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Article

Existence and Uniqueness of Positive Solutions for Hilfer–Hadamard-Type Fractional Differential Equations with γ-Concave and Sub-Homogeneous Operators

1
Department of Mathematics and Computer Science, Amirkabir University of Technology, Tehran 1591634311, Iran
2
Department of Mathematics, Faculty of Science, University of Bonab, Bonab 5551395133, Iran
3
Department of Mathematics and Statistics, Missouri S&T, Rolla, MO 65409, USA
*
Author to whom correspondence should be addressed.
Fractal Fract. 2026, 10(7), 449; https://doi.org/10.3390/fractalfract10070449
Submission received: 25 February 2026 / Revised: 20 June 2026 / Accepted: 23 June 2026 / Published: 30 June 2026
(This article belongs to the Special Issue Fractional Calculus and Nonlinear Analysis: Theory and Applications)

Abstract

In this research, we present necessary and sufficient conditions for the existence and uniqueness of positive solutions for a class of Hilfer–Hadamard-type fractional differential equations with boundary value problems, including those with integral boundary conditions. The obtained results are conditional on a specific set of strong assumptions, which substantially narrow the class of admissible nonlinearities, coefficients, and boundary data. Thus, the present work extends the Hadamard-type framework to the Hilfer–Hadamard setting only within this restrictive regime, rather than providing a full extension to all Hilfer–Hadamard systems. We utilize the properties of γ -concave and sub-homogeneous operators along with two Banach fixed point theorems to achieve our results. An illustrative example is also provided to demonstrate the applicability of the methodology and highlight the main findings. The novelty of our work in this paper is twofold: First, we have expanded the scope by considering Hilfer–Hadamard-type fractional differential equations. Second, we employed a new fixed point theorem to improve our results within this particular framework.

1. Introduction

In recent years, fractional differential equations (FDEs) have appeared as indispensable tools for modeling complex systems across disciplines, including mechanics, physics, chemistry, economics, engineering, and biology. For this reason, the study of FDEs has recently attracted the attention of many authors and modelers [1,2,3,4,5].
The Hilfer derivative, introduced by Hilfer the year 2000, generalizes the classical definition of derivatives to fractional orders, allowing for the incorporation of both fractional integral and derivatives aspects of functions. This increases the flexibility and accuracy of mathematical modeling, enabling a deeper understanding of complex dynamical systems. Hilfer’s work laid the groundwork for a rich avenue of exploration into the qualitative and quantitative analysis of fractional differential equations. For more information, one can refer to the references [6,7,8,9].
On the other hand, Hadamard fractional calculus, originating from the work of Hadamard, focuses on the properties and applications of fractional integrals and derivatives that exhibit nonlocal behavior. The synergy between Hadamard’s and Hilfer’s approaches has led to the development of equations that not only enhance theoretical understanding but also provide practical solutions to problems characterized by noninteger order dynamics [10,11,12].
Recently, some authors have introduced γ -concave and sub-homogeneous operators in order to study existence and uniqueness theorems of fixed points of mixed monotone operators. This contribution to the field has employed various methodologies, including topological techniques and fixed point theorems to explore the conditions of these equations. The interplay between increasing and decreasing operators and the formulation of γ -concave functions ( 0 γ < 1 ) have emerged as effective tools in proving the existence of solutions [13,14,15,16].
Previous studies [2,8,11,17] have explored the concept of Hilfer–Hadamard equations across a considerable body of the literature. In this paper, we extend the foundational contributions of Hilfer and Hadamard by presenting new results concerning uniqueness for this advanced case of fractional calculus, with the important caveat that our existence and uniqueness results are valid only under the restrictive assumptions ( A 1 )–( A 7 ) imposed on the nonlinearities, coefficients, and boundary data. Consequently, this is not a general result for all Hilfer–Hadamard systems, but rather a conditional extension of Hadamard-type models to a specific Hilfer–Hadamard setting. This approach further enriches the development of fractional differential equations and its practical applications.
In [14], the authors considered the Hadamard-type infinite boundary value problem
D 1 + β H ζ ( τ ) + ν ( τ ) φ ( τ , ζ ( τ ) ) + μ ( τ ) ψ ( τ , ζ ( τ ) ) = 0 , τ ( 1 , ) , ζ ( 1 ) = ζ ( 1 ) = 0 , D 1 + β 1 H ζ ( ) = i = 1 n i I 1 + ε i H ζ ( π ) + s j = 1 m ω j ζ ( s j ) ,
where D 1 + β H is the Hadamard derivative of order β ( 2 , 3 ] and I 1 + ε i H is the Hadamard integral of order ε i > 0 for i = 1 , , n .
Salamooni and coauthors in [7] considered the problem
D β , α HH ζ ( τ ) + φ ( τ , ζ ( τ ) ) = 0 , τ ( 1 , e ] , α ( 1 , 2 ] , β [ 0 , 1 ] , ζ ( 1 + δ ) = 0 , D 1 , 1 HH ζ ( e ) = μ D 1 , 1 HH ζ ( ν ) ,
where D β , α HH is the Hilfer–Hadamard-type derivative of order α ( 1 , 2 ] and type β [ 0 , 1 ] , μ [ 0 , 1 ) , ν [ 1 , e ) , δ [ 0 , 1 ) , D 1 , 1 HH = τ d d τ and φ : ( 1 , e ] × R + R + .
Here, we study the existence and uniqueness of positive solutions to a Hilfer–Hadamard equation under specified boundary conditions using the γ -concave and sub-homogeneous operators along with two Banach fixed point theorems. We consider the problem
D 1 + β , α HH ζ ( τ ) + ν ( τ ) φ ( τ , ζ ( τ ) ) + μ ( τ ) ψ ( τ , ζ ( τ ) ) = 0 , ζ ( 1 ) = 0 , τ [ 1 , e ] , ζ ( e ) = i = 1 n i I ε i H ζ ( π i ) + s j = 1 m ω j ζ ( s j ) = i = 1 n i Γ ( ε i ) 1 π i log π i ς ε i 1 ζ ( ς ) d ς ς + s j = 1 m ω j ζ ( s j ) ,
where D 1 + β , α HH is the Hilfer–Hadamard derivative and I ε i H the Hadamard integral of order β ( 1 , 2 ] and type α [ 0 , 1 ] , ε i > 0 , π i , s j ( 1 , e ) , i , ω j , s 0 for i = 1 , , n and j = 1 , , m , moreover ν , μ C ( J , R + ) and φ , ψ C ( J × R + , R + ) , for J = [ 1 , e ] .

