WEBVTT 00:00:00.000 --> 00:00:01.980 align:middle line:90% [SQUEAKING] 00:00:01.980 --> 00:00:03.465 align:middle line:90% [RUSTLING] 00:00:03.465 --> 00:00:06.435 align:middle line:90% [CLICKING] 00:00:06.435 --> 00:00:13.880 align:middle line:90% 00:00:13.880 --> 00:00:18.090 align:middle line:84% TOBIAS COLDING: Last time we talked about the Picard-Lindelöf 00:00:18.090 --> 00:00:18.590 align:middle line:90% theorem. 00:00:18.590 --> 00:00:25.520 align:middle line:90% 00:00:25.520 --> 00:00:35.090 align:middle line:84% And so this theorem is about solutions to ODEs. 00:00:35.090 --> 00:00:38.000 align:middle line:90% 00:00:38.000 --> 00:00:42.160 align:middle line:84% And so, specifically, it's asking that-- so you have 00:00:42.160 --> 00:00:47.270 align:middle line:90% a function f and a function g. 00:00:47.270 --> 00:00:50.680 align:middle line:90% 00:00:50.680 --> 00:00:56.020 align:middle line:90% And f here is a differentiable. 00:00:56.020 --> 00:01:00.730 align:middle line:90% 00:01:00.730 --> 00:01:06.710 align:middle line:84% And the derivative of f, as well as g, are continuous. 00:01:06.710 --> 00:01:12.250 align:middle line:90% 00:01:12.250 --> 00:01:17.790 align:middle line:84% And then the problem is that you're looking for solutions. 00:01:17.790 --> 00:01:19.550 align:middle line:90% So you want a solution. 00:01:19.550 --> 00:01:31.580 align:middle line:84% So looking for a solution y to the ODE. 00:01:31.580 --> 00:01:35.550 align:middle line:90% 00:01:35.550 --> 00:01:44.970 align:middle line:84% And the ODE is at the derivative should be equal to f of y. 00:01:44.970 --> 00:01:51.890 align:middle line:84% So at some point x, the derivative of y 00:01:51.890 --> 00:01:59.220 align:middle line:84% should be equal to f of y of x plus g of x. 00:01:59.220 --> 00:02:09.139 align:middle line:84% And you are looking for that this function, that y of 0, say, 00:02:09.139 --> 00:02:11.540 align:middle line:90% is equal to a. 00:02:11.540 --> 00:02:14.560 align:middle line:84% So this is what's called the initial condition. 00:02:14.560 --> 00:02:22.740 align:middle line:90% 00:02:22.740 --> 00:02:25.350 align:middle line:90% And this is so-- 00:02:25.350 --> 00:02:26.760 align:middle line:90% this is equal to that. 00:02:26.760 --> 00:02:29.740 align:middle line:90% 00:02:29.740 --> 00:02:32.820 align:middle line:90% And so in order to-- 00:02:32.820 --> 00:02:35.780 align:middle line:90% so we want existence-- 00:02:35.780 --> 00:02:43.000 align:middle line:84% so want existence and uniqueness. 00:02:43.000 --> 00:02:47.900 align:middle line:90% 00:02:47.900 --> 00:02:49.780 align:middle line:84% So existence just means that there 00:02:49.780 --> 00:02:52.100 align:middle line:90% is a solution for a given a. 00:02:52.100 --> 00:02:53.820 align:middle line:90% There is a solution. 00:02:53.820 --> 00:03:00.950 align:middle line:84% Uniqueness means that if you have two solutions to this ODE, 00:03:00.950 --> 00:03:04.310 align:middle line:90% then they will have to coincide. 00:03:04.310 --> 00:03:10.470 align:middle line:84% And so the way we were setting it up was that we were using-- 00:03:10.470 --> 00:03:15.120 align:middle line:84% so using the contraction mapping theorem. 00:03:15.120 --> 00:03:26.510 align:middle line:90% 00:03:26.510 --> 00:03:29.910 align:middle line:84% So the contraction mapping theorem is the following, 00:03:29.910 --> 00:03:36.870 align:middle line:84% that if you take a metric space x, and this metric space is x, 00:03:36.870 --> 00:03:54.920 align:middle line:84% d is a metric space, is a Cauchy complete metric space then 00:03:54.920 --> 00:04:10.760 align:middle line:84% a contracting map-- so a contracting map is a map g from 00:04:10.760 --> 00:04:16.600 align:middle line:84% this metric space into itself so that-- 00:04:16.600 --> 00:04:25.320 align:middle line:84% and a constant c strictly less than 1, 00:04:25.320 --> 00:04:33.040 align:middle line:84% so that if you measure the distance between the images, 00:04:33.040 --> 00:04:35.680 align:middle line:84% the images are squeezed together. 00:04:35.680 --> 00:04:40.000 align:middle line:84% So this means that the distance between the images 00:04:40.000 --> 00:04:42.120 align:middle line:84% is this constant that is strictly 00:04:42.120 --> 00:04:47.120 align:middle line:84% less than 1 times the distance between x and y. 00:04:47.120 --> 00:04:49.800 align:middle line:90% So this is a contracting map. 00:04:49.800 --> 00:04:52.330 align:middle line:84% And the contracting mapping theorem say-- 00:04:52.330 --> 00:05:03.930 align:middle line:90% 00:05:03.930 --> 00:05:08.460 align:middle line:84% so the contraction mapping theorem is the following. 00:05:08.460 --> 00:05:12.250 align:middle line:90% 00:05:12.250 --> 00:05:16.650 align:middle line:84% There exists if you have a Cauchy complete metric space 00:05:16.650 --> 00:05:19.970 align:middle line:84% and you have a contracting map like this, 00:05:19.970 --> 00:05:29.880 align:middle line:90% then T has a unique fixed point. 00:05:29.880 --> 00:05:33.570 align:middle line:90% 00:05:33.570 --> 00:05:36.032 align:middle line:84% And remember that a fixed point-- 00:05:36.032 --> 00:05:36.740 align:middle line:90% so a fixed point. 00:05:36.740 --> 00:05:40.810 align:middle line:90% 00:05:40.810 --> 00:05:47.550 align:middle line:84% This is x in this space so that T leaves it invariant. 00:05:47.550 --> 00:05:51.484 align:middle line:90% 00:05:51.484 --> 00:05:53.100 align:middle line:90% And so we proved this. 00:05:53.100 --> 00:05:57.780 align:middle line:84% We proved the contraction mapping theorem last time. 00:05:57.780 --> 00:06:03.620 align:middle line:84% Now, how were we going to use for this ODE? 00:06:03.620 --> 00:06:26.700 align:middle line:84% So we defined-- so if I here is an containing a 0, 00:06:26.700 --> 00:06:29.950 align:middle line:90% then we defined-- 00:06:29.950 --> 00:06:32.380 align:middle line:90% and u is a function. 00:06:32.380 --> 00:06:43.460 align:middle line:84% And u-- I'm sorry, y is a continuous function, 00:06:43.460 --> 00:06:46.580 align:middle line:90% then we defined g of y. 00:06:46.580 --> 00:06:54.230 align:middle line:84% So this is a continuous function on the interval on I. 00:06:54.230 --> 00:06:58.510 align:middle line:84% Then we defined another continuous function 00:06:58.510 --> 00:07:05.650 align:middle line:84% on the same interval as follows, that's the T of y. 00:07:05.650 --> 00:07:09.590 align:middle line:84% So this is going to be a continuous function. 00:07:09.590 --> 00:07:14.550 align:middle line:90% So in x, this thing here is-- 00:07:14.550 --> 00:07:15.690 align:middle line:90% this is a. 00:07:15.690 --> 00:07:18.390 align:middle line:90% 00:07:18.390 --> 00:07:22.310 align:middle line:84% Because we want it to have value at 0 a, and then 00:07:22.310 --> 00:07:39.710 align:middle line:84% we're integrating from 0 to x of f of y of s plus g of s ds. 00:07:39.710 --> 00:07:47.640 align:middle line:84% And so this-- this here gives you a map up from-- 00:07:47.640 --> 00:07:50.760 align:middle line:84% gives you a map from the continuous function 00:07:50.760 --> 00:07:53.960 align:middle line:84% on this interval into the continuous function 00:07:53.960 --> 00:07:56.560 align:middle line:90% on the interval. 00:07:56.560 --> 00:08:02.360 align:middle line:84% And it also has the property that if you take a function 00:08:02.360 --> 00:08:07.740 align:middle line:84% and you evaluate it at 0, then you get a. 00:08:07.740 --> 00:08:10.946 align:middle line:90% 00:08:10.946 --> 00:08:20.480 align:middle line:84% And now, we observe that, in fact, the image here-- 00:08:20.480 --> 00:08:24.120 align:middle line:84% the image is-- if you define-- when you define it 00:08:24.120 --> 00:08:27.500 align:middle line:84% like this, by the fundamental theorem of calculus, 00:08:27.500 --> 00:08:30.620 align:middle line:84% this function here is differentiable. 00:08:30.620 --> 00:08:36.679 align:middle line:90% 00:08:36.679 --> 00:08:38.640 align:middle line:90% So this function is-- 00:08:38.640 --> 00:08:40.919 align:middle line:90% so it actually takes-- 00:08:40.919 --> 00:08:48.895 align:middle line:84% so T actually takes the space of continuous functions 00:08:48.895 --> 00:08:51.850 align:middle line:84% into the space of functions that are not 00:08:51.850 --> 00:08:55.610 align:middle line:84% just continuous, but actually differentiable, 00:08:55.610 --> 00:08:58.930 align:middle line:84% and even the derivative is continuous. 00:08:58.930 --> 00:09:00.595 align:middle line:90% And the ones that are-- 00:09:00.595 --> 00:09:03.170 align:middle line:90% 00:09:03.170 --> 00:09:05.510 align:middle line:84% so this here is the space of functions. 00:09:05.510 --> 00:09:15.210 align:middle line:84% So this is functions that are where the functions y so that y 00:09:15.210 --> 00:09:21.670 align:middle line:90% prime exists and is continuous. 