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Available online at www.atnaa.org Research Article

A coupled non-separated system of Hadamard-type fractional dierential equations

Saleh S. Redhwana, Suad Y. Al-Mayyahib, Sadikali L. Shaikhc, Mohammed S. Abdod

aDepartment of Mathematics, Dr. Babasaheb Ambedkar Marathwada University, Aurangabad 431004 (M.S.), India.

bDepartment of Mathematics, Dept. of mathematics, college of education of pure science, University of Wasit, Iraq.

cDepartment of Mathematics, Maulana Azad College of arts, Science and Commerce, RozaBagh, Aurangabad 431004 (M.S.), India.

dDepartment of Mathematics, Department of Mathematics, Hodeidah University, Al-Hodeidah, Yemen.

Abstract

In this article, we discuss the existence and uniqueness of solutions of a coupled non-separated system for fractional dierential equations involving a Hadamard fractional derivative. The existence and uniqueness results obtained in the present study are not only new but also cover some results corresponding to special values of the parameters involved in the Caputo problems. These developed results are obtained by apply- ing Banach's xed point theorem and Leray-Schauder's nonlinear alternative. An example is presented to illustrate our main results.

Keywords: Hadamard-type fractional dierential equations xed point theorems boundary conditions.

2010 MSC: 26A33 34A12 30E25

1. Introduction

Fractional calculus (FC) is one of the sections of mathematics which is a popularization of classical calculus that include integrals and derivatives of non-integer order. FC has a substantial role in numerous

elds of science, engineering, and economics. FC tools have been found to support the development of mathematical methods which is more pragmatic to applied problems an expression of fractional dierential equations (FDEs). Recently, many versions of the fractional derivatives, such as Reimann-Liouville, Caputo,

Email addresses: saleh.redhwan909@gmail.com (Saleh S. Redhwan), Suadluck7@gmail.com (Suad Y. Al-Mayyahi), sad.math@gmail.com (Sadikali L. Shaikh), msabdo1977@gmail.com (Mohammed S. Abdo)

Received April 26, 2021, Accepted October 09, 2021, Online October 12, 2021

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Hadamard, Katugampola, Hilfer, ψ-Caputo, ψ-Hilfer, Atangana-Baleanu were presented, Here we refer to [2, 4, 15, 16, 19, 20, 21, 23, 24, 26, 27, 30].

Initial and boundary value problems for FDEs have won considerable signicance because of their many employment in applied sciences and engineering. Many authors have shown great interest in this topic and obtaining a variety of results for FDEs involving dierent kinds of the conditions and the fractional operators, see [1, 9, 28, 29, 34] and the references therein. Some numerical methods such as the Adomian decomposition method for solving the nonlinear FDEs and system of nonlinear FDEs were studied by Jafari and Daftardar-Gejji in [12, 17, 18].

Several authors studied the existence and uniqueness theory for FDEs involving Hadamard-type operator, see [2, 3, 5, 11, 14, 22, 25, 31] and references therein. Coupled systems of FDEs are of great value and an important eld. This type of system arises in business mathematics, management sciences, and other managerial sciences and so forth. To model such problems and some theoretical works on the coupled systems of FDEs, we refer the reader to some studied works [6, 7, 8, 10, 32, 33]. Motivated by the above discussion, in this article, we consider a coupled system of Hadamard-type FDEs:

 Dρ1ω(ϑ) = }1(ϑ, ω(ϑ), $(ϑ)), ϑ ∈ [1, e], 1 < ρ ≤ 2

Dσ1$(ϑ) = }2(ϑ, ω(ϑ), $(ϑ)), ϑ ∈ [1, e], 1 < σ ≤ 2 , (1) with the following non-separated coupled boundary conditions:

 ω(1) = ξ1$(e)

$(1) = `1ω(e)

 ω0(1) = ξ2$0(e)

$0(1) = `2ω0(e) , (2)

where ξ1, ξ2,`1, `2 are real constants, Dρ1, Dσ1 are the Hadamard fractional derivatives of order ρ and σ respectively, and }1, }2 :[1, e] × R × R → R are appropriate functions.

