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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

Multi-term fractional differential equations with nonlocal boundary conditions

Bashir Ahmad
  • Corresponding author
  • Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Jeddah Saudi Arabia
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  • De Gruyter OnlineGoogle Scholar
/ Najla Alghamdi
  • Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Jeddah Saudi Arabia
  • Department of Mathematics, Faculty of Science, University of Jeddah, P.O. Box 80327, Jeddah 21589, Jeddah Saudi Arabia
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/ Ahmed Alsaedi
  • Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Jeddah Saudi Arabia
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/ Sotiris K. Ntouyas
  • Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Jeddah Saudi Arabia
  • Department of Mathematics, University of Ioannina, 451 10 Ioannina, Ioannina Greece
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Published Online: 2018-12-31 | DOI: https://doi.org/10.1515/math-2018-0127

Abstract

We introduce and study a new kind of nonlocal boundary value problems of multi-term fractional differential equations. The existence and uniqueness results for the given problem are obtained by applying standard fixed point theorems. We also construct some examples for demonstrating the application of the main results.

Keywords: Caputo fractional derivative; multi-term fractional derivatives; existence; uniqueness; fixed point theorems

MSC 2010: 34A08; 34B10; 34B15

1 Introduction

Non-integer (arbitrary) order calculus has been extensively studied by many researchers in the recent years. The literature on the topic is now much enriched and contains a variety of results. The overwhelming interest in this branch of mathematical analysis results from its extensive applications in modeling several real world problems occurring in natural and social sciences. The mathematical models based on the tools of fractional calculus provide more insight into the characteristics of the associated phenomena in view of the nonlocal nature of fractional order operators in contrast to integer order operators. Examples include bioengineering [14], physics [9], thermoelasticity [15], etc. Boundary value problems of fractional order differential equations and inclusions have also attracted a significant attention and one can find a great deal of work on the topic involving different kinds of boundary conditions, for instance, see [1, 2, 3, 4, 5, 6, 7,] and the references cited therein.

Besides the equations involving only one differential operator, there are certain equations containing more than one differential operators. Such equations are called multi-term differential equations, see [5, 12, 13, 16].

In this paper we investigate a new type of boundary value problems of multi-term fractional differential equations and nonlocal three-point boundary conditions. Precisely, we consider the following problem:

(a2cDq+2+a1cDq+1+a0cDq)x(t)=f(t,x(t)),0<q<1,0<t<1,(1)

x(0)=0,x(η)=0,x(1)=0,0<η<1,(2)

where cDq denotes the Caputo fractional derivative of order q, f : [0, 1]×ℝ ➜ ℝ is a given continuous function and ai (i = 0, 1, 2) are real positive constants.

We prove the existence of solutions for the problem (1)-(2) by means of Krasnoselskii’s fixed point theorem and Leray-Schauder nonlinear alternative, while the uniqueness of solutions is established by Banach fixed point theorem. These results are presented in Section 3. An auxiliary lemma concerning the linear variant of (1)-(2) and some definitions are given in Section 2. Section 4 contains illustrative examples for the main results.

2 Basic results

We begin this Section with some definitions [10].

Definition 2.1

The Riemann-Liouville fractional integral of order τ > 0 of a function h : (0, ∞) ➝ ℝ is defined by

Iτh(u)=0u(uv)τ1Γ(τ)h(v)dv,u>0,

provided the right-hand side is point-wise defined on (0,∞), where Г is the Gamma function.

Definition 2.2

The Caputo derivative of order τ for a function h : [0, ∞) → R with h(x) 2 Cn[0, ∞) is defined by

cDτh(u)=1Γ(nτ)0uh(n)(v)(uv)τ+1ndv=Inτh(n)(u),t>0,n1<τ<n,

Property 2.1

With the given notations, the following equality holds:

Iτ(cDτh(u))=h(u)c0c1u...cn1un1,u>0,n1<τ<n,(3)

where ci (i = 1, ..., n − 1) are arbitrary constants.

The following lemma facilitates the transformation of the problem (1)-(2) into a fixed point problem.

Lemma 2.1

For any y ∈ C([0, 1], ℝ), the solution of linear multi-term fractional differential equation

(a2cDq+2+a1cDq+1+a0cDq)x(t)=y(t),0<q<1,0<t<1,(4)

supplemented with the boundary conditions (2) is given by

(i)x(t)=1a2(m2m1){0t0sΦ(t)(su)q1Γ(q)y(u)duds+σ1(t)010sΦ(1)(su)q1Γ(q)y(u)duds+σ2(t)0η0sΦ(η)(su)q1Γ(q)y(u)duds},ifa124a0a2>0,(5)

(ii)x(t)=1a2{0t0sΨ(t)(su)q1Γ(q)y(u)duds+ψ1(t)010sΨ(1)(su)q1Γ(q)y(u)duds+ψ2(t)0η0sΨ(η)(su)q1Γ(q)y(u)duds},ifa124a0a2=0,(6)

(iii)x(t)=1a2β{0t0sΩ(t)(su)q1Γ(q)y(u)duds+φ1(t)010sΩ(1)(su)q1Γ(q)y(u)duds+φ2(t)0η0sΩ(η)(su)q1Γ(q)y(u)duds},ifa124a0a2<0,(7)

where

Φ(κ)=em2(κs)em1(κs),κ=t,1,andη,m1=a1a124a0a22a2,m2=a1+a124a0a22a2,σ1(t)=γ2ρ2(t)γ4ρ1(t)μ,σ2(t)=γ3ρ1(t)γ1ρ2(t)μ,μ=γ1γ4γ2γ30,ρ1(t)=m2(1em1t)m1(1em2t)a2m1m2(m2m1),ρ2(t)=em1tem2t,γ1=m2(1em1)m1(1em2)a2m1m2(m2m1),γ2=m2(1em1η)m1(1em2η)a2m1m2(m2m1),γ3=em1em2,γ4=em1ηem2η,(8)