2. Preliminaries

In this section, we introduce the notations, definitions and lemmas used throughout the paper. We consider X = ζ C ( J , R ) : ζ X < with the norm
ζ X = sup τ J | ζ ( τ ) | ( log 10 ) γ 1 ( log τ ) γ 1 .
Then ( X , · X ) is a Banach space. Let θ be the zero element of X and consider a cone P = ζ X : ζ ( τ ) 0 X . For any ζ , ξ X , the property ζ ξ denotes that there are α > 0 and β > 0 such that α ζ ξ β ζ . For ρ > θ (i.e., ρ θ and ρ θ ), define P ρ = ζ X : ζ ρ . Clearly, P ρ P .
Definition 1
(Hadamard fractional integral [2]). Let β > 0 and ζ : [ a , b ] R be a continuous function. The left-sided and right-sided Hadamard integrals of order β for ζ are defined by
I a + β H ζ ( τ ) = 1 Γ ( β ) a τ log τ ς β 1 ζ ( ς ) d ς ς , a < τ
and
I b β H ζ ( τ ) = 1 Γ ( β ) τ b log ς τ β 1 ζ ( ς ) d ς ς , τ < b .
The Hadamard integral I a + β H for log τ a α 1 is defined by
I a + β H log τ a α 1 = Γ ( α ) Γ ( α + β ) log τ a α + β 1 , α , β > 0 , a < τ .
Let A C ( J ) denote the space of absolutely continuous functions on J . We define the space of differentiable absolutely continuous functions by
A C 1 ( J ) : = ζ : J R | d d τ ζ ( τ ) A C ( J ) .
Furthermore, we define the space of n-times ε -differentiable absolutely continuous functions as
A C ε n ( J ) : = ζ : J R | ε ( n 1 ) ζ ( τ ) A C ( J ) ,
where ε : = τ d d τ is the Stieltjes differential operator and n = [ β ] + 1 .
Definition 2
(Hadamard-type fractional derivative [2]). Let n 1 < β < n with n = [ β ] + 1 and the function ζ : [ a , b ] R be continuous. The left-sided and right-sided Hadamard derivatives of order β for ζ are defined by
D a + β H ζ ( τ ) = 1 Γ ( n β ) τ d d τ n a τ log τ ς n β 1 ζ ( ς ) d ς ς , a < τ
and
D b β H ζ ( τ ) = 1 Γ ( n β ) τ d d τ n τ b log ς τ n β 1 ζ ( ς ) d ς ς , τ < b ,
where log ( · ) = log e ( · ) . The Hadamard derivative D a + β H for log τ a α 1 is defined by
D a + β H log τ a α 1 = Γ ( α ) Γ ( α β ) log τ a α β 1 , α , β > 0 , a < τ .
Lemma 1
(See [12]). Let β > 0 and ζ C [ 1 , ) L 1 [ 1 , ) . Then the solution of the Hadamard-type differential equation D 1 + β H ζ ( τ ) = 0 is given by
ζ ( τ ) = i = 1 n c i log τ β i
with
I 1 + β H D 1 + β H ζ ( τ ) = ζ ( τ ) + i = 1 n c i log τ β i ,
where c i R ( i = 1 , 2 , , n ) and n 1 < β < n .
Definition 3
(Hilfer fractional derivative [7]). Let n 1 < β < n , 0 α 1 and ζ L 1 ( [ a , b ] ) . The left-sided Hilfer derivative of order β and type α for ζ is defined by
( D a + β , α RL ζ ) ( τ ) = I a + α ( n β ) RL τ d d τ n I a + ( 1 α ) ( n β ) RL ζ ( τ ) = I a + α ( n β ) RL τ d d τ n I a + ( n γ ) RL ζ ( τ ) = I a + α ( n β ) RL D a + γ RL ζ ( τ ) ,
where I a + RL and D a + RL are the Riemann–Liouville integral and derivative, respectively and γ = β + n α β α .
Theorem 1
(See [7]). Suppose that 0 < β < 1 , 0 α 1 , γ = β + α β α and ζ L 1 [ a , b ] . If D γ RL ζ L 1 [ a , b ] , then
I a + β RL D a + β , α RL ζ ( τ ) = I a + γ RL D a + γ RL ζ ( τ ) = ζ ( τ ) ( I a + 1 γ RL ζ ) ( a ) Γ ( γ ) τ a γ 1 .
Definition 4
(Hilfer–Hadamard derivative [8]). Let n 1 < β < n , 0 α 1 and ζ L 1 ( [ a , b ] ) . The left-sided Hilfer–Hadamard derivative of order β and type α for ζ is defined by
( D a + β , α HH ζ ) ( τ ) = I a + α ( n β ) H ε n I a + ( 1 α ) ( n β ) H ζ ( τ ) = I a + α ( n β ) H ε n I a + ( n γ ) H ζ ( τ ) = I a + α ( n β ) H D a + γ H ζ ( τ ) ,
where γ = β + n α α β , I a + ( · ) H and D a + ( · ) H are the Hadamard integral and derivative, respectively. Similarly, the right-sided Hilfer–Hadamard fractional derivative of order β and type α for ζ is defined by
( D b β , α HH ζ ) ( τ ) = I b α ( n β ) H ( ε ) n I b ( 1 α ) ( n β ) H ζ ( τ ) = I b α ( n β ) H ( ε ) n I b ( n γ ) H ζ ( τ ) = I b α ( n β ) H D b γ H ζ ( τ ) ,
where γ = β + n α α β , I b ( · ) H and D b ( · ) H are the Hadamard fractional integral and derivative, respectively.
Theorem 2
(See [7]). Suppose that n 1 < β < n with n = [ β ] + 1 . If ζ L 1 [ a , b ] , then
I a + β H D a + β H ζ ( τ ) = ζ ( τ ) i = 0 n 1 ( ε n i 1 ( I a + n β H ζ ) ) ( a ) Γ ( β i ) log τ a β i 1 .
Theorem 3
(See [6,7]). Suppose that n 1 < β < n , 0 α 1 , γ = β + n α β α and n 1 < γ n with n = [ β ] + 1 . If ζ L 1 [ a , b ] , then
I a + β H D a + β , α HH ζ ( τ ) = I a + γ H D a + γ H ζ ( τ ) = ζ ( τ ) i = 0 n 1 ( ε n i 1 ( I a + n γ H ζ ) ) ( a ) Γ ( γ i ) log τ a γ i 1 .
By Theorem 3, if α = 0 , then the formula reduces to the formula in Theorem 2.
Definition 5
(See [18]). The operator T 1 : X X is increasing if ζ ξ implies T 1 ζ T 1 ξ and the operator T 1 : X X is decreasing if ζ ξ implies T 1 ζ T 1 ξ .
Definition 6
(See [18]). Let 0 < γ < 1 . The operator T 1 : P P is γ -concave if T 1 ( κ ζ ) κ γ T 1 ζ for κ ( 0 , 1 ) and ζ P , and the operator T 2 : P P is sub-homogeneous if T 2 ( κ ζ ) κ T 2 ζ for κ > 0 and ζ P .
In [18], the authors studied the operator equation
T 1 ζ + T 2 ζ = ζ ,
where T 1 and T 2 are monotone operators. They gave the existence and uniqueness of positive solutions for (4).
Lemma 2
(See [18]). Let X be a real Banach space and P be a normal cone in X . Suppose that the operators T 1 , T 2 : P P are increasing, moreover T 1 is γ -concave and T 2 is sub-homogeneous. Assume that the following properties hold:
(i) 
There exists ρ > θ with T 1 ρ P ρ and T 2 ρ P ρ ;
(ii) 
There exists δ 0 > 0 with T 1 ζ δ 0 T 2 ζ for all ζ P .
Then (4) has a unique solution ζ * P ρ . Further, introducing the sequence
T 1 ζ n 1 + T 2 ζ n 1 = ζ n , n N ,
for any initial value ζ 0 P ρ , one has ζ n ζ * as n .
Lemma 3
(See [13]). Let X be a real Banach space and P be a normal cone in X . Suppose that the operator T 1 : P P is increasing, and the operator T 2 : P P is decreasing. Moreover, assume that the following properties hold:
(i) 
For ζ P and κ ( 0 , 1 ) , there exists f i ( κ ) ( κ , 1 ) , i = 1 , 2 , such that
T 1 ( κ ζ ) f 1 ( κ ) T 1 ζ , T 2 ( κ ζ ) 1 f 2 ( κ ) T 2 ζ ;
(ii) 
There exists ρ 0 P ρ such that T 1 ρ 0 + T 2 ρ 0 P ρ , where ρ > θ and P ρ = { ζ X : ζ ρ } .
Then (4) has a unique solution ζ * P ρ . Further, for any initial values ζ 0 , ξ 0 P ρ , defining the sequences
T 1 ζ n 1 + T 2 ξ n 1 = ζ n , T 1 ξ n 1 + T 2 ζ n 1 = ξ n , n N ,
one has ζ n ζ * , ξ n ζ * as n .

