modified variational iteration method for partial differential equations using ma’s transformation

Modified Variational Iteration Method for Partial Differential

Equations Using Ma’s Transformation

SYED TAUSEEF MOHYUD-DIN

Variational Iteration Techniques for Solving Initial and Boundary Value Problems

Introduction and History

Correction Functional

Conversion to a System of Equations

Restricted VariationSelection of Initial Value

Use of Initial andBoundary Conditions

Identification of Lagrange Multiplier Simpler

Variational Iteration Techniques for Solving Initial and Boundary Value Problems

Applications of Variational Iteration Method Modifications (VIMHP and VIMAP) Applications in Singular Problems (Use of New

Transformations)

Advantages of Variational Iteration Method

Use of Lagrange Multiplier (reduces the successive applications of integral operator)

Independent of the Complexities of Adomian’s Polynomials

Use of Initial Conditions only No Discretization or Linearization or Unrealistic

Assumptions Independent of the Small Parameter Assumption

Applications Boundary Value Problems of various-orders Boussinesq Equations Thomas-Fermi Model Unsteady Flow of Gas through Porous Medium Boundary Layer Flows Blasius Problem Goursat Problems Laplace Problems

Applications Heat and Wave Like Models Burger Equations Parabolic Equations KdVs of Third, Fourth and Seventh-orders Evolution Equations Higher-dimensional IBVPS Helmholtz Equations

Applications Fisher’s Equations Schrödinger Equations Sine-Gordon Equations Telegraph Equations Flierl Petviashivili Equations Lane-Emden Equations Emden-Fowler Equations

Variational Iteration Method ),(xguNuL

Correction functional

.))()(~)(()()(0

1 dssgsuNsuLxuxux

nnnn

.lim nnuu

Variational Iteration Method Using He’s Polynomials (VIMHP)

.)()()~()()()(00

)(

0

)(

0

00

)( dgduNpuLppxuupx

nn

n

nn

nx

nn

n

Modified Variational Iteration Method for Partial Differential

Equations Using Ma’s Transformation

Helmholtz Equation

2 2

2 2 2 2

, ,, 0,

u x y u x yu x y

x y

with initial conditions

0, , (0, ) cosh .xu y y u y y y

The exact solution

( , ) cosh .xu x y ye x y

Applying Ma’s transformation x t (by setting 1,k

2

22 0,d u

ud

with

, ( ) ,u A u B

)

0

1 .~2

1)()( dsuusBAu nnn

The correction functional

0

1 .2

1)()( dsuusBAu nnn

Applying modified variational iteration method (MVIM)

.2

12

2102

22

221

2

20

2

0

22

10 dsuppuud

udp

d

udp

d

udsBAuppuu

t

Comparing the co-efficient of like powers of p, following approximants are obtained

:0p 0 ( ) ,u A B

:1p 3 21

1 1( ) ,

12 4u A B B A

:2p 5 4 3 22

1 1 1 1( ) ,

480 96 12 4u A B B A B A

.

The series solution

.4

1

12

1

96

1

480

1)( 2345 AAAu

The inverse transformation

5 4 3 21 1 1 1( , ) ,

480 96 12 4u x y A B x y B x y A x y B x y A x y

the use of initial condition

2 23 2 2 2

4

2 22 2 2 3

4

2 6 24 24 12 122 ,

48

4 2 8 4 46 .

48

y y y y y y

y

y y y y y

y

y e y y e y e ye e e yA

e y

y e y y e y e ye eB

e y

The solution after two iterations is given by

3 2 3 3 2 3 3 4 2 5

4

2 2 3 3 2 2 4 2 2 2 5 2 3 2 3 3 2 4 3

1( , ) 96 8 48 4 6 96 2 4 2

2 48

2 24 48 2 4 2 ,

.

y y y y y y y y

y

y y y y y y y

u x y ye ye x x y e x y e x ye x y e x y e x y e xe y

y e x y e x y e x ye x e x y e e x y x y x y x x

Figure 3.1

Solution by Proposed Algorithm Exact solution

Helmholtz Equation

2 2

2 2 2 2

, ,8 , 0,

u x y u x yu x y

x y

with initial

conditions

0, sin 2 , (0, ) 0.xu y y u y

The exact solution for this problem is

( , ) cos 2 sin 2 .u x y x y

Applying Ma’s transformation x t (by setting 1,k

2

24 0,

d uu

d

with

, ( ) ,u A u B

The correction functional is given by

0

1 .~4)()( dsuusBAu nnn

0

1 .4)()( dsuusBAu nnn

Applying modified variational iteration method (MVIM)

