A Fast Numerical Method For Solving Calculus Of Variation Problems

AMO - Advanced Modeling and Optimization, Volume 15, Number 2, 2013
A fast numerical method for solving calculus
of variation problems
A. R. Nazemi1
, S. Hesam1
, A. Haghbin2
1
Department of Mathematics, School of Mathematical Sciences, Shahrood University of
Technology, P.O. Box 3619995161-316, Tel-Fax No:0098273-3392012, Shahrood, Iran.
2
Department of Mathematics, Gorgan branch, Islamic Azad University, Gorgan, Iran.
ABSTRACT In the modeling of a large class of problems in science and engineering, the
minimization of a functional is appeared. Finding the solution of these problems needs to solve
the corresponding ordinary differential equations which are generally nonlinear. In recent years,
differential transform method has been attracted a lot of attention of the researchers for solving
nonlinear problems. This method finds the solution of the problem without any discretization of
the equation. Since this method gives a closed form solution of the problem and avoids the round
off errors, it can be considered as an efficient method for solving various kinds of problems. In
this research, differential transform method (DTM) will be employed for solving some problems in
calculus of variations. Some examples are presented to show the efficiency of the proposed technique.
Keywords: Calculus of variations; Euler–Lagrange equation; differential transform method.
1. Introduction
The calculus of variations is concerned with finding the maxima and minima of a certain
functional [1, 2]. Functional minimization problems known as variational problems appear
in engineering and science where minimization of functionals, such as Lagrangian, strain,
potential, total energy, etc., give the laws governing the systems behavior. In optimal control
theory, minimization of certain functionals gives control functions for optimum performance
of the system [3]. The brachistochrone, geodesics and isoperimetric problems have played
an important role in the development of the calculus of variations [1, 2].
Several methods have been used to solve variational problems. For example, the direct
1(Corresponding author) email:nazemi20042003@yahoo.com
1email:taranome2009@yahoo.com
2email:Ahmadbin@yahoo.com
*AMO - Advanced Modeling and Optimization. ISSN: 1841-4311
133

A. R. Nazemi, S. Hesam, A. Haghbin
method of Ritz [1], Walsh functions [4], Laguerre series [5], shifted Legendre polynomial
series [6], shifted Chebyshev series [7], and Fourier series [8] have been applied to solve vari-
ational problems. Legendre and Walsh wavelet functions are also used to solve variational
problems in [9, 10], respectively. In [11, 12], rationalized Haar functions and Haar wavelets
are proposed to solve variational problems. Recently, Adomian decomposition method [13],
variational iteration method [14] and homotopy-perturbation method [15] have been in-
tensively developed to obtain exact and approximate analytical solutions of this kind of
variational problems.
Motivated by the above discussions, in the present work, we are concerned with the
application of the differential transform procedure, for calculus of variational problems. The
DTM was first proposed by Zhou [16] to solve nonlinear Genesio systems. It is a numerical
method based on the Taylor series expansion which constructs an analytical solution in the
form of a polynomial. The established high order Taylor series method requires only symbolic
computation. Another side, the DTM obtains a polynomial series solution by means of an
iterative procedure. The DTM is useful to obtain exact and approximate solutions of linear
and non-linear differential equation systems. No necessity to linearization, discretization
and large computational works. It has been used to solve efficiently, easily and accurately
a large class of nonlinear problems with approximations. These approximations converge
rapidly to exact solutions [17]-[45].
The paper is organized as follows. In Section 2, we introduce the general form of prob-
lems in calculus of variations, and their relations with ordinary differential equations are
highlighted. In Section 3, theoretical aspects of the differential transform are discussed. In
Section 4, effectiveness of the proposed approach is verified by solving several numerical
examples. A conclusion is presented in Section 5.
2. Statement of the problem
Let us consider the simplest form of the variational problems
η [y(x)] =
∫ x1
x0
F(x, y(x), y′
(x))dx, (1)
where η is the functional that its extremum must be found. To find the extreme value of η,
the boundary points of the admissible curves are known in the following form
y(x0) = θ, y(x1) = δ, (2)
where θ and δ are known. The necessary condition for the solution of the problem (1) is to
satisfy the Euler–Lagrange equation
Fy −
d
dx
Fy′ = 0, (3)
134

