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3 March 2026

26 Pages

Nonlinear Sequential Caputo Fractional Differential Systems: Existence and Hyers–Ulam Stability Under Coupled Mixed Boundary Constraints

,
and
1
Center for Nonlinear and Complex Networks, SRM TRP Engineering College, Tiruchirappalli 621105, Tamil Nadu, India
2
Center for Research, Easwari Engineering College, Chennai 600089, Tamil Nadu, India
3
Department of Mathematics, College of Science and Humanities in Alkharj, Prince Sattam bin Abdulaziz University, Alkharj 11942, Saudi Arabia
4
Department of Mathematics, Faculty of Science, Sakarya University, Sakarya 54050, Türkiye

Abstract

In this paper, we study a nonlinear system of sequential Caputo fractional differential equations equipped with coupled mixed multi-point boundary conditions. In particular, the boundary conditions involve the values of the unknown functions at the endpoints expressed as linear combinations of their values at several interior points, forming a closed system of relations. The existence of solutions is established by applying the Leray–Schauder alternative, while uniqueness is proved using Banach’s contraction principle. In addition, we investigate the Hyers–Ulam stability of the proposed system. Several examples are included to demonstrate the applicability of the theoretical results. Some special cases of the general problem are also discussed.

1. Introduction

Sequential Caputo fractional differential equations refer to systems in which derivatives of fractional order are applied successively, typically involving Caputo’s definition due to its compatibility with initial conditions of classical type. These equations are crucial in modeling complex dynamical systems that exhibit memory and hereditary properties, such as viscoelastic materials, diffusion processes, and biological systems. The sequential aspect enables the capture of layered temporal effects by combining multiple fractional orders, often leading to more accurate and flexible representations of real-world phenomena. Unlike classical integer-order models, sequential Caputo systems offer refined control over system dynamics, making them valuable for theoretical analysis and applications where multi-scale temporal behavior is significant [1,2,3,4,5,6].
Coupled systems consist of two or more interdependent differential equations in which the solution of one equation directly influences the others. These systems naturally arise in modeling phenomena involving multiple interacting components, such as in mechanical structures, electrical circuits, or biological networks. In the context of fractional calculus, coupling adds complexity by integrating memory effects across interconnected variables. Studying such systems is essential for understanding the joint dynamics of processes in which the isolation of individual components fails to capture the full behavior of the system [7,8,9,10,11,12,13].
Ulam–Hyers stability is a concept in the qualitative theory of differential equations that addresses the stability of approximate solutions. Specifically, it ensures that if a function nearly satisfies a given differential equation, then there exists an exact solution nearby. This type of stability is particularly important in applied mathematics, where exact solutions are often inaccessible, and numerical or approximate methods are used. In the context of fractional differential equations, Ulam–Hyers stability provides a robust framework for assessing the reliability and resilience of solutions under small perturbations or modeling inaccuracies [14,15,16,17].
In recent years, discrete fractional models involving the nabla fractional operator have received significant attention owing to their capability to effectively describe memory and hereditary characteristics in discrete-time dynamical systems. These nabla fractional difference equations serve as discrete analogues of continuous fractional differential equations and have been widely applied in the qualitative analysis of memory-driven processes [18]. In particular, several recent contributions have addressed coupled nabla fractional systems and investigated the existence, nonexistence, uniqueness, multiplicity, and stability of solutions for problems subject to various types of nonlocal boundary conditions, including multi-point summation boundary conditions [19,20]. The obtained results highlight the importance of discrete fractional operators in modeling systems where past states influence present dynamics. Motivated by these recent developments, the present study focuses on a coupled system of continuous fractional differential equations with multi-point summation boundary conditions, thereby extending the applicability of memory-dependent fractional models from discrete to continuous settings.
Let us now recall some works on closed boundary conditions. In ref. [21], Bashir Ahmad studied mixed nonlinearities with nonlocal closed boundary conditions.
D T − μ C Y ( τ ) + λ I T − ρ I 0 + σ K 1 ( τ , Y ( τ ) ) = K 2 ( τ , Y ( τ ) ) , τ ∈ J : = [ 0 , T ] ,
Y ( τ ) = p 1 Y ( ξ ) + p 2 T Y ′ ( ξ ) , T Y ′ ( ξ ) = q 1 Y ( ξ ) + q 2 T Y ′ ( ξ ) , 0 < ξ < T ,
where D T _ μ C represents the right Caputo fractional derivative operator of order μ ∈ ( 1 , 2 ] , I T − ρ and I 0 + σ , respectively, designate the right and left Riemann–Liouville operators of fractional orders ρ , σ > 0 . φ 1 , φ 2 : [ 0 , T ] × R → R are continuous functions and λ , p 1 , p 2 , q 1 , q 2 ∈ R . Recently, ref. [22] investigated the existence and uniqueness of solutions for a nonlinear system of fractional differential equations with coupled closed boundary conditions:
( c D α ) Y ( τ ) = K ( τ , Y ( τ ) , Z ( τ ) ) , ( c D β ) Z ( τ ) = L ( τ , Y ( τ ) , Z ( τ ) ) ,
with closed boundary conditions;
Y ( T ) = r 1 Z ( 0 ) + r 2 T Z ′ ( 0 ) , T Y ′ ( T ) = s 1 Z ( 0 ) + s 2 T Z ′ ( 0 ) , Z ( T ) = p 1 Y ( 0 ) + p 2 T Y ′ ( 0 ) , T Z ′ ( T ) = q 1 Y ( 0 ) + q 2 T Y ′ ( 0 ) ,
where D α c and D β c denote Caputo fractional derivatives of order α , β , 1 < α , β < 2 , respectively, r 1 , r 2 , s 1 , s 2 , p 1 , p 2 , q 1 , q 2 , ∈ R , T > 0 and K , L ∈ C ( J × R × R × R ) . Motivated by the mentioned work, in this paper, we propose a nonlocal variant of closed boundary conditions and study a coupled sequential fractional differential equation containing a right Caputo fractional derivative. Specifically, we explore the criteria that ensure the existence and stability of solutions for the following problem:
( c D α + λ 1 D α − 1 c ) Y ( τ ) = K ( τ , Y ( τ ) , Z ( τ ) ) , ( c D β + λ 2 D β − 1 c ) Z ( τ ) = L ( τ , Y ( τ ) , Z ( τ ) ) ,
with multi-point boundary conditions.
Y ( T ) = ∑ i = 1 m r 1 i Z ( k 1 ) + ∑ i = 1 m r 2 i T Z ′ ( k 2 ) , T Y ′ ( T ) = ∑ i = 1 m s 1 i Z ( u 1 ) + ∑ i = 1 m s 2 i T Z ′ ( u 2 ) , Z ( T ) = ∑ j = 1 n p 1 j Y ( ζ 1 ) + ∑ j = 1 n p 2 j T Y ′ ( ζ 2 ) , T Z ′ ( T ) = ∑ j = 1 n q 1 j Y ( v 1 ) + ∑ j = 1 n q 2 j T Y ′ ( v 2 ) ,
where D α c and D β c denote the Caputo fractional derivatives of order α , β , 1 < α , β < 2 , respectively, r i , s i , p i , q i , ∈ R ,   i ∈ { 1 , 2 } , and K , L ∈ C ( J × R × R × R ) and k i , u i , ζ i , v i ∈ [ 0 , T ] ,   i ∈ { 1 , 2 } . The boundary conditions (4) can be interpreted as the values of the unknown function and its derivative at the right end point T of the interval [ 0 , T ] are proportional to a linear combination of these values at arbitrary non-local positions k i , u i , ζ i , v i ∈ [ 0 , T ] ,   i ∈ { 1 , 2 } . Physically, the nonlocal multi-point closed boundary conditions provide a flexible mechanism to close the boundary at arbitrary positions in the given domain instead of the left end point of the domain.
The organization of this paper is summarized as follows:
  • Section 2 (Preliminaries): This section presents the fundamental definitions, auxiliary lemmas, and supporting results required for the subsequent analysis.
  • Section 3 (Existence and Uniqueness Results): In this section, we derive the integral equivalent formulation of the proposed problem and establish the main existence and uniqueness results using appropriate fixed-point techniques.
  • Section 4 (Hyers–Ulam Stability): This section is devoted to the study of the Hyers–Ulam stability of the considered system and provides sufficient conditions ensuring stability.
  • Section 5 (Examples): Two illustrative examples are presented to demonstrate the applicability of the theoretical results.
  • Section 6 (Conclusions): The final section summarizes the main findings and discusses possible directions for future research.

