Existence of the solutions for a class of nonlinear fractional order three-point boundary value problems with resonance

  • Zigen Ouyang1Email author and

    Affiliated with

    • Gangzhao Li1

      Affiliated with

      Boundary Value Problems20122012:68

      DOI: 10.1186/1687-2770-2012-68

      Received: 4 December 2011

      Accepted: 9 May 2012

      Published: 1 July 2012

      Abstract

      A class of nonlinear fractional order differential equation

      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equa_HTML.gif

      is investigated in this paper, where D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq1_HTML.gif is the standard Riemann-Liouville fractional derivative of order 1 < α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq2_HTML.gif, 0 < η < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq3_HTML.gif, f C ( [ 0 , 1 ] × R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq4_HTML.gif. Using intermediate value theorem, we obtain a sufficient condition for the existence of the solutions for the above fractional order differential equations.

      1 Introduction

      Consider the following boundary value problem
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ1_HTML.gif
      (1.1)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ2_HTML.gif
      (1.2)

      where D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq1_HTML.gif is the standard Riemann-Liouville fractional derivative of order 1 < α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq2_HTML.gif, 0 < η < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq3_HTML.gif and f C ( [ 0 , 1 ] × R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq4_HTML.gif.

      In the last few decades, many authors have investigated fractional differential equations which have been applied in many fields such as physics, mechanics, chemistry, engineering etc. (see references [1, 6, 10, 2123]). Especially, many works have been devoted to the study of initial value problems and bounded value problems for fractional order differential equations [12, 13, 15, 24].

      Recently, the existence of positive solutions of fractional differential equations has attracted many authors’ attention [25, 8, 9, 12, 14, 1720, 25, 26]. Using some fixed point theorems, they obtained some nice existence conditions for positive solutions.

      In more recent works, Jiang and Yuan [7] considered the following boundary value problem of fractional differential equations
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ3_HTML.gif
      (1.3)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ4_HTML.gif
      (1.4)

      where D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq1_HTML.gif is the standard Riemann-Liouville fractional derivative of order 1 < α < 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq5_HTML.gif and f : [ 0 , 1 ] × R + R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq6_HTML.gif is continuous. Using some properties of the Green function G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq7_HTML.gif, they obtain some new sufficient conditions for the existence of positive solutions for the above problem.

      Further, Li, Luo, and Zhou [4] investigated the following fractional order three-point boundary value problems
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ5_HTML.gif
      (1.5)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ6_HTML.gif
      (1.6)

      where D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq1_HTML.gif is the standard Riemann-Liouville fractional derivative of order 1 < α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq2_HTML.gif 0 β 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq8_HTML.gif 0 a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq9_HTML.gif ξ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq10_HTML.gif a ξ α β 2 1 β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq11_HTML.gif 0 α β 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq12_HTML.gif, and f : [ 0 , 1 ] × R + R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq6_HTML.gif is continuous.

      In this paper, we discuss the boundary value problem (1.1)-(1.2). Using some properties of the Green function G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq7_HTML.gif and intermediate value theorem, we establish some sufficient conditions for the existence of the positive solutions of the problem (1.1)-(1.2).

      The paper is arranged as follows: In Section 2, we introduce some definitions for fractional order differential equations and give our main results for the boundary value problem (1.1)-(1.2). We give some lemmas for our results in Section 3. In Section 4, we prove our main result; and finally, we give an example to illustrate our results.

      2 Main results

      In this section, we introduce some definitions and preliminary facts which are used in this paper.

      Definition 2.1 ([1, 10])

      The fractional integral of order α with the lower limit t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq13_HTML.gif for a function f is defined as
      I t 0 + α ( f ( t ) ) = 1 Γ ( α ) t 0 t f ( s ) ( t s ) 1 α d s , t > t 0 , α > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equb_HTML.gif

      provided that the integral on the right-hand side is point-wise defined on [ t 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq14_HTML.gif, where Γ is the Gamma function.

      Definition 2.2 ([1, 10])

      Riemann-Liouville derivative of order α with the lower limit t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq13_HTML.gif for a function f : [ 0 , ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq15_HTML.gif can be written as
      D t 0 + α ( f ( t ) ) = 1 Γ ( n α ) d n d t n t 0 t f ( s ) ( t s ) α + 1 n d s , t > t 0 , n 1 < α n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equc_HTML.gif

      where n is a positive integer.

