:11.355","Text":"This is the Laplace equation on an annulus and"},{"Start":"00:11.355 ","End":"00:17.730","Text":"u on the inner circle is this and u on the outer circle is this."},{"Start":"00:17.730 ","End":"00:21.165","Text":"We have both Dirichlet boundary conditions."},{"Start":"00:21.165 ","End":"00:25.680","Text":"To solve it, we use the usual formula in polar coordinates,"},{"Start":"00:25.680 ","End":"00:28.155","Text":"which is this series."},{"Start":"00:28.155 ","End":"00:35.110","Text":"This here is just another way of saying u (1 and Theta) is natural log of 2."},{"Start":"00:35.110 ","End":"00:39.135","Text":"If you substitute r=1 in this series,"},{"Start":"00:39.135 ","End":"00:42.484","Text":"recalling that natural log of 1 is 0,"},{"Start":"00:42.484 ","End":"00:47.400","Text":"what we\u0027re left with is a0 plus the sum of the series r is 1,"},{"Start":"00:47.400 ","End":"00:51.905","Text":"so the r to the n r to the minus n dropout and this is what we have."},{"Start":"00:51.905 ","End":"00:53.900","Text":"We can compare coefficients,"},{"Start":"00:53.900 ","End":"00:57.155","Text":"so we get that a naught is natural log of 2,"},{"Start":"00:57.155 ","End":"00:59.030","Text":"that\u0027s the constant coefficient."},{"Start":"00:59.030 ","End":"01:01.670","Text":"Then the cosine n Theta,"},{"Start":"01:01.670 ","End":"01:03.980","Text":"there\u0027s nothing on the left, 0."},{"Start":"01:03.980 ","End":"01:07.040","Text":"On the right it\u0027s a_n plus m minus n,"},{"Start":"01:07.040 ","End":"01:12.987","Text":"and similarly for bn b minus n. This is what happens when n is 0 and here n=1,"},{"Start":"01:12.987 ","End":"01:14.515","Text":"2, 3, etc."},{"Start":"01:14.515 ","End":"01:19.715","Text":"Next, we substitute r=2 and we are given this is natural log of 4."},{"Start":"01:19.715 ","End":"01:22.865","Text":"We get natural log of 4 is the series,"},{"Start":"01:22.865 ","End":"01:26.290","Text":"this one but r replaced by 2 this time."},{"Start":"01:26.290 ","End":"01:29.255","Text":"Again, we compare coefficients."},{"Start":"01:29.255 ","End":"01:33.500","Text":"The constant coefficient this time is both of these,"},{"Start":"01:33.500 ","End":"01:35.565","Text":"equal natural log of 4."},{"Start":"01:35.565 ","End":"01:41.640","Text":"Here we\u0027ve replaced r by 2,"},{"Start":"01:41.640 ","End":"01:45.330","Text":"so you get all these 2 to the n and 2 to the minus n."},{"Start":"01:45.330 ","End":"01:50.870","Text":"This is equal to 0 and this is equal to 0."},{"Start":"01:50.870 ","End":"01:55.670","Text":"Let\u0027s copy these here so we can have these equations alongside each other."},{"Start":"01:55.670 ","End":"01:58.145","Text":"From these we get all the a naught,"},{"Start":"01:58.145 ","End":"01:59.480","Text":"b naught, a_n,"},{"Start":"01:59.480 ","End":"02:00.740","Text":"b_n, a minus n,"},{"Start":"02:00.740 ","End":"02:04.185","Text":"b minus n. When n equals naught,"},{"Start":"02:04.185 ","End":"02:06.990","Text":"then we have this and this,"},{"Start":"02:06.990 ","End":"02:11.800","Text":"and taken together, this gives us a naught is natural log of 2."},{"Start":"02:11.800 ","End":"02:17.420","Text":"If you solve this, b naught natural log of 2 is log 4 minus log 2,"},{"Start":"02:17.420 ","End":"02:20.120","Text":"which is log 2, given that b naught is 1."},{"Start":"02:20.120 ","End":"02:21.650","Text":"When n is 1, 2, 3,"},{"Start":"02:21.650 ","End":"02:23.532","Text":"etc., or n bigger than 0,"},{"Start":"02:23.532 ","End":"02:27.395","Text":"first, let\u0027s take the a\u0027s from here and here."},{"Start":"02:27.395 ","End":"02:29.760","Text":"Solving this, we\u0027ve done this before,"},{"Start":"02:29.760 ","End":"02:32.765","Text":"it gives a_n and a minus n about 0."},{"Start":"02:32.765 ","End":"02:34.430","Text":"Similarly for the b_n,"},{"Start":"02:34.430 ","End":"02:37.175","Text":"is where n is bigger than 0."},{"Start":"02:37.175 ","End":"02:42.905","Text":"The only ones that are non-0 in all these coefficients are a naught and b naught."},{"Start":"02:42.905 ","End":"02:45.995","Text":"If we substitute these into the formula,"},{"Start":"02:45.995 ","End":"02:49.824","Text":"this one and this is what we\u0027re substituting,"},{"Start":"02:49.824 ","End":"02:52.730","Text":"of course, to remind you all the other coefficient is 0,"},{"Start":"02:52.730 ","End":"02:57.915","Text":"then what we get is u about Theta is a naught natural log 2."},{"Start":"02:57.915 ","End":"03:01.940","Text":"B naught is 1, 1 natural log of r and everything else is 0."},{"Start":"03:01.940 ","End":"03:05.480","Text":"We could say this solves the problem except that we were asked"},{"Start":"03:05.480 ","End":"03:08.594","Text":"to do it in Cartesian coordinates,"},{"Start":"03:08.594 ","End":"03:12.115","Text":"so just replace r by square root of x^2 plus y^2."},{"Start":"03:12.115 ","End":"03:13.695","Text":"Then we have u(x,"},{"Start":"03:13.695 ","End":"03:16.750","Text":"y) natural log of 2 plus natural log of the square root,"},{"Start":"03:16.750 ","End":"03:20.585","Text":"so we can forget the square root and put 1/2 out in front."},{"Start":"03:20.585 ","End":"03:24.720","Text":"Now this is the answer and we are done."}],"ID":30837},{"Watched":false,"Name":"Exercise 6a","Duration":"3m 52s","ChapterTopicVideoID":29262,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"In this exercise,"},{"Start":"00:01.770 ","End":"00:08.715","Text":"we have a boundary value problem which consists of the Laplace equation on an annulus,"},{"Start":"00:08.715 ","End":"00:13.380","Text":"and then we have a Neumann boundary condition on the inner circle,"},{"Start":"00:13.380 ","End":"00:17.175","Text":"and Dirichlet boundary condition on the outer circle."},{"Start":"00:17.175 ","End":"00:20.595","Text":"We have to solve the problem in polar coordinates,"},{"Start":"00:20.595 ","End":"00:26.175","Text":"and then in part B to prove that the solution is unique and we have a hint for part B."},{"Start":"00:26.175 ","End":"00:31.455","Text":"Now for part A we\u0027ll take out the usual formula, here it is."},{"Start":"00:31.455 ","End":"00:36.265","Text":"We\u0027ll also need the derivative because of this condition,"},{"Start":"00:36.265 ","End":"00:39.515","Text":"and just differentiate this and this is what we get."},{"Start":"00:39.515 ","End":"00:44.170","Text":"Let\u0027s just assume that the conditions for term by term differentiation are met,"},{"Start":"00:44.170 ","End":"00:47.565","Text":"substitute r equals 1 here,"},{"Start":"00:47.565 ","End":"00:52.705","Text":"and we get that cosine Theta plus sine Theta is the following,"},{"Start":"00:52.705 ","End":"01:00.120","Text":"when r is 1 the 1 over r disappears and so do all the r to the n whatever."},{"Start":"01:00.120 ","End":"01:02.385","Text":"This is what we\u0027re left with,"},{"Start":"01:02.385 ","End":"01:06.120","Text":"and if we compare coefficients, let\u0027s see what we get."},{"Start":"01:06.120 ","End":"01:12.590","Text":"For n equals 0, we have the b naught is 0 because there\u0027s no free coefficient."},{"Start":"01:12.590 ","End":"01:15.625","Text":"It\u0027s part of the cosine and Theta here."},{"Start":"01:15.625 ","End":"01:17.665","Text":"For n equals 1,"},{"Start":"01:17.665 ","End":"01:25.610","Text":"then we have that cosine Theta is this but for n equals 1 which is a_1 minus a minus 1,"},{"Start":"01:25.610 ","End":"01:29.465","Text":"and similarly for the b_1 and b minus 1,"},{"Start":"01:29.465 ","End":"01:31.145","Text":"and for the other n\u0027s,"},{"Start":"01:31.145 ","End":"01:35.870","Text":"we just have that this is 0 and this is 0."},{"Start":"01:35.870 ","End":"01:38.135","Text":"This is from 2 onwards."},{"Start":"01:38.135 ","End":"01:40.910","Text":"Next, we\u0027ll let r equals 2,"},{"Start":"01:40.910 ","End":"01:42.590","Text":"so when r is 2,"},{"Start":"01:42.590 ","End":"01:44.260","Text":"we get the following,"},{"Start":"01:44.260 ","End":"01:47.705","Text":"and again comparing coefficients,"},{"Start":"01:47.705 ","End":"01:50.690","Text":"this time the free coefficient is 0,"},{"Start":"01:50.690 ","End":"01:54.670","Text":"and here it\u0027s a naught plus b naught log 2, so that\u0027s that,"},{"Start":"01:54.670 ","End":"01:57.690","Text":"and all the other n\u0027s from 1 onwards,"},{"Start":"01:57.690 ","End":"02:01.310","Text":"we get that this plus this is 0,"},{"Start":"02:01.310 ","End":"02:03.860","Text":"and this plus this is 0."},{"Start":"02:03.860 ","End":"02:06.310","Text":"Let\u0027s copy these again over here,"},{"Start":"02:06.310 ","End":"02:08.270","Text":"so we\u0027ll have them together."},{"Start":"02:08.270 ","End":"02:10.205","Text":"Now from these,"},{"Start":"02:10.205 ","End":"02:12.590","Text":"we\u0027ll work out what all the a naught, b naught,"},{"Start":"02:12.590 ","End":"02:15.420","Text":"a_n, b_n, and so on, what they are."},{"Start":"02:15.420 ","End":"02:18.365","Text":"If we look at the n equals 0 part,"},{"Start":"02:18.365 ","End":"02:21.650","Text":"we have this is 0 and this is 0,"},{"Start":"02:21.650 ","End":"02:27.485","Text":"and that gives us that a naught and b naught are both 0."},{"Start":"02:27.485 ","End":"02:32.945","Text":"Then we can take this when n equals 1,"},{"Start":"02:32.945 ","End":"02:34.895","Text":"and that gives us this,"},{"Start":"02:34.895 ","End":"02:37.970","Text":"and then copy this over here,"},{"Start":"02:37.970 ","End":"02:41.810","Text":"so now we have 2 equations and 2 unknowns a_1 and minus 1,"},{"Start":"02:41.810 ","End":"02:43.415","Text":"and a bit of algebra."},{"Start":"02:43.415 ","End":"02:49.055","Text":"Solve it to get a_1 is a 1/5 and a minus 1 is minus 4/5."},{"Start":"02:49.055 ","End":"02:52.540","Text":"Actually, we get the same thing for b_1 and b minus 1,"},{"Start":"02:52.540 ","End":"02:55.770","Text":"same equation, same solution."},{"Start":"02:55.770 ","End":"02:59.385","Text":"Next, for all the other n\u0027s and from 2 onwards,"},{"Start":"02:59.385 ","End":"03:03.455","Text":"if you go back and see we get these equations for the a_n\u0027s,"},{"Start":"03:03.455 ","End":"03:06.610","Text":"and these equations for the b_n\u0027s and b minus n\u0027s,"},{"Start":"03:06.610 ","End":"03:08.225","Text":"and if we solve this,"},{"Start":"03:08.225 ","End":"03:11.255","Text":"we get that a_n and a minus n is 0,"},{"Start":"03:11.255 ","End":"03:14.680","Text":"and here we get that b_n and b minus n is 0."},{"Start":"03:14.680 ","End":"03:18.230","Text":"The only things left that are non-0 are these,"},{"Start":"03:18.230 ","End":"03:21.605","Text":"and also from the previous page a_1 and a minus 1."},{"Start":"03:21.605 ","End":"03:23.720","Text":"These are the non-0 ones."},{"Start":"03:23.720 ","End":"03:29.555","Text":"These are the only ones we have to substitute into this formula here."},{"Start":"03:29.555 ","End":"03:34.700","Text":"This and this are the same because b_n and a_n are the same,"},{"Start":"03:34.700 ","End":"03:36.635","Text":"and a minus n, b minus are the same."},{"Start":"03:36.635 ","End":"03:38.960","Text":"This is a 1/5 r to the 1,"},{"Start":"03:38.960 ","End":"03:42.680","Text":"minus 4/5 r to the minus 1 and same here so,"},{"Start":"03:42.680 ","End":"03:48.200","Text":"we\u0027re left with that times cosine 1 Theta plus sine 1 Theta."},{"Start":"03:48.200 ","End":"03:50.135","Text":"Anyway, this is the answer,"},{"Start":"03:50.135 ","End":"03:52.950","Text":"and we are done."}],"ID":30838},{"Watched":false,"Name":"Exercise 6b","Duration":"5m 3s","ChapterTopicVideoID":29263,"CourseChapterTopicPlaylistID":294437,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:04.259","Text":"Continuing, we come to Part b of the exercise."},{"Start":"00:04.259 ","End":"00:08.970","Text":"In Part a we solved this problem and in Part b we have to prove"},{"Start":"00:08.970 ","End":"00:14.175","Text":"that the solution is unique and the solution by the way is this."},{"Start":"00:14.175 ","End":"00:15.510","Text":"We\u0027re going to prove now this is"},{"Start":"00:15.510 ","End":"00:20.865","Text":"the only possible solution and we\u0027re given a hint to use Green\u0027s identity."},{"Start":"00:20.865 ","End":"00:25.180","Text":"The tactic is to show that if u_1 and u_2 are"},{"Start":"00:25.180 ","End":"00:29.280","Text":"solutions then necessarily you want is identically equal to u_2."},{"Start":"00:29.280 ","End":"00:32.900","Text":"We do this by proving that the difference; w,"},{"Start":"00:32.900 ","End":"00:37.707","Text":"which is u_1 minus u_2 must be 0."},{"Start":"00:37.707 ","End":"00:41.815","Text":"By linearity of the derivative and the Laplace operator,"},{"Start":"00:41.815 ","End":"00:48.474","Text":"we have that the Laplacian of w is Laplacian of u_1 minus Laplacian of u_2."},{"Start":"00:48.474 ","End":"00:51.710","Text":"W, well, we already said is u_1 minus u_2 and"},{"Start":"00:51.710 ","End":"00:56.720","Text":"also the derivative of w with respect to r is derivative of u_1"},{"Start":"00:56.720 ","End":"01:00.485","Text":"minus derivative of u_2 with respect to r and we need that"},{"Start":"01:00.485 ","End":"01:06.090","Text":"because one of the conditions here is anointment boundary condition."},{"Start":"01:06.560 ","End":"01:14.020","Text":"Back here we have that in the annulus this is 0 and this is 0,"},{"Start":"01:14.020 ","End":"01:15.995","Text":"so the difference is 0."},{"Start":"01:15.995 ","End":"01:23.887","Text":"On the outer radius w is u_1 minus u_2 and these are both equal, so the difference is 0."},{"Start":"01:23.887 ","End":"01:27.970","Text":"Similarly with the partial derivative with respect"},{"Start":"01:27.970 ","End":"01:32.530","Text":"to r each of these is equal to sine Theta plus cosine Theta."},{"Start":"01:32.530 ","End":"01:34.210","Text":"It doesn\u0027t matter what it is,"},{"Start":"01:34.210 ","End":"01:36.865","Text":"it\u0027s going be the same for both u_1 and u_2."},{"Start":"01:36.865 ","End":"01:41.000","Text":"If you need to subtract it, you\u0027re going to get 0."},{"Start":"01:41.030 ","End":"01:48.480","Text":"That means that the BVP that w satisfies is this one, all these are 0."},{"Start":"01:48.480 ","End":"01:51.890","Text":"These are just what it happens inside"},{"Start":"01:51.890 ","End":"01:57.175","Text":"the annulus on the inner boundary and on the outer boundary,"},{"Start":"01:57.175 ","End":"02:02.540","Text":"I will use the hint we were given about using Green\u0027s identity which is this."},{"Start":"02:02.540 ","End":"02:05.600","Text":"Now this involves two functions;"},{"Start":"02:05.600 ","End":"02:09.935","Text":"u and v, the same u as in the original problem."},{"Start":"02:09.935 ","End":"02:12.950","Text":"We\u0027ll take u equals v equals w,"},{"Start":"02:12.950 ","End":"02:16.840","Text":"and it will take w for both of these."},{"Start":"02:16.840 ","End":"02:26.225","Text":"If we do that the domain D will be the inside of the annulus."},{"Start":"02:26.225 ","End":"02:31.208","Text":"The boundary consists of the two circles where r=1 or r=2,"},{"Start":"02:31.208 ","End":"02:36.710","Text":"that\u0027s the boundary of D. Note that on"},{"Start":"02:36.710 ","End":"02:42.770","Text":"the inner boundary this is 0 because dw by dr is 0."},{"Start":"02:42.770 ","End":"02:45.890","Text":"The normal is parallel to the radius,"},{"Start":"02:45.890 ","End":"02:47.315","Text":"but it points inward."},{"Start":"02:47.315 ","End":"02:48.980","Text":"That\u0027s where we get a minus here,"},{"Start":"02:48.980 ","End":"02:52.778","Text":"but it\u0027s still 0 I guess I should mention."},{"Start":"02:52.778 ","End":"02:58.600","Text":"Of course dw by dr with this notation is the same as wr with this notation."},{"Start":"02:58.600 ","End":"03:05.735","Text":"On the outer circle we\u0027re given the condition that w is 0 just like this."},{"Start":"03:05.735 ","End":"03:09.230","Text":"If we look at this integral which like I said it\u0027s"},{"Start":"03:09.230 ","End":"03:13.735","Text":"gotten from this by replacing u and v each of them by w,"},{"Start":"03:13.735 ","End":"03:16.020","Text":"on the boundary,"},{"Start":"03:16.020 ","End":"03:17.605","Text":"either this is 0 and this is 0."},{"Start":"03:17.605 ","End":"03:19.790","Text":"The boundary has two parts and one of them,"},{"Start":"03:19.790 ","End":"03:22.586","Text":"this is 0 and the other this is 0."},{"Start":"03:22.586 ","End":"03:24.840","Text":"So the product is 0 in either case."},{"Start":"03:24.840 ","End":"03:27.345","Text":"That means that this integral is 0."},{"Start":"03:27.345 ","End":"03:30.890","Text":"This is equal to 0 because"},{"Start":"03:30.890 ","End":"03:38.925","Text":"Delta w is 0 in the whole annulus."},{"Start":"03:38.925 ","End":"03:45.559","Text":"These two are 0 and so this double integral is 0."},{"Start":"03:45.559 ","End":"03:53.810","Text":"Now a vector dot product with the vector itself is the norm of the vector squared,"},{"Start":"03:53.810 ","End":"03:58.010","Text":"so what we get here is the integral of the norm of"},{"Start":"03:58.010 ","End":"04:04.570","Text":"grad w squared or the norm squared is non-negative and continuous."},{"Start":"04:04.570 ","End":"04:10.805","Text":"When we have an integral which is 0 and the integrand is non-negative and continuous,"},{"Start":"04:10.805 ","End":"04:14.515","Text":"then the integrand is 0 on all of the domain."},{"Start":"04:14.515 ","End":"04:17.965","Text":"Grad w squared; the norm of,"},{"Start":"04:17.965 ","End":"04:21.650","Text":"is equal to derivative with respect to x squared plus"},{"Start":"04:21.650 ","End":"04:26.150","Text":"the derivative respect to y squared and that\u0027s identically 0"},{"Start":"04:26.150 ","End":"04:31.340","Text":"on D. If the sum of squares of two real numbers is"},{"Start":"04:31.340 ","End":"04:36.870","Text":"0 then each of them has to be 0,"},{"Start":"04:36.870 ","End":"04:40.903","Text":"and so the function itself is a constant."},{"Start":"04:40.903 ","End":"04:49.430","Text":"But w is 0 on the circle r=2 so if it\u0027s somewhere 0 and it\u0027s a constant,"},{"Start":"04:49.430 ","End":"04:51.575","Text":"it has to be everywhere 0."},{"Start":"04:51.575 ","End":"04:56.875","Text":"If w is 0 then u_1 minus u_2 is identically 0"},{"Start":"04:56.875 ","End":"04:59.620","Text":"which means that u_1 is identically equal to u_2 which is what we"},{"Start":"04:59.620 ","End":"05:03.470","Text":"wanted to show and that concludes this proof."}],"ID":30839}],"Thumbnail":null,"ID":294437},{"Name":"The Laplace Equation in a Wedge","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial + Exercise 1","Duration":"11m 19s","ChapterTopicVideoID":29258,"CourseChapterTopicPlaylistID":294438,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29258.jpeg","UploadDate":"2022-06-06T10:37:10.4000000","DurationForVideoObject":"PT11M19S","Description":null,"MetaTitle":"Tutorial + Exercise 1 - The Laplace Equation in a Wedge: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on The Laplace Equation in a Wedge practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/partial-differential-equations/the-laplace-equation/the-laplace-equation-in-a-wedge/vid30840","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.600","Text":"In this exercise, we\u0027re given a boundary value problem in polar coordinates."},{"Start":"00:06.600 ","End":"00:10.905","Text":"It\u0027s a Laplace equation here in a wedge."},{"Start":"00:10.905 ","End":"00:12.780","Text":"This is what a wedge looks like,"},{"Start":"00:12.780 ","End":"00:14.655","Text":"and it\u0027s like this,"},{"Start":"00:14.655 ","End":"00:17.940","Text":"r between 0 and 1 in this case,"},{"Start":"00:17.940 ","End":"00:21.000","Text":"could be any positive number."},{"Start":"00:21.000 ","End":"00:24.405","Text":"Theta goes between 0 and some Alpha."},{"Start":"00:24.405 ","End":"00:27.840","Text":"Alpha has to be up to 180 degrees."},{"Start":"00:27.840 ","End":"00:31.785","Text":"In this section, there\u0027ll be 3 boundary conditions;"},{"Start":"00:31.785 ","End":"00:33.720","Text":"one on here, one on here, and one on here,"},{"Start":"00:33.720 ","End":"00:36.300","Text":"but we require that these be 0."},{"Start":"00:36.300 ","End":"00:43.190","Text":"The ones where Theta is 0 or Alpha and r varies,"},{"Start":"00:43.190 ","End":"00:44.630","Text":"these got to be 0."},{"Start":"00:44.630 ","End":"00:48.965","Text":"Here on the arc part of the boundary,"},{"Start":"00:48.965 ","End":"00:53.030","Text":"u is equal to some function of Theta."},{"Start":"00:53.030 ","End":"00:55.550","Text":"Going to be like u of 1 and Theta,"},{"Start":"00:55.550 ","End":"00:57.230","Text":"but it\u0027s going to be some function of Theta."},{"Start":"00:57.230 ","End":"01:00.140","Text":"In this case, this would be f of Theta."},{"Start":"01:00.140 ","End":"01:03.815","Text":"This is going to be the setup in the problems in this section."},{"Start":"01:03.815 ","End":"01:08.835","Text":"Here\u0027s how we solve such problem starting with this particular example."},{"Start":"01:08.835 ","End":"01:11.580","Text":"We use separation of variables."},{"Start":"01:11.580 ","End":"01:15.530","Text":"It\u0027s similar to what we did in the wave equations on"},{"Start":"01:15.530 ","End":"01:19.325","Text":"a finite interval or the heat equation on the finite interval."},{"Start":"01:19.325 ","End":"01:21.700","Text":"Similar stages for solving."},{"Start":"01:21.700 ","End":"01:27.500","Text":"In the first stage, we find linearly independent solutions to the Laplace equation,"},{"Start":"01:27.500 ","End":"01:31.010","Text":"which are of the form u is equal to some function of"},{"Start":"01:31.010 ","End":"01:34.655","Text":"r times some function of Theta, the variables separated."},{"Start":"01:34.655 ","End":"01:38.410","Text":"Typically, there are countable number of them."},{"Start":"01:38.410 ","End":"01:41.520","Text":"U_n index like n equals 1,"},{"Start":"01:41.520 ","End":"01:43.410","Text":"2, 3, 4, 5, etc."},{"Start":"01:43.410 ","End":"01:46.285","Text":"We have this whole sequence of these."},{"Start":"01:46.285 ","End":"01:49.355","Text":"The general solution is going to be"},{"Start":"01:49.355 ","End":"01:55.415","Text":"a linear combination of these u_ns with coefficients A_n."},{"Start":"01:55.415 ","End":"01:57.880","Text":"It could be an infinite linear combination."},{"Start":"01:57.880 ","End":"02:03.890","Text":"As you\u0027ll see, getting a solution like this involves using the boundary conditions"},{"Start":"02:03.890 ","End":"02:10.565","Text":"here and here and involves using the solution to a Sturm–Liouville problem,"},{"Start":"02:10.565 ","End":"02:12.745","Text":"assuming you\u0027ve learned this."},{"Start":"02:12.745 ","End":"02:14.885","Text":"In the second stage,"},{"Start":"02:14.885 ","End":"02:20.795","Text":"we find these coefficients A_n by using the boundary condition on the arc part"},{"Start":"02:20.795 ","End":"02:27.460","Text":"that\u0027s this part here where r is 1 or in general might be something else,"},{"Start":"02:27.460 ","End":"02:31.465","Text":"and u is some function of Theta on this arc."},{"Start":"02:31.465 ","End":"02:36.190","Text":"We\u0027re about to start. Just a reminder that the Laplacian in polar coordinates,"},{"Start":"02:36.190 ","End":"02:37.930","Text":"this is one of the forms."},{"Start":"02:37.930 ","End":"02:43.205","Text":"Then we start with our Laplacian equation, Delta u."},{"Start":"02:43.205 ","End":"02:45.350","Text":"The Laplacian of u is 0."},{"Start":"02:45.350 ","End":"02:48.020","Text":"We translate that into polar coordinates,"},{"Start":"02:48.020 ","End":"02:50.440","Text":"so this is what it spells out."},{"Start":"02:50.440 ","End":"02:53.985","Text":"We\u0027re looking for solutions of this form."},{"Start":"02:53.985 ","End":"02:58.520","Text":"If we substitute u of this form in this equation,"},{"Start":"02:58.520 ","End":"03:03.865","Text":"u_rr is gotten simply by taking the second derivative of this that\u0027s here."},{"Start":"03:03.865 ","End":"03:05.685","Text":"1 over r is here."},{"Start":"03:05.685 ","End":"03:09.450","Text":"U_r just differentiate this once."},{"Start":"03:09.450 ","End":"03:11.130","Text":"1 over r-squared is here,"},{"Start":"03:11.130 ","End":"03:16.975","Text":"and u_ThetaTheta is just gotten by taking second derivative of big Theta."},{"Start":"03:16.975 ","End":"03:21.905","Text":"Now, multiply both sides by little r-squared and also"},{"Start":"03:21.905 ","End":"03:27.370","Text":"take this last term from the left and put it on the right with the minus."},{"Start":"03:27.370 ","End":"03:31.935","Text":"Note that here we have Theta of Theta, Theta of Theta,"},{"Start":"03:31.935 ","End":"03:35.615","Text":"and here we have R of r. I want to put this on the left"},{"Start":"03:35.615 ","End":"03:40.010","Text":"and Theta of Theta on the right in the denominators."},{"Start":"03:40.010 ","End":"03:42.295","Text":"This is what we get."},{"Start":"03:42.295 ","End":"03:47.195","Text":"Now, this is totally a function of just R, no Thetas here."},{"Start":"03:47.195 ","End":"03:52.895","Text":"And similarly here, we have just Theta but no R. In this case,"},{"Start":"03:52.895 ","End":"03:55.700","Text":"this is not an equality, it\u0027s an identity."},{"Start":"03:55.700 ","End":"03:59.645","Text":"It must be that each of these is a constant."},{"Start":"03:59.645 ","End":"04:02.005","Text":"Let\u0027s call that constant Lambda."},{"Start":"04:02.005 ","End":"04:03.930","Text":"From this, we can get 2 equations,"},{"Start":"04:03.930 ","End":"04:05.890","Text":"one in r and one in Theta."},{"Start":"04:05.890 ","End":"04:08.150","Text":"With that, this equals this."},{"Start":"04:08.150 ","End":"04:09.920","Text":"We get this first equation,"},{"Start":"04:09.920 ","End":"04:11.840","Text":"and then we let this equal Lambda,"},{"Start":"04:11.840 ","End":"04:14.900","Text":"we get this equation. Here we are."