Variational Iterative Method Applied to Variational Problems with Moving Boundaries

(1)

Variational Iterative Method Applied to Variational

Problems with Moving Boundaries

Fateme Ghomanjani, Sara Ghaderi

Department of Applied Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran Email: [email protected], [email protected]

Received January 3,2012; revised March 19, 2012; accepted March 26, 2012

ABSTRACT

In this paper, He’s variational iterative method has been applied to give exact solution of the Euler Lagrange equation which arises from the variational problems with moving boundaries and isoperimetric problems. In this method, general Lagrange multipliers are introduced to construct correction functional for the variational problems. The initial approxi-mations can be freely chosen with possible unknown constant, which can be determined by imposing the boundary con-ditions. Illustrative examples have been presented to demonstrate the efficiency and applicability of the variational it-erative method.

Keywords: Variational Iterative Method; Variational Problems; Moving Boundaries; Isoperimetric Problems

1. Introduction

In modeling a large class of problems arising in science, engineering and economics, it is necessary to minimize amounts of a certain functional. Because of the important role of this subject, special attention has been given to these problems. Such problems are called variational problems, see [1,2].

The simplest form of a variational problem can be considered as

 

1



   

0

, , d

x

v y x_ _



F x y x y x



x, (1)

where is the functional which its extremum must be found. Functional can be considered by two kinds of boundary conditions. In the fixed boundary problems, the admissible function

v

 

y x must satisfy following boun-

dary conditions

 

0 0,

 

1 1

y x y y x y (2)

In moving boundary problems, at least one of the boundary points of the admissible function is movable along a boundary curve. Further more many applications of the calculus of variations lead to problems in which not only boundary conditions, but also a quite different type of conditions known as constraints, are imposed on the admissible function. The necessary condition for the admissible solutions of such problems has to satisfy the Euler-Lagrange equation which is generally nonlinear.

In this work we consider He’s variational iterative method as a well known method for finding both analytic

and approximate solutions of differential equations. Here, the problem is initially approximated with possible un-knowns. Then a correction functional is constructed by a general Lagrange multiplier, which can be identified op-timally via the variational theory [3].

Variational iterative method is applied on various kinds of problems [4-31].

Author of [32] solved variational problems with mov-ing boundaries with Adomian decomposition method. Variational iterative method was applied to solve varia-tional problems with fixed boundaries (see [11,27,30]). In this work we obtain exact solution of variational problems with moving boundaries and isoperimetric problems by variational iterative method. In fact, varia-tional iterative method is applied to solve the Eu-ler-Lagrange equation with prescribed boundary condi-tions. To present a clear overview of the procedure sev-eral illustrative examples are included.

2. Variational Iterative Method

In variational iterative method which is stated by He [3], solutions of the problems are approximated by a set of functions that may include possible constants to be de-termined from the boundary conditions. In this method the problem is considered as

 

LyNyg x , (3)

where is a linear operator, and is a nonlinear

operator.

L N

 

g x is an inhomogeneos term. By using the

(2)

func-tional is taken into account

 

   



1 0

d , x

n n n n

y y 



 Ly s Ny s g s



s (4)

where  is Lagrange multiplier [5], the subscript n

denotes the n-th approximati yn is as a restricted

variation n 0

on,

i.e. y  [6-8]. Taking the variation from

both sides of the correct functional with respect to yn and

imposin n1 0, the stationary conditions are

ob-tained. By using the stationary conditions the optimal valu of the



 

g y

e  can be identified.



The successive approximation can be

es-tablished by determining a general lagrangian multiplier



1 k

y k

 and initial solution 0. Since this procedure avoids

the discretization of the problem, it is possible to find the closed form solution without any round off error.

y

In the case of m equations, the equations are rewritten in the form of:

 



1, ,



 

, 1, , ,

i i i m i

L y N y  y g x i  m

m

(5)

where _i is a linear with respect to _i, and _i is

nonlinear part of the ith equation. In this case the correct functionals are produced as

L y N

 







 



 





1

1 0

, , d ,

in i n

x

i i in n mn

y y

L y s N y s y s g s s



 



_

     (6)

and the optimal values of the  i,i 1, , are obtained by taking the variation from both sides of the correct functionals and finding stationary conditions using

