Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions

  • Bashir Ahmad1Email author and

    Affiliated with

    • Juan J Nieto1, 2

      Affiliated with

      Boundary Value Problems20112011:36

      DOI: 10.1186/1687-2770-2011-36

      Received: 20 July 2011

      Accepted: 17 October 2011

      Published: 17 October 2011

      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 fractional 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 ) , ( ϕ u ) ( t ) , ( ψ u ) ( t ) ) , t 0 , T , α 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ1_HTML.gif
      (1.1)
      subject to the boundary conditions of fractional order given by
      D α - 2 u ( 0 + ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ2_HTML.gif
      (1.2)
      D α - 1 u ( 0 + ) = ν I α - 1 u ( η ) , 0 < η < T , ν is a constant , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ3_HTML.gif
      (1.3)
      where D α denotes the Riemann-Liouville fractional derivative of order α, f: [0, T] × ℝ × ℝ × ℝ → ℝ is continuous, and
      ( ϕ x ) ( t ) = 0 t γ ( t , s ) x ( s ) d s , ( ψ x ) ( t ) = 0 t δ ( t , s ) x ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equa_HTML.gif

      with γ 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 integer-order models. Fractional differential equations arise in many engineering and scientific disciplines, such as physics, chemistry, biology, economics, control theory, signal and image processing, biophysics, blood flow phenomena, aerodynamics, fitting of experimental data, etc. (see [1, 2]). For some recent development on the topic, (see [319] and references therein).

      2 Preliminaries

      Let us recall some basic definitions (see [20, 21]).

      Definition 2.1 The Riemann-Liouville fractional integral of order α > 0 for a continuous function u: (0, ∞) → ℝ is defined as
      I α u ( t ) = 1 Γ ( α ) 0 t ( t - s ) α - 1 u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equb_HTML.gif

      provided the integral exists.

      Definition 2.2 For a continuous function u: (0, ∞) → ℝ, the Riemann-Liouville derivative of fractional order α > 0, n = [α] + 1 ([α] denotes the integer part of the real number α) is defined as
      D α u ( t ) = 1 Γ ( n - α ) 1 d t n 0 t ( t - s ) n - α - 1 u ( s ) d s = d d t n I n - α u ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equc_HTML.gif

      provided it exists.

      For α < 0, we use the convention that D α u = I u. Also for β ∈ [0, α), it is valid that D β I a u = I α-β u.

      Note that for λ >-1, λα - 1, α - 2,..., α - n, we have
      D α t λ = Γ ( λ + 1 ) Γ ( λ - α + 1 ) t λ - α , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equd_HTML.gif
      and
      D α t α - i = 0 , i = 1 , 2 , , n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Eque_HTML.gif
      In particular, for the constant function u(t) = 1, we obtain
      D α 1 = 1 Γ ( 1 - α ) t - α , α . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equf_HTML.gif

      For α ∈ ℕ, we get, of course, D α 1 = 0 because of the poles of the gamma function at the points 0, -1, -2,....

      For α > 0, the general solution of the homogeneous equation
      D α u ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equg_HTML.gif
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equh_HTML.gif

      where c i , i = 1, 2,..., n - 1, are arbitrary real constants.

      We always have D α I α 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equi_HTML.gif
      To define the solution for the nonlinear problem (1.1) and (1.2)-(1.3), we consider the following linear equation
      D α u ( t ) = σ ( t ) , α 1 , 2 , t 0 , T , T > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ4_HTML.gif
      (2.1)

      where σC[0, T].

      We define
      A = ν 0 η s α - 1 ( η - s ) α - 2 Γ ( α - 1 ) d s = ν Γ ( α ) η 2 α - 2 Γ ( 2 α - 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ5_HTML.gif
      (2.2)

      such that A ≠ Γ(α).

      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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ6_HTML.gif
      (2.3)

      with I α the usual Riemann-Liouville fractional integral of order α.

      From (2.3), we have
      D α - 1 u ( t ) = c 1 Γ ( α ) + I 1 σ ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ7_HTML.gif
      (2.4)
      D α - 2 u ( t ) = c 1 Γ ( α ) t + c 0 Γ ( α - 1 ) + I 2 σ ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ8_HTML.gif
      (2.5)
      Using the conditions (1.2) and (1.3) in (2.4) and (2.5), we find that c0 = 0 and
      c 1 = ν Γ ( α ) - A 0 η ( η - s ) α - 2 Γ ( α - 1 ) 0 s ( s - x ) α - 1 Γ ( α ) σ ( x ) d x d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equj_HTML.gif

      where A is defined by (2.2).