},{"Start":"04:14.900 ","End":"04:19.280","Text":"Now a reminder, u of r Theta is R of r Theta of Theta."},{"Start":"04:19.280 ","End":"04:23.885","Text":"Also, the 2 boundary conditions on the straight part,"},{"Start":"04:23.885 ","End":"04:26.535","Text":"not the arc, are these."},{"Start":"04:26.535 ","End":"04:32.145","Text":"If we substitute u as the product of R, big Theta,"},{"Start":"04:32.145 ","End":"04:34.260","Text":"what we get from this one,"},{"Start":"04:34.260 ","End":"04:35.925","Text":"we get that this is 0,"},{"Start":"04:35.925 ","End":"04:38.910","Text":"and from this one, we get that this is 0,"},{"Start":"04:38.910 ","End":"04:41.260","Text":"so this equals 0."},{"Start":"04:41.260 ","End":"04:45.860","Text":"Then we can divide by R of r. You might say,"},{"Start":"04:45.860 ","End":"04:48.500","Text":"hey, we might be dividing by 0."},{"Start":"04:48.500 ","End":"04:53.660","Text":"Well, the only way Theta of 0 is going to be non-zero is if"},{"Start":"04:53.660 ","End":"04:58.830","Text":"this is 0 everywhere for all r. Then not interesting."},{"Start":"04:58.830 ","End":"05:03.560","Text":"We\u0027re not looking for 0 solutions for R of r or Theta of Theta,"},{"Start":"05:03.560 ","End":"05:05.285","Text":"so leave that one out."},{"Start":"05:05.285 ","End":"05:08.300","Text":"Similarly, Theta of Alpha is 0."},{"Start":"05:08.300 ","End":"05:12.229","Text":"What we get if we take this together with this,"},{"Start":"05:12.229 ","End":"05:16.785","Text":"we have the following equation or problem for Theta."},{"Start":"05:16.785 ","End":"05:18.330","Text":"It\u0027s a Sturm–Liouville problem."},{"Start":"05:18.330 ","End":"05:24.000","Text":"This is the second-order differential equation and these are boundary conditions."},{"Start":"05:24.000 ","End":"05:26.400","Text":"When we studied Sturm–Liouville exercises,"},{"Start":"05:26.400 ","End":"05:30.050","Text":"we had a problem that\u0027s basically the same as this,"},{"Start":"05:30.050 ","End":"05:31.370","Text":"only with different letters."},{"Start":"05:31.370 ","End":"05:38.840","Text":"We had y\u0027\u0027 plus Lambda y equals 0 on the interval from 0-L and y is 0 at the endpoints."},{"Start":"05:38.840 ","End":"05:44.205","Text":"From this, we got eigenvalues n Pi over L all squared,"},{"Start":"05:44.205 ","End":"05:46.220","Text":"and there\u0027s a sequence of those,"},{"Start":"05:46.220 ","End":"05:48.335","Text":"n equals 1, 2, 3, etc,"},{"Start":"05:48.335 ","End":"05:51.350","Text":"and the eigenfunctions are Phi n,"},{"Start":"05:51.350 ","End":"05:55.655","Text":"which are sine of n Pi over L times x."},{"Start":"05:55.655 ","End":"05:58.910","Text":"All we have to do for our case is just switch letters,"},{"Start":"05:58.910 ","End":"06:01.250","Text":"replace y by Theta,"},{"Start":"06:01.250 ","End":"06:03.080","Text":"replace x by little Theta,"},{"Start":"06:03.080 ","End":"06:05.635","Text":"replace L by Alpha,"},{"Start":"06:05.635 ","End":"06:10.190","Text":"and then that gives us the solution that eigenfunctions Theta"},{"Start":"06:10.190 ","End":"06:16.115","Text":"n are sine of n Pi over Alpha Theta and eigenvalues Lambda n,"},{"Start":"06:16.115 ","End":"06:18.460","Text":"n Pi over Alpha squared."},{"Start":"06:18.460 ","End":"06:26.640","Text":"What we need next are the functions R_n of r. We ultimately want the Theta_n and the R_n."},{"Start":"06:26.640 ","End":"06:28.520","Text":"We don\u0027t need the Lambda_n at the end,"},{"Start":"06:28.520 ","End":"06:30.905","Text":"but we do need them meanwhile because the Lambda_n are"},{"Start":"06:30.905 ","End":"06:33.450","Text":"what connect the Theta_n with the R_n."},{"Start":"06:33.450 ","End":"06:35.355","Text":"They have the same Lambda_n."},{"Start":"06:35.355 ","End":"06:39.630","Text":"We substitute these Lambdas in this equation,"},{"Start":"06:39.630 ","End":"06:41.635","Text":"make Lambda equal to Lambda_n."},{"Start":"06:41.635 ","End":"06:43.415","Text":"That gives us the following,"},{"Start":"06:43.415 ","End":"06:47.555","Text":"just replacing Lambda_n equals n Pi over Alpha squared."},{"Start":"06:47.555 ","End":"06:51.260","Text":"This type of second-order equation is called an Euler"},{"Start":"06:51.260 ","End":"06:55.495","Text":"equation, sometimes Euler-Cauchy equation."},{"Start":"06:55.495 ","End":"07:01.080","Text":"I looked it up on the Wikipedia and here\u0027s what it says."},{"Start":"07:01.080 ","End":"07:05.000","Text":"Note in particular that one of the applications it"},{"Start":"07:05.000 ","End":"07:09.620","Text":"mentions is solving Laplace\u0027s equation in polar coordinates,"},{"Start":"07:09.620 ","End":"07:11.290","Text":"which is what we\u0027re doing."},{"Start":"07:11.290 ","End":"07:18.365","Text":"What they suggest is to use a trial solution, y equals x^m."},{"Start":"07:18.365 ","End":"07:22.460","Text":"We try y equals x^m and find some value of m for which it will work."},{"Start":"07:22.460 ","End":"07:25.580","Text":"Just copying this equation down here,"},{"Start":"07:25.580 ","End":"07:30.530","Text":"we substitute R of r equals r^m and see what value of m will make it work."},{"Start":"07:30.530 ","End":"07:34.730","Text":"In here, we have second derivative of r is"},{"Start":"07:34.730 ","End":"07:40.230","Text":"just m times m minus 1 and the index is lowered by 2."},{"Start":"07:40.230 ","End":"07:46.260","Text":"Similarly, here, r times m times r^m minus 1,"},{"Start":"07:46.260 ","End":"07:52.805","Text":"and here just r^m times this."},{"Start":"07:52.805 ","End":"07:57.925","Text":"Now I colored the powers of r. I forgot to color this 1."},{"Start":"07:57.925 ","End":"08:02.265","Text":"R squared times r^m minus 2 is r^m,"},{"Start":"08:02.265 ","End":"08:06.015","Text":"r times r^m minus 1 is also r^m,"},{"Start":"08:06.015 ","End":"08:07.665","Text":"and here we have r^m."},{"Start":"08:07.665 ","End":"08:15.630","Text":"It\u0027s the same power of r everywhere so we can just divide by r^m and get mm minus 1,"},{"Start":"08:15.630 ","End":"08:18.915","Text":"which is m squared minus m plus m,"},{"Start":"08:18.915 ","End":"08:23.025","Text":"and here minus n Pi over Alpha squared equals 0."},{"Start":"08:23.025 ","End":"08:25.650","Text":"Minus m and plus m cancel."},{"Start":"08:25.650 ","End":"08:27.270","Text":"This squared minus this squared is 0,"},{"Start":"08:27.270 ","End":"08:28.680","Text":"so this squared equals this squared,"},{"Start":"08:28.680 ","End":"08:33.330","Text":"so m is plus or minus n Pi over Alpha."},{"Start":"08:33.330 ","End":"08:37.380","Text":"Now put that m here in the r^m."},{"Start":"08:37.380 ","End":"08:38.600","Text":"Well, there\u0027s 2 solutions,"},{"Start":"08:38.600 ","End":"08:42.860","Text":"but really we can take any linear combination of the 2 solutions."},{"Start":"08:42.860 ","End":"08:45.930","Text":"We have r to the power of plus n Pi over"},{"Start":"08:45.930 ","End":"08:48.930","Text":"Alpha and r to the power of minus n Pi over Alpha,"},{"Start":"08:48.930 ","End":"08:51.410","Text":"and some combination of the 2 of them."},{"Start":"08:51.410 ","End":"08:56.130","Text":"If we take the minus r to the minus n Pi over Alpha,"},{"Start":"08:56.130 ","End":"09:03.035","Text":"this isn\u0027t defined when r equals 0 because we get 1 over 0 to something positive,"},{"Start":"09:03.035 ","End":"09:05.725","Text":"which is 1 over 0 not defined."},{"Start":"09:05.725 ","End":"09:09.800","Text":"To overcome that, we just have to take b_n equals 0 and just look"},{"Start":"09:09.800 ","End":"09:13.840","Text":"for solutions of the form A_n r^nPi over Alpha."},{"Start":"09:13.840 ","End":"09:16.268","Text":"But we don\u0027t need all the A_ns are different ones."},{"Start":"09:16.268 ","End":"09:17.930","Text":"Just choose 1 that\u0027s non-zero."},{"Start":"09:17.930 ","End":"09:20.255","Text":"They\u0027re all linearly dependent on each other."},{"Start":"09:20.255 ","End":"09:22.735","Text":"Easiest is to choose A_n equals 1."},{"Start":"09:22.735 ","End":"09:27.080","Text":"That means that R_n of r is just this bit because A_n is 1,"},{"Start":"09:27.080 ","End":"09:30.185","Text":"b_n is 0, then this is what we have,"},{"Start":"09:30.185 ","End":"09:32.450","Text":"and I wrote that in the summary."},{"Start":"09:32.450 ","End":"09:35.570","Text":"R_n is r^nPi over Alpha."},{"Start":"09:35.570 ","End":"09:37.430","Text":"Theta_n of Theta,"},{"Start":"09:37.430 ","End":"09:38.870","Text":"it should be Theta here,"},{"Start":"09:38.870 ","End":"09:43.000","Text":"is equal to sine of nPi over Alpha Theta,"},{"Start":"09:43.000 ","End":"09:47.440","Text":"and u is an infinite linear combination of 1 of these,"},{"Start":"09:47.440 ","End":"09:51.275","Text":"just 1 of these, the same n, R_n, Theta_n."},{"Start":"09:51.275 ","End":"09:54.650","Text":"Just replace R_n and Theta_n with this and this,"},{"Start":"09:54.650 ","End":"09:58.600","Text":"and this is what we have for the expression for u of r and Theta."},{"Start":"09:58.600 ","End":"10:02.240","Text":"What remains is to find the coefficients A_n."},{"Start":"10:02.240 ","End":"10:06.095","Text":"We haven\u0027t used the boundary condition on the arc part,"},{"Start":"10:06.095 ","End":"10:08.270","Text":"which says that u of 1,"},{"Start":"10:08.270 ","End":"10:11.370","Text":"Theta is this expression in Theta."},{"Start":"10:11.370 ","End":"10:16.145","Text":"On the other hand, u of 1 Theta can be gotten by replacing r equals 1 here."},{"Start":"10:16.145 ","End":"10:20.695","Text":"Note that this r^nPi over Alpha also becomes 1."},{"Start":"10:20.695 ","End":"10:24.570","Text":"We can also write 1 in front of the sign here and"},{"Start":"10:24.570 ","End":"10:28.935","Text":"here because we\u0027re going to do comparison of coefficients."},{"Start":"10:28.935 ","End":"10:35.340","Text":"What we\u0027ll get by comparing these is that A_2 is going to"},{"Start":"10:35.340 ","End":"10:42.840","Text":"equal 1 and A_3 is going to equal 1 and all the others are going to be 0."},{"Start":"10:42.840 ","End":"10:48.510","Text":"I just write it like that that A_n is 1 if n is 2 or 3 and 0 otherwise."},{"Start":"10:48.510 ","End":"10:51.380","Text":"Now returning to this expression,"},{"Start":"10:51.380 ","End":"10:53.540","Text":"which is an infinite sum,"},{"Start":"10:53.540 ","End":"10:56.330","Text":"but really it\u0027s only going to be a finite sum because we\u0027re only going"},{"Start":"10:56.330 ","End":"10:59.720","Text":"to take n to be 2 or 3."},{"Start":"10:59.720 ","End":"11:02.360","Text":"When n is 2, we get A_2,"},{"Start":"11:02.360 ","End":"11:09.780","Text":"which is 1^2Pi over Alpha sine 2Pi over Alpha Theta."},{"Start":"11:09.780 ","End":"11:13.889","Text":"Similarly, just replace 2 by 3 in the second term."},{"Start":"11:13.889 ","End":"11:16.290","Text":"This is the result we have,"},{"Start":"11:16.290 ","End":"11:19.870","Text":"and that concludes this exercise."}],"ID":30840},{"Watched":false,"Name":"Exercise 2","Duration":"7m 54s","ChapterTopicVideoID":29259,"CourseChapterTopicPlaylistID":294438,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.140","Text":"In this exercise, we\u0027re given a boundary value problem."},{"Start":"00:04.140 ","End":"00:08.025","Text":"We\u0027re going to solve it and express the solution as an infinite series."},{"Start":"00:08.025 ","End":"00:12.780","Text":"This is the Laplace equation and it\u0027s on a segment,"},{"Start":"00:12.780 ","End":"00:19.740","Text":"which is also a semicircle because the angle is 180 degrees at this Pi here."},{"Start":"00:19.740 ","End":"00:23.760","Text":"Then here are the boundary conditions on"},{"Start":"00:23.760 ","End":"00:26.336","Text":"the straight parts of the boundary and"},{"Start":"00:26.336 ","End":"00:29.320","Text":"this is the boundary condition on the circular part."},{"Start":"00:29.320 ","End":"00:32.660","Text":"We\u0027re going to use the method of separation of variables."},{"Start":"00:32.660 ","End":"00:40.320","Text":"These are the stages, first to find the independent solutions of the form R(r) Theta,"},{"Start":"00:40.320 ","End":"00:43.580","Text":"Theta and then look for a linear combination or"},{"Start":"00:43.580 ","End":"00:48.350","Text":"infinite linear combination of these and then using the boundary condition,"},{"Start":"00:48.350 ","End":"00:52.935","Text":"the one on the semicircle to find the coefficients A_n,"},{"Start":"00:52.935 ","End":"00:56.870","Text":"reminder of what the Laplacian is in polar coordinates."},{"Start":"00:56.870 ","End":"00:59.360","Text":"We\u0027ll start with the Laplace equation that u"},{"Start":"00:59.360 ","End":"01:02.810","Text":"satisfies and write this in polar coordinates."},{"Start":"01:02.810 ","End":"01:08.480","Text":"Now, method of separation of variables means we\u0027re going to look for u in this form."},{"Start":"01:08.480 ","End":"01:12.890","Text":"If we substitute that in here what we get is the following."},{"Start":"01:12.890 ","End":"01:17.165","Text":"Basically, u_rr means we have to differentiate r"},{"Start":"01:17.165 ","End":"01:21.860","Text":"twice and u_Theta Theta means we differentiate Theta twice and u_r,"},{"Start":"01:21.860 ","End":"01:23.630","Text":"we differentiate r once."},{"Start":"01:23.630 ","End":"01:25.250","Text":"This is what we get."},{"Start":"01:25.250 ","End":"01:29.510","Text":"Get rid of the fractions by multiplying everything by r squared."},{"Start":"01:29.510 ","End":"01:32.690","Text":"We\u0027ll also put this last term on the right-hand side."},{"Start":"01:32.690 ","End":"01:39.515","Text":"Now, some algebra, we can divide both sides by R(r) and also by Theta(Theta)."},{"Start":"01:39.515 ","End":"01:42.560","Text":"This will go into the denominator here and this will go into"},{"Start":"01:42.560 ","End":"01:46.885","Text":"the denominator here so this is what we get."},{"Start":"01:46.885 ","End":"01:53.200","Text":"Because this is a function of R and this is a function of Theta and they\u0027re equal,"},{"Start":"01:53.200 ","End":"01:55.850","Text":"that can only happen if they are both"},{"Start":"01:55.850 ","End":"01:59.330","Text":"a constant function and we\u0027ll call that constant Lambda."},{"Start":"01:59.330 ","End":"02:00.920","Text":"From this, we can get 2 equations,"},{"Start":"02:00.920 ","End":"02:02.990","Text":"one is by comparing this to this we get"},{"Start":"02:02.990 ","End":"02:05.810","Text":"an equation in R and one is comparing this to this"},{"Start":"02:05.810 ","End":"02:11.366","Text":"we get an equation in Theta and these are the 2 equations that we get."},{"Start":"02:11.366 ","End":"02:15.620","Text":"Continuing, we\u0027ll use a couple of the boundary conditions as 3."},{"Start":"02:15.620 ","End":"02:22.140","Text":"We\u0027ll use the 2 of them on the straight part that we have u(0) is 0 and u of our Pi is 0."},{"Start":"02:22.140 ","End":"02:24.660","Text":"We\u0027ll put that into this form of u,"},{"Start":"02:24.660 ","End":"02:26.850","Text":"which is R of our Theta(Theta)."},{"Start":"02:26.850 ","End":"02:34.325","Text":"What we get is R(r) Theta(0) is 0 and R of our Theta(Pi) is 0."},{"Start":"02:34.325 ","End":"02:36.840","Text":"We can cancel by R(r)."},{"Start":"02:36.840 ","End":"02:39.430","Text":"We\u0027ve got Theta(0) equals Theta of Pi equals 0,"},{"Start":"02:39.800 ","End":"02:45.905","Text":"and that together with this will give us the equations in Theta."},{"Start":"02:45.905 ","End":"02:49.490","Text":"Now this is a Sturm-Liouville problem and"},{"Start":"02:49.490 ","End":"02:53.000","Text":"we\u0027ve previously seen such a problem with different letters."},{"Start":"02:53.000 ","End":"02:54.740","Text":"We\u0027ve solved this problem."},{"Start":"02:54.740 ","End":"03:01.800","Text":"The solution is that the eigenvalues are (nPi/L)^2 for each n,"},{"Start":"03:01.800 ","End":"03:02.994","Text":"1, 2, 3, etc."},{"Start":"03:02.994 ","End":"03:09.645","Text":"and the corresponding eigenfunctions are sine(nPi/L times x)."},{"Start":"03:09.645 ","End":"03:12.440","Text":"In our case we just have to change the letters,"},{"Start":"03:12.440 ","End":"03:13.895","Text":"y is big Theta,"},{"Start":"03:13.895 ","End":"03:15.275","Text":"x is little Theta,"},{"Start":"03:15.275 ","End":"03:17.300","Text":"big L is Pi."},{"Start":"03:17.300 ","End":"03:21.365","Text":"Then we get the following eigenfunctions,"},{"Start":"03:21.365 ","End":"03:30.245","Text":"sine(nTheta) because here Pi/L is 1 because L is Pi and the eigenvalues are n^2."},{"Start":"03:30.245 ","End":"03:34.250","Text":"We found Theta_n now we want the corresponding R_n."},{"Start":"03:34.250 ","End":"03:37.215","Text":"We do this using the common Lambda_n."},{"Start":"03:37.215 ","End":"03:43.485","Text":"Ultimately we don\u0027t need the Lambda_n but the bridge from Theta_n to R_n."},{"Start":"03:43.485 ","End":"03:48.935","Text":"Using these Lambda_n\u0027s and substituting in this equation,"},{"Start":"03:48.935 ","End":"03:50.315","Text":"this is what we have,"},{"Start":"03:50.315 ","End":"03:54.775","Text":"but Lambda_n, we know is n^2 from here."},{"Start":"03:54.775 ","End":"04:01.930","Text":"This is the equation we get for R. This is an Euler Cauchy equation."},{"Start":"04:01.930 ","End":"04:04.390","Text":"Here\u0027s what the Wikipedia says."},{"Start":"04:04.390 ","End":"04:07.270","Text":"All we have to do is try a solution of"},{"Start":"04:07.270 ","End":"04:13.345","Text":"the form y=x^m and we\u0027ll find an appropriate m to make it work."},{"Start":"04:13.345 ","End":"04:22.775","Text":"Take this equation, substitute our big R is r^m and R\u0027\u0027 is this."},{"Start":"04:22.775 ","End":"04:24.955","Text":"R\u0027 is this."},{"Start":"04:24.955 ","End":"04:31.090","Text":"Note that the power of r in each of the terms is r^m so we can divide by r^m,"},{"Start":"04:31.090 ","End":"04:32.740","Text":"open the brackets here."},{"Start":"04:32.740 ","End":"04:36.960","Text":"This cancels with this and we\u0027re left with m^2=n^2,"},{"Start":"04:36.960 ","End":"04:42.005","Text":"which means that m is plus or minus n. Now R"},{"Start":"04:42.005 ","End":"04:47.090","Text":"is r^m so it could be r^n or r^minus n. In fact,"},{"Start":"04:47.090 ","End":"04:52.265","Text":"we can take a linear combination of both of them to get a more general solution."},{"Start":"04:52.265 ","End":"04:56.060","Text":"The r^minus n part isn\u0027t defined at the origin,"},{"Start":"04:56.060 ","End":"04:59.818","Text":"where r equals 0 so we\u0027ll take the b_n to be 0"},{"Start":"04:59.818 ","End":"05:03.890","Text":"and also we can take a_n to be anything non 0."},{"Start":"05:03.890 ","End":"05:06.475","Text":"The easiest thing is to take it to be equal to 1,"},{"Start":"05:06.475 ","End":"05:11.810","Text":"which means that our R_n(r) is just r^n."},{"Start":"05:11.810 ","End":"05:13.655","Text":"a_n, we take as 1,"},{"Start":"05:13.655 ","End":"05:15.940","Text":"b_n is 0 r^n."},{"Start":"05:15.940 ","End":"05:21.450","Text":"Summarizing, we found that Theta_n(Theta) is sine(nTheta),"},{"Start":"05:21.450 ","End":"05:25.520","Text":"R_n(r) is r^n and we want u to be"},{"Start":"05:25.520 ","End":"05:30.695","Text":"an infinite linear combination of the R_n and the Theta_n,"},{"Start":"05:30.695 ","End":"05:37.685","Text":"which means that it\u0027s equal to the sum of A_nr^n sine(nTheta)."},{"Start":"05:37.685 ","End":"05:40.670","Text":"We\u0027re just replacing R_n and Theta_n by this and"},{"Start":"05:40.670 ","End":"05:45.330","Text":"this and we need to find the coefficients A_n."},{"Start":"05:45.330 ","End":"05:48.140","Text":"We find these from the third boundary condition,"},{"Start":"05:48.140 ","End":"05:49.685","Text":"the one that we didn\u0027t use,"},{"Start":"05:49.685 ","End":"05:51.110","Text":"which is this one,"},{"Start":"05:51.110 ","End":"05:54.905","Text":"the one on the arc or semicircular part."},{"Start":"05:54.905 ","End":"05:57.470","Text":"Note that when r is 1,"},{"Start":"05:57.470 ","End":"06:01.900","Text":"r^n is 1 so the r^n part disappears and we have"},{"Start":"06:01.900 ","End":"06:06.955","Text":"the sum of A_n sine and Theta=Theta Pi minus Theta."},{"Start":"06:06.955 ","End":"06:12.220","Text":"This is a Fourier sine series and we can"},{"Start":"06:12.220 ","End":"06:17.770","Text":"find the coefficients using the formula that we learned,"},{"Start":"06:17.770 ","End":"06:20.929","Text":"the sine series of generally for Fourier series."},{"Start":"06:20.929 ","End":"06:24.635","Text":"The A_n is 2/Pi,"},{"Start":"06:24.635 ","End":"06:30.010","Text":"the integral of the function times sine(nTheta)dTheta."},{"Start":"06:30.010 ","End":"06:36.070","Text":"This integral can be done using the formula for integration by parts."},{"Start":"06:36.070 ","End":"06:38.080","Text":"We actually use it twice,"},{"Start":"06:38.080 ","End":"06:42.304","Text":"and I\u0027ll leave this to you as homework."},{"Start":"06:42.304 ","End":"06:46.406","Text":"You\u0027ve studied integration by parts in calculus,"},{"Start":"06:46.406 ","End":"06:48.740","Text":"details are not necessary here."},{"Start":"06:48.740 ","End":"06:51.470","Text":"The answer comes out to be this."},{"Start":"06:51.470 ","End":"06:58.790","Text":"Note that this expression here is either 2 or 0 depending on whether n is odd or even,"},{"Start":"06:58.790 ","End":"07:03.440","Text":"u of our Theta is this sum but hold the A_n\u0027s when"},{"Start":"07:03.440 ","End":"07:07.940","Text":"n is even come out to be 0 and the A_n\u0027s when n is odd,"},{"Start":"07:07.940 ","End":"07:10.000","Text":"come out to be 2."},{"Start":"07:10.000 ","End":"07:13.880","Text":"What we can do is instead of letting n go from 1 to infinity we"},{"Start":"07:13.880 ","End":"07:17.810","Text":"can let n just go along the odd numbers,"},{"Start":"07:17.810 ","End":"07:22.310","Text":"replace A_n by what it\u0027s equal to."},{"Start":"07:22.310 ","End":"07:26.255","Text":"Now, like we said, we just want to take odd numbers n,"},{"Start":"07:26.255 ","End":"07:31.950","Text":"take n=2k minus 1 and then this expression will be 2."},{"Start":"07:31.950 ","End":"07:34.965","Text":"What we\u0027ll get is 8/Pi."},{"Start":"07:34.965 ","End":"07:36.360","Text":"Bring that to the front,"},{"Start":"07:36.360 ","End":"07:39.390","Text":"n^3 is 2k minus 1^3,"},{"Start":"07:39.390 ","End":"07:44.295","Text":"r^n is r^2k minus 1 and here n is 2k minus 1."},{"Start":"07:44.295 ","End":"07:46.715","Text":"This is the answer."},{"Start":"07:46.715 ","End":"07:51.920","Text":"We\u0027ve expressed u(r Theta) in terms of an infinite series in R and Theta."},{"Start":"07:51.920 ","End":"07:55.140","Text":"That concludes this exercise."}],"ID":30841},{"Watched":false,"Name":"Exercise 3","Duration":"7m 59s","ChapterTopicVideoID":29260,"CourseChapterTopicPlaylistID":294438,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.865","Text":"In this exercise, we\u0027re given a boundary value problem in polar coordinates."},{"Start":"00:05.865 ","End":"00:10.050","Text":"We\u0027re going to solve it also in polar coordinates."},{"Start":"00:10.050 ","End":"00:14.910","Text":"This here is the Laplace equation and it satisfied,"},{"Start":"00:14.910 ","End":"00:18.540","Text":"you can call it a segment or you can call it a"},{"Start":"00:18.540 ","End":"00:22.805","Text":"half disk because the angle is 180 degrees or Pi."},{"Start":"00:22.805 ","End":"00:24.500","Text":"There are 3 boundary conditions,"},{"Start":"00:24.500 ","End":"00:26.815","Text":"this, this, and this."},{"Start":"00:26.815 ","End":"00:33.915","Text":"1 of them is here the other 1 is on the semicircle and the other ones here,"},{"Start":"00:33.915 ","End":"00:36.885","Text":"these two are 0, but this one is not."},{"Start":"00:36.885 ","End":"00:39.000","Text":"It\u0027s sine cubed Theta."},{"Start":"00:39.000 ","End":"00:42.725","Text":"When I solve it using separation of variables."},{"Start":"00:42.725 ","End":"00:44.419","Text":"These should be familiar,"},{"Start":"00:44.419 ","End":"00:48.020","Text":"so I won\u0027t read them out and we\u0027ll start straight away with"},{"Start":"00:48.020 ","End":"00:52.505","Text":"the first step where we take the Laplacian equation,"},{"Start":"00:52.505 ","End":"00:55.265","Text":"write it in polar coordinates,"},{"Start":"00:55.265 ","End":"01:02.165","Text":"and now look for you as function of r times function of Theta."},{"Start":"01:02.165 ","End":"01:03.725","Text":"If we substitute it here,"},{"Start":"01:03.725 ","End":"01:07.010","Text":"all these derivatives u_rr, u_r u_Theta Theta."},{"Start":"01:07.010 ","End":"01:10.465","Text":"They just have to be applied to the r part of the Theta part."},{"Start":"01:10.465 ","End":"01:13.115","Text":"We end up with is the following."},{"Start":"01:13.115 ","End":"01:15.395","Text":"For example, here it\u0027s 1 over r,"},{"Start":"01:15.395 ","End":"01:18.920","Text":"and d by dr is R\u0027(r) Theta of Theta is"},{"Start":"01:18.920 ","End":"01:23.570","Text":"a constant relative to r and similarly with the other 2 terms."},{"Start":"01:23.570 ","End":"01:28.985","Text":"Let\u0027s put a common denominator of r^2 to a multiply both sides,"},{"Start":"01:28.985 ","End":"01:33.805","Text":"also take the last term over to the right-hand side,"},{"Start":"01:33.805 ","End":"01:37.490","Text":"because we want to separate this R(r) and"},{"Start":"01:37.490 ","End":"01:41.690","Text":"this Theta of Theta just divide both sides by this times this,"},{"Start":"01:41.690 ","End":"01:43.280","Text":"this will go in the denominator here,"},{"Start":"01:43.280 ","End":"01:45.320","Text":"this will go into the denominator here."},{"Start":"01:45.320 ","End":"01:51.560","Text":"This is what will get and because this is a function of r purely,"},{"Start":"01:51.560 ","End":"01:53.660","Text":"this is a function of the Theta purely,"},{"Start":"01:53.660 ","End":"01:59.195","Text":"if they\u0027re identically equal must be a constant and we call that constant Lambda."