RESEARC H Open Access Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions Bashir Ahmad 1* and Juan J Nieto 1,2 * Correspondence: bashir_qau@yahoo.com 1 Department of Mathematics, Faculty of Science, King Abdulaziz University P.O. Box 80203, Jeddah 21589, Saudi Arabia Full list of author information is available at the end of the article Abstract This article investigates a boundary value problem of Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions. Some new existence results are obtained by applying standard fixed point theorems. 2010 Mathematics Subject Classification: 26A33; 34A34; 34B15. Keywords: Riemann-Liouville calculus, fractional integro-differential equations, frac- tional boundary conditions, fixed point theorems 1 Introduction In this article, we study the existence and uniqueness of solutions for the following nonlinear fractional integro-differential equation: D α u ( t ) = f ( t, u ( t ) , ( φu )( t ) , ( ψu )( t )) , t ∈ [0, T] , α ∈ ( 1,2] , (1:1) subject to the boundary conditions of fractional order given by D α−2 u ( 0 + ) =0 , (1:2) D α−1 u ( 0 + ) = νI α−1 u ( η ) ,0<η<T, ν is a constant , (1:3) where D a denotes the Riemann-Liouville fractional derivative of order a, f: [0, T]×ℝ × ℝ × ℝ ® ℝ is continuous, and (φx)(t)= t  0 γ (t, s)x(s)ds,(ψx)(t )= t  0 δ(t, s)x(s)ds , with g and δ being continuous functions on [0, T] × [0, T]. Boundary value problems for nonlinear fractional differential equations have recently been investigated by several researchers. As a matter of fact, fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes (see [1]) and make the fractional-order models more realistic and practical than the classical intege r-order models. Fractional differential equations arise in many engineering and scientifi c disciplines, such as physics, chemis- try, biology, economics, control theory, signal and image processing, biophysics, blood Ahmad and Nieto Boundary Value Problems 2011, 2011:36 http://www.boundaryvalueproblems.com/content/2011/1/36 © 2011 Ahmad and Nieto; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unre stricted use, distribution, and reproduction in any medium, provided the original work is properly cited. flow phenomena, aerodynamics, fitting of experimental data, etc. (see [1,2]). For some recent development on the topic, (see [3-19] and references therein). 2 Preliminaries Let us recall some basic definitions (see [20,21]). Definition 2.1 The Riemann-Liouville fractional integral of order a >0for a contin- uous function u: (0, ∞) ® ℝ is defined as I α u(t )= 1 (α) t  0 (t − s ) α−1 u(s)ds , provided the integral exists. Definition 2.2 For a continu ous function u:(0,∞) ® ℝ, the Riemann-Liouville deri- vative of fractional order a >0,n =[a]+1([a] denotes the integer part of the real number a) is defined as D α u(t )= 1 (n − α)  1 dt  n t  0 (t − s ) n−α−1 u(s)ds =  d dt  n I n−α u(t ) , provided it exists. For a < 0, we use the convention that D a u=I -a u. Also for b Î [0, a), it is valid that D b I a u=I a-b u. Note that for l >-1, l ≠ a -1,a - 2, , a - n, we have D α t λ = (λ +1)  ( λ − α +1 ) t λ−α , and D α t α−i =0 , i =1 , 2 , , n . In particular, for the constant function u(t) = 1, we obtain D α 1= 1  ( 1 − α ) t −α , α /∈ N . For aÎN, we get, of course, D a 1 = 0 because of the poles of the gamma function at the points 0, -1, -2, For a > 0, the general solution of the homogeneous equation D α u ( t ) = 0 in C(0, T) ∩ L(0, T)is u( t ) = c 0 t α−n + c 1 t α−n−1 + ···+ c n−2 t α−2 + c n−1 t α−1 , where c i , i = 1, 2, , n - 1, are arbitrary real constants. We always have D a I a u = u, and I α D α u ( t ) = u ( t ) + c 0 t α−n + c 1 t α−n−1 + ···+ c n−2 t α−2 + c n−1 t α−1 . To define the solution fo r the nonlinear problem (1.1) and (1.2)-(1.3), we consider the following linear equation Ahmad and Nieto Boundary Value Problems 2011, 2011:36 http://www.boundaryvalueproblems.com/content/2011/1/36 Page 2 of 9 D α u ( t ) = σ ( t ) , α ∈ ( 1,2] , t ∈ [0, T] , T > 0 , (2:1) where s Î C[0, T]. We define A = ν η  0 s α−1 (η − s) α−2 (α − 1) ds = ν(α) η 2α−2 (2α − 1) , (2:2) such that A ≠ Γ(a). The general solution of (2.1) is given by u( t ) = c 1 t α−1 + c 0 t α−2 + I α σ ( t ), (2:3) with I a the usual Riemann-Liouville fractional integral of order a. From (2.3), we have D α−1 u ( t ) = c 1  ( α ) + I 1 σ ( t ), (2:4) D α− 2 u ( t ) = c 1  ( α ) t + c 0  ( α − 1 ) + I 2 σ ( t ). (2:5) Using the conditions (1.2) and (1.3) in (2.4) and (2.5), we find that c 0 = 0 and c 1 = ν  (α) − A  η  0 (η − s) α−2 (α − 1) ⎛ ⎝ s  0 (s − x) α−1 (α) σ (x)dx ⎞ ⎠ ds , where A is defined by (2.2). Substituting the values of c 0 and c 1 in (2.3), the unique solution of (2.1) subject to the boundary conditions (1.2)-(1.3) is given by u (t )= t  0 (t − s ) α−1 (α) σ (s)ds + νt α−1  (α) − A  η  0 (η − s) α−2 (α − 1) ⎛ ⎝ s  0 (s − x) α−1 (α) σ (x)dx ⎞ ⎠ d s = t  0 (t − s ) α−1 (α) σ (s)ds + νt α−1  (α) − A  I 2α−1 σ (η). (2:6) 3 Main results Let C = C ( [0, T], R ) denotes the Banach space of all continuous functions from [0, T] ® ℝ endowed with the norm defined by ║u║ = sup{|u(t)|, t Î [0, T]}. If u is a solution of (1.1) and (1.2)-(1.3), then u (t )= t  0 (t − s ) α−1 (α) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds + ν 1 t α−1 η  0 (η − s) 2α−2 (2α − 1) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds, Ahmad and Nieto Boundary Value Problems 2011, 2011:36 http://www.boundaryvalueproblems.com/content/2011/1/36 Page 3 of 9 where ν 1 = ν  (α) − A  . Define an operator P : C → C as ( Pu ) (t )= t  0 (t − s ) α−1 (α) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds + ν 1 t α−1 η  0 (η − s) 2α−2 (2α − 1) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds, t ∈ [0, T] . Observe that the problem (1.1) and (1.2)-(1. 