\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{epic,amssymb}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 201, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/201\hfil Global and local behavior]
{Global and local behavior of the bifurcation diagrams
for semilinear problems}

\author[T. Shibata \hfil EJDE-2016/201\hfilneg]
{Tetsutaro Shibata}

\address{Tetsutaro Shibata \newline
Laboratory of Mathematics,
Institute of Engineering,
Hiroshima University,
Higashi-Hiroshima, 739-8527, Japan}
\email{shibata@amath.hiroshima-u.ac.jp}

\thanks{Submitted June 20, 2016. Published July 27, 2016.}
\subjclass[2010]{34F10}
\keywords{Asymptotic behavior; parabola-like bifurcation curves}

\begin{abstract}
 We consider the nonlinear eigenvalue problem
 \begin{gather*}
 u''(t) + \lambda (u(t)^p - u(t)^q) = 0, \quad u(t) > 0,\quad -1<t<1,\\
 u(1) = u(-1) = 0,
 \end{gather*}
 where $1 < p < q$ are constants and $\lambda > 0$ is a parameter.
 It is known in \cite{o1} that the bifurcation curve $\lambda(\alpha)$
 consists of two branches, which are denoted by $\lambda_\pm(\alpha)$.
 Here, $\alpha = \| u_\lambda\|_\infty$. We establish the
 asymptotic behavior of the turning point $\alpha_p$
 of $\lambda(\alpha)$, namely, the point which satisfies
 $d\lambda(\alpha_p)/d\alpha = 0$ as $p \to q$ and $p \to 1$.
 We also establish the asymptotic formulas for $\lambda_{+}(\alpha)$
 and $\lambda_{-}(\alpha)$ as $\alpha \to 1$ and $\alpha \to 0$, respectively.
\end{abstract}

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

\section{Introduction}

We consider the nonlinear eigenvalue problem
\begin{gather}
u''(t) + \lambda (u(t)^p - u(t)^q)
= 0, \quad t \in I := (-1,1), \label{e1.1}\\
u(t) > 0, \quad t \in I, \label{e1.2}\\
u(1) = u(-1) = 0, \label{e1.3}
\end{gather}
where $1 < p < q$ are constants and $\lambda > 0$ is a parameter.

Nonlinear elliptic eigenvalue problems
have been studied by many authors from the
standpoint of bifurcation analysis . We refer to 
\cite{b1,c1,c2,c3,c4,c5,c6,k1,l1,s1} and the
references therein. In particular, it is quite significant to
study the equations which have fine structures of the
bifurcation diagrams. Among other things,
\eqref{e1.1}--\eqref{e1.3} is well known as the model equation which has
the parabola-like bifurcation curve
(cf. Figure \ref{fig1} below). More precisely,
it has been
proved in  \cite[Theorem 2.15]{k2} and  \cite[Theorem 6.19]{o1} the
following basic properties of the structure
of bifurcation diagram for \eqref{e1.1}--\eqref{e1.3}.



\begin{theorem}[\cite{k2,o1}] \label{thm1.0}
 Assume that $1 < p < q$. Then there exists
a critical $\lambda_0$ such that \eqref{e1.1}--\eqref{e1.3} has no positive
solution for $0 < \lambda < \lambda_0$, exactly one positive
solution at $\lambda = \lambda_0$, and exactly two
positive solutions for $\lambda > \lambda_0$.
Furthermore, all solutions lie on a smooth solution curve, and $\lambda$ is
parameterized by $\alpha := \| u_\lambda\|_\infty$ as
$\lambda = \lambda(\alpha)$.
Further, $\lambda(\alpha)$ consists of two branches
$\lambda_\pm(\alpha)$ and
is a parabola-like curve with exactly one turn to the right
at $\alpha = \alpha_p \in (0, 1)$, namely, $\alpha_p$ is the
unique point which satisfies $\lambda'(\alpha_p) = 0$.
Furthermore,
\begin{gather}
\lambda_{-}(\alpha) \to \infty \quad \text{as }\alpha \to 0, \label{e1.4}\\
u_{-}(\lambda, 0) \to 0, \label{e1.5} \\
\lambda_{+}(\alpha) \to \infty \quad \text{as } \alpha \to 1, \label{e1.6} \\
u_{+}(\lambda,t) \to 1 \quad (t \in I) \quad
\text{as }\lambda \to \infty, \label{e1.7}
\end{gather}
where $u_{\pm}(\lambda, t)$ is a solution of \eqref{e1.1}--\eqref{e1.3}
corresponding to $\lambda = \lambda_\pm(\alpha)$.
\end{theorem}

