CUBO A Mathematical Journal
Vol.15, No¯ 02, (89–103). June 2013

Numerical solution of singular and non singular integral
equations

M.H.Saleh and D.Sh.Mohammed

Zagazig University,

Mathematics Department, Faculty of Science,

Zagazig, Egypt.

doaamohammedshokry@yahoo.com

ABSTRACT

This paper is devoted to study the approximate solution of singular and non singular in-

tegral equations by means of Chebyshev polynomial and shifted Chebyshev polynomial.

Some examples are presented to illustrate the mothed.

RESUMEN

Este art́ıculo se dedica al estudio de la solución aproximada de ecuaciones integrales

singulares y no singulares por medio de polinomios de Chebyshev con o sin corrimiento.

Se presentan algunos ejemplos para ilustrarlo.

Keywords and Phrases: Linear Hypersingular integral equations, nonlinear integral equations,

Chebyshev polynomial, shifted Chebyshev polynomial, Approximate solution.

2010 AMS Mathematics Subject Classification: 45E05; 65R20; 78W23.



90 M.H.Saleh and D.Sh.Mohammed CUBO
15, 2 (2013)

Introduction

Singular integral equations are usually difficult to solve analytically so it required to obtain

the approximate solution [7,8]. Chebyshev polynomials are of great importance in many areas of

mathematics particalarly approximation theory see ([1-3], [9-13] ). In this paper we analyze the

numerical solution of singular and non singular integral equations by using Chebyshev polynomial

and shifted Chebyshev polynomial.

This paper consists of two parts I and II. In part I, we study the approximate solution

of Hypersingular integral equations by means of Chebyshev polynomial. In part II, we study the

approximate solution of nonlinear integral equations by means of shifted Chebyshev polynomial.

Part I: Approximate solution of hypersingular integral equa-

tions

1. Formulation of the problem

Consider the following hypersingular integral equation:

=

∫1

−1

k(x, t)

(t − x)2
ϕ(t) dt +

∫1

−1

L(x, t) ϕ(t) dt = f(x) , − 1 ≤ x ≤ 1 (1.1)

where k(x, t) , L(x, t) and f(x) are given real-valued continuous functions defined on the set

[−1, 1] × [−1, 1], [−1, 1] × [−1, 1] and [−1, 1] respectively and ϕ(t) is unknown function satisfy the
following condition ϕ(±1) = 0.

The simplest hypersingular integral equation of the form (1.1) given by the following form:

=

∫1

−1

ϕ(t)

(t − x)2
dt = f(x) , (1.2)

where the finite-part integral in (1.2) can be defined as the derivative of a Cauchy principle value

integral as

=

∫1

−1

ϕ(t)

(t − x)2
dt =

d

dx

∫1

−1

ϕ(t)

t − x
dt. (1.3)



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Numerical solution of singular and non singular integral equations 91

Thus equation (1.2) can be written in the following form

d

dx

∫1

−1

ϕ(t)

t − x
dt = f(x) . (1.4)

Integrating both sides of equation (1.4) with respect to x , we obtain

∫1

−1

ϕ(t)

t − x
dt = F(x) , (1.5)

where

F(x) =

∫

f(x) dx . (1.6)

The exact solution of equation (1.5) given by the following form:

ϕ(x) = −

√
1 − x2

π2

∫1

−1

F(t)
√
1 − t2(t − x)

dt . (1.7)

The solution exists if and only if the function F(x) satisfies the following condition :

∫1

−1

F(t)
√
1 − t2

dt = 0 . (1.8)

2. The approximate solution

In this section we shall study the approximate solution of the hypersingular integral equation

(1.1).

The unknown function ϕ(x) satisfying the condition ϕ(±1) = 0, can be represented by the follow-
ing form:

ϕ(x) =
√

1 − x2 Φ(x), − 1 ≤ x ≤ 1 (2.1)

where Φ(x) is a well-behaved function of x in the interval x ∈ [−1, 1] . Let the unknown function
Φ(x) be approximated by means of a polynomial of degree n as the following :

Φ(x) ≈
n
∑

j=0

cj x
j , (2.2)

where cj (j = 0, 1, 2, ..., n) are unknown constants.



