Memors on Derental Equatons and Mathematcal Physcs Volume 11, 1997, 67{88 Guram Kharatshvl and Tamaz Tadumadze THE PROBLEM OF OPTIMAL CONTROL FOR NONLINEAR SYSTEMS WITH VARIABLE STRUCTURE, DELAYS AND PIECEWISE CONTINUOUS PREHISTORY
Abstract. Necessary condtons both for the optmalty of controls and ntal functons are proved n the form of an ntegral maxmum prncple and condtons of transversalty for nonlnear systems wth a varable structure, delays and a pecewse contnuous ntal functon n the case where values of the ntal functon (prehstory) and of the trajectory at a non-xed ntal moment and at a moment of varaton of the structure do not, generally speakng, concde. 1991 Mathematcs Subject Classcaton. 49K25. Key words and Phrases. Maxmum prncple, transversalty condtons, delays, varable structure systems. rezume. aracrpv dagvanebul argumentan cvladstruqturan sstemsatvs uban-uban uckvet sacks PunqcT damtkcebula martvs da sacks Punqcs optmalurobs auclebel probeb ntegralur maqsmums prncpsa da transversalobs probebs saxt m SemTxvevaS, roca sacks Punqcs (Cnastors) da traeqtors mnsvnelobeb arapqsrebul sacks moments da struqturs cvllebs moments, sazogadod, ar emtxvevan ertmanets.
69 1. Introducton Necessary condtons of optmalty for the problem gven below (see 2) are derved from a necessary condton of crtcalty [1,2,3] the bass of whch forms the noton of quas-convex set ntroduced by R.V. Gamkreldze [4]. Varaton of the structure of a system means that the system at some beforehand unknown moment may go over from one law of movement to another. After varaton of the structure, the values of the ntal functon of the system depend on ts prevous state. Ths jons them nto a sngle system wth varable structure. In concluson, t should be noted that some partcular cases of the problem under consderaton have been studed n [5]. 2. Statement of the Problem. Necessary Condtons of Optmalty Let O R n, G R r, = 1; : : : ; m, be open sets, R n be an n- dmensonal Eucldean space, J = [a; b] be a nte nterval; let the functons f : J O 2 G2! Rn, = 1; : : : ; m, be contnuously derentable wth respect to (x ; z ) 2 O 2, = 1; : : : ; m, respectvely, let : R 1! R 1, : R 1! R 1, = 1; : : : ; m, be contnuously derentable functons satsfyng the condtons (t) t, _ (t) > 0, (t) < t, _ (t) > 0; let q k (t 1 ; : : : ; t m+1 ; x 10 ; x 1 ; : : : ; x m0 ; x m ), k = 0; : : : ; l, be scalar functons contnuously derentable n all arguments t 2 J, = 1; : : : ; m + 1, (x 0 ; x ) 2 O 2, = 1; : : : ; m; let the functons g : J O 1! O, = 2; : : : ; m be contnuous and contnuously derentable wth respect to x 1 2 O 1, = 1; : : : ; m; respectvely, let = (J 1 ; N ) be the set of pecewse contnuous functons ' : J 1! N wth a nte number of ponts of dscontnuty, J 1 [ (a); b], N O be a convex bounded set, k' k = supf' (t)j t 2 J 1 g; let be the set of measurable functons u : J 2! U satsfyng the condton: clu (J 2 )-s a compactum lyng n G, J 2 = [ (a); b], U G s an arbtrary set. Consder the sets A = J 1+ Y p=1 O p Y p=1 Y p=1 p ; = 1; : : : ; m; wth the elements Q = (t 1 ; : : : ; t +1 ; x 10 ; : : : ; x 0 ; ' 1 ; : : : ; ' ; u 1 ; : : : ; u ), = 1; : : : ; m; O p=1 p = O 1 O, respectvely. To every element m 2 A m, t < t +1, = 1; : : : ; m, we assgn the derental equaton of varable structure _x (t) = f (t; x (t); x ( (t)); u (t); u ( (t))); t 2 [t ; t +1 ]; (1 ) x (t) = ' (t) + g (t; x 1 (t)); t 2 [ (t ); t ); x (t ) = x 0 ; (2 ) = 1; : : : ; m:
70 Here and n the sequel we assume that g 1 = 0, that s, x 1 (t) = ' 1 (t); t 2 [ 1 (t 1 ); t 1 ). Denton 1. The set of functons fx (t) = x (t; ); t 2 [ (t ); t +1 ]; = 1; : : : ; mg, s sad to be a soluton of the equaton wth varable structure whch correspond to an element m 2 A m, f the functon x (t) 2 O on the nterval [ (t ); t ] satses the condton (2 ), whle on the nterval [t ; t +1 ] t s absolutely contnuous and satses the equaton (1 ) almost everywhere. Denton 2. The element m 2 A m s sad to be admssble f the correspondng soluton fx (t); t 2 [ (t ); t +1 ]; = 1; : : : ; mg satses the condtons q k (t 1 ; : : : ; t m+1 ; x 10 ; x 1 (t 2 ); : : : ; s m0 ; x m (t m+1 )) = 0; k = 1; : : : ; l: The set of admssble elements wll be denoted by A 0 m. Denton 3. The element e m = (e t1 ; : : : ;e tm+1 ; ex 10 ; : : : ; ex m0 ; e' 1 ; : : : ; e' m ; eu 1 : : : ; eu m ) 2 A 0 m s sad to be locally optmal f there exst a number > 0 and a compact set K 0 ; = 1; : : : ; m, such that for arbtrary elements m 2 A 0 m satsfyng m+1 X mx je t t j + fjex 0 x 0 j + ke' ' k + ke f f k K g =1 the nequalty s fullled. Here =1 q 0 (e t1 ; : : : ;e tm+1 ; ex 10 ; ex 1 (e t2 ); : : : ; ex m0 ; ex m (e tm+1 )) q 0 (t 1 ; : : : ; t m+1 ; x 10 ; x 1 (t 2 ); : : : ; x m0 ; x m (t m+1 )) f e Z f K = J H t; f ; K dt; H t; f ; K = = sup f e t; x ; y f t; x ; y + @ f e() @x @f () + @ f e() @f () x ; y 2 K 2 @x @y @y ; ef t; x ; y = f t; x ; y ; eu (t); eu ( (t))); f (t; x ; y ) = = f (t; x ; y ; u (t); u ( (t)) ; ex = x (t; e ); x (t) = x (t; ): The problem of optmal control conssts n ndng a locally optmal element.
