Existence of solutions to nonlocal boundary value problems for fractional differential equations with impulses

Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 15, pp. 1–16. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu EXISTENCE OF SOLUTIONS TO NONLOCAL BOUNDARY VALUE PROBLEMS FOR FRACTIONAL DIFFERENTIAL EQUATIONS WITH IMPULSES DANIEL CAO LABORA, ROSANA RODR´ IGUEZ-L´ OPEZ, MOHAMMED BELMEKKI Abstract. In this work, through the application of fixed point theory, we consider the properties of the solutions to a nonlocal boundary value problem for fractional differential equations subject to impulses at fixed times. We compute the Green’s function related to the problem, which allows us to obtain an integral representation of the solution. This representation gives an explicit description of the solution when the source term does not depend on the solution. Nevertheless, when the description of the source term is implicit, we can not ensure the existence of a solution. In this case, we prove the existence of a solution for the integral problem via fixed point techniques. To do this, we develop a slight generalization of Arzel`a-Ascoli theorem that makes it suitable for piecewise uniformly continuous functions. 1. Introduction The investigation on fractional differential equations has experimented a huge expansion in the previous decades, and new applications have been proposed since then. Some examples of applications of fractional order equations can be found, for instance, in [1], where models of viscoplasticity are considered. The work [3] is focused on protein dynamics; [7] is devoted to continuum and statistical mechanics; and [8] in relaxation in filled polymers. On the other hand, the physical interpretation for fractional differential equations has been considered in [4] from the point of view of Riemann-Liouville derivatives, also in [11]. In this article, we consider a boundary value problem with integral conditions for a class of fractional differential equations subject to impulses. Some existence results for higher-order fractional differential equations with integral conditions can be found in [2], while some other results related to Pettis integral are included in [13]. We represent by Dδthe fractional derivative of Riemann-Liouville type and consider the following impulsive fractional differential equation with nonlocal boundary conditions: Dδ 0+u(t) + f(t, u(t)) = 0, t ∈(0, t1),(1.1) 2010 Mathematics Subject Classification. 