AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 540846 10.1155/2012/540846 540846 Research Article Multiplicity Result of Positive Solutions for Nonlinear Differential Equation of Fractional Order Liu Yang 1, 2 Weiguo Zhang 1 Staněk Svatoslav 1 College of Science University of Shanghai for Science and Technology Shanghai 200093 China usst.edu.cn 2 Department of Mathematics Hefei Normal University Anhui Hefei 230601 China hftc.edu.cn 2012 14 08 2012 2012 18 03 2012 18 06 2012 2012 Copyright © 2012 Yang Liu and Zhang Weiguo. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We investigate the existence of multiple positive solutions for a class of boundary value problems of nonlinear differential equation with Caputo’s fractional order derivative. The existence results are obtained by means of the Avery-Peterson fixed point theorem. It should be point out that this is the first time that this fixed point theorem is used to deal with the boundary value problem of differential equations with fractional order derivative.

1. Introduction

In this paper, we consider the existence and multiple existence of positive solutions for following boundary value problem of differential equation involving the Caputo's fractional order derivative (1.1)D0+αu(t)=f(t,u(t),D0+βu(t)),t(0,1),u(0)+u(0)=0,u(1)+u(1)=0, where 1<α<2, 0<β<α-1 and f:C([0,1]×R+×R,R+). Here D0+α is the Caputo's derivative of fractional order.

Due to the development of the theory of fractional calculus and its applications, such as in the fields of control theory, blood flow phenomena, Bode's analysis of feedback amplifiers, aerodynamics, and polymer rheology and many work on fractional calculus, fractional order differential equations has appeared . Recently, there have been many results concerning the solutions or positive solutions of boundary value problems for nonlinear fractional differential equations, see  and references along this line.

For example, Bai and Lü  considered the following Dirichlet boundary value problem of fractional differential equation: (1.2)D0+αu(t)+f(t,u(t))=0,t(0,1),u(0)=0=u(1),1<α2. By means of different fixed-point theorems on cone, some existence and multiplicity results of positive solutions were obtained. Jiang and Yuan  improved the results in  by discussing some new positive properties of the Green function for problem (1.2). By using the fixed point theorem on a cone due to Krasnoselskii, the authors established the existence results of positive solution for problem (1.2). Recently, Caballero et al.  obtained the existence and uniqueness of positive solution for singular boundary value problem (1.2). The existence results were established in the case that the nonlinear term f may be singular at t=0. As to positive solutions of problem (1.1), under the case that the nonlinear term was not involved with the derivative of the function u(t), Zhang  obtained the existence and multiplicity results of positive solutions by means of a fixed-point theorem on cones.

There are also some results concerning multipoint boundary value problems for differential equations of fractional order. Bai  investigated the existence and uniqueness of positive solution for three-point boundary value problem (1.3)D0+αu(t)+f(t,u(t))=0,t(0,1),u(0)=0,u(1)=βu(η), where 1<α2, η(0,1), 0<βηα-1<1. In , the uniqueness of positive solution was obtained by the use of contraction map principle and some existence results of positive solutions were established by means of the fixed point index theory. Very recently, Wang et al.  considered the boundary value problem of fractional differential equation with integral condition (1.4)D0+αu(t)+q(t)f(t,u(t))=0,t(0,1),n-1<α<n,u(0)=u(0)==u(n-2)(0)=0,u(1)=01u(s)dA(s), where α>2, 01u(s)dA(s) was given by Riemann-Stieltjes integral with a signed measure. By using the fixed point theorem, the existence of positive solution for this problem was established.