71 Theorem 1. (Necessary condtons of optmalty). Let e m 2 A 0 m be a locally optmal element; a < e t1 < < e tm+1 < b, e t < = (e t ) < +1 (e t+1 ) = 1; : : : ; m 1, e tm < m = m (e tm ) 6= e tm+1 ; let the functons (eu (t); eu ( (t))) = 1; : : : ; m, be contnuous at the ponts e t ; ;e t+1, respectvely, = 1; : : : ; m, and the functons (e' (t); e' ( (t))), = 1; : : : ; m be contnuous at the ponts e t, = 1; : : : ; m, respectvely. Then there exst a nonzero vector n = (n 0 ; : : : ; n l ), n 0 0, and a soluton (t), t 2 [e t ; (e t+1 )] of the conjugate equaton _ (t) = (t) @ e f (t) @x ((t)) @ e f ( (t)) @y (t) +1 ( +1 (t)) @ e f+1 ( +1 (t)) @y +1 _ (t) t 2 [e t ;e t+1 ]; (t) = 0; t 2 (e t+1 ; (e t+1 )]; = m; : : : ; 1; such that the followng condtons are fullled: 1) the ntegral maxmum prncple @eg +1 (t) @x _ +1 (t) (3 ) Zet (et ) ( (t)) @ e f ( (t)) @y _ (t)e' (t)dt Zet (et ) Z et +1 et ( (t)) @ e f ( (t)) @y _ (t)' (t)dt; 8' 2 ; = 1; : : : ; m; (4) (t) e f (t)dt et Z +1 et (t)f (t; ex (t); ex ( (t)u (t); u ( (t)))dt; 8' 2 ; = 1; : : : ; m: (5) 2) transversalty condtons @Q @t = a 1 (e t ) e f 1 (e t ) + b f (e t ) e f (e t ) + ( ) [ e f ( ; ex ( ); ex 0 ) e f ( ; ex ( ); e' (e t ) + eg (e t ))] _ (e t )g; @Q @x 0 = (e t ); = 1; : : : ; m + 1; (6) @Q @x = (e t+1 ); = 1; : : : ; m: (7)
72 Here ef (t) = f e @ f e(t) (t; ex (t); ex ( (t))); = @ f (t; ex (t); ex (t)); @x @x @eg +1 (t) eg (t) = g (t; ex 1 (t)); = @ g +1 (t; ex (t)); @x @x (t) s the characterstc functon of the segment [ (e t+1 );e t+1 )]; g m+1 = 0,.e., the last summand n the rght-hand sde of the equaton (3 m ) equals zero; (t) s the functon nverse to (t); a 1 = 0, a 2 = = a m+1 = 1, b 1 = = b m = 1, b m+1 = 0; tlde over Q denotes that the gradent s calculated at the pont (e t1 ; : : : ;e tm+1 ; ex 10 ; ex 1 (e t2 ); : : :, ex m0 ; ex m (e tm+1 ), then Remark 1. If rank @Q @t 1 ; : : : @Q @t m+1 ; @Q @x 10 ; : : : @Q ; @Q ; : : : ; @x m0 @x 1 mx maxfj (t)jjt 2 [e t ;e t+1 ]jg 6= 0: =1 @Q = 1 + l; @x m Remark 2. From the ntegral maxmum prncple 1) one can obtan n a standard way the pontwse maxmum prncple wth respect to the functons e' (t), = 1; : : : ; m: ( (t)) @ e f ( (t)) @y _ (t)e' (t) ( (t)) @ e f ( (t)) @y _ (t)' ; 8' 2 N ; a.e. on [ (e t );e t ]; = 1; : : : ; m; wth respect to the controls eu (t), = 1; : : : ; m, Xp p=1 _ p 1 Xp p=1 (t) ( p 1 _ p 1 (t) ( p 1 X p (t)) f e( p 1 (t)) + X p u ;p p+1; u ;p p) + p=1 p=1 (t))f ( p 1 (t); ex ( p 1 : _q p (t) (q p (t)) e f (q p (t)) (t)); ex ( ( p 1 (t))); _ p (t) ( p (t))f ( p (t); ex ( p (t)); ex ( ( p (t))); u ;p+p ; u ;p+p 1 ); 8(u 10 ; : : : ; u 1; ) 2 u 1+ ; a.e. on [ ;p ; ;p+1 ]; p = 1; : : : ; ; = 1; : : : ; m: Here ;p = ( ;p 1); p = 1; : : : ; ; ;0 = (e t ); ;p+1 = e t+1 ;
73 0(t) = t, p (t) = ( p 1 P (t)); (t) s the functon nverse to (t), 1 (t) = 0 (t); we assume that p=1 p = 0. Theorem 2. Let e m 2 A 0 m be a locally optmal element; a < e t1 < < e tm+1 < b, e t < < +1 e t+1, = 1; : : : ; m 1, e tm < m 6= e tm+1 ; let the functons (eu (t); eu ( (t))), = 1; : : : ; m, be left- (rght-) contnuous at the ponts e t ; ;e t+1, = 1; : : : ; m, respectvely. Then there exst a non-zero vector (+) = ( 0 (+); : : : ; q (+)) and a soluton (+)(t), t 2 [e t ; (e t+1 )] of the conjugate equaton (3 ), = 1; : : : ; m, such that the condtons n whch we have to substtute and (t) nstead of (+) and (+), respectvely, = 1; : : : ; m, are fullled. Moreover, the equalty (6) s replaced by the nequalty @ e @t a 1 ( e t ) e f 1 (e t ) + b f ( e t )e f (e t ) + ( )[ e f ( ; ex ( ); + @ e ex 10 ) e f ( ; ex ( ); ' (e t ) + eg (e t ))] _ (e t )g; = 1; : : : ; m + 1 a + @t (t) 1 f 1 