International Journal of Scientific & Engineering Research, Volume 5, Issue 6, June-2014 481
ISSN 2229-5518
A Simplified Method of Moment (MoM) Approach to solving nth Order Linear Differential Equations
H. M. El Misilmani, M. Abou Shahine, M. Al-Husseini, K. Y. Kabalan
Abstract—The Method of Moment (MoM) is a very successful tool in solving complex geometry electromagnetic (EM) problems, and is considered as the first computational approach in solving these problems. This paper approaches MoM in a very simple way and aims to produce a valuable procedure to solving boundary value problems, especially nth differential equations. Starting with the definition of the method, the problem is converted into its respective integral equation form, and then the unknown function is expanded into a sum of weighted basis functions, where the weight coefficients are to be found. The Galerkin method, which selects testing functions equal to the basic functions, is adopted. The problem then becomes a system of linear equations, which is solved analytically or numerically to find the needed weight coefficients. Two examples, a second- and a third-order differential equations, are considered to illustrate the application of the procedure, which can be used as well for the solution of other boundary-value problems. The considered examples show the detailed calculation process and make it easier to understand the solution procedure.
Index Terms—Basis functions, Differential equations, Method of Moment, Galerkin method, Weight coefficient
—————————— ——————————
EFORE digital computers, the design and analysis of elec- tromagnetic devices and structures were largely experi- mental. This changed after computers and programming languages appeared, where researchers began using them to challenge electromagnetic problems that could not be solved analytically. This led to a burst of development in a new field called computational electromagnetics (CEM). The Method of Moment (MoM) and other powerful numerical analysis tech- niques have been developed in this area in the last 50 years [1]. MoM, described in [2], is a simple numerical technique used to convert integro-differential equations into a linear system that can be solved numerically using a computer. When the order of the equation is small, MoM can analytically solve this prob-
lem in a general and very clear manner.
A large number of publications, including textbooks, grad- uate theses, and journal papers have been dedicated to MoM, but in most part these have been part of graduate courses on computational electromagnetics, or aimed to help profession- als apply MoM in their field problems. It is unusual to see a work on MoM addressed to undergraduate students to aid their understanding of this method, and help them apply it to solve problems they face in their undergraduate courses. This is reflected in the publications references hereafter. In [3], it is shown that the inner product involved in MoM is usually an integral, which is evaluated numerically by summing the inte- grand at certain discrete points. Three simple examples on the use of MoM in electromagnetics are presented in [4]. These examples deals with the input impedance of a short dipole, a plane wave scattering from a short dipole, and two coupled short dipoles. In [5], MoM is introduced in a way to give a deeper insight into electromagnetic phenomena. This is done by presenting examples and software programs, and also by giving the curriculum needed to quickly learn the basic con- cepts of numerical solutions.
This paper approaches MoM in a very simple way and aims
to produce a valuable procedure to solving boundary value problems, especially differential equations. Starting with the definition of the method, the problem is converted into its re- spective integral equation form, and then the unknown function is expanded into a sum of weighted basis functions, where the weight coefficients are to be found. The Galerkin method, which selects testing functions equal to the basis functions, is adopted. The problem then becomes a system of linear equations, which is solved analytically or numerically to find the needed weight coefficients. Two examples, a second- and a third-order differ- ential equations, are considered to illustrate the application of the procedure, which can be used as well for the solution of other boundary-value problems.
For students to be familiarized with MoM, it is good to intro- duce them to new terms such as expansion and testing. This is because the MoM method starts by expanding the unknown quantity, which is to be solved for, into a set of known func- tions with unknown coefficients. The resulting equation is then converted into a linear system of equations by enforcing the boundary conditions at a number of points. This resulting linear system is then solved analytically for the unknown coef- ficients. It is here to note that such an approach is very simple and quite interesting when applied to differential equation of order less than 3, but it is applicable for equations of higher order.
Accordingly, it is advisable to start with some basic mathemat- ical techniques for reducing functional equations to matrix equations. A deterministic problem is considered, which will be solved by reducing it to a suitable matrix equation, and hence the solution could be found by matrix inversion. The examples that we will choose are simple and easily illustrate the theory without any complicated mathematics. Linear spac- es and operators will be used in our solution. At first, it is rec- ommended to introduce MoM and define some terms related
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International Journal of Scientific & Engineering Research, Volume 5, Issue 6, June-2014 482
ISSN 2229-5518
to first order non-homogeneous differential equation. The
choice of this equation is important only for better under- standing of the solution.
