\documentclass[reqno]{amsart} %\pdfoutput=1\relax\pdfpagewidth=8.26in\pdfpageheight=11.69in\pdfcompresslevel=9 \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 37, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/37\hfil Asymptotic shape of solutions] {Asymptotic shape of solutions to nonlinear eigenvalue problems} \author[T. Shibata\hfil EJDE-2005/37\hfilneg] {Tetsutaro Shibata} \address{Tetsutaro Shibata \hfill\break Department of Applied Mathematics, Graduate School of Engineering, Hiroshima University, Higashi-Hiroshima, 739-8527, Japan} \email{shibata@amath.hiroshima-u.ac.jp} \date{} \thanks{Submitted January 11, 2005. Published March 29, 2005.} \subjclass[2000]{34B15} \keywords{Asymptotic formula; $L^1$-norm; simple pendulum; logistic equation} \begin{abstract} We consider the nonlinear eigenvalue problem $$-u''(t) = f(\lambda, u(t)), \quad u > 0, \quad u(0) = u(1) = 0,$$ where $\lambda > 0$ is a parameter. It is known that under some conditions on $f(\lambda, u)$, the shape of the solutions associated with $\lambda$ is almost box' when $\lambda \gg 1$. The purpose of this paper is to study precisely the asymptotic shape of the solutions as $\lambda \to \infty$ from a standpoint of $L^1$-framework. To do this, we establish the asymptotic formulas for $L^1$-norm of the solutions as $\lambda \to \infty$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \section{Introduction} We consider the nonlinear eigenvalue problem \begin{gather} -u''(t) = f(\lambda, u(t)), \quad t \in I:= (0, 1), \label{e1.1}\\ u(t) > 0, \quad t \in I, \\ u(0) = u(1) = 0, \label{e1.3} \end{gather} where $\lambda > 0$ is a parameter. The nonlinearities considered here are as follows: \begin{gather} f(\lambda, u) = \lambda\sin u, \label{e1.4}\\ f(\lambda, u) = \lambda\sin u - g(u), \label{e1.5}\\ f(\lambda, u) = \lambda(u - u^3). \label{e1.6} \end{gather} Equation \eqref{e1.1}--\eqref{e1.3} with the nonlinearities \eqref{e1.4} and \eqref{e1.5} are called the simple pendulum type equations (SPE), and that with \eqref{e1.6} is derived from the logistic equation of population dynamics (LEPD). Throughout this paper, in \eqref{e1.5}, we assume that $g(u)$ satisfies the following conditions: \begin{itemize} \item[(A1)] $g \in C^1(\mbox{\bf R})$ and $g(u) > 0$ for $u > 0$. \item[(A2)] $g(0) = g'(0) = 0$. \item[(A3)] $g(u)/u$ is strictly increasing for $0 \le u \le \pi$. \end{itemize} Nonlinear eigenvalue problems and singularly perturbed problems are intensively investigated by many authors. We refer to \cite{o2,s4} and the references therein. One of the most interesting problems to study in these fields is to clarify the asymptotic shapes of the solutions. We know (cf. \cite{c,f}) that for a given $\lambda > \pi^2$, there exists a unique solution $u \in C^2(I)$ to \eqref{e1.1}--\eqref{e1.3}. We denote by $u_{0,\lambda}, u_\lambda$ and $v_\lambda$ the solutions $u$ of \eqref{e1.1}--\eqref{e1.3} with \eqref{e1.4}, \eqref{e1.5} and \eqref{e1.6}, respectively. Let $m_\lambda := \min\{ m > 0: f(\lambda, m) = 0\}$. Then it is well known (cf. \cite{c,f} and Appendix) that $\Vert u_{0,\lambda}\Vert_\infty < m_\lambda$ (resp. $\Vert u_\lambda\Vert_\infty < m_\lambda$, $\Vert v_\lambda\Vert_\infty < m_\lambda$). Clearly, $m_\lambda = \pi$ and $m_\lambda = 1$ for \eqref{e1.4} and \eqref{e1.6}, respectively. By (A1), we see that $0 < m_\lambda < \pi$ for \eqref{e1.5}. Furthermore, it is known (cf. \cite{c,s1}) that $$\label{e1.7} u_{0,\lambda} \to \pi, \quad u_\lambda \to \pi, \quad v_\lambda \to 1$$ uniformly on any compact interval in $I$ as $\lambda \to \infty$. In other words, the asymptotic shape of these solutions are {\it almost boxes}. Therefore, a natural question we have to ask here is {\it how close to the boxes are the shape of the solutions $u_{0,\lambda}, u_\lambda$ and $v_\lambda$ asymptotically?} The purpose of this paper is to answer this question from a viewpoint of $L^1$-framework. More precisely, we restrict our attention to the typical nonlinearities \eqref{e1.4}--\eqref{e1.6}, and establish the precise asymptotic formulas for $L^1$-norm of $\Vert u_{0,\lambda}\Vert_1, \Vert u_{\lambda}\Vert_1$ and $\Vert v_{\lambda}\Vert_1$ as $\lambda \to \infty$. By these formulas, we understand well \begin{itemize} \item[(i)] The difference of the shape between $u_{0,\lambda}$ and $u_\lambda$ from a {\it non-local} point of view, and \item[(ii)] The difference between $\Vert v_\lambda\Vert_1$ and $\Vert v_\lambda\Vert_2$ when $\lambda \gg 1$. \end{itemize} The first approach to the study of the asymptotic shape of the solutions of (SPE) is to investigate the asymptotic behavior of the $L^\infty$-norm of the solutions as $\lambda \to \infty$ and the following results have been obtained in \cite{s1, s3}: \begin{gather} \Vert u_{0,\lambda}\Vert_\infty = \pi - 8 e^{-\sqrt{\lambda}/2} + o(e^{-\sqrt{\lambda}/2}), \label{e1.8} \\ \Vert u_\lambda \Vert_\infty = \pi - \frac{g(\pi)}{\lambda} + \frac{g(\pi)g'(\pi)}{\lambda^2} + o\big(\frac{1}{\lambda^2}\big). \label{e1.9} \end{gather} By \eqref{e1.8} and \eqref{e1.9}, we understand the difference between the {\it pointwise (local)} behavior of $u_{0,\lambda}$ and $u_\lambda$. However, we do not know the difference between the {\it total mass} of $u_{0,\lambda}$ and $u_\lambda$, which gives us the important information about the {\it non-local} property of $u_{0,\lambda}$ and $u_\lambda$. Therefore, it seems meaningful for us to establish the asymptotic formulas for $\Vert u_{0,\lambda}\Vert_1$ and $\Vert u_\lambda\Vert_1$, which give us the better understanding of the difference between the original (SPE) and the perturbed (SPE). We now state the results for (SPE). \begin{theorem} \label{thm1.1} As $\lambda \to \infty$ \begin{gather} \Vert u_{0,\lambda}\Vert_1 = \pi - C_1\frac{1}{\sqrt{\lambda}} + \frac{8}{\sqrt{\lambda}}e^{-\sqrt{\lambda}/2} + o\big(\frac{1}{\sqrt{\lambda}}e^{-\sqrt{\lambda}/2}\big), \label{e1.10} \\ \Vert u_\lambda\Vert_1 = \pi - C_1\frac{1}{\sqrt{\lambda}} - \frac{g(\pi)}{\lambda} + O\big(\frac{\log\lambda}{\lambda\sqrt{\lambda}}\big), \label{e1.11} \end{gather} where $C_1 = 8\int_0^{\pi/4} \log(\cot\theta)d\theta.$ \end{theorem} Roughly speaking, the second terms in \eqref{e1.10} and \eqref{e1.11} are derived from the width of the boundary layers of $u_{0,\lambda}$ and $u_\lambda$, while the third terms come directly from the second terms in \eqref{e1.8} and \eqref{e1.9}. We next consider (LEPD). The motivation to consider the asymptotic behavior of $\Vert v_\lambda\Vert_1$ as $\lambda \to \infty$ is as follows. Recently, from a viewpoint of $L^2$-bifurcation theory and nonlinear eigenvalue problems, the following formula for $\Vert v_\lambda\Vert_2$ as $\lambda \to \infty$ has been obtained. $$\label{e1.12} \Vert v_\lambda\Vert_2 = 1 - \sqrt{\frac{2}{\lambda}} - \frac{1}{\lambda} + o\big(\frac{1}{\lambda}\big).$$ This has been obtained by using \cite[Theorem 1]{s2}. On the other hand, since (LEPD) is a model equation of population density for some species, $\Vert v_\lambda\Vert_1$ stands for the total population of the species with $\lambda$, which describes the reciprocal number of its diffusion rate. Motivated by this biological background, it is also important to investigate the asymptotic behavior of $\Vert v_\lambda\Vert_1$ as $\lambda \to \infty$. Furthermore, since \eqref{e1.6} is a special nonlinearlity, we also obtain the asymptotic formula for $\Vert v_\lambda\Vert_\infty$ better than \eqref{e1.8} as $\lambda \to \infty$. Now we state our second results. \begin{theorem} \label{thm1.2} As $\lambda \to \infty$ \begin{gather} \Vert v_\lambda\Vert_1 = 1 - \frac{2\sqrt{2}\log2}{\sqrt{\lambda}} - 12e^{-\sqrt{2\lambda}} + o(e^{-\sqrt{2\lambda}}), \label{e1.13}\\ \Vert v_\lambda\Vert_\infty = 1 - 4 e^{-\sqrt{\lambda}/\sqrt{2}} - 8e^{-2\sqrt{\lambda}/\sqrt{2}} - 24\sqrt{2} \sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}} + o(\sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}}). \label{e1.14} \end{gather} \end{theorem} We see from \eqref{e1.12} and \eqref{e1.13} that the third term of $\Vert v_\lambda\Vert_1$ and $\Vert v_\lambda\Vert_2$ are totally different each other. The further direction of this study will be to treat more general nonlinear term $f(\lambda, u)$ and extend our results to PDE cases. The remainder of this paper is organized as follows. In Sections 2 and 3, we prove \eqref{e1.11} and \eqref{e1.10} in Theorem \ref{thm1.1}, respectively. By using the properties of complete elliptic integral, we prove Theorem \ref{thm1.2} in Section 4. \section{Proof of \eqref{e1.11} in Theorem \ref{thm1.1}} In this section, we consider \eqref{e1.1}--\eqref{e1.3} with \eqref{e1.5} and prove the formula \eqref{e1.11}. In what follows, the character $C$ denotes various positive constants independent of $\lambda \gg 1$. We know that $$\label{e2.1} \begin{gathered} u_\lambda(t) = u_\lambda(1-t) \quad \mbox{for } t \in \bar{I},\\ u_\lambda'(t) > 0 \quad \mbox{for } 0 \le t < 1/2, \\ \Vert u_\lambda\Vert_\infty = u_\lambda(1/2). \end{gathered}$$ We begin with the fundamental equality. Multiply \eqref{e1.1} by $u_\lambda'$. Then $$\{u_\lambda''(t) + \lambda\sin u_\lambda(t) - g(u_\lambda(t))\} u_\lambda'(t) = 0.$$ This implies that for $t \in \bar{I}$ $\frac{d}{dt}\big\{\frac12u_\lambda'(t)^2 - \lambda\cos u_\lambda(t) - G(u_\lambda(t))\big\} \equiv 0,$ where $G(u) = \int_0^u g(s)ds$. By this and \eqref{e2.1}, for $0 \le t \le 1$, $$\label{e2.2} \frac12u_\lambda'(t)^2 - \lambda\cos u_\lambda(t) - G(u_\lambda(t)) \equiv \mbox{constant} = -\lambda\cos\Vert u_\lambda\Vert_\infty - G(\Vert u_\lambda\Vert_\infty).$$ By this and \eqref{e2.1}, for $0 \le t \le 1/2$, $$\label{e2.3} u_\lambda'(t) = \sqrt{2\lambda (\cos u_\lambda(t) - \cos\Vert u_\lambda\Vert_\infty) + 2(G(u_\lambda(t)) - G(\Vert u_\lambda\Vert_\infty))}$$ We know from \eqref{e1.7} that as $\lambda \to \infty$ $$\label{e2.4} r_1(\lambda) := \Vert u_\lambda\Vert_\infty - \Vert u_\lambda\Vert_1 \to 0.