Adaptive Release of Natural Enemies in a Pest-Natural Enemy System with Pesticide Resistance

Abstract

Integrated pest management options such as combining chemical and biological control are optimal for combating pesticide resistance, but pose questions if a pest is to be controlled to extinction. These questions include (i) what is the relationship between the evolution of pesticide resistance and the number of natural enemies released? (ii) How does the cumulative number of natural enemies dying affect the number of natural enemies to be released? To address these questions, we developed two novel pest-natural enemy interaction models incorporating the evolution of pesticide resistance. We investigated the number of natural enemies to be released when threshold conditions for the extinction of the pest population in two different control tactics are reached. Our results show that the number of natural enemies to be released to ensure pest eradication in the presence of increasing pesticide resistance can be determined analytically and depends on the cumulative number of dead natural enemies before the next scheduled release time.

Appendices

Appendix A: The Deduction of Eq. (2) and Its Solution

Because P _s =ωP and P _r =(1−ω)P, and d ₁ is the mortality rate of susceptible pests, d ₂ is the mortality rate of resistant pests when the pesticide is sprayed, so we have

$$\left \{ \begin{aligned} &\frac{dP_s}{dt}=\omega rP(1-\eta P)-d_1P_s, \\ &\frac{dP_r}{dt}=(1-\omega) rP(1-\eta P)-d_2P_r. \end{aligned} \right . $$

However, for simplification we assume that the resistant pests display near-complete resistance to the pesticide, which means that d ₂≈0. Therefore,

$$\frac{dP}{dt}=\frac{dP_s}{dt}+\frac{dP_r}{dt}=rP(1-\eta P)-\omega d_1P. $$

Then we get

$$\frac{d\omega}{dt} =\frac{d}{dt} \biggl(\frac{P_s}{P} \biggr)=d_1\omega(\omega-1). $$

This resistance evolution equation has been widely used recently in different fields (Hall et al. 2004; Bonhoeffer and Nowak 1997; Gubbins and Gilligan 1999; Laxminarayan and Simpson 2000; Milgroom et al. 1989).

In practice, the evolution of pesticide resistance is dependent on the dosage of the pesticide applications, the frequency of applications or the pesticide application period. Although linking the evolution of pesticide resistance to the pest growth model is a great challenge, it is well known that the less frequent are the pesticide applications (i.e. the longer the period between them), the slower the development of the pest resistance. One possible way is to consider the effect of each pulse spraying of pesticides on the evolution of pesticide resistance as a perturbation constant, i.e. we have

$$\frac{d\omega(t)}{dt} =d_1\omega \bigl(\omega^{q_i}-1 \bigr), \quad \tau_{i-1}\leq t\leq \tau_i,\ i\in \mathcal{N}, $$

with initial value ω(τ _i−1) at each time interval t∈[τ _i−1,τ _i ] and ω(τ ₀)=ω(0)=ω ₀ is given. Where q _i denotes the perturbation constant of the ith pulse spraying of pesticides, which depends the number of pesticide applications, dosage X _i of the ith pesticide application and time interval Δτ _i . This equation resembles the classical Richards equation. As such, we anticipate the approach in resistant equation can be used to describe analytically the functional format of q _i .

The analytical solution of ω(t) can be determined as follows:

$$ \omega(t)= \bigl(1+e^{q_id_1(t-\tau_{i-1})} \bigl(\bigl(\omega( \tau_{i-1})\bigr)^{-q_i} -1 \bigr) \bigr)^{-\frac{1}{q_i}},\quad \tau_{i-1}\leq t\leq\tau_i, $$

(22)

which indicates that

$$ \omega(\tau_{i})= \bigl(1+e^{id_1} \bigl( \omega(\tau_{i-1})^{-q_i} -1 \bigr) \bigr)^{-1/q_i}. $$

(23)

For simplification, we assume that the same dosage of pesticides is applied each time, and without loss of generality we let X _i =1 for all $i\in \mathcal{N}$. It follows from the main text that one possible definition of q _i is as follows: q _i =i/Δτ _i , and then Δτ _i =T for all $i\in \mathcal{N}$. For this special case, the evolution of ω at each time point nT can be expressed as

$$ \omega(nT)= \bigl(1+e^{nd_1} \bigl(\omega\bigl((n-1)T \bigr)^{-n/T} -1 \bigr) \bigr)^{-T/n},\quad n\in \mathcal{N}. $$

(24)

Appendix B: The Proof of Theorem 2.1

For any given ε>0 (ε small enough), we will first show that there exists a τ(ε)>0 such that

$$ P(\tau)\leq \varepsilon e^{-\beta T}. $$

(25)

