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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 142, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/142\hfil Precise asymptotic behavior of solutions]
{Precise asymptotic behavior of solutions to
damped simple pendulum equations}

\author[T. Shibata\hfil EJDE-2009/142\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 April 21, 2009. Published November 7, 2009.}
\subjclass[2000]{34B15}
\keywords{Damped simple pendulum; asymptotic formula}

\begin{abstract}
 We consider the simple pendulum equation
 \begin{gather*}
 -u''(t) + \epsilon f(u'(t)) = \lambda\sin u(t), \quad t \in I:=(-1, 1),\\
 u(t) > 0, \quad t \in I, \quad u(\pm 1) = 0,
 \end{gather*}
 where $0 < \epsilon \le 1$, $\lambda > 0$, and the friction term
 is either $f(y) = \pm|y|$ or $f(y) = -y$.
 Note that when $f(y) = -y$ and $\epsilon = 1$, we have well known
 original damped simple pendulum equation.
 To understand the dependance of  solutions, to the damped simple
 pendulum equation with $\lambda \gg 1$,  upon the term $f(u'(t))$, 
 we  present asymptotic formulas for the maximum norm of the solutions.
 Also we present an asymptotic formula for the time at which
 maximum occurs, for the case $f(u) = -u$.
\end{abstract}

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

\section{Introduction}

We consider the damped simple pendulum equation
\begin{gather} \label{e1.1}
-u''(t) + \epsilon f(u'(t)) = \lambda\sin u(t), \quad t \in I := (-1, 1),\\
u(t) > 0, \quad t \in I,  \label{e1.2} \\
u(\pm 1) = 0, \label{e1.3}
\end{gather}
where $0 < \epsilon \le 1$, $\lambda > 0$, and the damping term
is either $f(y) = \pm| y |$ or $f(y) = -y$.
It is known that there exists a solution $u_{\epsilon,\lambda}$ to
 \eqref{e1.1}--\eqref{e1.3}
for $0 < \epsilon \le 1$ and $\lambda \gg 1$, with
$\| u_{\epsilon,\lambda}\|_\infty < \pi$; see for example \cite{a1}.

The purpose of this paper is to study the asymptotic behavior of
$u_{\epsilon,\lambda}(t)$ as $\lambda \to \infty$; this is useful for
understanding the effect of the damping term on the asymptotic
behavior of $u_{\epsilon,\lambda}$.
First, we recall some properties of the solution $u_{0,\lambda}$
for the  simple pendulum equation without friction
(i.e. the case where $\epsilon = 0$):
\begin{gather} \label{e1.4}
-u''(t) = \lambda\sin u(t), \quad t \in I,\\
u(t) > 0, \quad t \in I, \label{e1.5}\\
u(\pm 1) = 0. \label{e1.6}
\end{gather}
It is well known that $u_{0,\lambda} \to \pi$
locally uniformly in $I$ as $\lambda \to \infty$. Furthermore
(cf. Lemma \ref{lem2.1} in Section 2), as $\lambda \to \infty$,
\begin{equation} \label{e1.7}
\| u_{0,\lambda}\|_\infty = \pi - 8e^{-\sqrt{\lambda}}
- 32\sqrt{\lambda}e^{-3\sqrt{\lambda}} +
o(\sqrt{\lambda}e^{-3\sqrt{\lambda}}).
\end{equation}
It should be mentioned that the asymptotic behavior of solutions to
the original and perturbed simple pendulum problems have been studied
in \cite{s1,s2,s3}. We also refer the reader to \cite{f1} for
the basic properties of the solution to simple pendulum problems.
As far as the author knows,
there are only a few works concerning the precise
properties of solutions to \eqref{e1.1}--\eqref{e1.3}.
In particular, an asymptotic formulas such as
\eqref{e1.7} for $\| u_{\epsilon,\lambda}\|_\infty$ has
not been obtained yet.
Therefore, it seems worth considering the precise asymptotic
behavior of $\| u_{\epsilon,\lambda}\|_\infty$ as $\lambda \to \infty$,
for having a better understanding of the effect of the friction term.

Now we state our main results.
We denote by $u_{1,\epsilon,\lambda}$, $u_{2,\epsilon,\lambda}$ and
$u_{3,\epsilon,\lambda}$ the solutions of \eqref{e1.1}--\eqref{e1.3} with
$f(y) = -| y|$, $f(y) = | y|$ and $f(y) = -y$,
respectively.

\begin{theorem} \label{thm1.1}
Let $f(y) = -| y |$ and let
 $0 < \epsilon \le 1$ be fixed. Then, as $\lambda \to \infty$,
\begin{equation} \label{e1.8}
\| u_{1,\epsilon,\lambda}\|_\infty
= \pi - 8e^{-\epsilon}e^{-\sqrt{\lambda}}
+ O(\lambda^{-1/2}e^{-\sqrt{\lambda}}).
\end{equation}
\end{theorem}

Since $u_{1,\epsilon,\lambda}$ is a super-solution
of \eqref{e1.4}--\eqref{e1.6}, \eqref{e1.8} is well understood and
reasonable from a viewpoint of \eqref{e1.7}.
Moreover, the formula \eqref{e1.8} gives us the clear relationship between
$\| u_{0,\lambda}\|_\infty$
and $\| u_{1,\epsilon,\lambda}\|_\infty$.

The following result can be proved by the same arguments as those
used in the proof of Theorem \ref{thm1.1}.

\begin{theorem} \label{thm1.2}
Let  $f(y) = | y |$ and $0 < \epsilon \le 1$ be fixed.
Then, as $\lambda \to \infty$,
\begin{equation} \label{e1.9}
\| u_{2,\epsilon,\lambda}\|_\infty
= \pi - 8e^{\epsilon}e^{-\sqrt{\lambda}}
+ O(\lambda^{-1/4}e^{-\sqrt{\lambda}}).
\end{equation}
\end{theorem}

We also note that $u_{2,\epsilon,\lambda}$ is a sub-solution
of \eqref{e1.4}--\eqref{e1.6}, \eqref{e1.9} is also
reasonable result.

Now we consider the case $f(y) = -y$.
Let $0 < \epsilon \le 1$ be fixed.
Let $t_{\epsilon, \lambda} \in I$ be the unique point satisfying
$u_{3,\epsilon,\lambda}(t_{\epsilon,\lambda})
= \| u_{3,\epsilon,\lambda}\|_\infty$.
Then we know from \cite{b1} that $t_{\epsilon,\lambda} < 0$ for
$\lambda \gg 1$.

\begin{theorem} \label{thm1.3}
Let $f(y) = -y$. Then, as $\lambda \to \infty$,
\begin{gather}
t_{\epsilon,\lambda} = -\frac{\epsilon}{\sqrt{\lambda}}
+ O\big(\lambda^{-3/4}\big), \label{e1.10}\\
\| u_{3,\epsilon,\lambda}\|_\infty
=  \pi - 8e^{-\sqrt{\lambda}}
+ O(\lambda^{-1/4}e^{-\sqrt{\lambda}}). \label{e1.11}
\end{gather}
\end{theorem}

By \eqref{e1.10}, we obtain a precise asymptotic formula for
$t_{\epsilon,\lambda}$ as $\lambda \to \infty$. Moreover, since the
second term of \eqref{e1.11} is the same as that of \eqref{e1.7},
the friction term does not have any effect on the second term of
$\| u_{3,\epsilon,\lambda}\|_\infty$.

The rest of this paper is organized as follows.
In Section 2, we prove Theorem \ref{thm1.1} based on the crucial
tool Lemma \ref{lem2.2},
which will be proved in Section 3.
We prove Theorem \ref{thm1.2} in Section 4 
by almost the same argument as that to prove
Theorem \ref{thm1.1}.
We apply the modified argument for the proof of Theorem \ref{thm1.1} to
the proof of Theorem \ref{thm1.3} in Section 5.

