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\markboth{\hfil Existence of positive periodic solutions \hfil EJDE--2001/59}
{EJDE--2001/59\hfil Sui Sun Cheng \& Guang Zhang \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No. 59, pp. 1--8. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  Existence of positive periodic solutions for non-autonomous functional
  differential equations
 %
\thanks{ {\em Mathematics Subject Classifications:} 34B15.
\hfil\break\indent
{\em Key words:} positive periodic solutions, cone, delay.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted May 16, 2001. Published September 4, 2001. \hfil\break\indent
G. Zhang  was partially supported by the Natural Science Foundation of Shanxi
and by the \hfill\break\indent  Yanbei Normal Institute.} }
\date{}
%
\author{Sui Sun Cheng \& Guang Zhang}
\maketitle

\begin{abstract}
We establish the existence of positive periodic solutions for a
first-order differential equation with periodic delay. For this
purpose, we use the fixed point theorem proved by Krasnoselskii.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
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\newtheorem{corollary}[theorem]{Corollary}
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\section{Introduction}

In this article, we investigate the existence of positive periodic solutions
for the first-order functional differential equation
\begin{equation}
y'(t)=-a(t)y(t)+\lambda h(t)f(y(t-\tau (t)),  \label{1}
\end{equation}
where $a=a(t)$, $h=h(t)$ and $\tau =\tau (t)$ are continuous $T$-periodic
functions. We assume that $T,\lambda >0$, that $a=a(t)$, $f=f(t)$
and $h=h(t)$ are nonnegative, and that $a(t_{0})>0$ for some $t_{0}\in [0,T]$.

Functional differential equations with periodic delays appear in a number of
ecological models. In particular, our equation can be interpreted as the
standard Malthus population model $y'=-a(t)y$ subject to a
perturbation with periodical delay. One important question is whether these
equations can support positive periodic solutions. Such question has been
studied extensively by a number of authors; see for example
\cite{l1,j1,f1,f2,z1} and the
references therein. In this paper, we will obtain existence criteria for
$T$-periodic solutions of (\ref{1}) by means of a well known fixed
point theorem due to Krasnoselskii.

\begin{theorem} \label{thmA}
Let $E$ be a Banach space and let $P\subset E$ be a
cone. Assume $\Omega _1,\Omega _2$ are bounded open subsets of $E$ such
that $0\in \Omega _1\subset \overline{\Omega }_1\subset \Omega _2$.
Suppose that $T:P\cap ( \overline{\Omega }_2\backslash \Omega_1)
\to P$ is a completely continuous operator such that
\begin{enumerate}
\item $\| Tu\| \leq \| u\| $ for $u\in P\cap \partial \Omega
_1$ and $\| Tu\| \geq \| u\| $ for $P\cap \partial
\Omega _2$, or that
\item $\| Tu\| \geq \| u\| $ for $u\in
P\cap \partial \Omega _1$ and $\| Tu\| \leq \| u\| $
for $P\cap \partial \Omega _2$.
\end{enumerate}
Then $T$ has a fixed point in $P\cap
( \overline{\Omega }_2\backslash \Omega _1) $.
\end{theorem}

For the sake of convenience, the conditions needed for our
criteria are listed as follows:
\begin{enumerate}

\item[H1)] $f\in C([0,\infty ),[0,\infty ))$ and there are $x_{n}\to 0$
such that $f(x_{n})>0$ for $n=1,2,\dots$.

\item[H2)] $h(t)>0$ for $t\in R$.

\item[H3)] $\sup_{r>0}\min_{r\sigma \leq x\leq r}f(x)>0$, with $\sigma $
to be defined later.

\item[H4)] $f\in C([0,\infty ),[0,\infty ))$ and $f(x)>0$ for $x>0$.

