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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 269, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/269\hfil Solution branches]
{Solution branches for nonlinear problems with an 
asymptotic oscillation property}

\author[L. Gong, X. Li, B. Qin, X. Xu \hfil EJDE-2015/269\hfilneg]
{Lin Gong, Xiang Li, Baoxia Qin, Xian Xu}

\address{Lin Gong \newline
Department of Mathematics,
Jiangsu Normal University, Xuzhou,
Jiangsu 221116, China}
\email{gonglin812@163.com}

\address{Xiang Li \newline
Department of Mathematics,
Jiangsu Normal University, Xuzhou,
Jiangsu 221116, China}
\email{lxws3@126.com}

\address{Baoxia Qin \newline
School of Mathematics,
Qilu Normal  University, Jinan,
Shandong 250013,  China}
\email{qinbaoxia@126.com}

\address{Xian Xu\newline
Department of Mathematics,
Jiangsu Normal University, Xuzhou,
Jiangsu 221116, China}
\email{xuxian68@163.com}

\thanks{Submitted August 20, 2015. Published October 19, 2015.}
\subjclass[2010]{47H07, 47H10}
\keywords{Global solution branch; fixed point index;
\hfill\break\indent asymptotic oscillation property}

\begin{abstract}
 In this article we employ an oscillatory condition on the nonlinear term,
 to prove the existence of a connected component of solutions
 of a nonlinear problem, which  bifurcates from infinity and asymptotically
 oscillates over an interval of  parameter values.
 An interesting and immediate consequence of such oscillation property of
 the  connected component is the existence of infinitely many solutions 
 to the nonlinear  problem for all parameter values in that interval.
\end{abstract}

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

\section{Introduction}

 Rabinowtz \cite{r1} obtained  a well known result concerning the existence
of unbounded connected components bifurcating from infinity for
asymptotically linear operators, 1973. Since then, by using this result many
authors have studied the existence of  connected components of solutions
 of various   boundary value problems. By using different
methods-such as the  blow up method, the maximum principle, the moving
plane method, turning point theorem, eigenvalue theories and so on,
they tried to obtain much possible information on the  connected components.
An interesting question concerning the connected component bifurcating from
infinity is: in which manner the  connected component of solutions
approaches infinity? always from one side of a parameter in the 
parameter-norm plane, or oscillating infinitely about a parameter 
(even an interval of parameters)?

Let us first recall some results in the literature concerning the above problem.
 Schaaf and  Schmitt \cite{s1} studied the existence of solutions of nonlinear
Sturm Liouville problems whose linear part is at resonance.
By using  bifurcation methods, they studied the  one parameter  problem
\begin{equation}
\begin{gathered}
u''+\lambda u+g(u)=h(x), \quad 0\leq x\leq \pi;\\
u(0)=0=u(\pi).
\end{gathered}
\label{e1.1}
\end{equation}
They showed that \eqref{e1.1} has a connected component of solutions which
 bifurcates from infinity at $\lambda=1$, and showed that this connected
component must  cross the $\lambda=1$ parameter plane infinitely often.

 Davidson and  Rynne  \cite{d1} studied the
semilinear Sturm Liouville boundary value problem
\begin{equation}
\begin{gathered}
-u''=\lambda u+ f(u)\quad \text{in } (0, \pi),\\
u(0)=u(\pi)=0,
\end{gathered}  \label{e1.2}
\end{equation}
where $f:\mathbb{R}^+=[0,\infty)\to \mathbb{R}^1$ is Lipschitz continuous
and $\lambda$ is a real parameter. It is assumed in \cite{d1} that
$f(s)$ oscillates, as $s\to \infty$, in such a manner that the
problem $\eqref{e1.2}$ is not linearizable at $u=\infty$ but does have
a connected component $C$ of positive solutions bifurcating from infinity.
Then, they investigated the relationship between the oscillations of
$f$ and those of $C$ in the $\lambda-\|u\|$ plane at large $\|u\|$. They
obtained some  results about the oscillation properties of $C$
 over a single point $\lambda$, or over an
interval $I$ of $\lambda$ values. An immediate consequence of such
oscillations of $I$ is the existence of infinitely many solutions,
of arbitrarily large norm $\|u\|$, of problem $\eqref{e1.2}$ for all
values of $\lambda\in I$. Here, as defined as in \cite{d1}, a continuum
$C\subset\mathbb{R}^+\times E$ is said to
 oscillate over an interval $I=[\lambda_-,\lambda_+]$ if, for each
 $\nu\in\{+, -\}$, there exists a sequence of positive number
 $\{\zeta^\nu_n\}$,such that $\zeta^\nu_n\to \infty$ as
 $n\to \infty$, and any solution $(\lambda, u)\in C$ with
 $\|u||=\zeta^\nu_n$ must have $\nu(\lambda-\lambda_\nu)\geq 0$, and such
 solutions do exist for all sufficiently large $n$.
For other references concerning the connected component of solutions  with
asymptotic oscillation property one can refer to \cite{k1,m2,r2,s2,s3}.

Consider the  three-point boundary-value problem
\begin{equation}
\begin{gathered}
 -u''=\lambda u+ f(u)\quad \text{in } (0, 1),\\
u(0)=0,\quad u(1)=\alpha u(\eta),
\end{gathered}\label{e1.3}
\end{equation}
where $\eta\in (0,1)$, $\alpha\in [0,1)$,  $f:\mathbb{R}^+\to \mathbb{R}^1$
is Lipschitz continuous, $f(0)=0$ and $\lambda$ is
a real parameter.

During the past twenty years  the multi-point boundary value problems
have been studied extensively. Especially, some authors studied multi-boundary
value problems by using global bifurcation theories; see \cite{m1,r3,s5}.
The main purpose of this paper is to extend  some main results in
\cite{d1} to equation \eqref{e1.3}.
By employing an oscillatory condition on the nonlinear
term $f$ we will prove a result for the existence of a connected component
of solutions  of \eqref{e1.3}, which  bifurcates from infinity and
oscillates infinitely often over an interval of $\lambda$-values.
There is a main difficulty to extend the main results of \cite{d1} to the three
point boundary value problem \eqref{e1.3}. Obviously, the symmetric point
of every positive solution in \cite{d1} is known and this plays an important role
in the proof of \cite{d1}. For example, every positive solution of \eqref{e1.2}
is symmetric about $t_0= 1/2$ and has a single maximum occurring
at this point. However, the symmetric point of every positive solution
of \eqref{e1.3} is unknown and the positive solution of \eqref{e1.3}
may not be  symmetric about $t_0= 1/2$  when  $\alpha\neq 0$.
To overcome this difficulty in section 2 we will give a detailed  analysis
of positive solutions of \eqref{e1.3}. Note the nonlinearity $f$ may not be of
asymptotically linear type.  Consequently,  the corresponding nonlinear
operator may be non-differentiable when one converts \eqref{e1.3}
into an operator equation in $C[0,1]$, the methods in
Rabinowitz'  well known global bifurcation theorems from \cite{r1}
establishing existence results for unbounded connected components
bifurcating from infinity do not seem to work in our situation.
However,  due to the contributions of  Schmitt,  Berestycki et al.,
during the past forty years significant progress on the nonlinear
eigenvalue problems for non-differential mappings  has been achieved;
see \cite{b1,o1,p1,s4} and the references therein. By using the methods in
\cite{b1,o1,p1,s4} we can show the existence of connected component of solutions
 of \eqref{e1.3} bifurcating from infinity.

