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Nontrivial solutions for a fractional boundary value problem

Abstract

In this work, we discuss the existence of nontrivial solutions for the fractional boundary value problem

{ D 0 + α u = f ( t , u ) , t [ 0 , 1 ] , u ( 0 ) = u ( 0 ) = u ( 1 ) = 0 .

Here α(2,3] is a real number, D 0 + α is the standard Riemann-Liouville fractional derivative of order α. By virtue of some inequalities associated with Green’s function, without the assumption of the nonnegativity of f, we utilize topological degree theory to establish our main results.

MSC:26A33, 34B15, 34B18, 34B27.

1 Introduction

In this paper, we investigate nontrivial solutions for the boundary value problem of fractional order involving Riemann-Liouville’s derivative

{ D 0 + α u = f ( t , u ) , t [ 0 , 1 ] , u ( 0 ) = u ( 0 ) = u ( 1 ) = 0 ,
(1.1)

where α(2,3], f:[0,1]×RR (R:=(,+)) is continuous.

In view of a fractional differential equation’s modeling capabilities in engineering, science, economy and other fields, the last few decades have resulted in a rapid development of the theory of fractional differential equation; see the books [13]. This may explain the reason why the last few decades have witnessed an overgrowing interest in the research of such problems, with many papers in this direction published. We refer the interested reader to [421] and the references therein.

In [4], Bai and Lü studied the existence and multiplicity of positive solutions for the nonlinear fractional differential equation

{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 ,
(1.2)

where 1<α2 is a real number and f:[0,1]× R + R + is continuous. They obtained the existence of positive solutions by means of Guo-Krasnosel’skii fixed point theorem and Leggett-Williams fixed point theorem.

In [5], Jiang et al. discussed some positive properties of the Green function for boundary value problem (1.2), and as an application, they utilized the Guo-Krasnosel’skii fixed point theorem to obtain the existence of positive solutions for (1.2).

In [6], El-Shahed and Nieto investigated the existence of nontrivial solutions for a multi-point boundary value problem for fractional differential equations. Under certain growth conditions on the nonlinearity, several sufficient conditions for the existence of a nontrivial solution were obtained by using the Leray-Schauder nonlinear alternative.

In [7], Wang and Liu adopted the same methods in [6] to discuss the existence of solutions for nonlinear fractional differential equations with fractional anti-periodic boundary conditions

{ D α c x ( t ) = f ( t , x ( t ) , c D q x ( t ) ) , t [ 0 , T ] , x ( 0 ) = x ( T ) , c D p x ( 0 ) = c D p x ( T ) .
(1.3)

In [810], Ahmad et al. utilized fixed point theory to consider some fractional differential equations with fractional boundary conditions and obtained some new existence results. In particular, He and his coauthors [10] investigated the existence of solutions for the fractional nonlinear integro-differential equation of mixed type on a semi-infinite interval in a Banach space

{ D α u ( t ) + f ( t , u ( t ) , T u ( t ) , S u ( t ) ) = θ , n 1 < α n , n 2 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = θ , D α 1 u ( ) = u .
(1.4)

Meanwhile, we also note that they developed an explicit iterative sequence for approximating the solution together with an error estimate for the approximation.

In [22, 23], Sun and Zhang discussed a class of singular superlinear and sublinear Sturm-Liouville problems, respectively. In the two papers, the Sturm-Liouville problems are considered under some conditions concerning the first eigenvalues corresponding to the relevant linear operators, and the nonnegativity is not necessary to be nonnegative. The existence results of nontrivial solutions and positive solutions are given by means of topological degree theory.

Motivated by the works mentioned above, in our paper, we adopt the methods of [22, 23] to investigate the fractional problem (1.1). As we know, the eigenvalue and eigenfunction of an integer-order differential equation have been a very perfect theory; however, this work on fractional order differential equation has not appeared in the literature. In order to overcome the difficulty arising from it, we establish some inequalities associated with Green’s function; see Lemma 2.3 in Section 2. With the aid of these inequalities, the nonlinear term f can grow superlinearly and sublinearly, and we obtain that problem (1.1) has at least one nontrivial solution by topological degree theory. This means that both our methodology and results in this paper are different from those in [47, 1116].

