\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 74, pp. 1--10.\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 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/74\hfil Inverse spectral problems] {Inverse spectral problems for nonlinear Sturm-Liouville problems} \author[T. Shibata\hfil EJDE-2007/74\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 August 1, 2006. Published May 15, 2007.} \subjclass[2000]{34B15} \keywords{Inverse spectral problem; $L^2$-bifurcation diagram; logistic equations} \begin{abstract} This paper concerns the nonlinear Sturm-Liouville problem $$-u''(t) + f(u(t)) = \lambda u(t), \quad u(t) > 0, \quad t \in I := (0, 1), \quad u(0) = u(1) = 0,$$ where $\lambda$ is a positive parameter. We try to determine the nonlinear term $f(u)$ by means of the global behavior of the bifurcation branch of the positive solutions in $\mathbb{R}_+ \times L^2(I)$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} We consider the nonlinear Sturm-Liouville problem \begin{gather} -u''(t) + f(u(t)) = \lambda u(t), \quad t \in I := (0, 1), \label{e1.1}\\ u(t) > 0, \quad t \in I, \label{e1.2}\\ u(0) = u(1) = 0, \label{e1.3} \end{gather} where $\lambda$ is a positive parameter. We assume that $f(u)$ satisfies the following conditions: \begin{itemize} \item[(A1)] $f(u)$ is a function of $C^1$ for $u \ge 0$ satisfying $f(0) = f'(0) = 0$, \item[(A2)] $g(u) := f(u)/u$ is strictly increasing for $u \ge 0$ ($g(0) := 0$), \item[(A3)] $g(u) \to \infty$ as $u \to \infty$. \end{itemize} Then for each given $\alpha > 0$, there exists a unique solution $(\lambda, u) = (\lambda(\alpha), u_\alpha) \in \mbox{\bf R}_+ \times C^2(\bar{I})$ with $\Vert u_\alpha \Vert_2 = \alpha$. The set $\{(\lambda(\alpha), u_\alpha); \alpha > 0\}$ gives all solutions of \eqref{e1.1}--\eqref{e1.3} and is an unbounded curve of class $C^1$ in $\mbox{\bf R}_+ \times L^2(I)$ emanating from $(\pi^2, 0)$ (cf. \cite{b1,f1}). Typical examples of $f(u)$ are as follows: \begin{gather*} %e1.4 f(u) = u^p \quad (u \ge 0), \\ f(u) = u^p \big(1 - \frac{1}{1 + u^{q}}\big) \quad (u \ge 0), \end{gather*} where $p > 1$ and $q > 0$ are constants. The equation \eqref{e1.1}--\eqref{e1.3} is motivated by the logistic equation of population dynamics and vibration of string with self-interaction, and has been extensively investigated by many authors. We refer to \cite{b1,f1,h4,r1,r2} and the references therein for the works in $L^\infty$-framework from a viewpoint of local and global bifurcation theory. On the other hand, since \eqref{e1.1}--\eqref{e1.3} is regarded as an eigenvalue problem, it seems important to study \eqref{e1.1}--\eqref{e1.3} in $L^2$-framework. We refer to \cite{b2,c1,c2,c3,c4,h1,h2,h3,s1} for other works in this direction. In particular, the asymptotic formulas for $\lambda(\alpha)$ as $\alpha \to 0$ have been established in \cite{c2,c3}. Therefore, the problem we have to consider here is a global behavior of $u_\alpha$ as $\alpha \to \infty$, and it is known from \cite{b1} that $$%1.6 \frac{u_\alpha(t)}{g^{-1}(\lambda(\alpha))} \to 1$$ locally uniformly on $I$ as $\alpha \to \infty$. By this, it is easy to see that for $\alpha \gg 1$ $\alpha = \Vert u_{\alpha}\Vert_2 = (1 + o(1))g^{-1}(\lambda(\alpha)).