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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 141, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/141\hfil Bifurcation from infinity]
{Bifurcation from infinity and multiple solutions
for first-order periodic boundary-value problems}

\author[Z. Wang, C. Gao \hfil EJDE-2011/141\hfilneg]
{Zhenyan Wang, Chenghua Gao} 

\address{Zhenyan Wang \newline
Department of Mathematics, Northwest Normal University,
Lanzhou 730070, China}
\email{wangzhenyan86714@163.com}

\address{Chenghua Gao \newline
Department of Mathematics, Northwest Normal University,
Lanzhou 730070, China}
\email{gaokuguo@163.com}


\thanks{Submitted September 6, 2011. Published October 28, 2011.}
\subjclass[2000]{34B18}
\keywords{First-order periodic problems;
 Landsman-Lazer type condition; \hfill\break\indent
Leray-Schauder degree; bifurcation; existence}

\begin{abstract}
 In this article, we study the existence and multiplicity of
 solutions for the first-order periodic boundary-value problem
 \begin{gather*}
 u'(t)-a(t)u(t)=\lambda u(t)+g(u(t))-h(t), \quad  t\in (0, T),\\
 u(0)=u(T).
 \end{gather*}
\end{abstract}

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


\section{Introduction}

The first-order periodic differential equation
$$
u'(t)=a(t)u(t)-f(u(t-\tau(t)))
$$
has been proposed as models for a variety of physiological processes
and conditions including production of blood cells, respiration, and
cardiac arrhythmias, see \cite{GBN,MG,WL}.
Thus, the existence of periodic solutions of this periodic
differential equation has been discussed by several authors;
see for example \cite{AC,CZ,JW,MCC,M2,MS,PS1,PS2,WJX,W1,ZC}
and the references therein.

In these articles, the condition $\int_0^Ta(t)dt\neq0$ is
used for showing the existence of solutions. A natural question
is what would happen if $\int_0^Ta(t)dt=0$. It is easy to check that
if $\int_0^Ta(t)dt=0$, then the equation
$$
-u'(t)+a(t)u(t)=0, \quad u(0)=u(T)
$$
has nontrivial solutions. Thus, the operator $Lu=-u'(t)+a(t)u(t)$ is
not invertible.

In this article, using Leray-Schauder degree and bifurcation
techniques and under the condition that  $\int_0^Ta(t)dt=0$, we
discuss the existence and multiplicity of solutions for the
problem
\begin{gather}
u'(t)-a(t)u(t)=\lambda u(t)+g(u(t))-h(t), \quad
 t\in (0,T),\label{e1.1}\\
u(0)=u(T),\label{e1.2}
\end{gather}
where
$g:\mathbb{R}\to\mathbb{R}$ is continuous,
$h\in L^1(0, T)$, and the parameter $\lambda$ is close to $0$ which is
the eigenvalue of
$$
-u'(t)+a(t)u(t)=\lambda u(t), \quad u(0)=u(T).
$$

In this article, we use the following assumptions:
\begin{itemize}

\item[(H1)] $a(\cdot)\in C[0, T]$ and $\int^T_0a(t)dt=0$;

\item[(H2)] $g:\mathbb{R}\to\mathbb{R}$ is continuous, and there exist
$\alpha\in[0,1)$, $p,q\in(0,\infty)$, such that
$$
|g(u)|\leq p|u|^\alpha+q,\quad u\in\mathbb{R};
$$

\item[(H3)] There exist constants $A, a, R, r$ such that $r<0<R$ and
\begin{gather*}
g(u)\geq A,\quad\text{for  all } u\geq R,\\
g(u)\leq a,\quad\text{for all }u\leq r;
\end{gather*}

\item[(H3')] There exist constants $A, a, R, r$ such that $r<0<R$ and
\begin{gather*}
g(u)\leq A, \quad\text{for all }u\geq R,\\
g(u)\geq a,\quad\text{for all }u\leq r.
\end{gather*}