      Substituting the values of c0 and c1 in (2.3), the unique solution of (2.1) subject to the boundary conditions (1.2)-(1.3) is given by
      u ( t ) = 0 t ( t - s ) α - 1 Γ ( α ) σ ( s ) d s + ν t α - 1 Γ ( α ) - A 0 η ( η - s ) α - 2 Γ ( α - 1 ) 0 s ( s - x ) α - 1 Γ ( α ) σ ( x ) d x d s = 0 t ( t - s ) α - 1 Γ ( α ) σ ( s ) d s + ν t α - 1 Γ ( α ) - A I 2 α - 1 σ ( η ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ9_HTML.gif
      (2.6)

      3 Main results

      Let C = C ( [ 0 , T ] , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq1_HTML.gif denotes the Banach space of all continuous functions from [0, T] → ℝ endowed with the norm defined by ║u║ = sup{|u(t)|, t ∈ [0, T]}.

      If u is a solution of (1.1) and (1.2)-(1.3), then
      u ( t ) = 0 t ( t - s ) α - 1 Γ ( α ) f s , u s , ϕ u s , ψ u s d s + ν 1 t α - 1 0 η ( η - s ) 2 α - 2 Γ ( 2 α - 1 ) f s , u s , ϕ u s , ψ u s d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equk_HTML.gif
      where
      ν 1 = ν Γ ( α ) - A . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equl_HTML.gif
      Define an operator P : C C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq2_HTML.gif as
      P u ( t ) = 0 t ( t - s ) α - 1 Γ ( α ) f s , u s , ϕ u s , ψ u s d s + ν 1 t α - 1 0 η ( η - s ) 2 α - 2 Γ ( 2 α - 1 ) f s , u s , ϕ u s , ψ u s d s , t 0 , T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equm_HTML.gif

      Observe that the problem (1.1) and (1.2)-(1.3) has solutions if and only if the operator equation P u = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq3_HTML.gif has fixed points.

      Lemma 3.1 The operator P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq4_HTML.gif is compact.

      Proof
      1. (i)
        Let B be a bounded set in C[0, T]. Then, there exists a constant M such that |f(t,u(t), (φu)(t), (ψu)(t))| ≤ M, ∀u ∈ B, t∈[0, T]. Thus
        | P u ( t ) | M 0 t ( t - s ) α - 1 Γ ( α ) d s + M | ν 1 | t α - 1 0 η ( η - s ) 2 α - 2 Γ ( 2 α - 1 ) d s M T α - 1 T Γ ( α + 1 ) + | ν 1 | η 2 α - 1 Γ ( 2 α ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equn_HTML.gif
         
      which implies that
      | | P u | | M T α - 1 T Γ ( α + 1 ) + | ν 1 | η 2 α - 1 Γ ( 2 α ) < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equo_HTML.gif
      Hence, P ( B ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq5_HTML.gif is uniformly bounded.
      1. (ii)
        For any t 1, t 2 ∈ [0, T], uB, we have
        | P u t 1 - P u t 2 | = 0 t 1 t 1 - s α - 1 Γ ( α ) f s , u s , ϕ u s , ψ u s d s - 0 t 2 t 2 - s α - 1 Γ ( α ) f s , u s , ϕ u s , ψ u s d s + ν 1 t 1 α - 1 - t 2 α - 1 0 η η - s 2 α - s Γ 2 α - 1 f s , u s , ϕ u s , ψ u s d s M 1 Γ ( α ) 0 t 1 t 1 - s α - 1 - t 2 - s α - 1 d s - 1 Γ ( α ) t 1 t 2 t 2 - s α - 1 d s + ν 1 t 1 α - 1 - t 2 α - 1 0 η η - s 2 α - s Γ 2 α - 1 d s 0 as t 1 t 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equp_HTML.gif
         

      Thus, P ( B ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq5_HTML.gif is equicontinuous. Consequently, the operator P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq4_HTML.gif is compact. This completes the proof.   □

      We need the following known fixed point theorem to prove the existence of solutions for the problem at hand.

      Theorem 3.1 ([22]) Let E be a Banach space. Assume that T: EE be a completely continuous operator and the set V = {xE | 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 > 0 such that
      | f t , u t , ϕ u t , ψ u t | M , t 0 , T , u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equq_HTML.gif

      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 | u = μ P u , 0 < μ < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equr_HTML.gif
      and show that the set V is bounded. Let uV, then u = μ P u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq6_HTML.gif, 0 < μ < 1. For any t ∈ [0, T], we have
      | u t | μ 0 t t - s α - 1 Γ ( α ) | f s , u s , ϕ u s , ψ u s | d s + | ν 1 | t α - 1 0 η η - s 2 α - 2 Γ ( 2 α - 1 ) | f s , u s , ϕ u s , ψ u s | d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equs_HTML.gif
      As in part (i) of Lemma 3.1, we have
      P u M T α - 1 T Γ α + 1 + | ν 1 | η 2 α - 1 Γ ( 2 α ) < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equt_HTML.gif