92 M.H.Saleh and D.Sh.Mohammed CUBO
15, 2 (2013)

Substituting from form (2.1) and (2.2) into equation (1.1) we obtain:

n
∑

j=0

cj [=

∫1

−1

√
1 − t2 k(x, t) tj

(t − x)2
dt +

∫1

−1

√

1 − t2 L(x, t) tj dt] = f(x) , − 1 ≤ x ≤ 1. (2.3)

By using the following ” Chebyshev approximations” to the kernels k(x, t) and L(x, t)

k(x, t) ≈
∑m

p=0 kp(x) t
p ,

L(x, t) ≈
∑s

q=0 Lq(x) t
q ,























(2.4)

with known functions kp(x) and Lq(x) , then (2.3) takes

n
∑

j=0

cj [

m
∑

p=0

kp(x) up+j (x) +

s
∑

q=0

Lq(x) γq+j ] = f(x) , − 1 ≤ x ≤ 1 (2.5)

where

up+j (x) = =

∫1

−1

√
1 − t2 tp+j

(t − x)2
dt (2.6)

and

γq+j =

∫1

−1

√

1 − t2 tq+j dt . (2.7)

Using the zeros xk of the Chebyshev polynomial Tn+1 (x) into equation (2.5) we obtain the fol-

lowing system of linear equations with (n + 1) of the unknown constants cj (j = 0, 1, 2, ..., n)

n
∑

j=0

cj αj(xk) = f(xk) , (2.8)

where

xk = cos (
(2k − 1)

2(n + 1)
π) , k = 1, 2, 3, ..., n + 1 (2.9)

and

αj(xk) =

m
∑

p=0

kp(xk) up+j (xk) +

s
∑

q=0

Lq(xk) γq+j . (2.10)



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Numerical solution of singular and non singular integral equations 93

By solving the system of linear equations (2.8) we obtain the unknown constants cj and sub-

stituting into (2.1) and (2.2) we obtain the approximate solution of equation (1.1).

3. Numerical examples

In this section we shall give two examples to illustrate the above results .

Example 3.1

Consider the following hypersingular integral equation

=

∫1

−1

ϕ(t)

(t − x)2
dt +

∫1

−1

x t ϕ(t) dt = x + x3 , − 1 ≤ x ≤ 1 . (3.1)

Equation (3.1) can be written in the following form:

=

∫1

−1

ϕ(t)

(t − x)2
dt = (1 + µ) x + x3 , (3.2)

where

µ = −

∫1

−1

t ϕ(t) dt .

According to (1.3), (1.5), (1.6) and (1.7) it easy to show that the exact solution of equation

(3.2) given by:

ϕ(x) = −
1

40 π

√

1 − x2 [10 x3 + 27 x ] . (3.3)

Now, we study the approximate solution of equation (3.1). Since k(x, t) = 1 and L(x, t) = x t ,

then we have

k0(x) = 1 , k1(x) = k2(x) = ... = km(x) = 0

L1(x) = x , L0(x) = L2(x) = ... = Ls(x) = 0 .























(3.4)



94 M.H.Saleh and D.Sh.Mohammed CUBO
15, 2 (2013)

Substituting from (3.4) into (2.10) we obtain:

αj(xk) = uj (xk) + xk γj+1 , (j = 0, 1, 2, ..., n) (3.5)

By using the following formula

=

∫1

−1

√
1 − t2 Uj(t)

