Existence of solutions of infinite systems of integral equations in the Fréchet spaces

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1 Int. J. Nonlinear Anal. Al. 7 (216) No. 2, ISSN: (electronic) htt://dx.doi.org/1.2275/ijnaa Existence of solutions of infinite systems of integral equations in the Fréchet saces Reza Allahyari a, Reza Arab b,, Ali Shole Haghighi b a Deartment of Mathematics, Mashhad Branch, Islamic Azad University, Mashhad, Iran b Deartment of Mathematics, Sari Branch, Islamic Azad University, Sari, Iran (Communicated by A. Ebadian) Abstract In this aer we aly the technique of measures of noncomactness to the theory of infinite system of integral equations in the Fréchet saces. Our aim is to rovide a few generalization of Tychonoff fixed oint theorem and rove the existence of solutions for infinite systems of nonlinear integral equations with hel of the technique of measures of noncomactness and a generalization of Tychonoff fixed oint theorem. Also, we resent an examle of nonlinear integral equations to show the efficiency of our results. Our results extend several comarable results obtained in the revious literature. Keywords: Measure of noncomactness; Fréchet sace; Tychonoff fixed oint theorem; Infinite systems of equations. 21 MSC: Primary 47H9; Secondary 39B62, 34A Introduction The theory of infinite systems of integral equations is considered as an imortant branch of nonlinear analysis. In fact, infinite systems of integral equations are the natural generalization of infinite systems of differential equations which can arise in the theory of branching rocesses, the theory of neural nets, the theory of dissociation of olymers and real world roblems (cf. [29, 3, 31, 32, 33]). Also, infinite systems of integral equations are articular cases of integral equations in Banach saces which have been considered in many research aers [14, 15, 28, 32]. On the other hand, Measures of noncomactness are very useful tools in the theory of oerator equations in Banach saces. They are frequently used in the theory of functional equations, including Corresonding author addresses: rezaallahyari@mshdiau.ac.ir (Reza Allahyari), mathreza.arab@iausari.ac.ir (Reza Arab), ali.sholehaghighi@gmail.com (Ali Shole Haghighi) Received: December 215 Revised: Setember 216

2 26 Allahyari, Arab, Shole Haghighi ordinary differential equations, equations with artial derivatives, integral and integro-differential equations, otimal control theory, etc. In articular, the fixed oint theorems derived from them have many alications. There exists an enormous amount of considerable literature devoted to this subject (see for examle [2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 21, 22, 23, 24, 25, 26, 27, 28]). There have recently been many aers regarding the relationshi between the above concets, for examle, Arab et al. [11], Olszowy [27], Mursaleen and Mohiuddineb [23], Mursaleen and Alotaibi [24], Banaś and Lecko [15], Rzeka and Sadarangani [28] which discussed the solvability of infinite systems of differential and integral equations with the hel of measures of noncomactness. The aim of this aer is to give fixed oint theorems for condensing oerators in the Fréchet sace. Moreover, we study the roblem of the existence of solutions for infinite systems of integral equations of the form x n (t) = f n (t, x 1 (t),..., x n (t)) + k n (t, s)q n ((x i (s)) i= )ds. (1.1) We are going to show that Eq. (1.1) has solution that belongs to sace (L [, 1]) ω (denote the countable cartesian roduct of L [, 1] with itself). The obtained results extend several aers (see [3, 4, 5, 7, 8, 11, 12, 14], for examle). Finally, an examle is resented to show the efficiency of our results. 2. Preliminaries Here, we recall some basic facts concerning measures of noncomactness. Denote by R the set of real numbers and ut R + = [, + ). The symbol X, ConvX will denote the closure and closed convex hull of a subset X of E, resectively. Moreover, let N E indicate the family of all nonemty and relatively comact subsets of E. A toological vector sace (TVS) is a vector sace X over the field R which is endowed with a toology such that the mas (x, y) x + y and (α, x) αx are continuous from X X and R X to X. A toological vector sace is called locally convex if there is a basis for the toology consisting of convex sets (that is, sets A such that if x, y A then tx + (1 t)y A for < t < 1). Definition 2.1. [19] A Fréchet sace is a locally convex sace which is comlete with resect to a translation-invariant metric. Examle 2.2. Let E i be a Banach sace for all i N, then i N E i is a Fréchet sace by where x = (x 1, x 2,...), y = (y 1, y 2,...) i N E i. d(x, y) = su{ 1 2 i min{1, d i(x i, y i )} : i N}, Definition 2.3. [11] Let M be a class of subsets of a Fréchet sace E, we say M is an admissible set if N E M and if X M, then Conv(X), X M. Definition 2.4. [11] Let M be an admissible subset of a Fréchet sace E, we say that µ : M R + is a measure of noncomactness on Fréchet sace E if it satisfies the following conditions: (1 ) The family kerµ = {X M : µ(x) = } is nonemty and kerµ N E ;