Ψ(κ)=(κs)em(κs),κ=t,1,andη,ψ1(t)=(tη)em(t+η)temt+ηemηΛ,ψ2(t)=(1t)em(t+1)+temtemΛ,Λ=(η1)em(η+1)ηemη+em0,m=a12a2,(9)

Ω(κ)=eα(κs)sinβ(κs),κ=t,1,andη,α=a12a2,β=4a0a2a122a2,φ1(t)=ω4ϱ1(t)ω2ϱ2(t)Ω,φ2(t)=ω1ϱ2(t)ω3ϱ1(t)Ω,ϱ1(t)=ββeαtcosβtαeαtsinβtα2+β2,ϱ2(t)=a2βeαtsinβt,ω1=ββeαcosβαeαsinβα2+β2,ω2=ββeαηcosβηαeαηsinβηα2+β2,ω3=a2βeαsinβ,ω4=a2βeαηsinβη,Ω=ω2ω3ω1ω40.(10)

Proof. Case (i): a124a0a2>0.

Applying the operator Iq on (4) and using (3), we get

(a2D2+a1D+a0)x(t)=0t(ts)q1Γ(q)y(s)ds+c1,(11)

where c1 is an arbitrary constant. By the method of variation of parameters, the solution of (11) can be written as

x(t)=c2em1t+c3em2t1a2(m2m1)0tem1(ts)(0s(su)q1Γ(q)y(u)du+c1)ds+1a2(m2m1)0tem2(ts)(0s(su)q1Γ(q)y(u)du+c1)ds,(12)

where m1 and m2 are given by (8). Using x(0) = 0 in (12), we get

x(t)=c1[m2(1em1t)m1(1em2t)a2m1m2(m2m1)]+c2(em1tem2t)1a2(m2m1)[0t(em1(ts)em2(ts))(0s(su)q1Γ(q)y(u)du)ds],(13)

which together with the conditions x(1) = 0 and x(η) = 0 yields the following system of equations in the unknown constants c1 and c2:

c1γ1+c2γ3=1a2(m2m1)01(em1(1s)em2(1s))(0s(su)q1Γ(q)y(u)du)ds,c1γ2+c2γ4=1a2(m2m1)0η(em1(ηs)em2(ηs))(0s(su)q1Γ(q)y(u)du)ds.

Solving the above system together with the notations (8), we find that

c1=1a2μ(m2m1)[γ401(em1(1s)em2(1s))(0s(su)q1Γ(q)y(u)du)dsγ30η(em1(ηs)em2(ηs))(0s(su)q1Γ(q)y(u)du)ds],

and

c2=1a2μ(m2m1)[γ10η(em1(ηs)em2(ηs))(0s(su)q1Γ(q)y(u)du)dsγ201(em1(1s)em2(1s))(0s(su)q1Γ(q)y(u)du)ds].

Substituting the value of c1 and c2 in (13), we obtain the solution (5). The converse of the lemma follows by direct computation.

The other two cases can be treated in a similar manner. This completes the proof. □

3 Existence and Uniqueness Results

Let 𝒞 = C([0, 1], ℝ) denote the Banach space of all continuous functions from [0, 1] to ℝ equipped with the norm defined by ∥u∥ = sup {|u(t)| : t ∈ [0, 1]}.

By Lemma 2.1, we transform the problem (1)-(2) into equivalent fixed point problems according to the cases (i) − (iii) as follows.

  • (i) For a12 − 4a0a2 > 0, we define

x=Jx,(14)

where the operator 𝒥 : 𝒞 ➝ 𝒞 is given by

(Jx)(t)=1a2(m2m1){0t0sΦ(t)(su)q1Γ(q)f(u,x(u))duds+σ1(t)010sΦ(1)(su)q1Γ(q)f(u,x(u))duds+σ2(t)0η0sΦ(η)(su)q1Γ(q)f(u,x(u))duds},(15)

where Φ(·), σ1(t) and σ2(t) are defined by (8).

  • (ii) In case a12 − 4a0a2 = 0, we have

x=Hx,(16)

where the operator 𝓗 : 𝒞 ➝ 𝒞 is given by

(Hx)(t)=1a2{0t0sΨ(t)(su)q1Γ(q)f(u,x(u))duds+ψ1(t)010sΨ(1)(su)q1Γ(q)f(u,x(u))duds+ψ2(t)0η0sΨ(η)(su)q1Γ(q)f(u,x(u))duds},(17)

where Ѱ(·), ѱ1(t) and ѱ2(t) are given by (9).

  • (iii)When a12 − 4a0a2 < 0, let us define

x=Kx,(18)

where the operator 𝒦 : 𝒞 ➝ 𝒞 is given by

(Kx)(t)=1a2β{0t0sΩ(t)(su)q1Γ(q)f(u,x(u))duds+φ1(t)010sΩ(1)(su)q1Γ(q)f(u,x(u))duds+φ2(t)0η0sΩ(η)(su)q1Γ(q)f(u,x(u))duds},(19)

where Ω(·), φ1(t) and φ2(t) are defined by (10).

