1 00:00:00,000 --> 00:00:01,950 The following content is provided 2 00:00:01,950 --> 00:00:06,090 by MIT OpenCourseWare under a Creative Commons license. 3 00:00:06,090 --> 00:00:08,330 Additional information about our license 4 00:00:08,330 --> 00:00:10,490 and MIT OpenCourseWare in general 5 00:00:10,490 --> 00:00:11,790 is available at ocw.mit.edu. 6 00:00:16,410 --> 00:00:19,850 PROFESSOR: So this is the equation 7 00:00:19,850 --> 00:00:24,560 that we-- this inviscid Burgers' equation, no viscosity, 8 00:00:24,560 --> 00:00:30,110 is the one we've learned about for-- shocks appear, 9 00:00:30,110 --> 00:00:35,750 fans can appear and I want to do more 10 00:00:35,750 --> 00:00:38,830 today with exact solutions. 11 00:00:38,830 --> 00:00:41,400 So today is a little bit more analysis 12 00:00:41,400 --> 00:00:44,160 than numerical solutions. 13 00:00:44,160 --> 00:00:47,600 So while doing something, sort of PDE 14 00:00:47,600 --> 00:00:51,170 rather than numerical PDE, well, it's 15 00:00:51,170 --> 00:00:56,190 natural to include Burgers' equation with viscosity. 16 00:00:56,190 --> 00:00:58,580 So that's got the extra term there. 17 00:00:58,580 --> 00:01:01,490 And if I can, I would like to say something 18 00:01:01,490 --> 00:01:06,270 about two other important equations, nonlinear. 19 00:01:06,270 --> 00:01:11,040 One has a third derivative on the right-hand side, 20 00:01:11,040 --> 00:01:16,430 so it's a dispersion rather than a diffusion. 21 00:01:16,430 --> 00:01:21,840 And that equation is named after Korteweg-de Vries, 22 00:01:21,840 --> 00:01:24,040 KdV it's often called. 23 00:01:24,040 --> 00:01:28,100 I'll just put that down; KdV is the shorthand 24 00:01:28,100 --> 00:01:29,430 for that equation. 25 00:01:29,430 --> 00:01:36,400 And it comes up in water waves, one-way water waves 26 00:01:36,400 --> 00:01:40,890 actually, along a shallow channel. 27 00:01:40,890 --> 00:01:45,220 It was actually seen by somebody riding along next to the wave 28 00:01:45,220 --> 00:01:50,100 and the wave kept its shape. 29 00:01:50,100 --> 00:01:56,550 And so this equation has had a period 30 00:01:56,550 --> 00:02:00,320 of great fame in some way. 31 00:02:00,320 --> 00:02:03,020 It's nonlinear of course, but it was 32 00:02:03,020 --> 00:02:08,760 discovered that there was a really clever way to solve it. 33 00:02:08,760 --> 00:02:11,980 Actually, we will solve these equations 34 00:02:11,980 --> 00:02:16,500 even though they're nonlinear, that won't be too hard. 35 00:02:16,500 --> 00:02:23,200 The clever ideas that went into solving that one, 36 00:02:23,200 --> 00:02:27,870 I can describe a little bit, but not in full detail. 37 00:02:27,870 --> 00:02:33,930 And then we'll connect also to Hamilton-Jacobi type equations, 38 00:02:33,930 --> 00:02:37,830 which are very closely related to this in one dimension 39 00:02:37,830 --> 00:02:39,940 and in fact, they'll come into it. 40 00:02:39,940 --> 00:02:45,970 OK, so those are the equations and then, next time, 41 00:02:45,970 --> 00:02:48,320 will come a little more about finite differences 42 00:02:48,320 --> 00:02:52,570 for those equations and a lot more about the level set 43 00:02:52,570 --> 00:02:56,560 method, which is more connected to Hamilton-Jacobi 44 00:02:56,560 --> 00:03:03,020 and is quite a remarkable idea. 45 00:03:03,020 --> 00:03:05,300 OK, so that's for next time. 46 00:03:05,300 --> 00:03:06,510 This is for this time. 47 00:03:06,510 --> 00:03:11,820 So what I have in mind for this time is: to begin, 48 00:03:11,820 --> 00:03:16,690 can we solve our equation starting from a source, 49 00:03:16,690 --> 00:03:20,240 starting from a delta function, a point source? 50 00:03:20,240 --> 00:03:22,550 What's the solution look like? 51 00:03:22,550 --> 00:03:27,950 And I guess when I thought about it, 52 00:03:27,950 --> 00:03:30,440 I can see several ways to get at the answer 53 00:03:30,440 --> 00:03:38,970 and it's a very worth it example to do this from a point source. 54 00:03:38,970 --> 00:03:40,610 It has a natural interest and then 55 00:03:40,610 --> 00:03:42,790 it comes up in many other ways. 56 00:03:47,520 --> 00:03:49,950 So how we going to get the answer? 57 00:03:49,950 --> 00:03:54,990 Well, I can do it in terms of fans and shocks. 58 00:03:57,620 --> 00:04:00,900 Let me just proceed informally at first. 59 00:04:00,900 --> 00:04:06,660 The delta function is like a step up followed 60 00:04:06,660 --> 00:04:09,690 by a step down. 61 00:04:09,690 --> 00:04:14,490 Well, a very big step up and an immediate step down, right? 62 00:04:14,490 --> 00:04:18,060 So it's the extreme case of-- and remember 63 00:04:18,060 --> 00:04:22,100 that the step up-- for the step up, 64 00:04:22,100 --> 00:04:23,630 you remember the Riemann problem? 65 00:04:23,630 --> 00:04:27,830 Going from a lower to a higher one. 66 00:04:27,830 --> 00:04:32,245 That's the case where the characteristics separate, leave 67 00:04:32,245 --> 00:04:36,550 an open space that has to be filled in with a fan. 68 00:04:36,550 --> 00:04:40,860 So I would expect the solution to start with a fan 69 00:04:40,860 --> 00:04:46,300 and then I'm imagining that as the limit 70 00:04:46,300 --> 00:04:52,330 of an initial function like this. 71 00:04:52,330 --> 00:04:58,740 So that's step up, will start a fan going 72 00:04:58,740 --> 00:05:02,180 and then this step down will start a shock going. 73 00:05:05,390 --> 00:05:11,530 Because here, I have u is zero and here u is something big 74 00:05:11,530 --> 00:05:17,590 and that drop down means that the characteristics from here 75 00:05:17,590 --> 00:05:22,590 will immediately overtake-- the characteristics from the zero, 76 00:05:22,590 --> 00:05:28,810 starting from zero, have zero speed, in the xt-plane, 77 00:05:28,810 --> 00:05:30,290 they're just going up. 78 00:05:30,290 --> 00:05:35,350 The characteristics that start because of this thing 79 00:05:35,350 --> 00:05:37,700 are going to the right, so I'll have a shock. 80 00:05:37,700 --> 00:05:41,290 So I'm expecting a fan and then a shock 81 00:05:41,290 --> 00:05:43,930 based on the previous lectures. 82 00:05:43,930 --> 00:05:47,320 OK, and I'm going to figure out what fan it is 83 00:05:47,320 --> 00:05:51,490 and what shock it is and where the shock is. 84 00:05:51,490 --> 00:06:00,860 Now what I have on here is an exact formula for the solution. 85 00:06:00,860 --> 00:06:03,030 You might say, why didn't you tell me that before? 86 00:06:06,740 --> 00:06:11,170 And I will plug in the initial, it 87 00:06:11,170 --> 00:06:15,170 comes from the initial function and from a minimization process 88 00:06:15,170 --> 00:06:17,560 and it's not obvious. 89 00:06:17,560 --> 00:06:20,640 That's a funny way to describe a function of x and t, 90 00:06:20,640 --> 00:06:24,070 but of course it does. 91 00:06:24,070 --> 00:06:28,810 x and t appear here, y is the parameter 92 00:06:28,810 --> 00:06:31,290 and I'm minimizing with respect to y 93 00:06:31,290 --> 00:06:35,430 and that will give us-- well, it better give us the same answer. 