# New Perturbation Iteration Solutions for Fredholm and Volterra Integral Equations

Authors: İhsan Timuçin Dolapçı; Mehmet Şenol; Mehmet Pakdemirli
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

In this paper, recently developed perturbation iteration method is used to solve Fredholm and Volterra integral equations. The study shows that the new method can be applied to both types of integral equations. Some numerical examples are given, and results are compared with other studies to illustrate the efficiency of the method.


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## Body

## 1. Introduction

As one of the most important subjects of mathematics, differential and integral equations are widely used to model a variety of physical problems. Perturbation methods have been used in search of approximate analytical solutions for over a century [1–3]. Algebraic equations, integral-differential equations, and difference equations could be solved by these techniques approximately.However, a major difficulty in the implementation of perturbation methods is the requirement of a small parameter or inserting a small artificial parameter in the equation. Solutions obtained by these methods are therefore restricted by a validity range of physical parameters. To eliminate the small parameter assumption in regular perturbation analysis, iteration techniques are incorporated with perturbations. Many attempts in this issue appear in the literature recently [4–13].Recently, a new perturbation-iteration algorithm has been developed by Pakdemirli and his coworkers [14–16]. A preliminary study of developing root finding algorithms systematically [17–19] finally led to generalization of the method to differential equations also [14–16]. An iterative scheme is constituted over the perturbation expansion in the new technique. The method has been successfully implemented to first-order equations [15] and Bratu-type second-order equations [14].In this paper, this new technique is applied to integral equations for the first time. Fredholm and Volterra integral equations(1)y(t)=x(t)+∫01‍k(t,s)y(s)ds,y(t)=x(t)+∫0t‍k(t,s)y(s)ds
are considered, where x(t)∈L2[0,1), k(t,s)∈L2[0,1)×[0,1), and y(t) is the unknown function to be determined. Results are compared with some other studies.


## 2. Overview of the Method

In the present paper, the simplest perturbation-iteration algorithmPIA(1,1) is used by taking one correction term in the perturbation expansion and correction terms of only first derivatives in the Taylor series expansion, that is, n=1, m=1 [14–16]. Consider the Volterra integral equation
(2)y(t)=x(t)+∫0t‍k(t,s)y(s)ds
that has the form of
(3)F(u,∫‍u,ε)=0,
where
(4)F=y(t)-x(t)-ε∫0t‍k(t,s)y(s)ds
and ε is the artificially introduced perturbation parameter. In this method, we use only one correction term in the perturbation expansion:
(5)un+1=un+ε(uc)n.Substituting (5) into (3) and expanding in a Taylor series with first-order derivatives only yield
(6)F(un,∫‍un,0)+Fu(un,∫‍un,0)ε(uc)n+Fε(un,∫‍un,0)ε+F∫‍u(un,∫‍un,0)ε∫‍(uc)n=0
or
(7)(uc)n∂F∂u+(∫‍(uc)n)∂F∂(∫‍u)+∂F∂ε+Fε=0.
All derivatives are evaluated at ε=0.Starting with the initial conditionu0, first (uc)0 has been calculated by the help of (7). Then we substitute (uc)0 into (5) to find u1. Iteration process is repeated using (7) and (5) until we obtain a satisfactory result.