In the sequel, for the sake of computational convenience, we set

σ^1=maxt[0,1]|σ1(t)|,σ^2=maxt[0,1]|σ2(t)|,ε=maxt[0,1]|m2(1em1t)m1(1em2t)a2m1m2(m2m1)|,λ=1Γ(q+1){ε+σ^1γ1+ηqσ^2γ2},λ1=λεΓ(q+1),(20)

ψ^1=maxt[0,1]|ψ1(t)|,ψ^2=maxt[0,1]|ψ2(t)|,μ=1a2m2Γ(q+1){(1+ψ^1)((m1)em+1)+ψ^2ηq((mη1)emη+1)},μ1=μ((m1)em+1)a2m2Γ(q+1),(21)

φ^1=maxt[0,1]|φ1(t)|,φ^2=maxt[0,1]|φ2(t)|,ρ=1a2(α2+β2)Γ(q+1){(1+φ^1)(1eαcosβ(α/β)eαsinβ)+φ^2ηq(1eαηcosβη(α/β)eαηsinβη)},ρ1=ρ(1eαcosβ(α/β)eαsinβ)a2(α2+β2)Γ(q+1).(22)

Before presenting our first existence result for the problem (1)-(2), let us state Krasnoselskii’s fixed point theorem [11] that plays a key role in its proof.

Theorem 3.1

(Krasnoselskii’s fixed point theorem). Let Y be a bounded, closed, convex, and nonempty subset of a Banach space X. Let F1 and F 2 be the operators satisfying the conditions: (i) F1y1 + F 2y2 ϵ Y whenever y1, y2 ϵ Y; (ii) F1 is compact and continuous; (iii) F 2 is a contraction mapping. Then there exists y ∈ Y such that y = F1y + F2y.

In the forthcoming analysis, we need the following assumptions:

(A1) |f (t, x) − f (t, y)|ℓ|xy|, for all t ϵ [0, 1], x, y ϵ ℝ, > 0.

(A2) |f (t, x)|ϑ(t), for all (t, x) ϵ [0, 1] × ℝ and ϑ ϵ C([0, 1], ℝ+).

Theorem 3.2

Let f : [0, 1] ×ℝ ➝ ℝ be a continuous function satisfying the conditions (A1) and (A2). Then the problem (1)-(2) has at least one solution on [0, 1] provided that

  • (i) ℓλ1 < 1 for a12 − 4a0a2 > 0, where λ1 is given by (20);

  • (ii) ℓμ1 < 1 for a12 − 4a0a2 = 0, where μ1 is given by (21);

  • (iii) ℓρ1 < 1 for If a12 − 4a0a2 < 0, where ρ1 is given by (22).

Proof. (i) Setting supt∈[0,1]|ϑ(t)| = ∥ϑ∥ and choosing

r1ϑΓ(q+1){ε+σ^1γ1+ηqσ^2γ2},(23)

we consider a closed ball Br1 = {x ϵ 𝒞 : ∥x∥ ≤ r1g. Introduce the operators 𝒥1 and 𝒥2 on Br1 as follows:

(J1x)(t)=1a2(m2m1)0t0sΦ(t)(su)q1Γ(q)f(u,x(u))duds,(24)

(J2x)(t)=1a2(m2m1){σ1(t)010sΦ(1)(su)q1Γ(q)f(u,x(u))duds+σ2(t)0t0sΦ(η)(su)q1Γ(q)f(u,x(u))duds}.(25)

Observe that 𝒥 = 𝒥1 + 𝒥2. For x, y ϵ Br1 , we have

J1x+J2y=supt[0,1]|(J1x)(t)+(J2y)(t)|1a2(m2m1)supt[0,1]{0t0sΦ(t)(su)q1Γ(q)|f(u,x(u))|duds+|σ1(t)|010sΦ(1)(su)q1Γ(q)|f(u,y(u))|duds+|σ2(t)|0η0sΦ(η)(su)q1Γ(q)|f(u,y(u))|duds}ϑa2(m2m1)Γ(q+1)supt[0,1]{tq0t(em2(ts)em1(ts))ds+|σ1(t)|01(em2(1s)em1(1s))ds+ηq|σ2(t)|0η(em2(ηs)em1(ηs))ds}ϑΓ(q+1){ε+σ^1γ1+ηqσ^2γ2}r1,

where we used (23). Thus 𝒥 1x + 𝒥 2y ϵ Br1 . Using the assumption (A1) together with the condition ℓλ1 < 1, we can show that 𝒥2 is a contraction as follows:

J2xJ2y=supt[0,1]|(J2x)(t)(J2y)(t)|1a2(m2m1)supt[0,1]{|σ1(t)|010sΦ(1)(su)q1Γ(q)|f(u,x(u))f(u,y(u))|duds+|σ2(t)|0η0sΦ(η)(su)q1Γ(q)|f(u,x(u))f(u,y(u))|duds}a2(m2m1)Γ(q+1)supt[0,1]{|σ1(t)|01(em2(1s)em1(1s))ds+ηq|σ2(t)|0η(em2(ηs)em1(ηs))ds}xyΓ(q+1){σ^1γ1+ηqσ^2γ2}xy=λ1xy.

Note that continuity of f implies that the operator 𝒥1 is continuous. Also, 𝒥1 is uniformly bounded on Br1 as

J1x=1a2(m2m1)supt[0,1]|0t0sΦ(t)(su)q1Γ(q)f(u,x(u))duds|ϑεΓ(q+1).

