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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 107, pp. 1--37.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/107\hfil Bifurcation of positive solutions]
{Bifurcation of positive solutions for the one-dimensional
 $(p,q)$-Laplace equation}

\author[R. Kajikiya, M. Tanaka, S. Tanaka \hfil EJDE-2017/107\hfilneg]
{Ryuji Kajikiya, Mieko Tanaka, Satoshi Tanaka}

\address{Ryuji  Kajikiya \newline
Department of Mathematics,
Faculty of Science and Engineering,
Saga University, Saga 840-8502, Japan} 
\email{kajikiya@ms.saga-u.ac.jp}

\address{Mieko Tanaka \newline
Department of Mathematics,
Tokyo University of Science,
Kagurazaka 1-3, Shinjyuku-ku,
Tokyo 162-8601, Japan}
\email{miekotanaka@rs.tus.ac.jp}

\address{Satoshi Tanaka \newline
Department of Applied Mathematics,
Faculty of Science,
Okayama University of Science,
Okayama 700-005, Japan}
\email{tanaka@xmath.ous.ac.jp}

\dedicatory{Communicated by Raul Manasevich}

\thanks{Submitted February 8, 2017. Published April 21, 2017.}
\subjclass[2010]{34B09, 34B18, 34C23, 34L30}
\keywords{Bifurcation; positive solution; $(p,q)$-Laplace equation; time map; 
\hfill\break\indent multiple solutions}

\begin{abstract}
 In this article, we study the bifurcation of positive solutions for
 the one-dimensional $(p,q)$-Laplace equation under  Dirichlet boundary
 conditions. We investigate the shape of the bifurcation diagram
 and prove that there exist five different types of bifurcation diagrams.
 As a consequence, we prove the existence of multiple positive solutions and
 show the uniqueness of positive solutions for a bifurcation parameter in
 a certain range.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\allowdisplaybreaks


\section{Introduction}\label{section-1}

In this article, we study the bifurcation problem of positive solutions
for the one-dimensional $(p,q)$-Laplace
equation in the interval $(-L,L)$,
\begin{equation}
\begin{gathered}
(|u'|^{p-2}u')' + (|u'|^{q-2}u')' + \lambda(|u|^{p-2}u+|u|^{q-2}u)=0, \\
u(-L)=u(L)=0,
\end{gathered} \label{Plambda}
\end{equation}
where $L>0$, $1<q<p<\infty$ and $\lambda>0$ is a parameter.
We assume that $1<q<p$ throughout the paper.
We can deal with any interval $(a,b)$ instead of $(-L,L)$.
Indeed, putting $L:=(b-a)/2$ and using a translation, we can reduce $(a,b)$
to $(-L,L)$.
We call $(\lambda, u)$ a solution of \eqref{Plambda} if
$$
u \in C^1[-L, L], \quad |u'|^{p-2}u' + |u'|^{q-2}u' \in C^1[-L, L],
$$
and $(\lambda, u)$ satisfies \eqref{Plambda}.

There has been much interest in studying
autonomous equations with one-dimensional
$p$-Laplacian or one-dimensional Laplacian.
It is impossible to quote all of them.
Here, we only refer to \cite{AK,HW,Sc,Sh,T1}.
For the $(p,q)$-Laplace equation, the existence of positive solutions was studied in
\cite{FMM,YW,YY1,YY2}.
The problems in $\mathbb{R}^N$ were investigated in \cite{BB1,BB2,WY}.
The existence of nodal solutions was proved in \cite{MP}.
The one-dimensional $\Phi$-Laplacian problem
$$
(\Phi(u'))'+\lambda f(u) =0, \quad u(-L)=u(L)=0,
$$
which is a more general problem, is studied in \cite{FY,FN,GMS1,GMS2,GMZ,SS,Si},
where $\Phi$ is an increasing odd homeomorphism of $\mathbb{R}$ and
$f \in C(\mathbb{R})$.
However, it seems difficult to obtain precise results for
$\Phi$-Laplacian problems.
In this paper, we concentrate \eqref{Plambda} and investigate the structure of
solutions as precise as possible.

Several authors studied a more general problem than \eqref{Plambda},
\begin{gather*}
-\Delta_p u -\Delta_q u= \lambda(m_p(x)|u|^{p-2}u + m_q(x)|u|^{q-2}u) \quad
\text{in } \Omega, \\
u=0 \quad \text{on } \partial \Omega,
\end{gather*}
where $\Omega$ is a bounded smooth domain in $\mathbb{R}^N$.
Motreanu and Tanaka~\cite{MT} dealt with sign-indefinite weights
$m_p(x)$, $m_q(x)$ and
proved the existence and nonexistence of positive solutions for some
ranges of $\lambda$.
Applying the results in \cite{MT} to our problem, we obtain the following:
\begin{itemize}
 \item[(i)] if $0<\lambda<\min\{ \mu(p,L), \mu(q,L) \}$, then \eqref{Plambda} has
       no positive solutions;
 \item[(ii)] if $\min\{ \mu(p,L), \mu(q,L) \}<\lambda<\max\{ \mu(p,L), \mu(q,L) \}$,
       then \eqref{Plambda} has at least one positive solution,
\end{itemize}
where $\mu(p,L)$ is the first eigenvalue of the $p$-Laplacian, that is,
the following equation has a {\it positive} solution
if and only if $\mu=\mu(p,L)$:
\begin{equation}\label{eq:1.1}
 (|\phi'|^{p-2}\phi')' + \mu |\phi|^{p-2}\phi=0 \quad \text{in } (-L,L),
 \quad \phi(-L)=\phi(L)=0.
\end{equation}
However, when $\lambda \ge \max\{ \mu(p,L), \mu(q,L) \}$,
there is no information about the existence or nonexistence of positive solutions.
Moreover, it is unknown whether a positive solution is unique or not in the
case (ii). One of our purposes is to solve such problems.

The references mentioned above used mainly the variational method,
a minimizing method or the mountain pass lemma.
In this paper, we employ the bifurcation approach.
It seems to the authors that little is known about the bifurcation
diagram of \eqref{Plambda}.
Our problem \eqref{Plambda} seems simple, however the structure of the
solution space is very complicated.

The $p$-Laplace operator $\Delta_p$ is homogeneous with degree $p-1$,
that is, $\Delta_p(\lambda u)=\lambda^{p-1}\Delta_p u$ for $\lambda>0$.
However, the $(p,q)$-Laplace operator has no such homogeneity.
This is one of difficulties to our problem.

To explain another difficulty, we consider a large positive solution
of \eqref{Plambda}.
Then it approaches the first eigenfunction of the $p$-Laplacian \eqref{eq:1.1}.
Indeed, we rewrite \eqref{Plambda} as
$$
[(p-1)|u'|^{p-2}+(q-1)|u'|^{q-2}]u'' + \lambda(u^{p-2} + u^{q-2})u=0.
$$
(This expression is not rigorous because a solution does not have the
$C^2$-regularity. However we shall prove the convergence of solutions
to the eigenfunction  in the strict method.)
If $u$ and $|u'|$ are large enough, the equation above is nearly equal to
the following equation
$$
((p-1)|u'|^{p-2})u'' + \lambda u^{p-1}=0,
$$
which is exactly \eqref{eq:1.1}.
Thus a large positive solution converges to the first eigenfunction
of \eqref{eq:1.1} as the $L^\infty$ norm of $u$
diverges. This assertion will be proved in Section~\ref{section-4}.
On the other hand, a small positive solution is close to the first
eigenfunction of the $q$-Laplacian
because \eqref{Plambda} approaches the $q$-Laplace equation as the
$L^\infty$ norm of $u$ tends to zero.
It is well-known that the first eigenvalue $\mu(p,L)$ of \eqref{eq:1.1}
is represented as (see \cite[pp.4-5]{DR})
\begin{equation}\label{eq:1.2}
\mu(p,L) := (p-1) \left(\frac{\pi_p}{2L}\right)^p,
\end{equation}
where
\begin{equation}\label{eq:1.3}
\pi_p := 2 \int_0^1 (1-s^p)^{-1/p}\,ds = \frac{2\pi}{p\sin(\pi/p)}.
\end{equation}
For $\pi_p$, we also refer the reader to \cite{DEM} and \cite{T2}.
Therefore a small positive solution of \eqref{Plambda} is governed
by $\mu(q,L)$ and a large one by $\mu(p,L)$.
Hence the behavior of the first eigenvalue plays the key role in
the bifurcation problem.
If $L\leq 1$, then $\mu(p,L)$ is increasing on $p$.
However if $L>1$, then it is not increasing.
Indeed, we have the next theorem.

\begin{theorem}\label{th:1.1}
\begin{itemize}
\item[(i)] If $0<L\leq 1$, then $\mu(p,L)$ is strictly increasing with respect
to $p$, that is,       $\mu_p(p,L)>0$ for $1<p<\infty$, where $\mu_p(p,L)$
denotes the partial derivative with respect to $p$.
\item[(ii)] If $L>1$, then there exists a unique $p_*(L)>0$ such that
      $\mu_p(p,L)>0$ for $p\in (1,p_*(L))$ and $\mu_p(p,L)<0$ for
$p\in(p_*(L),\infty)$.
\end{itemize}
\end{theorem}

The theorem above may be known, however we can not find a  proof of it.
Therefore we give a proof in Section~\ref{section-2}.
The behavior of $\mu(p,L)$ as above makes the bifurcation problem complicated.
This is another difficulty in our problem.

We denote the $L^\infty(-L,L)$ norm of $u$ by $\|u\|_\infty$.
We shall prove that for any $\alpha>0$, there exists a unique positive
solution $(\lambda ,u)$ which satisfies $\|u\|_\infty=\alpha$.
We write it as $(\lambda, u)=(\lambda(\alpha), u(x,\alpha))$.
Then $(\lambda(\alpha), u(x,\alpha))$ with $\alpha>0$ represents all
positive solutions.
Since $u(x, \alpha)$ is uniquely determined by $\alpha$,
we can identify a curve $(\lambda(\alpha), u(x,\alpha))$ with $
(\lambda(\alpha),\alpha)$.
It will be shown that any positive solution $u$ of \eqref{Plambda} is even and
achieves its maximum at $x=0$ only.
See Lemma~\ref{le:3.2}.
Therefore $\alpha=\|u(\cdot, \alpha)\|_\infty=u(0,\alpha)$.
As mentioned before, if $(\lambda_n, u_n)$ is a sequence of positive solutions
and if $\|u_n\|_\infty\to \infty$,
then $\lambda_n\to \mu(p,L)$. This fact will be proved in Lemma~\ref{le:4.8}.
If $\|u_n\|_\infty \to 0$, then $\lambda_n\to \mu(q,L)$.
See Lemma~\ref{le:4.11} for the proof.
Since $(\lambda(\alpha), u(x,\alpha))$ is a positive solution satisfying
$\|u(\cdot, \alpha)\|_\infty=\alpha$,
it holds that $\lambda(\alpha) \to \mu(q,L)$ as $\alpha\to +0$ and
$\lambda(\alpha) \to \mu(p,L)$ as $\alpha\to \infty$.
Then the bifurcation curve $(\lambda(\alpha),\alpha)$
starts from the initial point $(\mu(q,L),0)$
and reaches the final point $(\mu(p,L),\infty)$.

Garc\'{i}a-Huidobro, Man\'{a}sevich and Schmitt in \cite{GMS1,GMS2}
considered $N$-dimensional $\Phi$-Laplacian problem
\begin{equation}
 (r^{N-1}\Phi(u'))'+\lambda r^{N-1} f(u)=0, \quad u'(0)=u(L)=0,
  \label{N}
\end{equation}
and they proved the following (i)--(iii):
(i) for each $\alpha>0$, there exists a positive solution $(\lambda,u)$
with $\|u\|_\infty=\alpha$;
(ii) there exists a constant $\lambda_0>0$ such that \eqref{N} has no
nontrivial solution if $0<\lambda<\lambda_0$;
(iii) there exists a connected component of positive solutions connecting
$(\mu(q,L),0)$ and $(\mu(p,L),\infty)$.
Moreover, they also showed the existence of sign-changing solutions.
See \cite[Theorem 1.1]{GMS1} and \cite[Theorem 5.1]{GMS2}.

The purpose of the present paper is to investigate the shape of the bifurcation
diagram. Moreover, observing the shape of the diagram, we study the number
 of positive solutions.
We draw the curve $(\lambda(\alpha),\alpha)$ in the $(\lambda,\alpha)$
plain, where $\lambda$-axis and $\alpha$-axis are chosen as axes of abscissa
and ordinate, respectively.
For simplicity, we write $\mu(p,L)$ as $\mu(p)$ if there is no
confusion.
In Section~\ref{section-4}, we shall show that the curve
$(\lambda(\alpha),\alpha)$ always stays in the right side of the line
$\lambda=\min\{\mu(p),\mu(q)\}$.
Using this fact, we classify all the bifurcation diagrams according to
whether the bifurcation curve always lies in the left side of the line
$\lambda=\max\{\mu(q),\mu(p)\}$
or protrudes from the line (see Figure 1):
\begin{itemize}
\item[(A)] $\mu(q) < \mu(p)$ and $\mu(q)<\lambda(\alpha) \leq \mu(p)$
for all $\alpha>0$.
\item[(B)] $\mu(q)<\mu(p)$ and $\mu(p)<\lambda(\alpha)$ at some $\alpha>0$.
\item[(C)] $\mu(p)<\mu(q)$ and $\mu(p)<\lambda(\alpha) \leq \mu(q)$ for all
$\alpha>0$.
\item[(D)] $\mu(p)<\mu(q)$ and $\mu(q)<\lambda(\alpha)$ at some $\alpha>0$.
\item[(E)] $\mu(p)=\mu(q)$ and $\mu(p)<\lambda(\alpha)$ for all $\alpha>0$.
\end{itemize}
The definitions above can be rewritten as below in terms of the number of positive
solutions.
\begin{itemize}
\item[(A)] $\mu(q)<\mu(p)$ and \eqref{Plambda} has at least one positive solution for
  $\lambda\in(\mu(q),\mu(p))$ and no positive solutions for
$\lambda\in(0,\mu(q)]\cup(\mu(p),\infty)$.

\item[(B)] $\mu(q)<\mu(p)$ and there exists $\lambda^*>\mu(p)$ such that
  \eqref{Plambda} has at least one positive solution for
 $\lambda\in(\mu(q),\mu(p)] \cup \{\lambda^*\}$,
  at least two positive solutions for $\lambda\in(\mu(p),\lambda^*)$ and
  no positive solutions for $\lambda\in(0,\mu(q)]\cup (\lambda^*,\infty)$.

\item[(C)] $\mu(p)<\mu(q)$ and \eqref{Plambda} has at least one positive
solution for   $\lambda\in(\mu(p),\mu(q))$ and no positive solutions for
$\lambda\in(0,\mu(p)]\cup (\mu(q),\infty)$.

\item[(D)] $\mu(p)<\mu(q)$ and there exists $\lambda^*>\mu(q)$ such that
  \eqref{Plambda} has at least one positive solution for
$\lambda\in(\mu(p),\mu(q)] \cup \{\lambda^*\}$,
  at least two positive solutions for $\lambda\in(\mu(q),\lambda^*)$ and
  no positive solutions for $\lambda\in(0,\mu(p)]\cup (\lambda^*,\infty)$.

\item[(E)] $\mu(p)=\mu(q)$ and there exists $\lambda^*>\mu(p)$ such that
  \eqref{Plambda} has at least one positive solution for $\lambda=\lambda^*$,
  at least two positive solutions for $\lambda\in(\mu(p),\lambda^*)$ and
  no positive solutions for $\lambda\in(0,\mu(p)]\cup (\lambda^*,\infty)$.
\end{itemize}

\begin{figure}[ht]
\setlength{\unitlength}{0.3pt}
\begin{center}
\begin{picture}(400,350)(0,0)
\put(0,0){\line(1,0){400}}
\multiput(100,0)(0,30){11}{\line(0,1){15}}
\multiput(310,0)(0,30){11}{\line(0,1){15}}
{\thicklines
\qbezier(100,0)(250,50)(300,300)
}
\put(60,-50){$\mu(q)$}
\put(260,-50){$\mu(p)$}
\put(150,250){\large (A)}
\end{picture}
\quad
\begin{picture}(400,350)(0,0)
\put(0,0){\line(1,0){400}}
\multiput(100,0)(0,30){11}{\line(0,1){15}}
\multiput(310,0)(0,30){11}{\line(0,1){15}}
{\thicklines
\qbezier(100,0)(450,50)(380,150)
\qbezier(380,150)(320,220)(320,300)
}
\put(60,-50){$\mu(q)$}
\put(260,-50){$\mu(p)$}
\put(150,250){\large (B)}
\end{picture}
\\[5ex]

\begin{picture}(400,350)(0,0)
\put(0,0){\line(1,0){400}}
\multiput(100,0)(0,30){11}{\line(0,1){15}}
\multiput(310,0)(0,30){11}{\line(0,1){15}}
{\thicklines
\qbezier(310,0)(160,50)(110,300)
}
\put(60,-50){$\mu(p)$}
\put(260,-50){$\mu(q)$}
\put(150,250){\large (C)}
\end{picture}
\quad
\begin{picture}(400,350)(0,0)
\put(0,0){\line(1,0){400}}
\multiput(100,0)(0,30){11}{\line(0,1){15}}
\multiput(310,0)(0,30){11}{\line(0,1){15}}
{\thicklines
\qbezier(310,0)(500,50)(320,100)
\qbezier(320,100)(120,150)(110,300)
}
\put(60,-50){$\mu(p)$}
\put(260,-50){$\mu(q)$}
\put(150,250){\large (D)}
\end{picture}
\\[5ex]

\begin{picture}(400,360)(0,-15)
\put(0,0){\line(1,0){400}}
\multiput(200,0)(0,30){11}{\line(0,1){15}}
{\thicklines
\qbezier(200,0)(400,50)(300,100)
\qbezier(300,100)(210,150)(210,300)
}
\put(100,-50){$\mu(p)=\mu(q)$}
\put(100,250){\large (E)}
\end{picture}
\end{center}
\caption{Bifurcation diagrams}
\label{fig1}
\end{figure}

We shall prove that the five types (A)--(E) of behaviors actually occur
for some $(p,q,L)$.
We note that if $\mu(p,L)=\mu(q,L)$, then (E) always occurs, by recalling
$(\lambda(\alpha),\alpha)$ always stays in the right side of the line
$\lambda=\min\{\mu(p,L),\mu(q,L)\}$.
Now we fix $p$ and $q$ and define
\begin{equation}\label{eq:1.4}
 L_* =L_*(p,q) := \frac{1}{2}\Big(\frac{(p-1)\pi_p^p}{(q-1)\pi_q^q} \Big)^{1/(p-q)}.
\end{equation}
Since $\mu(p,1)$ is increasing with respect to $p$ by Theorem \ref{th:1.1},
it holds that $\mu(q,1)<\mu(p,1)$ by $1<q<p$. This proves $L_*>1$.
From \eqref{eq:1.2} it follows that $\mu(p,L)=\mu(q,L)$ if and only if $L=L_*$.
If $L>L_*$, then $\mu(p,L)<\mu(q,L)$ by \eqref{eq:1.2}.
In Case (E), the bifurcation curve stays in the right side of the line
$\lambda=\mu(p,L)$.
Consider a small perturbation from (E).
Then, if $L$ is slightly greater than $L_*$, (D) occurs.
Conversely, (B) occurs when $L<L_*$ and $L$ is close to $L_*$.
In Section~\ref{section-6}, we shall show that if $L$ is small enough
or large enough, then $\lambda'(\alpha)>0$ or $\lambda'(\alpha)<0$
for all $\alpha>0$, respectively.
Therefore, if $L$ is small enough or large enough, then (A) or (C)
occurs respectively, and a positive solution is unique.
Consequently, we shall prove the existence of an $\varepsilon>0$
(see Theorem~\ref{th:7.14}) such that
(A), (B), (E), (D), (C) occur when $0<L<\varepsilon$,
$L_*-\varepsilon<L<L_*$, $L=L_*$, $L_*<L<L_*+\varepsilon$,
$1/\varepsilon<L<\infty$, respectively.
Furthermore, the types {\upshape (A)} with $0<L<\varepsilon$
and {\upshape (C)} with $1/\varepsilon<L<\infty$ have no turning points
and they are monotone.
 Therefore, the structure of solutions to \eqref{Plambda} changes
 depending on $L$.
 Such phenomena have been reported in \cite{CHW} and \cite{HCWC} for mean
 curvature equations.
\smallskip

\noindent\textbf{Contents of the article:}
In Section~\ref{section-2}, we give a proof of Theorem~\ref{th:1.1} and
study some properties of the
first eigenvalue, which play an important role for our proofs.

