Open Access

Nontrivial solutions for a higher fractional differential equation with fractional multi-point boundary conditions

Boundary Value Problems20122012:70

DOI: 10.1186/1687-2770-2012-70

Received: 7 April 2012

Accepted: 15 June 2012

Published: 3 July 2012

Abstract

This paper investigates the existence and uniqueness of nontrivial solutions to a class of fractional nonlocal multi-point boundary value problems of higher order fractional differential equation, this kind of problems arise from viscoelasticity, electrochemistry control, porous media, electromagnetic and signal processing of wireless communication system. Some sufficient conditions for the existence and uniqueness of nontrivial solutions are established under certain suitable growth conditions, our proof is based on Leray-Schauder nonlinear alternative and Schauder fixed point theorem.

MSC:34B15, 34B25.

Keywords

fractional differential equation nontrivial solution Green function Leray-Schauder nonlinear alternative

1 Introduction

The purpose of this paper is to establish the existence and uniqueness of nontrivial solutions to the following higher fractional differential equation:
{ D α x ( t ) = f ( t , x ( t ) , D μ 1 x ( t ) , D μ 2 x ( t ) , , D μ n 1 x ( t ) ) , 0 < t < 1 , x ( 0 ) = 0 , D μ i x ( 0 ) = 0 , D μ x ( 1 ) = j = 1 p 2 a j D μ x ( ξ j ) , 1 i n 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ1_HTML.gif
(1.1)

where n 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq1_HTML.gif, n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq2_HTML.gif, n 1 < α n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq3_HTML.gif, n l 1 < α μ l < n l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq4_HTML.gif, for l = 1 , 2 , , n 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq5_HTML.gif, and μ μ n 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq6_HTML.gif, α μ n 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq7_HTML.gif, α μ > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq8_HTML.gif, a j [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq9_HTML.gif, 0 < ξ 1 < ξ 2 < < ξ p 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq10_HTML.gif, j = 1 p 2 a j ξ j α μ 1 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq11_HTML.gif, D α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq12_HTML.gif is the standard Riemann-Liouville derivative, and f : [ 0 , 1 ] × R n R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq13_HTML.gif is continuous.

Differential equations of fractional order occur more frequently in different research areas such as engineering, physics, chemistry, economics, etc. Indeed, we can find numerous applications in viscoelasticity, electrochemistry control, porous media, electromagnetic and signal processing of wireless communication system, etc. [16].

For an extensive collection of results about this type of equations, we refer the reader to the monograph by Kilbas et al. [7], Miller and Ross [8], Podlubny [9], the papers [1024] and the references therein.

Recently, Salem [10] has investigated the existence of Pseudo solutions for the nonlinear m-point boundary value problem of a fractional type. In particular, he considered the following boundary value problem:
{ D α x ( t ) + q ( t ) f ( t , x ( t ) ) = 0 , 0 < t < 1 , α ( n 1 , n ] , n 2 , x ( 0 ) = x ( 0 ) = x ( 0 ) = = x ( n 2 ) ( 0 ) = 0 , x ( 1 ) = i = 1 m 2 ξ i x ( η i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ2_HTML.gif
(1.2)

where x takes values in a reflexive Banach space E 0 < η 1 < η 2 < < η m 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq14_HTML.gif and ξ i > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq15_HTML.gif with j = 1 m 2 ξ j η j α 1 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq16_HTML.gif. x ( k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq17_HTML.gif denotes the k th Pseudo-derivative of x and D α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq18_HTML.gif denotes the Pseudo fractional differential operator of order α. By means of the fixed point theorem attributed to O’Regan, a criterion was established for the existence of at least one Pseudo solution for the problem (1.2).

More recently, Zhang [11] has considered the following problem whose nonlinear term and boundary condition contain integer order derivatives of unknown functions:
{ D α x ( t ) + q ( t ) f ( x , x , , x ( n 2 ) ) = 0 , 0 < t < 1 , n 1 < α n , x ( 0 ) = x ( 0 ) = = x ( n 2 ) ( 0 ) = x ( n 2 ) ( 1 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ3_HTML.gif
(1.3)
where D α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq18_HTML.gif is the standard Riemann-Liouville fractional derivative of order α q may be singular at t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq19_HTML.gif and f may be singular at x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq20_HTML.gif x = 0 , , x ( n 2 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq21_HTML.gif. By using the fixed point theorem of a mixed monotone operator, a unique existence result of positive solution to the problem (1.3) was established. And then, Goodrich [12] was concerned with a partial extension of the problem (1.3) by extending boundary conditions
{ D α x ( t ) = f ( t , x ( t ) ) , 0 < t < 1 , n 1 < α n , n > 3 , x ( i ) ( 0 ) = 0 , 0 i n 2 , D α x ( 1 ) = 0 , 1 α n 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ4_HTML.gif
(1.4)

The author derived the Green’s function for the problem (1.4) and showed that it satisfies certain properties. Then, by using cone theoretic techniques, a general existence theorem for (1.4) was obtained when f ( t , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq22_HTML.gif satisfies some growth conditions.

