WEBVTT 00:00:00.000 --> 00:00:01.470 NARRATOR: The following content is 00:00:01.470 --> 00:00:04.966 provided by MIT OpenCourseWare under a Creative Commons 00:00:04.966 --> 00:00:06.060 license. 00:00:06.060 --> 00:00:08.230 Additional information about our license 00:00:08.230 --> 00:00:10.490 and MIT OpenCourseWare in general 00:00:10.490 --> 00:00:14.430 is available at ocw.mit.edu. 00:00:14.430 --> 00:00:19.220 PROFESSOR: Ready to go. 00:00:19.220 --> 00:00:23.290 So first of all, to say thank you to several people 00:00:23.290 --> 00:00:31.820 in the class who sent codes to find these wavefronts when 00:00:31.820 --> 00:00:39.230 a shock, a step function is moving along with velocity c. 00:00:39.230 --> 00:00:45.410 I had roughly drawn what these three methods produced, 00:00:45.410 --> 00:00:49.830 but it was much better to see it in Mr. Sekora's movie that's 00:00:49.830 --> 00:00:53.150 on 18.086 site. 00:00:53.150 --> 00:01:02.340 And sure enough, behind the moving front is an overshoot, 00:01:02.340 --> 00:01:06.860 that reminds us of the Gibbs phenomenon, from Lax-Wendroff, 00:01:06.860 --> 00:01:11.220 and I think it's sort of unavoidable if we have 00:01:11.220 --> 00:01:15.360 a negative -- because one of the three coefficients 00:01:15.360 --> 00:01:18.290 at the previous time is negative for Lax-Wendroff, 00:01:18.290 --> 00:01:21.170 that's how it gets its second order accuracy, 00:01:21.170 --> 00:01:24.160 and we see that better accuracy in a much steeper, 00:01:24.160 --> 00:01:26.580 much steeper profile. 00:01:26.580 --> 00:01:29.600 Then Lax-Friedrichs, if you've noticed, 00:01:29.600 --> 00:01:35.830 has a sort of jagged profile. 00:01:35.830 --> 00:01:38.600 Maybe you've realized why that would be. 00:01:38.600 --> 00:01:45.430 Somehow Lax-Friedrichs kind of operates on a staggered grid. 00:01:45.430 --> 00:01:49.140 We'll see, actually with -- I've drawn a staggered grid here, 00:01:49.140 --> 00:01:50.770 so I can say what I mean. 00:01:50.770 --> 00:01:54.350 Lax-Friedrichs computes this value 00:01:54.350 --> 00:01:59.770 from the two earlier values there and then 00:01:59.770 --> 00:02:01.900 at the next level again and so forth, 00:02:01.900 --> 00:02:09.690 so that actually, Lax-Friedrichs is using two staggered grids 00:02:09.690 --> 00:02:11.350 separately. 00:02:11.350 --> 00:02:16.460 I think -- normally, with Lax-Friedrichs, 00:02:16.460 --> 00:02:19.440 we call that n plus 1 and n plus 2 and n plus 3, 00:02:19.440 --> 00:02:23.350 but the picture is still right, so I think that's probably 00:02:23.350 --> 00:02:30.440 responsible for what you see, that there's two waves, 00:02:30.440 --> 00:02:36.360 two fronts, right, one delta x away from each other. 00:02:36.360 --> 00:02:40.070 And finally, upwind is also smeared out. 00:02:40.070 --> 00:02:46.680 I guess I was so pleased to see these actually showing 00:02:46.680 --> 00:02:51.670 on the website and that makes me think of more questions. 00:02:51.670 --> 00:02:54.670 Maybe you too. 00:02:54.670 --> 00:03:00.230 First of all, as time goes forward, 00:03:00.230 --> 00:03:03.500 we see this bump beginning to appear. 00:03:03.500 --> 00:03:06.970 So am I right in thinking that it reaches a certain height 00:03:06.970 --> 00:03:08.790 and stays there? 00:03:08.790 --> 00:03:13.170 So I guess I'm thinking about sort of further questions, 00:03:13.170 --> 00:03:15.820 because every time you do a numerical experiment, 00:03:15.820 --> 00:03:19.150 good additional question show up. 00:03:19.150 --> 00:03:23.990 So I'm wondering, as this wavefront forms, 00:03:23.990 --> 00:03:28.750 this characteristic profile for these three methods, 00:03:28.750 --> 00:03:31.900 first of all, does that profile settle down 00:03:31.900 --> 00:03:34.610 to a steady profile? 00:03:34.610 --> 00:03:37.110 Ultimately it'd be great to be able to predict 00:03:37.110 --> 00:03:40.330 what it would be. 00:03:40.330 --> 00:03:43.300 In particular, does that bump reach a certain height? 00:03:43.300 --> 00:03:45.770 Because you can see it just barely forming and then 00:03:45.770 --> 00:03:48.650 you see it growing to a certain point. 00:03:48.650 --> 00:03:52.200 And then, when the movie ends, fortunately it 00:03:52.200 --> 00:03:55.210 plots this picture and you really see the profile, 00:03:55.210 --> 00:03:57.420 and you can actually see, if you look closely, 00:03:57.420 --> 00:04:02.460 that it goes above 1 again and probably more. 00:04:05.630 --> 00:04:10.580 That would be, does it approach a steady profile? 00:04:10.580 --> 00:04:14.530 Does that height settle down? 00:04:14.530 --> 00:04:18.170 It'd be fantastic to be able to predict that height. 00:04:18.170 --> 00:04:20.540 For example, you always want to know, how do things 00:04:20.540 --> 00:04:22.780 depend on the parameter? 00:04:22.780 --> 00:04:26.510 The one parameter we have in this problem is r -- 00:04:26.510 --> 00:04:28.240 c delta t over delta x. 00:04:28.240 --> 00:04:32.180 If I change r, I'm changing all the coefficients. 00:04:32.180 --> 00:04:36.790 If I changed r to be exactly 1, then the profile 00:04:36.790 --> 00:04:38.500 would be perfect, right? 00:04:38.500 --> 00:04:43.370 That's the golden ratio, r equal 1. 00:04:43.370 --> 00:04:48.280 I would be computing that value exactly from this value, 00:04:48.280 --> 00:04:51.400 at exactly the right slope, on the characteristic, 00:04:51.400 --> 00:04:54.330 and perfection. 00:04:54.330 --> 00:04:57.600 Now as r goes away from 1, I presume -- 00:04:57.600 --> 00:04:58.980 it goes down below 1. 00:04:58.980 --> 00:05:02.620 Of course, if it goes above 1, catastrophe. 00:05:02.620 --> 00:05:08.140 If it goes below 1, I presume the bump begins to appear, 00:05:08.140 --> 00:05:09.890 goes higher. 00:05:09.890 --> 00:05:11.530 I don't know. 00:05:11.530 --> 00:05:16.130 Could you see how the dependence on r appears? 