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Existence theory for sequential fractional differential equations with anti-periodic type boundary conditions

  • Mohammed H. Aqlan , Ahmed Alsaedi , Bashir Ahmad EMAIL logo and Juan J. Nieto
Published/Copyright: October 16, 2016

Abstract

We develop the existence theory for sequential fractional differential equations involving Liouville-Caputo fractional derivative equipped with anti-periodic type (non-separated) and nonlocal integral boundary conditions. Several existence criteria depending on the nonlinearity involved in the problems are presented by means of a variety of tools of the fixed point theory. The applicability of the results is shown with the aid of examples. Our results are not only new in the given configuration but also yield some new special cases for specific choices of parameters involved in the problems.

MSC 2010: 34A08; 34B10; 34B15

1 Introduction

Recently, there has been an utterly great interest in developing theoretical analysis for boundary value problems of nonlinear fractional-order differential equations supplemented with a variety of boundary conditions. It has been mainly due to nonlocal nature of fractional-order differential operators which take into account memory and hereditary properties of some important and useful materials and processes. Fractional calculus has played a key role in improving the mathematical modelling of several phenomena occurring in engineering and scientific disciplines, such as blood flow problems, control theory, aerodynamics, nonlinear oscillation of earthquake, the fluid-dynamic traffic model, polymer rheology, regular variation in thermodynamics etc. For more details and explanation, see, for instance [13]. Some recent results on fractional-order boundary value problem can be found in a series of papers [412] and the references cited therein. Sequential fractional differential equations have also received considerable attention, for instance see [1317].

Anti-periodic boundary conditions are found to be quite significant and important in the mathematical modeling of certain physical processes and phenomena, for example trigonometric polynomials in the study of interpolation problems, wavelets, physics etc. For more details, see [18] and the references cited therein. For some recent works on fractional-order anti-periodic boundary value problems, we refer the reader to [1923]. However, the study of sequential fractional differential equations equipped with anti-periodic boundary conditions is yet to be initiated.

In this paper, we study new boundary value problems of Liouville-Caputo type sequential fractional differential equation:

(cDα+kcDα1)u(t)=f(t,u(t)),1<α2,0<t<T,T>0,(1)

subject to anti-periodic type (non-separated) boundary conditions of the form:

α1u(0)+ρ1u(T)=β1,  α2u(0)+ρ2u(T)=β2,(2)

and anti-periodic type (non-separated) nonlocal integral boundary conditions:

α1u(0)+ρ1u(T)=λ10nu(s)ds+λ2,α2u(0)+ρ2u(T)=μ1ξTu(s)ds+μ2,(3)

where cDα denotes the Liouville-Caputo fractional derivative of order α, k ∈ ℝ+, 0 < η < ξ < T, α1, α2, ρ1, ρ2, β1, β2, λ1, λ2, μ1μ2 ∈ ℝ with α1 + ρ1 ≠ 0, α2 + ρ2e-kT ≠ 0 and f : [0, T] × ℝ → ℝ is a given continuous function. Instead of writing the so-called “Caputo” derivative, we will call it “Liouville-Caputo” derivative as it was introduced by Liouville many decades ago.

The rest of the paper is organized as follows. In Section 2, we recall some basic concepts of fractional calculus and obtain the integral solution for the linear variants of the given problems. Section 3 contains the existence results for problem (1)-(2) obtained by applying Schaefer’s fixed point theorem, Leray-Schauder’s nonlinear alternative, Leray-Schauder’s degree theory, Banach’s contraction mapping principle and Krasnoselskii’s fixed point theorem. In Section 4, we provide the outline for the existence results of problem (1)-(3).

2 Preliminaries and auxiliary results

This section is devoted to some basic definitions of fractional calculus [1, 2] and auxiliary lemmas.

Definition 2.1

The fractional integral of order q with the lower limit zero for a function f is defined as

Iqf(t)=1Γ(q)0tf(s)(ts)1qds,  t > 0,  q > 0,

provided the right hand-side is point-wise defined on [0,∞), whereΓ()is the gamma function, which is defined byΓ(q)=0tq1etdt.

