Upper and Lower Solution Method for Nonlinear Fractional Differential Equations Boundary Value Problem

(1)

Upper and Lower Solution Method for Nonlinear

Fractional Differential Equations Boundary Value Problem

G. U. Pawar

^1*

and J. N. Salunke

²

*1

Department of Mathematics,

Mahatma Jyotiba Fule Jr. College,Pishor-431104,Dist. Aurangadad (M.S.), INDIA.

2

School of Mathematical Sciences,

Swami Ramanand Teerth Marathwada University, Nanded (M.S.), INDIA.

email:[email protected], [email protected].

(Received on: February 20, 2018)

ABSTRACT

In this article, we are concerned with the existence and uniqueness of positive solution of nonlinear fractional differential equations involving Riemann- Liouville fractional derivative with multipoint boundary condition. Existence and uniqueness result is obtained by using the upper and lower solution method and fixed point theorem on cone.

Keywords: Boundary value problem, existence and uniqueness, upper and lower solution.

1. INTRODUCTION

Fractional calculus is a generalization of ordinary derivative and integral to non- integer order. The theory of Fractional calculus is concerned with the n

^th

derivative and n-fold integral when n becomes an arbitrary parameter. But fractional orders of differentiation are more mysterious because they have no obvious geometric interpretation along the lines of the customary introduction to derivatives and integrals as slopes and areas. Fractional derivatives introduce an excellent instrument for the description of general properties of various materials and processes. The advantages of Fractional derivatives becomes apparent in modeling mechanical and electrical properties of real materials as well as in the description of properties of gases, liquids & rocks, and in many other fields

^1,2,3

.

Numerous research papers and monographs have appeared devoted to fractional

differential and fractional integral equations. It draws a great application in nonlinear

oscillations of earthquakes, many physical phenomena such as seepage monograph of Kilbas

et al.

¹

. Fractional differential equations arise in many engineering and scientific disciplines as

(2)

the mathematical modeling of systems and process in the field of chemistry, electrodynamics of complex medium, aerodynamics, polymer rheology etc involves derivatives of fractional order

^1-5

.

There has been much attention paid in developing the theory of existence and uniqueness of positive solutions for nonlinear fractional differential equations satisfying initial or boundary value conditions to mention few references

^2,3,4,11

. In consequence, many authors have investigated the existence of positive solutions for nonlinear fractional differential equation with multipoint boundary value condition.

In particular,

⁶

H. Jafari, V.D. Gejji, by using adomain decomposition method considered positive solutions of following nonlinear fractional boundary value problem D

₀₊^α

𝑢(𝑡) + 𝜇𝑓(𝑡, 𝑢(𝑡)) = 0 , 0 < 𝑡 < 1, 1 < 𝛼 ≤ 2

𝑢(0) = 0 , 𝑢(1) = 𝑐

Where D

₀₊^α

is the standard Riemann-Liouville fractional derivative of order α.

In 2012, by means of the fixed point index theory in cones,

⁷

Moustafa El-shahed and Wafa M.Shammakh, obtained the existence of multiple positive solutions for following nonlinear fractional eigen value problem with nonlocal conditions.

D

₀₊^α

𝑢(𝑡) + 𝜆𝑔(𝑡)𝑓(𝑡, 𝑢(𝑡)) = 0 , 0 < 𝑡 < 1, 𝑛 − 1 < 𝛼 ≤ 𝑛 𝑢(0) = 0 , 𝑢

^(𝑘)

(0) = 0 , 1 < 𝑘 ≤ 𝑛-2, 𝑢

^′′′

(1) = 𝜃[𝑢]

Where D

₀₊^α

is the standard Riemann-Liouville fractional derivative of order α and 𝜃[𝑢] =

∫ 𝑢(𝑠)𝑑𝐴(𝑠)

₀¹

is given by Riemann-Stieltjes integral with signed measure.

