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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 37, pp. 1--16.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/37\hfil Asymptotic shape of solutions]
{Asymptotic shape of solutions to nonlinear eigenvalue problems}
\author[T. Shibata\hfil EJDE-2005/37\hfilneg]
{Tetsutaro Shibata}

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

\date{}
\thanks{Submitted January 11, 2005. Published March 29, 2005.}
\subjclass[2000]{34B15}
\keywords{Asymptotic formula; $L^1$-norm; simple pendulum; logistic equation}

\begin{abstract}
 We consider the nonlinear eigenvalue problem
 $$
 -u''(t) =  f(\lambda, u(t)), \quad u > 0, \quad u(0) = u(1) = 0,
 $$
 where $\lambda > 0$ is a parameter. It is known that under some
 conditions on $f(\lambda, u)$, the shape of the solutions associated with
 $\lambda$ is almost `box' when $\lambda \gg 1$.
 The purpose of this paper is to study precisely the asymptotic shape of
 the solutions as $\lambda \to \infty$ from a standpoint of $L^1$-framework.
 To do this, we establish the asymptotic formulas for
 $L^1$-norm of the solutions as $\lambda \to \infty$.
\end{abstract}

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

\section{Introduction}

We consider the nonlinear eigenvalue problem
\begin{gather}
-u''(t) = f(\lambda, u(t)), \quad t \in I:= (0, 1), \label{e1.1}\\
 u(t) > 0, \quad t \in I, \\
u(0) = u(1) = 0, \label{e1.3}
\end{gather}
where $\lambda > 0$ is a parameter.
The nonlinearities considered here are as follows:
\begin{gather}
f(\lambda, u) = \lambda\sin u, \label{e1.4}\\
f(\lambda, u) = \lambda\sin u - g(u), \label{e1.5}\\
f(\lambda, u) = \lambda(u - u^3). \label{e1.6}
\end{gather}
Equation \eqref{e1.1}--\eqref{e1.3} with the
nonlinearities \eqref{e1.4} and \eqref{e1.5} are called the simple
pendulum type equations (SPE), and that with
\eqref{e1.6} is derived from the logistic equation of
population dynamics (LEPD). Throughout this paper, in \eqref{e1.5},
we assume that $g(u)$ satisfies the following conditions:
\begin{itemize}
\item[(A1)] $g \in C^1(\mbox{\bf R})$ and $g(u) > 0$ for $u > 0$.

\item[(A2)] $g(0) = g'(0) = 0$.

\item[(A3)] $g(u)/u$ is strictly increasing for $0 \le u \le \pi$.
\end{itemize}
Nonlinear eigenvalue problems and singularly perturbed problems
are intensively investigated by many authors. We refer to \cite{o2,s4}
and the references therein. One of the most interesting problems
to study in these fields is to clarify the asymptotic shapes of
the solutions.
We know (cf. \cite{c,f}) that for a given $\lambda > \pi^2$,
there exists a unique solution $u \in C^2(I)$ to \eqref{e1.1}--\eqref{e1.3}.
We denote by $u_{0,\lambda}, u_\lambda$ and $v_\lambda$ the solutions $u$
of \eqref{e1.1}--\eqref{e1.3} with \eqref{e1.4}, \eqref{e1.5} and \eqref{e1.6},
respectively.
Let
$m_\lambda := \min\{ m > 0: f(\lambda, m) = 0\}$. Then it is well
known (cf. \cite{c,f} and Appendix)
that $\Vert u_{0,\lambda}\Vert_\infty < m_\lambda$ (resp.
$\Vert u_\lambda\Vert_\infty < m_\lambda$,
$\Vert v_\lambda\Vert_\infty < m_\lambda$).
Clearly, $m_\lambda = \pi$ and $m_\lambda = 1$ for \eqref{e1.4} and \eqref{e1.6},
respectively.
By (A1), we see that $0 < m_\lambda < \pi$ for \eqref{e1.5}.
Furthermore, it is known (cf. \cite{c,s1}) that
\begin{equation} \label{e1.7}
u_{0,\lambda} \to \pi, \quad u_\lambda \to \pi, \quad v_\lambda \to 1
\end{equation}
uniformly on any compact interval in $I$ as $\lambda \to \infty$.
In other words, the asymptotic shape of these solutions are
{\it almost boxes}. Therefore, a natural question we have to ask
here is ``{\it how close to the boxes are the shape of the solutions
$u_{0,\lambda}, u_\lambda$ and $v_\lambda$ asymptotically?}

The purpose of this paper is to answer this question from a viewpoint
of $L^1$-framework. More precisely, we restrict our attention to
the typical nonlinearities \eqref{e1.4}--\eqref{e1.6}, and establish
the precise asymptotic formulas for $L^1$-norm of
$\Vert u_{0,\lambda}\Vert_1, \Vert u_{\lambda}\Vert_1$ and
$\Vert v_{\lambda}\Vert_1$
as $\lambda \to \infty$. By these formulas, we understand well
\begin{itemize}
\item[(i)] The difference of the shape between $u_{0,\lambda}$
and $u_\lambda$ from a {\it non-local} point of view, and

\item[(ii)] The difference between $\Vert v_\lambda\Vert_1$ and
$\Vert v_\lambda\Vert_2$ when $\lambda \gg 1$.
\end{itemize}
The first approach to the study of the asymptotic shape of the solutions
of (SPE) is to investigate the asymptotic behavior of
the $L^\infty$-norm of the solutions
as $\lambda \to \infty$ and the following results have been obtained in
\cite{s1, s3}:
\begin{gather}
\Vert u_{0,\lambda}\Vert_\infty = \pi - 8
e^{-\sqrt{\lambda}/2} + o(e^{-\sqrt{\lambda}/2}), \label{e1.8}
\\
\Vert u_\lambda \Vert_\infty = \pi - \frac{g(\pi)}{\lambda}
+ \frac{g(\pi)g'(\pi)}{\lambda^2} + o\big(\frac{1}{\lambda^2}\big).
\label{e1.9}
\end{gather}
By \eqref{e1.8} and \eqref{e1.9}, we understand the difference between
the {\it pointwise (local)}
behavior of $u_{0,\lambda}$ and $u_\lambda$.
However, we do not know the difference between the
{\it total mass} of
$u_{0,\lambda}$ and $u_\lambda$, which gives us the important
information about the
{\it non-local} property of $u_{0,\lambda}$ and $u_\lambda$.
Therefore, it seems meaningful for us to establish
the asymptotic formulas for $\Vert u_{0,\lambda}\Vert_1$
and $\Vert u_\lambda\Vert_1$, which give us the better understanding of the
difference between the original (SPE) and the perturbed (SPE).

We now state the results for (SPE).


\begin{theorem} \label{thm1.1}
As $\lambda \to \infty$
\begin{gather}
\Vert u_{0,\lambda}\Vert_1 = \pi - C_1\frac{1}{\sqrt{\lambda}}
+ \frac{8}{\sqrt{\lambda}}e^{-\sqrt{\lambda}/2}
+ o\big(\frac{1}{\sqrt{\lambda}}e^{-\sqrt{\lambda}/2}\big), \label{e1.10}
\\
\Vert u_\lambda\Vert_1 = \pi - C_1\frac{1}{\sqrt{\lambda}} -
\frac{g(\pi)}{\lambda} +
O\big(\frac{\log\lambda}{\lambda\sqrt{\lambda}}\big), \label{e1.11}
\end{gather}
where
\[
C_1 = 8\int_0^{\pi/4} \log(\cot\theta)d\theta.
\]
\end{theorem}

Roughly speaking, the second terms in \eqref{e1.10} and \eqref{e1.11} are
derived from the width of the boundary layers of $u_{0,\lambda}$ and $u_\lambda$,
while the third terms come directly from the second terms in \eqref{e1.8}
and \eqref{e1.9}.

