\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 $\lambda \to \infty$, $$\label{e1.7} \| u_{0,\lambda}\|_\infty = \pi - 8e^{-\sqrt{\lambda}} - 32\sqrt{\lambda}e^{-3\sqrt{\lambda}} + o(\sqrt{\lambda}e^{-3\sqrt{\lambda}}).$$ 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 $\lambda \to \infty$, $$\label{e1.8} \| u_{1,\epsilon,\lambda}\|_\infty = \pi - 8e^{-\epsilon}e^{-\sqrt{\lambda}} + O(\lambda^{-1/2}e^{-\sqrt{\lambda}}).$$ \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 $\lambda \to \infty$, $$\label{e1.9} \| u_{2,\epsilon,\lambda}\|_\infty = \pi - 8e^{\epsilon}e^{-\sqrt{\lambda}} + O(\lambda^{-1/4}e^{-\sqrt{\lambda}}).$$ \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 \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} For $-1 \le t \le 0$, we obtain $$\label{e2.6} \frac12u_{\epsilon,\lambda}'(t)^2 = \lambda(\cos u_{\epsilon,\lambda}(t) - \cos\| u_{\epsilon,\lambda}\|_\infty) + \epsilon\int_t^{0} | u_{\epsilon,\lambda}'(s)|^{2}ds.$$ 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 $-1 \le t \le 0$, $$\label{e2.9} u_{\epsilon,\lambda}'(t) = \sqrt{2(A(u_{\epsilon,\lambda}(t)) + \epsilon B(t))}.$$ Then $$\label{e2.10} 1 = \int_{-1}^{0} dt = \frac{1}{\sqrt{2}} \int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A (u_{\epsilon,\lambda}(t)) +\epsilon B(t)}}dt = \frac{1}{\sqrt{2}}(I + II),$$ 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 $\lambda \to \infty$, $$\label{e2.13} I = \sqrt{\frac{2}{\lambda}} \Big(\log\frac{4}{\sin (d_\lambda/2)} + \frac14(1 + o(1)) \big(\log\frac{4}{\sin (d_\lambda/2)}\big) \sin^2\frac{d_\lambda}{2}\Big).$$ \end{lemma} \begin{proof} Put $\theta = u_{\epsilon,\lambda}(t)$. Then \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} 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 $$\label{e2.15} K(k) = \log\frac{4}{k'} + \frac14\big(\log\frac{4}{k'}\big){k'}^2 (1 + o(1)),$$ 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 $$\label{e2.16} 1 < \frac{1}{\sqrt{2}}I \le \frac{1}{\sqrt{\lambda}} \big(1 + C\sin^2\frac{d_\lambda}{2}\big) \log\frac{4}{\sin(d_\lambda/2)}.$$ Then $$\label{e2.17} \sin\frac{d_\lambda}{2} \le 4(1 + o(1))e^{-\sqrt{\lambda}}, \enskip \frac{d_\lambda}{2} \le 4(1 + o(1))e^{-\sqrt{\lambda}}, \enskip \sin\| u_{\epsilon,\lambda}\|_\infty \le 8(1 + o(1))e^{-\sqrt{\lambda}}.$$ \begin{lemma}\label{lem2.2} As $\lambda \to \infty$, $$\label{e2.18} II = \frac{\sqrt{2}\epsilon} {\lambda}\log\big(\sin \frac{d_\lambda}{2}\big) + O(\lambda^{-1}).$$ \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 \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} This implies $$\label{e2.20} \big(1 - \frac{\epsilon}{\sqrt{\lambda}}\big) \log\sin\frac{d_\lambda}{2} = \log 4 - \sqrt{\lambda} + O(\lambda^{-1/2}).$$ By this, \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} 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 \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} To obtain Lemma \ref{lem2.2}, we estimate $II_1$ and $II_2$ by series of lemmas. \begin{lemma}\label{lem3.1} For $-1 \le t \le 0$, $$\label{e3.2} B(t) \le \sqrt{2A(u_{\epsilon,\lambda}(t))} (\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)) + 2\epsilon (\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t))^2.$$ \end{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 $-1 \le t \le 0$, \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} 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 \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} 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 $\lambda \gg 1$, $$\label{e3.7} | II_1 | \le -\frac{\sqrt{2}\epsilon}{\lambda} \log\sin\big(\frac{d_\lambda}{2}\big) + O(\lambda^{-1}).$$ \end{lemma} \begin{proof} By \eqref{e3.1} and Lemma \ref{lem3.1}, \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} We first calculate $X_1$. We put \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} We see that \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} Then by \eqref{e2.17}, \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} Next, \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} Now, we calculate $Q_2$. \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} By \eqref{e3.9}--\eqref{e3.13}, we obtain $$\label{e3.14} X_1 \leq -\frac{\sqrt{2}\epsilon}{\lambda} \log\sin\frac{d_\lambda}{2} + O(\lambda^{-1}).$$ 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 $t_\delta < t < 0$, \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{lemma} \begin{proof} We recall that for constants $a, b > 0$, \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} 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 $t_\delta \le t \le 0$, \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} where $$\label{e3.19} R_\lambda(u_{\epsilon,\lambda}(t)) =-2z\cos u_{\epsilon,\lambda}(t) + 2\sqrt{z^2\cos^2 u_{\epsilon,\lambda}(t) - 2z\cos u_{\epsilon,\lambda}(t) \sin\| u_{\epsilon,\lambda}\|_\infty}$$ 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 $\lambda \gg 1$, $$\label{e3.21} | II_1 | \ge -\frac{\sqrt{2}\epsilon}{\lambda}\log\sin \frac{d_\lambda}{2} + O(\lambda^{-1}).