26A33, 34B37, 34B05, 34B10, 34B27. Key words and phrases. Fractional differential equations; nonlocal boundary value problems; Riemann-Liouville fractional derivative; fixed point results. c 2020 Texas State University. Submitted May 9, 2018. Published February 10, 2020. 1 2 D. CAO LABORA, R. RODR´ IGUEZ-L ´ OPEZ, M. BELMEKKI EJDE-2020/15 Dδ t+ 1 u(t) + f(t, u(t)) = 0, t ∈(t1,1),(1.2) u(0) = 0, u(t+ 1) = 0, α0u(ξ0) + α1u(ξ1) = u(1), β Z1 0 u(s)ds =u(1),(1.3) where 1 < δ ≤2, 0 = t0< t1<1, ξi∈(ti, ti+1), i= 0,1, α0, α1, β ∈R\ {0}, and f: [0,1] ×R→Rcontinuous on [0, t1]×R. We also require fto be continuous on (t1,1] ×Rand that f|(t1,1]×Rcan be extended continuously to [t1,1] ×R, that is, the discontinuity at t1is of finite jump. The previous problem is a particular case of the following problem Dδ t+ k u(t) + f(t, u(t)) = 0, tk< t < tk+1, k = 0, . . . , m, (1.4) u(t+ k) = 0, k = 0, . . . , m, m X k=0 αku(ξk) = u(T), β ZT 0 u(s)ds =u(T),(1.5) where 1 < δ ≤2, T > 0, 0 = t0< t1<· · · < tk< tk+1 <· · · < tm+1 =T, ξi∈(ti, ti+1), i= 0, . . . , m,α0, α1, . . . , αm, β ∈R, and f: [0, T ]×R→Rcontinuous on [0, t1]×Rand on (tk, tk+1]×R, for k= 1, . . . , m, in such a way that f|(tk,tk+1]×R can be extended continuously to [tk, tk+1]×R, for k= 1, . . . , m. Here, if we consider T= 1 and m= 1, that is, 0 = t0< t1< t2= 1, then we obtain problem (1.1)–(1.3). In Section 2, we present some basic definitions and results. In Section 3, we consider a linear problem and obtain the corresponding Green’s function, and in Section 4, we give some existence results for the general nonlinear problem. 2. Basic definitions We introduce some basic concepts about fractional integrals and derivatives. Some relevant monographs on fractional calculus are [5, 6, 9, 12, 14]. Definition 2.1. The Riemann-Liouville fractional integral of order δ > 0 of a function f: (a, b]→Ris given by Iδ a+f(t) = 1 Γ(δ)Zt a (t−τ)δ−1f(τ)dτ, provided that the right-hand side is pointwise defined on (a, b], and where Γ denotes the Gamma function. Definition 2.2. For a continuous function f: (a, b]→R, the Riemann-Liouville derivative of fractional order δ > 0 is given by Dδ a+f(t) = 1 Γ(n−δ)d dtnZt a (t−τ)n−δ−1f(τ)dτ, where n=bδc+ 1, being bδcthe integer part of the real number δ. Lemma 2.3 ([14]).Given δ > 0, the solutions to the fractional differential equation Dδ 0+u(t) = 0 are the functions u(t) = c1tδ−1+c2tδ−2+· · · +cntδ−n, ci∈R, i = 1, . . . , n, where n=dδe, the largest integer less than equal to δ. EJDE-2020/15 EXISTENCE OF SOLUTIONS 3 Lemma 2.4. Given δ > 0, the solutions