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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

# 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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• Other articles by this author:
• 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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• Other articles by this author:
• De Gruyter OnlineGoogle Scholar
/ 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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• Other articles by this author:
• De Gruyter OnlineGoogle Scholar
/ 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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• De Gruyter OnlineGoogle Scholar
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.

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:

$(a2 cDq+2+a1 cDq+1+a0 cDq)x(t)=f(t,x(t)),0(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

$Iτh(u)=∫0u(u−v)τ−1Γ(τ)h(v)dv,u>0,$

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(n−τ)∫0uh(n)(v)(u−v)τ+1−ndv=In−τh(n)(u),t>0,n−1<τ

#### Property 2.1

With the given notations, the following equality holds:

$Iτ(cDτh(u))=h(u)−c0−c1u...−cn−1un−1,u>0,n−1<τ(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

$(a2 cDq+2+a1 cDq+1+a0 cDq)x(t)=y(t),0(4)

supplemented with the boundary conditions (2) is given by

$(i)x(t)=1a2(m2−m1){∫0t∫0sΦ(t)(s−u)q−1Γ(q)y(u)du ds+σ1(t)∫01∫0sΦ(1)(s−u)q−1Γ(q)y(u)du ds+σ2(t)∫0η∫0sΦ(η)(s−u)q−1Γ(q)y(u)du ds}, if a12−4a0a2>0,$(5)

$(ii)x(t)=1a2{∫0t∫0sΨ(t)(s−u)q−1Γ(q)y(u)du ds+ψ1(t)∫01∫0sΨ(1)(s−u)q−1Γ(q)y(u)du ds+ψ2(t)∫0η∫0sΨ(η)(s−u)q−1Γ(q)y(u)du ds}, if a12−4a0a2=0,$(6)

$(iii)x(t)=1a2β{∫0t∫0sΩ(t)(s−u)q−1Γ(q)y(u)du ds+φ1(t)∫01∫0sΩ(1)(s−u)q−1Γ(q)y(u)du ds+φ2(t)∫0η∫0sΩ(η)(s−u)q−1Γ(q)y(u)du ds}, if a12−4a0a2<0,$(7)

where

$Φ(κ)=em2(κ−s)−em1(κ−s),κ=t,1,andη,m1=−a1−a12−4a0a22a2,m2=−a1+a12−4a0a22a2,σ1(t)=γ2ρ2(t)−γ4ρ1(t)μ,σ2(t)=γ3ρ1(t)−γ1ρ2(t)μ,μ=γ1γ4−γ2γ3≠0,ρ1(t)=m2(1−em1t)−m1(1−em2t)a2m1m2(m2−m1),ρ2(t)=em1t−em2t,γ1=m2(1−em1)−m1(1−em2)a2m1m2(m2−m1),γ2=m2(1−em1η)−m1(1−em2η)a2m1m2(m2−m1),γ3=em1−em2,γ4=em1η−em2η,$(8)

$Ψ(κ)=(κ−s)em(κ−s),κ=t,1,andη,ψ1(t)=(t−η)em(t+η)−temt+ηemηΛ,ψ2(t)=(1−t)em(t+1)+temt−emΛ,Λ=(η−1)em(η+1)−ηemη+em≠0,m=−a12a2,$(9)

$Ω(κ)=e−α(κ−s)sinβ(κ−s),κ=t,1,andη,α=a12a2,β=4a0a2−a122a2,φ1(t)=ω4ϱ1(t)−ω2ϱ2(t)Ω,φ2(t)=ω1ϱ2(t)−ω3ϱ1(t)Ω,ϱ1(t)=β−βe−αtcosβt−αe−αtsinβtα2+β2,ϱ2(t)=a2βe−αtsinβt,ω1=β−βe−αcosβ−αe−αsinβα2+β2,ω2=β−βe−αηcosβη−αe−αηsinβηα2+β2,ω3=a2βe−αsinβ,ω4=a2βe−αηsinβη,Ω=ω2ω3−ω1ω4≠0.$(10)

Proof. Case (i): ${a}_{1}^{2}-4{a}_{0}{a}_{2}>0.$

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

$(a2D2+a1D+a0)x(t)=∫0t(t−s)q−1Γ(q)y(s)ds+c1,$(11)