      We call the function u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq16_HTML.gif a solution of (1.1)-(1.2) if u ( t ) C [ 0 , 1 ] L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq17_HTML.gif with a fractional derivative of order α belongs to C [ 0 , 1 ] L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq18_HTML.gif and satisfies Equation (1.1) and the boundary condition (1.2).

      We also need to introduce some lemmas as follows, which will be used in the proof of our main theorems.

      Lemma 2.1 ([26])

      Assume that h ( t ) C ( 0 , 1 ) L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq19_HTML.gifwith a fractional derivative of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq20_HTML.gifbelongs to C ( 0 , 1 ) L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq21_HTML.gif. Then, the fractional equation

      D t 0 + α ( h ( t ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ7_HTML.gif
      (2.1)
      has solutions
      h ( t ) = c 1 t α 1 + c 2 t α 2 + + c n t α n , c i R , i = 1 , 2 , , n , n = [ α ] + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ8_HTML.gif
      (2.2)

      Lemma 2.2 ([26])

      Assume that h ( t ) C ( 0 , 1 ) L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq19_HTML.gifwith a fractional derivative of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq20_HTML.gifbelongs to C ( 0 , 1 ) L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq21_HTML.gif. Then
      I t 0 + α D t 0 + α h ( t ) = h ( t ) + c 1 t α 1 + c 2 t α 2 + + c n t α n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ9_HTML.gif
      (2.3)

      for some c i R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq22_HTML.gif, i = 1 , 2 , , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq23_HTML.gif, n = [ α ] + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq24_HTML.gif.

      Lemma 2.3 ([16])

      Suppose that X be a Banach space, C X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq25_HTML.gifis closed and convex. Assume that U is a relatively open subset of C with 0 U http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq26_HTML.gif, and T : U ¯ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq27_HTML.gifis a completely continuous operator. Then, either
      1. (i)

        T has a fixed point in U ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq28_HTML.gif, or

         
      2. (ii)

        there exist u U http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq29_HTML.gif and γ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq30_HTML.gif with u = γ T u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq31_HTML.gif.

         

      Throughout this paper, we assume that f ( t , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq32_HTML.gif satisfies the following:

      (H) f ( t , u ) C ( [ 0 , 1 ] × R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq33_HTML.gif, and there exist two positive functions a ( t ) C ( [ 0 , 1 ] , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq34_HTML.gifand b ( t ) C ( [ 0 , 1 ] , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq35_HTML.gifsuch that
      | f ( t , t α 1 u ) | a ( t ) + b ( t ) | u | p , t [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ10_HTML.gif
      (2.4)
      where 0 p 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq36_HTML.gif. Furthermore,
      lim u ± f ( t , t α 1 u ) = ± http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ11_HTML.gif
      (2.5)

      for any t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq37_HTML.gif.

      We have our main results:

      Theorem 2.1 Suppose that (H) holds. If
      0 1 G ( s , s ) b ( s ) d s < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ12_HTML.gif
      (2.6)
      then the boundary value problem (1.1)-(1.2) has at least one solution, where
      G ( s , s ) = 1 Γ ( α ) ( 1 η α 1 ) { ( 1 s ) α 1 ( η s ) α 1 , 0 s η , ( 1 s ) α 1 , η s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equd_HTML.gif

      3 Some lemmas

      Let Ω = C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq38_HTML.gif, u Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq39_HTML.gif equipped the norm
      u = sup 0 t 1 | u ( t ) | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ13_HTML.gif
      (3.1)

      then Ω is a Banach space.

      We first give some lemmas as follows:

      Lemma 3.1 Problem (1.1)-(1.2) is equivalent to the following integral equation

      u ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s + u ( 1 ) t α 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ14_HTML.gif
      (3.2)
      where
      G ( t , s ) = { t α 1 ( 1 s ) α 1 t α 1 ( η s ) α 1 ( 1 η α 1 ) ( t s ) α 1 Γ ( α ) ( 1 η α 1 ) , 0 s min { t , η } 1 ; t α 1 ( 1 s ) α 1 t α 1 ( η s ) α 1 Γ ( α ) ( 1 η α 1 ) , 0 t s η 1 ; t α 1 ( 1 s ) α 1 ( 1 η α 1 ) ( t s ) α 1 Γ ( α ) ( 1 η α 1 ) , 0 η s t 1 ; t α 1 ( 1 s ) α 1 Γ ( α ) ( 1 η α 1 ) , 0 max { t , η } s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ15_HTML.gif
      (3.3)

      Proof The sufficiency is obvious, we only need to prove the necessity.