3 Existence of solutions of infinite systems... 7 (216) No. 2, (2 ) X Y = µ(x) µ(y ); (3 ) µ(x) = µ(x); (4 ) µ(convx) = µ(x); (5 ) µ(λx + (1 λ)y ) λµ(x) + (1 λ)µ(y ) for λ [, 1]; (6 ) If {X n } is a sequence of closed sets from M such that X n+1 X n for n = 1, 2,, and if lim n µ(x n) =, then X = n=1x n. Theorem 2.5. (Darbo [14]) Let C be a nonemty, closed, bounded, and convex subset of the Banach sace E and F : C C be a continuous maing. Assume that there exists a constant k [, 1) such that µ(f X) kµ(x), for any nonemty subset of C. Then F has a fixed oint in C. Theorem 2.6. (Tychonoff fixed oint theorem [1]) Let E be a Hausdorff locally convex linear toological sace, C be a convex subset of E and F : C E be a continuous maing such that F (C) A C, with A comact. Then F has at least one fixed oint. Theorem 2.7. ([11]) Suose µ i be a measure of noncomactness on Banach saces E i for all i N. If we define M = {C E i : su{µ i (π i (C))} < }, i where π i (C) denotes the natural rojection of E i into E i and µ : M R + by µ(c) = su{µ i (π i (C)) : i N}, (2.1) then M is an admissible set and µ is a measure of noncomactness on X = E i. 3. Main result In this section, we state some main results in Fréchet saces which generalize and imrove Darbo s fixed oint theorem, Tychonoff fixed oint theorem, the mentioned corresonding results of Arab et al. [11], Aghajani et al. [3] and several authors (see [4, 5, 7, 8, 12, 14]) Theorem 3.1. Let Ω be a nonemty, closed and convex subset of a Fréchet sace E, M be an admissible set such that Ω M and µ : M R + be a measure of noncomactness on E. Also, suose that F : Ω Ω is a continuous maing such that ψ(µ(f X)) ϕ(µ(x)), (3.1) and F (X) M for any nonemty subset X M where ψ, ϕ : R + R + are given functions such that ψ is lower semicontinuous and ϕ is uer semicontinuous on R. Moreover, ψ() = ϕ() = and ψ(t) > ϕ(t) > for t >. Then F has at least one fixed oint and the set of all fixed oints of F in Ω is comact.

4 28 Allahyari, Arab, Shole Haghighi Proof. By induction, we obtain a sequence {Ω n } such that Ω = Ω and Ω n = Conv(F Ω n 1 ), n 1. It is obvious that Ω n M for all n N. If there exists an integer N such that µ(ω N ) =, then Ω N is comact. Thus, Theorem 2.6 imlies that F has a fixed oint. Now assume that µ(ω n ) for n. Since {µ(ω n )} is a ositive decreasing sequence of real numbers, then there exists r such that µ(ω n ) r as n and by (3.1), we have ψ(r) lim n ψ(µ(ω n+1 )) = lim n ψ(µ(conv(f (Ω n )))) = lim n ψ(µ(f (Ω n ))) lim n ϕ(µ(ω n )) ϕ(r). This result, ψ() = ϕ() = and ψ(t) > ϕ(t) > for t > imly that r =. Hence we deduce that µ(ω n ) as n. Since the sequence (Ω n ) is nested, in view of axiom (6 ) of Definition 2.4 we deduce that the set Ω = Ω n is nonemty, closed and convex subset of the set Ω. Moreover, n=1 the set Ω is invariant under the oerator F and belongs to Kerµ. Thus, alying Tychonoff fixed oint theorem, F has a fixed oint. To comlete the roof it remains to verify that µ(f F ) = where F F = {x Ω : F x = x}. Since F (F F ) = F F and by (3.1), we have ψ(µ(f F )) = ψ(µ(f (F F ))) ϕ(µ(f F )). Moreover, ψ(t) > ϕ(t) > for t >, so µ(f F ) = and F F is relatively comact and since F is a continuous function so the set of fixed oints of F in Ω is comact. Corollary 3.2. ([11]) Let Ω be a nonemty, closed and convex subset of a Fréchet sace E, M an admissible set such that Ω M and µ : M R + is a measure of noncomactness on E. Let F : Ω Ω be a continuous maing such that µ(f X) ϕ(µ(x)), (3.2) and F (X) M for any nonemty subset X M where ϕ : R + R + is uer semicontinuous and nondecreasing function such that ϕ(t) < t for t > and ϕ() =. Then F has at least one fixed oint in the set Ω. Proof. Take ψ(t) = t in Theorem 3.1. We introduce the following useful corollary which will be used in Section 4 and extend recent result of Arab et al. [11] Corollary 3.3. Let Ω i (i N) be a nonemty, convex and closed subset of a Banach sace E i, µ i an arbitrary measure of noncomactness on E i and su i {µ i (Ω i )} <. Let F i : Ω i Ω i (i N) be a continuous oerator such that ψ(µ i (F i ( X i ))) ϕ(su µ i (X i )), (3.3) i for any subset X i of Ω i (i N) where ψ, ϕ : R + R + satisfies the hyotheses of Theorem 3.1 and ψ is nondecreasing. Then there exist (x j) j=1 Ω j such that for all i N j=1 F i ((x j) j=1) = x i. (3.4)

