On the Controllability of Non-Densely Defined Fractional Neutral Functional

Home , Functional derivative

International Journal of Pure and Applied Mathematics Volume 118 No. 11 2018, 257-276 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu doi: 10.12732/ijpam.v118i11.32 Special Issue ijpam.eu

On the controllability of non-densely deﬁned fractional neutral functional diﬀerential equations in Banach spaces

R. Subashini∗, S.Vimal Kumar†, S. Saranya‡ and C. Ravichandran§

Abstract In this paper, we investigates the controllability results for non-densely defined fractional neutral functional differential equations in Banach space by using fractional calculus theory and standard fixed point theorem. And also we showed the applications of abstract results.

Keywords: Fractional diﬀerential equations; Neutral equations; Controllability; Fixed point theorem. 2010 Mathematics Subject Classiﬁcation: 34A08

1 Introduction

In the recent years, the controllability plays an important role in the fractional calculus theory and has been extensively studied, the authors [3, 5, 13, 15, 20, 21, 25]. This is due to the fact that fractional differential equations have various applications in engineering and scientific disciplines, for example, fluid dynamics, fractal theory, diffusion in porous media, fractional biological neurons, traffic flow, polymer rheology, neural network modeling, viscoelastic panel in super sonic gas flow, real system characterized by power laws, electrodynamics of complex medium, sandwich system identification, nonlinear oscillation of earthquake, models of population growth, mathematical modeling of the diffusion of discrete particles in a turbulent fluid, nuclear reactors and theory of population dynamics. The details on the theory and its applications may be found in [8, 10, 11, 12, 14, 18, 22, 23, 24, 27, 28, 30]. Non-densely defined functional differential systems in abstract space with control system studied many authors [4, 6, 7]. Neutral fractional differential equations arise in many areas of

∗Research and Development Center, Bharathiar University, Coimbatore - 641046, Tamilnadu, India and Department of Mathematics, GTN Arts College, Dindigul - 624004, Tamilnadu, India. E. Mail-: sub- [email protected] †Department of Mathematics, RVS Technical Campus, Coimbatore-641 402, India. E. Mail-: svimalku- [email protected] ‡Department of Mathematics, Vignan’s Foundation For Science, Technology and Research, Guntur-522213, Andra Pradesh, India. E. Mail: [email protected] §PG & Research Department of Mathematics, Kongunadu Arts & Science College, Coimbatore - 641 029, Tamil Nadu, India. Email: [email protected]

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applied mathematics and for this reason these equations have received much attention in the last decades [1, 19, 26, 29]. But the literature related to neutral fractional diﬀerential equations with non-densely is very limited and we refer the reader to [9]. In this paper, we study the controllability of the integral solution for the fractional semilinear functional diﬀerential equations of the form

cDq[x(t) f(t, x )] = Ax(t) + Cu(t) + g(t, x ), t (0, b] (1.1) − t t ∈ x (s) + (h(x , x , ..., x ))(s) = φ(s), s [ r, 0]. 0 t1 t2 tn ∈ − where cDq is the Caputo fractional derivative of order 0 < q < 1, b > 0, r > 0. A : D(A) E ⊂ → E is a non-densely closed linear operator on E, E is a real Banach space with the norm . The |·| control function u( ) is given in L2([0, b],U), a Banach space of admissible control functions · with U as a Banach space. is a bounded linear operator from U to E. Let ([ r, 0],E) C C − denote the space of all continuous functions with the sup-norm φ c = maxs [ r,o] φ(s) , g, f : k k ∈ − | | [0, ) ([ r, 0],E) E and h :( ([ r, 0],E))n E are given functions are satisfying some ∞ × C − → C − → assumption, φ ([ r, 0],E) and deﬁne x ([ r, 0],E) by x (s) = x(t + s), s [ r, 0], for ∈ C − t ∈ C − t ∈ − t [0, b]. ∈

2 Back ground results

In this section, we introduce notations, deﬁnitions, and preliminary results which are used in the rest of the paper. We denoted by ([0, b],E) a Banach space of all continuous functions C from [0, b] into E with the norm x = sup y(t) : t [0, b] . B(E) denotes the Banach || || {| | ∈ } space of bounded linear operators from E into E, with the norm N = sup N(y) : y = 1 . || || {| | | | } Assume that J R, and 1 p . For measurable functions m : J R, deﬁne the norm ⊂ ≤ ≤ ∞ → 1 p p J m(t) dt , 1 p < m p = | | ≤ ∞ || ||L J  infR p(J)=0 supt J J¯ m(t) , p = ,  ∈ − | | ∞ where µ(J¯) is the Lebesgue measure on J¯. Let LP (J, R) be the Banach space of all Lebesgue

measurable functions m : J R with p < . → || · ||L J ∞ 1 1 r Lemma 2.1. (Holder’s inequality). Assume that r, p 1, and r + p = 1. If l L (J, R), then 1 ≥ ∈ for 1 p , lm L (J, R) and lm 1 l r m p . ≤ ≤ ∞ ∈ k kL J ≤ k kL J k kL J Lemma 2.2. (Bochner Theorem). A measurable function H : [0, b] E is Bochner’s integrable → if H is Lebesgue integrable. | | Lemma 2.3. (Schauder’s Fixed Point Theorem). If is a closed bounded and convex subset B of a Banach space E and F : is completely continuous, then F has a ﬁxed point in . B → B B Lemma 2.1 - 2.3 are classical, which can be found in many books.

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Deﬁnition 2.1. [2] Let E be a Banach space. An integrated semigroup is a family of operators

S(t) t 0 of bounded linear operators S(t) on E with the following properties: { } ≥ (i) S(0) = 0;

(ii) t S(t) is strongly continuous; → (iii) S(s)S(t) = s(S(t + r) S(r))dr for all t, s 0. 0 − ≥ Deﬁnition 2.2. R[10] An operator A is called a generator of an integrated semigroup, if there exists ω R such that (ω, + ) ρ(A) and there exists a strongly continuous exponentially ∈ ∞ ⊂ 1 bounded family (S(t))t 0 of linear bounded operator such that S(0) = 0 and (λI A)− = ≥ λt − λ 0∞ e− S(t)dt for all λ > ω.

