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Article

Coupled System of (k, ψ)-Hilfer and (k, ψ)-Caputo Sequential Fractional Differential Equations with Non-Separated Boundary Conditions

by
Furkan Erkan
1,
Nuket Aykut Hamal
1,
Sotiris K. Ntouyas
2,
Jessada Tariboon
3 and
Phollakrit Wongsantisuk
4,*
1
Department of Mathematics, Ege University, Bornova 35100, Izmir, Türkiye
2
Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece
3
Intelligent and Nonlinear Dynamic Innovations Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand
4
Department of Electronics Engineering Technology, College of Industrial Technology, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand
*
Author to whom correspondence should be addressed.
Axioms 2025, 14(9), 685; https://doi.org/10.3390/axioms14090685
Submission received: 13 July 2025 / Revised: 1 August 2025 / Accepted: 4 September 2025 / Published: 7 September 2025
(This article belongs to the Special Issue Advances in Nonlinear Analysis and Boundary Value Problems)

Abstract

This paper is concerned with the existence and uniqueness of solutions for a coupled system of ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo sequential fractional differential equations with non-separated boundary conditions. We make use of the Banach contraction mapping principle to obtain the uniqueness result, while two existence results are proved by using Leray–Schauder nonlinear alternative and Krasnosel’skiĭ’s fixed point theorem. The obtained results are illustrated by numerical examples.

1. Introduction

Fractional derivatives and integrals have become highly significant tools in a variety of fields, including chemistry, biology, physics, and finance. Owing to their wide range of applications, numerous new fractional derivative operators have been introduced and studied in the literature. Kilbas et al., in Ref. [1], introduced the Riemann–Liouville fractional derivative, the Caputo fractional derivative and the Riemann–Liouville fractional derivative with respect to another function. R. Hilfer, in Ref. [2], introduced the Hilfer fractional derivative, which is a combination of the Riemann–Liouville and Caputo fractional derivatives. For some applications involving the Hilfer fractional derivative, such as in control theory, the analysis of dynamical systems, and the modeling of anomalous diffusion processes, see [3,4,5,6]. Building on Kilbas’s idea of the Riemann–Liouville fractional derivative with respect to another function, Almeida introduced the Caputo fractional derivative with respect to another function in Ref. [7]. Similarly, Souza and De Oliveira, in Ref. [8], introduced the ψ -Hilfer fractional derivative, which is the derivative of the Hilfer fractional derivative with respect to another function.
Mubeen and Habibullah, in Ref. [9], introduced the k-Riemann–Liouville fractional integral by using the k-gamma function, which was introduced by Diaz and Pariguan in Ref. [10]. Similarly, Romero et al., in Ref. [11], introduced the k-Riemann–Liouville fractional derivative. Kwun et al., in Ref. [12], introduced the ( k , ψ ) -Riemann–Liouville fractional integral. Kucche and Mali, in Ref. [13], introduced the ( k , ψ ) -Riemann–Liouville, ( k , ψ ) -Caputo and ( k , ψ ) -Hilfer fractional derivative operators. In the literarture, ( k , ψ )-Hilfer nonlocal integro-multi-point boundary value problems were studied in Ref. [14], ( k , ψ ) )-Hilfer nonlocal fractional coupled systems in Ref. [15], ( k , ψ ) )-Hilfer Langevin fractional coupled systems in Ref. [16], ( k , ψ )-Hilfer variational problem in Ref. [17] and controllability of fractional dynamical systems with ( k , ψ )-Hilfer fractional derivative in Ref. [18].
The Hilfer fractional derivative, commonly encountered in the analysis of boundary value problems, generally necessitates a zero initial condition. This requirement considerably restricts its use in scenarios with more general boundary specifications. To overcome this limitation, one can employ a sequential combination of the Hilfer and Caputo fractional derivatives. This approach enables the study of boundary value problems with nonzero initial conditions. The combined use of the ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo derivatives offers a high degree of flexibility and generality in obtaining solutions to fractional differential equations. This representation provides a richer mathematical framework for modeling diverse physical and engineering systems. In recent years, several researchers have studied these fractional derivatives and explored their use in solving nonlinear fractional differential equations. In Ref. [19], the authors investigated the following mixed Hilfer and Caputo fractional Riemann–Stieltjes integral differential equations with non-separated boundary conditions:
D 0 + α , β ; ψ H D 0 + γ ; ψ C ϰ t = f t , ϰ t , I 0 + c , ψ ϰ t , 0 b ϰ s d ν ( s ) , t [ 0 , b ] , ϰ ( 0 ) + λ 1 ϰ ( b ) = 0 , D 0 + θ + γ 1 ; ψ C ϰ ( 0 ) + λ 2 D 0 + θ + γ 1 ; ψ C ϰ ( b ) = 0 ,
where D 0 + α , β ; ψ H and D 0 + γ ; ψ C , D 0 + θ + γ 1 ; ψ C , 0 < α , β , γ < 1 , θ = α + β ( 1 α ) , θ + γ > 1 are the ψ -Hilfer and ψ -Caputo fractional derivative operators, respectively. Moreover, λ 1 , λ 2 R , I 0 + c , ψ is the Riemann–Liouville fractional integral operator of order c > 0 with respect to a function ψ , f : [ 0 , b ] × R × R × R R is a nonlinear continuous function, 0 b ϰ s d ν ( s ) is the Riemann–Stieltjes integral and ν : [ 0 , b ] R is a function of bounded variation.
In Ref. [20], the authors investigated the following system:
D 0 + α 1 , β 1 ; ψ H D 0 + γ 1 ; ψ C ϰ t = f t , ϰ t , y t , I 0 + c 2 ; ψ y t , t [ 0 , b ] , D 0 + α 2 , β 2 ; ψ H D 0 + γ 2 ; ψ C y t = g t , ϰ t , I 0 + c 1 ; ψ ϰ t , y t , t [ 0 , b ] ϰ 0 + λ 1 y b = 0 , D 0 + θ 1 + γ 1 1 ; ψ C ϰ 0 + λ 2 D 0 + θ 2 + γ 2 1 ; ψ C y b = 0 , y 0 + δ 1 ϰ b = 0 , D 0 + θ 2 + γ 2 1 ; ψ C y 0 + δ 2 D 0 + θ 1 + γ 1 1 ; ψ C ϰ b = 0 ,
where D 0 + α ι ˙ , β ι ˙ ; ψ H and D 0 + γ ι ˙ ; ψ C , 0 < α ι ˙ , β ι ˙ , γ ι ˙ < 1 , θ ι ˙ = α ι ˙ + β ι ˙ α ι ˙ β ι ˙ , with α ι ˙ + γ ι ˙ > 1 , ι ˙ = 1 , 2 , are the ψ -Hilfer fractional derivative and ψ -Caputo fractional derivative, respectively. Moreover, λ ι ˙ , δ ι ˙ R with λ ι ˙ δ ι ˙ 1 , ι ˙ = 1 , 2 , I 0 + c ι ˙ ; ψ are the Riemann–Liouville fractional integral of order c ι ˙ > 0 , ι ˙ = 1 , 2 with respect to a function ψ , and f , g : [ 0 , b ] × R × R × R R are nonlinear continuous functions.
Recently, in Ref. [21], a sequential boundary value problems including both the ( k , ψ ) -Hilfer and the ( k , ψ ) -Caputo fractional derivatives supplemented with non-separated boundary conditions of the form
D 0 + ϑ , β ; ψ k , H D 0 + γ ; ψ k , C ϰ δ = φ δ , ϰ δ , δ [ 0 , b ] , λ 1 ϰ 0 + λ 2 ϰ b = 0 , λ 3 D 0 + ζ k + γ k ; ψ k , C ϰ 0 + λ 4 D 0 + ζ k + γ k ; ψ k , C ϰ b = 0 ,
were investigated, where the differential operator D 0 + ϑ , β ; ψ k , H is the ( k , ψ ) -Hilfer fractional derivative of order 0 < ϑ < 1 with the parameters 0 β 1 . D 0 + γ ; ψ k , C and D 0 + ζ k + γ k ; ψ k , C are the ( k , ψ ) -Caputo fractional derivative of orders 0 < γ < 1 and ζ k + γ k > 0 , respectively, where ζ k = ϑ + β ( k ϑ ) . Moreover, k > 0 , λ ι R ( ι = 1 , 2 , 3 , 4 ) and φ : [ 0 , b ] × R R is a continuous function. Existence and uniqueness of solutions are established through the application of fixed-point theorems by Banach, Schaefer, and Krasnosel’skiĭ, along with the Leray–Schauder nonlinear alternative.
In the present paper, we analyze the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo sequential fractional differential equations with non-separated boundary conditions of the following form:
D 0 + α 1 , β 1 ; ψ 1 k 1 , H D 0 + γ 1 ; ψ 1 k 1 , C ϰ t = f t , ϰ t , y t , I 0 + φ 2 ; ψ 2 k 2 y t , t [ 0 , b ] , D 0 + α 2 , β 2 ; ψ 2 k 2 , H D 0 + γ 2 ; ψ 2 k 2 , C y t = g t , ϰ t , I 0 + φ 1 ; ψ 1 k 1 ϰ t , y t , t [ 0 , b ] , λ 1 ϰ 0 + λ 2 y b = 0 , λ 3 D 0 + θ k 1 + γ 1 k 1 ; ψ 1 k 1 , C ϰ 0 + λ 4 D 0 + θ k 2 + γ 2 k 2 ; ψ 2 k 2 , C y b = 0 , δ 1 y 0 + δ 2 ϰ b = 0 , δ 3 D 0 + θ k 2 + γ 2 k 2 ; ψ 2 k 2 , C y 0 + δ 4 D 0 + θ k 1 + γ 1 k 1 ; ψ 1 k 1 , C ϰ b = 0 ,
where D 0 + α 1 , β 1 ; ψ 1 k 1 , H and D 0 + α 2 , β 2 ; ψ 2 k 2 , H are ( k 1 , ψ 1 ) -Hilfer and ( k 2 , ψ 2 ) -Hilfer fractional derivatives operator of orders 0 < α 1 , α 2 < 1 with the parameters 0 β 1 , β 2 1 . The differential operators D 0 + γ 1 ; ψ 1 k 1 , C and D 0 + θ k 1 + γ 1 k 1 ; ψ 1 k 1 , C are the ( k 1 , ψ 1 ) -Caputo fractional derivatives of orders 0 < γ 1 < 1 and θ k 1 + γ 1 > k 1 , respectively, where θ k 1 = α 1 + β 1 ( k 1 α 1 ) . Similarly, D 0 + γ 2 ; ψ 2 k 2 , C and D 0 + θ k 2 + γ 2 k 2 ; ψ 2 k 2 , C are the ( k 2 , ψ 2 ) -Caputo fractional derivatives of orders 0 < γ 2 < 1 and θ k 2 + γ 2 > k 2 , respectively, where θ k 2 = α 2 + β 2 ( k 2 α 2 ) . I 0 + φ ι ˙ ; ψ ι ˙ k ι ˙ are ( k ι ˙ , ψ ι ˙ ) -Riemann fractional integral of order φ ι ˙ > 0 , ( ι ˙ = 1 , 2 ) . Moreover, k 1 , k 2 > 0 , λ ι ˙ , δ ι ˙ R ( ι ˙ = 1 , 2 , 3 , 4 ) and f , g : [ 0 , b ] × R × R × R R are continuous functions.
It is worth noting that the present study is motivated by the generality of the ( k , ψ ) -Hilfer fractional derivative operator, which encompasses several well-known fractional derivative operators as special cases through appropriate choices of k , ψ and the parameter β . Specifically, the following apply:
(1)
When β = 0 , it reduces to the ( k , ψ ) -Riemann–Liouville fractional derivative;
In particular, for ψ ( t ) = t , it becomes the k-Riemann–Liouville fractional derivative.
(2)
When β = 1 , it becomes the ( k , ψ ) -Caputo fractional derivative;
Again, for ψ ( t ) = t , it corresponds to the k-Caputo fractional derivative.
(3)
For ψ ( t ) = t ρ : It yields the k-Hilfer–Katugampola fractional derivative;
Setting β = 0 gives the k-Katugampola fractional derivative;
Setting β = 1 gives the k-Caputo–Katugampola fractional derivative.
(4)
For ψ ( t ) = log t : It yields the k-Hilfer–Hadamard fractional derivative;
Setting β = 0 gives the k-Hadamard fractional derivative;
Setting β = 1 gives the k-Caputo–Hadamard fractional derivative.
Similarly, the ( k , ψ ) -Riemann–Liouville fractional integral operators, which appear in the fractional differential equations, specialize to the following:
  • The ψ -Riemann–Liouville fractional integral;
  • The k-Riemann–Liouville fractional integral;
  • The classical Riemann–Liouville fractional integral.
The above apply by taking k = 1 , ψ ( t ) = t , and both k = 1 and ψ ( t ) = t , respectively.
The combination of ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional derivative operators represents a novel approach in fractional calculus, which has not been extensively explored in the existing literature. Therefore, our contributions are expected to progress the ongoing development in this emerging area of research.
To the best of our knowledge, this work is the first to address coupled systems of boundary value problems of this particular form. Consequently, there are no directly comparable results available in the existing literature.
This paper is organized as follows: In Section 2, we provide the definitions and lemmas necessary for understanding the manuscript. In Section 3, we use fixed point theory to obtain our main results, proving the uniqueness of the solution using Banach’s contraction mapping principle, while two existence results are established through Leray–Schauder’s alternative and Krasnosel’skiĭ’s fixed point theorem. Furthermore, the obtained results are illustrated by numerical examples in Section 4.

