Solving pure Neumann problem enforcing B.C. with Lagrange Multiplier

Question

I want to solve the Laplace Equation with pure Neumann B.C. using Finite Element Method:

$- \Delta u = f \ $ in $ \ \Omega $

$- \partial u/\partial n = g \ $ on $ \ \Gamma = \partial \Omega$

With weak formulation

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

To obtain a unique solution, I was following the Lagrange Multiplier method, enforcing a constraint of the type

$\int_{\Omega} v \ dx= 0 $.

The following linear system would then be obtained

$\begin{pmatrix}A & B^T \\ B & 0\end{pmatrix} \begin{pmatrix}U \\ \lambda \end{pmatrix} = \begin{pmatrix}F \\ 0\end{pmatrix}$

Where $A$ corresponds to the LHS (stiffness matrix), $F$ to the RHS (load vector), $U$ to the solution vector and $\lambda$ to the Lagrange Multiplier, which can be discarded from the resulting solution vector.

According to Larson, Bengzon in the book "The Finite Element Method" (p. 95),

The Lagrangian multiplier $\lambda$ may be thought of as a force acting to enforce ̋the constraints. Because the zero mean value on $u_h$ is a constraint, which do not alter the solution to the underlying Neumann problem, the force should vanish or, at least, be very small.

I would like to know how small this value should be, and how to analyze the simulation results based on this (are the results valid if the multiplier is too large? Should I compare it based on the minimum value obtained from my solution? etc.).

Thanks

Answer 1

What is small is relative. I gave a few examples here to explain that "small" must be seen in the context of the other terms in your equation. For example, the total force exerted on the membrane that your equation describes is $\int_\Omega |f(x)| \; dx$, and the total boundary forces are $\int_{\partial\Omega} |g(x)| \; dx$. The Lagrange multiplier $\lambda$ should be small compared to these forces.