Wave equation on a compact Riemannian surface without boundary: no mass conservation? Consider a compact, smooth Riemmanian surface $\mathcal{S} \subset \mathbb{R}^3$ without boundary.
I would like to solve the wave equation:
$$u_{tt} + \Delta_{\mathcal{S}} u = 0$$
under the conditions:
$$u(0,x) = \delta_y(x)\\ u_t(0,x) = 0,$$
where $\Delta_{\mathcal{S}}$ is the Laplace-Beltrami operator on $\mathcal{S}$.
If I am not mistaken (please let me know if I am), the solution can be explicitly written down in terms of the Laplace-Beltrami eigenfuctions $\{\phi_k\}_{k\geq 1}$, which form a basis of $L^2(\mathcal{S})$, and the corresponding eigenvalues $\lambda_k$:
$$u(t,x) = \sum_{k=1}^{\infty} \cos(\sqrt{\lambda_k}t) \phi_k(y) \phi_k(x)$$
Now, what surprises me, is that "mass" is not conserved in the sense that
$$\frac{\partial}{\partial t} \int_{\mathcal{S}} u(x,t) dx = \sum_{k=1}^{\infty} -\sqrt{\lambda_k}\sin(\sqrt{\lambda_k}t) \phi_k(y) \int_{\mathcal{S}}\phi_k(x) dx$$
So "mass" changes over time depending on the $\lambda_k$ and thus on the geometry of the surface.
Question: Is there an intuitive explanation of why $\int_{\mathcal{S}} u(x,t) dx$ is not a conserved quantity?
 A: For generic initial data, the mass is never a conserved quantity for wave equations (except sometimes in 1+1 dimensions). Not even over Euclidean space. (In $\mathbb{R}^3$ for spherically symmetric, out-going initial data you can check explicitly that the mass grows at a rate that is about linearly in time.)
So there is absolutely no reason why you should expect that mass is conserved for general evolution on a Riemannian manifold. 
(Are you perhaps thinking of energy? (Or the mass in the $L^2$ sense?) That would be the natural conserved quantity for the wave equation on Euclidean space, and indeed, they are conserved also in the case you replace the space by a Riemannian manifold.)

For the specific case you are looking at, where the initial data has $u_t(0,x) = 0$, indeed you have conservation of total integral. And you've also proven it already in your question: eigenfunctions of the Laplace-Beltrami operator with non-zero eigenvalues have mean zero (integrate both sides of $-\triangle \phi_k = \lambda_k \phi_k$). 
