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Solvability for a system of Hadamard-type hybrid fractional differential inclusions

  • Keyu Zhang and Jiafa Xu EMAIL logo
Published/Copyright: June 15, 2023
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Abstract

In this article, a new system of Hadamard-type hybrid fractional differential inclusions equipped with Dirichlet boundary conditions was constructed. By virtue of a fixed-point theorem due to B. C. Dhage, (Existence results for neutral functional differential inclusions in Banach algebras, Nonlinear Anal. 64 (2006), no. 6, 1290–1306, doi: https://doi.org/10.1016/j.na.2005.06.036), the existence results of solutions for the considered problem are derived in a new norm space for multivalued maps. A numerical example is provided to illustrate our main results.

MSC 2010: 34A05; 34B18; 26A33

1 Introduction

The purpose of this article is to investigate the following system of Hadamard-type fractional hybrid differential inclusions given by:

(1) D α 1 x ( t ) f 1 ( t , x ( t ) , y ( t ) ) F 1 ( t , x ( t ) , y ( t ) ) , 1 < t < e , 1 < α 1 2 , D α 2 y ( t ) f 2 ( t , x ( t ) , y ( t ) ) F 2 ( t , x ( t ) , y ( t ) ) , 1 < t < e , 1 < α 2 2 , x ( 1 ) = x ( e ) = 0 , y ( 1 ) = y ( e ) = 0 ,

where D α 1 and D α 2 denote the Hadamard-type fractional derivatives of order α 1 and α 2 , f 1 , f 2 C ( [ 1 , e ] × R 2 , R \ { 0 } ) are the continuous functions, F 1 , F 2 : [ 1 , e ] × R 2 T ( R ) are the multi-valued maps, and T ( R ) is the family of all nonempty subsets of R .

Due to fractional-order derivative generalizes the classical integer-order derivative to an arbitrary-order case. Fractional-order differential equations can more accurately describe various phenomenon than integer-order differential equations in many complex and widespread fields such as physics, mechanics, chemistry, and engineering (see the books [15] and references cited therein). There has been a rapid increase in the number of fractional differential equations from both theoretical and applied perspectives (see [625]).

Note that most of the works on fractional differential equations involve either Riemann-Liouville-type derivative or Caputo-type derivative (see [810,12,15,16,19]). However, there is also another concept of Hadamard-type fractional derivative, which was first introduced by Hadamard in 1892 [26], which contains a logarithmic function of arbitrary exponent in the kernel of integral appearing in its definition. Hadmard-type integrals arise in the formulation of many problems in mechanics such as in fracture analysis. For details and applications of Hadamard-type fractional derivative and integral, see [2733].

The study of fractional differential inclusions also gained much attraction and interest as these tools of mathematical analysis are found to have a wide range of utility in stochastic modeling and optimal control problems [34]. Some recent achievement to the subject of fractional differential inclusions can be discovered in [3539] and references cited therein. In particular, the fractional differential inclusion for various types of single fractional differential equations with different boundary conditions is studied systematically in [28].

To our best knowledge, almost all fractional hybrid differential inclusions equipped with initial and boundary-value problem were focused on a single fractional hybrid differential equation in [28]. Is there a similar result for a system of coupled hybrid fractional differential inclusions of Hadamard-type? The main objective of this work is to obtain an existence result for system (1) under Lipschitz and Carathéodory conditions in virtue of a fixed-point theorem by Dhage [40] in a new Banach space. Inspired by the work mentioned earlier, this research (1) is also different from the recent results [3539]. This means that our work is new in the present configuration and contributes. We overcome some difficulties in proving operator convexity and closed graphs.

2 Preliminaries

In this section, we first recall some useful materials for Hadamard-type fractional derivatives and integrals.

Definition 1

[4,5] The Hadamard-type fractional derivative of order q > 0 for an integrable function g : [ 1 , + ) R is defined as:

D q g ( t ) = 1 Γ ( n q ) t d d t n 1 t log t s n q 1 g ( s ) s d s , n 1 < q < n ,

where n = [ q ] + 1 , [ q ] is the smallest integer greater than or equal to q and log ( ) = log e ( ) .

