Hopf Bifurcations in Delayed Rock–Paper–Scissors Replicator Dynamics

Abstract

We investigate the dynamics of three-strategy (rock–paper–scissors) replicator equations in which the fitness of each strategy is a function of the population frequencies delayed by a time interval $T$. Taking $T$ as a bifurcation parameter, we demonstrate the existence of (non-degenerate) Hopf bifurcations in these systems and present an analysis of the resulting limit cycles using Lindstedt’s method.

Appendices

Appendix 1: Derivation of replicator equation

Consider an exponential model of population growth,

$$\begin{aligned} \dot{\xi }_i = \xi _i g_i \quad (i=1,\ldots ,n) \end{aligned}$$

(109)

where $\xi _i$ is a real-valued function that approximates the population of strategy $i$ and $g_i(\xi _1,\ldots ,\xi _n)$ is the fitness of that strategy. The replicator Eq. [7] results from Eq. (109) by changing variables from the populations $\xi _i$ to the relative abundances, defined as $x_i \equiv \xi _i/p$ where $p$ is the total population:

$$\begin{aligned} p(t) = \sum _i \xi _i(t). \end{aligned}$$

(110)

We see that

$$\begin{aligned} \dot{p}&= \sum _i \dot{\xi }_i = \sum _i \xi _i g_i \end{aligned}$$

(111)

$$\begin{aligned}&= p \sum _i \frac{\xi _i}{p} g_i = p \sum _i x_i g_i \end{aligned}$$

(112)

$$\begin{aligned}&= p \phi \end{aligned}$$

(113)

where $\phi \equiv \sum _i x_i g_i$ is the average fitness of the whole population.

By the product rule,

$$\begin{aligned} \dot{x}_i&= \frac{\dot{\xi }_i}{p} - \frac{\xi _i \dot{p}}{p^2} \end{aligned}$$

(114)

$$\begin{aligned}&= \frac{\xi _i }{p}g_i - \frac{\xi _i}{p} \frac{\dot{p}}{p} \end{aligned}$$

(115)

$$\begin{aligned}&= x_i \left( g_i - \phi \right) . \end{aligned}$$

(116)

Therefore,

$$\begin{aligned} \sum _i \dot{x}_i&= \sum _i x_i g_i - \phi \sum _i x_i \end{aligned}$$

(117)

$$\begin{aligned}&= \sum _i x_i g_i - \sum _j x_j g_j \sum _i x_i. \end{aligned}$$

(118)

So, using the fact that

$$\begin{aligned} \sum _i x_i = \frac{\sum _i \xi _i}{p} = \frac{p}{p} \equiv 1 \end{aligned}$$

(119)

Equation (118) reduces to the identity

$$\begin{aligned} \sum _i \dot{x}_i = 0. \end{aligned}$$

(120)

The fitness of a strategy is assumed to depend only on the relative abundance of each strategy in the overall population, since the model only seeks to capture the effect of competition between strategies, not any environmental or other factors. Therefore, we assume that $g_i$ has the form

$$\begin{aligned} g_i(\xi _1,\ldots ,\xi _n) = f_i\left( \frac{\xi _1}{p},\ldots ,\frac{\xi _n}{p} \right) = f_i(x_1,\ldots ,x_n). \end{aligned}$$

(121)

Under this assumption, Eq. (116) is the replicator equation,

$$\begin{aligned} \dot{x}_i = x_i (f_i - \phi ), \end{aligned}$$

(122)

where $\phi$ is now expressed entirely in terms of the $x_i$, as

$$\begin{aligned} \phi = \sum _i x_i f_i. \end{aligned}$$

(123)

Mathematically, $\phi$ is a coupling term that introduces dependence on the abundance and fitness of other strategies.

