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Evolution of Cooperation in Prisoner’s Dilemma within Changeable External Environments

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Abstract

Rational individuals make a relatively dominant choice when faced with two opposing options of cooperation and betrayal in interactions. Without other interference, this often leads to a dilemma when social interests and personal interests are inconsistent, like the prisoner’s dilemma. However, in reality, an individual plays multiple games simultaneously. Members of a population may be influenced by other games to modify their strategic choices in the main game. In this paper, we consider intraspecific and interspecific games simultaneously from a multi-game perspective. How intraspecific cooperation evolves when different types of interspecific games as changeable environments is of great interest to us. We scrutinize the evolution of cooperation within the population in the prisoner’s dilemma by replicator dynamics with three different types of external environments of cooperation. The results show that the emergence and maintenance of cooperation in the intraspecific game are related to the relative size and the probability of cooperation in the external environment. Numerical experiences are provided to support the theoretical results.

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Funding

This work is supported by a grant of the National Natural Science Foundation of China (NNSF) (No. 10972011) and the Science Foundation of Hebei Normal University, China (No. L2018B01).

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GZ and HT developed ideas for the model. AX analysed the results and wrote the manuscript. HT corrected the manuscript.

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Correspondence to Haiyan Tian.

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Appendices

Appendix A The External Environment is Cooperation Type I

Proof of Theorem 1

To ensure \(\bar{\rho _{C} } \in (0,1)\), we get

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \theta (S_{2} +(1-S_{2}-T_{2})\varphi )+S_{1}>0,\\ \theta (S_{2} +(1-S_{2}-T_{2})\varphi )+S_{1}<S_{1}+T_{1}-1. \end{array}\right. } \Rightarrow {\left\{ \begin{array}{ll} \varphi < \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})},\\ \varphi > \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}. \end{array}\right. } \end{aligned} \end{aligned}$$
(A1)

Considering \(\varphi \in (0,1)\), the following conditions need to be satisfied:

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\ge 0,\\ \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\le 1. \end{array}\right. } \end{aligned} \Rightarrow \theta \ge \frac{T_{1}-1}{S_{2}}. \end{aligned}$$
(A2)

So, when \(\theta \ge \frac{T_{1}-1}{S_{2}}\) and \(\frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}<\varphi < \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\), internal equilibrium point \(\bar{\rho _{C}}\) exist. Considering

$$\begin{aligned} \begin{aligned}&{g}'(\bar{\rho _{C}})=\frac{1}{(1+\theta )^{2}} \left( S_{1} +\theta (S_{2}+(1-S_{2}-T_{2})\varphi )\right) \\&\qquad \qquad \quad \left( \frac{S_{1} +\theta (S_{2}+(1-S_{2}-T_{2})\varphi )}{S_{1}+T_{1}-1 }-1\right) <0,\\&{g}'(0)=\frac{1}{(1+\theta )^{2}} \left( S_{1} +\theta (S_{2}+(1-S_{2}-T_{2})\varphi )\right)>0,\\&{g}'(1)=-\frac{1}{(1+\theta )^{2}} \left( 1-T_{1} +\theta (S_{2}+(1-S_{2}-T_{2})\varphi )\right) >0. \end{aligned} \end{aligned}$$
(A3)

Therefore, internal equilibrium point \(\bar{\rho _{C}}\) is stable.

When \(\bar{\rho _{C}}\le 0\) or \(\bar{\rho _{C}}\ge 1\), internal equilibrium point \(\bar{\rho _{C}}\) doesn’t exist. We get

$$\begin{aligned} \begin{aligned}&S_{1} +\theta (S_{2}+(1-S_{2}-T_{2})\varphi )\le 0\\&\Rightarrow \theta > \frac{-S_{1}}{S_{2}}~~and~~\frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\le \varphi < 1. \end{aligned} \end{aligned}$$
(A4)

Now,

$$\begin{aligned} & {g}'(0)<0, {g}'(1)>0. \end{aligned}$$
(A5)
$$\begin{aligned} & \begin{aligned}&S_{1} +\theta (S_{2}+(1-S_{2}-T_{2})\varphi )\ge S_{1}+T_{1}-1\\&\Rightarrow \theta > \frac{T_{1}-1}{S_{2}}~~ and~~ 0< \varphi \le \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}. \end{aligned} \end{aligned}$$
(A6)

Now,

$$\begin{aligned} {g}'(1)<0, {g}'(0)>0. \end{aligned}$$
(A7)