},{"Start":"01:59.195 ","End":"02:01.230","Text":"The 2 equations here,"},{"Start":"02:01.230 ","End":"02:04.245","Text":"if we let this equal Lambda,"},{"Start":"02:04.245 ","End":"02:07.200","Text":"we get an equation in r and if we let this be Lambda,"},{"Start":"02:07.200 ","End":"02:09.345","Text":"we get an equation in Theta."},{"Start":"02:09.345 ","End":"02:11.705","Text":"We get these two equations,"},{"Start":"02:11.705 ","End":"02:13.130","Text":"simple algebra to check."},{"Start":"02:13.130 ","End":"02:15.965","Text":"Now, we\u0027ll use 2 of the 3 boundary conditions."},{"Start":"02:15.965 ","End":"02:19.425","Text":"We\u0027ll use u(r) naught is naught,"},{"Start":"02:19.425 ","End":"02:22.300","Text":"and u(r) Pi is naught."},{"Start":"02:22.670 ","End":"02:28.610","Text":"We\u0027ll use this form of view as R(r) Theta of Theta and if we do that,"},{"Start":"02:28.610 ","End":"02:31.790","Text":"this one becomes this equation is naught,"},{"Start":"02:31.790 ","End":"02:34.195","Text":"and this one becomes this one."},{"Start":"02:34.195 ","End":"02:43.640","Text":"We can cancel both sides by R(r) we end up with this equation or boundary conditions."},{"Start":"02:43.640 ","End":"02:46.175","Text":"Theta of naught equals Theta of Pi equals naught."},{"Start":"02:46.175 ","End":"02:50.030","Text":"This together with this equation here,"},{"Start":"02:50.030 ","End":"02:53.030","Text":"gives us a Sturm Liouville problem,"},{"Start":"02:53.030 ","End":"02:58.865","Text":"and in the section on Sturm Liouville problems basically solved this problem."},{"Start":"02:58.865 ","End":"03:00.305","Text":"If you look at this,"},{"Start":"03:00.305 ","End":"03:02.120","Text":"and this is practically the same,"},{"Start":"03:02.120 ","End":"03:04.385","Text":"just with different letters."},{"Start":"03:04.385 ","End":"03:09.560","Text":"We got the solution that the Eigenvalues Lambda_n are the following and"},{"Start":"03:09.560 ","End":"03:16.115","Text":"the corresponding eigenfunctions sine n pi over L times x lambda n,"},{"Start":"03:16.115 ","End":"03:19.595","Text":"n Pi over L squared, in our case,"},{"Start":"03:19.595 ","End":"03:21.620","Text":"is equal to Pi,"},{"Start":"03:21.620 ","End":"03:23.930","Text":"so pi over L is 1."},{"Start":"03:23.930 ","End":"03:26.605","Text":"It comes out a bit simpler."},{"Start":"03:26.605 ","End":"03:28.470","Text":"Here we just get n^2,"},{"Start":"03:28.470 ","End":"03:35.130","Text":"here we get nx and n Theta because here we have Theta."},{"Start":"03:35.130 ","End":"03:37.905","Text":"These are the eigenfunctions in Theta,"},{"Start":"03:37.905 ","End":"03:41.455","Text":"these are the eigenvalues n equals 1, 2, 3."},{"Start":"03:41.455 ","End":"03:44.785","Text":"That gives us the functions Theta n,"},{"Start":"03:44.785 ","End":"03:47.615","Text":"now we need the R_n."},{"Start":"03:47.615 ","End":"03:53.395","Text":"The R_n and the Theta_n are related by sharing a common Lambda_n."},{"Start":"03:53.395 ","End":"04:00.225","Text":"This is the equation we had earlier just replace Lambda by Lambda_n."},{"Start":"04:00.225 ","End":"04:02.170","Text":"We have a whole sequence of equations,"},{"Start":"04:02.170 ","End":"04:05.395","Text":"once for each n and we know that Lambda_n,"},{"Start":"04:05.395 ","End":"04:08.575","Text":"we just found it was equal to n^2."},{"Start":"04:08.575 ","End":"04:15.220","Text":"This is what we get what is called an Euler equation or Euler Cauchy equation."},{"Start":"04:15.220 ","End":"04:17.810","Text":"When we get a solution,"},{"Start":"04:17.810 ","End":"04:25.070","Text":"is to try y equals x^power of m and search for an m that will make it work."},{"Start":"04:25.070 ","End":"04:28.175","Text":"If R(r) is r^m,"},{"Start":"04:28.175 ","End":"04:33.455","Text":"then substituting in this differential equation,"},{"Start":"04:33.455 ","End":"04:35.075","Text":"we get the following."},{"Start":"04:35.075 ","End":"04:37.040","Text":"This and this is r^m,"},{"Start":"04:37.040 ","End":"04:38.675","Text":"this and this is r^m,"},{"Start":"04:38.675 ","End":"04:40.190","Text":"here we have r^m."},{"Start":"04:40.190 ","End":"04:42.400","Text":"Divided by r^m."},{"Start":"04:42.400 ","End":"04:49.695","Text":"Then m m minus 1 is m^2 minus m plus m minus n squared."},{"Start":"04:49.695 ","End":"04:53.975","Text":"These two cancel, we get that m squared equals n squared,"},{"Start":"04:53.975 ","End":"05:00.355","Text":"and hence that m is plus or minus n. We\u0027re looking for r equals r^m,"},{"Start":"05:00.355 ","End":"05:04.310","Text":"so we could take r^n or r to the minus n. Then in fact,"},{"Start":"05:04.310 ","End":"05:07.190","Text":"you can take a linear combination of the two."},{"Start":"05:07.190 ","End":"05:09.125","Text":"Because everything is linear."},{"Start":"05:09.125 ","End":"05:11.720","Text":"Whoever we\u0027re going to have to restrict this a bit,"},{"Start":"05:11.720 ","End":"05:15.350","Text":"the r to the minus n is no good because we want our solution"},{"Start":"05:15.350 ","End":"05:19.490","Text":"defined on the whole segment of semicircle,"},{"Start":"05:19.490 ","End":"05:22.656","Text":"including the boundary, including the origin but r to"},{"Start":"05:22.656 ","End":"05:27.455","Text":"the power of minus n isn\u0027t defined when r is 0,"},{"Start":"05:27.455 ","End":"05:29.095","Text":"because n is positive,"},{"Start":"05:29.095 ","End":"05:35.190","Text":"1 over 0 to something positive is an undefined is 1 over 0."},{"Start":"05:35.190 ","End":"05:40.970","Text":"We take the b_n equals 0 then won\u0027t have any r to the minus n as for a_n,"},{"Start":"05:40.970 ","End":"05:44.000","Text":"we can let it be whatever we want except 0."},{"Start":"05:44.000 ","End":"05:46.210","Text":"Choose a_n equals 1,"},{"Start":"05:46.210 ","End":"05:47.990","Text":"the non-zero will do,"},{"Start":"05:47.990 ","End":"05:51.860","Text":"but any 2 of these will be linearly dependent we don\u0027t need more than one,"},{"Start":"05:51.860 ","End":"05:54.255","Text":"just a_n equals 1 is the easiest."},{"Start":"05:54.255 ","End":"05:58.290","Text":"Then a_n is 1_b and 0 we just got that r_n is"},{"Start":"05:58.290 ","End":"06:04.085","Text":"little r^n so here is our n. Previously we had Theta_n let\u0027s summarize,"},{"Start":"06:04.085 ","End":"06:06.380","Text":"here is the Theta n we found,"},{"Start":"06:06.380 ","End":"06:08.755","Text":"here is the R_n we found."},{"Start":"06:08.755 ","End":"06:13.445","Text":"Looking for u of a form linear combination"},{"Start":"06:13.445 ","End":"06:18.910","Text":"infinite of R_n and Theta_n so it looks like this."},{"Start":"06:18.910 ","End":"06:22.910","Text":"Then replacing sine and Theta instead of Theta and"},{"Start":"06:22.910 ","End":"06:26.735","Text":"replacing r^n instead of R_n, we get this."},{"Start":"06:26.735 ","End":"06:29.900","Text":"We need to find the coefficients a_n and we\u0027ll get"},{"Start":"06:29.900 ","End":"06:33.365","Text":"those from the third boundary condition that we haven\u0027t used yet."},{"Start":"06:33.365 ","End":"06:36.710","Text":"You have 1 Theta equals sine cubed Theta."},{"Start":"06:36.710 ","End":"06:38.930","Text":"Substitute 1 in here,"},{"Start":"06:38.930 ","End":"06:43.205","Text":"and we get u of 1 Theta sine cubed Theta,"},{"Start":"06:43.205 ","End":"06:50.700","Text":"on the other hand on the right-hand side r^n is 1^n is just 1."},{"Start":"06:50.700 ","End":"06:54.585","Text":"We get the sum of a_n sine n_Theta."},{"Start":"06:54.585 ","End":"07:00.485","Text":"Now we have here at the sine series and here we have this function sine cubed Theta."},{"Start":"07:00.485 ","End":"07:04.310","Text":"We\u0027ll use the trigonometric identity that we\u0027ve used before so"},{"Start":"07:04.310 ","End":"07:09.710","Text":"sine cubed Theta is 3 quarters sine Theta minus a quarter sine 3 Theta."},{"Start":"07:09.710 ","End":"07:14.945","Text":"Now, these two belong in this series of sine n Theta."},{"Start":"07:14.945 ","End":"07:18.370","Text":"Here n is 1 and here n is 3."},{"Start":"07:18.370 ","End":"07:22.500","Text":"We get a_n is either 3 quarters when n is 1,"},{"Start":"07:22.500 ","End":"07:27.105","Text":"or minus a quarter when n is 3 and all the rest will be 0."},{"Start":"07:27.105 ","End":"07:32.285","Text":"Putting these coefficients back in this series,"},{"Start":"07:32.285 ","End":"07:35.075","Text":"we only need the 2 terms that are non-zero."},{"Start":"07:35.075 ","End":"07:37.405","Text":"When n is 1 and n is 3."},{"Start":"07:37.405 ","End":"07:39.375","Text":"When n is 1,"},{"Start":"07:39.375 ","End":"07:45.570","Text":"we have 3 quarters and then is 1 to r^1 which is r and sine 1 Theta which"},{"Start":"07:45.570 ","End":"07:52.005","Text":"is sine Theta in here a quarter r^3 sine of 3 Theta."},{"Start":"07:52.005 ","End":"07:56.200","Text":"This is the answer and we are done."}],"ID":30842},{"Watched":false,"Name":"Exercise 4","Duration":"6m 59s","ChapterTopicVideoID":29261,"CourseChapterTopicPlaylistID":294438,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.860","Text":"In this exercise, we\u0027re given a boundary value problem and it\u0027s on a wedge,"},{"Start":"00:06.860 ","End":"00:10.290","Text":"but it could also be seen as a half circle,"},{"Start":"00:10.290 ","End":"00:15.540","Text":"half disk because the angle is Pi, 180 degrees."},{"Start":"00:15.540 ","End":"00:18.374","Text":"This is the domain, the interior."},{"Start":"00:18.374 ","End":"00:24.000","Text":"This is the Laplace equation and there are 3 boundary conditions."},{"Start":"00:24.000 ","End":"00:29.100","Text":"There\u0027s this radius, this radius, and this semicircle."},{"Start":"00:29.100 ","End":"00:31.560","Text":"On these two radii,"},{"Start":"00:31.560 ","End":"00:35.805","Text":"it\u0027s 0 and here it\u0027s equal to the constant"},{"Start":"00:35.805 ","End":"00:40.260","Text":"1 should be familiar by now with the stages for solving so I\u0027ll leave them written,"},{"Start":"00:40.260 ","End":"00:41.670","Text":"but I won\u0027t read them out."},{"Start":"00:41.670 ","End":"00:45.860","Text":"We\u0027ll start with the first stage where we take the Laplace equation,"},{"Start":"00:45.860 ","End":"00:48.050","Text":"write it in polar coordinates,"},{"Start":"00:48.050 ","End":"00:51.620","Text":"and we\u0027re looking for u in the form of r(r"},{"Start":"00:51.620 ","End":"00:55.595","Text":"Theta)Theta) and if we substitute that in here,"},{"Start":"00:55.595 ","End":"00:58.320","Text":"we\u0027ve done this before this is what we get,"},{"Start":"00:58.320 ","End":"01:02.180","Text":"multiplying by r^2 we get the following,"},{"Start":"01:02.180 ","End":"01:06.110","Text":"and we bring this last term to the right-hand side with a minus."},{"Start":"01:06.110 ","End":"01:10.950","Text":"Now we divide both sides by r(r)and Theta(Theta) and we"},{"Start":"01:10.950 ","End":"01:16.190","Text":"get this and because this is equal to this identically function of r,"},{"Start":"01:16.190 ","End":"01:20.180","Text":"function of Theta must be a constant which we call Lambda."},{"Start":"01:20.180 ","End":"01:21.890","Text":"Then from this equals this,"},{"Start":"01:21.890 ","End":"01:25.220","Text":"we get an equation in R. From this equals this,"},{"Start":"01:25.220 ","End":"01:27.500","Text":"we get an equation in Theta,"},{"Start":"01:27.500 ","End":"01:29.825","Text":"two separate differential equations."},{"Start":"01:29.825 ","End":"01:34.460","Text":"We\u0027re going to first get some boundary conditions for the second equation in Theta."},{"Start":"01:34.460 ","End":"01:41.345","Text":"We do this by starting with the two boundary conditions along the radii, which are these."},{"Start":"01:41.345 ","End":"01:45.080","Text":"Substitute them into the general form of a separated u."},{"Start":"01:45.080 ","End":"01:48.950","Text":"What we get from here, u(r,0) = 0,"},{"Start":"01:48.950 ","End":"01:52.190","Text":"we get R(r) Theta(0) is 0 and similarly,"},{"Start":"01:52.190 ","End":"01:54.590","Text":"r(r) Theta(Pi) is 0."},{"Start":"01:54.590 ","End":"01:57.440","Text":"From this one, cancel the r(r)."},{"Start":"01:57.440 ","End":"02:00.140","Text":"Now we have boundary conditions for Theta,"},{"Start":"02:00.140 ","End":"02:01.909","Text":"which together with this,"},{"Start":"02:01.909 ","End":"02:06.470","Text":"give us a boundary value problem or initial value problem in Theta,"},{"Start":"02:06.470 ","End":"02:10.340","Text":"which you should recognize as the Sturm-Liouville problem."},{"Start":"02:10.340 ","End":"02:14.210","Text":"What this looks like is the following problem."},{"Start":"02:14.210 ","End":"02:20.180","Text":"It\u0027s essentially the same just with different letters has eigenvalues these,"},{"Start":"02:20.180 ","End":"02:24.005","Text":"and eigenfunctions these, and in our case,"},{"Start":"02:24.005 ","End":"02:27.740","Text":"we have the same thing, just with y replaced by Theta,"},{"Start":"02:27.740 ","End":"02:34.000","Text":"x replaced by little Theta and L replaced by Pi and so the eigenvalues and"},{"Start":"02:34.000 ","End":"02:42.315","Text":"eigenvectors become eigenvalues are n^2 because if L=Pi,"},{"Start":"02:42.315 ","End":"02:44.370","Text":"then Pi over L is 1,"},{"Start":"02:44.370 ","End":"02:48.270","Text":"so we just get here sine nx and here Pi over L is 1,"},{"Start":"02:48.270 ","End":"02:55.041","Text":"so we get n^2 so here sine n Theta and here n^2 and n is 1,"},{"Start":"02:55.041 ","End":"02:56.735","Text":"2, 3, 4, 5, etc."},{"Start":"02:56.735 ","End":"02:58.430","Text":"Now that we have the Theta n,"},{"Start":"02:58.430 ","End":"03:00.910","Text":"We want the corresponding R_n."},{"Start":"03:00.910 ","End":"03:03.750","Text":"We\u0027re going to find the R_n because"},{"Start":"03:03.750 ","End":"03:07.460","Text":"R_n(Theta n) share a common Lambda n. Just to remind you,"},{"Start":"03:07.460 ","End":"03:15.845","Text":"this is the equation we had for R(r) and if we just replace Lambda by Lambda n,"},{"Start":"03:15.845 ","End":"03:17.930","Text":"which is in fact n^2,"},{"Start":"03:17.930 ","End":"03:23.960","Text":"we get our familiar Euler equation and we know that the solution of this can be found by"},{"Start":"03:23.960 ","End":"03:30.035","Text":"substituting R=r^m and looking for m that satisfies this."},{"Start":"03:30.035 ","End":"03:36.515","Text":"Substituting r\u0027\u0027 is mm minus 1 r to the n minus 2, etc."},{"Start":"03:36.515 ","End":"03:41.230","Text":"Notice that the exponent of r in each of the terms is r^m."},{"Start":"03:41.230 ","End":"03:45.440","Text":"We can divide by r^m, expand these brackets,"},{"Start":"03:45.440 ","End":"03:54.055","Text":"and then simplify this to m^2=n^2 from which m equals plus or minus n. Now,"},{"Start":"03:54.055 ","End":"04:01.115","Text":"remember that R is r^m so either have r^n or r to the minus n,"},{"Start":"04:01.115 ","End":"04:05.285","Text":"or in general, a linear combination of both of them will do it."},{"Start":"04:05.285 ","End":"04:09.020","Text":"However, there\u0027s a problem with r to the minus n,"},{"Start":"04:09.020 ","End":"04:12.635","Text":"because when r goes to 0 or is 0,"},{"Start":"04:12.635 ","End":"04:16.210","Text":"then this isn\u0027t defined n is positive,"},{"Start":"04:16.210 ","End":"04:20.360","Text":"we can\u0027t have 0 to a negative exponent so we overcome this"},{"Start":"04:20.360 ","End":"04:25.250","Text":"by just taking b_n to be 0 and as for a_n,"},{"Start":"04:25.250 ","End":"04:27.230","Text":"we can choose any non-0 value."},{"Start":"04:27.230 ","End":"04:30.169","Text":"We\u0027re only looking for linearly independent solutions,"},{"Start":"04:30.169 ","End":"04:32.300","Text":"so b_n is 0,"},{"Start":"04:32.300 ","End":"04:37.205","Text":"a_n is 1. We have R_n=r^n."},{"Start":"04:37.205 ","End":"04:45.335","Text":"Summary so far, we\u0027ve found Theta_n and R_n as sine nTheta and r^n."},{"Start":"04:45.335 ","End":"04:48.200","Text":"We\u0027re looking for a general solution now."},{"Start":"04:48.200 ","End":"04:52.180","Text":"An infinite linear combination of the R_n Theta n"},{"Start":"04:52.180 ","End":"04:56.629","Text":"substituting r_n and Theta n from here and here,"},{"Start":"04:56.629 ","End":"04:58.400","Text":"what we get is"},{"Start":"04:58.400 ","End":"05:04.835","Text":"the following infinite linear combination and we need to find the coefficients a_n."},{"Start":"05:04.835 ","End":"05:09.410","Text":"We find these from the boundary condition that we haven\u0027t used yet,"},{"Start":"05:09.410 ","End":"05:12.755","Text":"the one on the semicircle part, the arc."},{"Start":"05:12.755 ","End":"05:16.405","Text":"u(1,Theta) is 1."},{"Start":"05:16.405 ","End":"05:19.214","Text":"We get on the one hand,"},{"Start":"05:19.214 ","End":"05:21.380","Text":"you have 1 Theta is 1 that\u0027s the right answer."},{"Start":"05:21.380 ","End":"05:26.538","Text":"On the other hand, it\u0027s equal to this when you replace r by 1,"},{"Start":"05:26.538 ","End":"05:33.825","Text":"1^n, here is just 1 and disappears so we have the sum of a_n sine nTheta=1."},{"Start":"05:33.825 ","End":"05:37.190","Text":"This is a sine series on the interval from"},{"Start":"05:37.190 ","End":"05:41.840","Text":"0-Pi and we know how to solve a Fourier sine series."},{"Start":"05:41.840 ","End":"05:46.715","Text":"The coefficients A_n are given as the following integral."},{"Start":"05:46.715 ","End":"05:48.710","Text":"This is a straightforward integral."},{"Start":"05:48.710 ","End":"05:56.870","Text":"The integral of sine and Theta is minus cosine n Theta over n. Take this from 0-Pi."},{"Start":"05:56.870 ","End":"06:01.550","Text":"We can forget about the minus here and assume it\u0027s from Pi-0."},{"Start":"06:01.550 ","End":"06:06.350","Text":"When we plug in 0, here, we get 1."},{"Start":"06:06.350 ","End":"06:07.970","Text":"When we plug in Pi,"},{"Start":"06:07.970 ","End":"06:09.470","Text":"we get cosine n Pi,"},{"Start":"06:09.470 ","End":"06:12.200","Text":"which is minus 1 to the n so altogether,"},{"Start":"06:12.200 ","End":"06:16.460","Text":"this is what we have now and returning to this form of the solution,"},{"Start":"06:16.460 ","End":"06:18.633","Text":"we now know what to substitute for A_n."},{"Start":"06:18.633 ","End":"06:19.940","Text":"A_n is equal to this,"},{"Start":"06:19.940 ","End":"06:24.965","Text":"but this is equal to 0 when n is even and when n is odd,"},{"Start":"06:24.965 ","End":"06:26.870","Text":"we get 1 minus minus 1, which is 2,"},{"Start":"06:26.870 ","End":"06:31.890","Text":"we get 4 over Pi n. If we let n=2k minus 1,"},{"Start":"06:31.890 ","End":"06:33.770","Text":"k goes from 1 to infinity,"},{"Start":"06:33.770 ","End":"06:37.760","Text":"we go over all the odd numbers and what we get is the following;"},{"Start":"06:37.760 ","End":"06:41.615","Text":"k goes from 1 to infinity instead of n in the denominator,"},{"Start":"06:41.615 ","End":"06:43.400","Text":"we have 2k minus 1."},{"Start":"06:43.400 ","End":"06:44.900","Text":"We already said that 1 minus,"},{"Start":"06:44.900 ","End":"06:50.760","Text":"minus 1^n is 2 when n is odd and the 2 with the 2 makes 4 over Pi and here again,"},{"Start":"06:50.760 ","End":"06:52.950","Text":"instead of n 2k minus 1,"},{"Start":"06:52.950 ","End":"06:55.585","Text":"instead of n 2k minus 1."},{"Start":"06:55.585 ","End":"07:00.030","Text":"This is the answer and we are done."}],"ID":30843}],"Thumbnail":null,"ID":294438},{"Name":"The Laplace Equation in a Rectangle","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"7m 8s","ChapterTopicVideoID":29271,"CourseChapterTopicPlaylistID":294439,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.695","Text":"In this exercise, we\u0027re given the following boundary value problem."},{"Start":"00:04.695 ","End":"00:08.160","Text":"This is a Laplace equation in the rectangle."},{"Start":"00:08.160 ","End":"00:10.628","Text":"This is the Laplace equation,"},{"Start":"00:10.628 ","End":"00:12.165","Text":"and this is the rectangle."},{"Start":"00:12.165 ","End":"00:14.879","Text":"Here\u0027s a picture bit more general."},{"Start":"00:14.879 ","End":"00:20.565","Text":"In general, the height is H and the length is L here will take L for both of them."},{"Start":"00:20.565 ","End":"00:25.620","Text":"These are the boundary conditions and this one I just called it f(x)."},{"Start":"00:25.620 ","End":"00:28.440","Text":"There\u0027s 4 sides to the rectangle and that\u0027s 1,"},{"Start":"00:28.440 ","End":"00:30.750","Text":"2, 3, 4 boundary conditions."},{"Start":"00:30.750 ","End":"00:33.900","Text":"The technique we use the separation of variables."},{"Start":"00:33.900 ","End":"00:37.950","Text":"At this point, I should say that I hope you remember"},{"Start":"00:37.950 ","End":"00:42.145","Text":"Sturm-Liouville differential equations stages for solving."},{"Start":"00:42.145 ","End":"00:44.570","Text":"First of all, we\u0027ll find solutions to"},{"Start":"00:44.570 ","End":"00:50.090","Text":"the Laplace equation of the form some function of x, now the function of y."},{"Start":"00:50.090 ","End":"00:54.035","Text":"Typically it turns out as a sequence of such solutions,"},{"Start":"00:54.035 ","End":"00:57.965","Text":"we get these using the boundary conditions,"},{"Start":"00:57.965 ","End":"01:01.220","Text":"this and this, one on this side, and the one on this side."},{"Start":"01:01.220 ","End":"01:04.820","Text":"When we found this sequence of functions u, n,"},{"Start":"01:04.820 ","End":"01:08.730","Text":"we take an infinite linear combination of them,"},{"Start":"01:08.730 ","End":"01:10.440","Text":"and that\u0027s the general solution."},{"Start":"01:10.440 ","End":"01:12.480","Text":"We find the coefficients a,"},{"Start":"01:12.480 ","End":"01:19.350","Text":"n from the value of u along this side and along this side."},{"Start":"01:19.350 ","End":"01:26.805","Text":"Let\u0027s start the Laplacian of u to write it out as u_xx plus u_yy, that\u0027s a 0."},{"Start":"01:26.805 ","End":"01:30.360","Text":"We\u0027re looking for a solution u of this form."},{"Start":"01:30.360 ","End":"01:36.090","Text":"Really we get this plus this equals 0 in 2 steps in 1."},{"Start":"01:36.090 ","End":"01:37.860","Text":"Then when this plus this is 0,"},{"Start":"01:37.860 ","End":"01:39.465","Text":"this is minus this."},{"Start":"01:39.465 ","End":"01:46.460","Text":"Then we can divide both sides by y(y) and x(x) and get the following."},{"Start":"01:46.460 ","End":"01:48.290","Text":"If these 2 are equal,"},{"Start":"01:48.290 ","End":"01:50.840","Text":"a function of x is identically equal to a function of y,"},{"Start":"01:50.840 ","End":"01:54.415","Text":"then there must be a constant that\u0027s called the constant minus Lambda."},{"Start":"01:54.415 ","End":"01:57.605","Text":"I just copying this down here."},{"Start":"01:57.605 ","End":"02:01.560","Text":"What we get if we substitute 0,"},{"Start":"02:01.560 ","End":"02:06.020","Text":"y or L, y we get 0 from the boundary conditions."},{"Start":"02:06.020 ","End":"02:07.280","Text":"On the other hand, from here,"},{"Start":"02:07.280 ","End":"02:10.670","Text":"what this says is that this is X(0),"},{"Start":"02:10.670 ","End":"02:13.165","Text":"Y(y), and this is X(L), Y(y)."},{"Start":"02:13.165 ","End":"02:19.155","Text":"This is 0 canceled by Y(y). We have X(0) = X(L) = 0."},{"Start":"02:19.155 ","End":"02:24.215","Text":"That will be a boundary value for the Sturm-Liouville problem,"},{"Start":"02:24.215 ","End":"02:27.350","Text":"which is taken from this equation here."},{"Start":"02:27.350 ","End":"02:34.085","Text":"Just the part with the x\u0027\u0027 over x equals minus Lambda gives us this."},{"Start":"02:34.085 ","End":"02:35.270","Text":"This one, like I said,"},{"Start":"02:35.270 ","End":"02:36.335","Text":"is just from this,"},{"Start":"02:36.335 ","End":"02:38.615","Text":"just throw out the Y(y)."},{"Start":"02:38.615 ","End":"02:41.060","Text":"Now we have a Sturm-Liouville problem."},{"Start":"02:41.060 ","End":"02:42.905","Text":"I\u0027ll just give you the solution."},{"Start":"02:42.905 ","End":"02:49.205","Text":"We have the eigenvalues Lambda n our n Pi over L^2."},{"Start":"02:49.205 ","End":"02:54.690","Text":"The eigenfunctions sin(n) Pi over L times x."},{"Start":"02:54.690 ","End":"02:56.960","Text":"We have the functions x, n,"},{"Start":"02:56.960 ","End":"02:59.000","Text":"and now we\u0027d like the functions y,"},{"Start":"02:59.000 ","End":"03:00.815","Text":"n that correspond to these."},{"Start":"03:00.815 ","End":"03:04.550","Text":"Well, they share the same Lambda n. Recall"},{"Start":"03:04.550 ","End":"03:10.205","Text":"that this was our equation this time we\u0027re not focusing on x we focusing on y,"},{"Start":"03:10.205 ","End":"03:11.990","Text":"and we want the same eigenvalues."},{"Start":"03:11.990 ","End":"03:16.625","Text":"What we get is y\u0027\u0027 over y,"},{"Start":"03:16.625 ","End":"03:22.265","Text":"which is Lambda n is n Pi over L^2."},{"Start":"03:22.265 ","End":"03:25.130","Text":"From this, we get the following equation,"},{"Start":"03:25.130 ","End":"03:29.240","Text":"which is similar to the one we had with x only here\u0027s a minus,"},{"Start":"03:29.240 ","End":"03:31.190","Text":"and the other was a plus, and this time,"},{"Start":"03:31.190 ","End":"03:34.055","Text":"instead of getting sine and cosine,"},{"Start":"03:34.055 ","End":"03:40.410","Text":"we get sine hyperbolic and cosine hyperbolic with arbitrary constants a,"},{"Start":"03:40.410 ","End":"03:47.175","Text":"n and b, n. Now we have the Y n(y) and we have the X n(x)."},{"Start":"03:47.175 ","End":"03:51.620","Text":"u is the infinite linear combination of the x and y n,"},{"Start":"03:51.620 ","End":"03:52.910","Text":"which we also called u,"},{"Start":"03:52.910 ","End":"03:56.040","Text":"n. We don\u0027t need these constants a,"},{"Start":"03:56.040 ","End":"03:58.835","Text":"n because we\u0027ve got enough constants with the a,"},{"Start":"03:58.835 ","End":"03:59.990","Text":"n, and b, n,"},{"Start":"03:59.990 ","End":"04:01.870","Text":"no need for this then."