00:09:21.670 --> 00:09:32.570 align:middle line:90% 00:09:32.570 --> 00:09:35.370 align:middle line:84% And so now, what we saw last time 00:09:35.370 --> 00:09:38.190 align:middle line:84% was that by the fundamental theorem of calculus-- 00:09:38.190 --> 00:09:50.940 align:middle line:90% 00:09:50.940 --> 00:09:58.160 align:middle line:90% of calculus, the derivative-- 00:09:58.160 --> 00:10:02.080 align:middle line:90% 00:10:02.080 --> 00:10:04.980 align:middle line:84% so by the fundamental theorem of calculus, 00:10:04.980 --> 00:10:11.540 align:middle line:84% the derivative of this function, if you take y 00:10:11.540 --> 00:10:15.300 align:middle line:84% and you take T of it, this is now a function, 00:10:15.300 --> 00:10:21.620 align:middle line:84% and it's differentiable, and the derivative is f of y. 00:10:21.620 --> 00:10:28.940 align:middle line:84% So at x, it's f of y of x plus g of x. 00:10:28.940 --> 00:10:31.100 align:middle line:90% And so this means that if-- 00:10:31.100 --> 00:10:40.430 align:middle line:84% so if y is a fixed point for the map g-- 00:10:40.430 --> 00:10:43.910 align:middle line:84% so if it's a fixed point, this just means that g of y 00:10:43.910 --> 00:10:45.670 align:middle line:90% is equal to y. 00:10:45.670 --> 00:10:51.950 align:middle line:84% Well then this thing here, this is the derivative of y. 00:10:51.950 --> 00:10:56.026 align:middle line:84% So then it means that y here solves the ODE. 00:10:56.026 --> 00:11:00.870 align:middle line:90% 00:11:00.870 --> 00:11:04.650 align:middle line:84% We have already checked that the initial value-- 00:11:04.650 --> 00:11:08.870 align:middle line:90% we did that somewhere here. 00:11:08.870 --> 00:11:10.630 align:middle line:84% We already checked that the initial value 00:11:10.630 --> 00:11:12.190 align:middle line:90% is what it should be. 00:11:12.190 --> 00:11:15.670 align:middle line:84% And if it's a fixed point, then the derivative 00:11:15.670 --> 00:11:19.670 align:middle line:90% is also what it should be. 00:11:19.670 --> 00:11:22.370 align:middle line:84% So we want to find a fixed point. 00:11:22.370 --> 00:11:29.470 align:middle line:84% And so what we did last time was that we saw that if we restrict 00:11:29.470 --> 00:11:31.270 align:middle line:90% the interval-- 00:11:31.270 --> 00:11:38.220 align:middle line:84% so last time, what we proved is this, 00:11:38.220 --> 00:11:40.140 align:middle line:84% that if we restrict the interval-- 00:11:40.140 --> 00:11:43.000 align:middle line:90% so last time we showed-- 00:11:43.000 --> 00:11:43.750 align:middle line:90% so we set. 00:11:43.750 --> 00:11:46.680 align:middle line:90% 00:11:46.680 --> 00:11:54.680 align:middle line:84% M here equal to a, absolute value of a plus 2, 00:11:54.680 --> 00:12:04.840 align:middle line:84% and then we saw that there exists a delta 0 at delta 0 00:12:04.840 --> 00:12:13.000 align:middle line:84% greater than 0, such that if you restrict to the interval, 00:12:13.000 --> 00:12:16.920 align:middle line:84% if you're looking at the interval of continuous function, 00:12:16.920 --> 00:12:18.800 align:middle line:84% the interval of continuous function 00:12:18.800 --> 00:12:21.520 align:middle line:90% on this little interval-- 00:12:21.520 --> 00:12:27.600 align:middle line:84% so this is an interval around 0 and you're looking at this map, 00:12:27.600 --> 00:12:30.000 align:middle line:90% then this map here-- 00:12:30.000 --> 00:12:31.800 align:middle line:84% well, first of all, we already know 00:12:31.800 --> 00:12:35.120 align:middle line:84% that it takes this the space of continuous function 00:12:35.120 --> 00:12:37.650 align:middle line:84% into the space of continuous function. 00:12:37.650 --> 00:12:40.170 align:middle line:90% But we have something more. 00:12:40.170 --> 00:12:45.370 align:middle line:84% We have that if you're looking at the subspace of this, which 00:12:45.370 --> 00:12:55.090 align:middle line:84% is the space of functions y that are here, 00:12:55.090 --> 00:12:58.690 align:middle line:84% and you also require that the norm here 00:12:58.690 --> 00:13:05.290 align:middle line:84% is less than or equal to M, then T takes this space here 00:13:05.290 --> 00:13:06.410 align:middle line:90% into itself. 00:13:06.410 --> 00:13:09.640 align:middle line:84% This is this lemma that we proved last time. 00:13:09.640 --> 00:13:15.690 align:middle line:90% 00:13:15.690 --> 00:13:21.690 align:middle line:84% So this space here, you can think about this space here, 00:13:21.690 --> 00:13:24.930 align:middle line:84% as we did last time, as-- so then we defined-- 00:13:24.930 --> 00:13:33.840 align:middle line:84% we defined-- we defined capital R equal to-- 00:13:33.840 --> 00:13:36.660 align:middle line:90% 00:13:36.660 --> 00:13:37.240 align:middle line:90% no, actually. 00:13:37.240 --> 00:13:37.740 align:middle line:90% Sorry. 00:13:37.740 --> 00:13:43.280 align:middle line:84% We defined-- we looked at the balls just on this. 00:13:43.280 --> 00:13:50.740 align:middle line:84% So we looked at the ball of radius capital N 00:13:50.740 --> 00:13:54.740 align:middle line:90% in this around this 0 function. 00:13:54.740 --> 00:13:58.420 align:middle line:84% So this thing here is the closed ball 00:13:58.420 --> 00:14:01.300 align:middle line:84% in the space of continuous function with the metric 00:14:01.300 --> 00:14:02.940 align:middle line:90% that we've been using. 00:14:02.940 --> 00:14:07.420 align:middle line:84% And this set here we can think about, if we want, as this set. 00:14:07.420 --> 00:14:15.900 align:middle line:90% 00:14:15.900 --> 00:14:18.420 align:middle line:90% So now what we have is-- 00:14:18.420 --> 00:14:23.100 align:middle line:84% so we have-- so the setup now is that we want to find a fixed 00:14:23.100 --> 00:14:29.820 align:middle line:84% point for this map T. We have a space that we want to look 00:14:29.820 --> 00:14:33.810 align:middle line:84% at, and the space is, again, now we're going to think about-- 00:14:33.810 --> 00:14:38.630 align:middle line:84% so just to-- so this is the space we're looking at. 00:14:38.630 --> 00:14:41.230 align:middle line:84% But this is what we before have denoted. 00:14:41.230 --> 00:14:44.110 align:middle line:84% Like, the ball in this metric space, 00:14:44.110 --> 00:14:47.910 align:middle line:84% it's centered at the function that is constant 0 00:14:47.910 --> 00:14:51.270 align:middle line:90% and this is the radius. 00:14:51.270 --> 00:14:57.530 align:middle line:84% And so I'm now looking at this, this map. 00:14:57.530 --> 00:15:01.910 align:middle line:84% So we have now shown that it takes this metric ball 00:15:01.910 --> 00:15:04.310 align:middle line:90% into itself. 00:15:04.310 --> 00:15:07.070 align:middle line:84% And I want to find a fixed point. 00:15:07.070 --> 00:15:08.490 align:middle line:90% I want to find a fixed point. 00:15:08.490 --> 00:15:16.270 align:middle line:90% 00:15:16.270 --> 00:15:23.070 align:middle line:90% Fixed point of T. 00:15:23.070 --> 00:15:25.110 align:middle line:90% So I need to-- 00:15:25.110 --> 00:15:26.670 align:middle line:90% I need to show-- 00:15:26.670 --> 00:15:29.430 align:middle line:90% so we want to show-- 00:15:29.430 --> 00:15:40.350 align:middle line:84% want to show that T is a contracting map. 00:15:40.350 --> 00:15:44.880 align:middle line:90% 00:15:44.880 --> 00:15:45.840 align:middle line:90% Why? 00:15:45.840 --> 00:15:48.440 align:middle line:84% Because if it is a contracting map, 00:15:48.440 --> 00:15:51.880 align:middle line:84% then we already know that if you take a closed 00:15:51.880 --> 00:16:00.160 align:middle line:84% interval like this, and you equip it with the usual distance 00:16:00.160 --> 00:16:04.190 align:middle line:84% that we have, this here is Cauchy complete. 00:16:04.190 --> 00:16:12.740 align:middle line:90% 00:16:12.740 --> 00:16:16.280 align:middle line:84% But now if you take a Cauchy complete space, 00:16:16.280 --> 00:16:18.560 align:middle line:84% and you're taking a closed subset-- 00:16:18.560 --> 00:16:20.880 align:middle line:84% and this here is a closed subset-- 00:16:20.880 --> 00:16:24.320 align:middle line:84% then this here is also Cauchy complete. 00:16:24.320 --> 00:16:25.530 align:middle line:90% This here is also. 00:16:25.530 --> 00:16:32.930 align:middle line:90% 00:16:32.930 --> 00:16:38.490 align:middle line:84% So if we could prove that T from this space into itself 00:16:38.490 --> 00:16:40.490 align:middle line:84% is a contracting map, so it means 00:16:40.490 --> 00:16:43.770 align:middle line:84% that there exists a C strictly less than 1. 00:16:43.770 --> 00:16:46.010 align:middle line:84% Then we would have a fixed point, 00:16:46.010 --> 00:16:49.382 align:middle line:84% and we would have a solution to the array. 00:16:49.382 --> 00:16:52.810 align:middle line:84% The thing is that this is not quite going to do it. 00:16:52.810 --> 00:16:57.090 align:middle line:84% It's not quite a contracting map. 00:16:57.090 --> 00:17:00.830 align:middle line:90% But so we will need-- 00:17:00.830 --> 00:17:13.650 align:middle line:84% so for it to be a contracting map, 00:17:13.650 --> 00:17:28.290 align:middle line:84% we need to restrict the integral further. 00:17:28.290 --> 00:17:32.550 align:middle line:90% 00:17:32.550 --> 00:17:44.350 align:middle line:84% So by that, I mean that we need to find-- we need delta to be-- 00:17:44.350 --> 00:17:46.910 align:middle line:84% we need to show that there's some delta. 