\begin{figure}[ht] 
\begin{center} 
\setlength{\unitlength}{1mm}
\begin{picture}(68,28)(0,0)
%\scriptsize
\put(0,5){\vector(1,0){65}}
\put(5,0){\vector(0,1){25}}
\qbezier(60,7)(-20,13)(60,19)
\dottedline{2}(5,21)(60,21)
\dottedline{2}(5,13)(20,13)
\dottedline{2}(20,5)(20,13)
\put(1,25){$\alpha$}
\put(2,20){1}
\put(1,13){$\alpha_p$}
\put(2,2){0}
\put(18,1){$\lambda_0$}
\put(63,1){$\lambda$}
\put(62,7){$\lambda_-(\alpha)$}
\put(62,18){$\lambda_+(\alpha)$}
\end{picture} 
\end{center}
\caption{Bifurcation curve $\lambda_{\pm}(\alpha)$}
\label{fig1}
\end{figure}

On the other hand, if $p = 1$, then we know from \cite{b1} that
the shape of the bifurcation curve looks as in Figure \ref{fig2}.

\begin{figure}[ht] 
\begin{center} 
\setlength{\unitlength}{1mm}
\begin{picture}(68,28)(0,0)
\put(0,5){\vector(1,0){65}}
\put(5,0){\vector(0,1){25}}
\qbezier(20,5)(20,15)(60,19)
\dottedline{2}(5,21)(60,21)
\put(1,25){$\alpha$}
\put(2,20){1}
\put(2,2){0}
\put(17,1){$\pi^2/4$}
\put(63,1){$\lambda$}
\put(62,18){$\lambda(\alpha)$}
\end{picture} 
\end{center}
\caption{Bifurcation curve for $p=1$}
\label{fig2}
\end{figure}


Consider now the relationship between Figures \ref{fig1} and \ref{fig2}.
It is quite natural to expect that if $p \to 1$, then the
shape of the graph in Figure \ref{fig1} will approach to the graph
in Figure \ref{fig2} in some sense.
To obtain the evidence of this expectation,
it is important to investigate
the asymptotic behavior of the graph in Figure \ref{fig1} as $p \to 1$.
Related to the observation above,
it is worth studying the asymptotic behavior of the graph in Figure \ref{fig1}
as $p \to q$.


Now we study the global behavior of $\lambda_{\pm}(\alpha)$.

\begin{theorem} \label{thm1.1}
 Let $1 < p < q$ be fixed constants.
Let an arbitrary $0 < \delta \ll 1$ be fixed.
(i) As $\alpha \to 1$,
\begin{equation} \label{e1.8}
\sqrt{\lambda_+(\alpha)} = \sqrt{\frac{1}{q-p}}
(1 + O(\delta))\alpha^{(1-p)/2} |\log(1-\alpha^{q-p})|
+ O(\delta^{-1}).
\end{equation}
(ii) As $\alpha \to 0$,
\begin{equation} \label{e1.9}
\sqrt{\lambda_{-}(\alpha)} = \sqrt{\frac{p+1}{2}}\alpha^{(1-p)/2}
(b_0 + b_1\alpha^{q-p} + O(\alpha^{2(q-p)}),
\end{equation}
where
\begin{gather} \label{e1.10}
b_0 = \int_0^1 \frac{1}{\sqrt{1-s^{p+1}}}ds, \\
b_1 = \frac{p+1}{2(q+1)}\int_0^1
\frac{1-s^{q+1}}{(1-s^{p+1})^{3/2}}ds. \label{e1.11}
\end{gather}
\end{theorem}

Our main purpose is to study the local behavior of $0 < \alpha_p < 1$,
which is the turning point of $\lambda(\alpha)$. We show how
Figure \ref{fig1} tends to Figure \ref{fig2} as $p \to 1$. To do this,
we establish the asymptotic formula for $\alpha_p$ as $p \to 1$.

\begin{theorem} \label{thm1.2}
 Let $q > 1$ be a fixed constant. Then as $p \to 1$,
\begin{equation} \label{e1.12}
\alpha_p^{q-p} = \frac{b_0(q+1)}{k_0(p+1)(q-p)}(p-1)+ O((p-1)^2),
\end{equation}
where
\begin{equation} \label{e1.13}
k_0 = \int_0^1 \frac{1-s^{q+1}}{(1-s^{p+1})^{3/2}}ds.
\end{equation}
Furthermore,
\begin{equation} \label{e1.14}
\sqrt{\lambda(\alpha_p)} = D_p(p-1)^{(1-p)/(2(q-p))}(b_0 + O(p-1)),
\end{equation}
where
\begin{equation} \label{e1.15}
D_p := \sqrt{\frac{p+1}{2}}\big[\frac{b_0(q+1)}{k_0(p+1)(q-p)}
\big]^{(1-p)/(2(q-p))}.
\end{equation}
\end{theorem}

