IHJPAS. 36(1)2023 

407 
 This work is licensed under a Creative Commons Attribution 4.0 International License 

 

   

 

 

Approximation Solution of Fuzzy Singular Volterra Integral                     

Equation by Non-Polynomial Spline 

 

 

 

 

 

 

 

 
 

 

 

Abstract 

        A non-polynomial spline (NPS) is an approximation method that relies on the triangular 
and polynomial parts, so the method has infinite derivatives of the triangular part of the NPS to 

compensate for the loss of smoothness inherited by the polynomial. In this paper, we propose 

polynomial-free linear and quadratic spline types to solve fuzzy Volterra integral equations (FVIE) 

of the 2nd kind with the weakly singular kernel (FVIEWSK) and Abel's type kernel. The linear 

type algorithm gives four parameters to form a linear spline. In comparison, the quadratic type 

algorithm gives five parameters to create a quadratic spline, which is more of a credit for the exact 

solution. These algorithms process kernel singularities with a simple technique. Illustrative 

examples use MathCad software to deal with upper and lower-bound solutions to fuzzy problems. 

The method provides a reliable way to ensure that an exact solution is approximated. Also, figures 

and tables show the potential of the method. 

 

Keywords: Fuzzy Volterra integral equation, weakly singular kernel, non-polynomial spline. 

1. Introduction 

     The fuzzy set theory is one of the essential theories introduced by' [1-2] Solving linear and 

nonlinear Abel fuzzy integral equations using Laplace transforms. [3] found the solution of the 

fuzzy Volterra integral equation of 2nd kind with Abel's type kernel by applying Laplace Adomain 

decomposition method. [4] proposed a piecewise spline collocations method with gradient meshes 

for FVIEWSK, showing that the gradient meshes are superior to uniform ones for this problem. 

[5] solved linear Volterra integral equations of the second kind with a weakly singular kernel by 

using the sixth order of non-polynomial spline functions. In this work, we drive linear and 

quadratic NPS to solve FVIEWSK and numerically drive linear NPS of FVIE with Abel's types 

kernel. This paper is organized as follows: in section 2, some preliminaries are given about basic 

definitions. Section 3 is about NPS' functions in linear and quadratic. Section 4 considers the linear 

doi.org/10.30526/36.1.2860 

Article history: Received 22 July 2022, Accepted 21 Augest 2022, Published in January 2023. 

 

 

 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

 

Zahraa A. Ibrahim 

Department of Mathematics, College of 

Sciences,Mustansiriyah University, 

Baghdad, Iraq. 

zara.ali1431@gmail.com 
 

Nabaa N. Hasan 
Department of Mathematics, College of 

Sciences,Mustansiriyah University, 

Baghdad, Iraq. 

alzaer1972@uomustansiriyah.edu.iq 

 
 

https://creativecommons.org/licenses/by/4.0/
mailto:zara.ali1431@gmail.com
mailto:alzaer1972@uomustansiriyah.edu.iq


IHJPAS. 36(1)2023 
 

408 
 

and quadratic methods of FVIEWSK and section 5 NPS of FVIE with Abel's type kernel. 

Furthermore, in section 6, some illustrated examples are given, showing the method's accuracy. 

Finally, conclusions are given in section 7. 

 2. Preliminaries concepts:  

       In this section some concepts and definitions of fuzzy set theory and kinds of singular kernels 

are given. 

Definition 2.1[4]: A fuzzy numbers is a fuzzy set 𝐴: 𝑅 → [0, 1] such that: 
(𝑖) 𝐴 is upper semi continuous, 
(𝑖𝑖) 𝐴(𝑥) = 0 outside some interval [𝑎, d], 
(𝑖𝑖𝑖) There are real' numbers 𝑏, : 𝑎 ≤ 𝑏 ≤ 𝑐 ≤ 𝑑 
for which 

