1
$\begingroup$

Let $\Omega$ be a finitely connected bounded domain in the complex plane bounded by $n+1$ analytic jordan curves. Letting $\partial \Omega$ denote the boundary of $\Omega$, we write $\partial \Omega = \{\Gamma_1 \cup \cdots \cup\Gamma_{n+1}\}$, where $\Gamma_i $ denote the boundary component of $\partial \Omega$ with the assumption that $\Gamma_{n+1}$ denote the outer boundary component.

Now consider the following $n$ harmonic harmonic function $u_1, u_2,\cdots,u_n$ defined on $\Omega$ which satisfy the following boundary value. \begin{align} u_j(z) &= 1\;\;\text{for}\;z\in \Gamma_j,\\ &= 0\;\;\text{for}\;z\in \partial \Omega /\Gamma_j \end{align} Existence of such harmonic function follows from the fact that $\partial \Omega$ consist of analytic jordan curve.

Now consider the following map $\phi:\Omega \to I^n $, $\big($where $I$ denote the unit open interval $(0,1) \subset \mathbb{R}\big)$ defined by $$\phi(z) =\big(u_1(z),u_2(z),\cdots,u_n(z)\big)$$ It is straight forward to see that for $n=1$, the map is onto (follows from connectedness of the domain $\Omega$). I believe that for $n\geq 2$ the map $\phi $ is never onto. The reason behind my belief is some dimension mismatch. But I am not able to prove that concretely.

So my question is whether the map $\phi$ is onto or not? Also I would like to know under what condition for a $n$ tuple of number $(a_1,\cdots,a_n) \in I^n$, there exist a $z\in \Omega$ so that $$\phi(z) = (a_1,\cdots,a_n)$$.

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

It's clear that $\phi$ is not onto when $n\ge 2$. Since $$\sum_{j=1}^{n+1}u_j(z)=1,$$ it follows that $$\sum_{j=1}^n u_j(z)<1.$$

I doubt that a suitable characterization of the range of $\phi$ exists.

$\endgroup$
2
  • $\begingroup$ can you say something about the range of $\phi$? will it be a sub manifold of $I^n$? $\endgroup$
    – Timon
    Commented Nov 17, 2015 at 19:29
  • $\begingroup$ I know nothing about the range. I doubt that it's a submanifold. For more or less the same reason as I thought it probably would be a submanifold a few seconds earlier: Say $p(z_1)=p(z_2)$. Then $\{p(w)\}$ for $|w-z_1|<\epsilon$ is probably something like what a neighborhood of $p(z_1)$ in a manifold would be. But there's also $|w-z_2|<\epsilon$, giving another "disk" about $p(z_j)$... $\endgroup$
    – David C. Ullrich
    Commented Nov 17, 2015 at 21:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .