chapter no 03

CHAPTER 3

LAPLACE ADOMIAN’S DECOMPOSITION METHOD (LADM)

3.1. Introduction

Many problems in mathematical physics, theoretical physics and chemical physics

are modelled by the so-called initial value and boundary value problems in the

second-order nonlinear ordinary differential equations. These equations are difficult to

be solved analytically and sometimes it is impossible then application must be made

to relevant numerical methods such as shooting method, finite difference etc. In recent

years, differential transform method has been used to solve this type of equations

[10,11,12]. In this work, the differential transform method is used to investigate the

numerical and analytical approximate solutions of the nonlinear singular initial value

problems of Emden-Fowler Type. Initial value problems in the second order is

considered which occur in applied mathematics, astrophysics and the numerical

solution of the Emden-Fowler equation, and the other linear and nonlinear singular

initial value problems, plays very important role because of the singularity behaviour

at the origin.

Laplace Adomian’s Decomposition Method (LADM) was first introduced by Suheil

A. Khuri [13], and has been successfully used to find the solution of differential

equations [14,15]. The Laplace Adomian’s Decomposition Method is a combination

of ADM and Laplace Transforms. This Method is successfully used to find the exact

solution of the Bratu and Duffing equation. The Significant advantage of this method

is its capability of combining the two powerful methods to obtain exact solution for

non-linear equation.

3.2. Analysis of Emden-Fowler Equation

Let us consider the Emden-Fowler equation

d2 yd x2

+ axd yd x

+ α xm−1 yr = 0 , n≠0 , n≠1 , m ,a ,α are parameters

which is used in mathematical physics, theoretical physics, and chemical physics.

Above equation has interesting mathematical and physical properties, and it has been

investigated from various points of view.

29

LAPLACE ADOMIAN DECOMPOSITION METHOD CH 3

3.3. Numerical Application

3.3.1 Problem 1

Consider Emden-Fowler equation of the second kind is

d2 yd x2

+ 2xd yd x

+ α xm y r = 0 , (3.1)

y (0 ) = y0 , y' (0 ) = 0 .

Throughout multiply Eq. (3.1) with x, we get

xd2 yd x2

+ 2d yd x

+ α xm + 1 yr = 0 .

(3.2)

Taking Laplace Transform of Eq. (3.2) on both sides, we have

L [ x d2 yd x2

+ 2d yd x ] +L [α xm + 1 yr ] = 0 ,

− s2 L' [ y ] − y (0 ) + L [α xm + 1 yr ] = 0 ,

− s2 L' [ y ] − 1 + L [ α xm + 1 yr ] = 0 . (3.3)

According to LADM, the solution y ( x ) of the given problem is defined by an infinite

series of the form

y ( x ) = ∑n= 0

∞

yn ( x ) , (3.4)

and the non-linear term yr

can be decompose into infinite series of the form

yr = ∑n = 0

∞

An , (3.5)

andAn ' s ,are called Adomian’s Polynomials,

Which yields that,

A0 = y0r,

A1 = r y1 y0r − 1 ,

A2 = r y2 y0r −1 + r (r − 1 )

y12

2 !y

0r − 2 ,

30


A3 = r (r − 1 ) (r − 2 )y

13

3 !y

0r −3 + r (r −1 ) y1 y2 y

0r − 2 + r y3 y

0r − 1 ,

⋮.

By substituting Eq.(3.4), (3.5) into (3.3)

− s2 L'[ ∑n = 0

∞

yn] − 1 + L [α xm + 1 ∑n= 0

∞

An ] = 0 ,

− s2 ∑n= 0

∞

L' [ yn] − 1 + ∑n = 0

∞

L [α xm + 1 An ] = 0 ,

∑n = 0

∞

L' [ yn ] =− s− 2 + s− 2 ∑n = 0

∞

L [α xm + 1 An ] . (3.6)

Its recursive relation is given by,

L' [ y0] =− s−2 ,

L' [ yn + 1] = s−2 α L [xm + 1 An] . n ≥ 0 (3.7)