In recent work [13], Rehman and Khan have investigated the multi-point boundary value problems for fractional differential equations of the form
{ D α y ( t ) = f ( t , y ( t ) , D β y ( t ) ) , t ( 0 , 1 ) , y ( 0 ) = 0 , D β y ( 1 ) i = 1 m 2 ζ i D β y ( ξ i ) = y 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ5_HTML.gif
(1.5)

where 1 < α 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq23_HTML.gif 0 < β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq24_HTML.gif 0 < ξ i < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq25_HTML.gif ζ i [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq26_HTML.gif with i = 1 m 2 ζ i ξ i α β 1 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq27_HTML.gif. By using the Schauder fixed point theorem and the contraction mapping principle, the authors established the existence and uniqueness of nontrivial solutions for BVP (1.5) provided that the nonlinear function f : [ 0 , 1 ] × R × R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq28_HTML.gif is continuous and satisfies certain growth conditions. However, Rehman and Khan only considered the case 1 < α 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq23_HTML.gif and the case of the nonlinear term f was not considered comprehensively.

Notice that the results dealing with the existence and uniqueness of solution for multi-point boundary value problems of fractional order differential equations are relatively scarce when the nonlinear term f and the boundary conditions all involve fractional derivatives of unknown functions. Thus, the aim of this paper is to establish the existence and uniqueness of nontrivial solutions for the higher nonlocal fractional differential equations (1.1) where nonlinear term f and the boundary conditions all involve fractional derivatives of unknown functions. In our study, the proof is based on the reduced order method as in [11] and the main tool is the Leray-Schauder nonlinear alternative and the Schauder fixed point theorem.

2 Basic definitions and preliminaries

Definition 2.1 A function x is said to be a solution of BVP (1.1) if x C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq29_HTML.gif and satisfies BVP (1.1). In addition, x is said to be a nontrivial solution if x 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq30_HTML.gif for t ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq31_HTML.gif and x is solution of BVP (1.1).

For the convenience of the reader, we present some definitions, lemmas, and basic results that will be used later. These and other related results and their proofs can be found, for example, in [69].

Definition 2.2 (see [8])

Let α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq32_HTML.gif with α R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq33_HTML.gif. Suppose that x : [ a , ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq34_HTML.gif then the α th Riemann-Liouville fractional integral is defined by
I α x ( t ) = 1 Γ ( α ) a t ( t s ) α 1 x ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equa_HTML.gif
whenever the right-hand side is defined. Similarly, with α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq32_HTML.gif with α R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq35_HTML.gif, we define the α th Riemann-Liouville fractional derivative to be
D α x ( t ) = 1 Γ ( n α ) ( d d t ) ( n ) a t ( t s ) n α 1 x ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equb_HTML.gif

where n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq36_HTML.gif is the unique positive integer satisfying n 1 α < n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq37_HTML.gif and t > a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq38_HTML.gif.

Remark 2.1 If x , y : ( 0 , + ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq39_HTML.gif with order α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq32_HTML.gif, then
D α ( x ( t ) + y ( t ) ) = D α x ( t ) + D α y ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equc_HTML.gif

Lemma 2.1 (see [7])

  1. (1)
    If x L 1 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq40_HTML.gif, ρ > σ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq41_HTML.gif, then
    I ρ I σ x ( t ) = I ρ + σ x ( t ) , D σ I ρ x ( t ) = I ρ σ x ( t ) , D σ I σ x ( t ) = x ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equd_HTML.gif
     
  2. (2)
    If ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq42_HTML.gif, ν > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq43_HTML.gif, then
    D ρ t ν 1 = Γ ( ν ) Γ ( ν ρ ) t ν ρ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Eque_HTML.gif
     

Lemma 2.2 (see [8])

Assume that x C ( 0 , 1 ) L 1 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq44_HTML.gifwith a fractional derivative of order α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq45_HTML.gif, then I α D α x ( t ) = x ( t ) + c 1 t α 1 + c 2 t α 2 + + c n t α n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq46_HTML.gif, where c i R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq47_HTML.gif, i = 1 , 2 , , n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq48_HTML.gif ( n = [ α ] + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq49_HTML.gif). Here I α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq50_HTML.gifstands for the standard Riemann-Liouville fractional integral of order α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq32_HTML.gifand D α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq18_HTML.gifdenotes the Riemann-Liouville fractional derivative as Definition 2.1.

