modified variational iteration method for partial differential equations using ma’s transformation
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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. Use of Initial and Boundary Conditions. - PowerPoint PPT PresentationTRANSCRIPT
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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