Georgian Mathematical Journal 1(94), No. 4, 419-427 ON THE INITIAL VALUE PROBLEM FOR FUNCTIONAL DIFFERENTIAL SYSTEMS V. ŠEDA AND J. ELIAŠ Astract. For a system of functional differential equations of an aritrary order the conditions are estalished for the initial value prolem to e solvale on an infinite interval, and the structure of the set of solutions to this prolem is studied. Introduction For the p-th order functional differential system x (p) (t) = f(t, x t,..., x (p 1) t ), t < +, (1) we consider the initial value prolem x (k) = ψ (k) (k = 0,..., p 1). (2) The invesigation is ased on Kuáček s theorem [2] asserting that under certain conditions the set of all fixed points of the compact map in the Fréchet space is a compact R δ -set. It is shown that some restrictions on the growth of the right-hand side of the functional differential system imply that the set of all solutions of the initial value prolem for that system is a compact R δ -set in the Fréchet space of C p 1 -functions. The result extends a similar theorem for first-order functional differential systems proved in [2] and the theorem for second-order functional differential systems proved in [3]. In the sequel we shall use the following notations and assumptions: Let h > 0, R, d N, and let e a norm in R d. Further, let H l = C l ([ h, 0], R d ) e provided with the norm { l x l = max x (k) (s) : h s 0 1991 Mathematics Suject Classification. 34K05, 34K25. 419 1072-947X/94/0700-0419$12.50/0 c 1994 Plenum Pulishing Corporation
420 V. ŠEDA AND J. ELIAŠ for each x H l and l = 0,..., p 1. For revity p 1 will e denoted y. Let X = C p 1 ([, ), R d ) e equipped with a topology of locally uniform convergence of the functions and of their p 1 derivatives on [, ). In the Fréchet space X the topology is given y the metric where p m (x) = sup d(x, y) = { p 1 m=1 1 2 m p m (x y) 1 + p m (x y), x (k) (t) : t + m, x, y X, m 1. Let X = C p 1 ([ h, ), R d ) e a Fréchet space whose topology is determined y seminorms p m(x) = sup { p 1 x (k) (t) : h t + m, x X, m 1. For x C([ h, ), R d ) we shall denote y x t H 0 the function x t (s) = x(t + s), s [ h, 0], t. Clearly (x t ) (k) (s) = (x (k) ) t (s), s [ h, 0], k = 0,..., p 1, and x X, t. It is assumed throughout the paper that f X([, ) H p 1 H 0, R d ), ψ H p 1. A solution x of (1), (2) is a function x X such that x [, ) C p ([, ), R d ) ad x satisfies (2) and the functional differential system (1) at each point t. 1. Auxiliary Propositions Now Kuáček s theorem in [2] will e stated as Lemma 1. In that lemma the compact R δ -set in the metric space (E, ρ) means a nonempty suset F of E which is homeomorphic to the intersection of a decreasing sequence of compact asolute retracts. By [1], p. 92, a metric space G is called an asolute retract when each continuous map f : K G has a continuous extension g : H G for each metric space H and each closed K H. For example, a nonempty convex suset of the Fréchet space is an asolute retract. Lemma 1. Let M e a nonempty closed set in the Fréchet space (E, ρ), T : M E a compact map (i.e., T is continuous and T (M) is a relatively compac set). Denote y S the map I T where I is the identity map on E. Let there exist a sequence {U n of closed convex sets in E fulfilling the conditions
ON THE INITIAL VALUE PROBLEM 421 (i) 0 U n for each n N; (ii) lim diam U n = 0, n and a sequence {T n of maps T n : M E fulfilling the conditions (iii) T (x) T n (x) U n for each x M and each n N; (iv) the map S n = I T n is a homeomorphism of the set S 1 n (U n ) onto U n. Then the set F of all fixed points of the map T is a compact R δ -set. In the special case E = X, ρ = d Lemma 1 implies Lemma 2. Let (X, d) e the Fréchet space given aove; let ϕ, ϕ n C([, ), [0, )), and let the following conditions e fulfilled: (v) For each t [, ) the sequence {ϕ n (t) is nonincreasing and lim ϕ n(t) = 0. Let r k R d, k = 0,..., p 1 and let n { M = x X : x (k) (t) r k ϕ(t), t, x (k) ()=r k, k =0,..., p 1. It is assumed that T : M X is a compact map with the property (T (x)) (k) () = r k, k = 0,..., p 1 for each x M and there