      Suppose that u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq16_HTML.gif is a solution of the problem (1.1)-(1.2). Integrating both sides of (1.1) of α order from 0 to t with respect to t, it follows that
      u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 f ( s , u ( s ) ) d s + c 1 t α 1 + c 2 t α 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ16_HTML.gif
      (3.4)
      According to (1.2) and (3.4), we have
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ17_HTML.gif
      (3.5)
      Combining (3.4) and (3.5), we obtain
      u ( t ) = 1 Γ ( α ) ( 1 η α 1 ) 0 t ( 1 η α 1 ) ( t s ) α 1 f ( s , u ( s ) ) d s + 1 Γ ( α ) ( 1 η α 1 ) { 0 1 t α 1 ( 1 s ) α 1 f ( s , u ( s ) ) d s 0 η t α 1 ( η s ) α 1 f ( s , u ( s ) ) d s } + u ( 1 ) t α 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Eque_HTML.gif

      According to (3.3), it is easy to show that (3.2) holds. The proof is completed. □

      Lemma 3.2 For any ( t , s ) [ 0 , 1 ] × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq40_HTML.gif, G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq7_HTML.gifis continuous, and G ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq41_HTML.giffor any ( t , s ) ( 0 , 1 ) × ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq42_HTML.gif.

      Proof The continuity of G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq7_HTML.gif for ( t , s ) [ 0 , 1 ] × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq40_HTML.gif is obvious.

      Let
      g 1 ( t , s ) = t α 1 ( 1 s ) α 1 t α 1 ( η s ) α 1 ( 1 η α 1 ) ( t s ) α 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equf_HTML.gif
      we only need to show that g 1 ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq43_HTML.gif for 0 s min { t , η } 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq44_HTML.gif, the rest of the proof is similar or obvious. From the definition of g 1 ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq45_HTML.gif, we have
      g 1 ( t , s ) = t α 1 { ( 1 s ) α 1 ( η s ) α 1 ( 1 η α 1 ) ( 1 s t ) α 1 } t α 1 { ( 1 s ) α 1 ( η s ) α 1 ( 1 η α 1 ) ( 1 s ) α 1 } t α 1 { η α 1 ( 1 s ) α 1 ( η s ) α 1 } t α 1 { ( η η s ) α 1 ( η s ) α 1 } > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equg_HTML.gif

      for 0 s min { t , η } 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq44_HTML.gif. The proof is completed. □

      Let
      G ( t , s ) = t α 1 G ( t , s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equh_HTML.gif
      then
      G ( t , s ) = { ( 1 s ) α 1 ( η s ) α 1 ( 1 η α 1 ) ( 1 s t ) α 1 Γ ( α ) ( 1 η α 1 ) , 0 s min { t , η } 1 ; ( 1 s ) α 1 ( η s ) α 1 Γ ( α ) ( 1 η α 1 ) , 0 t s η 1 ; ( 1 s ) α 1 ( 1 η α 1 ) ( 1 s t ) α 1 Γ ( α ) ( 1 η α 1 ) 0 η s t 1 ; ( 1 s ) α 1 Γ ( α ) ( 1 η α 1 ) , 0 max { t , η } s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ18_HTML.gif
      (3.6)

      The new Green’s function G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq46_HTML.gif has the following properties:

      Lemma 3.3 G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq46_HTML.gifis continuous for ( t , s ) ( 0 , 1 ) × ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq42_HTML.gif, and
      lim t 0 G ( t , s ) : = G ( 0 , s ) = { 1 Γ ( α ) ( 1 η α 1 ) { ( 1 s ) α 1 ( η s ) α 1 } , 0 s η , 1 Γ ( α ) ( 1 η α 1 ) ( 1 s ) α 1 , η s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equi_HTML.gif

      Furthermore, G ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq47_HTML.giffor ( t , s ) ( 0 , 1 ) × ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq42_HTML.gif.

