Laplacian on closed surface

Spectral Decomposition of the Laplacian

We have a Poisson problem setup as:

$$-\Delta u = f$$

where $\Delta$ is a Laplacian operator acting on a function $u$ in above. The spectral decomposition of $\Delta$ talks about functions where:

$$\Delta \phi_{i} = \lambda_{i} \phi_{i}$$

Here the $\phi_{i}$ are called as eigenmodes of the operator $\Delta$. These are functions on which the Laplacian acts as a linear scaling by $\lambda_{i}$.

Spectral Theorem

This theorem tell us that for a symmetric and self-adjoint operator, there exists real eigenvalues and eigenmodes. These eigenmodes are orthonormal under $L^2$ inner product and its eigenmode spans the entire space i.e., these eigenmodes are basis for the underlying space.

If the eigenvalues are real then we can naturally be ordered:

$$0\leq \lambda_{1} \leq \lambda_{2} \leq \lambda_{3} \leq \dots$$

A few interesting points:

• if $\lambda_{1}=0$ then the operator in non-invertable as it has some eigenmode going into the kernel of the operator. Which in-turn makes it non bijective.
• Higher values of $\lambda_{i}$ belongs to higher oscillatory eigenmodes on the domain.

Laplacian on Closed Surfaces

In our case, we often need to solve Poisson problem on a closed surface where:

$$\Delta u = f, \quad \partial\Omega = \{ \varnothing \}$$

Given that $\Delta$ is a symmetric and self-adjoint, spectral theorem tells us:

$$\Delta \phi_{i} = \lambda_{i}\phi_{i}$$

where $\phi_{i}$ and $\lambda_{i}$ are the eigenmodes and eigenvalues. Closed surface have null boundary. So integrating over the closed surface $\mathcal{M}$

$$\int_{\mathcal{M}} \Delta u = \int_{\mathcal{M}}f$$

but $\int_{\mathcal{M}}\Delta u = 0$ on a closed surface. (using integration by parts and then green’s theorem). Which forces a mean zero constrain on RHS $\int_{\mathcal{M}}f = 0$ for a valid solution to exist for the equation.

Mean-zero constraint only says solution exists but not a unique solution. As constant lies in the kernel of the Laplacian operator $\Delta$, there exists infinite number of solutions for $\int_{\mathcal{M}}\Delta (u + C) = \int_{\mathcal{M}}f$ for all constant $C$, as $\Delta C = 0$. For unique solution we need to add one more constraint like mean-zero $u$, i.e., $\int_{\mathcal{M}}u = 0$ to fix a gauge for a unique solution. Numerically, this mean-zero constraint on $u$ can be applied by Lagrange multiplier. Or by a Dirichlet pin.

Inverse using Eigenmodes

From spectral theorem we know that eigenmodes are basis for the underlying space. That means we can express both the solution $u$ and RHS $f$ using these modes as basis.

$$\begin{aligned} \Delta u = f \\ u = b_{i}\phi_{i},\quad f= c_{i}\phi_{i}, \quad \text{where $\phi_i$ is an eigenmode} \Delta \phi_{i} = \lambda_{i}\phi_{i} \\ \therefore \Delta(b_{i}\phi_{i}) = c_{i}\phi_{i} \\ b_{i}\lambda_{i}\phi_{i}= c_{i}\phi_{i} \\ \therefore \boxed{b_{i} = \frac{c_{i}}{\lambda_{i}}} \\ \end{aligned}$$

This give us a neat way of solving the Poisson problem. This also says that $\lambda_{i}>0$ else the inverse of $\Delta$ doesn’t exist on the whole space. We can still invert it on the subspace orthogonal to the kernel.