Applied Mathematics E-Notes, 13(2013), 109-119 c Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/ ISSN 1607-2510 On Mild Solutions Of Nonlocal Semilinear Impulsive Functional Integro-Di¤erential Equations Rupali Shikharchand Jainy, Machindra Baburao Dhaknez Received 4 June 2013 Abstract In the present paper, we investigate the existence, uniqueness and continuous dependence on initial data of mild solutions of …rst order nonlocal semilinear functional impulsive integro-di¤erential equations of more general type with …nite delay in Banach spaces. Our analysis is based on semigroup theory and Banach contraction theorem. 1 Introduction Impulsive equations arise in many di¤erent real processes and phenomena which appears in physics, population dynamics, medicine, economics etc. The study of impulsive functional di¤erential equations is linked to their utility in stimulating processes and phenomena subject to short time perturbations during their evolution. The perturbations are performed discretely and their duration is negligible in comparison with the total duration of the processes and phenomena. That is the reason for the perturbations to be considered as taking place instantaneously in the form of impulses. Impulsive di¤erential equations in recent years have been the object of investigation with increasing interest. For more information see the monographs, Lakshmikantham [14], Samoilenko and Perestyuk [14], the research papers [5, 16] and the references cited therein. Also, the problems of existence, uniqueness and other qualitative properties of solutions for semilinear di¤erential equations in Banach spaces has been studied extensively in the literature for last many years, see [1, 3, 4, 6, 7, 9–13, 15, 17]. On the other hand, di¤erential equations with nonlocal condition have been studied by many researchers as they are more precise to describe natural phenomena than di¤erential equations with classical initial condition, for example see [1, 4, 6, 9, 10, 12] and the references cited therein. Mathematics Subject Classi…cations: 45J05, 45N05, 47H10 ,47B38. of Mathematical Sciences, Swami Ramanand Teerth Marathwada University, Nanded431606, Maharashtra, India z Department of Mathematics, Dr. Babasaheb Ambedkar Marathwada University, Aurangabad431004, India y School 109 110 On Mild Solutions of Impulsive Functional Equations Recently, Li et. al [17] studied fractional integro-di¤erential equations of the type: q D x(t) = Ax(t) + f (t; x(t) + Z t k(t; s)h(t; s; x(s)ds); 0 x(0) = g(x) + x0 ; using measure of noncompactness and …xed point theorem of condensing maps. Also, Radhakrishnan and Bhalchandran [2] investigated controllability results for the impulsive integro-di¤erential evolution system with nonlocal condition of the form: Z t 0 x (t) = A(t)x(t) + Bu(t) + f t; x(t); a(t; s; x(s)ds ; t 2 J, t 6= ti ; 0 x(0) = g(x) + x0 ; xjt=ti = Ii x(ti ); i = 1; 2; ::; m: Ahmed et. al [8] obtained existence results of mild solutions of integro-di¤erential equations: Z t 0 u (t) = Au(t)) + f t; u(t); H(t; s)u(s)ds ; t 2 [0; K]; 0 u(0) = g(u) + u0 ; u(ti ) = Ii u(ti ); i = 1; 2; ::; m; using measure of noncompactness and new …xed point theorem. In the present paper we consider semilinear functional impulsive integro-di¤erential equation of …rst order of the type: Z t x0 (t) = Ax(t) + f t; xt ; k(t; s)h(s; xs )ds ; t 2 (0; T ]; t 6= k ; k = 1; 2; ::; m; (1) 0 x(t) + (g(xt1 ; :::; xtp ))(t) = (t); x( k) = Ik x( k ); r k = 1; 2; :::; m; t 0; (2) (3) where 0 < t1 < t2 < ::: < tp T , p 2 N, A is the in…nitesimal generator of strongly