      This implies that the set V is bounded independently of μ ∈ (0,1). Using Lemma 3.1 and Theorem 3.1, we obtain that the operator P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq4_HTML.gif has at least a fixed point, 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

      (A1) there exist positive functions L1(t), L2(t), L3(t) such that
      | f t , u t , ϕ u t , ψ u t - f t , v t , ϕ v t , ψ v t | L 1 t | u - v | + L 2 t | ϕ u - ϕ v | + L 3 t | ψ u - ψ v | , t 0 , 1 , u , v . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equu_HTML.gif
      (A2) Λ = (ξ1 + |ν1|Tα-1ξ2)(1 + γ0 + δ0) < 1, where
      γ 0 = sup t 0 , 1 | 0 t γ t , s d s | , δ 0 = sup t 0 , 1 | 0 t δ t , s d s | , 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 2 α - 1 L 1 η | , | I 2 α - 1 L 2 η | , | I 2 α - 1 L 3 η | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equv_HTML.gif

      Then the problem (1.1) and (1.2)-(1.3) has a unique solution on C[0, T].

      Proof Let us set supt∈[0, T]|f(t,0,0,0)| = M, and choose
      r ε M 1 - Λ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equw_HTML.gif
      Then we show that P B r B r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq7_HTML.gif, where B r = x C : u r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq8_HTML.gif. For x ∈ B r , we have
      P u t = sup t 0 , T | 0 t t - s α - 1 Γ α f s , u s , ϕ u s , ψ u s d s + ν 1 t α - 1 0 η η - s 2 α - 2 Γ 2 α - 1 f s , u s , ϕ u s , ψ u s d s | sup t 0 , T 0 t t - s α - 1 Γ α | f s , x s , ϕ x s , ψ x s - f s , 0 , 0 , 0 | + | f s , 0 , 0 , 0 | d s + | ν 1 | t α - 1 0 η η - s 2 α - s Γ 2 α - 1 | f s , x s , ϕ x s , ψ x s - f s , 0 , 0 , 0 + | f s , 0 , 0 , 0 | d s sup t 0 , T 0 t t - s α - 1 Γ α L 1 s | x s | + L 2 s | ϕ x s | + L 3 s | ψ x s | + M d s + | ν 1 | t α - 1 0 η η - s 2 α - 2 Γ 2 α - 1 L 1 s | x s | + L 2 s | ϕ x s | + L 3 s | ψ x s | + M d s sup t 0 , T 0 t t - s α - 1 Γ α L 1 s | x s | + γ 0 L 2 s | x s | + δ 0 L 3 s | x s | + M d s + | ν 1 | t α - 1 0 η η - s 2 α - 2 Γ 2 α - 1 L 1 s | x s | + γ 0 L 2 s | x s | + δ 0 L 3 s | x s | + M d s sup t 0 , T I α L 1 ( t ) + γ 0 I α L 2 t + δ 0 I α L 3 t r + M t q Γ q + 1 + | ν 1 | t α - 1 I ( 2 α - 1 ) L 1 ( η ) + γ 0 I ( 2 α - 1 ) L 2 ( η ) + δ 0 I ( 2 α - 1 ) L 3 ( η ) r + M η 2 α - 1 Γ ( 2 α ) ξ 1 + | ν 1 | T α - 1 ξ 2 1 + γ 0 + δ 0 r + M T α - 1 T Γ α + 1 + | ν 1 | η α - 1 Γ ( 2 α ) = Λ r + M ε r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equx_HTML.gif
      In view of (A1), for every t ∈ [0, T], we have
      | P u t - P v t | sup t 0 , T 0 t t - s α - 1 Γ α | f s , u s , ϕ u s , ψ u s - f s , v s , ϕ v s , ψ v s | d s + | ν 1 | t α - 1 0 η η - s 2 α - 2 Γ 2 α - 1 | f s , u s , ϕ u s , ψ u s - f s , υ s , ϕ v s , ψ v s | d s sup t 0 , T 0 t t - s α - 1 Γ α L 1 s | u - v | + L 2 s | ϕ v | + L 3 s | ψ u - ψ v | d s + | ν 1 | t α - 1 0 η η - s 2 α - 2 Γ 2 α - 1 | L 1 ( s ) | u - v | + L 2 ( s ) | ϕ u - ϕ v | + L 3 ( s ) | ψ u - ψ v | d s 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 ( 2 α - 1 ) L 1 ( η ) + γ 0 I ( 2 α - 1 ) L 2 ( η ) + δ 0 I ( 2 α - 1 ) L 3 ( η ) u - v ξ 1 + | ν 1 | T α - 1 ξ 2 1 + γ 0 + δ 0 u - v = Λ u - v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equy_HTML.gif