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

$x(t)=c2em1t+c3em2t−1a2(m2−m1)∫0tem1(t−s)(∫0s(s−u)q−1Γ(q)y(u)du+c1)ds+1a2(m2−m1)∫0tem2(t−s)(∫0s(s−u)q−1Γ(q)y(u)du+c1)ds,$(12)

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

$x(t)=c1[m2(1−em1t)−m1(1−em2t)a2m1m2(m2−m1)]+c2(em1t−em2t)−1a2(m2−m1)[∫0t(em1(t−s)−em2(t−s))(∫0s(s−u)q−1Γ(q)y(u)du)ds],$(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:

$c1γ1+c2γ3=1a2(m2−m1)∫01(em1(1−s)−em2(1−s))(∫0s(s−u)q−1Γ(q)y(u)du)ds,c1γ2+c2γ4=1a2(m2−m1)∫0η(em1(η−s)−em2(η−s))(∫0s(s−u)q−1Γ(q)y(u)du)ds.$

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

$c1=1a2μ(m2−m1)[γ4∫01(em1(1−s)−em2(1−s))(∫0s(s−u)q−1Γ(q)y(u)du)ds−γ3∫0η(em1(η−s)−em2(η−s))(∫0s(s−u)q−1Γ(q)y(u)du)ds],$

and

$c2=1a2μ(m2−m1)[γ1∫0η(em1(η−s)−em2(η−s))(∫0s(s−u)q−1Γ(q)y(u)du)ds−γ2∫01(em1(1−s)−em2(1−s))(∫0s(s−u)q−1Γ(q)y(u)du)ds].$

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

$(Jx)(t)=1a2(m2−m1){∫0t∫0sΦ(t)(s−u)q−1Γ(q)f(u,x(u))du ds+σ1(t)∫01∫0sΦ(1)(s−u)q−1Γ(q)f(u,x(u))du ds+σ2(t)∫0η∫0sΦ(η)(s−u)q−1Γ(q)f(u,x(u))du ds},$(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

$(Hx)(t)=1a2{∫0t∫0sΨ(t)(s−u)q−1Γ(q)f(u,x(u))du ds+ψ1(t)∫01∫0sΨ(1)(s−u)q−1Γ(q)f(u,x(u))du ds+ψ2(t)∫0η∫0sΨ(η)(s−u)q−1Γ(q)f(u,x(u))du ds},$(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

$(Kx)(t)=1a2β{∫0t∫0sΩ(t)(s−u)q−1Γ(q)f(u,x(u))du ds+φ1(t)∫01∫0sΩ(1)(s−u)q−1Γ(q)f(u,x(u))du ds+φ2(t)∫0η∫0sΩ(η)(s−u)q−1Γ(q)f(u,x(u))du ds},$(19)

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

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

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

$ψ^1=maxt∈[0,1]|ψ1(t)|,ψ^2=maxt∈[0,1]|ψ2(t)|,μ=1a2m2Γ(q+1){(1+ψ^1)((m−1)em+1)+ψ^2ηq((mη−1)emη+1)},μ1=μ−((m−1)em+1)a2m2Γ(q+1),$(21)

$φ^1=maxt∈[0,1]|φ1(t)|,φ^2=maxt∈[0,1]|φ2(t)|,ρ=1a2(α2+β2)Γ(q+1){(1+φ^1)(1−e−αcosβ−(α/β)e−αsin⁡β) +φ^2ηq(1−e−αηcosβη−(α/β)e−αηsinβη)},ρ1=ρ−(1−e−αcosβ−(α/β)e−αsin⁡β)a2(α2+β2)Γ(q+1).$(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

$r1≥∥ϑ∥Γ(q+1){ε+σ^1γ1+ηqσ^2γ2},$(23)

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

$(J1x)(t)=1a2(m2−m1)∫0t∫0sΦ(t)(s−u)q−1Γ(q)f(u,x(u))du ds,$(24)

$(J2x)(t)=1a2(m2−m1){σ1(t)∫01∫0sΦ(1)(s−u)q−1Γ(q)f(u,x(u))du ds+σ2(t)∫0t∫0sΦ(η)(s−u)q−1Γ(q)f(u,x(u))du ds}.$(25)

