WEBVTT - autoGenerated
00:00:00.000 --> 00:00:25.000
The second exercise concerns some calculation in one dimension, so if you assume that n is equal to 1 and uxt is v of x over square root of t for some vc2, then ut is equal to uxx, so u solves the heat equation in one dimension, if and only if v solves the equation, v2 plus z over 2v prime equal to 0.
00:00:25.000 --> 00:00:35.000
Then you have to show that the general solution is v of z equal to c, integral from 0 to z, e to the minus s squared over 4 in the s plus d.
00:00:35.000 --> 00:00:47.000
Then you have to differentiate this function u that you find with respect to x and select a proper constant c so that you get the fundamental solution in one dimension.
00:00:47.000 --> 00:00:56.000
And finally the optional part was to explain why this funny procedure actually gives you the fundamental solution.
00:00:56.000 --> 00:01:08.000
So part a was quite simple, I mean the derivative in time of uxt is simply v prime evaluated in x over square root of t times minus x over 2 t to the power 3 over 2.
00:01:08.000 --> 00:01:21.000
The first derivative in x of u is v prime of x over square root of t times 1 over square root of t and therefore the second derivative is simply the second derivative of v evaluated in x over square root of t times 1 over t.
00:01:21.000 --> 00:01:36.000
Therefore, ut is equal to uxx, if and only if, v prime of z times minus z over 2 t is equal to v second of z over t, where we set z to be x over square root of t, which is a real number so that makes sense.
00:01:36.000 --> 00:01:40.000
But this is the equation that we have here.
00:01:40.000 --> 00:01:59.000
Now, how to find the general solution, define v prime, I mean call v prime w, then w prime is equal to minus z over 2 w, but therefore the solution of this is simple, is some constant times e to the minus z squared over 4, the unique solution.
00:01:59.000 --> 00:02:06.000
And therefore v is simply given by the integration plus an arbitrary constant.
00:02:06.000 --> 00:02:18.000
As for point c, notice that uxt is a constant, the integral from 0 to x over square root of t e to the minus z squared over 4 in the s plus d, no?
00:02:18.000 --> 00:02:24.000
u is v of x over square root of t, so insert x over square root of t in the definition.
00:02:25.000 --> 00:02:40.000
Take a derivative, taking a derivative means we get c e to the minus x squared over 4t divided by square root of t, because we are taking a derivative in x, so we have to differentiate also this term over here, the end point of the integral.
00:02:40.000 --> 00:02:51.000
So that when you see that this is really the right form, it will be equal to the fundamental solution if we choose c to be 1 over square root 4 pi.
00:02:51.000 --> 00:02:58.000
Why? That's always the interesting question in math. Why do we get what we get?
00:02:58.000 --> 00:03:08.000
And the hint was, what is the initial condition for ux? And there are many ways to see it.
00:03:08.000 --> 00:03:12.000
One is, well, what is the initial condition for u?
00:03:12.000 --> 00:03:16.000
That means if I send t to 0+, what happens to uxt?
00:03:16.000 --> 00:03:24.000
Well, by definition, if x is positive, x over square root of t is going to plus infinity, the Gaussian function is clearly summable.
00:03:24.000 --> 00:03:32.000
So here I didn't specify, but everything works by Lebesgue, of course, because the Gaussian function is summable by one dimension, everything is fine.
00:03:32.000 --> 00:03:39.000
So here we get this number, c integral 0 to plus infinity e to the minus square over 4 in the s plus d for x positive.
00:03:39.000 --> 00:03:50.000
If x is 0, this term vanishes and there is only d. If x is negative, the opposite, so this goes to minus infinity, so reverse the order gives a minus sign.
00:03:50.000 --> 00:03:55.000
Now, but c is 1 over square root of 4 pi, we choose it.
00:03:55.000 --> 00:04:01.000
And this integral is equal to square root of pi, because the full integral is equal to 2 square root of pi.
00:04:01.000 --> 00:04:13.000
Therefore, we get that u of x is 0 will be some constant, plus 1 half the sine of x, which is 1 for x positive, 0 for x equal to 0, minus 1 for x, negative.
00:04:13.000 --> 00:04:22.000
But then the initial condition for ux is that ux is equal to the Dirac delta, in r times 0, in the sense of distributions, of course.
00:04:22.000 --> 00:04:26.000
The Dirac delta 0 is the Dirac delta centered in the origin.
00:04:26.000 --> 00:04:34.000
Now we go to explain why this works, but first let me point out immediately that this is the right explanation,
00:04:34.000 --> 00:04:43.000
because since ux solves the initial value problem, wt equals wxx, so the heat equation, with the initial datum given by the Dirac delta,
00:04:43.000 --> 00:04:52.000
this means that ux coincides with the fundamental solution, because remember, the fundamental solution can be held alternatively and equivalently,
00:04:52.000 --> 00:05:01.000
characterized as the solution for the heat equation with initial datum, the Dirac delta centered in the origin.
00:05:01.000 --> 00:05:08.000
By the way, how to check that ux solves also the heat equation, simply take another derivative.
00:05:08.000 --> 00:05:17.000
Now, since ut is equal to ux, u is a solution, if we take a derivative in x, nothing changes, and so well done.
00:05:17.000 --> 00:05:22.000
Because u is moved, this is plane from its definition.
00:05:22.000 --> 00:05:34.000
Very well, how to check that actually the derivative of this function over here, in the sense of distribution, is the Dirac delta centered in the origin?
00:05:34.000 --> 00:05:46.000
I mean, again, this was optional, so you don't have necessarily, you don't know, but it's good to get a little practice with this kind of calculations.
00:05:46.000 --> 00:06:00.000
This means that testing the sine of x against the derivative of a C1 function with compact support over R, we should get twice the Dirac delta centered in the origin, now because it's a factor of 1.
00:06:00.000 --> 00:06:07.000
And what is this integral? Well, it's an integral from 0 to plus infinity of phi prime minus the integral from minus infinity to 0 of phi prime, no?
00:06:07.000 --> 00:06:16.000
Because the sine is 1 or minus 1. But this is trivial because it's simply phi evaluated from 0 to plus infinity minus phi evaluated from minus infinity to 0.
00:06:16.000 --> 00:06:22.000
The first one gives me minus phi of 0 because phi has compact support, so it vanished at infinity.
00:06:22.000 --> 00:06:34.000
And the second gives me plus phi of 0, but then there is a minus, so minus and minus, minus 2, phi of 0, which is minus phi against twice the Dirac delta.
00:06:34.000 --> 00:06:36.000
And so will I.