1
$\begingroup$

Let $\Omega$ be an open connected subset of $\mathbb{R}^N$ satisfying the following property:

$$x=(x_1,\cdots,x_N)\in \Omega\implies (x_1,\cdots,-x_N)\in\Omega$$

Let $u$ be an harmonic function in

$$\Omega_+=\{x=(x_1,\cdots,x_N)\in\Omega: x_N>0\}$$

continuous in $\Omega_{-} = \{x=(x_1,\cdots,x_N)\in\Omega:x_N\ge 0\}$ and null when $x_N=0$. Show that there exists only one harmonic function $\overline{u}$ in $\Omega$ which coincides with $u$ in $\Omega_+$

Doesn't this follow from the unicity of the dirichlet problem? We have a function $u$ harmonic in the superior half plane and continuous in the superior + boundary. Suppose that there are two harmonic functions $u_1,u_2$ with such properties. Then $u_1-u_2$ is also harmonic. By the maximum principle, its max is attained in the border, but $u_1-u_2=0$ in the border. We can say the same for the minimum value. That is, the minimum is on the boundary and is $0$. Therefore $u_1-u_2=0$ everywhere, that is, $u_1=u_2$

The problem is that the hypothesis that $x=(x_1,\cdots,x_N)\in \Omega\implies (x_1,\cdots,-x_N)\in\Omega$ is useless. Maybe I'm missing something?

$\endgroup$
1
$\begingroup$

You are actually making two mistakes: First,the boundary of $\Omega_+$ need not be contained in the hyperplane $x_N=0.$ There's all the stuff above as well. For example, let $N=2$ and let $\Omega$ be the open unit disc. Then $\partial \Omega_+$ equals $\{e^{it}: t\in [0,\pi]\} \cup [-1,1].$ Secondly, even if all of $\partial \Omega_+$ lies in that hyperplane, as is true for $\Omega = \{x_N>0\},$ the unicity of the Dirichlet problem can fail. For example $u(x)\equiv 0$ and $u(x)=x_N$ both equal $0$ on $\partial \Omega_+$ in this case.

The key is boundedness: The solution to the Dirichlet problem for all bounded $\Omega$ is unique if it exists.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.