Boundary value problems of fractional differential equations with nonlocal and integral boundary con

International Journal of Advanced Engineering Research and Science (IJAERS) Peer-Reviewed Journal ISSN: 2349-6495(P) | 2456-1908(O) Vol-8, Issue-10; Oct, 2021 Journal Home Page Available: https://ijaers.com/ Article DOI: https://dx.doi.org/10.22161/ijaers.810.32

Boundary value problems of fractional differential equations with nonlocal and integral boundary conditions John Fiifi Essel, Lingling Zhang*, Emmanuel Addai Department of Mathematics, Taiyuan University of Technology, 030024, TaiYuan, Shanxi, China

Received: 29 Jul 2021, Received in revised form: 07 Oct 2021, Accepted: 13 Oct 2021, Available online: 28 Oct 2021 ©2021 The Author(s). Published by AI Publication. This is an open access article under the CC BY license (https://creativecommons.org/licenses/by/4.0/).

Abstract— The boundary value problems of fractional differential equations involving Caputo derivatives are examined in this article. The ultimate goal of our study is to institute the existence and uniqueness properties for our boundary problems with nonlocal and integral boundary conditions by applying the Banach and the Schaefer’s theorems of fixed point. We finally discuss two applicable questions to enhance the comprehension of our outcomes and conclude by summarizing our results and giving vital suggestions for further research works in relation to our study.

Keywords— boundary value problems, existence and uniqueness, fixed point theorems, fractional differential equations, nonlocal-integral boundary conditions.

I.

INTRODUCTION

The concept of fractional differentials and integrals date as far back as the middle of the nineteenth century by the works of some mathematicians such as Liouville, L’Hospital and Riemann. A century later, engineers and physicists found considerable applications to the concept in their respective fields of interest such as physics, identification of signals and processing of images, aerodynamics, blood flow phenomena and many other relevant fields of study[11,13]. Over the years, numerous concepts of fractional derivatives have been established such as the Riemann-Liouville and the Caputo derivative coupled with an extensive and significant advancements in the analysis of the solutions of FDEs, specifically of higher-order differentials with boundary restrictions.

The importance and applicability of fractional differential equations keep increasing extensively over the years due to its accuracy and objectiveness in describing nonlinear and or natural phenomenon such as stochastic and diffusion models, hydrology processes and finance. Lately, the advancement of fractional derivatives has not only portrayed in-depth research backgrounds but also a vast

www.ijaers.com

application to real life situations. The analysis of the multiplicity, uniqueness and existence of solutions and positive solutions of boundary value problems via the application of nonlinear techniques such as the theory of fixed points, the Leray-Schauder theorem, the technique of Upper and Lower solution among others has garnered much attention from researchers[3-13].

Benchohra et al[5], investigated the boundary problems of the FDE below; 𝐷0∝+ 𝑦(𝑡) = 𝑓(𝑡, 𝑦(𝑡)), 𝑡 ∈ 𝐽 = [0, 𝑇] , ∝ ∈ (1, 2]. 𝑦(0) = 𝑔(𝑦),

𝑦(𝑇) = 𝑦𝑇

where 𝐷0∝+ is a Caputo’s differential, 𝑓: [0, 𝑇] × 𝑅 → 𝑅 and𝑔: 𝑐(𝐽, 𝑅) → 𝑅 are continuous functions and 𝑦𝑇 ∈ 𝑅.

Cabada and Wang [6], examined the positive solutions of the FDE below with integral boundary conditions; 𝑐 𝛼 .𝐷 𝑢(𝑡)

+ 𝑓(𝑡, 𝑢(𝑡)) = 0,

0<𝑡<1 1

𝑢(0) = 𝑢′′ (0) = 0,

𝑢(1) = 𝜆 ∫ 𝑢(𝑠)𝑑𝑠 0

Page | 297

John Fiifi Essel et al.

International Journal of Advanced Engineering Research and Science, 8(10)-2021

where 𝛼 ∈ (2, 3), 𝜆 ∈ (0, 2), 𝑐.𝐷 𝛼 is a Caputo fractional differential, order 𝛼 and 𝑓: [0,1] × [0, ∞] → [0, ∞) is a continuous function.

Yan et al.[18], investigated the boundary value problems with integral conditions for the fractional differential equation; 𝑐 𝛼 .𝐷0+ 𝑥(𝑡)

𝑛 − 1 <∝ ≤ 2,

𝑇

𝑥(0) = 𝑦(𝑥), 𝑥(0) = 𝑦(𝑥),

= 𝑓(𝑡, 𝑥(𝑡)), 𝑡 ∈ [0, 𝑇],

∫0 𝑥(𝑡)𝑑𝑡 = 𝑚 𝑇

𝑥(𝑇) = ∫0 𝑔(𝑠)𝑥(𝑠)𝑑𝑠

in which 𝑐.𝐷0∝+ is a Caputo’s differential, 𝑓: [0, 𝑇] × 𝑅 → 𝑅,𝑦: 𝐶 2 ([0, 𝑇], 𝑅) → 𝑅 and 𝑔: [0, 𝑇] → 𝑅 are 𝐶 2 continuous function and 𝑚 ∈ 𝑅.

Inspired by the aforementioned articles, we examine the solutions of the FDE below involving Caputo derivatives; ; 𝑐 𝛾 .𝐷0+ 𝑢(𝑡)

= 𝑔(𝑡, 𝑢(𝑡), 𝑐.𝐷0𝛿+ 𝑢(𝑡)) = 0, 𝑡 ∈ [0, 𝜏], 𝛾 ∈ (𝑛 − 1, 𝑛] 𝑤ℎ𝑒𝑟𝑒 (𝑛 ≥ 2)

subject to the nonlocal and integral boundary conditions; τ

k 0 = μ(u), ∫0 u(t)dt = ρ k 0 = μ(u), 𝛾 Where 𝑐.𝐷0+ and

𝑐 𝛿 .𝐷0+

u(τ) = Uτ

(1.1)

are the Caputo fractional order differentials of 𝛾 and δ respectively, 𝑔: [0, 𝜏] × 𝑅 × 𝑅 → 𝑅 is a

continuous function, δ ∈ (0, 1) for all δ, ρ, Uτ ∈ R. As follows, we arrange the rest of the document. We stipulate our definitions and lemmas in section 2 to aid establish our main objectives. In section 3, by using the Banach and Schaefer's theorems of fixed points, we define some requirements for the existence of problem (1.1). Two examples to explain our key findings are discussed at the end.

II.

PRELIMINARIES

To commence the section, we stipulate some definitions and notations utilized in our study and proceed to prove our supporting lemmas before stating our key findings.

Definition 2.1([11]) The Caputo fractional differential of order α for a continuous function f(t) is defined by; 𝑐 𝛼 .𝐷0+ 𝑓(𝑡)

=

𝑡 1 ∫ (𝑡 𝛤(𝑛−𝛼) 0

− 𝑠)𝑛−𝛼−1 𝑓 (𝑛) (𝑠)𝑑𝑠,

where n = [𝛼] + 1, [𝛼] connotes the integer of ∝ and Γ is the gamma function.

Definition 2.2([18]). If 𝑓 ∈ (0, ∞)is a continuous function, the Riemann-Liuoville fractional integral of order𝛼 > 0 is defined as; 𝛼 𝐼0+ 𝑓(𝑡) =

𝑡 𝑓(𝑠) 1 𝑑𝑠, ∫ 𝛤(𝛼) 0 (𝑡−𝑠)1−𝛼

where Γ is the gamma function.

