IHJPAS. 53 (3)2022 

206 

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

 

  

 

 

Iterative Method for Solving a Nonlinear Fourth Order Integro-Differential 

Equation 

Areej Salah Mohammed 

Department of Mathematics, College of Education for Pure Science, Ibn Al Haitham,University 

of Baghdad, Iraq. 

areej.s.m@ihcoedu.uobaghdad.edu.iq 

 

                                                                                                             

 

 

Abstract  

     This study presents the execution of an iterative technique suggested by Temimi and Ansari 

(TA) method to approximate solutions to a boundary value problem of a 4th-order nonlinear 

integro-differential equation (4th-ONIDE) of the type Kirchhoff which appears in the study of 

transverse vibration of hinged shafts. This problem is difficult to solve because there is a non-

linear term under the integral sign, however, a number of authors have suggested iterative methods 

for solving this type of equation. The solution is obtained as a series that merges with the exact 

solution. Two examples are solved by TA method, the results showed that the proposed technique 

was effective, accurate, and reliable. Also, for greater reliability, the approximate solutions were 

compared with the classic Runge-Kutta method (RK4M) where good agreements were observed. 

For more accuracy the maximum error remainder was found, and the absolute error was computed 

between the semi-analytical method and the numerical method RK4M.  Mathematica® 11 was 

used as a program for calculations. 

Keywords an iterative technique, approximate solutions, fourth order integro-differential 

equation, maximum error remainder.  

1.Introduction 
     Many real-life phenomena engineering and physics are mathematically modeled into functional 

equations. Many mathematical formulas for physical phenomena and fluid dynamics contain 

differential and integral equations [1]. [2]. 

In this paper, the integro-differential equation of the fourth order of Kirchhoff type is considered 

as 

Ibn Al Haitham Journal for Pure and Applied Sciences 

Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index 

 

Doi: 10.30526/35.4.2776 

Article history: Received 16 January 2022, Accepted 3 February 2022, Published in October 2022. 

 

 

 

https://creativecommons.org/licenses/by/4.0/
mailto:areej.s.m@ihcoedu.uobaghdad.edu.iq


IHJPAS. 53 (3)2022 
 

207 

  
𝜐(4)(𝑡)−𝛿𝜐′′(𝑡)−

2

𝐵
∫ [𝜐′(𝑡)]

2
𝑑𝑡 

𝐵
0

𝜐′′(𝑡)=𝑘(𝑡),       0<𝑡<𝐵,

𝜐(0)=𝜐(𝐵)=𝜐′′(0)=𝜐′′(𝐵)=0,
]                              (1) 

where 𝜐(𝑡) represents the constant deviation of the shafts, 𝑘(𝑡) is a continuous function defined 

over [0, 𝐵]and 𝛿 and 𝐵 are positive constants. This type of problem represents a bending 

equilibrium model for expandable shafts that are simply supported on nonlinear bases. The force 

that the foundation applies to the shafts is represented by the function, 𝑘(𝑡). Small changes in 

shafts length and their effects are represented by : 
2

𝐵
∫ [𝜐′(𝑡)]2𝑑𝑡 

𝐵

0
[3]. It is obvious that it is difficult 

to find the exact solution to Equation (1), so many numerical and semi-analytical methods have 

been developed, such as [4] reduced the order of the equation to find the root of a nonlinear 

equation and solve it by Newton's method. [5] used the Optimal Homotopy Asymptotic strategy 

to solve equation (1) and proved the efficacy of the procedure by comparing it with a finite element 

technique. [6] proposed a numerical scheme for solving Equation (1) by transforming the problem 

into a nonlinear algebraic system and then solving it in an iterative method with the collocation 

technique.  

