I have to solve the following boundary value problem by using a series expansion.
$\Delta u(x,y) = 0 \ \ \ in \ \ \ \Omega = \left\{(x, y) \in R^2 :\ 0 \lt R_1^2 \lt x^2 + y^2 \lt R_2^2 \lt \infty\right\} \\ u(x,y) = f(x,y) \ \ \ on \ \ \ \Gamma_1 = \left\{(x, y) \in R^2 :\ x^2 + y^2 = R_1^2 \right\} \\ u(x,y) = g(x,y) \ \ \ on \ \ \ \Gamma_2 = \left\{(x, y) \in R^2 :\ x^2 + y^2 = R_2^2 \right\} \\$
What I've been thinking about so far:
Laplace operator in polar coordinates:
$\Delta u = 0 \ \Longleftrightarrow \ r^2 u_{rr} + ru_r + u_{\Phi\Phi} = 0$
$ u(r,\Phi) = w(r) \cdot v(\Phi)$
New PDG: $r^2w'' \cdot v + rw' \cdot v + w \cdot v'' = 0$
$v(r^2w'' + rw') = -w \cdot v''$
$\Longrightarrow \ \frac{r^2w'' + rw'}{w} = -\frac{v''}{v} = \lambda $
System ordinary PDG's:
$v''(\Phi) = -\lambda v (\Phi), \ \ \ r^2w''(r) + rw'(r) - \lambda_w = 0$
v is $2\pi$ periodic, therefore only $\lambda_k =k^2$ and
$v_k (\Phi) = a_k cos(k\Phi) + b_k sin(k\Phi) , \ k \in \mathbb{N}, \ v_0 (\Phi) = a_0$
are possible. Equation for matching $w_k$ is:
$r^2w''(r) + rw'(r) - k^2w = 0 $
$k=0 : \ \ \ \ w'' = - \frac{1}{r}w' \Longrightarrow w' = \frac{d_o}{r} \Longrightarrow w_o = c_0 + d_o ln(r)$
$ k \neq 0$ : Eulersche PDG: Substitution $r = e^t$ or $w(r) = r^\gamma $
$-k^2 \cdot r^\gamma + r \cdot \gamma \cdot r^{\gamma-1} + r^2 \cdot \gamma \cdot (\gamma - 1 ) \cdot r^{\gamma-2} = 0$
$\Longleftrightarrow r^\gamma (-k^2 + \gamma + \gamma^2 - \gamma) = 0 \Longleftrightarrow \gamma = \pm k$
and therefor $ \ w_k (r) = c_k r^-k + d_k r ^k$
Every function $w_k \cdot v_k$ solves the PDG. Because the PDG is linear, every linear combination is also a solution.
$u(r,\Phi) = c_0 + d_0 ln(r) + \sum (c_k r^-k + d_k r^k)(a_k cos(k\Phi) + b_k sin(k\Phi))$