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Existence of positive solutions for eigenvalue problem of nonlinear fractional differential equations

Advances in Difference Equations20122012:66

https://doi.org/10.1186/1687-1847-2012-66

  • Received: 14 December 2011
  • Accepted: 23 May 2012
  • Published:

Abstract

In this article, by using the fixed point theorem, existence of positive solutions for eigenvalue problem of nonlinear fractional differential equations

D 0 + α u ( t ) + λ a ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0

is considered, where 1 < α < 2 is a real number, D 0 + α is the standard Riemann-Liouville derivate, λ is a positive parameter and a(t) C([0, 1], [0, ∞)), f(t, u) C([0, 1] × [0, ∞), [0, ∞)).

MSC(2010): 34B18.

Keywords

  • fractional differential equation
  • positive solution
  • eigenvalue problem
  • fixed point

1 Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various sciences such as physics, mechanics, chemistry, engineering, etc. For details, see [16] and references therein.

Recently, many results were obtained dealing with the existence and multiplicity of solutions of nonlinear fractional differential equations by the use of techniques of nonlinear analysis, see [722] and the reference therein. Bai and Lu [7] studied the existence of positive solutions of nonlinear fractional differential equation
D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 ,
(1.1)

where 1 < α≤ 2 is a real number, D 0 + α is the standard Riemann-Liouville differentiation, and f : [0, 1] × [0, ∞) → [0, ∞) is continuous. They derived the corresponding Green function and obtained some properties as follows.

Proposition 1 The Green function G(t, s) satisfies the following conditions:

(R1) G(t, s) C([0,1] × [0,1]), and G(t, s) > 0 for t, s (0, 1);

(R2) There exists a positive function γ C(0, 1) such that
min 1 4 t 3 4 G ( t , s ) γ ( s ) max 0 t 1 G ( t , s ) γ ( s ) G ( s , s ) , s ( 0 , 1 ) ,
(1.2)
where
G ( t , s ) = [ t ( 1 - s ) ] α - 1 - ( t - s ) α - 1 Γ ( α ) , 0 s t 1 , [ t ( 1 - s ) ] α - 1 Γ ( α ) , 0 t s 1 .
(1.3)

It is well known that the cone plays a very important ... role in applying the Green function in research area. In [7], the authors cannot acquire a positive constant taken instead of the role of positive function γ(s) with 1 < α < 2 in (1.2). In [9], Jiang and Yuan obtained some new properties of the Green function and established a new cone. The results can be stated as follows.

Proposition 2 The Green function G(t, s) defined by (1.3) has the following properties: G(t, s) = G(1 - s, 1 - t) and
α - 1 Γ ( α ) t α - 1 ( 1 - t ) ( 1 - s ) α - 1 s G ( t , s ) 1 Γ ( α ) t α - 1 ( 1 - t ) ( 1 - s ) α - 2 t , s ( 0 , 1 ) .
(1.4)
Proposition 3 The function G*(t, s):= t2-αG(t, s) has the following properties:
q ( t ) Φ ( s ) G * ( t , s ) Φ ( s ) t , s [ 0 , 1 ] .
(1.5)

where q(t) = (α - 1)t(1 - t), Φ ( s ) = 1 Γ ( α ) s ( 1 - s ) α - 1 .

The purpose of this article is to establish the existence of positive solutions for eigenvalue problem of nonlinear fractional differential equations
D 0 + α u ( t ) + λ a ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0
(1.6)

where 1 < α < 2 is a real number, D 0 + α is the standard Riemann-Liouville derivate, λ is a positive parameter and a(t) C([0,1], [0, ∞)), f(t, u) C([0,1] × [0, ∞), [0, ∞)).

2 The preliminary lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature.

Definition 2.1 The fractional integral of order α > 0 of a function y : (0, ∞) → R is given by
I 0 + α y ( t ) = 1 Γ ( α ) 0 t ( t - s ) α - 1 y ( s ) d s

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

Definition 2.2 The fractional derivative of order α > 0 of a function y : (0, ∞) → R is given by
D 0 + α y ( t ) = 1 Γ ( n - α ) d d t n 0 t y ( s ) ( t - s ) α - n + 1 d s

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

Lemma 2.1 Let α > 0. If we assume u C(0, 1) ∩ L(0, 1), then the fractional differential equation
D 0 + α u ( t ) = 0

has u(t) = C1t α- 1 + C2t α- 2 + ... + C N t α-N , C i R, i = 1, 2, ..., N, where N is the smallest integer greater than or equal to α, as unique solutions.