Integrate Eq. (3.7) with respect to s,

L [ y0 ] =−∫ s−2 d s ,

L [ yn + 1 ] = α ∫ s−2 L [xm + 1 An ] d s . n ≥ 0

(3.8)

Now taking Inverse Laplace Transform of Eq. (3.8), it becomes

y0 = L−1 [−∫ s−2 d s ] ,

yn + 1 = L−1 [α ∫ s−2 L [ xm + 1 An] d s] . n ≥ 0 (3.9)

Therefore,

y0 = L− 1 [−∫ s−2 d s ] ,

= L− 1 [ s− 1] ,

= 1 ,

y1 = α L− 1 [∫ s− 2 L [xm + 1 A0] d s ] ,

31


= α L− 1 [∫ s− 2 L [xm + 1] d s ] ,

= α L− 1 [ (m + 1 ) ! s−m − 3

−3 −m ] ,=− α xm + 2

(m + 2 ) (m + 3 ), m >− 2 .

y2 = α L− 1 [∫ s− 2 L [xm + 1 A1] d s ] ,

= α L−1 [∫ s− 2 L [xm + 1 r y1 y0r − 1] d s ] ,

For r = 1 ,

y2 = α L− 1 [∫ s− 2 L [xm + 1 y1] d s ] ,

y2 = α L− 1 [∫ s− 2 L [ xm + 1 (− α xm + 2

(m + 2 ) (m + 3 ) )] d s ] ,y2 = α L− 1 [∫ s− 2 L [− α x2 m + 3

(m + 2 ) (m + 3 ) ] d s ] ,= α L− 1 [∫ s− 2 (− α (2 m + 3 ) !

(m + 2 ) (m + 3 ) s2 m + 4 ) d s] ,= α L− 1 [α (2 m + 3 ) !

(2 m + 5 ) (m + 2 ) (m + 3 ) s2 m + 5 ] ,= α

2 x2 m + 4

2 (2 m + 5 ) (m + 3 ) (m + 2 )2, m >− 2 .

Therefore,

y ( x ) = 1− α xm + 2

(m + 2 ) (m + 3 )+ α2 x2 m + 4

2 (2 m + 3 ) (m + 3 ) (m + 2 )2⋯ , m >− 2 ,

(3.10)

Particularly, we obtained the exact solution as

For m= 0 and r= 0 ,

32


y0 = 1 ,

y1 =− α x2

6,

y2 = 0 .

Therefore, the exact solution will be

y ( x ) = 1 − α x2

6.

For m= 0 and r= 1 ,

y0 = 1 ,

y1 =− α x2

6,

y2 = α2 x4

120,

y3=− α3 x6

5040,

⋮.

Therefore,

y ( x ) = 1 − α x2

6+ α

2 x4

120− α 3 x6

5040+⋯ ,

y ( x ) = 1x ( x − α x3

6+ α

2 x5

120− α3 x7

5040+⋯) ,

= 1√α x (√α x − α

32 x3

6+ α

52 x5

120− α

72 x7

5040+⋯) ,

y ( x ) = Sin (√α x )√α x

.

For m= 0 and r= 5 ,

33


y0 = 1 ,

y1 =− α x2

6,

y2 = α2 x4

24−3 α3 x6

70+ 17 α 4 x8

630− 59 α5 x10

11550+ α

6 x12

3510,

⋮.

y ( x ) = (1 + α x2

3 )− 12 .

