Existence of positive solutions for a system of semipositone fractional boundary value problems

Existence of positive solutions for a system of semipositone fractional boundary value problems

Johnny Henderson

and Rodica Luca

1Baylor University, One Bear Place 97328, Waco, Texas, 76798-7328 USA

2Gheorghe Asachi Technical University, 11 Blvd. Carol I, Ias,i 700506, Romania

Received 30 December 2015, appeared 2 May 2016 Communicated by Paul Eloe

Abstract. We investigate the existence of positive solutions for a system of nonlinear Riemann–Liouville fractional differential equations with sign-changing nonlinearities, subject to coupled integral boundary conditions.

Keywords: Riemann–Liouville fractional differential equations, coupled integral boundary conditions, positive solutions, sign-changing nonlinearities.

2010 Mathematics Subject Classification: 34A08, 45G15.

1 Introduction

We consider the system of nonlinear fractional differential equations (D₀^α₊u(t) +λf(t,u(t),v(t)) =0, t∈ (0, 1),

D₀^β₊v(t) +µg(t,u(t),v(t)) =0, t∈ (0, 1), (S) with the coupled integral boundary conditions











u(0) =u⁰(0) =· · · =u⁽ⁿ⁻²⁾(0) =0, u⁰(1) =

Z ₁

0 v(s)dH(s), v(0) =v⁰(0) =· · · =v^(m−2)(0) =0, v⁰(1) =

Z ₁

0

u(s)dK(s),

(BC)

where α∈(n−1,n], β∈ (m−1,m],n, m∈_N, n, m≥3, D^α₀₊and D₀^β₊denote the Riemann–

Liouville derivatives of orders α and β, respectively, the integrals from (BC) are Riemann–

Stieltjes integrals, and f and gare sign-changing continuous functions (that is, we have a so- called system of semipositone boundary value problems). These functions may be nonsingular or singular at t = 0 and/or t = 1. The boundary conditions above include multi-point and integral boundary conditions, as well as the sum of these in a single framework.

We present intervals for parameters λ and µ such that the above problem (S)–(BC) has at least one positive solution. By a positive solution of problem (S)–(BC) we mean a pair of

BCorresponding author. Email: rluca@math.tuiasi.ro

functions(u,v) ∈ C([0, 1])×C([0, 1]) satisfying (S) and (BC) withu(t) ≥ 0, v(t) ≥ 0 for all t ∈ [0, 1]andu(t)>0, v(t)>0 for all t ∈ (0, 1). The system (S) with the uncoupled integral boundary conditions











u(0) =u⁰(0) =· · · =u⁽ⁿ⁻²⁾(0) =0, u(1) =

Z ₁

0 u(s)dH(s), v(0) =v⁰(0) =· · ·=v^(m−2)(0) =0, v(1) =

Z ₁

0 v(s)dK(s),

(BC₁)

where the functions f andgare nonnegative, has been investigated in [6] and [11] by using the Guo–Krasnosel’skii fixed point theorem, and in [7] where in system (S) we haveλ=µ=1 and f(t,u,v)_andg(t,u,v)are replaced by fe(t,v)_andg_e(t,u), respectively, with efandgenonsingular or singular functions (denoted by (Se)). In [7] we used some theorems from the fixed point index theory and the Guo–Krasnosel’skii fixed point theorem. The semipositone case for problem (S)–(BC₁) was studied in [14] by using the nonlinear alternative of Leray–Schauder type. The systems (S) and (S) with coupled integral boundary conditionse











u(0) =u⁰(0) =· · ·=u⁽ⁿ⁻²⁾(0) =0, u(1) =

Z ₁

0 v(s)dH(s), v(0) =v⁰(0) =· · ·= v^(m−2)(0) =0, v(1) =

Z ₁

0 u(s)dK(s),

(BC₂)

have been investigated in [8] and [9] (problem (S)–(BC2) with f andgnonnegative functions), in [12] (problem (S)–(BCe ₂) with f and g nonnegative functions, singular or not), and in [10]

(problem (S)–(BC₂) with f, gsign-changing functions). We also mention the paper [20], where the authors studied the existence and multiplicity of positive solutions for system (S) with α= β,λ=µand the boundary conditionsu⁽ⁱ⁾(0) =v⁽ⁱ⁾(0) =0,i=0, . . . ,n−2,u(1) =av(ξ), v(1) = bu(η), ξ, η ∈ (0, 1), with ξ, η ∈ (0, 1), 0 < abξη < 1, and f and g are sign-changing nonsingular or singular functions.