A general nth order linear differential equation, defined over a domain D, has the form
[αn ] = [Lmn ]−1[Gm ] (10)
In our calculations, we will choose the test function wm equal
to the basis function 𝑓𝑛 , which is known as Garlekin method. The determination of matrix [Lmn ] is straightforward, and its
inverse is easy to obtain either analytically or numerically.
𝑎𝑛
𝑑𝑛 𝑓(𝑥)
𝑑𝑥 𝑛
+ 𝑎𝑛−1
𝑑𝑛−1 𝑓(𝑥)
𝑑𝑥 𝑛−1
+ ... + 𝑎1
𝑑𝑓(𝑥)
𝑑𝑥
+ 𝑎0 𝑓(𝑥) = 𝑔(𝑥) (1)
Once this is done, the αn coefficients are obtained, and the so- lution for 𝑓 is found.
In (1), the coefficients an, an−1,..., a1,a0 and 𝑔(𝑥) are known quantities, and 𝑓(𝑥) is the function whose solution is to be
determined. Equation (1) can be written in the form of an op-
erator equation
L�f(x)� = g(x), (2)
where 𝐿 is the operator equation, operating on 𝑓(𝑥), and given
by
It is good to note here that choosing the appropriate basis/test function is necessary to get fast to the accurate solution.
Considering the following second order differential equation defined by
d2 f(x) = x2 (11)
dx
L = an
dn ...
dxn
+ an−1
dn−1 ...
dxn−1
+ ... + a1
d...
dx
+ a0 ... (3)
defined over the domain 𝐷 = [0, 1] with the following bound- ary conditions 𝑓(0) = 𝑓(1) = 0. Starting by choosing the basis
function, let us choose
The solution of (1) is based on defining the inner product
< 𝑓, 𝑔 >, a scalar quantity valid over the domain of definition of 𝐿, which is given by
〈𝑓(𝑥), 𝑔(𝑥)〉 = ∫𝐷 𝑓 (𝑥)𝑔(𝑥)𝑑𝑥 (4)
Similarly, we define
〈L�f(x)�, g(x)〉 = ∫D L �f(x)�g(x)dx (5)
fn = x − xn+1 (12)
It is clear from (12) that the chosen basis functions meet the
boundary conditions and can be considered as a solution to
the problem. Substituting (12) into (6), the left-hand side ele-
ments of (7), which are the elements of the matrix [Lmn ], are
found to be
L = 〈w ,L(f )〉 = 1(x − xm+1 ) �
d2 �x−xn+1 �
2 � dx
dx
The first step in calculating the integral, using Method of Mo-
ments, is to expand 𝑓 into a sum of weighted basis functions
= 1(x − x
0
m+1
)(−n(n+1)x
−m
n−1
)dx
(13)
𝑓1 , 𝑓2 , 𝑓3 ; …, in the domain of 𝐿, as
f(x) = ∑n αn fn (6)
Testing functions denoted 𝑤1, 𝑤2, 𝑤3 ; …. are defined in the
L =
n+m+1
In the same manner, we compute the elements of the matrix
[Gm ], defined in (7), which are found to be
m
range of 𝐿. These testing functions are used for all values of 𝑛.
Using the inner product defined in (5), we obtain
Gm = 〈wm , g〉 = ∫(x − xm+1 ) x2 dx =
4(m+4)
(14)
∑n〈wm , L(fn )〉 = 〈wm , g(x)〉 form = 1,2,3, ... (7)
Expanding (7) over the values of m and n=1,2,3,…, the follow-
ing matrix equation is then obtained
Then, we start by choosing a value for n starting with 𝑛 = 1
until convergence of the solution. For the case when 𝑛 = 𝑚 =
1, 𝐿11 = 1/3; 𝐺1 = 1/5, 𝛼1 = 3/5, and f(x) is given by
f(x) = 3 (x − x2 ) (15)
5
〈𝑤1,𝐿(𝑓1 )〉 〈𝑤1,𝐿(𝑓2)〉 . . . . 〈𝑤1,𝐿(𝑓𝑛 )〉
𝛼1
〈𝑤1,𝑔〉
� 〈𝑤2,𝐿(𝑓1)〉 〈𝑤2,𝐿(𝑓2 )〉 . . . . 〈𝑤2,𝐿(𝑓𝑛 )〉 � �𝛼2 � = � 〈𝑤2,𝑔〉 �(8)
It is clear from (15), that the function 𝑓(𝑥) does not meet the
. . . . . . . . . . . . . . . .
〈𝑤𝑚 , 𝐿(𝑓1)〉 〈𝑤𝑚 , 𝐿(𝑓2 )〉 . . . . 〈𝑤𝑚 , 𝐿(𝑓𝑛 )〉
In a simpler form,
. . .
𝛼𝑛
. . .
