A continuous differential game with N players is a Nash differential game. Ordinary differential equations regulate how the game’s state is evaluated throughout time. Each participant affects the outcome of the game by selecting an acceptable strategy. The cost of a player is determined by all players’ strategies and the state’s related evaluation. Differential equations are used to express information about the players, such as the target set, all players’ cost functions, and the current space. It is assumed that each player’s opponents are rational and that each player employs an optimal strategy, in which each player seeks to maximize his own gain. The Nash equilibrium notion essentially states that if one person tries to change his approach unilaterally, he will not be able to better his own optimization.
Definition 1
If the cost functions for players \(1,2,\ldots,N\) are \(J_{1}(\nu _{1}, \nu _{2},\ldots, \nu _{N}),\ldots,J_{n}(\nu _{1},\nu _{2}, \ldots, \nu _{N})\), then \((\nu _{1}^{*},\nu _{2}^{*},\ldots,\nu _{N}^{*})\) is said to be a Nash equilibrium strategy if, for \(i=1,2,\ldots,N \),
$$ J_{i}\bigl(\nu _{1}^{*},\ldots,\nu _{i},\ldots,\nu _{n}^{*}\bigr)\geq J_{i}\bigl(\nu _{1}^{*},\ldots, \nu _{i}^{*},\ldots,\nu _{N}^{*}\bigr). $$
(6)
The Nash equilibrium solutions for two players, the government and the ITO, are investigated in this research.
Nash equilibrium
Take the next problem:
$$ \left. \begin{aligned} &\max_{\nu _{1}}J_{1}= \int _{0}^{\infty }e^{\zeta _{1}t} \bigl[ \omega _{1}~h(\nu _{1},\nu _{2})+q \kappa c Zk \nu _{2}\alpha _{1} \nu _{1} \bigr]\,dt, \\ &\max_{\nu _{2} } J_{2}= \int _{0}^{\infty }e^{\zeta _{2}t} [ \sigma _{1}~Z+\beta _{1} \nu _{2} \gamma _{1}~\kappa ]\,dt, \\ &\frac{d Z}{d t}=r~Z(t)h(\nu _{1},\nu _{2}),\qquad Z(0)=Z_{0}>0 ,\quad Z \geq 0, \\ &\frac{d \kappa }{d t}=\xi~\kappa + b_{1} \nu _{2}  a_{1} \nu _{1},\qquad \kappa (0)=\kappa _{0}>0 , \quad \kappa (t) \geq 0 . \end{aligned} \right\} $$
(7)
The Hamiltonian government’s \(H_{1}\) function is defined by
$$ \begin{aligned}H_{1}={}&\omega _{1}~ h(\nu _{1}, \nu _{2}) + q~\kappa (t) c~Z(t)k~\nu _{2}\alpha _{1} \nu _{1}+\chi _{1}(t)~\bigl( r~Z(t)h(\nu _{1}, \nu _{2})\bigr) \\ &{}+ \chi _{2}(t)~\bigl(\xi~\kappa (t) + b_{1} \nu _{2}  a_{1} \nu _{1}\bigr), \end{aligned} $$
where \(\chi _{1}\) and \(\chi _{2}\) are called the costate variables or the adjoint variables associated with the state variables \(Z(t)\) and \(\kappa (t)\) for ITO and the government, respectively.
The ITO’s Hamiltonian is defined as
$$\begin{aligned} H_{2} =&\sigma _{1} Z(t)+\beta _{1} \nu _{2} (t)\gamma _{1} \kappa (t)+ \Psi _{1}(t) \bigl(r Z(t)h(\nu _{1}, \nu _{2})\bigr) \\ &{}+\Psi _{2}(t) \bigl(\xi \kappa (t)+ b_{1} \nu _{2} a_{1} \nu _{1}\bigr), \end{aligned}$$
where \(\Psi _{1}\) and \(\Psi _{2}\) are the costate variables or adjoint variables for ITO and government, respectively, and are related with the state variables \(Z(t)\) and \(\kappa (t)\).
