Some existence results for a nonlinear fractional differential equation on partially ordered Banach spaces

  • Dumitru Baleanu1, 2, 3Email author,

    Affiliated with

    • Ravi P Agarwal4,

      Affiliated with

      • Hakimeh Mohammadi5 and

        Affiliated with

        • Shahram Rezapour5

          Affiliated with

          Boundary Value Problems20132013:112

          DOI: 10.1186/1687-2770-2013-112

          Received: 9 August 2012

          Accepted: 16 April 2013

          Published: 3 May 2013

          Abstract

          By using fixed point results on cones, we study the existence and uniqueness of positive solutions for some nonlinear fractional differential equations via given boundary value problems. Examples are presented in order to illustrate the obtained results.

          1 Introduction

          The field of fractional differential equations has been subjected to an intensive development of the theory and the applications (see, for example, [16] and the references therein). It should be noted that most of papers and books on fractional calculus are devoted to the solvability of linear initial fractional differential equations on terms of special functions. There are some papers dealing with the existence of solutions of nonlinear initial value problems of fractional differential equations by using the techniques of nonlinear analysis such as fixed point results, the Leray-Schauder theorem, stability, etc. (see, for example, [719] and the references therein). In fact, 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 and aerodynamics (see, for example, [2023] and the references therein). The main advantage of using the fractional nonlinear differential equations is related to the fact that we can describe the dynamics of complex non-local systems with memory. In this line of taught, the equations involving various fractional orders are important from both theoretical and applied view points. We need the following notions.

          Definition 1.1 ([1, 4])

          For a continuous function f : [ 0 , ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq1_HTML.gif, the Caputo derivative of fractional order α is defined by
          D α c f ( t ) = 1 Γ ( n α ) 0 t ( t s ) n α 1 f ( n ) ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equa_HTML.gif

          where n 1 < α < n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq2_HTML.gif, n = [ α ] + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq3_HTML.gif and [ α ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq4_HTML.gif denotes the integer part of α.

          Definition 1.2 ([1, 4])

          The Riemann-Liouville fractional derivative of order α for a continuous function f is defined by
          D α f ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t f ( s ) ( t s ) α n 1 d s ( n = [ α ] + 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equb_HTML.gif

          where the right-hand side is pointwise defined on ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq5_HTML.gif.

          Definition 1.3 ([1, 4])

          Let [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq6_HTML.gif be an interval in ℝ and α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq7_HTML.gif. The Riemann-Liouville fractional order integral of a function f L 1 ( [ a , b ] , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq8_HTML.gif is defined by
          I a α f ( t ) = 1 γ ( α ) a t f ( s ) ( t s ) 1 α d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equc_HTML.gif

          whenever the integral exists.

          Suppose that E is a Banach space which is partially ordered by a cone P E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq9_HTML.gif, that is, x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq10_HTML.gif if and only if y x P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq11_HTML.gif. We denote the zero element of E by θ. A cone P is called normal if there exists a constant N > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq12_HTML.gif such that θ x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq13_HTML.gif implies x N y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq14_HTML.gif (see [24]). Also, we define the order interval [ x 1 , x 2 ] = { x E | x 1 x x 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq15_HTML.gif for all x 1 , x 2 E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq16_HTML.gif [24]. We say that an operator A : E E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq17_HTML.gif is increasing whenever x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq10_HTML.gif implies A x A y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq18_HTML.gif. Also, x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq19_HTML.gif means that there exist λ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq20_HTML.gif and μ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq21_HTML.gif such that λ x y μ x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq22_HTML.gif (see [24]). Finally, put P h = { x E | x h } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq23_HTML.gif for all h > θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq24_HTML.gif. It is easy to see that P h P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq25_HTML.gif is convex and λ P h = P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq26_HTML.gif for all λ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq27_HTML.gif. We recall the following in our results. Let E be a real Banach space and let P be a cone in E. Let ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq28_HTML.gif be an interval and let τ and φ be two positive-valued functions such that φ ( t ) τ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq29_HTML.gif for all t ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq30_HTML.gif and τ : ( a , b ) ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq31_HTML.gif is a surjection. We say that an operator A : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq32_HTML.gif is τ-φ-concave whenever A ( τ ( t ) x ) φ ( t ) A x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq33_HTML.gif for all t ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq30_HTML.gif and x P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq34_HTML.gif [13]. We say that A is φ-concave whenever τ ( t ) = t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq35_HTML.gif for all t [13]. We recall the following result.