e (e t +) + b f + ( e t ) f e(e t +) + ( )[ f e( +; ex ( ); ex 10 ) f e( ; ex ( ); e' (e t +) + g (e t ))] _ (e t )g ; where ef (e t (+) ) = e f( e t (+) ; ex (e t ); ex (e (e t (+) )): Consder now for (1 ) and (2 ), = 1; : : : ; m, where g 2 = = g m = 0, the problem wth the boundary condtons q 1 (t 1 ; x 10 ) = 0; q (t ; x 1 (t )) x 0 = 0; = r; : : : ; m; and wth the functonal The functons q m+1 (t m+1 ; x m (t m+1 )) = 0; (8) q 0 (t m+1 ; x m (t m+1 ))! mn : (9) q 1 : J 0 1! R l1 ; q : J 0! R n ; = 2; : : : ; m; q m+1 : J 0 m! R l2 ; q 0 : J 0 m! R 1 are assumed to be contnuously derentable n all arguments. The functon Q for the problem under consderaton s of the form Q = (q 0 ; q 1 ; q 2 x 20 ; : : : ; q m x m0 ; q m+1 ): (10) Takng nto account (10), from Theorem 1 there follows
74 Theorem 3. Let m be a locally optmal element of the problem (1 ), (2 ), = 1; : : : ; m, (8), (9) and let the condtons of Theorem 1 be fullled. Then there exst a non-zero vector = ( 0 ; 1 ; : : : ; m+1 ), 0 0, 1 2 R l1, 2 R n, = 1; : : : ; m, m+1 2 R l2 and a soluton (t), t 2 [e t ; (e t+1 )], of the conjugate equaton (3 ) = 1; : : : ; m, where the last term n the rghthand sde equals zero, such that the condtons (4), (5) are fullled, whle the condtons of transversalty take the form @eq @t = a 1 (e t ) e f 1 (e t ) + b f (e t ) e f (e t ) + + ( )[ e f ( ; ex ( ); ex 0 ) e f ( ; ex ( ); e' (e t ))] (e t )g c 0 @eq 0 @t m+1 ; = 1; : : : ; m + 1; 1 @eq 1 @x 10 = 1 (e t1 ); (e t ) = ; = 2; : : : ; m; @q = 1 (e t ); = 2; : : : ; m; @x 1. Here 0 @eq 0 @x m + m+1 @eq m+1 @x m = m (e tm+1 ): c 1 = = c m = 0; c m+1=1 ; ex 0 = q (e t ; ex 1 (e t ); = 2; : : : ; m: 3. Proof of Theorem 1. The necessary condtons of optmalty are proved by the scheme gven n [1, 2, 3]. When applyng ths scheme, the prncpal moment s the constructon of a contnuous and derentable mappng whch plays an mportant role n dervng the necessary condtons of optmalty. To ths end we present below and prove (see 3.1 and 3.2) the approprate theorems. 3.1. Contnuous Dependence and Derentablty of the Soluton. Let O R n be an open set: E(J O 2 ; R n ) be a space of n-dmensonal functons f; J 0 2! R n satsfyng the followng condtons: 3) for a xed t 2 J the functon f(t; x; y) s contnuously derentable wth respect to (x; y) 2 O 2 ; 4) for a xed (x; y) 2 O 2 the functons f; f x ; f y are measurable wth respect to t; for an arbtrary compactum K O and an arbtrary functon f there exsts a functon m f;k (t) 2 L 1 (J; R 1 +), R 1 + = [0; 1) such that jf(t; x; y)j + jf x ( _ )j + jf y ( _ )j m f;k (t) 8(t; x; y) 2 J K 2 : In the space E(J O 2 ; R n ), let us ntroduce by means of the bases of neghborhoods of zero two locally convex separable topologes [1, 2, 3],
75 fv s (K; ) E(J 0 2 ; R n ) a compactum K and a number > 0 are arbtraryg, s = 1; 2, where V 1 (K; ) = ff 2 E(J O 2 ; R n )jh(f; K) g; (11) V 2 (K; ) = ff 2 E(J O 2 ; R n )j Zt 00 h(f; K) = sup H 0 (t; f; K) = sup Z J H 0 (t; f; K)dt g; (12) t 0 f(t; x; y)dtj(t 0 ; t 00 ; x; y) 2 J 2 K 2 ; n o jf(t; x; y)j + jf x ()j + jf y ()j (x; y) 2 K 2 : Consder the sets B = J 1+ Y p=1 O p Y p=1 (J 1 ; O ) Y p=1 E(J O 2 ; R n ); = 1; : : : ; m; wth the elements = (t 1 ; : : : ; t +1 ; x 10 ; : : : ; x 0 ; ' 1 ; : : : ; ' ; f 1 ; : : : ; f ), = 1; : : : ; m respectvely. In what follows we wll assume that the topologes T 1, T 2 are prescrbed n the spaces E(J O 2 ; Rn ) and = 1; : : : ; m (see (11) and (12)). To every element m 2 B m there corresponds the derental equaton of varable structure: _x (t) = f (t; x (t); x ((t))); t 2 [t ; t +1 ]; (13 ) x (t) = ' (t) + g (t; x 1 (t)); t 2 [ ; (t ); t +1 ); x (t ) = x 0 ; = 1; : : : ; m: (14 ) Denton 4. A set of functons fx (t) = x (t; ); t 2 [ (t ); t +1 ]; = 1; : : : ; mg s sad to be a soluton of the equaton of varable structure correspondng to the element m 2 B m, f the functon x (t) 2 0 on the segment [ (t ); t ] satses the condton (14 ) and on the nterval [t ; t +1 ] s absolutely contnuous and satses the equaton (13 ) a.e. Theorem 4. Let fex (t); t 2 [J (e t );e t+1 ]; = 1; : : : ; mg, a < e t1 < < e tm+1 < b, be a soluton correspondng to the element e m =(e t1 ; : : : ;e tm+1 ; ex 10 ; : : : ; ex m0 ; e' 1 ; : : : ; e' m ; e f1 ; : : : ; e fm ) 2 B m and let K 1 0 be a compactum contanng some neghborhood of the set K 0 = cl[fex (t)jt 2 [e t ;e t+1 ]g [ fe' (t)j 2 J 1 g]: Then for every " > 0 there exsts a number = (") > 0 such that to every element m 2 V (e m ; K 11 ; : : : ; K m1 ; ; c 0 ) = m+1 Y =1 V (e t ; ) my =1 V (e' ; )
76 my V (e' ; ) =1 =1 my V 1 (e f ; K 1 ) \ V 2 ( f e; K 1 ; c 0 )); there corresponds a soluton fx (t; ); t 2 [J (t ); t +1 ]; = 1; : : : ; mg. Moreover, the functon x (t; ) s dened on [J (t );e t+1 + ] (J (a); b), and satses almost everywhere on [t ;e t + ] the equaton (13 ). If then s m 2 V (e m ; K 11 ; : : : ; K m1 ; ; c 0 ); s = 1; 2; jx (t; 1 ) x (t; 2 )j "; t 2 [max(t 1 ; t 2 );e t+1 + ]; = 1; : : : ; m: (15) Here V (e t ; ), V (ex 0 ; ), V (e' ; ), are the -neghborhoods of the ponts e t,ex 0, e', n the spaces R 1, R n, respectvely; (J 1 ; R n ); V 1 ( e f ; K ; ) = ef + V 1 (K 1 ; ), V 2 ( e f ; K 1 c 0 ) = e f + V 2 (K 1 ; c 0 ), V 1 (K 0 ; ) E(J 0 2 ; R n, V 2 (K 1 ; c 0 ) E(J 0 2 ; Rn ; c 0 > 0 s a xed number. Theorem 4 can be proved by the method gven n [6] (see also [7]) and s used n provng the contnuty of the mappng (see 3.3). Remark 3. There exsts a number e 2 [0; ] (see (12)) such that V 2 ( e f ; K 1 ;e ) V1 ( e f ; K ; e ) \ V2 ( e f ; K 1 ; c 0 ); = 1; : : : ; m: Consequently, the nequalty (15) s the more so vald for m+1 Y s m 2 =1 V (e t ;e ) m Y =1 V (e' ; e ) m Y =1 V 2 ( e f ; K 1 ; e ): Ths fact s used n provng the openness of the set D 0 (see 3.2). Introduce the set V = f = (t 1 ; : : : ; t +1 ; x 10 ; : : : ; x 0 ; ' 1 ; : : : ; ' ; f 1 ; : : : ; f ) 2 2 B e jt j c ; sx x 0 c 1 ; k' k c ; f p = j=1 j j j c g; = 1; : : : ; m; j "fp j ; p = 1; : : : ; ; f j p 2 E(J 0 2 p; R np ), p = 1; : : : ;, j = 1; : : : ; s, are xed ponts and s, c > 0 are xed numbers. From Theorem 4 we have
77 Theorem 5. There exst numbers " 0 > 0, 0 > 0 such that for an arbtrary ("; m ) 2 [0; " 0 ]V m, to the element e m +" m there corresponds soluton fx (t; e +" ); t 2 [ (t ); t +1 ]; = 1; : : : ; mg; t = e t + "t : Moreover, x (t; e + " ) s dened on [ (t );e t+1 + 0 ]. Remark 4. Due to the unqueness, x (t; e ) on the nterval [ (e t );e t+1 + 0 ] s a contnuaton of ex (t). Therefore the functon ex (t) n the sequel s assumed to be dened on the whole nterval [ (e t );e t+1 + 0 ]. Usng the numbers 0 and " 0 (see Theorem 5), we ntroduce the notaton x (t; " ) = x (t; + " ) ex (t); t 2 [max(t ;e t );e t+1 + 0 ]; " 2 [0; " 0 ]: Theorem 6. Let a < e t1 < < e tm+1 < b, e t < = (e t ) < +1 (e t+1 ), + 1; : : : ; m 1, e tm < m = m (e tm ) 6= e tm+1 ; let the functons e f (t; x ; y ), = 1; : : : ; m, be contnuous respectvely at the ponts (e t ; ex 0 ; e' ( (e t ))), ( ; ex ( ); ex 0 ), ( ; ex ( ); e' (e t )), = 1; : : : ; m, and the functons e' (t), = 1; : : : ; m, be contnuous respectvely at the ponts e t, = 1; : : : ; m. Then there exst numbers " 1 2 [0; " 0 ], 1 2 [0; 0 ] such that the followng formula s vald: where x (t; " ) = "x (t; ) + o (t; " ); 8(t; "; ) 2 [ +1 (e t+1 ); e t+1 + 1 ] [0; " 1 ] V ; = 