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Intern ational Journ al of Scientific & E ngineering Research, Volume 5, Issue 6, June-2014 481
ISSN 2229-5518
IJSER © 2014
http://www.ijser.org
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
—————————— ——————————
1 INTRODUCTION
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.
2 METHOD OF MOMENTS SOLUTION APPROACH
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
B
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Intern ational Journ al of Scientific & E ngineering Research, Volume 5, Issue 6, June-2014 482
ISSN 2229-5518
IJSER © 2014
http://www.ijser.org
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
()
+()
+... +()
+()=() (1)
In (1), the coefficients a, a,..., a,a and () are known
quantities, and () is the function whose solution is to be
determined. Equation (1) can be written in the form of an op-
erator equation
Lf(x)= g(x), (2)
where is the operator equation, operating on (), and given
by
L = a...
+ a...
+... + a...
+ a... (3)
The solution of (1) is based on defining the inner product
<,>, a scalar quantity valid over the domain of definition
of , which is given by
(),()=
()() (4)
Similarly, we define
Lf(x), g(x)=L
f(x)g(x)dx (5)
The first step in calculating the integral, using Method of Mo-
ments, is to expand into a sum of weighted basis functions
,,; …, in the domain of , as
f(x)= f (6)
Testing functions denoted ,,; …. are defined in the
range of . These testing functions are used for all values of .
Using the inner product defined in (5), we obtain
w, L(f)
=w, 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
,() ,(). . . . ,()
,() ,(). . . . ,()
.... .... .... ....
,() ,(). . . . ,()
...
=,
,
...
,(8)
In a simpler form,
[L][]=[G], (9)
and the solution for the unknown coefficients is then
[]=[L][G] (10)
In our calculations, we will choose the test function wm equal
to the basis function , which is known as Garlekin method.
The determination of matrix [L] is straightforward, and its
inverse is easy to obtain either analytically or numerically.
Once this is done, the coefficients are obtained, and the so-
lution for is found.
It is good to note here that choosing the appropriate basis/test
function is necessary to get fast to the accurate solution.
2.1 Example 1
Considering the following second order differential equation
defined by
()
= x (11)
defined over the domain = [0, 1] with the following bound-
ary conditions (0) = (1) = 0. Starting by choosing the basis
function, let us choose
f= x x (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 [L], are
found to be
Lmn =w,L(f)=(xxm+)n+
dxdx
=(xxm+)(n(n+1)x)dx
Lmn =
n+m+
(13)
In the same manner, we compute the elements of the matrix
[G], defined in (7), which are found to be
G=w, g=(xx)xdx =
() (14)
Then, we start by choosing a value for n starting with = 1
until convergence of the solution. For the case when ==
1, = 1/3; = 1/5, = 3/5, and f(x) is given by
f(x)=
(xx) (15)
It is clear from (15), that the function () does not meet the
original differential equation defined in (11). Accordingly, we
need to increase the value of .
For == 2, the solution for () gives
f(x)=
(xx)
(xx) (16)
Again, (16) does not meet the original equation defined in (11).
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Intern ational Journ al of Scientific & E ngineering Research, Volume 5, Issue 6, June-2014 483
ISSN 2229-5518
IJSER © 2014
http://www.ijser.org
For == 3, the solution of () is given by
f(x)=
(xx) (17)
The function () given in (17) meets the boundary conditions
and (11), and accordingly it is the correct solution of the prob-
lem.
2.2 Example 2
Considering the following third order differential equation
defined by
2()
+ 4 ()
8f(x)= x6x14x (18)
defined over the domain = [0, 2] with the following bound-
ary conditions (0) = (2) = 0. Starting by choosing the basis
function, let us choose
f= 2
x (19)
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 [L], are
found to be
Lmn =w,L(f)
Lmn =2
x()()
+
2n(n1)
16
+8xdx
(20)
L= A + B + C + D
+
+
(21)
In (21),
A = ()()
for n + m 20
0 for n + m 2 = 0
,
B = 2n(n2) for n 1
0 for n = 1,
C = ()
for n + m 10
0 for n + m 1 = 0
,
D = 8(n2) for n 0, 1
0 for n = 0, 1 (22)
In the same manner, we compute the elements of the matrix
[G], defined in (7), which we found to be
G=w, g=2
x(x6x14x) (23)
G=
+
(24)
Then, we start by choosing a value for n starting with = 4
until convergence of the solution. The solution converges for
this value. In fact, for == 4, = 0, = 0, = 0, and
=1. For the case when == 5, = 0, = 0, =
0, =1, and = 0. Accordingly, () is given by
f(x)=2
+ x = x
(25)
It is clear from (25) that the function () does meet the origi-
nal differential equation defined in (18) and the two boundary
conditions of the problem.
3 CONCLUSION
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
could be easily used to solve mathematical problems and
equations. According to the type of the equation, the solution
of the Moment Method will vary to accommodate for the
change in the given problem.
REFERENCES
[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
Co., Inc., 1968.
[3] A. R. Djordjevic and T. K. Sarkar, “A Theorem on the Moment Methods,"
IEEE Transactions on Antennas and Propagation, vol. 35, no. 3, pp. 353-355,
March 1987.
[4] E. H. New man, “Simple Examples of the Method of Moments in Electromag-
netics," IEEE Transactions on Education, vol. 31 , no. 3, pp. 1 93-200, 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 Engineer ing, vol. 3, no. 3, pp. 115-118, 2011
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