2 Preliminaries

The Riemann-Liouville fractional derivative of order α>0 of a continuous function f:(0,+)(,+) is given by

D 0 + α f(t)= 1 Γ ( n α ) ( d d t ) n 0 t f ( s ) ( t s ) α n + 1 ds,

where n=[α]+1, [α] denotes the integer part of number α, provided that the right-hand side is pointwise defined on (0,+). For more details on fractional calculus, we refer the reader to the recent books; see [13]. Next, we present Green’s function of fractional differential equation boundary value problem (1.1).

Lemma 2.1 (See [[16], Lemma 2.7])

Let vC[0,1] and α(2,3]. Then D 0 + α u:=v, together with the boundary conditions u(0)= u (0)= u (1)=0, is equivalent to u(t)= 0 1 G(t,s)v(s)ds, where

G(t,s):= 1 Γ ( α ) { t α 1 ( 1 s ) α 2 ( t s ) α 1 , 0 s t 1 , t α 1 ( 1 s ) α 2 , 0 t s 1 .
(2.1)

Lemma 2.2 (See [[16], Lemma 2.8])

The functions G(t,s)C([0,1]×[0,1],[0,+)). Moreover, G(t,s) satisfies the following inequalities:

t α 1 s ( 1 s ) α 2 Γ(α)G(t,s)s ( 1 s ) α 2 ,t,s[0,1].
(2.2)

Lemma 2.3 Let φ(t)=t ( 1 t ) α 2 , t[0,1], and α Γ ( α 1 ) Γ ( 2 α ) (:= K 1 ) 1 α ( α 1 ) Γ ( α ) (:= K 2 ). Then

K 1 φ(s) 0 1 G(t,s)φ(t)dt K 2 φ(s),s[0,1].
(2.3)

Proof By (2.2), we arrive at the inequality (2.3) immediately. The proof is completed. □

By simple computation, we have max t [ 0 , 1 ] 0 1 G(t,s)ds= 0 1 φ(t)dt= 1 α ( α 1 ) = K 2 Γ(α). Let

E:=C[0,1],u:= max t [ 0 , 1 ] | u ( t ) | ,P:= { u E : u ( t ) 0 , t [ 0 , 1 ] } .

Then (E,) becomes a real Banach space and P is a cone on E. Now, note that u solves (1.1) if and only if u is a fixed point of the operator

(Au)(t):= 0 1 G(t,s)f ( s , u ( s ) ) ds,uE.
(2.4)

Clearly, fC([0,1]×R,R) implies A:EE is a completely continuous operator. Denote

(Lu)(t):= 0 1 G(t,s)u(s)ds,uE.

Then L:EE is a completely continuous linear operator, satisfying L(P)P. That is, L is a positive, completely continuous, linear operator. Let

P 0 := { u E : 0 1 u ( t ) φ ( t ) d t ω u , t [ 0 , 1 ] } ,

where φ(t) is defined by Lemma 2.3 and ω:= K 1 Γ(α)>0. By (2.2) and (2.3), we easily have the following result.

Lemma 2.4 L(P) P 0 .

Proof From (2.2), for uP, we have

(Lu)(t)= 0 1 G(t,s)u(s)ds 1 Γ ( α ) 0 1 φ(s)u(s)ds.

On the other hand, from (2.2) and (2.3), we find

0 1 (Lu)(t)φ(t)dt= 0 1 ( 0 1 G ( t , s ) u ( s ) d s ) φ(t)dt K 1 0 1 φ(s)u(s)ds K 1 Γ(α)Lu.

Therefore, L(P) P 0 . This completes the proof. □

Lemma 2.5 (See [24])

Let E be a Banach space and let ΩE be a bounded open set. Suppose that A: Ω ¯ E is a completely continuous operator. If there is u 0 0 such that uAu+μ u 0 , uΩ and μ0, then the topological degree deg(IA,Ω,0)=0.

Lemma 2.6 (See [24])

Let E be a Banach space and let ΩE be a bounded open set with 0Ω. Suppose that A: Ω ¯ E is a completely continuous operator. If Auμu, uΩ and μ1, then the topological degree deg(IA,Ω,0)=1.

3 Main results

We denote λ 1 := K 1 1 >0, λ 2 := K 2 1 >0 and B ρ :={uE:u<ρ} for ρ>0.

Theorem 3.1 If there exists a constant b0 such that

f(t,u)b,uR, lim inf u + f ( t , u ) u > λ 1 , lim sup u 0 | f ( t , u ) u |< λ 2 ,
(3.1)

then (1.1) has at least one nontrivial solution.

