0 & \mbox{if $02$ 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 \begin{equation} \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} \end{equation} and \begin{equation} \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} \end{equation} 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 \begin{equation} \int_{\Omega^\star} B(r) {u_1^\star}^2 dx \le \int_{S_1} B(r) z_1^2 dx \label{Bie2} \end{equation} and \begin{equation} \int_{\Omega^\star} {g(r)}^2 {u_1^\star}^2 dx \ge \int_{S_1} {g(r)}^2 z_1^2 dx \label{wie2} \end{equation} 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: \begin{equation} -\mu'(t) = \int_{u(x)=t} \frac{1}{|\nabla u|} H_{n-1}(dx). \label{coarea} \end{equation} 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 \begin{equation} -\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} \end{equation} 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 \begin{equation} \int_\Omega u_1^2 dx = \int_{\Omega^\star} {u_1^\star}^2 dx = \int_{S_1} z_1^2 dx. \label{normalization} \end{equation} 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 \begin{equation} \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{equation} \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. \begin{equation} \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} \end{equation} 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 \begin{equation} \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} \end{equation} 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 \begin{equation} I_{0}^{\tilde{s}_0}[f] = \int_{0}^{\tilde{s}_0} (\lambda_1 - V(w)) f(w) dw. \label{pickh} \end{equation} 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 \begin{equation} \lambda_1 < \frac{\int_{S_1} ({|\nabla F|}^2 + V F^2)dx} {\int_{S_1} F^2 dx} \le \lambda_1, \end{equation} 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}