3. The Green Function and Bounds

Lemma 4.
Let the function ϱ C ( J , R + ) satisfy 1 e ϱ ( ς ) d ς ς < and γ = β + 2 α β α , such that 1 < γ 2 . Then the function ζ is a solution of the Hilfer–Hadamard-type equation with the boundary conditions
D 1 + β , α HH ζ ( τ ) + ϱ ( τ ) = 0 , τ J , β ( 1 , 2 ] , α [ 0 , 1 ] , ζ ( 1 ) = 0 , ζ ( e ) = i = 1 n i I ε i H ζ ( π i ) + s j = 1 m ω j ζ ( s j )
if and only if
ζ ( τ ) = 1 e G ( τ , ς ) ϱ ( ς ) d ς ς , τ J ,
where
G ( τ , ς ) = G 1 ( τ , ς ) + G 2 ( τ , ς )
with
G 1 ( τ , ς ) : = G 1 ( τ , ς ) β = 1 Γ ( β ) ( log τ ) γ 1 ( log e ς ) β 1 log τ ς β 1 , 1 ς τ e , ( log τ ) γ 1 ( log e ς ) β 1 , 1 τ ς e
and
G 2 ( τ , ς ) = ( log τ ) γ 1 Θ i = 1 n i G 1 ( π i , ς ) ε i + β + ( log τ ) γ 1 Θ j = 1 m s ω j G 1 ( s j , ς ) β .
Here,
Θ : = 1 i = 1 n i I ε i H ( log π i ) γ 1 s j = 1 m ω j log s j γ 1 = 1 i = 1 n i Γ ( γ ) Γ ( γ + ε i ) ( log π i ) γ + ε i 1 s j = 1 m ω j log s j γ 1 0 .
Proof. 
Taking the Hadamard integral of order β in (6), we have
I 1 + β H ( D 1 + β , α HH ) ζ ( τ ) + I 1 + β H ϱ ( τ ) = 0 .
Using Theorem 3, (2), and (3), we obtain
ζ ( τ ) = I 1 + β H ϱ ( τ ) + ( ε I 1 + 2 γ H ζ ) ( 1 ) Γ ( γ ) log τ γ 1 + ( I 1 + 2 γ H ζ ) ( 1 ) Γ ( γ 1 ) log τ γ 2 .
Applying Lemma 1, we can write (9) as
ζ ( τ ) = 1 Γ ( β ) 1 τ log τ ς β 1 ϱ ( ς ) d ς ς + c 1 log τ γ 1 + c 2 log τ γ 2 .
By using the boundary condition (6), we have c 2 = 0 . Then (10) implies
ζ ( τ ) = 1 Γ ( β ) 1 τ log τ ς β 1 ϱ ( ς ) d ς ς + c 1 log τ γ 1 .
Letting τ = e in (11) and (6), we find
c 1 1 Γ ( β ) 1 e log e ς β 1 ϱ ( ς ) d ς ς = i = 1 n i I ε i H ζ ( π i ) + s j = 1 m ω j ζ ( s j ) = i = 1 n i I ε i H c 1 log π i γ 1 1 Γ ( β ) 1 π i log π i ς β 1 ϱ ( ς ) d ς ς + s j = 1 m ω j c 1 log s j γ 1 1 Γ ( β ) 1 s j log s j ς β 1 ϱ ( ς ) d ς ς = c 1 i = 1 n i I ε i H ( log π i ) γ 1 i = 1 n i Γ ( β ) I ε i H 1 π i log π i ς β 1 ϱ ( ς ) d ς ς s Γ ( β ) j = 1 m ω j 1 s j log s j ς β 1 ϱ ( ς ) d ς ς + c 1 j = 1 m s ω j log s j γ 1 .
By the above relations, we get
c 1 1 i = 1 n i I ε i H ( log π i ) γ 1 s j = 1 m ω j log s j γ 1 = i = 1 n i Γ ( β ) I ε i H 1 π i log π i ς β 1 ϱ ( ς ) d ς ς s Γ ( β ) j = 1 m ω j 1 s j log s j ς β 1 ϱ ( ς ) d ς ς + 1 Γ ( β ) 1 e log e ς β 1 ϱ ( ς ) d ς ς .
Therefore, by using (9), we have
c 1 = 1 Θ ( i = 1 n i Γ ( β ) I ε i H 1 π i log π i ς β 1 ϱ ( ς ) d ς ς s Γ ( β ) j = 1 m ω j 1 s j log s j ς β 1 ϱ ( ς ) d ς ς + 1 Γ ( β ) 1 e log e ς β 1 ϱ ( ς ) d ς ς ) ,
and hence
c 1 = 1 Θ ( i = 1 n i Γ ( ε i + β ) 1 π i log π i ς ε i + β 1 ϱ ( ς ) d ς ς + s Γ ( β ) j = 1 m ω j 1 s j log s j ς β 1 ϱ ( ς ) d ς ς 1 Γ ( β ) 1 e log e ς β 1 ϱ ( ς ) d ς ς ) = 1 Θ ( i = 1 n i Γ ( ε i + β ) 1 π i log π i ε i + β 1 log π i ς ε i + β 1 ϱ ( ς ) d ς ς + i = 1 n i Γ ( ε i + β ) π i e log π i ε i + β 1 ϱ ( ς ) d ς ς ) + 1 Θ ( j = 1 m s ω j Γ ( β ) 1 s j ( log s j ) β 1 log s j ς β 1 ϱ ( ς ) d ς ς + j = 1 m s ω j Γ ( β ) s j e ( log s j ) β 1 ϱ ( ς ) d ς ς ) + 1 Γ ( β ) 1 e log e ς β 1 ϱ ( ς ) d ς ς + 1 Θ Θ Γ ( β ) 1 e log e ς β 1 ϱ ( ς ) d ς ς = 1 Γ ( β ) 1 e log e ς β 1 ϱ ( ς ) d ς ς + 1 Θ i = 1 n i 1 e G 1 ( π i , ς ) ε i + β ϱ ( ς ) d ς ς + 1 Θ j = 1 m s ω j 1 e G 1 ( s j , ς ) β ϱ ( ς ) d ς ς .
So, by (11), we get
ζ ( τ ) = 1 Γ ( β ) 1 τ log τ ς β 1 ϱ ( ς ) d ς ς + ( log τ ) γ 1 Γ ( β ) 1 e log e ς β 1 ϱ ( ς ) d ς ς + ( log τ ) γ 1 Θ ( i = 1 n i 1 e G 1 ( π i , ς ) ε i + β ϱ ( ς ) d ς ς + j = 1 m s ω j 1 e G 1 ( s j , ς ) β ϱ ( ς ) d ς ς ) = 1 e G ( τ , ς ) ϱ ( ς ) d ς ς , τ J .
The proof is now completed. □
Lemma 5.
The Green functions G ( τ , ς ) and G 1 ( τ , ς ) have the following properties:
(1) 
G ( τ , ς ) is positive and continuous for ( τ , ς ) J × J ;
(2) 
For fixed β, G 1 ( τ , ς ) : = G 1 ( τ , ς ) β is positive and continuous for ( τ , ς ) J × J ;
(3) 
G 1 ( τ , ς ) is increasing with respect to τ;
(4) 
