VARIATIONAL ITERATION METHOD FOR VIBRATION PROBLEMS
Metin O. Kaya
Istanbul Technical University, Faculty of Aeronautics and Astronautics, 34469, Maslak,
Istanbul, Turkey
Abstract
In this study, linear/nonlinear, free/forced and damped/undamped vibrations of both one
degree of freedom and continous systems are discussed by using the Variational Iteration
Method. Additionally, common vibration problems are classified and Lagrange multipliers
are derived for each type of problem.
Keywords: Variational Iteration Method, VIM, Vibration, Lagrange Multiplier, Nonlinear
Vibration
1. Introduction
Vibration of dynamical systems can be divided into two main classes like discrete and
distributed. The variables in discrete systems depend on time only, whereas in distributed
systems such as beams, plates etc. variables depend on time and space. Therefore,
equations of motion of discrete systems are described by ordinary differential equations,
while equations of motion of distributed systems are described by partial differential
equations ( Meirovitch [1] ).
1
A considerable amount of studies have been made in the area of vibration problems.
Several techniques, such as finite element method, finite difference method, perturbation
techniques, series techniques, etc. have been used to handle vibration problems.
The Variational Iteration Method, VIM, was first proposed by He [2-11] and the method
has been applied to investigate many nonlinear partial differential equations, autonomous
and singular ordinary differential equations such that solitary wave solutions, rational
solutions, compacton solutions and other types of solution were found by Abdou and
Soliman [12, 13]. Additonally, He [4, 9] used VIM to solve linear/nonlinear vibration
problems.
In this study He’s studies are extended to cover vibration problems with damping, forced
vibration and vibration of beams. Therefore, here VIM is applied to various vibration
problems including vibration of linear/nonlinear, damped/undamped, free/forced vibrations
of one degree of freedom systems and beams as an example of continuous systems. The
procedure presented in this paper can be simply extended to solve more complex vibration
problems; such as aeroelasticity, random vibrations etc.
2. Variational Iteration Method
In order to illustrate the basic concepts of VIM, the following nonlinear partial differential
equation can be considered
Lu( x, t )  Ru( x, t )  Nu( x, t )  g ( x, t )
(1)
2
where R is a linear operator which has partial derivatives with respect to x , L is the linear
time derivative operator, Nu ( x, t ) is a nonlinear term and g ( x, t ) is an inhomogeneous term.
According to VIM, the following iteration formula can be constructed.
t
un1 ( x, t )  un ( x, t )     Lun  Run  Nun  g  d
(2)
0
where  is the general Lagrange multiplier which can be identified optimally via
variational theory, Run and Nun are considered as restricted variations, i.e.
 Run  0 ,  Nun  0 .
3. Different Form of L Operator
According to the L operator, a general classification of vibration problems can be made as
follows
3.1.
Case I:
Here the following form of the L operator is used
Lm
2
t 2
(3)
Considering Eq. (3), Eq. (1) can be expressed as follows
mu ( x, t )  Ru( x, t )  Nu( x, t )  g ( x, t )
(4)
3
The correction functional of Eq. (4) can be written as
t
un1 ( x, t )  un ( x, t )     mun  Run  Nun  g  d
(5)
0
Making the above correction functional stationary, and noticing that  u(0)  0 , the
following iteration can be written
t
 un1 ( x, t )   un ( x, t )      mun  Run  Nun  g  d
(6a)
0
t
  un (t )  m ( ) un ( )  t  m ( ) un ( )  t   m  un d  0
(6b)
0
which yields the following stationary conditions:
m   0
(7a)
1  m   t  0
(7b)
  t  0
(7c)
Therefore, in this case the Lagrange multiplier can be identified as follows

