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.


Example for first question:

$\Delta^2 u=0$, say on the unit disk, with boundary condition ${\partial\over\partial n}\Delta u=0$, $\Delta u-u=0$. The boundary conditions do not satisfy the Lopatinskii condition (note that the $u$ term is lower order). Nevertheless, it follows from the first boundary condition that $\Delta u$ is constant, and then the second boundary condition implies that $u$ is constant on the boundary. So the kernel is one-dimensional.

Example for second question:

ODE problems like $(x^2y')'-y=0$ with Dirichlet conditions $y(1)=y(-1)=0$.

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