Definition 2.2

The Riemann-Liouville fractional derivative of order q > 0, n — 1 < q < n, nN, is defined as

D0+qf(t)=1Γ(nq)(ddt)n0t(ts)nq1f(s)ds,

where the function f(t) has absolutely continuous derivative up to order (n — 1).

Definition 2.3

The Liouville-Caputo derivative of order q for a function f : [0,∞) → ℝ can be written as

cDqf(t)=Dq(f(t)k=0n1tkk!f(k)(0)),  t>0,  n1<q<n.
Remark 2.4

Iff(t)Cn[0,)then

cDqf(t)=1Γ(nq)0tf(n)(s)(ts)q+1nds=Inqf(n)(t),t>0,n1<q<n.

To define the fixed point problems associated with problems (1)-(2) and (1)-(3), we consider the following lemmas dealing with the linear variant of equation (1).

Lemma 2.5

Let hC([0, T], ℝ). Then the problem consisting of the equation

(cDα+kcDα1)u(t)=h(t)),1<α2, 0<t<T,T>0,(4)

and the boundary conditions (2) is equivalent to the integral equation

u(t)=v1(t)+0tek(ts)(0s(sx)α2Γ(α1)h(x)dx)ds+v2(t)0T(Ts)α2Γ(α1)h(s)ds                                                                 +v3(t)0Tek(Ts)(0s(sx)α2Γ(α1)h(x)dx)ds,(5)

where

v1(t)=β1(α1+ρ1)+((α1+ρ1ekT)(α1+ρ1)ekT)β2k(α1+ρ1)(α2+ρ2ekT),        v2(t)=ρ2((α1+ρ1)ekT(α1+ρ1ekT))k(α1+ρ1)(α2+ρ2ekT),       v3(t)=α1ρ2α2ρ1ρ2(α1+ρ1)ekt(α1+ρ1)(α2+ρ2ekT).(6)
Proof

As argued in [13], the general solution of (4) can be written as

u(t)=A0ekt+A1+0tek(ts)Iα1h(s)ds,(7)

where A0 and A1 are arbitrary constants and

Iα1h(t)=0t(tx)α2Γ(α1)h(x)dx.

Differentiating (7) with respect to t, we obtain

u(t)=kA0ektk0tek(ts)Iα1h(s)ds+Iα1h(t).(8)

Using the boundary conditions (2) in (7) and (8), we get

A0(α1+ρ1ekT)+A1(α1+ρ1)+ρ10Tek(Ts)Iα1h(s)ds=β1,(9)
kA0(α2+ρ2ekT)kρ20Tek(Ts)Iα1h(s)ds +ρ2Iα1h(T)=β2.(10)

Solving the system (9) and (10) for A0 and A1 we get

A0=β2k(α2+ρ2ekT)+ρ2k(α2+ρ2ekT){Iα1h(T)k0Tek(Ts)(0s(sx)α2Γ(α1)h(x)dx)ds},A1=β1(α1+ρ1)+(α1+ρ1ekT)β2k(α1+ρ1)(α2+ρ2ekT)                      ρ2(α1+ρ1ekT)k(α1+ρ1)(α2+ρ2ekT)0T(Ts)α2Γ(α1)h(s)                               +ρ2α1ρ1α2(α1+ρ1)(α2+ρ2ekT)0Tek(Ts)(0s(sx)α2Γ(α1)h(x)dx)ds.

Substituting the values of A0 and A1 in (7) yields the solution (5). Conversely, by direct computation, it can be established that (5) satisfies the equation (4) and boundary conditions (2). This completes the proof.