Now we prove the compactness of operator 𝒥1.We define sup(t,x)[0,1]×Br1|f(t,x)|=f¯.we have

|(J1x)(t2)(J1x)(t1)|=1a2(m2m1)|0t10s[Φ(t2)Φ(t1)](su)q1Γ(q)f(u,x(u))duds+t1t20sΦ(t2)(su)q1Γ(q)f(u,x(u))duds|f¯a2m1m2(m2m1)Γ(q+1){(t1qt2q)(m1(1em2(t2t1))m2(1em1(t2t1)))+t1q(m1(em2t2em2t1)m2(em1t2em1t1))}0ast2t10,

independent of x. Thus 𝒥1 is relatively compact on Br1 .Hence, by the Arzelá-Ascoli Theorem, 𝒥1 is compact on Br1 . Thus all the assumptions of Theorem 3.1 are satisfied. So, by the conclusion of Theorem 3.1, the problem (1)-(2) has at least one solution on [0, 1].

  • (ii) Let us consider Br2 = {x ϵ 𝒞 : ∥x∥ ≤ r2}, where supt2[0,1] (t)| = ∥ϑ∥ and

r2ϑa2m2Γ(q+1){(1+ψ^1)((m1)em+1)+ψ^2ηq((mη1)emη+1)}.(26)

Introduce the operators 𝓗1 and 𝓗2 on Br2 as follows:

(H1x)(t)=1a20t0sΨ(t)(su)q1Γ(q)f(u,x(u))duds,(27)

(H2x)(t)=1a2{ψ1(t)010sΨ(1)(su)q1Γ(q)f(u,x(u))duds+ψ2(t)0η0sΨ(η)(su)q1Γ(q)f(u,x(u))duds}.(28)

Observe that 𝓗 = 𝓗1 + 𝓗2. For x, y ϵ Br2 , we have

H1x+H2y=supt[0,1]|(H1x)(t)+(H2y)(t)|1a2supt[0,1]{0t0sΨ(t)(su)q1Γ(q)|f(u,x(u))|duds+ψ1(t)010sΨ(1)(su)q1Γ(q)|f(u,y(u))|duds+ψ2(t)0η0sΨ(η)(su)q1Γ(q)|f(u,y(u))|duds}ϑa2Γ(q+1)supt[0,1]{tq0t(ts)em(ts)ds+|ψ1(t)|01(1s)em(1s)ds+|ψ2(t)|ηq0η(ηs)em(ηs)ds}ϑa2m2Γ(q+1){(1+ψ^1)((m1)em+1)+ψ^2ηq((mη1)emη+1)}r2,

where we used (26). Thus 𝓗1x + 𝓗2y ϵ Br2 . Using the assumption (A1) together with ℓμ1 < 1, it is easy to show that 𝓗2 is a contraction. Note that continuity of f implies that the operator 𝓗1 is continuous. Also, 𝓗1 is uniformly bounded on Br2 as

H1x=supt[0,1]|(H1x)(t)|ϑa2m2Γ(q+1){(m1)em+1}.

Now we prove the compactness of operator 𝓗1. For that, let sup(t,x)[0,1]×Br2|f(t,x)|=f¯.Thus, for 0 < t1 < t2 < 1, we have

|(H1x)(t2)(H1x)(t1)|=1a2|0t10s(Ψ(t2)Ψ(t1))(su)q1Γ(q)f(u,x(u))duds+t1t20sΨ(t2)(su)q1Γ(q)f(u,x(u))duds|f¯a2m2Γ(q+1){t1q(mt2emt2mt1emt1+emt1emt2m(t2t1)em(t2t1)+em(t2t1)1))+t2q(m(t2t1)(em(t2t1)em(t2t1)+1)}0ast2t10,

independent of x. Thus, 𝓗1 is relatively compact on Br2 . Hence, by the Arzelá-Ascoli Theorem, 𝓗1 is compact on Br2 . Thus all the assumption of Theorem 3.1 are satisfied. So the conclusion of Theorem 3.1 applies and hence the problem (1)-(2) has at least one solution on [0, 1].

  • (iii) As before, letting supt2[0,1] (t)| = ∥ϑ∥ and

r3ϑa2(α2+β2)Γ(q+1){(1+φ^1)(1eαcosβ(α/β)eαsinβ)+φ^2ηq(1eαηcosβη(α/β)eαηsinβη)},(29)

we consider Br3 = {x ϵ 𝒞 : ∥x∥ ≤ r3}. Define the operators 𝒦1 and 𝒦2 on Br3 as follows:

(K1x)(t)=1a2β{0t0sΩ(t)(su)q1Γ(q)f(u,x(u))duds},(30)

(K2x)(t)=1a2β{φ1(t)010sΩ(1)(su)q1Γ(q)f(u,x(u))dusds+φ2(t)0η0sΩ(η)(su)q1Γ(q)f(u,x(u))duds}.(31)

Observe that 𝒦 = 𝒦1 + 𝒦2. For x, y ϵ Br3 , as before, it can be shown that 𝒦1x + 𝒦2y ϵ Br3 . Using the assumption (A1) together with ℓρ1 < 1, we can show that 𝒦2 is a contraction. Also, 𝒦1 is uniformly bounded on Br3 as

K1x=supt[0,1]|(K1x)(t)|ϑa2(α2+β2)Γ(q+1)(1eαcosβα/βeαsinβ).

