Nonlocal Initial Value Problems for Hybrid Caputo Fractional Integro-Differential Equations


Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3 (2025) 

  

985 
https://internationalpubls.com 

Nonlocal Initial Value Problems for Hybrid Caputo Fractional Integro-

Differential Equations 

 

Dr. Moffek Hamza 

Ecole Normale Supérieure de Ouargla,30000 Ouargla,Algeria , 

Email: moffek.hamza@ens-ouargla.dz 

 

Article History: 

Received: 19-09-2024 

Revised: 24-03-2025 

Accepted: 11-04-2025 

Abstract: This paper investigates nonlocal initial value problems for hybrid Caputo 

fractional integro-differential equations. By employing a fixed-point theorem due to 

Dhage, we establish the existence of solutions to these problems. The theoretical findings 

are illustrated through a concrete example, showcasing the applicability of our results. 

Keywords: The fractional Caputo derivative, The fractional integral, hybrid, Dhage fixed 

point. 

 

Introduction 

Fractional calculus, an extension of classical calculus, has garnered significant attention in recent years 

due to its ability to describe complex phenomena that integer-order derivatives and integrals fail to 

capture accurately. This branch of mathematics has proven its utility in various domains, including 

physics, engineering, biology, and economics. ([10],[13],[14],[16],[4],[12]). 

Nonlocal initial value problems (NIVPs) represent a category of problems in which the initial 

conditions depend on the values of the unknown function at multiple points rather than a single point. 

Such problems naturally arise in numerous real-world applications, such as heat transfer, viscoelastic 

material behavior, and control systems. 

Hybrid differential equations, which combine differential and integral operators, have emerged as a 

powerful tool for modeling complex systems. In recent years, there has been growing interest in 

studying hybrid fractional differential equations, which incorporate fractional derivatives and integrals 

into the hybrid structure. ([10],[13],[14],[16],[4],[12]). 

Lakshmikanthan and Dhage [7]  They initiated the study of hybrid equations by introducing a novel 

class of nonlinear differential equations known as ordinary hybrid differential equations. 

                                               {
𝑑

𝑑𝑡
(

𝑥(𝑡)

𝑓(𝑡,𝑥(𝑡))
) = 𝑔(𝑡, 𝑥(𝑡)),  𝑎. 𝑒.  𝑡 ∈ 𝐼0,

𝑥(𝑡0) = 𝑥0 ∈ ℝ
                           

They formulated essential hybrid differential inequalities that serve as key tools for proving the 

existence of extremal solutions. 

Zhao et al. [18]   extended Dhage’s work to the fractional-order case by examining boundary value 

problems involving fractional hybrid differential equations. 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3 (2025) 

  

986 
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                                                 {
𝐷0+
𝛼 (

𝑥(𝑡)

𝑓(𝑡,𝑥(𝑡))
) = 𝑔(𝑡, 𝑥(𝑡)),   𝑡 ∈ [0, 𝑇],

𝑥(𝑡0) = 0,
                       

where 𝐷0+
𝛼  is the Riemann-Liouville fractional derivative of order 0 < 𝛼 < 1. 

Hybrid fractional differential equations and inclusions have been the focus of considerable research in 

recent years. Prior to proceeding, we present a brief overview of some relevant contributions in this 

field. Ahmad et al. [2]   examined the existence of solutions for a hybrid inclusion problem involving 

nonlocal boundary conditions. 

                                      {
𝐶𝐷0+

𝛼 (
𝑥(𝑡)−∑ 𝐼

0+
𝛽𝑖𝑚

𝑖=1 ℎ𝑖(𝑡,𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡))
) ∈ 𝒢(𝑡, 𝑥(𝑡)),  𝑎. 𝑒.  𝑡 ∈ [0,1],

𝑥(0) = 𝜇(𝜉),   𝑥(1) = 𝑎 ∈ ℝ.

          

where 𝐶𝐷0+
𝛼  denotes the Caputo fractional derivative of order 1 < 𝛼 ≤ 2 and 𝐼

0+
𝛽𝑖  is the Riemann–

Liouville fractional integral of order 𝛽𝑖 > 0 with 𝑖 ∈ {1,2,3. . . , 𝑚}. In [6], Derbazi et al. confirmed the 

existence and uniqueness of solutions for a fractional hybrid boundary value problem. 

                               

{
 
 

 
 
𝐶𝐷0+

𝛼 (
𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡))
) = Θ(𝑡, 𝑥(𝑡)),  𝑎. 𝑒.  𝑡 ∈ [0, 𝑇],

𝑎1 (
𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡))
) |𝑡=0 + 𝑏1 (

𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡))
) |𝑡=𝑇 = 𝜆1,

𝑎2
𝐶𝐷

0+
𝛽
(
𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡))
) |𝑡=𝜂 + 𝑏2

𝐶𝐷
0+
𝛽
(
𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡))
) |𝑡=𝑇 = 𝜆2.

