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Physics Letters A 362 (2007) 401–406
www.elsevier.com/locate/pla
Solution of problems in calculus
of variations via He’s variational iteration method
MehdiTatari, Mehdi Dehghan∗
Department of Applied Mathematics, Faculty of Mathematics and Computer Science, Amirkabir University of Technology, No. 424, Hafez Ave., Tehran, Iran
Received 18 September 2006; accepted 18 September 2006
Available online 10 November 2006
Communicated by A.R. Bishop
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 He’s variational
iteration 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 He’s variational iteration method will be
employed for solving some problems in calculus of variations. Some examples are presented to show the efficiency of the proposed technique.
©2006Elsevier B.V. All rights reserved.
Keywords: Calculus of variations; Euler–Lagrange equation; He’s variational iteraction method; Symbolic computations; Extremum problem
1. Introduction In the large number of problems arising in analysis, mechan-
ics, geometry,itisnecessarytodeterminethemaximalandmin-
In this work we consider He’s variational iteration method imal of a certain functional. Because of the important role of
as a well known method for finding both analytic and approx- this subject in science and engineering, considerable attention
imate solutions of differential equations. The efficiency of this has been received on these kinds of problems. Such problems
method for solving various types of problems is shown for ex- are called variational problems [18].
ample in [1–9]. Employing this technique the exact solution of Onewell known method for solving variational problems is
a linear problem can be obtained by doing only one iteration direct method. In this technique the variational problem is re-
step. garded as a limiting case of a finite number of variables. This
This methodis used for solving autonomous ordinary differ- extremum problem of a function of a finite number of vari-
ential systems in [2]. Application of this method to Helmholtz ables is solved by ordinary methods, then a passage of limit
equation is investigated in [10]. This method is used for solv- yields the solution of the appropriate variational problem [19].
ing Burgers’ and coupled Burgers’ equations in [11].In[11] The direct method of Ritz and Galerkin has been investigated
the applications of the present method to coupled Schrödinger– for solving variational problems in [19,20]. Using Walsh se-
KdVequations and shallow water equation are provided. Also ries method a piecewise constant solution is obtained for varia-
the use of this method for solving linear fractional partial dif- tional methods [21]. Some orthogonal polynomials are applied
ferential equations arising from fluid mechanics is discussed in on variational problems to find continuous solutions for these
[12]. Other recent works in this field are found in [13–17]. problems [22–24]. Also Fourier series and Taylor series are ap-
plied to variational problems, respectively in [25] and [26] to
* Corresponding author. Fax: +9821 66497930. findacontinuoussolution for these kinds of problems.
E-mail addresses: mehditatari@aut.ac.ir (M. Tatari), mdehghan@aut.ac.ir More historical comments about variational problems are
(M. Dehghan). found in [19,20].
0375-9601/$ – see front matter © 2006 Elsevier B.V. All rights reserved.
doi:10.1016/j.physleta.2006.09.101
402 M.Tatari, M. Dehghan / Physics Letters A 362 (2007) 401–406
Theorganization of the rest of this Letter is as follows: where v is the functional that its extremum must be found. To
The well known He’s variational iteration method is re- findtheextremevalueofv,theboundarypointsoftheadmissi-
viewedinSection2.InSection3,weintroducethegeneralform ble curves are known in the following form
ofproblemsincalculusofvariations,andtheirrelationswithor-
y(x )=α, y(x )=β. (3.2)
dinary differential equations are highlighted. To present a clear 0 1
overview of the procedure, we select several examples in Sec- Thenecessaryconditionforthesolutionoftheproblem(3.1)
tion 3.1–3.4. A conclusion is presented in Section 4. is to satisfy the Euler–Lagrange equation
2. Variational iteration method d
Fy − dxFy′ =0, (3.3)
In this technique, the problem is initially approximated with with boundary conditions given in (3.2). The boundary value
possible unknowns. Then a correction functional is constructed problem (3.3) does not always have a solution and if the solu-
by a general Lagrange multiplier, which can be identified opti- tion exists, it may not be unique. Note that in many variational
mally via the variational theory [3]. In this method the problem problems the existence of a solution is obvious from the physi-
is considered as cal or geometrical meaningoftheproblemandifthesolutionof
Ly +Ny=g(x), (2.1) Euler’s equation satisfies the boundary conditions, it is unique.
where L is a linear operator, and N is a nonlinear operator, Also this unique extremal will be the solution of the given vari-
g(x)is an inhomogeneous term. Using the variational iteration ational problem [19].
