\documentclass[twoside]{article} \usepackage{amsthm, amssymb} \pagestyle{myheadings} \markboth{\hfil A second eigenvalue bound \hfil EJDE--2000/10} {EJDE--2000/10\hfil Craig Haile \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol.~{\bf 2000}(2000), No.~10, pp.~1--19. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % A second eigenvalue bound for the Dirichlet Schr\"{o}dinger equation wtih a radially \\ symmetric potential \thanks{ {\em 1991 Mathematics Subject Classifications:} 35J10, 35J15, 35J25, 35P15. \break\indent {\em Key words and phrases:} Schr\"{o}dinger eigenvalue equation, Dirichlet boundary conditions, \hfil\break\indent eigenvalue bounds, radially symmetric potential. \hfil\break\indent \copyright 2000 Southwest Texas State University and University of North Texas. \hfil\break\indent Submitted August 24, 1999. Published January 28, 2000.} } \date{} % \author{Craig Haile} \maketitle \begin{abstract} We study the time-independent Schr\"{o}dinger equation with radially symmetric potential $k|x|^\alpha$, $k \ge 0$, $k \in \mathbb{R}$, $\alpha \ge 2$ on a bounded domain $\Omega$ in $\mathbb{R}^n$, $(n \ge 2)$ with Dirichlet boundary conditions. In particular, we compare the eigenvalue $\lambda_2(\Omega)$ of the operator $-\Delta + k |x|^\alpha$ on $\Omega$ with the eigenvalue $\lambda_2(S_1)$ of the same operator $-\Delta +kr^\alpha$ on a ball $S_1$, where $S_1$ has radius such that the first eigenvalues are the same ($\lambda_1(\Omega) = \lambda_1(S_1)$). The main result is to show $\lambda_2(\Omega) \le \lambda_2(S_1)$. We also give an extension of the main result to the case of a more general elliptic eigenvalue problem on a bounded domain $\Omega$ with Dirichlet boundary conditions. \end{abstract} \newtheorem{thm}{Theorem}[section] \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \section{The Schr\"{o}dinger eigenvalue equation with radially symmetric potential} In this paper we consider the Schr\"{o}dinger eigenvalue equation with Dirichlet boundary conditions on a bounded domain $\Omega \subset \mathbb{R}^n$ ($n \ge 2$): \begin{eqnarray} \label{schroeq} &-\Delta u + k |x|^\alpha u = \lambda u \quad \mbox{on }\Omega \subset \mathbb{R}^n &\\ \label{schroeqb} &u=0 \quad \mbox{on } \partial \Omega & \end{eqnarray} with $\alpha \ge 2$, $k \ge 0$, $k \in \mathbb{R}$. Also of interest will be the generalization of the results to the case of a more general, second-order, elliptic, partial differential equation of the form \begin{eqnarray} \label{GEEa} &-\sum_{i,j=1}^n \frac{\partial}{\partial x_i}(a_{ij}(x) \frac{\partial u} {\partial x_j}) + k |x|^\alpha u = \lambda r(x) u \quad \mbox{on } \Omega \subset \mathbb{R}^n &\\ \label{GEEb} &u=0 \quad \mbox{on } \partial \Omega & \end{eqnarray} with the assumption that the equation is uniformly elliptic on $\Omega$. This means that there are positive numbers $a$ and $A$ such that the matrix $[a_{ij}]$ satisfies $0 < a \le [a_{ij}] \le A$ in the quadratic form sense throughout $\Omega$. Also required is that the weight function $r(x)$ satisfy $0 < c \le r(x) \le C$ for some positive constants $c$ and $C$ and all $x \in \Omega$. It is known that the set of eigenvalues ${\{ \lambda_k \}}_{k=1}^\infty$ corresponding to eigenfunctions $u_k$ of the problems (\ref{schroeq})-(\ref{schroeqb}) and (\ref{GEEa})-(\ref{GEEb}) are nonnegative and can be arranged in nondecreasing order as follows: $0 < \lambda_1 < \lambda_2 \le \lambda_3 \ldots \le \lambda_k \le \ldots \, \rightarrow \infty.$ The purpose of this paper is to compare the eigenvalues of (\ref{schroeq})-(\ref{schroeqb}) and (\ref{GEEa})-(\ref{GEEb}) to those of problems on a ball. We will obtain comparison results for $\lambda_2$ utilizing the fixed $\lambda_1$" technique of Ashbaugh and Benguria. This has been used to show for the Laplacian with no potential that $\lambda_2 (\Omega) \le \lambda_2 (S_1)$ where $S_1$ is the $n$-ball with the same first eigenvalue as $-\Delta$ on $\Omega$. Ashbaugh and Benguria then used the fact that the ratio of eigenvalues on balls for $-\Delta$ is constant to prove the Payne-P\'{o}lya-Weinberger conjecture (for $-\Delta$ the ratio of the first two eigenvalues is maximized for balls). \section{ New results} In this section a fixed $\lambda_1$" theorem for the Schr\"{o}dinger problem (\ref{schroeq})-(\ref{schroeqb}), is stated along with a generalization of that result to the general elliptic eigenvalue problem (\ref{GEEa})-(\ref{GEEb}). For the problem (\ref{schroeq})-(\ref{schroeqb}) the comparison eigenvalues considered will be those of the problem \begin{eqnarray} \label{neweqa} &-\Delta z + kr^\alpha z = \lambda z \quad\mbox{on } S_1 \in \mathbb{R}^n \\ \label{neweqb} &z=0 \quad\mbox{on } \partial S_1 \end{eqnarray} where $S_1$ is an $n$-dimensional ball of the appropriate radius so that we have \linebreak $\lambda_1 (\Omega) =\lambda_1(S_1)$. \paragraph{Note:} Here and throughout the paper we will use $k |x|^\alpha$ and $kr^\alpha$ somewhat interchangeably depending on whether we are talking about the potential function on a general domain or on a ball. It should be apparent that this only makes sense if the domain $\Omega$ allows us the make sense of the transformation of the potential function from $\Omega$ to $\Omega^\star$ (the $n$-ball having the same volume as $\Omega$) or to $S_1$. That is, the domain $\Omega$ should have its center of mass at the origin, i.e., at $r=0$, so that the origin of the potential on $\Omega$ and on $\Omega^\star$ or $S_1$ is unchanged. Throughout this paper we will assume this is the case. For the general elliptic problem (\ref{GEEa})-(\ref{GEEb}), the potential in (\ref{neweqa}) will be modified to $\frac{kr^\alpha}{a}$, where $a$ is the lower bound" of the matrix $[a_{ij}]$, and $S_1$ will be chosen so that $\lambda_1(S_1) = \frac{C}{a} \lambda_1(\Omega)$, where $C$ is the upper bound of the weight function $r(x)$. The reason for these modifications will be discussed later in the paper. The main result is now stated. \begin{thm} \label{mainthm} For equations (\ref{schroeq}) and (\ref{neweqa}), as specified above and with Dirichlet boundary conditions, the first two eigenvalues of the respective problems satisfy $$\label{mainresult2} \lambda_2 (\Omega) \le \lambda_2(S_1).$$ \end{thm} \noindent An extension of this theorem for equation (\ref{GEEa}) will also be derived: \begin{thm} \label{mainthm2} For equations (\ref{GEEa}) and (\ref{neweqa}) specified previously (note, in particular, that $S_1$ is specified by $\lambda_1 (S_1)= \frac{C}{a} \lambda_1(\Omega)$), and with Dirichlet boundary conditions, the first two eigenvalues of the respective problems satisfy $$\frac{\lambda_2}{\lambda_1}(\Omega) \le 1 + \frac{AC}{ac}\left (\frac{\lambda_2}{\lambda_1}(S_1)-1 \right). \label{mainresult1}$$ \end{thm} \section{Integral inequalities and rearrangements} We begin with the basic gap inequality $$\lambda_2 - \lambda_1 \le \frac{\int_\Omega {| \nabla P |}^2 u_1^2 dx}{\int_\Omega P^2 u_1^2 dx} \mbox{ provided that } \int_\Omega P u_1^2 dx = 0 \mbox{ and } P \not \equiv 0, \label{gap1}$$ derived using the Rayleigh-Ritz inequality and integration by parts. To make use of inequality (\ref{gap1}), we use not a single $P$ but rather $n$ different trial functions $P_i$, $i=1,2,\dots ,n$, where again $n$ is the dimension of the space. We define the $P_i$ by $P_i = g(r) \frac{x_i}{r}, \quad i=1,2,\dots ,n,$ where $g$ is chosen to be a nonnegative, nontrivial, continuous, differentiable, and bounded function of the radial variable $r=|x|$. The exact choice of $g$ will be made later. Following \cite{ash5} it can be shown that the side conditions in (\ref{gap1}) of $\int_\Omega P_i {u_1}^2 dx = 0$ for $i=1,2,\dots ,n$ can all be satisfied. Accepting this, rewrite (\ref{gap1}) as $(\lambda_2 - \lambda_1) \int_\Omega {P_i}^2u_1^2dx \le \int_\Omega {|\nabla P_i|}^2 u_1^2dx$ and sum on $i$, arriving at $\lambda_2 - \lambda_1 \le \frac{\int_\Omega (\sum_{i=1}^{n} {|\nabla P_i|}^2)u_1^2 dx} {\int_\Omega (\sum_{i=1}^{n} {P_i}^2)u_1^2dx}.