 
(1) 𝐴(𝑥) is monotonically' increasing on [𝑎, 𝑏], 
(2) 𝐴(𝑥) is 'monotonically decreasing on [𝑐, 𝑑], 
(3) A (𝑥) = 1, 𝑏 ≤ 𝑥 ≤ 𝑐 
Definition 2.2 [6]: Let X be a Cartesian' product of universes 𝑋1, 𝑋2, … , 𝑋𝑟 and �̃�1, �̃�2, . . . , �̃�𝑟  be 
r-fuzzy subsets of  𝑋1, 𝑋2, … , 𝑋𝑟 , respectively, 𝑓 = X → 𝑌,  
𝑦 = 𝑓(𝑥1, 𝑥2, … , 𝑥𝑟 ), then the extensions principles allow us to define a fuzzy set  �̃� in Y by: 
�̃� = {(𝑦, 𝜇�̃� (𝑦))| 𝑦 = 𝑓(𝑥1, 𝑥2, … , 𝑥𝑟 ), 𝑥1, 𝑥2, … , 𝑥𝑟 ∈ X}      
where 

𝜇�̃� (𝑦) = {

sup 𝑀𝑖𝑛{𝜇𝐴1̃ (𝑥1), … , 𝜇𝐴�̃� (𝑥𝑟 )},   𝑓
−1(𝑦) ≠ ∅ 

(𝑥1, 𝑥2, … , 𝑥𝑟 ) ∈ 𝑓
−1(𝑦)                                    

0 , 𝑜𝑡ℎ𝑒𝑟 𝑤𝑖𝑠𝑒                                                    

  

and 𝑓 −1 is the inverse image of 𝑓 
for  𝑟 = 1, the fuzzy extension principles, of course reduces to  
�̃� = 𝑓(�̃�) = 𝑓({(𝑦, 𝜇�̃� (𝑦))| 𝑦 = 𝑓(𝑥) , 𝑥 ∈ X}      
where    

𝜇�̃� (𝑦) = {
sup 𝜇�̃�(𝑥)     , 𝑓

−1(𝑦) ≠ ∅ 

𝑥 ∈ 𝑓 −1(𝑦)                           
0 ,      𝑜𝑡ℎ𝑒𝑟 𝑤𝑖𝑠𝑒                   

   

which is the definition of the fuzzy mapping. 

Definition 2.3 Crisp number [6]: 

A crisp number a is represented by: 

𝐴(𝑥) = {
1 𝑖𝑓 𝑥 = 𝑎
0 𝑖𝑓𝑥 ≠ 𝑎

      

A crisp interval [c, d] is represented by a fuzzy set 

𝐵(𝑥) = {
 1 if x ∈  [c, d]

0 if x ∉  [c, d]
  

Definition 2.4 some types of kernels [7]: 

i. Cauchy kernel  
   If the kernel 𝑘(𝑥, 𝑡) is of the form  

   𝑘(𝑥, 𝑡) =
𝐻(𝑥,𝑡)

𝑥−𝑡
  

   where 𝑘(𝑥, 𝑡) is differentiable function of (𝑥, 𝑡) with 𝐻(𝑥, 𝑡) ≠ 0, then the integral,               
.equation is said to be a singular equation with Cauchy kernels. 

ii. Weakly singular kernel 
     If the kernel 𝑘(𝑥, 𝑡) is of the form  

     𝑘(𝑥, 𝑡) =
𝐻(𝑥,𝑡)

|𝑥−𝑡|𝛼
  

  where 0 < 𝛼 < 1 and  𝐻(𝑥, 𝑡) is a differentiable and continuous function with 𝐻(𝑥, 𝑡) ≠ 0, 



IHJPAS. 36(1)2023 
 

409 
 

then the integral equation is said to be weakly singular kernel. 
 

iii. Strongly singular kernel 
If the kernel 𝑘(𝑥, 𝑡) is of the form  

𝑘(𝑥, 𝑡) =
𝐻(𝑥,𝑡)

|𝑥−𝑡|𝛼 
        𝛼 ≥ 2   

where 𝐻(𝑥, 𝑡) is a differentiable function of (𝑥, 𝑡) with 𝐻(𝑥, 𝑡) ≠ 0, then the integral equation 
is said to be strongly singular kernel. 
 

