\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil The first eigenvalue of the p-Laplacian\hfil EJDE--2001/35} {EJDE--2001/35\hfil Tilak Bhattacharya \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2001}(2001), No. 35, pp. 1--15. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Some observations on the first eigenvalue of the p-Laplacian and its connections with asymmetry % \thanks{ {\em Mathematics Subject Classifications:} 35J60, 35P30. \hfil\break\indent {\em Key words:} Asymmetry, De Giorgi perimeter, p-Laplacian, first eigenvalue, \hfil\break\indent Talenti's inequality. \hfil\break\indent \copyright 2001 Southwest Texas State University. \hfil\break\indent Submitted September 3, 2000. Published May 16, 2001.} } \date{} % \author{Tilak Bhattacharya} \maketitle \begin{abstract} In this work, we present a lower bound for the first eigenvalue of the $p$-Laplacian on bounded domains in $\mathbb{R}^2$. Let $\lambda_1$ be the first eigenvalue and $\lambda_1^*$ be the first eigenvalue for the ball of the same volume. Then we show that $\lambda_1\ge\lambda_1^*(1+C\alpha(\Omega)^{3})$, for some constant $C$, where $\alpha$ is the asymmetry of the domain $\Omega$. This provides a lower bound sharper than the bound in Faber-Krahn inequality. \end{abstract} \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode@=11 \@addtoreset{equation}{section} \catcode@=12 \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}{Lemma}[section] \section{Introduction} Let $\Omega \subset \mathbb{R}^{n}$, be a bounded domain. For $10$ and a non-zero minimizer, which we continue to call as $u=u(p, \Omega)$, exists and satisfies $$\mathop{\rm div}(|Du|^{p-2}Du) + \lambda_1 |u|^{p-2}u = 0, \quad \mbox {in } \Omega ,$$ where $u\in W^{1,p}_0(\Omega)$. The operator $\mathop{\rm div}(|Du|^{p-2}Du)$ is the $p$-Laplacian and this is the usual Laplacian when $p=2$. For $p\ne 2$, this is a quasi-linear and a degenerate elliptic operator. The equation in (1.1) is to be interpreted in the weak sense, i. e., $$\int_{\Omega}|Du|^{p-2}Du\cdot D\psi=\lambda_1\int_{\Omega}|u|^{p-2}u\psi,\;\; \forall\;\psi\in W^{1,p}_0(\Omega).$$ We refer to $\lambda_1$ as the first eigenvalue and $u$ as the first eigenfunction of the $p$-Laplacian on $\Omega$. It is well known that $\lambda_1$ is simple and $u$ has one sign \cite{b2,l1, l2}. The first eigenvalue is also known to be isolated [10]. Moreover, if $\Omega$ is a ball then $u$ is radial, decreasing and has only one critical point. It will be useful to note that $u$ is $L^{\infty}(\Omega)$ \cite{l2}. It is also quite well known that $u$ is $C^{1,\alpha}_{loc}$ in $\Omega$ \cite{d1,m1,t3}. See [2] for a more detailed discussion of matters related to the regularity of $u$. It should also be pointed out that unlike the case of the Laplacian, i. e., $p=2$, a complete characterization of the set of critical points of $u$, when $p\ne 2$, is still unknown. This fact or lack thereof becomes particularly important when working with level sets of $u$. The boundaries of such sets need not be smooth thus necessitating the