Fractional differential equations describe many phenomena in various fields of engineering and scientific disciplines such as physics, biophysics, chemistry, biology, economics, control theory, signal and image processing, aerodynamics, viscoelasticity, electromagnetics, and so on (see [2–4,13,15–19]). Integral boundary conditions arise in thermal conduction problems, semiconductor problems and hydrodynamic problems.

The paper is organized as follows. Section 2 contains some auxiliary results which inves- tigate a nonlocal boundary value problem for fractional differential equations. In Section 3, we prove several existence theorems for the positive solutions with respect to a cone for our problem (S)–(BC). Finally in Section 4 some examples are given to illustrate our main results.

2 Auxiliary results

We present here the definitions of the Riemann–Liouville fractional integral and the Riemann–

Liouville fractional derivative and then some auxiliary results that will be used to prove our main results.

Definition 2.1. The (left-sided) fractional integral of orderα>0 of a function f :(0,∞)→_R is given by

(I₀^α₊f)(t) = ¹ Γ(α)

Z _t

0

(t−s)^α−1f(s)ds, t >0,

provided the right-hand side is pointwise defined on (0,∞), whereΓ(α)is the Euler gamma function defined byΓ(α) =R_∞

0 t^α−1e^−tdt,α>0.

Definition 2.2. The Riemann–Liouville fractional derivative of order α ≥ 0 for a function f :(_0,∞)→_Ris given by

(D₀^α₊f)(t) = d

dt n

I₀ⁿ₊^−αf

(t) = ¹ Γ(n−α)

d dt

nZ _t

0

f(s)

(t−s)^{α−n+1ds, t} >0, wheren =bαc+1, provided that the right-hand side is pointwise defined on(0,∞).

The notation bαc stands for the largest integer not greater than α. If α = m ∈ _N then D^m₀₊f(t) = f^(m)(t)fort>0, and ifα=0 then D₀⁰₊f(t) = f(t)fort>0.

We consider now the fractional differential system

(D₀^α₊u(t) +_ex(t) =0, t∈(0, 1),

D₀^β₊v(t) +_ey(t) =0, t∈(0, 1), (2.1) with the coupled integral boundary conditions











u(0) =u⁰(0) =· · · =u⁽ⁿ⁻²⁾(0) =0, u⁰(1) =

Z ₁

0 v(s)dH(s), v(0) =v⁰(0) =· · · =v^(m−2)(0) =0, v⁰(1) =

Z ₁

0 u(s)dK(s),

(2.2)

where α∈ (n−1,n], β∈ (m−1,m], n, m∈_N, n, m≥3, and H,K : [0, 1]→_R are functions of bounded variation.

Lemma 2.3. If H, K : [_{0, 1}] → _R are functions of bounded variation, ∆ = (α−1)(β−1)− R1

0 τ^α−1dK(τ) R1

0 τ^β−1dH(τ) 6= 0 and x,e ey ∈ C(0, 1)∩L¹(0, 1), then the pair of functions (u,v)∈C([0, 1])×C([0, 1])given by

u(t) = − ¹ Γ(α)

Z _t

0

(t−s)^α−1x_e(s)ds+^t

α−1

∆

β−1 Γ(α−1)

Z ₁

0

(1−s)^α−2_ex(s)ds

− ¹ Γ(α)

_{Z 1}

0 s^β−1dH(s)

Z ₁

0

_{Z 1}

s

(τ−s)^α−1dK(τ)

xe(s)ds

+ ¹

Γ(β−1) _{Z 1}

0 s^β−1dH(s)

Z ₁

0

(1−s)^β−2y_e(s)ds

− ¹ Γ(β−1)

Z ₁

0

_{Z 1}

s

(τ−s)^β−1dH(τ)

ye(s)ds

, t∈ (0, 1], u(0) =0, v(t) = − ¹

Γ(β)

Z _t

0

(t−s)^β−1_ey(s)ds+ ^t

β−1

∆

α−1 Γ(β−1)

Z ₁

0

(1−s)^β−2_ey(s)ds

− ¹ Γ(β)

_{Z 1}

0 s^α−1dK(s)

Z ₁

0

_{Z 1}

s

(τ−s)^β−1dH(τ)

ey(s)ds

+ ¹

Γ(α−1) _{Z 1}

0 s^α−1dK(s)

Z ₁

0

(1−s)^α−2_ex(s)ds

− ¹ Γ(α−1)

Z ₁

0

Z ₁

s

(τ−s)^α−1dK(τ)

ex(s)ds

, t∈(0, 1], v(0) =0,

(2.3)

is solution of problem(2.1)–(2.2).