\item[(H4)]
$$
\int_0^T\frac{g^{-\infty}}{\psi(s)}ds
<\int_0^T\frac{h(s)}{\psi(s)}ds<\int_0^T\frac{g_{+\infty}}{\psi(s)}ds,
$$
where
$$
g^{-\infty}=\limsup_{s\to-\infty}g(s),\quad
g_{+\infty}=\liminf_{s\to+\infty}g(s),
$$
and $\psi(t)=e^{\int_0^ta(s)ds}$ is the solution of
$$
-u'+a(t)u=0, \quad u(0)=u(T).
$$

\item[(H4')]
$$
\int_0^T\frac{g^{+\infty}}{\psi(s)}ds
<\int_0^T\frac{h(s)}{\psi(s)}ds<\int_0^T\frac{g_{-\infty}}{\psi(s)}ds,
$$
where
$$
g^{+\infty}=\limsup_{s\to+\infty}g(s),\quad
g_{-\infty}=\liminf_{s\to-\infty}g(s).
$$
\end{itemize}

Our main results are as follows.

\begin{theorem} \label{thm1.1}
 Assume that {\rm (H1)--(H4)} hold. Then
there exists $\lambda_+,\lambda_-$ with
$\lambda_+>0>\lambda_-$ such that
\begin{itemize}
\item[(i)] \eqref{e1.1}, \eqref{e1.2} has at least one solution if
$\lambda\in[0,\lambda_+]$;

\item[(ii)] \eqref{e1.1}, \eqref{e1.2} has at least three solutions if
$\lambda\in[\lambda_-,0)$.

\end{itemize}
\end{theorem}


\begin{theorem} \label{thm1.2}
 Assume that {\rm (H1), (H2), (H3'),(H4')} hold. Then  there exists
$\lambda_+, \lambda_-$ with $\lambda_+>0>\lambda_-$ such that
\begin{itemize}
\item[(i)] \eqref{e1.1}, \eqref{e1.2} has at least one solution if
$\lambda\in[\lambda_-,0]$;

\item[(ii)] \eqref{e1.1}, \eqref{e1.2} has at least three solutions if
$\lambda\in(0,\lambda_+]$.

\end{itemize}
\end{theorem}

The rest of the paper is arranged as follows.
In section 2, we discuss the Lyapunov-Schmidt procedure
for \eqref{e1.1}, \eqref{e1.2}.  In
section 3, the existence of solutions of \eqref{e1.1}, \eqref{e1.2}
is discussed under `Landesman-Lazer' type conditions.

\section{Lyapunov-Schmidt procedure}

Let $X,Y$ be the Banach spaces $C[0, T]$, $L^1[0,T]$ with the norm
$\|x\|=\max\{|x(t)|:t\in [0,T]\}$,
$\|u\|_1=\int_0^T|u(s)|ds$, respectively. Define linear
operator $L: D(L)\subset X\to Y$ by
\begin{equation}
Lu=-u'+a(t)u, u\in D(L),\label{e2.1}
\end{equation}
where $D(L)=\{u\in W^{1,1}(0, T):u(0)=u(T)\}$. Let
$N: X\to X$ be the nonlinear operator defined by
\begin{equation}
(Nu)(t)=g(u(t)), \quad t\in[0, T],\; u\in D(L).\label{e2.2}
\end{equation}
It is easy to see that $N$ is continuous. Note that
\eqref{e1.1}, \eqref{e1.2} is equivalent to
\begin{equation}
Lu+\lambda u+Nu=h, u\in D(L).\label{e2.3}
\end{equation}



\begin{lemma} \label{lem2.1}
Let $L$ be defined by \eqref{e2.1}. Then
\begin{gather*}
\ker  L=\{x \in X:  x(t)=c\psi(t):c\in\mathbb{R}\},\\
\operatorname{Im} L=\{y\in Y:\int_0^T\frac{y(s)}{\psi(s)}ds=0\}.
\end{gather*}
\end{lemma}

\begin{proof}  It is easy to see that
$\ker L=\{c\psi(t):c\in\mathbb{R}\}$. The following will prove
that $\operatorname{Im}L=\{y\in Y:\int_0^T\frac{y(s)}{\psi(s)}ds=0\}$.

