On a System of Hadamard Fractional Differential Equations With Nonlocal Boundary Conditions on an Unbounded Domain

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On a System of Hadamard Fractional Differential Equations With Nonlocal Boundary Conditions on an Unbounded Domain

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Rodica Luca^*,

Alexandru Tudorache

Rodica Luca^*,

Alexandru Tudorache

This version is not peer-reviewed

Submitted:

10 May 2023

Posted:

11 May 2023

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Abstract

We study the existence of positive solutions for a system of Hadamard fractional differential equations on an infinite interval with nonnegative nonlinearities, subject to coupled nonlocal boundary conditions which contain Hadamard fractional derivatives and Riemann-Stieltjes integrals. In the proof of the main results, we use the Guo-Krasnosel'skii fixed point theorem and the Leggett-Williams fixed point theorem.

Keywords:

Subject: Computer Science and Mathematics - Analysis

MSC: 34A08; 34B10; 34B15; 34B18

1. Introduction

We consider the system of nonlinear Hadamard fractional differential equations

\{\begin{matrix} ^{H} D_{1 +}^{α} x (t) + a (t) f (x (t), y (t)) = 0, t \in (1, \infty), \\ ^{H} D_{1 +}^{β} y (t) + b (t) g (x (t), y (t)) = 0, t \in (1, \infty), \end{matrix}

(1)

supplemented with the nonlocal boundary conditions

\{\begin{matrix} x (1) = x^{'} (1) = \dots = x^{(n - 2)} (1) = 0, \\ ^{H} D_{1 +}^{α - 1} x (\infty) = \int_{1}^{\infty} x (s) d H_{1} (s) + \int_{1}^{\infty} y (s) d H_{2} (s), \\ y (1) = y^{'} (1) = \dots = y^{(m - 2)} (1) = 0, \\ ^{H} D_{1 +}^{β - 1} y (\infty) = \int_{1}^{\infty} x (s) d K_{1} (s) + \int_{1}^{\infty} y (s) d K_{2} (s), \end{matrix}

(2)

where

α \in (n - 1, n]

β \in (m - 1, m]

n, m \in N

n, m \geq 2

^{H} D_{1 +}^{κ}

denotes the Hadamard fractional derivative of order

κ

(for

κ = α, β, α - 1, β - 1

), the functions

a, b : (1, \infty) \to R_{+}

and

f, g : R_{+} \times R_{+} \to R_{+}

verify some assumptions, (

R_{+} = [0, \infty)

), and the integrals from (2) are Riemann-Stieltjes integrals with

H_{1}, H_{2}, K_{1}, K_{2} : [1, \infty) \to R

functions of bounded variation.

We will prove the existence of positive solutions of problem (1), (2) under some conditions on the data of problem, by using the Guo-Krasnosel’skii fixed point theorem and the Leggett-Williams fixed point theorem. By a positive solution of (1), (2) we mean a pair of functions

(x (t), y (t)), t \in [1, \infty)

satisfying (1) and (2), with

x (t) \geq 0

y (t) \geq 0

for all

t \in [1, \infty)

, and

x (t) > 0

for all

t \in (1, \infty)

y (t) > 0

for all

t \in (1, \infty)

. This problem is a generalization of the problem investigated in [13]. In the paper [13], the authors studied the system (1) with

α, β \in (1, 2]

(

n = m = 2

), subject to the boundary conditions

\{\begin{matrix} x (1) = 0,^{H} D_{1 +}^{α - 1} x (\infty) = \sum_{i = 1}^{m} λ_{i}^{H} I_{1 +}^{α_{i}} y (η), \\ y (1) = 0,^{H} D_{1 +}^{β - 1} y (\infty) = \sum_{j = 1}^{n} σ_{j}^{H} I_{1 +}^{β_{j}} x (ξ), \end{matrix}

(3)

where

λ_{i}, σ_{j} > 0

for

i = 1, \dots, m

j = 1, \dots, n

η > 1

ξ > 1

, and

^{H} I_{1 +}^{k}

is the Hadamard fractional integral of order k with lower limit 1 for

k = α_{1}, \dots, α_{m}, β_{1}, \dots, β_{n}

, defined by

^{H} I_{1 +}^{k} h (t) = \frac{1}{Γ (k)} \int_{1}^{t} {(ln \frac{t}{s})}^{k - 1} \frac{h (s)}{s} d s, t > 0,

(for a function

h : [1, \infty) \to R

), (see [7]). The last conditions for ∞ from (3) are particular cases of our conditions from (2). Indeed we have

\sum_{i = 1}^{m} λ_{i}^{H} I_{1 +}^{α_{i}} y (η) = \int_{1}^{\infty} y (s) d H_{2} (s), and \sum_{j = 1}^{n} σ_{j}^{H} I_{1 +}^{β_{j}} x (ξ) = \int_{1}^{\infty} x (s) d K_{1} (s),

with

H_{2} (s) = \sum_{i = 1}^{m} λ_{i} {\tilde{H}}_{i} (s)

, (

H_{1} \equiv 0

), and

K_{1} (s) = \sum_{j = 1}^{n} σ_{j} {\tilde{K}}_{j} (s)

, (

K_{2} \equiv 0

), where

{\tilde{H}}_{i} (s) = \{\begin{matrix} \frac{1}{Γ (α_{i} + 1)} ({(ln η)}^{α_{i}} - {(ln \frac{η}{s})}^{α_{i}}), if 0 \leq s \leq η, \\ \frac{1}{Γ (α_{i} + 1)} {(ln η)}^{α_{i}}, if s \geq η, \end{matrix}

and

{\tilde{K}}_{j} (s) = \{\begin{matrix} \frac{1}{Γ (β_{j} + 1)} ({(ln ξ)}^{β_{j}} - {(ln \frac{ξ}{s})}^{β_{j}}), if 0 \leq s \leq ξ, \\ \frac{1}{Γ (β_{j} + 1)} {(ln ξ)}^{β_{j}}, if s \geq ξ, \end{matrix}

for

i = 1, \dots, m

and

j = 1, \dots, n

. Therefore, in our paper, we consider general orders for the fractional derivatives in the equations of system (1). Besides, in the boundary conditions (2), the fractional derivatives

^{H} D_{1 +}^{α - 1} x

and

^{H} D_{1 +}^{β - 1} y

for ∞ are dependent on both functions x and y, in comparison with the condition (3) from [13], where the derivative of x is dependent only of y, and the derivative of y is dependent only of x. In addition, the conditions (2) with Riemann-Stieltjes integrals generalize, as we saw before, the conditions (3). These aspects represent the novelties for our problem (1), (2).

In what follows we will present other fractional boundary value problems studied in the last years, and related to problem (1), (2), namely, some Hadamard fractional differential equations subject to various nonlocal boundary conditions. In [14], the authors studied the existence of nonnegative multiple solutions for the Hadamard fractional differential equation with integral boundary conditions

\{\begin{matrix} ^{H} D_{1 +}^{α} u (t) + a (t) f (u (t)) = 0, t \in (0, \infty), \\ u (1) = 0,^{H} D_{1 +}^{α - 1} u (\infty) = \sum_{i = 1}^{m} λ_{i}^{H} I_{1 +}^{β_{i}} u (η), \end{matrix}

where

α \in (1, 2]

η \in (1, \infty)

β_{i} > 0

and

λ_{i} \geq 0

for all

i = 1, \dots, m

. In the proofs of the main results of [14], they applied Leggett-Williams and Guo-Krasnosel’skii fixed point theorems. In [16], by using the Banach fixed point theorem, the authors investigated the existence of the unique solution for the Hadamard fractional integro-differential equation subject to Hadamard fractional integral boundary conditions

\{\begin{matrix} ^{H} D_{1 +}^{γ} u (t) + f (t, u (t),^{H} I_{1 +}^{q} u (t)) = 0, t \in (1, \infty), \\ u (1) = u^{'} (1) = 0,^{H} D_{1 +}^{γ - 1} u (\infty) = \sum_{i = 1}^{m} λ_{i}^{H} I_{1 +}^{β_{i}} u (η), \end{matrix}

where

γ \in (2, 3)

η \in (1, \infty)

q > 0

β_{i} > 0

and

λ_{i} \geq 0

for all

i = 1, \dots, m

. In [15], the authors studied the existence of positive solutions for the nonlinear Hadamard fractional differential equation supplemented with Hadamard integral and multi-point boundary conditions

\{\begin{matrix} ^{H} D_{1 +}^{q} x (t) + σ (t) f (t, x (t)) = 0, t \in (1, \infty), \\ x (1) = x^{'} (1) = 0,^{H} D_{1 +}^{q - 1} x (\infty) = a^{H} I_{1 +}^{β} x (ξ) + b \sum_{i = 1}^{m - 2} α_{i} x (η_{i}), \end{matrix}

where

q \in (2, 3]

1 < ξ < η_{1} < \dots < η_{m - 2} < \infty

a, b \in R

, and

α_{i} \geq 0

for all

i = 1, \dots, m - 2

. In the main theorems they used the monotone iterative method to find two "twin" positive solutions of the problem, and then they presented monotone iterative schemes convergent to a unique positive solution. In [17], the authors investigated the nonlinear Hadamard fractional differential equation with nonlocal boundary conditions

\{\begin{matrix} ^{H} D_{1 +}^{α} x (t) + a (t) f (t, x (t)) = 0, t \in (1, \infty), \\ x (1) = x^{'} (1) = 0,^{H} D_{1 +}^{α - 1} x (\infty) = \sum_{i = 1}^{m} α_{i}^{H} I_{1 +}^{β_{i}} x (η) + b \sum_{j = 1}^{n} σ_{j} x (ξ_{j}), \end{matrix}

(4)

where

α \in (2, 3)

β_{i} > 0

and

α_{i} \geq 0

for all

i = 1, \dots, m

σ_{j} \geq 0

for all

j = 1, \dots, n

1 < η < ξ_{1} < \dots < ξ_{n} < \infty

. They studied the existence, uniqueness and multiplicity of positive solutions for problem (4), by using the Schauder fixed point theorem, the Banach contraction mapping principle, the monotone iterative method, and a fixed point theorem due to Avery and Peterson. In [18], the authors proved a generalization of a fixed point theorem due to Avery and Henderson, and then they applied it to problem (4) and proved the existence of at least three positive solutions of (4). We also mention the monographs [1,2,3,4,6,7,8,10,11,12,19] for various fractional differential equations and systems of fractional differential equations subject to diverse boundary conditions, and their applications in varied fields.

The paper is organized as follows. In Section 2 we present some preliminary results that will be used in the next section, including the solution for the corresponding linear problem, the Green functions and their properties. In Section 3 we give the main existence theorems for (1), (2), in Section 4 we present an example that illustrates the obtained results, and Section 5 contains the conclusions for our paper.

2. Auxiliary results

In this section we present some preliminary results that we will use in the next section.

Definition 1.

([7]) For a function

z : [a, \infty) \to R

, (

a \geq 0

), the left-sided Hadamard integral of fractional order

p > 0

with lower limit a is defined by

(^{H} I_{a +}^{p} z) (x) = \frac{1}{Γ (p)} \int_{a}^{x} {(ln \frac{x}{t})}^{p - 1} \frac{z (t)}{t} d t, x > 0 .

Definition 2.

([7]) For a function

z : [a, \infty) \to R

, (

a \geq 0

), the left-sided Hadamard fractional derivative of order

p \geq 0

with lower limit a is defined by

^{H} D_{a +}^{p} z (x) = {(x \frac{d}{d x})}^{n}^{H} I_{a +}^{n - p} z (x) = \frac{1}{Γ (n - p)} {(x \frac{d}{d x})}^{n} \int_{a}^{x} {(ln \frac{x}{t})}^{n - p - 1} \frac{z (t)}{t} d t,

where

n = [p] + 1

For

p = m \in N

, then

^{H} D_{a +}^{m} z (x) = (δ^{m} z) (x), x > a

, where

δ = x \frac{d}{d x}

is the

δ

-derivative.

Lemma 1.

([7]) If

α, β > 0

, and

a > 0

, then

\begin{matrix} (^{H} I_{a +}^{α} {(ln \frac{t}{a})}^{β - 1}) (x) = \frac{Γ (β)}{Γ (β + α)} {(ln \frac{x}{a})}^{β + α - 1}, \\ (^{H} D_{a +}^{α} {(ln \frac{t}{a})}^{β - 1}) (x) = \frac{Γ (β)}{Γ (β - α)} {(ln \frac{x}{a})}^{β - α - 1}, \end{matrix}

and in particular

^{H} D_{a +}^{p} {(ln \frac{t}{a})}^{p - j} (x) = 0

, for

j = 1, \dots, [p] + 1

Lemma 2.