Proof. We denote by c₁ = ¹

∆

β−₁ Γ(α−1)

Z ₁

0

(₁−s)^α−2_ex(s)ds− ¹ Γ(β−1)

Z ₁

0

Z ₁

s

(τ−s)^β−1dH(τ)

ye(s)ds

+ ¹

Γ(β−1) _{Z 1}

0

τ^β−1dH(τ)

Z ₁

0

(₁−s)^β−2_ey(s)ds

− ¹ Γ(α)

_{Z 1}

0 τ^β−1dH(τ)

Z ₁

0

_{Z 1}

s

(τ−s)^α−1dK(τ)

xe(s)ds

, and

d₁= ¹

∆

α−1 Γ(β−1)

Z ₁

0

(1−s)^β−2y_e(s)ds− ¹ Γ(α−1)

Z ₁

0

_{Z 1}

s

(τ−s)^α−1dK(τ)

xe(s)ds

+ ¹

Γ(α−1) _{Z 1}

0 τ^α−1dK(τ)

Z ₁

0

(1−s)^α−2_ex(s)ds

− ¹ Γ(β)

Z ₁

0

τ^α−1dK(τ)

Z ₁

0

Z ₁

s

(τ−s)^β−1dH(τ)

ey(s)ds

. Then the continuous functionsuandvfrom (2.3) can be written as











u(t) =c₁t^α−1− ¹ Γ(α)

Z _t

0

(t−s)^α−1_ex(s)ds=c₁t^α−1−I₀^α₊xe(t), t∈ (0, 1], u(0) =0, v(t) =d₁t^β−1− ¹

Γ(β)

Z _t

0

(t−s)^β−1_ey(s)ds=d₁t^β−1−I₀^β₊ye(t), t∈ (0, 1], v(0) =0.

Because D₀^α₊u(t) = c₁D^α₀₊(t^α−1)−D₀^α₊I₀^α₊xe(t) = −x_e(t) and D₀^β₊v(t) = d₁D₀^β₊(t^β−1)− D₀^β₊I₀^β₊ey(t) = −y_e(t) for all t ∈ (0, 1), we deduce that u and v satisfy the system (2.1). In addition, we haveu(0) =u⁰(0) =· · ·= u⁽ⁿ⁻²⁾(0) =0 andv(0) =v⁰(0) =· · · =v^(m−2)(0) =0.

A simple computation shows us thatu⁰(1) =R1

0 v(s)dH(s)andv⁰(1) =R1

0 u(s)dK(s), that is











c₁(α−1)− ¹ Γ(α−1)

Z ₁

0

(1−s)^α−2_ex(s)ds=

Z ₁

0

d₁s^β−1− ¹ Γ(β)

Z _s

0

(s−τ)^β−1_ey(τ)dτ

dH(s), d₁(β−1)− ¹

Γ(β−1)

Z ₁

0

(1−s)^β−2_ey(s)ds=

Z ₁

0

c₁s^α−1− ¹ Γ(α)

Z _s

0

(s−τ)^α−1_ex(τ)dτ

dK(s). Therefore we deduce that(u,v)is solution of problem (2.1)–(2.2).

Lemma 2.4. Under the assumptions of Lemma2.3, the solution(u,v)of problem(2.1)–(2.2)given by (2.3)can be written as









 u(t) =

Z ₁

0

G₁(t,s)x_e(s)ds+

Z ₁

0

G₂(t,s)_ey(s)ds, t∈[_{0, 1}]_, v(t) =

Z ₁

0 G3(t,s)y_e(s)ds+

Z ₁

0 G₄(t,s)_ex(s)ds, t∈[0, 1],

(2.4)

where 





























G₁(t,s) =g₁(t,s) + ^t

α−1

∆ _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0 g₁(τ,s)dK(τ)