Otherwise, P(t)≥εexp(−βT) for all t≥0. It follows from (b) that ∂F/∂P≤0 and then

$$F\bigl(s, P(s)\bigr)\leq F \bigl(s, \varepsilon e^{-\beta T} \bigr),\quad \mbox{for } s\geq 0. $$

Since

$$F \bigl(s,\varepsilon e^{-\beta T} \bigr)-F \biggl(s,\frac{\varepsilon}{2} e^{-\beta T} \biggr)=\frac{\varepsilon}{2} e^{-\beta T}\frac{\partial F}{\partial P} \bigl(s,\eta (s)\bigr), $$

where $\frac{\varepsilon}{2} \exp{(-\beta T)}<\eta (s)<\varepsilon \exp{(-\beta T)}$. Again assumption (b) implies that

$$\frac{\partial F}{\partial P}\bigl(s,\eta (s)\bigr)\leq -\varphi_0 \lambda(s), $$

where φ ₀=min{φ(P): exp(−βT)ε/2≤P≤εexp(−βT)}. Thus, we get

$$F\bigl(s, P(s)\bigr)\leq F\biggl(s,\frac{\varepsilon}{2} e^{-\beta T}\biggr)- \frac{\varepsilon}{2} e^{-\beta T}\varphi_0 \lambda(s)\leq F(s,0)- \frac{\varepsilon}{2} e^{-\beta T}\varphi_0 \lambda(s). $$

For any t≥0, there is a $h\in \mathcal{N}$, such that (h−1)T<t≤hT. Therefore,

$$\begin{aligned} \varepsilon e^{-\beta T} \leq&P(t) \\ =&\prod^{h-1}_{i=1}q(iT)P_0 \exp \biggl(\!\biggl(\int^{T}_{0}+\int ^{2T}_{T}+\cdots +\int^{(h-1)T}_{(h-2)T}+ \int^{t}_{(h-1)T}\biggr)F\bigl(s,P(s)\bigr)ds \biggr) \\ \leq& \prod^{h-1}_{i=1}q(iT)P_0 \exp \biggl(\biggl(\int^{T}_{0}+\cdots +\int ^{(h-1)T}_{(h-2)T}+\int^{t}_{(h-1)T} \biggr)\\ &{}\times\biggl(F(s,0)-\frac{\varepsilon}{2} e^{-\beta T}\varphi_0 \lambda(s) \biggr)ds \biggr) \\ =&\prod^{h-1}_{i=1}q(iT)P_0 \exp \biggl(\biggl(\int^{T}_{0}+\cdots +\int ^{t}_{(h-1)T}\biggr)F(s,0)ds \biggr)\\ &{}\times\exp \biggl(\int ^{t}_{0}\biggl(-\frac{\varepsilon}{2} e^{-\beta T}\varphi_0 \lambda(s)\biggr)ds \biggr). \end{aligned}$$

For ε>0 small enough and assumption (c) yields

$$q(iT)\exp{\int^{iT}_{(i-1)T}F \biggl(s, \frac{\varepsilon}{2} e^{-\beta T} \biggr)ds}\leq 1,\quad i=1,2,\ldots,h, $$

and F(s,0)≤β. Hence,

$$\varepsilon e^{-\beta T}\leq P(t)\leq P_0e^{\beta T}\exp \biggl(\int^{t}_{0}\biggl(-\frac{\varepsilon}{2} e^{-\beta T}\varphi_0 \lambda(s)\biggr)ds \biggr)\to 0 $$

as t→∞, in particular, P(t)<εexp(−βT) for t large enough. This contradiction establishes that (25) is true for some τ.

Now we will prove that

$$ P(t)\leq\varepsilon\quad \mbox{for } t\geq \tau. $$

(26)

Suppose that P(τ ₁)>ε≥εexp(−βT) for some τ ₁>τ and there is a $n\in \mathcal{N}$, such that (n−1)T≤τ ₁<nT. Then there exists a t ₁∈(τ,τ ₁), such that

$$P(t_1)=\varepsilon e^{-\beta T}, $$

and $P(t)>\varepsilon \exp{(-\beta T)},\ \rm{for}\ t\in (t_{1},\tau_{1}]$, which means that for any $k\in \mathcal{N}$, t ₁≠kT, due to 0≤q(kT)<1. Therefore, either t ₁>(n−1)T or t ₁<(n−1)T. If t ₁>(n−1)T, (this indicates τ ₁>(n−1)T), then τ ₁−t ₁<T and

