\documentclass[reqno]{amsart} \usepackage{graphicx} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 135, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/135\hfil Bifurcation of positive solutions] {Bifurcation of positive solutions for a semilinear equation with critical Sobolev exponent} \author[Y. Cheng\hfil EJDE-2006/135\hfilneg] {Yuanji Cheng} \address{Yuanji Cheng \newline School of Technology and Society, Malm\"o University, SE-205 06 Malm\"o, Sweden} \email{yuanji.cheng@ts.mah.se} \date{} \thanks{Submitted August 12, 2005. Published October 25, 2006.} \subjclass[2000]{49K20, 35J65, 34B15} \keywords{Critical Sobolev exponent; positive solutions; bifurcation} \begin{abstract} In this note we consider bifurcation of positive solutions to the semilinear elliptic boundary-value problem with critical Sobolev exponent \begin{gather*} -\Delta u = \lambda u - \alpha u^p+ u^{2^*-1}, \quad u >0 , \quad \hbox{in } \Omega,\\ u=0, \quad \hbox{on } \partial\Omega. \end{gather*} where $\Omega \subset \mathbb{R}^n$, $n\ge 3$ is a bounded $C^2$-domain $\lambda>\lambda_1$, $1 0$ is a bifurcation parameter. Brezis and Nirenberg \cite{b1} showed that a lower order (non-negative) perturbation can contribute to regain the compactness and whence yields existence of solutions. We study the equation with an indefinite perturbation and prove a bifurcation result of two solutions for this equation. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction and main result} It is well known that the following equation with a critical exponent has no solution on the star-shaped domains, \cite{p1}, $$\begin{gathered} -\Delta u =u^{\frac{n+2}{n-2}},\quad \hbox{in } \Omega, \\ u=0 , \quad \hbox{on } \partial \Omega, \end{gathered} \label{e1}$$ due to the lack of compactness in the embedding $H^1_0(\Omega)\hookrightarrow L_{\frac{2n}{n-2}}(\Omega)$. In their seminal work \cite{b1}, Brezis and Nirenberg show that perturbation by a lower order term suffices to regain the compactness and hence existence of a solution. Consider particularly for the following equation $$\begin{gathered} -\Delta u = \lambda u + u^{\frac{n+2}{n-2}},\quad u >0 \quad \hbox{in } \Omega, \\ u=0 , \quad \hbox{ on } \partial \Omega, \end{gathered} \label{e2}$$ where $\lambda$ is considered as a bifurcation parameter, let $\lambda_1 >0$ be the first eigenvalue of Laplacian with a Dirichlet boundary, then they show the following result. \begin{theorem}[\cite{b1}] \label{thm1} There is a constant $\lambda^* \in [0, \lambda_1)$, such that \eqref{e2} has a solution if $\lambda\in (\lambda^* , \lambda_1)$ and has no solution, if $\lambda \ge \lambda_1$. \end{theorem} Thereafter, there are many papers devoted to study of problems with critical Sobolev exponent (see \cite{g1,s1} and references therein). Effects of