However, in this work, the derivative of the unknown function u(t) was not involved in the nonlinear term explicitly. To our best knowledge, there are few papers considering the positive solution of boundary value problem of nonlinear fractional differential equations which the derivative of the unknown function u(t) is involved in the nonlinear term. In , Guo and Ge proved a new fixed point theorem, which can be regarded as an extension of Krasnoselskii's fixed point theorem in a cone. By applying this new theorem, Guo and Ge obtained the existence of positive solutions for second-order three-point boundary value problem (1.5)u′′+f(t,u(t),u(t))=0,t[0,1],u(0)=0,u(1)=αu(η),η(0,1), where f depended on the first order derivative of u. Very recently, Yang et al.  considered following boundary value problem (1.6)D0+αu(t)=f(t,u(t),u(t)),t(0,1),u(0)+u(0)=0,u(1)+u(1)=0. By means of Schauder's fixed point theorem and the fixed point theorem duo to Guo and Ge, some results on the existence of positive solutions were obtained.

In , Avery and Peterson gave an new triple fixed point theorem, which can be regarded as an extension of Leggett-Williams fixed point theorem. By using this method, many results concerning the existence of at least three positive solutions of boundary value problems of differential equation with integer order were established, see . For example, by using the Avery-Peterson fixed point theorem, Yang et al.  established the existence of at least three positive solutions of second-order multipoint boundary value problem (1.7)u′′+f(t,u(t),u(t))=0,t[0,1],u(0)=i=1m-2βiu(ξi),u(1)=i=1m-2αiu(ξi).

But by using the Avery-Peterson fixed point theorem, the nonlinear terms are often assumed to be nonnegative to ensure the concavity or convexity of the unknown function. When the differential equations of fractional order are considered, we cannot derive the concavity or convexity of function u(t) by the sign of its fractional order derivative. Thus the Avery-Peterson fixed point theorem cannot directly be used to consider the boundary value problem of nonlinear differential equation with fractional order where the derivative of the unknown function u(t) is involved in the nonlinear term explicitly.

In this paper, by obtaining some new inequalities of the unknown function and defining a special cone, we overcome the difficulties brought by the lack of the concavity or convexity of unknown function u(t). By an application of Avery-Peterson fixed point theorem, the existence of at least three positive solutions of problem (1.1) is established. It should be pointed out that it is the first time that the Avery-Peterson fixed point theorem is used to deal with the positive solutions of boundary value problem of differential equations with fractional order derivative.

2. Preliminary Results Definition 2.1.

The Riemann-Liouville fractional integral of order α>0 of a function u(t):(0,)R is given by (2.1)I0+αu(t)=1Γ(α)0t(t-s)α-1u(s)ds provided the right side is point-wise defined on (0,).

Definition 2.2.

The Caputo's fractional derivative of order α>0 of a continuous function u(t):(0,)R is given by (2.2)D0+αu(t)=1Γ(n-α)0tu(n)(s)(t-s)α-n+1ds, where n-1<αn, provided that the right side is point-wise defined on (0,).

Lemma 2.3.

Let α>0. Then the fractional differential equation D0+αu(t)=0 has solutions (2.3)        u(t)=c0+c1t+c2t2++cn-1tn-1,ciR,i=0,1,,n-1.

Lemma 2.4.

Let α>0. Then (2.4)I0+αD0+αu(t)=u(t)c0+c1t+c2t2++cn-1tn-1,ciR,i=1,2,,n-1.

Definition 2.5.

Let E be a real Banach space over R. A nonempty convex closed set PE is said to be a cone provided that

auP, for all uP, a0,

u,-uP implies u=0.

Definition 2.6.

An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

Definition 2.7.

The map α is said to be a continuous nonnegative convex functional on cone P of a real Banach space E provided that α:P[0,+) is continuous and (2.5)α(tx+(1-t)y)tα(x)+(1-t)α(y),  x,yP,t[0,1].

Definition 2.8.

The map β is said to be a continuous nonnegative concave functional on cone P of a real Banach space E provided that β:P[0,+) is continuous and (2.6)β(tx+(1-t)y)tβ(x)+(1-t)β(y),x,yP,t[0,1].