Clearly, as $p \to 1$,
\begin{equation} \label{e1.16}
D_p \to 1, \quad (p-1)^{(1-p)/(2(q-p))} \to 1, \quad b_0 \to \frac{\pi}{2}.
\end{equation}
Therefore, we see from \eqref{e1.14} that $\lambda(\alpha_p) \to \pi^2/4$ as
$p \to 1$, and the shape of the bifurcation curve when $0 < p-1 \ll 1$ is
as shown in Figure \ref{fig3}.

\begin{figure}[ht] 
\begin{center} 
\setlength{\unitlength}{1mm}
\begin{picture}(68,28)(0,0)
%\scriptsize
\put(0,5){\vector(1,0){65}}
\put(5,0){\vector(0,1){25}}
\qbezier(60,6.5)(-10,6.5)(60,19)
\dottedline{2}(5,21)(60,21)
\dottedline{2}(5,9.8)(25,9.8)
\dottedline{2}(25,5)(25,9.8)
\put(1,25){$\alpha$}
\put(2,20){1}
\put(1,9.4){$\alpha_p$}
\put(2,2){0}
\put(21,1){$\pi^2/4$}
\put(63,1){$\lambda$}
\put(62,7){$\lambda_-(\alpha)$}
\put(62,18){$\lambda_+(\alpha)$}
\end{picture} 
\end{center}
\caption{Bifurcation curve for $0<p-1\ll 1$}
\label{fig3}
\end{figure}

Finally, we establish the asymptotic formula for $\alpha_p$ as $p \to q$.

\begin{theorem} \label{thm1.3}
 Let $q > 1$ be fixed. Then 
\begin{equation} \label{e1.17}
\alpha_p^{q-p} = 1 - O\Big(\frac{q-p}{|\log (q-p)|^{2/3}}\Big)
\quad\text{as } p \to q.
\end{equation}
\end{theorem}

\begin{figure}[ht] 
\begin{center} 
\setlength{\unitlength}{1mm}
\begin{picture}(68,28)(0,0)
%\scriptsize
\put(0,5){\vector(1,0){65}}
\put(5,0){\vector(0,1){25}}
\qbezier(60,7)(-10,18)(60,19)
\dottedline{2}(5,21)(60,21)
\dottedline{2}(5,15.5)(25,15.5)
\dottedline{2}(25,5)(25,15.5)
\put(1,25){$\alpha$}
\put(2,20){1}
\put(1,16){$\alpha_p$}
\put(2,2){0}
\put(19,1){$\lambda_+(\alpha_p)$}
\put(63,1){$\lambda$}
\put(62,7){$\lambda_-(\alpha)$}
\put(62,18){$\lambda_+(\alpha)$}
\end{picture} 
\end{center}
\caption{Bifurcation curve for $0<q-p\ll 1$}
\label{fig4}
\end{figure}

It should be mentioned that as far as the author knows,
the results such as Theorems \ref{thm1.2} and \ref{thm1.3} seem to be new.

Our methods to prove Theorems \ref{thm1.1}--\ref{thm1.3} are based on the precise and
complicated calculation of the time map.