$$\begin{aligned} \varepsilon<P(\tau_1) =&P(t_1)\exp \biggl(\int ^{\tau_1}_{t_1}F\bigl(s,P(s)\bigr)ds \biggr) \\ \leq &\varepsilon e^{-\beta T}\exp \biggl(\int^{\tau_1}_{t_1}F \bigl(s,\varepsilon e^{-\beta T}\bigr)ds \biggr) \\ \leq &\varepsilon e^{-\beta T}e^{\beta T}=\varepsilon, \end{aligned}$$

this is a contradiction. So t ₁<(n−1)T, then there is a $m\in \mathcal{N}$, m<n such that (m−1)T<t ₁<mT (let t ₁=(m−1)T+l, 0<l<T) and τ ₁−t ₁<T, otherwise, τ ₁−t ₁≥T, i.e. τ ₁≥T+t ₁, and P(t)>εexp(−βT), t∈(t ₁,t ₁+T]. Therefore, (m−1)T<t ₁<mT<t ₁+T<(m+1)T, and solving (3) from t ₁ to t ₁+T, we get

$$\begin{aligned} \varepsilon e^{-\beta T} <&P(t_1+T) \\ =&q(mT)P(t_1)\exp \biggl(\int^{mT}_{t_1}F \bigl(s,P(s)\bigr)ds \biggr) \exp \biggl(\int_{mT}^{t_1+T}F \bigl(s,P(s)\bigr)ds \biggr) \\ \leq&q(mT)\varepsilon e^{-\beta T}\exp \biggl(\int^{t_1+T}_{t_1}F \bigl(s,\varepsilon e^{-\beta T}\bigr)ds \biggr) \\ =&q(mT)\varepsilon e^{-\beta T}\exp \biggl(\int^{mT+l}_{(m-1)T+l}F \bigl(s,\varepsilon e^{-\beta T}\bigr)ds \biggr) \\ \leq&\varepsilon e^{-\beta T}, \end{aligned}$$

this is a contradiction, thus τ ₁−t ₁<T. Therefore, if τ ₁>(n−1)T, then

$$\begin{aligned} \varepsilon <&P(\tau_1) \\ =&q\bigl((n-1)T\bigr)P(t_1)\exp \biggl(\int^{(n-1)T}_{t_1}F \bigl(s,P(s)\bigr)ds \biggr) \exp \biggl(\int_{(n-1)T}^{\tau_1}F \bigl(s,P(s)\bigr)ds \biggr) \\ =&q\bigl((n-1)T\bigr)P(t_1)\exp \biggl(\int_{t_1}^{\tau_1}F \bigl(s,P(s)\bigr)ds \biggr) \\ \leq&q\bigl((n-1)T\bigr)P(t_1)\exp \biggl(\int _{t_1}^{\tau_1}F\bigl(s,\varepsilon e^{-\beta T} \bigr)ds \biggr) \\ \leq&\varepsilon e^{-\beta T}e^{\beta T}=\varepsilon, \end{aligned}$$

it is a contradiction, if τ ₁=(n−1)T, then

$$\begin{aligned} \varepsilon <&P(\tau_1) \\ =&P(t_1)\exp \biggl(\int^{(n-1)T}_{t_1}F \bigl(s,P(s)\bigr)ds \biggr) \\ \leq&P(t_1)\exp \biggl(\int_{t_1}^{(n-1)T}F \bigl(s,\varepsilon e^{-\beta T}\bigr)ds \biggr) \\ \leq&\varepsilon e^{-\beta T}e^{\beta T}=\varepsilon. \end{aligned}$$

This is a contradiction. Hence, such τ ₁ does not exist. Therefore, P(t)≤ε for t≥τ, that is lim _t→∞ P(t)=0. The proof is complete.

Appendix C: The Proof of Theorem 3.1

It is seen from the second equation of system (4) that dN(t)/dt>−dN(t). Considering the following impulsive differential equation:

$$ \left \{ \begin{aligned} &\frac{dy(t)}{dt}=-dy(t), \quad t\neq nT, \\ &y\bigl(t^+\bigr)=y(t)+\delta_n,\quad t=nT, \\ &y\bigl(0^+\bigr)=N_0+\delta_0. \end{aligned} \right . $$

(27)

According to the comparison theorem on impulsive differential equations, B1 yields N(t)≥y(t)=N ^∗(t). It follows from the first equation of system (4) that

$$\frac{dP(t)}{dt}\leq rP(t) \bigl(1-\eta P(t)\bigr)-\beta P(t)N^*(t). $$

Now we consider the following impulsive differential equation

$$ \left \{ \begin{aligned} &\frac{dx(t)}{dt}= rx(t) \bigl(1-\eta x(t) \bigr)-\beta x(t)N^*(t),\quad t\neq nT, \\ &x\bigl(nT^{+}\bigr)=\bigl(1-\omega (nT) d_1\bigr)x(nT), \quad t= nT, \\ &\frac{d\omega(t)}{dt}=d_1\omega(t) \bigl(\omega(t)^{q_n}-1 \bigr), \\ &x\bigl(0^+\bigr)=P(0)\doteq P_0. \end{aligned} \right . $$

(28)

Again according to the comparison theorem on impulsive differential equations we have P(t)≤x(t).