By using the Leray-Schauder nonlinear alternative and a fixed point theorem on cones, Xinan Hao et.al.[8] obtained the existence of positive solutions to the following nonlinear fractional order singular and semipositone nonlocal boundary value problem

D

₀₊^𝛼

𝑢(𝑡) + 𝑓(𝑡, 𝑢(𝑡)) = 0 , 0 < 𝑡 < 1

𝑢(0) = 𝑢

^′

(0) = ⋯ = 𝑢

^(𝑛−2)

(0) = 0 , 𝑢(1) = 𝜇 ∫ 𝑢(𝑠)𝑑𝑠

1

0

Where 0 < 𝜇 < 𝛼 , 2 ≤ 𝑛 − 1 < 𝛼 ≤ 𝑛 , D

₀₊^𝛼

is the standard Riemann

Liouville derivative , and f(t, u) is semipositone and may be singular at u = 0.

In

⁹

Jiafa Xu, Zhongli Wei, Wei Dong, Considered uniqueness of positive solutions for a class of fractional following boundary value problem

D

₀₊^𝑣

𝑢(𝑡) + h(𝑡)𝑓(𝑡, 𝑢(𝑡)) = 0 , 0 < 𝑡 < 1, 𝑛 − 1 < 𝑣 ≤ 𝑛 𝑢(0) = 𝑢

^′

(0) = ⋯ = 𝑢

^(𝑛−2)

(0) = 0 , [D

₀₊^𝛼

𝑢(𝑡)]

_𝑡=0

= 0, 1 < 𝛼 ≤ 𝑛 − 2, where 𝑛 ∈ 𝑁 and 𝑛 > 3, D

₀₊^𝛼

is the standard Riemann Liouville derivative , 𝑓 ∈ 𝐶([0,1] × [0, ∞); (0, ∞))

and ℎ ∈ 𝐶(0,1) ∩ 𝐿(0,1) is nonnegative and may be singular at t = 0 and/or t = 1.

In

¹⁰

Yujun Cui, by using u-positive operator study the uniqueness of solution for boundary value problem for fractional differential equations

D

₀₊^𝑝

𝑥(𝑡) + p(𝑡)𝑓(𝑡, 𝑢(𝑡)) + 𝑞(𝑡) = 0 , 𝑡 ∈ (0,1)

𝑥(0) = 𝑥

^′

(0) = 0 , 𝑥(1) = 0,

(3)

where 2 < 𝑝 ≤ 3 is a real number.

Boundary value problem for nonlinear fractional differential equations have been studied by some above authors. But very few in this area have been done by using upper and lower solution

^12,13,[14

.

Motivated by all above work, in this paper, we study the existence and uniqueness of positive solutions of the following multipoint boundary value problem

𝐷

₀₊^𝛼

𝑥(𝑡) + p(𝑡)𝑓(𝑡, 𝑥(𝑡)) = 0 , 0 < 𝑡 < 1 (1) 𝑥(0) = 𝑥

^′

(0) = 𝑥

^′′

(0) = ⋯ = 𝑥

^(𝑛−2)

(0) = 0 , 𝑥

^(𝑞)

(1) = 0,

where 𝑛 − 1 < 𝛼

< 𝑛 is a real number and 𝐷

₀₊^𝛼

is the RiemannLiouville fractional derivative of order n − 1 < 𝛼 ≤ 𝑛.

As per our knowledge, no paper has considered the boundary value problem (1) using upper and lower solution. We establish the existence and uniqueness of positive solutions of boundary value problem. We shall use the upper and lower solution method and fixed point theorem on cone.

The rest of the paper is organized as follows: In section 2, we give preliminary facts and we construct Green’s function. Also we give some properties of Green’s function.

Consequently corresponding Fredholm integral equation of problem (1) is obtained. In section 3, we consider the existence and uniqueness of positive solution for equation (1) by using the upper and lower solution method.

