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Springerplus. 2016; 5(1): 957.
Published online 2016 July 2. doi:  10.1186/s40064-016-2665-8
PMCID: PMC4930441

On the existence of positive solutions for fractional differential inclusions at resonance

Abstract

In this paper, we discuss the existence of positive solutions for a boundary value problem of fractional differential inclusions with resonant boundary conditions. By using the Leggett–Williams theorem for coincidences of multi-valued operators due to O’Regan and Zima, results on the existence of positive solutions are established. An example is given to illustrate the efficiency of the main theorems.

Keywords: Fractional differential inclusions, Multi-valued operator, Positive solution, Resonance

Background

In this article, we investigate the existence of positive solutions of fractional differential inclusions with two-point boundary conditions:

D0+αu(t)ft,u(t),0t1,u(i)(0)=0,u(0)=u(1),i=1,2,,n-1,
1

where n - 1 < α < nn ≥ 2, D0+α denotes the Caputo fractional derivative, f:[0, 1] × ℝ → ℱ(ℝ), ℱ(ℝ) denotes the family of nonempty compact and convex subsets of .

Fractional calculus is a generalization of ordinary differentiation and integration to arbitrary order. The fractional differential equations play an important role in various fields of science and engineering, such as chemistry, biology, control theory, viscoelastic materials, signal processing, finance, life science and so on, see Kilbas et al. (2006), Samko et al. (1993), Podlubny (1999) and Orsingher and Beghin (2004).

During the last 10 years, boundary value problems for fractional differential equations are one of the most active fields in the researches of nonlinear differential equations theories. For further details, see Bai and Lü (2005), Zhang (2006), Caballero et al. (2011), Xu et al. (2009), Lin (2007) and Goodrich (2010). Meanwhile, fractional boundary value problems at resonance have been extensively studied. For some recent works on the topic, see Kosmatov (2008, 2010), Bai (2011), Bai and Zhang (2011) and Yang and Wang (2011) and references therein. It is well known that differential inclusions have proved to be valuable tools in the modeling of many realistic problems, such as economics, optimal control and so on. Recently, fractional differential inclusions have been investigated by several researchers, we refer the reader to Agarwal et al. (2010) and Chen et al. (2013).

As shown in the above mentioned works, we can see two facts. Firstly, although the boundary value problems for fractional differential equations at resonance have been studied by some authors, the existence of positive solutions to fractional differential equations at resonance are seldom considered. Secondly, there are few papers to deal with fractional differential inclusions under resonant conditions. The study of positive solutions for higher-order fractional differential inclusions under resonant conditions has yet to be initiated.

To fill this gap, we discuss the fractional differential inclusions (1) by using the Leggett–Williams theorem for coincidences of multi-valued operators due to O’Regan and Zima (2008).

The rest of this paper is organized as follows. “Preliminaries” section, we give some necessary notations, definitions and lemmas. In “Main results” section, we obtain the existence of positive solutions of (1) by Theorem 1. Finally, an example is given to illustrate our results in “Example” section.

Preliminaries

First of all, we present the necessary definitions and lemmas from fractional calculus theory. For more details, see Kilbas et al. (2006), Samko et al. (1993) and Podlubny (1999).

Definition 1

(Kilbas et al.2006) The Riemann–Liouville fractional integral of order α > 0 of a function f:(0, ) → ℝ is given by

I0+αf(t)=1Γ(α)0t(t-s)α-1f(s)ds,

provided that the right-hand side is pointwise defined on (0, ).

Definition 2

(Kilbas et al.2006) The Caputo fractional derivative of order α > 0 of a continuous function f:(0, ) → ℝ is given by

D0+αf(t)=1Γ(n-α)0t(t-s)n-α-1f(n)(s)ds,

where n - 1 < α ≤ n, provided that the right-hand side is pointwise defined on (0, ).

Lemma 1

(Kilbas et al. 2006) The fractional differential equation

D0+αy(t)=0

has solutiony(t) = c0c1t +  ⋯  + cn-1tn-1ci ∈ ℝ,i = 0, 1, …, n - 1,n = [α] + 1.