DefinitionR 2.3. [2] Let A be the generator of an integrated semigroup (S(t))t 0. Then for all ≥ x E and t 0, t S(s)xds D(A) and S(t)x = A t S(s)xds + tx. ∈ ≥ 0 ∈ 0 Definition 2.4. [10]R We say that linear operator A satisfiesR the Hille-Yosida condition if there exist M 0 and ω R such that (ω, + ) ρ(A) and sup (λ ω)n R(λ, A)n , n N, λ > ω ≥ ∈ ∞ ⊂ { − k k ∈ } M. Here and hereafter, we assume that A satisfies the Hille-Yosida condition. Let us intro- ≤ duce the part A0 of A in D(A): A0 = A on D(A0) = x D(A); Ax D(A) . Let (S(t))t 0 ∈ ∈ ≥ be the integrated semigroup generated by A. We note thatn (S0(t))t 0 is a C0-semigroupo on D(A) ≥ generated by A and S (t) Meωt, t 0, where M and ω are the constants considered in the 0 k 0 k ≤ ≥ Hille-Yosida condition. Semiflows generated by Lipschitz perturbations of non-densely defined operators. Let B = λR(λ, A) := λ(λI A) 1, Then for all x D(A), B x x as λ . Also from λ − − ∈ λ → → ∞ the Hille-Yosida condition it is easy to see that limλ Bλx M x . →∞ | | ≤ | | Definition 2.5. [18] The fractional integral of order α R+ with the lower limit zero for a ∈ function f : R+ E is defined as → 1 t f(s) Iαf(t) = ds, t > 0, 0 < α < 1, Γ(α) (t s)1 α Z0 − − provided the right side is point-wise defined on R+, where Γ( ) is the Gamma function. · Definition 2.6. [18] Riemann-Liouville derivative of order α R+ with the lower limit zero ∈ for a function f : R+ E is defined by → 1 d t f(s) LDαf(t) = ds, t > 0, 0 < α < 1, Γ(n α) dt (t s)α − Z0 − provided the right side is point-wise defined on R+, where Γ( ) is the Gamma function. · Definition 2.7. [18] The Caputo derivative of order 0 < α < 1 of a continuous function f : R+ E is defined by → 1 t f 0 (s) cDαf(t) = ds, t > 0. Γ(n α) (t s)α − Z0 − 3

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3 Main results

Deﬁnition 3.1. Let φ ([ r, 0],E), a function x :[ r, b] E be an integral solution of ∈ C − − → problem (1.1) if

(i) x is continuous on the interval [0,b];

(ii) 1 t(t s)q 1x(s)ds D(A) for t [0, b]; Γ(q) 0 − − ∈ ∈ (iii) R

φ(0) (h(x , x , ..., x ))(0) f(0, x ) + f(s, x ) − t1 t2 tn − 0 s x(t) =  + 1 A t(t s)q 1x(s)ds + 1 t(t s)q 1[Cu(s) + g(s, x )]ds, t [0, b],  Γ(q) 0 − Γ(q) 0 − s  − − ∈ φ(t), t R[ r, 0], R ∈ −   Lemma 3.1. If x is an integral solution of (1.1), then for all t [0, b], x(t) D(A). In ∈ ∈ particular, x(0) = φ(0) D(A). ∈ Proof. The proof is similar to the proof of the Lemma 2.7 in [17].

1 n 1 qn 1 Γ(nq+1) + Lemma 3.2. [16] Let ψ (θ) = ∞ ( 1) θ sin(nπq), θ R , then ψ (θ) is q π n=1 − − − − n! ∈ q a one-sided stable probability densityP function and its Laplace transform is given by

∞ pθ pq e− ψ (θ)dθ = e− , q (0, 1), p > 0. (3.1) q ∈ Z0 Lemma 3.3. For t [0, b], the integral solution in Deﬁnition 3.1 is given by ∈

∞ q x(t) = hq(θ)S0(t θ)[φ(0) (h(xt1 , xt2 , ..., xtn ))(0) f(0, x0)]dθ 0 − − Z t ∞ q 1 q + f(t, xt) + q θ(t s) − hq(θ)AS0((t s) θ)f(s, xs)dθds 0 0 − − t Z Z ∞ q 1 q + lim q θ(t s) − hq(θ)S0((t s) θ)Bλ[Cu(s) + g(s, xs)]dθds, λ − − →∞ Z0 Z0 1 1 1 1 q q + where hq(θ) = q θ− − ψq(θ− ) is the probability density function deﬁned on R .

Proof. From the deﬁnition, we have

x(t) =φ(0) (h(x , x , ..., x ))(0) f(0, x ) + f(s, x ) − t1 t2 tn − 0 s t t 1 q 1 1 q 1 + A (t s) − x(s)ds + (t s) − [Cu(s) + g(s, x )]ds, t [0, b] (3.2) Γ(q) − Γ(q) − s ∈ Z0 Z0

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Consider the Laplace transform

