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

\displaystyle\lambda (A) = \inf \left\{ {J(f):f \in H_0^1(A),f \ne 0} \right\} = \inf \left\{ {J(f):f \in C_0^\infty (A),f \ne 0} \right\}


\displaystyle J(f) = \frac{{\int_A {{{\left| {\nabla f} \right|}^2}dx} }}{{\int_A {{{\left| f \right|}^2}dx} }}.

We turn now to the proof of the theorem.


Part 1. There exist f \in C_0^\infty(A), g\in C_0^\infty(B) f,g \ne 0 such that

J(f)<\lambda(A)+\frac{\varepsilon}{2}, \quad J(g)<\lambda(B)+\frac{\varepsilon}{2}.

Here, we think of f as defined on all of \mathbb R^n with f(x)=0 for x\notin A and similarly for g. Without loss f and g can be assumed to be real valued. The normalization

\displaystyle\int f^2(x)dx = \int g^2(x)dx = 1

is imposed. The functions h_x given by

h_x(y) = f(y) g(y- x)

satisfy h_x\in C_0^\infty (A \cap B_x). Let

\displaystyle T(x) = \int {{{\left| {\nabla {h_x}} \right|}^2}(y)dy} , \quad D(x) = \int {{{\left| {{h_x}(y)} \right|}^2}dy}.


\displaystyle\int {D(x)dx} = \iint {f{{(y)}^2}g{{(y - x)}^2}dydx} = 1.

We now compute

\displaystyle {\left| {\nabla {h_x}} \right|^2}(y) = {\left| {\nabla f} \right|^2}(y){g^2}(y - x) + {f^2}(y){\left| {\nabla g} \right|^2}(y - x) + \frac{1}{2}(\nabla {f^2})(y) \cdot (\nabla {g^2})(y - x).

The last term can be written as

\displaystyle - \frac{1}{2}(\nabla {f^2})(y) \cdot {\nabla _x}{g^2}(y - x).

Thus the integral (over x) of this term vanishes and

\displaystyle\int {T(x)dx} = \int {{{\left| {\nabla f} \right|}^2}(x)dx} + \int {{{\left| {\nabla g} \right|}^2}(x)dx} < \lambda (A) + \lambda (B) + \varepsilon \equiv \Lambda .


\displaystyle\int {\left( {T(x) - \Lambda D(x)} \right)dx} < 0

and hence

\displaystyle\Lambda D(x) > T(x) \geqslant 0

on a set of positive measure. The proof then follows from the Rayleigh quotient.

Part 2. Since A and B are bounded there exist F \in H_0^1(A) and G \in H_0^1(B) such that

J(F)=\lambda (A), \quad J(G)=\lambda (B),


\displaystyle \int F^2=\int G^2 = 1.

This is a simple consequence of the Rellich-Kondrachov compactness theorem. (Again, we extend F to all of \mathbb R^n with F(x)=0, x \notin A, and similarly for G; it is easy to see that F, G \in H_0^1(\mathbb R^n).) Define


Since F,G \in L^2, H_x \in L^1. Likewise, since \nabla F and \nabla G are L^2 functions,

W_x(y)=(\nabla F)(y)G(y-x)+F(y) \nabla G(y-x) \in L^1.

It is easy to see (by approximating F, G by C_0^\infty functions) that \nabla H_x = W_x in the sense of distributions.

It is not a-priori clear that H_x or W_x \in L^2. (They are, in fact, in L^2 because F and G\in L^\infty. However, this latter fact is not elementary and we prefer to avoid using it. For our purpose it suffices to show that H_x and W_x \in L^2 for almost all x, and this can be done by the following elementary argument.) We note that (by Fubini) D(x)=\int H_x^2 satisfies \int D(x)dx=l. Thus, D(x)<\infty a.e. and H_x \in L^2 a.e. Likewise, repeating the argument for T(x)=\int |W_x|^2 we get

\displaystyle\int T(x) dx =\lambda(A)+\lambda(B) \equiv \Lambda.

(Here, one has to note as above that \nabla F(y)G(y-x) \in L^2(dy) a.e. and F(y)\nabla G(y-x) \in L^2(dy) a.e. By the Schwartz inequality,

Z_x(y) \equiv F(y)G(y-x)\nabla F(y)\cdot \nabla G(y-x) \in L^1(dy)

a.e. and

Z_x(y) \in L^1(dxdy).

Finally, 2G\nabla G = \nabla G^2 in the distributional sense, whence \iint (Z_x(y)dxdy=0.). Thus, for almost all x, H_x \in H_0^1(dy) and \nabla H_x = W_x. We have that \int (T-AD)=0. The remainder of the proof is as before, except that in order to prove the strict inequality, i.e. we must show that

T(x)=\Lambda D(x)

cannot hold a.e. To see this let K_a (resp. K_B) be the characteristic function of A (resp. B). Then

(K_A * K_B)(x) \equiv K(x)=|A \cap B_x|

is a continuous function of compact support. For \varepsilon>0 there is an open set C such that 0<K(x)<\varepsilon for x \in C. By the strong maximum principle, both F(y) and G(y-x) are positive in the open set A \cap B_x, x\in C. (Here we assume that A and B are connected; if not, it suffices to replace A and B by connected components on which F>0, G>0.) Thus, D(x)>0, x \in C. If T(x)=\Lambda D(x), x \in C, then \lambda (A\cap B_x) \geqslant \Lambda, but this is impossible for sufficiently small \varepsilon by the Faber-Krahn inequality.

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