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We consider a coupled PDE in a rectangle $\Omega=(-1,1)\times(-1,1)$. For the simplicity, we assume that the functions are periodic in $x_{1}$ direction. \begin{cases} \nabla\cdot u=f_{1},\ & \text{in }\Omega_{1}=(-1,1)\times(0,1), \\ -\Delta v=f_{2},\ & \text{in }\Omega_{2}=(-1,1)\times(-1,0), \\ u\cdot n=-\nabla v\cdot n+g,\ & \text{on } \Gamma=(-1,1)\times\{0\}, \\ u=0,\ & \text{on}\ \Gamma_{1}=(-1,1)\times\{1\}, \\ \partial_{x_{2}}v=0,\ & \text{on } \Gamma_{2}=(-1,1)\times\{-1\}. \end{cases}

Clearly, the existence of solution requires a compatiablity condition $$\int_{\Gamma}gdl=\int_{\Omega_{1}}f_{1}+\int_{\Omega_{2}}f_{2}.$$

How to solve for $(u,v)$ and prove the estimate $$\|u\|_{H^{1}(\Omega_{1})}+\|v\|_{H^{1}(\Omega_{2})}\leq C\big(\|g\|_{H^{\frac{1}{2}}(\Gamma)}+\|f_{1}\|_{L^{2}(\Omega_{1})}+\|f_{2}\|_{H^{-1}(\Omega_{2})}\big)\;?$$

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  • $\begingroup$ No boundary conditions on $\{-1,1\}\times (-1,1)$? $\endgroup$ Commented Dec 9, 2021 at 15:37
  • $\begingroup$ @WillieWong Sorry, I should assume that the functions are periodic in the horizontal direction. $\endgroup$ Commented Dec 9, 2021 at 15:39
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    $\begingroup$ This problem is underdetermined. Take u to be a divergence free vector field of compact support in $\Omega_1$. $\endgroup$ Commented Dec 9, 2021 at 18:32

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