             

where 1 < 𝛼 ≤ 2 , 0 < 𝛽 ≤ 1, 𝜂 ∈ [0, 𝑇] and 𝑎1, 𝑎2, 𝑏1, 𝑏2, 𝜆1, 𝜆2 are real constants. . Baleanu et al.  

[3] they employed a generalized version of Dhage’s hybrid fixed point theorem for the sum of three 

fractional operators to examine the existence of solutions to a fractional hybrid integro-differential 

equation subject to mixed hybrid integral boundary conditions. 

{
 
 
 
 
 
 
 

 
 
 
 
 
 
 𝐶𝐷0+

𝜔 (
𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡),𝐼

0+
𝛾1𝑥(𝑡),𝐼

0+
𝛾2𝑥(𝑡),...,𝐼

0+
𝛾𝑛𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡),𝐼
0+
𝜇1𝑥(𝑡),𝐼

0+
𝜇2𝑥(𝑡),...,𝐼

0+
𝜇𝑚𝑥(𝑡))

) = Υ(𝑡, 𝑥(𝑡)),  𝑎. 𝑒.  𝑡 ∈ [0,1],

𝜆1 ∫  𝐶
1

0
𝐷
0+
𝛽1 (

𝑥(𝑠)−ℎ(𝑠,𝑥(𝑠),𝐼𝛾1𝑥(𝑠),𝐼𝛾2𝑥(𝑠),...,𝐼𝛾𝑛𝑥(𝑠))

𝑔(𝑠,𝑥(𝑠),𝐼𝜇1𝑥(𝑠),𝐼𝜇2𝑥(𝑠),...,𝐼𝜇𝑚𝑥(𝑠))
)𝑑𝑠

+𝜆2 𝐶𝐷
0+
𝛼1 (

𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡),𝐼
0+
𝛾1𝑥(𝑡),𝐼

0+
𝛾2𝑥(𝑡),...,𝐼

0+
𝛾𝑛𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡),𝐼
0+
𝜇1𝑥(𝑡),𝐼

0+
𝜇2𝑥(𝑡),...,𝐼

0+
𝜇𝑚𝑥(𝑡))

) |𝑡=1

+𝜆3 (
𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡),𝐼

0+
𝛾1𝑥(𝑡),𝐼

0+
𝛾2𝑥(𝑡),...,𝐼

0+
𝛾𝑛𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡),𝐼
0+
𝜇1𝑥(𝑡),𝐼

0+
𝜇2𝑥(𝑡),...,𝐼

0+
𝜇𝑚𝑥(𝑡))

) |𝑡=0 = 0

𝜆4 ∫  𝐶
1

0
𝐷
0+
𝛽2 (

𝑥(𝑠)−ℎ(𝑠,𝑥(𝑠),𝐼𝛾1𝑥(𝑠),𝐼𝛾2𝑥(𝑠),...,𝐼𝛾𝑛𝑥(𝑠))

𝑔(𝑠,𝑥(𝑠),𝐼𝜇1𝑥(𝑠),𝐼𝜇2𝑥(𝑠),...,𝐼𝜇𝑚𝑥(𝑠))
)𝑑𝑠

+𝜆5 𝐶𝐷
0+
𝛼2 (

𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡),𝐼
0+
𝛾1𝑥(𝑡),𝐼

0+
𝛾2𝑥(𝑡),...,𝐼

0+
𝛾𝑛𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡),𝐼
0+
𝜇1𝑥(𝑡),𝐼

0+
𝜇2𝑥(𝑡),...,𝐼

0+
𝜇𝑚𝑥(𝑡))

) |𝑡=1

+𝜆6 (
𝑥(𝑡)−ℎ(𝑡,𝑥(𝑡),𝐼

0+
𝛾1𝑥(𝑡),𝐼

0+
𝛾2𝑥(𝑡),...,𝐼

0+
𝛾𝑛𝑥(𝑡))

𝑔(𝑡,𝑥(𝑡),𝐼
0+
𝜇1𝑥(𝑡),𝐼

0+
𝜇2𝑥(𝑡),...,𝐼

0+
𝜇𝑚𝑥(𝑡))

) |𝑡=0 = 0.

                 

 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3 (2025) 

  

987 
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where 1 < 𝜔 ≤ 2, 𝛽1, 𝛽2 ∈ (0,1], 𝛼1, 𝛼2 ∈ (0,1], 𝜆1, 𝜆2, 𝜆3, 𝜆4, 𝜆5, 𝜆6 ∈ ℝ
+ and 𝛾𝑖 > 0, 𝜇𝑗 > 0 with 

𝑖 ∈ {1,2, . . . 𝑛} and 𝑗 ∈ {1,2, . . . 𝑚}. 

Building upon the previous works, we establish an existence result for a class of fractional hybrid 

integro-differential problem.  