Definition 2

[4,5] The Hadamard-type fractional integral of order q > 0 for an integrable function g is defined as:

I q g ( t ) = 1 Γ ( q ) 1 t log t s q 1 g ( s ) s d s ,

provided that the integral exists.

In what follows, we provide some lemmas that are helpful to the proof of main theorems.

Lemma 1

Let ϕ 1 , ϕ 2 L 1 ( [ 1 , e ] , R ) be continuous functions. Then, the integral solution of the Hadamard-type fractional differential system

(2) D α 1 x ( t ) f 1 ( t , x ( t ) , y ( t ) ) = ϕ 1 ( t ) , 1 < t < e , 1 < α 1 2 , D α 2 y ( t ) f 2 ( t , x ( t ) , y ( t ) ) = ϕ 2 ( t ) , 1 < t < e , 1 < α 2 2 , x ( 1 ) = x ( e ) = 0 , y ( 1 ) = y ( e ) = 0 ,

is given by:

(3) x ( t ) = f 1 ( t , x ( t ) , y ( t ) ) 1 Γ ( α 1 ) 1 t log t s α 1 1 ϕ 1 ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 ϕ 1 ( s ) s d s , y ( t ) = f 2 ( t , x ( t ) , y ( t ) ) 1 Γ ( α 2 ) 1 t log t s α 2 1 ϕ 2 ( s ) s d s ( log t ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 ϕ 2 ( s ) s d s .

Proof

As argued in Chapter 9 in [28], the solution for (2) can be written as:

(4) x ( t ) = f 1 ( t , x ( t ) , y ( t ) ) 1 Γ ( α 1 ) 1 t log t s α 1 1 ϕ 1 ( s ) s d s + c 1 ( log t ) α 1 1 + c 2 ( log t ) α 1 2 , y ( t ) = f 2 ( t , x ( t ) , y ( t ) ) 1 Γ ( α 2 ) 1 t log t s α 2 1 ϕ 2 ( s ) s d s + d 1 ( log t ) α 2 1 + d 2 ( log t ) α 2 2 ,

where c 1 , c 2 , d 1 , d 2 R . Note that x ( 1 ) = x ( e ) = y ( 1 ) = y ( e ) = 0 in (2), we have

c 2 = 0 , c 1 = 1 Γ ( α 1 ) 1 e log e s α 1 1 ϕ 1 ( s ) s d s , d 2 = 0 , d 1 = 1 Γ ( α 2 ) 1 e log e s α 2 1 ϕ 2 ( s ) s d s .

Substituting these values into (4), we can obtain (3).□

Definition 3

A pair of function ( x , y ) is called the solution of system (1) if there exists a pair of function ( ϕ 1 , ϕ 2 ) L 1 ( [ 1 , e ] , R ) × L 1 ( [ 1 , e ] , R ) with ϕ 1 F 1 ( t , x ( t ) , y ( t ) ) and ϕ 2 F 2 ( t , x ( t ) , y ( t ) ) such that D α 1 x ( t ) f 1 ( t , x ( t ) , y ( t ) ) = ϕ 1 ( t ) , D α 2 x ( t ) f 2 ( t , x ( t ) , y ( t ) ) = ϕ 2 ( t ) almost every on [ 1 , e ] and x ( 1 ) = x ( e ) = y ( 1 ) = y ( e ) = 0 .

Next, we introduce some preliminary materials about norm spaces and multi-valued maps. Let X = C ( [ 1 , e ] , R ) denote a Banach space of continuous functions from [ 1 , e ] into R under the norm x = sup t [ 1 , e ] x ( t ) , which becomes a Banach algebra with respect to the multiplication “ ” defined by ( x , y ) ( t ) = x ( t ) y ( t ) for arbitrarily x , y X . The product space Ξ = X × X is also a Banach space under the norm ( x , y ) = x + y , which further becomes a Banach algebra with respect to the multiplication “ ” defined by ( ( x , y ) ( x ¯ , y ¯ ) ) ( t ) = ( x , y ) ( t ) ( x ¯ , y ¯ ) ( t ) = ( x ( t ) x ¯ ( t ) , y ( t ) y ¯ ( t ) ) for ( x , y ) , ( x ¯ , y ¯ ) X × X . For more details about the results concerning algebraic structure of the product space Ξ = X × X , please see [41,42].