The government’s and ITO’s best tactics must maximize the Hamiltonians \(H_{1}\) and \(H_{2}\); thus, the firstorder conditions are obtained:
$$ \frac{\partial H_{1}}{\partial \nu _{1}}=(\omega _{1}\chi _{1}) \frac{\partial h}{\partial {\nu _{1}}}\alpha _{1}a_{1} \chi _{2}=0 \quad \Rightarrow\quad \nu _{1}^{*}=\nu _{1}^{*}( \nu _{2}, \chi _{1}, \chi _{2} ) $$
(8)
and
$$ \frac{\partial H_{2}}{\partial \nu _{2}}=\beta _{1}\Psi _{1} { \frac{\partial h}{\partial {\nu _{2}}}}+b_{1}\Psi _{2}=0 \quad \Rightarrow\quad \nu _{2}^{*}=\nu _{2}^{*}(\nu _{1, \Psi _{1},\Psi _{2}}). $$
(9)
\(\chi _{1}\), \(\chi _{2}\), \(\Psi _{1}\), and \(\Psi _{2}\) are adjoint variables that satisfy the differential equations
$$ \begin{aligned} &\frac{{d\chi _{1}}}{dt}=\zeta _{1}\chi _{1} \frac{\partial H_{1}}{\partial Z}=(\zeta _{1}r)\chi _{1}+c, \\ &\frac{d\chi _{2}}{dt}=\zeta _{1}\chi _{2} \frac{\partial H_{1}}{\partial \kappa }=(\zeta _{1}\xi )\chi _{2}q, \\ &\frac{d\Psi _{1}}{dt}=\zeta _{2}\Psi _{1} \frac{\partial H_{2}}{\partial Z}=(\zeta _{2}r)\Psi _{1}\sigma _{1}, \\ &\frac{d\Psi _{2}}{dt}=\zeta _{2}\Psi _{2} \frac{\partial H_{2}}{\partial \kappa }=(\zeta _{2}\xi )\Psi _{2}+ \gamma _{1}. \end{aligned} $$
(10)
The conditions that limit transversality are as follows:
$$ \begin{aligned} &\lim_{t\to \infty } \int _{0}^{\infty }e^{(\zeta _{1}r)t} Z(t) \chi _{1}(t)\,dt =0, \\ &\lim_{t\to \infty } \int _{0}^{\infty }e^{(\zeta _{1}\xi )t} \kappa (t) \chi _{2}(t)\,dt =0, \\ &\lim_{t\to \infty } \int _{0}^{\infty }e^{(\zeta _{2}r)t} Z(t) \Psi _{1}(t)\,dt =0, \\ &\lim_{t\to \infty } \int _{0}^{\infty }e^{(\zeta _{2}\xi )t} \kappa (t) \Psi _{2}(t)\,dt =0. \end{aligned} $$
(11)
The adjoint equations’ solutions (10) are
$$ \begin{aligned} &\chi _{1}= \biggl(\chi _{10}+ \frac{c}{\zeta _{1}r} \biggr)e^{(\zeta _{1}r)t} \frac{c}{\zeta _{1}r}, \\ &\chi _{2}= \biggl(\chi _{20}\frac{q}{\zeta _{1}\xi } \biggr)e^{( \zeta _{1}\xi )t}+\frac{q}{\zeta _{1}\xi }, \\ &\Psi _{1}= \biggl(\Psi _{10}\frac{\sigma _{1}}{\zeta _{2}r} \biggr)e^{( \zeta _{2}r)t}+\frac{\sigma _{1}}{\zeta _{2}r}, \\ &\Psi _{2}= \biggl(\Psi _{20}+\frac{\gamma _{1}}{\zeta _{2}\xi } \biggr)e^{( \zeta _{2}\xi )t}\frac{\gamma _{1}}{\zeta _{2}\xi }. \end{aligned} $$
(12)
The adjoint variables diverge to ±∞ according to condition (5), violating the terms of the transversality, unless the values in the steady state are constant
$$ \begin{aligned} &\chi _{1}=\chi _{10}= \frac{c}{\zeta _{1}r},\qquad \chi _{2}=\chi _{20}= \frac{q}{\zeta _{1}\xi }, \\ &\Psi _{1}=\Psi _{10}=\frac{\sigma _{1}}{\zeta _{2}r},\qquad \Psi _{2}= \Psi _{20}=\frac{\gamma _{1}}{\zeta _{2}\xi }. \end{aligned} $$
(13)
With respect to the optimal strategies \(\nu_{1}\) and \(\nu_{2}\), the Hamiltonians \(H_{1} \) and \(H_{2}\) are concave, according to (13), where the harvest function is defined as follows:
$$ h(\nu _{1},\nu _{2})=\nu _{1}^{\tau _{1}}\nu _{2}^{\varrho },\quad \varrho >1, 0< \tau _{1} < 1. $$
Proposition 1
