Remarks on multifunction-based. dynamical systems. Bela Vizvari c. RRR , July, 2001

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1 R u t c o r Research R e p o r t Remarks on multifunction-based dynamical systems Gergely KOV ACS a Bela Vizvari c Marian MURESAN b RRR , July, 2001 RUTCOR Rutgers Center for Operations Research Rutgers University 640 Bartholomew Road Piscataway, New Jersey Telephone: Telefax: rrr@rutcor.rutgers.edu a Department of Operational Research, Eotvos Lorand University, Kecskemeti u , 1053 Budapest, HUNGARY, gkovacs@cs.elte.hu b Faculty of Mathematics and Computer Science, Babes-Bolyai University, M. Kogalniceanu 1, 3400 Cluj-Napoca, ROM ^ANIA, mmarian@math.ubbcluj.ro c Department of Operational Research, Eotvos Lorand University, Kecskemeti u , 1053 Budapest, HUNGARY, vizvari@cs.elte.hu

2 Rutcor Research Report RRR , July, 2001 Remarks on multifunction-based dynamical systems Gergely KOV ACS Marian MURESAN Bela Vizvari Abstract. If a dynamical system is dened by a multifunction, then there is no guarantee that a trajectory will reach an a priori given point. In many applications it is important to know whether or not the trajectories remain in a bounded region. In this paper a necessary and sucient condition is given so that no trajectory starting from a xed point leave a bounded set. Acknowledgements: The second and third authors express their gratitude to Hungarian Academy of Sciences (Grant No. AKP ,1) for supporting this research.

3 RRR Page 1 1 Introduction Let G : IR n! P(IR n ) (:= 2 IRn n ;) be a set-valued function. A discrete dynamical system is based on it if the system is dened by the relation x k+1 2 G(x k ); k = 0; 1; 2; :::; (1) and x 0 is a xed vector. Here we assume that the realization of x k+1 can be any element of G(x k ). If our knowledge on the behaviour of a dynamical system is restricted, then one way to model it is relation (1). An alternative possibility is to use stochastic variables. At the same time there are cases when we do not have enough information on the distribution or the distribution if it exists at all, is changing constantly. A typical situation of the second type is the cultivation of plants. It is easy to show by using moving averages of the yields that the technology is always improving. It is not possible to describe the joint eect of the changing technology and the random weather by stochastic variables. On stock markets the values of stocks have every day new distributions but the parameters are not known. Assume that system (1) has a xed point, i.e. there is a point y 2 IR n such that y 2 G(y): In economics the xed points are very important, as in mathematical models of economy the equilibrium used to be described by a xed point. Both Arrow and Debreu obtained the Nobel Prize for proving the existence of xed points. Debreu's model is a special case of (1). If the economy is in a xed point, i.e. for some k the equation x k = y holds, then unless G(y) = fyg, it is not necessary that the economy remains in the same state, i.e. the relation x k+1 = x k = y holds. As in the practice the equation x k = y; k = k 0 ; k 0 + 1; : : : never holds, where k 0 is a xed positive integer, it is important to know how far can move a trajectory from a particular xed point, especially whether or not the trajectories belong to a bounded set. Section 2 provides us with a necessary and sucient condition of boundedness. By some examples the sharpness of the bound is analyzed in section 3. In section 4 an example is provided for the stabilization of a dynamical system based on the set-valued function if the system has a control parameter. Some basic notions on set-valued analysis may be found, e.g., in [1], [3]. 2 Trajectories starting from a xed point The conditions (i) G(x) is a nonempty compact and convex set for every x 2 IR n ;