3. NPS Functions  

    The standard form of the 2nd FVIE is defined below [8]: 𝑢(𝑥, 𝑟) = 𝑓(𝑥, 𝑟) +

  𝜆; ∫ 𝑘(𝑥, 𝑡, 𝑢(𝑡, 𝑟), 𝑟)𝑑𝑡                                                                                
𝑥

0
                             (1) 

Consider the partition ∆= {𝑟0, 𝑟1, 𝑟2, … , 𝑟𝑛} of [𝑎, 𝑏] ⊂ 𝑅
′, let 𝑆(∆)′and indicate the arrangement of 

piecewise polynomial on subinterval [9]( 
𝐼𝑖 = [𝑟𝑖 , 𝑟𝑖+1] of segment ∆.  Let 𝑢(𝑟) be the exact solution. The NPS of n order 𝑆𝑖 (𝑟) at the form 
is: 

𝑆𝑖 (𝑟)′ = 𝑎𝑖 sin′ 𝑘(𝑟 − 𝑟𝑖 ) + 𝑏𝑖 cos′ 𝑘(𝑟 − 𝑟𝑖 ) + ⋯ + 𝑦𝑖 (𝑟 − 𝑟𝑖 )
𝑛−1 + 𝑧𝑖 .                          (2) 

Where 𝑎𝑖   , 𝑏𝑖  , … , 𝑦𝑖  and 𝑧𝑖  constants. 
 

In this section, we introduce different types of NPS functions, linear NPS functions and quadratic 

NPS functions. 

 

3.1 Linear Non-Polynomial Spline (LNPS):  

    We consider the LNPS method for finding an approximate solution of FVIE of the second kind 

[9]  

Consider the grid point 𝑟𝑖   on the interval  [𝑎, 𝑏],  as follows:   

𝑎 = 𝑟0 < 𝑟1 < 𝑟2 < ⋯ < 𝑟𝑛 = 𝑏                                                                                                              (3) 

𝑟𝑖 = 𝑟0 + 𝑖ℎ  , 𝑖 = 0,1,2, … , 𝑛                                                                                                        (4) 

ℎ =
  𝑏−𝑎 

𝑛
                                                                                                                                        (5) 

Where n is an appositive integer. Let 𝑢(𝑟) be the exact solution of equation (1) and 𝑆𝑖 (𝑟) be an 
approximation to 𝑢𝑖 = 𝑢(𝑟𝑖 ) obtained by the segment 𝑠𝑖 (𝑟) . Each NPS segment 𝑠𝑖 (𝑟)  has the 
form:  

𝑆𝑖 (𝑟)′ = 𝑎𝑖 sin′ 𝑘(𝑟 − 𝑟𝑖 ) + 𝑏𝑖 cos′ 𝑘(𝑟 − 𝑟𝑖 ) + 𝑐𝑖 (𝑟 − 𝑟𝑖 ) + 𝑑𝑖.                                                  (6) 

Where 𝑎𝑖 , 𝑏𝑖 , 𝑐𝑖 𝑎𝑛𝑑 𝑑𝑖 are constant and k is the frequency of the trigonometric function which will 
be used to raise the accuracy 'of the method. We consider the following relations: 

𝑆𝑖 (𝑟𝑖 ) = 𝑢(𝑟𝑖 )  
𝑆′𝑖′(𝑟𝑖 ) = 𝑘𝑏𝑖 + 𝑐𝑖 ≈ 𝑢

′
𝑖 (𝑟𝑖 )𝑠  

𝑆′′ 𝑖;(𝑟𝑖 ) = −𝑘
2𝑏𝑖 ≈ 𝑢

′′
𝑖 (𝑟𝑖 )𝑠   

𝑆′′′ 𝑖𝑠(𝑟𝑖 ) = −𝑘
3𝑎𝑖 ≈ 𝑢

′′′
𝑖 (𝑟𝑖 )𝑠  

Now, we can obtain the values of 𝑎𝑖,  𝑏𝑖 , 𝑐𝑖  and 𝑑𝑖   as follows: 

𝑎𝑖 =
 −1

𝑘3
𝑢′′′(𝑟𝑖 )𝑠                                                                                                                           (7) 

𝑏𝑖  =
−1

𝑘2  
u′′(𝑟𝑖 )s                                                                                                                            (8) 

𝑐𝑖 = 𝑢
′(𝑟𝑖) + 𝑠 

1

𝑘
u′′(𝑟𝑖 )𝑠                                                                                                                       (9) 

𝑑𝑖 = 𝑢(𝑟𝑖) +  
1

𝑘  
 u′′(𝑟𝑖 )𝑠                                                                                                                      (10) 

