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LINEAR INTEGRAL EQUATIONS AND ITS APPLICATIONS

Christy Tom Mathews, Bijesh P Biju Department of Mathematics, Deva Matha College Kuravilangad, Kerala, India

Department of Basic Science,Muslim Association College of Engineering,Venjaramoodu,Kerala,India

Email:[email protected]

[email protected]

Abstract

The theory of integral equations was first introduced by J. Fourier. In this study ,we go through

some basic definitions and classifications of integral equations. We will be concentrating more on

Volterra and Fredholm types of Linear integral equations, where we will see numerical methods of

solving them. Finally the study ends with some applications of integral equations to real world

problems.

Keywords: Integral equations, Linear integral equations ,Volterra integral equations ,Fredholm integral

equations , Numerical solutions

1. Introduction

Fourier (1768-1830) is the initiator of the theory of integral equations. The term integral

equation was first suggested by Du-Bois Reymond in 1888. The pioneering systematic investigations

goes back to late 19th and early 20th century works of Volterra, Fredholm & Hilbert. In 1887, Volterra

published a series of famous papers in which he singled out the notion of a functional and pioneered in

the development of a theory of functionals in the theory of linear integral equations of special types.

Fredholm presented the fundamentals of the Fredholm integral equation theory in a paper published in

1903 in the Acta Mathematica.

The Study here mainly points towards linear integral equations especially on Fredholm and

Volterra type of integral equations. One of the main relevance of this study arise from the idea of

conversion of differential equations in to integral equations and vice versa. It also includes some

numerical methods to solve these equations. Finally, the study establishes some real-world applications

of Linear integral equations.

2. Integral equation

Any equation in which the unknown function appears under the integral sign is known as an

integral equation. A typical form of an integral equation is

𝑢(𝑥) = 𝑓(𝑥) + 𝜆 ∫ 𝑘(𝑥, 𝑡)𝑢(𝑡)𝑑𝑡𝛽(𝑥)

𝛼(𝑥) (1)

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Volume XI, Issue VIII, August/2019

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Where, 𝑘(𝑥, 𝑡) is called kernel of the integral equation (1), 𝛼(𝑥), 𝛽(𝑥) are the limits of integration, 𝜆 is

a constant parameter. It can be easily observed that the unknown function 𝑢(𝑥) appears under integral

sign. It is to be noted that kernel 𝑘(𝑥, 𝑡) and the function 𝑓(𝑥) in the equation (1) are given functions.

Our prime objective is to find unknown function 𝑢(𝑥) that will satisfy equation (1) using a number of

solution techniques.

Eg: 𝑢(𝑥) = 𝑥 +1

2∫ 𝑥𝑡 𝑢(𝑡)𝑑𝑡

1

0

2.1. Classification of Integral Equations

An integral equation can be classified in to different categories as we have seen in ordinary and

partial differential equations. We can classify integral equations as Linear or Nonlinear,

Homogeneous or Nonhomogeneous. The most frequently used integral equations fall under two major

classes, namely Fredholm and Volterra integral equations.

An integral equation is called Linear if only linear operations are performed in it upon the

unknown function. For a linear integral equation, the unknown function 𝑢(𝑥) appearing under integral

sign is given in the functional form 𝐹(𝑢(𝑥)) such that the power of 𝑢(𝑥) is unity. The most general

type of linear integral equation is of the form,

ℎ(𝑥)𝑔(𝑥) = 𝑓(𝑥) + 𝜆 ∫ 𝑘(𝑥, 𝑡)𝑔(𝑡)𝑑𝑡𝑎

(2)

where, the upper limit may be either variable or fixed. The functions ℎ(𝑥), 𝑓(𝑥), 𝑘(𝑥, 𝑡) are known

functions while 𝑔(𝑥) is to be determined, 𝜆 is the non-zero real or complex parameter.