00:05:16.130 --> 00:05:23.030 And then I can even suggest one more thing, which is -- 00:05:23.030 --> 00:05:31.910 let's put a fourth guy onto this picture, 00:05:31.910 --> 00:05:33.850 which would be leapfrog. 00:05:33.850 --> 00:05:36.350 Leapfrog for first-order equations, 00:05:36.350 --> 00:05:40.210 so let me just put down below what that would be. 00:05:40.210 --> 00:05:44.110 So it would be on that kind of a staggered grid. 00:05:44.110 --> 00:05:46.800 A new value at time n plus 1 would 00:05:46.800 --> 00:05:51.650 come from these at time n, really, and this one, at time n 00:05:51.650 --> 00:05:52.370 minus 1. 00:05:52.370 --> 00:05:55.900 So leapfrog -- you'd think of it right away. 00:05:55.900 --> 00:06:04.070 I use a centered difference in time, U_(j, n+1) minus U_(j, 00:06:04.070 --> 00:06:09.850 n-1) over 2 delta t equals c times a centered difference 00:06:09.850 --> 00:06:20.510 in space, U_(j+1, n) minus U_(j-1, n) over 2 delta x. 00:06:20.510 --> 00:06:24.680 So you see right away, the 2's cancel. 00:06:24.680 --> 00:06:29.325 The delta x times c -- the delta t times c over delta x is 00:06:29.325 --> 00:06:36.110 our old ratio r, so this is easily written in terms of r 00:06:36.110 --> 00:06:37.300 again. 00:06:37.300 --> 00:06:44.140 It does have this two time step issue 00:06:44.140 --> 00:06:47.030 and of course, we have to think about stability. 00:06:47.030 --> 00:06:53.960 I'm writing down a fourth candidate for the equation u_t 00:06:53.960 --> 00:06:59.990 equals c*u_x, of course, before I come back to the second-order 00:06:59.990 --> 00:07:03.020 equations, or systems. 00:07:03.020 --> 00:07:05.220 I just wanted to put in these comments, 00:07:05.220 --> 00:07:11.500 partly because I'm thinking about projects that are -- 00:07:11.500 --> 00:07:20.120 and I hope you are -- so some people will have theses 00:07:20.120 --> 00:07:28.030 or problems from other courses which involve solving PDEs 00:07:28.030 --> 00:07:36.050 and that'd be perfectly natural to base some project 00:07:36.050 --> 00:07:42.720 on, but if you don't have some specific application, 00:07:42.720 --> 00:07:47.660 then here are a whole lot of interesting questions, 00:07:47.660 --> 00:07:49.790 to just go further with that. 00:07:49.790 --> 00:07:59.150 So the dependence on r would be a -- 00:07:59.150 --> 00:08:08.700 the profile dependence on r, c delta t over delta x, 00:08:08.700 --> 00:08:09.650 would be a question. 00:08:14.710 --> 00:08:18.240 Let me just emphasize, what's the goal? 00:08:18.240 --> 00:08:21.330 When I say that this upwind one is smeared out, 00:08:21.330 --> 00:08:26.190 I mean that it doesn't satisfy what you really hope for. 00:08:26.190 --> 00:08:31.860 You want to capture that shock, that discontinuity -- 00:08:31.860 --> 00:08:34.710 not perfectly, you can't expect that -- 00:08:34.710 --> 00:08:39.980 but you want to capture it, maybe in, let's say, 2 delta x. 00:08:39.980 --> 00:08:43.510 A good method captures the shock within 2 delta x 00:08:43.510 --> 00:08:50.960 and does it without much undesirable oscillation, 00:08:50.960 --> 00:08:54.250 physical oscillation like that. 00:08:54.250 --> 00:08:58.870 So these are methods where we're trading off a good shock 00:08:58.870 --> 00:09:03.400 capture versus a smeared one, which is not satisfactory, 00:09:03.400 --> 00:09:07.650 really, in a lot of applications. 00:09:07.650 --> 00:09:12.060 So capturing the shock within 2 delta 00:09:12.060 --> 00:09:19.200 x, roughly, is highly desirable, and you 00:09:19.200 --> 00:09:24.290 might have to go add a artificial viscosity that we'll 00:09:24.290 --> 00:09:28.300 discuss to get there. 00:09:28.300 --> 00:09:33.120 So that's past things, but still very much present, 00:09:33.120 --> 00:09:34.340 if I can say. 00:09:34.340 --> 00:09:38.720 Here I've introduced a new leapfrog, 00:09:38.720 --> 00:09:42.450 a one-way equation leapfrog, I'll 00:09:42.450 --> 00:09:50.230 call this, where the last lecture was 00:09:50.230 --> 00:09:52.520 about two-way leapfrog. 00:09:52.520 --> 00:09:56.040 Leapfrog for u_tt equals c squared 00:09:56.040 --> 00:09:58.510 u_xx, a two-way wave equation. 00:09:58.510 --> 00:10:04.480 I might just mention about this one-way leapfrog, 00:10:04.480 --> 00:10:10.170 once you put in e to the i*k*x as we always do, 00:10:10.170 --> 00:10:17.240 or e to the i*k*j delta x in the discrete case -- 00:10:17.240 --> 00:10:27.100 so you would put in an e to the i*j delta x times k, 00:10:27.100 --> 00:10:33.560 and then find the G, the growth factor in -- 00:10:33.560 --> 00:10:39.040 that'll depend on k and will tell us the time dependence. 00:10:39.040 --> 00:10:40.550 It'll be a G to the n-th. 00:10:43.330 --> 00:10:48.790 I think you'll find that for this guy, 00:10:48.790 --> 00:10:53.440 the absolute value of G is exactly 1 while it's stable. 00:10:53.440 --> 00:10:57.130 Of course, if the Courant condition is violated, 00:10:57.130 --> 00:10:59.670 then there's no way it could be stable 00:10:59.670 --> 00:11:04.010 and G's will grow, be bigger than 1. 00:11:04.010 --> 00:11:10.040 But, I think in the stable range, you will have -- 00:11:10.040 --> 00:11:13.290 this G will have absolute value exactly 1. 00:11:16.960 --> 00:11:18.620 So it'll be on the unit circle. 00:11:18.620 --> 00:11:23.390 You don't have much leeway, right? 00:11:23.390 --> 00:11:28.920 We found upwind was, like, well inside; 00:11:28.920 --> 00:11:32.840 Lax-Friedrichs was well inside; Lax-Wendroff, 00:11:32.840 --> 00:11:38.240 because it had higher accuracy for low frequencies, 00:11:38.240 --> 00:11:44.210 was closer; and I think this guy is right on. 00:11:44.210 --> 00:11:48.170 So this is the first-order methods. 00:11:48.170 --> 00:11:50.220 This would be Lax-Wendroff and this would 00:11:50.220 --> 00:11:53.150 be this second-order method. 