INTEGRAL EQUATIONS 8th Semester€¦ · Integral equations play an important role in the theory of...

INTEGRAL EQUATIONS8th Semester

DEPARTMENT OF MATHEMATICSUNIVERSITY OF MALAKAND

LECTURER 1

Preliminaries

• Integral Equations: An integral equation is an equation in which the unknown function u(x) appears under an integral sign as given by

Integro-Differential Equation:

Kinds of Integral Equations:There are several kinds integral equations in literature. Here we define some famous of them as:

1. Volterra Integral Equation: In Volterra integral equations, at least one of the limits of integration is avariable. For the first kind Volterra integral equations, the unknown function u(x) appears only inside integral sign in the form:

Similarly Volterra Integral Equations of second is given as:

2. Fredholm Integral Equations: The first kind integral of Fredholm is given

second kind Fredholm integral equation is:

Mixed Type Integral Equations: Some time both Volterra and Fredholm IEs are mixed as:

Singular Integral Equations :

• Volterra integral equations of the first kind :

Abel Type singular Integral Equations:

Exercise 1:

Lecturer No:2 On Applications of integral equations

Integral equations play an important role in the theory of ordinary and partial differential equationsand boundary value problems. The reduction of boundary value problems to integral equations allows forthe application of iteration and finite-difference methods of solving integral equations. These methodsare, as a rule, substantially simpler than those used for solving differential equations. Moreover, manydelicate proofs and qualitative results of the theory of differential equations have been obtained by theinvestigation of the corresponding integral equations.

0.1. Reduction of the Cauchy Problem for ODEs to Integral Equations. In this portion, wewill study the Cauchy Problem for First-Order ODEs and also the uniqueness and Existence Theorems.The Cauchy problem: find a solution of the equation

y′x= f(x, y), (0.1)

that satisfies the initial condition

y(x0) = y0 (0.2)

for given y0 and x0. Geometrical meaning of the Cauchy problem: find an integral curve of equation(0.1) passing through the point (x0, y0).

Theorem 0.1. (EXISTENCE). Let the function f(x, y) be continuous in an open domain D of thexy-plane. Then there is at least one integral curve of equation (0.1) that passes through each point(x0, y0) ∈ D, each of these curves can be extended at both ends up to the boundary of any closed domainD0 ⊂ D such that (x0, y0) belongs to the interior of D0.

Theorem 0.2. (UNIQUENESS). Let the function f(x, y) be continuous in an open domain D and havea bounded partial derivative in D with respect to y (or satisfy the Lipschitz condition: |f(x, y)–f(x, z)| ≤M |y–z|, where M > 0 is a constant). Then there is a unique solution of equation (0.1) satisfyingcondition (0.2).

0.2. Cauchy Problem for First-Order ODEs. Method of Successive Approximations. Inthis section, we will study the method of successive approximations (the Picard method) consists of twostages. On the first stage, the Cauchy problem (0.1)–(0.2) is reduced to the equivalent integral equation:

y(x) = y0 +

∫

x

x0

f(t, y(t))dt (0.3)

Then a solution of equation (0.3) is sought using the formula of successive approximations:

yn+1(x) = y0 +

∫

x

x0

f(t, yn(t))dt, n = 0, 1, 2, ...

The initial approximation y0(x) can be chosen arbitrarily the simplest way is to take y0 a constant. Theiterative process converges as n → ∞, provided the assumptions of the theorems in Subsection aboveare satisfied.

0.3. Cauchy Problem for Second-Order ODEs. Method of Successive Approximations. Inthis subsection, we will study the method of successive approximations is implemented in two steps.First, the Cauchy problem

y′′xx =f(x, y, y′x), (equation),

y(x0) =y0, y′x(x0) = y′0, (initial conditions),

1

Kamal Shah

Highlight

2

is reduced to an equivalent system of integral equations by the introduction of the new variable u(x) = y′x.These integral equations have the form

u(x) =y0 +

∫

x

x0

f(t, y(t), u(t))dt,

y(x) =y0 +

∫ x

x0

u(t)dt.

(0.4)

Then the solution of system (0.4) is sought by means of successive approximations defined by thefollowing recurrence formulas:

un+1(x) =y0 +

∫ x

x0

f(t, yn(t), un(t))dt,

yn+1(x) =y0 +

∫ x

x0

un(t)dt, n = 0, 1, 2, ...

As the initial approximation, one can take y0(x) = y0 and u0(x) = y′0. The iterative process convergesas n → ∞, under assumptions similar to those formulated in the theorems of Subsection 1.1.Remark. In a similar way, the Cauchy problem for an nth order ODE can be reduced to a system ofintegral equations. Consider the Cauchy problem for the following linear nth order ODE:

ynx + fn−1(x)yn−1x + ...+ f1(x)y

′ + f0(x)y = g(x) (0.5)

with the homogeneous initial conditions at the point x = a:

y(a) = y′x(a) = · · · = y(n–1)x (a) = 0. (0.6)

Introducing a new unknown function by

y(x) =1

(n− 1)!

∫

x

a

(x− t)n−1u(t)dt, (0.7)

and differentiating (0.7) n times, we get

y(k)x

(x) =1

(n− k − 1)!

∫

x

a

(x− t)n−k−1u(t)dt, k = 1, 2, ..., n− 1,

y(n)x (x) =u(x).

(0.8)

Obviously, the function (0.7) satisfies the initial conditions (0.6). Substituting (0.8) into the left-handside of equation (0.5), we obtain

u(x) +

∫ x

a

K(x, t)u(t)dt = g(x), (0.9)

where

K(x, t) = fn−1(x) + fn−2(x)x− t

1!+ · · ·+ f0(x)

(x − t)n−1

(n− 1)!. (0.10)

Thus, the Cauchy problem (0.5)–(0.6) has been reduced to the integral equation (0.9)–(0.10), which isa Volterra equation of the second kind. Finding the function u(x) from (0.9) and using formula (0.7)we obtain the desired solution y(x).Remark. The Cauchy problem for equation (0.5) with nonhomogeneous boundary conditions

y(a) = b0, y′

x(a) = b1, · · ·, y

n−1x

(a) = bn−1

can be reduced to a Cauchy problem with homogeneous boundary conditions for another function w(x)with the help of the substitution

y(x) = w(x) +

n−1∑

k=1

bk(x − a)k

k!.

3

1. Reduction of boundary value problems for odes to volterra integral equations.

Calculation of eigenvalues

In this section, we will study the reduction of Differential Equations to Volterra Integral Equations.Consider a linear nonhomogeneous ODE for the function y = y(x):

Ln[y] = h(x)(a < x < b), (1.1)

where

Ln[y] =n∑

k=0

fk(x)y(k)x

, fn(x) 6= 0. (1.2)

Let ϕ1(x), · · ·, ϕn(x) be a fundamental system of solutions of the truncated homogeneous equation

Ln[ϕ] = 0. (1.3)

Denote by W (x) its Wronskian determinant.