If $y\in\operatorname{Im}L$, then there exists $u\in D(L)$ such that
$-u'(t)+a(t)u(t)=y(t)$. So
$$
u(t)=u(0)\psi(t)-\int_0^ty(s)e^{\int_s^ta(\tau)d\tau}ds.
$$
Combining with $u(0)=u(T)$, we have
$$
\int_0^T\frac{y(s)}{\psi(s)}ds=0.
$$
On the other hand, if $y\in Y$ satisfies
$\int_0^T\frac{y(s)}{\psi(s)}ds=0$, then we set
$$
u(t):=-\int_0^ty(s)e^{\int_s^ta(\tau)d\tau}ds.
$$
It is not difficult to prove that $x\in D(L)$ and $Lu=y$.
\end{proof}

Define operator $P: X\to \ker L$,
\begin{equation}
(Pu)(t)=u(0)\psi(t), \quad u\in X. \label{e2.4}
\end{equation}
Let $Q: Y\to Y$ be such that
\begin{equation}
(Qy)(t)=\frac{1}{T}\psi(t)\int_0^T\frac{y(s)}{\psi(s)}ds.\label{e2.5}
\end{equation}
Denote $X_1=\{u\in X:u(0)=0\}$.


\begin{lemma} \label{lem2.2}
Let operators $P$ and $Q$ be defined by \eqref{e2.4} and \eqref{e2.5}.
Then
$$
X=X_1\oplus \ker L, \quad
Y=\operatorname{Im}L\oplus\operatorname{Im}Q.
$$
We define linear operator $K: \operatorname{Im}L\to D(L)\cap X_1$
\begin{equation}
(Ky)(t)=-\int_0^ty(s)e^{\int_s^ta(\tau)d\tau}ds, \quad
y\in \operatorname{Im}L,\label{e2.6}
\end{equation}
satisfying $K=L_p^{-1}$, where $L_p=L|_{D(L)\cap X_1}$.
\end{lemma}

\begin{proof}
 Let $y_1(t)=y(t)-(Qy)(t), y\in Y$, then it
is easy to verify that $y_1 \in\operatorname{Im}L$.
Thus $Y=\operatorname{Im}L + \operatorname{Im}Q$.
Also $\operatorname{Im}L\cap\operatorname{Im}Q= \{0\}$. Hence
$Y=\operatorname{Im}L\oplus\operatorname{Im}Q$.
If $u\in D(L)\cap X_1$, then
$$
(KL_pu)(t)=K\big(-u'(t)+a(t)u(t)\big)=u(t).
$$
If $y\in \operatorname{Im}L$, then
$$
(L_pKy)(t)=-\big(-\int_0^ty(s)e^{\int_s^ta(\tau)d\tau}ds\big)'-a(t)\int_0^ty(s)e^{\int_s^ta(\tau)d\tau}ds=y(t).
$$
This indicates $K=L_p^{-1}$.
\end{proof}

Therefore, for every $u\in X$ , we have a unique decomposition
$u(t) = \rho\psi(t) + v(t), t\in[0, T]$, where
$\rho\in\mathbb{R}, v\in X_1$. Similarly, for every $h\in Y$,
we have unique decomposition
$h(t)=\tau\psi(t)+\bar{h}(t), t\in[0, T]$, where
$\tau\in\mathbb{R}, \bar{h}\in \operatorname{Im}L$.
The operator $Q, K$ be defined as \eqref{e2.5}, \eqref{e2.6}.
Then $K(I-Q)N: X\to X$  is completely continuous, and \eqref{e2.3}
is equivalent to the system
\begin{gather}
v(t)+\lambda Kv(t)+K(I-Q)N(\rho\psi(t)+v(t))=K\bar{h}(t),\label{e2.7}\\
\lambda\rho\psi(t)+QN(\rho\psi(t)+v(t))=\tau\psi(t).\label{e2.8}
\end{gather}

