Deriving Variational Formulation

Question

Determine the variational formulation of

\begin{cases} -\Delta u+u=xy \quad& \text{in } \Omega\\ \nabla u\cdot \vec{n}+2u=3 \quad& \text{in } \partial\Omega \end{cases}

What I have tried:

\begin{align} \int_{\Omega}(-\Delta u+u)v &= \int_{\Omega}xyv \\ \int_{\Omega}uv-\int_{\Omega}\Delta uv &= \int_{\Omega}xyv \\ \int_{\Omega}uv-\int_{\Omega}\nabla u\cdot\nabla dv-\int_{\partial\Omega}\frac{\partial u}{\partial n}vds &= \int_{\Omega}vydx+vxdy \end{align}

How should I continue this? Is this correct?

Answer 1

The main idea is correct, note that there is two small mistakes

• In the source term, you cannot write $\int xy v dxdy$ as $\int vydx+vxdy$, as the second does not make sense.
• In the integration by part there is a sign error $$-\int_\Omega \Delta u v = {\color{red} +} \int_\Omega \nabla u \cdot \nabla v - \int_{\partial \Omega} \frac{\partial u}{\partial n} v ds.$$

In order to continue you must use the boundary condition $\partial_n u + 2u=3$ to write $$\int_{\partial \Omega} \frac{\partial u}{\partial n} v ds = -\int_{\partial \Omega}2 u v ds +\int_{\partial \Omega} 3v ds.$$

The variational formulation is then $\forall v \in H^1$ $$\underbrace{\int_\Omega \nabla u \cdot \nabla v + \int_\Omega u v +\int_{\partial \Omega} 2 u v ds}_{A(u,v)}= \underbrace{ \int_\Omega xy v + \int_{\partial \Omega} 3 v ds.}_{L(v)}$$

The last technical difference with Dirichlet or Neumann boundary conditions is to use trace theorems to check that these operators are continuous.