\section{Proof of Theorem \ref{thm1.1}}

In this section, we let $0 < p-1 \ll 1$.
We know that if
$(\lambda, u) \in \mathbb{R}_+ \times C^2(\bar{I})$ satisfies
\eqref{e1.1}--\eqref{e1.3}, then
\begin{gather}
u(t) = u(-t), \quad 0 \le t \le 1, \label{e2.1} \\
u(0) = \max_{-1\le t \le 1}u(t), \label{e2.2} \\
u'(t) < 0, \quad 0 < t \le 1. \label{e2.3}
\end{gather}
We parameterize the solution pair by using the $L^\infty$
norm of the solution $\alpha = \| u_\lambda\|_\infty$ such as
$(\lambda, u) = (\lambda(\alpha), u_\alpha)$ ($0 < \alpha < 1$).
By \eqref{e1.1}, for $t \in \bar{I}$,
\[
[u_\alpha''(t) + \lambda(u_\alpha(t)^p - u_\alpha(t)^q) ]
u_\alpha'(t) = 0.
\]
This implies that for $t \in \bar{I}$,
\[
\frac{d}{dt}\Big[ \frac12u_\alpha'(t)^2
+ \lambda \Big(\frac{1}{p+1}u_\alpha(t)^{p+1}
- \frac{1}{q+1}u_\alpha(t)^{q+1}\Big)\Big] = 0.
\]
By this, \eqref{e2.2} and putting $t = 0$, for $-1 \le t \le 1$, we obtain
\begin{align*}
&u_\alpha'(t)^2 + 2\lambda \Big(\frac{1}{p+1}u_\alpha(t)^{p+1}
- \frac{1}{q+1}u_\alpha(t)^{q+1}\Big) \\
&= \text{constant} = 2\lambda \Big(\frac{1}{p+1}\alpha^{p+1}
- \frac{1}{q+1}\alpha^{q+1}\Big).
\end{align*}
By \eqref{e2.3}, for $-1 \le t \le 0$, we obtain
\begin{equation} \label{e2.4}
u_\alpha'(t) = \sqrt{2\lambda}
\sqrt{\frac{1}{p+1}(\alpha^{p+1} -u_\alpha(t)^{p+1})
- \frac{1}{q+1}(\alpha^{q+1} - u_\alpha(t)^{q+1})}.
\end{equation}
By this and putting $\alpha s = u_\alpha(t)$, we obtain
\begin{equation} \label{e2.5}
\begin{aligned}
\sqrt{\lambda}
&= \frac{1}{\sqrt{2}}\int_{-1}^0 \frac{u_\alpha'(t)}
{\sqrt{(\alpha^{p+1} -u_\alpha(t)^{p+1})/(p+1)
- (q+1)(\alpha^{q+1} - u_\alpha(t))/(q+1)}}dt\\
&= \sqrt{\frac{p+1}{2}}\alpha^{(1-p)/2}
\int_{0}^1 \frac{1}
{\sqrt{1-s^{p+1}-(p+1)\alpha^{q-p}(1-s^{q+1})/(q+1)}}ds.
\end{aligned}
\end{equation}
We put
\begin{equation} \label{e2.6}
A_p(\alpha) := \int_{0}^1 \frac{1}
{\sqrt{1-s^{p+1}-(p+1)\alpha^{q-p}(1-s^{q+1})/(q+1)}}ds.
\end{equation}
Furthermore, we put for $0 \le s \le 1$,
\begin{equation} \label{e2.7}
h(s) := 1-s^{p+1}-\frac{p+1}{q+1}\alpha^{q-p}(1-s^{q+1}).
\end{equation}
It is clear that
\begin{gather}
h'(s) = -(p+1)s^p + (p+1)s^q\alpha^{q-p}, \label{e2.8}\\
h''(s) = (p+1) s^{p-1}(qs^{q-p}\alpha^{q-p} - p). \label{e2.9}
\end{gather}

\begin{lemma} \label{lem2.1}
Let an arbitrary $0 < \delta \ll 1$ be fixed.
Then as $\alpha \to 1$,
\begin{equation} \label{e2.10}
A_p(\alpha) = \sqrt{\frac{2}{(p+1)(q-p)}}
(1 + O(\delta)) | \log(1-\alpha^{q-p})|
+ O(\delta^{-1}).
\end{equation}
\end{lemma}

