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Article

Solvability for Two-Point Boundary Value Problems for Nonlinear Variable-Order Fractional Differential Systems

1
School of Mathematical Sciences, University of Jinan, Jinan 250022, China
2
School of Mathematics and Computer Science, Hunan City University, Yiyang 413000, China
*
Authors to whom correspondence should be addressed.
Fractal Fract. 2025, 9(9), 615; https://doi.org/10.3390/fractalfract9090615
Submission received: 21 August 2025 / Revised: 17 September 2025 / Accepted: 21 September 2025 / Published: 22 September 2025

Abstract

A class of boundary value problems for fractional differential systems involving variable-order derivatives is considered. Such problems can be transformed into some boundary value problems for nonlinear Caputo fractional differential systems. Here, the relations between linear Caputo fractional differential equations and their corresponding linear integral equations are investigated, and the results demonstrate that a proper Lipschitz-type condition is needed for studying nonlinear Caputo fractional differential equations. Then, an existence and uniqueness result is established in some vector subspaces by Banach’s fixed-point theorem and · e norm. In addition, two examples are presented to illustrate the theoretical conclusions.

1. Introduction

Fractional differential equations and fractional differential systems are widely used in a variety of fields, and research in these areas has attracted a lot of attention in recent decades; see [1,2,3]. Many interesting results on theoretical knowledge and practical applications have been established; see [4,5,6,7] and the references therein. In [7], Mahmudov and Al-Khateeb considered a coupled system of nonlinear fractional differential equations
D 0 + ζ C u ( χ ) = f ( χ , u ( χ ) , v ( χ ) ) , 0 χ T , D 0 + ξ C v ( χ ) = g ( χ , u ( χ ) , v ( χ ) ) , 0 χ T , u ( T ) = η 1 v ( ρ 1 ) , v ( T ) = η 2 u ( ρ 2 ) , u ( 0 ) = 0 , v ( 0 ) = 0 , ρ 1 , ρ 2 [ 0 , T ] ,
where D 0 + ζ C ,   D 0 + ξ C are the Caputo fractional derivatives, 1 < ζ ,   ξ 2 , f ,   g C ( [ 0 ,   T ]   ×   R 2 , R ) , η 1 ,   η 2 are real constants. By applying Leray–Schauder’s alternative and contraction mapping principle, they proved the existence and uniqueness of the solution. They also discussed the Hyers–Ulam stability of solutions.
Recently, a generalization of fractional operators which allows the order to be a function of certain variables has been discussed in [8,9]. With this variable order, the nonlocal properties of the fractional derivatives and integrals are more evident. In [10], Sun et al. considered that the variable-order fractional derivative is a better approach to represent the memory property.
Variable-order fractional operators have been used to represent numerous problems in science and engineering; see [11]. However, the existence of solutions for variable-order fractional differential equations and systems remains limited in the literature, primarily due to the lack of the semigroup property in variable-order operators; see [12,13,14]. To address the analytical challenges associated with such operators, Zhang [14] introduced piecewise constant functions and established the existence results for a class of initial value problem for fractional differential equations with variable-order derivatives. In [15], with the help of piecewise constant functions, Rezapour and Souid investigated the following boundary value problem (BVP) with a variable-order fractional derivative:
D χ ζ ( χ ) 0 + V u ( χ ) = f ( χ , u ( χ ) ) , 0 χ T , u ( 0 ) = u ( T ) ,
where D χ ζ ( χ ) 0 + V is the variable-order Caputo derivative, 0 < ζ ( χ ) 1 , f C ( [ 0 ,   T ]   ×   R , R ) . Based on Mawhin’s continuation theorem, the authors established some criteria for the existence of solutions.
Motivated by all the work above, in this paper, we consider the following BVP for a coupled system with variable-order fractional derivatives
D χ ζ ( χ ) 0 + V u ( χ ) = f ( χ , u ( χ ) , v ( χ ) ) , 0 < χ < T , D χ ξ ( χ ) 0 + V v ( χ ) = g ( χ , u ( χ ) , v ( χ ) ) , 0 < χ < T , u ( 0 ) = u ( T ) = v ( 0 ) = v ( T ) = 0 ,
where D χ ζ ( χ ) 0 + V ,   D χ ξ ( χ ) 0 + V are variable-order fractional derivatives, 1 < ζ ( χ ) ,   ξ ( χ ) < 2 . Because nonlinear behavior in constant-order differential equations can be modeled as linear behavior in a variable-order calculus framework, ref. [8] considered the variable-order differential system as a more reasonable model for complex physical phenomena. If ζ ( χ )   =   ζ ,   ξ ( χ ) = ξ , D χ ζ ( χ ) 0 + V ,   D χ ξ ( χ ) 0 + V coincides with the constant-order Caputo fractional derivatives, and the problem (1) recovers the classical BVP for fractional differential systems, see [7]. If the nonlinearities simplify to f ( χ ,   u ( χ ) ) and g ( χ ,   u ( χ ) ) , (1) becomes the BVPs for fractional differential equations, see [16]. In this context, the problem (1) stems from the mathematical modeling of anomalous diffusion, especially superdiffusion, in which the variance of the mean squares grows faster than in a Gaussian process [16].
To the best of our knowledge, the problem (1) has not yet been considered in the literature. Based on generalized intervals and piecewise constant functions, we investigate the existence of solutions for problem (1) by Banach’s fixed point theorem and the · e norm. In this paper, the relations between the linear Caputo fractional differential equations and their corresponding linear integral equations are also investigated. Notably, the results demonstrate that a proper Lipschitz-type condition is required to study nonlinear Caputo fractional differential equations.
The paper is organized as follows. In Section 2, some notations, definitions of fractional calculus, and useful lemmas are given. In Section 3, the existence and uniqueness result for the problem (1) is discussed. In Section 4, an example is presented to verify the theoretical conclusions.

