# Some formulations of domain coupling lead to saddle point problems. Is this merely an artifact?

Background I wanted to learn how to couple FEM and BEM (for the Poisson equation), because I wanted to better understand how open boundary conditions look like. Therefore I worked through the relevant sections of “Principles of Boundary Element Methods” by Martin Costabe, summer course lecture, 1986 (28 pages). The last section provides a formulation of the coupling, which leads to a symmetric saddle point problem. Of course, I would prefer a symmetric positive definite problem instead.

Even without BEM, some formulations of domain coupling lead to saddle point problems. Consider two domains $$\Omega_1$$ and $$\Omega_2$$ with common boundary $$\Gamma$$, and setup Poisson equations with homogeneous Neumann boundary conditions on $$\Gamma$$ in both domains. To couple the two domains, require that the potentials agree on the boundary, and add the transpose of this restriction by a Lagrange multiplier (to account for the fact that the normal derivative on both sides of $$\Gamma$$ is no longer restricted to 0, but must only agree with the normal derivative on the other side). The block structure of the matrix of the resulting linear system is

$$\pmatrix{A_1 & 0 & B_1 \\ 0 & A_2 & B_2 \\ C_1 & C_2 & 0}$$

Here $$C_1 = B_1^T$$ and $$C_2 = B_2^T$$.

In this case, $$C_1$$ and $$C_2$$ have a simple structure, like $$C_1=\pmatrix{0 & I}$$ and $$C_2 = \pmatrix{-I & 0}$$ if we order the nodes suitably. Therefore, it is often possible to eliminate the constraint and the Lagrange multiplier explicitly. My confusion is that resulting system seems to be easier to solve than directly trying to solve the initial saddle point problem.

What is different when I try to couple FEM and BEM is that the structure feels more complicated to me, and I know longer know how to explicitly eliminate the Lagrange multiplier. So now I try to understand why certain approaches to couple domains lead to saddle point problems in the first place. One of my guesses is that the Lagrange multiplier which occur in this formulation effectively correspond to the normal derivative of the potential on the inner interface. But why should a formulation where both the potential and (directional) derivatives of the potential occur lead to saddle point problems? (And is there a simple why to avoid saddle point problems, or at least a simple way to solve the resulting saddle point problems without sacrificing performance of iterative solvers like CG?)

• Regarding the solution of saddle point systems with positive subproblems, I think the Uzawa iteration is a good place to start for ideas. – rchilton1980 Jun 25 '20 at 12:59
• @rchilton1980 Part of my confusion is probably that the Uzawa iteration felt like a huge step back compared to CG to me. But that may be the wrong way to look at it. Since $A_1$ and $A_2$ are decoupled, I can solve them independently (and approximatively) by CG iterations if I want, and the resulting Schur complement which is effectively solved by the (inexact) Uzawa iteration might be well conditioned without any preconditioning. – Thomas Klimpel Jun 25 '20 at 13:44
• Yeah, I don't mean to imply that the Uzawa method is rocket science - just a structure-aware algorithm that transforms your indefinite/saddle system into 2 (or more) positive subsystems, where the last (schur complement) subsystem has been transformed with a well-placed minus sign, from negative definite to positive definite. It's basically CG for saddle systems. – rchilton1980 Jun 25 '20 at 14:28
• Don't use Uzawa. Use something like GMRES with a block-aware (e.g., block-triangular) preconditioner. – Wolfgang Bangerth Jun 26 '20 at 15:23