\documentclass[reqno]{amsart}
\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 119, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu (login: ftp)}
\thanks{\copyright 2003 Texas State University-San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE--2003/119\hfil
Positive solutions of a generalized logistic equation]
{Global positive solutions of a generalized logistic
equation with bounded and unbounded coefficients}
\author[G. N. Galanis \& P. K. Palamides\hfil EJDE--2003/119\hfilneg]
{George N. Galanis \& Panos K. Palamides}
\address{Naval Academy of Greece, Piraeus, 185 39, Greece}
\email[G. N. Galanis]{ggalanis@snd.edu.gr}
\email[P. K. Palamides]{ppalam@otenet.gr, ppalam@snd.edu.gr}
\date{}
\thanks{Submitted October 13, 2003. Published December 1, 2003.}
\subjclass[2000]{34B18, 34A12, 34B15}
\keywords{Generalized logistic equation, asymptotic behavior of solutions,
\hfill\break\indent
periodic solutions, Knesser's property, Consequent mapping, Continuum sets}
\begin{abstract}
In this paper we study the generalized logistic equation
$$
\frac{du}{dt}=a(t)u^{n}-b(t)u^{n+(2k+1)},\quad n,k\in \mathbb{N},
$$
which governs the population growth of a self-limiting specie,
with $a(t)$, $b(t)$ being continuous bounded functions.
We obtain a unique global, positive and bounded solution which,
further, plays the role of a frontier which clarifies the asymptotic
behavior or extensibility backwards and further it is an attractor
forward of all positive solutions. We prove also that the function
$$
\phi (t)=\sqrt[2k+1]{a(t)/b(t)}
$$
plays a fundamental role in the study of logistic equations since
if it is monotone, then it is an attractor of positive solutions
forward in time. Furthermore, we may relax the boundeness assumption
on $a(t)$ and $b(t)$ to a boundeness of it. An existence result of
a positive periodic solution is also given for the case where
$a(t)$ and $b(t)$ are also periodic (actually we derive a necessary
and sufficient condition for that). Our technique is a topological
one of Knesser's type (connecteness and compactness of the solutions
funnel).
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\section{Introduction}
One of the most popular differential equations with various applications in
economic and managerial sciences is the logistic equation:
\begin{equation}
\frac{du}{dt}=a(t)u(t)-b(t)u(t)^{2},\quad t\in\mathbb{R}. \label{1}
\end{equation}
While in the case where the coefficients $a(t),b(t)$ are constant, the
above-mentioned equation can be solved explicitly, by employing classical
techniques, and a stable equilibrium point of population may exists, when
$a(t)$ and $b(t)$ are variable the corresponding study becomes much more
complicated. As a matter of fact, no explicit solutions can be found in
general in this framework (see, among others, \cite{Am,Bo}) and the
equilibrium point may become unstable. However, it is clear that the
existence of stable periodic or stable bounded solutions is an essential
part of qualitative theory of differential equations. Furthermore, the
existence of a solution of such type is of fundamental importance
biologically, since it concerns the long time survival of species.
Similar problems appear also in the framework of partial differential
equations with logistic type nonlinearities (see e.g. \cite{Blat,Coh}) or
functional differential equations with discrete or continuous delays (see
\cite{TX} and \cite{JWZ} for some recent results).
A considerable number of authors have proposed different techniques in order
to determine non trivial solutions of the logistic equation and to study
their behavior. Among them J. Hale and H. Kocak in \cite{Ha} discuss the
time periodic case and N. Nkashama in his recent paper \cite{Nk} (published
in this journal) works on the study of a bounded solution of (\ref{1})
making ample use of classical techniques.
In this paper we study a generalized logistic equation. Namely, we assume
that the change of $u$ in time can be affected by higher order polynomials:
\begin{equation}
\frac{du}{dt}=a(t)u^{n}-b(t)u^{n+(2k+1)},\quad n,k\in \mathbb{N}, \label{2}
\end{equation}%
where the carrying capacity $a(t)$ and the self-limiting coefficient $b(t)$
are continuous and bounded functions:
\begin{equation}
00$ and a $\tau >t_{1}$\ such that
\begin{equation*}
G(x|[t_{1}-\varepsilon _{1},t_{1});P)\subseteq \omega ^{o}\quad \mbox{and}
\quad G(x|[t_{1},\tau ];P)\subseteq \partial {\omega }.
