Effects of Behavioural Strategy on the Exploitative Competition Dynamics

Abstract

We investigate a system of two species exploiting a common resource. We consider both abiotic (i.e. with a constant resource supply rate) and biotic (i.e. with resource reproduction and self-limitation) resources. We are interested in the asymmetric competition where a given consumer is the locally superior resource exploiter (LSE) and the other is the locally inferior resource exploiter (LIE). They also interact directly via interference competition in the sense that LIE individuals can use two opposite strategies to compete with LSE individuals: we assume, in the first case, that LIE uses an avoiding strategy, i.e. LIE individuals go to a non-competition patch to avoids competition with LSE individuals, and in the second one, LIE uses an aggressive strategy, i.e. being very aggressive so that LSE individuals have to go to a non-competition patch. We further assume that there is no resource in the non-competition patch so that individuals have to come back to the competition patch for their maintenance, and the migration process acts on a fast time scale in comparison with demography and competition processes. The models show that being aggressive is efficient for LIE’s survival and even provoke global extinction of the LSE and this result does not depend on the nature of resource.

Appendices

Appendix 1: Aggregation Method

Aggregation of variables methods was well performed in Auger et al. (2008a, b). Let us briefly describe the approximate aggregation procedure (see also in Marva et al. 2012). The considered models belong to a class of autonomous system of ordinary differential equations with two time scales can be expressed in the following form:

$$\begin{aligned} \dfrac{dn}{d\tau }=f(n) + \varepsilon s(n) \end{aligned}$$

(13)

with $n\in \mathbb {R}^m$, where maps f and s represent the fast and slow dynamics, respectively, and $\varepsilon$ is the small positive parameter measuring the time scales ratio when it is possible. To perform its approximate aggregation, system (13) is firstly converted into slow–fast form by means of an appropriate change of variables $n\in \mathbb {R}^m \rightarrow (x,y) \in \mathbb {R}^{m-k}\times \mathbb {R}^k:$

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{dx}{d\tau }=F(x,y) +\varepsilon S(x,y), \\ \dfrac{dy}{d\tau }=\varepsilon G(x,y), \end{array}\right. } \end{aligned}$$

(14)

where x represents the fast variable and y the slow variables. Finding the transformation $n \rightarrow (x,y)$ which yields the slow–fast form (14) of the system (13) could be a difficult task and the construction of general algorithms solving this problem is presently an active research line. On the other hand, in some applications for example in Auger et al. (2008a, b), Nguyen Ngoc et al. (2010), Nguyen-Ngoc et al. (2012), Nguyen-Ngoc et al. (2012) the context gives a natural way to define the so-called global variables y and thus to express the system (13) in a slow–fast form. The aggregation method now consists in different steps:

• Step 1 Taking $\varepsilon =0$ in the first equation of slow–fast form (14), i.e. $dx/d\tau =F(x,y)$. For constant y, finding asymptotically stable equilibrium $x^*(y)$ of this system.

• Step 2 Substituting $x^*(y)$ into the second equation of slow–fast form (14), obtaining the aggregated system:

  $$\begin{aligned} \dfrac{dy}{dt}=G(x^*(y),y), \end{aligned}$$

  (15)

  where $t=\varepsilon \tau$ represents the slow time variable.

• Step 3 Checking the two conditions: (H1) the system 15 is structurally stable and (H2) $\varepsilon$ is small enough, which ensures that the asymptotic behavior of the system (14) can be studied through the system (15).

The mathematical definition of the structural stability can be found in Kuznetsov (1998). The relationship between the asymptotic behaviors of the two systems is just an application of the classical Tikhonov theorem. This theorem and its applications were reviewed in Verhulst (2007).

Appendix 2: Equilibria and Local Stability Analysis of Model 2

1.1 Aggregated Model

$$\begin{aligned}&{\left\{ \begin{array}{ll} \dfrac{dR}{dt} = \gamma (R) - \dfrac{a_1m}{L(C_2)} RC_{1} - a_2 RC_{2} \\ \dfrac{dC_{1}}{dt} = \dfrac{C_1}{L(C_2)}\left[ -(d_{1C}m + d_{1N}\beta _{0}) - d_{1N} \beta C_{2} + a_1e_1m R \right] \\ \dfrac{dC_{2}}{dt} = C_{2}[-(d_{2} + l) + a_2e_2R] \end{array}\right. } \end{aligned}$$

(16)

$$\begin{aligned}&\Leftrightarrow {\left\{ \begin{array}{ll} \dfrac{dR}{dt} = x(R,C_1,C_2)\\ \dfrac{dC_{1}}{dt} = \dfrac{C_1}{L(C_{2})}y(R,C_2) \\ \dfrac{dC_{2}}{dt} = C_{2}z(R) \end{array}\right. } \end{aligned}$$

(17)