$ Again following Ashbaugh and Benguria \cite{ash5} the angular dependence of the $P_i$'s drops out, leading to the new gap inequality $$\lambda_2 - \lambda_1 \le \frac{\int_\Omega [g'(r)^2 + (n-1)g(r)^2/r^2]u_1^2dx} {\int_\Omega g(r)^2u_1^2dx}. \label{gap3}$$ The next step is to choose $g$ such that there is equality in (\ref{gap3}) if $\Omega$ is a ball. This implies the choice of $$g(r) = \left \{ \begin{array}{ll} \frac{y_2(r)}{y_1(r)} & \mbox{ for } 0 \le r < r^* \$3pt] \stackrel{\textstyle \lim}{{}_{t \nearrow r^*}} {}^{\textstyle g(t)} &\mbox{ for } r \ge r^*, \end{array} \right. \label{w}$$ where r^* is an appropriate radius (less than or equal to that of \Omega^*, with equality if \Omega = \Omega^\star) and y_1 and y_2 are solutions to certain one-dimensional radial eigenvalue problems obtained from (\ref{neweqa})-(\ref{neweqb}) via separation of variables. By standard separation of variables techniques in spherical coordinates and a result due to Baumgartner, Gross, and Martin for \mathbb{R}^3 \cite{bau}, extended by Ashbaugh-Benguria \cite{ashV} to all n and \Omega a ball, one finds that for radial potentials W(r) with [rW]'' \ge 0 (certainly satisfied in our case) the first two eigenvalues of -\Delta + W on a ball B of radius r^* are the respective first eigenvalues of certain one-dimensional problems. In our case these are $${y_1}'' + \frac{n-1}{r}{y_1}' + (\lambda^{\star}_{1}-kr^\alpha)y_1=0$$ with y_1(0) finite, y_1(r^*)=0, and $${y_2}'' + \frac{n-1}{r}{y_2}' + (\lambda^{\star}_{2}-\frac{n-1}{r^2} -kr^\alpha)y_2=0$$ with y_2(0)=y_2(r^*)=0. Here \lambda^{\star}_{1} and \lambda^{\star}_{2} are defined as the first eigenvalues of the respective problems. Note that by construction this makes g continuously differentiable on [0,\infty). Finally, by defining the function B(r) by $$B(r) \equiv {g'(r)}^2 + (n-1) {{g(r)}^2 \over r^2}. \label{B}$$ the gap inequality (\ref{gap3}) can be put into the form $$\lambda_2 - \lambda_1 \le \frac{ \int_\Omega B(r) u_1^2dx} {\int_\Omega {g(r)}^2 u_1^2 dx}. \label{gap4}$$ Next our attention is turned to proving two key properties of B and g. Both functions are nonnegative, and now two important facts about their respective derivatives need to be shown. Let r^* be the radius of the ball of definition for g and B. Then \begin{flushleft} (1) g(r) is increasing for r \in [0,r^*], and (2) B(r) is decreasing for r \in [0,r^*]. \end{flushleft} The idea of the proof is based on the properties of the function q(r) defined by $$q(r) = \frac{r g'(r)}{g(r)}. \label{q}$$ With q as above and B defined by (\ref{B}) an easy substitution shows B = [q^2 + (n-1)]{(g/r)}^2 and hence B'= 2[qq' + (q-1)(q^2 +(n-1))/r]{(g/r)}^2. It is clear that properties (1) and (2) will be met by showing that 0 \le q \le 1 and q' \le 0 for 0 \le r \le r^*. To prove these results one uses a Riccati differential equation satisfied by q on 0(\lambda^{\star}_{1} - \lambda^{\star}_{2})r_2 + (n-1)/{r_2} =q'(r_2) \ge 0, a contradiction. If r_2 = r^*, then using (\ref{q'}) and (\ref{qatr}) a similar argument to the above yields $$0 \ge q'(r_1)= (\lambda^{\star}_{1} - \lambda^{\star}_{2})r_1 + (n-1)/{r_1} > (\lambda^{\star}_{1} - \lambda^{\star}_{2})r^* + (n-1)/{r^*} = 3 q'(r^*) \ge 0$$ which is again a contradiction. \end{proof} Moving on, we make some further definitions to simplify our notation. Denote E = (\lambda^{\star}_{2} - \lambda^{\star}_{1})/2 and b = (n-2)/2; also let p= y_1'/y_1. Then from (\ref{q}) and (\ref{q'}) one has  q'=2T  with T defined to be $$T(r,q) = -pq - \frac{bq}{r} - \frac{q^2}{2r} + \frac{2b+1}{2r} - Er. \label{S}$$ To analyze the behavior of T for fixed values of q, we will study the behavior of \frac{\partial T}{\partial r} at points where T=0 (fixed q) and the behavior of T(r,\mbox{fixed q}) near r=0 and r=r^*. For r approaching zero, p is O(r), and thus from (\ref{S}) \[ T(r,q) = \frac{1}{2r} [(2b+1+q)(1-q)] + O(r).