\section{Proof of Theorem \ref{thm1.1}}
In the following two sections, we let $f(y) = -| y|$.
We fix $0 < \epsilon \le 1$. Further, we assume that $\lambda \gg 1$ and
we write $u_{\epsilon,\lambda} = u_{1,\epsilon,\lambda}$ for simplicity.
We consider the solution $u_{\epsilon,\lambda}(t)$
with $\| u_{\epsilon,\lambda}\|_\infty < \pi$.
We know
\begin{gather}
u_{\epsilon,\lambda}(t) = u_{\epsilon,\lambda}(-t),
\quad t \in I, \label{e2.1} \\
u_{\epsilon,\lambda}'(t) > 0, \quad t \in [-1, 0), \label{e2.2}\\
u_{\epsilon,\lambda}(0) = \| u_{\epsilon,\lambda}\|_\infty,
\label{e2.3}\\
u_{\epsilon,\lambda}(t) \to \pi \quad \mbox{as $\lambda \to \infty$},
\quad (t \in I). \label{e2.4}
\end{gather}
Note that \eqref{e2.1}-\eqref{e2.3} follow from \cite{b1}.
\eqref{e2.4} is a direct consequence of \eqref{e1.7}, \eqref{e2.3}
and \eqref{e2.6} below.

By \eqref{e1.1} and \eqref{e2.2},
for $-1 \le t \le 0$, we have
$$
\{u_{\epsilon,\lambda}''(t) + \epsilon u_{\epsilon,\lambda}'(t)
+ \lambda\sin u_{\epsilon,\lambda}(t)\}u_{\epsilon,\lambda}'(t) = 0.
$$
By this equality and \eqref{e2.3}, for $-1 \le t \le 0$, we have
\begin{equation} \label{e2.5}
\begin{aligned}
&\frac12u_{\epsilon,\lambda}'(t)^2 +
\epsilon\int_{-1}^t | u_{\epsilon,\lambda}'(s)|^{2}ds
- \lambda\cos u_{\epsilon,\lambda}(t)\\
&= \epsilon \int_{-1}^{0} | u_{\epsilon,\lambda}'(s)|^{2}ds
- \lambda\cos \| u_{\epsilon,\lambda}\|_\infty =\text{constant}.
\end{aligned}
\end{equation}
For $-1 \le t \le 0$, we obtain
\begin{equation} \label{e2.6}
\frac12u_{\epsilon,\lambda}'(t)^2 =
\lambda(\cos u_{\epsilon,\lambda}(t)
- \cos\| u_{\epsilon,\lambda}\|_\infty)
+ \epsilon\int_t^{0} | u_{\epsilon,\lambda}'(s)|^{2}ds.
\end{equation}
For $-1 \le t \le 0$, we put
\begin{gather} \label{e2.7}
A(\theta):= A_\lambda(\theta)
= \lambda(\cos \theta - \cos\| u_{\epsilon,\lambda}\|_\infty),
\\
B(t):= B_\lambda(t) = \int_t^{0} | u_{\epsilon,\lambda}'(s)|^{2}ds.
\label{e2.8}
\end{gather}
Then by \eqref{e2.2} and  \eqref{e2.6}--\eqref{e2.8}, for
$-1 \le t \le 0$,
\begin{equation} \label{e2.9}
u_{\epsilon,\lambda}'(t) = \sqrt{2(A(u_{\epsilon,\lambda}(t)) +
\epsilon B(t))}.
\end{equation}
Then
\begin{equation} \label{e2.10}
1 = \int_{-1}^{0} dt
= \frac{1}{\sqrt{2}}
\int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A
(u_{\epsilon,\lambda}(t)) +\epsilon B(t)}}dt
= \frac{1}{\sqrt{2}}(I + II),
\end{equation}
where
\begin{gather} \label{e2.11}
I = \int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A(u_{\epsilon,\lambda}(t))}}dt,
\\ \label{e2.12}
\begin{aligned}
II &= \int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t)) + \epsilon B(t)}}dt
- \int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A(u_{\epsilon,\lambda}(t))}}dt
\\
&= \int_{-1}^{0} \frac{-\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t))}\sqrt{A(u_{\epsilon,\lambda}(t))
+ \epsilon B(t)}}\\
&\quad\times \frac{1}{
\big(\sqrt{A(u_{\epsilon,\lambda}(t))}+\sqrt{A(u_{\epsilon,\lambda}(t))
+ \epsilon B(t)}\big)
}dt.
\end{aligned}
\end{gather}


\begin{lemma}\label{lem2.1}  Let $d_\lambda := \pi -
\| u_{\epsilon,\lambda}\|_\infty$.
Then, as $\lambda \to \infty$,
\begin{equation} \label{e2.13}
I = \sqrt{\frac{2}{\lambda}}
\Big(\log\frac{4}{\sin (d_\lambda/2)}
+ \frac14(1 + o(1))
\big(\log\frac{4}{\sin (d_\lambda/2)}\big)
\sin^2\frac{d_\lambda}{2}\Big).
\end{equation}
\end{lemma}

\begin{proof}
 Put $\theta = u_{\epsilon,\lambda}(t)$. Then
\begin{equation} \label{e2.14}
\begin{aligned}
I &= \frac{1}{\sqrt{\lambda}}
\int_0^{\| u_{\epsilon,\lambda}\|_\infty} \frac{1}
{\sqrt{\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty}}d\theta
\\
&= \frac{\sqrt{2}}{\sqrt{\lambda}\sin(\| u_{\epsilon,\lambda}\|_\infty/2)}
\int_0^{\| u_{\epsilon,\lambda}\|_\infty/2}
\frac{1}{\sqrt{1-\sin^2\varphi/\sin^2(\| u_{\epsilon,\lambda}\|_\infty/2)}}
d\varphi
\\
&= \sqrt{\frac{2}{\lambda}}\int_0^{\pi/2} \frac{1}
{\sqrt{1-\sin^2(\| u_{\epsilon,\lambda}\|_\infty/2)\sin^2\phi}}d\phi
\\
&= \sqrt{\frac{2}{\lambda}}K(k),
\end{aligned}
\end{equation}
where $K$ is the complete elliptic integral of the first kind and
$k = \sin(\| u_{\epsilon,\lambda}\|_\infty/2)$. Then by \cite{g1}, we have
\begin{equation} \label{e2.15}
K(k) = \log\frac{4}{k'} + \frac14\big(\log\frac{4}{k'}\big){k'}^2
(1 + o(1)),
\end{equation}
where $k' = \sqrt{1-k^2}= \cos(\| u_{\epsilon,\lambda}\|_\infty/2)
= \cos((\pi-d_\lambda)/2) = \sin(d_\lambda/2)$.
By this and \eqref{e2.14}, we obtain \eqref{e2.13}.
Thus the proof is complete.
\end{proof}

Since $II < 0$, by \eqref{e2.10}, \eqref{e2.15} and Lemma \ref{lem2.1},
we have
\begin{equation} \label{e2.16}
1 < \frac{1}{\sqrt{2}}I \le \frac{1}{\sqrt{\lambda}}
\big(1 + C\sin^2\frac{d_\lambda}{2}\big)
\log\frac{4}{\sin(d_\lambda/2)}.
\end{equation}
Then
\begin{equation} \label{e2.17}
\sin\frac{d_\lambda}{2} \le 4(1 + o(1))e^{-\sqrt{\lambda}},
\enskip \frac{d_\lambda}{2} \le 4(1 + o(1))e^{-\sqrt{\lambda}}, \enskip
\sin\| u_{\epsilon,\lambda}\|_\infty \le 8(1 + o(1))e^{-\sqrt{\lambda}}.
\end{equation}

\begin{lemma}\label{lem2.2}  As $\lambda \to \infty$,
\begin{equation} \label{e2.18}
II = \frac{\sqrt{2}\epsilon}
{\lambda}\log\big(\sin \frac{d_\lambda}{2}\big)
+ O(\lambda^{-1}).
\end{equation}
\end{lemma}