1.2 Jacobian Matrix

$$\begin{aligned} J(R, C_{1}, C_{2}) = \begin{pmatrix} x'_{R} &{} x'_{C_1} &{} x'_{C_2}\\ \dfrac{C_1}{L(C_2)}y'_{R} &{} \dfrac{y}{L(C_2)} &{} C_1\dfrac{y'_{C_2}L(C_2)-yL'_{C_2}}{L(C_2)^2}\\ C_2z'_{R} &{} 0 &{} z \end{pmatrix} \end{aligned}$$

(18)

1.3 Equilibria and Stability

• At (0, 0, 0) (in the biotic resource case):

  $$\begin{aligned} J(0,0,0) = \begin{pmatrix} r &{} 0 &{} 0\\ 0 &{} - \dfrac{d_{1C}m+d_{1N}\beta _0}{L(0)} &{} 0\\ 0 &{} 0 &{} -(d_2+l) \end{pmatrix}. \end{aligned}$$

  (19)

  The matrix has one positive eigenvalue r thus (0, 0, 0) is always unstable.

• At $(R^*,0,0)$:

  $$\begin{aligned} J(R^*,0,0) = \begin{pmatrix} -r &{} x'_{C_1} &{} x'_{C_2}\\ 0 &{} \dfrac{-(d_{1C}m+d_{1N}\beta _0)+a_1e_1mR^*}{L(0)}&{} 0\\ 0 &{} 0 &{} -(d_2+l)+a_2e_2R^* \end{pmatrix}. \end{aligned}$$

  The matrix has three eigenvalues: $\lambda _1=-r <0$, $\lambda _2=-(d_{1C}m+d_{1N}\beta _0)+a_1e_1mK$ and $\lambda _3=-(d_2+l)+a_2e_2K$. Hence, $(R^*,0,0)$ is stable if and only if $R^*<\text {min}\left\{ R_1^*, R_2^*\right\}$ where

  $$\begin{aligned} R_1^*:=\dfrac{d_{1C}m+d_{1N}\beta _0}{a_1e_1m} \,\, \text {and}\,\, R_2^*:= \dfrac{d_2+l}{a_2e_2} \end{aligned}$$

• At $(R_{1}^{*}, C_{1}^{*}, 0)$ where $C_1^*=\gamma (R_1^*)L(0)/a_1mR_1^*$: the condition for which this equilibrium is non-negative is given by $R_1^* < R^*$. In this case, the Jacobian matrix reads as follows:

  $$\begin{aligned} J(R_{1}^{*}, C_{1}^{*},0) = \begin{pmatrix} -r\theta (R_1^*) &{} -\dfrac{a_1mR_{1}^{*}}{L(0)} &{} -R_1^*a_2\\ \dfrac{a_1e_1mC_{1}^{*}}{L(0)} &{} 0 &{} -\dfrac{d_{1N}\beta C_1^*}{L(0)}\\ 0 &{} 0 &{} -(d_2+l) + a_2e_2R_1^* \end{pmatrix} \end{aligned}$$

  where $\theta (R)=R/K$ in the biotic resource case, and $\theta (R)=S/R$ in the abiotic resource case. The matrix has one eigenvalue

  $$\begin{aligned} \lambda _1= -(d_2+l) + a_2e_2R_1^* \end{aligned}$$

  and the others eigenvalues, $\lambda _2, \lambda _3$, are the solutions of the following equation:

  $$\begin{aligned} \lambda ^{2} + r\theta (R_{1}^{*})\lambda + \dfrac{a_1^{2}m^2e_1C_{1}^{*}R_{1}^{*}}{L(0)^2} = 0. \end{aligned}$$

  Since $\lambda _{2} + \lambda _{3} =-r\theta (R_1^*)< 0$ and $\lambda _{2}\lambda _{3}=a_1^{2}m^2e_1C_{1}^{*}R_{1}^{*}/L(0)^2 > 0$. Hence, these eigenvalues have negative real parts. Therefore, $(R_{1}^{*},C_{1}^{*},0)$ is stable provided

  $$\begin{aligned} -(d_2+l) + a_2e_2R_1^*<0 \Leftrightarrow R_1^* < R_2^*. \end{aligned}$$

  To summarize, $(R_{1}^{*},C_{1}^{*},0)$ is non-negative and stable provided

  $$\begin{aligned} R_1^* \,<\, \text {min} \left\{ R^*, R_2^*\right\} . \end{aligned}$$

• At $(R_{2}^{*}, 0, C_{2}^{*})$ where $C_2^*=\gamma (R_2^*)/a_2R_2^*$: the condition for which this equilibrium is non-negative is given by $R_2^*\,<\,R^*$. In this case, the Jacobian matrix reads as follows:

  $$\begin{aligned} J(R_{2}^{*},0, C_{2}^{*}) = \begin{pmatrix} -r\theta (R_2^*) &{} -\dfrac{a_1 m R_{2}^{*}}{L(C_2^*)} &{} -R_{2}^*a_2\\ 0 &{} \dfrac{-(d_{1C}m + d_{1N}\beta _{0}) - d_{1N}\beta C_{2}^* + a_1e_1mR_2^*}{L(C_2^*)}&{} 0\\ C_2^*a_2e_2 &{} 0 &{} 0 \end{pmatrix}. \end{aligned}$$