94 00:06:35,430 --> 00:06:39,250 So I'm taking this as an important case, which gives us 95 00:06:39,250 --> 00:06:44,520 a chance to review all the pieces that came into the-- all 96 00:06:44,520 --> 00:06:47,160 the rules that came into it. 97 00:06:47,160 --> 00:06:52,420 We had the rules for a fan, which was of the form x over t 98 00:06:52,420 --> 00:06:53,950 if it started at zero. 99 00:06:56,640 --> 00:07:01,190 We have the rules for a shock, which was that the shock speed 100 00:07:01,190 --> 00:07:08,300 should be the jump in f of u-- which is u squared over 2 101 00:07:08,300 --> 00:07:13,210 for Burgers'-- divided by the jump in u. 102 00:07:13,210 --> 00:07:15,630 And so that's a difference of squares, 103 00:07:15,630 --> 00:07:18,280 this is a difference of first powers. 104 00:07:18,280 --> 00:07:22,700 When I do that simple division, it factors out 105 00:07:22,700 --> 00:07:27,670 and I get a u_left plus a u_right over 2, 106 00:07:27,670 --> 00:07:31,460 the 2 coming from there, and the difference 107 00:07:31,460 --> 00:07:34,420 going into the difference of squares gives me that. 108 00:07:34,420 --> 00:07:39,470 OK, and then the other condition was the entropy condition 109 00:07:39,470 --> 00:07:44,180 that f prime-- remember, this is f prime of u-- 110 00:07:44,180 --> 00:07:48,850 so f prime of u, which is here, u. 111 00:07:48,850 --> 00:07:53,520 So u_left should be greater-- the shocks, 112 00:07:53,520 --> 00:07:56,150 what's the rule on the entropy condition? 113 00:07:56,150 --> 00:07:59,500 That the characteristic should run into the shock. 114 00:07:59,500 --> 00:08:02,370 So coming from the left, they should be faster. 115 00:08:02,370 --> 00:08:08,590 We have some shock line-- it's increasing with a speed 116 00:08:08,590 --> 00:08:13,500 s, which is d position of the shock-- dx/dt, 117 00:08:13,500 --> 00:08:17,340 and coming from the left, the shocks should be going faster 118 00:08:17,340 --> 00:08:18,660 and run into it. 119 00:08:18,660 --> 00:08:21,200 Coming from the right, the shocks-- 120 00:08:21,200 --> 00:08:24,350 the characteristics should be slower 121 00:08:24,350 --> 00:08:27,330 so that the shock line catches them 122 00:08:27,330 --> 00:08:32,410 and therefore, these also run into the shock. 123 00:08:32,410 --> 00:08:37,800 So this is f prime at u_right. 124 00:08:37,800 --> 00:08:40,580 And remember here, it's just u_left bigger 125 00:08:40,580 --> 00:08:45,860 than the shock speed, bigger than u_right. 126 00:08:45,860 --> 00:08:49,150 Now I'm going to go to the delta function 127 00:08:49,150 --> 00:08:52,290 itself, which will be cleaner and neater 128 00:08:52,290 --> 00:08:59,710 than this steep square wave. 129 00:08:59,710 --> 00:09:02,880 So we could pick any of those as the key 130 00:09:02,880 --> 00:09:08,660 to telling us-- this is what I'm expecting for a solution. 131 00:09:08,660 --> 00:09:09,730 Here is zero. 132 00:09:09,730 --> 00:09:12,300 Here is x. 133 00:09:12,300 --> 00:09:17,670 So I'm expecting that-- I guess I'll graph u. 134 00:09:17,670 --> 00:09:21,200 u of x at some time t. 135 00:09:21,200 --> 00:09:25,060 OK, so I'm graphing u at some later time t. 136 00:09:25,060 --> 00:09:29,500 OK, I'm expecting the solution to be u equal zero here, 137 00:09:29,500 --> 00:09:31,410 because it starts at zero-- well, 138 00:09:31,410 --> 00:09:35,110 I'm really-- now, when I point to that figure, what I really 139 00:09:35,110 --> 00:09:38,030 mean now is a delta function. 140 00:09:38,030 --> 00:09:41,400 So this is infinitely thin and infinitely tall. 141 00:09:44,100 --> 00:09:49,810 So characteristics are going to take off and well, who 142 00:09:49,810 --> 00:09:54,990 can say-- you might think the whole thing would break down 143 00:09:54,990 --> 00:09:59,510 because the delta function is in some sense, 144 00:09:59,510 --> 00:10:01,150 infinite at that point. 145 00:10:01,150 --> 00:10:02,160 But it doesn't. 146 00:10:02,160 --> 00:10:05,890 You'll see that a very sensible solution emerges here. 147 00:10:05,890 --> 00:10:12,300 And it's a fan, so u is x over t, I guess. 148 00:10:12,300 --> 00:10:21,290 So u starts at zero-- yeah, let me draw u at some later time t. 149 00:10:21,290 --> 00:10:25,880 It's linear and then it's a shock. 150 00:10:25,880 --> 00:10:28,160 So this is what I expect for the answer. 151 00:10:28,160 --> 00:10:34,980 I expect u to be zero, x over t up to some point, 152 00:10:34,980 --> 00:10:43,620 up to the shock, up to-- x is the shock position, 153 00:10:43,620 --> 00:10:50,880 and then after that it has the value zero 154 00:10:50,880 --> 00:10:57,410 because the initial function hasn't changed, 155 00:10:57,410 --> 00:10:59,710 hasn't got the message. 156 00:10:59,710 --> 00:11:08,270 So this was after x equal X of t. 157 00:11:08,270 --> 00:11:12,120 That's what it should look like. 158 00:11:12,120 --> 00:11:15,070 I'm proceeding here sort of on the basis of what 159 00:11:15,070 --> 00:11:17,610 we know about the solution. 160 00:11:17,610 --> 00:11:26,310 And maybe I can just mention that the text, the problems 161 00:11:26,310 --> 00:11:29,210 at the end of the section on conservation 162 00:11:29,210 --> 00:11:35,680 laws in the Introduction to Applied Math text 163 00:11:35,680 --> 00:11:38,380 ask about a few other initial functions. 164 00:11:38,380 --> 00:11:40,960 For example, minus the delta function. 165 00:11:44,690 --> 00:11:47,310 Maybe we should think through those questions. 166 00:11:47,310 --> 00:11:52,740 Is the solution with minus the delta function the negative 167 00:11:52,740 --> 00:11:54,170 of this solution? 168 00:11:57,170 --> 00:11:59,290 What do you think? 169 00:11:59,290 --> 00:12:01,340 No. 170 00:12:01,340 --> 00:12:04,110 This is a nonlinear equation and that fact 171 00:12:04,110 --> 00:12:11,600 shows up right away in the possibility 172 00:12:11,600 --> 00:12:15,090 that when you change the sign of the initial function, 173 00:12:15,090 --> 00:12:18,160 you don't just change the sign of the solution. 174 00:12:18,160 --> 00:12:20,460 And other things, I could ask well, 175 00:12:20,460 --> 00:12:22,120 suppose I have an initial function 176 00:12:22,120 --> 00:12:25,900 u_0 of x and I add 1 to it. 177 00:12:25,900 --> 00:12:28,880 Does that add 1 to the solution everywhere? 178 00:12:28,880 --> 00:12:30,760 That's a question to think about. 179 00:12:30,760 --> 00:12:35,430 So these are sort of questions that 180 00:12:35,430 --> 00:12:42,380 are not finite differences, but differential equations. 181 00:12:42,380 --> 00:12:48,340 All that remains here is to find the position of that shock. 182 00:12:48,340 --> 00:12:53,480 I guess one way to do it-- OK, I can think of several ways 183 00:12:53,480 --> 00:12:54,620 to do it. 184 00:12:54,620 --> 00:12:58,905 One way would be to use the fact that the total mass is 185 00:12:58,905 --> 00:13:00,610 conserved. 186 00:13:00,610 --> 00:13:05,100 So the integral of u at any time, 187 00:13:05,100 --> 00:13:10,650 the integral of that function from minus infinity 188 00:13:10,650 --> 00:13:14,660 to infinity-- that's the total mass-- 189 00:13:14,660 --> 00:13:18,670 has to agree with the initial total mass, which is 1, 190 00:13:18,670 --> 00:13:22,570 because the initial density distribution was 191 00:13:22,570 --> 00:13:24,670 that delta function at the origin 192 00:13:24,670 --> 00:13:26,960 and that we know integrates to 1. 193 00:13:26,960 --> 00:13:30,210 So one way to find this position would be the fact 194 00:13:30,210 --> 00:13:36,390 that the integral from zero to the place that the fan ends-- 195 00:13:36,390 --> 00:13:45,080 that's a capital X-- of x over t dx should be 1. 