The proof of the above lemma will be given in Section 3.
We accept Lemma \ref{lem2.2} tentatively to prove Theorem \ref{thm1.1}.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
 By Lemmas \ref{lem2.1} and \ref{lem2.2} and \eqref{e2.17}, we have
\begin{equation} \label{e2.19}
\begin{aligned}
1 &= \frac{1}{\sqrt{2}}(I + II) =
\frac{1}{\sqrt{\lambda}}
\Big(\log\frac{4}{\sin(d_\lambda/2)}
+ \frac14(1 + o(1))\sin^2\frac{d_\lambda}{2}
\log\frac{4}{\sin(d_\lambda/2)}\Big) \\
&\quad + \frac{\epsilon}{\lambda}\log\sin \frac{d_\lambda}{2} + O(\lambda^{-1})
\\
&= \frac{1}{\sqrt{\lambda}}
\big(\log 4 - \log\sin\frac{d_\lambda}{2}\big)
+ \frac{\epsilon}{\lambda}
\log\sin\frac{d_\lambda}{2} + O(\lambda^{-1}).
\end{aligned}
\end{equation}
This implies
\begin{equation} \label{e2.20}
\big(1 - \frac{\epsilon}{\sqrt{\lambda}}\big)
\log\sin\frac{d_\lambda}{2}
= \log 4 - \sqrt{\lambda} + O(\lambda^{-1/2}).
\end{equation}
By this,
\begin{equation} \label{e2.21}
\begin{aligned}
\log\sin\frac{d_\lambda}{2}
&= \big(1 + \frac{\epsilon}{\sqrt{\lambda}}
+ O(\lambda^{-1})\big)\big(\log 4 - \sqrt{\lambda}
+ O(\lambda^{-1/2})\big)\\
&= -\sqrt{\lambda} + \log 4 - \epsilon + O(\lambda^{-1/2}).
\end{aligned}
\end{equation}
By this and Taylor expansion,
\[ % \label{e2.21}
\sin\frac{d_\lambda}{2}
= \frac{d_\lambda}{2}\big(1 - \frac{d_\lambda^2}{3}
+ o(d_\lambda^2)\big) = 4e^{-\epsilon}
e^{-\sqrt{\lambda}}(1 + O(\lambda^{-1/2})).
\]
By this and \eqref{e2.17}, we obtain Theorem \ref{thm1.1}.
\end{proof}

\section{Proof of Lemma \ref{lem2.2}}

In this section, we focus our attention on the proof of Lemma \ref{lem2.2}.
Let $0 < \delta \ll 1$ be fixed. We define
$t_\delta := t_{\lambda,\delta} < 0$ by
$u_{\epsilon,\lambda}(t_\delta)
= \| u_{\epsilon,\lambda}\|_\infty - \delta$.
We set
\begin{equation}\label{e3.1}
\begin{aligned}
&II = II_1 + II_2
\\
&:= \int_{t_\delta}^0 \frac{-\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t))}\sqrt{A(u_{\epsilon,\lambda}(t))
+ \epsilon B(t)}
(\sqrt{A(u_{\epsilon,\lambda}(t))}
+\sqrt{A(u_{\epsilon,\lambda}(t)) + \epsilon B(t)})}dt
\\
& + \int_{-1}^{t_\delta} \frac{-\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t))}\sqrt{A(u_{\epsilon,\lambda}(t))
+ \epsilon B(t)}
\big(\sqrt{A(u_{\epsilon,\lambda}(t))}
+\sqrt{A(u_{\epsilon,\lambda}(t)) + \epsilon B(t)}\big)
}dt.
\end{aligned}
\end{equation}
To obtain Lemma \ref{lem2.2}, we estimate $II_1$ and $II_2$ by series of
lemmas.

\begin{lemma}\label{lem3.1}
For $-1 \le t \le 0$,
\begin{equation} \label{e3.2}
B(t) \le \sqrt{2A(u_{\epsilon,\lambda}(t))}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)) +
2\epsilon
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2.
\end{equation}
\end{lemma}

\begin{proof}
Since $\| u_{\epsilon,\lambda}\|_\infty < \pi$,
we see from \eqref{e1.1} that $u_{\epsilon,\lambda}''(t) \le 0$
for $t \in I$.
This along with \eqref{e2.8} and \eqref{e2.9} implies that
for $-1 \le t \le 0$,
\begin{equation} \label{e3.3}
\begin{aligned}
0 &< B(t) \\
&\le \max_{t \le s \le 0}| u_{\epsilon,\lambda}'(s)|
\int_t^0 u_{\epsilon,\lambda}'(s)ds\\
&= u_{\epsilon,\lambda}'(t)
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))
\\
&= \sqrt{2A(u_{\epsilon,\lambda}(t)) + 2\epsilon B(t)}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)).
\end{aligned}
\end{equation}
By \eqref{e3.3},
\[
B(t)^2 - 2\epsilon
B(t)(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2
- 2A(u_{\epsilon,\lambda}(t))(\| u_{\epsilon,\lambda}\|_\infty 
- u_{\epsilon,\lambda}(t))^2 \le 0.
\]
Since $\sqrt{a+b} \le \sqrt{a} + \sqrt{b}$ for $a, b \ge 0$,
by this, we obtain
\begin{equation}  \label{e3.4}
\begin{aligned}
B(t) &\le \epsilon
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2
\\
&\quad + \sqrt{\epsilon^2(\| u_{\epsilon,\lambda}\|_\infty
 - u_{\epsilon,\lambda}(t))^4 + 2A(u_{\epsilon,\lambda}(t))
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2}
\\
&\le 2\epsilon
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2
+ \sqrt{2A(u_{\epsilon,\lambda}(t))}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)).
\end{aligned}
\end{equation}
The proof is complete.
\end{proof}

By Taylor expansion, for $t_\delta \le t \le 0$ and $0 < \kappa \ll 1$,
\begin{gather} \label{e3.5}
\begin{aligned}
\cos u_{\epsilon,\lambda}(t)-\cos\| u_{\epsilon,\lambda}\|_\infty
&\le \sin\| u_{\epsilon,\lambda}\|_\infty
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)) \\
&\quad + \frac12
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2,
\end{aligned} \\
\label{e3.6}
\begin{aligned}
\cos u_{\epsilon,\lambda}(t)-\cos\| u_{\epsilon,\lambda}\|_\infty
&\ge \sin\| u_{\epsilon,\lambda}\|_\infty
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))\\
&\quad + \frac12(1-\kappa)
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2.
\end{aligned}
\end{gather}

\begin{lemma}\label{lem3.2}
 For $\lambda \gg 1$,
\begin{equation} \label{e3.7}
| II_1 | \le -\frac{\sqrt{2}\epsilon}{\lambda}
\log\sin\big(\frac{d_\lambda}{2}\big)
+  O(\lambda^{-1}).
\end{equation}
\end{lemma}