For i=0, 1, 2,…,n  

We differentiate equation (1) three times, with respect to r, then put  𝑟 = a to get:  

𝑢0 =𝑢(𝑎) = 𝑓(𝑎)                                                                                                                         (11) 



IHJPAS. 36(1)2023 
 

410 
 

𝑢0 =́ 𝑢 ́ (𝑎) = 𝑓 ́ (𝑎) + 𝑘(𝑎, 𝑎)𝑢(𝑎)                                                                                                   (12) 

Let 𝐸′(𝑥, 𝑡, 𝑢(𝑡, 𝑟); 𝑟) =  
𝜕𝑘(𝑥,𝑡 𝑢(𝑡,𝑟))′′

𝜕𝑟
 

𝑢′′(𝑟) = 𝑓0
′′(𝑟) + ∫

𝜕𝐸′(𝑥,𝑡,𝑢(𝑡,𝑟);𝑟)′′

𝜕𝑟

𝑥

0
 𝑑𝑡 + 2E(x, t, u(t, r): r)′′  

𝑢0
′′(𝑎) = 𝑢′′(𝑎) = 𝑓0

′′(𝑎) + 2𝐸′(𝑎, 𝑎, 𝑢(𝑎))                                                                                (13) 

Let F (x, t, u(t , r); r) =
𝜕𝐸(𝑥,𝑡,𝑢(𝑡,𝑟);𝑟)′

𝜕𝑟
 

u′′′(𝑟 ) = 𝑓
′′′(r) + ∫

∂F (x,t,u(t ,r);r)′

𝜕𝑟

𝑥

0
𝑑𝑡 + 3F (x, t, u(t , r); r)   𝑢0

′′′(a) = 𝑓 ′′′(𝑎) +

3F′ (a, a, a, u(a))                                                                                               (14) 
 

3.2 Quadratic Non-polynomial Splines (QNPS): 

      We consider the QNPS method for finding an approximate solution of FVIE of the second 

kind on the interval  [𝑎, 𝑏]  as the above equations (3),(4) and (5) 
The form of QNPS function [10] is 

 𝑄𝑖 (𝑥, 𝑟)′ = 𝑎𝑖 sin
′ 𝑘(𝑟 − 𝑟𝑖 ) + 𝑏𝑖 cos

′ 𝑘(𝑟 − 𝑟𝑖 ) + 𝑐𝑖 (𝑟 − 𝑟𝑖 ) + 𝑑𝑖 (𝑟 − 𝑟𝑖 )
2 + 𝑒𝑖                        (15)  

 where 𝑎𝑖, 𝑏𝑖, 𝑐𝑖, 𝑑𝑖 and 𝑒𝑖  are constants. We differentiate equation (15) four times with                                    
.respect to r and put r = a then replace r by 𝑟𝑖 in the relation equation (15) to yield: 
𝑄𝑖 (𝑥, 𝑟)

′ = 𝑎𝑖 + 𝑒𝑖    
𝑄′𝑖 (𝑥, 𝑟)

′ = 𝑘𝑏𝑖 +  𝑐𝑖    
𝑄′′𝑖 (𝑥, 𝑟)

′ = −𝑘2𝑎𝑖 + 2𝑑𝑖    
𝑄′′′𝑖 (𝑥, 𝑟)

′ = −𝑘3𝑏𝑖     

𝑄𝑖
(4)(𝑥, 𝑟) = 𝑘(4)𝑎𝑖   

We obtain the values of   𝑎𝑖, 𝑏𝑖, 𝑐𝑖, 𝑑𝑖 and 𝑒𝑖  from the above relation, as follows  

𝑎𝑖 =
𝑄𝑖

𝟒(𝑥,𝑟𝑖)

𝑘4
                                                                                                                                                (16) 

𝑏𝑖 = −
𝑄𝑖

′′′(𝑥,𝑟𝑖)

𝑘3
                                                                                                                                            (17) 

𝑐𝑖  = 𝑄′𝑖 (𝑥, 𝑟𝑖 )
′ − 𝑘𝑏𝑖                                                                                                                               (18) 

𝑑𝑖 =
𝑄′′𝑖(𝑥,𝑟𝑖)

