\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 161, pp. 1--7.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/161\hfil Asymptotic expansion formulas]
{Asymptotic expansion formulas for the maximum of
solutions to diffusive logistic equations}

\author[T. Shibata\hfil EJDE-2008/161\hfilneg]
{Tetsutaro Shibata}

\address{Tetsutaro Shibata \newline
Department of Applied Mathematics\\
Graduate School of Engineering\\
Hiroshima University, Higashi-Hiroshima, 739-8527, Japan}
\email{shibata@amath.hiroshima-u.ac.jp}

\thanks{Submitted May 7, 2008. Published December 9, 2008.}
\subjclass[2000]{34B15}
\keywords{Logistic equation; $L^\infty$-norm of solutions}

\begin{abstract}
 We consider the nonlinear eigenvalue problems
 \begin{gather*}
 -u''(t) + u(t)^p =  \lambda u(t),\\
 u(t) > 0, \quad t \in I := (0, 1), \quad u(0) = u(1) = 0,
 \end{gather*}
 where $p > 1$ is a constant and $\lambda > 0$ is a parameter.
 This equation is well known as the original logistic equation of
 population dynamics when $p = 2$.
 We establish the precise asymptotic formula for
 $L^\infty$-norm of the solution $u_\lambda$ as $\lambda \to \infty$
 when $p=2$ and $p=5$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

We consider the nonlinear eigenvalue problem
\begin{gather}
-u''(t) + u^p(t) = \lambda u(t), \quad t \in I:= (0, 1), \label{e1.1}\\
 u(t) > 0, \quad t \in I, \label{e1.2} \\
u(0) = u(1) = 0, \label{e1.3}
\end{gather}
where $p > 1$ is a constant and $\lambda > 0$ is a parameter.
The equation \eqref{e1.1}--\eqref{e1.3} is well known as the original
logistic equation of population dynamics when $p = 2$.
In this case, the equation
\eqref{e1.1}--\eqref{e1.3} describes the behavior of a species in $I$, where
$u$ and $\lambda$ imply the population density
and the growth rate, respectively.
Therefore, many authors has been investigated the properties of solutions to
\eqref{e1.1}--\eqref{e1.3} from not only
pure mathematical point of view but also an application to the
field of biology by computational analysis.

For the existence and uniqueness of the solutions,
we know from \cite{b1} that for a given $\lambda > \pi^2$,
there exists a unique solution $u = u_\lambda \in C^2(I)$ to \eqref{e1.1}--\eqref{e1.3}.
Furthermore, the set of solutions
$\{\lambda, u_\lambda\} \subset \text{\bf R}_+ \times C^2(I)$ of
\eqref{e1.1}--\eqref{e1.3} gives the clear picture of so called
bifurcation diagram. Therefore, there are many works which treated the
problem \eqref{e1.1}--\eqref{e1.3} by bifurcation theory of $L^\infty$-framework.

The purpose of this paper is to study more precisely
the asymptotic behavior of $\| u_\lambda\|_\infty$ as
$\lambda \to \infty$, which is certainly one of
the most important asymptotic properties of $u_\lambda$
to know the global structure of the bifurcation curve.
In spite of the simplicity of the equation, however, a few
information of global structure of bifurcation diagram in
$L^\infty$-framework have been known. Therefore, the problem here is
worth considering.

It is known that for general $p > 1$, the following asymptotic formula
for $\| u_\lambda\|_\infty$ holds as $\lambda \to \infty$ \cite{s2}.
\begin{equation} \label{e1.4}
\| u_\lambda\|_\infty
= \lambda^{1/(p-1)}(1 - e^{-\sqrt{(p-1)\lambda}(1 + o(1))/2}).
\end{equation}
In particular, if $p = 3$, then the following asymptotic
formula has been obtained in \cite{s1}.

\begin{theorem}[\cite{s1}] \label{thm1.1}
 Assume $p=3$ in \eqref{e1.1}. Then as $\lambda \to \infty$,
\begin{equation} \label{e1.5}
\| u_\lambda\|_\infty = \sqrt{\lambda}(1 - 4
e^{-\sqrt{\lambda}/\sqrt{2}}
- 8e^{-2\sqrt{\lambda}/\sqrt{2}}
- 24\sqrt{2}
\sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}}
+ o(\sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}})).
\end{equation}
\end{theorem}