\begin{lemma}[\cite{IN}] \label{lem2.3}
Assume that {\rm (H2), (H3)} hold. Then for each real number
$s > 0$, there exists a decomposition $g(u)=q_s(u) + g_s(u)$
of $g$ by $q_s$ and $g_s$ satisfying the conditions:
\begin{gather}
uq_s(u)\geq 0, u\in \mathbb{R},\label{e2.9} \\
|q_s(u)|\leq p|u|+q+s, u\geq 1,\label{e2.10},
\end{gather}
there exists $\sigma_s$ depending on $a, A$ and $g$ such that
\begin{equation}
|g_s(u)| \leq\sigma_s, u\in \mathbb{R}.\label{e2.11}
\end{equation}
\end{lemma}

\begin{lemma} \label{lem2.4}
 Assume that {\rm (H1)--(H4)} hold, and
$\lambda$ satisfies
\begin{equation}
0\leq\lambda\leq\eta_1:=\frac{1}{2\|K\|_{\operatorname{Im}L\to
X_1}}.\label{e2.12}
\end{equation}
Then there exists constant $R_0>0$ such that any solution $u$ of
\eqref{e1.1} \eqref{e1.2} satisfies
$\|u\|<R_0$.
\end{lemma}


\begin{proof}
 We divide the proof into several steps.

\textbf{Step 1.} By  assumption (H2), there exists a constant $b$ such
that
$$
|g(u)|\leq p|u|+b, u\in \mathbb{R},
$$
where $p=\eta_1/4$. Using Lemma \ref{lem2.3} with $s = 1$,
\eqref{e1.1}, \eqref{e1.2} is equivalent to
\begin{equation}
u'(t)-a(t)u(t)=\lambda
u(t)+g_1(u(t))+q_1(u(t))-h(t), t\in[0,T], u\in D(L),\label{e2.13}
\end{equation}
where $q_1$ and $g_1$ satisfying conditions \eqref{e2.9} and
\eqref{e2.11}.
Moreover, by \eqref{e2.10},
\begin{equation}
|q_1(u)|\leq p|u|+b+1.\label{e2.14}
\end{equation}
Let $ \bar{\delta}> 0$ and choose $B\in\mathbb{R}$ such that
\begin{equation}
(b+1)|\frac{1}{u}|\leq\frac{1}{4}\bar{\delta}\label{e2.15}
\end{equation}
for all $u\in\mathbb{R}$ with $|u|\geq B$. It follows from
\eqref{e2.14} and \eqref{e2.15} that
\begin{equation}
0\leq q_1(u)u^{-1}\leq p+\frac{1}{4}\bar{\delta}\label{e2.16}
\end{equation}
for all $u\in\mathbb{R}$ with $|u|\geq B$.


\textbf{Step 2.} Let us define $\gamma:\mathbb{R}\to\mathbb{R}$  by
\begin{equation}
\gamma(u)= \begin{cases}
u^{-1}q_1(u), &|u|\geq B;\\
B^{-1}q_1(B)(\frac{u}{B})+(1-\frac{u}{B})p, &0\leq u<B;\\
B^{-1}q_1(-B)(\frac{u}{B})+(1+\frac{u}{B})p, &-B< u\leq 0.
\end{cases}\label{e2.17}
\end{equation}
It is easy to see that $\gamma$ is continuous. Moreover,
 by \eqref{e2.16}, one has
\begin{equation}
0\leq\gamma(u)\leq p+\frac{1}{4}\bar{\delta}\label{e2.18}
\end{equation}
for all $u\in\mathbb{R}$.
Defining $f:\mathbb{R}\to\mathbb{R}$ by
\begin{equation}
f(u) = g_1(u) + q_1(u)-\gamma(u)u,\label{e2.19}
\end{equation}
it follows from \eqref{e2.16} that for some $\sigma\in\mathbb{R}$,
\begin{equation}
|f(u)|\leq\sigma\label{e2.20}
\end{equation}
for all $u\in\mathbb{R}$, where $\sigma$ depends only on $p$ and
$h$. Finally,  \eqref{e2.13} is equivalent to
$$
u'(t)-a(t)u(t)=\lambda
u(t)+f(u(t))+\gamma(u(t))u(t)-h(t), t\in[0,T], u\in D(L).
$$