 1 0, 1, ,

i n

y i m

 _    。

3. Statement of the Problem

3.1. Moving Boundary Problems

The necessary condition for the solution of problem (1) is to satisfy the Euler-Lagrange equation

,

d _0,

d

F F

y x y

 

 

  (7)

The general form of the variational problem (1) is





1



0

1, , ,2 , , , , , , , ,1 2 1 2 d ,

x

n n

x

v y y  y 



F x y y  y y y  yn



x (8)

Here, the necessary condition for the extermum of the functional (8) is to satisfy the following system of sec-ond-order differential equations

d

0, 1, 2, ,

d

i i

F F

i

y x y

 

  



   n

must be considered by the boundary conditions, but for

ith variable boundaries, we have tw

As the first case, those problems are consid-er

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In fixed boundary problems, Euler-Lagrange equation

the problems with variable boundaries, Euler-Lagrange equation must satisfy natural boundary conditions or transversality conditions which will be described in the following theorems.

For the problems w o cases:

Type 1:

ed for which at least one of the boundary points move freely along a line parallel to the y-axis. Indeed at this

point y x

 

is not specified. In this case all admissible

functions have the same domain



x x0, 1



and satisfy the

Euler-Lagrange equation in this al. Furthermore

such functions have to satisfy conditions called natural boundary conditions stated in the following theorem.

Theorem 3.1. Suppose the function interv

 

yy x in





1 0, 1

x x , yields a relative minimum of th nal

r which

C

(1) that fo

e functio

 

0 0

y x y , y x

 

1 y1 is arbitrary

(free right endpoint) and y x

 

0 ,

 

1 re arbitrary

(free endpoints). Then 0



y x a



y x the following

natural boundary conditio pectively:

satisfies, ns, res

   



x y1, 0 x1 ,y0 x1



0

y F

 _

 

 (10)

or

   





   



0 0 0 0 0

1 0 1 0 1

, ,

F

x y x y x y

F

x y x y x

y



0

 _

 

 _

 

 

(11)

Type 2: For the second case, the beginning and end points (or only one of them) can move freely on given curves y

 

x ,y

 

x . In this case, a function

 

y x is required, which emanates at some xx0

he curve

from t y

 

x and terminates for

1

some

xx on the curve y

 

x and minimizes the

func-1). In this p he points 0, 1

 

( roblem

tional , t x x are not

known, and they must satisfy the necessary conditions called transversality conditions, described in the follow-ing theorem.

Theorem 3 y y0

 

x C x x1



0,



, 0

.2. If the function 1

which emanates at some xx from the curve

 

1₍ _, ₎

y x C   and terminates for some xx1

on the curve

 

1



_,



C

y x    , yields a r

minimum for funct_{ional (1), where} 1

 

elative

FC R , R being

a domain in the



x y y, , 



space th all lineal

elements of

at contains

 

0

yy x , then it is necessary that

 

0

yy x to satisfy th Euler-Lagrange equation in the

interval

e



x x0, 1



and at the point of exit and the point of

entrance, llowing transversality conditions to be

satisfied: the fo

   







 



   





0 0 0 0 0 0 0 0

0 0 0 0 0

, ,

, , 0

F

x y x y x x y x

y

F x y x y x 

 _ _ _ _

 



 

(3)

   







 



   





1 0 1 0 1 1 0 1

1 0 1 0 1

, ,

, , 0

F

x y x y x x y x

y

F x y x y x 

 _ _ _ _

 



 

(13)

In the case that one of the points is fixed, then the transversality condition has to be held at the oth

One can consider transversality conditions for

lems with more than one unknown functions. For exam-pl

hich

er point. the

prob-e, in to minimize two dimensional casprob-e, a vector func-tion y x

 





y x1

   

,y2 x



is looked for such that





1





0

1, 2 , , , ' , ' d ,1 2 1 2 x

x

v y y 



F x y y y y x (14)

in w

 



₀



_2,

0 0

1 0 1,x , 2 y x and the end

y x y y x

a two-dimensional surface t



1 2



point

lies on hat is given by

,

xu y

1

y . Here the transversality conditions at

xx are:

 

0 0

1 2 1

1 0,

u

F y y F x

y y y y

 

 _{ } _  _

   

_ _ _ _ __ _  (15)

1

u u u

   

 

 

0 0

1 2 1

2 1 2 2

1 u u u 0,

F y y F x

y y y y

 

 _{ } _    _

   

_ _ _ _ __ _ 

  (16)

In which is an admissible vector

function.