𝐶𝐷0+
𝛼 (

𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) + 𝑔(𝑡, 𝐼

0+
𝜇1𝑢(𝑡), 𝐼

0+
𝜇2𝑢(𝑡), … , 𝐼

0+
𝜇𝑛𝑢(𝑡)) + ∫ 𝐾

𝑡

0
(𝑡, 𝑠, 𝑢(𝑠)𝑑𝑠 = 0, 𝑡 ∈ 𝐼      (1) 

                 𝑢(0) = ℎ(𝑢),   𝑎𝐷 (
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) |𝑡=0 + 𝑏 𝐶𝐷0+

𝛼 (
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) |𝑡=1 = 0.                     (2) 

where 𝛼 ∈ (1,2], 𝑎, 𝑏 ∈ ℝ, 𝐼 := [0,1]. Also, 𝐶𝐷0+
𝛼  denotes the fractional Caputo derivative of order 

𝛼, 𝐼
0+
𝜇𝑖  denotes the fractional Riemann–Liouville integral of order 𝜇𝑖 > 0 for all 𝑖 ∈ {1,2, . . . , 𝑛}, and 

the maps 𝑓: 𝐼 × ℝ → ℝ∗,  𝑔: 𝐼 × ℝ𝑛+1 → ℝ and 𝐾: 𝐼 × 𝐼 × ℝ → ℝ are continuous. 

This paper proceeds as follows: Some fundamental preliminaries are revisited in Section 2. 

In Section 3, we present the equivalent fractional integral equation corresponding to the linear part of 

the hybrid fractional differential equation ([1],[2]), and we prove the main existence result of this 

paper.One example is given in Section 4 to support the established findings. 

Preliminaries 

In this section, we present definitions and properties of fractional integration and differentiation, as 

well as the fixed point theorem employed in this work. For further details, the reader may refer to 

references [ [10], [13], [14] , [8]]. 

Definition 1.  Let 𝑔 be a real function defined on [0,1] and 𝛼 > 0. Then the left and right Riemann-

Liouville fractional integrals of order 𝛼 of 𝑔 are defined respectively by 𝐼0+
𝛼 𝑔(𝑡) =

1

Γ(𝛼)
∫

𝑔(𝑠)

(𝑡−𝑠)1−𝛼

𝑡

0
𝑑𝑠 

𝐼1−
𝛼 𝑔(𝑡) =

1

Γ(𝛼)
∫

𝑔(𝑠)

(𝑠−𝑡)1−𝛼

1

𝑡
𝑑𝑠 

Definition 2.  The left and the right Caputo fractional derivative of order 𝛼 > 0, of a function 𝑔 are, 

respectively 𝐶𝐷0+
𝛼 𝑔(𝑡) = (𝐼0+

𝑛−𝛼 𝑑𝑛

𝑑𝑡𝑛
𝑔(𝑡)) 𝐶𝐷1−

𝛼 𝑔(𝑡) = (−1)𝑛(𝐼1−
𝑛−𝛼 𝑑𝑛

𝑑𝑡𝑛
𝑔(𝑡)) where 𝑛 − 1 < 𝛼 < 𝑛. 

Proposition 3.  Let 𝑛 − 1 < 𝛼 < 𝑛 and 𝑓 ∈ 𝐿1[0 ,1]. Then (1)  𝐼0+
𝛼 𝐶𝐷0+

𝛼 𝑓(𝑡) = 𝑓(𝑡) − ∑
𝑓(𝑘)(0)

𝑘!
𝑛−1
𝑘=0 𝑡𝑘 

(2)  𝐼1−
𝛼 𝐶𝐷1−

𝛼 𝑓(𝑡) = 𝑓(𝑡) − ∑
(−1)𝑘𝑓(𝑘)(1)

𝑘!
𝑛−1
𝑘=0 (1 − 𝑡)𝑘 

Theorem 4.  (Dhage fixed point theorem)[8]) Let 𝑀 be a closed, bounded, convex and nonempty subset 

of a Banach algebra (𝐸, ∥ .  ∥) , and let 𝐴: 𝐸 → 𝐸 and 𝐵:𝑀 → 𝐸 be two operators such that 

(i) 𝐴 is Lipschitzian with Lipschitz constant 𝜆, 

(ii) 𝐵 is completely continuous,, 

(iii) 𝑥 = 𝐴𝑥𝐵𝑧 ⇒ 𝑥 ∈ 𝑀 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑧 ∈ 𝑀, 

(iv) 𝜆𝐿 < 1, where 𝐿 =∥ 𝐵(𝑀) ∥= 𝑠𝑢𝑝{∥ 𝐵(𝑥) ∥: 𝑥 ∈ 𝑀}. 

Then the operator equation 𝐴𝑦𝐵𝑦 = 𝑦 has a solution in 𝑀. 



Communications on Applied Nonlinear Analysis 

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Vol 32 No. 3 (2025) 

  

988 
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Main results 

Lemma 5.  Let 𝑦 ∈ 𝐴𝐶([0,1], ℝ) Then 𝑢 is a solution of the hybrid fractional integrodifferential 

problem 

{

𝐶𝐷0+
𝛼 (

𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) + 𝑦(𝑡) = 0, 𝑡 ∈ 𝐼 ≔ [0,1],

𝑢(0) = ℎ(𝑢),   𝑎𝐷 (
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) |𝑡=0 + 𝑏 𝐶𝐷0+

𝛼 (
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) |𝑡=1 = 0.