\end{equation*}
If, in addition, for any solution $x\in \mathcal{X}(P)$ there exists a point
$t_{2}\in Domx$ and a positive $\varepsilon _{2}>0$ such that
\begin{equation*}
G(x|[\tau ,t_{2}];P)\subseteq \partial {\omega }\quad \mbox{and}\quad
G(x|(t_{2},t_{2}+\varepsilon _{2}];P)\subseteq \Omega -\overline{\omega },
\end{equation*}%
then the point $P$ is called a point of \textit{strict semi-egress} of
$\omega $.
The set of all points of semi-egress of $\omega$ is denoted by $\omega^{s}$
and those of strict semi-egress by $\omega^{ss}$.
A second point now $Q=(\sigma ,\eta )\in \omega ^{s}$, with $\sigma \geq
\tau $, will be called a consequent of the initial one $P=(\tau ,\xi )$,
with respect to the set $\omega $ and the system (\ref{4}), if there exists
a solution $x\in \mathcal{X}(P,Q)=\mathcal{X}(P)\cap \mathcal{X}(Q)$ and a
point $t_{1}\in \lbrack \tau ,\sigma ]$ such that $G(x|_{[t_{1},\sigma
]})\subseteq \partial \omega $ and $G(x|_{(\tau ,t_{1})})\subseteq \omega
^{o}$, for $\tau \phi (s).
\end{gather*}%
Moreover, $\phi (t)$ is a positive and bounded function since for any $t\in
\mathbb{R}$:
\begin{equation*}
m:=\sqrt[2k+1]{a/B}\leq \phi (t)\leq \sqrt[2k+1]{A/b}:=M.
\end{equation*}
Under the previous notifications the following theorem holds true.
\begin{theorem} \label{Th1}
The generalized logistic equation (\ref{5}) admits exactly one
bounded solution which remains into the interval $I=[m,M]$ for all
$t\in\mathbb{R}$.
\end{theorem}
\begin{proof}
Let $\omega$ and $\omega_{0}$ be the sets given by
\begin{gather*}
\omega :=\{(s,v)\in\mathbb{R}\times\mathbb{R}:s\geq0,m\leq v\leq M\}, \\
\omega_{0} :=\{(s,v)\in\mathbb{\omega}:s=0\}
\end{gather*}
Then, every solution $v\in\mathcal{X}(P),\;P\in\omega_{0}$, of the
differential equation (\ref{7}), that reaches the boundary $\partial\omega
\; $of $\omega$, strictly egresses of it. In other words, using the
terminology defined in Section 2, $\omega^{s}=\omega^{ss}$. As a result, the
image $\mathcal{K}_{\omega}(\omega_{0})$ of the consequent mapping $\mathcal{%
K}_{\omega}$, with respect to (\ref{7}), has common points with both the
lines $v=m$, $v=M$. Therefore, based on the fact that the above-mentioned
image has to be a connected set (see Proposition \ref{Pr1}), we conclude
that there exists at least one solution $v=v(s)$ of (\ref{7}) that remains
into the interval $I=[m,M]$ for every $s\geq0$. Then, the corresponding
function $u(t)=v(-s)$ is the desired bounded solution of (\ref{5}), for $%
t\leq0$. On the other hand, if
\begin{equation*}
\omega_{1}:=\{(t,u)\in\mathbb{R}\times\mathbb{R}:t\geq0,m\leq u\leq M\},
\end{equation*}
then every point of $\partial{\omega_{1}}$ is not an egress one. As a
result, the solution $u=u(t)$ remains constantly into $I$ also for all
positive values of $t$.