\begin{proof}
 In what follows, $C$ denotes various
positive constants independent of $\alpha$ and $\delta$.
We apply the argument of  \cite[Theorem 1.1]{s2}
to the proof of \eqref{e2.10}.
By \eqref{e2.6}, we put
\begin{equation} \label{e2.11}
A_p(\alpha) = K_1 + K_2
:= \int_0^{1-\delta} \frac{1}{\sqrt{h(s)}}ds
+ \int_{1-\delta}^1 \frac{1}{\sqrt{h(s)}}ds.
\end{equation}
We first calculate $K_1$. For $0 \le s \le 1$, we have
\begin{equation} \label{e2.12}
h(s) > h_1(s) := 1-s^{p+1} - \frac{p+1}{q+1}(1-s^{q+1}).
\end{equation}
Then for $0 \le s \le 1$,
\begin{equation} \label{e2.13}
h_1'(s) = (p+1)s^p(-1+s^{q-p}) \le 0.
\end{equation}
This implies that for $0 \le s < 1 - \delta$, by Taylor expansion,
\begin{equation} \label{e2.14}
h(s) > h_1(s) > h_1(1-\delta) \ge C\delta^2.
\end{equation}
By this and \eqref{e2.11}, we obtain
\begin{equation} \label{e2.15}
K_1 \le C\int_0^{1-\delta} \frac{1}{\sqrt{\delta^2}}ds = O(\delta^{-1}).
\end{equation}
We next calculate $K_2$. By \eqref{e2.8}, \eqref{e2.9} and Taylor expansion,
for $1-\delta < s < 1$, there exists $s < \xi_s < 1$ such that
\begin{equation} \label{e2.16}
\begin{aligned}
h(s) &= h(1) + h'(1)(s-1) + \frac12h''(\xi_s)(s-1)^2 \\
&= (p+1)(1-\alpha^{q-p})(1-s) + \frac12(p+1)\xi_s^{p-1}
(q\xi_s^{q-p}\alpha^{q-p}-p)(1-s)^2\\
&= d_1(1-s)^2 + d_2(1-s),
\end{aligned}
\end{equation}
where
\begin{equation} \label{e2.17}
d_1 := \frac12(p+1)\xi_s^{p-1}(q\xi_s^{q-p}\alpha^{q-p}-p),
\quad d_2 := (p+1)(1-\alpha^{q-p}).
\end{equation}
By this, \eqref{e2.16} and putting $x = 1-s$, we obtain
\begin{equation} \label{e2.18}
\begin{aligned}
K_2 &= \int_0^\delta \frac{1}{\sqrt{d_1 x^2 + d_2x}}dx\\
&= \frac{1}{\sqrt{d_1}}
\big[\log| 2d_1x + d_2 + 2\sqrt{d_1(d_1x^2+d_2x)}|\big]_0^\delta\\
&= \frac{1}{\sqrt{d_1}}
\Big(\log(2d_1\delta + d_2 + 2\sqrt{d_1(d_1\delta^2+d_2\delta)})
- \log| d_2|\Big)\\
&=  \frac{1}{\sqrt{d_1}}(| \log(C\delta +
C(1-\alpha^{q-p}))|-\log(p+1) + | \log(1-\alpha^{q-p})|)\\
&= \frac{1}{\sqrt{d_1}}(O(| \log\delta|)
+ | \log(1-\alpha^{q-p})|).
\end{aligned}
\end{equation}
We have
\begin{equation} \label{e2.19}
\frac{1}{\sqrt{d_1}} = \sqrt{\frac{2}{(p+1)(q-p)}}
+ \sqrt{\frac{2}{p+1}}L_1,
\end{equation}
where
\begin{equation} \label{e2.20}
L_1 := \sqrt{\frac{1}{\xi_s^{p-1}
(q\xi_s^{q-p}\alpha_p^{q-p}-p)}} - \sqrt{\frac{1}{q-p}}.
\end{equation}
Then by direct calculation, as $\alpha \to 1$, we obtain
\begin{equation} \label{e2.21}
L_1 \le C(1-\alpha^{q-p} + \delta) \le C\delta.
\end{equation}
By this, \eqref{e2.11}, \eqref{e2.15}, \eqref{e2.18} and \eqref{e2.19},
we obtain \eqref{e2.10}. The proof is complete
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
By \eqref{e2.5}, \eqref{e2.6} and Lemma \ref{lem2.1}, we obtain \eqref{e1.8}.
We next prove \eqref{e1.9}. By \eqref{e2.6} and Taylor
expansion theorem, for $0 < \alpha \ll 1$, we obtain
\begin{equation} \label{e2.22}
\begin{aligned}
A_p(\alpha)
&= \int_0^1 \frac{1}{\sqrt{1-s^{p+1}}
\sqrt{1-\frac{p+1}{q+1}\alpha^{q-p}\frac{1-s^{q+1}}{1-s^{p+1}}}}ds
\\
&= \int_0^1 \frac{1}{\sqrt{1-s^{p+1}}}
\big\{1+\frac{p+1}{2(q+1)}\alpha^{q-p}
\frac{1-s^{q+1}}{1-s^{p+1}} + O(\alpha^{2(q-p)})\big\}ds
\\
&= b_0 + b_1\alpha^{q-p} + O(\alpha^{2(q-p)}).
\end{aligned}
\end{equation}
By this and \eqref{e2.5}, we obtain \eqref{e1.9}.
Thus the proof is complete.
\end{proof}

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

In this section, we study the asymptotic
behavior of $\alpha_p$ as $p \to 1$.
We put
\begin{equation} \label{e3.1}
B_p(\alpha) := \int_0^1 \frac{1-s^{q+1}}
{\{1-s^{p+1}-(p+1)\alpha^{q-p}(1-s^{q+1})/(q+1)\}^{3/2}}ds.
\end{equation}
By this, \eqref{e2.5} and \eqref{e2.6}, we have
\begin{equation} \label{e3.2}
(\sqrt{\lambda(\alpha)})' =
\sqrt{\frac{p+1}{2}}\alpha^{-(1+p)/2}
\big\{-\frac{p-1}{2}A_p(\alpha) + \frac{(p+1)(q-p)}{2(q+1)}B_p(\alpha)\alpha^{q-p}
\big\}.
\end{equation}

\begin{lemma} \label{lem3.1}
$\alpha_p \to 0$ as $p \to 1$.
\end{lemma}

\begin{proof}
Since $\lambda'(\alpha)
= 2\sqrt{\lambda(\alpha)}(\sqrt{\lambda(\alpha)})'$,
by \eqref{e3.2}, we consider the equation
\begin{equation} \label{e3.3}
-\frac{p-1}{2}A_p(\alpha_p) + \frac{(p+1)(q-p)}{2(q+1)}B_p(\alpha_p)
\alpha_p^{q-p} = 0.
\end{equation}
Then there are three cases to consider. Let an
arbitrary $0 < \delta \ll 1$ be fixed.
\smallskip

