Variational Iterative Method Applied to Variational Problems with Moving Boundaries ()
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)
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 
must satisfy following boundary conditions
(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 unknowns. Then a correction functional is constructed by a general Lagrange multiplier, which can be identified optimally via the variational theory [3].
Variational iterative method is applied on various kinds of problems [4-31].
Author of [32] solved variational problems with moving boundaries with Adomian decomposition method. Variational iterative method was applied to solve variational 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, variational iterative method is applied to solve the Euler-Lagrange equation with prescribed boundary conditions. To present a clear overview of the procedure several 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 determined from the boundary conditions. In this method the problem is considered as
, (3)
where 
is a linear operator, and 
is a nonlinear operator. 
is an inhomogeneos term. By using the variational iterative method, the following correct functional is taken into account
(4)
where 
is Lagrange multiplier [5], the subscript n denotes the n-th approximation, 
is as a restricted variation i.e. 
[6-8]. Taking the variation from both sides of the correct functional with respect to yn and imposing
, the stationary conditions are obtained. By using the stationary conditions the optimal value of the 
can be identified.
The successive approximation 
can be established by determining a general lagrangian multiplier 
and initial solution
. Since this procedure avoids the discretization of the problem, it is possible to find the closed form solution without any round off error.
In the case of m equations, the equations are rewritten in the form of:
(5)
where 
is a linear with respect to
, and 
is nonlinear part of the ith equation. In this case the correct functionals are produced as
(6)
and the optimal values of the 
are obtained by taking the variation from both sides of the correct functionals and finding stationary conditions using
。
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
(7)
The general form of the variational problem (1) is
(8)
Here, the necessary condition for the extermum of the functional (8) is to satisfy the following system of second-order differential equations
(9)
In fixed boundary problems, Euler-Lagrange equation must be considered by the boundary conditions, but for 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 with variable boundaries, we have two cases:
Type 1: As the first case, those problems are considered for which at least one of the boundary points move freely along a line parallel to the y-axis. Indeed at this point 
is not specified. In this case all admissible functions have the same domain 
and satisfy the Euler-Lagrange equation in this interval. Furthermore such functions have to satisfy conditions called natural boundary conditions stated in the following theorem.
Theorem 3.1. Suppose the function 
in
, yields a relative minimum of the functional (1) that for which
, 
is arbitrary (free right endpoint) and 
are arbitrary (free endpoints). Then 
satisfies, the following natural boundary conditions, respectively:
(10)
or
(11)
Type 2: For the second case, the beginning and end points (or only one of them) can move freely on given curves
. In this case, a function 
is required, which emanates at some 
from the curve 
and terminates for some 
on the curve 
and minimizes the functional (1). In this problem, the points 
are not known, and they must satisfy the necessary conditions called transversality conditions, described in the following theorem.
Theorem 3.2. If the function
, which emanates at some
from the curve 
and terminates for some 
on the curve 
, yields a relative minimum for functional (1), where
, R being a domain in the 
space that contains all lineal elements of
, then it is necessary that 
to satisfy the Euler-Lagrange equation in the interval 
and at the point of exit and the point of entrance, the following transversality conditions to be satisfied:
(12)
(13)
In the case that one of the points is fixed, then the transversality condition has to be held at the other point.
One can consider transversality conditions for the problems with more than one unknown functions. For example, in to minimize two dimensional case, a vector function 
is looked for such that
(14)
in which 
and the endpoint lies on a two-dimensional surface that is given by
. Here the transversality conditions at 
are:
(15)
(16)
In which 
is an admissible vector function.
For further information on transversality conditions, specially for the proofs of Theorems 3.1 and 3.2 and conditions (15), (16), see [2].
Example 3.1. Consider the following functional:
(17)
In which 
and 
is the amount of a capital at time t (see [1]).
Here, the capital stock 
at the initial time 
of the planning period is assumed to be known:
; on the other hand, the planner won’t wish to explain how large the capital would be at time
. Therefore, there is a variational problem with free right endpoint. Here we let
, and 
which has the analytical solution 
The corresponding Euler-Lagrange equation is:
Now natural boundary condition at 
is as following:
Therefore, the following boundary conditions are:
(18)
By using variational iterative method we consider the following functional is considered:
Taking the variation from both sides of the correct functional with respect to 
given:
For all variations 
and
. The following stationary conditions are obtained:
,
So that
. Therefore iterative formula can be found as:
If 
then
By imposing (18) 
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 
to the sphere
This problem is reduced to optimize the following functional:
(19)
where the point 
must lie on the sphere, with the exact solution
, see [33]. The corresponding Euler Lagrange equations for this problem
Figure 1. The graphs of approximated and exact solution for Example 3.1.
are:
So that
In above equations “e” and “f” are constant, so they can be rewritten as:
The transversality conditions are:
(20)
(21)
By using variational iteration method results:
and
The variation from both sides of above equations for finding the optimal value of 
is:
and
Therefore
.
and
which yields:
So that the following iterative formulas are obtained:
If 
then we have:
and
By choosing
,
Imposing (20) and (21) lead to, 
therefore:
which is the exact solution.
3.2. Isoperimetric Problems
Assume that two functions 
and 
are given. Among all curves 
along which the functional
assumes a given value l, determine the one for which the functional
Gives an extermal value. Suppose that F and G have continuous first and second partial derivatives for 
and for arbitrary values of the variables 
and
.
Euler’s theorem: If a curve 
extremizes the functional J 
under the conditions
and if 
is not an extremal of the functional K, there exists a constant 
such that the curve 
is an extremal of the functional
The necessary condition for the solution of this problem is to satisfy the Euler-Lagrange equation
with given boundary conditions in which 
for further information (see [2]).
Example 3.3. It is aimed to find the minimum of the functional
(22)
Such that
(23)
and
(24)
With exact solution 
[19]. According to the following auxiliary functional:
and the corresponding Euler-Lagrange equation:
so
By applying He’s variational iterative method results
To find the optimal value of 
following equation is required:
Therefore, the stationary conditions are obtained in the following form:
which yields
and the desired sequence is
By choosing 
Imposing (24) on this function given
(25)
If 
then from (24) 
but from (23) 
which is a contradiction.
Now imposing (24), we have: 
so 
and it is known that in this case imposing (24) on the Euler Lagrange equation yields
Hence:
and from (23)
. But y must be extremal when 
, therefore:
As it is observed that this solution is equal to exact solution (see Figure 2).
Example 3.4. The objective is to find an extremum of the functional
(26)
Such that
(27)
and
Figure 2. The graphs of approximated and exact solution for Example 3.3.
(28)
With exact solution
, 
, see [33]. By having the following auxiliary functional:
The system of Euler-Lagrange equations is in the form:
.
So
By using Homotopy variational iterative method gives:
Now
therefore
Hence
and
so
So 
is obtained as:
and the following iterative equations are obtained:
,
By choosing
:
And by imposing (28) on this functions:
from (27):
And consequently:
Figure 3. The graphs of approximated and exact solution for Example 3.4.
.
which is the exact solution (see Figure 3).
4. Conclusion
The He’s variational iterative method is an efficient method 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 discretize of the variables, there is no computational round off error. Moreover, only a few numbers of iterations are needed to obtain a satisfactory result.