FETI domain decomposition - kernel of local stiffness matrices Consider the differential equation
\begin{align*}
-\Delta u&=f\mathrm{\ in\ }\Omega \\
u & = 0 \mathrm{\ on\ }\partial\Omega
\end{align*}
with $\Omega=(0,1)^2$. We're splitting $\Omega$ into $N\times N$, $N\in\mathbb N$ compatible quadratic subdomains $\Omega_i$, $1\leq i\leq N^2$ and use $\mathcal P_1$-functions for finite element discretization. Let $K^{(N)}:=\sum_{i=1}^{N^2} K_i$, with $K_i$ the local stiffness matrix.
Can $K_i$, $1\leq i\leq N$, have a (non-trivial) kernel? Where do these kernels come from and are they depending on the choice of the finite element space $V^h$?
How do the the finite element space $V^h$ and the bilinear form $a(\cdot,\cdot):V\times V\to \mathbb R$ belonging to the variational formulation of the PDE have to be connected so one can see the kernel of the problem in the stiffness matrix.
I am kind of confused with these questions. I know what the kernel of a matrix is and I think the local stiffness matrices $K_i$ have a trivial kernel in case the subdomain $\Omega_i$ intersects with $\partial\Omega$ since our Dirichlet boundary gives rise to a unique solution here.
For subdomains in $\Omega$ I think we don't have this uniqueness. I read somewhere that this leads to $\ker K_i=\mathrm{span}\{1\}$, but I don't know why this is.
Can anyone tell me why one is so interested in these kernels? Is it because our finite element discretization has a uniquely solution if the kernel of the belonging system of equations is trivial?
And what does the finite element space used have to do with all this? I am really confused with this domain splitting thing.
I would be really grateful if someone could shed some light onto this. Thank you in advance!
 A: *

*Can $K_i$ have a non-trivial kernel? Yes, they do. In your example, if $\Omega_i$ is in the interior of the domain, the local stiffness will be the same as the discretization of your problem with Neumann conditions on $\partial \Omega_i$. This problem is not well-posed, since for any solution $u$, $u+c$ is also a solution (where $c \in \mathbb{R}$). So, the kernel is spanned by the constant functions ( $span{1}$ ).

*I would say that the kernel usually does not depend on the finite element space. They come from the restriction of the problem to the subdomain, which might be ill-posed depending on the way the restriction is done. Some domain decompositions methods can be defined at the continuous level, so the finite element plays no role (supposing that the finite element is suitable for the initial global problem).

*Finally, these kernels are interesting because, ideally, you would like to solve problems that only use the local stiffness (not the global one). But if you want to solve only local problems, you have to be sure that the solution is well-defined, i.e. that the kernel is trivial.


Hope it helps!
