Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$ with smooth boundary, and let $T>0$. Consider the non-homogeneous heat equation with Dirichlet boundary condition

$$\begin{aligned} u_t - \Delta u &= f & &\text{in }\Omega\times(0,T), \\ u &= 0 & &\text{on } \partial\Omega\times(0,T), \\ u(x,0) &= u_0(x) & &\text{for all } x \in \Omega.\end{aligned}$$

Suppose that $f \in L^\infty(\Omega\times(0,T))$ and $u_0 \in L^\infty(\Omega)$. Is it true that there exists a solution $u$ to the equation above such that $u \in C^{2,1}(\bar{\Omega}\times(0,T)) \cap L^\infty(\Omega\times(0,T))$ and $\lim_{t \rightarrow 0} u(x,t) = u_0(x)$ for a.e. $x\in \Omega$.

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First solution attempt. I tried to follow Section 2.3.1 in Evans book on pdes. But we only look at $\Omega \subset \mathbb{R}^n$, therefore we will use the Green function of the heat equation with Dirichlet boundary condition $G$ instead of the fundamental solution $\Phi$ that is used in the book. As in the book we split the problem into a homogeneous part with $u_0$ as initial data and a nonhomogeneous part with $0$ as initial data. Theorem 1 on page 47 in Evans should give us a solution for the homogeneous part. The proof for (i) and (ii) should still work for initial data in $L^\infty$, hence

$$u(x,t) = \int_\Omega G(x,y,t) u_0(y) \, \mathrm{d}y$$

is a smooth solution for the homogeneous part. Of course with noncontinuous initial data we can't expect (iii) to be true.

For the nonhomogeneous part we define

$$u(x,t) = \int_0^t\int_\Omega G(x,y,t-s) f(y,s) \, \mathrm{d}y \mathrm{d}s.$$

The problem is that in this case the regularity of $u$ doesn't follow straightforward from the regularity of $G$, because $G$ has a singularity at $t=0$, thus we can't differentiate under the integral. The proof of Theorem 2 on page 50 in Evans book assumes that $f \in C^{2,1}(\Omega\times(0,T))$ and that $f$ has compact support. The proof of Theorem 2 as presented by Evans doesn't work with $f \in L^\infty$. However, in Evans book it says that $f \in C^{2,1}(\Omega\times(0,T))$ with compact support is assumed for simplicity. The question is now if it is still possible to prove Theorem 2 with the assumption that $f \in L^\infty$ or does this approach just not work.


1 Answer 1


After writing down my first solution attempt I realized that it probably is not possible to find a regular $u \in C^{2,1}$ with just $f\in L^\infty$. The reason for this is that $f$ represents a heat source independent of $u$, thus we can't expect $u$ to be in $C^{2,1}$, if the outside heat source $f$ is non continuous.

Additionally, someone pointed out to me that if our solution $u$ is actually in $C^{2,1}$ then obviously $u_t - \Delta u \in C^0$. Hence, we need $f$ to be at least in $C^0$ if we want a solution $u \in C^{2,1}$. So we might be able to prove Evans Theorem 2 with $f \in C^0$ instead of $f\in C^{2,1}$. Nonetheless, we can say that the answer to my original question is no, because $u \in C^{2,1}$ implies $f \in C^0$ and hence $f \in L^\infty$ is not enough for the existence of a regular solution.


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