on a Caputo fractional differential equation

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DOI: 10.24193/subbmath.2019.4.06

Analysis of fractional boundary value problem with non local flux multi-point conditions

on a Caputo fractional differential equation

Muthaiah Subramanian, A Ramamurthy Vidhya Kumar and Thangaraj Nandha Gopal

Abstract. A brief analysis of boundary value problem of Caputo fractional differential equation with nonlocal flux multi-point boundary conditions has been done.

The investigation depends on the Banach fixed point theorem, Krasnoselskii- Schaefer fixed point theorem due to Burton and Kirk, fixed point theorem due to O’Regan. Relevant examples illustrating the main results are also constructed.

Mathematics Subject Classification (2010):34A08, 34A12, 34B10.

Keywords:Fractional differential equation, Caputo derivative, multi-point, nonlocal, integral conditions, existence, fixed point.

1. Introduction

In recent years, fractional differential equations are increasingly utilized to model many problems in biology, chemistry, engineering, physics, economic and other areas of applications. The fractional differential equations have become a useful tool for describing nonlinear phenomena of science and engineering models. Also, researchers found that fractional calculus was very suitable to describe long memory and hered- itary properties of various materials and processes. we refer the reader to the texts [16]-[14], [8], [9]-[6], and the references cited therein.

Fractional differential equations have attracted considerable interest because of their ability to model complex artefacts. These equations capture non local relations in space and time with memory essentials. Due to extensive applications of FDEs in engineering and science, research in this area has grown significantly all around the world., for instance, see [18], [11], [15] and the references cited therein. Recently, much interest has been created in establishing the existence of solutions for various types of

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boundary value problem of fractional order with nonlocal multi-point boundary conditions. Nonlocal multi-point conditions involving Liouville-Caputo derivative, first of its kind was explored by Agarwal et.al. [1] on nonlinear fractional order boundary value problem. Ahmad et.al. [2]-[5], [3], [7] profound the idea of new kind of nonlocal multi-point boundary value problem of fractional integro-differential equations involving multi-point strips integral boundary conditions.

In this paper the existence and uniqueness of solutions for the below fractional differential equations with nonlocal multi-point boundary conditions are discussed.

Consider the fractional differential equation

CD^δp(z) =k(z, p(z)), z∈J= [0,1], n−1< δ≤n, (1.1) supplemented with the nonlocal multi-point integral boundary conditions

p(0) =ψ(p), p⁰(0) =ρp⁰(ν), p⁰⁰(0) = 0, p⁰⁰⁰(0) = 0,· · ·, pⁿ⁻²(0) = 0, p(1) =λ

Z ς

0

p(σ)dσ+µ

m−2

X

j=1

ξjp(ζj), (1.2)

where^CD^δ denote the Caputo fractional derivative andk:J×RtoRandψ:C(J,R) to R, are given continuous functions, 0 < ν < ς < ζ₁ < ζ₂ < · · · < ζ_m−2 < 1, ξ_j, j = 1,2,· · ·, m−2, ρ, λ, µ are positive real constants. The rest of the paper is organised as follows: The preliminaries section is devoted to some fundamental concepts of fractional calculus with basic lemma related to the given problem. In section 3, the existence and uniqueness of solutions are obtained based on Banach fixed point theorem, Krasnoselskii-Schaefer fixed point theorem due to Burton and Kirk, and fixed point theorem due to O’Regan and also the validation of the results is done by providing examples.

2. Preliminaries

In this section, we introduce some notations and definitions of fractional calculus.

Definition 2.1. The fractional integral of orderδwith the lower limit zero for a function kis defined as

I^δk(z) = 1 Γ(δ)

Z z

0

k(σ)

(z−σ)^1−δdσ, z >0, δ >0,

provided the right hand-side is point-wise defined on [0,∞), where Γ(·) is the gamma function, which is defined by Γ(δ) =R∞

0 z^δ−1e^−zdz.

Definition 2.2. The Riemann-Liouville fractional derivative of orderδ > 0,n−1 <

δ < n,n∈Nis defined as

D^δ₀₊k(z) = 1 Γ(n−δ)

d dz

!ⁿ Z z

0

(z−σ)^n−δ−1k(σ)dσ,

where the functionk(z) has absolutely continuous derivative up to order (n−1).