      Lemma 3.4 For any s ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq48_HTML.gif, G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq46_HTML.gifis nonincreasing with respect to t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq49_HTML.gif. Especially, for any s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq50_HTML.gif, G t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq51_HTML.giffor t [ s , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq52_HTML.gif, and G t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq53_HTML.giffor t [ 0 , s ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq54_HTML.gif. That is G ( 1 , s ) G ( t , s ) G ( s , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq55_HTML.gif, where

      G ( 1 , s ) = 1 Γ ( α ) ( 1 η α 1 ) { η α 1 ( 1 s ) α 1 ( η s ) α 1 , 0 s η , η α 1 ( 1 s ) α 1 , η s 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ19_HTML.gif
      (3.7)
      and
      G ( s , s ) = 1 Γ ( α ) ( 1 η α 1 ) { ( 1 s ) α 1 ( η s ) α 1 , 0 s η , ( 1 s ) α 1 , η s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ20_HTML.gif
      (3.8)
      Let
      u ( t ) = t α 1 x ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ21_HTML.gif
      (3.9)
      Then, u ( 1 ) = x ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq56_HTML.gif, we have from Lemma 3.1, (3.6) and (3.9) that the integral Equation (3.2) can be rewritten as follows:
      x ( t ) = 0 1 G ( t , s ) f ( s , s α 1 x ( s ) ) d s + x ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ22_HTML.gif
      (3.10)
      Let
      y ( t ) = x ( t ) x ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ23_HTML.gif
      (3.11)
      Then, y ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq57_HTML.gif and (3.10) is equivalent to the following
      y ( t ) = 0 1 G ( t , s ) f ( s , s α 1 ( y ( s ) + x ( 1 ) ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ24_HTML.gif
      (3.12)

      We can divide our proof into the following two steps:

      First, we replace x ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq58_HTML.gif by any real number μ, then (3.12) can be rewritten as
      y ( t ) = 0 1 G ( t , s ) f ( s , s α 1 ( y ( s ) + μ ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ25_HTML.gif
      (3.13)

      It suffices to show that for any given real number μ, (3.13) has a solution y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq59_HTML.gif, which implies that Equation (1.1) has a solution u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq16_HTML.gif which satisfies the first boundary value condition u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq60_HTML.gif.

      Second, we show that there exists a μ such that the solution y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq59_HTML.gif of (3.13) satisfies y ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq57_HTML.gif, which implies that the solution u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq16_HTML.gif of (1.1) also satisfies the boundary value condition u ( 1 ) = 1 η α 1 u ( η ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq61_HTML.gif.

      In this section, we will prove the first step. For convenience sake, we define an operator T on the set Ω as follows:
      T y ( t ) = 0 1 G ( t , s ) f ( s , s α 1 ( y ( s ) + μ ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ26_HTML.gif
      (3.14)

      Lemma 3.5 Suppose that f C ( [ 0 , 1 ] × R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq4_HTML.gif, and (2.4) hold, then the operator T is completely continuous in Ω.

      Proof It is easy to show that the operator T maps Ω into itself. We divide the proof into the following three steps.

      Step 1. T y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq62_HTML.gif is continuous with respect to y ( t ) Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq63_HTML.gif.

      In fact, suppose that { y n ( t ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq64_HTML.gif is a sequence in Ω, and { y n ( t ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq64_HTML.gif converges to y ( t ) Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq65_HTML.gif. Since f ( t , t α 1 y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq66_HTML.gif is continuous with respect to y R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq67_HTML.gif, and it is obvious that G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq46_HTML.gif is uniformly continuous with respect to ( t , s ) [ 0 , 1 ] × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq68_HTML.gif from Lemma 3.3, then for any ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq69_HTML.gif, there exists an integer N, when n > N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq70_HTML.gif,
      f ( t , t α 1 ( y n ( t ) + μ ) ) f ( t , t α 1 ( y ( t ) + μ ) ) ε 0 1 G ( t , s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ27_HTML.gif
      (3.15)
      which follows from (3.14)-(3.15) that
      ( T y n ) ( t ) ( T y ) ( t ) = 0 1 G ( t , s ) { f ( s , s α 1 ( y n ( s ) + μ ) ) f ( s , s α 1 ( y ( s ) + μ ) ) } d s 0 1 G ( t , s ) d s f ( t , t α 1 ( y n ( t ) + μ ) ) f ( t , t α 1 ( y ( t ) + μ ) ) ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equj_HTML.gif

      Thus, the operator T is continuous in Ω.

      Step 2. T maps bounded set in Ω into bounded set.

      Suppose that B Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq71_HTML.gif is a bounded set with y ( t ) r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq72_HTML.gif for any y B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq73_HTML.gif. Then, we have from (2.4) and (3.14) that
      ( T y ) ( t ) = 0 1 G ( t , s ) f ( s , s α 1 ( y ( s ) + μ ) ) d s 0 1 G ( t , s ) a ( s ) d s + 0 1 G ( t , s ) b ( s ) | y ( s ) + μ | p d s 0 1 G ( t , s ) a ( s ) d s + 0 1 G ( t , s ) b ( s ) d s ( y ( t ) + μ ) p 0 1 G ( t , s ) a ( s ) d s + 0 1 G ( t , s ) b ( s ) d s ( r + μ ) p : = l . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equk_HTML.gif

      This gives that the operator T maps bounded set into bounded set in Ω.

