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Article

On Hybrid and Non-Hybrid Discrete Fractional Difference Inclusion Problems for the Elastic Beam Equation

1
Laboratory of Applied Mathematics, University of Kasdi Merbah, Ouargla 30000, Algeria
2
Department of Mathematics, College of Science, Qassim University, Buraydah 51452, Saudi Arabia
3
Department of Management Information Systems, College of Business and Economics, Qassim University, Buraydah 51452, Saudi Arabia
*
Author to whom correspondence should be addressed.
Fractal Fract. 2024, 8(8), 486; https://doi.org/10.3390/fractalfract8080486
Submission received: 13 July 2024 / Revised: 7 August 2024 / Accepted: 16 August 2024 / Published: 19 August 2024

Abstract

The results in this paper are related to the existence of solutions to hybrid and non-hybrid discrete fractional three-point boundary value inclusion problems for the elastic beam equation. The development of our results is attributed to the use of the Caputo and difference operators. The existence results for the non-hybrid discrete fractional inclusion problem are established by using fixed point theory for multi-valued upper semi-continuous maps, and the case of the hybrid discrete fractional inclusion problem is treated by Dhage’s fixed point theory. Additionally, we present two examples to illustrate our main results.

1. Introduction and Position of Problem

Fractional calculus is a generalized form of classical full-order calculus that examines the properties of fractional derivatives and integrals. This field of study is of considerable importance, due to its numerous applications in various scientific fields in recent years. Fractional calculus can be used in engineering science as well as aerodynamics, electrical circuits, fluid dynamics, heat transfer and physics. For detailed analysis, we refer to the extensive work in [1,2,3,4,5,6,7] for the latest trends in this direction. Researchers have studied various aspects of fractional difference equations. Obviously, the existence, uniqueness and stability analysis of solutions are very important characteristics of fractional difference equations. Various analytical methods and fixed point theory have been used to show the existence and stability of solutions. Several researchers have contributed, and several books and articles on the topic can be found in [8]. To find exactly solving nonlinear fractional difference equations is often very difficult; therefore, the stability analysis of the solution plays a vital role in such investigations. Various stabilities have been described in the literature: see B. Lyapunov’s stability [9], Mittag-Leffler’s stability [10] and exponential stability [11]. The stability presumed to be most reliable is called HU stabilize. This was modified to GHU stability; see [12]. In 1970, Rassias further generalized the stability described above. For fractional difference equations with different boundary conditions, Riemann–Liouville and Caputo operators, which solve existential and stable domains, the analytical equipment is well-equipped; see [13]. A new field called discrete fractional calculus is increasingly of interest to mathematicians and researchers. For discrete fraction operators, several practical problems have been studied; see [14]. Fractional difference equations, due to their applications, have recently piqued the interest of scientists. They can be applied in biology, ecology and applied sciences, in general; please see [15]. However, some research has been conducted on computations performed on discrete elastic beam equations of fractional order, which can be found in [16,17]. Fractional difference computation is a tool for explaining many phenomena in physics and control problems, modeling, chaotic dynamical systems and various fields of engineering and applications mathematics. Various types of methods and techniques are used in this direction, including numerical methods. Analytical methods are used by researchers to discuss discrete values of a specific proportion for continuous mathematical models and boundary value problems for elastic beam equations. For some recent developments on the existence, uniqueness and stability of solutions to fractional difference equations, see [1,17,18,19,20] and references therein. Discrete fractions and difference equations have opened up a new research context for mathematicians. For this reason, they have received increasing attention in recent years. This type of model analyzes real-life processes and phenomena by using discrete fractional operators, because such operators provide accurate tools for describing memory. In science and technology, the deflection of an elastic beam is a well-known phenomenon. Based on the relevance of their applications—such as in aircraft manufacturing, chemical sensors, material mechanics and medical diagnostics—physics three-point limit problems for the elastic beam equation have received much attention; see [21,22,23].
Many researchers have recently examined the elastic beam equation with various boundary conditions; see [24,25]. In [25], Gupta researched the fourth-order elastic beam equation with three-point boundary conditions:
v ( 4 ) ( κ ) = K ( κ , v ( κ ) ) , κ ( 0 , 1 ) v ( x = 0 ) = 0 , v ( x = 0 ) = 0 , v ( x = 1 ) = 0 , v ( x = 1 ) = 0 .
