consider the following approach to a solution of the one dimensional wave equation:
$-U''(x) = \lambda U(x), \quad \lambda := \frac{\omega^2}{c^2}, \quad 0 < x < 1, \quad U(0) = U(1) = 0$
I already know, that the solutions to the eigenvalue problem are the eigenvalues $\lambda_k = k^2 \pi^2$ with eigenfunctions $e_k(x) = \sin(m\pi x)$ with $m \in \mathbb{N}_0$.
What I am interested in is showing, that the eigenvalues of the discrete problem
$\frac{1}{h^2} [-u_{j-1} + 2 u_j - u_{j+1}] = \lambda u_j, \quad j=1,\dots,n, \quad v_0 = v_{n+1} = 0$
$h = \frac{1}{n+1},\quad u_j \approx U(x_j), \quad x_j = jh$
converge to the eigenvalues of the continuous problem, for $n \to \infty$.
I've tried to compute the eigenvalues of the coefficient matrix $A$ of the discrete problem, but that doesn't seem to lead anywhere, as I can't find a closed form for them.
Similarly using the estimate
$|\lambda_k(A) - \lambda_k(B)| \leq ||A-B||$
for the ordered eigenvalues $\lambda_k$ of the respective matrix and $B = diag(k^2\pi^2)_{k=0}^{n-1}$ doesn't help, as $||A-B||$ diverges to infinity for $n \to \infty$.
What is the best approach to this problem?
There is a closed form for the eigenvalues and eigenvectors of A:
$$\lambda_j = \frac{2}{h^2} (1-\cos(j\pi h)), \quad v_j = (\sin(kj\pi h))_{k=1}^n$$
which we can easily verify by computing $A v_j$ (using $\sin(x) + \sin(y) = \sin(\frac{x+y}{2}) \cos(\frac{x-y}{2})$).
As $n \to \infty$ is the same as $h \to 0$ we see that
$$\lim_{n \to \infty} \lambda_j = \lim_{h \to 0} 2 \frac{1 - \cos(j\pi h)}{h^2} = \lim_{h \to 0} -\frac{j\pi \sin(j\pi h)}{h} = \lim_{h \to 0} j^2\pi^2 \cos(j\pi h) = j^2 \pi^2$$