This wikipedia page https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative for the discrete second derivative ivative says that the eigenvalue problem $v_{k+1}-2v_k+v_{k-1}=\lambda v_k$ in the case of periodic boundary conditions $v_0=v_n$ ($k=1,\ldots,n$) has eigenvalues $$
\lambda_j=\begin{cases}-4\sin^2\left(\frac{\mathrm{\pi}(j-1)}{2n}\right) & j~\text{odd},\\-4\sin^2\left(\frac{\mathrm{\pi}j}{2n}\right)&j~\text{even}.\end{cases}
$$
How is this derived, or is there a reference anywhere? I wrote down the corresponding equations as:
\begin{align*}
-(2+\lambda_1)v_1+v_2+v_N &= 0, \\
v_{j-1}-(2+\lambda_j)v_j + v_{j+1} &= 0,\quad j=2,\ldots,N-1, \\
v_1+v_{N-1}-(2+\lambda_N)v_N &= 0.
\end{align*}
but how to actually solve this system of difference equations is not so clear to me.
Eigenvalues for discrete second derivative with periodic boundary conditions
circulant-matriceseigenvalues-eigenvectorsfinite differenceslinear algebrarecurrence-relations
Related Solutions
It's a litle confusing that you call ${\bf x_k}$ both the unnormalized and the normalized vector. Let's call ${\bf y_k} $ the normalized vector. We agree that, as $k\to \infty$
$${\bf x_k}=A^k {\bf x_0}=\lambda_1^k c_1 {\bf v_1}$$
Actually, this would correspond to the procedure without normalization. But the normalization would just multiply each ${\bf x_k}$ by some scalar, then we should actually write
$$ {\bf y_k} = \alpha \lambda_1^k c_1 {\bf v_1} = \beta {\bf v_1} $$
But this means that ${\bf y_k} $ is a multiple of the dominant eigenvector... then it's a (dominant) eigenvector. The same goes for ${\bf x_k} $.
Then $${\bf x_{k+1}} = A {\bf y_{k}} = \lambda_1 {\bf y_{k}} = \lambda_1 \frac{{\bf x_{k}}}{|{\bf x_{k}}|}$$
Then, because as $k$ grows ${\bf x_{k+1}} \approx {\bf x_{k}}$ we must have $\lambda_1 = |{\bf x_{k}}|$
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.
Best Answer
Great question!
Actually, solving the eigenvalue for circular boundary condition is surprisingly tractable. Consider the matrix $$ \begin{bmatrix} -2 & 1 & 0 &0 &...&1\\ 1 & -2 & 1 & 0&... &0\\ 0&1 & -2 & 1&... &0\\ &&&&...\\ 0&0 & 0 & 0&... &1\\ 1&0 & 0 & 0&... &-2\\ \end{bmatrix} $$ It's a circulant matrix, which could be diagonalized by a Discrete Fourier Transform matrix. Generally, if $$ C=\begin{bmatrix} c_0 & c_{n-1} & c_{n-2} & ... & c_2 & c_1\\ c_1 & c_{0} & c_{n-1} & ...& c_3 & c_2\\ &&...&&\\ c_{n-2} & c_{n-3} & c_{n-4} & ...& c_0& c_{n-1}\\ c_{n-1} & c_{n-2} & c_{n-3} & ... & c_1&c_0\\ \end{bmatrix} $$ Then, with $\omega = e^{2\pi i/n}$ $$ v_j=[1,\omega^j,\omega^{2j},...\omega^{(n-1)j}]^T\\ \lambda_j = c_0+c_1 \omega^j+c_2 \omega^{2j}+ ... + c_{n-1}\omega^{(n-1)j} $$ In this case, $c_0=-2,c_1=1,c_{n-1}=1$, then we can directly compute each eigenvalue $j=0,1,...n-1$ $$ \lambda_j = -2+\omega^j+\omega^{-j} $$ Then use the basic formula for $\cos$ double angle. $$ \lambda_j = -2+e^{2j\pi i/n}+e^{-2j\pi i/n}\\ =-2+2\cos(\frac{2\pi j}{n})\\ =-4\sin^2(\frac{\pi j}{n}) $$
Note that this looks different from your answer, but they are the same set of eigenvalues. Notice that $$ \sin^2(\frac{\pi j}{n})=\sin^2(\frac{\pi (n-j)}{n}) \text{ for } j=1,2...n-1 $$ So any $\lambda_j,j\neq0$ has multiplicity 2, $\lambda_0=0$ has multiplicity 1; if $n$ is even, $\lambda_{n/2}=-4$ has multiplicity 1 too.
So if you sort and re-oder the eigenvalues, they will be the same as your answer.