\section{Properties of positive solutions}

First let us recall some results concerning the linear eigenvalue problem
\begin{equation}
\begin{gathered}
-u''=\lambda u, \quad  t\in (0,1),\\
u(0)=0, \quad  u(1)=\alpha u(\eta). \\
\end{gathered}
\label{e2.1}
\end{equation}
According to \cite{x1}, there exists a sequence of eigenvalues
$$
0<\sqrt{\lambda_1(\alpha)}<\sqrt{\lambda_2(\alpha)}<\dots<\sqrt{\lambda_n(\alpha)}<\dots,
$$
where $\sqrt{\lambda_i(\alpha)}$ is the $i$-th positive solution of the elementary
equation $\sin x=\alpha \sin x\eta$. The corresponding eigenfunction to
$\sqrt{\lambda_n(\alpha)}$ is $\phi_{n,\alpha}(t)=\sin \sqrt{\lambda_n(\alpha)}t$.
In what follows, for brevity, denote $\phi_{1,\alpha}$ by $\phi_{\alpha}$,
for each $\alpha\in [0,1)$.
It is easy to show the following result concerning the principal
eigenvalue $\sqrt{\lambda_1(\alpha)}$.


\begin{lemma} \label{lem2.1}
Assume that $\alpha\in [0,1)$. Then
\begin{itemize}
\item[(1)] $\frac{\pi}{2}<\sqrt{\lambda_1(\alpha)}\leq \pi$;

\item[(2)]  $\sqrt{\lambda_1(\alpha)}$ is non-increasing with $\alpha\in [0,1)$;

\item[(3)] there exists an unique
$t'_\alpha\in[\frac{1}{2}, 1)$ such that $\|\phi_\alpha\|=\phi_\alpha(t'_\alpha)=1$, and
$\phi^\prime_\alpha(t)>0$ for $t\in [0,t'_\alpha)$, $\phi'_\alpha(t)<0$ for $t\in (t'_\alpha,1]$;

\item[(4)] $\phi_\alpha\to \phi_0=\sin\pi t$ in the $C^1$ norm on $[0,1]$ as $\alpha\to 0$.
\end{itemize}
\end{lemma}

For each $\alpha\in [0,\frac{1}{2}]$, let
\begin{align*}
S_\alpha=\big\{&(\lambda, u): \lambda\in \mathbb{R}^+, u\in C[0,1], u(t)>0
\text{ for $t\in (0,1)$, such that}\\
&\text{$u(t)$ is a solution of \eqref{e1.3}}\big\}.
\end{align*}
Then, for each $(\lambda, u)\in S_\alpha$, using  integration by parts and
the boundary condition, we have
\begin{equation}
(\lambda_1(\alpha)-\lambda)\int^1_0u\phi_{\alpha}dt=\int^1_0f(u)\phi_{\alpha}dt-W(\alpha),\label{e2.2}
\end{equation}
where
$$
W(\alpha)=\begin{vmatrix}
u(1),&u'(1)\\
\phi_\alpha(1), &\phi'_\alpha(1)
\end{vmatrix}
=\alpha\begin{vmatrix}
u(\eta),&u'(1)\\
\phi_\alpha(\eta), &\phi'_\alpha(1)
\end{vmatrix}.
$$
Let
$$
G(t,s)=\min\{t,s\}(1-\max\{t,s\}), \quad\forall t,s\in [0,1],
$$
and the operator $K_\alpha:C[0,1]\to C[0,1]$ be defined by
$$
K_\alpha x(t)=\int^1_0G(t,s)x(s)ds
+\frac{\alpha t}{1-\alpha\eta}\int^1_0G(\eta, s)x(s)ds, \quad t\in [0,1].
$$
 Let $P=\{x\in C[0,1]: x(t)\geq 0\ \text{for}\ t\in [0,1] \}$.
Then, for each $h\in C[0,1]$, $y=K_\alpha h$  if and only if
\begin{gather*}
-y''=h(t), \quad t\in (0,1),\\
y(0)=0, \quad y(1)=\alpha y(\eta).
\end{gather*}
Let
\begin{gather*}
e_\alpha(t)=\frac{1}{2}\eta(1-\eta)t[1-\alpha\eta-(1-\alpha)t],\quad  t\in [0,1],\\
e(t)=\frac{1}{4}\eta(1-\eta)t(1-t), \quad \forall t\in [0,1].
\end{gather*}
Then, we have $e_\alpha(t)\geq e(t)$ for $t\in [0,1]$ and
$\alpha\in [0,1/2]$.



\begin{lemma} \label{lem2.2}
For each $\alpha\in [0,1/2]$, let
$Q_\alpha=\{x\in P: x(t)\geq \|x\|e_\alpha(t)\text{ for }t\in [0,1] \}$.
 Then $K_\alpha:P\to Q_\alpha$ is completely continuous.
\end{lemma}

\begin{proof}
Obviously, $K_\alpha:P\to C[0,1]$ is completely continuous.
Take $h\in P$,  and let $y=K_\alpha h$.  Using the fact that $ y$ is
 a concave function on $[0,1]$, by \cite[Lemma 2]{x2}
 we  easily check that $K_\alpha:P\to Q_\alpha$. The proof is complete.
\end{proof}

In this paper we  use the following symbols.
\begin{gather*}
0 <\kappa^-=\liminf_{s\to+\infty}\frac{f(s)}{s}
\leq \limsup_{s\to+\infty}\frac{f(s)}{s}=\kappa^+<+\infty,\\
-\infty <\zeta^-=\liminf_{s\to+\infty}\frac{F(s)}{s^2}
\leq \limsup_{s\to+\infty}\frac{F(s)}{s^2}=\zeta^+<+\infty,
\end{gather*}
where $F(s)=\int^s_0f(t)dt$ for all $s\in [0,+\infty)$.

In this article we  always assume that
\begin{equation}
\frac{2-\eta}{3-\eta}>\kappa^+-\kappa^-. \label{e2.3}
\end{equation}

