Matematika. Vol. 48, No. 12, pp.43{54, 2004. UDC 519.873. ONE PURSUIT PROBLEM FOR DISCRETE GAMES. WITH SEVERAL PLAYERS. N.Yu. Satimov and ...
Russian Mathematics (Iz. VUZ) Vol. 48, No. 12, pp.43{54, 2004
Izvestiya VUZ. Matematika UDC 519.873
ONE PURSUIT PROBLEM FOR DISCRETE GAMES WITH SEVERAL PLAYERS N.Yu. Satimov and G.I. Ibragimov
1. Problem statement
Dierential games with integral constraints are studied in many works (e. g., 1]{9]). The basic results are obtained in 1]{4]. The games between a group of pursuers which act as one player and one escapee are of most interest. Earlier 5]{8] the su cient conditions of the termination of a game with integral constraints (and its discrete analog) were obtained. In this paper, we consider a linear discrete game with several players. We formulate the su cient condition of the termination of the pursuit from all points of a space. We prove that in the case of a single pursuer, this condition is also the necessary one. Assume that a discrete game is dened by the recurrent equations
zi (k + 1) = Ci zi (k) ; ui (k) + v(k) i = 1 : : : m (1) where zi ui v 2 Rn , n 1, k is the step number, k = 1 2 : : : Ci is a constant n n-matrix ui is a control parameter of the pursuit v is a control parameter of the escape. We choose the parameter ui as a sequence ui = ui () = (ui (1) ui (2) : : : ui (k) : : : ) from the closed ball Sp (i ) with the radius i centered at the space origin lp : 1=p 1 X kui ()k = jui (k)jp i : i=1
Here p, i , i = 1 : : : m, are known positive values, and the parameter v represents a sequence v = v() = (v(1) v(2) : : : v(k) : : : ) from the ball S () lp. Game (1) is said to be over if zi (k) = 0, for a certain value (j r) of the pair of indices (i k). De nition 1. We say that in game (1) from the initial state z0 = fz10 z20 : : : zm0 g it is possible to terminate the pursuit in N (z 0 ) steps if using any sequence v() 2 Sp (), one can construct sequences u1 () 2 Sp (1 ), u2 () 2 Sp (2 ) : : : um () 2 Sp (m ), such that for a certain value j of the index i, a solution zj = zj () = fzj (1) zj (2) : : : zj (k) : : : g of the equation zj (k + 1) = Cj zj (k) ; uj (k) + v(k) zj (1) = zj0 satises the condition zj (k) = 0 where 1 k N (z 0 ). In addition, in order to nd the values u1 (k) u2 (k) : : : um (k) of the parameters u1 u2 : : : um on the k-th step, k 1, one may use z1 (k) z2 (k) : : : zm(k) v(1) v(2) : : : v(k).
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