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


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