1
(  t )
m
(8)
4
3.2.
Case II:
Here the following form of the L operator is used
Lm
2

c
2
t
t
(9)
Considering Eq. (9), Eq. (1) can be expressed as follows
mu ( x, t )  cu( x, t )  Ru( x, t )  Nu( x, t )  g ( x, t )
(10)
Here Eq. (10) denotes vibration with damping where c is the damping coefficient and m is
the mass.
The correction functional of Eq. (10) can be written as
t
un1 ( x, t )  un ( x, t )     mun  Run  Nun  g  d
(11)
0
Making the above correction functional stationary, and noticing that  u(0)  0 , the
following iteration can be written
t
 un1 ( x, t )   un ( x, t )      mun  cu  Run  Nun  g  d
(12a)
0
t
  un (t )  m ( ) un ( )  t  [c  m ( )] un ( )  t   (m   c ) un d  0 (12b)
0
which yields the following stationary conditions
5
m   c  0
1  m   c

 t
 t
(13a)
0
(13b)
0
(13c)
Combining Eqs. (13b) and (13c), Eqs. (13a)-(13c) can be rewritten as follows
m   c  0
(14a)
1  m   t  0
(14b)

(14c)
 t
0
Therefore, in this case the Lagrange multiplier can be identified as follows
1 mc ( t )
  [e
 1]
c
(15)
Approximate Lagrange multiplier can be obtained simply by expanding Eq. (15) as follows
1 mc ( t )
1
c
c2
2
[e
 1]  (  t ) 
(


t
)

(  t )3
c
m
2! m2
3! m3
(16)
Hence,

1
c
c2
2
(  t ) 
(


t
)

(  t )3
2
3
m
2! m
3! m
(17)
6
3.3.
Case III:
Here the following form of the L operator is used
Lm
2
k
t 2
(18)
Considering Eq. (18), Eq. (1) can be expressed as follows
mu ( x, t )  ku( x, t )  Ru( x, t )  Nu( x, t )  g ( x, t )
(19)
whre k is the spring coefficient.
The correction functional of Eq. (19) can be written as
t
un1 ( x, t )  un ( x, t )     mun  kun  Run  Nun  g  d
(20)
0
Making the above correction functional stationary, and noticing that  u(0)  0 , the
following iteration can be written
t
 un1 ( x, t )   un ( x, t )      mun  ku  Run  Nun  g  d
(21a)
0
t
  un (t )  m ( ) un ( )  t  m ( ) un ( )  t   (m   k  ) un d  0
(21b)
0
which yields the following stationary conditions
m   k  0
(22a)
7
1  m   t  0
(22b)

(22c)
 t
0
Therefore, in this case the Lagrange multiplier can be identified as follows

1
sin  (  t ) 
m
(23)
where the circular frequency,  , is given by

k
.
m
(24)
Approximate Lagrange multiplier can be obtained simply by expanding Eq. (23) as follows
1
(  t )  2 (  t )3
sin  (  t )  

m
m
3! m
(25)
Hence,
(  t )  2 (  t )3


m
3! m
3.4.
(26)
Case IV:
Here the following form of the L operator is used
Lm
2

c k
2
t
t
(27)
8
Considering Eq. (27), Eq. (1) can be expressed as follows
mu ( x, t )  cu( x, t )  ku( x, t )  Ru( x, t )  Nu ( x, t )  g ( x, t )
(28)
The correction functional of Eq. (28) can be written as
t
un1 ( x, t )  un ( x, t )    mun  cu  ku  Run  Nun  g  d
(29)
0
Making the above correction functional stationary, and noticing that  u(0)  0 , the
following iteration can be written
t
 un1 ( x, t )   un ( x, t )      mun  cu  ku  Run  Nun  g  d
(30a)
0
t
  un (t )  m ( ) un ( )  t  [c  m ( )] un ( )  t   (m   c   k  ) un d  0
(30b)
0
which yields the following stationary conditions
m   c   k  0
(31a)
1  m   c
(31b)

 t
 t
0
0
(31c)
Combining Eqs. (31b) and (31c), Eqs. (31a)-(31c) can be rewritten as follows
m   c   k  0
(32a)
9
1  m   t  0
(32b)