 [7] suggested an iterative method namely Tamimi and Ansari (TA) method for solving 

nonlinear equations, and this technique was applied to solve numerous differential equations, for 

example, Korteweg-de Vries Equations [8], thin-film flow problems [9], Falkner-Skan problem 

[10]. In this work, the TA method is used to solve the 4th-ONIDE of the type Kirchhoff which 

appears in the study of transverse vibration of hinged shafts. The results acquired from the method 

showed that TA method is accurate, fast, and convenient. Two examples are solved and the results 

obtained are compared with the RK4M method, and the maximum error reminder is calculated. 

    This article is arranged as follows: Section 2 provides a clear formulation of the iterative method 

in general and specifically on Equation (1); Section 3 illustrates the effectiveness of the method 

with various examples before giving a brief conclusion in Section 4. 

1. The proposed method  
To clarify the fundamental thought of TA method. Let us consider the general equation  

          ℒ(υ(t)) + ℵ(υ(t)) + g(t) = 0,                                                                       (2)    

with boundary conditions     ℬ (υ,
d𝜐

𝑑𝑡
) = 0,    

where υ is unknown function, ℒ is the linear operator, ℵ is the nonlinear operator   and g is a known 
function. 

Assuming that υ0(𝑡) is a solution of Equation (3) of the initial condition  

         ℒ(υ0(𝑡)) + g(t) = 0, with ℬ (υ0,
d𝜐0

𝑑𝑡
) = 0 .                                                      (3)  

 To find the next iteration, we resolve the following equation: 

              ℒ(υ1(𝑡)) + ℵ(υ0(𝑡)) + g(t) = 0, ℬ (υ1,
d𝜐1

𝑑𝑡
) = 0.                                                             



IHJPAS. 53 (3)2022 
 

208 

Thus, an iterative procedure can be an effective solution to the following problem,  

 ℒ(υ𝑛+1(𝑡)) + ℵ(υ𝑛(𝑡)) + g(t) = 0, ℬ (υ𝑛+1,
d𝜐𝑛+1

𝑑𝑡
) = 0.                                   (4)               

Each of  𝜐𝑖 are solutions to Equation (1) [7].  

We will apply the steps of the method in solving equation (1), so it can be formulated as: 

  
ℒ(υ(t))−δR(υ(t))−

2

B
(∫ ℵ(υ(t))dt) 

B
0

R(υ(t))=k(t),       0<t<B,

υ(0)=υ(B)=υ′′(0)=υ′′(B)=0,
]                   (5)  

Where ℒ is the fourth derivative, R is the second derivative and ℵ the nonlinear term.  

υ0(𝑡) can be considered as a solution to the initial problem: 

ℒ(υ0(𝑡)) = k(t), with  𝜐0(0) = 𝜐0(𝐵) = 𝜐0
′′(0) = 𝜐0

′′(𝐵) = 0,                            (6)  

 υ1(𝑡) can be found by solving Equation (8): 

    
ℒ(υ1(𝑡))=δR(υ0(𝑡))+

2

B
(∫ ℵ(υ0(𝑡))dt) 

B
0

R(υ0(𝑡))+k(t),

υ1(0)=υ1(B)=υ1
′′(0)=υ1

′′(B)=0,
]                              (7)                         

Likewise, more solutions can be gained for the problems created by: 

    
ℒ(υ𝑛+1(𝑡))=δR(υ𝑛(𝑡))+

2

B
(∫ ℵ(υ𝑛(𝑡))dt) 

B
0

R(υ𝑛(𝑡))+k(t),

υ𝑛+1(0)=υ𝑛+1(B)=υ𝑛+1
′′(0)=υ𝑛+1

′′(B)=0,
]                      (8) 

2. Solution of the problem 
In this section, we will solve two problems using the method defined above 

Example 1. Consider the 4th-ONIDE [6] 

𝜐(4)(𝑡) − 𝜐′′(𝑡) − ∫ (𝜐′(𝑡))
21

0
𝑑𝑡 𝜐′′(𝑡) = 1   , 0 < 𝑡 < 1,                                             

𝜐(0) = 𝜐(1) = 𝜐′′(0) = 𝜐′′(1) = 0,                                                                           (9) 

we start with the assumption that  

 ℒ(𝜐(t)) = 𝜐(4)(𝑡) , ℵ(𝜐) = −𝜐′′(𝑡) − ∫ (𝜐′(𝑡))
21

0
𝑑𝑡 𝜐′′(𝑡), 𝑔(𝑡) = −1                  (10) 

