Fourier transforming the wave equation twice

Question

The wave equation $$\nabla^2 u(r,t)-\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(r,t)=0$$ can be Fourier transformed with respect to time, using $\frac{\partial}{\partial t}=i\omega$, to obtain the Helmholtz equation: $$\nabla^2 u(r,\omega)+\frac{\omega^2}{c^2}u(r,\omega)=0\,.$$ What I don't understand is what happens if we Fourier transform is subsequently with respect to space: $$-k^2 u(k,\omega)+\frac{\omega^2}{c^2}u(k,\omega)=0\,,$$ since we now get $$\left(-k^2+\frac{\omega^2}{c^2}\right)u(k,\omega)=0\,.$$ If this should hold for all $k$ and all $\omega$, the solution is the zero function, right?

Answer 1

When solving PDE we usually allow solutions to be distributions (aka generalised functions). I'll spare you the details; you'll have to read a book on PDE by a mathematician to understand that there is a deep connection between PDE and distributions. Long story short, the distributional solution of $$ (k^2-\omega^2)u(k,\omega)=0 $$ is $$ u(k,\omega)=f(k,\omega)\delta(k^2-\omega^2) $$ for an arbitrary function $f$. This function satisfies $(k^2-\omega^2)u=0$ but $u\neq 0$. (Note that I set $c=1$ to simplify the notation).

You can find some more details about this approach in this answer of mine, where I solve $(\partial^2+m^2)\phi=0$. Note that in that post, $\partial^2=\nabla^2-\partial_t^2$; to get the standard wave equation, you can take $m=0$.