Example 1: 𝑢(𝑥) = 𝑓(𝑥) + ∫ 𝑘(𝑥, 𝑡)𝑢(𝑡)𝑑𝑡𝑥

0

If the unknown function 𝑢(𝑥) appearing under integral sign is given in the functional form

𝐹(𝑢(𝑥)) such that the power of 𝑢(𝑥) is no longer unity, then such integral equations can be classified

as Nonlinear integral equations.(That is, 𝐹(𝑢(𝑥)) = 𝑢𝑛(𝑥); 𝑛 ≠ 1 𝑜𝑟 𝑠𝑖𝑛 𝑢(𝑥) 𝑒𝑡𝑐. )

Example 2: 𝑢(𝑥) = 𝑓(𝑥) + ∫ 𝑘(𝑥, 𝑡)𝑢2(𝑡)𝑑𝑡𝑥

0

Throughout this study we mainly focus on Linear Integral Equations.

2.1.1 Fredholm integral equations

The most standard form of Fredholm linear integral equations is given by the form

(𝑥)𝑢(𝑥) = 𝑓(𝑥) + ∫ k(𝑥, t)u(t)dtb

a (3)



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where the limits of integration a and b are constants and the unknown function 𝑢(𝑥) appears linearly

under the integral sign. K(𝑥,t) is the kernel of the equation

If (𝑥) 1, then (3) becomes

u(𝑥) = 𝑓(𝑥) + ∫ k(x, t)u(t)dtb

a (4)

and this equation is called Fredholm integral equation of second kind.

If (𝑥) 0, then (3) yields

𝑓(𝑥) + ∫ k(𝑥, t)u(t)dt = 0b

a (5)

which is called Fredholm integral equation of the first kind.

2.1.2 Volterra integral equations.

The most standard form of Volterra linear integral equations is of the form.

(𝑥)𝑢(𝑥) = 𝑓(𝑥) + ∫ k(𝑥, t)u(t)dt𝑥

a (6)

where the limits of integration are functions of 𝑥 and the unknown function 𝑢(𝑥) appears linearly under

the integral sign.

If the function (𝑥) 1, then (6) becomes

𝑢(𝑥) = 𝑓(𝑥) + ∫ k(𝑥, t)u(t)dt𝑥

a (7)

and this equation is known as the Volterra integral equation of the second kind.

If (𝑥) 0, then (6) becomes

𝑓(𝑥) + ∫ k(𝑥, t)u(t)dt = 0𝑥

a (8)

which is the Volterra integral equation of first kind.

2.1.3 Singular and Non – singular integral equations

A Singular integral equation is defined as an integral with the infinite limits when the Kernel

k(𝑥, t) of the integral becomes unbounded at a certain point in the interval.

Example 3: 𝑢(𝑥) = 𝑓(𝑥) + ∫ u(t)dt

−

If the Kernel k(𝑥, t)is bounded and continuous then the integral equation is said to be non – singular.

Remark;

If we set f(x)=0, in Volterra and Fredhlom integral equations then the resulting equation is

called a homogenous integral equation, otherwise it is called non-homogenous.

3. Relation Between Differential and Integral Equations

Integral and differential equations have a fundamental importance in functional analysis and

practice problems. But in many cases the resolution of differential equations with constant coefficients



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is easy but the resolution of these equations with variable coefficients is practically difficult or

impossible in more part of the cases. Here we present an analytical method which transforms a

differential equation into an integral equation. Also, it presents methods to convert an integral equation

into a differential equation.

3.1 Transformation of Differential Equations into Integral Equations.

A boundary value problem of the form 𝑦′′ + 𝑦′f1(𝑥) + 𝑦𝑓2(𝑥) = 𝑓3(𝑥) with 𝑦 (𝑎) =

𝑦0 & 𝑦 (𝑏) = 𝑦1 can be transformed in to a Fredholm equation. An initial value problem of the form

𝑦′′ + 𝑦′𝑓1(𝑥) + 𝑦𝑓2(𝑥) = 𝑓3(𝑥) with 𝑦 (𝑎) = 𝑦0 & 𝑦′ (𝑎) = 𝑦1 can be transformed in to a Volterra

equation.