We proceed now with the proof of the uniqueness of this bounded solution $%
u(t)$. Let $w(t)$ be another solution of (\ref{5}) with $w(t_{1})\neq
u(t_{1})$ for some $t_{1}\in\mathbb{R}$ and let us assume (with no loss of
generality) that $w(t_{1})__1$. As a result,
\begin{equation*}
(\frac{u^{1-n}}{1-n}-\frac{w^{1-n}}{1-n})'=b(t)(w^{2k+1}-u^{2k+1})
\end{equation*}
and, therefore, the function $u^{1-n}-w^{1-n}$ is increasing. This fact
ensures that
\begin{equation*}
\frac{1}{u(t)^{n-1}}-\frac{1}{w(t)^{n-1}}\leq\frac{1}{u(0)^{n-1}}-\frac {1}{%
w(0)^{n-1}}=c<0,\quad t\leq0.
\end{equation*}
Trivial calculations turns the previous result to
\begin{equation*}
(w(t)-u(t))\frac{w(t)^{n-2}+w(t)^{n-3}u(t)+\dots +w(t)u(t)^{n-3}+u(t)^{n-2}}{%
(u(t)w(t))^{n-1}}\leq c<0,
\end{equation*}
where the fraction emerged is a positive and bounded mapping. Hence, there
exists a real $\delta>0$ such that
\begin{equation*}
u(t)-w(t)\geq\delta,\quad\forall t\leq0.
\end{equation*}
Based on this we obtain
\begin{align*}
&(\frac{w^{1-n}}{1-n}-\frac{u^{1-n}}{1-n})'=b(t)(u-w)(u^{2k}
+u^{2k-1}w+\dots +uw^{2k-1}+w^{2k}) \\
&\geq b\delta(2k+1)m^{2k}=:\varepsilon_{0}>0,
\end{align*}
and by integration on the interval [t,0]:
\begin{equation*}
w(t)^{1-n}-u(t)^{1-n}>(w(0)^{1-n}-u(0)^{1-n})+(1-n)\varepsilon_{0}t.
\end{equation*}
Taking now the limits when $t\to-\infty$ we obtain that
\begin{equation*}
\lim_{t\to -\infty}(\frac{1}{w(t)^{n-1}}-\frac{1}{u(t)^{n-1}})=+\infty,
\end{equation*}
which cannot be true, since $w(t)$ has been assumed bounded.
In the case where $n=1$, we proceed in a similar way with only some
differences in the proof of the uniqueness of bounded solution. More
precisely, (\ref{5}) is now equivalent to
\begin{equation*}
(\ln u)'=a(t)-b(t)u(t)^{2k+1}
\end{equation*}
and assuming that $w(t)$ is a second solution of it with
\begin{equation*}
0\leq m\leq w(t)____0$, hence, $w(t)/u(t)$ is
increasing. As a result, $\frac{w(t)}{u(t)}\leq\frac{w(0)}{u(0)}=c<1$, for $%
t\leq0$, and
\begin{equation*}
u(t)-w(t)\geq(1-c)m.
\end{equation*}
Therefore,
\begin{align*}
(\ln w-\ln u)'&=b(t)(u^{2k+1}-w^{2k+1}) \\
&=b(t)(u-w)(u^{2k}+u^{2k-1}w+\dots +uw^{2k-1}+w^{2k}) \\
&\geq b(1-c)m(2k+1)m^{2k}=\varepsilon_{1}>0.
\end{align*}
Integrating on the interval [t,0] we obtain
\begin{equation*}
\ln (\frac{w(0)/u(0)}{w(t)/u(t)})>-\varepsilon_{1}t\Leftrightarrow \frac{w(t)%
}{u(t)}<\frac{w(0)}{u(0)}e^{\varepsilon_{1}t},\quad t\leq0.
\end{equation*}
Then,
\begin{equation*}
\lim_{t\to-\infty}(\frac{w(t)}{u(t)})=0
\end{equation*}
which is a contradiction to the fact that $\frac{w(t)}{u(t)}\geq\frac{m}{M}%
>0 $ and this concludes the proof.
\end{proof}
It is worthy to notice here that the boundeness conditions (3.2) can be
relaxed according to the next result.
\begin{theorem}\label{Th2}
Let $m^{\ast}$, $M^{\ast}$ be positive constants such that
\begin{equation}
0a-B(m-\varepsilon)^{2k+1}:=\varepsilon_{0}>0.