Proof The first two inequalities of (3.1) imply that there are ε>0 and b 1 >0 such that

f(t,u)( λ 1 +ε)u b 1 ,uR,t[0,1].
(3.2)

Take R>b K 2 Γ(α)+ ε 1 b( K 1 3 K 2 3 K 1 1 K 2 )+b K 1 2 K 2 3 + ε 1 b 1 K 1 2 K 2 2 . In what follows, we shall prove that

uAu+μ u ,uE,u=R,μ0,
(3.3)

where u P 0 . Indeed, if u 0 E, u 0 =R, and μ 0 0 such that

u 0 (t)=(A u 0 )(t)+ μ 0 u (t)= 0 1 G(t,s)f ( s , u 0 ( s ) ) ds+ μ 0 u (t).
(3.4)

Let u ˜ (t)=b 0 1 G(t,s)ds, then we have

u 0 (t)+ u ˜ (t)= 0 1 G(t,s) ( f ( s , u 0 ( s ) ) + b ) ds+ μ 0 u (t),

which leads to u 0 + u ˜ P 0 by Lemma 2.4. Combining this with (3.2), we find

0 1 ( A u 0 ) ( t ) φ ( t ) d t 0 1 u 0 ( t ) φ ( t ) d t = 0 1 φ ( t ) 0 1 G ( t , s ) f ( s , u 0 ( s ) ) d s d t 0 1 u 0 ( t ) φ ( t ) d t ( λ 1 + ε ) 0 1 φ ( t ) 0 1 G ( t , s ) u 0 ( s ) d s d t b 1 0 1 φ ( t ) 0 1 G ( t , s ) d s d t 0 1 u 0 ( t ) φ ( t ) d t = ( λ 1 + ε ) 0 1 φ ( t ) 0 1 G ( t , s ) ( u 0 ( s ) + u ˜ ( s ) ) d s d t ( λ 1 + ε ) 0 1 φ ( t ) 0 1 G ( t , s ) u ˜ ( s ) d s d t b 1 0 1 φ ( t ) 0 1 G ( t , s ) d s d t 0 1 u 0 ( t ) φ ( t ) d t ε λ 1 1 0 1 φ ( t ) ( u 0 ( t ) + u ˜ ( t ) ) d t + 0 1 u ˜ ( t ) φ ( t ) d t ( λ 1 + ε ) 0 1 φ ( t ) 0 1 G ( t , s ) u ˜ ( s ) d s d t b 1 0 1 φ ( t ) 0 1 G ( t , s ) d s d t ε ω λ 1 1 u 0 + u ˜ + b K 1 K 2 Γ ( α ) b ( λ 1 + ε ) K 2 3 Γ ( α ) b 1 K 2 2 Γ ( α ) ε ω λ 1 1 u 0 ε ω λ 1 1 u ˜ + b K 1 K 2 Γ ( α ) b ( λ 1 + ε ) K 2 3 Γ ( α ) b 1 K 2 2 Γ ( α ) ε ω λ 1 1 R ε ω λ 1 1 b K 2 Γ ( α ) + b K 1 K 2 Γ ( α ) b ( λ 1 + ε ) K 2 3 Γ ( α ) b 1 K 2 2 Γ ( α ) > 0 .

On the other hand, we have by (3.4)

0 1 u 0 (t)φ(t)dt 0 1 (A u 0 )(t)φ(t)dt= μ 0 0 1 u (t)φ(t)dt0.

That is a contradiction. As a result of this, (3.3) holds. Lemma 2.5 gives

deg(IA, B R ,0)=0.
(3.5)

It follows from the third inequality of (3.1) that there exists 0<r<R such that |f(t,u)| λ 2 |u|, |u|r, t[0,1]. In the following, we prove

Auμu,u B r ,μ1.
(3.6)

In fact, suppose that there exist u 1 B r , μ 1 1 such that A u 1 = μ 1 u 1 . We may suppose that μ 1 >1 (otherwise we are done). Thus

μ 1 | u 1 ( t ) | λ 2 0 1 G(t,s) | u 1 ( s ) | ds.
(3.7)

Multiply by φ(t) both sides of the preceding inequality and integrate over [0,1], and use (2.3) to obtain

μ 1 0 1 | u 1 ( t ) | φ(t)dt λ 2 0 1 φ(t) 0 1 G(t,s) | u 1 ( s ) | dsdt 0 1 | u 1 ( t ) | φ(t)dt.

