[{"Name":"Method of Characteristics","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Method of Characteristics","Duration":"15m 12s","ChapterTopicVideoID":29102,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29102.jpeg","UploadDate":"2022-06-06T07:35:27.3370000","DurationForVideoObject":"PT15M12S","Description":null,"MetaTitle":"Tutorial - Method of Characteristics: Video + Workbook | Proprep","MetaDescription":"First Order Equations - Method of Characteristics. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/partial-differential-equations/first-order-equations/method-of-characteristics/vid30663","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.110","Text":"To remind you, we\u0027re in the first-order partial differential equations."},{"Start":"00:07.110 ","End":"00:11.160","Text":"A specific kind called quasilinear as"},{"Start":"00:11.160 ","End":"00:16.350","Text":"a method of solution called the method of characteristics."},{"Start":"00:16.350 ","End":"00:20.310","Text":"Just to say what a quasilinear PDE is."},{"Start":"00:20.310 ","End":"00:26.145","Text":"It\u0027s of the form A u_x meaning derivative of u with respect to x"},{"Start":"00:26.145 ","End":"00:32.640","Text":"partial plus b partial of u with respect to y=C."},{"Start":"00:32.640 ","End":"00:34.890","Text":"The A, B, C are functions of x, y,"},{"Start":"00:34.890 ","End":"00:40.180","Text":"and u, and the solution we\u0027re looking for is of type u as a function of x and y."},{"Start":"00:40.720 ","End":"00:45.605","Text":"Now we need an initial condition in differential equations we always need,"},{"Start":"00:45.605 ","End":"00:46.819","Text":"and in this case,"},{"Start":"00:46.819 ","End":"00:48.755","Text":"we get it in the following form."},{"Start":"00:48.755 ","End":"00:55.970","Text":"Given a curve Gamma in the xy-plane and we\u0027re given the values u(x,"},{"Start":"00:55.970 ","End":"00:59.075","Text":"y) for (x,y) on this curve."},{"Start":"00:59.075 ","End":"01:06.110","Text":"Then we can build a curve in 3 dimensions such that the shadow of this,"},{"Start":"01:06.110 ","End":"01:11.225","Text":"the projection of this curve onto the xy-plane will be little Gamma."},{"Start":"01:11.225 ","End":"01:13.400","Text":"Big Gamma is x,"},{"Start":"01:13.400 ","End":"01:14.735","Text":"y, u of x,"},{"Start":"01:14.735 ","End":"01:20.345","Text":"y and the points on Gamma will be on the solution surface."},{"Start":"01:20.345 ","End":"01:24.040","Text":"A diagram at this point will help."},{"Start":"01:24.040 ","End":"01:27.970","Text":"We have in the plane some domain Omega,"},{"Start":"01:27.970 ","End":"01:30.610","Text":"it\u0027s an R2 in the xy-plane,"},{"Start":"01:30.610 ","End":"01:33.760","Text":"and we have a curve Gamma here,"},{"Start":"01:33.760 ","End":"01:38.715","Text":"and we\u0027re given the value of u along this Gamma,"},{"Start":"01:38.715 ","End":"01:43.390","Text":"which means that big Gamma would be this."},{"Start":"01:43.390 ","End":"01:48.820","Text":"We just take all the points on Gamma and raise them by u of x,y."},{"Start":"01:48.820 ","End":"01:51.427","Text":"This is just x, y;"},{"Start":"01:51.427 ","End":"01:52.540","Text":"x, y, u of x,"},{"Start":"01:52.540 ","End":"01:55.260","Text":"y, that\u0027s this curve."},{"Start":"01:55.260 ","End":"01:59.395","Text":"It\u0027s best to explain the method through the use of an example."},{"Start":"01:59.395 ","End":"02:01.210","Text":"Let\u0027s take the following,"},{"Start":"02:01.210 ","End":"02:04.645","Text":"which is of the quasilinear type."},{"Start":"02:04.645 ","End":"02:07.600","Text":"We\u0027ve got something u,"},{"Start":"02:07.600 ","End":"02:10.845","Text":"x plus something u, y equals something."},{"Start":"02:10.845 ","End":"02:12.990","Text":"In this case, A(x,"},{"Start":"02:12.990 ","End":"02:14.415","Text":"y, u) is 1,"},{"Start":"02:14.415 ","End":"02:16.200","Text":"B is 2,"},{"Start":"02:16.200 ","End":"02:18.915","Text":"and C is 3."},{"Start":"02:18.915 ","End":"02:23.645","Text":"This happens to be a linear partial differential equation with constant coefficients,"},{"Start":"02:23.645 ","End":"02:26.725","Text":"which is easier at least for starting off anyway."},{"Start":"02:26.725 ","End":"02:30.300","Text":"Gamma will give us basically,"},{"Start":"02:30.300 ","End":"02:32.040","Text":"it\u0027s the x-axis,"},{"Start":"02:32.040 ","End":"02:34.695","Text":"it\u0027s the set of all x, 0,"},{"Start":"02:34.695 ","End":"02:36.540","Text":"where x is the real number,"},{"Start":"02:36.540 ","End":"02:40.210","Text":"and the initial condition will be that u,"},{"Start":"02:40.210 ","End":"02:43.975","Text":"along this x-axis will just be x^2."},{"Start":"02:43.975 ","End":"02:46.795","Text":"Our curve Gamma will be x, 0,"},{"Start":"02:46.795 ","End":"02:51.675","Text":"x^2, that\u0027s u and I can illustrate that."},{"Start":"02:51.675 ","End":"02:56.235","Text":"This is the y-axis, x-axis."},{"Start":"02:56.235 ","End":"03:03.380","Text":"This is actually the little Gamma and big Gamma will be this,"},{"Start":"03:03.380 ","End":"03:07.255","Text":"all the points x, 0 but the height will be x^2."},{"Start":"03:07.255 ","End":"03:10.985","Text":"The solution surface has to pass through this."},{"Start":"03:10.985 ","End":"03:16.220","Text":"What we\u0027re looking for is a function u=u(x, y),"},{"Start":"03:16.220 ","End":"03:20.560","Text":"which on the one hand satisfies the partial differential equation,"},{"Start":"03:20.560 ","End":"03:23.660","Text":"on the other hand, it also satisfies the initial condition."},{"Start":"03:23.660 ","End":"03:31.175","Text":"The surface will pass through this and here\u0027s what it might look like."},{"Start":"03:31.175 ","End":"03:34.660","Text":"I used a 3D graphing tool."},{"Start":"03:34.660 ","End":"03:37.970","Text":"This is the surface. It passes through this."},{"Start":"03:37.970 ","End":"03:44.675","Text":"It\u0027s just to give you some idea to have something to picture in your mind if you need."},{"Start":"03:44.675 ","End":"03:48.070","Text":"You want to find the curve u(x,"},{"Start":"03:48.070 ","End":"03:51.695","Text":"y) which satisfies the partial differential equation"},{"Start":"03:51.695 ","End":"03:54.890","Text":"and passes through the initial curve Gamma,"},{"Start":"03:54.890 ","End":"03:57.065","Text":"which in our case is this."},{"Start":"03:57.065 ","End":"04:00.890","Text":"In this method, we\u0027re going to find the parametric form of the surface"},{"Start":"04:00.890 ","End":"04:04.940","Text":"u=u(x,y) but in terms of 2 parameters, t and Tau,"},{"Start":"04:04.940 ","End":"04:11.030","Text":"x will be x(t,τ) on y and u,"},{"Start":"04:11.030 ","End":"04:14.750","Text":"there will be functions of t and Tau, 2 parameters,"},{"Start":"04:14.750 ","End":"04:20.435","Text":"but in 3-dimensions it\u0027s a surface in 3D but like a 2D surface."},{"Start":"04:20.435 ","End":"04:24.350","Text":"For convenience, we\u0027re going to arrange it such that"},{"Start":"04:24.350 ","End":"04:29.020","Text":"the initial curve Gamma will be the curve where t=0,"},{"Start":"04:29.020 ","End":"04:32.705","Text":"Tau can vary but t=0."},{"Start":"04:32.705 ","End":"04:37.065","Text":"This will be Gamma when we replace the ts by 0,"},{"Start":"04:37.065 ","End":"04:40.605","Text":"the point on Gamma changes as we change Tau."},{"Start":"04:40.605 ","End":"04:44.520","Text":"It\u0027s really a 1 parameter curve, and that parameter is Tau."},{"Start":"04:44.520 ","End":"04:53.085","Text":"Now, we can describe Gamma as the sets of τ,0,τ^2."},{"Start":"04:53.085 ","End":"04:54.585","Text":"Gamma is x,0,x^2,"},{"Start":"04:54.585 ","End":"04:58.770","Text":"so if we just take x=Tau and we\u0027ll get this,"},{"Start":"04:58.770 ","End":"05:05.370","Text":"so x(0,τ) is Tau,"},{"Start":"05:05.370 ","End":"05:07.905","Text":"y(0,τ) is just 0,"},{"Start":"05:07.905 ","End":"05:14.960","Text":"the 0 here and that u is τ^2, x^2."},{"Start":"05:14.960 ","End":"05:18.545","Text":"The solution surface we can write as u(t,τ)="},{"Start":"05:18.545 ","End":"05:25.385","Text":"u(x,y) and x is x(t,τ) and y is y(t,τ)."},{"Start":"05:25.385 ","End":"05:27.770","Text":"We don\u0027t have these yet but when we have them,"},{"Start":"05:27.770 ","End":"05:31.460","Text":"we can write u as u(x,y) like this."},{"Start":"05:31.460 ","End":"05:34.400","Text":"Differentiating with respect to t,"},{"Start":"05:34.400 ","End":"05:41.990","Text":"we get u with respect to t. You can omit the (t,τ) when it\u0027s understood."},{"Start":"05:41.990 ","End":"05:44.750","Text":"In respect to t=u with respect to x,"},{"Start":"05:44.750 ","End":"05:48.515","Text":"x with respect to t plus u with respect to y,"},{"Start":"05:48.515 ","End":"05:49.830","Text":"y with respect to t,"},{"Start":"05:49.830 ","End":"05:51.635","Text":"that\u0027s the chain rule."},{"Start":"05:51.635 ","End":"05:56.970","Text":"Now note that this equation just write it with u_t on the right,"},{"Start":"05:56.970 ","End":"06:01.850","Text":"looks quite a bit like this."},{"Start":"06:01.850 ","End":"06:06.205","Text":"Which is the Au_x + Bu_y = C,"},{"Start":"06:06.205 ","End":"06:07.920","Text":"where in our case A is 1,"},{"Start":"06:07.920 ","End":"06:09.640","Text":"B is 2, and c is 3."},{"Start":"06:09.640 ","End":"06:14.315","Text":"What we can do is just require that x_t = A,"},{"Start":"06:14.315 ","End":"06:15.800","Text":"y_t = B,"},{"Start":"06:15.800 ","End":"06:21.560","Text":"and u_t = C. That gives us a system as follows,"},{"Start":"06:21.560 ","End":"06:27.500","Text":"we want x_t = 1,"},{"Start":"06:27.500 ","End":"06:31.295","Text":"and we want y_t = 2,"},{"Start":"06:31.295 ","End":"06:32.825","Text":"B in general,"},{"Start":"06:32.825 ","End":"06:36.230","Text":"and u_t = 3."},{"Start":"06:36.230 ","End":"06:40.100","Text":"We get these equations and these are called the"},{"Start":"06:40.100 ","End":"06:46.710","Text":"characteristic ordinary differential equations of the partial differential equation."},{"Start":"06:46.900 ","End":"06:50.900","Text":"Now the solution to this system is as"},{"Start":"06:50.900 ","End":"06:54.620","Text":"follows: since the derivative of x with respect to t is 1,"},{"Start":"06:54.620 ","End":"06:56.960","Text":"then x is the integral of 1dt,"},{"Start":"06:56.960 ","End":"06:59.000","Text":"which is t plus a constant,"},{"Start":"06:59.000 ","End":"07:05.150","Text":"but a constant really is an unknown function of Tau."},{"Start":"07:05.150 ","End":"07:08.165","Text":"If you differentiate this with respect to t, you get 0."},{"Start":"07:08.165 ","End":"07:09.890","Text":"Similarly with the others,"},{"Start":"07:09.890 ","End":"07:17.670","Text":"here we get the integral of 2 is 2t plus a function of Tau and here,"},{"Start":"07:17.670 ","End":"07:20.175","Text":"3t plus a function of Tau."},{"Start":"07:20.175 ","End":"07:27.125","Text":"The initial condition is the following that when t is 0,"},{"Start":"07:27.125 ","End":"07:30.995","Text":"then it goes through τ,0,τ^2,"},{"Start":"07:30.995 ","End":"07:33.390","Text":"which is the Gamma."},{"Start":"07:33.530 ","End":"07:41.195","Text":"From the above, if we let t = 0 here, here, and here,"},{"Start":"07:41.195 ","End":"07:42.920","Text":"this becomes 0, this is 0,"},{"Start":"07:42.920 ","End":"07:45.785","Text":"this is 0 so we just have C_1(τ),"},{"Start":"07:45.785 ","End":"07:48.755","Text":"and so we just compare this with this."},{"Start":"07:48.755 ","End":"07:51.920","Text":"We\u0027ve got C_1(τ) = τ,"},{"Start":"07:51.920 ","End":"07:55.855","Text":"C_2(τ) = 0, C_3(τ) = τ^2."},{"Start":"07:55.855 ","End":"07:58.760","Text":"Summarizing what we have so far,"},{"Start":"07:58.760 ","End":"08:01.025","Text":"we have that x, y,"},{"Start":"08:01.025 ","End":"08:10.282","Text":"and u as functions of t and Tau are given as follows and we even have c_1, c_2, c_3."},{"Start":"08:10.282 ","End":"08:14.845","Text":"We can just substitute these here."},{"Start":"08:14.845 ","End":"08:17.840","Text":"We get the solution,"},{"Start":"08:18.000 ","End":"08:22.810","Text":"x(t,Tau) is t plus Tau."},{"Start":"08:22.810 ","End":"08:25.690","Text":"Here we get 2t plus 0,"},{"Start":"08:25.690 ","End":"08:29.020","Text":"and here 3t plus Tau squared."},{"Start":"08:29.020 ","End":"08:30.580","Text":"We have x, y,"},{"Start":"08:30.580 ","End":"08:33.025","Text":"z in terms of t and Tau."},{"Start":"08:33.025 ","End":"08:35.110","Text":"This is in parametric form,"},{"Start":"08:35.110 ","End":"08:38.470","Text":"meaning x, y, and u in terms of t and Tau."},{"Start":"08:38.470 ","End":"08:41.260","Text":"We want it in explicit form,"},{"Start":"08:41.260 ","End":"08:45.190","Text":"meaning u as a function of x and y."},{"Start":"08:45.190 ","End":"08:49.750","Text":"What we have to do is take these 2 equations, hopefully,"},{"Start":"08:49.750 ","End":"08:56.035","Text":"reverse them to get Tau and t in terms of x and y."},{"Start":"08:56.035 ","End":"09:01.900","Text":"Then substitute that in here and will get u in terms of x and y."},{"Start":"09:01.900 ","End":"09:05.980","Text":"If we fix Tau and we let t vary,"},{"Start":"09:05.980 ","End":"09:15.580","Text":"we get a characteristic curve in t. The point where t equals 0 lies on Gamma."},{"Start":"09:15.580 ","End":"09:20.050","Text":"Together, letting Tau run over all possible values,"},{"Start":"09:20.050 ","End":"09:24.640","Text":"we get a bunch of curves that together form the surface."},{"Start":"09:24.640 ","End":"09:27.145","Text":"The diagram could help."},{"Start":"09:27.145 ","End":"09:31.930","Text":"Let\u0027s say that Tau equals 1."},{"Start":"09:31.930 ","End":"09:35.575","Text":"Here\u0027s the curve Gamma above it."},{"Start":"09:35.575 ","End":"09:38.980","Text":"There\u0027s the curve Gamma in 3D."},{"Start":"09:38.980 ","End":"09:41.695","Text":"This is the solution surface."},{"Start":"09:41.695 ","End":"09:45.835","Text":"Gamma is represented by T equals 0."},{"Start":"09:45.835 ","End":"09:52.105","Text":"T increases and decreases in this direction,"},{"Start":"09:52.105 ","End":"09:56.320","Text":"and Tau increases in this direction."},{"Start":"09:56.320 ","End":"10:00.340","Text":"This is like a Tau direction along here,"},{"Start":"10:00.340 ","End":"10:04.855","Text":"and T is width-wise."},{"Start":"10:04.855 ","End":"10:08.350","Text":"This would be where Tau equals 1,"},{"Start":"10:08.350 ","End":"10:11.230","Text":"just put 1 everywhere here."},{"Start":"10:11.230 ","End":"10:17.020","Text":"It has another characteristic curve letting Tau equals 2,"},{"Start":"10:17.020 ","End":"10:19.150","Text":"and here t is 0,"},{"Start":"10:19.150 ","End":"10:23.150","Text":"t is bigger or less than 0 here."},{"Start":"10:23.370 ","End":"10:26.890","Text":"Let Tau equal 3,"},{"Start":"10:26.890 ","End":"10:29.990","Text":"and we get another 1 of these."},{"Start":"10:31.050 ","End":"10:37.480","Text":"In general, just keep getting more of these as Tau varies."},{"Start":"10:37.480 ","End":"10:46.570","Text":"Now, remember we want to get back from u(t and Tau) to u(x and y)."},{"Start":"10:46.570 ","End":"10:48.250","Text":"Here are x, y,"},{"Start":"10:48.250 ","End":"10:51.925","Text":"and u in terms of t and Tau."},{"Start":"10:51.925 ","End":"10:54.490","Text":"Just consider the first 2."},{"Start":"10:54.490 ","End":"10:57.565","Text":"We have this set of equations,"},{"Start":"10:57.565 ","End":"11:00.160","Text":"x and y in terms of t and Tau."},{"Start":"11:00.160 ","End":"11:04.345","Text":"We want to reverse it to get t Tau in terms of x,y."},{"Start":"11:04.345 ","End":"11:05.650","Text":"To know if it\u0027s possible,"},{"Start":"11:05.650 ","End":"11:12.865","Text":"we have to check the Jacobian to make sure that the determinant of the Jacobian is not 0."},{"Start":"11:12.865 ","End":"11:16.750","Text":"The Jacobian is the matrix x_t,"},{"Start":"11:16.750 ","End":"11:19.375","Text":"x_Tau y_t, y_Tau."},{"Start":"11:19.375 ","End":"11:22.345","Text":"This is not absolute value, this is determinant."},{"Start":"11:22.345 ","End":"11:26.485","Text":"what we get, dx by dt is 1,"},{"Start":"11:26.485 ","End":"11:28.615","Text":"dx by dTau is 1,"},{"Start":"11:28.615 ","End":"11:32.305","Text":"dy by dt is 2,"},{"Start":"11:32.305 ","End":"11:34.900","Text":"and dy by dTau is 0."},{"Start":"11:34.900 ","End":"11:39.520","Text":"The determinant is 0 minus 2,"},{"Start":"11:39.520 ","End":"11:41.560","Text":"which is not 0,"},{"Start":"11:41.560 ","End":"11:43.885","Text":"and that\u0027s good. You want non-zero."},{"Start":"11:43.885 ","End":"11:46.165","Text":"Now let\u0027s start reversing it."},{"Start":"11:46.165 ","End":"11:51.250","Text":"We can get from here that t is y over 2."},{"Start":"11:51.250 ","End":"11:58.630","Text":"Then we can say from here that Tau is x minus t,"},{"Start":"11:58.630 ","End":"12:05.260","Text":"but t is y over 2. we have Tau in terms of x and y,"},{"Start":"12:05.260 ","End":"12:07.270","Text":"and here t in terms of x and y."},{"Start":"12:07.270 ","End":"12:09.565","Text":"Well, just in terms of y, but that\u0027s okay."},{"Start":"12:09.565 ","End":"12:14.875","Text":"Now we can substitute t and Tau into this third equation,"},{"Start":"12:14.875 ","End":"12:20.965","Text":"which gives us u(t and Tau), we get u(x,y)."},{"Start":"12:20.965 ","End":"12:25.990","Text":"3t is 3y over 2,"},{"Start":"12:25.990 ","End":"12:32.785","Text":"and Tau squared is x minus y over 2 squared."},{"Start":"12:32.785 ","End":"12:37.000","Text":"This gives us u in terms of x and y."},{"Start":"12:37.000 ","End":"12:39.865","Text":"This is the surface we\u0027re looking for."},{"Start":"12:39.865 ","End":"12:43.045","Text":"We\u0027re done, but don\u0027t go."},{"Start":"12:43.045 ","End":"12:45.940","Text":"Sometimes you might want to check the solution if we\u0027re"},{"Start":"12:45.940 ","End":"12:48.925","Text":"not sure if we have time on our hands."},{"Start":"12:48.925 ","End":"12:51.265","Text":"I\u0027ll show you how you would check this."},{"Start":"12:51.265 ","End":"12:54.730","Text":"It has to satisfy 2 things,"},{"Start":"12:54.730 ","End":"12:57.760","Text":"the initial condition and the differential equation."},{"Start":"12:57.760 ","End":"12:59.605","Text":"First, the initial condition,"},{"Start":"12:59.605 ","End":"13:03.235","Text":"we have to check that x,0,x^2,"},{"Start":"13:03.235 ","End":"13:05.560","Text":"which is our Gamma,"},{"Start":"13:05.560 ","End":"13:08.195","Text":"is on this surface."},{"Start":"13:08.195 ","End":"13:11.520","Text":"What we get, 3y over 2,"},{"Start":"13:11.520 ","End":"13:13.200","Text":"but y is 0,"},{"Start":"13:13.200 ","End":"13:15.390","Text":"so we get 3.0 over 2."},{"Start":"13:15.390 ","End":"13:20.370","Text":"Then here x and here y is 0,"},{"Start":"13:20.370 ","End":"13:24.565","Text":"so x minus 0 over 2 squared,"},{"Start":"13:24.565 ","End":"13:28.825","Text":"and u is x^2."},{"Start":"13:28.825 ","End":"13:31.690","Text":"We just basically put a question mark here."},{"Start":"13:31.690 ","End":"13:33.520","Text":"Does this equal this?"},{"Start":"13:33.520 ","End":"13:37.855","Text":"Well, yes, this is 0 and this part is 0."},{"Start":"13:37.855 ","End":"13:42.070","Text":"We\u0027re just left with x^2 equals x^2, which is true."},{"Start":"13:42.070 ","End":"13:43.870","Text":"That\u0027s the initial condition."},{"Start":"13:43.870 ","End":"13:52.765","Text":"Now, the PDE, is it true that u_x plus 2u_y=3? Let\u0027s see."},{"Start":"13:52.765 ","End":"13:55.255","Text":"Du by dx,"},{"Start":"13:55.255 ","End":"13:57.160","Text":"differentiate with respect to x,"},{"Start":"13:57.160 ","End":"14:01.150","Text":"this part is 0 because it doesn\u0027t have x in it."},{"Start":"14:01.150 ","End":"14:02.620","Text":"This with respect to x,"},{"Start":"14:02.620 ","End":"14:04.420","Text":"it\u0027s x minus a constant squared,"},{"Start":"14:04.420 ","End":"14:08.620","Text":"so it\u0027s twice x minus that constant."},{"Start":"14:08.620 ","End":"14:10.330","Text":"That\u0027s this first part now."},{"Start":"14:10.330 ","End":"14:13.315","Text":"Twice u with respect to y, so here\u0027s the 2."},{"Start":"14:13.315 ","End":"14:16.810","Text":"Du by dy, from 3y over 2,"},{"Start":"14:16.810 ","End":"14:18.880","Text":"we just get 3 over 2."},{"Start":"14:18.880 ","End":"14:23.605","Text":"From here we get twice x minus y over 2."},{"Start":"14:23.605 ","End":"14:25.630","Text":"The anti-derivative of this time is minus 1/2."},{"Start":"14:25.630 ","End":"14:28.570","Text":"Because it\u0027s with respect to y."},{"Start":"14:28.570 ","End":"14:31.180","Text":"Does this equal 3? Well, let\u0027s see."},{"Start":"14:31.180 ","End":"14:37.750","Text":"If we expand it, we get 2x minus y plus 3,"},{"Start":"14:37.750 ","End":"14:45.310","Text":"and then 2 with minus a half gives minus 1."},{"Start":"14:45.310 ","End":"14:47.170","Text":"Together with this 2,"},{"Start":"14:47.170 ","End":"14:48.895","Text":"we have minus 2."},{"Start":"14:48.895 ","End":"14:51.160","Text":"We have minus 2 times this,"},{"Start":"14:51.160 ","End":"14:55.510","Text":"which is minus 2x plus 1y."},{"Start":"14:55.510 ","End":"14:57.940","Text":"We got all this left-hand side."},{"Start":"14:57.940 ","End":"15:00.310","Text":"Now, 2x and 2x cancel,"},{"Start":"15:00.310 ","End":"15:02.140","Text":"minus y and y cancel."},{"Start":"15:02.140 ","End":"15:03.790","Text":"We\u0027re just left with 3."},{"Start":"15:03.790 ","End":"15:06.280","Text":"There we are. That\u0027s what we wanted to show."},{"Start":"15:06.280 ","End":"15:10.090","Text":"Yes, we have met both conditions."},{"Start":"15:10.090 ","End":"15:12.890","Text":"That concludes this exercise."}],"ID":30663},{"Watched":false,"Name":"Exercise 1","Duration":"6m 50s","ChapterTopicVideoID":29103,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"In this exercise, we have"},{"Start":"00:01.950 ","End":"00:08.475","Text":"a partial differential equation quasi-linear with an initial condition."},{"Start":"00:08.475 ","End":"00:10.725","Text":"This is the equation."},{"Start":"00:10.725 ","End":"00:14.475","Text":"This is the path in the x, y plane."},{"Start":"00:14.475 ","End":"00:16.635","Text":"This is the value of u along that path."},{"Start":"00:16.635 ","End":"00:18.420","Text":"This is the initial condition."},{"Start":"00:18.420 ","End":"00:23.700","Text":"Alpha here is just some constant not equal to 1/2."