As aforementioned, ( Ξ , ( , ) ) is a Banach space. Now, we redeclare some basic definitions on multi-valued maps (see [43]):

T cl ( Ξ ) = { F T ( Ξ ) : F is closed } , T b ( Ξ ) = { F T ( Ξ ) : F is bounded } , T cp ( Ξ ) = { F T ( Ξ ) : F is compact } , T cp , cv ( Ξ ) = { F T ( Ξ ) : F is compact and convex } .

Definition 4

A multi-valued map G : Ξ T ( Ξ ) is called convex (closed) valued if G ( x , y ) is convex (closed) for all ( x , y ) Ξ .

Definition 5

The map G is called bounded on bounded sets if G ( B ) = ( x , y ) B G ( x , y ) is bounded in Ξ for any bounded set B of Ξ (i.e., sup ( x , y ) B { ( u , v ) : ( u , v ) G ( x , y ) } < ).

Definition 6

The map G is called upper semi-continuous (u.s.c.) on Ξ if for each ( x , y ) Ξ , the set G ( x , y ) is a nonempty closed subset of Ξ , and if for each open set B of Ξ containing G ( x , y ) , there exists an open neighborhood N of ( x , y ) such that G ( N ) B .

Definition 7

The map G is called completely continuous if G ( B ) is relatively compact for every bounded subset B of Ξ .

Definition 8

A multi-valued map F : [ 1 , e ] × R 2 T ( R ) is called L 1 -Carathéodory if

  1. t F ( t , x , y ) is measurable for each ( x , y ) R × R ,

  2. ( x , y ) F ( t , x , y ) is upper semicontinuous for almost all t [ 1 , e ] ,

  3. there exists a function g r L 1 ( [ 1 , e ] , R + ) such that

    F ( t , x , y ) = sup { v : v F ( t , x , y ) g r ( t ) } ,

    for all x , y R with ( x , y ) r and for almost every t [ 1 , e ] .

For each ( x , y ) Ξ , define the set of selections of F x y = ( F 1 , x y , F 2 , x y ) by:

F 1 , x y { v 1 L 1 ( [ 1 , e ] , R ) : v 1 ( t ) F 1 ( t , x ( t ) , y ( t ) ) , for a.e. t [ 1 , e ] } , F 2 , x y { v 2 L 1 ( [ 1 , e ] , R ) : v 2 ( t ) F 2 ( t , x ( t ) , y ( t ) ) , for a.e. t [ 1 , e ] } .

Lemma 2

[40] Let X be a Banach algebra, A : X X a single-valued operator, and B : X T cp , cv ( X ) a multi-valued operator satisfying the following conditions:

  1. A is a single-valued Lipschitz with Lipschitz constant k ,

  2. B is compact and upper semicontinuous,

  3. 2 M k < 1 , where M = B ( X ) .

Then, either:
  1. the operator inclusion x A x B x has a solution,

or
  1. the set Φ = { u X μ u A u B u , μ > 1 } is unbounded.

Lemma 3

[43] If G : X T cl ( Y ) is upper semi-continuous, then G r ( G ) = { ( x , y ) X × Y , y G ( x ) } is a closed subset of X × Y ; i.e., for every sequence { x n } n N X and { y n } n N Y , if when n , x n x , y n y and y n G ( x n ) , then y G ( x ) . Conversely, if G is completely continuous and has a closed graph, then it is upper semi-continuous.

Lemma 4

[44] Let X be a Banach space, F : [ 1 , e ] × R T cp , cv ( X ) an L 1 -Carathéodory multi-valued map, and G a linear continuous mapping from L 1 ( [ 1 , e ] , X ) to C ( [ 1 , e ] , X ) . Then, the operator

G F : C ( [ 1 , e ] , X ) T cp , cv ( C ( [ 1 , e ] , X ) )

x ( G F ) ( x ) = G ( F ( x ) )

is a closed graph operator in C ( [ 1 , e ] , X ) × C ( [ 1 , e ] , X ) .

3 Main results

Let L 1 ( [ 1 , e ] , R ) be the Banach space of measurable functions x : [ 1 , e ] R , which is Lebesgue integrable and normed by x L 1 = 1 e x ( t ) d t . Now, we are in a position to show the existence of solutions for system (1).

