Dispersion Equation with Variable Wavenumber – Electromagnetic and Fluid Dynamics

Question

The wave equation
$$u_{tt}=c^2 u_{xx}$$ is known to have a simple wave solution $u(x,t)=Ae^{i(kx-\omega t)}$ where the dispersion equation is simply $c=\omega/k$. Yet, let the wavenumber be a function in $x$, then the independent variable $x$ will appear in the dispersion solution cause the first and second derivatives are functions in $x$ as the following:
$$ \dfrac{\partial{u}}{\partial x} = (ik+ixk_x) e^{i(kx-\omega t)}$$ and $$ \dfrac{\partial^2 u}{\partial x^2} = \left( (ik_x+ik_x+ixk_xx) + (ik+ixk_x)^2 \right) e^{i(kx-\omega t)} .$$
then $$c^2=\dfrac{-\omega^2}{(2ik_x+ixk_xx)+(ik+ixk_x)^2}$$
Did anyone encounter an independent variable as $x$ explicitly in the dispersion relationship as in the terms $ixk_x x$ and $ixk_x$?

Answer 1

Effectively, $u(x,t) = A e^{i(kx-\omega t}$ (with $c= \omega/k$) is a solution of the equation. The problem is in supposing that:

$$\tilde{u}(x,t) = A e^{i(k(x)x-\omega t}$$

is also a solution of the same equation, that step does not seem justified, in fact, for almost any choice of $k(x)$ we will not have a solution in the strict sense. The most general solution formed by plane waves would be a generalized linear combination of the type:

$$u(x,t) = \int_{-\infty}^{+\infty} A(k) e^{ik(x - ct)}\ \text{d}k$$

Given some function $k(x)$ there is no guarantee that you could found a function $A(k)$ such that:

$$e^{i(k(x)x-\omega t)} = \int_{-\infty}^{+\infty} A(k) e^{ik(x - ct)}\ \text{d}k$$