Fixing sup(t,x)ϵ[0,1]×Br3 |f (t, x)| = , we have

|(K1x)(t2)(K1x)(t1)|f¯a2(α2+β2)Γ(q+1){t1q((α/β)sinβ(t2t1)eα(t2t1)(α/β)sinβt2eαt2+(α/β)sinβt1eαt1+cosβ(t2t1)eα(t2t1)cosβt2eαt2+cosβt1eαt11)+t2q(1(α/β)sinβ(t2t1)eα(t2t1)cosβ(t2t1)eα(t2t1))},

which is independent of x and tends to zero as t2t1 ➝ 0 (0 < t1 < t2 < 1). Thus, employing the earlier arguments, 𝒦1 is compact on Br3 . In view of the foregoing arguments, it follows that the problem (1)-(2) has at least one solution on [0, 1]. The proof is completed. □

Remark 3.1

In the above theorem, we can interchange the roles of the operators

  • (1) 𝒥1 and 𝒥2 to obtain a second result by replacing ℓλ1 < 1 with the condition:

  • εΓ(q+1)<1;

  • (2) 𝓗1 and 𝓗2 to obtain a second result by replacing ℓμ1 < 1 with the condition:

  • {(m1)em+1}a2m2Γ(q+1)<1;

  • (3) 𝒦1 and 𝒦2 to obtain a second result by replacing ℓρ1 < 1 with the condition:

  • {1eαcosβ(α/β)eαsinβ}a2(α2+β2)Γ(q+1)<1.

Now we establish the uniqueness of solutions for the problem (1)-(2) by means of Banach’s contraction mapping principle.

Theorem 3.3

Assume that f : [0, 1] × ℝ ➝ ℝ is a continuous function such that (A1) is satisfied. Then the problem (1)-(2) has a unique solution on [0, 1] if

  • (i) ℓλ < 1 for a12 − 4a0a2 > 0, where λ is given by (20);

  • (ii) ℓμ < 1 for a12 − 4a0a2 = 0, where μ is given by (21);

  • (iii) ℓρ < 1 for If a12 − 4a0a2 < 0, where ρ is given by (22).

Proof. (i) Let us define supt2[0,1] |f (t, 0)| = M and select κ1λM1λshow that 𝒥Bk1 ⊂ Bk1 , where Bk1 = {x ϵ 𝒞 : ∥x∥ ≤ k1g and 𝒥 is defined by (15). Using the condition (A1), we have

|f(t,x)|=|f(t,x)f(t,0)+f(t,0)||f(t,x)f(t,0)|+|f(x,0)|x+Mκ1+M.(32)

Then, for x ϵ Bk1 , we obtain

J(x)=supt[0,1]|J(x)(t)|1a2(m2m1)supt[0,1]{0t0sΦ(t)(su)q1Γ(q)|f(u,x(u))|duds+|σ1(t)|010sΦ(1)(su)q1Γ(q)|f(u,x(u))|duds+|σ2(t)|0η0sΦ(η)(su)q1Γ(q)|f(u,x(u))|duds}(κ1+M)a2(m2m1)supt[0,1]{0t(em2(ts)em1(ts))sqΓ(q+1)ds+|σ1(t)|01(em2(1s)em1(1s))sqΓ(q+1)ds+|σ2(t)|0η(em2(ηs)em1(ηs))sqΓ(q+1)ds}(κ1+M)Γ(q+1){ε+σ^1γ1+ηqσ^2γ2}=(κ1+M)λκ1,

which clearly shows that 𝒥x ϵ Bk1 for any x ϵ Bk1 . Thus 𝒥Bk1 ⊂ Bk1 . Now, for x, y ϵ 𝒞 and for each t ϵ [0, 1], we have

(Jx)(Jy)1a2(m2m1)supt[0,1]{0t0sΦ(t)(su)q1Γ(q)|f(u,x(u))f(u,y(u))|duds+|σ1(t)|010sΦ(1)(su)q1Γ(q)|f(u,x(u))f(u,y(u))|duds+|σ2(t)|0η0sΦ(η)(su)q1Γ(q)|f(u,x(u))f(u,y(u))|duds}a2(m2m1)supt[0,1]{0t(em2(ts)em1(ts))sqΓ(q+1)ds+|σ1(t)|01(em2(1s)em1(1s))sqΓ(q+1)ds+|σ2(t)|0η(em2(ηs)em1(ηs))sqΓ(q+1)ds}xyΓ(q+1){ε+σ^1γ1+ηqσ^2γ2}xy=λxy,

where λ is given by (20) and depends only on the parameters involved in the problem. In view of the condition < 1/λ, it follows that J is a contraction. Thus, by the contraction mapping principle (Banach fixed point theorem), the problem (1)-(2) with a12 − 4a0a2 > 0 has a unique solution on [0, 1].

  • (ii) Let us define suptϵ[0,1] |f (t, 0)| = M and select κ2μM1μ.As in (i), one can show that 𝓗Bk2 ⊂ Bk2 , where H is defined by (17). Also, for x, y ϵ 𝒞 and for each t ϵ [0, 1], we can obtain

(Hx)(Hy)a2m2Γ(q+1){(1+ψ^1)((m1)em+1)+ψ^2ηq((mη1)emη+1)}xy=μxy

where μ is given by (21). By the condition < 1/μ, we deduce that the operator 𝓗 is a contraction. Thus, by the contraction mapping principle, the problem (1)-(2) with a12 − 4a0a2 = 0 has a unique solution on [0, 1].

  • (iii) Letting suptϵ[0,1] ϵf (t, 0)ϵ = M and selecting κ3ρM1ρ,it can be shown that KBk3 ⊂ Bk3 , where Bk3 = {x ϵ 𝒞 : ∥x∥ ≤ k3} and K is defined by (19). Moreover, for x, y ϵ 𝒞 and for each t ϵ [0, 1], we can find that

(Kx)(Ky)a2(α2+β2)Γ(q+1){(1+φ^1)(1eαcosβ(α/β)eαsinβ)+φ^2ηq(1eαηcosβη(α/β)eαηsinβη)}xy=ρxy,

where ρ is given by (22). Evidently, it follows by the condition < 1/ρ that K is a contraction. Thus, by the contraction mapping principle, the problem (1)-(2) with a12 − 4a0a2 < 0 has a unique solution on [0, 1]. This completes the proof. □

The next existence result is based on Leray-Schauder nonlinear alternative [8], which is stated below.

