# regularity of the Green's function vs. regularity of the solution

In Friedmans book there is analysis for the pde which is done via the fundamental solution. As I understand if we integrate that with initial data it gives us a solution of an equation. There are also results on regularity of the fundamental solution but I wonder if that has affect on the regularity of the solution itself? Is there a direct dependence? How smoothness of the fundamental solution affects the smoothness of a solution? thanks!

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Since the general solution of $Lf=g$ is the convolution of $g$ with the fundamental solution $\varphi$, we should expect that better regularity of $\varphi$ results in better regularity of $f=\varphi*g$. For example, if $\varphi\in L^p$ for some $p$, then $\varphi*g\in L^p$ for all $g\in L^1$, by Young's inequality. Of course, the more interesting question is the smoothness of solution rather than its integrability. Unfortunately, here the situation is complicated by the nature of $\varphi$: it is usually smooth except at one point where there is a singularity. One has to be careful estimating the contribution of the singularity to the solution. So, the relation between the smoothness of $\varphi$ and $\varphi*g$ cannot be described in broad terms that apply to all PDE.
@Leoned:singularity that you have mentioned occurs when $t=0$, since the density becomes a delta function, but for other $t>0$, would not partial derivative $u_x=\int \phi_xgdy$? Then the differentiability of $\phi$ would imply differentiability of $u$, would not it? –  Medan Jun 24 '12 at 23:58
@Medan We have $u(x)=\int\phi(x-t)g(x)\,dt=\int\phi(t)g(x-t)\,dt$. For every $x$ the integral involves contribution from the part of $\phi$ where it is singular. For example, the function $\phi(t)=\log|t|$ is $C^{\infty}$ outside of $0$, but its convolution with a continuous function $g$ is not even $C^2$ in general. –  user31373 Jun 25 '12 at 13:19
:I think it should be written as $u(x)=\int\phi(x-y)g(y)dy$? where the same for heat equation is $u(x,t)=\int\phi(x-y,t)g(y)dy$ which is singular only at $t=0$. Thus, we define it everywhere but $t=0$ to avoid those problems, then the regularity of $u(x,t)$ is the same as regularity of $\phi(x-y,t)$ w.r.t. to $x,t$. Where my logic is wrong? –  Medan Jun 25 '12 at 15:04