2. Preliminaries

Definition 1
([1]). If u A C n [ 0 , T ] , then the Caputo fractional derivative of order ζ ( n 1 ,   n ) for the function u is
D 0 + ζ C u ( χ ) = 0 χ ( χ s ) n ζ 1 Γ ( n ζ ) u ( n ) ( s ) d s , χ > 0 ,
where A C [ 0 ,   T ] is the absolutely continuous function space, A C n [ 0 ,   T ] denote the space of functions u which have continuous derivatives up to order n 1 on [ 0 ,   T ] such that u ( n 1 ) A C [ 0 ,   T ] .
Definition 2
([11]). For ζ ( χ ) ( 1 ,   2 ) , the variable-order fractional derivative is
D χ ζ ( χ ) 0 + V u ( χ )   =   0 χ ( χ     s ) 1     ζ ( χ ) Γ ( 2 ζ ( χ ) ) u ( s ) d s , χ > 0 .
Definition 2 shows that the memory effect of the system considered changes with time and is determined by the current state [10]. When ζ ( χ )   =   ζ is a constant, it reduces to a special case of Definition 1. Moreover, in (2), ζ ( · ) in the integral is a function of χ , which constitutes a fundamental difference from the definition used in [12,13].
Definition 3
([9]). Let ζ ( χ ) > 0 . Then, the variable-order fractional integral is
I χ ζ ( χ ) 0 + V u ( χ )   =   1 Γ ( ζ ( χ ) ) 0 χ ( χ s ) ζ ( χ ) 1 u ( s ) d s , χ > 0 .
Lemma 1
([14]). Let u ( χ ) , ζ ( χ ) , ξ ( χ ) be real functions on [ 0 , T ] , assume that variable-order fractional integrals I χ ζ ( χ ) 0 + V u ( χ ) , I χ ζ ( χ ) + ξ ( χ ) 0 + V u ( χ ) , I χ ζ ( χ ) 0 + V   I χ ξ ( χ ) 0 + V u ( χ ) defined by (3) exist. In the general case, we could claim that
I χ ζ ( χ ) 0 + V   I χ ξ ( χ ) 0 + V u ( χ ) I χ ζ ( χ ) + ξ ( χ ) 0 + V u ( χ ) , for some points in [ 0 ,   T ] .
Definition 4
([14]). An interval I R is termed a generalized interval if I is either an interval, or a point, or the empty set ∅. A partition of I is a finite set P of generalized intervals contained in I, such that every x I lies in exactly one of the generalized interval J in P. A function u : I R is said to be piecewise constant with respect to P, if for every J P , u is constant on J.
Zhang [14] proposed that the Lemma 1 is valid for piecewise constant functions ζ ( χ ) and ξ ( χ ) defined in the same partition of a finite interval [ 0 , T ] .
Lemma 2
([4]). The following assertions hold:
( i )
Let ϵ ( 0 ,   1 ) . Then, I 0 + 1 ϵ maps A C [ 0 ,   T ] to A C [ 0 ,   T ] ;
( i i )
Let ϵ   >   0 and let u L [ 0 ,   T ] or u C [ 0 ,   T ] . Then, D 0 + ϵ C I 0 + ϵ u ( χ ) = u ( χ ) .
Lemma 3.
For 1 < ζ k , ξ k < 2 , 0 < T k T and h 1 , h 2 A C [ 0 ,   T k ] , the solution of the linear system of fractional differential equations
D 0 + ζ k C u ( χ ) = h 1 ( χ ) , 0 < χ < T k , D 0 + ζ k C v ( χ ) = h 2 ( χ ) , 0 < χ < T k , u ( 0 ) = u ( T k ) = v ( 0 ) = v ( T k ) = 0 ,
is equivalent to functions u , v satisfying
u ( χ ) = 1 Γ ( ζ k ) 0 χ ( χ s ) ζ k 1 h 1 ( s ) d s χ T k Γ ( ζ k ) 0 T k ( T k s ) ζ k 1 h 1 ( s ) d s , v ( χ ) = 1 Γ ( ξ k ) 0 χ ( χ s ) ξ k 1 h 2 ( s ) d s χ T k Γ ( ξ k ) 0 T k ( T k s ) ξ k 1 h 2 ( s ) d s .
Proof. 
Solving (4) in a standard manner, it is easy to see that solutions u ( χ ) ,   v ( χ ) satisfy (5).
Conversely, let the functions u , v satisfy (5). Then, we have u ( 0 ) = u ( T k ) = v ( 0 ) = v ( T k ) = 0 and
u ( χ ) = I 0 + ζ k 1 h 1 ( χ ) 1 T k Γ ( ζ k ) 0 T k ( T k s ) ζ k 1 h 1 ( s ) d s , v ( χ ) = I 0 + ξ k 1 h 2 ( χ ) 1 T k Γ ( ξ k ) 0 T k ( T k s ) ξ k 1 h 2 ( s ) d s .
Since 1 < ζ k ,   ξ k < 2 and h 1 , h 2 A C [ 0 , T k ] , it follows from Lemma 2 ( i ) that
I 0 + ζ k 1 h 1 ( χ ) = I 0 + 1 ( 2 ζ k ) h 1 ( χ ) A C [ 0 , T k ] , I 0 + ξ k 1 h 2 ( χ ) = I 0 + 1 ( 2 ξ k ) h 2 ( χ ) A C [ 0 , T k ] .
Thus, u , v A C [ 0 ,   T k ] . By Definition 1 and Lemma 2 ( i i ) , we get D 0 + ζ k 1 C u , D 0 + ξ k 1 C v exist, and
D 0 + ζ k C u ( χ ) = I 0 + 2 ζ k u ( χ ) = I 0 + 1 ( ζ k 1 ) D u ( χ ) =   D 0 + ζ k 1 C u ( χ ) = D 0 + ζ k 1 C I 0 + ζ k 1 h 1 ( χ ) 1 T k Γ ( ζ k ) 0 T k ( T k s ) ζ k 1 h 1 ( s ) d s = h 1 ( χ ) ,
D 0 + ξ k C v ( χ ) = I 0 + 2 ξ k v ( χ ) = I 0 + 1 ( ξ k 1 ) D v ( χ ) = D 0 + ξ k 1 C v ( χ ) = D 0 + ξ k 1 C I 0 + ξ k 1 h 2 ( χ ) 1 T k Γ ( ξ k ) 0 T k ( T k s ) ξ k 1 h 2 ( s ) d s = h 2 ( χ ) .
Consequently, u ( χ ) ,   v ( χ ) is a solution of the problem (4). □
Remark 1.
As mentioned in [4], h 1 , h 2 C [ 0 ,   T k ] is not sufficient in the study of the equivalence between (4) and (5). From Definition 1, it can be seen that u , v A C 2 [ 0 ,   T k ] is an essential condition for the existence of D 0 + ζ k C u and D 0 + ξ k C v . Thus, we can prove that the solution of (4) satisfies (5) under the condition h 1 ,   h 2 C [ 0 ,   T k ] . However, we cannot prove that if h 1 ,   h 2 C [ 0 ,   T k ] , then u ( χ ) ,   v ( χ ) satisfies (5) as a solution of the problem (4).
Let C [ 0 ,   T ] with the norm u   =   max 0 χ T | u ( χ ) | . Then, C [ 0 ,   T ] is a Banach space. Denote
P = { u C [ 0 ,   T ] : u ( χ ) 0 , χ [ 0 ,   T ] } .
Obviously, P is a positive cone in C [ 0 , T ] . Thus, E = C [ 0 ,   T ]   ×   C [ 0 ,   T ] is a Banach space with the norm ( u ,   v ) 1 = max { u ,   v } , and P 1   =   P   ×   P is a cone in E.
Definition 5
([17]). Let P be a cone in real Banach space E and e P { θ } (θ denotes the zero element), set E e = { u E : λ > 0 , λ e u λ e } , and u e = inf { λ > 0 : λ e u λ e , u E e } .
Lemma 4
([17]). Let cone P be normal. Then, E e is a Banach space with e-norm u e .