Theorem 3.4

(Nonlinear alternative for single valued maps). Let Y be a closed, convex subset of a Banach space X and V be an open subset of Y with 0 ϵ V. Let G : VY be a continuous and compact (that is, G(V) is a relatively compact subset of Y) map. Then either G has a fixed point in V or there is a v ϵ ∂V (the boundary of V in Y) and ε ϵ (0, 1) with u = εG(u).

In order to establish our last result, we need the following conditions.

(H1)There exist a function g ϵ C([0, 1], ℝ+), and a nondecreasing function Q : ℝ+ ➝ ℝ+ such that |f (t, y)|g(t)Q(∥y∥), (t, y) ϵ [0, 1] × ℝ.

(H2)−(i, P) There exists a constant 𝒦i > 0 such that

KigQ(Ki)P>1,i=1,2,3,P{λ,μ,ρ}.

Theorem 3.5

Let f : [0, 1]×ℝ ➝ ℝ be a continuous function. Then the problem (1)-(2) has at least one solution on [0, 1] if

  • (a) (H1) and (H2)− (1, λ) are satisfied for a124a0a2>0;

  • (b) (H1) and (H2)− (2, μ) are satisfied for a124a0a2=0;

  • (c) (H1) and (H2)− (3, ρ) are satisfied for a124a0a2<0.

Proof. (a) Let us first show that the operator 𝒥 : 𝒞 ➝ 𝒞 defined by (15) maps bounded sets into bounded sets in 𝒥 = C([0, 1], ℝ). For a positive number ζ1, let 𝓑ζ1 = {x ϵ 𝒞 : ∥x∥ ≤ ζ1} be a bounded set in 𝒥. Then we have

J(x)=supt[0,1]|J(x)(t)|1a2(m2m1)supt[0,1]{0t0sΦ(t)(su)q1Γ(q)|f(u,x(u))|duds+|σ1(t)|010sΦ(1)(su)q1Γ(q)|f(u,x(u))|duds+|σ2(t)|0η0sΦ(η)(su)q1Γ(q)|f(u,x(u))|duds}gQ(ζ1)a2(m2m1)supt[0,1]{0t(em2(ts)em1(ts))sqΓ(q+1)ds+|σ1(t)|01(em2(1s)em1(1s))sqΓ(q+1)ds+|σ2(t)|0η(em2(ηs)em1(ηs))sqΓ(q+1)ds}gQ(ζ1)Γ(q+1){ε+σ^1γ1+ηqσ^2γ2},

which yields

JxgQ(ζ1)Γ(q+1){ε+σ^1γ1+ηqσ^2γ2}.

Next we show that 𝒥 maps bounded sets into equicontinuous sets of 𝒞. Let t1, t2 ϵ [0, 1] with t1 < t2 and y ϵ 𝓑ζ1 , where 𝓑ζ1 is a bounded set of 𝒞. Then we obtain

|(Jx)(t2)(Jx)(t1)|1a2(m2m1){|0t10s[Φ(t2)Φ(t1)](su)q1Γ(q)f(u,x(u))duds+t1t20sΦ(t2)(su)q1Γ(q)f(u,x(u))duds|+|σ1(t2)σ1(t1)|010sΦ(1)(su)q1Γ(q)|f(u,y(u))|duds+|σ2(t2)σ2(t1)|0η0sΦ(η)(su)q1Γ(q)|f(u,y(u))|duds}f¯a2m1m2(m2m1)Γ(q+1){(t1qt2q)(m1(1em2(t2t1))m2(1em1(t2t1)))+t1q(m1(em2t2em2t1)m2(em1t2em1t1))+|σ1(t2)σ1(t1)|(m2(1em1)m1(1em2))+|σ1(t2)σ1(t1)|ηq(m2(1em1η)m1(1em2η))},

which tends to zero as t2t1 ➝ 0 independently of x ϵ 𝓑ζ1 . From the foregoing arguments, it follows by the Arzelá-Ascoli theorem that 𝒥 : 𝒞 ➝ 𝒞 is completely continuous.

The proof will be complete by virtue of Theorem 3.4 once we establish that the set of all solutions to the equation x = θ𝒥x is bounded for θ ϵ [0, 1]. To do so, let x be a solution of x = θ𝒥x for θ ϵ [0, 1]. Then, for t ϵ [0, 1], we get

|x(t)|=|θJx(t)|1a2(m2m1)supt[0,1]{0t0sΦ(t)(su)q1Γ(q)|f(u,x(u))|duds+|σ1(t)|010sΦ(1)(su)q1Γ(q)|f(u,x(u))|duds+|σ2(t)|0η0sΦ(η)(su)q1Γ(q)|f(u,x(u))|duds}gQ(x)a2(m2m1)supt[0,1]{0t(em2(ts)em1(ts))sqΓ(q+1)ds+|σ1(t)|01(em2(1s)em1(1s))sqΓ(q+1)ds+|σ2(t)|0η(em2(ηs)em1(ηs))sqΓ(q+1)ds}gQ(x)Γ(q+1){ε+σ^1γ1+ηqσ^2γ2}=gQ(x)λ,

which on taking the norm for t ϵ [0, 1], yields

xgQ(x)λ1.