2. Auxiliary Results

In this section, we present several definitions and auxiliary results that will be used in the proofs of the main theorems.
Definition 1
([23]). Let α > 0 and let Y : [ 0 , ∞ ) → R be a function such that the following integral exists. The Riemann–Liouville fractional integral of order α is defined by
I α Y ( τ ) = 1 Γ ( α ) ∫ 0 τ ( τ − s ) α − 1 Y ( s ) d s , τ > 0 ,
where Γ ( · ) denotes the Gamma function given by
Γ ( p ) = ∫ 0 ∞ t p − 1 e − t d t , p > 0 .
Definition 2
([23]). Let n ∈ N and let α satisfy n − 1 < α < n . If Y : [ 0 , ∞ ) → R is ( n − 1 ) times continuously differentiable and its n-th derivative belongs to L loc 1 ( [ 0 , ∞ ) ) , then the Caputo fractional derivative of order α is defined by
D α c Y ( τ ) = 1 Γ ( n − α ) ∫ 0 τ ( τ − s ) n − α − 1 Y ( n ) ( s ) d s , τ > 0 .
In the following, we study the associated linear fractional system as a preliminary step toward the nonlinear analysis.
( c D α + λ 1 D α − 1 c ) Y ( τ ) = H 1 ( τ ) , ( c D β + λ 2 D β − 1 c ) Z ( τ ) = H 2 ( τ ) ,
supplemented with the boundary conditions (4), with H 1 , H 2 ∈ C [ 0 , T ] . We denote by
A 1 = e − λ 1 T , A 2 = ( 1 − e − λ 1 T ) λ 1 , A 3 = ∑ i = 1 m − r 1 e − λ 2 k 1 + r 2 T λ 2 e − λ 2 k 2 A 4 = ∑ i = 1 m − r 1 ( 1 − e − λ 2 k 1 ) λ 2 − r 2 T ( λ 2 e − λ 2 k 2 ) λ 2 , A 5 = T ( − λ 1 e − λ 1 T ) , A 6 = T ( λ 1 e − λ 1 T ) A 7 = ∑ i = 1 m − s 1 e − λ 2 u 1 + s 2 T λ 2 e − λ 2 u 2 , A 8 = ∑ i = 1 m − s 1 ( 1 − e − λ 2 u 1 ) λ 2 − s 2 T ( λ 2 e − λ 2 u 2 ) λ 2 A 9 = ∑ j = 1 n − p 1 e − λ 1 ζ 1 + p 2 T λ 1 e − λ 1 ζ 2 , A 10 = ∑ j = 1 n − p 1 1 λ 1 ( 1 − e − λ 1 ζ 1 ) − p 2 T 1 λ 1 ( λ 1 e − λ 1 ζ 2 ) A 11 = e − λ 2 T , A 12 = 1 λ 2 ( 1 − e − λ 2 T ) , A 13 = ∑ j = 1 n − q 1 e − λ 1 ( v 1 ) + q 2 T e − λ 1 v 2 A 14 = ∑ j = 1 n − q 1 1 λ 1 ( 1 − e − λ 1 v ) − q 2 T λ 1 ( λ 1 e − λ 1 v 2 ) , A 15 = − λ 2 T e − λ 2 T , A 16 = T λ 2 ( λ 2 e − λ 2 T ) .
E 1 = ∑ i = 1 m { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 H 2 ( s ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + r 2 T I β − 1 H 2 ( k 2 ) } − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s E 2 = ∑ i = 1 m { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + s 2 T I β − 1 H 2 ( u 2 ) − T I α − 1 H 1 ( T ) } + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s , E 3 = ∑ i = 1 n { p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + p 2 T I α − 1 H 1 ( ζ 2 ) } − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s E 4 = ∑ i = 1 n { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + q 2 T I α − 1 H 1 ( v 2 ) − T I β − 1 H 2 ( T ) } + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s , Δ 1 = ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 14 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) Δ 2 = ( A 4 A 5 − A 1 A 8 ) ( A 10 A 13 − A 9 A 14 ) ( A 11 A 13 − A 9 A 15 ) ( A 2 A 5 − A 1 A 6 ) − ( A 12 A 13 − A 9 A 16 ) ( A 2 A 5 − A 1 A 6 ) ( A 3 A 5 − A 1 A 7 ) ( A 10 A 13 − A 9 A 14 ) .
Y ( τ ) = [ 1 Δ 1 ( ( E 3 A 15 − E 4 A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( E 1 A 7 − E 2 A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ) ] e − λ 1 τ + [ 1 Δ 1 ( ( E 3 A 15 − E 4 A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( E 1 A 7 − E 2 A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ) ] 1 λ 1 ( 1 − e − λ 1 τ ) + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 H 1 ( s ) d s ,
Z ( τ ) = [ 1 Δ 2 ( ( E 1 A 5 − E 2 A 1 ) ( A 10 A 13 − A 9 A 14 ) ( A 12 A 13 − A 9 A 16 ) ( A 2 A 5 − A 1 A 6 ) − ( E 3 A 13 − E 4 A 9 ) ( A 2 A 5 − A 1 A 6 ) ( A 4 A 5 − A 1 A 8 ) ( A 10 A 13 − A 9 A 14 ) ) ] e − λ 2 τ + [ 1 Δ 2 ( ( E 1 A 5 − E 2 A 1 ) ( A 10 A 13 − A 9 A 14 ) ( A 11 A 13 − A 9 A 15 ) ( A 2 A 5 − A 1 A 6 ) − ( E 3 A 13 − E 4 A 9 ) ( A 2 A 5 − A 1 A 6 ) ( A 3 A 5 − A 1 A 7 ) ( A 10 A 13 − A 9 A 14 ) ) ] 1 λ 2 ( 1 − e − λ 2 τ ) + ∫ 0 τ e − λ 2 ( τ − s ) I β − 1 H 2 ( s ) d s ,
Lemma 1.
If Δ 1 , Δ 2 ≠ 0 , then the solution ( Y , Z ) ∈ ( C [ 0 , T ] ) 2 of the boundary value problem (3) and (4) is given by
Y ( τ ) = e − λ 1 τ Δ 1 [ ( { ∑ j = 1 n p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n p 2 T I α − 1 H 1 ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s } A 15 − { ∑ j = 1 n q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n q 2 T I α − 1 H 1 ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s − ∑ j = 1 n T I β − 1 H 2 ( T ) } A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( { ∑ i = 1 m r 1 ∫ 0 k 1 e − λ 1 ( k 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m r 2 T I β − 1 H 2 ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s } A 7 − { ∑ i = 1 m s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m s 2 T I β − 1 H 2 ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s − T I α − 1 H 1 ( T ) } A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 [ ( { ∑ j = 1 n