Generalized ’Lion & Man’ game of r. Rado Текст научной статьи по специальности «Математика»

Abdulla A.Azamov and Atamurot Sh.Kuchkarov

Institute of Mathematics and Information Technologies, Academyof Sciences Republic of Uzbekistan,

Durman yuli, 29, Tashkent, 100125, Uzbekistan E-mail: emailabdulla.azamov@gmail.com, kuchkarov1@yandex.ru

Abstract In the present work we study the game of degree for generalized ”Lion and Man” game where ”Lion” L moves on the plane while ”Man”

M must move along the given curve r . The case when r is circumference the problem was formulated by R.Rado and can be considered as the first example of dynamic games. By elementary but refined arguments R.Rado (Littlewood, 1957, Rado, 1973) proved that L can capture M if speeds of both points are equal. Further interesting results on ”Lion & Man” game concerning the case when both points moved inside a circle, was obtained by J.O.Flynn (1973, 1974).

Keywords: differential game, pursuit, evasion, strategy, ”Lion and Man”.

1. Introduction

We study the generalized game ”Lion and Man” in the case of R.Rado, whose work can be considered as the first meaningful example of the Theory of differential games.

In this generalized game the Lion L moves on the plane and tries to capture Man M who can’t leave the given curve r. Maximum speeds of this points are equal to p and a respectively, p > 0, a > 0. Rado analyzed the case when r is circumference of the radius R, R > 0, and p = a. By refined and quite elementary arguments he showed that the point L can capture the point L. Moreover, if the initial position of L is at the center O of the circumference r, the capture can be completed for the time T = Later J.Flint investigated the case where r is a circumference and a > p. He proved that there exists number l, l > 0, such that L can do |LM | = l approach, while M is able to ensure the inequality |LM| > l (Flynn, 1973, 1974).

If M moves in the circle G bounded by the circumference r, then S. Bezikovich proved that M can avoid of encounter with L, i.e. can ensure the relation M = L, thou L can approach the point M to any positive distance.

It should be noted that cases of the game ”Lion and Man” studied by R.Rado and S.Bezikovich are quite different in point of view of the theory of differential games: in the case of Rado states of the points can be defined by two coordinates in the polar system, while in the case of Bezicovich the phase space exactly tree dimensional (Lewin, 1986).Moreover, if we assume that L can move all over the plane (as we do in this paper), then the case of Rado relates to the games without phase constraints, while in the case of Bezicovich the condition M <G G is essential and makes a phase constraint.

Note also, arguments of Rado relayed on symmetry of circumference and they lose validity for curves different from the circumference.

In the present work the quantity problem for generalized ”Lion and Man” game of R.Rado is studied when L moves along the given curve r. Let r be given by the

absolutely continuous map 7 : R ^ Rn and parameterized by arc length. Motions of the points L and M are described respectively by the equations x = u, y = v with the phase constraint y(t) G r and control constraints |u| < p, |v| < a (x, y, u, v G Rn). The problem is interesting if r is embedded into Rn isometrically to R or S. (If r is isometric to half line [0, to) then problem can be reduced to the first case.)

In dependent on r, p and a we’ll prove either the pursuit problem or the evasion problem has a solution. Any of the existing approaches being applied to the pursuitevasion problems can be taken as the basis of the formalization (Friedman, 1971, Krasovskiy and Subbotin, 1988, Petrosyan, 1977). Exact statement of the problem requires definition of the concept of strategy U of the pursuer and that of the evader V. The strategy U relays on current information with discrimination of the evader: at each time t, t > 0, the value u(t) will be found by using information about x(s), y(s), 0 < s < t and v(t). According to the strategy V the value v(t) will be found by using information about x(s), y(s), 0 < s < t.

There is no necessity for description this concepts in detail, since each time we prove the solvability of the problems constructively. When we consider the pursuit problem, we construct a concrete strategy U*, such that the trajectories x(t), y(t), are generated uniquely by arbitrary measurable control v(-) of the evader and initial positions x(0) = x0, y(0) = y0 and prove that x(t) = y(t) for some t G [0, T]. In this case we’ll say that for the initial position (x0, y0) the pursuit problem is solvable on the time interval [0,T].

Similarly, for the evasion problem we construct a concrete strategy V* such that the trajectories x(t), y(t) are generated uniquely by arbitrary control of the pursuer u(-) and the initial position (x0, y0) and prove that x(t) = y(t) for all t G [0, +to) . In this case we say that for the initial position (x0, y0) evasion problem is solvable.