},{"Start":"00:23.700 ","End":"00:26.700","Text":"We\u0027ll follow the steps like in the tutorial."},{"Start":"00:26.700 ","End":"00:28.665","Text":"We find the initial curve,"},{"Start":"00:28.665 ","End":"00:34.335","Text":"big Gamma, which is u(xy)."},{"Start":"00:34.335 ","End":"00:36.960","Text":"It\u0027s a 3D curve, lies above Gamma."},{"Start":"00:36.960 ","End":"00:40.695","Text":"We parameterize it so that t equals 0,"},{"Start":"00:40.695 ","End":"00:44.065","Text":"then we have another parameter Tau."},{"Start":"00:44.065 ","End":"00:50.980","Text":"The easiest thing to do is just let x equals Tau and then y will equal Alpha Tau."},{"Start":"00:50.980 ","End":"00:54.335","Text":"Then u will equal x^2 plus y^2,"},{"Start":"00:54.335 ","End":"00:57.475","Text":"which is Tau^2 plus Alpha^2 Tau^2."},{"Start":"00:57.475 ","End":"01:01.570","Text":"The next step is to find what we call characteristic curves"},{"Start":"01:01.570 ","End":"01:06.070","Text":"by solving the characteristic ordinary differential equations."},{"Start":"01:06.070 ","End":"01:09.475","Text":"You could add the third one with u here,"},{"Start":"01:09.475 ","End":"01:12.580","Text":"but we don\u0027t need it now we\u0027ll use that one later."},{"Start":"01:12.580 ","End":"01:22.455","Text":"Now, this equation 2u_x plus u_y equals 0 gives us the a is 2, b is 1."},{"Start":"01:22.455 ","End":"01:24.970","Text":"Like I said, we don\u0027t need to c right now."},{"Start":"01:24.970 ","End":"01:29.020","Text":"This is comparing this to the template."},{"Start":"01:29.020 ","End":"01:32.820","Text":"x_t equals a, but a is 2."},{"Start":"01:32.820 ","End":"01:38.740","Text":"So x_t equals 2 and y_t will be 1."},{"Start":"01:38.740 ","End":"01:43.340","Text":"Partial derivative of y with respect to t. This one gives us"},{"Start":"01:43.340 ","End":"01:47.660","Text":"that x is 2t plus a constant."},{"Start":"01:47.660 ","End":"01:51.650","Text":"But a constant could be an arbitrary function of Tau."},{"Start":"01:51.650 ","End":"01:54.061","Text":"It\u0027s constant only as far as t goes."},{"Start":"01:54.061 ","End":"02:00.305","Text":"Similarly here, from this one we get y equals t plus some constant,"},{"Start":"02:00.305 ","End":"02:01.910","Text":"which is not really a constant,"},{"Start":"02:01.910 ","End":"02:04.180","Text":"it\u0027s a function of Tau."},{"Start":"02:04.180 ","End":"02:10.410","Text":"Then we can substitute t equals 0 so 2t is 0,"},{"Start":"02:10.410 ","End":"02:16.605","Text":"t is 0, we get this and compare these 2, these 2."},{"Start":"02:16.605 ","End":"02:24.980","Text":"That gives us that c_1 is Tau and c_2 is Alpha Tau."},{"Start":"02:24.980 ","End":"02:28.310","Text":"Now, plug these back in here."},{"Start":"02:28.310 ","End":"02:35.535","Text":"We have x(t) and Tau is 2t plus Tau and y(t) and Tau is t plus Alpha Tau."},{"Start":"02:35.535 ","End":"02:36.975","Text":"Now the next step,"},{"Start":"02:36.975 ","End":"02:44.180","Text":"we find u in terms of t and Tau from the third equation that we didn\u0027t use here,"},{"Start":"02:44.180 ","End":"02:46.310","Text":"du by dt is c,"},{"Start":"02:46.310 ","End":"02:52.039","Text":"and c in our case is 0. u by dt equals 0."},{"Start":"02:52.039 ","End":"02:54.260","Text":"But we also have our initial condition,"},{"Start":"02:54.260 ","End":"02:57.920","Text":"u of 0 and Tau is Tau^2 plus Alpha^2 Tau^."},{"Start":"02:57.920 ","End":"03:01.655","Text":"Yeah, we just scrolled it off, here it is."},{"Start":"03:01.655 ","End":"03:08.930","Text":"From this equation, we get that u(t) and Tau is a constant as far as t goes,"},{"Start":"03:08.930 ","End":"03:12.835","Text":"but it\u0027s an arbitrary function of Tau, call it c_3."},{"Start":"03:12.835 ","End":"03:18.555","Text":"That gives us that u(0) Tau is c_3 of Tau."},{"Start":"03:18.555 ","End":"03:21.780","Text":"Now, compare this to this,"},{"Start":"03:21.780 ","End":"03:27.155","Text":"and it gives us a c_3 of Tau is Tau^2 plus Alpha^2 Tau^2."},{"Start":"03:27.155 ","End":"03:32.870","Text":"Now we can plug this into here and we\u0027ve got what u is in terms of t and Tau,"},{"Start":"03:32.870 ","End":"03:36.310","Text":"but we want you in terms of x and y."},{"Start":"03:36.310 ","End":"03:40.250","Text":"Just copy this pair of equations here."},{"Start":"03:40.250 ","End":"03:43.355","Text":"This is x and y in terms of t and Tau."},{"Start":"03:43.355 ","End":"03:45.350","Text":"To make sure it\u0027s reversible,"},{"Start":"03:45.350 ","End":"03:48.245","Text":"we could check the determinant of the Jacobian,"},{"Start":"03:48.245 ","End":"03:51.488","Text":"which comes out to be 2, 1,"},{"Start":"03:51.488 ","End":"03:55.330","Text":"1 Alpha, which is 2 Alpha minus 1,"},{"Start":"03:55.330 ","End":"04:00.245","Text":"which is not 0, because we were given that Alpha is not equal to 1/2."},{"Start":"04:00.245 ","End":"04:07.820","Text":"Let\u0027s see if we can get t and Tau into the x and y. Subtract this minus twice this,"},{"Start":"04:07.820 ","End":"04:14.120","Text":"that will get rid of the t and we get x minus 2y equals Tau minus 2 Alpha Tau,"},{"Start":"04:14.120 ","End":"04:16.775","Text":"which gives us that Tau."},{"Start":"04:16.775 ","End":"04:18.620","Text":"Just take it out the brackets here,"},{"Start":"04:18.620 ","End":"04:19.880","Text":"we\u0027ve got one minus 2 Alpha."},{"Start":"04:19.880 ","End":"04:21.410","Text":"Bring that to the denominator,"},{"Start":"04:21.410 ","End":"04:23.450","Text":"so we have this."},{"Start":"04:23.450 ","End":"04:28.990","Text":"We just need Tau, we don\u0027t need t because u happens to be all in terms of Tau."},{"Start":"04:28.990 ","End":"04:35.825","Text":"So got that u of t and Tau is equal to 1 plus Alpha^2 Tau^2."},{"Start":"04:35.825 ","End":"04:37.700","Text":"Just factorizing this."},{"Start":"04:37.700 ","End":"04:42.050","Text":"Then we can substitute Tau from here and get that"},{"Start":"04:42.050 ","End":"04:48.465","Text":"u(xy) is 1 plus Alpha^2 times this squared."},{"Start":"04:48.465 ","End":"04:51.200","Text":"That\u0027s the answer. But don\u0027t go,"},{"Start":"04:51.200 ","End":"04:53.075","Text":"I\u0027d like to verify this solution."},{"Start":"04:53.075 ","End":"04:54.380","Text":"I\u0027ll do it in this exercise,"},{"Start":"04:54.380 ","End":"04:56.390","Text":"but not in the following ones."},{"Start":"04:56.390 ","End":"04:58.760","Text":"Here\u0027s the optional verification."},{"Start":"04:58.760 ","End":"05:00.989","Text":"This was what u was."},{"Start":"05:00.989 ","End":"05:04.160","Text":"Now, you have to check it against 2 conditions,"},{"Start":"05:04.160 ","End":"05:07.355","Text":"the initial condition, and the partial differential equation."},{"Start":"05:07.355 ","End":"05:10.828","Text":"The initial condition, which is the big Gamma,"},{"Start":"05:10.828 ","End":"05:12.250","Text":"we had it in terms of Tau,"},{"Start":"05:12.250 ","End":"05:19.385","Text":"but we can just as well put it in terms of x is on the surface of our solution at this."},{"Start":"05:19.385 ","End":"05:22.645","Text":"Let\u0027s check that this satisfies this."},{"Start":"05:22.645 ","End":"05:30.200","Text":"Substitute x and y. Y is Alpha x. we are working on the right-hand side,"},{"Start":"05:30.200 ","End":"05:32.240","Text":"take x out of the brackets,"},{"Start":"05:32.240 ","End":"05:35.450","Text":"so we have x^2 and 1 minus 2 Alpha^2."},{"Start":"05:35.450 ","End":"05:37.835","Text":"Now this cancels with this."},{"Start":"05:37.835 ","End":"05:42.655","Text":"We multiply this with this we\u0027ve got x^2 plus Alpha^2x^2,"},{"Start":"05:42.655 ","End":"05:44.645","Text":"which is exactly what we needed."},{"Start":"05:44.645 ","End":"05:49.120","Text":"This is u, so that\u0027s the initial condition satisfied."},{"Start":"05:49.120 ","End":"05:51.830","Text":"Now the partial differential equation,"},{"Start":"05:51.830 ","End":"05:54.830","Text":"which was this, let\u0027s see if it\u0027s true."},{"Start":"05:54.830 ","End":"05:57.260","Text":"Du by dx, we get from here,"},{"Start":"05:57.260 ","End":"06:00.440","Text":"this is a constant, just differentiate this with respect to x."},{"Start":"06:00.440 ","End":"06:07.430","Text":"We get twice x minus 2y and the internal derivative is 1 in the case of du by dy,"},{"Start":"06:07.430 ","End":"06:10.760","Text":"the derivative of this part is twice x minus 2y,"},{"Start":"06:10.760 ","End":"06:13.660","Text":"but there isn\u0027t in a derivative of minus 2."},{"Start":"06:13.660 ","End":"06:16.170","Text":"That\u0027s this minus 2 here."},{"Start":"06:16.170 ","End":"06:21.930","Text":"We have 1 plus Alpha^2 over 1 minus 2 Alpha^2 here."},{"Start":"06:21.930 ","End":"06:26.025","Text":"Here, we factorize it out so that\u0027s this."},{"Start":"06:26.025 ","End":"06:30.855","Text":"Then x minus 2y we have here and here."},{"Start":"06:30.855 ","End":"06:33.150","Text":"We put that here."},{"Start":"06:33.150 ","End":"06:39.750","Text":"All we\u0027re left with then is 2 times 2 plus 2 times minus 2,"},{"Start":"06:39.750 ","End":"06:43.575","Text":"which is 0 and that\u0027s what we wanted."},{"Start":"06:43.575 ","End":"06:45.880","Text":"That works out also."},{"Start":"06:45.880 ","End":"06:47.675","Text":"With this verification,"},{"Start":"06:47.675 ","End":"06:51.090","Text":"we\u0027ll conclude this exercise."}],"ID":30664},{"Watched":false,"Name":"Exercise 2","Duration":"5m 46s","ChapterTopicVideoID":29104,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.170","Text":"In this exercise, we have a partial differential equation, quasi-linear,"},{"Start":"00:07.170 ","End":"00:09.960","Text":"and we have an initial condition,"},{"Start":"00:09.960 ","End":"00:14.490","Text":"a path Gamma, which is half a parabola in the plane."},{"Start":"00:14.490 ","End":"00:18.000","Text":"We\u0027re given the value of u on this parabola,"},{"Start":"00:18.000 ","End":"00:21.195","Text":"that u(xy) is x minus y."},{"Start":"00:21.195 ","End":"00:25.695","Text":"We\u0027ll solve it using the method of characteristics like in the tutorial."},{"Start":"00:25.695 ","End":"00:30.465","Text":"The first step is to find a parametrization of the initial curve,"},{"Start":"00:30.465 ","End":"00:32.820","Text":"big Gamma, which is a set of all x,"},{"Start":"00:32.820 ","End":"00:36.030","Text":"y, and u(xy) where x and y are in Gamma."},{"Start":"00:36.030 ","End":"00:37.470","Text":"In this method,"},{"Start":"00:37.470 ","End":"00:43.010","Text":"we arrange it so that the curve Gamma is where t=0."},{"Start":"00:43.010 ","End":"00:44.420","Text":"We need 2 parameters;"},{"Start":"00:44.420 ","End":"00:45.785","Text":"t, and Tau,"},{"Start":"00:45.785 ","End":"00:49.955","Text":"and t=0 is the initial curve."},{"Start":"00:49.955 ","End":"00:54.080","Text":"Straightforward thing to do is to take x=Tau,"},{"Start":"00:54.080 ","End":"00:56.510","Text":"and then y is Tau squared,"},{"Start":"00:56.510 ","End":"00:59.195","Text":"and then u which is x minus y,"},{"Start":"00:59.195 ","End":"01:02.855","Text":"is Tau minus Tau squared as here."},{"Start":"01:02.855 ","End":"01:09.040","Text":"The condition x bigger or equal to 0 becomes Tau bigger or equal to 0."},{"Start":"01:09.040 ","End":"01:13.925","Text":"Then step 2 is to find the characteristic curves by"},{"Start":"01:13.925 ","End":"01:18.780","Text":"solving the pair of equations dx by dt is A,"},{"Start":"01:18.780 ","End":"01:20.925","Text":"and dy by dt is B."},{"Start":"01:20.925 ","End":"01:24.380","Text":"To our 3rd equation du by dt equals C,"},{"Start":"01:24.380 ","End":"01:27.265","Text":"but we don\u0027t need it just yet."},{"Start":"01:27.265 ","End":"01:35.240","Text":"A, B, and C we get them from this equation according to this template so that A is 3,"},{"Start":"01:35.240 ","End":"01:38.350","Text":"B is minus 2, C is 0."},{"Start":"01:38.350 ","End":"01:40.800","Text":"They\u0027re all constants it happens."},{"Start":"01:40.800 ","End":"01:43.560","Text":"This becomes x_t=3,"},{"Start":"01:43.560 ","End":"01:46.615","Text":"y_t equals minus 2."},{"Start":"01:46.615 ","End":"01:50.045","Text":"Integrating this with respect to t,"},{"Start":"01:50.045 ","End":"01:53.780","Text":"we have x is 3t plus,"},{"Start":"01:53.780 ","End":"01:58.730","Text":"not a constant but a function that only depends on Tau,"},{"Start":"01:58.730 ","End":"02:01.070","Text":"and not on t. Similarly,"},{"Start":"02:01.070 ","End":"02:07.070","Text":"for y we have minus 2t plus another constant which is really a function of Tau."},{"Start":"02:07.070 ","End":"02:12.510","Text":"If we substitute t=0 in these, we get x(0,"},{"Start":"02:12.510 ","End":"02:14.795","Text":"Tau) is C_1(Tau),"},{"Start":"02:14.795 ","End":"02:16.045","Text":"because this is 0,"},{"Start":"02:16.045 ","End":"02:20.390","Text":"and this is 0, so y(0, Tau) is C_2(Tau)."},{"Start":"02:20.390 ","End":"02:26.090","Text":"Now if we compare this pair of equations with the first 2 here,"},{"Start":"02:26.090 ","End":"02:29.735","Text":"it gives us that C_1(Tau) is Tau,"},{"Start":"02:29.735 ","End":"02:33.140","Text":"and C_2(Tau) is Tau squared."},{"Start":"02:33.140 ","End":"02:35.730","Text":"Yeah, that\u0027s what I just said,"},{"Start":"02:35.730 ","End":"02:40.035","Text":"and then by substituting these here, and here,"},{"Start":"02:40.035 ","End":"02:41.880","Text":"we\u0027ve got that x,"},{"Start":"02:41.880 ","End":"02:44.880","Text":"and y are as follows."},{"Start":"02:44.880 ","End":"02:47.915","Text":"We\u0027ve got x and y in terms of t and Tau."},{"Start":"02:47.915 ","End":"02:49.250","Text":"Now as I said,"},{"Start":"02:49.250 ","End":"02:52.910","Text":"we didn\u0027t put the 3rd equation in this series,"},{"Start":"02:52.910 ","End":"02:58.955","Text":"but the 3rd one is u with respect to t equals C. In our case,"},{"Start":"02:58.955 ","End":"03:02.340","Text":"C is 0 as here,"},{"Start":"03:02.340 ","End":"03:06.500","Text":"and what we get is that u with respect to t is 0,"},{"Start":"03:06.500 ","End":"03:10.835","Text":"but we also have an initial condition here."},{"Start":"03:10.835 ","End":"03:13.430","Text":"Integrating this with respect to t,"},{"Start":"03:13.430 ","End":"03:17.540","Text":"we have the u=C_3(Tau),"},{"Start":"03:17.540 ","End":"03:19.880","Text":"and if we put t=0,"},{"Start":"03:19.880 ","End":"03:23.495","Text":"we get u(0, Tau) equals C_3(Tau)."},{"Start":"03:23.495 ","End":"03:25.340","Text":"But we also have that u(0,"},{"Start":"03:25.340 ","End":"03:27.860","Text":"Tau) is Tau minus Tau squared."},{"Start":"03:27.860 ","End":"03:30.020","Text":"This has got to equal this,"},{"Start":"03:30.020 ","End":"03:33.695","Text":"and that gives us that C_3(Tau) is tau minus tau squared,"},{"Start":"03:33.695 ","End":"03:37.510","Text":"which we can then substitute in here,"},{"Start":"03:37.510 ","End":"03:42.770","Text":"and we have the u=Tau minus Tau squared."},{"Start":"03:42.770 ","End":"03:46.145","Text":"That\u0027s good, but we want u in terms of x and y,"},{"Start":"03:46.145 ","End":"03:48.250","Text":"so we go to the next step,"},{"Start":"03:48.250 ","End":"03:50.030","Text":"and to be safe,"},{"Start":"03:50.030 ","End":"03:55.385","Text":"we\u0027ll check the Jacobian to see that this transformation is reversible."},{"Start":"03:55.385 ","End":"03:58.475","Text":"You can get back t and Tau from x and y."},{"Start":"03:58.475 ","End":"04:01.070","Text":"The determinant of the Jacobian,"},{"Start":"04:01.070 ","End":"04:03.725","Text":"which is given by this expression is 3,"},{"Start":"04:03.725 ","End":"04:05.510","Text":"1 minus 2,"},{"Start":"04:05.510 ","End":"04:07.720","Text":"and here 2 Tau."},{"Start":"04:07.720 ","End":"04:12.035","Text":"That\u0027s equal 6 Tau plus 2, which is bigger than 0,"},{"Start":"04:12.035 ","End":"04:14.000","Text":"since Tau is bigger than 0,"},{"Start":"04:14.000 ","End":"04:16.235","Text":"and in particular it\u0027s not equal to 0,"},{"Start":"04:16.235 ","End":"04:18.205","Text":"which is what we need."},{"Start":"04:18.205 ","End":"04:20.715","Text":"We start reversing."},{"Start":"04:20.715 ","End":"04:23.720","Text":"What we can do here is multiply this one by 2,"},{"Start":"04:23.720 ","End":"04:25.955","Text":"and this one by 3,"},{"Start":"04:25.955 ","End":"04:29.675","Text":"then we have a 6t here and a minus 6t here."},{"Start":"04:29.675 ","End":"04:35.570","Text":"If we add them, the t disappears and we have 2 Tau plus 3Tau^2."},{"Start":"04:35.570 ","End":"04:37.850","Text":"If we rearrange this,"},{"Start":"04:37.850 ","End":"04:41.345","Text":"we get a quadratic equation in Tau."},{"Start":"04:41.345 ","End":"04:44.000","Text":"Get the 3Tau^2 plus 2 Tau,"},{"Start":"04:44.000 ","End":"04:46.940","Text":"and then this becomes minus equals 0,"},{"Start":"04:46.940 ","End":"04:49.760","Text":"so we can use the quadratic formula."},{"Start":"04:49.760 ","End":"04:52.115","Text":"This is what it comes out to be,"},{"Start":"04:52.115 ","End":"04:55.430","Text":"since Tau is bigger or equal to 0,"},{"Start":"04:55.430 ","End":"04:58.260","Text":"we\u0027d better take the plus here,"},{"Start":"04:58.480 ","End":"05:05.035","Text":"also dividing minus 2 over 6 is minus 1/3."},{"Start":"05:05.035 ","End":"05:06.890","Text":"Here in the square root,"},{"Start":"05:06.890 ","End":"05:10.025","Text":"we can put the 6 inside as 36,"},{"Start":"05:10.025 ","End":"05:12.755","Text":"4 over 36 is 1/9,"},{"Start":"05:12.755 ","End":"05:15.230","Text":"12 over 36 is a 1/3,"},{"Start":"05:15.230 ","End":"05:18.485","Text":"so this simplifies to this."},{"Start":"05:18.485 ","End":"05:21.815","Text":"Then we have Tau in terms of x and y,"},{"Start":"05:21.815 ","End":"05:26.965","Text":"and we don\u0027t need t because u just depends on Tau,"},{"Start":"05:26.965 ","End":"05:30.020","Text":"so u being Tau minus Tau squared,"},{"Start":"05:30.020 ","End":"05:33.445","Text":"just substitute what Tau is here."},{"Start":"05:33.445 ","End":"05:36.530","Text":"This is what we get, this part is Tau,"},{"Start":"05:36.530 ","End":"05:40.280","Text":"and minus this part is Tau squared."},{"Start":"05:40.280 ","End":"05:42.560","Text":"Maybe it simplifies, maybe not,"},{"Start":"05:42.560 ","End":"05:44.225","Text":"but just leave it like this,"},{"Start":"05:44.225 ","End":"05:47.250","Text":"and that concludes this exercise."}],"ID":30665},{"Watched":false,"Name":"Exercise 3","Duration":"5m 52s","ChapterTopicVideoID":29105,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"In this exercise, we\u0027re going to solve the following equation."},{"Start":"00:04.830 ","End":"00:08.760","Text":"It\u0027s a quasi-linear partial differential equation"},{"Start":"00:08.760 ","End":"00:12.360","Text":"and it has an initial condition like so."},{"Start":"00:12.360 ","End":"00:16.215","Text":"Gamma is a curve in the plane half a parabola,"},{"Start":"00:16.215 ","End":"00:22.620","Text":"and u is the value of the solution above the curve, little Gamma."},{"Start":"00:22.620 ","End":"00:30.878","Text":"To remind you, this is little Gamma in the plane and big Gamma is in the solution space,"},{"Start":"00:30.878 ","End":"00:33.315","Text":"and it\u0027s given by this formula."},{"Start":"00:33.315 ","End":"00:35.218","Text":"We don\u0027t need that,"},{"Start":"00:35.218 ","End":"00:37.754","Text":"and we\u0027ll be using the method of characteristics on"},{"Start":"00:37.754 ","End":"00:41.540","Text":"the first step in that is to find a parametrization of"},{"Start":"00:41.540 ","End":"00:48.325","Text":"big Gamma and we find one such that t equals 0."},{"Start":"00:48.325 ","End":"00:50.558","Text":"I should say its 2 parameters t,"},{"Start":"00:50.558 ","End":"00:54.450","Text":"and Tau that will be the solution surface."},{"Start":"00:54.450 ","End":"01:01.050","Text":"I brought the picture back again so that this is a two-dimensional surface,"},{"Start":"01:01.050 ","End":"01:02.870","Text":"it needs 2 parameters."},{"Start":"01:02.870 ","End":"01:08.875","Text":"We\u0027ll have t and Tau and this Gamma will be where t equals 0 and Tau varies."},{"Start":"01:08.875 ","End":"01:15.630","Text":"Later on, will vary t and get the whole surface that\u0027s the method of characteristics."},{"Start":"01:15.630 ","End":"01:18.630","Text":"We\u0027ll take x as just Tau,"},{"Start":"01:18.630 ","End":"01:22.740","Text":"it\u0027s the most obvious thing to do and then we need Tau less than or equal to 0."},{"Start":"01:22.740 ","End":"01:30.120","Text":"Y is x^2, so Tau^2 and u is x plus sine xy, xy is Tau^3."},{"Start":"01:30.120 ","End":"01:31.770","Text":"This is what we have."},{"Start":"01:31.770 ","End":"01:33.710","Text":"Now, we go to the second step,"},{"Start":"01:33.710 ","End":"01:40.445","Text":"which is to find these characteristic curves and we solve the characteristic ODEs."},{"Start":"01:40.445 ","End":"01:43.340","Text":"There are actually three of them, xt equals A,"},{"Start":"01:43.340 ","End":"01:47.075","Text":"yt equals B,"},{"Start":"01:47.075 ","End":"01:52.145","Text":"and ut equals C, but we don\u0027t need that third one just yet."},{"Start":"01:52.145 ","End":"01:57.240","Text":"The A, B, and the C come from this equation."},{"Start":"01:57.240 ","End":"02:01.050","Text":"In our case we have here the A is 1,"},{"Start":"02:01.050 ","End":"02:02.330","Text":"B is 2,"},{"Start":"02:02.330 ","End":"02:04.340","Text":"and c is 0."},{"Start":"02:04.340 ","End":"02:06.275","Text":"They\u0027re all constants here."},{"Start":"02:06.275 ","End":"02:10.115","Text":"Xt=A, so that\u0027s xt=1,"},{"Start":"02:10.115 ","End":"02:16.190","Text":"yt=B derivative of y with respect to t. We can integrate these with"},{"Start":"02:16.190 ","End":"02:22.730","Text":"respect to t. We get that x is t plus a constant,"},{"Start":"02:22.730 ","End":"02:24.340","Text":"but not a constant,"},{"Start":"02:24.340 ","End":"02:28.640","Text":"a function of Tau and similarly here,"},{"Start":"02:28.640 ","End":"02:33.730","Text":"integral of 2 is 2t plus a function of Tau."},{"Start":"02:33.730 ","End":"02:38.550","Text":"If we substitute t=0, this is 0,"},{"Start":"02:38.550 ","End":"02:46.610","Text":"this is 0 and we get this and compare these two equations to these two."},{"Start":"02:46.610 ","End":"02:51.500","Text":"Well, that gives us that C1 of Tau is Tau,"},{"Start":"02:51.500 ","End":"02:54.519","Text":"and C2 of Tau is Tau^2."