For ( τ , ς ) J × J with 1 < γ 2 and J = [ 1 , e ] , we have
( i ) G 1 ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1 1 , ( i i ) G ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1 1 Θ .
Proof. 
Parts (1) and (2) are obvious. Indeed, if ( log τ ) γ 1 ( log e ς ) β 1 log τ ς β 1 < 0 , then for τ < e , we have ( log τ ) γ 1 < log τ ς β 1 ( log e ς ) β 1 = log τ ς log e ς β 1 . Setting τ = e t and ς = e s , for some s and t ( 0 s < t < 1 ), we have
t γ 1 < t s 1 s β 1 .
Regarding as γ > β and t > t s 1 s , we get a contradiction. We prove (3). For fixed ς and β , we let
g 1 ( τ ) = 1 Γ ( β ) log τ γ 1 log e ς β 1 log τ ς β 1 , 1 ς τ e ,
g 2 ( τ ) = 1 Γ ( β ) ( log τ γ 1 log e ς β 1 , 1 τ ς e .
Clearly, g 1 is increasing on [ ς , e ] and g 2 is increasing on [ 1 , ς ] . Indeed, if τ 1 τ 2 , and we have
( log τ 2 ) γ 1 ( log e ς ) β 1 log τ 2 ς β 1 [ ( log τ 1 ) γ 1 ( log e ς ) β 1 log τ 1 ς β 1 ] < 0 ,
then we get
( ( log τ 2 ) γ 1 ( log τ 1 ) γ 1 ) log e ς β 1 < log τ 2 ς β 1 log τ 1 ς β 1 < 0 ,
which is a contradiction. Hence, g 1 is increasing. In a similar way, we can show that g 2 is increasing. Also, for any τ 1 , τ 2 with τ 1 τ 2 , we have the inequalities
g 1 ( τ 1 ) g 1 ( τ 2 ) , i . e . , G ( τ 1 , ς ) G ( τ 2 , ς ) for ς τ 1 , g 2 ( τ 1 ) g 2 ( τ 2 ) , i . e . , G ( τ 1 , ς ) G ( τ 2 , ς ) for τ 2 ς , g 2 ( τ 1 ) g 2 ( ς ) = g 1 ( ς ) g 1 ( τ 2 ) , i . e . , G ( τ 1 , ς ) G ( τ 2 , ς ) for τ 1 ς τ 2 .
Finally, we get
G ( τ 1 , ς ) G ( τ 2 , ς ) for 1 τ 1 τ 2 e .
We prove item (4). First, given that γ , β ( 1 , 2 ] , we have Γ ( β ) 1 and τ J , hence
( log τ ) γ 1 ( log 10 ) γ 1 ( log τ ) γ 1 1 .
So, log e ς β 1 1 . Therefore,
1 Γ ( β ) ( log τ ) γ 1 log e ς β 1 ( log 10 ) γ 1 ( log τ ) γ 1 1 ,
which shows (i). Next, for ( τ , ς ) J × J , by using (7) and (8), also according to (9), we know that Θ > 0 , and then we have
G ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1 = G 1 ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1 + ( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) i = 1 n i G 1 ( π i , ς ) ε i + β + ( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) j = 1 m s ω j G 1 ( s j , ς ) β 1 + 1 Θ i = 1 n i G 1 ( π i , ς ) ε i + β + j = 1 m s ω j G 1 ( s j , ς ) β 1 + i = 1 n i ( log π i ) γ 1 ( log e ς ) ε i + β 1 Θ Γ ( ε i + β ) + j = 1 m s ω j ( log s j ) γ 1 ( log e ς ) β 1 Θ Γ ( β ) = 1 + 1 Θ Γ ( β ) ( i = 1 n Γ ( β ) i Γ ( ε i + β ) ( log π i ) γ 1 log e ς ε i + β 1 + j = 1 m s ω j ( log s j ) γ 1 log e ς β 1 ) 1 + 1 Θ Γ ( γ ) i = 1 n Γ ( γ ) i Γ ( ε i + γ ) ( log π i ) γ + ε i 1 + j = 1 m s ω j ( log s j ) γ 1 1 + 1 Θ i = 1 n Γ ( γ ) i Γ ( ε i + γ ) ( log π i ) γ + ε i 1 + j = 1 m s ω j ( log s j ) γ 1 = 1 Θ ,
which shows (ii). □
To illustrate our result, we give the following numerical example.
Example 1.
We put α = 1 , β = 2 , γ = 2 and m , n = 1 and π i , s j = 2 and ε i , i , ω j , s = 1 and τ = e and ς = 1 and Θ 0.066 as a positive constant. Then we have
( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) = 0.769 Θ 11.651 , G 1 ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1 = 0.768
G ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1 = G 1 ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1   + ( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) i = 1 n i G 1 ( π i , ς ) ε i + β   + ( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) j = 1 m s ω j G 1 ( s j , ς ) β = 0.768 + 0.769 Θ [ i = 1 n i ( log π i ) γ 1 ( log e ς ) ε i + β 1 Γ ( ε i + β )   + j = 1 m s ω j ( log s j ) γ 1 ( log e ς ) β 1 Γ ( β ) ] = 0.768 + 11.651 ( log 2 ) ( log e ) 2 2 + ( log 2 ) ( log e ) = 0.768 + 11.651 0.693 2 + 0.693 = 12.880 .
According to the given constants and (9), we have
G ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1 12.880 , 1 Θ 15.1515 .
Finally,
G ( τ , ς ) ( log 10 ) γ 1 ( log τ ) γ 1 1 Θ .