$ It follows from this that, in many cases, as $\alpha \to \infty$ $$\label{e1.7} \lambda(\alpha) = g(\alpha) + o(g(\alpha)).$$ Note that if $f(u) = u^{p}$ ($p > 1$), then $g(\alpha) = \alpha^{p-1}$. Motivated by \eqref{e1.7}, the following asymptotic formula for $\lambda(\alpha)$ as $\alpha \to \infty$ has been given in \cite{s1}. \begin{theorem}[\cite{s1}] \label{thm1.1} Let $f(u) = u^p$ $(p > 1)$. Let $n$ be an arbitrary, fixed, number in $\mathbb{N}_0 = \{0, 1, 2, \dots\}$. Then the following asymptotic formula holds as $\alpha \to \infty$: $$\label{e1.8} \lambda(\alpha) = \alpha^{p-1} + C_0\alpha^{(p-1)/2} +\sum_{k=0}^n \frac{a_k(p)}{(p-1)^{k+1}}C_0^{k+2} \alpha^{k(1-p)/2} +o(\alpha^{n(1-p)/2}),$$ where $C_0 = (p+3)\int_0^1 \sqrt{\frac{p-1}{p+1} - \xi^2 + \frac{2}{p+1}\xi^{p+1}}d\xi$ and $a_k(p)$ is the polynomial ($\deg a_k(p) \le k +1$) which is determined by $a_0 = 1, a_1, \dots, a_{k-1}$. \end{theorem} Consider now the implication of Theorem \ref{thm1.1} and \eqref{e1.8} from the standpoint of inverse spectral problems. \subsection*{ Problem A} Assume that \eqref{e1.8} holds for any $n \in \mathbb{N}_0$. Then does $f(u) = u^p$ ($p > 1$) hold? For the first step to solve this problem, we simplified the problem as follows in \cite{s2}: \subsection*{Problem B} Assume that the following asymptotic formula is valid as $\alpha \to \infty$. $\lambda(\alpha) = \alpha^{p-1} + C_0\alpha^{(p-1)/2} + o(\alpha^{(p-1)/2}).$ Then does $f(u) = u^p$ hold? The answer to Problem B was given in \cite{s2}. \begin{theorem}[\cite{s2}] \label{thm1.2} Let $f(u) = u^p(1 - 1/(1 + u^2))$. Furthermore, assume that $1 < p < 5$. Then as $\alpha \to \infty$ $$\label{e1.11} \lambda(\alpha) = \alpha^{p-1} + C_0\alpha^{(p-1)/2} + o(\alpha^{(p-1)/2}).$$ \end{theorem} Therefore, we regret to say that the answer to the Problem B is negative. However, it seems worth considering the following problem on which it is imposed stronger conditions than those in Problem B. \subsection*{Problem C} Assume that the following asymptotic formula is valid as $\alpha \to \infty$. $$\label{e1.12} \lambda(\alpha) = \alpha^{p-1} + C_0\alpha^{(p-1)/2} + \frac{1}{p-1}C_0^2 + o(1).