Definition 2.3([19]). The gamma function, 𝛤(∙) is defined by, ∞

𝛤(∝) = ∫0 𝑒 −𝑡 𝑡 ∝−1 𝑑𝑡, where [𝛼] is the integer of ∝ 𝑎𝑛𝑑 Γ(∝ +1) =∝ 𝛤(∝).

Lemma 2.1([11]). If ∝ > 0, the FDE 𝐷0∝+ ℎ(𝑡) = 0 has the solution ℎ(𝑡) = 𝑘0 + 𝑘1 𝑡 + 𝑘2 𝑡 2 + ⋯ + 𝑘𝑛−1 𝑡 𝑛−1 , 𝑘𝑖 ∈ 𝑅, 𝑖 = 0,1, … , 𝑛 − 1.

Lemma 2.2 ([19]). If ∝ > 0, then 𝐼0∝+ 𝐷0∝+ ℎ(𝑡) = ℎ(𝑡) + 𝑘0 + 𝑘1 𝑡 + 𝑘2 𝑡 2 + ⋯ + 𝑘𝑛−1 𝑡 𝑛−1 for any 𝑘𝑖 ∈ 𝑅, 𝑖 = 0,1, … , 𝑛 − 1, 𝑛 = [∝] + 1.

Lemma 2.3. Let 𝛾 ∈ (1, 2] and 𝑞(𝑡) ∈ 𝑐 2 [0, 𝜏]. Then a function 𝑢 satisfies the BVP; {

𝑐 𝛾 .𝐷0+ 𝑢(𝑡)

= 𝑞(𝑠),

μ(u) = k 0 ,

www.ijaers.com

𝑡 ∈ [0, 𝜏]

τ ∫0 u(t)dt

=ρ

(2.1)

Page | 298

John Fiifi Essel et al.

International Journal of Advanced Engineering Research and Science, 8(10)-2021

Iff u satisfies the fractional integral equation; u(t) =

t 1 ∫ (t Γ(γ) 0

2t

2ρ

τ

τ2

− s)γ−1 q(s)ds + (1 − )μ(u) +

t−

τ 2t ∫ (τ τ2 Γ(γ+1) 0

− s)γ q(s)ds.

(2.2)

τ

Proof: From lemma 2.2 with the boundary conditions μ(u) = k 0 and ∫0 u(t)dt = ρ, then; γ

u(t) = I0+ q(t) + k 0 + k1 t where k 0 , k1 ∈ R, u(t) =

t 1 ∫ (t Γ(γ) 0

𝜏

− s)γ−1 q(s)ds + k 0 + k1 t τ

1

t

τ

τ

∫0 𝑢(𝑡)𝑑𝑡 = Γ(γ) ∫0 ∫0 (t − s)γ−1 q(s)ds + ∫0 k 0 dt + ∫0 k1 t dt 2ρ =

τ 2 ∫ (τ Γ(γ+1) 0 2𝜌

𝑘1 =

𝜏2

− s)γ q(s)ds + 2k 0 τ + τ2 k1 ;

2

− 𝜇(𝑢) − 𝜏

𝜏 2 ∫ (𝜏 𝜏2 𝛤(𝛾+1) 0

− 𝑠)𝛾 𝑞(𝑠)𝑑𝑠

Hence; u(t) =

t 1 ∫ (t Γ(γ) 0

2t

2ρ

τ

τ2

− s)γ−1 q(s)ds + (1 − )μ(u) +

t−

τ 2t ∫ (τ τ2 Γ(γ+1) 0

− s)γ q(s)ds.

Prove is completed.

Lemma 2.4 Let γ ∈ (1, 2] and q: [0, τ] → R be continuous. Then u satisfies the fractional integral equation; u(t) =

t 1 ∫ (t Γ(γ) 0

t

t

τ

τ

− s)γ−1 q(s)ds + (1 − )μ(u) + Uτ −

τ t ∫ (τ τΓ(γ) 0

− s)γ−1 q(s)ds

(2.3)

Iff u satisfies BVP (2.4); {

𝑐 𝛾 .𝐷0+ 𝑢(𝑡)

= 𝑞(𝑠), 𝑡 ∈ [0, 𝜏] (2.4)

μ(u) = k 0 , u(τ) = Uτ Proof: From lemma 2.2 and the boundary conditions μ(u) = k 0 , u(τ) = Uτ then; γ

u(t) = I0+ q(t) + k 0 + k1 t for some k 0 , k1 ∈ R, u(t) =

t 1 ∫ (t Γ(γ) 0

− s)γ−1 q(s)ds + k 0 + k1 t

this implies that; τ 1 ∫ (τ Γ(γ) 0

u(τ) = 𝑈𝜏 = 𝑘1 =

τ 1 ∫ (τ Γ(γ) 0 𝑈𝜏 𝜏

− s)γ−1 q(s)ds + k 0 + k1 τ

− s)γ−1 q(s)ds + μ(u) + k1 τ

1

− 𝜇(𝑢) − 𝜏

𝜏 1 ∫ (𝜏 𝜏𝛤(𝛾) 0

− 𝑠)𝛾−1 𝑞(𝑠)𝑑𝑠

Hence, u(t) =

t 1 ∫ (t Γ(γ) 0

− s)γ−1 q(s)ds −

τ t ∫ (τ τΓ(γ) 0

t

t

τ

τ

− s)γ−1 q(s)ds + (1 − )μ(u) + Uτ

Prove is completed.

III.

MAIN RESULTS

Now we find appropriate to establish our main findings.

Theorem 3.1 Suppose; (A1). A non-negative constant L exists and satisfies; |𝑔(𝑡, 𝑢1 , 𝑣1 ) − 𝑔(𝑡, 𝑢2 , 𝑣2 )| ≤ 𝐿(|𝑢1 − 𝑢2 | + |𝑣1 − 𝑣2 |) 𝑓𝑜𝑟 𝑎𝑛𝑦 t ∈ [0, τ] ∀ ui , vi ∈ R. (𝐴2 ) A non-negative constant 𝐿1 exists and satisfies; |𝜇(𝑡, 𝑢) − 𝜇(𝑡, 𝑣)| ≤ 𝐿1 |𝑢 − 𝑣|, ∀ 𝑡 ∈ [0, 𝜏] 𝑎𝑛𝑑 𝑢, 𝑣 ∈ 𝑅.

www.ijaers.com

Page | 299

John Fiifi Essel et al.

(𝐴3 ) If φ = max{(

International Journal of Advanced Engineering Research and Science, 8(10)-2021 τγ

Γ(γ+1)

+

2τγ−1

)2L + L1 ,

Γ(γ+2)

τ Γ(2−δ)

[(

τγ−1 Γ(γ)

+

2τγ−2

)2L + L1 ]} < 1.

Γ(γ+2)

Then the BVP (2.1) has a unique solution.

Proof: We convert BVP (2.1) to a problem with fixed point by considering the operator P ∶ U → U, where U ∈ C([0, τ], R) defined by; P(u)(t) =

t 1 ∫ (t Γ(γ) 0

− s)γ−1 g(s, u(s), c.Dδ0+ u(s))ds −

2t

2ρ

τ

τ2

+(1 − )μ(u) +

𝜏 2𝑡 ∫ (𝜏 𝜏2 𝛤(𝛾+1) 0

𝛿 − 𝑠)𝛾 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠))𝑑𝑠

t

(2.5)

Obviously, P has fixed points which satisfies problem (2.1). Now, from (𝐴1 ) we show that P has a fixed point and hence maps onto itself.