00:17:46.910 --> 00:17:49.070 align:middle line:90% It will actually be less than. 00:17:49.070 --> 00:17:52.430 align:middle line:84% So we need to show that there is some delta that is smaller 00:17:52.430 --> 00:17:55.550 align:middle line:84% than this delta zero, and so that if you restrict 00:17:55.550 --> 00:17:59.100 align:middle line:84% to this space here, now it's a continuous function. 00:17:59.100 --> 00:18:01.990 align:middle line:90% 00:18:01.990 --> 00:18:04.950 align:middle line:84% I will denote the ball by the same. 00:18:04.950 --> 00:18:09.330 align:middle line:84% But this here is actually a ball in a different metric space, 00:18:09.330 --> 00:18:13.750 align:middle line:84% because this is now a ball in the continuous functions 00:18:13.750 --> 00:18:15.900 align:middle line:90% on this smaller interval. 00:18:15.900 --> 00:18:19.536 align:middle line:90% 00:18:19.536 --> 00:18:21.970 align:middle line:90% But it's of the same radius. 00:18:21.970 --> 00:18:27.180 align:middle line:84% So if we restrict the map to this space here, then on this, 00:18:27.180 --> 00:18:29.220 align:middle line:90% it is a contracting map. 00:18:29.220 --> 00:18:31.910 align:middle line:84% So we need to cut down the integral. 00:18:31.910 --> 00:18:34.520 align:middle line:90% 00:18:34.520 --> 00:18:37.080 align:middle line:90% So now, how do we do that? 00:18:37.080 --> 00:18:44.000 align:middle line:84% So we need the contracting property again. 00:18:44.000 --> 00:18:47.420 align:middle line:84% And so suppose you take two functions. 00:18:47.420 --> 00:18:50.760 align:middle line:90% 00:18:50.760 --> 00:18:54.800 align:middle line:84% These are continuous maps on some interval, 00:18:54.800 --> 00:18:59.880 align:middle line:90% say, from right here. 00:18:59.880 --> 00:19:04.120 align:middle line:84% Again, delta is now-- that's a delta we are looking for. 00:19:04.120 --> 00:19:07.400 align:middle line:84% And it would be smaller than delta zero. 00:19:07.400 --> 00:19:14.820 align:middle line:84% So since it's smaller than delta zero, we already know that T-- 00:19:14.820 --> 00:19:17.280 align:middle line:90% so we have two maps like this. 00:19:17.280 --> 00:19:19.620 align:middle line:84% They are both supposed to be in the ball, 00:19:19.620 --> 00:19:23.930 align:middle line:84% so this means that this thing here is less than M. 00:19:23.930 --> 00:19:27.730 align:middle line:84% And we know we have already proven, just because delta is 00:19:27.730 --> 00:19:32.330 align:middle line:84% less than delta is zero, that T of this function 00:19:32.330 --> 00:19:38.410 align:middle line:84% here has norm less than or equal to M. 00:19:38.410 --> 00:19:41.590 align:middle line:84% But now, let's look at the difference. 00:19:41.590 --> 00:19:53.450 align:middle line:84% So T of y1 of x minus T of y2 of x. 00:19:53.450 --> 00:19:55.950 align:middle line:84% And we're interested in the norm here. 00:19:55.950 --> 00:19:58.530 align:middle line:90% 00:19:58.530 --> 00:20:02.520 align:middle line:84% And now x here is in this smaller interval. 00:20:02.520 --> 00:20:06.142 align:middle line:90% 00:20:06.142 --> 00:20:09.850 align:middle line:84% So now let's just write down what the map is. 00:20:09.850 --> 00:20:12.610 align:middle line:90% So the map here is the integral. 00:20:12.610 --> 00:20:13.450 align:middle line:90% So it's a. 00:20:13.450 --> 00:20:17.690 align:middle line:84% Sorry, it's a plus the integral from zero 00:20:17.690 --> 00:20:29.820 align:middle line:84% to x of f of y1 s plus g of s, ds, 00:20:29.820 --> 00:20:41.260 align:middle line:84% minus a plus the integral from zero to x of f of y2 s 00:20:41.260 --> 00:20:45.290 align:middle line:90% plus g of s, ds. 00:20:45.290 --> 00:20:48.260 align:middle line:90% 00:20:48.260 --> 00:20:51.380 align:middle line:84% And now you see that the a's cancel. 00:20:51.380 --> 00:20:54.500 align:middle line:90% The a's cancel. 00:20:54.500 --> 00:21:00.526 align:middle line:84% And the g of s, and those also cancel out. 00:21:00.526 --> 00:21:03.620 align:middle line:84% So you are left with that this thing here 00:21:03.620 --> 00:21:06.980 align:middle line:84% is the absolute value of the integral from zero 00:21:06.980 --> 00:21:25.710 align:middle line:84% to x of f of y1 of s minus f of y2 of s, ds. 00:21:25.710 --> 00:21:39.910 align:middle line:90% 00:21:39.910 --> 00:21:51.190 align:middle line:84% Now, so last time, when we proved 00:21:51.190 --> 00:22:00.350 align:middle line:84% that the map T takes this ball into the ball, we defined-- 00:22:00.350 --> 00:22:02.730 align:middle line:84% so let me just define that again. 00:22:02.730 --> 00:22:05.750 align:middle line:90% We defined L1. 00:22:05.750 --> 00:22:12.130 align:middle line:84% And L1 was the maximum of, actually-- 00:22:12.130 --> 00:22:22.320 align:middle line:90% 00:22:22.320 --> 00:22:24.360 align:middle line:84% I don't remember whether it was called L1. 00:22:24.360 --> 00:22:28.160 align:middle line:84% Let me just call it L to not call it. 00:22:28.160 --> 00:22:39.160 align:middle line:84% So L here is the maximum of f prime of z, 00:22:39.160 --> 00:22:47.560 align:middle line:84% where z here has norm less or equal to M. 00:22:47.560 --> 00:22:53.080 align:middle line:84% So this is so f prime exists and is continuous. 00:22:53.080 --> 00:22:58.000 align:middle line:84% So on this interval here, it has a maximum. 00:22:58.000 --> 00:23:00.320 align:middle line:90% So it's a nice finite number. 00:23:00.320 --> 00:23:02.400 align:middle line:90% And so now, we use the-- 00:23:02.400 --> 00:23:06.250 align:middle line:90% so by the mean value theorem-- 00:23:06.250 --> 00:23:07.250 align:middle line:90% the mean value theorem-- 00:23:07.250 --> 00:23:13.920 align:middle line:90% 00:23:13.920 --> 00:23:23.410 align:middle line:84% by the mean value theorem we have that f of z1 minus f of z2, 00:23:23.410 --> 00:23:33.930 align:middle line:84% as long as z1 and z2 are in the interval from minus M to M, 00:23:33.930 --> 00:23:36.530 align:middle line:84% then we have that this difference here 00:23:36.530 --> 00:23:42.010 align:middle line:84% is equal to the derivative in some intermediate point set 00:23:42.010 --> 00:23:43.622 align:middle line:90% to sigma-- 00:23:43.622 --> 00:23:45.010 align:middle line:90% to psi, I guess. 00:23:45.010 --> 00:23:48.930 align:middle line:84% Psi-- times the difference between the two. 00:23:48.930 --> 00:23:51.530 align:middle line:90% z1 minus z2. 00:23:51.530 --> 00:23:53.010 align:middle line:90% So we have this. 00:23:53.010 --> 00:23:55.890 align:middle line:90% This is the mean value theorem. 00:23:55.890 --> 00:23:58.450 align:middle line:90% And so we have that-- 00:23:58.450 --> 00:24:06.210 align:middle line:84% therefore, we have that the norm here of the difference 00:24:06.210 --> 00:24:08.850 align:middle line:90% is bounded by this here. 00:24:08.850 --> 00:24:10.690 align:middle line:84% When we take the absolute value of this, 00:24:10.690 --> 00:24:12.110 align:middle line:84% we're getting the absolute value. 00:24:12.110 --> 00:24:18.060 align:middle line:84% So this here is equal to the absolute value of f prime sigma 00:24:18.060 --> 00:24:20.150 align:middle line:90% times z1 minus z2. 00:24:20.150 --> 00:24:23.580 align:middle line:90% 00:24:23.580 --> 00:24:28.820 align:middle line:84% This thing here, because these two are in this interval, 00:24:28.820 --> 00:24:33.530 align:middle line:84% this is bounded by capital L. So you're getting this. 00:24:33.530 --> 00:24:37.660 align:middle line:90% 00:24:37.660 --> 00:24:43.540 align:middle line:84% And so now, if you use this factor, 00:24:43.540 --> 00:24:45.980 align:middle line:84% and you use this on this-- so this thing 00:24:45.980 --> 00:24:51.680 align:middle line:84% here is less than or equal to the integral from zero to x. 00:24:51.680 --> 00:24:53.380 align:middle line:84% And I can take the absolute value 00:24:53.380 --> 00:24:55.620 align:middle line:90% in under the integration sign. 00:24:55.620 --> 00:25:08.020 align:middle line:84% So I get f of y1s minus f of y2s, ds. 00:25:08.020 --> 00:25:09.820 align:middle line:90% Right? 00:25:09.820 --> 00:25:12.570 align:middle line:84% And so now I can continue this string. 00:25:12.570 --> 00:25:15.270 align:middle line:84% So I have what I'm interested in is T 00:25:15.270 --> 00:25:22.950 align:middle line:90% of y1 at x minus T of y2 of x. 00:25:22.950 --> 00:25:27.230 align:middle line:84% And I'm interested in the bound for the absolute value. 00:25:27.230 --> 00:25:32.190 align:middle line:84% And I have that this is less or equal to the integral from zero 00:25:32.190 --> 00:25:46.030 align:middle line:84% to x of f of y1s plus f of y2s, ds. 00:25:46.030 --> 00:25:50.310 align:middle line:84% And now I know that this integral here, x 00:25:50.310 --> 00:25:54.270 align:middle line:84% is supposed to lie in this interval here. 00:25:54.270 --> 00:25:59.630 align:middle line:84% So this means that the length of this interval from zero to-- 00:25:59.630 --> 00:26:01.350 align:middle line:90% is at most delta. 00:26:01.350 --> 00:26:03.030 align:middle line:90% So delta. 00:26:03.030 --> 00:26:07.390 align:middle line:84% And then I can take the maximum of this. 00:26:07.390 --> 00:26:10.500 align:middle line:90% Or I can take this one. 00:26:10.500 --> 00:26:13.560 align:middle line:84% I can take the maximum of this, but you 00:26:13.560 --> 00:26:16.560 align:middle line:84% see that this string here is bounded by. 00:26:16.560 --> 00:26:18.760 align:middle line:90% So this is the maximum. 