# Ngô Quốc Anh

## May 9, 2010

### The lowest eigenvalue of the Laplacian for the intersection of two domains

Filed under: PDEs — Tags: , , — Ngô Quốc Anh @ 5:59

Let us begin with a discussion of the geometric question. If $A$ is an open set in $\mathbb R^n$ (bounded or unbounded), let $\lambda(A)$ denote the lowest eigenvalue of $-\Delta$ in $A$ with Dirichlet boundary conditions. $\lambda(A)=-\infty$ if $A$ is empty. Intuitively, if $\lambda(A)$ is small then $A$ must be large in some sense. One well known result in this direction is the Faber-Krahn inequality which states that among all domains with a given volume $|A|$, the ball has the smallest $\lambda$. Thus,

$\displaystyle\lambda(A) \geqslant \beta_n\frac{1}{|A|^\frac{2}{n}}$

where $\beta_n$ is the lowest eigenvalue of a ball of unit volume. This inequality clearly does not tell the whole story. If $\lambda(A)$ is small then $A$ must not only have a large volume, it must also be “fat” in some sense.

Let us place here a very beautiful result due to Lieb among other big contributions. This result was published in Invent. Math. during 1983. The proof relies upon the Rayleigh quotient and a very clever choice of a trial function for the variational characterization of $\lambda(A\cap B_x)$. For the whole paper, we refer the reader to here.

Theorem. Let $A$ and $B$ be non-empty open sets in $\mathbb R^n$ ($n\geqslant 1$), and $\lambda(A)$ and $\lambda(B)$ be the lowest eigenvalue of $-\Delta$ with Dirichlet boundary conditions. Let $B_x$ denote $B$ translated by $x\in \mathbb R^n$. Let $\varepsilon>0$. Then there exists an $x$ such that

$\displaystyle\lambda(A \cap B_x)<\lambda(A)+\lambda(B) +\varepsilon$.

If $A$ and $B$ are both bounded then there is an $x$ such that

$\displaystyle\lambda(A \cap B_x)<\lambda(A)+\lambda(B)$.

Before proving the theorem, let us recall the so-called Rayleigh quotient. Precisely