to the fractional differential equation Dδ a+u(t)=0 are the functions u(t) = c1(t−a)δ−1+c2(t−a)δ−2+· · · +cn(t−a)δ−n, ci∈R, i = 1, . . . , n, where n=dδe. Proof. The result is well known for δ∈Z+. Consequently, we can assume that δ6∈ Z+. Then, the equation Dδ a+u(t) = 1 Γ(n−δ)d dtnZt a (t−τ)n−δ−1u(τ)dτ = 0 can be written, via the change of variable τ=a+s, as 1 Γ(n−δ)d dtnZt−a 0 (t−a−s)n−δ−1u(a+s)ds = 0, or 1 Γ(n−δ)d dtnZt−a 0 (t−a−s)n−δ−1v(s)ds = 0, where v(s) := u(a+s). Now, if we consider z=t−a, we obtain that 1 Γ(n−δ)d dz nZz 0 (z−s)n−δ−1v(s)ds = 0. Hence, the function vsatisfies the equation Dδ 0+v(z) = 0. From Lemma 2.3, we know that u(a+z) = v(z) = c1zδ−1+c2zδ−2+· · · +cnzδ−n, ci∈R, i = 1, . . . , n, or, equivalently, u(t) = v(t−a) = c1(t−a)δ−1+c2(t−a)δ−2+· · ·+cn(t−a)δ−n, ci∈R, i = 1, . . . , n.  It is straightforward to derive the following result, after a direct application of Lemma 2.4. Lemma 2.5. Given δ > 0, we have Iδ a+(Dδ a+u(t)) = u(t) + c1(t−a)δ−1+c2(t−a)δ−2+· · · +cn(t−a)δ−n, for ci∈R,i= 1, . . . , n. Proof. We write Iδ a+(Dδ a+u(t)) = u(t)+f(t) and we apply Dδ a+to both sides of this expression. Since Dδ a◦Iδ a= Id, we have that Dδ a+f= 0 and the conclusion follows from Lemma 2.4.  Remark 2.6. From Lemma 2.5, for a=tk,k= 0,1, . . . , m, and 1 < δ ≤2, we obtain Iδ t+ k (Dδ t+ k u(t)) = u(t) + c1,k(t−tk)δ−1+c2,k(t−tk)δ−2, c1,k, c2,k ∈R. 4 D. CAO LABORA, R. RODR´ IGUEZ-L ´ OPEZ, M. BELMEKKI EJDE-2020/15 3. Green’s function for a linear fractional differential equation In this section, we consider a related linear fractional differential equation with the same boundary conditions, to obtain the explicit expression of the Green’s function. First, we need to define the following space of functions. Definition 3.1. Given a partition 0 = t0< t1< t2= 1, the space of piecewise continuous functions from [0,1] to R, is defined as PC([0,1],R) = nu: [0,1] →R:u∈C([0, t1],R) and u∈C((t1,1],R) with lim t→t+ 1 u(t) finiteo. Lemma 3.2. Consider 1< δ ≤2,α0, α1, β ∈R\ {0},0 = t0< t1< t2= 1, and ξi∈(ti, ti+1)for i= 0,1. Assume also that σ∈P C([0,1],R). A function u∈PC([0,1],R)is a solution to the boundary value problem Dδ 0+u(t) + σ(t)=0, t ∈(0, t1),(3.1) Dδ t1+u(t) + σ(t) = 0, t ∈(t1,1),(3.2) u(0) = 0, u(t+ 1)=0, α0u(ξ0) + α1u(ξ1) = u(1), β Z1 0 u(s)ds =u(1) (3.3) if and only if it satisfies the integral equation u(t) = Z1 0 H(t, s)σ(s)ds, where H(t, s)is the Green’s function given in the proof, provided that ϕ(1) − 1 X k=0 αkϕ(ξk) = tδ 1β(1 −t1)1−δ(1 −t1)δ−1 δ−β(1 −t1)−α0ξδ−1 0 −α1 tδ 