Lemma 2.3 If 1 < α μ n 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq51_HTML.gif, α μ > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq52_HTML.gifand h L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq53_HTML.gif, then the boundary value problem
{ D α μ n 1 w ( t ) = h ( t ) , w ( 0 ) = 0 , D μ μ n 1 w ( 1 ) = j = 1 p 2 a j D μ μ n 1 w ( ξ j ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ6_HTML.gif
(2.1)
has the unique solution
w ( t ) = 0 1 K ( t , s ) h ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equf_HTML.gif
where K ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq54_HTML.gifis the Green function of BVP (2.1), and
K ( t , s ) = k 1 ( t , s ) + t α μ n 1 1 1 j = 1 p 2 a j ξ j α μ 1 j = 1 p 2 a j k 2 ( ξ j , s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ7_HTML.gif
(2.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ8_HTML.gif
(2.3)
Proof By applying Lemma 2.2, we may reduce (2.1) to an equivalent integral equation
w ( t ) = I α μ n 1 h ( t ) + c 1 t α μ n 1 1 + c 2 t α μ n 1 2 , c 1 , c 2 R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ9_HTML.gif
(2.4)
Note that w ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq55_HTML.gif and (2.4), we have c 2 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq56_HTML.gif. Consequently, a general solution of (2.3) is
w ( t ) = I α μ n 1 h ( t ) + c 1 t α μ n 1 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ10_HTML.gif
(2.5)
By (2.5) and Lemma 2.1, we have
D μ μ n 1 w ( t ) = D μ μ n 1 I α μ n 1 h ( t ) + c 1 D μ μ n 1 t α μ n 1 1 = I α μ h ( t ) + c 1 Γ ( α μ n 1 ) Γ ( α μ ) t α μ 1 = 0 t ( t s ) α μ 1 Γ ( α μ ) h ( s ) d s + c 1 Γ ( α μ n 1 ) Γ ( α μ ) t α μ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ11_HTML.gif
(2.6)
So, from (2.6), we have
D μ μ n 1 w ( 1 ) = 0 1 ( 1 s ) α μ 1 Γ ( α μ ) h ( s ) d s + c 1 Γ ( α μ n 1 ) Γ ( α μ ) , D μ μ n 1 w ( ξ j ) = 0 ξ j ( ξ j s ) α μ 1 Γ ( α μ ) h ( s ) d s + c 1 Γ ( α μ n 1 ) Γ ( α μ ) ξ j α μ 1 , for j = 1 , 2 , , p 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ12_HTML.gif
(2.7)
By D μ μ n 1 w ( 1 ) = j = 1 p 2 a j D μ μ n 1 w ( ξ j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq57_HTML.gif, combining with (2.7), we obtain
c 1 = 0 1 ( 1 s ) α μ 1 h ( s ) d s j = 1 p 2 a j 0 ξ j ( ξ j s ) α μ 1 h ( s ) d s Γ ( α μ n 1 ) ( 1 j = 1 p 2 a j ξ j α μ 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equg_HTML.gif
So, substituting c 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq58_HTML.gif into (2.5), the unique solution of the problem (2.1) is
w ( t ) = 0 t ( t s ) α μ n 1 1 Γ ( α μ n 1 ) h ( s ) d s + t α μ n 1 1 1 j = 1 p 2 a j ξ j α μ 1 × { 0 1 ( 1 s ) α μ 1 Γ ( α μ n 1 ) h ( s ) d s j = 1 p 2 a j 0 ξ j ( ξ j s ) α μ 1 Γ ( α μ n 1 ) h ( s ) d s } = 0 t ( t s ) α μ n 1 1 Γ ( α μ n 1 ) h ( s ) d s + ( 1 j = 1 p 2 a j ξ j α μ 1 + j = 1 p 2 a j ξ j α μ 1 ) t α μ n 1 1 1 j = 1 p 2 a j ξ j α μ 1 × 0 1 ( 1 s ) α μ 1 Γ ( α μ n 1 ) h ( s ) d s t α μ n 1 1 1 j = 1 p 2 a j ξ j α μ 1 j = 1 p 2 a j 0 ξ j ( ξ j s ) α μ 1 Γ ( α μ n 1 ) h ( s ) d s = 0 t ( t s ) α μ n 1 1 Γ ( α μ n 1 ) h ( s ) d s + 0 1 ( 1 s ) α μ 1 t α μ n 1 1 Γ ( α μ n 1 ) h ( s ) d s + t α μ n 1 1 1 j = 1 p 2 a j ξ j α μ 1 j = 1 p 2 a j 0 1 ( 1 s ) α μ 1 ξ j α μ 1 Γ ( α μ n 1 ) h ( s ) d s t α μ n 1 1 1 j = 1 p 2 a j ξ j α μ 1 j = 1 p 2 a j 0 ξ j ( ξ j s ) α μ 1 Γ ( α μ n 1 ) h ( s ) d s = 0 1 ( k 1 ( t , s ) + t α μ n 1 1 1 j = 1 p 2 a j ξ j α μ 1 j = 1 p 2 a j k 2 ( ξ j , s ) ) h ( s ) d s = 0 1 K ( t , s ) h ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equh_HTML.gif

