Linear Differential Game With Two Pursuers and One Evader

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Looking at Figure 8, case where τ 2 < τ 1 , the normalized zero-effort-miss at final time can be now written: z 1( 0) = z1( τ 1 ) + z1max ( τ 1 ) + Δz1( Δτ ) z = z τ − z τ + Δ 0 Since ( ) = Δ ( Δτ ) ( ) ( ) ( ) ( ) 2 0 2 2 2 max 2 z 2 Δ 0 z * z 1 1 because evader plays only against P1 ( v 1x1 = −1) during Δ τ (see Figure 8). Notice that in Figure 8 the x axis corresponds to τ 1 values ( τ 2 + Δτ ) . The third terms that * corresponds to the new change in due to v 2x1 is: Δz z 1 f And, thanks to the properties of integrals, H 21 * * ( Δτ ) = ε H ( Δτ → τ ) ( v − v ) 1 1 21 1 2x1 1x1 θ ⎛ ξ ⎞ ⎜ ⎟ ⎝ ε ⎠ 1 ( Δτ → τ ) = h dξ = H ( τ ) − H ( Δτ ) 1 ∫ Δτ 21 1 21 Expression becomes: Δz * [ ]( v 1) ( Δτ ) = ε H ( τ ) − ε H ( Δτ ) 1 1 21 1 1 21 2x1 + Substituting these expressions leads to: z * ( ) = z ( τ ) + μ H ( τ ) − ε H ( Δτ ) + [ ε H ( τ ) − ε H ( Δτ )] v 1 0 1 1 1 11 1 1 21 1 21 1 1 21 2x1 Moreover, the expression of z z 2 f remain the same as before when time to go coincided: * [ ] + ε H ( τ ) ( v 1) ( 0) z ( τ ) − μ H ( τ ) − ε H ( τ ) 2 = 2 2 2 12 2 2 22 2 2 22 2 2x1 − From which, using the initial equality, we can isolated v , in the case where τ 2 < τ 1 : * 2x1 v * 2× 1 ( z) = 2 [ − μ1 H 11( τ 1 ) − z1( τ 1 ) + ε 1H 21( Δτ )] τ P1 + [ μ 2 H 12 ( τ 2 ) − z 2 ( τ 2 )] 2 2 ε H ( τ ) τ + [ ε H ( τ ) − ε H ( Δτ )] τ 2 22 2 P2 1 21 1 1 21 P1 τ 2 P2 Due to the problem symmetry, the same calculus can be done switching the role of pursuers 1 and 2. Expression of v in case where τ 2 > τ 1 is then as follow: v * 2× 1 ( z) * 2x1 = 2 [ − μ1 H 11( τ 1 ) − z1( τ 1 )] τ P1 + [ μ 2 H 12 ( τ 2 ) − ε 2 H 22 ( Δτ ) − z 2 ( τ 2 )] 2 2 [ ε H ( τ ) − ε H ( Δτ )] τ + ε H ( τ ) τ 2 22 2 2 22 * Note that the particular case where Δτ = 0 leads to the expression of v 2x1 found in section 3.2 (same time to go). By the way, we compute the 2x1 NEZ extension when time to go differs. Figure 9 and Figure 10 show τ 2x1 limit remaining as a straight line (dash red line in Figure 9). The green crosses in Figure 10 represent an optimal trajectory on the 2x1 NEZ boundary at different time instants. P2 1 21 1 P1 τ 2 P2 - 8 -
12000 10000 8000 ZEM [m] 6000 4000 2000 0 0 2 4 6 8 10 12 τ go [s] −τ Figure 9 : 2x1 NEZ for 1 sec. τ 1 2 = 10000 10000 ZEM [m] 5000 5000 4000 0 0 5 10 4000 0 0 5 10 ZEM [m] 3000 2000 1000 1000 0 0 2 4 6 3000 2000 1000 1000 0 0 2 4 6 ZEM [m] 500 500 0 0 1 2 3 4 5 τ go [s] 0 0 1 2 3 4 5 τ go [s] Figure 10 : 2x1 NEZ sections at different time instants. 4. Conclusion We extend DGL/1 to the three players game involving two pursuers and one evader. We solve the game showing that for some initial conditions located between the pursuers the optimal evasion strategy is no more a maximum turn control. The optimal evasion consists then in a “trade off” strategy to drive the evader between the two pursuers. Considering 2x1 games we enlarge the interception area compared to solutions only involving two independent 1x1 NEZ. This approach could be applied when DGL/1 NEZ are bounded (closed) and when considering DGL/0 dynamics (linear differential game with first order for the pursuer and zero order for the evader). - 9 -
Page 1 and 2: Linear Differential Game With Two P
Page 3 and 4: The 2x1 game only makes a change in
Page 5 and 6: * ( μ 1H 11 − ε 1H 21 ) + ε 1H
Page 7: Moreover, in Figure 7 we draw an op

Linear Differential Game With Two Pursuers and One Evader

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