([7]) Let

p > 0

and

z, u \in C [1, \infty) \cap L^{1} (0, \infty)

. Then the Hadamard fractional differential equation

^{H} D_{1 +}^{p} z (t) = 0

has the solutions

z (t) = \sum_{i = 1}^{n} c_{i} {(ln t)}^{p - i},

and the following formula holds

^{H} I_{1 +}^{p}^{H} D_{1 +}^{p} u (t) = u (t) + \sum_{i = 1}^{n} d_{i} {(ln t)}^{p - i},

where

c_{i}, d_{i} \in R

i = 1, \dots, n

and

n = [p] + 1

We consider now the system of fractional differential equations

\{\begin{matrix} ^{H} D_{1 +}^{α} x (t) + h (t) = 0, t \in (1, \infty), \\ ^{H} D_{1 +}^{β} y (t) + k (t) = 0, t \in (1, \infty), \end{matrix}

(5)

where

h, k \in C ([1, \infty), R_{+})

, subject to the boundary conditions (2). We denote by

\begin{matrix} a = Γ (α) - \int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s), b = \int_{1}^{\infty} {(ln s)}^{β - 1} d H_{2} (s), \\ c = \int_{1}^{\infty} {(ln s)}^{α - 1} d K_{1} (s), d = Γ (β) - \int_{1}^{\infty} {(ln s)}^{β - 1} d K_{2} (s), \\ Δ = a d - b c . \end{matrix}

(6)

Lemma 3.

Assume that

a, b, c, d \in R

and the functions

h, k \in C ([1, \infty), R_{+})

satisfy the conditions

\int_{1}^{\infty} h (s) \frac{d s}{s} < \infty

and

\int_{1}^{\infty} k (s) \frac{d s}{s} < \infty

. If

Δ \neq 0

, then the solution of problem (5),(2) is given by

\begin{matrix} x (t) = - \frac{1}{Γ (α)} \int_{1}^{t} {(ln \frac{t}{s})}^{α - 1} h (s) \frac{d s}{s} \\ + \frac{{(ln t)}^{α - 1}}{Δ} [d \int_{1}^{\infty} h (s) \frac{d s}{s} - \frac{d}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \end{matrix}

\begin{matrix} - \frac{d}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s) \\ + b \int_{1}^{\infty} k (s) \frac{d s}{s} - \frac{b}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{b}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s)], t \in [1, \infty), \\ y (t) = - \frac{1}{Γ (β)} \int_{1}^{t} {(ln \frac{t}{s})}^{β - 1} k (s) \frac{d s}{s} \\ + \frac{{(ln t)}^{β - 1}}{Δ} [a \int_{1}^{\infty} k (s) \frac{d s}{s} - \frac{a}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{a}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s) \\ + c \int_{1}^{\infty} h (s) \frac{d s}{s} - \frac{c}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \\ - \frac{c}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s)], t \in [1, \infty) . \end{matrix}

(7)

Proof.

By Lemma 2 the solutions of system (5) are given by

\begin{matrix} x (t) = - \frac{1}{Γ (α)} \int_{1}^{t} {(ln \frac{t}{s})}^{α - 1} h (s) \frac{d s}{s} \\ + a_{1} {(ln t)}^{α - 1} + a_{2} {(ln t)}^{α - 2} + \dots + a_{n} {(ln t)}^{α - n}, t \in [1, \infty), \\ y (t) = - \frac{1}{Γ (β)} \int_{1}^{t} {(ln \frac{t}{s})}^{β - 1} k (s) \frac{d s}{s} \\ + b_{1} {(ln t)}^{β - 1} + b_{2} {(ln t)}^{β - 2} + \dots + b_{m} {(ln t)}^{β - m}, t \in [1, \infty), \end{matrix}

(8)

with

a_{i}, b_{j} \in R

for

i = 1, \dots, n

j = 1, \dots, m

. Because

x (1) = x^{'} (1) = \dots = x^{(n - 2)} (1) = 0

and

y (1) = y^{'} (1) = \dots = y^{(m - 2)} (1) = 0

we deduce that

a_{2} = \dots = a_{n} = 0

and

b_{2} = \dots = b_{m} = 0

. So, by (8) we find

\begin{matrix} x (t) = - \frac{1}{Γ (α)} \int_{1}^{t} {(ln \frac{t}{s})}^{α - 1} h (s) \frac{d s}{s} + a_{1} {(ln t)}^{α - 1}, t \in [1, \infty), \\ y (t) = - \frac{1}{Γ (β)} \int_{1}^{t} {(ln \frac{t}{s})}^{β - 1} k (s) \frac{d s}{s} + b_{1} {(ln t)}^{β - 1}, t \in [1, \infty) . \end{matrix}

(9)

We have

^{H} D_{1 +}^{α - 1} x (t) = - \int_{1}^{t} h (s) \frac{d s}{s} + a_{1} Γ (α)

and

^{H} D_{1 +}^{β - 1} y (t) = - \int_{1}^{t} k (s) \frac{d s}{s} + b_{1} Γ (β)

. Then the last conditions from (2), namely

^{H} D_{1 +}^{α - 1} x (\infty) = \int_{1}^{\infty} x (s) d H_{1} (s) + \int_{1}^{\infty} y (s) d H_{2} (s)

and

^{H} D_{1 +}^{β - 1} y (\infty) = \int_{1}^{\infty} x (s) d K_{1} (s) + \int_{1}^{\infty} y (s) d K_{2} (s)

, for the solution (9), give us

\{\begin{matrix} - \int_{1}^{\infty} h (s) \frac{d s}{s} + a_{1} Γ (α) = \int_{1}^{\infty} [- \frac{1}{Γ (α)} \int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ} + a_{1} {(ln s)}^{α - 1}] d H_{1} (s) \\ + \int_{1}^{\infty} [- \frac{1}{Γ (β)} \int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ} + b_{1} {(ln s)}^{β - 1}] d H_{2} (s), \\ - \int_{1}^{\infty} k (s) \frac{d s}{s} + b_{1} Γ (β) = \int_{1}^{\infty} [- \frac{1}{Γ (α)} \int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ} + a_{1} {(ln s)}^{α - 1}] d K_{1} (s) \\ + \int_{1}^{\infty} [- \frac{1}{Γ (β)} \int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ} + b_{1} {(ln s)}^{β - 1}] d K_{2} (s), \end{matrix}

\{\begin{matrix} a_{1} [Γ (α) - \int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s)] - b_{1} \int_{1}^{\infty} {(ln s)}^{β - 1} d H_{2} (s) \\ = \int_{1}^{\infty} h (s) \frac{d s}{s} - \frac{1}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \\ - \frac{1}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s), \\ - a_{1} \int_{1}^{\infty} {(ln s)}^{α - 1} d K_{1} (s) + b_{1} [Γ (β) - \int_{1}^{\infty} {(ln s)}^{β - 1} d H_{2} (s)] \\ = \int_{1}^{\infty} k (s) \frac{d s}{s} - \frac{1}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{1}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s) . \end{matrix}

The determinant of the above system in the unknowns

a_{1}

and

b_{1}

Δ

, which by the assumption of this lemma is different from zero. So the above system has a unique solution, namely

\begin{matrix} a_{1} = \frac{1}{Δ} [d \int_{1}^{\infty} h (s) \frac{d s}{s} - \frac{d}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \\ - \frac{d}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s) \\ + b \int_{1}^{\infty} k (s) \frac{d s}{s} - \frac{b}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{b}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s)], \\ b_{1} = \frac{1}{Δ} [a \int_{1}^{\infty} k (s) \frac{d s}{s} - \frac{a}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{a}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s) \\ + c \int_{1}^{\infty} h (s) \frac{d s}{s} - \frac{c}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \\ - \frac{c}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s)] . \end{matrix}

By replacing the above formulas for

a_{1}

and

b_{1}

in (9), we obtain the solution of problem (5), (2) given by (7). The converse follows by direct computation. □

Lemma 4.

Under the assumptions of Lemma 3, the solution of problem (5),(2) is

\{\begin{matrix} x (t) = \int_{1}^{\infty} G_{1} (t, s) h (s) \frac{d s}{s} + \int_{1}^{\infty} G_{2} (t, s) k (s) \frac{d s}{s}, t \in [1, \infty), \\ y (t) = \int_{1}^{\infty} G_{3} (t, s) h (s) \frac{d s}{s} + \int_{1}^{\infty} G_{4} (t, s) k (s) \frac{d s}{s}, t \in [1, \infty), \end{matrix}

(10)

where the Green functions

G_{i}, i = 1, \dots, 4

are given by

\begin{matrix} G_{1} (t, s) = g_{α} (t, s) + \frac{{(ln t)}^{α - 1}}{Δ} (d \int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ) + b \int_{1}^{\infty} g_{α} (τ, s) d K_{1} (τ)), \\ G_{2} (t, s) = \frac{{(ln t)}^{α - 1}}{Δ} (d \int_{1}^{\infty} g_{β} (τ, s) d H_{2} (τ) + b \int_{1}^{\infty} g_{β} (τ, s) d K_{2} (τ)), \\ G_{3} (t, s) = \frac{{(ln t)}^{β - 1}}{Δ} (c \int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ) + a \int_{1}^{\infty} g_{α} (τ, s) d K_{1} (τ)), \\ G_{4} (t, s) = g_{β} (t, s) + \frac{{(ln t)}^{β - 1}}{Δ} (c \int_{1}^{\infty} g_{β} (τ, s) d H_{2} (τ) + a \int_{1}^{\infty} g_{β} (τ, s) d K_{2} (τ)), \end{matrix}

(11)

with

\begin{matrix} g_{α} (t, s) = \frac{1}{Γ (α)} \{\begin{matrix} {(ln t)}^{α - 1} - {(ln \frac{t}{s})}^{α - 1}, 1 \leq s \leq t, \\ {(ln t)}^{α - 1}, 1 \leq t \leq s, \end{matrix} \\ g_{β} (t, s) = \frac{1}{Γ (β)} \{\begin{matrix} {(ln t)}^{β - 1} - {(ln \frac{t}{s})}^{β - 1}, 1 \leq s \leq t, \\ {(ln t)}^{β - 1}, 1 \leq t \leq s . \end{matrix} \end{matrix}

(12)

Proof.

By (7), we obtain

\begin{matrix} x (t) = \frac{1}{Γ (α)} \int_{1}^{\infty} {(ln t)}^{α - 1} h (s) \frac{d s}{s} - \frac{Δ}{Γ (α) Δ} \int_{1}^{\infty} {(ln t)}^{α - 1} h (s) \frac{d s}{s} \\ - \frac{1}{Γ (α)} \int_{1}^{t} {(ln \frac{t}{s})}^{α - 1} h (s) \frac{d s}{s} \\ + \frac{{(ln t)}^{α - 1}}{Δ} [d \int_{1}^{\infty} h (s) \frac{d s}{s} - \frac{d}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \\ - \frac{d}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s) \\ + b \int_{1}^{\infty} k (s) \frac{d s}{s} - \frac{b}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{b}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s)] \\ = \underset{A_{0}}{\underset{︸}{\frac{1}{Γ (α)} \int_{1}^{t} [{(ln t)}^{α - 1} - {(ln \frac{t}{s})}^{α - 1}] h (s) \frac{d s}{s} + \frac{1}{Γ (α)} \int_{t}^{\infty} {(ln t)}^{α - 1} h (s) \frac{d s}{s}}} \\ + \frac{{(ln t)}^{α - 1}}{Δ} [- \frac{a d - b c}{Γ (α)} \int_{1}^{\infty} h (s) \frac{d s}{s} + d \int_{1}^{\infty} h (s) \frac{d s}{s} \\ - \frac{d}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \\ - \frac{d}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s) + b \int_{1}^{\infty} k (s) \frac{d s}{s} \\ - \frac{b}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{b}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s)] \\ = A_{0} + \frac{{(ln t)}^{α - 1}}{Δ} \{- \frac{1}{Γ (α)} [Γ (α) Γ (β) - Γ (β) \int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s) \\ - Γ (α) \int_{1}^{\infty} {(ln s)}^{β - 1} d K_{2} (s) + (\int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s)) (\int_{1}^{\infty} {(ln s)}^{β - 1} d K_{2} (s)) \\ - (\int_{1}^{\infty} {(ln s)}^{α - 1} d K_{1} (s)) (\int_{1}^{\infty} {(ln s)}^{β - 1} H_{2} (s))] \int_{1}^{\infty} h (s) \frac{d s}{s} \\ + [Γ (β) - \int_{1}^{\infty} {(ln s)}^{β - 1} d K_{2} (s)] \int_{1}^{\infty} h (s) \frac{d s}{s} \\ - \frac{1}{Γ (α)} [Γ (β) - \int_{1}^{\infty} {(ln s)}^{β - 1} d K_{2} (s)] \int_{1}^{\infty} (\int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ)) h (s) \frac{d s}{s} \\ - \frac{1}{Γ (β)} [Γ (β) - \int_{1}^{\infty} {(ln s)}^{β - 1} d K_{2} (s)] \int_{1}^{\infty} (\int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ)) k (s) \frac{d s}{s} \\ + (\int_{1}^{\infty} {(ln s)}^{β - 1} d H_{2} (s)) \int_{1}^{\infty} k (s) \frac{d s}{s} \\ - \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln s)}^{β - 1} d H_{2} (s)) (\int_{1}^{\infty} (\int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ)) h (s) \frac{d s}{s}) \\ - \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln s)}^{β - 1} d H_{2} (s)) (\int_{1}^{\infty} (\int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ)) k (s) \frac{d s}{s})\} \end{matrix}