Let 𝑢1 , 𝑢2 ∈ 𝑢, then; 𝑡 1 ∫ (𝑡 𝛤(𝛾) 0

|𝑃(𝑢1 )(𝑡) − 𝑃(𝑢2 )(𝑡)| ≤ +

𝛿 𝛿 − 𝑠)𝛾−1 |𝑔(𝑠, 𝑢1 (𝑠), 𝑐.𝐷0+ 𝑢1 (𝑠)) − 𝑔(𝑠, 𝑢1 (𝑠), 𝑐.𝐷0+ 𝑢2 (𝑠))|𝑑𝑠

𝜏 2𝑡 𝛿 𝛿 ∫ (𝜏 − 𝑠)𝛾 |𝑔(𝑠, 𝑢1 (𝑠), 𝑐.𝐷0+ 𝑢1 (𝑠)) − 𝑔(𝑠, 𝑢1 (𝑠), 𝑐.𝐷0+ 𝑢2 (𝑠))𝑑𝑠| 𝜏 2 𝛤(𝛾 + 1) 0 2𝑡

+ |1 − | |𝜇(𝑢1 ) − 𝜇(𝑢2 )|.

(2.6)

𝜏

From (𝐴1 ), it implies that; 𝛿 𝛿 𝛿 𝛿 |𝑔(𝑠, 𝑢1 (𝑠), 𝑐.𝐷0+ 𝑢1 (𝑠)) − 𝑔(𝑠, 𝑢1 (𝑠), 𝑐.𝐷0+ 𝑢2 (𝑠))| ≤ L(|u1 (s) − u2 (s)| + | 𝑐.𝐷0+ 𝑢1 (𝑠)) − 𝑐.𝐷0+ 𝑢2 (𝑠))| 𝛿 𝛿 ≤ L(‖u1 − u2 ‖ + ‖ 𝑐.𝐷0+ 𝑢1 − 𝑐.𝐷0+ 𝑢2 ‖

≤ 2L‖u1 − u2 ‖∗

(2.7)

And from (𝐴2 ), |𝜇(𝑢1 ) − 𝜇(𝑢2 )| ≤ 𝐿1 ‖u1 − u2 ‖∗

(2.8)

Therefore from (2.6), |𝑃(𝑢1 )(𝑡) − 𝑃(𝑢2 )(𝑡)| ≤

𝑡 1 ∫ (𝑡 𝛤(𝛾) 0

+ ‖𝑃(𝑢1 ) − 𝑃(𝑢2 )‖ ≤

𝜏𝛾 𝛾𝛤(𝛾)

≤(

𝜏

𝜏 2𝑡 ∫ (𝜏 𝜏2 𝛤(𝛾+1) 0

− 𝑠)𝛾 𝑑𝑠. 2L‖u1 − u2 ‖∗

. 2L‖u1 − u2 ‖∗ + 𝐿1 ‖u1 − u2 ‖∗ +

2Lτγ

𝛤(𝛾+1)

≤ [(

2𝑡

− 𝑠)𝛾−1 𝑑𝑠. 2L‖u1 − u2 ‖∗ + |1 − | . 𝐿1 ‖u1 − u2 ‖∗

+

τγ 𝛤(𝛾+1)

4Lτγ−1 𝛤(𝛾+2)

+

2 τ(γ+1)Γ(γ+1)

τγ . 2L‖u1 − u2 ‖∗

). ‖u1 − u2 ‖∗ + 𝐿1 ‖u1 − u2 ‖∗

2τγ−1

)2L + 𝐿1 ]‖u1 − u2 ‖∗

𝛤(𝛾+2)

Again from (2.5); (Pu)′ (t) =

t 1 ∫ (t Γ(γ−1) 0

2

2ρ

τ

τ2

− s)γ−2 g(s, u(s), c.Dδ0+ u(s))ds − μ(u) +

−

𝜏 2 ∫ (𝜏 𝜏2 𝛤(𝛾+1) 0

𝛿 − 𝑠)𝛾 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠)𝑑𝑠

Then; |(𝑃𝑢1 )′ (𝑡) − (𝑃𝑢2 )′ (𝑡)| ≤

𝑡 1 ∫ (𝑡 𝛤(𝛾−1) 0

+

𝛿 𝛿 − 𝑠)𝛾−2 |𝑔(𝑠, 𝑢1 (𝑠), 𝑐.𝐷0+ 𝑢1 (𝑠)) − 𝑔(𝑠, 𝑢2 (𝑠), 𝑐.𝐷0+ 𝑢2 (𝑠))|𝑑𝑠

𝜏 2 𝛿 𝛿 ∫ (𝜏 − 𝑠)𝛾 |𝑔(𝑠, 𝑢1 (𝑠), 𝑐.𝐷0+ 𝑢1 (𝑠)) − 𝑔(𝑠, 𝑢2 (𝑠), 𝑐.𝐷0+ 𝑢2 (𝑠))|𝑑𝑠 𝜏 2 𝛤(𝛾 + 1) 0

+|𝜇(𝑢1 ) − 𝜇(𝑢2 )| From (2.7) and (2.8), |(𝑃𝑢1 )′ (𝑡) − (𝑃𝑢2 )′ (𝑡)| ≤

www.ijaers.com

𝑡 𝛾−1 𝛤(𝛾)

. 2L‖u1 − u2 ‖∗ + 𝐿1 ‖u1 − u2 ‖∗ +

2τγ−2 Γ(γ+2)

. 2L‖u1 − u2 ‖∗

Page | 300

John Fiifi Essel et al.

International Journal of Advanced Engineering Research and Science, 8(10)-2021

≤ [(

τγ−1 Γ(γ)

+

2τγ−2

)2L + 𝐿1 ] . ‖u1 − u2 ‖∗

Γ(γ+2)

But; 𝑐 𝛿 ′ .𝐷0+ (𝑃𝑢) (𝑡)

𝑡 1 ∫ (𝑡 𝛤(1−𝛿) 0

= = =

1

.

t1−δ

Γ(1−δ) 1−δ 1 Γ(2−δ)

− 𝑠)−𝛿 𝑑𝑠. (𝑃𝑢)′ (𝑡) (𝑃𝑢)′ (𝑡)

. (𝑃𝑢)′ (𝑡)

Therefore; 𝛿 𝛿 | 𝑐.𝐷0+ (𝑃𝑢1 )′ (𝑡) − 𝑐.𝐷0+ (𝑃𝑢2 )′ (𝑡)| ≤

≤

𝑡 1 ∫ (𝑡 𝛤(1−𝛿) 0 τ

Γ(2−δ)

− 𝑠)−𝛿 𝑑𝑠 |(𝑃𝑢1 )′ (𝑠) − (𝑃𝑢2 )′ (𝑠)|

τγ−1

[(

Γ(γ)

+

2τγ−2

)2L + 𝐿1 ]‖u1 − u2 ‖∗

Γ(γ+2)

Hence, 𝛿 𝛿 ‖𝑃𝑢1 − 𝑃𝑢2 ‖∗ = 𝑚𝑎𝑥{‖𝑃(𝑢1 ) − 𝑃(𝑢2 )‖, ‖ 𝑐.𝐷0+ (𝑃𝑢1 )′ (𝑡) − 𝑐.𝐷0+ (𝑃𝑢2 )′ (𝑡)‖} ≤ φ‖u1 − u2 ‖∗

Hence, P with φ ≤ 1 contracts. Therefore as a result of the contraction principle, we infer that P has a fixed point which is unique and satisfies (2.1). Consequently, the proof is completed.