\item[L1)] $\lim_{x\to 0}f(x)/x=\infty $
\item[L2)] $\lim_{x\to \infty }f(x)/x=\infty $
\item[L3)] $\lim_{x\to 0}f(x)/x=0$
\item[L4)] $\lim_{x\to \infty }f(x)/x=0$
\item[L5)] $\lim_{x\to 0}f(x)/x=l$ with $0<l<\infty $
\item[L6)] $\lim_{x\to \infty }f(x)/x=L$ with $0<L<\infty $.
\end{enumerate}

\section{Main Result}

We proceed formally from (\ref{1}) to obtain
\[
[ y(t)\exp ( \int_{-\infty }^{t}a(s)ds) ] '
=\lambda \exp ( \int_{-\infty }^{t}a(s)ds) h(t)f(y(t-\tau (t)).
\]
After integration from $t$ to $t+T$, we obtain
\begin{equation}
y(t)=\lambda \int_{t}^{t+T}G(t,s)h(s)f(y(s-\tau (s)))ds,  \label{2}
\end{equation}
where
\[
G(t,s)=\frac{\exp ( \int_{t}^{s}a(u)du) }{\exp (
\int_{0}^{T}a(u)du) -1}.
\]
Note that the denominator in $G(t,s)$ is not zero since we have assumed that
$a(t_{0})>0$ for some $t_{0}\in [0,T]$.
It is not difficult to check that any function $y(t)$ that satisfies (\ref{2})
is also a $T$-periodic solution of (\ref{1}). Note that
\[
N\equiv G(t,t)\leq G(t,s)\leq G(t,t+T)=G(0,T)\equiv M,\;t\leq s\leq t+T,
\]
and
\[
1\geq \frac{G(t,s)}{G(t,t+T)}\geq \frac{G(t,t)}{G(t,t+T)}=\frac{N}{M}>0.
\]
Now let $X$ be the set of all real $T$-periodic continuous functions,
endowed with the usual linear structure and the norm
\[
\| y\| =\sup_{t\in [0,T]}| y(t)| .
\]
Then $X$ is a Banach space with cone
\[
\Omega =\{ y(t):y(t)\geq \sigma \| y(t)\| ,\;t\in R\} ,
\]
where $\sigma =N/M$. Note that $a(t_{0})>0$ for some $t_{0}\in [0,T]$.
Clearly, $\sigma \in (0,1)$. Define a mapping $T:X\to X$ by
\[
( Ty) (t)=\lambda \int_{t}^{t+T}G(t,s)h(s)f(y(s-\tau (s)))ds.
\]
Then it is easily seen that $T$ is completely continuous on bounded subset
of $\Omega $, and for $y\in \Omega $,
\[
( Ty) (t)\leq \lambda M\int_{0}^{T}h(s)f(y(s-\tau (s)))ds
\]
so that
\[
( Ty) (t)\geq \lambda N\int_{0}^{T}h(s)f(y(s-\tau (s)))ds\geq
\sigma \| Ty\| .
\]
That is, $T\Omega $ is contained in $\Omega $.

\begin{lemma} \label{lm1}
With the above notation,  $T\Omega \subset \Omega $.
\end{lemma}

\begin{lemma} \label{lm2}
Assume that there exist two positive numbers $a$ and $b$ such that
$a\neq b$,
\begin{equation}
\max_{0\leq x\leq a}f(x)\leq \frac{a}{\lambda A},  \label{2.5}
\end{equation}
and
\begin{equation}
\min_{\sigma b\leq x\leq b}f(x)\geq \frac{b}{\lambda B}  \label{2.6}
\end{equation}
where
\begin{equation}
A=\max_{0\leq t\leq T}\int_{0}^{T}G(t,s)h(s)ds  \label{10}
\end{equation}
and
\begin{equation}
B=\min_{0\leq t\leq T}\int_{0}^{T}G(t,s)h(s)ds.  \label{11}
\end{equation}
Then there exists $\overline{y}\in \Omega $ which is a fixed point of $T$
and satisfies $\min \{ a,b\} \leq \| \overline{y}\|
\leq \max \{ a,b\} $.
\end{lemma}