Let γ and  θ be nonnegative continuous convex functionals on P, α a nonnegative continuous concave functional on P, and ψ a nonnegative continuous functional on P. Then for positive numbers a, b, c, and d, we define the following convex sets: (2.7)P(γ,d)={xPγ(x)<d},P(γ,α,b,d)={xPbα(x),γ(x)d},P(γ,θ,α,b,c,d)={xPbα(x),θ(x)c,γ(x)d}, and a closed set (2.8)      R(γ,ψ,a,d)={xPaψ(x),γ(x)d}.

Lemma 2.9 (see [<xref ref-type="bibr" rid="B31">31</xref>]).

Let P be a cone in Banach space E. Let γ, θ be nonnegative continuous convex functionals on P, α a nonnegative continuous concave functional on P, and ψ a nonnegative continuous functional on P satisfying (2.9)ψ(λx)λψ(x),for  0λ1, such that for some positive numbers l and d, (2.10)α(x)ψ(x),xlγ(x), for all xP(γ,d)¯. Suppose T:P(γ,d)¯P(γ,d)¯ is completely continuous and there exist positive numbers a, b, c with a<b such that

{xP(γ,θ,α,b,c,d)|α(x)>b} and α(Tx)>b for xP(γ,θ,α,b,c,d);

α(Tx)>b for xP(γ,α,b,d) with θ(Tx)>c;

0R(γ,ψ,a,d) and ψ(Tx)<a for xR(γ,ψ,a,d) with ψ(x)=a.

Then T has at least three fixed points x1,x2,x3P(γ,d)¯ such that (2.11)γ(xi)d,i=1,2,3;b<α(x1);a<ψ(x2),α(x2)<b;ψ(x3)<a.

3. Main Results Lemma 3.1 (see [<xref ref-type="bibr" rid="B30">30</xref>]).

Given y(t)C[0,1], then boundary value problem (3.1)        D0+αu(t)=y(t),u(0)+u(0)=0,u(1)+u(1)=0 is equivalent to (3.2)u(t)=0t((1-s)α-1(1-t)+(1-s)α-1Γ(α)+(1-s)α-2(1-t)Γ(α-1))y(s)ds+t1((1-s)α-1(1-t)Γ(α)+(1-s)α-2(1-t)Γ(α-1))y(s)ds=01G(t,s)y(s)ds, where (3.3)G(t,s)={(1-s)α-1(1-t)+(1-s)α-1Γ(α)+(1-s)α-2(1-t)Γ(α-1),0st1,(1-s)α-1(1-t)Γ(α)+(1-s)α-2(1-t)Γ(α-1),0ts1.

Lemma 3.2.

Given y(t)C[0,1], assume that u(t) is a solution of boundary value problem (3.4)        D0+αu(t)=y(t),u(0)+u(0)=0,u(1)+u(1)=0. Then (3.5)Dβu(t)=0t((t-s)α-β-1Γ(α-β)-t1-β(1-s)α-2(α-s)Γ(α)Γ(2-β))y(s)ds+t1-t1-β(1-s)α-2(α-s)Γ(α)Γ(2-β)y(s)ds.

Proof.

From Lemmas 2.3 and 2.4, we get that (3.6)u(t)=0t(t-s)α-1Γ(α)y(s)ds-C1-C2t,u(t)=0t(t-s)α-2Γ(α-1)y(s)ds-C2. The boundary condition u(0)+u(0)=0 implies that C1+C2=0. Considering the boundary condition u(1)+u(1)=0, we have (3.7)C2=-C1=1Γ(α)01(1-s)α-1y(s)ds+1Γ(α-1)01(1-s)α-2y(s)ds. From the definition of the Caputo derivative of fractional order, we see (3.8)Dβu(t)=Dβ(0t(1-s)α-1Γ(α)y(s)ds-C1-C2t)=0t(t-s)α-β-1Γ(α-β)y(s)ds-C2Γ(2-β)t1-β=0t((t-s)α-β-1Γ(α-β)-t1-β(1-s)α-2(α-s)Γ(α)Γ(2-β))y(s)ds+t1-t1-β(1-s)α-2(α-s)Γ(α)Γ(2-β)y(s)ds.

