Toward a Solution of the Active Target Defense Differential Game

Abstract

A novel pursuit-evasion differential game involving three agents is considered. An Attacker missile is pursuing a Target aircraft. The Target aircraft is aided by a Defender missile launched by, say, the wingman, to intercept the Attacker before it reaches the Target aircraft. Thus, a team is formed by the Target and the Defender which cooperate to maximize the separation between the Target aircraft and the point where the Attacker missile is intercepted by the Defender missile, while at the same time the Attacker tries to minimize said distance. A long-range Beyond Visual Range engagement which is in line with current CONcepts of OPeration is envisaged, and it is therefore assumed that the players have simple motion kinematics á la Isaacs. Also, the speed of the Attacker is equal to the speed of the Defender and the latter is interested in point capture. It is also assumed that at all time the Attacker is aware of the Defender’s position, i.e., it is a perfect information game. The analytic/closed-form solution of the target defense pursuit-evasion differential game delineates the state space region where the Attacker can reach the Target without being intercepted by the Defender, thus disposing of the Game of Kind. The target defense Game of Degree is played in the remaining state space. The analytic solution of the Game of Degree yields the agents’ optimal state feedback strategies, that is, the instantaneous heading angles for the Target and the Defender team to maximize the terminal separation between Target and Attacker at the instant of interception of the Attacker by the Defender, and also the instantaneous optimal heading for the Attacker to minimize said separation. Their calculation hinges on the real-time solution of a quartic equation. In this paper we contribute to the solution of a differential game with three states—an additional example to the, admittedly small, repertoire of pursuit-evasion differential games in 3-D which can be solved in closed form.

Appendix

Root locus analysis of the quartic equation

The problem parameter is $0\le \alpha \le 1$. When $\alpha =0$ the four roots of quartic equation (14) are: $y=y_T$, $y=y_T$, and $y=\pm ix_A$, for all $x_T\in \mathbb {R}^1$. When $\alpha =1$ we have a quadratic equation whose roots are $y=\frac{x_Ay_T}{x_A+x_T}$ and $y=\frac{x_Ay_T}{x_A-x_T}$. If $x_T=\pm x_A$, then one of the roots is given by $y=+\infty$. This root takes a finite value as soon as $\alpha$ decreases, that is, $y<\infty$ for $\alpha =1-\epsilon$ for some small $\epsilon >0$.

When the discriminant of the quartic equation ${\varDelta }<0$, the quartic equation has two real roots and a pair of complex roots. We already know that Eq. (14) has two real roots, $0<y<y_T$ and $y>y_T$ as is indeed required by our theory for the case where $x_T<0$ and $x_T>0$, respectively. We will investigate quartic equation (14) using the continuation method, starting out from $\alpha =0$ where we have a double real root $y=y_T$—we conduct a root locus investigation for $0\le \alpha <1$.

To this end, differentiate quartic equation (14) in $\alpha$

$$\begin{aligned} \begin{aligned}&\left\{ 2(1-\alpha ^2)y^3 - 3(1-\alpha ^2)y_Ty^2 + \left[ (1-\alpha ^2)y_T^2+x_A^2-\alpha ^2x_T^2 \right] y - x_A^2y_T\right\} \frac{\mathrm{d}y}{\mathrm{d}\alpha } \\&\quad - \alpha \left[ y^4-2y_Ty^3+\left( x_T^2+y_T^2\right) y^2\right] = 0. \end{aligned} \end{aligned}$$

(107)

Setting $\alpha =0$ and $y=y_T$, yields the relationship $0\cdot \frac{\mathrm{d}y}{\mathrm{d}\alpha }=0$ and we cannot calculate $\frac{\mathrm{d}y}{\mathrm{d}\alpha }\big |_{\alpha =0}$. Hence, differentiate Eq. (107) in $\alpha$ again and set $\alpha =0$

$$\begin{aligned} \begin{aligned}&\left[ 2y^3 - 3y_Ty^2 + \left( y_T^2+x_A^2\right) y - x_A^2y_T \right] \frac{\mathrm{d}^2y}{\mathrm{d}\alpha ^2} + \left( 6y^2 - 6y_Ty + x_A^2 +y_T^2\right) \left( \frac{\mathrm{d}y}{\mathrm{d}\alpha }\right) ^2 \\&\quad - \left[ y^4-2y_Ty^3+\left( x_T^2+y_T^2\right) y^2\right] = 0. \end{aligned} \end{aligned}$$