Since relation x = y doesn’t exclude possibility of approaching the point x to y to any small distance, it is required in the evasion problem (Pontryagin, 1970) that strategy V* has the following properties: there exists positive quantity 6 = 6(x0, y0) such that |x(t) — y(t)| > 6, for all t, t > 0 independent on control u(-).

We list the main results.

Theorem 1. Let p = a and r be closed. Then there exists a strategy of the pursuer that guaranties the capture, i.e. there exists T, T > 0, such that for any admissible control function v(-) of M x(t) = y(t) at some t G [0, T] for the corresponding trajectories.

Now suppose r is not closed, so r\y(0) consists of two components r+, .

Condition A : There exists a point 7(s±) G r± such that |x(0) — y(s±)| = |s±| respectively.

Obviously if Condition A is not occurred for r+ or then M easily avoids a capture. It turns out the converse also is true.

Theorem 2. Suppose p = a and r is not necessarily closed and Condition A holds for both arcs r±. Then there exists a strategy of the pursuer that guaranties the capture, i.e. there exists T, T > 0, such that for any admissible control function v(-) of M x(t) = y(t) at some t G [0,T] for the corresponding trajectories.

For the proof of this theorem we refer to (Kuchkarov, 2008). The following theorem improves and extends the main result of (Azamov, 1986), where was assumed r to be plane curve with class of smoothness C2.

Theorem 3. Let p < a and the function 7' satisfies Lipschitz condition i.e.

|Y'(si) — Y'(s2)| < A|si — S2I, A > 0.

Then the evasion problem is solvable.

2. Proof of the Theorems

2.1. The radial strategy of pursuit

It can be supposed

x(0) = 0, y(0) = Y(0) = 0. (1)

Let v = vv + vT (respectively u = uv + uT) be the decomposition such that

vv (uv) is directed to y (respectively x) and vT (respectively uT) is orthogonal to

vv (uv ) .

Definition 1. The function

Ur(x, y, v) = £vt + y(p2 — £2|vt|2)1/2 j|y| (2)

defined on the region |v| < a, |x| < |y| , y = 0 in R6n will be called Rado’s radial

strategy of pursuit (briefly R-strategy), where £ = |x|/|y|.

Lemma 1. The formula (2) representing the decomposition of the vector u = Ur has the following properties:

a) |u| = p(speed of the point L is maximal);

b) angular speeds of L and M equal:

I | ___ | |

c) |uv | > |vv | (velocity of L is directed, to the side of M ).

Proof is straightforward.

Here R— strategy of Rado will be used in the game in case of p = a. If this is the case, without any loss of generality we can assume that p = a = 1, since if it is necessary, we can change the scale.

2.2. Proof of Theorem 1.

Let the closed smooth curve r be given by the natural equation y = y(s) where s is arc-length and y (■) is a continuously differentiable periodical function with the period equal to the length of y of the curve r. Hence we can suppose |dY(s)/ds| = 1 and the point y(s) moves counter-clockwise along r when parameter s grows. Notice that r may have self-crossings.

We can write the equation of the motion of M as

ds(t)/dt = w(t), s(0) = 0 (3)

(see (1)) where w (■) is the control function of M satisfying the conditions: a) —1 < w(t) < 1 a.e.; b) w (■) is measurable.

Let W be the class of all control functions of M. In order to avoid cumbersome calculations we restrict ourselves considering the subclass W0 consisting of all functions w(-) G W satisfying the additionally condition

c) w(t) = 1 or w(t) = —1 a.e.

It should be noted that the last assumption will not restrict generality because of possibility to approximate any function w* (■) G W by sequence wn (■) G W0 so that the corresponding trajectories xn(t) ^ x* (t) uniformly on every finite interval.

The function y(t) = y(s(î)) describes the motion of M.

If dy/dt = v is its velocity, then w (t) = (7'(s(t)), v(t)) and v(t) = w(t)Y'(s(t)).

If r is a Jordanian curve then there exists the natural one-to-one correspondence between W (respectively W0 ) and the class of all admissible control functions v (■) : [0, to) ^ R2 such that |v(t) | < 1 (|v(t) | = 1) a.e. and

t

y(t) = yo + J v(t)dr G , t > 0. (4)

0

Lemma 2. For any control function w (■) G W0 the Cauchy problem

dx

— = uR(x, y(t), v(t)), x(0) = 0 (5)

dt

has a solution. It is unique on the interval J, where J = [0, +to), if |x(t)| < |y(t)| for all t > 0, in other case J = [0, 6], where 6 = min {t |x(t) = y(t) } .