\begin{matrix} = A_{0} + \frac{{(ln t)}^{α - 1}}{Δ} \{\int_{1}^{\infty} [\frac{Γ (β)}{Γ (α)} \int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) \\ - \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ)) (\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) \\ + \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) \\ - \frac{Γ (β)}{Γ (α)} \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ) \\ + \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ) \\ - \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) (\int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ))] h (s) \frac{d s}{s} \\ + \int_{1}^{\infty} [- \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ) \\ + \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ) \\ + (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) \\ - \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ)] k (s) \frac{d s}{s}\} \\ = A_{0} + \frac{{(ln t)}^{α - 1}}{Δ} \{\int_{1}^{\infty} \{\frac{Γ (β)}{Γ (α)} [\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ)] \\ - \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) [\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ)] \end{matrix}

\begin{matrix} + \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) \\ \times [\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ)]\} h (s) \frac{d s}{s} \\ + \int_{1}^{\infty} \{[\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ)] \\ + \frac{1}{Γ (β)} [(\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) (\int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ)) \\ - (\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) \\ + (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) (\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) \\ - (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) (\int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ))]\} k (s) \frac{d s}{s}\} \\ = A_{0} + \frac{{(ln t)}^{α - 1}}{Δ} \{\int_{1}^{\infty} \{\frac{d}{Γ (α)} [\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ)] \\ + \frac{b}{Γ (α)} [\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ)]\} h (s) \frac{d s}{s} \\ + \int_{1}^{\infty} \{\frac{d}{Γ (β)} [\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ)] \\ + \frac{b}{Γ (β)} [\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ)]\} k (s) \frac{d s}{s}\} \\ = \int_{1}^{\infty} g_{α} (t, s) h (s) \frac{d s}{s} + \frac{{(ln t)}^{α - 1}}{Δ} [d \int_{1}^{\infty} (\int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ)) h (s) \frac{d s}{s} \\ + b \int_{1}^{\infty} (\int_{1}^{\infty} g_{α} (τ, s) d K_{1} (τ)) h (s) \frac{d s}{s} \\ + d \int_{1}^{\infty} (\int_{1}^{\infty} g_{β} (τ, s) d H_{2} (τ)) k (s) \frac{d s}{s} \\ + b \int_{1}^{\infty} (\int_{1}^{\infty} g_{β} (τ, s) d K_{2} (τ)) k (s) \frac{d s}{s}] \\ = \int_{1}^{\infty} G_{1} (t, s) h (s) \frac{d s}{s} + \int_{1}^{\infty} G_{2} (t, s) k (s) \frac{d s}{s}, \end{matrix}

where

g_{α}

and

g_{β}

are given by (12), and

G_{1}

and

G_{2}

are given by (11). In the last relations above we used the equality

\begin{matrix} \int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ) \\ = \int_{1}^{s} {(ln τ)}^{α - 1} d H_{1} (τ) + \int_{s}^{\infty} [{(ln τ)}^{α - 1} - {(ln \frac{τ}{s})}^{α - 1}] d H_{1} (τ) \\ = Γ (α) \int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ), \end{matrix}

and similar equalities for

H_{2} (τ), K_{1} (τ)

and

K_{2} (τ)

In a similar manner, by (7) we find

\begin{matrix} y (t) = \frac{1}{Γ (β)} \int_{1}^{\infty} {(ln t)}^{β - 1} k (s) \frac{d s}{s} - \frac{Δ}{Γ (β) Δ} \int_{1}^{\infty} {(ln t)}^{β - 1} k (s) \frac{d s}{s} \\ - \frac{1}{Γ (β)} \int_{1}^{t} {(ln \frac{t}{s})}^{β - 1} k (s) \frac{d s}{s} \\ + \frac{{(ln t)}^{β - 1}}{Δ} [a \int_{1}^{\infty} k (s) \frac{d s}{s} - \frac{a}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{a}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s) \\ + c \int_{1}^{\infty} h (s) \frac{d s}{s} - \frac{c}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \\ - \frac{c}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s)] \\ = \underset{B_{0}}{\underset{︸}{\frac{1}{Γ (β)} \int_{1}^{t} [{(ln t)}^{β - 1} - {(ln \frac{t}{s})}^{β - 1}] k (s) \frac{d s}{s} + \frac{1}{Γ (β)} \int_{t}^{\infty} {(ln t)}^{β - 1} k (s) \frac{d s}{s}}} \\ + \frac{{(ln t)}^{β - 1}}{Δ} [- \frac{a d - b c}{Γ (β)} \int_{1}^{\infty} k (s) \frac{d s}{s} + a \int_{1}^{\infty} k (s) \frac{d s}{s} \\ - \frac{a}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{a}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s) \\ + c \int_{1}^{\infty} h (s) \frac{d s}{s} - \frac{c}{Γ (α)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s) \\ - \frac{c}{Γ (β)} \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s)] \\ = B_{0} + \frac{{(ln t)}^{β - 1}}{Δ} \{- \frac{1}{Γ (β)} [Γ (α) Γ (β) - Γ (β) \int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s) \\ - Γ (α) \int_{1}^{\infty} {(ln s)}^{β - 1} d K_{2} (s) + (\int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s)) (\int_{1}^{\infty} {(ln s)}^{β - 1} d K_{2} (s)) \\ - (\int_{1}^{\infty} {(ln s)}^{α - 1} d K_{1} (s)) (\int_{1}^{\infty} {(ln s)}^{β - 1} d H_{2} (s))] \int_{1}^{\infty} k (s) \frac{d s}{s} \\ + [Γ (α) - \int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s)] \int_{1}^{\infty} k (s) \frac{d s}{s} \\ - \frac{1}{Γ (α)} [Γ (α) - \int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s)] \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d K_{1} (s) \\ - \frac{1}{Γ (β)} [Γ (α) - \int_{1}^{\infty} {(ln s)}^{α - 1} d H_{1} (s)] \int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d K_{2} (s) \\ + (\int_{1}^{\infty} {(ln s)}^{α - 1} d K_{1} (s)) \int_{1}^{\infty} h (s) \frac{d s}{s} \\ - \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln s)}^{α - 1} d K_{1} (s)) (\int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{α - 1} h (τ) \frac{d τ}{τ}) d H_{1} (s)) \\ - \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln s)}^{α - 1} d K_{1} (s)) (\int_{1}^{\infty} (\int_{1}^{s} {(ln \frac{s}{τ})}^{β - 1} k (τ) \frac{d τ}{τ}) d H_{2} (s))\} \end{matrix}

\begin{matrix} = B_{0} + \frac{{(ln t)}^{β - 1}}{Δ} \{\int_{1}^{\infty} [- \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ) \\ + \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ)) \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ) \\ + \int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ) \\ - \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ)] h (s) \frac{d s}{s} \\ + \int_{1}^{\infty} [\frac{Γ (α)}{Γ (β)} \int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ) \\ - \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ)) (\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) \\ + \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) (\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ)) \\ - \frac{Γ (α)}{Γ (β)} \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ) \\ + \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ)) \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ) \\ - \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ)] k (s) \frac{d s}{s}\} \\ = B_{0} + \frac{{(ln t)}^{β - 1}}{Δ} \{\int_{1}^{\infty} \{[\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ)] \\ + \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ)) \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ) \\ - \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ)) (\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) \\ + \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ)) (\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) \\ - \frac{1}{Γ (α)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) (\int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ))\} h (s) \frac{d s}{s} \\ + \int_{1}^{\infty} \{\frac{Γ (α)}{Γ (β)} [\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ)] \\ - \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ)) [\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ)] \\ + \frac{1}{Γ (β)} (\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) [\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ) \\ - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ)]\} k (s) \frac{d s}{s}\} \\ = B_{0} + \frac{{(ln t)}^{β - 1}}{Δ} \{\int_{1}^{\infty} \{\frac{a}{Γ (α)} [\int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d K_{1} (τ)] \\ + \frac{c}{Γ (α)} [\int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{α - 1} d H_{1} (τ)]\} h (s) \frac{d s}{s} \\ + \int_{1}^{\infty} \{\frac{a}{Γ (β)} [\int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d K_{2} (τ)] \\ + \frac{c}{Γ (β)} [\int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ) - \int_{s}^{\infty} {(ln \frac{τ}{s})}^{β - 1} d H_{2} (τ)]\} k (s) \frac{d s}{s}\} \\ = \int_{1}^{\infty} g_{β} (t, s) k (s) \frac{d s}{s} + \frac{{(ln t)}^{β - 1}}{Δ} [a \int_{1}^{\infty} (\int_{1}^{\infty} g_{α} (τ, s) d K_{1} (τ)) h (s) \frac{d s}{s} \\ + c \int_{1}^{\infty} (\int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ)) h (s) \frac{d s}{s} \\ + a \int_{1}^{\infty} (\int_{1}^{\infty} g_{β} (τ, s) d K_{2} (τ)) k (s) \frac{d s}{s} \\ + c \int_{1}^{\infty} (\int_{1}^{\infty} g_{β} (τ, s) d H_{2} (τ)) k (s) \frac{d s}{s}] \\ = \int_{1}^{\infty} G_{3} (t, s) h (s) \frac{d s}{s} + \int_{1}^{\infty} G_{4} (t, s) k (s) \frac{d s}{s}, \end{matrix}

where

G_{3}

and

G_{4}

are given by (11). □

Based on the definitions of the functions

g_{α}, g_{β}, G_{i}

i = 1, \dots, 4

, we obtain easily the next lemma.

Lemma 5.

