\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2003(2003), No. 121, 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 (login: ftp)} \thanks{\copyright 2003 Texas State University-San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE--2003/121\hfil Optimal impulsive harvest policy] {Optimal impulsive harvest policy for time-dependent logistic equation with periodic coefficients} \author[Ling Bai \& Ke Wang\hfil EJDE--2003/121\hfilneg] { Ling Bai \& Ke Wang} % in alphabetical order \address{ Ling Bai\hfill\break Institute of Mathematics, Jilin University, Changchun, Jilin, 130061, China} \email{linglingbai@eyou.com} \address{ Ke Wang \hfill\break Department of Mathematics, Northeast Normal University, Changchun, Jilin, 130024, China} \email{shuaizs@hotmail.com} \date{} \thanks{Submitted March 4, 2003. Published December 9, 2003.} \thanks{Project Supported by grants 10171010 and 10201005 from the National Science Foundation \hfill\break\indent of China, and by grant 01061 from Major Project of Education Ministry of China.} \subjclass[2000]{92D25, 34A37} \keywords{Impulsive harvest equation, global attractor, \hfill\break\indent optimal impulsive harvest policy} \begin{abstract} We study a time-dependent logistic equation with periodic coefficients. First, we show that the impulsive harvest population equation has impulsive periodic solutions for constant effort harvest and for proportional harvest. Second, we investigate the optimal harvest effort that maximizes the sustainable yield per unit of time. Then we determine the corresponding optimal population levels. Our results generalize the results presented in \cite{z1}. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \section{Introduction} Most of the models for a single species dynamics have been derived from a differential equation of the form $$\dot{x}=xf(x,t)-g(t,x),\eqno{(1.1)}$$ where the solution $x=x(t)$ is the density (size, or biomass) of the resource population at time $t>0$, the function $f=f(t,x)$ is characterized by the population change at the moment $t$, the function $g=g(t,x)$ describes the continuous influences of outside factors, such as hunting, cutting down the space available, etc.. Various choices of the functions $f$ and $g$ lead us to various models. When we only consider an isolated population without any perturbations, namely $g(t,x)=0$, the classical model is the logistic equation $\begin{gathered} \dot{x}=rx(1-\frac{x}{K})\\ x(0)=x_0\,, \end{gathered} \eqno{(1.2)}$ or $\begin{gathered} \dot{x}=r(t)x(1-\frac{x}{K(t)})\\ x(t_0)=x_0\,, \end{gathered} \eqno{(1.3)}$ where (1.2) is an autonomous evolutionary model, and (1.3) is treated as the non-autonomous evolutionary model because the coefficients of (1.3) are dependent on the time $t$. In a real evolutionary processes of the population, the perturbation or the influence from outside occurs immediately" as impulses, and not continuously. The duration of these perturbations is negligible compared to the duration of the whole process. For instance, as we know, fisherman can not fish day and night in 24 hours. Instead, they only fish in some time every day. Furthermore, the seasons also decide the fishing period. So the problem of impulsive harvest is more practical and realistic compared to the