Open Access

Existence results for higher order fractional differential equation with multi-point boundary condition

Boundary Value Problems20122012:57

DOI: 10.1186/1687-2770-2012-57

Received: 23 October 2011

Accepted: 15 May 2012

Published: 15 May 2012

Abstract

The fixed point theorems on cones are used to investigate the existence of positive solution for higher order fractional differential equation with multi-point boundary condition.

MSC: 26A33; 34B15.

Keywords

fractional differential equation fixed point positive solution cone

1 Introduction

Recently, much attention has been paid to the fractional differential equations due to its wide application in physics, engineering, economics, aerodynamics, and polymer rheology etc. For the basic theory and development of the subject, we refer some contributions on fractional calculus, fractional differential equations, see Delbosco [1], Miller [2], and Lakshmikantham et al. [37]. Especially, there have been some articles dealing with the existence of solutions or positive solutions of boundary-value problems for nonlinear fractional differential equations (see [820] and references along this line). For examples, Jiang [16] obtained the existence of positive solution for boundary value problem of fractional differential equation
D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , u ( 0 ) = 0 , u ( 1 ) = 0 , 1 < α 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equa_HTML.gif

where D 0 + α u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq1_HTML.gif denotes the standard Riemann-Liouville fractional order derivative.

Agarwal et al. [17] investigated the existence of positive solution of singular problem
D 0 + α u ( t ) = f ( t , u ( t ) , D μ u ( t ) ) , u ( 0 ) = u ( 1 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equb_HTML.gif

where 1 < α < 2, 0 ≤ μα - 1 and f satisfies the Caratheodory conditions on [0,1] × [0, ∞) × R and f(t, x, y) is singular at x = 0. The existence results of positive solutions are established by using regularization and sequential techniques.

As to the nonlocal problem, Bai [18] established the existence of positive solution for three-point boundary value problem of fractional differential equation
D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , u ( 0 ) = 0 , u ( 1 ) = β u ( η ) , η ( 0 , 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equc_HTML.gif
By using the fixed point theorems on cones, Li et al. [19] established the existence of positive solutions for problem
D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = a D 0 + β u ( ξ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equd_HTML.gif

where 1 < α ≤ 2, 0 ≤ β ≤ 1, 0 ≤ a ≤ 1, ξ (0, 1) and α-β- 2 ≤ 1 - β, 0 ≤ α - β - 1 and f : [0,1] × [0, ∞) → [0, ∞) satisfies Caratheodory type conditions.

Very recently, Moustafa and Nieto [20] considered the nontrivial solution for following higher order multi-point problem
D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , n - 1 α n , n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equ1_HTML.gif
(1.1)
u ( 0 ) = u ( 0 ) = = u ( n - 2 ) = 0 , u ( 1 ) = i = 1 m - 2 β i u ( η i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equ2_HTML.gif
(1.2)

where n ≥ 2, 0 < η i < 1, β i > 0, i = 1, 2, . . . , m - 2, i = 1 m - 2 β i η i α - 1 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq2_HTML.gif, f C([0,1] × R, R). The existence of nontrivial solution was established by using the nonlinear alternative of Leray-Schauder. But, existence of positive solution for problem (1.1), (1.2), as far as we know, has not been considered before. Considering that problem (1.1) and (1.2) are more general than problems studied before, we believe that it is interesting to investigate the existence of positive solution for this problem.

In this article, we consider the existence and multiplicity of positive solutions for problem (1.1) and (1.2). We obtain some properties of the associated Green's function. By using these properties of Green's function and fixed point theorems on cones, we establish the existence and multiplicity of positive solutions.

2 Preliminaries

For the convenience of the reader, we present here the basic definitions and theory from fractional calculus theory. These definitions and theory can be founded in the literature [1].

Definition 2.1 The fractional integral of order α > 0 of a function u(t): (0, ∞) → R is given by
I 0 + α u ( t ) = 1 Γ ( α ) 0 t ( t - s ) α - 1 u ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Eque_HTML.gif

provided the right side is point-wise defined on (0, ∞).

