Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n=3$ ?

share|cite|improve this question
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. – user31373 Jun 2 '12 at 20:12
@LeonidKovalev nice example of a question worth a bounty. – user20266 Jun 6 '12 at 19:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.