The Nash problem’s best strategies can be found in
$$ \begin{aligned} &\nu _{1}= \biggl( \frac{\beta _{1} + b_{1} \Psi _{2}}{\varrho \Psi _{1}} \biggr)^{\frac{\varrho }{\tau _{1} +\varrho 1}} \biggl( \frac{\alpha _{1}+a_{1} \chi _{2}}{\tau _{1} (\omega _{1}\chi _{1})} \biggr)^{\frac{1\varrho }{\tau _{1}+\varrho 1}}, \\ &\nu _{2}= \biggl(\frac{\beta _{1}+b_{1} \Psi _{2}}{\varrho \Psi _{1}} \biggr)^{\frac{1\tau _{1}}{\tau _{1}+\varrho 1}} \biggl( \frac{\alpha _{1}+a_{1} \chi _{2}}{\tau _{1} (\omega _{1}\chi _{1})} \biggr)^{\frac{\tau _{1}}{\tau _{1}+\varrho 1}}, \\ & h(\nu _{1},\nu _{2})= \biggl( \frac{\beta _{1}+ b_{1} \Psi _{2}}{\varrho \Psi _{1}} \biggr)^{\frac{\varrho }{\tau _{1}+\varrho 1}} \biggl( \frac{\alpha _{1}+a_{1} \chi _{2}}{\tau _{1} (\omega _{1}\chi _{1})} \biggr)^{\frac{\tau _{1}}{\tau _{1}+\varrho 1}}. \end{aligned} $$
(14)
Proof
From the necessary conditions
$$\begin{aligned}& \frac{\partial H_{1}}{\partial \nu _{1}}=(\omega _{1}\chi _{1}) \frac{\partial h}{\partial {\nu _{1}}}\alpha _{1}a_{1}\chi _{2}=0, \\& \frac{\partial H_{2}}{\partial \nu _{2}}=\beta _{1}\Psi _{1} \frac{\partial h}{\partial {\nu _{2}}}+ b_{1} \Psi _{2}=0; \end{aligned}$$
then
$$ \frac{\partial {h}}{\partial {\nu _{1}}}= \frac{\alpha _{1}+ a_{1}\chi _{2}}{\omega _{1}\chi _{1}},\qquad \frac{\partial h}{\partial \nu _{2}}= \frac{\beta _{1}+b_{1}\Psi _{2}}{\Psi _{1}}. $$
(15)
We have obtained the following results from the harvest function \(h( \nu _{1},\nu _{2})\):
$$ \frac{\partial {h}}{\partial \nu _{1}} =\tau _{1} \nu _{1}^{\tau _{1}1} \nu _{2}^{\varrho }= \frac{\alpha _{1}+a_{1}\chi _{2}}{\omega _{1}\chi _{1}},\qquad \frac{\partial h}{\nu _{2}}= \varrho \nu _{1}^{\tau _{1}}\nu _{2}^{ \varrho 1}= \frac{\beta _{1}+ b_{1}\Psi _{2}}{\Psi _{1}}; $$
thus,
$$\begin{aligned}& \nu _{1}= \biggl( \frac{\alpha _{1}+a_{1}\chi _{2}}{\tau _{1}(\omega _{1}\chi _{1})} \biggr)^{\frac{1}{\tau _{1}1}}\nu _{2}^{\frac{\varrho }{1\tau _{1}}}, \end{aligned}$$
(16)
$$\begin{aligned}& \nu _{2}= \biggl(\frac{\beta _{1}+ b_{1}\Psi _{2}}{\varrho \Psi _{1}} \biggr)^{\frac{1}{\varrho 1}} \nu _{1}^{\frac{\tau _{1}}{1\varrho }}. \end{aligned}$$
(17)
By solving (16) and (17), we obtain
$$ \begin{aligned} &\nu _{1}= \biggl( \frac{\beta _{1}+b_{1} \Psi _{2}}{\varrho \Psi _{1}} \biggr)^{\frac{\varrho }{\tau _{1}+\varrho 1}} \biggl( \frac{\alpha _{1} + a_{1} \chi _{2}}{\tau _{1} (\omega _{1}\chi _{1})} \biggr)^{\frac{1\varrho }{\tau _{1}+\varrho 1}}, \\ &\nu _{2}= \biggl( \frac{\beta _{1} + b_{1} \Psi _{2}}{\varrho \Psi _{1}} \biggr)^{\frac{1\tau _{1}}{\tau _{1}+\varrho 1}} \biggl( \frac{\alpha _{1}+ a_{1} \chi _{2}}{\tau _{1} (\omega _{1}\chi _{1})} \biggr)^{\frac{\tau _{1}}{\tau _{1}+\varrho 1}}, \\ &h(\nu _{1},\nu _{2})= \biggl( \frac{\beta _{1} + b_{1} \Psi _{2}}{\varrho \Psi _{1}} \biggr)^{\frac{\varrho }{\tau _{1}+\varrho 1}} \biggl( \frac{\alpha _{1}+a_{1} \chi _{2}}{\tau _{1} (\omega _{1}\chi _{1})} \biggr)^{\frac{\tau _{1}}{\tau _{1}+\varrho 1}}. \end{aligned} $$
(18)
□
Remark

1.