$$ Then does $f(u) = u^p$ hold? The purpose of this paper is to answer this question. \begin{theorem} \label{thm1.3} Let $f(u) = u^p(1 - 1/(1 + u^q))$, where $p > 1$. \begin{itemize} \item[(i)] Assume that $(p-1)/2 < q < p+1$. Then \eqref{e1.11} holds as $\alpha \to \infty$. \item[(ii)] Assume that $p-1 < q < p+1$. Then \eqref{e1.12} holds as $\alpha \to \infty$. \end{itemize} \end{theorem} Therefore, unfortunately, the answer to the Problem C is not valid, either. It seems that the assumption in Problem C is still weak to solve the inverse spectral problem. However, it seems that Theorem \ref{thm1.3} certainly is the meaningful step to give the answer to Problem A. Our arguments here to prove Theorem \ref{thm1.3} are quite straightforward and are different from those of Theorem \ref{thm1.2}, which are variant of the proof of Theorem \ref{thm1.1}. Therefore, in \cite{s2}, it seems that the restriction $1 < p < 5$ for the case $q = 2$ is technical. However, it follows from Theorem \ref{thm1.3} (i) that this restriction is not technical but optimal. \section{Proof of Theorem \ref{thm1.3}} In this section, $C$ denotes various positive constants independent of $\lambda \gg 1$. We begin with the fundamental tools which play important roles in what follows. We denote by $(\lambda, u_\lambda)$ the solution pair of \eqref{e1.1}--\eqref{e1.3} for $\lambda > \pi^2$. We know from \cite{b1} that there exists a unique solution $u_\lambda \in C^2(\bar{I})$ of \eqref{e1.1}--\eqref{e1.3} for a given $\lambda > \pi^2$. We use this notation in what follows. Therefore, $\alpha = \Vert u_\lambda\Vert_2$. It is well known that \begin{gather} u_\lambda(t) = u_\lambda(1 - t), \quad t \in I, \label{e2.1}\\ u_\lambda'(t) > 0, \quad 0 \le t < \frac12, \label{e2.2}\\ \Vert u_\lambda\Vert_\infty = u_\lambda(\frac12). \label{e2.3} \end{gather} We know by \cite{b1} that $$\label{e2.4} \lambda = \Vert u_\lambda\Vert_\infty^{p-1}\frac {\Vert u_\lambda\Vert_\infty^q} {1 + \Vert u_\lambda\Vert_\infty^q} + \lambda_1,$$ where $\lambda_1 > 0$ is the remainder term of $\lambda$ with respect to $\Vert u_\lambda\Vert_\infty$ and depends on $\lambda$. For $\lambda \gg 1$, we have $$\label{e2.5} \lambda_1 \le C\lambda e^{-\sqrt{\kappa\lambda}/2}.$$ Here $\kappa > 0$ is a constant. For completeness, we give the proof of \eqref{e2.5} in Appendix. Now multiply \eqref{e1.1} by $u_\lambda'(t)$. Then $\Big(u_\lambda''(t) + \lambda u_\lambda(t) - u_\lambda(t)^p + \frac{u_\lambda^p(t)}{1 + u_\lambda^q(t)}\Big)u_\lambda'(t) = 0.$ This along with \eqref{e2.3} implies that \label{e2.7} \begin{aligned} &\frac12u_\lambda'(t)^2 + \frac12\lambda u_\lambda(t)^2 - \frac{1}{p+1}u_\lambda(t)^{p+1} + \int_0^{u_\lambda(t)}\frac{\xi^p}{1 + \xi^q}d\xi \equiv \mbox{constant} \\ &= \frac12\lambda \Vert u_\lambda\Vert_\infty^2 - \frac{1}{p+1}\Vert u_\lambda\Vert_\infty^{p+1} + \int_0^{\Vert u_\lambda\Vert_\infty}\frac{\xi^p}{1 + \xi^q}d\xi \quad (\mbox{put t = 1/2}). \end{aligned} Let $$\label{e2.8} L_\lambda(\theta) = \lambda (\Vert u_\lambda\Vert_\infty^2 - \theta^2) - \frac{2}{p+1}(\Vert u_\lambda\Vert_\infty^{p+1} - \theta^{p+1}) + 2 \int_\theta^{\Vert u_\lambda\Vert_\infty}\frac{\xi^p}{1 + \xi^q}d\xi.