We next consider (LEPD). The motivation to consider the
asymptotic behavior of
$\Vert v_\lambda\Vert_1$ as $\lambda \to \infty$ is as follows.
Recently, from a viewpoint of $L^2$-bifurcation theory and nonlinear
eigenvalue problems, the following
formula for $\Vert v_\lambda\Vert_2$ as $\lambda \to \infty$
has been obtained.
\begin{equation} \label{e1.12}
\Vert v_\lambda\Vert_2 = 1 - \sqrt{\frac{2}{\lambda}}
- \frac{1}{\lambda} + o\big(\frac{1}{\lambda}\big).
\end{equation}
This has been obtained by using \cite[Theorem 1]{s2}.
On the other hand, since (LEPD) is a
model equation of population density for some species,
$\Vert v_\lambda\Vert_1$
stands for the total population of the species
with $\lambda$, which describes
the reciprocal number of its diffusion rate.
Motivated by this biological background,
it is also important to investigate the asymptotic behavior of
$\Vert v_\lambda\Vert_1$ as $\lambda \to \infty$. Furthermore, since
\eqref{e1.6} is a special nonlinearlity, we also obtain the asymptotic
formula for $\Vert v_\lambda\Vert_\infty$
better than \eqref{e1.8} as $\lambda \to \infty$.

Now we state our second results.

\begin{theorem} \label{thm1.2}
As $\lambda \to \infty$
\begin{gather}
\Vert v_\lambda\Vert_1 = 1 - \frac{2\sqrt{2}\log2}{\sqrt{\lambda}}
- 12e^{-\sqrt{2\lambda}} + o(e^{-\sqrt{2\lambda}}), \label{e1.13}\\
\Vert v_\lambda\Vert_\infty = 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}}). \label{e1.14}
\end{gather}
\end{theorem}

We see from \eqref{e1.12} and \eqref{e1.13} that the third term of
$\Vert v_\lambda\Vert_1$ and $\Vert v_\lambda\Vert_2$
are totally different each other.
The further direction of this study will be to treat more general
nonlinear term $f(\lambda, u)$ and extend our results to PDE cases.

The remainder of this paper is organized as follows.
In Sections 2 and 3, we prove \eqref{e1.11} and \eqref{e1.10}
in Theorem \ref{thm1.1}, respectively. By using the properties of
complete elliptic integral, we prove Theorem \ref{thm1.2} in Section 4.

\section{Proof of \eqref{e1.11} in Theorem \ref{thm1.1}}

In this section, we consider \eqref{e1.1}--\eqref{e1.3} with \eqref{e1.5} and
prove the formula \eqref{e1.11}. In what follows, the character $C$ denotes
various positive constants independent of $\lambda \gg 1$.
We know that
\begin{equation} \label{e2.1}
\begin{gathered}
u_\lambda(t) = u_\lambda(1-t) \quad \mbox{for } t \in \bar{I},\\
u_\lambda'(t) > 0 \quad \mbox{for }  0 \le t < 1/2, \\
\Vert u_\lambda\Vert_\infty = u_\lambda(1/2).
\end{gathered}
\end{equation}
We begin with the fundamental equality. Multiply \eqref{e1.1} by $u_\lambda'$.
Then
$$
\{u_\lambda''(t) + \lambda\sin u_\lambda(t) - g(u_\lambda(t))\}
u_\lambda'(t) = 0.
$$
This implies that for $t \in \bar{I}$
\[
\frac{d}{dt}\big\{\frac12u_\lambda'(t)^2 - \lambda\cos u_\lambda(t)
- G(u_\lambda(t))\big\} \equiv 0,
\]
where $G(u) = \int_0^u g(s)ds$.
By this and \eqref{e2.1}, for $0 \le t \le 1$,
\begin{equation} \label{e2.2}
\frac12u_\lambda'(t)^2 - \lambda\cos u_\lambda(t)
- G(u_\lambda(t)) \equiv \mbox{constant}
= -\lambda\cos\Vert u_\lambda\Vert_\infty - G(\Vert u_\lambda\Vert_\infty).
\end{equation}
By this and \eqref{e2.1}, for $0 \le t \le 1/2$,
\begin{equation} \label{e2.3}
u_\lambda'(t) = \sqrt{2\lambda
(\cos u_\lambda(t) - \cos\Vert u_\lambda\Vert_\infty)
+ 2(G(u_\lambda(t)) - G(\Vert u_\lambda\Vert_\infty))}
\end{equation}
We know from \eqref{e1.7} that as $\lambda \to \infty$
\begin{equation}\label{e2.4}
r_1(\lambda) := \Vert u_\lambda\Vert_\infty - \Vert u_\lambda\Vert_1 \to 0.
\end{equation}
Since we have \eqref{e1.9}, we establish an asymptotic formula for $r_1(\lambda)$
as $\lambda \to \infty$ to obtain \eqref{e1.11}.
By \eqref{e2.1} and \eqref{e2.3}, for $\lambda \gg 1$
\begin{equation} \label{e2.5}
\begin{aligned}
r_1(\lambda) &= 2\int_0^{1/2} (\Vert u_\lambda\Vert_\infty - u_\lambda(t))dt
\\
&= 2\int_0^{1/2}
\frac{(\Vert u_\lambda\Vert_\infty - u_\lambda(t))u_\lambda'(t)dt}
{\sqrt{2\lambda
(\cos u_\lambda(t) - \cos\Vert u_\lambda\Vert_\infty)
+ 2(G(u_\lambda(t)) - G(\Vert u_\lambda\Vert_\infty))}}\\
&= 2\int_0 ^{\Vert u_\lambda\Vert_\infty}
\frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta}
{\sqrt{2\lambda
(\cos \theta - \cos\Vert u_\lambda\Vert_\infty)
+ 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))}}\\
&= K_1(\lambda) + K_2(\lambda),
\end{aligned}
\end{equation}
where
\begin{gather*}
K_1(\lambda)
:=2\int_{\Vert u_\lambda\Vert_\infty - 1/\lambda}
^{\Vert u_\lambda\Vert_\infty}
\frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta}
{\sqrt{2\lambda
(\cos \theta - \cos\Vert u_\lambda\Vert_\infty)
+ 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))}
},\\
K_2(\lambda) := 2\int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta}
{\sqrt{2\lambda
(\cos \theta - \cos\Vert u_\lambda\Vert_\infty)
+ 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))}
}.
\end{gather*}

\begin{lemma} \label{lem2.1}
$K_1(\lambda) = O(\lambda^{-3/2})$ for $\lambda \gg 1$.
\end{lemma}