method, the following correct functional is considered Thegeneral form of the variational problem (3.1) is
x v[y ,y ,...,y ]
1 2 n
x
yn+1 =yn+ λ Lyn(s)+Ny˜n(s)−g(s) ds, (2.2) 1
′ ′ ′
= F(x,y,y,...,y ,y,y,...,y )dx, (3.4)
0 1 2 n 1 2 n
x
where λ is Lagrange multiplier [9], the subscript n denotes the 0
nth approximation, y˜ is considered as a restricted variation i.e.
n with the given boundary conditions for all functions
δy˜n = 0 [4–6]. Taking the variation from both sides of the cor-
rect functional with respect to y and imposing δy =0,the y (x )=α ,y(x )=α ,...,y(x )=α , (3.5)
n n+1 1 0 1 2 0 2 n 0 n
stationary conditions are obtained. Using the stationary condi- y (x )=β ,y(x )=β ,...,y(x )=β . (3.6)
tions the optimal value of the λ can be identified. 1 1 1 2 1 2 n 1 n
Since this procedure avoids the discretization of the prob- Herethenecessaryconditionfortheextremumofthefunctional
lem, it is possible to find the closed form solution without any (3.4) is to satisfy the following system of second-order differ-
round off error. The use of symbolic computation is necessary ential equations
for finding the iterations. d
In the case of m equations, we rewrite equations in the form Fy − Fy′ =0,i=1,2,...,n, (3.7)
i dx i
L(y )+N (y ,...,y )=g (x), i =1,...,m, (2.3)
i i i 1 m i with boundary conditions given in (3.5) and (3.6).
where Li is linear with respect to yi, and Ni is the nonlinear The Euler–Lagrange equation is generally nonlinear. In this
part of the ith equation. In this case the correct functionals are Letter we apply the variational iteration method for solving
madeas Euler–Lagrange equations which arise from problems in cal-
x culus of variations. It is shown that this scheme is efficient for
yi(n+1) =yin+ λiLiyin(s) solving these kinds of problems.
0 3.1. Example 1
+N y˜1n(s),...,y˜mn(s) −g(s) ds, (2.4)
and the optimal values of λ , i = 1,...,mare obtained by tak- Asanelementary example we consider the following varia-
i
ing the variation from both sides of the correct functionals and tional problem
finding stationary conditions using 1
δyi(n+1) =0,i=1,...,m. ′ 2
minv= y(x)+y (x)−4exp(3x) dx, (3.8)
3. Statement of the problem 0
with given boundary conditions
The simplest form of a variational problem can be consid- 3
ered as y(0)=1,y(1)=e . (3.9)
x
1 Thecorresponding Euler–Lagrange equation is
′
v y(x) = F x,y(x),y (x) dx, (3.1) y′′ −y −8exp(3x)=0, (3.10)
x
0
M.Tatari, M. Dehghan / Physics Letters A 362 (2007) 401–406 403
withboundaryconditions(3.9).Theexactsolutionofthisprob- cycloid. Consider the following brachistochrone problem [27]
lem is y(x) = exp(3x). Using the He’s variational iteration 1
′ 2 1/2
methodwehavethefollowingcorrect functional 1+y (x)
minv= dx. (3.16)
x 1−y(x)
′′ 0
yn+1(x)=yn(x)+ λ yn(s)−y(s)−8exp(3s) ds, (3.11) Let the boundary conditions be
0
where λ is a general Lagrange multiplier [9], which can be y(0)=0,y(1)=−0.5. (3.17)
identified optimally via the variational theory, the subscript n In this case the Euler–Lagrange equation is written in the fol-
denotes the nth approximation, y˜n is considered as a restricted lowing form
variation, i.e. δy˜n = 0. Taking the variation from both sides of 11+y′2
(3.11) with respect to yn we have y′′ =− , (3.18)
x 2 y−1
′′ or equivalently
δyn+1(x)=δyn(x)+δ λ yn(s)−y(s)−8exp(3s) ds
0 y′′ −yy′′ − 1 − y′2 =0, (3.19)
=δy ′ ′
n(x)+λδyn(s) s=x −λ δyn(s) s=x 2 2
x with the given boundary conditions. The exact solution of this
+ (λ′′ −λ′)δynds=0, (3.12) problem in the implicit form is
0 2
for all variations δy and δy′. So we obtain the following sta- F(x,y)=− −y +0.381510869y+0.618489131
n n −0.8092445655
tionary conditions
√ y−0.1907554345
′′ ×arctan −y2+0.381510869y+0.618489131
λ (s)−λ(s)=0,λ(s)=0,
′ s=x −x+0.5938731505
1−λ(s)s=x =0. =0.