In (16), since \(\varrho > 1\); \(0 < \tau _{1} < 1 \), then the power of \(\nu _{2}\) is greater than 1 (\(\frac{\varrho }{1\tau _{1}}>1\))). This relation shows that the government is more aggressive when ITO increases their attacks and when the harvest function increases, as shown in Fig. 1. More discussion: Equation (16) showed the relationship between the strategy of the terrorist organization and the strategy of the government. We obtained this relationship from the condition \(\frac{\partial H_{1}}{\partial \nu _{1}}=0\). The government had neglected for short time the fight against terrorist organizations, after that it became aware and began to take measures and developed its strategies, this was shown in Fig. 1. We found that the government had played its role and taken control of the situation; for this, the activity of terrorist organizations stopped at a specific level, which was some light work. Also, we note that the government’s actions were a reaction to the terrorist organizations, but the reaction continued until the government took complete control of the situation.

2.
In (17), since \(\varrho > 1\), \(0 < \tau _{1} < 1 \), then the power of \(\xi _{1}\) is less than 1, (\(\frac{\tau _{1}}{1\varrho }<1\)) an increase in counterterror measures leads to a more cautious behavior by the terrorists and a lower harvest function, as shown in Fig. 2. Also, equation (17) showed the relationship between the strategies of terrorist organizations and the strategies of the government, which we obtained from the condition \(\frac{\partial H_{2}}{\partial \nu _{2}}=0\). Through this relationship, we noticed that when the government neglected the fight against terrorism, the organizations developed themselves and spread widely, this is clear from Fig. 2\(\nu _{1}\rightarrow 0\) as \(\nu _{2}\rightarrow \infty \); and when the government fulfilled its duty towards this problem, the activities of the organizations began to decline, and the government did not stop until it took control of the situation, which put things back in order, restricted terrorism, and the situation became stable \(\nu _{1}\rightarrow \infty \) as \(\nu _{2}\rightarrow 0\).
Lemma 1
For a constant harvest and constant strategies \(h(\nu _{1}, \nu _{2})\) and \(\nu _{1}\), \(\nu _{2}\) for the governments’ objective \(J_{1}\) and ITO’ objective \(J_{2}\) are
$$\begin{aligned}& J_{1}=\frac{\omega _{1} h}{\zeta _{1}}+\frac{q}{\zeta _{1}\xi } \biggl(\kappa _{0}\frac{a_{1} \nu _{1}b_{1} \nu _{2}}{\xi } \biggr)+ \frac{a _{1}\nu _{1}b_{1} \nu _{2}}{\xi \zeta _{1}} \frac{c}{\zeta _{1}r} \biggl( Z_{0}\frac{h}{r} \biggr) \frac{c h}{r\zeta _{1}}, \\& J_{2}=\frac{\sigma _{1}}{\zeta _{2}r} \biggl(Z_{0}\frac{h}{r} \biggr)+ \frac{\sigma _{1} h}{r\zeta _{2}}+ \frac{\beta _{1} \nu _{2}}{\zeta _{2}} \frac{\gamma _{1}}{\zeta _{2}\xi } \biggl(\kappa _{0} \frac{a_{1} \nu _{1}b_{1} \nu _{2}}{\xi } \biggr) \frac{\gamma _{1} (a_{1} \nu _{1}b_{1} \nu _{2})}{\xi \zeta _{2}}. \end{aligned}$$
Proof
The differential equation’s solution \(\frac{d \kappa }{d t}=\xi \kappa + b_{1} \nu _{2}a_{1} \nu _{1} \) is
$$ \kappa (t)= \biggl(\kappa _{0} \frac{a_{1} \nu _{1}b_{1} \nu _{2}}{\xi } \biggr)e^{\xi t}+ \frac{a_{1} \nu _{1}b_{1} \nu _{2}}{\xi }. $$
(19)
From (16) in (19), we have