We assume that the functions

H_{1}, H_{2}, K_{1}, K_{2}

are nondecreasing functions,

a, d \in R_{+}

b, c \in R

, and

Δ > 0

, and let

θ > 1

. Then the functions

g_{α}, g_{β},

and

G_{i}

i = 1, \dots, 4

have the following properties:

a) The functions

g_{α}

and

g_{β}

are continuous on

[1, \infty) \times [1, \infty)

;

0 \leq g_{α} (t, s) \leq \frac{1}{Γ (α)} {(ln t)}^{α - 1}, 0 \leq g_{β} (t, s) \leq \frac{1}{Γ (β)} {(ln t)}^{β - 1}, \forall t, s \in [1, \infty);

0 \leq \frac{g_{α} (t, s)}{1 + {(ln t)}^{α - 1}} \leq \frac{1}{Γ (α)}, 0 \leq \frac{g_{β} (t, s)}{1 + {(ln t)}^{β - 1}} \leq \frac{1}{Γ (β)}, \forall t, s \in [1, \infty);

d) The functions

G_{i}, i = 1, \dots, 4

are continuous on

[1, \infty) \times [1, \infty)

;

G_{i} (t, s) \geq 0

for all

(t, s) \in [1, \infty) \times [1, \infty)

and

i = 1, \dots, 4

;

\frac{G_{1} (t, s)}{1 + {(ln t)}^{α - 1}} \leq \frac{1}{Γ (α)} + \frac{1}{Δ Γ (α)} (d \int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) + b \int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ))

= \frac{d}{Δ} = : Λ_{1} > 0, \forall t, s \in [1, \infty)

;

\frac{G_{2} (t, s)}{1 + {(ln t)}^{α - 1}} \leq \frac{1}{Δ Γ (β)} (d \int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ) + b \int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ))

= \frac{b}{Δ} = : Λ_{2} \geq 0, \forall t, s \in [1, \infty)

;

\frac{G_{3} (t, s)}{1 + {(ln t)}^{β - 1}} \leq \frac{1}{Δ Γ (α)} (c \int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) + a \int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ))

= \frac{c}{Δ} = : Λ_{3} \geq 0, \forall t, s \in [1, \infty)

;

\frac{G_{4} (t, s)}{1 + {(ln t)}^{β - 1}} \leq \frac{1}{Γ (β)} + \frac{1}{Δ Γ (β)} (c \int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ) + a \int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ))

= \frac{a}{Δ} = : Λ_{4} > 0, \forall t, s \in [1, \infty)

;

min_{t \in [θ, \infty)} \frac{G_{1} (t, s)}{1 + {(ln t)}^{α - 1}} \geq \frac{{(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})}

\times (d \int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ) + b \int_{1}^{\infty} g_{α} (τ, s) d K_{1} (τ)),

\forall s \in [1, \infty)

;

min_{t \in [θ, \infty)} \frac{G_{2} (t, s)}{1 + {(ln t)}^{α - 1}} = \frac{{(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})}

\times (d \int_{1}^{\infty} g_{β} (τ, s) d H_{2} (τ) + b \int_{1}^{\infty} g_{β} (τ, s) d K_{2} (τ)),

\forall s \in [1, \infty)

;

min_{t \in [θ, \infty)} \frac{G_{3} (t, s)}{1 + {(ln t)}^{β - 1}} = \frac{{(ln θ)}^{β - 1}}{Δ (1 + {(ln θ)}^{β - 1})}

\times (c \int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ) + a \int_{1}^{\infty} g_{α} (τ, s) d K_{1} (τ)),

\forall s \in [1, \infty)

;

min_{t \in [θ, \infty)} \frac{G_{4} (t, s)}{1 + {(ln t)}^{β - 1}} \geq \frac{{(ln θ)}^{β - 1}}{Δ (1 + {(ln θ)}^{β - 1})}

\times (c \int_{1}^{\infty} g_{β} (τ, s) d H_{2} (τ) + a \int_{1}^{\infty} g_{β} (τ, s) d K_{2} (τ)),

\forall s \in [1, \infty)

Remark 1.

Under assumptions of Lemma 5 we find that

a, d > 0

and

b, c \geq 0

, so

Λ_{1}, Λ_{4} > 0

and

Λ_{2}, Λ_{3} \geq 0

We present now the fixed point theorems that we will use in the next section.

Let E be a real Banach space with the norm

∥ \cdot ∥

Definition 3.

A nonempty convex closed set

K \subset E

is a cone if it satisfies the conditions:

a) if

u \in K

and

α \geq 0

, then

α u \in K

;

b) if

u \in K

and

- u \in K

, then

u = 0

,where 0 is the zero element of E.

A cone

K \subset E

defines a partial ordering in E given by

x \leq y

if and only if

y - x \in K

Definition 4.

An operator

A : D (A) \subset E \to E

is compact if it maps bounded sets into relatively compact sets. An operator

A : D (A) \subset E \to E

is completely continuous if it is continuous and compact.

Theorem 1.

(Guo-Krasnosels’kii fixed point theorem - the fixed point theorem of cone expansion and compression of norm type, [5]). Let E be a real Banach space with the norm

∥ \cdot ∥

, and let

K \subset X

be a cone in E. Assume

Ω_{1}

and

Ω_{2}

are bounded open subsets of E with

0 \in Ω_{1}

{\bar{Ω}}_{1} \subset Ω_{2}

and let

A : K \cap ({\bar{Ω}}_{2} ∖ Ω_{1}) \to K

be a completely continuous operator such that, either

∥ A u ∥ \leq ∥ u ∥, \forall u \in K \cap \partial Ω_{1}

, and

∥ A u ∥ \geq ∥ u ∥, \forall u \in K \cap \partial Ω_{2}

, or

ii)

∥ A u ∥ \geq ∥ u ∥, \forall u \in K \cap \partial Ω_{1},

and

∥ A u ∥ \leq ∥ u ∥, \forall u \in K \cap \partial Ω_{2}

.Then

A

has at least one fixed point in

K \cap ({\bar{Ω}}_{2} ∖ Ω_{1})

We consider

Θ

a nonnegative continuous concave functional on the cone K, and

0 < c < d

. We define the convex sets

K_{c}

{\bar{K}}_{c}

K (Θ, c, d)

K_{c} = {u | u \in K, ∥ u ∥ < c}, {\bar{K}}_{c} = {u | u \in K, ∥ u ∥ \leq c},

and

K (Θ, c, d) = {u | u \in K, Θ (u) \geq c, ∥ u ∥ \leq d}

Theorem 2.

(Leggett-Williams fixed point theorem, [9]) Let K be a cone in a real Banach space E and

Θ (u)

a nonnegative continuous concave functional on K satisfying

Θ (u) \leq ∥ u ∥

for all

u \in {\bar{K}}_{l}

, (

l > 0

). Suppose that

A : {\bar{K}}_{l} \to {\bar{K}}_{l}

is a completely continuous operator and there exist positive numbers

0 < m < c < d < l

such that

{u | u \in K (Θ, c, d), Θ (u) > c} \neq \emptyset

and

Θ (A u) > c

for

u \in K (Θ, c, d)

;

ii)

∥ A u ∥ < m

for

u \in {\bar{K}}_{m}

;

iii)

Θ (A u) > c

for

u \in K (Θ, c, l)

and

∥ A u ∥ > d

Then A has at least three fixed points

u_{1}

u_{2}

and

u_{3}

satisfying

∥ u_{1} ∥ < m, Θ (u_{2}) > c, ∥ u_{3} ∥ > m and Θ (u_{3}) < c .

3. Main results

We introduce the space

X_{1} = \{x \in C (I, R), sup_{t \in I} \frac{| x (t) |}{1 + {(ln t)}^{α - 1}} < \infty\},

where

I = [1, \infty)

, with the norm

{∥ x ∥}_{1} = {sup}_{t \in I} \frac{| x (t) |}{1 + {(ln t)}^{α - 1}}

, the space

X_{2} = \{y \in C (I, R), sup_{t \in I} \frac{| y (t) |}{1 + {(ln t)}^{β - 1}} < \infty\},

with the norm

{∥ y ∥}_{2} = {sup}_{t \in I} \frac{| y (t) |}{1 + {(ln t)}^{β - 1}}

, and the space

X = X_{1} \times X_{2}

with the norm

∥ (x, y) ∥ = {∥ x ∥}_{1} + {∥ y ∥}_{2}

. The spaces

(X_{1}, ∥ \cdot ∥_{1})

(X_{2}, ∥ \cdot ∥_{2})

and

(X, ∥ (\cdot, \cdot) ∥)

are Banach spaces (see [14], Lemma 2.7).

Lemma 6.

([14], Lemma 2.8) Let

Ω \subset X_{1}

be a bounded set, which satisfy the following conditions:

(i) The functions

\frac{x (t)}{1 + {(ln t)}^{α - 1}}, x \in Ω

are equicontinuous on any compact interval of I;

(ii) For any

ϵ > 0

, there exists a constant

T = T (ϵ) > 0

such that

|\frac{x (t_{1})}{1 + {(ln t_{1})}^{α - 1}} - \frac{x (t_{2})}{1 + {(ln t_{2})}^{α - 1}}| < ϵ, \forall t_{1}, t_{2} \geq T, \forall x \in Ω .

Then Ω is relatively compact in

X_{1}

We define now the positive cone

P \subset X

P = {(x, y) \in X, x (t) \geq 0, y (t) \geq 0, \forall t \in [1, \infty)},

and the operator

A : P \to X

A (x, y) = (A_{1} (x, y), A_{2} (x, y)), (x, y) \in P

, where the operators

A_{1} : P \to X_{1}

and

A_{2} : P \to X_{2}

are defined by

\begin{matrix} A_{1} (x, y) (t) = \int_{1}^{\infty} G_{1} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} + \int_{1}^{\infty} G_{2} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s}, t \in I, \\ A_{2} (x, y) (t) = \int_{1}^{\infty} G_{3} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} + \int_{1}^{\infty} G_{4} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s}, t \in I, \end{matrix}

for

(x, y) \in P

In what follows we present the basic assumptions that we will use in our main results.

$(H 1)$: $α \in (n - 1, n]$ , $β \in (m - 1, m]$ , $n, m \in N$ , $n, m \geq 2$ , $H_{1}, H_{2}, K_{1}, K_{2} : [1, \infty) \to R$ are nondecreasing functions, $a, d \in R_{+}$ , $b, c \in R$ , and $Δ > 0$ .
$(H 2)$: The functions $f, g \in C (R_{+} \times R_{+}, R_{+})$ , $f, g \neq 0$ on any subinterval of $(0, \infty) \times (0, \infty)$ , and $f, g$ are bounded on $R_{+} \times R_{+}$ .
$(H 3)$: The functions $a, b : [1, \infty) \to R_{+}$ are not identical zero on any subinterval of $[1, \infty)$ , and $0 < \int_{1}^{\infty} \frac{a (s)}{s} d s < \infty$ , $0 < \int_{1}^{\infty} \frac{b (s)}{s} d s < \infty$ .

Lemma 7.

(H 1) - (H 3)

hold, then the operator

A : P \to P

is completely continuous.

Proof.

Under the assumptions of this lemma, we have

A (x, y) \in P

for all

(x, y) \in P

, that is

A : P \to P

. We will prove this lemma in four steps.

(I) We show firstly that the operator

A

is uniformly bounded on

P

. Let

S

be a bounded set of

P

. Then there exists

r > 0

such that

∥ (x, y) ∥ \leq r

, and so

{∥ x ∥}_{1} \leq r

and

{∥ y ∥}_{2} \leq r

for all

(x, y) \in S

. By

(H 2)

, there exist

M_{1} > 0

and

M_{2} > 0

such that

f (x, y) \leq M_{1}

and

g (x, y) \leq M_{2}

for all

x, y \in R_{+}

. Then by

(H 3)

, for any

(x, y) \in S

, we obtain

\begin{matrix} ∥ A_{1} {(x, y) ∥}_{1} = sup_{t \in I} \frac{| A_{1} (x, y) (t) |}{1 + {(ln t)}^{α - 1}} \\ = sup_{t \in I} \frac{1}{1 + {(ln t)}^{α - 1}} (\int_{1}^{\infty} G_{1} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} G_{2} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s}) \\ \leq M_{1} Λ_{1} \int_{1}^{\infty} a (s) \frac{d s}{s} + M_{2} Λ_{2} \int_{1}^{\infty} b (s) \frac{d s}{s} = : M_{3} < \infty, \\ ∥ A_{2} {(x, y) ∥}_{2} = sup_{t \in I} \frac{| A_{2} (x, y) (t) |}{1 + {(ln t)}^{β - 1}} \\ = sup_{t \in I} \frac{1}{1 + {(ln t)}^{β - 1}} (\int_{1}^{\infty} G_{3} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} G_{4} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s}) \\ \leq M_{1} Λ_{3} \int_{1}^{\infty} a (s) \frac{d s}{s} + M_{2} Λ_{4} \int_{1}^{\infty} b (s) \frac{d s}{s} = : M_{4} < \infty . \end{matrix}

∥ A (x, y) ∥ = ∥ A_{1} {(x, y) ∥}_{1} + {∥ A_{2} (x, y) ∥}_{2} \leq M_{3} + M_{4} < \infty, \forall (x, y) \in S,

that is, operator

A

is uniformly bounded.