196 00:13:45,080 --> 00:13:47,920 1 because it was initially 1. 197 00:13:47,920 --> 00:13:56,140 So that's gives us capital X squared over 2-- over 2t as 1. 198 00:13:56,140 --> 00:13:58,190 In fact, I guess I'm thinking now, 199 00:13:58,190 --> 00:14:01,200 this approach occurred to me sort of 200 00:14:01,200 --> 00:14:05,590 as I was walking to class, I'm thinking now that's unbeatable. 201 00:14:05,590 --> 00:14:09,500 That tells us where the shock position is, 202 00:14:09,500 --> 00:14:14,030 capital X squared over 2t is 1, so capital X 203 00:14:14,030 --> 00:14:16,730 is square root of 2t. 204 00:14:22,120 --> 00:14:30,630 OK, we could take that as a guess, if you like, 205 00:14:30,630 --> 00:14:38,000 and verify that it obeys all the nonlinear rules here. 206 00:14:38,000 --> 00:14:41,710 Right now it just obeys the fact that the total mass 207 00:14:41,710 --> 00:14:45,170 is conserved, which is like the minimal fact. 208 00:14:45,170 --> 00:14:49,480 But of course, we've built into this solution 209 00:14:49,480 --> 00:14:54,990 an expectation of fan followed by shock, which is now 210 00:14:54,990 --> 00:14:57,090 what we really want to verify. 211 00:14:57,090 --> 00:15:00,730 So I guess, what should I check next? 212 00:15:00,730 --> 00:15:03,360 I maybe should check the shock speed. 213 00:15:03,360 --> 00:15:06,170 Does this solution with that shock 214 00:15:06,170 --> 00:15:08,760 give me the correct shock speed? 215 00:15:08,760 --> 00:15:10,030 OK, let me check that. 216 00:15:14,120 --> 00:15:17,580 What's the shock speed? s is the speed 217 00:15:17,580 --> 00:15:25,020 of the shock, so it's the derivative of square root of 2t 218 00:15:25,020 --> 00:15:26,130 and what's that? 219 00:15:26,130 --> 00:15:30,360 That's square root of 2 t to the 1/2. 220 00:15:30,360 --> 00:15:33,540 So that's square root of 2 times t to the 1/2. 221 00:15:33,540 --> 00:15:38,310 The derivative of t to the 1/2 is 1/2 t to the minus 1/2. 222 00:15:42,190 --> 00:15:44,460 So it's 1 over square root of 2t. 223 00:15:49,900 --> 00:15:53,870 That's the shock speed for this proposed solution. 224 00:15:53,870 --> 00:15:56,980 If we decide that the shock should be at that point. 225 00:15:56,980 --> 00:15:58,550 Now what's supposed to happen? 226 00:15:58,550 --> 00:16:00,710 What is it that I want to check? 227 00:16:00,710 --> 00:16:05,490 I want to check that the jump condition is 228 00:16:05,490 --> 00:16:08,050 correct for this shock. 229 00:16:08,050 --> 00:16:12,190 That if a shock speed is that, so the question mark is, 230 00:16:12,190 --> 00:16:18,100 is this-- so that's the speed of the shock 231 00:16:18,100 --> 00:16:23,880 as we're sort of conjecturing it to be located. 232 00:16:23,880 --> 00:16:28,430 And now the question is whether that shock speed 233 00:16:28,430 --> 00:16:34,130 agrees with u on the left plus u on the right over 2. 234 00:16:34,130 --> 00:16:35,570 So what do I have to check? 235 00:16:35,570 --> 00:16:41,230 I have to check what is u on the left of the shock 236 00:16:41,230 --> 00:16:44,050 if this is my answer. 237 00:16:44,050 --> 00:16:49,380 Well, the left of the shock is the point when 238 00:16:49,380 --> 00:16:53,660 x equals capital X. That's where the shock happened, 239 00:16:53,660 --> 00:16:56,280 so it's capital X over t. 240 00:16:56,280 --> 00:16:59,440 Square root of 2t over t. 241 00:16:59,440 --> 00:17:06,840 So the value of u square root of 2t over t. 242 00:17:06,840 --> 00:17:09,770 That's where the fan is. 243 00:17:09,770 --> 00:17:12,640 That's this height and that height is zero. 244 00:17:12,640 --> 00:17:14,560 So it's the average of this and zero. 245 00:17:19,040 --> 00:17:20,770 So that's the jump condition. 246 00:17:20,770 --> 00:17:24,040 So this is the jump condition that we're 247 00:17:24,040 --> 00:17:32,510 checking and is it right? 248 00:17:32,510 --> 00:17:34,580 I hope so. 249 00:17:34,580 --> 00:17:36,620 Otherwise we're in mighty big trouble. 250 00:17:36,620 --> 00:17:40,400 OK yes, I have the same cancellation 251 00:17:40,400 --> 00:17:44,620 of square root of 2 into 2 and t to the 1/2 252 00:17:44,620 --> 00:17:47,480 leaves me with-- yes, I'm left with that. 253 00:17:47,480 --> 00:17:48,070 So yes. 254 00:17:53,830 --> 00:17:59,530 Jump condition satisfied by this conjectured solution. 255 00:17:59,530 --> 00:18:00,570 What else? 256 00:18:00,570 --> 00:18:07,100 Well, just having a shock with the right speed 257 00:18:07,100 --> 00:18:10,630 doesn't mean I've got the right answer. 258 00:18:10,630 --> 00:18:12,270 Doesn't mean I've got the right answer, 259 00:18:12,270 --> 00:18:17,790 can I just make that point because it's easy to slip by. 260 00:18:17,790 --> 00:18:25,740 Let me take an extreme case where I compare. 261 00:18:25,740 --> 00:18:29,460 So these comments are going to be 262 00:18:29,460 --> 00:18:35,270 to say that I really need to check the entropy condition. 263 00:18:38,360 --> 00:18:41,190 Because I could have two problems. 264 00:18:41,190 --> 00:18:47,800 One which started from initial value minus 1 jumping to 1. 265 00:18:47,800 --> 00:18:49,430 So that's a Riemann problem. 266 00:18:49,430 --> 00:18:54,080 And the other which starts from 1 and jumps to minus 1. 267 00:18:54,080 --> 00:18:59,090 And I guess in both cases, remember 268 00:18:59,090 --> 00:19:03,010 that f of u, which is u squared over 2, 269 00:19:03,010 --> 00:19:11,160 is 1/2 on both sides of that line x equals zero. 270 00:19:18,390 --> 00:19:20,250 In other words, could this solution 271 00:19:20,250 --> 00:19:26,030 stay at minus 1 and this solution stay at plus 1 272 00:19:26,030 --> 00:19:28,740 and have a jump between them? 273 00:19:28,740 --> 00:19:30,600 And same for this. 274 00:19:30,600 --> 00:19:33,870 This solution could stay at 1, this could stay at minus 1 275 00:19:33,870 --> 00:19:35,380 and have a jump. 276 00:19:35,380 --> 00:19:37,940 And now of course, in this case the jump 277 00:19:37,940 --> 00:19:41,060 will be upwards from minus 1 to 1. 278 00:19:41,060 --> 00:19:45,180 In this case it'll be a drop from 1 down to minus 1. 279 00:19:45,180 --> 00:19:53,740 And I guess I think this one is OK with the entropy condition 280 00:19:53,740 --> 00:19:55,190 and this one is not OK. 281 00:19:59,400 --> 00:20:02,310 Let me say again what I'm meaning. 282 00:20:02,310 --> 00:20:06,280 The fact that I've chosen this simple example 283 00:20:06,280 --> 00:20:10,420 with f of u equal u squared so that 1 squared 284 00:20:10,420 --> 00:20:14,010 and minus 1 squared give me the same answer. 285 00:20:14,010 --> 00:20:18,570 So the jump across this line is zero, the speed of the shock 286 00:20:18,570 --> 00:20:21,880 is zero, the jump condition is OK. 287 00:20:21,880 --> 00:20:30,440 So I'm saying that the jump condition is OK for both. 288 00:20:33,670 --> 00:20:36,540 And it's the right thing to have the jump in this one, 289 00:20:36,540 --> 00:20:37,790 but this is wrong. 290 00:20:37,790 --> 00:20:41,740 There should be a fan and I guess I've 291 00:20:41,740 --> 00:20:43,480 forgotten how the fan will go. 292 00:20:49,430 --> 00:20:54,230 We do have 1 here and we do have minus 1 there, 293 00:20:54,230 --> 00:20:59,180 but the value of u is connected not by a jump at a point, 294 00:20:59,180 --> 00:21:03,340 but by a fan that gets us from there to there. 