\textbf{Step 3.} It is to see that
$(L+\lambda I)|_{X_1\cap D(L)}:X_1\to\operatorname{Im}L$
is invertible. From \eqref{e2.12},
\begin{align*}
\|(L+\lambda I)|^{-1}_{X_1\cap D(L)}\|_{\operatorname{Im}L\to X_1}
&= \|L^{-1}|_{X_1\cap D(L)}(I+\lambda
K)^{-1}\|_{\operatorname{Im}L\to X_1}\\
&= \|K\|_{\operatorname{Im}L\to X_1}\|(I+\lambda
K)^{-1}\|_{\operatorname{Im}L\to X_1}\\
&\leq 2\|K\|_{\operatorname{Im}L\to X_1}.
\end{align*}
Let $u=\rho\psi(t)+v$ be a solution of \eqref{e2.13}, where
$\rho\in\mathbb{R}, v\in X_1$. Then from \eqref{e2.7},
\begin{align*}
\|v\|
&=  \|(L+\lambda I)|^{-1}_{X_1\cap D(L)}(I-Q)
 (\bar{h}-g(\rho\psi(t)+v(t)))\|\\
&\leq \|(L+\lambda I)|^{-1}_{X_1\cap D(L)}\|_{\operatorname{Im}L\to X_1}
 \|(I-Q)\|_{Y\to\operatorname{Im}L}[\|\bar{h}\|_1+p(|\rho|\cdot\|\psi\|+\|v\|)^\alpha+q]\\
&\leq
 2\|K\|_{\operatorname{Im}L\to X_1}\|(I-Q)\|_{Y\to\operatorname{Im}L}[\|\bar{h}\|_1+p(|\rho|\cdot\|\psi\|+\|v\|)^\alpha+q]\\
&= 2\|K\|_{\operatorname{Im}L\to X_1}\|(I-Q)\|_{Y\to\operatorname{Im}L}[\|\bar{h}\|_1+p(|\rho|\cdot\|\psi\|)^\alpha(1+\frac{\|v\|}{|\rho|\cdot\|\psi\|})^\alpha+q]\\
&\leq
2\|K\|_{\operatorname{Im}L\to X_1}\|(I-Q)\|_{Y\to\operatorname{Im}L}[\|\bar{h}\|_1+p(|\rho|\cdot\|\psi\|)^\alpha(1+\frac{\alpha\|v\|}{|\rho|\cdot\|\psi\|})+q]\\
&=  2\|K\|_{\operatorname{Im}L\to
X_1}\|(I-Q)\|_{Y\to\operatorname{Im}L}[\|\bar{h}\|_1
+p(|\rho|\cdot\|\psi\|)^\alpha\\
&\quad\times\big(1+\frac{\alpha}{(|\rho|\cdot\|\psi\|)^{1-\alpha}}
\cdot\frac{\|v\|}{(|\rho|\cdot\|\psi\|)^\alpha}\big)+q].
\end{align*}
Therefore,
$$
\frac{\|v\|}{(|\rho|\cdot\|\psi\|)^\alpha}\leq\frac{c_0}{(|\rho|\cdot\|\psi\|)^\alpha}+c_1+\frac{\alpha
c_1}{(|\rho|\cdot\|\psi\|)^{1-\alpha}}\cdot\frac{\|v\|}{(|\rho|\cdot\|
\psi\|)^\alpha},
$$
where
\begin{gather*}
c_0=2\|K\|_{\operatorname{Im}L\to
X_1}\|(I-Q)\|_{Y\to\operatorname{Im}L}(\|\bar{h}\|_1+q),\\
 c_1=2p\|K\|_{\operatorname{Im}L\to
X_1}\|(I-Q)\|_{Y\to\operatorname{Im}L}.
\end{gather*}
If
$$
|\rho|\geq \frac{(2\alpha
c_1)^{\frac{1}{1-\alpha}}}{\|\psi\|}:=\tilde{c},
$$
then
\begin{equation}
\frac{\|v\|}{(|\rho|\cdot\|\psi\|)^\alpha}
\leq\frac{2c_0}{(\tilde{c}\|\psi\|)^\alpha}+2c_1:=\bar{c}.\label{e2.21}
\end{equation}