′+𝑘2𝑎𝑖

𝟐
                                                                                                                                       (19) 

𝑒𝑖  = 𝑄𝑖 (𝑥, 𝑟𝑖 )
′ − 𝑎𝑖                                                                                                                                     (20) 

 
4. LNPS and QNPS Methods of FVIEWSK:  
     In this section, we use LNPS and QNPS methods to compute numerical solution of 
FVIEWSK which is 

𝑢(𝑥, 𝑟) − ∫  
𝑡𝜇−1

𝑥𝜇

𝑥

0
𝑢(𝑥, 𝑡)𝑑𝑡 = 𝑓(𝑥, 𝑟),           𝑥 ∈ [0, 𝑇]                                                                   (21)   

Where 0 < 𝜇 < 1 and f is a known function [11]. There is singularity at 𝑥 = 0 and 𝑡 = 0 for 
any positive value of t . 
To solve equation (21), we multiply both sides by 𝑥𝜇  to get 

𝑥𝜇  𝑢(𝑥, 𝑟) − ∫ 𝑡𝜇−1
𝑥

0
𝑢(𝑥, 𝑡)𝑑𝑡 = 𝑓(𝑥, 𝑟)𝑥𝜇                                                                                        (22)  

Hence, we differentiate equation (22) with respect to x, then we get  

𝑥𝜇  𝑢′(𝑥, 𝑟) + 𝜇𝑥𝜇−1𝑢(𝑥, 𝑟) −
1

𝑥1−𝜇
𝑢(𝑥, 𝑟) = 𝜇𝑥𝜇−1𝑓(𝑥, 𝑟) + 𝑥𝑓 ′(𝑥, 𝑟)𝑥𝜇                                 (23) 

And multiplying both sides by 𝑥1−𝜇 to get 
𝑥𝑢′(𝑥, 𝑟) + (1 − 𝜇)𝑢(𝑥, 𝑟) = 𝜇𝑓(𝑥, 𝑟) + 𝑥𝑓 ′(𝑥, 𝑟)                                                                          (24)  
Hence, we differentiate equation (24) four times with respect to 𝑥 and replace 𝑥 with the 
relation to yield: 

𝑢0 =
𝜇

𝜇−1
𝑓(𝑎, 𝑎)  

𝑢′0 =
𝜇+1

𝜇
𝑓 ′(𝑎, 𝑎)  



IHJPAS. 36(1)2023 
 

411 
 

𝑢′′0 =
𝜇+2

𝜇+1
𝑓′′(𝑎, 𝑎)                                                                                                                                    (25) 

𝑢0
′′′ =

𝜇+3

𝜇+2
𝑓 ′′′(𝑎, 𝑎)  

 𝑢0
4 =

𝜇+4

𝜇+3
𝑓 4(𝑎, 𝑎)  

To approximate the solution of linear FVIEWSK in (25) using linear NPS function, we present 
a method of solution following the algorithm.  

Algorithm of (LNPS) 

Step1:- input  ℎ =
𝑏−𝑎

𝑛
 , 𝑟𝑖 = 𝑟0 + 𝑖ℎ        , 𝑖 = 0,1, … , 𝑛  and 𝑢0 =

𝜇

𝜇−1
𝑓(𝑎, 𝑎)    

Step2:- Compute  𝑎0, 𝑏0, 𝑐0 𝑎𝑛𝑑 𝑑0  by ''substituting' the equations (11 – 14) into equations' 
             (7 – 10) 

Step3:- Evaluates S0(r) using step2 and equation (6) for i = 0" 

Step4:- Approximates"𝑢1 = 𝑢(𝑟1)  ≈ 𝑆0(𝑟1)𝑠 
Step5:- Do the following steps for 𝑖 = 1 to 𝑛 − 1 

Step6:- Compute 𝑎𝑖, 𝑏𝑖, 𝑐𝑖,𝑎𝑛𝑑  𝑑𝑖  by using equations (7 – 10) and replacing 

              𝑢0(𝑟𝑖), 𝑢′0(𝑟𝑖 ), 𝑢0
′′(𝑟𝑖 ) 𝑎𝑛𝑑 𝑢0

′′′(𝑟𝑖 ) 𝑖𝑛 𝑆(𝑟𝑖 ), S′(𝑟𝑖 ), 𝑆
′′(𝑟𝑖 )𝑎𝑛𝑑 𝑆

′′′(𝑟𝑖 )   

