On Some Applications of New Integral Transform “ELzaki Transform”

Global Journal of Mathematical Sciences: Theory and Practical.
ISSN 0974-3200 Volume 4, Number 1 (2012), pp. 15-23
© International Research Publication House
http://www.irphouse.com
On Some Applications of New Integral Transform
“ELzaki Transform”
1 Tarig M. Elzaki, 2 Salih M. Elzaki and 3 Elsayed A. Elnour
1 Math. Dept., Faculty of Sciences and Arts, Alkamil,
King Abdulaziz University, Jeddah, Saudi Arabia
1 Math. Dept., Sudan University of Science and Technology, Sudan
2 Math. Dept., Sudan University of Science and Technology, Sudan
3 Math. Dept., Faculty of Sciences and Arts, Khulais,
King Abdulaziz University, Jeddah, Saudi Arabia
3 Math. Dept., Alzaeim Alazhari University, Sudan
E-mail: 1 tarig.alzaki@gmail.com, 2 salih.alzaki@gmail.com,
3 sayedbayen@yahoo.com
Abstract
In this study a new integral transform, namely ELzaki transform was applied
to solve linear ordinary differential equations with constant coefficients. In
particular we apply this new transform technique to solve linear dynamic
systems and signals-delay differential equations and the renewal equation in
statistics.
Keywords: Elzaki Transform- Differential Equations--Applications.
Introduction
Many problems of physical interest are described by differential and integral
equations with appropriate initial or boundary conditions. These problems are usually
formulated as initial value problem, boundary value problems, or initial – boundary
value problem that seem to be mathematically more vigorous and physically realistic
in applied and engineering sciences. ELzaki transform method is very effective for
solution of differential and integral equations.
The technique that we used is ELzaki transform method which is based on Fourier
transform, it introduced by Tarig Elzaki (2010) see [1, 2, 3, 4, 5, 6].
In this study, ELzaki transform is applied to solve linear dynamic systems and
signals-delay differential equations and the renewal equation in statistics, which the
solution of these equations have a major role in the fields of science and engineering.

16 Tarig. M. Elzaki et al
When a physical system is modeled under the differential sense, if finally gives a
differential equation.
Recently .Tarig ELzaki introduced a new transform and named as ELzaki
transform [1] which is defined by:
∞
−
u
(), ⎤ ( ) () , ( , )
Ε ⎡⎣f t u⎦ = T u = u∫
e f t dt u∈
k k
0
t
1 2
Or for a function f ( t ) which is of exponential order,
f
( t)
⎧
⎪Me
− t k
< ⎨
t
⎪
k 2
Me t ≥
⎩
1
, t ≤ 0
, 0
∞
() ( ) 2 −t
, ⎤
( ) , ( , )
Ε ⎡⎣f t u⎦ = T u = u ∫ e f ut dt u∈
k k
Theorem (1)
Let T( u)
is the ELzaki transform of ⎡ ( )
Proof
0
1 2
( ) = ( ) ⎤.
⎣
E f t T u
⎦ then:
T( u)
T( u)
i Ε ⎡⎣f′ t ⎤⎦ = −u f ( 0) ( ii) Ε ⎡f ( t)
f 0 uf 0
2
u
⎣
′′ ⎤⎦
= − − ′
u
n−1
( n
( ) ) T( u) 2− n+
k ( k
iii Ε ⎡f () t ⎤
u f )
⎣ ⎦
= −
( 0
n ∑
)
u
() ()
() Ε ⎡ ′() ⎤ = ′()
∞
0
−t
u
k=
0
i ⎣f t ⎦ u∫ f t e dt Integrating by parts to find that:
( )
T u
Ε ⎡f () t ⎤
⎣
′
⎦
= − u f
u
( 0)
1
⎣ ⎦
u
⎣ ⎦
( ii ) Let g ( t) = f ′( t) , then: Ε ⎡g′
() t ⎤= Ε⎡g() t ⎤ −ug( 0)
We find that, by using( i ):
T( u)
Ε ⎡⎣f′′ () t ⎤⎦
= − f 0 −uf′
0
2
u
( ) ( )
( iii ) Can be proof by mathematical induction
( ) ( )