The existence and uniqueness of solutions for the system (1)-(2) are mainly investigated. To the best of our knowledge, a coupled system of Hadamard-type FDEs (1) with non-separated coupled boundary conditions (2) have were not widely studied. In consequence, the coupled system of FDEs with non-separated coupled boundary conditions will be studied by the Hadamard fractional derivative. Moreover, the main results are obtained by applying Banach xed point theorem and Leray-Schauder xed point theorem.

The outline of the paper is the following. In Section 2, we present some basic denitions and known results related to fractional calculus. Section 3 is devoted to proving the existence and uniqueness of the Hadamard coupled system (1)-(2). In the end, we present an illustrative example to justify our main results.

2. Preliminaries:

First of all, we present some denitions and properties from fractional calculus used throughout this article.

Denition 2.1. [20] For a continuous function ω : [1, +∞) → R, the Hadamard fractional integral of order ρ > 0 is dened by

Iρ1ω(ϑ) = 1 Γ(ρ)

Z ϑ 1

(logϑ

s)ρ−1ω(s)ds s , provided the right-hand side is point-wise dened on [1, +∞).

Denition 2.2. [16] Let n−1 < ρ < n, and ω(ϑ) has an absolutely continuous derivative up to order (n−1).

Then the Caputo-Hadamard fractional derivative of order ρ is dened as

Dρ1ω(ϑ) = 1 Γ(n − ρ)

Z ϑ 1

(logϑ

s)n−ρ−1δn(ω)(s)ds s , where δn= (ϑd )n, and n = [ρ] + 1.

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Lemma 2.3. [16] Let ρ > 0 and ω ∈ Cn[1, +∞) such that δ(n)(ω) exists a.e. on any bounded interval of [1, +∞). Then we have

Iρ1[Dρ1ω(ϑ)] = ω(ϑ) −

n−1

X

k=0

δ(k)ω(1)

Γ(k + 1)(log ϑ)k. In particular, if 0 < ρ < 1, then we have Iρ1[Dρ1ω(ϑ)] = ω(ϑ) − ω(1).

Lemma 2.4. [20] For all ` > 0 and ν > −1, then we have 1

Γ(`) Z ϑ

1

(logϑ

s)`−1(log s)νds

s = Γ(ν)

Γ(` + ν)(log ϑ)ν+`−1.

Theorem 2.5. [20] (Banach xed point theorem) Let (W, d) be a nonempty complete metric space with Π : W → W is a contraction mapping. Then map Π has a xed point.

Theorem 2.6. [13] (Leray-Schauder Nonlinear Alternative). Let W be a Banach space and S ⊆ W closed and convex. Assume that K is a relatively open subset of S with 0 ∈ K and Π : K −→ S is a compact and continuous mapping. Then ethier

1. Π has a xed point in K, or

2. there exists ω ∈ ∂K such that ω = λΠω for some λ ∈ (0, 1), where ∂K is boundary of K.

Now we present an auxiliary lemma which plays a key role in the sequel.

Lemma 2.7. Let u, v ∈ C([1, e], R). Then the solution of the linear fractional dierential system





Dρ1ω(ϑ) = u(ϑ), ϑ ∈ [1, e], 1 < ρ ≤ 2 Dσ1$(ϑ) = v(ϑ), ϑ ∈ [1, e], 1 < σ ≤ 2 ω(1) = ξ1$(e), ω0(1) = ξ2$0(e)

$(1) = `1ω(e), $0(1) = `2ω0(e)

, (3)

is equivalent to the system of integral equations ω(ϑ) = ξ1

η2 1 Γ(ρ)

Z e 1

(loge

s)ρ−1u(s)ds

s +`1ξ1

η2 1 Γ(σ)