Theorem 1

Suppose that

(H1) Functions f i : [ 1 , e ] × R × R R \ { 0 } are continuous and there exist constants L i > 0 such that

f i ( t , x , y ) f i ( t , x ¯ , y ¯ ) L i [ x ( t ) x ¯ ( t ) + y ( t ) y ¯ ( t ) ] , i = 1 , 2 ,

for almost every t [ 1 , e ] , x , y , x ¯ , y ¯ R ;

(H2) operators F i : [ 1 , e ] × R × R T ( R ) are L 1 -Carathéodory and have nonempty compact and convex values;

(H3) there exists a positive real number r such that

r > 2 F 10 g 1 r L 1 Γ ( α 1 ) + 2 F 20 g 2 r L 1 Γ ( α 2 ) 1 2 k g 1 r L 1 Γ ( α 1 ) 2 k g 2 r L 1 Γ ( α 2 ) ,

where 2 k g 1 r L 1 Γ ( α 1 ) + 2 k g 2 r L 1 Γ ( α 2 ) < 1 2 , k = L 1 + L 2 , F 10 = sup t [ 1 , e ] f 1 ( t , 0 , 0 ) , F 20 = sup t [ 1 , e ] f 2 ( t , 0 , 0 ) , here g 1 r ( t ) and g 2 r ( t ) have similar approach as Definition 8.

Then, system (1) has at least one solution on [ 1 , e ] × [ 1 , e ] .

Proof

Applying Lemma 1, we transform syetem (1) into an equivalent fixed-point problem. Define the operator T : Ξ T ( Ξ ) as T ( x , y ) ( t ) = ( T 1 ( x , y ) ( t ) , T 2 ( x , y ) ( t ) ) , where

(5) T 1 ( x , y ) ( t ) = h 1 C ( [ 1 , e ] , R ) : h 1 ( t ) = f 1 ( t , x ( t ) , y ( t ) ) 1 Γ ( α 1 ) 1 t log t s α 1 1 v 1 ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 v 1 ( s ) s d s , v 1 F 1 , x y

and

(6) T 2 ( x , y ) ( t ) = h 2 C ( [ 1 , e ] , R ) : h 2 ( t ) = f 2 ( t , x ( t ) , y ( t ) ) 1 Γ ( α 2 ) 1 t log t s α 2 1 v 2 ( s ) s d s ( log t ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 v 2 ( s ) s d s , v 2 F 2 , x y .

Next, we consider two operators A = ( A 1 , A 2 ) and B = ( B 1 , B 2 ) , where A i : Ξ Ξ are given by:

A i ( x , y ) ( t ) = f i ( t , x ( t ) , y ( t ) ) , t [ 1 , e ] , i = 1 , 2 ,

and B i : Ξ T ( Ξ ) are given by:

(7) B i ( x , y ) ( t ) = h i C ( [ 1 , e ] , R ) : h i ( t ) = 1 Γ ( α i ) 1 t log t s α i 1 v i ( s ) s d s ( log t ) α i 1 Γ ( α i ) 1 e log e s α i 1 v i ( s ) s d s , v i F i , x y , t [ 1 , e ] , i = 1 , 2 .

Note that T i ( x , y ) = A i ( x , y ) B i ( x , y ) , i = 1 , 2 , then T ( x , y ) = ( A 1 ( x , y ) B 1 ( x , y ) , A 2 ( x , y ) B 2 ( x , y ) ) . We next show that the operators A and B satisfy all the conditions of Lemma 2. For clarity, we display the proof in several steps.

Step 1. We first show that Lemma 2(a) holds, i.e., the single-valued operator A is a Lipschitz on Ξ . Let ( x , y ) , ( x ¯ , y ¯ ) Ξ , by (H1), we have

(8) A i ( x , y ) A i ( x ¯ , y ¯ ) = f i ( t , x , y ) f i ( t , x ¯ , y ¯ ) L i [ x ( t ) x ¯ ( t ) + y ( t ) y ¯ ( t ) ] L i [ x x ¯ + y y ¯ ] , t [ 1 , e ] , i = 1 , 2 .