Example 4:

Construct the linear integral equation corresponding to the differential equation 𝑦′′ + 𝑥𝑦 = 1

with initial conditions 𝑦(0) = 𝑦′(0) = 0

Solution

Given 𝑦′′ = 1 − 𝑥𝑦

Integrate w.r.t x from 0 to x

∫ 𝑦′′(𝑥)𝑑𝑥 = ∫ (1 − 𝑥𝑦)𝑑𝑥𝑥

0

𝑥

0

𝑦′(𝑥)]0𝑥 = 𝑥]0

𝑥 - ∫ 𝑥𝑦(𝑥)𝑑𝑥𝑥

0

Integrate w.r.t x from 0 to x

∫ [𝑦′(𝑥) − 𝑦′(0)]𝑑𝑥 = ∫ (𝑥 − 0)𝑑𝑥 − ∫ [∫ 𝑥𝑦(𝑥)𝑑𝑥]𝑑𝑥𝑥

0

𝑥

0

𝑥

0

𝑥

0

𝑦(𝑥)]0𝑥 = [

𝑥2

2]

0

𝑥

- 1

(2−1)!∫ (𝑥 − 𝑡)2−1𝑥

0ty(t)dt

y(x) - y (0) = 𝑥2

2 -∫ (𝑥 − 𝑡)𝑡𝑦(𝑡)𝑑𝑡

𝑥

0

y(x) = 𝑥2

2 -∫ (𝑥 − 𝑡)𝑡𝑦(𝑡)𝑑𝑡

𝑥

0

which is the required linear Volterra integral equation.

Example 5:

Consider the differential equation 𝑦′′(𝑥) = 𝑓(𝑥) with boundary conditions

𝑦(0) = 0, 𝑦(1) = 0

Solution

Given 𝑦′′(𝑥) = 𝑓(𝑥)

Integrate w.r.t 𝑥, ∫ 𝑦′′(𝑥)𝑥

0𝑑𝑥 = ∫ 𝑓(𝑥)

𝑥

0𝑑𝑥

⟹ 𝑦′(𝑥)]0𝑥 = ∫ 𝑓(𝑥)

𝑥

0𝑑𝑥

Again integrating, ∫ [𝑦′(𝑥)𝑥

0− 𝑦′(0)]𝑑𝑥 = ∫ ∫ 𝑓(𝑥)𝑑𝑥𝑑𝑥

𝑥

0



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=1

(2−1)!∫ ((𝑥 − 𝑡)2−1

𝑓(𝑡)𝑑𝑡𝑥

0

⟹ 𝑦(𝑥)]0𝑥 − 𝑥𝑦′(0)]0

𝑥 = ∫ (𝑥 − 𝑡)𝑥

0𝑓(𝑡)𝑑𝑡

⟹ 𝑦(𝑥) − 𝑦(0) − 𝑥𝑦′(0) = ∫ (𝑥 − 𝑡)𝑥

0𝑓(𝑡)𝑑𝑡

⟹ 𝑦(𝑥) = 𝑥𝑦′(0) + 𝑦(0) + ∫ (𝑥 − 𝑡)𝑥

0𝑓(𝑡)𝑑𝑡

Given that 𝑦(1) = 0

𝑦(1) = 𝑦′(0) + 𝑦(0) + ∫ (1 − 𝑡)𝑓(𝑡)𝑑𝑡1

0

⟹ 𝑦′(0) = 𝑦(1) − 𝑦(0) − ∫ (1 − 𝑡)𝑓(𝑡)𝑑𝑡1

0

⟹ 𝑦′(0) = 0 − 0 − ∫ (1 − 𝑡)𝑓(𝑡)𝑑𝑡1

0

⟹ 𝑦(𝑥) = −𝑥 ∫ (1 − 𝑡)𝑓(𝑡)𝑑𝑡1

0+ ∫ (𝑥 − 𝑡)𝑓(𝑡)𝑑𝑡

𝑥

0

= ∫ (𝑥 − 𝑡)𝑓(𝑡)𝑑𝑡 − 𝑥 [∫ (1 − 𝑡)𝑓(𝑡)𝑑𝑡 + ∫ (1 − 𝑡)𝑓(𝑡)𝑑𝑡1

𝑥

𝑥

0

]

