CONTACT:  

Faraj Y. Ishak             faraj.ishak@uod.ac         Department of Statistics, Collage of Administration & Economics 

University of Duhok, Iraq 

 

       The article can be accessed here. https://doi.org/10.15642/mantik.2022.8.2.89-98 

Jurnal Matematika MANTIK 

Vol. 8, No. 2, November 2022, pp.89-98 

 

ISSN: 2527-3159 (print) 2527-3167 (online) 

Mixed Boundary Value Problem for Nonlinear Fractional Volterra 
Integral Equation 

 

 

Faraj Y. Ishak 
 

University of Duhok Iraq, faraj.ishak@uod.ac 
 
 

 

Article history: 

Received Sep 9, 2022 

Revised, Dec 9, 2022 

Accepted, Dec 31, 2022 

 

Kata Kunci:  

Persamaan Volterra, 

teorema titik tetap 

Krasnoselskii, prinsip 

kontraksi Banach, 

teori Leray-Schauder 

degree. 

 

Abstrak. Artikel ini menyajikan hasil penelitian tentang eksistensi 

dari solusi persamaan integral fraksional nonlinier tipe Volterra 

dengan kondisi batas campuran, beberapa hipotesis yang 

diperlukan telah dikembangkan untuk membuktikan keberadaan 

solusi persamaan yang diusulkan. Teorema Krasnoselskii, prinsip 

Banach Contraction dan teori derajat Leray-Schauder adalah 

teorema dasar yang digunakan di sini untuk mencari solusi hasil. 

Dalam artikel ini juga diberikan contoh sederhana dari penerapan 

hasil persamaannya. 

 

 

Keywords:  

Volterra equation, 

Krasnoselskii theorem, 

Banach contraction 

principle, Leray-

Schauder degree theory. 

 

 

 

 

Abstract. In this paper we present the existence of solutions for a 

nonlinear fractional integral equation of Volterra type with mixed 

boundary conditions, some necessary hypotheses have been 
developed to prove the existence of solutions to the proposed 

equation. Krasnoselskii Theorem, Banach Contraction principle 

and Leray-Schauder degree theory are the basic theorems used here 

to find the results. A simple example of application of the main 

result is presented. 

 

  

How to cite: 

F. Y. Ishak, “Mixed Boundary Value Problem for Nonlinear Fractional Volterra Integral 

Equation”, J. Mat. Mantik, vol. 8, no. 2, pp. 89-98, December 2022. 

 

 

 

  

https://doi.org/10.15642/mantik.2021.7.1.9-19
http://u.lipi.go.id/1458103791


Jurnal Matematika MANTIK 

Vol. 8, No. 2, December 2022, pp. 89-98 

90 
 

1. Introduction 

Fractional calculus is the field of mathematical analysis which deals with the 

investigation and applications of integrals and derivatives of arbitrary order, the fractional 

calculus may be considered an old and yet novel topic. Recently, fractional differential 

equations have been of great interest. This is because of both the intensive development of 

the theory of fractional calculus itself and its applications in various sciences, such as 

physics, mechanics, chemistry, engineering.[1],[2],[3]. 

Integral equations appear in many engineering and scientific fields, such as unsteady 

aerodynamics, viscoelasticity, fluid dynamics, lots of population growth systems, neural 

network analysis, mathematical analysis of particle diffusion in an unstable fluid, heat 

conduction in memory resources, transmission lines, Population dynamics theory, nuclear 

reactors, inheritance systems. For details, see [4], [5], [6], [7], [8]. 

Boundary value problems have different applications. In addition to the previously 

mentioned fields, this type of problem appears in chemical engineering sciences, models 

of electromagnetic systems  and thermoelectric theory. For more detailed information on 

boundary conditions, see [9], [10]. For more details on local and non-local boundary 

conditions, see [11],[12],[13],[14],[15],[16]. 

Feng, Zhang and Yang [13] in 2011 studied the existence and multiplicity solution to 

the nonlocal boundary value problem, fixed-point theorems in the cone were the main tool 

to prove the solutions. In 2014 Nyamoradi and Alaei [17] employed the Guo – 

Krasnoselskii fixed point theorem in a cone to study the existence of solution to a new 

fractional nonlocal mixed boundary value problem. In 2022 Ishak [18] investigated the 

existence solution for a fractional BVP of the first sort with Hadamard type and three-point 

boundary conditions using Krasnoselskii Zabriko theorem and Banach contraction 

principle.  