3) has solutions if and only if the opera- tor equation P u = u has fixed points. Lemma 3.1 The operator P is compact. Proof (i) Let B be a bounded set in C[0, T]. Then, there exists a constant M s uch that |f (t,u(t), (u)(t), (ψu)(t))| ≤ M, ∀u Î B, tÎ[0, T]. Thus | ( Pu ) (t ) |≤M t  0 (t − s) α−1 (α) ds + M|ν 1 |t α−1 η  0 (η − s) 2α−2 (2α − 1) d s ≤ MT α−1  T  ( α +1 ) + |ν 1 |η 2α−1  ( 2α )  , which implies that | | ( Pu ) || ≤ MT α−1  T  ( α +1 ) + |ν 1 |η 2α−1  ( 2α )  < ∞ . Hence, P ( B ) is uniformly bounded. (ii) For any t 1 , t 2 Î [0, T], u Î B, we have | ( P u )( t 1 ) − ( P u )( t 2 ) | =       t 1  0 ( t 1 − s ) α−1 (α) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds − t 2  0 ( t 2 − s ) α−1 (α) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds +ν 1  t α−1 1 − t α−1 2  η  0 ( η − s ) 2α−s  ( 2α − 1 ) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds       ≤ M ⎛ ⎝       1 (α) t 1  0  ( t 1 − s ) α−1 − ( t 2 − s ) α−1  ds − 1 (α) t 2  t 1 ( t 2 − s ) α−1 ds       +       ν 1  t α−1 1 − t α−1 2  η  0 ( η − s ) 2α−s  ( 2α − 1 ) ds       ⎞ ⎠ → 0ast 1 → t 2 . Ahmad and Nieto Boundary Value Problems 2011, 2011:36 http://www.boundaryvalueproblems.com/content/2011/1/36 Page 4 of 9 Thus, P ( B ) is equicontinuous. Consequently, the operator P is compact. This com- pletes the proof. □ We need the following known fixed point theorem to prove the existence of solu- tions for the problem at hand. Theorem 3.1 ([22]) Let E be a Banach space. Assume that T: E ® E be a completely continuous operator and the set V = {x Î E | x=μTx,0<μ <1}be bounded. Then, T has a fixed point in E. Theorem 3.2 Assume that there exists a constant M >0such that |f ( t, u ( t ) , ( φu )( t ) , ( ψu )( t )) |≤M, ∀t ∈ [0, T] , u ∈ R . Then, the problem (1.1) and (1.2)-(1.3) has at least one solution on [0,T]. Proof We consider the set V = { u ∈ R|u = μ Pu,0< μ < 1 } , and show that the set V is bounded. Let u Î V,then u = μ P u ,0<μ < 1. For any t Î [0, T], we have | u ( t ) |≤μ ⎡ ⎣ t  0 ( t − s ) α−1 (α) |f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) |ds +|ν 1 |t α−1 η  0 ( η − s ) 2α−2 (2α − 1) |f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) |ds ⎤ ⎦ . As in part (i) of Lemma 3.1, we have  ( Pu ) ≤ MT α−1  T  ( α +1 ) + |ν 1 |η 2α−1  ( 2α )  < ∞ . This implies that the set V is bounded independently of μ Î (0,1). Using Lemma 3.1 and Theorem 3.1, we obtain that the operator P has at least a fixed p oint, which implies that the problem (1.1) and (1.2)- (1.3) has at least one solution. This completes the proof. Theorem 3.3 Assume that (A 1 ) there exist positive functions L 1 (t), L 2 (t), L 3 (t) such that |f ( t, u ( t ) , ( φu )( t ) , ( ψu )( t )) − f ( t, v ( t ) , ( φv )( t ) , ( ψv )( t )) | ≤ L 1 ( t ) |u − v| + L 2 ( t ) |φu − φv| + L 3 ( t ) | ψ u − ψ v|, ∀t ∈ [0, 1] , u, v ∈ R . (A 2 ) Λ =(ξ 1 +|ν 1 |T a-1 ξ 2 )(1 + g 0 + δ 0 )<1,where γ 0 =sup t∈[0,1] | t  0 γ ( t, s ) ds|, δ 0 =sup t∈[0,1] | t  0 δ ( t, s ) ds|, ξ 1 =sup t∈[0,T]  |I q L 1 ( t ) |, |I q L 2 ( t ) |, |I q L 3 ( t ) |  , ξ 2 =max  |I 2α−1 L 1 ( η ) |, |I 2α−1 L 2 ( η ) |, |I 2α−1 L 3 ( η ) |  , Then the problem (1.1) and (1.2)-(1.3) has a unique solution on C[0, T]. Proof Let us set sup tÎ[0, T] |f(t,0,0,0)| = M, and choose r ≥ εM 1 −  . Ahmad and Nieto Boundary Value Problems 2011, 2011:36 http://www.boundaryvalueproblems.com/content/2011/1/36 