Fareeha Transform Performance In Solving Fractional Differential
Telegraph Equations Combining Adomian Decomposition Method
NGUYEN MINH TUAN1ID , SANOE KOONPRASERT1,3ID ,
PHAYUNG MEESAD2ID
1Department of Mathematics, Faculty of Applied Science,
King Mongkut’s University of Technology North Bangkok,
1518 Pracharat 1 Road, Wongsawang, Bangsue, Bangkok 10800,
THAILAND
2Information Technology and Management Department,
King Mongkut’s University of Technology North Bangkok,
1518 Pracharat 1 Road, Wongsawang, Bangsue, Bangkok 10800,
THAILAND
3Centre of Excellence in Mathematics,
CHE, Si Ayutthaya Road, Bangkok 10400,
THAILAND
Abstract: Transformations have successfully outperformed a significant role in solving differential equations and
have been applied in large-scale aspects of science. Fareeha transform has been illustrated effectively in data com-
pression based on containing more information of the transform. In this paper, we expand the fractional Fareeha
transform in the Caputo derivative sense combining the Adomian Decomposition Method to seek the solutions of
fractional differential telegraph equations. The results of practical utilization have also been significantly shown
successful in solving fractional telegraph differential equations.
Key-Words: Fareeha Transform, ADM method, Laplace-typed transform, Time-fractional telegraph equations.
Received: March 26, 2023. Revised: January 22, 2024. Accepted: March 11, 2024. Published: April 16, 2024.
1 Introduction
Transformation is conveniently applied to perform
solutions of differential equations with constant-
coefficient equations, and usually combines other
suited methods for the coefficients containing poly-
nomial coefficients, for example, Laplace transform
is a powerful technique applied in solving mathemat-
ics and physics equations, especially in mechanics,
[1]. The application of Fourier and Laplace trans-
formation is varied such as managing the transfer
of electrical circuits, recognition systems, chemical
components, and computer programs, [2]. In recent
years, Fourier and Laplace transform has been used
more in analyzing and controlling digital signal pro-
cessing, [3]. The newfangled types of transforms
produced based on Fourier and Laplace transforms
to supply the needs of sciences have performed the
reliable approaches to solve particular problems, for
example, the Sumudu transform, [4]. Sumudu trans-
formation has been used in combination with homo-
topy analysis to solve the delay Fractional differential
Bagley-Torvik equation that does not need to calcu-
late the fractional derivative or integration terms, easy
and straightforward to get the exact solutions, [5].
The Sumudu transform created a new double inte-
gral Laplace-Sumudu transformation that has demon-
strated efficiency in solving the nonhomogeneous lin-
ear and nonlinear partial differential equations, [6].
Another transform such as the ELzaki transform de-
rived from the Fourier integral transform is a conve-
nient mathematical tool for solving differential equa-
tions, [7]. Elzaki transform is effective enough to
find the solution to the linear system expressed by an
ordinary differential equation associated with initial
or external disturbance condition, [8]. The associa-
tion of ELzaki transforms and the homotopy pertur-
bation method makes the Elzaki transform homotopy
perturbation method (ETHPM) used in physics and
engineering to solve linear and nonlinear differential
equations, [9].
Aboodh transform illustrated the main properties
to facilitate solving ordinary and partial differen-
tial equations, [10]. Aboodh transform connected
the new iterative method to perform a new approxi-
mate analytical method to solve fractional differential
equations in a biological population model, [11]. The
new alpha-integral Laplace transform supported the
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general features of the Laplace transform for the clas-
sical sense definition and theory, [12]. HY integral
transform, [13], is applied to solve Laguerre and Her-
mite differential equations by converting Lagoerre
and Hermite differential equations to first-order dif-
ferential equations. HY integral transform is useful in
finding Newtons law of cooling, [14], and expanded
for solving exponential growth and decay problems,
[15].
Another variation, the Mohand transform, is cre-
ated to efficiently apply in finding solutions of par-
tial differential equations, [16]. The Mohand trans-
form comes along with the Adomian decomposition
method to make the new path to solve the nonlinear
fractional evolution equations useful in mathemati-
cal models and engineering projects, [17]. Based on
the mathematical Fourier integral transform, the Sawi
transform was created providing the main properties
to demonstrate the fundamental properties for solv-
ing differential equations, [18]. The establishment
of the new transformation called the Kamal trans-
form supported new applications for ordinary and
partial differential equations, [19]. The duality rela-
tions make the Kamal transform visualize the full im-
portant sightseeing to other transformations, such as
Laplace, LaplaceCarson, Aboodh, Sumudu, Elzaki,
Mohand, and Sawi transforms, [20].