Theorem 3.2 Suppose g ∶ [0, τ] × R × R → R and μ ∶ C([0,1], R) → R are continuous functions and satisfy; (𝐵1 ) |𝑔(𝑡, 𝑢, 𝑣)| ≤ 𝑘 for any t ∈ [0, τ] ∀ u, v ∈ R, (𝐵2 ) |𝜇(𝑡, 𝑢) − 𝜇(𝑡, 𝑣)| ≤ 𝑘1 |𝑢 − 𝑣| for any t ∈ [0, τ] ∀ u, v ∈ R. (𝐵3 ) |𝜇(𝑢)| ≤ 𝑘∗ ∀ 𝜇 ∈ 𝐶([0, 𝜏], 𝑅). where k and 𝑘∗ are positive constants, then BVP (2.1) has a solution.

Proof: To ascertain our results, we use the Schaefer’s fixed point theorem. We divide the proof into four parts for simplicity and clarity.

Step 1: Firstly, we establish the continuity of P. Let 𝑢𝑛 ∈ 𝐶([0, 𝜏], 𝑅) be a sequence such that 𝑢𝑛 → 𝑢. Then ∀ t ∈ [0, τ], we derive; P(un )(t) − P(u)(t) ≤

𝑡 1 ∫ (𝑡 𝛤(𝛾) 0

+

𝛿 𝛿 − 𝑠)𝛾−1 |𝑔(𝑠, 𝑢𝑛 (𝑠), 𝑐.𝐷0+ 𝑢𝑛 (𝑠)) − 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠))|𝑑𝑠

𝜏 2 ∫ (𝜏 𝜏2 𝛤(𝛾+1) 0

𝛿 𝛿 − 𝑠)𝛾 |𝑔(𝑠, 𝑢𝑛 (𝑠), 𝑐.𝐷0+ 𝑢𝑛 (𝑠)) − 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠))|𝑑𝑠

+|𝜇(𝑢𝑛 ) − 𝜇(𝑢)| From (𝐵1 ) and (𝐵2 ); 𝛿 𝛿 |𝑔(𝑡, 𝑢𝑛 (𝑡), 𝑐.𝐷0+ 𝑢𝑛 (𝑡)) − 𝑔(𝑡, 𝑢(𝑡), 𝑐.𝐷0+ 𝑢(𝑡))| ≤ k(‖un (s) − u(s)‖ + 𝛿 𝛿 ‖ 𝑐.𝐷0+ 𝑢𝑛 (𝑠) − 𝑐.𝐷0+ 𝑢(𝑠)‖) 𝛿 𝛿 ≤ k(‖un − u‖ + ‖ 𝑐.𝐷0+ 𝑢𝑛 − 𝑐.𝐷0+ 𝑢‖)

≤ 2k‖un − u‖∗ and ‖𝜇(𝑢𝑛 ) − 𝜇(𝑢)‖ ≤ 𝑘1 ‖𝑢𝑛 − 𝑢‖ Therefore, |𝑃(𝑢𝑛 )(𝑡) − 𝑃(𝑢)(𝑡)| ≤

𝑡 1 ∫ (𝑡 𝛤(𝛾) 0

− 𝑠)𝛾−1 𝑑𝑠. 2k‖un − u‖∗ + 𝑘1 ‖𝑢𝑛 − 𝑢‖ +

𝑇 2 ∫ (𝑇 𝑇 2 𝛤(𝛾+1) 0

− 𝑠)𝛾 𝑑𝑠 .2k‖un − u‖∗

also, 𝛿 𝛿 | 𝑐.𝐷0+ 𝑃(𝑢𝑛 )(𝑡) − 𝑐.𝐷0+ 𝑃(𝑢)(𝑡)| ≤

www.ijaers.com

𝑡 1 ∫ (𝑡 𝛤(1−𝛿) 0

− 𝑠)−𝛿 |𝑃(𝑢𝑛 )(𝑠) − (𝑃𝑢)(𝑠)|𝑑𝑠

Page | 301

John Fiifi Essel et al.

International Journal of Advanced Engineering Research and Science, 8(10)-2021

𝛿 𝛿 Since the functions f and y are continuous, it implies that ‖𝑃(𝑢𝑛 ) − 𝑃(𝑢)‖ → 0 and ‖ 𝑐.𝐷0+ 𝑃(𝑢𝑛 ) − 𝑐.𝐷0+ 𝑃(𝑢)‖ → 0 as n → ∞. Therefore P is continuous.

Step 2: Now we prove the boundedness of P. By (𝐵1 ) 𝑎𝑛𝑑 (𝐵3 ) from (2.5) for any t ∈ [0, τ], we derive; |𝑃(𝑢)(𝑡)| ≤

𝑡 1 ∫ (𝑡 𝛤(𝛾) 0 t k ∫ (t Γ(γ) 0

≤

k

≤

γΓ(γ)

𝜏

Γ(γ+1)

𝜏 2𝑘 ∫ (𝜏 𝜏𝛤(𝛾+1) 0

2

− s)γ−1 ds + 2k ∗ + |𝜌| + 𝜏

. τγ +

k

≤

2

𝛿 − 𝑠)𝛾−1 |𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠))|𝑑𝑠 + 2|𝜇(𝑢)| + |𝜌| +

2k τ

. τγ +

.

τγ+1

𝛿 − 𝑠)𝛾 |𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠)|𝑑𝑠

− 𝑠)𝛾 𝑑𝑠

2

(γ+1)Γ(γ+1) 2k

𝜏 2 ∫ (𝜏 𝜏𝛤(𝛾+1) 0

+ 2k ∗ + |𝜌| 𝜏

2

Γ(γ+2)

. τγ + 2k ∗ + |𝜌| 𝜏

and, 𝛿 𝛿 |𝑃( 𝑐.𝐷0+ 𝑢(𝑡))| = | 𝑐.𝐷0+ 𝑃(𝑢)(𝑠)| ≤

≤

𝑡 1 ∫ (𝑡 𝛤(1−𝛿) 0 τ

− 𝑠)−𝛿 𝑑𝑠. |(𝑃𝑢)(𝑠)|

k

[

Γ(2−δ) Γ(γ+1)

. τγ +

2k

2

Γ(γ+2)

. τγ + 2k ∗ + |𝜌| 𝜏

Thus, P is evenly bounded.

Step 3: Next we establish that P is completely continuous.

Let 𝑡1 , 𝑡2 ∈ (0, 𝜏] where 𝑡1 < 𝑡2 , then; |𝑃(𝑢)(𝑡2 ) − 𝑃(𝑢)(𝑡1 )| = 𝑡1 1 𝛿 | ∫ [(𝑡2 − 𝑠)𝛾−1 − (𝑡1 − 𝑠)𝛾−1 ]𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠)) 𝑑𝑠 𝛤(𝛾) 0 𝑡2 1 𝛿 + ∫ [(𝑡2 − 𝑠)𝛾−1 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠))| 𝑑𝑠 𝛤(𝛾) 𝑡1

+

2(𝑡2 −𝑡1 )

≤

𝜏

𝑡1 k ∫ [(𝑡2 𝛤(𝛾) 0

+ ≤

|𝜇(𝑢)| +

k

𝜏2

|𝜌| +

2(𝑡2 −𝑡1 ) 𝜏 ∫ (𝜏 𝜏2 𝛤(𝛾+1) 0

− 𝑠)𝛾−1 − (𝑡1 − 𝑠)𝛾−1 ]𝑑𝑠 +

𝜏 2𝑘 ∫ (𝜏 𝑇 2 𝛤(𝛾+1) 0

Γ(γ+1)