5 Existence of solutions of infinite systems... 7 (216) No. 2, Proof. Assume that F : Ω i Ω i as follows for all (x j ) j=1 F ((x j ) j=1) = (F 1 ((x j ) j=1), F 2 ((x j ) j=1),..., F i ((x j ) j=1),...), Ω i. It is obvious that F is continuous. It suffices to show that the hyothesis (3.1) of Theorem 3.1 holds where µ is defined by Theorem 2.7. Take an arbitrary nonemty subset X of Ω i. Now, by (2 ) and (3.3) we obtain ψ(µ( F (X))) ψ(µ( F i ( π j (X)))) = su i su i su i j=1 ψ(µ i (F i (( π j (X))))) j=1 ϕ(su µ j (π j (X))) j ϕ(su(µ j (π j (X)))) j su ϕ(µ(x)) i ϕ(µ(x)). Therefore, all the conditions of Theorem 3.1 are satisfied, hence F has a fixed oint and there exist (x j) j=1 Ω j such that j=1 and that (3.4) holds. (x j) j=1 = F ((x j) j=1) = (F 1 ((x j) j=1), F 2 ((x j) j=1),..., F j ((x j) j=1),...) 4. Existence of solutions of infinite systems of integral equations In this section we are going to show how the result contained in section 3 can be alied to infinite systems of nonlinear integral equations. We will use a measure of noncomactness in the sace L [, 1]. In order to define this measure, take an arbitrary set X of M L [,1]. For x X and ε > let us ut ω(x, ε) = su{ τ h x x : h < ε}, ω(x, ε) = su{ω(x, ε) : x X} where τ h x(t) = { x(t + h) t + h 1 otherwise

6 21 Allahyari, Arab, Shole Haghighi for all t, h [, 1]. Moreover, ω (X) = lim ε ω(x, ε). It can be shown [14] that the maing ω = ω (X) is the measure of noncomactness in the sace L P [, 1]. Definition 4.1. A function f : R + R R is said to have the Carathéodory roerty if (i) For all x R the function t f(t, x) is measurable on R +. (ii) For almost all t R + the function x f(t, x) is continuous on R. Theorem 4.2. ( Minkowki s Inequality for Integrals) [6] Suose that (X, M, µ) and (Y, N, ν) are σ-finite measure saces, and let f be an (M N )-measurable function on X Y. If f and 1 <, then [ ( f(x, y)dν(y)) dµ(x) ] 1 ( f(x, y) dµ(x) Let us consider the Equation (1.1) under the following assumtions: dν(y). (a 1 ) f n : [, 1] R n R (n N) satisfies the Carathéodory conditions and f n (.,,..., ) L [, 1]. (a 2 ) There exist a non-decreasing, continuous and concave function φ : R + R + with φ(t) < t for all t >, φ() = and a L [, 1] such that f n (t, x 1,..., x n ) f n (s, y 1,..., y n ) a(t) a(s) + φ( max x i y i ), a.e. (4.1) for all n N (a 3 ) k n : [, 1] [, 1] R + (n N) is measurable if there exists g L [, 1] such that k n (t, s) g(t) for all t, s [, 1] and n N. (a 4 ) The oerator Q n (n N) acts continuously from the sace (L [, 1]) ω into L [, 1] and there exists a nondecreasing function ψ : R + R + such that Q n x ψ(su x i ) (4.2) for any x = (x i ) 1 L [, 1]) ω with su 1 i x i < and n N. (a 5 ) There exists a ositive solution r of the inequality φ(r ) + f n (., ) + ψ(r) K n r, (4.3) for all n N where (K n x)(t) = k n (t, s)x(s)ds. Theorem 4.3. Under assumtions (a 1 )-(a 5 ), the Equation (1.1) has at least a solution in the sace (L [, 1]) ω.