\begin{proof} For $j = 1, 2, \dots$, let
\[
I_j := \big[\Vert u_\lambda\Vert_\infty - \frac{1}{j\lambda},
\Vert u_\lambda\Vert_\infty - \frac{1}{(j+1)\lambda}\big].
\]
We put
\begin{equation} \label{e2.8}
J_j := 2\int_{I_j}
\frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta}
{\sqrt{2\lambda(\cos \theta - \cos\Vert u_\lambda\Vert_\infty)
+ 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))}}.
\end{equation}
We know by \cite{c} that
\begin{equation} \label{e2.9}
\lambda\sin\Vert u_\lambda\Vert_\infty > g(\Vert u_\lambda\Vert_\infty).
\end{equation}
Let an arbitrary $0 < \epsilon \ll 1$ be fixed. Let
$\eta_{\lambda,\epsilon}
:= \min_{\Vert u_\lambda\Vert_\infty-2\epsilon
\le u \le \Vert u_\lambda\Vert_\infty} g'(u)$.
Then by (A3), we see that $\eta_{\lambda,\epsilon} > 0$.
Then for
$\theta
\in [\Vert u_\lambda\Vert_\infty - 2\epsilon, \Vert u_\lambda\Vert_\infty]$,
by \eqref{e1.7}, \eqref{e2.9} and Taylor expansion, we have
\begin{equation} \label{e2.10}
\begin{aligned}
&2\lambda
(\cos \theta - \cos\Vert u_\lambda\Vert_\infty)
+ 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))
\\
&\ge 2\left(\lambda\sin\Vert u_\lambda\Vert_\infty
- g(\Vert u_\lambda\Vert_\infty)\right)(\Vert u_\lambda\Vert_\infty - \theta)
\\
&\quad + 2\left(-\frac{\lambda}{2}\cos
(\Vert u_\lambda\Vert_\infty - 2\epsilon)
+ \frac12\eta_{\lambda,\epsilon}\right)
(\Vert u_\lambda\Vert_\infty - \theta)^2
\\
&\ge C\lambda(\Vert u_\lambda\Vert_\infty - \theta)^2.
\end{aligned}
\end{equation}
By this and \eqref{e2.8}, for $\lambda > 1/\epsilon$,
%\begin{equation} \label{e2.11}
\begin{align*}
J_j &\le \int_{I_j} \frac{\Vert u_\lambda\Vert_\infty - \theta}
{\sqrt{C\lambda(\Vert u_\lambda\Vert_\infty - \theta)^2}}d\theta\\
&= \frac{1}{\sqrt{C\lambda}}\frac{1}{\lambda}
\big(\frac{1}{j} - \frac{1}{j+1}\big)
\le \frac{C}{\lambda^{3/2}}\frac{1}{j(j+1)}.
\end{align*}
By this,
\[
K_1(\lambda) = \sum_{j=1}^\infty J_j
\le \sum_{j=1}^\infty\frac{C}{\lambda^{3/2}}
\frac{1}{j(j+1)} \le \frac{C}{\lambda^{3/2}}.
\]
Thus the proof is complete.
 \end{proof}

The formula \eqref{e1.11} follows from \eqref{e1.9},
Lemma \ref{lem2.1} and the following Proposition.

\begin{proposition} \label{prop2.2}
For $\lambda \gg 1$
\begin{equation}
K_2(\lambda) = \frac{C_1}{\sqrt{\lambda}}
+ O\left(\frac{\log \lambda}{\lambda^{3/2}}\right).
\end{equation}
\end{proposition}

We prove this proposition using Lemmas \ref{lem2.3}--\ref{lem2.7} below.
 We put
\[ % \label{e2.12}
K_2(\lambda) := K_{2,1}(\lambda) + K_{2,2}(\lambda),
\]
where
\begin{gather*}
K_{2,1}(\lambda) := 2\int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta}
{\sqrt{2\lambda
(\cos \theta - \cos\Vert u_\lambda\Vert_\infty)}}, %\label{e2.14}
\\
K_{2,2}(\lambda) := K_2(\lambda) - K_{2,1}(\lambda). %\label{e2.15}
\end{gather*}
Furthermore, we put
\[ % \label{e2.16}
K_{2,1}(\lambda) := L_1(\lambda) + L_{2}(\lambda),
\]
where
\[ % \label{e2.17}
L_{1}(\lambda) := 2\int_0^{\pi - 1/\lambda} \frac{\pi - \theta}
{\sqrt{2\lambda(\cos\theta + 1)}}d\theta.
\]

\begin{lemma} \label{lem2.3}
For  $\lambda \gg 1$
\[ % \label{e2.18}
L_{1}(\lambda) = \frac{C_1}{\sqrt{\lambda}}
+ o\big(\frac{\log \lambda}{\lambda^{3/2}}\big).
\]
\end{lemma}

\begin{proof} For $\lambda \gg 1$
\begin{align*} %2.19
L_{1}(\lambda) &=
\frac{1}{\sqrt{\lambda}}
\int_0^{\pi-1/\lambda} \frac{\pi - \theta}{\cos(\theta/2)}d\theta\\
&=\frac{1}{\sqrt{\lambda}}
\int_{1/\lambda}^\pi \frac{t}{\sin(t/2)}dt
=\frac{4}{\sqrt{\lambda}}
\int_{1/(2\lambda)}^{\pi/2}
\frac{t}{\sin t}dt \quad (\mbox{put }\theta = \tan(t/2)) \\
&=\frac{8}{\sqrt{\lambda}}
\int_{\tan(1/(4\lambda))}^1 \frac{\tan^{-1}\theta}{\theta}d\theta\\
&= \frac{8}{\sqrt{\lambda}}\big\{
[\log\theta\tan^{-1}\theta]
_{\tan(1/(4\lambda))}^1
- \int_{\tan(1/(4\lambda))}^1 \frac{\log\theta}{1 + \theta^2}d\theta
\big\} \quad (\mbox{put } \theta = \tan t) \\
&= \frac{2 + o(1)}{\lambda^{3/2}}\log\lambda
-\frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\tan t)dt
+\frac{8}{\sqrt{\lambda}}\int_0^{1/(4\lambda)}\log(\tan t)dt
\\
&=\frac{2 + o(1)}{\lambda^{3/2}}\log\lambda
+ \frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\cot t)dt
+ \frac{8}{\sqrt{\lambda}}
(1 + o(1))\int_0^{1/(4\lambda)} \log t dt \\
&=\frac{2 + o(1)}{\lambda^{3/2}}\log\lambda
+ \frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\cot t)dt
+ \frac{8}{\sqrt{\lambda}}
(1 + o(1))\big(\frac{1}{4\lambda}\log\frac{1}{4\lambda}
- \frac{1}{4\lambda}\big) \\
&= \frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\cot t)dt
+ o\big(\frac{\log \lambda}{\lambda^{3/2}}\big).
\end{align*}
Thus the proof is complete.
\end{proof}

Next, we calculate $L_{2}(\lambda)$. To do this, we put
\begin{equation}
L_{2}(\lambda) = K_{2,1}(\lambda) - L_1(\lambda)
:= D_1(\lambda) + D_2(\lambda) + D_3(\lambda),
\end{equation}
where
\begin{gather*} %2.21
D_1(\lambda) :=
2\int_0^
{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{\Vert u_\lambda\Vert_\infty - \pi}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}d\theta,
\\
D_2(\lambda) :=
2\int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\Big(
\frac{\pi - \theta}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}
- \frac{\pi - \theta}
{\sqrt{2\lambda(\cos\theta + 1)}}
\Big)
d\theta,
\\
D_3(\lambda) := -2\int_ {\Vert u_\lambda\Vert_\infty - 1/\lambda}^{\pi - 1/\lambda}
\frac{\pi - \theta}
{\sqrt{2\lambda(\cos\theta + 1)}}d\theta.
\end{gather*}

\begin{lemma} \label{lem2.4}
$D_1(\lambda) = O(\lambda^{-3/2})$ as $\lambda \to \infty$.
\end{lemma}