2(𝑡2 −𝑡1 )

𝛿 − 𝑠)𝛾 |𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠))|𝑑𝑠

𝑡2 𝑘 ∫ (𝑡 𝛤(𝛾) 𝑡1 2

− 𝑠)𝛾−1 𝑑𝑠 +

2𝑘∗ 𝑇

(𝑡2 − 𝑡1 ) +

2|𝜌| 𝑇2

(𝑡2 − 𝑡1 )

− 𝑠)𝛾 𝑑𝑠. (𝑡2 − 𝑡1 ) γ

k

γ

[(t 2 − t1 )γ + t 2 − t1 ] +

Γ(γ+1)

(t 2 − t1 )γ +

2kτγ+1 τ2 Γ(γ+2)

(t 2 − t1 ) +

2𝑘∗ 𝜏

(t 2 − t1 ) +

2|𝜌| 𝜏2

(𝑡2 − 𝑡1 )

(2.9)

From (2.9), it can be easily seen that as 𝑡1 → 𝑡2 , the RHS tends to zero. Together with the Arzela - Ascoli theorem and the preceding steps, we conclude that P is completely continuous.

Step 4: Finally, we assume P maps onto itself, such that the set Φ = {u ∈ E: u = λPu for any λ ∈ (0, 1)} is bounded.

Now ∀ t ∈ [0, τ], P(u)(t) =

𝑡 λ ∫ (𝑡 Γ(γ) 0

−λ

𝑡

2𝑡

𝜏

𝜏2

𝛿 − 𝑠)𝛾−1 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠))𝑑𝑠 + 𝜆(1 − )𝜇(𝑢) + 𝜆

𝜏 2𝑡 ∫ (𝜏 𝜏2 𝛤(𝛾+1) 0

𝜌

𝛿 − 𝑠)𝛾 . 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷0+ 𝑢(𝑠))𝑑𝑠

By (B1 ) and (B3 ) for any t ∈ [0, τ]; |P(u)(t)| ≤ ≤

𝑡 𝑘 ∫ (𝑡 𝛤(𝛾) 0 𝑘 𝛤(𝛾+1)

2

− 𝑠)𝛾−1 𝑑𝑠 + 𝑘∗ + |𝜌| +

𝜏𝛾 +

𝜏

2𝑘 𝛤(𝛾+2)

𝜏 2𝑘 ∫ (𝜏 𝜏𝛤(𝛾+1) 0

− 𝑠)𝛾 𝑑𝑠

2

𝜏 𝛾+1 + 𝑘∗ + |𝜌| 𝜏

Hence the set Φ is bounded. Therefore P has a fixed point which is a solution to

www.ijaers.com

Page | 302

John Fiifi Essel et al.

International Journal of Advanced Engineering Research and Science, 8(10)-2021

problem(2.1). The end of prove.

At this point, we examine the solutions of (2.4) to ascertain its existence and uniqueness.

Theorem 3.3 Suppose; (𝐶1 ) A non-negative constant L exists and satisfies; |𝑔(𝑡, 𝑢1 , 𝑣1 ) − 𝑔(𝑡, 𝑢2 , 𝑣2 )| ≤ 𝐿(|𝑢1 − 𝑢2 | + |𝑣1 − 𝑣2 |) for each t ∈ J and all 𝑢𝑖 , 𝑣𝑖 ∈ 𝑅. (𝐶2 ) A constant 𝐿1 > 0 exists and satisfies condition; |𝜇(𝑢0 ) − 𝜇(𝑢1 )| < 𝐿1 |𝑢0 − 𝑢1 |, for any t ∈ J and all 𝑢0 , 𝑢1 ∈ 𝑈 (𝐶3 ) If φ = max{(

τγ

Γ(γ+1)

τγ−1

+

)2L + L1 ,

Γ(γ+1)

τγ−1

τ Γ(2−β)

[(

+

Γ(γ)

τγ−1

)2L + L1 ] } < 1.

Γ(γ+1)

Then the BVP (2.4) has a unique solution.

Proof: We convert BVP (2.4) to a problem with fixed point by considering the operator P ∶ U → U defined by; P(u)(t) =

t 1 ∫ (t Γ(γ) 0

−

t

t

τ

τ

− s)γ−1 g(s, u(s), c.Dδ0+ u(s))ds + (1 − )μ(u) + Uτ

τ t ∫ (τ τΓ(γ) 0

− s)γ−1 g(s, u(s), c.Dδ0+ u(s))ds

(2.10)

Now we show that P has a fixed point and hence is a contraction.

Let 𝑢1 , 𝑢2 ∈ 𝑈 ∀ 𝑡 ∈ [0, 𝜏]; |𝑃(𝑢1 )(𝑡) − 𝑃(𝑢2 )(𝑡)| ≤

t 1 ∫ (t Γ(γ) 0

+

− s)γ−1 |g(s, u1 (s), c.Dδ0+ u1 (s)) − g(s, u2 (s), c.Dδ0+ u2 (s))|ds

τ t ∫ (τ τΓ(γ) 0

− s)γ−1 |g(s, u1 (s), c.Dδ0+ u1 (s)) − g(s, u2 (s), c.Dδ0+ u2 (s))|ds

+|μ(u1 ) − μ(u2 )| From (𝐶1 ) and (𝐶2 ); |g(t, u1 (t), c.Dδ0+ u1 (t)) − g(t, u2 (t), c.Dδ0+ u2 (t))| ≤ L(|u1 (s) − u2 (s)| + | c.Dδ0+ u1 (s) − c.Dδ0+ u2 (s)|) ≤ L(‖u1 − u2 ‖ + ‖ c.Dδ0+ u1 − c.Dδ0+ u2 ‖) ≤ 2L‖u1 − u2 ‖1

(2.11)

And, |𝜇(𝑢1 )(𝑡) − 𝜇(𝑢2 )(𝑡)| ≤ 𝐿1 ‖u1 (s) − u2 (s)‖ ≤ 𝐿1 ‖𝑢1 − 𝑢2 ‖1

(2.12)

From (2.11) and (2.12), |𝑃(𝑢1 )(𝑡) − 𝑃(𝑢2 )(𝑡)| ≤

𝜏𝛾 𝛤(𝛾+1)

≤ [(

. 2𝐿‖u1 − u2 ‖1 +

𝜏𝛾 𝛤(𝛾+1)

+

𝜏𝛾−1

𝜏𝛾−1 𝛤(𝛾+1)

. 2L‖u1 − u2 ‖1 + L1 ‖u1 − u2 ‖1

)2𝐿 + 𝐿1 ]‖u1 − u2 ‖1

𝛤(𝛾+1)

Again from (2.10), (𝑃𝑢)′ (𝑡) =

t 1 ∫ (t Γ(γ−1) 0

1

1

τ

τ

− s)γ−2 g(s, u(s), c.Dδ0+ u(s))ds − μ(u) + Uτ −

τ 1 ∫ (τ τΓ(γ) 0

− s)γ−1 g(s, u(s), c.Dδ0+ u(s))ds

Then; |(𝑃𝑢1 )′ (𝑡) − (𝑃𝑢2 )′ (𝑡)| ≤ +

t 1 ∫ (t Γ(γ−1) 0

− s)γ−2 |g(s, u1 (s), c.Dδ0+ u1 (s)) − g(s, u2 (s), c.Dδ0+ u2 (s))|ds

τ 1 ∫ (τ − s)γ−1 |g(s, u1 (s), c.Dδ0+ u1 (s)) − g(s, u2 (s), c.Dδ0+ u2 (s))|ds τΓ(γ) 0 1

+ |μ(u1 ) − μ(u2 )| τ

www.ijaers.com

Page | 303

John Fiifi Essel et al.

