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Irreversible prey diapause as an optimal strategy of a physiologically extended Lotka–Volterra model

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We propose an optimal control framework to describe intra-seasonal predator–prey interactions, which are characterized by a continuous-time dynamical model comprising predator and prey density, as well as the energy budget of the prey over the length of a season. The model includes a time-dependent decision variable for the prey, representing the portion of the prey population in time that is active, as opposed to diapausing (a state of physiological rest). The predator follows autonomous dynamics and accordingly it remains active during the season. The proposed model is a generalization of the classical Lotka–Volterra predator–prey model towards non-autonomous dynamics that furthermore includes the effect of an energy variable. The model has been inspired by a specific biological system of predatory mites (Acari: Phytoseiidae) and prey mites (so-called fruit-tree red spider mites) (Acari: Tetranychidae) that feed on leaves of apple trees—its parameters have been instantiated based on laboratory and field studies. The goal of the work is to understand the decisions of the prey mites to enter diapause (a state of physiological rest) given the dynamics of the predatory mites: this is achieved by solving an optimization problem hinging on the maximization of the prey population contribution to the next season. The main features of the optimal strategy for the prey are shown to be that (1) once in diapause, the prey does not become active again within the same season and hence diapause is an irreversible process; (2) for the vast majority of parameter space, the portion of prey individuals entering diapause within the season does not decrease in time; (3) with an increased number of predators, the optimal population strategy for the prey is to start diapause earlier and to enter diapause more gradually. This optimal population strategy will be studied for its ESS properties in a sequel to the work presented in this article.

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  1. Mixed strategies are often referred to as singular or intermediate strategies.


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Correspondence to Kateřina Staňková.


Appendix A: Why energy has to be included in the model (quantitative argument from Sect. 2.3)

Let us consider the instantiated model in (3.2)–(3.4), and assume that the population-dependent environmental feedback is not explicit, namely \(E(t)=1, t\in [0,T]\). Then the optimization problem (3.1)–(3.4) simplifies to

$$\begin{aligned} u^{*}&= {\mathop {{\arg \sup }}\limits _{u}} \int \limits _0^T (1-u)r \mathrm{d}t^{\prime }; \\ \frac{\mathrm{d}p}{\mathrm{{d}}t}&= \left(\frac{1}{5}u r-\frac{1}{20}\right) p, \\ \frac{\mathrm{d}r}{\mathrm{{d}}t}&= \left(\frac{1}{5} u -\frac{1}{20} -u \, p\right) r. \end{aligned}$$

Introducing a value function \(W(p, r, t,u)=\int _{T-t}^T (1-u ) r \mathrm {d}t^{\prime }\) and the new variables \(b \stackrel{\mathrm{def}}{=}\frac{\partial W}{\partial p}\) and \(c \stackrel{\mathrm{def}}{=}\frac{\partial W}{\partial r}\), it is possible to show that the optimal control takes the form \(u^{*}= \mathrm{Heav}\, \fancyscript{C}\), where

$$\begin{aligned} \fancyscript{C} = \left(\frac{1}{5} p b- p c + \frac{1}{5} c - 1\right) r, \end{aligned}$$

which, as in the more general case, implies that the optimal behavior of the prey is again fully independent of the prey population level, and that \(u^{*}= 0\) at the end of the season.

The characteristic system can be expressed as follows:

$$\begin{aligned} \left\{ \begin{array}{l} p^{\prime } = \frac{1}{20} p - \frac{1}{5} u \, p\, r, \\ r^{\prime } = \frac{1}{20} r +u\, p\, r - \frac{1}{5} u \, r,\\ b^{\prime } = -\frac{1}{20} b + \frac{1}{5} u \, r b - u \, r c, \\ c^{\prime } = \frac{1}{5} u \, p b - \frac{1}{20} c - u \, p c+ \frac{1}{5} u \, c + 1-u. \end{array} \right. \end{aligned}$$

With reference to the reverse time \(\tau \), selecting a \(u=0\) and transversal conditions \(b(0)=c(0)=0\) yields:

$$\begin{aligned} p^{\prime }&= \frac{1}{20} p \Rightarrow p(\tau )=p(0)\,{\mathrm{e}^{\frac{\tau }{20}}},\\ r^{\prime }&= \frac{1}{20} r \Rightarrow r(\tau )=r(0)\,{\mathrm{e}^{\frac{\tau }{20}}}, \\ b^{\prime }&= -\frac{1}{20} b \Rightarrow b(\tau )={b(0)}\,{\mathrm{e}^{-\frac{\tau }{20}}}=0, \\ c^{\prime }&= 1- \frac{1}{20} c \Rightarrow c(\tau )= 20+{\mathrm{e}^{-\frac{\tau }{20}}} (c(0) -20)=20-20{\mathrm{e}^{-\frac{\tau }{20}}}. \end{aligned}$$

