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Assume $\Omega$ is a multi-connected bounded domain in $\mathbb{R}^2$, and function $v\in W^{1,2}_{0}(\Omega)$ satisfies:

$$\text{div} \; v=0$$

Then there exists a stream function $\psi\in W^{2,2}(\Omega)$ such that :

$$\nabla \psi=(-v_2,v_1)$$

How to prove this? I know the method when $v$ is continuous which use the Green Theorem and line integral.

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  • $\begingroup$ The fundamental theorem of calculus holds in for functions in $H^1$. You should be able to use the same argument as in the continuous case. $\endgroup$ – user113529 Nov 6 '15 at 8:21
  • $\begingroup$ @user43687 actually my only doubt is why $\psi$ is in $L^2$ ? $\endgroup$ – Kira Yamato Nov 6 '15 at 9:22
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Since $v$ is in $W_0^{1,2}(\Omega)$, extending it by zero outside of the domain gives a function in $W^{2,2}$. Thus, we can replace the multiply-connected domain $\Omega$ (with potentially bad boundary) by a ball $B$.

The existence of $\psi$ such that $\nabla \psi=(-v_2,v_1)$ is a form of the Poincaré lemma about closed form being exact. By definition, $\nabla \psi \in W^{1,2}(B)$. By the Poincaré inequality, this gives $\psi\in L^2(B)$, completing the proof of $\psi\in W^{2,2}(B)$.

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