(32c)
 t
0
Therefore, in this case the Lagrange multiplier can be identified as follows
Sin[  (  t )] 2cm( t )

e
m
(33)
where
4mk  c 2
2m

(34)
Approximate Lagrange multiplier can be obtained simply by expanding Eq. (34) as follows


Sin[  (  t )] 2cm ( t )  (  t ) c(  t ) 2
1  c2 k 
3
e



 2   (  t ) 
2
m
2m
3! m  m m 
 m

(35)
Hence,
 (  t ) c(  t ) 2

1  c2 k 
3


 2   (  t ) 
2
2m
3! m  m m 
 m


10
(36)
4. Illustrative Examples
EXAMPLE 1:
In this example, a simple mass-spring system that undergoes forced vibration is examined.
The differential equation of motion of this system is given by
my  ky  f (t )
(37)
where f (t )  F0 cos(t )
Dividing both sides by m , Eq.(37) can be rewritten as follows
y   2 y  F0 cos(t )
where F0 
(38)
F0
.
m
Here it is easily seen that the Lagrange multiplier of this problem is

1
sin  (  t ) 
m
(39)
Additionally, the iteration formula of this problem is
t
yn1 (t )  yn (t ) 
1
sin  (  t ) [ my ( )  ky( )  f ( )]d
m 0
The complementary solution of this problem is given by
11
(40)
y0  A cos  t   B sin t 
(41)
Using Eq. (41) as an initial approximation, we get
y1 (t )  A cos t   B sin t  
1
t
 0
sin  (  t ) [ F0 cos( )]d
(42)
or
y1 t   A cost   B sin t  
F0 cost  F0 cost 
 2
 2  2
  2
(43)
Since the last term in Eq. (43) automatically satisfies the complementary equation, this term
will not be used. Thus, Eq. (43) can be simplified to
y1 (t )  A cos t   B sin t  
F0 cos(t )
 2  2
(44)
which is the general solution of Eq. (37).
In order to point out the importance of using the exact Lagrange multiplier instead of the
approximate one, the following multiplier may be considered

(  t )
m
(45)
Considering Eq. (45), the following iteration expression can be written
t
yn1 (t )  yn (t )   (  t )[ y ( )   2 y( ) 
0
f ( )
]d
m
12
(46)
If we use the complementary solution given by Eq.(41) as an initial approximation, we get
t
y1 (t )  A cos t   B sin t    (  t )[ F0 cos( )]d
(47a)
0
or in compact form
y1  A cos  t   B sin t   F0 [
1  Cos(t )
]
2
(47b)
and
t

 2
 2Cos( ) 
1  Cos(t )

Cos
(


)

 F0 cos( )  d
y2  A cos t   B sin t   F0 [
]   (  t )  F0  2

2
2




0
 
(48a)
or in compact form
 1 2 
 1 2 
 2t 2
y2  A cos t   B sin t   F0  2  4   F0Cos(t )  2  4   F0
22
  
  
(48b)
In the same way, y 4 can be written as follows
 1 2 4 6 
 1 2 4 6 
y4  A cos t   B sin t   F0  2  4  6  8   F0 cost  2  4  6  8 


 
    


F0 2 t 2
2
 1  2  4  F0 4 t 4
 2  4  6  
24

 

Since,
13
 1  2  F0 6 t 6
 2  4  
720
 

 1 
 2
 
(49)
1
1 2 4 6





 2   2  2  4  6 8
(50)
The expression for y n can be written as follows
yn  A cos t   B sin t  
F0Cos(t )
F0   2t 2  4t 4 F0 6t 6

1



 2  2
 2  2 
2
24
720



(51)
As n   , the compact form of solution becomes
y(t )  A cos t   B sin t  
F0 [cos(t )  cos(t )]
 2  2
(52)
Note that Eq. (52) is the same as Eq. (47b).
If we use a different approximate form of the Lagrange multiplier,  , such as