Furthermore, fory ∈ ACn[0, 1],

I0+αD0+αy(t)=y(t)-k=0n-1y(k)(0)k!tk

and

D0+αI0+αy(t)=y(t).

Lemma 2

(Kilbas et al. 2006) The relation

Ia+αIa+βf(x)=Ia+α+βf(x),

is valid in following case:β > 0,  αβ > 0,  f ∈ L1(ab).

In the following, let us recall some definitions on Fredholm operators and cones in Banach space (see Mawhin 1979).

Let X, Y be real Banach spaces. Consider a linear mapping L:domL ⊂ X → Y and a nonlinear multivalued mapping N:X → 2Y. Assume that

  1. L is a Fredholm operator of index zero, that is, ImL is closed and dim (KerL) = codim(ImL) < ,
  2. N:X → 2Y is an upper semicontinuous mapping with nonempty compact convex values.

The assumption (A1) implies that there exist continuous projections P:X → X and Q:Y → Y such that ImP = KerL and KerQ = ImL. Moreover, since dim (ImQ) = codim (ImL), there exists an isomorphism J:ImQ → KerL. Denote by Lp the restriction of L to KerP ∩ domL. Clearly, Lp is an isomorphism from KerP ∩ domL to ImL, we denote its inverse by Kp:ImL → KerP ∩ domL. It is known that the inclusion Lx ∈ Nx is equivalent to

x ∈ (PJQN)xKP(IQ)Nx.

Let C be a cone in X such that

  1. μx ∈ C for all x ∈ C and μ ≥ 0,
  2. x,  - x ∈ C implies xθ.

It is well known that C induces a partial order in X by

x ⪯ y if and only if yx ∈ C.

The following property is valid for every cone in a Banach space X.

Lemma 3

Let C be a cone in X. Then for everyu ∈ C\{0}there exists a positive numberσ(u)such that

xu‖ ≥ σ(u)‖u‖ for all x ∈ C.

Let γ:X → C be a retraction, that is, a continuous mapping such that γ(x) = x for all x ∈ C. Set

Ψ: = PJQNKp(IQ)N and Ψγ: = Ψ ∘ γ.

We use the following result due to O’Regan and Zima.

Theorem 1

(O’Regan and Zima 2008) Let C be a cone in X and letΩ1,Ω2be open bounded subsets of X withΩ¯1Ω2andC(Ω¯2\Ω1). Assume that (A1), (A2) hold and the following assumptions hold:

  • (A3) QN:X → 2Yis bounded on bounded subsets ofC and Kp(IQ)N:X → 2Xbe compact on every bounded subset of C,
  • (A4) γmaps subsets ofΩ¯2into bounded subsets of C,
  • (A5) Lx ∉ λNxfor allx ∈ C ∩ Ω2 ∩ domLandλ ∈ (0, 1),
  • (A6)  deg {[I - (PJQN)γ]|kerL, kerL ∩ Ω2, 0} ≠ 0,
  • (A7) there existsu0 ∈ C\{0}such thatx‖ ≤ σ(u0)‖yforx ∈ C(u0) ∩ Ω1andy ∈ Ψx, whereC(u0) = {x ∈ C:μu0 ⪯ x for some μ > 0}andσ(u0)such thatxu0‖ ≥ σ(u0)‖xfor everyx ∈ C,
  • (A8) (PJQN)γ(Ω2) ⊂ C,
  • (A9) Ψγ(Ω¯2\Ω1)C,
  • (A10) x ∉ (PJQN)γx for x ∈ Ω2 ∩ KerL.

Then the equationLx ∈ Nxhas at least one solution in the setC(Ω¯2\Ω1).

Main results

In this section, we state our result on the existence of positive solutions for (1).

For simplicity of notation, we set

G(t,s)=α-Γ(α+1)Γ(2α)(1-s)α-αtαΓ(α+1)+αΓ(α+1)Γ(2α),0t<s1,α-Γ(α+1)Γ(2α)(1-s)α-αtαΓ(α+1)+αΓ(α+1)Γ(2α)+1Γ(α)t-s1-sα-1,0s<t1.