2. Preliminaries

In this section, we recall some definitions, lemmas, and a remark that will be used later. In the following, we suppose that ψ C n ( [ a , b ] , R ) and ψ is a positively continuous and increasing function satisfying the condition ψ ( t ) > 0 for each t [ a , b ] .
Definition 1
([12]). Let α , k R + and f L 1 ( [ a , b ] , R ) . Then the ( k , ψ ) -Riemann–Liouville fractional integral of order α for a function f is defined by
I k a + α ; ψ f ( t ) = 1 k Γ k ( α ) a t ψ ( s ) ψ ( t ) ψ ( s ) α k 1 f s d s .
Definition 2
([13,21]). Let α , k R + and f C ( [ a , b ] , R ) . Then the ( k , ψ ) -Caputo fractional derivative of order α for a function f is defined by
D a + α ; ψ k , C f ( t ) = I a + n k α ; ψ k k ψ ( t ) d d t n f ( t ) = 1 k Γ k ( n k α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) n α k 1 k ψ ( s ) d d s n f ( s ) d s .
where n = α k is the ceiling function of α k .
Definition 3
([13]). Let α , k R + , 0 β 1 and f C ( [ a , b ] , R ) . Then the ( k , ψ ) -Hilfer fractional derivative of order α and type β for a function f is defined by
D a + α , β ; ψ k , H f ( t ) = I a + β ( n k α ) ; ψ k k ψ ( t ) d d t n I a + ( 1 β ) ( n k α ) ; ψ k f ( t ) , n = α k .
Lemma 1
([21]). Let α , k R + and n = α k . Suppose that f C n ( [ a , b ] , R ) , Then
I a + α ; ψ k D a + α ; ψ k , C f ( t ) = f ( t ) j = 0 n 1 ψ ( t ) ψ ( a ) j Γ k ( j k + k ) k ψ ( t ) d d t j f ( t ) t = a .
Lemma 2
([13]). Let α , k R + , β [ 0 , 1 ] , θ k = α + β ( n k α ) and n = α k . Suppose that f C n ( [ a , b ] , R ) and I k a + n k θ k ; ψ f C n ( [ a , b ] , R ) . Then
I k a + α ; ψ D a + α , β ; ψ k , H f ( t ) = f ( t ) j = 1 n ψ ( t ) ψ ( a ) θ k k j Γ k ( θ k j k + k ) k ψ ( t ) d d t n j I k a + n k θ k ; ψ f ( t ) t = a .
Lemma 3
([13]). Let α 1 , α 2 , k R + . Then,
I k a + α 1 ; ψ I k a + α 2 ; ψ f ( t ) = I k a + α 1 + α 2 ; ψ f ( t )
Lemma 4
([21]). Let α 1 , α 2 , k R + with α 2 > α 1 . Then,
D a + α 1 ; ψ k , C I k a + α 2 ; ψ f ( t ) = I k a + α 2 α 1 ; ψ f ( t ) .
Lemma 5
([13,21]). Let α , k R + and μ R such that μ k > 1 . Then,
(i) 
I k a + α ; ψ ψ ( t ) ψ ( a ) μ k = Γ k ( μ + k ) Γ k ( μ + k + α ) ψ ( t ) ψ ( a ) μ + α k
(ii) 
D a + α ; ψ k , C ψ ( t ) ψ ( a ) μ k = Γ k ( μ + k ) Γ k ( μ + k α ) ψ ( t ) ψ ( a ) μ α k .
In the following lemma, a linear variant of the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional boundary value problem (2) is considered, which allows transforming the given nonlinear problem into an equivalent fixed-point formulation.
Lemma 6.
Assume that Δ 1 , Δ 2 0 and h 1 , h 2 C ( [ 0 , b ] , R ) . Then, the solution of the system
D 0 + α 1 , β 1 ; ψ 1 k 1 , H D 0 + γ 1 ; ψ 1 k 1 , C ϰ t = h 1 ( t ) , t [ 0 , b ] , D 0 + α 2 , β 2 ; ψ 2 k 2 , H D 0 + γ 2 ; ψ 2 k 2 , C y t = h 2 ( t ) , t [ 0 , b ] , λ 1 ϰ 0 + λ 2 y b = 0 , λ 3 D 0 + θ k 1 + γ 1 k 1 ; ψ 1 k 1 , C ϰ 0 + λ 4 D 0 + θ k 2 + γ 2 k 2 ; ψ 2 k 2 , C y b = 0 , δ 1 y 0 + δ 2 ϰ b = 0 , δ 3 D 0 + θ k 2 + γ 2 k 2 ; ψ 2 k 2 , C y 0 + δ 4 D 0 + θ k 1 + γ 1 k 1 ; ψ 1 k 1 , C ϰ b = 0 ,
is equivalent to the integral equations
ϰ ( t ) = I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 ( t ) + λ 2 δ 2 Δ 1 I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 ( b ) λ 2 δ 1 Δ 1 I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 ( b ) + Ω 1 ( t ) I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 ( b ) Ω 2 ( t ) I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 ( b ) , t [ 0 , b ] ,
and
y t = I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 ( t ) + λ 2 δ 2 Δ 1 I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 ( b ) λ 1 δ 2 Δ 1 I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 ( b ) Ω 3 ( t ) I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 ( b ) + Ω 4 ( t ) I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 ( b ) , t [ 0 , b ] ,
where
Δ 1 = λ 1 δ 1 λ 2 δ 2 , Δ 2 = λ 3 δ 3 λ 4 δ 4 , Φ ι ˙ η ( ζ ) = ( ψ ι ˙ ( ζ ) ψ ι ˙ ( 0 ) ) η k ι ˙ Γ k ι ˙ ( η + k ι ˙ ) , ( ι ˙ = 1 , 2 ) ,
and
Ω 1 ( t ) = δ 4 Δ 2 λ 2 Δ 1 λ 4 δ 2 Φ 1 θ k 1 + γ 1 k 1 ( b ) + λ 3 δ 1 Φ 2 θ k 2 + γ 2 k 2 ( b ) + λ 4 Φ 1 θ k 1 + γ 1 k 1 ( t ) ,
Ω 2 ( t ) = λ 4 Δ 2 λ 2 Δ 1 δ 2 δ 3 Φ 1 θ k 1 + γ 1 k 1 ( b ) + δ 1 δ 4 Φ 2 θ k 2 + γ 2 k 2 ( b ) + δ 3 Φ 1 θ k 1 + γ 1 k 1 ( t ) ,
Ω 3 ( t ) = δ 4 Δ 2 δ 2 Δ 1 λ 1 λ 4 Φ 1 θ k 1 + γ 1 k 1 ( b ) + λ 2 λ 3 Φ 2 θ k 2 + γ 2 k 2 ( b ) + λ 3 Φ 2 θ k 2 + γ 2 k 2 ( t ) ,
Ω 4 ( t ) = λ 4 Δ 2 δ 2 Δ 1 λ 1 δ 3 Φ 1 θ k 1 + γ 1 k 1 ( b ) + λ 2 δ 4 Φ 2 θ k 2 + γ 2 k 2 ( b ) + δ 4 Φ 2 θ k 2 + γ 2 k 2 ( t ) .
Proof. 
Assume that ( ϰ , y ) is a solution of the system (3). Operating I k 1 0 + α 1 ; ψ 1 and I k 2 0 + α 2 ; ψ 2 on both sides of the first and second equations in Equation (3), respectively, and using Lemma 2, we obtain for t [ 0 , b ] ,
D 0 + γ 1 ; ψ 1 k 1 , C ϰ t = I 0 + α 1 ; ψ 1 k 1 h 1 t + ( ψ 1 ( t ) ψ 1 ( 0 ) ) θ k 1 k 1 1 Γ k 1 ( θ k 1 ) c 1 ,
and
D 0 + γ 2 ; ψ 2 k 2 , C y t = I 0 + α 2 ; ψ 2 k 2 h 2 t + ( ψ 2 ( t ) ψ 2 ( 0 ) ) θ k 2 k 2 1 Γ k 2 ( θ k 2 ) c 2 ,
where
c 1 = I k 1 0 + k 1 θ k 1 ; ψ 1 h 1 ( t ) t = 0 and c 2 = I k 2 0 + k 2 θ k 2 ; ψ 2 h 2 ( t ) t = 0 .
Now, by taking the fractional integral I 0 + γ 1 ; ψ 1 k 1 and I 0 + γ 2 ; ψ 2 k 2 on both sides of Equations (6) and (7), respectively, and applying Lemma 1, we obtain
ϰ t = I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 t + ( ψ 1 ( t ) ψ 1 ( 0 ) ) θ k 1 + γ 1 k 1 1 Γ k 1 ( θ k 1 + γ 1 ) c 1 + d 1 = I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 t + Φ 1 θ k 1 + γ 1 k 1 ( t ) c 1 + d 1 ,
and
