Evolutionary dynamics of cooperation in the public goods game with pool exclusion strategies

Abstract

Social exclusion is widely used as a control mechanism to promote cooperative behavior in human societies. However, it remains unclear how such control strategies actually influence the evolutionary dynamics of cooperation. In this paper, we introduce two types of control strategies into a population of agents that play the public goods game, namely prosocial pool exclusion and antisocial pool exclusion, and we use the replicator equation to study the resulting evolutionary dynamics for infinite well-mixed populations. We show that the introduction of prosocial pool exclusion can stabilize the coexistence of cooperators and defectors by means of periodic oscillations, but only in the absence of second-order prosocial pool exclusion. When considering also antisocial pool exclusion, we show that the population exhibits a heteroclinic circle, where cooperators can coexist with other strategists. Moreover, when second-order exclusion is taken into account, we find that prosocial pool exclusion is the dominant strategy, regardless of whether the second-order exclusion is prosocial or antisocial. In comparison with punishment, we conclude that prosocial pool exclusion is a more effective control mechanism to curb free-riding.

Appendixes

1.1 Appendix A

In order to explore the evolutionary dynamics of $S_{4}$, which we discuss in Sect. 3, we first study the dynamics of each edge of $S_{4}$.

On the edge $D-$$ED$ ($x+z=0$ and $y+w=1$), we have $\dot{y}=y(1-y)(P_{D}-P_{ED})=y(1-y)\delta _{A}>0$, and thus, the direction of the dynamics goes from ED to D.

On the edge $C-$$EC$ ($y+w=0$ and $x+z=1$), we have $\dot{x}=x(1-x)(P_{C}-P_{EC})=x(1-x)\delta _{P}>0$; therefore, the direction of the evolution goes from EC to C.

On the edge $C-D$ ($z+w=0$ and $x+y=1$), we have $\dot{y}=y(1-y)(P_{D}-P_{C})=-y(1-y)(\frac{rc}{N}-c)>0$; therefore, the direction of the evolution goes from C to D.

On the edge $C-$$ED$ ($z+y=0$ and $x+w=1$), we have $\dot{w}=w(1-w)(P_{ED}-P_{C})=w(1-w)\{\frac{rc[1-w-(1-w)^{N-1}]}{w}-\delta _{A}+c\}$; therefore, when the antisocial exclusion cost is less than the cost of cooperation, namely $\delta _{A}<c$, the direction of the evolution goes from C to ED.

On the edge $D-$$EC$ ($x+w=0$ and $y+z=1$), we have $\dot{z}=z(1-z)(P_{EC}-P_{D})=z(1-z)(rc-c-\delta _{P})>0$, and thus, the direction of the dynamics goes from D to EC.

On the edge $ED-$$EC$ ($x+y=0$ and $w+z=1$), we have $\dot{w}=w(1-w)(P_{ED}-P_{EC})=w(1-w)[-\delta _{A}+c+\delta _{P}-(1-w)^{N-1}rc]$; thus, there exist an equilibrium, namely $w=1-[\frac{c+\delta _{P}-\delta _{A}}{rc}]^{\frac{1}{N-1}}$.

Next, we investigate the evolutionary dynamics of each face. We have discussed the dynamics on the face $C-D-$$EC$ in Sect. 2, and the three strategies can coexist in the population.

On the face $C-D-$$ED$ ($z=0$ and $x+y+w=1$), we have

$$\begin{aligned} P_{C}= & {} \sum _{N_{C}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{C}\end{array}}\right) x^{N_{C}}(1-x-w)^{N-N_{C}-1}\\&\times \, \frac{rc(N_{C}+1)}{N}-c\\= & {} (1-w)^{N-1}\frac{rc}{N}\frac{1-w+(N-1)x}{1-w}-c,\\ P_{ED}= & {} \sum _{N_{C}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{C}\end{array}}\right) x^{N_{C}}(1-x)^{N-N_{C}-1}\frac{rcN_{C}}{N-N_{C}}-\,\delta _{A}\\= & {} \frac{rcx(1-x^{N-1})}{1-x}-\,\delta _{A},\\ P_{D}= & {} \sum _{N_{C}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{C}\end{array}}\right) x^{N_{C}}(1-x-w)^{N-N_{C}-1}\frac{rcN_{C}}{N}\\&+\,\sum _{N_{C}=0}^{N-1}\sum _{N_{ED}=1}^{N-N_{C}-1}\left( {\begin{array}{c}N-1\\ N_{C}\end{array}}\right) \left( {\begin{array}{c}N-N_{C}-1\\ N_{ED}\end{array}}\right) x^{N_{C}}w^{N_{ED}}\\&\times \,(1-x-w)^{N-N_{ED}-N_{C}-1}\frac{rcN_{c}}{N-N_{C}}\\= & {} (1-w)^{N-1}\frac{rc}{N}\frac{(N-1)x}{1-w}\\&+\,\frac{rcx[(1-w-x)+wx^{N-1}-(1-w)^{N-1}(1-x)]}{(1-x)(1-w-x)}. \end{aligned}$$