$$ This along with \eqref{e2.2} and \eqref{e2.7} implies that $u_\lambda'(t) = \sqrt{L_\lambda(u_\lambda(t))}$ for $0 \le t \le 1/2$. By this and \eqref{e2.1}, we obtain \label{e2.10} \begin{aligned} \Vert u_\lambda\Vert_\infty^2 - \alpha^2 &= 2\int_0^{1/2} \frac{(\Vert u_\lambda\Vert_\infty^2 - u_\lambda^2(t))u_\lambda'(t)} {\sqrt{L_\lambda(u_\lambda(t))}}dt \\ &= 2\int_0^{\Vert u_\lambda\Vert_\infty} \frac{(\Vert u_\lambda\Vert_\infty^2 - \theta^2)} {\sqrt{L_\lambda(\theta)}}d\theta \\ &= \frac{2\Vert u_\lambda\Vert_\infty^2}{\sqrt{\lambda}} \int_0^1 \frac{1-s^2} {\sqrt{B_\lambda(s)}}ds \\ &= \frac{2\Vert u_\lambda\Vert_\infty^2}{\sqrt{\lambda}} \Big\{\int_0^1 \frac{1-s^2}{\sqrt{A(s)}}ds +\int_0^1 \Big(\frac{1-s^2} {\sqrt{B_\lambda(s)}} - \frac{1-s^2} {\sqrt{A(s)}}\Big)ds\Big\} \\ &= \frac{2\Vert u_\lambda\Vert_\infty^2}{\sqrt{\lambda}} (C_1 + M_\lambda), \end{aligned} where \begin{gather} A(s) := 1-s^2 - \frac{2}{p+1}(1 - s^{p+1}), \label{e2.11}\\ B_\lambda(s) := 1 - s^2 - \frac{2}{p+1} \frac{\Vert u_\lambda\Vert_\infty^{p-1}}{\lambda}(1 - s^{p+1}) + \frac{2}{\lambda\Vert u_\lambda\Vert_\infty^2} \int_{\Vert u_\lambda\Vert_\infty s}^{\Vert u_\lambda\Vert_\infty} \frac{\xi^p}{1 + \xi^q}d\xi, \label{e2.12} \\ C_1 := \int_0^1 \frac{1-s^2}{\sqrt{A(s)}}ds, \label{e2.13} \\ M_\lambda := \int_0^1 \Big( \frac{1-s^2} {\sqrt{B_\lambda(s)}}- \frac{1-s^2} {\sqrt{A(s)}}\Big)\,ds. \label{e2.14} \end{gather} By \eqref{e2.10}, we prove Theorem \ref{thm1.3}. Therefore, it is important to obtain the asymptotic formula for $M_\lambda$ as $\lambda \to \infty$. To this end, we first prove the following lemma. \begin{lemma} \label{lem2.1} Let $0 < \epsilon \ll 1$ be fixed. Then there exists a constant $0 < \delta \ll 1$ such that for $1 - \epsilon \le s \le 1$ and $\lambda \gg 1$, \begin{gather} A(s) = K_0(s)(1 - s)^2, \label{e2.15} \\ B_\lambda(s) = K_1(\lambda)(1 - s) + K_2(\lambda, s)(1 - s)^2, \label{e2.16} \end{gather} where \begin{gather} (p-1)(1-\delta) \le K_0(s) \le p-1\,, \label{e2.17} \\ K_1(\lambda) = \frac{2\lambda_1}{\lambda}\,, \label{e2.18}\\ (p-1)(1-\delta) \le K_2(\lambda, s) \le p-1\,. \label{e2.19} \end{gather} \end{lemma} \begin{proof} We have $A(1) = 0$. Furthermore, $$\label{e2.21} A'(s) = -2s + 2s^p, \quad A''(s) = -2 + 2ps^{p-1}.$$ By this and Taylor expansion, for $1 - \epsilon \le s \le 1$ and $\lambda \gg 1$, we obtain \begin{align*} A(s) &= A(1) + A'(1)(s-1) + \frac12A''(s_1)(s-1)^2 \\ &= \frac12A''(s_1)(s-1)^2, \end{align*} where $1 - \epsilon \le s < s_1 < 1$. This along with \eqref{e2.21} implies \eqref{e2.15}. Next, we have $B(1) = 0$. Furthermore, $B_\lambda'(s) = -2s + 2s^p\frac{\Vert u_\lambda\Vert_\infty^{p-1}}{\lambda} - \frac{2}{\lambda\Vert u_\lambda\Vert_\infty^2} \frac{\Vert u_\lambda\Vert_\infty^{p+1}s^p} {1 + \Vert u_\lambda\Vert_\infty^qs^q}.