3. Main Results

In the following, we assume that the condition ( H 1 ) and the Lipschitz-type condition ( H 2 ) hold.
( H 1 )
Let Λ be a partition of the finite interval [ 0 , T ] with Λ = { [ 0 , T 1 ] , ( T 1 , T 2 ] , ( T 2 , T 3 ] , , ( T N 1 , T ] } , and ζ ( χ ) , ξ ( χ ) : [ 0 , T ] ( 1 , 2 ) be piecewise constant functions with respect to Λ , i.e.,
ζ ( χ ) = ζ 1 , 0 χ T 1 , ζ 2 , T 1 < χ T 2 , , , ζ N , T N 1 < χ T , ξ ( χ ) = ξ 1 , 0 χ T 1 , ξ 2 , T 1 < χ T 2 , , , ξ N , T N 1 < χ T ,
where 1 < ζ k , ξ k < 2 ( k = 1 , 2 , , N ) are constants.
( H 2 )
f , g C ( [ 0 , T ] × R × R , R ) and there exist constants l i [ 0 , + ) , i = 0 , 1 , 2 , 3 , 4 , 5 such that
| f ( χ 1 , u 1 , v 1 ) f ( χ 2 , u 2 , v 2 ) |   l 0 | χ 1 χ 2 |   +   l 1 | u 1 u 2 |   +   l 2 | v 1 v 2 | ,
| g ( χ 1 , u 1 , v 1 ) f ( χ 2 , u 2 , v 2 ) |   l 3 | χ 1 χ 2 |   +   l 4 | u 1 u 2 |   +   l 5 | v 1 v 2 | ,
for any χ 1 , χ 2 [ 0 , T ] , and u 1 , u 2 , v 1 , v 2 R .
In view of Definition 2, the equations of the BVP (1) can be formulated as follows:
1 Γ ( 2 ζ ( χ ) ) 0 χ ( χ s ) 1 ζ ( χ ) u ( s ) d s = f ( χ , u ( χ ) , v ( χ ) ) , χ ( 0 , T ) , 1 Γ ( 2 ξ ( χ ) ) 0 χ ( χ s ) 1 ξ ( χ ) v ( s ) d s = g ( χ , u ( χ ) , v ( χ ) ) , χ ( 0 , T ) .
According to ( H 1 ) and Definition 1, Equation (6) in the interval χ ( 0 , T 1 ) can be written by
1 Γ ( 2 ζ 1 ) 0 χ ( χ s ) 1 ζ 1 u ( s ) d s = D 0 + ζ 1 C u ( χ ) = f ( χ , u ( χ ) , v ( χ ) ) , 1 Γ ( 2 ξ 1 ) 0 χ ( χ s ) 1 ξ 1 v ( s ) d s = D 0 + ξ 1 C v ( χ ) = g ( χ , u ( χ ) , v ( χ ) ) .
Equation (6) in the second interval χ ( T 1 , T 2 ) can be written by
1 Γ ( 2 ζ 2 ) 0 χ ( χ s ) 1 ζ 2 u ( s ) d s = f ( χ , u ( χ ) , v ( χ ) ) , 1 Γ ( 2 ξ 2 ) 0 χ ( χ s ) 1 ξ 2 v ( s ) d s = g ( χ , u ( χ ) , v ( χ ) ) .
Similarly to the above argument, Equation (6) in the interval χ ( T k 1 , T k ) , k = 3 , 4 , , N ( T N = T ) can be written by
1 Γ ( 2 ζ k ) 0 χ ( χ s ) 1 ζ k u ( s ) d s = f ( χ , u ( χ ) , v ( χ ) ) , 1 Γ ( 2 ξ k ) 0 χ ( χ s ) 1 ξ k v ( s ) d s = g ( χ , u ( χ ) , v ( χ ) ) .
Remark 2.
The above arguments are based on Definition 2 and Condition ( H 1 ) . They show that the equations of BVP (1) must be expressed through (7)–(9) in the different intervals. This fact affects the means of integrability for such systems and differentiates (1) from the BVP for constant-order fractional differential systems. Furthermore, it indicates that the numerical simulation of such variable-order systems should employ distinct expressions for different intervals.
Definition 6.
The BVP (1) has a solution, if there exist functions ( u k ( t ) , v k ( t ) ) ( k = 1 , 2 , , N ) such that u 1 , v 1 A C 2 [ 0 , T 1 ] satisfying Equation (7) and u 1 ( 0 ) = u 1 ( T 1 ) = v 1 ( 0 ) = v 1 ( T 1 ) = 0 ; u 2 , v 2 A C 2 [ 0 , T 2 ] satisfying Equation (8) and u 2 ( 0 ) = u 2 ( T 2 ) = v 2 ( 0 ) = v 2 ( T 2 ) = 0 ; u k , v k A C 2 [ 0 , T k ] satisfying Equation (9) and u k ( 0 ) = u k ( T k ) = v k ( 0 ) = v k ( T k ) = 0 ( k = 3 , 4 , , N ) .
Definition 7.
The BVP (1) has a unique solution, if functions ( u k ( t ) , v k ( t ) ) ( k = 1 , 2 , , N ) in Definition 6 are unique.
Theorem 1.
Assume that ( H 1 ) , ( H 2 ) hold, and