Z e 1

(loge

s)σ−1v(s)ds s + `1ξ1ξ2

η1η2 +`2ξ1ξ2

1η2 +`2ξ2

1 log ϑ

 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2u(s)ds s + `1`2ξ1ξ2

1η2 + `2ξ1 η1η2 + ξ2

η1 log ϑ

 1

Γ(σ − 1) Z e

1

(loge

s)σ−2v(s)ds s + 1

Γ(ρ) Z ϑ

1

(logϑ

s)ρ−1u(s)ds

s, (4)

and

$(ϑ) = `1

η2 1 Γ(σ)

Z e

1

(loge

s)σ−1v(s)ds

s +`1ξ1

η2 1 Γ(ρ)

Z e

1

(loge

s)ρ−1u(s)ds s + `1`2ξ1ξ2

1η2 + `1ξ2

η1η2 + `1

η1log ϑ

 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2u(s)ds s + `1`2ξ1

η1η2

+`1`2ξ21η2

+`2ξ21

log ϑ

 1

Γ(σ − 1) Z e

1

(loge

s)σ−2v(s)ds s + 1

Γ(σ) Z ϑ

1

(logϑ

s)σ−1v(s)ds

s , (5)

where η1= 1 − `2eξ22 6= 0, η2 = 1 − `1ξ1 6= 0.

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Proof. We know that the general solution of Hadamard coupled system in (3) can be written as

ω(ϑ) = c0+ c1(log ϑ) + 1 Γ(ρ)

Z ϑ 1

(logϑ

s)ρ−1u(s)ds

s , (6)

$(ϑ) = d0+ d1(log ϑ) + 1 Γ(σ)

Z ϑ 1

(logϑ

s)σ−1v(s)ds

s , (7)

where ci, di (i = 0, 1) are arbitrary real constants.

Using the non-separated coupled boundary conditions in (6) and (7), we have

ω(1) = ξ1$(e) ⇒ c0= ξ1



d0+ d1+ 1 Γ(σ)

Z e 1

(loge

s)σ−1v(s)ds s



, (8)

$(1) = `1ω(e) ⇒ ho = `1



c0+ c1+ 1 Γ(ρ)

Z e 1

(loge

s)ρ−1u(s)ds s



. (9)

Similarly, by using the non-separated coupled boundary conditions in (6) and (7), we have

ω0(1) = ξ2$0(e) ⇒ c1= ξ2 d1

e + 1

Γ(σ − 1) Z e

1

(loge

s)σ−2v(s)ds s



, (10)

$0(1) = `2ω0(e) ⇒ d1 = `2[c1

e + 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2u(s)ds

s ]. (11)

From the relations (10) and (11), we get c1 = `2ξ2

1

1 Γ(ρ − 1)

Z e 1

(loge

s)ρ−2u(s)ds s +ξ2

η1

1 Γ(σ − 1)

Z e 1

(loge

s)σ−2v(s)ds

s , (12)

d1 = `2ξ2

1 1 Γ(σ − 1)

Z e 1

(loge

s)σ−2v(s)ds s +`2

η1 1 Γ(ρ − 1)

Z e 1

(loge

s)ρ−2u(s)ds

s. (13)

Substituting c1 and d1 into (8) and (9), we get c0 = ξ1

η2 1 Γ(ρ)

Z e 1

(loge

s)ρ−1u(s)ds

s +`1ξ1 η2

1 Γ(σ)

Z e 1

(loge

s)σ−1v(s)ds s + `1ξ1ξ2

η1η2 +`2ξ1ξ21η2

 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2u(s)ds s + `1`2ξ1ξ2

1η2

+ `2ξ1 η1η2

 1

Γ(σ − 1) Z e

1

(loge

s)σ−2v(s)ds s , and

d0 = `1

η2 1 Γ(σ)

Z e 1

(loge

s)σ−1v(s)ds

s +`1ξ1

η2 1 Γ(ρ)

Z e 1

(loge

s)ρ−1u(s)ds s + `1`2ξ1ξ2

1η2 + `1ξ2

η1η2

 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2u(s)ds s + `1`2ξ1

η1η2 +`1`2ξ2

1η2

 1

Γ(σ − 1) Z e

1

(loge

s)σ−2v(s)ds s.

