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”
1Tarig M. Elzaki, 2Salih M. Elzaki and 3Elsayed A. Elnour
1Math. Dept., Faculty of Sciences and Arts, Alkamil, King Abdulaziz University, Jeddah, Saudi Arabia
1Math. Dept., Sudan University of Science and Technology, Sudan
2Math. Dept., Sudan University of Science and Technology, Sudan
3Math. Dept., Faculty of Sciences and Arts, Khulais, King Abdulaziz University, Jeddah, Saudi Arabia
3Math. Dept., Alzaeim Alazhari University, Sudan E-mail: 1[email protected], 2[email protected],
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.
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:
( ) ( ) ( ) (
1 2)
0
, , ,
t
f t u T u u e u f t dt u k k
∞ −
⎡ ⎤
Ε⎣ ⎦= =
∫
∈Or for a function f t
( )
which is of exponential order,( )
12
, 0
, 0
tk
tk
Me t
f t
Me t
⎧ − ≤
< ⎨⎪
⎪ ≥
⎩
( ) ( )
2( ) (
1 2)
0
, t , ,
f t u T u u e f ut dt u k k
∞
⎡ ⎤ −
Ε⎣ ⎦= =
∫
∈Theorem (1)
Let T u
( )
is the ELzaki transform of ⎡⎣E f t( ( ) )
=T u( )
⎤⎦. then:( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
( )( ) ( )
( )( )
2 1
2 0
0 0 0
0
n
n n k k
n k
T u T u
i f t u f ii f t f uf
u u
iii f t T u u f
u
− − +
=
′ ′′ ′
⎡ ⎤ ⎡ ⎤
Ε⎣ ⎦= − Ε⎣ ⎦= − −
⎡ ⎤
Ε⎣ ⎦= −
∑
Proof
( ) ( ) ( )
0
t
i f t u f t eu dt
∞ −
′ ′
⎡ ⎤
Ε⎣ ⎦=
∫
Integrating by parts to find that:( ) T u( ) ( )0
f t u f
⎡ ⎤ u
⎣ ′ ⎦
Ε = −
( )
ii Let g t( )
=f ′( )
t , then: g t( )
1 g t( )
ug( )
0′ u
⎡ ⎤ ⎡ ⎤
Ε⎣ ⎦= Ε⎣ ⎦ −
We find that, by using
( )
i :( )
T u( )
2( )
0( )
0f t f uf
′′ u ′
⎡ ⎤
Ε⎣ ⎦ = − −
( )
iii Can be proof by mathematical inductionTheorem (2)
Let f t
( )
∈ =A{
f t( )
∃M k k, ,1 2 >0,such that f t( )
<Met k/ i,if t∈ −( )
1 j×[ )
0,∞}
With Laplace transform F s
( )
,Then ELzaki transform T u( )
of f t( )
is given by( )
1T u u F u
= ⎛ ⎞⎜ ⎟⎝ ⎠
Proof
Let: f t
( )
∈A, then for − < <k1 u k2 20
( ) t ( )
T u u e f ut dt
∞
=
∫
−Let w=ut then we have:
( )
2( ) ( )
0 0
1 .
w w
u dw u
T u u e f w u e f w dw u F
u u
∞ − ∞ − ⎛ ⎞
=
∫
=∫
= ⎜ ⎟⎝ ⎠Also we have that T
( )
1 =F( )
1 so that both the ELzaki and Laplace transforms must coincide at u= =s 1Theorem (3)
Let f t
( )
and g t( )
be defined inA having Laplace transforms F s( )
and G s( )and ELzaki transforms M u
( )
andN u( ). Then ELzaki transform of the convolution of f and g( ) ( ) ( )
0
* ( )
f g t =∞
∫
f t g t−τ τd Is given by: E(
f *g)
( )t 1M u N u( ) ( )
⎡ ⎤ = u
⎣ ⎦
Proof
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 ⎤⎦=u L⎡⎣(
f *g)
( )t ⎤⎦,and since
( )
1 ,( )
1(
*)
( ) 1 . 1M v u F N u u G Then E f g t u F G
u u u u
⎡ ⎤
⎛ ⎞ ⎛ ⎞ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞
= ⎜ ⎟⎝ ⎠ = ⎜ ⎟⎝ ⎠ ⎣ ⎦= ⎢⎣ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎥⎦
( )
.( )
1( ) ( )
M u N u
u M u N u
u u u
⎡ ⎤
⎢ ⎥=
⎣ ⎦
Theorem (4)
If Ε⎡⎣f t
( )
⎤⎦=T u( )
, then:( ) (
1)
1
at u
e f t au T
au
− ⎛ ⎞
⎡ ⎤
Ε⎣ ⎦= + ⎜⎝ + ⎟⎠
Where a is a real constant.