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

$∥J1x+J2y∥=supt∈[0,1]|(J1x)(t)+(J2y)(t)|≤1a2(m2−m1)supt∈[0,1]{∫0t∫0sΦ(t)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|σ1(t)|∫01∫0sΦ(1)(s−u)q−1Γ(q)|f(u,y(u))|du ds+|σ2(t)|∫0η∫0sΦ(η)(s−u)q−1Γ(q)|f(u,y(u))|du ds}≤∥ϑ∥a2(m2−m1)Γ(q+1)supt∈[0,1]{tq∫0t(em2(t−s)−em1(t−s))ds+|σ1(t)|∫01(em2(1−s)−em1(1−s))ds+ηq|σ2(t)|∫0η(em2(η−s)−em1(η−s))ds}≤∥ϑ∥Γ(q+1){ε+σ^1γ1+ηqσ^2γ2}≤r1,$

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:

$∥J2x−J2y∥=supt∈[0,1]|(J2x)(t)−(J2y)(t)|≤1a2(m2−m1)supt∈[0,1]{|σ1(t)|∫01∫0sΦ(1)(s−u)q−1Γ(q)|f(u,x(u))−f(u,y(u))|du ds+|σ2(t)|∫0η∫0sΦ(η)(s−u)q−1Γ(q)|f(u,x(u))−f(u,y(u))|du ds}≤ℓa2(m2−m1)Γ(q+1)supt∈[0,1]{|σ1(t)|∫01(em2(1−s)−em1(1−s))ds+ηq|σ2(t)|∫0η(em2(η−s)−em1(η−s))ds}∥x−y∥≤ℓΓ(q+1){σ^1γ1+ηqσ^2γ2}∥x−y∥=ℓλ1∥x−y∥.$

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

$∥J1x∥=1a2(m2−m1)supt∈[0,1]|∫0t∫0sΦ(t)(s−u)q−1Γ(q)f(u,x(u))du ds|≤∥ϑ∥εΓ(q+1).$

Now we prove the compactness of operator 𝒥1.We define $\underset{\left(t,x\right)\in \left[0,1\right]×{B}_{{r}_{1}}}{sup}|f\left(t,x\right)|=\overline{f}.$we have

$|(J1x)(t2)−(J1x)(t1)|=1a2(m2−m1)|∫0t1∫0s[Φ(t2)−Φ(t1)](s−u)q−1Γ(q)f(u,x(u))du ds+∫t1t2∫0sΦ(t2)(s−u)q−1Γ(q)f(u,x(u))du ds|≤f¯a2m1m2(m2−m1)Γ(q+1){(t1q−t2q)(m1(1−em2(t2−t1))−m2(1−em1(t2−t1)))+t1q(m1(em2t2−em2t1)−m2(em1t2−em1t1))}→0 as t2−t1→0,$

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

$r2≥∥ϑ∥a2m2Γ(q+1){(1+ψ^1)((m−1)em+1)+ψ^2ηq((mη−1)emη+1)}.$(26)

Introduce the operators 𝓗1 and 𝓗2 on Br2 as follows:

$(H1x)(t)=1a2∫0t∫0sΨ(t)(s−u)q−1Γ(q)f(u,x(u))du ds,$(27)

$(H2x)(t)=1a2{ψ1(t)∫01∫0sΨ(1)(s−u)q−1Γ(q)f(u,x(u))du ds+ψ2(t)∫0η∫0sΨ(η)(s−u)q−1Γ(q)f(u,x(u))du ds}.$(28)

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

$∥H1x+H2y∥=supt∈[0,1]|(H1x)(t)+(H2y)(t)|≤1a2supt∈[0,1]{∫0t∫0sΨ(t)(s−u)q−1Γ(q)|f(u,x(u))|du ds+ψ1(t)∫01∫0sΨ(1)(s−u)q−1Γ(q)|f(u,y(u))|du ds+ψ2(t)∫0η∫0sΨ(η)(s−u)q−1Γ(q)|f(u,y(u))|du ds}≤∥ϑ∥a2Γ(q+1)supt∈[0,1]{tq∫0t(t−s)em(t−s)ds+|ψ1(t)|∫01(1−s)em(1−s)ds+|ψ2(t)|ηq∫0η(η−s)em(η−s)ds}≤∥ϑ∥a2m2Γ(q+1){(1+ψ^1)((m−1)em+1)+ψ^2ηq((mη−1)emη+1)}≤r2,$

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

$∥H1x∥=supt∈[0,1]|(H1x)(t)|≤∥ϑ∥a2m2Γ(q+1){(m−1)em+1}.$

Now we prove the compactness of operator 𝓗1. For that, let $\underset{\left(t,x\right)\in \left[0,1\right]×{B}_{{r}_{2}}}{sup}|f\left(t,x\right)|=\overline{f}.$Thus, for 0 < t1 < t2 < 1, we have