        (3) 

if and only if 𝑢 is a solution for the integral equation                                                  

                    𝑢(𝑡) = 𝑓(𝑡, 𝑢(𝑡)) [−∫
(𝑡−𝜏)𝛼−1

Γ(𝛼)

𝑡

0
𝑦(𝜏)𝑑𝜏 +

Γ(3−𝛼)𝑏𝑡

𝑎Γ(3−𝛼)+𝑏
∫ 𝑦
1

0
(𝜏)𝑑𝜏 +

ℎ(𝑢)

𝑓(0,ℎ(𝑢))
]    (4)  

Proof. we apply the right-hand side fractional integral 𝐼0+
𝛼  to equation ([3]).We get 

                                         
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
= −∫

(𝑡−𝜏)𝛼−1

Γ(𝛼)

𝑡

0
𝑦(𝜏)𝑑𝜏 + 𝑐1 + 𝑐2𝑡.                           (5)                             

Using the conditions nonlocal 𝑢(0) = ℎ(𝑢), so 𝑐1 =
ℎ(𝑢)

𝑓(0,ℎ(𝑢))
.and we have 

           𝐶𝐷0+
𝛼−1 (

𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) = −∫ 𝑦

𝑡

0
(𝜏)𝑑𝜏 + 𝑐2

𝑡2−𝛼

Γ(3−𝛼)
,                        

and 

                                             𝐷
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
= −∫

(𝑡−𝜏)𝛼−2

Γ(𝛼−1)

𝑡

0
𝑦(𝜏)𝑑𝜏 + 𝑐2,                            

so 

𝑎𝐷 (
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) |𝑡=0 + 𝑏 𝐶𝐷0+

𝛼 (
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) |𝑡=1 = 𝑎𝑐2 + 𝑏 (−∫ 𝑦

1

0
(𝜏)𝑑𝜏 +

𝑐2

Γ(3−𝛼)
)          

then we get 

                                                              𝑐2 =
Γ(3−𝛼)𝑏

𝑎Γ(3−𝛼)+𝑏
∫ 𝑦
1

0
(𝜏)𝑑𝜏                                     

Substituting the values of 𝑐1, 𝑐2 in ([5]), we get solution ([4]). The converse follows by direct 

computation. This completes the proof. 

 

In the sequel, we need the following assumptions. 

(H1)   The function 𝑓: 𝐼 × ℝ → ℝ− {0} is a continuous function satisfying the Lipschitz condition for 

a constant 𝜆𝑓 

∣ 𝑓(𝑡, 𝑢) − 𝑓(𝑡, 𝑣) ∣≤ 𝜆𝑓 ∣ 𝑢 − 𝑣 ∣. 

(H2)   The function ℎ:ℝ → ℝ is a continuous function and there exists a constant 𝑀0 > 0 such that : 

|
ℎ(𝑢)

𝑓(0,ℎ(𝑢))
| ≤ 𝑀0. 

(H3)   The function 𝑔: 𝐼 × ℝ𝑛+1 → ℝ is a continuous function and there exists a bounded mapping 

𝜃: 𝐼 ×→ ℝ+ such that for all 𝑢𝑖 , 𝑣𝑖 ∈ 𝐸, 



Communications on Applied Nonlinear Analysis 

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Vol 32 No. 3 (2025) 

  

989 
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∣ 𝑔(𝑡, 𝑢1(𝑡), 𝑢2(𝑡), . . . , 𝑢𝑛+1(𝑡)) − 𝑔(𝑡, 𝑣1(𝑡), 𝑣2(𝑡), . . . , 𝑣𝑛+1(𝑡)) ∣≤ 𝜃(𝑡)∑ ∣𝑛+1
𝑖=1 𝑢𝑖(𝑡) − 𝑣𝑖(𝑡) ∣,   

(H4)   The function 𝐾: 𝐼 × 𝐼 × ℝ → ℝ+ is a continuous function and there exists a constant 𝑀1 >

0such that : 

𝑀𝑎𝑥{𝐾(𝑡, 𝑠, 𝑢(𝑠)):  𝑡, 𝑠 ∈ 𝐼;  ∣ 𝑢(𝑠) ∣≤ 𝑅} ≤ 𝑀1 

where 𝑅 =
𝑀𝑓Υ

1−𝜆𝑓Υ
 and 

Υ = [
1

Γ(𝛼 + 1)
+

𝑏Γ(3 − 𝛼)

𝑎Γ(3 − 𝛼) + 𝑏
] [𝜃∗𝜉𝑅 + 𝐺∗ +𝑀1] + 𝑀0 

where 𝜃∗ = 𝑠𝑢𝑝𝑡∈𝐼𝜃(𝑡),   𝐺
∗ = 𝑠𝑢𝑝𝑡∈𝐼 ∣ 𝑔(𝑡, 0,0, . . . ,0) ∣,   𝑀𝑓 = 𝑠𝑢𝑝𝑡∈𝐼𝑓(𝑡, 0),   𝑎𝑛𝑑  𝜉 = 1 +