This, together with 0 1 | u 1 (t)|φ(t)dt>0, leads to μ 1 1, which is a contradiction. So, (3.6) holds. Lemma 2.6 implies

deg(IA, B r ,0)=1.
(3.8)

By (3.5) and (3.8), we have deg(IA, B R B ¯ r ,0)=deg(IA, B R ,0)deg(IA, B r ,0)=01=1. Then A has at least one fixed point on B R B ¯ r . This means that problem (1.1) has at least one nontrivial solution. □

In order to prove Theorem 3.2, we need the following result involving the spectral radius of L, denoted by r(L).

Lemma 3.1 0<r(L) K 2 .

Proof We easily obtain the result by Gelfand’s theorem and (2.2). This completes the proof. □

Theorem 3.2 If there exists a constant b0 such that

f(t,u)b,uR, lim inf u 0 f ( t , u ) | u | > λ 1 , lim sup u + f ( t , u ) u < λ 2 ,
(3.9)

then (1.1) has at least one nontrivial solution.

Proof By the second inequality of (3.9), there exist ε>0 and r 1 >0 such that

f(t,u)( λ 1 +ε)|u|,|u| r 1 ,t[0,1].
(3.10)

For every u B ¯ r 1 , we have from (3.10) that

(Au)(t)( λ 1 +ε) 0 1 G(t,s) | u ( s ) | ds,t[0,1]

and thus A( B ¯ r 1 )P. For all u B r 1 P, from (3.10), we know

(Au)(t)( λ 1 +ε) 0 1 G(t,s)u(s)ds,t[0,1].

We may suppose that A has no fixed point on B r 1 (otherwise, the proof is completed). Now we show that

uAu+μ u ,u B r 1 P,μ0,
(3.11)

where u P. Otherwise, there exist u 0 B r 1 P, μ 0 0 such that u 0 =A u 0 + μ 0 u A u 0 . Consequently,

u 0 (t)(A u 0 )(t)( λ 1 +ε) 0 1 G(t,s) u 0 (s)ds.

Multiply by φ(t) both sides of the preceding inequality and integrate over [0,1], and use (2.3) to obtain

0 1 u 0 (t)φ(t)dt( λ 1 +ε) 0 1 φ(t) 0 1 G(t,s) u 0 (s)dsdt( λ 1 +ε) λ 1 1 0 1 u 0 (t)φ(t)dt,

which implies 0 1 u 0 (t)φ(t)dt=0, and then u 0 (t)0, t[0,1]. It contradicts u 0 B r 1 P. Hence (3.11) is true. Since A( B ¯ r 1 )P, we have, from the permanence property of fixed point index and Lemma 2.5, that

deg(IA, B r 1 ,0)=i(A, B r 1 P,P)=0,
(3.12)

where i denotes fixed point index on P. Recall the definition of u ˜ . Clearly, u ˜ P and A:C[0,1]P u ˜ by (3.9). Define A ˜ u=A(u u ˜ )+ u ˜ , uC[0,1]. We easily find A ˜ :C[0,1]P. By the third inequality of (3.9), there exist r 2 > r 1 + u ˜ = r 1 +b K 2 Γ(α) and 0<σ<1 such that

f(t,u)σ λ 2 u,u r 2 ,t[0,1].
(3.13)

Let L 1 u=σ λ 2 Lu, uC[0,1]. Then L 1 :C[0,1]C[0,1] is a bounded linear operator and L 1 (P)P. Let

M=2max { sup u B ¯ r 2 0 1 G ( t , s ) | f ( s , u ( s ) ) | d s , 2 b K 2 Γ ( α ) } <+

and W:={uP:u=μ A ˜ u,0μ1}. In what follows, we will show that W is bounded. For all uW, let ψ ˜ (t)=min{u(t) u ˜ (t), r 2 } and e(u)={t[0,1]:u(t) u ˜ (t)> r 2 }. When u(t) u ˜ (t)<0, ψ ˜ (t)=u(t) u ˜ (t)u(t) r 2 r 2 , and so ψ ˜ r 2 . Consequently, for uW, we have from (3.13)