Lemma 2.6

Let hC([0, T], ℝ). Then the problem consisting of linear equation (4) equipped with boundary conditions (3) is equivalent to the integral equation

u(t)=B1(t){λ10n(0sek(sx)Iα1h(x)dx)dsρ10Tek(Ts)Iα1h(s)ds+λ2}                 +B2(t){μ1ξT(0sek(sx)Iα1h(x)dx)ds+kρ20Tek(Ts)Iα1h(s)ds                                                              ρ2Iα1h(T)+μ2}+0tek(ts)Iα1h(s)ds,(11)

where

B1(t)=(ϵ2ekt+δ2)Δ,   B2(t)=(ϵ1ektδ1)Δ,   Δ=δ1ϵ2+δ2ϵ1,                         δ1=α1+ρ1ekT+λ1k(ekn1),ϵ1=(α1+ρ1λ1η),                                       δ2=kα2kρ2ekT+μ1k(ekTekξ),  ϵ2=μ1(Tξ).(12)
Proof

Since the proof is similar to that of Lemma 2.5, we omit it.

3 Existence results for the problem (1)-(2)

In view of Lemma 2.5, we introduce a fixed point problem associated with the problem (1)-(2) as follows:

u=Hu,(13)

where the operator H:EE is

(Hu)(t)=v1(t)+0tek(ts)0s(sx)α2Γ(α1)f(x,u(x))dxds+v2(t)0T(Ts)α2Γ(α1)f(s,u(s))ds+v3(t)0Tek(Ts)0s(sx)α2Γ(α1)f(x,u(x))dxds.(14)

Here E=C([0,T],) denotes the Banach space of all continuous functions from [0, T] → ℝ endowed with the norm defined by u=sup{|u(t)|,t[0,T]}.

Observe that that problem (1)-(2) has solutions if the operator equation (13) has fixed points. For computational convenience, we set the notation:

Q=supt[0,T]{1ektkΓ(α)t1α+|v2(t)|Γ(α)T1α+|v3(t)|(1ekT)kΓ(α)T1α}.(15)

Now we are in a position to present our main results for the problem (1)-(2). The first one deals with Schaefer’s fixed point theorem [24].

Lemma 3.1

Let X be a Banach space. Assume thatT:XXis a completely continuous operator and the setY={uX|u=μTu,0<μ<1}is bounded. ThenThas a fixed point in X.

Theorem 3.2

Assume that there exists a positive constantL1such that|f(t,u(t))|L1fort[0,T],u. Then the boundary value problem (1)-(2) has at least one solution on [0, T].

Proof

In the first step, we show that the operator H defined by (14) is completely continuous. Observe that continuity of H follows from the continuity of f. For a positive constant r, let Br={uE:ur} be a bounded ball in E. Then for t ∈ [0, T], we have

|(Hu)(t)||v1(t)|+0tek(ts)(0s(sx)α2Γ(α1)|f(x,u(x))|dx)ds+|v2(t)|0T(Ts)α2Γ(α1)|f(s,u(s))|ds                                 +|v3(t)|0Tek(Ts)(0s(sx)α2Γ(α1)|f(x,u(x))|dx)ds.              |v1(t)|+L1{0tek(ts)(0s(sx)α2Γ(α1)dx)ds+|v2(t)|0T(Ts)α2Γ(α1)ds                                                 +|v3(t)|0Tek(Ts)(0s(sx)α2Γ(α1)dx)ds}L1Q+v1,

which consequently implies that

(Hu)L1Q+v1,

where Q is defined by (15).

Next we show that the operatorHmaps bounded sets into equicontinuous sets ofE. Let τ1, τ2 ∈ [0, T] with τ1 < τ2 and uBr. Then we have

(Hu)(τ2)(Hu)(τ1)v1(τ2)v1(τ1)+L1ekτ2ekτ10τ1eks0s(sx)α2Γ(α1)dxds+τ1τ2ek(τ2s)0s(sx)α2Γ(α1)dxds+v2(τ2)v2(τ1)0T(Ts)α2Γ(α1)ds+v3(τ2)v3(τ1)0Tek(Ts)0s(sx)α2Γ(α1)dxds.

As τ1τ2 → 0, the right-hand side of the above inequality tends to zero independently of uBr. Therefore, by the Arzelá-Ascoli theorem, the operator H:EE is completely continuous.