Lemma 3.3 (see [<xref ref-type="bibr" rid="B30">30</xref>]).

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

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

there exist a positive function γ(s),K(s)C(0,1) such that (3.9)max0t1K(s),s(0,1),mins[1/3,2/3]G(t,s)γ(s)K(s),0<s<1, where (3.10)K(s)=2(1-s)α-1Γ(α)+(1-s)α-2Γ(α-1),(3.11)γ(s)=13(1-s)α-1+(α-1)(1-s)α-22(1-s)α-1+(α-1)(1-s)α-2>16,s[0,  1).

Lemma 3.4.

Assume that y(t)>0 and u(t) is a solution of boundary value problem (3.1). There exists a positive constant γ0 such that (3.12)max0t1|u(t)|γ0max0t1Dβ|u(t)|.

Proof.

From Lemma 3.2,(3.13)max0t1|Dβu(t)|=max0t1|0t((t-s)α-β-1Γ(α-β)-t1-β(1-s)α-2(α-s)Γ(α)Γ(2-β))y(s)ds-t1t1-β(1-s)α-2(α-s)Γ(α)Γ(2-β)y(s)ds|max0t1|Dβu(1)|=01((1-s)α-2(α-s)Γ(α)Γ(2-β)-(1-s)α-β-1Γ(α-β))y(s)ds=01k(s)y(s)ds. Denote (3.14)h(s)=K(s)k(s)=2(1-s)α-1/Γ(α)+(1-s)α-2/Γ(α-1)(1-s)α-2(α-s)/Γ(α)Γ(2-β)-(1-s)α-β-1/Γ(α-β)=2(1-s)/Γ(α)+1/Γ(α-1)(α-s)/Γ(α)Γ(2-β)-(1-s)1-β/Γ(α-β). By a simple computation, we have (3.15)h(s)2/Γ(α)+1/Γ(α-1)(α-s0)/Γ(α)Γ(2-β)-(1-s0)1-β/Γ(α-β)=2/Γ(α)+1/Γ(α-1)((α-1)(1-β)-β((1-β)Γ(α)Γ(2-β)/Γ(α-β))1/β)/(1-β)Γ(α)Γ(α-β)=γ0>0. Thus, (3.16)max0t1|u(t)|=max0t101G(t,s)y(s)ds01K(s)y(s)ds01γ0k(s)y(s)dsγ0max0t1|Dβu(t)|.

Let the Banach space E={u(t)C[0,1], Dβu(t)C[0,1]} be endowed with the norm (3.17)u=max{max0t1|u(t)|,max0t1|Dβu(t)|},uE. We define the cone PE by (3.18)P={uEu(t)0,min1/3t2/3u(t)16max0t1u(t),max0t1u(t)γ0max0t1|Dβu(t)|}. Denote the positive constants (3.19)M=01K(s)ds=2Γ(α+1)+1Γ(α),N=1Γ(α-β+1)+1Γ(2-β)(1Γ(α)+1Γ(α+1)).

Lemma 3.5.

Let T:PE be the operator defined by (3.20)Tu(t):=01G(t,s)f(s,u(s),D0+βu(s))ds. Then T:PP is completely continuous.

Proof.