∑
1

Γ(1+𝜇𝑖)

𝑛
𝑖=1  

Theorem 6.  Assume that conditions (H1)–(H4) hold.and if 𝜆𝑓Υ < 1.then the problem ([1]-[2]) has 

at least one solution in 𝐸 = 𝐶([0,1]) 

Proof. We consider a subset Ω of 𝐸 given by Ω = {𝑢 ∈ 𝐸: ∥ 𝑢 ∥𝐸≤ 𝑅} and we define the operators 

𝐴:𝐸 → 𝐸 and 𝐵:Ω → 𝐸 as follows: 

                                                              𝐴𝑢(𝑡) = 𝑓(𝑡, 𝑢(𝑡)).  𝑡 ∈ 𝐼, 

               

𝐵𝑢(𝑡) = −∫
(𝑡−𝜏)𝛼−1

Γ(𝛼)

𝑡

0
[𝑔(𝜏, 𝐼

0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠]𝑑𝜏

+
𝑏Γ(3−𝛼)𝑡

𝑎Γ(3−𝛼)+𝑏
∫ [𝑔(𝜏, 𝐼0+

𝜇1𝑢(𝜏), 𝐼
0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠]

1

0
𝑑𝜏

+
ℎ(𝑢)

𝑓(0,ℎ(𝑢))
.

     

Obviously, problem ([1]-[2]) has a solution if and only if 𝐴𝑥𝐵𝑥 has a fixed point. 

Now, we show that the operators 𝐴 and 𝐵 satisfy all the conditions of theorem 4 in a series of steps. 

claim 1   𝐴 is a Lipschitz on 𝐸 Let 𝑢, 𝑣 ∈ 𝐸  for all  𝑡 ∈ 𝐼.  

Then in view of condition (H1), we get 

∣ 𝐴𝑢(𝑡) − 𝐴𝑣(𝑡) ∣=∣ 𝑓(𝑡, 𝑢(𝑡)) − 𝑓(𝑡, 𝑣(𝑡)) ∣≤ 𝜆𝑓 ∣ 𝑢 − 𝑣 ∣ 

Then, for each 𝑡 ∈ 𝐼 we obtain ∥ 𝐴𝑢 − 𝐴𝑣 ∥𝐸≤ 𝜆𝑓 ∥ 𝑢 − 𝑣 ∥𝐸 

claim 2 𝐵 is completely continuous on Ω. We firstly show that𝐵 is uniformly bounded for any 𝑢 ∈ Ω, 

by (H2)-(H5), we have. 



Communications on Applied Nonlinear Analysis 

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990 
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∣ 𝐵𝑢(𝑡) ∣≤ ∫
∣𝑡−𝜏∣𝛼−1

Γ(𝛼)

𝑡

0
[∣ 𝑔(𝜏, 𝑢(𝜏), 𝐼

0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) ∣ +∫ ∣

𝜏

0
𝐾(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠 ∣]𝑑𝜏

+
𝑏Γ(3−𝛼)𝑡

𝑎Γ(3−𝛼)+𝑏
∫ [∣ 𝑔(𝜏, 𝑢(𝜏), 𝐼0+

𝜇1𝑢(𝜏), 𝐼
0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) ∣ +∫ ∣

𝜏

0
𝐾(𝜏, 𝑠, 𝑢(𝑠) ∣ 𝑑𝑠]

1

0
𝑑𝜏

+|
ℎ(𝑢)

𝑓(0,ℎ(𝑢))
|,

≤ ∫
∣𝑡−𝜏∣𝛼−1

Γ(𝛼)

𝑡

0
[∣ 𝑔(𝜏, 𝑢(𝜏), 𝐼

0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) − 𝑔(𝑡, 0, . . . ,0) ∣

+∣ 𝑔(𝑡, 0, . . . ,0) ∣ + ∫ ∣
𝜏

0
𝐾(𝜏, 𝑠, 𝑢(𝑠) ∣ 𝑑𝑠]𝑑𝜏

+
𝑏Γ(3−𝛼)𝑡

𝑎Γ(3−𝛼)+𝑏
∫ [
1

0
∣ 𝑔(𝜏, 𝑢(𝜏), 𝐼

0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) − 𝑔(𝑡, 0, . . . ,0) ∣

+∣ 𝑔(𝑡, 0, . . . ,0) ∣ + ∫ ∣
𝜏

0
𝐾(𝜏, 𝑠, 𝑢(𝑠) ∣ 𝑑𝑠]𝑑𝜏

+|
ℎ(𝑢)

𝑓(0,ℎ(𝑢))
|,

≤
1

Γ(𝛼+1)
[𝜃∗ (1 +

1

Γ(𝜇1+1)
+

1

Γ(𝜇2+1)
+. . . +

1

Γ(𝜇𝑛+1)
) ∣ 𝑢 ∣ +𝐺∗ +𝑀1]