(  t )  2 (  t )3

m
3! m
(53)
the following iteration expression can be written
t
yn1 (t )  yn (t )   [(  t ) 
0
 2 (  t )3
3!
][ y ( )   2 y( ) 
f ( )
]d
m
(54)
Again using the complementary solution given by Eq.(41) as an initial approximation, we
get
t
y1 (t )  A cos t   B sin t    [(  t ) 
0
 2 (  t )3
3!
14
][ F0 cos( )]d
(55a)
or in compact form
 1 2 
 1 2 
 2t 2
y1  A cos t   B sin t   F0  2  4   F0Cos(t )  2  4   F0
22
  
  
(55b)
and
 1 2 4 6 
 1 2 4 6 
y2  A cos t   B sin t   F0  2  4  6  8   F0 cost  2  4  6  8 


 
    


F0 2 t 2
2
 1  2  4  F0 4 t 4
 2  4  6  
24

 

 1  2  F0 6 t 6
 2  4  
720
 

 1 
 2
 
(56)
(  t )  2 (  t )3

It can be easily seen that approximate Lagrange multiplier  
m
3! m
converges faster than  
(  t )
.
m
EXAMPLE 2:
In this example, a damped mass-spring system that undergoes forced vibration is examined.
The differential equation of motion of this system is given by
my  cy  ky  F0 cos(t )
(57)
The Lagrange multiplier of this problem is

Sin[  (  t )] 2cm( t )
e
m
(58)
15
where
4mk  c 2
2m

(59)
Hence using this Lagrange multiplier, the iteration formula can be written as
t
yn1 (t )  yn (t ) 
c
( t )
1
2m
e
Sin[  (  t )][my ( )  cy ( )  ky( )  f ( )]d

m 0
(60)
Dividing both sides of Eq. (57) by m , the following equation of motion is obtained
y  2 y   2 y  F0 cos(t )
(61)
where
F0 
F0
c
  .
and
m
2m
Therefore, Eq. (60) becomes
yn1 (t )  yn (t ) 
1

t
c
 e 2m
( t )
Sin[  (  t )] y ( )  2 y ( )  2 y ( )  F0 cos( )  d
(62)
0
and
  1   2
(63)
The complementary solution of this problem is given by
16
y0   A cos   t   B sin   t  et
(64)
Taking this complementary solution as an initial approximation, the following expression is
obtained
y1 (t )  y0 (t ) 
1

t
e
c
( t )
2m
Sin[  (  t )][ F0 cos( )]d
(65)
0
Integrating the second term of Eq. (65), we get
y1 (t )  y0 (t ) 
( 2  2 )Cos[t ]  2Sin[t ]
 4  2(2 2  1) 22  4
(2   2 )Cos[  t ]et
 (2   2 ) Sin[  t ]et
 4

  2(2 2  1) 22  4  4  2(2 2  1) 22  4  1   2


(66)
Since the last two terms of Eq. (66) automatically satisfy the homogeneous equation, they
will not be used. The second term of Eq. (66) can be written in a more compact form as
follows
( 2  2 )Cos[t ]  2Sin[t ]

 4  2(2 2  1) 22  4
Cos(t   )
1  
2
/
   2 /  
2 2
(67)
2
where
tan( ) 
2 / 
1  2 /  2
(68)
Hence, the exact solution of Eq. (57) is obtained after one iteration
17
y1   A cos   t   B sin   t  et 
F0Cos(t   )
1  
2
/
   2 /  
2 2
(69)
2
EXAMPLE 3:
In this example, transverse vibration of a uniform beam with simply supported ends is
examined. The differential equation of motion of this system is given by
4
2 y
2  y

c
 0, 0  x  L, t > 0
t 2
x 4
(70)
The initial and the boundary conditions for this problem are as follows:
y ( x,0)  sin
x
(71a)
L
y t ( x,0)  0
(71b)
y (0, t )  0
(71c)
y ( L, t )  0
(71d)
yxx (0, t )  0
(71e)
yxx ( L, t )  0
(71f)
y t ( x,0)  0 , yxx (0, t )  0
The Lagrange multiplier of this problem is
18
It is easily noticed that the Lagrange multiplier of this problem is