Consequently,

(9) A ( x , y ) A ( x ¯ , y ¯ ) = A 1 ( x , y ) A 1 ( x ¯ , y ¯ ) + A 2 ( x , y ) A 2 ( x ¯ , y ¯ ) L 1 [ x x ¯ + y y ¯ ] + L 2 [ x x ¯ + y y ¯ ] ( L 1 + L 2 ) ( x , y ) ( x ¯ , y ¯ ) .

So, the single-valued operator A is a Lipschitz on Ξ with Lipschitz constant k = L 1 + L 2 .

Step 2. We show that Lemma 2(b) holds, i.e., the multi-valued operator B is compact and upper semicontinuous on Ξ .

(i) We show that the operator B has convex values. Let u 11 , u 12 B 1 ( x , y ) , u 21 , u 22 B 2 ( x , y ) . Then, there are v 11 , v 12 F 1 , x y , v 21 , and v 22 F 2 , x y such that

(10) u 1 j ( t ) = 1 Γ ( α 1 ) 1 t log t s α 1 1 v 1 j ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 v 1 j ( s ) s d s , j = 1 , 2 , t [ 1 , e ] ,

(11) u 2 j ( t ) = 1 Γ ( α 2 ) 1 t log t s α 2 1 v 2 j ( s ) s d s ( log t ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 v 2 j ( s ) s d s , j = 1 , 2 , t [ 1 , e ] .

For any constant λ [ 0 , 1 ] , we have

(12) λ u 11 ( t ) + ( 1 λ ) u 12 ( t ) = 1 Γ ( α 1 ) 1 t log t s α 1 1 λ v 11 ( s ) + ( 1 λ ) v 12 ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 λ v 11 ( s ) + ( 1 λ ) v 12 ( s ) s d s ,

(13) λ u 21 ( t ) + ( 1 λ ) u 22 ( t ) = 1 Γ ( α 2 ) 1 t log t s α 2 1 λ v 21 ( s ) + ( 1 λ ) v 22 ( s ) s d s ( log t ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 λ v 21 ( s ) + ( 1 λ ) v 22 ( s ) s d s ,

where v ¯ 1 ( t ) = λ v 11 ( t ) + ( 1 λ ) v 12 ( t ) F 1 , x y , v ¯ 2 ( t ) = λ v 21 ( t ) + ( 1 λ ) v 22 ( t ) F 2 , x y for all t [ 1 , e ] .

Therefore, we have

λ u 11 ( t ) + ( 1 λ ) u 12 ( t ) B 1 ( x , y ) , λ u 21 ( t ) + ( 1 λ ) u 22 ( t ) B 2 ( x , y ) , B ( λ u 11 ( t ) + ( 1 λ ) u 12 ( t ) , λ u 21 ( t ) + ( 1 λ ) u 22 ( t ) ) = λ B ( u 11 ( t ) , u 21 ( t ) ) + ( 1 λ ) B ( u 12 ( t ) , u 22 ( t ) ) B ( x , y ) .

Then, we obtain the operator B ( x , y ) which is convex for each ( x , y ) Ξ . Then, operator B defines a multi-valued operator B : Ξ T cv ( Ξ ) .

(ii) We display that the operator B maps bounded sets into bounded sets in Ξ . Let Ω = { ( x , y ) ( x , y ) < r , ( x , y ) Ξ } . Then, for each h i B i ( x , y ) , i = 1 , 2 , there are v i F i , x y ( i = 1 , 2 ) such that

(14) h i ( t ) = 1 Γ ( α i ) 1 t log t s α i 1 v i ( s ) s d s ( log t ) α i 1 Γ ( α i ) 1 e log e s α i 1 v i ( s ) s d s , t [ 1 , e ] .

From (H2), we have

(15) B 1 ( x , y ) ( t ) = 1 Γ ( α 1 ) 1 t log t s α 1 1 v 1 ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 v 1 ( s ) s d s 1 Γ ( α 1 ) 1 t log t s α 1 1 g 1 r ( s ) s d s + ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 g 1 r ( s ) s d s 2 Γ ( α 1 + 1 ) g 1 r L 1

and

(16) B 2 ( x , y ) ( t ) 2 Γ ( α 2 + 1 ) g 2 r L 1 .

