I have the following problem:
Let $\mathcal{M}$ be a $2D$-manifold in $\mathbb{R}^3$ and let $g$ denote its metric. Furthermore it is known that $\mathcal{M}$ is a closed manifold (i.e. it has no boundary like a sphere, torus, etc.)
I believe that above information is sufficient to prove \begin{align} \int_{\mathcal{M}} W \nabla_\mathcal{M}^2 U \, \mathrm{d} \mu & = -\int_{\mathcal{M}} \langle \nabla_\mathcal{M} W,\nabla_\mathcal{M} U \rangle_g \, \mathrm{d}\mu, \end{align} for $W \in H^1(\mathcal{M})$ and $U \in C^2(\mathcal{M})$ (I need this for a finite element simulation). Here $\langle , \rangle_g$ denotes the inner product with respect to $g$.
According to Wikipedia, above equality holds for all compactly supported functions $W$ and $U$ (source: https://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator).
I am not quite sure whether I understand Wiki correctly. Suppose $\mathcal{M}$ is just a closed subset of the $xy$-plane, then the Beltrami operator would simply reduce to the ordinary $2D$-Laplacian and the right hand side would certainly contain an additional integral over $\partial \mathcal{M}$ (if $W$ and $U$ are nonzero there).
So to me the compactness of the supports seems insufficient for above equation to hold (could somebody tell me what I am getting wrong here ?). I believe, however, that if $\mathcal{M}$ has no boundary, the boundary integral should vanish and above equation should follow.
What I have done: let $S:\Omega \rightarrow \mathcal{M}$ parametrize $\mathcal{M}$, I carried out above integral in $\Omega$ utilizing $S$. Using standard calculus identities, after some steps I arrive at \begin{align} \int_{\mathcal{M}} W \nabla_\mathcal{M}^2 U \, \mathrm{d} \mu & = \int_{\partial \Omega} \left(w \sqrt{|g|} \nabla_\mathcal{M} u \right) \cdot \mathbf{n} \, \mathrm{d}s -\int_{\mathcal{M}} \langle \nabla_\mathcal{M} W,\nabla_\mathcal{M} U\rangle_g \, \mathrm{d}\mu \end{align} where $w \equiv W \circ S$ and $\nabla_\mathcal{M} u \equiv \nabla_\mathcal{M} U \circ S$. I believe that the boundary integral on the right hand side should vanish for closed $\mathcal{M}$. The reason is that the local counterparts $w$ and $u$ of $W$ and $U$ satisfy certain continuity constraints across $\partial \Omega$ which translates to $w$ and $\nabla_\mathcal{M} u$ having the same value on two segments of $\partial \Omega$ but with $\mathbf{n}$ pointing in the opposite direction so that the boundary integral vanishes. Of course my explanation lacks mathematical rigor and I was wondering what the best way to prove this is. I believe that there exist proves that avoid integrating over $\Omega$ all together but I don't know where to start looking, also I don't want to read an entire book to understand the concept of wedge-products etc so if the formal proof is complicated could someone refer me to a source that I can cite that proofs exactly above statement ? Thank you in advance.
Yes, you are correct. Wikipedia specifies compact support to ensure convergence and that the functions are zero on the boundary. Since you have no boundary, the latter does not apply (and notice that in fact the identity you want is used near the beginning of this section). The divergence operator on a manifold is defined so that $$ \operatorname{div}(fU) = \langle \operatorname{grad} f,U \rangle_g + f \operatorname{div}{U}, $$ and integrating this over the whole manifold, the divergence theorem forces the left-hand side to vanish, and you get the identity.