Effect of Hawk-Dove Game on the Dynamics of Two Competing Species

Abstract

Outcomes of interspecific competition, and especially the possibility of coexistence, have been extensively studied in theoretical ecology because of their implications in community assemblages. During the last decades, the influence of different time scales through the local/regional dynamics of animal communities has received an increasing attention. Nevertheless, different time scales involved in interspecific competition can result form other processes than spatial dynamics. Here, we envision and analyze a new theoretical framework that couples a game theory approach for competition with a demographic model. We take advantage of these two time scales to derive a reduced model governing the total densities of the two populations and we study how these two time scales interfere and influence outcomes of species competition. We find that a competition process occurring on a faster time scale than demography yields a “priority effect” where the first species introduced outcompetes the other one. We then confirm previous findings stipulating that species coexistence is favored by large difference in time scales because the extinction/recolonization process. Our results then highlight that an integration of demographic and competition time scales at both local and regional levels is mandatory to explain communities assemblages and should become a research priority.

Appendices

Appendix 1: Fast-Time System

We supply a quick analysis of the fast system in Eq. (11). System (11) is a set of two coupled ordinary differential equations. Our variables $y_{1}$ and $y_{2}$ lying in $[0,1]$. The biologically-relevant dynamics takes place within the square unit $[0,1]\times [0,1]$. The steady states are located at its corners $(0,0),(0,1),(1,0),(1,1)$, and another equilibrium, $\left( \frac{G_{2}}{C_{21}},\frac{G_{1}}{C_{12} }\right)$ inside the domain if $G_{2}<C_{21}$ and $G_{1}<C_{12}.$

To obtain the local stability results, we write the Jacobian matrix associated to system (11):

$$ J(y_{1},y_{2})=\left( \begin{array}{ll} {\frac{1}{2}\left( 1-2y_{1}\right)} \left( G_{1}-C_{12}y_{2}\right) &{} {-\frac{1}{2}C_{12}y_{1}\left( 1-y_{1}\right)} \\ {-\frac{1}{2}C_{21}y_{2}\left( 1-y_{2}\right)} &{} {\frac{1}{2}\left( 1-2y_{2}\right) \left( G_{2}-C_{21}y_{1}\right)} \end{array}\right) $$

(20)

• In the case of the origin the Jacobian matrix reads:

  $$ J(0,0)=\left( \begin{array}{ll} \frac{G_{1}}{2} &{} 0\\ 0 &{} \frac{G_{2}}{2} \end{array}\right) . $$

  The two eigenvalues are positive and the origin is an unstable node.

• In the case of the $(1,0)$ the Jacobian matrix reads:

  $$J(1,0)=\left( \begin{array}{ll} -\frac{G_{1}}{2} &{} 0\\ 0 &{} \frac{1}{2}\left( G_{2}-C_{21}\right) \end{array}\right) . $$

  Thus, $(1,0)$ is a stable node if both entries on the main diagonal are negative which can be expressed by $G_{2}<C_{21}.$

• In the case of the $(0,1)$ the Jacobian matrix reads:

  $$ J(1,0)=\left( \begin{array}{ll} \frac{1}{2}\left( G_{1}-C_{12}\right) &{} 0\\ 0 &{} -\frac{G_{2}}{2} \end{array}\right) $$

  Thus, $(0,1)$ is a stable node if $G_{1}<C_{12}.$

• Concerning the Jacobian matrix of $(1,1)$:

  $$J(1,1)=\left( \begin{array}{ll} -\frac{1}{2}\left( G_{1}-C_{12}\right) &{} 0\\ 0 &{} -\frac{1}{2}\left( G_{2}-C_{21}\right) \end{array}\right) .$$

  Its eigenvalues are negative which if $G_{1}>C_{12}$ and $G_{2}>C_{21}.$

• The last stationary state reads $\left( \frac{G_{2}}{C_{21}} ,\frac{G_{1}}{C_{12}}\right)$, after some algebra, the Jacobian matrix reads:

  $$J\left( \frac{G_{2}}{C_{21}},\frac{G_{1}}{C_{12}}\right) =\left( \begin{array}{ll} 0 &{} -\frac{1}{2}C_{12}\frac{G_{2}}{C_{21}}\left( 1-\frac{G_{2}}{C_{21} }\right) \\ -\frac{1}{2}C_{21}\frac{G_{1}}{C_{12}}\left( 1-\frac{G_{1}}{C_{12}}\right) &{} 0 \end{array}\right) $$

  The equilibrium point $\left( \frac{G_{2}}{C_{21}},\frac{G_{1}}{C_{12} }\right)$ must be located inside the unit square to make sense, i.e. $G_{1}<C_{12},$and $G_{2}<C_{21},$thus, this equilibrium is saddle.

Appendix 2: Slow-Time System

We sketch the analysis of the defined system (17), (18) and (19).