We know that $P_{C}<P_{D}$ is satisfied, and thus, there is no interior fixed point. In addition, the stability of three boundary points can be described as follow.

• (1) For $(x,y,w)=(0,0,1)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} -c+\delta _{A}&0\\ 0&\delta _{A} \end{bmatrix}; \end{aligned}$$

  thus, the fixed point is unstable.

• (2) For $(x,y,w)=(1,0,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} rc(N-2)+c-\delta _{A}&rc(N-2)-\delta _{A}+\frac{rc}{N}\\ 0&c-\frac{rc}{N} \end{bmatrix}; \end{aligned}$$

  thus, the fixed point is unstable.

• (3) For $(x,y,w)=(0,1,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} \frac{rc}{N}-c&0\\ -\frac{rc}{N}+c-\delta _{A}&-\delta _{A} \end{bmatrix}; \end{aligned}$$

  thus, the fixed point is stable.

On the face $D-$$EC$–ED ($x=0$ and $y+z+w=1$), the expected payoffs for these three strategies can be given by

$$\begin{aligned} P_{EC}= & {} \sum _{N_{EC}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{EC}\end{array}}\right) z^{N_{EC}}(1-z-w)^{N-N_{EC}-1}rc\nonumber \\&-\,c-\delta _{P},\nonumber \\= & {} (1-w)^{N-1}rc-c-\delta _{P} \end{aligned}$$

(39)

$$\begin{aligned} P_{D}= & {} 0,\end{aligned}$$

(40)

$$\begin{aligned} P_{ED}= & {} -\delta _{A}. \end{aligned}$$

(41)

Based on the above payoff expressions, we know that there is no interior fixed point. Furthermore, the stability of three boundary equilibria can be given as follows.

• (1) For $(y,z,w)=(0,0,1)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} \delta _{A}&0\\ 0&\delta _{A}-c-\delta _{P} \end{bmatrix}; \end{aligned}$$

  thus, this equilibrium is unstable.

• (2) For $(y,z,w)=(0,1,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} -rc+c+\delta _{P}&0\\ -\delta _{A}&-rc+c+\delta _{P}-\delta _{A} \end{bmatrix}; \end{aligned}$$

  thus, this equilibrium is stable.

• (3) For $(y,z,w)=(1,0,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} -\delta _{A}&-(rc-c-\delta _{P}+\delta _{A})\\ 0&rc-c-\delta _{P} \end{bmatrix}; \end{aligned}$$

  thus, this equilibrium is unstable.

• (4) For $(y,z,w)=(0, [\frac{c+\delta _{P}-\delta _{A}}{rc}]^{\frac{1}{N-1}},1-[\frac{c+\delta _{P}-\delta _{A}}{rc}]^{\frac{1}{N-1}})$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} \delta _{A}&0\\ a_{21}&(1-z)(N-1)z^{N-1}rc \end{bmatrix}, \end{aligned}$$

  where $a_{21}=(1-z)(N-1)z^{N-1}rc-z\delta _{A}$; thus, this equilibrium is unstable.