( H 3 ) max 4 T k ζ k Γ ( ζ k + 2 ) , 4 T k ξ k Γ ( ξ k + 2 ) l < 1 , l = max { l 1 , l 2 , l 4 , l 5 } , k = 1 , 2 , , N .
Then, the BVP (1) has a unique solution.
Proof. 
Equations of the BVP (1) can be formulated as (6). Equation (6) in the interval χ ( 0 , T 1 ) can be written as Equation (7). According to Definition 6, for χ [ 0 , T 1 ] , we consider the auxiliary BVP
D 0 + ζ 1 C u ( χ ) = f ( χ , u ( χ ) , v ( χ ) ) , 0 < χ < T 1 , D 0 + ξ 1 C v ( χ ) = g ( χ , u ( χ ) , v ( χ ) ) , 0 < χ < T 1 , u ( 0 ) = u ( T 1 ) = v ( 0 ) = v ( T 1 ) = 0 .
In order to discuss the existence results for the BVP (10), we define an operator as
F 1 ( u , v ) ( χ ) = ( A 1 ( u , v ) ( χ ) , A 2 ( u , v ) ( χ ) ) , ( u , v ) E 1 = C [ 0 , T 1 ] × C [ 0 , T 1 ] ,
where A 1 , A 2 : E 1 C [ 0 , T 1 ] are defined by
A 1 ( u , v ) ( χ ) = 1 Γ ( ζ 1 ) 0 χ ( χ s ) ζ 1 1 f ( s , u ( s ) , v ( s ) ) d s χ T 1 Γ ( ζ 1 ) 0 T 1 ( T 1 s ) ζ 1 1 f ( s , u ( s ) , v ( s ) ) d s , A 2 ( u , v ) ( χ ) = 1 Γ ( ξ 1 ) 0 χ ( χ s ) ξ 1 1 g ( s , u ( s ) , v ( s ) ) d s χ T 1 Γ ( ξ 1 ) 0 T 1 ( T 1 s ) ξ 1 1 g ( s , u ( s ) , v ( s ) ) d s .
Assume that ( u 1 , v 1 ) E 1 is a fixed point of F 1 . Then, u 1 ( χ ) = A 1 ( u 1 , v 1 ) ( χ ) and v 1 ( χ ) = A 2 ( u 1 , v 1 ) ( χ ) . Since f , g is continuous, we see that u 1 ( χ ) , v 1 ( χ ) exists for χ [ 0 , T 1 ] and max { | u 1 ( χ ) | } < , max { | v 1 ( χ ) | } < , χ [ 0 , T 1 ] . This, together with the condition ( H 2 ) , implies that for ( u 1 , v 1 ) E 1 and χ 1 , χ 2 [ 0 , T 1 ] ,
| f ( χ 1 , u 1 ( χ 1 ) , v 1 ( χ 1 ) ) f ( χ 2 , u 1 ( χ 2 ) , v 1 ( χ 2 ) ) | l 0 | χ 1 χ 2 |   +   l 1 | u 1 ( χ 1 ) u 1 ( χ 2 ) |   +   l 2 | v 1 ( χ 1 ) v 1 ( χ 2 ) | l 0 + l 1 max { | u 1 ( χ ) | } + l 2 max { | v 1 ( χ ) | } | χ 1 χ 2 |
and
| g ( χ 1 , u 1 ( χ 1 ) , v 1 ( χ 1 ) ) g ( χ 2 , u 1 ( χ 2 ) , v 1 ( χ 2 ) ) | l 3 | χ 1 χ 2 |   +   l 4 | u 1 ( χ 1 ) u 1 ( χ 2 ) |   +   l 5 | v 1 ( χ 1 ) v 1 ( χ 2 ) | l 3 + l 4 max { | u 1 ( χ ) | } + l 5 max { | v 1 ( χ ) | } | χ 1 χ 2 |
hold. It follows that f ( · , u 1 ( · ) , v 1 ( · ) ) , g ( · , u 1 ( · ) , v 1 ( · ) ) A C [ 0 , T 1 ] . By Lemma 3, we deduce that ( u 1 , v 1 ) is a solution of the problem (10), and u 1 , v 1 A C 2 [ 0 , T 1 ] . Obviously, u 1 , v 1 satisfies the Equation (7) and u 1 ( 0 ) = u 1 ( T 1 ) = v 1 ( 0 ) = v 1 ( T 1 ) = 0 .
In the next analysis, we will prove that F 1 has a fixed point. For ( u , v ) E 1 , by ( H 2 ) and (11), we have
| A 1 ( u , v ) ( χ ) |   1 Γ ( ζ 1 ) 0 χ ( χ s ) ζ 1 1 | f ( s , u ( s ) , v ( s ) ) f ( s , 0 , 0 ) | + | f ( s , 0 , 0 ) | d s + χ T 1 Γ ( ζ 1 ) 0 T 1 ( T 1 s ) ζ 1 1 | f ( s , u ( s ) , v ( s ) ) f ( s , 0 , 0 ) | + | f ( s , 0 , 0 ) | d s χ ζ 1 Γ ( ζ 1 + 1 ) + T 1 ζ 1 1 χ Γ ( ζ 1 + 1 ) l 1 u + l 2 v + M 1 2 T 1 ζ 1 Γ ( ζ 1 + 1 ) l ( u + v ) + M 1 χ T 1 ,
and
| A 2 ( u , v ) ( χ ) |   1 Γ ( ξ 1 ) 0 χ ( χ s ) ξ 1 1 | g ( s , u ( s ) , v ( s ) ) g ( s , 0 , 0 ) | + | g ( s , 0 , 0 ) | d s + χ T 1 Γ ( ξ 1 ) 0 T 1 ( T 1 s ) ξ 1 1 | g ( s , u ( s ) , v ( s ) ) g ( s , 0 , 0 ) | + | g ( s , 0 , 0 ) | d s χ ξ 1 Γ ( ξ 1 + 1 ) + T 1 ξ 1 1 χ Γ ( ξ 1 + 1 ) l 4 u + l 5 v + M 2 2 T 1 ξ 1 Γ ( ξ 1 + 1 ) l ( u + v ) + M 2 χ T 1 ,