Definition 2.2 The fractional derivative of order α > 0 of a continuous function u(t): (0, ∞) → R is given by
D 0 + α u ( t ) = 1 Γ ( n - α ) d d t n 0 t u ( s ) ( t - s ) α - n + 1 d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equf_HTML.gif

where n = [α] + 1, provided that the right side is point-wise defined on (0, ∞).

Lemma 2.1 Let α > 0. If we assume u C(0, 1) L(0, 1), then problem D 0 + α u ( t ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq3_HTML.gif has solution
u ( 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%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equg_HTML.gif

for some C i R, i = 1, 2, . . . , N, where N is the smallest integer greater than or equal to α.

Lemma 2.2 Assume that u C(0, 1) L(0, 1) with a fractional derivative of order α > 0 that belongs to C(0, 1) L(0, 1). Then
I 0 + α D 0 + α u ( t ) = u ( 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%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equh_HTML.gif

for some C i R, i = 1, 2, . . . , N.

Lemma 2.3 [21] Let E be a Banach space and K E be a cone. Assume Ω1, Ω2 are open bounded subsets of E with 0 Ω 1 Ω ̄ 1 Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq4_HTML.gif, and let
A : K Ω 2 \ Ω ̄ 1 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equi_HTML.gif
be a completely continuous operator such that
| | A u | | | | u | | , u K Ω 1 , a n d | | A u | | | | u | | , u K Ω 2 o r | | A u | | | | u | | , u K Ω 1 , a n d | | A u | | | | u | | , u K Ω 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equj_HTML.gif

then A has a fixed point in K ( Ω 2 \ Ω ̄ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq5_HTML.gif.

Let 0 < a < b be given and let ψ be a nonnegative continuous concave functional on the cone C. Define the convex sets C r and C(ψ , a, b) by
C r = { u C | | | u | | < r } C ( ψ , a , b ) = { u C | a ψ ( u ) , | | u | | b } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equk_HTML.gif

Lemma 2.4 [22] Let T : C ̄ r C ̄ r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq6_HTML.gif be a completely continuous operator and let ψ be a nonnegative continuous concave functional on C such that ψ(u) ≤ ||u|| for all u C ̄ r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq7_HTML.gif. Suppose that there exist 0 < a < b < dc such that

(S1) {u C(ψ , b, d)| ψ(u) > b}≠ and ψ(Tu) > b for u C(ψ , b, d),

(S2) ||Tu|| < a for ||u|| ≤ a and

(S3) ψ(Tu) > b for u C(ψ, b, c) with ||Tu|| ≥ d.