295 00:21:03,340 --> 00:21:08,360 A fan is also known as an expansion wave. 296 00:21:08,360 --> 00:21:12,660 That's the right thing to have for that problem. 297 00:21:12,660 --> 00:21:16,470 All that is just to say that I do 298 00:21:16,470 --> 00:21:18,990 have to check the entropy condition. 299 00:21:18,990 --> 00:21:22,350 But I hope it's not going to be too difficult. 300 00:21:22,350 --> 00:21:27,960 So my entropy condition-- remember, for this special case 301 00:21:27,960 --> 00:21:29,550 f prime is just u. 302 00:21:29,550 --> 00:21:31,560 So I'm really just checking, now, 303 00:21:31,560 --> 00:21:35,300 u_L greater than the shock speed, greater than u_R. 304 00:21:39,650 --> 00:21:42,020 u_R is zero of course. 305 00:21:42,020 --> 00:21:43,830 On this side of the shock it's zero. 306 00:21:46,720 --> 00:21:49,230 So the shock-- can you get that? 307 00:21:55,250 --> 00:21:58,490 We're going to be OK on the entropy condition. 308 00:21:58,490 --> 00:22:02,740 The shock speed is greater than zero and it's smaller than u_L. 309 00:22:02,740 --> 00:22:05,010 Well, how is that? 310 00:22:05,010 --> 00:22:09,110 Is this quantity, the shock speed-- 311 00:22:09,110 --> 00:22:12,860 no, greater than-- is this shock speed-- 312 00:22:12,860 --> 00:22:15,550 so checking that entropy condition means-- shall 313 00:22:15,550 --> 00:22:17,760 I put it up here? 314 00:22:17,760 --> 00:22:26,710 u_L bigger than-- so this was the characteristic speeds, 315 00:22:26,710 --> 00:22:29,530 the wave speeds on the left of the shock. 316 00:22:29,530 --> 00:22:31,610 They must run into the shock, so they must 317 00:22:31,610 --> 00:22:33,630 be going faster than the shock. 318 00:22:33,630 --> 00:22:37,130 OK, so let me just do this comparison. u_L 319 00:22:37,130 --> 00:22:42,720 at the left of the shock, on the left of the shock x is capital 320 00:22:42,720 --> 00:22:46,990 X, which we decided was root 2t. 321 00:22:46,990 --> 00:22:54,300 So this is root 2t over t. 322 00:22:54,300 --> 00:22:59,830 That's u_left because the shock starts 323 00:22:59,830 --> 00:23:05,690 when x is equal to root 2t and that's what u equals. 324 00:23:05,690 --> 00:23:11,510 And the shock speed we figured out here is 1 over. 325 00:23:11,510 --> 00:23:14,100 Now, so the question is, is this true? 326 00:23:14,100 --> 00:23:15,440 1 over root 2t. 327 00:23:18,370 --> 00:23:19,860 I guess I sure hope that it is. 328 00:23:22,980 --> 00:23:24,350 And it is, of course. 329 00:23:24,350 --> 00:23:29,910 If I multiply up by this, I have 2t over t, which is just 2 330 00:23:29,910 --> 00:23:32,050 is bigger than 1. 331 00:23:32,050 --> 00:23:35,470 So the answer again, is yes. 332 00:23:38,720 --> 00:23:46,870 So that completes checking that our proposed solution of fan 333 00:23:46,870 --> 00:23:50,780 then shock is OK. 334 00:23:50,780 --> 00:23:58,460 But, now let me come to this formula 335 00:23:58,460 --> 00:24:01,810 because it's quite a handy, useful formula. 336 00:24:01,810 --> 00:24:09,010 And what I have here is f of u divided by-- 337 00:24:09,010 --> 00:24:12,850 or f of x minus y divided by 2. 338 00:24:12,850 --> 00:24:16,520 That something squared over 2 divided by t. 339 00:24:16,520 --> 00:24:18,350 Sorry, divided by t. 340 00:24:18,350 --> 00:24:22,710 Well, what do I want to say about that going on? 341 00:24:26,950 --> 00:24:35,600 Well, first of all, if I do plug in the delta function for u_0, 342 00:24:35,600 --> 00:24:37,990 it gives the right answer. 343 00:24:37,990 --> 00:24:39,890 I can easily do this minimization. 344 00:24:39,890 --> 00:24:43,290 What what do I get when I plug in the delta function for u_0, 345 00:24:43,290 --> 00:24:47,830 just to start on trying this formula 346 00:24:47,830 --> 00:24:51,780 to be sure it gives the same answer that we just found? 347 00:24:51,780 --> 00:24:57,290 So the integral of u of x and zero, the integral of u 348 00:24:57,290 --> 00:24:59,580 is a step function. 349 00:24:59,580 --> 00:25:04,940 So this will be a step function of 1 for y 350 00:25:04,940 --> 00:25:08,140 greater than zero and zero for y less than zero. 351 00:25:11,020 --> 00:25:13,370 Minimization requires a little patience 352 00:25:13,370 --> 00:25:18,780 because a step function of y is coming in here, y is there. 353 00:25:18,780 --> 00:25:22,930 And the only way I would know to do that minimum 354 00:25:22,930 --> 00:25:25,940 would be the separate the different cases 355 00:25:25,940 --> 00:25:28,220 and do each one. 356 00:25:28,220 --> 00:25:29,890 Each one would be simple. 357 00:25:33,020 --> 00:25:35,560 I don't think we need to take our time to do that. 358 00:25:35,560 --> 00:25:37,000 It's going to give the same answer 359 00:25:37,000 --> 00:25:40,800 and the minimization is actually carried out 360 00:25:40,800 --> 00:25:42,880 in the applied math book. 361 00:25:45,410 --> 00:25:49,210 But I would like to say where it came from. 362 00:25:49,210 --> 00:25:53,400 Where did this formula-- so I'll go back to this formula. 363 00:25:53,400 --> 00:26:06,830 As a general formula for any starting value, 364 00:26:06,830 --> 00:26:09,630 starting function. 365 00:26:09,630 --> 00:26:14,690 And in some way derive the formula. 366 00:26:14,690 --> 00:26:19,500 Figure out where did that come from? 367 00:26:19,500 --> 00:26:24,660 It came from the Burgers' equation 368 00:26:24,660 --> 00:26:29,430 with-- so now I want to speak about this guy-- 369 00:26:29,430 --> 00:26:34,410 the Burgers' equation with viscosity. 370 00:26:34,410 --> 00:26:37,220 It turns out that that equation-- and this 371 00:26:37,220 --> 00:26:39,580 is very important. 372 00:26:39,580 --> 00:26:42,880 That equation has an exact solution, 373 00:26:42,880 --> 00:26:47,260 because I can change variables and make it linear. 374 00:26:47,260 --> 00:26:49,870 I can turn it into the heat equation. 375 00:26:49,870 --> 00:26:53,230 So that was a neat idea, which two people noticed 376 00:26:53,230 --> 00:26:55,960 at almost the same time. 377 00:26:55,960 --> 00:26:59,960 That very satisfying change of variables 378 00:26:59,960 --> 00:27:02,770 will turn this, will get rid of this nonlinearity 379 00:27:02,770 --> 00:27:05,740 and turn it into the heat equation. 380 00:27:05,740 --> 00:27:08,130 So it's a nonlinear change of variables of course, 381 00:27:08,130 --> 00:27:11,500 but not that hard to discover once you begin 382 00:27:11,500 --> 00:27:13,360 to hope that there is one. 383 00:27:13,360 --> 00:27:19,210 And then, once it's a heat equation we know the solution. 384 00:27:19,210 --> 00:27:22,050 The linear heat equation we know how to solve. 385 00:27:22,050 --> 00:27:25,790 So we solve it and then what? 386 00:27:25,790 --> 00:27:31,300 Let the viscosity nu go to zero. 387 00:27:31,300 --> 00:27:37,250 Watch what this solution is doing as nu goes to zero 388 00:27:37,250 --> 00:27:42,660 and it will-- of course, it'll be 389 00:27:42,660 --> 00:27:47,520 smooth for any positive value of new. 390 00:27:47,520 --> 00:27:50,950 The viscosity keeps things smooth. 391 00:27:50,950 --> 00:27:53,870 The shock appears in the limit. 392 00:27:53,870 --> 00:27:57,000 Maybe I should say something just physically about waves, 393 00:27:57,000 --> 00:27:58,040 breaking waves. 394 00:28:03,760 --> 00:28:07,820 If you look at waves in water, they're 395 00:28:07,820 --> 00:28:11,660 obeying a nonlinear wave equation, more complicated 396 00:28:11,660 --> 00:28:17,630 than ours, but nevertheless the same features. 