A fast numerical method for solving calculus of variation problems
with boundary conditions given in (2). The boundary value problem (3) does not always
have a solution and if the solution exists, it may not be unique. Note that in many variational
problems the existence of a solution is obvious from the physical or geometrical meaning of
the problem and if the solution of Euler’s equation satisfies the boundary conditions, it is
unique. Also this unique extremal will be the solution of the given variational problem [2].
The general form of the variational problem (1) is
η[y1, y2, ..., yn] =
∫ x1
x0
F(x, y1, y2, ..., yn, y′
1, y′
2, ..., y′
n)dx, (4)
with the given boundary conditions for all functions
y1(x0) = θ1, y2(x0) = θ2, ..., yn(x0) = θn, (5)
y1(x1) = δ1, y2(x1) = δ2, ..., yn(x1) = δn. (6)
Here the necessary condition for the extremum of the functional (4) is to satisfy the following
system of second-order differential equations
Fyi −
d
dx
Fy′
i
= 0, i = 1, 2, ..., n, (7)
with boundary conditions given in (5) and (6).
The Euler-Lagrange equation is generally nonlinear. In this manuscript we apply the
DTM for solving Euler–Lagrange equations which arise from problems in calculus of varia-
tions. It is shown that this scheme is efficient for solving these kinds of problems.
3. Basic idea of differential transform method
For convenience of the reader, we will present a review of the DTM.
As in [46]-[49], the differential transform of the function w(x) is in the form
W(k) =
1
k!
(
dk
w(x)
dxk
)
x=x0
, (8)
where w(x) is the original function and W(k) is the transformed function. The differential
inverse transform of W(k) is specified as follows
w(x) =
∞
∑
k=0
W(k)(x − x0)k
. (9)
From (8) and (9), we get
w(x) =
∞
∑
k=0
1
k!
(
dk
w(x)
dxk
)
x=x0
(x − x0)k
, (10)
which implies that the differential transform is derived from Taylor series expansion, but the
method does not evaluate derivatives symbolically. However, the corresponding derivatives
135

Table 1: The operations for the one-dimensional differential transform method.
Original function Transformed function
w(x) = u(x) ∓ v(x), W(k) = U(k) ∓ V (k)
w(x) = αu(x) W(k) = αU(k)
w(x) =
∂n
u(x)
∂xn W(k) =
(k+n)!
k!
U(k + n)
w(x) = u(x)v(x) W(k) =
k
∑
r=0
U(r)V (k − r)
w(x) = u(x)v(x)z(x) W(k) =
k
∑
r=0
k−r
∑
m=0
U(r)V (m)Z(k − r − m)
w(x) = xn W(k) = δ(k − n), where δ(k − n) =



1, k = n
0, otherwise
w(x) = eλt W(k) = δk
k!
w(x) = t W(k) = δ(k − 1)
are calculated recursively, and are defined by the transformed equations of the original
functions.
when x0 is taken as 0, then the function w(x) in (10) can be written as
w(x) =
∞
∑
k=0
W(k)xk
=
∞
∑
k=0
1
k!
[
dk
w(x)
dxk
]
x=0
xk
. (11)
In real applications, the function w(x) can be expressed by a finite series as
w(x) =
n
∑
k=0
W(k)xk
. (12)
The solution of the Euler-Lagrange equation (3) with boundary conditions (2) is given in a
series form that generally converges very rapidly in real physical problems. The fundamental
mathematical operations performed by differential transform can readily be obtained and
are listed in Table 1.
Here, we propose a new idea in order to use the DTM to solve optimization problem (1)
with boundary conditions (2). We consider the the following initial value problem (IVP) of
the Euler-Lagrange equation (3) as











