00:01
So in this question we're thinking about the error function, which has the form 2 over root pi times the integral from 0 to z, e to the minus u squared, du.
00:12
And we're asked to show that the following solves the heat equation.
00:16
U of x and t equals a plus b error function of x over.
00:21
And now it doesn't say this in the question, but i think it should.
00:25
I think it's a mis -a poorly formatted question.
00:31
There should be a square root over the 4kt.
00:35
And with that formatting issue fixed, we can go ahead and show that this solves the diffusion equation.
00:42
Now, let's have a look at the error function.
00:46
D by dz of the error function of z is going to be 2 over root pi times e to the minus z squared.
00:59
And that's just standard differentiation of an integral with respect to the endpoint.
01:05
So when we differentiate this error function, this heatless diffusion function here, what we're going to need to do is say that du by dx is going to be a, well, it's going to, well, the a is going to go away, but it's going to be b times d by dx of x over root 4kt times the derivative of the error function, which is 2 over root root pi, e to the minus x squared over 4kt, and that's just by the product rule.
01:44
So let's continue.
01:46
This is going to give us, well, the square root of 4 is 2, so we can take that 4 out of, well, no, let's just leave it there for now.
01:55
So differentiating this with respect to x, we're going to get 2b over the square root of 4 pi kt, and the 2 is going to cancel with the 4 inside the square root.
02:09
And then we've got this e to the minus x squared over 4kt.
02:14
Now let's take a second derivative, d2u by dx squared.
02:18
And this is going to give us, so we're going to get a factor of 2x over 4kt with a minus sign.
02:24
So minus 2bx over 4kt root pi kt, e to minus x squared over 4kt...