By using the formula of (24), we can easily have

$$\begin{aligned} q(nT) \doteq &1-\omega (nT)d_1 \\ =&1-\frac{d_1}{ (1+e^{nd_1} ((\omega((n-1)T))^{-\frac{n}{T}} -1 ) )^{\frac{T}{n}}} \end{aligned}$$

and

$$F(s,x)\doteq r-r\eta x(s)-\beta N^*(s). $$

Now we test and verify the conditions of Theorem 2.1. It is easy to see that condition (a) holds true naturally, and

$$F(s,0)=r-\beta N^*(s)\leq r $$

and

$$\frac{\partial F(s,x)}{\partial x}=-r\eta,\qquad \int^{\infty}_{0}\eta ds=\infty. $$

Therefore,

$$\begin{aligned} \exp \biggl(\int^{l+nT}_{l+(n-1)T}F(s,0)ds \biggr) =& \exp \biggl(\int^{l+nT}_{l+(n-1)T}r-\beta N^*(s)ds \biggr) \\ =&e^{rT}\exp \biggl(\int^{l+nT}_{l+(n-1)T} \bigl(-\beta N^*(s)\bigr)ds \biggr) \end{aligned}$$

with

$$\begin{aligned} &\exp \biggl(\int^{l+nT}_{l+(n-1)T}\bigl(-\beta N^*(s) \bigr)ds \biggr) \\ &\quad{}=\exp \Biggl( \Biggl(\frac{\beta N_0}{d}e^{-d(l+(n-1)T)}+\sum ^{n-1}_{i=0}\frac{\beta \delta_i}{d}e^{-d(l+(n-1-i)T)} \Biggr) \bigl(e^{-dT}-1 \bigr)\\ &\qquad{}+\frac{\beta \delta_n}{d}\bigl(e^{-dl}-1\bigr) \Biggr) \\ &\quad{}=\exp{ \bigl(M(n,l) \bigr)}. \end{aligned}$$

Thus,

$$\begin{aligned} &q(nT)\exp{\biggl(\int^{l+nT}_{l+(n-1)T}F(s,0)ds\biggr)} \\ &\quad{}=R_0(n,T)\exp{ \bigl(M(n,l) \bigr)} \\ &\quad{}\doteq R_0^{N}(n,T,l). \end{aligned}$$

According to Theorem 2.1, we can see that if $R_{0}^{N}(n,T,l)\leq 1$, then x(t)→0 as t→∞. Consequently, we have P(t)→0 as t→∞ provided $R_{0}^{N}(n,T,l)\leq1$.

Next, we prove that N(t)→N ^∗(t) as t→∞. For any 0<ε<d/(λβ), there exists a t ₁>0 such that 0<P(t)<ε for all t≥t ₁. Without loss of generality, we may assume that 0<P(t)<ε holds true for all t>0, then we have

$$-dN(t)\leq\frac{dN(t)}{dt}\leq(\lambda\beta\varepsilon-d)N(t). $$

For the left-hand inequality, it follows from impulsive differential equation (27) that N(t)≥y(t)=N ^∗(t). For the right-hand inequality, considering the following impulsive differential equation:

$$ \left \{ \begin{aligned} &\frac{dz(t)}{dt}=(\lambda\beta \varepsilon-d)z(t), \quad t\neq nT, \\ &z\bigl(t^+\bigr)=z(t)+\delta_n,\quad t=nT, \\ &z\bigl(0^+\bigr)=N_0+\delta_0. \end{aligned} \right . $$

(29)

The analytical solution of the above system at any impulsive interval ((n−1)T,nT] gives

$$ z^*(t)=N_0e^{(\lambda\beta\varepsilon-d)t}+\sum ^{n-1}_{i=0}\delta_ie^{(\lambda\beta\varepsilon-d)(t-iT)}, \quad (n-1)T<t\leq nT. $$

(30)

Therefore, for any ε ₁>0, there exists a t ₂>0 such that

$$N^*(t)-\varepsilon_1<N(t)<z^*(t)+\varepsilon_1 $$

for t>t ₂. Let ε→0, then we have

$$N^*(t)-\varepsilon_1<N(t)<N^*(t)+\varepsilon_1 $$

for t>t ₂, which indicates that N(t)→N ^∗(t) as t→∞. Therefore, the pest-free solution (7) is globally attractive if $R_{0}^{N}(n,T,l)\leq1$. The proof is complete.