      Step 3. T is equicontinuous in Ω.

      It suffices to show that for any y ( t ) B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq74_HTML.gif and any 0 < t 1 < t 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq75_HTML.gif, T y ( t 1 ) T y ( t 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq76_HTML.gif as t 1 t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq77_HTML.gif. We consider the following three cases:
      1. (i)

        0 < t 1 < t 2 < η http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq78_HTML.gif;

         
      2. (ii)

        0 < t 1 < η < t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq79_HTML.gif;

         
      3. (iii)

        0 < η < t 1 < t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq80_HTML.gif.

         
      We only prove the case (i), the rest two cases are similar. Since B is bounded, then there exists a M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq81_HTML.gif such that f M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq82_HTML.gif. According to (3.14), we have
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equl_HTML.gif

      According to Step 1-Step 3, the operator T is completely continuous in Ω. The proof is completed. □

      Further, we have

      Lemma 3.6 Suppose that f C ( [ 0 , 1 ] × R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq4_HTML.gif, and (2.4) and (2.6) holds, then, for any real number μ, the integral Equation (3.13) has at least one solution.

      Proof We only need to show that the operator T is priori bounded. Let
      r = max { 1 , 0 1 G ( s , s ) a ( s ) d s + 0 1 G ( s , s ) b ( s ) d s | μ | p 1 0 1 G ( s , s ) b ( s ) d s } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ28_HTML.gif
      (3.16)
      Define a set K Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq83_HTML.gif as follows
      K = { y Ω | y ( t ) r } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ29_HTML.gif
      (3.17)

      To show the existence of a fixed point of T by Lemma 2.3, we need to verify that the second possibility in Lemma 2.3 cannot happen.

      In fact, assume that there exists y K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq84_HTML.gif with y ( t ) = r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq85_HTML.gif and γ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq30_HTML.gif such that y = γ T y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq86_HTML.gif. It follows that
      y ( t ) = γ | ( T y ) ( t ) | = γ 0 1 G ( t , s ) f ( s , s α 1 ( y ( s ) + μ ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equm_HTML.gif
      and
      y ( t ) = γ 0 1 G ( t , s ) f ( s , s α 1 ( y ( s ) + μ ) ) d s γ 0 1 G ( s , s ) | f ( s , s α 1 ( y ( s ) + μ ) ) | d s < 0 1 G ( s , s ) a ( s ) d s + 0 1 G ( s , s ) b ( s ) d s y ( t ) + μ p 0 1 G ( s , s ) a ( s ) d s + 0 1 G ( s , s ) b ( s ) d s μ p + 0 1 G ( s , s ) b ( s ) d s r p 0 1 G ( s , s ) a ( s ) d s + 0 1 G ( s , s ) b ( s ) d s μ p + 0 1 G ( s , s ) b ( s ) d s r r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ30_HTML.gif
      (3.18)
      Here we have the use of the inequality
      ( a + b ) p a p + b p for a , b 0 , 0 p 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equn_HTML.gif

      It is obvious that (3.18) contradicts our assumption that y ( t ) = r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq85_HTML.gif. Therefore, by Lemma 2.3, it follows that T has a fixed point y K ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq87_HTML.gif. Hence, the integral Equation (3.14) has at least a solution y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq59_HTML.gif. The proof is completed. □

      4 The proof of the main results

      Now, we prove Theorem 2.1 by Lemma 3.4-3.5 and the intermediate value theorem.

      Proof of Theorem 2.1 It is obvious that the right-hand side of (3.14) is continuously dependent on the parameter μ, so we need to find a μ such that y ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq57_HTML.gif, which implies that u ( 1 ) = μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq88_HTML.gif.

      For any given real number μ, we rewrite (3.13) as follows:
      y μ ( t ) = 0 1 G ( t , s ) f ( s , s α 1 ( y μ ( s ) + μ ) ) d s , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ31_HTML.gif
      (4.1)
      From (4.1), it suffices to show that there exists a μ such that
      L ( μ ) : = y μ ( 1 ) = 0 1 G ( 1 , s ) f ( s , s α 1 ( y μ ( s ) + μ ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ32_HTML.gif
      (4.2)

      It is obvious that y μ ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq89_HTML.gif is continuously dependent on the parameter μ. In order to prove that there exists a μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq90_HTML.gif such that y μ ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq91_HTML.gif, we only need to show that lim μ L ( μ ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq92_HTML.gif, and lim μ L ( μ ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq93_HTML.gif.

