Suppose that I use Finite Difference Method to solve a stable PDE, which is 4th order in $x$ and 2nd order in $t$.
Unfortunately, the discretization in $x$ is irregular ($\Delta x_i$ values are different). Is there a mathematical result for checking the stability of the method (note, $\Delta t$ is uniform, i don't vary it)?
It seems that if I modify the problem to be uniformly discretized, with discretization step $min \Delta x_i$ (so I take the smallest step of the irregular discretization to be the step of the uniform one), then if the modified problem is stable, then the irregular one is stable too.
Is this statement true? Or it is not that simple, and this is dependent on the exact form of the PDE?
For a standard second-order wave equation in 1D with wave speed $c$, $$\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}-\frac{\partial^2 E}{\partial x^2}=0$$ time-discretized with a second-order accurate finite difference "leapfrog" method, the stability condition on a uniform grid is $$c\Delta_t<\Delta_x$$
On a non-uniform grid it becomes $$c\Delta_t<\frac{2}{\|D\|}$$ where $D$ is the matrix that represents the discretized spatial derivative operator on the non-uniform grid, and $\|D\|=\max_{|x|=1}|Dx|$ is the operator norm.
The proof is not super-complicated. See for example this thesis (eq. 2.113). The proof makes no assumptions on the nature of $D$, so could probably be easily extended to a fourth-order (in space) wave equation.