$$ Since we have \eqref{e1.9}, we establish an asymptotic formula for $r_1(\lambda)$ as $\lambda \to \infty$ to obtain \eqref{e1.11}. By \eqref{e2.1} and \eqref{e2.3}, for $\lambda \gg 1$ \label{e2.5} \begin{aligned} r_1(\lambda) &= 2\int_0^{1/2} (\Vert u_\lambda\Vert_\infty - u_\lambda(t))dt \\ &= 2\int_0^{1/2} \frac{(\Vert u_\lambda\Vert_\infty - u_\lambda(t))u_\lambda'(t)dt} {\sqrt{2\lambda (\cos u_\lambda(t) - \cos\Vert u_\lambda\Vert_\infty) + 2(G(u_\lambda(t)) - G(\Vert u_\lambda\Vert_\infty))}}\\ &= 2\int_0 ^{\Vert u_\lambda\Vert_\infty} \frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta} {\sqrt{2\lambda (\cos \theta - \cos\Vert u_\lambda\Vert_\infty) + 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))}}\\ &= K_1(\lambda) + K_2(\lambda), \end{aligned} where \begin{gather*} K_1(\lambda) :=2\int_{\Vert u_\lambda\Vert_\infty - 1/\lambda} ^{\Vert u_\lambda\Vert_\infty} \frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta} {\sqrt{2\lambda (\cos \theta - \cos\Vert u_\lambda\Vert_\infty) + 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))} },\\ K_2(\lambda) := 2\int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta} {\sqrt{2\lambda (\cos \theta - \cos\Vert u_\lambda\Vert_\infty) + 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))} }. \end{gather*} \begin{lemma} \label{lem2.1} $K_1(\lambda) = O(\lambda^{-3/2})$ for $\lambda \gg 1$. \end{lemma} \begin{proof} For $j = 1, 2, \dots$, let $I_j := \big[\Vert u_\lambda\Vert_\infty - \frac{1}{j\lambda}, \Vert u_\lambda\Vert_\infty - \frac{1}{(j+1)\lambda}\big].$ We put $$\label{e2.8} J_j := 2\int_{I_j} \frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta} {\sqrt{2\lambda(\cos \theta - \cos\Vert u_\lambda\Vert_\infty) + 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))}}.$$ We know by \cite{c} that $$\label{e2.9} \lambda\sin\Vert u_\lambda\Vert_\infty > g(\Vert u_\lambda\Vert_\infty).$$ Let an arbitrary $0 < \epsilon \ll 1$ be fixed. Let $\eta_{\lambda,\epsilon} := \min_{\Vert u_\lambda\Vert_\infty-2\epsilon \le u \le \Vert u_\lambda\Vert_\infty} g'(u)$. Then by (A3), we see that $\eta_{\lambda,\epsilon} > 0$. Then for $\theta \in [\Vert u_\lambda\Vert_\infty - 2\epsilon, \Vert u_\lambda\Vert_\infty]$, by \eqref{e1.7}, \eqref{e2.9} and Taylor expansion, we have \label{e2.10} \begin{aligned} &2\lambda (\cos \theta - \cos\Vert u_\lambda\Vert_\infty) + 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty)) \\ &\ge 2\left(\lambda\sin\Vert u_\lambda\Vert_\infty - g(\Vert u_\lambda\Vert_\infty)\right)(\Vert u_\lambda\Vert_\infty - \theta) \\ &\quad + 2\left(-\frac{\lambda}{2}\cos (\Vert u_\lambda\Vert_\infty - 2\epsilon) + \frac12\eta_{\lambda,\epsilon}\right) (\Vert u_\lambda\Vert_\infty - \theta)^2 \\ &\ge C\lambda(\Vert u_\lambda\Vert_\infty - \theta)^2. \end{aligned} By this and \eqref{e2.8}, for $\lambda > 1/\epsilon$, % \label{e2.11} \begin{align*} J_j &\le \int_{I_j} \frac{\Vert u_\lambda\Vert_\infty - \theta} {\sqrt{C\lambda(\Vert u_\lambda\Vert_\infty - \theta)^2}}d\theta\\ &= \frac{1}{\sqrt{C\lambda}}\frac{1}{\lambda} \big(\frac{1}{j} - \frac{1}{j+1}\big) \le \frac{C}{\lambda^{3/2}}\frac{1}{j(j+1)}. \end{align*} By this, $K_1(\lambda) = \sum_{j=1}^\infty J_j \le \sum_{j=1}^\infty\frac{C}{\lambda^{3/2}} \frac{1}{j(j+1)} \le \frac{C}{\lambda^{3/2}}.$ Thus the proof is complete. \end{proof} The formula \eqref{e1.11} follows from \eqref{e1.9}, Lemma \ref{lem2.1} and the following Proposition. \begin{proposition} \label{prop2.2} For $\lambda \gg 1$ K_2(\lambda) = \frac{C_1}{\sqrt{\lambda}} + O\left(\frac{\log \lambda}{\lambda^{3/2}}\right). \end{proposition} We prove this proposition using Lemmas \ref{lem2.3}--\ref{lem2.7} below. We put $% \label{e2.12} K_2(\lambda) := K_{2,1}(\lambda) + K_{2,2}(\lambda),$ where \begin{gather*} K_{2,1}(\lambda) := 2\int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{(\Vert u_\lambda\Vert_\infty - \theta)d\theta} {\sqrt{2\lambda (\cos \theta - \cos\Vert u_\lambda\Vert_\infty)}}, %\label{e2.14} \\ K_{2,2}(\lambda) := K_2(\lambda) - K_{2,1}(\lambda). %\label{e2.15} \end{gather*} Furthermore, we put $% \label{e2.16} K_{2,1}(\lambda) := L_1(\lambda) + L_{2}(\lambda),$ where $% \label{e2.17} L_{1}(\lambda) := 2\int_0^{\pi - 1/\lambda} \frac{\pi - \theta} {\sqrt{2\lambda(\cos\theta + 1)}}d\theta.$ \begin{lemma} \label{lem2.3} For $\lambda \gg 1$ $% \label{e2.18} L_{1}(\lambda) = \frac{C_1}{\sqrt{\lambda}} + o\big(\frac{\log \lambda}{\lambda^{3/2}}\big).$ \end{lemma} \begin{proof} For $\lambda \gg 1$ \begin{align*} %2.19 L_{1}(\lambda) &= \frac{1}{\sqrt{\lambda}} \int_0^{\pi-1/\lambda} \frac{\pi - \theta}{\cos(\theta/2)}d\theta\\ &=\frac{1}{\sqrt{\lambda}} \int_{1/\lambda}^\pi \frac{t}{\sin(t/2)}dt =\frac{4}{\sqrt{\lambda}} \int_{1/(2\lambda)}^{\pi/2} \frac{t}{\sin t}dt \quad (\mbox{put }\theta = \tan(t/2)) \\ &=\frac{8}{\sqrt{\lambda}} \int_{\tan(1/(4\lambda))}^1 \frac{\tan^{-1}\theta}{\theta}d\theta\\ &= \frac{8}{\sqrt{\lambda}}\big\{ [\log\theta\tan^{-1}\theta] _{\tan(1/(4\lambda))}^1 - \int_{\tan(1/(4\lambda))}^1 \frac{\log\theta}{1 + \theta^2}d\theta \big\} \quad (\mbox{put } \theta = \tan t) \\ &= \frac{2 + o(1)}{\lambda^{3/2}}\log\lambda -\frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\tan t)dt +\frac{8}{\sqrt{\lambda}}\int_0^{1/(4\lambda)}\log(\tan t)dt \\ &=\frac{2 + o(1)}{\lambda^{3/2}}\log\lambda + \frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\cot t)dt + \frac{8}{\sqrt{\lambda}} (1 + o(1))\int_0^{1/(4\lambda)} \log t dt \\ &=\frac{2 + o(1)}{\lambda^{3/2}}\log\lambda + \frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\cot t)dt + \frac{8}{\sqrt{\lambda}} (1 + o(1))\big(\frac{1}{4\lambda}\log\frac{1}{4\lambda} - \frac{1}{4\lambda}\big) \\ &= \frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\cot t)dt + o\big(\frac{\log \lambda}{\lambda^{3/2}}\big). \end{align*} Thus the proof is complete. \end{proof} Next, we calculate $L_{2}(\lambda)$. To do this, we put $$L_{2}(\lambda) = K_{2,1}(\lambda) - L_1(\lambda) := D_1(\lambda) + D_2(\lambda) + D_3(\lambda),$$ where \begin{gather*} %2.21 D_1(\lambda) := 2\int_0^ {\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{\Vert u_\lambda\Vert_\infty - \pi} {\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}d\theta, \\ D_2(\lambda) := 2\int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \Big( \frac{\pi - \theta} {\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}} - \frac{\pi - \theta} {\sqrt{2\lambda(\cos\theta + 1)}} \Big) d\theta, \\ D_3(\lambda) := -2\int_ {\Vert u_\lambda\Vert_\infty - 1/\lambda}^{\pi - 1/\lambda} \frac{\pi - \theta} {\sqrt{2\lambda(\cos\theta + 1)}}d\theta. \end{gather*} \begin{lemma} \label{lem2.4} $D_1(\lambda) = O(\lambda^{-3/2})$ as $\lambda \to \infty$. \end{lemma} \begin{proof} Let an arbitrary $0 < \epsilon \ll 1$ be fixed. Then for $\lambda \gg 1$ \begin{align*} %2.24 D_1(\lambda) &= D_{1,1}(\lambda) + D_{1,2}(\lambda) \\ &:= 2\int_{\Vert u_\lambda\Vert_\infty - \epsilon} ^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{\Vert u_\lambda\Vert_\infty - \pi} {\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}d\theta \\ &\quad + 2\int_0^{\Vert u_\lambda\Vert_\infty - \epsilon} \frac{\Vert u_\lambda\Vert_\infty - \pi} {\sqrt{2\lambda (\cos\theta - \cos\Vert u_\lambda\Vert_\infty) }}d\theta. \end{align*} For $0 \le \theta \le \Vert u_\lambda\Vert_\infty - \epsilon$, there exists a constant $C_\epsilon > 0$ such that for $\lambda \gg 1$ $$\label{e2.25} C_\epsilon \le \cos\theta - \cos\Vert u_\lambda\Vert_\infty.$$ By this and \eqref{e1.9}, for $\lambda \gg 1$, $$\label{e2.26} \vert D_{1,2}(\lambda) \vert \le \frac{2g(\pi)}{\lambda}(1 + o(1))\frac{1} {\sqrt{2C_\epsilon\lambda}}\pi \le C(\lambda^{-3/2}).$$ We next estimate $D_{1,1}(\lambda)$. For a given $\lambda \gg 1$, there exists $k_\lambda \in N$ satisfying $$\label{e2.27} \Vert u_\lambda\Vert_\infty - 2\epsilon \le \Vert u_\lambda\Vert_\infty - \frac{k_\lambda + 1}{\lambda} \le \Vert u_\lambda\Vert_\infty - \epsilon \le \Vert u_\lambda\Vert_\infty - \frac{k_\lambda}{\lambda}.$$ For $j = 1, 2, \dots, k_\lambda$, we define an interval $$\label{e2.28} M_j = \big[ \Vert u_\lambda\Vert_\infty - \frac{j + 1}{\lambda},\Vert u_\lambda\Vert_\infty - \frac{j}{\lambda} \big].$$ By \eqref{e2.27}, we see that $k_\lambda \le \epsilon\lambda$. By this, \eqref{e1.9} and \eqref{e2.10}, we obtain \begin{align*} %2.29 \vert D_{1,1}(\lambda) \vert &\le \frac{g(\pi)}{\lambda}(1 + o(1)) \sum_{j=1}^{k_\lambda} \int_{M_j} \frac{1} {\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}} d\theta \\ &\le \frac{g(\pi)}{\sqrt{2C}\lambda^{3/2}}(1 + o(1)) \sum_{j=1}^{k_\lambda} \int_{M_j} \frac{1}{\Vert u_\lambda\Vert_\infty - \theta}d\theta\\ &\le C\lambda^{-3/2} \sum_{j=1}^{k_\lambda}(\log(j + 1) - \log j)\\ &= C\lambda^{-3/2}\log(k_\lambda + 1) \le C\lambda^{-3/2}\log\lambda. \end{align*} By this and \eqref{e2.26}, we obtain our conclusion. Thus the proof is complete. \end{proof} \begin{lemma} \label{lem2.5} $D_2(\lambda) = O(\lambda^{-3/2})$ for $\lambda \gg 1$. \end{lemma} \begin{proof} We put $$\label{e2.30} A_\lambda(\theta) := 2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty), \enskip B_\lambda(\theta) := 2\lambda(\cos\theta + 1).$$ By \eqref{e1.9} and Taylor expansion, for $\lambda \gg 1$, we have $$\label{e2.31} 1 + \cos \Vert u_\lambda \Vert_\infty = \frac{g(\pi)^2}{2\lambda^2}(1 + o(1)).$$ Note that $A_\lambda(\theta) \le B_\lambda(\theta)$. By this, \eqref{e2.31} and Taylor expansion, for a fixed $0 < \epsilon \ll 1$ \label{e2.32} \begin{aligned} D_2(\lambda) &= 2\int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{2\lambda(\pi - \theta)(1 + \cos\Vert u_\lambda\Vert_\infty)} {\sqrt{A_\lambda(\theta)} \sqrt{B_\lambda(\theta)} (\sqrt{A_\lambda(\theta)} + \sqrt{B_\lambda(\theta)}) }d\theta\\ &\le \frac{2g(\pi)^2}{\lambda}(1 + o(1)) \Big[\int_{\Vert u_\lambda\Vert_\infty - \epsilon} ^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{\pi - \theta} {(2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty))^{3/2} }d\theta \\ &\quad + \int_0^{\Vert u_\lambda\Vert_\infty - \epsilon} \frac{\pi - \theta} {(2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty))^{3/2} }d\theta \Big]\\ &:= D_{2,1}(\lambda) + D_{2,2}(\lambda). \end{aligned} We know that $2\theta/\pi \le \sin\theta$ for $0 \le \theta \le \pi/2$. By this, \eqref{e1.9} and mean value theorem, for $\theta \in M_j$ defined by \eqref{e2.28}, we have \begin{align*} %.2.33 \cos\theta - \cos\Vert u_\lambda\Vert_\infty &\ge \sin\Big(\Vert u_\lambda \Vert_\infty - \frac{j}{\lambda}\Big) (\Vert u_\lambda\Vert_\infty - \theta)\\ &= \sin\Big(\pi - \frac{g(\pi)}{\lambda}(1 + o(1)) - \frac{j}{\lambda}\Big)(\Vert u_\lambda\Vert_\infty - \theta)\\ &=\sin\Big(\frac{g(\pi)}{\lambda}(1 + o(1)) + \frac{j}{\lambda}\Big) (\Vert u_\lambda\Vert_\infty - \theta)\\ &\ge \frac{2}{\pi} \Big(\frac{g(\pi)(1 + o(1)) + j}{\lambda}\Big) (\Vert u_\lambda\Vert_\infty - \theta). \end{align*} By this, \eqref{e1.9} and \eqref{e2.32}, \label{e2.34} \begin{aligned} D_{2,1}(\lambda) &\le \frac{C}{\lambda} \sum_{j=1}^{k_\lambda} \int_{M_j} \frac{(j + 1 + g(\pi) + o(1))/\lambda} {\{\frac{4}{\pi}(j + g(\pi) + o(1)) (\Vert u_\lambda \Vert_\infty - \theta)\}^{3/2}} d\theta\\ &\le \frac{C}{\lambda^2} \sum_{j=1}^{k_\lambda} (j + g(\pi) + 1 + o(1))(j + g(\pi) + o(1))^{-3/2} \int_{M_j} \!