$|(H1x)(t2)−(H1x)(t1)|=1a2|∫0t1∫0s(Ψ(t2)−Ψ(t1))(s−u)q−1Γ(q)f(u,x(u))du ds+∫t1t2∫0sΨ(t2)(s−u)q−1Γ(q)f(u,x(u))du ds|≤f¯a2m2Γ(q+1){t1q(mt2emt2−mt1emt1+emt1−emt2−m(t2−t1)em(t2−t1)+em(t2−t1)−1))+t2q(m(t2−t1)(em(t2−t1)−em(t2−t1)+1)}→0 as t2−t1→0,$

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

$r3≥∥ϑ∥a2(α2+β2)Γ(q+1){(1+φ^1)(1−e−αcosβ−(α/β)e−αsin⁡β)+φ^2ηq(1−e−αηcosβη−(α/β)e−αηsinβη)},$(29)

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

$(K1x)(t)=1a2β{∫0t∫0sΩ(t)(s−u)q−1Γ(q)f(u,x(u))du ds},$(30)

$(K2x)(t)=1a2β{φ1(t)∫01∫0sΩ(1)(s−u)q−1Γ(q)f(u,x(u))dus ds+φ2(t)∫0η∫0sΩ(η)(s−u)q−1Γ(q)f(u,x(u))du ds}.$(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

$∥K1x∥=supt∈[0,1]|(K1x)(t)|≤∥ϑ∥a2(α2+β2)Γ(q+1)(1−e−αcosβ−α/βe−αsin⁡β).$

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

$|(K1x)(t2)−(K1x)(t1)|≤f¯a2(α2+β2)Γ(q+1){t1q((α/β)sinβ(t2−t1)e−α(t2−t1)−(α/β)sinβt2e−αt2+(α/β)sinβt1e−αt1+cosβ(t2−t1)e−α(t2−t1)−cosβt2e−αt2+cosβt1e−αt1−1)+t2q(1−(α/β)sinβ(t2−t1)e−α(t2−t1)−cosβ(t2−t1)e−α(t2−t1))},$

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:

• $\frac{\ell \epsilon }{\mathrm{\Gamma }\left(q+1\right)}<1;$

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

• $\frac{\ell \left\{\left(m-1\right){e}^{m}+1\right\}}{{a}_{2}{m}^{2}\mathrm{\Gamma }\left(q+1\right)}<1;$

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

• $\frac{\ell \left\{1-{e}^{-\alpha }\mathrm{cos}\beta -\left(\alpha /\beta \right){e}^{-\alpha }\mathrm{sin}\beta \right\}}{{a}_{2}\left({\alpha }^{2}+{\beta }^{2}\right)\mathrm{\Gamma }\left(q+1\right)}<1.$

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 ${\kappa }_{1}\ge \frac{\lambda M}{1-\ell \lambda }$show that 𝒥Bk1 ⊂ Bk1 , where Bk1 = {x ϵ 𝒞 : ∥x∥ ≤ k1g and 𝒥 is defined by (15). Using the condition (A1), we have

$|f(t,x)|=|f(t,x)−f(t,0)+f(t,0)|≤|f(t,x)−f(t,0)|+|f(x,0)|≤ℓ∥x∥+M≤ℓκ1+M.$(32)

Then, for x ϵ Bk1 , we obtain

$∥J(x)∥=supt∈[0,1]|J(x)(t)|≤1a2(m2−m1)supt∈[0,1]{∫0t∫0sΦ(t)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|σ1(t)|∫01∫0sΦ(1)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|σ2(t)|∫0η∫0sΦ(η)(s−u)q−1Γ(q)|f(u,x(u))|du ds}≤(ℓκ1+M)a2(m2−m1)supt∈[0,1]{∫0t(em2(t−s)−em1(t−s))sqΓ(q+1)ds+|σ1(t)|∫01(em2(1−s)−em1(1−s))sqΓ(q+1)ds+|σ2(t)|∫0η(em2(η−s)−em1(η−s))sqΓ(q+1)ds}≤(ℓκ1+M)Γ(q+1){ε+σ^1γ1+ηqσ^2γ2}=(ℓκ1+M)λ≤κ1,$