$ This implies that for $q$ fixed, $0 0 & \mbox{if$0 2$this is seen to be an immediate contradiction of the necessary sign change of$Z$from negative to positive. If$\alpha=2$we consider$Z''=2 E^2-2k$. If$E^2-k$is less than zero we are done since that would make$Z''$negative. If$E^2-k$is zero then$Z \equiv Q$, which would mean$Z$has no sign changes, an obvious contradiction. If$E^2-k$is greater than zero then$Z'>0$, another contradiction. Thus$T$cannot have the required sign changes if we assume$q$increases away from one as$r$increases away from zero, so$q$must decrease from one as$r$increases from zero. \end{proof} With the end behavior of$q$established, the next step is to establish the condition that$q$is decreasing and hence$q \le 1$. \begin{lem} Given$q(0)=1$,$q$first decreases from one as$r$increases from zero,$q(r^*)=0$, and$q'(r^*) \le 0$, it follows that$q'(r) \le 0$for$0 \le r \le r^*$. \end{lem} \begin{cor} Given that$q'(r) \le 0$on$0 \le r \le r^*$and$q(0) = 1$, it follows that$q(r) \le 1$on$0 \le r \le r^*$. \end{cor} \begin{proof} (Of Lemma.) Suppose not. Then there exist three points$0 < r_1 < r_2 < r_3 < r^*$with$00$,$q'(r_3)<0$. Using the fixed value of$q<1$at these points in the function$T$, recall that$T$should be positive for small and large values of$r$. Also, since$T=q'/2$, we have$T(r_1) <0, T(r_2)>0$, and$T(r_3)<0$. Hence$T$must have at least four sign changes: positive-negative-positive-negative-positive. The value of$Z$at each of the roots of$T$must alternate in a similar way (positive to negative in$T$implies negative$Z$, etc.) Starting from$Z(0,q<1) = + \infty$, this shows that$Z$must also alternate positive-negative-positive-negative-positive. Recall from (\ref{Z}) that $Z(r,q<1) = \frac{M}{r^2} + \frac{E^2 r^2}{q} + Q - qkr^\alpha$ and thus \begin{eqnarray*} Z'(r,q<1) &= & \frac{-2M}{r^3} + \frac{2E^2r}{q} - q \alpha k r^{\alpha-1} \\ Z''(r,q<1) &=& \frac{6M}{r^4} + \frac{2E^2}{q} - q \alpha (\alpha-1) k r^{\alpha-2}. \end{eqnarray*} Since$q$less than one implies$M>0$and it follows that$Z''(r,q <1)$is strictly decreasing,$Z(r,q<1)$could not have the changes in concavity necessary for its changes in sign, a contradiction to the behavior of$T(r,q<1)$, and thus$q' \le 0$on$[0,r^*]$. This proves the lemma and thus the corollary. \end{proof} Combining our work, we have$0 \le q \le 1$and$q' \le 0$on$0 \le r \le r^*$, which in turn shows that$g(r)$is increasing and$B(r)$is decreasing. These results are fundamental in using the rearrangement techniques of Hardy, Littlewood, and P\'{o}lya \cite{har}. Rearrangements have several useful properties. In particular, if$f$and$h$are nonnegative functions, then we have the integral inequalities $\int_{\Omega^\star} f_\star h^\star dx \le \int_\Omega fh dx \le \int_{\Omega^\star} f^\star h^\star dx.$ Also, if$f$is a nonnegative, radially symmetric decreasing (resp. increasing) function on$\Omega$then$ f^\star (r) \le f(r) $and respectively$ f_\star (r) \ge f(r) $for$r$between 0 and the radius of$\Omega^\star$. The previous lemmas showed that$g(r)$is increasing and$B(r)$is decreasing (they are both nonnegative and radially symmetric), hence $$\int_\Omega B(r) u_1^2 dx \le \int_{\Omega^\star} {B(r)}^\star {u_1^\star}^2 dx \le \int_{\Omega^\star} B(r){u_1^\star}^2 dx, \label{Bie1}$$ and $$\int_\Omega {g(r)}^2 u_1^2 dx \ge \int_{\Omega^\star} g(r)_\star^2 {{u_1^\star}}^2 dx \ge \int_{\Omega^\star} {g(r)}^2 {{u_1^\star}}^2 dx. \label{wie1}$$ Note also, concerning our potential function, or any potential function$V(x)=V(r)$that is radially symmetric and increasing, that $\int_\Omega V(x) u_1^p dx \ge \int_{\Omega^\star} V_\star(r) {{u_1^\star}}^p dx \ge \int_{\Omega^\star} V(r) {{u_1^\star}}^p dx$ where$p=1,2.$With the inequalities (\ref{Bie1}), (\ref{wie1}) established, it is time to show the last set of inequalities necessary for completing the theorem. These are $$\int_{\Omega^\star} B(r) {u_1^\star}^2 dx \le \int_{S_1} B(r) z_1^2 dx \label{Bie2}$$ and $$\int_{\Omega^\star} {g(r)}^2 {u_1^\star}^2 dx \ge \int_{S_1} {g(r)}^2 z_1^2 dx \label{wie2}$$ where$z_1$is the first eigenfunction of the comparison problem (\ref{neweqa})-(\ref{neweqb}). In order to derive these two inequalities we establish a crossing result for$u_1^\star$and$z_1$similar to the type found in \cite{chi1,chi3,tal1,ash5}. The result will resemble the work in these papers, with the exception that the differential equation satisfied by$z_1$includes a potential term, whereas those of the previous works had no potential. \section{Chiti and Talenti type arguments} Since the results of this section apply to any potential function that is continuous, radially symmetric, increasing, and has its origin at the center of mass for$\Omega$, we will make our proof for a general potential function$V(x)=V(r)$, and keep in mind that our potential$V(x)= k |x|^\alpha=k r^\alpha=V(r)$is a special case. As a first step we follow Talenti \cite{tal1}, Chiti \cite{chi1}, Ashbaugh and Benguria \cite{ash5} in integrating both sides of$-\Delta u + Vu=\lambda u$over the level set$ \{x \in \Omega: u(x) > t \}$to get $-\int_{u(x)>t} \Delta u dx = \int_{u(x)>t}(\lambda - V(x)) u dx.