Lemma 2.2[7] Assume that u C(0, 1) L(0, 1) with a fractional derivative of order α > 0 that belongs to u 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

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

Lemma 2.3[7] Given y C[0,1] and 1 < α ≤ 2, the unique solution of
D 0 + α u ( t ) + y ( t ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 ,
(2.1)
is
u ( t ) = 0 1 G ( t , s ) y ( s ) d s ,
where
G ( t , s ) = [ t ( 1 - s ) ] α - 1 - ( t - s ) α - 1 Γ ( α ) , 0 s t 1 , [ t ( 1 - s ) ] α - 1 Γ ( α ) , 0 t s 1 .
(2.2)
Lemma 2.4[10] Let K be a cone in Banach space E. Suppose that T : K ¯ r K is a completely continuous operator.
  1. (i)

    If there exists u 0 K \ {θ} such that u - Tuμu 0 for any u K r and μ ≥ 0, then i(T, K r, K) = 0.

     
  2. (ii)

    If Tuμu for any u K r and μ ≥ 1, then i(T, K r, K) = 1.

     
Lemma 2.5[8] Let P be a cone in Banach space X. Suppose that T : PP is a completely continuous operator. If there exists a bounded open set Ω(P) such that each solution of
u = σ T u , u P , σ [ 0 , 1 ]

satisfies u Ω(P), then the fixed point index i(T, Ω(P), P) = 1.

3 The main results

Let E = C[0,1] be endowed with the ordering uv if u(t) ≤ v(t) for all t [0,1], and the maximum norm, u = max 0 t 1 u ( t ) . Define the cone P E by P = {u E|u(t) ≥ 0}, and
K = { u P | u ( t ) q ( t ) | | u | | } .

where q(t) is defined by (1.5).

It is easy to see that P and K are cones in E. For any 0 < r < R < +∞, let K r = {u K| ||u|| < r}, ∂K r = {u K| ||u|| = r}, K ¯ r = { u K | | | u | | r } and K ¯ R \ K r = { u K | r | | u | | R } .

For convenience, we introduce the following notations
g 0 = lim u 0 + g ( u ) u , g = lim u + g ( u ) u , 1 4 3 4 G * ( τ , s ) a ( s ) q 1 ( s ) d s = max 1 4 t 3 4 1 4 3 4 G * ( t , s ) a ( s ) q 1 ( s ) d s .

We assume the following conditions hold throughout the article:

(H1) a(t) C([0, 1], [0, ∞)), a(t) 0;

(H2) f(t, u) C([0, 1] × [0, ∞), [0, ∞)), and there exist g C([0, +∞), [0, +∞)), q1,q2 C((0, 1), (0, +∞)) such that
q 1 ( t ) g ( u ) f ( t , t α - 2 u ) q 2 ( t ) g ( u ) , t ( 0 , 1 ) , u [ 0 , + ) ,

where 0 1 q i ( s ) d s < + , i = 1 , 2 .

By similar arguments to Lemma 4.1 and Theorem 1.3 of [9], we obtain the following result.

Lemma 3.1 Assume that (H1)(H2) hold. Let
T u ( t ) : = λ 0 1 G * ( t , s ) a ( s ) f ( s , s α - 2 u ( s ) ) d s ,
(3.1)

then T : KK is completely continuous. Moreover, if u is a fixed point of T in K ¯ R \ K r , then y = t α- 2u is a positive solution of BVP (1.6).

By similar arguments to Theorems 2 and 3 of [18], we obtain the following result.

Theorem 3.1 Assume that (H1)(H2) hold. Then, for each λ satisfying
16 [ ( α - 1 ) 1 4 3 4 G * ( τ , s ) a ( s ) q 1 ( s ) d s ] g < λ < 1 0 1 Φ ( s ) a ( s ) q 2 ( s ) d s g 0 ,
(3.2)

there exists at least one positive solution of BVP (1.6) in K.

Theorem 3.2 Assume that (H1)(H2) hold. Then, for each λ satisfying
16 [ ( α - 1 ) 1 4 3 4 G * ( τ , s ) a ( s ) q 1 ( s ) d s ] g 0 < λ < 1 0 1 Φ ( s ) a ( s ) q 2 ( s ) d s g ,
(3.3)

there exists at least one positive solution of BVP (1.6) in K.

Set
L 1 u ( t ) : = 0 1 G * ( t , s ) a ( s ) q 1 ( s ) u ( s ) d s ,
(3.4)
L 2 u ( t ) : = 0 1 G * ( t , s ) a ( s ) q 2 ( s ) u ( s ) d s .
(3.5)

It is clear that L1, L2 is a completely continuous linear operator and L1(P) K, L2(P) K. By virtue of the Krein-Rutman theorem and Proposition 3, we have the following lemma.

