2

Let $\Omega$ be an open bounded double connected subset of $R^n$ with boundary $\Gamma=\Gamma_1\cup \Gamma_2$ with $\Gamma_1\cap \Gamma_2=\emptyset$. Let $u(x)$ ba harmonic in $\Omega$ and equal to $V$ on $\Gamma_1$ and equal to $-V$ on $\Gamma_2$. $V$ is a constant different from zero. Can I say that the gradient of $u(x)$ never vanish in $\Omega$. This is true if $n=2$. What happens for $n=3$ ?

giovanni cimatti
  • 21
  • 1
  • double connected ? – mike May 21 '12 at 21:17
  • In this context I interpret "doubly connected" as "homeomorphic to the spherical shell $1<|x|<2$." In two dimensions $\Omega$ is conformally equivalent to a circular annulus, in which we have an explicit formula $u=C\log|z|+C_1$. This is how we know that $\nabla u \ne 0$ in the case $n=2$. I think $\nabla u$ may vanish when $n=3$, but do not have a counterexample. –  Jun 02 '12 at 20:12
  • @LeonidKovalev nice example of a question worth a bounty. –  Jun 06 '12 at 19:31

0 Answers0