1; : : : ; m; (16) x (t; ) = Y (e t ; t) hx 0 f e(e t )t + h' (s) + @eg et +1 (s) x 1 (s) _ (s)ds + @x 1 Z Zet (et ) Y ( (s); t) @ e f ( (s)) @y ef ( ; ex ( ); ex 0 ) e f ( ; ex ( ); e' ; (e t ) + eg (e t )) _ (e t )t = et Y (s; t) e f ds Y ( ; t) = x 1 (t; ) x 2 (t; ); = 1; : : : ; m; (17) Y (s; t) s a matrx functon satsfyng the equaton @Y (s; t) @s = Y (s; t) @ e f (s) @x and also the condton next, Y ( (s); t) @ e f ( (s)) @y _ (s); s 2 [e t ;e t+1 ] Y (t; t) = E; Y (s; t) = 0; s > t; lm "!0 o (t; " )=" = 0; unformly wth respect to (t; ) 2 [ +1 (e t+1 );
78 e t+1 + 1 ] V ; e f (t) = f (t; ex (t); ex ( (t))); m+1 (e tm+1 ) = e tm+1 1 : The proof of Theorem 6 s conducted n a way descrbed n [2, 7]. In the same way one can prove more general Theorem 7. Let a < e t1 < < e tm+1 < b, e t < < +1 (e t+1 ), = 1; : : : ; m 1, e tm < m 6= e tm+1, and let the condtons (t; x )! (t (+); ex 0 ) lm e f (t; x ; e' ( (t))) = e f (e t (+)); ex 0 ; e'( (e t (+))) < 1; (t; x ; y )! ( (+); ex (); ex 0 ) lm e f (t; x ; y ) = e f (e (+); ex ( ); ex 0 ) < 1; be fullled. Then there exst numbers " 1 2 [0; " 0 ], 1 2 [0; 0 ], such that the formula (16) s vald for an arbtrary pont (t; "; ) 2 [ +1 (e t+1 );e t+1 + 1 ] [0; " 1 ] V (+), whle n the formula (17) before t there take place respectvely the expressons Here Y (e t ; t)e f (e t (+)) + Y ( ; t) e f ( ; (+); ex ( ); ex 0 ) ef ( (+); ex ( ); e' (e t (+)) + g(e t )) _ (e t ): V f 2 V j t K 0; K = 1; : : : ; + 1g; V + f 2 V j t K 0; K = 1; : : : ; + 1g: Theorem 8 ([The Cauchy formula [8]]). Let A(t), B(t), t 2 J 1 = [s 1 ; s 2 ] be summable matrx functons of dmenson n n; let F (t), t 2 J 1 be an n- dmensonal summable vector functon and (t) satsfy the same condtons as (t) do; let '(t), t 2 [(s 1 ); s 2 ] be a pecewse contnuous functon. Then the soluton of the equaton can be represented as _x(t) = A(t)x(t) + B(t)x((t)) + F (t); t 2 J 1 ; x(t) = '(t); t 2 [(s 1 ); s 1 ); x(s 1 ) = x 0 2 R n ; x(t) = Y (s 1 ; t)x 0 + Z s 1 Z t Y ((s); t)b((s)) _(s)'(s)ds + Y (s; t)f (s)ds; (s 1) t 2 J 1 ; s 1 where Y (s; t) s a matrx functon satsfyng the equaton @Y (s; t) @s = Y (s; t)a(s) Y ((s); t)b((s)) _(s); s 2 [s 1 ; t]
79 and also the condton (s) s the functon nverse to (s). Y (t; t) = E; Y (s; t) = 0; s > t; On the bass of Theorem 8 we can conclude that the functon x 1 (t; ) (see (17)) satses the equaton _x 1 (t) = @ e f (t) @x wth the ntal condton x 1 (t) + @ f e(t) x 1 ( (t)) + f e(t); @y t 2 [e t ;e t+1 ] (18) x 1 (t) = ' (t) + @eg (t) @x 1 x 1 (t); t 2 [ (e t );e t ); x 1 (e t ) = x 0 e f (e t1 )t ; (19) whle the functon x 2 (t; ) (see (17)) satses the equaton _x 2 (t) = @ f e(t) x 2 (t) + @ f e(t) x 2 ( (t)); t 2 [ ;e t+1 ]; (20) @x @y wth the ntal condton x 2 (t) = 0; t < x 2 ( ) = [e f ( ; ex ( ); ex 0 ) e f ( ; ex ( ); e' (e t ))] _ (e t )t : (21) For the sake of brevty we denote the functon, x (t; ), x 1 (t; ) and x 2 (t; ), respectvely by x (t), x 1 (t) and x 2 (t). Theorem 9. The followng formula s vald: where mx =1 @ e Q @x x (e t+1 ) = = Zet mx (Y (e t )x 1 (e t ) Y ( )x 2 ( ) + ); (22) =1 (et ) Y ( (t)) + et Z +1 @ e f ( (t)) @y _ (t)' (t)dt + et Y (t) e f (t)dt; (23)
80 Y (t) s a soluton of the matrx equaton _Y (t) = Y (t) @ e f (t) @x Y ( (t)) @ e f ( (t)) @y (t)y +1 ( +1 (t)) @eg +1(t) @x _ +1 (t); _ (t) t 2 [e t e t+1 ] (24) Y (e t+1 ) = @ e Q @x ; Y (t) = 0; t 2 (e t+1 ; (e t+1 )] = m; : : : ; 1: (25) Proof. Obvously (see (25)), @ Q e x 1 (e t+1 ) =Y (e t+1 )x 1 (e t+1 ) Y (e t )x 1 (e t ) + Y (e t )x 1 (e t ) = @x et Z +1 d = dt (Y (t)x 1 (t))dt + Y (e t )x 1 (e t ): (26) et From the fact that x 1 (t) satses the equaton (18), we obtan et Z +1 et d dt (Y (t)x 1 (t))dt = Z et +1 et h _Y 1 (t)x (t) + Y (t) + @ F e (t) x 1 ( (t)) + f e(t) @y @ e f (t) @x x 1 (t) + dt: (27) Now, takng nto account (19) and (25), we transform the ntegral Z et +1 et _ (t)x 1 (t)dt = Y (t) Zet @ e f (t) @y x 1 ( (t))dt = (et ) Y ( (t)) + Z et +1 et Y ( (t)) (et Z +1) (et ) @ e f ( (t)) @y ; @ f e( (t)) h ' (t) + @eg (t) x 1 (t) _ (t)dt + @y @x 1 @ e f ( (t)) @y x 1 (t) _(t)dt: (28)