The proof is completed. □

Lemma 2.4 | K ( t , s ) | M ( 1 s ) α μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq59_HTML.gif, for t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq60_HTML.gif, where
M = 1 + j = 1 p 2 a j | 1 j = 1 p 2 a j ξ j α μ 1 | Γ ( α μ n 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ13_HTML.gif
(2.8)
Proof Obviously, for t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq60_HTML.gif, we have k i ( t , s ) ( 1 s ) α μ 1 Γ ( α μ n 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq61_HTML.gif, i = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq62_HTML.gif. Thus
| K ( t , s ) | = | k 1 ( t , s ) + t α μ n 1 1 1 j = 1 p 2 a j ξ j α μ 1 j = 1 p 2 a j k 2 ( ξ j , s ) | ( 1 s ) α μ 1 Γ ( α μ n 1 ) + j = 1 p 2 a j ( 1 s ) α μ 1 Γ ( α μ n 1 ) | 1 j = 1 p 2 a j ξ j α μ 1 | ( 1 + j = 1 p 2 a j | 1 j = 1 p 2 a j ξ j α μ 1 | ) ( 1 s ) α μ 1 Γ ( α μ n 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equi_HTML.gif

This completes the proof. □

Now let us consider the following modified problem of BVP (1.1)
{ D α μ n 1 v ( t ) = f ( t , I μ n 1 v ( t ) , I μ n 1 μ 1 v ( t ) , , I μ n 1 μ n 2 v ( t ) , v ( t ) ) , v ( 0 ) = 0 , D μ μ n 1 v ( 1 ) = j = 1 p 2 a j D μ μ n 1 v ( ξ j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ14_HTML.gif
(2.9)

Lemma 2.5 Let x ( t ) = I μ n 1 v ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq63_HTML.gif, v ( t ) C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq64_HTML.gif. Then (2.9) can be transformed into (1.1). Moreover, if v C ( [ 0 , 1 ] , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq65_HTML.gifis a solution of the problem (2.9), then the function x ( t ) = I μ n 1 v ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq66_HTML.gifis a solution of the problem (1.1).

Proof Substituting x ( t ) = I μ n 1 v ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq63_HTML.gif into (1.1), by Lemmas 2.1 and 2.2, we can obtain that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ15_HTML.gif
(2.10)

and also D μ n 1 x ( 0 ) = v ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq67_HTML.gif. It follows from D μ x ( t ) = D μ I μ n 1 v ( t ) = d n d t n I n μ I μ n 1 v ( t ) = D μ μ n 1 v ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq68_HTML.gif that D μ μ n 1 v ( 1 ) = j = 1 p 2 a j D μ μ n 1 v ( ξ j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq69_HTML.gif. Using x ( t ) = I μ n 1 v ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq63_HTML.gif, v C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq70_HTML.gif, (2.9) is transformed into (1.1).

Now, let v C ( [ 0 , 1 ] , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq65_HTML.gif be a solution for the problem (2.9). Then, from Lemma 2.1, (2.9) and (2.10), one has
D α x ( t ) = d n d t n I n α x ( t ) = d n d t n I n α I μ n 1 v ( t ) = d n d t n I n α + μ n 1 v ( t ) = D α μ n 1 v ( t ) = f ( t , I μ n 1 v ( t ) , I μ n 1 μ 1 v ( t ) , , I μ n 1 μ n 2 v ( t ) , v ( t ) ) = f ( t , x ( t ) , D μ 1 x ( t ) , D μ 2 x ( t ) , , D μ n 1 x ( t ) ) , 0 < t < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equj_HTML.gif
Notice
I α v ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equk_HTML.gif
which implies that I α v ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq71_HTML.gif. Thus from (2.10), for i = 1 , 2 , , n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq72_HTML.gif, we have
x ( 0 ) = 0 , D μ i x ( 0 ) = 0 , D μ x ( 1 ) = j = 1 p 2 a j D μ x ( ξ j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equl_HTML.gif
Moreover, it follows from the monotonicity and property of I μ n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq73_HTML.gif that
I μ n 1 v C ( [ 0 , 1 ] , [ 0 , + ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equm_HTML.gif

Consequently, x ( t ) = I μ n 1 v ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq63_HTML.gif is a solution of the problem (1.1). □

Now let us define an operator T : C [ 0 , 1 ] C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq74_HTML.gif by
( T v ) ( t ) = 0 1 K ( t , s ) f ( s , I μ n 1 v ( s ) , I μ n 1 μ 1 v ( s ) , , I μ n 1 μ n 2 v ( s ) , v ( s ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ16_HTML.gif
(2.11)

Clearly, the fixed point of the operator T is a solution of BVP (2.9); and consequently is also a solution of BVP (1.1) from Lemma 2.5.

Lemma 2.6 T : C [ 0 , 1 ] C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq74_HTML.gifis a completely continuous operator.

Proof Noticing that f : [ 0 , 1 ] × R n R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq75_HTML.gif is continuous, by using the Ascoli-Arzela theorem and standard arguments, the result can easily be shown. □

Lemma 2.7 (see [25])

Let X be a real Banach space, Ω be a bounded open subset of X, where θ Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq76_HTML.gif, T : Ω ¯ X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq77_HTML.gifis a completely continuous operator. Then, either there exists x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq78_HTML.gif, λ > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq79_HTML.gifsuch that T ( x ) = λ x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq80_HTML.gif, or there exists a fixed point x Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq81_HTML.gif.

3 Main results

For the convenience of expression in rest of the paper, we let μ 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq82_HTML.gif.