$ Using Gauss's Divergence Theorem we have $- \int_{u(x)>t} \Delta u dx = \int_{u(x)=t} ({|\nabla u|}) H_{n-1}(dx),$ where$H_{n-1}$denotes ($n-1$)-dimensional measure. Now consider the distribution function$\mu(t)=\mbox{meas}\{x \in \Omega:|u(x)|>t\}$. Taking the derivative of$\mu(t)$we can derive the following special case of the coarea formula: $$-\mu'(t) = \int_{u(x)=t} \frac{1}{|\nabla u|} H_{n-1}(dx). \label{coarea}$$ Using (\ref{coarea}) and the Cauchy-Schwarz inequality we find \begin{eqnarray*} H_{n-1}\{x \in \Omega: u(x)=t \} &=& \int_{u=t} H_{n-1} (dx) \\ &=& \int_{u=t} \frac{{|\nabla u|}^{1/2}}{{|\nabla u|}^{1/2}} H_{n-1} (dx) \\ &\le& { \left (\int_{u=t} |\nabla u| H_{n-1} (dx) \right )}^{1/2} {\left (\int_{u=t} \frac{1}{|\nabla u|} H_{n-1} (dx) \right )}^{1/2} \\ &\le& {\left [- \mu'(t) \int_{u=t} |\nabla u| H_{n-1} (dx) \right ]}^{1/2}. \end{eqnarray*} Also, the$n$-dimensional isoperimetric inequality gives us $H_{n-1} \{x \in \Omega: u(x)=t\} \ge n {C_n}^{1/n} \mu(t)^{1-1/n}$ where$C_n$is the volume of the$n$-dimensional unit ball. Combining these two inequalities yields $\int_{u=t}|\nabla u| H_{n-1} (dx) \ge n^2 {C_n}^{2/n} \mu(t)^{2-2/n} (\frac{-1}{\mu'(t)}).$ Combining all these inequalities together for$u=u_1$and$\lambda=\lambda_1$, one finds $\int_{u_1(x) >t} [\lambda_1 - V(x)]u_1 dx \ge n^2 {C_n}^{2/n} \mu_1(t)^{2-2/n} (-1/\mu_1'(t)).$ We then use rearrangement properties to pass from$u_1$to$u_1^*$. Finally, considering the fact that$u_1^*$and$\mu_1$are essentially inverse functions of each other, we get the key inequality $$-\frac{du_1^*}{ds} \le { n^{-2} {C_n}^{-2/n} s^{-2+2/n}} \int_0^s (\lambda_1 u_1^* (w) - V(w) u_1^* (w)) dw, \label{derint}$$ which is similar to equations of Chiti \cite{chi1}, Ashbaugh and Benguria \cite{ash4}, and Talenti \cite{tal1}. Next consider the following lemma, an extension of the Faber-Krahn inequality \cite{fab,kra1,kra2}. \begin{lem}[Extension of Faber-Krahn] Suppose$V$is continuous, radially symmetric and increasing. Then the first eigenvalue for the operator$-\Delta + V$on$\Omega^\star$is less than or equal to$\lambda_1 (\Omega)$. Equivalently$|S_1| \le |\Omega|$, with equality if and only if$\Omega = \Omega^\star = S_1$. \end{lem} \begin{proof} By the Rayleigh-Ritz inequality and rearrangement properties we have \begin{eqnarray*} \lambda_1 &=& \frac{ \int_\Omega ( {|\nabla u_1|}^2 + Vu_1^2)dx} { \int_\Omega u_1^2 dx} \\ &\ge& \frac{ \int_{\Omega^\star} ( {|\nabla u_1^\star|}^2 + V{u_1^\star}^2)dx} { \int_{\Omega^\star} {u_1^\star}^2 dx} \\ &\ge& \mbox{[first eigenvalue of$-\Delta + V$on$\Omega^\star$]}. \end{eqnarray*} Thus ${\lambda_1} (S_1) = \lambda_1(\Omega) \ge \mbox{[first eigenvalue of -\Delta + V on \Omega^\star]} = \lambda_1(\Omega^\star).$ Since$\lambda_1$is exactly the first eigenvalue of$-\Delta + V$on$S_1$it follows by the monotonicity property of Dirichlet eigenvalues with respect to domains that$|S_1| \le |\Omega^\star| = |\Omega|$, and thus$S_1 \subseteq \Omega^\star$, with equality if and only if$\Omega=\Omega^*=S_1$. \end{proof} The groundwork has now been laid to prove a modified form of the Chiti comparison result. \begin{thm}[Modified Chiti Comparison Theorem] Let$V$be continuous, radially symmetric and increasing. Let$u_1$and$ z_1$be the first eigenfunctions of$-\Delta +V$on$\Omega$and$S_1$respectively, with Dirichlet Boundary conditions, assumed positive and normalized such that $$\int_\Omega u_1^2 dx = \int_{\Omega^\star} {u_1^\star}^2 dx = \int_{S_1} z_1^2 dx. \label{normalization}$$ Then, making a change of variables so that$z_1$and$u_1^\star$can be viewed as functions of$s=C_nr^n$, (so that we are actually looking