How to display Laplace operator as self-adjoint?

Question

How to display Laplace operator as self-adjoint?

I've read some hints about using Green's second identity and had some other ideas regarding constructing a matrix with the Laplace operator terms in the diagonal that's positive definite.

Answer 1

I'll just highlight the key points, there's a fair amount of details to fill in regarding function spaces. But at least formally, one can show

$$(-\Delta u, v)_{\Omega} = (\nabla u, \nabla v)_{\Omega} + \langle \nabla u \cdot \mathbf{n}, \gamma v \rangle = (u, -\Delta v)_{\Omega},$$

where $\mathbf{n}$ would be the normal vector to the domain $\Omega$, $\gamma$ is the trace, $\langle \cdot, \cdot \rangle$ is the duality/inner product on the boundary of $\Omega$ and $(\cdot, \cdot)_{\Omega}$ is the $L^2$ inner product.

The adjoint of an operator is basically the thing when you move it to the other side of the inner product. By using Green's theorem to move $-\Delta$ to the other side of the inner product, we see we get $-\Delta$ back. This means that the operator is self-adjoint.