Finally, we consider the set V={uE:u=μHu,0<μ<1} and show that V is bounded. For uV and t ∈ [0, T], we get

uL1Q+v1.

Therefore, V is bounded. Hence, by Lemma 3.1, the problem (1)-(2) has at least one solution on [0, T].

Our next existence result is based on Leray-Schauder’s nonlinear alternative.

Lemma 3.3

(Nonlinear alternative for single valued maps [25]). Let E be a Banach space, C a closed, convex subset of E, U an open subset of C and 0 ∈ U. Suppose thatA:U¯Cis a continuous, compact (that is, A(U¯)is a relatively compact subset of C ) map. Then either

  1. Ahas a fixed point in U¯, or

  2. there is a x ∈ ∂U(the boundary of U in C) andλ ∈ (0, 1) withx=λA(x).

Theorem 3.4

Assume that

  1. there exists a continuous nondecreasing functionχ:[0,)(0,)and a functionpC([0, T], ℝ+) such that

    |f(t,u)|p(t)χ(u)foreach(t,u)[0,T]×;
  2. there exists a constant N > 0 such that

    Nχ(N)pQ+v1>1,(16)

    where Q is given by (15).

Then the boundary value problem (1)-(2) has at least one solution on [0, T].

Proof

We complete the proof in different steps. We first show that the operator Hdefined by (14) maps bounded sets (balls) into bounded sets inE. For a positive constant r, let Br={uE:ur} be a bounded ball in E. Then, for t ∈ [0, T], we have

|(Hu)(t)||v1(t)|+χ(u)p{0tek(ts)(0s(sx)α2Γ(α1)dx)ds+|v2(t)|0T(Ts)α2Γ(α1)ds                                         +|v3(t)|0Tek(Ts)(0s(sx)α2Γ(α1)dx)ds}v1+χ(u)pQ,

which implies that (Hu)v1+χ(r)pQ.

In the second step, we establish that the operatorHmaps bounded sets into equicontinuous sets ofE. As in the proof of the previous result, for τ1, τ2 ∈ [0, T] with τ1 < τ2 and uBr, we can have

(Hu)(τ2)(Hu)(τ1)0 as τ2τ10,

independently of uBr. Therefore, it follows by the Arzela-Ascoli theorem that the operator H:EE is completely continuous.

Let u be a solution. Then, for t ∈ [0, T], we have that

uχ(u)pQ+v1.

In view of (E2), there exists N such that uN. Let us set

u={uE:u<N}.

We see that the operator H:U¯E is continuous and completely continuous. From the choice of U, there is no uU such that u=θHu for some θ ∈ (0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 3.3), we deduce that H has a fixed point uU¯ which is a solution of the problem (1)-(2). This completes the proof.

The next existence result is based on Leray-Schauder’s degree theory [25].

Theorem 3.5

Letf : [0, T] × ℝ → ℝ be a continuous function. Suppose that (E3) there exist constants 0 ≤ ω < Q-1, andM1 > 0 such that

|f(t,u)|ω|u|+M1forall (t,u)[0,T]×,

where Q is given by (15).

Then the boundary value problem (1)-(2) has at least one solution on [0, T].

Proof

We have to show the existence of at least one solution uE satisfying the fixed point problem

u=H,(17)

where the operator H:EE is defined by (14). Introduce a ball BRE as

BR={uE:u<R},

with a constant radius R > 0. Hence, we will show that the operator H:B¯RE satisfies the condition

uθHu,    uBR,    θ[0,1].(18)

Set

V(θ,u)=θHu,   uE,   θ[0,1].