at$u_1^*$), there will exist a point$s_1 \in (0,|S_1|)$such that$u_1^*(s_1)=z_1(s_1)$and $$\left\{ \begin{array}{ll} u_1^*(s) \le z_1(s) & \mbox{ for } 0 \le s \le s_1 \\ u_1^*(s) \ge z_1(s) & \mbox{ for } s_1 \le s \le |S_1|. \end{array} \right. \label{chiti}$$ \end{thm} \begin{proof} The arguments depend upon the continuity of$u_1^*$and$z_1$. (The absolute continuity of$u_1^*$is established in the arguments of Talenti \cite{tal1} or Chiti \cite{chi1}. See, in particular, Lemma 1 in \cite{chi1}.) It should also be noted that$z_1$is positive and decreasing on$[0,|S_1|]$. To show this, consider the Rayleigh quotient for$\lambda_1$and the related rearrangement inequality. $$\lambda_1 = \frac{\int_{S_1} ({|\nabla z_1|}^2 + V z_1^2)dx} {\int_{S_1} z_1^2 dx} \ge \frac{\int_{S_1} ({|\nabla z_1^\star|}^2 + V {z_1^\star}^2)dx} {\int_{S_1} {z_1^\star}^2 dx} \label{zdecr}$$ Since$V$is increasing, if$z_1$is not decreasing, then we have strict inequality in (\ref{zdecr}). However,$z_1^\star$is a valid trial function for$z_1$, so we have using Rayleigh-Ritz $$\lambda_1 \le \frac{\int_{S_1} ({|\nabla z_1^\star|}^2 + V {z_1^\star}^2)dx} {\int_{S_1} {z_1^\star}^2 dx}$$ which is a contradiction. The proof for (\ref{chiti}) then consists of basically two parts, each following the same line of reasoning. The first step is to show that if$z_1$is normalized so that$z_1(0) = \mbox{ess sup } u_1$, then$z_1(s) \le u_1^*(s)$for all$s \in [0,|S_1|]$. To prove this, suppose not. Assuming$|S_1| \neq |\Omega|$and$z_1 \not \equiv u_1^*$(otherwise it is trivial), there are two possibilities. The first is that$z_1$first starts out above$u_1^*$, and, since$|S_1| < |\Omega|$,$z_1$eventually drops below$u_1^*$. If this is the situation, multiply$u_1^*$by some constant$c>1$such that there is a point$s_0$sufficiently close to zero so that$ c u_1^*(s_0) = z_1(s_0), $and$\lambda_1 - V(s_0) \ge 0$. (This can be done since$\lambda_1 - V(0)$is positive and$V$is continuous at zero.) Now, define a new function$h(s)$such that $h(s) = \left\{ \begin{array}{ll} c u_1^*(s) &\mbox{ for } 0 \le s \le s_0 \\ z_1(s) &\mbox{ for } s_0 < s \le |S_1| \end{array} \right.$ and notice that$h$satisfies the inequality given in (\ref{derint}). This is true for$s \le s_0$by (\ref{derint}) and for$s \ge s_0$by the fact that$z_1(s)$satisfies (\ref{derint}) with equality, and replacing$z_1$by$c u_1^*$on the interval$(0,s_0$) will only increase the right-hand side since$cu_1^* \ge z_1$and$\lambda_1 - V(s) \ge 0$. (Note that with equality (\ref{derint}) is just an integrated form of the radial differential equation for$z$in the variable$s=C_nr^n$, hence the equality for$h \equiv z_1$, see for example Talenti \cite{tal1}.) Since$u_1^*(0) = z_1(0)$, we must have$cu_1^* > z_1$on some subinterval of$(0,s_0)$, and hence there is strict inequality in (\ref{derint}) beyond$s_0$. The other possibility is that$z_1$starts out below$u_1^*$, crosses above$u_1^*$at some point$\tilde{s}_0$, and then crosses back below$u_1^*$at some point$\tilde{s}_1$. If$\lambda_1 -V(\tilde{s}_0) \ge 0$, form$h(s)$exactly as above with$\tilde{s}_0$replacing$s_0$and the arguments that$h$satisfies the inequality in (\ref{derint}) hold as before. If$\lambda_1 - V(\tilde{s}_0) <0$, then on$[0,\tilde{s}_0]$pick$h$to be whichever function ($u_1^*$or$z_1$) yields the greater value in the functional $$I_{0}^{\tilde{s}_0}[f] = \int_{0}^{\tilde{s}_0} (\lambda_1 - V(w)) f(w) dw. \label{pickh}$$ Choose$h$on the whole interval$[0,|S_1|]$as follows: $h(s) = \left\{ \begin{array}{ll} \left\{ \begin{array}{ll} u_1^*(s) &\mbox{ if } I_{0}^{\tilde{s}_0}[u_1^*] \ge I_{0}^{\tilde{s}_0}[z_1] \\ z_1(s) &\mbox{ if } I_{0}^{\tilde{s}_0}[u_1^*] < I_{0}^{\tilde{s}_0}[z_1] \end{array} \right \} &\mbox{ for } s \in [0,\tilde{s}_0] \\ \min[u_1^*,z_1] &\mbox{ for } s \in (\tilde{s}_0, \tilde{s}_1) \\ z_1(s) &\mbox { for } s \in [\tilde{s}_1,|S_1|]. \end{array} \right.