W (x) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

ϕ1(x) · · · ϕn(x)ϕ′

1(x) · · · ϕ′

n(x)· · · · · · · · ·

ϕ(n−2)1 (x) · · · ϕ

(n−2)n (x)

ϕ(n−1)1 (x) · · · ϕ

(n−1)n (x)

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

,

and by Wν(x) the determinant obtained from W (x) by replacing the νth column by 0, · · ·, 0, h(x). Thegeneral solution of equation (1.1) can be written in the form

y(x) =

n∑

ν=1

ϕ(x)

∫

x

a

Wν(ξ)

fn(ξ)W (ξ)dξ +

n∑

ν=1

Cνϕν(x), (1.4)

where the first sum is a particular solution of equation (1.1), the second sum is the general solutionof the homogeneous equation (1.3), and Cν are arbitrary constants. For boundary value problems, theconstants Cν are found from the corresponding boundary conditions, and for the Cauchy problem, Cν

are obtained from the initial conditions.Consider the linear ODE for the function y = y(x) with a parameter λ:

Ln[y] = h(x)–λg(x)y (a < x < b), (1.5)

where Ln is the differential operator (1.2). Equation (1.5) differs from (1.1) only by an additional termin the right-hand side. Therefore, replacing the function h(x) by h(x)–λg(x)y(x) in the solution (1.4)and performing simple transformations, we come to the Volterra integral equation

y(x) + λ

∫

x

a

K(x, ξ)y(ξ)dξ = F (x). (1.6)

Where

K(x, ξ) =g(ξ)

fn(ξ)W (ξ)D(x, ξ), F (x) =

∫ x

a

h(ξ)

fn(ξ)W (ξ)D(x, ξ)dξ +

n∑

ν=1

Cνϕν(x), (1.7)

and

D(x, ξ) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

ϕ1(ξ) · · · ϕn(ξ)ϕ′

1(ξ) · · · ϕ′

n(ξ)· · · · · · · · ·

ϕ(n−2)1 (ξ) · · · ϕ

(n−2)n (ξ)

ϕ1(x) · · · ϕn(x)

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

. (1.8)

The Volterra integral equation (1.6)–(1.8) is equivalent to the differential equation (1.5).

4

1.1. Application of Volterra Equations to the Calculation of Eigenvalues. In this section, wewill study the application of Volterra equations to find the Eigenvalues.The Volterra integral equation (1.6) can be used for the calculation of the smallest eigenvalue and thecorresponding eigenfunction of various boundary value problems for the ODE (1.5). For this purpose,one utilizes the method of successive approximations: the first term y(x) of the integral equation isreplaced by yn(x) and y(ξ) in the integrand is replaced by yn−1(ξ). On each step, the parameter λ

is chosen such that the function yn(x) would satisfy the boundary conditions. This procedure can beillustrated by the following example.

Example 1.1. Consider the equation

y′′xx + λg(x)y = 0 (0 < x < 1) (1.9)

with the homogeneous boundary conditions of the first kind

y(0) = y(1) = 0. (1.10)

Equation (1.9) is a special case of (1.5) for n = 2, L2[y] = y′′xx, h(x) = 0, a = 0, b = 0. The fundamental

system of solutions of the truncated equation L2[ϕ] = 0 has the form

ϕ1(x) = 1, ϕ2(x) = x. (1.11)

Simple transformations with the help of (1.8) yield

W (x) = ϕ1(x)[ϕ2(x)]′

x − ϕ2(x)[ϕ1(x)]′

x = 1,

D(x, ξ) = ϕ1(ξ)ϕ2(x)–ϕ1(x)ϕ2(ξ) = x− ξ.(1.12)

Substituting (1.11)–(1.12) into (1.6)–(1.7), we come to the Volterra equation

y(x) = C1 + C2x− λ

∫ x

0

(x− ξ)g(ξ)y(ξ)dξ. (1.13)

From the first boundary condition in (1.10), we get C1 = 0. Since eigenfunctions are defined to withina constant coefficient, we can take C2 = 1 in (1.13). As a result we get

y(x) = x− λ

∫ x

0

(x − ξ)g(ξ)y(ξ)dξ. (1.14)

This equation can be solved by the method of successive approximations based on the formula

yn(x) = x− λ

∫

x

0

(x− ξ)g(ξ)yn−1(ξ)dξ, n = 1, 2.... (1.15)

Next, consider more closely the simplest case g(x) = 1. As the zero approximation, we take y0 = 1 andfind that

y1(x) = x− λ

∫

x

0

(x − ξ)dξ = x−1

2λx2. (1.16)

From the second boundary condition in (1.10), we get y1(1) = 0, and therefore, λ = λ1 = 2. It followsthat

y1(x) = x− x2.

Let us insert this function into the right-hand side of (1.16), where g(x) = 1. We have

y2(x) = x− λ

∫

x

0

(x− ξ)(ξ − ξ2)dξ = x− λ

(

1

6x3 −

1

12x4

)

.

Satisfying the second boundary condition in (1.10), i.e., y2(1) = 0, we obtain

λ2 = 12, y2(x) = x–2x3 + x4.

In a similar way, we find that

λ3 = 10, y3(x) = x−5

3x3 + 3x5 −

1

3x6. (1.17)

5

The exact smallest eigenvalue for g(x) = 1 is equal to λ = π2 ≈ 9.87, and the corresponding eigenfunctionhas the form

y(x) =1

πsin(πx) ≈ x− 1.6x3 + 0.8x5.

In the case under consideration, the choice of the initial approximation, y0(x) = 1, was not quitegood, since both boundary conditions in (1.10) do not hold for this function. The convergence ratemay be increased by taking the initial approximation of the form y0(x) = x–x2, in which case bothboundary conditions in (1.10) are satisfied. For an arbitrary continuous function g(x), in the absence ofinformation about eigenfunctions,it is convenient to take the initial approximation in (1.16) of the form

y0(x) = x–x2 or y0(x) =1

πsin(πx),

since these functions satisfy the boundary conditions (1.10). In the special case of g(x) = 1, the firstinitial function ensures fast convergence of the expression (1.16) to the exact result, and the secondyields the exact result immediately.

Lecturer 03: Conversion of IVP to Volterra Integral Equations

Integral equations appear in many forms. Two distinct ways that depend on the limits of integrationare used to characterize integral equations, namely:

(1) If the limits of integration are fixed, the integral equation is called a Fredholm integral equationgiven in the form:

u(x) = f(x) + λ∫ b

aK(x, t)u(t)dt

where a and b are constants.(2) If at least one limit is a variable, the equation is called a Volterra integral equation given in the

form:u(x) = f(x) + λ

∫ x

aK(x, t)u(t)dt

Moreover, two other distinct kinds, that depend on the appearance of the unknown function u(x), aredefined as follows:

(1) If the unknown function u(x) appears only under the integral sign of Fredholm or Volterraequation, the integral equation is called a first kind Fredholm or Volterra integral equationrespectively.

(2) If the unknown function u(x) appears both inside and outside the integral sign of Fredholm orVolterra equation, the integral equation is called a second kind Fredholm or Volterra integralequation respectively.

It is interesting to point out that any equation that includes both integrals and derivatives of theunknown function u(x) is called integro-differential equation. The Volterra integro-differential equationis of the form:

un(x) = f(x) + λ

∫ x

0

K(x, t)u(t)dt. (0.1)

where un(x) indicates the nth derivative of u(x). Volterra integro-differential equations appear when weconvert initial value problems to integral equations. The Volterra integro-differential equation containsthe unknown function u(x) and one of its derivatives un(x) inside and outside the integral sign. Itis important to note that initial conditions should be given for Volterra integro-differential equationsto determine the particular solutions. Examples of Volterra integral equation and Volterra integrodifferential equations are given by

Example 0.1.

xe−x =

∫ x

0

et−xu(t)dt, (0.2)

Example 0.2.