In Section~\ref{section-3}, we prove the uniqueness of solutions
for the initial value problem corresponding to our problem
\eqref{Plambda}. After that, we introduce a time map which denotes the
first zero in
$(0,\infty)$ of a solution for the initial value problem.

In Section~\ref{section-4},
we prove that all positive solutions are represented as a smooth curve
$(\lambda(\alpha), u(x,\alpha))$ with one parameter
$\alpha=u(0, \alpha)=\|u\|_\infty$
and study the properties of this bifurcation curve.

In Section~\ref{section-5}, to consider the behavior of
the bifurcation curve, we estimate a derivative of the time map,
which will be used in
Sections~\ref{section-6} and \ref{section-7}.

In Section~\ref{section-6}, we give sufficient conditions for which
the shape of the bifurcation curve is exactly type (A) or (C).

In Section~\ref{section-7}, we shall study the direction in which the
bifurcation curve moves near the initial point $(\mu(q,L),0)$ and near
the final point $(\mu(p,L),\infty)$.
Using these results, we shall construct types (B) and (D).
This is a different way from the perturbation method of type (E) as mentioned
after \eqref{eq:1.4}.
As a consequence, we shall obtain the existence of multiple positive
solutions and moreover the uniqueness of positive solutions in some
cases of $p$, $q$, and $L$.

\section{The first eigenvalue}\label{section-2}

In this section, we shall prove Theorem \ref{th:1.1} and
investigate the properties of the first eigenvalue of the $p$-Laplacian.

\begin{proof}[Proof of Theorem \ref{th:1.1}]
We put $g(x):=\log \mu(p,L)$ and $p:=\pi/x$. Then we find that
\begin{equation}\label{eq:2.1}
\begin{aligned}
g(x)
&=  \log \mu(p,L)=\log(p-1)+p\log[\pi/(Lp\sin(\pi/p))]  \\
&=  \log((\pi-x)/x) - (\pi/x)\log(\sin x/x) -(\pi/x)\log L.
\end{aligned}
\end{equation}
Differentiating it, we have
\begin{equation*}
g'(x) = \frac{-\pi}{x(\pi-x)}+ \pi x^{-2}\log(\sin x/x)
 -\pi x^{-1}\frac{\cos x}{\sin x} +\pi x^{-2}+ \pi x^{-2}\log L.
\end{equation*}
Putting $h(x)=\pi^{-1}x^2g'(x)$, we obtain
$$
h(x):=\pi^{-1}x^2g'(x)=
\frac{-x}{\pi-x} + \log(\sin x/x) -\frac{x\cos x}{\sin x} + \log L +1.
$$
Since $p\in(1,\infty)$, $x$ lies in $(0,\pi)$.
We compute $h'(x)$ as
\begin{equation}\label{eq:2.2}
\begin{aligned}
h'(x)
&=  \frac{-\pi}{(\pi-x)^2} + \frac{x^2-\sin^2 x}{x\sin^2x}  \\
&= \frac{x^2(\pi-x)^2-[(\pi-x)^2+\pi x]\sin^2 x}{(\pi-x)^2x\sin^2x}.
\end{aligned}
\end{equation}
We shall show that $h'(x)<0$ for $0<x<\pi$ in Lemma \ref{le:2.1} later on.
Thus $h(x)$ is strictly decreasing.

We shall show the assertion (i). Let $0<L \leq 1$.
Since $\lim_{x\to +0}h(x)=\log L\leq 0$ and $h(x)$ is decreasing,
it holds that $h(x)<0$ for $0<x<\pi$.
Accordingly, $g'(x)<0$.
Since $g(x)=\log \mu(p,L)$ and $p=\pi/x$, we find that
\begin{equation}\label{eq:2.3}
\mu(p,L)=\exp(g(x)), \quad \mu_p(p,L)=-\pi p^{-2}g'(x)\exp(g(x)).
\end{equation}
Consequently, $\mu_p(p,L)>0$ and the assertion (i) is obtained.

Let us show (ii). Let $L >1$.
Then $\lim_{x\to +0}h(x)=\log L> 0$.
Using L'Hospital's rule, we compute
\begin{equation}\label{eq:2.4}
\frac{-x}{\pi-x} -\frac{x\cos x}{\sin x} \to 0 \quad \text{as } x\to \pi-0.
\end{equation}
Hence $\lim_{x\to \pi-0}h(x)=-\infty$.
Since $h(x)$ is strictly decreasing, there exists a unique point $x_*\in(0,\pi)$ such that
\begin{equation}\label{eq:2.5}
h(x)>0 \quad \text{for } 0<x<x_*, \quad h(x)<0 \quad \text{for } x_*<x<\pi.
\end{equation}
Put $p_*=\pi/x_*$. When $x\in(0,x_*)$, $p$ lies in $(p_*,\infty)$.
Since $g'(x)>0$ for $x\in(0,x_*)$, $\mu_p(p,L)<0$ for $p\in(p_*,\infty)$
by \eqref{eq:2.3}.
In the same way, it holds that $\mu_p(p,L)>0$ for $p\in(0,p_*)$.
The proof is complete.
\end{proof}

We denote the numerator of $h'(x)$ in \eqref{eq:2.2} by $k(x)$, i.e.,
$$
k(x) := x^2(\pi-x)^2 -(x^2-\pi x+\pi^2)\sin^2x.
$$
We have used the next lemma in the proof of Theorem \ref{th:1.1}.

\begin{lemma}\label{le:2.1}
With the above notation,
$$
k(x)<0 \quad \text{for } x\in (0,\pi).
$$
\end{lemma}

\begin{proof}
Since $k(x)$ is symmetric with respect to the line $x=\pi/2$,
it is enough to show that $k(x)<0$ for $0<x \leq \pi/2$.
We use the inequality
$$
\sin x > x- \frac{1}{6}x^3>0 \quad \text{for } 0<x \leq \pi/2,
$$
to obtain
\begin{align*}
k(x)
&\leq  x^2(\pi-x)^2 - (x^2-\pi x+\pi^2)(x-x^3/6)^2 \\
&=  -\frac{x^3}{36}\left[x^5-\pi x^4 -(12-\pi^2)x^3 +12\pi x^2
 -12\pi^2 x+36\pi\right].
\end{align*}
Put
$$
K(x) :=x^5-\pi x^4 -(12-\pi^2)x^3 +12\pi x^2 -12\pi^2 x+36\pi.
$$
Then it has a derivative,
\begin{align*}
K'(x)
&=  5x^4-4\pi x^3 -3(12-\pi^2)x^2 +24\pi x -12 \pi^2 \\
&=  -(4\pi-5x)x^3 -3(12-\pi^2)x^2 -12\pi(\pi-2x)<0,
\end{align*}
because $4\pi-5x>0$ and $\pi-2x>0$ for $0<x<\pi/2$.
Hence $K(x)$ is decreasing. Moreover,
$$
K(\pi/2)=\frac{3}{32}\pi^5 -\frac{9}{2}\pi^3 +36\pi>0.
$$
Accordingly, $K(x)>0$ for $0<x \leq \pi/2$ and therefore $k(x)<0$ for $0<x \leq \pi/2$.
The proof is complete.
\end{proof}

To study the bifurcation problem, we need to investigate more properties
of the first eigenvalue $\mu(p,L)$ of
the $p$-Laplacian.
As proved in Theorem \ref{th:1.1}, for each $L>1$ fixed, $\mu(p,L)$ has a unique maximum point $p$ in $(1,\infty)$.
We denote it by $p_*(L)$.

\begin{proposition}\label{pr:2.2}
\begin{itemize}
\item[(i)]
  $\mu(p,L)$ has the following properties.
  \begin{gather}\label{eq:2.6}
  \lim_{p\to 1+0}\mu(p,L)=1/L \quad  \text{for all } L>0, \\
\label{eq:2.7}
  \lim_{p\to \infty}\mu(p,L)=\infty, \quad  \text{when } L\leq 1, \\
\label{eq:2.8}
  \lim_{p\to \infty}\mu(p,L)=0, \quad  \text{when } L>1.
  \end{gather}
\item[(ii)]
  The unique maximum point $p_*(L)$ of $\mu(p,L)$ is strictly decreasing
with respect to $L\in(1,\infty)$ and satisfies
  \begin{gather}\label{eq:2.9}
     \lim_{L\to 1+0}p_*(L)=\infty, \quad \lim_{L\to \infty}p_*(L)=1, \\
\label{eq:2.10}
  \lim_{L\to 1+0}\mu(p_*(L), L)=\infty, \quad \lim_{L\to \infty}\mu(p_*(L), L)=0.
  \end{gather}
\end{itemize}
\end{proposition}

\begin{proof}
We shall prove (i).
Since $\mu(p,L)=\mu(p,1)L^{-p}$ by \eqref{eq:1.2}, we have only to prove
 that $\lim_{p\to 1+0}\mu(p,1)=1$,
which ensures \eqref{eq:2.6}.
Let us show $\mu(p,1)^{1/p}\to 1$.
Put $x=1/p$. Then
\begin{align*}
\mu(p,1)^{1/p}
&=  (p-1)^{1/p}(\pi_p/2)=\Big(\frac{1-x}{x}\Big)^x\frac{\pi x}{\sin \pi x} \\
&=  x^{1-x}(1-x)^{x-1}\frac{\pi(1-x)}{\sin \pi x}.
\end{align*}
As $p\to 1+0$, $x$ converges to $1-0$.
We see readily that
$$
\lim_{x\to 1-0}(1-x)^{x-1}=1, \quad
\lim_{x\to 1-0}\frac{\pi(1-x)}{\sin \pi x}=1,
$$
which imply
$$
\lim_{x\to 1-0}x^{1-x}(1-x)^{x-1}\frac{\pi(1-x)}{\sin \pi x}=1.
$$
Therefore $\mu(p,1)^{1/p}\to 1$ and hence \eqref{eq:2.6} holds.

We shall show \eqref{eq:2.7} with $L=1$. Put
$$
y:=\mu(p,1)=(p-1)\left(\frac{\pi_p}{2}\right)^p, \quad x:=1/p.
$$
Then we see that
\begin{equation}\label{eq:2.11}
\log y =\log\Big(\frac{1-x}{x}\Big) + \frac{1}{x}
\log\Big(\frac{\pi x}{\sin \pi x}\Big).
\end{equation}
Since $p>1$, $x$ is in $(0,1)$.
Since $(\pi x)/\sin \pi x>1$, the second term on the right hand side
of \eqref{eq:2.11} is positive.
Accordingly, we have
$$
\log y \geq \log\Big(\frac{1-x}{x}\Big) \to \infty \quad \text{as } x\to +0.
$$
Therefore $\lim_{p\to\infty}\mu(p,1)=\infty$.
When $L\leq 1$, it follows that
$$
\mu(p,L)=\mu(p,1)L^{-p} \geq \mu(p,1) \to \infty \quad \text{as } p\to\infty.
$$
Consequently, \eqref{eq:2.7} holds.

Putting $x=1/p$, we have
$$
\pi_p/2 = \frac{\pi x}{\sin \pi x} \to 1 \quad \text{as } x\to +0.
$$
Therefore, when $L>1$, we obtain
$$
\mu(p,L)=(p-1)\left(\frac{\pi_p}{2L}\right)^p \to 0 \quad \text{as } p\to \infty,
$$
which ensures \eqref{eq:2.8}.

We shall show that $p_*=p_*(L)$ is decreasing.  We define
\begin{align*}
H(x)
:=& \frac{x}{\pi-x} - \log(\sin x/x) +\frac{x\cos x}{\sin x} \\
 =& -h(x)+1+\log L,
\end{align*}
where $h(x)$ is the function introduced in the proof of Theorem \ref{th:1.1}.
Since $h(x)$ is decreasing, $H(x)$ is increasing.
By recalling the proof of Theorem \ref{th:1.1}, there exists a unique
$x_*\in(0,\pi)$ such that \eqref{eq:2.5} holds and $p_*(L)=\pi/x_*$.
Since $h(x_*)=0$ (see \eqref{eq:2.5}), $x_*$ satisfies $H(x_*)=\log L +1$,
or equivalently $x_*=H^{-1}(\log L +1)$.
Therefore $x_*$ is an increasing function of $L$ and hence $p_*(L)$ is decreasing.
Since $h(x)\to\log L$ as $x\to+0$ and $h(x)\to-\infty$ as $x\to\pi-0$,
we have
$$
  \lim_{x\to1+0} H^{-1}(x) = +0, \quad
  \lim_{x\to\infty} H^{-1}(x) = \pi,
$$
 which show that $p_*(L)=\pi/H^{-1}(\log L +1) \to \infty$ as $L\to1+0$ and that
 $p_*(L)=\pi/H^{-1}(\log L +1) \to 1$ as $L\to\infty$.
Consequently we obtain \eqref{eq:2.9}.

We shall show \eqref{eq:2.10}.
Since $h(x_*)=0$, we have
$$
\log(\sin x_*/x_*) + \log L
=  \frac{x_*}{\pi-x_*} + \frac{x_*\cos x_*}{\sin x_*} -1.
$$
Let $g(x)$ be as in the proof of Theorem \ref{th:1.1}.
Substituting the equation above into $g(x_*)$ (see \eqref{eq:2.1}),
we find that
$$
g(x_*)=
\log((\pi-x_*)/x_*) - \frac{\pi}{\pi-x_*} - (\pi/x_*)
\Big(\frac{x_*\cos x_*}{\sin x_*} -1\Big).
$$
As $L\to 1+0$,
$x_*=H^{-1}(\log L +1) \to 0$.
An easy computation shows that
$$
(\pi/x_*)\Big(\frac{x_*\cos x_*}{\sin x_*} -1\Big) \to 0 \quad \text{as} \ x_*\to 0.
$$
Therefore $g(x_*)\to \infty$ as $x_*\to +0$, and hence
$\mu(p_*(L),L)=\exp(g(x_*))\to \infty$ as $L\to 1+0$.

As $L\to \infty$,
$x_*=H^{-1}(\log L +1) \to \pi-0$.
We rewrite $g(x_*)$ as
$$
 g(x_*)=\log((\pi-x_*)/x_*)
  + \frac{\pi}{x_*}\Big(\frac{-x_*}{\pi-x_*} - \frac{x_*\cos x_*}{\sin x_*} +1 \Big).
$$
Applying \eqref{eq:2.4}, we compute $g(x_*)\to -\infty$ as $L\to\infty$
and therefore $\mu(p_*(L),L)$ converges to $0$.
The proof is complete.
\end{proof}


\section{Initial value problem and time map}\label{section-3}

First we shall show the uniqueness of solutions for the initial value problem
\begin{equation}\label{eq:3.1}
\begin{gathered}
(|u'|^{p-2}u')' + (|u'|^{q-2}u')' + \lambda(|u|^{p-2}u+|u|^{q-2}u)=0, \\
u(x_0)=\alpha, \quad u^\prime(x_0)=\beta,
\end{gathered}
\end{equation}
because we shall use it several times.
Define
\begin{equation}\label{eq:3.2}
 f(t) :=|t|^{p-2}t + |t|^{q-2}t.
\end{equation}
Then the first equation of \eqref{eq:3.1} becomes
\begin{equation}\label{eq:3.3}
 f(u')' + \lambda f(u) = 0.
\end{equation}
Now, we define the energy $E(u)$ by
$$
E(u)(x):=\frac{p-1}{p}|u'(x)|^p + \frac{q-1}{q}|u'(x)|^q
+ \frac{\lambda}{p}|u(x)|^p + \frac{\lambda}{q}|u(x)|^q.
$$
Multiplying \eqref{eq:3.3} by $u'$, we see that $E(u)$ is constant on $x$
if $u$ is a solution of \eqref{eq:3.3}.
We put
\begin{equation}\label{eq:3.4}
\Phi(t) := \frac{p-1}{p} t^p + \frac{q-1}{q} t^q, \quad
F(t) := \frac{1}{p} t^p + \frac{1}{q} t^q.
\end{equation}
Then $E(u)$ is rewritten as
$$
E(u)(x)=\Phi(|u'(x)|)+\lambda F(|u(x)|).
$$

\begin{lemma}\label{le:3.1}
The initial value problem \eqref{eq:3.1} has a unique solution.
\end{lemma}

\begin{proof}
 Since \eqref{eq:3.1} is autonomous, we may assume that $x_0=0$, that is,
 we consider the initial value $u(0)=\alpha$ and $u'(0)=\beta$.
 By putting $v=f(u')$, problem \eqref{eq:3.1} is rewritten as
 $$
 \frac{d}{dx} \begin{pmatrix}
  u \\ v
 \end{pmatrix}
  =   \begin{pmatrix}
   f^{-1}(v) \\
   -\lambda f(u)
  \end{pmatrix}, \quad
 \begin{pmatrix}
  u(0) \\ v(0)
 \end{pmatrix}
  =  \begin{pmatrix}
   \alpha \\
   f(\beta)
  \end{pmatrix}.
$$
Then the existence of a local solution of \eqref{eq:3.1} is guaranteed by
the Peano existence theorem.

Next we shall show the uniqueness of local solutions.
Let $u$ be a local solution of \eqref{eq:3.1}.
Since the value $E(u)$ is constant on $x$, we conclude that
\begin{equation}\label{eq:3.5}
 E(u)(x) = E(u)(0) = \Phi(|\beta|) + \lambda F(|\alpha|).
\end{equation}
If $\alpha=\beta=0$, then $ E(u)(x)\equiv 0$, which means that $u(x)\equiv 0$
is a unique solution.
From \eqref{eq:3.5} it follows that
$$
 |u'(x)| = \Phi^{-1}\bigl(\Phi(|\beta|) + \lambda F(|\alpha|)
 - \lambda F(|u(x)|) \bigr),
$$
where $\Phi^{-1}$ is the inverse function of $\Phi$.
Now we assume that $\beta>0$.
Then $u'(x)>0$ near $x=0$ and hence
\begin{equation}\label{eq:3.6}
 u'(x) = \Phi^{-1}\bigl(\Phi(|\beta|) + \lambda F(|\alpha|)
- \lambda F(|u(x)|) \bigr),
\end{equation}
near $x=0$.
Since $\Phi^{-1}\bigl(\Phi(|\beta|) + \lambda F(|\alpha|) - \lambda F(|t|) \bigr)$
 is continuously differentiable near $t=\alpha$,
Picard's existence and uniqueness theorem
 implies that a solution of \eqref{eq:3.6} with the initial condition
 $u(0)=\alpha$ is unique.
In the same way, we can prove the uniqueness when $\beta<0$.
Finally, we assume that $\alpha\ne 0$ and $\beta=0$.
We consider the case $\alpha>0$ only, since the case $\alpha<0$ can be
treated similarly.
Then $u(x)>0$ for $x\in [-\varepsilon,\varepsilon]$ with a small
$\varepsilon>0$.
Since $f(u')'=-\lambda f(u)<0$ in $[-\varepsilon,\varepsilon]$,
$u'(x)$ is decreasing in this interval.
Since $u'(0)=\beta=0$, $u'(x)<0$ in $(0,\varepsilon]$.
Thus $u(x)>0$ and $u'(x)<0$ for $x \in (0,\varepsilon]$.
Using these facts with $\beta=0$ and computing as in \eqref{eq:3.6}, we have
$$
u'(t) = -\Phi^{-1}\bigl(\lambda F(\alpha) - \lambda F(u(t)) \bigr),
$$
or equivalently
$$
 \frac{-u'(t)}{\Phi^{-1}(\lambda[F(\alpha)-F(u(t))])}=1 \quad
  \text{for } t \in (0,\varepsilon].
$$
Integrating both sides over $(0,x)$, changing the variables $u(t)=\alpha s$
and using $u(0)=\alpha$,  we obtain
\begin{equation}\label{eq:3.7}
 {\mathcal T}(u(x)) = x
\end{equation}
for $x \in (0,\varepsilon]$, where
\begin{equation}\label{eq:3.8}
 {\mathcal T}(t)
 = \int_{t/\alpha}^1 \frac{\alpha}{\Phi^{-1}(\lambda[F(\alpha)-F(\alpha s)])}\,ds.
\end{equation}
 We note that ${\mathcal T}(t)$ is strictly decreasing in
$t \in (u(\varepsilon),\alpha)$  and it has an inverse function.
Consequently, any solution of \eqref{eq:3.1} is uniquely represented as
 $u(x)={\mathcal T}^{-1}(x)$,
which guarantees the uniqueness of solutions.