+
𝑏Γ(3−𝛼)

𝑎Γ(3−𝛼)+𝑏
[𝜃∗ (1 +

1

Γ(𝜇1+1)
+

1

Γ(𝜇2+1)
+. . . +

1

Γ(𝜇𝑛+1)
) ∣ 𝑢 ∣ +𝐺∗ +𝑀1] + 𝑀0

≤ [
1

Γ(𝛼+1)
+

𝑏Γ(3−𝛼)

𝑎Γ(3−𝛼)+𝑏
] [𝜃∗𝜉𝑅 + 𝐺∗ +𝑀1] + 𝑀0

≤ Υ

  

Thus, we get ∥ 𝐵𝑢 ∥𝐸≤ Υfor all 𝑢 ∈ Ω, this proves that 𝐵 is uniformly bounded in Ω. 

Next we show that 𝐵 is continuous on Ω. Let {𝑢𝑛}𝑛∈ℕ be a sequence in Ω converging to a point 𝑢 ∈

Ω.Then by (H2)-(H4), for all 𝑡 ∈ 𝐼, one has 

∣ 𝐵𝑢𝑛(𝑡) − 𝐵𝑢(𝑡) ∣≤ ∫
∣𝑡−𝜏∣𝛼−1

Γ(𝛼)

𝑡

0
[∣ 𝑔(𝜏, 𝑢𝑛, 𝐼0+

𝜇1𝑢𝑛(𝜏), 𝐼0+
𝜇2𝑢𝑛(𝜏), . . . , 𝐼0+

𝜇𝑛𝑢𝑛(𝜏))

−𝑔(𝜏, 𝑢(𝜏), 𝐼
0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), … , 𝐼

0+
𝜇𝑛𝑢(𝜏)) ∣ +∫ ∣

𝜏

0
𝐾(𝜏, 𝑠, 𝑢𝑛(𝑠) − 𝐾(𝜏, 𝑠, 𝑢(𝑠)

∣ 𝑑𝑠]𝑑𝜏

+
𝑏Γ(3−𝛼)𝑡

𝑎Γ(3−𝛼)+𝑏
∫ [
1

0
∣ 𝑔(𝜏, 𝑢𝑛, 𝐼0+

𝜇1𝑢𝑛(𝜏), 𝐼0+
𝜇2𝑢𝑛(𝜏), . . . , 𝐼0+

𝜇𝑛𝑢𝑛(𝜏))

−𝑔(𝜏, 𝑢(𝜏), 𝐼
0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), … , 𝐼

0+
𝜇𝑛𝑢(𝜏)) ∣ +∫ ∣

𝜏

0
𝐾(𝜏, 𝑠, 𝑢𝑛

(𝑠) − 𝐾(𝜏, 𝑠, 𝑢(𝑠) ∣ 𝑑𝑠]𝑑𝜏

+|
ℎ(𝑢𝑛)

𝑓(0,ℎ(𝑢𝑛))
−

ℎ(𝑢𝑛)

𝑓(0,ℎ(𝑢𝑛))
|,

≤
1

Γ(𝛼+1)
[𝜃∗ (1 +

1

Γ(𝜇1+1)
+

1

Γ(𝜇2+1)
+. . . +

1

Γ(𝜇𝑛+1)
) ∣ 𝑢𝑛 − 𝑢 ∣

+𝑘0 ∣ 𝑢𝑛 − 𝑢 ∣]

+
𝑏Γ(3−𝛼)

𝑎Γ(3−𝛼)+𝑏
[𝜃∗ (1 +

1

Γ(𝜇1+1)
+

1

Γ(𝜇2+1)
+. . . +

1

Γ(𝜇𝑛+1)
) ∣ 𝑢𝑛 − 𝑢 ∣

+𝑘0 ∣ 𝑢𝑛 − 𝑢 ∣]

+|
ℎ(𝑢𝑛)

𝑓(0,ℎ(𝑢𝑛))
−

ℎ(𝑢𝑛)

𝑓(0,ℎ(𝑢𝑛))
|,

≤ [
1

Γ(𝛼+1)
+

𝑏Γ(3−𝛼)

𝑎Γ(3−𝛼)+𝑏
] [𝜃∗𝜉 + 𝑘0] ∥ 𝑢𝑛 − 𝑢 ∥𝐸

+|
ℎ(𝑢𝑛)

𝑓(0,ℎ(𝑢𝑛))
−

ℎ(𝑢𝑛)

𝑓(0,ℎ(𝑢𝑛))
|,

  

Since that functions ℎ and 𝑓 are continuous, we deduce 



Communications on Applied Nonlinear Analysis 

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Vol 32 No. 3 (2025) 

  

991 
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∥ 𝐵𝑢𝑛 − 𝐵𝑢 ∥𝐸→ 0  𝑎𝑠  𝑛 → ∞. 

Then 𝐵 is continuous. 

