Problem: Solve $u_{xx} + u_{yy} = 0$ in the region ${α<θ<β, a<r<b}$ with the boundary conditions $u=0$ on the two sides $\theta=\alpha$ and $\theta=\beta$, $u=g(\theta)$ on the arc $r=a$, and $u=h(\theta)$ on the arc $r=b$
Attempt: $u_{xx} + u_{yy} = 0$
$u(r,α) = 0 = u(r,β)$
$u(a,θ) = g(θ)$, $u(b,θ) = h(θ)$
The textbook provided a general solution for these types of problems. But I feel like applying it directly in this case looks too simple. Am I doing anything wrong?
$u(x,y) = A_0/2 + \sum_{n=1}^{\infty}r_n(A_ncos(nθ) + B_nsin(nθ))$
$A_n = \frac{1}{a_n \pi} \int_{0}^{2\pi}g(\alpha)cos(n\alpha)d \alpha$
$B_n = \frac{1}{b_n \pi} \int_{0}^{2\pi}h(\beta)sin(n\beta)d\beta$
There are is one major issue: the Fourier series is flawed, because your variables don't make sense as written.
This is because we are working with $u(r,\theta)$ not $u(x,y)$. The Fourier series you provided is valid when you solve Laplace's Equation radially (using polar coordinates). Recall, this Fourier series is derived via the "separation of variables" method used commonly for linear PDEs on bounded domains. Write $u(r,\theta)=R(r)\Theta(\theta)$, get your system of ODEs (don't forget to rewrite Laplace's equation in polar coordinates when you do this).
However, the boundary conditions are well defined, and you can solve the PDE using that method, but be careful about $g(\theta)$ and $h(\theta)$. In the problem you are given, you are probably asked to assume some regularity of these two functions (i.e. they are smooth or at least $C^1$). At least, we need $g,h\in L^1_{loc}(D)$ where $D\subset \mathbb{R}^2$ is the bounded region given in the problem.