(\Vert u_\lambda\Vert_\infty - \theta)^{-3/2}d\theta\\ &\le \frac{C}{\lambda^{3/2}} \sum_{j=1}^{k_\lambda}j^{-1/2}\big( \frac{1}{\sqrt{j}} - \frac{1}{\sqrt{j+1}}\big)\\ &= \frac{C}{\lambda^{3/2}} \sum_{j=1}^{k_\lambda} \frac{1}{\sqrt{j}\sqrt{j+1}(\sqrt{j} + \sqrt{j+1})}j^{-1/2}\\ &\le \frac{C}{\lambda^{3/2}} \sum_{j=1}^{k_\lambda}\frac{1}{j^2} \le \frac{C}{\lambda^{3/2}}. \end{aligned} By \eqref{e2.25}, we have $%2.35 D_{2,2}(\lambda) \le \frac{2g(\pi)^2 + o(1)}{\lambda} \frac{1}{(2C_\epsilon\lambda)^{3/2}}\pi^2 \le C\lambda^{-5/2}.$ This along with \eqref{e2.32} and \eqref{e2.34} implies our conclusion. \end{proof} \begin{lemma} \label{lem2.6} $D_3(\lambda) = O(\lambda^{-3/2})$ for $\lambda \gg 1$. \end{lemma} \begin{proof} By \eqref{e1.9}, for $\lambda \gg 1$ \begin{align*} % 2.36 \vert D_3(\lambda)\vert &= \int_{\Vert u_\lambda\Vert_\infty - 1/\lambda} ^{\pi - 1/\lambda}\frac{\pi - \theta}{\sqrt{\lambda}\cos(\theta/2)}d\theta\\ &= \frac{1}{\sqrt{\lambda}} \int_{\pi-\Vert u_\lambda\Vert_\infty + 1/\lambda} ^{1/\lambda}\frac{-t}{\sin(t/2)}dt \\ &= \frac{4}{\sqrt{\lambda}} \int_{1/(2\lambda)}^{(\pi-\Vert u_\lambda\Vert_\infty + 1/\lambda)/2} \frac{\theta}{\sin\theta}d\theta \\ &= \frac{4}{\sqrt{\lambda}}(1 + o(1))\frac{\pi-\Vert u_\lambda\Vert_\infty}{2} = \frac{2g(\pi)}{\lambda^{3/2}}(1 + o(1)). \end{align*} Thus the proof is complete. \end{proof} By Lemmas \ref{lem2.3}--\ref{lem2.6}, we see that $$\label{e2.37} K_{2,1}(\lambda) = \frac{C_1}{\sqrt{\lambda}} + O\left(\frac{\log\lambda} {\lambda^{3/2}}\right).$$ Now we estimate $K_{2,2}(\lambda)$. \begin{lemma} \label{lem2.7} $K_{2,2}(\lambda) = O(\lambda^{-3/2}\log\lambda)$ for $\lambda \gg 1$. \end{lemma} \begin{proof} We put $E_\lambda(\theta) := 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))$. We recall $A_\lambda(\theta)$ defined in \eqref{e2.30}. Let an arbitrary $0 < \epsilon \ll 1$ be fixed. For $\lambda \gg 1$ \begin{align*} %2.38 &K_{2,2}(\lambda)\\ &= 2 \int_0^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \Big( \frac{\Vert u_\lambda\Vert_\infty - \theta} {\sqrt{A_\lambda(\theta) + E_\lambda(\theta)}} - \frac{\Vert u_\lambda\Vert_\infty - \theta} {\sqrt{A_\lambda(\theta)}} \Big)d\theta \\ &=2\int_0^{\Vert u_\lambda\Vert_\infty - \epsilon} \frac{(\Vert u_\lambda\Vert_\infty - \theta) (G(\Vert u_\lambda\Vert_\infty) - G(\theta))}{ \sqrt{A_\lambda(\theta) + E_\lambda(\theta)} \sqrt{A_\lambda(\theta)} (\sqrt{A_\lambda(\theta) + E_\lambda(\theta)} + \sqrt{A_\lambda(\theta)})}d\theta \\ &\quad + 2\int_{\Vert u_\lambda\Vert_\infty - \epsilon} ^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{(\Vert u_\lambda\Vert_\infty - \theta) (G(\Vert u_\lambda\Vert_\infty) - G(\theta))}{ \sqrt{A_\lambda(\theta) + E_\lambda(\theta)} \sqrt{A_\lambda(\theta)} (\sqrt{A_\lambda(\theta) + E_\lambda(\theta)} + \sqrt{A_\lambda(\theta)})}d\theta\\ &= H_1(\lambda) + H_2(\lambda). \end{align*} By \eqref{e2.25} we see that for $\lambda \gg 1$, $%39 H_1(\lambda) \le 2g(\pi)\pi^3(2C_\epsilon\lambda)^{-3/2}.$ Note that $A_\lambda(\theta) + E_\lambda(\theta) \le A_\lambda(\theta)$ for $\Vert u_\lambda\Vert_\infty - \epsilon \le \theta \le \Vert u_\lambda\Vert_\infty - 1/\lambda$. Then by \eqref{e2.10}, \begin{align*} %2.40 H_2(\lambda) &\le 2\int_{\Vert u_\lambda\Vert_\infty - \epsilon} ^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{g(\pi)(\Vert u_\lambda\Vert_\infty - \theta)^2} {(A_\lambda + E_\lambda)^{3/2}}d\theta\\ &\le \frac{C}{\lambda^{3/2}} \int_{\Vert u_\lambda\Vert_\infty - \epsilon} ^{\Vert u_\lambda\Vert_\infty - 1/\lambda} \frac{1}{\Vert u_\lambda\Vert_\infty - \theta}d\theta \\ &= \frac{C}{\lambda^{3/2}}(\log\epsilon - \log(1/\lambda)) \le \frac{C}{\lambda^{3/2}}\log\lambda. \end{align*} Thus the proof is complete. \end{proof} By \eqref{e2.37} and Lemma \ref{lem2.7}, we obtain Proposition \ref{prop2.2}. Now \eqref{e1.11} follows from \eqref{e1.9}, \eqref{e2.4}, \eqref{e2.5}, Lemma \ref{lem2.1} and Proposition \ref{prop2.2}. Thus the proof is complete. \section{Proof of \eqref{e1.10} in Theorem \ref{thm1.1}} To prove \eqref{e1.10}, we put $$\label{e3.1} Q(\lambda) := \pi - \Vert u_{0,\lambda}\Vert_1.