p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n p 2 T I α − 1 H 1 ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s } A 15 − { ∑ j = 1 n q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n q 2 T I α − 1 H 1 ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s − T I β − 1 H 2 ( T ) } A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( { ∑ i = 1 m r 1 ∫ 0 k 1 e − λ 1 ( k 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m r 2 T I β − 1 H 2 ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s } A 7 − { ∑ i = 1 m s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m s 2 T I β − 1 H 2 ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s − T I α − 1 H 1 ( T ) } A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ] + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 H 1 ( s ) d s , Z ( τ ) = e − λ 2 τ Δ 2 [ ( { ∑ i = 1 m r 1 ∫ 0 k 1 e − λ 1 ( k 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m r 2 T I β − 1 H 2 ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s } A 5 − { ∑ i = 1 m s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m s 2 T I β − 1 H 2 ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s − T I α − 1 H 1 ( T ) } A 1 ) × ( A 10 A 13 − A 9 A 14 ) ( A 12 A 13 − A 9 A 16 ) ( A 2 A 5 − A 1 A 6 ) − ( { ∑ j = 1 n p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n p 2 T I α − 1 H 1 ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s } A 13 − { ∑ j = 1 n q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n q 2 T I α − 1 H 1 ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s − T I β − 1 H 2 ( T ) } A 9 ) × ( A 2 A 5 − A 1 A 6 ) ( A 4 A 5 − A 1 A 8 ) ( A 10 A 13 − A 9 A 14 ) ] + 1 λ 2 ( 1 − e − λ 2 τ ) 1 Δ 2 [ ( { ∑ i = 1 m r 1 ∫ 0 k 1 e − λ 1 ( k 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m r 2 T I β − 1 H 2 ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s } A 5 − { ∑ i = 1 m s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m s 2 T I β − 1 H 2 ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s − T I α − 1 H 1 ( T ) } A 1 ) × ( A 10 A 13 − A 9 A 14 ) ( A 11 A 13 − A 9 A 15 ) ( A 2 A 5 − A 1 A 6 ) − ( { ∑ j = 1 n p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n p 2 T I α − 1 H 1 ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s } A 13 − { ∑ j = 1 n q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n q 2 T I α − 1 H 1 ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s − T I β − 1 H 2 ( T ) } A 9 ) × ( A 2 A 5 − A 1 A 6 ) ( A 3 A 5 − A 1 A 7 ) ( A 10 A 13 − A 9 A 14 ) ] + ∫ 0 τ e − λ 2 ( τ − s ) I β − 1 H 2 ( s ) d s .
Proof. 
The system (3) can be equivalently written as the following system:
( c D α + λ 1 D α − 1 c ) Y ( τ ) = H 1 ( τ ) , τ ∈ ( 0 , 1 ) , ( c D β + λ 2 D β − 1 c ) Z ( τ ) = H 2 ( τ ) , τ ∈ ( 0 , 1 ) ,
where D − 1 c is the integral operator I . The general solution of the system (9) is
Y ( τ ) = c 0 e − λ 1 τ + c 1 λ 1 ( 1 − e − λ 1 τ ) + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 H 1 ( s ) d s , Z ( τ ) = d 0 e − λ 2 τ + d 1 λ 2 ( 1 − e − λ 2 τ ) + ∫ 0 τ e − λ 2 ( τ − s ) I β − 1 H 2 ( s ) d s ,
we differentiate the system (10) and we find
Y ′ ( τ ) = − λ 1 c 0 e − λ 1 τ + c 1 λ 1 ( λ 1 e − λ 1 τ ) − λ 1 ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 H 1 ( s ) d s + I α − 1 H 1 ( τ ) , Z ′ ( τ ) = − λ 2 d 0 e − λ 2 τ + d 1 λ 2 ( λ 2 e − λ 2 τ ) − λ 2 ∫ 0 τ e − λ 2 ( τ − s ) I β − 1 H 2 ( s ) d s + I β − 1 H 2 ( τ ) ,
where c 0 , c 1 , d 0 and d 1 are unknown arbitrary constants. Using (11) in the nonlocal closed boundary conditions: Y ( T ) = r 1 Z ( k 1 ) + r 2 T Z ′ ( k 2 ) ,   T Y ′ ( T ) = s 1 Z ( u 1 ) + s 2 T Z ′ ( u 2 ) ,   Z ( T ) = p 1 Y ( ζ 1 ) + p 2 T Y ′ ( ζ 2 ) , and T Z ′ ( T ) = q 1 Y ( v 1 ) + q 2 T Y ′ ( v 2 ) , we find a system of algebraic equations in c 0 , c 1 , d 0 and d 1 :
A 1 c 0 + A 2 c 1 + A 3 d 0 + A 4 d 1 − E 1 = 0 , A 5 c 0 + A 6 c 1 + A 7 d 0 + A 8 d 1 − E 2 = 0 , A 9 c 0 + A 10 c 1 + A 11 d 0 + A 12 d 1 − E 3 = 0 , A 13 c 0 + A 14 c 1 + A 15 d 0 + A 16 d 1 − E 4 = 0 ,
c 0 = 1 Δ 1 [ ( E 3 A 15 − E 4 A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( E 1 A 7 − E 2 A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] , c 1 = 1 Δ 1 [ ( E 3 A 15 − E 4 A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( E 1 A 7 − E 2 A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ] , d 0 = 1 Δ 2 [ ( E 1 A 5 − E 2 A 1 ) ( A 10 A 13 − A 9 A 14 ) ( A 12 A 13 − A 9 A 16 ) ( A 2 A 5 − A 1 A 6 ) − ( E 3 A 13 − E 4 A 9 ) ( A 2 A 5 − A 1 A 6 ) ( A 4 A 5 − A 1 A 8 ) ( A 10 A 13 − A 9 A 14 ) ] , d 1 = 1 Δ 2 [ ( E 1 A 5 − E 2 A 1 ) ( A 10 A 13 − A 9 A 14 ) ( A 11 A 13 − A 9 A 15 ) ( A 2 A 5 − A 1 A 6 ) − ( E 3 A 13 − E 4 A 9 ) ( A 2 A 5 − A 1 A 6 ) ( A 3 A 5 − A 1 A 7 ) ( A 10 A 13 − A 9 A 14 ) ] .
Substituting the values of c 0 , c 1 , d 0 , and d 1 in (10) yields the solution. This completes the proof. □