\textbf{Step 4.}
 If we now assume that the conclusion of the lemma is false,
we obtain a sequence
$\{\lambda_n\}:0\leq\lambda_n\leq\eta_1, \lambda_n\to 0$ and
a sequence
$\{u_n\}:u_n=\rho_n\psi(t)+v_n, \rho_n\in\mathbb{R}, v_n\in X_1$
with $\|u_n\|\to\infty$ such that
\begin{equation}
\lambda_n\rho_n\psi(t)+Qg(\rho_n\psi(t)+v_n(t))=\tau\psi(t).
\label{e2.22}
\end{equation}
It follows immediately from \eqref{e2.21} that
\begin{equation}
|\rho_n|\to\infty, \|v_n\|(|\rho_n|\cdot\|\psi\|)^{-1}\to0,
\quad n\to\infty.\label{e2.23}
\end{equation}
So we infer that there exists sufficiently large $n_0\in \mathbb{N}$
such that for $n\geq n_0$
\begin{equation}
|v_n(t)|(|\rho_n|\psi(t))^{-1}\leq 1, \quad  t\in[0, T].\label{e2.24}
\end{equation}
Without loss of generality, let $\rho_n\to+\infty$  if
$n\to+\infty$ (the other case be proved by similar
method), then there exists sufficiently large $n_0\in\mathbb{N}$. If
$n\geq n_0$, $\lambda_n\rho_n\geq 0$; thus
\begin{equation}
\begin{gathered}
\tau-\frac{1}{T}\int_0^T\frac{g(\rho_n\psi(s)+v_n(s))}{\psi(s)}ds
\geq 0, \\
\tau\geq\frac{1}{T}\liminf_{n\to\infty}
\int_0^T\frac{g(\rho_n\psi(s)+v_n(s))}{\psi(s)}ds.
\end{gathered}\label{e2.25}
\end{equation}

To apply  Fatou's lemma to  \eqref{e2.25}, we need
a function $\hat{K}\in L^1[0,T]$ such that for
$s\in[0,T], \frac{g(u_n(s))}{\psi(s)}\geq\hat{K}(s)$. Indeed, from
the relation \eqref{e2.24}, one has that there exists nonnegative
function $k_1\in L^1[0,T]$ such that for $n\geq n_0$,
$$
|v_n(t)|(\rho_n\psi(t))^{-1}\leq k_1(t),\quad  t\in[0,T],
$$
and for every $s\in[0,T]$,
\begin{align*}
\gamma(u_n(s))u_n(s)+f(u_n(s))
&= \gamma(u_n(s))(\rho_n\psi(s)+v_n(s))+f(u_n(s))\\
&\geq \gamma(u_n(s))\frac{\rho_n\psi(s)+v_n(s)}{|\rho_n|\psi(s)}
 +f(u_n(s))\\
&\geq \gamma(u_n(s))(1-k_1(s))-|f(u_n(s))|\\
&\geq -(p+\frac{1}{4}\bar{\delta})(1-k_1(s))-\sigma:=\hat{K}(s).
\end{align*}
It follows from $\psi(s)>0$ that
$$
\frac{1}{\psi(s)}g(\rho_n\psi(s)+v_n(s))\geq
\frac{1}{\psi(s)}\hat{K}(s), s\in[0,T].
$$
Thus, applying Fatou's lemma to \eqref{e2.25}, we have
\begin{align*}
\tau&\geq \frac{1}{T}\liminf_{n\to\infty}\int_0^T
 \frac{g(\rho_n\psi(s)+v_n(s))}{\psi(s)}ds\\
&\geq \frac{1}{T}\int_0^T\liminf_{n\to\infty}
 \frac{g(\rho_n\psi(s)+v_n(s))}{\psi(s)}ds\\
&\geq \frac{1}{T}\int_0^T\frac{g_{+\infty}}{\psi(s)}ds.
\end{align*}
This contradicts with (H4).
\end{proof}