\noindent\textbf{Case 1.} Assume that
there exists a subsequence of $\{\alpha_p\}$, denoted by
$\{\alpha_p\}$ again, and constants $c_0$ and $c_1$ such that
$0 < c_0 < \alpha_p < c_1 < 1$
for $\lambda \gg 1$. Then for $0 < p-1 \ll 1$,
\begin{equation} \label{e3.4}
\begin{aligned}
B_p(\alpha_p)
&= \int_0^{1-\delta} \frac{1-s^{q+1}} {h(s)^{3/2}}ds + \int_{1-\delta}^{1}
\frac{1-s^{q+1}} {h(s)^{3/2}}ds\\
&=: I_1 + I_2 \\
&> I_1 > C\int_0^{1-\delta} \frac{(q+1)\delta}{[1-(p+1)c_0^{q-p}/(q+1)]^{3/2}}ds
\\
&= C_{1,\delta} > 0.
\end{aligned}
\end{equation}
By \eqref{e2.11}, we also obtain
\begin{equation} \label{e3.5}
\begin{aligned}
A_p(\alpha_p)
&= K_1 + K_2 := \int_{0}^{1-\delta} \frac{1}
{\sqrt{h(s)}}ds + \int_{1-\delta}^1 \frac{1}
{\sqrt{h(s)}}ds \\
&< C_{2,\delta} + K_2.
\end{aligned}
\end{equation}
By Taylor expansion and \eqref{e2.8}, for $1-\delta < s < 1$, we have
\begin{equation} \label{e3.6}
\begin{aligned}
h(s)
&= h(1) + h'(\tilde{s})(s-1)
= (p+1)\tilde{s}^p(1 - \alpha_p^{q-p}\tilde{s}^{q-p})(1-s)\\
&> (p+1)(1-\delta)^p(1-\alpha_p^{q-p})(1-s)\\
&> (p+1)(1-\delta)^p(1-c_1^{q-p})(1-s),
\end{aligned}
\end{equation}
where $s < \tilde{s} < 1$.
By this and \eqref{e3.5}, we obtain
\begin{equation} \label{e3.7}
\begin{aligned}
A_p &< C_{2,\delta} + K_2 \\
&< C_{2,\delta} + \int_{1-\delta}^1
 \frac{1}{\sqrt{(p+1)(1-\delta)^p(1-c_1^{q-p})(1-s)}}ds
= C_{3,\delta}.
\end{aligned}
\end{equation}
By this and \eqref{e3.3}, we obtain
\begin{equation} \label{e3.8}
\frac{p-1}{2}C_{3,\delta}
> \frac{p-1}{2}A_p
= \frac{(p+1)(q-p)}{2(q+1)}B_p\alpha_p^{q-p}
 > \frac{(p+1)(q-p)}{2(q+1)} C_{1,\delta}c_0^{q-p}.
\end{equation}
This is a contradiction, since $0 < p-1 \ll 1$.
\smallskip