(t − x)2
dt = −π (j + 1) Uj(x) , (3.7)

where Uj(x) is Chebyshev polynomial of the second kind. From (2.6) and (3.7) it easy to show

that

u0(x) = −π , u1(x) = −2π x, u2(x) = −π (3x
2 −

1

2
), u3(x) = −π (4x

3 − x) ,

u4(x) = −π (5x
4 −

3

2
x2 −

1

8
) , u5(x) = −π (6x

5 − 2x3 −
1

4
x) , ... (3.8)

from (2.7) we have

γ0 =
π

2
, γ2 =

π

8
, γ4 =

π

16
, γ6 =

5π

128
, γ8 =

7π

256
,

γ1 = 0 , γ3 = 0 , γ5 = 0 , γ7 = 0 , γ9 = 0 . (3.9)

Substituting from (3.8) and (3.9) into (2.10) and take n = 5 , we obtain the following system of

linear equations:

−π [c0 +
15

8
c1 xk + c2 (3x

2
k −

1

2
) + c3 (4x

3
k −

17

16
xk) + c4 (5x

4
k −

3

2
x2k −

1

8
) +

c5 (6x
5
k − 2x

3
k −

37

128
xk)] = xk + x

3
k , (k = 1, 2, ..., 6). (3.10)

Solving system (3.10) by using the zeros xk of Chebyshev polynomial Tn+1(x) , we obtain the

values of the constants as follows:

{ c0 = c2 = c4 = c5 = 0 , c1 =
−27

40π
and c3 =

−1

4π
} . (3.11)



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Numerical solution of singular and non singular integral equations 95

Substituting from (3.11) into (2.1) and (2.2) we obtain the approximate solution of equation (3.1)

which is the same as the exact solution which given by (3.3).

Example 3.2

Consider the following hypersingular integral equation

=

∫1

−1

ϕ(t)

(t − x)2
dt +

∫1

−1

(2x t + 4x3 t3) ϕ(t) dt = 4x3 − 2x + 1 . − 1 ≤ x ≤ 1 (3.12)

Equation (3.12) can be written in the following form:

=

∫1

−1

ϕ(t)

(t − x)2
dt = 2(µ1 − 1) x + 4(µ2 + 1)x

3 + 1 , (3.13)

µ1 = −

∫1

−1

t ϕ(t) dt and µ2 = −

∫1

−1

t3 ϕ(t) dt . (3.14)

It is easy to show that the exact solution of equation (3.13) given by:

ϕ(x) = −
1

π

√

1 − x2 [
832

825
x3 −

136

275
x + 1 ] . (3.15)

Similarly as in example 3.1 we study the approximate solution of equation (3.12). Since k(x, t) =

1 and L(x, t) = 2x t + 4x3 t3 , then we have

k0(x) = 1 , k1(x) = k2(x) = ... = km(x) = 0

L1(x) = 2x , L3(x) = 4x
3 , L0(x) = L2(x) = ... = Ls(x) = 0 .























(3.16)

Substituting from (3.16) into (2.10) we obtain:

αj(xk) = uj (xk) + 2xk γj+1 + 4x
3
k γj+3 . (j = 0, 1, 2, ..., n) (3.17)



96 M.H.Saleh and D.Sh.Mohammed CUBO
15, 2 (2013)

Substituting from (3.17) into (2.8) we obtain the following system of linear equations:

n
∑

j=0

cj (uj (xk) + 2xk γj+1 + 4x
3
k γj+3 ) = 4x

2
k − 2xk + 1. (j = 0, 1, 2, ..., n) (3.18)

Substituting from (3.8) and (3.9) into (3.18) and take n = 5 , we obtain the following system

of linear equations:

−π [c0 + c1 (
7

4
xk −

1

4
x3k) + c2 (3x

2
k −

1

2
) + c3 (

123

32
x3k −

9

8
xk) + c4 (5x

4
k −

3

2
x2k −

1

8
) +

c5 (6x
5
k −

135

64
x3k −

21

64
xk)] = 4x

3
k − 2xk + 1 , (k = 1, 2, ..., 6). (3.19)

Solving system (3.19) by using the zeros xk of Chebyshev polynomial Tn+1(x) , we obtain the

values of the constants as follows:

{ c2 = c4 = c5 = 0 , c0 = −
1

π
, c1 =

−136

275π
and c3 =

−832

825π
} . (3.20)

Substituting from (3.20) into (2.1) and (2.2) we obtain the approximate solution of equation (3.12)

which is the same as the exact solution which is given by (3.15).