$$ \end{lemma} \begin{proof} By \eqref{e3.1}, we have \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} By Lemma \ref{lem3.3}, we put \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} 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}, \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} Then \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} Further, by \eqref{e2.17}, $$\label{e3.28} \frac{1+\eta}{8\lambda}\epsilon \sin(\| u_{\epsilon,\lambda}\|_\infty - \delta) \int_0^{\delta} \frac{y^{3/2}} {(\sin\| u_{\epsilon,\lambda}\|_\infty + (1/2)y)^{3/2}}dy \le C\epsilon\lambda^{-1}.$$ By \eqref{e3.26}--\eqref{e3.28}, \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} Next, by \eqref{e3.6}, \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} Finally, by Lemma \ref{lem3.1} and \eqref{e3.22}, for $z_\lambda(u_{\epsilon,\lambda}(t)) := \| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)$, \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} Then by \eqref{e2.17}, \eqref{e3.6} and \eqref{e3.27}, \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} 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 $$\label{e4.1} \frac12u_{\epsilon,\lambda}'(t)^2 = \lambda(\cos u_{\epsilon,\lambda}(t) - \cos\| u_{\epsilon,\lambda}\|_\infty) - \epsilon\int_t^{0} | u_{\epsilon,\lambda}'(s)|^{2}ds.$$ Then by \eqref{e2.6}--\eqref{e2.8}, for $-1 \le t \le 0$, $$\label{e4.2} u_{\epsilon,\lambda}'(t) = \sqrt{2(A(u_{\epsilon,\lambda}(t)) - \epsilon B(t))}.$$ By this, $$\label{e4.3} 1 = \int_{-1}^{0} dt = \frac{1}{\sqrt{2}} \int_{-1}^{0} \frac{u_{\epsilon,\lambda}'(t)} {\sqrt{A(u_{\epsilon,\lambda}(t)) - \epsilon B(t)}}dt = \frac{1}{\sqrt{2}}(I + III),$$ 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 \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} \begin{lemma}\label{lem4.1} For $\lambda \gg 1$ and $-1 < t < 0$, $$\label{e4.7} B(t) \leq \sqrt{2A(u_{\epsilon,\lambda}(t))} (\| u_{\epsilon,\lambda}\|_\infty - u_{\epsilon,\lambda}(t)).$$ \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 $t_\delta \le t \le 0$, \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} 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 $$\label{e4.9} III_1 \le -\frac{\epsilon\sqrt{2}}{\lambda} \log\sin\frac{d_\lambda}{2} + O(\lambda^{-1}).$$ 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 $$\label{e4.10} \sin\frac{d_\lambda}{2} \le Ce^{-\sqrt{\lambda}}, \quad \sin\| u_{\epsilon,\lambda}\|_\infty \le Ce^{-\sqrt{\lambda}}.$$ \begin{lemma}\label{lem4.2} For $\lambda \gg 1$ and $t_\delta < t < 0$, \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{lemma} \begin{proof} By \eqref{e2.8}, \eqref{e4.2} and \eqref{e4.7}, \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} 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} \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} By \eqref{e4.8}, \eqref{e4.12} and Lemmas \ref{lem3.3} and \ref{lem3.4}, for $\lambda \gg 1$, $$\label{e4.14} \epsilon\int_{t_\delta}^0 \frac{W_1 u_{\epsilon,\lambda}'(t)} {2\sqrt{A(u_{\epsilon,\lambda}(t))}(A(u_{\epsilon,\lambda}(t)) - \epsilon B(t))}dt \le \frac{-\epsilon\sqrt{2}}{\lambda}\log\sin\frac{d_\lambda}{2} + O(\lambda^{-1}).$$ Further, by \eqref{e4.6}, \eqref{e4.8} and Lemma \ref{lem3.5}, for $\lambda \gg 1$, we obtain $$\label{e4.15} III_2 \le C\lambda^{-1}.$$ By \eqref{e4.6}, \eqref{e4.9} and \eqref{e4.13}--\eqref{e4.15}, we obtain $$\label{e4.16} III = \frac{-\epsilon\sqrt{2}}{\lambda}\log\sin\frac{d_\lambda}{2} + O(\lambda^{-3/4}).$$ 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 $$\label{e5.1} 1 + t_\lambda = \int_{-1}^{t_\lambda} dt = \frac{1}{\sqrt{2}} \int_{-1}^{t_\lambda} \frac{u_{\epsilon,\lambda}'(t)}{\sqrt{A(u_{\epsilon,\lambda}(t) + \epsilon B(t)}}dt = \frac{1}{\sqrt{2}}(P + Q),$$ 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 $$\label{e5.4} 1 - t_\lambda = \int_{t_\lambda}^{1} dt = \frac{1}{\sqrt{2}} \int_{t_\lambda}^1 \frac{-u_{\epsilon,\lambda}'(t)} {\sqrt{A(u_{\epsilon,\lambda}(t) - \epsilon B(t)}}dt = \frac{1}{\sqrt{2}}(R + S),$$ where $R = I$ and $S = III + O(\lambda^{-3/4})$. By \eqref{e5.1} and \eqref{e5.4}, we obtain $$\label{e5.5} 2t_\lambda = \sqrt{2}Q + O(\lambda^{-3/4}) = \sqrt{2}II + O(\lambda^{-3/4}) = \frac{2\epsilon}{\lambda} \log\big(\sin\frac{d_\lambda}{2}\big) + O(\lambda^{-3/4}),$$ which implies $$\label{e5.6} t_\lambda = \frac{\epsilon}{\lambda} \log\big(\sin\frac{d_\lambda}{2}\big) + O(\lambda^{-3/4}).$$ By this and \eqref{e5.1}, we obtain $$\label{e5.7} 1 = \frac{1}{\sqrt{2}}I + O(\lambda^{-3/4}).$$ 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}. 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