y t = I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 t + ( ψ 2 ( t ) ψ 2 ( 0 ) ) θ k 2 + γ 2 k 2 1 Γ k 2 ( θ k 2 + γ 2 ) c 2 + d 2 = I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 t + Φ 2 θ k 2 + γ 2 k 2 ( t ) c 2 + d 2 .
By Lemma 4, we have
D 0 + θ k 1 + γ 1 k 1 ; ψ 1 k 1 , C ϰ t = I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 t + c 1 ,
and
D 0 + θ k 2 + γ 2 k 2 ; ψ 2 k 2 , C y t = I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 t + c 2 .
From Equation (10) and applying the conditions λ 1 ϰ 0 + λ 2 y b = 0 , and λ 3 D 0 + θ k 1 + γ 1 k 1 ; ψ 1 k 1 , C ϰ 0 + λ 4 D 0 + θ k 2 + γ 2 k 2 ; ψ 2 k 2 , C y b = 0 , we obtain
λ 1 d 1 + λ 2 d 2 + λ 2 I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 b + λ 2 Φ 2 θ k 2 + γ 2 k 2 ( b ) c 2 = 0 ,
λ 3 c 1 + λ 4 c 2 + λ 4 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b = 0 .
From (11) and applying the conditions δ 1 y 0 + δ 2 ϰ b = 0 , δ 3 D 0 + θ k 2 + γ 2 k 2 ; ψ 2 k 2 , C y 0 + δ 4 D 0 + θ k 1 + γ 1 k 1 ; ψ 1 k 1 , C ϰ b = 0 , we obtain
δ 1 d 2 + δ 2 d 1 + δ 2 I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 b + δ 2 Φ 1 θ k 1 + γ 1 k 1 ( b ) c 1 = 0 ,
δ 3 c 2 + δ 4 c 1 + δ 4 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b = 0 .
From (13) and (15), we have
c 1 = λ 4 λ 3 δ 3 λ 4 δ 4 δ 4 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b δ 3 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b ,
c 2 = δ 4 λ 3 δ 3 λ 4 δ 4 λ 4 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b λ 3 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b .
From (12) and (14), we have
d 1 = λ 2 δ 2 λ 1 δ 1 λ 2 δ 2 ( I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 b + λ 4 Φ 1 θ k 1 + γ 1 k 1 ( b ) λ 3 δ 3 λ 4 δ 4 δ 4 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b δ 3 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b ) λ 2 δ 1 λ 1 δ 1 λ 2 δ 2 ( I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 b + δ 4 Φ 2 θ k 2 + γ 2 k 2 ( b ) λ 3 δ 3 λ 4 δ 4 λ 4 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b λ 3 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b ) ,
and
d 2 = λ 2 δ 2 λ 1 δ 1 λ 2 δ 2 ( I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 b + δ 4 Φ 2 θ k 2 + γ 2 k 2 ( b ) λ 3 δ 3 λ 4 δ 4 λ 4 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b λ 3 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b ) λ 1 δ 2 λ 1 δ 1 λ 2 δ 2 ( I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 b + λ 4 Φ 1 θ k 1 + γ 1 k 1 ( b ) λ 3 δ 3 λ 4 δ 4 δ 4 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b δ 3 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b ) .
Replacing the values c 1 , c 2 , d 1 , and d 2 in (8) and (9), we obtain
ϰ ( t ) = I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 t + λ 4 Φ 1 θ k 1 + γ 1 k 1 ( t ) Δ 2 δ 4 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b δ 3 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b + λ 2 δ 2 Δ 1 ( I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 b + λ 4 Φ 1 θ k 1 + γ 1 k 1 ( b ) Δ 2 δ 4 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b δ 3 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b ) λ 2 δ 1 Δ 1 ( I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 b + δ 4 Φ 2 θ k 2 + γ 2 k 2 ( b ) Δ 2 λ 4 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b λ 3 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b ) = I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 ( t ) + λ 2 δ 2 Δ 1 I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 ( b ) λ 2 δ 1 Δ 1 I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 ( b ) + Ω 1 ( t ) I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 ( b ) Ω 2 ( t ) I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 ( b ) , t [ 0 , b ] ,
and
y t = I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 t + δ 4 Φ 2 θ k 2 + γ 2 k 2 ( t ) Δ 2 λ 4 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b λ 3 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b + λ 2 δ 2 Δ 1 ( I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 b + δ 4 Φ 2 θ k 2 + γ 2 k 2 ( b ) Δ 2 λ 4 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b λ 3 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b ) λ 1 δ 2 Δ 1 ( I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 b + λ 4 Φ 1 θ k 1 + γ 1 k 1 ( b ) Δ 2 δ 4 I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 b δ 3 I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 b ) = I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 ( t ) + λ 2 δ 2 Δ 1 I 0 + α 2 + γ 2 ; ψ 2 k 2 h 2 ( b ) λ 1 δ 2 Δ 1 I 0 + α 1 + γ 1 ; ψ 1 k 1 h 1 ( b ) Ω 3 ( t ) I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 h 1 ( b ) + Ω 4 ( t ) I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 h 2 ( b ) , t [ 0 , b ] .
Conversely, by applying the ( k 1 , ψ 1 ) -Caputo fractional derivatives of orders γ i ( i = 1 , 2 ) to Equations (4) and (5), respectively, we obtain Equations (6) and (7), together with Equation (2). Applying the ( k 2 , ψ 2 ) -Hilfer fractional derivatives of orders α i ( i = 1 , 2 ) to Equations (6) and (7) together with Equation (2), respectively, we obtain the first two parts of Equation (3). To obtain the first conditions at lines 3 and 4 of Equation (3), we evaluate Equations (4) and (5) at t = 0 and t = b . Then, to obtain the second conditions at lines 3 and 4 of Equation (3), we applying the ( k 1 , ψ 1 ) -Caputo fractional derivatives of orders θ k i + γ i k i ( i = 1 , 2 ) to Equations (4) and (5), respectively, at t = 0 and t = b . Through direct computation, we verify that these second conditions are also satisfied. □