},{"Start":"04:01.870 ","End":"04:04.485","Text":"This is a general form of u."},{"Start":"04:04.485 ","End":"04:08.600","Text":"What remains now is to find the constants a, n and b,"},{"Start":"04:08.600 ","End":"04:14.059","Text":"n. These we find from the boundary conditions for the bottom of the rectangle,"},{"Start":"04:14.059 ","End":"04:16.460","Text":"on the top of the rectangle."},{"Start":"04:16.460 ","End":"04:21.725","Text":"This condition is u(x) naught = naught its whats given."},{"Start":"04:21.725 ","End":"04:26.000","Text":"This cosine hyperbolic of 0 is 1,"},{"Start":"04:26.000 ","End":"04:30.140","Text":"this stays, but sine hyperbolic of 0 is 0 is drops out."},{"Start":"04:30.140 ","End":"04:33.080","Text":"All we\u0027re left with from this square brackets is the a,"},{"Start":"04:33.080 ","End":"04:35.495","Text":"n, which we can put in front of the sine."},{"Start":"04:35.495 ","End":"04:41.045","Text":"This is what we get by comparison of coefficients for a Fourier sine series,"},{"Start":"04:41.045 ","End":"04:44.150","Text":"all the coefficients a, n = 0."},{"Start":"04:44.150 ","End":"04:45.920","Text":"The full a, n is 0,"},{"Start":"04:45.920 ","End":"04:47.885","Text":"this part drops off."},{"Start":"04:47.885 ","End":"04:49.700","Text":"We just have the b,"},{"Start":"04:49.700 ","End":"04:52.354","Text":"n sine hyperbolic sine,"},{"Start":"04:52.354 ","End":"04:54.470","Text":"that part like so."},{"Start":"04:54.470 ","End":"04:56.735","Text":"Now the second boundary condition,"},{"Start":"04:56.735 ","End":"04:59.915","Text":"this is the top edge of the rectangle."},{"Start":"04:59.915 ","End":"05:02.000","Text":"From this, we get,"},{"Start":"05:02.000 ","End":"05:04.550","Text":"just to remind you what it is, this is what it is."},{"Start":"05:04.550 ","End":"05:07.985","Text":"If we substitute y= L here,"},{"Start":"05:07.985 ","End":"05:10.090","Text":"we get the following."},{"Start":"05:10.090 ","End":"05:12.495","Text":"y is L, L over L cancels."},{"Start":"05:12.495 ","End":"05:14.059","Text":"We just have n Pi here,"},{"Start":"05:14.059 ","End":"05:19.440","Text":"and the rest of it\u0027s the same and left-hand side is just x L minus x."},{"Start":"05:19.440 ","End":"05:21.035","Text":"This is what we have now."},{"Start":"05:21.035 ","End":"05:26.765","Text":"This is a Fourier sine series and we can find the coefficients from the integral formula."},{"Start":"05:26.765 ","End":"05:29.915","Text":"Look at the b, n sine hyperbolic n Pi,"},{"Start":"05:29.915 ","End":"05:34.640","Text":"all the coefficient of the sine n pi over L x."},{"Start":"05:34.640 ","End":"05:39.035","Text":"This coefficient is equal by the formula,"},{"Start":"05:39.035 ","End":"05:43.930","Text":"the following integral to this function times the sine,"},{"Start":"05:43.930 ","End":"05:46.410","Text":"and the integral from 0-L,"},{"Start":"05:46.410 ","End":"05:48.545","Text":"and then 2 over L times that."},{"Start":"05:48.545 ","End":"05:51.620","Text":"That\u0027s the standard way of finding the coefficients of"},{"Start":"05:51.620 ","End":"05:54.990","Text":"Fourier series on a finite interval 0-L."},{"Start":"05:54.990 ","End":"05:59.435","Text":"This integral can be done with integration by parts."},{"Start":"05:59.435 ","End":"06:01.100","Text":"This is the formula."},{"Start":"06:01.100 ","End":"06:02.780","Text":"This is u\u0027,"},{"Start":"06:02.780 ","End":"06:05.615","Text":"this is v. Using this formula,"},{"Start":"06:05.615 ","End":"06:08.150","Text":"I\u0027m not going to get into the details."},{"Start":"06:08.150 ","End":"06:11.810","Text":"It\u0027s messy, and you know how to do integration by parts."},{"Start":"06:11.810 ","End":"06:14.284","Text":"I\u0027m going to give you the answer to this integral."},{"Start":"06:14.284 ","End":"06:16.525","Text":"It comes out to be the following."},{"Start":"06:16.525 ","End":"06:23.780","Text":"Then we can divide both sides by sine hyperbolic and Pi and get this numerator,"},{"Start":"06:23.780 ","End":"06:28.845","Text":"1 minus minus 1 to the n is either 0 or 2."},{"Start":"06:28.845 ","End":"06:33.185","Text":"0 if n is even and 2 if n is odd,"},{"Start":"06:33.185 ","End":"06:36.545","Text":"an odd number is typically 2k minus 1."},{"Start":"06:36.545 ","End":"06:39.680","Text":"We can write the coefficient b,"},{"Start":"06:39.680 ","End":"06:41.420","Text":"n is either b2 k minus 1,"},{"Start":"06:41.420 ","End":"06:45.590","Text":"meaning the odds, and b2 k the evens, and these are 0."},{"Start":"06:45.590 ","End":"06:50.720","Text":"Now when we go back to this formula and we substitute the b,"},{"Start":"06:50.720 ","End":"06:52.910","Text":"n we only need the odd n. In other words,"},{"Start":"06:52.910 ","End":"06:59.480","Text":"we take n equals 2k minus 1 if we do that n here and n here replaced by 2k minus 1,"},{"Start":"06:59.480 ","End":"07:02.470","Text":"and b, n is replaced by this."},{"Start":"07:02.470 ","End":"07:04.875","Text":"This is what we get."},{"Start":"07:04.875 ","End":"07:08.410","Text":"That\u0027s the answer, and we\u0027re done."}],"ID":30844},{"Watched":false,"Name":"Exercise 2","Duration":"7m 55s","ChapterTopicVideoID":29272,"CourseChapterTopicPlaylistID":294439,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.964","Text":"In this exercise, we\u0027re going to solve the following boundary value problem."},{"Start":"00:04.964 ","End":"00:12.435","Text":"It consists of the Laplace equation on a rectangle which happens to be a square."},{"Start":"00:12.435 ","End":"00:15.600","Text":"Sometimes the height is not the same as the length,"},{"Start":"00:15.600 ","End":"00:17.085","Text":"in this case it is."},{"Start":"00:17.085 ","End":"00:21.900","Text":"We\u0027re given boundary conditions on all 4 sides here at 0, here at 0,"},{"Start":"00:21.900 ","End":"00:30.390","Text":"here at 0 but here it\u0027s equal to f(x) which is just short for this function here."},{"Start":"00:30.390 ","End":"00:34.920","Text":"I will use the technique of separation of variables which consists"},{"Start":"00:34.920 ","End":"00:39.305","Text":"of several stages but we can sum them up in 2 stages."},{"Start":"00:39.305 ","End":"00:40.940","Text":"You know what, we\u0027ve done this before,"},{"Start":"00:40.940 ","End":"00:42.815","Text":"I\u0027ll just leave it written out here."},{"Start":"00:42.815 ","End":"00:48.305","Text":"You can pause. Let\u0027s jump straight into the solution."},{"Start":"00:48.305 ","End":"00:55.970","Text":"We start with the Laplace equation which is u_xx plus u_yy equals 0 in"},{"Start":"00:55.970 ","End":"01:00.380","Text":"cartesian coordinates which is what we\u0027re working in and we\u0027re looking for"},{"Start":"01:00.380 ","End":"01:06.045","Text":"a solution u of the form some function of X times some function of Y."},{"Start":"01:06.045 ","End":"01:12.260","Text":"If that\u0027s the case then this equation becomes u_xx is this with X"},{"Start":"01:12.260 ","End":"01:18.725","Text":"differentiated twice and u_yy is this with the Y differentiated twice."},{"Start":"01:18.725 ","End":"01:24.080","Text":"We can bring this term to the right-hand side with a minus and then we can"},{"Start":"01:24.080 ","End":"01:30.245","Text":"divide both sides by X(x) and Y(y) and get this equality."},{"Start":"01:30.245 ","End":"01:34.355","Text":"But when you have a function of X identically equal to a function of Y,"},{"Start":"01:34.355 ","End":"01:37.580","Text":"they have to both be a constant function,"},{"Start":"01:37.580 ","End":"01:41.375","Text":"a constant, call it minus Lambda."},{"Start":"01:41.375 ","End":"01:44.540","Text":"Just repeating this condition from here,"},{"Start":"01:44.540 ","End":"01:48.215","Text":"we\u0027re going to substitute 2 of the boundary conditions."},{"Start":"01:48.215 ","End":"01:50.180","Text":"The ones that are on the,"},{"Start":"01:50.180 ","End":"01:52.895","Text":"where are they, on the sides,"},{"Start":"01:52.895 ","End":"01:57.770","Text":"this one and this one which are these 2 boundary conditions,"},{"Start":"01:57.770 ","End":"01:59.560","Text":"u(0, y) and u(L,"},{"Start":"01:59.560 ","End":"02:01.355","Text":"y) are both 0."},{"Start":"02:01.355 ","End":"02:04.230","Text":"Let\u0027s get back here."},{"Start":"02:05.110 ","End":"02:09.970","Text":"These are the 2 conditions on the sides and if"},{"Start":"02:09.970 ","End":"02:14.170","Text":"you substitute them using this form of u we get,"},{"Start":"02:14.170 ","End":"02:17.530","Text":"for this one we get X(0), Y(y),"},{"Start":"02:17.530 ","End":"02:20.530","Text":"and for this one, X(L), Y(y),"},{"Start":"02:20.530 ","End":"02:24.605","Text":"both of these are 0, just cancel by Y(y)."},{"Start":"02:24.605 ","End":"02:27.865","Text":"We\u0027re left with is X(0) equals X(L) equals 0."},{"Start":"02:27.865 ","End":"02:34.678","Text":"This will become a boundary condition on an ordinary differential equation in X,"},{"Start":"02:34.678 ","End":"02:41.130","Text":"this is gotten from this equation but just the X part."},{"Start":"02:41.130 ","End":"02:46.120","Text":"This equals minus Lambda multiplied by X."},{"Start":"02:46.120 ","End":"02:48.880","Text":"Bring the minus Lambda x to the left hand side."},{"Start":"02:48.880 ","End":"02:55.120","Text":"This is the equation we get together with this Sturm-Liouville problem."},{"Start":"02:55.120 ","End":"03:01.740","Text":"Just write the solution for such a problem and this is what it is,"},{"Start":"03:01.740 ","End":"03:06.220","Text":"the eigenvalues are nPi over L^2."},{"Start":"03:06.220 ","End":"03:10.425","Text":"The eigenfunctions are sine nPi over L,"},{"Start":"03:10.425 ","End":"03:14.760","Text":"x and the whole sequence of these n goes from 1,"},{"Start":"03:14.760 ","End":"03:17.440","Text":"2, 3, 4, 5, etc, up to infinity."},{"Start":"03:17.440 ","End":"03:23.185","Text":"We have the solution for a sequence of functions X_n."},{"Start":"03:23.185 ","End":"03:26.575","Text":"We also want the corresponding sequence Y_n."},{"Start":"03:26.575 ","End":"03:30.555","Text":"Each X_n is paired with a Y_n."},{"Start":"03:30.555 ","End":"03:34.080","Text":"Now, thing in common are the eigenvalues,"},{"Start":"03:34.080 ","End":"03:38.435","Text":"so we can look with this eigenvalue Lambda n and try and find Y."},{"Start":"03:38.435 ","End":"03:44.220","Text":"Again, from this equation we just throw out the X part, we have this."},{"Start":"03:44.220 ","End":"03:47.715","Text":"Since the eigenvalues for Y are the same as for X,"},{"Start":"03:47.715 ","End":"03:53.840","Text":"throw out the minuses and replace Lambda by Lambda n which is Pi n over L^2."},{"Start":"03:53.840 ","End":"03:59.440","Text":"Then just rewriting this we get the following equation in Y,"},{"Start":"03:59.440 ","End":"04:03.260","Text":"this would sometimes be called minus Omega^2."},{"Start":"04:03.260 ","End":"04:08.435","Text":"We already had the case with plus Omega^2 and then we get sine and cosine,"},{"Start":"04:08.435 ","End":"04:13.685","Text":"with a minus its sine and cosine hyperbolic."},{"Start":"04:13.685 ","End":"04:20.695","Text":"We\u0027ve found Y_n and we have X_n,"},{"Start":"04:20.695 ","End":"04:23.780","Text":"and we need the product of these."},{"Start":"04:23.780 ","End":"04:28.505","Text":"Each product gives us a U_n but then we want to take a linear combination,"},{"Start":"04:28.505 ","End":"04:30.995","Text":"an infinite linear combination."},{"Start":"04:30.995 ","End":"04:33.830","Text":"Only we won\u0027t need these constants A_n because we\u0027ve got"},{"Start":"04:33.830 ","End":"04:37.145","Text":"plenty of constants already here and here."},{"Start":"04:37.145 ","End":"04:40.340","Text":"I mean, if you just throw in a constant A_n,"},{"Start":"04:40.340 ","End":"04:44.120","Text":"it just gets swallowed up with the other constants so we don\u0027t need this one."},{"Start":"04:44.120 ","End":"04:48.650","Text":"This is a general form of the solution u(x, y)."},{"Start":"04:48.650 ","End":"04:53.697","Text":"What we still have to do is find coefficients a_n and b_n."},{"Start":"04:53.697 ","End":"04:57.820","Text":"We\u0027ll find these from the 2 boundary conditions that we didn\u0027t use,"},{"Start":"04:57.820 ","End":"05:00.970","Text":"we used the ones at the sides of the rectangle,"},{"Start":"05:00.970 ","End":"05:04.550","Text":"we used the one for the bottom and the top of the rectangle."},{"Start":"05:04.550 ","End":"05:08.150","Text":"The one at the bottom is u(x, 0)=0."},{"Start":"05:08.150 ","End":"05:12.800","Text":"This will give us that 0 is equal to, well,"},{"Start":"05:12.800 ","End":"05:14.240","Text":"we\u0027re not replacing x,"},{"Start":"05:14.240 ","End":"05:15.845","Text":"we\u0027re only replacing y,"},{"Start":"05:15.845 ","End":"05:18.875","Text":"so this sine nPi over L, x stays."},{"Start":"05:18.875 ","End":"05:21.455","Text":"Now, here when y is 0,"},{"Start":"05:21.455 ","End":"05:26.225","Text":"this is all 0 and cosine hyperbolic of 0 is 1."},{"Start":"05:26.225 ","End":"05:29.945","Text":"Here we get sine hyperbolic of 0 and that\u0027s 0,"},{"Start":"05:29.945 ","End":"05:32.015","Text":"so this part drops off."},{"Start":"05:32.015 ","End":"05:35.510","Text":"This becomes just a_n because this part is 1,"},{"Start":"05:35.510 ","End":"05:38.485","Text":"so a_n we can bring to the front,"},{"Start":"05:38.485 ","End":"05:41.430","Text":"get a_n sine nPi over L,"},{"Start":"05:41.430 ","End":"05:44.100","Text":"x as a fourier series,"},{"Start":"05:44.100 ","End":"05:46.235","Text":"specifically a sine series."},{"Start":"05:46.235 ","End":"05:51.380","Text":"When we have a sine series we can compare coefficients on the left is 0,"},{"Start":"05:51.380 ","End":"05:56.100","Text":"all the coefficients is 0 so all the a_n\u0027s must be 0."},{"Start":"05:56.100 ","End":"05:59.295","Text":"If all the a_n\u0027s is 0 we can put a_n here,"},{"Start":"05:59.295 ","End":"06:01.980","Text":"so the a_n cosine hyperbolic this,"},{"Start":"06:01.980 ","End":"06:10.725","Text":"this part drops out and we have just b_n sine hyperbolic of nPi over L y. Yeah,"},{"Start":"06:10.725 ","End":"06:12.180","Text":"this is what we have left,"},{"Start":"06:12.180 ","End":"06:14.850","Text":"the b_n just put that in front."},{"Start":"06:14.850 ","End":"06:16.624","Text":"We found the a_n\u0027s,"},{"Start":"06:16.624 ","End":"06:18.645","Text":"we still have to find the b_n\u0027s,"},{"Start":"06:18.645 ","End":"06:21.590","Text":"but we didn\u0027t use all the boundary conditions,"},{"Start":"06:21.590 ","End":"06:22.970","Text":"we used this one."},{"Start":"06:22.970 ","End":"06:26.105","Text":"There\u0027s still the boundary condition at the top, u(x,"},{"Start":"06:26.105 ","End":"06:30.515","Text":"L) and that is equal to sine 3Pix over L."},{"Start":"06:30.515 ","End":"06:36.820","Text":"Notice that that is equal to one of these where n=3."},{"Start":"06:36.820 ","End":"06:43.055","Text":"We can call this one sine 3Pi over L x just to give it a coefficient."},{"Start":"06:43.055 ","End":"06:49.040","Text":"On the other side we have all this sum and the sum of various sine nPi over L x."},{"Start":"06:49.040 ","End":"06:51.380","Text":"What remains is the coefficient,"},{"Start":"06:51.380 ","End":"06:55.960","Text":"this b_n together with the sine hyperbolic and Pi is the coefficient."},{"Start":"06:55.960 ","End":"06:59.520","Text":"It must equal either 1,"},{"Start":"06:59.520 ","End":"07:02.355","Text":"if n is 3 and 0 otherwise,"},{"Start":"07:02.355 ","End":"07:08.475","Text":"so we get b_3 sine hyperbolic 3Pi is 1,"},{"Start":"07:08.475 ","End":"07:16.870","Text":"and if n is not 3 then b_n sine hyperbolic Pi n is 0 which gives us that"},{"Start":"07:16.870 ","End":"07:25.435","Text":"b_3 is 1 over sine hyperbolic 3Pi and b_n is 0 for the other n which is not 3."},{"Start":"07:25.435 ","End":"07:30.110","Text":"Now that we have the b_n we can substitute them in here."},{"Start":"07:30.110 ","End":"07:32.750","Text":"I could say that instead of taking the sum from 1 to"},{"Start":"07:32.750 ","End":"07:35.690","Text":"infinity we just take the 3 elements."},{"Start":"07:35.690 ","End":"07:37.550","Text":"I could say take the sum from 3-3,"},{"Start":"07:37.550 ","End":"07:42.560","Text":"meaning just substitute n=3 and take that one term only."},{"Start":"07:42.560 ","End":"07:48.025","Text":"When n=3 we know that b_n is 1 over sine hyperbolic 3Pi,"},{"Start":"07:48.025 ","End":"07:49.920","Text":"and here put n=3,"},{"Start":"07:49.920 ","End":"07:51.955","Text":"here put n=3,"},{"Start":"07:51.955 ","End":"07:55.560","Text":"and this is our solution and we\u0027re done."}],"ID":30845},{"Watched":false,"Name":"Exercise 3","Duration":"8m 56s","ChapterTopicVideoID":29273,"CourseChapterTopicPlaylistID":294439,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.649","Text":"In this exercise, we\u0027re going to solve the following boundary value problem."},{"Start":"00:04.649 ","End":"00:09.480","Text":"Here we have the Laplace equation on a rectangle."},{"Start":"00:09.480 ","End":"00:13.065","Text":"Well, in general, a rectangle with a height,"},{"Start":"00:13.065 ","End":"00:15.320","Text":"and a length, but here the height equals the length,"},{"Start":"00:15.320 ","End":"00:17.760","Text":"so actually it\u0027s a square."},{"Start":"00:17.810 ","End":"00:23.255","Text":"These are 4 boundary conditions on each of the 4 sides."},{"Start":"00:23.255 ","End":"00:25.820","Text":"Notice that on the left and on the right,"},{"Start":"00:25.820 ","End":"00:28.640","Text":"we have nonhomogeneous conditions."},{"Start":"00:28.640 ","End":"00:30.289","Text":"On the top and the bottom,"},{"Start":"00:30.289 ","End":"00:34.175","Text":"we have homogeneous Neumann boundary conditions."},{"Start":"00:34.175 ","End":"00:37.175","Text":"Going to use separation of variables."},{"Start":"00:37.175 ","End":"00:39.590","Text":"You\u0027ve seen these stages before."},{"Start":"00:39.590 ","End":"00:45.320","Text":"But what\u0027s different is the order of the boundary conditions."},{"Start":"00:45.320 ","End":"00:46.850","Text":"The first 1,"},{"Start":"00:46.850 ","End":"00:51.230","Text":"we use the homogeneous ones at the top and the bottom."},{"Start":"00:51.230 ","End":"00:55.300","Text":"Previously we used the left and the right edges."},{"Start":"00:55.300 ","End":"01:03.455","Text":"Now let\u0027s start. This is the Laplace equation and we want you to satisfy it."},{"Start":"01:03.455 ","End":"01:07.430","Text":"But such a u that has x and y separated as follows,"},{"Start":"01:07.430 ","End":"01:10.225","Text":"some function of x, some function of y."},{"Start":"01:10.225 ","End":"01:15.770","Text":"If we substitute this in this equation,"},{"Start":"01:15.770 ","End":"01:17.815","Text":"this is what we get."},{"Start":"01:17.815 ","End":"01:19.850","Text":"We can rearrange that,"},{"Start":"01:19.850 ","End":"01:26.755","Text":"move this to the right and then divide by Y(y) and X(x) and we get this."},{"Start":"01:26.755 ","End":"01:30.905","Text":"Let\u0027s first of all deal with the equation for y."},{"Start":"01:30.905 ","End":"01:37.930","Text":"So from here, we get the following second-order ODE for y."},{"Start":"01:37.930 ","End":"01:40.760","Text":"We want to use these boundary conditions,"},{"Start":"01:40.760 ","End":"01:42.905","Text":"and for that we first need u_y,"},{"Start":"01:42.905 ","End":"01:47.600","Text":"and we get this just by differentiating this with respect to y,"},{"Start":"01:47.600 ","End":"01:50.435","Text":"so we get a y\u0027 here."},{"Start":"01:50.435 ","End":"01:54.620","Text":"Substituting u_y(x) naught is this,"},{"Start":"01:54.620 ","End":"01:57.415","Text":"but with y=0,"},{"Start":"01:57.415 ","End":"02:00.000","Text":"and this is like this,"},{"Start":"02:00.000 ","End":"02:01.500","Text":"but with Y= L,"},{"Start":"02:01.500 ","End":"02:03.525","Text":"so both of these are 0."},{"Start":"02:03.525 ","End":"02:05.760","Text":"The X(x) cancels,"},{"Start":"02:05.760 ","End":"02:10.575","Text":"we take this together with this,"},{"Start":"02:10.575 ","End":"02:13.350","Text":"we get the following,"},{"Start":"02:13.350 ","End":"02:16.335","Text":"which is a Sturm-Liouville problem,"},{"Start":"02:16.335 ","End":"02:20.150","Text":"this time with Y\u0027 on the boundary condition,"},{"Start":"02:20.150 ","End":"02:24.380","Text":"whereas previously we\u0027d had Y in earlier exercises."},{"Start":"02:24.380 ","End":"02:27.880","Text":"That\u0027s because of the Neumann boundary condition."},{"Start":"02:27.880 ","End":"02:31.200","Text":"The solution to this Sturm-Liouville problem,"},{"Start":"02:31.200 ","End":"02:37.510","Text":"I\u0027m just going to give it assuming that you\u0027ve studied this when we did Sturm-Liouville."},{"Start":"02:37.510 ","End":"02:43.070","Text":"Here it is, the solution similar to the previous exercise."},{"Start":"02:43.070 ","End":"02:50.860","Text":"But we have a cosine instead of a sine that\u0027s because of the Y\u0027 instead of Y here,"},{"Start":"02:50.860 ","End":"02:55.125","Text":"and n begins from 0, not from 1."},{"Start":"02:55.125 ","End":"02:57.365","Text":"Now that we\u0027ve found the Y_n,"},{"Start":"02:57.365 ","End":"03:00.280","Text":"we want to get the corresponding X_n."},{"Start":"03:00.280 ","End":"03:03.790","Text":"Now X_n and Y_n share a common lambda n,"},{"Start":"03:03.790 ","End":"03:06.175","Text":"and that\u0027s how we\u0027ll get from 1 to the other."},{"Start":"03:06.175 ","End":"03:07.719","Text":"Now recall this equation,"},{"Start":"03:07.719 ","End":"03:10.990","Text":"we used it previously to figure out what Y is."},{"Start":"03:10.990 ","End":"03:13.655","Text":"This time we\u0027ll ignore the Y and we\u0027ll work on X,"},{"Start":"03:13.655 ","End":"03:16.180","Text":"so from this over this equals this,"},{"Start":"03:16.180 ","End":"03:22.435","Text":"we get that X\u0027\u0027 over X is equal to Lambda."},{"Start":"03:22.435 ","End":"03:24.300","Text":"We just get rid of the minus here and here,"},{"Start":"03:24.300 ","End":"03:29.170","Text":"and Lambda is Lambda n and Lambda n is what\u0027s written here,"},{"Start":"03:29.170 ","End":"03:31.085","Text":"n Pi over L^2."},{"Start":"03:31.085 ","End":"03:33.265","Text":"Rearranging a bit, just multiplying by"},{"Start":"03:33.265 ","End":"03:36.235","Text":"big X and then bringing this to the left-hand side,"},{"Start":"03:36.235 ","End":"03:37.795","Text":"we get the following."},{"Start":"03:37.795 ","End":"03:42.800","Text":"The solution to this depends on whether this quantity call it Omega,"},{"Start":"03:42.800 ","End":"03:45.815","Text":"where Omega is 0 or not."},{"Start":"03:45.815 ","End":"03:49.789","Text":"There was a plus here we know we get a solution with sine and cosine."},{"Start":"03:49.789 ","End":"03:51.320","Text":"If this is strictly negative,"},{"Start":"03:51.320 ","End":"03:54.380","Text":"we get a solution with sine hyperbolic and cosine hyperbolic."},{"Start":"03:54.380 ","End":"03:55.790","Text":"If this is 0,"},{"Start":"03:55.790 ","End":"03:58.235","Text":"we get a linear solution. Let\u0027s just write this."},{"Start":"03:58.235 ","End":"04:01.040","Text":"If n is not 0, then we have"},{"Start":"04:01.040 ","End":"04:05.330","Text":"a linear combination of cosine hyperbolic and sine hyperbolic."},{"Start":"04:05.330 ","End":"04:10.085","Text":"This is for X_n and Y_n just copying it from here is this."},{"Start":"04:10.085 ","End":"04:12.980","Text":"The other case where n is 0,"},{"Start":"04:12.980 ","End":"04:14.630","Text":"the solution for X is"},{"Start":"04:14.630 ","End":"04:23.240","Text":"just some linear polynomial because I mean the second derivative is 0,"},{"Start":"04:23.240 ","End":"04:26.090","Text":"so the first derivative is a constant and the function"},{"Start":"04:26.090 ","End":"04:29.660","Text":"itself is some constant times x plus another constant."},{"Start":"04:29.660 ","End":"04:32.485","Text":"In this case also Y_0,"},{"Start":"04:32.485 ","End":"04:35.325","Text":"when is naught is equal to 1,"},{"Start":"04:35.325 ","End":"04:38.925","Text":"because cosine of 0 is 1."},{"Start":"04:38.925 ","End":"04:43.860","Text":"Now u(x, y) is going to be the sum of u_0,"},{"Start":"04:43.860 ","End":"04:46.830","Text":"and the sum of u_n,"},{"Start":"04:46.830 ","End":"04:48.110","Text":"n goes from 1 to infinity."},{"Start":"04:48.110 ","End":"04:52.055","Text":"We separate the 0 from the rest of them because we have different expressions."},{"Start":"04:52.055 ","End":"04:56.805","Text":"We see from here that X0 and Y0 are like this,"},{"Start":"04:56.805 ","End":"04:59.360","Text":"and from here, x_n and y_n are like this."},{"Start":"04:59.360 ","End":"05:01.460","Text":"We just substitute those in here,"},{"Start":"05:01.460 ","End":"05:04.515","Text":"what we get is x_0,"},{"Start":"05:04.515 ","End":"05:10.480","Text":"y_0 plus the sum of this is y_n and this is x_n."},{"Start":"05:10.480 ","End":"05:14.090","Text":"It remains to find the coefficients a_n and b_n,"},{"Start":"05:14.090 ","End":"05:18.290","Text":"and we\u0027ll find these from the boundary conditions that we haven\u0027t used yet."},{"Start":"05:18.290 ","End":"05:24.155","Text":"These are the boundary conditions on the 2 sides of the rectangle."},{"Start":"05:24.155 ","End":"05:30.000","Text":"Let\u0027s first take the condition u(0, y) that\u0027s equal to 0."},{"Start":"05:30.140 ","End":"05:35.405","Text":"Now note that if we put x equals 0 here,"},{"Start":"05:35.405 ","End":"05:38.435","Text":"cosine hyperbolic of 0 is 1,"},{"Start":"05:38.435 ","End":"05:42.635","Text":"and sine hyperbolic of 0 is 0."},{"Start":"05:42.635 ","End":"05:47.135","Text":"All we get left from this square bracket is the a_n."},{"Start":"05:47.135 ","End":"05:52.860","Text":"We get that the 0 here is equal to x is 0,"},{"Start":"05:52.860 ","End":"05:54.915","Text":"so we\u0027re just left with the a0."