Solvability for fractional order boundary value problems at resonance

  • Zhigang Hu1Email author and

    Affiliated with

    • Wenbin Liu1

      Affiliated with

      Boundary Value Problems20112011:20

      DOI: 10.1186/1687-2770-2011-20

      Received: 10 May 2011

      Accepted: 5 September 2011

      Published: 5 September 2011

      Abstract

      In this paper, by using the coincidence degree theory, we consider the following boundary value problem for fractional differential equation

      D 0 + α x ( t ) = f ( t , x ( t ) , x ( t ) , x ( t ) ) , t [ 0 , 1 ] , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equa_HTML.gif

      where D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq1_HTML.gif denotes the Caputo fractional differential operator of order α, 2 < α ≤ 3. A new result on the existence of solutions for above fractional boundary value problem is obtained.

      Mathematics Subject Classification (2000): 34A08, 34B15.

      Keywords

      Fractional differential equations boundary value problems resonance coincidence degree theory

      1 Introduction

      Fractional calculus is a generalization of ordinary differentiation and integration on an arbitrary order that can be noninteger. This subject, as old as the problem of ordinary differential calculus, can go back to the times when Leibniz and Newton invented differential calculus. As is known to all, the problem for fractional derivative was originally raised by Leibniz in a letter, dated September 30, 1695.

      In recent years, the fractional differential equations have received more and more attention. The fractional derivative has been occurring in many physical applications such as a non-Markovian diffusion process with memory [1], charge transport in amorphous semiconductors [2], propagations of mechanical waves in viscoelastic media [3], etc. Phenomena in electromagnetics, acoustics, viscoelasticity, electrochemistry, and material science are also described by differential equations of fractional order (see [49]).

      Recently, boundary value problems (BVPs for short) for fractional differential equations at nonresonance have been studied in many papers (see [1016]). Moreover, Kosmatov studied the BVPs for fractional differential equations at resonance (see [17]). Motivated by the work above, in this paper, we consider the following BVP of fractional equation at resonance
      D 0 + α x ( t ) = f ( t , x ( t ) , x ( t ) , x ( t ) ) , t [ 0 , 1 ] , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ1_HTML.gif
      (1.1)

      where D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq1_HTML.gif denotes the Caputo fractional differential operator of order α, 2 < α ≤ 3. f : [0, 1] × ℝ3 → ×ℝ is continuous.

      The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions, and lemmas. In Section 3, we establish a theorem on existence of solutions for BVP (1.1) under nonlinear growth restriction of f, basing on the coincidence degree theory due to Mawhin (see [18]). Finally, in Section 4, an example is given to illustrate the main result.

      2 Preliminaries

      In this section, we will introduce notations, definitions, and preliminary facts that are used throughout this paper.

      Let X and Y be real Banach spaces and let L : domLXY be a Fredholm operator with index zero, and P : XX, Q : YY be projectors such that
      Im P = Ker L , Ker Q = Im L , X = Ker L Ker P , Y = Im L Im Q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equb_HTML.gif
      It follows that
      L | dom L Ker P : dom L Ker P Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equc_HTML.gif

      is invertible. We denote the inverse by K P .

      If Ω is an open bounded subset of X, and dom L Ω ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq2_HTML.gif, the map N : XY will be called L-compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq3_HTML.gif if Q N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq4_HTML.gif is bounded and K P ( I - Q ) N : Ω ¯ X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq5_HTML.gif is compact. Where I is identity operator.

      Lemma 2.1. ([18]) If Ω is an open bounded set, let L : domLXY be a Fredholm operator of index zero and N : XY L-compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq3_HTML.gif. Assume that the following conditions are satisfied
      1. (1)

        LxλNx for every (x, λ) ∈ [(domL\KerL)] ∩ ∂Ω × (0, 1);

         
      2. (2)

        Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω;

         
      3. (3)

        deg(QN|KerL, KerL ∩ Ω, 0) ≠ 0, where Q : YY is a projection such that ImL = KerQ.

         

      Then the equation Lx = Nx has at least one solution in dom L Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq6_HTML.gif.

      Definition 2.1. The Riemann-Liouville fractional integral operator of order α > 0 of a function x is given by
      I 0 + α x ( t ) = 1 Γ ( α ) 0 t ( t - s ) α - 1 x ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equd_HTML.gif

      provided that the right side integral is pointwise defined on (0, +∞).