This implies that

(17) B ( x , y ) = B 1 ( x , y ) + B 2 ( x , y ) 2 Γ ( α 1 + 1 ) g 1 r L 1 + 2 Γ ( α 2 + 1 ) g 2 r L 1 .

Thus, B ( Ξ ) is uniformly bounded. Then, B defines a multi-valued operator B : Ξ T b ( Ξ ) .

(iii) We show that the operator B maps bounded sets into equicontinuous sets. Let q i B i ( x , y ) ( i = 1 , 2 ) for some ( x , y ) Ω , where Ω is given as earlier. So, there exists u i F i , x y , such that

(18) q i ( t ) = 1 Γ ( α i ) 1 t log t s α i 1 u i ( s ) s d s ( log t ) α i 1 Γ ( α i ) 1 e log e s α i 1 u i ( s ) s d s , i = 1 , 2 .

For any τ 1 , τ 2 [ 1 , e ] and τ 1 < τ 2 , we have

(19) q 1 ( τ 1 ) q 1 ( τ 2 ) g 1 r L 1 Γ ( α i ) 1 τ 1 log τ 1 s α 1 1 1 s d s 1 τ 2 log τ 2 s α 1 1 1 s d s + g 1 r L 1 ( log τ 1 ) α 1 1 ( log τ 2 ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 1 s d s g 1 r L 1 Γ ( α 1 ) 1 τ 1 log τ 1 s α 1 1 log τ 2 s α 1 1 1 s d s + g 1 r L 1 Γ ( α 1 ) τ 1 τ 2 log τ 2 s α 1 1 1 s d s + g 1 r L 1 ( log τ 1 ) α 1 1 ( log τ 2 ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 1 s d s

and

(20) q 2 ( τ 1 ) q 2 ( τ 2 ) g 2 r L 1 Γ ( α 2 ) 1 τ 1 log τ 1 s α 2 1 log τ 2 s α 2 1 1 s d s + g 2 r L 1 Γ ( α 2 ) τ 1 τ 2 log τ 2 s α 2 1 1 s d s + g 2 r L 1 ( log τ 1 ) α 2 1 ( log τ 2 ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 1 s d s .

Note that the right-hand side of the two inequalities (19) and (20) go to zero for arbitrarily ( x , y ) Ω as τ 2 τ 1 . Therefore, B 1 and B 2 are equicontinuous sets. Also, note that B ( x , y ) = B 1 ( x , y ) + B 2 ( x , y ) , so B is equicontinuous sets.

From (ii)–(iii) and the Arzel á -Ascoli theorem, we have B : Ξ T ( Ξ ) is completely continuous. Thus, B defines a compact multi-valued operator B : Ξ T cp ( Ξ ) .

(iv) We claim that B has a closed graph. Let ( x n , y n ) ( x , y ) , ( h 1 n , h 2 n ) B ( x n , y n ) and ( h 1 n , h 2 n ) ( h 1 , h 2 ) . Then, we need to prove that ( h 1 , h 2 ) B ( x , y ) , i.e., h 1 B 1 ( x , y ) , h 2 B 2 ( x , y ) . Due to h 1 n B 1 ( x n , y n ) , h 2 n B 2 ( x n , y n ) , there are v 1 n F 1 , x y , v 2 n F 2 , x y such that

(21) h 1 n ( t ) = 1 Γ ( α 1 ) 1 t log t s α 1 1 v 1 n ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 v 1 n ( s ) s d s , t [ 1 , e ]

and

(22) h 2 n ( t ) = 1 Γ ( α 2 ) 1 t log t s α 2 1 v 2 n ( s ) s d s ( log t ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 v 2 n ( s ) s d s , t [ 1 , e ] .

Thus, it suffices to show that there are v 1 F 1 , x y , v 2 F 2 , x y such that for each t [ 1 , e ] ,

(23) h 1 = 1 Γ ( α i ) 1 t log t s α 1 1 v i ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 v 1 ( s ) s d s B 1 ( x , y )

and

(24) h 2 = 1 Γ ( α 2 ) 1 t log t s α 2 1 v 2 ( s ) s d s ( log t ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 v 2 ( s ) s d s B 2 ( x , y ) .