3.3.2. Problem 2

Consider another type of Emden-Fowler equation is

d2 yd x2

+ 2xd yd x

+ α xm e y = 0 ,

y (0 ) = 0 , y ' (0 ) = 0 . (3.11)


xd2 yd x2

+ 2d yd x

+ α xm + 1 e y = 0 . (3.12)


L [ x d2 yd x2

+ 2d yd x ] +L [α xm + 1 e y ] = 0 ,

− s2 L' [ y ] − y (0 ) + L [α xm + 1 e y ] = 0 ,

− s2 L' [ y ] − y (0 ) + L [α xm + 1 e y ] = 0 . (3.13)

According to LADM, the solution y ( x ) ,of the given problem is defined by an infinite

series of the form

y ( x ) = ∑n= 0

∞

yn ( x ) , (3.14)

and the non-linear term ey


e y = ∑n = 0

∞

An , (3.15)

34


andAn ' s are called Adomian’s Polynomials,

Which yields that,

A0 = ey0 ,

A1 = y 1 ey0 ,

A2 = y2 ey0 + 1

2 !y12 e

y0 ,

A3 = y3 ey0 + y1 y2 e

y0 + 13 !

y13 ey0 ,

⋮ .

By substituting Eq. (3.14), (3.15) into (3.13)

− s2 L'[ ∑n = 0

∞

yn] − y (0 ) + L [α xm + 1 ∑n = 0

∞

An] = 0 ,

− s2 ∑n= 0

∞

L' [ yn] − y (0 )+ ∑n= 0

∞

L [α xm + 1 An ] = 0 ,

∑n = 0

∞

L' [ yn ] =− s− 2 y (0 ) + s− 2 ∑n = 0

∞

L [α xm + 1 An ] . (3.16)


L' [ y0] =− s− 2 y (0 ) ,

L' [ yn + 1] = s−2 α L [xm + 1 An] . n ≥ 0 (3.17)


L [ y0 ] =−∫ s−2 y (0 ) d s ,

L [ yn + 1 ] = α ∫ s−2 L [xm + 1 An ] d s . n ≥ 0 (3.18)


y0 = L−1 [−∫ s− 2 y (0 ) d s] ,

yn + 1 = L−1 [α ∫ s−2 L [ xm + 1 An] d s] . n≥ 0 (3.19)

35


Therefore,

y0 = L− 1 [−∫ s− 2 y (0 ) d s ] ,

= L− 1 [0 ] = 0 ,

y1 = α L− 1 [∫ s− 2 L [xm + 1 A0] d s ] ,= α L− 1 [∫ s− 2 L [xm + 1] d s ] ,

= α L− 1 [ (m + 1 ) ! s−m − 3

−3 −m ] ,=− α xm + 2

(m + 2 ) (m + 3 ), m >− 2 .

y2 = α L− 1 [∫ s− 2 L [xm + 1 A1] d s ] ,

= α L−1 [∫ s− 2 L [ xm + 1 y1 ey0 ] d s ] ,

y2 = α L− 1 [∫ s− 2 L [xm + 1 y1] d s ] ,

y2 = α L− 1 [∫ s− 2 L [ xm + 1 (− α xm + 2

(m + 2 ) (m + 3 ) )] d s ] ,y2 = α L− 1 [∫ s− 2 L [− α x2 m + 3

(m + 2 ) (m + 3 ) ] d s ] ,= α L− 1 [∫ s− 2 (− α (2 m + 3 ) !

(m + 2 ) (m + 3 ) s2 m + 4 ) d s] ,= α L− 1 [α (2 m + 3 ) !

(2 m + 5 ) (m + 2 ) (m + 3 ) s2 m + 5 ] ,= α

2 x2 m + 4

2 (2 m + 5 ) (m + 3 ) (m + 2 )2, m >− 2 .

Therefore,

y ( x ) =− α xm + 2

(m + 2 ) (m + 3 )+ α 2 x2 m + 4

2 (2 m + 3 ) (m + 3 ) (m + 2 )2⋯ , m >− 2 ,

(3.20)

3.3.3. Problem 3

36


Consider nonlinear Emden-Fowler equation

d2 yd x2

+ 2xd yd x

+ α xm e− y = 0 ,

y (0 ) = 0 , y ' (0 ) = 0 . (3.21)

Throughout multiply Eq.(3.21) with x, we get

xd2 yd x2

+ 2d yd x

+ α xm + 1 e− y = 0 . (3.22)