Equation (1) describes an elastic beam model of dimension 1, which is governed by the displacement and bending moment at the left end to zero; the model can move freely at the right end with vanishing angles and shear forces.
Furthermore, Cianciaruso et al. [24] studied the cantilever beam equation model’s three-point boundary conditions:
v ( 4 ) ( κ ) = K ( κ , v ( κ ) ) , κ ( 0 , 1 ) v ( x = 0 ) = v ( x = 0 ) = v ( x = 1 ) = 0 , v ( x = 1 ) = h v ( ζ ) ,
where ζ ( 0 , 1 ) is a real constant. The model above shows the feedback mechanism model of the shear response ζ point displacement at the right end of the beam.
Jehad Alzabut et al., in [20], studied the existence and uniqueness of solutions with various types of Ulam stability results for the discrete fractional elastic beam equation subjected to the three-point boundary conditions, as follows:
Δ ϱ 4 ϱ v ( κ ) = K ( ϱ + κ 1 , v ( ϱ + κ 1 ) ) , κ N 0 β + 3 v ( ϱ 4 ) = Δ 2 v ( ϱ 4 ) = Δ v ( β + ϱ ) = 0 , Δ 3 v ( β + ϱ ) + v ( ζ ) = 0 ,
where ϱ ( 3 , 4 ] , and ζ N ϱ 1 β + ϱ + 2 , Δ ϱ 4 ϱ is the ϱ -order Riemann–Liouville fractional difference, and where
v : N ϱ 4 β + ϱ + 3 R
and
K : N ϱ 4 β + ϱ + 3 × R R
are continuous.
Hybrid fractional differential equations, also known as quadratic perturbations of nonlinear differential equations, are of significant interest to mathematicians and engineers, due to their ability to represent various dynamic models in specific cases. Recently, many researchers have examined hybrid fractional discrete problems. For example, Shammakh et al. [26] investigated the hybrid discrete pantograph equation, studied the existence of a unique solution and presented stability results.
In 2022, Shammakh et al. [27] investigated the stability results in the sense of Hyers and Ulam, utilizing the Mittag–Leffler function in a hybrid fractional order difference equation of the second type for the following problem:
Δ * ϱ ( w ( κ ) M ( κ , w ( κ ) ) ) = K ( ϱ + κ 1 , w ( ϱ + κ 1 ) ) , κ N 0 , 1 < ϱ < 2 w ( m 1 ) = g ( w ) ; Δ ( w ( m + n ) M ( m + n , w ( m + n ) ) ) ψ ( w ) = A ,
where Δ * ϱ is the fractional Caputo difference operator.
Non-hybrid discrete fractional inclusions problem: Many real-world problems that arise in applications, such as in economics and optimal control, can be modeled using differential inclusions. Consequently, non-hybrid discrete fractional inclusions have been the focus of many researchers (see [28,29]). In 2017, Lv [30] mainly studied the existence of solutions for a certain class of discrete fractional difference inclusions with boundary conditions:
Δ ϱ w ( κ ) K ( ϱ + κ 1 , w ( ϱ + κ 1 ) ) , κ N 0 b , 1 < ϱ < 2 w ( κ 2 ) = Δ ϱ 1 w ( b + 1 ) = 0 ,
Δ ϱ denotes the discrete Riemann–Liouville fractional difference of order ϱ .
Hybrid discrete fractional inclusions problem: To date, no paper has addressed the problem of hybrid discrete fractional inclusions using Dhag’s fixed point theorem for inclusion problems, which was published in 2006 [31].
This paper is a significant scientific contribution to the subject of hybrid discrete fractional inclusions. It provides new perspectives and a deeper understanding of fractional rank differential equations and their application in elastic beam equations, thereby helping to develop new solutions. The Caputo difference operator is used in this article as an important tool in fractional calculus, particularly in the numerical analysis of fractional differential equations. This operator provides a means to approximate the fractional derivative of a function, which is essential in modeling various physical and engineering processes that exhibit memory and hereditary properties.
We will address two generalized discrete problems that arise from the previously mentioned problems. Our primary objective is to investigate solutions to the non-hybrid discrete fractional inclusion problem for the elastic beam equation, including the Caputo-like difference operator governing the three-point boundary conditions given as follows:
Δ ϱ 4 ϱ w ( κ ) K ( ϱ + κ 1 , w ( ϱ + κ 1 ) ) , κ N 0 β + 3 w ( ϱ 4 ) = Δ 2 w ( ϱ 4 ) = Δ w ( β + ϱ ) = 0 , Δ α w ( β + ϱ ) + w ( ζ ) = 0 ,
where ϱ ( 3 , 4 ] , 2 < α 3 , ϱ α > 1 , ζ N ϱ 1 β + ϱ + 2 and Δ ϱ 4 ϱ is the ϱ -order Caputo fractional difference, and where
w : N ϱ 4 β + ϱ + 3 R
and
K : N ϱ 4 β + ϱ + 3 × R R
are continuous.
The second purpose is to investigate the existence of solutions to the hybrid discrete fractional inclusion problem for the elastic beam equation involving Caputo discrete derivatives,
Δ ϱ 4 ϱ w ( κ ) Φ ( κ , w ( κ ) ) K ( ϱ + κ 1 , w ( ϱ + κ 1 ) ) , κ N 0 β + 3 w ( ϱ 4 ) Φ ( ϱ 4 , w ( ϱ 4 ) ) = Δ 2 w ( ϱ 4 ) Φ ( ϱ 4 , w ( ϱ 4 ) ) = Δ w ( β + ϱ ) Φ ( β + ϱ , w ( β + ϱ ) ) = 0 , Δ α w ( β + ϱ ) Φ ( β + ϱ , w ( β + ϱ ) ) + w ( ζ ) Φ ( ζ , w ( ζ ) ) = 0 ,
where ϱ ( 3 , 4 ] , 2 < α 3 , ϱ α > 1 and ζ N ϱ 1 β + ϱ + 2 , and where
K : N ϱ 4 β + ϱ + 3 × R R
and
Φ : N ϱ 4 β + ϱ + 3 × R R / { 0 }
are continuous where β N 0 .
Below is a breakdown of the rest of this article. Section 2 introduces the basic tools and required definitions. Section 3 covers the existence results for the inclusion problem of the hybrid and non-hybrid discrete fractional elastic beam equations. In Section 4, we consider several examples to illustrate our results, and which could be applicable in real life.