〈𝑤𝑚 , 𝑔〉
original differential equation defined in (11). Accordingly, we
need to increase the value of 𝑛.
For 𝑛 = 𝑚 = 2, the solution for 𝑓(𝑥) gives
[Lmn ][αn ] = [Gm ], (9)
f(x) =
1 (x − x2 ) −
10
1 (x − x3 ) (16)
3
and the solution for the unknown coefficients is then
Again, (16) does not meet the original equation defined in (11).
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International Journal of Scientific & Engineering Research, Volume 5, Issue 6, June-2014 483
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For 𝑛 = 𝑚 = 3, the solution of 𝑓(𝑥) is given by
f(x) = 1 (x − x4 ) (17)
Then, we start by choosing a value for n starting with 𝑛 = 4
until convergence of the solution. The solution converges for
this value. In fact, for 𝑛 = 𝑚 = 4, 𝛼1 = 0, 𝛼2 = 0, 𝛼3 = 0, and
𝛼4
= −1. For the case when 𝑛 = 𝑚 = 5, 𝛼1
= 0, 𝛼2
= 0, 𝛼3 =
The function 𝑓(𝑥) given in (17) meets the boundary conditions
0, 𝛼4 = −1, and 𝛼5 = 0. Accordingly, 𝑓(𝑥) is given by
and (11), and accordingly it is the correct solution of the prob- lem.
f(x) = −2 �x�
+ x = x − x
(25)
2 8
Considering the following third order differential equation defined by
It is clear from (25) that the function 𝑓(𝑥) does meet the origi-
nal differential equation defined in (18) and the two boundary
conditions of the problem.
3 2
2 d f(x) + 4 d f(x) 4 2
dx3
dx2 − 8f(x) = x
− 6x
− 14x (18)
defined over the domain 𝐷 = [0, 2] with the following bound- ary conditions 𝑓(0) = 𝑓(2) = 0. Starting by choosing the basis
function, let us choose
x n
This paper presented, in simplified steps, the procedure of using the Method of Moments in solving simple differential equations of different orders. The considered examples show the detailed calculation process and make it easier to under- stand the solution procedure. As can be seen, this approach
f = 2 � �
2
− x (19)
could be easily used to solve mathematical problems and equations. According to the type of the equation, the solution
It is clear from (19) that the chosen basis functions meet the
boundary conditions and can be considered as a solution to
the problem. Substituting (19) into (6), the left-hand side ele-
ments of (7), which are the elements of the matrix [Lmn ], are
found to be
Lmn = 〈wm ,L(fn )〉
of the Moment Method will vary to accommodate for the
change in the given problem.
[1] W. C. Gibson, The Method of Moments in Electromagnetics, Chapman & Hall/ CRC, Taylor & Francis group, 2008.
[2] R. F. Harrington, Field Computation by Moment Methods, Krieger Publishing
n(n−1)(n−2)
x n−3
Co., Inc., 1968.
2 x m
� � +
2 2
(20)
[3] A. R. Djordjevic and T. K. Sarkar, “A Theorem on the Moment Methods,"
Lmn = ∫0 �2 �2�
− x� �
2n(n − 1) �x�
2
n−2
− 16 �x�
2
+ 8x
� dx
IEEE Transactions on Antennas and Propagation, vol. 35, no. 3, pp. 353-355,
March 1987.
[4] E. H. Newman, “Simple Examples of the Method of Moments in Electromag-
Lmn = A + B + C + D −
In (21),
64
n+m+1
+ 64
n+2
+ 64
m+2
− 63
3
(21)
netics," IEEE Transactions on Education, vol. 31, no. 3, pp. 193-200, August 1998.
[5] N.F.O. Serteller, A.G. Ak, G. Kocyigit, and T.C. Akinci, “Experimental Study
of Moment Method for Undergraduates in Electromagnetic”, Journal of Elec- tronics and Electrical Engineering, vol. 3, no. 3, pp. 115-118, 2011
A = �
2n(n−1)(n−2) for n + m − 2 ≠ 0
n+m−2 ,
0 for n + m − 2 = 0
B = � −2n(n − 2) for n ≠ 1 ,
0 for n = 1
C = �
8n(n−1) for n + m − 1 ≠ 0
n+m−1 ,
0 for n + m − 1 = 0
−8(n − 2) for n ≠ 0, 1
0 for n = 0, 1 (22)
In the same manner, we compute the elements of the matrix
[Gm ], defined in (7), which we found to be
G = 〈w , g〉 = �2 �x m
2
− x� (x4 − 6x2 − 14x) (23)
Gm =
64 m+5
− 96 −
m+3
112 m+2
+ 153
3
(24)
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