3. Main Results

Let us introduce the space X = { Y ( τ ) | Y ( τ ) ∈ C ( [ 0 , T ] ) } endowed with the norm | | Y | | = max { | Y ( τ ) , τ ∈ [ 0 , T ] } . Obviously ( Y , | | · | | ) is a Banach space. Also let Y = { Z ( τ ) | Z ( τ ) ∈ C ( [ 0 , T ] ) } endowed with the norm | | Z | | = max { | Z ( τ ) , τ ∈ [ 0 , T ] } .The product space ( X × Y , | | ( Y , Z ) | | ) is also a Banach space with norm | | Y , Z | | = | | Y | | + | | Z | | . Given Lemma 1, we define the operator T : X × Y → X × Y by
T ( Y , Z ) ( τ ) = T 1 ( Y , Z ) ( τ ) T 2 ( Y , Z ) ( τ ) ,
where
T 1 ( Y , Z ) ( τ ) = e − λ 1 τ Δ 1 [ ( { ∑ j = 1 n p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n p 2 T I α − 1 H 1 ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s } A 15 − { ∑ j = 1 n q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n q 2 T I α − 1 H 1 ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s − ∑ j = 1 n T I β − 1 H 2 ( T ) } A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( { ∑ i = 1 m r 1 ∫ 0 k 1 e − λ 1 ( k 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m r 2 T I β − 1 H 2 ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s } A 7 − { ∑ i = 1 m s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m s 2 T I β − 1 H 2 ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s − T I α − 1 H 1 ( T ) } A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 [ ( { ∑ j = 1 n p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n p 2 T I α − 1 H 1 ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s } A 15 − { ∑ j = 1 n q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n q 2 T I α − 1 H 1 ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s − T I β − 1 H 2 ( T ) } A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( { ∑ i = 1 m r 1 ∫ 0 k 1 e − λ 1 ( k 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m r 2 T I β − 1 H 2 ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s } A 7 − { ∑ i = 1 m s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m s 2 T I β − 1 H 2 ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s − T I α − 1 H 1 ( T ) } A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ] + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 H 1 ( s ) d s ,
T 2 ( Y , Z ) ( τ ) = e − λ 2 τ Δ 2 [ ( { ∑ i = 1 m r 1 ∫ 0 k 1 e − λ 1 ( k 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m r 2 T I β − 1 H 2 ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s } A 5 − { ∑ i = 1 m s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m s 2 T I β − 1 H 2 ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s − T I α − 1 H 1 ( T ) } A 1 ) × ( A 10 A 13 − A 9 A 14 ) ( A 12 A 13 − A 9 A 16 ) ( A 2 A 5 − A 1 A 6 ) − ( { ∑ j = 1 n p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n p 2 T I α − 1 H 1 ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s } A 13 − { ∑ j = 1 n q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n q 2 T I α − 1 H 1 ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s − T I β − 1 H 2 ( T ) } A 9 ) × ( A 2 A 5 − A 1 A 6 ) ( A 4 A 5 − A 1 A 8 ) ( A 10 A 13 − A 9 A 14 ) ] + 1 λ 2 ( 1 − e − λ 2 τ ) 1 Δ 2 [ ( { ∑ i = 1 m r 1 ∫ 0 k 1 e − λ 1 ( k 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m r 2 T I β − 1 H 2 ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s } A 5 − { ∑ i = 1 m s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 H 2 ( s ) d s − ∑ i = 1 m λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 H 2 ( s ) d s + ∑ i = 1 m s 2 T I β − 1 H 2 ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 H 1 ( s ) d s − T I α − 1 H 1 ( T ) } A 1 ) × ( A 10 A 13 − A 9 A 14 ) ( A 11 A 13 − A 9 A 15 ) ( A 2 A 5 − A 1 A 6 ) − ( { ∑ j = 1 n p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n p 2 T I α − 1 H 1 ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s } A 13 − { ∑ j = 1 n q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 H 1 ( s ) d s − ∑ j = 1 n λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 H 1 ( s ) d s + ∑ j = 1 n q 2 T I α − 1 H 1 ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 H 2 ( s ) d s − T I β − 1 H 2 ( T ) } A 9 ) × ( A 2 A 5 − A 1 A 6 ) ( A 3 A 5 − A 1 A 7 ) ( A 10 A 13 − A 9 A 14 ) ] + ∫ 0 τ e − λ 2 ( τ − s ) I β − 1 H 2 ( s ) d s .
We need the following hypotheses in the sequel:
  • ( H 1 ) Assume that there exist real constant m i , n i > 0 , i = 1 , 2 and m 0 > 0 , n 0 > 0 such that, for all τ ∈ [ 0 , T ] , U i ∈ R e , i = 1 , 2 ,
    | K ( τ , U 1 , U 2 ) | ≤ m 0 + m 1 | U 1 | + m 2 | U 2 | , | L ( τ , U 1 , U 2 ) | ≤ n 0 + n 1 | U 1 | + n 2 | U 2 | .
  • ( H 2 ) Assume that K , L : R e 2 → R e are continuous functions and there exist constants M i , N i , i = 1 , 2 such that for all τ ∈ [ 0 , T ] and Y i , Z i , i = 1 , 2 ,
    | K ( τ , Y 1 , Y 2 ) − K ( τ , Z 1 , Z 2 ) | ≤ M 1 | Y 1 − Z 1 | + M 2 | Y 2 − Z 2 | | L ( τ , Y 1 , Y 2 ) − L ( τ , Z 1 , Z 2 ) | ≤ N 1 | Y 1 − Z 1 | + N 2 | Y 2 − Z 2 | .
  • For convenience, we set
M 1 = e − λ 1 τ Δ 1 { ( ∑ j = 1 n p 1 ζ 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 1 ) + λ 1 p 2 T ζ 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 2 ) + p 2 T ζ 2 α − 1 Γ ( α ) A 15 + ∑ j = 1 n q 1 v 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 1 ) + λ 1 q 2 T v 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 2 ) + q 2 T v 2 α − 1 λ 1 Γ ( α ) A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) + T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) A 7 + λ 1 T T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) + T T α − 1 Γ ( α ) A 3 × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) } + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 { ( ∑ j = 1 n p 1 ζ 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 1 ) + λ 1 p 2 T ζ 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 2 ) + p 2 T ζ 2 α − 1 Γ ( α ) A 2 + ∑ j = 1 n q 1 v 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 1 ) + λ 1 q 2 T v 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 2 ) + q 2 T v 2 α − 1 λ 1 Γ ( α ) A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) + T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) A 7 + λ 1 T T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) + T T α − 1 Γ ( α ) A 3 × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) } + 1 λ 1 Γ ( α ) ( 1 − e − λ 1 ) ,
M 2 = e − λ 1 τ Δ 1 { T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) A 2 + λ 2 T T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) + T T β − 1 Γ ( β ) A 11 × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) + ( ∑ i = 1 m r 1 k 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 1 ) + λ 2 r 2 T k 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 2 ) + r 2 T k 2 β − 1 Γ ( β ) A 7 + ∑ i = 1 m s 1 u 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 1 ) + λ 2 s 2 T u 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 2 ) + s 2 T u 2 β − 1 Γ ( β ) A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) } + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 { T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) A 15 + λ 2 T T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) + T T β − 1 Γ ( β ) A 11 × ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) + ( ∑ i = 1 m r 1 k 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 1 ) + λ 1 r 2 T k 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 2 ) + r 2 T k 2 β − 1 Γ ( β ) A 7 + ∑ i = 1 m s 1 u 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 1 ) + λ 2 s 2 T u 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 2 ) + s 2 T u 2 β − 1 Γ ( β ) A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) } ,
M ^ 1 = e − λ 2 τ Δ 2 { T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) A 5 + λ 1 T T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) + T T α − 1 Γ ( α ) A 1 × ( A 10 A 13 − A 9 A 14 ) ( A 12 A 13 − A 9 A 16 ) ( A 2 A 5 − A 1 A 6 ) + ( ∑ j = 1 n p 1 ζ 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 1 ) + λ 1 p 2 T ζ 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 2 ) + p 2 T ζ 2 α − 1 Γ ( α ) A 13 + ∑ j = 1 n q 1 v 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 1 ) + λ 1 q 2 T v 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 2 ) + q 2 T v 2 α − 1 λ 1 Γ ( α ) A 9 ) × ( A 2 A 5 − A 1 A 6 ) ( A 4 A 5 − A 1 A 8 ) ( A 10 A 13 − A 9 A 14 ) } + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 { T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) A 5 + λ 1 T T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) + T T α − 1 Γ ( α ) A 1 × ( A 10 A 13 − A 9 A 14 ) ( A 11 A 13 − A 9 A 15 ) ( A 2 A 5 − A 1 A 6 ) + ( ∑ j = 1 n p 1 ζ 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 1 ) + λ 1 p 2 T ζ 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 2 ) + p 2 T ζ 2 α − 1 Γ ( α ) A 13 + ∑ j = 1 n q 1 v 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 1 ) + λ 1 q 2 T v 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 2 ) + q 2 T v 2 α − 1 λ 1 Γ ( α ) A 9 ) × ( A 2 A 5 − A 1 A 6 ) ( A 3 A 5 − A 1 A 7 ) ( A 10 A 13 − A 9 A 14 ) } + 1 λ 1 Γ ( α ) ( 1 − e − λ 1 ) ,
M ^ 2 = e − λ 2 τ Δ 2 { ( ∑ i = 1 m r 1 k 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 1 ) + λ 1 r 2 T k 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 2 ) + r 2 T k 2 β − 1 Γ ( β ) A 5 + ∑ i = 1 m s 1 u 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 1 ) + λ 2 s 2 T u 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 2 ) + s 2 T u 2 β − 1 Γ ( β ) A 1 ) × ( A 10 A 13 − A 9 A 14 ) ( A 12 A 13 − A 9 A 16 ) ( A 2 A 5 − A 1 A 6 ) + T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) A 13 + λ 2 T T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) + T T β − 1 Γ ( β ) A 9 × ( A 2 A 5 − A 1 A 6 ) ( A 4 A 5 − A 1 A 8 ) ( A 10 A 13 − A 9 A 14 ) } + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 { ( ∑ i = 1 m r 1 k 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 1 ) + λ 1 r 2 T k 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 2 ) + r 2 T k 2 β − 1 Γ ( β ) A 5 + ∑ i = 1 m s 1 u 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 1 ) + λ 2 s 2 T u 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 2 ) + s 2 T u 2 β − 1 Γ ( β ) A 1 ) × ( A 10 A 13 − A 9 A 14 ) ( A 11 A 13 − A 9 A 15 ) ( A 2 A 5 − A 1 A 6 ) + T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) A 13 + λ 2 T T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) + T T β − 1 Γ ( β ) A 9 × ( A 2 A 5 − A 1 A 6 ) ( A 3 A 5 − A 1 A 7 ) ( A 10 A 13 − A 9 A 14 ) } + 1 λ 2 Γ ( β ) ( 1 − e − λ 2 ) ,
Φ = min { 1 − [ ( M 1 + M 1 ^ ) κ 1 + ( M 2 + M 2 ^ ) κ 1 ^ ] , 1 − [ ( M 1 + M 1 ^ ) κ 2 + ( M 2 + M 2 ^ ) κ 2 ^ ] } .
We apply the following known result to prove the existence of solutions for the problems (3) and (4).
Lemma 2.
Leray–Schauder alternative. Let K ( X ) = { x ∈ D : x = k X ( x ) f o r   s o m e 0 < k < 1 } , where X : D → D is a completely continuous operator. Then either the set F ( X ) is unbounded or there exists at least one fixed point of the operator X.