With 𝑢(0) = 𝑢(1) = 𝑢′′(0) = 𝑢′′(1) = 0,                                                                  (11) 

Let us start with the first iteration 

 𝜐0
(4)(𝑡) = 1,                                                                                                                 (12) 

 with 𝜐0(0) = 𝜐0(1) = 𝜐0
′′(0) = 𝜐0

′′(1) = 0,                                                              (13) 

Then, we get  



IHJPAS. 53 (3)2022 
 

209 

 𝜐0(𝑡) = 0.04166(𝑡 − 2𝑡
3 + 𝑡4),                                                                                    (14) 

The second repetition can be performed as 

𝜐1
(4)(𝑡) = 1 + 𝜐0

′′(𝑡) + ∫ (𝜐0
′(𝑡))

21

0
𝑑𝑡 𝜐0

′′(𝑡),                                                              (15) 

with 𝜐1(0) = 𝜐1(1) = 𝜐1
′′(0) = 𝜐1

′′(1) = 0,                                                                    (16) 

Thus, we get  

 

𝜐1(𝑡) = 6.889329 × 10
−8(544269𝑡 − 1108715𝑡3 + 604800𝑡4 − 60531𝑡5 + 20177𝑡6)      (17)                                                                                                           

In the third repetition, the following equation must be solved: 

𝜐2
(4)(𝑡) = 1 + 𝜐1

′′(𝑡) + ∫ (𝜐1
′(𝑡))

21

0
𝑑𝑡 𝜐1

′′(𝑡),                                                               (18) 

Then, we get  

𝜐2(𝑡) = 3.7916 × 10
−26(1.00009 × 1024𝑡 − 2.032907 × 1024𝑡 3 + 1.0989 × 1024𝑡 4 − 1.00795 × 1023𝑡 5 +

3.6655 × 1022𝑡 6 − 2.6204 × 1021𝑡 7 + 6.5511 × 1020𝑡 8)                                                             (19)                                                             

We showed the first three solutions but we did not show the later ones because they get longer. 

By applying the (RK4M) using Mathematica® as a standard for evaluating TA method 

performance. In Figure 1, good agreement was observed between RK4M and TA method. 

 

Figure 1. The TA method and RK4M solutions for the problem 

The logarithmic plots for MER will be presented in Figure 2 for the approximate solution 

obtained by TA method for solving 4th-ONIDE for different values of n=1 to 5. 

 

Figure 2. Maximum Error Reminder for Solution by TA method 



IHJPAS. 53 (3)2022 
 

210 

Example 2: Consider the 4th-ONIDE [6] 

𝜐(4)(𝑡) − 𝜐′′(𝑡) − 2 ∫ (𝜐′(𝑡))
21

0
𝑑𝑡 𝜐′′(𝑡) = −𝑡   , 0 < 𝑡 < 1,                                           (20) 

𝜐(0) = 𝜐(1) = 𝑣′′(0) = 𝜐′′(1) = 0,                                                                                  (21) 

we start with the assumption that  

  ℒ(υ(t)) = 𝜐(4)(𝑡) , ℵ(𝜐) = −𝜐′′(𝑡) − 2 ∫ (𝜐′(𝑡))
21

0
𝑑𝑡 𝜐′′(𝑡), 𝑔(𝑡) = 𝑡                         (22) 

With  𝜐(0) = 𝜐(1) = 𝜐′′(0) = 𝜐′′(1) = 0,                                                                        (23) 

Let us start with the first iteration 

 𝜐0
(4)(𝑡) = −𝑡,                                                                                                                      (24) 

 with 𝜐0(0) = 𝜐0(1) = 𝜐0
′′(0) = 𝜐0

′′(1) = 0,                                                                      (25) 