By the monotonicity of the function, it is easy to verify that G(ts) > 0, ts ∈ [0, 1]. Here, we omit the proof. Moreover, κ is a constant which satisfies

0<κmin1,1maxt,s[0,1]G(t,s).
2

Thus, we get 1 - κG(ts) > 0, ts ∈ [0, 1].

Theorem 2

Assume that:

  1. f:[0, 1] × ℝ → ℱ(ℝ),f(t, u)is continuous for everyu ∈ ℝ,  t ∈ [0, 1],
  2. for eachr > 0, there existsαr ∈ L1[0, 1]such that|f(tu)| ≤ αr(t)for a.e.t ∈ [0, 1]and everyu ∈ [0, r], where|f(tu)| = sup{|w|:w ∈ f(tu)},
  3. there exist positive constantsb1b2b3c1c2Bwith
    B>c2c1+3b2c2αb1c1+3b3αb1,
    such that
    κx ≤ w ≤  - c1xc2 and w ≤  - b1|w| + b2xb3
    for allx ∈ [0, B]andw ∈ f(tx)witht ∈ [0, 1],
  4. there existb ∈ (0, B),t0 ∈ [0, 1],ρ ∈ (0, 1],δ ∈ (0, 1)and the functionq ∈ L1[0, 1],q(t) ≥ 0, t ∈ [0, 1],h ∈ C((0, b], ℝ+)such thatw(tu) ≥ q(t)h(u)for(tu) ∈ [0, 1] × (0, b]andw ∈ f(tu).h(u)uρis non-increasing on (0, b] with
    h(b)b01G(t0,s)(1-s)α-1q(s)ds1-δδρ.

Then the problem (1) has at least one positive solution on [0, 1].

Proof

We use the Banach space XYC[0, 1] with the supremum norm x‖ = maxt∈[0,1]|x(t)|.

Define L:domL → X and N:X → 2Y with domL=xX:D0+αx(t)C[0,1],x(i)(0)=0,x(0)=x(1),i=1,2,,n-1 by

Lu=D0+αu

and

Nu(t) = {y ∈ Y:y(t) ∈ f(tu(t)) a.e. on[0, 1]}.

Then the problem (1) can be written by

Lu ∈ Nuu ∈ domL.

By Lemma 1, D0+αu(t)=0 has solution

u(t) = c0c1t +  ⋯  + cn-1tn-1

where ci ∈ ℝ, i = 0, 1, …, n - 1. According to the boundary conditions of (1), we get ci = 0, i = 1, 2, …, n - 1. Thus, we obtain

KerL = {u ∈ domL:u(t) = c ∈ ℝ}.

Let y ∈ ImL, so there exists u ∈ domL which satisfies Luy. By Lemma 1, we have

u(t)=I0+αy(t)+c0+c1t++cn-1tn-1.

By the definition of domL, we have ci = 0,  i = 1, 2, …, n - 1. Hence,

u(t)=I0+αy(t)+c0.

Taking into account u(0) = u(1), we obtain

01(1-s)α-1y(s)ds=0.

On the other hand, suppose y satisfies the above equation. Let u(t)=I0+αy(t), and we can easily prove u(t) ∈ domL. Thus, we get

ImL=yY:01(1-s)α-1y(s)ds=0.

Define the linear continuous projector operator P:X → X by

Px(t)=α01(1-s)α-1x(s)ds,t[0,1].

Next, we define the operator Q:Y → Y by

Qy(t)=α01(1-s)α-1y(s)ds,t[0,1].

Noting that

PPx(t)=α01(1-s)α-1Px(t)ds=Px(t)·α01(1-s)α-1ds=Px(t),

then we have P2P. Similarly, we have Q2Q.

Then, one has ImP = KerL and KerQ = ImL. It follows from IndL = dim(kerL) - codim(ImL) = 0 that L is a Fredholm mapping of index zero. Then, (A1) holds.