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

$∥(Jx)−(Jy)∥≤1a2(m2−m1)supt∈[0,1]{∫0t∫0sΦ(t)(s−u)q−1Γ(q)|f(u,x(u))−f(u,y(u))|du ds+|σ1(t)|∫01∫0sΦ(1)(s−u)q−1Γ(q)|f(u,x(u))−f(u,y(u))|du ds+|σ2(t)|∫0η∫0sΦ(η)(s−u)q−1Γ(q)|f(u,x(u))−f(u,y(u))|du ds}≤ℓa2(m2−m1)supt∈[0,1]{∫0t(em2(t−s)−em1(t−s))sqΓ(q+1)ds+|σ1(t)|∫01(em2(1−s)−em1(1−s))sqΓ(q+1)ds+|σ2(t)|∫0η(em2(η−s)−em1(η−s))sqΓ(q+1)ds}∥x−y∥≤ℓΓ(q+1){ε+σ^1γ1+ηqσ^2γ2}∥x−y∥=ℓλ∥x−y∥,$

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 ${\kappa }_{2}\ge \frac{\mu M}{1-\ell \mu }.$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

$∥(Hx)−(Hy)∥≤ℓa2m2Γ(q+1){(1+ψ^1)((m−1)em+1)+ψ^2ηq((mη−1)emη+1)}∥x−y∥=ℓμ∥x−y∥$

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 ${\kappa }_{3}\ge \frac{\rho M}{1-\ell \rho },$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

$∥(Kx)−(Ky)∥≤ℓa2(α2+β2)Γ(q+1){(1+φ^1)(1−e−αcosβ−(α/β)e−αsin⁡β)+φ^2ηq(1−e−αηcosβη−(α/β)e−αηsinβη)}∥x−y∥=ℓρ∥x−y∥,$

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

$Ki∥g∥Q(Ki)P>1, i=1,2,3, P∈{λ,μ,ρ}.$

#### 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 ${a}_{1}^{2}-4{a}_{0}{a}_{2}>0;$

• (b) (H1) and (H2)− (2, μ) are satisfied for ${a}_{1}^{2}-4{a}_{0}{a}_{2}=0;$

• (c) (H1) and (H2)− (3, ρ) are satisfied for ${a}_{1}^{2}-4{a}_{0}{a}_{2}<0.$

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

$∥J(x)∥=supt∈[0,1]|J(x)(t)|≤1a2(m2−m1)supt∈[0,1]{∫0t∫0sΦ(t)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|σ1(t)|∫01∫0sΦ(1)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|σ2(t)|∫0η∫0sΦ(η)(s−u)q−1Γ(q)|f(u,x(u))|du ds}≤∥g∥Q(ζ1)a2(m2−m1)supt∈[0,1]{∫0t(em2(t−s)−em1(t−s))sqΓ(q+1)ds+|σ1(t)|∫01(em2(1−s)−em1(1−s))sqΓ(q+1)ds+|σ2(t)|∫0η(em2(η−s)−em1(η−s))sqΓ(q+1)ds}≤∥g∥Q(ζ1)Γ(q+1){ε+σ^1γ1+ηqσ^2γ2},$

which yields

$∥Jx∥≤∥g∥Q(ζ1)Γ(q+1){ε+σ^1γ1+ηqσ^2γ2}.$

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

$|(Jx)(t2)−(Jx)(t1)|≤1a2(m2−m1){|∫0t1∫0s[Φ(t2)−Φ(t1)](s−u)q−1Γ(q)f(u,x(u))du ds+∫t1t2∫0sΦ(t2)(s−u)q−1Γ(q)f(u,x(u))du ds|+|σ1(t2)−σ1(t1)|∫01∫0sΦ(1)(s−u)q−1Γ(q)|f(u,y(u))|du ds+|σ2(t2)−σ2(t1)|∫0η∫0sΦ(η)(s−u)q−1Γ(q)|f(u,y(u))|du ds}≤f¯a2m1m2(m2−m1)Γ(q+1){(t1q−t2q)(m1(1−em2(t2−t1))−m2(1−em1(t2−t1)))+t1q(m1(em2t2−em2t1)−m2(em1t2−em1t1))+|σ1(t2)−σ1(t1)|(m2(1−em1)−m1(1−em2))+|σ1(t2)−σ1(t1)|ηq(m2(1−em1η)−m1(1−em2η))},$