},{"Start":"02:54.519 ","End":"03:00.956","Text":"What we get, plugging these two here and here is we get x of t"},{"Start":"03:00.956 ","End":"03:07.295","Text":"and Tau is t plus Tau and y of t and Tau is 2t plus Tau^2."},{"Start":"03:07.295 ","End":"03:10.970","Text":"Now that we have x and y of t and Tau,"},{"Start":"03:10.970 ","End":"03:12.925","Text":"we also want u,"},{"Start":"03:12.925 ","End":"03:19.697","Text":"remember I said there\u0027s a third one here that u with respect to t is C. Yeah,"},{"Start":"03:19.697 ","End":"03:22.550","Text":"we get u with respect to t is C,"},{"Start":"03:22.550 ","End":"03:27.020","Text":"and in our case, c is 0 so we get u by dt is just"},{"Start":"03:27.020 ","End":"03:32.790","Text":"0 and there\u0027s also an initial condition, this one here."},{"Start":"03:32.790 ","End":"03:37.460","Text":"From this, we get that u is constant in t,"},{"Start":"03:37.460 ","End":"03:40.340","Text":"but really a function of Tau."},{"Start":"03:40.340 ","End":"03:42.335","Text":"This one I just copied."},{"Start":"03:42.335 ","End":"03:44.630","Text":"Now this gives us,"},{"Start":"03:44.630 ","End":"03:46.730","Text":"we plug in t equals 0,"},{"Start":"03:46.730 ","End":"03:50.765","Text":"we get u of 0 and Tau is C3 of Tau."},{"Start":"03:50.765 ","End":"03:57.865","Text":"But here we also have an expression u of 0 and Tau so C3 of tau is got to equal this."},{"Start":"03:57.865 ","End":"03:59.620","Text":"Now that we have C3,"},{"Start":"03:59.620 ","End":"04:06.370","Text":"we can put it here and get u of t and Tau is Tau plus sine of Tau^3."},{"Start":"04:06.370 ","End":"04:11.435","Text":"We have x and y and even u but in terms of t and Tau,"},{"Start":"04:11.435 ","End":"04:17.600","Text":"what we would like is to have u in terms of x and y and that brings us to Step 4."},{"Start":"04:17.600 ","End":"04:20.870","Text":"Now, here we have x and y in terms of t and Tau."},{"Start":"04:20.870 ","End":"04:24.260","Text":"I want to reverse it to get t and Tau in terms of x and y."},{"Start":"04:24.260 ","End":"04:26.015","Text":"To know if it\u0027s possible,"},{"Start":"04:26.015 ","End":"04:31.835","Text":"we can check the determinant of the Jacobian, which is this,"},{"Start":"04:31.835 ","End":"04:34.160","Text":"is 1, 1, 2,"},{"Start":"04:34.160 ","End":"04:40.010","Text":"2 Tau, this diagonal minus this diagonal is 2Tau minus 2."},{"Start":"04:40.010 ","End":"04:41.870","Text":"That is not 0. In fact,"},{"Start":"04:41.870 ","End":"04:45.905","Text":"it\u0027s less than 0 because Tau is less than or equal to 0."},{"Start":"04:45.905 ","End":"04:48.145","Text":"We\u0027re okay there."},{"Start":"04:48.145 ","End":"04:52.085","Text":"X and y, we have, from here,"},{"Start":"04:52.085 ","End":"04:57.665","Text":"we can double this row to get 2x equals 2t plus 2Tau,"},{"Start":"04:57.665 ","End":"05:03.205","Text":"and then if we subtract we get rid of t. We get y minus 2x"},{"Start":"05:03.205 ","End":"05:08.910","Text":"is Tau squared minus 2t and that\u0027s a quadratic equation in Tau."},{"Start":"05:08.910 ","End":"05:15.560","Text":"Just rearrange it like so and we can use the formula for a quadratic equation to get Tau."},{"Start":"05:15.560 ","End":"05:17.555","Text":"You know the formula,"},{"Start":"05:17.555 ","End":"05:18.950","Text":"this is what we get."},{"Start":"05:18.950 ","End":"05:21.040","Text":"Simplify it a bit."},{"Start":"05:21.040 ","End":"05:27.090","Text":"Also, we only need the minus because Tau is less than or equal to zeros,"},{"Start":"05:27.090 ","End":"05:29.295","Text":"we take a plus we\u0027ll get something positive."},{"Start":"05:29.295 ","End":"05:32.580","Text":"That gives us Tau in terms of x and y,"},{"Start":"05:32.580 ","End":"05:39.605","Text":"and we don\u0027t need t because u is completely in terms of Tau."},{"Start":"05:39.605 ","End":"05:46.505","Text":"Yeah, from here, just substitute Tau equals this and we get the following."},{"Start":"05:46.505 ","End":"05:48.950","Text":"This is Tau sine of Tau^3,"},{"Start":"05:48.950 ","End":"05:53.130","Text":"and that\u0027s the answer, we are done."}],"ID":30666},{"Watched":false,"Name":"Exercise 4","Duration":"5m 2s","ChapterTopicVideoID":29106,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"In this exercise, we\u0027re going to solve"},{"Start":"00:03.045 ","End":"00:09.150","Text":"the following quasi-linear partial differential equation with initial condition."},{"Start":"00:09.150 ","End":"00:11.835","Text":"This is the equation,"},{"Start":"00:11.835 ","End":"00:16.365","Text":"and we have a curve gamma in the plane,"},{"Start":"00:16.365 ","End":"00:23.640","Text":"which is actually just the x-axis and we have the value of the solution u on gamma,"},{"Start":"00:23.640 ","End":"00:26.670","Text":"which is x squared minus x to the fourth."},{"Start":"00:26.670 ","End":"00:31.800","Text":"First step is to parameterize the initial curve, big gamma."},{"Start":"00:31.800 ","End":"00:36.660","Text":"We parametrize the whole surface as dependent on t and"},{"Start":"00:36.660 ","End":"00:41.745","Text":"Tau and we arrange it such that when t equals 0,"},{"Start":"00:41.745 ","End":"00:45.510","Text":"it gives us our curve, big Gamma."},{"Start":"00:45.510 ","End":"00:53.404","Text":"The most straightforward is to take x as the parameter Tau, y."},{"Start":"00:53.404 ","End":"00:58.700","Text":"Well, it\u0027s forced y is equal to 0 and u would be,"},{"Start":"00:58.700 ","End":"01:03.380","Text":"since it\u0027s x squared minus x to the fourth would be Tau squared minus Tau to the fourth."},{"Start":"01:03.380 ","End":"01:06.300","Text":"We\u0027re using the method of characteristics."},{"Start":"01:06.300 ","End":"01:10.400","Text":"By the way, like in the tutorials or the next step will be to find"},{"Start":"01:10.400 ","End":"01:16.745","Text":"characteristic curves by solving these characteristic ODEs."},{"Start":"01:16.745 ","End":"01:20.275","Text":"We just need the first 2 at the moment."},{"Start":"01:20.275 ","End":"01:24.165","Text":"To remind you what big a and b are,"},{"Start":"01:24.165 ","End":"01:29.560","Text":"this is the template and a here would be 1,"},{"Start":"01:29.560 ","End":"01:32.095","Text":"b is minus 1,"},{"Start":"01:32.095 ","End":"01:35.105","Text":"and c is minus u."},{"Start":"01:35.105 ","End":"01:37.365","Text":"We just need these 2 for now,"},{"Start":"01:37.365 ","End":"01:38.720","Text":"the u will do later."},{"Start":"01:38.720 ","End":"01:41.285","Text":"So we get x t equals a,"},{"Start":"01:41.285 ","End":"01:43.070","Text":"which is 1, yt is b,"},{"Start":"01:43.070 ","End":"01:44.990","Text":"which is minus 1."},{"Start":"01:44.990 ","End":"01:48.245","Text":"We integrate each of these with respect to t,"},{"Start":"01:48.245 ","End":"01:53.300","Text":"we get that x is the integral of this plus an arbitrary function of"},{"Start":"01:53.300 ","End":"01:58.505","Text":"Tau and y here is minus t plus a function of Tau."},{"Start":"01:58.505 ","End":"02:03.230","Text":"If we substitute t equals 0 here and here these 2 become 0,"},{"Start":"02:03.230 ","End":"02:04.535","Text":"and we get this."},{"Start":"02:04.535 ","End":"02:08.375","Text":"Now we also have X of 0 and Tau here."},{"Start":"02:08.375 ","End":"02:11.410","Text":"X of 0 and Tau is Tau."},{"Start":"02:11.410 ","End":"02:17.265","Text":"That gives us that c1 of Tau is Tau and y of 0 Tau is 0."},{"Start":"02:17.265 ","End":"02:20.750","Text":"So this is going to equal 0."},{"Start":"02:20.750 ","End":"02:27.200","Text":"That means that x of t and Tau is t plus c1 of Tau is Tau,"},{"Start":"02:27.200 ","End":"02:29.960","Text":"and here we get minus d plus nothing."},{"Start":"02:29.960 ","End":"02:33.205","Text":"We have x and y in terms of t and Tau."},{"Start":"02:33.205 ","End":"02:37.535","Text":"Now we need to find what u is and we\u0027re going to use"},{"Start":"02:37.535 ","End":"02:43.145","Text":"this equation that u derivative with respect to t is c,"},{"Start":"02:43.145 ","End":"02:45.620","Text":"c is minus u,"},{"Start":"02:45.620 ","End":"02:49.085","Text":"so we get d by d t is minus u."},{"Start":"02:49.085 ","End":"02:51.815","Text":"Initial condition is this."},{"Start":"02:51.815 ","End":"02:57.370","Text":"This is a first-order linear constant coefficients in other solution to this,"},{"Start":"02:57.370 ","End":"03:00.890","Text":"u is c3 of Tau,"},{"Start":"03:00.890 ","End":"03:04.445","Text":"it like a constant e to the minus t. Now,"},{"Start":"03:04.445 ","End":"03:06.525","Text":"the initial condition is this."},{"Start":"03:06.525 ","End":"03:09.710","Text":"If we put t equals 0 here,"},{"Start":"03:09.710 ","End":"03:14.170","Text":"we get just the c3 because this becomes 1."},{"Start":"03:14.170 ","End":"03:15.980","Text":"From this and this,"},{"Start":"03:15.980 ","End":"03:19.055","Text":"we see that c3 has to equal this."},{"Start":"03:19.055 ","End":"03:26.945","Text":"Now that we have c3, we have u of t and Tau is Tau squared minus Tau to the fourth,"},{"Start":"03:26.945 ","End":"03:29.960","Text":"e to the minus t. That\u0027s good,"},{"Start":"03:29.960 ","End":"03:32.060","Text":"but we want u explicitly,"},{"Start":"03:32.060 ","End":"03:34.825","Text":"I mean, in terms of x and y."},{"Start":"03:34.825 ","End":"03:36.770","Text":"We want to reverse this."},{"Start":"03:36.770 ","End":"03:41.300","Text":"In other words, to get t and tau in terms of x and y."},{"Start":"03:41.300 ","End":"03:47.180","Text":"Let\u0027s first of all see if we can invert this using the Jacobian."},{"Start":"03:47.180 ","End":"03:51.695","Text":"The determinant of Jacobian is the determinant of x d x tau,"},{"Start":"03:51.695 ","End":"03:53.440","Text":"y t y tau."},{"Start":"03:53.440 ","End":"03:57.555","Text":"From here, when you see this is 1,1 minus 1,0."},{"Start":"03:57.555 ","End":"04:01.090","Text":"This determinant comes out to be 1,"},{"Start":"04:01.090 ","End":"04:02.825","Text":"which is certainly not 0,"},{"Start":"04:02.825 ","End":"04:04.910","Text":"which is good means that we can invert this."},{"Start":"04:04.910 ","End":"04:06.515","Text":"Now let\u0027s actually do it."},{"Start":"04:06.515 ","End":"04:13.445","Text":"We start off with x equals t plus Tau y equals minus t. Because t is minus y,"},{"Start":"04:13.445 ","End":"04:17.405","Text":"we can put x equals minus y plus tau,"},{"Start":"04:17.405 ","End":"04:19.580","Text":"and then we have t equals minus y."},{"Start":"04:19.580 ","End":"04:24.870","Text":"We can also bring y over to this side and get Tau equals x plus y."},{"Start":"04:24.870 ","End":"04:28.430","Text":"Now we have tau and t in terms of x and y."},{"Start":"04:28.430 ","End":"04:33.860","Text":"Now we can substitute t and Tau in here and we can get u in terms of"},{"Start":"04:33.860 ","End":"04:40.100","Text":"x and y. U equals Tau squared minus tau to the fourth,"},{"Start":"04:40.100 ","End":"04:43.325","Text":"e to the minus t. Just copying first,"},{"Start":"04:43.325 ","End":"04:45.880","Text":"now make this substitution."},{"Start":"04:45.880 ","End":"04:51.844","Text":"Tau is x plus y squared minus x plus y to the fourth,"},{"Start":"04:51.844 ","End":"04:54.880","Text":"e to the power of t is minus y,"},{"Start":"04:54.880 ","End":"04:57.355","Text":"so minus t is y."},{"Start":"04:57.355 ","End":"05:01.068","Text":"This is the solution,"},{"Start":"05:01.068 ","End":"05:03.140","Text":"and we are done."}],"ID":30667},{"Watched":false,"Name":"Exercise 5","Duration":"6m 2s","ChapterTopicVideoID":29107,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.505","Text":"In this exercise, we\u0027re going to solve the following equation with initial condition."},{"Start":"00:05.505 ","End":"00:09.495","Text":"This is a quasi-linear partial differential equation."},{"Start":"00:09.495 ","End":"00:14.684","Text":"You might wonder where the Gamma is that we usually define."},{"Start":"00:14.684 ","End":"00:19.545","Text":"Well, you can see that we are talking about x minus x."},{"Start":"00:19.545 ","End":"00:25.155","Text":"It could be the line y=minus x, this line here."},{"Start":"00:25.155 ","End":"00:30.150","Text":"Along this line, u is equal to this expression."},{"Start":"00:30.150 ","End":"00:31.650","Text":"It just depends on x,"},{"Start":"00:31.650 ","End":"00:33.900","Text":"not on y, but that\u0027s okay."},{"Start":"00:33.900 ","End":"00:37.930","Text":"General form is Au_ x plus Bu_ y=C."},{"Start":"00:37.930 ","End":"00:40.215","Text":"Well, you could bring the 2u to the other side."},{"Start":"00:40.215 ","End":"00:42.810","Text":"Then we have that A is 2,"},{"Start":"00:42.810 ","End":"00:45.360","Text":"B is minus 3,"},{"Start":"00:45.360 ","End":"00:48.600","Text":"and C is minus 2u."},{"Start":"00:48.600 ","End":"00:54.145","Text":"We\u0027re going to use the method of characteristics like a new tutorial."},{"Start":"00:54.145 ","End":"00:59.705","Text":"The first step is to parameterize the initial curve Gamma."},{"Start":"00:59.705 ","End":"01:03.125","Text":"Well, actually we\u0027re parameterizing the whole solution surface."},{"Start":"01:03.125 ","End":"01:07.510","Text":"Gamma will be the part where t=0."},{"Start":"01:07.510 ","End":"01:13.420","Text":"I should say it\u0027s going to be parameterized by 2 parameters, t and Tau."},{"Start":"01:13.670 ","End":"01:23.840","Text":"The way to get Gamma is we can say that x is Tau and then y being minus x is minus Tau,"},{"Start":"01:23.840 ","End":"01:28.950","Text":"and u from here just replace x by Tau,"},{"Start":"01:28.950 ","End":"01:31.350","Text":"and this is what it\u0027s equal to."},{"Start":"01:31.350 ","End":"01:32.710","Text":"Now step 2,"},{"Start":"01:32.710 ","End":"01:40.250","Text":"we find the characteristic curves by solving these 3 ordinary differential equations."},{"Start":"01:40.250 ","End":"01:43.160","Text":"We\u0027re just going to worry about the first 2,"},{"Start":"01:43.160 ","End":"01:45.994","Text":"and the third one comes in later."},{"Start":"01:45.994 ","End":"01:51.300","Text":"These 2 give us that x_t is A, y_t is B."},{"Start":"01:51.300 ","End":"01:53.805","Text":"We\u0027ve already figured out A and B,"},{"Start":"01:53.805 ","End":"01:56.955","Text":"the constants 2 and minus 3."},{"Start":"01:56.955 ","End":"02:00.500","Text":"By integrating this with respect to t,"},{"Start":"02:00.500 ","End":"02:05.205","Text":"we get that x is 2t plus a function of Tau,"},{"Start":"02:05.205 ","End":"02:08.925","Text":"and y is minus 3t plus another function of Tau,"},{"Start":"02:08.925 ","End":"02:10.800","Text":"c_1 and c_2,"},{"Start":"02:10.800 ","End":"02:13.815","Text":"like constants as far as t goes."},{"Start":"02:13.815 ","End":"02:17.850","Text":"If we substitute t=0,"},{"Start":"02:17.850 ","End":"02:19.560","Text":"then 2t is 0,"},{"Start":"02:19.560 ","End":"02:20.990","Text":"minus 3t is 0,"},{"Start":"02:20.990 ","End":"02:24.095","Text":"and we get this pair of equations."},{"Start":"02:24.095 ","End":"02:27.590","Text":"Now compare these to what we have here,"},{"Start":"02:27.590 ","End":"02:34.454","Text":"we have that x is Tau and y is minus Tau when t is 0."},{"Start":"02:34.454 ","End":"02:38.450","Text":"If we compare these 2 with these 2,"},{"Start":"02:38.450 ","End":"02:45.325","Text":"we get that c_1 of Tau is Tau and c_2 of Tau is minus Tau,"},{"Start":"02:45.325 ","End":"02:54.025","Text":"substituting c_1 and c_2 here and here we get this for x and y in terms of t and Tau."},{"Start":"02:54.025 ","End":"02:55.830","Text":"That was step 2."},{"Start":"02:55.830 ","End":"03:03.440","Text":"Now we need to find u in terms of t and Tau from the equation, this one,"},{"Start":"03:03.440 ","End":"03:09.785","Text":"u_t derivative of u with respect to t partial derivative is C. In our case,"},{"Start":"03:09.785 ","End":"03:13.040","Text":"C is minus 2u."},{"Start":"03:13.040 ","End":"03:18.750","Text":"What we get is du by dt is minus 2u,"},{"Start":"03:18.750 ","End":"03:21.870","Text":"that\u0027s the C. We have an initial condition,"},{"Start":"03:21.870 ","End":"03:25.035","Text":"u of 0, t. Let me scroll back."},{"Start":"03:25.035 ","End":"03:27.285","Text":"There it is. Tau plus 1,"},{"Start":"03:27.285 ","End":"03:29.175","Text":"e to the minus Tau."},{"Start":"03:29.175 ","End":"03:34.675","Text":"U of 0 Tau is Tau plus 1 e^minus Tau."},{"Start":"03:34.675 ","End":"03:41.045","Text":"Solving this differential equation as an ordinary differential equation,"},{"Start":"03:41.045 ","End":"03:45.380","Text":"it\u0027s like y prime =minus 2y."},{"Start":"03:45.380 ","End":"03:50.330","Text":"We know that the solution is exponential with the minus 2 in the exponent,"},{"Start":"03:50.330 ","End":"03:52.805","Text":"so it\u0027s e^minus 2t."},{"Start":"03:52.805 ","End":"03:55.355","Text":"We also put a constant in front."},{"Start":"03:55.355 ","End":"03:59.360","Text":"But here the constant is a function of Tau, call it c_3."},{"Start":"03:59.360 ","End":"04:02.275","Text":"If we put t=0 here,"},{"Start":"04:02.275 ","End":"04:06.600","Text":"then e^minus 2t is e^minus 0 is 1."},{"Start":"04:06.600 ","End":"04:13.095","Text":"We just get c_3 of Tau gives us that c_3 of Tau is this."},{"Start":"04:13.095 ","End":"04:16.670","Text":"If we substitute this back in here,"},{"Start":"04:16.670 ","End":"04:21.620","Text":"we get that u in terms of t and Tau is Tau plus 1,"},{"Start":"04:21.620 ","End":"04:26.255","Text":"e^minus Tau times e^minus 2t."},{"Start":"04:26.255 ","End":"04:28.445","Text":"Now, step 4,"},{"Start":"04:28.445 ","End":"04:31.280","Text":"we have to get back to x and y."},{"Start":"04:31.280 ","End":"04:33.020","Text":"We have u in terms of t and Tau,"},{"Start":"04:33.020 ","End":"04:35.270","Text":"we want it in terms of x and y."},{"Start":"04:35.270 ","End":"04:37.565","Text":"What we\u0027re going to do is try and reverse."},{"Start":"04:37.565 ","End":"04:39.290","Text":"We have x and y in terms of t and Tau."},{"Start":"04:39.290 ","End":"04:40.675","Text":"We want to reverse that."},{"Start":"04:40.675 ","End":"04:45.485","Text":"First, we\u0027ll check the Jacobian has a non-zero determinant."},{"Start":"04:45.485 ","End":"04:47.540","Text":"This is the formula."},{"Start":"04:47.540 ","End":"04:49.100","Text":"It\u0027s 2, 1,"},{"Start":"04:49.100 ","End":"04:51.494","Text":"minus 3, minus 1, let\u0027s see,"},{"Start":"04:51.494 ","End":"04:55.885","Text":"minus 2, minus minus 3 is plus 1."},{"Start":"04:55.885 ","End":"04:57.425","Text":"I accidentally wrote minus."},{"Start":"04:57.425 ","End":"05:00.050","Text":"Anyway, it\u0027s not 0, so that\u0027s the important thing."},{"Start":"05:00.050 ","End":"05:04.160","Text":"Now let\u0027s solve to get t and Tau in terms of x and y."},{"Start":"05:04.160 ","End":"05:05.600","Text":"Start with this."},{"Start":"05:05.600 ","End":"05:09.725","Text":"We could add these 2 equations and that will get rid of Tau."},{"Start":"05:09.725 ","End":"05:14.930","Text":"We get that x plus y is 2t minus 3t is"},{"Start":"05:14.930 ","End":"05:20.150","Text":"minus t. If we take 3 times this plus twice this,"},{"Start":"05:20.150 ","End":"05:26.330","Text":"we\u0027ll get rid of the t, 3 times Tau plus twice minus Tau is Tau."},{"Start":"05:26.330 ","End":"05:28.490","Text":"Just reverse sides,"},{"Start":"05:28.490 ","End":"05:30.350","Text":"and there\u0027s a minus here."},{"Start":"05:30.350 ","End":"05:34.685","Text":"We get t is minus x minus y and Tau is 3x plus 2y."},{"Start":"05:34.685 ","End":"05:39.065","Text":"Now we can substitute t and Tau in the expression for u,"},{"Start":"05:39.065 ","End":"05:41.495","Text":"which we had here."},{"Start":"05:41.495 ","End":"05:49.970","Text":"We get the answer which is that u of x and y is Tau is 3x plus 2y plus 1,"},{"Start":"05:49.970 ","End":"05:54.008","Text":"e^minus Tau is this,"},{"Start":"05:54.008 ","End":"05:59.745","Text":"and minus 2t would be plus 2x plus 2y."},{"Start":"05:59.745 ","End":"06:02.710","Text":"That\u0027s the end of this clip."}],"ID":30668},{"Watched":false,"Name":"Exercise 6","Duration":"6m 58s","ChapterTopicVideoID":29099,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.945","Text":"In this exercise, we\u0027re given that u(x,"},{"Start":"00:03.945 ","End":"00:11.625","Text":"y) is a solution to the following partial differential equation in the first quadrant and"},{"Start":"00:11.625 ","End":"00:20.790","Text":"satisfies the initial conditions that u(x,0)=1 and u(0, y)=0."},{"Start":"00:20.790 ","End":"00:26.220","Text":"We have to figure out what is u(1/5, 1)."},{"Start":"00:26.220 ","End":"00:29.220","Text":"This equation is quasi linear,"},{"Start":"00:29.220 ","End":"00:30.855","Text":"which is of this form."},{"Start":"00:30.855 ","End":"00:34.035","Text":"Where a, b, and c are functions of x, y, and u."},{"Start":"00:34.035 ","End":"00:36.720","Text":"A is y^2 from here,"},{"Start":"00:36.720 ","End":"00:38.145","Text":"B is 1,"},{"Start":"00:38.145 ","End":"00:40.130","Text":"C is minus u."},{"Start":"00:40.130 ","End":"00:43.700","Text":"Now let\u0027s get started with the steps in the tutorial."},{"Start":"00:43.700 ","End":"00:47.090","Text":"This is the method of characteristics."},{"Start":"00:47.090 ","End":"00:52.055","Text":"We first find the parameterization of the initial curve Gamma."},{"Start":"00:52.055 ","End":"00:54.770","Text":"Now the initial curve to corner,"},{"Start":"00:54.770 ","End":"00:59.295","Text":"it\u0027s like this, t=0 and Tau varies."},{"Start":"00:59.295 ","End":"01:01.200","Text":"Let Tau be 0 here,"},{"Start":"01:01.200 ","End":"01:03.980","Text":"negative Tau is along the x-axis,"},{"Start":"01:03.980 ","End":"01:08.255","Text":"and positive Tau is along the y-axis."},{"Start":"01:08.255 ","End":"01:12.690","Text":"u=1 here and 0 here."},{"Start":"01:12.690 ","End":"01:16.200","Text":"Has a formula, x(0,"},{"Start":"01:16.200 ","End":"01:20.490","Text":"Tau) is minus Tau when Tau is negative."},{"Start":"01:20.490 ","End":"01:22.560","Text":"Tau is minus 3,"},{"Start":"01:22.560 ","End":"01:29.085","Text":"then x is plus 3 and y=0 when Tau is negative."},{"Start":"01:29.085 ","End":"01:31.175","Text":"When Tau is positive,"},{"Start":"01:31.175 ","End":"01:35.450","Text":"then the x=0 and the y=Tau."},{"Start":"01:35.450 ","End":"01:37.310","Text":"That\u0027s along here."},{"Start":"01:37.310 ","End":"01:42.395","Text":"I said, u=1 when Tau is negative,"},{"Start":"01:42.395 ","End":"01:46.790","Text":"that\u0027s here and 0 when Tau is positive."