Variational formulation of a Elliptique -PDE and prove that the problem resultating is well-posed

Question

Determine the variational formulation of \begin{cases} -\Delta u(\mathbf{x})=1, & \mathbf{x}\in (0,1)\times (0,1) \\ -\partial_{x}u(\mathbf{x})+c(\mathbf{x})u(\mathbf{x})=0, & \mathbf{x}\in \{0\}\times (0,1)\\ \partial_{y}u(\mathbf{x})=0, & \mathbf{x} \in (0,1)\times \{0\}\cup (0,1)\times \{1\}\\ \partial_{x}u(\mathbf{x})=0, & \mathbf{x} \in\{1\}\times (0,1) \end{cases}

My approach: Let $v\in H^{1}(\Omega)$ and assuming that $u\in H^{2}(\Omega)$, so we have that \begin{eqnarray*} -\Delta u(\mathbf{x})&=&1\\ -\Delta u(\mathbf{x})v(\mathbf{x})&=& v(\mathbf{x})\\ \int_{\Omega}-\Delta u(\mathbf{x})v(\mathbf{x})d\mathbf{x}&=& \int_{\Omega}v(\mathbf{x})d\mathbf{x} \end{eqnarray*} Using Green's formula, we obtain that \begin{eqnarray*} \int_{\Omega}-\Delta u(\mathbf{x})v(\mathbf{x})d\mathbf{x}&=& \int_{\Omega}v(\mathbf{x})d\mathbf{x}\\ \int_{\Omega}\mathbf{\nabla u(x)}\cdot \mathbf{\nabla v(x)}d\mathbf{x}-\int_{\partial\Omega}\gamma_{1}u(\mathbf{x})\gamma_{0}v(\mathbf{x})ds_{\mathbf{x}}&=&\int_{\Omega}v(\mathbf{x})d\mathbf{x} \end{eqnarray*}

Thank you @VoB using your answer I was able to advance from where I was stuck: The cantity $\displaystyle \int_{\partial\Omega}\gamma_{1}u(\mathbf{x})\gamma_{0}v(\mathbf{x})ds_{\mathbf{x}}$ is equal to $\displaystyle \int_{\partial \Omega} v (\mathbf{\nabla u(\mathbf{x}) \cdot n})ds_\mathbf{x}$ we could find the last cantity in terms of the (BC), so the variational formulations is given by $$ \quad u\in H^{1}(\Omega): \quad a(u,v)=\ell(v), \quad \forall v\in V$$ where $$a(u,v):=\int_{\Omega} \mathbf{\nabla u(x)}\cdot \mathbf{\nabla v(x)}d\mathbf{x}-\int_{\partial \Omega} v (\mathbf{\nabla u(\mathbf{x}) \cdot n})ds_\mathbf{x}$$ and $$\ell(v):=\int_{\Omega} v(\mathbf{x})d\mathbf{x}$$

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Green's formula: Let $\Omega$ a bounded open set in $\mathbb{R}^{n}$ with border continous-Lipschitz $\partial \Omega$. Then, $\forall v\in H^{1}(\Omega)$ and $u\in H^{2}(\Omega)$ we have $$\int_{\Omega} v \Delta u=-\int_{\Omega}\mathbf{\nabla u\cdot \nabla v}+\int_{\partial \Omega}\gamma_{0}(v)\gamma_{0}(\mathbf{\nabla u)\cdot \nu}$$ where $\gamma_{0}(\mathbf{\nabla u})$ is the vector that is result of apply $\gamma_{0}$ to the components of $\mathbf{\nabla u}$.

Answer 1

After taking $v \in H^1(\Omega)$:

$$\int_{\Omega} \nabla u \cdot \nabla v dx- \int_{\partial \Omega} v (\nabla u \cdot n)ds_x = \int_{\Omega} vdx$$

Now, using the additivity of the integral on $\partial \Omega$ and the boundary conditions on the unit square:

$$\int_{\partial \Omega} v (\nabla u \cdot n)ds_x = \int_{(0,1) \times \{0\} \cup (0,1) \times \{1\}} v \partial_x u \cdot n ds_x + \int_{\{1\} \times(0,1)}v \partial_y u \cdot n_2 ds_x + \int_{\{0\} \times (0,1)} v (c(x) u(x) + \partial_y u \cdot n_2 )ds_x$$