On the face ED–C–EC ($y=0$ and $x+z+w=1$), the expected payoffs for these three strategies can be described by

$$\begin{aligned} P_{EC}= & {} \sum _{N_{EC}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{EC}\end{array}}\right) z^{N_{EC}}(1-z-w)^{N-N_{EC}-1}rc\nonumber \\&-\,c-\delta _{P}\nonumber \\= & {} (1-w)^{N-1}rc-c-\delta _{P}, \end{aligned}$$

(42)

$$\begin{aligned} P_{C}= & {} (1-w)^{N-1}rc-c, \end{aligned}$$

(43)

$$\begin{aligned} P_{ED}= & {} \sum _{N_{ED}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{ED}\end{array}}\right) x^{N-N_{ED}-1}(1-z-x)^{N_{ED}}\nonumber \\&\times \,\frac{rc(N-N_{ED}-1)}{N_{ED}+1}-\delta _{A}\nonumber \\= & {} \frac{rcx[(1-z)^{N-1}-x^{N-1}]}{1-x-z}-\delta _{A}, \end{aligned}$$

(44)

Based on the payoff expressions, we can get that $P_{C}>P_{EC}$. Therefore, there is no interior fixed point in $S_{3}$. And the stability of three boundary fixed points can be described as follows.

• (1) For $(x,z,w)=(0,0,1)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} \delta _{A}-c&0\\ 0&\delta _{A}-c-\delta _{P} \end{bmatrix}; \end{aligned}$$

  thus, this equilibrium is stable,

• (2) For $(x,z,w)=(0,1,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} \delta _{P}&0\\ -(rc-c+\delta _{A})&-(rc-c-\delta _{P}+\delta _{A}) \end{bmatrix}; \end{aligned}$$

  thus, this equilibrium is unstable.

• (3) for $(x,z,w)=(1,0,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} rc(N-2)+c-\delta _{A}&a_{12}\\ 0&-\delta _{P} \end{bmatrix}, \end{aligned}$$

  where $a_{12}=rc(N-2)+c+\delta _{P}-\delta _{A}$, and thus, this equilibrium is unstable.

• (4) For $(x,z,w)=(0, [\frac{c+\delta _{P}-\delta _{A}}{rc}]^{\frac{1}{N-1}},1-[\frac{c+\delta _{P}-\delta _{A}}{rc}]^{\frac{1}{N-1}})$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} \delta _{P}&0\\ a_{21}&rc(N-1)(1-z)z^{N-1} \end{bmatrix}, \end{aligned}$$

  where $a_{21}=rcz(1-z)[(N-1)z^{N-2}-(1-z)^{N-2}]-z\delta _{P}$; thus, this equilibrium is unstable.

1.2 Appendix B

Then, we explore the evolutionary dynamics of $S_{4}$, which we discuss in Sect. 5.

On the edge $C-$$EC$ ($y+w=0$ and $x+z=1$), we have $\dot{z}=z(1-z)(P_{EC}-P_{C})=z(1-z)\{\frac{rc[1-(1-z)^{N-1}]}{z}-\delta _{P}\}$ since $\frac{1-(1-z)^{N-1}}{z}$ decreases with increasing z. Thus, there is not interior equilibrium for $\delta _{P}<rc$. As a result, the direction of the evolution goes from C to EC. The dynamics of other edges of simplex $S_{4}$ are same to those in Sect. 3.

Next, we investigate the evolutionary dynamics on each face. We have discussed the situation on the face $C-D-$$EC$ in Sect. 4. Then, on the face $C-D-$$ED$ ($z=0$ and $x+y+w=1$), the expected payoffs for these three strategies can be given as

$$\begin{aligned} P_{C}= & {} \sum _{N_{C}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{C}\end{array}}\right) x^{N_{C}}(1-x-w)^{N-N_{C}-1}\nonumber \\&\times \,\frac{rc(N_{C}+1)}{N}-c\nonumber \\= & {} (1-w)^{N-1}\frac{rc[(N-1)x+(1-w)]}{N(1-w)}-c, \end{aligned}$$

(45)

$$\begin{aligned} P_{D}= & {} \sum _{N_{C}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{C}\end{array}}\right) x^{N_{C}}(1-x-w)^{N-N_{C}-1}\frac{rcN_{C}}{N}\nonumber \\= & {} (1-w)^{N-1}\frac{rc(N-1)x}{N(1-w)}, and\end{aligned}$$

(46)

$$\begin{aligned} P_{ED}= & {} \sum _{N_{ED}=0}^{N-1}\sum _{N_{D}=0}^{N-1-N_{ED}}\left( {\begin{array}{c}N-1\\ N_{ED}\end{array}}\right) \left( {\begin{array}{c}N-1-N_{ED}\\ N_{D}\end{array}}\right) \nonumber \\&\times \, x^{N-N_{ED}-N_{D}-1}(1-x-y)^{N_{ED}}y^{N_{D}}\nonumber \\\times & {} \frac{rc(N-N_{ED}-N_{D}-1)}{N_{ED}+1}-\delta _{A}\nonumber \\= & {} \frac{rcx[1-(x+y)^{N-1}]}{1-x-y}-\delta _{A}. \end{aligned}$$