5x2 + x3 =

∫ x

0

(5 + 3x− 3t)u(t)dt, (0.3)

Example 0.3.

u′(x) = −1 +1

2x2

− xex −

∫ x

0

tu(t)dt, u(0) = 0 (0.4)

Example 0.4.

u′′(x) + u′(x) = 1− x(sin(x) + cos(x)) −

∫ x

0

tu(t)dt, u′(0) = 0. (0.5)

0.1. Converting IVP to Volterra Integral Equation. we will study the technique that will con-vert an initial value problem (IVP) to an equivalent Volterra integral equation and Volterra integro-differential equation as well. Consider the general initial value problem:

yn(x) + a1(x)yn−1 + ...+ an−1(x)y

′ + an(x)y = g(x), (0.6)1

Kamal Shah

Highlight

2

subject to the initial conditions

y(0) = c0, y′(0) = c1, y

′′(0) = c2, ..., yn−1(0) = cn−1.

We assume that the functions ai(x), 1 ≤ i ≤ n are analytic at the origin, and the function g(x) iscontinuous through the interval of discussion. Let u(x) be a continuous function such that

yn(x) = u(x).

Integrating both sides with respect to x, we get

yn−1(x) = cn−1 +

∫ x

0

u(t)dt.

Integrating again both sides with respect to x, we get

yn−2(x) =cn−2 + cn−1x+

∫ x

0

∫ x

0

u(t)dtdt,

=cn−2 + cn−1(x) +

∫ x

0

(x− t)u(t)dt.

Proceeding by the same way, we have

yn−3(x) =cn−3 + cn−2x+1

2cn−1x

2 +

∫ x

0

∫ x

0

∫ x

0

u(t)dtdtdt,

=cn−3 + cn−2x+1

2cn−1x

2 +1

2

∫ x

0

(x− t)2u(t)dt.

Continuing the integration process leads to

y(x) =

n−1∑

k=0

ck

k!xk +

1

(n− 1)!

∫ x

0

(x− t)n−1u(t)dt.

Substituting the above values in (0.6), we get

u(x) = f(x)−

∫ x

0

K(x, t)u(t)dt, (0.7)

where

K(x, t) =n∑

k=1

an

(n− 1)!(x− t)k−1,+

1

(n− 1)!

∫ x

0

(x− t)n−1u(t)dt,

and

f(x) = g(x)−

n∑

j=1

aj

(

j∑

k=1

an−k

(j − k)!(x)j−k

)

.

Volterra integro-differential equation can be obtained by differentiating (0.7) as many times as we like.For simplicity, we will apply the above process to a second order initial value problem given by

y′′(x) + p(x)y′ + q(x)y(x) = g(x) (0.8)

subject to the initial conditions

y(0) = α, y′(0) = β,

where α and β are constants.The functions p(x) and q(x) are analytic functions, and g(x) is continuous.Let

y′′(x) = u(x),

where u(x) is a continuous function. Integrating both sides with respect to x, we get

y′(x)− y′(0) =

∫ x

0

u(t)dt, (0.9)

3

or equivalently

y′(x) = β +

∫ x

0

u(t)dt. (0.10)

Integrating it again with respect to x, we get

y(x)− y(0) = βx +

∫ x

0

∫ x

0

u(t)dtdt,

or equivalently

y(x) = α+ βx +

∫ x

0

(x − t)u(t)dt. (0.11)

Substituting (0.9), (0.10) and (0.11) in (0.8) yield the Volterra integral equation:

u(x) + p(x)

[

β +

∫ x

0

u(t)dt

]

+ q(x)

[

α+ βx+

∫ x

0

(x− t)u(t)dt

]

= g(x). (0.12)

The (0.12) can be written in standard Volterra integral equation form:

u(x) = f(x)−

∫ x

0

K(x, t)u(t)dt, (0.13)

whereK(x, t) = p(x) + q(x)(x − t),

andf(x) = g(x) − [βp(x) + αq(x) + βxq(x)].

In order to obtain Volterra integro-differential equation, differentiate (0.13) with respect to x, we willget the required Volterra integro-differential equation, which will be in the form of

u′(x) +K(x, x)u(x) = f ′(x)−

∫ x

0

∂K(x, t)

∂xu(t)dt, u0 = f(0) (0.14)

Example 0.5. Convert the following initial value problem to an equivalent Volterra integral equation:

y′(x)− 2xy(x) = ex2

, y0 = 1. (0.15)

We first sety′(x) = u(x). (0.16)

Integrating (0.15), we get

y(x)− y(0) =

∫ x

0

u(t)dt,

or equivalently

y(x) = 1 +

∫ x

0

u(t)dt, (0.17)

substituting (0.16) and (0.17) into (0.15) gives the equivalent Volterra integral equation

u(x) = 2x+ ex2

+ 2x

∫ x

0

u(t)dt.

0.2. Assignment. Convert each of the following IVPs to an equivalent Volterra integral equation:

(1) y′ − 4y = 0, y(0) = 0(2) y′′ + 4y = 0, y(0) = 0, y′(0) = 1(3) y′′ − 6y + 8y = 1, y(0) = 1, y′(0) = 1.

Convert each of the following Volterra integral equation to an equivalent IVP

(1) u(x) = x+∫ x

0u(t)dt

(2) u(x) = 1 + x2 +∫ x

0(x− t)u(t)dt

(3) u(x) = 1 + 2∫ x

0(x − t)3u(t)dt.

Lecture 04: Solving Volterra Integral Equations of the Second Kind by Laplace

transform method

0.1. Introduction. We will first study Volterra integral equations of the second kind given by

u(x) = f(x) + λ

∫x

a

K(x, t)u(t)dt (0.1)

depends on the difference x − t, then it is called a difference kernel. Examples of the difference kernelare ex−t ,cos(x− t), and x− t.

0.2. The Laplace Transform Method. The Laplace transform method is a powerful technique thatcan be used for solving initial value problems and integral equations as well.The integral equation canthus be expressed as

u(x) = f(x) + λ

∫x

a

K(x− t)u(t)dt (0.2)

we will examine specific Volterra integral equations where the kernel is a difference kerne By takingLaplace transform of both sides of (0.2), we get

U(x) = F (s) + λK(s)U(s), (0.3)

where U(s) = Lu(x), K(s) = LK(x), F (s) = Lf(x).Therefore(0.3) implies that

U(s) =F (s)

1− λK(s), λK(s) 6= 1. (0.4)

By taking the inverse transform of (0.4), we get

u(x) = L−1F (s)

1− λK(s)(0.5)

Solve the Volterra integral equation by using the Laplace transform method

Example 0.1.

u(x) = 1 +

∫x

0

u(t)dt. (0.6)

Here K(x− t) = 1, λ = 1. Taking laplace transform of both sides of (0.6), we get

L{u(x)} = L{1}+ L{1 ∗ u(x)dx},

so that

U(s) =1

s+

1

sU(s),

or equivalently

U(s) =1

s− 1(0.7)

By taking the inverse Laplace transform of both sides of (0.7) the exact solution is therefore given by

u(x) = ex

Example 0.2.

u(x) = 1−

∫x

0

(x− t)u(t)dt. (0.8)

Here K(x− t) = (x− t), λ = −1. Taking laplace transform of both sides of (0.8), we get

L{u(x)} = L{1} − L{(x− t) ∗ u(x)dx},

so that

U(s) =1

s−

1

s2U(s),

1

Kamal Shah

Highlight

2

or equivalently

U(s) =1

s2 + 1(0.9)


u(x) = cos(x).