\end{equation*}
By integration onto $[t,t_{3}]$ we obtain:
\begin{equation*}
u(t)^{1-n}>(n-1)\varepsilon_{0}(t_{3}-t)+u(t_{3})^{1-n},
\end{equation*}
which, in turn, gives that $\lim_{t\to-\infty}u(t)=0$. The same conclusion
is also reached in the case where $n=1$, with only some modifications in our
calculations since the function $\ln u$ is then replacing $\frac{u^{1-n}}{1-n%
}$.
\end{proof}
The previous proposition shows that there is a common behavior backward in
time, for all possible values of the indexes $n,k$, of the positive
solutions of the generalized logistic equation that remain bellow the unique
global solution $u_{b}$. This is not the case for those solutions of (\ref{5}%
) which are above $u_{b}$.
\begin{proposition}\label{Pr4}
Let $u$ be a solution of the generalized logistic equation
(\ref{5}) that lays above the unique bounded solution $u_{b}$ for some time
$t_{1}$. Then $u$ blows up backward in time and, more precisely,
\begin{itemize}
\item [(i)]There exists a $t_{0}\leq t_{1}$ such that $\lim_{t\to
t_{0}}u(t)=+\infty$, when $n>1$.
\item[(ii)] $\lim_{t\to-\infty}u(t)=+\infty$, when $n=1$.
\end{itemize}
\end{proposition}
\begin{proof}
We study first the case where the index $n$ is greater than $1$. Let $u(t)$
be a positive solution with $u(t_{1})>u_{b}(t_{1})$, for some time $t_{1}$.
Then, $u(t)>u_{b}(t)$, for every $t\in Dom(u)\cap Dom(u_{b})$, and
$u(t_{2})>M$ for some $t_{2}\leq t_{1}$ due to the uniqueness of $u_{b}$.
Taking into account that $u$ is then decreasing, we obtain $u(t)\geq
M+\delta $, for all $t\leq t_{2}$, for a positive $\delta$, and
\begin{equation*}
\frac{u'}{u^{n}}=a(t)-b(t)u^{2k+1}b\delta^{2k+1}(n-1)(t_{2}-t).
\end{equation*}
If we assume that the demand of $(i)$ does not fulfilled at any time, then
the function $u(t)$ would be defined for every $t_{n}0$ and
\begin{equation*}
\frac{1}{u(t_{2})^{n-1}}>\frac{1}{u(t_{2})^{n-1}}-\frac{1}{M_{n}^{n-1}}%
>b\delta^{2k+1}(n-1)(t_{2}-t_{n}),
\end{equation*}
which leads to a contradiction if we take $t_{n}\to-\infty$. As a result, a
point $t_{0}$ satisfying $\lim_{t\to t_{0}}u(t)=+\infty$ must exists.
In the case where $n=1$, equation (\ref{5}) takes the form
\begin{equation*}
(\ln u)^{\prime }=a(t)-b(t)u^{2k+1}
\end{equation*}%
and the existence of a positive solution $u>u_{b}$, as in previous, leads to
\begin{gather*}
\frac{u^{\prime }}{u}\leq -\delta _{1}<0\Rightarrow \int_{t}^{t_{2}}(\ln
u)^{\prime }dt\leq -\int_{t}^{t_{2}}\delta _{1}dt\Rightarrow \\
\ln (\frac{u(t)}{u(t_{2})})\geq \delta _{1}(t_{2}-t)\Rightarrow u(t)\geq
e^{\delta _{1}(t_{2}-t)}u(t_{2}),\quad t\leq t_{2}.
\end{gather*}%
As a result, $\lim_{t\rightarrow -\infty }u(t)=+\infty $.
\end{proof}
The next result describes the behavior of solutions of (\ref{5}) forward in
time. The unique bounded solution $u_{b}$ is, again, the key since it
attracts all such positive solutions.
\begin{proposition} \label{Pr5}
The unique bounded solution $u_{b}(t)$ is an attractor of all
positive solutions $w(t)$ of (\ref{5}) forward in time in the sense that
\[
\lim_{t\to+\infty}| u_{b}(t)-w(t)|=0.
