A viability result for second-order differential inclusions

Transcription

1 Electronic Journal of Differential Equations Vol. 00(00) No ISSN: URL: htt://ejde.math.swt.edu or htt://ejde.math.unt.edu ft ejde.math.swt.edu (login: ft) A viability result for second-order differential inclusions Vasile Luulescu Abstract We rove a viability result for the second-order differential inclusion F (x x ) (x(0) x (0)) = (x 0 y 0) Q := K Ω where K is a closed and Ω is an oen subsets of R m and is an uer semicontinuous set-valued ma with comact values such that F (x y) V (y) for some convex roer lower semicontinuous function V. 1 Introduction Bressan Cellina and Colombo [6] roved the existence of local solutions to the Cauchy roblem x F (x) x(0) = ξ K where F is an uer semicontinuous cyclically monotone and comact valued multifunction. While Rossi [15] roved a viability result for this roblem. On the other hand for the second order differential inclusion F (x x ) x(0) = x 0 x (0) = y 0 existence results were obtained by many authors [ ]). In [1] existence results are roven for the case when F (..) is an uer semicontinuous set-valued ma with comact values such that F (x y) V (y) for some convex roer lower semicontinuous function V. The aim of this aer is to rove a viability result for the second-order differential inclusion F (x x ) (x(0) x (0)) = (x 0 y 0 ) Q := K Ω where K is a closed and Ω is an oen subsets of R m and F : Q R m Rm is an uer semicontinuous set-valued ma with comact values such that F (x y) V (y) for some convex roer lower semicontinuous function V. Mathematics Subject Classifications: 34G0 47H0. Key words: second-order contingent set subdifferential viable solution. c 00 Southwest Texas State University. Submitted December Revised March Published August

2 A viability result EJDE 00/76 Preliminaries and statement of main result Let R m be the m-dimensional Euclidean sace with scalar roduct.. and norm.. For x R m and ε > 0 let B ε (x) = {y R m : x y < ε} be the oen ball centered at x with radius ε and let B ε (x) be its closure. Denote by B the oen unit ball B = {x R m : x < 1}. For x R m and for a closed subsets A R m we denote by d(x A) the distance from x to A given by d(x A) = inf{ x y : y A}. Let V : R m R be a roer lower semicontinuous convex function. The multifunction V : R m Rm defined by V (x) = {ξ R m : V (y) V (x) ξ y x y R m } is called subdifferential (in the sense of convex analysis) of the function V. We say that a multifunction F : R m Rm is uer semicontinuous if for every x R m and every ε > 0 there exists δ > 0 such that F (y) F (x) + B ε (0) y B δ (x). This definition of the uer semicontinuous multifunction is less restrictive than the usual (see Definition in [3] or Definition 1.1 in [11]). Actually such a roerty is called (ε δ)-uer semicontinuity (see Definition 1. in [11]) and it is only equivalent to the uer semicontinuity for comact-valued multifunctions (see Proosition 1.1 in [11]). For K R m and x K denote by T K (x) the Bouligand s contingent cone of K at x defined by T K (x) = { v R m : lim inf h 0+ d(x + hv K) h = 0 }. For K R m and (x y) K R m we denote by T () K (x y) the second-order contingent set of K at (x y) introduced by Ben-Tal [5] and defined by T () K (x y) = { v R m : lim inf h 0+ d(x + hy + h v K) h = 0 }. / We remar that if T () K (x y) is non-emty then necessarily y T K(x). Moreover (see [4] [10] [13]) if F is uer semicontinuous with comact convex values and if x : [0 T ] R m is a solution of the Cauchy roblem F (x x ) x(0) = x 0 x (0) = y 0