          Theorem 1.1 ([13])

          Let E be a Banach space, let P be a normal cone in E, and let A : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq32_HTML.gif be an increasing and τ-φ-concave operator. Suppose that there exists θ h P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq36_HTML.gif such that A h P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq37_HTML.gif. Then there are u 0 , v 0 P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq38_HTML.gif and r ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq39_HTML.gif such that r v 0 u 0 v 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq40_HTML.gif and u 0 A u 0 A v 0 v 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq41_HTML.gif, the operator A has a unique fixed point x [ u 0 , v 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq42_HTML.gif, and for x 0 P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq43_HTML.gif and the sequence { x n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq44_HTML.gif with x n = A x n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq45_HTML.gif, we have x n x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq46_HTML.gif.

          2 Main results

          We study the existence and uniqueness of a solution for the fractional differential equation
          D α u ( t ) + f ( t , u ( t ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equd_HTML.gif

          on partially ordered Banach spaces with two types of boundary conditions and two types of fractional derivatives, Riemann-Liouville and Caputo.

          2.1 Existence results for the fractional differential equation with the Riemann-Liouville fractional derivative

          First, we study the existence and uniqueness of a positive solution for the fractional differential equation
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equ1_HTML.gif
          (2.1)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equ2_HTML.gif
          (2.2)

          where D α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq47_HTML.gif is the Riemann-Liouville fractional derivative of order α. Let E = C [ ε , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq48_HTML.gif. Consider the Banach space of continuous functions on [ ε , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq49_HTML.gif with the sup norm and set P = { y C [ ε , T ] : min t [ ε , T ] y ( t ) 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq50_HTML.gif. Then P is a normal cone.

          Lemma 2.1 Let 0 < ε < T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq51_HTML.gif, T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq52_HTML.gif, t [ ε , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq53_HTML.gif, η ( ε , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq54_HTML.gif and 0 < α < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq55_HTML.gif. Then the problem D α u ( t ) + f ( t , u ( t ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq56_HTML.gif with the boundary value condition u ( η ) = u ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq57_HTML.gif has a solution u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq58_HTML.gif if and only if u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq58_HTML.gif is a solution of the fractional integral equation
          u ( t ) = ε T G ( t , s ) f ( s , u ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Eque_HTML.gif
          where
          G ( t , s ) = { t α 1 ( η s ) α 1 t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) ( t s ) α 1 Γ ( α ) , ε s η t T , t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) ( t s ) α 1 Γ ( α ) , ε η s t T , t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) , ε η t s T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equf_HTML.gif
          Proof From D α u ( t ) + f ( t , u ( t ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq56_HTML.gif and the boundary condition, it is easy to see that u ( t ) c 1 t α 1 = I ε α f ( t , u ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq59_HTML.gif. By the definition of a fractional integral, we get
          u ( t ) = c 1 t α 1 ε t ( t s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equg_HTML.gif
          Thus, u ( η ) = c 1 η α 1 ε η ( η s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq60_HTML.gif and
          u ( T ) = c 1 T α 1 ε T ( T s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equh_HTML.gif
          Since u ( η ) = u ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq61_HTML.gif, we obtain
          c 1 = 1 η α 1 T α 1 ε η ( η s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s 1 η α 1 T α 1 ε T ( T s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equi_HTML.gif
          Hence,
          u ( t ) = t α 1 η α 1 T α 1 ε η ( η s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s t α 1 η α 1 T α 1 ε T ( T s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s ε t ( t s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s = ε T G ( t , s ) f ( s , u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equj_HTML.gif

          This completes the proof. □

          Now, we are ready to state and prove our first main result.