81 Owng to (27) and (28), from (26) we obtan @ e Q @x x 1 (e t+1 ) = + Z (et Z et +1 et +Y ( (t)) @ e f ( (t)) @y (et ) Y ( (t)) + @ e f ( (t)) @y h _Y (t) + Y (t) _ (t) x 1 (t)dt + @ e f (t) @x + @eg (t) @x 1 _ (t)x 1 (t)dt + +Y (e t )x 1 (e t ): (29) Snce x 2 (t) satses equaton (20) and also the condton (21), we have analogously @ e Q @x x 2 (e t+1 ) = Thus + mx =1 Z et +1 et Z et +1 et @ Q e x (e t+1 ) = @x mx = =1 _Y (t) + Y (t) + Zet h _Y (t) + Y (t) @ e f (t) @x + Y ( (t)) @ e f ( (t)) @y _ (t) x 2 (t)dt + Y ( )x 2 ( ): (30) mx =1 h @ Q e x 1 (e t+1 ) @ Q e x 2 (e t+1 ) = @x @x (Y (e t )x 1 (e t ) Y ( )x 2 ( ) + + @ e f (t) @x (et ) Y ( (t)) We can easly see that mx =1 Zet (et ) Y ( (t)) + Y ( (t)) @ e f ( (t)) @y @ e f ( (t)) @y @ e f( (t)) @y _ (t) x (t)dt + @eg (t) @x 1 _ (t)x 1 (t)dt: (31) @eg (t) @x 1 _ (t)x 1 (t)dt =
82 = Zet mx =1et (t)y +1( +1(t)) @ f+1 e ( +1 (t))@eg +1 (t) _ +1 (t)x (t)dt: (32) @y +1 @x Because of (24) and (32), the equalty (31) provdes the formula (22). 3.2. Calculaton of the derental of the mappng. Consder the space E m = R 1+m my =1 R n my =1 (J 1 ; R n ) my =1 E(J 0 2 ; R n ) of the ponts m. Topologes T 1, T 2 specfy n E m two locally convex separable topologes; denote them respectvely by 1, 2. Denote by D 0 the set of elements m 2 B m to whch there corresponds a soluton of equaton of varable structure (14 ), (15 ), = 1; : : : ; m. Dene on a set D 0 the mappng by the formula S : D 0! my =1 R n ; (33) S( m ) = (S 1 ( 1 ); ; S m ( m )); :S ( ) = x (t +1 ; ); = 1; : : : ; m: Let e m 2 D 0 passng through the pont e m and L e m be a nte-dmensonal manfold L e m = m 2 E m j m = e m + m ; SX m = j j m ; j 2 R 1 ; j m 2 E m e m ; ; j=1 W m = = SX j=1 j j mj j j j c 2 be a bounded neghborhood of zero n L e m e m ; let c 2 > 0 e m be a xed number. The nte openness of the set D 0 mples the exstence of a number " 2 2 [0; " 1 ], such that for arbtrary ("; m ) 2 [0; " 2 ] m the pont e m +" m 2 D 0 \ L e m. " 2 s assumed to be so small that "jt j 2, = 2; : : : ; m + 1 for ("; m ) 2 [0; " 2 ] W m. Theorem 10. Let the condtons of Theorem 6 be fullled. Moreover, the functons e F (t; x ; y ), = 1; : : : ; m, are assumed to be contnuous respectvely at the pont (e t ; ex (e t); ex ( (e t ))), = 1; : : : ; m. Then the derental of the mappng (33) at the pont e m has the form ds e m ( m ) = ds 11 ( 1 ); : : : ; ds mm ( m ) ;
83 where ds e ( ) = x (e t+1 ; ) + e f (e t+1 )t +1 ; = 1; : : : ; m; (34) 2 W = f j ( ; 0; : : : ; 0) 2 W m g; = 1; : : : ; m 1: Proof. It s easly seen that the set V can be chosen n such a way that W V. Therefore for ("; ) 2 [0; " 2 ] W we have S (e + " ) S (e = x (e t+1 + "t +1 ; e + " ) ex (e t+1 ) = = x (e t+1 + "t +1 ; e + " ) + ex (e t+1 + "t ) ex (e t+1 ) = "x (e t + "t ; ) + et +1+"t Z +1 + et +1 e f (t)dt + 0 (e t+1; " ): (35) By assumpton, e f (t; x; y) s contnuous at the pont (e t ; ex (e t ); ex ( (e t ))), whch mples the contnuty of the functon e f (t) at the pont e t+1. Applyng now the theorem of the mean, we get Further, et +1+"t Z et +1 = " e f (e t+1)t + 0 (" ): (36) lm "!0 [x (e t+1 + "t +1 ; ) x (e t+1 ; )] = 0; unformly wth respect to 2 W : (37) Usng (36) and (37), from (35) we have where S (e + " ) S (e ) = "[x (e t ) + e f (e t )t ] + 0 (" ); 0 (" ) = 0 (e t+1 ; " ) + 0 (" ) + "[x (e t + "t ; ) x (e t )] and lm "!0 0 (" )=" = 0; unformly wth respect to 2 : Thus, the formula (34) s vald. Let us consder now the mappng P 1 : D 0! R 1+l (38) dened by the formula P 1 = QL, L( m ) = (t 1 ; : : : ; t m+1 ; x 10 ; : : : ; x m0 ; x 1 (t 1 ); : : : ; x m (t m+1 )).