      Definition 2.2. The Caputo fractional derivative of order α > 0 of a continuous function x is given by
      D 0 + α x ( t ) = I 0 + n - α d n x ( t ) d t n = 1 Γ ( n - α ) 0 t ( t - s ) n - α - 1 x ( n ) ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Eque_HTML.gif

      where n is the smallest integer greater than or equal to α, provided that the right side integral is pointwise defined on (0, +∞).

      Lemma 2.2. ([19]) For α > 0, the general solution of the Caputo fractional differential equation
      D 0 + α x ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equf_HTML.gif
      is given by
      x ( t ) = c 0 + c 1 t + c 2 t 2 + + c n - 1 t n - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equg_HTML.gif

      where c i ∈ ℝ, i = 0, 1, 2, . . ., n - 1; here, n is the smallest integer greater than or equal to α.

      Lemma 2.3. ([19]) Assume that xC(0, 1) ∩ L(0, 1) with a Caputo fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L(0, 1). Then,
      I 0 + α D 0 + α x ( t ) = x ( t ) + c 0 + c 1 t + c 2 t 2 + + c n - 1 t n - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equh_HTML.gif

      where c i ∈ ℝ, i = 0, 1, 2, . . ., n - 1; here, n is the smallest integer greater than or equal to α.

      In this paper, we denote X = C2[0, 1] with the norm ||x|| X = max{||x||, ||x'||, ||x"||} and Y = C[0, 1] with the norm ||y|| Y = ||y||, where ||x|| = max t ∈[0, 1] |x(t)|. Obviously, both X and Y are Banach spaces.

      Define the operator L : domLXY by
      L x = D 0 + α x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ2_HTML.gif
      (2.1)
      where
      dom L = { x X | D 0 + α x ( t ) Y , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 0 ) = 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equi_HTML.gif
      Let N : XY be the Nemytski operator
      N x ( t ) = f ( t , x ( t ) , x ( t ) , x ( t ) ) , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equj_HTML.gif
      Then, BVP (1.1) is equivalent to the operator equation
      L x = N x , x dom L . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equk_HTML.gif

      3 Main result

      In this section, a theorem on existence of solutions for BVP (1.1) will be given.

      Theorem 3.1. Let f : [0, 1] × ℝ3 → ℝ be continuous. Assume that

      (H1) there exist nonnegative functions p, q, r, sC[0, 1] with Γ(α - 1) - q1 - r1 - s1 > 0 such that
      | f ( t , u , v , w ) | p ( t ) + q ( t ) | u | + r ( t ) | v | + s ( t ) | w | , t [ 0 , 1 ] , ( u , v , w ) 3 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equl_HTML.gif

      where p1 = ||p||, q1 = ||q||, r1 = ||r||, s1 = ||s||.

      (H2) there exists a constant B > 0 such that for all u ∈ ℝ with |u| > B either
      u f ( t , u , v , w ) > 0 , t [ 0 , 1 ] , ( v , w ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equm_HTML.gif
      or
      u f ( t , u , v , w ) < 0 , t [ 0 , 1 ] , ( v , w ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equn_HTML.gif

      Then, BVP (1.1) has at leat one solution in X.

      Now, we begin with some lemmas below.

      Lemma 3.1. Let L be defined by (2.1), then
      Ker L = { x X | x ( t ) = c 0 , c 0 , t [ 0 , 1 ] } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ3_HTML.gif
      (3.1)
      Im L = { y Y | 0 1 ( 1 - s ) α - 1 y ( s ) d s = 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ4_HTML.gif
      (3.2)
      Proof. By Lemma 2.2, D 0 + α x ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq7_HTML.gif has solution
      x ( t ) = c 0 + c 1 t + c 2 t 2 , c 0 , c 1 , c 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equo_HTML.gif

      Combining with the boundary value condition of BVP (1.1), one has (3.1) hold.

      For y ∈ ImL, there exists x ∈ domL such that y = LxY. By Lemma 2.3, we have
      x ( t ) = 1 Γ ( α ) 0 t ( t - s ) α - 1 y ( s ) d s + c 0 + c 1 t + c 2 t 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equp_HTML.gif
      Then, we have
      x ( t ) = 1 Γ ( α - 1 ) 0 t ( t - s ) α - 2 y ( s ) d s + c 1 + 2 c 2 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equq_HTML.gif
      and
      x ( t ) = 1 Γ ( α - 2 ) 0 t ( t - s ) α - 3 y ( s ) d s + 2 c 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equr_HTML.gif
      By conditions of BVP (1.1), we can get that y satisfies
      0 1 ( 1 - s ) α - 1 y ( s ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equs_HTML.gif

      Thus, we get (3.2). On the other hand, suppose yY and satisfies 0 1 ( 1 - s ) α - 1 y ( s ) d s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq8_HTML.gif. Let x ( t ) = I 0 + α y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq9_HTML.gif, then x ∈ domL and D 0 + α x ( t ) = y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq10_HTML.gif. So that, y ∈ ImL. The proof is complete.