Part II: Approximate solution of nonlinear integral equations

In this part we transform the integral equation to a matrix equation which corresponds to a

system of nonlinear algebraic equations with unknown Chebyshev coefficients.

1. Formulation of the problem

Consider the following nonlinear integral equation:

φ(x) = f(x) + λ

∫1

0

K(x, t) [φ(t)]2 dt , (1.1)



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Numerical solution of singular and non singular integral equations 97

where f(x) , K(x, t) are given functions, λ is a real parameter and φ(x) is unknown function.

The unknown function φ(x) can be represented by truncated Chebyshev series as follows:

φ(x) =

N
∑

j=0

′

a∗j T
∗

j (x) , 0 ≤ x ≤ 1 (1.2)

where T∗j (x) denoted the shifted Chebyshev polynomial of the first kind, a
∗

j are the unknown

Chebyshev coefficients,
∑

′

is a sum whose first term is halved and N is any positive integer.

Suppose that the solution φ(x) of equation (1.1) and K(x, t) can be expressed as a truncated

Chebyshev series. Then (1.2) can be written in the following form

φ(x) = T∗(x) A∗ , (1.3)

where

T∗(x) = [ T∗0 (x) T
∗

1 (x) ... T
∗

N(x)] , A
∗ = [

a∗0
2

a∗1 ... a
∗

N]
T ,

and the function [φ(t)]2 can be written in the following matrix form [1]

[φ(t)]2 = T∗(t) B ∗ , (1.4)

where

T∗(t) = [T∗0 (x) T
∗

1 (x) ... T
∗

2N(x)] , B
∗ = [

b∗0
2

b∗1 ... b
∗

2N]
T ,

and the elements b∗i consists of a
∗

i and a
∗

−i = a
∗

i as follows:

b∗i =







































(a
∗

i/2
)
2

2
+
∑N−i/2

r=1 (a
∗

i
2
−r

) (a∗i
2
+r

) for even i

∑N− i−1
2

r=1 (a
∗

i+1
2

−r
) (a∗i−1

2
+r

) for odd i.



98 M.H.Saleh and D.Sh.Mohammed CUBO
15, 2 (2013)

Now, K(x, t) can be expanded by chebyshev series as follows:

K(xi, t) =

N
∑

r=0

′′

kr(xi) T
∗

r (t) ,

where
∑

′′

denotes a sum with first and last terms halved, xi are the chebyshev collocation points

defined by

xi =
1

2
[ 1 + cos (

iπ

N
)] , i = 0, 1, ..., N (1.5)

and Chebyshev coefficients kr(xi) are determined by the following relation:

kr(xi) =
2

N

N
∑

j=0

′′

K(xi, tj) T
∗

r (tj) , ti =
1

2
[ 1 + cos (

jπ

N
)]

which is given by [5 ].

Then the matrix representation of K(xi, t) given by

K(xi, t) = K(xi) T
∗(t)

T
. (1.6)

where

K(xi) = [
k0(xi)

2
k1(xi) ... kN−1(xi)

kN(xi)

2
] .

2. Solution of nonlinear integral equation

Our aim in this section to find the Chebyshev coefficients of (1.2), that is the matrix A∗. By

substituting from Chebyshev collocation points defined by (1.5) into equation (1.1) we obtain a

matrix equation of the form

Φ = F + λ I , (2.1)

where I(x) denotes the integral part of equation (1.1) and



CUBO
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Numerical solution of singular and non singular integral equations 99

Φ =







































φ(x0)

φ(x1)

.

.

.

φ(xN)







































, F =







































f(x0)

f(x1)

.

.