(II) Next we will prove that

A

is equicontinuous on any compact interval of I. We consider the interval

[1, T]

, where

T > 1

. Then for any

t_{1}, t_{2} \in [1, T]

, with

t_{1} < t_{2}

and

(x, y) \in S

, we have

\begin{matrix} Ξ_{1} = |\frac{A_{1} (x, y) (t_{2})}{1 + {(ln t_{2})}^{α - 1}} - \frac{A_{1} (x, y) (t_{1})}{1 + {(ln t_{1})}^{α - 1}}| \\ = |\int_{1}^{\infty} \frac{G_{1} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} a (s) f (x (s), y (s)) \frac{d s}{s} + \int_{1}^{\infty} \frac{G_{2} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} b (s) g (x (s), y (s)) \frac{d s}{s} \\ - \int_{1}^{\infty} \frac{G_{1} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}} a (s) f (x (s), y (s)) \frac{d s}{s} - \int_{1}^{\infty} \frac{G_{2} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}} b (s) g (x (s), y (s)) \frac{d s}{s}| \\ \leq |\int_{1}^{\infty} (\frac{G_{1} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{1} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}) a (s) f (x (s), y (s)) \frac{d s}{s}| \\ + |\int_{1}^{\infty} (\frac{G_{2} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{2} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}) b (s) g (x (s), y (s)) \frac{d s}{s}| \\ \leq \int_{1}^{\infty} |\frac{G_{1} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{1} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} |\frac{G_{2} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{2} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq \int_{1}^{\infty} [|\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| \\ + \frac{1}{Δ} (d \int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ) + b \int_{1}^{\infty} g_{α} (τ, s) d K_{1} (τ)) \\ \times (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}})] a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} \frac{1}{Δ} (d \int_{1}^{\infty} g_{β} (τ, s) d H_{2} (τ) + b \int_{1}^{\infty} g_{β} (τ, s) d K_{2} (τ)) \\ \times (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq M_{1} \underset{Ξ_{0}}{\underset{︸}{\int_{1}^{\infty} |\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) \frac{d s}{s}}} \\ + \frac{M_{1}}{Δ Γ (α)} (d \int_{1}^{\infty} {(ln τ)}^{α - 1} d H_{1} (τ) + b \int_{1}^{\infty} {(ln τ)}^{α - 1} d K_{1} (τ)) \\ \times (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) \int_{1}^{\infty} a (s) \frac{d s}{s} \\ + \frac{M_{2}}{Δ Γ (β)} (d \int_{1}^{\infty} {(ln τ)}^{β - 1} d H_{2} (τ) + b \int_{1}^{\infty} {(ln τ)}^{β - 1} d K_{2} (τ)) \\ \times (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) \int_{1}^{\infty} b (s) \frac{d s}{s} \\ = M_{1} Ξ_{0} + \frac{M_{1} (d Γ (α) - Δ)}{Δ Γ (α)} (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) \int_{1}^{\infty} a (s) \frac{d s}{s} \\ + \frac{M_{2} b}{Δ} (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) \int_{1}^{\infty} b (s) \frac{d s}{s} . \end{matrix}

For

Ξ_{0}

we deduce

\begin{matrix} Ξ_{0} = \int_{1}^{t_{1}} |\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) \frac{d s}{s} \\ + \int_{t_{1}}^{t_{2}} |\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) \frac{d s}{s} \\ + \int_{t_{2}}^{\infty} |\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) \frac{d s}{s} \end{matrix}

\begin{matrix} = \frac{1}{Γ (α)} \int_{1}^{t_{1}} |\frac{{(ln t_{2})}^{α - 1} - {(ln \frac{t_{2}}{s})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1} - {(ln \frac{t_{1}}{s})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}| a (s) \frac{d s}{s} \\ + \frac{1}{Γ (α)} \int_{t_{1}}^{t_{2}} |\frac{{(ln t_{2})}^{α - 1} - {(ln \frac{t_{2}}{s})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}| a (s) \frac{d s}{s} \\ + \frac{1}{Γ (α)} \int_{t_{2}}^{\infty} (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) a (s) \frac{d s}{s} \\ \leq \frac{1}{Γ (α)} (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) \int_{1}^{t_{1}} a (s) \frac{d s}{s} \\ + \frac{1}{Γ (α)} \int_{1}^{t_{1}} (\frac{{(ln \frac{t_{2}}{s})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln \frac{t_{1}}{s})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) a (s) \frac{d s}{s} \\ + \frac{1}{Γ (α)} (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) \int_{t_{1}}^{t_{2}} a (s) \frac{d s}{s} \\ + \frac{1}{Γ (α)} \int_{t_{1}}^{t_{2}} \frac{{(ln \frac{t_{2}}{s})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} a (s) \frac{d s}{s} \\ + \frac{1}{Γ (α)} (\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}) \int_{t_{2}}^{\infty} a (s) \frac{d s}{s} . \end{matrix}

Because the functions

\frac{{(ln t)}^{α - 1}}{1 + {(ln t)}^{α - 1}}

and

\frac{{(ln (t / s))}^{α - 1}}{1 + {(ln t)}^{α - 1}}

are uniformly continuous on

[1, T]

, respectively on

[1, T] \times [1, T]

, and the integrals

\int_{0}^{\infty} a (s) \frac{d s}{s}

\int_{0}^{\infty} b (s) \frac{d s}{s}

are convergent, we conclude that

Ξ_{0} \to 0

and

Ξ_{1} \to 0

t_{2} \to t_{1}

uniformly with respect to

(x, y) \in S

In a similar manner we obtain

|\frac{A_{2} (x, y) (t_{2})}{1 + {(ln t_{2})}^{β - 1}} - \frac{A_{2} (x, y) (t_{1})}{1 + {(ln t_{1})}^{β - 1}}| \to 0, as t_{2} \to t_{1},

uniformly with respect to

(x, y) \in S

. Then

A (S)

is equicontinuous on

[1, T]

(III) In what follows we will show that

A

is equiconvergent at ∞. We will prove firstly that

A_{1}

is equiconvergent at ∞, that is, for any

ϵ > 0

there exists

T_{0} > 0

such that for all

t_{1}, t_{2} \geq T_{0}

, and

(x, y) \in S

we have

|\frac{A_{1} (x, y) (t_{2})}{1 + {(ln t_{2})}^{α - 1}} - \frac{A_{1} (x, y) (t_{1})}{1 + {(ln t_{1})}^{α - 1}}| < Λ_{0} ϵ, (Λ_{0} > 0) .

For this, let

ϵ > 0

. Then there exists

δ_{1} > 1

such that

\int_{δ_{1}}^{\infty} a (s) \frac{d s}{s} < ϵ

and

\int_{δ_{1}}^{\infty} b (s) \frac{d s}{s} < ϵ

. Because

{lim}_{t \to \infty} \frac{{(ln t)}^{α - 1}}{1 + {(ln t)}^{α - 1}} = 1

and

{lim}_{t \to \infty} \frac{g_{α} (t, δ_{1})}{1 + {(ln t)}^{α - 1}} = 0

, we deduce that there exist

δ_{2} > 0

and

δ_{3} > δ_{1}

such that for any

t_{1}, t_{2} > δ_{2}

we have

|\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}| < ϵ,

and for any

t_{1}, t_{2} > δ_{3}

and

1 \leq s \leq δ_{1}

we have

\begin{matrix} |\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| \leq |\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}}| + |\frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| \\ \leq |\frac{g_{α} (t_{2}, δ_{1})}{1 + {(ln t_{2})}^{α - 1}}| + |\frac{g_{α} (t_{1}, δ_{1})}{1 + {(ln t_{1})}^{α - 1}}| < ϵ . \end{matrix}

Then we choose

T_{0} > max {δ_{2}, δ_{3}}

, and then for any

(x, y) \in S

and

t_{1}, t_{2} \geq T_{0}

we obtain

\begin{matrix} |\frac{A_{1} (x, y) (t_{2})}{1 + {(ln t_{2})}^{α - 1}} - \frac{A_{1} (x, y) (t_{1})}{1 + {(ln t_{1})}^{α - 1}}| \\ \leq \int_{1}^{\infty} |\frac{G_{1} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{1} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} |\frac{G_{2} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{2} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq \int_{1}^{δ_{1}} |\frac{G_{1} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{1} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{δ_{1}}^{\infty} |\frac{G_{1} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{1} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{δ_{1}} |\frac{G_{2} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{2} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| b (s) g (x (s), y (s)) \frac{d s}{s} \\ + \int_{δ_{1}}^{\infty} |\frac{G_{2} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{G_{2} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq M_{1} \int_{1}^{δ_{1}} |\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) \frac{d s}{s} \\ + M_{1} \int_{δ_{1}}^{\infty} |\frac{g_{α} (t_{2}, s)}{1 + {(ln t_{2})}^{α - 1}} - \frac{g_{α} (t_{1}, s)}{1 + {(ln t_{1})}^{α - 1}}| a (s) \frac{d s}{s} \\ + \frac{M_{1}}{Δ} |\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}| \\ \times (d \int_{1}^{\infty} \frac{1}{Γ (α)} τ^{α - 1} d H_{1} (τ) + b \int_{1}^{\infty} \frac{1}{Γ (α)} τ^{α - 1} d K_{1} (τ)) \int_{1}^{\infty} a (s) \frac{d s}{s} \\ + \frac{M_{2}}{Δ} |\frac{{(ln t_{2})}^{α - 1}}{1 + {(ln t_{2})}^{α - 1}} - \frac{{(ln t_{1})}^{α - 1}}{1 + {(ln t_{1})}^{α - 1}}| \\ \times (d \int_{1}^{\infty} \frac{1}{Γ (β)} τ^{β - 1} d H_{2} (τ) + b \int_{1}^{\infty} \frac{1}{Γ (β)} τ^{β - 1} d K_{2} (τ)) \int_{1}^{\infty} b (s) \frac{d s}{s} \\ \leq ϵ M_{1} \int_{1}^{δ_{1}} a (s) \frac{d s}{s} + \frac{2 ϵ M_{1}}{Γ (α)} + \frac{ϵ M_{1} (d Γ (α) - Δ)}{Δ Γ (α)} \int_{1}^{\infty} a (s) \frac{d s}{s} \\ + \frac{ϵ M_{2} b}{Δ} \int_{1}^{\infty} b (s) \frac{d s}{s} = Λ_{0} ϵ, (Λ_{0} > 0) . \end{matrix}

A_{1}

is equiconvergent at ∞. In a similar manner we show that

A_{2}

is equiconvergent at ∞, and then we deduce that

A

is equiconvergent at ∞.

(IV) In the last part of the proof we will prove that the operator

A

is continuous. Let

{((x_{n}, y_{n}))}_{n \in N}, (x, y) \in X

(x_{n}, y_{n}) \to (x, y)

X = X_{1} \times X_{2}

, for

n \to \infty

, that is

sup_{t \in I} \frac{| x_{n} (t) - x (t) |}{1 + {(ln t)}^{α - 1}} \to 0, and sup_{t \in I} \frac{| y_{n} (t) - y (t) |}{1 + {(ln t)}^{β - 1}} \to 0, as n \to \infty .

Then we deduce that for any

s \in I

x_{n} (s) - x (s) \to 0

and

y_{n} (s) - y (s) \to 0

, as

n \to \infty

Because

\begin{matrix} | f (x_{n} (s), y_{n} (s)) - f (x (s), y (s)) | \leq 2 M_{1}, \forall s \in I, n \in N, \\ | g (x_{n} (s), y_{n} (s)) - g (x (s), y (s)) | \leq 2 M_{2}, \forall s \in I, n \in N, \end{matrix}

we obtain by the Lebesgue convergence theorem that

\begin{matrix} \int_{1}^{\infty} a (s) | f (x_{n} (s), y_{n} (s)) - f (x (s), y (s)) | \frac{d s}{s} \to 0, as n \to \infty, \\ \int_{1}^{\infty} b (s) | g (x_{n} (s), y_{n} (s)) - g (x (s), y (s)) | \frac{d s}{s} \to 0, as n \to \infty . \end{matrix}

(13)

By using Lemma 5 and (13), we find

\begin{matrix} ∥ A_{1} (x_{n}, y_{n}) - A_{1} (x, y) ∥_{1} = sup_{t \in I} \frac{| A_{1} (x_{n}, y_{n}) (t) - A_{1} (x, y) (t) |}{1 + {(ln t)}^{α - 1}} \\ = sup_{t \in I} |\int_{1}^{\infty} \frac{G_{1} (t, s)}{1 + {(ln t)}^{α - 1}} a (s) f (x_{n} (s), y_{n} (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} \frac{G_{2} (t, s)}{1 + {(ln t)}^{α - 1}} b (s) g (x_{n} (s), y_{n} (s)) \frac{d s}{s} \\ - \int_{1}^{\infty} \frac{G_{1} (t, s)}{1 + {(ln t)}^{α - 1}} a (s) f (x (s), y (s)) \frac{d s}{s} - \int_{1}^{\infty} \frac{G_{2} (t, s)}{1 + {(ln t)}^{α - 1}} b (s) g (x (s), y (s)) \frac{d s}{s}| \\ \leq sup_{t \in I} (\int_{1}^{\infty} \frac{G_{1} (t, s)}{1 + {(ln t)}^{α - 1}} a (s) | f (x_{n} (s), y_{n} (s)) - f (x (s), y (s)) | \frac{d s}{s} \\ + \int_{1}^{\infty} \frac{G_{2} (t, s)}{1 + {(ln t)}^{α - 1}} b (s) | g (x_{n} (s), y_{n} (s)) - g (x (s), y (s)) | \frac{d s}{s}) \\ \leq Λ_{1} \int_{1}^{\infty} a (s) | f (x_{n} (s), y_{n} (s)) - f (x (s), y (s)) | \frac{d s}{s} \\ + Λ_{2} \int_{1}^{\infty} b (s) | g (x_{n} (s), y_{n} (s)) - g (x (s), y (s)) | \frac{d s}{s} \to 0, as n \to \infty . \end{matrix}

Then we deduce that

∥ A_{1} (x_{n}, y_{n}) - A_{1} (x, y) ∥_{1} \to 0,

n \to \infty

. In a similar manner we can prove that

∥ A_{2} (x_{n}, y_{n}) - A_{2} (x, y) ∥_{2} \to 0,

n \to \infty

. Therefore we obtain

∥ A (x_{n}, y_{n}) - A (x, y) ∥ \to 0

n \to \infty

. So the operator

A

is continuous.