We consider the mapping KP:ImL → domL ∩ KerP by

KPy(t)=01k(t,s)y(s)ds,t[0,1],

where

k(t,s):=1Γ(α)(t-s)α-1-Γ(1+α)Γ(2α)(1-s)2α-1,0st1,-Γ(1+α)Γ(2α)(1-s)2α-1,0t<s1,

Now, we will prove that is KP the inverse of L|domL ∩ KerP. In fact, for x ∈ domL ∩ KerP, we have D0+αx(t)=y(t)ImL and α01(1-s)α-1x(s)ds=0.

By Lemma 1, one has

(KPy)(t)=x(t)=I0+αy(t)+c0+c1t++cn-1tn-1.

According to the definition of domL, we get ci = 0, i = 1, 2, …, n - 1. Furthermore, by α01(1-s)α-1x(s)ds=0, we have c0=-Γ(1+α)(I0+2αy)(1).

Thus,

(KPy)(t)=I0+αy(t)+c0=I0+αy(t)-Γ(1+α)I0+2αy(1)=I0+αy(t)-Γ(1+α)·1Γ(2α)01(1-s)2α-1y(s)ds=01k(t,s)y(s)ds.

Obviously, LKPyy. Moreover, for x ∈ domL ∩ KerP, we get α01(1-s)α-1x(s)ds=0 and

KPLx=I0+αD0+αx(t)-Γ(1+α)I0+2αD0+αx(1)=x(t)-x(0)-Γ(1+α)I0+αI0+αD0+αx(1)=x(t)-x(0)-Γ(1+α)I0+αx(t)-x(0)|t=1=x(t)-x(0)-Γ(1+α)I0+αx(1)+Γ(1+α)I0+α(x(0))|t=1=x(t)-x(0)-Γ(1+α)Γ(α)01(1-s)α-1x(s)ds+Γ(1+α)Γ(α)01(1-s)α-1x(0)ds=x(t)-x(0)-α01(1-s)α-1x(s)ds+x(0)=x(t).

Thus, we know that KP = (L|domL∩KerP)-1. Moreover, it is easy to see that

|k(ts)| ≤ 3(1-s)α-1,  ∀ ts ∈ [0, 1].
3

Consider the cone

C = {x ∈ X:x(t) ≥ 0,  t ∈ [0, 1]}.

It is clear that (H1) and (H2) imply (A2) and (A3).

Let

Ω1=xX:δx<|x(t)|<b,t[0,1],Ω2=xX:x<B.

Clearly, Ω1 and Ω2 are bounded and open sets and

Ω¯1=xX:δx|x(t)|b,t[0,1]Ω2.

Moreover, C(Ω¯2\Ω1). Let JI and (γx)(t) = |x(t)| for x ∈ X, then γ is a retraction and maps subsets of Ω¯2 into bounded subsets of C, which means that (A4) holds.

Next, we will show (A5) holds. Suppose that there exist u0 ∈ Ω2 ∩ C ∩ domL and λ0 ∈ (0, 1) such that Lu0 ∈ λ0Nu0, then D0+αu0(t)λ0f(t,u0(t)) for all t ∈ [0, 1]. In view of (H3), we get that there exists w ∈ f(tu0(t)) such that

D0+αu0(t)=λ0w-λ0b1|w|+λ0b2u0(t)+λ0b3=-b1D0+αu0(t)+λ0b2u0(t)+λ0b3-b1D0+αu0(t)+b2u0(t)+b3,
4

and

D0+αu0(t)=λ0w-λ0c1u0(t)+λ0c2,
5

From (4), we obtain

0=u0(0)-u0(1)=I0+αD0+αu0(1)-b1Γ(α)01(1-s)α-1D0+αu0(s)ds+b2Γ(α)01(1-s)α-1u0(s)ds+b3Γ(α)01(1-s)α-1ds,

which gives

01(1-s)α-1D0+αu0(s)dsb2b101(1-s)α-1u0(s)ds+b3αb1.