      Lemma 3.2. Let L be defined by (2.1), then L is a Fredholm operator of index zero, and the linear continuous projector operators P : XX and Q : YY can be defined as
      P x ( t ) = x ( 0 ) , t [ 0 , 1 ] , Q y ( t ) = α 0 1 ( 1 - s ) α - 1 y ( s ) d s , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equt_HTML.gif
      Furthermore, the operator K P : ImL → domL ∩ KerP can be written by
      K P y ( t ) = 1 Γ ( α ) 0 t ( t - s ) α - 1 y ( s ) d s , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equu_HTML.gif
      Proof. Obviously, ImP = KerL and P2x = Px. It follows from x = (x - Px) + Px that X = KerP + KerL. By simple calculation, we can get that KerL ∩ KerP = {0}. Then, we get
      X = Ker L Ker P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equv_HTML.gif
      For yY, we have
      Q 2 y = Q ( Q y ) = Q y α 0 1 ( 1 - s ) α - 1 d s = Q y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equw_HTML.gif
      Let y = (y - Qy) + Qy, where y - Qy ∈ KerQ = ImL, Qy ∈ ImQ. It follows from KerQ = ImL and Q2y = Qy that ImQ ∩ ImL = {0}. Then, we have
      Y = Im L Im Q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equx_HTML.gif
      Thus,
      dim Ker L = dim Im Q = codim Im L = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equy_HTML.gif

      This means that L is a Fredholm operator of index zero.

      From the definitions of P, K P , it is easy to see that the generalized inverse of L is K P . In fact, for y ∈ ImL, we have
      L K P y = D 0 + α I 0 + α y = y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ5_HTML.gif
      (3.3)
      Moreover, for x ∈ domL ∩ KerP, we get x(0) = x'(0) = x"(0) = 0. By Lemma 2.3, we obtain that
      I 0 + α L x ( t ) = I 0 + α D 0 + α x ( t ) = x ( t ) + c 0 + c 1 t + c 2 t 2 , c 0 , c 1 , c 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equz_HTML.gif
      which together with x(0) = x'(0) = x"(0) = 0 yields that
      K P L x = x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ6_HTML.gif
      (3.4)

      Combining (3.3) with (3.4), we know that K P = (L|domL∩KerP)-1. The proof is complete.

      Lemma 3.3. Assume Ω ⊂ X is an open bounded subset such that dom L Ω ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq2_HTML.gif, then N is L-compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq3_HTML.gif.

      Proof. By the continuity of f, we can get that Q N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq4_HTML.gif and K P ( I - Q ) N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq11_HTML.gif are bounded. So, in view of the Arzelà -Ascoli theorem, we need only prove that K P ( I - Q ) N ( Ω ¯ ) X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq12_HTML.gif is equicontinuous.

      From the continuity of f, there exists constant A > 0 such that |(I - Q)Nx| ≤ A, x Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq13_HTML.gif, t ∈ [0, 1]. Furthermore, denote K P,Q = K P (I - Q)N and for 0 ≤ t1 < t2 ≤ 1, x Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq14_HTML.gif, we have
      ( K P , Q x ) ( t 2 ) - ( K P , Q x ) ( t 1 ) 1 Γ ( α ) 0 t 2 ( t 2 - s ) α - 1 ( I - Q ) N x ( s ) d s - 0 t 1 ( t 1 - s ) α - 1 ( I - Q ) N x ( s ) d s A Γ ( α ) 0 t 1 ( t 2 - s ) α - 1 - ( t 1 - s ) α - 1 d s + t 1 t 2 ( t 2 - s ) α - 1 d s = A Γ ( α + 1 ) ( t 2 α - t 1 α ) , | ( K P , Q x ) ( t 2 ) - ( K P , Q x ) ( t 1 ) | = α - 1 Γ ( α ) 0 t 2 ( t 2 - s ) α - 2 ( I - Q ) N x ( s ) d s - 0 t 1 ( t 1 - s ) α - 2 ( I - Q ) N x ( s ) d s A Γ ( α - 1 ) 0 t 1 ( t 2 - s ) α - 2 - ( t 1 - s ) α - 2 d s + t 1 t 2 ( t 2 - s ) α - 2 d s A Γ ( α ) ( t 2 α - 1 - t 1 α - 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equaa_HTML.gif
      and
      | ( K P , Q x ) ( t 2 ) ( K P , Q x ) ( t 1 ) | = ( α 2 ) ( α 1 ) Γ ( α ) | 0 t 2 ( t 2 s ) α 3 ( I Q ) N x ( s ) d s 0 t 1 ( t 1 s ) α 3 ( I Q ) N x ( s ) d s | A Γ ( α 2 ) [ 0 t 1 ( t 1 s ) α 3 ( t 2 s ) α 3 d s + t 1 t 2 ( t 2 s ) α 3 d s ] A Γ ( α 1 ) [ t 1 α 2 t 2 α 2 + 2 ( t 2 t 1 ) α 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equab_HTML.gif

