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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.

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  • $\begingroup$ Yes. The scalar wave equation is linear no matter what dimension it's in. $\endgroup$
    – andypea
    Commented Apr 26, 2019 at 9:18

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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.

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