From (5), we obtain

01(1-s)α-1u0(s)dsc2αc1.

From (3) and the equation

u0 = (IP)u0Pu0KPL(IP)u0Pu0Pu0KPLu0

we can get

u0=α01(1-s)α-1u0(s)ds+01k(t,s)D0+αu0(s)dsc2c1+01|k(t,s)|·D0+αu0(s)ds=c2c1+01|k(t,s)|(1-s)α-1·(1-s)α-1D0+αu0(s)dsc2c1+301(1-s)α-1D0+αu0(s)dsc2c1+3b2b101(1-s)α-1u0(s)ds+b3αb1c2c1+3b2c2αb1c1+3b3αb1.

Then, we have

B=u0c2c1+3b2c2αb1c1+3b3αb1,

which contradicts (H3). Hence (A5) holds.

To prove (A6), consider xKerLΩ¯2, then x(t) ≡ c on [0, 1]. Let

H(c,λ)=[I-λ(P+JQN)γ]c=c-λα01(1-s)α-1|c|ds-λα01(1-s)α-1f(s,|c|)ds=c-λ|c|-λα01(1-s)α-1f(s,|c|)ds,

for c ∈ [ - BB] and λ ∈ [0, 1]. It is easy to show that 0 ∈ H(cλ) implies c ≥ 0. Suppose 0 ∈ H(Bλ) for some λ ∈ (0, 1]. Then,

0=B-λB-λα01(1-s)α-1w(s,B)ds,

where w ∈ f(tB),  t ∈ [0, 1]. So (H3) leads to

0B(1-λ)=λα01(1-s)α-1w(s,B)dsλ(-c1B+c2)<0,

which is a contradiction. In addition, if λ = 0, then B = 0, which is impossible. Thus, H(xλ) ≠ 0 for x ∈ KerL ∩ Ω2, λ ∈ [0, 1]. As a result,

deg{[I-(P+JQN)γ]KerL,KerLΩ2,0}=deg{H(·,1),KerLΩ2,0}=deg{H(·,0),KerLΩ2,0}=deg{I,KerLΩ2,0}=10.

So (A6) holds.

Next, we prove (A7). Letting u0(t) ≡ 1, so we have u0 ∈ C\{0} and C(u0) = {x ∈ C:x(t) > 0, t ∈ [0, 1]}. We can take σ(u0) = 1. For x ∈ C(u0) ∩ Ω1, we get x(t) > 0, 0 < ‖x‖ ≤ b and x(t) ≥ δx‖,  t ∈ [0, 1].

By (H3) and (H4), for every x ∈ C(u0) ∩ Ω1 and v ∈ Ψx, there exits w ∈ Nx such that

v(t0)=α01(1-s)α-1x(s)ds+01G(t0,s)(1-s)α-1ws,x(s)dsδx+01G(t0,s)(1-s)α-1q(s)hx(s)ds=δx+01G(t0,s)(1-s)α-1q(s)·hx(s)xρ(s)xρ(s)dsδx+δρxρ01G(t0,s)(1-s)α-1q(s)·h(b)bρds=δx+δρx·b1-ρx1-ρ01G(t0,s)(1-s)α-1q(s)h(b)bdsδx+δρx·01G(t0,s)(1-s)α-1q(s)h(b)bdsδx+δρx·1-δδρ=x.

Thus, x‖ ≤ σ(u0)‖Ψx for all x ∈ C(u0) ∩ Ω1, i.e., (A7) holds.

Since for x ∈ Ω2 and w ∈ Nγx, from (H2) we have

(P+JQN)γx=α01(1-s)α-1|x(s)|ds+α01(1-s)α-1w(s,|x(s)|)dsα01(1-s)α-1(1-κ)|x(s)|ds0.

Thus, (PJQN)γx ⊂ C for x ∈ Ω2. Then (A8) holds.

Next, we prove (A9). Let xΩ¯2\Ω1

Ψγx(t)=vX:wNγxsuch thatv=α01(1-s)α-1|x|ds+01G(t,s)(1-s)α-1w(s,|x|)ds.