3 EJDE 00/76 Vasile Luulescu 3 such that x(t) K t [0 T ] then (x(t) x (t)) grah(t K ) t [0 T ) hence in articular (x 0 y 0 ) grah(t K ). For a multifunction F : Q := K Ω R m Rm and for any (x 0 y 0 ) grah(t K ) we consider the Cauchy roblem under the following assumtions: F (x x ) (x(0) x (0)) = (x 0 y 0 ) Q (.1) (H1) K is a closed and Ω and oen subset of R m such that Q := K Ω grah(t K ) (H) F is an uer semicontinuous comact valued multifunction such that F (x y) T () K (x y) (x y) Q; (H3) There exists a roer convex and lower semicontinuous function V : R m R such that F (x y) V (y) (x y) Q. Remar. A convex function V : R m R is continuous in the whole sace R m (Corollary in [14]) and almost everywhere differentiable (Theorem 5.5 in [14]). Therefore (H3) strongly restricts the multivaluedness of F. Definition. By viable solution of the roblem (.1) we mean any absolutely continuous function x : [0 T ] R m with absolutely continuous derivative x such that x(0) = x 0 x(0) = y 0 Our main result is the following: (t) F (x(t) x (t)) a.e. on [0 T ] (x(t) x (t)) Q t [0 T ]. Theorem.1 If F : Q R m Rm and V : R m R satisfy assumtions (H1) (H3) then then for every (x 0 y 0 ) Q there exist T > 0 and x : [0 T ] R m a viable solution of the roblem (.1). 3 Proof of the main result We start this section with the following technical result which will be used to rove the main result. Lemma 3.1 Assume Q = K Ω R m satisfies (H1) F : Q Rm satisfies (H) Q 0 Q is a comact subset and (x 0 y 0 ) Q 0. Then for every N there exist h 0 (0 1 ] and u0 Rm such that x 0 + h 0 y 0 + (h0 ) u 0 K (x 0 y 0 u 0 ) grah(f ) + 1 (B B B).

4 4 A viability result EJDE 00/76 Proof. Let (x y) Q be fixed. Since by (H) F (x y) T () K (x y) there exists v = v (xy) F (x y) such that lim inf h 0 + d(x + hy + h v K) h = 0. / Hence for every N there exists h = h (x y) (0 1 ] such that d(x + h y + h v K) < h 4. (3.1) By the continuity of the ma (a b) d(a + h b + h v K) it follows that N(x y) = { (a b) : d(a + h b + h v K) < h } 4 is an oen set and by (3.1) it contains (x y). Then there exists r := r(x y) (0 1 ) such that B r(x y) N(x y). Since Q 0 is comact there exists a finite subset {(x j y j ) Q : 1 j m} such that m Q 0 B rj (x j y j ). We set j=1 h 0 () := min{h (x j y j ) : j {1... m}}. Since (x 0 y 0 ) Q 0 there exists j 0 {1... m} such that (x 0 y 0 ) B rj0 (x j0 y j0 ) N(x j0 y j0 ). (3.) Denote by h 0 := h (x j0 y j0 ) and remar that by (3.1) and (3.) one has h 0 [h 0() 1 ] and there exists z 0 K such that we have that hence Let Then d(x 0 + h 0 y 0 + (h0 ) v 0 z 0 ) (h 0 ) / d(x 0 + h y 0 + (h 0 ) v 0 K) (h ) / < 1 z 0 x 0 h 0 y 0 (h 0 ) / u 0 := z 0 x 0 h 0 y 0 (h 0. ) / x 0 + h 0 y 0 + (h0 ) u 0 K. By (3.3) and (3.) we get successively: u v 0 < 1 d((x 0 y 0 ) (x j0 y j0 )) r j0 < 1 v 0 < 1. (3.3) hence (x 0 y 0 u 0 ) grah(f ) + 1 (B B B).

5 EJDE 00/76 Vasile Luulescu 5 Proof of Theorem.1 Let (x 0 y 0 ) Q grah(t K ). Since Ω R m is an oen subset there exist r > 0 such that B r (y 0 ) Ω. We set Q 0 := B r (x 0 y 0 ) (K B r (y 0 )). Since Q 0 is a comact set by the uer semicontinuity of F and Proosition in [3] we have that F (Q 0 ) := (xy) Q 0 F (x y) is a comact set hence there exists M > 0 such that: Let su{ v : v F (x y) (x y) Q 0 } M. T = min { r (M + 1) r M + 1 r }. (3.4) ( y 0 + 1) We shall rove the existence of a viable solution of the roblem (.1) defined on the interval [0 T ]. Since (x 0 y 0 ) Q 0 then by Lemma 3.1 there exist h 0 [h 0() 1 ] and u0 Rm such that x 0 + h 0 y (h0 ) u 0 K and (x 0 y 0 u 0 ) grah(f ) + 1 (B B B). Define x 1 :=x 0 + h 0 y (h0 ) u 0 ; y 1 :=y 0 + h 0 u 0. (3.5) We remar that if h 0 < T then x 1 x 0 h 0 y (h0 ) u 0 < h 0 y (h0 ) (M + 1) y 1 y 0 = h 0 u 0 < h 0 (M + 1) and by the choice of T we get x 1 x 0 < r y 1 y 0 < r. Therefore (x 1 y1 ) Q 0 and by Lemma 3.1 there exist h 1 u 1 Rm such that [h 0() 1 ] and x 1 + h 1 y (h1 ) u 1 K (x 1 y 1 u 1 ) grah(f ) + 1 (B B B). We claim that for each N there exist m() N and h x y u such that for every {... m() 1} we have that:

6 6 A viability result EJDE 00/76 (i) m() 1 j=0 h j T < m() j=0 hj (ii) (iii) (x y ) Q 0 x =x0 + ( 1 h i ) y (h i ) u i + 1 y =y0 + h i u i ; (iv) (x y u ) grah(f ) + 1 (B B B). 1 j=i+1 h i h j ui If h 0 + h1 T then we set m() = 1. Assume that h0 + h1 < T and define x :=x 1 + h 1 y (h1 ) u 1 y :=y 1 + h 1 u 1. Then by (3.5) and (3.6) we have that (3.6) x :=x 0 + (h 0 + h 1 )y (h0 ) u 1 + (h 1 ) u 1 + h 0 h 1 u 0 y :=y 0 + h 0 u 0 + h 1 u 1 and since h 0 + h1 T and it follows x x 0 (h 0 + h 1 ) y (h0 ) u (h1 ) u 1 + h 0 h 1 u 0 <(h 0 + h 1 ) y (h0 + h 1 ) (M + 1) x x 0 < r y y 0 < r hence (x y ) Q 0. Assume that h q xq yq uq have been constructed for q satisfying (ii) (iv) and that we construct h +1 x +1 y +1 u +1 satisfying such roerties. Since (x y ) Q 0 by lemma there exist h [h 0() 1 ] and u Rm such that x + h y + 1 (h ) u K (x y u ) grah(f ) + 1 (B B B). If h 0 +h1 + +h T then we set m() =. Assume that h0 +h1 + +h < T and define x +1 :=x + h y + 1 (h ) u (3.7) y +1 :=y + h u.

7 EJDE 00/76 Vasile Luulescu 7 Then by the above equations and (ii) we obtain that and x +1 = x + h y + 1 (h ) u = x0 + ( 1 h i ) y (h i ) u i Therefore and x (h i ) u i + = x 0 + ( y +1 h i ) y j=i+1 1 h i h j ui + h h i u i + 1 (h ) u 1 (h i ) u i + j=i+1 := y + h u = 1 y0 + h i u i + h u = y0 + x 0 ( y +1 ( h i ) y0 + 1 h i x 0 1 (h i ) u i + ) y0 + M + 1 ( h i ) h i h j ui h i u i. j=i+1 h i u i 0 (M + 1) ( h i ). Since hi < T one obtains that x +1 x 0 < r y +1 x 0 < r h i h j ui hence (x +1 y +1 ) Q 0. We remar that this iterative rocess is finite because h [h 0() 1 ] imlies the existence of an integer m() such that h 0 + h h m() 1 T < h 0 + h h m() 1 + h m(). By (iv) for every N and every { m()} there exists (a b v ) grah(f ) such that x a < 1 y b < 1 u v < 1 ; (3.8) hence x x x 0 + x x x 0 y y y 0 + y y y 0 (3.9) u u v + v 1 + M 1 + M.

8 8 A viability result EJDE 00/76 Let us set t = h0 + h h 1 t 0 = 0. We remar that for all N and all {1... m()} we have t t 1 < 1 and t m() 1 T < t m(). (3.10) For each 1 and for {1... m()} we set I = [t 1 t ] and for t I we define x (t) = x 1 + (t )y (t t 1 ) u 1. (3.11) Then x (t) = y 1 hence by (3.9) for all t [0 T ] we obtain + (t )u 1 t I (t) = u 1 t I (3.1) (t) u 1 < M + 1 x (t) y 1 + (t t ) u < y 0 + M + x (t) x 1 + (t x 0 + y 0 + M + 3. ) y (t t 1 ) u 1 (3.13) Moreover for all t [0 T ] we have that (x (t) x (t) (t)) (x y u ) + y 0 + M + B M + 1 B {0}; hence by (iv) we have (x (t) x (t) (t)) grah(f ) + ε()(b B {0}) (3.14) where ε() 0 when. Then by (3.11) (3.1) and (3.13) we obtain that ( ) is bounded in L ([0 T ] R m ) (x ) and (x ) are bounded in C([0 T ] R m ) and equi-lischitzian hence by Theorem in [3] there exist a subsequence (again denoted by (x ) ) and an absolutely continuous function x : [0 T ] R m such that (a) (x ) converge uniformly to x (b) (x ) converge uniformly to x (c) ( ) converge wealy in L ([0 T ] R m ) to. By (H3) and Theorem in [3] we get that (t) co F (x(t) x (t)) V (x (t)) a.e. on [0 T ]