The operator T is nonnegative and continuous in view of the nonnegativeness and continuity of functions G(t,s) and f(t,u(t),D0+βu(t)). Let ΩK be bounded. Then there exists a positive constant R1>0 such that uR1, uΩ. Denote (3.21)R=max0t1,uΩ|f(t,u(t),D0+βu(t))|+1. Then for uΩ, we have (3.22)|(Tu)(t)|01G(t,s)|f(s,u,D0+βu)|ds01K(s)|f(s,u,D0+βu)|dsR01K(s)ds=MR,|D0+β(Tu)(t)|=|1Γ(α-β)0t(t-s)α-β-1f(s,u(s),D0+βu(s))ds-t1-βΓ(2-β)(1Γ(α-1)01(1-s)α-2f(s,u,D0+βu)ds    +1Γ(α)01(1-s)α-1f(s,u,D0+βu)ds)|[1(α-β)Γ(α-β)+1Γ(2-β)(1Γ(α-1)01(1-s)α-2ds+1Γ(α)01(1-s)α-1ds)]×R=NR. Hence T(Ω) is bounded. On the other hand, for uΩ, t1,t2[0,  1], one has (3.23)|Tu(t2)-Tu(t1)|=|01G(t1,s)f(s,u(s),D0+βu(s))ds-01G(t2,s)f(s,u(s),D0+βu(s))ds|0t1|G(t2,s)-G(t1,s)||f(s,u(s),D0+βu(s))|ds+t1t2|G(t2,s)-G(t1,s)||f(s,u(s),D0+βu(s))|ds+t21|G(t2,s)-G(t1,s)||f(s,u(s),D0+βu(s))|ds[2(t2-t1)Γ(α)+t2α-t1αΓ(α+1)]×R,|D0+β(Tu(t2))-D0+β(Tu(t1))|=|1Γ(α-β)(0t2(t2-s)α-β-1f(s,u,D0+βu)ds-0t1(t1-s)α-β-1f(s,u,D0+βu)ds)+(t11-β-t21-β)Γ(2-β)(1Γ(α-1)01(1-s)α-2f(s,u,D0+βu)ds+1Γ(α)01(1-s)α-1f(s,u,D0+βu)ds)|R×|t2α-β-t1α-β|(α-β)Γ(α-β)+RΓ(2-β)(1Γ(α-1)01(1-s)α-2ds+1Γ(α)01(1-s)α-1)×|t21-β-t11-β|R(α-β)Γ(α-β)|t2α-β-t1α-β|+RΓ(2-β)(1Γ(α+1)+1Γ(α))|t21-β-t11-β|. Thus, (3.24)Tu(t2)-Tu(t1)0,for  t1t2. By means of the Arzela-Ascoli theorem, T is completely continuous. Furthermore, for uP, we have (3.25)min1/3t2/3Tu(t)=min1/3t2/301G(t,s)f(s,u(s),D0+βu(s))ds1601K(s)f(s,u(s),D0+βu(s))ds=16max0t1Tu(t),max0t1|u(t)|=01G(t,s)f(s,u(s),D0+βu(s))ds01K(s)f(s,u(s),D0+βu(s))ds01γ0k(s)f(s,u(s),D0+βu(s))ds=γ0max0t1|Dβu(t)|. Thus, T:PP is completely continuous.

Let the continuous nonnegative concave functional α, the the continuous nonnegative convex functionals γ, θ, and the continuous nonnegative functional ψ be defined on the cone by (3.26)γ(u)=max0t1|Dβu(t)|,θ(u)=ψ(u)=max0t1|u(t)|,α(u)=min1/3t2/3|u(t)|. By Lemmas 3.3 and 3.4, the functionals defined above satisfy that (3.27)16θ(u)α(u)θ(u)=ψ(u),uγ1γ(u),uP, where γ1=max{γ0,1}. Therefore condition (2.10) of Lemma 2.9 is satisfied.

Assume that there exist constants 0<a,b,d with a<b<(M/6N)d, c=6b such that

f(t,u,v)d/N, (t,u,v)[0,1]×[0,γ1d]×[-d,d],

f(t,u,v)>6b/M, (t,u,v)[0,1]×[b,6b]×[-d,d],

f(t,u,v)<a/M, (t,u,v)[0,1]×[0,a]×[-d,d].

Theorem 3.6.

Under assumptions (A1)–(A3), problem (1.1) has three positive solutions u1, u2, u3 satisfying (3.28)max0t1|Dβu(t)|d,i=1,2,3;b<min1/3t2/3|u1(t)|;a<max0t1|u2(t)|,min1/3t2/3|u2(t)|<b;max0t1|u3(t)|a.