397 00:28:17,630 --> 00:28:26,080 And a wave forms and as all water skiers know, 398 00:28:26,080 --> 00:28:28,770 surfers know-- surfers, I guess, know it best. 399 00:28:28,770 --> 00:28:35,090 The wave in time begins to break. 400 00:28:35,090 --> 00:28:40,150 Now, up to that time the viscosity 401 00:28:40,150 --> 00:28:43,110 is actually not so important. 402 00:28:43,110 --> 00:28:44,960 But as it begins to break there's 403 00:28:44,960 --> 00:28:52,540 a terrific u_xx now at the moments before breaking. 404 00:28:55,630 --> 00:29:01,020 And that u_xx then, that term produces viscosity. 405 00:29:01,020 --> 00:29:03,790 I mean, there is a viscosity there 406 00:29:03,790 --> 00:29:06,120 and suddenly it's important. 407 00:29:06,120 --> 00:29:11,470 And so the viscosity will prevent-- even though it's 408 00:29:11,470 --> 00:29:13,850 small, the viscosity will prevent 409 00:29:13,850 --> 00:29:18,140 the actual, complete break. 410 00:29:18,140 --> 00:29:22,050 I mean, what actually happens in that wave is, of course, 411 00:29:22,050 --> 00:29:24,360 a very difficult thing. 412 00:29:24,360 --> 00:29:27,100 The nonlinearity, turbulence-- everything's 413 00:29:27,100 --> 00:29:28,990 happening suddenly. 414 00:29:28,990 --> 00:29:36,550 Not easy, but the main point is that as in many problems, 415 00:29:36,550 --> 00:29:43,600 a term that's small in the case of smooth solutions, where u_xx 416 00:29:43,600 --> 00:29:49,480 has a normal size and nu is very small, so that term is small. 417 00:29:49,480 --> 00:29:53,180 Suddenly, u_xx has a giant size and that term 418 00:29:53,180 --> 00:29:56,890 is important, physically important. 419 00:29:56,890 --> 00:30:00,510 So Burgers' inviscid equation, this 420 00:30:00,510 --> 00:30:04,100 is the case with-- inviscid means nu equals zero. 421 00:30:04,100 --> 00:30:07,030 So Burgers' equation is dropping that term 422 00:30:07,030 --> 00:30:10,570 and it does have a shock, as we very well know. 423 00:30:15,100 --> 00:30:18,300 What are the steps that these two people 424 00:30:18,300 --> 00:30:21,670 noticed that take Burgers' equation 425 00:30:21,670 --> 00:30:23,680 and turn it into the heat equation? 426 00:30:27,300 --> 00:30:33,470 Let me see if I can remember the two steps. 427 00:30:33,470 --> 00:30:35,260 So I'm starting with Burgers' equation 428 00:30:35,260 --> 00:30:37,570 and I want to make it the heat equation. 429 00:30:37,570 --> 00:30:45,010 OK, so a first step is to introduce the integral of u. 430 00:30:45,010 --> 00:30:47,250 Let me call that capital U of x. 431 00:30:47,250 --> 00:30:50,540 So capital U of x will be the integral. 432 00:30:50,540 --> 00:31:02,130 So if I write it this way, I'm going to change u to capital U. 433 00:31:02,130 --> 00:31:04,870 So I'm introducing a new variable capital 434 00:31:04,870 --> 00:31:07,680 U. The linearity won't disappear yet, 435 00:31:07,680 --> 00:31:09,670 but it'll look a little different. 436 00:31:09,670 --> 00:31:19,100 So if I took this equation and I look at its-- well, 437 00:31:19,100 --> 00:31:20,730 let me just say, what's the equation 438 00:31:20,730 --> 00:31:23,050 going to look like for capital U? 439 00:31:23,050 --> 00:31:25,590 If I write it down, then you'll see that it's right. 440 00:31:25,590 --> 00:31:36,930 dU/dt plus-- instead of d by dx it'll be 1/2 dU/dx squared 441 00:31:36,930 --> 00:31:40,450 equals nu*U__xx. 442 00:31:43,420 --> 00:31:52,800 I claim that that's Burgers' equation written for capital 443 00:31:52,800 --> 00:31:56,850 U. Do you see that? 444 00:31:56,850 --> 00:31:59,160 Let me just see, well, how do I get from there 445 00:31:59,160 --> 00:32:02,870 to the equation with a star, Burgers' equation? 446 00:32:02,870 --> 00:32:05,290 I think I just take the x derivative. 447 00:32:05,290 --> 00:32:08,030 If I take the x derivative of every term, 448 00:32:08,030 --> 00:32:12,790 I'm taking the x derivative of U, so that's a little u. 449 00:32:12,790 --> 00:32:16,380 So I have d little u dt, as I want. 450 00:32:16,380 --> 00:32:19,620 If I'm taking the x derivative of this term 451 00:32:19,620 --> 00:32:22,870 I have little u_xx, as I want. 452 00:32:22,870 --> 00:32:26,960 And if I take the x derivative of this term 453 00:32:26,960 --> 00:32:29,390 it'll produce that. 454 00:32:29,390 --> 00:32:31,760 The x derivative, the 2 will come here, 455 00:32:31,760 --> 00:32:34,180 dU/dx will be the little u. 456 00:32:34,180 --> 00:32:36,130 The x derivative of that will be the u_x. 457 00:32:38,840 --> 00:32:41,370 It's straightforward. 458 00:32:41,370 --> 00:32:46,800 And so now I've got it in a slightly different, but still 459 00:32:46,800 --> 00:32:47,810 nonlinear form. 460 00:32:47,810 --> 00:32:48,630 Still quadratic. 461 00:32:52,230 --> 00:32:54,520 Now comes the nonlinear change of variables. 462 00:32:54,520 --> 00:32:59,720 I think it's u is minus 2*nu log-- 463 00:32:59,720 --> 00:33:03,930 so that's the nonlinear part-- of the new unknown that I 464 00:33:03,930 --> 00:33:08,800 finally want, which is w. 465 00:33:08,800 --> 00:33:12,480 So that's an exponential change of variable. 466 00:33:12,480 --> 00:33:19,180 Capital U is matching log w, so w is matching e to the minus 467 00:33:19,180 --> 00:33:21,150 an exponential of u. 468 00:33:21,150 --> 00:33:25,750 And I won't go through the steps, 469 00:33:25,750 --> 00:33:28,810 but when you introduce that in this equation, 470 00:33:28,810 --> 00:33:31,330 it turns into the heat equation. 471 00:33:35,900 --> 00:33:39,670 Of course, the initial function is different. 472 00:33:39,670 --> 00:33:46,290 The initial w is different from the initial u. 473 00:33:46,290 --> 00:33:51,270 Well, in fact we could see, at time zero-- 474 00:33:51,270 --> 00:33:53,500 so this was at all times. 475 00:33:53,500 --> 00:33:55,900 I didn't write the t explicitly. 476 00:33:55,900 --> 00:33:59,040 But at time zero, what is w of zero? 477 00:33:59,040 --> 00:34:00,970 I guess I just take the exponential. 478 00:34:00,970 --> 00:34:06,140 I put this over here and take its exponential. 479 00:34:06,140 --> 00:34:15,240 So w of x and zero will be u of x and zero, 480 00:34:15,240 --> 00:34:24,620 the starting value of u, divided by minus 2*nu, all exponential. 481 00:34:24,620 --> 00:34:28,810 This is the sort of expression, e to the something divided 482 00:34:28,810 --> 00:34:32,650 by-- with a minus sign-- divided by 2*nu, 483 00:34:32,650 --> 00:34:34,230 that we're going to meet. 484 00:34:34,230 --> 00:34:36,000 We meet it in the initial condition 485 00:34:36,000 --> 00:34:37,460 and we meet it in the solution. 486 00:34:40,150 --> 00:34:42,270 Well, actually we now know the solution. 487 00:34:42,270 --> 00:34:47,280 If we remember, what's the solution to the heat equation? 488 00:34:47,280 --> 00:34:48,570 So we have the heat equation. 489 00:34:51,260 --> 00:34:53,590 There's a constant in the heat equation, 490 00:34:53,590 --> 00:34:55,970 so we just have to remember, where does that constant go, 491 00:34:55,970 --> 00:34:59,060 and that's actually worth doing. 492 00:34:59,060 --> 00:35:03,060 And then we have to start it from this initial condition. 493 00:35:03,060 --> 00:35:05,590 Let me just write, over here, the solution to the heat 494 00:35:05,590 --> 00:35:06,090 equation. 