Hence, the condition \(\fancyscript{C} = 0\), related to mixed optimal strategies and \(u^{s}\in (0,1)\), can take place at time \(\tau _1\) if the following condition holds:

$$\begin{aligned}&-{p(0)} ( {\mathrm{e}^{\frac{\tau _1}{20}}}) ^{2}{r(0)}\, ( 20-20\,{\mathrm{e}^{-\frac{\tau _1}{20}}}) +\frac{1}{5}\,{r(0) \mathrm{e}^{\frac{\tau _1}{20}}} ( 20-20\,{\mathrm{e}^{-\frac{\tau _1}{20}}} ) -{ r(0)}\,{\mathrm{e}^{\frac{\tau _1}{20}}} \nonumber \\&\quad =r(0)( -20 {p(0)} {\mathrm{e}^{\frac{\tau _1}{10}}}+20\,{p(0)} {\mathrm{e}^{\frac{\tau _1}{20}}}+3\,{\mathrm{e}^{\frac{\tau _1}{20}}}-4 )=0 \end{aligned}$$

Assuming that \(r(0)>0\), the equality in (6.1) is satisfied if \({p(0)} = {\frac{-3+4\,{\mathrm{e}^{-\frac{\tau _1}{20}}}}{{20\,\mathrm{e}^{\frac{\tau _1}{20}}} ( -1+{\mathrm{e}^{-\frac{\tau }{20}}}) }}\). Assuming that \(p(0) \ge 0\), a time \(\tau _1\) such that \(u(\tau _1)\ne 0\) exists if \(\tau _1 \ge -20 \ln \frac{3}{4} \approx 5.75\). Furthermore, pairs \((\tau _1,p(0))\) related to possible non-zero optimal strategies for the prey are those corresponding to the curve depicted in Fig. 7. For all the values of \((\tau _1,p(0))\), such that \({p(0)} < {\frac{-3+4\,{\mathrm{e}^{-\frac{\tau _1}{20}}}}{{20\,\mathrm{e}^{\frac{\tau _1}{20}}}( -1+{\mathrm{e}^{-\frac{\tau }{20}}}) }}\) (below the curve in Fig. 7) the optimal strategy of the prey is to switch to an active state, namely \(u^{*}(\tau ) = 1\,\forall \tau \ge \tau _1\). For all the remaining values of \((\tau _1,p(0))\), which in practice means for \(p(0) > 0.05\) and any \(\tau _1\), the optimal strategy of the prey is to remain in diapause for the entire summer season: this in practice would deplete the energy of the prey and therefore would lead to its death. This leads to the conclusion that modeling the interactions of the considered system without including the energy variable leads to inconsistent outcomes.

Fig. 7
figure 7

If \(p(0)\) and \(\tau _1\) are situated above the red curve, then \(u^{*}=0\) for the entire summer. (The variable \(\tau _1\) has to be higher than \(-20 \ln \frac{3}{4} \approx 5.75\) as displayed in the left plot) (color figure online)

The argument can be generalized to models that are parameterized as in (3.3), (3.4), that is where \(\alpha \) and \(\gamma \) have not been fixed to the values \(1/20\) and \(1/5\), respectively. It can be algebraically shown that the maximal value of he curve \(p(0)\) is upper bounded by the quantity \((\gamma - \alpha )\) which, given the ranges of interest, is again a very small quantity.