4. Main Results

In this section, we will use Lemmas 2 and 3 to establish the uniqueness of the solution to (1). We define a partial ordering ⪯ by ζ ξ if and only if ξ ζ P ( P is defined in Section 2). This space is equipped with a partial order
ζ , ξ P , ζ ξ ζ ( τ ) ξ ( τ ) , τ J .
If 0 ζ ( τ ) ξ ( τ ) , then
sup τ J | ζ ( τ ) | ( log 10 ) γ 1 ( log τ ) γ 1 sup τ J | ξ ( τ ) | ( log 10 ) γ 1 ( log τ ) γ 1 ζ ξ ,
and, hence, P is a normal cone in X . To prove the results, we utilize the following assumptions:
( A 1
ν , μ : J R + are continuous and 0 < 1 e ν ( ς ) d ς ς , 1 e μ ( ς ) d ς ς < ;
( A 2
φ , ψ : J × R + R + are increasing with respect to the second argument, φ ( τ , 0 ) , ψ ( τ , 0 ) 0 , τ J ;
( A 3
For a bounded function ζ , the functions φ ( τ , ( ( log 10 ) γ 1 ( log τ ) γ 1 ) ζ ) and
ψ ( τ , ( ( log 10 ) γ 1 ( log τ ) γ 1 ) ζ ) are bounded with respect to τ for τ J ;
( A 4
ψ ( τ , κ ζ ) κ ψ ( τ , ζ ) for κ ( 0 , 1 ) , τ J , ζ R + , and there exists 0 < γ < 1 with condition φ ( τ , κ ζ ) κ γ φ ( τ , ζ ) for κ ( 0 , 1 ) , τ J , ζ R + ;
( A 5
There exists a fixed δ > 0 with ν ( τ ) φ ( τ , ζ ) δ μ ( τ ) ψ ( τ , ζ ) , τ J , ζ R + ;
( A 6
φ : J × R + R + is increasing in ζ and ψ : J × R + R + is decreasing in ζ ; also, φ ( τ , 0 ) , ψ ( τ , 0 ) 0 , τ J ;
( A 7
For κ ( 0 , 1 ) , there exist f i ( κ ) ( κ , 1 ) ( i = 1 , 2 ) with τ J , ζ R + , φ ( τ , κ ζ ) φ ( τ , ζ ) f 1 ( τ ) and ψ ( τ , κ ζ ) ψ ( τ , ζ ) f 2 ( κ ) .
Suppose that ρ ( τ ) = ( log τ ) γ 1 , τ J . As sup τ J | ρ ( τ ) | ( log 10 ) γ 1 ( log τ ) γ 1 < , we have ρ P . In the proof, we will consider P ρ = ζ X : ζ ρ .
From Lemma 4, we know that (1) has an integral formulation given by
ζ ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς + 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς , τ J ,
where G ( τ , ς ) is defined in (7). By using Lemma 2, we define T 1 : P X and T 2 : P X by
T 1 ζ ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς , T 2 ζ ( τ ) = 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς .
Then ζ is the solution of (1) if and only if
T 1 ζ + T 2 ζ = ζ .
Given Definitions 1 and 2 and using Lemmas 4 and 2, we introduce the operators T 1 , T 2 : P P associated with (1) as
T 1 ζ ( τ ) = 1 Γ ( β ) 1 τ log τ ς β 1 ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Γ ( β ) 1 e log e ς β 1 ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Θ i = 1 n i 1 e G 1 ( π i , ς ) ε i + β ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Θ j = 1 m s ω j 1 e G 1 ( s j , ς ) β ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς
and
T 2 ζ ( τ ) = 1 Γ ( β ) 1 τ log τ ς β 1 μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Γ ( β ) 1 e log e ς β 1 μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Θ i = 1 n i 1 e G 1 ( π i , ς ) ε i + β μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Θ j = 1 m s ω j 1 e G 1 ( s j , ς ) β μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς .
We have T 1 ( ζ ) ( τ ) + T 2 ( ζ ) ( τ ) = ζ ( τ ) .
Lemma 6.
If ( A 1 ) ( A 3 ) hold, then T 1 , T 2 : P P .
Proof. 
Let ζ P . Then ζ ( τ ) ( log 10 ) γ 1 ( log τ ) γ 1 < , τ J . From A 3 , ζ is bounded and therefore there exists M ζ > 0 such that
φ ς , ( ( log 10 ) γ 1 ( log ς ) γ 1 ) ζ ( ς ) ( log 10 ) γ 1 ( log ς ) γ 1 M ζ .
Moreover, from ( A 1 ) , ( A 2 ) and Lemma 5, we have
T 1 ζ ( τ ) ( log 10 ) γ 1 ( log τ ) γ 1 = 1 Γ ( β ) ( ( log 10 ) γ 1 ( log τ ) γ 1 ) 1 τ log τ ς β 1 ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Γ ( β ) ( ( log 10 ) γ 1 ( log τ ) γ 1 ) 1 e log e ς β 1 ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) i = 1 n i 1 e G 1 ( π i , ς ) ε i + β ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) j = 1 m s ω j 1 e G 1 ( s j , ς ) β ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς 1 e 1 Θ ν ( ς ) φ ς , ( ( log 10 ) γ 1 ( log ς ) γ 1 ) ζ ( ς ) ( log 10 ) γ 1 ( log ς ) γ 1 d ς ς M ζ Θ 1 e ν ( ς ) d ς ς < .
By Lemma 5, T 1 ζ ( τ ) > 0 on J , and, therefore, T 1 : P P . Similarly, if ζ P , then ζ ( τ ) ( log 10 ) γ 1 ( log τ ) γ 1 < for all τ J . From ( A 3 ), ζ is bounded, and hence there exists N ζ > 0 such that
ψ ς , ( ( log 10 ) γ 1 ( log ς ) γ 1 ) ζ ( ς ) ( log 10 ) γ 1 ( log ς ) γ 1 N ζ .
