TITLE: Where is the "wave information" hidden in this ODE
QUESTION [9 upvotes]: Consider the following ODE over $\mathbb{R}^+$ with $x(t)\in\mathbb{R}^n$ for large $n$:
\begin{equation}
\ddot x + K x = 0, \quad K=\begin{bmatrix} 2 & -1 &\dots \\ -1 & 2 &-1&\dots \\ \vdots & \ddots & \ddots & \ddots \\ 0 & \dots & -1 & 2 & -1 \\ 0  &\dots &\dots & -1 & 1\end{bmatrix}
\end{equation}
Physically, $x$ can be understood as the displacement of $n$ nodes of a spring-mass chain, fixed at one end (note that $K_{n,n}=1$, that's not a typo). This is what happens for an initial condition $x(0)=[0,\dots,0]$ and $\dot x(0)=[0,\dots,0,1]$, i.e. for a initial velocity of the mass at the end.

It appears that there is something looking like a propagating wave (going to the left), with a constant velocity. 
Question How to estimate this apparent velocity as a function of $n$?
Numerically, it is easy to estimate it, for example by plotting $(i,t,x_i(t))$ where $i\in \{ 1,\dots,n\}$, here for $n=100$:

We can clearly see a straight line in the plot $(i,t)$. On the left of this line, $x$ is (almost) zero: there is no displacement (information has not reached these points yet).
Note that the eigenvalues of $K$ are 
$$\omega_k=2\Big(1-\cos\Big(\dfrac{(2k-1)\pi}{2n+1}\Big)\Big)$$
and its eigenvectors are given by
$$\phi_k^{(i)}=\sin\Big(\dfrac{i(2k-1)\pi}{2n+1}\Big)$$
where $i$ is the component index and $k$ the index of the eigenvector.
The ODE can be solved in the first order form, posing $y^\top = [x,\ \dot x]$:
\begin{equation}
\dot y = Ay, \quad  A = \begin{bmatrix} 0 & I \\ -K &  0 \end{bmatrix}
\end{equation}
so the solutions are $y(t)=\exp(tA)y(0)$.
The question reduces to "extracting" the apparent velocity from $\exp(At)e_{2n}$ ($e_{2n}$ is the last vector of the canonical basis) but I can't see how. Digging in that direction, we can show that with
$$q_i(t)=\dfrac{1}{\omega_i} \phi_i^{(n)} \sin(\omega_i t)$$ and $q^\top =[q_1,\dots,q_n]$,
the solution for the initial velocity $y=e_{2n}$ is
$$ x(t)=\phi^{-1} q(t)$$
so the information in inside $\phi^{-1}q$.

REPLY [2 votes]: We may decompose $x(t)$ into a linear combination of eigenvectors, $x(t)=\sum_{k=1}^n\alpha_k(t)\phi_k$.
Each coefficient satisfies $\ddot\alpha_k(t)+\omega_k\alpha_k(t)=0$, so $\alpha_k(t)=a_k\exp(j\sqrt{\omega_k}t)+b_k\exp(-j\sqrt{\omega_k}t)$ for some complex $a_k,b_k$. I'm using $j=\sqrt{-1}$ because we are already using $i$ for the spatial index.
Similarly, each eigenvector can be written as $\phi_k^{(i)}=\frac1j\bigl(\exp(j\kappa_ki)-j\exp(-j\kappa_ki)\bigr)$, where $\kappa_k=\pi(2k-1)/(2n+1)$.
Multiplying them together, each eigenmode $\alpha_k(t)\phi_k^{(i)}$ can be expressed as a sum of terms of the form $c\exp\bigl(j(\pm\sqrt{\omega_k}t\pm\kappa_ki)\bigr)$, or equivalently, $c\exp\bigl(j\kappa_k\bigl(\pm i\pm(\sqrt{\omega_k}/\kappa_k)t\bigr)\bigr)$. Thus each eigenmode is a superposition of travelling waves of wavenumber $\kappa_k$ and velocity $\pm\sqrt{\omega_k}/\kappa_k$.
In particular, for $k\ll n$, we have $\sqrt{\omega_k}/\kappa_k\approx1$, consistent with the analysis from the wave equation. However, there is some dispersion at higher wavenumbers, which explains the trailing high-frequency oscillations.