, G₂(t,s) = (β−1)t^α−1

∆

Z ₁

0 g₂(τ,s)dH(τ), G3(t,s) =g2(t,s) + ^t

β−1

∆ Z ₁

0 τ^α−1dK(τ)

Z ₁

0 g2(τ,s)dH(τ)

, G₄(t,s) = (α−1)t^β−1

∆

Z ₁

0 g₁(τ,s)dK(τ), ∀t, s∈[0, 1],

(2.5)

and















g₁(t,s) = ¹ Γ(α)

(t^α−1(1−s)^α−2−(t−s)^α−1, 0≤s≤ t≤1, t^α−1(1−s)^α−2, ≤t ≤s≤1, g₂(t,s) = ¹

Γ(β)

(t^β−1(₁−s)^β−2−(t−s)^β−1_{, 0}≤s≤ t≤1, t^β−1(1−s)^β−2, 0≤t ≤s≤1.

(2.6)

Proof. By Lemma2.3and relation (2.3), we conclude

u(t) = ¹ Γ(α)

_{Z t}

0

[t^α−1(1−s)^α−2−(t−s)^α−1]x_e(s)ds+

Z ₁

t t^α−1(1−s)^α−2x_e(s)ds

−

Z ₁

0 t^α−1(1−s)^α−2_ex(s)ds

+(β−1)t^α−1

∆Γ(α−1)

Z ₁

0

(1−s)^α−2x_e(s)ds

− ^t

α−1

∆Γ(α) _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0

_{Z 1}

s

(τ−s)^α−1dK(τ)

ex(s)ds

+ ^t

α−1

∆Γ(β−1) _{Z 1}

0

_{Z 1}

0 τ^β−1(1−s)^β−2dH(τ)

ye(s)ds

−

Z ₁

0

_{Z 1}

s

(τ−s)^β−1dH(τ)

ey(s)ds

= ¹

Γ(α) _{Z t}

0

[t^α−1(1−s)^α−2−(t−s)^α−1]x_e(s)ds+

Z ₁

t t^α−1(1−s)^α−2x_e(s)ds

− ^β−1

∆Γ(α−1)

Z ₁

0 t^α−1(1−s)^α−2x_e(s)ds+ ¹

∆Γ(α) Z ₁

0 τ^α−1dK(τ)

Z ₁

0 τ^β−1dH(τ)

× _{Z 1}

0

t^α−1(₁−s)^α−2x_e(s)ds

+ ^β−1

∆Γ(α−1)

Z ₁

0

t^α−1(₁−s)^α−2x_e(s)ds

− ^t

α−1

∆Γ(α) _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0

_{Z 1}

s

(τ−s)^α−1dK(τ)

xe(s)ds

+ ^t

α−1

∆Γ(β−₁) _{Z 1}

0

_{Z 1}

0 τ^β−1(1−s)^β−2dH(τ)

ye(s)ds

−

Z ₁

0

_{Z 1}

s

(τ−s)^β−1dH(τ)

ye(s)ds

= ¹

Γ(α) _{Z t}

0

[t^α−1(1−s)^α−2−(t−s)^α−1]x_e(s)ds+

Z ₁

t t^α−1(1−s)^α−2x_e(s)ds

+ ^t

α−1

∆Γ(α) _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0

_{Z 1}

0 τ^α−1(1−s)^α−2dK(τ)

xe(s)ds

−

Z ₁

0

_{Z 1}

s

(τ−s)^α−1dK(τ)

xe(s)ds

+ ^t

α−1

∆Γ(β−1) Z ₁

0

Z ₁

0 τ^β−1(1−s)^β−2dH(τ)

ye(s)ds

−

Z ₁

0

Z ₁

s

(τ−s)^β−1dH(τ)

ye(s)ds

.