2. PRELIMINARIES

For the convenience of the reader, we present some neccessary definitions from fractional calculus

²

. Also we state some lemmas that will be used to prove our main results:

Definition (2.1) The Riemann-Liouville fractional integral of order α > 0 of a function f: (0 , ∞) → R is given by

I

₀₊^α

f(t) = 1

Γ(α) ∫ (t − s)

^α−1

f(s)ds

t

0

provided that the right-hand side is pointwise defined on (0 , ∞).

Definition (2.2) The Riemann-Liouville fractional derivative of order α > 0 of a function f: (0 , ∞) → R is given by

D

₀₊^α

f(t) = 1 Γ(n − α) ( d

dt )

n

∫ f(s)

(t − s)

^α−n+1

ds

t

0

where n − 1 < α ≤ n , provided that the right side is pointwise defined on (0 , ∞).

Let E = C[0,1] in which ‖x‖ =

_t∈[0,1]^max

|x(t)| .

we set P = {x ∈ C[0,1] | x(t) ≥ 0, ∀ t ∈ [0,1] }. Pis a positive cone in C[0,1] and P ⊂ E.

(4)

Lemma(2.1)[1] Let α > 0 and u ϵ C(0,1) ∩ L(0,1). Then the fractional differential equation

D

₀₊^α

u(t) = 0

has a unique solution

u(t) = c

1

t

^α−1

+ c

2

t

^α−2

+ … + c

N

t

^α−N

, c

i

∈ R , i = 1 , 2 , … , N , N = [α] + 1.

Lemma(2.2)[1] Assume that u ϵ C(0,1) ∩ L(0,1) with fractional derivative of order α ≥ 0 that belongs to C(0,1) ∩ L(0,1) . Then

I

₀₊^α

D

₀₊^α

u(t) = u(t) + c

₁

t

^α−1

+ c

₂

t

^α−2

+ … + c

_N

t

^α−N

, c

_i

∈ R , i = 1 , 2 , … , N .

In the following, we present the Green’s function for fractional differential equations with boundary value condition.

𝐓𝐡𝐞𝐨𝐫𝐞𝐦 (𝟐. 𝟏): Let n − 1 < α < n assume y(t) ∈ C[0,1] then the unique solution of the boundary value problem

D

₀₊^α

x(t) + y(t) = 0 , 0 < t < 1 (2)

x(0) = x

^′

(0) = ⋯ = x

⁽ⁿ⁻²⁾

(0) = 0 , x

^(q)

(1) = a is

x(t) = at

^α−1

∏

^q_i=1

(α − i) + ∫ G(t, s) y(s)ds

1 0

i. e x(t) = bt

^α−1

+ ∫ G(t, s) y(s)ds

₀¹

(3) where b = a

∏

^q_i=1

(α − i) and

G(t, s) = {

t

^α−1

(1 − s)

^α−1−q

− (t − s)

^α−1

Γ(α) 0 ≤ s ≤ t ≤ 1 t

^α−1

(1 − s)

^α−1−q

Γ(α) 0 ≤ t ≤ s ≤ 1

(4)

Proof: According to Lemma (2.2) , equation (1) is reduced into an equivalent integral equation as

𝑥(𝑡) = − 𝐼

₀₊^𝛼

𝑦(𝑡) + 𝑐

₁

𝑡

^𝛼−1

+ 𝑐

₂

𝑡

^𝛼−2

+ … + 𝑐

_𝑛

𝑡

^𝛼−𝑛

= − ∫

^{(𝑡−𝑠)}^𝛼−1

𝛤(𝛼)

𝑦(𝑠)𝑑𝑠

𝑡

0

+ 𝑐

₁

𝑡

^𝛼−1

+ 𝑐

₂

𝑡

^𝛼−2

+ … + 𝑐

_𝑛

𝑡

^𝛼−𝑛

(5) where 𝑐

₁

, 𝑐

₂

, … , 𝑐

_𝑛

∈ 𝑅.