The main tool in the proof of Theorem \ref{thm1.1} is an asymptotic
expansion formula for the {\it complete} elliptic integral of the
first kind. However, unfortunately, this approach can not be
applied to the general $p > 1$, in particular, to
the most important case $p = 2$.
We overcome this difficulty by using the asymptotic
expansion formula of the first elliptic integral in \cite{c1}
and obtain the following results.

\begin{theorem} \label{thm1.2}
Assume $p=2$ in \eqref{e1.1}. Then as $\lambda \to \infty$,
\begin{equation} \label{e1.6}
\| u_\lambda\|_\infty = \lambda
\big\{(1 -6(\sqrt{3}-1)^2e^{-\sqrt{\lambda}/2}-12(\sqrt{3}-1)^4
e^{-\sqrt{\lambda}}
+ O(\sqrt{\lambda} e^{-3\sqrt{\lambda}/2})\big\}.
\end{equation}
\end{theorem}

We find that almost the same idea, as that to prove Theorem \ref{thm1.2},
can be applied to the case $p = 5$.

\begin{theorem} \label{thm1.3}
Assume $p=5$ in \eqref{e1.1}. Then as $\lambda \to \infty$,
\begin{equation} \label{e1.7}
\| u_\lambda\|_\infty = \lambda^{1/4}
\big\{1 -6e^{-\sqrt{\lambda}} -30e^{-2\sqrt{\lambda}}
+ O(\sqrt{\lambda} e^{-3\sqrt{\lambda}})\big\}.
\end{equation}
\end{theorem}

The proofs of Theorems \ref{thm1.2} and \ref{thm1.3} are quite simple and
straightforward. The future direction of this study will be
to give the approach which can be applied to general $p > 1$.