Take $\delta>0$ small enough,  such that
$\frac{2-\eta}{3-\eta}-(\kappa^+-\kappa^-)>2\delta$ and $\kappa^--\delta>0$.
Let $\widehat{\kappa}^-=\kappa^--\delta$, $\widehat{\kappa}^+=\kappa^++\delta$, and
$$
\widetilde s_0=\min\big\{1,\frac{2-\eta}{2(3-\eta)}-\frac{1}{2}(\widehat{\kappa}^+-\widehat{\kappa}^-)\big\}.
$$
 Take $s_0>0$ large enough,  such that
\begin{equation}
\widehat{\kappa}^-s\leq f(s)\leq \widehat{\kappa}^+ s, \quad s\geq s_0.\label{e2.4}
\end{equation}
Let $M_0=\sup\{|f(s)|:s\in [0, s_0]\}$. By \eqref{e2.4} we have
\begin{equation}
\widehat{\kappa}^-s-M_0\leq f(s)\leq \widehat{\kappa}^+ s+M_0, \quad s\geq 0.\label{e2.5}
\end{equation}
Let
$$\gamma_{0,\alpha}=\frac{\|K_\alpha\|^{-1}-\widehat{\kappa}^+-\widehat{\kappa}^-}{2},\quad
 \gamma_{1,\alpha}=2\|K_\alpha e_\alpha\|^{-1}+\gamma_{0,\alpha}.
$$
Let
$$
k_0=\max_{t\in [0,1]}\int^1_0G(t,s)s(1-s)ds.
$$
For each  $t\in [0,1]$, we have
\begin{align*}
\frac{3-\eta}{2-\eta}
&\geq K_\alpha e_\alpha(t)
\geq \int^1_0 G(t,s)e_\alpha(s)ds
\geq \int^1_0 G(t,s)e(s)ds\\
&\geq \frac{1}{4}\eta(1-\eta)\int^1_0G(t,s)s(1-s)ds.
\end{align*}
and so,
$$
\frac{3-\eta}{2-\eta}\geq \|K_\alpha e_\alpha\|\geq \frac{1}{4}k_0\eta(1-\eta).
$$
On the other hand,  for $t\in [0,1]$ we have
$$
k_0\leq \|K_\alpha\|\leq 1+\frac{\alpha}{1-\alpha\eta}\leq \frac{3-\eta}{2-\eta}.
$$
Thus, we have
\begin{gather*}
a^-:=\frac{1}{2}\Big(\frac{2-\eta}{3-\eta}-\widehat{\kappa}^+-\widehat{\kappa}^-\Big)
\leq \gamma_{0,\alpha}\leq \frac{k_0^{-1}-\widehat{\kappa}^+-\widehat{\kappa}^-}{2},\\
a^+:=\frac{k_0^{-1}-\widehat{\kappa}^+-\widehat{\kappa}^-}{2}+\frac{8}{k_0\eta(1-\eta)}\geq \gamma_{1,\alpha}.
\end{gather*}
For each $\alpha\in [0,\frac{1}{2}]$, let
 $\Lambda_\alpha=\bar S_\alpha\cap \big([a^-, a^+]\times B^c(\theta,R_0)\big)$,
 where $B^c(\theta,R_0)=\{x\in C[0,1],\|x\|\geq R_0\}$ and
$R_0=\frac{8M_0}{\eta(1-\eta)}$.


\begin{lemma} \label{lem2.3}
For each $(\lambda, u)\in \Lambda_\alpha$, we have
\begin{equation}
u(t)\geq \frac{1}{2}\|u\| e_\alpha (t), \quad t\in [0,1].\label{e2.6}
\end{equation}
\end{lemma}


\begin{proof}
Define the operators $\widetilde F(\cdot,\cdot):[a^-, +\infty)\times C[0,1]\to C[0,1]$ by
$$
\widetilde F(\lambda, u)(t)=\lambda u(t)+f(u(t))+M_0, \quad t\in [0,1]
$$
and $B(\cdot,\cdot):[a^-, +\infty)\times C[0,1]\to C[0,1]$ by
$B(\lambda, u)(t)=K_\alpha\widetilde F(\lambda, u)(t)$ for  $t\in [0,1]$.
Let $v(t)=B(\lambda, u)(t)$ for all $t\in [0,1]$. By \eqref{e2.3} and
\eqref{e2.5} we have $\widetilde F(\lambda, u(t))\geq (\widehat{\kappa}^-+a^-) u(t)\geq 0$
for all $t\in [0,1]$. It follows from Lemma \ref{lem2.2} that $v\in Q_\alpha$.
Obviously, $u(t)=B(\lambda, u)(t)-M_0\omega_\alpha(t)= v(t)-M_0\omega_\alpha(t)$ for all
$t\in [0,1]$, where
$$
\omega_\alpha(t)=\int^1_0G(t,s)ds+\frac{\alpha t}{1-\alpha\eta}\int^1_0 G(\eta, s)ds,
\forall t\in [0,1].
$$
It is easy to see that
$$
\omega_\alpha(t)\leq \frac{1}{2}t\big[1-t+\frac{\alpha\eta(1-\eta)}{\eta(1-\alpha)}\big]
=\frac{e_\alpha(t)}{\eta(1-\eta)(1-\alpha)}\leq \frac{2e_\alpha(t)}{\eta(1-\eta)} , t\in [0,1].
$$
Then, we have
\begin{align*}
u(t)
&\geq \|v\|e_\alpha(t)-\frac{2M_0}{\eta(1-\eta)}e_\alpha(t)\\
&\geq \Big(\|u\|-M_0\|\omega_\alpha\|-\frac{2M_0}{\eta(1-\eta)}\Big) e_\alpha(t)\\
&\geq \frac{1}{2}\|u\|e_\alpha(t), \quad t\in [0,1].
\end{align*}
This implies that \eqref{e2.6} holds. The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.4}
 For each $(\lambda, u)\in \Lambda_\alpha$, there
exists an unique $t_\alpha\in (0,1)$ such that
\begin{itemize}
\item[(1)] $u(t_\alpha)=\|u\|$;

\item[(2)] $u'(t)>0$, $t\in (0, t_\alpha)$; $u'(t)<0$, $t\in (t_\alpha,1)$;

\item[(3)] $u(t_\alpha-s)=u(t_\alpha+s)$, $s\in [0, 1-t_\alpha]$;

\item[(4)] $1>t_\alpha\geq 1/2$;

\item[(5)] $t_\alpha=1/2$  as $\alpha=0$.
\end{itemize}
\end{lemma}

\begin{proof}
It follows from the theorem on the unique solutions of initial value problems
for differential equations (IVPU)  that $u'(0)>0$. From the boundary value
condition $u(1)=\alpha u(\eta), 0\leq \alpha<1/2$, there must exist
$t_\alpha\in (0,1)$ such that $u'(t_\alpha)=0$. Assuming that $t_\alpha<1/2$,
it follows from the fact that $f$  is independent of $t$
and IVPU that
$$
u(t_\alpha+s)=u(t_\alpha-s), \quad s\in [0, t_\alpha].
$$
Hence, $u(2t_\alpha)=u(0)=0$, $u'(2t_\alpha)=-u'(0)<0$, and so, there exists
$t'>2t_\alpha$ such that $u(t')<0$, which is a contradiction.
Hence, $1>t_\alpha\geq 1/2$,  and $u(t_\alpha-s)=u(t_\alpha+s)$ for
$s\in [0, 1-t_\alpha]$. Obviously,  (1) and (2) hold. When $\alpha=0$,
$u(1)=0$ and so $t_\alpha=1/2$.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.5}
 Assume that $\Lambda_\alpha\cap \big([a^-, a^+]\times B^c(\theta, R_1)\big)\neq \emptyset$,
where $R_1=R_0+24M_0c_1^{-1}$  and  $c_1=\frac{1}{8}\widetilde s_0\eta(1-\eta)$.
Let $\sigma_0(\alpha)=\frac{200\alpha}{\widetilde s_0\eta(1-\eta)}$ for
$\alpha\in [0, \frac{\widetilde s_0\eta(1-\eta)}{4000}]$.
For each $(\lambda, u)\in \Lambda_\alpha\cap \big([a^-, a^+]\times B^c(\theta, R_1)\big)$,
let $t_\alpha$ be defined as in Lemma \ref{lem2.4}. Then $\sigma_0(\alpha)\to 0$ as $\alpha \to 0$,
and for $\alpha\in [0, \widetilde s_0\eta(1-\eta)/4000]$,
\begin{equation}
0\leq t_\alpha-\frac{1}{2}\leq \sigma_0(\alpha).\label{e2.7}
\end{equation}
\end{lemma}