use of the DeGiorgi perimeter. We discuss this further in section 2. Also see \cite{b1,b2,t1,t2}. Let $D^*$ denote the symmetrized domain for an open set $D$, i .e., $D^{*}$ is the ball, centered at the origin, with volume equal to that of $D$. Let $\lambda ^*_1= \lambda_1 (\Omega^*)$; then the Faber-Krahn inequality states that $\lambda_1 \ge \lambda^{*}_1$, where equality holds if and only if $\Omega$ is a ball \cite{b2}. Our attempt, in this work, will be to characterize this lower bound for $\lambda_1$, for $10$, we will also assume throughout that $$\int _\Omega u^p = 1.$$ For $0\le t\le \sup u$, set \begin{eqnarray*} \Omega _t = \{ x \in \Omega : u (x) > t \},\;\;\mbox{and}\;\;\mu(t)=|\Omega_t|. \end{eqnarray*} Note that $\mu(t)$ is decreasing and right continuous. It is easy to show that $\mu(t)$ is continuous if and only if $|\{u=t\}|=0$. Clearly, $\mu(t)$ has at most countably many discontinuities. Since $u$ is only known to be $C^{1,\alpha}_{loc}$, it is not clear that $\mu(t)$ is continuous every where when $p\ne 2$ \cite{b1,b2,t1}. Let $u^*$ be the non-increasing rearrangement (Schwarz symmetrization) of $u$, defined as follows. First set $u^{\#}(a)=\inf \{t>0;\mu(t)0$, independent of $\Omega$, such that $$\lambda_1 (\Omega ) \ge \lambda_1 (\Omega ^*) (1 + C \alpha ^3) .$$ \end{theorem} We adapt the method developed in \cite{b1,b3,h1} to achieve our goal, i. e., we characterize the propagation of asymmetry $\alpha$ via the level sets of $u$. This is expressed in terms of the isoperimetric inequality. See Lemmas 3.3 and 3.5 for a more precise statement. Our result relies on several lemmas proven in Section 3 and the proof of the Theorem appears in Section 4. We mention that we make considerable use of the co-area formula in our work. In this context we refer to \cite{b4,f1}. The reader may find some overlap between this work and \cite{h3} however we believe some aspects of our work may be of independent interest. Lastly, we are unable to determine whether or not the third power appearing in (2.2) is optimal. However, in the case of the Laplacian it has been conjectured that the above Theorem holds with the second power and if true, it would then be optimal \cite{b3,n1}. \section {Preliminaries} In this section we present five lemmas which will lead to the proof of Theorem \ref{main-thm}. For compactness of our presentation, we take $|\Omega | = 1$. \begin{lemma} % Lemma 3.1 Let $u(x)$ be a solution of (1.1); set $h(t)= \int_{\Omega_t}|Du|^p$. Then $h(t)$ is convex in $t$ and for $0 < t < M$, $$\lambda_1 \Big(1 - t \int_\Omega u^{p-1}\Big) \le h(t) \le \lambda_1 \left(1 - t/M\right)\int_{\Omega_t}u^p.$$ \end{lemma} \paragraph{Proof.} A proof of this lemma can be worked by using the co-area formula. However, we will provide a proof which uses appropriate test functions (also see \cite{b2}). Recall the weak formulation in Section 1, i. e., $$\int_{\Omega}|Du|^{p-2}Du\cdot D\psi=\lambda_1\int_{\Omega}u^{p-1}\psi$$ where $\psi\in W^{1,p}_0(\Omega)$. Using the test function $(u-t)^+$ in (3.2), we find that $$h(t)=\lambda_1 \int_{\Omega_{t}}u^{p-1}(u-t).