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

$|x(t)|=|θJx(t)|≤1a2(m2−m1)supt∈[0,1]{∫0t∫0sΦ(t)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|σ1(t)|∫01∫0sΦ(1)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|σ2(t)|∫0η∫0sΦ(η)(s−u)q−1Γ(q)|f(u,x(u))|du ds}≤∥g∥Q(∥x∥)a2(m2−m1)supt∈[0,1]{∫0t(em2(t−s)−em1(t−s))sqΓ(q+1)ds+|σ1(t)|∫01(em2(1−s)−em1(1−s))sqΓ(q+1)ds+|σ2(t)|∫0η(em2(η−s)−em1(η−s))sqΓ(q+1)ds}≤∥g∥Q(∥x∥)Γ(q+1){ε+σ^1γ1+ηqσ^2γ2}=∥g∥Q(∥x∥)λ,$

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

$∥x∥∥g∥Q(∥x∥)λ≤1.$

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

$U1={x∈C:∥x∥

As the operator 𝒥 : ${\overline{U}}_{1}\to \mathcal{C}$is 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

$∥H(x)∥=supt∈[0,1]|H(x)(t)|≤1a2supt∈[0,1]{∫0t∫0sΨ(t)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|ψ1(t)|∫01∫0sΨ(1)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|ψ2(t)|∫0η∫0sΨ(η)(s−u)q−1Γ(q)|f(u,x(u))|du ds}≤∥g∥Q(ζ2)a2m2Γ(q+1){(1+ψ^1)((m−1)em+1)+ψ^2ηq((mη−1)emη+1)}.$

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

$|(Hx)(t2)−(Hx)(t1)|≤f¯a2m2Γ(q+1){t1q(mt2emt2−mt1emt1+emt1−emt2−m(t2−t1)em(t2−t1)+em(t2−t1)−1))+t2q(m(t2−t1)(em(t2−t1)−em(t2−t1)+1)}+f¯|ψ1(t2)−ψ1(t1)|a2∫01(1−s)em(η−s)sqΓ(q+1)ds+f¯|ψ2(t12)−ψ2(t1)|a2∫0η(η−s)em(η−s)sqΓ(q+1)ds,$

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

$|x(t)|=|θHx(t)|≤1a2supt∈[0,1]{∫0t∫0sΨ(t)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|ψ1(t)|∫01∫0sΨ(1)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|ψ2(t)|∫0η∫0sΨ(η)(s−u)q−1Γ(q)|f(u,x(u))|du ds}≤∥g∥Q(∥x∥)a2m2Γ(q+1){(1+ψ^1)((m−1)em+1)+ψ^2ηq((mη−1)emη+1)}=∥g∥Q(∥x∥)μ.$

Thus

$∥x∥∥g∥Q(∥x∥)μ≤1.$

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

$U2={x∈C:∥x∥

Since the operator $\mathcal{H}:{\overline{U}}_{2}\to \mathcal{C}$is 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

$|x(t)|=|θKx(t)|≤1a2β∫0t∫0sΩ(t)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|φ1(t)|∫01∫0sΩ(1)(s−u)q−1Γ(q)|f(u,x(u))|du ds+|φ2(t)|∫0η∫0sΩ(η)(s−u)q−1Γ(q)|f(u,x(u))|du ds≤∥g∥Q(∥x∥)a2(α2+β2)Γ(q+1){(1+φ^1)(1−e−αcosβ−(α/β)e−αsin⁡β)+φ^2ηq(1−e−αηcosβη−(α/β)e−αηsinβη)}=∥g∥Q(∥x∥)ρ,$

which yields

$∥x∥∥g∥Q(∥x∥)ρ≤1.$

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

$U3={x∈C:∥x∥

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

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

Here, $q=2/3,\eta =3/5,{a}_{2}=1,{a}_{1}=5,{a}_{0}=4,{a}_{1}^{2}-4{a}_{0}{a}_{2}=9>0,A$is a positive constant to be fixed later and

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

Clearly

$|f(t,x)−f(t,y)|≤0.04 A|x−y|,$

with = 0.04 A. Using the given values, we find that $\lambda =\left[\left(\epsilon +{\stackrel{^}{\sigma }}_{1}{\gamma }_{1}+{\eta }^{q}{\stackrel{^}{\sigma }}_{2}{\gamma }_{2}\right)/\mathrm{\Gamma }\left(q+1\right)\right]\approx 0.67232,$and λ1 = ${\lambda }_{1}=\lambda -\frac{\epsilon }{\mathrm{\Gamma }\left(q+1\right)}\approx 0.52953.$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+6 cD8/5+3 cD3/5)x(t)=Bt2+5(sint+tan−1x(t)),0(35)

supplemented with the boundary conditions

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

Here, $q=3/5,\eta =2/3,{a}_{2}=3,{a}_{1}=6,{a}_{0}=3,{a}_{1}^{2}-4{a}_{0}{a}_{2}=0,B$is a positive constant to be determined later and