\begin{proof} Let an arbitrary $0 < \epsilon \ll 1$ be fixed.
Then for $\lambda \gg 1$
\begin{align*} %2.24
D_1(\lambda) &= D_{1,1}(\lambda) + D_{1,2}(\lambda) \\
&:= 2\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{\Vert u_\lambda\Vert_\infty - \pi}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}d\theta
\\
&\quad +
2\int_0^{\Vert u_\lambda\Vert_\infty - \epsilon}
\frac{\Vert u_\lambda\Vert_\infty - \pi}
{\sqrt{2\lambda
(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)
}}d\theta.
\end{align*}
For $0 \le \theta \le \Vert u_\lambda\Vert_\infty - \epsilon$,
there exists a constant $C_\epsilon > 0$ such that for $\lambda \gg 1$
\begin{equation} \label{e2.25}
C_\epsilon \le \cos\theta - \cos\Vert u_\lambda\Vert_\infty.
\end{equation}
By this and \eqref{e1.9}, for $\lambda \gg 1$,
\begin{equation} \label{e2.26}
\vert D_{1,2}(\lambda) \vert \le \frac{2g(\pi)}{\lambda}(1 + o(1))\frac{1}
{\sqrt{2C_\epsilon\lambda}}\pi \le C(\lambda^{-3/2}).
\end{equation}
We next estimate $D_{1,1}(\lambda)$. For a given $\lambda \gg 1$, there
exists $k_\lambda \in N$ satisfying
\begin{equation} \label{e2.27}
\Vert u_\lambda\Vert_\infty - 2\epsilon \le
\Vert u_\lambda\Vert_\infty -
\frac{k_\lambda + 1}{\lambda}
\le
\Vert u_\lambda\Vert_\infty - \epsilon \le
\Vert u_\lambda\Vert_\infty - \frac{k_\lambda}{\lambda}.
\end{equation}
For $j = 1, 2, \dots, k_\lambda$, we define an interval
\begin{equation} \label{e2.28}
M_j = \big[
\Vert u_\lambda\Vert_\infty -
\frac{j + 1}{\lambda},\Vert u_\lambda\Vert_\infty - \frac{j}{\lambda}
\big].
\end{equation}
By \eqref{e2.27}, we see that $k_\lambda \le \epsilon\lambda$.
By this, \eqref{e1.9} and \eqref{e2.10}, we obtain
\begin{align*} %2.29
\vert D_{1,1}(\lambda) \vert
&\le \frac{g(\pi)}{\lambda}(1 + o(1))
\sum_{j=1}^{k_\lambda}
\int_{M_j} \frac{1}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}
d\theta \\
&\le
\frac{g(\pi)}{\sqrt{2C}\lambda^{3/2}}(1 + o(1))
\sum_{j=1}^{k_\lambda}
\int_{M_j} \frac{1}{\Vert u_\lambda\Vert_\infty
- \theta}d\theta\\
&\le C\lambda^{-3/2}
\sum_{j=1}^{k_\lambda}(\log(j + 1) - \log j)\\
&= C\lambda^{-3/2}\log(k_\lambda + 1) \le C\lambda^{-3/2}\log\lambda.
\end{align*}
By this and \eqref{e2.26}, we obtain our conclusion.
Thus the proof is complete.
\end{proof}

\begin{lemma} \label{lem2.5}
 $D_2(\lambda) = O(\lambda^{-3/2})$ for $\lambda \gg 1$.
\end{lemma}

\begin{proof} We put
\begin{equation} \label{e2.30}
A_\lambda(\theta)
:= 2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty), \enskip
B_\lambda(\theta) := 2\lambda(\cos\theta + 1).
\end{equation}
By \eqref{e1.9} and Taylor expansion, for $\lambda \gg 1$, we have
\begin{equation} \label{e2.31}
1 + \cos \Vert u_\lambda \Vert_\infty = \frac{g(\pi)^2}{2\lambda^2}(1 + o(1)).
\end{equation}
Note that $A_\lambda(\theta) \le B_\lambda(\theta)$. By this, \eqref{e2.31}
and Taylor expansion, for a fixed $0 < \epsilon \ll 1$
\begin{equation} \label{e2.32}
\begin{aligned}
D_2(\lambda) &=
2\int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{2\lambda(\pi - \theta)(1 + \cos\Vert u_\lambda\Vert_\infty)}
{\sqrt{A_\lambda(\theta)}
\sqrt{B_\lambda(\theta)}
(\sqrt{A_\lambda(\theta)} + \sqrt{B_\lambda(\theta)})
}d\theta\\
&\le \frac{2g(\pi)^2}{\lambda}(1 + o(1))
\Big[\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{\pi - \theta}
{(2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty))^{3/2}
}d\theta \\
&\quad +
\int_0^{\Vert u_\lambda\Vert_\infty - \epsilon}
\frac{\pi - \theta}
{(2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty))^{3/2}
}d\theta \Big]\\
&:= D_{2,1}(\lambda) + D_{2,2}(\lambda).
\end{aligned}
\end{equation}
We know that $2\theta/\pi \le \sin\theta$ for $0 \le \theta \le \pi/2$.
By this, \eqref{e1.9} and mean value theorem, for $\theta \in M_j$ defined by
\eqref{e2.28}, we have
\begin{align*} %.2.33
\cos\theta - \cos\Vert u_\lambda\Vert_\infty
&\ge \sin\Big(\Vert u_\lambda \Vert_\infty - \frac{j}{\lambda}\Big)
(\Vert u_\lambda\Vert_\infty - \theta)\\
&=
\sin\Big(\pi - \frac{g(\pi)}{\lambda}(1 + o(1))
- \frac{j}{\lambda}\Big)(\Vert u_\lambda\Vert_\infty - \theta)\\
&=\sin\Big(\frac{g(\pi)}{\lambda}(1 + o(1))
+ \frac{j}{\lambda}\Big)
(\Vert u_\lambda\Vert_\infty - \theta)\\
&\ge \frac{2}{\pi}
\Big(\frac{g(\pi)(1 + o(1)) + j}{\lambda}\Big)
(\Vert u_\lambda\Vert_\infty - \theta).
\end{align*}
By this, \eqref{e1.9} and \eqref{e2.32},
\begin{equation} \label{e2.34}
\begin{aligned}
D_{2,1}(\lambda) &\le \frac{C}{\lambda}
\sum_{j=1}^{k_\lambda}
\int_{M_j}
\frac{(j + 1 + g(\pi) + o(1))/\lambda}
{\{\frac{4}{\pi}(j + g(\pi) + o(1))
(\Vert u_\lambda \Vert_\infty - \theta)\}^{3/2}}
d\theta\\
&\le \frac{C}{\lambda^2}
\sum_{j=1}^{k_\lambda}
(j + g(\pi) + 1 + o(1))(j + g(\pi) + o(1))^{-3/2}
\int_{M_j} \!(\Vert u_\lambda\Vert_\infty - \theta)^{-3/2}d\theta\\
&\le \frac{C}{\lambda^{3/2}}
\sum_{j=1}^{k_\lambda}j^{-1/2}\big(
\frac{1}{\sqrt{j}} - \frac{1}{\sqrt{j+1}}\big)\\
&= \frac{C}{\lambda^{3/2}}
\sum_{j=1}^{k_\lambda}
\frac{1}{\sqrt{j}\sqrt{j+1}(\sqrt{j} + \sqrt{j+1})}j^{-1/2}\\
&\le
\frac{C}{\lambda^{3/2}}
\sum_{j=1}^{k_\lambda}\frac{1}{j^2} \le
\frac{C}{\lambda^{3/2}}.
\end{aligned}
\end{equation}
By \eqref{e2.25}, we have
\[ %2.35
D_{2,2}(\lambda) \le \frac{2g(\pi)^2 + o(1)}{\lambda}
\frac{1}{(2C_\epsilon\lambda)^{3/2}}\pi^2
\le C\lambda^{-5/2}.
\]
This along with \eqref{e2.32} and \eqref{e2.34} implies our conclusion.
\end{proof}

\begin{lemma} \label{lem2.6}
$D_3(\lambda) = O(\lambda^{-3/2})$ for $\lambda \gg 1$.
\end{lemma}

\begin{proof} By \eqref{e1.9}, for $\lambda \gg 1$
\begin{align*} % 2.36
\vert D_3(\lambda)\vert
&= \int_{\Vert u_\lambda\Vert_\infty - 1/\lambda}
^{\pi - 1/\lambda}\frac{\pi - \theta}{\sqrt{\lambda}\cos(\theta/2)}d\theta\\
&= \frac{1}{\sqrt{\lambda}}
\int_{\pi-\Vert u_\lambda\Vert_\infty + 1/\lambda}
^{1/\lambda}\frac{-t}{\sin(t/2)}dt \\
&= \frac{4}{\sqrt{\lambda}}
\int_{1/(2\lambda)}^{(\pi-\Vert u_\lambda\Vert_\infty + 1/\lambda)/2}
\frac{\theta}{\sin\theta}d\theta \\
&= \frac{4}{\sqrt{\lambda}}(1 + o(1))\frac{\pi-\Vert u_\lambda\Vert_\infty}{2}
= \frac{2g(\pi)}{\lambda^{3/2}}(1 + o(1)).
\end{align*}
Thus the proof is complete.
\end{proof}