Correctness of the statement follows from Caratheodory’s existence and uniqueness theorem for the Cauchy problem by virtue of condition |x(0) | =0 < |y(0)| (see (1)).

For the exact formulation it is required to reveal some numerical characteristics of the curve r.

Let

S+ = {s |<Y(s), Y'(s)> > 0} , S- = {s |<Y(s) Y'(s)> < 0} ,

S0 = {s KY(sX Y'(s)> = 0}

be the partition of some fixed segment I of the length y into three parts according to rising, descending and circular parts of r with respect to the initial position 0 of the point L. The segment I will be defined concretely below, in the Main Lemma. Notice that S+ and S_ are open. Obviously if mS0 = g, then <y(s), Y/(s)> = 0 a.e., which implies that r is a circumference, that is the case considered by R.Rado (mX denotes the Lebegue measure of X.)

So one can suppose that mS0 < g. Then periodicity of y(-) implies the inequalities y+ = mS+ > 0 and y- = mS— > 0. And the desired characteristics will be

6 = min {g+, g-} , K = max |y(s)| .

It’s clear ô > 0, M > 0.

Lemma 3. The solution of the Cauchy problem (5) may be represented, in the form x(t) = A(i)y(i), where A(-) is a nonnegative scalar function satisfying the conditions

dX y 1 — (î-vA)2 — vT\ dt \y(t)\

, A(0) = 0. (6)

Proof. If the function A = A(t) satisfies the conditions (6) then it can be seen by Lemma 1 that the product A(t)y(t) appears in the solution of the Cauchy problem (5). That’s why Lemma 1 is a corollary of uniqueness theorem for differential equations. □

Lemma 4. A' (t) > 0 a.e. whenever A(t) < 1. Moreover

d\(t) y/1 - A(t)

dt ~ K

as soon as <y(t), v(t)> < 0 (t G J).

Proof. The first part of the statement follows from (6). The second part is the

corollary of the inequality

y/l - A V - Ap> Vl - A, (7)

where the parameters A, p, q satisfy the conditions p < 0, p2 + q2 = 1, 0 < A < 1. The inequality (7) can easily be checked by raising to square. □

Thus A(t) increases. The task is to prove that A(t) reaches the value 1 meaning L = M.

Properties of the function s (■). In Lemma 5 and 6 the time parameter t varies

on the time interval [a, ft]. Further we put Y = s ([a, ft]). Notice that Y C

[s(a) — ft + a, s(a) + ft — a] in virtue of |s'(t)| < 1.

Lemma 5. If X is a measurable subset of Y then mX < m (s-1 (X)) (where s-1 (X) is preimage of the set X).

Proof easily follows from the condition |s'(t)| < 1.

The following notations are important for the future: if X C Y, then

T+(X) = {t|s(t) G X, s'(t) = 1} ,

T-(X) = {t|s(t) G X, s'(t) = —1} , (8)

T±(X) = {t|s(t) G X} .

Lemma 6. If X is open in the topology of Y and s(a) = s(ft) then

mT+(X) — mT-(X) = 0. (9)

Proof. If X = Y, then equality (9) follows from the property w(t) = +1 or w(t) = —1 and the equality s(a) = s(ft). Now consider the case X = (c, +to) n Y. Here the set {t|s(t) G X} will be union of a countable family of open intervals. The equality (9) is true for each of the components of the union as it has been noticed just above

- as in the situation X = Y. That’s why it is true for the entire union as well. □

For the further notice m {t|s(t) = c} = 0 because otherwise we would get

m {t|s'(t) = 0} > 0

notwithstanding with the preposition w(t) = ±1.

This implies that (9) holds for X = (c1, c2) n Y and X = [c1, c2) n Y. Finally (9) arises for any open set X in relative topology of Y from the properties above.

Corollary 1. If X is open in the topology of Y, then

mT+(X) — mT-(X)=sgn(s(ft) — s(a)) m (X n [s', s'']) (10)

where [s', s''] = [s(a), s(ft)] if s(a) < s(ft) and [s', s''] = [s(ft), s(a)] in the other case.

Proof. Suppose s(a) < s(ft). Set ft = ft + s(ft) — s(a) and w = w(t) when a < t < ft and w(t) = —1 when ft < t < ft.