When \(0< \theta \le \frac{-S_{1}}{S_{2}}\) and \(0<\varphi < 1\), then

$$\begin{aligned} (1-T_{2})\frac{-S_{1} }{S_{2}}+S_{1}<S_{1} +\theta (S_{2}+(1-S_{2}-T_{2})\varphi )<0. \end{aligned}$$
(A8)

So, \(\bar{\rho _{C}}<0\), internal equilibrium point \(\bar{\rho _{C}}\) doesn’t exist. Now,

$$\begin{aligned} {g}'(0)<0, {g}'(1)>0. \end{aligned}$$
(A9)

When \(\frac{-S_{1}}{S_{2}}< \theta < \frac{T_{1}-1}{S_{2}}\) and \(0< \varphi <\frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \), \(\bar{\rho _{C}}\) exist. Now,

$$\begin{aligned} {g}'(\bar{\rho _{C}})<0, {g}'(0)>0,{g}'(1)>0. \end{aligned}$$
(A10)

\(\square \)

Proof of Theorem 2

When \(S_{1}+\theta (S_{2} +(1-S_{2}-T_{2})\varphi ) < 0\), \({g}'(0)<0.\) Considering \(\varphi \in (0,1)\), we get \(\theta \ge \frac{-S_{1}}{S_{2}}\) and \(\frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}<\varphi <1\). The same reason can be used to obtain when \(S_{1}+\theta (S_{2} +(1-S_{2}-T_{2})\varphi ) > 0\). When \(0<\theta <\frac{-S_{1}}{S_{2}} \) and \(0<\varphi <1\), the proof is the same as in Case1. \(\square \)

Proof of Theorem 3

To ensure \(\bar{\rho _{C} } \in (0,1)\), we get

$$\begin{aligned} \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}< \varphi <\frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \end{aligned}$$
(A11)

Considering \(\varphi \in (0,1)\), the following conditions need to be satisfied:

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \ge 0\\ \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \le 1. \end{array}\right. } \end{aligned} \Rightarrow \theta \ge \frac{-S_{1}}{S_{2}}. \end{aligned}$$
(A12)

Thus, when \(\theta \ge \frac{-S_{1}}{S_{2}}\) and \( \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}< \varphi <\frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\), internal equilibrium point \(\bar{\rho _{C}}\) exist.Considering

$$\begin{aligned} \begin{aligned}&{g}'(\bar{\rho _{C}})> 0,\\&{g}'(0)< 0,\\&{g}'(1)< 0. \end{aligned} \end{aligned}$$
(A13)

Therefore, internal equilibrium point \(\bar{\rho _{C}}\) is not stable. Equilibrium point \(\rho _{C} =0\), \(\rho _{C} =1\) are stable.

When \(\bar{\rho _{C}}\le 0\) or \(\bar{\rho _{C}}\ge 1\), internal equilibrium point \(\bar{\rho _{C}}\) doesn’t exist. We get

$$\begin{aligned} & \theta > \frac{-S_{1}}{S_{2}}~~ and~~ 0< \varphi \le \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \Rightarrow {g}'(1)<0. \end{aligned}$$
(A14)
$$\begin{aligned} & \theta > \frac{T_{1}-1}{S_{2}}~~ and~~ \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\le \varphi< 1 \Rightarrow {g}'(0)<0. \end{aligned}$$
(A15)

When \(0< \theta \le \frac{T_{1}-1}{S_{2}}\) and \(0<\varphi < 1\), then \(\bar{\rho _{C}}<0\), internal equilibrium point \(\bar{\rho _{C}}\) doesn’t exist. Now,

$$\begin{aligned} {g}'(0)<0, {g}'(1)>0. \end{aligned}$$
(A16)

When \(\frac{T_{1}-1}{S_{2}}< \theta < \frac{-S_{1}}{S_{2}}\) and \(0< \varphi <\frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\), \(\bar{\rho _{C}}\) exist. We have

$$\begin{aligned} {g}'(\bar{\rho _{C}})>0, {g}'(0)<0,{g}'(1)<0. \end{aligned}$$
(A17)

\(\square \)

Appendix B The External Environment is Cooperation Type II

Proof of Theorem 4

To ensure \(\bar{\rho _{C} } \in (0,1)\), we have

$$\begin{aligned} \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}< \varphi < \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \end{aligned}$$
(B18)

Considering \(\varphi \in (0,1)\), the following conditions need to be satisfied:

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\ge 0,\\ \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\le 1. \end{array}\right. } \end{aligned} \Rightarrow \theta \ge \frac{T_{1}-1}{1-T_{2}}. \end{aligned}$$
(B19)

Then, \({g}'(0)>0\), \({g}'(1)>0\), \({g}'(\bar{\rho _{C}})<0\).