Let us take the linear operator Θ = ( Θ 1 , Θ 2 ) , where Θ i : L 1 ( [ 1 , e ] , R ) C ( [ 1 , e ] , R ) are given by:

(25) Θ i ( v i ) ( t ) = 1 Γ ( α i ) 1 t log t s α i 1 v i ( s ) s d s ( log t ) α i 1 Γ ( α i ) 1 e log e s α i 1 v i ( s ) s d s , i = 1 , 2 .

On the other hand,

(26) h 1 n ( t ) h 1 ( t ) = 1 Γ ( α 1 ) 1 t log t s α 1 1 v 1 n ( s ) v 1 ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 v 1 n ( s ) v 1 ( s ) s d s 0 , as n ,

and

(27) h 2 n ( t ) h 2 ( t ) = 1 Γ ( α 2 ) 1 t log t s α 2 1 v 2 n ( s ) v 2 ( s ) s d s ( log t ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 v 2 n ( s ) v 2 ( s ) s d s 0 , as n .

So it follows from Lemma 4 that Θ i F i , x y are the closed graph operators. This means that if h i n ( t ) Θ i ( F i , x y ) , then we obtain ( h 1 , h 2 ) B ( x , y ) since ( x n , y n ) ( x , y ) .

Note that B : Ξ T ( Ξ ) is completely continuous; it follows from Lemma 3 that the operator B is upper semicontinuous operator on Ξ .

Step 3. We show that Lemma 2(c) holds. From (H3), we have M = B ( Ξ ) = B ( x , y ) = B 1 ( x , y ) + B 2 ( x , y ) ( 2 / Γ ( α 1 ) ) g 1 r L 1 + ( 2 / Γ ( α 2 ) ) g 2 r L 1 , and k = L 1 + L 2 for ( x , y ) Ω .

Up to now, we have proved that all the conditions of Lemma 2 are satisfied and it means that either Lemma 2(i) or Lemma 2(ii) holds. Next, we display that Lemma 2(ii) is not satisfied.

Let Φ = { ( u , v ) Ξ μ ( u , v ) ( A 1 ( u , v ) B 1 ( u , v ) , A 2 ( u , v ) B 2 ( u , v ) ) } and ( u , v ) Φ be arbitrary. Then, for μ > 1 , μ ( u , v ) ( A 1 ( u , v ) B 1 ( u , v ) , A 2 ( u , v ) B 2 ( u , v ) ) , there exists ( ω 1 , ω 2 ) ( F 1 , x y , F 2 , x y ) such that, for any μ > 1 , we have

(28) u ( t ) = μ 1 f 1 ( t , u ( t ) , v ( t ) ) 1 Γ ( α 1 ) 1 t log t s α 1 1 ω 1 ( s ) s d s ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 ω 1 ( s ) s d s

and

(29) v ( t ) = μ 1 f 2 ( t , u ( t ) , v ( t ) ) 1 Γ ( α 2 ) 1 t log t s α 2 1 ω 2 ( s ) s d s ( log t ) α 2 1 Γ ( α 2 ) 1 e log e s α 2 1 ω 2 ( s ) s d s ,

for all t [ 1 , e ] . Therefore, we have

(30) u ( t ) μ 1 f 1 ( t , u ( t ) , v ( t ) ) 1 Γ ( α 1 ) 1 t log t s α 1 1 ω 1 ( s ) s d s + ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 ω 1 ( s ) s d s [ f 1 ( t , u ( t ) , v ( t ) ) f 1 ( t , 0 , 0 ) + f 1 ( t , 0 , 0 ) ] 1 Γ ( α 1 ) 1 t log t s α 1 1 ω 1 ( s ) s d s + ( log t ) α 1 1 Γ ( α 1 ) 1 e log e s α 1 1 [ L 1 r + F 10 ] 2 Γ ( α 1 ) g 1 r L 1

and

(31) v ( t ) [ L 2 r + F 20 ] 2 Γ ( α 2 ) g 2 r L 1 .