\section{Proof of Theorem \ref{thm1.2}}

In what follows, we assume that $\lambda \gg 1$. We put
\begin{equation} \label{e2.1}
\xi = \xi_\lambda = 1 - \frac{\| u_\lambda\|_\infty}{\lambda}.
\end{equation}
By \eqref{e1.4}, we know that $\xi_\lambda \to 0$ as $\lambda \to \infty$.
To prove Theorem \ref{thm1.2}, we establish the asymptotic formula for
$\xi_\lambda$. We know from \cite{b1} the fundamental properties of $u_\lambda$.
\begin{gather}
u_\lambda(t) = u_\lambda(1-t), \quad t \in I, \label{e2.2} \\
u_\lambda'(t) > 0, \quad t \in [0, 1/2), \label{e2.3}\\
\| u_\lambda\|_\infty = u_\lambda(1/2). \label{e2.4}
\end{gather}
Multiply \eqref{e1.1} by $u_\lambda'(t)$ to obtain
\[
\{u_\lambda''(t) + \lambda u_\lambda(t) - u_\lambda^2(t)\}u_\lambda'(t) = 0.
\]
This implies that
\[
\frac{d}{dt}\Big(
\frac12u_\lambda'(t)^2 + \frac12\lambda u_\lambda(t)^2
- \frac13u_\lambda(t)^3\Big) = 0.
\]
Namely, for $t \in \bar{I}$,
\begin{equation} \label{e2.5}
\frac12u_\lambda'(t)^2 + \frac12\lambda u_\lambda(t)^2
- \frac13u_\lambda(t)^3 \equiv \text{constant}
= \frac12\lambda\| u_\lambda\|_\infty^2
- \frac13\| u_\lambda\|_\infty^3.
\end{equation}
By this and \eqref{e2.3}, for $0 \le t \le 1/2$,
\begin{equation} \label{e2.6}
u_\lambda'(t) = \sqrt{S_\lambda(u_\lambda(t))},
\end{equation}
where
\begin{equation} \label{e2.7}
S_\lambda(w) = \lambda(\| u_\lambda\|_\infty^2 - w^2)
- \frac23(\| u_\lambda\|_\infty^3 - w^3).
\end{equation}
By this, we obtain
\begin{equation} \label{e2.8}
\frac12 = \int_0^{1/2} dt = \int_0^{1/2} \frac{u_\lambda'(t)}
{\sqrt{S_\lambda(u_\lambda(t))}}dt
= \frac{1}{\sqrt{\lambda}}
\int_0^1 \frac{1}{\sqrt{R_\lambda(s)}}ds.
\end{equation}
Here
\begin{gather}
R_\lambda(s) = 1 - s^2 - \eta_\lambda(1-s^3), \label{e2.9}\\
\eta = \eta_\lambda = \frac{2\| u_\lambda\|_\infty}{3\lambda}
= \frac{2}{3}(1-\xi). \label{e2.10}
\end{gather}
Furthermore, put $\theta = 1- s$. Then we obtain
\begin{equation} \label{e2.11}
R_\lambda(s) = U_\lambda(\theta) :=
\theta\eta(A_\lambda + B_\lambda\theta - \theta^2)
= \theta\eta(a_\lambda - \theta)(\theta - c_\lambda),
\end{equation}
where
\begin{gather}