2. Preliminaries

We now present several fundamental definitions and lemmas for discrete fractional calculus, which will be used throughout this article.
Definition 1 
([32]). We define the generalized falling function by
κ ϱ ̲ = Γ ( κ + 1 ) Γ ( κ + 1 ϱ )
for any κ and ϱ for which the right-hand side is defined. If κ + 1 ϱ is a pole of the gamma function and κ + 1 is not a pole then κ ϱ ̲ = 0 .
Lemma 1 
([15]). The factorial functions listed below are assumed to be well-defined:
1. 
( κ μ ) κ μ ̲ = κ μ + 1 ̲ , where μ R .
2. 
If κ r then κ α ̲ r α ̲ for any α > 0 .
3. 
κ α + ϱ ̲ = ( κ ϱ ) α ̲ κ ϱ ̲ .
Definition 2 
([16]). For ϱ > 0 and K defined on
N a = { a , a + 1 , }
the definition of the ϱth fractional Caputo difference operator is
Δ ϱ K ( κ ) = 1 Γ ( ϱ ) s = a κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( s ) ,
where κ N a + α and ϖ ( s ) = s + 1 .
Definition 3 
([15]). For ϱ > 0 and K defined on
N a = { a , a + 1 , } ,
the ϱ-order fractional sum of K is defined by
Δ α Δ ϱ K ( κ ) = Δ α ϱ K ( κ ) = Δ ( ϱ α ) K ( κ ) = 1 Γ ( ϱ α ) s = a κ ( ϱ α ) ( κ ϖ ( s ) ) ϱ α 1 ̲ K ( s ) ,
where κ N a + ϱ + n α and ϖ ( s ) = s + 1 .
Lemma 2 
([16]).
1. 
Assume that κ and ϱ are any numbers, such that κ ( ϱ ) and κ ( ϱ 1 ) are defined. Then, we have
Δ κ ϱ ̲ = ϱ κ ϱ 1 ̲ .
2. 
Assume that α > 0 and 0 N 1 < α N . Then,
Δ α Δ C α w ( κ ) = w ( κ ) + C 0 + C 1 κ 1 ̲ + C 2 κ 2 ̲ + + C N 1 κ N 1 ̲
for some C i R , 0 i N 1 .
Lemma 3 
([15]).
1. 
For κ and i, in which both ( κ ϖ ( i ) ) ϱ ̲ and ( κ 1 ϖ ( i ) ) ϱ ̲ are defined, then we have
Δ i ( κ ϖ ( i ) ) ϱ ̲ = ϱ ( κ 1 ϖ ( i ) ) ϱ 1 ̲ .
2. 
Let ϱ , v > 0 . Then,
Δ ϱ κ v ̲ = Γ ( v + 1 ) Γ ( v + ϱ + 1 ) κ v + ϱ ̲
and
Δ ϱ κ v ̲ = Γ ( v + 1 ) Γ ( v ϱ + 1 ) κ v ϱ ̲ .
The next relations in the fractional integrals are needed in proofs:
  • s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ = s = 0 κ ϱ ( κ s 1 ) = Γ ( κ + 1 ) ϱ Γ ( κ ϱ + 1 ) = 1 ϱ κ ϱ ̲ .
  • S = 0 T ( T + ϱ S 1 ) ϱ 1 ̲ = Γ ( T + ϱ + 1 ) ϱ Γ ( T + 1 ) = 1 ϱ ( T + ϱ ) ϱ ̲ .
  • S = 0 T + 1 ( T + ϱ S 1 ) ϱ 2 ̲ = Γ ( T + ϱ + 1 ) ( ϱ 1 ) Γ ( T + 2 ) = 1 ( ϱ 1 ) ( T + ϱ ) ϱ 1 ̲ ,
where T R + * , κ N a T is a variable of the function, and 3 < ϱ 4 is a fractional order.
For a normed space ( U * , | · | ) , the set of all subsets of U * , the set of all closed subsets of U * , the set of all bounded subsets of U * , the set of all compact subsets of U * and the set of all convex subsets of U * are represented by P ( U * ) , P c l ( U * ) , P b ( U * ) , P c p ( U * ) and P c v ( U * ) , respectively, where P ( U * ) denotes the family of all subsets of U * .
A multi-valued map,
G : U * P ( U * ) ,
is convex (closed, compact) valued if G ( κ ) is convex (closed, compact) for all κ U * .
Here, G is bounded on bounded sets if G ( Q ) = κ Q Q ( κ ) is bounded in U * for all Q P b ( U * ) ; that is,
sup κ Q { sup | z | : z N * ( κ ) } < .
The map N * is called upper semi-continuous on U * if for each κ 0 U * the set N * κ 0 is a non-empty, closed subset of U * and if for each open set O of U * containing N * κ 0 there exists an open neighborhood V 0 of κ 0 , such that N * V 0 O . N * is said to be completely continuous if N * ( B ) is relatively compact for every B P b ( U * ) . If a multi-valued map N * is completely continuous with non-empty compact values then N * is upper semi-continuous if and only if N * has a closed graph; that is,
κ n κ 0 , κ n κ 0 , κ n , N * κ n
imply that κ 0 N * κ 0 ; see [33].
At a point w C ( N ϱ 4 β + ϱ + 3 , R ) a set of selections of a set-valued map K is defined by
D H , w * = { K L 1 ( N ϱ 4 β + ϱ + 3 , R ) : K ( κ ) K ( κ , w ( κ ) ) , κ N ϱ 4 β + ϱ + 3 } .
Moreover, a Carathéodory set-valued map K : N ϱ 4 β + ϱ + 3 × R P ( R ) is said to be L 1 -Carathéodory whenever for each constant r > 0 there exists a function ψ r L ( N ϱ 4 β + ϱ + 3 , R + ) , such that
K ( κ , w ) = sup { | h | : h K ( κ , w ) } ψ r ( κ )
for each | w | r and κ N ϱ 4 β + ϱ + 3 .
We are ready now to present two fixed point theorems for multi-valued maps, which will be useful in the proofs of our main results.
Lemma 4 
([34]). Let B * be a non-empty closed convex subset of a Banach space E, and let
N * : B * P ( B * )
be an upper semi-continuous and compact map with non-empty, closed and convex values. Then, N * has a fixed point in B * .
Lemma 5 
([34]). Let T * be a Banach space, U be an open subset of T * and 0 U ¯ . Suppose
N * : U P T *
is an upper semi-continuous and compact map with non-empty, closed and convex values. Then, either
  • N * has a fixed point in U ¯ or
  • We have w U and λ 0 , 1 [ , such that w λ * N * ( w ) .
Theorem 1 
([35]). Suppose that X is a separable Banach space:
K : N ϱ 4 β + ϱ + 3 × X P c p , c v ( X ) ,
which is an L 1 -Carathéodory set-valued map and
: L 1 ( N ϱ 4 β + ϱ + 3 , X ) C ( N ϱ 4 β + ϱ + 3 , X )
is a linear continuous mapping. Then, the composition
D K * : C ( N ϱ 4 β + ϱ + 3 , X ) P c p , c v ( C ( N ϱ 4 β + ϱ + 3 , X ) )
is an operator in the product space
C ( N ϱ 4 β + ϱ + 3 , X ) × C ( N ϱ 4 β + ϱ + 3 , X ) ,
with
w D K * ( w ) = ( D K , w * )
having a closed-graph property.
Theorem 2 
([31]). Let U * be Banach algebra. Assume that there exists a single-valued map
A : U * U *
and a set-valued map
B : U * P c p , c l ( U * ) ,
such that
  • A is a Lipschitz operator with a Lipschitz constant M .
  • B is an upper semi-continuous operator with a compactness property.
  • 2 M μ < 1 is such that μ = B ( U * ) .
Then, either
(a)
There is a solution in U * for the operator inclusion w ( A w ) ( B w ) or
(b)
The set O = u * U * λ * u * A u * B u * , λ * > 1 is unbounded.