  The matrix has one eigenvalue

  $$\begin{aligned} \lambda _1= \dfrac{-(d_{1C}m + d_{1N}\beta _{0}) - d_{1N}\beta C_{2}^* + a_1e_1mR_2^*}{L(C_2^*)} \end{aligned}$$

  and the others eigenvalues, $\lambda _2, \lambda _3$, are the solutions of the following equation:

  $$\begin{aligned} \lambda ^{2} + r\theta (R_2^*)\lambda + a_2^{2}e_2C_{2}^{*}R_{2}^{*} = 0. \end{aligned}$$

  Since $\lambda _{2} + \lambda _{3} =-r\theta (R_2^*)< 0$ and $\lambda _{2}\lambda _{3}=a_2^{2}e_2C_{2}^{*}R_{2}^{*} > 0$. Hence, these eigenvalues have negative real parts. Therefore, $(R_{2}^{*},0, C_{2}^{*})$ is stable provided

  $$\begin{aligned}&\dfrac{-(d_{1C}m + d_{1N}\beta _{0}) - d_{1N}\beta C_{2}^* + a_1e_1mR_2^*}{L(C_2^*)}< 0\\&\quad \Leftrightarrow R_2^* < R_1^*+ \dfrac{d_{1N}\beta C_2^*}{a_1e_1m}=R_1^*+\dfrac{d_{1N}\beta }{a_1e_1m}.\dfrac{\gamma (R_2^*)}{a_2R_2^*}. \end{aligned}$$

  To summarize, $(R_{2}^{*},0, C_{2}^{*})$ is non-negative and stable provided

  $$\begin{aligned} R_2^* < \text {min} \left\{ R^*, R_1^*+\dfrac{d_{1N}\beta }{a_1e_1m}.\dfrac{\gamma (R_2^*)}{a_2R_2^*}\right\} . \end{aligned}$$

• At $(\hat{R}, \hat{C_1}, \hat{C_{2}})$ where $\hat{R}=R_2^*, \hat{C_1}=(\gamma (\hat{R})-a_2\hat{R}\hat{C_2})L(\hat{C_2})/a_1m\hat{R}, \hat{C_2}=a_1e_1m(R_2^* - R_1^*)/d_{1N}\beta$: the condition for which this equilibrium is non-negative is given by $R_2^* > R_1^*$ and $\gamma (\hat{R})-a_2\hat{R}\hat{C_2} > 0$. The second condition means that

  $$\begin{aligned} \gamma (R_2^*)-a_2R_2^*\dfrac{a_1e_1m(R_2^*-R_1^*)}{d_{1N}\beta } \end{aligned}$$

  or else

  $$\begin{aligned} R_2^* \,<\, R_1^*+ \dfrac{d_{1N}\beta }{a_1e_1m}.\dfrac{\gamma (R_2^*)}{a_2R_2^*}. \end{aligned}$$

  In this case, the jacobian matrix reads as follows:

  $$\begin{aligned} J(\hat{R},\hat{C_1}, \hat{C_2}) = \begin{pmatrix} -r\theta {\hat{R}} &{} -\dfrac{am}{L(\hat{C_2})}\hat{R} &{} \left( \dfrac{am\hat{C_1}\beta }{(L(\hat{C_2})^2} -b \right) \hat{R}\\ \dfrac{aem \hat{C_1}}{L(\hat{C_2})} &{}0&{} -\dfrac{d_{1N}\beta \hat{C_1}}{L(\hat{C_2})}\\ bf\hat{C_2}&{} 0 &{} 0 \end{pmatrix}. \end{aligned}$$

  The characteristic equation for the jacobian matrix $J(\hat{R},\hat{C_1}, \hat{C_2})$ is

  $$\begin{aligned} \lambda ^3 + A_1 \lambda ^2 + A_2 \lambda + A_3=0, \end{aligned}$$

  where

  $$\begin{aligned} A_1= & \, r\theta {\hat{R}},\\ A_2= & {} -\dfrac{a_1m\beta \hat{C_1}\hat{R}-a_1e_1m\hat{C_1}a_1m\hat{R}}{(L(\hat{(}C_2)))^2}+a_2\hat{R}, \\ A_3= & {} -\dfrac{a_2e_2\hat{C_2}a_1md_{1N}\beta \hat{C_1}\hat{R}}{(L(\hat{C_2}))^2} \end{aligned}$$

  Since $A_1 > 0$ and $A_3 < 0$ it implies that there exists at least one solution the equation whose real part is not negative (Routh-Hurwitz criteria). Therefore, $(\hat{R},\hat{C_1}, \hat{C_2})$ is unstable.