00:26:18.760 --> 00:26:21.800 align:middle line:90% It's the maximum. 00:26:21.800 --> 00:26:27.000 align:middle line:84% And if I use this inequality here, it's bounded by L, 00:26:27.000 --> 00:26:28.480 align:middle line:90% and then it's applied. 00:26:28.480 --> 00:26:31.600 align:middle line:84% This here you should think about as z1. 00:26:31.600 --> 00:26:34.840 align:middle line:84% And this here for each x, s, you can think 00:26:34.840 --> 00:26:37.600 align:middle line:90% about this as z1 one and z2. 00:26:37.600 --> 00:26:47.280 align:middle line:84% And so you're getting this thing here times y1s minus y2s. 00:26:47.280 --> 00:26:56.280 align:middle line:84% And here, s will now lie in the interval, in this interval here. 00:26:56.280 --> 00:26:59.120 align:middle line:90% Oops. 00:26:59.120 --> 00:27:00.400 align:middle line:90% Right. 00:27:00.400 --> 00:27:03.280 align:middle line:84% And so now you see that this is just a constant. 00:27:03.280 --> 00:27:09.890 align:middle line:84% So this is equal to delta L, and then this maximum 00:27:09.890 --> 00:27:20.090 align:middle line:84% s, these are equal to delta y1 minus y2, like this. 00:27:20.090 --> 00:27:27.290 align:middle line:84% This is evaluated in s, but I don't write that. 00:27:27.290 --> 00:27:30.530 align:middle line:84% And this thing here, but this is just 00:27:30.530 --> 00:27:34.050 align:middle line:84% delta L. This maximum here is just 00:27:34.050 --> 00:27:38.790 align:middle line:84% the distance between this function and that function. 00:27:38.790 --> 00:27:42.130 align:middle line:84% That's how we define this distance. 00:27:42.130 --> 00:27:48.530 align:middle line:84% And so you see that this right hand side now holds for all x. 00:27:48.530 --> 00:27:52.490 align:middle line:84% So this means that we have that the maximum of this thing 00:27:52.490 --> 00:27:58.530 align:middle line:84% here over x, where x is in the interval between zero and delta. 00:27:58.530 --> 00:28:05.140 align:middle line:84% So this means that we have that the distance between T of y1 00:28:05.140 --> 00:28:09.100 align:middle line:90% and T of y2-- 00:28:09.100 --> 00:28:12.880 align:middle line:84% this thing here is equal to the maximum. 00:28:12.880 --> 00:28:15.360 align:middle line:84% This is how we define that distance after all. 00:28:15.360 --> 00:28:19.380 align:middle line:90% 00:28:19.380 --> 00:28:21.560 align:middle line:84% And then it's of this thing here. 00:28:21.560 --> 00:28:26.140 align:middle line:90% 00:28:26.140 --> 00:28:28.460 align:middle line:90% Oh, sorry. 00:28:28.460 --> 00:28:38.500 align:middle line:90% x minus T of y2 in x. 00:28:38.500 --> 00:28:42.060 align:middle line:84% So we have that this and this inequality 00:28:42.060 --> 00:28:53.780 align:middle line:84% say that it's less than or equal to delta L, d, y1, comma y2. 00:28:53.780 --> 00:28:56.500 align:middle line:90% And so you see that now if you-- 00:28:56.500 --> 00:28:59.740 align:middle line:84% so L here was just a fixed constant. 00:28:59.740 --> 00:29:05.110 align:middle line:84% If I now set delta, just to make sure that there's no problem, 00:29:05.110 --> 00:29:08.850 align:middle line:84% I'm setting delta equal to 1 over 2L. 00:29:08.850 --> 00:29:11.310 align:middle line:84% I don't really need the one, but I'll just 00:29:11.310 --> 00:29:14.950 align:middle line:84% do it like this so I don't have zero in the denominator. 00:29:14.950 --> 00:29:20.630 align:middle line:84% If I choose the delta to be equal to this, 00:29:20.630 --> 00:29:25.190 align:middle line:84% well, then you see that this thing here, the inner product 00:29:25.190 --> 00:29:26.930 align:middle line:90% of these two is less-- 00:29:26.930 --> 00:29:31.390 align:middle line:84% or the product of these two is less than 1/2. 00:29:31.390 --> 00:29:33.070 align:middle line:84% So this is less than or equal to-- 00:29:33.070 --> 00:29:41.730 align:middle line:84% I can just replace it by 1/2, like that. 00:29:41.730 --> 00:29:44.590 align:middle line:84% And so you see this is exactly what it means 00:29:44.590 --> 00:29:46.250 align:middle line:90% to be a contracting mapping. 00:29:46.250 --> 00:29:48.330 align:middle line:84% In fact, it's a contracting mapping, 00:29:48.330 --> 00:29:53.830 align:middle line:84% which is pretty nice that the constant is not 00:29:53.830 --> 00:29:56.490 align:middle line:84% just strictly some constant strictly less than 1, 00:29:56.490 --> 00:29:58.290 align:middle line:90% but it's actually half. 00:29:58.290 --> 00:30:02.520 align:middle line:84% So it's a contracting mapping theorem in contracting map. 00:30:02.520 --> 00:30:05.240 align:middle line:84% And so now by the contracting mapping theorem, 00:30:05.240 --> 00:30:08.130 align:middle line:90% we have a fixed point. 00:30:08.130 --> 00:30:12.090 align:middle line:90% So this solves the array. 00:30:12.090 --> 00:30:16.760 align:middle line:84% And it does more than that, because the contracting mapping 00:30:16.760 --> 00:30:22.720 align:middle line:84% theorem says that it has a unique fixed point. 00:30:22.720 --> 00:30:27.820 align:middle line:84% So this means that-- and if you have a solution to this, 00:30:27.820 --> 00:30:30.920 align:middle line:84% if you have a solution to this, then 00:30:30.920 --> 00:30:34.420 align:middle line:84% we saw last time that if it is a solution, 00:30:34.420 --> 00:30:35.640 align:middle line:90% then it is a fixed point. 00:30:35.640 --> 00:30:38.332 align:middle line:90% 00:30:38.332 --> 00:30:43.720 align:middle line:84% So now we almost have existence and uniqueness. 00:30:43.720 --> 00:30:47.380 align:middle line:84% Let me explain what is missing to the existence and uniqueness. 00:30:47.380 --> 00:30:52.480 align:middle line:90% 00:30:52.480 --> 00:31:01.930 align:middle line:84% So for that, let me backtrack a little bit 00:31:01.930 --> 00:31:10.050 align:middle line:84% and talk about something about metric spaces, 00:31:10.050 --> 00:31:13.360 align:middle line:84% specifically about connected metric spaces. 00:31:13.360 --> 00:31:23.990 align:middle line:90% 00:31:23.990 --> 00:31:29.930 align:middle line:84% So suppose you have a metric space, 00:31:29.930 --> 00:31:38.510 align:middle line:84% then we say that it is connected. 00:31:38.510 --> 00:31:42.890 align:middle line:90% 00:31:42.890 --> 00:31:54.290 align:middle line:84% If x here is the only subset, so the whole 00:31:54.290 --> 00:32:04.260 align:middle line:84% set itself is the only subset of x, that is, 00:32:04.260 --> 00:32:06.090 align:middle line:90% it is both open and closed. 00:32:06.090 --> 00:32:13.540 align:middle line:90% 00:32:13.540 --> 00:32:17.960 align:middle line:84% Remember that if you take a metric space then the-- 00:32:17.960 --> 00:32:18.460 align:middle line:90% sorry. 00:32:18.460 --> 00:32:22.300 align:middle line:90% If x and-- sorry-- 00:32:22.300 --> 00:32:28.500 align:middle line:84% and the empty set are the only subset 00:32:28.500 --> 00:32:31.420 align:middle line:84% of x that are both open and closed. 00:32:31.420 --> 00:32:34.140 align:middle line:84% Remember that if you take a metric space, 00:32:34.140 --> 00:32:39.540 align:middle line:84% and if you take the whole set or the empty set, those are all-- 00:32:39.540 --> 00:32:43.100 align:middle line:84% they are both always closed and open. 00:32:43.100 --> 00:32:47.740 align:middle line:84% And so it's connected if those are the only two 00:32:47.740 --> 00:32:54.224 align:middle line:84% subsets that are at the same time open and closed. 00:32:54.224 --> 00:32:59.650 align:middle line:90% And so now, we have that. 00:32:59.650 --> 00:33:03.190 align:middle line:90% We have the following. 00:33:03.190 --> 00:33:04.920 align:middle line:90% We have the following lemma. 00:33:04.920 --> 00:33:07.450 align:middle line:90% 00:33:07.450 --> 00:33:07.950 align:middle line:90% Theorem. 00:33:07.950 --> 00:33:15.550 align:middle line:84% Let's call it a theorem that if you take an interval here 00:33:15.550 --> 00:33:19.490 align:middle line:90% with the usual metric here-- 00:33:19.490 --> 00:33:20.970 align:middle line:90% so I here is an interval-- 00:33:20.970 --> 00:33:28.430 align:middle line:90% 00:33:28.430 --> 00:33:37.930 align:middle line:84% then the theorem is that this metric space here is connected. 00:33:37.930 --> 00:33:42.430 align:middle line:90% 00:33:42.430 --> 00:33:44.370 align:middle line:90% So let's try to prove that. 00:33:44.370 --> 00:33:52.790 align:middle line:90% 00:33:52.790 --> 00:34:05.150 align:middle line:84% So I take now a subset here that is both open and closed. 00:34:05.150 --> 00:34:08.000 align:middle line:90% 00:34:08.000 --> 00:34:21.080 align:middle line:84% And we want to show that if A here is non-empty, 00:34:21.080 --> 00:34:26.500 align:middle line:84% then it implies that A is everything. 00:34:26.500 --> 00:34:28.600 align:middle line:84% So we're assuming that it's non-empty, 00:34:28.600 --> 00:34:36.239 align:middle line:84% and then we should try to prove that A is everything. 