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Definition 2.3. The Caputo derivative of orderδfor a functionk: [0,∞)→Rcan be written as

CD^δk(z) = D^δ₀₊ k(z)−

n−1

X

j=0

z^j j!k^(j)(0)

!

, z >0, n−1< δ < n.

Remark 2.4. Ifk(z)∈Cⁿ[0,∞), then

CD^δk(z) = 1 Γ(n−δ)

Z z

0

kⁿ(σ) (z−σ)^δ+1−ndσ

= I^n−δkⁿ(z), z >0, n−1< δ < n.

Lemma 2.5. For δ > 0, the general solution of the fractional differential equation

CD^δp(z) = 0 is given by

p(z) =a0+a1z+· · ·+an−1zⁿ⁻¹, whereai∈R,i= 1,2, . . . , n−1 (n= [δ] + 1).

In view of Lemma 2.5, it follows that

I^{δ C}D^δp(z) =p(z) +a0+a1z+· · ·+a_n−1zⁿ⁻¹, for somea_i∈R,i= 1,2, . . . , n−1 (n= [δ] + 1).

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

Lemma 2.6. For any ˆk ∈ C(J,R), the solution of the linear fractional differential equation

CD^δp(z) = ˆk(z), n−1< δ≤n, (2.1) supplemented with the boundary conditions (1.2) is given by

p(z) = Z z

0

(z−σ)^δ−1 Γ(δ)

ˆk(σ)dσ

+h

1 + (zυ1+zⁿ⁻¹$1)

ϑ (λδ+µ

m−2

X

j=1

ξ_j−1)i ψ(p)

+ρ(zυ2−zⁿ⁻¹$2) ϑ

hZ ν

0

(ν−σ)^δ−2 Γ(δ−1)

k(σ)dσˆ i

+(zυ₁+zⁿ⁻¹$₁) ϑ

h λ

Z ς

0

Z σ

0

(σ−θ)^δ−1 Γ(δ)

ˆk(θ)dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζj−σ)^δ−1 Γ(δ)

ˆk(σ)dσ− Z 1

0

(1−σ)^δ−1 Γ(δ)

ˆk(σ)dσi

(2.2)

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where

$1 = 1−ρ, $2 = 1−λδ² 2 −µ

m−2

X

j=1

ξjζj (2.3)

υ1 = (n−1)ρδⁿ⁻², υ2 = 1−λδⁿ n −µ

m−2

X

j=1

ξjζ_jⁿ⁻¹ (2.4) ϑ = $₁υ₂+$₂υ₁6= 0, (2.5) Proof.It is evident that the general solution of the fractional differential equations in (2.1) can be written as

p(z) = Z z

0

(z−σ)^δ−1 Γ(δ)

ˆk(σ)dσ+a0+a1z+a2z²+· · ·+an−1zⁿ⁻¹ (2.6)

where ai ∈ R, (i = 0,1,2, ...,(n−1)) are arbitrary constants. Using the boundary conditions given by (1.2) in (2.6), we get a0 = ψ(p). On using the notations (2.3)- (2.5) along with (1.2) in (2.6), we get

a₁$₁−a_n−1υ₁ = ρ Z ν

0

(ν−σ)^δ−2 Γ(δ−1)

k(σ)dσˆ (2.7)

a₁$₂+a_n−1υ₂ = λ Z ς

0

Z σ

0

(σ−θ)^δ−1 Γ(δ)

ˆk(θ)dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

ˆk(σ)dσ

− Z 1

0

(1−σ)^δ−1 Γ(δ)

ˆk(σ)dσ. (2.8)

Solving the system (2.7) and (2.8) fora1, a_n−1, we get

a1= 1 ϑ

"

υ2 ρ Z ν

0

(ν−σ)^δ−2 Γ(δ−1)

k(σ)dσˆ

!

+υ1 λ Z ς

0

Z σ

0

(σ−θ)^δ−1 Γ(δ)

ˆk(θ)dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζj

0

ˆk(σ)dσ+ψ(p) λδ+µ

m−2

X

j=1

ξj−1

− Z 1

0

(1−σ)^δ−1 Γ(δ)

ˆk(σ)dσ

!#

(2.9)

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a_n−1= −1 ϑ

"

$₂ ρ Z ν

0

(ν−σ)^δ−2 Γ(δ−1)

k(σ)dσˆ

!