𝑥

0

= ∫ (𝑥 − 𝑡 − 𝑥 + 𝑥𝑡)𝑓(𝑡)𝑑𝑡 − 𝑥 ∫ (1 − 𝑡)𝑓(𝑡)𝑑𝑡1

𝑥

𝑥

0

= ∫ 𝑡(𝑥 − 1)𝑓(𝑡)𝑑𝑡 + ∫ 𝑥(𝑡 − 1)𝑓(𝑡)𝑑𝑡1

𝑥

𝑥

0

= ∫ [𝑡(𝑥 − 1) + (𝑡 − 1)𝑥]𝑓(𝑡)𝑑𝑡1

0

= ∫ 𝑘(𝑥, 𝑡)𝑓(𝑡)𝑑𝑡1

0

; 𝑘(𝑥, 𝑡) = 𝑓(𝑥) = {(𝑥 − 𝑡)𝑡; 𝑡 < 𝑥(𝑡 − 1)𝑥; 𝑥 < 𝑡

3.2 Transformation of Integral Equations into Differential Equations

If the upper limit is 𝑥 & Kernel doesn’t depend on 𝑥, then Volterra equation is

equivalent to an ordinary differential equation.

(𝑥) = 𝑓(𝑥) + ∫ 𝑘(𝑥, 𝑡)𝑥

0(𝑡)𝑑𝑡

⟹ ′(𝑥) = 𝑓′(𝑥) + 𝑘(𝑥)(𝑥)

If the upper limit is 𝑥 and kernel depends on 𝑥 then it is a possible to reduce the integral

equation to an ordinary differential equation by differentiating several times w.r.t. 𝑥

Example 6:

Convert 𝑦(𝑥) = − ∫ (𝑥 − 𝑡)𝑦(𝑡)𝑑𝑡𝑥

0 into an initial value problem.



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Solution

𝑦(𝑥) = − ∫ (𝑥 − 𝑡)𝑦(𝑡)𝑑𝑡𝑥

0 (9)

Differentiating equation (9) w.r.t 𝑥,

𝑦′(𝑥) = −𝑑

𝑑𝑥∫ (𝑥 − 𝑡)𝑦(𝑡)𝑑𝑡

𝑥

0

𝑦′(𝑥) = − ∫ 𝑦(𝑡)𝑑𝑡𝑥

0 (10)

Again, differentiating equation (10) w.r.t 𝑥 will give

𝑦′′(𝑥) = −𝑑

𝑑𝑥∫ 𝑦(𝑡)𝑑𝑡

𝑥

0

𝑦′′(𝑥) = −𝑦(𝑥)

𝑦′′(𝑥) + 𝑦(𝑥) = 0 (11)

Putting the lower limit 𝑥 = 0 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒) the equations (9) & (10) will give respectively the

following.

𝑦(0) = − ∫ (0 − 𝑡)0

0𝑦(𝑡)𝑑𝑡

𝑦(0) = 0 (12)

𝑦′(0) = − ∫ 𝑦0

0(𝑡)𝑑𝑡

𝑦′(0) = 0 (13)

(11), (12) & (13) form the ordinary differential form of given integral equation.

4. Numerical Solutions of Integral Equations

Now we discuss about the different methods to solve integral equations. First, we shall discuss

the various methods of solutions of the Fredholm integral equation. Then we’ll study the different

methods to solve Volterra integral equations.

4.1 Numerical Methods for Fredholm Equations

There exist several methods for the numerical solutions of Fredholm equations like the method of

degenerate Kernels, methods of successive approximations, method of successive substitutions etc.

4.1.1 Method of Degenerate Kernels

We consider the integral equation

𝑓(𝑥) − ∫ 𝑘(𝑥, 𝑡)𝑏

𝑎𝑓(𝑡)𝑑𝑡 = (𝑥) (14)

A Kernel 𝑘(𝑥, 𝑡) is said to be degenerate it can be expressed in the form.

𝑘(𝑥, 𝑡) = ∑ ui(x)vi(t)ni=1 (15)

Substituting this in (9) we obtain



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𝑓(𝑥) − ∑ ∫ ui(x)vi(t)𝑓(𝑡)b

a𝑑𝑡 = (𝑥)𝑛

𝑖=1 (16)

Setting ∫ 𝑣𝑖(𝑡)𝑓(𝑡)𝑑𝑡 =𝑏

𝑎𝐴𝑖 (17)

equation (11) gives

𝑓(𝑥) = ∑ 𝐴𝑖 𝑢𝑖 (𝑥) + (𝑥)𝑛𝑖=1 (18)