\]
\end{proposition}
\begin{proof}
Let us consider the case $n>1$ and let $w(t)\;$be an arbitrarily chosen
positive solution of (\ref{5}). If we assume that $w(t)\hat{M}(t-t_{1})+k,
\end{equation*}
where $\hat{M}$ is a positive constant. This directly gives that
$\lim_{t\rightarrow +\infty }(\frac{1}{u_{b}^{n-1}(t)})=+\infty $ which
obviously contradicts the fact that the solution $u_{b}$ remains into the
interval $[m,M]$. Therefore, $\lim_{t\rightarrow +\infty }(\frac{1}{
u_{b}^{n-1}(t)}-\frac{1}{w^{n-1}(t)})=0$ which, in turns, gives rise to the
desired $\lim_{t\rightarrow +\infty }|u_{b}(t)-w(t)|=0$ .
Analogously we work in the case where the solution $w$ lays above $u_{b}$.
In the special case where $n=1$ we proceed similarly just adopting the
formalism presented in Theorem \ref{Th1}.
\end{proof}
We conclude this section with a detailed study of the asymptotic behavior of
positive solutions of the generalized Logistic Equation (3.1) in the special
case where $\phi (t)=\sqrt[2k+1]{a(t)/b(t)}$ is monotone. First, we outline
in next Remark the behavior of the unique bounded solution $u_{b}$.
\subsection*{Remark}
(i)\quad If $\phi(t)$ is always decreasing, then the following choices are
possible for $u_{b}$:
\begin{itemize}
\item[(A)] $u_{b}(t)$ lays always over $\phi(t)$ being constantly
decreasing. In this case there is no possibility of $u_{b}$ and $\phi$ to
intersect since if such an incident occurs at a time $t_{0}$, then we would
have
\begin{equation*}
u_{b}'(t_{0})=0,u_{b}(t_{0}+h)<\phi(t_{0}+h),\quad \mbox{for }h>0.
\end{equation*}
As a result,
\begin{equation*}
\phi'(t_{0})=\lim_{h\to0+}\frac{\phi(t_{0}+h)-\phi(t_{0})}{h} \geq
\lim_{h\to0+}\frac{u_{b}(t_{0}+h)-u_{b}(t_{0})}{h}=u_{b}^{\prime }(t_{0})=0
\end{equation*}
which contradicts the fact that $\phi$ is decreasing.
\item[(B)] $u_{b}$ begins bellow $\phi$. Then either $u_{b}(t)<\phi(t)$, for
all $t\in\mathbb{R}$, and $u_{b}$ is increasing or $u_{b}$ intersects $\phi$
at a unique point $t_{0}$ and then follows the behavior described in case
(A). Moreover it is obvious that $\max\left\{ u_{b}(t):t\in\mathbb{R}%
\right\} =u_{b}(t_{0})$.
\end{itemize}
\noindent(ii)\quad If $\phi(t)$ is always increasing in $\mathbb{R}$, then $u_{b}(t)$
has an analogous behavior with three possible choices:
\begin{itemize}
\item[(C)] Remains constantly increasing bellow $\phi$,
\item[(D)] Remains above $\phi $ for every $t\in \mathbb{R}$ and approaching
it being constantly decreasing.
\item[(E)] Begins above $\phi$ decreasing until they intersect and falling
in case (C) thereafter. Then clearly $\min\left\{ u_{b}(t):t\in\mathbb{R}%
\right\} =u_{b}(t_{0})$.
\end{itemize}
(iii)\quad In the special case where $\phi(t)$ is constant, it coincides
with the unique bounded solution $u_{b}$. \medskip
In view of the previous thoughts, we are now in a position to prove the next
basic result which illustrates the asymptotic behavior of all positive
solutions of the generalized Logistic Equation (3.1) under the assumptions
(3.2)\ in the case where the function $\phi (t)=\sqrt[2k+1]{a(t)/b(t)}$ is
monotone.
\begin{proposition} \label{prop6}
If $\phi(t)$ is monotone, then it is an attractor of all positive solutions
$w(t)$ of (3.1) forward in time in the sense that
\[
\lim_{t\to+\infty}|\phi(t)-w(t)|=0.
