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

The general form of the variational problem (1) is

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

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:

or

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:

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

in which and the endpoint lies on a two-dimensional surface that is given by. Here the transversality conditions at are:

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:

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:

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:

where the point must lie on the sphere, with the exact solution, see [33]. The corresponding Euler Lagrange equations for this problem

are:

So that

In above equations “e” and “f” are constant, so they can be rewritten as:

The transversality conditions are:

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.

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

Such that

and

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

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

Such that

and

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:

.

which is the exact solution (see Figure 3).