Example 0.3.

u(x) = sin(x) + cos(x) +

∫x

0

sin(x− t)u(t)dt. (0.10)

Applying the linear property of the Laplace transforms, we have

L{u(x)} = L{sin(x) + cos(x)} + 2L{sin(x − t) ∗ u(x)dx},

so that

U(s) =1

s2 + 1+

s

s2 + 1+

2

s2 + 1U(s),

or equivalently

U(s) =1

s− 1(0.11)


u(x) = ex.

0.3. Assignment. Use the Laplace transform method to solve the Volterra integral equations:

(1) u(x) = x+∫x

0(x− t)u(t)dt

(2) u(x) = x− 1 +∫x

0(x− t)u(t)dt

(3) u(x) = sin(x) + cosh(x) + sinh(x) − 2∫x

0cos(x− t)u(t)dt

(4) u(x) = sinh(x) + cosh(x) + cos(x) − 2∫x

0cos(x − t)u(t)dt

(5) u(x) = sin(x)− cos(x) + cosh(x) − 2∫x

0cos(x − t)u(t)dt

Lecture 05: Solving Volterra Integral Equations of the First Kind by Laplace

transform method

0.1. Introduction. The standard form of the Volterra integral equations of the first kind is given by

f(x) =

∫

x

0

K(x, t)u(t)dt (0.1)

where the kernel K(x, t) and the function f(x) are given real-valued functions, and u(x) is the functionto be determined. In this section we will discuss Laplace transform method that are commonly used forhandling the Volterra integral equations of the first kind.

0.2. The Laplace Transform Method. The Laplace transform method is a powerful technique thatwe used before for solving initial value problems and Volterra integral equations of the second kind

f(x) =

∫

x

0

K(x, t)u(t)dt (0.2)

By taking Laplace transform of both sides of (0.2), we get

F (s) = K(s)U(s), (0.3)

where U(s) = Lu(x), K(s) = LK(x), F (s) = Lf(x).Therefore(0.3) implies that

U(s) =F (s)

K(s),K(s) 6= 0. (0.4)

By taking the inverse transform of (0.4), we get

u(x) = L−1F (s)

K(s)(0.5)

Solve the Volterra integral equation of the first kind by using the Laplace transform method

Example 0.1.

ex − sin(x) − cos(x) =

∫

x

0

2ex−tu(t)dt. (0.6)

Taking laplace transform of both sides of (0.6), we get

1

s− 1−

1

s2 − 1−

s

s2 − 1=

2

s− 1U(s),

or equivalently2

(s− 1)(s2 + 1)=

2

s− 1U(s).

This gives us

U(s) =1

s2 + 1.

Taking the inverse Laplace transform of both sides gives the exact solution

u(x) = sin(x). (0.7)

Solve the Volterra integral equation of the first kind by using the Laplace transform method

Example 0.2.

− 1 + x2 +1

6x3 + 2 sinh(x) + cosh(x) =

∫

x

0

(x− t+ 2)u(t)dt. (0.8)

Taking laplace transform of both sides of (0.8), we get

s3 + s2 − 1

s2(s2 − 1)=

(

1

s2+

2

s

)

U(s),

1

Kamal Shah

Highlight

2

or equivalently

U(s) =1

s2+

s

s2 − 1. (0.9)

Taking the inverse Laplace transform of both sides gives the exact solution

u(x) = x+ cosh(x). (0.10)

0.3. Assignment. Use the Laplace transform method to solve the Volterra integral equations of thefirst kind:

(1) x− sin(x) =∫

x

0(x − t)u(t)dt

(2) 1 + x− sin(x)− cos(x) =∫

x

0(x− t)u(t)dt

(3) sinh(x) =∫

x

0ex−tu(t)dt

(4) 1− cos(x) =∫

x

0cos(x− t)u(t)dt

(5) x =∫

x

0cos(x− t+ 1)u(t)dt

Lecturer 06: Study on Fredholm Integral Equations

0.1. Introduction. Integral equations appear in many forms. If the limits of integration are fixed, theintegral equation is called a Fredholm integral equation.

0.2. Fredholm integral equation of the first kind. If the unknown function u(x) appears onlyunder the integral sign of Fredholm equation, the integral equation is called a first kind Fredholmintegral equation. The first kind is represented by the form:

f(x) =

∫

b

a

K(x, t)u(t)dt (0.1)

0.3. Fredholm integral equation of the second kind. If the unknown function u(x) appears bothinside and outside the integral sign of Fredholm equation, the integral equation is called a second kindFredholm integral equation.The second kind is represented by the form:

u(x) = f(x) + λ

∫ b

a

K(x, t)u(t)dt. (0.2)

Examples of the two kinds are given by

Example 0.1.

sin(x)− x cos(x)

x2=

∫ 1

0

sin(xt)u(t)dt, (0.3)

Example 0.2.

u(x) = x+1

2

∫ 1

−1

(x− t)u(t)dt, (0.4)

0.4. Converting BVP to Fredholm Integral Equation. we will present a method that will converta boundary value problem to an equivalent Fredholm integral equation. The technique requires morework if compared with the initial value problems when converted to Volterra integral equations. Wewill present two specific distinct boundary value problems (BVPs) to derive two distinct formulas thatcan be used for converting BVP to an equivalent Fredholm integral equation.Type 1. We first consider the following boundary value problem:

y′′(x) + g(x)y(x) = h(x), 0 < x < 1, (0.5)

with the boundary conditions:

y(0) = α, y(1) = β.

Consider

y′′(x) = u(x), (0.6)

integrate (0.6) with respect to x, we obtain∫

x

0

y′′(t)dt =

∫

x

0

u(t)dt,

that gives

y′(x) = y′(0) +

∫

x

0

u(t)dt, (0.7)

where the condition y′(0)is not given in a boundary value problem. The condition y′(0) will be deter-mined later by using the boundary condition at x = 1. Integrating both sides of (0.7), we get

y(x) = y(0) + xy′(0) +

∫ x

0

∫ x

0

u(t)dtdt,

1

Kamal Shah

Highlight

2

or equivalently

y(x) = α+ xy′(0) +

∫ x

0

(x− t)u(t)dt, (0.8)

now we have to find y′(0), for this substitute x = 1 in (0.8), we get

y(1) = α+ y′(0) +

∫ 1

0

(1 − t)u(t)dt, (0.9)

which gives us

β = α+ y′(0) +

∫ 1

0

(1− t)u(t)dt.