$ Again we must check that (\ref{derint}) is satisfied. Certainly it is on$(0,\tilde{s_0})$for whichever function is$h$. For$s \in (\tilde{s_0}, \tilde{s_1})$, if$h(s)=u_1^*(s)$we have \begin{eqnarray*} -\frac{dh}{ds} &=& -\frac{du_1^*}{ds} \\ &\le& n^{-2} {C_n}^{-2/n} s^{-2+2/n} \left (\int_0^{\tilde{s_0}} (\lambda_1 - V)u_1^* dw + \int_{\tilde{s_0}}^{s} (\lambda_1 - V)u_1^* dw \right ) \\ & \le & n^{-2} {C_n}^{-2/n} s^{-2+2/n} \left ( \int_0^{\tilde{s_0}} (\lambda_1 - V)h dw + \int_{\tilde{s_0}}^{s} (\lambda_1 - V)h dw \right ) \end{eqnarray*} the first replacement on$(0,\tilde{s_0})$true by the construction that$h$would give the greater integral for (\ref{pickh}) and the second replacement true since$h \le u_1^*$and$\lambda_1 - V(s) < 0$on$(\tilde{s_0}, \tilde{s_1})$. (Note that$\lambda_1 - V(s)$is a decreasing function.) Similarly, if$h(s)=z_1(s)$for some$s \in (\tilde{s_0}, \tilde{s_1})$, then at this value of$s$we have $-\frac{dh}{ds} = -\frac{dz_1}{ds} \le n^{-2} {C_n}^{-2/n} s^{-2+2/n} \left ( \int_0^{\tilde{s_0}} (\lambda_1 - V)h dw + \int_{\tilde{s_0}}^{s} (\lambda_1 - V)h dw \right ).$ Finally, note that$h=u_1^* u_1^*(0)$, then by the extension of the Faber-Krahn inequality$|S_1| < |\Omega|$, and, by the Dirichlet boundary conditions$u_1^*(|\Omega|) = z_1(|S_1|) = 0$, it is clear that there is at least one point$s \in (0,|S_1|)$at which$u_1^*(s) = z_1(s)$. Take$s_1$to be the largest$s \in (0,|S_1|)$such that the condition$u_1^*(w) \le z_1(w)$for all$0 \le w \le s$holds. Then it remains to be proven that${u_1}^*(s) > z_1(s)$for all$s \in (s_1,|S_1|]$. If not, there is a point$s_2$, with$s_1 < s_2 < |S_1|$, which we define to be the largest$s$such that$u_1^*(w) \ge z_1(w)$for all$s_1 \le w \le s$. If$s_2$exists, there also exists a point$s_3$,$s_2 < s_3 < |S_1|$, defined to be the largest$s$such that$u_1^*(w) \le z_1(w)$for all$s_2 \le w \le s$. Now, as before, we piece together a function$v(s)$from$u_1^*$and$z_1$which will satisfy the key integral inequality (\ref{derint}). This construction is done again based upon the interval where the decreasing function$\lambda_1 - V(s)$changes sign from positive to negative. The possibilities are as follows: {\bf Case 1.}$\lambda_1 - V(s)$changes sign in$[s_2,|S_1|]$or not at all. Then define the trial function$v$as $v(s) = \left\{ \begin{array}{ll} z_1(s) &\mbox{ for } s \in [0,s_1]\cup[s_2,|S_1|] \\ \max [u_1^*,z_1] &\mbox{ for } s \in (s_1,s_2) \end{array} \right.$ {\bf Case 2.}$\lambda_1 - V(s)$changes sign in$(s_1,s_2)$. Then $v(s) = \left\{ \begin{array}{ll} z_1(s) &\mbox{ for } s \in [0,s_1] \\ \left\{ \begin{array}{ll} u_1^*(s) &\mbox{ if } I_{s_1}^{s_2}[u_1^*] \ge I_{s_1}^{s_2}[z_1] \\ z_1(s) &\mbox{ if } I_{s_1}^{s_2}[u_1^*] < I_{s_1}^{s_2}[z_1] \end{array} \right \} &\mbox{ for } s \in (s_1,s_2) \\ \min [u_1^*,z_1] &\mbox{ for } s \in (s_2,s_3) \\ z_1(s) &\mbox{ for } s \in [s_3,|S_1|]. \end{array} \right.$ {\bf Case 3.}$\lambda_1 - V(s)$changes sign in$[0,s_1]$. Then $v(s) = \left\{ \begin{array}{ll} \left\{ \begin{array}{ll} u_1^*(s) &\mbox{ if } I_{0}^{s_1}[u_1^*] \ge I_{0}^{s_1}[z_1] \\ z_1(s) &\mbox{ if } I_{0}^{s_1}[u_1^*] < I_{0}^{s_1}[z_1] \end{array} \right \} &\mbox{ for } s \in [0,s_1] \\ \min [u_1^*,z_1] &\mbox{ for } s \in (s_1,s_3) \\ z_1(s) &\mbox{ for } s \in [s_3,|S_1|]. \end{array} \right.$ As before, let$F(x)=v(C_n|x|^n)$for whichever$v$is applicable. Then as long as$s_2$exists$F$cannot be the ground-state eigenfunction on$|S_1|$, which gives the left-hand side of the inequality below. However, using the same methods that allowed us to arrive at (\ref{contraine}) for$\Phi$, one can derive the right-hand side of the inequality $$\lambda_1 < \frac{\int_{S_1} ({|\nabla F|}^2 + V F^2)dx} {\int_{S_1} F^2 dx} \le \lambda_1,$$ a contradiction. \end{proof} The fact that$g(r)$is increasing and the modified Chiti Comparison Theorem can be used to prove the inequality (\ref{wie2}). Keeping in mind that$g$and$u_1^\star$are functions of the radial variable only, let$r^*$be the radius of$S_1$. Also, let$r_1$correspond to$s_1$from the Chiti theorem via the change of variable$s_1=C_n r_1^n$, and let$R$be the radius of$\Omega^\star$. Then we have \begin{eqnarray*} \lefteqn{ \int_{S_1} {g(r) }^2 z_1^2 dx - \int_{\Omega^\star} {g(r)}^2 {u_1^\star}^2 dx}\\ &=& nC_n \bigg[ \int_0^{r_1} g(r)^2 (z_1^2 - {u_1^*}^2)r^{n-1} dr + \int_{r_1}^{r^*} g(r)^2 (z_1^2 \\ &&- {u_1^*}^2)r^{n-1} dr - \int_{r^*}^{R} g(r)^2 {u_1^*}^2 r^{n-1} dr \bigg] \\ &\le& nC_n \left [ \int_0^{r_1} g(r_1)^2 (z_1^2 - {u_1^*}^2)r^{n-1} dr \right. \\ & &\left.