International Journal of Advanced Engineering Research and Science, 8(10)-2021

From (2.11) and (2.12), ‖(𝑃𝑢1 )′ − (𝑃𝑢2 )′ ‖ ≤

𝜏𝛾−1 𝛤(𝛾)

≤ [(

2𝐿‖u1 − u2 ‖1 + L1 ‖u1 − u2 ‖1 +

𝜏𝛾−1 𝛤(𝛾)

+

𝜏𝛾

𝜏𝛾 𝛤(𝛾+1)

2𝐿‖u1 − u2 ‖1

)2𝐿 + 𝐿1 ]‖u1 − u2 ‖1

𝛤(𝛾+1)

and 𝑡 1 ∫ (𝑡 𝛤(1−𝛿) 0

| c.Dδ0+ (Pu1 )(t) − c.Dδ0+ (Pu2 )(t)| ≤ ≤

𝜏 (2−𝛿)

𝜏𝛾−1

[(

𝛤(𝛾)

− 𝑠)−𝛿 |(𝑃𝑢1 )′ (𝑡) − (𝑃𝑢2 )′ (𝑡)|𝑑𝑠 +

𝜏𝛾

)2𝐿 + 𝐿1 ]‖u1 − u2 ‖1

𝛤(𝛾+1)

Therefore; ‖𝑃(𝑢1 ) − 𝑃(𝑢2 )‖1 = 𝑚𝑎𝑥{‖𝑃(𝑢1 ) − 𝑃(𝑢2 )‖, ‖ 𝑐.𝐷𝛿 𝑃(𝑢1 ) − 𝑐.𝐷 𝛿 𝑃(𝑢2 )‖} ≤ 𝜑‖u1 − u2 ‖1 Therefore with φ < 1, P is said to contract and hence uniquely satisfies the BVP (2.4). Thus, the end of proof..

Theorem 3.4 Suppose g ∶ [0, τ] × R × R → R and μ ∶ C([0, 1], R) → R are continuous functions and satisfy; (𝐷1 ) |𝑔(𝑡, 𝑢, 𝑣)| ≤ 𝑘 for any t ∈ J ∀ u, v ∈ R. (𝐷2 ) |𝜇(𝑢)| ≤ 𝑘1 ∀ 𝑢 ∈ 𝐶([0, 𝜏], 𝑅). where k and k1 are positive constants, then BVP (2.4) has a solution.

Proof: Similarly, we apply the Schaefer’s theorem of fixed points in proving our results.

Step 1: We prove that P is continuous. Let the sequence 𝑢𝑛 exist in such a way that 𝑢𝑛 → 𝑢 ∈ 𝑈. For any t ∈ [0, τ]; |𝑃(𝑢𝑛 )(𝑡) − 𝑃(𝑢)(𝑡)| ≤

𝑡 1 ∫ (𝑡 𝛤(𝛾) 0

+

− 𝑠)𝛾−1 |𝑔(𝑠, 𝑢𝑛 (𝑠), 𝑐.𝐷𝛿 𝑢𝑛 (𝑠)) − 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷𝛿 𝑢(𝑠))|𝑑𝑠 + |𝜇(𝑢𝑛 ) − 𝜇(𝑢)|

𝜏 1 ∫ (𝜏 𝜏𝛤(𝛾) 0

− 𝑠)𝛾−1 |𝑔(𝑠, 𝑢𝑛 (𝑠), 𝑐.𝐷𝛿 𝑢𝑛 (𝑠)) − 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷𝛿 𝑢(𝑠))|𝑑𝑠

From assumptions (𝐷2 ) and (D3 ); |𝑔(𝑠, 𝑢𝑛 (𝑠), 𝑐.𝐷𝛿 𝑢𝑛 (𝑠)) − 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷𝛿 𝑢(𝑠))| ≤ 2k‖un − u‖2 and |𝜇(𝑢𝑛 ) − 𝜇(𝑢)| ≤ 𝑘1 ‖un − u‖2 Therefore; |𝑃(𝑢𝑛 )(𝑡) − 𝑃(𝑢)(𝑡)| ≤

𝜏𝛾 𝛤(𝛾+1)

2k‖un − u‖2 + 𝑘1 ‖un − u‖2 +

τγ−1 Γ(γ+1)

2k‖un − u‖2

also, | 𝑐.𝐷𝛿 𝑃(𝑢𝑛 )(𝑡) − 𝑐.𝐷𝛿 𝑃(𝑢)(𝑡)| ≤

𝑡 1 ∫ (𝑡 𝛤(1−𝛿) 0

− 𝑠)−𝛿 |𝑃(𝑢𝑛 )(𝑠) − 𝑃(𝑢)(𝑠)|𝑑𝑠

From the above, we conclude that P is continuous since the functions g and μ are continuous which implies that ‖𝑃(𝑢𝑛 ) − 𝑃(𝑢)‖ → 0 and ‖ 𝑐.𝐷𝛿 𝑃(𝑢𝑛 )(𝑡) − 𝑐.𝐷𝛿 𝑃(𝑢)(𝑡)‖ → 0 as n approaches infinity.

Step 2: P transforms bounded sets into another sets bounded in U. A non-negative constant r exists and satisfies ‖𝑃(𝑢)‖ ≤ 𝑟.

Hence, for t ∈ [0, τ] we derive; By (𝐷1 ) and (D2 ); |𝑃(𝑢)(𝑡)| ≤ ≤ ≤

𝑡 1 ∫ (𝑡 𝛤(𝛾) 0 t k ∫ (t Γ(γ) 0 k Γ(γ+1)

− 𝑠)𝛾−1 |𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷𝛿 𝑢(𝑠))|𝑑𝑠 + |𝜇(𝑢)| + |𝑈𝜏 | +

− s)γ−1 ds + k1 + |Uτ | +

τγ +

k Γ(γ+1)

τ k ∫ (τ Γ(γ) 0

𝜏 1 ∫ (𝜏 𝛤(𝛾) 0

− 𝑠)𝛾−1 |𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷𝛿 𝑢(𝑠))|𝑑𝑠

− s)γ−1 ds

τγ + k1 + |Uτ |

Therefore;

www.ijaers.com

Page | 304

John Fiifi Essel et al.