We have shown that \eqref{eq:3.1} has a unique local solution $u$.
The energy identity \eqref{eq:3.5} implies that both $u$ and $u'$ are
 bounded as far as $u$ exists.
Thus, by a standard argument, we conclude that $u$ exists on
 $\mathbb{R}$ and is unique.
The proof is complete.
\end{proof}

We summarize the properties of positive solutions for \eqref{Plambda}
in the next lemma.

\begin{lemma}\label{le:3.2}
Any positive solution $u(x)$ of \eqref{Plambda} is concave, even and
$u'(x)<0$ in $(0,L]$.
\end{lemma}

\begin{proof}
Let $u$ be any positive solution of \eqref{Plambda}.
Since $u>0$, the first equation of \eqref{Plambda} ensures that $f(u')'<0$
in $(-L,L)$, where $f$ is the function defined by \eqref{eq:3.2}.
Therefore $u'$ is decreasing and hence $u$ is concave.
Since $u$ is concave, it has a unique critical point $x_0$, i.e., $u'(x_0)=0$.
Put $v(x):=u(2x_0-x)$, which satisfies the first equation of \eqref{Plambda}.
Since $v(x_0)=u(x_0)$ and $v'(x_0)=u'(x_0)=0$,
$v$ is identically equal to $u$ because of Lemma~\ref{le:3.1}.
Accordingly, $u(x)$ is symmetric with respect to the line $x=x_0$.
Since $u(L)=u(-L)=0$, it holds that $x_0=0$.
Thus $u(x)$ is even.
Since $u$ is even and concave, $u'(x)<0$ in $(0,L]$.
\end{proof}

Lemma \ref{le:3.2} implies that, to find all positive solutions, it is
sufficient to consider the initial value problem
\begin{equation}\label{eq:3.9}
 \begin{gathered}
  (|u'|^{p-2}u')' + (|u'|^{q-2}u')'
   + \lambda ( |u|^{p-2}u + |u|^{q-2}u ) = 0, \\
   u'(0)=0, \quad u(0)=\alpha>0.
   \end{gathered}
\end{equation}
Then Lemma~\ref{le:3.1} implies that \eqref{eq:3.9}
has a unique solution, which is periodic and has zeros
by a standard argument.
Since the energy $E(u)$ is constant on $x$, the solution of \eqref{eq:3.9} satisfies
\begin{equation*}
E(u) =\Phi(|u'(x)|) + \lambda F(|u(x)|) = \lambda F(\alpha)
\quad \text{for all } x \in \mathbb{R}.
\end{equation*}
Denote the smallest zero in $(0,\infty)$ of the solution of \eqref{eq:3.9}
by $T(\lambda, \alpha)$.
Then we have the following lemma.

\begin{lemma}\label{le:3.3}
\begin{equation}\label{eq:3.10}
T(\lambda, \alpha)=\int_0^1 \frac{\alpha}{\Phi^{-1}(\xi)}\,ds,
\end{equation}
where $\Phi^{-1}$ is the inverse function of $\Phi$ given by \eqref{eq:3.4}
and $\xi$ is defined by
\begin{equation}\label{eq:3.11}
\xi:= \frac{\lambda}{p}(1-s^p)\alpha^p+ \frac{\lambda}{q}(1-s^q)\alpha^q.
\end{equation}
\end{lemma}

\begin{proof}
 By the same argument as in the proof of Lemma \ref{le:3.1}, we have
 \eqref{eq:3.7} for $x \in (0,T(\lambda,\alpha)]$, where
 ${\mathcal T}$ is given by \eqref{eq:3.8}.
 Letting $x=T(\lambda,\alpha)$ in \eqref{eq:3.7}, we have
 \begin{equation*}
 \int_0^1 \frac{\alpha}{\Phi^{-1}(\lambda[F(\alpha)-F(\alpha s)])}\,ds
  = T(\lambda,\alpha).
 \end{equation*}
 We put $\xi := \lambda(F(\alpha)-F(\alpha s))$, which is reduced to
 \eqref{eq:3.11}. The proof is complete.
\end{proof}


\section{Existence and nonexistence of positive solutions}\label{section-4}

In this section, we investigate the range of $\lambda$ for which \eqref{Plambda}
 has a positive solution by using the time map and the energy.
Moreover, we prove that all positive solutions are represented as a smooth curve
$(\lambda(\alpha), u(x,\alpha))$ with one parameter $\alpha=u(0, \alpha)=\|u\|_\infty$
and study the properties of this bifurcation curve.
Recall that $1<q<p$ is assumed throughout the paper.
By Theorem \ref{th:1.1}, if $L\leq 1$, then $\mu(q,L)<\mu(p,L)$.
However, if $L>1$, then all cases $\mu(q,L)<\mu(p,L)$, $\mu(q,L)>\mu(p,L)$
and $\mu(q,L)=\mu(p,L)$ can occur.
We state one of the main results.

\begin{theorem}\label{th:4.1}
Let $1<q<p$ and $L>0$. Then the following assertions hold.
\begin{itemize}
\item[(i)]
  For any $\alpha>0$, there exists a unique positive solution
$(\lambda, u)$ of \eqref{Plambda} which satisfies $u(0)=\alpha$.
  Denote it by $(\lambda(\alpha), u(x,\alpha))$.
\item[(ii)]
  The set of all positive solutions consists only of
$(\lambda(\alpha), u(x,\alpha))$ with $\alpha>0$.
\item[(iii)]
  $\lim_{\alpha\to +0}\lambda(\alpha)=\mu(q,L)$ and
$\lim_{\alpha\to \infty}\lambda(\alpha)=\mu(p,L)$.
\item[(iv)]
  Let $\phi(x)$ and $\psi(x)$ be the first eigenfunctions of the $p$ and $q$ Laplacian,
  respectively, which satisfy $\phi(0)=\psi(0)=1$. Then it holds that
  $$
  \lim_{\alpha\to +0}u(x,\alpha)/\|u(\cdot,\alpha)\|_\infty=\psi(x), \quad
  \lim_{\alpha\to \infty}u(x,\alpha)/\|u(\cdot,\alpha)\|_\infty=\phi(x),
  $$
  where the convergence is the strong topology in $C^1[-L,L]$.
\item[(v)]
  $\lambda(\alpha)$ is a $C^1(0,\infty)$ function which satisfies, for $\alpha>0$
  \begin{equation}\label{eq:4.1}
\begin{aligned}
    & \min\{\mu(p,L), \mu(q,L)\}  \\
    & < \lambda(\alpha) < \max\{(p-1)(\pi_q/(2L))^p, (q-1)(\pi_q/(2L))^q\}.
\end{aligned}
  \end{equation}
\end{itemize}
\end{theorem}

Garc\'{i}a-Huidobro, Man\'{a}sevich and Schmitt \cite[Theorem 5.1]{GMS2}
proved the existence part of (i) and (iii) of Theorem \ref{th:4.1}
for more general problem \eqref{N}.
By using the theorem above, we obtain the next corollary.

\begin{corollary}\label{co:4.2}
Let $1<q<p$ and $L>0$.\\
(i)   Suppose that $\mu(p,L)\neq \mu(q,L)$.
  Then \eqref{Plambda} has a positive solution when
  $$
  \min\{\mu(p,L), \mu(q,L)\} < \lambda < \max\{\mu(p,L), \mu(q,L)\}.
  $$
(ii)  Suppose that $L>1$ and $\mu(p,L)=\mu(q,L)$.
  Then there exists a $\lambda^*>\mu(p,L)$ such that
  \eqref{Plambda} has no positive solutions when $\lambda \leq \mu(p,L)$,
  at least two positive solutions when $\mu(p,L)<\lambda <\lambda^*$,
  at least one positive solution when $\lambda =\lambda^*$,
  and no positive solutions when $\lambda >\lambda^*$.
\end{corollary}

Corollary \ref{co:4.2} (ii) gives the bifurcation type (E) stated
in Section~\ref{section-1}. The assertion (i) has been proved in
 Motreanu and Tanaka~\cite{MT} also by using the variational method.
We shall prove the theorem by applying the bifurcation method.

We shall show the nonexistence of a positive solution when $\lambda$
does not satisfy \eqref{eq:4.1}.
To this end, we shall show the next lemma.

\begin{lemma}\label{le:4.3}
Suppose that $L>1$, $1<q<p$ and $\mu(p,L)=\mu(q,L)$.
Then no first eigenfunction of the $p$-Laplacian is equal to that of the
$q$-Laplacian.
\end{lemma}

\begin{proof}
Suppose on the contrary that $u$ is a first eigenfunction of both
$p$-Laplacian and $q$-Laplacian.
By the scalar multiplication, we may assume that $u(0)=1$.
Then $u$ is positive, even, concave and $0<u(x)<u(0)=1$ for $x\in(0,L)$.
Put $\lambda:=\mu(p,L)=\mu(q,L)$.
Since $u$ is a first eigenfunction of the $p$-Laplacian and $u'(0)=0$
and $u(0)=1$, we have the energy identity
$$
\frac{p-1}{p}|u'(x)|^p + \frac{\lambda}{p} u(x)^p=\frac{\lambda}{p},
$$
or equivalently,
$$
|u'(x)|=(\lambda/(p-1))^{1/p}(1-u(x)^p)^{1/p} \quad \text{for all } x\in [0,L].
$$
Similarly, we have
$$
|u'(x)|=(\lambda/(q-1))^{1/q}(1-u(x)^q)^{1/q}.
$$
Therefore it follows that for $x\in[0,L]$,
$$
(\lambda/(p-1))^{1/p}(1-u(x)^p)^{1/p} =(\lambda/(q-1))^{1/q}(1-u(x)^q)^{1/q}.
$$
As $x$ varies on $[0,L]$, $u(x)$ takes all values on $[0,1]$.
Accordingly, we see that
$$
(\lambda/(p-1))^{1/p}(1-t^p)^{1/p} =(\lambda/(q-1))^{1/q}(1-t^q)^{1/q} \quad
\text{for all } t\in[0,1].
$$
This causes a contradiction.
Indeed, differentiating the equation above with respect to $t$ and dividing
it by $t^{q-1}$, we obtain
$$
(\lambda/(p-1))^{1/p}(1-t^p)^{-(p-1)/p}t^{p-q}
= (\lambda/(q-1))^{1/q}(1-t^q)^{-(q-1)/q}.
$$
As $t\to +0$, we find a contradiction. The proof is complete.
\end{proof}

For simplicity, we write $\mu(p,L)$ as $\mu(p)$ if there is no confusion.

\begin{lemma}\label{le:4.4}
If $\lambda \leq \min\{\mu(p), \mu(q)\}$, then \eqref{Plambda} has no
positive solutions.
\end{lemma}

\begin{proof}
Hereafter $W^{1,p}_0(-L,L)$ denotes the Sobolev space and $\|u\|_p$ denotes
the $L^p(-L,L)$ norm of $u$.
Since the first eigenvalue is the minimum of the Rayleigh quotient,
we have
\begin{equation}\label{eq:4.2}
\mu(p)\|u\|_p^p \leq \|u'\|_p^p \quad \text{for } u\in W^{1,p}_0(-L,L).
\end{equation}
Here the equality occurs if and only if $u\equiv 0$ or it is a first eigenfunction of the $p$-Laplacian.
Let $\lambda \leq \min\{\mu(p), \mu(q)\}$ but assume that \eqref{Plambda} has a positive solution $u$.
We divide the proof into three cases.
\smallskip

\noindent\textbf{Case 1.} Assume that $\lambda<\min\{\mu(p), \mu(q)\}$.
Multiplying \eqref{Plambda} by $u$, integrating it on $(-L,L)$ and using
\eqref{eq:4.2}, we have
\begin{equation}\label{eq:4.3}
\begin{aligned}
\|u'\|_p^p + \|u'\|_q^q
&=  \lambda(\|u\|_p^p + \|u\|_q^q)  \\
& <  \mu(p)\|u\|_p^p + \mu(q)\|u\|_q^q  \\
&\leq  \|u'\|_p^p + \|u'\|_q^q.
\end{aligned}
\end{equation}
This is impossible.
\smallskip

\noindent\textbf{Case 2.}
 Assume that $\lambda=\min\{\mu(p), \mu(q)\}$ and $\mu(p)\neq \mu(q)$.
Then \eqref{eq:4.3} is still true. This is a contradiction.
\smallskip

\noindent\textbf{Case 3.} Assume that $\lambda=\mu(p)= \mu(q)$.
Then \eqref{eq:4.3} remains valid.
Indeed, if the equality holds, then we have
$$
\|u'\|_p^p + \|u'\|_q^q =\lambda(\|u\|_p^p + \|u\|_q^q)
= \mu(p)\|u\|_p^p + \mu(q)\|u\|_q^q,
$$
which with \eqref{eq:4.2} shows that
$$
\|u'\|_p^p = \mu(p)\|u\|_p^p \quad \text{and } \quad \|u'\|_q^q = \mu(q)\|u\|_q^q.
$$
Hence $u$ is a first eigenfunction of both $p$-Laplacian and $q$-Laplacian.
This contradicts Lemma \ref{le:4.3}.
The proof is complete.
\end{proof}

By using the unique solution of the initial value problem \eqref{eq:3.9}
introduced in Section~\ref{section-3}, we shall solve our problem
\eqref{Plambda}.
Thus,
we denote the unique solution of \eqref{eq:3.9} by $U(x, \lambda, \alpha)$.
Let $T(\lambda, \alpha)$ be as in \eqref{eq:3.10}.
Consider the case where $U(x,\lambda, \alpha)$ becomes a positive solution of \eqref{Plambda}, that is,
$$
U(x,\lambda, \alpha)>0 \quad \text{in } [0,L), \quad
U(L,\lambda, \alpha)=0.
$$
Then the next lemma readily follows.

\begin{lemma}\label{le:4.5}
Problem \eqref{Plambda} has a positive solution at $\lambda$ if and only if
$T(\lambda, \alpha)=L$ at some $\alpha>0$.
In this case, $U(x,\lambda, \alpha)$ is a positive solution.
\end{lemma}

\begin{lemma}\label{le:4.6}
Problem \eqref{Plambda} has no positive solutions if
\begin{equation}\label{eq:4.4}
\lambda \geq \max\{(p-1)(\pi_q/(2L))^p, (q-1)(\pi_q/(2L))^q\}.
\end{equation}
\end{lemma}

\begin{proof}
Let $\lambda$ satisfy the inequality above.
We shall show that $T(\lambda, \alpha)<L$ for all $\alpha>0$.
Then the conclusion follows from Lemma \ref{le:4.5}.
We shall estimate the time map $T(\lambda, \alpha)$.
Let $\xi$ be as in \eqref{eq:3.11}.
For $s \in [0,1]$, we have
\begin{align*}
\xi
&=  \frac{\lambda}{p}(1-s^p)\alpha^p+ \frac{\lambda}{q}(1-s^q)\alpha^q \\
&=  \frac{p-1}{p}
    \Big[ \lambda^{1/p} \frac{(1-s^p)^{1/p}}{(p-1)^{1/p}} \alpha \Big]^p
    + \frac{q-1}{q}
    \Big[ \lambda^{1/q} \frac{(1-s^q)^{1/q}}{(q-1)^{1/q}} \alpha \Big]^q.
\end{align*}
Define
$$
m(\lambda) := \min \Big\{ \frac{\lambda^{1/p}}{(p-1)^{1/p}},
\frac{\lambda^{1/q}}{(q-1)^{1/q}} \Big\}.
$$
Since $(1-s^p)^{1/p} > (1-s^p)^{1/q} > (1-s^q)^{1/q}$ for $s \in (0,1)$, we have
\begin{align*}
\xi
& >  \frac{p-1}{p} \big[ m(\lambda) (1-s^q)^{1/q} \alpha \big]^p
    + \frac{q-1}{q} \big[ m(\lambda) (1-s^q)^{1/q} \alpha \big]^q \\
&=  \Phi( m(\lambda) (1-s^q)^{1/q} \alpha),
\end{align*}
which is rewritten as $\Phi^{-1}(\xi)>m(\lambda)(1-s^q)^{1/q}\alpha$.
Therefore we see that
$$
T(\lambda, \alpha)=\int_0^1 \frac{\alpha}{\Phi^{-1}(\xi)}ds
< \frac{1}{m(\lambda)}\int_0^1(1-s^q)^{-1/q}ds =\frac{\pi_q}{2m(\lambda)}.
$$
Condition \eqref{eq:4.4} is equivalent to the inequality
$\pi_q/(2m(\lambda)) \leq L$.
Consequently, $T(\lambda, \alpha)<L$ for all $\alpha>0$.
The proof is complete.
\end{proof}

Combining Lemmas \ref{le:4.4} and \ref{le:4.6}, we have

\begin{lemma}\label{le:4.7}
If $\lambda$ satisfies either $\lambda \leq \min\{\mu(p), \mu(q)\}$
or \eqref{eq:4.4},
then \eqref{Plambda} has no positive solutions.
\end{lemma}

We shall investigate the behavior of positive solution $(\lambda, u)$
 when $\|u\|_\infty$ diverges to infinity.

\begin{lemma}\label{le:4.8}
Let $(\lambda_n,u_n)$ be a sequence of positive solutions to \eqref{Plambda}
which satisfies $\|u_n\|_\infty\to \infty$.
Put $v_n:=u_n/\|u_n\|_\infty$.
Then $(\lambda_n, v_n)$ converges to $(\mu(p,L), \phi)$ in
$\mathbb{R}\times C^1[-L,L]$,
where $\phi(x)$ is the first eigenfunction of the $p$-Laplacian which satisfies
$\phi(0)=1$.
\end{lemma}

To show the lemma above, we need an a priori estimate of positive solutions.