+$₁ λ Z ς

0

Z σ

0

(σ−θ)^δ−1 Γ(δ)

ˆk(θ)dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

ˆk(σ)dσ+ψ(p) λδ+µ

m−2

X

j=1

ξj−1

− Z 1

0

(1−σ)^δ−1 Γ(δ)

k(σ)dσˆ

!#

. (2.10)

Substituting the values of a0, a1, a_n−1 in (2.6), we get the solution (2.2). This completes the proof.

3. Main results

We denote byG=C(J,R) be the Banach space of all continuous functions from J→R, equipped with the norm defined by

kpk= sup

z∈J

|p(z)|, z∈J}.

Also by L¹(J,R), we denote the Banach space of measurable functions p : J → R which are Lebesgue integral and normed by

kpk_L1 = Z 1

0

|p(z)|dz.

In view of Lemma 2.6, we define an operator T : G → G associated with problem (1.1) as

(Tp)(z) = Z z

0

(z−σ)^δ−1

Γ(δ) k(σ, p(σ))dσ +h

1 + (zυ1+zⁿ⁻¹$1)

ϑ (λδ+µ

m−2

X

j=1

ξj−1)i ψ(p)

+ρ(zυ2−zⁿ⁻¹$2) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) k(σ, p(σ))dσi +(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

(σ−θ)^δ−1

Γ(δ) k(θ, p(θ))dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζ_j−σ)^δ−1

Γ(δ) k(σ, p(σ))dσ

− Z 1

0

(1−σ)^δ−1

Γ(δ) k(σ, p(σ))dσi

(3.1)

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Let us defineT1,T2 :G→Gby (T₁p)(z) =

Z z

0

(z−σ)^δ−1

Γ(δ) k(σ, p(σ))dσ +ρ(zυ2−zⁿ⁻¹$2)

ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) k(σ, p(σ))dσi +(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

(σ−θ)^δ−1

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζj−σ)^δ−1

− Z 1

0

(1−σ)^δ−1

(3.2) and

(T2p)(z) =h

1 + (zυ1+zⁿ⁻¹$1)

ϑ (λδ+µ

m−2

X

j=1

ξj−1)i

ψ(p) (3.3)

In the sequel, we use the notations:

bη= 1 Γ(δ+ 1)

"

1 +ρ|(υ2−$2)|ν^δ−1

ϑδ +(υ1+$1) ϑ

λς^δ+1 δ+ 1 +µ

m−2

X

j=1

ξ_jζ_j^δ+ 1

# (3.4) and

ωb= 1 +(υ1+$1) ϑ

λδ+µ

m−2

X

j=1

ξ_j+ 1

(3.5) Theorem 3.1. The continuous function k defined from J×R toR. Let us speculate that

(E₁) |k(z, p)−k(z, q)| ≤ Skp−qk, ∀z∈J,S>0, p, q∈R.

(E₂) The continuous function ψ defined fromC(J,R)→Rsatisfying the condition:

|ψ(v)−ψ(w)| ≤εkv−wk, εω <b 1, ∀ v, w∈C(J,R), ε >0.

(E₃) Θ := Sηb+εω <b 1. Then the boundary value problem (1.1)-(1.2) has unique solution onJ.

Proof. Forp, q∈Gand for eachz∈J, from the definition ofTand assumptions (E₁) and (E2). We obtain

|(Tp)(z)−(Tq)(z)| ≤sup

z∈J

(Z z

0

(z−σ)^δ−1

Γ(δ) |k(σ, p(σ))−k(σ, q(σ))|dσ +

h

1 +(zυ1+zⁿ⁻¹$1)

ϑ (λδ+µ

m−2

X

j=1

ξj−1)i

|ψ(p)−ψ(q)|

+

ρ(zυ2−zⁿ⁻¹$2) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) |k(σ, p(σ))−k(σ, q(σ))|dσi

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+

(zυ₁+zⁿ⁻¹$₁) ϑ

hλ Z ς

0

Z σ

0

(σ−θ)^δ−1

Γ(δ) |k(θ, p(θ))−k(θ, q(θ))|dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζj

0

(ζj−σ)^δ−1

Γ(δ) |k(σ, p(σ))−k(σ, q(σ))|dσ +

Z 1

0

(1−σ)^δ−1

Γ(δ) |k(σ, p(σ))−k(σ, q(σ))|dσi )

≤ Z z

0

(z−σ)^δ−1

Γ(δ) (Skp−qk)dσ +

h1 +(zυ₁+zⁿ⁻¹$₁)

ϑ (λδ+µ

m−2

X

j=1

ξ_j−1)i

|ψ(p)−ψ(q)|

+

ρ(zυ₂−zⁿ⁻¹$₂) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) (Skp−qk)dσi +

(zυ1+zⁿ⁻¹$1) ϑ

h

λ Z ς

0

Z σ

0

(σ−θ)^δ−1

Γ(δ) (Skp−qk)dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζ_j−σ)^δ−1

Γ(δ) (Skp−qk)dσ+ Z 1

0

(1−σ)^δ−1

Γ(δ) (Skp−qk)dσi

≤ S

Γ(δ+ 1)

"

1 +ρ|(υ2−$2)|ν^δ−1

ϑδ +(υ1+$1) ϑ

λς^δ+1 δ+ 1 +µ

m−2

X

j=1

ξjζ_j^δ+ 1

# kp−qk

+

"

1 + (υ1+$1) ϑ

λδ+µ

m−2

X

j=1

ξj+ 1

#

εkp−qk ≤(Sbη+εω)kpb −qk.

Hence

k(Tp)−(Tq)k ≤ Θkp−qk.

As Θ<1 by (E₃), the operator T:G→Gis a contraction. Hence the conclusion of the theorem follows by the Banach fixed point theorem.

Example 3.2. Consider the fractional differential equation given by

CD

7

3p(z) = sinz+e^−zsinp(z) 4√

z⁶+ 16, z∈J, (3.6)

subject to the boundary conditions p(0) = 1

10p(z), p⁰(0) = 1 4x⁰1

5

p(1) = Z ¹₃

0

p(σ)dσ+

4

X

j=1

ξjp(ζj). (3.7) Here

2< δ ≤3, λ=µ= 1, ρ=1

4, ν=1 5, ς= 1

3, ξ1=1

5, ξ2=1

7, ξ3=1

6, ξ4= 1 8,

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ζ1=1

2, ζ2=1

4, ζ3=1

3, ζ4= 1 5. Using the given data, we find that

|k(z, p(z))|= sinz+e^−zsinp(z) 4√

z⁶+ 16, ψ(p) = 1 10p(z).

Since

|k(z, p)−k(z, q)≤ 1

16kp−qk,

|ψ(v)−ψ(w)| ≤ 1

10kv−wk,

therefore, (E₁) and (E₂) are respectively satisfied withS=₁₆¹ andε= ₁₀¹. With the given data, we find thatbη= 5.18462,bω= 2.62014, it is found that

Θ :=Sηb+εωb∼= 0.586053<1.

Thus, the assumptions of Theorem 3.1 hold and the problem (3.6)-(3.7) has at most one solution onJ.

Theorem 3.3. Let Y be a Banach space, and H1,H2 :Y→Y be two operators such that H1 is a contraction and H2 is completely continuous. Then either

(i) the operator equation u=H₁(u) +H₂(u)has a solution, or

(ii) the setF={w∈Y:κH₁(^w_κ) +κH₂(w) =w} is unbounded forκ∈(0,1).

Theorem 3.4. The continuous functionkdefined fromJ×RtoRand condition(E2) hold. Also let us understand that:

(E4) ψ(0) = 0.

(E5) there exists a function x ∈ L¹(J,R+) such that |k(z, v)| ≤ x(z), for almost everywhere eachz∈J, and each v∈R.

Then the problem (1.1)-(1.2) has at least one solution onJ.

Proof. To transform the problem (1.1)-(1.2) into a fixed point problem. we consider the mapT:G→Ggiven by (Tp)(z) = (T1p)(z) + (T2p)(z), z∈J, whereT1andT2

are defined by (3.2) and (3.3) respectively.

We shall show that the operatorsT₁andT₂satisfy all the conditions of Theorem 3.3.

Step 1. The operatorT1 defined by (3.2) is continuous.

Letpn ⊂Bθ={p∈G:kpk ≤θ}withkpn−pk →0.

