Maxwell's equations solutions by means of the Natural transform

Abstract

The Natural transform is applied to Maxwell’s Equations describing transient electromagnetic planar waves propagating in lossy medium (TEMP), in order to obtain its electric and magnetic fields solutions. To achieve this task, many of the basic Natural transform properties are initially
investigated, and then pragmatically used.

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MATHEMATICS IN ENGINEERING, SCIENCE AND AEROSPACE
MESA - www.journalmesa.com
Vol. 3, No. 3, pp. 313-323, 2012
c
⃝CSP - Cambridge, UK; I&S - Florida, USA, 2012
Maxwell’s equations solutions by means of the natural
transform
Fethi Bin Muhammed Belgacem1,⋆, Rathinavel Silambarasan2
1Department of Mathematics, Faculty of Basic Education, PAAET, Shaamyia, Kuwait.
2M. S. Software Engg., School of Information Technology, V. I. T. University, Vellore, Tamilnadu,
India.
⋆Corresponding Author. fbmbelgacem@gmail.com.
Abstract. The Natural transform is applied to Maxwell’s Equations describing transient electromag-
netic planar waves propagating in lossy medium (TEMP), in order to obtain its electric and magnetic
fields solutions. To achieve this task, many of the basic Natural transform properties are initially
investigated, and then pragmatically used.
1 Introduction
The Natural transform of the function f(t)defined for time variable tin the domain (−∞,∞)is defined
by
N[f(t)] = R(s,u) = ∫∞
−∞
e−st f(ut)dt ;s,u∈(−∞,∞)(1.1)
The integral equation (1.1) is the Two-sided Natural transformation of the (time) function f(t), and
the variables sand uare the Natural transform variables. Integral equation (1.1) can be now re-defined
as
N[f(t)] = ∫∞
−∞
e−st f(ut)dt ;s,u∈(−∞,∞)
=∫0
−∞
e−st f(ut)dt ;s,u∈(−∞,0)+∫∞
0
e−st f(ut)dt ;s,u∈(0,∞)
=N−[f(t)] + N+[f(t)]
=N[f(t)H(−t)] + N[f(t)H(t)]
=R−(s,u) + R+(s,u)
2010 Mathematics Subject Classification 35Q61, 44A10, 44A15, 44A20, 44A30, 44A35, 81V10
Keywords: Fourier transform, Laplace transform, Maxwell’s equations, natural, Sumudu transform

314 F.B.M. Belgacem and R. Silambarasan
where H(.) is the Heaviside function. To define formally, for the function f(t)H(t)defined in the
positive real axis t∈(0,∞), and in the set
A={f(t)|∃M,τ1,τ2>0,|f(t)|<Me |t|
τj,if t∈(−1)j×[0,∞)}(1.2)
the Natural transform (as N transform in [16]) is defined by
N[f(t)H(t)] = N+[f(t)] = R+(s,u) = ∫∞
0
e−st f(ut)dt ;s,u∈(0,∞)(1.3)
The detailed theory of Natural transform with the Multiple Shift theorems, etc are given in [10]. As
explained in [16] the integral equation (1.3) reduces to Laplace transform [11, 19] when u≡1 in
(1.3), and to Sumudu transform [21, 22] when s≡1 in (1.3). More sources of Sumudu transform are
explained in [4, 5, 6, 7] and for the introductory properties see [1, 2, 3, 23, 24]. Natural transform is
the Linear operator [16]. Moreover
N+[a f (t)±bg(t)] = aN+[f(t)] ±bN+[g(t)]
=aF+(s,u)±bG+(s,u)(1.4)
In (1.4) aand bare non-zero constants and F+(s,u)and G+(s,u)are respective Natural transform
of f(t)and g(t)both defined in set the A. The relation between Natural transform to Laplace and
Sumudu transforms are proved in next two theorems.
Theorem 1.1. If R+(s,u)is Natural transform of f(t)∈Aand F(s)is Laplace transform [11, 19] of
f(t)∈Athen,
N+[f(t)] = R+(s,u) = 1
u∫∞
0
e−st
uf(t)dt =1
uFs
u(1.5)
Proof. Substituting ω=ut in eqn (1.3) and replacing limits and dummy variable the result (1.5)
obtained.
Theorem 1.2. If R+(s,u)and G(u)are respective Natural and Sumudu transforms [4, 5] of function
f(t)∈Athen,
N+[f(t)] = R+(s,u) = 1
s∫∞
0
e−tfut
sdt =1
sGu
s(1.6)
Proof. Substituting ω=st in eqn (1.3) and performing similar operations as theorem 1. the result
(1.6) obtained.
Thus either from the Laplace transform [11, 19] (or) from the Sumudu transform [4, 5], the Natural
transform can be derived. We call the relations (1.5) and (1.6) as Natural-Laplace-Duality (NLD) and
Natural-Sumudu-Duality (NSD) respectively. For instance the Natural transform of first kind Bessel’s
function of order zero and first kind Modified Bessel’s function of order zero are given respectively
by [10],
N+[J0(αt)] = 1
√s2+α2u2; Re(s),Re(u)>0 (1.7)
N+[I0(αt)] = 1
√s2−α2u2; Re(s)>|αu|,Re(u)>0 (1.8)

