I am trying to derive the correct variational form for the Poisson equation with pure Neumann boundary conditions, and an additional contraint $\int_{\Omega} u \, {\rm d} x = 0$, as described in this link.

As I understand it, the problem can be defined as:

Find $(u, c) \in V \times \mathbb{R}$ such that

$F(u, c) = \frac1{2} \int \nabla u \cdot \nabla u - \int f u - \int_{\partial \Omega}g u + c \int u$

is minimized wrt $u$ and $c$

To minimize with respect to $u$, I started writing

$\left. \frac{\partial}{\partial \epsilon}F(u + \epsilon v, c)\right|_{\epsilon=0} = 0$

which led to

$\int \nabla u \cdot \nabla v + \int c v = \int f v + \int_{\partial \Omega} g v $.

With respect to $c$ this gives $\int u = 0$.

Just adding these two equalities leads to the variational form mentioned in the link (where $d \int u = 0$ for all $d \in \mathbb{R}$ ensures that the two equations remain true separatly). Is this correct so far?

Now, in the finite element description, we have one more unknown, $c$, i.e. the Lagrange multiplier, no? To go to a set of equations, the variational form of the equation must be true for all $N$ testfunctions (shapefunctions) $v$. Am I right that this will lead to a $N \times (N + 1)$ matrix equation? Am I also correct that, since the expression should be valid for all $d \in \mathbb{R}$ we can pick a random value in the actual evaluation of these forms?

  • $\begingroup$ Notice in the link that they add the two equations together, but include also two unknowns - the solution + Lagrange multiplier. There are also two sets of test functions - one for the solution and one for the Lagrange multiplier, so they end up with a square system. $\endgroup$
    – Jesse Chan
    Commented Nov 7, 2014 at 4:09
  • 1
    $\begingroup$ In your question, you mention $\lambda$, but that doesn't appear in the formulation. You should replace it by $c$. Furthermore, the statement "is minimized with respect to $u$ and $c$" is wrong. What you are looking for is not a minimizer but a stationary point of $F$. The stationary point will turn out to be a saddle point for your problem. $\endgroup$ Commented Nov 7, 2014 at 4:42
  • $\begingroup$ @Wolfgang The $\lambda$ indeed was a leftover of my own notation, I corrected it. $\endgroup$
    – Maarten
    Commented Nov 7, 2014 at 13:33
  • $\begingroup$ @JLC Am I correct in believing that to build this matrix, the expression $\int \nabla u \cdot \nabla v + \int c v = \int f v + \int_{\partial \Omega} g v $ leads to $N$ equations, and the constraint $\int u = 0$ accounts for the last equation? That would not require a testfunction $d$ (I have no idea what else is meant with a 'test function' for the space of real numbers). $\endgroup$
    – Maarten
    Commented Nov 7, 2014 at 13:40
  • $\begingroup$ I call it a test function for the Lagrange multiplier, but it's really just adding an extra equation (the test function for a real number is a constant). Notice that in their link, they have $a((u,c),(v,d)) = \int_\Omega \nabla u\nabla v + \int_\Omega c v + \int_\Omega d u$; the last term is what you're missing in your equation (and what leads to the "saddle point" Wolfgang mentioned). $\endgroup$
    – Jesse Chan
    Commented Nov 7, 2014 at 13:58


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