concave and convex combination on bifurcation have been studied in \cite{a1,c1,c2,c3,t1}. In this paper we consider the equation with an indefinite lower order perturbation. For simplicity consider the prototype equation $$\begin{gathered} -\Delta u = \lambda u -\alpha u^p + u^{\frac{n+2}{n-2}},\quad u >0 , \quad \hbox{in } \Omega, \\ u=0 , \quad \hbox{on } \partial \Omega, \end{gathered} \label{e3}$$ where $\lambda$ is a fixed positive constant, and $\alpha >0$ is considered as a bifurcation parameter. The main result of this note is the the following theorem showing the existence of two solutions. \begin{theorem} \label{thm2} If $\lambda >\lambda_1$ and $3\le n \le 5$, $1< p < 4/(n-2)$, then there is a constant $\alpha_0 >0$ such that \eqref{e3} has at least two solutions for $\alpha > \alpha_0$ and has no solution if $\alpha < \alpha_0$ \end{theorem} \begin{figure}[ht] \begin{center} \includegraphics[width=0.7\textwidth]{fig1} \end{center} \caption{Bifurcation diagram of (1.3)} \end{figure} \section{Auxiliary lemmas} In this section we establish some estimates which are needed in the proof of Theorem \ref{thm2}. Without loss of generality, we assume that the domain $\Omega$ contains the origin and choose $R>0$ small enough so that $\{x: |x| \le 2 R\} \subset \Omega$. Let $\psi (x)$ be a cut-off function such that $\psi(x)\equiv\begin{cases} 1, & |x|\le R, \\ 0, & |x|\ge 2R, \end{cases}$ and $N=\sqrt{n(n-2)}$. Also let $$u_{\varepsilon}(x) =\psi(x) u_{0\varepsilon}(x), \quad u_{0\varepsilon}(x)=\big( \frac{N\varepsilon}{\varepsilon^2+|x|^2}\big)^{(n-2)/2}.$$ Then $\|\nabla u_{0\varepsilon}\|_2^2 = S^{n/2} =\|u_{0\varepsilon}\|_{2^*}^{2^*}$ for all $\varepsilon >0$. The following estimates will be needed in the proof of Theorem \ref{thm2}. \begin{lemma} \label{lem1} The following estimates hold for some constant $K=K(q) >0$ \begin{itemize} \item[(a)] $\|\nabla u_{\varepsilon}\|_2^2 = S^{n/2} + O( \varepsilon^{n-2})$ \item[(b)] $\|u_{\varepsilon}\|_{2^*}^{2^*} = S^{n/2} + O( \varepsilon^{n})$ \item[(c)] $1\le q< 2^*$, $\|u_{\varepsilon}\|_q^q \begin{cases} =K \varepsilon^{\frac{2n-(n-2)q}{2}}+ O(\varepsilon^{\frac{(n-2)q}{2}}) , & q > n/(n-2)\\ =\varepsilon^{n/2}(K|\ln \varepsilon|+O(1)) , & q=n/(n-2) \\ \approx \varepsilon^{(n-2)q/2} , & q0  such that the following inequalities hold for all a, b \ge 0 \begin{itemize} \item[(1)] p \ge 2, \beta_1(a^{p-1}b+ ab^{p-1}) \ge (a+b)^p-a^p-b^p \ge \beta_2(a^{p-1}b+ab^{p-1}). \item[(2)]  p \in(1, 2), (a+b)^p-a^p-b^p \le \beta a^{p-1}b. \end{itemize} \end{lemma} \begin{proof} The inequalities follow from the facts that h(t) = \frac{(1+t)^p-1-t^p}{t+ t^{p-1}}\to p as either t\to 0+  or  t\to +\infty; h_0(t) =\frac{(1+t)^p-1-t^p}{t}\to p  as  t \to 0+ and h_0(t) \to 0  as  t \to +\infty. \end{proof} We would like to point out here that if 10  such that the following estimate holds for all a, b \ge 0,  (a+b)^p\ge a^p+b^p +\beta a^{p-1}b.  \section{Proof of Theorem \ref{thm2}} Now we consider $$\begin{gathered} -\Delta u = \lambda u -\alpha u^p + u^{\frac{n+2}{n-2}},\quad u >0 , \quad \hbox{in } \Omega, \\ u=0 , \quad \hbox{on } \partial \Omega, \end{gathered} \label{e3alpha}$$ We first observe that for small \alpha>0 there is no solution for \eqref{e3alpha} by comparison, because f(u):=\lambda u -\alpha u^p + u^{\frac{n+2}{n-2}}  satisfies the inequality  f(u) > \lambda_1u on ( 0, \infty). On the other hand, if  \alpha  is big enough, then f(u)  vanishes somewhere on (0, \infty)  and whence a constant u_+(x)= M  suffices for a super-solution. To find a sub-solution, we can take u_-(x) = t\phi_1(x)>0, where \phi_1(x)>0 is the normalized eigenfunction associated to \lambda_1, because $$-\Delta (t\phi_1) -\lambda (t\phi_1) +\alpha(t\phi_1)^p -(t\phi_1)^{2^*-1} = t(\lambda_1-\lambda)\phi_1 +\alpha (t\phi_1)^{p}-(t\phi_1)^{2^*-1} < 0.\label{e4}$$ Thus by the sub- and super-solution method, there is a solution for \eqref{e3alpha}. Furthermore for given \alpha_0 > 0 if the problem (3.1) has a solution u_{ \alpha_0}, we shall show then for any \alpha>\alpha_0  the problem \eqref{e3alpha} has also a solution. Clearly u_{ \alpha_0} is a super-solution for \eqref{e3alpha}, because $$-\Delta u_{\alpha_0}-\lambda u_{\alpha_0} + \alpha u^p_{ \alpha_0} -u_{\alpha_0}^{2^*-1} = ( \alpha - \alpha_0) u_{ \alpha_0}^p >0, \label{e5}$$ and moreover t\phi_1(x)  still suffices as a sub-solution. Further, by the Hopf's lemma \frac{\partial u_{\alpha_0}}{\partial \nu}>0 on \partial \Omega, we deduce that t\phi_1(x) 0  large then (\lambda +k)u -\alpha u^p + u^\frac{n+2}{n-2} will be increasing on (0 , \infty)  and whence we deduce from \cite[Theorem 2]{b2} that u_{\alpha} is a local minimizer for J  in H^1_0(\Omega)-topology. We now define \alpha_0  to be the infimum of all \alpha>0 such that \eqref{e3alpha} has a solution, then we infer that \alpha_0>0 is an finite number, and it remains to show that for all  \alpha >\alpha_0  there are two solutions for \eqref{e3alpha}. Let  \alpha > \alpha_0 be given, and u_{\alpha} be the solution of \eqref{e3alpha} obtained by the sub- and super-solution method. To establish the second solution we exploit the truncation and translation technique and define  v = u -u_{\alpha} and \[ g(x, v) = \begin{cases} \lambda v -\alpha(( v+ u_{\alpha})^p -u_{\alpha}^p) + (v+u_{\alpha})^{2^*-1}-u_{\alpha}^{2^*-1} & v \ge 0 \\ 0 & v < 0.