3. Main Results

Now, we look into the existence of solutions to the discrete fractional hybrid and non-hybrid inclusion issues (5) and (6). To this end, we denote
U * = Y : N ϱ 4 β + ϱ + 3 R , Y continuous ,
and we observe that U * becomes a Banach space when equipped with the usual maximum norm; that is, for any κ U * ,
Y = max | Y ( κ ) | : κ N ϱ 4 β + ϱ + 3 .
Definition 4. 
A function w U * is said to be a solution to problem (5) if there exists a corresponding function:
K ( κ ) : N ϱ 4 β + ϱ + 3 × R R ,
such that K ( κ ) K ( κ , w ( κ ) ) on N ϱ 4 β + ϱ + 3 and
Δ ϱ 4 ϱ w ( κ ) = K ( ϱ + κ 1 ) , κ N 0 β + 3 w ( ϱ 4 ) = Δ 2 w ( ϱ 4 ) = Δ w ( β + ϱ ) = 0 , Δ α w ( β + ϱ ) + w ( ζ ) = 0 .
For the discrete fractional order three-point boundary value problem, we have the next result.
Lemma 6. 
Let K : N ϱ 4 β + ϱ + 3 R be given. Then, the discrete fractional elastic beam equation with three-point boundary conditions,
Δ ϱ 4 ϱ w ( κ ) = K ( ϱ + κ 1 ) , κ N 0 β + 3 , ϱ ( 3 , 4 ] w ( ϱ 4 ) = Δ 2 w ( ϱ 4 ) = Δ w ( β + ϱ ) = 0 , Δ α w ( β + ϱ ) + w ( ζ ) = 0 , 2 < α 3 ,
has the unique solution, for κ N ϱ 4 β + ϱ + 3 ,
w ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ,
such that
e 1 = ( ϱ 4 ) 1 ̲ , e 2 = ( ϱ 4 ) 2 ̲ , e 3 = ( ϱ 4 ) 3 ̲ , e 4 = 6 ( ϱ + n ) ( ϱ 4 ) 3 ( ϱ + n ) 2 ̲ , f 0 = 1 Γ ( 1 α ) ( β + ϱ ) α ̲ + 1 , f 1 = 1 Γ ( 2 α ) ( β + ϱ ) 1 α ̲ + ζ , f 2 = 2 Γ ( 3 α ) ( β + ϱ ) 2 α ̲ + ζ 2 ̲ , f 3 = 6 Γ ( 4 α ) ( β + ϱ ) 3 α ̲ + ζ 3 ̲ , E = f 0 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 + f 1 e 4 3 ( ϱ 4 ) f 2 + f 3 , E 1 ( κ ) = ( 3 ( ϱ 4 ) κ 2 ̲ ( 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 ) e 4 κ 1 ̲ κ 3 ̲ , E 2 ( κ ) = E 1 ( κ ) ( f 0 e 1 f 1 ) + E ( e 1 κ ) .
Proof. 
From (11), we have
Δ ϱ 4 ϱ w ( κ ) = K ( ϱ + κ 1 ) .
By applying the fractional Δ ϱ of order ϱ ( 3 , 4 ] along with (7) to (13), we have
w ( κ ) = C 0 + C 1 κ 1 ̲ + C 2 κ 2 ̲ + C 3 κ 3 ̲ + 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ,
for κ N ϱ 4 β + ϱ + 3 and some constants C j R , where j = 0 , 1 , 2 , 3 . By applying the first boundary condition w ( ϱ 4 ) = 0 in (14), we obtain
C 0 + C 1 e 1 + C 2 e 2 + C 3 e 3 = 0 .
Considering the following statements,
e 1 = ( ϱ 4 ) 1 ̲ , e 2 = ( ϱ 4 ) 2 ̲ , e 3 = ( ϱ 4 ) 3 ̲ ,
using Lemma 2, Definition 3 and taking the operator Δ on both sides of (15), we obtain
Δ w ( κ ) = 1 Γ ( ϱ 1 ) s = 0 κ ϱ + 1 ( κ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + C 1 + 2 C 2 κ 1 ̲ + 3 C 3 κ 2 ̲ .
The operator Δ is applied on both sides of (16), and we obtain
Δ 2 w ( κ ) = 2 C 2 + 6 C 3 κ 1 ̲ + 1 Γ ( ϱ 2 ) s = 0 κ ϱ + 2 ( κ ϖ ( s ) ) ϱ 3 ̲ K ( 1 + ϱ + s ) .
From the boundary condition, using (16) and (17) implies
C 2 = 3 C 3 ( ϱ 4 )
and
C 1 = 6 ( ϱ + n ) ( ϱ 4 ) 3 ( ϱ + n ) 2 ̲ C 3 1 Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) .
We put
e 4 = 6 ( ϱ + n ) ( ϱ 4 ) 3 ( ϱ + n ) 2 ̲ .
By using the fractional difference of order 2 < α 3 , we have
Δ α w ( κ ) = C 0 Γ ( 1 ) Γ ( 1 α ) κ α ̲ + C 1 Γ ( 2 ) Γ ( 2 α ) κ 1 α ̲ + C 2 Γ ( 3 ) Γ ( 3 α ) κ 2 α ̲ + C 3 Γ ( 4 ) Γ ( 4 α ) κ 3 α ̲ + 1 Γ ( ϱ α ) s = 0 κ ϱ + α ( κ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) .
From the second condition of (11), we obtain
C 0 f 0 + C 1 f 1 + C 2 f 2 + C 3 f 3 + 1 Γ ( ϱ α ) s = 0 α + β ( ( β + ϱ ) ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + 1 Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) = 0 ,
where
f 0 = 1 Γ ( 1 α ) ( β + ϱ ) α ̲ + 1 , f 1 = 1 Γ ( 2 α ) ( β + ϱ ) 1 α ̲ + ζ f 2 = 2 Γ ( 3 α ) ( β + ϱ ) 2 α ̲ + ζ 2 ̲ , f 3 = 6 Γ ( 4 α ) ( β + ϱ ) 3 α ̲ + ζ 3 ̲ .
From Equations (15), (18), (19) and (21) we find
C 3 = f 1 f 0 e 1 E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) 1 E Γ ( ϱ α ) s = 0 α + β ( ( β + ϱ ) ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) 1 E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s )
and
C 2 = 3 ( ϱ 4 ) E Γ ( ϱ α ) s = 0 α + β ( ( β + ϱ ) ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + 3 ( ϱ 4 ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) 3 ( ϱ 4 ) ( f 1 f 0 e 1 ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s )
and
C 1 = e 4 ( f 1 f 0 e 1 ) E E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) e 4 E Γ ( ϱ α ) s = 0 α + β ( ( β + ϱ ) ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) e 4 E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s )
and
C 0 = ( 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 ) ( f 1 f 0 e 1 ) E e 1 E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) ( 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 ) E Γ ( ϱ α ) s = 0 α + β ( ( β + ϱ ) ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) ( 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s )
with
E = f 0 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 + f 1 e 4 3 ( ϱ 4 ) f 2 + f 3 .
We put
E 1 ( κ ) = ( 3 ( ϱ 4 ) κ 2 ̲ 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 ) e 4 κ 1 ̲ κ 3 ̲
and
E 2 ( κ ) = E 1 ( κ ) ( f 0 e 1 f 1 ) + E ( e 1 κ ) .
Substituting the value of C 0 , C 1 , C 2 , C 3 in (14), we find
w ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) .
Remark 1. 
From the definition of w ( κ ) given in (12), we can easily see that
w ( κ ) ( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) × s = 0 β + ϱ + 3 K ( 1 + ϱ + s ) ,
where
E 1 * = max | E 1 ( κ ) | κ N ϱ 4 β + ϱ + 3 , E 2 * = max | E 2 ( κ ) | κ N ϱ 4 β + ϱ + 3 , E * = f 0 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 + f 1 e 4 3 ( ϱ 4 ) f 2 + f 3 .
Define an operator
F : U * P ( U * )
by
F ( w ) = { g U * : g ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E [ 1 Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + 1 Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ] , K D K , w } ,
where
D K , w = K U * , K ( κ ) K ( κ , w ( κ ) ) , κ N ϱ 4 β + ϱ + 3 .
Some essential hypotheses are presented as follows:
  • ( I 1 )  K : N ϱ 4 β + ϱ + 3 × R P ( R ) is a non-empty, compact and convex multi-valued map.
  • ( I 2 )   y K ( κ , y ) is upper semi-continuous for each κ N ϱ 4 β + ϱ + 3 .
  • ( I 3 )  For each r 1 > 0 there exists a function
    ψ r 1 : N ϱ 4 β + ϱ + 3 [ 0 , + [ ,
    such that
    K ( κ , y ) = sup { | h | : h K ( κ , y ) } ψ r 1 ( κ )