3. Main Results

Let X = C ( [ 0 , b ] , R ) be the Banach space of all continuous functions from [ 0 , b ] to R equipped with the norm ϰ X = sup { | ϰ ( t ) | : t [ 0 , b ] } . The product space X × Y , · X × Y is a Banach space with norm ( ϰ , y ) X × Y = ϰ X + y Y for ( ϰ , y ) X × Y .
In view of Lemma 6, we define an operator T : X × Y X × Y by
T ϰ , y ( t ) = T 1 ϰ , y ( t ) T 2 ϰ , y ( t ) ,
where
T 1 ϰ , y ( t ) = I 0 + α 1 + γ 1 ; ψ 1 k 1 f ϰ , y ( t ) + λ 2 δ 2 Δ 1 I 0 + α 1 + γ 1 ; ψ 1 k 1 f ϰ , y ( b ) λ 2 δ 1 Δ 1 I 0 + α 2 + γ 2 ; ψ 2 k 2 g ϰ , y ( b ) + Ω 1 ( t ) I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 f ϰ , y ( b ) Ω 2 ( t ) I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 g ϰ , y ( b ) , t [ 0 , b ] ,
and
T 2 ϰ , y ( t ) = I 0 + α 2 + γ 2 ; ψ 2 k 2 g ϰ , y ( t ) + λ 2 δ 2 Δ 1 I 0 + α 2 + γ 2 ; ψ 2 k 2 g ϰ , y ( b ) λ 1 δ 2 Δ 1 I 0 + α 1 + γ 1 ; ψ 1 k 1 f ϰ , y ( b ) Ω 3 ( t ) I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 f ϰ , y ( b ) + Ω 4 ( t ) I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 g ϰ , y ( b ) , t [ 0 , b ] ,
where we used the notations
f ϰ , y ( ξ ) = f ξ , ϰ ξ , y ξ , I 0 + φ 2 ; ψ 2 k 2 y ξ , ξ = t , b , g ϰ , y ( ξ ) = g ξ , ϰ ξ , I 0 + φ 1 ; ψ 1 k 1 ϰ ξ , y ξ , ξ = t , b .
For computational convenience, we set
Θ ι ˙ = 1 + Φ ι ˙ φ ι ˙ ( b ) , ι ˙ = 1 , 2 , Λ 1 = Φ 1 α 1 + γ 1 ( b ) 1 + λ 2 δ 2 Δ 1 + Ω 1 ( b ) Φ 1 α 1 θ k 1 + k 1 ( b ) , Λ 2 = λ 2 δ 1 Φ 2 α 2 + γ 2 ( b ) Δ 1 + Ω 2 ( b ) Φ 2 α 2 θ k 2 + k 2 ( b ) , Λ 3 = λ 1 δ 2 Φ 1 α 1 + γ 1 ( b ) Δ 1 + Ω 3 ( b ) Φ 1 α 1 θ k 1 + k 1 ( b ) , Λ 4 = Φ 2 α 2 + γ 2 ( b ) 1 + λ 2 δ 2 Δ 1 + Ω 4 ( b ) Φ 2 α 2 θ k 2 + k 2 ( b ) , Λ 1 * = Λ 1 Φ 1 α 1 + γ 1 ( b ) and Λ 4 * = Λ 4 Φ 2 α 2 + γ 2 ( b ) .
In the first result, Banach’s contraction mapping principle is used to prove the existence and uniqueness of solutions for the system in Equation (2).
Theorem 1.
Let f , g : [ 0 , b ] × R × R × R R be continuous functions. Assume that the following condition is satisfied:
  • (H1) There exists constants ν , ω > 0 such that
    f ( t , ϰ 1 , y 1 , z 1 ) f ( t , ϰ 2 , y 2 , z 2 ) ν ϰ 2 ϰ 1 + y 2 y 1 + z 2 z 1 , g ( t , ϰ 1 , y 1 , z 1 ) g ( t , ϰ 2 , y 2 , z 2 ) ω ϰ 2 ϰ 1 + y 2 y 1 + z 2 z 1 ,
for all t [ 0 , b ] and ϰ ι ˙ , y ι ˙ , z ι ˙ R , ι ˙ = 1 , 2 .
If
( Λ 1 + Λ 3 ) ν Θ 2 + ( Λ 2 + Λ 4 ) ω Θ 1 < 1 ,
where Λ i , i = 1 , 2 , 3 , 4 are given by Equation (16), then the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions in Equation (2) has a unique solution ( ϰ , y ) on [ 0 , b ] .
Proof. 
Let B ρ = { ( ϰ , y ) X × Y : ( ϰ , y ) X × Y ρ } be a closed and bounded ball with
ρ Λ 1 + Λ 3 M + Λ 2 + Λ 4 N 1 Λ 1 + Λ 3 ) ν Θ 2 + Λ 2 + Λ 4 ) ω Θ 1 ,
where M = sup t [ 0 , b ] f ( t , 0 , 0 , 0 ) : t [ 0 , b ] < and N = sup t [ 0 , b ] g ( t , 0 , 0 , 0 ) : t [ 0 , b ] < .
By assumption ( H 1 ) , it follows that
f ϰ , y ( ξ ) = f ξ , ϰ ξ , y ξ , I 0 + φ 2 ; ψ 2 k 2 y ξ f ξ , ϰ ξ , y ξ , I 0 + φ 2 ; ψ 2 k 2 y ξ f ( ξ , 0 , 0 , 0 ) + f ( ξ , 0 , 0 , 0 ) ν ϰ ( ξ ) + y ( ξ ) + I 0 + φ 2 ; ψ 2 k 2 y ( ξ ) + M ν ϰ X + y Y + Φ 2 φ 2 ( b ) y Y + M ν ϰ X + y Y + Φ 2 φ 2 ( b ) y Y + Φ 2 φ 2 ( b ) ϰ X + M = ν 1 + Φ 2 φ 2 ( b ) ϰ X + y Y + M ν Θ 2 ρ + M ,
and
g ϰ , y ( ξ ) = g ξ , ϰ ξ , I 0 + φ 1 ; ψ 1 k 1 ϰ ξ , y ξ ω ϰ ( ξ ) + I 0 + φ 1 ; ψ 1 k 1 ϰ ( ξ ) + y ( ξ ) + N ω ϰ X + Φ 1 φ 1 ( b ) ϰ X + y Y + N ω Θ 1 ρ + N .
Let us first show that T ( B ρ ) B ρ . For each ( ϰ , y ) B ρ , we have
T 1 ϰ , y ( t ) 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ , y ( s ) d s + λ 2 δ 2 Δ 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ , y ( s ) d s + λ 2 δ 1 Δ 1 k 2 Γ k 2 ( α 2 + γ 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 + γ 2 k 2 1 g ϰ , y ( s ) d s + Ω 1 ( b ) k 1 Γ k 1 ( α 1 θ k 1 + k 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 θ k 1 k 1 f ϰ , y ( s ) d s + Ω 2 ( b ) k 2 Γ k 2 ( α 2 θ k 2 + k 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 θ k 2 k 2 g ϰ , y ( s ) d s ν Θ 2 ρ + M k 1 Γ k 1 ( α 1 + γ 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 + γ 1 k 1 1 d s + λ 2 δ 2 ( ν Θ 2 ρ + M ) Δ 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 + γ 1 k 1 1 d s + λ 2 δ 1 ( ω Θ 1 ρ + N ) Δ 1 k 2 Γ k 2 ( α 2 + γ 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 + γ 2 k 2 1 d s + Ω 1 ( b ) ( ν Θ 2 ρ + M ) k 1 Γ k 1 ( α 1 θ k 1 + k 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 θ k 1 k 1 d s + Ω 2 ( b ) ( ω Θ 1 ρ + N ) k 2 Γ k 2 ( α 2 θ k 2 + k 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 θ k 2 k 2 d s = Φ 1 α 1 + γ 1 ( b ) 1 + λ 2 δ 2 Δ 1 + Ω 1 ( b ) Φ 1 α 1 θ k 1 + k 1 ( b ) ν Θ 2 ρ + M + λ 2 δ 1 Φ 2 φ 2 ( b ) Δ 1 + Ω 2 ( b ) Φ 2 α 2 θ k 2 + k 2 ( b ) ω Θ 1 ρ + N = Λ 1 ν Θ 2 + Λ 2 ω Θ 1 ρ + Λ 1 M + Λ 2 N .
Therefore, we deduce that
T 1 ϰ , y X Λ 1 ν Θ 2 + Λ 2 ω Θ 1 ρ + Λ 1 M + Λ 2 N .
In a similar way of computation, we get
T 2 ϰ , y Y Λ 3 ν Θ 2 + Λ 4 ω Θ 1 ρ + Λ 1 M + Λ 2 N .
From the two inequalities in Equations (18) and (19) above, we can conclude that