495 00:35:11,770 --> 00:35:16,390 So this board will follow that board 496 00:35:16,390 --> 00:35:19,370 and remember-- so do you remember the solution 497 00:35:19,370 --> 00:35:20,410 to the heat equation? 498 00:35:24,360 --> 00:35:26,480 You remember, we know the solution that 499 00:35:26,480 --> 00:35:31,000 starts from a delta function, that fundamental solution 500 00:35:31,000 --> 00:35:39,050 is that e to the minus whatever over 2t, is it? 501 00:35:39,050 --> 00:35:39,970 Yeah. 502 00:35:39,970 --> 00:35:45,040 So what is w of x and t? 503 00:35:45,040 --> 00:35:46,400 Let's see. 504 00:35:46,400 --> 00:35:49,130 So we have the heat equation. 505 00:35:49,130 --> 00:35:54,120 There is a square root of 4*pi and now it's 4*pi*nu*t. 506 00:35:57,650 --> 00:35:59,930 When nu was 1 we didn't know that, 507 00:35:59,930 --> 00:36:05,620 but now, essentially this nu, I can change the time variable 508 00:36:05,620 --> 00:36:07,500 to nu times t. 509 00:36:07,500 --> 00:36:12,050 So everywhere I saw a t, I now will see nu times t. 510 00:36:12,050 --> 00:36:15,090 And then, do you remember what went in here? 511 00:36:15,090 --> 00:36:20,950 The initial value, w of x and zero, 512 00:36:20,950 --> 00:36:27,200 times the fundamental solution, that bell curve 513 00:36:27,200 --> 00:36:35,280 that is centered at x, and grows-- 514 00:36:35,280 --> 00:36:38,570 so it's e to the minus-- do you remember what it looked like? 515 00:36:38,570 --> 00:36:41,280 I think there's an e to the minus-- centered 516 00:36:41,280 --> 00:36:48,760 at x, so it's x minus t squared over-- was it 2? 517 00:36:48,760 --> 00:36:52,840 Over 2t, it was, and now there's a nu*t. 518 00:36:58,360 --> 00:37:02,050 x minus-- oh, wait a minute. 519 00:37:02,050 --> 00:37:04,240 I can't have x's. 520 00:37:04,240 --> 00:37:09,900 Let me take w of s and zero, e to the x minus s squared. 521 00:37:12,760 --> 00:37:15,490 This is from minus infinity to infinity. 522 00:37:15,490 --> 00:37:19,850 And that, finally, is going to give me w of x and t. 523 00:37:24,980 --> 00:37:26,750 So I integrate ds. 524 00:37:30,360 --> 00:37:34,850 It's actually good to go back to and see something 525 00:37:34,850 --> 00:37:38,000 that we derived before but we haven't 526 00:37:38,000 --> 00:37:40,760 thought about it for a while. 527 00:37:45,760 --> 00:37:50,360 This is the fundamental solution starting at the point s. 528 00:37:50,360 --> 00:37:54,960 This is the amount of initial function 529 00:37:54,960 --> 00:37:56,930 there is at the point s. 530 00:37:56,930 --> 00:38:06,190 And I integrate over all those little starts of point sources. 531 00:38:06,190 --> 00:38:10,000 I integrate over all the sources, all the way from s 532 00:38:10,000 --> 00:38:11,740 equal minus infinity to infinity, 533 00:38:11,740 --> 00:38:12,990 and I get that answer. 534 00:38:12,990 --> 00:38:15,910 And now I know what w of s and zero is. 535 00:38:15,910 --> 00:38:17,570 That's what I just found. 536 00:38:17,570 --> 00:38:24,660 This is e to the minus, as we've just said, 537 00:38:24,660 --> 00:38:32,460 capital U of x and zero-- U_0 I'll call it-- over 2*nu. 538 00:38:32,460 --> 00:38:35,550 So for this, substitute this. 539 00:38:42,790 --> 00:38:48,690 It's requiring patience, but not genius I'd say, to do this. 540 00:38:48,690 --> 00:38:53,350 Now the question is, what happens as nu goes to zero 541 00:38:53,350 --> 00:38:57,910 and that the thing that makes this worth presenting. 542 00:38:57,910 --> 00:39:01,680 What happens to integrals like this? 543 00:39:01,680 --> 00:39:06,700 I have an integral of something and the exponentials 544 00:39:06,700 --> 00:39:07,890 are both negative. 545 00:39:07,890 --> 00:39:10,060 I've somehow an integral like this. 546 00:39:10,060 --> 00:39:14,130 I have the integral of some e to the minus something-- 547 00:39:14,130 --> 00:39:21,070 I'll say B-- over nu ds. 548 00:39:21,070 --> 00:39:27,480 Do you see that the B involves the U_0 over 2? 549 00:39:27,480 --> 00:39:32,160 Well, let me put a 2*nu. 550 00:39:32,160 --> 00:39:38,650 This is really what I have in this formula. 551 00:39:38,650 --> 00:39:42,570 Well, divided by this thing, so let me put that 1 over square 552 00:39:42,570 --> 00:39:45,630 root outside. 553 00:39:45,630 --> 00:39:47,780 And remember, my question is, what 554 00:39:47,780 --> 00:39:49,850 happens as nu goes to zero? 555 00:39:49,850 --> 00:39:53,180 Do you have any idea-- I mean, it's a mess. 556 00:39:53,180 --> 00:39:55,340 We do not want to do this integration. 557 00:39:59,890 --> 00:40:03,750 It's an exact formula, but not one I want to deal with. 558 00:40:03,750 --> 00:40:09,450 What I do want is to know what happens as nu goes to zero. 559 00:40:09,450 --> 00:40:13,730 And the main point is that nu is showing up in the denominator. 560 00:40:13,730 --> 00:40:15,820 We have some quantity in the numerator, 561 00:40:15,820 --> 00:40:17,380 what is that quantity? 562 00:40:17,380 --> 00:40:28,400 That quantity is U_0 plus this x minus x squared over 2*nu*t-- 563 00:40:28,400 --> 00:40:35,220 over t, because-- in fact, if we're lucky, 564 00:40:35,220 --> 00:40:39,270 this B is exactly this quantity. 565 00:40:39,270 --> 00:40:43,940 This is exactly this bracket. 566 00:40:43,940 --> 00:40:50,490 That's, I think, my B, if I change letter from s to y. 567 00:40:50,490 --> 00:40:53,780 If I'd been like prepared in advance, 568 00:40:53,780 --> 00:40:56,090 I wouldn't have introduced y and s 569 00:40:56,090 --> 00:41:01,140 because they're both just dummy variables 570 00:41:01,140 --> 00:41:03,130 and they should have been the same. 571 00:41:03,130 --> 00:41:07,960 You see that this guy-- because, you 572 00:41:07,960 --> 00:41:12,450 remember the change between little u and capital U? 573 00:41:12,450 --> 00:41:15,200 Capital U is just the integral. 574 00:41:15,200 --> 00:41:17,390 So that's U_0. 575 00:41:17,390 --> 00:41:24,200 And this is this x minus y squared over 2t. 576 00:41:28,600 --> 00:41:32,010 I'm not sure if I've got a 2 somewhere lost, 577 00:41:32,010 --> 00:41:36,000 but we won't worry about that. 578 00:41:36,000 --> 00:41:38,060 The main point that I want to make 579 00:41:38,060 --> 00:41:40,580 is, how do you deal with an integral like this? 580 00:41:40,580 --> 00:41:46,280 Because this is like a major topic in a more advanced course 581 00:41:46,280 --> 00:41:52,800 on applied math, a more advanced course on the analysis of 582 00:41:52,800 --> 00:41:56,080 applied math, not the numerical analysis. 583 00:41:58,850 --> 00:42:07,130 What's happening as nu goes to zero= Well, as nu goes to zero, 584 00:42:07,130 --> 00:42:09,300 this thing is getting very large. 585 00:42:11,870 --> 00:42:15,270 e to the minus a very large amount. 586 00:42:15,270 --> 00:42:20,320 So it's as near nothing as you can imagine. 587 00:42:20,320 --> 00:42:22,470 e to the minus a very negative amount. 588 00:42:22,470 --> 00:42:25,550 But still, the total is considered somehow, 589 00:42:25,550 --> 00:42:29,920 and the question is, what's the main leading term 590 00:42:29,920 --> 00:42:34,500 in this quantity, and the answer is, you look at the point 591 00:42:34,500 --> 00:42:40,400 where B is a minimum. 