Substituting the values of ci, di(i = 0, 1)in (6) and (7), we get solutions (4) and (5). The converse follows by direct computation. This completes the proof.

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3. Main results:

Let us introduce the space

W = {ω(ϑ)|ω(ϑ) ∈ C([1, e], R)}, endowed with the norm

kωk = sup{|ω(ϑ)| , ϑ ∈ [1, e]}.

Clearly, (W, k·k) is a Banach space. Then the product space W × G = {ω × $ : kω, $k = sup

ϑ∈[1,e]

|(ω, $)(ϑ)|}, is also a Banach space equipped with the norm

k(ω, $)k = kωk + k$k .

In view of Lemma (2.7), we dene the operator Π : W × G → W × G by Π(ω, $)(ϑ) =Π1(ω, $)(ϑ)

Π2(ω, $)(ϑ)

 , where

Π1(ω, $)(ϑ) = ξ1

η2 1 Γ(ρ)

Z e 1

(loge

s)ρ−1}1(s, ω(s), $(s))ds s +`1ξ1

η2 1 Γ(σ)

Z e 1

(loge

s)σ−1}2(s, ω(s), $(s))ds s + `1ξ1ξ2

η1η2 +`2ξ1ξ2

1η2 +`2ξ2

1 log ϑ



×

 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2}1(s, ω(s), $(s))ds s



+ `1`2ξ1ξ21η2

+ `2ξ1 η1η2

2 η1

log ϑ



×

 1

Γ(σ − 1) Z e

1

(loge

s)σ−2}2(s, ω(s), $(s))ds s



+ 1 Γ(ρ)

Z ϑ 1

(logϑ

s)ρ−1}1(s, ω(s), $(s))ds s , and

Π2(ω, $)(ϑ) = `1 η2

1 Γ(σ)

Z e 1

(loge

s)σ−1}2(s, ω(s), $(s))ds s +`1ξ1

η2 1 Γ(ρ)

Z e

1

(loge

s)ρ−1}1(s, ω(s), $(s))ds s + `1`2ξ1ξ2

1η2 + `1ξ2 η1η2 +`1

η1log ϑ



×

 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2}1(s, ω(s), $(s))ds s



+ `1`2ξ1

η1η2

+`1`2ξ2

1η2

+`2ξ2

1

log ϑ



×

 1

Γ(σ − 1) Z e

1

(loge

s)σ−2}2(s, ω(s), $(s))ds s



+ 1 Γ(σ)

Z ϑ 1

(logϑ

s)σ−1}2(s, ω(s), $(s))ds s .

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To simplify, we put

N1= 1 Γ(ρ)

 |`1| |ξ1| |ξ2|

1| |η2| + |`2| |ξ1| |ξ2|

e |η1| |η2| +|`2| |ξ2| e |η1|



+ 1

Γ(ρ + 1)

 |ξ1|

2|+ 1



, (14)

N2 = 1 Γ(σ)

 |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`2| |ξ1|

1| |η2|+ |ξ2|

1|



+ 1

Γ(σ + 1)

 |`1| |ξ1|

2|



, (15)

N3= 1 Γ(ρ)

 |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`1| |ξ2|

1| |η2|+ |`1|

1|



+ 1

Γ(ρ + 1)

 |`1| |ξ1|

2|



, (16)

N4= 1 Γ(σ)

 |`1| |`2| |ξ1|

1| |η2| + |`1| |`2| |ξ2|

e |η1| |η2| +|`2| |ξ2| e |η1|



+ 1

Γ(σ + 1)

 |`1|

2|+ 1



. (17)

The rst result is based on Banach's xed point theorem. To this end, we need the following hypotheses:

(H1) }1, }2 : [1, e] × R × R → R are continuous and there exist two positive constants κ1 and κ2 such that

|}1(ϑ, ω1, ω2) − }1(ϑ, $1, $2)| ≤ κ1(|ω1− $1| + |ω2− $2|) ,

|}2(ϑ, ω1, ω2) − }2(ϑ, $1, $2)| ≤ κ2(|ω1− $1| + |ω2− $2|) , for all ϑ ∈ [1, e] and ωi, $i ∈ R i = 1, 2.