On Some Applications of New Integral Transform “ELzaki Transform” 17
Theorem (2)
{ }
t / k
j
i
Let f () t ∈ A= f () t ∃ M, k1, k2
> 0, such that f () t < Me , if t∈( − 1) × [ 0, ∞ )
With Laplace transform F ( s ), Then ELzaki transform T( u)
of f ( )
⎛1
⎞
T( u)
= uF⎜ ⎟
⎝ u ⎠
Proof
Let: f () t
∞
( ) =
2 −t
∫ ( )
0
∈ A, then for − k1< u<
k2
Tu u e futdt
∞
t is given by
w
w
−
2 dw − 1
u
u
⎛ ⎞
Let w= ut then we have: T( u) = u ∫e f ( w) = u e f ( w)
dw uF .
u
∫
= ⎜ ⎟
u
0 0
⎝ ⎠
Also we have that T () 1 = F ( 1)
so that both the ELzaki and Laplace transforms
must coincide at u= s=
1
Theorem (3)
Let () ( )
F s and Gs ()
and ELzaki transforms M ( u ) and Nu. ( ) Then ELzaki transform of the convolution
f t and g t be defined in A having Laplace transforms ( )
of f and g
Proof
∞
( f * g) ( t)
= ∫ f ( t) g( t−τ
) dτ
Is given by: E ⎡⎣( f * g) ( t)
⎤ ⎦ = M ( u) N( u)
0
The Laplace transform of ( f * g ) is given by: L ⎡⎣( f * g) ⎤ ⎦ = F( s) G( s)
By the duality relation (Theorem (2)), we have: ( ) = ( )
E ⎡⎣ f * g ( t) ⎤⎦ uL⎡⎣ f * g ( t)
⎤⎦ ,
and since
⎛1⎞ ⎛1⎞ ⎡ ⎛1⎞ ⎛1⎞⎤
M( v) = uF⎜ ⎟ , N( u) = uG⎜ ⎟ Then E⎡( f * g)
( t) ⎤ = u F . G
u u
⎣ ⎦ ⎢ ⎜ ⎟ ⎜ ⎟
u u
⎥
⎝ ⎠ ⎝ ⎠ ⎣ ⎝ ⎠ ⎝ ⎠⎦
⎡M ( u) N( u)
⎤ 1
u⎢
. ⎥ = M ( u) N( u)
⎣ u u ⎦ u
Theorem (4)
u
Ε
⎣
e − f t
⎦
= + au T
⎛ ⎜ ⎞
⎟
⎝1+
au ⎠
at
If Ε ⎡⎣f ( t) ⎤⎦ = T ( u),
then: ⎡ ( ) ⎤ ( 1 )
Where a is a real constant.
∞
1
u

18 Tarig. M. Elzaki et al
Proof
−at
We have, by definition ⎡ ( )
∞
0
( 1+
)
au t
u
( )
Ε ⎣e f t ⎤
⎦ = u∫ e f t dt . Let
−
u
w = or
1 + au
w
u = , we have:
1 − aw
∞ a
∞ t
−
w −
1 1
u
w
⎡ u ⎤
u∫e f () t dt = e f () t dt = T ( w)
= T and
1−aw ∫
1− aw au ⎢1
0 0
1
au ⎥
− ⎣ + ⎦
1+
au
−at
u
Ε ⎡
⎣e f () t ⎤
⎦ = ( 1+
au)
T
1+
au
Theorem (5)
−
u
If Ε ⎡⎣f ( t) ⎤⎦ = T ( u),
then: Ε⎡⎣f ( t −a) H ( t − a) ⎤⎦
= e T ( u)
Where H ( t − a)
is Heaviside unit step function.
Proof
It follows from the definition that:
∞
a
−
−
u
u
( ) ( ) ⎤ ( ) ( ) ( )
Ε⎡f t−a H t− a = u e f t−a H t− a dt = u e f t−a dt
⎣ ⎦ ∫ ∫
Let t a τ ,
0 0
a ∞ t a
− − −
u
u
τ
− = then we have: e u e f ( τ) dτ
= e T( u)
∫
0
2
In particular if f ( t ) = 1, then: H ( t a) u e − u
Ε⎡⎣ − ⎤⎦
=
Also we can prove by mathematical induction that:
n−1
⎡
a
( t−
a)
⎤
−
n+
1 u
Ε⎢
H ( t− a)
⎥ = u e
⎢ Γ( n
⎣
) ⎥⎦
Example (1) (Linear dynamical systems and signals)
In physical and engineering sciences, a large number of linear dynamical systems
with a time dependent input signal f ( t ) that generates an output signal x ( t ) can be
described by the ordinary differential equation with constant coefficients.
n n−
( 1 m m−
D + a ) ( ) ( 1
n−1D + ... + a0 x t = D + bm−
1D + .... + b0) f ( t)
(1)
d
Where D = , a0, a1,... an−
1
, b0, b1,...
bm−
1
are constants
dx
We apply ELzaki transform to find the output x ( t ) so that (1) becomes.
P ( u) x ( u) R ( u) q ( u) f ( u) s ( u)
− = − (2)
n n−1 m m−1
a
a
∞
t

