Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.