2𝜏𝛾

‖𝑃(𝑢)‖ ≤

𝛤(𝛾+1)

International Journal of Advanced Engineering Research and Science, 8(10)-2021

𝑘 + 𝑘1 + |Uτ | = r

Step 3: P maps any bounded sets into an equi-continuous sets in U. Let 𝑡1 , 𝑡2 ∈ [0, 𝜏], 𝑡1 ≤ 𝑡2 . Then from (2.10) we have; |𝑃(𝑢)(𝑡2 ) − 𝑃(𝑢)(𝑡1 )| =

+ + ≤

|∫0 1[(𝑡2 − 𝑠)𝛾−1 − (𝑡1 − 𝑠)𝛾−1 ] 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷𝛿 𝑢(𝑠))| 𝑑𝑠 1

𝛤(𝛾)

𝑡

(𝑡2 −𝑡1 )

1

𝜏

|∫𝑡 2(𝑡 − 𝑠)𝛾−1 𝑔(𝑠, 𝑢 (𝑠), 𝑐.𝐷𝛿 𝑢(𝑠))| 𝑑𝑠 +

(𝑡2 −𝑡1 ) 𝜏

|Uτ | +

𝑡1 k ∫ [(𝑡2 Γ(γ) 0

+ ≤

𝑡

1 𝛤(𝛾)

(𝑡2 −𝑡1 ) 𝜏

M Γ(γ+1)

(𝑡2 −𝑡1 ) 𝜏𝛤(𝛾)

𝜏

∫0 (𝜏 − 𝑠)𝛾−1 |𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷 𝛿 𝑢(𝑠))|𝑑𝑠

− 𝑠)𝛾−1 − (𝑡1 − 𝑠)𝛾−1 ] 𝑑𝑠 +

|Uτ | +

𝑘(𝑡2 −𝑡1 ) 𝜏𝛤(𝛾) γ

|𝜇(𝑢)|

𝑡2 𝑘 ∫ (𝑡 𝛤(𝛾) 𝑡1

− 𝑠)𝛾−1 𝑑𝑠 +

(𝑡2 −𝑡1 ) 𝜏

𝑘1

𝜏

∫0 (𝜏 − 𝑠)𝛾−1 𝑑𝑠 k

γ

[(t 2 − t1 )γ + t 2 − t1 ] +

Γ(γ+1)

(t 2 − t1 )γ +

(t2 −t1 ) τ

k1 +

(𝑡2 −𝑡1 ) 𝜏

|Uτ | +

𝑘𝜏𝛾−1 𝛤(𝛾+1)

(𝑡2 − 𝑡1 )

(2.13) As 𝑡1 → 𝑡2 , the RHS of (2.13) tends to zero. Therefore P is completely continuous as a consequence of the above steps and the Arzela-Ascoli theorem.

Step 4: Finally, we prove the set ϕ = {u ∈ E; u = λPu for any λ ∈ (0, 1)} Let u ∈ ϕ, then u = λP(u) ∀ λ ∈ (0, 1) and 0 ≤ t ≤ τ, 𝑡 𝜆 ∫ (𝑡 𝛤(𝛾) 0

P(u)(t) =

𝑡

𝜆𝑡

𝜏

𝜏

− 𝑠)𝛾−1 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷𝛿 𝑢(𝑠)) + 𝜆(1 − )𝜇(𝑢) −

𝑈𝜏 −

𝜏 𝜆𝑡 ∫ (𝜏 𝜏𝛤(𝛾) 0

− 𝑠)𝛾−1 𝑔(𝑠, 𝑢(𝑠), 𝑐.𝐷𝛿 𝑢(𝑠))𝑑𝑠

From (𝐷2 ) and (D3 ); |𝑃(𝑢)(𝑡)| ≤ ≤ ‖𝑃(𝑢)‖ ≤

𝑡 𝑘 ∫ (𝑡 𝛤(𝛾) 0 k Γ(γ+1) 𝜏𝛾

𝛤(𝛾+1)

− 𝑠)𝛾−1 𝑑𝑠 + 𝑘1 + |𝑈𝜏 | +

. τγ +

k Γ(γ+1)

𝜏 𝑘 ∫ (𝜏 𝛤(𝛾) 0

− 𝑠)𝛾−1 𝑑𝑠

. τγ + 𝑘1 + |𝑈𝜏 |

. 2𝑘 + 𝑘1 + |𝑈𝜏 |

This shows that the set ϕ is bounded and hence satisfies problem (2.10).

IV.

APPLICATION

At this point, we apply our obtained results to some selected examples.

Example 4.1 Consider the fractional boundary value problem; 𝑐 1.5 .𝐷0+ 𝑢(𝑡)

𝑘0 =

0.5 = 𝑡 + 0.1𝑡𝑢(𝑡) + (0.1𝑡 2 ) 𝑐.𝐷0+ 𝑢(𝑡);

∑𝑛𝑖=1 𝑐𝑖 𝑢(𝑡𝑖 ),

𝜏 ∫0 𝑢(𝑡)𝑑𝑡

𝑡 ∈ (0, 1],

𝛾 ∈ (1, 2],

=1

Where 𝑡𝑖 ∈ (0, 1), 𝑐𝑖 , 𝑖 = 1, 2, . . . , 𝑛 − 1, 𝑛 are non-negative constants with 1

∑𝑛𝑖=1 𝑐𝑖 < . Set γ = 1.5 (n ≥ 2), δ = 0.5, μ(u) = ∑ni=1 ci u(𝑡𝑖 ); 𝜏 = 1, 𝜌 = 1 and 2

𝛿 𝑔(𝑡, 𝑢(𝑡), 𝑐.𝐷0+ 𝑢(𝑡))

0.5 = 𝑡 + 0.1𝑡𝑢(𝑡) + (0.1𝑡 2 ) 𝑐.𝐷0+ 𝑢(𝑡).

Let t ∈ [0, τ] and 𝑢𝑖 , 𝑣𝑖 ∈ 𝑅, where i = 1, 2.

Then we have; |𝑔(𝑡, 𝑢1 , 𝑢2 ) − 𝑔(𝑡, 𝑣1 , 𝑣2 )| = |0.1𝑡(𝑢1 ) − 0.1𝑡(𝑢2 )| + |0.1𝑡 2 (𝑣1 ) − 0.1𝑡 2 (𝑣2 )| ≤ 0.1(|u1 − u2 | + |v1 − v2 |)

www.ijaers.com

Page | 305

John Fiifi Essel et al.

International Journal of Advanced Engineering Research and Science, 8(10)-2021

Hence, satisfies condition (A1 ) with L = 0.1. Also, |𝜇(𝑢1 ) − 𝜇(𝑢2 )| = |∑𝑛𝑖=1 𝑐𝑖 𝑢1 (𝑡𝑖 ) − ∑𝑛𝑖=1 𝑐𝑖 𝑢2 (𝑡𝑖 )| ≤ ∑ni=1 ci ‖u1 − u2 ‖ Hence, (𝐴1 ) is satisfied with 𝐿1 = ∑𝑛𝑖=1 𝑐𝑖 <

1 2

Finally, we show that (A3 ) holds. τγ

φ=(

Γ(γ+1)

2τγ−1

+

1

)2L + L1 = (

Γ(γ+2)

+

Γ(2.5)

2

)2(0.1) + 0.5 ≅ 0.771 < 1.

Γ(3.5)

and 𝜏 𝛤(2−𝛿)

𝜏𝛾−1

[(

𝛤(𝛾)

+

2𝜏𝛾−2

)2𝐿 + 𝐿1 =

𝛤(𝛾+2)

1 𝛤(1.5)

[(

1 𝛤(1.5)

+

2

)] 2(0.1) + 0.5 ≅ 0.954.

𝛤(3.5)

(A 3 ) is satisfied with φ < 1.

Hence, from theorem 3.1 we conclude that BVP (2.1) has a unique solution.

