I'm stuck at proving the coercivity of the bilinear functional in the variational formulation of the problem:

\begin{array}{c} -\nabla^{2} u=f \quad \text { in } \Omega \\ u=g_{D} \text { on } \partial \Omega_{D} \quad \text { and } \quad \frac{\partial u}{\partial n}=g_{N} \text { on } \partial \Omega_{N} \end{array}

where $\partial \Omega_{D} \cup \partial \Omega_{N}=\partial \Omega$ and $\partial \Omega_{D}$ and $\partial \Omega_{N}$ are distinct.

The variational formulation is :

Find $u \in \mathcal{H}_{E}^{1}$ such that $$ \int_{\Omega} \nabla u \cdot \nabla v=\int_{\Omega} v f+\int_{\partial \Omega_{N}} v g_{N} \quad \text { for all } v \in \mathcal{H}_{E_{0}}^{1} $$ where: \begin{aligned} \mathcal{H}_{E}^{1} &:=\left\{u \in \mathcal{H}^{1}(\Omega) | u=g_{D} \text { on } \partial \Omega_{D}\right\} \\ \mathcal{H}_{E_{0}}^{1} &:=\left\{v \in \mathcal{H}^{1}(\Omega) | v=0 \text { on } \partial \Omega_{D}\right\} \end{aligned}

The proof by contradiction holds for $a(u, v)=\int_{\Omega} \nabla u \cdot \nabla v+\int_{\partial \Omega} uv$, but here, our bilinear form doesn't contain the term $\int_{\partial \Omega} uv$

Thank you for your help

  • 3
    $\begingroup$ Have you looked into any of the usual finite element books where this is proved? You will need the Poincare inequality. $\endgroup$ – Wolfgang Bangerth May 13 '20 at 14:55
  • $\begingroup$ I think Poincare inequality work just for $H_{0}^{1}(\Omega)$, could you please suggest for me a reference, I read some of them but they don't explain why the solution exists and unique $\endgroup$ – Almendrof66 May 14 '20 at 4:02
  • $\begingroup$ Poincaré inequality can surely be extended to any situation where the trace of the function is constrained on a $d-1$ dimensional subset of $\Omega \in \mathbb{R}^d$ to some given values - in this case $u |_{\partial \Omega_D} = g_D$. See e.g. the classical technique for proving Poincaré inequality in Evans - Partial Differential Equations. $\endgroup$ – knl May 15 '20 at 7:27
  • $\begingroup$ Also, your space ${\cal H}^1_{E_0}$ is precisely $H^1_0$. $\endgroup$ – Wolfgang Bangerth May 15 '20 at 15:43
  • 1
    $\begingroup$ @Almendrof66 Not directly, but basically everything that holds for $H^1_0$ also holds in some modified form for functions that are zero on only parts of the boundary. In particular, what you need to show here is that the only functions that are in the kernel (null space) of the gradient are the constant functions (obvious) and that the only constant function that is in your space is the zero function (because it needs to be zero at least at one point of the boundary. That's all you need. $\endgroup$ – Wolfgang Bangerth May 16 '20 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.