∞ pt ∞ pt v(p) = e− Bλx(t)dt, χ(p) = e− Bλf(s, xs)dt, Z0 Z0 ∞ pt w(p) = e− Bλ[Cu(s) + g(s, xs)]dt, p > 0. (3.3) Z0 Note that for each t > 0, B [Cu(s) + g(s, x )] D(A), then we have λ s ∈ 1 1 1 v(p) = [B φ(0) (h(x , x , ..., x ))(0) f(0, x )] + χ(p) + Av(p) + w(p) p λ − t1 t2 tn − 0 pq pq q 1 q 1 = p − (p I A)− [B φ(0) (h(x , x , ..., x ))(0) f(0, x )] − λ − t1 t2 tn − 0 q q 1 q 1 + p (p I A)− χ(p) + (p I A)− w(p) − − q 1 ∞ pqs = p − e− S0(s)B [φ(0) (h(x , x , ..., x ))(0) f(0, x )]ds λ − t1 t2 tn − 0 Z0 q ∞ pqs ∞ pqs + p e− S0(s)χ(p)ds + e− S0(s)w(p)ds (3.4) Z0 Z0 where I is the identity operator deﬁned on E. From (3.1), we get

q 1 ∞ pqs p − e− S0(s)B [φ(0) (h(x , x , ..., x ))(0) f(0, x )] ds λ − t1 t2 tn − 0 Z0 ∞ q 1 (pt)q q = q(pt) − e− S0(t )B [φ(0) (h(x , x , ..., x ))(0) f(0, x )] dt λ − t1 t2 tn − 0 Z0 ∞ 1 d (pt)q q = − [e− S0(t )B [φ(0) (h(x , x , ..., x ))(0) f(0, x )] dt p dt λ − t1 t2 tn − 0 Z0 ∞ ∞ ptθ q = θψq(θ)e− S0(t )Bλ [φ(0) (h(xt1 , xt2 , ..., xtn ))(0) f(0, x0)] dθdt 0 0 − − Z Z q ∞ pt ∞ t = e− ψ (θ)S0 B [φ(0) (h(x , x , ..., x ))(0) f(0, x )] dθ dt, q θ λ − t1 t2 tn − 0 Z0 Z0 (3.5)

∞ pqs e− S0(s)w(p)ds Z0 ∞ ∞ pqs pt = e− e− S0(s)Bλ[Cu(s) + g(s, xs)]dtds Z0 Z0 ∞ ∞ q 1 (ps)q pt q = qs − e− e− S0(s )Bλ[Cu(s) + g(s, xs)]dtds Z0 Z0 ∞ ∞ ∞ psθ pt q = qψq(θ)e− e− S0(s )Bλ[Cu(s) + g(s, xs)]dθdtds Z0 Z0 Z0 q 1 q ∞ ∞ ∞ p(s+t) s − s = qψq(θ)e− q S0 Bλ[Cu(s) + g(s, xs)]dθdtds 0 0 0 θ θ Z Z Z t q 1 q ∞ pt ∞ (t s) − (t s) = e− q ψ (θ) − S0 − B [Cu(s) + g(s, x )]dθdsdt q θq θq λ s Z0 Z0 Z0 (3.6)

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and

q ∞ pqs p e− S0(s)χ(p)ds Z0 (pt)q ∞ q (ps) ∞ ∞ ∞ q 1 (λt)q q ps = e− S0(t )e− f(s, x )ds + qt − e− AS0(t )e− f(s, x )dsdt − s s Z0 t=0 Z0 Z0 t q q ∞ pt ∞ (t s) (t s) = e− f(t, x ) + q ψ (θ)AS0 − f(s, x ) − dθds dt. (3.7) t q θq s θq Z0 Z0 Z0 According to (3.4)-(3.7), we have

q ∞ pt ∞ t v(p) = e− ψ (θ)S0 B [φ(0) (h(x , x , ..., x ))(0) f(0, x )] dθ dt q θ λ − t1 t2 tn − 0 Z0 Z0 t q 1 q ∞ pt ∞ (t s) − (t s) + e− q ψ (θ) − S0 − B [Cu(s) + g(s, x )]dθdsdt q θq θq λ s Z0 Z0 Z0 t q q ∞ pt ∞ (t s) (t s) + e− f(t, x ) + q ψ (θ)AS0 − f(s, x ) − dθds dt. t q θq s θq Z0 Z0 Z0 Inverting the last Laplace transform, we obtain

q ∞ t B x(t) = ψ (θ)S0 B [φ(0) (h(x , x , ..., x ))(0) f(0, x )] dθ λ q θ λ − t1 t2 tn − 0 Z0 t q q ∞ (t s) (t s) + f(t, x ) + q ψ (θ)AS0 − f(s, x ) − dθds t q θq s θq Z0 Z0 t q 1 q ∞ (t s) − (t s) + q ψ (θ) − S0 − B [Cu(s) + g(s, x )]dθds q θq θq λ s Z0 Z0 ∞ q = hq(θ)S0(t θ)Bλ [φ(0) (h(xt1 , xt2 , ..., xtn ))(0) f(0, x0)] dθ 0 − − Z t ∞ q 1 q + f(t, xt) + q θ(t s) − hq(θ)AS0((t s) θ)f(s, xs)dθds 0 0 − − t Z Z ∞ q 1 q + q θ(t s) − h (θ)S0((t s) θ)B [Cu(s) + g(s, x )]dθds. − q − λ s Z0 Z0

In view of limλ Bλx = x for x D(A) and Lemma 3.1, we have for t [0, b]. →∞ ∈ ∈

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Remark 3.1. One can verify that

1 1 ∞ θh (θ)dθ = ∞ ψ (θ)dθ = (3.8) q θq q Γ(1 + q) Z0 Z0 Lemma 3.4. [29] For any x E, β (0, 1) and η (0, 1], we have AT (t)x = A1 βT (t)Aβx, ∈ ∈ ∈ q − q qCn Γ(2 η 0 t a, and ATq(t) tqn Γ(1+q(1− η)) , 0 < t a. ≤ ≤ | | ≤ − ≤ Deﬁnition 3.2. The problem (1.1) is said to be controllable on the interval [0, b] if for every initial function π ([ r, 0],E) with π(0) D(A) and x D(A), there exists a control ∈ C − ∈ 1 ∈ u L2([0, b],U) such that the integral solution x(t) of (1.1) satisﬁes x(b) = x . we are now ∈ 1 in a position to state and prove our main results for the controllability of solutions of problem (1.1). Let us list the following assumptions:

(A1) A satisﬁes the Hille-Yosida condition.