\begin{lemma}\label{le:4.9}
Let $(\lambda, u)$ be a positive solution of \eqref{Plambda}.
Then there exists a constant $C>0$ independent of $u$, $\lambda$ and $L$ such that
$$
|u'(x)| \leq C\lambda^{1/(p-1)}L^{1/(p-1)}(\|u\|_\infty +1) \quad \text{for } x\in [-L,L].
$$
\end{lemma}

\begin{proof}
Since $u$ is even, it is enough to show the inequality above for $x\in [0,L]$.
Since $1<q<p$, there exists a constant $C>0$ such that
\begin{equation}\label{eq:4.5}
|f(t)| \leq C(|t|^{p-1}+1) \quad \text{for } t\in \mathbb{R},
\end{equation}
where $f(t)$ is given by \eqref{eq:3.2}.
Since $f(t)\geq t^{p-1}$ for $t\geq 0$, the inverse function $f^{-1}$ satisfies
\begin{equation}\label{eq:4.6}
f^{-1}(t) \leq t^{1/(p-1)} \quad \text{for } t \geq 0.
\end{equation}
We rewrite \eqref{Plambda} as
$$
f(u')'+\lambda f(u)=0.
$$
Integrating this equation on $[0,x]$, using $u'(0)=0$ and operating $f^{-1}$,
we obtain
\begin{equation}\label{eq:4.7}
u'(x) = -f^{-1}\Big(\lambda \int_0^x f(u(t))dt \Big).
\end{equation}
Using \eqref{eq:4.5} and \eqref{eq:4.6}, we obtain, for $x\in[0,L]$
\begin{align*}
|u'(x)|
&\leq  \Big(\lambda \int_0^x f(u(t))dt \Big)^{1/(p-1)} \\
&\leq  \left(\lambda LC(\|u\|_\infty^{p-1}+1) \right)^{1/(p-1)} \\
&\leq  \lambda^{1/(p-1)}L^{1/(p-1)}C'(\|u\|_\infty+1).
\end{align*}
\end{proof}

\begin{proof}[Proof of Lemma \ref{le:4.8}]
Let $(\lambda_n, u_n)$ be as in Lemma \ref{le:4.8}.
Since $\lambda_n$ is bounded and bounded away from zero by Lemma \ref{le:4.7},
a subsequence of $\lambda_n$ (again denoted by $\lambda_n$) converges
to a limit $\lambda>0$.
Integrating \eqref{Plambda} on $(0,L)$ and using $u'(0)=0$, we have
\begin{equation}\label{eq:4.8}
|u'_n(x)|^{p-2}u'_n(x) + |u'_n(x)|^{q-2}u'_n(x)
= -\lambda_n \int_0^x(u_n(t)^{p-1} +u_n(t)^{q-1})dt.
\end{equation}
Put $v_n:=u_n/\|u_n\|_\infty$.
Dividing the both sides by $\|u_n\|_\infty^{p-1}$, we have
\begin{equation}\label{eq:4.9}
|v'_n|^{p-2}v'_n
= -\lambda_n \int_0^x(v_n(t)^{p-1} + u_n(t)^{q-1}/\|u_n\|_\infty^{p-1})dt
- |u'_n|^{q-2}u'_n/\|u_n\|_\infty^{p-1}.
\end{equation}
The second term $u_n(t)^{q-1}/\|u_n\|_\infty^{p-1}$ of the integrand
uniformly converges to zero.
By Lemma \ref{le:4.9}, we estimate the last term as
$$
\|u'_n\|_\infty^{q-1}/\|u_n\|_\infty^{p-1}
\leq C^{q-1}(\|u_n\|_\infty +1)^{q-1}/\|u_n\|_\infty^{p-1} \to 0.
$$
Since $\|v_n\|_\infty=1$ by the definition of $v_n$, the right-hand side
of \eqref{eq:4.9} is uniformly bounded.
Thus $\|v'_n\|_\infty$ is bounded.
By the Ascoli-Arzela theorem, a subsequence of $v_n$ uniformly converges
to a limit $v$.
Denote the right hand side of \eqref{eq:4.9} by $w_n(x)$.
Then $w_n(x)$ uniformly converges to
\begin{equation}\label{eq:4.10}
w(x):=-\lambda \int_0^x v(t)^{p-1}dt.
\end{equation}
Since $|v'_n|^{p-2}v'_n=w_n$ by \eqref{eq:4.9}, we rewrite this equation as
$$
v_n(x)= -\int_x^L \text{\rm sgn}(w_n)|w_n(t)|^{1/(p-1)}dt.
$$
Since $v_n$ and $w_n$ uniformly converge to $v$ and $w$, respectively, we find that
$$
v(x)= -\int_x^L \text{\rm sgn}(w)|w(t)|^{1/(p-1)}dt.
$$
Differentiating it and using \eqref{eq:4.10}, we obtain
\begin{gather*}
(|v'|^{p-2}v')' + \lambda {v}^{p-1}=0, \quad v \geq 0, \quad \text{in } (-L,L),\\
v(-L)=v(L)=0.
\end{gather*}
Since $v_n(0)=1$, it holds that $v(0)=1$. Hence $v(x)>0$ for $x\in(-L,L)$.
Therefore $v$ is a positive eigenfunction and so $\lambda$ must be the
first eigenvalue.
Consequently, $\lambda=\mu(p,L)$ and $v=\phi$.
From the uniqueness of the limit, $(\lambda_n, v_n)$ itself
(without extracting a subsequence) converges.
\end{proof}

We next consider the case where a sequence of positive solutions converges to zero.
To this end, we prepare another a priori estimate of positive solutions.

\begin{lemma}\label{le:4.10}
For any positive solution $(\lambda, u)$ satisfying $\|u\|_\infty\leq 1$,
it holds 
$$
|u'(x)|\leq (2\lambda L)^{1/(q-1)}\|u\|_\infty \quad \text{for } x\in [-L,L].
$$
\end{lemma}

\begin{proof}
It is sufficient to show the inequality above for $x\in [0,L]$, because $u$ is even.
Let $f(t)$ be as in \eqref{eq:3.2}. Then we have
$$
|f(t)|\leq 2|t|^{q-1} \text{ for } |t|\leq 1, \quad f^{-1}(t) \leq t^{1/(q-1)} 
\text{ for } t\geq 0.
$$
Using these inequalities, we estimate \eqref{eq:4.7} for $x\in[0,L]$ as
$$
|u'(x)| \leq \Big(\lambda \int_0^x f(u(t))dt \Big)^{1/(q-1)} 
\leq (2 \lambda L)^{1/(q-1)}\|u\|_\infty.
$$
\end{proof}

We shall show that a sequence of positive solutions converging to zero 
approaches the first eigenfunction of the $q$-Laplacian.

\begin{lemma}\label{le:4.11}
Let $(\lambda_n,u_n)$ be a sequence of positive solutions of \eqref{Plambda} 
which satisfies $\|u_n\|_\infty\to 0$.
Put $v_n:=u_n/\|u_n\|_\infty$.
Then $(\lambda_n, v_n)$ converges to $(\mu(q,L), \psi)$ in 
$\mathbb{R}\times C^1[-L,L]$, where $\psi(x)$ denotes the first eigenfunction 
of the $q$-Laplacian satisfying $\psi(0)=1$.
\end{lemma}

\begin{proof}
We use the same method as in the proof of Lemma \ref{le:4.8}.
By Lemma \ref{le:4.7}, $\lambda_n$ is bounded and bounded away from zero.
A subsequence of $\lambda_n$ converges to a limit $\lambda>0$.
Dividing \eqref{eq:4.8} by $\|u_n\|_\infty^{q-1}$, we have
\begin{equation}\label{eq:4.11}
\begin{aligned}
&|v'_n|^{q-2}v'_n \\
&= -\lambda_n \int_0^x(v_n(t)^{q-1} + u_n(t)^{p-1}/\|u_n\|_\infty^{q-1})dt 
- |u'_n|^{p-2}u'_n/\|u_n\|_\infty^{q-1}.
\end{aligned}
\end{equation}
Since $1<q<p$ and $\|u_n\|_\infty\to0$, we use Lemma \ref{le:4.10} to get
$$
\|u_n\|_\infty^{p-1}/\|u_n\|_\infty^{q-1}\to 0, \quad 
\|u'_n\|_\infty^{p-1}/\|u_n\|_\infty^{q-1}\to 0.
$$
Therefore the right hand side of \eqref{eq:4.11} is uniformly bounded and hence
 $\|v_n'\|_\infty$ is bounded.
By the Ascoli-Arzela theorem, a subsequence of $v_n$ uniformly converges to a 
limit $v$. The rest of proof is the same as in the proof of Lemma \ref{le:4.8}.
\end{proof}

We are now in a position to prove Theorem \ref{th:4.1}.

\begin{proof}[Proof of Theorem \ref{th:4.1}]
It is easy to verify that for any $\alpha>0$ fixed, $T(\lambda, \alpha)$ 
is strictly decreasing on $\lambda$ and satisfies
$$
\lim_{\lambda\to +0}T(\lambda, \alpha)=\infty, \quad 
\lim_{\lambda\to \infty}T(\lambda, \alpha)=0.
$$
Therefore for each $\alpha>0$, the equation $T(\lambda, \alpha)=L$ has a 
unique solution $\lambda$.
We denote it by $\lambda(\alpha)$, that is,
\begin{equation}\label{eq:4.12}
T(\lambda(\alpha),\alpha)=L \quad \text{for } \alpha>0.
\end{equation}
Recall that $U(x,\lambda, \alpha)$ denotes a solution of \eqref{eq:3.9}.
We define
$$
u(x,\alpha):=U(x,\lambda(\alpha),\alpha).
$$
Then $(\lambda(\alpha), u(x,\alpha))$ is a positive solution of \eqref{Plambda}
with $\lambda$ replaced by $\lambda(\alpha)$, which satisfies $u(0,\alpha)=\alpha$ .
We shall show the uniqueness of such a solution.
Let $\alpha>0$. If $(\mu,v(x))$ is a positive solution satisfying $v(0)=\alpha$,
then $T(\mu,\alpha)=L$ by the definition of $T$.
From the uniqueness of $\lambda(\alpha)$ satisfying \eqref{eq:4.12},
it follows that $\lambda(\alpha)=\mu$.
Since $u(0,\alpha)=\alpha=v(0)$ and $u'(0,\alpha)=0=v'(0)$, it holds that 
$u(x,\alpha)=v(x)$ from Lemma~\ref{le:3.1}.
Therefore $(\mu,v(x))=(\lambda(\alpha), u(x,\alpha))$ and we obtain the assertion 
(i). By definition, $(\lambda(\alpha),u(x,\alpha))$ with $\alpha>0$ represents 
all positive solutions. Thus (ii) is valid.
The integrand
$$
\alpha/\Phi^{-1}(\xi)= \alpha/\Phi^{-1}((\lambda/p)(1-s^p)\alpha^p+(\lambda/q)(1-s^q)\alpha^q)
$$
in \eqref{eq:3.10}
is differentiable with respect to $\lambda$ and $\alpha$ and,
it holds:
\begin{gather*}
\frac{\partial}{\partial \lambda} (\alpha/\Phi^{-1}(\xi))
=-\frac{\alpha d\xi/d\lambda}{(\Phi^{-1}(\xi))^2\Phi'(\Phi^{-1}(\xi))},
\\
\frac{\partial}{\partial \alpha} (\alpha/\Phi^{-1}(\xi))
=\frac{\Phi^{-1}(\xi)\Phi'(\Phi^{-1}(\xi))
 -\alpha d\xi/d\alpha}{(\Phi^{-1}(\xi))^2\Phi'(\Phi^{-1}(\xi))},
\end{gather*}
where
\begin{gather*}
\Phi'(\Phi^{-1}(\xi))
=(p-1)\Phi^{-1}(\xi)^{p-1}+ (q-1)\Phi^{-1}(\xi)^{q-1}, \\
\frac{d\xi}{d\lambda}=(1-s^p)\alpha^{p}/p + (1-s^q)\alpha^{q}/q, \\
\frac{d\xi}{d\alpha}=\lambda(1-s^p)\alpha^{p-1} + \lambda(1-s^q)\alpha^{q-1}.
\end{gather*}
Hence, for each compact set $D$ in $(0,\infty)\times(0,\infty)$, there exists 
a constant $C>0$ such that
 \[
  \big|\frac{\partial}{\partial \lambda} (\alpha/\Phi^{-1}(\xi))\big|
    \le C (1-s^q)^{-1/q}
  \quad \text{and} \quad
  \big|\frac{\partial}{\partial \alpha} (\alpha/\Phi^{-1}(\xi))\big|
   \le C (1-s^q)^{-1/q}
 \]
 for $s \in (0,1)$ and $(\alpha,\lambda) \in D$.
See also the proof of Proposition \ref{pr:5.1}.
Therefore, since $(1-s^q)^{-1/q} \in L^1(0,1)$, $T(\lambda,\alpha)$ 
has partial derivatives.
Denote them by $T_\lambda$ and $T_\alpha$.
It is easy to verify that $T_\lambda<0$.
Applying the implicit function theorem to \eqref{eq:4.12}, we find that
$\lambda(\alpha)$ is a $C^1$ function which satisfies
\begin{equation}\label{eq:4.13}
\lambda'(\alpha)=-\frac{T_\alpha(\lambda(\alpha), \alpha)}{T_\lambda(\lambda(\alpha),
 \alpha)}.
\end{equation}
Combining Lemmas \ref{le:4.8} and \ref{le:4.11}, we have (iii) and (iv).
Lemma \ref{le:4.7} ensures (v).
The proof is complete.
\end{proof}

Let $u(x,\alpha)$ be as in Theorem \ref{th:4.1}.
Since $u(x,\alpha)$ is uniquely determined by $\alpha$, we can identify the 
bifurcation curve $(\lambda(\alpha), u(x,\alpha))$ with $(\lambda(\alpha),\alpha)$.
Then Theorem \ref{th:4.1} (iii) means that the bifurcation curve 
$(\lambda(\alpha),\alpha)$
starts from the initial point $(\mu(q),0)$ and goes to the final point
 $(\mu(p),\infty)$. Using these results, we shall prove Corollary \ref{co:4.2}.

\begin{proof}[Proof of Corollary \ref{co:4.2}]
By Theorem \ref{th:4.1} (i) and (ii), \eqref{Plambda} has a positive solution 
at $\lambda$ if and only if
$\lambda$ is in the range of $\lambda(\alpha)$, i.e.,
$\lambda \in \{\lambda(\alpha): \alpha>0\}$.
This fact with (iii) in Theorem \ref{th:4.1} shows the assertion (i).
We shall show (ii). Let $\mu(p)=\mu(q)$.
We draw the curve $(\lambda(\alpha),\alpha)$ in the plain, where
we choose $\lambda$-axis and $\alpha$-axis as axes of abscissa and ordinate, 
respectively.
By \eqref{eq:4.1}, $\lambda(\alpha)>\mu(p)=\mu(q)$ for all $\alpha$.
Thus the bifurcation curve stays in the right side of the line $\lambda=\mu(p)$.
We define
\begin{equation}\label{eq:4.14}
\begin{aligned}
\lambda^*
:= & \sup\{\lambda(\alpha): \alpha>0\}  \\
=&  \sup\{\lambda: \eqref{Plambda} \text{ has a positive solution} \}.
\end{aligned}
\end{equation}
Then $\lambda^*>\mu(p)$ and Theorem \ref{th:4.1} (iii) ensures the assertion (ii).
\end{proof}

Since $\lambda(\alpha)$ depends on $p$, $q$ and $L$ also, we write it as 
$\lambda(\alpha,p,q,L)$. Then $\lambda^*$ given by \eqref{eq:4.14} depends 
on $p$, $q$ and $L$ and we denote it by $\lambda^*(p,q,L)$.

\begin{lemma}\label{le:4.12}
The function $\lambda^*(p,q,L)$ is continuous.
\end{lemma}

\begin{proof}
Let $1<q_0<p_0$, $0<L_0$ and let $(p_n, q_n,L_n)$ be a sequence converging to 
$(p_0,q_0,L_0)$. We shall show that along a subsequence of $(p_n,q_n,L_n)$,
\begin{equation}\label{eq:4.15}
\lim_{n\to\infty}\lambda^*(p_n,q_n,L_n)  \leq \lambda^*(p_0,q_0,L_0).
\end{equation}
Choose $\alpha_n>0$ such that
\begin{equation}\label{eq:4.16}
|\lambda(\alpha_n,p_n,q_n,L_n) - \lambda^*(p_n,q_n,L_n)|<1/n.
\end{equation}
After extracting a subsequence, we may assume that
$\alpha_n\to \infty$, $\alpha_n\to 0$ or $\alpha_n\to \alpha_\infty>0$.
By Theorem \ref{th:4.1} (iii), we have
$\mu(p_0,L_0), \mu(q_0,L_0) \leq \lambda^*(p_0,q_0,L_0)$.
Observing the proof of Lemma \ref{le:4.8}, we can prove that if $\alpha_n\to\infty$,
then $\lambda(\alpha_n,p_n,q_n,L_n)$ converges to $\mu(p_0,L_0)$. Hence
$$
\lim_{n\to\infty}\lambda^*(p_n,q_n,L_n)=\mu(p_0,L_0)\leq \lambda^*(p_0,q_0,L_0).
$$
In the same way as in the proof of Lemma \ref{le:4.11},
we can show that if $\alpha_n\to 0$, then
$\lambda(\alpha_n,p_n,q_n,L_n)$ converges to $\mu(q_0,L_0)$. Thus
$$
\lim_{n\to\infty}\lambda^*(p_n,q_n,L_n)=\mu(q_0,L_0)\leq \lambda^*(p_0,q_0,L_0).
$$
If $\alpha_n$ converges to a limit $\alpha_\infty>0$, the continuity of $\lambda$ 
implies that
$$
\lim_{n\to\infty}\lambda(\alpha_n,p_n,q_n,L_n) 
=\lambda(\alpha_\infty,p_0,q_0,L_0) \leq \lambda^*(p_0,q_0,L_0),
$$
which with \eqref{eq:4.16} yields \eqref{eq:4.15}. Consequently, all the cases 
of $\alpha_n$ satisfy \eqref{eq:4.15}.