Step7:- calculate 𝑆𝑖 (𝑟) using step 6 and equations (6) 

Step8:- Approximate 𝑢𝑖+1 = 𝑆𝑖 (𝑟𝑖+1) 

 

Remark: To approximate solution of QNPS by the above. algorithm and replace step 6 by 

(Compute 𝑎𝑖, 𝑏𝑖,𝑐𝑖, 𝑑𝑖 and 𝑒𝑖  using equations  ( 16 – .20 ) and replacing 𝑄𝑖 (𝑟𝑖 ),

Q′𝑖 (𝑟𝑖), Q′′𝑖 (𝑟𝑖 ) , 𝑄0
′′′(𝑟𝑖 ) 𝑎𝑛𝑑 𝑄𝑖

(𝟒)(𝑟𝑖 )  𝑖𝑛  𝑆(𝑟𝑖 ), S
′(𝑟𝑖), 𝑆′′(𝑟𝑖 ), 𝑆

′′′(𝑟𝑖 )𝑎𝑛𝑑 𝑆𝑖
(𝟒)(𝑟𝑖 )   

and replace step7 by (calculate 𝑆𝑖 (𝑟) using step 6 and equation (15)). 
 

5. NPS of FVIE with Abel's type kernel: 

      Singular FVIE with Abel's type kernel [3] can be written in a general form as 

�̃�(𝑥, 𝑟) = 𝑓(𝑥, 𝑟) + ∫
𝑢(𝑡,𝑟)

√𝑥−𝑡

𝑥

0
𝑑𝑡.                                                                                                       (26) 

Where 𝑟, 𝑥 ∈ [0,1] 
To solve equation (26), applying Laplace transformation to both sides, we have  

   ℒ�̃�(𝑥, 𝑟) = ℒ𝑓(𝑥, 𝑟) + ℒ [∫
𝑢(𝑡,𝑟)

√𝑥−𝑡

𝑥

0
𝑑𝑡]                                                                                      (27) 

By convolution theorem, equation (27) yields 

ℒ�̃�(𝑥, 𝑟) = ℒ 𝑓(𝑥, 𝑟) + ℒ[𝑥 −1 2⁄ ] ℒ [�̃�(𝑥, 𝑟)                                                                             (28) 
Then, we have   

  ℒ�̃�(𝑥, 𝑟) = ℒ 𝑓(𝑥, 𝑟) + √
𝜋

𝑠
 ℒ [�̃�(𝑥, 𝑟)                                                                                             (29) 

Upon using the inverse of the Laplace transformation to both sides of equation (29), we obtain  

�̃�(𝑥, 𝑟) =  𝑓(𝑥, 𝑟) + ℒ−1 [√
𝜋

𝑠
 ℒ [�̃�(𝑥, 𝑟)]                                                                                 (30) 

Let the solution of equation (30) be in the form of the spline  

where  



IHJPAS. 36(1)2023 
 

412 
 

�̃�(𝑥, 𝑟) = [𝑢(𝑥, 𝑟), 𝑢(𝑥, 𝑟)] = 𝑎 cos(𝑥) + 𝑏 sin(𝑥) + 𝑐(𝑥) + 𝑑  

𝑓(𝑥, 𝑟) = [𝑓(𝑥, 𝑟), 𝑓(𝑥, 𝑟)]  

by substituting the above two equations into equation (30) to get: 

𝑎 cos(𝑥) + 𝑏 sin(𝑥) + 𝑐(𝑥) + 𝑑 = 𝑓(𝑥, 𝑟) + ℒ −1 [√
𝜋

𝑠
 ℒ[𝑎 cos(𝑥) + 𝑏 sin(𝑥) + 𝑐(𝑥) + 𝑑] ] (31)   

Now, by simplifying equation (31) to obtain: 

[𝑎(cos(𝑥) − √𝜋 cos(𝑥) (√2 − 1) − √𝜋 sin(𝑥) (2 − √2) + 𝑏 (sin(𝑥) − √𝜋 cos(𝑥) (−1 +

1

√2
) − √𝜋 sin(𝑥) (2 −

1

√2
) + 𝑐 (𝑥 −

4𝑥3 2⁄

3
) + 𝑑(1 − 2√𝑥)] = 𝑓 (𝑥, 𝑟)                                     (32)                                                                          

Let ∆   be a partition for the x, s.t  ∆: 0 = 𝑥0 < 𝑥1 < 𝑥2 < 𝑥3 = 1.  
Where ℎ = 1 3⁄ then 𝑥0 = 0, 𝑥1 = 1 3⁄ , 𝑥2 = 1 3⁄  and 𝑥3 = 1. 