Then the limitkpn(z)−p(z)k →0 is uniformly valid onJ. From the uniform continuity ofk(z, p) on the compact setJ×[−θ, θ], it follows thatkk(z, pn(z))−k(z, p(z))k →0 uniformly onJ. HencekT1pn−T1pk →0 asn→ ∞which implies that the operator T1 is continuous.

Step 2. The operatorT1 maps bounded sets into bounded sets inG.

It is indeed enough to show that for anyθ >0 there exists a positive constantSsuch that for each

p∈Bθ={p∈G:kpk ≤θ}, we have

kT1pk ≤Q.

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Letp∈Bθ. Then kT1pk ≤

Z z

0

(z−σ)^δ−1

Γ(δ) |k(σ, p(σ))|dσ +ρ(zυ2−zⁿ⁻¹$2)

ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) |k(σ, p(σ))|dσi +(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

(σ−θ)^δ−1

Γ(δ) |k(θ, p(θ))|dθ

! dσ

+µ

m−2

X

j=1

ξ_j Z ζj

0

(ζ_j−σ)^δ−1

Γ(δ) |k(σ, p(σ))|dσ +

Z 1

0

(1−σ)^δ−1

Γ(δ) |k(σ, p(σ))|dσi

≤ Z z

0

(z−σ)^δ−1 Γ(δ) x(σ)dσ +ρ(zυ₂−zⁿ⁻¹$₂)

ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) x(σ)dσi +(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

(σ−θ)^δ−1 Γ(δ) x(θ)dθ

! dσ

+µ

m−2

X

j=1

ξ_j Z ζj

0

(ζj−σ)^δ−1

Γ(δ) x(σ)dσ+ Z 1

0

(1−σ)^δ−1

Γ(δ) x(σ)dσi

≤ kxk Γ(δ+ 1)

"

1 +ρ|(υ2−$₂)|ν^δ−1 ϑδ +(υ1+$1)

ϑ

λς^δ+1 δ+ 1 +µ

m−2

X

j=1

ξjζ_j^δ+ 1

# := Q

Step 3. The operatorT₁ maps bounded sets into equicontinuous sets inG.

Let%₁, %₂∈Jwith%₁< %₂ andp∈B_θ, we obtain

|(T1p)(%₂)−(T₁p)(%₁)| ≤

Z %₁

0

[(%₂−σ)^δ−1−(%₁−σ)^δ−1]

Γ(δ) ×k(σ, p(σ))dσ

+

Z %₂

%1

(%2−σ)^δ−1

+

ρ((%2−%1)υ2−(%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$2) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) |k(σ, p(σ))|dσi +((%₂−%₁)υ₁+ (%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$₁)

ϑ

hλ Z ς

0

Z σ

0

(σ−θ)^δ−1

! dσ

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+µ

m−2

X

j=1

ξj

Z ζj

0

(ζj−σ)^δ−1

Γ(δ) |k(σ, p(σ))|dσ+ Z 1

0

(1−σ)^δ−1

≤

Z %₁

0

[(%₂−σ)^δ−1−(%₁−σ)^δ−1]

Γ(δ) ×x(σ)dσ

+

Z %₂

%₁

(%₂−σ)^δ−1 Γ(δ) x(σ)dσ

+

ρ((%₂−%₁)υ₂−(%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$₂) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) |x(σ)|dσi

+((%₂−%₁)υ₁+ (%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$₁) ϑ

hλ Z ς

0

Z σ

0

(σ−θ)^δ−1

Γ(δ) |x(θ)|dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζj

0

(ζj−σ)^δ−1

Γ(δ) |x(σ)|dσ+ Z 1

0

(1−σ)^δ−1

Γ(δ) |x(σ)|dσi

≤ kxk Γ(δ+ 1)

"

[2(%2−%1)^δ+ (%^δ₂−%^δ₁)] +ρ((%₂−%₁)υ₂−(%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$₂)ν^δ−1 ϑδ

+((%₂−%₁)υ₁+ (%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$₁) ϑ

λς^δ+1 δ+ 1 +µ

m−2

X

j=1

ξ_jζ_j^δ+ 1

#

which is independent ofpand tends to zero as%₂−%₁→0. Thus,T₁is equicontinuous.