\paragraph{Proof.}  Let $\Omega _{\xi }=\{ w\in \Omega |\| w\| <\xi\} $.
Assume that $a<b$. Then, for any $y\in \Omega $ which satisfies $\| y\| =a$,
in view of (\ref{2.5}), we have
\begin{equation}
( Ty) (t)\leq \{ \lambda \int_{t}^{t+T}G(t,s)h(s)ds\}
\cdot \frac{a}{\lambda A}\leq \lambda A\cdot \frac{a}{\lambda A}=a.
\label{11.5}
\end{equation}
That is, $\| Ty\| \leq \| y\| $ for $y\in \partial
\Omega _{a}$. For any $y\in \Omega $ which satisfies $\| y\| =b$,
we have
\begin{equation}
( Ty) (t)\geq \{ \lambda \int_{t}^{t+T}G(t,s)h(s)ds\}
\cdot \frac{b}{\lambda B}\geq \lambda B\cdot \frac{b}{\lambda B}.
\label{11.6}
\end{equation}
That is, we have $\| Ty\| \geq \| y\| $ for $y\in
\partial \Omega _{b}$. In view of Theorem \ref{thmA}, there exists $\overline{y}\in
\Omega $ which satisfies $a\leq \| \overline{y}\| \leq b$ such
that $T\overline{y}=\overline{y}$. If $a>b$, (\ref{11.5}) is replaced by
$(Ty)(t)\geq b$ inview of (\ref{2.6}), and (\ref{11.6}) is replaced by
$(Ty)(t)\leq a$ in view of (\ref{2.5}). The same conclusion then follows. The
proof is complete.

\begin{theorem} \label{thm1}
Suppose (H1), (H2), (L1) and (L2) hold. Then for any
$\lambda \in (0,\lambda ^{*})$, equation (\ref{1}) has at least two positive
periodic solutions, where
\[
\lambda ^{*}=\frac{1}{A}\sup_{r>0}\frac{r}{\max_{0\leq x\leq r}f(x)},
\]
and $A$ is defined by (\ref{10}).
\end{theorem}

\paragraph{Proof.} Let $q(r)=r/( A\max_{0\leq x\leq r}f(x)) $. In view of
(H1), we have that $q\in C( (0,\infty ),(0,\infty )) $. In view of
(L1) and (L2), we see further that $\lim_{r\to
0}q(r)=\lim_{r\to \infty }q(r)=0$. Thus, there exists $r_{0}>0$ such
that $q(r_{0})=\max_{r>0}q(r)=\lambda ^{*}$. For any $\lambda \in (0,\lambda
^{*})$, by the intermediate value theorem, there exist $a_1\in (0,r_{0})$
and $a_2\in (r_{0},\infty )$ such that $q(a_1)=q(a_2)=\lambda $. Thus,
we have $f(x)\leq a_1/( \lambda A) $ for $x\in [0,a_1]$ and
$f(x)\leq a_2/( \lambda A) $ for $x\in [0,a_2]$. On the other
hand, in view of (L1) and (L2), we see that there exist $b_1\in (0,a_1)$
and $b_2\in (a_2,\infty )$ such that $f(x)/x\geq 1/(\lambda \sigma B)$
for $x\in (0,b_1]\cup [b_2\sigma ,\infty )$. That is, $f(x)\geq
b_1/(\lambda B)$ for $x\in [b_1\sigma ,b_1]$ and $f(x)\geq
b_2/(\lambda B)$ for $x\in [b_2\sigma ,b_2]$. An application of Lemma
\ref{lm2} leads to two distinct solutions of (\ref{1}).
\smallskip

We remark that the arguments in the above proof actually yield the
following result: If (H1) and (H2) hold, and if either (L1) or (L2) holds,
then for any $0<\lambda <\lambda ^{*}$, equation (\ref{1}) has at least one
positive periodic solution.

\begin{theorem} \label{thm2}
Suppose (H2), (H4), (L3) and (L4) hold. Then for any $\lambda>\lambda ^{**}$,
equation (\ref{1}) has at least two positive periodic solutions, where
\[
\lambda ^{**}=\frac{1}{B}\inf_{r>0}\frac{r}{\min_{\sigma r\leq x\leq r}f(x)},
\]
and $B$ is defined by (\ref{11}).
\end{theorem}

