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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @JochenGlueck In principle you're right, but both are die-hard spectral theoretical conventions. $\endgroup$ – Delio Mugnolo Oct 8 '18 at 15:51

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