\begin{proof}
 By \eqref{e3.1} and Lemma \ref{lem3.1},
\begin{equation} \label{e3.8}
\begin{aligned}
| II_1 | & \le \epsilon\int_{t_\delta}^0 \frac{B(t)u_{\epsilon,\lambda}'(t)}
{2A(u_{\epsilon,\lambda}(t))^{3/2}}dt
= X_1 + X_2
\\
&:=
\frac{\epsilon}{\sqrt{2}\lambda}
\int_{t_\delta}^0 \frac{\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)}
{\cos u_{\epsilon,\lambda}(t) - \cos\| u_{\epsilon,\lambda}\|_\infty}
u_{\epsilon,\lambda}'(t)dt
\\
&\quad + \epsilon^2
\int_{t_\delta}^0 \frac{(\| u_{\epsilon,\lambda}\|_\infty
- u_{\epsilon,\lambda}(t))^2}
{(\lambda(\cos u_{\epsilon,\lambda}(t)
- \cos\| u_{\epsilon,\lambda}\|_\infty)^{3/2}}
u_{\epsilon,\lambda}'(t)dt
\\
&= \frac{\epsilon}{\sqrt{2}\lambda}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{\| u_{\epsilon,\lambda}\|_\infty-\theta}
{\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty}d\theta
\\
&\quad + \frac{\epsilon^2}{\lambda^{3/2}}
\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{(\| u_{\epsilon,\lambda}\|_\infty-\theta)^2}
{(\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty)^{3/2}}d\theta.
\end{aligned}
\end{equation}
We first calculate $X_1$. We put
\begin{equation} \label{e3.9}
\begin{aligned}
X_1 &= Q_1 + Q_2 \\
&:=\frac{\epsilon}{\sqrt{2}\lambda}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{\| u_{\epsilon,\lambda}\|_\infty-\theta}
{\cos\theta - \cos\pi}d\theta
\\
&\quad + \frac{\epsilon}{\sqrt{2}\lambda}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\Big(
\frac{\| u_{\epsilon,\lambda}\|_\infty-\theta}
{\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty}
- \frac{\| u_{\epsilon,\lambda}\|_\infty-\theta}
{\cos\theta - \cos\pi}
\Big)
d\theta.
\end{aligned}
\end{equation}
We see that
\begin{equation} \label{e3.10}
\begin{aligned}
Q_1 &= Q_{11} + Q_{12} \\
&:=\frac{\epsilon}{\sqrt{2}\lambda}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{\| u_{\epsilon,\lambda}\|_\infty-\pi}
{\cos\theta + 1}d\theta
+\frac{\epsilon}{\sqrt{2}\lambda}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{\pi - \theta}
{\cos\theta + 1}d\theta.
\end{aligned}
\end{equation}
Then by \eqref{e2.17},
\begin{equation} \label{e3.11}
\begin{aligned}
Q_{11} &= \frac{-d_\lambda\epsilon}{\sqrt{2}\lambda}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{1}{\cos\theta + 1}d\theta \\
&= \frac{-d_\lambda\epsilon}{\sqrt{2}\lambda}\big[
\tan\frac{\theta}{2}
\big]_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}\\
&= \frac{-d_\lambda\epsilon}{\sqrt{2}\lambda}\big[
\frac{\cos(d_\lambda/2)}{\sin(d_\lambda/2)}
- \frac{\sin(\pi - d_\lambda - \delta)/2)}
{\cos(\pi - d_\lambda - \delta)/2)} \big]
= O(\lambda^{-1}).
\end{aligned}
\end{equation}
Next,
\begin{equation} \label{e3.12}
\begin{aligned}
Q_{12} &= \frac{\epsilon}{\sqrt{2}\lambda}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{\pi - \theta}{\cos\theta + 1}d\theta\\
&= \frac{\epsilon}{\sqrt{2}\lambda}
\int^{d_\lambda + \delta}
_{d_\lambda}\frac{y}{1-\cos y}dy\\
&= \frac{\epsilon}{\sqrt{2}\lambda}
\int^{d_\lambda + \delta}
_{d_\lambda} y\left(-\cot\frac{y}{2}\right)'dy
\\
&= \frac{\epsilon}{\sqrt{2}\lambda}
\Big(-(d_\lambda + \delta)\cot\frac{d_\lambda+\delta}{2}
+ d_\lambda\cot\frac{d_\lambda}{2}\\
&\quad + 2\log\sin\big(\frac{d_\lambda + \delta}{2}\big)
- 2\log\sin\big(\frac{d_\lambda}{2}\big)\Big) \\
&= -\frac{\sqrt{2}\epsilon}{\lambda}\log\sin\frac{d_\lambda}{2}
+ O(\lambda^{-1}).
\end{aligned}
\end{equation}
Now, we calculate $Q_2$.
\begin{equation} \label{e3.13}
\begin{aligned}
Q_2 &= \frac{\epsilon}{\sqrt{2}\lambda}
(1 + \cos\| u_{\epsilon,\lambda}\|_\infty)
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{1}{\cos\theta + 1}
\cdot\frac{\| u_{\epsilon,\lambda}\|_\infty - \theta}
{\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty}d\theta
\\
&\le \frac{\epsilon}{\sqrt{2}\lambda}(1 - \cos d_\lambda)
\frac{1}{\sin \| u_{\epsilon,\lambda}\|_\infty}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{1}{\cos\theta + 1}
d\theta
\\
&\leq C\epsilon d_\lambda\lambda^{-1}\big[\tan\frac{\theta}{2}\big]
_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}\\
&= C\epsilon
\lambda^{-1}d_\lambda\left(\frac{\cos(d_\lambda/2)}{\sin(d_\lambda/2)}
- \tan\frac{\pi - d_\lambda-\delta}{2} \right)
=  O(\lambda^{-1}).
\end{aligned}
\end{equation}
By \eqref{e3.9}--\eqref{e3.13}, we obtain
\begin{equation} \label{e3.14}
X_1 \leq  -\frac{\sqrt{2}\epsilon}{\lambda}
\log\sin\frac{d_\lambda}{2}
+  O(\lambda^{-1}).
\end{equation}
Finally, we calculate $X_2$. By \eqref{e2.17} and \eqref{e3.6},
\begin{align*}
X_2 &= \epsilon^2\lambda^{-3/2}\\
&\times\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{(\| u_{\epsilon,\lambda}\|_\infty - \theta)^2}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + (1/2)(1 -\kappa)
(\| u_{\epsilon,\lambda}\|_\infty - \theta))^{3/2}
(\| u_{\epsilon,\lambda}\|_\infty - \theta)^{3/2}}d\theta
\\
&\leq C\epsilon^2\lambda^{-3/2}
\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{(\| u_{\epsilon,\lambda}\|_\infty - \theta)^{1/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty +
(\| u_{\epsilon,\lambda}\|_\infty - \theta))^{3/2}}d\theta
\\
&= C\epsilon^2\lambda^{-3/2}\int_0^\delta \frac{y^{1/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + y)^{3/2}}dy
\\
&\le C\epsilon^2\lambda^{-3/2}\int_0^\delta \frac{1}
{{\sin\| u_{\epsilon,\lambda}\|_\infty + y}}dy\\
&\le C\epsilon^2
\lambda^{-3/2}| \log\sin\| u_{\epsilon,\lambda}\|_\infty|
= O(\lambda^{-1}).
\end{align*}
By this and \eqref{e3.14}, we obtain \eqref{e3.7}.
Thus the proof is complete.
\end{proof}

We estimate $II_1$ from below. To do this, we need the following lemma.