Existence Results via Leray–Schauder Alternative

In this section, we prove an existence result based on the Leray–Schauder alternative.
Theorem 1.
Assume that ( H 1 ) holds. In addition, it is assumed that
( M 1 + M 2 ) κ 1 + ( M 1 ^ + M 2 ^ ) κ 1 ^ < 1 ,
( M 1 + M 2 ) κ 2 + ( M 1 ^ + M 2 ^ ) κ 2 ^ < 1 ,
then system (3) and (4) has at least one solution on [ 0 , T ] .
Proof. 
First, we show that T is continuous, continuity of functions Y and Z . Let Ω ⊂ 𝒳 × 𝒴 be bounded. Then there exist positive constants L 1 and L 2 such that
| K ( τ , Y ( τ ) , Z ( τ ) ) | ≤ L 1 , | L ( τ , Y ( τ ) , Z ( τ ) ) | ≤ L 2 ,
∀ ( Y , Z ) ∈ Ω . Then for any ( Y , Z ) ∈ Ω , we have
∥ T 1 ( Y , Z ) ( τ ) ∥ ≤ | e − λ 1 τ Δ 1 [ ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( K ( s , Y ( s ) , Z ( s ) ) ) ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + q 2 T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + r 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + s 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 ( ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( K ( s , Y ( s ) , Z ( s ) ) ) ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + q 2 T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + r 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + s 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ) + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s | ,
≤ e − λ 1 τ Δ 1 { ( ∑ j = 1 n p 1 ζ 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 1 ) + λ 1 p 2 T ζ 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 2 ) + p 2 T ζ 2 α − 1 Γ ( α ) A 15 + ∑ j = 1 n q 1 v 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 1 ) + λ 1 q 2 T v 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 2 ) + q 2 T v 2 α − 1 λ 1 Γ ( α ) A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) + T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) A 7 + λ 1 T T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) + T T α − 1 Γ ( α ) A 3 × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) } + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 { ( ∑ j = 1 n p 1 ζ 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 1 ) + λ 1 p 2 T ζ 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 2 ) + p 2 T ζ 2 α − 1 Γ ( α ) A 2 + ∑ j = 1 n q 1 v 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 1 ) + λ 1 q 2 T v 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 2 ) + q 2 T v 2 α − 1 λ 1 Γ ( α ) A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) + T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) A 7 + λ 1 T T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) + T T α − 1 Γ ( α ) A 3 × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) } + 1 λ 1 Γ ( α ) ( 1 − e − λ 1 ) + e − λ 1 τ Δ 1 { T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) A 2 + λ 2 T T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) + T T β − 1 Γ ( β ) A 11 × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) + ( ∑ i = 1 m r 1 k 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 1 ) + λ 2 r 2 T k 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 2 ) + r 2 T k 2 β − 1 Γ ( β ) A 7 + ∑ i = 1 m s 1 u 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 1 ) + λ 2 s 2 T u 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 2 ) + s 2 T u 2 β − 1 Γ ( β ) A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) } + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 { T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) A 15 + λ 2 T T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) + T T β − 1 Γ ( β ) A 11 × ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) + ( ∑ i = 1 m r 1 k 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 1 ) + λ 1 r 2 T k 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 2 ) + r 2 T k 2 β − 1 Γ ( β ) A 7 + ∑ i = 1 m s 1 u 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 1 ) + λ 2 s 2 T u 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 2 ) + s 2 T u 2 β − 1 Γ ( β ) A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) } ,
which implies that
| 𝒯 1 ( Y , Z ) | = M 1 L 1 + M 2 L 2 .
Similarly, we get
| 𝒯 2 ( Y , Z ) | = L 1 M 3 + L 2 M 4 ,
which, in view of the notation (15) and (16) yields
| | 𝒯 1 ( Y , Z ) | | ≤ M 1 L 1 + M 2 L 2 .
Likewise, using the notation (17) and (18), we have
| | 𝒯 2 ( Y , Z ) | | ≤ M 1 ^ L 1 + M 2 ^ L 2 .
Then it follows from (23) and (24) that, from the above inequalities, we conclude that the operator 𝒯 is uniformly bounded, since
| | 𝒯 ( Y , Z ) | | ≤ ( M 1 + M 1 ^ ) L 1 + ( M 2 + M 2 ^ ) L 2 ,
which shows that the operator 𝒯 is uniformly bounded.
Next, we show that 𝒯 is equi-continuous. Let τ 1 , τ 2 ∈ [ 0 , T ] with τ 1 < τ 2 . Then we have
| T 1 ( Y ( τ 2 ) , Y ( τ 2 ) ) − T 1 ( Y ( τ 1 ) , Y ( τ 1 ) ) | ≤ | e − λ 1 τ Δ 1 [ ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( K ( s , Y ( s ) , Z ( s ) ) ) ( ζ 2 ) + ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s A 15 + { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + q 2 T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) + ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + r 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( k 2 ) + ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s } A 7 + { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + s 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 ( ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( K ( s , Y ( s ) , Z ( s ) ) ) ( ζ 2 ) + ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s A 15 + { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + q 2 T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) + ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + r 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( k 2 ) + ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s } A 7 + { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + s 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ) | + | ∫ 0 τ [ e − λ 1 ( τ 2 − s ) − e − λ 1 ( τ 1 − s ) ] I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s | + | ∫ τ 2 τ 1 e − λ 1 ( τ 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s | → 0 ,
clearly, | T 1 ( Y ( τ 2 ) , Y ( τ 2 ) ) − T 1 ( Y ( τ 1 ) , Y ( τ 1 ) ) | → 0 , and similarly | T 2 ( Y ( τ 2 ) , Y ( τ 2 ) ) − T 2 ( Y ( τ 1 ) , Y ( τ 1 ) ) | → 0 as τ 2 → τ 1 independent of ( Y , Z ) ∈ Ω . In consequence, the operator T ( Y , Z ) is equi-continuous. Hence, it follows, according to Arzelá Ascoli theorem, that T ( Y , Z ) is completely continuous.
Finally, it will be established that the set W = { ( Y , Z ) ∈ 𝒳 × 𝒴 | ( Y , Z ) = σ T ( Y , Z ) , 0 < σ < T } and show that it is bounded. Let ( Y , Z ) ∈ W , then ( Y , Z ) = σ T ( Y , Z ) and for any τ ∈ [ 0 , T ] , we have
( Y , Z ) ( τ ) = σ T 1 ( Y , Z ) ( τ ) , σ T 2 ( Y , Z ) ( τ ) .
Then, given the assumption, we obtain