    for each ( κ , y ) N ϱ 4 β + ϱ + 3 × [ r 1 , r 1 ] and lim r 1 + sup 1 r 1 s = 0 β + ϱ + 3 ψ r 1 ( κ ) = R .
  • ( I 4 )  There exists a non-decreasing function
    ϕ : [ 0 , + [ ] 0 , + [
    and a function
    υ : N ϱ 4 β + ϱ + 3 ] 0 , + [
    satisfying s = 0 β + ϱ + 3 υ ( 1 + ϱ + s ) 0 and a positive number δ, such that
    K ( κ , y ) = sup { | h | : h K ( κ , y ) } υ ( κ ) ϕ ( | y | ) ,
    for each ( κ , y ) N ϱ 4 β + ϱ + 3 × R and
    ( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) × ϕ ( δ ) s = 0 β + ϱ + 3 υ ( 1 + ϱ + s ) < δ .
  • ( I 5 )  There exists a constant > 0 , such that
    | Φ ( κ , y 1 ) Φ ( κ , y 2 ) | < | y 1 y 2 | .
  • ( I 6 )   There exists r ¯ > 0 , such that
    r ¯ > M * Δ 1 K Δ ,
    where
    K Δ < 1 , M * = sup { | Φ ( κ , 0 ) | , κ N ϱ 4 β + ϱ + 3 }
    and
    K ( κ , y ) = sup { | h | : h K ( κ , y ) } ϕ ( κ ) sup κ N ϱ 4 β + ϱ + 3 ϕ ( κ ) = R *
    and
    Δ = ϱ E 2 * ( β + ϱ ) ϱ 1 ̲ + E * ( β + ϱ + 3 ) ϱ ̲ + E 1 * ϱ α ̲ β + ϱ ϱ α ̲ + ( ζ ) ϱ ̲ E * Γ ( ϱ + 1 ) R * ,
    where
    Δ 1 2 .
Remark 2. 
If K satisfies hypotheses ( I 1 ) and ( I 3 ) then
K : N ϱ 4 , β + ϱ + 3 × R P c p , c v ( R )
is in an L 1 -Carathéodory set-valued map.
Now, we state and prove our first result regarding the non-hybrid inclusion problem (5).
Theorem 3. 
If assumptions ( I 1 ) and ( I 3 ) are satisfied and
( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) R < 1
then problem (5) has at least one solution in U * .
Proof. 
We divide the proof into three steps, to show that the operator F, defined by (29), satisfies all the assumptions of Lemma 4.
  • Step1
We show that the operator F ( w ) is convex for any w U * . Indeed, if g 1 , g 2 F ( w ) then there exists K 1 , K 2 D K , w , such that for each κ N ϱ 4 β + ϱ + 3 we have
g j ( x ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K j ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K j ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K j ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K j ( 1 + ϱ + s ) ,
where j = 1 , 2 . Let λ * [ 0 , 1 ] ; hence, for each κ N ϱ 4 β + ϱ + 3 we have
λ * g 1 + 1 λ * g 2 = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ λ * K 1 ( 1 + ϱ + s ) + 1 λ * K 2 ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ λ * K 1 ( 1 + ϱ + s ) + 1 λ * K 2 ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ λ * K 1 ( 1 + ϱ + s ) + 1 λ * K 2 ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ λ * K 1 ( 1 + ϱ + s ) + 1 λ * K 2 ( 1 + ϱ + s ) .
Since D K , w is convex (because K is assumed to have convex values) we have λ * g 1 + 1 λ * g 2 F ( w ) , implying that F ( w ) is convex for any w U * .
  • Step2
We show that there exists a positive number r 2 , such that
F : B r 2 P B r 2
is a completely continuous map with compact values, where
B r 2 = { w U * ; w r 2 } .
Indeed, if the claimed assertion is not true then for each positive number r 1 a function w r 1 B r 1 can be found, such that g r 1 F ( w r 1 ) and g r 1 > r 1 with
g r 1 = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K r 1 ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K r 1 ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K r 1 ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K r 1 ( 1 + ϱ + s )
for some K r 1 D K , w r 1 . On the other hand, from (30) in ( I 3 ) and Remark 1, we have
r 1 g r 1 = max κ N ϱ 4 , β + ϱ + 3 | 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K r 1 ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K r 1 ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K r 1 ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K r 1 ( 1 + ϱ + s ) |
( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) × s = 0 β + ϱ + 3 | K r 1 ( 1 + ϱ + s ) | ( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) × s = 0 β + ϱ + 3 ψ r 1 ( 1 + ϱ + s )
Dividing both sides, taking the limit as r 1 + , we conclude that
( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) > 1 ,
which contradicts (34). Hence, the existence of a positive number r 2 is proved. Furthermore, since E is a finite dimension space, the operator
F : B r 2 P B r 2
is completely continuous with compact values.
  • Step3
We show that F is an upper semi-continuous map. For this purpose, we first define the linear continuous operator,
J : X X K ( J K ) ( κ ) ,
such that
( J K ) ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ,
where κ N ϱ 4 β + ϱ + 3 . Let lim n + w n w * , g n F w n and lim n + g n g * . It is easy to see that we only have to show that g * F w * . To this end, observe first that the relation g n F w n means that there exists K n D K , w n , such that g n = J K n . Since w n n N 0 is bounded, from its convergence and the condition (30) in ( I 3 ) it follows that there exists a compact set O of U * with K n n N 0 O ; see [36,37]. Therefore, there exists a convergent sub-sequence K n p p N 0 of K n n N 0 , such that lim p + K n p K * . Now, let lim p + K n p K * and K n p ( x ) H κ , w n p ( κ ) for all κ N ϱ + 4 β + ϱ + 3 . Then, since H ( κ , . ) is upper semi-continuous for all κ N ϱ + 4 β + ϱ + 3 , we can conclude that K * ( κ ) H κ , w * ( κ ) , implying that K * D K , w * . Next, since lim p + K n p K * and J is continuous, we see that g n p = lim p + J K w n p = J K * and, hence, g * = J K * J D K , w * = F w * . As a result, F is an upper semi-continuous map.
Thus,
F : B r 2 P B r 2
is a compact multi-valued map and is upper semi-continuous with convex closed values. Hence, we can apply Lemma 4 to conclude that F has at least one fixed point, w B r 2 U * , which is a solution to the problem (5). Theorem 3 is proved. □
Theorem 4. 
If the assumptions ( I 1 ) , ( I 2 ) and ( I 4 ) are fulfilled, then problem (5) has at least one solution.
Proof. 
In order to use Lemma 5, we first show that the set
W = { w U * : w α F ( w ) , for some θ ( 0 , 1 ) }
is bounded, where F is defined by Equation (29). Let w W ; then, there exists a function K D K , w , such that for each κ N ϱ 4 β + ϱ + 3 we have
w ( κ ) = θ { 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) } ,
for some θ ( 0 , 1 ) . Hence, in view of Remark 1 and ( I 4 ) , for any κ N ϱ 4 , β + ϱ + 3 we can write