      Now, we show that lim μ L ( μ ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq92_HTML.gif. On the contrary, we assume that lim ̲ μ L ( μ ) < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq94_HTML.gif. Then, there exists a sequence { μ n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq95_HTML.gif, lim n μ n = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq96_HTML.gif such that lim μ n L ( μ n ) < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq97_HTML.gif, which implies that the sequence { L ( μ n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq98_HTML.gif is bounded from above. Notice that the function f ( t , t α 1 y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq66_HTML.gif is continuous with respect to t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq49_HTML.gif and y R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq67_HTML.gif. We first claim that it is impossible to have
      f ( t , t α 1 ( y μ n ( t ) + μ n ) ) 0 for all t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ33_HTML.gif
      (4.3)
      as μ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq99_HTML.gif is large enough. Indeed, assume that (4.3) is true. Then, by (4.1), we have
      y μ n ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ34_HTML.gif
      (4.4)
      for all t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq49_HTML.gif. Thus, we get
      lim μ n ( y μ n ( t ) + μ n ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ35_HTML.gif
      (4.5)
      for all t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq49_HTML.gif. Since we have assumed in (H) that
      lim u f ( t , t α 1 u ) = , t ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ36_HTML.gif
      (4.6)
      by (4.2), (4.5)-(4.6), we have
      lim μ n y μ n ( 1 ) = lim μ n 0 1 G ( 1 , s ) f ( s , s α 1 ( y μ n ( s ) + μ n ) ) d s = lim μ n 1 4 3 4 G ( 1 , s ) f ( s , s α 1 ( y μ n ( s ) + μ n ) ) d s = , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ37_HTML.gif
      (4.7)

      which contradicts our assumption.

      Now, for large μ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq99_HTML.gif, we define
      I n = { t [ 0 , 1 ] | f ( t , t α 1 ( y μ n ( t ) + μ n ) ) < 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equo_HTML.gif

      Then, I n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq100_HTML.gif is not empty.

      Further, we divide the set I n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq100_HTML.gif into two sets I ˜ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq101_HTML.gif and I ˆ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq102_HTML.gif as follows:
      I ˜ n = { t I n | y μ n ( t ) + μ n > 0 } , I ˆ n = { t I n | y μ n ( t ) + μ n 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equp_HTML.gif

      It is easy to know that I ˜ n I ˆ n = ϕ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq103_HTML.gif, and I ˜ n I ˆ n = I n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq104_HTML.gif, and we have from (H) that I ˆ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq102_HTML.gif is not empty.

      From (H) again, the function f ( t , t α 1 u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq105_HTML.gif is bounded below by a constant for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq49_HTML.gif and u [ 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq106_HTML.gif. Thus, there exists a constant M (<0), independent of t and μ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq99_HTML.gif, such that
      f ( t , t α 1 ( y μ n ( t ) + μ n ) ) M , t I ˜ n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ38_HTML.gif
      (4.8)
      Let
      m ¯ ( μ n ) = min t I n y μ n ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equq_HTML.gif
      From the definitions of I ˜ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq101_HTML.gif and I ˆ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq102_HTML.gif, we have
      m ¯ ( μ n ) = min t I ˆ n y μ n ( t ) = y μ n ( t ) I ˆ n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equr_HTML.gif
      and it follows that m ¯ ( μ n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq107_HTML.gif as μ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq108_HTML.gif (since if m ¯ ( μ n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq109_HTML.gif is bounded below by a constant as μ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq108_HTML.gif, then (4.7) holds). Therefore, we can choose μ n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq110_HTML.gif large enough so that
      m ¯ ( μ n ) < min { 1 , M 0 1 G ( s , s ) d s 0 1 G ( s , s ) a ( s ) d s 1 0 1 G ( s , s ) b ( s ) d s } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ39_HTML.gif
      (4.9)
      for n > n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq111_HTML.gif. From (H), (4.1), (4.8)-(4.9), and the definitions of I ˜ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq101_HTML.gif and I ˆ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq102_HTML.gif, for any μ n > μ n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq112_HTML.gif, we have
      y μ n ( t ) = 0 1 G ( t , s ) f ( s , s α 1 ( y μ n ( s ) + μ n ) ) d s I n G ( s , s ) f ( s , s α 1 ( y μ n ( s ) + μ n ) ) d s I ˜ n G ( s , s ) f ( s , s α 1 ( y μ n ( s ) + μ n ) ) d s + I ˆ n G ( s , s ) ( a ( s ) b ( s ) | y μ n ( s ) + μ n | p ) d s ( M I ˜ n G ( s , s ) d s I ˆ n G ( s , s ) a ( s ) d s ) I ˆ n G ( s , s ) b ( s ) d s y μ n ( t ) + μ n p , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equs_HTML.gif
      from which it follows that
      y μ n ( t ) M 0 1 G ( s , s ) d s 0 1 G ( s , s ) a ( s ) d s 0 1 G ( s , s ) b ( s ) d s y μ n ( t ) I n p M 0 1 G ( s , s ) d s 0 1 G ( s , s ) a ( s ) d s + 0 1 G ( s , s ) b ( s ) d s m ¯ ( μ n ) , t I n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equt_HTML.gif
      which implies that
      m ¯ ( μ n ) M 0 1 G ( s , s ) d s 0 1 G ( s , s ) a ( s ) d s 1 0 1 G ( s , s ) b ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equu_HTML.gif