\\ \end{cases}$ In the sequel we shall study the boundary-value problem $$\begin{gathered} -\Delta v = g(x, v) \quad \hbox{in } \Omega \\ v=0 \quad \hbox{on } \partial \Omega. \end{gathered} \label{e6}$$ First we notice that any nontrivial solution $v$ of \eqref{e6} must be non-negative and then by the strong maximal principle it should be strictly positive on $\Omega$. Whence if $v\ne 0$ is a solution of \eqref{e6}, then $u= v+ u_{\alpha}$ will be a positive solution to the problem \eqref{e3alpha}, which is bigger than $u_{\alpha}$. We will exploit the critical point method and whence will study the associated functional to the problem \eqref{e6}, $$E(v) = \int_{\Omega}\frac{1}{2}|\nabla v|^2-G(x, v), \quad G(x, v)= \int_0^v g(x, t)\, dx.$$ Given any $v \in H$, decomposed into positive part $v_+$, and negative part $v_-$, then we test the equation \eqref{e3alpha} for the solution $u_{\alpha}$ by $v_+$ and obtain $$\int_{\Omega}\nabla u_{\alpha} \cdot \nabla v_+ = \int_{\Omega} ( \lambda u_{\alpha} -\alpha u_{\alpha}^p + u_{\alpha}^{\frac{n+2}{n-2}})v_+\,.$$ Furthermore we obtain the relation $$E(v) = J(v_++u_{\alpha}) -J(u_{\alpha})+ \frac{1}{2} \|v_-\|^2, \label{e7}$$ which shows that zero is even a local minimizer for $E$. \begin{lemma} \label{lem3} The equation \eqref{e6} satisfies the Palais-Smale condition (P.S.)$_c$ for any $c \in (0, \frac{1}{n}S^{n/2})$. \end{lemma} \begin{proof} Arguments in \cite[Lemma 2.3]{s1} works also here. \end{proof} By the min-max principle, if we can find $v>0$ such that $$c= \inf_{\phi\in \Gamma} \max \{ E(\phi (t)): t\in [0, 1]\}$$ is finite and $E$ satisfies the local Palais-Smale condition (P.S.)$_c$, where $$\Gamma = \{\phi \in C([0, 1], H): \phi(0)=0, \phi(1) = v \} \label{e8}$$ then there is a critical point $u$ of $E$ at level $c$. It follows from \eqref{e7} that $c\ge 0$. If $c> 0$, then we will have a nontrivial solution $u$. If $c=0$, then by \cite[Theorem 5.10]{f1}, see also \cite{f2,g2}, we deduce that there is a continua of minimizers $u^{\varepsilon}(x), \varepsilon \in (0, \varepsilon_0)$ such that $E(u^{\varepsilon}) = E(u_{\alpha})$. So we are also done even in this case. To find the function $v$ in \eqref{e8}, we shall test $v= t u_{\varepsilon}$. For $n=3$, we may assume $p\in( 2, 3)$ then we have $2^*= 6$, $\frac{n}{n-2} =3$ and by Lemma \ref{lem2} we obtain \begin{gather*} (v+u_{\alpha})^{2^*-1} -u_{\alpha}^{2^*-1} \ge v^5 + 4 v^4 u_{\alpha},\\ ( v+ u_{\alpha})^p -u_{\alpha}^p\le v^p+\beta_1( v^{p-1} u_{\alpha} + vu_{\alpha}^{p-1}) \end{gather*} and consequently $$G(x, v )\ge \frac{\lambda}{2} v^2 -\alpha ( \frac{1}{p+1}v^{p+1} + \beta (\frac{1}{2} v^2u^{p-1}_{\alpha}+ \frac{1}{p} v^pu_{\alpha}))+\frac{1}{6}v^{6} + \frac{\beta_2 }{5}v^{5}u_{\alpha}.$$ Since $u_{\alpha}$ is strictly positive on $\Omega$, so there are constants $C_1\ge C_2 >0$ such that $C_1\ge u_{\alpha}(x)\ge C_2$, for all $x\in \Omega, |x|\le 2R$. We deduce that for some constants $C_3, C_4 >0$, $$E(t u_{\varepsilon}) \le \int_{\Omega}\frac{t^2}{2}|\nabla u_{\varepsilon} |^2+C_4( t^2 u_{\varepsilon}^2 + t^{p} u_{\varepsilon}^{p}+ t^{p+1} u_{\varepsilon}^{p+1}) -C_3t^{5} u_{\varepsilon}^{5}- \frac{t^{6}}{6} u_{\varepsilon}^{6}.