T ϰ , y X × Y = T 1 ϰ , y X + T 2 ϰ , y Y Λ 1 ν Θ 2 + Λ 2 ω Θ 1 ρ + Λ 1 M + Λ 2 N + Λ 3 ν Θ 2 + Λ 4 ω Θ 1 ρ + Λ 3 M + Λ 4 N = Λ 1 + Λ 3 ν Θ 2 + Λ 2 + Λ 4 ω Θ 1 ρ + Λ 1 + Λ 3 M + Λ 2 + Λ 4 N ρ ,
which yields that T ( B ρ ) B ρ .
By assumption ( H 1 ) , it follows that
f ϰ 2 , y 2 ( ξ ) f ϰ 1 , y 1 ( ξ ) = f ξ , ϰ 2 ξ , y 2 ξ , I 0 + φ 2 ; ψ 2 k 2 y 2 ξ f ξ , ϰ 1 ξ , y 1 ξ , I 0 + φ 2 ; ψ 2 k 2 y 1 ξ ν ϰ 2 ϰ 1 + y 2 y 1 + I 0 + φ 2 ; ψ 2 k 2 y 2 I 0 + φ 2 ; ψ 2 k 2 y 1 ν ϰ 2 ϰ 1 X + y 2 y 1 Y + Φ 2 φ 2 ( b ) y 2 y 1 Y ν ϰ 2 ϰ 1 X + y 2 y 1 Y + Φ 2 φ 2 ( b ) y 2 y 1 Y + Φ 2 φ 2 ( b ) ϰ 2 ϰ 1 X = ν 1 + Φ 2 φ 2 ( b ) ϰ 2 ϰ 1 X + y 2 y 1 Y ν Θ 2 ϰ 2 ϰ 1 X + y 2 y 1 Y
and, similarly,
g ϰ 2 , y 2 ( ξ ) g ϰ 1 , y 1 ( ξ ) = g ξ , ϰ 2 ξ , I 0 + φ 1 ; ψ 1 k 1 ϰ 2 ξ , y 2 ξ g ξ , ϰ 1 ξ , I 0 + φ 1 ; ψ 1 k 1 ϰ 1 ξ , y 1 ξ ω ϰ 2 ϰ 1 + I 0 + φ 1 ; ψ 1 k 1 ϰ 2 I 0 + φ 1 ; ψ 1 k 1 ϰ 1 + y 2 y 1 ω ϰ 2 ϰ 1 X + Φ 1 φ 1 ( b ) ϰ 2 ϰ 1 X + y 2 y 1 Y ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y .
Now, we will show that the operator T is a contraction. For each ( ϰ 1 , y 1 ) , ( ϰ 2 , y 2 ) B ρ and for any t [ 0 , b ] , we have
T 1 ( ϰ 2 , y 2 ) ( t ) T 1 ( ϰ 1 , y 1 ) ( t ) 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ 2 , y 2 ( s ) f ϰ 1 , y 1 ( s ) d s + λ 2 δ 2 Δ 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ 2 , y 2 ( s ) f ϰ 1 , y 1 ( s ) d s + λ 2 δ 1 Δ 1 k 2 Γ k 2 ( α 2 + γ 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 + γ 2 k 2 1 g ϰ 2 , y 2 ( s ) g ϰ 1 , y 1 ( s ) d s + Ω 1 ( b ) k 1 Γ k 1 ( α 1 θ k 1 + k 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 θ k 1 k 1 f ϰ 2 , y 2 ( s ) f ϰ 1 , y 1 ( s ) d s + Ω 2 ( b ) k 2 Γ k 2 ( α 2 θ k 2 + k 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 θ k 2 k 2 g ϰ 2 , y 2 ( s ) g ϰ 1 , y 1 ( s ) d s Φ 1 α 1 + γ 1 ( b ) 1 + λ 2 δ 2 Δ 1 + Ω 1 ( b ) Φ 1 α 1 θ k 1 + k 1 ( b ) ν Θ 2 ϰ 2 ϰ 1 X + y 2 y 1 Y + λ 2 δ 1 Φ 2 α 2 + γ 2 ( b ) Δ 1 + Ω 2 ( b ) Φ 2 α 2 θ k 2 + k 2 ( b ) ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y = Λ 1 ν Θ 2 ϰ 2 ϰ 1 X + y 2 y 1 Y + Λ 2 ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y ,
and consequently, we obtain
T 1 ( ϰ 2 , y 2 ) T 1 ( ϰ 1 , y 1 ) X Λ 1 ν Θ 2 + Λ 2 ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y .
Similarly, we can find that
T 2 ( ϰ 2 , y 2 ) T 2 ( ϰ 1 , y 1 ) Y Λ 3 ν Θ 2 + Λ 4 ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y .
From the inequalities in Equations (20) and (21), we conclude that
T ( ϰ 2 , y 2 ) T ( ϰ 1 , y 1 ) X × Y ( Λ 1 + Λ 3 ) ν Θ 2 + ( Λ 2 + Λ 4 ) ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y .
Since ( Λ 1 + Λ 3 ) ν Θ 2 + ( Λ 2 + Λ 4 ) ω Θ 1 < 1 , the operator T is a contraction. Thus, by Banach’s contraction mapping principle, the operator T has a unique fixed point. Consequently, the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions in Equation (2) has a unique solution on [ 0 , b ] . This completes the proof. □
Lemma 7
([22] Leray–Schauder alternative). Let X be a Banach space, and T : X X be a completely continuous operator (i.e., a map restricted to any bounded set in X is compact). Let Ω ( T ) = { ϰ T : ϰ = μ T ( ϰ ) for some 0 < μ < 1 } . Then either the set Ω ( T ) is unbounded, or T has at least one fixed point.
Theorem 2.
Let f , g : [ 0 , b ] × R × R × R R be continuous functions. Assume that the following condition is satisfied:
  • (H2) There exist constants p ι ˙ , q ι ˙ 0 ( ι ˙ = 1 , 2 ) and p 0 > 0 , q 0 > 0 , such that
    f ( t , ϰ , y , z ) p 0 + p 1 ϰ + p 2 y + p 3 z , g ( t , ϰ , y , z ) q 0 + q 1 ϰ + q 2 y + q 3 z ,
    for all t [ 0 , b ] and ϰ , y , z R .
If
( Λ 1 + Λ 3 ) p 1 + ( Λ 2 + Λ 4 ) ( q 1 + q 2 Φ 1 φ 1 ( b ) ) < 1 , ( Λ 1 + Λ 3 ) ( p 2 + p 3 Φ 2 φ 2 ( b ) ) + ( Λ 2 + Λ 4 ) q 3 < 1 ,
where Φ i , i = 1 , 2 are given by Equation (6) and Λ i , i = 1 , 2 , 3 , 4 are given by Equation (16), then the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions in Equation (2) has at least one solution on [ 0 , b ] .
Proof. 
In view of the continuity of functions f and g, the operator T is continuous. Now, we will show that T maps bounded set into bounded set in X × Y . For a positive r, let
B r = { ( ϰ , y ) X × Y : ( ϰ , y ) X × Y r } ,
be a bounded set in X × Y .
By assumption ( H 2 ) , it follows that
f ϰ , y ( ξ ) = f ξ , ϰ ξ , y ξ , I 0 + φ 2 ; ψ 2 k 2 y ξ p 0 + p 1 ϰ ( ξ ) + p 2 y ( ξ ) + p 3 I 0 + φ 2 ; ψ 2 k 2 y ( ξ ) p 0 + p 1 ϰ X + p 2 y Y + p 3 Φ 2 φ 2 ( b ) y Y = p 0 + p 1 ϰ X + p 2 + p 3 Φ 2 φ 2 ( b ) y Y p 0 + p 1 r + p 2 + p 3 Φ 2 φ 2 ( b ) r = p 0 + p 1 + p 2 + p 3 Φ 2 φ 2 ( b ) r : = L 1 ,
and
g ϰ , y ( ξ ) = g ξ , ϰ ξ , I 0 + φ 1 ; ψ 1 k 1 ϰ ξ , y ξ q 0 + q 1 + q 2 Φ 1 φ 1 ( b ) + q 3 r : = L 2 .
For any ( ϰ , y ) B r , we have
T 1 ( ϰ , y ) ( t ) Φ 1 α 1 + γ 1 ( b ) 1 + λ 2 δ 2 Δ 1 + Ω 1 ( b ) Φ 1 α 1 θ k 1 + k 1 ( b ) L 1 + λ 2 δ 1 Φ 2 α 2 + γ 2 ( b ) Δ 1 + Ω 2 ( b ) Φ 2 α 2 θ k 2 + k 2 ( b ) L 2 ,
which leads to