00:34:36.239 --> 00:34:42.639 align:middle line:84% So we let x zero be an element in A, but we're assuming 00:34:42.639 --> 00:34:48.120 align:middle line:84% it's non-empty, so we can let x zero be an element in A. And now 00:34:48.120 --> 00:34:49.299 align:middle line:90% I want to define-- 00:34:49.299 --> 00:34:59.410 align:middle line:90% 00:34:59.410 --> 00:35:01.050 align:middle line:90% So I have here-- 00:35:01.050 --> 00:35:03.010 align:middle line:90% I have here my interval. 00:35:03.010 --> 00:35:07.730 align:middle line:84% This is my interval I. I have here x zero, 00:35:07.730 --> 00:35:28.490 align:middle line:84% and I'm looking at the y in A, so that if you take everything 00:35:28.490 --> 00:35:29.350 align:middle line:90% so that-- 00:35:29.350 --> 00:35:37.410 align:middle line:84% so y is an A. And I'm going to look at y with y 00:35:37.410 --> 00:35:39.450 align:middle line:90% bigger or equal to x zero. 00:35:39.450 --> 00:35:45.850 align:middle line:84% And I'm going to look at the y so that this whole interval here 00:35:45.850 --> 00:35:55.220 align:middle line:84% from x zero to y is contained in A. 00:35:55.220 --> 00:36:16.860 align:middle line:84% So define as subset here, B of I, so that it's the y. 00:36:16.860 --> 00:36:23.020 align:middle line:84% The subset B is the y in A with the property 00:36:23.020 --> 00:36:29.300 align:middle line:84% that it is that y here is bigger or equal to x zero, 00:36:29.300 --> 00:36:32.740 align:middle line:90% and you have-- 00:36:32.740 --> 00:36:46.820 align:middle line:84% and this interval here is contained in A. 00:36:46.820 --> 00:36:49.090 align:middle line:90% And now I want to show that. 00:36:49.090 --> 00:36:52.310 align:middle line:84% So let's say for the sake of argument, 00:36:52.310 --> 00:36:58.050 align:middle line:84% let me assume that this integral is a closed interval, like this. 00:36:58.050 --> 00:36:59.550 align:middle line:90% It works equally well. 00:36:59.550 --> 00:37:02.270 align:middle line:84% The argument obvious modification 00:37:02.270 --> 00:37:08.580 align:middle line:84% work equally well if it's an open or half open interval. 00:37:08.580 --> 00:37:16.470 align:middle line:90% 00:37:16.470 --> 00:37:21.030 align:middle line:84% So I take some y here, and I have-- 00:37:21.030 --> 00:37:27.710 align:middle line:84% so I have a y, and y is-- so it's 00:37:27.710 --> 00:37:33.360 align:middle line:84% a set so that this whole interval is contained in A. 00:37:33.360 --> 00:37:46.522 align:middle line:84% And what I want to show is, so I want to show that y is in A, 00:37:46.522 --> 00:37:49.920 align:middle line:84% and this is the first thing I want to show. 00:37:49.920 --> 00:37:54.960 align:middle line:90% One is that y is in A. 00:37:54.960 --> 00:37:57.800 align:middle line:84% And then the second thing I want to show 00:37:57.800 --> 00:38:03.980 align:middle line:84% is that y is equal to B. You see, 00:38:03.980 --> 00:38:07.880 align:middle line:84% if I had proven both of these properties, 00:38:07.880 --> 00:38:10.540 align:middle line:90% then it would be that-- 00:38:10.540 --> 00:38:19.120 align:middle line:84% so then it would mean that everything here 00:38:19.120 --> 00:38:24.130 align:middle line:84% would be in A, including this last point. 00:38:24.130 --> 00:38:28.520 align:middle line:84% And so this means that we have this entire interval is in A-- 00:38:28.520 --> 00:38:29.760 align:middle line:90% in an I, sorry. 00:38:29.760 --> 00:38:32.360 align:middle line:84% And then likewise, we could argue on the other side. 00:38:32.360 --> 00:38:34.950 align:middle line:90% 00:38:34.950 --> 00:38:37.920 align:middle line:90% So now, how do we prove this? 00:38:37.920 --> 00:38:51.590 align:middle line:84% So well, the first thing is that if we take-- so if we set-- 00:38:51.590 --> 00:38:52.090 align:middle line:90% sorry. 00:38:52.090 --> 00:38:54.810 align:middle line:90% 00:38:54.810 --> 00:38:56.010 align:middle line:90% Sorry about this. 00:38:56.010 --> 00:38:58.770 align:middle line:90% This here should have been-- 00:38:58.770 --> 00:39:02.730 align:middle line:84% This here should have been y zero. 00:39:02.730 --> 00:39:06.890 align:middle line:84% And this should have been the soup of y 00:39:06.890 --> 00:39:13.850 align:middle line:90% in A. Soup of these guys. 00:39:13.850 --> 00:39:18.450 align:middle line:84% And then this here, and we want to prove that this thing here 00:39:18.450 --> 00:39:24.730 align:middle line:84% is in A. And the second thing is we want 00:39:24.730 --> 00:39:28.302 align:middle line:84% to prove that this soup is actually equal to B. 00:39:28.302 --> 00:39:29.510 align:middle line:90% That's what we want to prove. 00:39:29.510 --> 00:39:33.236 align:middle line:90% 00:39:33.236 --> 00:39:40.790 align:middle line:84% So the soup of these, well, if y zero is soup of elements, 00:39:40.790 --> 00:39:45.940 align:middle line:84% then you have a sequence, so here you have y zero. 00:39:45.940 --> 00:39:50.540 align:middle line:84% And this means that you have a sequence of elements, which 00:39:50.540 --> 00:39:54.220 align:middle line:84% are then in elements in the set B, 00:39:54.220 --> 00:40:01.980 align:middle line:84% this is soup of B. This is the same as soup of B. Sorry. 00:40:01.980 --> 00:40:07.900 align:middle line:84% Soup of B. I could write it like this. 00:40:07.900 --> 00:40:13.020 align:middle line:84% So I take this set, and these are the y so 00:40:13.020 --> 00:40:17.900 align:middle line:84% that this entire interval is contained in A. 00:40:17.900 --> 00:40:21.300 align:middle line:84% And then I take the soup of this set, 00:40:21.300 --> 00:40:25.860 align:middle line:84% and I want to prove that this soup is in A, 00:40:25.860 --> 00:40:33.380 align:middle line:84% and then I want to prove that, in fact, the soup is B. So 00:40:33.380 --> 00:40:38.700 align:middle line:84% that is a soup of such elements in capital 00:40:38.700 --> 00:40:48.230 align:middle line:84% B means that there exists a sequence of elements in B, 00:40:48.230 --> 00:40:58.030 align:middle line:84% so that y here, y zero, is the soup of B. 00:40:58.030 --> 00:41:03.630 align:middle line:84% This here implies that there exists a sequence yn. 00:41:03.630 --> 00:41:12.230 align:middle line:84% And this yn are in A, and you have that x zero, 00:41:12.230 --> 00:41:17.270 align:middle line:84% yn is contained in A. And then you 00:41:17.270 --> 00:41:23.070 align:middle line:84% have that these guys converge to y zero. 00:41:23.070 --> 00:41:26.330 align:middle line:84% This is what you get from that it's a soup. 00:41:26.330 --> 00:41:31.550 align:middle line:84% These two things are that the elements are in-- 00:41:31.550 --> 00:41:43.415 align:middle line:84% that these elements, so this here is just that the yn on B. 00:41:43.415 --> 00:41:46.320 align:middle line:84% So now, what you have, so this here means-- 00:41:46.320 --> 00:41:49.360 align:middle line:90% so here I have x zero. 00:41:49.360 --> 00:41:51.480 align:middle line:90% Here I have y zero. 00:41:51.480 --> 00:41:57.840 align:middle line:90% And I have these elements yn. 00:41:57.840 --> 00:42:04.780 align:middle line:84% And in particular, they are in A. But A is closed. 00:42:04.780 --> 00:42:18.200 align:middle line:84% Since A is closed, since yn is in A, and yn and A is closed, 00:42:18.200 --> 00:42:27.800 align:middle line:84% it's a closed subset, this means that the limits here-- 00:42:27.800 --> 00:42:30.800 align:middle line:84% the limit y zero, which is the limit, 00:42:30.800 --> 00:42:37.130 align:middle line:84% this here must also be in A. So we 00:42:37.130 --> 00:42:39.863 align:middle line:90% have proven the first property. 00:42:39.863 --> 00:42:46.170 align:middle line:90% 00:42:46.170 --> 00:42:48.670 align:middle line:90% You've proven that. 00:42:48.670 --> 00:42:52.090 align:middle line:84% So we've proven that y zero is an A. 00:42:52.090 --> 00:42:58.330 align:middle line:84% But now you see that we have that for each of the yn, we have 00:42:58.330 --> 00:43:05.250 align:middle line:84% that this interval here lie in A. And they converge to this. 00:43:05.250 --> 00:43:10.450 align:middle line:84% So it means that if you take anything to the left of y zero, 00:43:10.450 --> 00:43:15.330 align:middle line:84% well then eventually, there's some yn to the right of that. 00:43:15.330 --> 00:43:19.730 align:middle line:84% And so this whole interval is in A. So you see that this-- 00:43:19.730 --> 00:43:25.850 align:middle line:84% it now means that not only is y zero in A, but y zero 00:43:25.850 --> 00:43:33.610 align:middle line:90% is really in the set B. 00:43:33.610 --> 00:43:39.440 align:middle line:84% This here only used that A was closed. 00:43:39.440 --> 00:43:41.830 align:middle line:84% So now we need to use that A is open. 00:43:41.830 --> 00:43:48.686 align:middle line:90% 00:43:48.686 --> 00:43:54.620 align:middle line:84% So now, suppose we take a element. 