Since $\lambda(\alpha,p_n,q_n,L_n) \leq \lambda^*(p_n,q_n,L_n)$ for $\alpha>0$,
 we pass to the limit to obtain
$$
\lambda(\alpha,p_0,q_0,L_0)\leq \liminf_{n\to\infty}\lambda^*(p_n,q_n,L_n).
$$
Taking the supremum on $\alpha$, we have
$$
\lambda^*(p_0,q_0,L_0)\leq \liminf_{n\to\infty}\lambda^*(p_n,q_n,L_n).
$$
By this inequality and \eqref{eq:4.15}, 
$\lambda^*(p_n,q_n,L_n)$ has a subsequence that converges to $\lambda^*(p_0,q_0,L_0)$.
From the uniqueness of the limit, $\lambda^*(p_n,q_n,L_n)$ itself converges.
Therefore $\lambda^*(p,q,L)$ is continuous.
\end{proof}

We define
$$
m(p,q,L) := \min\{\mu(p,L), \mu(q,L)\}, \quad 
M(p,q,L) := \max\{\mu(p,L), \mu(q,L)\}.
$$
Consider a small perturbation of $(p,q,L)$ satisfying Corollary \ref{co:4.2} (ii).
Then we have the next result.

\begin{theorem}\label{th:4.13}
Let $p_0$, $q_0$ and $L_0$ satisfy that $1<L_0$, $1<q_0<p_0$ and 
$\mu(p_0,L_0)=\mu(q_0,L_0)$.
If $(p,q,L)$ is sufficiently close to $(p_0,q_0,L_0)$ and satisfies 
$\mu(p, L)\neq \mu(q,L)$,
then $m(p,q,L)< M(p,q,L) < \lambda^*(p,q,L)$.
Moreover \eqref{Plambda} has no positive solutions if $\lambda \leq m(p,q,L)$,
at least one positive solution if $m(p,q,L)<\lambda \leq M(p,q,L)$,
at least two positive solutions if $M(p,q,L)<\lambda < \lambda^*(p,q,L)$,
at least one positive solution if $\lambda = \lambda^*(p,q,L)$,
and no positive solutions if $\lambda > \lambda^*(p,q,L)$.
\end{theorem}

\begin{proof}
Note that $\mu(p,L)$ and $\mu(q,L)$ are continuous and so is 
$\lambda^*(p,q,L)$ by Lemma \ref{le:4.12}.
It follows from Corollary \ref{co:4.2} (ii) that 
$\lambda^*(p_0,q_0,L_0)> \mu(p_0,L_0)$.
Therefore $\lambda^*(p,q,L)$ is greater than $M(p,q,L)$
when $(p,q,L)$ is sufficiently close to $(p_0,q_0,L_0)$.
This shows the theorem.
\end{proof}

\begin{remark}\label{re:4.14} \rm
Recall that the five types (A)--(E) of the bifurcation diagrams have been 
defined in Section~\ref{section-1}.
Let $L_*=L_*(p,q)$ be defined by \eqref{eq:1.4}.
As mentioned in Section~\ref{section-1}, $\mu(p,L)=\mu(q,L)$ if and only 
if $L=L_*$.
By \eqref{eq:1.2}, $\mu(p,L)>\mu(q,L)$ when $L<L_*$ and $\mu(p,L)<\mu(q,L)$ 
when $L>L_*$.
This fact with Theorem~\ref{th:4.13} shows that if $L$ is slightly less than
 $L_*$, then (B) occurs, and if it is slightly greater than $L_*$, 
then (D) occurs. Moreover, we shall prove in Section~\ref{section-6} that
if $L$ is small enough, then (A) appears and if it is large enough, (C) occurs.
Therefore, as $L$ increases, the bifurcation diagram changes in order of 
(A), (B), (E), (D), (C).
\end{remark}

\section{Derivative of the time map}\label{section-5}

Let $T(\lambda, \alpha)$ be defined by \eqref{eq:3.10}.
We denote by $T_\alpha$ and $T_\lambda$ the partial derivatives of 
$T(\lambda, \alpha)$ with respect to $\alpha$ and $\lambda$, respectively.
Since $T_\lambda<0$, \eqref{eq:4.13} implies that
$\lambda'(\alpha)$ and $T_\alpha$ have the same sign.
To investigate the sign of $\lambda'(\alpha)$, we estimate $T_\alpha$.
In particular, we compute the limit of $T_\alpha$ as $\alpha \to \infty$ 
or as $\alpha \to +0$.

\begin{proposition}\label{pr:5.1}
\begin{equation}\label{eq:5.1}
\lim_{\alpha\to\infty}\alpha^{p+1-q}T_\alpha(\lambda, \alpha)= T_\infty(\lambda),
\end{equation}
where $T_\infty(\lambda)$ is given by
\begin{gather*}
T_\infty(\lambda) = T_\infty(\lambda,p,q) := c_1(p,q)\lambda^{-1/p} 
- c_2(p,q)\lambda^{(q-p-1)/p}, \\
c_1(p,q):=q^{-1}(p-q)(p-1)^{1/p}\int_0^1(1-s^q)(1-s^p)^{-(p+1)/p}ds, \\
c_2(p,q):=q^{-1}(p-q)(q-1)(p-1)^{-(q-1)/p}\int_0^1(1-s^p)^{(q-p-1)/p}ds.
\end{gather*}
Furthermore, for any compact subset $K$ in $(0,\infty)$,
the convergence of \eqref{eq:5.1} is uniform on $\lambda \in K$.
\end{proposition}

\begin{proof}
Let $\xi$ be defined by \eqref{eq:3.11}.
We differentiate $T(\lambda, \alpha)$ given by \eqref{eq:3.10} with respect 
to $\alpha$ to obtain
\begin{equation*}
T_\alpha(\lambda, \alpha)
= \int_0^1 \frac{\Phi^{-1}(\xi)\Phi'(\Phi^{-1}(\xi))-\alpha d\xi/d\alpha}
{(\Phi^{-1}(\xi))^2\Phi'(\Phi^{-1}(\xi))}ds.
\end{equation*}
We denote the numerator and the denominator of the integrand by $P$ and $Q$, 
respectively, i.e.,
\begin{equation*}
T_\alpha(\lambda, \alpha) = \int_0^1 \frac{P(s, \lambda, \alpha)}
{Q(s, \lambda, \alpha)}ds.
\end{equation*}
We observe that
\begin{gather*}
\Phi'(\Phi^{-1}(\xi))=(p-1)\Phi^{-1}(\xi)^{p-1}+ (q-1)\Phi^{-1}(\xi)^{q-1}, \\
\frac{d\xi}{d\alpha}=\lambda(1-s^p)\alpha^{p-1} + \lambda(1-s^q)\alpha^{q-1}.
\end{gather*}
Therefore,
\begin{gather}\label{eq:5.2}
P:=(p-1)\Phi^{-1}(\xi)^p + (q-1)\Phi^{-1}(\xi)^q 
 - \lambda[(1-s^p)\alpha^p+(1-s^q)\alpha^q], \\
\label{eq:5.3}
Q:=(p-1)\Phi^{-1}(\xi)^{p+1}+ (q-1)\Phi^{-1}(\xi)^{q+1}.
\end{gather}
Put $\eta:=\Phi^{-1}(\xi)$. Then we see that
\begin{equation*}
\xi=\Phi(\eta)=\frac{p-1}{p}\eta^p + \frac{q-1}{q}\eta^q.
\end{equation*}
We rewrite the relation above as
$$
(p-1)\eta^p + (q-1)\eta^q= p\xi -\frac{(p-q)(q-1)}{q}\eta^q.
$$
Substituting $\eta=\Phi^{-1}(\xi)$ into the relation above, we obtain
$$
(p-1)\Phi^{-1}(\xi)^p + (q-1)\Phi^{-1}(\xi)^q
= p\xi -\frac{(p-q)(q-1)}{q}\Phi^{-1}(\xi)^q.
$$
Using this relation in \eqref{eq:5.2} and replacing $\xi$ by \eqref{eq:3.11}, 
we obtain
\begin{equation*}
P=\frac{p-q}{q}\lambda(1-s^q)\alpha^q - \frac{(p-q)(q-1)}{q}\Phi^{-1}(\xi)^q.
\end{equation*}
Denote the first and the second terms on the right hand side by $I$ and $J$, 
respectively, i.e.,
\begin{equation}\label{eq:5.4}
I:=\frac{p-q}{q}\lambda(1-s^q)\alpha^q, \quad
J:=\frac{(p-q)(q-1)}{q}\Phi^{-1}(\xi)^q.
\end{equation}
Then $P=I-J$. We compute the limits of $\alpha^{p+1-q}I/Q$ and $\alpha^{p+1-q}J/Q$ 
as $\alpha\to \infty$.
For functions $f(t)$ and $g(t)$, we define the notation
$$
f(t)\sim g(t) \quad \text{as } t\to \infty,
$$
if $\lim_{t\to \infty}f(t)/g(t)=1$.
Fix $0<s<1$ arbitrarily. Then we see that
\begin{gather}\label{eq:5.5}
\Phi^{-1}(\xi)\sim \Big(\frac{p}{p-1}\Big)^{1/p}\xi^{1/p}
\quad \text{as } \xi\to\infty, \\
\label{eq:5.6}
\xi \sim \frac{\lambda}{p}(1-s^p)\alpha^p \quad \text{as } \alpha\to\infty.
\end{gather}
Therefore as $\alpha\to\infty$,
\begin{equation}\label{eq:5.7}
\begin{aligned}
Q
&=  (p-1)\Phi^{-1}(\xi)^{p+1}+ (q-1)\Phi^{-1}(\xi)^{q+1}  \\
& \sim  (p-1)\Phi^{-1}(\xi)^{p+1}  \\
& \sim  (p-1)(p/(p-1))^{(p+1)/p} \xi^{(p+1)/p}  \\
& \sim  (p-1)^{-1/p}\lambda^{(p+1)/p}(1-s^p)^{(p+1)/p}\alpha^{p+1}.
\end{aligned}
\end{equation}
Accordingly, we have
\begin{equation}\label{eq:5.8}
\lim_{\alpha\to \infty} \frac{\alpha^{p+1-q}I}{Q}=
\frac{p-q}{q}(p-1)^{1/p}\lambda^{-1/p}(1-s^q)(1-s^p)^{-(p+1)/p}.
\end{equation}
By \eqref{eq:5.4}--\eqref{eq:5.6}, $J$ has an asymptotic formula
\begin{equation*}
J \sim q^{-1}(p-q)(q-1)(p-1)^{-q/p}\lambda^{q/p}(1-s^p)^{q/p}\alpha^q.
\end{equation*}
The relation above with \eqref{eq:5.7} implies 
\begin{equation}\label{eq:5.9}
\begin{aligned}
&\lim_{\alpha\to \infty} \frac{\alpha^{p+1-q}J}{Q} \\
&= q^{-1}(p-q)(q-1)(p-1)^{-(q-1)/p}\lambda^{(q-p-1)/p}(1-s^p)^{(q-p-1)/p}.
\end{aligned}
\end{equation}
We shall prove in the next Lemma~\ref{le:5.2} that there exists a constant 
$C>0$ independent of $\alpha$ and $s$
such that for $s\in(0,1)$ and $\alpha\geq 1$,
\begin{gather}\label{eq:5.10}
0\leq \alpha^{p+1-q}I/Q \leq C(1-s^p)^{-1/q}, \\
\label{eq:5.11}
0\leq \alpha^{p+1-q}J/Q \leq C(1-s^p)^{(q-p-1)/p}.
\end{gather}
The right hand sides of \eqref{eq:5.10} and \eqref{eq:5.11} are integrable 
on $(0,1)$. By the Lebesgue dominated convergence theorem with 
\eqref{eq:5.8}--\eqref{eq:5.11}, we obtain
\begin{align*}
\lim_{\alpha\to \infty}\alpha^{p+1-q}\int_0^1 \frac{I}{Q}\,ds
&=  \frac{p-q}{q}(p-1)^{1/p}\lambda^{-1/p} \int_0^1 (1-s^q)(1-s^p)^{-(p+1)/p}\,ds  \\
&=  c_1(p,q)\lambda^{-1/p},
\end{align*}
and
\begin{align*}
&\lim_{\alpha\to \infty}\alpha^{p+1-q}\int_0^1 \frac{J}{Q}\,ds\\
&=  \frac{p-q}{q}(q-1)(p-1)^{-(q-1)/p} \lambda^{-(p+1-q)/p} 
  \int_0^1(1-s^p)^{(q-p-1)/p}ds \\
&=  c_2(p,q)\lambda^{-(p+1-q)/p}.
\end{align*}
Here $c_1(p,q)$ and $c_2(p,q)$ have been defined in the statement of
 Proposition \ref{pr:5.1}.
Thus $\alpha^{p+1-q}T_\alpha$ converges to $T_\infty$.

Let $K$ be any compact subset of $(0,\infty)$.
Then \eqref{eq:5.10} and \eqref{eq:5.11} are valid for all $\lambda \in K$ and
the constant $C$ depends only on $K$.
Therefore the convergence of \eqref{eq:5.1} is uniform on $\lambda\in K$.
The proof is complete.
\end{proof}

Let $I$, $J$ and $Q$ be defined in the proof above.
We shall give estimates of $I/Q$ and $J/Q$, which imply \eqref{eq:5.10} 
and \eqref{eq:5.11}.

\begin{lemma}\label{le:5.2}
There exists a positive constant $C$ such that
\begin{gather}\label{eq:5.12}
\lambda^{1/p}\alpha^{p+1-q}I/Q \leq C(1-s^p)^{-1/q}
(1+\lambda^{-(p-q)/pq}\alpha^{-(p-q)/q}), \\
\label{eq:5.13}
\lambda^{(p+1-q)/p}\alpha^{p+1-q}J/Q \leq C(1-s^p)^{(q-p-1)/p}
(1+\alpha^{q-p})
\end{gather}
for every $\lambda>0$, $\alpha>0$ and $s\in (0,1)$.
\end{lemma}

\begin{proof}
In this proof, we denote various positive constants independent of $\alpha$, $s$, $\lambda$ by $c$ or $C$.
From an easy computation, it follows that for all $t\geq 0$,
\begin{equation}\label{eq:5.14}
\Phi^{-1}(t) \leq \left(\frac{pt}{p-1}\right)^{1/p}, \quad
\Phi^{-1}(t) \leq \left(\frac{qt}{q-1}\right)^{1/q}.
\end{equation}
Note that
\begin{gather*}
\Phi^{-1}(t) \sim \left(\frac{q}{q-1}\right)^{1/q}t^{1/q} \quad \text{as } t\to +0,\\
\Phi^{-1}(t) \sim \Big(\frac{p}{p-1}\Big)^{1/p}t^{1/p} \quad \text{as } t\to \infty.
\end{gather*}
By the relations above, there exists a constant $c>0$ such that
\begin{equation}\label{eq:5.15}
\Phi^{-1}(t) \geq ct^{1/q} \quad \text{when } 0\leq t \leq 1, \quad
\Phi^{-1}(t) \geq ct^{1/p} \quad \text{when } t \geq 1.
\end{equation}
Furthermore, for $s\in (0,1)$, we have
\begin{equation}\label{eq:5.16}
\xi=\frac{\lambda}{p}(1-s^p)\alpha^p + \frac{\lambda}{q}(1-s^q)\alpha^q
\leq C\lambda(1-s^p)(\alpha^p+\alpha^q).
\end{equation}
We divide the proof into two cases: $0\leq \xi \leq 1$ and $\xi>1$.
\smallskip

\noindent\textbf{Case 1.} Let $0\leq \xi \leq 1$.
Using the first inequality in \eqref{eq:5.15}, we have
\begin{equation}\label{eq:5.17}
Q\geq (q-1)\Phi^{-1}(\xi)^{q+1} \geq c\xi^{(q+1)/q} 
\geq c\lambda^{(q+1)/q}(1-s^p)^{(q+1)/q}\alpha^{p(q+1)/q}.
\end{equation}
This inequality and \eqref{eq:5.4} imply
\begin{align*}
\frac{\lambda^{1/p}\alpha^{p+1-q}I}{Q}
&\leq  C\lambda^{-(p-q)/pq} \alpha^{-(p-q)/q}(1-s^q)(1-s^p)^{-(q+1)/q} \\
&\leq  C\lambda^{-(p-q)/pq} \alpha^{-(p-q)/q}(1-s^p)^{-1/q}.
\end{align*}
Thus we have \eqref{eq:5.12}.
Let us estimate $J$.
Using \eqref{eq:5.4}, the second inequality in \eqref{eq:5.14} and 
\eqref{eq:5.16}, we have
$$
J \leq C\Phi^{-1}(\xi)^q \leq C\xi\leq C\lambda (1-s^p)(\alpha^p+\alpha^q),
$$
which with \eqref{eq:5.17} implies that
\begin{equation}\label{eq:5.18}
\frac{\lambda^{(p+1-q)/p}\alpha^{p+1-q}J}{Q}
 \leq C(\lambda \alpha^p)^{(p-q)(q-1)/pq}(1-s^p)^{-1/q} (1+\alpha^{q-p}).
\end{equation}
Since $\xi\leq 1$, it holds that $(\lambda/p)(1-s^p)\alpha^p\leq \xi \leq1$,
and so $\lambda\alpha^p\leq p(1-s^p)^{-1}$.
Substituting this inequality into the right hand side of \eqref{eq:5.18}, we obtain
$$
\frac{\lambda^{(p+1-q)/p}\alpha^{p+1-q}J}{Q}\leq C(1-s^p)^{(q-p-1)/p}
(1+\alpha^{q-p}).
$$
Consequently, \eqref{eq:5.13} holds.
\smallskip

\noindent\textbf{Case 2.} Let $\xi>1$.
By the second inequality in \eqref{eq:5.15}, we have
\begin{equation}\label{eq:5.19}
Q\geq (p-1)\Phi^{-1}(\xi)^{p+1}\geq c\xi^{(p+1)/p} 
\geq c\lambda^{(p+1)/p}(1-s^p)^{(p+1)/p}\alpha^{p+1}.
\end{equation}
Therefore
$$
\frac{\lambda^{1/p}\alpha^{p+1-q}I}{Q}
\leq C(1-s^q)(1-s^p)^{-(p+1)/p} \leq C(1-s^p)^{-1/p} \leq C(1-s^p)^{-1/q},
$$
which shows \eqref{eq:5.12}.
We use the first inequality in \eqref{eq:5.14} and \eqref{eq:5.16} to obtain
$$
J\leq C\Phi^{-1}(\xi)^q \leq C\xi^{q/p} \leq C\lambda^{q/p}(1-s^p)^{q/p}(\alpha^p+\alpha^q)^{q/p}.
$$
This inequality and \eqref{eq:5.19} give us
\begin{align*}
\frac{\lambda^{(p+1-q)/p}\alpha^{p+1-q}J}{Q}
&\leq  C(1-s^p)^{(q-p-1)/p} (1+\alpha^{q-p})^{q/p} \\
&\leq  C(1-s^p)^{(q-p-1)/p} (1+\alpha^{q-p}).
\end{align*}
Thus \eqref{eq:5.13} holds and the proof is complete.
\end{proof}

In the following proposition, we compute the limit of $T_\alpha$ as $\alpha\to+0$.

\begin{proposition}\label{pr:5.3}
\begin{equation}\label{eq:5.20}
\lim_{\alpha\to +0}\alpha^{q+1-p}T_\alpha(\lambda, \alpha)=T_0(\lambda),
\end{equation}
where 
\begin{gather*}
T_0(\lambda):=T_0(\lambda,p,q) := d_2(p,q)\lambda^{(p-q-1)/q} 
- d_1(p,q)\lambda^{-1/q}, \\
d_1(p,q):=p^{-1}(p-q)(q-1)^{1/q}\int_0^1 (1-s^p)(1-s^q)^{-(q+1)/q}\,ds, \\
d_2(p,q):=p^{-1}(p-q)(p-1)(q-1)^{-(p-1)/q}\int_0^1 (1-s^q)^{(p-q-1)/q}\,ds.
\end{gather*}
Moreover, for any compact subset $K$ of $(0,\infty)$,
the convergence of \eqref{eq:5.20} is uniform on $\lambda \in K$.
\end{proposition}

\begin{remark}\label{re:5.4} \rm
We point out the relation between $T_0(\lambda,p,q)$ and
$T_\infty(\lambda,p,q)$ as in Proposition~\ref{pr:5.1}.
First, we note that
\begin{equation}\label{eq:5.21}
c_1(p,q)=-d_1(q,p) \quad \text{and}\quad c_2(p,q)=-d_2(q,p),
\end{equation}
where $c_i$ and $d_i$ ($i=1, 2$)
are constants as in Proposition~\ref{pr:5.1}
or Proposition~\ref{pr:5.3}, respectively.
This leads to
$$
T_0(\lambda,p,q)=-c_2(q,p)\lambda^{(p-q-1)/q} + c_1(q,p)\lambda^{-1/q}
=T_\infty(\lambda,q,p),
$$
whence $T_0$ is obtained by replacing
$p$ and $q$ each other in $T_\infty$
as a matter of form.
\end{remark}

\begin{proof}[Proof of Proposition~\ref{pr:5.3}]
We use the same way as in the proof of Proposition \ref{pr:5.1}.
Replacing $p$ and $q$ each other in the proof of Proposition \ref{pr:5.1},
we obtain
\begin{equation*}
P = \frac{(p-q)(p-1)}{p}\Phi^{-1}(\xi)^p
- \frac{p-q}{p}\lambda(1-s^p)\alpha^p.
\end{equation*}
Denote the first and the second terms by $M$ and $N$, respectively, that is,
$$
M:=\frac{(p-q)(p-1)}{p}\Phi^{-1}(\xi)^p, \quad
N:=\frac{p-q}{p}\lambda(1-s^p)\alpha^p.
$$
Then $P=M-N$.
We compute the limits of $\alpha^{q+1-p}M/Q$ and $\alpha^{q+1-p}N/Q$ as 
$\alpha\to +0$.
Fix $0<s<1$ arbitrarily. Observe that
\begin{gather}\label{eq:5.22}
\Phi^{-1}(\xi)\sim \Big(\frac{q}{q-1}\Big)^{1/q}\xi^{1/q}, \quad \xi\to +0, \\
\label{eq:5.23}
\xi \sim \frac{\lambda}{q}(1-s^q)\alpha^q, \quad \alpha\to +0.
\end{gather}
Therefore we find that as $\alpha\to +0$,
\begin{equation}\label{eq:5.24}
\begin{aligned}
Q
&=  (p-1)\Phi^{-1}(\xi)^{p+1}+ (q-1)\Phi^{-1}(\xi)^{q+1}  \\
& \sim  (q-1)\Phi^{-1}(\xi)^{q+1}  \\
& \sim  (q-1)(q/(q-1))^{(q+1)/q} \xi^{(q+1)/q}  \\
& \sim  (q-1)^{-1/q}\lambda^{(q+1)/q}(1-s^q)^{(q+1)/q}\alpha^{q+1}.
\end{aligned}
\end{equation}
Accordingly,
\begin{equation*}
\lim_{\alpha\to +0} \frac{\alpha^{q+1-p}N}{Q}=
\frac{p-q}{p}(q-1)^{1/q}\lambda^{-1/q}(1-s^p)(1-s^q)^{-(q+1)/q}.
\end{equation*}
By \eqref{eq:5.22} and \eqref{eq:5.23}, the asymptotic formula of $M$ is computed as
\[
M  \sim  p^{-1}(p-q)(p-1)(q-1)^{-p/q}\lambda^{p/q}(1-s^q)^{p/q}\alpha^p.
\]
This expression and \eqref{eq:5.24} imply
\[
\lim_{\alpha\to +0} \frac{\alpha^{q+1-p}M}{Q}
=  p^{-1}(p-q)(p-1)(q-1)^{-(p-1)/q}  \lambda^{(p-q-1)/q}(1-s^q)^{(p-q-1)/q}.
\]