},{"Start":"01:46.790 ","End":"01:48.905","Text":"That\u0027s the first step."},{"Start":"01:48.905 ","End":"01:54.740","Text":"Now, step 2 is to find the characteristic curves by solving these equations,"},{"Start":"01:54.740 ","End":"01:56.330","Text":"this is actually a third 1,"},{"Start":"01:56.330 ","End":"01:58.640","Text":"u with respect to t is c,"},{"Start":"01:58.640 ","End":"02:01.055","Text":"but we don\u0027t need that just yet."},{"Start":"02:01.055 ","End":"02:04.375","Text":"Dx by dt is y^2,"},{"Start":"02:04.375 ","End":"02:09.364","Text":"that\u0027s A and dy by dt is 1, that\u0027s B."},{"Start":"02:09.364 ","End":"02:11.780","Text":"This one, if we integrate with respect to t,"},{"Start":"02:11.780 ","End":"02:16.085","Text":"gives us that y is t plus a constant in t,"},{"Start":"02:16.085 ","End":"02:18.830","Text":"but really an arbitrary function in Tau."},{"Start":"02:18.830 ","End":"02:22.970","Text":"We can\u0027t integrate this because we don\u0027t know how y depends on t,"},{"Start":"02:22.970 ","End":"02:24.520","Text":"so let\u0027s leave that."},{"Start":"02:24.520 ","End":"02:27.180","Text":"From this one if we let t=0,"},{"Start":"02:27.180 ","End":"02:30.675","Text":"we get y(0, Tau) is c_2(Tau)."},{"Start":"02:30.675 ","End":"02:35.205","Text":"But we also have y(0, Tau) from here."},{"Start":"02:35.205 ","End":"02:39.810","Text":"Just copied it. That gives us that c_2(Tau)"},{"Start":"02:39.810 ","End":"02:45.370","Text":"is 0 or Tau according to whether t is negative or positive."},{"Start":"02:45.370 ","End":"02:49.195","Text":"We have y in terms of t and Tau."},{"Start":"02:49.195 ","End":"02:54.565","Text":"We can say that xt is this squared."},{"Start":"02:54.565 ","End":"02:57.985","Text":"By integrating this with respect to"},{"Start":"02:57.985 ","End":"03:01.285","Text":"t. This is just a function of Tau, it\u0027s like a constant."},{"Start":"03:01.285 ","End":"03:04.340","Text":"We get raised the power by 1,"},{"Start":"03:04.340 ","End":"03:07.110","Text":"(t+c_2)^3 and divide by 3."},{"Start":"03:07.110 ","End":"03:08.170","Text":"We have to add c_1,"},{"Start":"03:08.170 ","End":"03:10.160","Text":"which is also a function of Tau."},{"Start":"03:10.160 ","End":"03:14.310","Text":"We have an initial condition we had this just copied from before,"},{"Start":"03:14.310 ","End":"03:18.240","Text":"x(0, Tau) is this."},{"Start":"03:18.240 ","End":"03:21.390","Text":"If we let t=0 in this equation,"},{"Start":"03:21.390 ","End":"03:28.305","Text":"we get 1/3(c_2)^3 +c_1. That\u0027s x(0, Tau)."},{"Start":"03:28.305 ","End":"03:30.000","Text":"On the other hand is equal to this,"},{"Start":"03:30.000 ","End":"03:32.120","Text":"so it\u0027s equal to this."},{"Start":"03:32.120 ","End":"03:34.745","Text":"Just comparing these 2 when t=0."},{"Start":"03:34.745 ","End":"03:38.855","Text":"If we look at these 2 that I\u0027ve colored in blue,"},{"Start":"03:38.855 ","End":"03:41.630","Text":"we get that if Tau is negative,"},{"Start":"03:41.630 ","End":"03:45.040","Text":"then c_2 is 0."},{"Start":"03:45.040 ","End":"03:49.185","Text":"That c_1 has to equal minus Tau."},{"Start":"03:49.185 ","End":"03:51.510","Text":"If Tau is positive,"},{"Start":"03:51.510 ","End":"03:54.765","Text":"then c_2 is Tau."},{"Start":"03:54.765 ","End":"04:02.070","Text":"We have that 1/3(Tau^3) + c_1=0."},{"Start":"04:02.070 ","End":"04:06.795","Text":"Which gives us that c_1 is minus 1/3 Tau^3."},{"Start":"04:06.795 ","End":"04:12.880","Text":"Just summarizing that we have c_2 and c_1 as functions of Tau,"},{"Start":"04:12.880 ","End":"04:14.225","Text":"what they\u0027re equal to."},{"Start":"04:14.225 ","End":"04:15.680","Text":"Now we come to step 3."},{"Start":"04:15.680 ","End":"04:17.270","Text":"We\u0027re going to find u(t,"},{"Start":"04:17.270 ","End":"04:22.850","Text":"Tau) from the equation du by dt equals c. In our case,"},{"Start":"04:22.850 ","End":"04:24.785","Text":"c is minus u,"},{"Start":"04:24.785 ","End":"04:27.440","Text":"so ut is minus u."},{"Start":"04:27.440 ","End":"04:30.335","Text":"The solution to this differential equation"},{"Start":"04:30.335 ","End":"04:34.730","Text":"is u equals some constant times e to the minus t,"},{"Start":"04:34.730 ","End":"04:37.925","Text":"but it\u0027s not a constant, it\u0027s a function of Tau."},{"Start":"04:37.925 ","End":"04:41.660","Text":"Earlier we had the initial condition that u(0,"},{"Start":"04:41.660 ","End":"04:44.485","Text":"Tau) is 1 or 0."},{"Start":"04:44.485 ","End":"04:49.745","Text":"We can compare because you put t=0 here we get c3(Tau)."},{"Start":"04:49.745 ","End":"04:53.080","Text":"C3(Tau) must equal this."},{"Start":"04:53.080 ","End":"04:55.620","Text":"Multiplying c_3 by e^ minus t,"},{"Start":"04:55.620 ","End":"04:57.260","Text":"you get that u(t,"},{"Start":"04:57.260 ","End":"04:59.610","Text":"Tau) is the following."},{"Start":"04:59.610 ","End":"05:02.340","Text":"We found u, but in terms of t and Tau."},{"Start":"05:02.340 ","End":"05:06.245","Text":"We want u expressed in terms of x and y,"},{"Start":"05:06.245 ","End":"05:08.210","Text":"which brings us to step 4."},{"Start":"05:08.210 ","End":"05:12.610","Text":"Recall that we found y and x."},{"Start":"05:12.610 ","End":"05:17.600","Text":"We also found that c_2 is this and c_1 is this."},{"Start":"05:17.600 ","End":"05:19.425","Text":"This is all stuff we had from before."},{"Start":"05:19.425 ","End":"05:20.660","Text":"Let\u0027s just rearrange it,"},{"Start":"05:20.660 ","End":"05:23.225","Text":"tidy it up into 2 cases."},{"Start":"05:23.225 ","End":"05:26.590","Text":"When Tau is negative or when Tau is positive."},{"Start":"05:26.590 ","End":"05:28.520","Text":"If Tau is negative,"},{"Start":"05:28.520 ","End":"05:35.015","Text":"then we get that c_2 is 0, so is Y=t."},{"Start":"05:35.015 ","End":"05:38.555","Text":"It\u0027s just collecting it all and arranging it."},{"Start":"05:38.555 ","End":"05:45.880","Text":"We have y and x in 2 cases when Tau is positive and Tau is negative."},{"Start":"05:45.880 ","End":"05:52.515","Text":"From this one, we can let t+Tau=y. We get this."},{"Start":"05:52.515 ","End":"05:58.150","Text":"Here we can say that t is y so we have this."},{"Start":"05:58.150 ","End":"06:07.125","Text":"We can collect these 2 together and say that x equals the common part is 1/3 y^3."},{"Start":"06:07.125 ","End":"06:09.330","Text":"It\u0027s either minus Tau or"},{"Start":"06:09.330 ","End":"06:14.110","Text":"minus 1/3 Tau^3 according to whether Tau is positive or negative."},{"Start":"06:14.110 ","End":"06:21.280","Text":"Recall the original task was to find what u is when x is 1/5 and y is 1."},{"Start":"06:21.280 ","End":"06:22.870","Text":"But we don\u0027t have u(x,"},{"Start":"06:22.870 ","End":"06:24.845","Text":"y), we have u(t, Tau)."},{"Start":"06:24.845 ","End":"06:27.825","Text":"We\u0027d like to know what t equals and Tau equals."},{"Start":"06:27.825 ","End":"06:31.550","Text":"Maybe we don\u0027t have to know this precisely. You\u0027ll see."},{"Start":"06:31.550 ","End":"06:35.845","Text":"Anyway, recall also that u(t, Tau) is this."},{"Start":"06:35.845 ","End":"06:41.390","Text":"Put x=1/5 here and y=1."},{"Start":"06:41.390 ","End":"06:46.175","Text":"We get 1/5=1/3 minus this expression."},{"Start":"06:46.175 ","End":"06:49.785","Text":"1/5-1/3= 2/15."},{"Start":"06:49.785 ","End":"06:52.605","Text":"We get that 2/15 is equal to this."},{"Start":"06:52.605 ","End":"06:55.515","Text":"Now, 2/15 is positive."},{"Start":"06:55.515 ","End":"06:59.075","Text":"We can eliminate the case where Tau is negative."},{"Start":"06:59.075 ","End":"07:02.165","Text":"We have the 2/5 is 1/3 Tau^3."},{"Start":"07:02.165 ","End":"07:03.935","Text":"We can get Tau from this."},{"Start":"07:03.935 ","End":"07:10.425","Text":"We get that Tau^3 is 3 times 2/15, which is 2/5."},{"Start":"07:10.425 ","End":"07:14.565","Text":"Tau is the cube root of 2/5."},{"Start":"07:14.565 ","End":"07:20.020","Text":"From this, we could get t because we have y= t+Tau."},{"Start":"07:20.020 ","End":"07:22.775","Text":"But we don\u0027t actually need to find t,"},{"Start":"07:22.775 ","End":"07:27.320","Text":"because from this we have that u(t,"},{"Start":"07:27.320 ","End":"07:28.840","Text":"Tau) is e to the minus t(Tau)."},{"Start":"07:28.840 ","End":"07:30.785","Text":"Negative 0 is Tau is positive,"},{"Start":"07:30.785 ","End":"07:34.555","Text":"but in our case, Tau is positive."},{"Start":"07:34.555 ","End":"07:42.050","Text":"We just have to take it from the bottom line and get that u=0."},{"Start":"07:42.050 ","End":"07:46.385","Text":"In conclusion, u(1/5,1) is 0,"},{"Start":"07:46.385 ","End":"07:49.680","Text":"and that completes this exercise."}],"ID":30670},{"Watched":false,"Name":"Exercise 7","Duration":"7m 49s","ChapterTopicVideoID":29100,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.975","Text":"This exercise we have the following problem,"},{"Start":"00:03.975 ","End":"00:09.045","Text":"which is partial differential equation with initial conditions."},{"Start":"00:09.045 ","End":"00:12.345","Text":"This is the region it\u0027s defined on."},{"Start":"00:12.345 ","End":"00:16.290","Text":"The a here is a real constant."},{"Start":"00:16.290 ","End":"00:18.855","Text":"It\u0027s a quasi-linear equation,"},{"Start":"00:18.855 ","End":"00:22.140","Text":"which we\u0027ll solve using the method of characteristics."},{"Start":"00:22.140 ","End":"00:27.462","Text":"In Step 1, we find the parameterization of the initial curve Gamma."},{"Start":"00:27.462 ","End":"00:31.935","Text":"We\u0027ll have it as a function of t and tau."},{"Start":"00:31.935 ","End":"00:34.826","Text":"t=0 is the curve Gamma,"},{"Start":"00:34.826 ","End":"00:39.500","Text":"t and tau together will give the solution surface."},{"Start":"00:39.500 ","End":"00:44.570","Text":"What we\u0027ll do is we\u0027ll arrange the boundary such that t=0,"},{"Start":"00:44.570 ","End":"00:50.390","Text":"of course, and tau will be positive here and negative here."},{"Start":"00:50.390 ","End":"00:53.675","Text":"Here it starts with 0 and tau increases."},{"Start":"00:53.675 ","End":"00:56.755","Text":"Tau equals whatever x is."},{"Start":"00:56.755 ","End":"01:00.950","Text":"Here tau will be minus x or minus y."},{"Start":"01:00.950 ","End":"01:03.980","Text":"Same thing here. Let\u0027s write that."},{"Start":"01:03.980 ","End":"01:07.220","Text":"So x is a function of tau,"},{"Start":"01:07.220 ","End":"01:11.075","Text":"where t is 0 is tau or minus tau."},{"Start":"01:11.075 ","End":"01:15.125","Text":"Here x is tau here x is minus tau."},{"Start":"01:15.125 ","End":"01:22.430","Text":"Then here y is 0 and here y equals minus tau."},{"Start":"01:22.430 ","End":"01:25.415","Text":"Here, when tau is positive,"},{"Start":"01:25.415 ","End":"01:30.595","Text":"then x is positive from here."},{"Start":"01:30.595 ","End":"01:35.165","Text":"You will equal this and since x=tau here,"},{"Start":"01:35.165 ","End":"01:41.300","Text":"then this is what we get and here it\u0027s 0 because u(y),"},{"Start":"01:41.300 ","End":"01:46.235","Text":"y is 0 and that corresponds to tau negative and tau negative,"},{"Start":"01:46.235 ","End":"01:50.240","Text":"then y is minus tau, which is positive."},{"Start":"01:50.240 ","End":"01:53.760","Text":"Yeah, so 0,0."},{"Start":"01:53.760 ","End":"01:55.320","Text":"We have x, y,"},{"Start":"01:55.320 ","End":"01:59.385","Text":"and u when t=0 as dependent on tau,"},{"Start":"01:59.385 ","End":"02:01.425","Text":"and that\u0027s Step 1."},{"Start":"02:01.425 ","End":"02:03.350","Text":"Now Step 2 is to find"},{"Start":"02:03.350 ","End":"02:09.815","Text":"the characteristic curves which are given as dx by dt=a, dy by dt=b."},{"Start":"02:09.815 ","End":"02:14.300","Text":"There\u0027s a third one, d u by dt=c but we don\u0027t need that just yet."},{"Start":"02:14.300 ","End":"02:19.245","Text":"Another reminder this was the differential equation."},{"Start":"02:19.245 ","End":"02:22.355","Text":"This is A, B,"},{"Start":"02:22.355 ","End":"02:28.210","Text":"it\u0027s not written B is 1 and C is minus u."},{"Start":"02:28.210 ","End":"02:30.625","Text":"For these 2,"},{"Start":"02:30.625 ","End":"02:35.370","Text":"we get this is equal to a which is 2."},{"Start":"02:35.370 ","End":"02:40.360","Text":"Then y is B, which is 1."},{"Start":"02:40.360 ","End":"02:45.850","Text":"Later we\u0027ll say that UT is minus u."},{"Start":"02:45.850 ","End":"02:51.985","Text":"Now, integrating this with respect to t gives us that x=2t plus"},{"Start":"02:51.985 ","End":"02:59.855","Text":"some function of tau and y=1t plus another function of tau."},{"Start":"02:59.855 ","End":"03:06.275","Text":"Now we have initial conditions for x0 tau and y is 0 tau."},{"Start":"03:06.275 ","End":"03:08.550","Text":"First of all, let\u0027s deal with the x,"},{"Start":"03:08.550 ","End":"03:14.020","Text":"x0 tau is just copying it from here."},{"Start":"03:14.210 ","End":"03:16.290","Text":"On the other hand,"},{"Start":"03:16.290 ","End":"03:24.870","Text":"x0tau= C_1 of tau because substitute t=0 here,"},{"Start":"03:24.870 ","End":"03:27.870","Text":"and we\u0027re just left with the C_1 of tau."},{"Start":"03:27.870 ","End":"03:33.520","Text":"Now, on the other hand, take 2."},{"Start":"03:33.800 ","End":"03:36.450","Text":"Now we have 2,"},{"Start":"03:36.450 ","End":"03:40.540","Text":"and we also have another form for x=0 and tau,"},{"Start":"03:40.540 ","End":"03:44.185","Text":"which is what you get if you substitute t=0 here."},{"Start":"03:44.185 ","End":"03:47.650","Text":"This is equal to C1 of tau."},{"Start":"03:47.650 ","End":"03:51.280","Text":"Now, these 2 things are equal to the same thing,"},{"Start":"03:51.280 ","End":"03:52.480","Text":"so they\u0027re equal to each other."},{"Start":"03:52.480 ","End":"03:56.755","Text":"So C_1 of tau is this split function."},{"Start":"03:56.755 ","End":"04:00.385","Text":"We can substitute that in"},{"Start":"04:00.385 ","End":"04:06.230","Text":"here and get that x of t tau is the 2t from here and the C_1 of t,"},{"Start":"04:06.230 ","End":"04:07.850","Text":"which is equal to this."},{"Start":"04:07.850 ","End":"04:13.940","Text":"That\u0027s x. Now let\u0027s take care of y. y(0,𝜏 ) is like this."},{"Start":"04:13.940 ","End":"04:16.290","Text":"Just copy that."},{"Start":"04:17.900 ","End":"04:21.875","Text":"The other hand, y(0,𝜏) = c_2(𝜏),"},{"Start":"04:21.875 ","End":"04:25.160","Text":"which we get from here by putting t=0."},{"Start":"04:25.160 ","End":"04:30.630","Text":"We have these 2 are equal so c_2(𝜏) is equal to this."},{"Start":"04:30.800 ","End":"04:34.955","Text":"Then substitute c_2(𝜏) here,"},{"Start":"04:34.955 ","End":"04:39.305","Text":"replace that by this split function."},{"Start":"04:39.305 ","End":"04:44.210","Text":"We get that y(t,𝜏) is the following."},{"Start":"04:44.210 ","End":"04:50.370","Text":"Now we have x(t,𝜏) and y(t,𝜏) explicitly in terms of t and tau."},{"Start":"04:51.800 ","End":"05:00.290","Text":"We can rewrite these 2 piecewise-defined functions just slightly differently."},{"Start":"05:00.290 ","End":"05:05.105","Text":"We can write them like so to put everything inside the curly brace."},{"Start":"05:05.105 ","End":"05:08.255","Text":"Here we have 2t plus tau 2t minus tau,"},{"Start":"05:08.255 ","End":"05:11.340","Text":"t, t minus tau."},{"Start":"05:11.380 ","End":"05:19.170","Text":"The next step is to find u in terms of t and tau from the equation du by dt=c."},{"Start":"05:19.170 ","End":"05:21.770","Text":"In our case, c is minus u."},{"Start":"05:21.770 ","End":"05:26.985","Text":"We get the equation du by dt equals minus u."},{"Start":"05:26.985 ","End":"05:31.400","Text":"It\u0027s a straightforward linear differential equation"},{"Start":"05:31.400 ","End":"05:33.380","Text":"of first order with constant coefficients."},{"Start":"05:33.380 ","End":"05:35.075","Text":"We know how to solve this."},{"Start":"05:35.075 ","End":"05:39.275","Text":"The solution is that u is some constant,"},{"Start":"05:39.275 ","End":"05:46.470","Text":"which is really a function of tau times e to the minus t. Now Step 3,"},{"Start":"05:46.470 ","End":"05:51.980","Text":"we\u0027re going to find what u is from the derivative of u with respect to t,"},{"Start":"05:51.980 ","End":"05:56.100","Text":"which is equal to C. In our case, remember that c=-u."},{"Start":"05:56.770 ","End":"06:01.770","Text":"From this you get the u by dt=-u and look"},{"Start":"06:01.770 ","End":"06:06.943","Text":"on tau as a constant so it\u0027s a differential equation in the variable t,"},{"Start":"06:06.943 ","End":"06:08.615","Text":"and we know how to solve this."},{"Start":"06:08.615 ","End":"06:12.830","Text":"The solution is that u equals some constant times e to the"},{"Start":"06:12.830 ","End":"06:17.000","Text":"minus t. Only the constant here is a function of tau."},{"Start":"06:17.000 ","End":"06:20.030","Text":"Certainly, you can check if you differentiate this with respect to t,"},{"Start":"06:20.030 ","End":"06:22.910","Text":"you just get minus the same thing."},{"Start":"06:22.910 ","End":"06:25.055","Text":"So u by ut is minus u."},{"Start":"06:25.055 ","End":"06:30.050","Text":"Now, earlier we had the initial condition that u(0,𝜏)"},{"Start":"06:30.050 ","End":"06:35.510","Text":"is a cosine tau plus sine tau when tau is positive and 0 otherwise."},{"Start":"06:35.510 ","End":"06:36.785","Text":"On the other hand,"},{"Start":"06:36.785 ","End":"06:42.310","Text":"we can substitute t=0 here and get c_3(𝜏)."},{"Start":"06:42.310 ","End":"06:47.915","Text":"You have 0 tau here and here means that c_3(𝜏) is equal to this."},{"Start":"06:47.915 ","End":"06:51.950","Text":"Then you can put this in here and then we get"},{"Start":"06:51.950 ","End":"06:56.075","Text":"that u of t and tau is e to the minus t times this,"},{"Start":"06:56.075 ","End":"06:58.400","Text":"which is what\u0027s written here."},{"Start":"06:58.400 ","End":"07:02.675","Text":"That gives us u in terms of the parameters t and tab."},{"Start":"07:02.675 ","End":"07:03.860","Text":"We want to explicitly,"},{"Start":"07:03.860 ","End":"07:07.130","Text":"in other words, u in terms of x and y."},{"Start":"07:07.130 ","End":"07:09.185","Text":"That brings us to step 4."},{"Start":"07:09.185 ","End":"07:12.560","Text":"Now, recall these we had earlier,"},{"Start":"07:12.560 ","End":"07:15.545","Text":"we have x as a function of t and tau and y."},{"Start":"07:15.545 ","End":"07:19.100","Text":"Let\u0027s add this one to this list so we have x,"},{"Start":"07:19.100 ","End":"07:22.885","Text":"y, and u all in terms of t and tau."},{"Start":"07:22.885 ","End":"07:26.705","Text":"They all depend on whether tau is positive or negative."},{"Start":"07:26.705 ","End":"07:28.235","Text":"Now in this exercise,"},{"Start":"07:28.235 ","End":"07:33.475","Text":"we don\u0027t have to go back to get t and tau in terms of x and y,"},{"Start":"07:33.475 ","End":"07:36.225","Text":"there is a bit of a shortcut we can do."},{"Start":"07:36.225 ","End":"07:41.525","Text":"Notice that x minus 2y is tau. Let\u0027s see."},{"Start":"07:41.525 ","End":"07:43.610","Text":"If tau is bigger than 0,"},{"Start":"07:43.610 ","End":"07:45.395","Text":"like here and here,"},{"Start":"07:45.395 ","End":"07:48.860","Text":"then x minus 2y certainly tau."},{"Start":"07:48.860 ","End":"07:52.325","Text":"2t plus tau minus twice t. That\u0027s that."},{"Start":"07:52.325 ","End":"07:53.945","Text":"If tau is negative,"},{"Start":"07:53.945 ","End":"07:56.925","Text":"we have this minus twice this."},{"Start":"07:56.925 ","End":"07:59.355","Text":"Again, the 2t minus the 2t cancels,"},{"Start":"07:59.355 ","End":"08:04.335","Text":"and minus tau minus twice minus tau is plus tau."},{"Start":"08:04.335 ","End":"08:06.890","Text":"This is true in either case,"},{"Start":"08:06.890 ","End":"08:14.195","Text":"and that means that in place of tau here and here we can put x minus 2y."},{"Start":"08:14.195 ","End":"08:16.570","Text":"Then we\u0027ll still have the t here,"},{"Start":"08:16.570 ","End":"08:20.375","Text":"but this t only occurs in the case where tau is bigger than 0,"},{"Start":"08:20.375 ","End":"08:23.350","Text":"and in that case, t is equal to y."},{"Start":"08:23.350 ","End":"08:26.510","Text":"Doing a little bit of substitutions, minus t,"},{"Start":"08:26.510 ","End":"08:29.150","Text":"we have minus y and here,"},{"Start":"08:29.150 ","End":"08:33.245","Text":"instead of tau, we have x minus 2y here and here."},{"Start":"08:33.245 ","End":"08:36.215","Text":"Also in the inequalities here,"},{"Start":"08:36.215 ","End":"08:40.805","Text":"this is x minus 2y bigger than 0 and x minus 2y less than 0."},{"Start":"08:40.805 ","End":"08:43.714","Text":"We\u0027re almost there. Just some fine-tuning."},{"Start":"08:43.714 ","End":"08:49.490","Text":"Remember the original domain was 0 less than or equal to y, less than or equal to x."},{"Start":"08:49.490 ","End":"08:52.385","Text":"Remember that was an eighth of the plane,"},{"Start":"08:52.385 ","End":"08:54.460","Text":"where we shaded it in the very beginning."},{"Start":"08:54.460 ","End":"08:57.180","Text":"Here\u0027s a picture, what I\u0027m talking about."},{"Start":"08:57.180 ","End":"09:02.825","Text":"We have to sharpen this inequality a bit because we want it to be in original domain."},{"Start":"09:02.825 ","End":"09:06.050","Text":"What we get is this is equal to this."},{"Start":"09:06.050 ","End":"09:07.310","Text":"That\u0027s the same thing here."},{"Start":"09:07.310 ","End":"09:13.040","Text":"Here. We can say that x is bigger than 2y."},{"Start":"09:13.040 ","End":"09:17.305","Text":"From here y is bigger or equal to 0 so 2y is bigger or equal to 0."},{"Start":"09:17.305 ","End":"09:21.065","Text":"In this case, 2y is bigger than x,"},{"Start":"09:21.065 ","End":"09:24.755","Text":"but still x is bigger or equal to y, bigger or equal to 0."},{"Start":"09:24.755 ","End":"09:30.110","Text":"If I draw these 2 domains, sketch them."},{"Start":"09:30.110 ","End":"09:32.555","Text":"This was our original sector,"},{"Start":"09:32.555 ","End":"09:37.550","Text":"eighth of a plane where zero-sum or equal to y, less than or equal to x."},{"Start":"09:37.550 ","End":"09:40.055","Text":"This is a dividing line."},{"Start":"09:40.055 ","End":"09:43.040","Text":"This is where x=y."},{"Start":"09:43.040 ","End":"09:44.810","Text":"Here 2y is bigger than x,"},{"Start":"09:44.810 ","End":"09:47.600","Text":"bigger than y, and here x bigger than 2y."