Therefore, we obtain u(t) = ¹

Γ(α) _{Z t}

0

[t^α−1(1−s)^α−2−(t−s)^α−1]_ex(s)ds+

Z ₁

t t^α−1(1−s)^α−2_ex(s)ds

+ ^t

α−1

∆Γ(α) _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0

_{Z s}

0 τ^α−1(1−s)^α−2dK(τ)

xe(s)ds +

Z ₁

0

Z ₁

s

τ^α−1(1−s)^α−2dK(τ)

xe(s)ds−

Z ₁

0

Z ₁

s

(τ−s)^α−1dK(τ)

xe(s)ds

+ ^t

α−1

∆Γ(β−1) _{Z 1}

0

_{Z s}

0

τ^β−1(₁−s)^β−2dH(τ)

ey(s)ds +

Z ₁

0

_{Z 1}

s τ^β−1(1−s)^β−2dH(τ)

ye(s)ds−

Z ₁

0

_{Z 1}

s

(τ−s)^β−1dH(τ)

ye(s)ds

= ¹

Γ(α) _{Z t}

0

[t^α−1(1−s)^α−2−(t−s)^α−1]_ex(s)ds+

Z ₁

t t^α−1(1−s)^α−2_ex(s)ds

+ ^t

α−1

∆Γ(α) _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0

_{Z s}

0 τ^α−1(1−s)^α−2dK(τ)

xe(s)ds +

Z ₁

0

Z ₁

s

[τ^α−1(1−s)^α−2−(τ−s)^α−1]dK(τ)

xe(s)ds

+ ^t

α−1

∆Γ(β−1) Z ₁

0

Z _s

0

τ^β−1(₁−s)^β−2dH(τ)

ey(s)ds +

Z ₁

0

_{Z 1}

s

[τ^β−1(₁−s)^β−2−(τ−s)^β−1]dH(τ)

ye(s)ds

=

Z ₁

0 g₁(t,s)x_e(s)ds+^t

α−1

∆ _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0

_{Z 1}

0 g₁(τ,s)dK(τ)

xe(s)ds

+(β−1)t^α−1

∆

_{Z 1}

0

_{Z 1}

0 g₂(τ,s)dH(τ)

ye(s)ds

=

Z ₁

0 G₁(t,s)x_e(s)ds+

Z ₁

0 G₂(t,s)_ey(s)ds.

In a similar manner, we deduce v(t) =

Z ₁

0 g₂(t,s)y_e(s)ds+ ^t

β−1

∆ _{Z 1}

0 τ^α−1dK(τ)

Z ₁

0

_{Z 1}

0 g₂(τ,s)dH(τ)

ye(s)ds

+(α−1)t^β−1

∆

_{Z 1}

0

_{Z 1}

0 g₁(τ,s)dK(τ)

xe(s)ds

=

Z ₁

0 G3(_t,s)_ye(_s)_ds+

Z ₁

0 G4(_t,s)_ex(_s)_ds.

Therefore, we obtain the expression (2.4) for the solution(u,v)of problem (2.1)–(2.2) given by relations (2.3).

Lemma 2.5. The functions g₁, g₂ given by(2.6)have the properties

a) g₁, g₂ : [0, 1]×[0, 1] → _R+ are continuous functions, and g₁(t,s) > 0, g₂(t,s) > 0 for all (_t,s)∈(_{0, 1}]×(_{0, 1})_.

b) g₁(t,s) ≤ h₁(s), g₂(t,s) ≤ h₂(s)for all (t,s) ∈ [0, 1]×[0, 1], where h₁(s) = ^s(1_Γ⁻₍^s)α−2

α) and h₂(s) = ^s(1_Γ⁻₍^s)β−2

β) for all s∈[0, 1].

c) g₁(t,s)≥t^α−1h₁(s), g₂(t,s)≥t^β−1h₂(s)for all(t,s)∈[0, 1]×[0, 1]. d) g₁(t,s)≤ _Γ^tα₍⁻¹

α), g2(t,s)≤ _Γ^tβ₍⁻¹

β), for all(t,s)∈[0, 1]×[0, 1]. Proof. Part a) of this lemma is evident.

b) The functiong₁ is nondecreasing in the first variable. Indeed, fors≤ t, we have

∂g₁

∂t (t,s) = ¹

Γ(α)[(α−1)t^α−2(1−s)^α−2−(α−1)(t−s)^α−2]

= ¹

Γ(α−1)[(t−ts)^α−2−(t−s)^α−2]≥0.

Then, g₁(t,s)≤g₁(1,s)for all(t,s)∈ [0, 1]×[0, 1]withs ≤t.

Fors≥ t, we obtain

∂g₁

∂t (t,s) = ¹ Γ(α−1)^t

α−2(1−s)^α−2 ≥0.

Hence, g₁(t,s)≤ g₁(s,s)_{for all}(t,s)∈[_{0, 1}]×[_{0, 1}]_withs≥ t.