(A2) The operator S0(t) is compact D(A) whenever t > 0 and satisﬁes supt [0, ] S0(t) = ∈ ∞ || || M < , where M is a constant. 0 ∞ 0 (A ) For each t [0, b], the function g(t, ): ([ r, 0],E) E is continuous and for each 3 ∈ · C − → v ([ r, 0],E), the function g( , v) : [0, b] E is strongly measurable. ∈ C − · → (A ) There exists a constant q [0, q) and a function m L1/q1 ([0, b],R+) such that 4 1 ∈ ∈ g(t, v) m(t) for all v ([ r, 0],E) and almost all t [0, b]. | | ≤ ∈ C − ∈ (A ) The function h : [0, b] E is continuous and there exists a constant β (0, 1) and 5 × C → ∈ H,H > 0 such that h D(Aβ) and for any x, y , the function Aβf( , x) is strongly 1 ∈ ∈ C · measurable and Aβf(t, ) satisﬁes the Lipschitz condition Aβf(t, x) Aβf(t, y) H x y β · − ≤ k − k and the inequality A f(t, x) H1( x + 1). ≤ k k

(A6) There exists a constant L > 0 such that h(xt1 , xt2 , ..., xtn ) h(yt1 , yt2 , ..., ytn ) k − k ≤ L x y , for x, y ([ r, b],E). k − k ∈ C − (A ) The linear operator W : L2([0, b],U) D(A) deﬁne by 7 → b ∞ q 1 q W u = lim q θ(b s) − hq(θ)S0((b s) θ)Bλ u(s)dθds λ − − C →∞ Z0 Z0 1 2 has an induced inverse operator W − which takes values in L ([0, b],U)/kerW and there 1 exists positive constants M1 and M2 such that M1 and W − M2. For each f ||C|| ≤ || || ≤ positive constant k, Let Bk = x ([0, b], D(A); x k), then Bk is clearly a bounded ∈ C || || ≤ f closed and convex subset in ([0, b],E). C Theorem 3.1. Let φ ([ r, 0],E) with φ(0) D(A). If the assumption (A ) - (A ) are ∈ C − ∈ 1 7 satisﬁed, then the problem (1.1) is controllable on the interval [0,b].

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Proof. Using assumption (A ), for arbitrary function x( ) and x D(A), we deﬁne the control 5 · 1 ∈

1 ∞ q ux(t) = W − x1 hq(θ)S0(b θ)[φ(0) (h(xt1 , xt2 , ..., xtn ))(0) f(0, x0)]dθ − 0 − − h Z b f ∞ q 1 q f(t, xt) q θ(b s) − hq(θ)AS0((b s) θ)f(s, xs)dθds − − 0 0 − − b Z Z ∞ q 1 q lim q θ(b s) − hq(θ)S0((b s) θ)Bλg(s, xs)dθds (t). (3.9) − λ 0 0 − − →∞ Z Z i Using this control we will prove that the operator F : B ([0, b], D(A)) deﬁned by k → C

∞ q (F x)(t) = hq(θ)S0(t θ)[φ(0) (h(xt1 , xt2 , ..., xtn ))(0) f(0, x0)]dθ 0 − − Z t ∞ q 1 q + f(t, xt) + q θ(t s) − hq(θ)AS0((t s) θ)f(s, xs)dθds 0 0 − − t Z Z ∞ q 1 q + lim q θ(t s) − hq(θ)S0((t s) θ)Bλ[Cus(s) + g(s, xs)]dθds, t [0, b] λ − − ∈ →∞ Z0 Z0 has a ﬁxed point x( ). Then it is easy to check that x(b) = (F x)(b) = x , which implies that the · 1 system (1.1) is controllable. We will ﬁrstly show that for a positive constant k, the operator

F is well deﬁned on Bk. In fact, in view of supt [0, ] S0(t) = M0 < , for any positive ∈ ∞ || || ∞ constant k and x B , we have ∈ k

∞ q h (θ)S0(t θ)[φ(0) (h(x , x , ..., x ))(0) f(0, x )]dθ q − t1 t2 tn − 0 Z0 β β M φ(0) + L x v + h(x , x , ..., x )(0) + A− A f(0, x ) ≤ 0 | | k − k | t1 t2 tn | | 0 | β M φ(0) + Lk + h(x , x , ..., x )(0) + A− H (k + 1) . (3.10) ≤ 0 | | | t1 t2 tn | | | 1 For x Bk, since x(t) is continuous in t, according to (A2), g(t, xt) is a measurable function ∈ 1 q 1 1 q on [0, b]. Direct calculation show that (t s) − L − 1 [0, t], for t [0, b] and q1 [0, q). Let q 1 − ∈ ∈ ∈ a = − ( 1, 0), M = m 1 . 1 q1 3 − ∈ − k kL q1 [0,t] By using Lemma 2.1 and (A ), for t [0, b], we have 4 ∈

1 q1 t t q 1 − q 1 1 −q (t s) − g(s, xs) ds (t s) − 1 ds m 1 − ≤ − k kL q1 [0,t] Z0 Z0

M3 (1+a)(1 q1) b − . (3.11) 1 q1 ≤ (1 + a) −

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From (A ), (3.8),(3.10) and by the fact that B M, we get 2 k λk ≤ t ∞ q 1 q θ(t s) − h (θ)S0((t s) θ)B g(s, x )dθ ds − q − λ s Z0 Z0 t ∞ q 1 MM θh (θ) (t s) − g(s, x )dθ ds ≤ 0 q − s Z0 Z0 t MM0 q 1 (t s) − g(s, x ) ds ≤ Γ(1 + q) − s Z0

