In reviewing complex analysis, I stumbled upon the following problem:
Find a function $g(x,y)$ harmonic on $\{ 1<x^2+y^2<16\}$ such that $g(x,y)=3$ when $x^2+y^2=1$ and $g(x,y)=8$ when $x^2+y^2=16$.
In my notebook, I saw some notes on the side that $g(x,y)$ should have the form $a \cdot \ln(x^2+y^2)+b$, but not only am I unsure of how to derive this equation- I'm not certain it's correct to begin with.
I know that in the standard Cartesian plane, a Harmonic function is one that satisfies Laplace's equation, that is, that the sum of the second partial derivatives is $0$.
Where to proceed from here, however, I am unsure.
In cylindrical coordinates, let $ g(\rho ,\phi)$ be harmonic in the region $1<\rho<4$. Then, $g(\rho,\phi)$ satisfies Laplace's Equation
$$\frac{1}{\rho}\frac{\partial }{\partial \rho}\left(\rho \frac{\partial g}{\partial \rho}\right)+\frac{1}{\rho^2}\frac{\partial^2 g}{\partial \phi^2}=0$$
with $g(1,\phi)=3$ and $g(4,\phi)=8$.
Symmetry suggests that $g$ is independent of $\phi$ and the problem reduces to an ordinary differential equation
$$\frac{1}{\rho}\frac{d }{d\rho}\left(\rho \frac{d g}{d \rho}\right)=0 \tag 1$$
Integrating $(1)$, we find that $g(\rho)=A\log(\rho)+B$. Enforcing the boundary conditions, we have $g(1)=B=3$ and $g(4)=A\log(4)+3=8\implies A=\frac{5}{\log(4)}$. Therefore, the solution is
$$g(\rho)=\frac{5\log(\rho)}{\log(4)}+3$$