\paragraph{Proof.}
Let $p(r)=r/( B\min_{\sigma r\leq x\leq r}f(x)) $.
Clearly, $q\in C( (0,\infty ),(0,\infty )) $. From (L3)
and (L4), we see that $\lim_{r\to 0}p(r)=\lim_{r\to \infty
}p(r)=\infty $. Thus, there exists $r_{0}>0$ such that
$p(r_{0})=\min_{r>0}p(r)=\lambda ^{**}$. For any $\lambda >\lambda ^{**}$,
there exist $b_1\in (0,r_{0})$ and $b_2\in (r_{0},\infty )$ such that
$p(b_1)=p(b_2)=\lambda $. Thus, we have $f(x)\geq b_1/( \lambda
B) $ for $x\in [\sigma b_1,b_1]$ and $f(x)\geq b_2/(
\lambda B) $ for $x\in [\sigma b_2,b_2]$. On the other hand, in
view of (L3), we see that $f(0)=0$ and that there exists $a_1\in (0,b_1)$
such that $f(x)/x\leq 1/( \lambda A) $ for $x\in (0,a_1]$.
Thus, we have $f(x)\leq a_1/( \lambda A) $. In view of (L4), we
see that there exists $a\in (b_2,\infty )$ such that $f(x)/x\leq 1/(
\lambda A) $ for $x\in [a,\infty )$. Let $\delta =\max_{0\leq x\leq
a}f(x)$. Then we have $f(x)\leq a_2/( \lambda A) $ for $x\in
[0,a_2]$, where $a_2>a$ and $a_2\geq \lambda \delta A$. An application
of Lemma \ref{lm2} leads to two distinct solutions of (\ref{1}).
\smallskip

Again, we remark that the proof of Theorem \ref{thm2} shows the following: If (H1),
(H2) and (H3) hold, and if (L3) or (L4) holds, then for any
$\lambda>\lambda ^{**}$, equation (\ref{1}) has a positive periodic solution.

\begin{theorem} \label{thm3}
Assume that (H1), (H2), (L5) and (L6) hold. Then, for each $\lambda $
satisfying
\begin{equation}
\frac{1}{\sigma BL}<\lambda <\frac{1}{Al}  \label{100}
\end{equation}
or
\[
\frac{1}{\sigma Bl}<\lambda <\frac{1}{AL},
\]
equation (\ref{1}) has a positive periodic solution.
\end{theorem}

\paragraph{Proof.}  Suppose (\ref{100}) holds. Let $\varepsilon >0$ be such that
\[
\frac{1}{\sigma B(L-\varepsilon )}\leq \lambda \leq \frac{1}{A(l+\varepsilon
)}.
\]
Note that $l>0$, thus there exists $H_1>0$ such that $f(x)\leq
(l+\varepsilon )x$ for $0<x\leq H_1$. So, for $y\in \Omega $ with $\|
y\| =H_1$, we have
\begin{eqnarray*}
( Ty) (t) &\leq &\lambda (l+\varepsilon
)\int_{t}^{t+T}G(t,s)h(s)y(s-\tau (s))ds \\
&\leq &\lambda (l+\varepsilon )\| y\| \int_{0}^{T}G(t,s)h(s)ds \\
&\leq &\lambda A(l+\varepsilon )\leq \| y\| .
\end{eqnarray*}
Next, since $L>0$, there exists a $\overline{H}_2>0$ such that $f(x)\geq
(L-\varepsilon )x$ for $x\geq \overline{H}_2$. Let $H_2=\max \{
2H_1,\sigma \overline{H}_2\} $, then for $y\in \Omega $ with
$\| y\| =H_2$, \begin{eqnarray*}
( Ty) (t) &\geq &\lambda (L-\varepsilon
)\int_{t}^{t+T}G(t,s)h(s)y(s-\tau (s))ds \\
&\geq &\lambda (L-\varepsilon )\sigma \| y\|
\int_{0}^{T}G(t,s)h(s)ds \\
&\geq &\lambda (L-\varepsilon )\sigma B\| y\| \geq \|
y\| .
\end{eqnarray*}
In view of Lemma \ref{lm2}, we see that equation (\ref{1}) has a positive
periodic solution. The other case is similarly proved.