It is also well known that fixed-point theorems have been applied to various boundary 

value problems to show the existence of solutions; for example, see [3],[5]. However, this 

researcher’s  remains not enough compared to the broad applications of this type of 

equations. The aim of this paper is to fill this gap. in this study we will investigate the 

existence and uniqueness solutions for the boundary value problem: 

          𝑫 
𝒄 𝜶𝒙(𝒕) = 𝒈(𝒕) + ∫ 𝝍(𝒕, 𝒔)𝝓(𝒕, 𝒔, 𝒙(𝒔))𝒅𝒔      

𝒕

−∞

 

                             𝑥(0) = 𝑎𝑥(𝛾) + 𝑏,   𝑥 ́(0) = ω ,   0 ≤ 𝑡 ≤ 1 , 1 < 𝛼 ≤ 2                  (1) 

 

Where  𝑫 
𝒄  indicates the Caputo fractional operator 𝝓: [𝟎, 𝟏]𝘅[𝟎, 𝟏]𝘅𝐗 → 𝐗 is a given 

continuous function in Banach space (𝐗 ,...) and 𝐂 = 𝐂([𝟎, 𝟏], 𝐗)  is Banach space of all 

continuous functions from [0,1]→ 𝐗  endowed with the norm denoted by  ‖. ‖,

𝒂 , 𝒃, 𝛚 [𝟎, +∞), 𝝍: [𝟎, 𝟏]𝘅[𝟎, 𝟏] → 𝐗  is a given kernel , 𝒈: [𝟎, 𝟏] → [𝟎, +∞) is known 
continuous function. 

 

2. Preliminaries   

In this section we will mention some basic definitions in fractional calculus. 

Definition 2.1: [19] The fractional integral of order q is defined by 

𝐼𝑞 𝑓(𝑡) =
1

Γ(𝑞)
∫ (𝑡 − 𝑠)𝑞−1

𝑡

0

𝑓(𝑠)𝑑𝑠,       𝑞 > 0 

provided the integral exists. 



Faraj Y. Ishak 

Mixed Boundary Value Problem for Nonlinear Fractional Volterra Integral Equation 

91 
 

Definition 2.2: [19] The fractional derivative of order q is defined by 

𝐷𝑞 𝑓(𝑡) =
1

Γ(𝑛 − 𝑞)
(

𝑑

𝑑𝑡
)

𝑛

∫ (𝑡 −  𝑠)𝑛−𝑞−1
𝑡

0

𝑓(𝑠)𝑑𝑠,    

𝑛 − 1 < 𝑞 ≤ 𝑛, 𝑞 > 0, Provided the right-hand side is pointwise defined on (0, +∞). 

Lemma 2.1: [18] For α, β > 0, then the following relation hold: 

𝐷𝛼 𝑡𝛽 =
Γ(𝛽 + 1)

Γ(𝛽 + 1 − 𝛼)
𝑡𝛽−𝛼−1, 𝛽 > 𝑛 𝑎𝑛𝑑 𝐷𝛼 𝑡𝑘 = 0, 

𝑘 = 0,1, … , 𝑛 − 1 

 

Lemma 2.2: [18] Let 𝛼 > 0, then the differential equation    

 𝑐 𝐷0+
𝛼 𝑥(𝑡) = 0 

Has a unique solution  𝑥(𝑡) = 𝑐0 + 𝑐1𝑡 + ⋯ 𝑐𝑛−1𝑡
𝑛−1, 𝑐𝑖 ∈ 𝑅, 𝑖 = 1,2, … , 𝑛, where 𝑛 −

1 < 𝛼 ≤ 𝑛 

In view of Lemma 2.2, it follows that 

𝐼𝑞  𝑐 𝐷𝑞 𝑥(𝑡) = 𝑥(𝑡) + 𝑐0 + 𝑐1𝑡 + ⋯ 𝑐𝑛−1𝑡
𝑛−1,  

     𝑐𝑖 ∈ 𝑅, 𝑖 = 1,2, … , 𝑛 

 

Theorem 2.1: [5] (Krasnoselskii fixed point theorem) Let M be a closed convex and 

nonempty subset of a Banach space X. Let A, B be the operators such that (i) Ax + By ∈ M 

whenever x, y ∈ M (ii) A is compact and continuous (iii) B is a contraction mapping. Then 

there exists z ∈ M such that z = Az + Bz. 

 

Theorem 2.2: (Arzela -Ascoli theorem) Let 𝛺 be a compact Hausdorff metric space. Then 

𝑀 ⊂ 𝐶(𝛺) is relatively compact ⟺ 𝑀 is uniformly bounded and uniformly equicontinuous.  