\]
\end{proposition}
\begin{proof}
We study the case where $\phi (t)$ is decreasing. The second option ($\phi
(t)$ increasing) can be developed analogously. Since $\phi (t)$ is bounded,
it will converge (when $t\rightarrow +\infty $) to its infimum: $%
\lim_{t\rightarrow +\infty }\phi (t)=m_{1}>0$. On the other hand, if the
unique bounded solution $u_{b}$ of (3.1) lays above $\phi (t)$ (case A of
Remark 1), then it will also be decreasing and bounded converging to a
positive $m_{2}$. If we assume that $m_{2}>m_{1}$, then there will exists a $%
t_{1}>0$ such that $m_{1}\leq \phi (t)m_{1}$. Therefore,
\begin{equation*}
u_{b}(t)-\phi (t)>\frac{\delta }{2},\quad t\geq t_{1}.
\end{equation*}%
Taking into account that the generalized Logistic Equation (3.1) is
equivalent to
\begin{align*}
(\frac{u(t)^{1-n}}{1-n})^{\prime }& =b(t)(\phi (t)^{2k+1}-u(t)^{2k+1}) \\
& =b(t)[\phi (t)-u(t)][\phi (t)^{2k}+\phi (t)^{2k-1}u(t)+\dots +\phi
(t)u(t)^{2k-1}+u(t)^{2k}],
\end{align*}%
we obtain $(\frac{u_{b}(t)^{1-n}}{1-n})^{\prime }<-b\frac{\delta }{2}%
m_{1}^{2k}$ which, by integration on $[t_{1},t]$ gives
\begin{equation*}
\frac{1}{u_{b}(t)^{n-1}}>\frac{1}{u_{b}(t_{1})^{n-1}}+(n-1)b\frac{\delta }{2}%
m_{1}^{2k}(t-t_{1}).
\end{equation*}%
Taking the limits when $t\rightarrow +\infty $, we obtain $%
\lim_{t\rightarrow +\infty }\frac{1}{u_{b}(t)^{n-1}}=+\infty $, which
contradicts the fact that $u_{b}(t)$ remains into the interval $[m,M]$ for
every $t\in \mathbb{R}$ (Theorem \ref{Th1}). Therefore, the limits $m_{1}$, $%
m_{2}$ have to coincide and $\lim_{t\rightarrow +\infty }|\phi
(t)-u_{b}(t)|=0$.
If we assume now that $u_{b}$ is bellow $\phi $ for some time (case B of
Remark 1), then either it intersects $\phi $ at a unique point and then
follows the behavior described above, approaching $\phi $ when $t\rightarrow
+\infty $, or it remains bellow $\phi $ for all $t\in \mathbb{R}$. If this
is the case, then $u_{b}$ is constantly increasing and bounded, so that $%
\lim_{t\rightarrow +\infty }u_{b}(t)=m_{2}>0$. If $m_{2}$ does not coincides
with $m_{1}$, then $m_{2}=m_{1}-\delta $, $\delta >0$, and
\begin{equation*}
u_{b}(t)\leq m_{2}\leq \phi (t)-\delta ,\quad t\in \mathbb{R}.
\end{equation*}%
Following now similar thoughts as before, we are leading to the
contradiction $\lim_{t\rightarrow +\infty }u_{b}(t)=0$. As a result, in this
case too, $m_{2}$ and $m_{1}$ have to be equal and $\lim_{t\rightarrow
+\infty }|\phi (t)-u_{b}(t)|=0$.
Taking in mind that $u_{b}(t)$ is an attractor of all positive solutions of
(2.1) (Proposition \ref{Pr5}) we reach the desired result of the proposition.
\end{proof}
\section{The Periodic Problem}
In this section we consider again the differential equation
\begin{equation}
\frac{du}{dt}=a(t)u^{n}-b(t)u^{n+(2k+1)},\quad n,k\in\mathbb{N}, \label{10}
\end{equation}
under the assumption
\begin{equation}
00$ and $\tau\in\mathbb{R}\,$\ the generalized logistic
equation (\ref{10}) admits a positive solution $x=x(t),\;t\in\lbrack\tau
,\tau+T]$, which satisfies the periodic condition
\[
x(\tau)=x(\tau+T).