      This contradicts (4.9).

      Now, we have proved that lim μ L ( μ ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq92_HTML.gif. By a similar method, we can prove that lim μ L ( μ ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq93_HTML.gif. The detail is omitted.

      Notice that L ( μ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq113_HTML.gif is continuous with respect to μ ( , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq114_HTML.gif. It follows from the intermediate value theorem [11] that there exists a μ ( , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq115_HTML.gif such that L ( μ ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq116_HTML.gif, that is y ( 1 ) = y μ ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_IEq117_HTML.gif, which satisfies the second boundary value condition of (1.2). The proof is completed. □

      5 Examples

      Example 5.1 Consider the following boundary value problem
      { D 3 / 2 u ( t ) + t 2 + u 2 = 0 , t [ 0 , 1 ] , u ( 0 ) = 0 , u ( 1 ) = 2 1 2 u ( 1 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equ40_HTML.gif
      (5.1)
      where
      α = 3 / 2 , η = 1 2 , 2 1 2 ( 1 2 ) 3 2 1 = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equv_HTML.gif
      and
      f ( t , u ) = t 2 + u 2 , f ( t , t 1 2 u ) = t 2 + t 1 2 u 2 , b ( t ) = t 1 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equw_HTML.gif
      It is easy to show that
      lim u ± f ( t , t 1 2 u ) = ± , t ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equx_HTML.gif
      and
      0 1 G ( s , s ) b ( s ) d s 1 2 1 Γ ( 3 2 ) ( 1 ( 1 2 ) 1 2 ) 0 1 ( 1 s ) 1 2 s 1 2 d s = 1 2 1 Γ ( 3 2 ) ( 1 ( 1 2 ) 1 2 ) Γ ( 3 2 ) Γ ( 3 2 ) Γ ( 3 ) 0.756 < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-68/MediaObjects/13661_2011_Article_166_Equy_HTML.gif

      Thus, the conditions of Theorem 2.1 are satisfied. Therefore, the problem (5.1) has at least a nontrivial solution.

      Author’s contributions

      Each of the authors, ZO and GL contributed to each part of this study equally and read and approved the final version of the mnanuscript.

      Declarations

      Acknowledgements

      Supported partially by China Postdoctoral Science Foundation under Grant No.20110491280 and the Subject Lead Foundation of University of South China No. 2007XQD13.

      Authors’ Affiliations

      (1)
      School of Mathematics and Physics, School of Nuclear Science and Technology, University of South China