where M 1 = max χ [ 0 , T 1 ] | f ( χ , 0 , 0 ) | , M 2 = max χ [ 0 , T 1 ] | g ( χ , 0 , 0 ) | . Consequently, F 1 maps all of E 1 into the following vector space:
E 1 , e = { ( u , v ) E 1 : λ > 0 , λ e ( χ ) ( u ( χ ) , v ( χ ) ) λ e ( χ ) , χ [ 0 , T 1 ] } ,
where e ( χ ) = χ T 1 , χ T 1 . In view of Definition 5 and Lemma 4, E 1 , e is a subspace of E 1 and E 1 , e is a Banach space with the norm
( u , v ) 1 , e = inf { λ > 0 : λ e ( χ ) ( u ( χ ) , v ( χ ) ) λ e ( χ ) , χ [ 0 , T 1 ] } .
Hence, it suffices to consider the fixed point of F 1 in E 1 , e . Let ( u , v ) , ( u ˜ , v ˜ ) E 1 , e . Then, we have
| A 1 ( u , v ) ( χ ) A 1 ( u ˜ , v ˜ ) ( χ ) | 1 Γ ( ζ 1 ) 0 χ ( χ s ) ζ 1 1 | f ( s , u ( s ) , v ( s ) ) f ( s , u ˜ ( s ) , v ˜ ( s ) ) | d s + χ T 1 Γ ( ζ 1 ) 0 T 1 ( T 1 s ) ζ 1 1 | f ( s , u ( s ) , v ( s ) ) f ( s , u ˜ ( s ) , v ˜ ( s ) ) | d s 1 Γ ( ζ 1 ) 0 χ ( χ s ) ζ 1 1 l 1 | u ( s ) u ˜ ( s ) | + l 2 | v ( s ) v ˜ ( s ) | d s + χ T 1 Γ ( ζ 1 ) 0 T 1 ( T 1 s ) ζ 1 1 l 1 | u ( s ) u ˜ ( s ) | + l 2 | v ( s ) v ˜ ( s ) | d s 1 Γ ( ζ 1 ) 0 χ ( χ s ) ζ 1 1 ( l 1 + l 2 ) ( u , v ) ( u ˜ , v ˜ ) 1 , e s T 1 d s + χ T 1 Γ ( ζ 1 ) 0 T 1 ( T 1 s ) ζ 1 1 ( l 1 + l 2 ) ( u , v ) ( u ˜ , v ˜ ) 1 , e s T 1 d s 1 Γ ( ζ 1 ) 0 χ ( χ s ) ζ 1 1 s T 1 d s + χ T 1 Γ ( ζ 1 ) 0 T 1 ( T 1 s ) ζ 1 1 s T 1 d s 2 l ( u , v ) ( u ˜ , v ˜ ) 1 , e 4 T 1 ζ 1 Γ ( ζ 1 + 2 ) l ( u , v ) ( u ˜ , v ˜ ) 1 , e χ T 1
and
| A 2 ( u , v ) ( χ ) A 2 ( u ˜ , v ˜ ) ( χ ) | 1 Γ ( ξ 1 ) 0 χ ( χ s ) ξ 1 1 | g ( s , u ( s ) , v ( s ) ) g ( s , u ˜ ( s ) , v ˜ ( s ) ) | d s + χ T 1 Γ ( ξ 1 ) 0 T 1 ( T 1 s ) ξ 1 1 | g ( s , u ( s ) , v ( s ) ) g ( s , u ˜ ( s ) , v ˜ ( s ) ) | d s 1 Γ ( ξ 1 ) 0 χ ( χ s ) ξ 1 1 ( l 4 + l 5 ) ( u , v ) ( u ˜ , v ˜ ) 1 , e s T 1 d s + χ T 1 Γ ( ξ 1 ) 0 T 1 ( T 1 s ) ξ 1 1 ( l 4 + l 5 ) ( u , v ) ( u ˜ , v ˜ ) 1 , e s T 1 d s 4 T 1 ξ 1 Γ ( ξ 1 + 2 ) l ( u , v ) ( u ˜ , v ˜ ) 1 , e χ T 1 .
The above inequality implies that
F 1 ( u , v ) F 1 ( u ˜ , v ˜ ) 1 , e max 4 T 1 ζ 1 Γ ( ζ 1 + 2 ) , 4 T 1 ξ 1 Γ ( ξ 1 + 2 ) l ( u , v ) ( u ˜ , v ˜ ) 1 , e .
As condition ( H 3 ) holds, the operator F 1 is a contraction. Hence, F 1 has a unique fixed point ( u 1 , v 1 ) E 1 , e , which is a solution of the problem (10).
We now prove the uniqueness of the solution of the problem (10) in E 1 , e . Suppose that ( u ˜ 1 , v ˜ 1 ) E 1 , e is another solution of (10). From Remark 1, we deduce that ( u ˜ 1 , v ˜ 1 ) satisfies
u ˜ 1 ( χ ) = 1 Γ ( ζ 1 ) 0 χ ( χ s ) ζ 1 1 f ( s , u ˜ 1 ( s ) , v ˜ 1 ( s ) ) d s χ T 1 Γ ( ζ 1 ) 0 T 1 ( T 1 s ) ζ 1 1 f ( s , u ˜ 1 ( s ) , v ˜ 1 ( s ) ) d s , v ˜ 1 ( χ ) = 1 Γ ( ξ 1 ) 0 χ ( χ s ) ξ 1 1 g ( s , u ˜ 1 ( s ) , v ˜ 1 ( s ) ) d s χ T 1 Γ ( ξ 1 ) 0 T 1 ( T 1 s ) ξ 1 1 g ( s , u ˜ 1 ( s ) , v ˜ 1 ( s ) ) d s .
Thus, ( u ˜ 1 , v ˜ 1 ) E 1 , e is a fixed point of the operator F 1 ( u , v ) . It follows that ( u ˜ 1 , v ˜ 1 ) = ( u 1 , v 1 ) . Hence, ( u 1 , v 1 ) is the unique solution of the problem (10).
In addition, the Equation (6) in the interval χ ( T 1 , T 2 ) can be written as Equation (8). In order to consider the existence result of a solution to (8), we consider the following BVP in the interval χ [ 0 , T 2 ]