Step 4. The operatorT₂ defined by (3.3) is continuous and Θ- contractive.

To show the continuity ofT2forz∈J, let us consider a sequencepn converging top.

Then we have

kT2p_n−T₂pk ≤h

1 +(zυ₁+zⁿ⁻¹$₁)

ϑ (λδ+µ

m−2

X

j=1

ξ_j−1)i

|ψ(pn)−ψ(p)|

≤h

1 +(υ₁+$₁) ϑ

λδ+µ

m−2

X

j=1

ξ_j+ 1i

εkpn−pk,

which, in view ofE2, implies thatT2is continuous. Also isT2is Θ- contractive, since

Θ =h

1 +(υ1+$1) ϑ

λδ+µ

m−2

X

j=1

ξj+ 1i

ε=ωε <b 1.

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Step 5. It remains to show that the set F is bounded for every κ!. Let p ∈ F be a solution of the integral equation

p(z) = Z z

0

κ(z−σ)^δ−1

Γ(δ) k(σ, p(σ))dσ +κh

1 +(zυ1+zⁿ⁻¹$1)

ϑ (λδ+µ

m−2

X

j=1

ξj−1)i ψ(p)

+κρ(zυ2−zⁿ⁻¹$2) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) k(σ, p(σ))dσi +κ(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

(σ−θ)^δ−1

! dσ

+µ

m−2

X

j=1

ξ_j Z ζj

0

(ζ_j−σ)^δ−1

− Z 1

0

(1−σ)^δ−1

, z∈J Then, for eachz∈J, we have

|p(z)| ≤ Z z

0

(z−σ)^δ−1

Γ(δ) x(σ)dσ+κh

1 +(zυ1+zⁿ⁻¹$1)

ϑ (λδ+µ

m−2

X

j=1

ξj−1)i

× ψp(σ)

κ

−ψ(0)

+|ψ(0)|

!

+ρ(zυ2−zⁿ⁻¹$2) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) x(σ)dσi +(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

! dσ

+µ

m−2

X

j=1

ξ_j Z ζj

0

(ζ_j−σ)^δ−1

0

(1−σ)^δ−1

Γ(δ) x(σ)dσi

≤ Z z

0

(z−σ)^δ−1

Γ(δ) x(σ)dσ+ρ(zυ₂−zⁿ⁻¹$₂) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) x(σ)dσi +(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζ_j−σ)^δ−1

0

(1−σ)^δ−1

Γ(δ) x(σ)dσi +h

1 +(υ1+$1)

ϑ (λδ+µ

m−2

X

j=1

ξj+ 1)i εkpk or

(1−ωε)kpk ≤b Z z

0

(z−σ)^δ−1

Γ(δ) x(σ)dσ+ρ(zυ2−zⁿ⁻¹$2) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) x(σ)dσi

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+(zυ₁+zⁿ⁻¹$₁) ϑ

hλ Z ς

0

Z σ

0

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζ_j−σ)^δ−1

0

(1−σ)^δ−1

Γ(δ) x(σ)dσi . Consequently, we have

kpk ≤ V:= 1 (1−ωε)b

(Z z

0

(z−σ)^δ−1 Γ(δ) x(σ)dσ +ρ(zυ2−zⁿ⁻¹$2)

ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) x(σ)dσi +(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζj−σ)^δ−1 Γ(δ) x(σ)dσ +

Z 1

0

(1−σ)^δ−1

Γ(δ) x(σ)dσi )

which shows that the set F is bounded, since ωε <b 1. Hence, T has a fixed point in J by Theorem 3.3, and consequently the problem (1.1)-(1.2) has a solution. This

completes the proof.

Finally, we show that the existence of solutions for the boundary value problem (1.1)-(1.2) by applying a fixed poin theorem due to O’Regan.

Lemma 3.5. Denote by X an open set in a closed, convex set A of a Banach space H. Assume 0 ∈ X. Also assume that T(bX) is bounded and that T: Xb →A is given by T=T1+T2, in which T1:Xb →His a nonlinear contraction (i.e., there exists a nonnegative nondecreasing functionϕ: [0,∞)→[0,∞)satisfyingϑ(y)< y fory >0, such that kT2(p)−T₂(q)k ≤ϑ(kp−qk)∀p, q∈X. Then, eitherb

(W₁) Thas a fixed pointx∈X; orb

(W₂) there exist a point x∈∂X and κ∈(0,1) with x=κT(x), whereXb and ∂X, respectively, represent the closure and boundary of X.