Attaining solutions of dynamical systems by us-
ing Laplace-Carson, fractional order associated with
the terms in the differential equations were demon-
strated to get the exact solution, [21]. Another per-
formance of Laplace-Carson is to raise the develop-
ment of the gradient flow of a viscoplastic medium,
[22]. A Laplace-typed transform called G_Trasform
established is applied to engineering problems, [23].
Anuj transforms illustrated new properties of linear-
ity, scaling, translation, convolution, and application
in solving ordinary differential equations, [21, 24].
The differential equations performance was signifi-
cantly solved by using Laplace transform, [25], and
ELzaki Transform, [26]. Another useful and effective
application of the Anuj transform is to find the solu-
tion of Volterra integral equations, [27]. The general
form of the transform can be seen in Table 1.
In this paper, the Fareeha transform will be ex-
tended for fractional differential derivative defined as
the following, [28]:
Gf(u(t)) = F(sn) = ∞
0
u(t)e−sntdt.(1)
where t>0,sn>0,n=2k+1,k=0,1,2,···. The Fa-
reeha transform has been built for the general form of
some following specific cases shown in Table 2. The
inverse Fareeha transform is defined as the following:
u(t) = G−1
fF(sn).(2)
Table 1: Demographic performance of recent trans-
forms.
Cases of transform Transform
sm∞
0u(t)e−sntdt
m=0,n=1 Laplace transform
m=n=−1 Sumudu transform
m=1,n=−1 Elzaki transform
m=−1,n=1 Aboodh transform
m=0,n=
α
−1,
α
-Integral Laplace-
α
∈R+
0Transform
m=1,n=2 HY integral transform
m=2,n=1 Mohand transform
m=−2,n=−1 Sawi transform
m=0,n=−1 Kamal transform
m=
α
,n=−1 G_transform
m=2,n=−1 Anuj transform
m=1,n=1 LaplaceCarson transform
In this paper, the Fareeha transform will be consid-
ered to solve the general fractional telegraph equation
of the form
D
α
tu(x,t) + L[u(x,t)] + N[u(x,t)] = q(x,t).(3)
subject to the initial and boundary conditions
u(x,0) = g(x),ut(x,0) = h(x),0<x<1,
t>0,1<
α
≤2.(4)
and D
α
tu(x,t)represents Caputo derivative in terms
of tof order
α
,L[u(x,t)],N[u(x,t)] respectively
perform the linear and nonlinear terms of function
u(x,t).
2 Basic Functions Using Fareeha
Transform
Definition 1 (Gamma function).(Shown in [29])
Given a value z ∈C, and Re(z)>0, the integration
presented as following is valid
Γ(z) = ∞
0
tz−1e−tdt.(5)
We collect some particular results related to the
gamma function:
Γ(z+1) = zΓ(z),Γ(n+1) = n!.(6)
Definition 2 (Caputo’s Fractional Derivatives).(See
in [30], chapter 2) Given a continuous function
y=f(x), and an arbitrary value m −1<
α
≤m, the
Caputo’s fractional derivative of order
α
is defined
by:
C
aD
α
xf(x) = 1
Γ(m−
α
)x
a
(x−s)m−
α
−1f(m)(s)ds.
(7)
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In this section, we will expand the transform for
the polynomial-coefficient functions and fractional
derivatives that are performed in Table 2.
Proposition 1. Using the Fareeha transform, we ap-
ply some basic function derivatives, [28]:
Gf(u′) = F(sn) = snF(sn)−u(0).(8)
Gf(u′′) = F(sn) = s2nF(sn)−snu(0)−u′(0).(9)
Gf(u′′′) = s3nF(sn)−s2nu(0)−snu′(0)−u′′(0).
(10)
Gf(u(k)(t)) = sknGf(u(t)) −
k−1
∑
m=0
s(k−m−1)nf(m)(0).
(11)
Proposition 2. Fareeha transform of the function
f(t) = t
α
:
Gf(t
α
) = Γ(
α
+1)
s(
α
+1)n.(12)
Proof.
Using Fareeha’s definition, [28], Equation (1), and
Gamma function (5), we have
Gf(t
α
) = ∞
0
t
α
e−sntdt,(substituting u =snt)
=1
s(
α
+1)n∞
0
u
α
e−udu =Γ(
α
+1)
s(
α
+1)n.(13)
Proposition 3. Now, we will extend the Fareeha
transform for the Caputo derivative which is a main
point applied to this work:
GfC
0D
α
tu(t)=sn
α
F(sn)−
m−1
∑
k=0
sn(
α
−k−1)u(k)(0).
(14)
Proof.
Using convolution property of f(t)and g(t)in Table
2,
f∗g=t
0
f(u)g(t−u)du.(15)
From definition 2 of the Caputo derivative, we have
C
0D
α
tu(t) = 1
Γ(m−
α
)t
0
(t−
τ
)m−
α
−1x(m)(
τ
)d
τ
=1
Γ(m−
α
)tm−
α
−1∗x(m)(t).(16)
Using Fareeha definition, [28], Equation (11), we
Table 2: The Fareeha transform of basic functions.