exists a sequence {T n of compact maps T n : M X such that (T n (x) (k) () = r k, k = 0,..., p 1 for each x M and (vi) ( T n (x) ) (k) ( ) (k)(t) (t) T (x) ϕ n (t), x M, t ; (vii) for each n N there exists a function ϕ n C([, ), [0, )) such that ϕ n + ϕ n ϕ on [, ) and ( T n (x) ) (k) (t) rk ϕ n (t), x M, t ; (viii) the map S n = I T n is injective on M, where I is the identity on X. Then the set F of all fixed points of the map T is a compact R δ. Proof. The set U n = { x X : x (k) (t) ϕ n (t), t, x (k) () = 0, k = 0,..., p 1 is convex and closed on X for each n N. We shall show that the sequence {U n satisfies all conditions of Kuáček s theorem when E = X, ρ = d and thus Lemma 2 will follow from Lemma 1. Clearly, (i) is fulfilled. As to the condition (ii), we choose an aritrary ε > 0. Then there is an m 0 N such that m=m [ +1 (1/2m ) < ε/2. The
422 V. ŠEDA AND J. ELIAŠ condition (v) and the Dini theorem imply that the sequence {ϕ n converges on [, ) locally uniformly to 0. Therefore for ε > 0 and m 0 N there is an n 0 N such that g m (ϕ n ) = sup{ ϕ n (t) : t + m ε/4 m 0 for n n 0 and m = 1, 2,..., m 0. Hence for n n 0 and x, y U n we have d(x, y) = m=1 m 0 m0 1 p m (x y) 2 m 1 + p m (x y) 2g m (ϕ n ) + m=1 m=m 0 +1 m=1 p m (x y) + m=m 0 +1 1 2 m 2m ε 0 + ε 4m 0 2 = ε. 1 2 m This implies that (ii) is satisfied. The assumption (iii) follows from (vi) and from the definition of T, T n, n N. Now we show the inclusion U n S n (M). The condition (viii) then implies that S n is the ijection of Sn 1 (U n ) onto U n and is fulfilled, since the continuity of Sn 1 Un is the consequence of the compactness of T n. Thus we have to prove that for each y U n there is an x y M such that x y T n (x y ) = y. This means that the map P n (x) = y + T n (x) has a fixed point for each y U n. The condition (vii) implies y (k) (t) + ( T n (x) ) (k) (t) rk y (k) (t) + + ( T n (x) ) (k) (t) rk ϕ n (t) + ϕ n (t) ϕ(t), t, and (P n (t)) (k) () = r k, k = 0,..., p 1, for each x M. Therefore P n (M) M. As M is a closed convex ounded set and P n (M) M is a compact map, y Tikhonov s fixed point theorem, P n has a fixed point. Now the function ϕ in Lemma 2 will e given as a solution of an integral equation. The existence and some properties of the solution to that equation will e discussed in the following two lemmas. Lemma 3. Let ψ H p 1, ω C([, ),[0, )), g C([0, ),[0, )) e a nondecreasing function. Further, let p 2 ( p 1 σ(t) = σ(t) 0, t <, if p = 1, and k=l+1 K(t, s) = ψ (k) (0) (k l)! (t )k l, t <, if p 2, (t s) p 1 l (p 1 l)! Then the following statements hold: (3), s t. (4)
ON THE INITIAL VALUE PROBLEM 423 1. A solution ϕ C([, ), [0, )) of the integral equation ϕ(t) = σ(t) + exists and satisfies K(t, s)ω(s)g( ψ + ϕ(s)) ds, t <, (5) 0 ϕ(t) λ(t), t <, (6) if and only if there exists a function λ C([, ), [0, )) such that λ(t) σ(t) + K(t, s)ω(s)g( ψ + λ(s)) ds, t <, (7) i.e., if and only if there exists an upper solution λ of (5). 2. A solution ϕ of (5) (whenever it exists) is a nondecreasing function in [, ). 3. If (5) has a solution ϕ and ω 1 C([, ), (0, )) satisfies 0 ω 1 (t) ω(t) for t <, then the equation ϕ(t) = σ(t) + K(t, s)ω 1 (s)g( ψ + ϕ(s)) ds (8) has a solution ϕ 1 such that 0 ϕ 1 (t) ϕ(t), t <. Proof. 1. The necessity is clear. To prove the sufficiency we shall proceed y the method of steps. Hence we prove y mathematical induction that for each m = 1, 2,... : (a) There exists a solution y m C([, + m], [0, )) of (5) satisfying the inequalities 0 y m (t) λ(t), t + m, and () y m+1 (t) = y m (t), t + m. Consider the partially ordered Banach space X 1 = C([, +1], R) with the sup-norm where z 1 z 2 if and only if z 1 (t) z 2 (t) for each t [, +1] and each pair z 1, z 2 from that space. Then, y definition, the interval z 1, z 2 = {y X 1 : z 1 (t) y(t) z 2 (t), t + 1. The operator U 1 : X 1 X 1 defined y U 1 (t) = σ(t) + K(t, s)ω(s)g( ψ + y(s)) ds, t + 1, is completely continuous, nondecreasing, and, in view of (7), maps the interval 0, λ [,+1] into itself. Hence y Schauder s fixed point theorem U 1 has a fixed point Y 1 satisfying (6) on [, + 1].