1β(1 −t1)1−δ(ξ1−t1)δ−1 δ−β(1 −t1)6= 0, where ϕ(t) =    tδ−1, t ∈(0, t1), tδ 1β(1−t1)1−δ(t−t1)δ−1 δ−β(1−t1), t ∈(t1,1]. Proof. Integrating (3.1) and (3.2), we obtain Iδ t+ k Dδ t+ k u(t) = −Iδ t+ k σ(t), t ∈(tk, tk+1), k = 0,1. From Remark 2.6, equations (3.1) and (3.2) can be rewritten as u(t) + c1,k(t−tk)δ−1+c2,k(t−tk)δ−2=−1 Γ(δ)Zt tk (t−τ)δ−1σ(τ)dτ, for t∈(tk, tk+1), where c1,k, c2,k ∈R,k= 0,1, or equivalently, by renaming the constants used, u(t) = −1 Γ(δ)Zt tk (t−τ)δ−1σ(τ)dτ +c1,k(t−tk)δ−1+c2,k(t−tk)δ−2, for t∈(tk, tk+1), where c1,k, c2,k ∈R, for k= 0,1. EJDE-2020/15 EXISTENCE OF SOLUTIONS 5 Note that the conditions u(t+ k) = 0, for k= 0,1, imply that c2,k = 0, for k= 0,1, so that u(t) = −1 Γ(δ)Zt tk (t−τ)δ−1σ(τ)dτ +c1,k(t−tk)δ−1, t ∈(tk, tk+1), where c1,k ∈R, for k= 0,1. Using the integral condition βR1 0u(s)ds =u(1), we obtain c1,1(1 −t1)δ−1=βZ1 0 u(s)ds +1 Γ(δ)Z1 t1 (1 −s)δ−1σ(s)ds. The previous calculations lead, for k= 0, to u(t) = −1 Γ(δ)Zt 0 (t−τ)δ−1σ(τ)dτ +c1,0tδ−1, t ∈(0, t1),(3.4) and, for k= 1, to u(t) = −1 Γ(δ)Zt t1 (t−τ)δ−1σ(τ)dτ +β(1 −t1)1−δ(t−t1)δ−1Z1 0 u(s)ds +(1 −t1)1−δ Γ(δ)(t−t1)δ−1Z1 t1 (1 −s)δ−1σ(s)ds, t ∈(t1,1). (3.5) From these expressions, we obtain Z1 0 u(s)ds =Zt1 0 u(s)ds +Z1 t1 u(s)ds =−1 Γ(δ)Zt1 0Zt 0 (t−s)δ−1σ(s)ds dt +Zt1 0 c1,0tδ−1dt −1 Γ(δ)Z1 t1Zt t1 (t−s)δ−1σ(s)ds dt +β(1 −t1)1−δZ1 t1 (t−t1)δ−1dt Z1 0 u(s)ds +(1 −t1)1−δ Γ(δ)Z1 t1 (t−t1)δ−1dt Z1 t1 (1 −s)δ−1σ(s)ds =−1 δΓ(δ)Zt1 0 (t1−s)δσ(s)ds +c1,0 δtδ 1 −1 δΓ(δ)Z1 t1 (1 −s)δσ(s)ds +β1−t1 δZ1 0 u(s)ds +1−t1 δΓ(δ)Z1 t1 (1 −s)δ−1σ(s)ds. Then Z1 0 u(s)ds =−1 (δ−β(1 −t1))Γ(δ)Zt1 0 (t1−s)δσ(s)ds +c1,0 (δ−β(1 −t1))tδ 1 −1 (δ−β(1 −t1))Γ(δ)Z1 t1 (1 −s)δσ(s)ds +(1 −t1) (δ−β(1 −t1))Γ(δ)Z1 t1 (1 −s)δ−1σ(s)ds. (3.6) 6 D. CAO LABORA, R. RODR´ IGUEZ-L ´ OPEZ, M. BELMEKKI EJDE-2020/15 Therefore, we can find a piecewise defined integral kernel G(t, s) that allows us to express uin a simple way. On the one hand, for t∈(0, t1), it follows from (3.4) that u(t) = −1 Γ(δ)Zt 0 (t−s)δ−1σ(s)ds +c1,0tδ−1 =Z1 0 G(t, s)σ(s)ds +c1,0tδ−1 =Z1 0 G(t, s)σ(s)ds +c1,0ϕ(t), t ∈(0, t1). On the other hand, if we replace expression (3.6) in (3.5), we can describe u(t), for t∈(t1,1), as u(t) = −1 Γ(δ)Zt t1 (t−s)δ−1σ(s)ds −β(1 −t1)1−δ(t−t1)δ−1 (δ−β(1 −t1))Γ(δ)Zt1 0 (t1−s)δσ(s)ds +c1,0 δ−β(1 −t1)tδ 1β(1 −t1)1−δ(t−t1)δ−1 −β(1 −t1)1−δ(t−t1)δ−1 (δ−β(1 −t1))Γ(δ)Z1 t1 (1 −s)δσ(s)ds +β(1 −t1)2−δ(t−t1)δ−1 (δ−β(1 −t1))Γ(δ)Z1 t1 (1 −s)δ−1σ(s)ds +(1 −t1)1−δ Γ(δ)(t−t1)δ−1Z1 t1 (1 −s)δ−1σ(s)ds =−1 Γ(δ)Zt1 0 β(1 −t1)1−δ(t−t1)δ−1 δ−β(1 −t1)(t1−s)δσ(s)ds +1 Γ(δ)Zt t1−(t−s)δ−1+ (1 −t1)1−δ(t−t1)δ−1(1 −s)δ−1σ(s)ds −1 Γ(δ)Zt t1 β(1 −t1)1−δ(t−t1)δ−1 δ−β(1 −t1)(1 −s)δσ(s)ds +1 Γ(δ)Zt t1 β(1 −t1)2−δ(t−t1)δ−1 δ−β(1 −t1)(1 −s)δ−1σ(s)ds −1 Γ(δ)Z1 t β(1 −t1)1−δ(t−t1)δ−1 δ−β(1 −t1)(1 −s)δσ(s)ds +1 Γ(δ)Z1 t β(1 −t1)2−δ(t−t1)δ−1 δ−β(1 −t1)(1 −s)δ−1σ(s)ds +1 Γ(δ)Z1 t (1 −t1)1−δ(t−t1)δ−1(1 −s)δ−1σ(s)ds +c1,0 δ−β(1 −t1)tδ 1β(1 −t1)1−δ(t−t1)δ−1 =Z1 0 G(t, s)σ(s)ds +c1,0 δ−β(1 −t1)tδ 1β(1 −t1)1−δ(t−t1)δ−1 EJDE-2020/15 EXISTENCE OF SOLUTIONS 7 =Z1 0 G(t, s)σ(s)ds +c1,0ϕ(t), t ∈(t1,1]. In summary, u(t) = Z1 0 G(t, s)σ(s)ds +c1,0ϕ(t), t ∈(0,1],(3.7) where G(t, s) = 1 Γ(δ)                                          −(t−s)δ−1if 0 ≤s≤t<t1, 0 if 0 ≤t≤s≤1,0≤t<t1, −β(1−t1)1−δ(t−t1)δ−1 δ−β(1−t1)(t1−s)δif 0 ≤s≤t1≤t, −(t−s)δ−1+ (1 −t1)1−δ(t−t1)δ−1(1 −s)δ−1 −β(1−t1)1−δ(t−t1)δ−1 δ−β(1−t1)(1 −s)δ +β(1−t1)2−δ(t−t1)δ−1 δ−β(1−t1)(1 −s)δ−1if t1≤s≤t≤1, −β(1−t1)1−δ(t−t1)δ−1 δ−β(1−t1)(1 −s)δ +β(1−t1)2−δ(t−t1)δ−1 δ−β(1−t1)(1 −s)δ−1 +(1 −t1)1−δ(t−t1)δ−1(1 −s)δ−1if t1≤t≤s≤1; that is, G(t, s) = 1 Γ(δ)                              −(t−s)δ−1if 0 ≤s≤t<t1, 0 if 0 ≤t≤s≤1,0≤t < t1, −β(1−t1)1−δ(t−t1)δ−1 δ−β(1−t1)(t1−s)δif 0 ≤s≤t1≤t, −(t−s)δ−1+ (1 −t1)1−δ(t−t1)δ−1(1 −s)δ−1 ×1−β(1−s) δ−β(1−t1)+β(1−t1) δ−β(1−t1)if t1≤s≤t≤1, (1 −t1)1−δ(t−t1)δ−1(1 −s)δ−1 ×1−β(1−s) δ−β(1−t1)+β(1−t1) δ−β(1−t1)if t1≤t≤s≤1, or, equivalently, G(t, s) =1 Γ(δ)(δ−β(1 −t1)) ×                          −(t−s)δ−1(δ−β(1 −t1)) if 0 ≤s≤t < t1, 0 if 0 ≤t≤s≤1,0≤t<t1, −β(1 −t1)1−δ(t−t1)δ−1(t1−s)δif 0 ≤s≤t1≤t, −(t−s)δ−1(δ−β(1 −t1)) +(1 −t1)1−δ(t−t1)δ−1(1 −s)δ−1(δ−β(1 −s)) if t1≤s≤t≤1, (1 −t1)1−δ(t−t1)δ−1(1 −s)δ−1(δ−β(1 −s)) if t1≤t≤s≤1. (3.8) Finally, if we consider the condition α0u(ξ0) + α1u(ξ1) = u(1), we have u(1) = Z1 0 G(1, s)σ(s)ds +c1,0ϕ(1) 8 D. CAO LABORA, R. RODR´ IGUEZ-L ´ OPEZ, M. BELMEKKI EJDE-2020/15 =α0Z1 0 G(ξ0, s)σ(s)ds +α0c1,0ϕ(ξ0) + α1Z1 0 G(ξ1, s)σ(s)ds +α1c1,0ϕ(ξ1) =Z1 0 [α0G(ξ0, s) + α1G(ξ1, s)]σ(s)ds + (α0ϕ(ξ0) + α1ϕ(ξ1))c1,0 =Z1 0 1 X k=0 αkG(ξk, s)σ(s)ds +c1,0 1 X k=0 αkϕ(ξk); that is, Z1 0h1 X k=0 αkG(ξk, s)−G(1, s)iσ(s)ds =c1,0hϕ(1) − 1 X k=0 αkϕ(ξk)i. Since, by hypothesis, ϕ(1) −P1 k=0 αkϕ(ξk)6= 0, we have c1,0=1 ϕ(1) −P1 k=0 αkϕ(ξk)Z1 0h1 X k=0 αkG(ξk, s)−G(1, s)iσ(s)ds. We know that G(1, s) = 1 Γ(δ)(δ−β(1 −t1)) (−β(t1−s)δ,0≤s≤t1, (1 −s)δ−1β(s−t1), t1≤s≤1. Moreover, for ξ0∈(0, t1), we have G(ξ0, s) = 1 Γ(δ)(δ−β(1 −t1)) (−(ξ0−s)δ−1(δ−β(1 −t1)),0≤s≤ξ0, 0, ξ0≤s≤1, and, for ξ1∈(t1,1), G(ξ1, s) = 1 Γ(δ)(δ−β(1 −t1)) ×           −β(1 −t1)1−δ(ξ1−t1)δ−1(t1−s)δ,0≤s≤t1, −(ξ1−s)δ−1(δ−β(1 −t1)) +(1 −t1)1−δ(ξ1−t1)δ−1(1 −s)δ−1(δ−β(1 −s)), t1≤s≤ξ1, (1 −t1)1−δ(ξ1−t1)δ−1(1 −s)δ−1(δ−β(1 −s)), ξ1≤s≤1. If we take A=1 ϕ(1)−P1 k=0 αkϕ(ξk)and B=A Γ(δ)(δ−β(1−t1)) , we have c1,0=AZ1 0h1 X k=0 αkG(ξk, s)−G(1, s)iσ(s)ds =AZ1 0 [α0G(ξ0, s) + α1G(ξ1, s)−G(1, s)]σ(s)ds =BZξ0 0−α0(ξ0−s)δ−1(δ−β(1 −t1)) −α1β(1 −t1)1−δ(ξ1−t1)δ−1(t1−s)δ+β(t1−s)δσ(s)ds +BZt1 ξ0 [−α1β(1 −t1)1−δ(ξ1−t1)δ−1(t1−s)δ+β(t1−s)δ]σ(s)ds EJDE-2020/15 EXISTENCE OF SOLUTIONS 9 +BZξ1 t1 −α1(ξ1−s)δ−1(δ−β(1 −t1))σ(s)ds +BZξ1 t1 α1(1 −t1)1−δ(ξ1−t1)δ−1(1 −s)δ−1(δ−β(1 −s))σ(s)ds +BZξ1 t1 [−(1 −s)δ−1β(s−t1)]σ(s)ds +BZ1 ξ1α1(1 −t1)1−δ(ξ1−t1)δ−1(1 −s)δ−1(δ−β(1 −s)) −(1 −s)δ−1β(s−t1)σ(s)ds =Z1 0 K(s)ds, where K(s) =                                B−α0(ξ0−s)δ−1(δ−β(1 −t1)) −α1β(1 −t1)1−δ(ξ1−t1)δ−1(t1−s)δ+Bβ(t1−s)δ,0≤s≤ξ0, B[−α1β(1 −t1)1−δ(ξ1−t1)δ−1(t1−s)δ+β(t1−s)δ],0≤ξ0≤s≤t1, −Bα1(ξ1−s)δ−1(δ−β(1 −t1)) +Bα1(1 −t1)1−δ(ξ1−t1)δ−1(1 −s)δ−1(δ−β(1 −s)) +B[−(1 −s)δ−1β(s−t1)], t1≤s≤ξ1, Bα1(1 −t1)1−δ(ξ1−t1)δ−1(1 −s)δ−1(δ−β(1 −s)) −B(1 −s)δ−1β(s−t1), ξ1< s ≤1. Then, for t∈(0, t1], we can write u(t) = −1 Γ(δ)Zt 0 (t−s)δ−1σ(s)ds +c1,0tδ−1=Z1 0 H(t, s)σ(s)ds, where His defined in the following way. Denoting C=A δ−β(1−t1), for t∈(0, ξ0], we have H(t, s) = tδ−1 Γ(δ)                                                            −(t−s)δ−1t1−δ+C−α0(ξ0−s)δ−1 −α1β(1 −t1)1−δ(ξ1−t1)δ−1(t1−s)δ+Cβ(t1−s)δ if 0 ≤s≤t≤ξ0, C−α0(ξ0−s)δ−1−α1β(1 −t1)1−δ(ξ1−t1)δ−1(t1−s)δ +Cβ(t1−s)δif 0 ≤t<s≤ξ0, C−α1β(1 −t1)1−δ(ξ1−t1)δ−1(t1−s)δ+β(t1−s)δ if 0 ≤t≤ξ0< s ≤t1, −Cα1(ξ1−s)δ−1(δ−β(1 −t1)) +Cα1(1 −t1)1−δ(ξ1−t1)δ−1(1 −s)δ−1(δ−β(1 −s)) +C−(1 −s)δ−1β(s−t1) if 0 ≤t≤ξ0, t1≤s≤ξ1, Cα1(1 −t1)1−δ(ξ1−t1)δ−1(1 −s)δ−1(δ−β(1 −s)) −C(1 −s)δ−1β(s−t1) if 0 ≤t≤ξ0, ξ1< s ≤1, 16 D. CAO LABORA, R. RODR´ IGUEZ-L ´ OPEZ, M. BELMEKKI EJDE-2020/15 [2] M. Feng, X. Zhang, W. Ge; New existence results for higher-order nonlinear fractional differential equations with integral boundary conditions, Bound. Value Probl. 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Rudin,; Principles of Mathematical Analysis. McGraw-Hill, New York, 1976. Daniel Cao Labora Departamento de Estad´ ıstica, An´ alisis Matem´ atico y Optimizaci´ on, Universidade de Santiago de Compostela, 15782, Spain Email address:[email protected] Rosana Rodr´ ıguez-L´ opez Departamento de Estad´ ıstica, An´ alisis Matem´ atico y Optimizaci´ on, Universidade de Santiago de Compostela, 15782, Spain Email address:[email protected] Mohammed Belmekki Ecole Sup´ erieure en Sciences Appliqu´ ees, BP. 165 RP, Bel Horizon, Tlemcen,13000, Algeria Email address:[email protected]