MM0M3 (1+a)(1 q1) b − , for t [0, b]. (3.12) 1 q1 ≤ Γ(1 + q)(1 + a) − ∈

For x B , according to (A ), (A ) and (3.9), we have ∈ k 2 5 t ∞ q 1 q θ(t s) − h (θ)S0((t s) θ)B Cu (s)dθ ds − q − λ x Z0 Z0 t MM0M1 q 1 (t s) − ux(s) U ds ≤ Γ(1 + q) 0 − | | Z q MM0M1M2b qMM0M3 (1+a)(1 q1) x1 + M0 φ(0) + b − , (3.13) ≤ qΓ(1 + q) | | | | Γ(1 + q)(1 + a)1 q1 − where symbol denotes the norm. Therefore | · |U

∞ q 1 q θ(t s) − h (θ)S0((t s) θ)B [Cu (s) + g(s, x )]dθ − q − λ x s Z0

is Lebesgue integrable with respect to s [0, t] for all t [0, b]. From Lemma 2.2, we get ∈ ∈

∞ q 1 q θ(t s) − h (θ)S0((t s) θ)B [Cu (s) + g(s, x )]dθ − q − λ x s Z0 is Bochner integrable with respect to s [0, t] for all t [0, b]. Note that ∈ ∈ t q 1 q 1 (t s) − (λI A)S0((t s) θ)(λI A)− [Cu (s) + g(s, x )]ds D(A), − − − − x s ∈ Z0 we get

t q 1 q 1 lim Bλ (t s) − (λI A)S0((t s) θ)(λI A)− [Cux(s) + g(s, xs)]ds λ − − − − →∞ 0 Z t q 1 q 1 = lim Bλ(λI A) (t s) − S0((t s) θ)(λI A)− [Cux(s) + g(s, xs)]ds λ − − − − →∞ 0 t Z q 1 q = lim (t s) − S0((t s) θ)Bλ[Cux(s) + g(s, xs)]ds λ − − →∞ Z0

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exists. Thus,

t ∞ q 1 q q θhq(θ) lim (t s) − S0((t s) θ)Bλ[Cux(s) + g(s, xs)]dsdθ λ − − 0 →∞ 0 Z Z t ∞ q 1 q = lim q θhq(θ) (t s) − S0((t s) θ)Bλ[Cux(s) + g(s, xs)]dsdθ λ − − →∞ 0 0 Z t Z ∞ q 1 q = lim q θ(t s) − hq(θ)S0((t s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ − − →∞ Z0 Z0

exists. Using above the equation and (3.10) we conclude that F is well deﬁned on Bk. According

to [29] and hypothesis (A5), we obtain the following relation

t ∞ q 1 q θ(t s) − h (θ)S0((t s) θ)f(s, x )dθ ds − q − s Z0 Z0 t 1 q 1 1 β q β (t s) − A − S0((t s) θ)A f (s, xs) ds ≤ Γ(1 + q) 0 − − Z 1 Γ(1 + β) qβ C1 βH1(k + 1)b . (3.14) ≤ Γ(1 + q) βΓ(1 + qβ) −

Next, we will prove F x = x has a ﬁxed point on Bk, which is equivalent to determining a

positive constant k0 such that F has a ﬁxed point on Bk0 . In fact, by choosing

β β k =M φ(0) + Lk + h(x , x , ..., x )(0) + A− H (k + 1) + A− H (k + 1) 0 0 | | | t1 t2 tn | | | 1 | | 1 1 Γ(1 + β) qβ + C1 βH1(k + 1)b Γ(1 + q) βΓ(1 + qβ) − q MM0M1M2b qMM0M3 (1+a)(1 q1) + x1 + M0 φ(0) + b − qΓ(1 + q) | | | | Γ(1 + q)(1 + a)1 q1 − MM0M3 (1+a)(1 q1) + b − , 1 q1 Γ(1 + q)(1 + a) −

we can prove that F has a ﬁxed point on Bk0 . The proof will be divided into ﬁve steps. Step 1. For x B , F x k . ∈ k0 || || ≤ 0 Let x B and t [0, b], according to (3.10), (3.12) and (3.13), we obtain ∈ k0 ∈

β β (F x)(t) M φ(0) + Lk + h(x , x , ..., x )(0) + A− H (k + 1) + A− H (k + 1) | | ≤ 0 | | | t1 t2 tn | | | 1 | | 1 1 Γ(1 + β) qβ + C1 βH1(k + 1)b Γ(1 + q) βΓ(1 + qβ) − q MM0M1M2b qMM0M3 (1+a)(1 q1) + x1 + M0 φ(0) + b − qΓ(1 + q) | | | | Γ(1 + q)(1 + a)1 q1 − MM0M3 (1+a)(1 q1) + b − 1 q1 Γ(1 + q)(1 + a) − k . ≤ 0 Hence, F x k for every pair x B . || || ≤ 0 ∈ k0

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Step 2. F is continuous on Bk0 . n n Let x , x Bk (n = 1, 2, ...), such that limn x x = 0. From the deﬁnition of , we 0 →∞ ⊂ n || − || n || · || have limn xt = xt, t [0, b]. Then from (H3), we have limn g(t, xt ) = g(t, xt), t [0, b]. →∞ →∞ ∈ n ∈ It follows that sups [0,b] g(s, xs ) g(s, xs) 0, as n . On the other hand, for t [0, b], ∈ | − | → → ∞ ∈

