What does the Von Neumann's stability analysis tell us about non-linear finite difference equations?

Question

I am reading a paper [1] where they solve the following non-linear equation \begin{equation} u_t + u_x + uu_x - u_{xxt} = 0 \end{equation} using finite difference methods. They also analyse the stability of the schemes using the Von Neumann's stability analysis. However, as the authors realize, this is only applicable to linear PDE's. So the authors work around this by "freezing" the non-linear term, i.e. they replace the $uu_x$ term with $Uu_x$, where $U$ is "considered to represent locally constant values of $u$."

So my question is two-fold:

1: how to interpret this method and why does it (not) work?

2: could we also replace the $uu_x$ term with the $uU_x$ term, where $U_x$ is "considered to represent locally constant values of $u_x$"?

References

1. Eilbeck, J. C., and G. R. McGuire. "Numerical study of the regularized long-wave equation I: numerical methods." Journal of Computational Physics 19.1 (1975): 43-57.

Answer 1

What you are saying is referred to as linearization. It is a common technique used in the analysis of non-linear PDE's. What is done is to cast equations in the format,

$u_t+Au=0$

Here A is a matrix resulting from the linearization of the equation.

Now to your questions,

1. As you are thinking, it works to some extent, but does not to some other extent. The utility is that stability can be proven for linear systems but not readily for non-linear systems. So the linear results are extended to the non-linear systems. Often, different methods are adopted for particular cases. For example,

$uu_x = \frac{1}{2}(u^2)_x$

which is the conservation form. So,

$u_t+\frac{1}{2}(u^2)_x = 0$

when represented in a finite volume sense gives limits on the evolution of u.

2. What is the utility of doing the replacement. You will remove the equation from a wave equation form. Which would mean that the solutions would not behave as a wave equation. So in the stability analysis, the test solutions would have to be completely different and un-physical as well.

Answer 2

To elaborate on the linearization argument, in uu_x you want to assume u is locally constant, not u_x, for two reasons: a) u varies more slowly than its derivative, and b) in this particular case, if you assume u_x is locally constant, by definition you also assume u is locally linear, which means higher space derivatives are zero, and this not only introduces additional approximation error, but it may imply you may be throwing out the baby with the bathwater, depending on your equation.