u ( t ) = μ ( A ˜ u ) ( t ) 0 1 G ( t , s ) f ( s , u ( s ) u ˜ ( s ) ) d s + u ˜ ( t ) = e ( u ) G ( t , s ) f ( s , u ( s ) u ˜ ( s ) ) d s + [ 0 , 1 ] e ( u ) G ( t , s ) f ( s , u ( s ) u ˜ ( s ) ) d s + u ˜ ( t ) σ λ 2 0 1 G ( t , s ) u ( s ) d s + 0 1 G ( t , s ) f ( s , ψ ˜ ( s ) ) d s + 2 u ˜ ( t ) σ λ 2 0 1 G ( t , s ) u ( s ) d s + M = ( L 1 u ) ( t ) + M

and then ((I L 1 )u)(t)M, t[0,1]. By Lemma 3.1 and 0<σ<1, r( L 1 )=σ λ 2 r(L)σ λ 2 K 2 <1. Therefore, the inverse operator ( I L 1 ) 1 exists and ( I L 1 ) 1 =I+ L 1 + L 1 2 ++ L 1 n + . It follows from L 1 (P)P that ( I L 1 ) 1 (P)P. So, we have u(t) ( I L 1 ) 1 M, t[0,1] and W is bounded.

Select r 3 >max{ r 2 ,supW+b K 2 Γ(α)} and thus A ˜ has no fixed point on B r 3 . Indeed, if there exists u 1 B r 3 such that A ˜ u 1 = u 1 , then u 1 W and u 1 = r 3 >supW, which is a contradiction. Then we have from the permanence property and the homotopy invariance property of fixed point index that

deg(I A ˜ , B r 3 ,0)=i( A ˜ , B r 3 P,P)=i(0, B r 3 P,P)=1.
(3.14)

Set the completely continuous homotopy H(t,u)=A(ut u ˜ )+t u ˜ , (t,u)[0,1]× B ¯ r 3 . If there exists ( t 0 , u 2 )[0,1]× B r 3 such that H( t 0 , u 2 )= u 2 , and then A( u 2 t 0 u ˜ )= u 2 t 0 u ˜ and A ˜ ( u 2 t 0 u ˜ + u ˜ )= u 2 t 0 u ˜ + u ˜ . Thus u 2 t 0 u ˜ + u ˜ W and u 2 t 0 u ˜ + u ˜ u 2 (1 t 0 ) u ˜ r 3 u ˜ >supW, which is a contradiction. From the homotopy invariance of topological degree and (3.14), we have

deg(IA, B r 3 ,0)=deg(I A ˜ , B r 3 ,0)=1.
(3.15)

By (3.12) and (3.15), we get deg(IA, B r 3 B ¯ r 1 ,0)=deg(IA, B r 3 ,0)deg(IA, B r 1 ,0)=1, which implies that A has at least one fixed point on B r 3 B ¯ r 1 . This means that the problem (1.1) has at least one nontrivial solution. □

Two examples 1. Let

f(t,u)= a 1 u+ a 2 u 2 ++ a n u n ,(t,u)[0,1]×R,

where n is a positive even number, a i R (i=1,2,,n1), | a 1 |< λ 2 , a n >0. It is easy to see that f(t,u) is bounded below and usually sign-changing for u0. In addition, lim sup u 0 | f ( t , u ) u |=| a 1 |< λ 2 and lim inf u + f ( t , u ) u =+. Thus by Theorem 3.1, we can obtain the existence of a nontrivial solution of (1.1).

  1. 2.

    Let

    f(t,u)= 1 u 2 1 + u 2 ,(t,u)[0,1]×R.

It is easy to see that f(t,u) is bounded below and usually sign-changing for u0. In addition, lim sup u + f ( t , u ) u =0< λ 2 and lim inf u 0 f ( t , u ) | u | =+. Thus, by Theorem 3.2, we can obtain the existence of a nontrivial solution of (1.1).

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Acknowledgements

Research is supported by the NNSF-China (10971046), Shandong and Hebei Provincial Natural Science Foundation (ZR2012AQ007, A2012402036), GIIFSDU (yzc12063), IIFSDU (2012TS020).

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Correspondence to Keyu Zhang.

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KZ and JX gave the proof for the main result together. All authors read and approved the final manuscript.

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Zhang, K., Xu, J. Nontrivial solutions for a fractional boundary value problem. Adv Differ Equ 2013, 171 (2013). https://doi.org/10.1186/1687-1847-2013-171

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Keywords

  • fractional boundary value problem
  • nontrivial solution
  • topological degree
  • Riemann-Liouville derivative