(108)

Set $y=y_T$ in Eq. (108)

$$\begin{aligned} 0\cdot \frac{d^2y}{d\alpha ^2} + \left( x_A^2 +y_T^2\right) \left( \frac{\mathrm{d}y}{\mathrm{d}\alpha }\right) ^2 = x_T^2y_T^2 \ \ \Rightarrow \ \ \frac{\mathrm{d}y}{\mathrm{d}\alpha }\big |_{\alpha =0,y=y_T} = \pm \frac{x_Ty_T}{\sqrt{x_A^2+y_T^2}} . \end{aligned}$$

(109)

Also, setting $\alpha =0$ and $y=\pm ix_A$ in Eq. (107) yields

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}\alpha }\big |_{\alpha =0,y=\pm ix_A} = 0 . \end{aligned}$$

Note that setting $y=\pm ix_A$ in Eq. (108) yields

$$\begin{aligned} \left[ 2x_A^2y_T + \left( y_T^2-x_A^2\right) y\right] \frac{\mathrm{d}^2y}{\mathrm{d}\alpha ^2}=\left( x_A^2-x_T^2-y_T^2+2y_Ty\right) x_A^2 \end{aligned}$$

Fig. 19

Root locus of quartic equation (14)

Let $\frac{\mathrm{d}^2y}{\mathrm{d}\alpha ^2}=a+ib$. We calculate

$$\begin{aligned}&\left[ 2x_A^2y_T + ix_A\left( y_T^2-x_A^2\right) \right] (a+ib)=(x_A^2-x_T^2-y_T^2+2ix_Ay_T)x_A^2 \\&\quad \Rightarrow \ \begin{bmatrix} 2x_Ay_T&x_A^2-y_T^2 \\ -\left( x_A^2-y_T^2\right)&2x_Ay_T \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}=x_A \begin{bmatrix} x_A^2-x_T^2 -y_T^2 \\ 2x_Ay_T \end{bmatrix} \\&\quad \Rightarrow \ \begin{bmatrix} a \\ b \end{bmatrix}= \frac{1}{\left( x_A^2+y_T^2\right) ^2} \begin{bmatrix} 2x_Ay_T&-\left( x_A^2-y_T^2\right) \\ x_A^2-y_T^2&2x_Ay_T \end{bmatrix} \begin{bmatrix} x_A^2-x_T^2 -y_T^2 \\ 2x_Ay_T \end{bmatrix} x_A \\&\quad \Rightarrow \\&\quad \quad \quad a=-2\left( \frac{x_Ax_T}{x_A^2+y_T^2}\right) ^2 y_T <0 \\ \\&\quad \quad \quad b= \frac{\left( x_A^2+y_T^2\right) ^2 - \left( x_A^2-y_T^2\right) x_T^2}{(x_A^2+y_T^2)^2}x_A \end{aligned}$$

Similarly,

$$\begin{aligned}&\left[ 2x_A^2y_T - ix_A\left( y_T^2-x_A^2\right) \right] (a+ib)=\left( x_A^2-x_T^2-y_T^2-2ix_Ay_T\right) x_A^2 \\&\quad \Rightarrow \ \begin{bmatrix} 2x_Ay_T&-\left( x_A^2-y_T^2\right) \\ x_A^2-y_T^2&2x_Ay_T \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix}=x_A \begin{bmatrix} x_A^2-x_T^2 -y_T^2 \\ -2x_Ay_T \end{bmatrix} \\&\quad \Rightarrow \\&\quad \quad \quad a=-2\left( \frac{x_Ax_T}{x_A^2+y_T^2}\right) ^2 y_T <0 \ \ \ \text {as before!} \\&\quad \quad \quad b= -\frac{\left( x_A^2+y_T^2\right) ^2 - \left( x_A^2-y_T^2\right) x_T^2}{\left( x_A^2+y_T^2\right) ^2}x_A ; \ \ \ \text {note sign inversion!} \end{aligned}$$

So the two imaginary roots wander as a pair into the LHS of the complex plane, whereas the two real roots split off at $y=y_T$ and move in opposite direction toward $\frac{x_Ay_T}{x_A\pm x_T}$. The root locus picture of the roots of quartic equation (14) as a function of the problem parameter $0\le \alpha <1$ is shown in Fig. 19. Importantly, our quartic Eq. (14) has two complex roots and two real roots $y_1\ge y_T$ and $y_2\le y_T$, as required by the theory of the ATDDG.