As argued in Theorem 3.2, the operator H is continuous, uniformly bounded and equicontinuous. Thus, by the Arzelá-Ascoli theorem, a continuous map hθ defined by hθ(u)=uV(θ,u)=uθHu is completely continuous. If (18) holds, then the following Leray-Schauder degrees are well defined and by the homotopy invariance of topological degree, it follows that

deg(hθ,BR,0)=deg(IθH,BR,0)=deg(h1,BR,0)                     =deg(h0,BR,0)=deg(I,BR,0)=10,   0BR,

where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, we have h1(u)=uHu=0 for at least one uBR. Let us assume that u=θHu for some θ ∈ [0, 1] and for all t ∈ [0, T]. Then, using the assumption (E3), it is easy to find that

|u(t)|=|θHu(t)|(ωu+M1)Q+v1,

which implies that

uM1Q+v11ωQ

If R=M1Q+v11ωQ+1 the inequality (18) holds. This completes the proof.

Next we show the existence of a unique solution of the problem (1)-(2) by applying Banach’s contraction mapping principle.

Theorem 3.6

Assume thatf : [0, T] × ℝ → ℝ is a continuous function satisfying the Lipschitz condition: (E4) there exists a positive number ℓ such that|f(t,u)f(t,υ)||uυ|,t[0,T],u,υ.

Then the boundary value problem (1)-(2) has a unique solution on [0, T] ifQ < 1/, whereQis given by (15).

Proof

Consider a set Br={uE:ur} with rQM+v11Q, where M=supt[0,T]|f(t,0)| and Q is given by (15). In the first step, we show that HBrBr, where the operator H is defined by (14). For any uBr, t ∈ [0, T], observe that

|f(t,u(t))|=|f(t,u(t))f(t,0)+f(t,0)||f(t,u(t))f(t,0)|+|f(t,0)|u+Mr+M,

where we have used the assumption (E4). Then, for uBr, we obtain

(Hu)supt[0,T]{|v1(t)+|0tek(ts)(0s(sx)α2Γ(α1)|f(x,u(x))|dx)ds+|v2(t)|0T(Ts)α2Γ(α1)|f(s,u(s))|ds+|v3(t)|0Tek(Ts)(0s(sx)α2Γ(α1)|f(x,u(x))|dx)ds.}(r+M)supt[0,T]{0tek(ts)(0s(sx)α2Γ(α1)dx)ds+|v2(t)|0T(Ts)α2Γ(α1)ds+|v3(t)|0Tek(Ts)(0s(sx)α2Γ(α1)dx)ds}+v1(r+M)Q+v1r,

which implies that HuBr. Thus HBrBr. Next we show that the operator H is a contraction. Using the assumption (E4) and (15), we get

HuHvsupt[0,T]{0tek(ts)(0s(sx)α2Γ(α1)|f(x,u(x))f(x,v(x))|dx)ds+|v2(t)|0T(Ts)α2Γ(α1)|f(s,u(s))f(s,v(s))|ds+|v3(t)|0Tek(Ts)(0s(sx)α2Γ(α1)|f(x,u(x))f(x,v(x))|dx)ds}uvsupt[0,T]{0tek(ts)(0s(sx)α2Γ(α1)dx)ds+|v2(t)|0T(Ts)α2Γ(α1)ds+|v3(t)|0Tek(Ts)(0s(sx)α2Γ(α1)dx)ds}Quv.

In view of the assumption: Q < 1/ℓ, it follows that the operator H is a contraction. Thus, by Banach’s contraction mapping principle, we deduce that the operator H has a fixed point, which in turn implies that there exists a unique solution for the problem (1)-(2) on [0, T].

In the following theorem, we show the existence of solutions for the problem (1)-(2) by applying Krasnoselskii’s fixed point theorem.

Lemma 3.7

(Krasnoselskii’s fixed point theorem [24]). Let Y be a closed bounded, convex and nonempty subset of a Banach space X. LetB1, B2be the operators such that (i)B1y1 + B2y2Ywhenevery1,y2Y;(ii)B1is compact and continuous and (iii)B2is a contraction mapping. Then there existszYsuch thatz = B1z + B2z.