Let $\lambda_0$ and $\lambda_1$ be any numbers satisfying $0<\lambda_0<\lambda_1$.
We choose an $\alpha_0\in(0,1)$ so small that $\xi\leq 1$
for $\alpha\in(0,\alpha_0)$, $\lambda\in[\lambda_0, \lambda_1]$ and $s\in(0,1)$.
According to the next Lemma~\ref{le:5.5},
we can prove the existence of a constant $C>0$ depending only on $\lambda_0$ 
and $\lambda_1$ such that for all $\alpha\in(0,\alpha_0)$,
 $\lambda\in[\lambda_0, \lambda_1]$ and $s\in(0,1)$,
\begin{equation}\label{eq:5.25}
0\leq \alpha^{q+1-p}M/Q \leq C(1-s^q)^{(p-q-1)/q}(1+\alpha_0^{p-q})^{p/q},
\end{equation}
\begin{equation}\label{eq:5.26}
0\leq \alpha^{q+1-p}N/Q \leq C(1-s^q)^{-1/q}.
\end{equation}
Therefore, by the Lebesgue dominated convergence theorem, we obtain the 
assertion of the proposition.
\end{proof}

In Section~\ref{section-6}, we need the next lemma, which ensures \eqref{eq:5.25} 
and \eqref{eq:5.26} also.
Let $M$, $N$ and $Q$ be functions as in Proposition~\ref{pr:5.3}.

\begin{lemma}\label{le:5.5}
There exists a positive constant $C$ such that
\begin{gather}\label{eq:5.27}
\lambda^{1/p}\alpha^{q/p}M/Q \leq C(1-s^q)^{-1/p}(1+\alpha^{p-q})^{p/q}, \\
\label{eq:5.28}
\lambda^{1/p}\alpha^{-p+q(p+1)/p}N/Q \leq C(1-s^q)^{-1/q}(1+(\lambda\alpha^q)^{-(p-q)/pq}),
\end{gather}
for every $\lambda>0$, $\alpha>0$ and $s\in (0,1)$ and moreover,
\begin{gather}\label{eq:5.29}
\lambda^{(q+1-p)/q}\alpha^{q+1-p}M/Q \leq C(1-s^q)^{(p-q-1)/q}(1+\alpha^{p-q})^{p/q},\\
\label{eq:5.30}
\lambda^{1/q}\alpha^{q+1-p}N/Q \leq C(1-s^q)^{-1/q}
\end{gather}
if $\lambda>0$, $\alpha>0$, $s\in (0,1)$ and $\xi\leq 1$, where $\xi$ is
 defined by \eqref{eq:3.11}.
\end{lemma}

\begin{proof}
Observe that
\begin{equation}\label{eq:5.31}
1-s^p<(p/q)(1-s^q) \quad \text{for } 0<s<1.
\end{equation}
By \eqref{eq:5.14} and the inequality above, we have
\begin{gather}\label{eq:5.32}
M \leq C\Phi^{-1}(\xi)^p\leq C\xi^{p/q} \leq C\lambda^{p/q}(1-s^q)^{p/q}\alpha^p 
(1+\alpha^{p-q})^{p/q}, \\
\label{eq:5.33}
M \leq C\Phi^{-1}(\xi)^p\leq C\xi \leq C\lambda (1-s^q)\alpha^q(1+\alpha^{p-q}).
\end{gather}
In the same way as in the proof of Lemma~\ref{le:5.2},
we divide the proof into two cases: $0\leq \xi\leq 1$ and $\xi>1$.
\smallskip

\noindent\textbf{Case 1.} Let $0\leq \xi\leq 1$.
We use the first inequality in \eqref{eq:5.15} to obtain
\begin{equation}\label{eq:5.34}
Q \geq (q-1)\Phi^{-1}(\xi)^{q+1} \geq c\xi^{(q+1)/q} 
\geq c\lambda^{(q+1)/q}(1-s^q)^{(q+1)/q}\alpha^{q+1}.
\end{equation}
This inequality and \eqref{eq:5.32} show that
$$
\lambda^{1/p}\alpha^{q/p}M/Q \leq C(\lambda \alpha^q)^{(p-1)(p-q)/pq}
(1+\alpha^{p-q})^{p/q}(1-s^q)^{(p-q-1)/q}.
$$
Since $(\lambda/q)(1-s^q)\alpha^q \leq \xi\leq 1$, we have 
$\lambda\alpha^q \leq q(1-s^q)^{-1}$.
Substituting this inequality into the right hand side of the inequality above, 
we obtain
$$
\lambda^{1/p}\alpha^{q/p}M/Q \leq C(1-s^q)^{-1/p}(1+\alpha^{p-q})^{p/q}.
$$
Therefore \eqref{eq:5.27} holds.
We use the definition of $N$ and \eqref{eq:5.34} to get
$$
\lambda^{1/p}\alpha^{-p+q(p+1)/p}N/Q 
\leq C(\lambda\alpha^q)^{-(p-q)/pq}(1-s^p)(1-s^q)^{-(q+1)/q},
$$
which with \eqref{eq:5.31} proves \eqref{eq:5.28}.
Using \eqref{eq:5.32}, \eqref{eq:5.34} and the definition of $N$, we have
\begin{gather*}
\lambda^{(q+1-p)/q}\alpha^{q+1-p}M/Q \leq C(1-s^q)^{(p-q-1)/q}(1+\alpha^{p-q})^{p/q},\\
\lambda^{1/q}\alpha^{q+1-p}N/Q \leq C(1-s^p)(1-s^q)^{-(q+1)/q}.
\end{gather*}
These two inequalities with \eqref{eq:5.31} prove \eqref{eq:5.29} and
 \eqref{eq:5.30}.
\smallskip

\noindent\textbf{Case 2.} Let $\xi>1$.
We have only to prove \eqref{eq:5.27} and \eqref{eq:5.28}.
We use the second inequality in \eqref{eq:5.15} to get
\begin{equation*}
Q \geq (p-1)\Phi^{-1}(\xi)^{p+1} \geq c\xi^{(p+1)/p} 
\geq c\lambda^{(p+1)/p}(1-s^q)^{(p+1)/p}\alpha^{q(p+1)/p}.
\end{equation*}
This inequality, \eqref{eq:5.33} and \eqref{eq:5.31} show that
$$
\lambda^{1/p}\alpha^{q/p}M/Q \leq C(1-s^q)^{-1/p}(1+\alpha^{p-q})
\leq C(1-s^q)^{-1/p}(1+\alpha^{p-q})^{p/q},
$$
and
\begin{align*}
\lambda^{1/p}\alpha^{-p+q(p+1)/p}N/Q
&\leq  C(1-s^p)(1-s^q)^{-(p+1)/p} \\
&\leq  C(1-s^q)^{-1/p} \leq C(1-s^q)^{-1/q}.
\end{align*}
Therefore we have \eqref{eq:5.27} and \eqref{eq:5.28}. The proof is complete.
\end{proof}

\section{Exact shape of the bifurcation curve}\label{section-6}

In this section, we investigate the shape of the bifurcation curve 
$(\lambda(\alpha),\alpha)$. The next theorem draws a whole bifurcation diagram 
$(\lambda(\alpha),\alpha)$ under the assumption,
\begin{equation}\label{eq:6.1}
\mu(q,L) \leq (p-1)^{-q/(p-q)}(q-1)^{p/(p-q)}.
\end{equation}
If $p$ and $q$ are fixed and $L$ is large enough, the inequality above is fulfilled.

\begin{theorem}\label{th:6.1}
Assume \eqref{eq:6.1}. Then it holds that $\mu(p,L)<\lambda(\alpha)<\mu(q,L)$ 
and $\lambda'(\alpha)<0$ for all $\alpha>0$.
Therefore \eqref{Plambda} possesses a unique positive solution if and only 
if $\mu(p,L)<\lambda<\mu(q,L)$.
\end{theorem}

The theorem above says that the bifurcation curve starts from the initial point 
$(\mu(q), 0)$ and goes monotonically to the left and reaches the final point 
$(\mu(p), \infty)$. This is of type (C) stated in Section~\ref{section-1}.
In the next theorem, we consider the opposite case where $L$ is small enough and
$\mu(q,L)<\mu(p,L)$.

\begin{theorem}\label{th:6.2}
Fix $1<q<p$. If $L>0$ is small enough, then $\mu(q,L)<\lambda(\alpha)<\mu(p,L)$ 
and $\lambda'(\alpha)>0$ for all $\alpha>0$.
Therefore \eqref{Plambda} possesses a unique positive solution if and only 
if $\mu(q,L)<\lambda<\mu(p,L)$.
\end{theorem}

The theorem above shows that the bifurcation curve starts from $(\mu(q), 0)$,
goes monotonically to the right and reaches the final point $(\mu(p), \infty)$.
This behavior is of type (A).

\begin{proof}[Proof of Theorem \ref{th:6.1}]
Assumption \eqref{eq:6.1} is equivalent to the inequality
\begin{equation}\label{eq:6.2}
(p-1)(\pi_q/(2L))^p \leq (q-1)(\pi_q/(2L))^q=\mu(q).
\end{equation}
This inequality implies that the right hand side of \eqref{eq:4.1} 
is equal to $\mu(q)$.

On the other hand, observing \eqref{eq:1.3}, we see that $\pi_p$ is 
decreasing on $p$. Then we use \eqref{eq:6.2} to obtain
$$
\mu(p)=(p-1)(\pi_p/(2L))^p <  (p-1)(\pi_q/(2L))^p \leq \mu(q).
$$
Hence $\mu(p)<\mu(q)$. By Theorem \ref{th:4.1} (v),
$\mu(p) < \lambda(\alpha) < \mu(q)$ for all $\alpha>0$.

We shall show that $\lambda'(\alpha)<0$.
Let $I$ and $J$ be as in \eqref{eq:5.4}.
We shall prove that $I<J$, i.e.,
$$
\frac{p-q}{q}\lambda(1-s^q)\alpha^q < \frac{(p-q)(q-1)}{q}\Phi^{-1}(\xi)^q
$$
for all $\alpha>0$ and $s\in (0,1)$, where
$\Phi$ and $\xi$ are the functions defined by
\eqref{eq:3.4} and \eqref{eq:3.11}, respectively.
Since $\Phi$ is increasing, this inequality is equivalent to
$$
\Phi\left(\left[\lambda(q-1)^{-1}(1-s^q)\right]^{1/q}\alpha \right) < \xi.
$$
By the definitions of $\Phi$ and $\xi$, this is rewritten as
$$
\frac{p-1}{p}\left(\lambda(q-1)^{-1}(1-s^q)\right)^{p/q}\alpha^p 
< \frac{\lambda}{p}(1-s^p)\alpha^p,
$$
or equivalently,
\begin{equation}\label{eq:6.3}
\lambda^{(p-q)/q}(1-s^q)^{p/q} < (p-1)^{-1}(q-1)^{p/q}(1-s^p).
\end{equation}
Hence we have only to show the inequality above with $\lambda=\lambda(\alpha)$.
Since $\lambda(\alpha)<\mu(q)$, \eqref{eq:6.1} ensures that
$$
\lambda(\alpha)^{(p-q)/q} < \mu(q)^{(p-q)/q} \leq (p-1)^{-1}(q-1)^{p/q}.
$$
Observe that $(1-s^q)^{p/q} < 1-s^q< 1-s^p$.
Multiplying these inequalities, we obtain \eqref{eq:6.3}.
Therefore $I<J$, and so
$$
T_\alpha(\lambda(\alpha), \alpha) = \int_0^1 \frac{I-J}{Q}\,ds <0.
$$
Since $T_\lambda<0$, we conclude from \eqref{eq:4.13} that $\lambda'(\alpha)<0$ 
for all $\alpha$.
The proof is complete.
\end{proof}

Since $T(\lambda, \alpha)$ depends on $p$ and $q$ also, we denote it by
$T(\lambda, \alpha, p, q)$.
To prove Theorem \ref{th:6.2}, we need the next lemma.

\begin{lemma}\label{le:6.3}
For any $1<q<p$, there exists a $\Lambda(p,q)>0$ such that
$$
T_\alpha(\lambda, \alpha, p,q)>0 \quad \text{for $\alpha>0$ and 
$\lambda>\Lambda(p,q)$.}
$$
\end{lemma}

\begin{proof}
Fix $1<q<p$ and denote $T_\alpha(\lambda,\alpha,p,q)$ by $T_\alpha(\lambda,\alpha)$ 
for simplicity.
Suppose on the contrary that $T_\alpha(\lambda_n,\alpha_n)\leq 0$ along some 
sequences $\alpha_n$ and $\lambda_n$,
where $\lambda_n\to \infty$ as $n\to \infty$.
After choosing a subsequence of $\alpha_n$, we divide the proof into five cases below:
(i) $\alpha_n\to \infty$, (ii) $\alpha_n\to \alpha_0>0$,
(iii) $\alpha_n\to 0$ and $\lambda_n\alpha_n^q \to\infty$,
(iv) $\alpha_n\to 0$ and $\lambda_n\alpha_n^q \to c_0>0$,
(v) $\alpha_n\to 0$ and $\lambda_n\alpha_n^q \to 0$.
We shall prove that all the cases lead to a contradiction.
Let $\xi_n$, $P_n$, $Q_n$, $I_n$ and $J_n$ be defined by 
\eqref{eq:3.11}, \eqref{eq:5.2}, \eqref{eq:5.3}
and \eqref{eq:5.4} with $\lambda$ and $\alpha$ replaced by $\lambda_n$ and 
$\alpha_n$, respectively.
Since $T_\alpha(\lambda_n,\alpha_n)\leq 0$, we have
\begin{equation}\label{eq:6.4}
\int_0^1 P_n/Q_n\,ds \leq 0 \quad\text{for all } n\in \mathbb{N}.
\end{equation}

(i) Assume that $\alpha_n\to \infty$.
Fix $0<s<1$ arbitrarily.
Since $\xi_n\to \infty$, the same computation as in \eqref{eq:5.7} shows that
$$
Q_n \sim (p-1)^{-1/p}\lambda_n^{(p+1)/p}\alpha_n^{p+1}(1-s^p)^{(p+1)/p}
\quad \text{as } n\to \infty.
$$
Therefore 
$$
\lim_{n\to \infty}\lambda_n^{1/p} \alpha_n^{p+1-q}I_n/Q_n 
= q^{-1}(p-q)(p-1)^{1/p}(1-s^q)(1-s^p)^{-(p+1)/p}.
$$
By \eqref{eq:5.5} and \eqref{eq:5.6}, we see that
$$
J_n \sim q^{-1}(p-q)(q-1)(p-1)^{-q/p}\lambda_n^{q/p}\alpha_n^q(1-s^p)^{q/p}.
$$
Hence
\begin{align*}
&\lim_{n\to\infty}\lambda_n^{(p+1-q)/p}\alpha_n^{p+1-q}J_n/Q_n\\
&=  q^{-1}(p-q)(q-1)(p-1)^{-(q-1)/p} 
 (1-s^p)^{(q-p-1)/p}.
\end{align*}
According to Lemma~\ref{le:5.2} and noting $\lambda_n, \alpha_n\to \infty$,
we obtain
\begin{gather*}
0 \leq \lambda_n^{1/p}\alpha_n^{p+1-q}I_n/Q_n \leq C(1-s^p)^{-1/q}, \\
0 \leq \lambda_n^{(p+1-q)/p}\alpha_n^{p+1-q}J_n/Q_n \leq C(1-s^p)^{(q-p-1)/p},
\end{gather*}
for $0<s<1$, where $C>0$ is independent of $s$ and $n$.
From the Lebesgue dominated convergence theorem, it follows that
$$
\lim_{n\to \infty}\lambda_n^{1/p} \alpha_n^{p+1-q} \int_0^1 I_n/Q_n\,ds = c, \quad
\lim_{n\to \infty}\lambda_n^{(p+1-q)/p}\alpha_n^{p+1-q} \int_0^1 J_n/Q_n\,ds = d,
$$
where
\begin{gather*}
c:= q^{-1}(p-q)(p-1)^{1/p} \int_0^1 (1-s^q)(1-s^p)^{-(p+1)/p}ds, \\
d:= q^{-1}(p-q)(q-1)(p-1)^{-(q-1)/p} \int_0^1 (1-s^p)^{(q-p-1)/p}ds.
\end{gather*}
Using $P_n=I_n-J_n$ and combining the relations above, we obtain
$$
\lim_{n\to\infty}\lambda_n^{1/p} \alpha_n^{p+1-q} \int_0^1 P_n/Q_n\,ds =c>0.
$$
This contradicts \eqref{eq:6.4}.