| w ( κ ) | = θ | 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) | max κ N ϱ 4 , β + ϱ + 3 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ | K ( 1 + ϱ + s ) | + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ | K ( 1 + ϱ + s ) | + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ | K ( 1 + ϱ + s ) | + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ | K ( 1 + ϱ + s ) |
( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) × s = 0 β + ϱ + 3 | K ( 1 + ϱ + s ) | ( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) × s = 0 β + ϱ + 3 υ ( 1 + ϱ + s ) ϕ ( | w ( 1 + ϱ + s ) | ) ( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) × ϕ ( | w ) s = 0 β + ϱ + 3 υ ( 1 + ϱ + s ) .
This shows that the set W is bounded and that the inequality
w ( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) × ϕ ( | w ) s = 0 β + ϱ + 3 υ ( 1 + ϱ + s )
holds for each w W . Then, by ( I 4 ) there exists δ , such that w δ .
We define
U = { w U * : w < δ } .
As in the proof of Theorem 3, we can show that the operator
F : U ¯ P ( U * )
is a compact multi-valued map and is upper semi-continuous with convex closed values. From the choice of U , we can easily find that the second possibility given in Lemma 5 is ruled out. So, the first possibility of Lemma 5 holds and we conclude that F has at least one fixed point, w U ¯ U * , which is a solution to problem (5). Theorem 4 is now proved. □
Next, we investigate the existence of solutions to the hybrid inclusion problem (6).
Lemma 7. 
Let ϱ ( 3 , 4 ] , 2 < α 3 and w U * , and we have
K : N ϱ 4 β + ϱ + 3 R
and
Φ : N ϱ 4 β + ϱ + 3 × R R / { 0 } .
Then, the function w is a solution to the discrete EBE
Δ ϱ 4 ϱ w ( κ ) Φ ( κ , w ( κ ) ) = K ( ϱ + κ 1 ) , κ N 0 β + 3 w ( ϱ 4 ) Φ ( ϱ 4 , w ( ϱ 4 ) ) = Δ 2 w ( ϱ 4 ) Φ ( ϱ 4 , w ( ϱ 4 ) ) = Δ w ( β + ϱ ) Φ ( β + ϱ , w ( β + ϱ ) ) = 0 , Δ α w ( β + ϱ ) Φ ( β + ϱ , w ( β + ϱ ) ) + w ( ζ ) Φ ( ζ , w ( ζ ) ) = 0 ,
if and only if w ( κ ) for κ N ϱ 4 β + ϱ + 3 is a solution to the following fractional difference formula:
w ( κ ) = Φ ( κ , w ( κ ) ) { 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) } ,
such that
e 1 = ( ϱ 4 ) 1 ̲ , e 2 = ( ϱ 4 ) 2 ̲ , e 3 = ( ϱ 4 ) 3 ̲ , e 4 = 6 ( ϱ + n ) ( ϱ 4 ) 3 ( ϱ + n ) 2 ̲ f 0 = 1 Γ ( 1 α ) ( β + ϱ ) α ̲ + 1 , f 1 = 1 Γ ( 2 α ) ( β + ϱ ) 1 α ̲ + ζ f 2 = 2 Γ ( 3 α ) ( β + ϱ ) 2 α ̲ + ζ 2 ̲ , f 3 = 6 Γ ( 4 α ) ( β + ϱ ) 3 α ̲ + ζ 3 ̲ E = f 0 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 + f 1 e 4 3 ( ϱ 4 ) f 2 + f 3 E 1 ( κ ) = 3 ( ϱ 4 ) κ 2 ̲ ( 3 ( ϱ 4 ) e 2 e 3 e 4 e 1 ) e 4 κ 1 ̲ κ 3 ̲ E 2 ( κ ) = E 1 ( κ ) ( f 0 e 1 f 1 ) + E ( e 1 κ ) .
Proof. 
We put
G ( κ ) = w ( κ ) Φ ( κ , w ( κ ) ) .
Using the same steps as Lemma 6, we find
G ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) .
By substituting (36) in (37), we obtain
w ( κ ) = Φ ( κ , w ( κ ) ) { 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s )
+ E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) } .
Now, we prove our result in relation to the inclusion problem (6).
Theorem 5. 
If the assumptions ( I 1 ) ( I 5 ) ( I 6 ) are fulfilled then the hybrid inclusion problem (6) has a solution.
Proof. 
For every w U * we define the set of selections of the operator D K , w by
D K , w = { K U * , K ( κ ) K ( κ , w ( κ ) ) , κ N ϱ 4 β + ϱ + 3 } .
It is clear that
D K , w D K , w * .
We consider the set-valued map
F 1 : U * P ( U * )
by
F 1 ( w ) = C U * : C ( κ ) = Φ ( κ , w ( κ ) ) { 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E [ 1 Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + 1 Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ] } , K D K , w } ,
for all κ N ϱ 4 β + ϱ + 3 . The function w is a solution for the hybrid inclusion problem (6) if and only if w is a fixed point of the operator F 1 .
Indeed, ( )
If the function w is a solution for the hybrid inclusion problem (6) then there exists K H , such that
Δ ϱ 4 ϱ w ( κ ) Φ ( κ , w ( κ ) ) = K ( ϱ + κ 1 ) .
According to Lemma 7, we have
w ( κ ) = Φ ( κ , w ( κ ) ) { 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) } .
and K K , implying that if K D K , w then we have w in the set-valued map F 1 .
( )
If the function w is in the set-valued map F 1 then w is given as follows:
w ( κ ) = Φ ( κ , w ( κ ) ) { 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E [ 1 Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + 1 Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ] } , K D K , w .
According to Lemma 7, we have
Δ ϱ 4 ϱ w ( κ ) Φ ( κ , w ( κ ) ) = K ( ϱ + κ 1 ) ,
with K D K , w , implying
Δ ϱ 4 ϱ w ( κ ) Φ ( κ , w ( κ ) ) D K , w .
According to (38),
Δ ϱ 4 ϱ w ( κ ) Φ ( κ , w ( κ ) ) K ( κ , w ( κ ) ) .
Then, w is a solution for the hybrid inclusion problem (6).
Now, we define the single-valued mapping
A : U * U *
by ( A w ) ( κ ) = Φ ( κ , w ( κ ) ) , and we define the set-valued map
B : U * P ( U * )
by
( B w ) ( κ ) = χ U * : χ ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E [ 1 Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + 1 Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ] , K D K , w }
for all κ N ϱ 4 β + ϱ + 3 . Then, we obtain F 1 ( w ) = ( A w ) ( B w ) . We show that both operators A and B satisfy the assumptions of Theorem 5. We first prove that the operator A is Lipschitz. Let w 1 , w 2 E ; thus, assumption ( K 5 ) implies that
A w 1 ( κ ) A w 2 ( κ ) = Φ κ , w 1 ( κ ) Φ κ , w 2 ( κ ) K w 1 ( κ ) w 2 ( κ ) K w 1 w 2 ,
for all κ N ϱ 4 β + ϱ + 3 . Hence, we obtain A w 1 A w 2 K w 1 w 2 for all w 1 , w 2 U * . This shows that operator A is Lipschitz with a Lipschitz constant K > 0 .
In this step, we prove that the set-valued map B has convex values. Let χ 1 , χ 2 B ( w ) ; then, we have K 1 , K 2 D K , w , such that for each κ N ϱ 4 β + ϱ + 3 we have
χ j ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K j ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K j ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K j ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K j ( 1 + ϱ + s ) ,
where j = 1 , 2 .
Let λ * [ 0 , 1 ] ; then, for each κ N ϱ 4 β + ϱ + 3 we have
λ * χ 1 + 1 λ * χ 2 ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ λ * K 1 ( 1 + ϱ + s ) + 1 λ * K 2 ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ λ * K 1 ( 1 + ϱ + s ) + 1 λ * K 2 ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ λ * K 1 ( 1 + ϱ + s ) + 1 λ * K 2 ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ λ * K 1 ( 1 + ϱ + s ) + 1 λ * K 2 ( 1 + ϱ + s ) .