\begin{lemma}\label{lem3.3}
 For $\lambda \gg 1$ and $t_\delta < t < 0$,
\begin{equation} \label{e3.15}
\begin{aligned}
B(t) &\geq  \frac{\sqrt{\lambda}}{2}
\sqrt{-\cos u_{\epsilon,\lambda}(t)}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2\\
&\quad
- \frac12\sqrt{\lambda}\frac{\sin\| u_{\epsilon,\lambda}\|_\infty}
{\sqrt{-\cos u_{\epsilon,\lambda}(t)}}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)).
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
 We recall that for constants $a, b > 0$,
\begin{equation} \label{e3.16}
\begin{aligned}
&\int \sqrt{ax^2 + bx} dx \\
&= \frac{2ax+b}{4a}\sqrt{ax^2+bx}
- \frac{b^2}{8a}\frac{1}{\sqrt{a}}\log
\big| 2ax + b + 2\sqrt{a(ax^2+bx)}\big|.
\end{aligned}
\end{equation}
By Taylor expansion, for
$\| u_{\epsilon,\lambda}\|_\infty - \delta \le u_{\epsilon,\lambda}(t)
\le \theta \le \| u_{\epsilon,\lambda}\|_\infty$, we have
\[ % \label{e3.17}
\cos\theta -\cos\| u_{\epsilon,\lambda}\|_\infty
\ge \sin \| u_{\epsilon,\lambda}\|_\infty
(\| u_{\epsilon,\lambda}\|_\infty - \theta)
- \frac12\cos u_{\epsilon,\lambda}(t)
(\| u_{\epsilon,\lambda}\|_\infty - \theta)^2.
\]
By this, \eqref{e2.7}--\eqref{e2.9} and \eqref{e3.16}, for
 $t_\delta \le t \le 0$,
\begin{equation} \label{e3.18}
\begin{aligned}
B(t) &=  \int_t^0
\sqrt{2A(u_{\epsilon,\lambda}(t)) + 2\epsilon B(t)}u_{\epsilon,\lambda}'(t)dt
\\
&\geq  \sqrt{2\lambda}\int_{u_{\epsilon,\lambda}(t)}^{\| u_{\epsilon,\lambda}\|_\infty}
\sqrt{\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty}d\theta
\\
&\geq  \sqrt{\lambda}\int_{0}^{\| u_{\epsilon,\lambda}\|_\infty 
 - u_{\epsilon,\lambda}(t)}
\sqrt{-x^2\cos u_{\epsilon,\lambda}(t) 
+ 2x\sin\| u_{\epsilon,\lambda}\|_\infty}dx
\\
&=
\sqrt{\lambda}\Big(\frac{1}{2}z
- \frac{\sin\| u_{\epsilon,\lambda}\|_\infty}
{2\cos u_{\epsilon,\lambda}(t)}
\Big)
\sqrt{-z^2\cos u_{\epsilon,\lambda}(t)
+ 2z\sin \| u_{\epsilon,\lambda}\|_\infty}
\\
&\quad - \frac{\sqrt{\lambda}}{2}
\frac{\sin^2\| u_{\epsilon,\lambda}\|_\infty}
{(-\cos u_{\epsilon,\lambda}(t))^{3/2}}
\big\{\log(R_\lambda(u_{\epsilon,\lambda}(t))\\
&\quad + 2\sin\| u_{\epsilon,\lambda}\|_\infty)
- \log(2\sin\| u_{\epsilon,\lambda}\|_\infty)\big\},
\end{aligned}
\end{equation}
where
\begin{equation} \label{e3.19}
R_\lambda(u_{\epsilon,\lambda}(t))
=-2z\cos u_{\epsilon,\lambda}(t)
+ 2\sqrt{z^2\cos^2 u_{\epsilon,\lambda}(t)
- 2z\cos u_{\epsilon,\lambda}(t)
\sin\| u_{\epsilon,\lambda}\|_\infty}
\end{equation}
and
$z := \| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)$.
We know that $\log(1 + x) \le x$
for $x \ge 0$. By this,
\begin{align*} % \label{e}
&\log(R_\lambda(u_{\epsilon,\lambda}(t))
+ 2\sin\| u_{\epsilon,\lambda}\|_\infty)
- \log(2\sin\| u_{\epsilon,\lambda}\|_\infty)
\\
&= \log\Big(1 + \frac
{-2z\cos u_{\epsilon,\lambda}(t) + 2\sqrt{z^2\cos^2 u_{\epsilon,\lambda}(t)
- 2z\cos u_{\epsilon,\lambda}(t)\sin\| u_{\epsilon,\lambda}\|_\infty}}
{2\sin\| u_{\epsilon,\lambda}\|_\infty}
\Big)
\\
&\le \frac{-z\cos u_{\epsilon,\lambda}(t) + \sqrt{-\cos u_{\epsilon,\lambda}(t)}
\sqrt{-z^2\cos u_{\epsilon,\lambda}(t)
+ 2z\sin\| u_{\epsilon,\lambda}\|_\infty}}
{\sin\| u_{\epsilon,\lambda}\|_\infty}.
\end{align*}
By this and \eqref{e3.18}, we obtain \eqref{e3.15}.
Thus the proof is complete.
\end{proof}

\begin{lemma}\label{lem3.4}
For $\lambda \gg 1$,
\begin{equation} \label{e3.21}
| II_1 | \ge  -\frac{\sqrt{2}\epsilon}{\lambda}\log\sin
\frac{d_\lambda}{2}
+ O(\lambda^{-1}).
\end{equation}
\end{lemma}