Next we prove that the operator𝐵 equicontinuous. Let 𝑢 ∈ Ω and 𝑡1, 𝑡2 ∈ 𝐼 with 𝑡1 < 𝑡2 Then we have 

      

∣ 𝐵𝑢(𝑡2) − 𝐵𝑢(𝑡1) ∣≤ | ∫
(𝑡2−𝜏)

𝛼−1

Γ(𝛼)

𝑡2
0

[𝑔(𝜏, 𝑢(𝜏), 𝐼
0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠]𝑑𝜏

+
𝑏Γ(3−𝛼)𝑡2

𝑎Γ(3−𝛼)+𝑏
∫ [𝑔(𝜏, 𝑢(𝜏), 𝐼0+

𝜇1𝑢(𝜏), 𝐼
0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠]

1

0
𝑑𝜏

−∫
(𝑡1−𝜏)

𝛼−1

Γ(𝛼)

𝑡1
0

[𝑔(𝜏, 𝑢(𝜏), 𝐼
0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ ∣

𝜏

0
𝐾(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠]𝑑𝜏

+
𝑏Γ(3−𝛼)𝑡1

𝑎Γ(3−𝛼)+𝑏
∫ [𝑔(𝜏, 𝑢(𝜏), 𝐼0+

𝜇1𝑢(𝜏), 𝐼
0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠]

1

0
𝑑𝜏|

  

          

≤ ∫ |
𝑡1
0

(𝑡2−𝜏)
𝛼−1−(𝑡1−𝜏)

𝛼−1

Γ(𝛼)
|[ |𝑔(𝜏, 𝑢(𝜏), 𝐼

0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠|]𝑑𝜏

+∫ |
𝑡2
𝑡1

(𝑡2−𝜏)
𝛼−1

Γ(𝛼)
|[ |𝑔(𝜏, 𝑢(𝜏), 𝐼

0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠|]𝑑𝜏

+
𝑏Γ(3−𝛼)∣𝑡2−𝑡1∣

𝑎Γ(3−𝛼)+𝑏
∫ [ |
1

0
𝑔(𝜏, 𝑢(𝜏), 𝐼

0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏))| + ∫ |

𝜏

0
𝐾(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠|]𝑑𝜏

≤ [
𝜃∗𝜉𝑅+𝐺∗+𝑀1

Γ(𝛼+1)
] [2 ∣ 𝑡2 − 𝑡1 ∣

𝛼+ 𝑡2
𝛼 − 𝑡1

𝛼]

+
𝑏Γ(3−𝛼)∣𝑡2−𝑡1∣

𝑎Γ(3−𝛼)+𝑏
[𝜃∗𝜉𝑅 + 𝐺∗ +𝑀1]

  

which is independent of 𝑢 ∈ Ω. As 𝑡1 → 𝑡2, the right-hand side of the above inequality tends to zero. 

Therefore, it follows from the Arzel´a-Ascoli theorem that 𝐵 is a completely continuous operator on 

Ω. 

claim 3   Now we show that the (iii) hypothesis of theorem 4 is satisfied. Let 𝑢 ∈ 𝐸 and 𝑣 ∈ Ω such 

that 𝑢 = 𝐴𝑢𝐵𝑣 Then, for 𝑡 ∈ 𝐼 we have 

        

∣ 𝑢(𝑡) ∣≤∣ 𝐴𝑢(𝑡) ∣∣ 𝐵𝑣(𝑡) ∣

≤∣ 𝑓(𝑡, 𝑢(𝑡)) ∣ [ | − ∫
(𝑡−𝜏)𝛼−1

Γ(𝛼)

𝑡

0
[𝑔(𝜏, 𝐼

0+
𝜇1𝑢(𝜏), 𝐼

0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠]𝑑𝜏

+
𝑏Γ(3−𝛼)𝑡

𝑎Γ(3−𝛼)+𝑏
∫ [𝑔(𝜏, 𝐼0+

𝜇1𝑢(𝜏), 𝐼
0+
𝜇2𝑢(𝜏), . . . , 𝐼

0+
𝜇𝑛𝑢(𝜏)) + ∫ 𝐾

𝜏

0
(𝜏, 𝑠, 𝑢(𝑠)𝑑𝑠]

1

0
𝑑𝜏

+
ℎ(𝑢)

𝑓(0,ℎ(𝑢))
|]

≤ [∣ 𝑓(𝑡, 𝑢(𝑡)) − 𝑓(𝑡, 0) ∣ +∣ 𝑓(𝑡, 0) ∣] ([
1

Γ(𝛼+1)
+

𝑏Γ(3−𝛼)

𝑎Γ(3−𝛼)+𝑏
] [𝜃∗𝜉𝑅 + 𝐺∗ +𝑀1] + 𝑀0)

≤ [𝜆𝑓 ∣ 𝑢 ∣ +𝑀𝑓]Υ.