By Lemmas \ref{lem2.3}--\ref{lem2.6}, we see that
\begin{equation} \label{e2.37}
K_{2,1}(\lambda) = \frac{C_1}{\sqrt{\lambda}} + O\left(\frac{\log\lambda}
{\lambda^{3/2}}\right).
\end{equation}
Now we estimate $K_{2,2}(\lambda)$.

\begin{lemma} \label{lem2.7}
$K_{2,2}(\lambda) = O(\lambda^{-3/2}\log\lambda)$
for $\lambda \gg 1$.
\end{lemma}

\begin{proof}  We put
$E_\lambda(\theta) := 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))$.
We recall $A_\lambda(\theta)$ defined in \eqref{e2.30}.
Let an arbitrary $0 < \epsilon \ll 1$ be fixed. For $\lambda \gg 1$
\begin{align*} %2.38
&K_{2,2}(\lambda)\\
&= 2 \int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\Big(
\frac{\Vert u_\lambda\Vert_\infty - \theta}
{\sqrt{A_\lambda(\theta) + E_\lambda(\theta)}} -
\frac{\Vert u_\lambda\Vert_\infty - \theta}
{\sqrt{A_\lambda(\theta)}}
\Big)d\theta \\
&=2\int_0^{\Vert u_\lambda\Vert_\infty - \epsilon}
\frac{(\Vert u_\lambda\Vert_\infty - \theta)
(G(\Vert u_\lambda\Vert_\infty) - G(\theta))}{
\sqrt{A_\lambda(\theta) + E_\lambda(\theta)}
\sqrt{A_\lambda(\theta)}
(\sqrt{A_\lambda(\theta) + E_\lambda(\theta)}
+ \sqrt{A_\lambda(\theta)})}d\theta \\
&\quad  +
2\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{(\Vert u_\lambda\Vert_\infty - \theta)
(G(\Vert u_\lambda\Vert_\infty) - G(\theta))}{
\sqrt{A_\lambda(\theta) + E_\lambda(\theta)}
\sqrt{A_\lambda(\theta)}
(\sqrt{A_\lambda(\theta) + E_\lambda(\theta)}
+ \sqrt{A_\lambda(\theta)})}d\theta\\
&= H_1(\lambda) + H_2(\lambda).
\end{align*}
By \eqref{e2.25} we see that for $\lambda \gg 1$,
\[ %39
H_1(\lambda) \le 2g(\pi)\pi^3(2C_\epsilon\lambda)^{-3/2}.
\]
Note that $A_\lambda(\theta) + E_\lambda(\theta)
\le A_\lambda(\theta)$ for
$
\Vert u_\lambda\Vert_\infty - \epsilon
\le \theta \le \Vert u_\lambda\Vert_\infty - 1/\lambda$.
Then by \eqref{e2.10},
\begin{align*} %2.40
H_2(\lambda) &\le
2\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{g(\pi)(\Vert u_\lambda\Vert_\infty - \theta)^2}
{(A_\lambda + E_\lambda)^{3/2}}d\theta\\
&\le
\frac{C}{\lambda^{3/2}}
\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - 1/\lambda}
\frac{1}{\Vert u_\lambda\Vert_\infty - \theta}d\theta
\\
&=
\frac{C}{\lambda^{3/2}}(\log\epsilon - \log(1/\lambda))
\le \frac{C}{\lambda^{3/2}}\log\lambda.
\end{align*}
Thus the proof is complete.
\end{proof}

By \eqref{e2.37} and Lemma \ref{lem2.7}, we obtain Proposition \ref{prop2.2}.
Now \eqref{e1.11} follows from
\eqref{e1.9}, \eqref{e2.4}, \eqref{e2.5}, Lemma \ref{lem2.1} and
Proposition \ref{prop2.2}.
Thus the proof is complete.

\section{Proof of \eqref{e1.10} in Theorem \ref{thm1.1}}

To prove \eqref{e1.10}, we put
\begin{equation} \label{e3.1}
Q(\lambda) := \pi - \Vert u_{0,\lambda}\Vert_1.
\end{equation}
By the similar calculation to that in \eqref{e2.5}, we have
\begin{gather} \label{e3.2}
Q(\lambda) = 2\int_0^{\Vert u_{0,\lambda}\Vert_\infty}
\frac{\pi - \theta}{\sqrt{2\lambda(\cos\theta
- \cos\Vert u_{0,\lambda}\Vert_\infty)}} d\theta
= Q_{1}(\lambda) + Q_{2}(\lambda),
\\
Q_{2}(\lambda) := Q(\lambda) - Q_1(\lambda), \label{e3.3}
\end{gather}
where
\begin{equation} \label{e3.4}
Q_1(\lambda) := 2\int_0^\pi \frac{\pi - \theta}
{\sqrt{2\lambda(\cos\theta + 1)}}d\theta.
\end{equation}

\begin{lemma} \label{lem3.1}
$Q_{1}(\lambda) = C_1\lambda^{-1/2}$.
\end{lemma}

\begin{proof}
\begin{align*}
Q_{1}(\lambda) &=
\frac{1}{\sqrt{\lambda}}
\int_0^\pi \frac{\pi - \theta}{\cos(\theta/2)}d\theta\\
&=\frac{1}{\sqrt{\lambda}}
\int_0^\pi \frac{t}{\sin(t/2)}dt
=\frac{4}{\sqrt{\lambda}}
\int_0^{\pi/2} \frac{t}{\sin t}dt \quad (\mbox{put } \theta = \tan(t/2))\\
&=\frac{8}{\sqrt{\lambda}}
\int_0^1 \frac{\tan^{-1}\theta}{\theta}d\theta \\
&= \frac{8}{\sqrt{\lambda}}\big\{[\log\theta\tan^{-1}\theta
]_0^1 - \int_0^1 \frac{\log\theta}{1 + \theta^2}d\theta
\big\} \quad (\mbox{put }\theta = \tan t) \\
&=-\frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\tan t)dt
= \frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\cot t)dt
\end{align*}
Thus the proof is complete.
\end{proof}

\begin{lemma} \label{lem3.2}
$Q_{2}(\lambda) = -\frac{8}{\sqrt{\lambda}}(1 + o(1))e^{-\sqrt{\lambda}/2}$
as $\lambda \to \infty$.
\end{lemma}