continuous harvest. However, to the best of our knowledge, there no results on impulsive harvest for renewable resources in the literature. In this paper, we research optimal impulsive harvest policy for a single population resource. The organization of this article is as follows: In the next section, we establish the mathematical model for impulsive time harvest for the well known logistic equation. We also obtain the maximum of increasing density of population per unit time. In subsequent portions of this paper, the main results on the existence and the stability of impulsive periodic solution for the impulsive equation are proved. Then the optimal impulsive harvest policies are determined for both constant effort harvest and for harvest proportional to the size of the population. \section{The impulsive harvest model} Considering the feasible operation, we suppose that we harvest once every time $T$ for the population $X$ which obeys the logistic growth law. We shall establish the mathematical model of impulsive time harvest for the logistic equation: $\begin{gathered} \frac{dN}{dt}=r(t)N\big( 1-\frac{N}{K(t)}\big)-\delta(s(t))Eh(N(t)) \\ N(t_0)=N_0\,. \end{gathered}\eqno{(2.1)}$ Here, assume that $r$ and $K$ are both positive $T$-periodic functions with respect to $t$. $h(N(t))$ is the function of general harvest; $\delta$ is the Dirac impulsive function, which satisfies $\delta(0)=\infty$ and $\delta(s)=0$ for $s\neq 0$ and $\int_{-\infty}^{\infty}\delta(s)ds=1$, and $$s(t)=\begin{cases} 0 &\mbox{if } t=nT,\; n\in N,\\ -1 &\mbox{otherwise.} \end{cases}$$ From this explanation, it is obvious that the population $X$ will increase according to logistic increasing curve without exploitation and the management of the resource will harvest $Eh(N(t))$ every time $T$. For explaining the latter, we discuss the impulse function $\delta$ deliberately. As is well known, the Heaviside function satisfies $$\theta(t)=\begin{cases} 1 &\mbox{if }t\geq 0,\\ 0 &\mbox{if }t<0.\end{cases}$$ Using generalized derivatives, $\theta' =\delta$. Thus, if $t\neq nT$, $s(t)=-1$ and $\theta(s(t))=0$, namely, the management does not harvest. If $t=nT$, $s(t)=0$ and $\theta(s(t))=1$, namely, in $nT$, the management harvests $Q(nT)$, which satisfies $$Q(nT)=\int_{-\infty}^{nT}\delta(s(t))Eh(x(t))dt -\int_{-\infty}^{(n-1)T}\delta(s(t))Eh(x(t))dt = Eh(x(nT)).$$ Clearly, the general solution of (1.3) may be written in the form $$x(t,t_0,x_0)=\Big(\frac{1}{x_0}\exp\big\{ - \int_{t_0}^{t} r(s)ds \big\} + \int_{t_0}^{t} \frac{r(s)}{K(s)}\exp \big\{- \int_{s}^{t}r(\tau)d\tau \big\} ds \Big)^{-1}.$$ For convenience, denote $x(t,t_0,x_0)=\frac{1}{\frac{1}{x_0}A(t)+B(t)}=\frac{x_0}{A(t)+B(t)x_0}$, where $$A(t)=\exp\big\{ - \int_{t_0}^{t} r(s)ds \big\},\quad B(t)=\int_{t_0}^{t} \frac{r(s)}{K(s)} \exp \big\{ - \int_{s}^{t}r(\tau)d\tau \big\} ds. \eqno{(2.2)}$$ For biological considerations, we are interested only in positive solutions. In this paper, we always need $x_0>0$. After time $T$, the increase of population in (1.3) without harvest is $x(T,0,x_0)-x_0=:f(x_0)$. Then $f(x_0)=\frac{x_0}{A(T)+B(T)x_0}-x_0\,. \eqno{(2.3)}$ In the following, our objective is to find an $x_0$ such that $f(x_0)$ reaches its maximum at $\bar x_0$. Let $f'(x_0)=0$, so we have $$x_0^{1}=\frac{-A(T)+\sqrt{A(T)}}{B(T)}>0\,,\quad x_0^{2}=\frac{-A(T)-\sqrt{A(T)}}{B(T)}<0\,.