Page 5 of 9 Then we show that P B r ⊂ B r , where B r = { x ∈ C :  u  ≤ r } . For x Î B r , we have  ( Pu )( t )  =sup t∈[0,T] | t  0 ( t − s ) α−1  ( α ) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds + ν 1 t α−1 η  0 ( η − s ) 2α−2  ( 2α − 1 ) f ( s, u ( s ) , ( φu )( s ) , ( ψu )( s )) ds| ≤ sup t∈[0,T] ⎛ ⎝ t  0 ( t − s ) α−1  ( α )  |f ( s, x ( s ) , ( φx )( s ) , ( ψx )( s )) − f ( s,0,0,0 ) | + |f ( s,0,0,0 ) |  ds + |ν 1 |t α−1 η  0 ( η − s ) 2α−s  ( 2α − 1 )  |f ( s, x ( s ) , ( φx )( s ) , ( ψx )( s )) − f ( s,0,0,0 ) + |f ( s,0,0,0 ) |  ds  ≤ sup t∈[0,T] ⎛ ⎝ t  0 ( t − s ) α−1  ( α ) ( L 1 ( s ) |x ( s ) | + L 2 ( s ) | ( φx )( s ) | + L 3 ( s ) | ( ψx )( s ) | + M ) ds + |ν 1 |t α−1 η  0 ( η − s ) 2α−2  ( 2α − 1 ) ( L 1 ( s ) |x ( s ) | + L 2 ( s ) | ( φx )( s ) | + L 3 ( s ) | ( ψx )( s ) | + M ) ds ⎞ ⎠ ≤ sup t∈[0,T] ⎛ ⎝ t  0 ( t − s ) α−1  ( α ) ( L 1 ( s ) |x ( s ) | + γ 0 L 2 ( s ) |x ( s ) | + δ 0 L 3 ( s ) |x ( s ) | + M ) ds + |ν 1 |t α−1 η  0 ( η − s ) 2α−2  ( 2α − 1 ) ( L 1 ( s ) |x ( s ) | + γ 0 L 2 ( s ) |x ( s ) | + δ 0 L 3 ( s ) |x ( s ) | + M ) ds ) ≤ sup t∈[0,T]   I α L 1 (t)+γ 0 I α L 2 ( t ) + δ 0 I α L 3 ( t )  r + Mt q   q +1  + |ν 1 |t α−1  I (2α−1) L 1 (η)+γ 0 I (2α−1) L 2 (η)+δ 0 I (2α−1) L 3 (η)  r + Mη 2α−1 (2α)  ≤  ξ 1 + |ν 1 |T α−1 ξ 2  ( 1+γ 0 + δ 0 ) r + MT α−1  T  ( α +1 ) + |ν 1 |η α−1 (2α)  = r + Mε ≤ r In view of (A 1 ), for every t Î [0, T], we have | ( P u )( t ) − ( P v )( t ) | ≤ sup t∈[0,T] ⎛ ⎝ t  0 ( t − s ) α−1  ( α ) |f  s, u ( s ) , ( φu )( s ) , ( ψu )( s ) − f ( s, v ( s ) , ( φv )( s ) , ( ψv )( s )) |ds +|ν 1 |t α−1 η  0 ( η − s ) 2α−2  ( 2α − 1 ) |f  s, u ( s ) , ( φu )( s ) , ( ψu )( s ) − f ( s, υ ( s ) , ( φv )( s ) , ( ψv )( s )) |ds  ≤ sup t∈[0,T] ⎛ ⎝ t  0 ( t − s ) α−1  ( α ) ( L 1 ( s ) |u − v| + L 2 ( s ) |φv| + L 3 ( s ) |ψu − ψv| ) ds +|ν 1 |t α−1 η  0 ( η − s ) 2α−2  ( 2α − 1 ) |  L 1 (s)|u − v| + L 2 (s)|φu − φv| + L 3 (s)|ψu − ψv|   ds  ≤ sup t∈[0,T]  I α L 1 (t)+γ 0 I α L 2 (t)+δ 0 I α L 3 (t)   u − v  +|ν 1 |T α−1  I (2α−1) L 1 (η)+γ 0 I (2α−1) L 2 (η)+δ 0 I (2α−1) L 3 (η)   u − v  ≤  ξ 1 + |ν 1 |T α−1 ξ 2  ( 1+γ 0 + δ 0 )  u − v =   u − v  Ahmad and Nieto Boundary Value Problems 2011, 2011:36 http://www.boundaryvalueproblems.com/content/2011/1/36 Page 6 of 9 By assumption (A 2 ), Λ < 1, therefore, the operator P is a contraction. Hence, by Banach fixed point theorem, we deduce that P has a unique fixed point which in fact is a unique solution of problem (1.1) and (1.2)-(1.3). This completes the proof. □ Theorem 3.4 (Krasnoselskii’ s fixed point theorem [22]). Let M be a closed convex and nonempty subset of a Banach space X. Let A, B be the operators such that (i) x, y ∈ M whenever x, y ∈ M ; (ii) A is compact and continuous; (iii) B is a contraction mapping. Then, there exists z ∈ M such that z = Az + Bz. Theorem 3.5 Assume that f:[0,T]×ℝ × ℝ × ℝ ® ℝ is a continuous function and the following assumptions hold: (H 1 ) |f ( t, u ( t ) , ( φu )( t ) , ( ψu )( t )) − f ( t, v ( t ) , ( φv )( t ) , ( ψv )( t )) | ≤ L 1 ( t ) |u − v| + L 2 ( t ) |φu − φv| + L 3 ( t ) |ψu − ψv|, ∀t ∈ [0, T] , u, v ∈ R . (H 2 )|f (t,u)| ≤ μ(t), ∀(t,u)Î[0, T]×ℝ, and μ Î C([0, T],ℝ + ). If |ν 1 |T α−1 η 2α−1  ( 2α ) < 1 , (3:1) then the boundary value problem (1.1) and (1.2)-(1.3) has at least one solution on [0, T]. Proof Letting sup tÎ[0, T] |μ(t)| = ||μ||, we fix ¯ r ≥ μ  T α−1  T  ( α +1 ) + |ν 1 |η 2α−1  ( 2α )  , and consider B ¯ r = { u ∈ C :  u  ≤ ¯ r } . We define the operators P 1 and P 2 on B ¯ r as ( P 1 u ) (t )= t  0 ( t − s ) α−1 (α) f  s, u(s), (φu)(s), (ψu)(s)  ds, ( P 2 u ) (t )=ν 1 t α−1 η  0 ( η − s ) 2α−s  ( 2α − 1 ) f  s, u ( s ) , ( φu ) (s), (ψu(s))ds  . For u , v ∈ B ¯ r , we find that  P 1 u + P 2 v ≤ μ  T α−1  T  ( α +1 ) + |ν 1 |η 2α−1  ( 2α )  ≤ ¯ r . Thus, P 1 u + P 2 v ∈ B ¯ r . It follows from the assumption (H 1 ) together with (3.1) that P 2 is a contraction mapping. Continuity of f implies that the operator P 1 is continuous. Also, P 1 is uniformly bounded on B ¯ r as  P 1 u ≤  u  T α  ( α +1 ) . Now we prove the compactness of the operator P 1 . In view of (H 1 ), we define sup ( t,x,φx,ψx ) ∈[0,T]×B r ×B r ×B r |f (t, x, φx, ψx)| = ¯ f , and conse- quently we have Ahmad and Nieto Boundary Value Problems 2011, 2011:36 http://www.boundaryvalueproblems.com/content/2011/1/36 Page 7 of 9 | ( P 1 u )( t 1 ) − ( P 2 u ) (t 2 )| =     1 (α) t 1  0  (t 1 − s) α−1 − (t 2 − s) α−1   s, u(s), (φu)(s), (ψu)(s)  d s − 1 (α) t 2  t 1 (t 2 − s) α−1 f  s, u(s), (φu)(s), (ψu)(s)  ds       ≤ ¯ f  ( α +1 ) |2(t 2 − t 1 ) α + t α 1 − t α 2 |, which is i ndependent of u and tends to zero as t 2 ® t 1 .So, P 1 is relatively compact on B ¯ r . Hence, by the Arzelá-Ascoli Theorem, P 1 is compact on B ¯ r . Thus, all the assumptions of Theorem 3.4 are satisfied. So the conclusion of Theorem 3.4 implies that the boundary value problem (1.1) and (1.2)-(1.3) has at least one solution on [0, T]. 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Cambridge University Press, Cambridge (1980) doi:10.1186/1687-2770-2011-36 Cite this article as: Ahmad and Nieto: Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions. Boundary Value Problems 2011 2011:36. Submit your manuscript to a journal and benefi t from: 7 Convenient online submission 7 Rigorous peer review 7 Immediate publication on acceptance 7 Open access: articles freely available online 7 High visibility within the fi eld 7 Retaining the copyright to your article Submit your next manuscript at 7 springeropen.com Ahmad and Nieto Boundary Value Problems 2011, 2011:36 http://www.boundaryvalueproblems.com/content/2011/1/36 Page 9 of 9 . article Abstract This article investigates a boundary value problem of Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions. Some new existence. (1980) doi:10.1186/1687-2770-2011-36 Cite this article as: Ahmad and Nieto: Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions. Boundary Value Problems 2011 2011:36. Submit. RESEARC H Open Access Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions Bashir Ahmad 1* and Juan J Nieto 1,2 *