592 00:42:40,400 --> 00:42:44,311 If B at one point is 1, than we have an e to the minus 1 over 593 00:42:44,311 --> 00:42:44,810 2*nu. 594 00:42:48,280 --> 00:42:52,720 And compared to e to the minus 100 over 2*nu, 595 00:42:52,720 --> 00:42:55,980 the e to the minus 1 over 2*nu is way bigger. 596 00:42:55,980 --> 00:42:59,170 So that in the end, the contribution all 597 00:42:59,170 --> 00:43:03,740 comes from the point where B is a minimum. 598 00:43:03,740 --> 00:43:07,460 And that's what this formula says we should look for. 599 00:43:07,460 --> 00:43:09,910 And if you figure out what that contribution is, 600 00:43:09,910 --> 00:43:15,440 and taking into account the fact that we do have a 4*pi*nu*t 601 00:43:15,440 --> 00:43:20,140 here-- without this, the whole thing would just be going 602 00:43:20,140 --> 00:43:25,000 to zero as nu goes to zero; this e to the minus very negative 603 00:43:25,000 --> 00:43:25,610 stuff is zero. 604 00:43:25,610 --> 00:43:33,000 But we have a nu here, so in the end something nonzero emerges, 605 00:43:33,000 --> 00:43:37,810 but it emerges totally at this point where B is a minimum 606 00:43:37,810 --> 00:43:42,090 and that's what this formula has found. 607 00:43:42,090 --> 00:43:45,300 This is the minimum value of b. 608 00:43:45,300 --> 00:43:49,530 OK, I won't track down further. 609 00:43:49,530 --> 00:43:51,850 Do I remember the right word? 610 00:43:51,850 --> 00:43:55,260 Is it stationary phase? 611 00:43:55,260 --> 00:43:57,870 Did anybody meet these words? 612 00:43:57,870 --> 00:44:05,280 I'm speaking now about something in the history of applied math. 613 00:44:05,280 --> 00:44:08,500 Still, those tools that were created 614 00:44:08,500 --> 00:44:12,460 over the pre-computer years are still 615 00:44:12,460 --> 00:44:16,890 extremely valuable to understand what's going on 616 00:44:16,890 --> 00:44:19,840 and this is just an illustration. 617 00:44:19,840 --> 00:44:31,800 OK, that's my comments on the limiting case of Burgers', 618 00:44:31,800 --> 00:44:36,590 which is inviscid Burgers', and we actually 619 00:44:36,590 --> 00:44:43,440 used this form of the equation as the first step 620 00:44:43,440 --> 00:44:44,640 in linearization. 621 00:44:50,680 --> 00:44:54,760 As I say, it's quite a nice exercise, now, 622 00:44:54,760 --> 00:44:58,050 to use this formula. 623 00:44:58,050 --> 00:45:02,870 An explicit formula, which we, last time, did not suspect. 624 00:45:02,870 --> 00:45:06,640 We were tracking down-- we knew the problem was nonlinear 625 00:45:06,640 --> 00:45:11,240 and were tracking down shocks and fans. 626 00:45:11,240 --> 00:45:15,310 Here's a formula that just plain gives us the answer. 627 00:45:15,310 --> 00:45:19,710 So this gives us a shock or a fan, whichever. 628 00:45:19,710 --> 00:45:26,010 This has the right shock speeds if there's a shock 629 00:45:26,010 --> 00:45:29,390 and it has the right-- the entropy condition is satisfied 630 00:45:29,390 --> 00:45:34,330 because it was satisfied all along as nu was going to zero. 631 00:45:34,330 --> 00:45:38,060 OK, so now do I want to make some final comments 632 00:45:38,060 --> 00:45:42,390 on this Korteweg-de Vries-- KdV-- equation. 633 00:45:52,720 --> 00:45:58,020 I mean, that would be an ideal numerical example. 634 00:45:58,020 --> 00:46:03,260 In fact, its importance was discovered numerically. 635 00:46:03,260 --> 00:46:05,510 One could say that this is one of the biggest 636 00:46:05,510 --> 00:46:13,810 contributions of numerical simulation to the theory 637 00:46:13,810 --> 00:46:16,020 of nonlinear PDEs. 638 00:46:16,020 --> 00:46:23,280 The discovery-- and it just came out of the blue-- the discovery 639 00:46:23,280 --> 00:46:28,360 that that particular nonlinear equation was 640 00:46:28,360 --> 00:46:31,320 what we might call integrable. 641 00:46:31,320 --> 00:46:38,220 That there is some twist that produce 642 00:46:38,220 --> 00:46:42,860 linear equations whose solution give the answer 643 00:46:42,860 --> 00:46:45,770 to this nonlinear equation. 644 00:46:45,770 --> 00:46:48,160 Maybe what I ought to say is, first 645 00:46:48,160 --> 00:46:51,770 to tell you what those numerical experiments were like. 646 00:46:51,770 --> 00:46:56,310 And people just like looked at the answers and said, 647 00:46:56,310 --> 00:46:59,430 can this be? 648 00:46:59,430 --> 00:47:00,990 OK, what were they like? 649 00:47:00,990 --> 00:47:04,960 So we could find-- without much work, 650 00:47:04,960 --> 00:47:15,070 we could find solutions of the form, of a traveling wave form. 651 00:47:15,070 --> 00:47:20,720 So we could find solutions that have the form some function 652 00:47:20,720 --> 00:47:24,690 of x minus c*t. 653 00:47:24,690 --> 00:47:34,320 If you plug that in, you have then derivatives of f. 654 00:47:34,320 --> 00:47:36,370 You have an ordinary differential equation for f, 655 00:47:36,370 --> 00:47:41,220 and you can actually find shapes, waves that 656 00:47:41,220 --> 00:47:44,760 travel following that equation. 657 00:47:44,760 --> 00:47:49,430 So of course, what you can't do is add two of them 658 00:47:49,430 --> 00:47:53,100 together and still have a solution because the equation 659 00:47:53,100 --> 00:47:54,430 is not linear. 660 00:47:54,430 --> 00:47:58,520 So everybody assumed, OK, people had found these traveling waves 661 00:47:58,520 --> 00:48:02,250 before and the computer will-- if you 662 00:48:02,250 --> 00:48:04,010 start with the right initial condition, 663 00:48:04,010 --> 00:48:06,290 the computer will produce a traveling wave. 664 00:48:06,290 --> 00:48:09,190 If you start with the right initial condition, 665 00:48:09,190 --> 00:48:12,550 the computer will show two traveling waves 666 00:48:12,550 --> 00:48:14,570 at different speeds. 667 00:48:14,570 --> 00:48:19,480 This c will be different from one wave to the other. 668 00:48:19,480 --> 00:48:21,590 So what will happen as time goes forward? 669 00:48:21,590 --> 00:48:27,540 One wave will catch the other one because it's going faster. 670 00:48:27,540 --> 00:48:31,100 So the shapes of each wave are known. 671 00:48:31,100 --> 00:48:34,330 I think it's 1 over hyperbolic cosine squared. 672 00:48:36,870 --> 00:48:41,660 So we have one wave going with a different c, 673 00:48:41,660 --> 00:48:44,640 going faster than another wave. 674 00:48:44,640 --> 00:48:45,800 We're computing along. 675 00:48:48,740 --> 00:48:53,070 The fast wave catches the slow wave and they interact. 676 00:48:53,070 --> 00:48:56,420 Of course they interact because the problem's nonlinear, 677 00:48:56,420 --> 00:49:00,150 and you can't see what's happening at the time 678 00:49:00,150 --> 00:49:03,090 when the big one catches the small one. 679 00:49:03,090 --> 00:49:06,480 But you can keep computing. 680 00:49:06,480 --> 00:49:10,380 And what happened was that after this jumble 681 00:49:10,380 --> 00:49:15,350 the fast wave emerged with the same shape. 682 00:49:15,350 --> 00:49:18,940 And the slow wave emerge behind it with the same shape. 683 00:49:18,940 --> 00:49:24,160 And there was some lag of course, some delay moment, 684 00:49:24,160 --> 00:49:26,830 which had been used up in the jumble 685 00:49:26,830 --> 00:49:28,510 as the one passed the other. 686 00:49:28,510 --> 00:49:31,750 But they emerged looking the same. 687 00:49:35,670 --> 00:49:37,980 I mean, that's sort of like a linear property. 