Then T has at least three fixed points u1, u2, and u3 such that
| | u 1 | | < a , b < ψ ( u 2 ) , | | u 3 | | > a , ψ ( u 3 ) < b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equl_HTML.gif
Lemma 2.5 Denote η0 = 0, η m- 1 = 1 and β0 = β m- 1 = 0. Given y(t) C[0,1]. The problem
D 0 + α u ( t ) + y ( t ) = 0 , u ( 0 ) = u ( 0 ) = = u ( n - 2 ) = 0 , u ( 1 ) = i = 1 m - 2 β i u ( η i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equ3_HTML.gif
(3.1)
is equivalent to
u ( t ) = 0 1 G ( t , s ) y ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equm_HTML.gif
where
G ( t , s ) = t α - 1 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 t s - ( t - s ) α - 1 Γ ( α ) + t α - 1 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 t s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equn_HTML.gif
Furthermore, the function G(t, s) is continuous on [0,1] × [0,1] and satisfies the condition
G ( t , s ) > 0 , t , s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equo_HTML.gif
Proof. From Lemma 2.1, we get that problem (3.1) is equivalent to
u ( t ) = - 0 t ( t - s ) α - 1 Γ ( α ) y ( s ) d s + C 1 t α - 1 + C 2 t α - 2 + + C n t α - n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equp_HTML.gif
The boundary conditions u(0) = u'(0) = ... = u(n-2) = 0 induce that C2 = C3 = ... = C n = 0.. Considering the boundary condition u ( 1 ) = i = 0 m - 1 β i u ( η i ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq8_HTML.gif, we get
C 1 = 1 1 - i = 0 m - 1 β i η i α - 1 Γ ( α ) 0 1 ( 1 - s ) α - 1 y ( s ) d s - i = 0 m - 1 β i 0 η i ( η i - s ) α - 1 y ( s ) d s . u ( t ) = - 1 Γ ( α ) 0 t ( t - s ) α - 1 y ( s ) d s + t α - 1 1 - i = 0 m - 1 β i η i α - 1 Γ ( α ) × 0 1 ( 1 - s ) α - 1 y ( s ) d s - i = 0 m - 1 β i 0 η i ( η i - s ) α - 1 y ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equq_HTML.gif
Then for η i- 1 < t < η i , i = 1, 2,. . . , m - 1,
u ( t ) = - 1 Γ ( α ) 0 t ( t - s ) α - 1 y ( s ) d s + t α - 1 1 - i = 1 m - 2 β i η i α - 1 Γ ( α ) 0 1 ( 1 - s ) α - 1 y ( s ) d s - i = 1 m - 2 β i 0 η i ( η i - s ) α - 1 y ( s ) d s = k = 1 i - 1 η k - 1 η k - 1 Γ ( α ) ( t - s ) α - 1 + t α - 1 1 - i = 1 m - 2 β i η i α - 1 Γ ( α ) ( 1 - s ) α - 1 - j = k m - 1 β j ( η j - s ) α - 1 y ( s ) d s + η i - 1 t - 1 Γ ( α ) ( t - s ) α - 1 + t α - 1 1 - i = 1 m - 2 β i η i α - 1 Γ ( α ) ( 1 - s ) α - 1 - j = i i - 1 β j ( η j - s ) α - 1 y ( s ) d s + t η i t α - 1 1 - i = 1 m - 2 β i η i α - 1 Γ ( α ) ( 1 - s ) α - 1 - j = i m - 1 β j ( η j - s ) α - 1 y ( s ) d s + k = i m - 1 η k - 1 η k t α - 1 1 - i = 1 m - 2 β i η i α - 1 Γ ( α ) ( 1 - s ) α - 1 - j = k m - 1 β j ( η j - s ) α - 1 y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equr_HTML.gif
Furthermore, for η i -1sη i , i = 1, 2, . . . , m-1 and ts
Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 G ( t , s ) t α - 1 k = i m - 2 1 - s α - 1 - η k - s α - 1 > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equs_HTML.gif
For η i- 1s ≤ η i , i = 1, 2, . . . , m-1 and ts
Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 G ( t , s ) t α - 1 ( 1 - s ) α - 1 - 1 - s t α - 1 + t α - 1 k = 0 i - 1 β k η k α - 1 1 - s t α - 1 - 1 - s η k α - 1 > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equt_HTML.gif

Lemma 2.6 The function G(t, s) satisfies the following conditions:
  1. (1)

    G(t, s) ≤ G(s, s), t, s [0, 1],

     
  2. (2)

    There exists function γ(s) such that min η m - 2 s 1 G ( t , s ) γ ( s ) G ( s , s ) , 0 < s < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq9_HTML.gif.