4 Page 2 RRR (ii) G(x) is continuous with respect to the Hausdor metric on every x 2 IR n ; (iii) dim G(x) = n for every x 2 IR n ; will be called the regularity conditions of the multifunction G: The assumptions of compactness and convexity of the values to a set-valued function G are commonly used in mathematical economics. The well-known Hausdor metric is called sometimes Hausdor-Pompeiu metric (see [2], [5, p. 111]). Paper [4] contains the following result Theorem 1 If the regularity conditions are satised and there is a point x 0 such that x 0 2 int G(x 0 ); then there exists a neighborhood V (x 0 ) of x 0 such that each x 2 V (x 0 ) is a xed point too, i.e. x 2 G(x). Denition 1 Assume that the relation x 0 2 int G(x 0 ) holds and V (x 0 ) is the neighborhood of x 0 supplied by the theorem 1. Let and S 1 := [ x2s0 S 0 := V (x 0 ); G(x) = G(S 0 ) S 0 i.e., the set of the points that are reachable [ from the set S 0 by exactly one step. Consider S k := G(x); k = 1; :::; and x2sk?1 S := i.e., the set of the points that are reachable from the set S 0 by any step. 1[ 0 We remark that S 0 S 1 and this implies that S k S k+1 : Thus we may write that S k 8n 2 IN S := Let's denote :=j S 0 j= sup x2s0 jxj: If there is a point x 2 S such that j G(x) j= 1; then S is unbounded. Similarly, if there is a sequence (x n ) n ; x n 2 S; such that sup n 1[ k=n j G(x n ) j= 1; then j S j= 1: Thus, if the boundedness of S is investigated, then these cases must be excluded. Therefore we assume that the following conditions hold S k :

5 RRR Page 3 (a) the set S 0 is bounded; (b) there exists a bounded set H such that for every point x 2 S there exists a point y x so that G(x) y x + H. Take :=j H j. Theorem 2 Suppose that the regularity conditions and the assumptions (a) and (b) are satised. Then the following two statements are equivalent (1) the set S is bounded; (2) there exists a xed 0 < < 1 so that j y x j j x j for any x 2 S: Proof. (1)=)(2). There holds G(S) = G( Thus if we choose H := S; then 1[ 1 S k ) = 1[ 1 G(S k ) = 1[ 2 S k = S: i.e. y x = 0 proves the statement. (2)=)(1). First we show that j S 1 j +. j S 1 j = j [ x2s0 G(x) j j [ x2s0 G(x) 0 + S (y x + H) j sup j y x j + j H j + (2) x2s0 Now we show that 1? : Since S 1 S 0 it follows that j S 1 jj S 0 j and + : Thus our claim is proved. For all positive integer n j S n j n + 1? n 1? : We prove this statement by induction. If n = 1; then by (2) it follows that j S 1 j 1 +. Assume that the statement is true for n = k? 1 and we show that it is true for n = k; too. Then [! j S n j = j G(x) j n?1 + 1? n?1 + = n + 1? n 1? 1? : x2sn?1

6 Page 4 RRR Let us remark that the sequence n + 1? n 1? is increasing. Finally we prove that the set S is bounded. From the denition and the above results we have j S j sup n! n + 1? n = lim 1? n! n2in! n + 1? n = 1? 1? ; hence the set S is bounded. Q.E.D. Remark. Let us emphasize that the proof of theorem 2 is based drastically on the existence and the properties of the set S 0 : At the same time this set S 0 is supplied us by theorem 1 which is based on the regularity conditions. A question arises: is it necessary condition (ii)? The answer is no. Theorem 3 Suppose that we have a multifunction G : IR n! P(IR n ) satisfying the regularity conditions except the Hausdor continuity. Instead of (ii) we assume that there exists a continuity point x 0 so that x 0 2 int G(x 0 ) and (ii) 0 9 an open neighborhood N of x 0 such that there is no discontinuity point of G in N. Then there exists a neighborhood V (x 0 ) of x 0 such that x 2 G(x) for all x 2 V (x 0 ): Proof. If G has no point of discontinuity, then theorem 1 provides the conclusion. Suppose we have discontinuity points. By hypothesis x 0 is a point of continuity and there exists a neighborhood W := x 0 + B(0; ") of x 0 such that it does not contain any point of discontinuity. Consider the following closed neighborhood of x 0 U := x 0 + B(0; "=2): Then for each point y 2 IR n n U there is precisely one point z y from the boundary of U such that the points x 0 ; z y and y lie on the same straight line. Let us dene a new multifunction F by ( G(y) if y 2 U; F (y) = G(z y ) if y =2 U: Then multifunction F satises all the hypotheses of theorem 1. Hence there exists a neighborhood N of x 0 such that x 2 F (x); for all x 2 N: Then the neighborhood V (x 0 ) := N \ U of x 0 satises the theorem. Q.E.D. Remark. The conclusion of the theorem 3 is false if x 0 coincides with a point of discontinuity. Consider for x 2 IR The only xed point is x = 1. G(x) = 8 >< >: (2; 3) x < 1; (1=2; 3=2) x = 1; (0; 1=2) x > 1:

7 RRR Page 5 Theorem 4 Suppose that the conditions (i); (ii) 0 ; and (iii) and the assumptions (a) and (b) are satised. Then the following two statements are equivalent (1) the set S is bounded; (2) there exists a xed 0 < < 1 so that j y x j j x j for any x 2 S: Proof. It is similar to the proof of theorem 2 with the only change that instead of theorem 1 we use theorem 3. Q.E.D. 3 Examples 1. For a > 0 and x 2 IR let be G(x) := f(x) + B[0; ] where B[0; ] denotes the [?; ] interval. Consider ( 0 x; j x j a f(x) = 1 x; j x j> a with 1 > 0 > 1 > 0. Suppose j S 0 j< a. If 1? 0 a; then j S j= 1? 0 : If 1? 0 > a; then there exists an index k such that j S k?1 j< a and j S k j> a. If j S k+1 j>j S k+2 j then j S j=j S k+1 j< k 0? k : 1? 0 The right-hand side of the above relation is less then =(1? 0 ). If j S k+1 j<j S k+2 j; then j S j= =(1? 1 ) since in this case j S k+l+2 j< k+2 0 l 1 + 1? k+2 0 l 1 + 1? l 1! : 1? 0 1? 1 1? 1 We remark that =(1? 1 ) < =(1? 0 ). 2. Let us note that if = 1; theorem 2 fails. Let G(x) = x + B[0; r]; where r > 0: Then j S j= 1:

8 Page 6 RRR An example for control In many applications it is important to stabilize the system, i.e. to keep the trajectory in a certain region. E.g. if there are many small companies in a market, then high uctuation in the prices might be very disadvantageous because a lot of small rms cannot survive the low-price periods. Thus the government aim should be to keep the prices around a realistic value if it is possible. In other cases it might be necessary that the trajectories avoid certain domains. In this section it is supposed that there is a parameter which can be chosen, i.e. the system can be controlled in this way. In the example the set of the possible values of the parameter is such that both aforementioned problems can be solved. Let G(x) = px + B(0; ); x 2 IR; (3) where p 2 [p 1 ; p 2 ] and 0 < p 1 < 1 < p 2. We can choose the value of the parameter p from this interval in every step. Theorem 5 For every x 0 2 IR we can choose the parameter p such that lim n!1 d x n; B 0;!! = 0: (4) Proof. Let p = p 1. If x n > ; (5) then x n > p 1 x n + x n+1 ) x n > x n+1 : (6) Hence the following sequence of inequalities is obtained by induction: j x 1 j j x 0 j p 1 + < j x 0 j; j x n j j x 0 j p n 1 + p n?1 1 + ::: + = p n 1 j x 0 j + 1? pn 1 : Here the rst term converges to 0 and the second one converges to if n goes to +1. :::

9 RRR Page 7 If then x n ; (7) x n+1 p 1 x n + : (8) Q.E.D. Theorem 6 If x 0 =2 B 0;! p 2? 1 (9) and ( ) A > max? ; p 2? 1 + ; (10) then we can choose the parameter p such that lim d(x n; B(A; )) = 0: (11) n!1 Proof. If x n > (12) and p = p 1 ; then x n+1 < x n. If x n > p 2? 1 ; (13) then x n+1 > x n by p = p 2, because x n > p 2? 1 ) < x n(p 2? 1) ) x n < p 2 x n? x n+1 : (14) If x n > A and p 1 x n > A; then let p = p 1. If x n > A and p 1 x n < A then let p = A=x n. If x n < A and p 1 x n < A; then let p = p 2. If x n < A and p 1 x n > A; then let p = A=x n : From the estimation of the absolute value of x n similar to the proof of the previous theorem it is easy to see that this strategy of choice of p is correct. Q.E.D.

10 Page 8 RRR References [1] J. P. Aubin, H. Frankowska, Set-Valued Analysis, Birkhauser, Basel, [2] Y. G. Borisovich, B. D. Gel'man, A. D. Myshkis, B. B. Obukhovski, Multivalued functions, Itogi nauki i tekhniki, 19(1982), (Russian). [3] M. Muresan, An Introduction to Set-Valued Analysis, Cluj University Press, Cluj- Napoca, [4] M. Muresan, B. Vizvari, On the stabilization of economic processes, Mathematica (Cluj) (1)42 (2000). [5] T. R. Rockafellar, R. J. B. Wets, Variational Analysis, GMW, Springer, Berlin, 1998.