\begin{proof}
For each $(\lambda, u)\in \Lambda_\alpha\cap \big([a^-, a^+]\times B^c(\theta, R_1)\big)$,
let $v(t)=u(t)/\|u\|$ for $t\in [0,1]$. Then
\begin{equation}
\begin{gathered}
-v''(t)=\lambda v(t)+\frac{1}{\|u\|}f(\|u\| v(t)),\quad t\in (0,1),\\
v(0)=0, \quad v(1)=\alpha v(\eta).
\end{gathered}\label{e2.8}
\end{equation}
It follows from Lemma \ref{lem2.4} that $v(t_\alpha)=1$,
$t_\alpha\geq  1/2$ and
$$
v(t_\alpha-s)=v(t_\alpha+s), \quad \forall s\in [0,1-t_\alpha].
$$
Hence, $v(2t_\alpha-1)=v(1)=\alpha v(\eta)\leq \alpha$.
It follows from Lemma \ref{lem2.4} that $v'(t)>0$ for $ t\in (0, t_\alpha)$.
Thus, if there exits a  $t'\in [0,t_\alpha)$ such that
$v(t')\geq \alpha\geq v(2 t_\alpha-1)$, then we have $t'\geq 2 t_\alpha-1$ or
$t_\alpha\leq \frac{1}{2}(1+t')$.

By Lemma \ref{lem2.3},  for $t\in [0, t_\alpha)$, we have
\begin{equation}
\begin{aligned}
u'(t)&=-\int^{t_\alpha}_t  u''(s)ds\\
&\geq \frac{1}{2}\widetilde s_0\|u\|\int^{t_\alpha}_t e_\alpha(s)ds-M_0(t_\alpha-t)\\
&\geq \frac{1}{2}\widetilde s_0\|u\|\int^{t_\alpha}_t e(s)ds-M_0(t_\alpha-t),
\end{aligned} \label{e2.9}
\end{equation}
and so
$$
v'(t)\geq \frac{1}{2}\widetilde s_0\int^{t_\alpha}_t e(s)ds-\frac{M_0}{\|u\|}(t_\alpha-t).
$$
 Then, we have
\begin{align*}
v(t)
&=v(0)+\int^{t}_0 v'(s)ds=\int^{t}_0v'(s)ds\\
&\geq \frac{1}{2}\widetilde s_0\int^{t}_0ds\int^{t_\alpha}_s e(\tau)d\tau
-\frac{M_0}{\|u\|}\int^{t}_0(t_\alpha-s)ds\\
&\geq c_1\Big(t\int^{t_\alpha}_ts(1-s)ds+\int^t_0s^2(1-s)ds\Big)
-\frac{M_0}{\|u\|}t\\
&=c_1\Big(t\int^{t_\alpha}_0s(1-s)ds-t\int^t_0s(1-s)ds
 +\int^t_0s^2(1-s)ds\Big)-\frac{M_0}{\|u\|}t\\
&\geq c_1\Big(t\int^{1/2}_0s(1-s)ds-t\big(\frac{1}{2}t^2
 -\frac{1}{3}t^3\big)+\big(\frac{1}{3}t^3-\frac{1}{4}t^4\big)\Big)
 -\frac{M_0}{\|u\|}t\\
&\geq \Big(\frac{1}{12}c_1-\frac{M_0}{\|u\|}\Big)t-\frac{1}{6}c_1t^2\\
&\geq \frac{1}{24}c_1t-\frac{1}{6}c_1t^2.
\end{align*}
Solving the inequality
$$
\frac{1}{24}c_1t-\frac{1}{6}c_1t^2\geq \alpha,
$$
we obtain
$$
\frac{1-\sqrt{1-384\alpha c_1^{-1}}}{8}
\leq t\leq \frac{1+\sqrt{1-384\alpha c_1^{-1}}}{8}.
$$
Let
$$
t'=\frac{1-\sqrt{1-384\alpha c_1^{-1}}}{8}.
$$
Then we have
$$
0\leq t_\alpha-\frac{1}{2}\leq \frac{1-\sqrt{1-384\alpha c_1^{-1}}}{16}
\leq \frac{24\alpha}{c_1}\leq \sigma_0(\alpha).
$$
Obviously, $\sigma_0(\alpha)\to 0$ as $\alpha\to 0$. The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.6}
Let $\lambda_1(\alpha)$, $t'_\alpha$ be defined as in Lemma \ref{lem2.1} and
$\sigma_1(\alpha)=\frac{100\alpha}{\eta(1-\eta)}$ for each
$\alpha\in [0, \frac{\widetilde s_0\eta(1-\eta)}{4000}]$.
Then for each $\alpha\in [0, \frac{\widetilde s_0\eta(1-\eta)}{4000}]$,
$$
0\leq t'_\alpha-\frac{1}{2}\leq \sigma_1(\alpha).
$$
\end{lemma}

\begin{proof}
It follows from (2) in Lemma \ref{lem2.1} that
$\pi^2=\lambda_1(0)\geq \lambda_1(\alpha)\geq \lambda_1(\frac{1}{2})\geq \frac{\pi^2}{4}$
for each $\alpha\in [0,1/2]$. According to Lemma \ref{lem2.2} we have
$\phi_\alpha\in Q_\alpha$  for $\alpha\in [0,1/2]$, and so
$$
\phi_\alpha(t)\geq \|\phi_\alpha\|e_\alpha(t)=e_\alpha(t)\geq e(t),\quad
 \forall t\in [0,1].
$$
Note that $\frac{\widetilde s_0\eta(1-\eta)}{4000}\leq  1/2$.
As in the proof Lemma \ref{lem2.5}, for $t\in [0,t'_\alpha]$, we have
\begin{align*}
\phi_\alpha(t)
&=\phi_\alpha (0)+\int^t_0\phi_\alpha'(s)ds=\int^t_0\phi_\alpha'(s)ds\\
&\geq \lambda_1(\frac{1}{2})\int^t_0ds\int^{t'_\alpha}_s\phi_\alpha(\tau)d\tau\\
&\geq \frac{\eta(1-\eta)}{4}\lambda_1(\frac{1}{2})
 \int^t_0ds\int^{t'_\alpha}_s\tau(1-\tau)d\tau\\
&\geq \frac{\pi^2\eta(1-\eta)}{16}\Big(t\int^{1/2}_0s(1-s)ds
 -t\int^t_0s(1-s)ds+\int^t_0s^2(1-s)ds\Big)\\
&\geq \frac{\pi^2\eta(1-\eta)}{16}\big(\frac{1}{12}t-\frac{1}{6}t^2\big).
\end{align*}
Solving the inequality
$$
\frac{\pi^2\eta(1-\eta)}{16}\Big(\frac{1}{12}t-\frac{1}{6}t^2\Big)\geq \alpha,
$$
we have
$$
\frac{1-\sqrt{1-\frac{1536\alpha}{\pi^2\eta(1-\eta)}}}{4}
\leq t\leq \frac{1+\sqrt{1-\frac{1536\alpha}{\pi^2\eta(1-\eta)}}}{4}.
$$
Then we have
$$
0\leq t'_\alpha-\frac{1}{2}\leq \frac{1-\sqrt{1-\frac{1536\alpha}{\pi^2\eta(1-\eta)}}}{8}
\leq \sigma_1(\alpha).
$$
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.7}
 Assume that $\Lambda_\alpha\cap \big([a^-, a^+]\times B^c(\theta, R_2)\big)\neq \emptyset$,
where $R_2=R_1+\frac{200M_0}{\widetilde s_0\eta(1-\eta)}$. Let $c_1$ and
$\sigma_0(\alpha)$ be defined as in Lemma \ref{lem2.5}, $c_2=\frac{c_1}{24}$.
Assume that $\alpha\in [0, \frac{\widetilde s_0 \eta(1-\eta)}{4000}]$.
For each $(\lambda, u)\in \Lambda_\alpha\cap \big([a^-, a^+]\times B^c(\theta, R_2)\big)$,
let $t_\alpha$ be defined as in Lemma \ref{lem2.4}. Then we have
\begin{equation}
|u'(t)|\geq c_2\|u\||t_\alpha-t|, t\in [0,1]. \label{e2.10}
\end{equation}
\end{lemma}