T 1 ( ϰ , y ) X Λ 1 L 1 + Λ 2 L 2 .
In the same way, we have
T 2 ( ϰ , y ) Y Λ 3 L 1 + Λ 4 L 2 .
Hence,
T ( ϰ , y ) X × Y = T 1 ( ϰ , y ) X + T 2 ( ϰ , y ) Y ( Λ 1 + Λ 3 ) L 1 + ( Λ 2 + Λ 4 ) L 2 ,
which implies the uniformly boundedness property of the operator T.
For the equicontinuity of T, we set t 1 , t 2 [ 0 , b ] with t 1 < t 2 and ( ϰ , y ) B r . Then we have
T 1 ( ϰ , y ) ( t 2 ) T 1 ( ϰ , y ) ( t 1 ) 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 t 1 ψ 1 ( s ) ψ 1 ( t 2 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 ψ 1 ( t 1 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ , y ( s ) d s + 1 k 1 Γ k 1 ( α 1 + γ 1 ) t 1 t 2 ψ 1 ( s ) ψ 1 ( t 2 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ , y ( s ) d s + Ω 1 ( t 2 ) Ω 1 ( t 1 ) k 1 Γ k 1 ( α 1 θ k 1 + k 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 θ k 1 k 1 f ϰ , y ( s ) d s + Ω 2 ( t 2 ) Ω 2 ( t 1 ) k 2 Γ k 2 ( α 2 θ k 2 + k 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 θ k 2 k 2 g ϰ , y ( s ) d s L 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 t 1 ψ 1 ( s ) ψ 1 ( t 2 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 ψ 1 ( t 1 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 d s + L 1 k 1 Γ k 1 ( α 1 + γ 1 ) t 1 t 2 ψ 1 ( s ) ψ 1 ( t 2 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 d s + L 1 Ω 1 ( t 2 ) Ω 1 ( t 1 ) k 1 Γ k 1 ( α 1 θ k 1 + k 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 θ k 1 k 1 d s + L 2 Ω 2 ( t 2 ) Ω 2 ( t 1 ) k 2 Γ k 2 ( α 2 θ k 2 + k 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 θ k 2 k 2 d s = L 1 Φ 1 α 1 + γ 1 ( t 2 ) Φ 1 α 1 + γ 1 ( t 1 ) + Φ 1 α 1 θ k 1 + k 1 ( b ) Ω 1 ( t 2 ) Ω 1 ( t 1 ) + L 2 Φ 2 α 2 θ k 2 + k 2 ( b ) Ω 2 ( t 2 ) Ω 2 ( t 1 ) ,
which implies
T 1 ( ϰ , y ) ( t 2 ) T 1 ( ϰ , y ) ( t 1 ) 0 , as t 1 t 2 independently   of ( ϰ , y ) B r .
In addition, we obtain
T 2 ( ϰ , y ) ( t 2 ) T 2 ( ϰ , y ) ( t 1 ) L 1 Φ 2 α 2 + γ 2 ( t 2 ) Φ 2 α 2 + γ 2 ( t 1 ) + Φ 1 α 2 θ k 1 + k 1 ( b ) Ω 3 ( t 2 ) Ω 3 ( t 1 ) + L 2 Φ 2 α 2 θ k 2 + k 2 ( b ) Ω 4 ( t 2 ) Ω 4 ( t 1 ) .
Then,
T 2 ( ϰ , y ) ( t 2 ) T 2 ( ϰ , y ) ( t 1 ) 0 , as t 1 t 2 independently   of ( ϰ , y ) B r .
Thus, the set T ( B r ) is equicontinuous. By taking into account the Arzelá–Ascoli theorem, T ( B r ) is relatively compact. Then, the operator T is completely continuous.
Finally, we show that the set
Υ = { ( ϰ , y ) X × X : ( ϰ , y ) = μ T ( ϰ , y ) , 0 < μ < 1 } ,
is bounded. For any ( ϰ , y ) Υ , then ( ϰ , y ) = μ T ( ϰ , y ) for some μ ( 0 , 1 ) . Hence, for t [ 0 , b ] , we have
ϰ ( t ) = μ T 1 ( ϰ , y ) ( t ) and y ( t ) = μ T 2 ( ϰ , y ) ( t ) .
Then, we can compute that
ϰ ( t ) = μ T 1 ( ϰ , y ) ( t ) Λ 1 p 0 + p 1 ϰ X + ( p 2 + p 3 Φ 2 φ 2 ( b ) ) y Y + Λ 2 q 0 + ( q 1 + q 2 Φ 1 φ 1 ( b ) ) ϰ X + q 3 y Y ,
and
y ( t ) = μ T 2 ( ϰ , y ) ( t ) Λ 3 p 0 + p 1 ϰ X + ( p 2 + p 3 Φ 2 φ 2 ( b ) ) y Y + Λ 4 q 0 + ( q 1 + q 2 Φ 1 φ 1 ( b ) ) ϰ X + q 3 y Y .
Therefore, we obtain
ϰ X Λ 1 p 0 + p 1 ϰ X + ( p 2 + p 3 Φ 2 φ 2 ( b ) ) y Y + Λ 2 q 0 + ( q 1 + q 2 Φ 1 φ 1 ( b ) ) ϰ X + q 3 y Y ,
and
y Y Λ 3 p 0 + p 1 ϰ X + ( p 2 + p 3 Φ 2 φ 2 ( b ) ) y Y + Λ 4 q 0 + ( q 1 + q 2 Φ 1 φ 1 ( b ) ) ϰ X + q 3 y Y ,
which yield
ϰ X + y Y ( Λ 1 + Λ 3 ) p 0 + ( Λ 2 + Λ 4 ) q 0 + ( Λ 1 + Λ 3 ) p 1 + ( Λ 2 + Λ 4 ) ( q 1 + q 2 Φ 1 φ 1 ( b ) ) ϰ X + ( Λ 1 + Λ 3 ) ( p 2 + p 3 Φ 2 φ 2 ( b ) ) + ( Λ 2 + Λ 4 ) q 3 y Y .
Therefore
( ϰ , y ) X × Y ( Λ 1 + Λ 3 ) p 0 + ( Λ 2 + Λ 4 ) q 0 K * ,
where
K * = min 1 ( Λ 1 + Λ 3 ) p 1 + ( Λ 2 + Λ 4 ) ( q 1 + q 2 Φ 1 φ 1 ( b ) ) , 1 ( Λ 1 + Λ 3 ) ( p 2 + p 3 Φ 2 φ 2 ( b ) ) + ( Λ 2 + Λ 4 ) q 3 ,
which shows that Υ is bounded. By using Leray–Schauder’s alternative, we conclude that the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions in Equation (2) has at least one solution on 0 , b . This completes the proof. □
The last existence theorem is based on the following Krasnosel’skiĭ’s fixed point theorem.
Theorem 3
([23]). Let B be a bounded, closed, convex and nonempty subset of a Banach space X with operators T 1 and T 2 be operators such that
(i) 
T 1 ( ϰ ) + T 2 ( y ) B where ϰ , y B ,
(ii) 
T 1 is compact and continuous,
(iii) 
T 2 is a contraction mapping.
Then, there exists z B such that z = T 1 ( z ) + T 2 ( z ) .
Theorem 4.
Let f , g : [ 0 , b ] × R × R × R R be continuous functions satisfying the assumption ( H 1 ) . Moreover, we assume that:
  • (H3) There exist continuous functions ϕ , σ C ( [ 0 , b ] , R + ) such that
    f ( t , ϰ , y , z ) ϕ ( t ) a n d g ( t , ϰ , y , z ) σ ( t ) ,
    for each ( t , ϰ , y , z ) [ 0 , b ] × R × R × R .
If
( Λ 1 * + Λ 3 ) ν Θ 2 + ( Λ 2 + Λ 4 * ) ω Θ 1 < 1 ,
where Λ 1 * , Λ 2 , Λ 3 and Λ 4 * are given by Equation (16), then the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions in Equation (2) has at least one solution on [ 0 , b ] .
Proof. 
First, we separate the operator T as
T 1 ϰ , y ( t ) = T 1 , 1 ϰ , y ( t ) + T 1 , 2 ϰ , y ( t ) , T 2 ϰ , y ( t ) = T 2 , 1 ϰ , y ( t ) + T 2 , 2 ϰ , y ( t ) ,