  

Thus, we obtain 

∥ 𝑢(𝑡) ∥𝐸≤
𝑀𝑓Υ

1 − 𝜆𝑓Υ
= 𝑅 



Communications on Applied Nonlinear Analysis 

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992 
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Then 𝑢 ∈ Ω, thus the (iii) hypothesis of theorem 4 is satisfied. 

claim 4   Now, we show that 𝜆𝑓𝐿 < 1, where 𝐿 =∥ 𝐵(Ω) ∥𝐸= sup{∥ 𝐵𝑢 ∥𝐸: 𝑢 ∈ Ω} 

Since 𝐿 = sup𝑢∈Ω{sup𝑡∈𝐼 ∣ 𝐵𝑢(𝑡) ∣} ≤ Υ, then 𝜆𝑓Υ < 1, 

Thus all the conditions of theorem 4 are satisfied and hence the operator equation 𝑢 = 𝐴𝑢𝐵𝑢 has a 

solution in Ω. In consequence, problem ([1]-[2]) has a solution on 𝐼. This completes the proof  

An Example 

This section includes an example that showcases how Theorem 6 can be applied. 

Let us consider the following boundary value problem: 

      𝐶𝐷0+
1.7 [

𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
] + 𝑔(𝑡, 𝑢(𝑡), 𝐼0+

0.7𝑢(𝑡), 𝐼0+
0.2𝑢(𝑡)) + ∫ 𝐾

𝑡

0
(𝑡, 𝑠, 𝑢(𝑠)))𝑑𝑠 = 0 ,  𝑡 ∈ 𝐼 = [0,  1]. (𝟔)  

                          𝑢(0) = ℎ(𝑢),    100𝐷 (
𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) |𝑡=0 +

𝐶𝐷0+
0.7 (

𝑢(𝑡)

𝑓(𝑡,𝑢(𝑡))
) |𝑡=1 = 0.    (𝟕) 

here 𝛼 = 1.7,   𝜇1 = 0.7,   𝜇2 = 0.2,   𝑎 = 100 and 𝑏 = 1. 

                                Where 𝑓(𝑡, 𝑢(𝑡)) =
7𝑒−3𝑡

15(𝑡2+2)

|𝑢|+1

|𝑢|+2
, and ℎ(𝑢) =

sin(𝑢)

100+𝑢2
, 

 

and  

             𝑔(𝑡, 𝑢(𝑡), 𝐼0+
0.7𝑢(𝑡), 𝐼0+

0.2𝑢(𝑡)) =
3

100(𝑡2+1)
[𝑢(𝑡) + |cos(𝐼0+

0.7𝑢(𝑡)) − 𝑐ℎ(𝐼0+
0.2𝑢(𝑡))|] +

𝑡

100
,  

and 

𝐾(𝑡, 𝑠, 𝑢(𝑠)) =
cos(𝑢2)𝑒−𝑡(𝑠2+1)

100
 Note that 𝑀𝑓 = sup𝑡∈𝐼 ∣ 𝑓(𝑡, 0) ∣=

7𝑒−3𝑡

30(𝑡2+2)
=

7

60
 and 𝑀0 =

  |
ℎ(𝑢(𝑡)

𝑓(0,ℎ(𝑢(𝑡))
| =

60

70
  

Setting 𝑀1 =
2

100
, 𝐺∗ = sup𝑡∈𝐼 ∣ 𝑔(𝑡, 0, . . . ,0) ∣= 0.01, and we have 

                                             
∣ 𝑓(𝑡, 𝑢) − 𝑓(𝑡, 𝑣) ∣≤

7𝑒−3𝑡

15(𝑡2+2)
|
𝑢+1

𝑢+2
−
𝑣+1

𝑣+2
|

≤
7

30
|

∣𝑣−𝑢∣

∣𝑣+2∣∣𝑢+2∣
≤

7

30
∣ 𝑣 − 𝑢 ∣.

  

then 𝜆𝑓 =
7

30
, for 𝑢, 𝑣 ∈ ℝ, we have 

                                    

∣ 𝑔(𝑡, 𝑣(𝑡), 𝐼0+
0.7𝑣(𝑡), 𝐼0+

0.2𝑣(𝑡)) − 𝑔(𝑡, 𝑢(𝑡), 𝐼0+
0.7𝑢(𝑡), 𝐼0+

0.2𝑢(𝑡)) ∣

≤
3

100(𝑡2+1)
(1 +

𝑡0.2

Γ(1.2)
+

𝑡0.7

Γ(1.7)
) ∣ 𝑣 − 𝑢 ∣

≤
3

100(𝑡2+1)
(1 +

1

Γ(1.2)
+

1

Γ(1.7)
) ∣ 𝑣 − 𝑢 ∣

    

Thus, the assumption (A2) holds true with 𝜃(𝑡) =
3

100(𝑡2+1)
 and 𝜃∗ = 0.03 and 𝜉 = 3.189672. 

By the above data, we get Υ = 0.939755 and 𝜆𝑓Υ = 0.219276 < 1.. Thus, we can choose 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3 (2025) 

  

993 
https://internationalpubls.com 

0.935712 < 𝑅 < 54.19965. Accordingly, all the conditions of Theorem 6 are fulfilled, the hybrid 

fractional problem (6-7) has at least one solution on [0, 1]. 

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