I am trying to find the dispersion relation for a system of linear ODEs.
I can do this for a single linear PDE, for example $$u_x = u_t$$ by substituting $u = Ae^{i(kx-wt)}$, here $w = w(k)$ where $k$ is the wavenumber. Then we get
$$(ik) = -iw $$$$\Rightarrow w = -k$$
However, now I am trying to find the dispersion relation for a system of linear ODEs. For example $$\dot{u_n} =u_{n-1}+u_{n+1}$$ and $n = 1,2,3,...,2N$ when $u_0 = u_{2N}, u_1 = u_{2N+1}$
In the first example, $w$ was a function of a continuous variable $k$. But since there are no $x$ derivatives now, $w$ is a function of a discrete variable $w = w_k$.
My question: what do I use to substitute into equation $(1)$ to find the dispersion relation. Would I use the same substitution as the first example? What's confusing me is that we have the term $u_{n-1}$ and I am not sure how we are meant to reflect that in the substitution.
Thanks a lot.
In electrodynamics it is common to derive dispersion relations for Maxwell's equations, which are a system of linear PDEs, somewhat like your example (but with spatial derivatives). See for example section 8.3.1 here.
You can only find dispersion relations for (systems of) differential equations that actually have wavelike solutions, with a wavelength (or wave vector) and a frequency, such that you can find a relation between the two.
Certainly you can insert $u_n=A_n\exp(-i\omega t)$ into your equations. Demanding that a non-trivial solution exist leads to an equation of this form $$M=\left[\begin{matrix}-i\omega & -1 & 0 & 0 & \ldots & 0 & -1\\ -1 & -i\omega & -1 & 0 & \ldots & 0 & 0\\ 0 & -1 & -i\omega & -1 & \ldots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{matrix}\right]\\ \left|M\right|=0$$ which gives some finite number (probably $2N$) of possible solutions $\omega$. You might call this a "dispersion relation", in the sense that it describes the possible frequencies supported in your "material", but given that your equations are not wave equations this is a bit of a stretch. (If you actually want to calculate this determinant you should look into the theory of circulant matrices)
Now suppose you find an $\omega_i$ that makes $|M|=0$, then you have a solution for the $A_n=\text{ker}(M)$. It is likely so that this $A_n$ is itself of the form $\exp(i k_i n)$ which would let you associate a discrete $k_i$ with $\omega_i$.
OK, in fact we can work this out explicitly. Because $M$ is circulant we have $$M=W \text{diag}(-i\omega -\Omega_0 -\Omega_0^{2N-1},-i\omega -\Omega_1 -\Omega_1^{2N-1},\ldots) W^{\dagger}$$ where $W$ is a DFT matrix, and $\Omega_j=\exp(\frac{\pi i j}{N})$. The discrete solutions $\omega_j$ thus obey $$-i\omega_j -\Omega_j -\Omega_j^{2N-1}=0$$ $$\omega_j =\frac{\Omega_j +\Omega_j^{2N-1}}{-i}$$ $$\omega_j =\frac{\exp(\frac{\pi i j}{N}) +\exp(\frac{\pi i j}{N})^{2N-1}}{-i}$$ $$\omega_j =2 i \cos\left(\frac{j\pi}{N}\right)$$ The corresponding eigenvector in the nullspace is a column of $W$: $$\left[\begin{matrix}A_1 \\ A_2 \\ \vdots \\A_{2N}\end{matrix}\right]\propto\left[\begin{matrix}\exp\left(\frac{0 \pi i j}{N}\right) \\ \exp\left(\frac{1 \pi i j}{N}\right) \\ \vdots \\ \exp\left(\frac{(2N-1) \pi i j}{N}\right) \end{matrix}\right]$$ which has $k=\pi j/N$.
Thus, your dispersion relation is $(\omega_j,k_j)=\left(2 i \cos\left(\frac{j\pi}{N}\right),\pi j/N\right)$