On Some Applications of New Integral Transform “ELzaki Transform” 19
Where,
1 an−
1 b
Pn( u) = + + ... + a , q ( u)
= + + ... + b
u u u u
1 m−1
n n−1 0 m
m m−1
0
n−1 n−2
∑
2− n+ k ( k
( ) ) 3− n+
k ( k
= ( 0) + )
( 0 ) + ... + ( 0)
R u u x a u x ux
n−1 n−1
k= 0 k=
0
m−1 m−2
∑
2− m+ k ( k
( ) ) 3− m+
k ( k
= ( 0) + )
( 0 ) + ... + ( 0)
S u u f b u f uf
∑
∑
m−1 m−1
k= 0 k=
0
It is convenient to express (2) in the form
x u = h u f u + g u
(3)
( ) ( ) ( ) ( )
Where
( )
( ) = qm
u
p ( u)
And g ( u )
h u
n
( ) −
m−
( )
P ( u)
R u S u
n−1 1
= (4)
n
And h ( u ) is usually called the transfer function. The inverse ELzaki transform
combined with the convolution theorem leads to the formal solution,
t
( ) = ( − τ) ( τ) τ + ( )
x t ∫ f t h d g t
(5)
0
With zero initial data, g ( u ) = 0. the transfer function takes the simple form.
x ( u )
h ( u)
= f ( u )
or x ( u ) h ( u ) f ( u )
t
If f ( t) = δ ( t)
and h( t) e
( )
x t
3
−1
⎡ u ⎤
t
=Ε ⎢ = e −1
1+
u
⎥
⎣ ⎦
= (6)
= then, the output function is
h( t ) is known as the impulse response.
Example (2) (Delay Differential Equations)
In many problems the derivatives of the unknown function x ( t ) are related to its
value at different times t −τ.
this leads us to consider differential equations of the
form:
dx
+ ax ( t − τ ) = f ( t )
(7)
dt
Where a is a constant and f ( t ) is a given function. Equations of this type are
called delay differential equations. In general , initial value problems for these
equations involve the specification of x ( t ) in the interval t 0
−τ
≤t≤ t 0
and this

20 Tarig. M. Elzaki et al
information combined with the equation it self, is sufficient to determine x ( t ) for
t > t . We show how equation (7) can be solved by ELzaki transform when 0
t
0
= 0
and x ( t) = x
0
for t ≤ 0. In view of the initial condition , we can write.
x ( t − τ ) = x ( t −τ) H ( t − τ)
So equation (7) is equivalent to,
dx
+ ax ( t −τ) H ( t − τ) = f ( t )
(8)
dt
Applying ELzaki transform to (8) gives,
( )
x u
u
τ
−
u
( 0) ( ) ( )
− ux + ae x u = f u
Or ( )
And
2
( ) + u x( 0)
τ
uf u
2
⎡
− ⎤
u
x u = = u f ( u) u x( 0)
1 aue
τ
−
⎣
⎡ + ⎤
⎦⎢
+ ⎥
u
1+
aue
⎣ ⎦
2
n
( ) = ⎡ ( ) + ( 0) ⎤ ( −1) ( )
∞
n = 0
n
x u
⎣
uf u u x
⎦∑ au e
(9)
nτ
−
u
The inverse ELzaki transform gives the formal solution:
−1 2
n
() =Ε ⎡ ( ) + ( 0) ⎤ ( −1) ( )
∞
x t
⎣
u f u u x
⎦∑ au e
(10)
n=
0
In order to write an explicit solution, we choose x
0
= 0 and f ( t)
= t and hence
(10) be comes.
nτ
n+
2
∞
1 4
n n −
∞
−
⎡ ⎤ n n ( t−
nτ
)
u
xt () =Ε ⎢u∑( − 1)( au) e ⎥= ∑ ( −1 ) a Ht ( − nτ
)
⎣ n= 0 ⎦ n=
0 ( n + 2! )
, t>
0
Example (3) (Renewal Equation in statistics)
The random function x ( t ) of time t represents the number of times some event has
occurred between time 0 and time t , and usually referred to as a counting Process. A
random variable X
n
that recodes the time it assumes for X to get the value n from
the n − 1 is referred to as an inter-arrival time .If the random variables X
1, X
2....
are
independent and identically distributed, then the counting process X ( t ) is called a
renewal process. We denote their common Probability distribution function by F ( t )
and the density function by f ( t ) so that F ( t) f ( t).
n
nτ
−
u
−1
′ = Then renewal function is
defined by the expected number of time the event being counted occurs by time t and
is denoted by r( t ) so that .