Further let

t

3(t) = s(a) + J w(t)dt

a

and T+(X), T_(X) be a partition of the segment [a, ft] for the function 3 (■) analogical to the partition (8) for s (■) . Notice s(a) = 3 (a) = 3(ft) . Hence we have

mT+(X) = mT+(X) and

mT^X) = mT-(X) + m{t G [ft, ft]|5(t) G X} =

= mT-(X) + m{s|s G [s(/3), s(ft)], s G X} =

= mT-(X) + m(X n [s', s'']).

The second equality is based on the property s'(t) = —1 for t G [ft, ft] in according with the definition.

The proof can be provided in the same way if s(a) > s(ft). Q.E.D. □

During the following lemma t changes in the time interval [0, g]. Let t* ( respectively t* ) be a point of the maximum (minimum) of s(t) on the interval [0, g]. Notice that s ([0, g]) = [s(t*), s(t*)] C [s(t*) — g, s(t*)].

Below we take the segment [s(t*) — y, s(t*)] as the interval I and deal with the sets: T+ (S+), T-(S+), T+(S-), T-(S-), T±($,).

The main Lemma 7.

mT-(S+)+mT+(S-)+mT±(So) >

Proof. By virtue of analogy it’s enough to consider the situation s(g) > 0.

Thus s ([0, g]) C [s(t*), s(t*)] C [s(t*) — g, s(t*)] .

It will be considered two cases separately.

The 1st case: s(t*) < g — g-/2. Let

S£ = S£ n [s(t*),g] (10)

where e replaces one of the signs +, — or 0. Here we have mT+(i>-) = mT-(S-) because the intersection S- n [s', s''] has the measure equal 0 (see, (10)).

Similarly mT+(S-) = mT-(S-).

Applying this two equalities, the notation (10) and the assumption s(t*) < g — g-/2 we get

mT-(S+) + mT+(S-) + mT±(S0) >

> }-[mT+(S+)+mT-(S+)+mT+(S-)+mT-(S-)+mT±(S0)] >

2

-1-1 jr

> —m[s(t*), g] = -(g - s(t*j) > ^ > -.

The 2nd case: s(t*) < g—g-/2. Here in virtue of Lemma 6 mT+(S-) = mT_(S_). Further using Lemma 5 and the condition for s(t*) < g — g-/2 we get

2mT+(S-) > mT+(S-) + mT-(S-) = m{t|s(t) G S-} >

> m{t|s(t) G fl [0, s(t*)]} =

= ms-1{S- O [0, s(t*)]} > m{S— O [0, s(t*)]} = mS— + m[0, s(t*)] — m{S— U [0, s(t*)]} >

g- + s(i*) — g > g-/2.

Here the relation S_ O [0, s(t*)] C O [s(t*), s(t*)] O ([0, g]) is used in order to apply Lemma 5.

Consequently, here also

mT—(S+) + mT+ (S_) + mT0 > g-/4.

Q.E.D. □

Ending of the proof of theorem 1. Let N = [8K/¿] + 1. It should be shown

for any w (•) G W0 there exists t G [0, Ng] such that A(t) = 1. Assume the contrary

i.e. A(t) < 1 for all t G [0, Ng]. Then Lemma 4 implies

x/l - A(g) - 1 = ~\J

<_1 / w 1 *, f w _ 1 ^

2 \Jt-(s+) dt - A(t) Jt+(S-) dt \]l - A(i) y

1 / dA(t) 1

1 W

2Jt±(s0) dt \/l ~ A(i)

Let each integral in the last expression be estimated. If t G T_(S+) i.e. s(t) G S+ and w(t) = —1 then

<Y'(s(t)) Y(s(t))) > 0, v(t) = —7/(s(t))-These relations and y(t) = Y(s(t)) imply <v(t), z(t)) < 0.

Similarly it t G T+(S_) i.e. s(t) G S— and w(t) = +1 then v(t) = 7'(s(t)) that’s why <v(t), z(t)) < 0 again.

At last it t G T±(So) then (v(t), z(t)) = 0. Thus A'(i) > \/l — A/K in all cases in virtue of Lemma 4. Applying this inequality and the main Lemma we get

y/T^ÀS)-1 < _mT*- + 'nT-* + mT« < - A.