Proof
We have, by definition
( )
(1 )( )
0
au t
at u
e f t u e f t dt
∞ − +
⎡ − ⎤
Ε⎣ ⎦=
∫
. Let w =1+uau or1 , u w
= aw
− we have:
( ) ( ) ( )
( ) ( )
0 0
1 1
1 1 1
1 1
1 1
a t
u w
a t
w u
u e f t dt e f t dt T w T and
aw aw au au
au e f t au T u
au
∞ − ∞ −
−
⎡ ⎤
= − = − = − ⎢⎣ + ⎥⎦
+
⎡ ⎤
Ε⎣ ⎦= + +
∫ ∫
Theorem (5)
If Ε⎡⎣f t
( )
⎤⎦=T u( )
, then: Ε⎡⎣f t(
−a H t) (
−a)
⎤⎦= e T u−ua( )
Where H t
(
−a)
is Heaviside unit step function.Proof
It follows from the definition that:
( ) ( ) ( ) ( ) ( )
0 0
a t
u u
f t a H t a u e f t a H t a dt u e f t a dt
∞ − ∞ −
⎡ ⎤
Ε⎣ − − ⎦=
∫
− − =∫
−Let t − =a τ, then we have:
( ) ( )
0
a t a
u u
e− u e∞
∫
−τ f τ τd =e T u− In particular if f t( )
=1, then: Ε⎡⎣H t(
−a)
⎤⎦= u e2 −uaAlso we can prove by mathematical induction that:
( )
( )
n 1( )
n 1 aut a
H t a u e
n
− + −
⎡ − ⎤
Ε⎢ − ⎥=
⎢ Γ ⎥
⎣ ⎦
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 bedescribed by the ordinary differential equation with constant coefficients.
(
Dn +an−1Dn−1+ +... a x t0) ( )
=(
Dm +bm−1Dm−1+....+b0)
f t( )
(1) Where d , 0, 1,... n 1 , 0, ,...1 m1D a a a b b b
dx − −
= are constants
We apply ELzaki transform to find the output x t
( )
so that (1) becomes.( ) ( )
1( ) ( ) ( )
1( )
n n m m
P u x u −R − u =q u f u −s − u (2)
Where,
( )
11 0( )
11 01 1
... , ...
n m
n n n m m m
a b
P u a q u b
u u u u
− −
− −
= + + + = + + +
( )
( )( )
( )( ) ( )
( )
( )( )
( )( ) ( )
1 2
2 3
1 1
0 0
1 2
2 3
1 1
0 0
0 0 ... 0
0 0 ... 0
n n
k k
n k n k
n n
k k
m m
k k
m k m k
m m
k k
R u u x a u x ux
S u u f b u f uf
− −
− + − +
− −
= =
− −
− + − +
− −
= =
= + + +
= + + +
∑ ∑
∑ ∑
It is convenient to express (2) in the form
( ) ( ) ( ) ( )
x u =h u f u +g u (3)
Where
( ) ( )
m
( )
n
q u
h u = p u And
( ) ( ) ( )
1
( )
1n m
n
R u S u
g u P u
− − −
= (4)
And h u
( )
is usually called the transfer function. The inverse ELzaki transform combined with the convolution theorem leads to the formal solution,( ) ( ) ( ) ( )
0 t
x t =
∫
f t −τ h τ dτ+g t (5)With zero initial data, g u
( )
=0. the transfer function takes the simple form.( ) ( ) ( )
h u x u
=f u or x u
( )
=h u( ) ( )
f u (6)If f t
( )
=δ( )
t and h t( )
=et then, the output function is( )
1 3 11 u t
x t e
u
− ⎡ ⎤
= Ε ⎢⎣ + ⎥⎦= − h t
( )
is known as the impulse response.Example (2) (Delay Differential Equations)
In many problems the derivatives of the unknown functionx t
( )
are related to its value at different times t −τ.this leads us to consider differential equations of the form:( ) ( )
dx ax t f t
dt + −τ = (7)
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 t0− ≤ ≤τ t t0 and thisinformation combined with the equation it self, is sufficient to determine x t
( )
for0.