      By assumption (A2), Λ < 1, therefore, the operator P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq4_HTML.gif is a contraction. Hence, by Banach fixed point theorem, we deduce that P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq4_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq9_HTML.gifbe a closed convex and nonempty subset of a Banach space X. Let A, B be the operators such that (i) A x + B y M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq10_HTML.gifwhenever x , y M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq11_HTML.gif; (ii) A is compact and continuous; (iii) B is a contraction mapping. Then, there exists z M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq12_HTML.gifsuch 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 , ϕ u t , ψ u t - f t , v t , ϕ v t , ψ v t | L 1 ( t ) | u - v | + L 2 ( t ) | ϕ u - ϕ v | + L 3 ( t ) | ψ u - ψ v | , t 0 , T , u , v . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equz_HTML.gif

      (H 2 ) |f (t,u)| ≤ μ(t), ∀(t,u)∈[0, T] × ℝ, and μC([0, T],ℝ+).

      If
      | ν 1 | T α - 1 η 2 α - 1 Γ 2 α < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equ10_HTML.gif
      (3.1)

      then the boundary value problem (1.1) and (1.2)-(1.3) has at least one solution on [0, T].

      Proof Letting supt∈[0, T]|μ(t)| = ||μ||, we fix
      r ̄ μ T α - 1 T Γ ( α + 1 ) + | ν 1 | η 2 α - 1 Γ ( 2 α ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equaa_HTML.gif
      and consider B r ̄ = u C : u r ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq13_HTML.gif. We define the operators P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq14_HTML.gif and P 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq15_HTML.gif on B r ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq16_HTML.gif as
      P 1 u ( t ) = 0 t t - s α - 1 Γ ( α ) f s , u ( s ) , ( ϕ u ) ( s ) , ( ψ u ) ( s ) d s , P 2 u ( t ) = ν 1 t α - 1 0 η η - s 2 α - s Γ 2 α - 1 f s , u s , ϕ u ( s ) , ( ψ u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equab_HTML.gif
      For u , v B r ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq17_HTML.gif, we find that
      P 1 u + P 2 v μ T α - 1 T Γ ( α + 1 ) + | ν 1 | η 2 α - 1 Γ ( 2 α ) r ̄ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equac_HTML.gif

      Thus, P 1 u + P 2 v B r ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq18_HTML.gif. It follows from the assumption (H1) together with (3.1) that P 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq15_HTML.gif is a contraction mapping. Continuity of f implies that the operator P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq14_HTML.gif is continuous.

      Also, P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq14_HTML.gif is uniformly bounded on B r ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq16_HTML.gif as
      P 1 u u T α Γ α + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equad_HTML.gif

      Now we prove the compactness of the operator P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq14_HTML.gif.

      In view of (H1), we define sup ( t , x , ϕ x , ψ x ) 0 , T × B r × B r × B r | f ( t , x , ϕ x , ψ x ) | = f ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq19_HTML.gif, and consequently we have
      | P 1 u t 1 - P 2 u ( t 2 ) | = 1 Γ ( α ) 0 t 1 ( t 1 - s ) α - 1 - ( t 2 - s ) α - 1 s , u ( s ) , ( ϕ u ) ( s ) , ( ψ u ) ( s ) d s - 1 Γ ( α ) t 1 t 2 ( t 2 - s ) α - 1 f s , u ( s ) , ( ϕ u ) ( s ) , ( ψ u ) ( s ) d s f ̄ Γ ( α + 1 ) | 2 ( t 2 - t 1 ) α + t 1 α - t 2 α | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_Equae_HTML.gif

      which is independent of u and tends to zero as t2t1. So, P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq14_HTML.gif is relatively compact on B r ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq16_HTML.gif. Hence, by the Arzelá-Ascoli Theorem, P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq14_HTML.gif is compact on B r ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-36/MediaObjects/13661_2011_Article_82_IEq16_HTML.gif. 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]. This completes the proof.   □

      Declarations

      Acknowledgements

      This study was partially supported by Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia.

      Authors’ Affiliations

      (1)
      Department of Mathematics, Faculty of Science, King Abdulaziz University
      (2)
      Departamento de Análisis Matemático, Facultad de Matemáticas Universidad de Santiago de Compostela

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      © Ahmad and Nieto; licensee Springer. 2011

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