V vy; - 2if - 8if

All reasonings done above for the interval [0, g] are correct for each of the

time intervals [(k — 1)g, kg], k = 2, 3, ..., N of the length g. Obtained in this way

estimations being summarized will lead to the following

y/1 - A(JVA) < 1 - ^ < 0

the last inequality contradicts to the assumption has been made in the beginning of the proof. Q.E.D.

Thus, in the generalized Rado’s game R-strategy is winning on the time interval [0, 0] where 0 = 8Kg/£. □

Remark 1. As obvious from the proof, Theorem is also true for curves r given

by the absolutely continuous function 7 (■) if piecewise constant functions w (■) are taken as admissible.

Remark 2. Another more simple proof of Theorem 1 follows from the proof of the Theorem 2. The last will be published in [9].

2.3. Proof of the theorem 3.

Let p < a and the function 7 ' satisfies Lipschitz condition i.e.

|Y'(si) — Y'(s2)| < A|si — S21, A > 0. (11)

This condition allows localize the evasion problem.

Let so and are positive roots of the equations

a — P A2s2

1 - - As = 0, (12)

2a 3Aso

I — a2 6 — A2Sq ’

(13)

respectively ( their existence and uniqueness are clear).

Lemma 8. Let <x0,y'(0)> < 0 (<x0,y'(0)> > 0). Then

|x0 — Y(s0)| — ps0/ct > As0 (|x0 — Y(—s0)| — ps0/ff > As0 ) . (14)

Furthermore, if |x01 > Asq, then

|xo — Y(s)| — ps/a > ao |xo| ( |xo — y( — s)|— ps/a > ao |xo|) , 0 < s < so. (15)

Proof. From (11) we obtain |y'(0) — y'(s)| < As, s > 0. Squaring this inequality we have 2 — 2y1 (s) < (As)2 , s > 0. Here and in the sequel we denote

Yi(s) = <Y(0),Y(s)> , Y2(s) = Y(s) — <Y(0),Y(s)> Y(0).

From here according to Yi(0) = 0 we obtain yi (s) > s — A2s3/6, s > 0. Then

|Y2(s)| < y s2 — (yi(s)) <y s2 — (s — A2s3/6) = As2y 3 1 — A2s2/36 < As2

at least on the interval s <G [0, s0]. Hence, by definition of s0 (see, 12), we have

Yi(s) — |Y2(s)| — ps/ct > s — A2s3/6 — As2 — ps/ct >

> s ^--------------------As^j > 0, 0 < s < so. (16)

First, we prove the inequality (15). It is not difficult to verify that using (12) and (13) yields

2\1/2 /, ,2, 2, ^1/2

|x0 — Y(s)| > (yi (s) + (|x0| — |Y2(s)|)^ > (|x0|2 + Yi2(s))

— |Y2(s)| > a0 |x0| + |Yi(s)| — |Y2(s)|.

Therefore, 16) implies (15). We proceed to show the inequality (14). In accordance with (12), we have

|x0 — Y(s0)| — ps0/ct > |Yi(s0)| — ps0/ct >

> s0 — A2s0/6 — Ps0/ct —s0 — s0 (A2sQ/6 — P/ct) —

= s0 — s0 (1 — As0) = As0 This completes the proof of the lemma. □

Proof of the theorem 3. Let u(-) be a control of the pursuer L and x(-) is the solution of the Cauchy problem x = u(t), x(0) = x0. We define the strategy V* and corresponding trajectory y(-) of the evader by stepwise process. Let A = {t0, ti, t2, ...} is the partition of the interval [0, to) , ij = is0/<r, i = 0, 1, ....

We assume that the trajectory y(-) is defined up to the instant of time ij. We set x(ij) = xj, y(ij) = yj, (i = 1,2, .... ) We continue the trajectory y(-) on the time interval [tj,tj+i] in the following way: if <xj — yj,y' (s(yj))> < 0, then y(-) is defined as the solution of the Cauchy problem y = cty'(s(y)), , y(tj) = yj, and if <x — y,Y' (s(y))> > 0, then y(-) is defined as the solution of the Cauchy problem y = —cty'(s(y)) , y(tj) = yj. Note that in both cases |y(t)| = 1.

Let <x0 — y0, y' (s(y0))> < 0. Then at 0 < t < s0/ct we have s(t) = CTt and

|x(t) — y(t)| >

t

x0 — Y(s(t)) + J u(t)dt 0

> |x0 — Y(s(t))| —

—pt = |x0 — Y(s(t)) | — Ps(t)/CT.