Then, we get  

 𝜐0(𝑡) = 0.0027(−7𝑡 + 10𝑡
3 − 3𝑡5 )                                                                                    (26) 

The second repetition can be performed as 

𝜐1
(4)(𝑡) = −𝑡 + 𝑣0

′′(𝑡) + 2 ∫ (𝜐0
′(𝑡))

21

0
𝑑𝑡 𝜐0

′′(𝑡),                                                  (27) 

with 𝜐1(0) = 𝜐1(1) = 𝜐1
′′(0) = 𝜐1

′′(1) = 0,                                                                         (28) 

Thus, we get  

 

𝜐1(𝑡) = 1.3997 × 10
−8(−1242613𝑡 + 1752877𝑡3 − 496083𝑡5 − 14181𝑡7)                  (29) 

                                                                                                                                                                                            

In the third repetition, the following equation must be solved: 

𝜐2
(4)(𝑡) = −𝑡 + 𝜐1

′′(𝑡) + 2 ∫ (𝜐1
′(𝑡))

21

0
𝑑𝑡 𝜐1

′′(𝑡),                                                            (30) 

Then, we get  

𝜐2(𝑡) = 8.1826 × 10
−27(−2.1513 × 1024𝑡 + 3.0403 × 1024𝑡3 − 8.6844 × 1023𝑡5 −

2.0212 × 1022𝑡7 − 3.3703 × 1020𝑡9))                                                                                (31)  

We showed the first three solutions but we did not show the later ones because they get longer. 

By applying the (RK4M) using MATHEMATICA as a standard for evaluating TA method 

performance. In Figure 3, good agreement was observed between RK4M and TA method. 



IHJPAS. 53 (3)2022 
 

211 

 

Figure 3. The TA method and RK4M solutions for the problem 

The logarithmic plots for MER will be presented in Figure 2 for the approximate solution 

obtained by TA method for solving 4th-ONIDE for different values of n=1 to 5. 

 

Figure 4. Maximum Error Reminder for Solution by TA method 

The absolute error was calculated by RK4M and TA method, we note a good approximation 

as shown in Table 1 for both examples. 

Table 1. The absolute error for RK4M and TA method 

 

 

x Absolute error for example 1 Absolute error for example 2 

 |𝜐𝑇𝐴𝑀 (𝑡) − 𝜐𝑅𝐾4(𝑡)| |𝜐𝑇𝐴𝑀 (𝑡) − 𝜐𝑅𝐾4 (𝑡)| 

0 1.796565911 × 10−7 2.506032872×10-8 

0.1 3.351539886 × 10−7 4.987247103 × 10−8 

0.2 4.575623109 × 10−7 7.253721456 × 10−8 

0.3 5.365609015 × 10−7 9.058700125" × 10−8 

0.4 5.636405136 × 10−7 1.013928555 × 10−7 

0.5 5.35561508 × 10−7 1.028496388 × 10−7 

0.6 4.558146015 × 10−7 9.384872716 × 10−8 

0.7 3.331583074 × 10−7 7.409719157 × 10−8 

0.8 1.781605395 × 10−7 4.326019289 × 10−8 

0.9 0 1.734723476 × 10−18 



IHJPAS. 53 (3)2022 
 

212 

3. Conclusion  
 In this work, a proposed iterative technique is presented for solving a class of 4th-ONIDE 

Kirchhoff type that emerges in the investigation of transversal vibrations of hinge shafts. The TA 

method is an effective method for solving various kinds of linear and non-linear problems 

accurately. The fundamental benefit of this technique is that it does not require the estimation of 

variables. It very well may be seen that when the number of repetition increases, the maximum 

error reminder have decreased. In comparison with the RK4M, we find a great agreement in the 

results through the above figures and table. 

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https://www.aimsciences.org/article/doi/10.3934/proc.2007.2007.694