| T 1 ( Y , Z ) ( τ ) | ≤ e − λ 1 τ Δ 1 ( ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s − λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s + q 2 T I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s − T I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s + r 2 T I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s + s 2 T I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s − T I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 ( ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s − ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s + q 2 T I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s − T I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s + r 2 T I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) d s + s 2 T I β − 1 ( n 0 + n 1 | U 1 | + n 2 | U 2 | ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s − T I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ) + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 ( m 0 + m 1 | U 1 | + m 2 | U 2 | ) d s .
Hence, we have
| | Y | | ≤ ( m 0 + m 1 | | U 1 | | + m 2 | | U 2 | | ) M 1 + ( n 0 + n 1 | | U 1 | | + n 2 | | U 2 | | ) M 2 .
Similarly, one can obtain
| | Z | | ≤ ( m 0 + m 1 | | U 1 | | + m 2 | | U 2 | | ) M 3 + ( n 0 + n 1 | | U 1 | | + n 2 | | U 2 | | ) M 4 .
From (26) and (27), we obtain
| | Y | | + | | Z | | ≤ ( m 0 + m 1 | | U 1 | | + m 2 | | U 2 | | ) M 1 + ( n 0 + n 1 | | U 1 | | + n 2 | | U 2 | | ) M 2 + ( m 0 + m 1 | | U 1 | | + m 2 | | U 2 | | ) M 3 + ( n 0 + n 1 | | U 1 | | + n 2 | | U 2 | | ) M 4 ,
Consequently, we get
| | ( Y , Z ) | | ≤ ( m 0 ( M 1 + M 3 ) ) + ( n 0 ( M 2 + M 4 ) ) Φ
which shows that the set W is bounded. Thus, by Lemma 2, the operator T has at least one fixed point. In consequence, the problem (3) and (4) has at least one solution on [ 0 , T ] . The proof is complete. □
We make use of Banach’s contraction mapping principle to establish the existence and uniqueness of solutions for the problem (3) and (4).
Theorem 2.
Assume that Y , Z : [ 0 , T ] × R 2 → R are continuous functions and there exist constants M i , N , i = 1 , 2 such that for all τ ∈ [ 0 , T ] and Y i , Z i ∈ R , i = 1 , 2 .
| K ( τ , Y 1 , Y 2 ) − K ( τ , Z 1 , Z 2 ) | ≤ M 1 | Y 1 − Z 1 | + M 2 | Y 2 − Y 2 | | L ( τ , Y 1 , Y 2 ) − L ( τ , Z 1 , Z 2 ) | ≤ N 1 | Y 1 − Z 1 | + N 2 | Y 2 − Z 2 | .
In addition, assume that
( M 1 + M 3 ) ( M 1 + M 2 ) + ( M 2 + M 4 ) ( N 1 + N 2 ) < 1 ,
where M 1 , M 2 , M 31 , M 4 are given by (15)–(18). Then the boundary value problem (4) has a unique solution.
Proof. 
Fixing sup τ ∈ [ 0 , T ] K ( τ , 0 , 0 ) = Q 1 < ∞ and sup τ ∈ [ 0 , T ] L ( τ , 0 , 0 ) = Q 2 < ∞ and using the assumption ( H 2 ) , we obtain the following:
| K ( τ , Y ( τ ) , Z ( τ ) ) | = | K ( τ , Y ( τ ) , Z ( τ ) ) − K ( τ , 0 , 0 ) + K ( τ , 0 , 0 ) | ≤ M 1 | | Y | | + M 2 | | Z | | + Q 1 , | L ( τ , Y ( τ ) , Z ( τ ) ) | = | L ( τ , Y ( τ ) , Z ( τ ) ) − L ( τ , 0 , 0 ) + L ( τ , 0 , 0 ) | ≤ N 1 | | Y | | + N 2 | | Z | | + Q 2 .
Now, we consider the closed ball B r = { ( Y , Z ) ∈ 𝒳 × 𝒴 : | | ( Y , Z ) | | ≤ r } and show that T ( B r ) ⊂ B r , where
Q 1 ( M 1 + M 3 ) + Q 2 ( M 2 + M 4 ) 1 − [ ( M 1 + M 3 ) ( M 1 + M 2 ) + ( M 2 + M 4 ) ( N 1 + N 2 ) ] < r .
Let ( Y , Z ) ∈ B r . We obtain,
| T 1 ( Y , Z ) ( τ ) | ≤ e − λ 1 τ Δ 1 ( ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s − λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s + q 2 T I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s − T I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s + r 2 T I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s + s 2 T I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s − T I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 ( ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s − ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s + q 2 T I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s − T I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s + r 2 T I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) d s + s 2 T I β − 1 ( N 1 | | Y | | + N 2 | | Z | | + Q 2 ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s − T I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ) + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 ( M 1 | | Y | | + M 2 | | Z | | + Q 1 ) d s ,
which, on taking the norm for τ ∈ [ 0 , T ] , yields
| | ( T 1 ( Y , Z ) ) | | ≤ [ ( M 1 + M 2 ) r + Q 1 ] M 1 + [ ( N 1 + N 2 ) r + Q 2 ] M 2 .
In the same way, we obtain
| | ( T 2 ( Y , Z ) ) | | ≤ [ ( M 1 + M 2 ) r + Q 1 ] M 3 + [ ( N 1 + N 2 ) r + Q 2 ] M 4 .
Consequently, it follows that | | T ( Y , Z ) | | ≤ r , that is, ( Y , Z ) ∈ B r . Hence, T ( B r ) ⊂ B r . Next, we show that the operator T is a contraction. For that, let ( Y 1 , Z 1 ) , ( Y 2 , Z 2 ) ∈ X × Y . Then, for any τ ∈ [ 0 , T ] , we get
| | T 1 ( Y 2 , Z 2 ) ( τ ) − T 1 ( Y 1 , Z 1 ) | | ≤ e − λ 1 τ Δ 1 [ ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s + ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s } A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s + q 2 T I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − T I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( T ) } A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s + r 2 T I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s + s 2 T I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − T I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( T ) } A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 ( ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s + ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s } A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s + q 2 T I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − T I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( T ) } A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s + r 2 T I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s + s 2 T I β − 1 | L ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − L ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s − T I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | ( T ) } A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ) + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 | K ( τ , Y 2 ( τ ) , Z 2 ( τ ) ) − K ( τ , Y 1 ( τ ) , Z 1 ( τ ) ) | d s , ≤ ( M 1 | | Y 2 − Y 1 | | + M 2 | | Z 2 − Z 1 | | ) M 1 + ( N 1 | | Y 2 − Y 1 | | + N 2 | | Z 2 − Z 1 | | ) M 2 .
In consequence, we get
| | T 1 ( Y 2 , Z 2 ) ( τ ) − T 1 ( Y 1 , Z 1 ) ( τ ) | | ≤ [ ( M 1 + M 2 ) M 1 + ( N 1 + N 2 ) M 2 ] ( | | Y 2 − Y 1 | | + | | Z 2 − Z 1 | | ) .
Similarly,
| | T 2 ( Y 2 , Z 2 ) ( τ ) − T 2 ( Y 1 , Z 1 ) ( τ ) | | ≤ [ ( M 1 + M 2 ) M 3 + ( N 1 + N 2 ) M 4 ] ( | | Y 2 − Y 1 | | + | | Z 2 − Z 1 | | ) .
It follows from (28) and (29) that
| | T ( Y 2 , Z 2 ) ( τ ) − T ( Y 1 , Z 1 ) ( τ ) | | ≤ [ ( M 1 + M 2 ) ( M 1 + M 3 ) + ( N 1 + N 2 ) ( M 2 + M 4 ) ] ( | | Y 2 − Y 1 | | + | | Z 2 − Z 1 | | ) .
which, given (19), shows that the operator T is a contraction. Hence, it follows by Banach’s contraction mapping principle that the operator T has a unique fixed point, which corresponds to the unique solution of the problem (3) and (4). This completes the proof. □