 

Lemma 2.3: Given 𝑓 ∈ 𝐶(0,1) ∩ 𝐿(0,1), the unique solution of (1) is: 

𝑥(𝑡) = ∫
(𝑡 − 𝑠)𝛼−1

Γ(𝛼)

𝑡

0

𝑓(𝑠)𝑑𝑠 +  
𝑎

1 − 𝑎
∫

(𝛾 − 𝑠)𝛼−1

Γ(𝛼)
𝑓(𝑠)𝑑𝑠

𝛾

0

+
𝑎𝜔𝛾 + 𝑏

1 − 𝑎
+ 𝜔𝑡        (2) 

Where,                     

𝑓(𝑡) = 𝑔(𝑡) + ∫ 𝜓(𝑡, 𝑠)𝜙(𝑡, 𝑠, 𝑥(𝑠))𝑑𝑠
𝑡

−∞

 

Proof: In view of lemma (2.2) the fractional differential equation (1) is equivalent to the 

integral equation: 

𝑥(𝑡) = 𝐼0+
𝛼 𝑓(𝑡) + 𝑐0 + 𝑐1𝑡    

𝑥(𝑡) = ∫
(𝑡 − 𝑠)𝛼−1

Γ(𝛼)

𝑡

0

𝑓(𝑠)𝑑𝑠 + 𝑐0 + 𝑐1𝑡 

Where 𝑐0, 𝑐1 ∈ 𝑅. From the boundary conditions (1), we have   𝑐1 = 𝜔 and 

𝑐0 =
𝑎

1 − 𝑎
∫

(𝛾 − 𝑠)𝛼−1

Γ(𝛼)
𝑓(𝑠)𝑑𝑠

𝛾

0

+
𝑎𝜔𝛾 + 𝑏

1 − 𝑎
 

 



Jurnal Matematika MANTIK 

Vol. 8, No. 2, December 2022, pp. 89-98 

92 
 

By substituting 𝑐0, 𝑐1 in 𝑥(𝑡), the proof will be completed . 

We will need the following hypotheses: 

(H1) for 𝐾 ∈ 𝑅+ the inequality holds: 

‖𝜙(𝑡, 𝑠, 𝑥2) − 𝜙(𝑡, 𝑠, 𝑥1)‖ ≤ 𝐾‖𝑥2 − 𝑥1‖,     for all  𝑡, 𝑠 ∈  [0,1],   𝑥1, 𝑥2 ∈ 𝑋 

 

(H2) ‖𝜙(𝑡, 𝑠, 𝑥)‖ ≤ 𝐿    for all 𝑡, 𝑠, 𝑥 ∈ [0,1]𝗑[0,1]𝗑X,   L ∈ 𝑅+ Further 

‖𝑔(𝑡)‖ ≤ 𝜉   , ‖𝜓(𝑡, 𝑠)‖ ≤ 𝛿𝑒 −𝜆(𝑡−𝑠)     for all  𝜉, 𝛿, 𝜆 ∈ 𝑅+ 

 

3. Main Result 

Proves of theorems of existence and uniqueness solution for equation (1) will be given in 

this section. 

 

Theorem 3.1: Suppose that 𝜙: [0,1]𝗑[0,1]𝗑 X → X is continuous and fulfilled H1and H2, if: 

𝑎𝛿 𝐾𝛾𝛼

𝜆(1 − 𝑎)Γ(𝛼 + 1)
< 1                                                                                                                   (3) 

Then equation (1) has at least one solution. 

Proof: Let 𝜑𝑟 = {𝑥 ∈ 𝐶: ‖𝑥‖ ≤ 𝑟} where:  

𝑎𝛾𝛼 (𝜆𝜉 + 𝛿𝐿)

𝜆(1 − 𝑎)Γ(𝛼 + 1)
+

𝜆𝜉 + 𝛿𝐿

𝜆Γ(𝛼 + 1)
+

𝑎𝜔𝛾 + 𝑏

1 − 𝑎
+ 𝜔 ≤ 𝑟 

define tow mapping F, G on 𝜑𝑟  s.t. 

(𝐹𝑥)(𝑡) = ∫
(𝑡 − 𝑠)𝛼−1

Γ(𝛼)

𝑡

0

𝑓(𝑠)𝑑𝑠 

(𝐺𝑥)(𝑡) =
𝑎

1 − 𝑎
∫

(𝛾 − 𝑠)𝛼−1

Γ(𝛼)
𝑓(𝑠)𝑑𝑠

𝛾

0

+
𝑎𝜔𝛾 + 𝑏

1 − 𝑎
+ 𝜔𝑡 

For 𝑥, 𝑦 ∈ 𝜑𝑟, by (H2) we obtain: 