D 0 + ζ 2 C u ( χ ) = f ( χ , u ( χ ) , v ( χ ) ) , 0 < χ < T 2 , D 0 + ξ 2 C v ( χ ) = g ( χ , u ( χ ) , v ( χ ) ) , 0 < χ < T 2 , u ( 0 ) = u ( T 2 ) = v ( 0 ) = v ( T 2 ) = 0 .
Define an operator as
F 2 ( u , v ) ( χ ) = ( A 3 ( u , v ) ( χ ) , A 4 ( u , v ) ( χ ) ) , ( u , v ) E 2 = C [ 0 , T 2 ] × C [ 0 , T 2 ] ,
where A 3 , A 4 : E 2 C [ 0 , T 2 ] are defined by
A 3 ( u , v ) ( χ = 1 Γ ( ζ 2 ) 0 χ ( χ s ) ζ 2 1 f ( s , u ( s ) , v ( s ) ) d s χ T 2 Γ ( ζ 2 ) 0 T 2 ( T 2 s ) ζ 2 1 f ( s , u ( s ) , v ( s ) ) d s , A 4 ( u , v ) ( χ ) = 1 Γ ( ξ 2 ) 0 χ ( χ s ) ξ 2 1 g ( s , u ( s ) , v ( s ) ) d s χ T 2 Γ ( ξ 2 ) 0 T 2 ( T 2 s ) ξ 2 1 g ( s , u ( s ) , v ( s ) ) d s .
Assume that ( u 2 , v 2 ) E 2 is a fixed point of F 2 . Then, it follows from the continuous of f , g that u 2 ( χ ) , v 2 ( χ ) exists for χ [ 0 , T 2 ] and
max { | u 2 ( χ ) | } = max | I 0 + ζ 2 1 f ( χ , u 2 ( χ ) , v 2 ( χ ) ) 1 T 2 I 0 + ζ 2 f ( T 2 , u 2 ( T 2 ) , v 2 ( T 2 ) ) | < , max { | v 2 ( χ ) | } = max | I 0 + ξ 2 1 g ( χ , u 2 ( χ ) , v 2 ( χ ) ) 1 T 2 I 0 + ξ 2 g ( T 2 , u 2 ( T 2 ) , v 2 ( T 2 ) ) | < ,
Combining with ( H 2 ) , for ( u 2 , v 2 ) E 2 and χ 1 , χ 2 [ 0 , T 2 ] , we have
| f ( χ 1 , u 2 ( χ 1 ) , v 2 ( χ 1 ) ) f ( χ 2 , u 2 ( χ 2 ) , v 2 ( χ 2 ) ) | l 0 + l 1 max { | u 2 ( χ ) | } + l 2 max { | v 2 ( χ ) | } | χ 1 χ 2 |
and
| g ( χ 1 , u 2 ( χ 1 ) , v 2 ( χ 1 ) ) g ( χ 2 , u 2 ( χ 2 ) , v 2 ( χ 2 ) ) | l 3 + l 4 max { | u 2 ( χ ) | } + l 5 max { | v 2 ( χ ) | } | χ 1 χ 2 |
hold. Thus, f ( · , u 2 ( · ) , v 2 ( · ) ) , g ( · , u 2 ( · ) , v 2 ( · ) ) A C [ 0 , T 2 ] . By Lemma 3, it implies that ( u 2 , v 2 ) is a solution of the problem (12), and u 2 , v 2 A C 2 [ 0 , T 2 ] . Moreover, from (13), we get
1 Γ ( 2 ζ 2 ) 0 χ ( χ s ) 1 ζ 2 u 2 ( s ) d s = 1 Γ ( 2 ζ 2 ) 0 χ ( χ s ) 1 ζ 2 ( s ζ 2 2 Γ ( ζ 2 1 ) f ( 0 , u 2 ( 0 ) , v 2 ( 0 ) ) + 1 Γ ( ζ 2 1 ) 0 s ( s τ ) ζ 2 2 d f ( τ , u 2 ( τ ) , v 2 ( τ ) ) ) = f ( 0 , u 2 ( 0 ) , v 2 ( 0 ) ) + 0 χ d f ( s , u 2 ( s ) , v 2 ( s ) ) = f ( χ , u 2 ( χ ) , v 2 ( χ ) ) , 0 < χ < T 2 ,
and
1 Γ ( 2 ξ 2 ) 0 χ ( χ s ) 1 ξ 2 v 2 ( s ) d s = 1 Γ ( 2 ξ 2 ) 0 χ ( χ s ) 1 ξ 2 ( s ξ 2 2 Γ ( ξ 2 1 ) g ( 0 , u 2 ( 0 ) , v 2 ( 0 ) ) + 1 Γ ( ξ 2 1 ) 0 s ( s τ ) ξ 2 2 d g ( τ , u 2 ( τ ) , v 2 ( τ ) ) ) = g ( 0 , u 2 ( 0 ) , v 2 ( 0 ) ) + 0 χ d g ( s , u 2 ( s ) , v 2 ( s ) ) = g ( χ , u 2 ( χ ) , v 2 ( χ ) ) , 0 < χ < T 2 .
It is clear that u 2 , v 2 satisfy
1 Γ ( 2 ζ 2 ) 0 χ ( χ s ) 1 ζ 2 u 2 ( s ) d s = f ( χ , u 2 ( χ ) , v 2 ( χ ) ) , T 1 < χ < T 2 , 1 Γ ( 2 ξ 2 ) 0 χ ( χ s ) 1 ξ 2 v 2 ( s ) d s = g ( χ , u 2 ( χ ) , v 2 ( χ ) ) , T 1 < χ < T 2 ,
which means that the function u 2 , v 2 is a solution of (8) with u ( 0 ) = u ( T 2 ) = v ( 0 ) = v ( T 2 ) = 0 .
Next, we verify that F 2 has a fixed point. According to ( H 2 ) , for ( u , v ) E 2 , we have
| A 3 ( u , v ) ( χ ) | 1 Γ ( ζ 2 ) 0 χ ( χ s ) ζ 2 1 | f ( s , u ( s ) , v ( s ) ) f ( s , 0 , 0 ) | + | f ( s , 0 , 0 ) | d s + χ T 2 Γ ( ζ 2 ) 0 T 2 ( T 2 s ) ζ 2 1 | f ( s , u ( s ) , v ( s ) ) f ( s , 0 , 0 ) | + | f ( s , 0 , 0 ) | d s 2 T 2 ζ 2 Γ ( ζ 2 + 1 ) l ( u + v ) + M 3 χ T 2 ,
and