Moreover, from ( A 1 ) , ( A 2 ) , and Lemma 5, we have
T 2 ζ ( τ ) ( log 10 ) γ 1 ( log τ ) γ 1 = 1 Γ ( β ) ( ( log 10 ) γ 1 ( log τ ) γ 1 ) 1 τ log τ ς β 1 μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Γ ( β ) ( ( log 10 ) γ 1 ( log τ ) γ 1 ) 1 e log e ς β 1 μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) i = 1 n i 1 e G 1 ( π i , ς ) ε i + β μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς + ( log τ ) γ 1 Θ ( ( log 10 ) γ 1 ( log τ ) γ 1 ) j = 1 m s ω j 1 e G 1 ( s j , ς ) β μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς . 1 e 1 Θ μ ( ς ) ψ ς , ( ( log 10 ) γ 1 ( log ς ) γ 1 ) ζ ( ς ) ( log 10 ) γ 1 ( log ς ) γ 1 d ς ς N ζ Θ 1 e μ ( ς ) d ς ς < .
Also, T 2 ζ ( τ ) > 0 on J , and therefore, T 2 : P P . □
Lemma 7.
Suppose that φ , ψ satisfy ( A 1 ) , ( A 2 ) and ( A 4 ) . Then T 1 : P P is an increasing γ -concave operator and T 2 : P P is increasing sub-homogeneous.
Proof. 
First, we show T 1 and T 2 are two increasing operators. Note that
ζ , ξ P , ζ ξ ζ ( τ ) ξ ( τ ) , τ J .
By ( A 1 ) , ( A 2 ) and Lemma 5, we have
T 1 ζ ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ξ ( ς ) ) d ς ς = T 1 ξ ( τ ) .
Therefore, T 1 ζ T 1 ξ . We have
T 2 ζ ( τ ) = 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ξ ( ς ) ) d ς ς = T 2 ξ ( τ ) .
So, T 2 ζ T 2 ξ .
Secondly, we show that T 1 is a γ -concave operator. For any κ , γ ( 0 , 1 ) and ζ P , from ( A 1 ) , ( A 2 ) , ( A 4 ) and Lemma 5, we obtain
T 1 ( κ ζ ) ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , κ ζ ( ς ) ) d ς ς κ γ 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς = κ γ T 1 ζ ( τ ) .
Hence, T 1 ( κ ζ ) κ γ T 1 ζ , κ , γ ( 0 , 1 ) and ζ P .
We show that T 2 is sub-homogeneous. For κ ( 0 , 1 ) and ζ P , from ( A 1 ) , ( A 2 ) , ( A 4 ) and Lemma 5, we obtain
T 2 ( κ ζ ) ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , κ ζ ( ς ) ) d ς ς κ 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς = κ T 2 ζ ( τ ) ,
that is, T 2 ( κ ζ ) κ T 2 ζ , κ ( 0 , 1 ) and ζ P . □
Lemma 8.
Let ( A 1 ) ( A 3 ) be satisfied. Then T 1 ρ P ρ and T 2 ρ P ρ .
Proof. 
If ρ P , then ρ ( τ ) ( log 10 ) γ 1 ( log τ ) γ 1 < . From ( A 3 ) , there exists M ρ > 0 with
φ ς , ( ( log 10 ) γ 1 ( log ς ) γ 1 ) ρ ( ς ) ( log 10 ) γ 1 ( log ς ) γ 1 M ρ .
Let
g 1 = j = 1 m s ω j Θ 1 s n G 1 ( s j , ς ) ν ( ς ) φ ( ς , 0 ) d ς ς , g 2 = { M ρ ( log e ς ) β 1 Γ ( β ) 1 e ν ( ς ) d ς ς + M ρ Θ i = 1 n i 1 e G 1 ( π i , ς ) ε i + β ν ( ς ) d ς ς + M ρ Θ j = 1 m s ω j 1 e G 1 ( s j , ς ) β ν ( ς ) d ς ς } .
From ( A 1 ) , ( A 2 ) and Lemmas 4 and 5, we get
T 1 ρ ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ( log ς ) γ 1 ) d ς ς 1 e G ( τ , ς ) ν ( ς ) φ ( ς , 0 ) d ς ς 1 e G 2 ( τ , ς ) ν ( ς ) φ ( ς , 0 ) d ς ς = ( log τ ) γ 1 Θ 1 e i = 1 n i G 1 ( π i , ς ) ε i + β + j = 1 m s ω j G 1 ( s j , ς ) β ν ( ς ) φ ( ς , 0 ) d ς ς ( log τ ) γ 1 Θ 1 e j = 1 m s ω j G 1 ( s j , ς ) β ν ( ς ) φ ( ς , 0 ) d ς ς = ( log τ ) γ 1 Θ j = 1 m s ω j 1 e G 1 ( s j , ς ) ν ( ς ) φ ( ς , 0 ) d ς ς j = 1 m s ω j Θ 1 s n G 1 ( s j , ς ) ν ( ς ) φ ( ς , 0 ) d ς ς ( log τ ) γ 1 = g 1 ( log τ ) γ 1 = g 1 ρ ( τ ) .
From ( A 3 ),
T 1 ρ ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ρ ( ς ) ) d ς ς = 1 e G ( τ , ς ) ν ( ς ) φ ς , ( ( log 10 ) γ 1 ( log ς ) γ 1 ) ρ ( ς ) ( log 10 ) γ 1 ( log ς ) γ 1 d ς ς 1 e G ( τ , ς ) ν ( ς ) M ρ d ς ς 1 e M ρ ( log τ ) γ 1 ( log e ς ) β 1 Γ ( β ) ν ( ς ) d ς ς + 1 e M ρ ( log τ ) γ 1 Θ i = 1 n i G 1 ( π i , ς ) ε i + β ν ( ς ) d ς ς + 1 e M ρ ( log τ ) γ 1 Θ j = 1 m s ω j G 1 ( s j , ς ) β ν ( ς ) d ς ς = { M ρ ( log e ς ) β 1 Γ ( β ) 1 e ν ( ς ) d ς ς + M ρ Θ 1 e i = 1 n i G 1 ( π i , ς ) ε i + β ν ( ς ) d ς ς + M ρ Θ 1 e j = 1 m s ω j G 1 ( s j , ς ) β ν ( ς ) d ς ς } ( log τ ) γ 1 = g 2 ( log τ ) γ 1 = g 2 ρ ( τ ) .
Note φ ( ς , 0 ) 0 , G 1 ( s j , ς ) β 0 and 1 e ν ( ς ) d ς ς > 0 . From G 1 ( s j , ς ) β ν ( ς ) φ ( ς , 0 ) 0 , we obtain
1 s n G 1 ( s j , ς ) β ν ( ς ) φ ( ς , 0 ) d ς ς > 0 .
From ( A 2 ) and ( A 3 ), we have φ ( ς , 0 ) φ ( ς , ρ ( ς ) ) M ρ . Then, considering 1 e ν ( ς ) d ς ς > 0 , we get