(H2) There exist real constants δi, βi ≥ 0, i = 0, 1, 2,such that ∀ωj ∈ R(j = 1, 2), we have

|}1(ϑ, ω1, ω2)| ≤ δ0+ δ11| + δ22| ,

|}2(ϑ, ω1, ω2)| ≤ β0+ β11| + β22| . Theorem 3.1. Assume that (H1) holds. If

φ1 := [(N1+ N3) κ1+ (N2+ N4) κ2] < 1, (18) then system (1)-(2) has a unique solution, where Ni, i = 1, 2, 3, 4 are given by (14)-(17).

Proof. Let supϑ∈[1,e]}1(ϑ, 0, 0) = 1 < ∞and supϑ∈[1,e]}2(ϑ, 0, 0) = 2 < ∞and r > 0, we show that ΠBr⊂ Br, where

Br = {(ω, $) ∈ W × G : k(ω, $)k < r}, with r ≥ 1−φφ21,where φ1< 1 and

φ2:= (N1+ N3) 1+ (N2+ N4) 2. By hypotheses (H1) and for (ω, $) ∈ Br, ϑ ∈ [1, e], we have

|}1(ϑ, ω(ϑ), $(ϑ))| ≤ |}1(ϑ, ω(ϑ), $(ϑ)) − }1(ϑ, 0, 0)| + |}1(ϑ, 0, 0)|

≤ κ1(|ω(ϑ)| + |$(ϑ)|) + 1

≤ κ1r + 1. As same way, we have

|}2(ϑ, ω(ϑ), $(ϑ))| ≤ κ2r + 2,

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which give to

1(ω, $)(ϑ)| ≤  |ξ1|

2|

 1 Γ(ρ)

Z e 1

(loge

s)ρ−1|}1(s, ω(s), $(s))|ds s + |`1| |ξ1|

2|

 1 Γ(σ)

Z e 1

(loge

s)σ−1|}2(s, ω(s), $(s))|ds s + |`1| |ξ1| |ξ2|

1| |η2| + |`2| |ξ1| |ξ2|

e |η1| |η2| +|`2| |ξ2| e |η1| |log ϑ|



×

 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2|}1(s, ω(s), $(s))|ds s



+ |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`2| |ξ1|

1| |η2|+ |ξ2|

1||log ϑ|



×

 1

Γ(σ − 1) Z e

1

(loge

s)σ−2|}2(s, ω(s), $(s))|ds s



≤  |ξ1|

2|+ 1 (κ1r + 1)

Γ(ρ + 1) + |`1| |ξ1|

2|

 (κ2r + 2) Γ(σ + 1) + |`1| |ξ1| |ξ2|

1| |η2| + |`2| |ξ1| |ξ2|

e |η1| |η2| +|`2| |ξ2| e |η1|

 (κ1r + 1) Γ(ρ) + |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`2| |ξ1|

1| |η2|+ |ξ2|

1|

 (κ2r + 2) Γ(σ)

≤ (κ1N1+ κ2N2)r + (1N1+ 2N2).

Hence

1(ω, $)(ϑ)k ≤ (κ1N1+ κ2N2)r + (1N1+ 2N2).