This in turn gives

y′(0) = (β − α)−

∫

1

0

(1 − t)u(t)dt, (0.10)

putting (0.10) in (0.8), we get

y(x) = α+ (β − α)x−

∫

1

0

x(1 − t)u(t)dt+

∫

x

0

(1− t)u(t)dt, (0.11)

substituting (0.6) and (0.11) in (0.5), we get

u(x) + αg(x) + (β − α)xg(x) −

∫

1

0

xg(x)(1 − t)u(t)dt+

∫

1

0

g(x)(x − t)u(t)dt = h(x), (0.12)

which implies that

u(x) = h(x)−αg(x)−(β−α)xg(x)−g(x)

∫

x

0

(x−t)u(t)dt+xg(x)

[∫

x

0

(1− t)u(t)dt+

∫

1

0

(1− t)u(t)dt

]

,

which gives

u(x) = f(x) +

∫ x

0

t(1− x)g(x)u(t)dt +

∫ 1

x

x(1 − t)g(x)u(t)dt,

which gives us the required Fredholm integral equation

u(x) = f(x) +

∫

1

0

K(x, t)u(t)dt,

where

f(x) = h(x) − αg(x)− (β − α)xg(x)

and K(x, t) is given by

K(x, t) =

{

t(1− x)g(x), 0 ≤ t ≤ x,

x(1 − t)g(x), x ≤ t ≤ 1,(0.13)

The techniques presented above will be illustrated by the following two examples

Example 0.3. Convert the following BVP to an equivalent Fredholm integral equation:

y′′(x) + 9y(x) = cos(x), y(0) = y1 = 0. (0.14)

Here α = β = 0, g(x) = 9 and h(x) = cos(x), which gives

f(x) = cos(x). (0.15)

Therefore the required Fredholm integral equation will be

u(x) = cos(x) +

∫ 1

0

K(x− t)u(t)dt, (0.16)

3

where

K(x, t) =

{

9t(1− x), 0 ≤ t ≤ x,

9x(1− t), x ≤ t ≤ 1,(0.17)

Type 2. We next consider the following boundary value problem:

y′′(x) + g(x)y(x) = h(x), 0 < x < 1, (0.18)

with the boundary conditions:y(0) = α1, y′(1) = β1.

we set againy′′(x) = u(x), (0.19)

integrating both sides of (0.19), we get∫ x

0

y′′(t)dt =

∫ x

0

u(t)dt,

which gives

y′(x) = y′(0)

∫

x

0

u(t)dt. (0.20)

The condition y′ is not given, it will be derived later by using the boundary condition at y′(1) = β1.Integrating both sides of (0.20), we get

y(x) = y(0) + xy′(0) +

∫

x

0

∫

x

0

u(t)dtdt,

or equivalently

y(x) = α1 + xy′(0) +

∫ x

0

(x− t)u(t)dt. (0.21)

To determined y′(0), we first differentiate (0.21), with respect to x, we get

y′(x) = y′(0) +

∫

x

0

u(t)dt,

using boundary condition y′(1) = β1, we get

y′(0) = β1 −

∫

1

0

u(t)dt, (0.22)

using (0.22) into (0.21) gives

u(x) + α1g(x) + β1xg(x)−

∫

1

0

xg(x)u(t)dt+

∫

x

0

g(x)(x − t)u(t)dt = h(x), (0.23)

which implies that

u(x) = h(x)− (α1 + β1x)g(x) + xg(x)

[∫

x

0

u(t)dt+

∫

1

x

u(t)dt

]

− g(x)

∫

x

0

(x− t)u(t)dt, (0.24)

The (0.24) can be written as

u(x) = f(x) +

∫

x

0

tg(x)u(t)dt +

∫

1

0

xg(x)u(t)dt,

which gives the Fredholm integral equation

u(x) = f(x) +

∫

1

0

K(x, t)u(t)dt,

wheref(x) = h(x)− (α1 + β1x)g(x),

4

and

K(x, t) =

{

tg(x), 0 ≤ t ≤ x,

xg(x), x ≤ t ≤ 1,(0.25)

The second type of conversion that was presented above will be illustrated by the following example.


y′′(x) + y(x) = 0, y(0) = y′(1) = 0. (0.26)

Here α1β1 = 0, g(x) = 1 and h(x) = 0. This in turn gives

f(x) = 0.

Substituting these value in the general form of Fredholm integral equation, we get

u(x) =

∫

1

0

K(x, t)u(t)dt,

where

K(x, t) =

{

t, 0 ≤ t ≤ x,

x, x ≤ t ≤ 1,(0.27)


y′′(x) + 2y(x) = 0, y(0) = 0, y′(1) = 1. (0.28)

Here α1 = 0, β1 = 1, g(x) = 2 and h(x) = 4. This in turn gives

f(x) = 4− 2x.

Substituting these value in the general form of Fredholm integral equation, we get

u(x) = 4− 2x+

∫ 1

0

K(x, t)u(t)dt,

where

K(x, t) =

{

2t, 0 ≤ t ≤ x,

2x, x ≤ t ≤ 1,(0.29)

0.5. Assignment. Convert each of the following BVPs to an equivalent Fredholm integral equation:

(1) y′′ + 4y = 0, y(0) = y(1) = 0(2) y′′ + 2y = x, 0 < x < 1, y(0) = 1, y(1) = 0(3) y′′ + 4xy = 2, 0 < x < 1, y(0) = 1, y′(1) = 0

Lecture 07: Solving the Fredholm Integral Equations by Adomian Decomposition

Method

0.1. Introduction. We will first study Fredholm integral equations of the second kind given by

u(x) = f(x) + λ

∫ b

a

K(x, t)u(t)dt. (0.1)

The unknown function u(x), that will be determined, occurs inside and outside the integral sign.Thekernel K(x, t) and the function f(x) are given real-valued functions, and λ is a parameter.

The Adomian decomposition method consists of decomposing the unknown function u(x) of anyequation into a sum of an infinite number of components defined by the decomposition series

u(x) =

∞∑

n=0

un(x), (0.2)

or equivalently

u(x) = u0(x) + u1(x) + u2(x) + ..., (0.3)

where the components un(x), n ≥ 0 will be determined recurrently. The Adomian decomposition methodconcerns itself with finding the components u0(x) + u1(x) + u2(x) + ... individually. Determination ofthese components can be achieved in an easy way through a recurrence relation that usually involvessimple integrals that can be easily evaluated.

To establish the recurrence relation, we substitute (0.4) into the Fredholm integral equation, hence

∞∑

n=0

un(x) = f(x) + λ

∫ b

a

K(x, t)

(

∞∑

n=0

un(x)

)

dt, (0.4)

or equivalently

u0(x) + u1(x) + u2(x) + ... = f(x)λ

∫ b

a

K(x, t)[u0(x) + u1(x) + u2(x) + ...]dt. (0.5)

The zeroth component u0(x) is identified by all terms that are not included under the integral sign.This means that the components uj(x), j ≥ 0 of the unknown function u(x) are completely determinedby setting the recurrence relation

u0(x) = f(x), un+1(x) = λ

∫ b

a

K(x, t)un(t)dt, n ≥ 0,

or equivalently

u0(x) = f(x),

u1(x) = λ

∫ b

a

K(x, t)u0(t)dt,

u2(x) = λ

∫ b

a

K(x, t)u1(t)dt,

u3(x) = λ

∫ b

a

K(x, t)u2(t)dt,

(0.6)

and so on for other components.In view of (0.6), the components u0(x), u1(x), u2(x), ... are completely determined. As a result, the

solution u(x) of the Fredholm integral equation (0.1) is readily obtained in a series form by using theseries assumption.