00:11:53.150 --> 00:11:56.710 So you're really playing with fire there 00:11:56.710 --> 00:12:00.630 and maybe this is not a favorite, just 00:12:00.630 --> 00:12:04.920 because it's too close to the edge, in fact, right 00:12:04.920 --> 00:12:06.530 on the edge. 00:12:06.530 --> 00:12:10.960 So those are some comments really generated 00:12:10.960 --> 00:12:17.890 by your numerical experiments and I think that's the way this 00:12:17.890 --> 00:12:23.490 course should -- homeworks and projects should develop out 00:12:23.490 --> 00:12:26.740 of numerical experiment. 00:12:26.740 --> 00:12:31.730 Now I'll come back to complete my lecture 00:12:31.730 --> 00:12:34.110 on the second-order wave equation. 00:12:37.260 --> 00:12:43.110 So we studied leapfrog for that. 00:12:43.110 --> 00:12:47.000 I won't repeat that, but then with second-order wave 00:12:47.000 --> 00:12:49.730 equation, we have another option, 00:12:49.730 --> 00:12:56.080 which is to reduce a second-order equation with two 00:12:56.080 --> 00:13:01.260 time derivatives to a system of two equations that are 00:13:01.260 --> 00:13:05.570 first order in time and space. 00:13:05.570 --> 00:13:12.310 I've chosen to use the letters H and E as a kind of hint 00:13:12.310 --> 00:13:17.350 that we're touching a really big application, which 00:13:17.350 --> 00:13:20.430 is electromagnetism. 00:13:20.430 --> 00:13:26.370 So E for the electric field, H for the magnetic field but here 00:13:26.370 --> 00:13:35.400 in 1-D -- so it allows me just to mention that in 3-D, 00:13:35.400 --> 00:13:39.160 when E has three components, H has three components, 00:13:39.160 --> 00:13:45.310 this is really the curl and this is really minus the curl. 00:13:45.310 --> 00:13:56.920 And the notes will show the beautiful Yee mesh, 00:13:56.920 --> 00:14:05.790 staggered mesh that copies for finite difference methods, 00:14:05.790 --> 00:14:11.700 the properties of the curl, our interpretation 00:14:11.700 --> 00:14:16.210 of and the properties of -- that Maxwell discovered the physics 00:14:16.210 --> 00:14:23.099 of, you know, current, magnetic fields going around a wire that 00:14:23.099 --> 00:14:24.890 along which one an electric field is going. 00:14:28.200 --> 00:14:31.670 But let me stay here with one dimension. 00:14:31.670 --> 00:14:37.180 So first of all, can we recover the wave equation from that? 00:14:37.180 --> 00:14:41.070 Let's just see that we haven't left behind the wave equation. 00:14:45.190 --> 00:14:48.810 So I just take the second derivative -- 00:14:48.810 --> 00:14:51.820 so the second derivative -- I just want to recover the wave 00:14:51.820 --> 00:14:56.360 equation so that I haven't lost it here -- so this will be, 00:14:56.360 --> 00:14:58.810 if I -- I'm taking the time derivative of the first 00:14:58.810 --> 00:15:07.220 equation, so I get d second h dt dx, right? 00:15:07.220 --> 00:15:13.730 But by the useful, extremely useful fact that this is 00:15:13.730 --> 00:15:20.060 the same as taking the t derivative first and the x 00:15:20.060 --> 00:15:24.860 derivative second, now I have the x derivative of this 00:15:24.860 --> 00:15:32.240 equation, so this is c -- now I have the x derivative, 00:15:32.240 --> 00:15:35.310 or c times it, so now I'm up to c squared, 00:15:35.310 --> 00:15:39.920 the x derivative of the x derivative, right? 00:15:43.310 --> 00:15:49.246 So we eliminated H, got back to one equation, second order, 00:15:49.246 --> 00:15:50.370 and it's the wave equation. 00:15:53.660 --> 00:15:58.560 I want to show the same thing now for finite differences. 00:15:58.560 --> 00:16:05.060 It's just -- it's such a simple and useful device and our point 00:16:05.060 --> 00:16:09.610 will be that it works for -- just as well for differences. 00:16:09.610 --> 00:16:14.980 That allows us to do this staggered grid idea. 00:16:14.980 --> 00:16:21.480 So the staggered grid idea is, I'll do E on this grid, 00:16:21.480 --> 00:16:26.170 at these grid points, H at the half points, 00:16:26.170 --> 00:16:32.750 E again at the integer points, H again at the half points. 00:16:32.750 --> 00:16:41.790 Now, so my difference method, leapfrog, 00:16:41.790 --> 00:16:46.390 on this staggered mesh, will copy -- 00:16:46.390 --> 00:16:54.900 so the first equation will give E at the new time from 00:16:54.900 --> 00:16:59.770 the difference in H at this time. 00:16:59.770 --> 00:17:02.250 So, you see it. 00:17:02.250 --> 00:17:08.810 dE/dt is going to be replaced by that time difference. 00:17:08.810 --> 00:17:14.180 dH/dx is going to be replaced by this space difference, 00:17:14.180 --> 00:17:16.420 and that will determine E at the new time. 00:17:16.420 --> 00:17:21.280 So I need time 0 and time 1/2 to get started. 00:17:21.280 --> 00:17:23.610 Maybe I put those down here, somewhere. 00:17:23.610 --> 00:17:26.330 I have to use the initial condition, 00:17:26.330 --> 00:17:33.710 the initial velocity to get started on E and H 00:17:33.710 --> 00:17:35.990 and then I go onwards. 00:17:35.990 --> 00:17:38.470 I've got to this point and this point. 00:17:38.470 --> 00:17:43.230 Then this guy comes from this one, this one and this one. 00:17:46.520 --> 00:17:52.410 Of course, that similarly comes from the same equation 00:17:52.410 --> 00:17:55.810 shifted over. 00:17:55.810 --> 00:17:56.980 Now I know these. 00:17:56.980 --> 00:17:59.540 Now I'm ready for this one. 00:17:59.540 --> 00:18:03.150 Coming from -- what am I approximating now? 00:18:03.150 --> 00:18:09.270 I'm on the staggered part of the grid, the half steps. 00:18:09.270 --> 00:18:14.220 I'm approximating this on the half steps, where H is defined. 00:18:14.220 --> 00:18:18.990 So that minus that is the time difference. 00:18:18.990 --> 00:18:23.380 This minus this -- which, I know these now -- 00:18:23.380 --> 00:18:26.190 is the space difference in this equation, 00:18:26.190 --> 00:18:30.060 and that's what allows me to compute the new one. 00:18:30.060 --> 00:18:33.940 You see, it's a simple idea. 00:18:33.940 --> 00:18:41.710 It's close but different from my scalar 00:18:41.710 --> 00:18:44.580 leapfrog, one-way leapfrog here. 00:18:44.580 --> 00:18:49.140 Here it was the same U at all the points. 