A = A_\lambda = \frac{2-3\eta}{\eta} = \frac{3\xi}{1-\xi}
  = 3\xi(1 + \xi + O(\xi^2)), \label{e2.12}\\
B = B_\lambda = \frac{3\eta-1}{\eta} = \frac{3(1-2\xi)}{2(1-\xi)}
  = \frac{3}{2}(1 - \xi - \xi^2 + O(\xi^3)), \label{e2.13}\\
a = a_\lambda = \frac{B + \sqrt{B^2 + 4A}}{2}
  = \frac32 + \frac{1}{2}\xi - \frac{1}{6}\xi^2
  + O(\xi^3), \label{e2.14}\\
c = c_\lambda = \frac{B - \sqrt{B^2 + 4A}}{2}
= -2\xi - \frac{4}{3}\xi^2 + O(\xi^3). \label{e2.15}
\end{gather}
We obtain \eqref{e2.12}--\eqref{e2.15} by Taylor expansion.
By \eqref{e2.8} and \eqref{e2.11}, we obtain
\begin{equation} \label{e2.16}
\frac{\sqrt{\lambda}}{2} = \frac{1}{\sqrt{\eta}}
\int_0^1 \frac{1}{\sqrt{\theta(a-\theta)(\theta-c)}}d\theta.
\end{equation}
We set
\begin{equation} \label{e2.17}
\phi = \sin^{-1}\sqrt{\frac{a-c}{a(1-c)}}, \quad k = \sqrt{\frac{a}{a-c}}.
\end{equation}
By \cite[pp. 266]{g1}, we know
\begin{equation} \label{e2.18}
\int_0^1 \frac{1}{\sqrt{\theta(a-\theta)(\theta-c)}}d\theta
=
\frac{2}{\sqrt{a-c}}F(\phi,k)
= \frac{2}{\sqrt{a-c}}\int_0^\phi \frac{1}{\sqrt{1-k^2\sin^2 t}}dt.
\end{equation}
By this, \eqref{e2.10} and \eqref{e2.16}, we obtain
\begin{equation} \label{e2.19}
\frac{\sqrt{\lambda}}{2} = \frac{1}{\sqrt{\eta}}
\frac{2}{\sqrt{a-c}}F(\phi,k) =
\sqrt{\frac{6}{(1-\xi)}}\sqrt{\frac{1}{a-c}}F(\phi,k).
\end{equation}
Therefore, to prove Theorem \ref{thm1.2}, we calculate $F(\phi,k)$ precisely.
By \cite[Eq. (1.16)]{c1}, we know that as $k \to 1-0$ and
$\phi \to \pi/2 - 0$,
\begin{equation} \label{e2.20}
F(\phi, k) = \frac{\sin \phi}{1-\beta}
\big[\big(1 + \frac{\zeta^2 + \cos^2\phi}{4}\big)
\log\frac{4}{\zeta + \cos\phi} - \frac{\zeta^2 + \cos^2\phi - \zeta\cos\phi}
{4}\big],
\end{equation}
where
\begin{equation} \label{e2.21}
\zeta = (1 - k^2\sin^2\phi)^{1/2}, \quad 0 < \beta < \frac38\zeta^4.
\end{equation}

\begin{lemma} \label{lem2.1}
For $\lambda \gg 1$,
\begin{eqnarray} \label{e2.22}
\log(\zeta + \cos\phi)
&=& \frac12\log\xi + \frac12\log 2 - \frac12\log 3 + \log(\sqrt{3}+1)
 \\
 &&\mbox{}- \frac{\sqrt{3}}{1 + \sqrt{3}}
\big(\frac23 + \frac{\sqrt{3}}{9}\big)
\xi + O(\xi^2).
\nonumber
\end{eqnarray}
\end{lemma}