4. Hyers–Ulam Stability of the System

In this section, we investigate the Hyers–Ulam stability of the proposed coupled fractional system. Consider the following perturbed system:
D α c + λ 1 D α − 1 c Y ( τ ) − K ( τ , Y ( τ ) , Z ( τ ) ) ≤ ϵ 1 , D β c + λ 2 D β − 1 c Z ( τ ) − L ( τ , Y ( τ ) , Z ( τ ) ) ≤ ϵ 2 , τ ∈ [ 0 , T ] ,
where ϵ 1 , ϵ 2 > 0 are given constants.
Definition 3
([24]). The boundary value problem (3) is said to be Hyers–Ulam stable if there exist positive constants M 1 , M 2 , M 3 , M 4 such that for every ϵ 1 , ϵ 2 > 0 , and for every pair ( Y , Z ) ∈ C ( [ 0 , T ] , R ) × C ( [ 0 , T ] , R ) satisfying (30), there exists an exact solution ( Y * , Z * ) of problem (3) such that
| Y ( τ ) − Y * ( τ ) | ≤ M 1 ϵ 1 + M 2 ϵ 2 , | Z ( τ ) − Z * ( τ ) | ≤ M 3 ϵ 1 + M 4 ϵ 2 ,
for all τ ∈ [ 0 , T ] .
Remark 1.
A pair ( Y , Z ) is said to satisfy the inequality (30) if there exist functions Z 1 , Z 2 ∈ C ( [ 0 , T ] , R ) such that
D α c + λ 1 D α − 1 c Y ( τ ) = K ( τ , Y ( τ ) , Z ( τ ) ) + Z 1 ( τ ) , D β c + λ 2 D β − 1 c Z ( τ ) = L ( τ , Y ( τ ) , Z ( τ ) ) + Z 2 ( τ ) ,
for all τ ∈ [ 0 , T ] , where
| Z 1 ( τ ) | ≤ ϵ 1 , | Z 2 ( τ ) | ≤ ϵ 2 , τ ∈ [ 0 , T ] .
Remark 2.
If ( Y , Z ) represent a solution of inequality (30), then ( Y , Z ) is a solution of the following inequality
| Y ( τ ) − Y * ( τ ) | ≤ M 1 ϵ 1 + M 2 ϵ 2 , τ ∈ [ 0 , T ] , | Z ( τ ) − Z * ( τ ) | ≤ M 3 ϵ 1 + M 4 ϵ 2 , τ ∈ [ 0 , T ] ,
as from Remark 1, we have
( c D α + λ 1 D α − 1 c ) Y ( τ ) = K ( τ , Y ( τ ) , Z ( τ ) ) + Z 1 ( τ ) , τ ∈ [ 0 , T ] , ( c D β + λ 2 D β − 1 c ) Z ( τ ) = L ( τ , Y ( τ ) , Z ( τ ) ) + Z 2 ( τ ) , τ ∈ [ 0 , T ] ,
with the help of Definition 3 and Remark 1, we verified Remark 2, in the following lines
| Y ( τ ) − Y * ( τ ) | ≤ e − λ 1 τ Δ 1 [ ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( K ( s , Y ( s ) , Z ( s ) ) ) ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + q 2 T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + r 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + s 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) ] + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 ( ( { ∑ j = 1 n ( p 1 ∫ 0 ζ 1 e − λ 1 ( ζ 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − ∑ j = 1 n ( λ 1 p 2 T ∫ 0 ζ 2 e − λ 1 ( ζ 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + ∑ j = 1 n λ 1 p 2 T ∫ 0 ζ 2 ( K ( s , Y ( s ) , Z ( s ) ) ) ( ζ 2 ) − ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s A 15 − { q 1 ∫ 0 v 1 e − λ 1 ( v 1 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − ∑ j = 1 n ( λ 1 q 2 T ∫ 0 v 2 e − λ 1 ( v 2 − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s + q 2 T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( v 2 ) + λ 2 T ∫ 0 T e − λ 2 ( T − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) − ( { r 1 ∫ 0 k 1 e − λ 2 ( k 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − λ 2 r 2 T ∫ 0 k 2 e − λ 2 ( k 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + r 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( k 2 ) − ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s } A 7 − { s 1 ∫ 0 u 1 e − λ 2 ( u 1 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s − λ 2 s 2 T ∫ 0 u 2 e − λ 2 ( u 2 − s ) I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) d s + s 2 T I β − 1 ( L ( s , Y ( s ) , Z ( s ) ) ) ( u 2 ) + λ 1 T ∫ 0 T e − λ 1 ( T − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s − T I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) ( T ) } A 3 ) ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) ) + ∫ 0 τ e − λ 1 ( τ − s ) I α − 1 ( K ( s , Y ( s ) , Z ( s ) ) ) d s ,
≤ ϵ 1 [ e − λ 1 τ Δ 1 { ( p 1 ζ 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 1 ) + λ 1 p 2 T ζ 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 2 ) + p 2 T ζ 2 α − 1 Γ ( α ) A 2 + q 1 v 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 1 ) + λ 1 q 2 T v 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 2 ) + q 2 T v 2 α − 1 λ 1 Γ ( α ) A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) + T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) A 7 + λ 1 T T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) + T T α − 1 Γ ( α ) A 3 × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) } + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 { ( p 1 ζ 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 1 ) + λ 1 p 2 T ζ 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 ζ 2 ) + p 2 T ζ 2 α − 1 Γ ( α ) A 2 + q 1 v 1 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 1 ) + λ 1 q 2 T v 2 α λ 1 Γ ( α ) ( 1 − e − λ 1 v 2 ) + q 2 T v 2 α − 1 λ 1 Γ ( α ) A 11 ) × ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) + T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) A 7 + λ 1 T T α λ 1 Γ ( α ) ( 1 − e − λ 1 T ) + T T α − 1 Γ ( α ) A 3 × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) } + 1 λ 1 Γ ( α ) ( 1 − e − λ 1 ) ] + ϵ 2 [ e − λ 1 τ Δ 1 { T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) A 2 + λ 2 T T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) + T T β − 1 Γ ( β ) A 11 × ( A 4 A 7 − A 3 A 8 ) ( A 2 A 7 − A 3 A 6 ) ( A 12 A 15 − A 11 A 16 ) + ( r 1 k 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 1 ) + λ 2 r 2 T k 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 2 ) + r 2 T k 2 β − 1 Γ ( β ) A 7 + s 1 u 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 1 ) + λ 2 s 2 T u 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 2 ) + s 2 T u 2 β − 1 Γ ( β ) A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 4 A 7 − A 3 A 8 ) ( A 10 A 15 − A 11 A 16 ) } + 1 λ 1 ( 1 − e − λ 1 τ ) 1 Δ 1 { ( T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) A 2 + λ 2 T T β λ 2 Γ ( β ) ( 1 − e − λ 2 T ) + T T β − 1 Γ ( β ) A 11 ) ( A 4 A 7 − A 3 A 8 ) ( A 1 A 7 − A 3 A 5 ) ( A 12 A 15 − A 11 A 16 ) + ( r 1 k 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 1 ) + λ 1 r 2 T k 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 k 2 ) + r 2 T k 2 β − 1 Γ ( β ) A 7 + s 1 u 1 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 1 ) + λ 2 s 2 T u 2 β λ 2 Γ ( β ) ( 1 − e − λ 2 u 2 ) + s 2 T u 2 β − 1 Γ ( β ) A 3 ) × ( A 12 A 15 − A 11 A 16 ) ( A 9 A 15 − A 11 A 13 ) ( A 4 A 7 − A 3 A 8 ) } ] ≤ M 1 ϵ 1 + M 2 ϵ 2 .
By the same method, we can obtain that
| Z ( τ ) − Z * ( τ ) | ≤ M 3 ϵ 1 + M 4 ϵ 2 , τ ∈ [ 0 , T ] ,
where M 1 , M 2 , M 3 , M 4 are given by (15)–(18). Hence, Remark 2 is verified, with the help of (31) and (32). Thus, the nonlinear sequential coupled system of Caputo fractional differential equations is Hyers–Ulam stable, and consequently, the system (3) is Hyers–Ulam stable.