Lemma 3.2 Assume that (H1)(H2) hold. Then the spectral radius r(L1) > 0, r(L2) > 0 and L1, L2, respectively, has a positive eigenfunction φ1, φ2 corresponding to its first eigenvalue λ1 = (r(L1))-1, λ2 = (r(L1))-1, that is φ1 = λ1L1φ1, φ2 = λ2L2φ2.

Obviously, L1φ1 K, L2φ2 K, so φ1 K, φ2 K.

In the following, we will obtain some existence results under some conditions concerning the first eigenvalue with respect to linear operator L1, L2.

Assume that

(H3) λ 1 g 0 < λ < λ 2 g ;

(H3) λ 1 g < λ < λ 2 g 0 .

Theorem 3.3 Assume (H1)(H2)(H3) hold. Then the BVP (1.6) has at least one positive solution.

Proof. First, by Lemma 3.1, we know that T : KK is completely continuous.

By (H3), we have g 0 > λ 1 λ , there exists r1> 0 such that
g ( u ) λ 1 λ u , 0 < u r 1 .
Then for u K r 1 , we have
( T u ) ( t ) = λ 0 1 G * ( t , s ) a ( s ) f ( s , s α - 2 u ( s ) ) d s λ 0 1 G * ( t , s ) a ( s ) q 1 ( s ) g ( u ( s ) ) d s λ 1 0 1 G * ( t , s ) a ( s ) q 1 ( s ) u ( s ) d s = λ 1 ( L 1 u ) ( t ) .
By Lemma 3.2, φ1 = λ1L1φ1. We may suppose that T has no fixed points on K r 1 (otherwise, the proof is finished). Now we show that
u - T u μ φ 1 , u K r 1 , μ 0 .
(3.6)
Assume by contradicts that there exist u K r 1 and μ1 ≥ 0 such that u1 - Tu1 = μ1φ1, then μ1> 0 and u1 = Tu1 + μ1φ1μ1φ1. Let μ ̄ = sup { μ | u 1 μ φ 1 } , then μ ̄ μ 1 , u 1 μ ̄ φ 1 and T u 1 λ 1 μ ̄ L 1 φ 1 = μ ̄ φ 1 . Thus,
u 1 = T u 1 + μ 1 φ 1 μ ̄ φ 1 + μ 1 φ 1 = ( μ ̄ + μ 1 ) φ 1 ,
which contradicts the definition of μ ̄ So (3.6) is true and by lemma 2.4 we have
i ( T , K r 1 , K ) = 0 .
(3.7)
On the other hand, by (H3), we have g < λ 2 λ , there exists 0 < σ < 1 and r2> r1> 0 such that
g ( u ) σ λ 2 λ u , u r 2 .
Let = σλ2L2φ, φ C[0,1]. Then L : C[0,1] → C[0,1] is a bounded linear operator and L(K) K. Denote
M = ( max 0 t , s 1 G * ( t , s ) ) sup φ K r 2 0 1 λ a ( s ) q 2 ( s ) g ( φ ( s ) ) d s .
It is clear that M < +∞ Let
W = { φ K | φ = μ T φ , 0 μ 1 } .

In the following, we prove that W is bounded.

For any φ W , set φ ̄ ( t ) = min { φ ( t ) , r 2 } and denote E(φ) = {t [0,1]|φ(t) > r2}, then
φ ( t ) = μ ( T φ ) ( t ) ( T φ ) ( t ) = λ 0 1 G * ( t , s ) a ( s ) f ( s , s α - 2 φ ( s ) ) d s λ 0 1 G * ( t , s ) a ( s ) q 2 ( s ) g ( φ ( s ) ) d s = λ E ( φ ) G * ( t , s ) a ( s ) q 2 ( s ) g ( φ ( s ) ) d s + λ [ 0 , 1 ] \ E ( φ ) G * ( t , s ) a ( s ) q 2 ( s ) g ( φ ̄ ( s ) ) d s σ λ 2 0 1 G * ( t , s ) a ( s ) q 2 ( s ) φ ( s ) d s + λ 0 1 G * ( t , s ) a ( s ) q 2 ( s ) g ( φ ̄ ( s ) ) d s ( L φ ) ( t ) + M , t [ 0 , 1 ] .
Thus ((I - L)φ)(t) ≤ M, t [0,1]: Since λ2 is the first eigenvalue of L2 and 0 < σ < 1, the first eigenvalue of L, (r(L))-1> 1. Therefore, the inverse operator (I - L)-1 exists and
( I - L ) - 1 = I + L + L 2 + + L n .

It follows from L(K) K that (I - L)-1(K) K. So we have φ(t) ≤ (I - L)-1M, t [0,1] and W is bounded.

