Solve initial value two-dimensional heat equation inside a circle $\frac{\partial u}{\partial t} = k \, \nabla^2 u$ with $u(a, \theta, t) = g(\theta)$ 
Q: Solve the initial value problem for a two-dimensional heat equation inside a circle (of radius $a$) with time-independent boundary conditions:
  \begin{align*}
  \frac{\partial u}{\partial t} &= k \,\nabla^2 u \\
  u(a, \theta, t) &= g(\theta) \\
  u(r, \theta, 0) &= f(r, \theta) \\
\end{align*}

I'm pretty sure that the first step is to transform this problem into a problem with homogeneous boundary conditions. I can do that. I'm having problems solving it from there, however.
EDIT: Fixed the equilibrium solution.
Identify equilibrium solution (omitting the steps):
\begin{align*}
  \nabla^2 u_E(r, \theta) &= 0 \\
\end{align*}
\begin{align*}
  u_E(r,\theta) &= \sum\limits_{n=0}^\infty A_n r^n \cos (n \theta) + \sum\limits_{n=1}^\infty B_n r^n \sin (n \theta) \\
\end{align*}
Then we use the orthogonality equations to solve for the coefficients:
\begin{align*}
  A_0 &= \frac{1}{2\pi} \int_{-\pi}^{\pi} g(\theta) \, d\theta \\
  A_{n\ge 1} &= \frac{1}{\pi a^n} \int_{-\pi}^{\pi} g(\theta) \cos (n \theta) \, d\theta \\
  B_{n \ge 1} &= \frac{1}{\pi a^n} \int_{-\pi}^{\pi} g(\theta) \sin (n \theta) \, d\theta \\
\end{align*}
Transform to displacement problem with homogeneous boundary conditions:
\begin{align*}
  v(r, \theta, t) &= u(r, \theta, t) - u_E(r, \theta) \\
  u(r, \theta, t) &= v(r, \theta, t) + u_E(r, \theta) \\
  \frac{\partial v}{\partial t} &= k \, \nabla^2 v \\
  v(a, \theta, t) &= 0 \\
  v(r, \theta, 0) &= f(r, \theta) - u_E(r, \theta) \\
\end{align*}
Solve the homogeneous displacement problem.
Periodicity:
\begin{align*}
  v(r, \pi, t) &= v(r, -\pi, t) \\
  \frac{dv}{d\theta}(r, \pi, t) &= \frac{dv}{d\theta}(r, -\pi, t) \\
\end{align*}
Separate:
\begin{align*}
  v(r, \theta, t) &= w(r, \theta) h(t) \\
  \frac{\partial v}{\partial t} &= k \,\nabla^2 v \\  
  \frac{1}{k} \frac{1}{h} \frac{dh}{dt} &= \frac{1}{w} \, \nabla^2 w = -\lambda \\
  w(a, \theta) &= 0 \\
\end{align*}
separate again:
\begin{align*}
  w(r, \theta) &= \phi(\theta)G(r) \\
  G(a) &= 0 \\
  \nabla^2 w &= -\lambda w \\
  \nabla^2 w &= \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2} = -\lambda w \\
  \frac{1}{r} \frac{1}{G} \frac{d}{dr} \left( r \frac{dG}{dr} \right) + \frac{1}{r^2} \frac{1}{\phi} \frac{d^2 \phi}{d\theta^2} &= -\lambda \\  
  \frac{1}{r} \frac{1}{G} \frac{d}{dr} \left( r \frac{dG}{dr} \right) &= -\lambda - \frac{1}{r^2} \frac{1}{\phi} \frac{d^2 \phi}{d\theta^2} = -\mu \\  
%  \nabla^2 w &= \frac{\partial^2 w}{\partial r^2} + \frac{1}{r} \frac{\partial w}{\partial r} + \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2} &= -\lambda w \\
%  \frac{1}{G} \frac{d^2 G}{dr^2} + \frac{1}{G} \frac{1}{r} \frac{dG}{dr} + \frac{1}{\phi} \frac{1}{r^2} \frac{d^2 \phi}{d\theta^2} &= -\lambda \\
\end{align*}
\begin{align*}
  \frac{d}{dr} \left( r \frac{dG}{dr} \right) &= -\mu rG \\
\end{align*}
I'm not sure what to do next and I suspect I didn't take the right route. Any suggestions?
 A: Separation of variables gives eigenfunctions of the Laplacian $\phi_n(r,\theta)=r^{|n|} e^{in\theta}$ for $n \in \mathbb{Z}$. Expanding $g$ in a Fourier series $\sum c_n e^{in\theta}$, you then have that $u_E(r,\theta)=\sum a^{-|n|} c_n \phi_n(r,\theta)$ is a steady state solution. You can now reduce the desired heat equation to one with homogeneous Dirichlet boundary conditions by subtracting off this steady state from the initial condition, i.e. you are left to solve 
$$\frac{\partial v}{\partial t}=k \nabla^2 v \\
v(a,\theta,t)=0 \\
v(r,\theta,0)=f(r,\theta) - u_E(r,\theta).$$
This can again be done by separation of variables; if $f(r,\theta)=\sum b_n \phi_n(r,\theta)$ then the initial condition is now $\sum (b_n - a^{-|n|} c_n) \phi_n(r,\theta)$ and now you can solve the heat equation when the initial condition is an eigenfunction of the Laplacian and then apply superposition. Solving the heat equation when the initial condition, say $h$, is an eigenfunction of the Laplacian is very simple: if $h$ is an eigenfunction of the Laplacian which corresponds to the boundary condition and with eigenvalue $\lambda$, then the heat equation will just give you $v=e^{\lambda k t} h$. Here the eigenvalues are $-n^2$, so you end up with 
$$u(r,\theta,t)=\sum a^{-|n|} c_n \phi_n(r,\theta) + \sum e^{-n^2 kt}(b_n-
 a^{-|n|} c_n) \phi_n(r,\theta).$$
Note that one must have $\lim_{r \to a^-} f(r,\theta)=g(\theta)$ for this to all make sense.
