# A free boundary problem

Do there exist Jordan analytic curves $J$ in the complex plane $C$, other than circles, with the following property:

There exists a harmonic function $u$ in the unbounded component of $C\backslash J$, $u(z)=0$ and $|\mathrm{grad}\, u(z)|=1$ for $z\in J$.

If we impose mild restrictions on $u$ near $\infty$, for example that $u$ is bounded from below or from above, then such things do not exist.

• A little remark about reducing the problem to the circle: Let $U(J)$ be the unbounded component of $C\setminus J$ and let $S$ be the unit circle. By Riemann's mapping theorem $U(J)$ and $U(S)$ are conformally equivalent. Harmonicity is conformally invariant in 2D. Thus if the function on $U(J)$ you seek exists, there is a harmonic function on $U(S)$ vanishing on $S$. The gradient condition seems trickier. – Joonas Ilmavirta Aug 14 '14 at 19:22
• @ChristianRemling, yes, but to find a noncircular example via Riemann's theorem requires finding a function with nonconstant gradient norm on the circle and a matching conformal map. – Joonas Ilmavirta Aug 14 '14 at 19:32

Sorry, I solved it few hours after posting.

Consider two functions $$\phi(z)=z+\frac{2a^2}{z}-\frac{a^4}{3z^3},$$ and $$v(z)=(1+a^4)\log|z|-\frac{a^2}{2}\Re(z^2-z^{-2}).$$ They have the following properties.

(i) When $a>0$ is sufficiently small, $\phi$ is univalent in $\Delta=\{ z:|z|\geq 1\}$, and $\phi'(z)\neq 0$ in $\Delta$.

(ii) $v$ is harmonic in $\Delta$ and $v(z)=0$ on the unit circle.

(iii) For $z$ on the unit circle, we have $|\mathrm{grad}\, v(z)|=|\phi'(z)|$.

First two properties are evident, and (iii) is verified by direct calculation.

We choose $J=\phi(\{ z:|z|=1\})$, and $u=v\circ\phi^{-1}$ in the outer component of $J$. Evidently So $u$ is harmonic and $u(z)=0$ on $J$. Finally for $z\in J$ we have $$|\mathrm{grad}\, u|=|\mathrm{grad}\, v||(\phi^{-1})'|=1,$$ in view of (iii).