For further information on transversality con

specially for the proofs of Theorems 3.1 and 3.2 and conditions (15), (16), see [2].

1. Consider the u

 

   



0 0



1 , 2 y x y x

ditions,

Example 3. following functional:

 



 

_*



2

0

d , T

J y 



a by t y t c t (17)

In which _, _0, * ₀

a b c  and y t

 

is the amount of

a capital at tim

Here, the capital stock at the initial

e t (see [1]).

 

0

y time t0

of the planning period is assumed to be known:

 

0 0

y y ; o nd, the nner won’t wish to

explain how large the capital would be at time t T

n the other ha pla

 .

Therefore, there is a varia l problem with free right

endpoint. Here we let * _1, ₁

a b c  T , and y0 2

tiona



whic the analytical solution

 

1 t.

y t  e The cor-

responding Euler-Lagrange equation is:

   

1 0.

y t y t  

Now natural boundary conditi is as

fol-lowing: h has

on at t1

   



1, 1 , 1



2



 

1



₁ 0

t f

y y y t y t

y 

 _ _{ } _ _ _ _

 

itio

Therefore, the following boundary cond ns are:

 

0 2, 1

 

1 1 0

y  y y   . (18)

By using variational iterative method we consid he

fo

sides of the correct

functional with respect to given:



)d

er t

llowing functional is considered:

 



 



1

t

n n n

n

y t y t 



 y z y z  0

Taking the variation from both

1 d ,z

n y

 



 





 





0

( 1

d 0

t

n

   







 

1

0

n n n n

n n _{z t} _z

t

n

y t y t y z y z z

z z y z

y z z

   



  



   



  



t

For all variations

y t z y

   



 

 

  

and  yn n

y

 . The following

sta-tionary conditions are obtained:

   

z z 0

   ,

 





0

z t z

 _ 

 



1



0

z t z

 _

 

 

1 1

2 2

z t t

z e e

  z

So that   . Therefore iterative for-

mula can be found as:

 





1

0 2 2

n t

z t

e e

 _ 





1 1 1 d ,

n t z

n n

y t y t

y z y z z



 



 _{  } _

 

If ₀ t t,

y Ae Be then

 





1

0

1 1 _{1 d}

2 2

2

1 1

1

2 2

t

t t z t t z

t

t t z t t z

t t

y t Ae Be e e z

Ae B e e

A e B e

  

 



 

   _  _ 

 

  

   

_  _ _  _ 

   



By imposing (18) 1

e 

3 1

,

2 2

A B are resulted. Which

yields the exact solutions of the problem (see Figure 1). Example 3.2. We want to find the shortest distance from the point A(1,1,1) to the sphere

2 2 2 ₁

x y z 

This problem is reduced to optimize the following functional:

 

_, 1 ₁ 2

 

2_{( )d}

J

1

x

y z 



y x z x x (19)

point



where the B x y z



1, ,1 1



act solution y x,

must lie on the sphere,

with the ex 1 z1x

equations

, see [33]. The cor-

(4)

approximate solution exact solution

–3 –2 –1 0 1 2 3

t

20

18

16

14

12

10

8

6

4

[image:4.595.65.287.85.307.2]

2

Figure 1. The graphs of approximated and exact solution for Example 3.1.

are:

 

2 2

dx_ ₁_y _x _z _x _

2 2

d

0,

d _0.

d ₁

y

z

x _y _x _z _x

 _ 

 _{ }

 _ 

 _{ }

 _ _ _ _ 

 

So that

 

2 2 , 2 2 .

1 1

y z

e f

y x z x y x z x

 

 

   

   

In above equations “e” and “f” are constant, so they

can be rewritten as:

 

2 2 2 2

1 0

y e y x x

f y x z

z

z x

 

    

 

   

,

.