L [ x d2 yd x2

+ 2d yd x ] +L [α xm + 1 e− y ] = 0 ,

− s2 L' [ y ] − y (0 ) + L [α xm + 1 e− y ] = 0 ,

− s2 L' [ y ] − y (0 ) + L [α xm + 1 e− y ] = 0 . (3.23)


series of the form

y ( x ) = ∑n= 0

∞

yn ( x ) , (3.24)

and the non-linear term e− y


e− y= ∑n= 0

∞

An , (2.25)


Which yields that,

37


A0 = e− y0 ,

A1 =− y 1 e− y0 ,

A2 =− y2 e− y0 + 1

2 !y12 e

− y0 ,

A3 =− y3 e− y0 + y1 y2 e

− y0 − 13 !

y13 e− y0 ,

⋮ .


− s2 L'[ ∑n = 0

∞

yn] − y (0 ) + L [α xm + 1 ∑n = 0

∞

An] = 0 ,

− s2 ∑n= 0

∞

L' [ yn] − y (0 )+ ∑n= 0

∞

L [α xm + 1 An ] = 0 ,

∑n = 0

∞

L' [ yn ] =− s− 2 y (0 ) + s− 2 ∑n = 0

∞

L [α xm + 1 An ] . (3.26)


L' [ y0] =− s− 2 y (0 ) ,

L' [ yn + 1] = s−2 α L [xm + 1 An] . n ≥ 0 (3.27)


L [ y0 ] =−∫ s−2 y (0 ) d s ,

L [ yn + 1 ] = α ∫ s−2 L [xm + 1 An ] d s . n ≥ 0 (3.28)


y0 = L−1 [−∫ s− 2 y (0 ) d s] ,

yn + 1 = L−1 [α ∫ s−2 L [ xm + 1 An] d s] . n≥ 0 (3.29)

Therefore,

38


y0 = L− 1 [−∫ s− 2 y (0 ) d s ] ,

= L− 1 [0 ] = 0 ,

y1 = α L− 1 [∫ s− 2 L [xm + 1 A0] d s ] ,= α L− 1 [∫ s− 2 L [xm + 1] d s ] ,

= α L− 1 [ (m + 1 ) ! s−m − 3

−3 −m ] ,=− α xm + 2

(m + 2 ) (m + 3 ), m >− 2 .

y2 = α L− 1 [∫ s− 2 L [xm + 1 A1] d s ] ,

= α L−1 [∫ s− 2 L [ xm + 1 (− y1 e− y0)] d s ] ,

y2 =− α L− 1 [∫ s− 2 L [xm + 1 y1] d s ] ,

y2 =− α L− 1 [∫ s− 2 L [ xm + 1 (− α xm + 2

(m + 2 ) (m + 3 ) )] d s ] ,y2 =− α L− 1 [∫ s− 2 L [− α x2 m + 3

(m + 2 ) (m + 3 ) ] d s ] ,=− α L− 1 [∫ s− 2 (− α (2 m + 3 ) !

(m + 2 ) (m + 3 ) s2 m + 4 ) d s] ,=− α L− 1 [α (2 m + 3 ) !

(2 m + 5 ) (m + 2 ) (m + 3 ) s2 m + 5 ] ,=− α

2 x2 m + 4

2 (2 m + 5 ) (m + 3 ) (m + 2 )2, m >− 2 .

Therefore,

y ( x ) =− α xm + 2

(m + 2 ) (m + 3 )− α2 x2 m + 4

2 (2 m + 3 ) (m + 3 ) (m + 2 )2⋯ , m >− 2 ,

(3.30)

For m= 0 ,Eq. (3.30) becomes,

39


y ( x ) =− α x2

6− α2 x4

120− 8 α3 x6

15120⋯ ,

3.3.4. Problem 4

Consider Emden-Fowler equation of the second kind is

d2 yd x2

+ 2xd yd x

+ α xm ln x yr = 0 ,

y (0 ) = 1 , y ' (0 ) = 0 . (3.31)


xd2 yd x2

+ 2d yd x

+ α xm + 1 ln x yr = 0 . (3.32)