Hence,

|x(t) — y(t)| > min (|x0 — y(s(t))| — ps(t)/CT) > 0, 0 < t < ti. (17)

0<t<ti

Moreover, by using the inequality (14) we have

|x(ti) — y(ti)| > |x0 — Y(s(ti))| — ps(ti)/ct = |x0 — Y(s0)| — ps0/ct > AsQ.

(In case of <x0 — y0, y' (s(y0))> > 0 arguments are the same.)

The last relation allows us to continue the evasion process to the next time interval [ti,t2] = [s0/ct, 2s0/ct ] , while preserving the estimate

|x(t) — y(t)| > a0As0, |x(t2) — y(t2)| > AsQ.

This follows from lemma 8 if we take y(ti), x(ti) as the initial position of players and apply the reasoning above. This process can be continued infinitely. As the evader each time passes the path of length s0, then the estimate |x(t) — y(t)| > a0As0 is true for all t > ti.

Thus, the point M can act in the game in a such way that the distance between M and L estimates by

|x(t) — y(t)| > min (|x0 — y(s(t))| — ps(t)/CT) > 0, 0 < t < s0/ct,

0<t<s0/ct

|x(t) — y(t)| > a0As0, t > s0/ct.

The proof of the theorem is complete. □

Remark 3. Let r be an arbitrary smooth curve with unbounded curvature. As the curvature K(s) of the curve r is a continuous function, then it is bounded on every interval [si, s2] and, hence, the function Y'(s) satisfies the Lifschitz condition on each bounded interval. By using this fact, it can be shown as it was done in proof of theorem 3 that the evasion problem is solvable. However, the evader not always can ensure the estimate for the distance below by positive constant (for example, if r is graph of the function y = xsin(1/x) ).

3. Evasion from Many Pursuers Along the Smooth Surface

We consider the differential game with many pursuers moving all over the space and evader moving along given surface. We’ll construct a positional evasion strategy ensuring the estimate for the distance between players below by positive constant, provided the number of pursuers doesn’t exceed dimension of the surface. We are given n dimensional surface r from the class of smoothness C2 in the space Rn+i. At each point principal curvature of the surface r is bounded in absolute value by number 1/r, r > 0 and any geodesic on the surface r is either closed or from each its point in both direction has infinite length.

Movement of the pursuers Lj and the evader M are described by

Li : xi = u, xi(0) = Xio, M : y = v, y(0) = yo,

in the space Rn+i, where xjo = y0; uj, v are control vectors that satisfy conditions |uj| < pj, |v| < ct, i = 1, 2, ..., k. During the game the evader M can’t leave the surface r: y(t) G r, t > 0.

Aim of the present section is to construct the strategy of the evader ensuring estimate for the distance between evader and closest to pursuer. Control and strategy of players are defined as well as in section 1.

Theorem 4. Let n > 2, k < n. Then evasion problem is solvable. Furthermore, there exist a positive constant ^ ^ (x0,y0,p, ct, r) and a positional strategy of M

such that for any admissible control of L = {Li,Lq, ..., L&} the inequality

min |xj (t) — y(t) I > ^

i<j<k

holds for all t, t > 0.

Proof. Let K(z, m) be the normal curvature of the surface r in direction of the unit tangent vector m at z, and yZ (■, m) be a geodesic (parameterized by the length of the arc) on r such that yZ(0, m) = z, yZ (0, m) = m |yZ(z, m)| = 1. Then Y''(s, m) = Kzm(s)Nzx(s), where N is unit oriented normal vector field in r, and Nzm(s) is its value at yZ(s,m). Hence, |y''(s, m)| < 1/r and all the functions of the form y'(■, m) satisfy the Lifschitz condition with the same constant A > 0 :

Wz (s^ m) — Y' (s2 , m)| < A |si — s2 | . (18)

At t = 0 we define the unit vector v0 from the tangent space Ty(0) so that

<V0,xjo — y0> = min max <v,xj0 — y0> ,

|v| = i i<j<k

where v G Ty(0) and construct the geodesic Yy(0) (s, v0), s > 0. It is not difficult to verify by using k < n that <v0,xjo — y0> < 0. As (18) holds, then according to Theorem 3 the evader M preserves the inequality

|xj(t) — y(t)| > min (|y(s) — xj01— ps/ct) > 0

0<t<so

at 0 < s < s0/ct. while moving along geodesic Yy(0)(s, v0), s > 0, with the speed ct (see,(17))