00:18:49.140 --> 00:18:56.400 Here it is E at the integer time levels and H at the half 00:18:56.400 --> 00:18:59.700 integer time levels. 00:18:59.700 --> 00:19:03.630 It's natural to say -- I'm calling this leapfrog. 00:19:06.990 --> 00:19:12.180 Could I do this for the difference equation? 00:19:12.180 --> 00:19:17.760 I just want to see that if I eliminate H -- 00:19:17.760 --> 00:19:19.750 you saw what happened there. 00:19:19.750 --> 00:19:24.640 By using the identity that the second derivative in t and x 00:19:24.640 --> 00:19:28.640 is always the same as the second derivative in x and t, 00:19:28.640 --> 00:19:32.730 I eliminated H and got an equation that involved E only. 00:19:36.550 --> 00:19:44.030 What's going to be the corresponding identity? 00:19:44.030 --> 00:19:47.630 Trivial, but just -- trivial things are not bad things 00:19:47.630 --> 00:19:51.820 to notice -- for the different method. 00:19:54.760 --> 00:19:58.570 This is what I wanted to be sure of, for the H part. 00:19:58.570 --> 00:20:05.460 This H -- so the H were -- do you mind if I draw yet another 00:20:05.460 --> 00:20:06.650 symbol? 00:20:06.650 --> 00:20:11.840 I want to use this idea to eliminate H. So 00:20:11.840 --> 00:20:15.270 let me put two H's there. 00:20:15.270 --> 00:20:16.700 Is that right? 00:20:16.700 --> 00:20:18.020 Yeah. 00:20:18.020 --> 00:20:25.000 I think what I'd like to know -- let me just say what I believe. 00:20:25.000 --> 00:20:31.090 I believe that the time difference of the space 00:20:31.090 --> 00:20:35.990 difference of H at mesh points is 00:20:35.990 --> 00:20:44.490 equal to the space difference of the time difference of H 00:20:44.490 --> 00:20:45.560 at the same mesh points. 00:20:45.560 --> 00:20:47.370 I meant to put these as subscripts. 00:20:51.290 --> 00:20:52.670 It's not delta x. 00:20:52.670 --> 00:20:57.180 It's delta sub x. 00:20:57.180 --> 00:21:01.130 Difference in the time direction, difference 00:21:01.130 --> 00:21:02.950 in the space direction. 00:21:02.950 --> 00:21:05.520 Let's just see what this means. 00:21:05.520 --> 00:21:08.870 Let me redraw these four points. 00:21:08.870 --> 00:21:13.140 These are H. H at all those points. 00:21:13.140 --> 00:21:19.480 So this says, figure out the space differences -- so this, 00:21:19.480 --> 00:21:22.920 this minus this and this minus this -- 00:21:22.920 --> 00:21:27.790 and then take the time difference of those. 00:21:27.790 --> 00:21:30.470 Can I see what I'll get if I do that? 00:21:30.470 --> 00:21:33.890 So this means I take the space differences, 00:21:33.890 --> 00:21:39.190 so I have 1 minus 1 and that would be 1 minus 1, 00:21:39.190 --> 00:21:42.870 but now I'm going to take the time difference of those. 00:21:42.870 --> 00:21:45.340 So that'll be this one minus this one 00:21:45.340 --> 00:21:48.610 and this one plus this one. 00:21:48.610 --> 00:21:51.630 Do you see that? 00:21:51.630 --> 00:21:58.490 What we have here is H at this point minus H 00:21:58.490 --> 00:22:00.740 at this point minus H at this point 00:22:00.740 --> 00:22:05.740 plus H at this point plus 1. 00:22:05.740 --> 00:22:09.170 I claim that we have the same thing here, 00:22:09.170 --> 00:22:12.640 that if I take the time differences first -- 00:22:12.640 --> 00:22:14.020 can I do that? 00:22:14.020 --> 00:22:16.340 Instead of taking the space differences first, 00:22:16.340 --> 00:22:19.740 I'll take this one, this time difference, 00:22:19.740 --> 00:22:23.780 this minus this and I'll subtract -- 00:22:23.780 --> 00:22:27.510 I'm taking now the space difference -- I'll subtract -- 00:22:27.510 --> 00:22:32.160 are you with me? 00:22:32.160 --> 00:22:35.960 You see, it's just the same four values -- 00:22:35.960 --> 00:22:42.840 two 1's and two minus 1's, whether we go first -- in fact, 00:22:42.840 --> 00:22:46.640 this actually is a -- I think maybe throws a little light 00:22:46.640 --> 00:22:56.740 on the calculus identity, that we can exchange the order 00:22:56.740 --> 00:23:00.820 of the derivatives, because we know we can exchange the order 00:23:00.820 --> 00:23:04.120 of the differences, because we just got numbers like 1, 00:23:04.120 --> 00:23:09.190 minus 1, minus 1, and plus 1. 00:23:12.940 --> 00:23:19.180 If we let the spacing, the mesh, go to 0, this -- 00:23:19.180 --> 00:23:24.960 and divided by the delta x and the delta t, 00:23:24.960 --> 00:23:31.840 I suppose we would prove the continuous case. 00:23:31.840 --> 00:23:36.030 So in a way, this throws light on something that's probably 00:23:36.030 --> 00:23:38.110 something that we take for granted, 00:23:38.110 --> 00:23:43.010 but we wouldn't want to be pressed on too closely. 00:23:46.360 --> 00:23:51.880 So having that identity allows me 00:23:51.880 --> 00:23:57.000 to do exactly what I did with the continuous case there 00:23:57.000 --> 00:24:03.270 and show that this, when I eliminate the half levels, 00:24:03.270 --> 00:24:08.030 I've got ordinary leapfrog. 00:24:08.030 --> 00:24:11.670 So when I eliminate half levels, I will be -- 00:24:11.670 --> 00:24:15.310 when I eliminate this half level, let's say, 00:24:15.310 --> 00:24:20.320 then I'll be connecting -- when I eliminate half levels, 00:24:20.320 --> 00:24:25.020 I'm going to be connecting these at three levels -- one, two, 00:24:25.020 --> 00:24:28.310 three. 00:24:28.310 --> 00:24:30.790 It'll be leapfrog. 00:24:30.790 --> 00:24:37.890 So since it's been a few days since we wrote down 00:24:37.890 --> 00:24:44.220 two-way leapfrog or second-order equation leapfrog, let me just 00:24:44.220 --> 00:24:47.410 write it again, and that's what we would get. 00:24:47.410 --> 00:24:55.100 The time difference squared of E will 00:24:55.100 --> 00:25:01.250 be equal to c squared times the space difference squared of E. 00:25:01.250 --> 00:25:05.070 And of course, here we have a delta x squared 00:25:05.070 --> 00:25:08.760 and here a delta t squared. 