      Since t α , t α -1 and t α -2 are uniformly continuous on [0, 1], we can get that K P , Q ( Ω ¯ ) C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq15_HTML.gif, ( K P , Q ) ( Ω ¯ ) C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq16_HTML.gif and ( K P , Q ) ( Ω ¯ ) C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq17_HTML.gif are equicontinuous. Thus, we get that K P , Q : Ω ¯ X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq18_HTML.gif is compact. The proof is completed.

      Lemma 3.4. Suppose (H1), (H2) hold, then the set
      Ω 1 = { x dom L \ Ker L | L x = λ N x , λ ( 0 , 1 ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equac_HTML.gif

      is bounded.

      Proof. Take x ∈ Ω1, then Nx ∈ ImL. By (3.2), we have
      0 1 ( 1 - s ) α - 1 f ( s , x ( s ) , x ( s ) , x ( s ) ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equad_HTML.gif

      Then, by the integral mean value theorem, there exists a constant ξ ∈ (0, 1) such that f(ξ, x(ξ), x'(ξ), x"(ξ)) = 0. Then from (H2), we have |x(ξ)| ≤ B.

      Then, we have
      | x ( t ) | = x ( ξ ) + ξ t x ( s ) d s B + x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equae_HTML.gif
      That is
      x B + x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ7_HTML.gif
      (3.5)
      From x ∈ domL, we get x'(0) = 0. Therefore,
      | x ( t ) | = x ( 0 ) + 0 t x ( s ) d s x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equaf_HTML.gif
      That is
      x x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ8_HTML.gif
      (3.6)
      By Lx = λNx and x ∈ domL, we have
      x ( t ) = λ Γ ( α ) 0 t ( t - s ) α - 1 f ( s , x ( s ) , x ( s ) , x ( s ) ) d s + x ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equag_HTML.gif
      Then we get
      x ( t ) = λ Γ ( α - 1 ) 0 t ( t - s ) α - 2 f ( s , x ( s ) , x ( s ) , x ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equah_HTML.gif
      and
      x ( t ) = λ Γ ( α - 2 ) 0 t ( t - s ) α - 3 f ( s , x ( s ) , x ( s ) , x ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equai_HTML.gif
      From (3.5),(3.6), and (H1), we have
      x 1 Γ ( α - 2 ) 0 t ( t - s ) α - 3 | f ( s , x ( s ) , x ( s ) , x ( s ) ) | d s 1 Γ ( α - 2 ) 0 t ( t - s ) α - 3 [ p ( s ) + q ( s ) | x ( s ) | + r ( s ) | x ( s ) | + s ( s ) | x ( s ) | ] d s 1 Γ ( α - 2 ) 0 t ( t - s ) α - 3 ( p 1 + q 1 x + r 1 x + s 1 x ) d s 1 Γ ( α - 2 ) 0 t ( t - s ) α - 3 [ p 1 + q 1 B + ( q 1 + r 1 + s 1 ) x ] d s 1 Γ ( α - 1 ) [ p 1 + q 1 B + ( q 1 + r 1 + s 1 ) x ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equaj_HTML.gif
      Thus, from Γ(α - 1) - q1 - r1 - s1 > 0, we obtain that
      x p 1 + q 1 B Γ ( α - 1 ) - q 1 - r 1 - s 1 : = M 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equak_HTML.gif
      Thus, we get
      x x M 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equal_HTML.gif
      and
      x B + x B + M 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equam_HTML.gif
      Therefore,
      x X max { M 1 , B + M 1 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equan_HTML.gif

      So Ω1 is bounded. The proof is complete.