Appendix B: Proof of Proposition 3.1

As portrayed in Fig. 8, let us assume that there exists a time \(\tau _{3}: u^{*}(\tau _3) = 1\) and there exists a \(\delta >0: \forall \tau \in (\tau _3, \tau _3 +\delta ]: u^{*}(\tau ) \in [0,1)\). Let variable \(E_2\stackrel{\mathrm{def}}{=}E(\tau _2)\). Setting \(u^{*}= 1\) for \(\tau >\tau _2\), the energy \(E(\tau )\) satisfies

$$\begin{aligned} E(\tau ) = 1+ (E_2-1)\mathrm{e}^{d\,(\tau -\tau _{2})}, \end{aligned}$$

therefore \(E_3\stackrel{\mathrm{def}}{=}E(\tau _3)=1+ (E_2-1)\mathrm{e}^{d\,(\tau _3-\tau _{2})}\). Moreover, let us introduce the following variables: \(a_3 \stackrel{\mathrm{def}}{=}a(\tau _3), c_3 \stackrel{\mathrm{def}}{=}a(\tau _3), p_3 \stackrel{\mathrm{def}}{=}p(\tau _3)\), and \(r_3\stackrel{\mathrm{def}}{=}r(\tau _3)\).

Fig. 8
figure 8

Study of the existence of a second event in reverse time: can the optimal strategy \(u^{*}\) again admit values within the interval \([0,1)\), after being equal to \(1\)?

At time \(\tau =\tau _3\), the condition \(\fancyscript{S}=0\) has to be satisfied. Substituting expression for \(E_3\) into equation \(\fancyscript{S}=0\) leads to

$$\begin{aligned}&{\frac{1}{250}} (1+{\mathrm{e}^{d(\tau _3-\tau _2)}} ( {E_2}-1 ) ) a_3 +d\,a_3 -d ( 1+{\mathrm{e}^{d(\tau _3-\tau _2)}} ( {E_2}-1)) a_3 + \frac{1}{5} \,p_3 \, r_3 \, b_3\\&\quad -p_3 r_3 c_3 + \frac{1}{5} ( 1+{\mathrm{e}^{d(\tau _3-\tau _2)}} ({E_2}-1)) r_3 c_3 - ( 1+{\mathrm{e}^{d(\tau _3-\tau _2)}} ( {E_2}-1)) r_3=0. \end{aligned}$$

From this equation we can express \(E_2\) as:

$$\begin{aligned} E_2 =1+ {\frac{ a_3 + 50\,p_3 r_3 \, b_3 - 250\,p_3 r_3 c_3 + 50 \,r_3 \, c_3 - 250 \,r_3}{{\mathrm{e}^{d\,\Delta }} ( 250\,d\,a_3 -\,a_3 - 50 \,r_3 \,c_3 + 250\,r_3) }}, \end{aligned}$$

where we have set \(\Delta =\tau _3-\tau _2>0\).

Since \(E_2\in (0,1]\), the inequality (7.1) is satisfied only if

$$\begin{aligned} 1 > {\frac{ 250\,p_3 r_3 c_3 -a_3 - 50\,p_3 r_3 \, b_3 - 50 \,r_3 \, c_3 + 250 \,r_3}{{\mathrm{e}^{d\,\Delta }} ( 250\,d\,a_3 -\,a_3 - 50 \,r_3 \,c_3 + 250\,r_3) }} \ge 0. \end{aligned}$$

Recall that if there is a second event at time \(\tau _3\), then \(\fancyscript{S}(\tau _3)=0,\) but also \(\fancyscript{S}^{\prime }(\tau _3)=\fancyscript{S}^{\prime \prime }(\tau _3)= \cdots =0\). From the equation \(\fancyscript{S}^{\prime }=0\) the parameter \(a_3\) can be expressed in terms of the other variables:

$$\begin{aligned} \nonumber a_3&= \frac{r_3}{{10\,d}} ( 2 \,c_3 - 2 \,c_3 {\mathrm{e}^{d\,\Delta }} + 875 \,{\mathrm{e}^{d\Delta }}{E_2}- 2{,}500\, p_3 + 375- 125\,p_3 \, c_3 \\&+ 2{,}500 \,{\mathrm{e}^{d\Delta }}p_3 + 25\, p_3\, b_3 + 500\,{\mathrm{e}^{ 2\,d \,\Delta }}{(E_2)}^{2}- 1{,}000\,{\mathrm{e}^{ 2\,d\,\Delta }}{E_2}+ 2 \,c_3 {\mathrm{e}^{d\,\Delta }}{E_2}\nonumber \\&- 2{,}500\,{\mathrm{e}^{d \,\Delta }}{E_2}\,p_3\,a_3 + 2{,}500 \,d{\mathrm{e}^{d\,\Delta _3}}- 2{,}500\,d{\mathrm{e}^{d\Delta }}{E_2}\nonumber \\&+ 500\,{\mathrm{e}^{ 2\,d\Delta }}- 875\,{\mathrm{e}^{d\Delta }}). \end{aligned}$$

Furthermore, \(b_3\) can be expressed from the equation \(\fancyscript{S}^{\prime \prime }=0\) (after substituting (7.3) into this same equation), and likewise \(c_3\) can be expressed from the condition \(\fancyscript{S}^{\prime \prime \prime }=0\) (after substituting expressions for \(a_3\) and \(b_3\) into this equation)—we omit reporting the expressions for \(b_3\) and \(c_3\), as their computation is straightforward.