Fy − d
dx Fy′ = 0,
y(x0) = θ,
y′
(x0) = α,
(13)
where α ∈ I
R is an unknown parameter. Using the DTM, we find the series solution of y(x)
consists of an unknown constant α. To find this constant, we impose the boundary condition
y(x1) = δ to the obtained approximate solution (12) which results in an equation in α. By
solving this equation, we find α and then the optimal solution y(.) is immediately given.
136

A similar procedure is done to solve problem (4) with respect to (5) and (6), where the
imposed boundary condition is given by (6).
According to the above discussions, the following theorem can be stated:
Theorem 3.1 Consider the calculus of variation problem (1) with boundary conditions
(2). Employing the DTM, the optimal solution is given as
y∗
(x) =
∞
∑
k=0
Y (k)xk
, x ∈ [x0, x1]. (14)
A similar theorem can be concluded for problem (4) with boundary conditions (5) and (6).
It is clearly impossible to obtain the optimal trajectory law as in (14), since it contains
infinite series. In practice, the Nth order suboptimal trajectory is obtained by replacing ∞
with a finite positive integer N in (14) as follows:
y(x) ≃
N
∑
k=0
Y (k)xk
, x ∈ [x0, x1]. (15)
4. Simulation results
These examples are chosen such that there exist analytical solutions for them to give an
obvious overview of the DTM.
Example 1: We consider the following variational problem [13]
min η[y(x)] =
∫ 1
0
(y(x) + y′
(x) − 4 exp(3x))
2
dx, (16)
with given boundary conditions
y(0) = 1, y(1) = e3
, (17)
which has the following analytical solution
y(x) = exp(3x). (18)
The corresponding Euler-Lagrange equation is











y′′
− y − 8 exp(3x) = 0,
y(0) = 1,
y′
(0) = α.
(19)
Using the DTM we have
(k + 1)(k + 2)Y (k + 2) − Y (k) − 8
3k
k!
= 0, (20)
Y (0) = 1, Y (1) = α. (21)
137

Substituting (21) into (20) and by an iterative procedure, we achieve
Y (2) =
9
2
, Y (3) =
24 + α
6
, Y (4) =
27
8
, Y (5) =
1
20
(
36 +
24 + α
6
)
,
Y (6) =
81
80
, Y (7) =
1
42
(
81
5
+
1
20
(36 +
24 + α
6
)
)
, Y (8) =
729
4480
,
Y (9) =
1
72
(
243
70
+
1
42
(
81
5
+
1
20
(36 +
24 + α
6
))
)
, Y (10) =
729
44800
, ....
Substituting all Y (k) into (15), the 11-term of the DTM series solution of y(x) can be
given by
y(x) ≃
10
∑
k=0
Y (k)xk
= 1 + αx +
9
2
x2
+
24 + α
6
x3
+
27
8
x4
+
1
20
(
36 +
24 + α
6
)
x5
+
81
80
x6
+
1
42
x7
(
81
5
+
1
20
(36 +
24 + α
6
)
)
+
729
4480
x8
+
1
72
x9
(
243
70
+
1
42
(
81
5
+
1
20
(36 +
24 + α
6
))
)
+
729
44800
x10
. (22)
This gives the approximation of the y(x) in a series form. Now to find the constant α, the
boundary condition at x = 1 is imposed on the approximate solution of y(x) in (22). We
have
y(1) = e3
, (23)
which results in
α = 3.0049963740455447321. (24)
Replacing α into y(x) in (22), an approximate solution is obtained for y(x). Higher ac-
curacy is also obtained using more components of y(x); for example if n = 20, we get
α = 3.0000000002015122890. An absolute error between different values term of DTM
solution y(x) in (22) and the exact solution (18) is also depicted in Figure 1.
Example 2: In this example, consider the following variational problem [13]:
min η[y(x)] =
∫ 1
0
1 + y2
(x)
y′2(x)
dx, (25)
with given boundary conditions
y(0) = 0, y(1) = 0.5. (26)
The exact solution of this problem is
y(x) = sinh(0.4812118250x). (27)
138