‖(𝐹𝑥)(𝑡) + (𝐺𝑥)(𝑡)‖ ≤

∫
(𝑡−𝑠)𝛼−1

Γ(𝛼)
‖𝑔(𝑠)‖𝑑𝑠 +

𝑡

0
∫

(𝑡−𝑠)𝛼−1

Γ(𝛼)
(∫ ‖𝜓(𝑡, 𝑠)‖

𝑡

−∞

𝑡

0
‖𝜙(𝑡, 𝑠, 𝑥(𝑠))‖𝑑𝑠)𝑑𝑠 +

|
𝑎

1−𝑎
| ∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
‖𝑔(𝑠)‖𝑑𝑠 +

𝑎

1−𝑎

𝛾

0
∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
(

𝛾

0
∫ ‖𝜓(𝑡, 𝑠)‖

 𝑡

−∞ 
‖𝜙(𝑡, 𝑠, 𝑥(𝑠))‖𝑑𝑠)𝑑𝑠 +

|
𝑎𝜔𝛾+𝑏

1−𝑎
| + |𝜔𝑡|  

≤
𝜉 

Γ(𝛼+1)
+  

𝛿

𝜆
∫

(𝑡−𝑠)𝛼−1

Γ(𝛼)
𝐿𝑑𝑠

𝑡

0
+

𝑎𝜉𝛾𝛼

(1−𝑎)Γ(𝛼+1)
+

𝑎𝛿

𝜆(1−𝑎)
∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
𝐿𝑑𝑠

𝛾

0
+

𝑎𝜔𝛾+𝑏

1−𝑎
+ 𝜔𝑡  

≤
𝜉 

Γ(𝛼+1)
+  

𝛿𝐿

𝜆Γ(𝛼+1) 
+

𝑎𝜉𝛾𝛼

(1−𝑎)Γ(𝛼+1)
+   

𝑎𝛿𝐿𝛾𝛼

𝜆(1−𝑎)Γ(𝛼+1) 
+

𝑎𝜔𝛾+𝑏

1−𝑎
+ 𝜔  

≤
𝑎𝛾𝛼(𝜆𝜉+𝛿𝐿)

𝜆(1−𝑎)Γ(𝛼+1)
+

𝜆𝜉+𝛿𝐿

𝜆Γ(𝛼+1)
+

𝑎𝜔𝛾+𝑏

1−𝑎
+ 𝜔 ≤ 𝑟  

Which means 𝐹𝑥 + 𝐺𝑥 ∈ 𝜑𝑟 . 



Faraj Y. Ishak 

Mixed Boundary Value Problem for Nonlinear Fractional Volterra Integral Equation 

93 
 

Since F is continuous because ϕ is continuous, we have to prove that 𝐹 is compact. (𝐹𝑥)(t) 

is uniformly bounded on 𝜑𝑟 , as: 

‖(𝐹𝑥)(𝑡)‖ ≤
𝜆𝜉 + 𝛿𝐿

𝜆Γ(𝛼 + 1)
   

Since 𝜙 is bounded on [0,1]𝗑[0,1]𝗑𝜑𝑟 let: 

𝜙𝑚𝑎𝑥 = 𝑠𝑢𝑝⏟
𝑠,𝑡,𝑥(𝑡)∈[0,1]𝗑[0,1]𝗑𝜑𝑟

‖𝜙(𝑡, 𝑠, 𝑥(𝑡))‖ 

Then for 𝑡1, 𝑡2 ∈ [0,1] we get 

‖(𝐹𝑥)(𝑡2) − (𝐹𝑥)(𝑡1)‖ ≤
1

Γ(𝛼)
∫ ‖((𝑡2 − 𝑠)

𝛼−1 − (𝑡1 − 𝑠)
𝛼−1)𝑓(𝑠)𝑑𝑠‖

𝑡1
0

+

1

Γ(𝛼)
∫ ‖(𝑡2 − 𝑠)

𝛼−1𝑓(𝑠)𝑑𝑠‖
𝑡2

𝑡1
  

              ≤
1

Γ(𝛼)
∫ ‖((𝑡2 − 𝑠)

𝛼−1 − (𝑡1 − 𝑠)
𝛼−1) (𝑔(𝑠) +

𝑡1
0

∫ 𝜓(𝑡, 𝑠)𝜙(𝑡, 𝑠, 𝑥(𝑠))
𝑡

−∞
𝑑𝑠)‖ 𝑑𝑠  +  

1

Γ(𝛼)
∫ ‖(𝑡2 − 𝑠)

𝛼−1 (𝑔(𝑠) +
𝑡2

𝑡1

∫ 𝜓(𝑡, 𝑠)𝜙(𝑡, 𝑠, 𝑥(𝑠))
𝑡

−∞
𝑑𝑠)‖ 𝑑𝑠  

 

              ≤
𝜉|(𝑡2 − 𝑡1)