},{"Start":"05:54.915 ","End":"06:00.910","Text":"Like I said here we just get the a_n together with the cosine n Pi over L_y."},{"Start":"06:00.910 ","End":"06:07.250","Text":"Now if we compare coefficients of this Fourier cosine series,"},{"Start":"06:07.250 ","End":"06:09.980","Text":"we get that all the a_n\u0027s are 0,"},{"Start":"06:09.980 ","End":"06:14.520","Text":"including n=0 and from 1 to infinity."},{"Start":"06:14.520 ","End":"06:18.755","Text":"If we put a_n=0 here and here,"},{"Start":"06:18.755 ","End":"06:21.005","Text":"then what we get is that u(x,"},{"Start":"06:21.005 ","End":"06:25.620","Text":"y) equals b_0 x, that\u0027s this b_0."},{"Start":"06:25.700 ","End":"06:28.800","Text":"From here we only take the b_n\u0027s,"},{"Start":"06:28.800 ","End":"06:33.735","Text":"we get b_n sine hyperbolic n Pi over L_x."},{"Start":"06:33.735 ","End":"06:37.320","Text":"From here, cosine n Pi over L_y."},{"Start":"06:37.320 ","End":"06:42.980","Text":"I just put the b_n in front and then the cosine and then the sine hyperbolic."},{"Start":"06:42.980 ","End":"06:44.310","Text":"Now we have u(x,"},{"Start":"06:44.310 ","End":"06:47.420","Text":"y) with just coefficients b_n."},{"Start":"06:47.420 ","End":"06:53.290","Text":"We can find those by the boundary condition that we haven\u0027t used yet,"},{"Start":"06:53.290 ","End":"06:55.870","Text":"that\u0027s this 1, u(L, y),"},{"Start":"06:55.870 ","End":"07:01.795","Text":"which is equal to 1 plus cosine 2 Pi y over L. If we do that,"},{"Start":"07:01.795 ","End":"07:04.000","Text":"we get that this,"},{"Start":"07:04.000 ","End":"07:06.190","Text":"which just slightly rewrite,"},{"Start":"07:06.190 ","End":"07:10.645","Text":"put the y to the side and put a 1 in front of the cosine,"},{"Start":"07:10.645 ","End":"07:12.340","Text":"so that its coefficient,"},{"Start":"07:12.340 ","End":"07:17.405","Text":"equals b_0, and then x is L everywhere."},{"Start":"07:17.405 ","End":"07:21.090","Text":"Here x is L,"},{"Start":"07:21.090 ","End":"07:23.520","Text":"so L over L is 1,"},{"Start":"07:23.520 ","End":"07:26.280","Text":"so we just get sine hyperbolic n Pi."},{"Start":"07:26.280 ","End":"07:28.170","Text":"For the sine hyperbolic we get n Pi,"},{"Start":"07:28.170 ","End":"07:31.755","Text":"for the cosine we have the same thing, n Pi over L_y."},{"Start":"07:31.755 ","End":"07:36.605","Text":"A comparing coefficients is only 2 things that match when n is 0 or 2."},{"Start":"07:36.605 ","End":"07:39.440","Text":"When n is 0, we get b_0 L=1."},{"Start":"07:39.440 ","End":"07:41.135","Text":"When n is 2,"},{"Start":"07:41.135 ","End":"07:45.530","Text":"here we have 2 Pi over L_y cosine and here also,"},{"Start":"07:45.530 ","End":"07:49.250","Text":"so we\u0027d get that b_n =1,"},{"Start":"07:49.250 ","End":"07:53.125","Text":"so we get 1=b_0 L,"},{"Start":"07:53.125 ","End":"07:55.400","Text":"we just want the coefficient part, not the part with the Y,"},{"Start":"07:55.400 ","End":"07:59.105","Text":"so b_2 sine hyperbolic 2 Pi."},{"Start":"07:59.105 ","End":"08:01.345","Text":"If n is not 0 or 2,"},{"Start":"08:01.345 ","End":"08:04.020","Text":"then we get that this expression,"},{"Start":"08:04.020 ","End":"08:06.815","Text":"b_n sine hyperbolic n Pi is 0."},{"Start":"08:06.815 ","End":"08:09.740","Text":"This will give us that b naught is 1/L,"},{"Start":"08:09.740 ","End":"08:15.895","Text":"b_2 is 1 over sine hyperbolic 2 Pi and b_n is 0."},{"Start":"08:15.895 ","End":"08:17.640","Text":"This is what we said."},{"Start":"08:17.640 ","End":"08:19.770","Text":"Now we can say that u(x,"},{"Start":"08:19.770 ","End":"08:22.050","Text":"y) is the sum of just 2 things,"},{"Start":"08:22.050 ","End":"08:27.035","Text":"b_0x, plus the 2 term from the sum."},{"Start":"08:27.035 ","End":"08:31.190","Text":"I just wrote it by saying that the limit goes from 2-2,"},{"Start":"08:31.190 ","End":"08:34.310","Text":"meaning just the term when n equals 2,"},{"Start":"08:34.310 ","End":"08:40.160","Text":"which comes out to b. B_0 is 1/L times x,"},{"Start":"08:40.160 ","End":"08:47.925","Text":"and this b_2 is 1 over sine hyperbolic 2 Pi cosine n is 2,"},{"Start":"08:47.925 ","End":"08:51.660","Text":"so 2 Pi over L_y and here sine hyperbolic n again is 2,"},{"Start":"08:51.660 ","End":"08:52.945","Text":"2P i over L_x."},{"Start":"08:52.945 ","End":"08:54.305","Text":"Anyway, this is the answer,"},{"Start":"08:54.305 ","End":"08:56.430","Text":"and we are done."}],"ID":30846},{"Watched":false,"Name":"Exercise 4","Duration":"9m 4s","ChapterTopicVideoID":29274,"CourseChapterTopicPlaylistID":294439,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.960","Text":"In this exercise, we\u0027re given a boundary value problem,"},{"Start":"00:03.960 ","End":"00:07.380","Text":"the Laplace equation on a rectangle,"},{"Start":"00:07.380 ","End":"00:08.985","Text":"which is actually a square,"},{"Start":"00:08.985 ","End":"00:11.775","Text":"because the height and the length are the same."},{"Start":"00:11.775 ","End":"00:14.939","Text":"We have the 4 boundary conditions,"},{"Start":"00:14.939 ","End":"00:19.095","Text":"one for each of the sides of the rectangle."},{"Start":"00:19.095 ","End":"00:23.945","Text":"These two are Neumann boundary conditions."},{"Start":"00:23.945 ","End":"00:30.390","Text":"The ones on the top and the bottom and the two-side ones are Dirichlet conditions."},{"Start":"00:30.470 ","End":"00:34.250","Text":"These two are homogeneous."},{"Start":"00:34.250 ","End":"00:37.670","Text":"Here are the stages for solving,"},{"Start":"00:37.670 ","End":"00:39.695","Text":"which we\u0027ve done already."},{"Start":"00:39.695 ","End":"00:43.495","Text":"So I\u0027m just giving them as a review."},{"Start":"00:43.495 ","End":"00:47.600","Text":"Let\u0027s start with the Laplace equation,"},{"Start":"00:47.600 ","End":"00:51.320","Text":"which can be written in Cartesian coordinates as ux,"},{"Start":"00:51.320 ","End":"00:53.510","Text":"x plus uy, y equals 0."},{"Start":"00:53.510 ","End":"00:57.170","Text":"We\u0027re looking for a solution of the type,"},{"Start":"00:57.170 ","End":"01:02.160","Text":"some function of x, some function of y product."},{"Start":"01:02.330 ","End":"01:05.990","Text":"If we differentiate twice with respect to x,"},{"Start":"01:05.990 ","End":"01:13.390","Text":"we just get x\u0027\u0027 and uy y will give us y\u0027\u0027 here."},{"Start":"01:13.390 ","End":"01:20.030","Text":"This is what happens to Laplace equation for a separated function."},{"Start":"01:20.030 ","End":"01:27.245","Text":"Bring this to the right-hand side and divide both sides by y(y),"},{"Start":"01:27.245 ","End":"01:30.770","Text":"x(x) and this is what we get."},{"Start":"01:30.770 ","End":"01:34.156","Text":"Let\u0027s just concentrate on the y part,"},{"Start":"01:34.156 ","End":"01:36.500","Text":"get an equation from that."},{"Start":"01:36.500 ","End":"01:43.470","Text":"This gives us the following second-order differential equation in y,"},{"Start":"01:43.670 ","End":"01:47.715","Text":"and what we need is some boundary conditions."},{"Start":"01:47.715 ","End":"01:55.040","Text":"We can use these two boundary conditions."},{"Start":"01:55.040 ","End":"01:56.960","Text":"We substitute them into here."},{"Start":"01:56.960 ","End":"01:58.985","Text":"This is just uy,"},{"Start":"01:58.985 ","End":"02:03.304","Text":"which we get from differentiating this with respect to y."},{"Start":"02:03.304 ","End":"02:06.965","Text":"We plug in these two into here."},{"Start":"02:06.965 ","End":"02:10.565","Text":"We get x(x),"},{"Start":"02:10.565 ","End":"02:15.335","Text":"y\u0027(0) equals x(x),"},{"Start":"02:15.335 ","End":"02:20.645","Text":"y\u0027(l) equals 0 and we can divide by x(x)."},{"Start":"02:20.645 ","End":"02:24.770","Text":"We have y\u0027(0) equals y\u0027(l) equals 0."},{"Start":"02:24.770 ","End":"02:26.615","Text":"Now, if we take"},{"Start":"02:26.615 ","End":"02:33.920","Text":"this ordinary differential equation together with these boundary conditions,"},{"Start":"02:33.920 ","End":"02:36.424","Text":"then we get this problem,"},{"Start":"02:36.424 ","End":"02:39.335","Text":"which is a Sturm-Louiville problem."},{"Start":"02:39.335 ","End":"02:41.765","Text":"We know how to solve these."},{"Start":"02:41.765 ","End":"02:44.840","Text":"Note that there\u0027s a y\u0027 here and here."},{"Start":"02:44.840 ","End":"02:49.520","Text":"So our solution will be with a cosign not a sine,"},{"Start":"02:49.520 ","End":"02:52.885","Text":"and n will start from 0."},{"Start":"02:52.885 ","End":"02:57.930","Text":"Eigenvalues are these eigenfunctions are these."},{"Start":"02:57.930 ","End":"03:00.800","Text":"That\u0027s why in our next,"},{"Start":"03:00.800 ","End":"03:03.835","Text":"you want to find the corresponding xn."},{"Start":"03:03.835 ","End":"03:07.010","Text":"We have the eigenfunctions for yn,"},{"Start":"03:07.010 ","End":"03:12.400","Text":"which are these, and the corresponding lambda n, which are these."},{"Start":"03:12.400 ","End":"03:15.265","Text":"Next we want the corresponding xn."},{"Start":"03:15.265 ","End":"03:23.330","Text":"Now return to this equation where last time we compared y\u0027\u0027 over y with minus lambda."},{"Start":"03:23.330 ","End":"03:25.490","Text":"This time we\u0027ll use the x part."},{"Start":"03:25.490 ","End":"03:34.460","Text":"Minus x\u0027\u0027 over x is minus lambda throw out the minuses and we get that x\u0027\u0027/x is lambda."},{"Start":"03:34.460 ","End":"03:37.070","Text":"But lambda we can take as lambda n,"},{"Start":"03:37.070 ","End":"03:39.645","Text":"which is n pi over l squared,"},{"Start":"03:39.645 ","End":"03:42.295","Text":"and x here will be xn."},{"Start":"03:42.295 ","End":"03:44.765","Text":"Rewrite this a little bit throughout the lambda n,"},{"Start":"03:44.765 ","End":"03:49.220","Text":"bring this to the left-hand side and multiply also by x(x)."},{"Start":"03:49.220 ","End":"03:50.785","Text":"This is what we get,"},{"Start":"03:50.785 ","End":"03:59.000","Text":"and the solution to this ODE depends on whether this is 0 or non-zero."},{"Start":"03:59.000 ","End":"04:04.190","Text":"If it\u0027s non-zero we get that xn is a linear combination of cosine"},{"Start":"04:04.190 ","End":"04:10.265","Text":"hyperbolic n pi over lx and sine hyperbolic n pi over lx."},{"Start":"04:10.265 ","End":"04:17.495","Text":"If this is 0, then we get that the second-order derivative of x is 0,"},{"Start":"04:17.495 ","End":"04:22.250","Text":"so x multiply x is a linear polynomial in x,"},{"Start":"04:22.250 ","End":"04:25.145","Text":"like a naught plus v naught x."},{"Start":"04:25.145 ","End":"04:29.565","Text":"When n is naught, then y naught is 1,"},{"Start":"04:29.565 ","End":"04:32.130","Text":"and when n is not 0,"},{"Start":"04:32.130 ","End":"04:35.335","Text":"yn(y) is cosine n pi over ly."},{"Start":"04:35.335 ","End":"04:39.425","Text":"This, we can get from this by letting n equals"},{"Start":"04:39.425 ","End":"04:44.375","Text":"0 and cosine of 0 is 1 so that\u0027s where we get the 1."},{"Start":"04:44.375 ","End":"04:49.085","Text":"Next, we want the sum of the products of nx with the y."},{"Start":"04:49.085 ","End":"04:52.390","Text":"Separately, we\u0027ll take x naught, with y naught."},{"Start":"04:52.390 ","End":"04:54.210","Text":"Then each xn,"},{"Start":"04:54.210 ","End":"04:55.815","Text":"with each yn,"},{"Start":"04:55.815 ","End":"04:57.915","Text":"we multiply and then sum."},{"Start":"04:57.915 ","End":"05:02.510","Text":"We don\u0027t need constants here because we\u0027re going to get enough constants from the a,"},{"Start":"05:02.510 ","End":"05:06.515","Text":"n and b, n. We get x naught of x is this,"},{"Start":"05:06.515 ","End":"05:08.255","Text":"y naught of y is this."},{"Start":"05:08.255 ","End":"05:15.245","Text":"Similarly, xn(x) is all this square bracket and yn(y) is this."},{"Start":"05:15.245 ","End":"05:18.140","Text":"What we would like is to find the a, n and the b,"},{"Start":"05:18.140 ","End":"05:21.380","Text":"n. I will get these from the boundary conditions,"},{"Start":"05:21.380 ","End":"05:23.450","Text":"the pair that we haven\u0027t used yet,"},{"Start":"05:23.450 ","End":"05:28.655","Text":"that\u0027s the two on the sides, U0Y and ULY."},{"Start":"05:28.655 ","End":"05:30.620","Text":"We\u0027ll start with the left one,"},{"Start":"05:30.620 ","End":"05:34.925","Text":"which is u naught(y) equals cosine pi y over L."},{"Start":"05:34.925 ","End":"05:39.715","Text":"That means that we put x equals 0 in here."},{"Start":"05:39.715 ","End":"05:43.145","Text":"B0, x drops off and also"},{"Start":"05:43.145 ","End":"05:49.940","Text":"the signs all disappear all this because when x is 0 sine hyperbolic is 0."},{"Start":"05:49.940 ","End":"05:53.075","Text":"But when x is 0, the cosine hyperbolic is 1,"},{"Start":"05:53.075 ","End":"05:54.695","Text":"so we\u0027re left with the an."},{"Start":"05:54.695 ","End":"05:58.400","Text":"We just have an together with this cosine,"},{"Start":"05:58.400 ","End":"06:01.880","Text":"and that gives us the following sum."},{"Start":"06:01.880 ","End":"06:04.820","Text":"Now we can compare coefficients."},{"Start":"06:04.820 ","End":"06:09.770","Text":"The only non-zero coefficient in this cosine series is 1,"},{"Start":"06:09.770 ","End":"06:14.660","Text":"which goes with cosine of 1 pi y over l. Only when n equals 1,"},{"Start":"06:14.660 ","End":"06:15.905","Text":"do we get something."},{"Start":"06:15.905 ","End":"06:17.255","Text":"When n equals 1,"},{"Start":"06:17.255 ","End":"06:19.720","Text":"here we get A1."},{"Start":"06:19.720 ","End":"06:23.614","Text":"A1 is 1 and all the other an\u0027s are 0."},{"Start":"06:23.614 ","End":"06:27.590","Text":"Now we can get closer to what u(xy) is by"},{"Start":"06:27.590 ","End":"06:32.750","Text":"substituting the an\u0027s in this equation from the first part,"},{"Start":"06:32.750 ","End":"06:36.654","Text":"a naught is 0, so we just get v naught x."},{"Start":"06:36.654 ","End":"06:38.355","Text":"Then from this part,"},{"Start":"06:38.355 ","End":"06:40.530","Text":"let\u0027s take first the an\u0027s and the bn\u0027s."},{"Start":"06:40.530 ","End":"06:44.030","Text":"The an is only one that\u0027s non-zero, that\u0027s a1."},{"Start":"06:44.030 ","End":"06:48.875","Text":"We get cosine 1 pi over ly,"},{"Start":"06:48.875 ","End":"06:52.775","Text":"cosine hyperbolic 1 pi over lx,"},{"Start":"06:52.775 ","End":"06:55.850","Text":"and then the bn we get them all,"},{"Start":"06:55.850 ","End":"06:58.295","Text":"but they\u0027re just unknowns as bn."},{"Start":"06:58.295 ","End":"07:01.025","Text":"Now we just have to find the bn\u0027s."},{"Start":"07:01.025 ","End":"07:05.640","Text":"For that, we use the other boundary condition,"},{"Start":"07:05.650 ","End":"07:09.980","Text":"u(ly) equals cosine 3 pi y over L. We can"},{"Start":"07:09.980 ","End":"07:13.985","Text":"substitute x equals L here and again compare coefficients."},{"Start":"07:13.985 ","End":"07:17.300","Text":"So we get, on the one hand cosine 3 pi over ly,"},{"Start":"07:17.300 ","End":"07:20.975","Text":"on the other hand, p naught L cosine pi over L,"},{"Start":"07:20.975 ","End":"07:29.540","Text":"y cosine hyperbolic pi over L times L, which is just Pi."},{"Start":"07:29.540 ","End":"07:33.530","Text":"Then bn. Just put x equals l here again,"},{"Start":"07:33.530 ","End":"07:36.665","Text":"L/ l is 1, so this is also n Pi."},{"Start":"07:36.665 ","End":"07:40.114","Text":"Take this term and move it over to the left."},{"Start":"07:40.114 ","End":"07:43.375","Text":"Now we can compare coefficients."},{"Start":"07:43.375 ","End":"07:51.550","Text":"When n equals 3 on this side we have b3 sine hyperbolic 3Pi and when n is 1,"},{"Start":"07:51.550 ","End":"07:57.010","Text":"the coefficient here is b1 sine hyperbolic 1 pi."},{"Start":"07:57.010 ","End":"07:58.570","Text":"On the left-hand side,"},{"Start":"07:58.570 ","End":"07:59.860","Text":"the coefficient is 1,"},{"Start":"07:59.860 ","End":"08:01.135","Text":"so the 1 equals this,"},{"Start":"08:01.135 ","End":"08:06.400","Text":"and here the coefficient is 1 times cosine hyperbolic pi."},{"Start":"08:06.400 ","End":"08:07.975","Text":"That\u0027s equal to this."},{"Start":"08:07.975 ","End":"08:12.250","Text":"Well, the others are 0. B naught is 0 because there\u0027s no free coefficient here."},{"Start":"08:12.250 ","End":"08:13.450","Text":"B naught l is 0,"},{"Start":"08:13.450 ","End":"08:17.690","Text":"so V naught is 0, and all the other bn are also 0."},{"Start":"08:17.690 ","End":"08:21.760","Text":"We get b3 is one over sine hyperbolic 3 pi,"},{"Start":"08:21.760 ","End":"08:27.380","Text":"b1 cosine hyperbolic sine hyperbolic Pi."},{"Start":"08:27.380 ","End":"08:32.885","Text":"The other bn\u0027s is 0. Going back to this expression which we had earlier,"},{"Start":"08:32.885 ","End":"08:35.225","Text":"substitute all the b n\u0027s."},{"Start":"08:35.225 ","End":"08:36.800","Text":"B naught is 0,"},{"Start":"08:36.800 ","End":"08:37.820","Text":"so this drops off."},{"Start":"08:37.820 ","End":"08:39.320","Text":"This doesn\u0027t contain any b,"},{"Start":"08:39.320 ","End":"08:41.275","Text":"so this stays as is."},{"Start":"08:41.275 ","End":"08:45.600","Text":"For this part we get n equals 1 and n equals 3."},{"Start":"08:45.600 ","End":"08:49.160","Text":"This plus n equals 1 gives us this expression,"},{"Start":"08:49.160 ","End":"08:52.265","Text":"and n equals 3 here gives us this expression."},{"Start":"08:52.265 ","End":"08:56.600","Text":"But we have b1 and b3 here and here."},{"Start":"08:56.600 ","End":"08:58.325","Text":"So if we put those in,"},{"Start":"08:58.325 ","End":"09:00.440","Text":"we get the following."},{"Start":"09:00.440 ","End":"09:04.920","Text":"That\u0027s the answer, and we are done."}],"ID":30847},{"Watched":false,"Name":"Exercise 5a","Duration":"4m 50s","ChapterTopicVideoID":29275,"CourseChapterTopicPlaylistID":294439,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.180","Text":"In this exercise, we\u0027re given a boundary value problem and we\u0027ll see why it\u0027s"},{"Start":"00:06.180 ","End":"00:09.690","Text":"anointment boundary value problem in the moment and"},{"Start":"00:09.690 ","End":"00:13.800","Text":"involves a parameter K. K appears over here."},{"Start":"00:13.800 ","End":"00:16.245","Text":"This is the Laplace equation."},{"Start":"00:16.245 ","End":"00:18.885","Text":"This is the rectangle."},{"Start":"00:18.885 ","End":"00:22.665","Text":"These are 4 boundary conditions on"},{"Start":"00:22.665 ","End":"00:27.330","Text":"each of the 4 sides of the rectangle. Three of them are 0."},{"Start":"00:27.330 ","End":"00:30.270","Text":"The one that\u0027s not 0 is the one here."},{"Start":"00:30.270 ","End":"00:32.772","Text":"We\u0027ll just do Part A now,"},{"Start":"00:32.772 ","End":"00:38.625","Text":"but we have to find which value of K makes the necessary condition hold."},{"Start":"00:38.625 ","End":"00:41.795","Text":"We studied this in a previous clip in this section."},{"Start":"00:41.795 ","End":"00:46.240","Text":"That when you have a Laplace equation with Neumann boundary conditions,"},{"Start":"00:46.240 ","End":"00:50.360","Text":"there\u0027s a very particular necessary condition. Here it is."},{"Start":"00:50.360 ","End":"00:55.130","Text":"It\u0027s the integral along the border of"},{"Start":"00:55.130 ","End":"01:01.040","Text":"the directional derivative of u in the direction of the unit outer normal."},{"Start":"01:01.040 ","End":"01:03.875","Text":"This integral has to be 0."},{"Start":"01:03.875 ","End":"01:10.819","Text":"Now Neumann boundary condition is when we\u0027re given du by dn on the boundary,"},{"Start":"01:10.819 ","End":"01:15.656","Text":"but it doesn\u0027t seem that we are given to du by dy or du by dx,"},{"Start":"01:15.656 ","End":"01:18.785","Text":"but it turns out that it\u0027s almost the same thing."},{"Start":"01:18.785 ","End":"01:21.005","Text":"Give or take a plus or minus."},{"Start":"01:21.005 ","End":"01:26.105","Text":"For example, on this lower edge of the rectangle,"},{"Start":"01:26.105 ","End":"01:32.210","Text":"the directional derivative is minus the derivative with respect to y,"},{"Start":"01:32.210 ","End":"01:34.714","Text":"because y increases upwards,"},{"Start":"01:34.714 ","End":"01:36.935","Text":"but this increases downwards."},{"Start":"01:36.935 ","End":"01:41.510","Text":"Here, the directional derivative is the same as du by dx,"},{"Start":"01:41.510 ","End":"01:43.655","Text":"here it\u0027s du by dy,"},{"Start":"01:43.655 ","End":"01:46.420","Text":"and here it\u0027s minus du by dx."},{"Start":"01:46.420 ","End":"01:51.680","Text":"We can break up this integral into 4 separate integrals."},{"Start":"01:51.680 ","End":"01:55.610","Text":"Like I said, for this one we need a minus."},{"Start":"01:55.610 ","End":"01:58.170","Text":"For this, we need a plus, for this,"},{"Start":"01:58.170 ","End":"01:59.280","Text":"we need a plus,"},{"Start":"01:59.280 ","End":"02:03.460","Text":"that\u0027s this one here and here, a minus again."},{"Start":"02:03.590 ","End":"02:08.495","Text":"Now this is equal to 0 and so is this,"},{"Start":"02:08.495 ","End":"02:09.995","Text":"and so is this,"},{"Start":"02:09.995 ","End":"02:14.390","Text":"this 1 not referring to the 4 boundary conditions."},{"Start":"02:14.390 ","End":"02:17.305","Text":"Will get 0 for 3 of them,"},{"Start":"02:17.305 ","End":"02:21.875","Text":"and we\u0027re just left with this integral from 0 to Pi"},{"Start":"02:21.875 ","End":"02:26.810","Text":"of du by dx along this right-hand side."},{"Start":"02:26.810 ","End":"02:30.170","Text":"Going to do it in more detail than we need."},{"Start":"02:30.170 ","End":"02:38.546","Text":"You will actually just replace this by what it\u0027s equal to with over here and ds is dy,"},{"Start":"02:38.546 ","End":"02:41.705","Text":"but let\u0027s do it with parameterization."},{"Start":"02:41.705 ","End":"02:48.455","Text":"We can parameterize the edge from here to here as follows."},{"Start":"02:48.455 ","End":"02:50.405","Text":"X is always 2Pi,"},{"Start":"02:50.405 ","End":"02:55.885","Text":"y is t, and t goes from 0 to Pi."},{"Start":"02:55.885 ","End":"03:00.015","Text":"Dx is 0 because it\u0027s a constant,"},{"Start":"03:00.015 ","End":"03:02.695","Text":"dy is just dt,"},{"Start":"03:02.695 ","End":"03:06.590","Text":"and ds, which is this expression is also dt."},{"Start":"03:06.590 ","End":"03:15.500","Text":"That means that we can get this integral as du by dx is this function."},{"Start":"03:15.500 ","End":"03:23.860","Text":"Then we substitute x=2Pi and y=t."},{"Start":"03:23.860 ","End":"03:26.385","Text":"Well, there is no x here."},{"Start":"03:26.385 ","End":"03:30.695","Text":"Then put t here and here instead of the y."},{"Start":"03:30.695 ","End":"03:35.525","Text":"Now, cosine squared 2y there\u0027s a formula we can use."},{"Start":"03:35.525 ","End":"03:41.340","Text":"It\u0027s 1 plus cosine twice 2t that\u0027s 4t over"},{"Start":"03:41.340 ","End":"03:48.335","Text":"2 plus 2 plus t cosine t. The integral of this is here."},{"Start":"03:48.335 ","End":"03:52.100","Text":"K stays 0.5 becomes 0.5t,"},{"Start":"03:52.100 ","End":"04:01.845","Text":"cosine 4t over 2 becomes 1/8 sine 4t and the k carries to here 2 integral of 2 is 2t,"},{"Start":"04:01.845 ","End":"04:05.355","Text":"integral of 3 cosine t is 3 sine t,"},{"Start":"04:05.355 ","End":"04:09.955","Text":"and all this between the limits of 0 and Pi."},{"Start":"04:09.955 ","End":"04:16.070","Text":"This is 0 because sine of multiples of Pi is 0 and so is this."},{"Start":"04:16.070 ","End":"04:19.200","Text":"We just have these 2 terms."},{"Start":"04:19.630 ","End":"04:24.575","Text":"This 1 becomes 0.5KPi,"},{"Start":"04:24.575 ","End":"04:27.455","Text":"which is Pi over 2K."},{"Start":"04:27.455 ","End":"04:30.170","Text":"This 1 becomes 2Pi."},{"Start":"04:30.170 ","End":"04:35.195","Text":"The question is, under what condition is this equal to 0?"},{"Start":"04:35.195 ","End":"04:39.905","Text":"Well, multiply everything by 2 and divide by Pi,"},{"Start":"04:39.905 ","End":"04:44.440","Text":"and we get that K plus 4 equals 0."},{"Start":"04:44.440 ","End":"04:47.760","Text":"K equals minus 4."},{"Start":"04:47.760 ","End":"04:51.460","Text":"That\u0027s the answer to Part A."}],"ID":30848},{"Watched":false,"Name":"Exercise 5b","Duration":"7m 45s","ChapterTopicVideoID":29276,"CourseChapterTopicPlaylistID":294439,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.065","Text":"We\u0027re continuing the exercise from Part a."},{"Start":"00:04.065 ","End":"00:05.595","Text":"In part b,"},{"Start":"00:05.595 ","End":"00:10.860","Text":"we have to solve the boundary value problem with the value of K from Part"},{"Start":"00:10.860 ","End":"00:17.910","Text":"a and the additional information that u(0, 0) is 10."},{"Start":"00:17.910 ","End":"00:22.440","Text":"In Part a, we got the result that K=minus 4."},{"Start":"00:22.440 ","End":"00:28.785","Text":"Our reminder of the stages for solving this problem,"},{"Start":"00:28.785 ","End":"00:34.845","Text":"Laplace equation on a rectangle using separation of variables method."},{"Start":"00:34.845 ","End":"00:35.990","Text":"We\u0027ve done this before."},{"Start":"00:35.990 ","End":"00:38.060","Text":"I\u0027ll just let you read it if you want,"},{"Start":"00:38.060 ","End":"00:40.820","Text":"skip over it, and let\u0027s actually start."},{"Start":"00:40.820 ","End":"00:43.655","Text":"This is the Laplace equation."},{"Start":"00:43.655 ","End":"00:49.625","Text":"We want to find u of the form some function of x times some function of y."},{"Start":"00:49.625 ","End":"00:53.240","Text":"If we substitute this u in this equation,"},{"Start":"00:53.240 ","End":"00:54.545","Text":"we get the following."