Dear students, hello, I hope you had a nice Christmas vacation.
So let us come to the solutions to the seventh exercise sheet.
We start by recalling some usual basic definition of notation.
The only important thing is that in our, according to our notation, the parabolic domain is open,
so it's simply the cross product between u and the interval 0, capital T,
and the parabolic boundary is defined as usual, so basically it's the boundary
without the upper layer u cross capital T, time t.
And this is important because in the maximum principle, the only part of the chain,
the maximum is actually achieved on this part of the boundary.
It can be achieved also inside, but the maximum only on this part of the boundary is equal to the maximum
of the closure of the full set, ut.
The important thing is that in this way, when we are talking about a function in C2, 1, ut,
we are talking about a function which is twice differentiable in x and 1s in t on an open set.
Evans' book uses another notation where here we are considering time t,
so here there is a square bracket. It's a little weird to me.
I understand that in this way gamma t is defined, a little simpler, but I prefer to have
differentiable functions defined on open sets to avoid the complicancies with the boundary values.
So this is the heat equation in ln, and the fundamental solutions for the heat equation
are given by the function phi xt, which is 1 over 4 pi t to the n over 2 e to the minus
norm of x squared over 40.
And so the function which is given by the convolution between phi time t and the initial
d2f is a smooth solution of the initial value problem for the heat equation.
First exercise basically considers some nice scaling property of the heat equation.
So suppose u is smooth and solves the heat equation in a range times 0 plus infinity.
Then you have to show that the function u ln x t, defined as u of ln x ln squared t,
is also a solution for all lambda non-zero, because 0 doesn't make sense because time
must be positive. And then you can use this to prove that the function vxt, defined as
x scalar product u plus 2t u t, is also a solution.
And of course one can verify this, but the task was used point A.
How to do it? Well, it's obvious that the derivative in time of u ln is simply ln squared
u t evaluated in ln x ln squared t, the gradient of u ln is ln, the gradient of u evaluated
in ln x ln squared t and the Laplacian therefore gives us ln squared, this is a chain rule.
And therefore u ln t minus Laplacian of u ln is equal to ln squared u t ln, because this
if we define this operation of taking the lambda of a function as taking this type
of parabolic scaling, minus lambda squared Laplacian u lambda. And so this is clearly
linear, so well done. Thanks to point A, we know that this equation is satisfied for all
lambda positive. But then u is moved, and therefore we can differentiate in lambda.
The derivative in lambda of u lambda evaluated in xt is x scalar product u, because again,
chain rule first differentiate in the first term. Remember that x is a vector, but one
could also do component by component, plus twice lambda t u t evaluated in lambda x,
lambda squared t, because the derivative of lambda squared is true lambda. But therefore
what is v is simply the derivative of u in lambda, of u lambda, sorry, in lambda evaluated
lambda equal to 1. No, simple. Therefore the trick is simply that you consider this equation,
the heat equation for u lambda is satisfied for every lambda, therefore it's satisfied
also for the incremental ratio because, in lambda, because it's a linear equation. So
we can differentiate in lambda, set lambda equal to 1, and we're done.