Functions u(t)F(sn)(s) = ∞
0u(t)e−sntdt
1s−n
t s−2n
eat 1
sn−a
t
α
Γ(
α
+1)
s(
α
+1)n
sinat a
s2n+a2
tsinat 2asn
(s2n+a2)2
cosat sn
s2n+a2
tcosat s2n−a2
(s2n+a2)2
sinhat a
s2n−a2
coshat sn
s2n−a2
tsinhat 2asn
s2n−a2
tcoshat s2n+a2
s2n−a2
eat sin
ω
t
ω
(sn−a2)2+
ω
2
eat cos
ω
tsn−a
(sn−a2)2+
ω
2
eat sinh
ω
t
ω
(sn−a2)2−
ω
2
eat cosh
ω
tsn−a
(sn−a2)2−
ω
2
sintcost1
s2n+4
u(t)∗v(t)U(sn)×V(sn)
(convolution of u(t)and v(t),
have
GfC
0D
α
tu(t)=1
Γ(m−
α
)Gftm−
α
−1×Gfx(m)(t)
=1
Γ(m−
α
)
Γ(m−
α
)
s(m−
α
)n
×snmF(sn)−
m−1
∑
k=0
sn(m−k−1)u(k)(0)
=1
s(m−
α
)n
×snmF(sn)−
m−1
∑
k=0
sn(m−k−1)u(k)(0)
=sn
α
F(sn)−
m−1
∑
k=0
sn(
α
−k−1)u(k)(0).
(17)
3 Methodology
In this section, we will combine the transform of
the fractional derivative using Proposition 3 and the
Adomian decomposition method for solving the fac-
tional differential telegraph equations. First, we re-
mind the Adomian decomposition method deployed
in this work
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3.1 Adomian decomposition method
The Adomian decomposition method is an effective
procedure for finding the analytic solutions intro-
duced to solve linear and nonlinear differential equa-
tions, [31]. The Adomain decomposition method
was summarized to determine the exact solutions of
Bratu-type equations, [32], and has been used in most
system theories to simplify the disjoint state spaces.
The differential equations, [33], have been developed
enormously and expanded in some models of real life
such as the circuit model, Spring-mass system, dif-
fusion system, and population model, [25]. In this
paper, the newfangled Fareeha transform, [28], com-
bines the Adomian decomposition method to perform
the effectiveness of the transform for solving frac-
tional telegraph equations. The Adomain decompo-
sition method, [34], expresses a function u(x)in the
form of
u(x) =
∞
∑
n=0
un(x),(18)
where the term un(x)be iteratively defined. Espe-
cially, the nonlinear term N(u)can be decomposition
into an infinite series of polynomials in the form of
N(u) =
∞
∑
n=0
An(u0,u1,u2,···,un),(19)
where Anbe the so-called Adomian polynomial terms
of u0,u1, ..., undefined by
An=1
n!
dn
d
λ
n[F(
n
∑
i=0
λ
iui)]
λ
=0,n=0,1,2,··· (20)
where the terms are determined as the following
A0=F(u0),
A1=u1F′(u0),
A2=u2F′(u0) + 1
2u2
1F′′(u0),
A3=u3F′(u0) + u1u2F′′(u0) + 1
3u3
1F′′′(u0,
A4=u4F′(u0) + (u1u3+1
2u2
2)F′′(u0)
+1
2u2
1u2F′′′(u0) + 1
24u4
1F(4)u0,
···
3.2 Using Fareeha transform for the
fractional telegraph equations
In this section, we deploy the Fareeha transform
for general fractional telegraph equations as the
following:
D
α
tu(x,t) + L[u(x,t)] + N[u(x,t)] = q(x,t),
(21)
satisfy the condition
u(x,0) = g(x),ut(x,0) = h(x),0<
α
≤2.(22)
Step 1: Taking Fareeha transform for the equa-
tion (21), and using property (14) including condition
(22):
Gf[D
α
tu(x,t) + L[u(x,t)] + N[u(x,t)]] = Gf[q(x,t)].
(23)
Step 2: Simplify the equation in the form:
u(x,t) = G−1
fg(x)s−n+h(x)s−2n+1
sn
α
Gf[q(x,t)]
+G−1
f1
sn
α
Gf{−L[u(x,t)] −N[u(x,t)]}.
(24)
Step 3: Setting u(x,t) = ∑∞
k=0uk(x,t),N(u) =
−L[u(x,t)−N[u(x,t)] then calculate the terms:
u0=G−1
fg(x)s−n+h(x)s−2n+1
sn
α
Gf[q(x,t)]
uk+1=G−1
f1
sn
α
Gf(Ak),k=0,1,2,···
where Akis the Adomian terms given by the Equation
(20).