          Theorem 2.2 Let 0 < ε < T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq51_HTML.gif be given and let τ and φ be two functions on ( ε , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq62_HTML.gif such that φ ( t ) τ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq63_HTML.gif for all t ( ε , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq64_HTML.gif. Suppose that τ : ( ε , T ) ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq65_HTML.gif is a surjection and f ( t , u ( t ) ) C ( [ ε , T ] × [ 0 , ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq66_HTML.gif is increasing in u for each fixed t, f ( t , u ( t ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq67_HTML.gif and f ( t , τ ( λ ) u ( t ) ) φ ( λ ) f ( t , u ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq68_HTML.gif for all t , λ ( ε , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq69_HTML.gif and u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq70_HTML.gif. Assume that there exist M 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq71_HTML.gif, M 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq72_HTML.gif and θ h P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq73_HTML.gif such that
          M 1 h ( t ) ε T G ( t , s ) f ( s , h ( s ) ) d s M 2 h ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equk_HTML.gif

          for all t [ ε , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq53_HTML.gif, where G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq74_HTML.gif is the green function defined in Lemma 2.1. Then the problem (2.1) with the boundary value condition (2.2) has a unique positive solution u P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq75_HTML.gif. Moreover, for the sequence u n + 1 = ε T G ( t , s ) f ( s , u n ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq76_HTML.gif, we have u n u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq77_HTML.gif for all u 0 P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq78_HTML.gif.

          Proof By using Lemma 2.1, the problem is equivalent to the integral equation
          u ( t ) = ε T G ( t , s ) f ( s , u ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equl_HTML.gif
          where
          G ( t , s ) = { t α 1 ( η s ) α 1 t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) ( t s ) α 1 Γ ( α ) , ε s η t T , t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) ( t s ) α 1 Γ ( α ) , ε η s t T , t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) , ε η t s T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equm_HTML.gif
          Define the operator A : P E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq79_HTML.gif by A u ( t ) = ε T G ( t , s ) f ( s , u ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq80_HTML.gif. Then u is a solution for the problem if and only if u = A u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq81_HTML.gif. It is easy to check that the operator A is increasing on P. On the other hand,
          A ( τ ( λ ) u ) ( t ) = ε T G ( t , s ) f ( s , τ ( λ ) u ( s ) ) d s φ ( λ ) ε T G ( t , s ) f ( s , u ( s ) ) d s = φ ( λ ) A u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equn_HTML.gif
          for all λ [ ε , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq82_HTML.gif and u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq70_HTML.gif. Thus, the operator A is τ-φ-concave. Since
          M 1 h ( t ) A h ( t ) = ε T G ( t , s ) f ( s , h ( s ) ) d s M 2 h ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equo_HTML.gif

          for all t [ ε , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq53_HTML.gif, we get A h P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq83_HTML.gif. Now, by using Theorem 1.1, the operator A has a unique positive solution u P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq75_HTML.gif. This completes the proof. □

          Here, we give the following example to illustrate Theorem 2.2.