(47)

Since $P_{D}>P_{C}$, there is no interior point. And the stability of three boundary points can be described as follows

• (1) For $(x,y,w)=(0,0,1)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} -c+\delta _{A}&0\\ 0&\delta _{A} \end{bmatrix}; \end{aligned}$$

  thus, this fixed point is unstable.

• (2) For $(x,y,w)=(1,0,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} rc(N-2)+c-\delta _{A}&rc(N-2)-\delta _{A}+\frac{rc}{N}\\ 0&c-\frac{rc}{N} \end{bmatrix}; \end{aligned}$$

  thus, the fixed point is unstable.

• (3) For $(x,y,w)=(0,1,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} \frac{rc}{N}-c&0\\ -\frac{rc}{N}+c-\delta _{A}&-\delta _{A} \end{bmatrix}; \end{aligned}$$

  thus, this fixed point is stable.

The dynamic on the face $D-$$EC$–ED is the same to those in the situation without second-order exclusion. On the face $C-$$EC$–ED ($y=0$ and $x+z+w=1$), the expected payoffs of these three strategies can be given by

$$\begin{aligned} P_{C}= & {} (1-z-w)^{N-1}rc-c,\end{aligned}$$

(48)

$$\begin{aligned} P_{EC}= & {} \sum _{N_{C}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{C}\end{array}}\right) x^{N_{C}}(1-x-w)^{N-N_{C}-1}\nonumber \\&\quad \frac{rcN}{N-N_{C}}-c-\delta _{P},\nonumber \\= & {} \frac{rc[(1-w)^{N}-(1-w-z)^{N}]}{z}-c-\delta _{P},\nonumber \\\end{aligned}$$

(49)

$$\begin{aligned} P_{ED}= & {} \sum _{N_{ED}=0}^{N-1}\left( {\begin{array}{c}N-1\\ N_{ED}\end{array}}\right) (1-z-w)^{N-N_{ED}-1}w^{N_{ED}}\nonumber \\&\times \,\frac{rc(N-N_{ED}-1)}{N_{ED}+1}-\delta _{A},\nonumber \\= & {} \frac{rcx[(1-z)^{N-1}-x^{N-1}]}{w}-\delta _{A}. \end{aligned}$$

(50)

Solving $P_{C}=P_{ED}$ yields $w{=}\frac{rcx(1-z)[(1-z)^{N-2}-x^{N-2}]}{\delta _{A}-c}$; therefore, there is not interior fixed point for $\delta _{A}<c$. In addition, the stability of these four boundary points can be described as follows.

• (1) For $(x,z,w)=(0,0,1)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} -c+\delta _{A}&0\\ 0&\delta _{A}-c-\delta _{P} \end{bmatrix}; \end{aligned}$$

  thus, this fixed point is stable.

• (2) For $(x,z,w)=(1,0,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} rc(N-2)+c-\delta _{A}&-(rc-c+\delta _{A}-\delta _{P})\\ 0&rc(N-1)-\delta _{P} \end{bmatrix}; \end{aligned}$$

  thus, this fixed point is unstable.

• (3) For $(x,z,w)=(0,1,0)$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} -rc+\delta _{P}&0\\ c-\delta _{A}&-(rc-c-\delta _{P}+\delta _{A}) \end{bmatrix}; \end{aligned}$$

  thus, this fixed point is stable.

• (4) For $(x,z,w)=(0,[\frac{c+\delta _{P}-\delta _{A}}{rc}]^{\frac{1}{N-1}},1-[\frac{c+\delta _{P}-\delta _{A}}{rc}]^{\frac{1}{N-1}})$, the Jacobian is

  $$\begin{aligned} J=\begin{bmatrix} \delta _{A}-c&0\\ a_{21}&rcz^{N-1}(1-z)(N-1) \end{bmatrix}, \end{aligned}$$

  where $a_{21}=rcz(1-z)[Nz^{N-2}-(1-z)^{N-2}]-z(\delta _{A}-c)$; thus, this equilibrium is unstable.