From the above steps and Lemma 6, we conclude that the operator

A

is completely continuous. □

Now for

θ > 1

we introduce the following constants

\begin{matrix} Q_{1} = \int_{1}^{\infty} a (s) \frac{d s}{s}, Q_{2} = \int_{1}^{\infty} b (s) \frac{d s}{s}, Q_{3} = \int_{θ}^{\infty} a (s) \frac{d s}{s}, Q_{4} = \int_{θ}^{\infty} b (s) \frac{d s}{s}, \\ {\tilde{L}}_{1} = \frac{d}{Γ (α)} \int_{1}^{θ} {(ln τ)}^{α - 1} d H_{1} (τ) + \frac{b}{Γ (α)} \int_{1}^{θ} {(ln τ)}^{α - 1} d K_{1} (τ), \\ {\tilde{L}}_{2} = \frac{d}{Γ (β)} \int_{1}^{θ} {(ln τ)}^{β - 1} d H_{2} (τ) + \frac{b}{Γ (β)} \int_{1}^{θ} {(ln τ)}^{β - 1} d K_{2} (τ), \\ {\tilde{L}}_{3} = \frac{c}{Γ (α)} \int_{1}^{θ} {(ln τ)}^{α - 1} d H_{1} (τ) + \frac{a}{Γ (α)} \int_{1}^{θ} {(ln τ)}^{α - 1} d K_{1} (τ), \\ {\tilde{L}}_{4} = \frac{c}{Γ (β)} \int_{1}^{θ} {(ln τ)}^{β - 1} d H_{2} (τ) + \frac{a}{Γ (β)} \int_{1}^{θ} {(ln τ)}^{β - 1} d K_{2} (τ), \\ L_{1} = \frac{{\tilde{L}}_{1} {(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})}, L_{2} = \frac{{\tilde{L}}_{2} {(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})}, \\ L_{3} = \frac{{\tilde{L}}_{3} {(ln θ)}^{β - 1}}{Δ (1 + {(ln θ)}^{β - 1})}, L_{4} = \frac{{\tilde{L}}_{4} {(ln θ)}^{β - 1}}{Δ (1 + {(ln θ)}^{β - 1})}, \\ Υ_{1} = Q_{3} (L_{1} + L_{3}), Υ_{2} = Q_{4} (L_{2} + L_{4}), Υ_{3} = Q_{1} (Λ_{1} + Λ_{3}), Υ_{4} = Q_{2} (Λ_{2} + Λ_{4}) . \end{matrix}

(14)

In our first main theorem we will prove the existence of at least one positive solution for problem (1), (2) by using the Guo-Krasnosel’skii fixed point theorem (Theorem 1).

Theorem 3.

We assume that

(H 1) - (H 3)

hold, and there exists

θ > 1

such that

Q_{j} > 0

j = 3, 4

, and

{\tilde{L}}_{i} > 0

i = 1, \dots, 4

. In addition, we suppose that there exist positive constants

r_{1}, r_{2}

with

r_{1} < r_{2}

, and

σ_{1} \in [Υ_{1}^{- 1}, \infty)

σ_{2} \in [Υ_{2}^{- 1}, \infty)

σ_{3} \in (0, Υ_{3}^{- 1}]

and

σ_{4} \in (0, Υ_{4}^{- 1}]

such that

$(H 4)$: $f ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \geq \frac{σ_{1} r_{1}}{2}, \forall t \in [θ, \infty), (x, y) \in [0, r_{1}] \times [0, r_{1}],$

$g ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \geq \frac{σ_{2} r_{1}}{2}, \forall t \in [θ, \infty), (x, y) \in [0, r_{1}] \times [0, r_{1}];$
$(H 5)$: $f ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \leq \frac{σ_{3} r_{2}}{2}, \forall t \in I, (x, y) \in [0, r_{2}] \times [0, r_{2}],$

$g ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \leq \frac{σ_{4} r_{2}}{2}, \forall t \in I, (x, y) \in [0, r_{2}] \times [0, r_{2}] .$

Then problem (1), (2) has at least one positive solution

(x, y)

such that

r_{1} \leq ∥ (x, y) ∥ \leq r_{2} .

Proof.

By Lemma 7, the operator

A

is completely continuous. We introduce the set

Ω_{1} = {(x, y) \in X, ∥ (x, y) ∥ < r_{1}}

. Then for

(x, y) \in P \cap \partial Ω_{1}

, we have

{∥ x ∥}_{1} + {∥ y ∥}_{2} = r_{1}

, so

{∥ x ∥}_{1} \leq r_{1}

and

{∥ y ∥}_{2} \leq r_{1}

, that is

0 \leq \frac{x (t)}{1 + {(ln t)}^{α - 1}} \leq r_{1}

and

0 \leq \frac{y (t)}{1 + {(ln t)}^{β - 1}} \leq r_{1}

for all

t \in I

Therefore by

(H 4)

and Lemma 5, we obtain

\begin{matrix} ∥ A_{1} {(x, y) ∥}_{1} = sup_{t \in I} \frac{1}{1 + {(ln t)}^{α - 1}} (\int_{1}^{\infty} G_{1} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} G_{2} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s}) \\ \geq inf_{t \in [θ, \infty)} \frac{1}{1 + {(ln t)}^{α - 1}} (\int_{1}^{\infty} G_{1} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} G_{2} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s}) \\ \geq inf_{t \in [θ, \infty)} \frac{1}{1 + {(ln t)}^{α - 1}} \int_{1}^{\infty} G_{1} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + inf_{t \in [θ, \infty)} \frac{1}{1 + {(ln t)}^{α - 1}} \int_{1}^{\infty} G_{2} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s} \\ \geq \int_{1}^{\infty} inf_{t \in [θ, \infty)} \frac{G_{1} (t, s)}{1 + {(ln t)}^{α - 1}} a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} inf_{t \in [θ, \infty)} \frac{G_{2} (t, s)}{1 + {(ln t)}^{α - 1}} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \geq \frac{{(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})} \int_{θ}^{\infty} (d \int_{1}^{\infty} g_{α} (τ, s) d H_{1} (τ) \\ + b \int_{1}^{\infty} g_{α} (τ, s) d K_{1} (τ)) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \frac{{(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})} \int_{θ}^{\infty} (d \int_{1}^{\infty} g_{β} (τ, s) d H_{2} (τ) \\ + b \int_{1}^{\infty} g_{β} (τ, s) d K_{2} (τ)) b (s) g (x (s), y (s)) \frac{d s}{s} \\ \geq \frac{{(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})} \int_{θ}^{\infty} (d \int_{1}^{θ} g_{α} (τ, s) d H_{1} (τ) \\ + b \int_{1}^{θ} g_{α} (τ, s) d K_{1} (τ)) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \frac{{(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})} \int_{θ}^{\infty} (d \int_{1}^{θ} g_{β} (τ, s) d H_{2} (τ) \\ + b \int_{1}^{θ} g_{β} (τ, s) d K_{2} (τ)) b (s) g (x (s), y (s)) \frac{d s}{s} \\ = \frac{{(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})} \int_{θ}^{\infty} a (s) f (x (s), y (s)) \frac{d s}{s} \\ \times (\frac{d}{Γ (α)} \int_{1}^{θ} {(ln τ)}^{α - 1} d H_{1} (τ) + \frac{b}{Γ (α)} \int_{1}^{θ} {(ln τ)}^{α - 1} d K_{1} (τ)) \\ + \frac{{(ln θ)}^{α - 1}}{Δ (1 + {(ln θ)}^{α - 1})} \int_{θ}^{\infty} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \times (\frac{d}{Γ (β)} \int_{1}^{θ} {(ln τ)}^{β - 1} d H_{2} (τ) + \frac{b}{Γ (β)} \int_{1}^{θ} {(ln τ)}^{β - 1} d K_{2} (τ)) \\ = \frac{{(ln θ)}^{α - 1} {\tilde{L}}_{1}}{Δ (1 + {(ln θ)}^{α - 1})} \frac{σ_{1} r_{1}}{2} \int_{θ}^{\infty} a (s) \frac{d s}{s} + \frac{{(ln θ)}^{α - 1} {\tilde{L}}_{2}}{Δ (1 + {(ln θ)}^{α - 1})} \frac{σ_{2} r_{1}}{2} \int_{θ}^{\infty} b (s) \frac{d s}{s} \\ = \frac{σ_{1} r_{1} L_{1} Q_{3}}{2} + \frac{σ_{2} r_{1} L_{2} Q_{4}}{2} = (\frac{σ_{1} L_{1} Q_{3}}{2} + \frac{σ_{2} L_{2} Q_{4}}{2}) r_{1} . \end{matrix}

In a similar manner we find

\begin{matrix} ∥ A_{2} {(x, y) ∥}_{2} \geq \frac{{(ln θ)}^{β - 1} {\tilde{L}}_{3}}{Δ (1 + {(ln θ)}^{β - 1})} \frac{σ_{1} r_{1}}{2} \int_{θ}^{\infty} a (s) \frac{d s}{s} + \frac{{(ln θ)}^{β - 1} {\tilde{L}}_{4}}{Δ (1 + {(ln θ)}^{β - 1})} \frac{σ_{2} r_{1}}{2} \int_{θ}^{\infty} b (s) \frac{d s}{s} \\ = \frac{σ_{1} r_{1} L_{3} Q_{3}}{2} + \frac{σ_{2} r_{1} L_{4} Q_{4}}{2} = (\frac{σ_{1} L_{3} Q_{3}}{2} + \frac{σ_{2} L_{4} Q_{4}}{2}) r_{1} . \end{matrix}

Therefore we deduce

\begin{matrix} ∥ A (x, y) ∥ = ∥ A_{1} {(x, y) ∥}_{1} + {∥ A_{2} (x, y) ∥}_{2} \\ \geq [\frac{(L_{1} + L_{3}) σ_{1} Q_{3}}{2} + \frac{(L_{2} + L_{4}) σ_{2} Q_{4}}{2}] r_{1} = (\frac{σ_{1} Υ_{1}}{2} + \frac{σ_{2} Υ_{2}}{2}) r_{1} \geq r_{1}, \end{matrix}

which gives us

∥ A (x, y) ∥ \geq ∥ (x, y) ∥, \forall (x, y) \in P \cap \partial Ω_{1} .

