Is there a harmonic function which satisfies the following conditions?

Question

Let $\Omega\subset R^2$ be a simply connected domain with smooth boundary $\partial \Omega$. Let $\Gamma_1$ be a subset of $\partial \Omega$ such that $\partial\Omega\subset\overline \Gamma_1\neq\emptyset$. Does there exist a harmonic function $u\in C^\infty(\overline\Omega)$ such that $|\nabla u|\neq 0$ in $\overline\Omega$ and $\frac{\partial u}{\partial\nu} = 0$ on $\Gamma_1$? Here $\nu$ denotes the outward normal vector of $\Omega$.

Answer 1

I assume that "$\partial \Omega\subset \overline{\Gamma_1}\ne \emptyset$" should be "$\partial \Omega\setminus \overline{\Gamma_1}\ne \emptyset$".

The answer is yes, such $u$ exists. Let $\Omega'$ be a $C^\infty$-smooth domain such that $\partial \Omega$ contains a horizontal line segment $L$. You can conformally map $\Omega$ onto $\Omega'$ so that the image of $\overline{\Gamma_1}$ is contained in $L$. The conformal map is $C^\infty$ smooth on the closure of domain, in both directions.

The function $z\mapsto \operatorname{Re}z$ is harmonic in $\Omega'$, has nowhere vanishing gradient, and has zero normal derivative at every point of $L$. Transferring this function back to $\Omega$, you get the required $u$, since harmonicity is preserved under composition with conformal maps, and the normal derivative is merely scaled by the derivative of conformal map.