          Example 2.1 Let 0 < ε < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq84_HTML.gif be given. Consider the periodic boundary value problem
          D 1 3 u ( t ) + { g ( t ) + [ u ( t ) ] α } = 0 ( t [ ε , 1 ] ) , u ( η ) = u ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equp_HTML.gif
          where η ( ε , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq85_HTML.gif, g is continuous on [ ε , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq86_HTML.gif and min t [ ε , 1 ] g ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq87_HTML.gif. Put
          G ( t , s ) = { t 2 / 3 ( η s ) 2 / 3 t 2 / 3 ( 1 s ) 2 / 3 ( η 2 / 3 1 2 / 3 ) Γ ( 1 / 3 ) ( t s ) 2 / 3 Γ ( 1 / 3 ) , ε s η t 1 , t 2 / 3 ( 1 s ) 2 / 3 ( η 2 / 3 1 2 / 3 ) Γ ( 1 / 3 ) ( t s ) 2 / 3 Γ ( 1 / 3 ) , ε η s t 1 , t 2 / 3 ( 1 s ) 2 / 3 ( η 2 / 3 1 2 / 3 ) Γ ( 1 / 3 ) , ε η t s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equq_HTML.gif
          Then ε 1 G ( t , s ) d s = t 2 / 3 ( η ε ) 1 / 3 t 2 / 3 ( 1 ε ) 1 / 3 ( t ε ) 1 / 3 ( η 2 / 3 1 ) Γ ( 4 / 3 ) ( η 2 / 3 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq88_HTML.gif. Now, define τ ( t ) = t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq35_HTML.gif, φ ( t ) = t 1 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq89_HTML.gif, γ 1 = min t [ ε , 1 ] g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq90_HTML.gif, γ 2 = max t [ ε , 1 ] g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq91_HTML.gif and also f ( t , u ) = g ( t ) + u 1 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq92_HTML.gif for all t. Then τ : ( 0 , 1 ) ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq93_HTML.gif is a surjection and φ ( t ) > τ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq94_HTML.gif for all t ( ε , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq95_HTML.gif. For each u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq96_HTML.gif, we have
          f ( t , τ ( λ ) u ( t ) ) = f ( t , λ u ( t ) ) = g ( t ) + λ 1 / 3 [ u ( t ) ] 1 / 3 λ 1 / 3 ( g ( t ) + [ u ( t ) ] 1 / 3 ) = φ ( λ ) f ( t , u ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equr_HTML.gif
          Now, put h 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq97_HTML.gif, M 1 = ( γ 1 + 1 ) min t [ ε , 1 ] , η [ ε , 1 ] t 2 / 3 ( 1 ε ) 1 / 3 ( t ε ) 1 / 3 ( η 2 / 3 1 ) Γ ( 4 / 3 ) ( η 2 / 3 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq98_HTML.gif and M 2 = ( γ 2 + 1 ) max η [ ε , 1 ] ε 2 / 3 η 1 / 3 Γ ( 4 / 3 ) ( η 2 / 3 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq99_HTML.gif. Then we get
          ε 1 G ( t , s ) { g ( s ) + [ h ( s ) ] 1 / 3 } d s ε 1 G ( t , s ) ( γ 2 + 1 ) d s ( γ 2 + 1 ) max t [ ε , 1 ] ε 1 G ( t , s ) d s ( γ 2 + 1 ) ( max η [ ε , 1 ] ε 2 / 3 η 1 / 3 Γ ( 4 / 3 ) ( η 2 / 3 1 ) ) = M 2 h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equs_HTML.gif
          and
          ε 1 G ( t , s ) { g ( s ) + [ h ( s ) ] 1 / 3 } d s ( γ 1 + 1 ) min t [ ε , 1 ] ε 1 G ( t , s ) d s ( γ 1 + 1 ) min t [ ε , 1 ] , η [ ε , 1 ] t 2 / 3 ( 1 ε ) 1 / 3 ( t ε ) 1 / 3 ( η 2 / 3 1 ) Γ ( 4 / 3 ) ( η 2 / 3 1 ) = M 1 h . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equt_HTML.gif

          Thus, by using Theorem 2.2, the problem has a unique solution in P h = P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq100_HTML.gif.