Denoting the nominator and denominator of the fraction in Eq. (7.2) as “\(\mathrm{Nom}\)” and “\(\mathrm{Den}\)”, respectively, there are two cases characterizing the necessary conditions in (7.2), for the existence of a \(\tau _3 > \tau _2\) such that \(u(\tau )=1\) for \(\tau \in [\tau _2,\tau _3]\) and of a \(\delta >0: \forall \tau \in (\tau _3,\tau _3+\delta ], u(\tau )\in (0,1)\):

  • Case 1

    $$\begin{aligned}&0 \le \mathrm{Nom} ,\end{aligned}$$
    $$\begin{aligned}&0 < \mathrm{Den} ,\end{aligned}$$
    $$\begin{aligned}&0 < \mathrm{Den}-\mathrm{Nom}. \end{aligned}$$
  • Case 2

    $$\begin{aligned}&0 \ge \mathrm{Nom}, \end{aligned}$$
    $$\begin{aligned}&0 > \mathrm{Den}, \end{aligned}$$
    $$\begin{aligned}&0 > \mathrm{Den}- \mathrm{Nom}. \end{aligned}$$

The quantities \(\mathrm{Den}\) and \(\mathrm{Nom}\) can be then written with \(E_2\) expressed by (7.1).

In the following, the two cases are considered in detail. Case 1: Condition (7.4) implies

$$\begin{aligned} 250\,r_3-50\,p_3\, r_3\, b_3+250\,p_3\,r_3\,c_3-50\,r_3\,c_3 \le a_3, \end{aligned}$$

whereas condition (7.5) implies

$$\begin{aligned} a_3< \frac{50\,r_3 ( c_3- 5) }{250\,d- 1} \end{aligned}$$

and condition (7.6) implies

$$\begin{aligned} a_3< \frac{50\,r_3 ( \mathrm{e}^{d\,\Delta }\,c_3- 5 \,{\mathrm{e}^{d\Delta }}- \,b_3\,p_3+ 5\,p_3 \,c_3- c_3+ 5) }{- {\mathrm{e}^{d\Delta }}+ 250\,{\mathrm{e}^{d\,\Delta }}d+ 1}. \end{aligned}$$

Note that inequalities (7.10) and (7.11) imply either

$$\begin{aligned} 0<- 50\,r_3\,b_3+ 250\,r_3\,c_3 \end{aligned}$$


$$\begin{aligned} p_3<-{\frac{ 250\,d ( c_3- 5 ) }{( 250\,d - 1) ( b_3- 5\,c_3) }}, \end{aligned}$$


$$\begin{aligned} - 50\,r_3\,b_3+ 250\,r_3\,c_3<0 \end{aligned}$$


$$\begin{aligned} {\frac{- 250\,d (c_3- 5) }{(250\,d -1) ( b_3- 5\,c_3 ) }} < p_3. \end{aligned}$$

We have substituted the expressions for \(a_3, b_3,\) and \(c_3\) into inequalities (7.13), (7.14), (7.15) and (7.16), respectively.

Assuming that \(p_3 > 0, r_3 > 0, E^{f}\in (0,1]\), and \(d>\frac{1}{250}\), in both cases it can be shown that

$$\begin{aligned} a_3> {\frac{50\,r_3 ({\mathrm{e}^{d\Delta }}c_3- 5 \,{\mathrm{e}^{d\Delta }}- b_3\,p_3+ 5\,p_3 \,c_3-c_3+ 5) }{- \mathrm{e}^{d\Delta }+ 250\,\mathrm{e}^{d\,\Delta }\,d+ 1}}, \end{aligned}$$

which contradicts Eq. (7.12) and therefore also Eq. (7.6).