\begin{proof} We put
\begin{equation} \label{e3.5}
Q_{2}(\lambda) := R(\lambda) + S(\lambda),
\end{equation}
where
\begin{gather*} %3.6, 3.7
R(\lambda) = \sqrt{\frac{2}{\lambda}}
\int_0^{\Vert u_{0,\lambda}\Vert_\infty}
\Big(
\frac{\pi-\theta}
{\sqrt{\cos\theta-\cos\Vert u_{0,\lambda}\Vert_\infty}}
- \frac{\pi-\theta}{\sqrt{\cos\theta + 1}}
\Big) d\theta \\
S(\lambda) = - \sqrt{\frac{2}{\lambda}}
\int_{\Vert u_{0,\lambda}\Vert_\infty}^\pi
\frac{\pi-\theta}{\sqrt{\cos\theta + 1}}d\theta.
\end{gather*}
For $\lambda \gg 1$
\begin{equation} \label{e3.8}
\begin{aligned}
R(\lambda) &= \sqrt{\frac{2}{\lambda}}
\int_0^{\Vert u_{0,\lambda}\Vert_\infty}
\frac{(\pi-\theta) (\sqrt{\cos\theta + 1}
-\sqrt{\cos\theta-\cos\Vert u_{0,\lambda}\Vert_\infty})}
{\sqrt{\cos\theta + 1}
\sqrt{\cos\theta-\cos\Vert u_{0,\lambda}\Vert_\infty}}d\theta
\\
&=\sqrt{\frac{2}{\lambda}}(1 + o(1))
\int_0^{\Vert u_{0,\lambda}\Vert_\infty}
\frac{(\pi-\theta)
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)}
{(\cos\theta + 1)
(\sqrt{\cos\theta-\cos\Vert u_{0,\lambda}\Vert_\infty}
+ \sqrt{\cos\theta + 1})}d\theta
\\
&=\sqrt{\frac{2}{\lambda}}
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1))
\int_0^{\Vert u_{0,\lambda}\Vert_\infty}
\frac{\pi-\theta}{2(\cos\theta + 1)^{3/2}}d\theta
\\
&=\sqrt{\frac{2}{\lambda}}
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1))
\int_0^{\Vert u_{0,\lambda}\Vert_\infty}
\frac{\pi-\theta}{4\sqrt{2}\cos^3(\theta/2)}d\theta
\\
&=\sqrt{\frac{2}{\lambda}}
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1))
\int_\pi^{\pi-\Vert u_{0,\lambda}\Vert_\infty}
\frac{-\theta}{4\sqrt{2}\sin^3(\theta/2)}d\theta
\\
&=\sqrt{\frac{1}{\lambda}}
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1))
\int_{(\pi-\Vert u_{0,\lambda}\Vert_\infty)/2}^{\pi/2}
\frac{\theta}{\sin^3\theta}d\theta
\\
&=\sqrt{\frac{1}{\lambda}}
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1))
\int
_{(\pi-\Vert u_{0,\lambda}\Vert_\infty)/2}^{\pi/2}
\frac{1}{\sin^2\theta}d\theta
\\
&=
\sqrt{\frac{1}{\lambda}}
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1))
[-\cot \theta]_{(\pi-\Vert u_{0,\lambda}\Vert_\infty)/2}^{\pi/2}
\\
&=\sqrt{\frac{1}{\lambda}}
\frac{\cos((\pi-\Vert u_{0,\lambda}\Vert_\infty)/2)}
{\sin((\pi-\Vert u_{0,\lambda}\Vert_\infty)/2)}
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1))
\\
&=\sqrt{\frac{1}{\lambda}}
\frac{2}{\pi-\Vert u_{0,\lambda}\Vert_\infty}
(\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)).
\end{aligned}
\end{equation}
By \eqref{e1.8} and Taylor expansion, for $\lambda \gg 1$
\begin{equation} \label{e3.9}
\begin{aligned}
\cos\Vert u_{0,\lambda}\Vert_\infty
&=\cos\big(\pi - 8(1 + o(1))e^{-\sqrt{\lambda}/2}\big)
\\
&= -\cos\big(8(1 + o(1))e^{-\sqrt{\lambda}/2}\big)\\
&= -1 + 32(1 + o(1))e^{-\sqrt{\lambda}}.
\end{aligned}
\end{equation}
By this, \eqref{e1.8} and \eqref{e3.8}, for $\lambda \gg 1$
\begin{equation} \label{e3.10}
R(\lambda) = \frac{8}{\sqrt{\lambda}}(1 + o(1))
e^{-\sqrt{\lambda}/2}.
\end{equation}
Next, we calculate $S(\lambda)$.
By \eqref{e1.8}, for $\lambda \gg 1$
\begin{align*} % 3.11
S(\lambda) &=
- \sqrt{\frac{2}{\lambda}}
\int_{\pi - \Vert u_{0,\lambda}\Vert_\infty}^0
\frac{-t}{{\sqrt{2}\cos((\pi-t)/2)}}dt\\
&= - \sqrt{\frac{1}{\lambda}}
\int_0^{\pi - \Vert u_{0,\lambda}\Vert_\infty}
\frac{t}{\sin(t/2)}dt\\
&= - \frac{4}{\sqrt{\lambda}}
\int_0^{(\pi - \Vert u_{0,\lambda}\Vert_\infty)/2}
\frac{\theta}{\sin\theta}d\theta\\
&=- \frac{4}{\sqrt{\lambda}}\frac{\pi - \Vert u_{0,\lambda}\Vert_\infty}{2}
(1 + o(1))\\
&= -\frac{16}{\sqrt{\lambda}}(1 + o(1))e^{-\sqrt{\lambda}/2}.
\end{align*}
By this, \eqref{e3.5} and \eqref{e3.10}, we obtain our conclusion. Thus the proof
is complete.
\end{proof}

Now \eqref{e1.10} follows from \eqref{e1.8}, \eqref{e3.1}, \eqref{e3.2}
and Lemmas \ref{lem3.1} and \ref{lem3.2}.
Thus the proof is complete.

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

In this section, we consider \eqref{e1.1}--\eqref{e1.3} with \eqref{e1.6}.
By \eqref{e1.1}, we have
\[
(v_\lambda''(t) + \lambda(v_\lambda(t) - v_\lambda^3(t)))v_\lambda'(t) = 0.
\]
This implies that for $t \in \bar{I}$
\[
\frac{d}{dt}\big(
\frac12v_\lambda'(t)^2 + \frac12\lambda v_\lambda^2(t)
- \frac14\lambda v_\lambda^4(t)
\big) = 0.
\]
This implies that for $t \in \bar{I}$,
\begin{equation} \label{e4.1}
\frac12v_\lambda'(t)^2 + \frac12\lambda v_\lambda^2(t)
- \frac14\lambda v_\lambda^4(t) \equiv \mbox{constant}
= \frac12\lambda \Vert v_\lambda\Vert_\infty^2
- \frac14\lambda\Vert v_\lambda\Vert_\infty^4.
\end{equation}
We know that
\begin{equation} \label{e4.2}
v_\lambda'(t) \ge 0, \quad 0 \le t \le 1/2,
\quad v_\lambda(t) = v_\lambda(1-t),
\quad t \in I.
\end{equation}
Therefore, by \eqref{e4.1} and \eqref{e4.2}, for $0 \le t \le 1/2$,
\begin{equation} \label{e4.3}
v_\lambda'(t) =
\sqrt{\lambda\{
(\Vert v_\lambda\Vert_\infty^2 - v_\lambda(t)^2) - \frac12
(\Vert v_\lambda\Vert_\infty^4 - v_\lambda(t)^4)\}}.
\end{equation}
The following Lemma \ref{lem4.1} implies \eqref{e1.14} in Theorem \ref{thm1.2}.