with
T 1 , 1 ϰ , y ( t ) = I 0 + α 1 + γ 1 ; ψ 1 k 1 f ϰ , y ( t ) , t [ 0 , b ] , T 1 , 2 ϰ , y ( t ) = λ 2 δ 2 Δ 1 I 0 + α 1 + γ 1 ; ψ 1 k 1 f ϰ , y ( b ) λ 2 δ 1 Δ 1 I 0 + α 2 + γ 2 ; ψ 2 k 2 g ϰ , y ( b ) + Ω 1 ( t ) I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 f ϰ , y ( b ) Ω 2 ( t ) I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 g ϰ , y ( b ) , t [ 0 , b ] , T 2 , 1 ϰ , y ( t ) = I 0 + α 2 + γ 2 ; ψ 2 k 2 g ϰ , y ( t ) , t [ 0 , b ] , T 2 , 2 ϰ , y ( t ) = λ 2 δ 2 Δ 1 I 0 + α 2 + γ 2 ; ψ 2 k 2 g ϰ , y ( b ) λ 1 δ 2 Δ 1 I 0 + α 1 + γ 1 ; ψ 1 k 1 f ϰ , y ( b ) Ω 3 ( t ) I 0 + α 1 θ k 1 + k 1 ; ψ 1 k 1 f ϰ , y ( b ) + Ω 4 ( t ) I 0 + α 2 θ k 2 + k 2 ; ψ 2 k 2 g ϰ , y ( b ) , t [ 0 , b ] .
It is clear that T = T 1 , 1 + T 1 , 2 + T 2 , 1 + T 2 , 2 . Let B R = { ( ϰ , y ) X × Y : ( ϰ , y ) X × Y R } , be a closed and bounded ball with
R ( Λ 1 + Λ 3 ) ϕ X + ( Λ 2 + Λ 4 ) σ Y
where ϕ X = sup t [ 0 , b ] ϕ ( t ) < and σ Y = sup t [ 0 , b ] σ ( t ) < .
For any ( ϰ 1 , y 1 ) , ( ϰ 2 , y 2 ) B R , we find that
T 1 , 1 ( ϰ 1 , y 1 ) ( t ) + T 1 , 2 ( ϰ 2 , y 2 ) ( t ) Λ 1 ϕ X + Λ 2 σ Y .
Similarly, one can get
T 2 , 1 ( ϰ 1 , y 1 ) ( t ) + T 2 , 2 ( ϰ 2 , y 2 ) ( t ) Λ 3 ϕ X + Λ 4 σ Y .
Thus, we obtain
T 1 ( ϰ 1 , y 1 ) + T 2 ( ϰ 2 , y 2 ) X × Y ( Λ 1 + Λ 3 ) ϕ X + ( Λ 2 + Λ 4 ) σ Y R ,
which shows that T 1 ( ϰ 1 , y 1 ) + T 2 ( ϰ 2 , y 2 ) B R .
Using assumption (H1) along with Equation (23), we show that ( T 1 , 2 , T 2 , 2 ) is a contraction mapping. For ( ϰ 1 , y 1 ) , ( ϰ 2 , y 2 ) B R , and for any t [ 0 , b ] , we have
T 1 , 2 ( ϰ 2 , y 2 ) ( t ) T 1 , 2 ( ϰ 1 , y 1 ) ( t ) λ 2 δ 2 Δ 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ 2 , y 2 ( s ) f ϰ 1 , y 1 ( s ) d s + λ 2 δ 1 Δ 1 k 2 Γ k 2 ( α 2 + γ 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 + γ 2 k 2 1 g ϰ 2 , y 2 ( s ) g ϰ 1 , y 1 ( s ) d s + Ω 1 ( b ) k 1 Γ k 1 ( α 1 θ k 1 + k 1 ) 0 b ψ 1 ( s ) ψ 1 ( b ) ψ 1 ( s ) α 1 θ k 1 k 1 f ϰ 2 , y 2 ( s ) f ϰ 1 , y 1 ( s ) d s + Ω 2 ( b ) k 2 Γ k 2 ( α 2 θ k 2 + k 2 ) 0 b ψ 2 ( s ) ψ 2 ( b ) ψ 2 ( s ) α 2 θ k 2 k 2 g ϰ 2 , y 2 ( s ) g ϰ 1 , y 1 ( s ) d s λ 2 δ 2 Φ 1 α 1 + γ 1 ( b ) Δ 1 + Ω 1 ( b ) Φ 1 α 1 θ k 1 + k 1 ( b ) ν Θ 2 ϰ 2 ϰ 1 X + y 2 y 1 Y + λ 2 δ 1 Φ 2 α 2 + γ 2 ( b ) Δ 1 + Ω 2 ( b ) Φ 2 α 2 θ k 2 + k 2 ( b ) ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y = Λ 1 * ν Θ 2 ϰ 2 ϰ 1 X + y 2 y 1 Y + Λ 2 ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y ,
and consequently, we obtain
T 1 , 2 ( ϰ 2 , y 2 ) T 1 , 2 ( ϰ 1 , y 1 ) X Λ 1 * ν Θ 2 + Λ 2 ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y .
Similarly, we can find that
T 2 , 2 ( ϰ 2 , y 2 ) T 2 , 2 ( ϰ 1 , y 1 ) Y Λ 1 ν Θ 2 + Λ 4 * ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y .
From the inequalities in Equations (24) and (25), we conclude that
( T 1 , 2 , T 2 , 2 ) ( ϰ 2 , y 2 ) ( T 1 , 2 , T 2 , 2 ) ( ϰ 1 , y 1 ) X × Y ( Λ 1 * + Λ 3 ) ν Θ 2 + ( Λ 2 + Λ 4 * ) ω Θ 1 ϰ 2 ϰ 1 X + y 2 y 1 Y .
Since ( Λ 1 * + Λ 3 ) ν Θ 2 + ( Λ 2 + Λ 4 * ) ω Θ 1 < 1 , the operator ( T 1 , 2 , T 2 , 2 ) is a contraction.
Continuity of f and g implies that the operator ( T 1 , 1 , T 2 , 1 ) is continuous. Also, ( T 1 , 1 , T 2 , 1 ) is uniformly bounded on B R as
T 1 , 1 ( ϰ , y ) X ϕ X Φ 1 α 1 + γ 1 ( b ) and T 2 , 1 ( ϰ , y ) Y σ Y Φ 2 α 2 + γ 2 ( b ) .
Consequently, we obtain
( T 1 , 1 , T 2 , 1 ) X × Y ϕ X Φ 1 α 1 + γ 1 ( b ) + σ Y Φ 2 α 2 + γ 2 ( b ) .
Hence, ( T 1 , 1 , T 2 , 1 ) is uniformly bounded. Lastly, we will show that the set ( T 1 , 1 , T 2 , 1 ) is equicontinuous. For t 1 , t 2 [ 0 , b ] with t 1 < t 2 and ( ϰ , y ) B R , we have
T 1 , 1 ( ϰ , y ) ( t 2 ) T 1 , 1 ( ϰ , y ) ( t 1 ) 1 k 1 Γ k 1 ( α 1 + γ 1 ) 0 t 1 ψ 1 ( s ) ψ 1 ( t 2 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 ψ 1 ( t 1 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ , y ( s ) d s + 1 k 1 Γ k 1 ( α 1 + γ 1 ) t 1 t 2 ψ 1 ( s ) ψ 1 ( t 2 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 f ϰ , y ( s ) d s ϕ X k 1 Γ k 1 ( α 1 + γ 1 ) 0 t 1 ψ 1 ( s ) ψ 1 ( t 2 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 ψ 1 ( t 1 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 d s + ϕ X k 1 Γ k 1 ( α 1 + γ 1 ) t 1 t 2 ψ 1 ( s ) ψ 1 ( t 2 ) ψ 1 ( s ) α 1 + γ 1 k 1 1 d s = ϕ X Φ 1 α 1 + γ 1 ( t 2 ) Φ 1 α 1 + γ 1 ( t 1 ) .
Similarly, we have
T 2 , 1 ( ϰ , y ) ( t 2 ) T 2 , 1 ( ϰ , y ) ( t 1 ) σ Y Φ 2 α 2 + γ 2 ( t 2 ) Φ 2 α 2 + γ 2 ( t 1 ) .
From the inequalities in Equations (26) and (27), we conclude that
T 1 , 1 ( ϰ , y ) ( t 2 ) T 1 , 1 ( ϰ , y ) ( t 1 ) 0 and T 2 , 1 ( ϰ , y ) ( t 2 ) T 2 , 1 ( ϰ , y ) ( t 1 ) 0 ,
as t 1 t 2 independently of ( ϰ , y ) B R . Therefore the operator ( T 1 , 1 , T 2 , 1 ) is equicontinuous. Hence, by the Arzelá–Ascoli Theorem, ( T 1 , 1 , T 2 , 1 ) is compact on B R . Therefore, by the conclusion of Krasnosel’skiĭ’s fixed point theorem, Equation (2) has at least one solution on [ 0 , b ] . This completes the proof. □