j = 1 m s ω j Θ 1 s n G 1 ( s j , ς ) ν ( ς ) φ ( ς , 0 ) d ς ς M ρ ( log e ς ) β 1 Γ ( β ) 1 e ν ( ς ) d ς ς + M ρ Θ i = 1 n i 1 e G 1 ( π i , ς ) ε i + β ν ( ς ) d ς ς + M ρ Θ j = 1 m s ω j 1 e G 1 ( s j , ς ) ν ( ς ) d ς ς > 0 .
We can assume 0 < g 1 g 2 . So, g 1 ρ ( τ ) T 1 ρ ( τ ) g 2 ρ ( τ ) . Hence, T 1 ρ P ρ .
Also,
T 2 ρ ( τ ) = 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ρ ( ς ) ) d ς ς j = 1 m s ω j Θ 1 s n G 1 ( s j , ς ) μ ( ς ) ψ ( ς , 0 ) d ς ς ρ ( τ ) ,
T 2 ρ ( τ ) = 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ρ ( ς ) ) d ς ς { M ρ ( log e ς ) β 1 Γ ( β ) 1 e μ ( ς ) d ς ς + M ρ Θ i = 1 n i 1 e G 1 ( π i , ς ) ε i + β μ ( ς ) d ς ς + M ρ Θ j = 1 m s ω j 1 e G 1 ( s j , ς ) β μ ( ς ) d ς ς } ρ ( τ ) .
On the basis of ψ ( ς , 0 ) 0 and 1 e μ ( ς ) d ς ς > 0 , ψ ( ς , 0 ) ψ ( ς , ρ ( ς ) ) M ρ , we deduce T 2 ρ P ρ . □
By using Lemmas 6–8, and assuming ( A 1 ) ( A 7 ) , we can establish the existence and uniqueness of a positive solution for (1).
Theorem 4.
Let ( A 1 ) ( A 5 ) hold. Then (1) has a unique solution ζ * P ρ . For ζ 0 P ρ , defining a sequence by
ζ n + 1 ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ n ( ς ) ) d ς ς + 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ n ( ς ) ) d ς ς ,
with n = 0 , 1 , 2 , , we have ζ n ( τ ) ζ * ( τ ) as n , where G ( τ , ς ) is given in (7).
Proof. 
From Lemmas 6–8, we prove item (ii) of Lemma 2. For ζ P , by ( A 1 ) , ( A 2 ) , ( A 5 ) , and Lemma 5, we have
T 1 ζ ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς δ 0 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς = δ 0 T 2 ζ ( τ ) .
So, we achieve T 1 ζ δ 0 T 2 ζ for ζ P . By using Lemma 5, ζ * P ρ is a unique solution of T 1 ζ + T 2 ζ = ζ . Therefore,
ζ * ( τ ) = 1 e G ( τ , ς ) [ ν ( ς ) φ ( ς , ζ * ( ς ) ) + μ ( ς ) ψ ( ς , ζ * ( ς ) ) ] d ς ς ,
and it is the unique positive solution of (1) in P ρ . Moreover, for every ζ 0 P ρ , the sequence ζ n = T 1 ζ n 1 + T 2 ζ n 1 satisfies ζ n ( τ ) ζ * ( τ ) , that is,
ζ n + 1 ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ n ( ς ) ) d ς ς + 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ n ( ς ) ) d ς ς .
Hence, the proof is complete. □
Lemma 9
(See [18]). Assume that φ, ψ satisfy ( A 1 ) , ( A 3 ) and ( A 6 ) . Then T 1 ρ + T 2 ρ P ρ .
Theorem 5.
Suppose that ( A 1 ) , ( A 3 ) , ( A 6 ) , ( A 7 ) hold. Then (1) has a unique positive solution ζ * P ρ . For the given initial values ζ 0 , ξ 0 P ρ , constructing the sequences
ζ n + 1 ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ n ( ς ) ) d ς ς + 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ξ n ( ς ) ) d ς ς
and
ξ n + 1 ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ξ n ( ς ) ) d ς ς + 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ n ( ς ) ) d ς ς ,
with n = 0 , 1 , 2 , and τ J , we get ζ n ( τ ) ζ * ( τ ) , ξ n ( τ ) ζ * ( τ ) , n , where G ( τ , ς ) is defined in (7).
Proof. 
From ( A 1 ) , ( A 6 ) , and Lemma 5, T 1 : P P is increasing and T 2 : P P is decreasing. Moreover, by ( A 7 ) , we have
T 1 ( κ ζ ) ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , κ ζ ( ς ) ) d ς ς f 1 ( κ ) 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ ( ς ) ) d ς ς = f 1 ( κ ) T 1 ζ ( τ ) ,
T 2 ( κ ζ ) ( τ ) = 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , κ ζ ( ς ) ) d ς ς 1 f 2 ( κ ) 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ ( ς ) ) d ς ς = 1 f 2 ( κ ) T 2 ζ ( τ ) .
We can show that T 1 and T 1 satisfy (5). By Lemma 9, and considering that item (ii) of Lemma 3 holds, the operator equation T 1 ζ + T 2 ζ = ζ has a unique solution ζ * P ρ . For the given initial values ζ 0 , ξ 0 P ρ , getting the sequences
ζ n = T 1 ζ n 1 + T 2 ξ n 1 , ξ n = T 1 ξ n 1 + T 2 ζ n 1 , n = 1 , 2 , ,
we get ζ n ( τ ) ζ * ( τ ) , ξ n ( τ ) ζ * ( τ ) as n . Evidently, ζ * is the unique positive solution for (1) in P ρ . For the given initial values ζ 0 , ξ 0 P ρ , making the sequences
ζ n + 1 ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ζ n ( ς ) ) d ς ς + 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ξ n ( ς ) ) d ς ς
and
ξ n + 1 ( τ ) = 1 e G ( τ , ς ) ν ( ς ) φ ( ς , ξ n ( ς ) ) d ς ς + 1 e G ( τ , ς ) μ ( ς ) ψ ( ς , ζ n ( ς ) ) d ς ς , τ J ,
we have ζ n ( τ ) ζ * ( τ ) and ξ n ( τ ) ζ * ( τ ) . □