00:25:08.760 --> 00:25:11.120 So altogether, an r squared. 00:25:11.120 --> 00:25:13.540 That's the formula we know. 00:25:13.540 --> 00:25:14.740 That's the formula we know. 00:25:19.320 --> 00:25:25.480 So if I simplify this to make it look good 00:25:25.480 --> 00:25:27.910 and put the delta t squared up there, 00:25:27.910 --> 00:25:29.870 you see that it's just r squared. 00:25:37.720 --> 00:25:47.040 Maybe it's worth realizing that, again, the magic r equal 1 -- 00:25:47.040 --> 00:25:58.950 The magic ratio r equal 1 gives the exact U, 00:25:58.950 --> 00:26:02.900 agreeing exactly with u. 00:26:02.900 --> 00:26:08.030 This equation, if r is 1, would be satisfied exactly 00:26:08.030 --> 00:26:14.920 by the continuous solution and so it's the same as this one. 00:26:18.520 --> 00:26:22.040 So that's leapfrog. 00:26:22.040 --> 00:26:24.650 So can I summarize leapfrog? 00:26:24.650 --> 00:26:29.090 Leapfrog was, first of all, that led us to these -- 00:26:29.090 --> 00:26:32.270 it worked for second-order equations. 00:26:32.270 --> 00:26:38.200 It led us to the same stability condition 00:26:38.200 --> 00:26:42.130 that r had to be less or equal to 1, 00:26:42.130 --> 00:26:54.730 but it took a quadratic equation for G. Remember, 00:26:54.730 --> 00:27:01.070 we had G squared minus 2G times some quantity 00:27:01.070 --> 00:27:07.830 a that involved r's and cosines plus 1 equals 0. 00:27:07.830 --> 00:27:09.782 The reason it was a quadratic was there 00:27:09.782 --> 00:27:10.990 were two time steps involved. 00:27:14.810 --> 00:27:17.100 We handled that. 00:27:17.100 --> 00:27:21.070 We got stability, if r was less than or equal to 1. 00:27:21.070 --> 00:27:25.330 That led us to a, in magnitude, less or equal 1. 00:27:25.330 --> 00:27:30.100 And that led us to G -- I'll just remember those steps -- 00:27:30.100 --> 00:27:34.955 a less or equal 1, and that let us to G less than or equal 00:27:34.955 --> 00:27:38.230 to 1. 00:27:38.230 --> 00:27:47.480 Yes, the notes, including more about Maxwell's equation 00:27:47.480 --> 00:27:54.290 and Yee's beautiful method, with a figure that shows his mesh, 00:27:54.290 --> 00:28:01.130 will go up -- I mean, every section is getting upgraded 00:28:01.130 --> 00:28:05.040 and this one is the next one to be upgraded. 00:28:05.040 --> 00:28:10.650 Probably by tomorrow, you'll have a good section 00:28:10.650 --> 00:28:14.050 on second-order equations. 00:28:17.070 --> 00:28:24.220 For me, that's what I wanted to say about the wave 00:28:24.220 --> 00:28:31.510 equation alone, with no diffusion 00:28:31.510 --> 00:28:36.810 and now I'm ready to move to the heat equation. 00:28:36.810 --> 00:28:40.400 So what do we have coming up? 00:28:40.400 --> 00:28:42.160 Maybe important to say what's coming up. 00:28:47.120 --> 00:28:50.010 So we've done one-way wave. 00:28:54.240 --> 00:28:56.280 Now we've done two-way waves. 00:29:00.740 --> 00:29:10.820 Next comes the heat equation, heat/diffusion, 00:29:10.820 --> 00:29:14.080 and then will come a mixture of the two, 00:29:14.080 --> 00:29:26.810 convection-diffusion -- and by the way, 00:29:26.810 --> 00:29:34.560 so I'll go straight to these -- in a couple of weeks, 00:29:34.560 --> 00:29:38.810 we'll have a guest lecturer who will do the applications 00:29:38.810 --> 00:29:44.980 to financial mathematics, mathematical finance. 00:29:44.980 --> 00:29:49.280 Specifically, the Black-Scholes equation. 00:29:53.920 --> 00:30:00.040 So you probably know that a lot of effort and a lot of money 00:30:00.040 --> 00:30:04.730 are going into the valuation of options 00:30:04.730 --> 00:30:13.330 and other financial derivatives and that this Black-Scholes 00:30:13.330 --> 00:30:18.590 equation beautifully produced the heat equation, 00:30:18.590 --> 00:30:23.330 and additional correction terms will 00:30:23.330 --> 00:30:26.220 produce a convection-diffusion equation. 00:30:26.220 --> 00:30:29.200 So this is an application that we 00:30:29.200 --> 00:30:34.060 wouldn't have made some years ago, 00:30:34.060 --> 00:30:37.970 but now it's worth focusing on that. 00:30:37.970 --> 00:30:41.300 I think it's interesting in itself. 00:30:41.300 --> 00:30:45.500 So this is what's coming and then 00:30:45.500 --> 00:30:48.780 you could say one-way wave equations, but nonlinear. 00:30:48.780 --> 00:30:53.870 So nonlinear wave equations and those 00:30:53.870 --> 00:30:56.040 are called conservation laws. 00:31:05.740 --> 00:31:09.500 That's after these linear cases. 00:31:09.500 --> 00:31:10.840 All right. 00:31:10.840 --> 00:31:17.170 So I'm ready to tackle the heat equation, starting 00:31:17.170 --> 00:31:22.100 then on to the next section of the notes. 00:31:22.100 --> 00:31:25.510 So let me tackle it here. 00:31:25.510 --> 00:31:33.010 So the heat equation. 00:31:33.010 --> 00:31:38.010 Let me take the constant to be 1; u_t equals u_xx. 00:31:41.310 --> 00:31:43.060 Because we already see the big difference. 00:31:43.060 --> 00:31:45.970 Now, there's a second x derivative, but only 00:31:45.970 --> 00:31:49.840 a first derivative in time. 00:31:49.840 --> 00:31:53.510 The dimensions of t and x are no longer the same. 00:31:53.510 --> 00:31:58.810 We won't have delta t and delta x comparable. 00:31:58.810 --> 00:32:03.515 Now it will be delta t comparable with delta x 00:32:03.515 --> 00:32:06.930 squared, and that's a very significant difference, 00:32:06.930 --> 00:32:09.680 because that's -- if delta x is small, 00:32:09.680 --> 00:32:12.110 delta x squared is extremely small. 00:32:12.110 --> 00:32:19.060 So when delta t is constrained by delta x squared, 00:32:19.060 --> 00:32:22.650 we're looking at small time steps. 00:32:22.650 --> 00:32:27.610 We're looking at stiff systems, absolutely. 00:32:27.610 --> 00:32:29.370 This is going to be a stiff problem, 00:32:29.370 --> 00:32:36.220 because the low frequencies and the high frequencies 00:32:36.220 --> 00:32:39.100 differ enormously in the decay rate. 00:32:39.100 --> 00:32:41.550 In fact, why don't we see it right away, 00:32:41.550 --> 00:32:44.530 starting as always with exponentials. 