4. Illustrative Examples

Consider the following coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions of the form
D 0 + 1 4 , 1 3 ; tan 1 t 1 2 , H D 0 + 2 3 ; tan 1 t 1 2 , C ϰ t = f t , ϰ t , y t , I 0 + 13 12 ; sin t 5 6 y t , t 0 , 3 2 , D 0 + 3 4 , 3 5 ; sin t 5 6 , H D 0 + 1 3 ; sin t 5 6 , C y t = g t , ϰ t , I 0 + 11 12 ; tan 1 t 1 2 ϰ t , y t , t 0 , 3 2 , 9 5 ϰ 0 + 2 y 3 2 = 0 , 11 3 D 0 + 1 2 ; tan 1 t 1 2 , C ϰ 0 + 13 25 D 0 + 3 10 ; sin t 5 6 , C y 3 2 = 0 , 5 2 y 0 + 1 7 ϰ 3 2 = 0 , D 0 + 3 10 ; sin t 5 6 , C y 0 + 6 17 D 0 + 1 2 ; tan 1 t 1 2 , C ϰ 3 2 = 0 .
Here, k 1 = 1 2 , k 2 = 5 6 , α 1 = 1 4 , α 2 = 3 4 , β 1 = 1 3 , β 2 = 3 5 , γ 1 = 2 3 , γ 2 = 1 3 , φ 1 = 11 12 , φ 2 = 13 12 , b = 3 2 λ 1 = 9 5 , λ 2 = 2 , λ 3 = 11 3 , λ 4 = 13 25 , δ 1 = 5 2 , δ 2 = 1 7 , δ 3 = 1 , δ 4 = 6 17 , ψ 1 ( t ) = tan 1 t and ψ 2 ( t ) = sin t . Using the given values, we find that Δ 1 4.21428 , Δ 2 3.48313 , Φ 1 θ k 1 + γ 1 k 1 ( b ) 1.99498 , Φ 2 θ k 2 + γ 2 k 2 ( b ) 1.19849 , Φ 1 α 1 + γ 1 ( b ) 2.05693 , Φ 2 α 2 + γ 2 ( b ) 1.08282 , Φ 1 α 1 θ k 1 + k 1 ( b ) 1.89026 , Φ 2 α 2 θ k 2 + k 2 ( b ) 1.21315 , Ω 1 ( b ) 0.64054 , Ω 2 ( b ) 0.39294 , Ω 3 ( b ) 0.48188 , Ω 4 ( b ) 0.08560 , Θ 1 3.05693 , Θ 2 2.08282 , Λ 1 3.40715 , Λ 2 1.76139 , Λ 3 1.03638 , Λ 4 1.26007 , Λ 1 * 1.35022 , Λ 4 * 0.17725 .
Example 1.
We consider the functions f , g : [ 0 , 3 2 ] × R × R × R R defined on [ 0 , 3 2 ] , as
f t , ϰ , y , I 0 + 13 12 ; sin t 5 6 y = e 2 t 8 t + 25 ϰ 2 + 2 ϰ 1 + ϰ + 1 7 ( t + 3 ) sin 2 y + 1 21 tan 1 I 0 + 13 12 ; sin t 5 6 y + 1 8 t 3 + 27 ,
and
g t , ϰ , I 0 + 11 12 ; tan 1 t 1 2 ϰ , y = 1 4 ( t + 5 ) cos 2 ϰ + 1 41 sin I 0 + 11 12 ; tan 1 t 1 2 ϰ + 1 t 2 + 40 y 2 + 2 y 1 + y + 1 4 t + 1 .
Clearly f and g satisfy the Lipschitz condition, since
f ( t , ϰ 1 , y 1 , z 1 ) f ( t , ϰ 2 , y 2 , z 2 ) 1 20 ϰ 2 ϰ 1 + y 2 y 1 + z 2 z 1 ,
and
g ( t , ϰ 1 , y 1 , z 1 ) g ( t , ϰ 2 , y 2 , z 2 ) 1 19 ϰ 2 ϰ 1 + y 2 y 1 + z 2 z 1 ,
with Lipschitz constants ν = 1 20 , ω = 1 19 . Therefore, the functions f and g satisfy condition ( H 1 ) in Theorem 1. In addition, we can find that
( Λ 1 + Λ 3 ) ν Θ 2 + ( Λ 2 + Λ 4 ) ω Θ 1 0.94887 < 1 ,
which implies that the inequality in Equation (17) is satisfied. Therefore, we deduce by Theorem 1, the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions in Equation (28) with f and g given by Equations (29) and (30), respectively, has a unique solution on [ 0 , 3 2 ] .
Example 2.
We consider the functions f , g : [ 0 , 3 2 ] × R × R × R R defined on [ 0 , 3 2 ] , as
f t , ϰ , y , I 0 + 13 12 ; sin t 5 6 y = 1 7 t + 14 + 4 e y t 18 ( t + 2 ) ϰ 2025 1 + ϰ 2024 + y ( 2 t + 1 ) 84 sin ϰ + cos 4 ϰ t 2 t 2 + 400 I 0 + 13 12 ; sin t 5 6 y ,
and
g t , ϰ , I 0 + 11 12 ; tan 1 t 1 2 ϰ , y = 1 12 t + 24 + ϰ t + 1 2 26 π tan 1 y + sin 4 y t 3 t 4 + 49 I 0 + 11 12 ; tan 1 t 1 2 ϰ + e 4 ϰ t 27 ( t + 3 ) y 2 1 + y .
Then, we have
f ( t , ϰ , y , z ) 1 14 + 1 9 ϰ + 1 21 y + 1 20 z ,
and
g ( t , ϰ , y , z ) 1 24 + 1 26 ϰ + 1 81 y + 1 7 z .
By setting p 0 = 1 14 , p 1 = 1 9 , p 2 = 1 21 , p 3 = 1 20 , q 0 = 1 24 , q 1 = 1 26 , q 2 = 1 81 , q 3 = 1 7 we obtain
( Λ 1 + Λ 3 ) p 1 + ( Λ 2 + Λ 4 ) ( q 1 + q 2 Φ 1 φ 1 ( b ) ) 0.68666 < 1 ,
and
( Λ 1 + Λ 3 ) ( p 2 + p 3 Φ 2 φ 2 ( b ) ) + ( Λ 2 + Λ 4 ) q 3 0.88381 < 1 ,
which implies that the inequalities in Equation (22) are satisfied. Therefore, we deduce by Theorem 2, the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions in Equation (28) with f and g given by Equations (31) and (32), respectively, has at least one solution in [ 0 , 3 2 ] .
Example 3.
We consider the functions f , g : [ 0 , 3 2 ] × R × R × R R defined on [ 0 , 3 2 ] , as
f t , ϰ , y , I 0 + 13 12 ; sin t 5 6 y = 1 6 t + 20 sin ϰ + 1 t + 100 ln ( 1 + y ) 1 + ln ( 1 + y ) + cos 2 π t 11 ( t + 1 ) | I 0 + 13 12 ; sin t 5 6 y | 1 + | I 0 + 13 12 ; sin t 5 6 y | + e t 2 + 1 + 7 ,
and
g t , ϰ , I 0 + 11 12 ; tan 1 t 1 2 ϰ , y = e 4 t 2 4 ( t + 3 ) ln ( 1 + ϰ ) 1 + ln ( 1 + ϰ ) + sin 2 π t t + 25 | I 0 + 11 12 ; tan 1 t 1 2 ϰ | 1 + | I 0 + 11 12 ; tan 1 t 1 2 ϰ | + 1 3 t 3 + 64 3 tan 1 | y | + 40 t + 14 .
Then, we have
f ( t , ϰ , y , z ) 1 6 t + 20 + 1 t + 100 + cos 2 π t 11 ( t + 1 ) + e t 2 + 1 + 7 ,
and
g ( t , ϰ , y , z ) e 4 t 2 4 ( t + 3 ) + sin 2 π t t + 25 + 1 3 t 3 + 64 3 + 40 t + 14 .
Moreover, f and g satisfy the Lipschitz condition, since
f ( t , ϰ 1 , y 1 , z 1 ) f ( t , ϰ 2 , y 2 , z 2 ) 1 10 ϰ 2 ϰ 1 + y 2 y 1 + z 2 z 1 ,
and
g ( t , ϰ 1 , y 1 , z 1 ) g ( t , ϰ 2 , y 2 , z 2 ) 1 12 ϰ 2 ϰ 1 + y 2 y 1 + z 2 z 1 ,
with Lipschitz constants ν = 1 10 , ω = 1 12 . Therefore, the functions f and g satisfy condition ( H 1 ) in Theorem 1. In addition, we can find that
( Λ 1 * + Λ 3 ) ν Θ 2 + ( Λ 2 + Λ 4 * ) ω Θ 1 0.99093 < 1 ,
which implies that the inequality in Equation (23) is satisfied. Therefore, we deduce by Theorem 4, the coupled system of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions (28) with f and g given by (33) and (34), respectively, has at least one solution in [ 0 , 3 2 ] .

5. Conclusions

In this paper, we have established the existence and uniqueness results for a new class of coupled systems of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations with non-separated boundary conditions. The uniqueness result depends on the Banach contraction mapping principle, while the existence results are established using the Leray–Schauder nonlinear alternative and Krasnosel’skii’s fixed point theorem. All the obtained results are well supported and illustrated by carefully constructed numerical examples. Our results are novel and make a significant contribution to enriching the existing results in the literature on coupled systems of sequential ( k , ψ ) -Hilfer and ( k , ψ ) -Caputo fractional differential equations.

Author Contributions

Conceptualization, S.K.N. and J.T.; methodology, F.E., N.A.H., S.K.N., J.T. and P.W.; formal analysis, F.E., N.A.H., S.K.N., J.T. and P.W.; writing—original draft preparation, F.E., N.A.H., S.K.N., J.T. and P.W.; funding acquisition, J.T. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by King Mongkut’s University of Technology North Bangkok, Contract No. KMUTNB-67-KNOW-16.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Data are contained within the article.

Conflicts of Interest

The authors declare no conflicts of interest.

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Erkan, F.; Hamal, N.A.; Ntouyas, S.K.; Tariboon, J.; Wongsantisuk, P. Coupled System of (k, ψ)-Hilfer and (k, ψ)-Caputo Sequential Fractional Differential Equations with Non-Separated Boundary Conditions. Axioms 2025, 14, 685. https://doi.org/10.3390/axioms14090685

AMA Style

Erkan F, Hamal NA, Ntouyas SK, Tariboon J, Wongsantisuk P. Coupled System of (k, ψ)-Hilfer and (k, ψ)-Caputo Sequential Fractional Differential Equations with Non-Separated Boundary Conditions. Axioms. 2025; 14(9):685. https://doi.org/10.3390/axioms14090685

Chicago/Turabian Style

Erkan, Furkan, Nuket Aykut Hamal, Sotiris K. Ntouyas, Jessada Tariboon, and Phollakrit Wongsantisuk. 2025. "Coupled System of (k, ψ)-Hilfer and (k, ψ)-Caputo Sequential Fractional Differential Equations with Non-Separated Boundary Conditions" Axioms 14, no. 9: 685. https://doi.org/10.3390/axioms14090685

APA Style

Erkan, F., Hamal, N. A., Ntouyas, S. K., Tariboon, J., & Wongsantisuk, P. (2025). Coupled System of (k, ψ)-Hilfer and (k, ψ)-Caputo Sequential Fractional Differential Equations with Non-Separated Boundary Conditions. Axioms, 14(9), 685. https://doi.org/10.3390/axioms14090685

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