L [ x d2 yd x2

+ 2d yd x ] +L [α xm + 1 ln x y r ] = 0 ,

− s2 L' [ y ] − y (0 ) + L [α xm + 1 ln x y r ] = 0 . (3.33)


series of the form

y ( x ) = ∑n= 0

∞

yn ( x ) , (3.34)

and the non-linear term yr


yr = ∑n = 0

∞

An , (3.35)


Which yields that,

A0 = y0r,

A1 = r y1 y0r − 1 ,

A2 = r y2 y0r −1 + r (r − 1 )

y12

2 !y

0r − 2 ,

40


A3 = r (r − 1 ) (r − 2 )y

13

3 !y

0r −3 + r (r −1 ) y1 y2 y

0r − 2 + r y3 y

0r − 1 ,

⋮.


− s2 L'[ ∑n = 0

∞

yn] − y (0 ) + L [α xm + 1 ln x ∑n= 0

∞

An] = 0 ,

− s2 ∑n= 0

∞

L' [ yn] − y (0 )+ ∑n= 0

∞

L [α xm + 1 ln x An ] = 0 ,

∑n = 0

∞

L' [ yn ] =− s− 2 y (0 ) + s− 2 ∑n = 0

∞

L [α xm + 1 ln x An ] . (3.36)


L' [ y0] =− s− 2 y (0 ) ,

L' [ yn + 1] = s−2 α L [xm + 1 ln x An] . n ≥ 0 (3.37)


L [ y0 ] =−∫ s−2 y (0 ) d s ,

L [ yn + 1 ] = α ∫ s−2 L [xm + 1 ln x An] d s . n ≥ 0 (3.38)


y0 = L−1 [−∫ s− 2 y (0 ) d s] ,

yn + 1 = L−1 [α ∫ s−2 L [ xm + 1 ln x An ] d s] . n ≥ 0 (3.39)

Therefore, for r = 0 and m=−1 ,Eq. (3.39) becomes,

y0 = L− 1 [−∫ s− 2 y (0 ) d s ] ,

= L− 1 [− s− 1 ] = 1 ,

y1 = α L− 1 [∫ s− 2 L [xm + 1 ln x A0] d s ] ,= α L− 1 [∫ s− 2 L [ xm + 1 ln x y

0r ] d s ] ,

41


= α L− 1 [∫ s− 2 L [xm + 1 ln x ] d s ] ,

= α L− 1 [∫ s− 2 (− Euler Gamma + ln ss ) d s ] ,

= α L− 1 [14 s2 +Euler Gamma

2 s2+

ln s

s2 ] ,= α x

2 ( ln x − 32 ) ,

y2 = 0 .

So the exact solution will be,

y ( x ) = 1 − α x2 ( ln x − 3

2 ) .

Similarly

y ( x ) = 1 − α x2

6 ( ln x − 56 ) .

y ( x ) = 1 − α x3

12 ( ln x − 712 ) .

y ( x ) = 1 − α x4

20 ( ln x − 920 ) .

y ( x ) = 1 − α x5

30 ( ln x − 1130 ) .

3.3.5. Problem 5 [67]

Consider a Time-Dependent Lane-Emden equation

∂2 y∂ x2

+2x

∂ y∂ x

− (6 + 4 x2 − cos t ) y = ∂ y∂ t

,

y (0 , t ) = esin t , yx (0 , t ) = 0 . (3.40)


42


x∂2 y∂ x2

+ 2∂ y∂ x

− (6 + 4 x2 − cos t ) x y = x∂ y∂ t

. (3.41)


L [ x ∂2 y∂ x2

+ 2∂ y∂ x ] = L [ (6 + 4 x2 − cos t ) x y + x ∂ y

∂ t ] ,− s2 ∂

∂ sL [ y ( x ,t ) ] − y (0 , t ) = L [ (6 + 4 x2 − cos t ) x y + x ∂ y

∂ t ] ,∂

∂ sL [ y ( x , t ) ] =− s− 2 y (0 , t ) − s− 2 L [ (6 + 4 x2 − cos t ) x y + x ∂ y

∂ t ] . (3.42)

According to LADM, the solution y ( x , t ) of the given problem is defined by an

infinite series of the form

y ( x , t ) = ∑n = 0

∞

yn ( x , t ) , (3.43)

also it’s a linear problem, so we need not to find the Adomian’s Polynomials.

