Can any solution to the three-dimensional wave equation be written as a superposition of plane waves?

Can any solution to the three-dimensional wave equation, $$\nabla^2f = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2},$$ be written as a superposition of sinusoidal plane waves? In "Introduction to Electrodynamics" Griffiths points out that sinusoidal waves are of interest because any solution to the one-dimensional wave equation, $$\frac{\partial^2 f}{\partial z^2} = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2},$$ is a superposition of sinusoidal waves, but that doesn't necessarily mean that any solution to the three-dimensional wave equation can be written as a superposition of sinusoidal plane waves.

• Yes. The scalar wave equation is linear no matter what dimension it's in. Apr 26, 2019 at 9:18

In a simplified effort let's focus on integrable functions $$L^1$$ and assume solutions of the wave equation $$f\in L^1$$: $$f(r,t) = \int dk d\omega \exp(\text{i}( k\cdot r + \omega t)) \hat{f}(k,\omega)$$ $$(\Delta - v^{-2}\partial^2_t)f = 0$$ Putting the fourier-integral definition of $$f$$ into the wave equation yields $$0 = \int dk d\omega (v^{-2}\omega^2 - k^2)\exp(\text{i}( k\cdot r + \omega t)) \hat{f}(k,\omega) \forall r,t$$ Now the only way this may work if the only nonvanishing $$\hat{f}(k,\omega)$$ are the ones with $$(v^{-2}\omega^2 - k^2) = 0$$, since $$\exp(\text{i}( k\cdot r + \omega t))$$ are linearly independent, and this characterizes exactly the plane waves.