Appendix 2: Coefficients generated in the RPS problem

The entries of the matrix $J$ from Eq. (34) are

$$\begin{aligned} \alpha&= x^* \left( (a_1-b_1)(x^*-1)-(a_2+b_1+b_3)y^* \right) \end{aligned}$$

(124)

$$\begin{aligned} \beta&= x^* \left( (a_1+a_3+b_2)(x^*-1)+(a_3-b_3)y^* \right) \end{aligned}$$

(125)

$$\begin{aligned} \gamma&= y^* \left( (a_1-b_1)x^*-(a_2+b_1+b_3)(y^*-1) \right) \end{aligned}$$

(126)

$$\begin{aligned} \delta&= y^* \left( (a_1+a_3+b_2)x^*+(a_3-b_3)(y^*-1) \right) \end{aligned}$$

(127)

where $x^*$ and $y^*$ are the coordinates of the interior equilibrium point,

$$\begin{aligned} (x^*,y^*)&= \left( \frac{b_3 \left( a_3+b_2\right) +a_1 a_3}{a_1\left( a_2+a_3+b_1\right) +a_2 \left( a_3+b_2\right) +b_3\left( a_3+b_1+b_2\right) +b_1 b_2},\right. \nonumber \\&\quad \left. \frac{a_1 \left( a_2+b_1\right) +b_1 b_3}{a_1 \left( a_2+a_3+b_1\right) +a_2 \left( a_3+b_2\right) +b_3 \left( a_3+b_1+b_2\right) +b_1 b_2}\right) . \end{aligned}$$

(128)

The coefficients in Eqs. (52) and (53) are

$$\begin{aligned} c_1&= (a_1-b_1)(2x^*-1)-(a_2+b_1+b_3)y^* \end{aligned}$$

(129)

$$\begin{aligned} c_2&= (a_1+a_3+b_2)(2x^*-1)+(a_3-b_3)y^* \end{aligned}$$

(130)

$$\begin{aligned} c_3&= -(a_2+b_1+b_3)x^* \end{aligned}$$

(131)

$$\begin{aligned} c_4&= (a_3-b_3)x^* \end{aligned}$$

(132)

$$\begin{aligned} d_1&= a_1-b_1 \end{aligned}$$

(133)

$$\begin{aligned} d_2&= a_1+a_3+b_2 \end{aligned}$$

(134)

$$\begin{aligned} d_3&= -(a_2+b_1+b_3) \end{aligned}$$

(135)

$$\begin{aligned} d_4&= a_3-b_3 \end{aligned}$$

(136)

$$\begin{aligned} h_1&= (a_1-b_1)y^* \end{aligned}$$

(137)

$$\begin{aligned} h_2&= (a_1+a_3+b_2)y^* \end{aligned}$$

(138)

$$\begin{aligned} h_3&= (a_1-b_1)x^*-(a_2+b_1+b_3)(2y^*-1) \end{aligned}$$

(139)

$$\begin{aligned} h_4&= (a_1+a_3+b_2)x^*-(a_3-b_3)(2y^*-1) \end{aligned}$$

(140)

$$\begin{aligned} j_1&= -(a_2+b_1+b_3) \end{aligned}$$

(141)

$$\begin{aligned} j_2&= a_3-b_3 \end{aligned}$$

(142)

$$\begin{aligned} j_3&= a_1-b_1 \end{aligned}$$

(143)

$$\begin{aligned} j_4&= a_1+a_3+b_2. \end{aligned}$$

(144)

The coefficients $B_1,\ldots ,B_4$ in Eqs. (77) and (78) are

$$\begin{aligned} B_1&= \frac{1}{2} \left[ s \left( 2 c_4 r+c_2+c_3\right) \cos (\omega _0 T_0) \right. \nonumber \\&\quad \left. -\left( c_4 (r-s) (r+s)+\left( c_2+c_3\right) r+c_1\right) \sin (\omega _0 T_0)\right] \end{aligned}$$

(145)

$$\begin{aligned} B_2&= \frac{1}{2} \left[ -s \left( 2 c_4 r+c_2+c_3\right) \sin (\omega _0 T_0) \right. \nonumber \\&\quad \left. -\left( c_4 (r-s) (r+s)+\left( c_2+c_3\right) r+c_1\right) \cos (\omega _0 T_0)\right] \end{aligned}$$

(146)

$$\begin{aligned} B_3&= \frac{1}{2} \left[ s \left( 2 h_4 r+h_2+h_3\right) \cos (\omega _0 T_0) \right. \nonumber \\&\quad \left. -\left( h_4 (r-s) (r+s)+\left( h_2+h_3\right) r+h_1\right) \sin (\omega _0 T_0)\right] \end{aligned}$$

(147)

$$\begin{aligned} B_4&= \frac{1}{2} \left[ -s \left( 2 h_4 r+h_2+h_3\right) \sin (\omega _0 T_0) \right. \nonumber \\&\quad \left. -\left( h_4 (r-s) (r+s)+\left( h_2+h_3\right) r+h_1\right) \cos (\omega _0 T_0)\right] \end{aligned}$$

(148)

where $r$ and $s$ are as in Eq. (74).