00:43:54.620 --> 00:43:59.780 align:middle line:84% Suppose so now here we have x zero. 00:43:59.780 --> 00:44:03.780 align:middle line:84% We have here B, and we want to show 00:44:03.780 --> 00:44:09.100 align:middle line:84% that this y zero we defined to be the soup of this set capital 00:44:09.100 --> 00:44:16.980 align:middle line:84% B, we want to prove that this thing here is, in fact, y zero. 00:44:16.980 --> 00:44:19.460 align:middle line:90% Suppose not. 00:44:19.460 --> 00:44:20.280 align:middle line:90% Suppose not. 00:44:20.280 --> 00:44:25.100 align:middle line:84% So suppose that you have here, strictly between these two, 00:44:25.100 --> 00:44:26.580 align:middle line:90% this here is y zero. 00:44:26.580 --> 00:44:29.180 align:middle line:90% 00:44:29.180 --> 00:44:35.230 align:middle line:84% Now, we already know that y zero is in A. 00:44:35.230 --> 00:44:39.070 align:middle line:90% This is what we proved here. 00:44:39.070 --> 00:44:48.670 align:middle line:84% But since A here is open, there exists 00:44:48.670 --> 00:44:50.250 align:middle line:90% some tiny little interval. 00:44:50.250 --> 00:44:59.230 align:middle line:84% There exists some interval here around this y zero. 00:44:59.230 --> 00:45:04.630 align:middle line:84% So if you take y zero, and you subtract this epsilon, 00:45:04.630 --> 00:45:06.770 align:middle line:84% so there exists some epsilon greater than zero, 00:45:06.770 --> 00:45:09.310 align:middle line:84% so that if you're looking at this tiny little interval 00:45:09.310 --> 00:45:16.410 align:middle line:84% centered at y zero, then this here is contained in A. 00:45:16.410 --> 00:45:19.950 align:middle line:90% This is just because A was open. 00:45:19.950 --> 00:45:23.970 align:middle line:84% But this means that now you have that-- 00:45:23.970 --> 00:45:27.150 align:middle line:84% so this here is you assume this is y zero. 00:45:27.150 --> 00:45:32.440 align:middle line:84% So you have that this thing here lie in the set capital A. 00:45:32.440 --> 00:45:35.040 align:middle line:84% And this I guarantee you that you can actually 00:45:35.040 --> 00:45:38.360 align:middle line:84% go a little bit past it, that even if you 00:45:38.360 --> 00:45:43.400 align:middle line:84% go a little bit past it, if you go a little bit past it here, 00:45:43.400 --> 00:45:47.800 align:middle line:84% then these are still in A. So this means that this could not 00:45:47.800 --> 00:45:50.000 align:middle line:90% be the soup. 00:45:50.000 --> 00:45:53.280 align:middle line:84% So therefore, this point here had 00:45:53.280 --> 00:45:57.460 align:middle line:84% to be B. It had to be the final point. 00:45:57.460 --> 00:46:00.280 align:middle line:84% So this proved these two properties. 00:46:00.280 --> 00:46:02.440 align:middle line:84% And so once you have these two properties, 00:46:02.440 --> 00:46:05.680 align:middle line:84% you have that the interval is connected, 00:46:05.680 --> 00:46:10.880 align:middle line:84% that the only non-empty subset of the interval that is 00:46:10.880 --> 00:46:13.320 align:middle line:90% both open and closed is itself. 00:46:13.320 --> 00:46:15.960 align:middle line:90% 00:46:15.960 --> 00:46:20.520 align:middle line:84% Now, why is this related to what we want? 00:46:20.520 --> 00:46:24.440 align:middle line:90% So this is just so we have-- 00:46:24.440 --> 00:46:26.540 align:middle line:90% suppose you take two functions. 00:46:26.540 --> 00:46:34.130 align:middle line:90% 00:46:34.130 --> 00:46:38.850 align:middle line:84% So why is this related to the uniqueness for the array? 00:46:38.850 --> 00:46:40.540 align:middle line:90% So this is the following lemma. 00:46:40.540 --> 00:46:43.850 align:middle line:90% 00:46:43.850 --> 00:46:45.710 align:middle line:90% Suppose you have an interval. 00:46:45.710 --> 00:46:50.050 align:middle line:84% So this is a interval in R. And suppose you 00:46:50.050 --> 00:46:53.370 align:middle line:90% have two functions, U1 and U2. 00:46:53.370 --> 00:46:57.250 align:middle line:84% They're functions on the interval, 00:46:57.250 --> 00:46:59.310 align:middle line:90% and they are both continuous. 00:46:59.310 --> 00:47:03.770 align:middle line:90% 00:47:03.770 --> 00:47:06.340 align:middle line:90% And then assume that for some-- 00:47:06.340 --> 00:47:21.690 align:middle line:84% assume that for some x zero in the interval U1 of x zero 00:47:21.690 --> 00:47:26.840 align:middle line:84% is equal to U2 of x zero, so they coincide in one point. 00:47:26.840 --> 00:47:31.300 align:middle line:90% 00:47:31.300 --> 00:47:33.520 align:middle line:84% And assume you have a second condition. 00:47:33.520 --> 00:47:46.260 align:middle line:84% And the second condition is that if U1 of x 00:47:46.260 --> 00:48:00.420 align:middle line:84% is equal to U2 of x, then there exists a neighborhood, so 00:48:00.420 --> 00:48:05.700 align:middle line:84% epsilon, where this is allowed to depend on x, 00:48:05.700 --> 00:48:14.460 align:middle line:84% so that U1 is equal to U2 in not just at x, 00:48:14.460 --> 00:48:23.180 align:middle line:84% but actually, on this little interval x minus epsilon, x 00:48:23.180 --> 00:48:26.490 align:middle line:90% plus epsilon. 00:48:26.490 --> 00:48:28.130 align:middle line:84% So you have these two properties. 00:48:28.130 --> 00:48:32.230 align:middle line:84% So you have two continuous functions on an interval. 00:48:32.230 --> 00:48:34.750 align:middle line:90% They agree at one point. 00:48:34.750 --> 00:48:38.910 align:middle line:84% And if they agree at any point in this interval, 00:48:38.910 --> 00:48:42.350 align:middle line:84% then they agree on a little neighborhood of that point. 00:48:42.350 --> 00:48:45.990 align:middle line:84% But the neighborhood may depend on the point. 00:48:45.990 --> 00:48:48.950 align:middle line:84% And then the statement is-- so the claim 00:48:48.950 --> 00:49:05.590 align:middle line:84% is that U1 is actually equal to U2 everywhere in the interval 00:49:05.590 --> 00:49:08.670 align:middle line:90% where they're defined. 00:49:08.670 --> 00:49:10.776 align:middle line:90% That's the claim. 00:49:10.776 --> 00:49:14.480 align:middle line:84% And so now, you see that if you take-- 00:49:14.480 --> 00:49:23.110 align:middle line:90% 00:49:23.110 --> 00:49:25.660 align:middle line:84% if you're looking at the proof of this 00:49:25.660 --> 00:49:29.750 align:middle line:90% is that suppose you prove-- 00:49:29.750 --> 00:49:32.760 align:middle line:90% 00:49:32.760 --> 00:49:39.880 align:middle line:84% suppose you define a function, and this function here 00:49:39.880 --> 00:49:45.280 align:middle line:90% is U1 of x minus U2 of x. 00:49:45.280 --> 00:49:49.360 align:middle line:84% If you define it like this, U1 and U2 are continuous. 00:49:49.360 --> 00:49:51.060 align:middle line:90% So f here is continuous. 00:49:51.060 --> 00:49:57.000 align:middle line:90% 00:49:57.000 --> 00:50:00.600 align:middle line:84% You also have that f of x zero, where 00:50:00.600 --> 00:50:05.800 align:middle line:84% this is U1 of x zero minus U2 of x zero, 00:50:05.800 --> 00:50:08.500 align:middle line:84% and these two functions coincided at x zero, 00:50:08.500 --> 00:50:11.720 align:middle line:90% so this is zero. 00:50:11.720 --> 00:50:15.760 align:middle line:90% You also have that if f of x-- 00:50:15.760 --> 00:50:26.700 align:middle line:84% so if f of x is equal to zero, well, right now, 00:50:26.700 --> 00:50:32.300 align:middle line:84% then you have by this second property here, this property 00:50:32.300 --> 00:50:33.540 align:middle line:90% here-- 00:50:33.540 --> 00:50:38.140 align:middle line:84% so if f is zero, this is just meaning that the two function 00:50:38.140 --> 00:50:39.980 align:middle line:90% coincided at that x. 00:50:39.980 --> 00:50:44.940 align:middle line:84% So this means that they exist, and they exist at delta-- 00:50:44.940 --> 00:50:54.940 align:middle line:84% epsilon, sorry-- epsilon greater than zero, such that f of y 00:50:54.940 --> 00:51:05.060 align:middle line:84% is equal to 0 in this little interval around x. 00:51:05.060 --> 00:51:07.380 align:middle line:90% So you have that. 00:51:07.380 --> 00:51:09.440 align:middle line:90% And now I claim that-- 00:51:09.440 --> 00:51:13.860 align:middle line:84% so now the claim is that the zero set, so the set 00:51:13.860 --> 00:51:20.750 align:middle line:84% where the function f is zero, this is both open and closed. 00:51:20.750 --> 00:51:32.550 align:middle line:84% So the claim here is That the set capital A, 00:51:32.550 --> 00:51:39.590 align:middle line:84% this is the set of x in the interval where f of x 00:51:39.590 --> 00:51:47.500 align:middle line:84% is equal to zero, that this here is both open and closed. 00:51:47.500 --> 00:51:53.270 align:middle line:90% 00:51:53.270 --> 00:51:57.750 align:middle line:84% We already know that this set, this is followup here, 00:51:57.750 --> 00:52:00.150 align:middle line:90% that x zero is-- 00:52:00.150 --> 00:52:06.650 align:middle line:84% so since x zero is in this set, then we have-- 00:52:06.650 --> 00:52:16.270 align:middle line:84% and the interval is connected, and I is connected, 00:52:16.270 --> 00:52:18.700 align:middle line:84% if this set is both open and closed, 00:52:18.700 --> 00:52:24.840 align:middle line:84% then we have that then A is equal to the whole interval. 00:52:24.840 --> 00:52:30.440 align:middle line:84% So this means that f is constant zero, which means that the two 00:52:30.440 --> 00:52:32.460 align:middle line:90% functions coincided everywhere. 00:52:32.460 --> 00:52:41.160 align:middle line:84% And so we just need to prove that f is both open and closed. 