From 𝑥(0) = 𝑥

^′

(0) = ⋯ = 𝑥

^(𝑛−2)

(0) = 0 we have 𝑐

₂

= 𝑐

₃

= … = 𝑐

_𝑛

= 0 Then the general solution of (2) is

𝑥(𝑡) = − ∫

₀^𝑡^{(𝑡−𝑠)}_𝛤(𝛼)^𝛼−1

𝑦(𝑠)𝑑𝑠 + 𝑐

₁

𝑡

^𝛼−1

(6) 𝑥

^(𝑞)

(𝑡) = − ∏(𝛼 − 𝑖)

𝑞

𝑖=1

∫ (𝑡 − 𝑠)

^{𝛼−1−𝑞}

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

𝑡 0

+ 𝑐

₁

∏(𝛼 − 𝑖)

𝑞

𝑖=1

𝑡

^{𝛼−1−𝑞}

(5)

𝑥

^(𝑞)

(1) = − ∏(𝛼 − 𝑖)

𝑞

𝑖=1

∫ (1 − 𝑠)

^{𝛼−1−𝑞}

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

1 0

+ 𝑐

₁

∏(𝛼 − 𝑖)

𝑞

𝑖=1

(1)

^{𝛼−1−𝑞}

From 𝑥

^(𝑞)

(1) = 𝑎 , we have

𝑎 = − ∏(𝛼 − 𝑖)

𝑞

𝑖=1

∫ (1 − 𝑠)

^{𝛼−1−𝑞}

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

1 0

+ 𝑐

₁

∏(𝛼 − 𝑖)

𝑞

𝑖=1

𝑐

₁

∏(𝛼 − 𝑖) =

𝑞

𝑖=1

𝑎

+ ∏(𝛼 − 𝑖)

𝑞

𝑖=1

∫ (1 − 𝑠)

^{𝛼−1−𝑞}

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

1

0

Therefore, 𝑐

₁

=

^𝑎

∏^𝑞_𝑖=1(𝛼−𝑖)

+ ∫

^(1−𝑠)^{𝛼−1−𝑞}

𝛤(𝛼)

𝑦(𝑠)𝑑𝑠

1

0

(7)

Put value of 𝑐

₁

from (7)into (6) 𝑥(𝑡) = − ∫ (𝑡 − 𝑠)

^𝛼−1

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

𝑡 0

+ [ 𝑎

∏

^𝑞_𝑖=1

(𝛼 − 𝑖) + ∫ (1 − 𝑠)

^{𝛼−1−𝑞}

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

1 0

] 𝑡

^𝛼−1

𝑥(𝑡) = 𝑎𝑡

^𝛼−1

∏

^𝑞_𝑖=1

(𝛼 − 𝑖) − ∫ (𝑡 − 𝑠)

^𝛼−1

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

𝑡 0

+ ∫ (1 − 𝑠)

^{𝛼−1−𝑞}

𝑡

^𝛼−1

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

1

0 𝑥(𝑡) = 𝑎𝑡^𝛼−1

∏^𝑞_𝑖=1(𝛼 − 𝑖)+ ∫ (1 − 𝑠)^{𝛼−1−𝑞}𝑡^𝛼−1− (𝑡 − 𝑠)^𝛼−1

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

𝑡 0

+ ∫ (1 − 𝑠)^{𝛼−1−𝑞}𝑡^𝛼−1

𝛤(𝛼) 𝑦(𝑠)𝑑𝑠

1

𝑡

𝑥(𝑡) = 𝑎𝑡

^𝛼−1

∏

^𝑞_𝑖=1

(𝛼 − 𝑖) + ∫ 𝐺(𝑡, 𝑠) y(s)𝑑𝑠

1

0

𝑥(𝑡) = 𝑏𝑡

^𝛼−1

+ ∫ 𝐺(𝑡, 𝑠) y(s)𝑑𝑠

1 0

where 𝑏 = 𝑎

∏

^𝑞_𝑖=1

(𝛼 − 𝑖) and 𝐺(𝑡, 𝑠) is given in (4).