The transversality conditions are:

1

2 2

2

2 2

1

0

x x

y y

y

x z

z



 _



   



  





 _ 



 



(20)

2 2 2 2

1 x y 1 y

z



 

 

 _ _  _ _ _ _ _

  

1

2 2 2

2 2 . 0

1 1 1

x x

z

z z

y y

y y x y _

 _  _

 

    

 _ 

 

 

 

  

 



 

2

 

2

 



1 1

0

1 n n d

n x

n n

y_ x y x 



 y s e y s z s s

and

 



 

2

 

2

 



1 2

0

1 n n d

x

n n n

z x z x 



 z s f y s z s s

The variation from both sides of above equations for finding the optimal value of  is:

 

1 1

0

n n n

 

1

 

1

 

0 d

d 0

x

n n _{s x} n

y x y x y s s

y x y s y s s

   

   _  



 

  _ _ 

_





and

 

1 2

0

d n x

n n

s x

z x z x z s s

 _   





 

 

2

 

2

 

0

d 0

x

n n n

z x z s z s s

    

 _ _ 

_





Therefore

 

1 1

1 0,

s x s x

s s

 _  _ 0

  

   

    .

and

 

2 2

1 0,

s x s x

s s

 _  _ 0

  

   

   

which yields:

 

1 s 1, 2 s 1.

    

So that the following iterative formulas are obtained:

 



 



1

2 2 0

1 ( ) 1 n n( ) d

n n

x n

y x y x

y s e y s z s s

 

  

 

_

  

 





1

2 2 0

( 1) 1 n n

n n

x n

z x z x

z s f y s z s s

 

  

 

_

   

If

d

 

0 , 0

y x ax b z x cxd then we have:



2 2



1 1 d

x

y ax b 



a e a c s 0

2 2 1

e a c x b

   

and



2 2



1

0

2 2

1 d

1 x

z cx d c f a c s

f a c x d

     

   



2 (21)

(5)







1 1 , 1 1



.

y   b x b z  d xd

1 3 0,

Imposing (20) and (21) lead to, 0, .

3

b d x 

therefore:

which is the exact solution.

3.2. Isoperimetric Problems

Assume that two functions 1 , 1 . y x z x



, ,



G x y y and F x y y



, , 



are given. Among all curves

 

1





C x x0, 1 along yy x 

which the functional

 

1



, ,



d x

0

x

K y 



G x y y x

assumes a given value l, determine the one for which the

functional

 

0

x





1

, , d

x

J y 



F x y y x

ose that F and G have

partial derivatives for Gives an extermal value. Supp

continuous first and second

0 1

x  x x and for arbitrary les y

and y.

Euler’s theorem: I

values of the variab

f a curve yy x

 

d

y x

extremizes the

functional J

 





1

0

, , x

x

y 



F x y  unde tr he conditions

 

1





0

, , d ,

x x

K y 



G x y y xl

 

0 0,

 

1 1

y x y y x y

and if yy x

 

is not an extremal of the functional K,

there exists a constant  such that the curve yy x

 

is an e al of the functional

sary condition for th

lem is to satisfy the Euler-Lagrange equation xtrem





1

, , d

x

L



_F x y y   _ x

The neces e solution of this

prob-

0

, , x

G x y y

d ₀

d

H H

y x y

 

 

  

with given boundary conditions in which H F G

for further information (see [2]). ple 3.3.

Exam It is aimed to find the minimum of the

functional

π

 

2

 

0 d

J y 



y x x (22)

Such that

and

 

π

2

0

d 1

y x x



(23)

 

0 0,

 

π 0

y  y  (24)

With exact solution y x

 

  2sinx [19].

Accord-

x

an sponding Euler-Lagrange equation:

π

ing to the following auxiliary functional:



1

2 2

0

d

L



y y

d the corre

 

y 0 d

2 2

d

y x

   

so

0

y y

By applying He’s variational iterative method results

To find the optimal value of

 





0

d

y 1 x y





 

y s x

n n x 



 yn s 

 following equation is required:

 



 



1

n n n _{s x}

x

y x y x  y s _

 





 

0

[ yn s ]s x yn s yn s ds

   

  _    

  

 

   0

 

_

 

herefore, the stationary conditions are obtained in the following form:

T













( ) 0,

0. s x

s x

s   

 

 

1_{s x}_ 0,

 

which yields

s x   

and the desired sequence is

 

  

  



 



1

0

d

x

n n n n

y_ x y x 



sx y s y s s

By choosing y₀ a sin

 

cx bcos

 

cx

 





 





 





 

1

2 2

0

2 2 2

cos

( ) sin cos d

sin cos

x

x b cx

sin

y x a c

s x ac a cs b b cs s

a cx b cx b ax

b acx

c

c c

 

   

      

    



Imposing (24) on this function given

c

 

 

 

1 2

si

a

 n

0, cx ax

b y x acx

c c



(6)

0

 

If then from (24) ac0, but from (23)

3 3 ,

π

ac which is a contra

Now imposing (24), we have: diction.