\begin{proof}
 Now, \eqref{e2.9} holds for each
$(\lambda, u)\in \Lambda_\alpha\cap \big([a^-, a^+]\times B^c(\theta, R_2)\big)$.
 From $\alpha\in [0, \frac{\widetilde s_0 \eta(1-\eta)}{4000}]$, we have
$\sigma_0(\alpha)<\frac{1}{10}$, and so,  by \eqref{e2.7} we have
$t_\alpha\in [1/2, 3/5]$ .
On the other hand,  we have for $t\in [0,t_\alpha)$,
\begin{equation}
\int^{t_\alpha}_t e(s)ds=\frac{\eta(1-\eta)}{4}
\big[\frac{1}{2}(t_\alpha+t)-\frac{1}{3}(t_\alpha^2+t_\alpha t+t^2)\big](t_\alpha-t).
\label{e2.11}
\end{equation}
Let
$$
g(t)=\frac{1}{2}(t_\alpha+t)-\frac{1}{3}(t_\alpha^2+t_\alpha t+t^2), \quad \forall t\in [0,1].
$$
It is easy to see that $g(t)\geq \min\{g(0), g(t_\alpha)\}\geq \frac{1}{8}$ for
$t\in [0,t_\alpha]$.
It follows from \eqref{e2.9} and \eqref{e2.11} that for $ t\in (0, t_\alpha)$,
\begin{equation}
u'(t)\geq \big[\frac{1}{64}\widetilde s_0\eta(1-\eta)\|u\|-M_0\big](t_\alpha-t)
\geq c_2\|u\|(t_\alpha-t). \label{e2.12}
\end{equation}
Similarly,  for $t\in [t_\alpha, 1]$, we have
\begin{equation}
\begin{aligned}
-u'(t)&=-\int^t_{t_\alpha}u''(s)ds\\
&\geq \frac{1}{2}\widetilde s_0\|u\|\int^t_{t_\alpha}e_\alpha(s)ds-M_0(t-t_\alpha)\\
&\geq \frac{1}{2}\widetilde s_0\|u\|\int^t_{t_\alpha}e(s)ds-M_0(t-t_\alpha)\\
&\geq \big( c_1\|u\|g(t)-M_0\big)(t-t_\alpha)\\
&\geq \big(\frac{1}{12}c_1\|u\|-M_0\big)(t-t_\alpha)\\
&\geq c_2\|u\|(t-t_\alpha).
\end{aligned} \label{e2.13}
\end{equation}
Now \eqref{e2.10} follows from \eqref{e2.12} and \eqref{e2.13}.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.8}
 Assume that $\Lambda_\alpha\cap \big([a^-, a^+]\times B^c(\theta, R_3)\big)\neq \emptyset$,
 where $R_3=R_2+M_0$.  For each
$(\lambda, u)\in \Lambda_\alpha\cap \big([a^-, a^+]\times B^c(\theta, R_2)\big)$,
let $t_\alpha$ be defined as in Lemma \ref{lem2.4}. Then the following
inequalities also hold:
\begin{gather}
(a^-+\widehat{\kappa}^-)u(t)\leq -u''(t)\leq (|a^+|+\widehat{\kappa}^+)u(t), \quad
  \forall t\in E^+(u;s_0),\label{e2.14}\\
|u''(t)|\leq (|a^+|+\widehat{\kappa}^+)s_0+3M_0,  \quad \forall t\in E^-(u;s_0),\label{e2.15}\\
|u'(t)|\leq c_3 \|u\||t-t_\alpha|, \forall t\in [0,1].\label{e2.16}
\end{gather}
where $c_3=|a^+|+\widehat{\kappa}^++1$, $E^+(u;s_0)=\{s\in [0,1]:u(s)\geq s_0\}$ and
$E^-(u;s_0)=\{s\in [0,1]:u(s)< s_0\}$.
\end{lemma}


\begin{proof}
It follows from \eqref{e2.4} and \eqref{e2.5} that \eqref{e2.14}
and \eqref{e2.15} hold. Also, by \eqref{e2.5} we have
\begin{equation}
\begin{aligned}
u'(t)&=-\int^{t_\alpha}_t u''(s)ds=\int^{t_\alpha}_t(\lambda u(s)+f(u(s)))ds\\
&\leq \int^{t_\alpha}_t(\lambda u(s)+\widehat{\kappa}^+ u(s)+M_0]ds\\
&\leq (|a^+|+\widehat{\kappa}^++1)\|u\|(t_\alpha-t), \ \ t\in (0, t_\alpha).
\end{aligned}\label{e2.17}
\end{equation}
Similarly, we have
\begin{equation}
-u'(t)\leq (|a^+|+\widehat{\kappa}^++1)\|u\| (t-t_\alpha), \forall t\in (t_\alpha, 1).\label{e2.18}
\end{equation}
Now \eqref{e2.16} follows from \eqref{e2.17} and \eqref{e2.18}.
The proof is complete.
\end{proof}

Recall that if $\{\Sigma_n\}$ is a sequence of sets then
\begin{align*}
\liminf _{n\to \infty}\Sigma_n=\big\{& x: \text{ there exists a positive integer
$N_0$  such that every}\\
&\text{neighborhood of $x$ intersects $\Sigma_n$ for }n\geq N_0\big\},
 \end{align*}
\begin{align*}
\limsup_{n\to \infty}\Sigma_n=\big\{& x:  \text{ every neighborhood of
$x$ intersects $\Sigma_n$ }\\
&\text{for infinitely many integers } n\big\}.
 \end{align*}

The proof of the next lemma can be found in \cite{w1}.

\begin{lemma} \label{lem2.9}
 Let $\{\Sigma_n\}$ be a sequence of connected sets in
a complete metric space $M$. Assume that
\begin{itemize}
\item[(i)] $\cup_{n=1}^\infty \Sigma_n$ is precompact in $M$;

\item[(ii)] $\liminf_{n\to \infty}\Sigma_n \neq\emptyset$.
\end{itemize}
Then $\limsup_{n\to \infty}\Sigma_n$ is non-empty, closed and
connected.
\end{lemma}