Step 4: The analytic solution is depicted in the form
that could approach to the exact solution:
u(x,t) = u0(x,t) + u1(x,t) + u2(x,t) + ··· (25)
4 Application
The Fareeha transform seems well suited to solutions
of ordinary differential equations when they are lin-
early combined with constant coefficients, and when
initial conditions create the solutions. We will apply
the Fareeha transform for the fractional differential
telegraph equations using the Adomian decomposi-
tion method. We consider the first example as the
following
Example 4.1.Consider the fractional hyperbolic tele-
graph equation as follows, [35]:
D
α
tu(x,t) = u(x,t)−2ut(x,t)−uxx(x,t),(26)
u(x,0) = ex,ut(x,0) = −2ex,0<
α
≤2.(27)
Taking Fareeha transform on both sides from
Equation (26), using equation (14) including condi-
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tion (27)
Gf[D
α
tu(x,t)] = Gfu(x,t)−2ut(x,t)−uxx(x,t)
sn
α
F(x,sn)−sn(
α
−1)ex+2exsn(
α
−2)
=Gf[u(x,t)−2ut(x,t)−uxx(x,t)].
F(x,sn) = ex(sn(
α
−1)−2sn(
α
−2))
sn
α
+1
sn
α
Gfu(x,t)−2ut(x,t)−uxx(x,t).
(28)
Simplify the equations (28) and taking converse
Laplace on both sides, we can get
u(x,t) = G−1
fex(s−n−2s−2n)
+1
sn
α
Gf[u(x,t)−2ut(x,t)−uxx(x,t)]
=G−1
fex(s−n)−2s−2n)
+G−1
f1
sn
α
Gf[u(x,t)−2ut(x,t)−uxx(x,t)].
(29)
Taking F[u] = u−2ut−uxx, and setting the terms of
ADM F[u] = ∑∞
k=0Ak, where
Ak=1
k!
dk
d
λ
kFk
∑
i=0
λ
iui
λ
=0
;k=0,1,2,3,···
(30)
Assume the solution u(x,t) = ∑∞
i=0uk(x,t), so
∞
∑
i=0
uk(x,t) = G−1
fex(s−n)−2s−2n)
+G−1
f1
sn
α
Gf[Ak],k=0,1,2,3,···
(31)
Construct the recursion form of the Adomian decom-
position method (29), we calculate the terms as fol-
lows:
u0=G−1
fex(s−n−2s−2n)=ex(1−2t)(32)
uk+1=G−1
f1
sn
α
Gf[Ak],k=0,1,2,3,··· (33)
Using (32), we can find the term A0as follows:
A0=1
0!
d0
d
λ
0F0
∑
i=0
λ
0u0
λ
=0
=u0−2u0t−u0xx
=ex(1−2t)−2(−2ex)−ex(1−2t) = 4ex.
(34)
From equation (33), we evaluate the terms of solution
u1using (34) as follows
u1=G−1
f1
sn
α
Gf(A0)=G−1
f1
sn
α
Gf(4ex)
=4ext
α
Γ(
α
+1).(35)
Now we calculate the term A1using equation (35)
A1=1
1!
d1
d
λ
1F1
∑
i=0
λ
1ui
λ
=0
=F(u1)
=u1−2u1t−u1xx =−8ex
α
t
α
−1
Γ(
α
+1).(36)
From equation (36), A1, we can find the value of u2
u2=G−1
f1
snGf(A1)=G−1
f1
sn
α
Gf−8ex
α
t
α
−1
Γ(
α
+1)
=−8exG−1
f1
sn
α
Gf
α
t
α
−1
Γ(
α
+1)=−8ext2
α
−1
Γ(2
α
−1).
(37)
Using equation (37), u2, we can calculate A2:
A2=1
2!
d2
d
λ
2F2
∑
i=0
λ
2uixx
λ
=0
=1
2!
d2
d
λ
2
λ
0u0+
λ
u1+
λ
2u2
λ
=0
=1
22F[u2] = u2−2u2t−u2xx =16ext2
α
−2
Γ(2
α
−1).
(38)
Using equation (38), A2, we can calculate u3:
u3=G−1
f1
sn
α
Gf(A2)=16ex
Γ(2
α
−1)G−1
f1
sn
α
Gft2
α
−2
=16ex
Γ(2
α
−1)G−1
f1
sn
α
Γ(2
α
−1)
s(2
α
−1)n=16ex
Γ(3
α
−1)t3
α
−2.