Similarly, we obtain

2(ω, $)(ϑ)| ≤  |`1|

2|

 (κ2r + 2)

Γ(σ + 1) + |`1| |ξ1|

2|

 (κ1r + 1) Γ(ρ + 1) + |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`1| |ξ2|

1| |η2|+ |`1|

1|

 (κ1r + 1) Γ(ρ) + |`1| |`2| |ξ1|

1| |η2| +|`1| |`2| |ξ2|

e |η1| |η2| + |`2| |ξ2| e |η1|

 (κ2r + 2)

Γ(σ) +(κ2r + 2) Γ(σ + 1)

=

 1 Γ(ρ)

 |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`1| |ξ2|

1| |η2|+ |`1|

1|



+ 1

Γ(ρ + 1)

 |`1| |ξ1|

2|



1r + 1) +

 1 Γ(σ)

 |`1| |`2| |ξ1|

1| |η2| +|`1| |`2| |ξ2|

e |η1| |η2| +|`2| |ξ2| e |η1|



+ 1

Γ(σ + 1)

 |`1|

2|+ 1



2r + 2)

= (κ2N3+ κ1N4)r + (2N3+ 1N4).

Hence

2(ω, $)(ϑ)k ≤ (κ1N3+ κ2N4)r + (1N3+ 2N4).

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Consequently,

kΠ(ω, $)(ϑ)k ≤ kΠ1(ω, $)(ϑ)k + kΠ2(ω, $)(ϑ)k

≤ [(N1+ N31+ (N2+ N42] r [N1+ N3] 1+ [N2+ N4] 2

≤ φ1r + φ2= r.

Now, for (ω1, $1), (ω2, $2) ∈ W × Gand for any ϑ ∈ [1, e], we get

12, $2)(ϑ) − Π11, $1)(ϑ)|

 1 Γ(ρ)

 |`1| |ξ1| |ξ2|

1| |η2| +|`2| |ξ1| |ξ2|

e |η1| |η2| +|`2| |ξ2| e |η1|



+ 1

Γ(ρ + 1)

 |ξ1|

2|+ 1



× κ1(kω2− ω1k + k$2− $1k) +

 1 Γ(σ)

 |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`2| |ξ1|

1| |η2|+|ξ2|

1|



+ 1

Γ(σ + 1)

 |`1| |ξ1|

2|



× κ2(kω2− ω1k + k$2− $1k)

≤ (N1κ1+ N2κ2) (kω2− ω1k + k$2− $1k) . Hence

12, $2)(ϑ) − Π11, $1)(ϑ)k ≤ (N1κ1+ N2κ2) (kω2− ω1k + k$2− $1k) . (19) In the same way,

22, $2)(ϑ) − Π21, $1)(ϑ)k ≤ (N3κ1+ N4κ2) (kω2− ω1k + k$2− $1k) . (20) It follows from (19) and (20) that

kΠ(ω2, $2)(ϑ) − Π(ω1, $1)(ϑ)k

≤ [(N1+ N3) κ1+ (N2+ N4) κ2] (kω2− ω1k + k$2− $1k) .

Since φ1 < 1, the operator Π is a contraction mapping. So, we conclude that the Hadamard coupled system (1)-(2) has a unique solution due to Banach xed point theorem. The proof is completed.

The next result is based on Leray-Schauder's xed point theorem.

Theorem 3.2. Assume that (H1) and (H2) holds. If

[(N1+ N3) δ1+ (N2+ N4) β1] < 1, [(N1+ N3) δ2+ (N2+ N4) β2] < 1, and

[(N1+ N3) µ1+ (N2+ N4) µ2] < 1,

where µ1 > 0, µ2 > 0 and Ni, i = 1, 2, 3, 4, are given by (14)-(17), then the Hadamard coupled system (1)-(2) has at least one solution.

Proof. Firstly, we will prove that the operator Π : W × G → W × G is completely continuous. Since }1 and }2 are continuous functions on [1, e], it is obvious that Π is continuous too. Let S ⊂ W × G be bounded.

Then there exist two positive constants µ1 > 0 and µ2 > 0 such that

|}1(ϑ, ω(ϑ), $(ϑ))| < µ1, for (ϑ, ω, $) ∈ [1, e] × S,

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and

|}2(ϑ, ω(ϑ), $(ϑ))| < µ2, for (ϑ, ω, $) ∈ [1, e] × S.