\begin{proof} By \eqref{e3.1}, we have
\begin{equation}
\begin{aligned} \label{e3.22}
| II_1 |
&\geq  \epsilon
\int_{t_\delta}^0 \frac{B(t)u_{\epsilon,\lambda}'(t)}
{2\sqrt{A(u_{\epsilon,\lambda}(t)}
(A(u_{\epsilon,\lambda}(t))+\epsilon B(t))}dt
:= II_{1,1} - II_{1,2}\\
&= \epsilon
\int_{t_\delta}^0 \frac{B(t)u_{\epsilon,\lambda}'(t)}
{2A(u_{\epsilon,\lambda}(t))^{3/2}}dt\\
&\quad +
\epsilon\Big(\int_{t_\delta}^0
\frac{B(t)u_{\epsilon,\lambda}'(t)}{2\sqrt{A(u_{\epsilon,\lambda}(t))}
(A(u_{\epsilon,\lambda}(t))+\epsilon B(t))}dt
- \int_{t_\delta}^0 \frac{B(t)u_{\epsilon,\lambda}'(t)}
{2A(u_{\epsilon,\lambda}(t))^{3/2}}dt\Big).
\end{aligned}
\end{equation}
By Lemma \ref{lem3.3}, we put
\begin{equation} \label{e3.23}
\begin{aligned}
II_{1,1} &=  II_{1,2,1} - II_{1,2,2}
\\
&=  \frac{\epsilon}{2}\sqrt{\lambda}
\int_{t_\delta}^0
\frac{\sqrt{-\cos u_{\epsilon,\lambda}(t)}
(\| u_{\epsilon,\lambda}\|_\infty - u _\lambda(t))^2u_{\epsilon,\lambda}'(t)}
{2\lambda^{3/2}(\cos u_{\epsilon,\lambda}(t)
 - \cos\| u_{\epsilon,\lambda}\|_\infty)^{3/2}}dt
\\
&\quad- \frac{\sqrt{\lambda}\epsilon}{2}
\sin\| u_{\epsilon,\lambda}\|_\infty
\int_{t_\delta}^0
\frac{(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))
u_{\epsilon,\lambda}'(t)}
{2\sqrt{-\cos u_{\epsilon,\lambda}(t)}\lambda^{3/2}
(\cos u_{\epsilon,\lambda}(t) - \cos\| u_{\epsilon,\lambda}\|_\infty)^{3/2}}dt.
\end{aligned}
\end{equation}
By Taylor expansion, for $t_\delta \le t \le 0$ and $0 < \eta \ll 1$,
\begin{gather} \label{e3.24}
\sqrt{-\cos\| u_{\epsilon,\lambda}\|_\infty}
- \sqrt{-\cos u_{\epsilon,\lambda}(t)} \le \frac{1 + \eta}{2}
\sin u_{\epsilon,\lambda}(t)
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)),
\\  \label{e3.25}
\sqrt{-\cos\| u_{\epsilon,\lambda}\|_\infty} =
(\cos d_\lambda)^{1/2} = 1 - \frac{1}{4}(1 + o(1))d_\lambda^2.
\end{gather}
By \eqref{e3.5}, \eqref{e3.23} and \eqref{e3.24},
\begin{equation} \label{e3.26}
\begin{aligned}
II_{1,2,1}
&\geq \frac{\epsilon}
{4\lambda}\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{\sqrt{-\cos \theta}
(\| u_{\epsilon,\lambda}\|_\infty - \theta)^{1/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty
+ (1/2)(\| u_{\epsilon,\lambda}\|_\infty - \theta))^{3/2}}d\theta
\\
&= \frac{\epsilon}
{4\lambda}\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{\sqrt{-\cos\| u_{\epsilon,\lambda}\|_\infty}
(\| u_{\epsilon,\lambda}\|_\infty - \theta)^{1/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + (1/2)(\| u_{\epsilon,\lambda}\|_\infty - \theta))
^{3/2}} d\theta
\\
&\quad - \frac{\epsilon}
{4\lambda}\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{(\sqrt{-\cos\| u_{\epsilon,\lambda}\|_\infty} - \sqrt{-\cos\theta})
 (\| u_{\epsilon,\lambda}\|_\infty - \theta)^{1/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty
+ (1/2)(\| u_{\epsilon,\lambda}\|_\infty - \theta))
^{3/2}}d\theta
\\
&\geq
\frac{\epsilon}{4\lambda}\sqrt{-\cos\| u_{\epsilon,\lambda}\|_\infty}
\int_0^{\delta} \frac{y^{1/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + (1/2)y)^{3/2}}dy
\\
&\quad - \frac{1+\eta}{8\lambda}\epsilon
\sin(\| u_{\epsilon,\lambda}\|_\infty - \delta)
\int_0^{\delta} \frac{y^{3/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + (1/2)y)^{3/2}}dy.
\end{aligned}
\end{equation}
Then
\begin{equation} \label{e3.27}
\begin{aligned}
\frac{\epsilon}{4\lambda}\int_0^{\delta} \frac{y^{1/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + (1/2)y)^{3/2}}dy
&=  \frac{\epsilon}{\sqrt{2}\lambda}\int_0^{\delta} \frac{y^{1/2}}
{(2\sin\| u_{\epsilon,\lambda}\|_\infty + y)^{3/2}}dy
\\
&=  \frac{\sqrt{2}\epsilon}{\lambda}
\int_0^{\sqrt{\delta/(2\sin\| u_{\epsilon,\lambda}\|_\infty)}}
\frac{z^2}{(1 + z^2)^{3/2}}dz
\\
&=  -\frac{\sqrt{2}\epsilon}
{\lambda}\log\sin\| u_{\epsilon,\lambda}\|_\infty
+ O(\lambda^{-1}).
\end{aligned}
\end{equation}
Further, by \eqref{e2.17},
\begin{equation} \label{e3.28}
\frac{1+\eta}{8\lambda}\epsilon
\sin(\| u_{\epsilon,\lambda}\|_\infty - \delta)
\int_0^{\delta} \frac{y^{3/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + (1/2)y)^{3/2}}dy
\le C\epsilon\lambda^{-1}.
\end{equation}
By \eqref{e3.26}--\eqref{e3.28},
\begin{equation} \label{e3.29}
\begin{aligned}
II_{1,2,1} &\geq  -\frac{\sqrt{2}\epsilon}{\lambda}
\log\sin\| u_{\epsilon,\lambda}\|_\infty
+ O(\lambda^{-1})
\\
&=  -\frac{\sqrt{2}\epsilon}{\lambda}\log\sin d_\lambda
+ O(\lambda^{-1})
\\
&=  -\frac{\sqrt{2}\epsilon}{\lambda}
\Big(\log 2 + \sin\frac{d_\lambda}{2}
+ \cos\frac{d_\lambda}{2}\Big)
+ O(\lambda^{-1})
\\
&= -\frac{\sqrt{2}\epsilon}{\lambda}\log\sin\frac{d_\lambda}{2}
+ O(\lambda^{-1}).
\end{aligned}
\end{equation}
Next, by \eqref{e3.6},
\begin{equation} \label{e3.30}
\begin{aligned}
II_{1,2,2} &\leq  C\epsilon\lambda^{-1}
\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{\sin\| u_{\epsilon,\lambda}\|_\infty}
{(\sin\| u_{\epsilon,\lambda}\|_\infty +
(\| u_{\epsilon,\lambda}\|_\infty - \theta))^{3/2}
(\| u_{\epsilon,\lambda}\|_\infty - \theta)^{1/2}}d\theta
\\
&\leq  C\epsilon\lambda^{-1}
\int_0^{\sqrt{\delta/(\sin\| u_{\epsilon,\lambda}\|_\infty)}}
\frac{1}{(1 + z^2)^{3/2}}dz \le C\epsilon\lambda^{-1}.
\end{aligned}
\end{equation}
Finally, by Lemma \ref{lem3.1} and \eqref{e3.22}, for
$z_\lambda(u_{\epsilon,\lambda}(t))
:= \| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)$,
\begin{equation} \label{e3.31}
\begin{aligned}
| II_{1,2}|
&=  \epsilon^2\int_{t_\delta}^0
\frac{B(t)^2u_{\epsilon,\lambda}'(t)}{A(u_{\epsilon,\lambda}(t))^{5/2}}dt
\\
&\leq  C\epsilon^2\int_{t_\delta}^0
\frac{A(u_{\epsilon,\lambda}(t))z_\lambda(u_{\epsilon,\lambda}(t))^2
+ \epsilon^2z_\lambda(u_{\epsilon,\lambda}(t))^4}
{A(u_{\epsilon,\lambda}(t))^{5/2}}u_{\epsilon,\lambda}'(t)dt
\\
&=  C\epsilon^2\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty} \frac{z_\lambda(\theta)^2}
{A(\theta)^{3/2}}d\theta +
C\epsilon^4\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty} \frac{z_\lambda(\theta)^4}
{A(\theta)^{5/2}}d\theta
\\
&:= Y_1 + Y_2.
\end{aligned}
\end{equation}
Then by \eqref{e2.17}, \eqref{e3.6} and \eqref{e3.27},
\begin{equation} \label{e3.32}
\begin{aligned}
Y_1 &=  C\epsilon^2\lambda^{-3/2}
\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2}
{(\cos\theta - \cos| u_{\epsilon,\lambda}\|_\infty)^{3/2}}d\theta
\\
&\leq  C\epsilon^2\lambda^{-3/2}
\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^{1/2}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + (1/2)(1 -\kappa)
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)))^{3/2}}
d\theta
\\
&\leq  C\epsilon^2\lambda^{-3/2}\int_0^\delta
\frac{y^{1/2}}{(\sin\| u_{\epsilon,\lambda}\|_\infty + y)^{3/2}}dy
\\
&\leq  \epsilon^2
\lambda^{-3/2}(C + \log\sin\| u_{\epsilon,\lambda}\|_\infty)
= O(\lambda^{-1}).
\end{aligned}
\end{equation}
By the same argument as that just above, by \eqref{e2.17} and \eqref{e3.6}, 
we obtain
\begin{align*} % \label{e}
Y_2 &=  C\epsilon^4\lambda^{-5/2}
\int_{\| u_{\epsilon,\lambda}\|_\infty -\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^4}
{(\cos\theta - \cos| u_{\epsilon,\lambda}\|_\infty)^{5/2}}d\theta
\\
&\leq  C\epsilon^4\lambda^{-5/2}
\Big(\int_0^{\delta}\frac{1}
{\sin\| u_{\epsilon,\lambda}\|_\infty + y}dy + C\Big)
\\
&\leq  C\epsilon^4
\lambda^{-5/2}
\left(| \log\sin\| u_{\epsilon,\lambda}\|_\infty| + C\right)
=  O(\lambda^{-3/2}).
\end{align*}
Thus the proof is complete.
\end{proof}

\begin{lemma}\label{lem3.5}
For $\lambda \gg 1$, we have $ | II_2| \leq  C\lambda^{-1}$.
\end{lemma}

\begin{proof} By \eqref{e3.1} and Lemma \ref{lem3.1},
\begin{align*}
&| II_2| \\
&= \int_{-1}^{t_\delta}
\frac{\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t)) + \epsilon B(t)}
\sqrt{A(u_{\epsilon,\lambda}(t))}
(\sqrt{A(u_{\epsilon,\lambda}(t)) + \epsilon B(t)}
+ \sqrt{A(u_{\epsilon,\lambda}(t))})}dt
\\
&\le C\epsilon\int_{-1}^{t_\delta}
\frac{B(t)u_{\epsilon,\lambda}'(t)}{2A(u_{\epsilon,\lambda}(t))^{3/2}}dt
\\
&\le C\epsilon\int_0^{\| u_{\epsilon,\lambda}\|_\infty-\delta}
\frac{2\epsilon(\| u_{\epsilon,\lambda}\|_\infty - \theta)^2 +
\sqrt{2\lambda(\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty)}
(\| u_{\epsilon,\lambda}\|_\infty - \theta)}
{(\lambda(\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty))
^{3/2}}d\theta
\\
&\le C\epsilon(\lambda^{-3/2} + \lambda^{-1}).
\end{align*}
The proof is complete.
\end{proof}

Now Lemma \ref{lem2.2} follows from Lemmas \ref{lem3.2}--\ref{lem3.5}.
The proof is complete.