1
(  t )
m
(72)
Using this Lagrange multiplier, the iteration formula can be written as
t
yn1 (t )  yn (t ) 
1
(  t )[ y  c 2 yxxxx ]d

m0
(73)
The complementary solution of this problem is given by
y0 (t )  sin
x
(74)
L
Using the iteration formula given by Eq. (73) and taking the complementary solution as an
initial approximation, we get
y1 (t )  sin
x
L

t
 c2 4
 x c 2 4t 2  x
1
x

sin

sin
(


t
)
sin
d

 L4
L
2 L4
L
m 0
L 

(75)
and
y2 (t )  sin
t
 c2 4  x 2   4
 x c2 4t 2  x 1
 x c 2 8 2
 x 

c
sin

sin

sin

(


t
)

sin
 4
  d

8
4
4

L
2L
L 
L
2L
L m0
L
L
 L
 sin
x
L

c 2 4t 2
 x c 4 8t 4  x
sin

sin
2! L4
L
4! L8
L
Hence the expression for y n can be written as follows
19
(76)
yn (t )  sin
x
c2 4t 2 c4 8t 4
1




L 
2! L4
4! L8
(1) n
c2 n 4 nt 2 n 

(2n)! L4 n 
(77)
Thus we have,
lim yn (t )  sin
x
n
L
cos
 2 ct
(78)
L2
which is the exact solution.
EXAMPLE 4:
In this example, transverse vibration of a variable coefficient beam that was studied by
Wazwaz [14] is considered. The differential equation of motion of this system is given by
2 y
4 y


(
x
)
 f ( x, t ) ,  ( x )  0 , 0  x  L , t > 0
t 2
x 4
(79)
Here it can be that the equation of motion of the beam is as follows:
2 y
4 y
6

(1

x
)
 ( x 4  x 3  )cos t
2
4
t
x
7!
(80)
The initial and the boundary conditions for this problem are as follows:
y ( x,0) 
6 7
x ,
7!
y t ( x,0)  0 ,
0  x 1
(81a)
0  x 1
(81b)
20
y (0, t )  0 .
y (1, t ) 
6
cos t ,
7!
yxx (0, t )  0 ,
y xx (1, t ) 
1
cos t ,
20
t>0
(81c)
t>0
(81d)
t>0
(81e)
t>0
(81f)
The Lagrange multiplier of this problem is

1
(  t )
m
and m  1 for this problem. Hence the iteration formula can be written as
t
yn1 (t )  yn (t )   (  t )[ y  (1  x) y xxxx  ( x 4  x 3 
0
6
)cos ]d
7!
(82)
The complementary solution of this problem that is used as an initial approximation is
given by
y0 ( t ) 
6 7
x cos t  ( x 3  x 4 )(1  cos t )
7!
(83)
By using the iteration expression given by Eq. (82) and the initial approximation given by
Eq. (83), we get
y1 (t ) 
6 7
x cos t  ( x 3  x 4 )(1  cos t )  (1  x)(24  12t 2  x 3  (24  x 3 )cos t
7!
21
(84a)
or in compact form
y1 (t )  12(t 2  2)(1  x)  (
x7
 24 x  24)cos t
7!
(84b)
and
y2 (t )  12(t 2  2)(1  x)  (
x7
 24 x  24)cos t  12(1  x)(t 2  2  2cos t )
7!
(85)
Eq. (85) can be simplified to the following expression
y2 ( t ) 
x7
cos t
7!
(86)
As it is seen, the exact solution of this problem is obtained quickly in two iteration.
EXAMPLE 5:
In this example, Mathematical Pendulum that was studied by He [4, 9] is considered.
The differential equation of motion of the undamped mathematical pendulum is given by,
y   2 sin y  0
(87)
The initial conditions for this problem are as follows:
y (0)  A
(88a)
y (0)  0
(88b)
22
The Siny term in Eq. (87) is a nonlinear term and it can be expanded as
sin y  y 
1 3
y
6
(89)
Substituting Eq. (89) into Eq. (87) gives
y  y 
2
2
6
y3  0
(90)
A more detailed form of this mathematical pendulum was investigated by He [5, 7].
The Lagrange multiplier of this problem is