Appendix D: Determining the New Number of Natural Enemies to be Released for Case 3.2

In this case $R^{N}_{0_{\max}}(n,T,l)=R_{0}^{N}(n,T,T)= R_{C}$. It follows from (8) that we have M(n,T)=ln(R _c /R ₀(n,T)), that is

$$\Biggl(\frac{\beta N_0}{d}e^{-dnT}+\sum^{n}_{i=0} \frac{\beta \delta_i}{d}e^{-d(n-i)T} \Biggr) \bigl(e^{-dT}-1 \bigr)=\ln \biggl(\frac{R_C}{R_0(n,T)} \biggr), $$

this indicates that

$$ G_{n^{'}}+\sum^{n}_{i=n^\prime+1} \delta_i e^{diT}= -\frac{de^{dnT}}{\beta (1-e^{-dT} )}\ln \biggl( \frac{R_C}{R_0(n,T)} \biggr). $$

(31)

That is

$$\sum^{n}_{i=n^\prime+1} \delta_i e^{diT}= -\frac{de^{dnT}}{\beta (1-e^{-dT} )}\ln \biggl(\frac{R_C}{R_0(n,T)} \biggr)-G_{n^{'}}\doteq A_n. $$

Therefore, when n=n′+1 we get

$$\delta_{n^\prime+1}=A_{n^\prime+1}e^{-d(n^\prime+1)T}. $$

When n=n′+2, we have

$$\delta_{n^\prime+1}e^{d(n^\prime+1)T} +\delta_{n^\prime+2}e^{d(n^\prime+2)T}=A_{n^\prime+2}, $$

that is

$$\delta_{n^\prime+2}=(A_{n^\prime+2}-A_{n^\prime+1})e^{-d(n^\prime+2)T}. $$

Similarly, when n=n′+3 we get

$$\delta_{n^\prime+1}e^{d(n^\prime+1)T} +\delta_{n^\prime+2}e^{d(n^\prime+2)T}+ \delta_{n^\prime+3}e^{d(n^\prime+3)T}=A_{n^\prime+3}, $$

that is

$$\delta_{n^\prime+3}=(A_{n^\prime+3}-A_{n^\prime+2})e^{d(n^\prime+3)T}. $$

By induction, the new number of natural enemies to be released δ _n can be determined as follows:

$$\delta_n=\left \{\begin{array}{l} \delta_c, \quad \mbox{if } n\leq n^{\prime},\\ A_{n^{\prime}+1}e^{-d(n^\prime+1)T},\quad n=n^{\prime}+1,\\ (A_{n}-A_{n-1})e^{-dnT},\quad \mbox{if } n>n^{\prime}+1. \end{array} \right . $$

Appendix E: The Proof of Theorem 4.1

It is seen from the second equation of system (15) that dN(t)/dt>−dN(t). Considering the following impulsive differential equation:

$$ \left \{ \begin{aligned} &\frac{dy(t)}{dt}=-dy(t), \quad t \neq hT_N, \\ &y\bigl(t^+\bigr)=y(t)+\delta_h,\quad t=hT_N, \\ &y\bigl(0^+\bigr)=N_0.\end{aligned} \right . $$

(32)

According to the comparison theorem on impulsive differential equations, (E1) yields N(t)≥y(t)=N ^∗(t). It follows from the first equation of system (15) that

$$\frac{dP(t)}{dt}\leq rP(t) \bigl(1-\eta P(t)\bigr)-\beta P(t)N^*(t). $$

Now we consider the following impulsive differential equation:

$$ \left \{ \begin{aligned} &\frac{dx(t)}{dt}= rx(t) \bigl(1-\eta x(t) \bigr)-\beta x(t)N^*(t),\quad t\neq \tau_n, \\ &x\bigl(hT_p^{+}\bigr)=\bigl(1-\omega (hT_p) d_1\bigr)x(hT_p),\quad t= \tau_n, \\ &\frac{d\omega(t)}{dt}=d_1\omega(t) \bigl(\omega(t)^{q_n}-1 \bigr), \\ &x\bigl(0^+\bigr)=P(0)\doteq P_0.\end{aligned} \right . $$

(33)

Again according to the comparison theorem on impulsive differential equations, we have P(t)≤x(t).