continuous semigroup of bounded linear operators fT (t)gt 0 and Ik (k = 1; 2; :::; m) are the linear operators acting in a Banach space X. The functions f; h; g; k and are given functions satisfying some assumptions. The impulsive moments k are such that 0 T , m 2 N and x( k ) = x( k + 0) x( k 0) 0 < 1 < 2 < ::: < m < m+1 where x( k + 0) and x( k 0) are, respectively, the right and the left limits of x at k . Equations of the form (1)-(3) or their special forms arise in some physical applications as a natural generalization of the classical initial value problems. Our aim of the present paper is to generalize and extend results reported in [5, 16]. The models in this paper are studied for impulse e¤ect. We study the existence, uniqueness and continuous dependence of mild solution of nonlocal IVP problem for an impulsive functional integro-di¤erential equation. The main tool used in our analysis is based on an application of the Banach contraction theorem and semigroup theory. R. S. Jain and M. B. Dhakne 111 As usual, in the theory of impulsive di¤erential equations at the points of discontinuity i of the solution t ! x(t), we assume that x( i ) x( i 0). It is clear that, in general the derivatives x0 ( i ) do not exist. On the other hand, from (1), there exist the limits x0 ( i 0). According to the above convention, we assume x0 ( i ) = x0 ( i 0). This paper is organized as follows. Section 2 presents the preliminaries and hypotheses. In Section 3, we prove existence and uniqueness of mild solution. In section 4, we prove continuous dependence of solutions on initial data. Finally, in section 5, we give application based on our result . 2 Preliminaries and Hypotheses Let X be a Banach space with the norm k k. Let C = C([ r; 0]; X); 0 < r < 1; be the Banach space of all continuous functions : [ r; 0] ! X endowed with supremum norm k kC = supfk (t)k : r t 0g and B denote the set fx : [ r; T ] ! X : x(t) is continuous at t 6= k , left continuous at t = k , and the right limit x( k + 0) exists for k = 1; 2; :::; mg. Clearly, B is a Banach space with the supremum norm kxkB = supfkx(t)k : t 2 [ r; T ]nf 1 ; 2 ; :::; m gg. For any x 2 B and t 2 [0; T ]nf 1 ; 2 ; :::; m g, we denote xt the element of C given by xt ( ) = x(t + ) for 2 [ r; 0] and is a given element of C. In this paper, we assume that, there exist positive constant K 1 such that kT (t)k K for every t 2 [0; T ]. Also assume k : [0; T ] [0; T ] ! R is continuous function and as the set [0; T ] [0; T ] is compact, there exists a constant L > 0 such that jk(t; s)j L for 0 s t T . DEFINITION 2.1. A function x 2 B satisfying the equations x(t) = T (t) (0) T (t)(g(xt1 ; :::; xtp ))(0) Z t Z s + T (t s)f s; xs ; k(s; )h( ; x )d 0 0 X + T (t k )Ik x( k ) 0< ds k <t for t 2 (0; T ], and x(t) + (g(xt1 ; :::; xtp ))(t) = (t) for r t 0; is said to be the mild solution of the initial value problem (1)-(3). REMARK. A mild solution of equations (1)-(3) satis…es (2) and (3). However, a mild solution may not be di¤erentiable at zero. The following inequality will be useful while proving our result. 