The constants 𝐴𝑖 are still to be determined, but substituting from (18) in (17) we get

∫ 𝑣𝑖(𝑡)[∑ 𝐴𝑗 𝑢𝑗 (𝑡) + (𝑡)𝑛𝑗=1 ]𝑑𝑡 =

𝑏

𝑎𝐴𝑖

Or

∑ 𝐴𝑗 ∫ 𝑣𝑖(𝑡)𝑏

𝑎𝑢𝑗 (𝑡)𝑑𝑡 +𝑛

𝑗=1 ∫ 𝑣𝑖(𝑡)𝑏

𝑎(𝑡)𝑑𝑡 = 𝐴𝑖

which represent a system of n equations in the n unknowns 𝐴1,𝐴2 … … … 𝐴𝑛. When the 𝐴𝑖 𝑠 are

determined, equation (18) then gives 𝑓(𝑥).

Example 7:

We consider the equation 𝑓(𝑥) − ∫ sin 𝑥 𝑐𝑜𝑠𝑡𝜋

20

𝑓(𝑡)𝑑𝑡 = sin 𝑥

Setting ∫ 𝑐𝑜𝑠𝑡𝜋

20

𝑓(𝑡)𝑑𝑡 = 𝐴, the integral equation becomes

𝑓(𝑥) = 𝐴 𝑠𝑖𝑛 𝑥 + sin 𝑥 = (𝐴 + 1) sin 𝑥

Substituting this in above equation we get

∫ 𝑐𝑜𝑠𝑡(𝐴 + 1)𝑠𝑖𝑛𝑡 𝑑𝑡 = 𝐴𝜋/2

0

⟹ ∫ 𝑢𝑑𝑢 =𝐴

𝐴+1

1

0 ; 𝑢 = 𝑠𝑖𝑛𝑡

⟹ 𝑢2

2]

0

1

= 𝐴

𝐴+1

⟹ 1

2 =

𝐴

𝐴+1

⟹ 𝐴 = 1

2−

Hence 𝑓(𝑥) = (

2−+ 1) sin 𝑥

Therefore, the solution of the integral equation is

𝑓(𝑥) =2

2−sin 𝑥 ( ≠ 2)

4.1.2 Method of Successive Substitutions

Let 𝑢 (𝑥) = 𝑓(𝑥) + ∫ 𝑘(𝑥, 𝑡)𝑢 (𝑡)𝑑𝑡𝑏

𝑎 (19)

In equation (19) as usual k(𝑥, 𝑡) ≠ 0 , is real & continuous in the rectangle R, for which 𝑎 ≤ 𝑥 ≤ 𝑏,

𝑎 ≤ 𝑡 ≤ 𝑏; 𝑓(𝑥) ≠ 0 is real and continuous in the interval I for which 𝑎 ≤ 𝑥 ≤ 𝑏 𝑎𝑛𝑑 , a constant

parameter. Substituting in the 2nd member of (19), in place of 𝑢 (𝑡), its value as given by the equation

itself yields



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u(x) = 𝑓(𝑥) + ∫ k(x, t)𝑏

𝑎

𝑓(𝑡)𝑑𝑡 + 2 ∫ k(x, t)

𝑏

𝑎

∫ k(t, 𝑡1)u(t1)𝑏

𝑎

𝑑𝑡1𝑑𝑡

Here, again we substitute for u(𝑡1) its value as given in equation (19). Thus, we get,

u(x) = 𝑓(𝑥) + ∫ k(x, t)𝑏

𝑎

𝑓(𝑡)𝑑𝑡 + 2 ∫ k(x, t)

𝑏

𝑎

∫ k(t, 𝑡1)𝑏

𝑎

𝑓(𝑡1)𝑑𝑡1𝑑𝑡 +

3 ∫ k(x, t)

𝑏

𝑎

∫ k(t, 𝑡1) ∫ k(𝑡1, 𝑡2)u𝑏

𝑎

𝑏

𝑎

(𝑡2)𝑑𝑡2𝑑𝑡1𝑑𝑡

Proceeding like this we get

u(x) = 𝑓(𝑥) + ∫ k(x, t)𝑏

𝑎

𝑓(𝑡)𝑑𝑡 + 2 ∫ k(x, t)