Maxwells Equations Solutions by Means of the Natural Transform 315
Now as given in [16], u≡1 in (7) and (8) gives the Laplace transform of J0(αt)and I0(αt)respectively
[19] and simillarly s≡1 in (7) and (8) gives the Sumudu transform of J0(αt)and I0(αt)respectively
[5]. More instances of Natural transform of elementary functions are given in [10]. In the next section
some of the Natural transform properties are proved, which will be applied for solving Maxwell’s
Equations.
2 Natural transform properties
We start with the Second Shifting theorem.
Theorem 2.1. If Ha(t)is the unit step Heaviside function for any real number ’a’ defined
Ha(t) = 1 for t≥a
0 for t<a(2.1)
Then the Natural transform of shifted function f(t−a) = f(t−a)Ha(t)) is given by
N+[f(t−a)Ha(t)] = e−as
uR+(s,u)(2.2)
Proof. The Sumudu transform of f(t−a)Ha(t)is given by [4].
S+[f(t−a)Ha(t)] = e−a
uG(u)(2.3)
Now applying Natural-Sumudu-Dual relation (1.6) to equation (2.3), i.e.,
N+[f(t−a)Ha(t)] = 1
se−a
u
sGu
s
=1
se−as
uGu
s
=e−as
uR+(s,u)
where the proof ends.
The Natural transform of derivative of function f(t)∈A, w. r. t. ’t’ is given in the succeeding
theorem.
Theorem 2.2. If fn(t)∈Ais the n−th derivative of function f(t)∈A, w. r. t. ’t’ then the Natural
transform of fn(t)∈Ais given by
N+[fn(t)] = sn
unR+(s,u)−
n−1
∑
k=0
sn−(k+1)
un−kfk(0)(2.4)
Proof. The Sumudu transform of fn(t)∈Ais given by [4, 5, 6]
S+[fn(t)] = G(u)
un−
n−1
∑
k=0
fk(0)
un−k(2.5)
Using the Natural-Sumudu-Duality (1.6) to equation (2.5),