      References

      1. Kilbas AA, Srivastava HM, Trujillo JJ: Theory and Applications of Fractional Differential Equations. Elsevier, Amstadam; 2006.MATH
      2. Bakhani A, Daftardar-Gejji V: Existence of positive solutions of nonlinear fractional differential equations. J. Math. Anal. Appl. 2003, 278: 434-442. 10.1016/S0022-247X(02)00716-3MathSciNetView Article
      3. Bakhani A, Daftardar-Gejji V: Positive solutions of a system of non-autonomous fractional differential equations. J. Math. Anal. Appl. 2005, 302: 56-64. 10.1016/j.jmaa.2004.08.007MathSciNetView Article
      4. Li CF, Luo XN, Zhou Y: Existence of positive solutions of the boundary value problem for nonlinear fractional differential equations. Comput. Math. Appl. 2010, 59: 1363-1375. 10.1016/j.camwa.2009.06.029MATHMathSciNetView Article
      5. Bai CZ: Positive solutions for nonlinear fractional differential equations with coefficient that changes sign. Nonlinear Anal. 2006, 64: 677-685. 10.1016/j.na.2005.04.047MATHMathSciNetView Article
      6. Delbosco D: Fractional calculus and function spaces. J. Fract. Calc. 1994, 6: 45-53.MATHMathSciNet
      7. Jiang DQ, Yuan CJ: The positive properties of the Green function for Dirichlet-type boundary value problems of nonlinear fractional differential equations and its application. Nonlinear Anal. 2010, 72: 710-719. 10.1016/j.na.2009.07.012MATHMathSciNetView Article
      8. Kaufmann ER, Mboumi E: Positive solutions of a boundary value problem for a nonlocal fractional differential equations. Electron. J. Qual. Theory Differ. Equ. 2008, 3: 1-11.MathSciNetView Article
      9. Jafari H, Gejji VD: Positive solutions fractional nonlinear boundary value problems using Adomian decomposition method. Appl. Math. Comput. 2006, 180: 700-706. 10.1016/j.amc.2006.01.007MATHMathSciNetView Article
      10. Podlubny I: Fractional Differential Equations. Academic Press, New York; 1993.
      11. Liao KR, Li ZY: Mathematical Analysis. Higher Education Press, Beijing; 1986. in Chinese
      12. Belmekki M, Benchohra M: Existence results for fractional order semilinear functional differential equations with nondense domain. Nonlinear Anal. 2010, 72: 925-932. 10.1016/j.na.2009.07.034MATHMathSciNetView Article
      13. 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.2009.01.073MATHMathSciNetView Article
      14. EI-Shahed M: Positive solutions for boundary value problem of nonlinear fractional differential equations. Abstr. Appl. Anal. 2007., 2007:
      15. Kosmatov N: Integral equations and initial value problems for nonlinear differential equations of fractional order. Nonlinear Anal. 2009, 70: 2521-2529. 10.1016/j.na.2008.03.037MATHMathSciNetView Article
      16. Agarmal RP, Meehan M, O’Regan D: Fixed Point Theory and Applications. Cambridge University Press, Cambridge; 2001.View Article
      17. Agarwal RP, Zhou Y, He Y: Existence of fractional neutral functional differential equations. Comput. Math. Appl. 2010, 59: 1095-1110. 10.1016/j.camwa.2009.05.010MATHMathSciNetView Article
      18. Liang SH, Zhang JH: Positive solutions for boundary value problems of nonlinear fractional differential equation. Nonlinear Anal. 2009, 71: 5545-5550. 10.1016/j.na.2009.04.045MATHMathSciNetView Article
      19. Zhang SQ: Positive solutions for boundary value problems of nonlinear fractional differential equations. Electron. J. Differ. Equ. 2006, 2006: 1-12.View Article
      20. Gejji VD: Positive solutions of a system of non-autonomous fractional differential equations. J. Math. Anal. Appl. 2005, 302: 56-64. 10.1016/j.jmaa.2004.08.007MATHMathSciNetView Article
      21. Lakshmikantham V: Theory of fractional function differential equations. Nonlinear Anal. 2008, 69: 3337-3343. 10.1016/j.na.2007.09.025MATHMathSciNetView Article
      22. Lakshmikantham V, Vatsala AS: Basic theory of fractional differential equations. Nonlinear Anal. 2008, 69: 2677-2682. 10.1016/j.na.2007.08.042MATHMathSciNetView Article
      23. Lakshmikantham V, Leela S, Vasundhara Devi J: Theory of Fractional Dynamic Systems. Cambridge Scientific Publishers, Cambridge; 2009.MATH
      24. Zhou Y, Jiao F, Li J: Existence and uniqueness of fractional neutral differential equations with infinite delay. Nonlinear Anal. 2009, 71: 3249-3256. 10.1016/j.na.2009.01.202MATHMathSciNetView Article
      25. Bai ZB: On positive solutions of a nonlocal fractional boundary value problem. Nonlinear Anal. 2010, 72: 916-924. 10.1016/j.na.2009.07.033MATHMathSciNetView Article
      26. Bai ZB, Lu HS: Positive solutions for boundary value problem of nonlinear fractional differential equation. J. Math. Anal. Appl. 2005, 311: 495-505. 10.1016/j.jmaa.2005.02.052MATHMathSciNetView Article

      Copyright

      © Ouyang and Li; licensee Springer 2012

      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.