\begin{lemma} \label{lem2.4'}
 Assume that {\rm (H1), (H2), (H3'), (H4')} hold, and
$\lambda$ satisfies
$$
0\leq\lambda\leq\eta_1:=\frac{1}{2\|K\|_{\operatorname{Im}L\to
X_1}}.
$$
Then there exists constant $R_0>0$ such that any solution $u$ of
\eqref{e1.1} \eqref{e1.2} satisfy
$\|u\|<R_0$.
\end{lemma}

\section{The Proof of the Main Result}

\begin{lemma} \label{lem3.1}
 Assume that {\rm (H1)--(H4)} hold. Then there exists
$R_1: R_1\geq R_0$ such that for $0\leq\lambda\leq\delta$,
and $R\geq R_1$ one has
$$
\deg(L+\lambda I+N-h, B(R), 0)=\deg(L+\delta I, B(R), 0)=\pm 1,
$$
where $B(R)=\{u\in C[0, T]: \|u\|<R\}$, and the $\deg$ denotes
Leray-Schauder degree when $\lambda\neq 0$ and coincidence degree
when $\lambda=0$. Then \eqref{e1.1},\eqref{e1.2} has a
solution in $\bar{B}(R)$
for $0\leq\lambda\leq\delta$.
\end{lemma}

\begin{proof} From Lemma \ref{lem2.4} and the definition of $L$, if
$\lambda\in[0, \delta]$,
$$
\deg(L+\delta I, B(R), 0)
$$
is defined and depends on $\lambda$. Let
$(\mu, u)\in[0, 1]\times X$ be a solution of \eqref{e2.3}. Then
$$
Lu+\delta u+\mu(Nu-h)=0.
$$
So
\[
\|u\|= \mu\|(L+\delta)^{-1}(h-Nu)\|
\leq  \|(L+\delta)^{-1}\|_{Y\to
X}(\|h\|_1+p\|u\|^\alpha+q).
\]
Therefore there exists $R'_0>0$ such that
$\|u\|<R'_0$.
Choosing $R_1=\max\{R'_0, R_0\}$, then for arbitrary $R>R_1$,
$$
\deg(L+\lambda I+N-h, B(R), 0)=\deg(L+\delta I, B(R), 0)=\pm 1.
$$
\end{proof}

\begin{lemma} \label{lem3.1'}
 Assume that {\rm (H1), (H2), (H3'),(H4')} hold. Then
there exists $R_1: R_1\geq R_0$
such that for $0\leq\lambda\leq\delta$, and $R\geq R_1$ one has
$$
\deg(L+\lambda I+N-h, B(R), 0)=\deg(L+\delta I, B(R), 0)=\pm 1,
$$
where $B(R)=\{u\in C[0, T]: \|u\|<R\}$.
\end{lemma}