112 On Mild Solutions of Impulsive Functional Equations LEMMA 2.2 ([15, p.12]). Let a nonnegative piecewise continuous function u(t) satis…es the inequality u(t) Z D+ t v(s)u(s)ds + t0 X t0 < i u( i ) i <t for t t0 where D 0, i 0, v(t) > 0 and i are the …rst kind discontinuity points of the function u(t). Then the following estimate holds for the function u(t), u(t) D Y t0 < (1 + i ) exp Z t v(s)ds : to i <t We list the following hypotheses for our convenience. (H1 ) Let f : [0; T ] C X ! X such that for every w 2 B, x 2 X and t 2 [0; T ], f ( ; wt ; x) 2 B and there exists a constant F > 0 such that kf (t; ; x) f (t; ; y)k F (k kC + kx yk); ; 2 C; x; y 2 X: (H2 ) Let h : [0; T ] C ! X such that for every w 2 B and t 2 [0; T ], h( ; wt ) 2 B and there exists a constant H > 0 such that kh(t; ) h(t; )k Hk kC ; (H3 ) Let g : C p ! C such that there exists a constant G k(g(xt1 ; xt2 ; :::; xtp ))(t) (g(yt1 ; yt2 ; :::; ytp ))(t)k ; 2 C: 0 satisfying Gkx ykB ; t 2 [ r; 0]: (H4 ) Let Ik : X ! X are functions such that there exists constants Lk satisfying kIk (v)k 3 Lk kvk; v 2 X; k = 1; 2; :::; m: Existence and Uniqueness We have the following THEOREM 3.1. Suppose that the hypotheses (H1 )–(H4 ) are satis…ed and where X = KG + KF [1 + LHT ]T + K Lk : 0< k <t Then the initial-value problem (1)-(3) has a unique mild solution x on [ r; T ]. PROOF. We introduce an operator F on a Banach space B as follows <1 R. S. Jain and M. B. Dhakne 113 8 > < (t) (g(xt1 ; :::; xtp ))(t) if r R t 0; Rs t (Fx)(t) = T (t)[ (0) g(xt1 ; :::; xtp )(0)] + 0 T (t s)f s; xs ; 0 k(s; )h( ; s )d > : P + 0< k <t T (t k )Ik x( k ) if t 2 (0; T ]: ds It is easy to see that F : B ! B. Now we will show that F is a contraction on B. Let x; y 2 B. Then k(Fx)(t) (Fy)(t)k = kg(xt1 ; :::; xtp )(t) g(yt1 ; :::; ytp )(t)k Gkx ykB (4) for t 2 [ r; 0] and k(Fx)(t) (Fy)(t)k T (t) g(xt1 ; :::; xtp )(0) g(yt1 ; :::; ytp )(0) Z s t + T (t s)f s; xs ; k(s; )h( ; x )d 0 0 Z s f s; ys ; k(s; )h( ; y )d ds 0 X + T (t Ik y( k )]k k )[Ik x( k ) Z 0< k <t kT (t)kkg(xt1 ; :::; xtp )(0) g(yt1 ; :::; ytp )(0)k Z t Z s + kT (t s)k f s; xs ; k(s; )h( ; x )d 0 0 Z s ds f s; ys ; k(s; )h( ; y )d 0 X + kT (t Ik y( k )]k k )kk[Ik x( k ) 0< k <t KGkx for t 2 (0; T ] where Z t J1 = kT (t 0 f K s; ys ; Z s)kk f Z s Z [kxs [kx Z Z k(s; )h( ; x )d ds s 0 ys kC + L t 0 KF ys kC + t 0 KF s k(s; )h( ; y )d F [kxs Z (5) 0 0 KF Z s; xs ; 0 t ykB + J1 + J2 Z jk(s; )jkh( ; x ) h( ; y )kd ]ds s Hkx 0 ykB + LHT kx y kC d ]ds ykB ]ds t [1 + LHT ]kx ykB ds 0 KF [1 + LHT ]T kx ykB (6) 114 On Mild Solutions of Impulsive Functional Equations and J2 = X 0< k <t 0< k <t X kT (t k )kkIk x( k ) KkIk x( k) X K 0< k <t 0< k <t K X Lk kx( Lk kx k) Ik y( y( Ik y( k )k k )k k )k ykB : (7) Using (6)-(7), inequality (5) becomes k(Fx)(t) (Fy)(t)k KGkx ykB + KF [1 + LHT ]T + K X 0< Lk k <t ! kx ykB (8) for t 2 [0; T ]: As K 1, in view of inequality (4) and (8), we can say that inequality (8) holds good for t 2 [ r; T ]. Therefore, for t 2 [ r; T ], ! X k(Fx)(t) (Fy)(t)k KGkx ykB + KF [1 + LHT ]T + K Lk kx ykB 0< X KG + KF [1 + LHT ]T + K 0< k <t Lk ! k <t kx ykB ; which implies kFx where FykB kx ykB = KG + KF [1 + LHT ]T + K X 0< Lk : k <t Since < 1, the operator F satis…es all the assumptions of Banach contraction theorem and therefore F has unique …xed point in the space B and clearly it is the mild solution of nonlocal IVP problem (1)-(3) with impulse e¤ect. This completes the proof. 4 Continuous Dependence on Initial Data We have the following THEOREM 4.1. Suppose that hypotheses (H1 )-(H4 ) are satis…ed and < 1. Then for each 1 ; 2 2 C and for the corresponding mild solutions x1 and x2 of the problems Z t x0 (t) = Ax(t) + f t; xt ; k(t; s)h(s; xs )ds ; t 2 (0; T ]; (9) 0 x( k ) = Ik x( x(t) + g(xt1 ; :::; xtp )(t) = i (t); k ); k = 1; 2; :::; m; i = 1; 2; t 2 [ r; 0]; (10) (11) R. S. Jain and M. B. Dhakne 115 the following inequality holds Q K kx1 0< x2 kB (1 + KLk ) exp(KF T ) k <t Q 1 0< where k (1 + KLk ) exp(KF T ) 2 kC 1 (12) k <t = GK + KF LHT 2 : PROOF. Let 1 ; 2 2 B be arbitrary functions