𝛼 − 𝑡2
𝛼 |

Γ(𝛼 + 1)
+

𝛿𝜙𝑚𝑎𝑥 |(𝑡2 − 𝑡1)
𝛼 − 𝑡2

𝛼 |

𝜆Γ(𝛼 + 1)
+

𝜉𝑡1
𝛼

Γ(𝛼 + 1)
+

𝛿𝜙𝑚𝑎𝑥 𝑡1
𝛼

𝜆Γ(𝛼 + 1)

−  
𝜉|(𝑡2 − 𝑡1)

𝛼 |

Γ(𝛼 + 1)
−

𝛿𝜙𝑚𝑎𝑥 |(𝑡2 − 𝑡1)
𝛼 |

𝜆Γ(𝛼 + 1)
 

                          ≤ (
𝜉

Γ(𝛼 + 1)
+

𝛿𝜙𝑚𝑎𝑥
𝜆Γ(𝛼 + 1)

)|𝑡1
𝛼 − 𝑡2

𝛼 | 

Which is independent of 𝑥  there for (𝐹𝑥)(𝑡 )  is relatively compact on 𝜑𝑟  ,by Arzela-

Ascoli’s theorem (𝐹𝑥)(𝑡 ) is compact in 𝜑𝑟 . 

For 𝑥, 𝑦 ∈ 𝜑𝑟  and 𝑡 ∈ [0,1], by H1  we have: 

 

‖(𝐺𝑥)(𝑡) − (𝐺𝑦)(𝑡)‖ ≤
𝑎

1−𝑎
∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
(

𝛾

0
∫ ‖𝜓(𝑡, 𝑠)‖

 𝑡

−∞ 
‖𝜙(𝑡, 𝑠, 𝑥(𝑠))‖𝑑𝑠)𝑑𝑠 −

  
𝑎

1−𝑎
∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
(

𝛾

0
∫ ‖𝜓(𝑡, 𝑠)‖

 𝑡

−∞ 
‖𝜙(𝑡, 𝑠, 𝑦(𝑠))‖𝑑𝑠)𝑑𝑠  

 

≤
𝑎

1 − 𝑎
∫

(𝛾 − 𝑠)𝛼−1

Γ(𝛼)
(

𝛾

0

∫ 𝛿𝑒 −𝜆(𝑡−𝑠)‖𝜙(𝑡, 𝑠, 𝑥(𝑠)) − 𝜙(𝑡, 𝑠, 𝑦(𝑠))‖
 𝑡

−∞ 

𝑑𝑠)𝑑𝑠 

          ≤
𝑎𝛿

𝜆(1−𝑎)
∫

(𝛾−𝑠)𝛼−1𝐾

Γ(𝛼)
‖𝑥 − 𝑦‖𝑑𝑠 

𝛾

0
  

  ≤
𝑎𝛿 𝐾𝛾𝛼

𝜆(1−𝑎)Γ(𝛼+1)
‖𝑥 − 𝑦‖                           (4) 



Jurnal Matematika MANTIK 

Vol. 8, No. 2, December 2022, pp. 89-98 

94 
 

It follows from (4) that (𝐺𝑥)(𝑡) is contraction mapping, this completes the prove. 

Theorem 3.2: Suppose that 𝜙: [0,1]𝗑[0,1]𝗑X → X is continuous and fulfilled (H1). If: 

Ω =
𝛿𝐾 + 𝑎𝛿𝐾(𝛾𝛼 − 1)

𝜆(1 − 𝑎)𝛤(𝛼 + 1) 
< 1                                                                                                       (5) 

Then equation (1) has a unique solution. 

Proof: Let Φ: 𝐶 → 𝐶 is defined as: 

(Φ𝑥)(𝑡) = ∫
(𝑡−𝑠)𝛼−1

Γ(𝛼)

𝑡

0
𝑓(𝑠)𝑑𝑠 +

𝑎

1−𝑎
∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
𝑓(𝑠)𝑑𝑠

𝛾

0
+

𝑎𝜔𝛾+𝑏

1−𝑎
+  𝜔𝑡,         𝑡 ∈ [0,1]       

let 𝑠𝑢𝑝𝑡∈[0,1]|𝜙(𝑡, 𝑠, 0)| = 𝑀 and choose: 

 

𝑟 ≥
𝜆𝜉+𝛿𝑀

(1−𝛺)λΓ(𝛼+1)
+

𝑎𝛾𝛼(𝜆𝜉+𝛿𝑀)

(1−𝛺)λ𝑎Γ(𝛼+1)
+

𝑎𝜔𝛾+𝑏

(1−𝛺)(1−𝑎)
+

𝜔𝑡

(1−𝛺)
                         (6) 

It is claimed that Φ𝜑𝑟 ⊂ 𝜑𝑟  where: 

