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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.

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1 Answer 1

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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