Theorem 3.8

Letf : [0,T] × ℝ → ℝ be a continuous function satisfying the condition (E4) and that|f(t,x)|g(t)(t,x) ∈ [0, T] × ℝ withgC([0, T], ℝ+) andsupt[0,T]|g(t)|=g . In addition, it is assumed thatQ1<1/, where

Q1=supt[0,T]{|v2(t)Γ(α)T1α+|v3(t)|(1ekT)kΓ(α)T1α}.(19)

Then problem (1)-(2) has at least one solution on [0, T].

Proof

Consider Ba={uE:ua}, where aQg+v1 with Q given by (15). We define the operators H1 and H2 on Ba as

       (H1u)(t)=0tek(ts)(0s(sx)α2Γ(α1)f(x,u(x))dx)ds(H2u)(t)=v1(t)+v2(t)0T(Ts)α2Γ(α1)f(s,u(s))ds            +v3(t)0Tek(Ts)(0s(sx)α2Γ(α1)f(x,u(x))dx)ds.

For u, υBa, it is easy to verify that H1u+H2υQg+v1, where Q is given by (15). Thus, H1u+H2υBa . Using the assumption (E4) and (19), we can get H1u+H2υQ1uυ, which implies that H2 is a contraction in view of the given condition: Q1<1/.

Notice that continuity of f implies that the operator H1 is continuous. Also, H1 is uniformly bounded on Ba as

H1u(1ekT)Tα1gkΓ(α).

Next, it will be shown that the operator H1 is compact. Fixing sup(t,u)[0,T]×Ba|f(t,u)|=fa and for t1, t2 ∈ [0, T] (t1 < t2), consider

(H1u)(t2)(H1u)(t1)fa(|ekt2ekt1|0t1eks(0s(sx)α2Γ(α1)dx)ds+t1t2ek(t2s)(0s(sx)α2Γ(α1)dx)ds)0 as t2t10,

independent of u. This implies that H1 is relatively compact on Ba. Hence, by the Arzelá-Ascoli Theorem, the operator H1 is compact on Ba. Thus all the assumptions of Lemma (3.7) are satisfied. In consequence, by the conclusion of Lemma (3.7), the problem (1)-(2) has at least one solution on [0, T].

Example 3.9

Consider the following anti-periodic fractional boundary value problem:

{(cD3/2+1.5cD1/2)u(t)=f(t,u(t)),t[0,2],1.25u(0)+4u(2)=1,0.5u(0)2u(2)=2.5.(20)

Here T = 2, k = 1.5, α1 = 1.25, ρ1 = 4, β1 = —1, α2 = 0.5, ρ2 = —2, β2 = 2.5. With the given values, we find that Q ≈ 7.742915 (Q is given by (15)).

(a) Let

f(t,u)=eu2(cos2(3u+2)t2+9+tsint1+u2+2t+3).(21)

Clearly |f(t,u(t))|3=L1 for all t ∈ [0, 2], u ∈ ℝ. Thus, by Theorem 3.2, the problem (20) with f(t,u) given by (21) has at least one solution on [0, 2].

(b) Letting

f(t,u)=et27(|u|31+|u|3+|u|1+|u|+1t+1),(22)

we have |f(t,u)|et/9=p(t)χ(u) . Selecting χ(u)=1 and p(t)=et/9(p=1/9), we find that the assumption (E2) holds true for N > 4.064141. As all the conditions of Theorem 3.4 are satisfied, there exists at least one solution of the problem (20) with f(t, u) given by (22) on [0, 2].

(c) Let us take

f(t,u)=1t2+100sinu+1t+2.(23)

Then |f(t,u)|(1/10)u+1/2 implies that ω = 1/10, M1 = 1/2. Clearly ω < 1/Q(Q ≈ 7.742915). In consequence, the conclusion of Theorem 3.5 applies and the problem (20) with f(t, u) given by (23) has a solution on [0, T].

(d) Let us choose

f(t,u)=110tan1u(t)+cost.(24)

Clearly =1/10 as |f(t,u)f(t,υ)|110|uυ| and Q0.774292<1 . Thus all the conditions of Theorem 3.6 are satisfied. Hence we deduce by the conclusion of Theorem 3.6 that there exists a unique solution for the problem (20) with f(t, u) given by (24).

