Why a solution for $c^2 \Delta u = u_{tt}$ must have eigenfunctions as its series terms expansion?

Question

I'm reading this which explains how to arrive at a solution for $u$ as a series expansion involving $J_0, J_1,\cdots$ which are Bessel Functions.

It concludes at page 5 saying that $u$ is the following:

and that the terms $J_0$ and $J_i$ for $i=1,\cdots$ are eigenfunctions of the Laplacian Operator. But why?

It isn't clear to me why the process of separating variables and arriving at a soluton will give me an expansion in terms of eigenfunctions. Maybe it's a theorem? Are the eigenvalues the terms multiplying $J_0, J_1,\cdots$?

UPDATE:

Can you explain it better why the equation is $u_{tt}-c^2\nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-\nabla u = 0 $?

Answer 1

It arises from the Neumann spectral theorem which guarantees that Laplacian operator has basis of eigenfunctions and that basis is full and orthogonal. So every function in Hilbert space can be represented as a linear sum of eigenfunctions and that representation is unique. And since the time operator is linear too, all eigenfunctions evolve independently.

So the logic is the following.

1. At each point $t$ the solution $\tilde u$ can be uniquely represented as a sum of series of eigenfunctions $f_i(\mathbf{r})$ with coefficients $a_i$.
2. Thus, coefficients are some functions of $t$: $a_i = a_i(t)$ and the solution can be uniquely represented as a sum $$ \tilde u(\mathbf r, t) = \sum_i a_i(t)f_i(\mathbf r) $$
3. Since both Laplacian and second time derivative are linear, then all terms $a_i(t)f_i(\mathbf r)$ must satisfy original equation.

The eigenvalues are those $\lambda_{n,m}$.