𝑏

𝑎

∫ k(t, 𝑡1)𝑏

𝑎

𝑓(𝑡1)𝑑𝑡1𝑑𝑡 +

3

∫ k(x, t)𝑏

𝑎 ∫ k(t, 𝑡1) ∫ k(𝑡1, 𝑡2)𝑏

𝑎

𝑏

𝑎𝑓(𝑡2)𝑑𝑡2𝑑𝑡1𝑑𝑡 + ⋯ (20)

The series solution given in (20) converges uniformly in the interval [a, b] if

𝑀(b − a) < 1 where |k (x, t) |≤ 𝑀

Example 8:

Consider the Fredholm equation 𝑢(𝑥) = cos 𝑥 +1

2∫ sin 𝑥

𝜋

20

u(t)dt

Here =1

2, f(x) = cos 𝑥 & k(x, t) = sin 𝑥 and substituting these values in equation (20) yields.

𝑢(𝑥) = cos 𝑥 +1

2∫ sin 𝑥

𝜋

2

0

costdt

+1

4∫ sin 𝑥

𝜋

2

0

∫ sin 𝑡 cos 𝑡1𝑑𝑡1𝑑𝑡

𝜋

2

0

+ 1

8∫ sin 𝑥

𝜋

2

0

∫ sin 𝑡 ∫ sin 𝑡1 cos 𝑡2𝑑𝑡2𝑑𝑡1𝑑𝑡 +

𝜋

2

0

…

𝜋

2

0

= cos 𝑥 +1

2𝑠𝑖𝑛 𝑥 +

1

4𝑠𝑖𝑛𝑥 +

1

8𝑠𝑖𝑛𝑥 + ⋯

= cos 𝑥 + sin 𝑥

4.2 Numerical Method for Volterra Equations

Here we discuss three methods: series solutions method, differentiation method and method of

resolvant kernels.

4.2.1 Series Solutions of Volterra Equations

Consider a simple Volterra equation (𝑥) = 𝑥 + ∫ 𝑥

0(𝑡)𝑑𝑡. For small ,

0

(𝑥) = 𝑥 is a first approximation. Insert this in the integral term to get a better approximation.



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1

(𝑥) = 𝑥 + ∫ 𝑡𝑑𝑡 = 𝑥 +𝑥2

2

𝑥

0

Again, put this in to the integral to get

2

(𝑥) = 𝑥 + ∫ (𝑡 +1

2𝑡2)𝑑𝑡

𝑥

0

= 𝑥 + [t2

2]

0

x

+

2

2[

𝑡3

3]

0

𝑥

= 𝑥 +𝑥2

2+

2𝑥3

6

Continuing like this we get,

𝑛

(𝑥) = 𝑥 +𝑥2

2+ ⋯

1

𝑛!

𝑛−1𝑥𝑛

We let 𝑛 → , the series converges to 𝑒𝑥−1

. Substituting into the equation verifies that this is

the correct solution. So, for this Volterra equation the technique of expansion in series gives a result

which is convergent for all .

We now show that this works for all Volterra equations, subject to some fairly general

conditions. Suppose that we look for a solution with 𝑥 in some finite interval [0,1]& and that on this

interval f(x) is bounded with [𝑎, 𝑏]x[𝑎, 𝑏]; 𝑘[𝑥, 𝑠] < 𝑚

Then |0(𝑥)| < |𝑓(𝑥)| < 𝑀

|1(𝑥)| = |𝑓(𝑥) + ∫ 𝑘(𝑥, 𝑠)𝑓(𝑠)𝑑𝑠

𝑥

0

|

< 𝑚 + ||𝑚𝑀(𝑥 − 𝑎)

|2(𝑥)| < |𝑓(𝑥) + ∫ 𝑘(𝑥, 𝑠)

1(𝑥)𝑑𝑠

𝑥

0

|

< 𝑚 + ∫ 𝑀(𝑚 + 𝑚𝑀(𝑠 − 𝑎)𝑥

0

𝑑𝑠

= 𝑚(1 + ||𝑀(𝑥 − 𝑎) + |2|𝑀2 (𝑥−𝑎)2

2)