316 F.B.M. Belgacem and R. Silambarasan
N+[fn(t)] = 1
sG(u
s)
un
sn−
n−1
∑
k=0
fk(0)
un−k
sn−k
=1
ssn
unGu
s−
n−1
∑
k=0
sn−k
un−kfk(0)
=sn
un
1
sGu
s−
n−1
∑
k=0
sn−k−1
un−kfk(0)
=sn
unR+(s,u)−
n−1
∑
k=0
sn−(k+1)
un−kfk(0)
which gives the relation (2.4).
Now n=1 and n=2 in equation (2.4) gives the respective Natural transform of first and second
derivative of function f(t)∈A, w. r. t. ’t’,
N+[f′(t)] = s
uR+(s,u)−f(0)
u(2.6)
N+[f′′(t)] = s2R+(s,u)−s f (0)
u2−f′(0)
u(2.7)
We give the following corollaries of Natural transform of partial derivatives of f(z,t)∈A, w. r. t. ’t’
and ’z’ respectively.
Corollary 2.1. If N+[f(z,t)] = R+(z,s,u)then The Natural transform of n−th partial derivative of
f(z,t)∈A, w. r. t. ’t’ is defined by
N+∂nf(z,t)
∂tn=sn
unR+(z,s,u)−
n−1
∑
k=0
sn−(k+1)
un−klim
t→0
∂kf(z,t)
∂tk(2.8)
Corollary 2.2. If N+[f(z,t)] = R+(z,s,u)then The Natural transform of n−th partial derivative of
f(z,t)∈A, w. r. t. ’z’ is defined by
N+∂nf(z,t)
∂zn=dnR+(z,s,u)
dzn(2.9)
The proof of Corollaries 1 and 2 are similar to Theorem 4. When n=1 and n=2 in equation (2.8)
gives the Natural transform of first and second partial derivative of f(z,t)∈A, w. r. t. ’t’ respectively
by
N+∂f(z,t)
∂t=s
uR+(z,s,u)−1
ulim
t→0f(z,t)(2.10)
N+∂2f(z,t)
∂t2=s2
u2R+(z,s,u)−s
u2lim
t→0f(z,t)−1
ulim
t→0
∂f(z,t)
∂t(2.11)
When n=1 and n=2 in equation (2.9) gives the Natural transform of first and second partial deriva-
tive of f(z,t)∈A, w. r. t. ’z’ respectively by
N+∂f(z,t)
∂z=d
dz R+(z,s,u)(2.12)
N+∂2f(z,t)
∂t2=d2
dz2R+(z,s,u)(2.13)

Maxwells Equations Solutions by Means of the Natural Transform 317
Theorem 2.3. If R+(s,u)is the Natural transform of function f(t)in A, then its inverse Natural trans-
form [10] is defined by
N−1[R+(s,u)] = f(t) = lim
T→∞
1
2πi∫γ+iT
γ−iT
est
uR+(s,u)ds (2.14)
In equation (2.14), γis real constant and the integral is taken along s=γin the complex plane
s=x+iy. The real number γis chosen so that s=γlies on right of all (finite (or) countably infinite)
singularities. Suppose when T→∞then the integral in (2.14) over Γtends to zero, then by Cauchy
residue theorem, eqn (2.14) is defined by
N−1[R+(s,u)] = f(t) = lim
T→∞
1
2πi∫γ+iT
γ−iT
est
uR+(s,u)ds
f(t) = ∑Residues of est
uR+(s,u)at the poles of R+(s,u)
The detailed derivation and explanation of equation (2.14) is given in [10]. The convolution of two
functions f(t),g(t)∈Ais defined by
(f∗g)(t) = ∫t
0
f(τ)g(t−τ)dτ(2.15)
Theorem 2.4. The Natural transform of Convolution of two functions f(t),g(t)both defined in set
the Awith N+[f(t)] = F+(s,u)and N+[g(t)] = G+(s,u), is given by
N+[( f∗g)(t)] = uF+(s,u)G+(s,u)(2.16)
Proof. The Sumudu transform of Convolution of two functions (2.15) is defined by [4, 5, 8, 15]
S+[( f∗g)(t)] = uF(u)G(u)(2.17)
Applying the Natural-Sumudu-Duality (1.6) to the equation (2.17)
N+[( f∗g)(t)] = 1
su
sFu
sGu
s
=u1
sFu
s1
sGu
s
=uF+(s,u)G+(s,u)
which gives the result (2.16).
3 Maxwell’s equation solutions
James Clerk Maxwell (1831-1879) integrated the electric, magnetic and electro-magnetic induction
theories and formed the set of differential equations, this integration is after called Maxwell’s equa-
tions. In electromagnetics there are four relationships to describe the response of the medium for
various input. The four, are relation between electric field Ewith conductive current Jand electric
displacement D. And the relation between magnetic field Hwith magnetic induction Band magnetic
polarization M.