This completes the proof.

3. MAIN RESULTS

In this section, we consider the existence and uniqueness of a positive solution for equation(1).

Let A: P → P be the operator defined as Ax(t) = bt

^α−1

+ ∫ G(t, s) p(s)y(s)ds

1 0

𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 (𝟑. 𝟏): The function v(t) ∈ E is called a lower solution of operator A if

(6)

v(t) ≤ Av(t) (D

^α

v(t) ≤ p(t) f(t, v(t))) , 0 < t < 1

and the function w(t) ∈ E is called a upper solution of operator A if w(t) ≥ Aw(t) (D

^α

w(t) ≥ p(t)f(t, w(t))) , 0 < t < 1

If the strict inequalities hold, v(t)and w(t)are strict lower and upper solutions.

We need the following abstract theorem

𝐓𝐡𝐞𝐨𝐫𝐞𝐦(𝟑. 𝟏): [𝟏𝟓]Let D be a subset of the cone P of semi − order Banach space E , T: D → E be nondecreasing. If there exists x

₀

, y

₀

∈ D such that x

₀

≤ y

₀

, 〈x

₀

, y

₀

〉 ⊂ D and x

₀

, y

₀

are the lower and upper solutions of equation x − T(x) = 0 ,

then the equation x − T(x) = 0 has maximum solution and minimum

solution x

^∗

, y

^∗

in〈x

₀

, y

₀

〉 such that x

^∗

≤ y

^∗

, when one of the following conditions hold (1) P is normal and T is compact continuous

(2) P is regular and T is continuous

(3) E is reflexive, P is normal and T is continuous or weak continuous.

𝐓𝐡𝐞𝐨𝐫𝐞𝐦(𝟑. 𝟐): [𝐄𝐱𝐢𝐬𝐭𝐞𝐧𝐜𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦]

Assume that

(𝐇

_𝟏

) f: [0,1] × [0, +∞) → [0, +∞) is continuous , f(t,∙) is nondecreasing for each t

∈ [0,1],

and there exists a positive constant K such that f(t,∙) is strictly increasing on [0, K] for each t ∈ [0,1].

(𝐇

_𝟐

) 0 < lim

x→+∞

f(t, x(t)) < +∞ for each t ∈ [0,1].

(𝐇

_𝟑

) p(t) ∈ C(0,1) ∩ L(0,1)is nonnegative and may be singular at t = 0 and(or)t = 1.

p: (0,1) → [0, +∞) is continuous and dose not vanish identically on any subinterval of (0,1) such that 0 < ∫ p(s)ds <

1 0

+ ∞ Then the equation (1) has a positive solution.

Proof: We will prove the theorem in the following four steps

Step1. The operator A ∶ P → P is completely contionuous. The operator A ∶ P → P is continuous in view of non-negativeness and continuity of 𝐺(𝑡, 𝑠) 𝑎𝑛𝑑 𝑓(𝑡, 𝑥) 𝑎𝑛𝑑 𝑝(𝑡).

Let Ω ⊂ 𝑃 be bounded which to say there exists a positive constant 𝑀 > 0 such that ‖𝑥‖ ≤ 𝑀; ∀𝑥 ∈ Ω

Let 𝐿 = 𝑚𝑎𝑥

0≤𝑡≤1,0≤𝑥≤𝑀

|𝑓(𝑡, 𝑥(𝑡))|𝑝(𝑡) + 1 .Then , for all 𝑥 ∈ Ω, we have

|𝐴𝑥(𝑡)| ≤ ∫ 𝐺(𝑡, 𝑠) 𝑝(𝑠)𝑓(𝑠, 𝑥(𝑠))𝑑𝑠

1 0

≤ 𝐿 ∫ 𝐺(𝑡, 𝑠)𝑑𝑠

1

0

Hence 𝐴(Ω) is bounded.