Since D K , w , is convex (because K assumes convex values), we have λ * χ 1 + 1 λ * χ 2 B ( w ) , implying that B ( w ) is convex for any w U * . In order to prove that the operator B is completely continuous, we must first show that it maps all bounded sets into bounded subsets of U * . To do this, we will need to demonstrate two equicontinuity and uniform boundedness properties for the set B ( U * ) . We demonstrate that B ( U * ) maps all bounded sets into bounded subsets of U * . For a positive number r * > 0 , consider the bounded ball
V r * = w U * ; w r * .
For every w V r * and χ B w , there exists a function K D K , w , such that
χ ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s )
+ E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s )
for each κ N ϱ 4 β + ϱ + 3 . On the other hand, from (30) in ( I 6 ) , we have
χ = max κ N ϱ 4 β + ϱ + 3 | 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) | max κ N ϱ 4 β + ϱ + 3 | 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ ϕ ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ ϕ ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ ϕ ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ ϕ ( 1 + ϱ + s ) | [ 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ + E 2 * E * Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ + E 1 * E * 1 Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ + 1 Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ ] R * [ ϱ E 2 * ( β + ϱ ) ϱ 1 ̲ + E * ( β + ϱ + 3 ) ϱ ̲ + E 1 * ϱ α ̲ ( β + ϱ ) ϱ α ̲ + ( ζ ) ϱ ̲ E * Γ ( ϱ + 1 ) ] R * .
Then, the set B ( U * ) is uniformly bounded. Now, we prove that the operator B transforms bounded sets into equicontinuous sets. Let w V r * and χ B ( w ) . Choose K D K , w , such that
χ ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) .
For ϱ 4 κ 1 < κ 2 β + ϱ + 3 , we have
B w ( κ 1 ) B w ( κ 2 ) = | 1 Γ ( ϱ ) s = 0 κ 1 ϱ ( κ 1 ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) 1 Γ ( ϱ ) s = 0 κ 2 ϱ ( κ 2 ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ 1 ) E 2 ( κ 2 ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ 1 ) E 1 ( κ 2 ) E [ 1 Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + 1 Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ] | R * | κ 1 ϱ ̲ κ 2 ϱ ̲ | Γ ( ϱ + 1 ) + R * ( β + ϱ ) ϱ 1 ̲ E Γ ( ϱ ) | E 2 ( κ 1 ) E 2 ( κ 2 ) | + R * ( ζ ) ϱ ̲ + ϱ α ̲ ( β + ϱ ) ϱ α ̲ E Γ ( ϱ + 1 ) | E 1 ( κ 1 ) E 1 ( κ 2 ) | < ϵ .
Notice that the right-hand side tends to zero when κ 2 κ 1 . By the Arzela Ascoli Theorem, we ascertain that the operator B is a complete continuity. In the following, we prove that B has a closed graph, which, in turn, implies that the operator B is upper semi-continuous.
We assume that w n V r * and χ n B w n with lim n + w n w * and lim n + χ n χ * . We claim that χ * B w * . For every n N 1 and χ n B w n , we choose K n D K , w n , such that
χ n ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K n ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K n ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K n ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K n ( 1 + ϱ + s ) .
According to (8), it is enough to demonstrate the existence of a function K * D K , w * , such that
χ * ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K * ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K * ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K * ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K * ( 1 + ϱ + s )
for each κ N ϱ 4 β + ϱ + 3 , for which we define the continuous linear operator
: U * L 1 N ϱ 4 β + ϱ + 3 , R U *
where
( K ) ( κ ) = w ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) ,
for all κ N ϱ 4 β + ϱ + 3 . Hence,
χ n ( κ ) χ * ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ ( K n ( 1 + ϱ + s ) K * ( 1 + ϱ + s ) ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ ( K n ( 1 + ϱ + s ) K * ( 1 + ϱ + s ) ) + E 1 ( κ ) E 1 Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ ( K n ( 1 + ϱ + s ) K * ( 1 + ϱ + s ) ) + 1 Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ ( K n ( 1 + ϱ + s ) K * ( 1 + ϱ + s ) ) 0 .
Thus, using Remark 2, according to Theorem 1 it is deduced that the operator D H has a closed graph. Also, since χ n D K , w n and lim n + w n w * , since D H has a closed graph, by (41) and (8) there exists K * D K , w * , such that
χ * ( κ ) = 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K * ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K * ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K * ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K * ( 1 + ϱ + s ) .
Hence, χ * B w * , and then B has a closed graph. This means that the operator B is upper semi-continuous. On the other hand, since the operator B has compact values then B is a compact and upper semi-continuous operator. By using assumption ( I 6 ) , we have
B ( U * ) = sup κ N ϱ 4 β + ϱ + 3 { | B w | : w U * } ϱ E 2 * ( β + ϱ ) ϱ 1 ̲ + E * ( β + ϱ + 3 ) ϱ ̲ + E 1 * ϱ α ̲ β + ϱ ϱ α ̲ + ( ζ ) ϱ ̲ E * Γ ( ϱ + 1 ) R * = Δ .
Then, Δ 1 2 , and it remains the case that it is proved that claim (ii) of Theorem 2 cannot be realized. Let w U * and λ * > 1 be such that w = r ¯ and λ * w ( x ) ( A w ) ( κ ) ( B w ) ( κ ) and K D K , w * . It follows that
| w ( κ ) | = 1 λ * | Φ ( κ , w ( κ ) ) { 1 Γ ( ϱ ) s = 0 κ ϱ ( κ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) + E 2 ( κ ) E Γ ( ϱ 1 ) s = 0 1 + β ( β + ϱ ϖ ( s ) ) ϱ 2 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ α ) s = 0 α + β ( β + ϱ ϖ ( s ) ) ϱ α 1 ̲ K ( 1 + ϱ + s ) + E 1 ( κ ) E Γ ( ϱ ) s = 0 ζ ϱ ( ζ ϖ ( s ) ) ϱ 1 ̲ K ( 1 + ϱ + s ) } | 1 λ * ( | w | + | Φ ( κ , 0 ) | ) × ϱ E 2 * ( β + ϱ ) ϱ 1 ̲ + E * ( β + ϱ + 3 ) ϱ ̲ + E 1 * ϱ α ̲ ( β + ϱ ) ϱ α ̲ + ( ζ ) ϱ ̲ ] E * Γ ( ϱ + 1 ) R * 1 λ * ( | w | + | Φ ( κ , 0 ) | ) Δ ( | w | + | Φ ( κ , 0 ) | ) Δ
for all κ N ϱ 4 , β + ϱ + 3 . Hence, we obtain
r ¯ M * Δ 1 Δ .
According to condition I 6 and (32), we see that condition ( b ) of Theorem 2 is impossible. Thus, w ( A w ) ( B w ) . Hence, the operator F 1 has a fixed point, and so the hybrid inclusion problem (6) has a solution. □