In view of (H2)− (1, λ), there is no solution x such that ∥x∥ 6= K1. Let us set

U1={xC:x<K1}.

As the operator 𝒥 : U¯1Cis continuous and completely continuous, we infer from the choice of U1 that there is no u ϵ ∂U1 such that u = θ𝒥(u) for some θ ϵ (0, 1). Hence, by Theorem 3.4, we deduce that 𝒥 has a fixed point u ϵ U1 which is a solution of the problem (1)-(2).

  • (b) As in part (a), it can be shown that the operator H : 𝒞 ➝ 𝒞 defined by (17) maps bounded sets into bounded sets in 𝒞 = C([0, 1], ℝ). For that, let ζ2 be a positive number and let 𝓑ζ2 = {x ϵ 𝒞 : ∥x∥ ≤ ζ2} be a bounded set in 𝒞. Then we have

H(x)=supt[0,1]|H(x)(t)|1a2supt[0,1]{0t0sΨ(t)(su)q1Γ(q)|f(u,x(u))|duds+|ψ1(t)|010sΨ(1)(su)q1Γ(q)|f(u,x(u))|duds+|ψ2(t)|0η0sΨ(η)(su)q1Γ(q)|f(u,x(u))|duds}gQ(ζ2)a2m2Γ(q+1){(1+ψ^1)((m1)em+1)+ψ^2ηq((mη1)emη+1)}.

In order to show that 𝓗 maps bounded sets into equicontinuous sets of 𝒞, let t1, t2 ϵ [0, 1] with t1 < t2 and ϵ 2 𝓑ζ2 , where 𝓑ζ2 is a bounded set of 𝒞. Then we get

|(Hx)(t2)(Hx)(t1)|f¯a2m2Γ(q+1){t1q(mt2emt2mt1emt1+emt1emt2m(t2t1)em(t2t1)+em(t2t1)1))+t2q(m(t2t1)(em(t2t1)em(t2t1)+1)}+f¯|ψ1(t2)ψ1(t1)|a201(1s)em(ηs)sqΓ(q+1)ds+f¯|ψ2(t12)ψ2(t1)|a20η(ηs)em(ηs)sqΓ(q+1)ds,

which tends to zero as t2t1 ➝ 0 independently of x ϵ 𝓑ζ2 . As argued before, 𝓗 : 𝒞 ➝ 𝒞 is completely continuous. To show that the set of all solutions to the equation x = θ𝓗x is bounded for θ ϵ [0, 1], let x be a solution of x = θ𝓗x for θ ϵ [0, 1]. Then, for t ϵ [0, 1], we find that

|x(t)|=|θHx(t)|1a2supt[0,1]{0t0sΨ(t)(su)q1Γ(q)|f(u,x(u))|duds+|ψ1(t)|010sΨ(1)(su)q1Γ(q)|f(u,x(u))|duds+|ψ2(t)|0η0sΨ(η)(su)q1Γ(q)|f(u,x(u))|duds}gQ(x)a2m2Γ(q+1){(1+ψ^1)((m1)em+1)+ψ^2ηq((mη1)emη+1)}=gQ(x)μ.

Thus

xgQ(x)μ1.

In view of (H2)− (2, μ), there is no solution x such that ∥x∥ 6= K 2. Let us define

U2={xC:x<K2}.

Since the operator H:U¯2Cis continuous and completely continuous, there is no u ϵ ∂U2 such that u = θ𝓗(u) for some θ ϵ (0, 1) by the choice of U2. In consequence, by Theorem 3.4, we deduce that 𝓗 has a fixed point u ϵ U which is a solution of the problem (1)-(2).

  • (c) As in the preceding cases, one can show that the operator 𝒦 defined by (19) is continuous and completely continuous. We only provide the outline for the last part (a-priori bounds) of the proof. Let x be a solution of x = θKx for θ ϵ [0, 1], where K is defined by (19). Then, for t ϵ [0, 1], we have

|x(t)|=|θKx(t)|1a2β0t0sΩ(t)(su)q1Γ(q)|f(u,x(u))|duds+|φ1(t)|010sΩ(1)(su)q1Γ(q)|f(u,x(u))|duds+|φ2(t)|0η0sΩ(η)(su)q1Γ(q)|f(u,x(u))|dudsgQ(x)a2(α2+β2)Γ(q+1){(1+φ^1)(1eαcosβ(α/β)eαsinβ)+φ^2ηq(1eαηcosβη(α/β)eαηsinβη)}=gQ(x)ρ,

which yields

xgQ(x)ρ1.

In view of (H2)− (3, ρ), there is no solution x such that ∥x∥ 6= K3. Let us set

U3={xC:x<K3}.

As before, one can show that the operator K has a fixed point u ϵ Ū3, which is a solution of the problem (1)-(2). This completes the proof. □

4 Examples

Example 4.1

Consider the following boundary value problem

(cD8/3+5cD5/3+4cD2/3)x(t)=A(t+5)2(|x|1+|x|+cost),0<t<1,(33)

x(0)=0,x(3/5)=0,x(1)=0,(34)

Here, q=2/3,η=3/5,a2=1,a1=5,a0=4,a124a0a2=9>0,Ais a positive constant to be fixed later and

f(t,x)=A(t+5)2(|x|1+|x|+cost).

Clearly

|f(t,x)f(t,y)|0.04A|xy|,

with = 0.04 A. Using the given values, we find that λ=[(ε+σ^1γ1+ηqσ^2γ2)/Γ(q+1)]0.67232,and λ1 = λ1=λεΓ(q+1)0.52953.It is easy to check that |f (t, x)| ≤ 2A/(t + 5)2 = ϑ(tγ) and ℓλ1 < 1 when A < 47.211678.