7 Existence of solutions of infinite systems... 7 (216) No. 2, Remark 4.4. Under the assumtions (a 3 ) and (a 5 ) the linear Fredholm integral oerator K n : L [, 1] L [, 1] (n N) is a continuous oerator. Proof. Let us fix arbitrarily n N. F n : (L [, 1]) ω L [, 1] (n N) is defined by F n ((x j ) j=1)(t) = f n (t, x 1 (t),, x n (t)) + k n (t, s)q n ((x i (s)) i= )ds. (4.4) By using conditions (a 1 ) (a 5 ) and since f n is concave, for arbitrary fixed t [, 1], we have Thus F n ((x j ) j=1)(t) f n (t, x 1 (t),, x n (t)) f n (t,,..., ) + f n (t,,..., ) + φ( max x i(t) ) + f n (t,,..., ) + k n (t, s)q n ((x i (s)) i= )ds a.e. k n (t, s)q n ((x i (s)) i= )ds F n ((x j ) j=1) φ( max x i ) + f n (., ) + K n ψ( (x i ) i= ). (4.5) Hence F n ((x j ) j=1) L [, 1] for any (x j ) j=1 (L [, 1]) ω and F n is well defined and from (4.5), we have F n ((B r ) ω ) B r, where r is a constant aearing in assumtion (4.3). Also, F n is continuous in (L [, 1]) ω because f n (t,.), K n and Q n are continuous for almost all t [, 1]. If we define k n,s : [, 1] R + by k n,s (t) := k n (t, s) for all s [, 1], then we show that ω ({k n,s : s [, 1]}) =. To do this, fix arbitrary ε >. We define the function ϑ : [, 1] R as follows ϑ(s) = k n (t, s) dt. (4.6) Since there exists g L [, 1] such that k n (t, s) g(t) for all t, s [, 1], so ϑ is continuous and there exists δ 1 > such that ϑ(s) ϑ(t) < ε for all s, t [, 1] with s t < δ. Moreover, there exist s 1,..., s m and δ 2 > such that [, 1] m B δ1 (s i ) and τ h k n,si k n,si ε where h δ 2. Since {k n,s1,..., k n,sm } is a comact subset of L [, 1] and ω ({k n,s1,..., k n,sm }) = for every s [, 1] and h δ 2, there exists s i such that s s i ε and ( τ h k n,s k n,s = k n (t, s) k n (t + h, s) dt So, we have ( k n (t, s) k n (t, s i ) dt + ( ( + k n (t + h, s) k n (t + h, s i ) dt 2 ϑ(s) ϑ(s i ) + τ h k n,si k n,si 2ε + ε. ω(k n,s, δ 2 ) 2ε + ε, ω({k n,s : s [, 1]}, δ 2 ) 2ε + ε, k n (t, s i ) k n (t + h, s i ) dt

8 212 Allahyari, Arab, Shole Haghighi and ω ({k n,s : s [, 1]}) =. Now, For any nonemty subset X j of B r for all j N, we claim that [ω (F n X)] φ([ω (X)] ). To verify this, let ε >, (x j ) j=1 X and t, h [, 1] with h ε, thus we have So, (F n (x j ) j=1)(t + h) (F n (x j ) j=1)(t) f n (t + h, x 1 (t + h),, x n (t + h)) + f n (t, x 1 (t),, x n (t)) k n (t + h, s)q n ((x i (s)) i= )ds k n (t, s)q n ((x i (s)) i= )ds f n (t + h, x 1 (t + h),, x n (t + h)) f n (t, x 1 (t),, x n (t)) + k n (t + h, s)q n ((x i (s)) i= )ds k n (t, s)q n ((x i (s)) i= )ds a(t + h) a(t) + φ[ max x i(t + h) x i (t) ] + k n (t + h, s) k n (t, s) Q n ((x i (s)) i= ) ds. Indeed, τ h F n (x j ) j=1 F n (x j ) j=1 ( = (F n (x j ) j=1)(t + h) (F n (x j ) j=1)(t) dt ( ( + a(t + h) a(t) dt + ( ( φ[ max x i(t + h) x i (t) ]) dt k n (t + h, s) k n (t, s) Q n ((x i (s)) i= ) ds dt ω(a, ε) + φ( max τ hx i x i + ω({kn,s : s [, 1]}, ε)ψ( (x i ) i= ). (4.7) ( ( φ[ max x i(t + h) x i (t) ]) dt ( = φ[ max x i(t + h) x i (t) ]dt ( = φ[ max x i(t + h) x i (t) dt] = [φ( max τ hx i x i )] 1,