u n (t) u (t) | x − x |U b 1 ∞ q 1 q n = W − lim q θ(b s) − hq(θ)S0((b s) θ)Bλ (g(s, xs ) g(s, xs)) dθds (t) λ − − − →∞ Z0 Z0 U b f ∞ q 1 q n M 2 lim q θ(b s) − hq(θ)S0((b s) θ)Bλ (g(s, xs ) g(s, xs)) dθds ≤ λ − − − →∞ Z0 Z0 q MM 0M2b n sup g(s, xs ) g(s, xs) . ≤ Γ(1 + q) s [0,b] | − | ∈ Therefore, for t [0, b], ∈ (F xn)(t) (F x)(t) | − | t ∞ q 1 q n = lim q θ(t s) − hq(θ)S0((t s) θ)Bλ (g(s, xs ) g(s, xs)) dθds λ − − − →∞ 0 0 Z Z t ∞ q 1 q + lim q θ(t s) − hq(θ)S0((t s) θ)Bλ (Cuxn (s) Cux(s)) dθds λ − − − →∞ 0 0 Zt Z qMM 0 q 1 n (t s) − g(s, xs ) g(s, xs) ds ≤ Γ(1 + q) 0 − | − | Z t qMM0M1 q 1 + (t s) − uxn (s) ux(s) ds Γ(1 + q) 0 − | − | q Z q MM0b n MM0M1b sup g(s, xs ) g(s, xs) + sup uxn (s) ux(s) , ≤ Γ(1 + q) s [0,b] | − | Γ(1 + q) s [0,b] | − | ∈ ∈ which implies F xn F x 0, as n . This is show that F is continuous. k − k → → ∞ Step 3. The family F x, x B is equicontinuous. { ∈ k0 } For any x B and 0 t t a, we have ∈ k0 ≤ 1 ≤ 2 ≤ (F x)(t ) (F x)(t ) 2 − 1

∞ q = hq(θ)S 0(t2θ)[φ(0) (h(xt1 , xt2 , ..., xtn ))(0) f(0, x0)]dθ 0 − − Z ∞ q hq(θ)S0(t θ)[φ(0) (h(xt , xt , ..., xt ))(0) f(0, x0)]dθ − 1 − 1 2 n − Z0 + f(t , x ) f(t , x ) 2 t − 1 t t2 ∞ q 1 q + q θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds 2 − q 2 − s Z0 Z0 t1 ∞ q 1 q q θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds − 1 − q 1 − s Z0 Z0

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t2 ∞ q 1 q + lim q θ(t2 s) − hq(θ)S0((t2 s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ − − →∞ Z0 Z0 t1 ∞ q 1 q lim q θ(t1 s) − hq(θ)S0((t1 s) θ)Bλ[Cux(s) + g(s, xs)]dθds − λ 0 0 − − →∞ Z Z ∞ q q hq(θ) S0(t θ) S0(t θ) [φ(0) (h(xt , xt , ..., xt ))(0) f(0, x0)]dθ ≤ 2 − 1 − 1 2 n − Z0 t2 ∞ q 1 q + q θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds 2 − q 2 − s Zt1 Z0 t1 ∞ q 1 q + q θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds 2 − q 2 − s Z0 Z0 t1 ∞ q 1 q q θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds − 1 − q 2 − s Z0 Z0 t1 ∞ q 1 q + q θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds 1 − q 2 − s Z0 Z0 t1 ∞ q 1 q q θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds − 1 − q 1 − s Z0 Z0 t2 ∞ q 1 q + lim q θ(t2 s) − hq(θ)S0((t2 s) θ)Bλ[Cux (s) + g(s, xs)]dθds λ − − →∞ Zt1 Z0 t1 ∞ q 1 q + lim q θ(t2 s) − hq(θ)S0((t2 s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ − − →∞ Z0 Z0 t1 ∞ q 1 q lim q θ(t1 s) − hq(θ)S0((t2 s) θ)Bλ[Cux(s) + g(s, xs)]dθds − λ − − →∞ Z0 Z0 t1 ∞ q 1 q + lim q θ(t1 s) − hq(θ)S0((t2 s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ − − →∞ Z0 Z0 t1 ∞ q 1 q lim q θ(t1 s) − hq(θ)S0((t1 s) θ)Bλ[Cux(s) + g(s, xs)]dθds − λ 0 0 − − →∞ Z Z + f(t , x ) f(t , x ) | 2 t − 1 t | ∞ q q = h (θ) S0(t θ) S0(t θ) [φ(0) (h(x , x , ..., x ))(0) f(0, x )]dθ q 2 − 1 − t1 t2 tn − 0 Z0 t2 ∞ q 1 q + q θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds 2 − q 2 − s Zt1 Z0 t1 ∞ q 1 q 1 q + q θ (t s) − (t s) − h (θ)AS0((t s ) θ)f(s, x )dθds 2 − − 1 − q 2 − s Z0 Z0 t1 ∞ q 1 q q + q θ(t s) − h (θ)A S0((t s) θ) S0((t s) θ) f(s, x )dθds 1 − q 2 − − 1 − s Z0 Z0 t2 ∞ q 1 q + lim q θ(t2 s) − hq(θ)S0((t2 s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ − − →∞ Zt1 Z0 t1 ∞ q 1 q 1 q + lim q θ (t2 s) − (t1 s) − hq(θ)S0((t2 s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ − − − − →∞ Z0 Z0 t1 ∞ q 1 q q + lim q θ(t1 s) − hq(θ) S0((t2 s) θ) S0((t1 s) θ) Bλ[Cux(s) + g(s, xs)]dθds λ 0 0 − − − − →∞ Z Z + f(t , x ) f(t , x ) | 2 t − 1 t |