According to (H3) and (2), for xΩ¯2\Ω1 and v ∈ Ψγx, there exits w ∈ Nγx such that

v(t)=α01(1-s)α-1|x(s)|ds+01G(t,s)(1-s)α-1ws,|x(s)|ds>01(1-s)α-1|x(s)|1-κG(t,s)ds0.

Hence, ΨγΩ¯2\Ω1C; i.e., (A9) holds.

To prove (A10), suppose there exists u0 ∈ Ω2 ∩ KerL, i.e., u0c ∈ ℝ and |c| = B such that c ∈ (PJQN)γu. For w ∈ Nγc, we have

c=α01(1-s)α-1|c|ds+α01(1-s)α-1w(s,|c|)dsα01(1-s)α-1(1-κ)|c|ds0.

Hence, we get c ∈ (PJQN)γu implies c ≥ 0. Then for cB and w ∈ NγB, we have

B=α01(1-s)α-1Bds+α01(1-s)α-1w(s,B)ds.

Hence,

α01(1-s)α-1w(s,B)ds=0.

On the other hand, from (H3), we have

0=α01(1-s)α-1w(s,B)ds-c1B+c2<0.

This contradiction implies (A10) holds.

Hence, applying Theorem 1, BVP (1) has a positive solution u on [0, 1] with b ≤ ‖u‖ ≤ B. This completes the proof.

Example

To illustrate how our main result can be used in practice, we present here an example.

Let us consider the following fractional differential inclusion at resonance

D0+1.5u(t)f(t,u),0t1,u(0)=0,u(0)=u(1),
6

where f(t,u)=w(t,u)+125v:v[0,1], w(t,u)=13001+2t-2t2u2-4u+3u.

Corresponding to BVP (1), we have that α = 1.5 and

G(t,s)=32-Γ(2.5)Γ(3)(1-s)1.5-1.5t1.5Γ(2.5)+1.5Γ(2.5)Γ(3),0t<s1,32-Γ(2.5)Γ(3)(1-s)1.5-1.5t1.5Γ(2.5)+1.5Γ(2.5)Γ(3)+1Γ(1.5)t-s1-s0.5,0s<t1,

It is easy to see that G(ts) ≥ 0 for ts ∈ [0, 1].

Let κ = 0.003 and B = 2. By the monotonicity of the function, for x ∈ [0, 2] and w ∈ f(tx), t ∈ [0, 1], we can prove that

-31000xw(t,x)-130x+117

and

w(t,x)-83|w|+130x+14.

Then, we can choose c1=130, c2=117, b1=83, b2=130, b3 = ¼. By calculation, we have

c2c1+3b2c2αb1c1+3b3αb11.764+0.044+0.187=1.808<2=B.

Take q(t)=12401+2t-t2 and h(x) = x. We see that q ∈ L1[0, 1], q(t) ≥ 0 and h ∈ C((0, b], ℝ+),  where b = 1/2 ∈ (0, B) = (0, 2). Furthermore, for (tu) ∈ [0, 1] × (0, 1/2] and w ∈ f(tu), by a simple computation, we get that

w(tu) ≥ q(t)h(u).

Choose ρ = 1, so we have h(u)uρ1 which is non-increasing on (0, b]. By Choosing t0 = 0, δ = 0.997, with simple calculations, we can get

h(b)b01G(t0,s)(1-s)α-1q(s)ds1-δδρ.

Therefore, (H1)–(H4) of Theorem 2 are satisfied. Then BVP (6) has a positive solution on [0, 1].

Conclusions

In this paper, we have obtained the existence of positive solutions for a boundary value problem of fractional differential inclusions at resonance. By using the Leggett–Williams theorem for coincidences of multi-valued operators due to O’Regan and Zima, we have found the existence results. Our results are new in the context of fractional differential inclusions and positive solutions. As applications, an example is presented to illustrate the main results. In the future, we will consider the the uniqueness of positive solutions for the fractional differential equations at resonance.