     
Proof (1) For η i -1 < s < η i , i = 1, 2, . . . , m-1, Denote
g 1 ( t , s ) = t α - 1 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 , g 2 ( t , s ) = - ( t - s ) α - 1 Γ ( α ) + t α - 1 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equu_HTML.gif
The facts that
g 1 ( t , s ) t = ( α - 1 ) t α - 2 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 > 0 , g 2 ( t , s ) t = - ( α - 1 ) ( t - s ) α - 2 Γ ( α ) + ( α - 1 ) t α - 2 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equv_HTML.gif
imply that g1(t, s) is decreasing with respect to t for [η i- 1, s] and g2 (t, s) is increasing with respect to t for [s, η i ], i = 1, 2, . . . , m - 1. Thus one can easily check that
G ( t , s ) G ( s , s ) , t , s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equw_HTML.gif
  1. (2)
    For η m -2 < t < 1, denote
    γ i ( t , s ) = - ( t - s ) α - 1 Γ ( α ) + t α - 1 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 , γ ( s ) = min γ i ( η m - 2 , s ) , 1 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 , i = 1 , 2 , , m - 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equx_HTML.gif
    γ i ( t , s ) = - ( t - s ) α - 1 Γ ( α ) + t α - 1 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 , γ ( s ) = min γ i ( η m - 2 , s ) , 1 Γ ( α ) 1 - k = 0 m - 1 β k η k α - 1 ( 1 - s ) α - 1 - k = i m - 1 β k ( η k - s ) α - 1 , i = 1 , 2 , , m - 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equy_HTML.gif
     
Thus we have
γ ( s ) > 0 , min η m - 2 < t < 1 G ( t , s ) γ ( s ) G ( s , s ) = γ ( s ) max 0 t 1 G ( t , s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equz_HTML.gif

3 Main results

Let X = C[0,1] be a Banach space endowed with the norm
| | u | | = max 0 t 1 | u ( t ) | , u X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equaa_HTML.gif

Define the cone P E by P = {u X | u(t) ≥ 0}.

Theorem 3.1 Define the operator T : PX,
T u ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equab_HTML.gif

then T : PP is completely continuous.

Proof From the nonnegative and continuous properties of function f and G(t, s), one can obtain easily that the operator T : PP and T is continuous. Let Ω be a bounded subset of cone P. That is, there exists a positive constant M1 > 0 such that ||u|| ≤ M1 for all u Ω. Thus for each u Ω, t1, t2 [0,1], one has
| T u ( t 1 ) - T u ( t 2 ) | = 0 1 ( G ( t 1 , s ) - G ( t 2 , s ) ) f ( s , u ( s ) ) d s 0 1 G ( t 1 , s ) - G ( t 2 , s ) f ( s , u ( s ) ) d s M 2 0 1 G ( t 1 , s ) - G ( t 2 , s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equac_HTML.gif
Then the continuity of function G(t, s) implies that T is equicontinuity on the bounded subset of P . On the other hand, for u Ω, there exist constant M2 > 0 such that
f ( t , u ) M 2 , t [ 0 , 1 ] , u Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equad_HTML.gif
Then
T u ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s M 2 0 1 G ( s , s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equae_HTML.gif

which implies that T is uniformly bounded on the bounded subset of P . Then an application of Ascoli-Arezela ensures that T : PP is completely continuous.

Theorem 3.2 Assume that there exist two positive constant r 2 > N M r 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq10_HTML.gif such that

(A1) f(t, u) ≤ Mr2, (t, u) [0, 1] × [0, r2]

(A2) f(t, u) ≥ Nr1, (t, u) [0, 1] × [0, r1]

where
M = 0 1 G ( s , s ) d s - 1 , N = η m - 2 1 γ ( s ) G ( s , s ) d s - 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equaf_HTML.gif

Then problem (1.1) and (1.2) has at least one positive solution u such that r1 ≤ ||u|| ≤ r2.

Proof Let Ω2 = {u P | ||u|| ≤ r2}. For u ∂Ω2, considering assumption (A 1), we have
0 u ( t ) r 1 , a n d f ( t , u ) M r 2 , t [ 0 , 1 ] , T u ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s M r 2 0 1 G ( s , s ) d s r 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equag_HTML.gif

Thus ||Tu|| ≤ ||u||, u ∂Ω2.