9 Existence of solutions of infinite systems... 7 (216) No. 2, and by Minkowski s Inequality for Integrals, we have ( k n (t + h, s) k n (t, s) Q n ((x i (s)) i= ) ds dt = ( k n (t + h, s) k n (t, s) Q n ((x i (s)) i= ) dt ω(k n,s, ε) Q n ((x i (s)) i= ) ds ω({k n,s : s [, 1]}, ε) Q n (x i ) i= ω({k n,s : s [, 1]}, ε)ψ( (x i ) i= ). Therefore τ h F n (x j ) j=1 F n (x j ) j=1 ω(a, ε) + [φ( max τ hx i x i )] 1 + ω({kn,s : s [, 1]}, ε)ψ( (x i ) i= ) ( ω(a, ε) + φ( max [ω(x i, ɛ)] ) + ω({k n,s : s [, 1]}, ε)ψ( (x i ) i= ). By using the above estimate we have ω(f n ( X i ), ε) ω(a, ε) + ( φ( max 1 j n [ω(x j, ɛ)] ) ds + ω({k n,s : s [, 1]}, ε)ψ(r ). Since {a} is a comact set and ω ({k n,s : s [, 1]}) =, so we have ω(a, ε) and ω({k n,s : s [, 1]}, ε) as ε. Then we obtain [ω (F n ( X i ))] φ([ max ω (X i )] ). Now, by considering the functions ψ, ϕ : [, ) [, ) defined by we get ψ(t) = t, and ϕ(t) = φ(t ), ψ(ω (F n ( X i ))) ϕ( max ω (X i )). Thus from Corollary 3.3 the functional integral equation (1.1) has at least a solution in (L [, 1]) ω. Examle 4.5. Consider the following integral equation of the form x n (t) = 1 3 t + 1 i=n 1 t ln( x i (t) + 1) + (t s) 2 x i (s) ds n 2 i e x i(s) + n t 1 (4.8) In this examle, we have k n (t, s) = (t s) 2 χ E (t, s) (E = {(t, s) : s t 1}) f n (t, x 1, x 2,..., x n ) = 1 3 t + 1 i=n ln( x i + 1), n Q n (x i ) i= = 1 2 i x i e x i + n.

10 214 Allahyari, Arab, Shole Haghighi It can readily be seen that f n satisfies assumtion a 1 and hyothesis (a 2 ) with < 3, a(t) = 1 3 t and φ(t) = (ln t + 1), indeed, f n (t, x 1,..., x n ) f n (s, y 1,..., y n ) = 1 3 t 1 3 s + 1 i=n ln( x i + 1) ln( y i + 1) n 1 3 t 1 3 s + 1 n 1 3 t 1 3 s + 1 n i=n i=n ln( x i + 1 y i + 1 ) ln(1 + x i y i y i + 1 ) 1 3 t 1 + ln(1 + ( max 3 s x i y i ) = a(t) a(s) + φ( max x i y i ). Also, k n (t, s) (n N) satisfies hyothesis (a 3 ) with g(t) = t 2, k n (., s) L [, 1] for each s [, 1] and g L [, 1] and K n 1. Now, we show that Q n is continuous oerator of (L [, 1]) ω into L [, 1]. To establish this claim, let us fix x (L [, 1]) ω, n N and ε = 1 2 n and take arbitrary ((y j ) j=1 ) (L [, 1]) ω, such that su{ 1 2 i min{1, x i y i } : i N} < ε. Then we have Thus, where Q n (x i ) i= Q n (y i ) i= n 4ε + 4ε + Q n (x i ) i= Q n (y i ) i= 4ε x i y i + i e x i + n e y i + n n 1 2 ( x i y i + i e x i + n n i=n x i y i i e x i + n e y i + n y i e x i e y i (e x i + n)(e y i ) + n) 1 2 i ( x i y i + e x i e y i ). n 6ε + 2ϑ(ε), ϑ(ε) = su{ e x i e y i : 1 i n, y i b + ε}, 1 2 i ( x i y i + e x i e y i ) with b = su i { x i + ε}. Hence, we obtain ϑ(ε) as ε and Q n is a continuous oerator. Moreover, for all x (L [, 1]) ω we deduce Q n (x i ) i= 1 2 x i 1, i e x i + n and this oerator satisfies hyothesis (a 4 ) with ψ(t) = 1. On the other hand, with choosing [1, 3) we can comute r which satisfies the following inequality: 3 φ(r ) + f n (., ) + ψ(r) K n ln(r + 1) r.

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