1

Kamal Shah

Highlight

2

Example 0.1. Solve the following Fredholm integral equation

u(x) = x+ ex −4

3+

∫ 1

0

tu(t)dt. (0.7)

Substituting the decomposition series into both sides of (0.7) gives∞∑

n=0

un(x) = x+ ex −4

3+

∫ 1

0

t

∞∑

n=0

un(x)dt,

or equivalently

u0(x) + u1(x) + u2(x) + ... = x+ ex −4

3+

∫ 1

0

t[u0(t) + u1(t) + u2(t) + ...]dt. (0.8)

Proceeding as before, we set the following recurrence relation

u0(x) = x+ ex −4

3, uk+1(x) =

∫ 1

0

tuk(t)dt, k ≥ 0

Consequently, we obtain

u0(x) = x+ ex −4

3,

u1(x) =

∫ 1

0

tu0(t)dt =2

3,

u2(x) =

∫ 1

0

tu1(t)dt =1

3,

u3(x) =

∫ 1

0

tu2(t)dt =1

6,

u4(x) =

∫ 1

0

tu3(t)dt =1

12,

(0.9)

and so on. Therefore the required solution will be

u(x) = x+ ex −4

3+

2

3

(

1 +1

2+

1

4+

1

8+ ...

)

. (0.10)

0.2. Assignment. Solve the following Fredholm integral equations by using the Adomian decomposi-tion method

(1) u(x) = ex + 1− e+∫ 1

0u(t)dt

(2) u(x) = cos(x) + 2x+∫ π

0xtu(t)dt

(3) u(x) = x cos(x) + 1 + 1

2

∫ 1

0u(t)dt

Lecturer 08: APPLICATION OF LAPLACE-ADOMIAN DECOMPOSITION METHEOD

In this section, we will study LADM on linear and non linear system of Partial differential equations(PDEs). Adomian Decomposition Method (ADM) was first introduced by Gorge Adomian in 1980. Itwas used very effectively on a wide range of physical models of partial differential equations, such asBurger’s equation is a non linear PDE of second order, which have many applications in sciences andtechnology.

0.1. Laplace Adomian-Decomposition Methed. In this section, we present a Laplace-Adomiandecomposition method for solving of partial differential equations written in an operator form

{

Ltu+R1(u, v) +N1(u, v) = g1,

Ltv +Rt(u, v) +N2(u, v) = g2.(0.1)

With initial data{

u(x, 0) = f1(x),

v(x, 0) = f2(x).(0.2)

where Lt is considered a first-order partial differential operator, R1, R2 and N1, N2 are linear andnonlinear operators, respectively. And g1 and g2 are source terms. The method consists of first applyingthe Laplace transform to both sides of equations in system (0.1) and then by using initial conditions(0.2), we have:

{

L[Ltu] + L[R1(u, v)] + L[N1(u, v)] = L[g1],

L[Ltv] + L[Rt(u, v)] + L[N2(u, v)] = L[g2].(0.3)

Using the differentiation property of Laplace transform, we get{

L[u] = f1(x)s

+ 1sL[g1]−

1sL[R1(u, v)] +

1sL[N1(u, v)],

L[v] = f2(x)s

+ 1sL[g2]−

1sL[R2(u, v)] +

1sL[N2(u, v)].

(0.4)

The LADM defines the solutions u(x, t) and v(x, t) by the infinite series

u(x, t) =∞∑

n=0

un, v(x, t) =∞∑

n=0

vn. (0.5)

The nonlinear terms N1, N2 are usually represented by an infinite series of the so-called Adomianpolynomials.

N1(u, v) =∞∑

n=0

An, N2(u, v) =∞∑

n=0

A∗

n. (0.6)

The Adomian polynomials can be generated for all forms of nonlinearity. They are determined by thefollowing relations:

An =1

n!

[

dn

dλn

[

N1

∞∑

i=0

(λiui)

]]

λ=0

, (0.7)

and

A∗

n =1

n!

[

dn

dλn

[

N2

∞∑

i=0

(λivi)

]]

λ=0

. (0.8)

Substituting (0.5) and (0.6) into (0.4), gives{

L[∑

∞

n=0 un] =f1(x)

s+ 1

sL[g1]−

1sL[R1(

∑

∞

n=0 un,∑

∞

n=0 vn)] +1sL[∑

∞

n=0 An],

L[∑

∞

n=0 vn] =f2(x)

s+ 1

sL[g2]−

1sL[R2(

∑

∞

n=0 un,∑

∞

n=0 vn)] +1sL[∑

∞

n=0 A∗

n].(0.9)

1

Kamal Shah

Highlight

2

Applying the linearity of the Laplace transform, we define the following recursively formula:{

L[u0] =f1(x)

s+ 1

sL[g1],

L[v0] =f2(x)

s+ 1

sL[g2].

(0.10)

{

L[u1] = − 1sL[R1(u0, v0)] +

1sL[A0],

L[v1] = − 1sL[R2(u0, v0)] +

1sL[A∗

0].(0.11)

In general, for k ≥ 1, the recursive relations are given by{

L[uk+1] = − 1sL[R1(uk, vk)] +

1sL[Ak],

L[vk+1] = − 1sL[R2(uk, vk)] +

1sL[A∗

k].(0.12)

Applying the inverse Laplace transform, we can evaluate uk and vk (k ≥ 0). In some cases the exactsolution in the closed form may also be obtained.

1. Applications

In this section, we use the LADM to solve homogeneous and inhomogeneous linear system of par-tial differential equations and homogeneous and inhomogeneous nonlinear system of partial differentialequations. The algorithms are performed by Matlab.

1.1. The Homogenous linear system. Consider the homogeneous linear system of PDEs:{

ut − vx + (u+ v) = 0,

vt − ux + (u+ v) = 0.(1.1)

With initial conditionsu(x, 0) = sinhx, v(x, 0) = coshx. (1.2)

Taking the Laplace transform on both sides of Eq. (1.1) then, by using the differentiation property ofLaplace transform and initial conditions (1.2) gives

{

L[u] = 1ssinhx+ 1

sL[vx]−

1sL[u+ v],

L[v] = 1scoshx+ 1

sL[ux]−

1sL[u+ v].

(1.3)

The LADM defines the solutions u(x, t), v(x, t) by the series

u(x, t) =

∞∑

n=0

un, v(x, t) =

∞∑

n=0

vn, (1.4)

and the terms vx and ux by an infinite series

ux(x, t) =

∞∑

n=0

unx, vx(x, t) =

∞∑

n=0

vnx. (1.5)

Substituting series in (1.4) and (1.5) into both sides of Eq. (1.3) yields{

L[∑

∞

n=0 un] =1ssinhx+ 1

sL[∑

∞

n=0 vnx]− 1

sL[∑

∞

n=0 un +∑

∞

n=0 vn],

L[∑

∞

n=0 vn] =1scoshx+ 1

sL[∑

∞

n=0 unx]− 1

sL[∑

∞

n=0 un +∑

∞

n=0 vn].(1.6)

Now we define the following recursively formula:{

L[u0] =1ssinhx,

L[v0] =1scoshx,

(1.7)

{

L[un+1] =1sL[vnx

]− 1sL[un + vn], n ≥ 0,

L[vn+1] =1sL[unx

]− 1sL[un + vn], n ≥ 0.

(1.8)

3

Taking the inverse Laplace transform of both sides of the (1.7) and (1.8), we have{

[u0] = sinhx,

[v0] = coshx,(1.9)

System of PDEs{

[u1] = −t coshx,

[v1] = −t sinhx,(1.10)

{

[u2] =t2

2! sinhx,

[v2] =t2

2! coshx,(1.11)

and so on for other components. Using (1.4), the series solutions are therefore given by{

[u(x, t)] = sinhx(1 + t2

2! +t4

4! ...)− coshx(t+ t3

3! +t5

5! ...),

[v(x, t)] = coshx(1 + t2

2! +t4

4! ...)− sinhx(t+ t3

3! +t5

5! ...),(1.12)

using the Taylor expansion for sinh t and cosh t, we can find the exact solutions

u(x, t) = sinh(x− t), v(x, t) = cosh(x− t).