\]
Furthermore (\ref{10}) has exactly one (classical) global $T$-periodic
solution $u=u(t),\;t\in\mathbb{R}$, provided that both functions $a(t)$\ and
$b(t)$ are also $T$-periodic.
\end{theorem}
\begin{proof}
We consider the set
\begin{equation*}
\Omega :=\left\{ \left( t,u\right) \in \mathbb{R}^{2}:m^{\ast }\leq u\leq
M^{\ast }\right\}
\end{equation*}%
and, for a fixed time $t=\tau $, its subset
\begin{equation*}
\Omega \lbrack \tau ,\tau +T]:=\left\{ \left( t,u\right) \in \Omega :\tau
\leq t\leq \tau +T\right\} .
\end{equation*}%
Let also
\begin{equation*}
\Omega (\tau ):=\left\{ \left( t,u\right) \in \Omega :t=\tau \right\} \text{%
\ \ and \ \ }\Omega (\tau +T):=\left\{ \left( t,u\right) \in \Omega :t=\tau
+T\right\}
\end{equation*}%
be the cross-sections of $\Omega $ at the time $t=\tau $ and $t=\tau +T$
respectively, where we notice that
\begin{equation*}
\Omega \lbrack \tau ]=\Omega \lbrack \tau +T]=I^{\ast }=[m^{\ast },M^{\ast
}].
\end{equation*}%
Taking into account the sign of the nonlinearity of the\ differential
equation (4.1), it is clear that any solution $x\in \mathcal{X}(I^{\ast })$
egresses strictly from $\Omega \lbrack \tau ,\tau +T]$, through the face $%
\Omega (\tau +T)$. Thus, the consequent mapping
\begin{equation*}
\mathcal{K}:I^{\ast }\rightarrow I^{\ast }
\end{equation*}%
is well defined and continuous and $\mathcal{K}$ admits a fixed point $P_{0}$%
. In other words, there exists a solution $x\in \mathcal{X}(I^{\ast })$ of (%
\ref{10}) remaining in $\Omega \lbrack \tau ,\tau +T]$ such that
\begin{equation*}
x(\tau )=x(\tau +T),\;\tau \in \mathbb{R}.
\end{equation*}%
Now we extend periodically\ the obtaining solution $x=x(t)$, $t\in \lbrack
\tau ,\tau +T]$. More precisely, for any integer $n$ we set
\begin{equation*}
u(t):=x(t-nT)=x(s),\quad t\in \left[ \tau +nT,\tau +(n+1)T\right] .
\end{equation*}%
Then, clearly $u=u(t)$, $t\in \mathbb{R}$, is a periodic function.
Furthermore (notice that $s=t-nT\in \lbrack \tau ,\tau +T]$) by the
periodicity of $a(t)$ and $b(t)$, we obtain
\begin{align*}
u^{\prime }(t)& =x^{\prime }(t-nT)=x^{\prime }(s) \\
& =a(s)x^{n}(s)-b(s)x^{n+(2k+1)}(s) \\
& =a(t-nT)u^{n}(t)-b(t-nT)u^{n+(2k+1)}(t) \\
& =a(t)u^{n}(t)-b(t)u^{n+(2k+1)}(t).
\end{align*}%
As a result, $u=u(t)$ is also\ a solution of equation (\ref{10}) remaining
in $\Omega $ for all $t\in \mathbb{R}$. The uniqueness of the obtained
periodic solution follows by Theorem \ref{Th1}.
\end{proof}
The assumption of $T$-periodicity of $a(t)$\ and $b(t)$ is essential at the
above Theorem. Indeed, this is clarified in the next Example given by the
referee.
\subsection{Example}
Let us consider the equation
\begin{equation}
\frac{du}{dt}=u-b(t)u^{4}, \label{13}
\end{equation}%
where $b(t)$ is converging to a positive number when $t\rightarrow +\infty $%
. When $u\neq 0$, we have from (\ref{13})
\begin{equation*}
u^{-4}\frac{du}{dt}=u^{-3}-b(t).