0 0.2 0.4 0.6 0.8 1
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
x
|y
DTM
(x)−y
Exact
(x)|
n=8
n=9
n=10
Figure 1: Error functions |yDT M (x) − yExact(x)| with different values of n for 0 ≤ x ≤ 1 in
Example 1.
The Euler-Lagrange equation of this problem is











y′′
+ y′′
y2
− yy′2
= 0,
y(0) = 0,
y′
(0) = α.
(28)
Utilizing the DTM we get
(1 + k)(2 + k)Y (2 + k) +
k
∑
s=0
k−s
∑
m=0
(1 + k − m − s)(2 + k − m − s)Y (2 + k − m − s)Y (m)Y (s) −
k
∑
s=0
k−s
∑
m=0
(1 + m)(1 + k − m − s)Y (1 + m)Y (1 + k − m − s)Y (s) = 0, (29)
Y (0) = 0, Y (1) = α. (30)
Substituting (30) into (29) we acquire
Y (k) = 0, ∀ k = 2, 4, ...
Y (3) =
α3
6
, Y (5) =
α5
120
, , Y (7) =
α7
5040
, Y (9) =
α9
362880
, ...
139

Substituting all Y (k) into (15), the 5-term DTM series solution of y(x) is obtained as
following
y(x) ≃
9
∑
k=0
Y (k)xk
= xα +
x3
α3
6
+
x5
α5
120
+
x7
α7
5040
+
x9
α9
362880
. (31)
Implementing the boundary condition y(1) = 0.5 on y(x) in (31), we obtain α = 0.48121182506679352167.
In Figure 2, the |y′′
+ y′′
y2
− yy′2
| is plotted.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
1.2
x 10
−9
x
|y
′
′
+y
′
′
y
2
−yy
′2
|
Figure 2: Error function |y′′
+ y′′
y2
− yy′2
| for 0 ≤ x ≤ 1 in Example 2.
Example 3: We consider the following brachistochrone problem [13]
min η[y(x)] =
∫ 1
0
√
1 + y′2
(x)
1 − y(x)
dx, (32)
subject to the boundary conditions
y(0) = 0, y(1) = −0.5. (33)
The analytical solution of this problem in the implicit form is
F(x, y(x)) = −
√
−y2 + 0.381510869y + 0.618489131 −
0.8092445655 × arctan
(
y − 0.1907554345
√
−y2 + 0.381510869y + 0.618489131
)
−x + 0.5938731505 = 0.
140

The corresponding Euler-Lagrange equation is given by











y′′
− yy′′
− 1
2 − y′2
2 = 0,
y(0) = 0,
y′
(0) = α.
(34)
According to the DTM, we have
2(k + 1)(k + 2)Y (k + 2) = 2
k
∑
s=0
Y (s)(k − s + 2)(k − s + 1)Y (k − s + 2) +
k
∑
s=0
(s + 1)Y (s + 1)(k − s + 1)Y (k − s + 1) + δ(k), (35)
Y (0) = 0, Y (1) = α. (36)
Substituting (36) into (35) we get
Y (2) =
1
4
(1 + α2
),
Y (3) =
1
6
α(1 + α2
),
Y (4) =
1
48
(1 + 8α2
+ 7α4
)
Y (5) =
1
240
t5
α(11 + 46α2
+ 35α4
),
Y (6) =
11 + 237α2
+ 681α4
+ 455α6
2880
,
Y (7) =
α(73 + 696α2
+ 1533α4
+ 910α6
)
5040
,
....
Substituting all Y (k) into (15), the series solution of y(x) is
y(x) ≃
7
∑
k=0
Y (k)xk
= αx +
1
4
x2
(1 + α2
) +
1
6
x3
α(1 + α2
) +
1
48
x4
(1 + 8α2
+ 7α4
) +
1
240
x5
α(11 + 46α2
+ 35α4
) +
x6
(11 + 237α2
+ 681α4
+ 455α6
)
2880
+
x7
α(73 + 696α2
+ 1533α4
+ 910α6
)
5040
. (37)
Imposing the boundary condition y(1) = −0.5 on the DTM solution y(x) in (37), we obtain
α = −0.707749337327525455524125800640. In Figure 3, the error function |F(x, yn)| is
plotted for n = 1, 3, 5. The convergence of the iteration formula is clear in this figure.
Example 4: We consider the problem of finding the extremals of the functional [13]
η[y(x), z(x)] =
∫ π
2
0
[
y′2
(x) + z′2
(x) + 2y(x)z(x)
]
dx, (38)
141