Therefore we have According to He’s variational iteration method we have
λ(s)= −1exp(s−x)+ 1exp(x−s), (3.13)
2 2 y (x)
n+1
x
and we find the following iteration formula 1 y′2(s)
=y (x)+ λ y′′(s)−y (s)y′′(s)− − n ds.
x n n n n
yn+1(x)=yn(x)+ −1exp(s−x)+ 1exp(x−s) 0 2 2 (3.20)
2 2
0 For finding the optimal value of λ we have
× y′′(s)−y(s)−8exp(3s) ds. (3.14)
n δy (x)
If we set n+1
x
1 y′2(s)
=δy (x)+δ λ y′′(s)−y (s)y′′(s)− − n ds,
y0 =Aexp(3x)+Bexp(−3x), (3.15) n n n n 2 2
and find y ,wehave 0 (3.21)
1
A=1, which gives
x
B=0. δyn+1(x)=δyn(x)+δ λy′′(s)ds
Byimposing the boundary conditions on y , the exact solution n
1 0
of the problem will be found. ′ ′
=δyn(x)+λδyn(s)s=x −λδyn(s)s=x
x
3.2. Example 2 + (λ′′)δynds
The problem of brachistochrone is proposed in 1696 by Jo- 0
hannBernoulliwhichisrequiredtofindthelineconnectingtwo =0. (3.22)
certain points A and B that do not lie on a vectorial line and Therefore we have
possessing the property that a moving particle slides down this
line from A to B in the shortest time. This problem is solved ′′
λ (s)=0,λ(s)=0,
by Johann Bernoulli, Jacob Bernoulli, Leibnitz, Newton and ′ s=x
L’Hospital. It is shown that the solution of this problem is a 1−λ(s)s=x =0.
404 M.Tatari, M. Dehghan / Physics Letters A 362 (2007) 401–406
These equations yield
λ(s)=s−x, (3.23)
and the following iteration formula is obtained
yn+1(x)
x
1 y′2(s)
=y (x)+ (s−x) y′′(s)−y (s)y′′(s)− − n ds.
n n n n 2 2
0 (3.24)
By choosing y = Ax +B, and computing only two itera-
0
tions and imposing the boundary conditions we have
A=−0.7460779052,B=0.
Fig. 1. Error functions |F(x,y )| (1), |F(x,y )| (2) and |F(x,y )| (3) for
Astheresult we have 1 2 3
0x1.
v(y ) −v(y )=−0.3570300e−3.
min 2
More accurate solution is obtained easily using more iter-
ations. In this work we have used the well known software
Maple 8 because of its nice ability in symbolic computation
for computing the iterations.
In Fig. 1, the error function |F(x,y )| is plotted for m =
m
1,2,3. The convergence of the iteration formula is clear in this
figure.
In Fig. 2, the exact solution of the problem is plotted.
3.3. Example 3
In this example, consider the following variational prob-
lem[19] Fig. 2. The exact solution of the brachistochrone problem (3.16).
1 2
minv= 1+y (x)dx. (3.25) Similar to Example 2, we have λ = s − x. Therefore, the
y′2(x) following variational iteration method will be obtained.
0
x
Let the boundary conditions be ′′ 2 ′′
δyn+1(x)=δyn(x)+δ (s −x) yn(s)+yn(s)yn(s)
y(0)=0,y(1)=0.5. (3.26) 0
Theexact solution of this problem is −y′2(s)yn(s) ds. (3.31)
n
Choosingy =Ax+B,theothertermsofthesequence(y )
y =sinh(0.4812118250x). (3.27) 0 n
are computed easily using symbolic computations.
TheEuler–Lagrange equation of this problem can be written in In Fig. 3,the|y −y |isplotted for 0x 1.
min 3
the following form Imposing the boundary conditions on y3 we have
y′′ +y′′y2 −yy′2 =0. (3.28) A=0.4812127078,B=0.
Let the boundary conditions be In Fig. 4,the|y′′ +y′′y2 −y3y′2| is plotted for 0 x 1.
3 3 3 3
y(0)=0,y(1)=0.5. (3.29) 3.4. Example 4
Applying the He’s variational iteration method on this prob-
lem we have In this example we consider the problem of finding the ex-
x tremals of the functional [19].
′′ 2 ′′
δyn+1(x)=δyn(x)+δ λ yn(s)+yn(s)yn(s) π/2
0 ′2 ′ 2
−y′2 (3.30) v y(x),z(x) = y (x)+z (x)+2y(x)z(x) dx, (3.32)
n (s)yn(s) ds. 0
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