(39)
The solution collected using (25) is performed as fol-
lows
u(x,t) = ex(1−2t) + 4ext
α
Γ(
α
+1)−8ext2
α
−1
Γ(2
α
)
+16ex
Γ(3
α
−1)t3
α
−2+···
=ex1−2t+4t
α
Γ(
α
+1)−8t2
α
−1
Γ(2
α
)
+16
Γ(3
α
−1)t3
α
−2+···,(40)
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when
α
=2, the solution (40) becomes as the follow-
ing
u(x,t) = ex1−2t+(2t)2
2! −(2t)3
3! +(2t)4
4! − ···
=ex−2t.(41)
and their graphs have been depicted in Fig. 1, and
Fig. 2, respectively (Appendix). The comparison of
the solutions by different values of
α
is shown in Ta-
ble 3 (Appendix).
Example 4.2.Consider a non-linear time-fractional
hyperbolic telegraph equation, [35]:
D
α
tu(x,t) = uxx(x,t) + ut(x,t)−u2(x,t) + xu(x,t).ux(x,t),
(42)
u(x,0) = x;ut(x,0) = x,0<
α
≤2.(43)
Let F[u] = uxx(x,t) + ut(x,t)−u2(x,t) +
xu(x,t).ux(x,t), and taking Fareeha transform on
both sides of equation (42), using equation (14)
consisting of condition (43)
Gf[D
α
tu(x,t)] = Gf[N(u)] (44)
sn
α
F(x,sn)−sn(
α
−1)x−sn(
α
−2)x=Gf[N(u)].(45)
Simplify equation (45) and collect the terms F(sn),
we have
F(x,sn) = s−n
α
[x(sn(
α
−1)−sn(
α
−2))] + s−n
α
Gf[N(u)].
(46)
Deriving the solution by dividing sn
α
on both sides of
equation (46) by inverse Fareeha transform
u(x,t) = G−1
fs−n
α
[x(sn(
α
−1)−sn(
α
−2))]
+G−1
f1
sn
α
Gf[N(u)].(47)
Adopting N(u) = ∑∞
k=0Ak,u(x,t) = ∑∞
k=0uk(x,t),
where
Ak=1
k!
dk
d
λ
kFk
∑
i=0
λ
iui
λ
=0
;n=0,1,2,···
(48)
the equation (47) could be rewritten as follows
∞
∑
k=0
uk(x,t) = G−1
fs−n
α
[x(sn(
α
−1)−sn(
α
−2))]
+G−1
f1
sn
α
Gf[N(u)],(49)
and we calculate the iterative form of ADM using
equation (47) as the following
u0=G−1
fs−n
α
[x(sn(
α
−1)−sn(
α
−2))]=x(1+t).
(50)
uk+1=G−1
f1
sn
α
Gf[Ak],k=0,1,2,··· (51)
We can calculate the terms A0using equation (50) as
follows
A0=1
0!
d0
d
λ
0F0
∑
i=0
λ
0u0
λ
=0
=u0xx +u0t−u2
0+xu0.u0x
=x−[x(1+t)]2+x2(1+t)(1+t) = x.(52)
Considering the terms ukfrom equation (51), and
equation (52) as follows:
u1=G−1
f1
sn
α
Gf(A0)=G−1
f1
sn
α
Gf(x)=xt
α
Γ(
α
+1).
(53)
Using equation (52), u1, we can calculate A1
A1=1
1!
d1
d
λ
1F1
∑
i=0
λ
1uixx
λ
=0
=d
d
λ
λ
0u0+
λ
u1
λ
=0
=u1xx +u1t−u2
1+xu1.u1x.
=x
α
t
α
−1
Γ(
α
+1)−x2t2
α
(Γ(
α
+1))2+x2t
α
Γ(
α
+1)
t
α
Γ(
α
+1)
=x
α
t
α
−1
Γ(
α
+1).(54)
Using equation (54), A1, we can calculate u2
u2=G−1
f1
snGf(A1)=G−1
f1
sn
α
Gfx
α
t
α
−1
Γ(
α
+1)
=x
α
Γ(
α
+1)G−1
f1
sn
α
Γ(
α
)
s
α
n=xt2
α
−1
Γ(2
α
−1).
(55)
Using equation (55), u2, we calculate A2:
A2=1
2!
d2
d
λ
2F2
∑
i=0
λ
2uixx
λ
=0
=1
2!
d2
d
λ
2
λ
0u0+
λ
u1+
λ
2u2
λ
=0
=u2xx +u2t−u2
2+xu2.u2x=xt2
α
−2
Γ(2
α
−1).(56)
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Now, we can calculate u3using equation (56)
u3=G−1
f1
sn
α
Gf(A2)
=x
Γ(2
α
−1)G−1
f1
sn
α
Gft2
α
−2
=x
α
Γ(
α
)
Γ(
α
+1)
(2
α
−1)
Γ(2
α
)G−1
f1
sn
α
Γ(2
α
−1)
s(2
α
−1)n
=xt3
α
−2
Γ(3
α
−1).(57)
Solution collected using equation (25) is performed
as follows
u(x,t) = u0(x,t) + u1(x,t) + u2(x,t) + u3(x,t) + ···
=x(1+t) + xt
α
Γ(
α
+1)+xt2
α
−1
Γ(2
α
)
+xt3
α
−2
Γ(3
α
−1)+··· (58)
When
α
=2, the solution from equation (58) be-
comes
u(x,t) = x1+t+t2
2! +t3
3! +t4
4! +···=xet.(59)
and their graphs has been respectively portrayed in
Fig. 3, and Fig. 4 (Appendix). The comparison of
different values of
α
is shown in Table 4 (Appendix).