$$ \kappa (t)=\kappa _{0} e^{\xi t}\frac{b_{1}}{\xi } \bigl(1e^{\xi t}\bigr) \nu _{2} + \frac{a_{1}}{\xi } \bigl( \bigl(1e^{\xi t}\bigr) \bigr) \biggl( \frac{\alpha _{1}+a_{1}\chi _{2}}{\tau _{1}(\omega _{1}\chi _{1})} \biggr)^{\frac{1}{\tau _{1}1}}\nu _{2}^{\frac{\varrho }{1\tau _{1}}}, $$
(20)
and from (17) in (19), we have
$$ \kappa (t)= \kappa _{0}e^{\xi t}+\frac{a_{1}}{\xi }\bigl(1e^{\xi t}\bigr) \nu _{1} \frac{b_{1}}{\xi } \bigl(1+ e^{\xi t} \bigr) \biggl( \frac{\beta _{1} + b_{1}\Psi _{2}}{\varrho \Psi _{1}} \biggr)^{\frac{1}{1\varrho }}\nu _{1}^{\frac{\tau _{1}}{1\varrho }}, $$
(21)
and the differential equation’s solution \(\frac{d Z}{d t} = r Zh( \nu _{1},\nu _{2})\) is
$$ Z(t)= \biggl(Z_{0}\frac{h}{r} \biggr)e^{r t}+ \frac{h}{r}. $$
(22)
In the government objective \(J_{1}\) and the ITO objective \(J_{2}\), from (19) and (22),
$$\begin{aligned} J_{1} =& \int _{0}^{\infty }e^{\zeta _{1}t} \biggl[\omega _{1}~h + q~\biggl(\kappa _{0}\frac{a_{1} \nu _{1}b_{1} \nu _{2}}{\xi } \biggr)~e^{ \xi t}+\frac{q~(~a_{1}\nu _{1}b_{1} \nu _{2})}{\xi } \\ &{}c \biggl(Z_{0} \frac{h}{r} \biggr)~e^{r t} \frac{c h}{r}k \nu _{2}\alpha _{1} \nu _{1} \biggr]\,dt \\ =&\frac{\omega _{1} h}{\zeta _{1}}+\frac{q}{\zeta _{1}\xi } \biggl( \kappa _{0} \frac{a_{1} \nu _{1}b_{1} \nu _{2}}{\xi } \biggr)+ \frac{q~(a_{1} \nu _{1}b_{1}\nu _{2})}{\xi~\zeta _{1}} \\ &{} \frac{c}{\zeta _{1}r} \biggl(Z_{0}\frac{h}{r} \biggr) \frac{c~h}{r~\zeta _{1}}\frac{ k~\nu _{2}}{\zeta _{1}} \frac{\alpha _{1} \nu _{1}}{\zeta _{1}} \end{aligned}$$
and
$$\begin{aligned} J_{2} =& \int _{0}^{\infty }e^{\zeta _{2}t} \biggl[\sigma _{1} \biggl( Z_{0} \frac{ h}{r} \biggr)~e^{rt}+\frac{\sigma _{1}~h}{r}\gamma _{1} \biggl( \kappa _{0}\frac{a_{1}\nu _{1}b_{1}\nu _{2}}{\xi } \biggr)e^{\xi t} \\ &{}+ \frac{\gamma _{1}~(a~\nu _{1}b_{1} \nu _{2})}{\xi }+\beta _{1} \nu _{2} \biggr]\,dt \\ =&\frac{\sigma _{1}}{\zeta _{2}r} \biggl(\kappa _{0}\frac{ h}{r} \biggr)+ \frac{\sigma _{1}~h}{r~\zeta _{2}} \frac{\gamma _{1}}{\zeta _{2}\xi } \biggl(\kappa _{0} \frac{a_{1} \nu _{1}b_{1} \nu _{2}}{\xi } \biggr) \\ &{} \frac{\gamma _{1}~(a_{1} \nu _{1}b_{1}\nu _{2})}{\xi~\zeta _{2}}+ \frac{\beta _{1} \nu _{2}}{\zeta _{2}}, \end{aligned}$$
where \(\nu _{1}\), \(\nu _{2}\) as well as \(h(\nu _{1},\nu _{2})\) are given in(18). □
Remark

1.
Equation (20) represents the relationship between government activity \(\kappa (t)\) and the strategies of the terrorist organization \(\nu _{2}\). When \(\nu _{2}\rightarrow 0\), the government had a clear activity, and the relationship became increasing until the maximum value. After that the relationship became decreasing as the organizations increased their activity and developed their strategies, which affected the government that abandoned some of its duties until the organization took control of the state’s capabilities as shown in Fig. 3.

2.
Equation (21) represents the relationship between the government’s activity \(\kappa (t)\) and its strategies \(\nu _{1}\). When the strategies were weak, this reflected on the government’s performance, which also became weak; when the government developed its strategy, this led to an increase in its activity, and this activity continued with the development of government strategies until it reached to the stability, this is evident in Fig. 4.