Let $\Omega \subset \mathbb R^n$, $n \in \mathbb N$, be a smooth bounded domain, $\mathcal A$ an elliptic operator (for instance $\mathcal A = - \Delta$), $u_0 \in C^0(\overline \Omega)$ and $g \in C^\alpha(\partial \Omega)$ for some $\alpha \in (0, 1)$. What is known about the equation

$$\begin{cases} u_t + \mathcal A u = 0 & \text{in $(0, \infty) \times \Omega$}, \\ u = g & \text{in $(0, \infty) \times \partial \Omega$}, \\ u(\cdot, t) = u_0 & \text{in $\Omega$} \end{cases}$$

(or also about the corresponding elliptic equation)?

If $g \in C^{2+\beta}(\partial \Omega)$ for some $\beta \in (0, 1)$, then one can extend $g$ to a function $\tilde g \in C^{2+\beta}(\overline \Omega)$ and consider $u-\tilde g$ instead of $u$. Of course, this is no longer possible for less regular $g$.

However, I would still expect that the problem above (or at least the elliptic version) has a (unique?) classical solution – which may even be $C^\alpha$ up to the boundary (but of course not more).

I guess this has been treated somewhere and hence I am thankful for any references.


1 Answer 1


I think Theorem 9, Chapter 3.4, in Avner Friedman's book, Partial Differential Equations of Parabolic Type, might help you. Basically it says that with suitable assumptions on $\mathcal{A}$ and the boundary of $\Omega$, you get a unique solution that is in $C^{2+\alpha}$ for a $0<\alpha<1$, but only in the interior of $\Omega\times(0,T)$.

For the proof you don't even need that $g\in C^\alpha$. It is enough that $u=\psi$ on the parabolic boundary $\Omega\times\{t=0\} \cup \partial\Omega\times(0,T)$, where $\psi$ is a continuous function on that boundary. For a bounded domanin $\Omega$ we can for example apply Tietze's extension theorem to get a $\psi\in C(\bar{\Omega}\times[0,T]$). Let $N$ be the rectangle in $\mathbb{R}^{d+1}$ that contains $\bar{\Omega}\times[0,T]$. By the Weierstrass approximation theorem there exists a sequence of polynomilas $\psi_k$ that approximate $\psi$ uniformly in $N$. As you already suggested, you can now apply interior parabolic schauder estimates to $u-\psi_k$. You get a sequence of solutions $u_k$, by interior parabolic Schauder estimates you can show that $u_k$ converges uniformly to a $u$. $u$ is continuous in $\bar{\Omega}\times[0,T]$ and $u\in C^{2+\alpha}$ locally in $\Omega\times(0,T)$.

Tl;dr, you got the right idea. You consider $u-g$, but since $g$ is not differentiable, you approximate $g$ by a sequence of polynomials $g_k$. Interior Schauder estimates show that the corresponding sequence $u_k$ converges uniformly to your solution $u$.

If you want $u\in C^{2+\alpha}(\bar{\Omega}\times[0,T])$, i.e. differentiability up to the boundary, then you need that the boundary condition $\psi$ is in $C^{2+\alpha}$, see for instance Ladyzenskaja-Solonnikov-Ural'ceva's book Linear and Quasi-linear Equations of Parabolic Type, Theorem 5.2,Chapter IV Section 5.

  • $\begingroup$ That makes a lot of sense, thanks! Which of his books do you refer to? “Partial differential equations“ or “Partial differential equations of parabolic type”? $\endgroup$
    – Keba
    Commented Nov 2, 2020 at 15:12
  • $\begingroup$ And, secondly, is there any hope for a uniqueness theorem? $\endgroup$
    – Keba
    Commented Nov 2, 2020 at 15:12
  • $\begingroup$ Ah, you named the title of the book, sorry. $\endgroup$
    – Keba
    Commented Nov 2, 2020 at 15:23
  • $\begingroup$ @Keba It is "Partial differential equations of parabolic type". Yes, the theorem that I am referring to states that $u$ is unique. $\endgroup$
    – SC2020
    Commented Nov 2, 2020 at 15:24
  • $\begingroup$ Thanks again. Will try to get the book then. $\endgroup$
    – Keba
    Commented Nov 2, 2020 at 15:33

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