In the next result, we use the terminology:

∆θ={p∈G:kpk< θ}, Vθ=max{|k(z, p)|: (z, p)∈J×[θ,−θ]}.

Theorem 3.6. The continuous function k defined from J×R to R and conditions (E₁),(E₂),(E₄)hold. Also let us understand that:

(E₆) there exists a nonnegative function x∈ C(J,R) and a nondecreasing function φ: [0,∞)→[0,∞)such that |k(z, v)| ≤x(z)φ(kvk)for any (z, v)∈J×R;

(E7) sup

θ∈(0,∞)

θ

bηφ(θ)kxk > 1

1−ωεb , where bη and bω are defined in (3.4) and (3.5) respectively. Then the problem (1.1)-(1.2) has at least one solution onJ.

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Proof. By the assumption (E7), there exists a numberθ >b 0 such that θb

ηφ(bb θ)kxk > 1

1−ωεb (3.8)

We shall show that the operators T1 andT2 defined by (3.2) and (3.3) respectively, satisfy all the conditions of Lemma 3.5.

Step 1. The operatorT1is continuous and completely continuous. We first show that T1(∆

θb) is bounded. For anyp∈∆

θb, we have kT1pk ≤

Z z

0

(z−σ)^δ−1

Γ(δ) |k(σ, p(σ))|dσ +ρ(zυ₂−zⁿ⁻¹$₂)

ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) |k(σ, p(σ))|dσi +(zυ₁+zⁿ⁻¹$₁)

ϑ

hλ Z ς

0

Z σ

0

(σ−θ)^δ−1

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζj−σ)^δ−1

Γ(δ) |k(σ, p(σ))|dσ +

Z 1

0

(1−σ)^δ−1

≤ V_θ Z z

0

(z−σ)^δ−1 Γ(δ) x(σ)dσ +Vθρ(zυ2−zⁿ⁻¹$2)

ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) x(σ)dσi +Vθ(zυ1+zⁿ⁻¹$1)

ϑ

h λ

Z ς

0

Z σ

0

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζj−σ)^δ−1

0

(1−σ)^δ−1

Γ(δ) x(σ)dσi

≤ kxkVθ

Γ(δ+ 1)

"

1 +ρ|(υ2−$2)|ν^δ−1 ϑδ +(υ₁+$₁)

ϑ

λς^δ+1 δ+ 1 +µ

m−2

X

j=1

ξjζ_j^δ+ 1

#

= Vθkpkη.b Thus the operator T₁(V

bθ) is uniformly bounded. Let %₁, %₂ ∈ J with %₁ < %₂ and p∈B_θ. Then

|(T1p)(%₂)−(T₁p)(%₁)| ≤V_θ

Z %1

0

[(%₂−σ)^δ−1−(%₁−σ)^δ−1]

Γ(δ) ×x(σ)dσ

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+V_θ

Z %2

%₁

(%₂−σ)^δ−1 Γ(δ) x(σ)dσ

+

Vθρ((%2−%1)υ2−(%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$2) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) |x(σ)|dσi +V_θ((%₂−%₁)υ₁+ (%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$₁)

ϑ

hλ Z ς

0

Z σ

0

(σ−θ)^δ−1

Γ(δ) |x(θ)|dθ

! dσ

+µ

m−2

X

j=1

ξj

Z ζ_j

0

(ζj−σ)^δ−1

Γ(δ) |x(σ)|dσ+ Z 1

0

(1−σ)^δ−1

Γ(δ) |x(σ)|dσi

≤ kxkVθ

Γ(δ+ 1)

"

[2(%2−%1)^δ+ (%^δ₂−%^δ₁)] +ρ((%2−%1)υ2−(%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$2)ν^δ−1 ϑδ

+((%2−%1)υ1+ (%ⁿ⁻¹₂ −%ⁿ⁻¹₁ )$1) ϑ

λς^δ+1 δ+ 1 +µ

m−2

X

j=1

ξjζ_j^δ+ 1

#

which is independent ofpand tends to zero as%₂−%₁→0. Thus,T₁is equicontinuous.