Example 4.3.Taking into account 2D time-fractional
telegraph equation given as follows, [35]:
D2
α
tu(x,y,t) + 3D
α
tu(x,y,t) + 2u(x,y,t) = uxx(x,y,t)
+uyy(x,y,t),(60)
satisfy the initial conditions
u(x,y,0) = ex+y;ut(x,y,0) = −3ex+y,0<
α
≤1.
(61)
Let F[u] = −3D
α
tu−2u+uxx +uyy, and apply the
Fareeha transform on both sides of equation (60) in-
cluding (61)
GfD2
α
tu(x,y,t)=Gf[N(u)].
s2n
α
F(x,y,sn)−sn(2
α
−1)u(x,y,0)−sn(2
α
−2)ut(x,y,0)
=Gf[N(u)].(62)
Taking inverse Fareeha transform from equation (62),
we have
u(x,y,t) = G−1
fsn(2
α
−1)u(x,y,0)−sn(2
α
−2)ut(x,y,0)
+G−1
f1
s2n
α
Gf[N(u)].(63)
By setting N(u) = ∑∞
k=0Ak, and u(x,y,t) =
∑∞
k=0uk(x,y,t), where
Ak=1
k!
dk
d
λ
kFk
∑
i=0
λ
iui
λ
=0
,k=0,1,2,3,···
(64)
the equation (65) is rewritten as follows
∞
∑
k=0
uk(x,y,t) = G−1
fsn(2
α
−1)u(x,y,0)−sn(2
α
−2)ut(x,y,0)
+G−1
f1
s2n
α
Gf[N(u)].(65)
Using equation (65), and setting reiterated terms as
the following
u0=G−1
fsn(2
α
−1)u(x,y,0)−sn(2
α
−2)ut(x,y,0).
=u(x,y,0) + tut(x,y,0) = ex+y(1−3t)(66)
ur+1=G−1
fs−2n
α
Gf[−3D
α
tu−2u+uxx +uyy],
r=1,2,··· (67)
We calculate the terms of ADM, A0
A0=1
0!
d0
d
λ
0F0
∑
i=0
λ
0u0
λ
=0
=−3D
α
tu0−2u0+u0xx +u0yy =−3D
α
tu0.(68)
Using an equation (68), we calculate the term u1
u1=G−1
f1
s2n
α
Gf(A0)=G−1
f1
s2n
α
Gf(−3D
α
tu0)
=−3G−1
f1
s2n
α
sn
α
Gf[u0]−sn(
α
−1)u0(x,y,0)
=−3G−1
f1
s2n
α
ex+ysn
α
1
sn−3
s2n−sn(
α
−1)
=−3G−1
f1
s2n
α
−3
s(2−
α
)nex+y
=−3G−1
f1
s2n
α
Gf(D
α
tu0)
=−3G−1
fs−2n
α
−3
s(2−
α
)nex+y
=9ex+yt
α
+1
Γ(
α
+2).(69)
Using equation (69), we can calculate the term A1
A1=1
1!
d1
d
λ
1F1
∑
i=0
λ
1uixx
λ
=0
=d
d
λ
λ
0u0+
λ
u1
λ
=0
=−3D
α
tu1−2u1+u1xx +u1yy.(70)
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Using equation (69), and equation (70), we have
u2=G−1
f1
s2n
α
Gf(A1)
=G−1
f1
s2n
α
Gf(−3D
α
tu1−2u1+u1xx +u1yy)
=−3G−1
f1
s2n
α
sn
α
Gf[u1]−sn(
α
−1)u1(x,y,0)
=−3G−1
f1
s2n
α
sn
α
Gf9ex+yt
α
+1
Γ(
α
)
=−27ex+yt2
α
+1
Γ(2
α
+2).(71)
Using equation (71), we have
A2=1
2!
d2
d
λ
2F2
∑
i=0
λ
2uixx
λ
=0
=1
2!
d2
d
λ
2
λ
0u0+
λ
u1+
λ
2u2
λ
=0
=−3D
α
tu2−2u2+u2xx +u2yy.(72)
u3=G−1
f1
sn
α
Gf(A2)
=G−1
f1
sn
α
Gf(−3D
α
tu2−2u2+u2xx +u2yy)
=G−1
f1
sn
α
Gf(−3D
α
tu2)=G−1
f1
sn
α
Gf(−3D
α
tu2)
=−3G−1
f1
sn
α
(s
α
nGf[u2]−s
α
−1u2(x,y,0))
=−3G−1
f1
sn
α
−27ex+y
s(
α
+2)n=81ex+yt3
α
+1
Γ(3
α
+2).
