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Consider a Neumann-Poisson problem. \begin{align*} -\Delta u & = f \ \mathrm{in}\ \Omega\\ \frac{\partial u}{\partial\nu} & = g \ \mathrm{on}\ \partial\Omega \end{align*} Then $u$ is uniquely defined only up to an additive constant, i.e. if $u$ is a solution, $\tilde u = u+c$, $c\in\mathbb R$ is a solution, too.

To get a unique solution we introduce the constraint $\int_\Omega u dx =0$. Now I want to do a Finite Element discretization for this problem and I have problems introducing the constraint.

The variational formulation of the PDE reads:

Find $u\in H^1(\Omega)$ such that $$\int\nabla u\nabla\varphi d x = \int_\Omega f\varphi dx + \int_{\partial\Omega} g\varphi d\sigma\qquad \forall \varphi\in H^1(\Omega)$$

I read that using Lagrange multipliers I can introduce the constraint by

Find $\lambda\in\mathbb R$ such that $$c\int_\Omega udx = \lambda\int_\Omega \varphi dx \qquad\forall \varphi\in H^1(\Omega), c\in\mathbb R$$

but I am completely clueless as to why this is a variational formulation of the constraint.

Can anyone help me with this? Or point me in another direction on how to introduce the constraint? Thanks!

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    $\begingroup$ According to this paper, "some practitioners" constrain $u$ by prescribing its value at a particular node. This gives a unique solution, which you can integrate and subtract a suitable constant to get the constraint you wanted. $\endgroup$ – user147263 May 16 '14 at 20:23
  • $\begingroup$ I will give this a read, thank you! $\endgroup$ – dinosaur May 16 '14 at 20:53
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The standard way to impose the mean-value zero condition is to modify the weak formulation as follows: $$(\nabla u, \nabla v) + \epsilon (u,v) = (f,v) + \langle g,v \rangle,$$ where $\epsilon$ is a small positive number, $$(a,b):= \int_\Omega a \cdot b \,\mathrm{d}x,$$ and $$\langle a,b \rangle:= \int_{\partial \Omega} a \cdot b \,\mathrm{d}s.$$ To see that this gives mean-value zero, substitute $v=1$ to get $$\epsilon (u,1) = (f,1) + \langle g,1 \rangle = 0,$$ where the right-hand side is zero due to the combatibility condition of $f$ and $g$ (i.e., Newton's second law in every physically-interpretable problem). Thus, you get $$(u,1)=0,$$ which is exactly what you want. In practice, you want to try out different values for $\epsilon$. In theory, it should be chosen such that $\epsilon = Ch$ where $C$ is some positive constant.

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  • $\begingroup$ This is very helpful, thank you! $\endgroup$ – dinosaur May 26 '14 at 17:19

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