},{"Start":"00:54.545 ","End":"00:57.200","Text":"For u_xx, we just differentiate x twice,"},{"Start":"00:57.200 ","End":"00:58.310","Text":"and for u_yy,"},{"Start":"00:58.310 ","End":"01:00.635","Text":"we just differentiate y twice."},{"Start":"01:00.635 ","End":"01:02.720","Text":"Bring this over to the other side."},{"Start":"01:02.720 ","End":"01:07.205","Text":"Now, divide by Y (y) X (x) we get that this equals this."},{"Start":"01:07.205 ","End":"01:10.370","Text":"When a function of X is identically equal to a function of Y,"},{"Start":"01:10.370 ","End":"01:14.125","Text":"then there must be a constant and we call this constant minus Lambda."},{"Start":"01:14.125 ","End":"01:17.090","Text":"If we just look at the Y part,"},{"Start":"01:17.090 ","End":"01:21.785","Text":"then we get the following second-order ODE for y,"},{"Start":"01:21.785 ","End":"01:25.070","Text":"but we need some boundary conditions on that."},{"Start":"01:25.070 ","End":"01:28.535","Text":"We\u0027ll get those from the boundary conditions on u."},{"Start":"01:28.535 ","End":"01:30.785","Text":"If we differentiate this,"},{"Start":"01:30.785 ","End":"01:35.510","Text":"we get that u_y is X(x) Y\u0027(y)."},{"Start":"01:35.510 ","End":"01:37.390","Text":"If we substitute x,"},{"Start":"01:37.390 ","End":"01:39.530","Text":"0 or x, Pi,"},{"Start":"01:39.530 ","End":"01:42.085","Text":"we get 0. This was given."},{"Start":"01:42.085 ","End":"01:48.620","Text":"The first equality that this equals 0 gives us that X(x) Y\u0027(0)."},{"Start":"01:48.620 ","End":"01:50.945","Text":"This one gives us that this is 0."},{"Start":"01:50.945 ","End":"01:58.990","Text":"We can cancel X(x) and get that Y\u0027(0)=Y\u0027(Pi)=0."},{"Start":"01:58.990 ","End":"02:02.150","Text":"This boundary condition together with this equation"},{"Start":"02:02.150 ","End":"02:05.315","Text":"gives us this pair and now we have this problem,"},{"Start":"02:05.315 ","End":"02:08.375","Text":"which is a Sturm–Liouville problem."},{"Start":"02:08.375 ","End":"02:15.680","Text":"We\u0027ve done this problem before and the answer comes out to be the following."},{"Start":"02:15.680 ","End":"02:19.370","Text":"These are the eigenfunctions and eigenvalues,"},{"Start":"02:19.370 ","End":"02:21.895","Text":"n=0,1, 2, 3, etc."},{"Start":"02:21.895 ","End":"02:26.945","Text":"Recall the equation we had this one previous time we used it for Y,"},{"Start":"02:26.945 ","End":"02:33.610","Text":"now we\u0027ll use it for X throughout the minus and we get X\u0027\u0027/X is Lambda,"},{"Start":"02:33.610 ","End":"02:36.005","Text":"but Lambda is n^2."},{"Start":"02:36.005 ","End":"02:38.675","Text":"This is what we have, X is X_n."},{"Start":"02:38.675 ","End":"02:42.575","Text":"Moving this to this side and then bringing the right side to the left side,"},{"Start":"02:42.575 ","End":"02:44.060","Text":"we get the following."},{"Start":"02:44.060 ","End":"02:45.454","Text":"There are 2 cases."},{"Start":"02:45.454 ","End":"02:48.410","Text":"If n is not 0, then the general solution is"},{"Start":"02:48.410 ","End":"02:52.490","Text":"a linear combination of cosine hyperbolic nx and sine hyperbolic nx."},{"Start":"02:52.490 ","End":"02:58.655","Text":"If n is 0, we just get the solution to X\u0027\u0027=0."},{"Start":"02:58.655 ","End":"03:01.805","Text":"It\u0027s a linear polynomial in x,"},{"Start":"03:01.805 ","End":"03:04.225","Text":"a naught plus b naught x."},{"Start":"03:04.225 ","End":"03:09.245","Text":"Then u(x, y) is an infinite linear combination of these,"},{"Start":"03:09.245 ","End":"03:11.000","Text":"X_n times Y_ n,"},{"Start":"03:11.000 ","End":"03:12.200","Text":"but X naught,"},{"Start":"03:12.200 ","End":"03:15.170","Text":"Y naught separately because it\u0027s a different formula."},{"Start":"03:15.170 ","End":"03:19.625","Text":"I forgot to say to bring forward the Y_n that we had previously."},{"Start":"03:19.625 ","End":"03:25.565","Text":"When n is 0, this gives us Y_0(y) is cosine 0, which is 1."},{"Start":"03:25.565 ","End":"03:28.325","Text":"Now we put this Y naught of y=1,"},{"Start":"03:28.325 ","End":"03:35.805","Text":"Y_n is cosine n y and the X_n and X naught are here and here."},{"Start":"03:35.805 ","End":"03:39.340","Text":"This is what we have now for u(x, y)."},{"Start":"03:39.340 ","End":"03:42.300","Text":"We need to find the coefficients a_n, b_n."},{"Start":"03:42.300 ","End":"03:46.680","Text":"U(x, y) is a naught plus b naught x times 1."},{"Start":"03:46.680 ","End":"03:53.495","Text":"That\u0027s this times this plus the sum of this times this."},{"Start":"03:53.495 ","End":"03:57.510","Text":"We need u_x, so differentiate with respect to x."},{"Start":"03:57.510 ","End":"03:59.345","Text":"This is what we get."},{"Start":"03:59.345 ","End":"04:03.710","Text":"Now we have the boundary condition u_x (0, y)= 0."},{"Start":"04:03.710 ","End":"04:09.020","Text":"Also, we\u0027ll use these so that when x is 0,"},{"Start":"04:09.020 ","End":"04:11.405","Text":"this comes out to be 0,"},{"Start":"04:11.405 ","End":"04:14.075","Text":"this comes out to be 1."},{"Start":"04:14.075 ","End":"04:15.755","Text":"From all these brackets,"},{"Start":"04:15.755 ","End":"04:19.910","Text":"we just get nb_n and then times the cosine ny."},{"Start":"04:19.910 ","End":"04:24.050","Text":"If we compare the coefficients of this cosine series,"},{"Start":"04:24.050 ","End":"04:27.515","Text":"we get that b naught is 0."},{"Start":"04:27.515 ","End":"04:30.425","Text":"Also, nb_n is 0,"},{"Start":"04:30.425 ","End":"04:32.180","Text":"but here n is not 0,"},{"Start":"04:32.180 ","End":"04:33.290","Text":"so we can divide by it,"},{"Start":"04:33.290 ","End":"04:34.715","Text":"so b_n is 0."},{"Start":"04:34.715 ","End":"04:36.770","Text":"In short b_n is 0 for all the n,"},{"Start":"04:36.770 ","End":"04:38.495","Text":"0, 1, 2, 3, etc."},{"Start":"04:38.495 ","End":"04:43.385","Text":"Now, if we put all the b_n\u0027s to be equal to 0 in this equation,"},{"Start":"04:43.385 ","End":"04:45.799","Text":"what we get is the following."},{"Start":"04:45.799 ","End":"04:51.065","Text":"I move the cosine ny in front of the cosine hyperbolic nx."},{"Start":"04:51.065 ","End":"04:54.065","Text":"Now differentiating this with respect to x,"},{"Start":"04:54.065 ","End":"04:56.335","Text":"we get the following."},{"Start":"04:56.335 ","End":"05:00.675","Text":"We need this for this boundary condition."},{"Start":"05:00.675 ","End":"05:04.600","Text":"If we substitute x=2PI,"},{"Start":"05:04.600 ","End":"05:10.320","Text":"on the right-hand side here we change this x to 2Pi, we get this."},{"Start":"05:10.320 ","End":"05:12.910","Text":"I colored the cosine ny,"},{"Start":"05:12.910 ","End":"05:14.815","Text":"so it\u0027s easier, you\u0027ll see."},{"Start":"05:14.815 ","End":"05:16.720","Text":"The right-hand side we can simplify,"},{"Start":"05:16.720 ","End":"05:18.805","Text":"use a bit of trigonometry."},{"Start":"05:18.805 ","End":"05:23.420","Text":"Cosine squared 2y is 1 plus cosine 4y/2."},{"Start":"05:23.420 ","End":"05:27.670","Text":"A bit of simplification gives us the following."},{"Start":"05:27.670 ","End":"05:29.425","Text":"Now you see why I colored them."},{"Start":"05:29.425 ","End":"05:35.045","Text":"I want to compare cosine ny ones with cosine 4y and ones with cosine 1y."},{"Start":"05:35.045 ","End":"05:38.175","Text":"Comparing the coefficient of cosine 4y,"},{"Start":"05:38.175 ","End":"05:41.508","Text":"we get 4a_4,"},{"Start":"05:41.508 ","End":"05:46.680","Text":"sine hyperbolic 2 times 4 times Pi, which is 8Pi."},{"Start":"05:46.680 ","End":"05:53.220","Text":"If we compare the cosine 1y here with n=1, we get 1,"},{"Start":"05:53.220 ","End":"05:57.120","Text":"a1 sine hyperbolic 2Pi,"},{"Start":"05:57.120 ","End":"06:00.360","Text":"and that\u0027s equal to 3."},{"Start":"06:00.360 ","End":"06:03.770","Text":"For all the other n\u0027s except 1 and 4,"},{"Start":"06:03.770 ","End":"06:05.555","Text":"we get the following is 0,"},{"Start":"06:05.555 ","End":"06:08.660","Text":"which actually gives us that a n is 0,"},{"Start":"06:08.660 ","End":"06:10.820","Text":"because sine hyperbolic is not 0,"},{"Start":"06:10.820 ","End":"06:14.455","Text":"it\u0027s only 0 when the argument is 0."},{"Start":"06:14.455 ","End":"06:20.450","Text":"We get that a_4 is this over this,"},{"Start":"06:20.450 ","End":"06:23.390","Text":"so it\u0027s minus a half, just as written,"},{"Start":"06:23.390 ","End":"06:28.785","Text":"and a_n is 0 if n is not 1 or 4."},{"Start":"06:28.785 ","End":"06:33.210","Text":"But note that this sum is from 1 to infinity,"},{"Start":"06:33.210 ","End":"06:34.940","Text":"a naught doesn\u0027t appear here,"},{"Start":"06:34.940 ","End":"06:38.525","Text":"so we have no extra information about a naught."},{"Start":"06:38.525 ","End":"06:41.450","Text":"In this equation for u(x,"},{"Start":"06:41.450 ","End":"06:45.470","Text":"y) we can substitute all the a_n\u0027s from 1 to infinity,"},{"Start":"06:45.470 ","End":"06:46.880","Text":"but a naught has to stay."},{"Start":"06:46.880 ","End":"06:48.420","Text":"We don\u0027t know anything about it."},{"Start":"06:48.420 ","End":"06:52.910","Text":"We get a naught, then the term where n=1 we get this,"},{"Start":"06:52.910 ","End":"06:54.530","Text":"where n=4 we get this."},{"Start":"06:54.530 ","End":"06:56.690","Text":"The others are 0, so they don\u0027t appear."},{"Start":"06:56.690 ","End":"07:00.950","Text":"We can replace now a_1 and a_4 with what they are from here,"},{"Start":"07:00.950 ","End":"07:02.315","Text":"and we get this."},{"Start":"07:02.315 ","End":"07:08.130","Text":"Now we\u0027re going to use the extra information that u naught equals 10."},{"Start":"07:08.130 ","End":"07:13.325","Text":"We get a naught plus cosine(0) and cosine hyperbolic of 0,"},{"Start":"07:13.325 ","End":"07:14.390","Text":"are both of them 1,"},{"Start":"07:14.390 ","End":"07:17.095","Text":"so this is 1 and this is 1."},{"Start":"07:17.095 ","End":"07:20.115","Text":"These just disappear and we get the following."},{"Start":"07:20.115 ","End":"07:22.455","Text":"We get that 10 equals this."},{"Start":"07:22.455 ","End":"07:28.400","Text":"That gives us the a naught is 10 minus this plus this or plus this minus this."},{"Start":"07:28.400 ","End":"07:30.140","Text":"Now, slightly rewriting this,"},{"Start":"07:30.140 ","End":"07:33.920","Text":"putting this in the numerator and this in the numerator, we get this."},{"Start":"07:33.920 ","End":"07:37.670","Text":"Now, just put a naught as to what it\u0027s equal from"},{"Start":"07:37.670 ","End":"07:41.970","Text":"here and we get the answer to the problem u(x,"},{"Start":"07:41.970 ","End":"07:45.420","Text":"y) is this, and we are done."}],"ID":30849}],"Thumbnail":null,"ID":294439},{"Name":"The Mean Value Property","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - The Mean Value Property","Duration":"2m 49s","ChapterTopicVideoID":29243,"CourseChapterTopicPlaylistID":294440,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.085","Text":"We come to a new topic."},{"Start":"00:02.085 ","End":"00:05.925","Text":"The mean value property of harmonic functions."},{"Start":"00:05.925 ","End":"00:10.500","Text":"In case you don\u0027t remember what harmonic functions are we\u0027ll define them again."},{"Start":"00:10.500 ","End":"00:14.490","Text":"A real function of 2 variables is called harmonic."},{"Start":"00:14.490 ","End":"00:17.610","Text":"If it\u0027s twice continuously differentiable,"},{"Start":"00:17.610 ","End":"00:20.645","Text":"meaning it has continuous second derivatives."},{"Start":"00:20.645 ","End":"00:23.860","Text":"It satisfies Laplace\u0027s equation,"},{"Start":"00:23.860 ","End":"00:25.710","Text":"which can be written like this,"},{"Start":"00:25.710 ","End":"00:31.490","Text":"or u_xx plus u_yy=0 is also a polar definition."},{"Start":"00:31.490 ","End":"00:35.959","Text":"The mean value property of harmonic functions is as follows."},{"Start":"00:35.959 ","End":"00:38.180","Text":"Suppose we have a disk D,"},{"Start":"00:38.180 ","End":"00:44.550","Text":"which has radius R centered at some point x-naught, y-naught."},{"Start":"00:44.550 ","End":"00:51.470","Text":"Suppose that u is harmonic in the open disk D and continuous in the closed disk,"},{"Start":"00:51.470 ","End":"00:55.835","Text":"and is continuous in the open disk and on the boundary."},{"Start":"00:55.835 ","End":"00:59.540","Text":"Then we have the following equality."},{"Start":"00:59.540 ","End":"01:06.395","Text":"The value of u at the center of the circle is equal to the following."},{"Start":"01:06.395 ","End":"01:12.700","Text":"This is the x and the y are parameterized by an angle Theta."},{"Start":"01:12.700 ","End":"01:17.630","Text":"It\u0027s the points along this circle of radius rho."},{"Start":"01:17.630 ","End":"01:21.200","Text":"We take the value of u along these points,"},{"Start":"01:21.200 ","End":"01:24.650","Text":"integrate, and then divide by 2 Pi."},{"Start":"01:24.650 ","End":"01:29.420","Text":"What this gives us is the average value this, by the way,"},{"Start":"01:29.420 ","End":"01:33.800","Text":"is true for all rho between 0 and R,"},{"Start":"01:33.800 ","End":"01:35.090","Text":"like in the picture."},{"Start":"01:35.090 ","End":"01:37.840","Text":"This property is called the mean value property,"},{"Start":"01:37.840 ","End":"01:39.429","Text":"and it means that,"},{"Start":"01:39.429 ","End":"01:43.880","Text":"the value of u at the center is the average of the values of"},{"Start":"01:43.880 ","End":"01:48.500","Text":"view on the circle centered at this point with radius rho,"},{"Start":"01:48.500 ","End":"01:53.315","Text":"where rho can be anywhere from 0 to r. What this means is that"},{"Start":"01:53.315 ","End":"01:58.700","Text":"if we know the value of a harmonic function you along a circle,"},{"Start":"01:58.700 ","End":"02:01.940","Text":"then we can compute the value at the center using"},{"Start":"02:01.940 ","End":"02:05.780","Text":"this formula without solving the Laplace equation."},{"Start":"02:05.780 ","End":"02:07.040","Text":"I mean it\u0027s one way to do it, solve"},{"Start":"02:07.040 ","End":"02:09.650","Text":"the Laplace equation and substitute x naught, y naught."},{"Start":"02:09.650 ","End":"02:11.845","Text":"But we don\u0027t need to do all that."},{"Start":"02:11.845 ","End":"02:17.610","Text":"This will be shown in the exercises that follow this tutorial."},{"Start":"02:17.610 ","End":"02:23.030","Text":"The remark that instead of taking the average along the circle,"},{"Start":"02:23.030 ","End":"02:28.550","Text":"we can take the average along the disk to two-dimensional average."},{"Start":"02:28.550 ","End":"02:32.960","Text":"If D is the disk centered at x-naught, y-naught,"},{"Start":"02:32.960 ","End":"02:37.070","Text":"then you at the center is 1 over the area of"},{"Start":"02:37.070 ","End":"02:42.125","Text":"the disk times the integral of u on the disk."},{"Start":"02:42.125 ","End":"02:44.360","Text":"But we won\u0027t use this version."},{"Start":"02:44.360 ","End":"02:47.225","Text":"We\u0027ll just use the average of the circle."},{"Start":"02:47.225 ","End":"02:49.890","Text":"That concludes this clip."}],"ID":30850},{"Watched":false,"Name":"Exercise 1","Duration":"2m 42s","ChapterTopicVideoID":29244,"CourseChapterTopicPlaylistID":294440,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.630","Text":"This exercise, we\u0027re given a boundary value problem consisting of a Laplace equation"},{"Start":"00:06.630 ","End":"00:10.020","Text":"on a disk and its Direchlet because we\u0027re given"},{"Start":"00:10.020 ","End":"00:14.415","Text":"the value of the function along the boundary of the disk, which is the circle."},{"Start":"00:14.415 ","End":"00:20.835","Text":"We have to compute u(0,0) without actually solving the boundary value problem."},{"Start":"00:20.835 ","End":"00:26.460","Text":"The way we do this is by using the mean value property of this function."},{"Start":"00:26.460 ","End":"00:29.775","Text":"If it satisfies the Laplace\u0027s equation,"},{"Start":"00:29.775 ","End":"00:31.860","Text":"then it\u0027s a harmonic function,"},{"Start":"00:31.860 ","End":"00:35.550","Text":"and so the value at a point is the average of"},{"Start":"00:35.550 ","End":"00:40.125","Text":"the value along a circle at which the point is centered."},{"Start":"00:40.125 ","End":"00:41.835","Text":"In our case,"},{"Start":"00:41.835 ","End":"00:45.560","Text":"(0,0) is the center of a circle of radius 2,"},{"Start":"00:45.560 ","End":"00:47.225","Text":"row is 2,"},{"Start":"00:47.225 ","End":"00:51.205","Text":"and x_naught and y_naught are both 0."},{"Start":"00:51.205 ","End":"00:55.890","Text":"We get u of 0,"},{"Start":"00:55.890 ","End":"01:00.500","Text":"row is 2, so 2 cosine Theta and here 2 sine Theta, d Theta."},{"Start":"01:00.500 ","End":"01:04.820","Text":"Now the function u(x,y) is given as absolute value of"},{"Start":"01:04.820 ","End":"01:10.235","Text":"x on the circle and these points are all on the circle of radius 2."},{"Start":"01:10.235 ","End":"01:14.180","Text":"This is equal to the integral of absolute value of x,"},{"Start":"01:14.180 ","End":"01:17.470","Text":"which is absolute value of 2 cosine Theta, d Theta."},{"Start":"01:17.470 ","End":"01:20.120","Text":"We can take the 2 out of the absolute value and"},{"Start":"01:20.120 ","End":"01:23.080","Text":"out of the integral and get 1 over Pi here."},{"Start":"01:23.080 ","End":"01:29.180","Text":"The absolute value of cosine Theta will either be cosine Theta or minus cosine Theta,"},{"Start":"01:29.180 ","End":"01:31.715","Text":"depending if x is positive,"},{"Start":"01:31.715 ","End":"01:39.135","Text":"which means that we\u0027re in the 1st quadrant or the 4th quadrant as we go around,"},{"Start":"01:39.135 ","End":"01:40.485","Text":"it\u0027s x is positive,"},{"Start":"01:40.485 ","End":"01:45.175","Text":"and when we\u0027re in the 2nd and 3rd quadrant then x is negative."},{"Start":"01:45.175 ","End":"01:48.500","Text":"What we can do is split this integral up into 3."},{"Start":"01:48.500 ","End":"01:49.925","Text":"From here to here,"},{"Start":"01:49.925 ","End":"01:53.555","Text":"absolute value of cosine Theta is cosine Theta."},{"Start":"01:53.555 ","End":"01:55.775","Text":"From here to here,"},{"Start":"01:55.775 ","End":"01:58.925","Text":"absolute value is minus cosine theta."},{"Start":"01:58.925 ","End":"02:01.480","Text":"That\u0027s where this minus comes in."},{"Start":"02:01.480 ","End":"02:05.100","Text":"Then here it\u0027s also equal to cosine Theta."},{"Start":"02:05.100 ","End":"02:07.290","Text":"We\u0027ve gotten rid of the absolute value now."},{"Start":"02:07.290 ","End":"02:09.245","Text":"Now it\u0027s a straightforward integral."},{"Start":"02:09.245 ","End":"02:12.020","Text":"The integral of cosine is sine."},{"Start":"02:12.020 ","End":"02:15.295","Text":"We have sine between these 2 limits,"},{"Start":"02:15.295 ","End":"02:17.940","Text":"minus sine between these 2 limits,"},{"Start":"02:17.940 ","End":"02:21.240","Text":"plus sine between these 2 limits."},{"Start":"02:21.240 ","End":"02:26.255","Text":"Just substituting sine of Pi over 2 is 1, etc."},{"Start":"02:26.255 ","End":"02:27.995","Text":"We get the following,"},{"Start":"02:27.995 ","End":"02:29.590","Text":"which comes out to be,"},{"Start":"02:29.590 ","End":"02:34.635","Text":"this is 1 minus minus 2 is plus 2 plus 1 over Pi."},{"Start":"02:34.635 ","End":"02:38.205","Text":"1 plus 2 plus 1 is 4,"},{"Start":"02:38.205 ","End":"02:40.530","Text":"so the answer is 4 over Pi,"},{"Start":"02:40.530 ","End":"02:43.240","Text":"and that concludes this exercise."}],"ID":30851},{"Watched":false,"Name":"Exercise 2","Duration":"2m 8s","ChapterTopicVideoID":29245,"CourseChapterTopicPlaylistID":294440,"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":"In this exercise, we\u0027re given that u(x,"},{"Start":"00:02.925 ","End":"00:07.665","Text":"y) satisfies the following boundary value problem."},{"Start":"00:07.665 ","End":"00:14.265","Text":"It\u0027s a Laplace PDE with Dirichlet boundary conditions."},{"Start":"00:14.265 ","End":"00:20.010","Text":"The domain is the disk of radius 2, 4 is 2^2."},{"Start":"00:20.010 ","End":"00:25.560","Text":"We have to compute the value of u at 0,0 without actually solving the BVP."},{"Start":"00:25.560 ","End":"00:32.115","Text":"What we\u0027ll use is the formula for the mean value property of harmonic functions."},{"Start":"00:32.115 ","End":"00:33.315","Text":"u is harmonic,"},{"Start":"00:33.315 ","End":"00:36.570","Text":"and this is what the mean value property says."},{"Start":"00:36.570 ","End":"00:42.285","Text":"The value at a point is equal to the average of values in a circle around the point."},{"Start":"00:42.285 ","End":"00:43.560","Text":"In our case x_naught,"},{"Start":"00:43.560 ","End":"00:49.720","Text":"y_naught is the origin and the radius is 2 because this has to be Rho^2."},{"Start":"00:49.720 ","End":"00:53.585","Text":"Replacing x_naught, y_naught and Rho here,"},{"Start":"00:53.585 ","End":"00:55.880","Text":"we get the following integral."},{"Start":"00:55.880 ","End":"01:00.395","Text":"Now, u(x) and y is y plus y^2,"},{"Start":"01:00.395 ","End":"01:03.635","Text":"so it\u0027s going to be this plus this squared,"},{"Start":"01:03.635 ","End":"01:05.440","Text":"which is the following."},{"Start":"01:05.440 ","End":"01:08.855","Text":"2 sine Theta, 2 sine Theta^2."},{"Start":"01:08.855 ","End":"01:11.730","Text":"Now it\u0027s just a straightforward integral."},{"Start":"01:11.730 ","End":"01:14.375","Text":"We use some trigonometry."},{"Start":"01:14.375 ","End":"01:18.545","Text":"We know that sine^2 Theta is 1 minus cosine 2 Theta over 2."},{"Start":"01:18.545 ","End":"01:21.710","Text":"This 2 goes into this 4 twice."},{"Start":"01:21.710 ","End":"01:28.610","Text":"What we can do now is cancel the 2 here with the 2 here and the 2 here,"},{"Start":"01:28.610 ","End":"01:35.470","Text":"then we get just 1 over Pi and then sine Theta plus 1 minus cosine 2 Theta."},{"Start":"01:35.470 ","End":"01:37.005","Text":"Straightforward integral."},{"Start":"01:37.005 ","End":"01:38.430","Text":"Integral of sine is cosine,"},{"Start":"01:38.430 ","End":"01:39.840","Text":"integral of 1 is Theta,"},{"Start":"01:39.840 ","End":"01:43.650","Text":"integral of cosine 2 Theta is sine 2 Theta over 2."},{"Start":"01:43.650 ","End":"01:46.455","Text":"We want all this between 0 and 2Pi."},{"Start":"01:46.455 ","End":"01:49.890","Text":"The first one, substitute 0 and 2Pi,"},{"Start":"01:49.890 ","End":"01:52.155","Text":"we get 1 minus 1."},{"Start":"01:52.155 ","End":"01:54.885","Text":"Here 2Pi minus 0."},{"Start":"01:54.885 ","End":"01:59.280","Text":"Well, sine 4Pi and sine 0 are both 0."},{"Start":"01:59.280 ","End":"02:01.095","Text":"Anyway, we get this,"},{"Start":"02:01.095 ","End":"02:02.360","Text":"1 minus 1 cancels,"},{"Start":"02:02.360 ","End":"02:03.590","Text":"0 minus 0 cancels."},{"Start":"02:03.590 ","End":"02:07.070","Text":"We\u0027re just left with 2Pi, 2Pi over Pi is 2,"},{"Start":"02:07.070 ","End":"02:09.300","Text":"and that\u0027s the answer."}],"ID":30852},{"Watched":false,"Name":"Exercise 3","Duration":"1m 42s","ChapterTopicVideoID":29246,"CourseChapterTopicPlaylistID":294440,"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\u0027re given that u satisfies the following boundary value problem,"},{"Start":"00:06.045 ","End":"00:11.880","Text":"it consists of a Laplace equation with a Dirichlet boundary condition."},{"Start":"00:11.880 ","End":"00:18.720","Text":"Our task is to compute the value of u at the origin without actually solving the BVP."},{"Start":"00:18.720 ","End":"00:24.285","Text":"We\u0027re going to do this by noting that if u satisfies a Laplace equation,"},{"Start":"00:24.285 ","End":"00:26.970","Text":"then u is harmonic and if it\u0027s harmonic,"},{"Start":"00:26.970 ","End":"00:29.325","Text":"it has the mean value property,"},{"Start":"00:29.325 ","End":"00:32.355","Text":"which is given by the following formula."},{"Start":"00:32.355 ","End":"00:33.930","Text":"In our case, x naught,"},{"Start":"00:33.930 ","End":"00:37.410","Text":"y naught will be 0,0 and rho,"},{"Start":"00:37.410 ","End":"00:39.330","Text":"which is the radius of the circle,"},{"Start":"00:39.330 ","End":"00:42.605","Text":"is going to be 3 because 9 is 3^2."},{"Start":"00:42.605 ","End":"00:44.300","Text":"If we do that,"},{"Start":"00:44.300 ","End":"00:47.090","Text":"instead of x-naught plus rho cosine Theta,"},{"Start":"00:47.090 ","End":"00:50.795","Text":"we\u0027ll get just 3 cosine Theta and here 3 sine Theta."},{"Start":"00:50.795 ","End":"00:53.293","Text":"All we have to do is compute this integral."},{"Start":"00:53.293 ","End":"00:56.360","Text":"But first let\u0027s replace u by what the function is."},{"Start":"00:56.360 ","End":"00:57.880","Text":"It\u0027s x^2 + x + y,"},{"Start":"00:57.880 ","End":"01:03.110","Text":"so here we get this squared plus this, plus this."},{"Start":"01:03.110 ","End":"01:08.300","Text":"We can take the coefficients in front of the integral and submit"},{"Start":"01:08.300 ","End":"01:13.285","Text":"this up into 3 so we get 9 over 2 Pi the integral of cosine squared,"},{"Start":"01:13.285 ","End":"01:18.349","Text":"3 over 2 Pi the integral of cosine 3 over 2 Pi integral of sine."},{"Start":"01:18.349 ","End":"01:21.913","Text":"Now these integrals, we know are 0."},{"Start":"01:21.913 ","End":"01:24.815","Text":"The integral of cosine squared is a simple problem,"},{"Start":"01:24.815 ","End":"01:27.920","Text":"turns out to be Pi and you could do it if you use"},{"Start":"01:27.920 ","End":"01:32.690","Text":"this trigonometric identity and if this is equal to Pi,"},{"Start":"01:32.690 ","End":"01:37.170","Text":"then 9 over 2 Pi times Pi is 9 over 2 and"},{"Start":"01:37.170 ","End":"01:42.930","Text":"the others are 0 so this is the answer and that completes this exercise."