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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 $\lambda \in H^{-\frac12}(\gamma)$.

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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.

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  • $\begingroup$ 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 $-$. $\endgroup$ – Felix Nov 1 '17 at 14:20
  • $\begingroup$ Just curious: What is your end goal? Are you planning to use this method in practice? $\endgroup$ – knl Nov 1 '17 at 14:25
  • $\begingroup$ I use the Nitsche method in practice. I just want to compare Nitsche to Lagrangian and to Penalty in my master thesis. The goal is to find a fictitious domain method which can be calculated very fast. It will be used in cancer therapy. $\endgroup$ – Felix Nov 1 '17 at 14:44
  • $\begingroup$ Maybe you are already aware of the connection between Nitsche's method and Lagrange multiplier method, but nevertheless see the paper by R. Stenberg on the subject. Especially page 146 where he demonstrates that the Nitsche's method is a special case of the stabilized Lagrange multiplier method. $\endgroup$ – knl Nov 2 '17 at 10:31
  • $\begingroup$ Sounds interesting. I will have a look at it. $\endgroup$ – Felix Nov 2 '17 at 10:50

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