A solution to the nonhomogeneous heat equation with $0$ initial data is $u:\Bbb R^n\times [0,\infty)\to \Bbb R$ solving
$$\begin{align}
\partial_t u(x,t) – \Delta_xu(x,t) &= f(x,t) \\
u(x,0) &\equiv 0.
\end{align}$$
The regularity of $f$ is unspecified for now.Recall the heat kernel $\Phi(x,t)=\frac 1{(4\pi t)^{n/2}}e^{-|x|^2/4t}$. The Duhamel's principle states that
$$
u(x,t)=\int_0^t\int_{\Bbb R^n} \Phi(x-y,t-s)f(y,s)dyds
$$
solves our equation.
In books I've read, e.g. Evans' PDE, assume quite a strict condition on the regularity of $f$. For example, in Evans' book he assumed that $f\in C^2_1(\Bbb R^n\times[0,\infty))$ and is compactly supported. Then the proof that $u$ really solves the heat equation or the smoothness of $u$ is shown by rewriting the formula as
$$
u(x,t)=\int_0^t\int_{\Bbb R^n} \Phi(y,s)f(x-y,t-s)dyds
$$
and differentiate under integral sign, e.g.
$$
\partial_t u(x,t)=\int_0^t\int_{\Bbb R^n} \Phi(y,s)\partial_t f(x-y,t-s)dyds+\int_{\Bbb R^n}\Phi(y,t)f(x-y,0)dy.
$$
To show smoothness of $u$, some even assume $f$ to be smooth. I can't help but feeling that this is a massive overkill. Some integrability condition on $f$ should suffice.
What are some reasonable integrability conditions we can put on $f$ so that
$$
u(x,t)=\int_0^t\int_{\Bbb R^n} \Phi(x-y,t-s)f(y,s)dyds
$$
works? What should we impose further to get higher regularity of $u$?
Of course, if $f$ is really irregular we cannot expect $u$ to be regular. I just want to know how much regularity do we gain from the fact that $u$ solves heat equation.
Best Answer
In Folland's Introduction to PDE's, Theorem 4.7 states that if $f\in L^1(\Bbb R^n\times\Bbb R)$, then $u$ is a solution in the distribution sense. In a remark following the proof of the theorem, it is stated that if $f\in C^\alpha$ for some $\alpha>0$, then $u$ is a classical solution. No proof is given, only a comment saying that the proof is similar to the one of a previous theorem.