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= I1 + q (I2 + I3 + I4 + I5 + I6 + I7) + I8

where

∞ q q I = h (θ) S0(t θ) S0(t θ) [φ(0) (h(x , x , ..., x ))(0) f(0, x )]dθ , 1 q 2 − 1 − t1 t2 tn − 0 Z0 t2 ∞ q 1 q I = θ(t s) − h (θ)AS0((t s) θ)f(s, x )dθds , 2 2 − q 2 − s Zt1 Z0 t1 ∞ q 1 q 1 q I = θ (t s) − (t s) − h (θ)AS0((t s ) θ)f(s, x )dθds , 3 2 − − 1 − q 2 − s Z0 Z0 t1 ∞ q 1 q q I = θ(t s) − h (θ)A S0((t s) θ) S0((t s) θ) f(s, x )dθds , 4 1 − q 2 − − 1 − s Z0 Z0 t2 ∞ q 1 q I5 = lim θ(t2 s) − hq(θ)S0((t2 s) θ)Bλ[Cux(s) + g(s, xs)]dθds , λ − − →∞ Zt1 Z0 t1 ∞ q 1 q 1 q I6 = lim θ (t2 s) − (t1 s) − hq(θ)S0((t2 s) θ)Bλ[Cux(s) + g(s, xs)]dθds , λ − − − − →∞ Z0 Z0 t1 ∞ q 1 q q I7 = lim θ(t1 s) − hq(θ) S0((t2 s) θ) S0((t1 s) θ) Bλ[Cux(s) + g(s, xs)]dθds . λ 0 0 − − − − →∞ Z Z I = f(t , x ) f(t , x ) 8 | 2 t − 1 t | By using analogous argument performed in (3.12) and (3.13), we can conclude that

q MM0M3 (1+a)(1 q1) MM0M1M2(t2 t1) I5 b − + − x1 + M0 φ(0) 1 q1 ≤ Γ(1 + q)(1 + a) − qΓ(1 + q) | | | | qMM0M3 (1+a)(1 q1) q + b − (t2 t1) , 1 q1 Γ(1 + q)(1 + a) − − t MM 1 1 (1 q1) 0 q 1 q 1 1 q − I ((t s) − (t s) − ) − 1 ds m 1 6 1 2 q ≤ Γ(1 + q) − − − || ||L 1 [0,t1] Z0 t1 MM0M1M2 q 1 q 1 + ((t1 s) − (t2 s) − )ds x1 + M0 φ(0) Γ(1 + q) 0 − − − | | | | Z qMM0M3 (1+a)(1 q1) + b − 1 q1 Γ(1 + q)(1 + a) − t MM M 1 (1 q1) 0 3 ((t s)a (t s)a)ds − ≤ Γ(1 + q) 1 − − 2 − Z0 t1 MM0M1M2 q 1 q 1 + ((t1 s) − (t2 s) − )ds x1 + M0 φ(0) Γ(1 + q) 0 − − − | | | | Z qMM0M3 (1+a)(1 q1) + b − 1 q1 Γ(1 + q)(1 + a) −

Hence limt2 t1 I5 = 0, and limt2 t1 I6 = 0. Using the similar argument and (A5), we can → → conclude that limt2 t1 I2 = 0, limt2 t1 I3 = 0 and limt2 t1 I8 = 0. → → → On the other hand, (A2) implies that S0(t) for t > 0 is continuous in the uniform operator

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topology, then from the Lebesgue domainated convergence theorem, we get

∞ q q lim I1 = lim hq(θ) S0(t2θ) S0(t1θ) [φ(0) (h(xt1 , xt2 , ..., xtn ))(0) f(0, x0)]dθ , t2 t1 t2 t1 0 − − − → → Z ∞ q q = h q(θ) lim S0(t2θ) S0(t1θ) φ(0) (h(xt1 , xt2 , ..., xtn ))(0) f(0, x0) dθ t2 t1 || − ||| − − | Z0 → = 0,

t1 ∞ q 1 q q lim I4 = lim θ(t1 s) − hq(θ)A S0((t2 s) θ) S0((t1 s) θ) f(s, xs)dθds , t2 t1 t2 t1 − − − − → → Z0 Z0 t1 ∞ q 1 q q = θ(t1 s) − hq(θ)A lim S0((t2 s) θ) S0((t1 s) θ) f(s, xs) dθds, − t2 t1 || − − − ||| | Z0 Z0 → = 0,

t1 ∞ q 1 q q lim I7 = lim M θ(t1 s) − hq(θ) S0((t2 s) θ) S0((t1 s) θ) Cux(s) t2 t1 t2 t1 − || − − − ||| → → Z0 Z0 + g(s, x ) dθds s | t1 ∞ q 1 q q = M θ(t1 s) − hq(θ) lim S0((t2 s) θ) S0((t1 s) θ) Cux(s) − t2 t1 || − − − ||| Z0 Z0 → + g(s, x ) dθds s | = 0.

Consequently, (F x)(t ) (F x)(t ) 0 independently of x B as t t , which mean that | 2 − 1 | → ∈ k0 2 → 1 F is continuous on Bk0 . Step 4. For each t [0, b],V (t) = (F x)(t), x B is relatively compact in E. ∈ { ∈ k0 } Obviously, V(0) is relatively compact in E. Let 0 < t < b be ﬁxed. For arbitrary (0, t) and ∈ δ > 0, deﬁne an operator F,δ on Bk0 by

∞ q (F,δx)(t) = hq(θ)S0(t θ)[φ(0) (h(xt1 , xt2 , ..., xtn ))(0) + f(0, x0)]dθ δ − Z t − ∞ q 1 q + f(t, xt) + q θ(t s) − hq(θ)AS0((t s) θ)f(s, xs)dθds 0 δ − − t Z Z − ∞ q 1 q + lim q θ(t s) − hq(θ)S0((t s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ − − →∞ Z0 Zδ ∞ q q q = hq(θ)S0(t δ)S0(t θ t δ)[φ(0) (h(xt1 , xt2 , ..., xtn ))(0) + f(0, x0)]dθ δ − − Z t − ∞ q 1 q q q + f(t, xt) + q θ(t s) − hq(θ)AS0( δ)S0((t s) θ δ)f(s, xs)dθds 0 δ − − − t Z Z − ∞ q 1 q q q + lim q θ(t s) − hq(θ)S0( δ)S0((t s) θ δ)Bλ[Cux(s) λ − − − →∞ Z0 Zδ + g(s, xs)]dθds

where x B . Then from the compactness of S (tqδ) and S (qδ), we get that the set V (t) = ∈ k0 0 0 ,δ (F x)(t), x B is relatively compact in E for each (0, t) and δ > 0. Moreover, for { ,δ ∈ k0 } ∈