and let x1 and x2 be the mild solutions of the problem(9)-(11). Then we have h i x1 (t) x2 (t) = T (t)[ 1 (0) (0)] T (t) g(x ; :::; x )(0) g(x ; :::; x )(0) 1t1 1tp 2t1 2tp 2 Z t Z s + T (t s) f s; x1s ; k(s; )h(s; x1s )d 0 0 Z s f s; x2s ; k(s; )h(s; x2s )d ds 0 X + T (t Ik x2 ( k )) (13) k )(Ik x1 ( k ) 0< k <t for t 2 (0; T ] and x1 (t) x2 (t) = 1 (t) 2 (t) [g(x1t1 ; :::; x1tp )(t) g(x2t1 ; :::; x2tp )(t)] (14) for t 2 [ r; 0]. From (13) and using hypothesis (H1 )-(H4 ), we get kx1 (t) x2 (t)k Kk + HL 2 kC + GKkx1 1 Z kx1 x2 k C d 1 X 0< 0 0< k <t F kx1s Lk kx1 ( k) x2s kC x2 ( k )k Z t x2 k B + K F kx1s x2s kC ds 0 X x2 k B + K Lk kx1 ( k ) x2 ( k )k + F KLHT 2 kx1 +K ds + K t 2 kC + GKkx1 1 Kk Z X s 0 Kk x2 k B + K k <t 2 kC 0< + kx1 Lk kx1 ( k) k <t x2 kB + KF x2 ( Z 0 t kx1s x2s kC ds k )k: (15) Simultaneously, by (14) and hypothesis (H3 ),we get kx1 (t) x2 (t)k k 1 2 kC + Gkx1 x2 kB ; t 2 [ r; 0]: (16) 116 On Mild Solutions of Impulsive Functional Equations Since K 1, the inequalities (15) and(16) imply, for t 2 [ r; T ] Z t kx1 (t) x2 (t)k Kk 1 k + kx x k + KF kx1s 1 2 B 2 C 0 X +K Lk kx1 ( k ) x2 ( k )k; 0 < k t; 0< x2s kC ds t 2 [0; T ]: k <t (17) De…ne the function z : [ r; T ] ! R by z(t) = supfkx1 (s) x2 (s)k : r s tg; t 2 [0; T ]. Let t 2 [ r; t] be such that z(t) = kx1 (t ) x2 (t )k. If t 2 [0; t], then from inequality (17), we have z(t) = kx1 (t ) Kk +K k z(t) 2 kC 1 X 0< for 0 < x2 (t )k k <t + kx1 Lk kx1 ( k) x2 kB + KF x2 ( Z t 0 kx1s x2s kC ds k )k t and Kk 2 kC + kx1 1 x2 kB + KF Z t z(s)ds + K 0 X 0< Lk z( k) (18) k <t Now applying Lemma 2.2 to the inequality (18), we get Y z(t) (Kk 1 x2 kB ) (1 + KLk ) exp(KF T ): 2 kC + kx1 0< k <t Hence, we get kx1 x2 kB (Kk 1 2 kC + kx1 x2 kB ) Y 0< (1 + KLk ) exp(KF T ): k <t The inequality given by (12) is the immediate consequence of the above inequality. This completes the proof. 5 Application To illustrate the application of our result proved in section 3, consider the following semilinear partial functional di¤erential equation of the form Z t @ @2 w(u; t) = w(u; t) + H t; w(u; t r); k(t; s)P (s; w(s r))ds ; (19) @t @u2 0 w(0; t) = w( ; t) = 0; 0 w(u; ) + p X t T; w(u; ti + ) = (u; t); (20) (21) i=1 w(u; k) = Ik (w(u; k )); k = 1; 2; :::; m; (22) R. S. Jain and M. B. Dhakne 117 for 0 u ; 0 t T and r 0 where 0 < t1 t2 tp T , the functions H : [0; T ] R R ! R, P : [0; T ] R ! R and Ik : R ! R are continuous . We assume that the functions H, P and Ik satisfy the following conditions: For every t 2 [0; T ] and u; v; x; y 2 R, there exists positive constants l, p, ck and d such that jH(t; u; x) H(t; v; y)j jP (t; u) P (t; v)j jIk (x)j p X l(ju vj + jx p(ju vj); ck jxj ; k = 1; 2; :::; m; jw(u; ti + t)j d; Kd + Kl(1 + LpT )T + K X i=1 and yj); 0< ck < 1: k <t 00 Let us take X = L2 [0; ]. De…ne the operator A : X ! X by Az = z with domain 0 00 D(A) = fz 2 X : z; z are absolutely continuous, z 2 X and z(0) = z( ) = 0g. Then the operator A can be written as Az = 1 X n2 (z; zn )zn ; z 2 D(A) 1 X exp( n2 t)(z; zn )zn ; z 2 X: n=1 p where zn (u) = ( 2= ) sin nu; n = 1; 2; ::: is the orthogonal set of eigenvectors of A and A is the in…nitesimal generator of an analytic semigroup T (t); t 0 and is given by T (t)z = n=1 Now, the analytic semigroup T (t) being compact, there exists constant K such that jT (t)j De…ne the functions f : [0; T ] C K; for each t 2 [0; T ]: X ! X, h : [0; T ] C ! X, Ik : X ! X as follows f (t; ; x)(u) = H(t; ( r)u; x(u)); h(t; )(u) = P (t; ( r)u); for t 2 [0; T ]; ; 2 C, x 2 X and 0 u . 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