00:52:41.160 --> 00:52:45.080 align:middle line:84% But the zero set, if you take a continuous function, so 00:52:45.080 --> 00:52:55.860 align:middle line:84% if f here is continuous, and you have a sequence 00:52:55.860 --> 00:52:59.880 align:middle line:84% so that each element here in the sequence is zero, 00:52:59.880 --> 00:53:05.900 align:middle line:84% and this xn converges to some x, well, then by continuity, 00:53:05.900 --> 00:53:10.740 align:middle line:84% by continuity, f of x would also be zero. 00:53:10.740 --> 00:53:12.160 align:middle line:90% This is by continuity. 00:53:12.160 --> 00:53:17.730 align:middle line:90% 00:53:17.730 --> 00:53:22.990 align:middle line:84% So this means that this set here is definitely closed. 00:53:22.990 --> 00:53:24.530 align:middle line:90% And the open thing-- 00:53:24.530 --> 00:53:28.370 align:middle line:84% well, the open is just exactly this property 00:53:28.370 --> 00:53:32.530 align:middle line:84% that we're assuming that if you have an element in it, 00:53:32.530 --> 00:53:35.930 align:middle line:84% each element is interior, which means that there is 00:53:35.930 --> 00:53:38.770 align:middle line:90% this little interval around it. 00:53:38.770 --> 00:53:43.950 align:middle line:84% So we have proven that it's both closed. 00:53:43.950 --> 00:53:46.530 align:middle line:84% We prove that first, and also open. 00:53:46.530 --> 00:53:53.010 align:middle line:84% And so we have that if you have two such functions defined 00:53:53.010 --> 00:53:56.250 align:middle line:84% both continuous, and defined on an interval, 00:53:56.250 --> 00:53:58.610 align:middle line:90% they coincide at one point. 00:53:58.610 --> 00:54:00.990 align:middle line:84% And if they coincide at one point, 00:54:00.990 --> 00:54:02.990 align:middle line:84% then they coincide in a little neighborhood. 00:54:02.990 --> 00:54:05.830 align:middle line:84% If you have two functions with this property, 00:54:05.830 --> 00:54:08.950 align:middle line:84% then they must be equal in the whole interval. 00:54:08.950 --> 00:54:13.650 align:middle line:84% So now, how does this relate to the uniqueness 00:54:13.650 --> 00:54:15.000 align:middle line:90% that we wanted to prove? 00:54:15.000 --> 00:54:18.420 align:middle line:90% 00:54:18.420 --> 00:54:20.120 align:middle line:90% So we're starting again. 00:54:20.120 --> 00:54:22.680 align:middle line:90% So we have the-- 00:54:22.680 --> 00:54:25.360 align:middle line:84% so this is, again, the Picard-Lindelof theorem, 00:54:25.360 --> 00:54:26.930 align:middle line:90% and this is the uniqueness part. 00:54:26.930 --> 00:54:33.620 align:middle line:90% 00:54:33.620 --> 00:54:36.820 align:middle line:84% And so you're looking at, again, you're 00:54:36.820 --> 00:54:41.840 align:middle line:84% looking at solutions to this array. 00:54:41.840 --> 00:54:49.820 align:middle line:90% 00:54:49.820 --> 00:54:55.420 align:middle line:84% And we have that now, we had that inner-- 00:54:55.420 --> 00:54:59.460 align:middle line:84% we have that inner, so and then we let so-- 00:54:59.460 --> 00:55:06.120 align:middle line:90% and let y1 and y2 be solutions. 00:55:06.120 --> 00:55:11.950 align:middle line:90% 00:55:11.950 --> 00:55:14.550 align:middle line:84% And let's say that these two, they 00:55:14.550 --> 00:55:16.910 align:middle line:84% may be defined on different intervals. 00:55:16.910 --> 00:55:20.750 align:middle line:84% Of course, but you just check the intersection of the two 00:55:20.750 --> 00:55:22.630 align:middle line:90% intervals that they're defined. 00:55:22.630 --> 00:55:24.910 align:middle line:84% So suppose that they're both defined, 00:55:24.910 --> 00:55:35.830 align:middle line:84% say, on the same interval, like this. 00:55:35.830 --> 00:55:38.930 align:middle line:84% And this interval here and I, of course, 00:55:38.930 --> 00:55:44.990 align:middle line:84% has to include the zero where we want this condition. 00:55:44.990 --> 00:55:47.270 align:middle line:90% Well, then what do we have? 00:55:47.270 --> 00:55:50.563 align:middle line:84% So we have an interval, and we have two continuous functions. 00:55:50.563 --> 00:55:52.230 align:middle line:84% They're actually better than continuous. 00:55:52.230 --> 00:55:54.590 align:middle line:90% They're differentiable. 00:55:54.590 --> 00:55:56.770 align:middle line:84% And even the derivative is continuous. 00:55:56.770 --> 00:55:58.378 align:middle line:90% But never mind that. 00:55:58.378 --> 00:56:01.990 align:middle line:90% 00:56:01.990 --> 00:56:06.090 align:middle line:84% We have that y1 of zero is equal to A, 00:56:06.090 --> 00:56:09.550 align:middle line:90% and that's equal to y2 of zero. 00:56:09.550 --> 00:56:12.800 align:middle line:90% So they coincide at one point. 00:56:12.800 --> 00:56:16.560 align:middle line:84% And now, the Picard-Lindelof theorem 00:56:16.560 --> 00:56:20.020 align:middle line:84% said that because there were two solutions, 00:56:20.020 --> 00:56:25.000 align:middle line:84% if you restrict the integral to looking 00:56:25.000 --> 00:56:27.520 align:middle line:84% at this little interval around zero, 00:56:27.520 --> 00:56:30.400 align:middle line:84% Well, then there was a unique fixed point. 00:56:30.400 --> 00:56:33.560 align:middle line:84% So this means that in a small neighborhood 00:56:33.560 --> 00:56:50.360 align:middle line:84% of the origin, so two in a small neighborhood of the origin. 00:56:50.360 --> 00:56:56.040 align:middle line:84% And again, this is thanks to that in the contraction mapping 00:56:56.040 --> 00:56:57.840 align:middle line:84% theorem, you had a unique fixed point 00:56:57.840 --> 00:57:05.240 align:middle line:84% in a small neighborhood of the origin, y1 is equal to y2. 00:57:05.240 --> 00:57:08.690 align:middle line:84% But actually, the Picard-Lindelof-- the argument 00:57:08.690 --> 00:57:12.210 align:middle line:90% would work at any other point. 00:57:12.210 --> 00:57:16.850 align:middle line:84% Suppose you take any other point. 00:57:16.850 --> 00:57:19.450 align:middle line:84% And you take a solution to the array. 00:57:19.450 --> 00:57:23.830 align:middle line:84% Well, if you take another point, then the value-- 00:57:23.830 --> 00:57:26.650 align:middle line:90% so if you take-- 00:57:26.650 --> 00:57:28.790 align:middle line:90% so take any other point. 00:57:28.790 --> 00:57:31.570 align:middle line:90% So here you have zero. 00:57:31.570 --> 00:57:33.630 align:middle line:90% And so now take any other point. 00:57:33.630 --> 00:57:36.690 align:middle line:90% Let's call it x zero. 00:57:36.690 --> 00:57:41.650 align:middle line:84% And if you assume that these two functions coincide 00:57:41.650 --> 00:57:43.490 align:middle line:90% at this point. 00:57:43.490 --> 00:57:53.210 align:middle line:84% So suppose that y1 of x zero is equal to y2 of x zero. 00:57:53.210 --> 00:57:57.010 align:middle line:84% If you now restrict this interval, 00:57:57.010 --> 00:58:02.530 align:middle line:84% make it sufficiently small, then the argument of Picard-Lindelof 00:58:02.530 --> 00:58:05.470 align:middle line:84% tells you that you had this unique fixed point. 00:58:05.470 --> 00:58:09.220 align:middle line:84% So you had a unique solution on this little interval. 00:58:09.220 --> 00:58:11.200 align:middle line:84% But since they coincide at this point, 00:58:11.200 --> 00:58:14.620 align:middle line:84% it means that they must be the same in this little interval. 00:58:14.620 --> 00:58:21.500 align:middle line:84% So you see that now we have the condition for the lemma that 00:58:21.500 --> 00:58:24.420 align:middle line:84% says that so you have these two continuous functions, 00:58:24.420 --> 00:58:26.440 align:middle line:84% and they were equal at one point. 00:58:26.440 --> 00:58:28.660 align:middle line:84% And if they're equal at a point, then they're 00:58:28.660 --> 00:58:31.420 align:middle line:90% equal in a whole neighborhood. 00:58:31.420 --> 00:58:34.940 align:middle line:84% So these two functions, these two solutions 00:58:34.940 --> 00:58:42.340 align:middle line:84% are equal on the interval where they're both defined. 00:58:42.340 --> 00:58:45.300 align:middle line:90% Now, so now you might-- 00:58:45.300 --> 00:58:48.040 align:middle line:84% and then I want to do a little bit of review at the end 00:58:48.040 --> 00:58:51.200 align:middle line:84% and at least say something about the final. 00:58:51.200 --> 00:58:53.420 align:middle line:84% But before we get to that, let me just 00:58:53.420 --> 00:59:00.020 align:middle line:84% make one more observation about the Picard-Lindelof. 00:59:00.020 --> 00:59:06.150 align:middle line:84% So you may say that in the proof of the Picard-Lindelof, 00:59:06.150 --> 00:59:09.030 align:middle line:84% we kind of had to restrict the integral. 00:59:09.030 --> 00:59:10.910 align:middle line:90% Is that really necessary? 00:59:10.910 --> 00:59:12.370 align:middle line:90% Why should we restrict? 00:59:12.370 --> 00:59:15.150 align:middle line:84% Why should we only be able to say 00:59:15.150 --> 00:59:18.470 align:middle line:84% that we had a solution in a very small interval? 