$$ In view of lemma \ref{lem1}, we obtain \begin{gather*} \|u_{\varepsilon}\|_2^2 \le A \varepsilon, \quad \|u_{\varepsilon}\|_p^p \le A \varepsilon^{p/2}, \quad \|u_{\varepsilon}\|_{p+1}^{p+1} = K(p+1) \varepsilon^{(5-p)/2}+O(\varepsilon^{(p+1)/2}), \\ \|u_{\varepsilon}\|_5^5 = K\eqref{e5} \sqrt{\varepsilon}+O(\varepsilon^{5/2}), \quad \|u_{\varepsilon}\|_6^6 = S^{3/2} +O(\varepsilon^{3}) \end{gather*} thus \begin{align*} E(t u_{\varepsilon}) &\le \frac{t^2}{2}(S^{3/2}+O(\varepsilon))+C_4( t^2 A\varepsilon+t^pA\varepsilon^{p/2} + t^{p+1} (K(p+1)\varepsilon^{\frac{5-p}{2}} +O(\varepsilon^{\frac{p+1}{2}}) ))\\ \\ &\quad -t^{5}C_3 (K(5)\sqrt{\varepsilon}+ O(\varepsilon^{5/2}))-\frac{t^6}{6} (S^{3/2} +O(\varepsilon^3)) :=h_3(t). \end{align*} The function $h_3(t)$ attains its maximum on $(0, \infty)$ at $t_{max3} :=1-\frac{5K\eqref{e5}C_3}{4S^{3/2}}\sqrt{\varepsilon} +o(\sqrt{\varepsilon})$. Moreover $h_3( t_{max3})= \frac{1}{3}S^{3/2} - C_3K\eqref{e5} \sqrt{\varepsilon} +o(\sqrt{\varepsilon})$. Therefore, we deduce that for $\varepsilon > 0$ enough small $$c = \inf_{\phi\in \Gamma} \max \{ E(\phi (t)): t\in [0, 1]\}\le h_3( t_{max3}) <\frac{1}{3}S^{3/2}$$ and obtain via the mountain pass theorem that \eqref{e6} admits a positive solution $u$. The proof is complete for the case of dimension $3$. If $n=4$ or 5, then by the assumption $p < 4/(n-2) \le 2$ and thus it follows from the lemma \ref{lem2} that \begin{gather*} ( v+ u_{\alpha})^p -u_{\alpha}^p\le v^p+\beta v u_{\alpha}; \quad (v+u_{\alpha})^{2^*-1}-u_{\alpha}^{2^*-1} \ge v^{2^*-1} + \beta_2 v^{2^*-2}u_{\alpha},\\ g(x, v )\ge \lambda v -\alpha ( v^p +\beta vu^{p-1}_{\alpha})+v^{2^*-1} + \beta_2 v^{2^*-2}u_{\alpha} \end{gather*} and consequently \begin{gather*} G(x, v )\ge \frac{\lambda}{2} v^2 -\alpha ( \frac{1}{p+1}v^{p+1} + \frac{\beta}{2} v^2u^{p-1}_{\alpha})+\frac{1}{2^*}v^{2^*} + \frac{\beta_2 }{2^*-1}v^{2^*-1}u_{\alpha}, \\ E(v) \le \int_{\Omega}\frac{1}{2}|\nabla v|^2-(\frac{\lambda}{2} v^2 -\alpha ( \frac{1}{p+1}v^{p+1} + \frac{\beta}{2} v^2u^{p-1}_{\alpha})+\frac{1}{2^*}v^{2^*} + \frac{\beta_2 }{2^*-1}v^{2^*-1}u_{\alpha}). \end{gather*} In analogy as the case $n=3$, we deduce that for some constants $C_3, C_4>0$. $$E(t u_{\varepsilon}) \le \int_{\Omega}\frac{t^2}{2}|\nabla u_{\varepsilon} |^2+C_4( t^2 u_{\varepsilon}^2 + t^{p+1} u_{\varepsilon}^{p+1}) -C_3t^{2^*-1} u_{\varepsilon}^{2^*-1}- \frac{t^{2^*}}{2^*} u_{\varepsilon}^{2^*}.