(73)
So the solution is collected as the following
u(x,y,t) = ex+y1−3t+9t
α
+1
Γ(
α
+2)
−27t2
α
+1
Γ(2
α
+2)+81t3
α
+1
Γ(3
α
+2)···.(74)
When
α
=1, the solution from the equation (74)
shown as follows
u(x,y,t) = ex+y1−3t+(3t)2
2! −(3t)3
3! +···=ex+y−3t.
(75)
and the solutions have been respectively shown in
Fig. 7, and Fig. 8 (Appendix). The comparison of
the solutions of arbitrary values of
α
is shown in Ta-
ble 5 (Appendix).
Example 4.4.To perform the effectiveness of the
Fareeha transform, we take into account the space-
differential telegraph equation as follows, [36]:
D
α
xu(x,t) = utt (x,t) + ut(x,t) + u(x,t),(76)
u(0,t) = e−t,ux(0,t) = e−t,0<
α
≤2,t≥0.(77)
Considering F[u] = utt +ut+u, taking Fareeha
transform on both sides equation (76), using equation
(14) including equation (77):
Gf(D
α
xu(x,t)) = Gf[N(u)].(78)
sn
α
F(x,sn)−sn(
α
−1)e−t−sn(
α
−2)e−t=Gf[N(u)].
F(x,sn) = [sn(
α
−1)−sn(
α
−2)]e−t
sn
α
+1
sn
α
Gf[(N(u)].
(79)
Taking inverse Fareeha transform of equation (79) by
sn
α
, and we have
u(x,t) = G−1
f[sn(
α
−1)−sn(
α
−2)]e−t
sn
α
+1
sn
α
Gf[N(u)]
=G−1
f[sn(
α
−1)−sn(
α
−2)]e−t
sn
α
.
+G−1
f1
sn
α
Gf[N(u)](80)
Setting N(u) = ∑∞
k=0Ak,u(x,t) = ∑∞
k=0uk(x,t), where
Ak=1
n!
dk
d
λ
kFk
∑
i=0
λ
iui
λ
=0
;k=0,1,2,3,···
(81)
and the equation (80) is rewritten as follows
∞
∑
k=0
uk(x,t) = G−1
f[sn(
α
−1)−sn(
α
−2)]e−t
sn
α
+G−1
f1
sn
α
Gf[N(u)].(82)
Now, we will establish repeated terms of ADM using
(82)
u0=G−1
f(sn(
α
−1)−sn(
α
−2))e−t
sn
α
=e−t+xe−t=e−t(1+x).(83)
uk+1=G−1
f1
sn
α
Gf(utt +ut+u)=G−1
f1
sn
α
Gf(An).
(84)
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Using equation (83), we evaluate A0as follows
A0=1
0!
d0
d
λ
0F0
∑
i=0
λ
0u0
λ
=0
=u0tt +u0t+u0
= (1+x)e−t+ (1+x)(−e−t)+(1+x)e−t= (1+x)e−t.
(85)
Based on equation (85), and equation (84), we can
calculate the terms of the solution as follows:
u1=G−1
f1
sn
α
Gf(A0)=G−1
f1
sn
α
Gf(1+x)e−t
=e−tG−1
f1
sn
α
(1
sn+1
s2n)=e−tx
α
Γ(
α
+1)+x
α
+1
Γ(
α
).
(86)
The following steps are to calculate the rest of ADM
and solution, for example, using equation (86), we
can calculate A1
A1=1
1!
d1
d
λ
1F1
∑
i=0
λ
1uixx
λ
=0
=d
d
λ
λ
0u0+
λ
u1
λ
=0
=u1tt +u1t+u1=e−tx
α
Γ(
α
+1)+x
α
+1
Γ(
α
).
(87)
Using equation (87), we calculate u2
u2=G−1
f1
snGf(A1)
=G−1
f1
sn
α
Gfe−tx
α
Γ(
α
+1)+x
α
+1
(
α
+1)!
=e−tG−1
f1
sn
α
1
s(
α
+1)n+1
s(
α
+2)n
=e−tx2
α
Γ(2
α
+1)+x2
α
+1
Γ(2
α
+2).(88)
From equation (88), we can calculate A2
A2=1
2!
d2
d
λ
2F2
∑
i=0
λ
2uixx
λ
=0
=1
2!
d2
d
λ
2
λ
0u0+
λ
u1+
λ
2u2
λ
=0
=u2tt +u2t+u2=e−tx2
α
Γ(2
α
+1)+x2
α
+1
Γ(2
α
+2).