1.2. The Inhomogenous linear system. Consider the inhomogeneous linear system{

ut − vx − (u− v) = 2,

vt − ux − (u− v) = 2.(1.13)

With initial conditionsu(x, 0) = 1 + ex, v(x, 0) = −1 + ex. (1.14)


{

L[u] = 1s+ ex

s− 2

s2+ 1

sL[vx] +

1sL[u − v],

L[v] = −1s

+ ex

s− 2

s2− 1

sL[ux] +

1sL[u− v].

(1.15)

Using the decomposition series (1.4) and (1.5) for the linear terms u(x, t), v(x, t) and ux, vx, we obtain{

L[∑

∞

n=0 un] =1s+ ex

s− 2

s2+ 1

sL[∑

∞

n=0 vnx] + 1

sL[∑

∞

n=0 un − v∑

∞

n=0 vn],

L[∑

∞

n=0 vn] =−1s

+ ex

s− 2

s2− 1

sL[∑

∞

n=0 unx] + 1

sL[∑

∞

n=0 un −∑

∞

n=0 vn].(1.16)

The LADM presents the recursive relations{

L[u0] =1s+ ex

s− 2

s2,

L[v0] =−1s

+ ex

s− 2

s2.

(1.17)

{

L[un+1] =1sL[vnx

] + 1sL[un − vn],

L[vn+1] = − 1sL[unx

] + 1sL[un − vn].

(1.18)

Taking the inverse Laplace transform of both sides of the (1.17) and (1.18), we have{

u0 = 1 + ex − 2t,

v0 = −1 + ex − 2t,(1.19)

{

u1 = tex + 2t,

v1 = −tex + 2t,(1.20)

{

u2 = t2ex

2! ,

v2 = t2ex

2! ,(1.21)

4

and so on for other components. Using (1.4), the series solutions are therefore given by

u(x, t) = 1 + ex(

1 + t+ t2

2! +t3

3! ...

)

,

v(x, t) = −1 + ex(

1− t+ t2

2! −t3

3! ...

)

,

(1.22)

that converges to the exact solutions

u(x, t) = 1 + ex+t, v(x, t) = −1 + ex−t.

1.3. The Inhomogenous Nonlinear system.{

ut + vux + u = 1,

vt − uvx − v = 1.(1.23)

with initial conditionsu(x, 0) = ex, v(x, 0) = e−x. (1.24)


{

L[u] = ex

s− 1

s2− 1

sL[vux]−

1sL[u],

L[v] = e−x

s+ 1

s2+ 1

sL[uvx] +

1sL[v].

(1.25)

We represent u(x, t) and v(x, t) by the infinite series (1.4) then, inserting these series into both sides ofEq. (1.25) yields

{

L[∑

∞

n=0 un] =ex

s− 1

s2− 1

sL[∑

∞

n=0 An]−1sL[∑

∞

n=0 un],

L[∑

∞

n=0 vn] =e−x

s+ 1

s2+ 1

sL[∑

∞

n=0 A∗

n] +1sL[∑

∞

n=0 vn].(1.26)

Where An and A∗

n are the so-called Adomian polynomials by (0.12) that represent the nonlinear termsvux and uvx, respectively. We have a few terms of the Adomian polynomials for vux and uvx, whichare given by

A0 = v0u0x ,

A1 = v1u0x + v0u1x ,

A2 = v2u0x + v1u1x + v0u2x ,

A3 = v3u0x + v2u1x + v1u2x + v0u3x ,

· · ·

and

A∗

0 = u0v0x ,

A∗

1 = u1v0x + u0v1x ,

A∗

2 = u2v0x + u1v1x + u0v2x ,

A∗

3 = u3v0x + u2v1x + u1v2x + u0v3x ,

· · ·.

Now we define the following recursively formula:{

L[u] = 1s+ ex

s,

L[v] = 1s+ e−x

s,

(1.27)

{

L[un+1] =1s2

− 1sL[An]−

1sL[un],

L[vn+1] =1s2

+ 1sL[A∗

n] +1sL[vn],

(1.28)

5

Operating with Laplace inverse on both sides of (1.27) and (1.28) gives{

L[u0] = t+ ex,

L[v0] = t+ e−x,(1.29)

{

L[u1] = −t− t2

2! − tex − t2

2! ex,

L[v1] = −t− t2

2! + te−x − t2

2! e−x,

(1.30)

{

L[u2] =t2

2! + t2ex...,

L[v2] =t2

2! + t2e−x....(1.31)

Similarly, we can find other components. Using (1.4), the series solutions are therefore given by

u(x, t) = ex(

1− t+ t2

2! −t3

3! ...

)

,

v(x, t) = e−x

(

1 + t+ t2

2! +t3

3! ...

)

.

(1.32)

By using the Taylor expansion for et and e−t we can find the exact solutions

u(x, t) = ex−t, v(x, t) = e−x+t.

Lecture 09: Solution of Integro -Differential Equations

0.1. Introduction. The Adomian decomposition method was introduced and developed by GeorgeAdomian in 1980. A considerable amount of research work has been invested recently in applyingthis method to a wide class of linear and nonlinear ordinary differential equations, partial differentialequations and integral equations as well. This method generates a solution in the form of a series whoseterms are determined by a recursive relation using the Adomian polynomials. Mathematicians work tomodify Adomian decomposition method.

The modified form of Laplace decomposition method has been introduced by Khuri. Then Hus-sain and Khan used modified Laplace decomposition method to solve fractional differentia equations.Recently, the authors have used several methods for the numerical or the analytical solution of linearand non linear integro-differential equations of the second kind. But in this lecture we use modifiedLaplace decomposition method for the integro-differential equations. Because this is simple method forsolving the integro-differential equation. The main advantage of this method is the fact that it givesthe analytical solution. It is also note that the advantage of the decomposition methodology displays afast convergence of the solutions.

For this, consider the general form of second order nonlinear partial differential equations with initialconditions in the form

Lu(x, t) +Ru(x, t) +Nu(x, t) = h(x, t), u(x, 0) = f(x), ut(x, 0) = g(x), (0.1)

where L is the second order differential operator Lxx = ∂n

∂xn, R is the remaining linear operator, N

represents a general non-linear differential operator and h(x, t) is the source term. Applying Laplacetransform on both sides of (0.1) we have

L{Lu(x, t) +Ru(x, t) +Nu(x, t)} = L{h(x, t)},

which implies that

L{Lu(x, t)}+ L{Ru(x, t)}+ L{Nu(x, t)} = L{h(x, t)}. (0.2)

Now using the differentiation property of Laplace transform, we have

s2L{u(x, t)} − sf(x)− g(x) + L{Ru(x, t)}+ L{Nu(x, t)} = L{h(x, t)},

by rearranging, we have

L{u(x, t)} =f(x)

s+

g(x)

s2−

1

s2L{Ru(x, t)} −

1

s2L{Nu(x, t)}+

1

s2L{h(x, t)}. (0.3)

The next step in Laplace decomposition method is representing the solution as an infinite series givenbelow

u(x, t) =∞∑

n=0

un(x, t), (0.4)

The nonlinear operator is decomposed as

Nu(x, t) =

∞∑

n=0

An(x, t), (0.5)

where for every n ∈ N, An is the Adomian polynomial given below

An =1

n!

