# Poisson Problem, Fictitious Domain Method, Lagrange Multiplier

Consider the problem \begin{align} - \Delta u &= f & in \text{ } \omega \\ u &= g & on \text{ } \gamma \end{align} $\omega$ is a bounded domain of $\mathbb{R}^d$ $(d\ge0)$ and $\gamma$ its boundary $\delta \omega$. We assume that $u \in H^1(\omega)$, $f \in L^2(\omega)$ and $g \in H^{\frac{1}{2}}(\gamma)$.

Let us consider a 'box' $\Omega$ which is a domain in $\mathbb{R}^d$ such that $\omega \subseteq \Omega$. Now I want to derive the weak form by the Lagrange Multiplier Method. The corresponding minimization problem: $$u = \arg\min_{v \in H^1(\Omega),\\ v = g \text{ on } \gamma } \frac12 \int_\Omega |\nabla v|^2\,\mathrm{d}x - \int_\Omega fv\,\mathrm{d}x.$$ We define a Lagrangian functional $\mathscr{L}: H^1(\Omega) \times H^{-\frac12}(\gamma)\rightarrow \mathbb{R}$ $$\mathscr{L} (v, \mu) = \frac12 \int_\Omega |\nabla v|^2\,\mathrm{d}x - \int_\Omega \tilde{f}v\,\mathrm{d}x- <\mu, v - g>$$ where $\tilde{f} \in L^2(\Omega)$ and $\tilde{f}|_{\omega}=f$, $<.,.>$ denotes the duality pairing between $H^{-\frac12}(\gamma)$ and $H^{\frac12}(\gamma)$. How can I derive this two equations: \begin{aligned} \int_\Omega \nabla \tilde{u} \cdot \nabla v \,\mathrm{d}x&= \int_\Omega \tilde{f}v\,\mathrm{d}x + <\lambda, v> \quad &\forall v \in H^1(\Omega),\\ <\mu, \tilde{u}-g> &= 0 \quad &\forall \mu \in H^{-\frac12}(\gamma),\\ \end{aligned} $\tilde{u} \in H^1(\Omega)$ and $\lambda \in H^{-\frac12}(\gamma)$.

The first (or the second) equation is equivalent to setting the Fréchet derivative of the Lagrangian with respect $v$ (or with respect to $\mu$) to zero.
An example how to obtain the first equation: $$0 = \frac{\partial \mathscr{L}(u + \epsilon v, \lambda)}{\partial \epsilon}\Bigg|_{\epsilon = 0} = \int_\Omega \nabla (u+\epsilon v) \cdot \nabla v \,\mathrm{d}x - \int_\Omega f v \,\mathrm{d}x + \langle \lambda, v \rangle \Bigg|_{\epsilon = 0}$$ This holds for any direction $v \in H^1(\Omega)$. Remark: It seems to me that your sign for $\langle \lambda, v \rangle$ is wrong.
• Thank you, that's what I was looking for. Yes you are right, the sign in front of the $<\mu, v - g>$ in $\mathscr{L} (v, \mu)$ was wrong. I changed it now from $+$ to $-$. – Felix Nov 1 '17 at 14:20