4
$\begingroup$

Under what conditions on $a(x)$ and domain $D$, the spectral gap of the elliptic operator $ \nabla \cdot(a(x)\cdot \nabla)$ defined on $D$, can be controlled?

The boundary condition is that the solution at the boundary is zero. Assume that $D$ is a unit ball in $R^{d}$. Since the eigenvalues of this operator are countable and nonnegative, the spectral gap is the difference between its smallest nonzero eigenvalue and zero.

$\endgroup$
  • 11
    $\begingroup$ Do you impose any boundary conditions? What is the dimension of $D$? Please define precisely what you mean by spectral gap. $\endgroup$ – Liviu Nicolaescu Oct 5 '18 at 18:39
  • $\begingroup$ @LiviuNicolaescu OP has updated the question. $\endgroup$ – Minkov Oct 8 '18 at 4:05
  • 1
    $\begingroup$ Well, by the minimax principle the spectral gap of $\nabla\cdot(a\nabla)$ is not smaller than the spectral gap of $\alpha\Delta$, where $\alpha$ is the essential infimum of $M:=\{\xi^* a(x)\xi:\xi \in {\mathbb C}^d\}$; and not larger than $A\Delta$, where $A$ is the essential supremum of $M$. Now, you can control the spectral gap of $\alpha \Delta$ (resp., $A\Delta$) by invoking the literature devoted to spectral geometry of $\Delta$; for instance, the most classical lower bound for the spectral gap of $\Delta$ in terms of the volume of $\Omega$ is the Faber-Krahn inequality. $\endgroup$ – Delio Mugnolo Oct 8 '18 at 11:23
  • $\begingroup$ @DelioMugnolo: This raises a question as to what happens if the operator is (pointwise) elliptic but not uniformly elliptic. $\endgroup$ – Nate Eldredge Oct 8 '18 at 13:57
  • 1
    $\begingroup$ @JochenGlueck In principle you're right, but both are die-hard spectral theoretical conventions. $\endgroup$ – Delio Mugnolo Oct 8 '18 at 15:51

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.