.4 22

1022

22

21

2

20

2

0

22

10 dsuppuud

udp

d

udp

d

udsBAuppuu

t

Comparing the co-efficient of like powers of p, following approximants are obtained

:0p 0 ( ) ,u A B

:1p 3 21

2( ) 2 ,

3u A B B A

:2p 5 4 3 22

2 2 2( ) 2 ,

15 3 3u A B B A B A

.

The series solution is given by

5 4 3 22 2 2( ) 2 ,

15 3 3u A B B A B A

the inverse transformation will yield

5 4 3 22 2 2( , ) 2 ,

15 3 3u x y A B x y B x y A x y B x y A x y

The use of initial condition gives

4 2

6 8

2

6 8

sin 2 2 6 315 ,

16 4 45

sin 2 3 260 .

16 4 45

y y yA

y y

y y yB

y y

The solution after two iterations is given by

54 4 2 4 3 3 4 2 3 56 8

4 4 5 3 6 2 2 6 8

sin 2( , ) 45 30 2 24 60 80 60 16

16 4 45

60 80 40 90 16 4 ,

.

yu x y x y yx y x y x y x y x

y y

y x y x y x x y y

Table 1Table 1 (Error estimates at .1y )

x Exact solution Approx solution *Errors

-1.0 -.0744491770 -.082675613 8.22E-03

-0.8 -.0039143995 -.0058010496 1.88E-03

-0.6 .0722477834 .0719893726 2.58E-04

-0.4 .1384269365 .1384142557 1.26E-05

-0.2 .1829867759 .1829865713 2.04E-07

0 .1986693308 .1986693308 0.000000

0.2 .1829991064 .1829865713 1.25E-05

0.6 .1386872460 .1384142557 2.72E-04

0.8 .0740356935 .0719893726 2.04E-03

1.0 .0033413560 -.0058010496 9.14E-03

1.0 -.0526997339 -.0826756135 2.99E-02

*Error = Exact solution – Approximate solution

Homogeneous Telegraph Equation.

2 2

2 2 2 2

, , ,, ,

u x t u x t u x tu x y

x t t

with initial and boundary

conditions

2 2. 0, , (0, ) ,

. ,0 , ( ,0) 2 .

t tx

x xx

BC u t e u t e

I C u x e u x e

The exact solution for this problem is

2( , ) .x tu x t e

Applying Ma’s transformation x t (by setting 1, 2,k

2

23 2 0,d u du

ud d

with

, ( ) ,u A u B

0

1 .~3

1~3

2)()( dsuuusBAu nnnn

0

1 .3

1

3

2)()( dsuuusBAu nnnn

Applying modified variational iteration method (MVIM)

.3

1

3

210

1021

2

20

2

0

10 dspuud

dup

d

du

d

udp

d

udsBApuu

t

Comparing the co-efficient of like powers of p, following approximants are obtained

:0p 0 ( ) ,u A B

:1p 3 2 21

1 1 1( ) ,

18 6 3u A B B A B

The series solution is given by

5 4 4 3 3 2 22 1 1 7 1 1 1( ) ,

1080 216 54 54 27 6 3u A B B A B B A A B

The inverse transformation would yield

5 4 4 3

3 2 2

2 1 1 7( , ) 2 2 2 2 2

1080 216 54 541 1 1

2 2 2 ,27 6 3

u x t A B x t B x t A x t B x t B x t

A x t A x t B x t

and use of initial condition gives

2 2 3

4

2 2

4

9 6 6 43 ,

4 27 36

3 2 29 .

4 27 36

t

t

e t t tA

t t

e t tB

t t

The solution after two iterations is given by

2 3 2 3 2 2 2 3 3 2 4 4

4

54 3 72 27 2 6 2 8 8 54 72 81( , )

2 4 27 36

te x tx x tx t x t x t x t x x t tu x t

t t

.

Solution by Proposed Algorithm Exact solution

CONCLUSION

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