For each 𝑥 ∈ Ω , ∀𝑡

₁

, 𝑡

₂

∈ [0,1], satisfy 𝑡

₁

≤ 𝑡

₂

then we have

(7)

|𝐴𝑥(𝑡

₂

) − 𝐴𝑥(𝑡

₁

)| = |∫ 𝐺(𝑡

₂

, 𝑠) 𝑝(𝑠)𝑓(𝑠, 𝑥(𝑠))𝑑𝑠

1 0

− ∫ 𝐺(𝑡

1

, 𝑠) 𝑝(𝑠)𝑓(𝑠, 𝑥(𝑠))𝑑𝑠

1 0

|

= ∫ |𝐺(𝑡

₂

, 𝑠) − 𝐺(𝑡

₁

, 𝑠)|

𝑡₁ 0

𝑝(𝑠) 𝑓(𝑠, 𝑣(𝑠))𝑑𝑠 + ∫ |𝐺(𝑡

₂

, 𝑠) − 𝐺(𝑡

₁

, 𝑠)| 𝑝(𝑠)

𝑡₂ 𝑡₁

𝑓(𝑠, 𝑣(𝑠))𝑑𝑠

+ ∫ |𝐺(𝑡

₂

, 𝑠) − 𝐺(𝑡

₁

, 𝑠)|

1 𝑡₂

𝑝(𝑠) 𝑓(𝑠, 𝑣(𝑠))𝑑𝑠

≤ ∫ | 𝑡

₂^𝛼−1

(1 − s)

^{𝛼−1−𝑞}

− (𝑡

₂

− s)

^α−1

Γ(α)

𝑡₁

0

− 𝑡

₁^𝛼−1

(1 − s)

^{𝛼−1−𝑞}

− (𝑡

₁

− s)

^α−1

Γ(α) | 𝑝(𝑠) 𝑓(𝑠, 𝑥(𝑠))𝑑𝑠

+ ∫ | 𝑡

₂^𝛼−1

(1 − s)

^{𝛼−1−𝑞}

− (𝑡

₂

− s)

^α−1

Γ(α)

𝑡₂

𝑡₁

− 𝑡

₁^𝛼−1

(1 − s)

^{𝛼−1−𝑞}

Γ(α) | 𝑝(𝑠) 𝑓(𝑠, 𝑥(𝑠))𝑑𝑠 + ∫ | 𝑡

₂^𝛼−1

(1 − s)

^{𝛼−1−𝑞}

Γ(α) − 𝑡

₁^𝛼−1

(1 − s)

^{𝛼−1−𝑞}

Γ(α) |

1 𝑡₂

𝑝(𝑠) 𝑓(𝑠, 𝑥(𝑠))𝑑𝑠

≤ 𝑏(𝑡

₂^𝛼−1

− 𝑡

₁^𝛼−1

) + 𝐿

𝛤(𝛼) ∫ (1 − 𝑠)

^{𝛼−1−𝑞}

1 0

𝑑𝑠 (𝑡

₂^𝛼−1

− 𝑡

₁^𝛼−1

)

≤ 𝑏(𝑡

₂^𝛼−1

− 𝑡

₁^𝛼−1

) + 2𝐿

(𝑝 − 𝛼)𝛤(𝛼) (𝑡

₂^𝛼−1

− 𝑡

₁^𝛼−1

)

≤ (𝑏 + 2𝐿

(𝑝 − 𝛼)𝛤(𝛼) ) (𝑡

2𝛼−1

− 𝑡

1𝛼−1

)

Since 𝑡

^𝛼−1

is uniformly continuous when 0 ≤ 𝑡 ≤ 1 𝑎𝑛𝑑 1 < 𝛼 ≤ 2 , it is easy to prove 𝐴(Ω)is

equicontinuous . The Arzele − Ascoli Theorem implies that 𝐴(Ω) ̅̅̅̅̅̅̅ is compact.

Therefore A ∶ P → P is completely contionuous.