}],"ID":30853},{"Watched":false,"Name":"Exercise 4","Duration":"1m 42s","ChapterTopicVideoID":29247,"CourseChapterTopicPlaylistID":294440,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.760","Text":"In this exercise, we\u0027re given that u satisfies the following boundary value problem."},{"Start":"00:05.760 ","End":"00:11.205","Text":"It consists of a Laplace equation and the Dirichlet boundary condition."},{"Start":"00:11.205 ","End":"00:14.745","Text":"In this case, u( x,y) equals e^x."},{"Start":"00:14.745 ","End":"00:18.540","Text":"All this is on the unit disk, the unit circle."},{"Start":"00:18.540 ","End":"00:24.495","Text":"Our task is to prove that u(0,0) is bigger or equal to 1 over e,"},{"Start":"00:24.495 ","End":"00:27.380","Text":"and we\u0027re not going to do this by solving the equation,"},{"Start":"00:27.380 ","End":"00:31.519","Text":"we\u0027re going to use the mean value property of harmonic functions."},{"Start":"00:31.519 ","End":"00:34.430","Text":"Since u satisfies the Laplace equation,"},{"Start":"00:34.430 ","End":"00:35.695","Text":"u is harmonic,"},{"Start":"00:35.695 ","End":"00:36.950","Text":"and if u is harmonic,"},{"Start":"00:36.950 ","End":"00:39.410","Text":"it satisfies the mean value property,"},{"Start":"00:39.410 ","End":"00:41.435","Text":"which in general is this."},{"Start":"00:41.435 ","End":"00:44.580","Text":"But in our case, x_naught,"},{"Start":"00:44.580 ","End":"00:51.360","Text":"y_naught is 0,0 and Rho is 1 because 1 is 1^2."},{"Start":"00:51.360 ","End":"00:54.000","Text":"What we get is the following,"},{"Start":"00:54.000 ","End":"00:57.645","Text":"and then we know that u is e^x,"},{"Start":"00:57.645 ","End":"01:00.004","Text":"so we get e^cosine Theta."},{"Start":"01:00.004 ","End":"01:05.630","Text":"Now, the cosine of Theta is always bigger or equal to minus 1."},{"Start":"01:05.630 ","End":"01:07.805","Text":"It\u0027s between 1 and minus 1."},{"Start":"01:07.805 ","End":"01:10.670","Text":"If this is bigger or equal to minus 1,"},{"Start":"01:10.670 ","End":"01:14.090","Text":"because the exponential is an increasing function,"},{"Start":"01:14.090 ","End":"01:16.670","Text":"e^cosine Theta is bigger or equal to"},{"Start":"01:16.670 ","End":"01:20.360","Text":"e^minus 1 and the bigger or equal to is preserved by the integral,"},{"Start":"01:20.360 ","End":"01:22.685","Text":"so we get this bigger or equal to."},{"Start":"01:22.685 ","End":"01:26.600","Text":"This integral is equal to 2Pi,"},{"Start":"01:26.600 ","End":"01:30.635","Text":"which is the length of the interval times the constant 2Pi e^minus 1."},{"Start":"01:30.635 ","End":"01:32.780","Text":"The 2Pi cancels with the 2Pi,"},{"Start":"01:32.780 ","End":"01:36.220","Text":"and all we\u0027re left with is the e^minus 1,"},{"Start":"01:36.220 ","End":"01:38.310","Text":"which is 1 over e,"},{"Start":"01:38.310 ","End":"01:40.170","Text":"and that\u0027s what we had to prove,"},{"Start":"01:40.170 ","End":"01:43.000","Text":"and so we are done."}],"ID":30854},{"Watched":false,"Name":"Exercise 5","Duration":"2m 3s","ChapterTopicVideoID":29248,"CourseChapterTopicPlaylistID":294440,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.395","Text":"In this exercise, we have 2 functions, u and v,"},{"Start":"00:04.395 ","End":"00:09.165","Text":"and both of them satisfy a BVP as given."},{"Start":"00:09.165 ","End":"00:12.030","Text":"They\u0027re both on the unit disk."},{"Start":"00:12.030 ","End":"00:15.345","Text":"They both satisfy the Laplace\u0027s equation,"},{"Start":"00:15.345 ","End":"00:20.055","Text":"but the difference is that the Dirichlet boundary condition is different."},{"Start":"00:20.055 ","End":"00:24.690","Text":"For u it\u0027s equal to x^2 -2xy and for v it\u0027s equal to -y^2."},{"Start":"00:24.690 ","End":"00:29.520","Text":"Our task is to prove that u is bigger than v at the origin."},{"Start":"00:29.520 ","End":"00:31.260","Text":"A common thing to do when we\u0027re comparing"},{"Start":"00:31.260 ","End":"00:34.015","Text":"2 different functions is to consider their difference."},{"Start":"00:34.015 ","End":"00:42.425","Text":"Let w be defined as u minus v. W also satisfies the Laplace\u0027s equation on the unit disk."},{"Start":"00:42.425 ","End":"00:45.560","Text":"The boundary condition will be simply the difference."},{"Start":"00:45.560 ","End":"00:50.300","Text":"This minus this, which is x ^2 - 2xy+y^2,"},{"Start":"00:50.300 ","End":"00:53.035","Text":"which is (x-y )^2."},{"Start":"00:53.035 ","End":"00:55.700","Text":"Because w satisfies the Laplace\u0027s equation,"},{"Start":"00:55.700 ","End":"01:00.575","Text":"it\u0027s harmonic and therefore has the mean value property, which is this."},{"Start":"01:00.575 ","End":"01:04.970","Text":"In our case, x-naught y-naught is the origin (0,"},{"Start":"01:04.970 ","End":"01:06.540","Text":"0) and Rho,"},{"Start":"01:06.540 ","End":"01:08.880","Text":"the radius is equal to 1."},{"Start":"01:08.880 ","End":"01:11.105","Text":"If we substitute that in here,"},{"Start":"01:11.105 ","End":"01:13.430","Text":"we get the following."},{"Start":"01:13.430 ","End":"01:16.040","Text":"In our case it\u0027s w and not u,"},{"Start":"01:16.040 ","End":"01:18.080","Text":"but it\u0027s a just letter."},{"Start":"01:18.080 ","End":"01:21.050","Text":"Now we have the definition of w here."},{"Start":"01:21.050 ","End":"01:22.910","Text":"Just substitute cosine Theta,"},{"Start":"01:22.910 ","End":"01:26.630","Text":"sine Theta for x and y and we get this."},{"Start":"01:26.630 ","End":"01:32.105","Text":"Note that the integrand is always bigger or equal to 0 because it\u0027s something squared."},{"Start":"01:32.105 ","End":"01:34.670","Text":"But at sometimes strictly bigger than 0 because"},{"Start":"01:34.670 ","End":"01:37.955","Text":"cosine Theta is not identically equal to sine Theta."},{"Start":"01:37.955 ","End":"01:41.330","Text":"If we take the integral of something that\u0027s"},{"Start":"01:41.330 ","End":"01:45.750","Text":"continuous and sometimes bigger than 0 and always bigger or equal to 0,"},{"Start":"01:45.750 ","End":"01:48.890","Text":"this has to be strictly bigger than 0,"},{"Start":"01:48.890 ","End":"01:51.774","Text":"which gives us that u(0,"},{"Start":"01:51.774 ","End":"01:53.905","Text":"0) is bigger than v(0, 0)."},{"Start":"01:53.905 ","End":"01:57.000","Text":"If w is bigger than 0 and w is u minus v,"},{"Start":"01:57.000 ","End":"02:03.580","Text":"then u is bigger than v. That\u0027s what we had to prove and that concludes this exercise."}],"ID":30855},{"Watched":false,"Name":"Exercise 6","Duration":"2m 25s","ChapterTopicVideoID":29249,"CourseChapterTopicPlaylistID":294440,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.690","Text":"In this exercise, we\u0027re given that the function u"},{"Start":"00:03.690 ","End":"00:07.140","Text":"satisfies the following boundary value problem."},{"Start":"00:07.140 ","End":"00:10.245","Text":"This is a Poisson equation."},{"Start":"00:10.245 ","End":"00:14.880","Text":"It\u0027s not Laplace because there\u0027s no 0 here,"},{"Start":"00:14.880 ","End":"00:17.505","Text":"and this is a Dirichlet boundary condition."},{"Start":"00:17.505 ","End":"00:21.810","Text":"This is on the unit disk whose boundaries the unit circle."},{"Start":"00:21.810 ","End":"00:25.709","Text":"We have to prove that at the origin u is equal to 0."},{"Start":"00:25.709 ","End":"00:27.780","Text":"U is not harmonic."},{"Start":"00:27.780 ","End":"00:31.515","Text":"We can\u0027t use the mean value property for u."},{"Start":"00:31.515 ","End":"00:33.935","Text":"But we can try and fix it somehow."},{"Start":"00:33.935 ","End":"00:40.445","Text":"We\u0027ll look for a function v such that the Laplacian of v is equal to 2."},{"Start":"00:40.445 ","End":"00:43.175","Text":"One possibility for v is x^2."},{"Start":"00:43.175 ","End":"00:47.930","Text":"The second derivative with respect to x is 2 and with respect to y,"},{"Start":"00:47.930 ","End":"00:49.100","Text":"it\u0027s equal to 0,"},{"Start":"00:49.100 ","End":"00:50.495","Text":"so 2 plus 0 is 2."},{"Start":"00:50.495 ","End":"00:54.065","Text":"You could also take y^2, other possibilities."},{"Start":"00:54.065 ","End":"00:58.030","Text":"Now we define w to be equal to u minus v,"},{"Start":"00:58.030 ","End":"01:05.795","Text":"and w will satisfy the Laplacian is equal to 0 because it\u0027s this 2 minus this 2,"},{"Start":"01:05.795 ","End":"01:07.805","Text":"and w of x, y,"},{"Start":"01:07.805 ","End":"01:16.260","Text":"the boundary value will equal x^2 plus xy minus the x^2 from here."},{"Start":"01:16.260 ","End":"01:18.700","Text":"That is equal to just xy."},{"Start":"01:18.700 ","End":"01:25.330","Text":"We can use the mean value property on w because w is harmonic."},{"Start":"01:25.330 ","End":"01:27.470","Text":"We can use this formula,"},{"Start":"01:27.470 ","End":"01:29.750","Text":"but in our case, x naught,"},{"Start":"01:29.750 ","End":"01:31.715","Text":"y naught is 0, 0,"},{"Start":"01:31.715 ","End":"01:34.765","Text":"and Rho is equal to 1."},{"Start":"01:34.765 ","End":"01:37.365","Text":"When instead of view we have w also,"},{"Start":"01:37.365 ","End":"01:38.670","Text":"so we have the w of 0,"},{"Start":"01:38.670 ","End":"01:40.470","Text":"0 is the integral."},{"Start":"01:40.470 ","End":"01:44.825","Text":"Well, what\u0027s written, and we have the formula for w here, it\u0027s just x, y."},{"Start":"01:44.825 ","End":"01:47.965","Text":"Here it\u0027s cosine Theta times Sine Theta."},{"Start":"01:47.965 ","End":"01:54.470","Text":"We can use a trigonometric identity that sine 2 Theta is 2 cosine Theta sine Theta."},{"Start":"01:54.470 ","End":"01:55.940","Text":"We put the half in here,"},{"Start":"01:55.940 ","End":"01:57.560","Text":"that will be okay."},{"Start":"01:57.560 ","End":"02:02.250","Text":"The integral from 0-2 Pi of sine 2 Theta is 0."},{"Start":"02:02.250 ","End":"02:04.575","Text":"This is equal to 0."},{"Start":"02:04.575 ","End":"02:06.720","Text":"Since w is u minus v,"},{"Start":"02:06.720 ","End":"02:09.015","Text":"u is w plus v,"},{"Start":"02:09.015 ","End":"02:11.010","Text":"so u of 0, 0, w of 0,"},{"Start":"02:11.010 ","End":"02:13.080","Text":"0 plus V of 0, 0,"},{"Start":"02:13.080 ","End":"02:18.375","Text":"and that\u0027s equal to 0 plus v of x, y is x^2."},{"Start":"02:18.375 ","End":"02:20.325","Text":"Here it\u0027s 0^2."},{"Start":"02:20.325 ","End":"02:22.920","Text":"That is equal to 0."},{"Start":"02:22.920 ","End":"02:26.170","Text":"That\u0027s what we had to show. We are done."}],"ID":30856},{"Watched":false,"Name":"Exercise 7","Duration":"2m 12s","ChapterTopicVideoID":29242,"CourseChapterTopicPlaylistID":294440,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.145","Text":"In this exercise, we\u0027re given that u is harmonic on the whole plane,"},{"Start":"00:05.145 ","End":"00:08.100","Text":"and at the origin u is positive."},{"Start":"00:08.100 ","End":"00:14.220","Text":"We have to prove that for every positive Epsilon, there is some x,"},{"Start":"00:14.220 ","End":"00:21.230","Text":"y on the circle of radius Epsilon centered at the origin such that u(x,"},{"Start":"00:21.230 ","End":"00:23.630","Text":"y) is strictly positive."},{"Start":"00:23.630 ","End":"00:25.640","Text":"We will do a proof by contradiction,"},{"Start":"00:25.640 ","End":"00:28.640","Text":"so first we have to invert this property."},{"Start":"00:28.640 ","End":"00:33.980","Text":"The opposite of that is that there exists an Epsilon bigger than naught,"},{"Start":"00:33.980 ","End":"00:36.680","Text":"such that for all x,"},{"Start":"00:36.680 ","End":"00:39.590","Text":"y satisfying this property,"},{"Start":"00:39.590 ","End":"00:41.960","Text":"u(x, y) is not bigger than 0,"},{"Start":"00:41.960 ","End":"00:43.970","Text":"meaning less than or equal to 0."},{"Start":"00:43.970 ","End":"00:46.280","Text":"We want to reach a contradiction from this,"},{"Start":"00:46.280 ","End":"00:48.980","Text":"and then that will prove our claim."},{"Start":"00:48.980 ","End":"00:51.170","Text":"Now, u is harmonic given,"},{"Start":"00:51.170 ","End":"00:53.965","Text":"so it has the mean value property,"},{"Start":"00:53.965 ","End":"00:56.415","Text":"and here\u0027s the formula for that."},{"Start":"00:56.415 ","End":"00:59.600","Text":"In our case, we\u0027ll take x naught,"},{"Start":"00:59.600 ","End":"01:01.340","Text":"y naught to be 0,"},{"Start":"01:01.340 ","End":"01:06.065","Text":"0 and Rho will be Epsilon and the picture could help."},{"Start":"01:06.065 ","End":"01:10.775","Text":"This is the origin radius Epsilon, and here\u0027s x, y."},{"Start":"01:10.775 ","End":"01:15.900","Text":"Then what this says if we make this substitution is that 1 over 2 Pi,"},{"Start":"01:15.900 ","End":"01:19.835","Text":"the integral of u of this is Epsilon cosine Theta."},{"Start":"01:19.835 ","End":"01:23.500","Text":"This is Epsilon sine Theta d Theta."},{"Start":"01:23.500 ","End":"01:26.100","Text":"Now for every Theta,"},{"Start":"01:26.100 ","End":"01:28.950","Text":"this point, Epsilon cosine Theta,"},{"Start":"01:28.950 ","End":"01:34.430","Text":"Epsilon sine Theta satisfies the following because cosine squared plus sine squared"},{"Start":"01:34.430 ","End":"01:40.280","Text":"equals 1 and so by the given I mean by the supposition of the contrary,"},{"Start":"01:40.280 ","End":"01:44.995","Text":"u of Epsilon cosine theta Epsilon sine Theta is less than or equal to 0."},{"Start":"01:44.995 ","End":"01:48.585","Text":"This is for all Theta between 0 and 2 Pi,"},{"Start":"01:48.585 ","End":"01:50.330","Text":"and so u(0,"},{"Start":"01:50.330 ","End":"01:54.665","Text":"0) which is the integral here,"},{"Start":"01:54.665 ","End":"01:57.410","Text":"is going to be less than or equal to 0 because"},{"Start":"01:57.410 ","End":"02:01.946","Text":"the integrant is less than or equal to 0 for each Theta,"},{"Start":"02:01.946 ","End":"02:04.310","Text":"so the integral will be less than or equal to 0."},{"Start":"02:04.310 ","End":"02:07.970","Text":"But this contradicts given that u(0,0) is bigger than"},{"Start":"02:07.970 ","End":"02:12.990","Text":"0 and this contradiction proves our claim, and we\u0027re done."}],"ID":30857}],"Thumbnail":null,"ID":294440},{"Name":"The Maximum and Minimum Principles","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Maximum and Minimum Principles","Duration":"2m 34s","ChapterTopicVideoID":29251,"CourseChapterTopicPlaylistID":294441,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.910","Text":"A new topic, the maximum principle for"},{"Start":"00:02.910 ","End":"00:08.265","Text":"harmonic functions and they\u0027ll also be a minimum principle which is very similar."},{"Start":"00:08.265 ","End":"00:09.930","Text":"Let\u0027s start with the maximum."},{"Start":"00:09.930 ","End":"00:12.105","Text":"The setup is this."},{"Start":"00:12.105 ","End":"00:14.610","Text":"We have a domain Omega,"},{"Start":"00:14.610 ","End":"00:18.765","Text":"and domain means open and connected, and non-empty."},{"Start":"00:18.765 ","End":"00:25.710","Text":"Let u(x,y) be harmonic in Omega and continuous on the closure of Omega,"},{"Start":"00:25.710 ","End":"00:28.695","Text":"which is Omega together with its boundary."},{"Start":"00:28.695 ","End":"00:32.505","Text":"Then the maximum of u,"},{"Start":"00:32.505 ","End":"00:36.070","Text":"and there is a maximum because Omega bar is compact,"},{"Start":"00:36.070 ","End":"00:41.065","Text":"so the maximum is achieved on the boundary, diagram might help."},{"Start":"00:41.065 ","End":"00:47.675","Text":"This is omega and this is the boundary of Omega and Omega bar is the whole thing."},{"Start":"00:47.675 ","End":"00:51.440","Text":"What we\u0027re saying in symbols is that there is some point on"},{"Start":"00:51.440 ","End":"00:56.915","Text":"the boundary such that u has a maximum at this point."},{"Start":"00:56.915 ","End":"00:58.760","Text":"Which means that for all x,"},{"Start":"00:58.760 ","End":"01:04.155","Text":"y in Omega bar u(x,y) is less than or equal to the maximum u."},{"Start":"01:04.155 ","End":"01:09.680","Text":"Now, there could be another maximum point inside and then the maximum would be equal."},{"Start":"01:09.680 ","End":"01:13.740","Text":"It certainly has at least 1 point on the boundary,"},{"Start":"01:13.740 ","End":"01:15.650","Text":"which is the maximum and also could have"},{"Start":"01:15.650 ","End":"01:18.140","Text":"other points on the boundary which are the maximum."},{"Start":"01:18.140 ","End":"01:22.520","Text":"At any rate, we can get the maximum value on the boundary."},{"Start":"01:22.520 ","End":"01:28.820","Text":"The second conclusion from this setup is that if u have a local maximum in Omega,"},{"Start":"01:28.820 ","End":"01:31.220","Text":"then u has to be a constant function."},{"Start":"01:31.220 ","End":"01:36.290","Text":"If there\u0027s a point inside which is bigger than all the points nearby,"},{"Start":"01:36.290 ","End":"01:38.815","Text":"then the function has to be constant."},{"Start":"01:38.815 ","End":"01:41.795","Text":"Then every point is going to be a maximum."},{"Start":"01:41.795 ","End":"01:45.910","Text":"It also follows that if you have the global maximum inside then"},{"Start":"01:45.910 ","End":"01:50.185","Text":"u is a constant because the global maximum is also a local maximum."},{"Start":"01:50.185 ","End":"01:54.365","Text":"That is all I want to say in this introduction to the maximum principle."},{"Start":"01:54.365 ","End":"01:56.835","Text":"Now, the minimum principle."},{"Start":"01:56.835 ","End":"02:00.470","Text":"The minimum principle is exactly the same as the maximum principle,"},{"Start":"02:00.470 ","End":"02:03.140","Text":"except that we replace maximum with minimum,"},{"Start":"02:03.140 ","End":"02:06.665","Text":"so we have minimum here instead of maximum."},{"Start":"02:06.665 ","End":"02:11.060","Text":"Also, here the word minimum instead of maximum."},{"Start":"02:11.060 ","End":"02:12.620","Text":"In fact, that\u0027s what I did."},{"Start":"02:12.620 ","End":"02:15.050","Text":"I did a copy-paste and then I changed maximum to"},{"Start":"02:15.050 ","End":"02:17.810","Text":"minimum and also the sign of the inequality."},{"Start":"02:17.810 ","End":"02:26.120","Text":"In short, the minimum of u on the domain with its closure is achieved on the boundary."},{"Start":"02:26.120 ","End":"02:32.408","Text":"You can\u0027t have a local or global minimum inside except if the function is constant."},{"Start":"02:32.408 ","End":"02:34.500","Text":"That\u0027s it."}],"ID":30858},{"Watched":false,"Name":"Exercise 1","Duration":"4m 25s","ChapterTopicVideoID":29252,"CourseChapterTopicPlaylistID":294441,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.690","Text":"In this exercise, we have 2 functions, u and v,"},{"Start":"00:03.690 ","End":"00:06.405","Text":"each of them defined on the annulus,"},{"Start":"00:06.405 ","End":"00:08.415","Text":"which is shown in the picture."},{"Start":"00:08.415 ","End":"00:11.130","Text":"It\u0027s between radius 1/2 and radius"},{"Start":"00:11.130 ","End":"00:16.320","Text":"1 1/2 and each of them satisfies its own boundary value problem."},{"Start":"00:16.320 ","End":"00:18.090","Text":"This is the one for u."},{"Start":"00:18.090 ","End":"00:23.885","Text":"This is the one for v. Note that the boundary consists of the 2 circles,"},{"Start":"00:23.885 ","End":"00:28.380","Text":"which are where r = 1/2 and r ="},{"Start":"00:28.380 ","End":"00:33.345","Text":"3/2 and data could be anything and similarly here 1/2 and 3/2."},{"Start":"00:33.345 ","End":"00:36.350","Text":"This is a Dirichlet problem for each of them."},{"Start":"00:36.350 ","End":"00:40.135","Text":"We\u0027re given the value of the function on the boundary."},{"Start":"00:40.135 ","End":"00:43.805","Text":"Now the question we have to answer is, which is greater?"},{"Start":"00:43.805 ","End":"00:46.895","Text":"u(1, 2 Pi/3),"},{"Start":"00:46.895 ","End":"00:50.200","Text":"or v(1, 2 Pi/3)?"},{"Start":"00:50.200 ","End":"00:51.785","Text":"It\u0027s the same point here."},{"Start":"00:51.785 ","End":"00:55.580","Text":"See radius 1 is here,"},{"Start":"00:55.580 ","End":"01:00.390","Text":"2 Pi/3 is 120 degrees, somewhere about here."},{"Start":"01:00.680 ","End":"01:05.855","Text":"The usual trick when we\u0027re comparing 2 functions is to subtract them."},{"Start":"01:05.855 ","End":"01:10.310","Text":"Let\u0027s define a function w to be u minus"},{"Start":"01:10.310 ","End":"01:16.580","Text":"v. Then if we check if w at this point is positive,"},{"Start":"01:16.580 ","End":"01:21.425","Text":"negative or 0, then we\u0027ll get respectively that u is bigger than v,"},{"Start":"01:21.425 ","End":"01:22.550","Text":"less than v,"},{"Start":"01:22.550 ","End":"01:29.700","Text":"or equal to v according to these outcome."},{"Start":"01:29.700 ","End":"01:35.930","Text":"Note that neither u nor v satisfies the Laplace equation."},{"Start":"01:35.930 ","End":"01:38.180","Text":"You can call this a Poisson equation,"},{"Start":"01:38.180 ","End":"01:42.410","Text":"which is like a nonhomogeneous Laplace equation but we want it to be"},{"Start":"01:42.410 ","End":"01:48.740","Text":"homogeneous in order for the u or v or whatever to be harmonic."},{"Start":"01:48.740 ","End":"01:52.190","Text":"But we\u0027ll get lucky with w as you see."},{"Start":"01:52.190 ","End":"01:57.020","Text":"The Laplacian of w is the Laplacian of u minus the Laplacian of v. This is"},{"Start":"01:57.020 ","End":"02:01.500","Text":"equal to 1 minus sine^2 Theta minus cosine^2 Theta,"},{"Start":"02:01.500 ","End":"02:05.420","Text":"it\u0027s 0, so w satisfies the Laplace equation."},{"Start":"02:05.420 ","End":"02:12.290","Text":"Now let\u0027s check the 2 boundary conditions for w at the inner circle and the outer circle."},{"Start":"02:12.290 ","End":"02:16.850","Text":"Again, just subtraction w on the circle of radius 1/2"},{"Start":"02:16.850 ","End":"02:22.310","Text":"is u minus v. This is the expression for u there."},{"Start":"02:22.370 ","End":"02:25.460","Text":"This is the expression for v there,"},{"Start":"02:25.460 ","End":"02:26.510","Text":"but with a minus,"},{"Start":"02:26.510 ","End":"02:31.130","Text":"but the minus minus makes it a plus and clearly this expression is bigger"},{"Start":"02:31.130 ","End":"02:35.780","Text":"or equal to 0. w is bigger or equal to 0 on the inner circle."},{"Start":"02:35.780 ","End":"02:39.185","Text":"Let\u0027s check the outer circle 3 over 2."},{"Start":"02:39.185 ","End":"02:41.135","Text":"Similar calculation."},{"Start":"02:41.135 ","End":"02:45.805","Text":"We get 9 over 4 minus 9 over 4 cosine^2 Theta."},{"Start":"02:45.805 ","End":"02:48.680","Text":"Since cosine squared Theta is less than or equal to 1,"},{"Start":"02:48.680 ","End":"02:51.340","Text":"this minus this is bigger or equal to 0."},{"Start":"02:51.340 ","End":"02:52.950","Text":"Here bigger or equal to 0,"},{"Start":"02:52.950 ","End":"02:54.540","Text":"here bigger or equal to 0,"},{"Start":"02:54.540 ","End":"02:55.889","Text":"so on all the boundary,"},{"Start":"02:55.889 ","End":"02:59.915","Text":"w is non-negative and here\u0027s a picture."},{"Start":"02:59.915 ","End":"03:03.830","Text":"The inner circle, the outer circle on each of them,"},{"Start":"03:03.830 ","End":"03:06.110","Text":"w is bigger or equal to 0."},{"Start":"03:06.110 ","End":"03:08.600","Text":"We can\u0027t say immediately what the minimum is,"},{"Start":"03:08.600 ","End":"03:10.865","Text":"but in fact, the minimum is 0."},{"Start":"03:10.865 ","End":"03:12.395","Text":"If you look at this point,"},{"Start":"03:12.395 ","End":"03:14.775","Text":"the value of Theta here is 0."},{"Start":"03:14.775 ","End":"03:18.510","Text":"If we put Theta = 0, well,"},{"Start":"03:18.510 ","End":"03:25.475","Text":"we get exactly 0 here because cosine^2 is 1 and 9/4 minus 9/4 is 0."},{"Start":"03:25.475 ","End":"03:29.135","Text":"This is the minimum, and this is where w=0."},{"Start":"03:29.135 ","End":"03:30.980","Text":"Let\u0027s just write some of this down."},{"Start":"03:30.980 ","End":"03:33.980","Text":"Omega is the gray area between"},{"Start":"03:33.980 ","End":"03:40.790","Text":"the 2 circles and the boundary is the union of both circles."},{"Start":"03:40.790 ","End":"03:45.170","Text":"We\u0027ve shown that w is bigger or equal to 0 on the boundary,"},{"Start":"03:45.170 ","End":"03:48.315","Text":"and w is not a constant,"},{"Start":"03:48.315 ","End":"03:53.000","Text":"so it can\u0027t have a minimum point inside in the gray area."},{"Start":"03:53.000 ","End":"03:56.930","Text":"In Omega, w is strictly bigger than 0."},{"Start":"03:56.930 ","End":"04:00.005","Text":"In particular, w at this point,"},{"Start":"04:00.005 ","End":"04:01.650","Text":"which is 1,"},{"Start":"04:01.650 ","End":"04:05.850","Text":"2 Pi/3 is strictly bigger than 0,"},{"Start":"04:05.850 ","End":"04:12.290","Text":"but w is u minus v. If u minus v is bigger than 0,"},{"Start":"04:12.290 ","End":"04:16.700","Text":"then u is bigger than v. The value of u at this point is"},{"Start":"04:16.700 ","End":"04:21.665","Text":"bigger than the value of v at this point and that\u0027s what we had to check."},{"Start":"04:21.665 ","End":"04:26.010","Text":"This is the answer and we are done."}],"ID":30859},{"Watched":false,"Name":"Exercise 2","Duration":"2m 12s","ChapterTopicVideoID":29253,"CourseChapterTopicPlaylistID":294441,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.425","Text":"In this exercise, we have a boundary value problem,"},{"Start":"00:04.425 ","End":"00:11.130","Text":"which consists of a Laplace\u0027s equation on the disk of radius 2,"},{"Start":"00:11.130 ","End":"00:16.890","Text":"and we have Dirichlet boundary conditions on the circle of radius 2."},{"Start":"00:16.890 ","End":"00:24.100","Text":"We have to compute the minimum of u(x,y) on the closed disk of radius 2."},{"Start":"00:24.100 ","End":"00:27.240","Text":"We\u0027ll be using the minimum principle."