Carrying on like this we get

|𝑛(𝑥)| < 𝑚(1 + ||)𝑀(𝑥 − 𝑎) + |2|𝑀2 (𝑥−𝑎)2

2+ ⋯

1

𝑛!| |𝑛𝑀𝑛(𝑥 − 𝑎)𝑛)

Since this series here is of exponential type, we get convergence for all values of

Example 9:

(𝑥) = ∫ 1

0(𝑠)𝑑𝑠

Now we get 0

(𝑥) = 𝑥

1

(𝑥) = 𝑥 + ∫ 1

0(𝑠)𝑑𝑠



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= 𝑥 + ∫ 𝑠𝑑𝑠1

0

= 𝑥 + [𝑠2

2]

0

1

= 𝑥 +

2

2

(𝑥) = 𝑥 + ∫ (𝑠 +

2)𝑑𝑠

1

0

= 𝑥 + [𝑠2

2+s

2]

0

1

= 𝑥 + (1

2+

2)

= 𝑥 +

2+2

2

Continuing this way, 𝑛

(𝑥) = 𝑥 +1

2(+

2 + ⋯ + 𝑛)

4.2.2 Method of Differentiation

A Volterra equation with a simple separable Kernel can be solved by reducing it to a differential

equation.

Example 10:

Consider (𝑥) = 𝑥5 + ∫ 𝑥𝑠2𝑥

0(𝑠)𝑑𝑠

We can divide this through by 𝑥, so that the integral term does not depend on 𝑥, getting

(𝑥)

𝑥= 𝑥4 + ∫

𝑥

0

(𝑠)𝑠2𝑑𝑠

Differentiating w.r.t x gives

𝑑

𝑑𝑥((𝑥)

𝑥) = 4𝑥3 + 𝑥2(𝑥)

= 4𝑥3 + 𝑥3((𝑥)

𝑥)

which is a simple linear differential equation. We get

𝑑

𝑑𝑥((𝑥)

𝑥) 𝑒−

1

4𝑥4

= 4𝑥3𝑒−1

4𝑥4

And so (𝑥) = −4𝑥 + 𝑐𝑥𝑒−1

4𝑥4

This involves an arbitrary constant, whereas a Volterra integral equation has an unique solution. We

can evaluate the constant by going back to the integral equation. The condition that (𝑥) = 0 when 𝑥 =

0 tells us nothing, but if we use the fact that ∅(𝑥)

𝑥→ as 𝑥 → 0 we see that the solution is

(𝑥) = −4𝑥 + 4𝑥𝑒−1

4𝑥4



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5. Applications of Integral Equations

The study of integral equations has a significant role in both pure and applied field of mathematics.

The development of science has led to the formation of many physical laws, which when restated in

mathematical form, often appear as differential equations. Engineering problems can be mathematically

described by differential equations and thus differential equations play very important roles in the

solution of practical problems. For example, Newton’s law, stating that the rate of change of the

momentum of a particle is equal to the force acting on it, can be translated into mathematical language

as a differential equation. Similarly, problems arising in electric circuits, chemical kinetics and transfer

of heat in a medium can all be represented mathematically as differential equations. These differential

equations can be transformed to the equivalent integral equations of Volterra and Fredholm types. There

are many physical problems that are governed by integral equations and these equations can be easily

transformed to the differential equation.

Applications of some special integral equations are tested below.

5.1 Abel’s integral equation (𝑓(𝑥) = ∫ k(𝑥, 𝑡)(𝑡)𝑑𝑡𝑥

0)

This equation arises in the problem of finding the path of a particle which is constrained to

move under gravity in a vertical plane.

5.2 Vandrey’s equation (𝑉(𝑠) = (𝑠) −

𝜋∫ k(𝑠, 𝜎)𝑣(𝜎)𝑑𝜎; 0 ≤ 𝑠 ≤ 𝐿

𝐿

0)

This equation occurs in fluid dynamics while calculating the pressure distribution on the surface of the

body of revolution moving in a fluid.

5.3 Seismic Response of Dams

In order to analyze the safety and stability of an earth dam during an earthquake we need to

know the response of the dam to earthquake ground motion so that the inertia forces that will be

generated in the dam by the earthquake can be derived. Once the inertia forces are known, the safety

and the stability of the structure can be determined.

