The unique existence of a harmonic function $u$ that agrees with a continuous function $h$ on the smooth boundary of a domain

Question

I'm reading below statement at page 36 of these notes, i.e.,

Fact. Let $D$ be an open and bounded domain in $\mathbb{R}^n$ and $\partial D$ be its (smooth) boundary. Let $h \in \mathcal{C}(\partial D)$. Then there exists a unique function $u \in \mathcal{C}^2(D)$ such that $$ \begin{cases} \Delta u(\underline{x})=0, & \forall \underline{x} \in D, \\ u(\underline{x})=h(\underline{x}), & \forall \underline{x} \in \partial D . \end{cases} $$

Remark. The function $u$ takes values in $\mathbb{R}$ (not in $\mathbb{R}^n$ ).

Could you confirm if my below understanding is fine?

$h:\partial D \to \mathbb R$ is continuous.
$u: \overline D \to \mathbb R$ is twice continuously differentiable. Here $\overline D$ is the closure of $D$ in $\mathbb R^n$.

score 2 · Accepted Answer · answered Mar 13 '23 at 23:11

2

Yes.
No, $u$ is twice continuously differentiable in $D$ (i.e. in its interior), but only continuous on the boundary $\partial D$. Thus, one usually writes $u\in C^2(D)\cap C(\overline{D})$.

Of course, you could assume $u\in C^2(\overline{D})$, but it is not necessary for the theory of this type of PDEs.

answered Mar 13 '23 at 23:11

Symplectic Witch

3,106

1

I think you meant $u\in C(D)\cap C^2(\overline{D})$ rather than $u\in C^2(D)\cap C(\overline{D})$. – Akira Mar 13 '23 at 23:26
@Akira No, that would be a trivial statement: $C^2(\overline{ D})$ is a subset of $C(D)$, so their intersection would be just $C^2(\overline{ D})$. – Symplectic Witch Mar 13 '23 at 23:29
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Ah, I'm sorry for being silly. – Akira Mar 13 '23 at 23:31
Could you have a look at my related question here? – Akira Mar 14 '23 at 01:01

The unique existence of a harmonic function $u$ that agrees with a continuous function $h$ on the smooth boundary of a domain

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