},{"Start":"00:27.240 ","End":"00:29.400","Text":"The domain Omega,"},{"Start":"00:29.400 ","End":"00:31.980","Text":"which is the open disk of radius 2,"},{"Start":"00:31.980 ","End":"00:35.520","Text":"2 because 4 is 2^2 is the following."},{"Start":"00:35.520 ","End":"00:40.715","Text":"The closure of Omega is where it\u0027s less than or equal to 4,"},{"Start":"00:40.715 ","End":"00:44.795","Text":"and the boundary of Omega is the circle."},{"Start":"00:44.795 ","End":"00:47.015","Text":"By the minimum principle,"},{"Start":"00:47.015 ","End":"00:54.875","Text":"the minimum on the closed disk Omega bar is the same as the minimum on just the boundary."},{"Start":"00:54.875 ","End":"01:00.245","Text":"The boundary we can express as x^2 plus y^2 equals 4,"},{"Start":"01:00.245 ","End":"01:04.865","Text":"and u(x,y) from here is y plus y^2."},{"Start":"01:04.865 ","End":"01:06.800","Text":"X doesn\u0027t appear here."},{"Start":"01:06.800 ","End":"01:12.230","Text":"But from here, we can get that y goes between minus 2 and 2."},{"Start":"01:12.230 ","End":"01:15.680","Text":"It\u0027s clear, imagine a circle of radius 2,"},{"Start":"01:15.680 ","End":"01:18.785","Text":"y goes from minus 2 to plus 2."},{"Start":"01:18.785 ","End":"01:27.170","Text":"This is a regular Calculus 1 basic problem finding the minimum of a parabola."},{"Start":"01:27.170 ","End":"01:30.740","Text":"We can actually use a formula, not use calculus."},{"Start":"01:30.740 ","End":"01:36.649","Text":"There\u0027s a formula, c minus b^2 over 4a or something like that."},{"Start":"01:36.649 ","End":"01:45.335","Text":"It comes out to be minus 1/4 and it occurs when y is minus 1/2."},{"Start":"01:45.335 ","End":"01:48.950","Text":"Maybe the easiest ways to differentiate 2y plus 1,"},{"Start":"01:48.950 ","End":"01:50.510","Text":"set that equal to 0,"},{"Start":"01:50.510 ","End":"01:52.430","Text":"y equals minus 1/2,"},{"Start":"01:52.430 ","End":"01:59.360","Text":"substitute minus 1/2 in here and we get minus 1/2 plus 1/4, which is minus 1/4."},{"Start":"01:59.360 ","End":"02:03.185","Text":"It is a minimum because the second derivative,"},{"Start":"02:03.185 ","End":"02:06.290","Text":"which is 2 is positive,"},{"Start":"02:06.290 ","End":"02:07.640","Text":"so it\u0027s a minimum."},{"Start":"02:07.640 ","End":"02:09.920","Text":"Anyway, the answer is minus 1/4,"},{"Start":"02:09.920 ","End":"02:12.720","Text":"and that concludes this exercise."}],"ID":30860},{"Watched":false,"Name":"Exercise 3","Duration":"4m 9s","ChapterTopicVideoID":29254,"CourseChapterTopicPlaylistID":294441,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.780","Text":"In this exercise, u is harmonic in the square from 0 to Pi here,"},{"Start":"00:06.780 ","End":"00:08.865","Text":"from 0 to Pi here."},{"Start":"00:08.865 ","End":"00:13.365","Text":"We\u0027re given the value of u on the boundary."},{"Start":"00:13.365 ","End":"00:15.824","Text":"The boundary has 4 parts,"},{"Start":"00:15.824 ","End":"00:18.945","Text":"and we\u0027re given it as follows."},{"Start":"00:18.945 ","End":"00:21.900","Text":"U(x, naught) equals u(x) Pi=0,"},{"Start":"00:21.900 ","End":"00:26.730","Text":"which means that on this side and on this side, it\u0027s 0."},{"Start":"00:26.730 ","End":"00:30.840","Text":"Then the 2 vertical sides here and here,"},{"Start":"00:30.840 ","End":"00:34.455","Text":"it\u0027s equal to y^4th minus Pi^3 y."},{"Start":"00:34.455 ","End":"00:36.855","Text":"We have to prove that u(x,"},{"Start":"00:36.855 ","End":"00:41.030","Text":"y) is bigger than this expression for all x,"},{"Start":"00:41.030 ","End":"00:44.480","Text":"y inside the square,"},{"Start":"00:44.480 ","End":"00:46.265","Text":"not on the boundary."},{"Start":"00:46.265 ","End":"00:49.910","Text":"Note that u is well-defined at the corners."},{"Start":"00:49.910 ","End":"00:53.540","Text":"It seems to be doubly defined because this corner belongs to this side,"},{"Start":"00:53.540 ","End":"00:54.950","Text":"and to this side."},{"Start":"00:54.950 ","End":"00:56.960","Text":"But it\u0027s 0 here,"},{"Start":"00:56.960 ","End":"00:59.045","Text":"here, here, and here."},{"Start":"00:59.045 ","End":"01:01.220","Text":"U is continuous on the boundary,"},{"Start":"01:01.220 ","End":"01:06.810","Text":"it\u0027s continuous in each of the 4 sides and it matches up where they meet, so continuous."},{"Start":"01:06.810 ","End":"01:11.615","Text":"That\u0027s what we need for the minimum principle for harmonic function."},{"Start":"01:11.615 ","End":"01:16.460","Text":"This means that the minimum on the closure of"},{"Start":"01:16.460 ","End":"01:23.480","Text":"D of the function u is the same as the minimum on the boundary of D, of u."},{"Start":"01:23.480 ","End":"01:26.690","Text":"The minimum along all these 4 sides,"},{"Start":"01:26.690 ","End":"01:32.210","Text":"it\u0027s just going to be the minimum along 1 of these vertical sides, choose 1."},{"Start":"01:32.210 ","End":"01:37.250","Text":"Otherwise, the minimum of y^4th minus Pi^3 y,"},{"Start":"01:37.250 ","End":"01:40.640","Text":"where y goes from 0 to Pi."},{"Start":"01:40.640 ","End":"01:42.395","Text":"Let\u0027s write that down."},{"Start":"01:42.395 ","End":"01:44.120","Text":"Because like on this side,"},{"Start":"01:44.120 ","End":"01:48.125","Text":"it\u0027s 0 at the endpoints and it could be less than 0 in the middle,"},{"Start":"01:48.125 ","End":"01:50.450","Text":"maybe not, and then the minimum will be 0,"},{"Start":"01:50.450 ","End":"01:53.240","Text":"but actually dips below 0 here."},{"Start":"01:53.240 ","End":"01:54.690","Text":"Let\u0027s give it a name."},{"Start":"01:54.690 ","End":"01:57.285","Text":"This function we\u0027ll call it g(y)."},{"Start":"01:57.285 ","End":"02:00.680","Text":"If it\u0027s convenient, we can write it factorized."},{"Start":"02:00.680 ","End":"02:03.805","Text":"We find the minimum using straightforward calculus."},{"Start":"02:03.805 ","End":"02:07.265","Text":"We differentiate and we get the following."},{"Start":"02:07.265 ","End":"02:09.965","Text":"Let me look for where the derivative is 0."},{"Start":"02:09.965 ","End":"02:13.100","Text":"Look for say, y_naught where the minimum occurs."},{"Start":"02:13.100 ","End":"02:20.150","Text":"There\u0027s only 1 place where y_naught is Pi over the cube root of 4 because we\u0027re at 0,"},{"Start":"02:20.150 ","End":"02:22.160","Text":"4y^3 equals Pi^3,"},{"Start":"02:22.160 ","End":"02:23.930","Text":"so y^3 is Pi^3 over 4."},{"Start":"02:23.930 ","End":"02:26.360","Text":"Take the cube root, this is what we get."},{"Start":"02:26.360 ","End":"02:29.060","Text":"Let\u0027s see if it\u0027s a minimum or a maximum."},{"Start":"02:29.060 ","End":"02:31.550","Text":"Well, the second derivative is positive,"},{"Start":"02:31.550 ","End":"02:32.960","Text":"so we have a minimum."},{"Start":"02:32.960 ","End":"02:35.465","Text":"Now what is this minimum value?"},{"Start":"02:35.465 ","End":"02:38.470","Text":"It\u0027s equal to g of this y_naught,"},{"Start":"02:38.470 ","End":"02:40.845","Text":"which is this value,"},{"Start":"02:40.845 ","End":"02:43.595","Text":"y_naught, use this expression here,"},{"Start":"02:43.595 ","End":"02:48.290","Text":"y^3 minus Pi^3 and here\u0027s a graph that I"},{"Start":"02:48.290 ","End":"02:54.875","Text":"made of g of y between 0 and Pi."},{"Start":"02:54.875 ","End":"02:59.165","Text":"This here is a numerical answer the graph program gave."},{"Start":"02:59.165 ","End":"03:00.500","Text":"Anyway, we see,"},{"Start":"03:00.500 ","End":"03:02.045","Text":"and because it has a minimum,"},{"Start":"03:02.045 ","End":"03:04.325","Text":"this really is the minimum over the whole interval."},{"Start":"03:04.325 ","End":"03:07.130","Text":"Minimum could be a local minimum or at the endpoints,"},{"Start":"03:07.130 ","End":"03:08.720","Text":"but at the endpoints is 0."},{"Start":"03:08.720 ","End":"03:14.060","Text":"This is going to be the global minimum on this side and this interval."},{"Start":"03:14.060 ","End":"03:16.820","Text":"Continuing with this calculation,"},{"Start":"03:16.820 ","End":"03:18.230","Text":"this is what we get."},{"Start":"03:18.230 ","End":"03:24.295","Text":"This thing cubed is Pi^3 over 4 minus Pi^3 and then times y_naught."},{"Start":"03:24.295 ","End":"03:26.270","Text":"This comes out to be,"},{"Start":"03:26.270 ","End":"03:30.440","Text":"you simplify it, this is minus 3/4 Pi^3."},{"Start":"03:30.440 ","End":"03:34.580","Text":"Together with this, this is what we get for the minimum value."},{"Start":"03:34.580 ","End":"03:38.600","Text":"Now, this is the minimum over the closed squares,"},{"Start":"03:38.600 ","End":"03:40.535","Text":"square with the boundaries."},{"Start":"03:40.535 ","End":"03:43.715","Text":"But u is not a constant function."},{"Start":"03:43.715 ","End":"03:45.144","Text":"Quite clearly."},{"Start":"03:45.144 ","End":"03:51.400","Text":"Along one of the side it goes from 0 to minus this expression, not a constant."},{"Start":"03:51.400 ","End":"03:52.925","Text":"If it\u0027s not a constant,"},{"Start":"03:52.925 ","End":"03:57.645","Text":"then it can\u0027t have a minimum inside the square."},{"Start":"03:57.645 ","End":"04:05.000","Text":"Inside the square, it has to be always strictly bigger than this minimum value."},{"Start":"04:05.000 ","End":"04:06.725","Text":"This is what we had to prove,"},{"Start":"04:06.725 ","End":"04:09.840","Text":"and that concludes this exercise."}],"ID":30861},{"Watched":false,"Name":"Exercise 4","Duration":"3m 11s","ChapterTopicVideoID":29255,"CourseChapterTopicPlaylistID":294441,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.564","Text":"In this exercise, u is harmonic in the disk,"},{"Start":"00:05.564 ","End":"00:10.650","Text":"which is centered at the origin and has radius square root of 8."},{"Start":"00:10.650 ","End":"00:13.229","Text":"It has a Dirichlet boundary condition,"},{"Start":"00:13.229 ","End":"00:17.895","Text":"meaning we are given the value of the function on the boundary."},{"Start":"00:17.895 ","End":"00:19.455","Text":"There it\u0027s 2xy."},{"Start":"00:19.455 ","End":"00:25.020","Text":"The boundary of course is the circle where x squared plus y squared is 8."},{"Start":"00:25.020 ","End":"00:27.294","Text":"Our task is to prove that u(x,"},{"Start":"00:27.294 ","End":"00:35.165","Text":"y) is strictly between minus 8 and 8 on all of the interior of the disk."},{"Start":"00:35.165 ","End":"00:38.015","Text":"That\u0027s the original D with less than."},{"Start":"00:38.015 ","End":"00:43.625","Text":"Now u is not a constant function because 2xy is not a constant."},{"Start":"00:43.625 ","End":"00:45.800","Text":"It\u0027s not constant even on the boundary."},{"Start":"00:45.800 ","End":"00:47.570","Text":"We can use the minimum principle and"},{"Start":"00:47.570 ","End":"00:52.100","Text":"the maximum principle and get that the value of u at any point in"},{"Start":"00:52.100 ","End":"00:55.220","Text":"the interior is strictly greater than the minimum on"},{"Start":"00:55.220 ","End":"00:58.985","Text":"the boundary and strictly less than the maximum on the boundary."},{"Start":"00:58.985 ","End":"01:02.510","Text":"Now we\u0027ll figure out what is this minimum and what is this maximum,"},{"Start":"01:02.510 ","End":"01:06.235","Text":"and hopefully it will turn out to be minus 8 and 8."},{"Start":"01:06.235 ","End":"01:10.275","Text":"Now, we could use Lagrange Multipliers,"},{"Start":"01:10.275 ","End":"01:16.700","Text":"because what we have is a maximum of a function under a constraint."},{"Start":"01:16.700 ","End":"01:21.850","Text":"But it turns out to be easier just to use polar coordinates in this case."},{"Start":"01:21.850 ","End":"01:27.945","Text":"We have that u(x, y) is 2xy on the circle."},{"Start":"01:27.945 ","End":"01:31.645","Text":"In polar, that comes out to be u( r,"},{"Start":"01:31.645 ","End":"01:34.700","Text":"Theta) equals x is r cosine Theta,"},{"Start":"01:34.700 ","End":"01:36.200","Text":"y is r sine Theta."},{"Start":"01:36.200 ","End":"01:38.980","Text":"We have 2 r squared cosine Theta sine Theta."},{"Start":"01:38.980 ","End":"01:44.730","Text":"This in polar is r squared equals 8 or r equals square root of 8."},{"Start":"01:44.730 ","End":"01:48.665","Text":"Just to summarize, on r equals square root of 8,"},{"Start":"01:48.665 ","End":"01:50.230","Text":"Theta can be whatever,"},{"Start":"01:50.230 ","End":"01:56.100","Text":"u(r, Theta) is r^2 times 2 cosine Theta sine Theta."},{"Start":"01:56.100 ","End":"02:02.320","Text":"Put the r-squared in front because 2 cosine Theta sine Theta is sine of 2 Theta."},{"Start":"02:02.320 ","End":"02:04.140","Text":"On our particular boundary,"},{"Start":"02:04.140 ","End":"02:05.160","Text":"r^2 is 8,"},{"Start":"02:05.160 ","End":"02:08.180","Text":"so we have 8 sine of 2 Theta."},{"Start":"02:08.180 ","End":"02:10.760","Text":"Now we can say what the maximum and minimum are,"},{"Start":"02:10.760 ","End":"02:12.755","Text":"because the sine 2 Theta,"},{"Start":"02:12.755 ","End":"02:16.405","Text":"the minimum is minus 1 and the maximum is 1."},{"Start":"02:16.405 ","End":"02:20.450","Text":"Sine 2 Theta goes between minus 1 and 1 and we"},{"Start":"02:20.450 ","End":"02:24.740","Text":"have more than a whole period between 0 and 2 Pi,"},{"Start":"02:24.740 ","End":"02:26.389","Text":"in fact 2 whole periods."},{"Start":"02:26.389 ","End":"02:28.160","Text":"Yeah, 1 and minus 1,"},{"Start":"02:28.160 ","End":"02:30.610","Text":"now multiply by 8."},{"Start":"02:30.610 ","End":"02:34.500","Text":"We have that the maximum of 8 sine 2 Theta is 8."},{"Start":"02:34.500 ","End":"02:37.720","Text":"The minimum of 8 sine 2 Theta is minus 8,"},{"Start":"02:37.720 ","End":"02:44.420","Text":"which means that the maximum on the circle of radius root 8 of 2 x y is 8."},{"Start":"02:44.420 ","End":"02:49.480","Text":"We\u0027re returning now to Cartesian coordinates and the minimum is minus 8."},{"Start":"02:49.480 ","End":"02:52.535","Text":"Let\u0027s just copy this line down here."},{"Start":"02:52.535 ","End":"02:57.140","Text":"We had this, and now we found this minimum that\u0027s minus 8."},{"Start":"02:57.140 ","End":"02:59.845","Text":"We found this maximum that\u0027s 8."},{"Start":"02:59.845 ","End":"03:05.280","Text":"We have minus 8 is less than u is less than 8 for all points x,"},{"Start":"03:05.280 ","End":"03:07.275","Text":"y in D,"},{"Start":"03:07.275 ","End":"03:08.955","Text":"that\u0027s inside the disk."},{"Start":"03:08.955 ","End":"03:12.490","Text":"This is what we had to show. We are done."}],"ID":30862},{"Watched":false,"Name":"Exercise 5","Duration":"3m 5s","ChapterTopicVideoID":29256,"CourseChapterTopicPlaylistID":294441,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.125","Text":"In this exercise, D is the annulus,"},{"Start":"00:04.125 ","End":"00:05.820","Text":"the one that\u0027s in the picture."},{"Start":"00:05.820 ","End":"00:06.990","Text":"This is Radius 1,"},{"Start":"00:06.990 ","End":"00:08.805","Text":"this is Radius 2."},{"Start":"00:08.805 ","End":"00:12.735","Text":"U and v are 2 functions each satisfies the BVP,"},{"Start":"00:12.735 ","End":"00:14.445","Text":"this is the one for u."},{"Start":"00:14.445 ","End":"00:16.290","Text":"This is the one for v we\u0027re"},{"Start":"00:16.290 ","End":"00:24.540","Text":"given a Poisson equation because it\u0027s not homogeneous here and here."},{"Start":"00:24.540 ","End":"00:27.630","Text":"We\u0027re given the value on the boundary."},{"Start":"00:27.630 ","End":"00:31.305","Text":"The boundary consists of 2 parts where r =1 and r =2."},{"Start":"00:31.305 ","End":"00:35.205","Text":"Similarly here we have v when r =1 and when r =2."},{"Start":"00:35.205 ","End":"00:41.690","Text":"What we have to prove is that v is less than u and u is less than 1 plus v"},{"Start":"00:41.690 ","End":"00:48.680","Text":"everywhere on the interior of D. This whole exercise will be just in polar coordinates."},{"Start":"00:48.680 ","End":"00:53.299","Text":"The usual procedure when we want to compare 2 things is to subtract them,"},{"Start":"00:53.299 ","End":"01:01.880","Text":"so we\u0027ll let w be identically equal to u minus v and then we can see what w satisfies."},{"Start":"01:01.880 ","End":"01:04.340","Text":"The Laplace operator is linear,"},{"Start":"01:04.340 ","End":"01:11.465","Text":"so Delta w is Delta u minus Delta v. We get 1 minus 1 is 0,"},{"Start":"01:11.465 ","End":"01:20.795","Text":"the same for the boundary values we also subtract cos^2 minus minus sin^2 is 1."},{"Start":"01:20.795 ","End":"01:24.435","Text":"Sin^2 minus 0 is sin^2."},{"Start":"01:24.435 ","End":"01:27.650","Text":"Note that w is not a constant and"},{"Start":"01:27.650 ","End":"01:31.069","Text":"it\u0027s harmonic because it satisfies the Laplace equation,"},{"Start":"01:31.069 ","End":"01:35.750","Text":"so we can use the minimum and maximum principles and get strict inequalities."},{"Start":"01:35.750 ","End":"01:42.830","Text":"In this case, the minimum on the boundary of w is strictly less than the value of"},{"Start":"01:42.830 ","End":"01:50.630","Text":"w at any point inside D and that\u0027s less strictly than the maximum on the boundary."},{"Start":"01:50.630 ","End":"01:54.260","Text":"Note that the boundary consists of 2 separate circles,"},{"Start":"01:54.260 ","End":"01:56.540","Text":"r = 1 and r = 2."},{"Start":"01:56.540 ","End":"02:00.290","Text":"Let\u0027s see what the maximum and minimum are on each of these circles and then we can"},{"Start":"02:00.290 ","End":"02:04.450","Text":"take the smaller of the 2 minima and the larger of the 2 maxima."},{"Start":"02:04.450 ","End":"02:08.270","Text":"W equals 1 and r equals 1 and since it\u0027s constant,"},{"Start":"02:08.270 ","End":"02:10.985","Text":"the minimum and the maximum are both 1,"},{"Start":"02:10.985 ","End":"02:14.030","Text":"when r equals 1 and r equals 2,"},{"Start":"02:14.030 ","End":"02:17.330","Text":"we have that w is sin^2 Theta,"},{"Start":"02:17.330 ","End":"02:24.665","Text":"and that ranges from 0 to 1 and so the minimum is 0 and the maximum is 1."},{"Start":"02:24.665 ","End":"02:27.395","Text":"Now if we take the smallest of the 2 minima,"},{"Start":"02:27.395 ","End":"02:30.005","Text":"it will be 0, and the largest of the 2 maxima,"},{"Start":"02:30.005 ","End":"02:31.820","Text":"it will be 1."},{"Start":"02:31.820 ","End":"02:36.880","Text":"The minimum is 0 and the maximum is 1 on the boundary."},{"Start":"02:36.880 ","End":"02:41.165","Text":"That means that for all our Theta inside the annulus,"},{"Start":"02:41.165 ","End":"02:45.860","Text":"we have at the minimum is less than w is less than the maximum,"},{"Start":"02:45.860 ","End":"02:51.175","Text":"and the minimum is 0 and the maximum is 1 so we get that w is between 0 and 1, strictly."},{"Start":"02:51.175 ","End":"02:58.040","Text":"W is equal to u minus v add v to everything and we get that v"},{"Start":"02:58.040 ","End":"03:05.730","Text":"is less than u is less than 1 plus v. That is what we had to show and we are done."}],"ID":30863},{"Watched":false,"Name":"Exercise 6","Duration":"2m 9s","ChapterTopicVideoID":29257,"CourseChapterTopicPlaylistID":294441,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.274","Text":"In this exercise, D is the open unit disc so that the boundary is the unit circle."},{"Start":"00:07.274 ","End":"00:11.910","Text":"Let u and v satisfy the following boundary value problems."},{"Start":"00:11.910 ","End":"00:16.260","Text":"That Laplacian of u is x^2 plus y^2 in"},{"Start":"00:16.260 ","End":"00:21.770","Text":"the disk and u is equal to cosine^2 x on the boundary."},{"Start":"00:21.770 ","End":"00:26.965","Text":"For v, the same Laplacian x^2 plus y^2."},{"Start":"00:26.965 ","End":"00:31.925","Text":"But the value on the boundary is x^2 minus sine^2 x."},{"Start":"00:31.925 ","End":"00:40.160","Text":"We have to prove that u is strictly between v and 1 plus v for all x,"},{"Start":"00:40.160 ","End":"00:43.040","Text":"y in the open unit disc."},{"Start":"00:43.040 ","End":"00:50.900","Text":"The usual trick we subtract u minus v and call it w. The PDE for w,"},{"Start":"00:50.900 ","End":"00:53.959","Text":"since the Laplacian is linear,"},{"Start":"00:53.959 ","End":"00:57.680","Text":"is x^2 plus y^2 minus x^2 plus y^2,"},{"Start":"00:57.680 ","End":"00:58.930","Text":"in other words, 0,"},{"Start":"00:58.930 ","End":"01:02.585","Text":"so that w is harmonic and on the boundary,"},{"Start":"01:02.585 ","End":"01:11.015","Text":"w is cosine^2 x minus x^2 minus sine^2 x and cosine^2 minus minus sine^2 is 1."},{"Start":"01:11.015 ","End":"01:15.245","Text":"We get 1 minus x^2 on the boundary."},{"Start":"01:15.245 ","End":"01:22.545","Text":"W is harmonic and obviously non-constant because it\u0027s non-constant on the boundary."},{"Start":"01:22.545 ","End":"01:26.000","Text":"Using the maximum principle and the minimum principle,"},{"Start":"01:26.000 ","End":"01:30.605","Text":"we have that w(x,y) inside the disk"},{"Start":"01:30.605 ","End":"01:36.305","Text":"is strictly between the minimum and the maximum on the boundary."},{"Start":"01:36.305 ","End":"01:40.550","Text":"Now, 1 minus x^2 looks like this."},{"Start":"01:40.550 ","End":"01:46.640","Text":"The maximum is 1 and the minimum is 0 achieved here and here."},{"Start":"01:46.640 ","End":"01:50.870","Text":"We get that w(x,y) is between 0 and 1,"},{"Start":"01:50.870 ","End":"01:59.265","Text":"where x and y as the berberine D. Now we remember that w is u minus v,"},{"Start":"01:59.265 ","End":"02:04.445","Text":"so add v to everything and we get that v is less than u,"},{"Start":"02:04.445 ","End":"02:06.350","Text":"less than 1 plus v,"},{"Start":"02:06.350 ","End":"02:07.820","Text":"that\u0027s what we had to show,"},{"Start":"02:07.820 ","End":"02:10.110","Text":"so we are done."}],"ID":30864},{"Watched":false,"Name":"Exercise 7","Duration":"2m 42s","ChapterTopicVideoID":29250,"CourseChapterTopicPlaylistID":294441,"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"},{"Start":"00:02.850 ","End":"00:06.660","Text":"the function u satisfies the following boundary value problem."},{"Start":"00:06.660 ","End":"00:08.925","Text":"This is a Poisson equation."},{"Start":"00:08.925 ","End":"00:12.555","Text":"The Laplacian is not equal to 0, it\u0027s equal to 2."},{"Start":"00:12.555 ","End":"00:15.480","Text":"That\u0027s on the unit disk open."},{"Start":"00:15.480 ","End":"00:19.950","Text":"On the boundary, u is given to be 2x^2."},{"Start":"00:19.950 ","End":"00:23.190","Text":"We have to prove that u(x,"},{"Start":"00:23.190 ","End":"00:28.350","Text":"y) is strictly between x^2 and 1+x^2 for all x,"},{"Start":"00:28.350 ","End":"00:30.750","Text":"y inside the unit disc,"},{"Start":"00:30.750 ","End":"00:37.650","Text":"and the problem is that u is not harmonic because Laplacian of u is not 0."},{"Start":"00:37.650 ","End":"00:44.690","Text":"We can try and fix it by looking for a function v such that the Laplacian of v is 2,"},{"Start":"00:44.690 ","End":"00:51.945","Text":"and then subtract v from u to get w. One possibility,"},{"Start":"00:51.945 ","End":"00:55.970","Text":"and there are many, is v (x, y) = x^2."},{"Start":"00:55.970 ","End":"01:00.065","Text":"If you take the second derivative with respect to x, it\u0027s 2,"},{"Start":"01:00.065 ","End":"01:02.330","Text":"second derivative with respect to y is 0,"},{"Start":"01:02.330 ","End":"01:03.530","Text":"2+0 is 2,"},{"Start":"01:03.530 ","End":"01:05.045","Text":"so we\u0027re all right."},{"Start":"01:05.045 ","End":"01:06.830","Text":"Then u - v,"},{"Start":"01:06.830 ","End":"01:11.465","Text":"which is w, satisfies this BVP."},{"Start":"01:11.465 ","End":"01:14.000","Text":"The Laplacian now is 0."},{"Start":"01:14.000 ","End":"01:15.200","Text":"The Laplacian is linear,"},{"Start":"01:15.200 ","End":"01:19.570","Text":"so it\u0027s 2 minus the 2 from here is 0."},{"Start":"01:19.570 ","End":"01:28.130","Text":"On the boundary, w is u - v. U is 2x^2 on the boundary and v is x^2 everywhere,"},{"Start":"01:28.130 ","End":"01:29.410","Text":"in particular on the boundary,"},{"Start":"01:29.410 ","End":"01:31.969","Text":"so we get this minus this, which is x^2."},{"Start":"01:31.969 ","End":"01:36.650","Text":"The boundary is the circle x^2 + y^2 = 1."},{"Start":"01:36.650 ","End":"01:38.765","Text":"Note that on the boundary,"},{"Start":"01:38.765 ","End":"01:43.815","Text":"x is between -1 and 1 and w is x^2,"},{"Start":"01:43.815 ","End":"01:45.845","Text":"2x^2 minus x squared."},{"Start":"01:45.845 ","End":"01:49.550","Text":"But the minimum and maximum principles which we can"},{"Start":"01:49.550 ","End":"01:54.364","Text":"apply to w because it is harmonic and also it\u0027s non-constant."},{"Start":"01:54.364 ","End":"01:57.425","Text":"For all x, y in the open unit disc,"},{"Start":"01:57.425 ","End":"02:03.380","Text":"the minimum on the boundary of w is less than w on the disk,"},{"Start":"02:03.380 ","End":"02:07.880","Text":"which is less than the maximum of w on the boundary."},{"Start":"02:07.880 ","End":"02:10.070","Text":"In other words, w in the disk is strictly"},{"Start":"02:10.070 ","End":"02:13.205","Text":"between the minimum and maximum on the boundary."},{"Start":"02:13.205 ","End":"02:19.550","Text":"The minimum of x^2 is 0 when x is between -1 and 1,"},{"Start":"02:19.550 ","End":"02:24.150","Text":"and the maximum is 1, that\u0027s fairly clear."},{"Start":"02:24.350 ","End":"02:27.240","Text":"W is between 0 and 1,"},{"Start":"02:27.240 ","End":"02:30.540","Text":"but w is u - x^2,"},{"Start":"02:30.540 ","End":"02:34.580","Text":"so all we have to do is add the x^2 to all the sides and we get"},{"Start":"02:34.580 ","End":"02:38.840","Text":"that u is between x^2 and 1 plus x^2,"},{"Start":"02:38.840 ","End":"02:40.220","Text":"and that\u0027s what we had to show,"},{"Start":"02:40.220 ","End":"02:42.810","Text":"and so we are done."}],"ID":30865}],"Thumbnail":null,"ID":294441}]

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1.1

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