Positive harmonic functions vanishing on the boundary for the Laplacian in unbounded horn-shaped domains

Abstract:

Denote points $\bar x \in {R^{d + 1}}$, $d \geq 2$, by $\bar x = (\rho ,\theta ,z)$, where $\rho > 0$, $\theta \in {S^{d - 1}}$, and $z \in R$. Let $a:[0,\infty ) \to (0,\infty )$ be a nondecreasing ${C^2}$-function and define the "horn-shaped" domain $\Omega = \{ \bar x = (\rho ,\theta ,z):|z| < a(\rho )\}$ and its unit "cylinder" $D = \{ \bar x = (\rho ,\theta ,z) \in \Omega :\rho < 1\}$. Under appropriate regularity conditions on a, we prove the following theorem: (i) If ${\smallint ^\infty }a(\rho )/{\rho ^2}d\rho = \infty$, then the Martin boundary at infinity for $\frac {1}{2}\Delta$ in $\Omega$ is a single point, (ii) If ${\smallint ^\infty }a(\rho )/{\rho ^2}d\rho < \infty$, then the Martin boundary at infinity for $\frac {1}{2}\Delta$ in $\Omega$ is homeomorphic to ${S^{d - 1}}$. More specifically, a sequence $\{ ({\rho _n},{\theta _n},{z_n})\} _{n = 1}^\infty \subset \Omega$ satisfying ${\lim _{n \to \infty }}{\rho _n} = \infty$ is a Martin sequence if and only if ${\lim _{n \to \infty }}{\theta _n}$ exists on ${S^{d - 1}}$. From (i), it follows that the cone of positive harmonic functions in $\Omega$ vanishing continuously on $\partial \Omega$ is one-dimensional. From (ii), it follows easily that the cone of positive harmonic functions on $\Omega$ vanishing continuously on $\partial \Omega$ is generated by a collection of minimal elements which is homeomorphic to ${S^{d - 1}}$. In particular, the above result solves a problem stated by Kesten, who asked what the Martin boundary is for $\frac {1}{2}\Delta$ in $\Omega$ in the case $a(\rho ) = 1 + {\rho ^\gamma }$, $0 < \gamma < 1$. Our method of proof involves an analysis as $\rho \to \infty$ of the exit distribution on $\partial D$ for Brownian motion starting from $(\rho ,\theta ,z) \in \Omega$ and conditioned to hit D before exiting $\Omega$.