Theorem 3.1 Suppose f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq83_HTML.giffor any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq84_HTML.gif. Moreover, there exist nonnegative functions p 1 , p 2 , , p n , q L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq85_HTML.gifsuch that
| f ( t , u 1 , u 2 , , u n ) | i = 1 n p i ( t ) | u i | + q ( t ) , a.e. ( t , u 1 , u 2 , , u n ) [ 0 , 1 ] × R n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ17_HTML.gif
(3.1)
and
M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s < ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ18_HTML.gif
(3.2)

where M is defined by (2.8). Then BVP (1.1) has at least one nontrivial solution.

Proof Since f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq86_HTML.gif, there exists [ σ , τ ] [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq87_HTML.gif such that
min t [ σ , τ ] | f ( t , 0 , , 0 ) | > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equn_HTML.gif
By condition (3.1), we have q ( t ) | f ( t , 0 , , 0 ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq88_HTML.gif, a.e. t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq89_HTML.gif, thus
0 1 ( 1 s ) α μ 1 q ( s ) d s > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equo_HTML.gif
On the other hand, from (3.2), we know
( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equp_HTML.gif
Take
r = M 0 1 ( 1 s ) α μ 1 q ( s ) d s 1 ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equq_HTML.gif

then r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq90_HTML.gif.

Now let Ω r = { v C [ 0 , 1 ] : x < r } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq91_HTML.gif, suppose v Ω r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq92_HTML.gif, λ > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq93_HTML.gif such that T v = λ v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq94_HTML.gif. Then
λ r = λ v = T v = max t [ 0 , 1 ] | T v ( t ) | M 0 1 ( 1 s ) α μ 1 f ( s , I μ n 1 v ( s ) , I μ n 1 μ 1 v ( s ) , , I μ n 1 μ n 2 v ( s ) , v ( s ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ19_HTML.gif
(3.3)
Moreover, for i = 0 , 1 , 2 , , n 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq95_HTML.gif,
| I μ n 1 μ i v ( t ) | = | 0 t ( t s ) μ n 1 μ i 1 v ( s ) Γ ( μ n 1 μ i ) d s | v Γ ( μ n 1 μ i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equr_HTML.gif
thus we have, by hypothesis (3.1),
| f ( s , I μ n 1 v ( s ) , I μ n 1 μ 1 v ( s ) , , I μ n 1 μ n 2 v ( s ) , v ( s ) ) | p 1 ( s ) | I μ n 1 v ( s ) | + p 2 ( s ) | I μ n 1 μ 1 v ( s ) | + + p n 1 ( s ) | I μ n 1 μ n 2 v ( s ) | + p n ( s ) | v ( s ) | + q ( s ) v Γ ( μ n 1 ) p 1 ( s ) + v Γ ( μ n 1 μ 1 ) p 2 ( s ) + + v Γ ( μ n 1 μ n 2 ) p n 1 ( s ) + v p n ( s ) + q ( s ) ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) v [ p 1 ( s ) + p 2 ( s ) + + p n 1 ( s ) + p n ( s ) ] + q ( s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equs_HTML.gif
Consequently, from (3.3), we have
λ r ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s v + M 0 1 ( 1 s ) α μ 1 q ( s ) d s = r ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s + M 0 1 ( 1 s ) α μ 1 q ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equt_HTML.gif
Therefore,
λ ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s + M 0 1 ( 1 s ) α μ 1 q ( s ) d s r = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equu_HTML.gif

This contradicts λ > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq79_HTML.gif. By Lemma 2.7, T has a fixed point v Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq96_HTML.gif, since f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq83_HTML.gif; so then, by Lemma 2.5, BVP (1.1) has a nontrivial solution v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq97_HTML.gif. This completes the proof. □

Theorem 3.2 Suppose f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq98_HTML.giffor any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq84_HTML.gif. Moreover, there exist nonnegative functions p 1 , p 2 , , p n , q L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq85_HTML.gifsuch that
| f ( t , u 1 , u 2 , , u n ) | i = 1 n p i ( t ) | u i | σ i + q ( t ) , a.e. ( t , u 1 , u 2 , , u n ) [ 0 , 1 ] × R n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ20_HTML.gif
(3.4)

where 0 < σ 1 , σ 2 , , σ n < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq99_HTML.gifare nonnegative constants. Then BVP (1.1) has at least one nontrivial solution.

Proof By Lemma 2.6, we know T : C [ 0 , 1 ] C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq100_HTML.gif is a completely continuous operator.