Acknowledgements

The research was supported by the Science Foundation of Shandong Jiaotong University (Z201429).

Competing interests

The author declares that he has no competing interests.

References

  • Agarwal RP, Benchohra M, Hamani S. A survey on existence results for boundary value problems of nonlinear fractional differential equations and inclusions. Acta Appl Math. 2010;109:973–1033. doi: 10.1007/s10440-008-9356-6. [Cross Ref]
  • Bai Z. Solvability for a class of fractional m-point boundary value problem at resonance. Comput Math Appl. 2011;62:1292–1302. doi: 10.1016/j.camwa.2011.03.003. [Cross Ref]
  • Bai Z, Lü H. Positive solutions for boundary value problem of nonlinear fractional differential equation. J Math Anal Appl. 2005;311:495–505. doi: 10.1016/j.jmaa.2005.02.052. [Cross Ref]
  • Bai Z, Zhang Y. Solvability of fractional three-point boundary value problems with nonlinear growth. Appl Math Comput. 2011;218:1719–1725.
  • Caballero J, Harjani J, Sadarangani K. Positive solutions for a class of singular fractional boundary value problems. Comput Math Appl. 2011;63:1325–1332. doi: 10.1016/j.camwa.2011.04.013. [Cross Ref]
  • Chen Y, Tang X, He X. Positive solutions of fractional differential inclusion at resonace. Mediterr J Math. 2013;10:1207–1220. doi: 10.1007/s00009-013-0273-1. [Cross Ref]
  • Goodrich CS. Existence of a positive solution to a class of fractional differential equation. Appl Math Lett. 2010;23:1050–1055. doi: 10.1016/j.aml.2010.04.035. [Cross Ref]
  • Kilbas AA, Srivastava HM, Trujillo JJ. North-Holland mathematics studies. Amsterdam: Elsevier Science B.V.; 2006. Theory and applications of fractional differential equations.
  • Kosmatov N. Multi-point boundary value problems on an unbounded domain at resonance. Nonlinear Anal. 2008;68:2158–2171. doi: 10.1016/j.na.2007.01.038. [Cross Ref]
  • Kosmatov N. A boundary value problem of fractional order at resonance. Electron J Differ Equ. 2010;135:1–10.
  • Lin W. Global existence theory and chaos control of fractional differential equations. J Math Anal Appl. 2007;332:709–726. doi: 10.1016/j.jmaa.2006.10.040. [Cross Ref]
  • Mawhin J. NSFCBMS regional conference series in mathematics. Providence: American Mathematical Society; 1979. Topological degree methods in nonlinear boundary value problems.
  • O’Regan D, Zima M. Leggett–Williams theorems for coincidences of multivalued operators. Nonlinear Anal. 2008;68:2879–2888. doi: 10.1016/j.na.2007.02.034. [Cross Ref]
  • Orsingher E, Beghin L. Time-fractional telegraph equations and telegraph processes with brownian time. Probab Theory Relat Fields. 2004;128:141–160. doi: 10.1007/s00440-003-0309-8. [Cross Ref]
  • Podlubny I. Fractional differential equations. New York: Academic Press; 1999.
  • Samko SG, Kilbas AA, Marichev OI. Theory and applications. Yverdon: Gordon and Breach; 1993. Fractional integrals and derivatives.
  • Xu X, Jiang D, Yuan C. Multiple positive solutions for the boundary value problem of a nonlinear fractional differential equation. Nonlinear Anal. 2009;71:4676–4688. doi: 10.1016/j.na.2009.03.030. [Cross Ref]
  • Yang A, Wang H. Positive solutions for two-point boundary value problems of nonlinear fractional differential equation at resonace. Electron J Qual Theory Differ Equ. 2011;71:1–15.
  • Zhang S. Positive solutions for boundary-value problems of nonlinear fractional differential equations. Electron J Differ Equ. 2006;36:1–12. doi: 10.1155/ADE/2006/90479. [Cross Ref]

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