When \(\bar{\rho _{C}}\le 0\) or \(\bar{\rho _{C}}\ge 1\), internal equilibrium point \(\bar{\rho _{C}}\) doesn’t exist. We have

$$\begin{aligned} & \begin{aligned} \theta > \frac{-S_{1}}{1-T_{2}}~~ and~~ 0< \varphi \le \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \end{aligned} \Rightarrow {g}'(0)<0. \end{aligned}$$
(B20)
$$\begin{aligned} & \begin{aligned} \theta > \frac{T_{1}-1}{1-T_{2}}~~ and~~ \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \le \varphi< 1 \end{aligned} \Rightarrow {g}'(1)<0. \end{aligned}$$
(B21)

When \(0< \theta \le \frac{-S_{1}}{1-T_{2}}\) and \(0<\varphi < 1\), then \(\bar{\rho _{C}}<0\), internal equilibrium point \(\bar{\rho _{C}}\) doesn’t exist. Now,

$$\begin{aligned} {g}'(0)<0, {g}'(1)>0. \end{aligned}$$
(B22)

When \(\frac{-S_{1}}{1-T_{2}}< \theta < \frac{T_{1}-1}{1-T_{2}}\) and \(\frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}< \varphi < 1\), \(\bar{\rho _{C}}\) exist. Now,

$$\begin{aligned} {g}'(\bar{\rho _{C}})<0, {g}'(0)>0, {g}'(1)>0. \end{aligned}$$
(B23)

\(\square \)

Proof of Theorem 5

When \(S_{1}+\theta (S_{2} +(1-S_{2}-T_{2})\varphi ) < 0\), \({g}'(0)<0.\) Considering \(\varphi \in (0,1)\), we get \(\theta \ge \frac{-S_{1}}{1-T_{2}}\) and \(0<\varphi <\frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\). The same reason can be used to obtain when \(S_{1}+\theta (S_{2} +(1-S_{2}-T_{2})\varphi ) > 0\). When \(0< \theta < \frac{-S_{1}}{1-T_{2}} \) and \(0<\varphi < 1\), the proof is the same as in Case1. \(\square \)

Proof of Theorem 6

To ensure \(\bar{\rho _{C} } \in (0,1)\), we have

$$\begin{aligned} \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}< \varphi < \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}. \end{aligned}$$
(B24)

Considering \(\varphi \in (0,1)\), the following conditions need to be satisfied:

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\ge 0\\ \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})}\le 1. \end{array}\right. } \end{aligned} \Rightarrow \theta \ge \frac{-S_{1}}{1-T_{2}}. \end{aligned}$$
(B25)

Then, \({g}'(0)<0\), \({g}'(1)<0\), \({g}'(\bar{\rho _{C}})>0\).

When \(\bar{\rho _{C}}\le 0\) or \(\bar{\rho _{C}}\ge 1\), internal equilibrium point \(\bar{\rho _{C}}\) doesn’t exist. We have

$$\begin{aligned} & \theta > \frac{-S_{1}}{1-T_{2}}~~ and~~ \frac{-S_{1}-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \le \varphi< 1 \Rightarrow {g}'(1)<0. \end{aligned}$$
(B26)
$$\begin{aligned} & \theta > \frac{T_{1}-1}{1-T_{2}}~~ and~~ 0< \varphi \le \frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})} \Rightarrow {g}'(0)<0. \end{aligned}$$
(B27)

When \(0< \theta \le \frac{T_{1}-1}{1-T_{2}}\) and \(0<\varphi < 1\), then \(\bar{\rho _{C}}<0\), internal equilibrium point \(\bar{\rho _{C}}\) doesn’t exist. Now,

$$\begin{aligned} {g}'(0)<0, {g}'(1)>0. \end{aligned}$$
(B28)

When \(\frac{T_{1}-1}{1-T_{2}}< \theta < \frac{-S_{1}}{1-T_{2}}\) and \(\frac{T_{1}-1-\theta S_{2} }{\theta (1-S_{2}-T_{2})}< \varphi <1\), \(\bar{\rho _{C}}\) exist. We have

$$\begin{aligned} {g}'(\bar{\rho _{C}})>0, {g}'(0)<0,{g}'(1)<0. \end{aligned}$$
(B29)

\(\square \)

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Xing, A., Zhang, G. & Tian, H. Evolution of Cooperation in Prisoner’s Dilemma within Changeable External Environments. Dyn Games Appl (2024). https://doi.org/10.1007/s13235-024-00607-1

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