      Lemma 3.5. Suppose (H2) holds, then the set
      Ω 2 = { x | x Ker L , N x Im L } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equao_HTML.gif

      is bounded.

      Proof. For x ∈ Ω2, we have x(t) = c, c ∈ ℝ, and Nx ∈ ImL. Then, we get
      0 1 ( 1 - s ) α - 1 f ( s , c , 0 , 0 ) d s = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equap_HTML.gif
      which together with (H2) implies |c| ≤ B. Thus, we have
      x X B . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equaq_HTML.gif

      Hence, Ω2 is bounded. The proof is complete.

      Lemma 3.6. Suppose the first part of (H2) holds, then the set
      Ω 3 = { x | x Ker L , λ x + ( 1 - λ ) Q N x = 0 , λ [ 0 , 1 ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equar_HTML.gif

      is bounded.

      Proof. For x ∈ Ω3, we have x(t) = c, c ∈ ℝ, and
      λ c + ( 1 - λ ) α 0 1 ( 1 - s ) α - 1 f ( s , c , 0 , 0 ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ9_HTML.gif
      (3.7)
      If λ = 0, then |c| ≤ B because of the first part of (H2). If λ ∈ (0, 1], we can also obtain |c| ≤ B. Otherwise, if |c| > B, in view of the first part of (H2), one has
      λ c 2 + ( 1 - λ ) α 0 1 ( 1 - s ) α - 1 c f ( s , c , 0 , 0 ) d s > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equas_HTML.gif

      which contradicts to (3.7).

      Therefore, Ω3 is bounded. The proof is complete.

      Remark 3.1. Suppose the second part of (H2) hold, then the set
      Ω 3 = { x | x Ker L , - λ x + ( 1 - λ ) Q N x = 0 , λ [ 0 , 1 ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equat_HTML.gif

      is bounded.

      The proof of Theorem 3.1. Set Ω = {xX | ||x|| X < max{M1, B, B + M1} + 1}. It follows from Lemma 3.2 and 3.3 that L is a Fredholm operator of index zero and N is L-compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq3_HTML.gif. By Lemma 3.4 and 3.5, we get that the following two conditions are satisfied
      1. (1)

        LxλNx for every (x, λ) ∈ [(domL\KerL) ∩ ∂Ω] × (0, 1);

         
      2. (2)

        Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω.

         
      Take
      H ( x , λ ) = ± λ x + ( 1 - λ ) Q N x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equau_HTML.gif
      According to Lemma 3.6 (or Remark 3.1), we know that H(x, λ) ≠ 0 for x ∈ KerL ∩ ∂Ω. Therefore,
      deg ( Q N Ker L , Ω Ker L , 0 ) = deg ( H ( , 0 ) , Ω Ker L , 0 ) = deg ( H ( , 1 ) , Ω Ker L , 0 ) = deg ( ± I , Ω Ker L , 0 ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equav_HTML.gif

      So that, the condition (3) of Lemma 2.1 is satisfied. By Lemma 2.1, we can get that Lx = Nx has at least one solution in dom L Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq6_HTML.gif. Therefore, BVP (1.1) has at least one solution. The proof is complete.

      4 An example

      Example 4.1. Consider the following BVP
      { D 0 + 5 2 x ( t ) = t 16 ( x 10 ) + t 2 16 e | x | + t 3 16 sin [ ( x ) 2 ] , t [ 0,1 ] x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 0 ) = 0. http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equ10_HTML.gif
      (4.1)
      where
      f ( t , u , v , w ) = t 1 6 ( u - 1 0 ) + t 2 1 6 e - | v | + t 3 1 6 sin ( w 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equaw_HTML.gif
      Choose p ( t ) = 1 0 t + 2 1 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq19_HTML.gif, q ( t ) = t 1 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq20_HTML.gif, r(t) = 0, s(t) = 0, B = 10. We can get that q 1 = 1 1 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_IEq21_HTML.gif, r1 = 0, s1 = 0 and
      Γ 5 2 - 1 - q 1 - r 1 - s 1 > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-20/MediaObjects/13661_2011_Article_12_Equax_HTML.gif

      Then, all conditions of Theorem 3.1 hold, so BVP (4.1) has at least one solution.

      Declarations

      Acknowledgements

      The authors would like to thank the referees very much for their helpful comments and suggestions. This research was supported by the Fundamental Research Funds for the Central Universities (2010LKSX09) and the Science Foundation of China University of Mining and Technology (2008A037).