Let Ω1 = {u P | ||u|| ≤ r1}. For u ∂Ω1, considering assumption (A 2), we have
0 u ( t ) r 1 , a n d f ( t , u ) N r 2 , t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equah_HTML.gif
Thus for t [η m- 2, 1], we get
T u ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s η m - 2 1 γ ( s ) G ( s , s ) f ( s , u ( s ) ) d s r 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equai_HTML.gif

which gives that ||Tu|| ≥ ||u||, u ∂Ω1. An application of Lemma (2.5) ensures the existence of positive solution u(t) of problem (1.1) and (1.2).

Theorem 3.3 Suppose that there exist constants 0 < a < b < c such that

(A 3) f(t, u) < Ma, for (t, u) [0,1] × [0, a],

(A 4) f(t, u) ≥ Nb, for (t, u) [η m- 2, 1] × [b, c],

(A 5) f(t, u) ≤ Mc, for (t, u) [0,1] × [0, c],

then problem (1.1) and (1.2) has at least three positive solution u1, u2, u3 with
max 0 t 1 | u 1 | a , b < min η m - 2 t 1 | u 2 | < max 0 t 1 | u 2 | c , a < max 0 t 1 | u 3 | c , min η m - 2 t 1 | u 3 | < b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equaj_HTML.gif
Proof Let the nonnegative continuous concave functional θ on the cone P defined by
θ ( u ) = min η m - 2 t 1 | u ( t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equak_HTML.gif
If u P ̄ c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq11_HTML.gif, then ||u|| ≤ c. Then by condition (A5), we have
f ( t , u ) M c , for ( t , u ) [ 0 , 1 ] × [ 0 , c ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equal_HTML.gif
Thus
| T ( u ) ( t ) | = 0 1 G ( t , s ) f ( s , u ( s ) ) d s M c 0 1 G ( s , s ) d s = c . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equam_HTML.gif
which yields that T : P ̄ c P ̄ c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq12_HTML.gif. In the same way, we get that
T u < a , f o r u a . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equan_HTML.gif
We chose the function u ( t ) = b + c 2 t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq13_HTML.gif. We claim that b + c 2 { u P ( θ , b , c ) | θ ( u ) > b } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_IEq14_HTML.gif, which ensures that {u P (θ, b, c)|θ(u) > b} ≠ . And for u P (θ, b, c), we have
f ( t , u ( t ) ) N b , t [ η m - 2 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equao_HTML.gif
Then
θ ( T u ) = min η m - 2 t 1 0 1 G ( t , s ) f ( s , u ( s ) ) d s > N b η m - 2 1 γ ( s ) G ( s , s ) d s = b , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equap_HTML.gif

which yields that θ(Tu) > b, for u P (θ, b, c).

An application of Lemma (2.6) ensures that problem (1.1) and (1.2) has at least three positive solutions with
max 0 t 1 | u 1 | a , b < min η m - 2 t 1 | u 2 | < max 0 t 1 | u 2 | c , a < max 0 t 1 | u 3 | c , min η m - 2 t 1 | u 3 | < b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-57/MediaObjects/13661_2011_Article_151_Equaq_HTML.gif

Declarations

Acknowledgements

This study was supported by the Anhui Provincial Natural Science Foundation (10040606Q50) and the Natural Science Foundation of Anhui Department of Education (KJ2010A285).