9 EJDE 00/76 Vasile Luulescu 9 where co stands for the closed convex hull; hence by Lemma 3.3 in [7] we obtain that d dt V (x (t)) = (t) a.e. on [0 T ]; hence On the other hand since B such that z 1 1 u 1 F (x 1 V (x (T )) V (x (0)) = a 1 (t) = u 1 t I y 1 b 1 0 (t) dt. (3.15) by (iv) there exist a 1 b 1 ) V (y 1 b 1 ) N (3.16) and so the roerties of the subdifferential of a convex function imly that for every < m() and for every N we have hence V (x (t ) b ) V (x ( u 1 = u 1 ) b 1 ) x (t ) x ( = (t) dt z 1 (t)dt + u 1 V (x (t ) b ) V (x ( ) b 1 (t) dt z 1 ) ) + b 1 b = b 1 b = (t)dt + u 1 (t)dt + u 1 Analogously if T I m() then by (3.10) we have b 1 b ; b 1 b. (3.17) V (x (T )) V (x (t m() 1 ) b m() 1 u m() 1 z m() 1 t m() 1 ) = (t) dt z m() 1 t m() 1 + u m() 1 z m() 1 b m() 1. (t)dt + b m() 1 t m() 1 (t)dt (3.18) By adding the m() 1 inequalities from (3.17) and the inequality from (3.18) we get V (x (T )) V (y 0 b 0 ) 0 (t) dt + α() (3.19)

10 10 A viability result EJDE 00/76 where Since m() 1 α() = =1 z m() 1 m() 1 α() =1 + z m() m() 1 =1 z 1 t m() 1 m() 1 (t)dt + =1 (t)dt + u m() 1 m() 1 z 1 (t)dt + t m() 1 z 1 + z m() t m() 1 (M + )(3m() 1) =1 (t)dt + u m() 1 m() 1 (t)dt + =1 (t)dt + u m() 1 u 1 u 1 b 1 b z m() 1 b m() 1. b 1 b + z m() 1 b m() 1 u 1 b 1 b z m() 1 b m() 1 it following that α() 0 when ; hence by (3.19) we assing to the limit for we obtain V (x (T )) V (y 0 ) lim su Therefore by (3.15) and (3.0) 0 (t) dt lim su 0 0 (t) dt. (3.0) (t) dt and since ( ) converges wealy in L ([0 T ] R m ) to by alying Proosition III.30 in [8] we obtain that ( ) converge strongly in L ([0 T ] R m ) to hence a subsequence again denoted by ( ) converge oinwise a.e. to. Since by (3.14) lim d((x (t) x (t) (t)) grah(f )) = 0 and since by (H) the grah of F is closed ([3] Proosition 1.1.) we have that (t) F (x(t) x (t)) a.e. on [0 T ]. It remains to rove that (x(t) x (t)) Q t [0 T ]. Indeed by (3.11) (3.1) and (3.13) we have that x (t) x < y 0 + M + x (t) y < M + 1

11 EJDE 00/76 Vasile Luulescu 11 hence lim d((x (t) x (t)) (x y )) = 0. Since (x y ) Q 0 N by (a) and (b) we have that On the other hand d((x(t) x (t)) Q 0 ) lim d((x(t) x (t)) (x (t) x (t))) = 0. d((x(t) x (t)) (x (t) x (t))) + d((x (t) x (t)) (x y )) + d((x y ) Q 0); (3.1) hence by assing to the limit we obtain that d((x(t) x (t)) Q 0 ) = 0 t [0 T ]. Since Q 0 is closed we obtain that (x(t) x (t)) Q 0 for all t [0 T ] which comletes the roof. Acnowledgements. This research was done while the author was visiting the Research unit Mathematics and Alications (Control Theory Grou) of the Deartment of Mathematics of Aveiro University where the author was suorted by a ost-doctoral fellowshi. The author wishes to than Vasile Staicu for having introduced him to the subject of the aer; The author also wants to thans the referee for his/her remars that imroved the results and their resentation. References [1] B. Aghezaaf and S. Sajid On the second order contingent set and differential inclusions. Journal of Convex Analysis 7: [] J. P. Aubin and H. Franowsa Set-Valued Analysis Biräuser [3] J. P. Aubin and A. Cellina Differential Inclusions. Sriger-Verlag Berlin [4] A. Auslender and J. Mechler Second order viability roblems for differential inclusions. J. Math. Anal. Al. 181: [5] A. Ben-Tal Second order theory of extremum roblems in extremal methods and system analysis. In A. V. Fiaco and Kortane editors Lecture Notes in Economics and Mathematical Systems Vol. 179 volume 179. Sringer Verlag New Yor [6] A. Bressan A. Cellina and G. Colombo. Uer semicontinuous differential inclusions without convexity. Proc. Amer. Math. Soc. 106:

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