\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: $$\frac{du}{dt}=a(t)u(t)-b(t)u(t)^{2},\quad t\in\mathbb{R}. \label{1}$$ 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: $$\frac{du}{dt}=a(t)u^{n}-b(t)u^{n+(2k+1)},\quad n,k\in \mathbb{N}, \label{2}$$% where the carrying capacity $a(t)$ and the self-limiting coefficient $b(t)$ are continuous and bounded functions: 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}$be positive constants such that 0a-B(m-\varepsilon)^{2k+1}:=\varepsilon_{0}>0. \end{equation*} By integration onto$[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 $$\frac{du}{dt}=a(t)u^{n}-b(t)u^{n+(2k+1)},\quad n,k\in\mathbb{N}, \label{10}$$ under the assumption 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 $$\frac{du}{dt}=u-b(t)u^{4}, \label{13}$$% 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 $$\frac{dx}{dt}=-3x+3b(t). \label{14}$$% 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. \begin{thebibliography}{9} \bibitem{Am} H. Amann, \emph{Ordinary Differential Equations: An Introduction to Nonlinear Analysis}, Walter de Gruyter and Co., New York, 1990. \bibitem{Blat} J. Blat and K. J. 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