318 F.B.M. Belgacem and R. Silambarasan
In this section we apply the theorems and properties of preceding section to solve Maxwell’s
equations describing planar transverse electromagnetic wave (TEMP) propagating in lossy medium.
The Laplace transform method and mathematical models were used to solve Maxwell’s partial
differential equations in [12, 13, 14, 17, 18, 20, 25]. Other than Laplace transform, F. B. M. Belgacem
applied the new Sumudu transform [21, 22] to the Maxwell’s equation in [8]. M. G. M. Hussian. et.
al. solved the Maxwell’s equations using Sumudu transform and obtained the transient electric field
solution in [15].
The planar transverse electromagnetic wave (TEMP) propagate in zdirection in lossy medium with
constant permittivity ε, permeability µand conductivity σ. The electric field vector Eand magnetic
field vector Hare related by [8, 15, 17, 20]
▽×E=−µ∂H
∂t(3.1)
▽×H=ε∂E
∂t+σE(3.2)
When the electric field vector is polarized along xdirection thus Ex(z,t)and magnetic field along y
direction Hy(z,t).
The Maxwell’s equations (3.1) and (3.2) are expressed in differential equation as
∂Ex(z,t)
∂z+µ∂Hy(z,t)
∂t=0 (3.3)
∂Hy(z,t)
∂z+ε∂Ex(z,t)
∂t+σEx(z,t) = 0 (3.4)
Applying Natural transform to equations (3.3) and (3.4), In view of equations (2.10) and (2.12), the
partial differential equations (2.10) and (2.11) are transformed into
∂F(z,s,u)
∂z+sµ
uG(z,s,u)−µ
uHy(z,0) = 0 (3.5)
∂G(z,s,u)
∂z+sε
uF(z,s,u)−ε
uEx(z,0) + σF(z,s,u) = 0 (3.6)
where F(z,s,u) = N+[Ex(z,t)] and G(z,s,u) = N+[Hy(z,t)]. Differentiating the eqn (3.5) partially w.
r. t. ’z’ gives
∂2F(z,s,u)
∂z2+sµ
u
∂G(z,s,u)
∂z=µ
u
∂Hy(z,0)
∂z(3.7)
Next writting eqn (3.6) as
∂G(z,s,u)
∂z=−sε
uF(z,s,u) + ε
uEx(z,0)−σF(z,s,u)(3.8)
Substituting equation (3.7) in (3.8), simplifying and re-arranging gives the following equation,
∂2F(z,s,u)
∂z2−s2µε
u2+sµσ
uF(z,s,u) = µ
u∂Hy(z,t)
∂zt=0−sµε
u2Ex(z,0)(3.9)
Hence the differential equation (3.9) is only in F(z,s,u). Now re-writing eqn (3.4) as

Maxwells Equations Solutions by Means of the Natural Transform 319
∂Hy(z,t)
∂zt=0
=−ε∂Ex(z,t)
∂tt=0−σ[Ex(z,t)]t=0(3.10)
Now substituting eqn (3.9) in eqn (3.10), simplifying and re-arranging, results in
∂2F(z,s,u)
∂z2−s2µε
u2+sµσ
uF(z,s,u) = −sµε
u2+µσ
uEx(z,0)−µε
u∂Ex(z,t)
∂tt=0
(3.11)
The equation (3.11) is transformed Maxwell’s differential equation of transient electric field Ex(z,t).
Also at this stage it is worth to check, as given in [16] when u≡1 in eqn (3.11) gives the Laplace
transform application of Maxwell’s equation [20]. And s≡1 in eqn (3.11) is the Sumudu transform
of the Maxwell’s equation [8, 15].
Now we have to solve the non-homogeneous differential equation (3.11) for the transient electric
field. We consider the initial condition [8, 15]
[Ex(z,t)]t=0=f0(z)(3.12)
∂Ex(z,t)
∂tt=0
=f′
0(z)(3.13)
and the boundary condition [8, 15]
lim
z→0Ex(z,t) = f(t)if t≥0
0 if t<0(3.14)
Substituting the initial conditions (3.12) and (3.13) in equation (3.11) gives
d2F(z,s,u)
dz2−s2µε
u2+sµσ
uF(z,s,u) = −sµε
u2+µσ
uf0(z)−µε
uf′
0(z)(3.15)
Substituting γ2=sµε
u2+µσ
uin (3.15) leads to
d2F(z,s,u)
dz2−sγ2F(z,s,u) = −γ2f0(z)−µε
uf′
0(z)(3.16)
Let
P(z,s,u) = −γ2f0(z)−µε
uf′
0(z)(3.17)
So that equation (3.16) results in
d2F(z,s,u)
dz2−sγ2F(z,s,u) = P(z,s,u)(3.18)
The Complementary solution and Particular integral solution (by method of variation of parameter)
of non-homogeneous equation (3.18) are respectively given by
Fc(z,s,u) = A(s,u)e−γ√sz +B(s,u)eγ√sz (3.19)
Fp(z,s,u) = eγ√sz
2γ√s∫e−γ√szP(z,s,u)d z +e−γ√sz
2γ√s∫eγ√szP(z,s,u)dz (3.20)