0 0.2 0.4 0.6 0.8 1
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
x
|F(x,y
n
(x))| n=1
n=3
n=5
Figure 3: Error functions F(x, y1(x)), F(x, y2(x)) and F(x, y3(x)) for 0 ≤ x ≤ 1 in Example
3.
with the given boundary conditions as follows:
y(0) = 0, y(
π
2
) = 1, (39)
z(0) = 0, z(
π
2
) = −1, (40)
which has the following analytical solution





y(x) = sin(x),
z(x) = − sin(x).
(41)
The system of Euler’s differential equations is of the form































y′′
− z = 0,
z′′
− y = 0,
y(0) = 0,
y′
(0) = α,
z(0) = 0,
z′
(0) = β.
(42)
142

Implementing the DTM we have
(k + 1)(k + 2)Y (k + 2) − Z(k) = 0, (43)
(k + 1)(k + 2)Z(k + 2) − Y (k) = 0, (44)
Y (0) = 0, Y (1) = α, Z(0) = 0, Z(1) = β. (45)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
0
1
2
3
4
5
6
7
x 10
−8
x
|y
DTM
(x)−y
Exact
(x)|
Figure 4: Comparison of the exact solution with the DTM solution.
Substituting (45) into (43) and (44) we get
Y (k) = 0, ∀ k = 2, 4, ...
Y (3) =
β
6
, Y (5) =
α
120
, Y (7) =
β
5040
, ...
Z(k) = 0, ∀ k = 2, 4, ...
Z(3) =
α
6
, Z(5) =
β
120
, Z(7) =
α
5040
, ...
Substituting all Y (k) into (15), the 6-term DTM series solutions of y(x) and z(x) are given
143

by
y(x) ≃
11
∑
k=0
Y (k)xk
= αx +
βx3
6
+
αx5
120
+
βx7
5040
+
αx9
362880
+
βx11
39916800
, (46)
z(x) ≃
11
∑
k=0
Z(k)xk
= βx +
αx3
6
+
βx5
120
+
αx7
5040
+
βx9
362880
+
αx11
39916800
. (47)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
−1
−0.9
−0.8
−0.7
−0.6
−0.5
−0.4
−0.3
−0.2
−0.1
0
x
z
DTM
(x)
Figure 5: The DTM solution z(t) in Example 4.
In order to find the unknown constants α and β, we use the boundary condition y(π
2 ) = 1
and z(π
2 ) = −1; we get





y(π
2 ) = 1 =⇒ α = 1.0000000562589522947,
z(π
2 ) = −1 =⇒ β = −1.0000000562589522947.
(48)
An absolute error between y(x) in (46) and the corresponding exact solution in (41) is
depicted in Figures 4. The DTM solution of z(x) in (47) is also shown in Figure 5. These
graphs show that the proposed method has an appropriate convergence rate.
144

5. Conclusion
The DTM is employed for finding the solution of the ordinary differential equations which
arise from problems of calculus of variations. The present study has confirmed that the
DTM offers great advantages of straightforward applicability, computational efficiency and
high accuracy. The DTM needs less work in comparison with the traditional methods.
Therefore, this method can be applied to many complicated linear and non-linear problems
and does not require linearization, discretization or perturbation. Mathematica and Matlab
have been used for computations and simulations in this paper.
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