Choose r3> max{r2, || (I - L)-1M||}. Then by lemma 2.5, we have
i ( T , K r 3 , K ) = 1 .
(3.8)
By (3.7) and (3.8), one has
i ( T , K r 3 \ K ̄ r 1 , K ) = i ( T , K r 3 , K ) - i ( T , K r 1 , K ) = 1 .

Then T has at least one fixed point on K r 3 \ K ̄ r 1 By Lemma 3.1, this means that problem (1.6) has at least one positive solution. The proof is complete.

Theorem 3.4 Assume (H1)(H2)(H4) hold. Then the BVP (1.6) has at least one positive solution.

Theorem 3.5 Assume (H1)(H2) hold and λ 1 λ < g 0 , λ 1 λ < g . Moreover, the following condition holds:

(H5) there exists a p > 0 such that 0 ≤ up implies g(u) < M1p, where M 1 = [ λ 0 1 Φ ( s ) a ( s ) q 2 ( s ) d s ] - 1 .

Then the BVP (1.6) has at least two positive solution.

Theorem 3.6 Assume (H1)(H2) hold and 0 g 0 < λ 2 λ , 0 g < λ 2 λ . Moreover, the following condition holds:

(H6) there exists a p > 0 such that p ( α - 1 ) 16 u p implies g(u) > M2p, where M 2 = [ λ 1 4 3 4 G ( 1 2 , s ) a ( s ) q 1 ( s ) d s ] 1 . Then the BVP (1.6) has at least two positive solution.

Remark 3.1 Theorems 3.3, 3.4 extend and improve Theorem 1.5 in [9], Theorems 3.5 and 3.6 extend and improve Theorems 1.3 and 1.4 in [9], respectively.

4 Example

Consider the boundary value problem
D 0 + α u ( t ) + λ μ ( u a ( t ) + u b ( t ) ) = 0 , 0 < t < 1 , 0 < a < 1 < b < 1 2 - α , 1 < α < 2 , u ( 0 ) = u ( 1 ) = 0 .
(3.9)

Then (3.9) have at least two positive solutions u1 and u2, for each 0 < μ < μ*, where μ* is some positive constant.

Proof. We will apply Theorem 3.5. To this end we take a(t) ≡ 1, f(t, u) = μ(u a (t) + u b (t)), then f(t, t α- 2y) = μ(t a (α-2)y a (t) + t b (α-2) yb (t)).

Let q1(t) = t a (α-2), q2(t) = t b (α-2) and g(y) = μ(y a + y b ), then q1(t)g(y) ≤ f(t, t α- 2y) ≤ q2(t)g(y) and q1, q2 L1(0, 1), g C([0, +∞), [0, +∞)). Thus (H1)(H2)is satisfied, and it is to see that λ 1 λ < g 0 , λ 1 λ < g .

Since
1 Γ ( α ) 0 1 s ( 1 - s ) α - 1 q 2 ( s ) d s = 1 Γ ( α ) 0 1 s ( 1 - s ) α - 1 s b ( α - 2 ) d s = 1 Γ ( α ) 0 1 s ( 2 + b ( α - 2 ) ) - 1 ( 1 - s ) α - 1 d s = 1 Γ ( α ) Γ ( 2 + b ( α - 2 ) ) Γ ( α ) Γ ( 2 + b ( α - 2 ) + α ) = Γ ( 2 + b ( α - 2 ) ) Γ ( 2 + b ( α - 2 ) + α ) ,

then M 1 = ( λ Γ ( 2 + b ( α - 2 ) ) Γ ( 2 + b ( α - 2 ) + α ) ) - 1 .

Let
μ * = F ( p ) = sup y > 0 { F ( y ) : F ( y ) = M 1 y y a + y b } = F ( ( 1 - a b - 1 ) 1 b - a ) ,

where p = ( 1 - a b - 1 ) 1 b - a .

Then, for μ < μ*, we have
g ( y ) = μ ( y a + y b ) μ ( p a + p b ) < M 1 p , 0 y p ,

thus (H5) holds. Therefore, (3.9) have at least two positive solutions u1 and u2, for each 0 < μ < μ*.

Declarations

Acknowledgements

The authors are grateful to the anonymous referee for their valuable suggestions. X. Han was supported by the NSFC (No. 11101335), the Gansu Provincial Department of Education Fund (No. 1101-02), and the Project of Science and Technology Bureau of Lanzhou (2011-2-72).

Authors’ Affiliations

(1)
Department of Mathematics, Northwest Normal University, Lanzhou, 730070, P. R. China

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© Han and Gao; 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.

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