| A 4 ( u , v ) ( χ ) | 1 Γ ( ξ 2 ) 0 χ ( χ s ) ξ 2 1 | g ( s , u ( s ) , v ( s ) ) g ( s , 0 , 0 ) | + | g ( s , 0 , 0 ) | d s + χ T 2 Γ ( ξ 2 ) 0 T 2 ( T 2 s ) ξ 2 1 | g ( s , u ( s ) , v ( s ) ) g ( s , 0 , 0 ) | + | g ( s , 0 , 0 ) | d s 2 T 2 ξ 2 Γ ( ξ 2 + 1 ) l ( u + v ) + M 4 χ T 2 ,
where M 3 = max χ [ 0 , T 2 ] | f ( χ , 0 , 0 ) | , M 4 = max χ [ 0 , T 2 ] | g ( χ , 0 , 0 ) | . Consequently, F 2 maps all of E 2 into
E 2 , e = { u E 2 : λ > 0 , λ e ( χ ) ( u ( χ ) , v ( χ ) ) λ e ( χ ) , χ [ 0 , T 2 ] } ,
where e ( χ ) = χ T 2 , χ T 2 . Let P 1 = { u C [ 0 , T 2 ] : u ( χ ) 0 , χ [ 0 , T 2 ] } × { u C [ 0 , T 2 ] : u ( χ ) 0 , χ [ 0 , T 2 ] } . Then, P 1 is a normal cone in a real Banach space E 2 and e ( χ ) P 1 { θ } . From Definition 5 and Lemma 4, E 2 , e is a subspace of E 2 and E 2 , e is a Banach space with the norm
( u , v ) 2 , e = inf { λ > 0 : λ e ( χ ) ( u ( χ ) , v ( χ ) ) λ e ( χ ) , χ [ 0 , T 2 ] } .
Let ( u , v ) , ( u ˜ , v ˜ ) E 2 , e . Then, we have
| A 3 ( u , v ) ( χ ) A 3 ( u ˜ , v ˜ ) ( χ ) | 1 Γ ( ζ 2 ) 0 χ ( χ s ) ζ 2 1 l 1 | u ( s ) u ˜ ( s ) | + l 2 | v ( s ) v ˜ ( s ) | d s + χ T 2 Γ ( ζ 2 ) 0 T 2 ( T 2 s ) ζ 2 1 l 1 | u ( s ) u ˜ ( s ) | + l 2 | v ( s ) v ˜ ( s ) | d s 1 Γ ( ζ 2 ) 0 χ ( χ s ) ζ 2 1 ( l 1 + l 2 ) ( u , v ) ( u ˜ , v ˜ ) 2 , e s T 2 d s + χ T 2 Γ ( ζ 2 ) 0 T 2 ( T 2 s ) ζ 2 1 ( l 1 + l 2 ) ( u , v ) ( u ˜ , v ˜ ) 2 , e s T 2 d s 4 T 2 ζ 2 Γ ( ζ 2 + 2 ) l ( u , v ) ( u ˜ , v ˜ ) 2 , e χ T 2
and
| A 4 ( u , v ) ( χ ) A 4 ( u ˜ , v ˜ ) ( χ ) | 1 Γ ( ξ 2 ) 0 χ ( χ s ) ξ 2 1 l 4 | u ( s ) u ˜ ( s ) | + l 5 | v ( s ) v ˜ ( s ) | d s + χ T 2 Γ ( ξ 2 ) 0 T 2 ( T 2 s ) ξ 2 1 l 4 | u ( s ) u ˜ ( s ) | + l 5 | v ( s ) v ˜ ( s ) | d s 1 Γ ( ξ 2 ) 0 χ ( χ s ) ξ 2 1 ( l 4 + l 5 ) ( u , v ) ( u ˜ , v ˜ ) 2 , e s T 2 d s + χ T 2 Γ ( ξ 2 ) 0 T 2 ( T 2 s ) ξ 2 1 ( l 4 + l 5 ) ( u , v ) ( u ˜ , v ˜ ) 2 , e s T 2 d s 4 T 2 ξ 2 Γ ( ξ 2 + 2 ) l ( u , v ) ( u ˜ , v ˜ ) 2 , e χ T 2 .
Hence,
F 2 ( u , v ) F 2 ( u ˜ , v ˜ ) 2 , e max 4 T 2 ζ 2 Γ ( ζ 2 + 2 ) , 4 T 2 ξ 2 Γ ( ξ 2 + 2 ) l ( u , v ) ( u ˜ , v ˜ ) 2 , e .
In view of ( H 3 ) , we find that F 2 has a unique fixed point ( u 2 , v 2 ) E 2 , e , which is a solution of the problem (12). The uniqueness proof for the solution to problem (12) follows the same approach as that used for problem (10), so it is omitted. Consequently, ( u 2 , v 2 ) is the unique solution of (12).
Similarly, for k = 3 , 4 , , N , we can get that (9) has a unique solution u k , v k A C 2 [ 0 , T k ] with u k ( 0 ) = u k ( T k ) = v k ( 0 ) = v k ( T k ) = 0 .
As a result, we find that the BVP (1) has a unique solution. □
Remark 3.
It may be favorable to consider the fixed point of F k in E k , e ( k = 1 , 2 , , N ) . If we consider it in E k , the result of Theorem 1 remains true except that the condition ( H 3 ) is replaced by the following condition:
max 4 T k ζ k Γ ( ζ k + 1 ) , 4 T k ξ k Γ ( ξ k + 1 ) l < 1 , l = max { l 1 , l 2 , l 4 , l 5 } , k = 1 , 2 , , N .
According to 1 < ζ k , ξ k < 2 , we have
1 Γ ( ζ k + 2 ) = 1 Γ ( ζ k + 1 ) · 1 ζ k + 1 < 1 Γ ( ζ k + 1 ) ,
1 Γ ( ξ k + 2 ) = 1 Γ ( ξ k + 1 ) · 1 ξ k + 1 < 1 Γ ( ξ k + 1 ) ,
which yield Theorem 1 which provides the same result with weaker condition.