As all the conditions of Theorem 3.2 (i) are satisfied, the problem (33)-(34) has at least one solution on [0, 1]. On the other hand, ℓλ < 1 whenever A < 37.184674 and thus there exists a unique solution for the problem (33)-(34) on [0, 1] by Theorem 3.3 (i).

Example 4.2

Consider the multi-term fractional differential equation

(3cD13/5+6cD8/5+3cD3/5)x(t)=Bt2+5(sint+tan1x(t)),0<t<1,(35)

supplemented with the boundary conditions

x(0)=0,x(2/3)=0,x(1)=0,(36)

Here, q=3/5,η=2/3,a2=3,a1=6,a0=3,a124a0a2=0,Bis a positive constant to be determined later and

f(t,x)=Bt2+5(sint+tan1x(t)).

Clearly

|f(t,x)f(t,y)|0.2B|xy|,

where = 0.2 B. Using the given values, it is found that μ=[(1+ψ^1)((m1)em+1)+ψ^2ηq((mη1)emη+1)/a2m2Γ(q+1)]0.59146, and μ1=μ((m1)em+1)a2m2Γ(q+1)0.29573.Further, |f (t, x)|B(2 + π)/2(t2 + 5) = (t) and ℓμ1 < 1 when B < 16.907314. As all the conditions of Theorem 3.2 (ii) are satisfied, the problem (35)-(36) has at least one solution on [0, 1]. On the other hand, ℓμ < 1 whenever B < 8.453657. Thus there exists a unique solution for the problem (35)-(36) on [0, 1] by Theorem 3.3 (ii).

Example 4.3

consider the multi-term fractional boundary value problem given by

(cD8/3+cD5/3+cD2/3)x(t)=C36+t2(|x|1+|x|+12),0<t<1,(37)

x(0)=0,x(3/5)=0,x(1)=0,(38)

where, q=2/3,η=3/5,a2=1,a1=1,a0=1,a124a0a2=3<0,Cis a positive constant to be fixed later and

f(t,x)=C36+t2(|x|1+|x|+12).

Clearly

|f(t,x)f(t,y)|(C/6)|xy|,

with = C /6. Using the given values, we have

ρ=1a2(α2+β2)Γ(q+1){(1+φ^1)(1eαcosβα/βeαsinβ)+φ^2ηq(1eαηcosβηα/βeαηsinβη)}0.75392,

and ρ1=ρ(1eαcosβα/βeαsinβ)a2(α2+β2)Γ(q+1)0.37696.Also |f(t,x)|3C/236+t2=ϑ(t)and ℓρ1 < 1 when C < 15.916808. Clearly all the conditions of Theorem 3.2 (iii) hold true. Thus the problem (37)-(38) has at least one solution on [0, 1]. On the other hand, ℓρ < 1 whenever C < 7.958404. Thus there exists a unique solution for the problem (37)-(38) on [0, 1] by Theorem 3.3 (iii).

Example 4.4

Consider the following nonlocal boundary value problem of multi-term fractional differential equation

(cD8/3+5cD5/3+4cD2/3)x(t)=Ptt+16(|x|8(1+|x|)+sinx+18),0<t<1,(39)

x(0)=0,x(3/5)=0,x(1)=0,(40)

where, q=2/3,η=3/5,a124a0a2=9>0,P is a positive constant and

f(t,x)=Ptt+16(|x|8(1+|x|)+sinx+18).

Clearly

|f(t,x)|Ptt+16(14+x)=g(t)Q(x),

with g(t)=Ptt+16,Q(x)=14+x.Letting P = 2 and using the condition (H2)− (1, ), we find that K1 > 0.10102 (we have used λ = 0.67232). Thus, the conclusion of Theorem 3.5 (a) applies to the problem (39)-(40).

Example 4.5

Consider the following boundary value problem

(3cD13/5+6cD8/5+3cD3/5)x(t)=12t+9(|x|31+|x|3+et),0<t<1,(41)

x(0)=0,x(2/3)=0,x(1)=0,(42)

where, q=3/5,η=2/3,a124a0a2=0,

f(t,x)=12t+9(|x|31+|x|3+et).

Clearly

|f(t,x)|1+et2t+9=g(t)Q(x),

with g(t)=1+et2t+9,Q(x)=1.Using the condition (H2)− (2, μ), we find that K2 > 0.3943 (with μ = 0.59146). Thus, the conclusion of Theorem 3.5 (b) applies to the problem (41)-(42).

Example 4.6

consider the following problem

(cD8/3+cD5/3+cD2/3)x(t)=3tt2+6(12+sinx),0<t<1,(43)

x(0)=0,x(3/5)=0,x(1)=0,(44)

Here, q=2/3,η=3/5,a124a0a2=3<0,and

f(t,x)=3tt2+6(12+sinx).

Clearly

|f(t,x)|3tt2+6(12+x),

with g(t)=3tt2+6,Q(x)=12+x.By the condition (H2)− (3, ρ), it is find that K3 > 0.30252 (with ρ = 0.75392). Thus, the conclusion of Theorem 3.5 (c) applies to the problem (43)-(44).

Acknowledgement

This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia under grant no. (KEP-PhD-11-130-39). The authors, therefore, acknowledge with thanks DSR technical and financial support.

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About the article

Received: 2018-01-16

Accepted: 2018-08-01

Published Online: 2018-12-31


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1519–1536, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0127.

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© 2018 Ahmad et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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