},{"Start":"09:47.600 ","End":"09:49.940","Text":"That\u0027s just for interest\u0027s sake. It doesn\u0027t really matter."},{"Start":"09:49.940 ","End":"09:53.060","Text":"You could just leave it like this without the picture."},{"Start":"09:53.060 ","End":"09:55.805","Text":"We found that u is 0 in this part,"},{"Start":"09:55.805 ","End":"10:00.780","Text":"and here it\u0027s equal to this expression. We are done."}],"ID":30671},{"Watched":false,"Name":"Exercise 8","Duration":"10m ","ChapterTopicVideoID":29101,"CourseChapterTopicPlaylistID":294427,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":30672}],"Thumbnail":null,"ID":294427},{"Name":"Lagrange\u0027s Method","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Lagrange\u0026#39;s Method","Duration":"2m 7s","ChapterTopicVideoID":29085,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.280","Text":"We return to quasilinear first-order partial differential equations."},{"Start":"00:05.280 ","End":"00:08.445","Text":"Previously, we learned the method of characteristics."},{"Start":"00:08.445 ","End":"00:11.265","Text":"Now, we\u0027re going to learn Lagrange\u0027s method."},{"Start":"00:11.265 ","End":"00:15.705","Text":"Let\u0027s start by reminding you what a quasilinear PDE is."},{"Start":"00:15.705 ","End":"00:17.790","Text":"It\u0027s this form."},{"Start":"00:17.790 ","End":"00:20.430","Text":"We\u0027re looking for u,"},{"Start":"00:20.430 ","End":"00:23.160","Text":"which is a function of x and y."},{"Start":"00:23.160 ","End":"00:27.330","Text":"In Lagrange\u0027s method, we consider Lagrange\u0027s auxiliary equations."},{"Start":"00:27.330 ","End":"00:29.895","Text":"This is the basis of the method."},{"Start":"00:29.895 ","End":"00:31.950","Text":"Now, there\u0027s really 2 equations here,"},{"Start":"00:31.950 ","End":"00:33.450","Text":"you might think there\u0027s 3."},{"Start":"00:33.450 ","End":"00:35.325","Text":"This equals this, this equals this,"},{"Start":"00:35.325 ","End":"00:36.795","Text":"and this equals this, but really,"},{"Start":"00:36.795 ","End":"00:38.595","Text":"once you have 2, you have the 3rd."},{"Start":"00:38.595 ","End":"00:42.000","Text":"Choose 2 of these. We found a solution for each."},{"Start":"00:42.000 ","End":"00:46.475","Text":"The solution typically involves an arbitrary constant which you put on the right."},{"Start":"00:46.475 ","End":"00:50.240","Text":"We get one solution in the general form Phi_1 (x, y,"},{"Start":"00:50.240 ","End":"00:53.705","Text":"u) is C_1 and the other in the form Phi_2 (x,"},{"Start":"00:53.705 ","End":"00:55.385","Text":"y, u) is C_2."},{"Start":"00:55.385 ","End":"00:58.535","Text":"There\u0027s a condition that these must be linearly independent."},{"Start":"00:58.535 ","End":"01:02.090","Text":"You can\u0027t just double this one and get this one, for example."},{"Start":"01:02.090 ","End":"01:04.880","Text":"At least one of them has to contain u explicitly."},{"Start":"01:04.880 ","End":"01:06.770","Text":"They can\u0027t just involve x and y."},{"Start":"01:06.770 ","End":"01:09.440","Text":"This will be true in all our examples."},{"Start":"01:09.440 ","End":"01:11.660","Text":"But just thought I\u0027d state it."},{"Start":"01:11.660 ","End":"01:16.160","Text":"The general solution of the PDE is the following."},{"Start":"01:16.160 ","End":"01:17.750","Text":"An arbitrary function f,"},{"Start":"01:17.750 ","End":"01:23.824","Text":"which is differentiable of Phi_1 and Phi_2 of x, y, u is 0."},{"Start":"01:23.824 ","End":"01:26.520","Text":"Now some important remarks,"},{"Start":"01:26.520 ","End":"01:29.495","Text":"it won\u0027t really be understood until we get to the examples,"},{"Start":"01:29.495 ","End":"01:31.895","Text":"but I\u0027ll just have them recorded here."},{"Start":"01:31.895 ","End":"01:37.100","Text":"If Phi_1 is a function just of x and y not u,"},{"Start":"01:37.100 ","End":"01:42.220","Text":"then we can write big f of Phi_1 equals Phi_2."},{"Start":"01:42.220 ","End":"01:45.360","Text":"Like I said, it depend in the examples."},{"Start":"01:45.360 ","End":"01:48.635","Text":"Similarly, if Phi_ 2 is function just of x y,"},{"Start":"01:48.635 ","End":"01:52.285","Text":"then we can write F(Phi_2) is Phi_1."},{"Start":"01:52.285 ","End":"01:55.280","Text":"Lastly, if C,"},{"Start":"01:55.280 ","End":"01:57.830","Text":"the C here is 0,"},{"Start":"01:57.830 ","End":"02:02.090","Text":"then we can write u equals C_1 or u equals C_2."},{"Start":"02:02.090 ","End":"02:05.239","Text":"Following this, there will be some examples,"},{"Start":"02:05.239 ","End":"02:07.680","Text":"but that\u0027s it for the tutorial."}],"ID":30673},{"Watched":false,"Name":"Example 1","Duration":"3m 12s","ChapterTopicVideoID":29086,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.780","Text":"Here\u0027s an example of how to use Lagrange\u0027s method to"},{"Start":"00:03.780 ","End":"00:07.860","Text":"solve a partial differential equation, this one."},{"Start":"00:07.860 ","End":"00:11.730","Text":"We\u0027re also given an initial condition."},{"Start":"00:11.730 ","End":"00:13.530","Text":"First, we find the general solution,"},{"Start":"00:13.530 ","End":"00:16.634","Text":"then we find the solution that satisfies this condition."},{"Start":"00:16.634 ","End":"00:21.195","Text":"Let\u0027s start by stating the auxiliary equations."},{"Start":"00:21.195 ","End":"00:23.340","Text":"Only in this case,"},{"Start":"00:23.340 ","End":"00:28.305","Text":"C is equal to 0. a is y,"},{"Start":"00:28.305 ","End":"00:29.970","Text":"b is minus x,"},{"Start":"00:29.970 ","End":"00:32.130","Text":"and c is 0."},{"Start":"00:32.130 ","End":"00:34.620","Text":"We just take the one equation,"},{"Start":"00:34.620 ","End":"00:41.450","Text":"this equals this and we can solve it using separation of variables,"},{"Start":"00:41.450 ","End":"00:45.305","Text":"just cross multiply and we get xdx equals minus ydy."},{"Start":"00:45.305 ","End":"00:51.360","Text":"Integrate 1/2 x^2 equals minus 1/2 y^2 plus a constant and"},{"Start":"00:51.360 ","End":"00:57.065","Text":"then we can double that and get x^2 plus y^2 equals another constant."},{"Start":"00:57.065 ","End":"01:01.100","Text":"Let\u0027s call the left-hand side Phi 1 of x and y."},{"Start":"01:01.100 ","End":"01:03.625","Text":"Notice it doesn\u0027t involve u."},{"Start":"01:03.625 ","End":"01:08.210","Text":"Now, I want to remind you of the remarks we had in the tutorial."},{"Start":"01:08.210 ","End":"01:09.890","Text":"Note that in our case,"},{"Start":"01:09.890 ","End":"01:13.690","Text":"Phi 1 depends just on x and y."},{"Start":"01:13.690 ","End":"01:17.565","Text":"We can write f (Phi 1) equals Phi 2."},{"Start":"01:17.565 ","End":"01:24.780","Text":"Also, we have that C equals 0 so we\u0027re going to write u equals C_2."},{"Start":"01:24.780 ","End":"01:27.675","Text":"u is C_2,"},{"Start":"01:27.675 ","End":"01:30.030","Text":"F( Phi 1) is Phi 2,"},{"Start":"01:30.030 ","End":"01:31.520","Text":"Phi 2 of x,"},{"Start":"01:31.520 ","End":"01:34.625","Text":"y and u is equal to u."},{"Start":"01:34.625 ","End":"01:40.110","Text":"We get that f(x)^2 plus y^2 equals u."},{"Start":"01:40.110 ","End":"01:43.325","Text":"Now just switch sides and write u as u(x,"},{"Start":"01:43.325 ","End":"01:47.075","Text":"y) and that gives us our answer to the equation."},{"Start":"01:47.075 ","End":"01:48.575","Text":"Let\u0027s do a check."},{"Start":"01:48.575 ","End":"01:50.510","Text":"Won\u0027t always do this just this time."},{"Start":"01:50.510 ","End":"01:54.410","Text":"I\u0027ll do a check that this really satisfies this PDE."},{"Start":"01:54.410 ","End":"01:57.770","Text":"Let\u0027s check if the left-hand side equals the right-hand side."},{"Start":"01:57.770 ","End":"02:00.920","Text":"yu_x minus xu_y,"},{"Start":"02:00.920 ","End":"02:07.385","Text":"ux from here is a derivative of f times the anti-derivative with respect to x,"},{"Start":"02:07.385 ","End":"02:16.445","Text":"which is 2x minus x times uy is f\u0027 times the derivative of this with respect to y."},{"Start":"02:16.445 ","End":"02:19.175","Text":"Notice that here we have y and x."},{"Start":"02:19.175 ","End":"02:20.630","Text":"Here we have x and y."},{"Start":"02:20.630 ","End":"02:22.025","Text":"In other words, it\u0027s the same,"},{"Start":"02:22.025 ","End":"02:26.080","Text":"so that this equals this so the difference really is equal to 0."},{"Start":"02:26.080 ","End":"02:28.370","Text":"Now, we come to part b,"},{"Start":"02:28.370 ","End":"02:33.960","Text":"where we\u0027re given the initial condition that u(x) naught equals x^2,"},{"Start":"02:33.960 ","End":"02:36.170","Text":"you have to find a specific solution from"},{"Start":"02:36.170 ","End":"02:39.715","Text":"the general solution such that this is satisfied."},{"Start":"02:39.715 ","End":"02:43.235","Text":"If we let y equals naught here,"},{"Start":"02:43.235 ","End":"02:47.165","Text":"we get f(x)^2 plus naught squared equals x squared."},{"Start":"02:47.165 ","End":"02:50.930","Text":"Make a substitution that t equals x squared and we get that"},{"Start":"02:50.930 ","End":"02:56.495","Text":"f(t) equals t. That means that F is the identity function."},{"Start":"02:56.495 ","End":"03:00.335","Text":"If you have x, y equals f(x) squared plus y squared,"},{"Start":"03:00.335 ","End":"03:05.870","Text":"then it\u0027s just equal to x squared plus y squared because we said that f is the identity."},{"Start":"03:05.870 ","End":"03:08.029","Text":"This is the answer to the PDE,"},{"Start":"03:08.029 ","End":"03:13.410","Text":"which also satisfies the initial condition and we\u0027re done."}],"ID":30674},{"Watched":false,"Name":"Example 2","Duration":"2m 56s","ChapterTopicVideoID":29087,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.700","Text":"Here\u0027s another example of a PDE using Lagrange\u0027s method."},{"Start":"00:05.700 ","End":"00:09.780","Text":"We\u0027re also given an initial condition."},{"Start":"00:09.780 ","End":"00:11.850","Text":"First, you find the general solution,"},{"Start":"00:11.850 ","End":"00:15.450","Text":"then we find a solution which satisfies this."},{"Start":"00:15.450 ","End":"00:19.890","Text":"The auxiliary equations are as follows,"},{"Start":"00:19.890 ","End":"00:23.985","Text":"where this is A and this is B,"},{"Start":"00:23.985 ","End":"00:29.535","Text":"and this is C. In this case, C is 0."},{"Start":"00:29.535 ","End":"00:31.710","Text":"Forget about the last bit."},{"Start":"00:31.710 ","End":"00:33.000","Text":"From this equals this,"},{"Start":"00:33.000 ","End":"00:36.930","Text":"we get the following and we can use separation of variables,"},{"Start":"00:36.930 ","End":"00:39.840","Text":"just bring the x to this side."},{"Start":"00:39.840 ","End":"00:42.890","Text":"Also doubled both sides because then we\u0027ll have 2x"},{"Start":"00:42.890 ","End":"00:46.205","Text":"which is the derivative of the denominator and the numerator."},{"Start":"00:46.205 ","End":"00:48.335","Text":"Now integrate both sides."},{"Start":"00:48.335 ","End":"00:54.440","Text":"Natural log of 1 plus x equals twice natural log of y plus a constant."},{"Start":"00:54.440 ","End":"01:00.575","Text":"Bring this over to the other side and also put the 2 inside, so it\u0027s y^2."},{"Start":"01:00.575 ","End":"01:07.190","Text":"Then we have natural log of this divided by this is c. Take the reverse logarithm here,"},{"Start":"01:07.190 ","End":"01:09.440","Text":"the exponentiation of both sides."},{"Start":"01:09.440 ","End":"01:12.980","Text":"We have this is equal to e^c."},{"Start":"01:12.980 ","End":"01:15.425","Text":"But it\u0027s another constant."},{"Start":"01:15.425 ","End":"01:19.670","Text":"We call this phi_1(x, y), and u,"},{"Start":"01:19.670 ","End":"01:24.925","Text":"but it really only depends on x and y and because C is 0,"},{"Start":"01:24.925 ","End":"01:30.995","Text":"the second equation becomes u equals c_2."},{"Start":"01:30.995 ","End":"01:34.415","Text":"That phi_2(x, y) and u is just u."},{"Start":"01:34.415 ","End":"01:40.445","Text":"This is from the remarks in the tutorial that was this one."},{"Start":"01:40.445 ","End":"01:44.615","Text":"Now we\u0027re going to use this because phi_1 just depends on x and y."},{"Start":"01:44.615 ","End":"01:50.585","Text":"So we can write F of phi_1 is phi_2 and that becomes in our case,"},{"Start":"01:50.585 ","End":"01:52.865","Text":"F of this equals this."},{"Start":"01:52.865 ","End":"01:54.140","Text":"I just rearranged a bit,"},{"Start":"01:54.140 ","End":"01:55.445","Text":"bring you to the left."},{"Start":"01:55.445 ","End":"01:57.230","Text":"Remember that it\u0027s u(x,"},{"Start":"01:57.230 ","End":"01:59.870","Text":"y) is f of 1 plus x^2 over y^2."},{"Start":"01:59.870 ","End":"02:01.775","Text":"That is our solution."},{"Start":"02:01.775 ","End":"02:03.865","Text":"That\u0027s Part A."},{"Start":"02:03.865 ","End":"02:06.860","Text":"This was a solution to Part A and now we want"},{"Start":"02:06.860 ","End":"02:12.905","Text":"a specific solution which satisfies u(0, y) equals y^2."},{"Start":"02:12.905 ","End":"02:15.440","Text":"If we let x equals 0, here,"},{"Start":"02:15.440 ","End":"02:20.975","Text":"we get F (1 /y^2)equals u(0, y)."},{"Start":"02:20.975 ","End":"02:27.590","Text":"That means that F (1 /y^2) is equal to the y^2 from here."},{"Start":"02:27.590 ","End":"02:31.140","Text":"If you let t equals (1/y^2),"},{"Start":"02:31.140 ","End":"02:38.105","Text":"then y^2 is 1 over t. We get F (t) is 1 over t and then we can write u,"},{"Start":"02:38.105 ","End":"02:43.920","Text":"which is F of 1 plus x^2 over y^2 like we saw here."},{"Start":"02:44.600 ","End":"02:49.628","Text":"F is the reciprocal function so this is equal to,"},{"Start":"02:49.628 ","End":"02:51.050","Text":"just reverse the fraction,"},{"Start":"02:51.050 ","End":"02:53.435","Text":"y^2 over 1 plus x^2."},{"Start":"02:53.435 ","End":"02:56.820","Text":"That\u0027s the answer and we\u0027re done."}],"ID":30675},{"Watched":false,"Name":"Example 3","Duration":"1m 53s","ChapterTopicVideoID":29088,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.545","Text":"Another example of a PDE,"},{"Start":"00:04.545 ","End":"00:09.300","Text":"quasi-linear that we\u0027re going to solve with Lagrange\u0027s method so this is"},{"Start":"00:09.300 ","End":"00:14.580","Text":"our equation and we start off with the auxiliary equations."},{"Start":"00:14.580 ","End":"00:15.945","Text":"Just to remind you,"},{"Start":"00:15.945 ","End":"00:17.685","Text":"this is A,"},{"Start":"00:17.685 ","End":"00:23.520","Text":"this is B, this is C, and these denominators."},{"Start":"00:23.520 ","End":"00:26.055","Text":"That becomes the following."},{"Start":"00:26.055 ","End":"00:28.500","Text":"Then we want to take 2 of these equations."},{"Start":"00:28.500 ","End":"00:30.765","Text":"We\u0027ll take this equals this,"},{"Start":"00:30.765 ","End":"00:33.240","Text":"and this equals this."},{"Start":"00:33.240 ","End":"00:35.520","Text":"We start off with this one, this one later."},{"Start":"00:35.520 ","End":"00:40.380","Text":"This one gives us natural log of y equals natural log of x plus c."},{"Start":"00:40.380 ","End":"00:45.695","Text":"Bring the natural log of y to the other side and we get natural log of a quotient."},{"Start":"00:45.695 ","End":"00:48.680","Text":"Then take the inverse log of both sides,"},{"Start":"00:48.680 ","End":"00:53.180","Text":"and we get that x over y equals each of the power of c, which is another constant,"},{"Start":"00:53.180 ","End":"00:59.090","Text":"call it c_1 and that gives us our function Phi_1(x,"},{"Start":"00:59.090 ","End":"01:02.720","Text":"y, and u), which is really just dependent on x and y."},{"Start":"01:02.720 ","End":"01:06.530","Text":"Now we go to the other equation, this one."},{"Start":"01:06.530 ","End":"01:10.895","Text":"This is also separated the variables are,"},{"Start":"01:10.895 ","End":"01:13.115","Text":"and we take the integral of both sides,"},{"Start":"01:13.115 ","End":"01:16.855","Text":"natural log of x is u plus c_2."},{"Start":"01:16.855 ","End":"01:22.470","Text":"Bring this to the other side and this one we can call it Phi_2."},{"Start":"01:22.470 ","End":"01:25.230","Text":"Now we have Phi_1 and Phi_2."},{"Start":"01:25.230 ","End":"01:30.590","Text":"Notice that Phi_1 depends just on x and y and in the tutorial,"},{"Start":"01:30.590 ","End":"01:35.300","Text":"we had the following remarks that if Phi_1 just depends on x and y,"},{"Start":"01:35.300 ","End":"01:38.645","Text":"we can write F(Phi_1) equals Phi_2,"},{"Start":"01:38.645 ","End":"01:44.840","Text":"which gives us F(x over y) is natural log of x minus u. I just rearrange it a bit,"},{"Start":"01:44.840 ","End":"01:45.950","Text":"put u on this side,"},{"Start":"01:45.950 ","End":"01:49.650","Text":"put the f on this side and we have the following"},{"Start":"01:49.650 ","End":"01:54.820","Text":"is the solution and that concludes this exercise."}],"ID":30676},{"Watched":false,"Name":"Example 4","Duration":"2m 4s","ChapterTopicVideoID":29089,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.500","Text":"Now yet another example of how to solve a PDE"},{"Start":"00:04.500 ","End":"00:08.190","Text":"that\u0027s quasi-linear using Lagrange\u0027s method."},{"Start":"00:08.190 ","End":"00:09.800","Text":"This is the equation."},{"Start":"00:09.800 ","End":"00:15.270","Text":"We start with the auxiliary equations where this is A,"},{"Start":"00:15.270 ","End":"00:21.195","Text":"this is B, and this is C. This becomes the following."},{"Start":"00:21.195 ","End":"00:23.602","Text":"Then we want a pair of these,"},{"Start":"00:23.602 ","End":"00:26.655","Text":"so we\u0027ll take this equals this,"},{"Start":"00:26.655 ","End":"00:28.215","Text":"and this equals this."},{"Start":"00:28.215 ","End":"00:30.480","Text":"First of all, we\u0027ll solve this one,"},{"Start":"00:30.480 ","End":"00:33.315","Text":"and this we\u0027ll do later."},{"Start":"00:33.315 ","End":"00:38.880","Text":"This one gives us natural log(u) equals natural log(y) plus a constant."},{"Start":"00:38.880 ","End":"00:40.439","Text":"Bring this to the other side."},{"Start":"00:40.439 ","End":"00:43.355","Text":"Natural log of a quotient equals c,"},{"Start":"00:43.355 ","End":"00:46.475","Text":"and so u over y is e^c,"},{"Start":"00:46.475 ","End":"00:48.005","Text":"which is c_ 1."},{"Start":"00:48.005 ","End":"00:50.270","Text":"Note that this we could write as this,"},{"Start":"00:50.270 ","End":"00:51.980","Text":"we\u0027ll need this in a moment."},{"Start":"00:51.980 ","End":"00:55.220","Text":"The other equation is this one,"},{"Start":"00:55.220 ","End":"00:58.130","Text":"but here we don\u0027t have just x and y,"},{"Start":"00:58.130 ","End":"01:01.325","Text":"we have this u here and that\u0027s where we\u0027ll use this."},{"Start":"01:01.325 ","End":"01:03.495","Text":"This equals c_ 1 y."},{"Start":"01:03.495 ","End":"01:05.025","Text":"We get the following."},{"Start":"01:05.025 ","End":"01:09.065","Text":"I guess I missed a step where we multiply both sides by y,"},{"Start":"01:09.065 ","End":"01:12.510","Text":"so instead of y^2 minus c_ 1 y,"},{"Start":"01:12.510 ","End":"01:14.310","Text":"we just have y minus c_ 1,"},{"Start":"01:14.310 ","End":"01:16.600","Text":"and here we just get dy."},{"Start":"01:16.700 ","End":"01:21.255","Text":"That gives us that dx is y minus c_ 1 dy."},{"Start":"01:21.255 ","End":"01:27.835","Text":"Then integrating, we get that x=1/2y^2 minus c_ 1 plus c_ 2."},{"Start":"01:27.835 ","End":"01:30.955","Text":"Now we can use this one again."},{"Start":"01:30.955 ","End":"01:32.850","Text":"Just rearranging a bit,"},{"Start":"01:32.850 ","End":"01:35.900","Text":"leaves c_ 2 on one side and all the rest on the other side,"},{"Start":"01:35.900 ","End":"01:39.155","Text":"and c_ 1 y is x so we get this."},{"Start":"01:39.155 ","End":"01:43.475","Text":"I forgot earlier to say that when we have this,"},{"Start":"01:43.475 ","End":"01:45.410","Text":"we can call Phi_ 1(x, y,"},{"Start":"01:45.410 ","End":"01:47.645","Text":"u) to be u over y."},{"Start":"01:47.645 ","End":"01:51.260","Text":"From here we got Phi_ 2 to be this."},{"Start":"01:51.260 ","End":"01:56.450","Text":"Then remember that the general solution is F(Phi_ 1,"},{"Start":"01:56.450 ","End":"02:01.445","Text":"Phi_ 2)=0, which in our case gives us the following."},{"Start":"02:01.445 ","End":"02:04.830","Text":"That concludes this exercise."}],"ID":30677},{"Watched":false,"Name":"Exercise 1","Duration":"2m 24s","ChapterTopicVideoID":29090,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.575","Text":"In this exercise, we\u0027re given a quasilinear PDE,"},{"Start":"00:04.575 ","End":"00:10.005","Text":"and we have to find the general solution using the Lagrange\u0027s method."},{"Start":"00:10.005 ","End":"00:14.790","Text":"We\u0027ll do it a bit shorter than in the tutorial."},{"Start":"00:14.790 ","End":"00:17.100","Text":"We can skip certain things."},{"Start":"00:17.100 ","End":"00:19.650","Text":"We don\u0027t have to say a equals x,"},{"Start":"00:19.650 ","End":"00:22.050","Text":"b equals y, u etc."},{"Start":"00:22.050 ","End":"00:23.670","Text":"We also don\u0027t need Phi 1,"},{"Start":"00:23.670 ","End":"00:25.500","Text":"Phi 2. Anyway, you\u0027ll see."},{"Start":"00:25.500 ","End":"00:27.650","Text":"We start with the auxiliary equations."},{"Start":"00:27.650 ","End":"00:31.010","Text":"We can straight away write dx over what we call a,"},{"Start":"00:31.010 ","End":"00:35.910","Text":"we take straight away from here and dy over what\u0027s in"},{"Start":"00:35.910 ","End":"00:41.660","Text":"front of the uy equals du over what\u0027s on the right-hand side."},{"Start":"00:41.660 ","End":"00:46.595","Text":"Then we choose two pairs of equations from here."},{"Start":"00:46.595 ","End":"00:51.535","Text":"Let\u0027s say we take the first equals the last,"},{"Start":"00:51.535 ","End":"00:55.265","Text":"and the other one will be this equals this,"},{"Start":"00:55.265 ","End":"00:56.795","Text":"the second and the third."},{"Start":"00:56.795 ","End":"00:58.460","Text":"Work on this one first,"},{"Start":"00:58.460 ","End":"00:59.965","Text":"then come back to this."},{"Start":"00:59.965 ","End":"01:04.850","Text":"From here, we can straight away integrate and get"},{"Start":"01:04.850 ","End":"01:09.695","Text":"natural log of x equals natural log of u plus a constant."