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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.

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    $\begingroup$ Generally $u$ gains 2 degrees more differentiability than $f$. If $f \in H^k$ then $u\in H^{k+2}$, and if $f \in C^{k,\alpha}$ then $u \in C^{k+2,\alpha}$. The correct place to look for the Sobolev space theory is Chapter 7 in Evans (regularity for parabolic equations). You can't do any better than this (If $u \in C^k$ then clearly $f \in C^{k-2}$, so you can't have a situation where $f$ has low regularity and $u$ is smooth). $\endgroup$ – Jeff Jan 18 '18 at 22:31
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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.

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  • $\begingroup$ Very interesting, thank you. Just to be sure, did you mean $f\in C^{\alpha}$ as a function in $\Bbb R^{n+1}$? $\endgroup$ – BigbearZzz Jan 18 '18 at 19:37
  • $\begingroup$ The book does not say. $\endgroup$ – Julián Aguirre Jan 19 '18 at 18:44
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    $\begingroup$ @BigbearZzz it is enough that $f\in C^{\alpha,0}_{x,t}$ meaning Holder wrt $x$ and continuous in $(x,t)$ (plus in $L_1$ or something for integral to converge). $\endgroup$ – Andrew Feb 2 '18 at 11:29

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