Step2. A is an increasing operator.

Let 𝑥

₁

, 𝑥

₂

∈ P and 𝑥

₁

≤ 𝑥

₂

then by (H

₁

), we have that 𝐴𝑥

1

(t) = 𝑏𝑡

^𝛼−1

+ ∫ 𝐺(𝑡, 𝑠) 𝑝(𝑠)𝑓(𝑠, 𝑥

1

(𝑠))𝑑𝑠

1

0

≤ 𝑏𝑡

^𝛼−1

+ ∫ 𝐺(𝑡, 𝑠) 𝑝(𝑠)𝑓(𝑠, 𝑥

₂

(𝑠))𝑑𝑠

1

0

≤ 𝐴𝑥

2

(t)

Therefore A is an increasing operator.

Applying the definition of lower and upper solution , we have 𝐴𝑥

1

≥ 𝑥

1

𝑎𝑛𝑑 𝐴𝑥

2

≤ 𝑥

2

.

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Hence A: 〈x

₁

, x

₂

〉 → 〈x

₁

, x

₂

〉 is a compact continuous operator.

Step 3. By (H

₃

),there exists positive constants M

₁

and N such that x ≥ N it holds p(t)f(t, x(t)) ≤ M

₁

:

On the other hand, by (H

₁

), f: [0,1] × [0, N] is continuous, ∃M

₂

> 0 such that, it holds p(t)f(t, x(t)) ≤ M

₂

.

Let M = max{M

1

, M

2

} , then we have p(t)f(t, x(t)) ≤ M , ∀x ≥ 0.

Now, we consider the following equation

D

^α

w(t) + M = 0 , 0 < t < 1, n − 1 < α ≤ n (8) w(0) = w

^′

(0) = ⋯ = w

⁽ⁿ⁻²⁾

(0) = 0 , w

^(q)

(1) = a

From the theorem (2.1), we have the solution (8) is w(t) = bt

^α−1

+ ∫ G(t, s) Mds

1

0

≥ bt

^α−1

+ ∫ G(t, s) p(s)f(s, w(s))ds

1

0

= Aw(t)

which implies that 𝑤(𝑡) is an upper solution of the operator A.

Obviously, 𝑤(𝑡) ≡ 0 is a upper solution of the operator A. On the other hand, it is obvious that 𝑣(𝑡) ≡ 0

is a lower solution of the operator A, we have 𝑣(𝑡) ≤ 𝑤(𝑡).

Step 4.Since P is a normal cone, Theorem(3.1) implies that A has a fixed point 𝑥(𝑡) ∈

〈0, 𝑤(𝑡)〉.

Therefore, the equation (1) has a positive solution. This completes the proof.

𝐓𝐡𝐞𝐨𝐫𝐞𝐦(𝟑. 𝟐): [𝐔𝐧𝐢𝐪𝐮𝐞𝐧𝐞𝐬𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦]

Assume that f satisfies

|f(t, x) − f(t, y)| ≤ a(t)|x − y|, (9) where t ∈ [0,1], x, y ∈ [0, ∞) , a: [0,1] → [0, ∞) is a continuous function. If

∫ s

₀¹ ^α−1

(1 − s)

^α−1−q

𝑎(𝑠)𝑑𝑠 < (𝑞 − 𝛼)𝛤(𝛼) (10) 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (1)ℎ𝑎𝑠 𝑢𝑛𝑖𝑞𝑢𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

𝐏𝐫𝐨𝐨𝐟: If A

ⁿ

is a contraction operator for n sufficiently large , then the equation (1) has a unique positive solution.