00:32:44.530 --> 00:32:55.870 So if I plug in u of x and t is some G of t e to the i*k*x, 00:32:55.870 --> 00:32:57.860 separating variables as always. 00:33:01.100 --> 00:33:03.860 Look for a pure exponential solution, 00:33:03.860 --> 00:33:12.870 plug that in and we get dG/dt times e to the i*k*x 00:33:12.870 --> 00:33:16.970 on the left, and now I have to take two x derivatives, 00:33:16.970 --> 00:33:23.750 so that brings i*k down twice, so that's minus k squared -- 00:33:23.750 --> 00:33:29.610 i squared giving the minus 1 -- times G, times e to the i*k*x. 00:33:29.610 --> 00:33:40.870 And now, of course, I can cancel that, which is never 0, 00:33:40.870 --> 00:33:45.010 and I've separated out the time part, 00:33:45.010 --> 00:33:51.050 the G part, dG/dt is minus k squared G. So G 00:33:51.050 --> 00:33:55.880 is e to the minus k squared t. 00:33:55.880 --> 00:33:59.580 Notice that it's k squared, and notice 00:33:59.580 --> 00:34:02.750 that it's real and negative. 00:34:02.750 --> 00:34:06.320 See, that's a very big difference. 00:34:06.320 --> 00:34:16.010 Compare e to the i*k*c*t, which it was for the one-way wave 00:34:16.010 --> 00:34:17.490 equation. 00:34:17.490 --> 00:34:21.450 The difference is enormous here. 00:34:21.450 --> 00:34:26.920 That's magnitude 1, energy conserved. 00:34:26.920 --> 00:34:30.180 This is magnitude smaller than 1, energy dissipated. 00:34:36.080 --> 00:34:40.810 We had a ratio delta t over delta x. 00:34:40.810 --> 00:34:43.390 Now we're going to see delta t over delta x squared 00:34:43.390 --> 00:34:52.010 because somehow x and t are -- now have -- 00:34:52.010 --> 00:34:56.950 it's x squared that matches t now. 00:34:56.950 --> 00:35:00.890 So now, when I put that into here -- let me just do it -- 00:35:00.890 --> 00:35:06.270 G of t is e to the minus k squared t. 00:35:06.270 --> 00:35:10.100 So there is the pure exponential. 00:35:10.100 --> 00:35:13.120 Simple as always, because we have constant coefficients, 00:35:13.120 --> 00:35:16.770 but highly informative. 00:35:16.770 --> 00:35:20.750 I guess another thing that I read off of that is, 00:35:20.750 --> 00:35:24.620 what frequencies are decaying fast? 00:35:24.620 --> 00:35:26.220 High frequencies? 00:35:26.220 --> 00:35:30.750 If k is large, k squared is very large 00:35:30.750 --> 00:35:34.690 and e to the minus k squared t is decaying very quickly. 00:35:34.690 --> 00:35:39.840 So high-frequency noise in this system, 00:35:39.840 --> 00:35:46.970 roughness, discontinuities are going to get smoothed out, 00:35:46.970 --> 00:35:51.070 strongly, by moving forward in time. 00:35:51.070 --> 00:35:52.750 Physically, what's happening is, we 00:35:52.750 --> 00:36:00.020 have our wavefront, which stayed a perfect step 00:36:00.020 --> 00:36:07.680 function in the wave equation, will instantly smear. 00:36:07.680 --> 00:36:10.990 So diffusion smears discontinuity. 00:36:14.420 --> 00:36:15.550 Heat travels. 00:36:15.550 --> 00:36:22.940 If we were back over here, if the temperature starts out as 1 00:36:22.940 --> 00:36:28.540 on this side and 0 on this side, then in an instant, 00:36:28.540 --> 00:36:33.870 some heat flows from right to left. 00:36:33.870 --> 00:36:39.790 In fact, it flows with infinite speed. 00:36:39.790 --> 00:36:44.220 In a short delta t, there's a little heat all the way out 00:36:44.220 --> 00:36:48.930 there -- very little, of course, because it's -- 00:36:48.930 --> 00:36:54.690 I guess we want to see what that behavior is. 00:36:54.690 --> 00:36:59.160 So I now want to write down -- staying with the differential 00:36:59.160 --> 00:37:03.380 equation, what do we want to do? 00:37:03.380 --> 00:37:08.590 We did the exponential case and learned a good bit 00:37:08.590 --> 00:37:12.110 from that, the fast decay of high frequencies. 00:37:16.640 --> 00:37:20.680 The next step would be to put different frequencies, 00:37:20.680 --> 00:37:22.550 different k's together. 00:37:25.080 --> 00:37:33.120 By linearity, we can add these solutions, 00:37:33.120 --> 00:37:36.830 we can multiply that by any numbers we want, 00:37:36.830 --> 00:37:40.400 depending on k; we still have solutions. 00:37:40.400 --> 00:37:43.980 And we can integrate over k, so we will, actually. 00:37:43.980 --> 00:37:47.160 That that will be the next step. 00:37:47.160 --> 00:37:50.710 Let me write down what that does. 00:37:50.710 --> 00:37:55.780 I'm going to jump, take a little jump here 00:37:55.780 --> 00:37:58.630 to put on the board what I just said in words. 00:37:58.630 --> 00:38:03.806 I can multiply this, e to the minus k squared t e 00:38:03.806 --> 00:38:09.930 to the i*k*x -- that's what's happening to the pure frequency 00:38:09.930 --> 00:38:11.350 k. 00:38:11.350 --> 00:38:23.610 If I multiply that by the amount of that frequency which is 00:38:23.610 --> 00:38:26.220 in the initial condition -- so here is the initial condition, 00:38:26.220 --> 00:38:30.730 u_0, and I've taken its Fourier transform. 00:38:30.730 --> 00:38:36.020 This tells me how much of frequency k is in the problem, 00:38:36.020 --> 00:38:38.930 is in the initial problem. 00:38:38.930 --> 00:38:41.140 This tells me how much of frequency k 00:38:41.140 --> 00:38:43.430 is in the initial problem. 00:38:43.430 --> 00:38:48.220 That amount multiplies this, with this rapid decay as time 00:38:48.220 --> 00:38:53.020 goes forward, but at any later time, I'm just adding up -- 00:38:53.020 --> 00:38:55.930 and since k -- we're on the whole line -- 00:38:55.930 --> 00:39:03.380 k is a continuous -- so we add up from k equal minus infinity 00:39:03.380 --> 00:39:09.530 to infinity and I think we have to remember the 2*pi that 00:39:09.530 --> 00:39:14.100 depends on our particular definition of that transform, 00:39:14.100 --> 00:39:17.020 but I'm going to stay with this one. 00:39:17.020 --> 00:39:18.470 That's the answer. 