Moreover, as in proof of theorem 3 applying lemma 7 at t = s0/ct we have

|xj(s0/ct) — y(s0/CT)| > AsQ.

his allows to repeat the process infinitely. Every time starting from the second, applying lemma 7 as in proof of theorem 3, it can be shown preservation of the inequality |xj(t) — y(t)| > a0As0 for all t > s0/ct and i = 1, 2, ..., k. The proof of the theorem is complete. □

Remark 4. Let r be any smooth surface of unbounded curvature. As the principal curvatures K(s) of the surface r is continuous function, then it is bounded on each interval [si, s2]. By using this fact it can be shown as in proof of theorem 4 that evasion problem is solvable. If the surface r is not smooth (for example, r is a cone), then theorem 4, in general, is not valid (see, appendix 2)

Appendix

1. Evasion along the Graph

Let r is a finite graph. Finite graph is a set consisting of finite number of points (called vertexes) and arcs of curves (called edges) connecting pairs of these points, each point being connected at least one another point.

We assume that r is a singly connected and each edge of the graph is image of a mapping y : [0,a] ^ Rn from the class of smoothness C2. Evader moves along the edges of the graph, and pursuer moves all over the space.

Let A be a vertex of the graph r and Yi(■), i = 1, 2,..., m is the set of all edges of the graph r with one end at A and A = y»(0) for all i = 1, 2, ..., m. We say that vertex A is convenient for evasion, if convex hull of all vectors y(0) contains A. By using lemma 7 it can be shown as in proof of theorem 3 that if p < ct and all vertices of the graph r are convenient for evasion, then in this version of the game it is solvable the evasion problem by preserving estimates for distances between players below. If p = ct, then arguing as in proof of theorem 1 one can verified that point L by using R-strategy will be able to complete pursuit from any initial position.

2. The Case when the evader moves along piecewise smooth curve

Theorem 5. Let r be a closed piecewise smooth Jordan curve and 2a is the least angle between tangent rays at breaking points, a G [0,n/2]. If p > CTsina, then there exists an initial point x0, y0, for that pursuit problem is solvable. If here r is convex or p = ct, then pursuit problem is solvable for any initial positions.

Proof of both parts of theorem are derived from the following statement. Let A be isosceles triangle with the base BC and vertex A, ZA = 2a, a G (0,n/2). Let x0 G BC, y0 G ABU AC and segment with ends x0 ,y0 are perpendicular to BC. Then strategy of parallel approach (Petrosyan, 1977, Azamov and Samatov, 2000) ensures solvability of the pursuit problem for the time h/ \J p2 — a2 sin2a, where h is hight of A.

The following statement can be also proved on the basis of the same consideration.

Theorem 6. Let r be an arbitrary finite graph in the space Rn with straight edges. Then there exists k G (0,1) such that at p > kCT pursuit problem is solvable for any initial position.

References

Littlewood, F. I. (1957). A Mathematician’s Miscellany. Methuen Co., London.

Rado, R. (1973). How the leon tames was saved. Math. Spectrum, 6(1), 14-18.

Flynn, J. O. (1973). Lion and Man: the boundari constraint. SIAM J. Control Optim., 11(3), 397-411.

Flynn, J. O. (1974). Pursuit in the circle: lion versus man. Different. Games and Cont. Theory, 99-124.

Lewin, J. (1986). The Lion and Man Problem Revisited. JOTA, 49(3), 411-430. Friedman, A. (1971). Differential games. Wilev, New York.

Krasovskiy, N. N. and A. I. Subbotin (1988). Game-theoretical control problems. Springer, Verlag.

Petrosyan, L. A. (1977). Differential games of pursuit. Leningrad University Press: Leningrad.(in Russian)

Pontryagin, L. S. (1970). Linear differential dames of evasion. Dokladi Akademii Nauk SSSR, 191(2), 283-285.(in Russian).

Kuchkarov, A. Sh. (2008). Solusion of Simple pursuit-Evasion Problem when Evader Moves on Given Curve. IGTR, to appear, www.worldscinet.com/iqtr/edioterial/paper. Azamov, A. A. (1986). On a Probem of Escape Along a Prescribed Curve. J. App. Math.

and Mash.,46(4), 553-555.

Azamov, A. A. and B. T. Samatov (2000). n-strategy . Press of the Nashional University of Uzbekistan, Tashkent.