84 Theorem 11. Let the condtons of Theorem 9 be fullled. Then the dfferental of the mappng (38) at the pont e m s of the form dp 1 e m ( m ) = mx =1 @ e Q @t + a Y 1 (e t ) e f 1 b Y (e t ) e f (e t ) + +Y ( ) f e( ; ex ( ); ex 0 ) f e( ; ex ( ); e' (e t ) + eg (e t )) _ (e t ) t + mx h @ e Q + + Y (e t ) x 0 + @x 0 =1 where a 1 = 0, a 2 = = a m+1, b 1 = = b m = 1, b m+1 = 0. Proof. It s easy to see that dp 1 e m ( m ) = m+1 X =1 @ e Q @t t + @ e Q @x 0 t + @ e Q @x ds e ( ) ; (39) + @ e Q @t m+ t m+1 : (40) Takng nto account (34), (22), (19) and (21), we obtan m+1 X =1 @ e Q @t t + mx =1 @ e Q @x ds e ( ) = m+1 X =1 + @ e Q @t + a Y 1 (e t ) e f 1 (e t ) b hy (e t )e f (e t ) + Y ( ) F e ( ; ex ( ); ex 0 ) f e( ; ex ( ); e' (e mx t ) + eg (t )) _ (e t ) t + Y (e t )x 0 + =1 : (41) Owng to (41), from (40) we get the formula (39). 3.3. Necessary Condtons of Crtcalty. Dervng the necessary condton of crtcalty. Consder the space E z = R 1 E m. Let D = fz = (s; m )js 2 R 1 g: The nte openness of the set D 0 mples the nte openness of D. Dene on the set D 0 the mappng P : D! R 1+l (42) by usng the formula P (z) = P 1 ( m ) + (s; 0; : : : ; 0). The topology 1 denes n E 2 a locally convex separable topology whch we denote by.
85 In E 2 we prescrbe a lter ez, ez = (0; e 0), wth elements Y m+1 er = R1 \ + V 0 V et =1 my =1 V ex 0 my =1 e' my =1 ef ; where V 0, V et, V ex 0 are arbtrary neghborhoods of the ponts 0 2 R 1, e t 2 (a; b), ex 0 2 0 ; e' = \ V e', V e' (J 1 ; R n ) s an arbtrary neghborhood of the pont e' ; ef s an arbtrary element of the lter ef. Elements of the lter ef are ef = f \ V ef ; where V E(J O ef 2 ; Rn ) s an arbtrary neghborhood of the pont ef = f e(t; x ; y ) = f (t; x ; y ; eu (t); eu ( (t))) n the topology T 2, F = ff (t; x ; y ; u (t); u ( (t)))ju 2 g: Beng a drect product of convex lters by quas-convex lters [1, 2, ef 3], the lter ez s -quas-convex. The proof of quas-convexty of s ef gven n [9]. The crtcalty of the mappng P onto ez follows n a standard way from the local optmalty of e m [1, 2, 3]. From Remark 3 and Theorem 4 there follows the exstence of an element W ez, such that c 0w ez D, and the mappng (42) onto c 0w ez s contnuous n the topology. The assumptons of Theorem 1 ensure the fulllment of the condtons of Theorem 10 whch n ts turn ensures derentablty of the mappng (42) at the pont ez: dp ez (z) = dp e m ( m ) + (s; 0; : : : ; 0); z = (s; m ): Thus all the premses for the necessary condton of crtcalty are fullled [1, 2, 3]. Consequently, there exsts a non-zero vector = ( 0 ; : : : ; l ) and the element Wz f 2 ez such that dp ez (z) 0; 8z 2 K(c 0 ew ez ez); (43) K(M) denotes a convex cone stretched on the set M. Below from the necessary condton of crtcalty (43) we wll derve the necessary condtons of optmalty. The condton z 2 K(c 0 ew ez ez) s equvalent to the condton s 2 R1 +, t 2 R 1, t 0 2 R, ' 2 K(ew e' e' ), f 2 K(c 0 ew ef e F ). Usng the expresson (39), lettng f = ' = 0, = 1; : : : ; m, s = 0 n (43) and takng nto account that t, = 1; : : : ; m+1 x = 0, = 1; : : : ; m,