By using the formula of (24), we can easily have

$$\begin{aligned} q(nT) \doteq &1-\omega (nT)d_1 \\ =&1-\frac{d_1}{ (1+e^{nd_1} ((\omega((n-1)T))^{-\frac{n}{T}} -1 ) )^{\frac{T}{n}}} \end{aligned}$$

and

$$F(s,x)\doteq r-r\eta x(s)-\beta N^*(s). $$

Now we test and verify the conditions of Theorem 2.1. It is easy to see that condition (a) holds true naturally, and

$$F(s,0)=r-\beta N^*(s)\leq r $$

and

$$\frac{\partial F(s,x)}{\partial x}=-r\eta,\qquad \int^{\infty}_{0} \eta ds=\infty. $$

Therefore,

$$\begin{aligned} \exp \biggl(\int^{l+(h-1)T_N+\tau_k}_{l+(h-1)T_N+\tau_{k-1}}F(s,0)ds \biggr) =& \exp \biggl(\int^{l+(h-1)T_N+\tau_k}_{l+(h-1)T_N+\tau_{k-1}}r-\beta N^*(s)ds \biggr) \\ =&e^{r\tau}\exp \biggl(\int^{l+(h-1)T_N+\tau_k}_{l+(h-1)T_N+\tau_{k-1}} \bigl(-\beta N^*(s)\bigr)ds \biggr), \end{aligned}$$

where k≥1, τ _k =kτ, and

$$\begin{aligned} &\int^{l+(h-1)T_N+\tau_k}_{l+(h-1)T_N+\tau_{k-1}}\bigl(-\beta N^*(s)\bigr)ds \\ &\quad{}=-\beta\int^{l+(h-1)T_N+\tau_k}_{l+(h-1)T_N+\tau_{k-1}} \Biggl(N_0+\sum^{h-1}_{i=1} \delta_ie^{idT_N} \Biggr)e^{-ds}ds \\ &\quad{}=\frac{\beta}{d}e^{-d(l+(h-1)T_N+\tau_{k-1})} \Biggl(N_0+\sum ^{h-1}_{i=1}\delta_ie^{idT_N} \Biggr) \bigl(e^{-d\tau}-1 \bigr) \end{aligned}$$

for l∈(0,τ] and 1≤k≤k _p , thus

$$\begin{aligned} &\hat{R}_{0}^{T_N}(h,k-1,T_N,l) \\ &\quad{}\doteq q \bigl((h-1)T_N+\tau_k \bigr)\exp \biggl(\int^{l+(h-1)T_N+\tau_k}_{l+(h-1)T_N+\tau_{k-1}}F(s,0)ds \biggr) \\ &\quad{}=R_0 \bigl((h-1)T_N+\tau_k \bigr) \\ &\qquad{}\times\exp{ \Biggl(\frac{\beta}{d}e^{-d(l+(h-1)T_N+\tau_{k-1})} \Biggl(N_0+\sum^{h-1}_{i=1} \delta_ie^{idT_N}\Biggr) \bigl(e^{-d\tau}-1\bigr) \Biggr)}, \end{aligned}$$

where R ₀((h−1)T _N +τ _k ,τ)=(1−d ₁ ω((h−1)T _N +τ _k ))exp(rτ).

Moreover,

$$\begin{aligned} &\int^{l+hT_N}_{l+(h-1)T_N+\tau_{k_p}}\bigl(-\beta N^*(s)\bigr)ds \\ &\quad{}=\int^{hT_N}_{l+(h-1)T_N+\tau_{k_p}}\bigl(-\beta N^*(s) \bigr)ds+\int^{l+hT_N}_{hT_N}\bigl(-\beta N^*(s)\bigr)ds \\ &\quad{}=-\beta\int^{hT_N}_{l+(h-1)T_N+\tau_{k_p}} \Biggl(N_0+\sum^{h-1}_{i=1} \delta_ie^{idT_N}\Biggr)e^{-ds}ds \\ &\qquad{}-\beta\int^{l+hT_N}_{hT_N} \Biggl(N_0+\sum^{h}_{i=1} \delta_ie^{idT_N}\Biggr)e^{-ds}ds \\ &\quad{}=\frac{\beta}{d}e^{-d(l+(h-1)T_N+\tau_{k_p})} \Biggl(N_0+\sum ^{h-1}_{i=1}\delta_ie^{idT_N} \Biggr) \bigl(e^{-d\tau}-1\bigr)+\frac{\beta\delta_{h}}{d}\bigl(e^{-dl}-1 \bigr) \\ &\quad{}\doteq \Psi(l), \end{aligned}$$

therefore,

$$\begin{aligned} \hat{R}_{0}^{T_N}(h,k_p,T_N,l) \doteq & q(hT)\exp{ \biggl(r-\int^{l+hT_N}_{l+(h-1)T_N+\tau_{k_p}}\beta N^*(s)ds \biggr)} \\ =& R_0(hT_N,\tau)e^{\Psi(l)}, \end{aligned}$$

for l∈(0,τ].