          2.2 Existence results for the fractional differential equation with the Caputo fractional derivative

          Here, we study the existence and uniqueness of a positive solution for the fractional differential equation
          D α c u ( t ) + f ( t , u ( t ) ) = 0 ( t [ 0 , T ] , T 1 , 1 < α < 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equ3_HTML.gif
          (2.3)
          u ( 0 ) = β 1 u ( η ) , u ( T ) = β 2 u ( η ) ( η ( 0 , t ) , 0 < β 1 < β 2 < 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equ4_HTML.gif
          (2.4)
          where D α c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq101_HTML.gif is the Caputo fractional derivative of order α. Let E = C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq102_HTML.gif be the Banach space of continuous functions on [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq103_HTML.gif with the sup norm and
          P = { y C [ 0 , T ] : min t [ 0 , T ] y ( t ) 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equu_HTML.gif

          It is known that P is a normal cone. Similar to the proof of Lemma 2.1, we can prove the following result.

          Lemma 2.3 Let 1 < α < 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq104_HTML.gif, T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq52_HTML.gif, t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq105_HTML.gif, η ( 0 , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq106_HTML.gif and 0 < β 1 < β 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq107_HTML.gif. Then the problem D α c u ( t ) + f ( t , u ( t ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq108_HTML.gif with the boundary value conditions u ( 0 ) = β 1 u ( η ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq109_HTML.gif and u ( T ) = β 2 u ( η ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq110_HTML.gif has a solution u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq58_HTML.gif if and only if u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq58_HTML.gif is a solution of the fractional integral equation u ( t ) = 0 T G ( t , s ) f ( s , u ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq111_HTML.gif, where
          G ( t , s ) = { [ β 1 T + t ( β 2 β 1 ) ] ( η s ) α 1 + t ( T s ) α 1 T ( t s ) α 1 T Γ ( α ) , 0 s η t T , t ( T s ) α 1 T ( t s ) α 1 T Γ ( α ) , 0 η s t T , t ( T s ) α 1 T Γ ( α ) , 0 η t s T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equv_HTML.gif
          Theorem 2.4 Let T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq52_HTML.gif be given and let τ and φ be two positive-valued functions on ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq112_HTML.gif such that φ ( t ) τ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq113_HTML.gif for all t ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq114_HTML.gif. Suppose that τ : ( 0 , T ) ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq115_HTML.gif is a surjection and f ( t , u ( t ) ) C ( [ ε , T ] × [ 0 , ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq66_HTML.gif is increasing in u for each fixed t, f ( t , u ( t ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq116_HTML.gif whenever 0 < η < s < t < T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq117_HTML.gif and f ( t , u ( t ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq118_HTML.gif otherwise, and also f ( t , τ ( λ ) u ( t ) ) φ ( λ ) f ( t , u ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq119_HTML.gif for all t , λ ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq120_HTML.gif and u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq70_HTML.gif. Assume that there exist M 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq71_HTML.gif, M 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq72_HTML.gif and θ h P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq73_HTML.gif such that
          M 1 h ( t ) 0 T G ( t , s ) f ( s , h ( s ) ) d s M 2 h ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equw_HTML.gif

          for all t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq105_HTML.gif, where G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq74_HTML.gif is the green function defined in Lemma 2.3. Then the problem (2.3) with the boundary value conditions (2.4) has a unique positive solution u P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq75_HTML.gif. Moreover, for the sequence u n + 1 = ε T G ( t , s ) f ( s , u n ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq76_HTML.gif, we have u n u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq77_HTML.gif for all u 0 P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq78_HTML.gif.

          Proof It is sufficient to define the operator A : P E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq79_HTML.gif by
          A u ( t ) = 0 T G ( t , s ) f ( s , u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equx_HTML.gif

          Now, by using a similar proof of Theorem 2.2, one can show that A u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq121_HTML.gif for all u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq122_HTML.gif and t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq105_HTML.gif, and also the operator A is τ-φ-concave. By using Theorem 1.1, the operator A has a unique positive solution u P h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq75_HTML.gif. This completes the proof by using Lemma 2.3. □

          Below we present an example to illustrate Theorem 2.4.

