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

Question

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.

Answer 1

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.