.

f(xN)







































, I =







































I(x0)

I(x1)

.

.

.

I(xN)







































, T∗ =







































T∗(x0)

T∗(x1)

.

.

.

T∗(xN)







































.

When we substitute from Chebyshev collocation points (1.5) into (1.3), the matrix Φ becomes

Φ = T∗ A∗ . (2.2)

Substituting from (1.4) and (1.6) in I(xi) for i = 0, 1, ..., N , i = 0, 1, ..., 2N and using the following

relation [6 ],

Z =

∫1

0

T∗(t)
T
T∗(t) dt = [

∫1

0

T∗i (t) T
∗

j (t) dt ] =
1

2
[zij] ,

where

zij =



































1
1−(i+j)2

+ 1
1−(i−j)2

for even i+j

0 for odd i+j ,

we obtain

I(xi) = K(xi) Z B
∗ . (2.3)

Therefore, we obtain the matrix I in terms of Chebyshev coefficients matrix in the following form:

I = K Z B∗ , (2.4)



100 M.H.Saleh and D.Sh.Mohammed CUBO
15, 2 (2013)

where

K = [ k(x0) k(x1) ... k(xN)]
T .

Now, by using the relation (2.2) and (2.4), the integral equation (1.1) transform to a matrix equa-

tion which is given by:

T∗ A∗ − λ K Z B∗ = F . (2.5)

The matrix equation (2.5) corresponds to a system of (N + 1) nonlinear algebraic equations with

(N + 1) unknown Chebyshev coefficients. Thus the unknown coefficients a∗j can be computed from

this equation and substituting from these coefficients into (1.2) we obtain the approximate solution.

3. Numerical examples

In this section we shall give two examples to illustrate the above results .

Example 3.1

Consider the following nonlinear integral equation

φ(x) = x −
1

3
+

∫1

0

[φ(t)]2 dt . (3.1)

From (3.1) we have

f(x) = x −
1

3
, K(x, t) = 1 and λ = 1 .

For N = 2 , the Chebyshev collocation points on [0, 1] can be found from (1.5) as

x0 = 1 , x1 =
1

2
, x2 = 0

and the matrix equation corresponds to the integral equation (3.1) given by

T∗ A∗ − K Z B∗ = F , (3.2)

where



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Numerical solution of singular and non singular integral equations 101

T∗ =























1 1 1

1 0 −1

1 −1 1























, F =























2
3

1
6

−1
3























, Z =























1 0 −1
3

0 − 1
15

0 1
3

0 −1
5

0

−1
3

0 7
15

0 − 19
105























K =























1 0 0

1 0 0

1 0 0























, A∗ =























a∗0/2

a∗1

a∗2























, B∗ =













































1
2
(
a

∗

0
2

2
+ a∗1

2
+ a∗2

2
)

a∗0 a
∗

1 + a
∗

1 a
∗

2

a
∗

1
2

2
+ a∗0 a

∗

2

a∗1 a
∗

2

a
∗

2
2

2













































.

Substituting from these matrices into (3.2), we obtain a system of nonlinear algebraic equations,

the solution of this given by:

(a∗0 = 1 , a
∗

1 =
1

2
, a∗2 = 0) .

Substituting from these values into (1.2) when N = 2 we obtain the approximate solution φ(x) = x ,

which is the exact solution.

Example 3.2

Consider the following nonlinear integral equation

φ(x) = x2 −
x

6
− 1 +

∫1

0

x t [φ(t)]2 dt . (3.3)



102 M.H.Saleh and D.Sh.Mohammed CUBO
15, 2 (2013)

Similarly as in example 3.1 it is easy to show that the values of a∗j (for N = 2 , j = 0, 1, 2)

given by

(a∗0 = −
5

4
, a∗1 =

1

2
, a∗2 =

1

8
) ,

and the approximate solution of the integral equation (3.3) given by φ(x) = x2 − 1 which is the

exact solution.

Received: October 2011. Accepted: September 2012.

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