$$ Furthermore, $f''(x_0^{1})<0$, then $\bar{x}_0=x_0^1$. Thus the maximum of increasing density of population is $\omega=:\max f(x_0)=f(\bar{x}_0) =\frac{\big(1-\sqrt{A(T)}\big)^2}{B(T)},\eqno{(2.4)}$ and the maximum of increasing density of population per unit of time is $\max \frac{f(x_0)}{T}=\frac{f(\bar{x}_0)}{T} =\frac{\big(1-\sqrt{A(T)}\big)^2}{B(T)T}.\eqno{(2.5)}$ \section{Optimal impulsive harvest policy for constant effort harvest} Now, we consider single population $X$ of size $N(t)$, which obeys the logistic growth law, is impulsively harvested by means of a constant effort, $h(N)\equiv 1$, namely, every time $T$, the management harvest constant is $E$. Equation of the impulsively harvested population reads $\begin{gathered} \frac{dN}{dt}=r(t)N\left( 1-\frac{N}{K(t)}\right)-\delta(s(t))E\,,\\ N(t_0)=N_0\,. \end{gathered} \eqno{(3.1)}$ We always denote the solution of (3.1) by $N(t,t_0,N_0)$, while represent $x(t,t_0,x_0)$ as the solution of (1.3) without harvest. It is known that the solution of a nonautomated system with T-periodic coefficients has the property of periodic translation, we can denote $x(t,t_0,x_0)$ and $x(t-nT,t_0-nT,x_0)$ as the same solution of a system. \begin{theorem} \label{thm3.1} (1) If 0 0,meanwhile, it is easy to see that the equation F(y)=0 has two roots, that is \begin{gather*} y_1=\frac{1-A(T)-EB(T)- \sqrt{(1-A(T)-EB(T))^2-4EA(T)B(T)}}{2B(T)},\; n \in N, \\ y_2=\frac{1-A(T)-EB(T)+ \sqrt{(1-A(T)-EB(T))^2-4EA(T)B(T)}}{2B(T)},\; n \in N. \end{gather*} It follows that y_2>y_1>0. Next, we prove that N(t,0,y_1) and N(t,0,y_2) are T-periodic solutions. It is obvious that \begin{align*} N(T,0,y_1)&=x(T,0,y_1)-E=x(T,0,y_1)-y_1-E+y_1\\ &=f(y_1)-E+y_1=F(y_1)+y_1=y_1=N(0,0,y_1), \end{align*} and N(2T,0,y_1)=N(2T,T,N(T,0,y_1))=x(2T,T,y_1)-E=x(T,0,y_1)-E=y_1\,. $$Therefore. we obtain inductively$$ N(nT,0,y_1)=y_1\quad \hbox{for all } n \in N. $$Similarly, we have$$ N(nT,0,y_2)=y_2=N(0,0,y_2)\quad\hbox{for all } n \in N. $$Let N(t,0,y_1)=\xi_1 (t), N(t,0,y_2)=\xi_2 (t). Then \xi_1 (t) and \xi_2 (t) are impulsive periodic solutions of (3.1) with \xi_1 (nT)=y_1, \xi_2 (nT)=y_2 for all n \in N. If E=\omega, then F(y)=0 has one and only one root with y_3=\frac{1-A(T)-EB(T)}{2B(T)}, so (3.1) has only one impulsive periodic solution \xi_3 (t) with$$ \xi_3(nT)=\frac{1-A(T)-EB(T)}{2B(T)},~~\forall n \in N. $$The proof is completed. \end{proof} \begin{theorem} \label{thm3.2} (1) If E < \omega, then N(t,0,N_0) \to \xi_2 (t) as t \to +\infty for N_0 >y_1 and N(t,0,N_0) \to 0 for 0y_3 and N(t,0,N_0) \to 0 for 0\omega, then N(t,0,N_0) \to 0 as t \to +\infty for all N_0 >0. \end{theorem} \begin{proof} First, we know that F(y)>0 for y_1y_2. Suppose E< \omega and N_0>y_2. For convenience, denote N_n=N(nT,0,N_0). We can write$$ N_1=N(T,0,N_0)=x(T,0,N_0)-E=f(N_0)+N_0-E=F(N_0)+N_0y_2 implies $$N_1=x(T,0,N_0)-E>x(T,0,y_2)-E=N(T,0,y_2)=\xi_2 (T)=y_2.