$ Then by \eqref{e2.4}, \begin{align*} B_\lambda'(1) &= -2 + 2\frac{\Vert u_\lambda\Vert_\infty^{p-1}}{\lambda} - \frac{2}{\lambda\Vert u_\lambda\Vert_\infty^2} \frac{\Vert u_\lambda\Vert_\infty^{p+1}} {1 + \Vert u_\lambda\Vert_\infty^q} \\ &= -2\Big(1 - \frac{\Vert u_\lambda\Vert_\infty^{p+q-1}} {\lambda(1 + \Vert u_\lambda\Vert_\infty^q)}\Big) \\ &= -\frac{2\lambda_1}{\lambda}. \end{align*} Furthermore, $B_\lambda''(s) = \frac{2(p\Vert u_\lambda\Vert_\infty^{p-1}s^{p-1}-\lambda)}{\lambda} - \frac{2(p-q)s^{p+q-1}\Vert u_\lambda\Vert_\infty^{p+q-1} + ps^{p-1}\Vert u_\lambda\Vert_\infty^{p-1}} {\lambda(1 + \Vert u_\lambda\Vert_\infty^{q}s^q)^2}.$ Therefore, by \eqref{e2.4}, $2(p-1)(1-\delta) \le B_\lambda''(s) \le 2(p-1)$ for $1-\epsilon \le s \le 1$ and $\lambda \gg 1$. By this and Taylor expansion, we obtain \begin{align*} B_\lambda(s) &= B_\lambda(1) + B_\lambda'(s)(s-1) + \frac12B_\lambda''(s_2)(s-1)^2 \\ &= K_1(\lambda)(1-s) + K_2(\lambda, s)(1-s)^2, \end{align*} where $1 - \epsilon < s < s_2 < 1$. Thus the proof is complete. \end{proof} \begin{lemma} \label{lem2.2} For $\lambda \gg 1$, $$\label{e2.25} M_\lambda = C_2(q)\Vert u_\lambda\Vert_\infty^{-q}(1 + o(1)).$$ Here $C_2(q) = \int_0^1 \frac{(1-s^2)\{(1-s^{p+1})/(p+1) - (1-s^{p-q+1})/(p-q+1)\}} {A(s)^{3/2}}ds.$ \end{lemma} \begin{proof} It is easy to see that $M_\lambda = \int_0^1 \frac{(1-s^2) (A(s) - B_\lambda(s))} {\sqrt{A(s)}\sqrt{B_\lambda(s)}(\sqrt{A(s)} + \sqrt{B_\lambda(s)})}ds.$ Then by \eqref{e2.4}, \eqref{e2.11} and \eqref{e2.12}, for $0 < s \le 1$ and $\lambda \gg 1$, \label{e2.28} \begin{aligned} &A(s) - B_\lambda(s)\\ &= \frac{2}{p+1} \big(\frac{\Vert u_\lambda\Vert_\infty^{p-1}}{\lambda} - 1\big) (1 - s^{p+1})-\frac{2}{\lambda\Vert u_\lambda\Vert_\infty^2} \int_{\Vert u_\lambda\Vert_\infty s}^{\Vert u_\lambda\Vert_\infty} \frac{\xi^p}{1 + \xi^q}d\xi \\ &= \frac{2}{p+1}(1 + o(1)) \Vert u_\lambda\Vert_\infty^{-q}(1 - s^{p+1}) - \frac{2}{p-q+1}(1 + o(1)) \frac{\Vert u_\lambda\Vert_\infty^{p-q-1}} {\lambda}(1 - s^{p-q+1}) \\ &=2\Vert u_\lambda\Vert_\infty^{-q}(1 + o(1)) \big\{\frac{1}{p+1}(1 - s^{p+1}) - \frac{1}{p-q+1}(1 - s^{p-q+1})\big\}. \end{aligned} Furthermore, since $q < p+ 1$, as $\lambda \to \infty$, $$\label{e2.29} \vert A(0) - B_\lambda(0)\vert \le \big\vert \frac{2}{p+1}\big(\frac{\Vert u_\lambda\Vert_\infty^{p-1}}{\lambda} - 1\big) - \frac{2}{\lambda\Vert u_\lambda\Vert_\infty^2} \int_{0}^{\Vert u_\lambda\Vert_\infty} \frac{\xi^p}{1 + \xi^q}d\xi \big\vert \\ \le C\Vert u_\lambda\Vert_\infty^{-q}.