Authors’ Affiliations

(1)
Department of Mathematics, Anhui Normal University

References

  1. Delbosco D: Fractional calculus and function spaces. J Fract Calc [J] 1994, 6: 45-53.MATHMathSciNet
  2. Miller KS, Ross B: An Introduction to the Fractional calculus and Fractional Differential Equations. Wiley, New York (1993) Lakshmikantham (Author), S.Leela (Author), J.Vasundhara Devi (Author);MATH
  3. Leela S, Devi J: Theory of Fractional Dynamic Systems. Cambridge Academic, Cambridge; 2009.MATH
  4. Lakshmikantham V, Devi J: Theory of fractional differential equations in a Banach space. Eur J Pure Appl Math 2008, 1: 38-45.MATHMathSciNet
  5. Lakshmikantham V, Leela S: Nagumo-type uniqueness result for fractional differential equations. Nonlinear Anal TMA 2009, 71: 2886-2889. 10.1016/j.na.2009.01.169MATHMathSciNetView Article
  6. Lakshmikantham V, Leela S: A Krasnoselskii-Krein-type uniqueness result for fractional differential equations. Nonlinear Anal TMA 2009, 71: 3421-3424. 10.1016/j.na.2009.02.008MATHMathSciNetView Article
  7. Lakshmikantham V: Theory of fractional differential equations. Nonlinear Anal TMA 2008, 69: 3337-3343. 10.1016/j.na.2007.09.025MATHMathSciNetView Article
  8. Zhang S: The existence of a positive solution for a nonlinear fractional differential equation. J Math Anal Appl 2000, 252: 804-812. 10.1006/jmaa.2000.7123MATHMathSciNetView Article
  9. Zhang S: Existence of positive solution for some class of nonlinear fractional differential equation. J Math Anal Appl 2003, 278: 136-148. 10.1016/S0022-247X(02)00583-8MATHMathSciNetView Article
  10. Babakhani A, Gejji VD: Existence and positive solutions of nonlinear Fractional differential equations. J Math Anal Appl 2003, 278: 434-442. 10.1016/S0022-247X(02)00716-3MATHMathSciNetView Article
  11. Xu XJ, Jiang DQ, Yuan CJ: Multiple positive solutions for the boundary value problem of a nonlinear fractional differential equations. Nonlinear Anal TMA 2009, 71: 4676-4688. 10.1016/j.na.2009.03.030MATHMathSciNetView Article
  12. Liang S, Zhang J: Positive solutions for boundary value problems of nonlinear fractional differential equation. Nonlinear Anal 2009, 71: 5545-5550. 10.1016/j.na.2009.04.045MATHMathSciNetView Article
  13. Ahmad B, Nieto J: Existence results for a coupled system of nonlinear fractional differential equations with three-point boundary conditions. Comput Math Appl 2009, 58: 1838-1843. 10.1016/j.camwa.2009.07.091MATHMathSciNetView Article
  14. Bai Z, Lü H: Positive solutions for boundary value problem of nonlinear fractional differential equation. J Math Anal Appl [J] 2005, 311: 495-505.MATHView Article
  15. Kosmatov N: A singular boundary value problem for nonlinear differential equations of fractional order. J Appl Math Comput 2010, 29: 125-135.MathSciNetView Article
  16. Jiang D, Yuan C: The positive properties of the Green function for Direchlet-type boundary value problems of nonlinear fractional differential equations and its application. Nonlinear Anal 2010, 15: 710-719.MathSciNetView Article
  17. Agarwal RP, O'Regan D, Stanêk Svatoslav: Positive solutions for Dirichlet problems of sigular nonlinear fractional differential equations. J Math Anal Appl 2010, 371: 57-68. 10.1016/j.jmaa.2010.04.034MATHMathSciNetView Article
  18. Bai Z: On positive solutions of a nonlocal fractional boundary value problem. Nonlinear Anal TMA [J] 2010, 72: 916-924. 10.1016/j.na.2009.07.033MATHView Article
  19. Li C, Luo X, 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
  20. Shahed M, Nieto J: Nontrivial solution for a nonlinear multi-point boundary value problem of fractional order. Comput Math Appl 2010, 59: 3438-3443. 10.1016/j.camwa.2010.03.031MATHMathSciNetView Article
  21. Krasnosel'skii M: Topological methods in the theory of nonlinear integral equations. Pergamon, Elmsford; 1964.
  22. Leggett R, Williams L: Multiple positive fixed points of nonlinear operators on ordered Banach spaces. Indiana Univ Math J 1979, 28: 673-688. 10.1512/iumj.1979.28.28046MATHMathSciNetView Article

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