By substitution Eq. (3.43) in Eq. (3.42), we get

∂∂ s

L [ ∑n = 0

∞

yn ( x , t )]=− s− 2 y (0 , t ) − s− 2 L [ (6 + 4 x2 − cos t ) x ∑n = 0

∞

y n + x∂∂ t ∑

n = 0

∞

yn] ,∑n = 0

∞ ∂∂ s

L [ yn ( x , t ) ] =− s− 2 y (0 , t ) − s− 2 ∑n =0

∞L [ (6 + 4 x2 − cos t ) x yn + x

∂ yn∂ t ] .


∂∂ x

L [ y0 ( x , t ) ] =− s− 2 y (0 , t ) ,

∂∂ x

L [ yn + 1 (x , t ) ] =−s− 2 L [ (6 + 4 x2 − cos t ) x yn + x∂ y n∂ t ] . n ≥ 0

(3.44)


L [ y0 ( x , t ) ] =−∫ s− 2 y (0 , t ) d s ,

L [ yn + 1 ( x , t ) ] =−∫s− 2 L [( 6 + 4 x2 − cos t ) x yn + x∂ yn∂ t ] d s . n ≥ 0

(3.45)

Now taking the Inverse Laplace Transform of the Eq. (3.45),

43


y0 (x , t ) = L−1 [−∫ s− 2 y (0 , t ) d s ] ,

yn + 1 ( x , t ) = L− 1 [−∫s− 2 L [( 6 + 4 x2 − cos t ) x yn + x∂ yn∂ t ] d s ] . n ≥ 0

(3.46)

Therefore,

y0 (x , t ) =− L−1 [∫ s− 2 y (0 , t ) d s ] ,

=− L− 1 [− esin t s−1 ] ,

= esin t ,

y1 ( x , t ) = L− 1 [−∫ s− 2 L [( 6 + 4 x2 − cos t ) x y0 + x∂ y0

∂ t ] d s ] ,= L− 1 [−∫ s− 2 L [ (6 + 4 x2 − cos t ) x esin t + x ∂ esin t

∂ t ] d s ] ,= L−1 [−∫ s− 2 L [ (6 + 4 x2 − cos t ) x esin t + x cos t esin t ] d s ] ,= L−1 [−∫ s− 2 L [ (6 + 4 x2 ) x esin t ] d s ] ,

= esin t L− 1 [−∫ s− 2 (24s4

+ 6s2 ) d s] ,

= esin t L−1 [24

5 s5+

2

s3 ] ,= esin t ( x2 +

x4

5 ) .

y2 ( x , t ) = L− 1 [−∫ s− 2 L [( 6 + 4 x2 − cos t ) x y1 + x∂ y1

∂ t ] d s] ,= esin t (310

x4 + 13105

x6 + x8

90 ) .

44


y3 ( x , t ) = L− 1 [−∫ s− 2 L [(6 + 4 x2 − cos t ) x y2 + x∂ y2

∂ t ] d s] ,= esin t (3 x6

70+ 17 x8

630+ 54 x10

11550+ 1

3510x12) ,

⋮.

Thus from Eq. (3.43) solution will be,

y ( x , t ) = y0 + y1 + y2 + y3 +⋯ ,

y ( x , t ) = esin t + esin t (x2 + x4

5 ) + esin t ( 310

x 4 + 13105

x6 + x8

90 ) + esin t ( 3 x6

70+ 17 x8

630+ 54 x10

11550+ 1

3510x12) +⋯,

y ( x , t ) = esin t (1 + x2 + x4

2 !+ x

6

3 !+ x

8

4 !+⋯ ,) ,

= esin t (ex 2 ) ,

= ex2 + sin t . (3.47)

This is exact solution.

45

chapter no 03

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