The inertia forces generated during an earthquake will depend on the geometry of the dam, the material

properties, the earthquake time history.

The reality of the problem is that an earth dam is a three dimensional structure. The material

properties are nonlinear inelastic and the earthquake time history is a time varying phenomenon. The

problem is therefore very complex and a proper solution requires the use of a finite element program,

which can deal with nonlinear inelastic material properties. The earthquake is the travelling wave

phenomenon, which arrives at the base of the dam through the foundation rock. Since the foundation is

not rigid, part of the energy, which vibrates the dam is lost through the foundation causing radiation

damping.

The analytical solution to such a problem is not at all possible. However, it is often necessary

to have approximate solutions that can be used to understand the behaviour of the dam during



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earthquakes. In order to make the problem amenable to analytical solutions, some approximations are

made to create a mathematical model. Such a model in this case is known as the shear beam model (SB)

of earth dams. Here we formulate a differential equation using the ideas from physics such as shear

modulus, strain, inertia, acceleration, momentum and so on. And we convert this differential equation

to an integral equation to get solution in an easier way.

5.5 Transverse Oscillations of a Homogenous Elastic Bar

Consider a homogeneous elastic bar with linear mass density d. Its axis coincides

with the segment (0, 𝑙) of the s axis when the bar is in its state of rest. It is clamped at

the end 𝑠 = 0, free at the end 𝑠 = 𝑙, and is forced to perform simple harmonic

oscillations with period 2𝜋𝜔⁄ , 𝜔 is the angular frequency.

s

𝑦(𝑠)

L

y

Figure 1: Transverse oscillations of a homogenous elastic bar

The problem is to find the deflection 𝑦(𝑠) that is parallel to the y axis and satisfies the system

of equations

𝑑4𝑦

𝑑𝑠4− 𝑘4𝑦 = 0, 𝑘4 =

𝜔2𝑑

𝐸𝐼

𝑦(0) = 𝑦 ,(0) = 0, 𝑦 ,,(𝑙) = 𝑦 ,,,(𝑙) = 0

Where 𝐸𝐼 is the bending rigidity of the bar

The above differential equation with initial conditions can be reduced to 𝑦 ,,(0) = 𝑐2 𝑦 ,,,(0) = 𝑐3, then

we obtain the required integral equation as

𝑔(𝑠) = 𝑘4 (𝑠2

2!𝑐2 +

𝑠3

3!𝑐3) + 𝑘4 ∫

(𝑠 − 𝑡)3

3!

𝑠

0

𝑔(𝑡)𝑑𝑡

Where, 𝑔(𝑠) =𝑑4𝑦

𝑑𝑠4

Then the solution 𝑦(𝑠) of the above differential equation is given by

𝑦(𝑠) = ∫(𝑠 − 𝑡)3

3!

𝑠

0

𝑔(𝑡)𝑑𝑡 +𝑠3

3!𝑐3 +

𝑠2

2!𝑐2



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Here we can find the constants 𝑐2 and 𝑐3 from given conditions. Hence we can transform given

differential equation with initial conditions in to an integral equation and obtain the solution easily.

6. Conclusion

This study briefly explains the different concepts associated with an integral equation,

especially linear integral equation. It mainly points towards linear integral equations such as Fredholm

and Volterra integral equations. There are a number of methods to solve the two types of integral

equations. We have discussed a few among them in this study. One obvious reason for using integral

equations rather than differential equations is that all of the conditions specifying the initial value

problem or boundary value problem for a differential equation can often be condensed into a single

integral equation.

Linear integral equations have undoubtedly a wide range of applications in all fields of science.

In this work, we have discussed its importance in the study of seismic response of dams and transverse

oscillations of a homogeneous elastic bar. Although the idea of linear integral equations can be extended

to a wide range of real-world problems such as to study interaction of ocean waves, Flow of heat in a

metal bar etc.

References;

1. Rahman M., Integral equation and their Applications, WIT press Southampton, Boston, 2007

2. Ram P. Kanwal, Linear Integral Equations Theory and Technique, Academic press New

York,1971.

3. James S. Scarborough, Numerical Mathematical Analysis, 6th edition.

4. Kreyzig Erwin, Introductory functional analysis with applications, John Wiley & Sons, 1978.



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