},{"Start":"01:09.695 ","End":"01:16.560","Text":"That gives us that x equals u times e^c,"},{"Start":"01:16.560 ","End":"01:20.860","Text":"but e^c is not a constant, so it\u0027s uc_1."},{"Start":"01:20.890 ","End":"01:25.480","Text":"This gives us that c_1 is x/u."},{"Start":"01:25.480 ","End":"01:28.080","Text":"Let\u0027s go to the other one now."},{"Start":"01:28.080 ","End":"01:31.655","Text":"We\u0027re going to get c_2 equals something from the other one."},{"Start":"01:31.655 ","End":"01:38.235","Text":"Here, the obvious thing to do is multiply both sides by u."},{"Start":"01:38.235 ","End":"01:42.405","Text":"This u and this u will cancel each other,"},{"Start":"01:42.405 ","End":"01:46.395","Text":"and what we\u0027re left with is dy/y=du."},{"Start":"01:46.395 ","End":"01:53.615","Text":"Integrate that natural log of y=u plus another constant, c_2."},{"Start":"01:53.615 ","End":"01:58.550","Text":"That c_2 equals natural log of y minus u."},{"Start":"01:58.550 ","End":"02:00.960","Text":"Now we have these two,"},{"Start":"02:00.960 ","End":"02:05.775","Text":"these are what we call in the tutorial Phi 1 and Phi 2."},{"Start":"02:05.775 ","End":"02:12.455","Text":"The general solution is some function of this and this equals 0."},{"Start":"02:12.455 ","End":"02:16.035","Text":"Any arbitrary differentiable function,"},{"Start":"02:16.035 ","End":"02:21.815","Text":"F(x/u) and natural log y minus u equals 0."},{"Start":"02:21.815 ","End":"02:24.600","Text":"That concludes this clip."}],"ID":30678},{"Watched":false,"Name":"Exercise 2","Duration":"2m 2s","ChapterTopicVideoID":29091,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.610","Text":"In this exercise, we have a partial differential equation of"},{"Start":"00:05.610 ","End":"00:09.900","Text":"the quasi-linear type and we\u0027re going to solve"},{"Start":"00:09.900 ","End":"00:14.400","Text":"it using Lagrange\u0027s method to find the general solution."},{"Start":"00:14.400 ","End":"00:17.490","Text":"First, we write the auxiliary equations."},{"Start":"00:17.490 ","End":"00:19.655","Text":"We have dx, dy,"},{"Start":"00:19.655 ","End":"00:23.490","Text":"and du in the numerators then here we have what we call a,"},{"Start":"00:23.490 ","End":"00:26.340","Text":"b, and c. We have x^2 here,"},{"Start":"00:26.340 ","End":"00:29.520","Text":"y^2 here, and u^2 here."},{"Start":"00:29.520 ","End":"00:33.375","Text":"We choose 2 of the equations from these 3."},{"Start":"00:33.375 ","End":"00:37.395","Text":"For the first one, we\u0027ll take this equals this,"},{"Start":"00:37.395 ","End":"00:39.315","Text":"and for the second one,"},{"Start":"00:39.315 ","End":"00:41.640","Text":"we\u0027ll take this equals this."},{"Start":"00:41.640 ","End":"00:43.350","Text":"Let\u0027s start with this one,"},{"Start":"00:43.350 ","End":"00:44.795","Text":"return to this later."},{"Start":"00:44.795 ","End":"00:47.420","Text":"This already has the variables separated,"},{"Start":"00:47.420 ","End":"00:49.805","Text":"we just have to integrate each side."},{"Start":"00:49.805 ","End":"00:54.310","Text":"Integral of 1/x squared is minus 1/x and similarly with y,"},{"Start":"00:54.310 ","End":"00:57.795","Text":"and we have a constant of integration."},{"Start":"00:57.795 ","End":"01:03.415","Text":"Rearrange, and we have 1/y minus 1/x is C_1."},{"Start":"01:03.415 ","End":"01:07.325","Text":"Now the other one, it also has its variable separated,"},{"Start":"01:07.325 ","End":"01:09.110","Text":"just have to integrate both sides."},{"Start":"01:09.110 ","End":"01:11.435","Text":"It will look very similar to this."},{"Start":"01:11.435 ","End":"01:13.805","Text":"Only we have a constant C_2,"},{"Start":"01:13.805 ","End":"01:17.905","Text":"again rearrange, and we have this equals C_2."},{"Start":"01:17.905 ","End":"01:20.240","Text":"This is in the tutorial,"},{"Start":"01:20.240 ","End":"01:22.915","Text":"what we called Phi 1 and Phi 2."},{"Start":"01:22.915 ","End":"01:27.675","Text":"This function depends only on x and y and not on u."},{"Start":"01:27.675 ","End":"01:29.795","Text":"We said that in such a case,"},{"Start":"01:29.795 ","End":"01:36.485","Text":"what we can write is that this one is a function of this one."},{"Start":"01:36.485 ","End":"01:40.220","Text":"Here we put a function of the one that doesn\u0027t have u in it."},{"Start":"01:40.220 ","End":"01:44.315","Text":"That\u0027s not the end because we wanted to get u isolated."},{"Start":"01:44.315 ","End":"01:49.220","Text":"What we\u0027ll do is bring the 1/x to"},{"Start":"01:49.220 ","End":"01:51.005","Text":"the other side and that will be plus"},{"Start":"01:51.005 ","End":"01:54.365","Text":"1/x and then take the reciprocal and that will give us u."},{"Start":"01:54.365 ","End":"02:00.470","Text":"We have 1/f of 1/y minus 1/x plus 1/x,"},{"Start":"02:00.470 ","End":"02:03.270","Text":"and that\u0027s the answer."}],"ID":30679},{"Watched":false,"Name":"Exercise 3","Duration":"3m 33s","ChapterTopicVideoID":29092,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.370","Text":"In this exercise, we have"},{"Start":"00:02.370 ","End":"00:08.910","Text":"the following quasi-linear PDE as written here and we want to find"},{"Start":"00:08.910 ","End":"00:17.850","Text":"the general solution using Lagrange\u0027s method so we start with the auxiliary equations,"},{"Start":"00:17.850 ","End":"00:21.690","Text":"dx over this part,"},{"Start":"00:21.690 ","End":"00:28.720","Text":"dy over this and du over this, they\u0027re all equal."},{"Start":"00:28.720 ","End":"00:34.085","Text":"We have to choose 2 of these to solve."},{"Start":"00:34.085 ","End":"00:43.220","Text":"We\u0027ll go with this equals this and the other one we\u0027ll use differentials."},{"Start":"00:43.220 ","End":"00:47.510","Text":"We just multiply everything by x,"},{"Start":"00:47.510 ","End":"00:51.530","Text":"y, u and then here we have a y in the numerator."},{"Start":"00:51.530 ","End":"00:53.090","Text":"Here the missing one is x,"},{"Start":"00:53.090 ","End":"00:55.430","Text":"so it\u0027s x, dy and here the new thing, what is u?"},{"Start":"00:55.430 ","End":"00:56.870","Text":"So it\u0027s u, du,"},{"Start":"00:56.870 ","End":"00:59.250","Text":"and there\u0027s a minus here."},{"Start":"00:59.560 ","End":"01:05.690","Text":"There\u0027s a clip on the use of differentials in this section."},{"Start":"01:05.690 ","End":"01:08.705","Text":"Just look for it."},{"Start":"01:08.705 ","End":"01:12.210","Text":"We\u0027re going to solve this one first,"},{"Start":"01:12.210 ","End":"01:16.405","Text":"multiply both sides by u so we get rid of the u here and"},{"Start":"01:16.405 ","End":"01:22.989","Text":"here and this has its variable separated we just have to integrate both sides."},{"Start":"01:22.989 ","End":"01:27.265","Text":"Natural log of x equals natural log of y plus c"},{"Start":"01:27.265 ","End":"01:32.650","Text":"brings us to the other side and use the difference of logs is the log of"},{"Start":"01:32.650 ","End":"01:37.585","Text":"the quotient and we get the natural log of x over y equals"},{"Start":"01:37.585 ","End":"01:44.290","Text":"c and if natural log of x over y is constant,"},{"Start":"01:44.290 ","End":"01:46.570","Text":"then so is x over y,"},{"Start":"01:46.570 ","End":"01:52.580","Text":"it will equal each of the power of this constant anyway, call it c_1."},{"Start":"01:52.880 ","End":"01:56.585","Text":"Now, returning here,"},{"Start":"01:56.585 ","End":"02:01.710","Text":"these 2 are equal amongst themselves and each one of them is equal to this,"},{"Start":"02:01.710 ","End":"02:06.335","Text":"if we add this plus this will get twice this."},{"Start":"02:06.335 ","End":"02:11.770","Text":"This plus this is twice this that\u0027s what I wrote here."},{"Start":"02:11.770 ","End":"02:13.720","Text":"Now with the left-hand side,"},{"Start":"02:13.720 ","End":"02:19.600","Text":"what we can do is say that this is the differential of xy."},{"Start":"02:19.600 ","End":"02:23.290","Text":"When we have the differential of a function of x and y,"},{"Start":"02:23.290 ","End":"02:27.040","Text":"the differential is the derivative of this with respect to x,"},{"Start":"02:27.040 ","End":"02:29.395","Text":"which is y times dx,"},{"Start":"02:29.395 ","End":"02:32.515","Text":"plus the derivative of this with respect to y,"},{"Start":"02:32.515 ","End":"02:34.525","Text":"which is x times dy."},{"Start":"02:34.525 ","End":"02:39.695","Text":"Bring this to the other side and now we integrate both sides."},{"Start":"02:39.695 ","End":"02:42.500","Text":"The integral of d of xy is just xy,"},{"Start":"02:42.500 ","End":"02:45.770","Text":"and this is d of u squared,"},{"Start":"02:45.770 ","End":"02:49.790","Text":"d of u squared is 2u times du,"},{"Start":"02:49.790 ","End":"02:51.470","Text":"like I say later on,"},{"Start":"02:51.470 ","End":"02:58.100","Text":"there\u0027s a clip on differentials so now we have x over y equals c_1,"},{"Start":"02:58.100 ","End":"03:08.390","Text":"xy plus u squared equals c_2 and because this depends only on x and y and not on u,"},{"Start":"03:08.390 ","End":"03:15.055","Text":"we can say that this is f of this or f of this is equal to this."},{"Start":"03:15.055 ","End":"03:21.680","Text":"What have to do now is isolate u so we bring xy to the other side,"},{"Start":"03:21.680 ","End":"03:23.990","Text":"f(x ) over y minus xy,"},{"Start":"03:23.990 ","End":"03:27.865","Text":"and then we get u squared so the square root of that will give us"},{"Start":"03:27.865 ","End":"03:33.460","Text":"u and this is the answer so that\u0027s the end of the clip."}],"ID":30680},{"Watched":false,"Name":"Exercise 4","Duration":"4m 8s","ChapterTopicVideoID":29093,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:06.045","Text":"In this exercise, we have a partial differential equation itself,"},{"Start":"00:06.045 ","End":"00:09.270","Text":"the quasi-linear type and we\u0027re going to find"},{"Start":"00:09.270 ","End":"00:13.290","Text":"its general solution using Lagrange\u0027s method."},{"Start":"00:13.290 ","End":"00:17.820","Text":"The first step is to write Lagrange\u0027s auxiliary equations."},{"Start":"00:17.820 ","End":"00:24.240","Text":"dx over the part that comes in front of ux equals dy over"},{"Start":"00:24.240 ","End":"00:31.949","Text":"the part that goes with uy equals du over the bit that\u0027s on the right-hand side."},{"Start":"00:31.949 ","End":"00:33.650","Text":"We call them a,"},{"Start":"00:33.650 ","End":"00:35.855","Text":"b, and c in the tutorial."},{"Start":"00:35.855 ","End":"00:38.240","Text":"This is really a double equation,"},{"Start":"00:38.240 ","End":"00:41.360","Text":"2 equations, but there really are 3 forms."},{"Start":"00:41.360 ","End":"00:42.590","Text":"You see the first is the second,"},{"Start":"00:42.590 ","End":"00:44.960","Text":"the second equals the third, the first equal the third."},{"Start":"00:44.960 ","End":"00:46.450","Text":"We choose 2 of them,"},{"Start":"00:46.450 ","End":"00:49.760","Text":"we\u0027ll take second equals the third,"},{"Start":"00:49.760 ","End":"00:55.320","Text":"but simplify it by multiplying by minus x and we\u0027ll also,"},{"Start":"00:55.320 ","End":"00:56.830","Text":"well, you\u0027ll see,"},{"Start":"00:56.830 ","End":"00:58.130","Text":"we have to do one at a time."},{"Start":"00:58.130 ","End":"01:00.260","Text":"We\u0027ll start with this one and then do this."},{"Start":"01:00.260 ","End":"01:03.800","Text":"From here, we get by integration natural log of"},{"Start":"01:03.800 ","End":"01:07.775","Text":"y equals natural log of u and we have to add a constant of integration,"},{"Start":"01:07.775 ","End":"01:10.220","Text":"bring the natural log to the other side,"},{"Start":"01:10.220 ","End":"01:11.630","Text":"you have a difference,"},{"Start":"01:11.630 ","End":"01:16.250","Text":"make it the logarithm of a quotient and then we can inverse log"},{"Start":"01:16.250 ","End":"01:21.455","Text":"or exponentiate both sides and we get that y over u is c1,"},{"Start":"01:21.455 ","End":"01:26.110","Text":"which is really e to the c and that\u0027s the first one."},{"Start":"01:26.110 ","End":"01:30.510","Text":"Now here, we\u0027re going to use a bit of a trick,"},{"Start":"01:30.510 ","End":"01:35.030","Text":"if we multiply everything by x, this x disappears,"},{"Start":"01:35.030 ","End":"01:39.340","Text":"this x disappears, and we have an x appearing up here"},{"Start":"01:39.340 ","End":"01:44.040","Text":"and this we can write as ud over minus u^2."},{"Start":"01:44.040 ","End":"01:46.605","Text":"You\u0027ll see why this is more convenient."},{"Start":"01:46.605 ","End":"01:53.390","Text":"We have this situation where dy over minus y is equal to both this and this."},{"Start":"01:53.390 ","End":"01:56.750","Text":"Now we\u0027re going to use a result from algebra."},{"Start":"01:56.750 ","End":"01:58.670","Text":"This will be like Lambda,"},{"Start":"01:58.670 ","End":"02:00.080","Text":"this will be a over b,"},{"Start":"02:00.080 ","End":"02:06.079","Text":"this will be c over d. If Lambda\u0027s equal to both a over b and c over d,"},{"Start":"02:06.079 ","End":"02:11.390","Text":"then it\u0027s also equal to a plus c over b plus d,"},{"Start":"02:11.390 ","End":"02:16.055","Text":"so that this and this and this are all the same as is Lambda."},{"Start":"02:16.055 ","End":"02:18.785","Text":"I\u0027ll give you a proof in a moment, but let\u0027s continue."},{"Start":"02:18.785 ","End":"02:23.870","Text":"In our case, what we get is that this plus this,"},{"Start":"02:23.870 ","End":"02:28.220","Text":"over this plus this is also equal to dy over minus"},{"Start":"02:28.220 ","End":"02:35.840","Text":"y. u^2 minus u^2 cancels and we\u0027re left with this."},{"Start":"02:35.840 ","End":"02:41.555","Text":"Multiply both sides by y^2 and we have this."},{"Start":"02:41.555 ","End":"02:45.245","Text":"Bring this to the other side, like so."},{"Start":"02:45.245 ","End":"02:51.290","Text":"Now we can integrate each of these pieces so we have half x^2 plus a"},{"Start":"02:51.290 ","End":"02:59.125","Text":"half y^2 plus a half u^2 equals a constant and we can double this."},{"Start":"02:59.125 ","End":"03:02.190","Text":"It\u0027s still a constant, call it c2."},{"Start":"03:02.190 ","End":"03:05.180","Text":"Now we have this and we had this."},{"Start":"03:05.180 ","End":"03:06.600","Text":"Well, let\u0027s just write it again here,"},{"Start":"03:06.600 ","End":"03:09.170","Text":"so we have these 2 function of x,"},{"Start":"03:09.170 ","End":"03:10.410","Text":"y and u equals c2,"},{"Start":"03:10.410 ","End":"03:12.775","Text":"function of x, y and u equals c1."},{"Start":"03:12.775 ","End":"03:18.920","Text":"Neither of these is a function of just x and y otherwise we could have a simpler form."},{"Start":"03:18.920 ","End":"03:26.480","Text":"Just go for the general form than an arbitrary function of this and this equals 0."},{"Start":"03:26.480 ","End":"03:32.630","Text":"That\u0027s the most general solution for our differential equation. I almost forgot."},{"Start":"03:32.630 ","End":"03:37.715","Text":"I still owe you the proof of this, here it is."},{"Start":"03:37.715 ","End":"03:42.465","Text":"Lambda is a over b and c over d. Well from this,"},{"Start":"03:42.465 ","End":"03:44.535","Text":"we get that a is b Lambda."},{"Start":"03:44.535 ","End":"03:47.520","Text":"From this we get that c is d Lambda."},{"Start":"03:47.520 ","End":"03:49.515","Text":"If we add a plus c,"},{"Start":"03:49.515 ","End":"03:52.410","Text":"that\u0027s b Lambda plus d Lambda."},{"Start":"03:52.410 ","End":"03:54.435","Text":"Of course you can take Lambda out,"},{"Start":"03:54.435 ","End":"03:57.029","Text":"it\u0027s b plus d times Lambda."},{"Start":"03:57.029 ","End":"04:00.780","Text":"The b plus d, we can bring on the other side in the denominator and we get Lambda"},{"Start":"04:00.780 ","End":"04:05.030","Text":"a plus c over b plus d. That\u0027s what we had to show."},{"Start":"04:05.030 ","End":"04:09.330","Text":"Okay, this completes our exercise."}],"ID":30681},{"Watched":false,"Name":"Exercise 5","Duration":"3m 25s","ChapterTopicVideoID":29094,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"In this exercise, we\u0027re given"},{"Start":"00:01.980 ","End":"00:07.140","Text":"the following quasi-linear partial differential equation and there are two parts."},{"Start":"00:07.140 ","End":"00:09.210","Text":"First, we have to find the general solution of"},{"Start":"00:09.210 ","End":"00:12.825","Text":"the equation using Lagrange\u0027s method and the other part,"},{"Start":"00:12.825 ","End":"00:16.265","Text":"we\u0027ll find a particular solution which satisfies u(x,"},{"Start":"00:16.265 ","End":"00:20.020","Text":"1) = x for all x."},{"Start":"00:20.020 ","End":"00:25.640","Text":"Let\u0027s start. We start by writing the auxiliary equations."},{"Start":"00:25.640 ","End":"00:29.580","Text":"D_x over a equals d_y over b,"},{"Start":"00:29.580 ","End":"00:32.120","Text":"equals d_u over c, where this is a,"},{"Start":"00:32.120 ","End":"00:37.230","Text":"this is b and this is c. We\u0027ll take this pair of equations,"},{"Start":"00:37.230 ","End":"00:40.700","Text":"this is a separated differential equation,"},{"Start":"00:40.700 ","End":"00:43.235","Text":"in fact, we can integrate it already."},{"Start":"00:43.235 ","End":"00:45.805","Text":"I don\u0027t know why I wrote it twice anyway."},{"Start":"00:45.805 ","End":"00:49.620","Text":"Natural log of x equals natural log of y plus the constant of"},{"Start":"00:49.620 ","End":"00:52.470","Text":"integration and then we"},{"Start":"00:52.470 ","End":"00:55.450","Text":"can bring the natural log of y to the other side. So it\u0027s a minus."},{"Start":"00:55.450 ","End":"00:59.150","Text":"It\u0027s the log of the quotient equals this constant."},{"Start":"00:59.150 ","End":"01:01.730","Text":"If natural log of this is this,"},{"Start":"01:01.730 ","End":"01:04.546","Text":"then e to the power of this is this."},{"Start":"01:04.546 ","End":"01:07.475","Text":"So x over y is e^c."},{"Start":"01:07.475 ","End":"01:09.330","Text":"We\u0027ll just call that c_1."},{"Start":"01:09.330 ","End":"01:12.020","Text":"Now we\u0027ll use the little trick."},{"Start":"01:12.020 ","End":"01:19.835","Text":"We\u0027ll multiply all three sides of this equation by xy and we get the following."},{"Start":"01:19.835 ","End":"01:22.055","Text":"Now, we can do is say,"},{"Start":"01:22.055 ","End":"01:24.125","Text":"if this is equal to this,"},{"Start":"01:24.125 ","End":"01:26.255","Text":"and this is equal to this,"},{"Start":"01:26.255 ","End":"01:30.750","Text":"then this plus this is twice this."},{"Start":"01:30.890 ","End":"01:33.980","Text":"Each of these two is half of d_u."},{"Start":"01:33.980 ","End":"01:36.485","Text":"So together they make a whole d_u."},{"Start":"01:36.485 ","End":"01:37.820","Text":"Now we\u0027ve seen this before."},{"Start":"01:37.820 ","End":"01:44.600","Text":"This is the differential of xy and there is a section on differentials or review."},{"Start":"01:44.600 ","End":"01:51.514","Text":"Now we can integrate both sides and get xy equals u plus another constant of integration."},{"Start":"01:51.514 ","End":"01:55.005","Text":"So that c_2 is xy minus u."},{"Start":"01:55.005 ","End":"01:57.670","Text":"Now we have c_1 and c_2."},{"Start":"01:57.670 ","End":"01:59.929","Text":"We could use the general equation,"},{"Start":"01:59.929 ","End":"02:03.500","Text":"that big F of this and this equals 0,"},{"Start":"02:03.500 ","End":"02:07.625","Text":"but here, x over y doesn\u0027t contain u."},{"Start":"02:07.625 ","End":"02:10.465","Text":"We can use the simpler formula,"},{"Start":"02:10.465 ","End":"02:13.265","Text":"that F of this equals this."},{"Start":"02:13.265 ","End":"02:15.500","Text":"F is an arbitrary function,"},{"Start":"02:15.500 ","End":"02:19.880","Text":"has to be differentiable and now we just have to extract u from this."},{"Start":"02:19.880 ","End":"02:23.000","Text":"We get that u, which is u(x,"},{"Start":"02:23.000 ","End":"02:30.150","Text":"y) is xy minus F of x over y and that completes Part A."},{"Start":"02:30.150 ","End":"02:32.160","Text":"Now let\u0027s do Part b."},{"Start":"02:32.160 ","End":"02:38.040","Text":"We want to find a particular solution that satisfies u(x,1)= x."},{"Start":"02:39.520 ","End":"02:45.725","Text":"We take this equation from here and substitute y equals 1,"},{"Start":"02:45.725 ","End":"02:48.925","Text":"then we get x times y is x times 1,"},{"Start":"02:48.925 ","End":"02:51.630","Text":"and x over y is just x."},{"Start":"02:51.630 ","End":"02:57.635","Text":"Now, this equals this because they\u0027re both equal to u(x,1)."},{"Start":"02:57.635 ","End":"03:02.990","Text":"We can compare these two and get x minus F(x)= x,"},{"Start":"03:02.990 ","End":"03:07.010","Text":"or F(x)=0, it\u0027s identically 0."},{"Start":"03:07.010 ","End":"03:09.710","Text":"If F is the 0 function applied to anything,"},{"Start":"03:09.710 ","End":"03:12.500","Text":"it gives 0, even to x over y."},{"Start":"03:12.500 ","End":"03:17.655","Text":"This F of x over y drops and we just have here that u(x,"},{"Start":"03:17.655 ","End":"03:26.170","Text":"y)= xy and that is the answer and that completes this exercise."}],"ID":30682},{"Watched":false,"Name":"Exercise 6","Duration":"3m 35s","ChapterTopicVideoID":29095,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.600","Text":"In this exercise, we have a quasi-linear partial differential equation as follows"},{"Start":"00:06.600 ","End":"00:14.250","Text":"and we want to solve it under the condition that u(x, 0) equals 1."},{"Start":"00:14.250 ","End":"00:19.490","Text":"As usual, we start by writing the auxiliary equations,"},{"Start":"00:19.490 ","End":"00:24.020","Text":"dx over this equals dy over this,"},{"Start":"00:24.020 ","End":"00:26.315","Text":"equals du over this."},{"Start":"00:26.315 ","End":"00:29.760","Text":"We\u0027ll take this pair first."},{"Start":"00:29.760 ","End":"00:32.630","Text":"Later, we\u0027ll take this pair."},{"Start":"00:32.630 ","End":"00:34.640","Text":"Let\u0027s start with this one."},{"Start":"00:34.640 ","End":"00:39.500","Text":"Cross multiply, you get e^x dx equals minus e^y dy,"},{"Start":"00:39.500 ","End":"00:42.448","Text":"integrate integral of e^x is e^x,"},{"Start":"00:42.448 ","End":"00:47.870","Text":"the integral of e^y is e^y and we need a constant of integration."