$$ By the similar calculation to that in \eqref{e2.5}, we have \begin{gather} \label{e3.2} Q(\lambda) = 2\int_0^{\Vert u_{0,\lambda}\Vert_\infty} \frac{\pi - \theta}{\sqrt{2\lambda(\cos\theta - \cos\Vert u_{0,\lambda}\Vert_\infty)}} d\theta = Q_{1}(\lambda) + Q_{2}(\lambda), \\ Q_{2}(\lambda) := Q(\lambda) - Q_1(\lambda), \label{e3.3} \end{gather} where $$\label{e3.4} Q_1(\lambda) := 2\int_0^\pi \frac{\pi - \theta} {\sqrt{2\lambda(\cos\theta + 1)}}d\theta.$$ \begin{lemma} \label{lem3.1} $Q_{1}(\lambda) = C_1\lambda^{-1/2}$. \end{lemma} \begin{proof} \begin{align*} Q_{1}(\lambda) &= \frac{1}{\sqrt{\lambda}} \int_0^\pi \frac{\pi - \theta}{\cos(\theta/2)}d\theta\\ &=\frac{1}{\sqrt{\lambda}} \int_0^\pi \frac{t}{\sin(t/2)}dt =\frac{4}{\sqrt{\lambda}} \int_0^{\pi/2} \frac{t}{\sin t}dt \quad (\mbox{put } \theta = \tan(t/2))\\ &=\frac{8}{\sqrt{\lambda}} \int_0^1 \frac{\tan^{-1}\theta}{\theta}d\theta \\ &= \frac{8}{\sqrt{\lambda}}\big\{[\log\theta\tan^{-1}\theta ]_0^1 - \int_0^1 \frac{\log\theta}{1 + \theta^2}d\theta \big\} \quad (\mbox{put }\theta = \tan t) \\ &=-\frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\tan t)dt = \frac{8}{\sqrt{\lambda}}\int_0^{\pi/4}\log(\cot t)dt \end{align*} Thus the proof is complete. \end{proof} \begin{lemma} \label{lem3.2} $Q_{2}(\lambda) = -\frac{8}{\sqrt{\lambda}}(1 + o(1))e^{-\sqrt{\lambda}/2}$ as $\lambda \to \infty$. \end{lemma} \begin{proof} We put $$\label{e3.5} Q_{2}(\lambda) := R(\lambda) + S(\lambda),$$ where \begin{gather*} %3.6, 3.7 R(\lambda) = \sqrt{\frac{2}{\lambda}} \int_0^{\Vert u_{0,\lambda}\Vert_\infty} \Big( \frac{\pi-\theta} {\sqrt{\cos\theta-\cos\Vert u_{0,\lambda}\Vert_\infty}} - \frac{\pi-\theta}{\sqrt{\cos\theta + 1}} \Big) d\theta \\ S(\lambda) = - \sqrt{\frac{2}{\lambda}} \int_{\Vert u_{0,\lambda}\Vert_\infty}^\pi \frac{\pi-\theta}{\sqrt{\cos\theta + 1}}d\theta. \end{gather*} For $\lambda \gg 1$ \label{e3.8} \begin{aligned} R(\lambda) &= \sqrt{\frac{2}{\lambda}} \int_0^{\Vert u_{0,\lambda}\Vert_\infty} \frac{(\pi-\theta) (\sqrt{\cos\theta + 1} -\sqrt{\cos\theta-\cos\Vert u_{0,\lambda}\Vert_\infty})} {\sqrt{\cos\theta + 1} \sqrt{\cos\theta-\cos\Vert u_{0,\lambda}\Vert_\infty}}d\theta \\ &=\sqrt{\frac{2}{\lambda}}(1 + o(1)) \int_0^{\Vert u_{0,\lambda}\Vert_\infty} \frac{(\pi-\theta) (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)} {(\cos\theta + 1) (\sqrt{\cos\theta-\cos\Vert u_{0,\lambda}\Vert_\infty} + \sqrt{\cos\theta + 1})}d\theta \\ &=\sqrt{\frac{2}{\lambda}} (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)) \int_0^{\Vert u_{0,\lambda}\Vert_\infty} \frac{\pi-\theta}{2(\cos\theta + 1)^{3/2}}d\theta \\ &=\sqrt{\frac{2}{\lambda}} (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)) \int_0^{\Vert u_{0,\lambda}\Vert_\infty} \frac{\pi-\theta}{4\sqrt{2}\cos^3(\theta/2)}d\theta \\ &=\sqrt{\frac{2}{\lambda}} (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)) \int_\pi^{\pi-\Vert u_{0,\lambda}\Vert_\infty} \frac{-\theta}{4\sqrt{2}\sin^3(\theta/2)}d\theta \\ &=\sqrt{\frac{1}{\lambda}} (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)) \int_{(\pi-\Vert u_{0,\lambda}\Vert_\infty)/2}^{\pi/2} \frac{\theta}{\sin^3\theta}d\theta \\ &=\sqrt{\frac{1}{\lambda}} (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)) \int _{(\pi-\Vert u_{0,\lambda}\Vert_\infty)/2}^{\pi/2} \frac{1}{\sin^2\theta}d\theta \\ &= \sqrt{\frac{1}{\lambda}} (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)) [-\cot \theta]_{(\pi-\Vert u_{0,\lambda}\Vert_\infty)/2}^{\pi/2} \\ &=\sqrt{\frac{1}{\lambda}} \frac{\cos((\pi-\Vert u_{0,\lambda}\Vert_\infty)/2)} {\sin((\pi-\Vert u_{0,\lambda}\Vert_\infty)/2)} (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)) \\ &=\sqrt{\frac{1}{\lambda}} \frac{2}{\pi-\Vert u_{0,\lambda}\Vert_\infty} (\cos\Vert u_{0,\lambda}\Vert_\infty + 1)(1 + o(1)). \end{aligned} By \eqref{e1.8} and Taylor expansion, for $\lambda \gg 1$ \label{e3.9} \begin{aligned} \cos\Vert u_{0,\lambda}\Vert_\infty &=\cos\big(\pi - 8(1 + o(1))e^{-\sqrt{\lambda}/2}\big) \\ &= -\cos\big(8(1 + o(1))e^{-\sqrt{\lambda}/2}\big)\\ &= -1 + 32(1 + o(1))e^{-\sqrt{\lambda}}. \end{aligned} By this, \eqref{e1.8} and \eqref{e3.8}, for $\lambda \gg 1$ $$\label{e3.10} R(\lambda) = \frac{8}{\sqrt{\lambda}}(1 + o(1)) e^{-\sqrt{\lambda}/2}.$$ Next, we calculate $S(\lambda)$. By \eqref{e1.8}, for $\lambda \gg 1$ \begin{align*} % 3.11 S(\lambda) &= - \sqrt{\frac{2}{\lambda}} \int_{\pi - \Vert u_{0,\lambda}\Vert_\infty}^0 \frac{-t}{{\sqrt{2}\cos((\pi-t)/2)}}dt\\ &= - \sqrt{\frac{1}{\lambda}} \int_0^{\pi - \Vert u_{0,\lambda}\Vert_\infty} \frac{t}{\sin(t/2)}dt\\ &= - \frac{4}{\sqrt{\lambda}} \int_0^{(\pi - \Vert u_{0,\lambda}\Vert_\infty)/2} \frac{\theta}{\sin\theta}d\theta\\ &=- \frac{4}{\sqrt{\lambda}}\frac{\pi - \Vert u_{0,\lambda}\Vert_\infty}{2} (1 + o(1))\\ &= -\frac{16}{\sqrt{\lambda}}(1 + o(1))e^{-\sqrt{\lambda}/2}. \end{align*} By this, \eqref{e3.5} and \eqref{e3.10}, we obtain our conclusion. Thus the proof is complete. \end{proof} Now \eqref{e1.10} follows from \eqref{e1.8}, \eqref{e3.1}, \eqref{e3.2} and Lemmas \ref{lem3.1} and \ref{lem3.2}. Thus the proof is complete. \section{Proof of Theorem \ref{thm1.2}} In this section, we consider \eqref{e1.1}--\eqref{e1.3} with \eqref{e1.6}. By \eqref{e1.1}, we have $(v_\lambda''(t) + \lambda(v_\lambda(t) - v_\lambda^3(t)))v_\lambda'(t) = 0.