$$ By this and \eqref{e2.28}, for $0 \le s \le 1$, as $\lambda \to \infty$, $B_\lambda(s) \to A(s)$. We apply Lebesgue's convergence theorem to our situation. Let an arbitrary $0 < \epsilon \ll 1$ be fixed. Then \label{e2.31} \begin{aligned} M_\lambda &= \int_0^{1-\epsilon}\frac{(1-s^2) (A(s) - B_\lambda(s))}{\sqrt{A(s)}\sqrt{B_\lambda(s)}(\sqrt{A(s)} + \sqrt{B_\lambda(s)})}ds \\ &\quad + \int_{1-\epsilon}^1 \frac{(1-s^2) (A(s) - B_\lambda(s))} {\sqrt{A(s)}\sqrt{B_\lambda(s)}(\sqrt{A(s)} + \sqrt{B_\lambda(s)})}ds\\ &:= M_{1,\lambda} + M_{2,\lambda}. \end{aligned} We know that $A(s)$ and $B_\lambda(s)$ is strictly decreasing for $0 \le s \le 1$ and $B_\lambda(1) = 0$ (cf. Appendix). So we see from \eqref{e2.16} that for $0 \le s \le 1 - \epsilon$ and $\lambda \gg 1$, \begin{gather*} A(s) \ge A(1-\epsilon) \ge (p-1)(1 - \delta)\epsilon^2, \\ B_\lambda(s) \ge B_\lambda(1-\epsilon) \ge (p-1)(1 - \delta)\epsilon^2 > 0. \end{gather*} By this, there exists a constant $C_\epsilon > 0$ such that for $0 \le s \le 1 - \epsilon$ and $\lambda \gg 1$, $\big\vert\frac{(1-s^2)(A(s) - B_\lambda(s))} {\sqrt{A(s)}\sqrt{B_\lambda(s)}\big(\sqrt{A(s)} + \sqrt{B_\lambda(s)}\big)} \big\vert \le C_\epsilon.$ Therefore, by \eqref{e2.4}, \eqref{e2.28}, \eqref{e2.29}, \eqref{e2.31} and Lebesgue's convergence theorem, we obtain \label{e2.34} \begin{aligned} M_{1,\lambda} &= \int_0^{1-\epsilon} \frac{(1-s^2)(A(s) - B_\lambda(s))} {\sqrt{A(s)}\sqrt{B_\lambda(s)}(\sqrt{A(s)} + \sqrt{B_\lambda(s)})}ds \\ &\to \int_0^{1-\epsilon} \frac{(1-s^2)\{(1-s^{p+1})/(p+1) - (1-s^{p-q+1})/(p-q+1)\}} {A(s)^{3/2}}ds. \end{aligned} By Lemma \ref{lem2.1}, for $1-\epsilon \le s \le 1$ and $\lambda \gg 1$, \begin{align*} &\big\vert\frac{(1-s^2)(A(s) - B_\lambda(s))} {\sqrt{A(s)}\sqrt{B_\lambda(s)}(\sqrt{A(s)} + \sqrt{B_\lambda(s)})}\big\vert \\ &\le \frac{(1-s^2)\{K_1(\lambda)(1-s) + \vert K_0(s) - K_2(\lambda,s)\vert(1-s)^2\}} {(K_1(\lambda)(1 - s) + K_2(\lambda, s)(1 - s)^2) \sqrt{K_0(s)(1 - s)^2}} \\ &\le 2\frac{(K_1(\lambda)(1-s) + \vert K_0(s) - K_2(\lambda,s)\vert(1-s)^2)} {\{K_1(\lambda)(1 - s) + K_2(\lambda, s)(1 - s)^2\} \sqrt{K_0(s)}} \\ &\le 2\frac{K_1(\lambda) + \vert K_0(s) - K_2(\lambda,s)\vert(1-s)} {\{K_1(\lambda) + K_2(\lambda, s)(1 - s)\} \sqrt{K_0(s)}} \le C. \end{align*} By this, we apply Lebesgue's convergence theorem to $M_{2,\lambda}$ to obtain \begin{align*} M_{2,\lambda} &= \int_{1-\epsilon}^1 \frac{(1-s^2)(A(s) - B_\lambda(s))} {\sqrt{A(s)}\sqrt{B_\lambda(s)}(\sqrt{A(s)} + \sqrt{B_\lambda(s)})}ds \\ &\to \int_{1-\epsilon}^1 \frac{(1-s^2)\{(1-s^{p+1})/(p+1) - (1-s^{p-q+1})/(p-q+1)\}} {A(s)^{3/2}}ds. \end{align*} By this and \eqref{e2.34}, we obtain \eqref{e2.25}. Thus the proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.3}] By \eqref{e2.10} and Lemma \ref{lem2.2}, we obtain $\Vert u_\lambda\Vert_\infty^2 - \alpha^2 = \frac{2\Vert u_\lambda\Vert_\infty^2}{\sqrt{\lambda}} (C_1 + C_2\Vert u_\lambda\Vert_\infty^{-q}(1 + o(1))).