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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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