Harmonic functions constant on circumferences

Question

I want to find all the harmonic functions in $\mathbb{R}^{2}-\{(0,0)\}$ which are constant on circumferences with center in $(0,0)$.

$\mathbb{R}^{2}-\{(0,0)\}$ isn't simply connected so we can't apply the theorem about harmonic coniugates. Any hint ?

Answer 1

If $u$ is constant on circles centered at $(0,0)$ then $u$ is a radial function, that is, its values depend only 0n $r=\sqrt{x^2+y^2}$, the distance from $(x,y)$ to $(0,0)$. If $u=U(r)$, then $$ \Delta u=\frac{d^2u}{dr^2}+\frac{1}{r}\,\frac{du}{dr}=0. $$ Solving the differential equation we get $$ \frac{du}{dr}=\frac{A}{r}\text{ and }u=A\,\log r+B $$ for some constants $A$ and $B$.

Answer 2

Identifying $\mathbb{R}^2$ with $\mathbb{C}$ as usual, if we compose a harmonic function $h$ on $\mathbb{C}\setminus \{0\}$ with the exponential function, we obtain a harmonic $2\pi i$-periodic function $g$ on $\mathbb{C}$.

If $h$ is constant on circles with centre $0$, then $g$ is constant on the lines $\operatorname{Re} z = \operatorname{const}$, so $g$ is an entire harmonic function depending only on $\operatorname{Re} z$. That means

$$g(z) = a + b\operatorname{Re} z$$

for some constants $a,b$. It remains to write $\operatorname{Re} z$ as a function of $e^z$ to find $h$.