   The approximate equation �̃�(𝑥, 𝑟) = 𝑓 (𝑥, 𝑟)[𝑎 cos(𝑥) + 𝑏 sin(𝑥) + 𝑐(𝑥) + 𝑑]                       (33) 

Now, we need to solve equation (33) to find the constants (𝑎 , 𝑏, 𝑐 𝑎𝑛𝑑 𝑑) using the system: 

𝑀𝐶 = 𝐹  to find 𝐶 calculate 𝐶 = 𝑀−1𝐹 and substitute this solution in equation (33) to find 
approximation solution. 

6. Illustrative Examples  

    In this section, two test examples are illustrated below to solve ( FVIEWSK ) and FVIE with 

Abel's types kernel in upper and lower solution, 𝑠(𝑥, 𝑟) the approximate solution by the proposed 
method and error = |𝑢(𝑥, 𝑟) − 𝑠(𝑥, 𝑟)| where 𝑢(𝑥, 𝑟) the exact solution. 
 

Example (1):  

Consider LFVIEs with weakly singular 

𝑢(𝑥, 𝑟) − ∫  
𝑡𝛽−1

𝑥𝛽

𝑥

0
𝑢(𝑥, 𝑡)𝑑𝑡 = 𝑓(𝑥, 𝑟),  where 

𝑓(𝑥, 𝑟) = [(0.71428571𝑥3 − 0.6𝑥2)(𝑟 − 1) ;  (0.71428571𝑥3 −   0.6𝑥2)(1 − 𝑟)]  
The exact solution is 𝑢(𝑥, 𝑟) = [(𝑥3 − 𝑥2)(𝑟 − 1) ; (𝑥3 − 𝑥2)(1 − 𝑟)] [11]     

 
Table 1. shows the results of LFVIEs with weakly singular in lower solution 

 
𝒙  𝒖(𝒙, 𝒓)  𝒔 (𝒙, 𝒓) in Error in 

Linear Quadratic Linear Quadratic 

0 0.000000000 0.000000000 -0.000000000 0.000000000 0.000000000 

0.1 −8.1 × 10−3 -0.008092924 −8.1 × 10−3 7.004 × 10−6  4.499 × 10−7  
0.2 -0.02900000 -0.028694546 -0.028814338 1.055 × 10−4  1.439 × 10−5  
0.3 -0.05700000 -0.056203437 -0.056809116 4.966 × 10−4  1.091 × 10−4  
0.4 -0.08600000 -0.849492630 -0.086859048 1.451 × 10−3  4.590 × 10−4  
0.5 -0.11300000 -0.109249304 -0.113897909 3.251 × 10−3  1.398 × 10−3  
0.6 -0.13200000 -0.123465262 -0.133069357 6.135 × 10−3  3.469 × 10−3  
0.7 -0.13200000 -0.122059594 -0.139775512 0.010000000 7.476 × 10−3 

0.8 -0.11500000 -0.099650843 -0.129722893 0.016000000 0.015000000 

0.9 -0.07300000 -0.051067411 -0.098965315 0.022000000 0.026000000 

1 0.000000000 0.02860077 -0.043943323 0.029000000 0.044000000 

 
Table 2. Shows the results of LFVIEs with weakly singular in upper solution 

 

𝒙 𝒖(𝒙, 𝒓)  𝒔(𝒙, 𝒓) in Error in 

Linear Quadratic Linear Quadratic 

0 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 

0.1 8.1 × 10−3  0.008092924 8.10 × 10−3 7.004 × 10−6 4.499 × 10−7 
0.2 0.029000000 0.028694546 0.028814338 1.055 × 10−4 1.439 × 10−5 