$$ For $n=4$, we have \begin{align*} E(t u_{\varepsilon}) &\le \frac{t^2}{2}(S^{2}+0(\varepsilon^2))+C_4( t^2 (\varepsilon^2(K(2)|\ln\varepsilon|+O(1)) + t^{p+1} (K(p+1)\varepsilon^{3-p} \\ &\quad +O(\varepsilon^{p+1}) )) -t^{3}C_3 (K(3)\varepsilon+ O(\varepsilon^3))-\frac{t^{4}}{4} (S^2 +O(\varepsilon^4)) :=h_4(t)\,. \end{align*} Then $h_4(t)$ attains its maximum on $(0, \infty)$ at $t_{max4} :=1-\frac{3K(3)C_3}{2S^2}\varepsilon+o(\varepsilon)$, which satisfies \begin{align*} &S^{2}+O(\varepsilon^2)+C_4( 2\varepsilon^2(K(2)|\ln\varepsilon| +O(1)) +t^{p-1} (p+1) (K(p+1)\varepsilon^{3-p} +O(\varepsilon^{p+1}) )) \\ &=t3C_3 (K(3)\varepsilon+ O(\varepsilon^3))+t^2 (S^2 +O(\varepsilon^4)) \end{align*} and moreover $h_4(t_{max4}) = \frac{1}{4}S^2 -C_3K(3) \varepsilon +o(\varepsilon) <\frac{1}{4}S^2$, for sufficient small $\varepsilon >0$. So we are done in this case. If $n=5$, we obtain in a similar way that \begin{align*} E(t u_{\varepsilon}) &\le \frac{t^2}{2}(S^{5/2}+O(\varepsilon^3))+C_4( t^2 (\varepsilon^2K(2)+O(\varepsilon^3)) + t^{p+1} (K(p+1)\varepsilon^{(7-3p)/2} \\ &\quad +O(\varepsilon^{\frac{3(p+1)}{2}}) )) -t^{\frac{7}{3}}C_3 (K(\frac{7}{3})\varepsilon^{\frac{3}{2}} + O(\varepsilon^{\frac{7}{2}}))-\frac{3t^{\frac{10}{3}}}{10} (S^{\frac{5}{2}} +O(\varepsilon^5)) :=h_5(t). \end{align*} Because $p < 4/3$, we see that $(7-3p)/2 > 3/2$ and whence $h_5(t)$ attends its maximum on $(0, \infty)$ at $t_{max5} :=1-\frac{7K(7/3)C_3}{4S^{5/2}}\varepsilon^{3/2} +o(\varepsilon^{3/2})$, which satisfies \begin{align*} &S^{5/2}+C_4( 2\varepsilon^2K(2)+O(\varepsilon^3) + (p+1)t^{p-1} (K(p+1)\varepsilon^{3-p} +O(\varepsilon^{p+1}) )) \\ &=\frac{7}{3}C_3t^{1/3} (K(7/3)\varepsilon^{3/2} + O(\varepsilon^{7/2}))+t^{4/3} (S^{5/2} +O(\varepsilon^5))\,. \end{align*} Moreover $h_5(t_{max5}) = \frac{1}{5}S^{5/2} -C_3K(7/3) \varepsilon^{3/2} +o(\varepsilon^{3/2}) <\frac{1}{5}S^{5/2}$, for sufficient small $\varepsilon >0$. So the proof is complete in this case. \section{ An example} In this part we show a numerical result of solutions for an equation on the the unite ball in $\mathbb{R}^3$. we consider an equation with a critical exponent $\Omega=\{x\in \mathbb{R}^3:\|x\|< 1\}$, \begin{gather*} -\Delta u(x) = 4\pi u(x) -\alpha u^2(x) + u^5(x), \quad \|x\|<1, \\ u(x)=0, \quad \|x\| =1. \end{gather*} By Gidas, Ni and Nirenberg \cite{g3}, any positive solution must be radial symmetric, i.e. $u(x)=u(r)$, $r= \|x\|$ and thus satisfies ordinary differential equation \begin{gather*} -(r^2u'(r))' = r^2(4\pi u(r) -\alpha u^2(r) + u^5(r)), \quad r \in (0, 1),\\ u'(0)=0, \quad u(1)=0. \end{gather*} By a numerical simulation for $\alpha =7.5$, we find two positive solutions, where their maxima of the solutions are $u_1(0)=0.575$ and $u_2(0) = 3.44$. \begin{figure}[ht] \begin{center} \includegraphics[width=0.6\textwidth]{fig2} \end{center} \caption{Numerical simulation of solutions on unit ball in $\mathbb{R}^3$} \end{figure} \begin{thebibliography}{00} \bibitem{a1} A. 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