## 3. Application

Example 1.
Consider the Fredholm integral equation of the second kind(8)u(x)=∫01‍(-13e2x-(5/3)t)u(t)dt+e2x+(1/3)
with exact solution
(9)u(x)=e2x.
Equation (8) can be rewritten in the following form:
(10)F(u,ε)=u(x)-e2x+(1/3)-ε∫01‍(-13e2x-(5/3)t)u(t)dt,
where ε is a small parameter. The terms in (7) are
(11)F=un(x)-e2x+(1/3),Fu=1,Fε=-∫01‍(-13e2x-(5t/3)un(t))dt,F∫‍u=0.
Note that introducing the small parameter ε as a coefficient of the integral term simplifies (7) and makes it solvable. For this specific example (7) reads
(12)e1/3+2x+∫01‍-13e-5t/3+2xun(t)dt=(uc)n(x)+un(x).
When applying the iteration formula (5), we select an initial guess appropriate to the boundary condition and at each step we determine coefficients from the boundary condition. Starting with the initial function
(13)u0=1
and using the formula, the approximate solutions at each step are
(14)u1=-e2x5+15e-5/3+2x+e1/3+2x,u2=-15e-5/3+2x(-1+e1/3+e5/3-11e2+5e7/3),u3=15e-(5/3)+2x×(1+e1/3(-2+e1/30000000000×(1+e(-1+17e1/3-16e2/3+5e)))).
Higher iterations are not given here for brevity. Using a symbolic manipulation software, iterations could be calculated up to any order. In Table1, some of our iterations are compared with the exact solution and the error between the exact solution, and u20 are given which are of order 10-8.Table 1
Numerical result of Example1.
x
u
5
u
10
u
15
u
20
Exact solution
Error (u20)
0.0
1.00572
0.99995
1.00000
1.00000
1.00000
5.20782
×
1
0
-
9
0.1
1.22839
1.22134
1.22140
1.22140
1.22140
6.36084
×
1
0
-
9
0.2
1.50035
1.49174
1.49183
1.49182
1.49182
7.76915
×
1
0
-
9
0.3
1.83254
1.82202
1.82212
1.82212
1.82212
9.48926
×
1
0
-
9
0.4
2.23826
2.22542
2.22554
2.22554
2.22554
1.15902
×
1
0
-
8
0.5
2.73382
2.71813
2.71828
2.71828
2.71828
1.41563
×
1
0
-
8
0.6
3.33910
3.31993
3.32012
3.32012
3.32012
1.72906
×
1
0
-
8
0.7
4.07838
4.05498
4.05520
4.05520
4.05520
2.11187
×
1
0
-
8
0.8
4.98135
4.95276
4.95304
4.95303
4.95303
2.57945
×
1
0
-
8
0.9
6.08423
6.04931
6.04965
6.04965
6.04965
3.15055
×
1
0
-
8
1.0
7.43130
7.38865
7.38906
7.38906
7.38906
3.84809
×
1
0
-
8Example 2.
Consider the following integral equation:(15)u(x)=cosx-∫0x‍(x-t)cos(x-t)u(t)dt.
The exact solution of the problem is
(16)u(x)=13(2cos3x+1).
Equation (15) can be rewritten in the following form:
(17)F(u,ε)=u(x)-cosx-ε∫0x‍(x-t)cos(x-t)u(t)dt,
where ε is a small artificial parameter. The terms in (7) are
(18)F=un(x)-cosx,Fu=1,Fε=∫0x‍(x-t)cos(x-t)un(t)dt,F∫‍u=0.
Equation (7) reduces to
(19)∫0x‍(x-t)cos(t-x)undt+(uc)n+un=cosx.
Choosing the initial guess(20)u0=1
and using the formula, the approximate solutions at each step are
(21)u1=1-xsinx,u2=1+112x(-12+x2)sinx,u3=1+1480x(15xcosx-(495-45x2+x4)sinx).
Higher iterations are not given for brevity. In Table2, some of our iterations are compared with the exact solution, and the errors between the exact solution and u10 are given which are of order 10-16.Table 2
Numerical result of Example2.
x
u
5
u
10
Exact solution
Error (u10)
0.0
1
1
1
0
0.1
0.990025
0.990025
0.990025
0
0.2
0.960398
0.960398
0.960398
1.11022
×
1
0
-
16
0.3
0.912007
0.912007
0.912007
0
0.4
0.846298
0.846298
0.846298
1.11022
×
1
0
-
16
0.5
0.765240
0.765240
0.765240
0
0.6
0.671256
0.671256
0.671256
2.22045
×
1
0
-
16
0.7
0.567160
0.567160
0.567160
0
0.8
0.456066
0.456066
0.456066
1.11022
×
1
0
