Let $\Omega \subset \mathbb{R}$ in a bounded polygon domain and $f:\Omega \to \mathbb{R}$ known function.We split the boundary into two parts $\partial \Omega_{1}$ and $\partial \Omega_{2}$ such that $\partial \Omega =\partial \Omega_{1} \cup \partial \Omega_{2}.$ The elliptic problem of Dirichlet & Neumann conditions is :
\begin{align*} -\Delta u +u &=f \quad in \quad \Omega \subset\mathbb{R}^2 \\ u&=0 \quad in \quad \partial \Omega_{1} \\ \frac{\partial u}{\partial \mathbf{n}}&=1 \quad in \quad \partial \Omega_{2} \end{align*}
It was asked to write the problem in a weak form in a suitable solution space $\mathcal{V}$ and the bilinear form $a(\cdot,\cdot):\mathcal{V} \times \mathcal{V} \to \mathbb{R}$.
My thought :
\begin{align*} -\Delta u+u &=f \\ -\Delta u v+uv &=fv \\ -\int_{\Omega}\Delta u v dV+ \int_{\Omega}uv dV&=\int_{\Omega}fv dV \\ \int_{\Omega}\nabla u \cdot \nabla v dV- \int_{\partial \Omega_{2}}v \nabla u \cdot \vec{n} dS + \int_{\Omega}uv dV&=\int_{\Omega}fv dV \\ \int_{\Omega}\nabla u \cdot \nabla v dV- 1 + \int_{\Omega}uv dV&=\int_{\Omega}fv dV \\ \end{align*} Is -1 right ?
Or :
$$\int_{\Omega}\nabla u \cdot \nabla v dV- \int_{\Omega} v dx + \int_{\Omega}uv dV=\int_{\Omega}fv dV$$
because $$\int_{\partial \Omega}v \nabla u \cdot \vec{n} dS = \int_{\Omega_{2}} v \cdot 1 dx$$
Is that correct ? How you calculate the integral in a polygon domain such this one with these boundary conditions ?
Solution \begin{align*} \int_{\partial \Omega}v \nabla u \cdot \vec{n} dS &=\int_{\partial\Omega_{1}} v \cdot \frac{\partial u}{\partial n} - \int_{\partial\Omega_{2}} v \cdot \frac{\partial u}{\partial n} dx \\ &=0 - \int_{\partial\Omega_{2}} v \cdot 1 dx \\ &=- \int_{\partial\Omega_{2}} v dx \end{align*}
$$\int_{\Omega}\nabla u \cdot \nabla v dV +\int_{\partial\Omega_{2}} v dx + \int_{\Omega}uv dV=\int_{\Omega}fv dV$$