$$ We now make some observations which will prove useful later. For $\delta>0$, $t\le u1$ it is clear that $p^2-2p+2>0$ and $-\mu(t)^{(p^2-2p+2)/(2p(p-1))}$ is increasing and right continuous. Integrating from $0$ to $t$, we find that \begin{eqnarray*} t\le \left(\frac{2p(p-1)}{ p^2-2p+2}\right)\left(\frac{\lambda_1}{(4\pi)^{p/2}}\right)^{1/(p-1)} \left(1-\mu(t)^{(p^2-2p+2)/(2p(p-1))}\right) \end{eqnarray*} Thus we get the estimate in the statement of the lemma. \hfill$\diamondsuit$ \paragraph{Remark.} Lemma 3.2 leads to an upper bound under the assumption $\lambda_1 \le 2\lambda ^*_1$.\medskip The next three lemmas are crucial to proving the Theorem, they indicate how asymmetry enters into the calculations. In Lemmas 3.3 and 3.5, $k$ stands for a positive constant in $(0,1/100)$, whose exact value will determined in Section 4. The basic approach to proving the Theorem is along the lines of \cite{b1,b3,h1} and this motivates the following lemma. \begin{lemma} \label{lm3.3} Let $u$ solve (2.1), $\alpha$ be the asymmetry of $\Omega$. Assume that there exists a $T$ with $0 < T < M$ such that for a. e. $t\in [0,T]$, there exists a constant $k, 0 < k < 1/100$, with the property that $$L(\partial \Omega _t)^2 \ge 4\pi (1 + k\alpha ^2 )\mu (t) .$$ Then $$\lambda_1 (\Omega )\ge \lambda_1 (\Omega ^*)\left(1 + \frac {Tp}{8M}k\alpha^2\right).$$ \end{lemma} \paragraph{Proof.} Our starting point will be Lemma 2 in \cite{t1} and the outline of the proof is quite similar to the method used in Lemma 1 in \cite{t2}. From \cite[Lemma 2]{t1} for a. e. $t$, $$L(\partial\Omega_t)^{p/(p-1)}\le -\mu(t)' \left(-\frac{d}{dt}\int_{\Omega_t} |Du|^p\right)^{1/(p-1)},$$ where $L(\partial\Omega_t)$ is the De Giorgi perimeter. Employing the isoperimetric inequality, $$(4\pi\mu(t))^{p/2(p-1)}\le -\mu(t)' \left(-\frac{d}{dt}\int_{\Omega_t} |Du|^p\right)^{1/(p-1)}\;\;\mbox{a. e. t.}$$ Employing the hypothesis of the lemma, we get for a. e. $t$, \begin{eqnarray*} \left(4\pi(1+k\alpha^2)\mu(t)\right)^{p/2(p-1)}\le -\mu(t)' \left(-\frac{d}{dt}\int_{\Omega_t}|Du|^p\right)^{1/(p-1)} \end{eqnarray*} Therefore, $$-\frac{d}{dt}\int_{\Omega_t}|Du|^p \ge \frac{(4\pi(1+k\alpha^2)\mu(t))^{p/2}} {(-\mu(t)')^{p-1}}.$$ Recall that from Lemma 3.1, $\int_{\Omega_{t}}|Du|^{p}$ is convex and hence continuous on $[0,M]$. Since $-\int_{\Omega_{t}}|Du|^{p}$ is non-decreasing integrating (3.12) from $0$ to $T$, we obtain \begin{eqnarray*} \int_{\Omega}|Du|^p -\int_{\Omega_T}|Du|^p \ge (1+k\alpha^2)^{p/2} \int^T_0 \frac{4\pi\mu(s))^{p/2}}{(-\mu(s)')^{p-1}}ds \end{eqnarray*} Rewriting and using that $p$-Dirichlet integrals diminish under symmetrization, we obtain \begin{eqnarray} \lambda_1(\Omega)&\ge& \int_{\Omega_T^*}|Du^{*}|^p + (1+k\alpha^2)^{p/2} \int^T_0 \frac{(4\pi\mu(s))^{p/2}}{(-\mu(s)')^{p-1}}ds \end{eqnarray} where $u^*$ is the Schwartz non-increasing radial rearrangement of $u$. Recall that $u$ is $C^{1,\alpha}_{loc}(\Omega)$, hence $u^*$ is locally Lipschitz continuous. Since $u^*(x)=u^*(|x|)$, we define $r(t)=\sqrt{\mu(t)/\pi}$. Clearly, $u^*(r(t))=t$ where $r=|x|$. Thus the co-area formula yields $$\int_{\Omega_t^*}|Du^*|^p=\int_t^{\infty} \left(\int_{\partial\Omega_s^{*}}|Du^*|^{p-1} \right)ds=\int_{t}^{\infty} |Du^{*}|^{p-1}(r(s))L(\partial\Omega_{s}^{*})ds,$$ where $r(s)=\sqrt{\mu(s)/\pi}$. Thus for a. e. $t$, $$\frac{d}{dt}\int_{\Omega_t^*}|Du^*|^p= -|Du^*|^{p-1}(r(t))L(\partial\Omega^*_t),$$ where $r=\sqrt{|\Omega^*_t|/\pi}=\sqrt{\mu(t)\pi}$. Note that the above also shows that $\int_{\Omega_{t}^*}|Du^{*}|^{p}$ is Lipschtiz continuous in $t$. However, using polar coordinates we may express \begin{eqnarray*} \int_{\Omega^*_t}|Du^*|^p =2\pi \int_0^{\sqrt{\mu(t)/\pi}}|Du^*|^p r\,dr. \end{eqnarray*} Differentiating the above and using (3.15) \begin{eqnarray*} \frac{d}{dt}\int_{\Omega^*_t}|Du^*|^p&=&\sqrt{\pi}|Du^{*}|^{p}(t) r(t)\mu(t)^{\prime}/\sqrt{\mu(t)}\\ &=& |Du^*|^p\mu(t)'=-|Du^*|^{p-1} L(\partial\Omega^*_t). \end{eqnarray*} Simplifying we obtain for a. e. $t$ that $$|Du^*|=-\frac{L(\Omega^*_t)}{\mu(t)'}.$$ Employing (3.16) in the co-area formula (3.14) results in $$\int_{\Omega^*_t}|Du^*|^p=\int^{\infty}_t \frac{(4\pi\mu(s))^{p/2}}{ (-\mu(s)')^{p-1}}ds.$$ Now since $k<1/100$ and $\alpha\le 1$, clearly $(1+k\alpha^2)^{p/2} \ge 1+ kp\alpha^2/4,$ for all $1& 4\pi \Big\{|F| + |H| + 2 \{|F||H|\}^{1/2}\Big\}\\ &=& 4\pi \Big\{|\Omega _t| + 2|H| +2[(|\Omega _t|+|H|)|H|]^{1/2}\Big\}. \end{eqnarray*} Simplifying, $$2\Big\{|H|(|\Omega _t| + |H|)\Big\}^{1/2} \le k\alpha^2 |\Omega _t|.$$ One then easily obtains $$|H| \le \frac {k^2\alpha ^4}{4} |\Omega _t |.$$ (B) Our second step is to show that of the$C_i$'s all but one have small areas. In order to simplify our computations, we set$R_i = \sqrt {|C_i|/\pi}, \quad i = 1,2,\dots .$Label$R_i$'s such that$R_1 = \sup \{ R_i, i = 1,\dots \}$. This supremum is attained since$\Sigma|C_i| = \pi \Sigma R^2_i = |\Omega _t|< \infty$. Also note that$L(\partial C_i) \ge 2\pi R_i, \; \forall \;\; i$. Thus \begin{eqnarray*} 4\pi ^2 (\Sigma R_i)^2 \le (\Sigma L(\partial C_i)) ^2=(L(\partial\Omega_t))^2 < 4\pi ^2 (1 + k\alpha ^2 )(\Sigma R^2_i ) . \end{eqnarray*} Set$\varepsilon _i = R_i/R_1$,$i = 1,2,\dots$, then $$\Big(1 + {\sum _{i > 1}} \varepsilon _i\Big)^2 \le (1 + k\alpha ^2)\Big(1 + {\sum_{i>1} \varepsilon ^2_i} \Big).$$ Thus, $$1 + 2 {\sum _{i > 1}} \varepsilon _i + {\sum _{i > 1}} \varepsilon ^2_i \le (1 + k\alpha ^2) (1 + {\sum _{i > 1}} \varepsilon ^2_i ),$$ hence, $$2 {\sum _{i > 1}} \varepsilon _i \le k\alpha ^2 (1 + {\sum _{i > 1}} \varepsilon _i ^2);$$ now, together with the fact$ {\sum _{i > 1}} \varepsilon _i^2 \le {\sum _{i > 1}} \varepsilon _i $, we get $${\sum _{i > 1}} \varepsilon _i \le \frac {k\alpha ^2}{2-k\alpha ^2} \le k\alpha ^2 .$$ Thus, $${\sum _{i > 1}} \varepsilon _i^2 \le ( {\sum _{i > 1}} \varepsilon _i)^2 \le k^2 \alpha ^4,$$ implying that \begin{eqnarray*} {\sum _{i>1}} |C_i| \le k^2\alpha ^4|C_1| \le k^2\alpha ^4 |\Omega_t|. \end{eqnarray*} It n easy to see that $$|C_1| \ge \frac {|\Omega _t|}{1 + k^2\alpha ^4} \ge |\Omega _t| (1 - k^2\alpha^4).