
\documentclass[reqno]{amsart}
\usepackage{graphicx}
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\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<p < 2^* -1= \frac{n+2}{n-2} $
 and $\alpha >0$ 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{equation}
\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}
\end{equation}
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{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}
\end{equation}
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{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}
\end{equation}
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)] $1\le q< 2^*$,
\[
 \|u_{\varepsilon}\|_q^q  \begin{cases}
 =K \varepsilon^{\frac{2n-(n-2)q}{2}}+ O(\varepsilon^{\frac{(n-2)q}{2}}) ,
&   q >  n/(n-2)\\
=\varepsilon^{n/2}(K|\ln \varepsilon|+O(1)) , &  q=n/(n-2) \\
 \approx \varepsilon^{(n-2)q/2} , &  q<n/(n-2).
\end{cases}
\]
\end{itemize}
\end{lemma}


\begin{proof} The estimate in (a) and (b)  are known. Estimate (c)
 can be shown similarly as in \cite{g1,s1}.
\end{proof}


\begin{lemma} \label{lem2}
 There are  constants $\beta, \beta_1, \beta_2 >0 $ 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 $1<p<2$ then there is no
constant $\beta>0 $  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{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{e3alpha}
\end{equation}
 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
\begin{equation}
-\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}
\end{equation}
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
\begin{equation}
-\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}
\end{equation}
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 $t\phi_1(x) <u_{ \alpha_0}(x)$ on
the whole domain $\Omega$  and thus again via sub- and
super-solution method  we obtain a solution $ u_{ \alpha}(x) $ for
\eqref{e3alpha}, where $u_{ \alpha}(x)$ is a  minimizer of
$$
J(u) =  \int_{\Omega}\frac{1}{2}|\nabla u|^2-\frac{\lambda}{2}
u^2+ \frac{\alpha}{p+1}| u|^{p+1}+\frac{1}{2^*}|u|^{2^*}\,dx
$$
over the convex set
$ K= \{u\in H^1_0(\Omega): t\phi_1(x) \le u(x) \le u_{\alpha_0}(x)
\text{ a. e. in  }\Omega \}$. Furthermore, since
 $t\phi_1, u_{\alpha_0}$ are not solutions of (3)$_\alpha$, we
 deduce that $t\phi_1(x) < u(x) < u_{\alpha_0}(x) $ on $
 \Omega$. If we choose $k >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{equation}
\begin{gathered}
-\Delta v = g(x, v)
\quad    \hbox{in } \Omega  \\
v=0 \quad \hbox{on } \partial \Omega.
\end{gathered} \label{e6}
\end{equation}
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
\begin{equation}
E(v) = J(v_++u_{\alpha}) -J(u_{\alpha})+ \frac{1}{2} \|v_-\|^2, \label{e7}
\end{equation}
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
\begin{equation}
\Gamma = \{\phi \in C([0, 1], H):
\phi(0)=0, \phi(1) = v \} \label{e8}
\end{equation}
 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}


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