 \section{Oscillatory bifurcation from infinity}

Let $\zeta_0=\zeta^++\zeta^-$,
$\zeta=\frac{\zeta^+-\zeta^-}{2}$, $c_4=c_3(2\widehat{\kappa}^++\zeta_0)+4\zeta$,
 $$
c_5=\frac{\eta(1-\eta)\zeta^2}{64c_4}\Big(1-\frac{\zeta}{4c_4}\Big),
$$
and
$$
\vartheta=\min\Big\{\frac{\widetilde s_0 \eta(1-\eta)}{4000},
\frac{c_2c_5\widetilde s_0\eta(1-\eta)}{6400 c_4},\frac{c_5}{16(\pi+c_3)} \Big\}.
$$
Recall the definition of  $S_\alpha$  in section 2. Now we have the
following main result.

\begin{theorem} \label{thm3.1}
 Suppose that \eqref{e2.3} holds, and $\zeta^-<\zeta^+$. Then,
for $\alpha\in [0, \vartheta/2]$, $\bar S_\alpha$ possesses at least
one connected component $C_{\alpha,\infty}$ bifurcating from infinity and
oscillating over an interval $I_\alpha:=[d_-(\alpha), d_+(\alpha)]$, where
$$
d_-(\alpha)=\lambda_1(\alpha)-\frac{3}{4}c_5-\zeta_0, d_+(\alpha)
=\lambda_1(\alpha)+\frac{3}{4}c_5-\zeta_0.
$$
\end{theorem}

\begin{proof}
Take  $\tau_n\in (0,1)$  such that $\tau_n\to 1$ as $n\to \infty$. Let
\begin{gather*}
c_5^{(\tau_n)}=\frac{\tau_n(2-\tau_n)\eta(1-\eta)\zeta^2}{64c_3c_4}
\Big(1-\frac{(2-\tau_n)\zeta}{4c_4}\Big),\\
\vartheta^{(\tau_n)}=\min\Big\{\frac{\widetilde s_0 \eta(1-\eta)}{4000},
\frac{c_2c_5^{(\tau_n)}\widetilde s_0\eta(1-\eta)}{6400 c_4},
\frac{c_5^{(\tau_n)}}{16(\pi+c_3)} \Big\}.
\end{gather*}
Except for the last paragraph, in what follows of the proof we always assume
that $\alpha\in [0,\vartheta^{(\tau_n)}]$.

 By using the methods showing the main results in \cite{b1,o1,p1,s4}
we can prove that $\bar S_\alpha$ possesses at least one connected component
bifurcating from infinity. For brevity, we will omit the process.
 Let $\widetilde f(s)=f(s)-\zeta_0 s$, and
$\widetilde F(s)=F(s)-\frac{1}{2}\zeta_0 s^2$ for $s\geq 0$.  Then  we have
$$
\limsup_{s\to+\infty}\frac{\widetilde F(s)}{s^2}=\zeta,\quad
\liminf_{s\to+\infty}\frac{\widetilde F(s)}{s^2}=-\zeta.
$$
Thus, there exists two sequences of positive numbers
$\{\rho_n^+\}$ and $\{\rho_n^-\}$, with $\rho_n^+\to \infty$ and
$\rho_n^-\to \infty$, such that
\begin{equation}
\widetilde F(\rho_n^-)\geq \frac{\zeta}{2}(\rho_n^-)^2, \quad
\widetilde F(\rho_n^+)\leq -\frac{\zeta}{2}(\rho_n^+)^2.\label{e3.1}
\end{equation}
For $n=1,2,\dots$, let
\begin{gather*}
\varsigma^-_n=\inf\big\{s\geq 0: \widetilde F(s)\geq \frac{\zeta}{2}(\rho_n^-)^2\big\},\\
\varsigma^+_n=\inf\big\{s\geq 0: \widetilde F(s)\leq - \frac{\zeta}{2}(\rho_n^+)^2\big\}.
\end{gather*}
Assume without loss of generality that $\{\varsigma^+_n\}$ and $\{\varsigma^-_n\}$ are
strictly increasing, and $R_4^{(\tau_n)}\leq \varsigma_n^+<\varsigma_n^-<\varsigma_{n+1}^+$
for $n\in \mathbb{N}$, where
\begin{gather*}
R_4^{(\tau_n)}=R_3+3M_0(\widehat{\kappa}^+)^{-1}+32s_0(\eta(1-\eta))^{-1}
+1+1024s_0c_6(\eta(1-\eta)c_5^{(\tau_n)})^{-1},\\
c_6=\frac{ c_4[(|a^+|+|\widehat{\kappa}^+|)s_0+3M_0]}{2c_2^2},
\end{gather*}
 and $R_3$ is defined as in Lemma \ref{lem2.8}.
Let $(\lambda, u)\in \bar S_\alpha$, with $\|u\|=\varsigma^-_n$ and $\lambda\in [a^-,a^+]$.
Then, $(\lambda, u)$ satisfies \eqref{e2.2}.
Writing $Z(u;t)=\widetilde F(\|u\|)-\widetilde F(u(t))$ for $t\in [0,1]$. Obviously,
$Z(u;t)\geq 0$ for $t\in [0,1]$.
By a direct computation, we have
\begin{equation}
\int^1_0\widetilde f(u)\phi_\alpha dt=\int^1_0Z(u; t)\frac{u'\phi_\alpha'
-\phi_\alpha u''}{(u')^2}dt-Z(u;1)\frac{\phi_\alpha(1)}{u'(1)}.\label{e3.2}
\end{equation}
By \eqref{e2.5} and using the fact that $\|u\|\geq 3M_0(\widehat{\kappa}^+)^{-1}$, we have
\begin{equation}
|f(u(t))|\leq \widehat{\kappa} ^+u(t)+3M_0\leq 2\widehat{\kappa}^+\|u\|,\quad t\in [0,1].\label{e3.3}
\end{equation}
Let $t_\alpha\in [1/2, 1)$ be such that $u(t_\alpha)=\|u\|$, and
 $u'(t_\alpha)=0$.  It follows from Lemma \ref{lem2.5} that
$t_\alpha\in [\frac{1}{2}, \frac{1}{2}+\sigma_0(\alpha)]$.
 Since $Z'_t(u;t)=-\widetilde f(u(t))u'(t)$, by \eqref{e2.16} and \eqref{e3.3} we have
\begin{equation}
|Z_t'(u;t)|\leq c_4\|u\|^2|t-t_\alpha|, \quad \forall t\in [0,1],\label{e3.4}
\end{equation}
and so
\begin{align*}
|Z(u;t)|
&=\big|Z(u;t_\alpha)+\int^t_{t_\alpha} Z'_s(u; s)ds\big|\\
&\leq \int^t_{t_\alpha} |Z'_s(u; s)|ds\\
&\leq \frac{1}{2}c_4\|u\|^2(t-t_\alpha)^2
\end{align*}
for each $t\in [t_\alpha, 1]$. Similarly,  we also have
$$
|Z(u;t)|\leq \frac{1}{2}c_4\|u\|^2(t-t_\alpha)^2, \quad \forall t\in [0,t_\alpha].
$$
Thus, we have
\begin{equation}
|Z(u;t)|\leq \frac{1}{2}c_4\|u\|^2(t-t_\alpha)^2, \quad \forall t\in [0,1].\label{e3.5}
\end{equation}

Now we give estimates for each part of the right side of the equality \eqref{e3.2},
 and for $W(\alpha)$.