\begin{lemma} \label{lem4.1}
As $\lambda \to \infty$
\begin{equation} \label{e4.4}
\Vert v_\lambda\Vert_\infty = 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{lemma}

\begin{proof}
By \eqref{e4.3},
\begin{equation} \label{e4.5}
\begin{aligned}
\frac12 &= \int_0^{1/2} dt
= \int_0^{1/2}\frac{v_\lambda'(t)}
{\sqrt{\lambda
\{(\Vert v_\lambda\Vert_\infty^2 - v_\lambda(t)^2) - \frac12
(\Vert v_\lambda\Vert_\infty^4 - v_\lambda(t)^4)
\}}}dt
\\
&= \frac{1}{\sqrt{\lambda}}
\int_0^{\Vert v_\lambda\Vert_\infty} \frac{1}
{\sqrt{(\Vert v_\lambda\Vert_\infty^2 - \theta^2)
- \frac12(\Vert v_\lambda\Vert_\infty^4 - \theta^4)}
}d\theta \quad (\mbox{put} \enskip \theta = \Vert v_\lambda\Vert_\infty s)
\\
&= \frac{1}{\sqrt{\lambda}}\frac{\sqrt{2}}
{\sqrt{2-\Vert v_\lambda\Vert_\infty^2}}
\int_0^1 \frac{1}{\sqrt{(1-s^2)(1-k^2s^2)}}ds
\\
&= \frac{1}{\sqrt{\lambda}}\frac{\sqrt{2}}
{\sqrt{2-\Vert v_\lambda\Vert_\infty^2}}K(k),
\end{aligned}
\end{equation}
where
$k = \Vert v_\lambda\Vert_\infty/\sqrt{2-\Vert v_\lambda\Vert_\infty^2}$
and
\[ %4.6
K(k) := \int_0^{\pi/2} \frac{1}{\sqrt{1-k^2\sin^2\theta}}d\theta.
\]
It is known (cf. \cite[p.909, 8.113]{g}) that as $k \to 1$
\begin{equation} \label{e4.7}
\begin{aligned}
K(k) &= -\frac12\log(1-k^2) + 2\log 2 - \frac{1-k^2}{8}
\log(1-k^2) \\
&\quad + \big(\frac12\log2 - \frac14\big)(1 - k^2)
- \frac{9}{128}(1-k^2)^2\log(1-k^2)\\
&\quad + o\big((1-k^2)^2\log(1-k^2)\big).
\end{aligned}
\end{equation}
We put $\xi_\lambda := 1 - \Vert v_\lambda\Vert_\infty^2$.
Then $\xi_\lambda > 0$ and $\xi_\lambda \to 0$ as $\lambda \to \infty$
by \eqref{e1.7}. Then
\begin{equation} \label{e4.8}
1-k^2
= \frac{2(1-\Vert v_\lambda\Vert_\infty^2)}{2-\Vert v_\lambda\Vert_\infty^2}
= \frac{2\xi_\lambda}{1 + \xi_\lambda}.
\end{equation}
By this, the Taylor expansion, and \eqref{e4.7},
\begin{equation} \label{e4.9}
\begin{aligned}
K(k) &=
-\frac12
\big(\log 2 + \log \xi_\lambda - \log(1 + \xi_\lambda)\big)
\\
&\quad + 2\log 2 - \frac{\xi_\lambda}{4(1 + \xi_\lambda)}
\big(\log 2 + \log \xi_\lambda - \log(1 + \xi_\lambda)\big)
\\
&\quad + \big(\frac12\log 2 - \frac14\big)
\frac{2\xi_\lambda}{1 + \xi_\lambda}
-\frac{9}{32}\frac{\xi_\lambda^2}{(1 + \xi_\lambda)^2}
\big(\log 2 + \log \xi_\lambda - \log(1 + \xi_\lambda)\big)
\\
&\quad +o(\xi_\lambda^2\log\xi_\lambda)
\\
&= -\frac12
\big(\log 2 + \log \xi_\lambda - \xi_\lambda + O(\xi_\lambda^2)\big)
+ 2\log 2
\\
&\quad - \frac14\xi_\lambda(1 - \xi_\lambda + O(\xi_\lambda^2))
\big(\log 2 + \log \xi_\lambda - \xi_\lambda + O(\xi_\lambda^2)\big)
\\
&\quad + \big(\log 2 - \frac12\big)
\xi_\lambda(1 - \xi_\lambda + O(\xi_\lambda^2))
-\frac{9}{32}
\xi_\lambda^2\log\xi_\lambda
+ o(\xi_\lambda^2\log\xi_\lambda)
\\
&= -\frac12\log\xi_\lambda + \frac32\log 2
- \frac14\xi_\lambda\log\xi_\lambda
+ \frac{3\log2}{4}\xi_\lambda
- \frac{1}{32}\xi_\lambda^2\log\xi_\lambda
+ o(\xi_\lambda^2\log\xi_\lambda).
\end{aligned}
\end{equation}
Furthermore, by Taylor expansion, for $\lambda \gg 1$,
\[ %4.10
\frac{1}{\sqrt{2-\Vert v_\lambda\Vert_\infty^2}}
= (1 + \xi_\lambda)^{-1/2} = 1 - \frac12\xi_\lambda
+ \frac{3}{8}\xi_\lambda^2 + o(\xi_\lambda^2).
\]
This along with \eqref{e4.5} and \eqref{e4.9} implies that
\begin{equation} \label{e4.11}
\frac{\sqrt{\lambda}}{2\sqrt{2}}
=-\frac12\log\xi_\lambda + \frac32\log 2
-\frac{3}{32}\xi_\lambda^2\log\xi_\lambda
+ o(\xi_\lambda^2\log\xi_\lambda).
\end{equation}
By this, for $\lambda \gg 1$, we have
\[
\frac{\sqrt{\lambda}}{\sqrt{2}} = -\log\xi_\lambda + \log 8 + \log(1 + o(1))
= \log\frac{8(1 + o(1))}{\xi_\lambda}.
\]
This implies that for $\lambda \gg 1$,
$\xi_\lambda = 8(1 +o(1))e^{-\sqrt{\lambda}/\sqrt{2}}$. %4.12
Then for $\lambda \gg 1$,
\[
-\frac{3}{32}\xi_\lambda^2\log\xi_\lambda
= 3\sqrt{2}(1 + o(1))\sqrt{\lambda}e^{-\sqrt{2\lambda}}.
\]
By this, \eqref{e4.11} and Taylor expansion, for $\lambda \gg 1$
\begin{equation} \label{e4.13}
\begin{aligned}
\xi_\lambda &=8e^{-\sqrt{\lambda}/\sqrt{2}}\cdot
e^{6\sqrt{2}(1 + o(1))\sqrt{\lambda}e^{-\sqrt{2\lambda}}}
\\
&= 8e^{-\sqrt{\lambda}/\sqrt{2}}(1 + 6\sqrt{2}(1 + o(1))
\sqrt{\lambda}e^{-\sqrt{2\lambda}}))
\\
&= 8e^{-\sqrt{\lambda}/\sqrt{2}}
+ 48\sqrt{2}(1 + o(1))
\sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}}.
\end{aligned}
\end{equation}
By this and Taylor expansion, for $\lambda \gg 1$,
\begin{align*}
\Vert v_\lambda\Vert_\infty
&= \sqrt{1 - \xi_\lambda}\\
&=\Big(1- 8e^{-\sqrt{\lambda}/\sqrt{2}}
- 48\sqrt{2}(1 + o(1))\sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}})
\Big)^{1/2}
\\
&= 1 + \frac12\Big(- 8e^{-\sqrt{\lambda}/\sqrt{2}}
-48\sqrt{2}(1 + o(1))\sqrt{\lambda}
e^{-3\sqrt{\lambda}/\sqrt{2}})\Big)
\\
&\quad- \frac{1}{8}(- 8e^{-\sqrt{\lambda}/\sqrt{2}}
- 48\sqrt{2}(1 + o(1))
\sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}}))^2
+ O(e^{-3\sqrt{\lambda}/\sqrt{2}}).
\end{align*}
By this, we obtain \eqref{e4.4}.
\end{proof}


The following implies \eqref{e1.13} in Theorem \ref{thm1.2}.