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every x B , we have ∈ k0 (F x)(t) (F x)(t) | − ,δ | ∞ q hq(θ)S0(t θ)[φ(0) (h(xt1 , xt2 , ..., xtn ))(0) + f(0, x0)]dθ ≤ 0 − Z t δ q 1 q + q θ(t s) − hq(θ)AS0((t s) θ)f(s, xs)dθds 0 0 − − Z t Z ∞ q 1 q + q θ(t s) − hq(θ)AS0((t s) θ)f(s, xs)dθds t δ − − Z − Z t δ q 1 q + q lim θ(t s) − hq(θ)S0((t s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ − − →∞ 0 0 Z t Z ∞ q 1 q + q lim θ(t s) − hq(θ)S0((t s) θ)Bλ[Cux(s) + g(s, xs)]dθds λ t δ − − →∞ Z − Z δ β M0 φ(0) + Lk + h(xt1 , xt2 , ..., xtn )(0) + A− H1(k + 1) θhq(θ)dθ ≤ | | | | | | 0 δ Z Γ(1 + β) qβ + C1 βH1(k + 1)b θhq(θ)dθ βΓ(1 + qβ) − Z0 Γ(1 + β) qβ ∞ + C1 βH1(k + 1)b θhq(θ)dθ βΓ(1 + qβ) − Zδ 1 q1 t q 1 − δ 1 −q + qMM0 (t s) − 1 ds m 1 θhq(θ)dθ − || ||L q1 [0,t] Z0 Z0 δ q qMM0M3 (1+a)(1 q1) + MM0M1M2b x1 + M0 φ(0) + b − θhq(θ)dθ | | | | Γ(1 + q)(1 + a)1 q1 − Z0 1 q1 t q 1 − 1 −q ∞ + qMM (t s) − 1 ds m 1 θh (θ)dθ 0 q q t − || ||L 1 [t ,t] 0 Z − − Z MM0M1M2 qMM0M3 (1+a)(1 q1) q + x1 + M0 φ(0) + b − 0 as , δ 0. qΓ(1 + q) | | | | Γ(1 + q)(1 + a)1 q1 → → − Therefore, there are relatively compact sets arbitrarily close to the set V (t), t > 0. Hence the set V (t), t > 0 is also relatively compact in E. As a consequence of the Arzela-Ascoli theorem it is concluded that F x, x B is a completely continuous operator. Then all the conditions ∈ k0 of Lemma 2.3 are satisﬁed, hence F has a ﬁxed point on Bk0 . Therefore, the problem (1.1) is controllable. The proof is completed.

4 Example

As an application of our results we consider the following fractional neutral functional differential equations of the form

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π ∂2 cDq[u(t, z) U(z, y)u (s, y)dy] = (u(t, z) + Cu(t) + g(t, u(t r, z)), z [0, π], t (0, b], − t ∂z2 − ∈ ∈ Z0 u(t, 0) = u(t, π) = 0, t [0, b], (4.1) ∈ n π

u(s, z) + k(z, y)uti (s, y)dy = (φ(s))(z), s [ r, 0] 0 ∈ − Xi=0 Z where cDq is a Caputo fractional neutral functional derivative of order 0 < q < 1, b > 0, f : [0, b] R R is a given function, n is a positive integer, 0 < t < t < ... < t < b, × → 0 1 n φ ([ r, 0],E), that is φ(z) E. Consider E = L2([0, π]; R) endowed with the uniform ∈ C − ∈ topology and the operator A : D(A) E E defined by ⊂ → ∂2 D(A) = u L2([0, π]; R): u(t, 0) = u(t, π) = 0 , Au = (u(t, z)). { ∈ } ∂z2 Now, we have D(A) = u E : u(t, 0) = u(t, π) = 0 = E. { ∈ } 6 As we know from [4], R(λ, A) 1 , A satisfies the Hille-Yosida condition with (0, + ) ρ(A) | | ≤ λ ∞ ⊆ and λ > 0. Hence, operator A satisfies (A1), (A2) and M = M0 = 1. The system(4.1) can be reformulated as the following fractional neutral functional differential equations

cDq[x(t) f(t, x )] = Ax(t) + Cu(t) + g(t, x ), t (0, b] (1.1) − t t ∈ x (s) + (h(x , x , ..., x ))(s) = φ(s), s [ r, 0]. 0 t1 t2 tn ∈ − where x = u (s, ), that is (x(t + s))(z) = u(t + s, z), t [0, b], z [0, π], s [ r, 0]. The t t · ∈ ∈ ∈ − function h : [0, b] E is given by × C → π f(t, x )z = U(z, y)u t(s, y)dy. t − Z0 The function g : n = is given by C C n

(h(xt1 , xt2 , ..., xtn ))(s) = Kgxti (s), Xi=0 where (K v)(z) = π k(z, y)v(y)dy, for v E, z [0, π]. Take g 0 ∈ ∈ R 1 t u t r, x g(t, u(t r, x)) = e− | − | + sin t, t [0, b], − 6√π 1 + u t r, x ∈ | − | then we have f : [0, b] R R is continuous functions. Moreover, × →

1 t 1/q1 + g(t, u) = e− + sin t := m(t) L ([0, b],R ). | | 6√π | | ∈

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Therefore, assumptions (A ), (A ) are satisﬁed W : L2([0, b],U) D(A) deﬁned by 3 4 → b ∞ q 1 q W u = lim q θ(b s) − hq(θ)S0((b s) θ)BλCu(s)dθds λ − − →∞ Z0 Z0

satisﬁes assumption (A7). Then all conditions of Theorem 3.1 are satisﬁed and we deduce the problem 4.1 is controllable.

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