\begin{lemma} \label{lem3.2}
 Assume that {\rm (H1)--(H4)} hold. Then there exists
$\mu\geq 0$ such that for $-\mu\leq\lambda\leq 0$ one has
$$
\deg(L+\lambda I+N-h, B(R), 0)=\deg(L+\delta I, B(R), 0)=\pm 1,
$$
where $R$ be defined in Lemma \ref{lem3.1}.
Then \eqref{e1.1},\eqref{e1.2} has a solution
in $B(R)$ for $-\mu\leq\lambda\leq\delta$.
\end{lemma}


\begin{proof}
 Let
$$
\tau_0=\inf_{u\in\partial B(R)\cap X}\|Lu+Nu-h\|.
$$
It is easy to verify that $\tau_0>0$. Choosing sufficiently small
$\mu>0$ such that $\mu R<\tau_0$, then if $\lambda\in[-\mu, \mu]$,
$$
\deg(L+\lambda I+N-h, B(R), 0)=\deg(L+N-h, B(R), 0).
$$
Combined with Lemma \ref{lem3.1}, the result can be proved. That is to see
that if $\lambda\in[-\mu, \delta]$, \eqref{e2.3} has at least
one solution in $\bar{B}(R)$.
\end{proof}

\begin{lemma} \label{lem3.2'}
Assume that {\rm (H1), (H2), (H3')(H4')} hold.
Then there exists $\mu\geq 0$ such
that for $-\mu\leq\lambda\leq 0$, one has
$$
\deg(L+\lambda I+N-h, B(R), 0)=\deg(L+\delta I, B(R), 0)=\pm 1,
$$
where $R$ be defined in Lemma \ref{lem3.1}.
Then \eqref{e1.1}, \eqref{e1.2} has a solution
in $B(R)$ for $-\mu\leq\lambda\leq\delta$.
\end{lemma}

\begin{remark} \label{rmk1} \rm
Since $g$ is L-completely
continuous and satisfies (H2) and since $\lambda=0$ is a simple
eigenvalue of L, it follows from bifurcation results of
\cite{IN} that
there exist two connected sets
$\mathcal{C}_+, \mathcal{C}_-\subset
\mathbb{R}\times X$ of solutions of \eqref{e1.1}, \eqref{e1.2}
such that for all sufficiently small $\epsilon>0$,
$$
\mathcal{C}_+\cap U_\epsilon\neq\emptyset,\quad
\mathcal{C}_-\cap U_\epsilon\neq\emptyset,
$$
where $U_\epsilon:=\{(\lambda, u)\in\mathbb{R}\times
X, |\lambda|<\epsilon, \|u\|>1/\epsilon\}$.
\end{remark}


\begin{proof}[Proof of Theorem \ref{thm1.1}]
Set $\lambda^+=\delta$, then it follows from Lemma \ref{lem3.1} and
Lemma \ref{lem3.2} that \eqref{e1.1}, \eqref{e1.2} has at
least one solution in $B(R)$ for $\lambda\in[-\mu, \lambda_+]$. On
the other hand,  Remark \ref{rmk1} shows that there exists two connected sets
$\mathcal{C}+$ and $\mathcal{C}-$ of solutions of \eqref{e1.1},
\eqref{e1.2}
bifurcating from infinity at $\lambda=0$. Hence by Lemma \ref{lem2.4}, the
connected sets $\mathcal{C}+$ and $\mathcal{C}-$ of Remark
\ref{rmk1} must satisfy
$$
\mathcal{C}_+, \mathcal{C}_-\subset\{(\lambda, u):
\|u\|\geq1/\epsilon, -\mu<\lambda<0\}.
$$
and hence, if $1/\epsilon \geq R$; i.e.,
$\epsilon\leq 1/k$. Choosing
$\lambda_{-}=\max\{-\mu, -1/k\}$,  we obtain two solutions
$u_1, u_2: u_1\in\mathcal{C}_+, u_2\in\mathcal{C}_-$, and
$\|u_i\|\geq R$ ($i=1, 2$).
\end{proof}

Theorem \ref{thm1.2} can be proved by a similar method.


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\end{document}