$$ Similarly, we have \begin{align*} N_2&=N(2T,0,N_0)=N(2T,T,N_1)\\ &=x(2T,T,N_1)-E=x(T,0,N_1)-E\\ &=f(N_1)+N_1-E=F(N_1)+N_1x(T,0,y_2)-E=\xi_2 (T)=y_2\,. Therefore, by the same arguments we can obtain a monotone decreasing sequence \{N_n\} with a lower bound y_2. It is obvious that the sequence \{N_n\} has a limit, suppose it is \beta, then \beta \geq y_2. If \beta >y_2, then \begin{align*} N_{n+1}-N_n&=N((n+1)T,0,N_0)-N_n=N((n+1)T,nT,N_n)-N_n\\ &=x((n+1)T,nT,N_n)-N_n=x(T,0,N_n)-E-N_n=F(N_n)\,. \end{align*} Therefore, F(\beta)=0 as n \to \infty. Because F(y)=0 has only two roots y_1 or y_2, we get a contradiction. Thus \beta=y_2, that is \lim_{n \to \infty}N_n=\beta=y_2. According to the continuous dependence of solution on initial value in finite time, for any given \epsilon >0 there is a \delta \in (0,\epsilon), such that |x_0-y_2|< \delta  implies |x(t,0,x_0)-x(t,0,y_2)|<\epsilon for t \in [0,T). In addition, we know that \lim_{n \to \infty}N_n=\beta, for the previous \delta, there exists a natural number \bar{N} such that n \geq \bar{N} implies that 0y_2. If y_1y_2 or y_1 y_2 or y_1\omega, the population approaches 0 for any initial level N_0 in a finite time. If E=\omega, there exists a unique positive impulsive periodic solution \xi_3 (t) of (3.1) with \xi_3 (nT)=\frac{1-A(T)-EB(T)}{2B(T)}, which is semi-stable" in the sense that N(t,0,N_0) approaches \xi_3 (t) if N_0> y_3=\xi_3 (T), but N(t,0,N_0) approaches 0 if N_00 satisfies: \lim_{t \to +\infty}|N(t,0,N_0)-\xi(t)|=0. $$\begin{theorem} \label{thm4.1} If 00 for 0\tilde{y}. Next we prove that N(t,0,\tilde{y}) is impulsive periodic solution of (4.1). It is easy to see that$$ N(T,0,\tilde{y})=(1-E)x(T,0,\tilde{y})=G(\tilde{y})+\tilde{y}=\tilde{y} and \begin{align*} N(2T,0,\tilde{y})&=N(2T,T,N(T,0,\tilde{y}))=N(2T,T,\tilde{y})\\ &=(1-E)x(2T,T,\tilde{y})=(1-E)x(T,0,\tilde{y})=\tilde{y}. \end{align*} Inductively, we prove that N(nT,0,\tilde{y})=\tilde{y} for all n \in N. Therefore, (4.1) has unique impulsive periodic solution N(t,0,\tilde{y}):=\xi(t) with \xi(nT)= \tilde{y} for \forall n \in N. Next, we prove the global attractiveness of \xi(t). Suppose that N_0>\tilde{y}, and N_n:=N(nT,0,N_0), n \in N. We have N_1=N(T,0,N_0)=(1-E)x(T,0,N_0)=G(N_0)+N_0(1-E)x(T,0,\tilde{y})=N(T,0,\tilde{y})=\tilde{y}. Similarly, we can prove that \tilde{y}\tilde{y}, then \begin{align*} N_{n+1}-N_n&=N((n+1)T,0,N_0)-N_n\\ &=N((n+1)T,nT,N_n)-N_n\\ &=(1-E)x((n+1)T,nT,N_n)-N_n\\ &=(1-E)x(T,0,N_n)-N_n=G(N_n), \end{align*} which implies that G(\tilde{\beta})=0, this contradicts with the fact that the equation G(y)=0 has a unique root \tilde{y}. Thus \tilde{\beta}=\tilde{y}  and we have proved that \lim_{n \to +\infty}N_n=\tilde{\beta}=\tilde{y}. $$Therefore, for any given \epsilon >0, there is a \delta \in (0,\epsilon) such that n>\tilde{N} implies 0\tilde{y}.$$ By a similar argument, we can prove $$\lim_{t \to \infty}|N(t,0,N_0)-\xi(t)|=0 \quad\hbox{for } 00, the size of population X tends to extinction. \end{theorem} In real life, fishers would like to make a decision how to obtain maximum harvest. From Theorem \ref{thm4.1}, when T is a fixed constant, the sustainable yield per unit time is$$ Y(E)=E\frac{1-E-A(T)}{B(T)T(1-E)}.\eqno{(4.3)} $$Our objective is to find an E^{*} such that Y(E) reaches its maximum at E=E^{*}. This is the optimization of a function. The derivative of Y(E) is written as$$ Y'(E)=\frac{E^2-2E+1-A(T)}{TB(T)(1-E)^{2}}, $$then the equation E^{2}-2E-A(T)+1=0 has two roots, which are E_1=1+ \sqrt{A(T)}>1 and E_2=1- \sqrt{A(T)}<1. Furthermore, we can obtain$$ Y'' (E)= \frac{2A(T)}{TB(T)(-1+E)^3}<0,\quad \forall 0