Example 4.2 Consider the fractional boundary value problem; 𝑐 1.5 .𝐷0+ 𝑢(𝑡)

−1 𝑡 𝑒 2 𝑢(𝑡)

−1

=𝑒2𝑡+

𝑒 −𝑡

+(

10𝑒 𝑡 (1+𝑢(𝑡))

3

10𝑒 𝑡

) 𝑐.𝐷00.5 + 𝑢(𝑡)

u(0) = ∑ni=1 ci u(t i ); u(1) = 0 Where 𝑡𝑖 ∈ (0, 1)

𝑐𝑖, 𝑖 = 1,2, . . . , 𝑛 − 1, 𝑛 are non-negative constants with

1

∑𝑛𝑖=1 𝑐𝑖 < . From BVP (3.2), γ = 1.5, δ = 0.5, τ = 1, Uτ = 0, 6

𝑔(𝑡, 𝑢(𝑡), 𝑐.𝐷𝛿 𝑢(𝑡))

=𝑒

−1 𝑡 2

+

−1 𝑡 𝑒 2 𝑢(𝑡)

𝑒 −𝑡

3

+(

10𝑒 𝑡 (1+𝑢(𝑡))

10𝑒 𝑡

n ) 𝑐.𝐷00.5 + 𝑢(𝑡) , μ(u) = ∑i=1 ci u(t i );

Let 𝑢𝑖 , 𝑣𝑖 ∈ 𝑅, 𝑖 = 1,2 𝑎𝑛𝑑 𝑡 ∈ [0, 1]. Then; |𝑔(𝑡, 𝑢1 , 𝑣1 ) − 𝑔(𝑡, 𝑢2 , 𝑣2 )| = | =

𝑒

−1⁄ 𝑡2 2

1 10

10𝑒 𝑡

𝑢1 (𝑡) −

𝑒

−1⁄ 𝑡2 2

10𝑒 𝑡

𝑢2 (𝑡)| + |

𝑒

−1⁄ 𝑡2 2

10𝑒 𝑡

𝑣1 (𝑡) −

𝑒

−1⁄ 𝑡2 2

10𝑒 𝑡

𝑣2 (𝑡)|

(|u1 − u2 | + |v1 − v2 |)

Hence, satisfies condition (C1 ) with L =

1 10

.

Also, |𝜇(𝑢) − 𝜇(𝑣)| ≤ ∑𝑛𝑖=1 𝑐𝑖 |𝑢 − 𝑣| Now, we verify that (𝐶3 ) is satisfied with τ = 1. (

τγ Γ(γ+1)

+

τγ−1

)2L + L1 = (

Γ(γ+1)

2

1

) × 2(0.1) + ≅ 0.4675 < 1.

Γ(2.5)

6

and, 𝜏 𝛤(2−𝛽)

[(

𝜏𝛾−1 𝛤(𝛾)

+

𝜏𝛾

)2𝐿 + 𝐿1 ] =

𝛤(𝛾+1)

1 𝛤(1.5)

[(

1 𝛤(1.5)

+

1

1

) × 2(0.1) + ] ≅ 0.6123 < 1

𝛤(2.5)

6

which satisfies (𝐶3 ) with φ < 1 for any γ ∈ (1, 2]. Hence, we establish that BVP (2.4) has a unique solution.

V.

CONCLUSION

This article examines the existence and uniqueness of solutions using the principle of contraction mapping and the Schaefer’s fixed point theorem to ascertain our results for fractional differential equations involving Caputo derivatives with nonlocal and integral boundary conditions.

Boundary problems for fractional derivatives with multiple

www.ijaers.com

boundary conditions or eigenvalue problems may be considered for further research works.

REFERENCES [1] R. Agarwal, P. O’Regan, D. Stanek, Positive solutions for Dirichlet problems of singular nonlinear fractional differential equations. J. Math. Anal. Appl.

371, 57-68, 2010.

Page | 306

John Fiifi Essel et al.

International Journal of Advanced Engineering Research and Science, 8(10)-2021

[2] Z. Bai, H. L, Positive solutions for boundary value problem of nonlinear fractional differential equation.

J. Math. Anal. Appl. 311, 495-505, 2005.

[3] K. Balanchandra, The nonlocal Cauchy problem for nonlinear fractional integro differential equations in Banach spaces.Nonlinear Analysis. 72,4587-4593, 2010.

[4] K. Balachandran, S. Kiruthika,J.J. Trujillo, Existence of solutions of nonlinear fractional pantograph equations. Acta Mathematica Scientia.33, 712-720, 2013.

[5] M. Benchohra, S. Hamani, S. K. Ntouyas, Boundary value problems for differential equations with fractional order and nonlocal conditions. Nonlinear Anal. 71, 2391-2396, 2009.

[6] A. Cabada, G. Wang, Positive solutions of nonlinear fractional differential equations with integral boundary value conditions. J. Math. Anal. Appl. 389, 403-411, 2012.

[7] M. El-shahed, J.J. Nieto, Nontrivial solutions for a nonlinear multi-point boundary value problem of fractional order. Comput. Math. Appl. 59, 3438-3443, 2010.

[8] A.M.A. El-Sayed, Nonlinear functional differential equations of arbitrary orders, Nonlinear Anal. 33, 181186, 1998.

[9] W. Feng, S. Sun, Z. Han, Y. Zhao, Existence of solutions for a singular system of nonlinear fractional differential equations. Comput. Math. Appl. 62, 1370-1378, 2011.

[10] S. Hamani, J. Henderson, Boundary value problems for fractional differential inclusions with nonlocal conditions. Mediterranean Journal of Mathematics.28, 147164, 2015.

[11] A. A. Kilbas, H. H. Srivastava, J. J. Trujillo, Theory and Applications of Fractional Differential Equations.

Elsevier, Amsterdam, 2006.

[12] X. Liu, Multiple solutions for fractional differential equations with nonlinear boundary conditions.

Computers and Mathematics with Applications.59, 2880-2886, 2010.

[13] K. B. Oldham, J. Spanier, The Fractional Calculus.

Academic Press, New York, 1974.

[14] M. U. Rehman,R. A. Khan, Existence and uniqueness of solutions for multi-point boundary value problems for fractional differential equations. Appl. Math. Lett.

23, 1038-1044, 2010.

[15] S. Sun, Y. Zhao, Z. Han, Y. Li, The existence of solutions for boundary value problem of fractional hybrid differential equations. Commun. Nonlinear Sci. Numer. Simul. 17, 4961-4967, 2012.

www.ijaers.com

[16] S. Sun, Y. Zhao, Z. Han, M. Xu, Uniqueness of positive solutions for boundary value problems of singular fractional differential equations. Inverse Probl. Sci. Eng. 20, 299-309, 2012.

[17] Y. Tian, Existence results for the three-point impulsive boundary value problem involving fractional differential equations. Computers and Mathematics with Applications.59, 2601-2609, 2010.

[18] R. A. Yan, S. R. Sun, D. W. Yang, The existence of solutions for boundary value problems of two types fractional perturbation differential equations. J. Appl.

Math. Comput. 48,187203, 2015.

[19] R. Yan, S. Sun, Y. Sun, Boundary value problems for fractional differential equations with nonlocal boundary conditions. Adv Differ Equ 2013, 176, 2013.

[20] S. Zhang, Positive solutions for boundary-value problems of nonlinear fractional differential equations. Electron. J. Differ. Equ. 36, 1-12, 2006.

[21] Y. Zhao, S. Sun, Z. Han, M. Zhang, Positive solutions for boundary value problems of nonlinear fractional differential equations. Appl. Math. Comput.

217, 6950-6958, 2011.

[22] Y. Zhao, S. Sun, Z. Han, Q. Li, Positive solutions to boundary value problems of nonlinear fractional differential equations. Abstr. Appl. Anal. 2011, Article ID 390543, 2011.

[23] Y. Zhao, S. Sun, Z. Han, Q. Li, The existence of multiple positive solutions for boundary value problems of nonlinear fractional differential equations. Commun. Nonlinear Sci. Numer. Simul.

16, 2086-2097, 2011.

[24] Y. Zhao, S. Sun, Z. Han, Q. Li, Theory of fractional hybrid differential equations. Comput. Math. Appl.

62, 1312-1324, 2011.

Page | 307