$ By this, \eqref{e2.4} and the Taylor expansion, for $\lambda \gg 1$, \label{e2.38} \begin{aligned} &\Vert u_\lambda\Vert_\infty^2- \alpha^2\\ &= 2\Vert u_\lambda\Vert_\infty^2 \Big(\Vert u_\lambda\Vert_\infty^{p-1} - \Vert u_\lambda\Vert_\infty^{p-q-1}(1 + o(1)) \Big)^{-1/2} \big(C_1 + C_2\Vert u_\lambda\Vert_\infty^{-q}(1 + o(1))\big) \\ &= 2\Vert u_\lambda\Vert_\infty^{(5-p)/2} (C_1 + (C_1/2 + C_2)\Vert u_\lambda\Vert_\infty^{-q}(1 + o(1))). \end{aligned} By \eqref{e2.4} and direct calculation, we obtain $$\label{e2.39} \Vert u_\lambda\Vert_\infty = \lambda^{1/(p-1)}\Big(1 + \frac{1}{p-1} \lambda^{-q/(p-1)} + o\big(\lambda^{-q/(p-1)}\big)\Big).$$ By this, \eqref{e2.38} and Taylor expansion, \label{e2.40} \begin{aligned} \alpha^2 &= \Vert u_\lambda\Vert_\infty^2 - 2\Vert u_\lambda\Vert_\infty^{(5-p)/2} (C_1 + (C_1/2 + C_2)\Vert u_\lambda\Vert_\infty^{-q}(1 + o(1))) \\ &= \lambda^{2/(p-1)} + \frac{2}{p-1}\lambda^{(2-q)/(p-1)} + o\big(\lambda^{(2-q)/(p-1)} \big) \\ &\quad -2\lambda^{(5-p)/(2(p-1))} \big\{C_1 + \big(\frac{5-p}{2(p-1)}C_1 + \frac12C_1 + C_2 \big)\lambda^{-q/(p-1)} (1 + o(1))\big\}. \end{aligned} (i) Assume that $q > (p-1)/2$. Then for $\lambda \gg 1$, $\lambda^{(5-p)/(2(p-1)) - q/(p-1)} \ll \lambda^{(2-q)/(p-1)} \ll \lambda^{(5-p)/(2(p-1))}.$ By this and \eqref{e2.40}, we obtain $\alpha^2 = \lambda^{2/(p-1)} -2C_1\lambda^{(5-p)/(2(p-1))} + \frac{2}{p-1}\lambda^{(2-q)/(p-1)}(1 + o(1)).$ Now we put $\lambda = \alpha^{p-1}(1 +C_3\alpha^{-\eta_1}(1 + o(1))).$ Then by direct calculation, we obtain $C_3 = (p-1)C_1, \quad \eta_1 = \frac{p-1}2.$ Since $(p-1)C_1 = C_0$, this implies Theorem \ref{thm1.3} (i). \noindent (ii) Furthermore, assume that $q > p-1$. We put $\lambda = \alpha^{p-1}(1 + (p-1)C_1\alpha^{-(p-1)/2} + C_4\alpha^{-\eta_2}(1 + o(1))).$ Then by a straightforward calculation, we obtain $C_4 = (p-1)C_1^2, \quad \eta_2 = p-1.$ This implies that for $\lambda \gg 1$ $\lambda = \alpha^{p-1} + (p-1)C_1\alpha^{(p-1)/2} + (p-1)C_1^2 + o(1).$ Since $(p-1)C_1 = C_0$, we obtain Theorem \ref{thm1.3} (ii). Thus the proof is complete. \end{proof} \section{Appendix} We first prove \eqref{e2.5}. We consider \eqref{e1.1}--\eqref{e1.3} with $f(u) = u^{p+q}/(1 + u^q)$ for $p > 1$ and $q > 0$. We put $F(u) := \int_0^u f(s)ds$. Furthermore, let $$\label{e3.1} Q_\lambda(\theta) = \lambda(\Vert u_\lambda\Vert_\infty^2 - \theta^2) - 2(F(\Vert u_\lambda\Vert_\infty) - F(\theta)).$$ For $0 \le t \le 1$, \eqref{e3.1} is equivalent to \eqref{e2.8}. Then for $0 \le t \le 1/2$, we obtain $$\label{e3.2} u_\lambda'(t) = \sqrt{Q(u_\lambda(t))}.