\begin{proof}
By \eqref{e2.15}, \eqref{e2.17} and Taylor expansion,
\begin{equation} \label{e2.23}
k^2\sin^2 \phi = \frac{1}{1-c}
= 1 - 2\xi + \frac{8}{3}\xi^2 + O(\xi^3).
\end{equation}
By this, \eqref{e2.21} and Taylor expansion,
\begin{equation} \label{e2.24}
\zeta = (1 - k^2\sin^2\phi)^{1/2}
= \sqrt{2\xi}\big(1 - \frac{2}{3}\xi + O(\xi^2)\big).
\end{equation}
By \eqref{e2.14}, \eqref{e2.15}, \eqref{e2.17} and \eqref{e2.23},
\begin{equation} \label{e2.25}
\sin^2\phi = \frac{a-c}{a}\frac{1}{1-c}
= 1 - \frac{2}{3}\xi + \frac{4}{9}\xi^2 + O(\xi^3).
\end{equation}
By this and Taylor expansion,
\begin{equation} \label{e2.26}
\cos\phi = \sqrt{1 - \sin^2\phi}
= \sqrt{2\xi}\big(\frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}}\xi + O(\xi^2)
\big).
\end{equation}
Then by \eqref{e2.24} and \eqref{e2.26},
\begin{eqnarray} \label{e2.27}
\log(\zeta + \cos\phi) &=&
\frac12\log\xi + \frac12\log 2 - \frac12\log 3 + \log(\sqrt{3}+1)
\\
&&\mbox{} - \frac{\sqrt{3}}{1 + \sqrt{3}}
\big(\frac{2}{3} + \frac{\sqrt{3}}{9}\big)\xi
+ O(\xi^2).
\nonumber
\end{eqnarray}
Thus the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
 By \eqref{e2.24} and \eqref{e2.26}, we have
\begin{equation} \label{e2.28}
\frac{\zeta^2 + \cos^2\phi}{4} = \frac23\xi + O(\xi^2).
\end{equation}
Further, by \eqref{e2.24}, \eqref{e2.25} and \eqref{e2.26},
\begin{gather}
\sin\phi = 1 - \frac13\xi + O(\xi^2), \label{e2.29}\\
\zeta\cos\phi = \frac{2}{\sqrt{3}}\xi + O(\xi^2),
\label{e2.30}\\
\zeta^4 = O(\xi^2). \label{e2.31}
\end{gather}
By \eqref{e2.14} and \eqref{e2.15},
\begin{equation} \label{e2.32}
\frac{1}{\sqrt{a-c}}
= \sqrt{\frac{2}{3}}\Big(1 - \frac56\xi + O(\xi^2)\Big).
\end{equation}
Then by \eqref{e2.20}, \eqref{e2.22} and  \eqref{e2.28}--\eqref{e2.32},
\begin{equation} \label{e2.33}
\begin{aligned}
F(\phi,k) &= (1 + O(\xi^2))(1 - \frac13\xi + O(\xi^2))\\
&\quad \times\big\{(1 + \frac23\xi + O(\xi^2))
(\log 4 - \log(\zeta + \cos\phi)) - \frac{4-\sqrt{3}}{6}\xi + O(\xi^2))\big\}
\\
&= -\frac12\log\xi + D - \frac16\xi\log\xi
+ \big(\frac16+\frac13 D\big)\xi + O(\xi^2\log\xi),
\end{aligned}
\end{equation}
where
\begin{equation} \label{e2.34}
D = \frac32\log 2 + \frac12\log 3 - \log(\sqrt{3}+1)
= \frac12\log 6(\sqrt{3}-1)^2.
\end{equation}
By this, \eqref{e2.19}, \eqref{e2.32} and Lemma \ref{lem2.1},
\begin{equation} \label{e2.35}
\begin{aligned}
\frac{\sqrt{\lambda}}{2}
&= \sqrt{6}(1-\xi)^{-1/2}\sqrt{\frac23}
\Big(1 - \frac56\xi + O(\xi^2)\Big)F(\phi,k) \\
&= 2(1 - \frac13\xi + O(\xi^2))F(\phi,k) \\
&= -\log\xi + \log 6(\sqrt{3}-1)^2 + \frac13\xi + O(\xi^2\log\xi) \\
&= -\log\xi + \log 6(\sqrt{3}-1)^2 + \frac13\xi
   + O(\sqrt{\lambda}e^{-\sqrt{\lambda}}).
\end{aligned}
\end{equation}
By this and Taylor expansion,
\begin{equation} \label{e2.36}
\begin{aligned}
\xi &= e^{-\sqrt{\lambda}/2}
\cdot e^{\log 6(\sqrt{3}-1)^2}
\cdot e^{\xi/3}\cdot e^{O(\sqrt{\lambda}e^{-\sqrt{\lambda}})}
\\
&= 6(\sqrt{3}-1)^2e^{-\sqrt{\lambda}/2}
\Big(1 + \frac13\xi + O(\xi^2)\Big)
\left(1 + O(\sqrt{\lambda}e^{-\sqrt{\lambda}})\right)
\\
&= 6(\sqrt{3}-1)^2e^{-\sqrt{\lambda}/2}
(1 + O(\sqrt{\lambda}e^{-\sqrt{\lambda}}))
+ 2(\sqrt{3}-1)^2\xi e^{-\sqrt{\lambda}/2}.
\end{aligned}
\end{equation}
By this, we obtain
\begin{equation} \label{e2.37}
\xi\left(1 - 2(\sqrt{3}-1)^2e^{-\sqrt{\lambda}/2}\right)
= 6(\sqrt{3}-1)^2e^{-\sqrt{\lambda}/2}
(1 + O(\sqrt{\lambda}e^{-\sqrt{\lambda}})).
\end{equation}
This implies
\begin{align*}
\xi &= 6(\sqrt{3}-1)^2
e^{-\sqrt{\lambda}/2}\left(1 + O(\sqrt{\lambda}e^{-\sqrt{\lambda}})\right)
\left(1 + 2(\sqrt{3}-1)^2e^{-\sqrt{\lambda}/2} + O(e^{-\sqrt{\lambda}})\right)
\\
&= 6(\sqrt{3}-1)^2 e^{-\sqrt{\lambda}/2} + 12(\sqrt{3}-1)^4
e^{-\sqrt{\lambda}} + O(\sqrt{\lambda}e^{-3\sqrt{\lambda}/2}).
\end{align*}
Thus, the proof is complete.
\end{proof}