For x, y ∈ P , we have the estimate, by the definition of Greens function |𝐴𝑥(𝑡) − 𝐴𝑦(𝑡)|

= |𝑏𝑡

^𝛼−1

+ ∫ 𝐺(𝑡, 𝑠) 𝑝(𝑠)𝑓(𝑠, 𝑥(𝑠))𝑑𝑠

1

0

− (𝑏𝑡

^𝛼−1

+ ∫ 𝐺(𝑡, 𝑠) 𝑝(𝑠)𝑓(𝑠, 𝑦(𝑠))𝑑𝑠

1 0

)|

(9)

≤ ∫ 𝐺(𝑡, 𝑠)𝑝(𝑠)𝑎(𝑠) |𝑓(𝑠, 𝑥(𝑠)) − 𝑓(𝑠, 𝑦(𝑠))|𝑑𝑠

1

0

≤ ∫ 𝐺(𝑡, 𝑠)𝐾𝑎(𝑠)|𝑥(𝑠) − 𝑦(𝑠)|𝑑𝑠

1

0

≤ ‖𝑥 − 𝑦‖𝑡

^𝛼−1

𝛤(𝛼) ∫ (1 − 𝑠)

^{𝛼−1−𝑞}

1 0

N𝑎(𝑠)𝑑𝑠 Denote 𝐻 = ∫ (1 − 𝑠)

^{𝛼−1−𝑞}

1 0

𝑁𝑎(𝑠) ≤ 𝐻𝑡

^𝛼−1

𝛤(𝛼) ‖𝑥 − 𝑦‖

Similarly, |𝐴

²

𝑥(𝑡) − 𝐴

²

𝑦(𝑡)| = ∫ 𝐺(𝑡, 𝑠)𝑝(𝑠) |𝑓(𝑠, 𝐴𝑥(𝑠)) − 𝑓(𝑠, 𝐴𝑦(𝑠))|𝑑𝑠

1

0

≤ ∫ 𝐺(𝑡, 𝑠)𝑎(𝑠)𝑝(𝑠)|𝐴𝑥(𝑠) − 𝐴𝑦(𝑠)|𝑑𝑠

1

0

≤ ∫ 𝐺(𝑡, 𝑠)𝑎(𝑠) 𝐻𝑡

^𝛼−1

𝛤(𝛼) ‖𝑥 − 𝑦‖𝑑𝑠

1

0

≤ ∫ 𝑡

^𝛼−1

(1 − 𝑠)

^{𝛼−1−𝑞}

𝛤(𝛼)

𝐻𝑡

^𝛼−1

𝛤(𝛼) ‖𝑥 − 𝑦‖𝑎(𝑠)𝑑𝑠

1

0

≤ 𝐻‖𝑥 − 𝑦‖𝑡

^𝛼−1

𝛤

²

(𝛼) ∫ 𝑠

^𝛼−1

(1 − 𝑠)

^{𝛼−1−𝑞}

1 0

𝑎(𝑠)𝑑𝑠 ≤ 𝐻𝐾𝑡

^𝛼−1

𝛤

²

(𝛼) ‖𝑥 − 𝑦‖

𝑤ℎ𝑒𝑟𝑒 𝐾 = ∫ 𝑠

₀¹ ^𝛼−1

(1 − 𝑠)

^{𝛼−1−𝑞}

𝑎(𝑠)𝑑𝑠By mathematical induction, it follows that

|𝐴

^𝑛

𝑥(𝑡) − 𝐴

^𝑛

𝑦(𝑡)| ≤

^𝐾𝐻^𝑛−1^𝑡^𝛼−1

𝛤^𝑛(𝛼)

‖𝑥 − 𝑦‖, (11) By (11), for n large enough, we have

𝐻𝐾

^𝑛−1

𝛤

^𝑛

(𝛼) = 𝐻

𝛤(𝛼) ( 𝐾 𝛤(𝛼) )

𝑛−1

< 1.

Hence , it holds

‖𝐴

^𝑛

𝑥(𝑡) − 𝐴

^𝑛

𝑦(𝑡)‖ ≤ ‖𝑥 − 𝑦‖

which implies A

ⁿ

is a contraction operator for n sufficiently large, then the equation (1) has a unique positive solution.

This completes the proof.

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