Then, for any (ω, $) ∈ S, we have

1(ω, $)(ϑ)| ≤

 1 Γ(ρ)

 |`1| |ξ1| |ξ2|

1| |η2| +|`2| |ξ1| |ξ2|

e |η1| |η2| + |`2| |ξ2| e |η1|



+ 1

Γ(ρ + 1)

 |ξ1|

2|+ 1



µ1 +

 1 Γ(σ)

 |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`2| |ξ1|

1| |η2|+ |ξ2|

1|



+ 1

Γ(σ + 1)

 |`1| |ξ1|

2|



µ2

≤ N1µ1+ N2µ2,

which gives kΠ1(ω, $)k ≤ N1µ1+ N2µ2.Analogously, we have kΠ2(ω, $)k ≤ N3µ1+ N4µ2. Thus, it follows from the above inequalities that the operator Π is uniformly bounded, since

kΠ(ω, $)k ≤ [(N1+ N3) µ1+ (N2+ N4) µ2] < 1.

Next, we show that Π is equicontinuous, i.e. we prove that a bounded set S is mapped into an equicontinuous set of W × G by Π. Let ϑ1, ϑ2 ∈ [1, e]with ϑ1 < ϑ2. Then we have

1(ω(ϑ2), $(ϑ2)) − Π1(ω(ϑ1), $(ϑ1)|

≤ 1

Γ(ρ) Z ϑ2

1

(logϑ2

s )ρ−1|}1(s, ω(s), $(s))|ds s + 1

Γ(ρ) Z ϑ1

1

(logϑ1

s )ρ−1|}1(s, ω(s), $(s))|ds s

≤ µ1

 1 Γ(ρ)

Z ϑ1

1

 (logϑ2

s )ρ−1− (logϑ1

s )ρ−1 ds s + 1

Γ(ρ) Z ϑ2

ϑ1

(logϑ2

s )ρ−1ds s



≤ µ1

Γ(ρ + 1){2 (log ϑ2− log ϑ1)ρ+ (log ϑ2)ρ− (log ϑ1)ρ} . (21) In a similar way, we can easily get

2(ω(ϑ2), $(ϑ2)) − Π2(ω(ϑ1), $(ϑ1)|

≤ µ2

Γ(σ + 1){2 (log ϑ2− log ϑ1)σ+ (log ϑ2)σ− (log ϑ1)σ} . (22) Since log(ϑ) is uniformly continuous on [1, e], the right-hand sides of the inequalities (21) and (22) tend to zero as ϑ2 → ϑ1. Therefore, the operator Π(ω, $) is equicontinuous. The Arzela-Ascoli theorem along with the above steps shows that Π is completely continuous mapping.

Finally, we shall verify that the set

S = {(ω, $) ∈ W × G : (ω, $) = λΠ(ω, $), 0 ≤ λ ≤ 1}, is bounded. Indeed, Let (ω, $) ∈ S with (ω, $) = λΠ(ω, $). For any ϑ ∈ [1, e], we have

ω(ϑ) = λΠ1(ω, $)(ϑ), $(ϑ) = λΠ2(ω, $)(ϑ).

(10)

Then

|ω(ϑ)| < |Π1(ω, $)(ϑ)|

≤  |ξ1|

2|

 1 Γ(ρ)

Z e 1

(loge

s)ρ−1|}1(s, ω(s), $(s))|ds s + |`1| |ξ1|

2|

 1 Γ(σ)

Z e 1

(loge

s)σ−1|}2(s, ω(s), $(s))|ds s + |`1| |ξ1| |ξ2|

1| |η2| +|`2| |ξ1| |ξ2|

e |η1| |η2| +|`2| |ξ2| e |η1| |log ϑ|



×

 1

Γ(ρ − 1) Z e

1

(loge

s)ρ−2|}1(s, ω(s), $(s))|ds s



+ |`1| |`2| |ξ1| |ξ2|

e |η1| |η2| + |`2| |ξ1|

1| |η2|+|ξ2|

1||log ϑ|



×

 1

Γ(σ − 1) Z e

1

(loge

s)σ−2|}2(s, ω(s), $(s))|ds s



+ 1 Γ(ρ)