(15)

Now we introduce the set

Ω_{2} = {(x, y) \in X, ∥ (x, y) ∥ < r_{2}}

. Then for any

(x, y) \in P \cap \partial Ω_{2}

, we have

{∥ x ∥}_{1} + {∥ y ∥}_{2} = r_{2}

, and then

{∥ x ∥}_{1} \leq r_{2}

and

{∥ y ∥}_{2} \leq r_{2}

, that is

0 \leq \frac{x (t)}{1 + {(ln t)}^{α - 1}} \leq r_{2}

and

0 \leq \frac{y (t)}{1 + {(ln t)}^{β - 1}} \leq r_{2}

for all

t \in I

Then by

(H 5)

and Lemma 5, we obtain

\begin{matrix} ∥ A_{1} {(x, y) ∥}_{1} = sup_{t \in I} \frac{A_{1} (x, y)}{1 + {(ln t)}^{α - 1}} \\ \leq sup_{t \in I} \frac{1}{1 + {(ln t)}^{α - 1}} \int_{1}^{\infty} G_{1} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + sup_{t \in I} \frac{1}{1 + {(ln t)}^{α - 1}} \int_{1}^{\infty} G_{2} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq \int_{1}^{\infty} sup_{t \in I} \frac{G_{1} (t, s)}{1 + {(ln t)}^{α - 1}} a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} sup_{t \in I} \frac{G_{2} (t, s)}{1 + {(ln t)}^{α - 1}} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq Λ_{1} \int_{1}^{\infty} a (s) f (x (s), y (s)) \frac{d s}{s} + Λ_{2} \int_{1}^{\infty} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq \frac{σ_{3} r_{2} Λ_{1}}{2} \int_{1}^{\infty} a (s) \frac{d s}{s} + \frac{σ_{4} r_{2} Λ_{2}}{2} \int_{1}^{\infty} b (s) \frac{d s}{s} \\ = (\frac{σ_{3} Q_{1} Λ_{1}}{2} + \frac{σ_{4} Q_{2} Λ_{2}}{2}) r_{2}, \end{matrix}

and

\begin{matrix} ∥ A_{2} {(x, y) ∥}_{2} = sup_{t \in I} \frac{A_{2} (x, y)}{1 + {(ln t)}^{β - 1}} \\ \leq sup_{t \in I} \frac{1}{1 + {(ln t)}^{β - 1}} \int_{1}^{\infty} G_{3} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + sup_{t \in I} \frac{1}{1 + {(ln t)}^{β - 1}} \int_{1}^{\infty} G_{4} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq \int_{1}^{\infty} sup_{t \in I} \frac{G_{3} (t, s)}{1 + {(ln t)}^{β - 1}} a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} sup_{t \in I} \frac{G_{4} (t, s)}{1 + {(ln t)}^{β - 1}} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq Λ_{3} \int_{1}^{\infty} a (s) f (x (s), y (s)) \frac{d s}{s} + Λ_{4} \int_{1}^{\infty} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq \frac{σ_{3} r_{2} Λ_{3}}{2} \int_{1}^{\infty} a (s) \frac{d s}{s} + \frac{σ_{4} r_{2} Λ_{4}}{2} \int_{1}^{\infty} b (s) \frac{d s}{s} \\ = (\frac{σ_{3} Q_{1} Λ_{3}}{2} + \frac{σ_{4} Q_{2} Λ_{4}}{2}) r_{2} . \end{matrix}

Then we deduce

\begin{matrix} ∥ A (x, y) ∥ \leq [\frac{(Λ_{1} + Λ_{3}) σ_{3} Q_{1}}{2} + \frac{(Λ_{2} + Λ_{4}) σ_{4} Q_{2}}{2}] r_{2} \\ = (\frac{σ_{3} Υ_{3}}{2} + \frac{σ_{4} Υ_{4}}{2}) r_{2} \leq r_{2}, \end{matrix}

that is

∥ A (x, y) ∥ \leq ∥ (x, y) ∥, \forall (x, y) \in P \cap \partial Ω_{2} .

(16)

Therefore by (15), (16) and Theorem 1 ii), we conclude that the operator

A

has a fixed point in

P \cap ({\bar{Ω}}_{2} ∖ Ω_{1})

, with

r_{1} \leq ∥ (x, y) ∥ \leq r_{2}

. So

x (t) \geq 0

and

y (t) \geq 0

for all

t \in I

, and by

(H 2)

(H 3)

, we obtain that

x (t) > 0

for all

t \in (1, \infty)

y (t) > 0

for all

t \in (1, \infty)

, that is

(x, y)

is a positive solution of problem (1), (2). □

In a similar manner as we proved Theorem 3, we can obtain the following theorem.

Theorem 4.

We assume that

(H 1) - (H 3)

hold, and there exists

θ > 1

such that

Q_{j} > 0

j = 3, 4

, and

{\tilde{L}}_{i} > 0

i = 1, \dots, 4

. In addition, we suppose that there exist positive constants

r_{1}, r_{2}

with

r_{1} < r_{2}

, and

σ_{1} \in [Υ_{1}^{- 1}, \infty)

σ_{2} \in [Υ_{2}^{- 1}, \infty)

σ_{3} \in (0, Υ_{3}^{- 1}]

and

σ_{4} \in (0, Υ_{4}^{- 1}]

such that

$(H 6)$: $f ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \leq \frac{σ_{3} r_{1}}{2}, \forall t \in I, (x, y) \in [0, r_{1}] \times [0, r_{1}],$

$g ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \leq \frac{σ_{4} r_{1}}{2}, \forall t \in I, (x, y) \in [0, r_{1}] \times [0, r_{1}];$
$(H 7)$: $f ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \geq \frac{σ_{1} r_{2}}{2}, \forall t \in [θ, \infty), (x, y) \in [0, r_{2}] \times [0, r_{2}],$

$g ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \geq \frac{σ_{2} r_{2}}{2}, \forall t \in [θ, \infty), (x, y) \in [0, r_{2}] \times [0, r_{2}] .$

Then problem (1), (2) has at least one positive solution

(x, y)

such that

r_{1} \leq ∥ (x, y) ∥ \leq r_{2} .

In what follows, we will prove the existence of at least three positive solutions for problem (1), (2) by applying the Leggett-Williams fixed point theorem (Theorem 2).

Theorem 5.

We assume that

(H 1) - (H 3)

hold, and there exists

θ > 1

such that

Q_{j} > 0

j = 3, 4

, and

{\tilde{L}}_{i} > 0

i = 1, \dots, 4

. In addition, we suppose that there exist positive constants

a_{0} < b_{0} < c_{0}

such that

$(H 8)$: $f ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) < \frac{a_{0}}{2 Υ_{3}}, \forall t \in I, (x, y) \in [0, a_{0}] \times [0, a_{0}],$

$g ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) < \frac{a_{0}}{2 Υ_{4}}, \forall t \in I, (x, y) \in [0, a_{0}] \times [0, a_{0}];$
$(H 9)$: $f ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) > \frac{b_{0}}{2 Υ_{1}}, \forall t \in [θ, \infty), x, y \geq 0, b_{0} \leq x + y \leq c_{0},$

$g ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) > \frac{b_{0}}{2 Υ_{2}}, \forall t \in [θ, \infty), x, y \geq 0, b_{0} \leq x + y \leq c_{0};$
$(H 10)$: $f ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \leq \frac{c_{0}}{2 Υ_{3}}, \forall t \in I, (x, y) \in [0, c_{0}] \times [0, c_{0}],$

$g ((1 + {(ln t)}^{α - 1}) x, (1 + {(ln t)}^{β - 1}) y) \leq \frac{c_{0}}{2 Υ_{4}}, \forall t \in I, (x, y) \in [0, c_{0}] \times [0, c_{0}] .$

Then problem (1),(2) has at least three positive solutions

(x_{1}, y_{1})

(x_{2}, y_{2})

and

(x_{3}, y_{3})

such that

∥ (x_{1}, y_{1}) ∥ < a_{0}

∥ (x_{3}, y_{3}) ∥ > a_{0}

, and

inf_{t \in [θ, \infty)} (\frac{x_{2} (t)}{1 + {(ln t)}^{α - 1}} + \frac{y_{2} (t)}{1 + {(ln t)}^{β - 1}}) > b_{0}, inf_{t \in [θ, \infty)} (\frac{x_{3} (t)}{1 + {(ln t)}^{α - 1}} + \frac{y_{3} (t)}{1 + {(ln t)}^{β - 1}}) < b_{0} .

Proof.

We show firstly that operator

A : {\bar{P}}_{c_{0}} \to {\bar{P}}_{c_{0}}

. For any

(x, y) \in {\bar{P}}_{c_{0}}

, we have

∥ (x, y) ∥ \leq c_{0}

, and so

{∥ x ∥}_{1} \leq c_{0}

{∥ y ∥}_{2} \leq c_{0}

. Using the assumption

(H 10)

and Lemma 5, we find

\begin{matrix} ∥ A_{1} {(x, y) ∥}_{1} \leq sup_{t \in I} \frac{1}{1 + {(ln t)}^{α - 1}} \int_{1}^{\infty} G_{1} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + sup_{t \in I} \frac{1}{1 + {(ln t)}^{α - 1}} \int_{1}^{\infty} G_{2} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s} \end{matrix}

\begin{matrix} \leq \int_{1}^{\infty} sup_{t \in I} \frac{G_{1} (t, s)}{1 + {(ln t)}^{α - 1}} a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} sup_{t \in I} \frac{G_{2} (t, s)}{1 + {(ln t)}^{α - 1}} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq Λ_{1} \int_{1}^{\infty} a (s) f (x (s), y (s)) \frac{d s}{s} + Λ_{2} \int_{1}^{\infty} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq Λ_{1} \frac{c_{0}}{2 Υ_{3}} \int_{1}^{\infty} a (s) \frac{d s}{s} + Λ_{2} \frac{c_{0}}{2 Υ_{4}} \int_{1}^{\infty} b (s) \frac{d s}{s} \\ = (\frac{Λ_{1} Q_{1}}{Υ_{3}} + \frac{Λ_{2} Q_{2}}{Υ_{4}}) \frac{c_{0}}{2}, \end{matrix}

and

\begin{matrix} ∥ A_{2} {(x, y) ∥}_{2} \leq sup_{t \in I} \frac{1}{1 + {(ln t)}^{β - 1}} \int_{1}^{\infty} G_{3} (t, s) a (s) f (x (s), y (s)) \frac{d s}{s} \\ + sup_{t \in I} \frac{1}{1 + {(ln t)}^{β - 1}} \int_{1}^{\infty} G_{4} (t, s) b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq \int_{1}^{\infty} sup_{t \in I} \frac{G_{3} (t, s)}{1 + {(ln t)}^{β - 1}} a (s) f (x (s), y (s)) \frac{d s}{s} \\ + \int_{1}^{\infty} sup_{t \in I} \frac{G_{4} (t, s)}{1 + {(ln t)}^{β - 1}} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq Λ_{3} \int_{1}^{\infty} a (s) f (x (s), y (s)) \frac{d s}{s} + Λ_{4} \int_{1}^{\infty} b (s) g (x (s), y (s)) \frac{d s}{s} \\ \leq Λ_{3} \frac{c_{0}}{2 Υ_{3}} \int_{1}^{\infty} a (s) \frac{d s}{s} + Λ_{4} \frac{c_{0}}{2 Υ_{4}} \int_{1}^{\infty} b (s) \frac{d s}{s} \\ = (\frac{Λ_{3} Q_{1}}{Υ_{3}} + \frac{Λ_{4} Q_{2}}{Υ_{4}}) \frac{c_{0}}{2} . \end{matrix}

Then we obtain

∥ A (x, y) ∥ \leq [\frac{(Λ_{1} + Λ_{3}) Q_{1}}{Υ_{3}} + \frac{(Λ_{2} + Λ_{4}) Q_{2}}{Υ_{4}}] \frac{c_{0}}{2} = c_{0},

A : {\bar{P}}_{c_{0}} \to {\bar{P}}_{c_{0}}

We consider

d_{0} \in (b_{0}, c_{0})

, and we define the concave nonnegative continuous functional w on

P

w (x, y) = inf_{t \in [θ, \infty)} (\frac{x (t)}{1 + {(ln t)}^{α - 1}} + \frac{y (t)}{1 + {(ln t)}^{β - 1}}), (x, y) \in P .

We see easily that

w (x, y) \leq ∥ (x, y) ∥

for all

(x, y) \in {\bar{P}}_{c_{0}}

Next we will verify the conditions of Theorem 2, with

E = X

K = P

A = A

m = a_{0}

c = b_{0}

l = c_{0}

d = d_{0}

Θ = w

. We verify first condition ii). For

(x, y) \in {\bar{P}}_{a_{0}}

we will show that

∥ A (x, y) ∥ < a_{0}

. For this, let

(x, y) \in {\bar{P}}_{a_{0}}

. We obtain as in the above inequalities that

∥ A_{1} {(x, y) ∥}_{1} < (\frac{Λ_{1} Q_{1}}{Υ_{3}} + \frac{Λ_{2} Q_{2}}{Υ_{4}}) \frac{a_{0}}{2}, {∥ A_{2} (x, y) ∥}_{2} < (\frac{Λ_{3} Q_{1}}{Υ_{3}} + \frac{Λ_{4} Q_{2}}{Υ_{4}}) \frac{a_{0}}{2},

and so

∥ A (x, y) ∥ < a_{0}

. Then we have assumption ii) of Theorem 2.