t >t We show how equation (7) can be solved by ELzaki transform when t0 =0 and x t
( )
=x0 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 H t f t
dt + −τ −τ = (8)
Applying ELzaki transform to (8) gives, x u
( )
ux( )
0 ae x uu( )
f u( )
u
−τ
− + =
Or
( ) ( )
2( )
0( )
2( )
0 1 11
u u
uf u u x
x u u f u u x a u e
a u e
τ τ
− −
−
⎡ ⎤
+ ⎡ ⎤
= =⎣ + ⎦⎢ + ⎥
⎣ ⎦
+ And
( ) ( )
2( ) ( ) ( )
0
0 1
n n n
u n
x u u f u u x au e
∞ − τ
=
⎡ ⎤
=⎣ + ⎦
∑
− (9)The inverse ELzaki transform gives the formal solution:
( )
1( )
2( ) ( ) ( )
0
0 1
n n n
u n
x t u f u u x au e
∞ −τ
−
=
⎡ ⎤
= Ε ⎣ + ⎦
∑
− (10)In order to write an explicit solution, we choose x0 =0 and f t
( )
=t and hence (10) be comes.( ) ( ) ( ) ( ) ( )
( )
2( )
1 4
0 0
1 1 , 0
2 !
n n
n n u n n
n n
t n
x t u au e a H t n t
n
τ τ
τ
∞ − ∞ +
−
= =
⎡ ⎤ −
= Ε ⎢⎣
∑
− ⎥⎦=∑
− + − >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 variableXn 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 X1, 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( )
.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 .( ) { ( )
1} ( )
0
( ) :
r t X t x t X x f x d x
∞
⎡ ⎤ ⎡ ⎤
= Ε⎣ ⎦= Ε
∫
⎣ ⎦ = (11)Where Ε
{
X t( )
:X1=x}
is the conditional expected value of X t( )
under the condition that X1=x and has the value.{
X t( )
:X1 x}
⎡1 r t(
x)
⎤H t(
x)
Ε = = +⎣ − ⎦ − (12)
Thus
( ) ( ) ( )
0
1
t
r t =
∫
⎡⎣ +r t −x ⎤⎦f x dx Or( ) ( ) ( ) ( )
0 t
r t =F t +
∫
r t −x f x dx (13)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( ) ( )
r u F u r u f u
= +u Or
( ) ( )
( )
r u uF u
u f u
= −
The inverse transform gives the formal solution r t( ) of renewal function.
(i) If F t
( )
=et and f t( )
=et , then:( )
1 3 2 1 sin h 21 2 2
r t u t
u
− ⎡ ⎤
= Ε ⎢⎣ − ⎥⎦= (ii) If F t
( )
=t and f t( )
=1, then:( )
1 2 2 11 u t
r t E u e
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
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 second Edition, Chapman & Hall /CRC (2006).
[8] A.Kilicman and H.E.Gadain. An application of double Laplace transform and Sumudu transform, Lobachevskii J. Math.30 (3) (2009), pp.214-223.
[9] J. Zhang, A Sumudu based algorithm m for solving differential equations, Comp. Sci. J. Moldova 15(3) (2007), pp – 303-313.
[10] Hassan Eltayeb and Adem kilicman, A Note on the Sumudu Transforms and differential Equations, Applied Mathematical Sciences, VOL, 4,2010, no.22,1089-1098
[11] Kilicman A. & H. ELtayeb. A note on Integral Transform and Partial Differential Equation, Applied Mathematical Sciences, 4(3) (2010), PP.109- 118.
[12] Hassan ELtayeh and Adem kilicman, on Some Applications of a new Integral Transform, Int. Journal of Math. Analysis, Vol, 4, 2010, no.3, 123-132.
Appendix
ELzaki Transform of some Functions
( )
f t Ε⎡⎣f t
( )
⎤⎦=T u( )
1 u2
t u3
tn n u! n+2
( )
1
, 0 ta
a a
−
Γ >
1
ua+
eat 2
1 u
−au teat
( )
3
1 2
u
−au
( )
1
, 1, 2,....
1 !
n at
t e n n
− =
−
( )
1
1
n n
u au
+
−
sinat 3
2 2
1 au +a u
cosat 2
2 2
1 u +a u
sinhat 3
2 2
1 au
−a u
cos hat 2
2 2
1 au
−a u
atsin e bt
( )
3
2 2 2
1 bu au b u
− +
atcos
e bt
( )
( )
2
2 2 2
1 1
au u au b u
−
− +
sin
t at 4
2 2
2 1
au +a u
0
( )
J at 2
1 2
u +au
( )
H t −a 2 au
u e−
(
t a)
δ − ua
u e−