4. Numerical Applications

To illustrate our results, we give numerical examples in this section.
Example 1. 
Δ 1 4 15 4 w ( κ ) w ( κ ) ( sin ( w ( κ ) ) ) 2 w ( κ ) ( sin ( w ( κ ) ) ) 2 + 5 + κ 10 , w ( κ ) ( cos ( w ( κ ) ) ) 4 w ( κ ) ( cos ( w ( κ ) ) ) 4 + 3 + 4 κ 39.997 [ w ( 1 4 ) = Δ 2 w ( 1 4 ) = Δ w ( 23 4 ) = 0 , Δ 9 4 w ( 23 4 ) + w ( 27 4 ) = 0 , κ N 0 5 . .
Set ϱ = 16 5 , α = 14 5 , ζ = 16 5 , β = 3 .
Consider a set-valued
K : N 1 4 39 4 × R P ( R ) ,
defined by
K ( κ , w ( κ ) ) = sin ( w ( κ ) ) 2 sin ( w ( κ ) ) 2 + 5 + κ 10 , w ( κ ) ( cos ( w ( κ ) ) ) 4 w ( κ ) ( cos ( w ( κ ) ) ) 4 + 3 + 4 κ 18.999 ; κ N 1 4 39 4 .
This obviously satisfies assumption ( I 1 ) ; then, for any κ N 1 4 39 4 and r 1 > 0 ,
K max sin w ( κ ) 2 ( sin w ( κ ) ) 2 + 5 + κ 10 , cos w ( κ ) 4 cos ( w ( κ ) ) 4 + 3 + 4 κ 18.999 1 + 4 κ 18.999 < ( 1 + 4 κ 18.999 ) ( r 1 + 1 ) = ψ r 1 ( κ ) ,
for each ( κ . y ) N 1 4 39 4 × [ r 1 , r 1 ] . Then,
lim r 1 + sup 1 r 1 s = 0 46 5 ψ r 1 ( κ ) = 0.01 = R ,
and then, we obtain
( ( 4 + β + ϱ ) ϱ 1 ̲ Γ ( ϱ ) + E 2 * ( 1 + β + ϱ ) ϱ 2 ̲ E * Γ ( ϱ 1 ) + E 1 * E * ( 1 + β + ϱ ) ϱ α 1 ̲ Γ ( ϱ α ) + ( ζ + 1 ) ϱ 1 ̲ Γ ( ϱ ) ) R = 0.5955 < 1 .
This shows that assumption ( I 3 ) and (34) are satisfied. It is clear that K satisfies assumption ( I 1 ) . So, by Theorem 3, we conclude the existence of a solution for problem (42).
We can use Theorem 4 to prove that there is a solution to problem (42). It is clear that ( I 1 ) and ( I 2 ) are satisfied for
K max sin w ( κ ) 2 ( sin w ( κ ) ) 2 + 5 + κ 10 , cos w ( κ ) 4 cos ( w ( κ ) ) 4 + 3 + 4 κ 18.999 | 1 + 4 κ 18.999 | | 1 + 4 κ 18.999 | 2 + 1 333 2 | w ( κ ) | = υ ( κ ) ϕ ( | w ( κ ) | ) .
From (31), we have
59.5497 × ( 2 + δ ) × 88.0100 < δ .
We can take δ > 11002 , showing that assumption ( I 1 ) is satisfied. So, by Theorem 4, we conclude that problem (42) has at least one solution.
Example 2. 
Let the system
Δ 1 3 11 3 w ( κ ) Φ ( κ , w ( κ ) ) K ( κ + 8 3 , w ( κ + 8 3 ) ) , κ N 0 5 w ( 1 3 ) Φ ( 1 3 , w ( 1 3 ) ) = Δ 2 w ( 1 3 ) Φ ( 1 3 , w ( 1 3 ) ) = Δ w ( 17 3 ) Φ ( 17 3 , w ( 17 3 ) ) = 0 Δ 5 2 w ( 17 3 ) Φ ( 17 3 , w ( 17 3 ) ) + w ( 14 3 ) Φ ( 14 3 , w ( 14 3 ) ) = 0 .
Set ϱ = 11 3 , α = 5 2 , ζ = 14 3 , β = 2 . Consider a set-valued map,
K : N 1 3 26 3 × R P ( R ) ,
where
K ( κ , w ( κ ) ) = 7 9 , ( κ + 2 ) | w ( κ ) | log ( cos 2 ( κ ) + 2 π ) ( | w ( κ ) | + 1 ) ( κ 2 + 65 17 ) 2 κ N 1 3 26 3 ,
and a set-valued map
Φ : N 1 3 26 3 × R R ,
where
Φ ( κ , w ( κ ) ) = | w ( κ ) | a r c t a n 3 ( κ + π ) ( | w ( κ ) | + π ) ( 55 κ ) 4 κ N 1 3 26 3 .
Noting that
M * = sup N 1 3 26 3 ( | Φ ( κ , 0 ) | ) = 0 .
Since the map Φ is Lipschitz, for each w 1 , w 2 R , by Hypothesis ( K 5 ) we have
Φ κ , w 1 ( κ ) Φ κ , w 2 ( κ ) ( π 2 ) 3 ( 55 26 3 ) 4 | w 1 ( κ ) w 2 ( κ ) | K = ( π 2 ) 3 ( 55 26 3 ) 4 > 0
for κ N 1 3 26 3 , and, since
υ max 7 9 , ( κ + 2 ) | w ( κ ) | log ( c o s 2 ( κ ) + 2 π ) ( | w ( κ ) | + 1 ) ( ( κ ) 2 + 65 17 ) 2 ( κ + 2 ) log ( 1 + 2 π ) ( 1 9 + 289 ) ,
for all υ K ( κ , w ( κ ) ) we find
K ( κ , w ( κ ) ) = sup | K | , K K ( κ , w ( κ ) ) ϕ ( κ ) = ( κ + 2 ) log ( 1 + 2 π ) ( 1 9 + 289 ) .
Then,
R * = ϕ ( κ ) 9.4130 × 10 04 .
Hence,
= ϱ E 2 * ( β + ϱ ) ϱ 1 ̲ + E * ( β + ϱ + 3 ) ϱ ̲ + E 1 * ϱ α ̲ β + ϱ ϱ α ̲ + ( ζ ) ϱ ̲ E * Γ ( ϱ + 1 ) R * 0.1788 .
Now, we choose r ¯ > 0 with r ¯ > M * Δ 1 Δ ; then, we have
Δ 1.6830 × 10 04 < 1 2 .
By using Theorem 5, the hybrid inclusion problem (43) has a solution.

5. Conclusions and Perspectives

This work was conceived, firstly, to study the existence of solutions to the hybrid discrete fractional inclusions problem, using Dhag’s fixed point theorem for inclusion problems. We studied the existence of solutions to hybrid and non-hybrid discrete fractional three-point boundary value inclusion problems for the elastic beam equation. The results were illustrated by suitable examples. The existence of a solution to the elastic beam equation indicates that the mathematical model used to represent the beam is accurate and capable of providing reliable results. Solutions to this equation offer valuable insights into the behavior of beams under various loads. This information is crucial for the design and analysis of engineering structures, such as bridges and buildings, to ensure their safety and sustainability. Additionally, these solutions help us to understand how the beam will behave over time under varying loads, assisting in predicting and responding to dynamic forces such as wind and earthquakes. In future works, we plan to extend our approach to other types of discrete differential inclusions or fully hybrid discrete fractional differential equations. It will be very interesting to extend our approach to other types of discrete differential inclusions. We will also seek numerical solutions, using approximate methods, and we will conduct numerical simulations to validate the theoretical results.

Author Contributions

Writing—original draft preparation, F.A. and A.A.; writing—review and editing, T.R.; supervision, K.Z. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

The data presented in this study are available on request from the corresponding author.

Acknowledgments

The researchers would like to thank the Deanship of Graduate Studies and Scientific Research at Qassim University for financial support (QU-APC-2024-9/1).

Conflicts of Interest

The authors declare no conflicts of interest.

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MDPI and ACS Style

Alili, F.; Amara, A.; Zennir, K.; Radwan, T. On Hybrid and Non-Hybrid Discrete Fractional Difference Inclusion Problems for the Elastic Beam Equation. Fractal Fract. 2024, 8, 486. https://doi.org/10.3390/fractalfract8080486

AMA Style

Alili F, Amara A, Zennir K, Radwan T. On Hybrid and Non-Hybrid Discrete Fractional Difference Inclusion Problems for the Elastic Beam Equation. Fractal and Fractional. 2024; 8(8):486. https://doi.org/10.3390/fractalfract8080486

Chicago/Turabian Style

Alili, Faycal, Abdelkader Amara, Khaled Zennir, and Taha Radwan. 2024. "On Hybrid and Non-Hybrid Discrete Fractional Difference Inclusion Problems for the Elastic Beam Equation" Fractal and Fractional 8, no. 8: 486. https://doi.org/10.3390/fractalfract8080486

APA Style

Alili, F., Amara, A., Zennir, K., & Radwan, T. (2024). On Hybrid and Non-Hybrid Discrete Fractional Difference Inclusion Problems for the Elastic Beam Equation. Fractal and Fractional, 8(8), 486. https://doi.org/10.3390/fractalfract8080486

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