\end{equation*}%
Let $x=u^{-3}$, then $\frac{dx}{dt}=-3u^{-4}\frac{du}{dt}$. Hence, equation (%
\ref{13}) becomes
\begin{equation}
\frac{dx}{dt}=-3x+3b(t). \label{14}
\end{equation}%
The solution of equation \ (\ref{14}) with initial value $x(0)=x^{0}$ is
\begin{equation*}
x(t)=e^{-3t}[x_{0}+\int_{0}^{t}b(s)e^{3s}ds],
\end{equation*}%
which implies
\begin{equation*}
\lim_{t\rightarrow +\infty }x(t)=\lim_{t\rightarrow +\infty
}x_{0}e^{-3t}+\lim_{t\rightarrow +\infty }\frac{\int_{0}^{t}b(s)e^{3s}ds}{%
e^{3t}}=\lim_{t\rightarrow +\infty }\frac{b(t)}{3}.
\end{equation*}%
This shows that (\ref{14}), and consequently (\ref{13}), has no nonconstant
periodic solution. In fact, zero is always a constant periodic solution of (%
\ref{13}). \medskip
We conclude this paper by proving, further, that the periodicity of the
coefficients $a(t)$ and $b(t)$ is an almost necessary and sufficient
condition for the existence of a global periodic solution of the generalized
logistic equation (\ref{10}). Namely:
\begin{theorem}
The generalized logistic equation (\ref{10}) admits a positive $T$-periodic
solution, for any $T>0$, if and only if both functions $a(t)$ and $b(t)$ are
$T$-periodic, provided that the function $u_{p}(t):=\sqrt[2k+1]{\frac
{a(t+T)-a(t)}{b(t+T)-b(t)}}$ is not a periodic solution. In the case where
$u_{p}$ is a solution of (\ref{10}) and the function $\phi=\phi(t)$ is
nonconstant and periodic, then $u_{p}$ is the unique (particular) periodic
solution of the logistic equation.
\end{theorem}
\begin{proof}
The sufficiency of the $T$-periodicity of both $a(t)$ and $b(t)$ has been
proven in Theorem \ref{Th3}.
If we assume now that a $T$-periodic solution $u$ exists then, taking into
account the uniqueness of the bounded solution $u_{b}$ of (\ref{10}), we
obtain that $u=u_{b}$ and $u^{\prime }(t+T)=u^{\prime }(t)$. Consequently,
since $u(t)$ satisfies the equation (\ref{10}), we easily get%
\begin{equation*}
a(t+T)-a(t)=[b(t+T)-b(t)]u^{2k+1}(t).
\end{equation*}%
As a result, if $u=u_{p}(t)$ is not a solution of (\ref{10}) or it is not a
periodic function, then necessarily both the coefficients $a(t)\;$and $b(t)$
must be $T$-periodic functions. Hence the first part of the theorem is
established.
Suppose now that $a(t)$ or $b(t)$ is not $T$-periodic and further that the
map $u=u_{p}(t)$ is a solution of (\ref{10}). In order to finish the proof,
it is enough to show that $u_{p}(t)\;$is a $T$-periodic function. However,
the $T$-periodicity of $u_{p}(t)\;$is equivalent to:
\begin{equation*}
\frac{a(t)-a(t-T)}{b(t)-b(t-T)}=\frac{a(t+T)-a(t)}{b(t+T)-b(t)}
\end{equation*}
or
\begin{equation*}
(a(t)-a(t-T))(b(t+T)-b(t))=(a(t+T)-a(t))(b(t)-b(t-T)).
\end{equation*}
The last equality holds if the function $\phi(t)$ (or simply the map $%
a(t)/b(t)$) is $T$-periodic, since it is equivalent to
\begin{align*}
&\phi^{2k+1}(t)-\phi^{2k+1}(t-T)\frac{b(t-T)}{b(t)}+\phi^{2k+1}(t-T)\frac {%
b(t-T)}{b(t+T)} \\
&=\phi^{2k+1}(t+T)-\phi^{2k+1}(t+T)\frac{b(t-T)}{b(t)} +\phi^{2k+1}(t)\frac{%
b(t-T)}{b(t+T)}
\end{align*}
\end{proof}
\subsection*{Acknowledgement}
We would like to thank the anonymous referee for his/her valuable comments
which led to the clarification of our results in Section 4.
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\end{document}
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