Let
a = ( 1 + i = 0 n 2 1 Γ σ i + 1 ( μ n 1 μ i ) ) M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s , b = M 0 1 ( 1 s ) α μ 1 q ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equv_HTML.gif
Choose
R { ( n + 1 ) b , [ ( n + 1 ) a ] 1 1 σ 1 , [ ( n + 1 ) a ] 1 1 σ 2 , , [ ( n + 1 ) a ] 1 1 σ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equw_HTML.gif
and define a ball M = { v C [ 0 , 1 ] : v R , t [ 0 , 1 ] } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq101_HTML.gif. For every v M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq102_HTML.gif, we have
| T v ( t ) | 0 1 K ( t , s ) | f ( s , I μ n 1 v ( s ) , I μ n 1 μ 1 v ( s ) , , I μ n 1 μ n 2 v ( s ) , v ( s ) ) | d s M 0 1 ( 1 s ) α μ 1 | f ( s , I μ n 1 v ( s ) , I μ n 1 μ 1 v ( s ) , , I μ n 1 μ n 2 v ( s ) , v ( s ) ) | d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equx_HTML.gif
On the other hand, it follows from (3.4) that
| f ( s , I μ n 1 v ( s ) , I μ n 1 μ 1 v ( s ) , , I μ n 1 μ n 2 v ( s ) , v ( s ) ) | p 1 ( s ) | I μ n 1 v ( s ) | σ 1 + p 2 ( s ) | I μ n 1 μ 1 v ( s ) | σ 2 + + p n 1 ( s ) | I μ n 1 μ n 2 v ( s ) | σ n 1 + p n ( s ) | v ( s ) | σ n + q ( s ) v σ 1 Γ σ 1 ( μ n 1 ) p 1 ( s ) + v σ 2 Γ σ 2 ( μ n 1 μ 1 ) p 2 ( s ) + + v σ n 1 Γ σ n 1 ( μ n 1 μ n 2 ) p n 1 ( s ) + v σ n p n ( s ) + q ( s ) ( v σ n + i = 0 n 2 v σ i + 1 Γ σ i + 1 ( μ n 1 μ i ) ) [ p 1 ( s ) + p 2 ( s ) + + p n 1 ( s ) + p n ( s ) ] + q ( s ) ( 1 + i = 0 n 2 1 Γ σ i + 1 ( μ n 1 μ i ) ) i = 1 n v σ i i = 1 n p i ( s ) + q ( s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ21_HTML.gif
(3.5)
In view of (3.5), we have the following estimate:
| T v ( t ) | ( 1 + i = 0 n 2 1 Γ σ i + 1 ( μ n 1 μ i ) ) M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s i = 1 n v σ i + M 0 1 ( 1 s ) α μ 1 q ( s ) d s = a i = 1 n v σ i + b n R n + 1 + R n + 1 = R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equy_HTML.gif

Therefore, T v R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq103_HTML.gif. Thus we have T : M M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq104_HTML.gif. Hence the Schauder fixed point theorem implies the existence of a solution in M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq105_HTML.gif for BVP (2.9). Since f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq106_HTML.gif, then by Lemma 2.5, BVP (1.1) has a nontrivial solution v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq97_HTML.gif. This completes the proof. □

Theorem 3.3 Suppose f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq107_HTML.giffor any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq84_HTML.gif. Moreover, there exist nonnegative functions p 1 , p 2 , , p n , q L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq108_HTML.gifsuch that
| f ( t , u 1 , u 2 , , u n ) | i = 1 n p i ( t ) | u i | σ i + q ( t ) , a.e. ( t , u 1 , u 2 , , u n ) [ 0 , 1 ] × R n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ22_HTML.gif
(3.6)

where σ 1 , σ 2 , , σ n > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq109_HTML.gifare nonnegative constants. Then BVP (1.1) has at least one nontrivial solution.

Proof The proof is similar to that of Theorem 3.2, so it is omitted. □

Remark 3.1 In [13], the authors studied the cases 1 < α 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq110_HTML.gif μ 1 = μ 2 = = μ n 1 = β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq111_HTML.gif 0 < β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq24_HTML.gif, but the case of σ i = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq112_HTML.gif i = 1 , 2 , , n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq48_HTML.gif was not considered. Here we extend the results of [13] and fill the case σ i = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq112_HTML.gif i = 1 , 2 , , n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq113_HTML.gif.

Theorem 3.4 Suppose f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq107_HTML.giffor any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq84_HTML.gif. Moreover, there exist nonnegative functions p 1 , p 2 , , p n L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq114_HTML.gifsuch that
| f ( t , u 1 , u 2 , , u n ) f ( t , v 1 , v 2 , , v n ) | i = 1 n p i ( t ) | u i v i | , a.e. ( t , u 1 , u 2 , , u n ) , ( t , v 1 , v 2 , , v n ) [ 0 , 1 ] × R n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ23_HTML.gif
(3.7)

and (3.2) holds. Then BVP (1.1) has a unique nontrivial solution.

Proof In fact, if v 1 = v 2 = = v n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq115_HTML.gif, then we have
| f ( t , u 1 , u 2 , , u n ) | i = 1 n p i ( t ) | u i | + | f ( t , 0 , 0 , , 0 ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equz_HTML.gif

From Theorem 3.1, we know BVP (1.1) has a nontrivial solution.

