Eigenvalues and eigenvectors of the second derivative Consider the $1$-dimensional eigenvalue problem $$\begin{cases}\frac{d^2u}{dt^2}=\lambda u \\ u(0)=u(1)=0\end{cases}$$ I want to compute the eigenvalues and eigenfunctions. I found something related in wikipedia. Here is the link: https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative#The_discrete_case
In wiki, there are two cases. In the continuous case, the $j$th eigenvalue and eigenvector are $$\begin{align} &\lambda_j=-j^2 \pi^2 ~~(\text{In my problem $L=1$}) \\ &v_j=\sqrt{2}\sin(j\pi t)\end{align}$$
In the discrete case, the $j$th eigenvalue and eigenvector are 
$$\begin{align} &\lambda_j=-\frac{4}{h^2}\sin^2\left(\frac{\pi j}{2(n+1)}\right) \\ &v_{i,j}=\sqrt{\frac{2}{n+2}}\sin\left(\frac{ij \pi}{n+1}\right)\end{align} $$(the index i represents the $i$th component of the eigenvector). I think $n$ should be number of grid points and $h$ should be the length of a subinterval in the partition.
What's the relation of between the two $\lambda_j~'s$?  Why do the eigenvalues in continuous and discrete cases look very different?
 A: The two eigenvalue expressions are approximately equal when the frequency $j$ is small or moderate compared to the number of gridpoints $n$, and differ when the frequency $j$ is large compared to $n$. Roughly, the expressions begin to differ from each other when the number of gridpoints per wavelenth becomes too small for the discrete eigenvector to accurately resolve the continuous sinusoidal eigenvector.

To see this, first rescale/changing variables to map the domain of the discrete problem to $[0,1]$ by setting 
$$t_i := \frac{i}{n+1}.$$
Then with respect to the variables $t_i$, the eigenvectors of the discrete operator are evenly spaced samples of the eigenvectors for the continuous problem (up to a constant scaling factor):
$$v_{i,j} = C \sin\left(j \pi t_i\right).$$
Now apply the small angle approximation $\sin(\theta) \approx \theta$ to the eigenvalue expression:
\begin{align*}
\lambda_j =& -\frac{4}{h^2}\sin^2\left(\frac{\pi j}{2(n+1)}\right) \\
=& -4 n^2\sin^2\left(\frac{\pi j}{2(n+1)}\right) \\
\approx& -4 n^2\left(\frac{\pi j}{2 (n+1)}\right)^2 \\
\approx& -j^2 \pi^2,
\end{align*}
which is the corresponding eigenvalue of the continuous problem. The small angle approximation we made is valid when $\frac{\pi j}{2(n+1)} \sim \frac{j}{n}$ is small. Since $j$ is the number of "bumps" in the sinusoid, and $n$ is the number of gridpoints, the approximation is valid whenever the number of "gridpoints per bump" (alternatively, gridpoints per wavelength) is sufficiently small. I.e., the frequency is small enough that discrete approximation accurately resolves the continuous eigenvector.
The small angle approximation breaks down when there are few gridpoints per bump, which makes sense since the continuous eigenvectors are not well approximated by discrete samples anymore in this regime.