86 may take arbtrary values, we obtan @ Q e h = a Y 1 (e t ) f 1 e (e t ) + b Y (e t ) + Y ( ) f e( ; ex ( ); ex 0 ) @t e f ( ; ex ( ); e' (e t ) _ (e t ) ; = 1; : : : ; m + 1; (44) @ e Q @t = Y (e t ); = 1; : : : ; m: (45) Lettng t = 0, = 1; : : : ; m + 1, x 0 = f = 0, = 1; : : : ; m, s = 0 n (43) and takng nto consderaton that f W e' e', = 1; : : : ; m, contans a non-zero element, we get Zet (et ) Y (t) @ e f((t)) @y _ (t)' (t)dt 0; 8' 2 K f W e' e' ; = 1; : : : ; m: (46) For t = 0, = 1; : : : ; m + 1, x 0 = ' = 0, = 1; : : : ; m, s = 0 we have et Z +1 t Y (t)f t; ex(t); ex( (t)) dt 0; 8f 2 K c 0 f W ef e F ; = 1; : : : ; m: (47) If t = 0, = 1; : : : ; m+1, x 0 = ' = f = 0, = 1; : : : ; m, we obtan that 0. Introduce the notaton (t) = Y (t); t 2 [e t ; (e t+1 )]; = 1; : : : ; m: (48) Obvously, (t) on the nterval [e t ;e t+1 ] satses equaton (3 ) and also the condton et+1 @ Q e = ; (t) = 0; t 2 et+1; (e @x t+1) : (49) Condtons (44), (45) and (49) on account of (48) provde the condton of transversalty (2). From the convexty and boundedness of the set N follows the ncluson e' K(f W e' e' ): (50) Condton (46), due to (48), (50) and the fact that ' = ' e', results wth respect to e' n the maxmum prncple (4).
87 Prove now the condton (5). The mappng f! et Z +1 t Y (t)f t; ex (t); ex ( (t)) dt; f 2 V 2 K1 ; c 0 ; (51) s contnuous n the topology T 1 (see [9]). The contnuty of the mappng (51) makes t possble to conclude that nequalty (47) s vald for an arbtrary where [f W ef ] 1 M f 2 K([f W ef ] 1 M e f ); M = V 2 (K 1 ; c 0 ); s a closure wth respect to M of the set W f \ M ef n the topology T. Usng now the proven n [9] ncluson F e f K([f W ef ] 1 M e f ); we can say that the nequalty (47) s vald for arbtrary f 2 f e F. From (47), due to (48) as well as f = f t; x ; y ; u (t); u ( (t)) e f t; x ; y ; eu (t); eu ( (t)) 2 F e f we obtan wth respect to the controls the maxmum prncple (5). Thus Theorem 1 s proved completely. Fnally we note that Theorem 2 s proved analogously to Theorem 1. In ths case for calculaton of the derental we use Theorem 7, whle n the lter ez we replace respectvely the neghborhoods V et, = 1; : : : ; m + 1, by R et \ V et (R + et ), where R et = ( 1;e t ], R et = [e t ; 1). References 1. R. V. Gamkreldze and G. L. Kharatshvl, Extremal problems n lnear topologcal spaces 1. Math. Systems Theory,1(1967), No. 3, 229{ 256. 2. R. V. Gamkreldze and G. L. Kharatshvl, Extremal problems n lnear topologcal spaces. (Russan) Izv. Akad. Nauk SSSR, Ser. Mat., 33(1969), No. 4, 781{839. 3. R. V. Gamkreldze, The necessary condtons of rst order and axomatcs of extremal problems. (Russan) Trudy Mat. Inst. Steklova 62(1971), 152{180. 4. R. V. Gamkreldze, On some extremal problems n the theory of derental equatons wth applcatons to the theory of optmal control. SIAM J. Control, 3(1965), 106{128.
88 5. N. M. Avalshvl, Maxmum prncple for the optmal problem wth a varable structure and delays. (Russan) In Optmal problems wth a varable structure. Tbls Unversty Press, Tbls, 1985, 48{79. 6. R. V. Gamkreldze, Fundamentals of the optmal control. (Russan) Tbls State Unversty Press, Tbls, 1977. 7. G. L. Kharatshvl, Z.A. Machadze, N.I. Markozashvl and T. A. Tadumadze, Abstract varatonal theory and ts applcaton to the optmal problems wth delays. (Russan) Metsnereba, Tbls, 1973. 8. R. Gabasov and F. M. Krllova, Qualtatve theory of optmal processes. (Russan)Nauka, Moscow, 1975. 9. G. L. Kharatshvl and T. A. Tadumadze, Nonlnear optmal control systems wth varable delays. (Russan)Mat. Sb., 107(149)(1978), No. 4(12), 613{633. Authors' addresses: (Receved 24.12.96) G. Kharatshvl Cybernetcs Insttute of the Georgan Academy of Scences 5, S.Eul St., Tbls 380086 Georga T. Tadumadze I.N. Vekua Insttute of Appled Mathematcs of Tbls State Unversty 2, Unversty St., Tbls 380043 Georga