# Failure of Fredholm property of elliptic PDE systems

Roughly speaking, a PDE operator satisfies the Fredholm property if its principal symbol is elliptic and the information provided on the boundary satisfies the Shapiro-Lopatinskii condition.

What can happen when one of these conditions is not met, but the other holds:

• In case of failure of the Shapiro-Lopatinskii boundary condition (while having ellipticity). Does one automatically have an infinite dimensional kernel?

• What happens when the operator is elliptic everywhere but at some point, while the Lopatinskii condition is still satisfied? Can the PDE problem still be well-posed?

Is there a clear picture of what happens in these situations, or are there counterexamples of different type, depending on regularity of coefficients and boundary?

Thank you for any information.

-

$\Delta u=0$, say on the unit disk, with boundary condition $u_{xx}=0$. The kernel is given by linear combinations of $1$, $x$, $y$ and $xy$, so it is four-dimensional.
ODE problems like $(x^2y')'-y=0$ with Dirichlet conditions $y(1)=y(-1)=0$.