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

Let $f(y) = | y|$ in this section.
We fix $0 < \epsilon \le 1$. Further, we assume that $\lambda \gg 1$ and
we write $u_{\epsilon,\lambda} = u_{2,\epsilon,\lambda}$ for simplicity.
We consider the solution $u_{\epsilon,\lambda}(t)$
with $\| u_{\epsilon,\lambda}\|_\infty < \pi$.
We know that the properties \eqref{e2.1}--\eqref{e2.4} are also valid.
By the same argument as that to obtain \eqref{e2.5}, we have
\begin{equation} \label{e4.1}
\frac12u_{\epsilon,\lambda}'(t)^2 =
\lambda(\cos u_{\epsilon,\lambda}(t)
- \cos\| u_{\epsilon,\lambda}\|_\infty)
- \epsilon\int_t^{0} | u_{\epsilon,\lambda}'(s)|^{2}ds.
\end{equation}
Then by \eqref{e2.6}--\eqref{e2.8}, for $-1 \le t \le 0$,
\begin{equation} \label{e4.2}
u_{\epsilon,\lambda}'(t) = \sqrt{2(A(u_{\epsilon,\lambda}(t)) -
\epsilon B(t))}.
\end{equation}
By this,
\begin{equation} \label{e4.3}
1 =  \int_{-1}^{0} dt
= \frac{1}{\sqrt{2}}
\int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t))
 - \epsilon B(t)}}dt
= \frac{1}{\sqrt{2}}(I + III),
\end{equation}
where
\begin{gather} \label{e4.4}
I =  \int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t))}}dt,
\\ \label{e4.5}
\begin{aligned}
III &=  \int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t)) - \epsilon B(t)}}dt
- \int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A(u_{\epsilon,\lambda}(t))}}dt
\\
&=  \int_{-1}^{0} \frac{\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t))}\sqrt{A(u_{\epsilon,\lambda}(t))
- \epsilon B(t)}}\\
&\quad\times\frac{1}{
\big(\sqrt{A(u_{\epsilon,\lambda}(t))}+\sqrt{A(u_{\epsilon,\lambda}(t))
- \epsilon B(t)}\big)}dt.
\end{aligned}
\end{gather}

Let $0 < \delta \ll 1$ be fixed. Further, let
$-1 < t_\delta < 0$ satisfy $u_{\epsilon,\lambda}(t_\delta)
= \| u_{\epsilon,\lambda}\|_\infty$. We put
\begin{equation} \label{e4.6}
\begin{aligned}
&III = III_1 + III_2
\\
&:= \int_{t_\delta}^0 \frac{\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t))}\sqrt{A(u_{\epsilon,\lambda}(t))
- \epsilon B(t)}
(\sqrt{A(u_{\epsilon,\lambda}(t))}
+\sqrt{A(u_{\epsilon,\lambda}(t)) - \epsilon B(t)})}dt
\\
&+
\int_{-1}^{t_\delta} \frac{\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t))}\sqrt{A(u_{\epsilon,\lambda}(t))
- \epsilon B(t)}
\big(\sqrt{A(u_{\epsilon,\lambda}(t))}
+\sqrt{A(u_{\epsilon,\lambda}(t)) - \epsilon B(t)}\big)}dt.
\end{aligned}
\end{equation}

\begin{lemma}\label{lem4.1}
For $\lambda \gg 1$ and $-1 < t < 0$,
\begin{equation} \label{e4.7}
B(t) \leq  \sqrt{2A(u_{\epsilon,\lambda}(t))}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)).
\end{equation}
\end{lemma}

\begin{proof}
By \eqref{e2.2}, \eqref{e2.8} and \eqref{e4.2},
\begin{align*}
B(t) &=  \int_t^0 u_{\epsilon,\lambda}'(s)^2ds\\
&\le \int_t^0 \sqrt{2A(u_{\epsilon,\lambda}(s))}u_{\epsilon,\lambda}'(s)ds
\\
&\leq  \sqrt{2A(u_{\epsilon,\lambda}(t))}
\int_t^0 u_{\epsilon,\lambda}'(s)ds \\
&= \sqrt{2A(u_{\epsilon,\lambda}(t))}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)).
\end{align*}
Thus the proof if complete.
\end{proof}


By \eqref{e3.6} and Lemma \ref{lem4.1}, we obtain the estimate of
$III_1$ from above as follows. Indeed, for $t_\delta \le t \le 0$,
\begin{equation} \label{e4.8}
\begin{aligned}
A(u_{\epsilon,\lambda}(t)) - \epsilon B(t)
&\geq  A(u_{\epsilon,\lambda}(t)) - \epsilon\sqrt{2A(u_{\epsilon,\lambda}(t)}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))
\\
&= A(u_{\epsilon,\lambda}(t))\Big(1 - \frac{\sqrt{2}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))}
{\sqrt{A(u_{\epsilon,\lambda}(t))}}
\Big)
\\
&\geq  A(u_{\epsilon,\lambda}(t))
\Big(1 -\frac
{2(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))}
{\sqrt{(\lambda(1-\kappa)/2)}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))}
\Big)
\\
&\geq  A(u_{\epsilon,\lambda}(t))
\Big(1 -\frac{C}{\sqrt{\lambda}}\Big).
\end{aligned}
\end{equation}
By this and \eqref{e4.6},
\begin{align*}
III_1 &\leq
\int_{t_\delta}^{0} \frac{\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{2\sqrt{A(u_{\epsilon,\lambda}(t))}(A(u_{\epsilon,\lambda}(t))
- \epsilon B(t))}
\\
&\leq
\int_{t_\delta}^{0} \frac{\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{2\sqrt{A(u_{\epsilon,\lambda}(t))}
A(u_{\epsilon,\lambda}(t))(1 - C/\sqrt{\lambda})}dt
\\
&\leq
\Big(1 + \frac{C}{\sqrt{\lambda}}\Big)
\int_{t_\delta}^{0} \frac{\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{2\sqrt{A(u_{\epsilon,\lambda}(t))}A(u_{\epsilon,\lambda}(t))}dt.
\end{align*}
By this, Lemma \ref{lem4.1} and the same argument as that used
in Lemma \ref{lem3.2},
for $\lambda \gg 1$, we obtain
\begin{equation} \label{e4.9}
III_1 \le -\frac{\epsilon\sqrt{2}}{\lambda}
\log\sin\frac{d_\lambda}{2} + O(\lambda^{-1}).
\end{equation}
Furthermore, by \eqref{e4.3} and \eqref{e4.8},
\begin{align*}
1 &\leq  \frac{1}{\sqrt{2}}\int_{-1}^0
\frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A(u_{\epsilon,\lambda}(t))}
(1 - C/\sqrt{\lambda})}dt
\\
&\leq  \frac{1}{\sqrt{2}}\Big(1 + \frac{C}{\sqrt{\lambda}}\Big)
\int_{-1}^0
\frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A(u_{\epsilon,\lambda}(t))}}dt.
\end{align*}
By this and \eqref{e2.14}--\eqref{e2.16}, we obtain
\begin{equation} \label{e4.10}
\sin\frac{d_\lambda}{2} \le Ce^{-\sqrt{\lambda}},
\quad
\sin\| u_{\epsilon,\lambda}\|_\infty \le Ce^{-\sqrt{\lambda}}.
\end{equation}

\begin{lemma}\label{lem4.2}
 For $\lambda \gg 1$ and $t_\delta < t < 0$,
\begin{equation} \label{e4.11}
\begin{aligned}
B(t) &\geq  \sqrt{2}\int_{u_{\epsilon,\lambda}(t)}
^{\| u_{\epsilon,\lambda}\|_\infty}
\sqrt{\lambda(\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty)}d\theta
\\
&\quad - 2\sqrt{\epsilon}A(u_{\epsilon,\lambda}(t))^{1/4}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^{3/2}.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof} By \eqref{e2.8}, \eqref{e4.2} and \eqref{e4.7},
\begin{equation}  \label{e4.12}
\begin{aligned}
B(t) &=  \int_t^0 u_{\epsilon,\lambda}'(s)^2ds
\\
&\geq \int_t^0 \sqrt{2A(u_{\epsilon,\lambda}(s))}u_{\epsilon,\lambda}'(s)ds
- \int_t^0\sqrt{2\epsilon B(s)}u_{\epsilon,\lambda}'(s)ds
\\
&\geq \int_t^0 \sqrt{2A(u_{\epsilon,\lambda}(s))}u_{\epsilon,\lambda}'(s)ds
- \sqrt{2\epsilon B(t)}\int_t^0u_{\epsilon,\lambda}'(s)ds
\\
&\geq
\int_t^0 \sqrt{2A(u_{\epsilon,\lambda}(s))}u_{\epsilon,\lambda}'(s)ds
- \sqrt{2\epsilon B(t)}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))
\\
&\geq
\int_t^0 \sqrt{2A(u_{\epsilon,\lambda}(s))}u_{\epsilon,\lambda}'(s)ds
- 2\sqrt{\epsilon}A(u_{\epsilon,\lambda}(t))^{1/4}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^{3/2}
\\
&:= W_1 - W_2.
\end{aligned}
\end{equation}
The proof is complete.
\end{proof}