For the applicability of Theorem 3.8, we find that |f(t,u)|g(t)=π/20+cost with g=(20+π)/20 and Q1 ≈ 6.732035 (Q1 is given by (19)). Obviously Q10.673203<1 . Thus all the conditions of Theorem 3.8 are satisfied. Hence the conclusion of Theorem 3.8 implies that the problem (20) with f(t, u) given by (24) has at least one solution on [0, 2].

Remark 3.10

By fixing the parameters involved in the boundary conditions(2), we can obtain some new special results for different problems arising from the problem (1)-(2). For instance, forα1 = α2 = ρ1 = ρ2 = 1, β1 = β2 = 0, we obtain the existence results for sequential fractional differential equation (1) with anti-periodic boundary condition: u(0) + u(T) = 0, u′(0) + u′(T) = 0. Our results correspond to the ones obtα1ned in [14] foru(0) = a, u′(0) = u′(1) by takingα1 = 1 = α2, ρ1 = 0, ρ2 = —1, β1 = a, β2 = 0.

4 Existence results for the problem (1)-(3)

In this section, we present some existence results for the problem (1)-(3). We omit the proofs as the method of proof is similar to the one employed in the previous section. First of all, by Lemma 2.6, we define a fixed point operator G:EE associated with the problem (1)-(3) as

(Gu)(t)=B1(t){λ10η(0sek(sx)Iα1h(x)dx)dsρ10Tek(Ts)Iα1h(s)ds+λ2}+B2(t){μ1ξT(0sek(sx)Iα1h(x)dx)ds+kρ20Tek(Ts)Iα1h(s)dsρ2Iα1h(T)+μ2}+0tek(ts)Iα1h(s)ds(25)

where B1(t) and B2(t) are given by (12).

Using the operator (25) and the method of proof for the results obtained in the last section, we can establish the following results for the problem (1)-(3).

Theorem 4.1

Letf : [0,T] × ℝ → ℝ be a continuous function satisfying the assumption (E4). Then the boundary value problem (1)-(3) has a unique solution on [0, T] ifQ¯<1/, where

Q¯=supt[0,T]{|B1(t)[λ1kΓ(α)(ηαα+ηα1(ekη1)k)ρ1Tα1(1ekT)kΓ(α)]|+|B2(t)[μ1kΓ(α)(Tαξαα+Tα1(ekTekξ)k)ρ2Tα1Γ(α)]|+(1ekt)tα1kΓ(α)}.(26)
Theorem 4.2

Letf : [0,T] × ℝ → ℝ be a continuous functions satisfying the condition (E4) and that|f(t,x)|g¯(t), (t,x)[0,T]×withg¯C([0,T],+) .In addition, it is assumed thatQ¯<1/, where

Q¯1=supt[0,T]{|B1(t)[λ1kΓ(α)(ηαα+ηα1(ekη1)k)ρ1Tα1(1ekT)kΓ(α)]|+|B2(t)[μ1kΓ(α)(Tαξαα+Tα1(ekTekξ)k)ρ2Tα1Γ(α)]|}.(27)

Then problem (1)-(3) has at least one solution on [0,T].

Theorem 4.3

Letf : [0,T] × ℝ → ℝ be a jointly continuous function. Assume that (E5) there exists a continuous nondecreasing functionψ : [0,) → (0,) and a functionφ ∈ C([0, T], ℝ+) such that

|f(t,u)|φ(t)ψ(u)  foreach(t,u)[0,T]×;

(E6) there exists a constantN1 > 0 such that

N1ψ(N1)φQ¯+Q^>1,(28)

whereQ¯is given by (26),

Q^=supt[0,T]{|λ2B1(t)+μ2B2(t)|}.(29)

Then the boundary value problem (1)-(3) has at least one solution on [0,T].