dn

dλn

[

N

(

∞∑

n=0

λiui

)]

λ=0

. (0.6)

1

Kamal Shah

Highlight

2

putting the values of (0.4) and (0.5) in (0.3), we get

∞∑

n=0

L{un(x, t)} =f(x)

s+

g(x)

s2−

1

s2L{Ru(x, t)} −

1

s2L

[

∞∑

n=0

An(x, t)

]

+1

s2L{h(x, t)}. (0.7)

Comparing both sides of (0.7), we get

L{u0(x, t)} = k1(x, s),

L{u1(x, t)} = k2(x, s)−1

s2L{R0u(x, t)} −

1

s2L{A0(x, t)},

L{un+1(x, t)} = −1

s2L{Rnu(x, t)} −

1

s2L{An(x, t)}, n ≥ 1,

(0.8)

Applying the inverse Laplace transform to (0.8) gives our required recursive relation as follows

u0(x, t) = k1(x, t),

u1(x, t) = k2(x, t)− L−1

[

1

s2L{R0u(x, t)} −

1

s2L{A0(x, t)}

]

,

un+1(x, t) = L−1

[

1

s2L{Rnu(x, t)} −

1

s2L{An(x, t)}

]

, n ≥ 1,

(0.9)

The solution through the modified Adomian decomposition method highly depends upon the choice ofk0(x, t) and k1(x, t), where k0(x, t) and k1(x, t) represent the terms arising from the source term andprescribed initial conditions.

We give some examples to illustrate this method for the integro-differential equations

Example 0.1. Let us first consider the nonlinear integro-differential equation

u′(x) = −1 +

∫ x

0

u2(t)dt, u(0) = 0. (0.10)

Applying the Laplace transform and by using the initial condition we have

sU(s) = −1

s+ L

[∫ x

0

u2(t)dt

]

,

which implies that

U(s) = −1

s2+

1

sL

[∫ x

0

u2(t)dt

]

.

Applying the inverse Laplace transform we get

u(x) = −x+ L−1

[

1

sL

(∫ x

0

u2(t)dt

)]

. (0.11)

We decompose the solution as an infinite sum given below

u(x) =

∞∑

n=0

un(x), (0.12)


∞∑

n=0

un(x) = −x+ L−1

[

1

sL

(

∫ x

0

∞∑

n=0

An(t)dt

)]

, (0.13)

3

here An =∑n

j=0ujun−j . Thus recursive relation is given below

u0(x) = −x,

u1(x) =x4

12,

u2(x) =−x7

252,

(0.14)

and so on

Example 0.2. Consider the second order nonlinear integro-differential equation

u′′(x) = sinh(x) + x−

∫ x

0

x(cosh2(t)− u2(t))dt, u(0) = 0, u′(0) = 1. (0.15)

Applying the Laplace transform and by using the initial conditions we obtain

s2U(s)− 1 =1

s2 − 1+

1

s2− L

[∫ x

0

x(cosh2(t)− u2(t))dt

]

,

which implies that

U(s) =1

s2+

1

s2(s2 − 1)+

1

s4−

1

s2L

[∫ x

0


]

,


u(x) = sinh(x) +x3

6− L−1

[

1

s2L

(∫ x

0


)]

, (0.16)

We decompose the solution as an infinite sum given below

u(x) =

∞∑

n=0

un(x), (0.17)


∞∑

n=0

un(x) = sinh(x) +x3

6− L−1

[

1

s2L

(

∫ x

0

∞∑

n=0

An(t)dt

)]

, (0.18)

here An =∑n

j=0ujun−j . Thus recursive relation is given below

u0(x) = sinh(x),

u1(x) =x6

6− L−1

[

1

s2L

(

x

∫ x

0

(cosh2(t)− u20(t))dt

)]

= 0,

un+1(x) = −L−1

[

1

s2L

(

x

∫ x

0

(cosh2(t)− u2n(t))dt

)]

= 0, n ≥ 1,

(0.19)

for every n ≥ 1, An = 0, Thus, the exact solution is

u(x) = sinh(x),

Example 0.3. Consider the third-order linear integro-differential equation

u′′′(x) = sin(x)− x−

∫ π

2

0

xtu′(t))dt, u(0) = 1, u′(0) = 0, u′′(0) = −1. (0.20)

4

Applying the Laplace transform and by using the initial conditions we obtain

s3U(s)− s2 + 1 =1

s2 + 1−

1

s2− L

[

∫ π

2

0

xtu′(t))dt

]

,

which implies that

U(s) =1

s−

1

s3+

1

s3(s2 + 1)−

1

s5−

1

s3L

[

∫ π

2

0

xtu′(t)dt

]

,


u(x) = cos(x)−1

24x4− L−1

[

1

s3L

(

∫ π

2

0

xtu′(t)dt

)]

, (0.21)

Applying the same procedure as in the previous example we arrive the modified recursive relation givenbelow

u0(x) = cos(x),

u1(x) = −1

24x4− L−1

[

1

s3L

(

x

∫ π

2

0

tu′

0(t)dt

)]

= 0,

un+1(x) = −L−1

[

1

s3L

(

∫ π

2

0

tu′

n(t)dt

)]

= 0,

(0.22)

and so the total solution of the above problem is given as

u(x) = cos(x).

Example 0.4. Consider the second-order linear integro-differential equation

u′′(x) + xu′(x)− xu(x) = ex − 2 sin(x) +

∫ 1

−1

sin(x)e−tu(t))dt, u(0) = 1, u′(0) = 1. (0.23)

We Applying the Laplace transform and by using the initial conditions we obtain

(s2 − 1)U(s)− (s− 1)d

dsU(s)− s− 1 =

1

s− 1−

2

s2 + 1+ L

[∫ 1

−1

sin(x)e−tu(t))dt

]

,

which implies that

U(s) =1

s+ 1

d

dsU(s) +

1

s− 1+

1

(s− 1)2(s+ 1)−

2

(s2 − 1)(s2 + 1)

−1

s2 − 1L

[∫ 1

−1

sin(x)e−tu(t))dt

]

,


u(x) = ex + L−1

[

1

s+ 1

d

dsU(s) +

1

(s− 1)2(s+ 1)−

2

(s2 − 1)(s2 + 1)

−1

(s2 − 1)(s2 + 1)L

(∫ 1

−1

e−tu(t))dt

)]

,

5

Applying the same procedure as in the previous example we arrive the modified recursive relation givenas

u0(x) = ex,

u1(x) = L−1

[

1

s+ 1

d

dsU(s) +

1

(s− 1)2(s+ 1)−

2

(s2 − 1)(s2 + 1)

−1

(s2 − 1)(s2 + 1)L

(∫ 1

−1

e−te−t)dt

)]

= 0,

un+1(x) = 0, n ≥ 1.

(0.24)

Thus, the exact solution isu(x) = ex.

In the above examples we observed that the modified Laplace Adomian decomposition method withthe initial approximation obtained from initial conditions yield a good approximation to the exactsolution only in a few iterations.

0.2. Assignment. Solve the nth-order linear integro-differential equation by modified Laplace Adomiandecomposition method.

INTEGRAL EQUATIONS 8th Semester€¦ · Integral equations play an important role in the theory of...

Documents

Transcript of INTEGRAL EQUATIONS 8th Semester€¦ · Integral equations play an important role in the theory of...