I have read a few questions on StackExchange about radial solutions for Laplace's equation. However, I do not feel like they answer my specific question. If you know of a previously asked question that may answer what I am wondering here, please let me know. Otherwise, my question is the following:
Suppose $\phi_{xx}+\phi_{yy}=0$ in the domain between $x^2+y^2=1$ and $(x-1)^2+y^2=\big(\frac{5}{2}\big)^2$, and assume the boundary conditions $\phi(x,y)=a$ on $x^2+y^2=1$ and $\phi(x,y)=b$ on $(x-1)^2+y^2=\big(\frac{5}{2}\big)^2$. If I conformally map this domain and the boundaries to two concentric circles and consider now the polar Laplacian, why is it that $\phi(r,\theta)$ is independent of $\theta$? In other words, why do constant boundary conditions imply that the polar Laplacian of the harmonic function $\phi$ is independent of $\theta$?
Thank you for your responses!
$\phi$ must be independent of $\theta$ since otherwise it cannot be constant on the boundary. Put it this way, suppose $\phi = \phi(r,\theta)$. Say the boundary is the circle $r=a$, imposing the boundary condition says that $\phi(a,\theta) = \textrm{constant}$ for all $\theta\in [0,2\pi]$. The only way this can be true is only if $\phi$ is independent of $\theta$.