      Authors’ Affiliations

      (1)
      Department of Mathematics, China University of Mining and Technology

      References

      1. Metzler R, Klafter J: Boundary value problems for fractional diffusion equations. Phys A 2000, 278: 107-125. 10.1016/S0378-4371(99)00503-8View ArticleMathSciNet
      2. Scher H, Montroll E: Anomalous transit-time dispersion in amorphous solids. Phys Rev B 1975, 12: 2455-2477. 10.1103/PhysRevB.12.2455View Article
      3. Mainardi F: Fractional diffusive waves in viscoelastic solids. In Nonlinear Waves in Solids. Edited by: Wegner JL, Norwood FR. ASME/AMR, Fairfield NJ; 1995:93-97.
      4. Diethelm K, Freed AD: On the solution of nonlinear fractional order differential equations used in the modeling of viscoplasticity. In Scientific Computing in Chemical Engineering II-Computational Fluid Dynamics, Reaction Engineering and Molecular Properties. Edited by: Keil F, Mackens W, Voss H, Werther J. Springer, Heidelberg; 1999:217-224.
      5. Gaul L, Klein P, Kempfle S: Damping description involving fractional operators. Mech Syst Signal Process 1991, 5: 81-88. 10.1016/0888-3270(91)90016-XView Article
      6. Glockle WG, Nonnenmacher TF: A fractional calculus approach of self-similar protein dynamics. Biophys J 1995, 68: 46-53. 10.1016/S0006-3495(95)80157-8View Article
      7. Mainardi F: Fractional calculus: some basic problems in continuum and statistical mechanics. In Fractals and Fractional Calculus in Continuum Mechanics. Edited by: Carpinteri A, Mainardi F. Springer, Wien; 1997:291-348.View Article
      8. Metzler F, Schick W, Kilian HG, Nonnenmacher TF: Relaxation in filled polymers: a fractional calculus approach. J Chem Phys 1995, 103: 7180-7186. 10.1063/1.470346View Article
      9. Oldham KB, Spanier J: The Fractional Calculus. Academic Press, New York, London; 1974.
      10. Agarwal RP, ORegan D, Stanek S: Positive solutions for Dirichlet problems of singular nonlinear fractional differential equations. J Math Anal Appl 2010, 371:, 57-68.View ArticleMathSciNetMATH
      11. Bai Z, Hu L: Positive solutions for boundary value problem of nonlinear fractional differential equation. J Math Anal Appl 2005, 311: 495-505. 10.1016/j.jmaa.2005.02.052View ArticleMathSciNetMATH
      12. Kaufmann ER, Mboumi E: Positive solutions of a boundary value problem for a nonlinear fractional differential equation. Electron J Qual Theory Differ Equ 2008, 3: 1-11.View ArticleMathSciNet
      13. Jafari H, Gejji VD: Positive solutions of nonlinear fractional boundary value problems using Adomian decomposition method. Appl Math Comput 2006, 180: 700-706. 10.1016/j.amc.2006.01.007View ArticleMathSciNetMATH
      14. Benchohra M, Hamani S, Ntouyas SK: Boundary value problems for differential equations with fractional order and nonlocal conditions. Nonlinear Anal 2009, 71: 2391-2396. 10.1016/j.na.2009.01.073View ArticleMathSciNetMATH
      15. Liang S, Zhang J: Positive solutions for boundary value problems of nonlinear fractional differential equation. Nonlinear Anal 2009, 71: 5545-5550. 10.1016/j.na.2009.04.045View ArticleMathSciNetMATH
      16. Zhang S: Positive solutions for boundary-value problems of nonlinear fractional differential equations. Electron J Differ Equ 2006, 36: 1-12.View Article
      17. Kosmatov N: A boundary value problem of fractional order at resonance. Electron J Differ Equ 2010, 135: 1-10.MathSciNet
      18. Mawhin J: Topological degree and boundary value problems for nonlinear differential equations in topological methods for ordinary differential equations. Lect Notes Math 1993, 1537: 74-142. 10.1007/BFb0085076View ArticleMathSciNet
      19. Lakshmikantham V, Leela S, Vasundhara Devi J: Theory of Fractional Dynamic Systems. Cambridge Academic Publishers, Cambridge; 2009.

      Copyright

      © Hu and Liu; licensee Springer. 2011

      This article is published under license to BioMed Central Ltd. 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 unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.