},{"Start":"00:47.870 ","End":"00:52.445","Text":"Then we can write it as e^x plus e^y equals constant number 1."},{"Start":"00:52.445 ","End":"00:54.865","Text":"Now let\u0027s take this pair,"},{"Start":"00:54.865 ","End":"00:56.070","Text":"this equals this,"},{"Start":"00:56.070 ","End":"01:00.245","Text":"multiply both sides by minus e x plus y"},{"Start":"01:00.245 ","End":"01:06.890","Text":"minus e^x will cancel with minus e^x and we still have times e^y so we get e^y,"},{"Start":"01:06.890 ","End":"01:09.890","Text":"dy equals du over u."},{"Start":"01:09.890 ","End":"01:14.480","Text":"Integrate e^y equals natural log of u plus"},{"Start":"01:14.480 ","End":"01:19.605","Text":"another constant c_2 so we can write this as this minus this equals c_2."},{"Start":"01:19.605 ","End":"01:22.420","Text":"Now we have c_1 and c_2."},{"Start":"01:22.420 ","End":"01:26.210","Text":"This left-hand side does not contain u."},{"Start":"01:26.210 ","End":"01:28.850","Text":"This is what we call Phi 1."},{"Start":"01:28.850 ","End":"01:37.245","Text":"This is Phi 2 and we have that Phi 2 equals F of Phi 1 or F of Phi 1 is"},{"Start":"01:37.245 ","End":"01:41.750","Text":"Phi 2 rearrange to bring natural log of u on the left"},{"Start":"01:41.750 ","End":"01:46.400","Text":"and the rest on the right recall that we have our initial condition,"},{"Start":"01:46.400 ","End":"01:48.770","Text":"u of x naught equals 1,"},{"Start":"01:48.770 ","End":"01:52.190","Text":"which means that when y equals 0,"},{"Start":"01:52.190 ","End":"01:58.360","Text":"u equals 1 so let\u0027s substitute u equals 1 so that\u0027s natural log of 1,"},{"Start":"01:58.360 ","End":"02:01.340","Text":"e to the 0, and here e to the 0."},{"Start":"02:01.340 ","End":"02:03.815","Text":"Anyway, this is 0, this is 1 this is 1."},{"Start":"02:03.815 ","End":"02:10.980","Text":"Then we get by rearranging F of e^x plus 1 equals 1."},{"Start":"02:10.980 ","End":"02:14.625","Text":"Just bring this to the other side or the 0 there."},{"Start":"02:14.625 ","End":"02:20.915","Text":"Now e^x plus 1 is a pretty much arbitrary number not quite arbitrary because"},{"Start":"02:20.915 ","End":"02:27.615","Text":"has to be bigger than 1 because e^x is positive so we can say that F of t equals 1,"},{"Start":"02:27.615 ","End":"02:29.905","Text":"at least for t bigger than 1."},{"Start":"02:29.905 ","End":"02:33.965","Text":"Because also any t is bigger than 1 can be written like this."},{"Start":"02:33.965 ","End":"02:37.190","Text":"Subtract 1, take the natural logarithm, you get x."},{"Start":"02:37.190 ","End":"02:45.150","Text":"Natural log of u from here is equal to e^y minus F of e^x plus"},{"Start":"02:45.150 ","End":"02:48.935","Text":"e^y that\u0027s just copying it under the condition that"},{"Start":"02:48.935 ","End":"02:54.670","Text":"the function F is identically 1 or strictly for t bigger than 1."},{"Start":"02:54.670 ","End":"02:57.570","Text":"Then this is equal to 1,"},{"Start":"02:57.570 ","End":"03:00.500","Text":"so we can just say this is e^y minus 1."},{"Start":"03:00.500 ","End":"03:03.860","Text":"Not quite sure why this is bigger than 1 each of these"},{"Start":"03:03.860 ","End":"03:07.760","Text":"is bigger than 0 so let\u0027s just say that x is bigger than 0,"},{"Start":"03:07.760 ","End":"03:11.900","Text":"then this is bigger than 1 and this is bigger than 0 so satisfies"},{"Start":"03:11.900 ","End":"03:17.540","Text":"this condition might be a way around this anyway and then take the inverse log,"},{"Start":"03:17.540 ","End":"03:20.060","Text":"which is the exponent of each side and we have that u"},{"Start":"03:20.060 ","End":"03:23.765","Text":"equals e to the power of this if you check it,"},{"Start":"03:23.765 ","End":"03:28.810","Text":"we don\u0027t have to have x bigger than 0 if x isn\u0027t even involved in this."},{"Start":"03:28.810 ","End":"03:31.715","Text":"I guess that was okay retroactively."},{"Start":"03:31.715 ","End":"03:35.820","Text":"Anyway, this is the answer and so we are done."}],"ID":30683},{"Watched":false,"Name":"Exercise 7a","Duration":"5m 23s","ChapterTopicVideoID":29096,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.385","Text":"In this exercise, we have the following partial differential equation."},{"Start":"00:05.385 ","End":"00:10.560","Text":"In Part A, we have to find the general solution using Lagrange\u0027s method,"},{"Start":"00:10.560 ","End":"00:12.030","Text":"and in Part B,"},{"Start":"00:12.030 ","End":"00:17.265","Text":"to verify that the solution we found in A really does satisfy the PDE."},{"Start":"00:17.265 ","End":"00:19.710","Text":"Let\u0027s start with Part A."},{"Start":"00:19.710 ","End":"00:22.725","Text":"We start with the auxiliary equations."},{"Start":"00:22.725 ","End":"00:25.890","Text":"It\u0027s dx over,"},{"Start":"00:25.890 ","End":"00:30.314","Text":"divided by this means this float up to the numerator,"},{"Start":"00:30.314 ","End":"00:34.880","Text":"plus dy over y=du over y^2"},{"Start":"00:34.880 ","End":"00:40.850","Text":"u. I will break this up into two equations."},{"Start":"00:40.850 ","End":"00:44.255","Text":"The first one we\u0027ll take is this equals this and later,"},{"Start":"00:44.255 ","End":"00:46.525","Text":"we\u0027ll do this equals this."},{"Start":"00:46.525 ","End":"00:48.200","Text":"It\u0027s just telling you what\u0027s coming,"},{"Start":"00:48.200 ","End":"00:50.600","Text":"but for now we start with this one."},{"Start":"00:50.600 ","End":"00:52.640","Text":"We can rearrange this,"},{"Start":"00:52.640 ","End":"00:53.870","Text":"put the dx down here,"},{"Start":"00:53.870 ","End":"00:55.010","Text":"put the y up here,"},{"Start":"00:55.010 ","End":"00:57.325","Text":"switch sides. We get this."},{"Start":"00:57.325 ","End":"01:02.035","Text":"dy by dx we can call it y\u0027 is equal to e^x,"},{"Start":"01:02.035 ","End":"01:03.770","Text":"y times square root of y,"},{"Start":"01:03.770 ","End":"01:07.480","Text":"which is y^3/2 plus y."},{"Start":"01:07.480 ","End":"01:13.260","Text":"Then we can write it as y\u0027 minus y equals e^x y^3/2."},{"Start":"01:13.260 ","End":"01:15.470","Text":"If you\u0027ve done ordinary differential equations,"},{"Start":"01:15.470 ","End":"01:18.995","Text":"you should recognize this as a Bernoulli equation."},{"Start":"01:18.995 ","End":"01:22.475","Text":"The general form of Bernoulli\u0027s equation is this,"},{"Start":"01:22.475 ","End":"01:24.545","Text":"and here n is 3/2."},{"Start":"01:24.545 ","End":"01:28.880","Text":"The solution or the strategy for solving a Bernoulli equation is to"},{"Start":"01:28.880 ","End":"01:33.950","Text":"substitute z=y^(1 minus n)."},{"Start":"01:33.950 ","End":"01:39.692","Text":"In our case, z=y^(1 minus 3/2)."},{"Start":"01:39.692 ","End":"01:41.765","Text":"One minus 3/2 is minus a half."},{"Start":"01:41.765 ","End":"01:44.980","Text":"We\u0027ll write it as z(x) equals y^-1/2 of"},{"Start":"01:44.980 ","End":"01:48.275","Text":"x to remind you that z and y are functions of x here."},{"Start":"01:48.275 ","End":"01:50.450","Text":"Differentiate both sides,"},{"Start":"01:50.450 ","End":"01:57.330","Text":"z\u0027 minus a 1/2y^(-1/2) times inner derivative is y\u0027."},{"Start":"01:57.330 ","End":"02:00.690","Text":"Now let\u0027s extract y\u0027 from here."},{"Start":"02:00.690 ","End":"02:02.150","Text":"Take all of this,"},{"Start":"02:02.150 ","End":"02:03.320","Text":"move it to the other side,"},{"Start":"02:03.320 ","End":"02:06.455","Text":"and then swap sides, and we get,"},{"Start":"02:06.455 ","End":"02:08.690","Text":"-1/2 becomes -2,"},{"Start":"02:08.690 ","End":"02:11.120","Text":"the reciprocal of y^(-3/2),"},{"Start":"02:11.120 ","End":"02:13.820","Text":"becomes y^(3/2) on the other side."},{"Start":"02:13.820 ","End":"02:18.830","Text":"Anyway, we get this. Now let\u0027s just copy this for convenience down here"},{"Start":"02:18.830 ","End":"02:24.740","Text":"and then substitute y\u0027 in here, we\u0027ll need y."},{"Start":"02:24.740 ","End":"02:30.160","Text":"From here, y=z^-2."},{"Start":"02:30.160 ","End":"02:32.535","Text":"Now substituting y\u0027,"},{"Start":"02:32.535 ","End":"02:34.395","Text":"copy it from here,"},{"Start":"02:34.395 ","End":"02:39.900","Text":"and y like we said is z^-2, equals this."},{"Start":"02:39.900 ","End":"02:46.740","Text":"y^3/2 is (z^-2)^3/2. The same here."},{"Start":"02:46.740 ","End":"02:52.425","Text":"(z^-2)^3/2 would be z^-3,"},{"Start":"02:52.425 ","End":"02:54.875","Text":"so we get this,"},{"Start":"02:54.875 ","End":"02:59.195","Text":"multiply both sides by z^3,"},{"Start":"02:59.195 ","End":"03:00.500","Text":"and we get this."},{"Start":"03:00.500 ","End":"03:04.560","Text":"z^-2, z^3 is just z. Rearrange it,"},{"Start":"03:04.560 ","End":"03:10.285","Text":"and we get z\u0027 plus a 1/2(z)= -1/2(e^x )."},{"Start":"03:10.285 ","End":"03:15.853","Text":"This is a linear equation of the first-degree with constant coefficients,"},{"Start":"03:15.853 ","End":"03:17.705","Text":"and we know how to solve this."},{"Start":"03:17.705 ","End":"03:21.695","Text":"One way is to multiply both sides by an integrating factor."},{"Start":"03:21.695 ","End":"03:25.520","Text":"This is a 1/2, the integrating factor is going to be e^(1/2 x),"},{"Start":"03:25.520 ","End":"03:27.545","Text":"will see that it works."},{"Start":"03:27.545 ","End":"03:30.575","Text":"Multiply everything by the integrating factor."},{"Start":"03:30.575 ","End":"03:37.550","Text":"Now notice that the left-hand side is the derivative of the product, e^(1/2 x) times z."},{"Start":"03:37.550 ","End":"03:40.430","Text":"It\u0027s this times the derivative of this,"},{"Start":"03:40.430 ","End":"03:42.575","Text":"plus the derivative of this here,"},{"Start":"03:42.575 ","End":"03:45.890","Text":"times this is equal to the right-hand side."},{"Start":"03:45.890 ","End":"03:51.365","Text":"Now we can integrate both sides and get that e^(1/2) z is the integral of this,"},{"Start":"03:51.365 ","End":"03:54.755","Text":"which is this plus a constant."},{"Start":"03:54.755 ","End":"03:59.330","Text":"Just rewrite this 3/2 and the denominator is 2/3,"},{"Start":"03:59.330 ","End":"04:00.960","Text":"2/3 and 1/2 is a 1/3."},{"Start":"04:00.960 ","End":"04:02.655","Text":"We\u0027ve got this."},{"Start":"04:02.655 ","End":"04:07.635","Text":"Remember that z is y^-1/2."},{"Start":"04:07.635 ","End":"04:13.470","Text":"We want to get back to y. z is y^-1/2, equals this."},{"Start":"04:13.470 ","End":"04:15.405","Text":"Write it as the following,"},{"Start":"04:15.405 ","End":"04:18.500","Text":"y^-1/2 is 1 over the square root of y."},{"Start":"04:18.500 ","End":"04:20.735","Text":"Bring this to the other side and make it plus."},{"Start":"04:20.735 ","End":"04:22.685","Text":"Now we have c_1."},{"Start":"04:22.685 ","End":"04:25.790","Text":"We want to work on another equation to get c_2."},{"Start":"04:25.790 ","End":"04:28.010","Text":"Remember, we said we\u0027re going to work on this one."},{"Start":"04:28.010 ","End":"04:31.085","Text":"This can be solved with separation of variables."},{"Start":"04:31.085 ","End":"04:33.150","Text":"Bring the y^2 to the other side."},{"Start":"04:33.150 ","End":"04:36.885","Text":"We\u0027ve got ydy=du over u."},{"Start":"04:36.885 ","End":"04:38.810","Text":"The integral of this is 1/2(y^2)."},{"Start":"04:38.810 ","End":"04:42.155","Text":"The integral of this is natural log u plus constant,"},{"Start":"04:42.155 ","End":"04:47.085","Text":"and so c_2 is equal to this minus this."},{"Start":"04:47.085 ","End":"04:50.550","Text":"We have c_1 and we have c_2."},{"Start":"04:50.550 ","End":"04:55.125","Text":"But c_1 just contains x and y without u."},{"Start":"04:55.125 ","End":"04:59.855","Text":"What we do in this case is take this equals F of this."},{"Start":"04:59.855 ","End":"05:03.320","Text":"This is the general solution to the equation."},{"Start":"05:03.320 ","End":"05:08.570","Text":"Let\u0027s just bring it in the form of u equals something depending on x and y."},{"Start":"05:08.570 ","End":"05:11.915","Text":"Bringing the natural log of u to the left-hand side,"},{"Start":"05:11.915 ","End":"05:15.125","Text":"now extract u by taking the exponent of both sides,"},{"Start":"05:15.125 ","End":"05:16.950","Text":"so u which is u(x,"},{"Start":"05:16.950 ","End":"05:19.757","Text":"y) equals e to the power of all of this."},{"Start":"05:19.757 ","End":"05:24.270","Text":"This is the solution to Part A."}],"ID":30684},{"Watched":false,"Name":"Exercise 7b","Duration":"4m 48s","ChapterTopicVideoID":29097,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.550","Text":"Now we come to part b."},{"Start":"00:02.550 ","End":"00:06.555","Text":"In part a, we solve this partial differential equation."},{"Start":"00:06.555 ","End":"00:10.440","Text":"In part b, we have to verify that the solution works."},{"Start":"00:10.440 ","End":"00:14.865","Text":"A reminder, solution from Part a was this,"},{"Start":"00:14.865 ","End":"00:18.165","Text":"and just copying the PDE here."},{"Start":"00:18.165 ","End":"00:22.555","Text":"We have to show that if we substitute this in here, we get equality."},{"Start":"00:22.555 ","End":"00:26.010","Text":"What we\u0027ll do is we\u0027ll compute ux, we\u0027ll compute uy,"},{"Start":"00:26.010 ","End":"00:27.885","Text":"and substitute in the left-hand side,"},{"Start":"00:27.885 ","End":"00:31.350","Text":"and see that it gives us y^2 u."},{"Start":"00:31.350 ","End":"00:34.800","Text":"From here, the derivative is e to the power of"},{"Start":"00:34.800 ","End":"00:39.390","Text":"the same thing times the derivative of this exponent,"},{"Start":"00:39.390 ","End":"00:42.135","Text":"which is with respect to x,"},{"Start":"00:42.135 ","End":"00:44.405","Text":"1/2y^2 is a constant, so it disappears."},{"Start":"00:44.405 ","End":"00:48.875","Text":"We have minus f\u0027 of this,"},{"Start":"00:48.875 ","End":"00:52.985","Text":"and we still have the anti-derivative of this with respect to x."},{"Start":"00:52.985 ","End":"00:56.680","Text":"So the 1 over square root y is a constant stays."},{"Start":"00:56.680 ","End":"01:00.015","Text":"Derivative of this is a 1/2 e to the 1/2x."},{"Start":"01:00.015 ","End":"01:04.715","Text":"From here, we get another 3/2 in front of the exponent."},{"Start":"01:04.715 ","End":"01:09.535","Text":"Now, we can simplify this minus with this a 1/2,"},{"Start":"01:09.535 ","End":"01:11.205","Text":"and this also is a 1/2."},{"Start":"01:11.205 ","End":"01:19.065","Text":"We can pull it out front and get minus a 1/2 here and this is equal to u."},{"Start":"01:19.065 ","End":"01:21.240","Text":"Just u is equal to this,"},{"Start":"01:21.240 ","End":"01:22.710","Text":"so we can replace this by u,"},{"Start":"01:22.710 ","End":"01:24.950","Text":"then we have the minus a 1/2 here,"},{"Start":"01:24.950 ","End":"01:30.150","Text":"then this bit, and this bit without the blue numbers."},{"Start":"01:30.230 ","End":"01:32.820","Text":"That is du by dx."},{"Start":"01:32.820 ","End":"01:34.240","Text":"Now you want du by dy."},{"Start":"01:34.240 ","End":"01:37.355","Text":"So first let\u0027s just copy again what u is."},{"Start":"01:37.355 ","End":"01:40.759","Text":"Now differentiate with respect to y."},{"Start":"01:40.759 ","End":"01:43.505","Text":"Again, we get e to the power of this."},{"Start":"01:43.505 ","End":"01:46.760","Text":"The derivative of this with respect to y,"},{"Start":"01:46.760 ","End":"01:51.185","Text":"1/2 y^2 gives us y, f gives us f\u0027,"},{"Start":"01:51.185 ","End":"01:54.110","Text":"and the anti-derivative this time,"},{"Start":"01:54.110 ","End":"01:56.690","Text":"this part gives us 0, and here,"},{"Start":"01:56.690 ","End":"01:59.150","Text":"with respect to y,"},{"Start":"01:59.150 ","End":"02:04.930","Text":"we just differentiate the 1/y to give us minus 1/2y root y."},{"Start":"02:04.930 ","End":"02:06.765","Text":"It\u0027s y to the minus a 1/2,"},{"Start":"02:06.765 ","End":"02:09.865","Text":"so minus a 1/2 y to the minus 3/2."},{"Start":"02:09.865 ","End":"02:16.385","Text":"Once again, we can replace this by u and the rest of it as is."},{"Start":"02:16.385 ","End":"02:19.474","Text":"Recall that this was the differential equation."},{"Start":"02:19.474 ","End":"02:26.660","Text":"Now we want to substitute du by dx and du by dy into here on the left-hand side,"},{"Start":"02:26.660 ","End":"02:29.830","Text":"we\u0027ll do the left-hand side and see if we can reach the right."},{"Start":"02:29.830 ","End":"02:31.750","Text":"We have this times,"},{"Start":"02:31.750 ","End":"02:34.240","Text":"this is ux,"},{"Start":"02:34.240 ","End":"02:39.575","Text":"plus the other part now plus y times this part,"},{"Start":"02:39.575 ","End":"02:42.775","Text":"which is due with respect to y."},{"Start":"02:42.775 ","End":"02:45.530","Text":"We can tidy this up a bit by letting t"},{"Start":"02:45.530 ","End":"02:51.270","Text":"equals what\u0027s written in here and also what\u0027s written in here."},{"Start":"02:52.250 ","End":"02:57.860","Text":"We\u0027ll also do some other tidying up like putting this over a common denominator,"},{"Start":"02:57.860 ","End":"03:01.580","Text":"square root of y and bringing the minus to the front."},{"Start":"03:01.580 ","End":"03:06.625","Text":"Then from here, we get uy times y is y^2 u."},{"Start":"03:06.625 ","End":"03:08.525","Text":"Then we have minus,"},{"Start":"03:08.525 ","End":"03:10.880","Text":"which goes with the minus here."},{"Start":"03:10.880 ","End":"03:14.560","Text":"It\u0027s a plus times u,"},{"Start":"03:14.560 ","End":"03:21.020","Text":"and we don\u0027t need the y because the y will cancel with the y here."},{"Start":"03:21.020 ","End":"03:23.645","Text":"This part is t,"},{"Start":"03:23.645 ","End":"03:26.895","Text":"and then this part here,"},{"Start":"03:26.895 ","End":"03:30.340","Text":"which is just this."},{"Start":"03:30.560 ","End":"03:34.575","Text":"Continuing, this is what we had."},{"Start":"03:34.575 ","End":"03:39.720","Text":"Let\u0027s take e to the 1/2 x out of here,"},{"Start":"03:39.720 ","End":"03:45.810","Text":"and from here, that leaves just 1 plus e to the x root y."},{"Start":"03:45.810 ","End":"03:49.290","Text":"Here\u0027s the e to the 1/2 x that we took out,"},{"Start":"03:49.290 ","End":"03:51.920","Text":"and everything else is the same."},{"Start":"03:51.920 ","End":"03:54.545","Text":"Notice that the denominator here,"},{"Start":"03:54.545 ","End":"04:00.815","Text":"e to the x root y plus 1 is the same as 1 plus e to dx root y, that will cancel."},{"Start":"04:00.815 ","End":"04:04.570","Text":"What will be left with is minus,"},{"Start":"04:04.570 ","End":"04:08.270","Text":"then e to the 1/2 x,"},{"Start":"04:08.270 ","End":"04:11.390","Text":"then u then f\u0027(t)."},{"Start":"04:11.390 ","End":"04:15.700","Text":"Then the 2 with the root y is 1/2 root y."},{"Start":"04:15.700 ","End":"04:17.190","Text":"That\u0027s up to here."},{"Start":"04:17.190 ","End":"04:19.560","Text":"Then plus y^2 u."},{"Start":"04:19.560 ","End":"04:22.685","Text":"Then this as is."},{"Start":"04:22.685 ","End":"04:26.915","Text":"Now notice this bit colored in blue is the same as this bit."},{"Start":"04:26.915 ","End":"04:33.670","Text":"I mean, all this part up to here is the same as this part here."},{"Start":"04:33.670 ","End":"04:36.285","Text":"But here there\u0027s a minus also,"},{"Start":"04:36.285 ","End":"04:41.800","Text":"so this will cancel this and we\u0027re just left with y^2 u,"},{"Start":"04:41.800 ","End":"04:44.165","Text":"and that\u0027s what we had to get to."},{"Start":"04:44.165 ","End":"04:45.905","Text":"That means that we\u0027re done."},{"Start":"04:45.905 ","End":"04:49.410","Text":"We\u0027re okay. That concludes this exercise."}],"ID":30685},{"Watched":false,"Name":"Exercise 8","Duration":"3m 28s","ChapterTopicVideoID":29081,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.500","Text":"In this exercise, we have a partial differential equation,"},{"Start":"00:04.500 ","End":"00:10.718","Text":"cosine y du by dx plus sine y du by dy equals e to the y sine y times u"},{"Start":"00:10.718 ","End":"00:17.265","Text":"and it\u0027s restricted to the domain where u is positive and y is between 0 and Pi over 2."},{"Start":"00:17.265 ","End":"00:21.105","Text":"We have to find the general solution using Lagrange\u0027s method."},{"Start":"00:21.105 ","End":"00:25.530","Text":"In the Lagrange\u0027s method, we start with Lagrange\u0027s auxiliary equations,"},{"Start":"00:25.530 ","End":"00:32.925","Text":"which is dx over this part equals dy over the bit goes front of the uy,"},{"Start":"00:32.925 ","End":"00:36.705","Text":"and du over the part that\u0027s on the right hand side."},{"Start":"00:36.705 ","End":"00:40.185","Text":"Call these A, B, and C in the tutorial."},{"Start":"00:40.185 ","End":"00:43.353","Text":"This is really 2 equations."},{"Start":"00:43.353 ","End":"00:45.315","Text":"So we split it up into 2."},{"Start":"00:45.315 ","End":"00:49.005","Text":"The first one we\u0027ll take as this equals this,"},{"Start":"00:49.005 ","End":"00:53.090","Text":"and we\u0027ll solve this now and later we\u0027ll take this equals this."},{"Start":"00:53.090 ","End":"00:56.215","Text":"But leave that for now we\u0027ll work on this one."},{"Start":"00:56.215 ","End":"01:02.330","Text":"This we can separate the variable that we"},{"Start":"01:02.330 ","End":"01:09.725","Text":"bring cosine y to this side and then maybe change sides so we get the following."},{"Start":"01:09.725 ","End":"01:16.150","Text":"Note that the numerator cosine y is the derivative of sine y."},{"Start":"01:16.150 ","End":"01:18.590","Text":"When we integrate this,"},{"Start":"01:18.590 ","End":"01:22.160","Text":"we get the natural logarithm of sine y. Yeah,"},{"Start":"01:22.160 ","End":"01:23.899","Text":"it should be absolute value,"},{"Start":"01:23.899 ","End":"01:26.855","Text":"but sine y is positive in"},{"Start":"01:26.855 ","End":"01:34.615","Text":"this domain as equal to integral of this as x plus constant of integration."},{"Start":"01:34.615 ","End":"01:39.830","Text":"From this, we can extract sine y by taking e^x"},{"Start":"01:39.830 ","End":"01:46.310","Text":"plus c, e^x, e^c."},{"Start":"01:46.310 ","End":"01:50.280","Text":"E^c is c1. Another constant."