Case 2: Condition (7.7) implies

$$\begin{aligned} 250\,r_3-50\,p_3\, r_3b_3+250\,p_3\,r_3\,c_3-50\,r_3\,c_3 \ge a_3, \end{aligned}$$

whereas condition (7.8) implies

$$\begin{aligned} a_3> {\frac{50\,r_3 (c_3- 5) }{250\,d- 1 }} \end{aligned}$$

and condition (7.9) implies

$$\begin{aligned} a_3> {\frac{50\,r_3 ( {\mathrm{e}^{d\Delta }}c_3- 5 \,{\mathrm{e}^{d\Delta }}- \,b_3\,p_3+ 5\,p_3 \,c_3- c_3+ 5) }{- {\mathrm{e}^{d\Delta }}+ 250\, \mathrm{e}^{d \Delta }\,d+ 1}}. \end{aligned}$$

Note that the inequalities in (7.17) and (7.18) imply either

$$\begin{aligned} 0<- 50\,r_3 b_3+ 250\,r_3 c_3 \end{aligned}$$


$$\begin{aligned} p_3>{\frac{- 250\,d ( c_3- 5) }{( - 1 + 250\,d) (b_3- 5\,c_3) }}, \end{aligned}$$


$$\begin{aligned} - 50\,r_3b_3+ 250\,r_3 c_3<0 \end{aligned}$$


$$\begin{aligned} {\frac{- 250\,d ( c_3- 5) }{( - 1+ 250 \,d) ( b_3- 5\,c_3) }}>p_3. \end{aligned}$$

We have substituted expressions for \(a_3, b_3,\) and \(c_3\) into inequalities (7.20), (7.21), (7.22) and (7.23), respectively. Assuming that \(p_3 > 0, r_3 > 0, E^{f}\in (0,1]\), and \(d>\frac{1}{250}\), in both cases it can be shown that

$$\begin{aligned} a_3< {\frac{50\,r_3 ( {\mathrm{e}^{d\Delta }}c_3- 5 \,{\mathrm{e}^{d\Delta }}- \,b_3\,p_3+ 5\,p_3 \,c_3- c_3+ 5) }{- \,{\mathrm{e}^{d\Delta }}+ 250\,{\mathrm{e}^{d \Delta }}d+ 1}}, \end{aligned}$$

which contradicts Eq. (7.19) and therefore also Eq. (7.9). \(\square \)

Appendix C: Proof of Proposition 3.2

If \(p=0,\) the characteristic system takes the following form in reverse time:

$$\begin{aligned} \left\{ \begin{array}{l} E^{\prime } = \frac{1}{250} (1- u)E- d \,u + d \,u \,E,\\ r^{\prime } = \frac{1}{20} r - \frac{1}{5} u \,E \, r,\\ a^{\prime } = -\frac{1}{250}(1-u) a - d u a + \frac{1}{5} u \, r\,c +(1-u) r, \\ c^{\prime } = - \frac{1}{20} c + \frac{1}{5} u\, E\, c + (1-u) E. \end{array} \right. \end{aligned}$$

Furthermore, the surface \(\fancyscript{S}\) can be expressed as

$$\begin{aligned} \fancyscript{S}=\frac{1}{250} E a + d a - d E a + \frac{1}{5} E \,r\, c - E\, r, \end{aligned}$$

It is again easy to check that \(u^{s}(0) = 0\) in reverse time, and that \(\tau _1=500 \ln w\), with \(w\) being the smallest root of the following polynomial

$$\begin{aligned}&100\,(E^{f})^{2}{w}^{31}+ (-25\,{E^{f}}-500\,d{E^{f}}) {w}^{29}+500\,d{w}^{27} \\&\quad -100\,(E^{f})^{2}{w}^{4}+ (-2\,{E^{f}}+500\,d{ Ef}) {w}^{2}-500\,d. \end{aligned}$$

From (8.1), the energy level \(E^1\) of the prey entering diapause is \(E(\tau _1)=E^{f}e^{\frac{\tau _1}{250}}\). Figure 2 represents the values of \(\tau _1\) as functions of \(d\) and \(E^{f}\). Notice that \(E^1\ge 1\) for \(E^{f}\ge 0.9775\), therefore in the following we will assume that \(E^{f}\in (0,0.9775)\). (Moreover, recall that \(d>1/250\).)