But in this case, we prefer to concentrate on the uniqueness of a nontrivial solution for BVP (1.1). Let T be given in (2.11), we shall show that T is a contraction. In fact, by (3.7), a similar method to Theorem 3.1, we have
| f ( s , I μ n 1 u ( s ) , I μ n 1 μ 1 u ( s ) , , I μ n 1 μ n 2 u ( s ) , u ( s ) ) f ( s , I μ n 1 v ( s ) , I μ n 1 μ 1 v ( s ) , , I μ n 1 μ n 2 v ( s ) , v ( s ) ) | ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) u v i = 1 n p i ( s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equaa_HTML.gif
And then
T u T v M 0 1 ( 1 s ) α μ 1 | f ( s , I μ n 1 u ( s ) , I μ n 1 μ 1 u ( s ) , , I μ n 1 μ n 2 u ( s ) , u ( s ) ) f ( s , I μ n 1 v ( s ) , I μ n 1 μ 1 v ( s ) , , I μ n 1 μ n 2 v ( s ) , v ( s ) ) | d s ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s u v . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equab_HTML.gif

Then (3.2) implies that T is indeed a contraction. Finally, we use the Banach fixed point theorem to deduce the existence of a unique nontrivial solution to BVP (1.1). □

Corollary 3.1 Suppose f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq116_HTML.giffor any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq84_HTML.gif, and (3.1) holds. Then BVP (1.1) has at least one nontrivial solution if one of the following conditions holds
  1. (1)
    There exists a constant p > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq117_HTML.gif such that
    0 1 [ i = 1 n p i ( s ) ] p d s < ( p ( α μ 1 ) p 1 + 1 ) p 1 ( M + i = 0 n 2 M Γ ( μ n 1 μ i ) ) p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ24_HTML.gif
    (3.8)
     
  2. (2)
    There exists a constant λ > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq118_HTML.gif such that
    i = 1 n p i ( s ) < Γ ( α + λ μ 1 ) Γ ( α μ ) Γ ( λ + 1 ) ( M + i = 0 n 2 M Γ ( μ n 1 μ i ) ) 1 s λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ25_HTML.gif
    (3.9)
     
  3. (3)
    There exists a constant λ > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq118_HTML.gif such that
    i = 1 n p i ( s ) < ( α + λ μ ) ( M + i = 0 n 2 M Γ ( μ n 1 μ i ) ) 1 ( 1 s ) λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ26_HTML.gif
    (3.10)
     
  4. (4)
    p i ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq119_HTML.gif ( i = 1 , 2 , , n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq48_HTML.gif) satisfy
    i = 1 n p i ( s ) < ( α μ ) ( M + i = 0 n 2 M Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ27_HTML.gif
    (3.11)
     
Proof Let
R = M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equac_HTML.gif
From the proof of Theorem 3.1, we only need to prove
R < ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equad_HTML.gif
  1. (1)
    If (3.8) holds, let 1 p + 1 q = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq120_HTML.gif, and by using H o ¨ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq121_HTML.giflder inequality,
    R M ( 0 1 [ i = 1 n p i ( s ) ] p d s ) 1 p ( 0 1 ( 1 s ) q ( α μ 1 ) d s ) 1 q = M [ q ( α μ 1 ) + 1 ] 1 q ( 0 1 [ i = 1 n p i ( s ) ] p d s ) 1 p = M [ p ( α μ 1 ) p 1 + 1 ] p 1 p ( 0 1 [ i = 1 n p i ( s ) ] p d s ) 1 p < ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equae_HTML.gif
     
  2. (2)
    In this case, it follows from (3.9) that
    R < M Γ ( α + λ μ 1 ) Γ ( α μ ) Γ ( λ + 1 ) ( M + i = 0 n 2 M Γ ( μ n 1 μ i ) ) 1 0 1 ( 1 s ) α μ 1 s λ d s = ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equaf_HTML.gif
     
  3. (3)
    In this case, it follows from (3.10) that
    R < M ( α + λ μ ) ( M + i = 0 n 2 M Γ ( μ n 1 μ i ) ) 1 0 1 ( 1 s ) α μ 1 ( 1 s ) λ d s = ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equag_HTML.gif
     
  4. (4)
    If (3.11) is satisfied, we have
    R < M ( α μ ) ( M + i = 0 n 2 M Γ ( μ n 1 μ i ) ) 1 0 1 ( 1 s ) α μ 1 d s = ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equah_HTML.gif
     

This completes the proof of Corollary 3.1. □

Corollary 3.2 Suppose f ( t , 0 , , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq122_HTML.giffor any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq84_HTML.gif. Moreover,
0 lim sup i = 1 n | u i | + max t [ 0 , 1 ] | f ( t , u 1 , u 2 , , u n ) | i = 1 n | u i | < α μ M ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ28_HTML.gif
(3.12)

Then BVP (1.1) has at least one nontrivial solution.