424 V. ŠEDA AND J. ELIAŠ Suppose now that there exists a solution y m of (5) on [, +m]. Consider the space X m+1 = C([, + m + 1], R) with the sup-norm and with the natural ordering. Let U m+1 e the operator given y the right-hand side of (5) on [, + m + 1]. U m+1 is completely continuous, nondecreasing, and maps the interval 0, λ [,+m+1] = {y X m+1 : 0 y(t) λ(t), t + m + 1 into itself. Similarly, U m+1 maps the closed and convex set Y m+1 ( = {x X m+1 : x(t) ) = y m (t), t + m into itself. Hence U m+1 0, λ [,+m+1] Y m+1 0, λ [,+m+1] Y m+1 and there exists a fixed point y m+1 of U m+1 in 0, λ [,+m+1] Y m+1. This is the sought-for function y m+1 with the properties (a) and (). Then the function ϕ(t) = y m (t) for t + m, m = 1, 2,..., is a solution of (5) in [, ), satisfying (6). 2. The statement follows from (3), (4) and (5). 3. Since each solution ϕ of (5) is an upper solution of (8), Statement 1 implies Statement 3. The existence of an upper solution λ of (5) is provided y Lemma 4. Let ψ, σ, and K have the same meaning as in Lemma 3 and let g C([0, ), (0, )) e a nondecreasing function. Then λ C([,, [0, )) is an upper solution of (5) if there is a function ρ C([, ), [0, )) such that and λ(t) = σ(t) + 0 ω(t) K(t, s)ρ(s) ds, t <, (9) ρ(t), t <. (10) g( ψ + λ(t)) Proof. By comining (9), (10) we get that λ determined y (9) is an upper solution of (5) in [, ). Remark 1. Similarly, the necessary and sufficient condition for λ C([,, [0, )) to e a solution of (5) is that there exists a function ρ C([,, [0, )) such that (9) is fulfilled and ω(t) = ρ(t) g( ψ + λ(t)), t <.
ON THE INITIAL VALUE PROBLEM 425 2. The Main Theorem The main theorem reads as follows. Theorem 1. Let ψ H p 1, f C([, ) H p 1 h 0, R d ). Let, further, ω C([, ), [0,, )), g C([0, ), [0, )) e a nondecreasing function, and let (ix) f(t, χ t,..., χ (p 1) t ω(t)g( χ t ) for each (t, χ) [, ) M, where M = { x X : x (k) (t) ψ (k) (0) ϕ(t) for t x (k) = ψ (k), k = 0,..., p 1, and and ϕ is a solution of equation (5) where the functions σ and K are determined y (3) and (4), respectively. Then the prolem (1), (2) has a solution x lying on M and the set F is a compact R δ -set in the space X. Proof. Consider the set M = { x X : x (k) (t) ψ (k) (0) ϕ(t) for t x (k) () = ψ (k) (0), k = 0,..., p 1. and Clearly, the restriction P : X X determined y P (x) = x [, ) is a homeomorphism of M onto M. Let the map T : m X e determined y + T (x)(t) = ψ (k) (0) (t ) k + k! (t s) p 1 f[s, (P 1 x) s,..., (P 1 x) (p 1) s ] ds, x M, t, (11) (p 1)! where P 1 is the inverse of P M. Then F = P 1 (F ), where F is the set of all fixed points of the map T. Since the homeomorphic image of the compact R δ -set is again a compact R δ -set, it is sufficient to prove that F is a compact R δ -set in the space X. This will e done y using Lemma 2 where