1

sin  (  t ) 
(91)
Hence the iteration formula is
yn1 (t )  yn (t ) 
1

t
2
 sin (  t )[ y( )   y( ) 
0
2
6
y 3 ( )]d
(92)
The complementary solution of this problem that is used as an initial approximation is
given by
y0 (t )  A cos(t )
(93)
where  is an unknown constant.
Substituting the initial approximation into Eq. (90), the following residual is obtained
23
R0 (t )  y   2 y 
2
1
1 3 2
y 3  A(1  A2   2 ) 2 cos(t ) 
A  cos(3t )
6
8
24
(94)
The coefficient of the cos( t ) term is set to zero in order to eliminate the secular term
which may occur in the next iteration. Doing so, the expression of  is found as follows
A2
  1
8
(95)
Hence,
A3
y1 (t )  A cost 
(cos3t  cost )
24(9 2  1) 2
(96)
with  defined in Eq. (95).
The period can be expressed as follows
T
2
1
 1  A2
8
If A 

2
(97)
, then T  1.20T0 . On the other hand He’s [4, 9] approximation gives T  1.17T0 ,
while the exact period is Tex  1.16T0 , where T0  2 /  .
24
EXAMPLE 6:
In this example, the problem that was studied by Nayfeh and Mook [15] is considered.
The differential equation of motion is given by,
u   2u   u 2u  0
(98)
The initial conditions for this problem are as follows:
y (0)  A
(99a)
y (0)  0
(99b)
The Lagrange multiplier of this problem is

1

sin  (  t ) 
(100)
The iteration formula is given by
un1 (t )  un (t ) 
1
t
sin  (  t ) [u( )   u( )   u ( )u( )]d

2
2
(101)
0
The complementary solution of this problem that is used as an initial approximation is
given by
u0 (t )  A cos(t )
(102)
where  is an unknown constant.
25
Substituting the initial approximation given by Eq. (102), the following residual is obtained
as follows
R0 (t )  u   2u   u 2u  A 2 (1   2 
3 2
1
A  )cos(t )   A3 2 2 cos(3 t )
4
4
(103)
The coefficient of the cos( t ) term is set to zero in order to eliminate the secular term
which may occur in the next iteration. Doing so, the expression for  is obtained as
follows

2
(104)
4  3 A2
Hence,
y1 (t )  A cost 
 A3 2
(cost  cos3t )
4(9 2  1)
(105)
with  defined in Eq. (104).
The new frequency is defined as follows
1  

1 
2
4  3 A2

(106)
The frequency that is obtained by Nayfeh and Mook [15] using the perturbation method is
3
8
1   (1   A2 )
(107)
26
Note that Eq. (107) is valid only for small  values. However, the frequency expression
given by Eq. (106) is valid for all  values and takes the following form for small  values
3
8
1  1   A2 
27 2 4
 A 
128
(108)
EXAMPLE 7:
In this example, the Duffing-harmonic oscillator that was studied by Lim and Wu [16] and
Mickens [17] is considered.
The differential equation of motion is given by,
d2 y
y3