\noindent\textbf{Case 2.} We assume that
$\alpha_p \to 1$ and derive a contradiction.
By \eqref{e2.18}, \eqref{e3.4}, Taylor expansion and putting $x = 1-s$, we obtain
\begin{equation} \label{e3.9}
\begin{aligned}
I_2 &\ge C\int_0^\delta \frac{(q+1)x}{(d_1x^2+d_2x)^{3/2}}dx\\
&= C\Big[\frac{2d_2(q+1)x}{d_2^2\sqrt{d_1x^2+d_2x}}\Big]_0^\delta \\
&= \frac{2C(q+1)\delta}{d_2\sqrt{d_1\delta^2 + d_2\delta}}.
\end{aligned}
\end{equation}
By this, \eqref{e2.17} and \eqref{e3.2}, as $\alpha_p \to 1$, we obtain
\begin{equation} \label{e3.10}
0 = (\sqrt{\lambda(\alpha_p)})'
\ge -C| \log(1-\alpha_p^{q-p})| + C_\delta\frac{1}{1-\alpha_p^{q-p}} \to \infty.
\end{equation}
This is a contradiction. Thus the proof is complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.2}]
By \eqref{e2.7}, Lemma \ref{lem3.1} and Taylor
expansion, we obtain
\begin{equation} \label{e3.11}
\begin{aligned}
&B_p(\alpha_p) \\
&= \int_0^1 \frac{1-s^{q+1}}
{(1-s^{p+1})^{3/2}
(1-\frac{p+1}{q+1}\alpha_p^{q-p}\frac{1-s^{q+1}}{1-s^{p+1}})^{3/2}}ds
\\
&= \int_0^1 \frac{1-s^{q+1}}{(1-s^{p+1})^{3/2}}
\big\{1+\frac{3(p+1)}{2(q+1)}\alpha_p^{q-p}
\frac{1-s^{q+1}}{1-s^{p+1}} + O(\alpha^{2(q-p)})\big\}ds\\
&= k_0 + k_1\alpha^{q-p} + O(\alpha^{2(q-p)}),
\end{aligned}
\end{equation}
where $k_0$ is a constant defined in \eqref{e1.13} and
\begin{equation} \label{e3.12}
k_1 = \frac{3(p+1)}{2(q+1)}\int_0^1
\frac{(1-s^{q+1})^2}{(1-s^{p+1})^{5/2}}ds.
\end{equation}
By this, \eqref{e2.22} and \eqref{e3.3}, for $0 < p-1 \ll 1$, we obtain
\begin{equation} \label{e3.13}
\alpha_p^{q-p} = \frac{b_0(q+1)}{k_0(p+1)(q-p)}(p-1)
+ O((p-1)^2).
\end{equation}
This implies \eqref{e1.12}.
Finally, we show \eqref{e1.14}.
By \eqref{e3.13}, as $p \to 1$, we obtain
\begin{equation} \label{e3.14}
\begin{aligned}
\log\alpha_p^{(1-p)/2}
&=\frac{1-p}{2}\log\alpha_p\\
&= \frac{1-p}{2(q-p)}
\Big(\log(p-1) + \log\Big(\frac{b_0(q+1)(1 + o(1))}
{k_0(p+1)(q-p)}\Big)\Big) \to 0.
\end{aligned}
\end{equation}
This implies that $\alpha_p^{(1-p)/2} \to 1$ as $p \to 1$.
By this, \eqref{e1.9}, \eqref{e2.5} and \eqref{e3.12}, we obtain
\begin{equation} \label{e3.15}
\begin{aligned}
\sqrt{\lambda(\alpha_p)}
&= \sqrt{\lambda_{-}(\alpha_p)} \\
&=\sqrt{\frac{p+1}{2}}\Big[\frac{b_0(q+1)}{k_0(p+1)(q-p)}(p-1)
+ O((p-1)^2)\Big]^{(1-p)/(2(q-p))} \\
&\quad\times (b_0 + b_1\alpha_p^{q-p} + O(\alpha_p^{2(q-p)})\\
&= \sqrt{\frac{p+1}{2}}\Big[\frac{b_0(q+1)}{k_0(p+1)(q-p)}
\Big]^{(1-p)/(2(q-p))} \\
&\quad\times (p-1)^{(1-p)/(2(q-p))}(b_0 + O(p-1)).
\end{aligned}
\end{equation}
The proof is complete.
\end{proof}

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

In this section, we prove Theorems \ref{thm1.3} by following the strategy of
the proof of Theorems \ref{thm1.1} and \ref{thm1.2}.

\begin{lemma} \label{lem4.1} 
 $\alpha_p \to 1$ as $p \to q$.
\end{lemma}

\begin{proof} As in the proof of Lemma \ref{lem3.1}, we  consider \eqref{e3.3}.
There are two cases to be considered.
\smallskip

\noindent\textbf{Case 1.} Assume that
there exists a subsequence of $\{a_p\}$, denoted by
$\{a_p\}$ again, and constants $c_0$ and $c_1$ such that
$0 < c_0 < \alpha_p < c_1 < 1$
for $\lambda \gg 1$. Let $0 < \delta \ll 1$ be a fixed constant.
By \eqref{e3.1}, we obtain
\begin{equation} \label{e4.1}
B_p(\alpha_p) < \int_{0}^1
\frac{1-s^{q+1}}{(1-s^{p+1} - (p+1)c_1^{q-p}(1-s^{q+1})/(q+1))^{3/2}}ds < C.
\end{equation}
Since $A_p(\alpha_p)> C_{4,\delta}$ by \eqref{e3.5}, by \eqref{e3.3}
and \eqref{e4.1}, we have
\begin{equation} \label{e4.2}
\begin{aligned}
&-\frac{p-1}{2}A_p(\alpha_p)
+ \frac{(p+1)(q-p)}{2(q+1)}B_p(\alpha_p)\alpha_p^{q-p} \\
&< -\frac{p-1}{2}C_{4,\delta} + \frac{(p+1)(q-p)}{2(q+1)}C < 0.
\end{aligned}
\end{equation}
This is a contradiction.
\smallskip