$ This implies that for $t \in \bar{I}$ $\frac{d}{dt}\big( \frac12v_\lambda'(t)^2 + \frac12\lambda v_\lambda^2(t) - \frac14\lambda v_\lambda^4(t) \big) = 0.$ This implies that for $t \in \bar{I}$, $$\label{e4.1} \frac12v_\lambda'(t)^2 + \frac12\lambda v_\lambda^2(t) - \frac14\lambda v_\lambda^4(t) \equiv \mbox{constant} = \frac12\lambda \Vert v_\lambda\Vert_\infty^2 - \frac14\lambda\Vert v_\lambda\Vert_\infty^4.$$ We know that $$\label{e4.2} v_\lambda'(t) \ge 0, \quad 0 \le t \le 1/2, \quad v_\lambda(t) = v_\lambda(1-t), \quad t \in I.$$ Therefore, by \eqref{e4.1} and \eqref{e4.2}, for $0 \le t \le 1/2$, $$\label{e4.3} v_\lambda'(t) = \sqrt{\lambda\{ (\Vert v_\lambda\Vert_\infty^2 - v_\lambda(t)^2) - \frac12 (\Vert v_\lambda\Vert_\infty^4 - v_\lambda(t)^4)\}}.$$ The following Lemma \ref{lem4.1} implies \eqref{e1.14} in Theorem \ref{thm1.2}. \begin{lemma} \label{lem4.1} As $\lambda \to \infty$ $$\label{e4.4} \Vert v_\lambda\Vert_\infty = 1 - 4 e^{-\sqrt{\lambda}/\sqrt{2}} - 8e^{-2\sqrt{\lambda}/\sqrt{2}}- 24\sqrt{2} \sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}} + o(\sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}}).$$ \end{lemma} \begin{proof} By \eqref{e4.3}, \label{e4.5} \begin{aligned} \frac12 &= \int_0^{1/2} dt = \int_0^{1/2}\frac{v_\lambda'(t)} {\sqrt{\lambda \{(\Vert v_\lambda\Vert_\infty^2 - v_\lambda(t)^2) - \frac12 (\Vert v_\lambda\Vert_\infty^4 - v_\lambda(t)^4) \}}}dt \\ &= \frac{1}{\sqrt{\lambda}} \int_0^{\Vert v_\lambda\Vert_\infty} \frac{1} {\sqrt{(\Vert v_\lambda\Vert_\infty^2 - \theta^2) - \frac12(\Vert v_\lambda\Vert_\infty^4 - \theta^4)} }d\theta \quad (\mbox{put} \enskip \theta = \Vert v_\lambda\Vert_\infty s) \\ &= \frac{1}{\sqrt{\lambda}}\frac{\sqrt{2}} {\sqrt{2-\Vert v_\lambda\Vert_\infty^2}} \int_0^1 \frac{1}{\sqrt{(1-s^2)(1-k^2s^2)}}ds \\ &= \frac{1}{\sqrt{\lambda}}\frac{\sqrt{2}} {\sqrt{2-\Vert v_\lambda\Vert_\infty^2}}K(k), \end{aligned} where $k = \Vert v_\lambda\Vert_\infty/\sqrt{2-\Vert v_\lambda\Vert_\infty^2}$ and $%4.6 K(k) := \int_0^{\pi/2} \frac{1}{\sqrt{1-k^2\sin^2\theta}}d\theta.$ It is known (cf. \cite[p.909, 8.113]{g}) that as $k \to 1$ \label{e4.7} \begin{aligned} K(k) &= -\frac12\log(1-k^2) + 2\log 2 - \frac{1-k^2}{8} \log(1-k^2) \\ &\quad + \big(\frac12\log2 - \frac14\big)(1 - k^2) - \frac{9}{128}(1-k^2)^2\log(1-k^2)\\ &\quad + o\big((1-k^2)^2\log(1-k^2)\big). \end{aligned} We put $\xi_\lambda := 1 - \Vert v_\lambda\Vert_\infty^2$. Then $\xi_\lambda > 0$ and $\xi_\lambda \to 0$ as $\lambda \to \infty$ by \eqref{e1.7}. Then $$\label{e4.8} 1-k^2 = \frac{2(1-\Vert v_\lambda\Vert_\infty^2)}{2-\Vert v_\lambda\Vert_\infty^2} = \frac{2\xi_\lambda}{1 + \xi_\lambda}.$$ By this, the Taylor expansion, and \eqref{e4.7}, \label{e4.9} \begin{aligned} K(k) &= -\frac12 \big(\log 2 + \log \xi_\lambda - \log(1 + \xi_\lambda)\big) \\ &\quad + 2\log 2 - \frac{\xi_\lambda}{4(1 + \xi_\lambda)} \big(\log 2 + \log \xi_\lambda - \log(1 + \xi_\lambda)\big) \\ &\quad + \big(\frac12\log 2 - \frac14\big) \frac{2\xi_\lambda}{1 + \xi_\lambda} -\frac{9}{32}\frac{\xi_\lambda^2}{(1 + \xi_\lambda)^2} \big(\log 2 + \log \xi_\lambda - \log(1 + \xi_\lambda)\big) \\ &\quad +o(\xi_\lambda^2\log\xi_\lambda) \\ &= -\frac12 \big(\log 2 + \log \xi_\lambda - \xi_\lambda + O(\xi_\lambda^2)\big) + 2\log 2 \\ &\quad - \frac14\xi_\lambda(1 - \xi_\lambda + O(\xi_\lambda^2)) \big(\log 2 + \log \xi_\lambda - \xi_\lambda + O(\xi_\lambda^2)\big) \\ &\quad + \big(\log 2 - \frac12\big) \xi_\lambda(1 - \xi_\lambda + O(\xi_\lambda^2)) -\frac{9}{32} \xi_\lambda^2\log\xi_\lambda + o(\xi_\lambda^2\log\xi_\lambda) \\ &= -\frac12\log\xi_\lambda + \frac32\log 2 - \frac14\xi_\lambda\log\xi_\lambda + \frac{3\log2}{4}\xi_\lambda - \frac{1}{32}\xi_\lambda^2\log\xi_\lambda + o(\xi_\lambda^2\log\xi_\lambda). \end{aligned} Furthermore, by Taylor expansion, for $\lambda \gg 1$, $%4.10 \frac{1}{\sqrt{2-\Vert v_\lambda\Vert_\infty^2}} = (1 + \xi_\lambda)^{-1/2} = 1 - \frac12\xi_\lambda + \frac{3}{8}\xi_\lambda^2 + o(\xi_\lambda^2).$ This along with \eqref{e4.5} and \eqref{e4.9} implies that $$\label{e4.11} \frac{\sqrt{\lambda}}{2\sqrt{2}} =-\frac12\log\xi_\lambda + \frac32\log 2 -\frac{3}{32}\xi_\lambda^2\log\xi_\lambda + o(\xi_\lambda^2\log\xi_\lambda).$$ By this, for $\lambda \gg 1$, we have $\frac{\sqrt{\lambda}}{\sqrt{2}} = -\log\xi_\lambda + \log 8 + \log(1 + o(1)) = \log\frac{8(1 + o(1))}{\xi_\lambda}.$ This implies that for $\lambda \gg 1$, $\xi_\lambda = 8(1 +o(1))e^{-\sqrt{\lambda}/\sqrt{2}}$. %4.12 Then for $\lambda \gg 1$, $-\frac{3}{32}\xi_\lambda^2\log\xi_\lambda = 3\sqrt{2}(1 + o(1))\sqrt{\lambda}e^{-\sqrt{2\lambda}}.