## Derivative of certain integrals - Evans PDE

Hi All,

In PDE by Evans, he gives the fundamental solution to the heat equation for t>0

$\displaystyle \Phi(x,t)=\frac{1}{(4\pi t)^\frac{n}{2}}e^{-\frac{|x|^2}{4t}}$

Now he reasons that the convolution

$\displaystyle u(x,t)=\int_{\mathbb{R}^n}\Phi(x-y,t)g(y)\,dy$ should solve the IVP with $\displaystyle u=g$ on $\displaystyle \mathbb{R}^n \times \{t=0\}$.

A lot in chapter two he makes a big fuss (a justified one, I'm sure...) about bringing derivatives underneath the integral sign (for example with the solution to Laplace's equation earlier on in the chapter...)

In the case I'm talking about however, he has no trouble down the page saying that

$\displaystyle u_t(x,t)-\Delta u(x,t) = \int_{\mathbb{R}^n}[(\Phi_t - \Delta_x \Phi)(x-y,t)]g(y)\,dy$

He says that $\displaystyle \Phi$ is infinitely differentiable with uniformly bounded derivatives of all orders... so is this the reason why he can bring the derivative wrt $\displaystyle t$ and the laplacian with respect to $\displaystyle x$ inside the integral in that case? I know there are theorems from measure theory that guarantee you can do this - if that is his reasoning, I am fine with it.

However, I get confused when he then goes on to show

$\displaystyle u(x,t):=\int_0^t \int_{\mathbb{R}^n}\Phi(x-y,t-s)f(y,s)\,dy\,ds$ solves the non-homogenous problem. He says:

"Since $\displaystyle \Phi$ has a singularity at (0,0), we cannot directly justify differentiating under the integral sign. We instead proceed somewhat as in the proof of theorem 1 in 2.2.1..."

Okay, first of all, which derivative is he talking about?! Have we run in to problems now because we are integrating wrt to $\displaystyle s$ from 0 to $\displaystyle t$ as well?

I don't really understand this... is anyone able to shed any light for me on the situation?

Thanks!