Theorem 4.4

Letf : [0,T] × ℝ → ℝ be a continuous function. Suppose that (E7) there exist constants0ω<1/Q¯, andM2 > 0 such that

|f(t,u)|ω1|u|+M2  forall(t,u)[0,T]×,

whereQ¯is given by (26).

Then the boundary value problem (1)-(3) has at least one solution on [0, T].

Theorem 4.5

Letf : [0,T] × ℝ → ℝ be a continuous function. Assume that there exists a positive constantL2suchthat|f(t,u(t))|L2for t ∈ [0,T], u ∈ ℝ. Then the boundary value problem (1)-(3) has at least one solution on [0, T].

Example 4.6

Consider the following anti-periodic fractional boundary value problem:

{(cD3/2+1.5cD1/2)u(t)=1t+25((5t)sinu30+etcost),t[0,2],1.25u(0)+4u(2)=01u(s)ds1,0.5.u(0)2u(2)=21.252u(s)ds+2.5,(30)

wheref(t,u(t))=1t+25((5t)sinu15+etcost), T = 2, η = 1, ξ = 1.25, k = 1.5, α1 = 1.25, ρ1 = 4, λ1 = —1, λ2 = —1, λ2 = 0.5, ρ2 = —2, μ1 = 2, μ2 = 2.5.

With the given values, we find thatQ¯14.422595 ( Q¯is given by(26)) and=1/30as|f(t,u)f(t,v)|130|uυ| . Clearly, Q¯<1/ .Thus all the conditions of Theorem 4.1 are satisfied. Hence we deduce by the conclusion of Theorem 4.1 that there exists a unique solution for the problem (30).

For the applicability of Theorem 4.2, we find that|f(t,u)|g¯(t)=1t+25(5t30+1)withg¯=730andQ¯113.411715 ( Q¯1is given by(27)). ObviouslyQ¯1<1/ .Thus all the conditions of Theorem 4.2 are satisfied. Hence the conclusion of Theorem 4.2 applies to the problem (30).

To illustrate Theorem 4.3, we take|f(t,u)|φ(t)ψ(u)1t+25(5t30+1), φ(t)=1t+25(5t30+1), ψ(u)=1withφ=730andN>ψ(N)φQ¯+Q^=9.333636(Q¯andQ^are given by(26)and(29)respectively). Thus all the conditions of Theorem 4.3 are satisfied. Hence the conclusion of Theorem 4.3 implies that the problem (30) has at least one solution on [0, T].

Remark 4.7

Several special cases of the existence results for the problem (1)-(3) follow by fixing the values of the parameters involved in the problem. For example, by takingα1= 1 = α2, ρ1 = 0 = ρ2, λ1 = 1 = μ1, λ2 = 0 = μ2, the results of this section correspond to the conditions: u(0)=0nu(s)ds, u(0)=ξTu(s)ds . In case we takeα1 = 0 = α2, ρ1 = 1 = ρ2, λ1 = 1 = μ1, λ2 = 0 = μ2, we obtain the results for terminal-point conditions: u(T)=0nu(s)ds, u(T)=ξTu(s)ds . Lettingα1 = 1 = α2, ρ1 = 1 = ρ2, λ1 = 1/eta, μ1 = 1/((Tξ), λ2 = 0 = μ2, we get the results for the average valued (integral) conditions: u(0)+u(T)=(1/η)0nu(s)ds, u(0)+u(T)=1/(Tξ)ξTu(s)ds. By takingα = 2, our results correspond to the equation: (D2 + kD)u(t) = f (t, u(t)), 0 < t < T, T > 0, which are also new.

Acknowledgement

The research of J.J. Nieto was partially supported by the Ministerio de Economia y Competi-tividad of Spain under grant and MTM2013-43014-P, and Xunta de Galicia under grant GRC 2015/004.

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Received: 2016-4-12
Accepted: 2016-7-27
Published Online: 2016-10-16
Published in Print: 2016-1-1

© 2016 Aqlan et al., published by De Gruyter Open

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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