\noindent\textbf{Case 2.}
Assume that there exists a subsequence of $\{\alpha_p\}$, denoted by
$\{\alpha_p\}$ again, such that $\alpha_p \to 0$. Then clearly, we have
\begin{equation} \label{e4.3}
A_p \to \int_0^1 \frac{1}{\sqrt{1-s^{p+1}}}ds, \quad
B_p \to \int_0^1 \frac{1-s^{q+1}}{(1-s^{p+1})^{3/2}}ds.
\end{equation}
This implies
\begin{equation} \label{e4.4}
-\frac{p-1}{2}A_p(\alpha_p)
+ \frac{(p+1)(q-p)}{2(q+1)}B_p(\alpha_p)\alpha_p^{q-p}
< 0.
\end{equation}
This is a contradiction that completes the proof.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.3}]
We calculate $B_p(\alpha_p)$ by using \eqref{e3.4}.
Let $0 < \delta_0 \ll 1$ be a fixed constant. Then
for $0 < \delta < \delta_0$, we put
\begin{equation} \label{e4.5}
I_1 := I_{1,1} + I_{1,2} =
\int_0^{1-\delta_0} \frac{1-s^{q+1}}{h(s)^{3/2}}ds
+ \int_{1-\delta_0}^{1-\delta} \frac{1-s^{q+1}}{h(s)^{3/2}}ds.
\end{equation}
It is cleat that $I_{1,1} = O(1)$. By \eqref{e2.16}, \eqref{e2.17}, \eqref{e2.18} and
Taylor expansion, we have
\begin{equation} \label{e4.6}
\begin{aligned}
I_{1,2}
&\le C\int_{\delta}^{\delta_0} \frac{(q+1)x}{(d_1x^2 + d_2x)^{3/2}}dx\\
&\le C \int_{\delta}^{\delta_0} \frac{(q+1)x}{d_1x^2 + d_2x}
\frac{1}{\sqrt{d_1x^2 + d_2x}}dx \le C\delta^{-1}
 \int_{\delta}^{\delta_0} \frac{1}{\sqrt{d_1x^2 + d_2x}}dx\\
&\le C\delta^{-1}| \log \delta|.
\end{aligned}
\end{equation}
By mean value theorem, for $1-\delta < s < 1$, we have
\begin{equation} \label{e4.7}
s^{q+1} = 1 + (q+1)\eta_s^q(s-1),
\end{equation}
where $1-\delta < \eta_s < 1$.
By this, \eqref{e2.16} and \eqref{e2.17}, we obtain
\begin{equation} \label{e4.8}
\begin{aligned}
I_2 &= \int_{1-\delta}^1 \frac{(q+1)\eta_s^q(1-s)}
{(d_1(1-s)^2 + d_2(1-s))^{3/2}}ds\\
&= \int_0^\delta \frac{(q+1)\eta_s^qx}
{(d_1 x^2 + d_2x)^{3/2}}dx\\
&= (q+1)(1 + O(\delta))\Big[\frac{2d_2x}{d_2^2\sqrt{d_1x^2+d_2x}}
\Big]_0^\delta \\
&= 2(q+1)(1 + O(\delta))\frac{\delta}{d_2\sqrt{d_1\delta^2 + d_2\delta}}\\
&=\frac{2(q+1)}{(p+1)(1-\alpha_p^{q-p})}(1 + O(\delta))M_\delta,
\end{aligned}
\end{equation}
where
\begin{equation} \label{e4.9}
M_\delta := \frac{\delta}{\sqrt{d_1\delta^2 + d_2\delta}}.
\end{equation}
By this and \eqref{e3.3}, we obtain
\begin{equation} \label{e4.10}
\frac{p-1}{2}A_p(\alpha_p) = (q-p)
\Big(M_\delta\frac{\alpha_p^{q-p}}{1-\alpha_p^{q-p}} 
+ O(\delta^{-1}| \log \delta|)
\Big).
\end{equation}
This implies
\begin{equation} \label{e4.11}
\alpha_p^{q-p} = \frac{1}{1 + O((q-p)M_\delta/A_p)}
= 1 - O\Big(\frac{(q-p)M_\delta}{A_p}\Big).
\end{equation}
By \eqref{e2.17} and \eqref{e4.9}, we have
\begin{equation} \label{e4.12}
M_\delta \le \sqrt{\frac{\delta}{d_2}} \le C\frac{1}{\sqrt{1-\alpha_p^{q-p}}}.
\end{equation}
By this, Lemma \ref{lem2.1}, \eqref{e2.17} and \eqref{e4.11}, we obtain
\begin{equation} \label{e4.13}
1-\alpha_p^{q-p} \le C\frac{(q-p)^{3/2}}{
| \log(1-\alpha_p^{q-p})|\sqrt{1-\alpha_p^{q-p}}}.
\end{equation}
This implies
$1-\alpha_p^{q-p} \le C(q-p) \ll 1$,
namely,
\begin{equation} \label{e4.14}
\frac{1}{| \log(1-\alpha_p^{q-p})|} \le
C\frac{1}{| \log (q-p) |}.
\end{equation}
By this and \eqref{e4.13}, we obtain Theorem \ref{thm1.3}. Thus the proof is complete.
\end{proof}

\subsection*{Acknowledgments}
This research was partially supported by a Grant-in-Aid for Scientific Research
(C) (No.25400167) of Japan Society for the Promotion of Science.



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