688 00:49:37,980 --> 00:49:41,550 Not exactly, because you couldn't just add them, 689 00:49:41,550 --> 00:49:44,710 it wasn't perfect, but the shapes were perfect. 690 00:49:44,710 --> 00:49:51,480 And that suggested that this equation might have hidden 691 00:49:51,480 --> 00:49:55,180 linearity that nobody noticed. 692 00:49:55,180 --> 00:49:58,830 And then finally that was discovered. 693 00:49:58,830 --> 00:50:00,170 It was a search. 694 00:50:00,170 --> 00:50:03,230 People found conservation laws, it's 695 00:50:03,230 --> 00:50:05,120 easy to show that the integral of u 696 00:50:05,120 --> 00:50:07,110 is conserved by this equation. 697 00:50:07,110 --> 00:50:10,680 We've done that a thousand times. 698 00:50:10,680 --> 00:50:13,160 Then other integrals are conserved 699 00:50:13,160 --> 00:50:16,910 and if you just like to do algebra, 700 00:50:16,910 --> 00:50:20,120 a short list of conserved quantities: integral of u, 701 00:50:20,120 --> 00:50:24,910 integral of u squared and a few others, not just 702 00:50:24,910 --> 00:50:27,900 u cubed and u fourth, of course. 703 00:50:27,900 --> 00:50:29,960 People were creating a list. 704 00:50:29,960 --> 00:50:33,090 And then somebody saw how to get an infinite number 705 00:50:33,090 --> 00:50:34,420 of integrals. 706 00:50:34,420 --> 00:50:40,470 So that was a suspicion that the problem was integrable. 707 00:50:40,470 --> 00:50:42,700 That if you could find an infinite list 708 00:50:42,700 --> 00:50:47,130 of conserved quantities, conserved even 709 00:50:47,130 --> 00:50:50,570 under this nonlinear interaction term, 710 00:50:50,570 --> 00:50:52,710 that somewhere something was linear. 711 00:50:52,710 --> 00:50:56,670 And then eventually, a change of variables-- well, 712 00:50:56,670 --> 00:50:59,270 more than a change of variables, a total twist 713 00:50:59,270 --> 00:51:06,670 was found to integrate it and get an explicit answer. 714 00:51:06,670 --> 00:51:09,520 That's more than I can do, but I want to say, 715 00:51:09,520 --> 00:51:12,070 of course, that immediately set out 716 00:51:12,070 --> 00:51:18,630 a mad search for other nonlinear problems 717 00:51:18,630 --> 00:51:22,980 that were in this sense completely integrable. 718 00:51:22,980 --> 00:51:26,480 And now, am I going to be able to list some of the ones that 719 00:51:26,480 --> 00:51:27,920 were discovered? 720 00:51:27,920 --> 00:51:32,440 I think I can-- whoops-- not on that board. 721 00:51:32,440 --> 00:51:38,280 So I'll just end with the completely integrable nonlinear 722 00:51:38,280 --> 00:51:40,240 equations. 723 00:51:40,240 --> 00:51:45,860 And I don't know if the list is complete; 724 00:51:45,860 --> 00:51:50,860 the discovery of new ones is like a major thing. 725 00:51:50,860 --> 00:51:53,490 So we have Korteweg-de Vries. 726 00:51:53,490 --> 00:51:55,370 That was this the first. 727 00:51:55,370 --> 00:51:58,430 Then there was a nonlinear Schrodinger equation. 728 00:51:58,430 --> 00:52:02,450 And the solutions, I have to say what the solutions are called. 729 00:52:02,450 --> 00:52:06,480 Those waves that I mentioned, which-- 730 00:52:06,480 --> 00:52:08,480 they're not quite that bad-- are called 731 00:52:08,480 --> 00:52:13,970 solitons, solitary waves. 732 00:52:13,970 --> 00:52:16,870 And it was the interaction of solitons and the fact 733 00:52:16,870 --> 00:52:19,310 that they emerged looking unchanged 734 00:52:19,310 --> 00:52:24,600 that set off this horrific effort to figure out 735 00:52:24,600 --> 00:52:25,730 what was going on. 736 00:52:25,730 --> 00:52:29,530 So I was going to write nonlinear Schrodinger in here, 737 00:52:29,530 --> 00:52:37,100 and without saying exactly where the nonlinearity appears, 738 00:52:37,100 --> 00:52:38,830 but you take Schrodinger equation 739 00:52:38,830 --> 00:52:45,360 and-- then oh, I'm blanking out on other-- 740 00:52:45,360 --> 00:52:49,070 there are just a couple more famous ones. 741 00:52:51,960 --> 00:52:55,890 I'll put those names into an e-mail 742 00:52:55,890 --> 00:52:58,210 rather than stumble here. 743 00:52:58,210 --> 00:53:02,880 OK, so that's a little history, with no details, 744 00:53:02,880 --> 00:53:07,770 of an incredibly special class of nonlinear PDEs. 745 00:53:07,770 --> 00:53:12,320 Well of course, the Navier-Stokes equations 746 00:53:12,320 --> 00:53:14,190 aren't on that list. 747 00:53:14,190 --> 00:53:19,560 And we still don't know, especially in 3D, 748 00:53:19,560 --> 00:53:23,010 I guess that's one of the million dollar Clay prizes, 749 00:53:23,010 --> 00:53:28,690 is to know whether the Navier-Stokes equations have 750 00:53:28,690 --> 00:53:30,580 a solution for all time. 751 00:53:30,580 --> 00:53:33,490 So that's one of these seven famous prizes 752 00:53:33,490 --> 00:53:38,260 of which one of the seven, geometry analysis 753 00:53:38,260 --> 00:53:42,230 prize, apparently that problem has been solved. 754 00:53:42,230 --> 00:53:44,490 So there are six to go and one of them 755 00:53:44,490 --> 00:53:49,230 is whether the Navier-Stokes equations have solutions. 756 00:53:49,230 --> 00:53:52,300 Actually, I'll just take one more minute of your time. 757 00:53:52,300 --> 00:53:54,430 Thank you. 758 00:53:54,430 --> 00:53:58,070 People had, or somebody had conjectured 759 00:53:58,070 --> 00:54:01,160 some possible vortex blow-up in which 760 00:54:01,160 --> 00:54:07,570 maybe that was a case, that was an example of nonexistence. 761 00:54:07,570 --> 00:54:12,110 We could find a particular initial function 762 00:54:12,110 --> 00:54:16,170 that created such a blow up in a small space 763 00:54:16,170 --> 00:54:22,210 that the Navier-Stokes equations couldn't be continued. 764 00:54:22,210 --> 00:54:25,820 But big numerical experiments, just reported in the last 765 00:54:25,820 --> 00:54:30,400 weeks, show that that particular example, 766 00:54:30,400 --> 00:54:37,310 that somebody thought might prove nonexistence after 767 00:54:37,310 --> 00:54:41,210 a finite time, doesn't. 768 00:54:41,210 --> 00:54:45,500 So it's back to the drawing board, really. 769 00:54:45,500 --> 00:54:47,500 I don't see how numerically we're ever 770 00:54:47,500 --> 00:54:53,550 going to see that it does exist, but this numerical experiment 771 00:54:53,550 --> 00:54:59,130 was sufficiently clear to say that that particular phenomenon 772 00:54:59,130 --> 00:55:02,930 that looked dangerous didn't go wrong. 773 00:55:02,930 --> 00:55:06,870 OK, so next time, it's a little more 774 00:55:06,870 --> 00:55:11,440 about numerical solutions of these equations 775 00:55:11,440 --> 00:55:13,660 and then about the level set method. 776 00:55:13,660 --> 00:55:20,150 And then, looking ahead, we have a guest lecture 777 00:55:20,150 --> 00:55:23,830 on financial mathematics, the PDEs of financial math, 778 00:55:23,830 --> 00:55:28,570 for everybody who wants to go to New York and make a billion. 779 00:55:28,570 --> 00:55:33,940 And then we're going to very quickly 780 00:55:33,940 --> 00:55:37,410 be in the second part of the course, 781 00:55:37,410 --> 00:55:41,750 on solving large linear systems. 782 00:55:41,750 --> 00:55:43,920 OK, I'll see you Friday then for level sets. 783 00:55:43,920 --> 00:55:44,420 Good. 784 00:55:44,420 --> 00:55:45,680 Thanks for your patience. 785 00:55:54,970 --> 00:55:55,770 Hi. 786 00:55:55,770 --> 00:55:56,690 AUDIENCE: [INAUDIBLE] 787 00:55:56,690 --> 00:55:58,500 PROFESSOR: Of course I have.