$$ \noindent (C) We now work with$F_1$; by hypothesis of the lemma and (3.27) \begin{eqnarray} L(\partial F_1)^2 \le L(\partial C_1)^2 \le L(\partial \Omega _t)^2 &\le & 4\pi (1 + k\alpha ^2)(1 + k^2\alpha ^4)|C_1|\nonumber\\ & \le & 4\pi \left(1 + \frac {10}{9} k\alpha ^2\right) |F_1| . \end{eqnarray} The last inequality follows from noting that$k < 1/100$. Since$F_1$is simply connected, we may calculate the inradius$I$(see \cite{o1}) using (3.28), \begin{eqnarray*} I & \ge & \frac{L (\partial F_1) - \sqrt {L( \partial F_1)^2 - 4\pi |F_1|}}{2\pi} \\ &\ge& \frac{\sqrt {4\pi |F_1|} - \sqrt {4\pi (1 + \frac {10}{9} k\alpha^2) |F_1|-4\pi |F_1|}}{2\pi} \\ & \ge & \sqrt {\frac {|F_1|}{\pi}} \left ( 1 - \frac {11}{10} \sqrt {k} \alpha \right ) \ge \sqrt {\frac {|\Omega _t|}{\pi}} \frac {\left ( 1 - \frac {11}{10} \sqrt {k} \; \alpha \right )}{\sqrt {1 + k^2\alpha ^4}} \\ & \ge & \sqrt {\frac {|\Omega _t|}{\pi}} \left(1 - \frac {12}{10} \sqrt {k} \alpha \right) = R. \end{eqnarray*} Clearly the ball$B_R$with an appropriate center lies in$F_1$, and so$B_R\backslash H$lies in$C_1$. We now estimate$\Omega_t$by using the properties of the in-ball, the definition of asymmetry$\alpha $(see (1.2)) and (3.26), \begin{eqnarray*} \alpha |\Omega | & \le & |\Omega \backslash (B_R\backslash H)| = |\Omega | - |B_R \backslash H| \\ &\le& 1 - \Big[ \big(1 - \frac {12}{10} \sqrt {k} \alpha\big)^2 - \frac {k^2\alpha ^4}{4}\Big]|\Omega _t| \\ &\le& 1 - \Big(1 - \frac {5}{2} \sqrt {k}\alpha \Big)|\Omega _t | \end{eqnarray*} Thus, $$|\Omega _t | \le \frac {1-\alpha}{1-5\sqrt {k}\alpha /2}.$$ We now recall the discussion at the beginning of our proof. The inequality in (3.29) is derived for$S_i$with$\alpha=\alpha_i=\alpha(S_i)$. As pointed out, taking the limit$i\to\infty$provides justification for validity of (3.29) for$\Omega_t$. \hfill$\diamondsuit$\paragraph{Remark 3.3} If$t$satisfies the conditions of Lemma 3.5, thenLemma 3.4 implies $$\lambda_1 (\Omega ) \ge \lambda_1 (\Omega ^*) \left(\frac {1-5\sqrt {k}\alpha/2} {1-\alpha }\right)^{p/2} (1-t)^{p-1} .$$ \section {Proof of the main result} We take$k =1/625$; we recapitulate the above results as follows: \\ (a) If asymmetry propagates over a t'' interval$[0,T]$, i.e.,$L( \partial \Omega _t)^2 \ge 4 \pi (1 + k\alpha ^2)|\Omega _t|$a. e.$t\in[0,T]$, then $$\lambda_1 (\Omega ) \ge \lambda_1 (\Omega ^* ) \left(1 + \frac {kT}{8M} \alpha ^2 \right)\;\;. \eqno {(*)}$$ This follows from Lemma 3.3, and Remark 3.1. Note that it is enough toassume that$\lambda_1 \le 2\lambda ^*_1$, for otherwise the theorem follows. \\ (b) If not, i.e., for some$t$in$[0, T]$, we have$L(\partial \Omega _t)^2 < 4\pi (1 + k\alpha ^2 )|\Omega _t |, $then via Lemma 3.4, and Remark 3.3 with$k=1/625$, we have \begin{eqnarray} \lambda_1 (\Omega ) &\ge& \lambda_1 (\Omega ^*) \Big(\frac {1-5\sqrt {k}\; \alpha/2}{1-\alpha}\Big)^{p/2} (1-t)^{p-1} \\ &\ge& \lambda_1(\Omega^*) \left(1+\frac{9\alpha}{10}\right)^{p/2}(1-t)^{p-1} \nonumber \end{eqnarray} We make the following simple observations keeping in mind that$0\le \alpha \le 1$. Firstly \left( 1+9\alpha/10\right)^{p/2} \ge 1+9p\alpha/40\; \mbox{when$1