Thus, we have

( u , v ) k r 2 Γ ( α 1 ) g 1 r L 1 + 2 Γ ( α 2 ) g 2 r L 1 + 2 F 10 Γ ( α 1 ) g 1 r L 1 + 2 F 20 Γ ( α 2 ) g 2 r L 1 ,

where F i 0 and g i r ( i = 1 , 2 ) are defined in (H3). Then, note that ( u , v ) = R , we have

r 2 F 10 g 1 r L 1 Γ ( α 1 ) + 2 F 20 g 2 r L 1 Γ ( α 2 ) 1 2 k g 1 r L 1 Γ ( α 1 ) 2 k g 2 r L 1 Γ ( α 2 ) .

Therefore, Lemma 2(ii) is not satisfied by (H3). Then, there exists ( x , y ) Ξ such that

( x , y ) = ( A 1 ( x , y ) B 1 ( x , y ) , A 2 ( x , y ) B 2 ( x , y ) ) ,

i.e., system (1) at least one solution on [ 1 , e ] × [ 1 , e ] . This completes the proof.□

Example 1

Consider the following system of Hadamard-type fractional hybrid differential inclusions

(32) D 1.5 x ( t ) 0.1 e 1 t ( sin x + sin y + 2 ) F 1 ( t , x ( t ) , y ( t ) ) , 1 < t < e , D 1.25 y ( t ) 0.15 ( arctan x + arctan y + 3 ) F 2 ( t , x ( t ) , y ( t ) ) , 1 < t < e , x ( 1 ) = x ( e ) = 0 , y ( 1 ) = y ( e ) = 0 ,

where α 1 = 1.5 , α 2 = 1.25 , F i : [ 1 , e ] × R 2 R ( i = 1 , 2 ) are multi-valued maps given by:

t F 1 ( t , x ( t ) , y ( t ) ) = x 3 10 ( x 3 + y 3 + 3 ) , sin x 16 ( sin x + sin y + 1 ) + 1 16

and

t F 2 ( t , x ( t ) , y ( t ) ) = x 3 12 ( x 3 + y 3 + 2 ) + 1 12 , sin x 10 ( sin x + sin y + 3 ) .

By (H1), L 1 = 0.1 , L 2 = 0.15 with k = 0.25 . For f 1 F 1 , f 2 F 2 , we have

f 1 max x 3 10 ( x 3 + y 3 + 3 ) , sin x 16 ( sin x + sin y + 1 ) + 1 16 1 8 , ( x , y ) R 2 ,

and

f 2 max x 3 12 ( x 3 + y 3 + 2 ) + 1 12 , sin x 10 ( sin x + sin y + 3 ) 1 6 , ( x , y ) R 2 .

Then,

F 1 ( t , x , y ) = sup { v 1 : v 1 F 1 ( t , x , y ) } 1 8 = g 1 r ( t ) , ( x , y ) R 2 , F 2 ( t , x , y ) = sup { v 2 : v 2 F 2 ( t , x , y ) } 1 6 = g 2 r ( t ) , ( x , y ) R 2 .

Clearly, g 1 r L 1 = e 1 8 , g 2 r L 1 = e 1 6 , F 10 = 1 16 , F 10 = 1 12 . Hence, 2 g 1 r L 1 Γ ( α 1 ) + 2 g 2 r L 1 Γ ( α 2 ) k 0.279156 < 1 2 and r > 2 F 10 g 1 r L 1 Γ ( α 1 ) + 2 F 20 g 2 r L 1 Γ ( α 2 ) / 1 2 k g 1 r L 1 Γ ( α 1 ) 2 k g 2 r L 1 Γ ( α 2 ) 0.115079 . Consequently, all the conditions of Theorem 1 are satisfied, and system (32) has at least one solution on [ 1 , e ] × [ 1 , e ] .

Acknowledgement

The authors thank the reviewers for their constructive remarks on their work.

  1. Funding information: This research was supported by Natural Science Foundation of Chongqing (cstc2020jcyj-msxmX0123) and Technology Research Foundation of Chongqing Educational Committee (KJQN202000528).

  2. Author contributions: All authors accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors declare no conflicts of interest.

  4. Ethical approval: The conducted research is not related to either human or animal use.

  5. Informed consent: Informed consent was obtained from all individuals included in this study.

  6. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2022-11-17
Revised: 2023-03-13
Accepted: 2023-03-23
Published Online: 2023-06-15

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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