# Using the maximum principle to show that a harmonic function will be identically zero if it vanishes on the boundary

Let $$(M,g)$$ be a compact Riemannian manifold with nontrivial boundary and suppose $$u$$ is a harmonic function on $$M$$ that vanishes on $$\partial M$$. I'd like to determine whether or not $$u$$ is identically zero on $$M$$. My guess is a yes, and I'm trying to find a proof. Now, for some reasons, I'd like the maximum principle to be the cornerstone of the proof. Here is the deal.

Let $$X$$ be a smooth vector field on a Riemannian manifold $$M$$ and let $$f$$ be a smooth function on $$M$$. We define a second-order elliptic operator $$L$$ acting on functions on $$M$$ by $$L(u)=-\Delta u+\langle X,\nabla u\rangle+fu,$$ where $$\Delta$$ is the Laplacian. Then the strong maximum principle states that

Theorem. Let $$(M,g)$$ be a connected Riemannian manifold with or without boundary, and consider the operator $$L$$ with $$f\geq 0$$. Let $$u\in C^2(\mathrm{int}M)\cap C^0(M)$$ and assume that $$L(u)\leq 0$$ on $$\mathrm{int}M$$. If $$u$$ attains a nonnegative maximum value on $$\mathrm{int}M$$, or if $$u$$ attains a nonnegative maximum value at a point in $$\partial M$$, where $$\frac{\partial u}{\partial\nu}=0$$ with $$\nu$$ denoting the unit outer normal, then $$u$$ must be constant on $$M$$.

In my DG book, a smooth manifold $$M$$ is a disjoint union of $$\mathrm{int}M$$ and $$\partial M$$. So, to justify my guess mentioned earlier, I have to show that $$u$$ is zero at each interior point of $$M$$. Proceeding with proof by contradiction, I assume $$u$$ is, say, positive at some interior point $$x$$. Now, $$u$$ is continuous, so I can choose a neighborhood $$U$$ of $$x$$ on which $$u$$ is positive and attains its maximum. Applying the maximum principle shows that $$u$$ is constant on $$U$$. Then I can find a cover of $$M$$ by overlapping neighborhoods and use a patching argument to show $$u$$ is constant on all of $$M$$. Will my strategy stand a chance? Thank you.

• If $\Delta u = 0$, then by the maximum principle, $\sup_{\overline{M}}|u| = \sup_{\partial M}|u|$. So if $\Delta u = 0$ and $u|_{\partial M} = 0$, then $u = 0$. Commented Aug 7, 2022 at 18:48

I found something! My guess can really be justified by using the maximum principle. In fact, it is the weak version of this principle that completes the justification. The theorem I cited in the post is actually the strong maximum principle, from which we can deduce the weak one through an argument that requires a decomposition of the manifold into connected pieces, like I sketched in the latter part of my post.

Anyway, the weak maximum principle says that the maximum value of $$u$$ can be achieved on $$\partial M$$, where $$u$$ is, by hypothesis, identically zero. This establishes that $$u\leq 0$$. On the other hand, we can use the weak minimum principle to establish that $$u\geq 0$$, yielding the result that $$u$$ is null everywhere on $$M$$.

Honestly, I don't have an adequate knowledge of topology to carry out the decomposition into connected pieces, and this answer may not meet the standard of a complete proof. So, if someone is willing to give a piece of advice, that will be dearly great.