Appendix 3: Removal of secular terms in Lindstedt’s method with delay

Consider a system of differential delay equations of the form

$$\begin{aligned} \omega \frac{\mathrm{d}u}{\mathrm{d}t}&= \alpha \bar{u} + \beta \bar{v} + K_1 \sin t + K_2 \cos t \end{aligned}$$

(149)

$$\begin{aligned} \omega \frac{\mathrm{d}v}{\mathrm{d}t}&= \gamma \bar{u} + \delta \bar{v} + K_3 \sin t + K_4 \cos t. \end{aligned}$$

(150)

where $\bar{u} = u(t-\omega T)$ and $\bar{v} = v(t-\omega T)$, and where $\omega$ and $T$ are such that the associated homogeneous problem,

$$\begin{aligned} \omega \frac{\mathrm{d}u}{\mathrm{d}t}&= \alpha \bar{u} + \beta \bar{v} \end{aligned}$$

(151)

$$\begin{aligned} \omega \frac{\mathrm{d}v}{\mathrm{d}t}&= \gamma \bar{u} + \delta \bar{v} \end{aligned}$$

(152)

admits solutions of the form $\sin t$ and $\cos t$, or equivalently $\mathrm{e}^{i t}$.

Substituting $u=r \mathrm{e}^{i t}$ and $v=s \mathrm{e}^{i t}$ into Eqs. (151) and (152), we obtain the characteristic equations

$$\begin{aligned} i r \omega&= \mathrm{e}^{-i \omega T}(\alpha r +\beta s ) \end{aligned}$$

(153)

$$\begin{aligned} i s \omega&= \mathrm{e}^{-i \omega T}(\gamma r +\delta s ). \end{aligned}$$

(154)

Rearranging, these become

$$\begin{aligned} \left( \begin{array}{cc} \alpha \mathrm{e}^{-i\omega T}-i\omega &{}\quad \beta \mathrm{e}^{-i\omega T} \\ \gamma \mathrm{e}^{-i\omega T} &{}\quad \delta \mathrm{e}^{-i\omega T} - i\omega \end{array}\right) \left( \begin{array}{c} r \\ s \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \end{array}\right) . \end{aligned}$$

(155)

Define

$$\begin{aligned} R \equiv \left( \begin{array}{cc} \alpha \mathrm{e}^{-i\omega T}-i\omega &{}\quad \beta \mathrm{e}^{-i\omega T} \\ \gamma \mathrm{e}^{-i\omega T} &{}\quad \delta \mathrm{e}^{-i\omega T} - i\omega \end{array}\right) . \end{aligned}$$

(156)

A non-trivial solution for $r$ and $s$ requires that $\det R = 0$. Separating the real and imaginary parts, this means that

$$\begin{aligned} Re(\det R)&= \cos (2 \omega T) (\alpha \delta -\beta \gamma )-\omega ((\alpha +\delta ) \sin (\omega T)+\omega ) =0 \end{aligned}$$

(157)

$$\begin{aligned} Im(\det R)&= -\cos (\omega T) (\sin (\omega T) (2 \alpha \delta -2 \beta \gamma )+\omega (\alpha +\delta )) = 0. \end{aligned}$$

(158)

Equation (158) tells us that

$$\begin{aligned} \sin (\omega T) = \frac{\omega (\alpha +\delta )}{2 (\beta \gamma -\alpha \delta )}. \end{aligned}$$

(159)

(We neglect the alternate possibility that $\cos (\omega T)=0$.) Then, we substitute this back into Eq. (157) to obtain

$$\begin{aligned} \omega ^2 = \alpha \delta -\beta \gamma . \end{aligned}$$

(160)

Under the conditions (159) and (160), the solutions to Eqs. (149) and (150) will in general have secular terms:

$$\begin{aligned} u&= m_1\cos t + m_2 \sin t + n_1 t \cos t + n_2 t \sin t \end{aligned}$$

(161)

$$\begin{aligned} v&= m_3\cos t + m_4 \sin t + n_3 t \cos t + n_4 t \sin t. \end{aligned}$$

(162)

We wish to derive conditions on the $K_i$ in Eqs. (149) and (150) such that the $n_i$ are all equal to 0.

