OpenYear of origin: 1974
Posted online: 2018-06-20 13:00:08Z by Lavi Karp85
Cite as: P-180620.1
In the complex plane the Schwarz function of an arc $\Gamma$ is a holomorphic function $S(z) $ near $\Gamma$ and such that $S(z)=\bar{z}$ on the arc $\Gamma$. The Schwarz function of a straight line is a linear function $az+b$. The presuppose is that the Schwarz function of non-linear arcs develops singularities. And indeed, a beautiful theorem of Davis confirms this hypotheses [3]. Namely, suppose the Schwarz function $S(z)$ of an arc $\Gamma$ is an entire function in the complex plane, then $\Gamma$ is a straight line.
One may consider the Schwarz function as the complex derivative of the solution to the Cauchy problem $\Delta U_\Gamma =0$ near $\Gamma$ and $u(z)=z\bar{z}$ on $\Gamma$, here $\Delta$ is the Laplacian in $\mathbb{R^2}\sim\mathbb{C}$. We call $U_\Gamma$ the Schwarz potential of the arc $\Gamma$. Obviously this definition can be extended to higher dimensions. So let $\Gamma$ be a real analytic hyper-surface in $\mathbb{R}^d$, the Schwarz potential $U_\Gamma$ of $\Gamma$ is the solution to the Cauchy problem $\Delta U_\Gamma=0$ near $\Gamma$ and $U_\Gamma(x)=|x|^2$ on $\Gamma$. For example, if $\Gamma$ is the hyper--plane $\{{\bf{a}}\cdot x=0\}$, where $\bf{a}$ is a unit vector, then $U_\Gamma(x)=|x|^2-2d({\bf{a}}\cdot x)^2$ is its Schwarz potential.
Now, given a real analytic hyper--surface $\Gamma$ in $\mathbb{R}^d$, then the existence of the Schwarz potential follows from the Cauchy--Kowalevski theorem. Khavinson and Shapiro formed the conjecture that if the Schwarz potential of $\Gamma$ is an entire harmonic function, then $\Gamma$ is a hyperplane [6].
The conjecture was confirmed under several additional restrictions. In the case $\Gamma$ is an algebraic surface, Khavinson and Shapiro proved that $\Gamma$ is a hyperplane [6]. Another approach is so called one side strategy: Suppose $\mathbb{R}^d=\Omega\cup \Gamma\cup D$ is a disjoint union, we set $V_\Gamma(x)=(2d)^{-1}\left(|x|^2-U_\Gamma(x)\right)$ for $ x\in \Omega$, and $V_\Gamma(x)=0$ for $ x\in D$, that is, it is a solution to the free boundary problem $V_\Gamma(\Delta V_\Gamma-1)\equiv 0$. So suppose $u$ satisfies the global free boundary problem $u(\Delta u-1)\equiv 0$, $\Omega=\{\Delta u =1\}$, $\Gamma$ is the boundary of $\Omega$ and $D=\mathbb{R^d}\setminus \Omega$. The question is under which conditions we can conclude that $\Omega$ is a half space. So if $D$ is bounded there are many types of solutions, see e.g. [4,7,9]. In case $D$ is unbounded and no growth conditions at infinity are assumed, then there is an example of a solution to this free boundary problem in which $\Omega$ is not a half space, but then $u(x)$ has an exponential growth at infinity (see [9]). So the restriction to a polynomial growth is an indispensable demand in the one side approach. It was proved by Shahgholian [8] that if $u(x)$ has a polynomial growth and $D$ contains balls of arbitrary large radius, then $u(x)\sim |x|^2\log|x|$ near infinity. This type of growth is insufficient to derive further geometrical properties, one has to get rid of the logarithmic term. This was done in [4]. Having the quadratic growth, it follows from [1,2] that $D$ is convex. In that case there are also solutions of the free boundary problem for which $\Omega$ is a not half space. For example, when $\Gamma$ is a paraboloid. However, if we require that $D$ is thick at infinity, $\limsup_{\rho\to\infty}\frac{|D\cap B_\rho|}{|B_\rho}>0$, then $\Omega$ is a half space [2,4,5]. Here $B_\rho$ is a ball with radius $\rho$ and $ |\ \cdot \ |$.
The conclusions are that if either $\Gamma$ is an algebraic surface, or one side of $\Gamma$ is thick at infinity and the Schwarz potential has a polynomial growth, then Davis theorem holds in $\mathbb{R}^d$, otherwise it is an open problem.
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Created at: 2018-06-20 13:00:08Z
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