# Spectral Gap of Elliptic Operator

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.

• Do you impose any boundary conditions? What is the dimension of $D$? Please define precisely what you mean by spectral gap. – Liviu Nicolaescu Oct 5 '18 at 18:39
• @LiviuNicolaescu OP has updated the question. – Minkov Oct 8 '18 at 4:05
• 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. – Delio Mugnolo Oct 8 '18 at 11:23
• @DelioMugnolo: This raises a question as to what happens if the operator is (pointwise) elliptic but not uniformly elliptic. – Nate Eldredge Oct 8 '18 at 13:57
• @JochenGlueck In principle you're right, but both are die-hard spectral theoretical conventions. – Delio Mugnolo Oct 8 '18 at 15:51