IHJPAS. 36(1)2023 
 

413 
 

0.3 0.057000000 0.056203437 0.056809116 4.966 × 10−4 1.091 × 10−4 
0.4 0.086000000 0.084949263 0.086859048 1.451 × 10−3 4.590 × 10−4 
0.5 0.113000000 0.109249304 0.113897909 3.251 × 10−3 1.398 × 10−3 
0.6 0.132000000 0.123465262 0.133069357 6.135 × 10−3 3.469 × 10−3 
0.7 0.132000000 0.122059594 0.139775512 0.010000000 7.476 × 10−3 
0.8 0.115000000 0.099650843 0.129722893 0.016000000 0.015000000 

0.9 0.073000000 0.051067411 0.098965315 0.022000000 0.026000000 

1 0.000000000 0.02860077 0.043943323 0.029000000 0.044000000 

 

Example (2): 

 Consider the FVIE with Abel's type kernel  

{
𝑢(𝑥, 𝑟) = (𝑥 +

4

3
𝑥 3 2⁄ ) (4 + 𝑟) − ∫

𝑢(𝑡,𝑟)

√𝑥−𝑡

𝑥

0
𝑑𝑡                          

𝑢(𝑥, 𝑟) = (𝑥 +
4

3
𝑥 3 2⁄ ) (𝑟 − 6) − ∫

𝑢(𝑡,𝑟)

√𝑥−𝑡

𝑥

0
𝑑𝑡                           

                 

where the exact solution is   𝑢(𝑥, 𝑟) = [(4 + 𝑟)𝑥 , (6 − 𝑟)𝑥][3]  

Table 3. FVIE with Abel's type kernel in the parametric form in upper and lower solution. 

𝒙 𝒖(𝒙, 𝒓) 𝒔(𝒙, 𝒓) Error 𝒖(𝒙, 𝒓) 𝒔(𝒙, 𝒓) Error 

0 0.00000000 0.000000000 0.000000000 0.0000000000 0.0000000000 0.000000000 

0.1 -0.59000000 -0.590000000 0.000000000 0.4100000000 0.4100000000 0.000000000 

0.2 -1.18000000 -1.180000000 0.000000000 0.8200000000 0.8200000000 0.000000000 

0.3 -1.77000000 -1.770000000 0.000000000 1.2300000000 1.2300000000 0.000000000 

0.4 -2.36000000 -2.360000000 0.000000000 1.6400000000 1.6400000000 0.000000000 

0.5 -2.95000000 -2.950000000 0.000000000 2.0500000000 2.0500000000 0.000000000 

0.6 -3.54000000 -3.540000000 0.000000000 2.4600000000 2.4600000000 0.000000000 

0.7 -4.13000000 -4.130000000 0.000000000 2.8700000000 2.8700000000 0.000000000 

0.8 -4.72000000 -4.720000000 0.000000000 3.2800000000 3.2800000000 0.000000000 

0.9 -5.31000000 -5.310000000 0.000000000 3.6900000000 3.6900000000 0.000000000 

1 -5.90000000 -5.900000000 0.000000000 4.1000000000 4.1000000000 0.000000000 

 

 
            Figure 1. Results of example (1)                                                Figure 2.Results of example (2)    

                                                                                       

In Figures 1, 2, we plot the graphs of the approximate ((+) for upper and (+) for lower) solution 

and exact ((×) for upper and (×) for lower) solution among different values of x with fuzzy 

parameter 𝑟 = 0.1, 0 ≤ 𝑥 ≤ 1, where s(x) .is approximate, and e(x) is exact for the upper 
parametric form and  s1(x) .is approximate, and  e1(x) is exact for the lower parametric form. 
 

7. Conclusions  



IHJPAS. 36(1)2023 
 

414 
 

                The examples show that the results of the method are convergent to the exact solution. 

The non-polynomial spline has been successfully used to obtain the .approximate solutions; the 

trigonometric term of this spline has infinite derivatives, which agree with the exact solution. 

Given the results, tables and figures show that the proposed technique is a powerful mathematical 

tool for solving FVIEWSK with MathCad programming implementation. We will consider 

FVIEWSK, algorithm, and applied fractional order problems for future work.                                

 

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