},{"Start":"01:50.690 ","End":"01:57.725","Text":"Yeah, from here we\u0027d like to extract c1,"},{"Start":"01:57.725 ","End":"02:03.865","Text":"so just bring e^x over to the other side as e^ minus x and we have c1."},{"Start":"02:03.865 ","End":"02:07.945","Text":"We\u0027ll work on another equation and try and get C2."},{"Start":"02:07.945 ","End":"02:12.550","Text":"From here, sine y with sine y cancels."},{"Start":"02:12.550 ","End":"02:15.065","Text":"The e^y goes over here."},{"Start":"02:15.065 ","End":"02:16.959","Text":"Again, we separated the variables,"},{"Start":"02:16.959 ","End":"02:18.925","Text":"so we integrate both sides."},{"Start":"02:18.925 ","End":"02:20.780","Text":"This is just e^y,"},{"Start":"02:20.780 ","End":"02:24.114","Text":"and this is natural log of u because u is positive,"},{"Start":"02:24.114 ","End":"02:25.995","Text":"we don\u0027t need an absolute value,"},{"Start":"02:25.995 ","End":"02:28.375","Text":"and plus a constant of integration."},{"Start":"02:28.375 ","End":"02:32.175","Text":"Use a different symbol on c1, call it c2."},{"Start":"02:32.175 ","End":"02:37.975","Text":"Then we\u0027re going to get c2 on its own by just rearranging,"},{"Start":"02:37.975 ","End":"02:40.615","Text":"bringing the natural log over to the other side."},{"Start":"02:40.615 ","End":"02:43.750","Text":"Now we have c2 and c1."},{"Start":"02:43.750 ","End":"02:47.080","Text":"This in the tutorial we called Phi."},{"Start":"02:47.080 ","End":"02:49.600","Text":"Phi 1(x) and y,"},{"Start":"02:49.600 ","End":"02:53.285","Text":"and this was Phi(x), y, u."},{"Start":"02:53.285 ","End":"02:55.830","Text":"This one doesn\u0027t depend on u,"},{"Start":"02:55.830 ","End":"02:57.285","Text":"just on x and y,"},{"Start":"02:57.285 ","End":"03:02.565","Text":"so the idea is to write this is some function of this,"},{"Start":"03:02.565 ","End":"03:05.030","Text":"so e^y minus natural log of u is"},{"Start":"03:05.030 ","End":"03:10.715","Text":"some function has to be differentiable of e to the minus x sine y."},{"Start":"03:10.715 ","End":"03:13.025","Text":"We want to extract u from this."},{"Start":"03:13.025 ","End":"03:19.270","Text":"First of all, isolate natural log of u and then take the exponent of both sides."},{"Start":"03:19.270 ","End":"03:22.910","Text":"We have that u is e to the power of what\u0027s written here,"},{"Start":"03:22.910 ","End":"03:29.310","Text":"and this is the answer and that concludes this exercise."}],"ID":30686},{"Watched":false,"Name":"Exercise 9","Duration":"2m 12s","ChapterTopicVideoID":29082,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.540","Text":"In this exercise, we have a quasi-linear partial differential equation as follows."},{"Start":"00:06.540 ","End":"00:10.410","Text":"Under the conditions xy are positive and we have to"},{"Start":"00:10.410 ","End":"00:14.520","Text":"find the general solution using Lagrange\u0027s method."},{"Start":"00:14.520 ","End":"00:16.140","Text":"Lagrange\u0027s method starts with"},{"Start":"00:16.140 ","End":"00:20.610","Text":"the Lagrange\u0027s auxiliary equations where we have dx over a,"},{"Start":"00:20.610 ","End":"00:22.020","Text":"a is this part."},{"Start":"00:22.020 ","End":"00:27.330","Text":"Dividing by 1 over y is like multiplying by y equals dy over this,"},{"Start":"00:27.330 ","End":"00:30.915","Text":"which is xdy equals du over this,"},{"Start":"00:30.915 ","End":"00:32.640","Text":"which is du over 2."},{"Start":"00:32.640 ","End":"00:34.785","Text":"This is really 2 equations."},{"Start":"00:34.785 ","End":"00:37.670","Text":"We\u0027ll take one of the equations first,"},{"Start":"00:37.670 ","End":"00:40.810","Text":"that will be this equals this."},{"Start":"00:40.810 ","End":"00:44.315","Text":"This we can solve with separation of variables,"},{"Start":"00:44.315 ","End":"00:47.090","Text":"dx over x equals dy over y."},{"Start":"00:47.090 ","End":"00:51.425","Text":"Integrate natural log of x equals natural log of y plus a constant."},{"Start":"00:51.425 ","End":"00:53.085","Text":"Bring this to the other side."},{"Start":"00:53.085 ","End":"00:59.525","Text":"It\u0027s natural log(x) over y equals c and then take the exponent of both sides,"},{"Start":"00:59.525 ","End":"01:01.340","Text":"which cancels with the log."},{"Start":"01:01.340 ","End":"01:06.980","Text":"We get x over y equals a different c e to the c, c1."},{"Start":"01:06.980 ","End":"01:08.930","Text":"We\u0027ll take another,"},{"Start":"01:08.930 ","End":"01:10.880","Text":"a little trick that we\u0027ve used before."},{"Start":"01:10.880 ","End":"01:14.135","Text":"We\u0027ll take this equals this equals this."},{"Start":"01:14.135 ","End":"01:19.205","Text":"Then say that if this equals this and this equals this,"},{"Start":"01:19.205 ","End":"01:22.850","Text":"then the sum of these two is going to be twice this."},{"Start":"01:22.850 ","End":"01:28.685","Text":"We get xdy plus ydx equals twice d over 2, which is du."},{"Start":"01:28.685 ","End":"01:32.435","Text":"This we know is the differential of x times y."},{"Start":"01:32.435 ","End":"01:36.530","Text":"Now integrate both and here we get xy equals u plus a different"},{"Start":"01:36.530 ","End":"01:41.375","Text":"constant of integration and write it as something equals c2."},{"Start":"01:41.375 ","End":"01:45.200","Text":"Now we have one function of x and y is c1."},{"Start":"01:45.200 ","End":"01:48.050","Text":"Another function of xy and u is c2."},{"Start":"01:48.050 ","End":"01:51.755","Text":"This one is a function of x and y not containing u."},{"Start":"01:51.755 ","End":"01:56.765","Text":"In that case, the general solution is this equals a function of this,"},{"Start":"01:56.765 ","End":"02:00.695","Text":"xy minus u is sum f(x) over y."},{"Start":"02:00.695 ","End":"02:02.590","Text":"How you want to extract u?"},{"Start":"02:02.590 ","End":"02:05.600","Text":"That\u0027s easy, just bring u over here, f over here,"},{"Start":"02:05.600 ","End":"02:08.900","Text":"u is xy minus f(x) over y."},{"Start":"02:08.900 ","End":"02:12.720","Text":"That is the general solution. And we\u0027re done."}],"ID":30687},{"Watched":false,"Name":"Exercise 10","Duration":"4m 58s","ChapterTopicVideoID":29083,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.520","Text":"In this exercise, we\u0027re given a quasi-linear PDE as follows."},{"Start":"00:05.520 ","End":"00:10.065","Text":"We have to find the general solution using Lagrange\u0027s method."},{"Start":"00:10.065 ","End":"00:17.430","Text":"We start off with the auxiliary equations dx over this equals dy over this,"},{"Start":"00:17.430 ","End":"00:19.860","Text":"equals du over this."},{"Start":"00:19.860 ","End":"00:21.840","Text":"This will give us 2 equations."},{"Start":"00:21.840 ","End":"00:23.670","Text":"We\u0027ll start off with this equals this,"},{"Start":"00:23.670 ","End":"00:25.590","Text":"and then we\u0027ll do another one."},{"Start":"00:25.590 ","End":"00:30.230","Text":"Yeah, dx over x plus y equal to y over y minus x,"},{"Start":"00:30.230 ","End":"00:34.700","Text":"dy over dx equals y minus x over y plus x,"},{"Start":"00:34.700 ","End":"00:37.220","Text":"divide top and bottom by x,"},{"Start":"00:37.220 ","End":"00:39.635","Text":"and we get that y\u0027 is the following."},{"Start":"00:39.635 ","End":"00:42.245","Text":"Now, this is a homogeneous equation."},{"Start":"00:42.245 ","End":"00:45.720","Text":"If you replace x and y by Lambda x,"},{"Start":"00:45.720 ","End":"00:48.380","Text":"Lambda y, we get the same thing as"},{"Start":"00:48.380 ","End":"00:51.350","Text":"a more general definition of homogeneous, but this is good enough."},{"Start":"00:51.350 ","End":"00:53.570","Text":"When we have a homogeneous equation,"},{"Start":"00:53.570 ","End":"00:55.310","Text":"there is a standard trick."},{"Start":"00:55.310 ","End":"00:58.685","Text":"Replace y over x by v. In other words,"},{"Start":"00:58.685 ","End":"01:00.830","Text":"y equals x v,"},{"Start":"01:00.830 ","End":"01:03.440","Text":"differentiate both sides with respect to x."},{"Start":"01:03.440 ","End":"01:07.360","Text":"We get y\u0027 equals v plus x v\u0027."},{"Start":"01:07.360 ","End":"01:11.945","Text":"Y\u0027 is this, and y\u0027 equals this."},{"Start":"01:11.945 ","End":"01:15.320","Text":"This is v minus 1 over v plus 1."},{"Start":"01:15.320 ","End":"01:18.470","Text":"We get that this equals this."},{"Start":"01:18.470 ","End":"01:26.725","Text":"Then again, rearranging, we get x v\u0027 equals this minus v common denominator."},{"Start":"01:26.725 ","End":"01:28.440","Text":"Simplify, I did the work for you."},{"Start":"01:28.440 ","End":"01:37.470","Text":"You can compute it v times v plus 1 is v^2 plus v. If we subtract v^2 plus v from this,"},{"Start":"01:37.470 ","End":"01:39.410","Text":"we get minus 1 minus v^2,"},{"Start":"01:39.410 ","End":"01:41.240","Text":"so it\u0027s minus v^2 plus 1."},{"Start":"01:41.240 ","End":"01:42.740","Text":"Anyway, so what we get."},{"Start":"01:42.740 ","End":"01:45.770","Text":"Then v\u0027 is dv by dx,"},{"Start":"01:45.770 ","End":"01:49.825","Text":"and now we can separate variables as follows."},{"Start":"01:49.825 ","End":"01:53.180","Text":"Now before we integrate, we\u0027ll rewrite this."},{"Start":"01:53.180 ","End":"01:57.350","Text":"This, we can write it as v over this plus 1 over this."},{"Start":"01:57.350 ","End":"02:00.190","Text":"V is 2v times 1/2."},{"Start":"02:00.190 ","End":"02:03.345","Text":"We want 2v because it\u0027s the derivative of the denominator."},{"Start":"02:03.345 ","End":"02:07.100","Text":"Then plus 1 over e^2 plus 1 equals this."},{"Start":"02:07.100 ","End":"02:09.605","Text":"Now we can integrate and get for this."},{"Start":"02:09.605 ","End":"02:13.330","Text":"A half natural log of v^2 plus 1."},{"Start":"02:13.330 ","End":"02:15.885","Text":"The integral of this is arctangent."},{"Start":"02:15.885 ","End":"02:19.625","Text":"Here minus natural log of x plus a constant."},{"Start":"02:19.625 ","End":"02:23.460","Text":"Remember that v is y over x."},{"Start":"02:23.660 ","End":"02:27.030","Text":"Substituting v equals y over x,"},{"Start":"02:27.030 ","End":"02:28.680","Text":"we get the following."},{"Start":"02:28.680 ","End":"02:36.934","Text":"Continuing, we can reorganize this part here is y^2 plus x^2 over x^2."},{"Start":"02:36.934 ","End":"02:41.780","Text":"Now the half we can put inside the natural logarithm and make it square root."},{"Start":"02:41.780 ","End":"02:44.945","Text":"We have the square root of y^2 plus x^2."},{"Start":"02:44.945 ","End":"02:49.250","Text":"Just change the order here over the square root of x^2, which is x."},{"Start":"02:49.250 ","End":"02:51.730","Text":"The rest is the same."},{"Start":"02:51.730 ","End":"02:57.515","Text":"This is logarithm of the numerator minus logarithm of the denominator."},{"Start":"02:57.515 ","End":"03:02.360","Text":"At this point, we can cancel minus log x with minus log x,"},{"Start":"03:02.360 ","End":"03:08.480","Text":"we just get log of x^2 plus y^2 square root plus arc tangent of y over x equals c_1."},{"Start":"03:08.480 ","End":"03:11.255","Text":"Now we want another equation and get c_2."},{"Start":"03:11.255 ","End":"03:12.985","Text":"Let\u0027s go back,"},{"Start":"03:12.985 ","End":"03:16.250","Text":"we\u0027ll compare this to this and get the following."},{"Start":"03:16.250 ","End":"03:19.970","Text":"Now multiply both sides by x^2 minus y^2."},{"Start":"03:19.970 ","End":"03:24.650","Text":"We\u0027ll get this because x^2 minus y^2 over x plus y is x minus y."},{"Start":"03:24.650 ","End":"03:27.970","Text":"Now, we\u0027ll also compare this with this."},{"Start":"03:27.970 ","End":"03:32.150","Text":"Just put a minus here and put it as x minus y."},{"Start":"03:32.150 ","End":"03:35.570","Text":"Now if we multiply out by this product,"},{"Start":"03:35.570 ","End":"03:37.340","Text":"we get something similar."},{"Start":"03:37.340 ","End":"03:40.175","Text":"We get minus x plus y dy equals du."},{"Start":"03:40.175 ","End":"03:41.795","Text":"Now from these two,"},{"Start":"03:41.795 ","End":"03:46.825","Text":"we can get this plus this equals twice du."},{"Start":"03:46.825 ","End":"03:50.435","Text":"Open this up, we have the following."},{"Start":"03:50.435 ","End":"03:52.220","Text":"Now organize it,"},{"Start":"03:52.220 ","End":"03:56.794","Text":"xdx and ydy separately and the 2 middle ones together,"},{"Start":"03:56.794 ","End":"03:59.825","Text":"we get in the middle of ydx plus xdy,"},{"Start":"03:59.825 ","End":"04:03.155","Text":"which we know is equal to d(x, y)."},{"Start":"04:03.155 ","End":"04:05.210","Text":"Now integrate everything."},{"Start":"04:05.210 ","End":"04:11.270","Text":"Have 1/2 x^2 minus xy minus 1/2 y^2 equals 2u plus a constant."},{"Start":"04:11.270 ","End":"04:16.775","Text":"Then we can double everything get 4u plus a different constant, c_2."},{"Start":"04:16.775 ","End":"04:19.685","Text":"Then we can write that this equals c_2."},{"Start":"04:19.685 ","End":"04:25.220","Text":"Let\u0027s take this and return here and put the c_2 equation next to it."},{"Start":"04:25.220 ","End":"04:27.575","Text":"Now that we have these two together,"},{"Start":"04:27.575 ","End":"04:31.430","Text":"we see that this one just contains x and y,"},{"Start":"04:31.430 ","End":"04:33.080","Text":"and this one contains x, y, and u."},{"Start":"04:33.080 ","End":"04:36.655","Text":"The solution will be that this is a function of this."},{"Start":"04:36.655 ","End":"04:45.415","Text":"We can write it as f of this part equals x^2 minus 2xy minus y^2 minus 4u."},{"Start":"04:45.415 ","End":"04:47.475","Text":"Now we want to extract u."},{"Start":"04:47.475 ","End":"04:49.340","Text":"Bring the 4u to this side,"},{"Start":"04:49.340 ","End":"04:52.130","Text":"this to the other side, and then divide by 4."},{"Start":"04:52.130 ","End":"04:55.460","Text":"We get that u is the following expression."},{"Start":"04:55.460 ","End":"04:58.830","Text":"That concludes this exercise."}],"ID":30688},{"Watched":false,"Name":"Exercise 11","Duration":"6m ","ChapterTopicVideoID":29084,"CourseChapterTopicPlaylistID":294428,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.640","Text":"This exercise is not like the others it\u0027s kind of a reverse exercise."},{"Start":"00:05.640 ","End":"00:14.220","Text":"We\u0027re given that this is the general solution to a PDE that\u0027s quasi linear."},{"Start":"00:14.220 ","End":"00:17.730","Text":"We have to find the PDE, in other words,"},{"Start":"00:17.730 ","End":"00:19.560","Text":"to find the functions A, B,"},{"Start":"00:19.560 ","End":"00:24.015","Text":"and C. That\u0027s part A, and in part B,"},{"Start":"00:24.015 ","End":"00:26.895","Text":"we have to find a particular solution,"},{"Start":"00:26.895 ","End":"00:29.220","Text":"particular as opposed to general,"},{"Start":"00:29.220 ","End":"00:37.140","Text":"the particular solution which satisfies the condition that u(0, y) equals y^2."},{"Start":"00:37.140 ","End":"00:40.385","Text":"Just copy this over here,"},{"Start":"00:40.385 ","End":"00:44.540","Text":"and let\u0027s compute du by dx and du by dy."},{"Start":"00:44.540 ","End":"00:46.310","Text":"Well, not exactly first,"},{"Start":"00:46.310 ","End":"00:48.785","Text":"we\u0027ll multiply by y plus 1,"},{"Start":"00:48.785 ","End":"00:50.930","Text":"more convenient this way."},{"Start":"00:50.930 ","End":"00:57.830","Text":"Now, let\u0027s differentiate both sides with respect to x. Y plus 1 doesn\u0027t depend on x,"},{"Start":"00:57.830 ","End":"01:01.310","Text":"so we just have to differentiate u with respect to x."},{"Start":"01:01.310 ","End":"01:03.620","Text":"On this side also,"},{"Start":"01:03.620 ","End":"01:09.425","Text":"e^y is a constant so we differentiate this by taking F prime,"},{"Start":"01:09.425 ","End":"01:14.250","Text":"and now the antiderivative with respect to x. Y plus 1 is a"},{"Start":"01:14.250 ","End":"01:22.435","Text":"constant and 1 over x^2 plus 1 has a derivative of minus 1 over x^2 plus 1^2."},{"Start":"01:22.435 ","End":"01:28.250","Text":"Understood another in a derivative of the x^2 plus 1 so that\u0027s times 2x,"},{"Start":"01:28.250 ","End":"01:32.105","Text":"now we\u0027re going to differentiate this with respect to y."},{"Start":"01:32.105 ","End":"01:33.785","Text":"Here we have a product."},{"Start":"01:33.785 ","End":"01:38.135","Text":"Differentiate u with respect to y and the y plus 1 as is,"},{"Start":"01:38.135 ","End":"01:44.460","Text":"and then u as is times derivative of y plus 1 is 1 equals,"},{"Start":"01:44.460 ","End":"01:47.049","Text":"the right-hand side we have a product,"},{"Start":"01:47.049 ","End":"01:51.900","Text":"derivative of e^y is e^y times this as is,"},{"Start":"01:51.900 ","End":"01:55.490","Text":"and then e^y times the derivative of this."},{"Start":"01:55.490 ","End":"02:00.515","Text":"It\u0027s F\u0027 of this and then the anti-derivative,"},{"Start":"02:00.515 ","End":"02:03.860","Text":"x^2 plus 1 and the denominator is a constant so we"},{"Start":"02:03.860 ","End":"02:07.385","Text":"just differentiate the y plus 1 and get 1."},{"Start":"02:07.385 ","End":"02:14.210","Text":"Note that this is equal to u times y plus 1, because, well,"},{"Start":"02:14.210 ","End":"02:18.375","Text":"it\u0027s written here, u times y plus 1,"},{"Start":"02:18.375 ","End":"02:20.321","Text":"z^y, F of this,"},{"Start":"02:20.321 ","End":"02:22.585","Text":"so yeah this is u times y plus 1."},{"Start":"02:22.585 ","End":"02:29.770","Text":"Then we can bring this term to the other side and we get u times 1 minus y plus 1."},{"Start":"02:29.770 ","End":"02:32.160","Text":"It\u0027s just minus y,"},{"Start":"02:32.160 ","End":"02:36.890","Text":"so this and this becomes this and the rest as is."},{"Start":"02:36.890 ","End":"02:41.356","Text":"Now we can see the way to get rid of this F\u0027 here and here."},{"Start":"02:41.356 ","End":"02:44.255","Text":"Just divide the two equations."},{"Start":"02:44.255 ","End":"02:50.405","Text":"We\u0027ll divide this by this so we get this over this,"},{"Start":"02:50.405 ","End":"02:54.980","Text":"the left-hand sides and then equals this,"},{"Start":"02:54.980 ","End":"03:00.170","Text":"the right-hand side here over the right hand side from here."},{"Start":"03:00.170 ","End":"03:02.525","Text":"Now I\u0027ve colored the stuff that\u0027ll cancel."},{"Start":"03:02.525 ","End":"03:05.660","Text":"The y plus 1 on the left will cancel with the y plus 1 on"},{"Start":"03:05.660 ","End":"03:10.070","Text":"the right side and here we have something in the numerator and denominator,"},{"Start":"03:10.070 ","End":"03:11.695","Text":"so that will cancel."},{"Start":"03:11.695 ","End":"03:19.310","Text":"Then all that\u0027s left on the right is this minus 2x and x^2"},{"Start":"03:19.310 ","End":"03:27.345","Text":"plus 1^2 over 1 over x^2 plus 1 makes it just x^2 plus 1 to the power of 1."},{"Start":"03:27.345 ","End":"03:35.970","Text":"Next, we\u0027ll cross multiply so we get ux times x^2 plus 1 equals this times this."},{"Start":"03:35.970 ","End":"03:40.245","Text":"It\u0027s minus 2x times uy(y plus 1),"},{"Start":"03:40.245 ","End":"03:42.135","Text":"then a minus and the minus is a plus,"},{"Start":"03:42.135 ","End":"03:45.460","Text":"plus 2x times uy."},{"Start":"03:46.880 ","End":"03:50.175","Text":"Bring this term over to the left,"},{"Start":"03:50.175 ","End":"03:52.010","Text":"and then we have our solution."},{"Start":"03:52.010 ","End":"03:55.040","Text":"We have that x^2 plus 1 is A,"},{"Start":"03:55.040 ","End":"03:57.470","Text":"2xy plus 1 is B,"},{"Start":"03:57.470 ","End":"04:02.790","Text":"and 2xyu is C. That concludes part A."},{"Start":"04:02.790 ","End":"04:05.400","Text":"Now we come to part B."},{"Start":"04:05.400 ","End":"04:09.485","Text":"Remember this is the general solution to the PDE,"},{"Start":"04:09.485 ","End":"04:15.365","Text":"and we want to find this particular solution that satisfies this."},{"Start":"04:15.365 ","End":"04:21.350","Text":"The first step will be to find the function F. Substitute"},{"Start":"04:21.350 ","End":"04:26.905","Text":"x equals 0 here and we get u(0, y)."},{"Start":"04:26.905 ","End":"04:30.334","Text":"The only place a substitute x is here, it\u0027s 0."},{"Start":"04:30.334 ","End":"04:33.500","Text":"Now u(0, y) equals this and u(0,"},{"Start":"04:33.500 ","End":"04:35.300","Text":"y) equals this,"},{"Start":"04:35.300 ","End":"04:36.995","Text":"so this equals this,"},{"Start":"04:36.995 ","End":"04:41.155","Text":"and now we can bring this over to the other side and invert it."},{"Start":"04:41.155 ","End":"04:48.810","Text":"We have F(y plus 1) equals y^2 over e^y times y plus 1."},{"Start":"04:48.830 ","End":"04:51.541","Text":"Now, we can make a substitution,"},{"Start":"04:51.541 ","End":"04:54.310","Text":"we want to know what F of a variable is."},{"Start":"04:54.310 ","End":"04:58.495","Text":"Let y plus 1 equals s. If s is y plus 1,"},{"Start":"04:58.495 ","End":"05:02.845","Text":"we\u0027ll also need the inverse y equals s minus 1."},{"Start":"05:02.845 ","End":"05:10.295","Text":"That gives us that F(s) equals y is s minus 1^2."},{"Start":"05:10.295 ","End":"05:14.520","Text":"Again, why is s minus 1 and y plus 1 is"},{"Start":"05:14.520 ","End":"05:20.145","Text":"s. That gives us the function F of whatever, F(s)."},{"Start":"05:20.145 ","End":"05:21.370","Text":"Now that we have F,"},{"Start":"05:21.370 ","End":"05:26.080","Text":"we can substitute F in this expression for the general solution."},{"Start":"05:26.080 ","End":"05:29.065","Text":"Just replace F by this definition,"},{"Start":"05:29.065 ","End":"05:31.700","Text":"so wherever we see s,"},{"Start":"05:31.700 ","End":"05:36.150","Text":"we can put y plus 1 over x^2 plus 1,"},{"Start":"05:36.150 ","End":"05:39.030","Text":"and I\u0027ve shaded it in blue,"},{"Start":"05:39.030 ","End":"05:40.650","Text":"so you see it better."},{"Start":"05:40.650 ","End":"05:47.505","Text":"S here is this and then the s here is this,"},{"Start":"05:47.505 ","End":"05:50.520","Text":"the s here is this,"},{"Start":"05:50.520 ","End":"05:52.595","Text":"and s here is this."},{"Start":"05:52.595 ","End":"05:57.215","Text":"This could possibly be simplified but let\u0027s just leave it at that."},{"Start":"05:57.215 ","End":"06:00.330","Text":"That concludes this exercise."}],"ID":30689}],"Thumbnail":null,"ID":294428}]

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1.1

1