Therefore, we deduce that g₁(t,s) ≤ h₁(s) for all (t,s) ∈ [0, 1]×[0, 1], where h₁(s) = g₁(1,s) = _Γ(¹

α)s(1−s)^α−2 fors ∈[0, 1]. c) Fors≤t, we have

g₁(t,s) = ¹

Γ(α)[t^α−1(1−s)^α−2−(t−s)^α−1]

≥ ¹

Γ(α)[t^α−1(1−s)^α−2−(t−ts)^α−1] =t^α−1h₁(s). Fors≥ t, we obtain

g₁(t,s) = ¹ Γ(α)^t

α−1(₁−s)^α−2 ≥t^α−1h₁(s)_.

Hence, we conclude thatg₁(t,s)≥ t^α−1h₁(s)for all (t,s)∈[0, 1]×[0, 1]. d) For all(t,s)∈[0, 1]×[0, 1]we have

g₁(t,s)≤ ¹ Γ(α)^t

α−1(1−s)^α−2≤ ^t

α−1

Γ(α)^.

In a similar manner we obtain the corresponding inequalities forg₂, withh₂(s) = ^s(1_Γ⁻₍^s)β−2

β)

fors∈ [0, 1].

Lemma 2.6. If H, K:[0, 1]→_Rare nondecreasing functions, and∆>0, then G_i, i=1, . . . , 4,given by(2.5)are continuous functions on[0, 1]×[0, 1]and satisfy G_i(t,s)≥0for all(t,s)∈ [0, 1]×[0, 1], i= 1, . . . , 4. Moreover, ifex, ye∈C(0, 1)∩L¹(0, 1)satisfyxe(t)≥0,ye(t)≥0for all t ∈ (0, 1), then the solution(u,v)of problem(2.1)–(2.2)given by(2.4)satisfies u(t)≥0, v(t)≥0for all t ∈[0, 1]. Proof. By using the assumptions of this lemma, we haveG_i(t,s)≥0 for all(t,s)∈[0, 1]×[0, 1], i=1, . . . , 4, and sou(t)≥0,v(t)≥0 for allt ∈[0, 1].

Lemma 2.7. Assume that H, K : [0, 1] → _R are nondecreasing functions, ∆ > 0, and that R1

0 τ^α−1dK(τ) > 0, R1

0 τ^β−1dH(τ) > 0. Then the functions G_i, i = 1, . . . , 4satisfy the inequal- ities

a₁)G₁(t,s)≤σ₁h₁(s), ∀(t,s)∈[0, 1]×[0, 1], where σ₁ =1+ ¹

∆(K(1)−K(0))

Z ₁

0 τ^β−1dH(τ)>0.

a₂)G₁(t,s)≤δ₁t^α−1, ∀(t,s)∈[0, 1]×[0, 1], where δ₁ = ¹

Γ(α)

1+ ¹

∆ Z ₁

0

τ^β−1dH(τ)

Z ₁

0

τ^α−1dK(τ)

>_0.

a₃)G₁(t,s)≥$₁t^α−1h₁(s), (t,s)∈[0, 1]×[0, 1], where

$₁ =1+ ¹

∆ Z ₁

0 τ^β−1dH(τ)

Z ₁

0 τ^α−1dK(τ)

>0.

b₁)G2(t,s)≤σ₂h2(s), ∀(t,s)∈[0, 1]×[0, 1], whereσ₂ = ^β−_∆¹(H(1)−H(0))>0.

b₂)G₂(t,s)≤δ₂t^α−1, ∀(t,s)∈[0, 1]×[0, 1], whereδ₂= _∆Γ(¹

β−1)

R1

0 τ^β−1dH(τ)>0.

b3)G2(t,s)≥$2t^α−1h2(s), ∀(t,s)∈[0, 1]×[0, 1], where$2= ^β−_∆¹R1

0 τ^β−1dH(τ)>0.

c₁)G₃(t,s)≤σ₃h₂(s), ∀(t,s)∈[0, 1]×[0, 1], where σ₃ =1+ ¹

∆(H(1)−H(0))

Z ₁

0 τ^α−1dK(τ)>0.

c₂)G₃(t,s)≤δ₃t^β−1, ∀(t,s)∈ [0, 1]×[0, 1], where δ₃= ¹

Γ(β)

