Suppose $\Omega$ is a bounded open simply connected region in $\mathbb{R}^2$, and let $u:\overline{\Omega} \to \mathbb{R}^2$. Suppose $u$ is divergence free. Then there exists a stream function $\psi$ such that $u = (-\partial_2 \psi, \partial_1 \psi)$.
I would like to know how regular $\psi$ is in terms of the regularity of $u$. Also, does the regularity of the boundary of $\Omega$ play a role here, and does $u$ need to be regular up to and including the boundary, or just on the interior?
In particular, I'm interested in the case $u \in C^1(\overline{\Omega})$. Can I say $\psi \in C^2(\overline{\Omega})$? Feel free to assuming $\partial \Omega$ is $C^2$ as well.
Since $u$ is divergence free, the rotated field $u^\perp := (-u_2,u_1)$ is conservative. Therefore, there exists $\psi$ such that $\nabla \psi = u^\perp$. Since $ u \in C^1(\overline{\Omega})$, we see that the first and second derivatives of $\psi$ continuously extend to $\overline{\Omega}$. To complete the proof of $\psi\in C^2(\overline{\Omega})$, it remains to show that $\psi$ itself extends continuously to $\overline{\Omega}$. This is where the geometry of the domain becomes relevant.
Since the boundary of $\Omega$ is smooth, it follows that $\Omega$ is quasiconvex: any two points $a,b\in \Omega$ can be joined by a curve of length at most $C|a-b|$. Integrating $\nabla \psi$ along this curve, we find that $\psi$ is Lipschitz in $\Omega$. Thus, it indeed extends continuously to the closure.