\begin{corollary} \label{coro1}
Assume that (H1) and (H2) hold. Assume further that either (L1)
and (L4) hold, or, (L2) and (L3) hold, then for any $\lambda >0$, equation
(\ref{1}) has a positive periodic solution.
\end{corollary}

\paragraph{Proof.}
Suppose first that (L1) and (L4) hold. If $\sup_{0\leq x<\infty
}f(x)=D<\infty $, then $\lambda ^{*}\geq ( 1/A) \sup_{r>0}(
r/D) =\infty $. If $f(x)$ is unbounded, then there exist a sequence $%
\{ r_{n}\} $ such that $f(r_{n})=\max_{0\leq x\leq r_{n}}f(x)$
and $\lim_{n\to \infty }r_{n}=\infty $. In view of (L4), we have $%
\lambda ^{*}\geq ( 1/A) \sup ( r_{n}/f( r_{n})
) =\infty $. Thus, we have proved $\lambda ^{*}=\infty $. In this
case, our assertion follows from the remark following Theorem \ref{thm1}.
If (L2) and (L3) hold, then we have $\lim_{x\to \infty }f(x)=\infty $. Thus,
(H3) holds. Let $\{ r_{n}\} $ satisfy $\lim_{n\to \infty
}r_{n}=\infty $ and $f(\sigma r_{n})=\min_{\sigma r_{n}\leq x\leq r_{n}}f(x)$.
 In view of (L2), we have $\lambda ^{**}\leq ( 1/B) \inf
(r_{n}/f(\sigma r_{n}))=0$. Thus, $\lambda ^{**}=0$. In this case, our
assertion follows from the remark following Theorem \ref{thm2}.

\begin{corollary} \label{coro2}
Assume that (H1) and (H2) hold. Assume further that either (L1)
and (L6) hold, or, (L2) and (L5) hold. Then for any $0<\lambda <1/(
Al) $ or $0<\lambda <1/( AL) $ equation (\ref{1}) has a
positive periodic solution.
\end{corollary}

\begin{corollary} \label{coro3}
Assume that (H1) and (H2) hold. Assume further that either (L3)
and (L6) hold, or, (L4) and (L5) hold. Then for any $1/(\sigma LB)<\lambda
<\infty $ or $1/(\sigma lB)<\lambda <\infty $ equation (\ref{1}) has a
positive periodic solution.
\end{corollary}

Similarly, we can also discuss the equation
\begin{equation}
x'(t)=a(t)x(t)-\lambda h(t)f(x(t-\tau (t)).  \label{3}
\end{equation}
where $a=a(t)$, $h=h(t)$ and $f=f(t)$ satisfy the same assumptions stated
for equation (\ref{1}). By (\ref{3}), we have
\[
x(t)=\int_{t}^{t+T}H(t,s)h(s)f(x(s-\tau (s))ds,
\]
where
\[
H(t,s) =\frac{\exp ( -\int_{t}^{s}a(u)du) }{1-\exp (-\int_{0}^{T}a(u)du) }
=\frac{\exp ( \int_{s}^{t+T}a(u)du) }{\exp (\int_{0}^{T}a(u)du-1) }
\]
which satisfies
\[
M=G(0,T)=H(t,t)\geq H(t,s)\geq H(t,t+T)=H(0,T)=G(t,t)=N
\]
and
\[
1\geq \frac{H(t,s)}{H(t,t)}\geq \frac{H(t,t+T)}{H(t,t)}=\frac{N}{M}=\sigma .
\]
Let
\[
A'=\max_{0\leq t\leq T}\int_{0}^{T}H(t,s)h(s)ds
\]
and
\[
B'=\min_{0\leq t\leq T}\int_{0}^{T}H(t,s)h(s)ds.
\]
Then we have the following results.