\begin{lemma} \label{lem4.2}
 As $\lambda \to \infty$
\begin{equation} \label{e4.15}
\Vert v_\lambda\Vert_1 = 1 - \frac{2\sqrt{2}\log 2}{\sqrt{\lambda}}
- 12e^{-\sqrt{2\lambda}} + o(e^{-\sqrt{2\lambda}}).
\end{equation}
\end{lemma}

\begin{proof} By \eqref{e4.3}, for $\lambda \gg 1$
\begin{equation} \label{e4.16}
\begin{aligned}
\Vert v_\lambda\Vert_1
&= 2\int_0^{1/2} v_\lambda(t)\,dt \\
&= 2 \int_0^{1/2}\frac{v_\lambda(t)v_\lambda'(t)}
{\sqrt{\lambda\{
(\Vert v_\lambda\Vert_\infty^2 - v_\lambda(t)^2) - \frac12
(\Vert v_\lambda\Vert_\infty^4 - v_\lambda(t)^4)\}}}\,dt
\\
&= \frac{2}{\sqrt{\lambda}}
\int_0^{\Vert v_\lambda\Vert_\infty} \frac{\theta}
{\sqrt{\Vert v_\lambda\Vert_\infty^2 - \theta^2
- \frac12(\Vert v_\lambda\Vert_\infty^4 - \theta^4)}}\,d\theta
\\
&=\frac{2\Vert v_\lambda\Vert_\infty}{\sqrt{\lambda}}
\int_0^1 \frac{s}
{\sqrt{(1-s^2)-\frac12\Vert v_\lambda\Vert_\infty^2(1-s^4)}}\,ds
\\
&=\frac{\Vert v_\lambda\Vert_\infty}{\sqrt{\lambda}}
\frac{\sqrt{2}}{\sqrt{2-\Vert v_\lambda\Vert_\infty^2}}
\int_0^1\frac{1}{(\sqrt{1-t)(1-k^2t)}}\,dt
\\
&= \sqrt{\frac{2}{\lambda}}k\int_0^1
\frac{1}{\sqrt{(1-t)(1-k^2t)}}\,dt.
\end{aligned}
\end{equation}
By putting $s = \sqrt{(1-k^2t)/(1-t)}$, we obtain easily
\[ %4.17
\int_0^1 \frac{1}{\sqrt{(1-t)(1-k^2t)}}dt = \frac{1}{k}
\log\big(\frac{1+k}{1-k}\big).
\]
This along with \eqref{e4.16} implies that
\begin{equation} \label{e4.18}
\Vert v_\lambda\Vert_1
=\sqrt{\frac{2}{\lambda}}\log\frac{1+k}{1-k}
=\sqrt{\frac{2}{\lambda}}\log\frac{(1+k)^2}{1-k^2}
=\sqrt{\frac{2}{\lambda}}(2\log(1 + k) - \log(1-k^2)).
\end{equation}
By \eqref{e4.8}, \eqref{e4.13} and Taylor expansion, for $\lambda \gg 1$
\begin{equation} \label{e4.19}
\begin{aligned}
\log(1 - k^2) &= \log\frac{2\xi_\lambda}{1 + \xi_\lambda}\\
&= \log 2 + \log\xi_\lambda - \log(1 + \xi_\lambda)\\
&= 4\log2 - \sqrt{\frac{\lambda}{2}}
+ 6\sqrt{2}\sqrt{\lambda}e^{-\sqrt{2\lambda}}
- \left(\xi_\lambda + O(\xi_\lambda^2)\right)\\
&= 4\log2 - \sqrt{\frac{\lambda}{2}} - 8e^{-\sqrt{\lambda/2}}
+ 6\sqrt{2}\sqrt{\lambda}e^{-\sqrt{2\lambda}}
+ o(\sqrt{\lambda}e^{-\sqrt{2\lambda}}).
\end{aligned}
\end{equation}
By Lemma \ref{lem4.1}, \eqref{e4.13} and Taylor expansion, for $\lambda \gg 1$,
\begin{align*} % 4.20
k &= \frac{\Vert v_\lambda\Vert_\infty}{\sqrt{2-\Vert v_\lambda\Vert_\infty}}
= \Vert v_\lambda\Vert_\infty(1 + \xi_\lambda)^{-1/2}
\\
&= \Vert v_\lambda\Vert_\infty\Big(1 - \frac12\xi_\lambda
+ \frac{3}{8}\xi_\lambda^2(1 + o(1))\Big)
\\
&= (1 - 4e^{-\sqrt{\lambda/2}} - 8e^{-2\sqrt{\lambda/2}}(1 + o(1)))
(1 - 4e^{-\sqrt{\lambda/2}} + 24e^{-2\sqrt{\lambda/2}}(1 + o(1)))
\\
&= 1 - 8e^{-\sqrt{\lambda/2}} + 32e^{-2\sqrt{\lambda/2}}(1 + o(1)).
\end{align*}
By this and Taylor expansion, for $\lambda \gg 1$,
\begin{align*} %4.21
\log(1 + k) &= \log(2 - (1-k)) = \log 2 - \frac12(1-k) - \frac18(1-k)^2
+ o((1-k)^2) \\
&= \log 2 - 4e^{-\sqrt{\lambda/2}} + 8(1 + o(1))e^{-2\sqrt{\lambda/2}}.
\end{align*}
By this and \eqref{e4.19}, we obtain
\[
\log\big(\frac{1+k}{1-k}\big) = -2\log 2 + \sqrt{\frac{\lambda}{2}}
-6\sqrt{2}(1 + o(1))\sqrt{\lambda}e^{-\sqrt{2\lambda}}.
\]
By this and \eqref{e4.18} and \eqref{e4.19}, we obtain \eqref{e4.15}.
Thus the proof is complete.
\end{proof}

\section{Appendix}

We show that $\Vert u_\lambda\Vert_\infty < \pi$ for completeness.
By \eqref{e2.1},
$$
-u_\lambda''(1/2) = \lambda\sin\Vert u_\lambda\Vert_\infty
- g(\Vert u_\lambda\Vert_\infty) \ge 0.
$$
This along with (A1) implies that
there exists a non-negative integer $k$ such that
\begin{equation} \label{e5.1}
2k\pi < \Vert u_\lambda\Vert_\infty < (2k+1)\pi.
\end{equation}
Assume that $k \ge 1$. Then by \eqref{e2.1}, there exists a unique
$t_\lambda \in (0, 1/2)$ such that
$u_\lambda(t_\lambda) = \Vert u_\lambda \Vert_\infty - 2k\pi$.
Then by \eqref{e2.2},
\begin{equation} \label{e5.2}
\begin{aligned}
&\frac12 u_\lambda'(t_\lambda)^2 - \lambda \cos u_\lambda(t_\lambda)
- G(u_\lambda(t_\lambda))\\
& = \frac12 u_\lambda'(t_\lambda)^2
- \lambda \cos \Vert u_\lambda\Vert_\infty
- G(\Vert u_\lambda\Vert_\infty - 2k\pi) \\
& = - \lambda \cos \Vert u_\lambda\Vert_\infty
- G(\Vert u_\lambda\Vert_\infty).
\end{aligned}
\end{equation}
Since $G(u)$ is strictly increasing for $u \ge 0$ by (A1),
by \eqref{e5.2}, we obtain
$$
\frac12 u_\lambda'(t_\lambda)^2 = G(\Vert u_\lambda\Vert_\infty - 2k\pi)
- G(\Vert u_\lambda\Vert_\infty) < 0.
$$
This is a contradiction. Thus $k = 0$ in \eqref{e5.1} and we get our assertion.

\subsection*{Acknowledgment}
The author thanks the anonymous referee for his/her helpful comments.

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\end{document}