(1)\ Estimate for $Z(u;1)\frac{\phi_\alpha(1)}{u'(1)}$. It follows from \eqref{e3.5}
and Lemma \ref{lem2.7} that for $\alpha\in [0, \vartheta^{(\tau_n)}]$,
\begin{equation}
\big|Z(u;1)\frac{\phi_\alpha(1)}{u'(1)}\big|
\leq \frac{1}{2}c_4\|u\|^2(1-t_\alpha)^2\frac{\alpha\phi_\alpha(\eta)}{c_2\|u\|(1-t_\alpha)}
\leq \frac{ c_4\|u\|\alpha}{2c_2}\leq \frac{1}{16}c_5^{(\tau_n)}\|u\|.
\label{e3.6}
\end{equation}

(2) Estimate for $\int^1_0Z(u; t)\frac{-\phi_\alpha u''}{(u')^2}dt$.
 Since $Z(u;t)\geq 0$ for $t\in [0,1]$, and $-u''(t)\geq 0$ for
 $t\in E^+(u;s_0)$, we have
\begin{equation}
\begin{aligned}
\int^1_0Z(u; t)\frac{-\phi_\alpha u''}{(u')^2}dt
&=\Big(\int_{E^+(u;s_0)}+\int_{E^-(u;s_0)}\Big)Z(u; t)\frac{-\phi_\alpha u''}{(u')^2}dt\\
&\geq \int_{E^-(u;s_0)}Z(u; t)\frac{-\phi_\alpha u''}{(u')^2}dt.
\end{aligned} \label{e3.7}
\end{equation}
On the other hand, by \eqref{e3.5}, Lemmas \ref{lem2.7} and \ref{lem2.8} we have
\begin{align*}
&\big|\int_{E^-(u;s_0)}Z(u; t)\frac{-\phi_\alpha u''}{(u')^2}dt\big|\\
&\leq \int_{E^-(u;s_0)}|Z(u; t)|
 \big|\frac{\phi_\alpha u''}{(u')^2}\big|dt\\
&\leq \int_{E^-(u;s_0)}|Z(u; t)|\big|\frac{u''}{(u')^2}\big|dt\\
&\leq \int_{E^-(u;s_0)} \frac{1}{2}c_4\|u\|^2(t-t_\alpha)^2
 \frac{ |u''|}{c_2^2\|u\|^2(t-t_\alpha)^2}dt\\
&=\int_{E^-(u;s_0)} \frac{ c_4 |u''|}{2c_2^2}dt\\
&\leq \int_{E^-(u;s_0)} \frac{ c_4[(|a^+|+|\widehat{\kappa}^+|)s_0+3M_0]}{2c_2^2}dt\\
&\leq c_6\cdot\text{meas} (E^-(u;s_0)).
\end{align*}
from Lemma \ref{lem2.3}, it follows that for $\alpha\in [0,\vartheta^{(\tau_n)}]$,
$$
E^-(u;s_0)\subset\Big[0, \frac{1-\sqrt{1-\frac{32s_0}{\eta(1-\eta)\|u\|}}}{2}\Big]
\cup \Big[\frac{1+\sqrt{1-\frac{32s_0}{\eta(1-\eta)\|u\|}}}{2},1\Big].
$$
and so
\begin{equation}
\begin{aligned}
\big|\int_{E^-(u;s_0)}Z(u; t)\frac{-\phi_\alpha u''}{(u')^2}dt\big|
&\leq c_6\cdot\text{mes} (E^-(u;s_0))\\
&\leq \frac{64s_0c_6}{\eta(1-\eta)\|u\|}
\leq \frac{1}{16}c_5^{(\tau_n)}\|u\|.
\end{aligned}\label{e3.8}
\end{equation}
It follows from \eqref{e3.7} and \eqref{e3.8} that
\begin{equation}
\int^1_0Z(u; t)\frac{-\phi_\alpha u''}{(u')^2}dt
\geq -\frac{1}{16}c_5^{(\tau_n)}\|u\|. \label{e3.9}
\end{equation}


(3) Estimate for $\int^1_0Z(u; t)\frac{ \phi_\alpha'}{u'}dt$.
According to Lemmas \ref{lem2.5} and \ref{lem2.6}, there exist $t_\alpha, t'_\alpha\geq 1/2$
 such that $\|u\|=u(t_\alpha)$, $\|\phi_\alpha\|=\phi_\alpha(t'_\alpha)$, and
$0\leq t_\alpha-\frac{1}{2}\leq \sigma_0(\alpha)$,
$0\leq t'_\alpha-\frac{1}{2}\leq \sigma_1(\alpha)$.
Let $\sigma(\alpha)=\max\{\sigma_0(\alpha),\sigma_1(\alpha)\}$.  Obviously, we have
$\sigma(\alpha)=\sigma_0(\alpha)$. Now we have
\begin{equation}
\int^1_0Z(u; t)\frac{ \phi_\alpha'}{u'}dt=\Big(\int^{1/2}_0
+\int_{\frac{1}{2}}^{\frac{1}{2}+\sigma(\alpha)}
+\int_{\frac{1}{2}+\sigma(\alpha)}^1\Big)Z(u; t)\frac{ \phi_\alpha'}{u'}dt.\label{e3.10}
\end{equation}
It follows from Lemmas \ref{lem2.1} and \ref{lem2.4}
 that $\phi_\alpha'(t)<0$ and $u'(t)<0$ for
$t\in [\frac{1}{2}+\sigma(\alpha),1]$, and so
\begin{equation}
\int_{\frac{1}{2}+\sigma(\alpha)}^1 Z(u; t)\frac{ \phi_\alpha'}{u'}dt\geq 0.\label{e3.11}
\end{equation}
 From Lemma \ref{lem2.1} we have $\pi/2<|\sqrt{\lambda_1(\alpha)}|\leq \pi$, and so
$\pi/2\leq |\phi'_\alpha|\leq \pi$. It follows from \eqref{e3.5},
 Lemmas \ref{lem2.7} and \ref{lem2.8} that
\begin{equation}
\begin{aligned}
\big|\int_{\frac{1}{2}}^{\frac{1}{2}+\sigma(\alpha)}Z(u; t)\frac{\phi_\alpha'}{u'}dt\big|
&\leq\int_{\frac{1}{2}}^{\frac{1}{2}+\sigma(\alpha)} |Z(u; t)
\frac{\phi_\alpha'}{u'}|dt\\
&\leq \int_{\frac{1}{2}}^{\frac{1}{2}
 +\sigma(\alpha)}\frac{1}{2}c_4\|u\|^2(t-t_\alpha)^2\frac{\pi}{c_2\|u\|(t-t_\alpha)}dt\\
&\leq \frac{c_4\pi\|u\|}{2c_2}\sigma(\alpha)\\
&\leq \frac{2c_4\|u\|}{c_2}\sigma_0(\alpha)
 \leq \frac{1}{16}c_5^{(\tau_n)}\|u\|.
\end{aligned} \label{e3.12}
\end{equation}
Obviously,   by \eqref{e3.1} we have
\begin{equation}
Z(u;0)=\widetilde F(\|u\|)-\widetilde F(u(0))
=\widetilde F(\|u\|)=\frac{\zeta}{2}(\rho_n^-)^2\geq \frac{\zeta}{2}(\varsigma^-_n)^2
= \frac{\zeta}{2}\|u\|^2. \label{e3.13}
\end{equation}
By \eqref{e3.4} and \eqref{e3.13},
 for $t\in \Big[0, 1-\sqrt{1-\frac{(2-\tau_n)\zeta}{2c_4}}\Big]$, we have
\begin{equation}
\begin{aligned}
Z(u;t)&=Z(u;0)+\int^t_0Z'_s(u;s)ds\\
&\geq \frac{\zeta}{2}\|u\|^2-\int^t_0c_4\|u\|^2(t_\alpha-s)ds\\
&\geq \frac{\zeta}{2}\|u\|^2-\int^t_0c_4\|u\|^2(1-s)ds\\
&=\frac{\|u\|^2}{2}(\zeta+c_4t^2-2c_4t)
\geq \frac{\tau_n\zeta}{4}\|u\|^2.
\end{aligned}\label{e3.14}
\end{equation}
It follows from Lemmas \ref{lem2.1} and \ref{lem2.2} that
$\phi_\alpha=\lambda_1(\alpha)K\phi_\alpha\in Q_\alpha$, and so
$$
\phi_\alpha(t)\geq \|\phi_\alpha\|e_\alpha(t)= e_\alpha(t)\geq e(t), \quad
 \forall t\in [0,1].
$$
It follows from Lemmas \ref{lem2.1} and \ref{lem2.4} that $\phi_\alpha'(t)>0$ and
 $u'(t)>0$ for $t\in [0, 1/2]$. By Lemma \ref{lem2.7} and \eqref{e3.14} we have
\begin{equation}
\begin{aligned}
\int^{1/2}_0Z(u; t)\frac{\phi_\alpha'}{u'}dt
&=\Big(\int^{1-\sqrt{1-\frac{(2-\tau_n)\zeta}{2c_4}}}_0
+\int_{1-\sqrt{1-\frac{(2-\tau_n)\zeta}{2c_4}}}^{1/2}\Big)Z(u; t)
 \frac{ \phi_\alpha'}{u'}dt\\
&\geq \int^{1-\sqrt{1-\frac{(2-\tau_n)\zeta}{2c_4}}}_0Z(u; t)
 \frac{ \phi_\alpha'}{u'}dt\\
&\geq \int^{1-\sqrt{1-\frac{(2-\tau_n)\zeta}{2c_4}}}_0
 \frac{\tau_n\zeta}{4}\|u\|^2\frac{\phi_\alpha'}{c_3\|u\|}dt\\
&\geq \frac{\tau_n\zeta}{4c_3}\phi_\alpha
 \big(1-\sqrt{1-\frac{(2-\tau_n)\zeta}{2c_4}}\big)\|u\|\\
&\geq \frac{\tau_n\zeta}{4c_3}\phi_\alpha
 \big(\frac{(2-\tau_n)\zeta}{4c_4}\big)\|u\|\\
&\geq \frac{\tau_n\eta(1-\eta)\zeta}{64c_3}
 \frac{(2-\tau_n)\zeta}{c_4}\big(1-\frac{(2-\tau_n)\zeta}{4c_4}\big)\|u\|\\
&= c_5^{(\tau_n)}\|u\|.
\end{aligned} \label{e3.15}
\end{equation}