4. Example

Example 1.
Consider
D χ ζ ( χ ) 0 + V u ( χ ) = 1 30 900 + χ | u ( χ ) | 1 + | u ( χ ) | + sin v ( χ ) , 0 < χ < 1 , D χ ξ ( χ ) 0 + V v ( χ ) = 1 40 1600 + χ 2 | v ( χ ) | 1 + | v ( χ ) | + cos u ( χ ) , 0 < χ < 1 , u ( 0 ) = u ( 1 ) = v ( 0 ) = v ( 1 ) = 0 ,
where ζ , ξ : [ 0 , 1 ] ( 1 , 2 ) defined by
ζ ( χ ) = 1.2 , 0 χ 1 2 , 1.6 , 1 2 < χ 1 , ξ ( χ ) = 1.1 , 0 χ 1 2 , 1.5 , 1 2 < χ 1 ,
and
f ( χ , u ( χ ) , v ( χ ) ) = 1 30 900 + χ | u ( χ ) | 1 + | u ( χ ) | + sin v ( χ ) ,
g ( χ , u ( χ ) , v ( χ ) ) = 1 40 1600 + χ 2 | v ( χ ) | 1 + | v ( χ ) | + cos u ( χ ) .
Clearly, ζ ( χ ) , ξ ( χ ) are piecewise constant functions with respect to the partition [ 0 , 1 2 ] , ( 1 2 , 1 ] of [ 0 , 1 ] , and
| f ( χ 1 , u 1 , v 1 ) f ( χ 2 , u 2 , v 2 ) | = | 1 30 900 + χ 1 | u 1 | 1 + | u 1 | + sin v 1 1 30 900 + χ 2 | u 2 | 1 + | u 2 | + sin v 2 | = | 1 30 900 + χ 1 | u 1 | 1 + | u 1 | + sin v 1 | u 2 | 1 + | u 2 | sin v 2 . + . 1 30 900 + χ 1 1 30 900 + χ 2 | u 2 | 1 + | u 2 | + sin v 2 | 1 900 | | u 1 | 1 + | u 1 | | u 2 | 1 + | u 2 | | + | sin v 1 sin v 2 | + 2 | 1 30 900 + χ 1 1 30 900 + χ 2 | 1 900 ( | | u 1 | | u 2 | | + | v 1 v 2 | ) + 2 | χ 2 χ 1 | 30 900 + χ 1 900 + χ 2 ( 900 + χ 2 + 900 + χ 1 ) 1 900 ( | u 1 u 2 | + | v 1 v 2 | ) + 1 900 2 | χ 1 χ 2 | .
and
| g ( χ 1 , u 1 , v 1 ) g ( χ 2 , u 2 , v 2 ) | = | 1 40 1600 + χ 1 2 | v 1 | 1 + | v 1 | + cos u 1 1 40 1600 + χ 2 2 | v 2 | 1 + | v 2 | + cos u 2 | 1 1600 ( | v 1 v 2 | + | u 1 u 2 | ) + 1 1600 × 800 | χ 1 χ 2 | .
Thus, we have
ζ 1 = 1.2 , ζ 2 = 1.6 , ξ 1 = 1.1 , ξ 2 = 1.5 , T 1 = 1 2 , T 2 = 1 ,
l 0 = 1 900 2 , l 1 = l 2 = 1 900 , l 3 = 1 1600 × 800 , l 4 = l 5 = 1 1600 , l = 1 900 ,
and
max 4 T 1 ζ 1 Γ ( ζ 1 + 2 ) , 4 T 1 ξ 1 Γ ( ξ 1 + 2 ) l = 9.4344 × 10 4 < max 4 T 1 ζ 1 Γ ( ζ 1 + 1 ) , 4 T 1 ξ 1 Γ ( ξ 1 + 1 ) l = 1.9813 × 10 3 < 1 ,
max 4 T 2 ζ 2 Γ ( ζ 2 + 2 ) , 4 T 2 ξ 2 Γ ( ξ 2 + 2 ) l = 1.3373 × 10 3 < max 4 T 2 ζ 2 Γ ( ζ 2 + 1 ) , 4 T 2 ξ 2 Γ ( ξ 2 + 1 ) l = 3.3433 × 10 3 < 1 .
Hence, by Theorem 1 or Remark 3, the BVP (14) has a unique solution. Moreover, Theorem 1 provides the same result with a weaker condition.
Example 2.
Consider
D χ ζ ( χ ) 0 + V u ( χ ) = χ + c 1 20 u ( χ ) + v ( χ ) , 0 < χ < 3 , D χ ξ ( χ ) 0 + V v ( χ ) = χ 2 + c 2 30 ( sin u ( χ ) + sin v ( χ ) ) , 0 < χ < 3 , u ( 0 ) = u ( 3 ) = v ( 0 ) = v ( 3 ) = 0 ,
where 0 c 1 , c 2 1 are real constants, ζ , ξ : [ 0 , 3 ] ( 1 , 2 ) are piecewise constant functions with respect to the partition [ 0 , 1 ] , ( 1 , 2 ] , ( 2 , 3 ] of [ 0 , 3 ] , and ζ , ξ defined by
ζ ( χ ) = 1.2 , 0 χ 1 , 1.5 , 1 < χ 2 , 1.8 , 2 < χ 3 , ξ ( χ ) = 1.1 , 0 χ 1 , 1.4 , 1 < χ 2 , 1.7 , 2 < χ 3 .
Let χ 1 , χ 2 [ 0 , 3 ] , and u 1 , u 2 , v 1 , v 2 R . Then,
| f ( χ 1 , u 1 , v 1 ) f ( χ 2 , u 2 , v 2 ) | | χ 1 χ 2 | + 1 20 | u 1 u 2 | + 1 20 | v 1 v 2 | ,
and
| g ( χ 1 , u 1 , v 1 ) g ( χ 2 , u 2 , v 2 ) | 6 | χ 1 χ 2 | + 1 30 | u 1 u 2 | + 1 30 | v 1 v 2 | .
Using the data given, we find that
ζ 1 = 1.2 , ζ 2 = 1.5 , ζ 3 = 1.8 , ξ 1 = 1.1 , ξ 2 = 1.4 , ξ 3 = 1.7 , T 1 = 1 , T 2 = 2 , T 3 = 3 ,
l 0 = 1 , l 1 = l 2 = 1 20 , l 3 = 6 , l 4 = l 5 = 1 30 , l = 1 20 ,
and
max 4 T 1 ζ 1 Γ ( ζ 1 + 2 ) , 4 T 1 ξ 1 Γ ( ξ 1 + 2 ) l = 0.0910 < 1 , max 4 T 2 ζ 2 Γ ( ζ 2 + 2 ) , 4 T 2 ξ 2 Γ ( ξ 2 + 2 ) l = 0.1770 < 1 ,
max 4 T 3 ζ 3 Γ ( ζ 3 + 2 ) , 4 T 3 ξ 3 Γ ( ξ 3 + 2 ) l = 0.3104 < 1 .
Hence, all the conditions of Theorem 1 are satisfied and, consequently, the BVP (15) has a unique solution.

5. Conclusions

This paper has discussed the existence and uniqueness of a solution for the BVP (1) for a variable-order fractional differential system. Due to the essential difference about the variable-order fractional calculus and the constant-order fractional calculus, we introduce generalized intervals and piecewise constant functions to establish the main result. Based on these definitions, we give the definition of solutions to the BVP (1), and convert the variable-order system to some constant-order systems. Then, we establish some criteria of existence for (1) by using Banach’s fixed point theorem and the · e norm. Two examples are provided to validate the accuracy of these results. Because the variable-order fractional derivatives in BVP (1) show that memory changes over time and is determined by the current state, our results can enrich the theoretical knowledge and applications of the system whose memory property changes over time. For future research directions, we plan to deal with variable-order fractional differential systems with continuous functions characterizing the orders of fractional derivatives, and analyze the behaviors of solutions such as existence, uniqueness, stability and numerical solutions.

Author Contributions

Conceptualization, Y.Z. and R.Y.; methodology, R.Y.; validation, Y.Z. and R.Y.; formal analysis, Y.Z. and R.Y.; investigation, Y.Z. and R.Y.; writing—original draft preparation, R.Y.; writing—review and editing, Y.Z.; funding acquisition, Y.Z. and R.Y. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by National Natural Science Foundation of China (12201199, 61703180), and Science and Technology Program of University of Jinan (1008399).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

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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Zhao, Y.; Yan, R. Solvability for Two-Point Boundary Value Problems for Nonlinear Variable-Order Fractional Differential Systems. Fractal Fract. 2025, 9, 615. https://doi.org/10.3390/fractalfract9090615

AMA Style

Zhao Y, Yan R. Solvability for Two-Point Boundary Value Problems for Nonlinear Variable-Order Fractional Differential Systems. Fractal and Fractional. 2025; 9(9):615. https://doi.org/10.3390/fractalfract9090615

Chicago/Turabian Style

Zhao, Yige, and Rian Yan. 2025. "Solvability for Two-Point Boundary Value Problems for Nonlinear Variable-Order Fractional Differential Systems" Fractal and Fractional 9, no. 9: 615. https://doi.org/10.3390/fractalfract9090615

APA Style

Zhao, Y., & Yan, R. (2025). Solvability for Two-Point Boundary Value Problems for Nonlinear Variable-Order Fractional Differential Systems. Fractal and Fractional, 9(9), 615. https://doi.org/10.3390/fractalfract9090615

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