According to Theorem 2.1, we can see that if $\hat{R}_{0}^{T_{N}}(h,k-1,T_{N},l)\leq 1$, then x(t)→0 as t→∞. Consequently, we have P(t)→0 as t→∞ provided $\hat{R}_{0}^{T_{N}}(h,k-1,T_{N},l)\leq 1$, for k=1,2,…,k _p +1.

Because $\hat{R}_{0}^{T_{N}}(h,k-1,T_{N},l)$ increases with respect to k and l, for k=2,3,…,k _p , we have

$$\begin{aligned} &\hat{R}_{0_{\max}}^{T_N}(h,k-1,T_N,l) \\ &\quad{}=\hat{R}_{0}^{T_N}(h,k_p-1,T_N, \tau) \\ &\quad{}=R_0\bigl((h-1)T_N+\tau_{k_p},\tau \bigr) \\ &\qquad{}\times\exp \Biggl(\frac{\beta}{d}e^{-d((h-1)T_N+\tau_{k_p})} \Biggl(N_0+ \sum^{h-1}_{i=1} \delta_ie^{idT_N}\Biggr) \bigl(e^{-d\tau}-1\bigr) \Biggr). \end{aligned}$$

From

$$\begin{aligned} \frac{\partial \hat{R}_{0}^{T_N}(h,k_p,T_N,l)}{\partial l} =&\hat{R}_{0}^{T_N}(h,k_p,T_N,l) \frac{\partial \Psi(l)}{\partial l} \\ =&\hat{R}_{0}^{T_N}(h,k_p,T_N,l) \beta e^{-dl}(D_{h-1}-\delta_{h}), \end{aligned}$$

we conclude that if D _h−1>δ _h then $\hat{R}_{0}^{T_{N}}(h,k_{p},T_{N},l)$ increases with respect to l, so

$$\begin{aligned} &\hat{R}_{0_{\max}}^{T_N}(h,k_p,T_N,l) \\ &\quad{}=\hat{R}_{0}^{T_N}(h,k_p,T_N, \tau) \\ &\quad{}=R_0 (hT_N,\tau )\exp{ \Biggl( \frac{\beta}{d}e^{-dhT_N} \Biggl(N_0+\sum ^{h}_{i=1}\delta_ie^{idT_N} \Biggr) \bigl(e^{-d\tau}-1 \bigr) \Biggr)}. \end{aligned}$$

We can easily see that $\hat{R}_{0_{\max}}^{T_{N}}(h,k-1,T_{N},l)<\hat{R}_{0_{\max}}^{T_{N}}(h,k_{p},T_{N},l)$. Therefore,

$$\begin{aligned} R^{T_N}_{0}(h,T_N) =&\max \bigl\{ \hat{R}_{0_{\max}}^{T_N}(h,k-1,T_N,l), \hat{R}_{0_{\max}}^{T_N}(h,k_p,T_N,l) \bigr\} \\ =&\max \bigl\{\hat{R}_{0}^{T_N}(h,k_p-1,T_N, \tau),\hat{R}_{0}^{T_N}(h,k_p,T_N, \tau) \bigr\} \\ =&R_0 (hT_N,\tau )\exp{ \Biggl( \frac{\beta}{d}e^{-dhT_N} \Biggl(N_0+\sum^{h}_{i=1} \delta_ie^{idT_N} \Biggr) \bigl(e^{-d\tau}-1 \bigr) \Biggr)}. \end{aligned}$$

If D _h−1≤δ _h then $\hat{R}_{0}^{T_{N}}(h,k_{p},T_{N},l)$ decreases with respect to l, so

$$\begin{aligned} &\hat{R}_{0_{\max}}^{T_N}(h,k_p,T_N,l) \\ &\quad{}=\hat{R}_{0}^{T_N}(h,k_p,T_N,0) \\ &\quad{}=R_0 (hT_N,\tau )\exp{ \Biggl( \frac{\beta}{d}e^{-d((h-1)T_N+\tau_{k_p})}\Biggl(N_0+\sum ^{h-1}_{i=1}\delta_ie^{idT_N} \Biggr) \bigl(e^{-d\tau}-1\bigr) \Biggr)}. \end{aligned}$$