320 F.B.M. Belgacem and R. Silambarasan
As z→∞,Ex(z,t)→finite so that B(s,u) = 0. And using the boundary condition equation (3.14)
N+lim
z→0Ex(z,t)=N+[f(t)] = F(s,u) = A(s,u)(3.21)
Thus
F(z,s,u) = F(s,u)e−γ√sz (3.22)
Now expanding e−γ√sz
γ√s, as in [8, 15, 20]. (By applying the Natural-Sumudu-Duality property equation
(1.6) to the equation (3.20) through equation (3.22) of [15] ) gives
e−γ√sz
γ√s=a∫∞
z/a
e−bt J0b
az2−a2t2e−st
udt (3.23)
where in equation (3.23), a=1
√µεand b=σ
2εand J0(.)is the first kind Bessel’s function of order zero.
Next differentiating equation (3.23) w. r. t.’z’.
e−γ√sz =e−b
aze−s
au z−a∫∞
z/a
e−bt ∂
∂zJ0b
az2−a2t2e−st
udt (3.24)
In equation (3.24), substituting v=st
uso that t=uv
sand dt =udv
sand noting, as t→z
a,v=sz
au and as
t→∞,v→∞. Hence equation (3.24) becomes
e−γ√sz =e−b
aze−s
au z−au∫∞
zs/au
1
se−b(uv
s)∂
∂zJ0b
az2−a2uv
s2e−vdv (3.25)
Now assuming ’v’ as time variable, from theorem 2. equation (1.6), the Natural transform of f(v)is
N+[f(v)] = R(s,u) = 1
s∫∞
0
e−vfuv
s(3.26)
In view of equation (3.26), the definite integral in equation (3.25) is Natural transform of Φ(z,v),
where Φ(z,v)is defined by [8, 15].
Φ(z,v) = e−bv ∂
∂zJ0b
a√z2−av2for v≥z
a
0 for 0 <v<z
a
(3.27)
It is important to note that Φ(z,v)is non-periodic function. So we expanded e−γ√sz into
e−γ√sz =e−b
aze−s
au z−auN+[Φ(z,v)] (3.28)
Substituting equation (3.28) in equation (3.22) for e−γ√sz,
F(z,s,u) = F(s,u)e−b
aze−s
au z−auF(s,u)N+[Φ(z,v)] (3.29)
Now using the second shifting property theorem 3 equation (2.2) and Convolution theorem 8 equa-
tion (2.16) and noting N−1[F(z,s,u)] = Ex(z,t), inverting (Inverse Natural transform) equation (3.29)
finally results transient electric field
Ex(z,t) = e−b
azft−z
aH(z
a)(t)−a∫∞
z/a
f(t−τ)e−bτ∂
∂zJ0b
az2−(aτ)2dτ(3.30)
The equation (3.30) is the transient electric field solution of TEMP waves and it is exactly true with
the solution in [8, 15, 20]. When using the different initial condition, we have the following result.