Hence, by the Arzela-Ascoli Theorem. T1(V

bθ) is a relatively compact set. Now, let pn ⊂V

θbwithkpn−pk →0. Then thekpn(z)−p(z)k →0 is uniformly valid onJ.

From the uniform continuity ofk(z, p) on the compact set J×[bθ,−bθ], it follows that kk(z, pn(z))−k(z, p(z))| →0

uniformly on J. Hence kT1pn−T1pk →0 asn→ ∞ which proves the continuity of T1. This completes the proof Step 1.

Step 2. The operatorT2:V

bθ→C(J,R) is contractive. This is a consequence of (E2).

Step 3. The setT(V

θb) is bounded. The assumptions (E2) and (E4) imply that kT2pk ≤ωεbb θ,

for any p∈ V

θb. This, with the boundedness of the set T1(V

θb) implies that the set T(Vθb) is bounded.

Step 4. Finally, it will be shown that the case W2 in Lemma 3.5 does not hold. On the contrary, we suppose that W2 holds. Then, we have that there exist κ ∈ (0,1) andp∈∂V

θbsuch thatp=κTp.

So, we havekpk=θband p(z) =

Z z

0

κ(z−σ)^δ−1

Γ(δ) k(σ, p(σ))dσ +κh

1 +(zυ1+zⁿ⁻¹$1)

ϑ (λδ+µ

m−2

X

j=1

ξj−1)i ψ(p)

+κρ(zυ₂−zⁿ⁻¹$₂) ϑ

hZ ν

0

(ν−σ)^δ−2

Γ(δ−1) k(σ, p(σ))dσi +κ(zυ₁+zⁿ⁻¹$₁)

ϑ

hλ Z ς

0

Z σ

0

(σ−θ)^δ−1

! dσ

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+µ

m−2

X

j=1

ξj

Z ζj

0

(ζj−σ)^δ−1

Γ(δ) k(σ, p(σ))dσ− Z 1

0

(1−σ)^δ−1

Γ(δ) k(σ, p(σ))dσi z∈J.

Using the assumptions (E4)-(E6), we get θb ≤ φ(bθ)kxk

Γ(δ+ 1)

"

1 + ρ|(υ2−$2)|ν^δ−1 ϑδ +(υ₁+$₁)

ϑ

λς^δ+1 δ+ 1 +µ

m−2

X

j=1

ξ_jζ_j^δ+ 1

#

+bθεh

1 + (υ₁+$₁) ϑ

λδ+µ

m−2

X

j=1

ξ_j+ 1i .

which yields

bθ ≤ bηφ(bθ)kxk+ωε.b Thus, we get a contradiction :

θb

bηφ(bθ)kxk ≤ 1 1−ωεb .

Thus, the operators T₁ and T₂ satisfy all the conditions of Lemma 3.5. Hence, the operator T has at least one fixed point p∈ V

θb, which is a solution of the problem

(1.1)-(1.2). This completes the proof.

Example 3.7. Consider the fractional differential equation given by

CD

5

2p(z) = e^−z 2√

z²+ 16 1

2 +ztan⁻¹(z)

, z∈J, (3.9)

supplemented with the boundary conditions of Example 3.2.

Observe that|k(z, p)| ≤x(z)φ(|p|) with x(z) = e^−z

4√

z²+ 16, φ(|p|) = 1 +|p|

andψ(0) = 0,ε= ₁₀¹ as |ψ(v)−ψ(w)| ≤ ₁₀¹|v−w|. With φ(θ) = 1 +θ, kxk= 1

16, bη∼= 1.0683, bω∼= 0.36416, we have that (E7) holds, since

θb bηφ(bθ)kxk

∼= 14.9771> 1.03779∼= 1 1−ωεb .

Thus, all the conditions of Theorem 3.6 is satisfied and here the problem (3.9) with (3.7) has at least one solution onJ.

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Muthaiah Subramanian Department of Mathematics

Sri Ramakrishna Mission Vidyalaya College of Arts and Science Coimbatore - 641 020, Tamilnadu, India

e-mail:[email protected] A Ramamurthy Vidhya Kumar Department of Mathematics

e-mail:[email protected] Thangaraj Nandha Gopal Department of Mathematics

e-mail:[email protected]