00:39:18.470 --> 00:39:19.760 That's u of x and t. 00:39:29.400 --> 00:39:32.910 So I've said that in words, but actually, we 00:39:32.910 --> 00:39:36.170 could just check it. 00:39:36.170 --> 00:39:41.350 We would like to know that this left-hand side solves the heat 00:39:41.350 --> 00:39:43.990 equation, and we would like to know 00:39:43.990 --> 00:39:49.970 that it has the correct start, the correct u of x and 0. 00:39:49.970 --> 00:39:50.960 Let's check that. 00:39:50.960 --> 00:39:56.060 At t equals 0, at the start, this is 1. 00:39:56.060 --> 00:40:00.460 So at the start, this formula, without that thing 00:40:00.460 --> 00:40:04.910 there, is just the inverse Fourier transform 00:40:04.910 --> 00:40:07.780 that takes the Fourier transform, multiplies by this, 00:40:07.780 --> 00:40:12.380 integrates to recover u_0. 00:40:12.380 --> 00:40:14.010 So it does start out correctly. 00:40:17.300 --> 00:40:19.240 Should I say that again? 00:40:19.240 --> 00:40:22.520 It starts correctly because at t equal 0, 00:40:22.520 --> 00:40:25.415 this is just the inverse Fourier transform of u_0, 00:40:25.415 --> 00:40:29.830 so it produces u_0. 00:40:29.830 --> 00:40:34.040 It inverts this transform that we took. 00:40:34.040 --> 00:40:37.110 Secondly, does it solve the heat equation? 00:40:37.110 --> 00:40:41.770 Sure, because for each k, we've just 00:40:41.770 --> 00:40:43.720 found that that solves the heat equation 00:40:43.720 --> 00:40:47.520 and now we're just putting different k's together. 00:40:47.520 --> 00:40:50.740 So that's the answer. 00:40:50.740 --> 00:40:52.740 So you might say, OK, perfection. 00:40:59.360 --> 00:41:02.000 Do we need finite differences? 00:41:04.670 --> 00:41:05.770 I guess we do. 00:41:05.770 --> 00:41:13.700 This is typical of formulas that give the answer exactly, 00:41:13.700 --> 00:41:17.630 but in terms that are not numerically convenient. 00:41:17.630 --> 00:41:21.660 Because, to use his formula, first of all, 00:41:21.660 --> 00:41:29.400 this this requires an integral, an infinite integral to find -- 00:41:29.400 --> 00:41:37.130 integral to find -- for u_0, u hat. 00:41:37.130 --> 00:41:42.240 To use this formula, we'd have to find this transform, 00:41:42.240 --> 00:41:45.340 and then we have to do this integral 00:41:45.340 --> 00:41:57.730 for the inverse transform, this integral, for the answer, u. 00:41:57.730 --> 00:41:58.880 Two infinite integrals. 00:42:03.710 --> 00:42:09.580 And, not to mention the fact that on a finite interval, 00:42:09.580 --> 00:42:12.160 when x doesn't go all the way from minus infinity 00:42:12.160 --> 00:42:19.180 to infinity, we have to think all over again. 00:42:19.180 --> 00:42:24.440 So we have a nice formula for the answer, but -- 00:42:24.440 --> 00:42:28.240 it looks nice, but it's not numerically terrific. 00:42:31.390 --> 00:42:34.330 We can get a lot of information out of this 00:42:34.330 --> 00:42:39.660 and there's one other special solution that's 00:42:39.660 --> 00:42:42.220 also extremely informative. 00:42:42.220 --> 00:42:47.320 What if -- so I'm going to do a special case, 00:42:47.320 --> 00:42:49.350 but it's the big case. 00:42:49.350 --> 00:42:57.240 What if the initial conditions is a delta function? 00:42:57.240 --> 00:43:02.870 What if the initial condition is a delta function? 00:43:02.870 --> 00:43:14.430 An impulse at time 0, x equal 0, an instant source of heat 00:43:14.430 --> 00:43:18.372 at a point -- a point source with finite strength, 00:43:18.372 --> 00:43:19.280 you could say. 00:43:19.280 --> 00:43:22.580 An impulse at times 0. 00:43:27.450 --> 00:43:29.980 For the wave equation, what happened? 00:43:29.980 --> 00:43:32.150 That was like turning on a flash of light 00:43:32.150 --> 00:43:35.920 and it went along the characteristics. 00:43:35.920 --> 00:43:42.600 For the heat equation, it'll be quite different. 00:43:42.600 --> 00:43:47.000 There aren't characteristics here, 00:43:47.000 --> 00:43:49.530 but this is an extremely important solution, 00:43:49.530 --> 00:43:54.255 and I think you could call it the fundamental solution 00:43:54.255 --> 00:43:55.130 to the heat equation. 00:44:01.390 --> 00:44:02.570 You may know what it is. 00:44:02.570 --> 00:44:05.230 You may have seen this formula. 00:44:05.230 --> 00:44:11.720 It's fantastic that we get a formula for this one. 00:44:11.720 --> 00:44:14.930 You could say, we've got a formula. 00:44:14.930 --> 00:44:18.170 What's the Fourier transform, what's u_0 hat of k 00:44:18.170 --> 00:44:19.510 for the delta function? 00:44:22.280 --> 00:44:24.120 1, right? 00:44:24.120 --> 00:44:30.010 The Fourier transform of the delta function is a constant 1. 00:44:30.010 --> 00:44:34.280 All frequencies are there in equal amounts. 00:44:34.280 --> 00:44:38.310 This is a constant, so I can remove it. 00:44:38.310 --> 00:44:39.710 It's 1. 00:44:39.710 --> 00:44:43.320 I'm left with a specific integral to do. 00:44:43.320 --> 00:44:46.500 And the question is, can I do it? 00:44:46.500 --> 00:44:49.410 And the answer turns out to be yes. 00:44:49.410 --> 00:44:53.100 You can do that integral when this is a constant 00:44:53.100 --> 00:44:55.030 and that will give you the answer 00:44:55.030 --> 00:44:58.130 with that starting value. 00:44:58.130 --> 00:45:00.920 Let me just say what that answer is. 00:45:00.920 --> 00:45:04.970 u of x and t -- this is the fundamental solution, 00:45:04.970 --> 00:45:09.800 u fundamental, maybe I should say, 00:45:09.800 --> 00:45:14.940 because it's such an important case -- is, turns out to be -- 00:45:14.940 --> 00:45:19.960 it's not obvious what that integral is. 00:45:19.960 --> 00:45:37.030 By the way, normally -- let's see -- normally it's -- 00:45:37.030 --> 00:45:39.050 to get the answer that I'm going to write down, 00:45:39.050 --> 00:45:43.510 we take a kind of end run on the integral, 00:45:43.510 --> 00:45:48.450 and I'm going to take a serious end run and write down 00:45:48.450 --> 00:45:49.580 the answer exactly. 