0
dt 2 1  y 2
(109)
The initial conditions for this problem are as follows:
y (0)  A
(110a)
y (0)  0
(110b)
with initial conditions y (0)  A and y (0)  0 .
For small y values, Eq. (109) reduces to
d2 y
 y3
dt 2
0
(111a)
On the other hand, for large y values, Eq. (109) reduces to
27
d2 y
y
dt 2
(111b)
0
Considering Eqs. (111a) and (111b) respectively, it is noticed that for small y values, Eq.
(109) reduces to the equation of motion of the Duffing-type nonlinear oscillator while for
large y values, it reduces to the equation of motion of a linear harmonic oscillator.
Therefore, Eq. (109) is called as Duffing-harmonic oscillator equation of motion.
The following form of Eq. (109) is going to be studied in this example
(1  y 2 )
d2 y
 y3  0
2
dt
(112)
He’s technique is going to be used to overcome seculer terms that appear in the iterations.
The initial approximation is,
y0 (t )  A cos( t )
(113)
where  is an unknown constant.
Substituting the initial approximation into Eq. (112), the following residual is obtained
3
3
A3
R0 (t )  (1  y 2 ) y  y 3  ( A2   2  A2 2 )cos( t )  (1   2 )cos(3 t )
4
4
4
(114)
In oreder to discard the seculer terms, the coefficient of cos( t ) is set to zero which gives
the expression of  as follows
28
3 2
A
4

3
1  A2
4
(115)
Hence the new frequency is defined as follows
3 2
A
4

3
1  A2
4
(116)
which is the same with the one found by Lim and Wu [16] and Mickens [17].
The iteration formula is given by
t
u1 (t )  u0 (t )   (  t )[
0
A3
(1   2 )cos(3 )]d
4
(117)
Hence,
y1 (t )  cos t 
A
(cos3t  1)
27
(118)
with  defined in Eq. (116).
For small values of amplitude A, the frequency expression given in Eq. (116) is expressed
as follows

3
A
4
(119a)
29
Additionally, for large values of amplitude A, the frequency expression given in Eq. (116)
is expressed as follows
 1
(119b)
which agree with the approximations made for the equations of motion given in Eqs. (111a)
and (111b).
5. Conclusion
In thispaper, which extends He’s studies , various types of vibration problems including
vibration of linear/nonlinear, damped/undamped, free/forced vibrations of one degree of
freedom systems and beams as an example of continuous systems are solved using the
Variational Iteration Method. Lagrange multipliers which arise from different types of
vibration problems are presented and the solutions are made in a detailed way.
Additionally, the procedure presented in this paper can be simply extended to solve more
complex vibration problems; such as aeroelasticity, random vibrations etc.
30
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[3] J.H. He, Approximate analytical solution for seepage flow with fractional derivatives in
porous media, Comput. Methods Appl. Mech. Eng. 167 (1998) 57–68.
[4] J.H. He, Approximate solution of nonlinear differential equations with convolution
product nonlinearities, Comput. Methods Appl. Mech. Eng. 167 (1998) 69–73.
[5] J.H. He, Variational iteration method—a kind of non-linear analytical technique: some
examples, Internat. J. Nonlinear Mech. 34 (1999) 699–708.
[6] J.H. He, Variational iteration method for autonomous ordinary differential systems,
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[9] J.H. He, A generalized variational principle in micromorphic thermoelasticity, Mech.
Res. Comm. 32 (1) (2005) 93–98.
31
[10] J.H. He, Some asymptotic methods for strongly nonlinearly equations, Internat. J.
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[11] J.H. He, Variational iteration method—Some recent results and new interpretations,
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[12] M.A. Abdou, A.A. Soliman, New applications of variational iteration method, Physica
D 211 (2005) 1–8.
[13] M.A. Abdou, A.A. Soliman, Variational iteration method for solving Burger’s and
coupled Burger’s equations J. Comput. Appl. Math. 181 (2005) 245–251.
[14] A. M. Wazwaz, Analytic treatment for variable coefficient fourth-order parabolic
partial differential equation, Appl. Math. Comput. 123 (2001) 219-227.
[15] A. H. Nayfeh and D. T. Mook, Nonlinear Oscillation, John Wiley & Sons, 1979
[16] C.W. Lim and B.S. Wu, A new analytical approach to the Duffing-harmonic oscillator,
Physics Letters A 311 (2003) 365–373
[17] R. E. Mickens, Mathematical and Numerical Study of the Duffing Harmonic
Oscillator, Journal of Sound and Vibration (2001) 244(3), 563-567.
32