We verify condition i) from Theorem 2. We choose the element

(x_{0} (t), y_{0} (t)) = (\frac{b_{0} + d_{0}}{4} (1 + {(ln t)}^{α - 1}), \frac{b_{0} + d_{0}}{4} (1 + {(ln t)}^{β - 1})), t \in I .

Because

x_{0} (t) \geq 0

y_{0} (t) \geq 0

for all

t \in I

∥ (x_{0}, y_{0}) ∥ = \frac{b_{0} + d_{0}}{2} < d_{0}

and

w (x_{0}, y_{0}) = \frac{b_{0} + d_{0}}{2} > b_{0}

, we deduce that

(x_{0}, y_{0}) \in {(x, y) | (x, y) \in P (w, b_{0}, d_{0}), w (x, y) > b_{0}}

. Now, let

(x, y) \in P (w, b_{0}, d_{0})

, that is

(x, y) \in P

w (x, y) \geq b_{0}

and

∥ (x, y) ∥ \leq d_{0}

. So we have

\frac{x (t)}{1 + {(ln t)}^{α - 1}} + \frac{y (t)}{1 + {(ln t)}^{β - 1}} \leq d_{0}

for all

t \in I

, and

{inf}_{t \in [θ, \infty)} (\frac{x (t)}{1 + {(ln t)}^{α - 1}} + \frac{y (t)}{1 + {(ln t)}^{β - 1}}) \geq b_{0}

. Then by

(H 9)

and Lemma 5, and similar computations as those from the first part of the proof of Theorem 3, we find

\begin{matrix} w (A (x, y)) = inf_{t \in [θ, \infty)} (\frac{A_{1} (x, y)}{1 + {(ln t)}^{α - 1}} + \frac{A_{2} (x, y)}{1 + {(ln t)}^{β - 1}}) \\ \geq L_{1} \int_{θ}^{\infty} a (s) f (x (s), y (s)) \frac{d s}{s} + L_{2} \int_{θ}^{\infty} b (s) g (x (s), y (s)) \frac{d s}{s} \\ + L_{3} \int_{θ}^{\infty} a (s) f (x (s), y (s)) \frac{d s}{s} + L_{4} \int_{θ}^{\infty} b (s) g (x (s), y (s)) \frac{d s}{s} \\ > L_{1} \frac{b_{0} Q_{3}}{2 Υ_{1}} + L_{2} \frac{b_{0} Q_{4}}{2 Υ_{2}} + L_{3} \frac{b_{0} Q_{3}}{2 Υ_{1}} + L_{4} \frac{b_{0} Q_{4}}{2 Υ_{2}} \\ = (L_{1} + L_{3}) \frac{b_{0} Q_{3}}{2 Υ_{1}} + (L_{2} + L_{4}) \frac{b_{0} Q_{4}}{2 Υ_{2}} = b_{0} . \end{matrix}

Therefore

w (A (x, y)) > b_{0}

, and we have assumption i) of Theorem 2.

Now we verify condition iii) of Theorem 2, namely

w (A (x, y)) > b_{0}

for

(x, y) \in P (w, b_{0}, c_{0})

and

∥ A (x, y) ∥ > d_{0}

. So let

(x, y) \in P (w, b_{0}, c_{0})

and

∥ A (x, y) ∥ > d_{0}

. By similar arguments used before we deduce that

w (A (x, y)) > b_{0}

, that is assumption iii) is satisfied.

Then by Theorem 2 and

(H 2)

(H 3)

we deduce that problem (1), (2) has at least three positive solutions

(x_{1}, y_{1})

(x_{2}, y_{2})

and

(x_{3}, y_{3})

with

∥ (x_{1}, y_{1}) ∥ < a_{0}

∥ (x_{3}, y_{3}) ∥ > a_{0}

w (x_{2}, y_{2}) > b_{0}

and

w (x_{3}, y_{3}) < b_{0}

. □

4. An example

Let

α = \frac{5}{2}

β = \frac{10}{3}

n = 3

m = 4

a (t) = \frac{1}{{(t - 1 / 2)}^{1 / 3}}, t \in [1, \infty)

b (t) = \frac{t + 2}{{(t - 1 / 3)}^{5 / 4}}, t \in [1, \infty)

f (x, y) = 6 + e^{- | x - y |} {sin}^{2} (x y), x, y \geq 0

g (x, y) = \frac{2}{3} + 95 e^{- x y} {cos}^{4} (x + y), x, y \geq 0

H_{1} (s) = {\frac{19}{110}, s \in [1, 3); \frac{s}{11} - \frac{1}{10}, s \in [3, 7); \frac{59}{110}, s \in [7, 10); \frac{116}{165}, s \in [10, \infty)}

H_{2} (s) = {\frac{5}{136} {(s - 1)}^{17 / 5}, s \in [1, 2); \frac{5}{136}, s \in (2, \infty)}

K_{1} (s) = {1, s \in [1, \frac{5}{3}); \frac{14}{13}, s \in [\frac{5}{3}, 2); \frac{7}{290} {(s - 2)}^{29 / 7} + \frac{14}{13}, s \in [2, \frac{15}{6}); \frac{7}{290} {(\frac{1}{2})}^{29 / 7} + \frac{14}{13}, s \in [\frac{15}{6}, \infty)}

K_{2} (s) = {3, s \in [1, 4); \frac{286}{95}, s \in [4, 9); \frac{s^{2}}{146} + \frac{34061}{13870}, s \in [9, 11); \frac{45556}{13870}, s \in [11, \infty)}

We consider the system of fractional differential equations

\{\begin{matrix} ^{H} D_{1 +}^{5 / 2} x (t) + \frac{1}{{(t - 1 / 2)}^{1 / 3}} (6 + e^{- | x (t) - y (t) |} {sin}^{2} (x (t) y (t))) = 0, t \in (1, \infty), \\ ^{H} D_{1 +}^{10 / 3} y (t) + \frac{t + 2}{{(t - 1 / 3)}^{5 / 4}} (\frac{2}{3} + 95 e^{- x (t) y (t)} {cos}^{4} (x (t) + y (t))) = 0, t \in (1, \infty), \end{matrix}

(17)

subject to the boundary conditions

\{\begin{matrix} x (1) = x^{'} (1) = 0, y (1) = y^{'} (1) = y^{''} (1) = 0, \\ ^{H} D_{1 +}^{3 / 2} x (\infty) = \frac{1}{11} \int_{3}^{7} x (s) d s + \frac{1}{6} x (10) + \frac{1}{8} \int_{1}^{2} {(s - 1)}^{12 / 5} y (s) d s, \\ ^{H} D_{1 +}^{7 / 3} y (\infty) = \frac{1}{13} x (\frac{5}{3}) + \frac{1}{10} \int_{2}^{15 / 6} {(s - 2)}^{22 / 7} x (s) d s + \frac{1}{95} y (4) + \frac{1}{73} \int_{9}^{11} s y (s) d s . \end{matrix}

(18)

By using Mathematica program, we obtain

a \approx 0.01753901

b \approx 0.01031779

c \approx 0.02920547

d \approx 0.83235873

Δ \approx 0.01429742 > 0

\int_{1}^{\infty} a (s) \frac{d s}{s} \approx 3.15855472

\int_{1}^{\infty} b (s) \frac{d s}{s} \approx 6.53061854

. So assumptions

(H 1) - (H 3)

are satisfied. In addition, we take

θ = 2

, and we deduce

Λ_{1} = \frac{d}{Δ} \approx 58.21742336

Λ_{2} = \frac{b}{Δ} \approx 0.72165469

Λ_{3} = \frac{c}{Δ} \approx 2.04270988

Λ_{4} = \frac{a}{Δ} \approx 1.22672621

{\tilde{L}}_{1} \approx 0.00021797

{\tilde{L}}_{2} \approx 0.00309129

{\tilde{L}}_{3} \approx 0.00037053

{\tilde{L}}_{4} \approx 0.00010846

L_{1} \approx 0.00557882

L_{2} \approx 0.07911642

L_{3} \approx 0.00773204

L_{4} \approx 0.00226336

Q_{1} \approx 3.15855472

Q_{2} \approx 6.53061854

Q_{3} \approx 2.43620073

Q_{4} \approx 4.28276191

Υ_{1} \approx 0.03242794

Υ_{2} \approx 0.34853023

Υ_{3} \approx 190.33492843

Υ_{4} \approx 12.72413242

We also consider

r_{1} = \frac{1}{3}

r_{2} = 2800

σ_{1} = 31 > Υ_{1}^{- 1}

σ_{2} = 3 > Υ_{2}^{- 1}

σ_{3} = 0.005 < Υ_{3}^{- 1}

and

σ_{4} = 0.07 < Υ_{4}^{- 1}

. Then we obtain the inequalities

\begin{matrix} f ((1 + {(ln t)}^{3 / 2}) x, (1 + {(ln t)}^{7 / 3}) y) \geq 6 > \frac{σ_{1} r_{1}}{2} = \frac{31}{6}, \forall t \in [2, \infty), x, y \in [0, \frac{1}{3}], \\ g ((1 + {(ln t)}^{3 / 2}) x, (1 + {(ln t)}^{7 / 3}) y) \geq \frac{2}{3} > \frac{σ_{2} r_{1}}{2} = \frac{1}{2}, \forall t \in [2, \infty), x, y \in [0, \frac{1}{3}], \end{matrix}

so assumption

(H 4)

is satisfied. In addition we have

\begin{matrix} f ((1 + {(ln t)}^{3 / 2}) x, (1 + {(ln t)}^{7 / 3}) y) \leq 7 = \frac{σ_{3} r_{2}}{2}, \forall t \in [1, \infty), x, y \in [0, 2800], \\ g ((1 + {(ln t)}^{3 / 2}) x, (1 + {(ln t)}^{7 / 3}) y) \leq \frac{287}{3} < \frac{σ_{4} r_{2}}{2} = 98, \forall t \in [1, \infty), x, y \in [0, 2800], \end{matrix}

hence assumption

(H 5)

is also satisfied. Therefore by Theorem 3 we conclude that problem (17), (18) has at least one positive solution

(x (t), y (t)), t \in [1, \infty)

, with

\frac{1}{3} \leq {sup}_{t \in [1, \infty)} \frac{x (t)}{1 + {(ln t)}^{3 / 2}} + {sup}_{t \in [1, \infty)} \frac{y (t)}{1 + {(ln t)}^{7 / 3}} \leq 2800

5. Conclusions

In this paper we investigate the system of nonlinear fractional differential equations (1) with Hadamard derivatives of various orders

α \in (n - 1, n]

and

β \in (m - 1, m]

, respectively, and nonnegative nonlinearities on the infinite interval

(1, \infty)

. The system (1) is supplemented with general nonlocal boundary conditions (2), where the unknown functions x and y in the point 1 and their derivatives until orders

n - 2

and

m - 2

, respectively, are all 0, and the Hadamard derivatives of x and y of order

n - 1

and

m - 1

at ∞ are dependent on both Riemann-Liouville integrals of x and y. Our problem generalizes the problem studied in [13], by considering here different orders for the fractional derivatives in the equations of system (1), and also a general form of the boundary conditions from (2) at ∞. Under some assumptions on the data of this problem, we give firstly the solution of the associated linear boundary value problem, and the corresponding Green functions with their properties. Then in the main section of paper we prove the existence of positive solutions of (1), (2) by applying the Guo-Krasnosl’skii fixed point theorem and the Leggett-Williams fixed point theorem. We also present finally an example for illustrating our results.

Author Contributions

Conceptualization, R.L.; Formal analysis, R.L. and A.T.; Methodology, R.L. and A.T. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Conflicts of Interest

The authors declare no conflict of interest.

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On a System of Hadamard Fractional Differential Equations With Nonlocal Boundary Conditions on an Unbounded Domain

Abstract

1. Introduction

2. Auxiliary results

3. Main results

4. An example

5. Conclusions

Author Contributions

Funding

Conflicts of Interest

References

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