𝜑𝑟 = {𝑥 ∈ 𝐶: ‖𝑥‖ ≤ 𝑟} 

In fact, for 𝑥 ∈ 𝜑𝑟 , by (4), (5) and H1 we obtain: 

‖(Φ𝑥)(𝑡)‖ ≤ ∫
(𝑡 − 𝑠)𝛼−1

Γ(𝛼)

𝑡

0

‖𝑔(𝑠)‖𝑑𝑠 

+ ∫
(𝑡−𝑠)𝛼−1

Γ(𝛼)

𝑡

0
(∫ ‖𝜓(𝑡, 𝑠)‖

 𝑡

−∞ 
‖𝜙(𝑡, 𝑠, 𝑥(𝑠))‖𝑑𝑠) 𝑑𝑠 +  

𝑎

1−𝑎
∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
‖𝑔(𝑠)‖𝑑𝑠

𝛾

0
+  

 

𝑎

1−𝑎
∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
(∫ ‖𝜓(𝑡, 𝑠)‖

 𝑡

−∞ 
‖𝜙(𝑡, 𝑠, 𝑥(𝑠))‖𝑑𝑠) 𝑑𝑠

𝛾

0
+ |

𝑎𝜔𝛾+𝑏

1−𝑎
| + |𝜔𝑡|  

 

≤
𝜉

𝛤(𝛼+1)
+ ∫

(𝑡−𝑠)𝛼−1

𝛤(𝛼)
∫ 𝛿𝑒 −𝜆

(𝑡−𝑠) 𝑡

−∞ 
(‖𝜙(𝑡, 𝑠, 𝑥(𝑠)) − 𝜙(𝑡, 𝑠, 0)‖ +

𝑡

0

‖𝜙(𝑡, 𝑠, 0)‖)𝑑𝑠 𝑑𝑠 +
𝑎𝜉𝛾𝛼

(1−𝑎)𝛤(𝛼+1)
+

𝑎

1−𝑎
∫

(𝛾−𝑠)𝛼−1

𝛤(𝛼)
(∫ 𝛿𝑒 −𝜆

(𝑡−𝑠) 𝑡

−∞ 
‖𝜙(𝑡, 𝑠, 𝑥(𝑠)) −

𝛾

0

𝜙(𝑡, 𝑠, 0)‖ + ‖𝜙(𝑡, 𝑠, 0)‖𝑑𝑠)𝑑𝑠 +
𝑎𝜔𝛾+𝑏

1−𝑎
+ 𝜔𝑡         

 

≤
𝜉

Γ(𝛼+1)
+

𝑟𝛿𝐾

λΓ(𝛼+1)
+

𝛿𝑀

λΓ(𝛼+1)
+

𝑎𝜉𝛾𝛼

(1−𝑎)Γ(𝛼+1)
+

𝑟𝑎𝛿𝐾𝛾𝛼

𝜆(1−𝑎)𝛤(𝛼+1)
+

𝑎𝛿𝑀𝛾𝛼

𝜆(1−𝑎)𝛤(𝛼+1)
+

𝑎𝜔𝛾+𝑏

1−𝑎
+ 𝜔𝑡  

≤ 𝑟 (
𝛿𝐾 + 𝑎𝛿𝐾(𝛾𝛼 − 1)

𝜆(1 − 𝑎)𝛤(𝛼 + 1) 
) +

𝜆𝜉 + 𝛿𝑀

λΓ(𝛼 + 1)
+

𝑎𝛾𝛼 (𝜆𝜉 + 𝛿𝑀)

𝜆(1 − 𝑎)𝛤(𝛼 + 1)
+

𝑎𝜔𝛾 + 𝑏

1 − 𝑎
+ 𝜔𝑡 

 

≤ 𝑟𝛺 + (1 − 𝛺)𝑟 = 𝑟 

Now we have to prove that the function Φ is contraction. For 𝑥, 𝑦 ∈ 𝐶 and 𝑡 ∈ [0,1], by 

(5) and H1 we have:  

 

‖(Φ𝑥)(𝑡) − (Φ𝑦)(𝑡)‖ 



Faraj Y. Ishak 

Mixed Boundary Value Problem for Nonlinear Fractional Volterra Integral Equation 

95 
 

≤ ∫
(𝑡−𝑠)𝛼−1

Γ(𝛼)
(∫ ‖𝜓(𝑡, 𝑠)‖

 𝑡

−∞ 
‖𝜙(𝑡, 𝑠, 𝑥(𝑠)) − 𝜙(𝑡, 𝑠, 𝑦(𝑠))‖𝑑𝑠)