Z ϑ 1

(logϑ

s)ρ−1|}1(s, ω(s), $(s))|ds s . Assumption (H2) gives

|ω(ϑ)| ≤ N10+ δ1|ω| + δ2|$|) + N20+ β1|ω| + β2|$|)

= N1δ0+ N2β0+ (N1δ1+ N2β1) |ω| + (N1δ2+ N2β2) |$| , and

|$(ϑ)| ≤ N30+ δ1|ω| + δ2|$|) + N40+ β1|ω| + β2|$|)

≤ N3δ0+ N4β0+ (N3δ1+ N4β1) |ω| + (N3δ2+ N4β2) |$| . Hence

kωk ≤ N1δ0+ N2β0+ (N1δ1+ N2β1) kωk + (N1δ2+ N2β2) k$k , and

k$k ≤ N3δ0+ N4β0+ (N3δ1+ N4β1) kωk + (N3δ2+ N4β2) k$k , which imply that

kωk + k$k ≤ (N1+ N3) δ0+ (N2+ N4) β0+ [(N1+ N31+ (N2+ N41] kωk + [(N1+ N32+ (N2+ N42] k$k . Consequently,

kω + $k ≤ (N1+ N3) δ0+ (N2+ N4) β0

N0 ,

where N0= min{1 − (N1+ N31+ (N2+ N41, 1 − (N1+ N32+ (N2+ N42},which shows that the set S is bounded. Thus, as a consequence Theorem 2.6, the Hadamard coupled system (1)-(2) has at least one solution. The proof is complete.

Example 3.3. Let ρ = σ = 43, ξ1= 34, ξ2 = 12, `1 = 32, `2 = 12,

}1(ϑ, ω, $) = 1 16(ϑ + 1)2

|ω(ϑ)|

1 + |ω(ϑ)| + 1 + 1

128sin2($(ϑ)) + 1

ϑ2+ 1, and

}2(ϑ, ω, $) = 1

128πsin(2πω(ϑ)) + |$(ϑ)|

64 (1 + |$(ϑ)|)+1 2.

(11)

Consider the Hadamard-type coupled system (

D

4 3

1ω(ϑ) = }1(ϑ, ω, $), ϑ ∈ [1, e], D

4 3

1$(ϑ) = }2(ϑ, ω, $), ϑ ∈ [1, e], (23)

with the non-separated coupled boundary conditions

 ω(1) = 34$(e)

$(1) = 32ω(e) ,

 ω0(1) = 12$0(e)

$0(1) = 12ω0(e) . (24)

For ϑ ∈ [1, e] and ω, ω, $, $∈ R+, we have

|}1(ϑ, ω, ω) − }1(ϑ, $, $)| ≤ 1

64|ω − $| + 1

64|ω− $| and

|}2(ϑ, ω, ω) − }2(ϑ, $, $)| ≤ 1

64|ω − $| + 1

64|ω− $| .

Hence the condition (H1)holds with κ1 = κ2= 641, η1= 0.966 17 6= 0, η2 = −0.125 6= 0, supϑ∈[1,e]}1(ϑ, 0, 0) = 1 +12 = 1 < ∞ and supϑ∈[1,e]}2(ϑ, 0, 0) = 12 = 2 < ∞. We shall check that condition (18) holds. Indeed, by some simple calculations we nd that N1= 11. 841, N2 = 12. 575, N3 = 17. 211 and N4 = 17. 52 With the given data, we see that φ1= 0.924 17 < 1.

Therefore, by Theorem 3.1, we conclude that problem (23)-(24) has a unique solution.

4. conclusion:

In this article, we have established the existence and uniqueness results of a new type of Hadamard-type fractional dierential equations with coupled non-separated boundary conditions. Our analysis is based on the reduction of convert dierential equations to integral equations and using some xed point theorems which are completely general and eective. We are condent the reported results here will have a favorable impact on the expansion of further applications in applied sciences and engineering.

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