$f(t,x)=Bt2+5(sint+tan−1x(t)).$

Clearly

$|f\left(t,x\right)-f\left(t,y\right)|\le 0.2\text{\hspace{0.17em}}B|x-y|,$

where = 0.2 B. Using the given values, it is found that $\mu =\left[\left(1+{\stackrel{^}{\psi }}_{1}\right)\left(\left(m-1\right){e}^{m}+1\right)+{\stackrel{^}{\psi }}_{2}{\eta }^{q}\left(\left(m\eta -1\right){e}^{m\eta }+1\right)/{a}_{2}{m}^{2}\mathrm{\Gamma }\left(q+1\right)\right]\approx 0.59146,$ and ${\mu }_{1}=\mu -\frac{\left(\left(m-1\right){e}^{m}+1\right)}{{a}_{2}{m}^{2}\mathrm{\Gamma }\left(q+1\right)}\approx 0.29573.$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(37)

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

where, $q=2/3,\eta =3/5,{a}_{2}=1,{a}_{1}=1,{a}_{0}=1,{a}_{1}^{2}-4{a}_{0}{a}_{2}=-3<0,C$is a positive constant to be fixed later and

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

Clearly

$|f(t,x)−f(t,y)|≤(C/6) |x−y|,$

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

$ρ=1a2(α2+β2)Γ(q+1){(1+φ^1)(1−e−αcosβ−α/βe−αsin⁡β)+φ^2ηq(1−e−αηcosβη−α/βe−αηsinβη)}≈0.75392,$

and ${\rho }_{1}=\rho -\frac{\left(1-{e}^{-\alpha }\mathrm{cos}\beta -\alpha /\beta {e}^{-\alpha }\mathrm{sin}\beta \right)}{{a}_{2}\left({\alpha }^{2}+{\beta }^{2}\right)\mathrm{\Gamma }\left(q+1\right)}\approx 0.37696.$Also $|f\left(t,x\right)|\le 3C/2\sqrt{36+{t}^{2}}=\vartheta \left(t\right)$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+5 cD5/3+4 cD2/3)x(t)=Ptt+16(|x|8(1+|x|)+sin⁡x+18),0(39)

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

where, $q=2/3,\eta =3/5,{a}_{1}^{2}-4{a}_{0}{a}_{2}=9>0,$P is a positive constant and

$f(t,x)=Ptt+16(|x|8(1+|x|)+sin⁡x+18).$

Clearly

$|f(t,x)|≤Ptt+16(14+∥x∥)=g(t)Q(∥x∥),$

with $g\left(t\right)=\frac{Pt}{\sqrt{t+16}},\phantom{\rule{thinmathspace}{0ex}}Q\left(\parallel x\parallel \right)=\frac{1}{4}+\parallel x\parallel .$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+6 cD8/5+3 cD3/5)x(t)=12t+9(|x|31+|x|3+e−t),0(41)

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

where, $q=3/5,\eta =2/3,{a}_{1}^{2}-4{a}_{0}{a}_{2}=0,$

$f(t,x)=12t+9(|x|31+|x|3+e−t).$

Clearly

$|f(t,x)|≤1+e−t2t+9=g(t)Q(∥x∥),$

with $g\left(t\right)=\frac{1+{e}^{-t}}{2\sqrt{t+9}},\phantom{\rule{thinmathspace}{0ex}}Q\left(\parallel x\parallel \right)=1.$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+sin⁡x),0(43)

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

Here, $q=2/3,\eta =3/5,{a}_{1}^{2}-4{a}_{0}{a}_{2}=-3<0,$and

$f(t,x)=3tt2+6(12+sin⁡x).$

Clearly

$|f(t,x)|≤3tt2+6(12+∥x∥),$

with $g\left(t\right)=\frac{3t}{{t}^{2}+6},\phantom{\rule{thinmathspace}{0ex}}Q\left(\parallel x\parallel \right)=\frac{1}{2}+\parallel x\parallel .$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,

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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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