\begin{proof}[Proof of Theorem \ref{thm1.2}]
We calculate the estimate of $III_1$ from below.
By \eqref{e3.6}, \eqref{e4.6}, \eqref{e4.8}, \eqref{e4.10} and
\eqref{e4.12}
\begin{equation} \label{e4.13}
\begin{aligned}
&\epsilon\int_{t_\delta}^0 \frac{W_2 u_{\epsilon,\lambda}'(t)}
{2\sqrt{A(u_{\epsilon,\lambda}(t))}(A(u_{\epsilon,\lambda}(t))
- \epsilon B(t))}dt
\\
&\le C\int_{t_\delta}^0
\frac{A(u_{\epsilon,\lambda}(t))^{1/4}
(\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^{3/2}}
{\sqrt{A(u_{\epsilon,\lambda}(t))}
(A(u_{\epsilon,\lambda}(t))(1 - C/\sqrt{\lambda})}u_{\epsilon,\lambda}'(t)dt
\\
&\le C
\Big(1 + \frac{C}{\sqrt{\lambda}}\Big)
\lambda^{-5/4}\int_{\| u_{\epsilon,\lambda}\|_\infty-\delta}
^{\| u_{\epsilon,\lambda}\|_\infty}
\frac{(\| u_{\epsilon,\lambda}\|_\infty - \theta)^{3/2}}
{(\cos\theta - \cos\| u_{\epsilon,\lambda}\|_\infty)^{5/4}}d\theta
\\
&\le C\Big(1 + \frac{C}{\sqrt{\lambda}}\Big)
\lambda^{-5/4}\int_0^\delta \frac{y^{1/4}}
{(\sin\| u_{\epsilon,\lambda}\|_\infty + y)^{5/4}}dy
\\
&\le C\Big(1 + \frac{C}{\sqrt{\lambda}}\Big)
\lambda^{-5/4}\int_0^{\delta/\sin\| u_{\epsilon,\lambda}\|_\infty}
\frac{z^{1/4}}{(1+z)^{5/4}}dz
\\
&\le C\Big(1 + \frac{C}{\sqrt{\lambda}}\Big)
\lambda^{-5/4}
| \log \sin\| u_{\epsilon,\lambda}\|_\infty |
\\
&\le C\lambda^{-3/4}.
\end{aligned}
\end{equation}
By \eqref{e4.8}, \eqref{e4.12} and Lemmas \ref{lem3.3} and \ref{lem3.4},
for $\lambda \gg 1$,
\begin{equation} \label{e4.14}
\epsilon\int_{t_\delta}^0 \frac{W_1 u_{\epsilon,\lambda}'(t)}
{2\sqrt{A(u_{\epsilon,\lambda}(t))}(A(u_{\epsilon,\lambda}(t))
- \epsilon B(t))}dt
\le \frac{-\epsilon\sqrt{2}}{\lambda}\log\sin\frac{d_\lambda}{2}
+ O(\lambda^{-1}).
\end{equation}
Further, by \eqref{e4.6}, \eqref{e4.8} and Lemma \ref{lem3.5},
for $\lambda \gg 1$, we obtain
\begin{equation} \label{e4.15}
III_2 \le C\lambda^{-1}.
\end{equation}
By \eqref{e4.6}, \eqref{e4.9} and \eqref{e4.13}--\eqref{e4.15}, we obtain
\begin{equation} \label{e4.16}
III = \frac{-\epsilon\sqrt{2}}{\lambda}\log\sin\frac{d_\lambda}{2}
+ O(\lambda^{-3/4}).
\end{equation}
By this, \eqref{e2.14} and the same argument as
\eqref{e2.18}--\eqref{e2.20},
we obtain \eqref{e1.9}. The proof is complete.
\end{proof}

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

In this section, let $f(y) = -y$. We write
$t_\lambda = t_{\epsilon,\lambda} < 0$ for simplicity.
The proof of the Theorem \ref{thm1.3} is almost the same as
those of Theorems \ref{thm1.1} and \ref{thm1.2}.
We begin with the fundamental properties of $u_{\epsilon,\lambda}$.


\begin{lemma}\label{lem5.1}
\begin{itemize}

\item[(i)]
$u_{\epsilon,\lambda}'(t) > 0$ for $-1 \le t < t_\lambda$ and
$u_{\epsilon,\lambda}'(t) > 0$ for $t_\lambda < t < 1$.

\item[(ii)] $u_{\epsilon,\lambda}(t) \to \pi$ locally uniformly in $I$ as
$\lambda \to \infty$.

\item[(iii)] $t_\lambda < 0$ and $t_\lambda \to 0$ as $\lambda \to \infty$.
\end{itemize}
\end{lemma}

Since the proof of Lemma \ref{lem5.1} is quite easy, we omit it. To prove
Theorem \ref{thm1.3}, we repeat the same arguments as those in Sections 3 and 4.
We see that
\begin{equation} \label{e5.1}
1 + t_\lambda =  \int_{-1}^{t_\lambda} dt
= \frac{1}{\sqrt{2}}
\int_{-1}^{t_\lambda}
\frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A(u_{\epsilon,\lambda}(t) +
\epsilon B(t)}}dt
= \frac{1}{\sqrt{2}}(P + Q),
\end{equation}
where
\begin{gather} \label{e5.2}
P =  \int_{-1}^{t_\lambda}
\frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A(u_{\epsilon,\lambda}(t)}}dt,
\\  \label{e5.3}
\begin{aligned}
Q &=  \int_{-1}^{t_\lambda} \frac{u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t) + \epsilon B(t)}}dt
- \int_{-1}^{t_\lambda} \frac{u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t)}}dt
\\
&=  \int_{-1}^{t_\lambda} \frac{-\epsilon B(t)u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t)}
\sqrt{A(u_{\epsilon,\lambda}(t) + \epsilon B(t)}
(\sqrt{A(u_{\epsilon,\lambda}(t)}
+\sqrt{A(u_{\epsilon,\lambda}(t) + \epsilon B(t)})
}dt.
\end{aligned}
\end{gather}

Then it is clear that $P = I$ in \eqref{e2.11}.
Furthermore, by using the same argument as that in Section 3,
it is easy to show that $Q = II + O(\lambda^{-1})$. We also find that
\begin{equation} \label{e5.4}
1 - t_\lambda =  \int_{t_\lambda}^{1} dt
= \frac{1}{\sqrt{2}}
\int_{t_\lambda}^1 \frac{-u_{\epsilon,\lambda}'(t)}
{\sqrt{A(u_{\epsilon,\lambda}(t) - \epsilon B(t)}}dt
= \frac{1}{\sqrt{2}}(R + S),
\end{equation}
where $R = I$ and $S = III + O(\lambda^{-3/4})$.
By \eqref{e5.1} and \eqref{e5.4}, we obtain
\begin{equation} \label{e5.5}
2t_\lambda = \sqrt{2}Q + O(\lambda^{-3/4})
 = \sqrt{2}II + O(\lambda^{-3/4}) =
\frac{2\epsilon}{\lambda}
\log\big(\sin\frac{d_\lambda}{2}\big) + O(\lambda^{-3/4}),
\end{equation}
which implies
\begin{equation} \label{e5.6}
t_\lambda = \frac{\epsilon}{\lambda}
\log\big(\sin\frac{d_\lambda}{2}\big) + O(\lambda^{-3/4}).
\end{equation}
By this and \eqref{e5.1}, we obtain
\begin{equation} \label{e5.7}
1 = \frac{1}{\sqrt{2}}I + O(\lambda^{-3/4}).
\end{equation}
By this and Lemma \ref{lem2.1}, we obtain \eqref{e1.11}.
By \eqref{e1.11} and \eqref{e5.6}, we obtain
\eqref{e1.10}. The proof is complete.


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\end{document}