Proof Take ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq123_HTML.gif such that
α μ M ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 ϵ > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equai_HTML.gif
by (3.12), there exists a large enough constant R 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq124_HTML.gif such that for any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq125_HTML.gif, i = 1 n | u i | R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq126_HTML.gif, one has
| f ( t , u 1 , u 2 , , u n ) | ( α μ M ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 ϵ ) i = 1 n | u i | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equaj_HTML.gif
Let
ħ = max t [ 0 , 1 ] , i = 1 n | u i | R 0 | f ( t , u 1 , u 2 , , u n ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equak_HTML.gif
Then for any ( t , u 1 , u 2 , , u n ) [ 0 , 1 ] × R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq127_HTML.gif, we have
| f ( t , u 1 , u 2 , , u n ) | ħ + ( α μ M ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 ϵ ) i = 1 n | u i | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equal_HTML.gif
Let
i = 1 n p i ( s ) = ( α μ M ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 ϵ ) , q ( s ) = ħ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equam_HTML.gif
we prove
R < ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equan_HTML.gif
In fact,
R = M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s M ( α μ M ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 ϵ ) 0 1 ( 1 s ) α μ 1 d s < ( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equao_HTML.gif

Then it follows from Theorem 3.1 that BVP (1.1) has at least one nontrivial solution. □

4 Examples

Example 4.1 Consider the boundary value problem
{ D 5 2 x ( t ) = t sin x ( t ) 100 π + | x ( t ) | D 9 8 x ( t ) 10 2 + | D 5 4 x ( t ) | D 5 2 x ( t ) = + ( 1 + t 2 ) D 5 4 x ( t ) 100 + t 3 2 + cos t , t ( 0 , 1 ) , x ( 0 ) = D 9 8 x ( 0 ) = D 5 4 x ( 0 ) = 0 , D 11 8 x ( 1 ) = 2 2 D 11 8 ( 1 4 ) + 1 5 D 11 8 ( 1 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ29_HTML.gif
(4.1)
Proof Let α = 5 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq128_HTML.gif, μ 1 = 9 8 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq129_HTML.gif, μ 2 = 5 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq130_HTML.gif, μ = 11 8 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_IEq131_HTML.gif, and set
f ( t , u 1 , u 2 , u 3 ) = t sin u 1 100 π + | u 1 | u 2 10 2 + | u 3 | + ( 1 + t 2 ) u 3 100 + t 3 2 + cos t , p 1 ( t ) = t 100 π , p 2 ( t ) = 1 10 2 , p 3 ( t ) = 1 + t 2 100 , q ( t ) = t 3 2 + cos t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equap_HTML.gif
Then
| f ( t , u 1 , u 2 , u 3 ) | p 1 ( t ) | u 1 | + p 2 ( t ) | u 2 | + p 3 ( t ) | u 3 | + q ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equaq_HTML.gif
and
( 1 + i = 0 n 2 1 Γ ( μ n 1 μ i ) ) 1 = ( 1 Γ ( 5 4 ) + 1 Γ ( 1 8 ) + 1 ) 1 0.4472 , M = 1 + j = 1 p 2 a j | 1 j = 1 p 2 a j ξ j α μ 1 | Γ ( α μ n 1 ) 2.3934 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equar_HTML.gif
Thus we have
M 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s = 2.3934 0 1 ( 1 s ) α μ 1 i = 1 n p i ( s ) d s 0.07679 × 2.3934 0.1838 < 0.4472 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equas_HTML.gif

Thus the condition (3.2) in Theorem 3.1 is satisfied, and from Theorem 3.1, BVP (4.1) has a nontrivial solution. □

Example 4.2 Consider the boundary value problem
{ D 8 3 x ( t ) = 1 2 ( t t 2 ) x 5 ( t ) ( sin t + e t ) [ D 7 6 x ( t ) ] 9 8 D 8 3 x ( t ) = + 2 t 3 [ D 4 3 x ( t ) ] 3 + e t + t , t ( 0 , 1 ) , x ( 0 ) = D 7 6 x ( 0 ) = D 4 3 x ( 0 ) = 0 , D 3 2 x ( 1 ) = 1 π D 3 2 ( 1 3 ) 2 D 3 2 ( 2 3 ) + 1 2 D 3 2 ( 3 4 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equ30_HTML.gif
(4.2)
Proof Let
f ( t , u 1 , u 2 , u 3 ) = 1 2 ( t t 2 ) | u 1 | 5 + ( sin t + e t ) | u 2 | 9 8 + 2 t 3 | u 3 | 3 + e t + t , p 1 ( t ) = 1 2 ( t t 2 ) , p 2 ( t ) = sin t + e t , p 3 ( t ) = 2 t 3 , q ( t ) = e t + t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equat_HTML.gif
Then
| f ( t , u 1 , u 2 , u 3 ) | p 1 ( t ) | u 1 | 5 + p 2 ( t ) | u 2 | 9 8 + p 3 ( t ) | u 3 | 3 + q ( t ) , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-70/MediaObjects/13661_2012_Article_174_Equau_HTML.gif

Thus Theorem 3.4 guarantees a nontrivial solution for BVP (4.2). □

Declarations

Acknowledgement

The authors thank the referee for helpful comments and suggestions which led to an improvement of the paper. The authors were supported financially by the National Natural Science Foundation of China (11071141) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017) and the Fundamental Research Funds for the Central Universities (Grant No. HIT. NSRIF. 2010091), the National Science Foundation for Post-doctoral Scientists of China (Grant No. 2012M510956).

Authors’ Affiliations

(1)
Communication Research Center, Harbin Institute of Technology
(2)
School of Mathematical and Informational Sciences, Yantai University

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