00:59:18.470 --> 00:59:21.570 align:middle line:84% And so if you're looking at the example, 00:59:21.570 --> 00:59:28.390 align:middle line:84% suppose you're looking at the example where f-- 00:59:28.390 --> 00:59:37.350 align:middle line:84% so f of y is equal to y squared, g of x is equal to zero. 00:59:37.350 --> 00:59:41.270 align:middle line:84% And so you're looking at the array solutions here. 00:59:41.270 --> 00:59:42.270 align:middle line:90% Why? 00:59:42.270 --> 00:59:48.510 align:middle line:84% So that y prime is equal to y squared, 00:59:48.510 --> 01:00:01.330 align:middle line:84% and let's say equal to a for x equal to x zero. 01:00:01.330 --> 01:00:05.800 align:middle line:84% So suppose you're looking at an array like this. 01:00:05.800 --> 01:00:11.960 align:middle line:84% So at this x zero, you want the value of the function to be-- 01:00:11.960 --> 01:00:20.320 align:middle line:84% so you want-- sorry-- y of x zero equal to-- 01:00:20.320 --> 01:00:22.860 align:middle line:84% maybe this is a little bit better notation. 01:00:22.860 --> 01:00:29.760 align:middle line:84% So I want a function so that y prime is equal to y squared. 01:00:29.760 --> 01:00:33.880 align:middle line:84% And I want y of x zero equal to a. 01:00:33.880 --> 01:00:38.840 align:middle line:90% 01:00:38.840 --> 01:00:45.960 align:middle line:84% Now, let's say that x zero and a are both positive. 01:00:45.960 --> 01:00:54.400 align:middle line:84% So if you're looking at this array, then you can just-- 01:00:54.400 --> 01:01:00.690 align:middle line:84% this one here is easy to solve, because you have a function. 01:01:00.690 --> 01:01:05.440 align:middle line:84% You have a function y prime equal to y squared. 01:01:05.440 --> 01:01:09.010 align:middle line:90% 01:01:09.010 --> 01:01:11.250 align:middle line:84% And now let's try to find a solution 01:01:11.250 --> 01:01:14.210 align:middle line:90% in a neighborhood of x zero. 01:01:14.210 --> 01:01:16.770 align:middle line:84% But there it's supposed to be positive. 01:01:16.770 --> 01:01:19.290 align:middle line:84% So in a small neighborhood, it's positive. 01:01:19.290 --> 01:01:21.710 align:middle line:90% So I can divide by y squared. 01:01:21.710 --> 01:01:24.370 align:middle line:84% So I can looking at it so this thing here 01:01:24.370 --> 01:01:31.570 align:middle line:84% is equivalent to that y prime over y squared is equal to 1. 01:01:31.570 --> 01:01:40.050 align:middle line:84% And this thing here I can write as 1 over y prime times minus 1, 01:01:40.050 --> 01:01:41.450 align:middle line:90% this thing here. 01:01:41.450 --> 01:01:44.930 align:middle line:84% If you're looking at this function here by the chain rule, 01:01:44.930 --> 01:01:47.130 align:middle line:90% it's equal to that. 01:01:47.130 --> 01:01:48.810 align:middle line:84% And so this means that now you can 01:01:48.810 --> 01:01:51.650 align:middle line:84% apply the fundamental theorem of calculus. 01:01:51.650 --> 01:01:54.650 align:middle line:84% And if you're integrating this function here 01:01:54.650 --> 01:01:59.900 align:middle line:84% and you're integrating it from its two values 01:01:59.900 --> 01:02:08.420 align:middle line:90% x zero to x, d of s-- 01:02:08.420 --> 01:02:09.670 align:middle line:90% sorry, this is the prime-- 01:02:09.670 --> 01:02:12.940 align:middle line:90% 01:02:12.940 --> 01:02:17.600 align:middle line:84% then what you get is that this thing here is equal to, well, 01:02:17.600 --> 01:02:20.240 align:middle line:84% on one hand, by the fundamental theorem of calculus, 01:02:20.240 --> 01:02:23.950 align:middle line:90% it's 1 over y of x minus-- 01:02:23.950 --> 01:02:25.200 align:middle line:90% and then there's a minus sign. 01:02:25.200 --> 01:02:28.180 align:middle line:84% Sorry, this here, because of this minus sign. 01:02:28.180 --> 01:02:34.780 align:middle line:84% So it's minus this plus 1 over y of x zero. 01:02:34.780 --> 01:02:37.340 align:middle line:84% And on the other hand, it's equal to this. 01:02:37.340 --> 01:02:41.240 align:middle line:84% And when you're integrating this on this interval here, 01:02:41.240 --> 01:02:47.784 align:middle line:84% so it's also equal to x zero x, and then 1ds. 01:02:47.784 --> 01:02:50.340 align:middle line:90% 01:02:50.340 --> 01:02:54.180 align:middle line:84% This thing here is just x minus x zero. 01:02:54.180 --> 01:02:57.710 align:middle line:90% x minus x zero. 01:02:57.710 --> 01:03:02.150 align:middle line:84% So if you solve it, then let me just write the result. 01:03:02.150 --> 01:03:05.510 align:middle line:84% If you check what this gives you, 01:03:05.510 --> 01:03:11.230 align:middle line:84% this gives you that when you solve it like this, 01:03:11.230 --> 01:03:14.230 align:middle line:90% you're getting-- 01:03:14.230 --> 01:03:17.110 align:middle line:90% you're getting that y here is-- 01:03:17.110 --> 01:03:20.790 align:middle line:90% so you have this array here. 01:03:20.790 --> 01:03:23.910 align:middle line:84% And you're assuming that these things here are positive, 01:03:23.910 --> 01:03:30.910 align:middle line:84% and then the solutions are of the form y of x equal to a, 01:03:30.910 --> 01:03:36.730 align:middle line:90% over 1 plus ax zero minus x. 01:03:36.730 --> 01:03:40.340 align:middle line:90% 01:03:40.340 --> 01:03:43.870 align:middle line:84% And so what you should observe here is that this thing 01:03:43.870 --> 01:03:53.350 align:middle line:84% here, as this is not defined, so this is note 01:03:53.350 --> 01:04:05.440 align:middle line:84% that y is not defined on all of R. 01:04:05.440 --> 01:04:14.560 align:middle line:84% But instead, as x here goes to x zero plus 1 01:04:14.560 --> 01:04:21.640 align:middle line:84% over a, as x zero goes to this, this thing here goes to minus 1, 01:04:21.640 --> 01:04:26.520 align:middle line:84% so the denominator goes to zero, so the function blows up. 01:04:26.520 --> 01:04:31.320 align:middle line:84% So this here shows you that it does, 01:04:31.320 --> 01:04:35.040 align:middle line:84% in a neighborhood of this point, it has a solution. 01:04:35.040 --> 01:04:37.300 align:middle line:90% But that neighborhood is not-- 01:04:37.300 --> 01:04:40.480 align:middle line:84% it's some relatively small neighborhood, and the function, 01:04:40.480 --> 01:04:45.960 align:middle line:84% you cannot extend it past s, past s. 01:04:45.960 --> 01:04:49.560 align:middle line:84% A certain value, you can't extend it past. 01:04:49.560 --> 01:04:51.960 align:middle line:84% In this case, you can't extend it past this point 01:04:51.960 --> 01:04:58.218 align:middle line:84% here, so to the left of this point 01:04:58.218 --> 01:05:01.640 align:middle line:84% to the-- sorry-- to the right of that point. 01:05:01.640 --> 01:05:05.830 align:middle line:90% 01:05:05.830 --> 01:05:10.010 align:middle line:84% So now, I want to say a little bit about the final. 01:05:10.010 --> 01:05:11.170 align:middle line:90% Yeah, sorry. 01:05:11.170 --> 01:05:13.130 align:middle line:84% AUDIENCE: Why is it that you can't-- 01:05:13.130 --> 01:05:16.370 align:middle line:84% once you reach that point where it blows up, 01:05:16.370 --> 01:05:19.150 align:middle line:84% it never solves the differential equation past that point? 01:05:19.150 --> 01:05:20.608 align:middle line:84% TOBIAS COLDING: Yeah, that's right. 01:05:20.608 --> 01:05:23.870 align:middle line:90% So the thing is, so right. 01:05:23.870 --> 01:05:28.970 align:middle line:84% So this here, this array is going to force it to-- 01:05:28.970 --> 01:05:32.850 align:middle line:84% once this here, once y become large, 01:05:32.850 --> 01:05:35.190 align:middle line:84% then y become really, really large. 01:05:35.190 --> 01:05:38.470 align:middle line:84% So it becomes large too quickly, so it becomes quickly. 01:05:38.470 --> 01:05:42.470 align:middle line:84% So it will give you some of array like this 01:05:42.470 --> 01:05:46.090 align:middle line:84% that you have some critical value where you couldn't extend 01:05:46.090 --> 01:05:52.340 align:middle line:84% it past that, because it will have a tangent that 01:05:52.340 --> 01:05:54.940 align:middle line:90% is going vertical like this. 01:05:54.940 --> 01:05:58.120 align:middle line:90% And so you can't-- 01:05:58.120 --> 01:06:01.260 align:middle line:90% so this means that there's-- 01:06:01.260 --> 01:06:01.880 align:middle line:90% that's right. 01:06:01.880 --> 01:06:08.740 align:middle line:84% So you have a solution that exists on some interval, 01:06:08.740 --> 01:06:10.680 align:middle line:84% but it doesn't exist past this point. 01:06:10.680 --> 01:06:15.940 align:middle line:84% So you can't have a solution on some interval 01:06:15.940 --> 01:06:19.460 align:middle line:90% larger than this to the left. 01:06:19.460 --> 01:06:22.640 align:middle line:84% And there's probably a similar phenomenon on the other side. 01:06:22.640 --> 01:06:27.580 align:middle line:90% 01:06:27.580 --> 01:06:33.260 align:middle line:84% And it's a kind of standard problem, standard issue, 01:06:33.260 --> 01:06:38.060 align:middle line:84% that for differential equations, you can get blowups like this. 01:06:38.060 --> 01:06:40.340 align:middle line:84% That is a particular simple example. 01:06:40.340 --> 01:06:41.480 align:middle line:90% Any other questions? 01:06:41.480 --> 01:06:45.700 align:middle line:90% 01:06:45.700 --> 01:06:46.620 align:middle line:90% OK. 01:06:46.620 --> 01:06:48.410 align:middle line:90% OK, great. 01:06:48.410 --> 01:06:50.000 align:middle line:90%