1.1 Equilibria and Local Stability Analysis of Model 17: The Case When Species 1 is Pure Hawk

There are four feasible equilibria. $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1-b)$ The feasibility conditions for the last equilibrium require that $b<1$. We can analyze the stability of equilibrium point by the following Jacobian matrix

$$ \begin{gathered} J(u,v) = \left( {\begin{array}{*{20}l} {r_{1} \left( {1 - 2u} \right)} \hfill & 0 \hfill \\ { - r_{2} bvr_{2} } \hfill & {(1 - 2v - bu)} \hfill \\ \end{array} } \right) \hfill \\ \hfill \\ \end{gathered} $$

• The Jacobian matrix $J(0,0)$ is a diagonal matrix; its eigenvalues are $r_{1}$ and $r_{2}$. This implies the equilibrium $\left( 0,0\right)$ is unstable.

• The Jacobian matrix $J(1,0)$ is a diagonal matrix. This matrix has two eigenvalues: one is negative number $\lambda _{1}=-r_{1}<0$ and another eigenvalue $\lambda _{2}=r_{2}(1-b).$ Thus the equilibrium $\left( 1,0\right)$ is stable if $b>1.$

• The Jacobian matrix $J(0,1)$ is a lower triangular matrix; its eigenvalues are $\lambda _{1}=r_{1}$ and $\lambda _{2}=-r_{2}$ Thus,$\left( 0,1\right)$ is saddle.

• The last stationary state reads $(1,1-b)$, After some algebra one gets that as soon as the Jacobian matrix $J(1,1-b)$ is an lower triangular matrix; its two eigenvalues are $\lambda _{1}=-r_{1}$ and $\lambda _{2}=-r_{2}(1-b)$. This implies the equilibrium $\left( 1,1-b\right)$ if it exists, is stable.

1.2 Equilibria and Local Stability Analysis of Model 18: The Case When Species 2 is Pure Hawk

A similar analysis yields the local stability of the stationary state of system (18), We can write the Jacobian matrix as following:

$$ J(u,v)=\left( \begin{array}{ll} r_{1}\left( 1-2u-av\right) &{} -ar_{1}u\\ 0 &{} r_{2}\left( 1-2v\right) \end{array}\right) $$

• For the first equilibrium $(0,0)$ we find $\lambda _{1}=r_{1}$ and $\lambda _{2}=r_{2}$. Thus the equilibrium $(0,0)$ is unstable.

• For $(1,0)$ one gets again two eigenvalues, $\lambda _{1}=-r_{1}$ and $\lambda _{2}=r_{2}.$ Thus $\left( 1,0\right)$ is saddle point.

• For $(0,1)$ the first eigenvalue is $\lambda _{2}=-r_{2}$. The other eigenvalue is $\lambda _{1}=r_{1}(1-a)$. Thus if $a>1$, the equilibrium is stable, else it is saddle.

• In the remaining case of the equilibrium point $(1-a,1)$. (the condition for its existence is that $a<1$) after some algebra, one gets that as soon as the Jacobian matrix $J(1-a,1)$ is an upper triangular matrix; its eigenvalues are $\lambda _{1}=-r_{1}(1-a)$ and $\lambda _{2}=-r_{2}$ yielding local stability of $(1-a,1)$ as soon as $a<1$ hold.

1.3 Equilibria and Local Stability Analysis of Model 19: The Case of All Aggressive Individuals

• The Jacobian matrix of (19) computed at $(0,0)$ is given by

  $$ J\left( 0,0\right) =\left( \begin{array}{ll} r_{1} &{} 0\\ 0 &{} r_{2} \end{array}\right)$$

  For any value of the parameters the origin is unstable.

• The Jacobian matrix of (19) computed at $(1,0)$ is given by

  $$ J\left( {1,0} \right) = \left( {\begin{array}{*{20}l} { - r_{1} } \hfill & { - ar_{1} } \hfill \\ 0 \hfill & {r_{2} (1 - b)} \hfill \\ \end{array} } \right) $$

  Thus, the equilibrium $(1,0)$ is stable if $b>1$ otherwise, it is saddle node.

• The Jacobian matrix of (19) computed at $(0,1)$ is given by

  $$ J\left( {0,1} \right) = \left( {\begin{array}{*{20}l} {r_{1} \left( {1 - a} \right)} \hfill & 0 \hfill \\ { - r_{2} b} \hfill & { - r_{2} } \hfill \\ \end{array} } \right) $$

  This Jacobian matrix has two eigenvalues: $-r_{2}$ and $r_{1}\left( 1-a\right)$. Thus if $a>1$, the equilibrium is stable, else it is saddle.

• For the last equilibrium $(u^{*},v^{*})$. The Jacobian matrix $J\left( u^{*},v^{*}\right)$ becomes:

  $$ J\left( u^{*},v^{*}\right) =\left( \begin{array}{ll} -r_{1}u^{*} &{} -ar_{1}u^{*}\\ -r_{2}bv^{*} &{} -r_{2}v^{*} \end{array}\right) $$

  Since, the trace of this previous 2-dimensional matrix is negative. To be a sink, its determinant must be positive leading to the condition $ab<1.$ otherwise, it is saddle.