# 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

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

where

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

We turn now to the proof of the theorem.

Proof.

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

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

Clearly,

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

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

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

Therefore,

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

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

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

with

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

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

$H_x(y)=F(y)G(y-x)$.

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 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.