On Some Applications of New Integral Transform “ELzaki Transform” 21
∞
() ()
0
{ 1 } ( )
rt () =Ε ⎡⎣X t⎤⎦ = ∫ Ε ⎡⎣xt ⎤⎦:
X = x f x dx
(11)
{ 1 }
Where Ε X ( t)
: X = x is the conditional expected value of X ( )
condition that X
1
= x and has the value.
{ X ( t)
: X
1
x} ⎡1
r( t x ) ⎤H ( t x )
t under the
Ε = = ⎣ + − ⎦ −
(12)
Thus
t
( ) = ⎡1
+ ( − ) ⎤ ( )
∫
r t ⎣ r t x ⎦f x dx
Or
0
t
( ) = ( ) + ( − ) ( )
r t F t ∫ r t x f x dx
(13)
0
This is called the renewal equation in mathematical statistics .We solve the
equation by taking ELzaki transform with respect to t , and ELzaki transformed
equation is,
1
uF ( u )
r ( u) = F( u) + r ( u) f ( u)
Or r ( u)
=
u
u − f u
The inverse transform gives the formal solution rt () of renewal function.
3
t
t
−1
⎡ u ⎤ 1
(i) If F ( t) = e and f ( t) = e , then: r( t)
=Ε ⎢ =
2 sin h 2 t
1 2u
⎥
⎣ − ⎦ 2
(ii) If F ( t)
= t and ( ) 1,
f t = then: ( )
( )
2
−1⎡ u 2⎤
t
r t = E ⎢ − u = e −1
1−u
⎥
⎣ ⎦
Conclusion
In this study, we introduced new integral transform called ELzaki transform to solve
linear dynamical systems and signals – delay differential equations and the renewal
equation in statistics. It has been shown that ELzaki transform is a very efficient tool
for solving these equations

22 Tarig. M. Elzaki et al
References
[1] Tarig M. Elzaki, The New Integral Transform “Elzaki Transform” Global
Journal of Pure and Applied Mathematics, ISSN 0973-1768,Number 1(2011),
pp. 57-64.
[2] Tarig M. Elzaki & Salih M. Elzaki, Application of New Transform “Elzaki
Transform” to Partial Differential Equations, Global Journal of Pure and
Applied Mathematics, ISSN 0973-1768,Number 1(2011), pp. 65-70.
[3] Tarig M. Elzaki & Salih M. Elzaki, On the Connections Between Laplace and
Elzaki transforms, Advances in Theoretical and Applied Mathematics, ISSN
0973-4554 Volume 6, Number 1(2011),pp. 1-11.
[4] Tarig M. Elzaki & Salih M. Elzaki, On the Elzaki Transform and Ordinary
Differential Equation With Variable Coefficients, Advances in Theoretical and
Applied Mathematics. ISSN 0973-4554 Volume 6, Number 1(2011), pp. 13-18.
[5] Tarig M. Elzaki, Adem Kilicman, Hassan Eltayeb. On Existence and
Uniqueness of Generalized Solutions for a Mixed-Type Differential Equation,
Journal of Mathematics Research, Vol. 2, No. 4 (2010) pp. 88-92.
[6] Tarig M. Elzaki, Existence and Uniqueness of Solutions for Composite Type
Equation, Journal of Science and Technology, (2009). pp. 214-219.
[7] Lokenath Debnath and D. Bhatta. Integral transform and their Application
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On Some Applications of New Integral Transform “ELzaki Transform” 23
Appendix
ELzaki Transform of some Functions
f ( t )
Ε ⎡f ( t) ⎤ = T ( u)
2
1
t
3
⎣
u
u
n
nu !
+
n
2
t
t
a−1
( a)
Γ , a>
0
a 1
u +
at
2
e
u
1−au
3
u
2
( 1−au
)
n−1
at
n + 1
t e
u
, n = 1,2,....
n
( n − 1! )
( 1−au
)
sinat
3
au
2 2
1+ au
cosat 2
u
2 2
1+ au
sinhat
3
au
2 2
1−a u
cos hat 2
au
2 2
1−a u
3
sinbt
bu
1− au + b u
at
te
at
e
( )
at
e cos bt ( 1−
)
( 1− au ) +
⎦
2 2 2
2
au u
b u
t sinat 4
2au
2 2
1+ au
J
0 ( at )
2
u
2
1+ au
H t a
a
( )
( t a)
δ −
− 2
ue − u
a
ue − u
2 2 2