426 V. ŠEDA AND J. ELIAŠ we put r k = ψ (k) (0), k = 0,..., p 1. Due to (ix) the maps T n : M X determined y p 1 ψ (k) (0) k! (t ) k for t + 1 n, p 1 ψ (k) (0) T n (x)(t) = k! (t ) k + 1 n (t 1 n s)p 1 (p 1)! (12) f[s, (P 1 x) s,..., (P 1 x) (p 1) s ] ds for + 1 n t < and x M are, together with T, compact and, again y (ix), ( T n (x) ) (l) ( ) (l)(t) (t) T (x) p 1 (t s) p 1 l { p 1 1 n (t s) p 1 l (t 1 n s)p 1 l (p 1 l)! + t 1 n (p 1 l)! ω(s)g( (p 1 x) s )ds, t + 1 n, (t s) p 1 l (p 1 l)! ω(s)g( (p 1 x) s ) ω(s)g( (p 1 x) s )ds+ ds, + 1 n t <. By Lemma 3 ϕ is nondecreasing in [, ) and therefore (P 1 x) s ψ + ϕ(s) for each s <, x M. Hence, using also (4), we get ( T n (x) ) (l) ( ) (l)(t) (t) T (x) ϕn (t), t, x M, where ϕ n (t) = K(t, s)ω(s)g( ψ + ϕ(s))ds, t + 1 n, t K(t, s)ω(s)g( ψ + ϕ(s))ds 1 n K(t 1 n, s)ω(s) ( ψ + ϕ(s))ds, + 1 n t <, n N. Clearly, ϕ n C([, ), [0, )) and the relations ϕ n+1 (t) ϕ n (t), lim n ϕ n(t) = 0 can e proved for each t. Therefore these functions satisfy the assumptipons (v), (vi) of Lemma 2. Further, where ( T n (x) ) (l) (t) ϕ (l) (0) ϕ n (t), t, x M, ϕ n = σ(t), t + 1 n, σ(t) + 1 n K(t 1 n, s)ω(s)g( ψ + ϕ(s))ds, + 1 n t <,
ON THE INITIAL VALUE PROBLEM 427 n N, σ and K are the functions introduced y (3) and (4). The functions ϕ n are nonnegative continuous functions on [, ) and, y virtue of (5) we otain ϕ n (t) + ϕ n (t) = ϕ(t), t, n N. Hence the assumption (vii) of Lemma 2 is satisfied, too. Thus it remains for us to show that the assumption (viii) holds, and then Lemma 2 will imply the statement of Theorem 1. Let n N e aritrary ut fixed. If x, y M, x y, then there exists a t 0 [, ) such that x(t 0 ) = y(t 0 ). If t 0 + 1 n, then, taking into account (12), we otain x(t 0 ) T n (x)(t 0 ) y(t 0 ) T n (y)(t 0 ). In the other case there is a t 1 + 1 n such that t 1 = sup{τ > : x(t) = y(t) for t < τ. Then there exists t 0 (t 1, t 1 + 1 n ) such that x(t 0) = y(t 0 ). By (12) we now have + T n (x)(t 0 ) = p=1 ψ (k) t 0 1 n (0) (t 0 ) k + k! f[s, (P 1 x) s,..., (P 1 x) s (p 1) ]ds = t 0 1 n (t 0 1 n s)p 1 (p 1)! ψ (k) (0) (t 0 ) k + k! (t 0 1 n s)p 1 f[s, (P 1 y) s,..., (P 1 y) (p 1) s ]ds = T n (y)(t 0 ) (p 1)! and thus x(t 0 ) T n (x)(t 0 ) y(t 0 ) T n (y)(t 0 ), which we were to prove. References 1. J. Dugundji and A. Granas, Fixed point theory. PWN, Warszawa, 1982. 2. Z. Kuáček, Remarks on the paper of K. Czarnowski and T. Pruszko On the structure of fixed point sets.... Preprint. 3. V. Šeda and Š. Belohorec, A remark on second-order functional differential system (sumitted to Arch. Math.). (Received 15.04.1993) Authors addresses: Valter Šeda Jozef Eliaš Comenius University Slovak Technical University Department of Mathematical Analysis Department of Mathematics Mlynská dolina Mám. Sloody 17 842 15 Bratislava, Slovakia 812 31 Bratislava, Slovakia