(ii) Assume that $\alpha_n$ converges to a positive limit $\alpha_0$.
Fix $0<s<1$ arbitrarily.
Then $\xi_n \sim \lambda_n\xi_0(s)$ as $n\to\infty$, where $\xi_0(s)$ is defined by
$$
\xi_0(s):=\frac{1}{p}(1-s^p)\alpha_0^p + \frac{1}{q}(1-s^q)\alpha_0^q.
$$
Since $\xi_n\to \infty$, we have
\begin{gather*}
\Phi^{-1}(\xi_n) \sim (p\xi_n/(p-1))^{1/p} 
\sim (p/(p-1))^{1/p}\lambda_n^{1/p}\xi_0(s)^{1/p}, \\
J_n \sim q^{-1}(p-q)(q-1)(p/(p-1))^{q/p}\lambda_n^{q/p}\xi_0(s)^{q/p}, \\
Q_n \sim (p-1)\Phi^{-1}(\xi_n)^{p+1} 
 \sim p^{(p+1)/p}(p-1)^{-1/p}\lambda_n^{(p+1)/p}\xi_0(s)^{(p+1)/p}.
\end{gather*}
Moreover, due to Lemma~\ref{le:5.2} and by noting $\lambda_n\to\infty$ and 
$\alpha_n\to \alpha_0>0$,
there is a constant $C>0$ independent of $n$ and $s$ such that
$$
\lambda_n^{1/p}I_n/Q_n \leq C(1-s^p)^{-1/q}, \quad
\lambda_n^{(p+1-q)/p}J_n/Q_n \leq C(1-s^p)^{(q-p-1)/p},
$$
for $0<s<1$.
The Lebesgue dominated convergence theorem shows that
$$
\lim_{n\to\infty} \lambda_n^{1/p}\int_0^1 I_n/Q_n\,ds = c, \quad
\lim_{n\to\infty} \lambda_n^{(p+1-q)/p}\int_0^1 J_n/Q_n\,ds = d,
$$
where
\begin{gather*}
c:=q^{-1}(p-q)p^{-(p+1)/p}(p-1)^{1/p}\alpha_0^q \int_0^1 (1-s^q)\xi_0(s)^{-(p+1)/p}ds,\\
d:=q^{-1}(p-q)(q-1)p^{(q-p-1)/p}(p-1)^{-(q-1)/p} \int_0^1 \xi_0(s)^{(q-p-1)/p}ds.
\end{gather*}
By the relations above, we have
$$
\lim_{n\to\infty} \lambda_n^{1/p}\int_0^1 P_n/Q_n\,ds = c>0,
$$
which contradicts \eqref{eq:6.4}.

(iii) Assume that $\alpha_n\to 0$ and $\lambda_n\alpha_n^q \to\infty$.
We rewrite $M$ and $N$ given in the proof of Proposition \ref{pr:5.3} as $M_n$ 
and $N_n$, after replacing
$\lambda$ and $\alpha$ by $\lambda_n$ and $\alpha_n$, respectively.
Since $\alpha_n\to 0$ and $\lambda_n\alpha_n^q \to\infty$, we have, as $n\to\infty$
\begin{gather*}
\xi_n \sim \frac{\lambda_n}{q}(1-s^q)\alpha_n^q, \\
\Phi^{-1}(\xi_n)^p \sim p\xi_n/(p-1) 
 \sim \frac{p}{(p-1)q}\lambda_n(1-s^q)\alpha_n^q, \\
\begin{aligned}
Q_n
& \sim  (p-1)\Phi^{-1}(\xi_n)^{p+1} \\
& \sim  (p/q)^{(p+1)/p}(p-1)^{-1/p}\lambda_n^{(p+1)/p}\alpha_n^{q(p+1)/p} 
(1-s^q)^{(p+1)/p},
\end{aligned}
\end{gather*}
which shows that
$$
M_n \sim q^{-1}(p-q)\lambda_n\alpha_n^q (1-s^q).
$$
By Lemma~\ref{le:5.5} with the facts that $p>q$, $\alpha_n\to 0$ and 
$\lambda_n\alpha_n^q\to \infty$,
there exists a constant $C>0$ independent of $n$ such that
\begin{gather*}
\lambda_n^{1/p}\alpha_n^{q/p}M_n/Q_n \leq C(1-s^q)^{-1/p}, \\
\lambda_n^{1/p}\alpha_n^{-p+q(p+1)/p}N_n/Q_n \leq C(1-s^q)^{-1/q}.
\end{gather*}
Therefore we obtain constants $c,d>0$ such that
\begin{gather*}
\lim_{n\to\infty} \lambda_n^{1/p}\alpha_n^{q/p}\int_0^1M_n/Q_n\,ds =c>0,\\
\lim_{n\to\infty} \lambda_n^{1/p}\alpha_n^{-p+q(p+1)/p}\int_0^1 N_n/Q_n\,ds =d>0.
\end{gather*}
Combining these identities, we obtain
$$
\lim_{n\to\infty} \lambda_n^{1/p}\alpha_n^{q/p}\int_0^1 P_n/Q_n\,ds =c>0.
$$
This contradicts \eqref{eq:6.4}.

(iv) Assume that $\alpha_n\to 0$ and $\lambda_n\alpha_n^q$ converges to a positive 
limit $c_0$. From this assumption, it follows that $\lambda_n\alpha_n^p \to 0$. 
Therefore, as $n\to\infty$,
\begin{gather*}
\xi_n \to \xi_0(s):=\frac{c_0}{q}(1-s^q), \\
Q_n \to (p-1)\Phi^{-1}(\xi_0(s))^{p+1} + (q-1)\Phi^{-1}(\xi_0(s))^{q+1}, \\
M_n \to p^{-1}(p-q)(p-1)\Phi^{-1}(\xi_0(s))^p, \\
N_n=\frac{p-q}{p}\lambda_n(1-s^p)\alpha_n^p \to 0.
\end{gather*}
All the convergences above are uniform on $s$.
Hence we see that
$$
\lim_{n\to\infty} \int_0^1 P_n/Q_n\,ds =c>0,
$$
with some $c>0$. This contradicts \eqref{eq:6.4}.

(v) Assume that $\alpha_n\to 0$ and $\lambda_n\alpha_n^q \to 0$.
Then $\xi_n \sim q^{-1}\lambda_n(1-s^q)\alpha_n^q$.
Since $\xi_n$ converges to $0$ uniformly on $s$, we use \eqref{eq:5.22} to obtain
$$
\Phi^{-1}(\xi_n) \sim (q\xi_n/(q-1))^{1/q}
 \sim (q-1)^{-1/q}\lambda_n^{1/q}\alpha_n(1-s^q)^{1/q}.
$$
Moreover,
\begin{gather*}
M_n \sim p^{-1}(p-q)(p-1)(q-1)^{-p/q}\lambda_n^{p/q}\alpha_n^p(1-s^q)^{p/q}, \\
Q_n \sim (q-1)\Phi^{-1}(\xi_n)^{q+1} 
\sim (q-1)^{-1/q}\lambda_n^{(q+1)/q}\alpha_n^{q+1}(1-s^q)^{(q+1)/q}.
\end{gather*}
Note that $p>q$, $\alpha_n\to 0$ and $\xi_n\to 0$ uniformly on $s$.
Then by Lemma~\ref{le:5.5}, there exists a constant $C>0$ independent of $n$ 
such that
\begin{gather*}
\lambda_n^{(q+1-p)/q}\alpha_n^{q+1-p}M_n/Q_n \leq C(1-s^q)^{(p-q-1)/q}, \\
\lambda_n^{1/q}\alpha_n^{q+1-p}N_n/Q_n \leq C(1-s^q)^{-1/q},
\end{gather*}
for sufficiently large $n$.
By the Lebesgue dominated convergence theorem, we have constants $c ,d>0$ such that
\begin{gather*}
\lim_{n\to\infty} \lambda_n^{(q+1-p)/q}\alpha_n^{q+1-p}\int_0^1 M_n/Q_n\,ds =c,\\
\lim_{n\to\infty} \lambda_n^{1/q}\alpha_n^{q+1-p}\int_0^1 N_n/Q_n\,ds =d.
\end{gather*}
From these identities, we obtain
$$
\lim_{n\to\infty} \lambda_n^{(q+1-p)/q}\alpha_n^{q+1-p}\int_0^1 P_n/Q_n\,ds =c>0,
$$
which contradicts \eqref{eq:6.4}. The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th:6.2}]
Fix $1<q<p$.
If $L\leq 1$, then $\mu(q,L)<\mu(p,L)$ by Theorem \ref{th:1.1}.
Recall that $\lambda(\alpha)>\min\{\mu(p,L),\mu(q,L)\}$ for all $\alpha>0$.
Therefore $\mu(q,L)<\lambda(\alpha)$ for $\alpha>0$.
Let $\Lambda(p,q)$ be as in Lemma \ref{le:6.3}.
Observe that $\mu(q,L)\to \infty$ as $L\to +0$ by the definition of $\mu(q,L)$.
Thus, if $L$ is small enough, then $\mu(q,L)>\Lambda(p,q)$ and hence 
$\lambda(\alpha)>\Lambda(p,q)$.
Lemma \ref{le:6.3} with \eqref{eq:4.13} ensures that
$$
\lambda'(\alpha)=-\frac{T_\alpha(\lambda(\alpha), \alpha)}{T_\lambda(\lambda(\alpha), \alpha)}>0
\quad \text{for all } \alpha>0.
$$
Since $\lambda(\alpha)$ converges to $\mu(p,L)$ as $\alpha\to\infty$,
it holds that $\mu(q,L)<\lambda(\alpha)<\mu(p,L)$ for $\alpha>0$.
The proof is complete.
\end{proof}


\section{Bifurcation curve near the initial and final points}\label{section-7}

In this section, we investigate the direction in which the bifurcation curve 
moves near the initial point and near
the final point. Using these results, we shall construct
the bifurcation diagrams (B) and (D) stated in Section~\ref{section-1} .

Let $c_i(p,q)$ and $d_i(p,q)$ with $i=1,2$ be the constants given in
 Propositions \ref{pr:5.1} and \ref{pr:5.3}.
We define
\begin{gather}\label{eq:7.1}
z_\infty(p,q):=(c_2(p,q)/c_1(p,q))^{p/(p-q)}, \\
\label{eq:7.2}
z_0(p,q):=(d_1(p,q)/d_2(p,q))^{q/(p-q)}.
\end{gather}
Then $z_\infty$ and $z_0$ are unique zeros of $T_\infty(\lambda)$ and 
$T_0(\lambda)$ in $(0,\infty)$, respectively.
Moreover, we easily see that $z_\infty(p,q)=z_0(q,p)$ as a matter of form.
Indeed, it follows from \eqref{eq:5.21} in Remark~\ref{re:5.4} that
\begin{align*}
z_\infty(p,q)
&=  \Big(\frac{c_2(p,q)}{c_1(p,q)}\Big)^{p/(p-q)}
       =\Big(\frac{d_2(q,p)}{d_1(q,p)}\Big)^{p/(p-q)} \\
&=  \Big(\frac{d_1(q,p)}{d_2(q,p)}\Big)^{p/(q-p)} =z_0(q,p).
\end{align*}
By the definitions of $T_\infty$ and $T_0$, we have

\begin{lemma}\label{le:7.1}
\begin{itemize}
\item[(i)] $T_\infty(\lambda)>0$ when $\lambda>z_\infty$ and $T_\infty(\lambda)<0$ 
when $\lambda<z_\infty$.
\item[(ii)] $T_0(\lambda)>0$ when $\lambda>z_0$ and $T_0(\lambda)<0$ when $\lambda<z_0$.
\end{itemize}
\end{lemma}

In the next proposition, we shall prove that
the sign of $\lambda'(\alpha)$ for $\alpha$ large enough (or small enough) 
is determined by the order relation between the first eigenvalue $\mu(p,L)$ 
(or $\mu(q,L)$) and the zero $z_\infty$ (or $z_0$, respectively).

\begin{proposition}\label{pr:7.2}
\begin{itemize}
\item[(i)] If $\mu(p,L)>z_\infty(p,q)$, then $\lambda'(\alpha)>0$ for 
 $\alpha>0$ large enough.
\item[(ii)] If $\mu(p,L)<z_\infty(p,q)$, then $\lambda'(\alpha)<0$ for $\alpha>0$ 
 large enough.
\item[(iii)] If $\mu(q,L)>z_0(p,q)$, then $\lambda'(\alpha)>0$ for $\alpha>0$ 
 small enough.
\item[(iv)] If $\mu(q,L)<z_0(p,q)$, then $\lambda'(\alpha)<0$ for $\alpha>0$ 
 small enough.
\end{itemize}
\end{proposition}

\begin{proof}
Since $\lim_{\alpha\to \infty}\lambda(\alpha)=\mu(p,L)$, we use Proposition 
\ref{pr:5.1} to obtain
$$
\lim_{\alpha\to \infty}\alpha^{p+1-q}T_\alpha(\lambda(\alpha), \alpha)
=T_\infty(\mu(p,L)).
$$
Assume that $\mu(p,L)>z_\infty(p,q)$. Then $T_\infty(\mu(p,L))>0$ by Lemma 
\ref{le:7.1} (i).
For $\alpha>0$ large enough, we use \eqref{eq:4.13} to obtain
$$
\alpha^{p+1-q}\lambda'(\alpha)
= -\frac{\alpha^{p+1-q}T_\alpha(\lambda(\alpha), \alpha)}{T_\lambda(\lambda(\alpha),
 \alpha)}>0,
$$
because $T_\lambda<0$.
Therefore the assertion (i) holds. The other assertions can be proved in the 
same method.
\end{proof}


The proposition above indicates which direction the bifurcation curve goes
 to near the initial point $(\mu(q), 0)$
and near the final point $(\mu(p), \infty)$.
If $\lambda'(\alpha)>0$, then the curve
$(\lambda(\alpha),\alpha)$ goes to the right in the $(\lambda, \alpha)$ plain.
If $\lambda'(\alpha)<0$, then it moves to the left.

\begin{theorem}\label{th:7.3}
Let $q>1$ and $L>0$.
\begin{itemize}
\item[(i)]
  If $\pi_q>2L$, then for $p$ large enough, $\lambda'(\alpha)$ is positive for $\alpha>0$ small enough.
\item[(ii)]
  If $\pi_q < 2L$, then for $p$ large enough, it holds that $\mu(p,L)<\mu(q,L)$ and
  $\lambda'(\alpha)$ is negative for $\alpha>0$ small enough.
\end{itemize}
\end{theorem}

The value $2L$ is the length of the interval $[-L,L]$.
The theorem above means that the relation between the length of the interval and 
$\pi_q$ determines the direction in which the bifurcation curve starts up.
If the length is less (or greater) than $\pi_q$, the curve grows to the right 
(or left, respectively).

Using Theorem~\ref{th:7.3}, we show the uniqueness of positive solutions for 
$\lambda$ slightly greater than $\mu(q,L)$ in the next corollary.

\begin{corollary}\label{co:7.4}
Let $1<q$ and $0<L\leq 1$. If $p$ is large enough, there exists an $\varepsilon>0$ 
such that a positive solution of \eqref{Plambda} is unique when 
$\mu(q,L)<\lambda < \mu(q,L)+\varepsilon$.
\end{corollary}

\begin{proof}
Let $1<q<p$ and $0<L \leq 1$.
Then Theorem~\ref{th:1.1} ensures that $\mu(q,L)<\mu(p,L)$.
It follows easily from \eqref{eq:1.3} that $\pi_q$ is strictly decreasing in $q$,
$\lim_{q\to 1+0}\pi_q=\infty$ and $\lim_{q\to \infty}\pi_q=2$.
Therefore $\pi_q>2 \geq 2L$.
If $p$ is large enough, then Theorem~\ref{th:7.3} (i) shows that 
$\lambda'(\alpha)>0$ for $\alpha\in (0,a)$
with an $a>0$ small enough.
We define $\lambda_0:=\inf_{a<\alpha<\infty}\lambda(\alpha)$.
Then $\lambda_0>\mu(q,L)$
because $\lambda(\alpha)>\mu(q,L)$ for any $\alpha>0$
and $\lambda(\alpha) \to \mu(p,L)$ as $\alpha\to \infty$
by Theorem~\ref{th:4.1} (iii) and (v) (note $\mu(q,L)<\mu(p,L)$ also).
For each $\mu \in (\mu(q,L), \lambda_0)$,
there exists a unique $\alpha$ which satisfies $\mu=\lambda(\alpha)$,
that is, a positive solution is unique.
The proof is complete.
\end{proof}

Using Theorem~\ref{th:7.3}, we have another corollary.

\begin{corollary}\label{co:7.5}
Let $q$ and $L$ satisfy that $\pi_q>2L>2$.
If $p$ is large enough, then $\mu(p,L)<\mu(q,L)<\lambda^*$, where 
$\lambda^*$ is given by \eqref{eq:4.14}.
Moreover, the following assertions hold.
\begin{itemize}
\item[(i)] If $\lambda \leq \mu(p,L)$ or $\lambda > \lambda^*$, \eqref{Plambda} 
 has no positive solutions.
\item[(ii)] If $\mu(p,L)<\lambda \leq \mu(q,L)$ or 
 $\lambda = \lambda^*$, \eqref{Plambda} has at least one positive solution.
\item[(iii)] If $\mu(q,L)<\lambda < \lambda^*$, \eqref{Plambda} has at least 
two positive solutions.
\end{itemize}
\end{corollary}

\begin{proof}
Since $L>1$, $\mu(p,L)\to 0$ as $p\to\infty$ by \eqref{eq:2.8}.
Therefore $\mu(p,L)<\mu(q,L)$ for $p$ large enough.
Since $\pi_q>2L$, the bifurcation curve $(\lambda(\alpha),\alpha)$
grows to the right from the initial point $(\mu(q,L),0)$ by Theorem~\ref{th:7.3} (i).
Therefore $\mu(p,L)<\mu(q,L)<\lambda^*$ and the assertions (i)--(iii) follow.
\end{proof}

Corollary~\ref{co:7.5} gives an example of type (D).
In theorem \ref{th:7.3}, we fixed $q$ and then took $p$ large enough.
In the next theorem, we consider the opposite case where $p$ is fixed and then $q$ is sufficiently close to $1$.

\begin{theorem}\label{th:7.6}
Let $p>1$ and $L>0$.
If $q\in(1,\infty)$ is sufficiently close to $1$,
then $\lambda'(\alpha)>0$ for $\alpha>0$ small enough.
Therefore the bifurcation curve $(\lambda(\alpha),\alpha)$
grows to the right from the initial point $(\mu(q),0)$.
\end{theorem}

Since $\pi_q\to\infty$ as $q\to1+0$, it holds that $\pi_q>2L$ for $q$ sufficiently 
close to $1$.
Hence Theorem \ref{th:7.6} perhaps follows from Theorem \ref{th:7.3} (i).
However this is not true.
Indeed, in Theorem \ref{th:7.3} (i) we need to choose $p$ large enough,
but in Theorem \ref{th:7.6} we can take $p$ as any number greater than $1$
and then choose $q$ sufficiently close to $1$.
Therefore non of these theorems follows from another.