\section{Proof of Theorem \ref{thm1.3}}

In this section, let $p = 5$. We put
\begin{equation} \label{e3.1}
r = r_\lambda = 1 - \frac{\| u_\lambda\|_\infty^4}{\lambda}.
\end{equation}
To prove Theorem \ref{thm1.3}, we calculate $r$ precisely.
Let
\begin{equation} \label{e3.2}
\sigma = \sigma_\lambda = \frac{1}{3}
\frac{\| u_\lambda\|_\infty^4}{\lambda}
= \frac{1-r}{3}
\end{equation}
and
\begin{equation} \label{e3.3}
Y_\lambda(s) := 1 - s^2 - \sigma(1 - s^6)
= (1 - s^2)\{1 - \sigma(1 + s^2 + s^4)\}.
\end{equation}
Then by the same calculation as that to obtain \eqref{e2.8}, we have
\begin{equation} \label{e3.4}
\frac12 = \frac{1}{\sqrt{\lambda}}\int_0^1 \frac{1}{\sqrt{Y_\lambda(s)}}ds.
\end{equation}
This along with \eqref{e3.3} implies
\begin{equation} \label{e3.5}
\begin{aligned}
\frac{\sqrt{\lambda}}{2}
&=\int_0^1 \frac{1}{\sqrt{(1-s^2)\{1-\sigma(1+s^2+s^4)\}}}ds \\
&= \frac12 \int_0^1 \frac{2sds}{\sqrt{s^2(1-s^2)\{1-\sigma(1+s^2+s^4)\}}}ds
\\
&=\frac12\int_0^1 \frac{dt}{\sqrt{t(1-t)\{1-\sigma(1+t+t^2)\}}}ds
\\
&=\frac{1}{2\sqrt{\sigma}}\int_0^1\frac{1}{\sqrt{t(1-t)(t-d)(a-t)}}dt,
\end{aligned}
\end{equation}
where
\begin{gather}
a = \frac{-1 + \sqrt{1-4\delta}}{2} = 1 + r + \frac23r^2 + O(r^3),
 \label{e3.6} \\
d = \frac{-1 - \sqrt{1-4\delta}}{2} = -2 -r - \frac23r^2 + O(r^3),
 \label{e3.7}\\
\delta = 1 - \frac{1}{\sigma} = -2 - 3r - 3r^2 + O(r^3). \label{e3.8}
\end{gather}
By \eqref{e3.5} and \cite[p. 290]{g1}, we know
\begin{equation} \label{e3.9}
\sqrt{\lambda} = \frac{1}{\sqrt{\sigma}}\frac{2}{\sqrt{a(1-d)}}
F\big(\frac{\pi}{2}, k\big),
\end{equation}
where
\begin{equation} \label{e3.10}
k = \sqrt{\frac{a-d}{a(1-d)}}.
\end{equation}
By \eqref{e2.20}, as $k \to 1-0$,
\begin{equation} \label{e3.11}
F\left(\frac{\pi}{2},k\right) = \frac{1}{1-\beta}
\big[
\big(1 + \frac{\zeta^2}{4}\big)
\log\frac{4}{\zeta}- \frac{\zeta^2}{4}
\big],
\end{equation}
where (by \eqref{e3.6}, \eqref{e3.7} and \eqref{e3.10},
\begin{gather}
\zeta^2 = 1 - k^2 = \frac{d(1-a)}{a(1-d)}
= \frac{2}{3}r\big(1 - \frac16r + O(r^2)\big), \label{e3.12}
\\
0 < \beta < \frac{3}{8}\zeta^4 = O(r^2). \label{e3.13}
\end{gather}
By this, \eqref{e3.11}--\eqref{e3.13} and Taylor expansion,
\begin{equation} \label{e3.14}
\begin{aligned}
F\left(\frac{\pi}{2},k\right)
&= (1 + O(r^2))\big[(1 + \frac{1}{6}r + O(r^2))(\log 4
- \log\zeta) - \frac16r + O(r^2)\big] \\
&= -\frac12\log r + \frac12\log 24 - \frac{1}{12}r\log r
+ \big(\frac{1}{12}\log 24 - \frac{1}{12}\big)r + O(r^2).
\end{aligned}
\end{equation}
Further, by \eqref{e3.2}, \eqref{e3.6}, \eqref{e3.7} and Taylor expansion,
\begin{gather}
\frac{1}{\sqrt{\sigma}} = \sqrt{3}\big(1 + \frac12r + \frac38r^2
+ O(r^3)\big), \label{e3.15}
\\
\frac{1}{\sqrt{a(1-d)}} = \frac{1}{\sqrt{3}}
\big(1 - \frac23 r + O(r^2)\big). \label{e3.16}
\end{gather}
By \eqref{e3.9} and \eqref{e3.14}--\eqref{e3.16}, we obtain
\begin{equation} \label{e3.17}
\begin{aligned}
\sqrt{\lambda}
&= \big(1 - \frac16 r + O(r^2)\big)
 (-\log r + \log 24 - \frac{1}{6}r\log r
 + \frac{1}{6}(\log 24 - 1)r + O(r^2)) \\
&= -\log r + \log 24 - \frac{1}{6}r + O(r^2\log r).
\end{aligned}
\end{equation}
By this, we obtain
\begin{equation} \label{e3.18}
re^{r/6} = 24e^{-\sqrt{\lambda}}e^{O(r^2\log r)}.
\end{equation}
By this and Taylor expansion, we obtain
\begin{equation} \label{e3.19}
\begin{aligned}
r &= 24\Big(1 - \frac{1}{6}r + O(r^2)\Big)(1 + O(r^2\log r))
e^{-\sqrt{\lambda}} \\
&= 24e^{-\sqrt{\lambda}} - 96e^{-2\sqrt{\lambda}} + o(e^{-2\sqrt{\lambda}}).
\end{aligned}
\end{equation}
By substituting this into \eqref{e3.17} again, we obtain
\[
r = 24e^{-\sqrt{\lambda}} - 96e^{-2\sqrt{\lambda}}
+ O(\sqrt{\lambda} e^{-3\sqrt{\lambda}}).
\]
By this and \eqref{e3.1}, we obtain
\begin{align*}
\| u_\lambda\|_\infty
&= \lambda^{1/4} (1 - 24e^{-\sqrt{\lambda}} + 96e^{-2\sqrt{\lambda}}
+ O(\sqrt{\lambda} e^{-3\sqrt{\lambda}}))^{1/4} \\
&= \lambda^{1/4}(1 - 6e^{-\sqrt{\lambda}} -30e^{-2\sqrt{\lambda}}
+ O(\sqrt{\lambda} e^{-3\sqrt{\lambda}})).
\end{align*}
Thus, the proof is complete.

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