General solution of 3d wave equation as a superposition of plane/spherical waves Is the general solution to the 3d wave equation a superposition of plane/spherical waves? Can this be shown?
 A: You can solve it using Fourier transforms. If you have never seen them, the following could be a bit confusing. However, basically what one can do is write your unknown function as a infinite sum of waves with a spatial frequency $k$, solve it for each value of $k$ and then go sum it back to the original function. This procedure, as the wave equation is relatively easy to solve for a single wave, often really helps!
If you have the wave equation
$${\partial^2 y \over \partial t^2} = c \nabla ^2 y$$
where $\nabla^2={\partial^2  \over \partial x^2}+{\partial^2 \over \partial y^2}+{\partial^2 \over \partial z^2}$
you can use a Fourier transform to attempt solving it, i.e. a sort of infinite sum of plane waves.
In case you don't know Fourier transforms, briefly you can write every function in 3D $y(\textbf{r}=(x, y, z))$ (provided they satisfy some very general conditions) as a sum of plane waves of the form $e^{i\textbf{k}\cdot\textbf{r}}$, with $\textbf{k}=(k_x, k_y, k_z)$, as
$$y(\textbf{r}) = A \int \hat{y}(\textbf{k}) e^{i\textbf{k}\cdot\textbf{r}}d\textbf{k}$$
where $A$ is a constant and $\hat{y}(\textbf{k})$ is the so called Fourier transform and represents the continous coefficients of the series.
By substituting this expression in the wave equation we get an equation for $\hat{y}(\textbf{k})$ which is
$${\partial^2 \hat{y}(\textbf{k}) \over \partial t^2} = -c k^2 \hat{y}(\textbf{k})$$
To prove it, just plug the expression I wrote up there in the wave equation and use the fact that $i^2=-1$ and ${\partial e^{i\textbf{k}\cdot\textbf{r}} \over \partial x} = i k_x e^{i\textbf{k}\cdot\textbf{r}}$ and similar.
This equations is often easier to solve (it is very close the equation for a harmonic oscillator) but still requires the boundary conditions for $y$ to be completely solvable. Once it is solved, using the expression for $y(\textbf{r})$, you can "invert" the Fourier transform to get the solution of your original functions.
You could do the same for spherical waves by finding an appropriate expression for $\nabla$ and for the Fourier transform.
