We consider a coupled PDE in a rectangle $\Omega=(-1,1)\times(-1,1)$. \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 $$\|u\|_{H^{1}(\Omega_{1})}+\|v\|_{H^{1}(\Omega_{2})}\leq C\big(\|g\|_{H^{\frac{1}{2}}(\Gamma)}+\|f_{1}\|_{L^{2}(\Omega_{1})}+\|f_{2}\|_{H^{-1}(\Omega_{2})}\big)\;?$$

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 $$\|u\|_{H^{1}(\Omega_{1})}+\|v\|_{H^{1}(\Omega_{2})}\leq C\big(\|g\|_{H^{\frac{1}{2}}(\Gamma)}+\|f_{1}\|_{L^{2}(\Omega_{1})}+\|f_{2}\|_{H^{-1}(\Omega_{2})}\big)\;?$$

We consider a coupled PDE in a rectangle $\Omega=(-1,1)\times(-1,1)$. \begin{equation*} \left\lbrace \begin{split} \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{split} \right. \end{equation*}

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 $(u,v)$ and prove the estimate $$||u||_{H^{1}(\Omega_{1})}+||v||_{H^{1}(\Omega_{2})}\leq C(||g||_{H^{\frac{1}{2}}(\Gamma)}+||f_{1}||_{L^{2}(\Omega_{1})}+||f_{2}||_{H^{-1}(\Omega_{2})}).$$

We consider a coupled PDE in a rectangle $\Omega=(-1,1)\times(-1,1)$. \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 $$\|u\|_{H^{1}(\Omega_{1})}+\|v\|_{H^{1}(\Omega_{2})}\leq C\big(\|g\|_{H^{\frac{1}{2}}(\Gamma)}+\|f_{1}\|_{L^{2}(\Omega_{1})}+\|f_{2}\|_{H^{-1}(\Omega_{2})}\big)\;?$$