Maxwells Equations Solutions by Means of the Natural Transform 321
Theorem 3.1. The transient electric field Ex(z,t)solution and magnetic field Hy(z,t)solution of
Maxwell’s equations (3.3) and (3.4) using the initial conditions
lim
t→0[Ex(z,t)] = 0=lim
t→0[Hy(z,t)] (3.31)
lim
t→0∂Ex(z,t)
∂t=0=lim
t→0∂Hy(z,t)
∂t(3.32)
and the boundary conditions, equation (3.14) and equation (4.4) and also, both Ex(z,t)and Hy(z,t)
are finite as z→∞, are given respectively by
Ex(z,t) = ∫t
0f′(t−τ) + f(0)1
2πi
1
s∫λ+i∞
λ−i∞
est
ue−γ√szdsdτ(3.33)
Hy(z,t) = ∫t
0g′(t−ζ) + g(0)1
2πi
1
s∫λ+i∞
λ−i∞
est
ucosγ√szdsdζ(3.34)
Appendix
4 General natural transform properties
Property Definition
Definition N+[f(t)∈A] = ∫∞
0
e−st f(ut)dt ;s,u∈(0,∞)
First Shifting N+[e±tf(t)] = s
s∓uR+su
s∓u
First Scaling N+[f(at)] = 1
aR+s
a,u
Integration N+∫t
0
.. .∫t
0
f(τ)(dτ)n=un
snR+(s,u)
Product shift N+[tnf(t)] = un
sn
dn
dununR+(s,u)
Product shift and function derivative N+[tnfn(t)] = undn
dunR+(s,u)
Division shift N+f(t)
tn=1
un∫∞
s
.. .∫∞
s
R+(s,u)(ds)n
Function anti-derivative and division shift N+fn(t)
tn=1
(−u)n∫∞
u
...∫∞
u
R+(s,u)(du)n
Initial Value lim
t→0f(t) = lim
u→0
s→∞
sR+(s,u)
Final Value lim
t→∞f(t) = lim
u→∞
s→0
sR+(s,u)
Heaviside’s expansion f(t) =
n
∑
i=1
F+(αi,u)
G+′(αi,u)e
αit
u

322 F.B.M. Belgacem and R. Silambarasan
5 Conclusion and future work
In this paper, by defining some properties, we applied the Natural transform to Maxwell’s equations
and obtained the transient electric and magnetic field solution.
Similarly the transformed Maxwell’s differential equation of transient Magnetic field Hy(z,t)is
given by
d2G(z,s,u)
dz2+s2µε
u2+sµσ
uG(z,s,u) = −sµε
u2+µσ
uHy(z,0)−µε
u∂Hy(z,t)
∂tt=0
(5.1)
Now applying the initial conditions
[Hy(z,t)]t=0=g0(z)(5.2)
∂Hy(z,t)
∂tt=0
=g′
0(z)(5.3)
and the boundary condition
lim
z→0Ex(z,t) = g(t)if t≥0
0 if t<0(5.4)
with the substitution γ2=sµε
u2+µσ
u, the equation (4.1) is re-written as
d2G(z,s,u)
dz2+sγ2G(z,s,u) = −γ2g0(z)−µε
ug′
0(z)(5.5)
Let
Q(z,s,u) = −γ2g0(z)−µε
ug′
0(z)(5.6)
So that equation (4.5) is
d2G(z,s,u)
dz2+sγ2G(z,s,u) = Q(z,s,u)(5.7)
The respective complementary and particular (by using method of variation of parameter) solutions
of non-homogeneous equation (4.7) are
Gc(z,s,u) = C(s,u)cos γ√sz +D(s,u)sinγ√sz (5.8)
Gp(z,s,u) = sin γ√sz
2γ√s∫cosγ√szQ(z,s,u)d z +cosγ√sz
2γ√s∫sinγ√szQ(z,s,u)dz (5.9)
The finiteness of Hy(z,t)requires D(s,u) = 0 as z→∞, and from the boundary condition equation
(4.4),
G(z,s,u) = G(s,u)cosγ√sz (5.10)
Now expanding cos γ√sz
γ√sof equation (4.10) and performing inverse Natural transform operation, the
transient magnetic field Hy(z,t)solution [12, 13] of Maxwell’s equations (3.3) and (3.4) with the
initial conditions (4.2), (4.3) and boundary condition (4.4) is obtained. This will be our future work.

Maxwells Equations Solutions by Means of the Natural Transform 323
Acknowledgements
The Authors wish to acknowledge, and thank Professor Ganquan Xie, for comments that helped
improve the flow of the paper, during the PIERS2011, Suzhou, China.
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