+ \int_{r_1}^{r^*} g(r_1)^2 (z_1^2 - {u_1^*}^2)r^{n-1} dr - \int_{r^*}^{R} g(r_1)^2 {u_1^*}^2 r^{n-1} dr \right ] \\ &=& g(r_1)^2 nC_n \left [\int_0^{r^*} z_1^2 r^{n-1}dr - \int_0^{R} {u_1^*}^2 r^{n-1} dr \right ] \\ &=& g(r_1)^2 \left [\int_{S_1} z_1^2 dx - \int_{\Omega^\star} {u_1^\star}^2 dx \right ] \\ &=& 0 \end{eqnarray*} the last line by virtue of the normalization hypothesis (\ref{normalization}). By a similar calculation, using the fact that$B(r)$is radially decreasing and the Chiti Theorem, we can show the inequality (\ref{Bie2}): $\int_{\Omega^\star} B(r) {u_1^\star}^2 dx \le \int_{S_1} B(r) z_1^2 dx.$ \section{Conclusion of the proof of the main theorem} Combining the results (\ref{Bie1}), (\ref{wie1}), (\ref{Bie2}), (\ref{wie2}) yields the string of inequalities $\int_\Omega B(r) u_1^2 dx \le \int_{\Omega^\star} {B(r)}^\star {{u_1^\star}}^2 dx \le \int_{\Omega^\star} B(r){{u_1^\star}}^2 dx \le \int_{S_1} B(r) z_1^2 dx$ and $\int_\Omega {g(r)}^2 u_1^2 dx \ge \int_{\Omega^\star} g(r)_\star^2 {{u_1^\star}}^2 dx \ge \int_{\Omega^\star} {g(r)}^2 {{u_1^\star}}^2 dx \ge \int_{S_1} {g(r) }^2 z_1^2 dx.$ Together with (\ref{gap4}) this gives $\lambda_2 (\Omega) - \lambda_1 (\Omega) \le \frac{\int_{S_1} B(r) z_1^2 dx}{\int_{S_1} {g(r)}^2 z_1^2 dx}.$ Since$B$and$g$were defined in such a way as to give equality in the case of$\Omega$equal to a ball, the right hand side yields ${\lambda_2}(S_1) - {\lambda_1}(S_1) = \frac{\int_{S_1} B(r) z_1^2 dx}{\int_{S_1} {g(r)}^2 z_1^2 dx}$ and so this reduces to the main inequality $\lambda_2 (\Omega) \le {\lambda_2}(S_1),$ since$S_1$was chosen so that$\lambda_1(S_1)=\lambda_1(\Omega)$. \section{Extension to general elliptic equation} This section gives an outline of how to extend the bound given in Theorem \ref{mainthm} for the Schr\"{o}dinger problem to the bound (\ref{mainresult1}) referring to the general elliptic problem of Theorem \ref{mainthm2}. Proceeding as in the case of the Schr\"{o}dinger operator we find$B$and$g$are defined as before with the exception that$kr^\alpha$becomes$\frac{kr^\alpha}{a}$. Equality to$\lambda_2 - \lambda_1$in the integral ratio $\frac{ \int_\Omega B(r) u_1^2dx} { \int_\Omega {g(r)}^2 u_1^2 dx}$ again comes for the Schr\"{o}dinger operator on a ball. Since$B$and$g$differ only by a constant in$kr^\alpha$, their properties of$g$increasing and$B$decreasing remain unchanged. The Chiti and Talenti type arguments proceed with little change, the key differences being that$S_1$is chosen so that $\left\{ \begin{array}{ll} -\Delta z + {\frac{kr^\alpha}{a}} z = \lambda z & \mbox{ on } S_1 \\ z=0 & \mbox{ on } \partial S_1, \end{array} \right.$ has$\frac{C}{a} \lambda_1(\Omega)$as its first eigenvalue as opposed to$\lambda_1(\Omega)$. Another extension of the Faber-Krahn result shows that$|S_1| \le |\Omega^*| = |\Omega|$. Corresponding to the change from$\lambda_1$to$\frac{C}{a} \lambda_1$, the key integral inequality (\ref{derint}) changes to $-\frac{du_1^*}{ds} \le \frac{{ n^{-2} {C_n}^{-2/n} s^{-2+2/n}}}{a} \int_0^s (C \lambda_1 u_1^* (w) - kw^\alpha u_1^* (w)) dw.$ (In this inequality we see the necessity for the change in potential and first eigenvalue.) The Chiti comparison result follows as before, leading to $\lambda_2 (\Omega) - \lambda_1 (\Omega) \le \frac{A}{c} \frac{\int_{S_1} B(r) z_1^2 dx}{\int_{S_1} {g(r)}^2 z_1^2 dx} = \frac{A}{c} (\lambda_2 (S_1) - \lambda_1 (S_1) )$ which simplifies to the ratio inequality $\frac{\lambda_2}{\lambda_1}(\Omega) \le 1 + \frac{AC}{ac}\left (\frac{\lambda_2}{\lambda_1}(S_1)-1 \right )$ which is the result stated in Theorem \ref{mainthm2}. \paragraph{Acknowledgments.} I wish to express my gratitude to Professor Mark Ashbaugh for his guidance. 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(6) {\bf 4-B} (1985), pp. 917-949. \end{thebibliography} \medskip \noindent{\sc Craig Haile }\\ Department of Mathematics and Physics\\ College of the Ozarks\\ Point Lookout, MO 65726-0017, USA\\ e-mail haile@cofo.edu \end{document}