5. Examples

Consider the following coupled system of sequential Caputo fractional differential equations:
D α c + λ 1 D α − 1 c Y ( τ ) = K ( τ , Y ( τ ) , Z ( τ ) ) , D β c + λ 2 D β − 1 c Z ( τ ) = L ( τ , Y ( τ ) , Z ( τ ) ) .
The system is supplemented with the coupled multi-point boundary conditions:
Y ( T ) = ∑ i = 1 m r 1 i Z ( k 1 ) + ∑ i = 1 m r 2 i T Z ′ ( k 2 ) , T Y ′ ( T ) = ∑ i = 1 m s 1 i Z ( u 1 ) + ∑ i = 1 m s 2 i T Z ′ ( u 2 ) , Z ( T ) = ∑ j = 1 n p 1 j Y ( ζ 1 ) + ∑ j = 1 n p 2 j T Y ′ ( ζ 2 ) , T Z ′ ( T ) = ∑ j = 1 n q 1 j Y ( v 1 ) + ∑ j = 1 n q 2 j T Y ′ ( v 2 ) ,
for the above system, the associated constants were computed as
A 1 = 0.090718 , A 2 = 0.757735 , A 3 = 0.636377 , A 4 = 0.856517 ,
A 5 = 0.217723 , A 6 = 0.108862 , A 7 = 0.00633549 , A 8 = 0.472758 ,
A 9 = 0.399548 , A 10 = 0.832956 , A 11 = 0.0333733 , A 12 = 0.568604 ,
A 13 = 0.146026 , A 14 = 0.786147 , A 15 = 0.113469 , A 16 = 0.0667465 .
Moreover,
M 1 ≈ 0.472758 , M 2 ≈ 0.000125368 , M ^ 1 ≈ 0.000405607 , M ^ 2 = 0.354819 .
Example 1.
To illustrate Theorem 1, consider the nonlinear functions
K ( τ , Y ( τ ) , Z ( τ ) ) = 1 3 ( τ 2 + 7 ) | Y ( τ ) | 2 1 + | Y ( τ ) | + 1 27 sin Z ( τ ) + 1 8 2 + 9 ,
L ( τ , Y ( τ ) , Z ( τ ) ) = 1 30 ( 1 + | Z ( τ ) | ) | Y ( τ ) | 2 ( 9 + τ 2 ) 2 + Z ( τ ) cos Y ( τ ) 2 ( τ + 4 ) 2 .
From (35), the constants were determined as
κ 0 = 1 24 , κ 1 = 1 21 , κ 2 = 1 27 ,
κ ^ 0 = 1 32 , κ ^ 1 = 1 30 , κ ^ 2 = 1 81 .
Furthermore,
( M 1 + M 2 ) κ 1 + ( M ^ 1 + M ^ 2 ) κ ^ 1 ≈ 0.034359075 < 1 , ( M 1 + M 2 ) κ 2 + ( M ^ 1 + M ^ 2 ) κ ^ 2 ≈ 0.02189968 < 1 .
Therefore, all the assumptions of Theorem 1 were satisfied. Consequently, the boundary value problem (33) and (34), with nonlinearities defined in (35), admitted at least one solution on the interval [ 0 , 2 ] .
Example 2.
To illustrate the applicability of Theorem 2, we consider the nonlinear functions defined by
K ( τ , Y ( τ ) , Z ( τ ) ) = e − τ 2 τ 2 + 40 tan − 1 Y ( τ ) + 1 ( 35 + τ 3 ) 1 + | Z ( τ ) | + cos τ 8 τ 2 + 1 , L ( τ , Y ( τ ) , Z ( τ ) ) = 1 ( τ 4 + 36 ) 1 + | Y ( τ ) | + | Y ( τ ) | τ 2 + 625 cos Z ( τ ) + e τ 2 ( sin 2 τ + 3 ) .
From (37), we obtained the constants
M 1 = 1 40 , M 2 = 1 35 , N 1 = 1 36 , N 2 = 1 25 .
A direct computation yielded
( M 1 + M 3 ) ( M 1 + M 2 ) + ( M 2 + M 4 ) ( N 1 + N 2 ) ≈ 0.0494939148 < 1 .
Therefore, all the hypotheses of Theorem 2 were satisfied. Consequently, the boundary value problem (33), with the functions K and L defined in (37), admitted a unique solution on the interval [ 0 , 2 ] .

6. Conclusions

In this paper, we investigated a nonlinear system of sequential Caputo fractional differential equations subject to coupled mixed multi-point boundary conditions. We established the existence of solutions by applying the Leray–Schauder alternative and proved uniqueness using Banach’s contraction principle. Furthermore, we examined the Hyers–Ulam stability of the system and derived sufficient conditions ensuring its stability.
The results showed that fixed-point techniques provided an effective framework for analyzing coupled fractional systems with nonlocal boundary constraints. The theoretical findings were supported by illustrative examples, which confirmed the validity and applicability of the main results. Several special cases were also presented to demonstrate the flexibility of the proposed framework.
Future research could have addressed more general nonlinear terms or extended the analysis to broader classes of multi-point boundary conditions. The construction of appropriate numerical methods for approximating solutions could also have offered further insight into the qualitative behavior and potential applications of such coupled fractional systems.

Author Contributions

Conceptualization: M.M., S.F.A.; Methodology: S.F.A., H.G.; Software: M.M., H.G.; Validation: S.F.A.; Formal analysis: M.M., S.F.A., H.G.; Investigation: H.G.; Writing—original draft preparation: M.M., S.F.A.; Writing—review and editing: M.M., H.G.; Supervision: S.F.A.; Project administration: H.G. All authors have read and agreed to the published version of the manuscript.

Funding

M. Manigandan would like to thank SRM TRP Engineering College, India, for their financial support, vide number SRM/TRP/RI/005. The authors extend their appreciation to Prince Sattam bin Abdulaziz University for funding this research work through the project number (PSAU/2025/01/32463).

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Conflicts of Interest

The authors declare no conflicts of interest.

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