\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