1+ ¹

∆ Z ₁

0

τ^α−1dK(τ)

Z ₁

0

τ^β−1dH(τ)

>_0.

c₃)G₃(t,s)≥$₃t^β−1h₂(s), ∀(t,s)∈[0, 1]×[0, 1], where

$₃ =1+ ¹

∆ _{Z 1}

0 τ^α−1dK(τ)

Z ₁

0 τ^β−1dH(τ)

= $₁ >0.

d₁)G₄(t,s)≤ σ₄h₁(s), ∀(t,s)∈[0, 1]×[0, 1], whereσ₄ = ^α−_∆¹(K(1)−K(0))>0.

d₂)G₄(t,s)≤ δ₄t^β−1, ∀(t,s)∈[_{0, 1}]×[_{0, 1}], whereδ₄ = _∆Γ(¹

α−1)

R1

0 τ^α−1dK(τ)>_0.

d3)G₄(t,s)≥ $₄t^β−1h₁(s), ∀(t,s)∈ [0, 1]×[0, 1], where $₄ = ^α−_∆¹R1

0 τ^α−1dK(τ)>0.

Proof. From the assumptions of this lemma, we obtain K(1)−K(0) =

Z ₁

0 dK(τ)≥

Z ₁

0 τ^α−1dK(τ)>0, H(1)−H(0) =

Z ₁

0 dH(τ)≥

Z ₁

0 τ^β−1dH(τ)>0.

By using Lemma2.4 and Lemma2.5, we deduce for all (t,s)∈[_{0, 1}]×[_{0, 1}]

a₁)

G₁(t,s) =g₁(t,s) + ^t

α−1

∆ Z ₁

0

τ^β−1dH(τ)

Z ₁

0

g₁(_τ,s)dK(τ)

≤h₁(s) + ¹

∆ _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0 h₁(s)dK(τ)

=h₁(s)

1+ ¹

∆(K(1)−K(0))

Z ₁

0 τ^β−1dH(τ)

= σ₁h₁(s). a₂)

G₁(t,s)≤ ^t

α−1

Γ(α)+ ^t

α−1

∆ _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0

τ^α−1 Γ(α)^dK(τ)

= t^α−1 1 Γ(α)

1+ ¹

∆ _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0 τ^α−1dK(τ)

=δ₁t^α−1. a3)

G₁(t,s)≥t^α−1h₁(s) + ^t

α−1

∆ _{Z 1}

0 τ^β−1dH(τ)

Z ₁

0 τ^α−1h₁(s)dK(τ)

=t^α−1h₁(s)

1+ ¹

∆ Z ₁

0 τ^β−1dH(τ)

Z ₁

0 τ^α−1dK(τ)

= $₁t^α−1h₁(s). b₁)

G₂(t,s) = (β−1)t^α−1

∆

Z ₁

0 g₂(τ,s)dH(τ)≤ ^β−1

∆ Z ₁

0 h₂(s)dH(τ)

= ^β−1

∆ (H(1)−H(0))h2(s) =σ2h2(s). b₂)

G2(t,s)≤ (β−1)t^α−1

∆

Z ₁

0

τ^β−1

Γ(β)^dH(τ) =δ2t^α−1. b₃)

G₂(t,s)≥ (β−1)t^α−1

∆

Z ₁

0 τ^β−1h₂(s)dH(τ) =$₂t^α−1h₂(s). c₁)

G3(t,s) = g2(t,s) + ^t

β−1

∆ Z ₁

0 τ^α−1dK(τ)

Z ₁

0 g2(τ,s)dH(τ)

≤h₂(s) + ¹

∆ _{Z 1}

0

τ^α−1dK(τ)

Z ₁

0

h₂(s)dH(τ)

=h₂(s)

1+ ¹

∆(H(1)−H(0))

Z ₁

0 τ^α−1dK(τ)

= σ₃h₂(s). c₂)

G₃(t,s)≤ ^t

β−1

Γ(β)+ ^t

β−1

∆ Z ₁

0

τ^α−1dK(τ)

Z ₁

0

τ^β−1 Γ(β)^dH(τ)

= ^t

β−1

Γ(β)

1+ ¹

∆ Z ₁

0 τ^α−1dK(τ)

Z ₁

0 τ^β−1dH(τ)

=δ₃t^β−1.