𝑡

0
𝑑𝑠 +

   
𝑎

1−𝑎
∫

(𝛾−𝑠)𝛼−1

Γ(𝛼)
(∫ ‖𝜓(𝑡, 𝑠)‖

 𝑡

−∞ 
‖𝜙(𝑡, 𝑠, 𝑥(𝑠)) − 𝜙(𝑡, 𝑠, 𝑦(𝑠))‖𝑑𝑠)𝑑𝑠

𝛾

0
  

≤ ∫
(𝑡 − 𝑠)𝛼−1

Γ(𝛼)
𝐾‖𝑥 − 𝑦‖(∫ 𝛿𝑒 −𝜆(𝑡−𝑠)

 𝑡

−∞ 

𝑑𝑠
𝑡

0

+
𝑎

1 − 𝑎
∫

(𝛾 − 𝑠)𝛼−1

Γ(𝛼)
𝐾‖𝑥 − 𝑦‖(∫ 𝛿𝑒 −𝜆(𝑡−𝑠)

 𝑡

−∞ 

𝑑𝑠)𝑑𝑠
𝛾

0

 

 

≤
𝛿𝐾‖𝑥 − 𝑦‖

λΓ(𝛼 + 1) 
+

𝑎𝛿𝛾𝛼 𝐾‖𝑥 − 𝑦‖

𝜆(1 − 𝑎)Γ(𝛼 + 1)
 

 

≤
𝛿𝐾 + 𝑎𝛿𝐾(𝛾𝛼 − 1)

𝜆(1 − 𝑎)𝛤(𝛼 + 1) 
‖𝑥 − 𝑦‖ ≤ Ω‖𝑥 − 𝑦‖ 

Ω < 1 ensure that (Φ𝑥)(𝑡) is contractive. Therefor the conclusion of the theorem follows 

from the contraction mapping principle. 

Theorem 3.3: Let 𝜙: [0, 𝑇] × [0, 𝑇] × 𝑅 → 𝑅 , and let  𝑘 ∈ 𝑅 such that  0 ≤ 𝑘 <
1

Ω
, where  

Ω =
𝑡𝛼

Γ(𝛼 + 1)
+

𝑎𝛾𝛼

(1 − 𝑎)Γ(𝛼 + 1)
 

 

and M > 0 such that |𝜙(𝑡, 𝑠, 𝑥(𝑡))| ≤ 𝑘|𝑥| + 𝑀 for all 𝑡 ∈ [0, 𝑇], 𝑥 ∈ 𝑅 then problem (1) 

has at least one solution. 

Proof: Define an operator Ψ: Λ → Λ as: 

 

Ψ(𝑡) = ∫
(𝑡 − 𝑠)𝛼−1

Γ(𝛼)

𝑡

0

𝑓(𝑠)𝑑𝑠 +  
𝑎

1 − 𝑎
∫

(𝛾 − 𝑠)𝛼−1

Γ(𝛼)
𝑓(𝑠)𝑑𝑠

𝛾

0

+
𝑎𝜔𝛾 + 𝑏

1 − 𝑎
+  𝜔𝑡 

 

Where   Λ = 𝐶([0,1], 𝑅)  denote to the Bansch space of all continuous functions from 

[0,1] → 𝑅 endowed with the norm defined by  ‖𝑥‖ = sup{|𝑥(𝑡)|, 𝑡 ∈ [0,1]}. Let us define 

a fixed-point problem by: 

𝑥 = Ψ𝑥               (7) 

Now we need to prove the existence of at least one solution 𝑥 ∈ {0, 𝑇] satisfying (7). Define 

a ball ℬ𝑟 ⊂ 𝐶[0, 𝑇] with 𝑟 > 0 as: 

ℬ𝑟 = {𝑥 ∈ 𝐶[0, 𝑇]: max
𝑡∈[0,𝑇]

|𝑥(𝑡)| < 𝑟} 

Where 𝑟 well be given later, then it’s enough to show that  ℬ𝑟̅̅̅̅ → 𝐶[0, 𝑇]satisfies: 

𝑥 ≠ 𝜎Ψ𝑥, ∀𝑥 ∈ 𝜕ℬ𝑟  𝑎𝑛𝑑 ∀𝜎 ∈ [0, 𝑇] (8) 

Let us define 𝐻(𝜎, 𝑥) = 𝜎Ψ𝑥, 𝑥 ∈ 𝐶(ℝ), 𝜎 ∈ [0, 𝑇]   Then by Arzesla’-Ascoli theorem 

ℎ𝜎 (𝑥) = 𝑥 − 𝐻(𝜎, 𝑥) = 𝑥 − 𝜎Ψ𝑥 is completely continuous if (8) is true then the following 