Consider the case where $L>1$ and $\mu(p,L)<1/L$.
For example, fix $L>1$ and then choose $p$ large enough. Then this inequality holds.
Recall \eqref{eq:2.6}, i.e., $\mu(q,L) \to 1/L$ as $q\to 1+0$.
Therefore if $q$ is sufficiently close to $1$, then $\mu(p,L)<\mu(q,L)$.
By Theorem \ref{th:7.6}, the bifurcation curve $(\lambda(\alpha),\alpha)$ grows 
to the right from $(\mu(q,L),0)$, and hence $\lambda^*>\mu(q,L)>\mu(p,L)$.
This case gives an example of type (D). Furthermore, we have the next corollary.

\begin{corollary}\label{co:7.7}
Let $L$ and $p$ satisfy that $L>1$ and $\mu(p,L)<1/L$.
If $q$ is sufficiently close to $1$, then $\mu(p,L)<\mu(q,L)<\lambda^*$ and
the assertions of Corollary \ref{co:7.5} are valid.
\end{corollary}

To prove Theorems \ref{th:7.3} and \ref{th:7.6}, we need the two lemmas below.

\begin{lemma}\label{le:7.8}
Define
$$
K(p,q):=\int_0^1 (1-s^q)^p\,ds.
$$
Then for each $q>1$,
$$
\lim_{p\to\infty} p^{1/q}q K(p,q)=\int_0^\infty t^{-(q-1)/q}e^{-t} dt.
$$
\end{lemma}

\begin{proof}
We use the change of variables $s^q=t/p$. Since
$$
s=(t/p)^{1/q}, \quad\,ds = p^{-1/q}q^{-1}t^{-(q-1)/q}dt,
$$
$K$ is rewritten as
$$
K=p^{-1/q}q^{-1}\int_0^p (1-t/p)^pt^{-(q-1)/q}dt,
$$
or equivalently,
$$
p^{1/q}qK=\int_0^p (1-t/p)^pt^{-(q-1)/q}dt.
$$
We denote the integrand by $g(t,p)$.
Then
$$
\lim_{p\to\infty}g(t,p)=t^{-(q-1)/q}e^{-t},
$$
$$
0\leq g(t,p)\leq t^{-(q-1)/q}e^{-t} \quad \text{for $p>1$ and $t\in(0,p)$}.
$$
By the Lebesgue dominated convergence theorem, as $p\to \infty$,
$$
\int_0^p g(t,p)dt \to \int_0^\infty t^{-(q-1)/q}e^{-t}dt.
$$
This completes the proof.
\end{proof}

\begin{lemma}\label{le:7.9}
Let $z_0(p,q)$ be the constant as in \eqref{eq:7.2}. Then
\begin{gather}\label{eq:7.3}
\lim_{p\to \infty}z_0(p,q)=q-1, \\
\label{eq:7.4}
\lim_{q\to 1+0}z_0(p,q)=0.
\end{gather}
\end{lemma}

\begin{proof}
By the definitions of $d_1$ and $d_2$ in Proposition \ref{pr:5.3}, we have
\begin{equation}\label{eq:7.5}
\frac{d_1(p,q)}{d_2(p,q)} =(p-1)^{-1}(q-1)^{p/q}(V/W),
\end{equation}
where
\begin{gather*}
V=V(p,q) :=\int_0^1(1-s^p)(1-s^q)^{-(q+1)/q}ds, \\
W=W(p,q) :=\int_0^1 (1-s^q)^{(p-q-1)/q}ds.
\end{gather*}
Using $K(p,q)$ in Lemma \ref{le:7.8}, we write
$W(p,q)=K((p-q-1)/q,q)$ and obtain
\begin{align*}
\lim_{p\to\infty}p^{1/q}W(p,q)
&=  \lim_{p\to\infty}q^{1/q}((p-q-1)/q)^{1/q}K((p-q-1)/q,q) \\
&=  q^{-(q-1)/q}\int_0^\infty t^{-(q-1)/q}e^{-t} dt.
\end{align*}
We denote the right hand side by $W_q$.
For $p$ large enough, it holds that
\begin{equation}\label{eq:7.6}
(1/2)p^{-1/q}W_q \leq W(p,q) \leq 2p^{-1/q}W_q.
\end{equation}
Since $1-s<1-s^p<p(1-s)$ for $s\in(0,1)$, it follows that
$$
\int_0^1(1-s)(1-s^q)^{-(q+1)/q}ds \leq V \leq p\int_0^1(1-s)(1-s^q)^{-(q+1)/q}ds.
$$
Writing the left hand side as $V_q$,
we have
\begin{equation}\label{eq:7.7}
V_q\leq V(p,q) \leq pV_q.
\end{equation}
Note that the constants $V_q$ and $W_q$ depend only on $q$.
By \eqref{eq:7.5}--\eqref{eq:7.7}, we obtain
\begin{align*}
& (V_q W_q^{-1}/2)^{q/(p-q)}
(p-1)^{-q/(p-q)}p^{1/(p-q)}(q-1)^{p/(p-q)} \\
& \leq (d_1/d_2)^{q/(p-q)} \\
&\leq (2 V_q W_q^{-1})^{q/(p-q)}
(p-1)^{-q/(p-q)}p^{(q+1)/(p-q)}(q-1)^{p/(p-q)}.
\end{align*}
Recall that $z_0(p,q)=(d_1/d_2)^{q/(p-q)}$. Letting $p\to \infty$, 
we obtain \eqref{eq:7.3}.

We shall show \eqref{eq:7.4}.
Since $1-s^p<p(1-s)$ and $1-s^q>1-s$ for $s\in (0,1)$, we have
$$
V(p,q) \leq p \int_0^1 (1-s)^{-1/q}ds = pq/(q-1).
$$
Hence
\begin{equation}\label{eq:7.8}
\limsup_{q\to 1+0}(q-1)V(p,q)\leq p.
\end{equation}
By the Lebesgue dominated convergence theorem, we
see easily that
\begin{equation}\label{eq:7.9}
\lim_{q\to 1+0}W(p,q)
= \int_0^1 (1-s)^{p-2}ds =\frac{1}{p-1}.
\end{equation}
From \eqref{eq:7.5}, \eqref{eq:7.8} and \eqref{eq:7.9}, it follows that
$$
\limsup_{q\to 1+0}(d_1/d_2)=\limsup_{q\to1+0} (p-1)^{-1}(q-1)^{p/q}(V/W)=0.
$$
Since $z_0=(d_1/d_2)^{q/(p-q)}$, \eqref{eq:7.4} holds.
\end{proof}

Using the two lemmas above, we shall prove Theorems \ref{th:7.3} and \ref{th:7.6}.

\begin{proof}[Proof of Theorem \ref{th:7.3}]
Assume that $\pi_q>2L$.
We use Lemma \ref{le:7.9} to get
$$
\mu(q,L)=(q-1)(\pi_q/(2L))^q>q-1=\lim_{p\to\infty}z_0(p,q).
$$
Therefore $\mu(q,L)>z_0(p,q)$ for $p$ large enough.
By Proposition \ref{pr:7.2} (iii),  $\lambda'(\alpha)>0$ for $\alpha>0$ small enough.
Thus the assertion (i) holds.

Let us show (ii).
As stated in the proof of Corollary~\ref{co:7.4}, $\pi_q$ is greater than $2$.
Therefore if $\pi_q \leq 2L$, then $L>1$.
Hence
\eqref{eq:2.8} implies that
$\mu(p,L)< \mu(q,L)$ for $p$ large enough.
The negativity of $\lambda'(\alpha)$ can be proved in the same way as in (i).
\end{proof}

\begin{proof}[Proof of Theorem \ref{th:7.6}]
By \eqref{eq:2.6} and \eqref{eq:7.4}, $\mu(q,L)>z_0(p,q)$ for $q$ sufficiently 
close to $1$. From Proposition \ref{pr:7.2} (iii), the conclusion follows.
\end{proof}

In Theorems \ref{th:7.3} and \ref{th:7.6}, we studied the behavior of the 
bifurcation curve $(\lambda(\alpha),\alpha)$ near the initial point $(\mu(q),0)$.
We shall investigate it near the final point $(\mu(p),\infty)$.
To this end, we compute $z_\infty(p,q)$ in the next lemma.

\begin{lemma}\label{le:7.10}
Let $z_\infty(p,q)$ be the constant as in \eqref{eq:7.1}.
Then
\begin{gather}\label{eq:7.10}
\lim_{p\to \infty}z_\infty(p,q)=q+1, \\
\label{eq:7.11}
\lim_{q\to 1+0}z_\infty(p,q)
=(p-1)^{-1/(p-1)}\Big(\int_0^1 (1-s)(1-s^p)^{-(p+1)/p}ds \Big)^{-p/(p-1)}.
\end{gather}
\end{lemma}

\begin{proof}
We define
\begin{equation} \label{eq:7.12}
\begin{gathered}
X:=X(p,q) :=\int_0^1 (1-s^q)(1-s^p)^{-(p+1)/p}ds, \\
Y:=Y(p,q) :=\int_0^1 (1-s^p)^{(q-p-1)/p}ds.
\end{gathered}
\end{equation}
Fix $q>1$ arbitrarily.
We apply the Lebesgue dominated convergence theorem to obtain
\begin{equation}\label{eq:7.13}
\lim_{p\to\infty}X(p,q) =\int_0^1(1-s^q)ds=\frac{q}{q+1}.
\end{equation}
We shall compute the limit of $Y(p,q)$ as $p\to \infty$.
Changing the variables $s^p=t$, we have
$$
Y(p,q) = (1/p)B(1/p,(q-1)/p),
$$
where $B(x,y):=\int_0^1t^{x-1}(1-t)^{y-1}dt$ is the beta function.
Using the relation
$B(x,y) = \Gamma(x)\Gamma(y)/\Gamma(x+y)$, where $\Gamma(x)$ is the gamma function,
we rewrite $Y(p,q)$ as
$$
Y(p,q)= \frac{\Gamma(1/p)\Gamma((q-1)/p)}{p\Gamma(q/p)}.
$$
Since $\lim_{x\to+0}x\Gamma(x)=\lim_{x\to+0}\Gamma(x+1)=\Gamma(1)=1$, we have
\begin{equation}\label{eq:7.14}
\begin{aligned}
 \lim_{p\to\infty} Y(p,q)
 & = \lim_{p\to\infty} \frac{\Gamma(1/p)\Gamma((q-1)/p)}{p\Gamma(q/p)}   \\
 & = \lim_{p\to\infty} \frac{[(1/p)\Gamma(1/p)][((q-1)/p)\Gamma((q-1)/p)]}{(q/p)
 \Gamma(q/p)} \frac{q}{q-1}  \\
 & = \frac{q}{q-1}.
\end{aligned}
\end{equation}
By the definitions of $z_\infty$, $c_1$ and $c_2$, we find that
$$
z_\infty(p,q) = (c_2/c_1)^{p/(p-q)} = (q-1)^{p/(p-q)} (p-1)^{-q/(p-q)} (Y/X)^{p/(p-q)}.
$$
Using \eqref{eq:7.13} and \eqref{eq:7.14}, we obtain $\lim_{p\to\infty}z_\infty(p,q)=q+1$.
Thus \eqref{eq:7.10} holds.

Let us prove \eqref{eq:7.11}.
In the same way as in \eqref{eq:7.14}, we have
\begin{align*}
 \lim_{q\to 1+0}(q-1)Y(p,q)
 & = \lim_{q\to1+0} \frac{(q-1)\Gamma(1/p)\Gamma((q-1)/p)}{p\Gamma(q/p)} \\
 & = \lim_{q\to1+0} \frac{\Gamma(1/p)[((q-1)/p)\Gamma((q-1)/p)]}
                            {\Gamma(q/p)} 
  = 1.
\end{align*}
As $q\to 1+0$, we find easily that
$$
X(p,q) =\int_0^1 (1-s^q)(1-s^p)^{-(p+1)/p}ds \to \int_0^1 (1-s)(1-s^p)^{-(p+1)/p}ds.
$$
Therefore 
\begin{align*}
\frac{c_2}{c_1}
&=  (q-1)(p-1)^{-q/p}Y(p,q)X(p,q)^{-1} \\
& \to   (p-1)^{-1/p}\Big(\int_0^1 (1-s)(1-s^p)^{-(p+1)/p}\,ds\Big)^{-1}
\quad \text{as}\ q\to 1+0.
\end{align*}
Since $z_\infty=(c_2/c_1)^{p/(p-q)}$, we obtain \eqref{eq:7.11}.
The proof is complete.
\end{proof}

In the next theorem, we shall investigate how the bifurcation curve behaves
as it approaches the final point $(\mu(p), \infty)$.

\begin{theorem}\label{th:7.11}
\begin{itemize}
\item[(i)]
  Let $q>1$ and $L\leq 1$.
  If $p$ is large enough, then $\mu(q,L)<\mu(p,L)$ and $\lambda'(\alpha)>0$ 
 for $\alpha$ large enough.
\item[(ii)]
  Let $q>1$ and $L>1$.
  If $p$ is large enough, then $\mu(p,L)<\mu(q,L)$ and $\lambda'(\alpha)<0$ 
 for $\alpha$ large enough.
  Moreover, a positive solution of \eqref{Plambda} is unique when
  $\mu(p,L)<\lambda<\mu(p,L)+\varepsilon$
  with a small $\varepsilon>0$.
\end{itemize}
\end{theorem}

The theorem above says that if $L\leq 1$ and $p$ is large enough,
the bifurcation curve $(\lambda(\alpha),\alpha)$ approaches the final point 
$(\mu(p), \infty)$ from the left.
On the other hand, if $L>1$ and $p$ is large enough,
then the curve approaches the final point from the right.
If $\pi_q>2L$, we combine Theorem \ref{th:7.3} (i) with Theorem \ref{th:7.11} (ii).
Then this gives Type (D).

\begin{proof}[Proof of Theorem \ref{th:7.11}]
Fix $q>1$ and $L\leq 1$ arbitrarily.
We use \eqref{eq:7.10} with the fact that $\mu(p,L)\to \infty$ as $p\to\infty$ 
by \eqref{eq:2.7}.
Then $\mu(p,L)>z_\infty(p,q)$ for $p$ large enough.
By Proposition \ref{pr:7.2} (i), $\lambda'(\alpha)>0$ for $\alpha>0$ large enough.

Let us show (ii). Fix $q>1$ and $L>1$.
Since $\lim_{p\to\infty}\mu(p,L)=0$ by \eqref{eq:2.8},
it follows from \eqref{eq:7.10} that
$\mu(p,L)<\min\{z_\infty(p,q),\mu(q,L)\}$ for $p$ large enough.
By Proposition \ref{pr:7.2} (ii), there exists an $A>0$ such that 
$\lambda'(\alpha)<0$ for $\alpha \geq A$.
Consequently, $\lambda(\alpha)$ is decreasing and converges to $\mu(p,L)$
as $\alpha\to \infty$.
We put
$\lambda_0:=\inf_{0<\alpha\leq A}\lambda(\alpha)$.
Then $\lambda_0>\mu(p,L)$ holds because $\lim_{\alpha\to +0}\lambda(\alpha)=\mu(q,L)$ and
$\lambda(\alpha)>\mu(p,L)$ for any $\alpha>0$.
For each $\mu \in (\mu(p,L), \lambda_0)$,
there exists a unique $\alpha$ which satisfies $\mu=\lambda(\alpha)$.
This shows the uniqueness of positive solutions.
The proof is complete.
\end{proof}

Let us consider type (B), which has been obtained as a small perturbation 
from type (E) in Remark~\ref{re:4.14}.
We shall construct type (B) by a different method in the next theorem.

\begin{theorem}\label{th:7.12}
If $L\in(1,\infty)$ is sufficiently close to $1$,
there exist $p$ and $q$ such that $1<q<p$ and
\begin{equation}\label{eq:7.15}
\mu(q,L)<\mu(p,L)<z_\infty(p,q).
\end{equation}
Therefore $\lambda'(\alpha)<0$ for $\alpha>0$ large enough and 
$\mu(q,L)< \mu(p,L)<\lambda^*$.
\end{theorem}

The theorem above says that the bifurcation curve reaches the final point from 
the right and gives an example of type (B).
The next corollary follows immediately from the theorem above.

\begin{corollary}\label{co:7.13}
Let $p$, $q$ and $L$ satisfy $L>1$ and \eqref{eq:7.15}.
Then the following assertions hold.
\begin{itemize}
\item[(i)] If $\lambda \leq \mu(q,L)$ or $\lambda > \lambda^*$, 
 there exist no positive solutions.
\item[(ii)] If $\mu(q,L)<\lambda \leq \mu(p,L)$ or $\lambda = \lambda^*$, 
 there exists at least one positive solution.
\item[(iii)] If $\mu(p,L)<\lambda < \lambda^*$, there exist at least two 
positive solutions.
\end{itemize}
\end{corollary}

In Lemma~\ref{le:7.10}, we investigated the behavior of $z_\infty(p,q)$ as $q\to 1+0$.
Using this result, we prove Theorem \ref{th:7.12}.

\begin{proof}[Proof of Theorem \ref{th:7.12}]
Denote the right hand side of \eqref{eq:7.11} by $Z_\infty(p)$, i.e.,
$$
Z_\infty(p):=(p-1)^{-1/(p-1)}\Big(\int_0^1 (1-s)(1-s^p)^{-(p+1)/p}ds 
\Big)^{-p/(p-1)}.
$$
A simple computation shows that $\lim_{p\to \infty}Z_\infty(p)=2$.
Fix $a$ satisfying $1<a<2$.
Then there exists a $p(a)>1$ such that if $p>p(a)$, then $Z_\infty(p)>a$.
By \eqref{eq:2.9} and \eqref{eq:2.10}, we observe that
$$
\lim_{L\to 1+0}p_*(L)= \infty, \quad \lim_{L\to 1+0}\mu(p_*(L),L)= \infty.
$$
Accordingly, if $L$ is slightly greater than $1$, then
\begin{equation}\label{eq:7.16}
\mu(p_*(L),L)>a, \quad p_*(L) > p(a).
\end{equation}
We fix such an $L>1$.
Since $\mu(p,L)$ is strictly decreasing in $p\in [p_*(L),\infty)$ and converges 
to $0$ as $p\to \infty$, we can choose $p\in(p_*(L), \infty)$ satisfying 
$1<\mu(p,L)<a$.
We fix such a $p$. Since $\lim_{q\to 1+0}\mu(q,L)=1/L<1$,
$\mu(q,L)$ is less than $1$ for $q$ sufficiently close to $1$. Thus we have
$$
\mu(q,L)<1<\mu(p,L)<a.
$$
Since $p>p_*(L)$, \eqref{eq:7.16} ensures that $p>p(a)$. Hence $Z_\infty(p)>a$. Therefore we obtain
$$
\lim_{q\to 1+0}z_\infty(p,q)=Z_\infty(p)>a.
$$
Consequently, if $q$ is sufficiently close to $1$, it holds that
$$
z_\infty(p,q)>a>\mu(p,L)>1>\mu(q,L).
$$
Thus \eqref{eq:7.15} is obtained.
By Proposition \ref{pr:7.2}, $\lambda'(\alpha)<0$ for $\alpha$ large enough.
Since $\lambda(\alpha)$ is decreasing and converges to $\mu(p,L)$
as $\alpha\to\infty$, it holds that $\lambda^* >\mu(p,L)$.
The proof is complete.
\end{proof}

Observing all our results and using $L_*$ defined by \eqref{eq:1.4}, we propose the next conjecture.

\begin{conjecture} \rm
Fix $1<q<p$. Then there exist constants $\underline{L}, \overline{L}$ satisfying
$0<\underline{L}<L_*<\overline{L}$ such that
the types  (A), (B), (E), (D) and  (C) occur when
$0<L \leq \underline{L}$, $\underline{L}<L<L_*$,
$L=L_*$, $L_*<L<\overline{L}$ and $\overline{L} \leq L<\infty$, respectively.
The types {\upshape (A)} and {\upshape (C)} have no turning points and
 (B), (D), (E) have exactly one turning point.
\end{conjecture}

The conjecture above is partially solved in the present paper.
Indeed, we have already proved the following theorem
(see Remark~\ref{re:4.14} and Theorems~\ref{th:6.1} and \ref{th:6.2}).

\begin{theorem}\label{th:7.14}
There exists a constant $\varepsilon>0$ such that
{\upshape (A), (B), (E), (D), (C)} occur when $0<L<\varepsilon$,
$L_*-\varepsilon<L<L_*$, $L=L_*$, $L_*<L<L_*+\varepsilon$, $1/\varepsilon<L<\infty$, 
respectively,
and there exist no turning points for {\upshape (A)} with $0<L<\varepsilon$
and {\upshape (C)} with $1/\varepsilon<L<\infty$.
\end{theorem}

\subsection*{Acknowledgments}
R.  Kajikiya was supported by the JSPS KAKENHI Grant  16K05236.
M. Tanaka was supported by the JSPS KAKENHI Grant 15K17577.
S. Tanaka was supported by the JSPS KAKENHI Grant  26400182.

The authors would like to thank the referee for carefully reading our manuscript
and for giving constructive comments which improved the quality of this paper.

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