$$ By this, we obtain $$\label{e3.3} \frac12 = \int_0^{1/2} \frac{u_\lambda'(t)} {\sqrt{Q(u_\lambda(t)}}dt = \int_0^{\Vert u_\lambda\Vert_\infty} \frac{1}{\sqrt{Q_\lambda(\theta)}}d\theta = \frac{1}{\sqrt{\lambda}} \int_0^1 \frac{1}{\sqrt{R_\lambda(s)}} ds,$$ where $$\label{e3.4} R_\lambda(s) := 1 - s^2 - \frac{2} {\lambda\Vert u_\lambda\Vert_\infty^2} (F(\Vert u_\lambda\Vert_\infty) - F(\Vert u_\lambda\Vert_\infty s))$$ Let an arbitrary $0 < \epsilon \ll 1$ be fixed. Then for $1 - \epsilon \le s \le 1$, by Taylor expansion, $F(\Vert u_\lambda\Vert_\infty s) = F(\Vert u_\lambda\Vert_\infty) + f(\Vert u_\lambda\Vert_\infty)\Vert u_\lambda\Vert_\infty(s-1) + \frac12f'(\Vert u_\lambda\Vert_\infty s_1) \Vert u_\lambda\Vert_\infty^2(s-1)^2,$ where $s < s_1 < 1$ and $s_1$ depends on $s$. Since $f'(u) = pu^{p-1}(1 + o(1))$ for $u \gg 1$, there exists a constant $\delta > 0$ such that for $1 - \epsilon \le s \le 1$ \label{e3.6} \begin{aligned} R_\lambda(s) &=2\big(1 - \frac{f(\Vert u_\lambda\Vert_\infty)} {\lambda\Vert u_\lambda\Vert_\infty} \big)(1 - s) + \big(\frac{f'(\Vert u_\lambda\Vert_\infty s_1)}{\lambda} - 1 \big) (1 - s)^2 \\ &\ge 2\xi(1 - s) + \delta(1 - s)^2, \end{aligned} where $$\label{e3.7} \xi := 1 - \frac{f(\Vert u_\lambda\Vert_\infty)} {\lambda\Vert u_\lambda\Vert_\infty} > 0.$$ We know that $\xi \to 0$ as $\lambda \to \infty$. By this, \eqref{e3.3} and \eqref{e3.6}, we obtain \begin{align*} \frac{\sqrt{\lambda}}{2} &\le \int_0^{1-\epsilon}\frac{1}{\sqrt{R_\lambda(s)}}ds + \int_{1-\epsilon}^1 \frac{1}{\sqrt{2\xi(1-s) + \delta(1-s)^2}}ds \\ &\le C + \int_0^\epsilon\frac{1}{\sqrt{2\xi v + \delta v^2}}dv \\ &= C + \delta^{-1} \left[\log \vert 2\delta v + 2\xi + 2\sqrt{\delta(\delta v^2 + 2\xi v)} \vert \right]_0^\epsilon \\ &= \delta^{-1}(\log C - \log 2\xi). \end{align*} By this, we obtain $2\xi \le Ce^{-\delta\sqrt{\lambda}/2},$ which along with \eqref{e3.7} implies $\lambda \le \frac{f(\Vert u_\lambda\Vert_\infty)} {\Vert u_\lambda\Vert_\infty} + \frac{C}{2}\lambda e^{-\delta\sqrt{\lambda}/2}.$ Thus the proof of \eqref{e2.5} is complete. We next prove that $B_\lambda(s)$ is decreasing for $0 \le s \le 1$. Indeed, since $B_\lambda(s) = R_\lambda(s)$ in \eqref{e3.4}, by (A2) and \eqref{e2.4}, \begin{align*} B_\lambda'(s) &= -2s + \frac{2s}{\lambda} \frac{f(\Vert u_\lambda\Vert_\infty s)}{\Vert u_\lambda\Vert_\infty s} \\ &\le -2s + \frac{2s}{\lambda} \frac{f(\Vert u_\lambda\Vert_\infty)}{\Vert u_\lambda\Vert_\infty} < 0. \end{align*} Thus the proof is complete. \begin{thebibliography}{00} \bibitem{b1} H. Berestycki, {\it Le nombre de solutions de certains probl\`emes semi-lin\'eares elliptiques}, J. Funct. Anal. {\bf 40} (1981), 1--29. \bibitem{b2} A. Bongers, H.-P. Heinz and T. K\"upper, {\it Existence and bifurcation theorems for nonlinear elliptic eigenvalue problems on unbounded domains}, J. Differential Equations {\bf 47} (1983), 327--357. \bibitem{c1} J. Chabrowski, {\it On nonlinear eigenvalue problems}, Forum Math. {\bf 4} (1992), 359--375. \bibitem{c2} R. Chiappinelli, {\it Remarks on bifurcation for elliptic operators with odd nonlinearity}, Israel J. 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