(4) Estimate for $W(\alpha)$. It follows from Lemma \ref{lem2.8} that
\begin{equation}
\begin{aligned}
|W(\alpha)|
&=\big|\alpha\big[u(\eta)\phi'_\alpha(1)-u'(1)\phi_\alpha(\eta)\big]\big|\\
&\leq \alpha\big(\|u\|\|\phi'_\alpha\|+c_3\|u\|(1-t_\alpha)\|\phi_\alpha\|\big)\\
&\leq \alpha\|u\|\big(\pi+c_3\big)\leq \frac{1}{16}c_5^{(\tau_n)}\|u\|.
\end{aligned} \label{e3.16}
\end{equation}

From Lemmas \ref{lem2.2} and \ref{lem2.3} it follows that
\begin{equation}
0<\frac{1}{2}\|u\|\int^1_0 e^2(t) dt
\leq \frac{1}{2}\|u\|\int^1_0e^2_\alpha(t) dt
\leq \int^1_0 u\phi_\alpha dt\leq \|u\|.\label{e3.17}
\end{equation}
From \eqref{e3.2}, \eqref{e3.6}, \eqref{e3.9}--\eqref{e3.12},
\eqref{e3.15}--\eqref{e3.17} it follows that
\begin{equation}
\int^1_0\widetilde f(u)\phi_\alpha dt-W(\alpha)
\geq \frac{3}{4}c_5^{(\tau_n)}\|u\|
\geq \frac{3}{4}c_5^{(\tau_n)}\int^1_0 u\phi_\alpha dt.\label{e3.18}
\end{equation}
 Thus, by \eqref{e2.2},\eqref{e3.17} and \eqref{e3.18}, we have
$$
\lambda\leq \lambda_1(\alpha)-\frac{3}{4}c_5^{(\tau_n)}-\zeta_0:=d^{(\tau_n)}_-(\alpha).
$$
Similarly, for each $(\lambda, u)\in \bar S_\alpha$ with
$\|u\|=\varsigma^+_n\geq R_4^{(\tau_n)}$ and $\lambda\in [a^-, a^+]$, we have
$$
\lambda\geq \lambda_1(\alpha)+\frac{3}{4}c_5^{(\tau_n)}-\zeta_0
:=d^{(\tau_n)}_+(\alpha).
$$

Assume that $\vartheta^{(\tau_n)}\to \vartheta$ as $n\to \infty$.
Obviously, $ \vartheta>0$.  Assume  without loss of generality that
$\vartheta^{(\tau_n)}\geq \frac{\vartheta}{2}$ for all $n\in \mathbb{N}$.
From the proof above we see that, for each $\alpha\in [0,\vartheta/2]$
and $n\in \mathbb{N}$, there exists at least one connected component
$C_{\alpha,\infty}^{(\tau_n)}$ of $\bar S_\alpha$ oscillating over the
interval $[d^{(\tau_n)}_-(\alpha), d^{(\tau_n)}_+(\alpha)]$.
Note $d^{(\tau_n)}_-(\alpha)\to d_-(\alpha)$ and
$d^{(\tau_n)}_+(\alpha)\to d_+(\alpha)$ as $n\to \infty$.
Applying Lemma \ref{lem2.9}, we see that for each $\alpha\in [0,\vartheta/2]$,
there exists at least one connected component $C_{\alpha,\infty}$ of
$\bar S_\alpha$ such that $C_{\alpha,\infty}$ contains
$\limsup_{n\to \infty}C_{\alpha,\infty}^{(\tau_n)}$  and oscillates
over the interval $[d_-(\alpha), d_+(\alpha)]$. The proof is complete.
\end{proof}

\begin{corollary} \label{coro3.1}
 Suppose that all condition of Theorem \ref{thm3.1} hold. Then for
$\alpha\in [0,\vartheta/2]$ and $\lambda\in [d_-(\alpha), d_+(\alpha)]$,
\eqref{e1.3} has infinitely many solutions.
\end{corollary}

\subsection*{Acknowledgments}
This research was supported by NSFC 11501260.
The authors would like to  thank  Prof. K. Schmitt for his many helpful
suggestions about improving the article.

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\end{document}