Proof.

Problem (1.1) has a solution u=u(t) if and only if u solves the operator equation (3.29)u(t)=01G(t,s)f(s,u(s),Dβu(s))ds=(Tu)(t). For uP(γ,d)¯, we have γ(u)=max0t1|Dβu(t)|<d. From assumption (A1), we obtain (3.30)f(t,u(t),Dβu(t))dN. Thus (3.31)γ(Tu)=max0t1|Dβ(Tu)|=|1Γ(α-β)0t(t-s)α-β-1f(s,u,D0+βu)ds-t1-βΓ(2-β)(1Γ(α-1)01(1-s)α-2f(s,u,D0+βu)ds+1Γ(α)01(1-s)α-1f(s,u,D0+βu)ds)|[1(α-β)Γ(α-β)+1Γ(2-β)(1Γ(α-1)01(1-s)α-2ds+1Γ(α)01(1-s)α-1ds)]×dN=d. Hence, T:P(γ,d)¯P(γ,d)¯.

The fact that the constant function u(t)=6bP(γ,θ,α,b,c,d) and α(6b)>b implies that {uP(γ,θ,α,b,c,d|α(u)>b)}.

For uP(γ,θ,α,b,c,d), we have bu(t)6b and |Dβu(t)|<d for 0t1. From assumption (A2), (3.32)f(t,u(t),Dβu(t))>6bM. Thus (3.33)α(Tu)=min1/3t2/301G(t,s)f(s,u(s),D0+βu(s))ds1601K(s)f(s,u(s),Dβu(s))ds1601K(s)ds×6bM=b, which means α(Tu)>b, for  all uP(γ,θ,α,b,6b,d). These ensure that condition (S1) of Lemma 2.9 is satisfied. Secondly,  for all uP(γ,α,b,d) with θ(Tu)>6b, (3.34)α(Tu)16θ(Tu)>16×c=16×6b=b. Thus, condition (S2) of Lemma 2.3 holds. Finally we show that (S3) also holds. We see that ψ(0)=0<a and 0R(γ,ψ,a,d). Suppose that uR(γ,ψ,a,d) with ψ(x)=a. Then by assumption (A3), (3.35)ψ(Tu)=max0t1|(Tu)(t)|=01G(t,s)f(s,u(s),Dβu(s))ds01K(s)ds×aM=a. Thus, all conditions of Lemma 2.9 are satisfied. Hence problem (1.1) has at least three positive concave solutions u1, u2, u3 satisfying (3.28).

4. Example

Consider the nonlinear FBVPs (4.1)D0+αu(t)=f(t,u(t),D0+βu(t)),t(0,1),u(0)+u(0)=0,u(1)+u(1)=0, where α=1.6, β=0.5, n=2 and (4.2)f(t,u,v)={1π4et+110u4+1100sin(v10000),0u10,1π4et+1000+1100sin(v10000),u>10. Choose a=1, b=3, d=10000. By a simple computation, we obtain that (4.3)γ08.7149,γ18.7149,M2.5181,N2.8672. We can check that the nonlinear term f(t,u,v) satisfies

f(t,u,v)<d/N, (t,u,v)[0,1]×[0,87149]×[-10000,10000],

f(t,u,v)>6b/M, (t,u,v)[0,1]×[3,18]×[-10000,10000],

f(t,u,v)<a/M, (t,u,v)[0,1]×[0,1]×[-10000,10000].

Then all assumptions of Theorem 3.6 are satisfied. Thus problem (4.1) has at least three positive solutions u1(t), u2(t), u3(t) satisfying (4.4)max0t1|D0+βu(t)|10000,i=1,2,3;3<min1/2t1|u1(t)|,1<max0t1|u2(t)|,min1/2t1|u2(t)|<3,max0t1|u3(t)|1.

Acknowledgments

This work was supported by the Anhui Provincial Natural Science Foundation (10040606Q50) and Innovation Program of Shanghai Municipal Education Commission (13zz118).

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