          Example 2.2 Let α = 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq123_HTML.gif. Consider the periodic boundary value problem
          D α c u ( t ) + g ( t ) + [ u ( t ) ] α = 0 ( t [ 0 , 1 ] ) , u ( 0 ) = 1 3 u ( 1 2 ) u ( 1 ) = 1 2 u ( 1 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equy_HTML.gif
          where g is a continuous function on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq124_HTML.gif with min t [ 0 , 1 ] g ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq125_HTML.gif. Put β 2 = η = 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq126_HTML.gif, β 1 = 1 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq127_HTML.gif and
          G ( t , s ) = { [ 1 3 + 1 6 t ] ( 1 2 s ) 1 / 2 + t ( 1 s ) 1 / 2 ( t s ) 1 / 2 Γ ( 3 / 2 ) , 0 s η t 1 , t ( 1 s ) 1 / 2 ( t s ) 1 / 2 Γ ( 3 / 2 ) , 0 η s t 1 , t ( 1 s ) 1 / 2 Γ ( 3 / 2 ) , 0 η t s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equz_HTML.gif
          Then 0 1 G ( t , s ) d s = [ 1 3 + 1 6 t ] ( 1 2 ) 3 / 2 + t t 3 / 2 Γ ( 5 / 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq128_HTML.gif. Now, define τ ( t ) = t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq129_HTML.gif, φ ( t ) = t α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq130_HTML.gif, γ 1 = min t [ 0 , 1 ] g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq131_HTML.gif, γ 2 = max t [ 0 , 1 ] g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq132_HTML.gif and f ( t , u ) = g ( t ) + u α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq133_HTML.gif. Then it is easy to see that τ : ( 0 , 1 ) ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq134_HTML.gif is a surjection map and φ ( t ) > τ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq135_HTML.gif for t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq136_HTML.gif. Also, we have
          f ( t , τ ( λ ) u ( t ) ) = f ( t , λ u ( t ) ) = g ( t ) + λ α [ u ( t ) ] α λ α ( g ( t ) + [ u ( t ) ] α ) = φ ( λ ) f ( t , u ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equaa_HTML.gif
          for all u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq96_HTML.gif. Now, put h 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq97_HTML.gif, M 1 = ( γ 1 + 1 ) min t [ 0 , 1 ] 1 3 t ( 1 2 ) 3 / 2 t 3 / 2 Γ ( 5 / 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq137_HTML.gif and also M 2 = ( γ 2 + 1 ) 5 6 ( 1 2 ) 3 / 2 + 1 Γ ( 5 / 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq138_HTML.gif. Then we have
          0 1 G ( t , s ) { g ( s ) + [ h ( s ) ] 3 / 2 } d s 0 1 G ( t , s ) ( γ 2 + 1 ) d s ( γ 2 + 1 ) max t [ 0 , 1 ] 0 1 G ( t , s ) d s ( γ 2 + 1 ) 5 6 ( 1 2 ) 3 / 2 + 1 Γ ( 5 / 2 ) = M 2 h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equab_HTML.gif
          and
          0 1 G ( t , s ) { g ( s ) + [ h ( s ) ] 3 / 2 } d s ( γ 1 + 1 ) min t [ 0 , 1 ] 0 1 G ( t , s ) d s ( γ 1 + 1 ) min t [ 0 , 1 ] 1 3 t ( 1 2 ) 3 / 2 t 3 / 2 Γ ( 5 / 2 ) = M 1 h . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_Equac_HTML.gif

          Thus, by using Theorem 2.4, the problem has a unique solution in P h = P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-112/MediaObjects/13661_2012_Article_363_IEq100_HTML.gif.

          Declarations

          Acknowledgements

          This work is partially supported by the Scientific and Technical Research Council of Turkey. Research of the third and forth authors was supported by Azarbaidjan Shahid Madani University. Also, the authors express their gratitude to the referees for their helpful suggestions which improved final version of this paper.

          Authors’ Affiliations

          (1)
          Department of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University
          (2)
          Department of Mathematics, Cankaya University
          (3)
          Institute of Space Sciences
          (4)
          Department of Mathematics, Texas A&M University
          (5)
          Department of Mathematics, Azarbaidjan Shahid Madani University

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          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.