-
16
0.9
0.341300
0.341300
0.341300
5.55112
×
1
0
-
17
1.0
0.226296
0.226296
0.226296
5.55112
×
1
0
-
17Example 3.
Consider the equation(22)u(x)=e3x-19(2e3+1)x+∫01‍xtu(t)dt
with the exact solution
(23)u(x)=e3x.
Equation (22) is rewritten in the following form:
(24)F(u,ε)=u(x)-e3x+19(2e3+1)x-ε∫01‍xtu(t)dt,
where ε is an artificially introduced small parameter. The terms in (7) are
(25)F=un(x)-e3x+19(2e3+1)x,Fu=1,Fε=-∫01‍xtun(t)dt,F∫‍u=0.
Equation (7) reduces to
(26)9(e3x+∫01‍txundt)=x+2e3x+9(uc)n+9un.
Choosing the initial guess
(27)u0=1
and using the formula, the approximate solutions at each step are
(28)u1=e3x+7x18-2e3x9,u2=e3x-154(-7+4e3)x,u3=e3x-1162(-7+4e3)x.
Higher iterations are not given for brevity. In Table 3, some of our iterations are compared with the exact solution, and the errors between the exact solution and u20 are given which are of order 10-8.Table 3
Numerical result of Example3.
x
u
5
u
10
u
15
u
20
Exact solution
Error (u20)
0
1
1
1
1
1
0
0.1
1.34483
1.34984
1.34986
1.34986
1.34986
2.45400
×
1
0
-
9
0.2
1.81206
1.82208
1.82212
1.82212
1.82212
4.90801
×
1
0
-
9
0.3
2.44451
2.45954
2.45960
2.45960
2.45960
7.36201
×
1
0
-
9
0.4
3.299996
3.32003
3.32011
3.32012
3.32012
9.81602
×
1
0
-
9
0.5
4.45654
4.48159
4.48169
4.48169
4.48169
1.22700
×
1
0
-
8
0.6
6.01947
6.04952
6.04964
6.04965
6.04965
1.47240
×
1
0
-
8
0.7
8.13096
8.16603
8.16617
8.16617
8.16617
1.71780
×
1
0
-
8
0.8
10.98293
11.02301
11.02317
11.02318
11.02318
1.96320
×
1
0
-
8
0.9
14.83446
14.87955
14.87973
14.87973
14.87973
2.20860
×
1
0
-
8
1.0
20.03523
20.08533
20.08553
20.08554
20.08554
2.45400
×
1
0
-
8Example 4.
Consider the following integral equation:(29)u(x)=910x2+∫01‍12x2t2u(t)dt.
The exact solution of the problem is
(30)u(x)=x2.
Equation (29) is rewritten in the following form:
(31)F(u,ε)=u(x)-910x2-ε∫01‍12x2t2u(t)dt,
and proceeding in a similar way yields the following iteration algorithm:
(32)(uc)n+un=9x210+∫01‍12t2x2undt.
One may select the initial guess as u0=0. The successive approximations are
(33)u0=0,u1=9x210,u2=99x2100,u3=999x21000.
Higher iterations are not given for brevity. In Table4, some of our iterations are compared with the exact solution, and the errors between the exact solution and u20 are given which are of order 10-17.Table 4
Numerical result of Example4.
x
u
5
u
10
u
15
u
20
Exact solution
Error (u20)
0.0
0
0
0
0
0
0
0.1
0.0099999
0.00999
0.01000
0.01000
0.01000
0
0.2
0.0399996
0.03999
0.04000
0.04000
0.04000
0
0.3
0.0899991
0.08999
0.09000
0.09000
0.09000
1.38778
×
1
0
-
17
0.4
0.159998
0.15999
0.16000
0.16000
0.16000
0
0.5
0.249998
0.24999
0.25000
0.25
0.25
0
0.6
0.359996
0.35999
0.36000
0.36000
0.36000
5.55112
×
1
0
-
17
0.7
0.489995
0.48999
0.49000
0.49000
0.49000
5.55112
×
1
0
-
17
0.8
0.639994
0.63999
0.64000
0.64000
0.64000
0
0.9
0.809992
0.80999
0.81000
0.81000
0.81
1.11022
×
1
0
-
16
1.0
0.99999
0.99999
1.00000
1.0
1.0
0


## 4. Conclusion

In this paper, we have applied the newly developed Perturbation Iteration AlgorithmPIA(1,1) to some Fredholm and Volterra type integral equations for the first time. Numerical results show that method PIA(1,1) is an effective perturbation-iteration technique producing successful analytical results for integral equations.



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*Source: 682537-2013-05-23.xml*