 

3_π

sin π π

c

c c

 

 

so 0. and it is known that in this case imposing (24) on the Euler Lagrange equation yields





2

, 1,

c     k k 2,

Hence:

 

sin

y x a kx

and from (23) 2

π

a  . But y must be extremal when

, therefore:

0 x π

 

2sin

π

y x   x

lution is equal to e ct

4 d

As it is observed that this so xa

solution (see Figure 2).

Example 3.4. The objective is to find an extremum of the functional

   

1



2 2



0

, 4

x z x y z xz z

J y 



     x (26)

Such that

and

2

0

)d 2

z x

    (27) 1

2 (y xy



x

–3 –2 –1 0 1 2 3 0.6

 

0 0, 0

 

0, 1

 

1, 1

 

1

y  z  y  z  . (28)

With exact solution

 

2

7 5

2

x x

x   , z x

 

x, see

[33]. By having the following auxiliary functional:

The system of Euler-Lagrange equations is in the form:









1

2 2

0

2 ₄ ₄ 2 _d

L



y z  xz z y xy z x









d d

2 2 0, 4 2 4 2

dx y yx  dx z x z 0.

So



2 2 



y  0, 2 2



 



y0.

By using Homotopy variational iterative method gives:

Now

 



  



 



  



1 1

1 2

0

2 2 d ,

2 2 d .

x

n n n

x

n n n

y x y x y s s

x z s s

  

 





   



 



 

0

z x z

 









 









 





 





1 1

1

2 2

2 2 d 0.

n n n

0

s x

y x y x y

  

   n

s x x

n

s

y s

y s s

 

  

 _



   

 _{ } _

  

_  _





_

 

therefore





1

1 2 2 0,

s x

  _









1

2 2 0,

2 2 0.

s x

 



  

 

 

 

 

 

  

 

 

Hence

0.4

0.2

–0.2

–0.4

[image:6.595.55.532.71.736.2]

–0.6

1 _{2 2}

s x 

  



and

 





 







 









 





1 2

2

2 0

2 2

2 2 d 0

n n _{s x}

n _{s x}

n n

x

z x z x z s

z s

z s s

    

  

 _



  

 _  _

  

_  _





_

 

so













2

1 2 2 0,

2 2 0,

2 2 0.

s x

 





  

 

 

 

 

 

  

 

(7)

So 2 is obtained as:

2

2 2

s x 

  



and the following iterative equations are obtained:

 



  



1

0

2 2 d

2 2

x

n n n

s x

y x y x  y s 





 _

   





s,

 



  



1

0

2 2 d .

2 2 x

n n n

s x

z x z x  z s s





 _

  





By choosing y0ax b z , 0cxd:

 

1

0 2

1

d 2 2

,

4(1 )

x

s x

y x ax b s

x ax b

z cx d

  

 

   



  



 



And by imposing (28) on this functions:





 

_

_

_

2

_

1

1 ,

4 1

a

  

 0, 1, 0,

1 ,

4 1 4 1

b c d

x

y x x



 

  

 

 _ _

 

from (27):

z x

 





1 2

10 12

,

11 11

     

And consequently:

x

–10

–20

–30

[image:7.595.57.298.118.708.2]

–3 –2 –1 0 1 2 3

 

7 5 2,

 

2

x x

y x   z x x.

which is the exact solution (see Figure 3).

4. Conclusion

The He’s variational iterative method is an efficient me-thod for solving various kinds of problems. In this paper variational iterative method is employed for finding the minimum of a functional with moving boundaries and isoperimetric problems. Using He’s variational iterative method the solution of the problem is provided in a closed form. Since this method does not need to the dis-cretize of the variables, there is no computational round off error. Moreover, only a few numbers of iterations are needed to obtain a satisfactory result.

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