$ By this, \eqref{e4.11} and Taylor expansion, for $\lambda \gg 1$ \label{e4.13} \begin{aligned} \xi_\lambda &=8e^{-\sqrt{\lambda}/\sqrt{2}}\cdot e^{6\sqrt{2}(1 + o(1))\sqrt{\lambda}e^{-\sqrt{2\lambda}}} \\ &= 8e^{-\sqrt{\lambda}/\sqrt{2}}(1 + 6\sqrt{2}(1 + o(1)) \sqrt{\lambda}e^{-\sqrt{2\lambda}})) \\ &= 8e^{-\sqrt{\lambda}/\sqrt{2}} + 48\sqrt{2}(1 + o(1)) \sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}}. \end{aligned} By this and Taylor expansion, for $\lambda \gg 1$, \begin{align*} \Vert v_\lambda\Vert_\infty &= \sqrt{1 - \xi_\lambda}\\ &=\Big(1- 8e^{-\sqrt{\lambda}/\sqrt{2}} - 48\sqrt{2}(1 + o(1))\sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}}) \Big)^{1/2} \\ &= 1 + \frac12\Big(- 8e^{-\sqrt{\lambda}/\sqrt{2}} -48\sqrt{2}(1 + o(1))\sqrt{\lambda} e^{-3\sqrt{\lambda}/\sqrt{2}})\Big) \\ &\quad- \frac{1}{8}(- 8e^{-\sqrt{\lambda}/\sqrt{2}} - 48\sqrt{2}(1 + o(1)) \sqrt{\lambda}e^{-3\sqrt{\lambda}/\sqrt{2}}))^2 + O(e^{-3\sqrt{\lambda}/\sqrt{2}}). \end{align*} By this, we obtain \eqref{e4.4}. \end{proof} The following implies \eqref{e1.13} in Theorem \ref{thm1.2}. \begin{lemma} \label{lem4.2} As $\lambda \to \infty$ $$\label{e4.15} \Vert v_\lambda\Vert_1 = 1 - \frac{2\sqrt{2}\log 2}{\sqrt{\lambda}} - 12e^{-\sqrt{2\lambda}} + o(e^{-\sqrt{2\lambda}}).$$ \end{lemma} \begin{proof} By \eqref{e4.3}, for $\lambda \gg 1$ \label{e4.16} \begin{aligned} \Vert v_\lambda\Vert_1 &= 2\int_0^{1/2} v_\lambda(t)\,dt \\ &= 2 \int_0^{1/2}\frac{v_\lambda(t)v_\lambda'(t)} {\sqrt{\lambda\{ (\Vert v_\lambda\Vert_\infty^2 - v_\lambda(t)^2) - \frac12 (\Vert v_\lambda\Vert_\infty^4 - v_\lambda(t)^4)\}}}\,dt \\ &= \frac{2}{\sqrt{\lambda}} \int_0^{\Vert v_\lambda\Vert_\infty} \frac{\theta} {\sqrt{\Vert v_\lambda\Vert_\infty^2 - \theta^2 - \frac12(\Vert v_\lambda\Vert_\infty^4 - \theta^4)}}\,d\theta \\ &=\frac{2\Vert v_\lambda\Vert_\infty}{\sqrt{\lambda}} \int_0^1 \frac{s} {\sqrt{(1-s^2)-\frac12\Vert v_\lambda\Vert_\infty^2(1-s^4)}}\,ds \\ &=\frac{\Vert v_\lambda\Vert_\infty}{\sqrt{\lambda}} \frac{\sqrt{2}}{\sqrt{2-\Vert v_\lambda\Vert_\infty^2}} \int_0^1\frac{1}{(\sqrt{1-t)(1-k^2t)}}\,dt \\ &= \sqrt{\frac{2}{\lambda}}k\int_0^1 \frac{1}{\sqrt{(1-t)(1-k^2t)}}\,dt. \end{aligned} By putting $s = \sqrt{(1-k^2t)/(1-t)}$, we obtain easily $%4.17 \int_0^1 \frac{1}{\sqrt{(1-t)(1-k^2t)}}dt = \frac{1}{k} \log\big(\frac{1+k}{1-k}\big).$ This along with \eqref{e4.16} implies that $$\label{e4.18} \Vert v_\lambda\Vert_1 =\sqrt{\frac{2}{\lambda}}\log\frac{1+k}{1-k} =\sqrt{\frac{2}{\lambda}}\log\frac{(1+k)^2}{1-k^2} =\sqrt{\frac{2}{\lambda}}(2\log(1 + k) - \log(1-k^2)).$$ By \eqref{e4.8}, \eqref{e4.13} and Taylor expansion, for $\lambda \gg 1$ \label{e4.19} \begin{aligned} \log(1 - k^2) &= \log\frac{2\xi_\lambda}{1 + \xi_\lambda}\\ &= \log 2 + \log\xi_\lambda - \log(1 + \xi_\lambda)\\ &= 4\log2 - \sqrt{\frac{\lambda}{2}} + 6\sqrt{2}\sqrt{\lambda}e^{-\sqrt{2\lambda}} - \left(\xi_\lambda + O(\xi_\lambda^2)\right)\\ &= 4\log2 - \sqrt{\frac{\lambda}{2}} - 8e^{-\sqrt{\lambda/2}} + 6\sqrt{2}\sqrt{\lambda}e^{-\sqrt{2\lambda}} + o(\sqrt{\lambda}e^{-\sqrt{2\lambda}}). \end{aligned} By Lemma \ref{lem4.1}, \eqref{e4.13} and Taylor expansion, for $\lambda \gg 1$, \begin{align*} % 4.20 k &= \frac{\Vert v_\lambda\Vert_\infty}{\sqrt{2-\Vert v_\lambda\Vert_\infty}} = \Vert v_\lambda\Vert_\infty(1 + \xi_\lambda)^{-1/2} \\ &= \Vert v_\lambda\Vert_\infty\Big(1 - \frac12\xi_\lambda + \frac{3}{8}\xi_\lambda^2(1 + o(1))\Big) \\ &= (1 - 4e^{-\sqrt{\lambda/2}} - 8e^{-2\sqrt{\lambda/2}}(1 + o(1))) (1 - 4e^{-\sqrt{\lambda/2}} + 24e^{-2\sqrt{\lambda/2}}(1 + o(1))) \\ &= 1 - 8e^{-\sqrt{\lambda/2}} + 32e^{-2\sqrt{\lambda/2}}(1 + o(1)). \end{align*} By this and Taylor expansion, for $\lambda \gg 1$, \begin{align*} %4.21 \log(1 + k) &= \log(2 - (1-k)) = \log 2 - \frac12(1-k) - \frac18(1-k)^2 + o((1-k)^2) \\ &= \log 2 - 4e^{-\sqrt{\lambda/2}} + 8(1 + o(1))e^{-2\sqrt{\lambda/2}}. \end{align*} By this and \eqref{e4.19}, we obtain $\log\big(\frac{1+k}{1-k}\big) = -2\log 2 + \sqrt{\frac{\lambda}{2}} -6\sqrt{2}(1 + o(1))\sqrt{\lambda}e^{-\sqrt{2\lambda}}.$ By this and \eqref{e4.18} and \eqref{e4.19}, we obtain \eqref{e4.15}. Thus the proof is complete. \end{proof} \section{Appendix} We show that $\Vert u_\lambda\Vert_\infty < \pi$ for completeness. By \eqref{e2.1}, $$-u_\lambda''(1/2) = \lambda\sin\Vert u_\lambda\Vert_\infty - g(\Vert u_\lambda\Vert_\infty) \ge 0.$$ This along with (A1) implies that there exists a non-negative integer $k$ such that $$\label{e5.1} 2k\pi < \Vert u_\lambda\Vert_\infty < (2k+1)\pi.$$ Assume that $k \ge 1$. Then by \eqref{e2.1}, there exists a unique $t_\lambda \in (0, 1/2)$ such that $u_\lambda(t_\lambda) = \Vert u_\lambda \Vert_\infty - 2k\pi$. Then by \eqref{e2.2}, \label{e5.2} \begin{aligned} &\frac12 u_\lambda'(t_\lambda)^2 - \lambda \cos u_\lambda(t_\lambda) - G(u_\lambda(t_\lambda))\\ & = \frac12 u_\lambda'(t_\lambda)^2 - \lambda \cos \Vert u_\lambda\Vert_\infty - G(\Vert u_\lambda\Vert_\infty - 2k\pi) \\ & = - \lambda \cos \Vert u_\lambda\Vert_\infty - G(\Vert u_\lambda\Vert_\infty). \end{aligned} Since $G(u)$ is strictly increasing for $u \ge 0$ by (A1), by \eqref{e5.2}, we obtain $$\frac12 u_\lambda'(t_\lambda)^2 = G(\Vert u_\lambda\Vert_\infty - 2k\pi) - G(\Vert u_\lambda\Vert_\infty) < 0.$$ This is a contradiction. Thus $k = 0$ in \eqref{e5.1} and we get our assertion. \subsection*{Acknowledgment} The author thanks the anonymous referee for his/her helpful comments. \begin{thebibliography}{00} \bibitem{b} H. Berestycki; {\it Le nombre de solutions de certains probl\emes semi-lin\'eares elliptiques}, J. Functional Analysis {40} (1981), 1--29. \bibitem{c} P. Cl\'ement and G. 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