We substitute the solutions (161) and (162) into Eqs. (149) and (150), and set the coefficients of $\sin t$, $\cos t$, $t\sin t$ and $t\cos t$ separately equal to 0 in both equations.

The coefficients of $\sin t$ and $\cos t$ give us a system of linear equations on the $m_i$ and $n_i$, of the form

$$\begin{aligned} M\cdot \mathbf m + N\cdot \mathbf n = - \mathbf k \end{aligned}$$

(163)

where $\mathbf m = (m_1,\ldots ,m_4)^\mathrm{T}$, $\mathbf n = (n_1,\ldots ,n_4)^\mathrm{T}$ and $\mathbf k =(K_1,\ldots ,K_4)^\mathrm{T}$.

Similarly, the coefficients of $t\sin {t}$ and $t\cos {t}$ give us a system of linear equations on the $n_i$, of the form

$$\begin{aligned} S\cdot \mathbf n = \mathbf 0 . \end{aligned}$$

(164)

By row reducing in Mathematica, we find that both $M$ and $S$ have rank 2. To eliminate the $n_i$, we proceed as follows:

• Without loss of generality, set $m_3=m_4=0$.

• Solve any two independent rows of Eq. (164) for $n_3$ and $n_4$ in terms of $n_1$ and $n_2$. The result is

  $$\begin{aligned} n_3&=\frac{n_2 \omega \cos (\omega T)-n_1 (\alpha +\omega \sin (\omega T))}{\beta } \end{aligned}$$

  (165)
  $$\begin{aligned} n_4&= -\frac{n_1 \omega \cos (\omega T)+n_2(\alpha + \omega \sin (\omega T))}{\beta } \end{aligned}$$

  (166)

• Substitute these expressions for $n_3$ and $n_4$ into Eq. (163). This is now a full-rank linear system of equations on $m_1,m_2,n_1$ and $n_2$. Solve this system to obtain expressions for $m_1,m_2,n_1$ and $n_2$ in terms of the $K_i$.

• Substitute the expressions for $n_1$ and $n_2$ from the previous step into Eqs. (165) and (166). Now we have all the $n_i$ in terms of the $K_i$.

• Set the $n_i$ expressions equal to 0. This gives a rank-2 system of equations on the $K_i$, so it is possible to solve for $K_3$ and $K_4$ in terms of $K_1$ and $K_2$. The result is

  $$\begin{aligned} K_3&= \frac{\gamma (K_1 (\alpha +\omega \sin (\omega T))+K_2 \omega \cos (\omega T))}{\alpha ^2+2 \alpha \omega \sin (\omega T)+\omega ^2} \end{aligned}$$

  (167)
  $$\begin{aligned} K_4&= \frac{\gamma (K_2 (\alpha +\omega \sin (\omega T))-K_1 \omega \cos (\omega T))}{\alpha ^2+2 \alpha \omega \sin (\omega T)+\omega ^2}. \end{aligned}$$

  (168)

  Using Eqs. (159) and (160), these reduce to

  $$\begin{aligned} K_3&= \frac{K_1 (\delta -\alpha )-K_2 \sqrt{-(\alpha -\delta )^2-4 \beta \gamma }}{2 \beta } \end{aligned}$$

  (169)
  $$\begin{aligned} K_4&= \frac{K_1 \sqrt{-(\alpha -\delta )^2-4 \beta \gamma }+K_2 (\delta -\alpha )}{2 \beta }. \end{aligned}$$

  (170)

If Eqs. (169) and (170) hold, then there are solutions of Eqs. (149) and (150) with no secular terms.