Jurnal Matematika MANTIK 

Vol. 8, No. 2, December 2022, pp. 89-98 

96 
 

Leray-Schauder degree are well define and by the homotopy 

invariance of topological degree it follows that: 

deg(ℎ𝜎 , ℬ𝑟 , 0) = deg(𝐼 − 𝜎Ψ𝑥, ℬ𝑟 , 0) = deg(ℎ1, ℬ𝑟 , 0) = deg(ℎ0, ℬ𝑟 , 0) = 

deg(𝐼, ℬ𝑟 , 0) = 1 ≠ 0,0 ∈ ℬ𝑟. Where I denote the unit operator, by non-zero property of 

the Leray-Schauder degree ℎ1(𝑥) = 𝑥 − 𝜎Ψ𝑥 = 0 for at least one 𝑥 ∈ ℬ𝑟  in order to prove 

(8) we assume that 𝑥 = 𝜎Ψ𝑥 for some 𝜎 ∈ [0, 𝑇]and for all 𝑡 ∈ [0, 𝑇] such that: 

 

|𝑥(𝑡)| = |𝜎Ψ𝑥 | = ∫
(𝑡 − 𝑠)𝛼−1

Γ(𝛼)

𝑡

0

(|𝑔(𝑡)| + ∫ |𝜓(𝑡, 𝑠)||𝜙(𝑡, 𝑠, 𝑥(𝑠))|𝑑𝑠
𝑡

−∞

)𝑑𝑠

+  
𝑎

1 − 𝑎
∫

(𝛾 − 𝑠)𝛼−1

Γ(𝛼)
(|𝑔(𝑡)| + ∫ |𝜓(𝑡, 𝑠)||𝜙(𝑡, 𝑠, 𝑥(𝑠))|𝑑𝑠

𝑡

−∞

)𝑑𝑠
𝛾

0

+ |
𝑎𝜔𝛾 + 𝑏

1 − 𝑎
| + |𝜔𝑡 |      

≤ (𝜉 +
𝛿(𝑘|𝑥| + 𝑀

𝜆
)(

𝑡𝛼

Γ(𝛼 + 1)
+

𝑎𝛾𝛼

(1 − 𝑎)Γ(𝛼 + 1)
) + 𝑀1 

 

≤ (𝜉 +
𝛿(𝑘|𝑥| + 𝑀

𝜆
)Ω + 𝑀1 

Which on taking norm (𝑠𝑢𝑝𝑡∈[0,𝑇]|𝑥| = ‖𝑥‖) and solving for ‖𝑥‖ yields: 

 

‖𝑥‖ =
ξΩ(Mσ + 1) + 𝑀1

𝜎(𝜎 − Ω𝛿𝑘)
 

Where 𝑀1 = |
𝑎𝜔𝛾+𝑏

1−𝑎
| + |𝜔𝑡 | , letting  

𝑟 =
ξΩ(Mσ + 1) + 𝑀1

𝜎(𝜎 − Ω𝛿𝑘)
+ 1 

(8) hold, this completes the proof. 

 

Example: Consider the following fractional integrodifferential equation 

𝐷3 2⁄ 𝑥(𝑡) = sin(𝑡) + ∫ (2𝑡 + 𝑠)(𝑡 + 𝑥(𝑠))𝑑𝑠
𝑡

−∞

 

                              𝑥(0) = 0.5𝑥(1.25) + 3,      �́�(0) = 1.5                        (9) 

 

Comparing (9) and (1), we see that 𝛼 = 3 2⁄
 
, 𝑔(𝑡) = sin(𝑡) , 𝜓(𝑡, 𝑠) = 2𝑡 +

𝑠, 𝜙(𝑡, 𝑠, 𝑥(𝑡)) = 𝑡 + 𝑥(𝑠),   𝛾 = 1.25, 𝑎 = 0.5, 𝑏 = 3, 𝜔 = 1.5. If we choose 𝜉 = 1,

𝛿 = 1, 𝜆 = 5 , 𝐾 = 2, then H1 holds, and 

𝛿𝐾 + 𝑎𝛿𝐾(𝛾𝛼 − 1)

𝜆(1 − 𝑎)𝛤(𝛼 + 1) 
=

(1)(2) + (0.5)(1)(2)(1.253 2⁄ − 1)

(5)(1 − 0.5)( 1.3293)
< 1  

      

That is, (5) holds. Thus, by theorem 3.2, equation (9) has a unique solution. 



Faraj Y. Ishak 

Mixed Boundary Value Problem for Nonlinear Fractional Volterra Integral Equation 

97 
 

4. Conclusion 

Based on the results and discussion, it can be concluded that the idea of Krasnoselskii 

fixed point theorem was very effective to proof the existence solution for the proposed 

equation. Also, under some suitable hypotheses and conditions we were able to complete 

the proof of existence solution for the equation proposed in this paper smoothly. The present 

work can be extended to the nonlocal and non-separable fractional boundary value problem. 

 

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