Equations \(\fancyscript{S}=0, \fancyscript{S}^{\prime }=0,\) and \(\fancyscript{S}^{\prime \prime }=0\) allow expressing \(a(\tau ), c(\tau ),\) and \(u^{s}(\tau )\) in terms of \(r(\tau )\) and \(E(\tau ),\) respectively. Of interest to this proof, the expression for the mixed strategy \(u^{s}(\tau )\) reads as the ratio of two polynomials:

\((E ( -25\, E^{2}-29\cdot 10^3 \,d E^{3}+125\cdot 10^3 \,{d}^{2} E^{2}+116\, E^{3 }-13\cdot 10^3 \,d\,E +230\cdot 10^3 \,{d}^{2}+67{,}750\,d E^{2}-375\cdot 10^3 \,{d}^{2}E))/( 4\,(-125\,d \,E^{2} -5\cdot 10^3 \,{d}^{2}E -21\, E^{4}-1{,}250\cdot 10^3 \,{d}^{3}+100\,E^{5}-9\cdot 10^3 \,d E^ {3}+54{,}250\,d E^{4}+531{,}250\,{d}^{2 } E^{2}-875\cdot 10^3 \,{d}^{2} E^{3}+1{,}250\cdot 10^3 \,{d}^{3}E +250\cdot 10^3 \,{d}^{2} E^{4}- 25\cdot 10^3 \, E^{5}d))\).

Remarkably, the expression is independent of \(r\), which aligns to earlier outcomes on the independence of the prey population density.

Since \(u^{s}\) cannot by definition take values that are lower than \(0\) or greater than \(1\), we denote the values for \((d,E)\) for which \(u^{s}\in [0,1]\) as “feasible” and we will call those for which \(u^{s}\ne [0,1]\) “unfeasible.” Figure 9 represents the feasibility regions for \((d,E)\), assuming \(d\in (\frac{1}{250},1]\) and \(E\in (0,0.9775]\). The feasible region for the given parameters corresponds to possible trajectories that have mixed strategies, whereas the unfeasible region relates to trajectories that stay always in diapause mode (\(u^{s}= 0\)), or discontinuously switch to \(u^{s}= 1\). In either case, trajectories for the optimal strategy will be non-decreasing in reverse time.

Fig. 9
figure 9

Regions of \(d\) and \(E\) for which \(u^{s}\in (0,1].\) The purple bold line corresponds to \(d\) and \(E\) for which \(u^{s}=1\) and coincides with the boundary of the feasibility region (color figure online)

From \(\fancyscript{S}^{\prime \prime \prime }=0\) we can derive the expression of \(\frac{\mathrm{d}u^{s}}{\mathrm{d}\tau }\). Figure 10 plots the parts of the feasibility region for which \(u^{s}\) is increasing and decreasing, respectively.

Fig. 10
figure 10

Regions of \(d\) and \(E\) for which \(\frac{\mathrm{d}u^{s}}{\mathrm{d}\tau }>0\) (denoted with a \(+\)) and for which \(\frac{\mathrm{d}u^{s}}{\mathrm{d}\tau }<0\) (denoted by a \(-\)). The green dashed line corresponds to the \((d,E)\) values for which \(\frac{\mathrm{d}u^{s}}{\mathrm{d}\tau }=0\) (color figure online)

With focus on the feasible region, the following observations can be made:

  • For \(d>0.15\), no mixed strategy takes place, since the parameter space corresponds to the unfeasible region.

  • For \(\frac{1}{250}<d<0.15\), either

    • no mixed strategy takes place if \(E\) is small, else

    • a mixed strategy \(u^{s}\) takes place.

Let us further elaborate on the latter case (presence of mixed strategies) with the help of Fig. 10. Recall that

$$\begin{aligned} E^{\prime } = \frac{E}{250} (1- u) - d\,u(1-E), \end{aligned}$$

and assume the dynamics at time \(\tau _1\) land in a region where \(u^{s}\) has negative derivative. It can be numerically shown that the values of \(u^{s}\) are quite small in this region. Because of the values of \(E>0.5\) and \(1/250 < d < 0.15, E^{\prime } > 0\) persistently, given that \(d u^{s}/dt < 0\). This regime will be sustained until \(E = 1\), which will force \(u^{s}\) to switch to \(u^{*}= 1\). Given the values of the quantities of interest, this will happen for a short interval—if \(d = \frac{1}{250}\), the interval will be approximately less than \(0.5\), whereas if \(d \sim 0.15\), the interval will be even smaller. \(\square \)

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Staňková, K., Abate, A. & Sabelis, M.W. Irreversible prey diapause as an optimal strategy of a physiologically extended Lotka–Volterra model. J. Math. Biol. 66, 767–794 (2013).

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