\begin{theorem} \label{thm4}
Assume that (H1) and (H2) hold. Suppose further that either (L1)
or (L2) holds. Then for any $\lambda \in (0,\overline{\lambda })$, equation (%
\ref{3}) has a positive periodic solution, where
\[
\overline{\lambda }=\frac{1}{A'}\sup_{r>0}\frac{r}{\max_{0\leq
x\leq r}f(x)}.
\]
\end{theorem}

\begin{theorem} \label{thm5}
Suppose (H1), (H2), (L1) and (L2) hold. Then for any $%
\lambda \in (0,\overline{\lambda })$, equation (\ref{3}) has at least two
positive periodic solutions.
\end{theorem}

\begin{theorem} \label{thm6}
Assume that (H1), (H2) and (H3). Suppose further that either (L3)
or (L4) holds. Then for any $\lambda >\underline{\lambda }$, equation (\ref
{3}) has a positive periodic solution, where
\[
\underline{\lambda }=\frac{1}{B'}\inf_{r>0}\frac{r}{\min_{\sigma
r\leq x\leq r}f(x)}.
\]
\end{theorem}

\begin{theorem} \label{thm7}
 Suppose (H2), (H4), (L3) and (L4) hold. Then for any
 $\lambda >\underline{\lambda }$, equation (\ref{3}) has at least two positive
 periodic solutions.
\end{theorem}

\begin{theorem} \label{thm8}
Assume that (H1), (H2), (L5) and (L6) hold. Then, for each $\lambda $ satisfying
\[
\frac{1}{\sigma B'L}<\lambda <\frac{1}{A'l}
\]
or
\[
\frac{1}{\sigma B'l}<\lambda <\frac{1}{A'L},
\]
equation (\ref{3}) has a positive periodic solution.
\end{theorem}

\begin{corollary} \label{coro4}
Assume that (H1) and (H2) hold. Suppose further that either
(L1) and (L4) hold, or, (L2) and (L3) hold. Then for any $\lambda >0$, equation (\ref{3}) has a positive periodic solution.
\end{corollary}

\begin{corollary} \label{coro5}
Assume that (H1) and (H2) hold. Suppose further that either
(L1) and (L6) hold, or, (L2) and (L5) hold. Then for any $0<\lambda
<1/( A'L) $ or $0<\lambda <1/( A'l) $
equation (\ref{3}) has a positive periodic solution.
\end{corollary}

\begin{corollary} \label{coro6}
Assume that (H1) and (H2) hold. Suppose further that either
(L3) and (L6) hold, or, (L4) and (L5) hold. Then for any $1/(\sigma
LB')<\lambda <\infty $ or $1/(\sigma lB')<\lambda <\infty $
equation (\ref{3}) has a positive periodic solution:
\end{corollary}

\begin{thebibliography}{9} \frenchspacing

\bibitem{f1}  M. Fan and K. Wang, \emph{Optimal harvesting policy for single
population with periodic coefficients}, Math. Biosci., 152(1998), 165-177.

\bibitem{f2}  M. Fan and K. Wang, \emph{Uniform ultimate boundedness and periodic
solutions of functional differential equations with infinite delay}, J. Sys.
Sci. Math. Sci., 17(3)(1999), 323-327 (in Chinese).

\bibitem{j1}  D. Q. Jiang and J. J. Wei, \emph{Existence of positive periodic
solutions of nonautonomous functional differential equations}, Chinese Ann.
Math., A20(6)(1999), 715-720 (in Chinese).

\bibitem{l1}  Y. K. Li, \emph{Existence and global attractivity of positive
periodic solutions for a class of delay differential equations}, Chinese
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\bibitem{z1}  S. N. Zhang, \emph{Periodicity in functional differential
equations}, Ann. Diff. Eqns., 12(2)(1996), 252-257.

\end{thebibliography}


\noindent\textsc{Sui Sun Cheng} \\
Department of Mathematics,
National Tsing Hua University\\
Hsinchu, Taiwan 30043 \\
e-mail: sscheng@math2.math.nthu.edu.tw
\medskip

\noindent\textsc{Guang Zhang}\\
Department of Mathematics,
Yanbei Normal Institute\\
Datong, Shanxi 037000, P.R.China\\
e-mail: dtgzhang@yahoo.com.cn

\end{document}