Solving a wave equation with boundary conditions on a circle? How might one solve a wave equation which had boundary conditions on a circle?
i.e. given $$(\partial^2_x+\partial^2_y)\phi(x,y)=0$$
And known values, $f$ on a circle:
$$\phi(\cos(\theta),\sin(\theta)) = f(\theta)$$
How would one find a solution for $\phi$ in terms of $f$?
 A: There are a few techniques for this, but the most plug-and-chug one is separation of variables in polar coordinates.   Most books on PDEs or mathematical methods for scientists will devote a good chunk of a chapter to explaining this, but here's a summary that will fit in a StackExchange answer:

*

*"Guess" a solution of the form
$$
\phi(r, \theta) = R(r) \Theta(\theta).
$$
If we rewrite the Laplacian in terms of $r$ and $\theta$, it becomes
$$
\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \phi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} = 0,
$$
and after some algebra this can be rearranged to read
$$
\frac{r}{R} \frac{\partial}{\partial r} \left( r \frac{\partial R}{\partial r} \right) + \frac{1}{\Theta}\frac{\partial^2 \Theta}{\partial \theta^2} = 0.
$$


*Realize that since $R$ doesn't depend on $\theta$, and $\Theta$ doesn't depend on $r$, the two terms in the above equations must be constants.  Moreover, since $\Theta(0) = \Theta(2\pi)$, the solution for $\Theta$ must be sinusoidal rather than exponential, with an integer number of periods in $2 \pi$ radians.  Taken together, this implies that
$$
\frac{r}{R} \frac{\partial}{\partial r} \left( r \frac{\partial R}{\partial r} \right) = n^2 \qquad \frac{1}{\Theta} \frac{\partial^2 \Theta}{\partial \theta^2} = -n^2.
$$
for some $n \in \mathbb{Z}$.


*Solve the above two ODEs (not PDEs any more!) to obtain $R$ and $\Theta$ for a fixed value of $n$.  These solutions turn out to be
$$
R(r) = A_n r^n + B_n r^{-n} \qquad \Theta(\theta) = C_n \sin (n \theta) + D_n \cos (n \theta).
$$
If the solution is to be valid at $r = 0$, then we cannot have $|R| \to \infty$ there, and so we must have $B_n = 0$.  This then implies that the solution is of the form
$$
\phi(r, \theta) = r^n [C_n \sin (n \theta) + D_n \cos (n \theta)].
$$


*Use the linearity of the PDE to write a more general solution as the sum of these individual solutions:
$$
\phi(r, \theta) = \sum_{n = 0}^\infty r^n \left[C_n \sin (n \theta) + D_n \cos (n \theta)\right] \tag{1}
$$


*Note that on the boundary $r = 1$, this ansatz becomes
$$
\phi(1, \theta) = f(\theta) = \sum_{n = 0}^\infty \left[C_n \sin (n \theta) + D_n \cos (n \theta) \right]
$$
and thus the coefficients $C_n$ and $D_n$ can be found by expanding the function $f(\theta)$ in a Fourier series.  The full solution to Laplace's equation with the given boundary conditions is then just Eq. (1) above with the coefficients $C_n$ and $D_n$ plugged in appropriately.
Depending on the function $f(\theta)$, the solution so obtained may be a finite series or an infinite series.  In particular, if $\phi(x,y)$ on the boundary can be written in terms of a polynomial in $x$ and $y$, then $f(\theta)$ will be expressible as a set of powers of $\cos \theta$ and $\sin \theta$, and so the Fourier series will have a finite number of terms.  IIRC this then leads to $\phi(x,y)$ being a polynomial everywhere on the disc.
An infinite series solution, on the other hand, is often not as useful.  It can sometimes (but not always) be re-summed into an explicit form—but when it can, this is often a signal that a different technique (e.g., conformal mapping) could also have been used to find the solution.
A: This is a suplimentary answer.
Taking Michael's answer above and feeding back in the Fourier components of $f(\theta)$ using e.g.
$$C_n = \frac{1}{\pi}\int f(\theta) \sin(n\theta) d\theta$$
$$D_n = \frac{1}{\pi}\int f(\theta) \cos(n\theta) d\theta$$
Then doing the summation, (which the main part is a simple power series in:
$$2\Re[ \sum_{n=0}^{\infty} (r \exp(i(\theta-\beta))^n]-1$$
I find that I can write the solution as:
$$\phi(r,\theta) =  \frac{1}{2\pi}\int\limits_0^{2\pi}\left(\frac{1 -  r^2 }{1-2r \cos(\theta-\beta)+ r^2}\right)f(\beta)d \beta$$
which works everywhere except at the boundary where you must take the limit as $r\rightarrow 1$.