Due to the development of pest resistance, we can see that $R_{0}(hT_{N},\tau)> R_{0}((h-1)T_{N}+\tau_{k_{p}},\tau)$, thus $\hat{R}_{0}^{T_{N}}(h,k_{p},T_{N},0)>\hat{R}_{0}^{T_{N}}(h,k_{p}-1,T_{N},\tau)$, that is $\hat{R}_{0_{\max}}^{T_{N}}(h,k-1,T_{N},l)<\hat{R}_{0_{\max}}^{T_{N}}(h,k_{p},T_{N},l)$, therefore,

$$\begin{aligned} &R^{T_N}_{0}(h,T_N)\\ &\quad{}=\max \bigl\{ \hat{R}_{0_{\max}}^{T_N}(h,k-1,T_N,l), \hat{R}_{0_{\max}}^{T_N}(h,k_p,T_N,l) \bigr\} \\ &\quad{}=\max \bigl\{\hat{R}_{0}^{T_N}(h,k_p-1,T_N, \tau),\hat{R}_{0}^{T_N}(h,k_p,T_N,0) \bigr\} \\ &\quad{}=R_0 (hT_N,\tau )\exp{ \Biggl( \frac{\beta}{d}e^{-d((h-1)T_N+\tau_{k_p})} \Biggl(N_0+\sum^{h-1}_{i=1} \delta_ie^{idT_N}\Biggr) \bigl(e^{-d\tau}-1\bigr) \Biggr)}. \end{aligned}$$

thus, if $R^{T_{N}}_{0}(h,T_{N})<1$, then P(t)→0 as t→∞.

The following proof is the same as Theorem 3.1. The proof is complete.

Appendix F: Determining the New Number of Natural Enemies to Be Released for Case 4.2

In this case, solving equation

$$R^{T_N}_{0}(h,T_N)=R_C $$

with respect to δ _i , yields

$$ \sum^{h}_{i=1} \delta_ie^{idT_N}=-\frac{de^{dhT_N}}{\beta(1-e^{-d\tau})} \ln \biggl( \frac{R_C}{R_0(hT_N,\tau)} \biggr)-N_0,\quad h\in \mathcal{N}. $$

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Due to δ _i =δ _c , when i≤h′, we have

$$N_0+\delta_c\sum^{h^\prime}_{i=1}e^{idT_N}+ \sum^{h}_{i=h^\prime+1}\delta_ie^{idT_N} =-\frac{de^{dhT_N}}{\beta(1-e^{-d\tau})} \ln \biggl(\frac{R_C}{R_0(hT_N,\tau)} \biggr), $$

that is

$$\sum^{h}_{i=h^\prime+1}\delta_ie^{idT_N} =-\frac{de^{dhT_N}}{\beta(1-e^{-d\tau})} \ln \biggl(\frac{R_C}{R_0(hT_N,\tau)} \biggr) -N_0- \delta_c\sum^{h^\prime}_{i=1}e^{idT_N} \doteq A^\prime_{h}. $$

Therefore, when h=h′+1, we have

$$\delta_{h^\prime+1}=A^\prime_{h^\prime+1}e^{-d(h^\prime+1)T_N}. $$

When h=h′+2, we get

$$\delta_{h^\prime+1}e^{d(h^\prime+1)T_N}+\delta_{h^\prime+2}e^{d(h^\prime+2)T_N}=A^\prime_{h^\prime+2}, $$

that is

$$\delta_{h^\prime+2}=\bigl(A^\prime_{h^\prime+2}-A^\prime_{h^\prime+1} \bigr)e^{-d(h^\prime+2)T_N}. $$

Similarly, when h=h′+3, we get

$$\delta_{h^\prime+3}=\bigl(A^\prime_{h^\prime+3}-A^\prime_{h^\prime+2} \bigr)e^{-d(h^\prime+3)T_N}. $$

By induction, the new number of natural enemies to be released δ _h can be determined as follows:

$$\delta_h=\left \{\begin{array}{l} \delta_c, \quad \mbox{if } h\leq h^{\prime},\\ A^\prime_{h^\prime+1}e^{-d(h^\prime+1)T_N},\quad h=h^{\prime}+1,\\ (A^\prime_{h}-A^\prime_{h-1})e^{-dhT_N},\quad \mbox{if } h>h^{\prime}+1. \end{array} \right . $$