00:45:54.340 --> 00:45:55.940 Here's the key part of the answer. 00:46:00.270 --> 00:46:03.880 It's a solution which starts from a delta function, spreads 00:46:03.880 --> 00:46:10.620 of course, and the shape as it spreads is a bell. 00:46:10.620 --> 00:46:15.665 It's the famous bell-shaped Gaussian distribution of e 00:46:15.665 --> 00:46:23.230 to the minus -- so is it e to the minus x squared, I guess? 00:46:23.230 --> 00:46:29.880 And then, always here comes -- that scalar tells us the width 00:46:29.880 --> 00:46:32.020 of the bell. 00:46:32.020 --> 00:46:34.900 How far has it spread at time t? 00:46:34.900 --> 00:46:37.210 And what goes there is 4t. 00:46:40.230 --> 00:46:42.250 That's the key observation. 00:46:42.250 --> 00:46:45.960 And notice how x squared and t are coming in. 00:46:45.960 --> 00:46:50.380 It's that ratio, x squared to t, that's crucial. 00:46:50.380 --> 00:46:52.480 That's the parameter again. 00:46:56.250 --> 00:46:59.550 Of course, we need to multiply by a constant. 00:47:03.630 --> 00:47:06.790 How do I know I need a constant? 00:47:06.790 --> 00:47:12.450 Because the total heat is conserved. 00:47:16.140 --> 00:47:19.210 In other words, the integral of the delta function, 00:47:19.210 --> 00:47:24.010 the original source of heat, was 1. 00:47:24.010 --> 00:47:26.280 The integral from minus infinity to infinity 00:47:26.280 --> 00:47:31.150 of the delta function, which is all totally concentrated 00:47:31.150 --> 00:47:33.290 at one point, is 1. 00:47:33.290 --> 00:47:35.480 So I need to put on whatever constant 00:47:35.480 --> 00:47:37.360 it takes so that the integral of this thing 00:47:37.360 --> 00:47:44.590 shall stay 1 for later times. 00:47:44.590 --> 00:47:47.940 This is one integral from minus infinity to infinity that we 00:47:47.940 --> 00:47:51.740 can do and it turns out to need 1 over the square root 00:47:51.740 --> 00:47:53.220 of 4*pi*t. 00:47:59.520 --> 00:48:01.180 That solves the heat equation. 00:48:05.610 --> 00:48:11.770 It's not a lot of fun to plug that into the heat equation, 00:48:11.770 --> 00:48:13.910 but it's possible, right? 00:48:13.910 --> 00:48:17.380 The x derivative, the time derivative 00:48:17.380 --> 00:48:19.010 is going to be messy. 00:48:19.010 --> 00:48:20.260 It'll have two terms. 00:48:20.260 --> 00:48:23.400 The x derivative will have two terms and they'll agree. 00:48:27.550 --> 00:48:31.150 The notes will give one way to do the integral 00:48:31.150 --> 00:48:31.990 and get that answer. 00:48:37.370 --> 00:48:40.390 You see that that integral, it's -- 00:48:40.390 --> 00:48:52.510 integrals with e to the minus k squared in them are impossible 00:48:52.510 --> 00:48:57.360 on a finite interval, but here we're going all the way from 00:48:57.360 --> 00:49:00.050 minus infinity to infinity, which you might think makes 00:49:00.050 --> 00:49:04.640 the problem harder, but actually it makes it a great deal easier 00:49:04.640 --> 00:49:09.810 and we can get an explicit answer, a beautiful answer. 00:49:09.810 --> 00:49:15.340 So we learn that the solution to the heat equation, 00:49:15.340 --> 00:49:22.180 coming from a point source, is a bell-shaped Gaussian curve that 00:49:22.180 --> 00:49:26.570 gets wider and wider as t increases, 00:49:26.570 --> 00:49:29.010 but I guess you would want to look to see, 00:49:29.010 --> 00:49:32.890 what happens if t comes down to 0? 00:49:32.890 --> 00:49:38.220 Does that really approximate -- I meant to say, 00:49:38.220 --> 00:49:44.130 does it really converge to -- this should, as t goes to 0, 00:49:44.130 --> 00:49:52.030 this should go to the delta function, and in some way, 00:49:52.030 --> 00:49:52.960 it does. 00:49:52.960 --> 00:49:54.600 In some way, it does. 00:49:54.600 --> 00:49:57.810 It's wonderful. 00:49:57.810 --> 00:50:05.520 The delta function, of course, is 0 away from the origin. 00:50:05.520 --> 00:50:08.840 So why does this approach 0 away from the origin? 00:50:08.840 --> 00:50:11.070 Suppose x is 1. 00:50:11.070 --> 00:50:12.310 Suppose x is 1. 00:50:12.310 --> 00:50:19.890 Then do you see this thing going to 0 as t goes to 0? 00:50:19.890 --> 00:50:20.910 Look what's happening. 00:50:20.910 --> 00:50:26.600 As t goes to 0, this part is blowing up, but kind of weakly. 00:50:26.600 --> 00:50:29.630 That's not a disastrous blowup. 00:50:29.630 --> 00:50:34.290 Compared to this e to the minus -- let's say 1 over 4t. 00:50:34.290 --> 00:50:40.450 As t goes to 0, that's e to a very negative exponent. 00:50:40.450 --> 00:50:42.630 That's going to 0 very fast, much 00:50:42.630 --> 00:50:45.200 faster than this is going to 0. 00:50:45.200 --> 00:50:48.640 If you like to think of l'Hopital's rule or ratio 00:50:48.640 --> 00:50:53.180 or something, this quantity is going to 0 way 00:50:53.180 --> 00:50:56.500 faster than this one is, and the result 00:50:56.500 --> 00:51:03.100 is 0, except at the one point x equals 0, where this is 1. 00:51:08.140 --> 00:51:12.040 The fact that the total amount of heat stays at 1 00:51:12.040 --> 00:51:17.590 tells us that the delta function is the limit there. 00:51:20.350 --> 00:51:27.180 Today was the heat equation, continuous case, 00:51:27.180 --> 00:51:30.950 with these explicit answers. 00:51:30.950 --> 00:51:36.570 Then next time is -- tomorrow, I guess -- is the heat equation, 00:51:36.570 --> 00:51:39.400 finite differences. 00:51:39.400 --> 00:51:43.050 Again, we'll have several difference methods 00:51:43.050 --> 00:51:46.330 and to compare them would be terrific. 00:51:49.080 --> 00:51:53.200 Please think about that, comparison with the heat 00:51:53.200 --> 00:51:57.290 equation and/or what I was saying 00:51:57.290 --> 00:52:00.960 at the beginning of class, a deeper look 00:52:00.960 --> 00:52:04.970 at the comparison for the wave equation. 00:52:04.970 --> 00:52:11.150 See you tomorrow for difference equations for this problem. 00:52:11.150 --> 00:52:12.400 Thanks.