5. Examples

Example 2.
Let τ J and consider the problem
D 1 + 3 2 , 1 2 HH ζ ( τ ) + τ e τ 2 + ζ ( τ ) 1 2 ( log 10 ) 3 4 ( log τ ) 3 4 + τ e 2 τ 1 + ζ ( τ ) 1 3 ( log 10 ) 3 4 ( log τ ) 3 4 = 0 , ζ ( 1 ) = 0 , ζ ( e ) = i = 1 2 i I ε i H ζ ( π i ) + Γ ( 5 2 ) 6 j = 1 3 ω j ζ ( s j ) ,
where
β = 3 2 , α = 1 2 , γ = 7 4 , n = 2 , m = 3 , ε 1 = 1 2 , ε 2 = 3 2 , π 1 = e 1 2 , π 2 = e 1 3 , s = Γ ( 5 2 ) 6 , ω 1 = 1 4 , ω 2 = 1 3 , ω 3 = 1 2 , s 1 = e 1 3 , s 2 = e 1 2 , s 3 = 2 , 1 = 1 , 2 = 6 .
We take
r ν ( τ ) = τ e τ , μ ( τ ) = τ e 2 τ , φ ( τ , ζ ( τ ) ) = 2 + ζ ( τ ) 1 2 ( log 10 ) 3 4 ( log τ ) 3 4 , ψ ( τ , ζ ( τ ) ) = 1 + ζ ( τ ) 1 3 ( log 10 ) 3 4 ( log τ ) 3 4 ,
Θ = 1 i = 1 2 i Γ ( 7 4 ) Γ ( 7 4 + ε i ) ( log π i ) 3 4 + ε i Γ ( 5 2 ) 6 j = 1 3 ω j log s j 3 4 0.388 > 0 = 1 0.919 1.225 ( 0.420 ) + 5.514 3.078 ( 0.084 ) 0.221 0.109 + 0.178 + 0.379 .
We know that Θ 0.388 . Finally, ν and μ are continuous functions in τ J , and
1 e ν ( ς ) d ς ς = 1 e ς e ς d ς ς = e 1 e e < , 1 e μ ( ς ) d ς ς = 1 e ς e 2 ς d ς ς = 1 2 e 2 1 2 e 2 e < .
Evidently, φ, ψ C ( J × R + , R + ) are continuous and increasing functions with respect to the second argument, so φ ( τ , 0 ) > 0 , ψ ( τ , 0 ) > 0 and the items ( A 1 ) and ( A 2 ) are satisfied, where 0 < ζ < M ζ ,
φ ( τ , ( ( log 10 ) γ 1 ( log τ ) γ 1 ) ζ ) = 2 ( log 10 ) 3 4 ( log τ ) 3 4 + ( ( log 10 ) 3 4 ( log τ ) 3 4 ) 1 2 1 ζ ( τ ) 1 2 2 ( log 10 ) 3 4 ( log e ) 3 4 + M ζ , ψ ( τ , ( ( log 10 ) γ 1 ( log τ ) γ 1 ) ζ ) = 1 ( log 10 ) 3 4 ( log τ ) 3 4 + ( ( log 10 ) 3 4 ( log τ ) 3 4 ) 1 3 1 ζ ( τ ) 1 3 1 ( log 10 ) 3 4 ( log e ) 3 4 + M ζ 3 .
Therefore, the item ( A 3 ) is satisfied. Taking γ = 1 2 for τ J , κ ( 0 , 1 ) and ζ R + , we have
φ ( τ , κ ζ ) = 2 + κ 1 2 ζ ( τ ) 1 2 ( log 10 ) 3 4 ( log τ ) 3 4 κ 1 2 2 + ζ ( τ ) 1 2 ( log 10 ) 3 4 ( log τ ) 3 4 = κ 1 2 φ ( τ , ζ ) , ψ ( τ , κ ζ ) = 1 + κ 1 3 ζ ( τ ) 1 3 ( log 10 ) 3 4 ( log τ ) 3 4 κ 1 + ζ ( τ ) 1 3 ( log 10 ) 3 4 ( log τ ) 3 4 = κ ψ ( τ , ζ ) .
Therefore, the item ( A 4 ) is satisfied. Moreover, if we take δ ( 0 , 1 ] , for τ J , ζ R + , then we obtain
ν ( τ ) φ ( τ , ζ ) = τ e τ ( 2 + ζ ( τ ) 1 2 ) ( log 10 ) 3 4 ( log τ ) 3 4 τ e 2 τ ( 1 + ζ ( τ ) 1 3 ) ( log 10 ) 3 4 ( log τ ) 3 4 δ μ ( τ ) ψ ( τ , ζ ) .
All the items of Theorem 4 hold. Hence, the mentioned problem has a unique positive solution ζ * P ρ , where ρ ( τ ) = ( log τ ) 3 2 , τ J . Letting ζ 0 P ρ and making the sequence
ζ n + 1 ( τ ) = 1 e G ( τ , ς ) ς e ς 2 + ( ζ n ( ς ) ) 1 2 ( log 10 ) 3 4 ( log ς ) 3 4 + ς e 2 ς 1 + ( ζ n ( ς ) ) 1 3 ( log 10 ) 3 4 ( log ς ) 3 4 d ς ς , n = 0 , 1 , ,
we have ζ n ( τ ) ζ * ( τ ) , where
G ( τ , ς ) = G 1 ( τ , ς ) + G 2 ( τ , ς )
and
G 1 ( τ , ς ) : = G 1 τ , ς , 3 2 = 1 0.886 ( log τ ) 3 4 ( log e ς ) 1 2 log τ ς 1 2 , 1 ς τ e , ( log τ ) 3 4 ( log e ς ) 1 2 , 1 τ ς e
and
G 2 ( τ , ς ) = ( log τ ) 3 4 0.388 i = 1 2 i G 1 ( π i , ς ) ε i + β + ( log τ ) 3 4 0.388 j = 1 3 s ω j G 1 ( s j , ς ) β = ( log τ ) 3 4 0.388 G 1 ( e 1 2 , ς , 2 ) + 6 G 1 ( e 1 3 , ς , 3 ) + ( log τ ) 3 4 0.388 Γ ( 5 2 ) 6 1 4 G 1 e 1 3 , ς , 3 2 + 1 3 G 1 e 1 2 , ς , 3 2 + 1 2 G 1 2 , ς , 3 2 .

6. Conclusions

In this paper, we have investigated the existence and uniqueness of positive solutions for a class of Hilfer–Hadamard-type fractional differential equations subject to boundary value problems, including integral boundary conditions. The results obtained in this work extend and complement several earlier contributions in the literature by generalizing Hadamard-type fractional models to the broader Hilfer–Hadamard framework. However, this generalization is valid only under the strong assumptions ( A 1 )–( A 7 ) detailed in Section 4. These hypotheses significantly restrict the admissible nonlinearities, coefficients, and boundary data, and therefore limit the class of problems for which the main theorems are applicable. In this sense, the present work should be viewed as a conditional extension rather than a full generalization to all Hilfer–Hadamard systems. An illustrative example was presented to validate the theoretical findings. This confirms the effectiveness of the developed methodology in addressing nonlinear fractional differential equations with complex boundary conditions within the prescribed assumptions. Overall, the results of this study contribute to the ongoing development of fractional calculus, particularly in the analysis of boundary value problems involving nonlocal operators. Future research may focus on extending these techniques to more general classes of fractional operators, investigating numerical methods for approximating solutions, or exploring applications in real-world models arising in science and engineering.

Author Contributions

Conceptualization, H.R.; methodology, H.R. and H.A.; formal analysis, H.R., H.A. and M.B.; investigation, M.B.; writing—original draft preparation, H.R. and H.A.; writing—review and editing, M.B. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Data sharing is not applicable to this paper as no data sets were generated or analyzed during the current research.

Conflicts of Interest

The authors declare no conflicts of interest.

References

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MDPI and ACS Style

Rasouli, H.; Afshari, H.; Bohner, M. Existence and Uniqueness of Positive Solutions for Hilfer–Hadamard-Type Fractional Differential Equations with γ-Concave and Sub-Homogeneous Operators. Fractal Fract. 2026, 10, 449. https://doi.org/10.3390/fractalfract10070449

AMA Style

Rasouli H, Afshari H, Bohner M. Existence and Uniqueness of Positive Solutions for Hilfer–Hadamard-Type Fractional Differential Equations with γ-Concave and Sub-Homogeneous Operators. Fractal and Fractional. 2026; 10(7):449. https://doi.org/10.3390/fractalfract10070449

Chicago/Turabian Style

Rasouli, Hasan, Hojjat Afshari, and Martin Bohner. 2026. "Existence and Uniqueness of Positive Solutions for Hilfer–Hadamard-Type Fractional Differential Equations with γ-Concave and Sub-Homogeneous Operators" Fractal and Fractional 10, no. 7: 449. https://doi.org/10.3390/fractalfract10070449

APA Style

Rasouli, H., Afshari, H., & Bohner, M. (2026). Existence and Uniqueness of Positive Solutions for Hilfer–Hadamard-Type Fractional Differential Equations with γ-Concave and Sub-Homogeneous Operators. Fractal and Fractional, 10(7), 449. https://doi.org/10.3390/fractalfract10070449

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