(89)
Using equation (89), we can find the term u3
u3=G−1
f1
sn
α
Gf(A2)
=e−tG−1
f1
sn
α
Gfe−tx2
α
Γ(2
α
+1)+x2
α
+1
Γ(2
α
+2)
=e−tG−1
f1
sn
α
1
s(2
α
+1)n+1
s(2
α
+2)n
=e−t1
s(3
α
+1)n+1
s(3
α
+2)n
=e−tx3
α
Γ(3
α
+1)+x3
α
+1
Γ(3
α
+2).(90)
Then the solution was derived using (25) as follows
u(x,t) = e−t1+x+x
α
Γ(
α
+1)
+x
α
+1
Γ(2
α
+2)+x2
α
Γ(2
α
+1)+x2
α
+1
Γ(2
α
+2)
+x3
α
Γ(3
α
+1)+x3
α
+1
Γ(2
α
+2)+···.(91)
When
α
=2 the solution gained from equation (91)
as the following
u(x,t) = ex−t.(92)
and the solutions have been respectively depicted us-
ing the Matlab application in Fig. 5 and Fig. 6 (Ap-
pendix). The comparison of the solutions by different
values of
α
is shown in Table 6 (Appendix).
5 Conclusion
In this paper, some extensive properties for the Fa-
reeha transform have been created, especially the
fractional derivative in the Caputo sense. Fareeha
transform has also performed successfully by solv-
ing time-fractional telegraph equations. To show the
effectiveness of the transform, a space-fractional tele-
graph equation was conducted briefly to get the an-
alytic solution. Using series expansion, the close
forms have been derived that approach exact solu-
tions played an essential role in physics [37], and
their graphs have been performed to show the nec-
essary role of the methods. Like other Laplace-typed
transforms, the inverse Fareeha transform is derived
from the direct Fareeha transform summarized in Ta-
ble 2. The transform is conducted by taking Fareeha
transform for fractional terms and other linear and
nonlinear terms are included in Adomian decomposi-
tion method terms. In general, the Fareeha transform
has effectively contributed to finding the solution to
fractional telegraph equations.
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Acknowledgment:
The authors acknowledge the anonymous reviewer’s
comments, which improve the manuscript’s quality,
and kindly provide advice and valuable discussions.
References:
[1] Schiff, J. L. (1999). The Laplace Transform:
Theory and Applications. 245.
[2] Holbrook, J. G. (1973). Laplace-Transformation.
Vieweg+Teubner Verlag.
https://doi.org/10.1007/978-3-663-01882-7.
[3] Beerends, R. J. (2003). Fourier and Laplace
Transforms. 459.
[4] Watugala, G. K. (1993). Sumudu transform: A
new integral transform to solve differential
equations and control engineering problems.
International Journal of Mathematical Education
in Science and Technology, 24(1), 3543.
https://doi.org/10.1080/0020739930240105.
[5] Alomari, A. K., Syam, M. I., Anakira, N. R., &
Jameel, A. F. (2020). Homotopy Sumudu
transform method for solving applications in
physics. Results in Physics, 18, 103265.
https://doi.org/10.1016/j.rinp.2020.103265.
[6] Ahmed, S. A., Elzaki, T. M., Elbadri, M., &
Mohamed, M. Z. (2021). Solution of partial
differential equations by new double integral
transform (LaplaceSumudu transform). Ain
Shams Engineering Journal, 12(4), 40454049.
https://doi.org/10.1016/j.asej.2021.02.032.
[7] Elzaki, T. M. (2011). The New Integral
Transform ELzaki Transform. 9.
[8] Kim, H. (2017b). On The Form and Properties of
an Integral Transform with Strength in Integral
Transforms. Far East Journal of Mathematical
Sciences (FJMS), 102(11), 28312844.
https://doi.org/10.17654/MS102112831
[9] Elzaki, T. M., & Biazar, J. (2013). Homotopy
Perturbation Method and Elzaki Transform for
Solving System of Nonlinear Partial Differential
Equations.
[10] Aboodh, K. S. (2013). The New Integral
Transform Aboodh Transform. 11.
[11] Ojo, G. O., & Mahmudov, N. I. (2021).
Aboodh Transform Iterative Method for Spatial
Diffusion of a Biological Population with
Fractional-Order. Mathematics, 9(2), 155.
https://doi.org/10.3390/math9020155.
[12] Ojeda, N., & Romero, L. (2016). A NEW
α
Iintegral Laplace Transform. 5, 5962.
https://doi.org/10.15520/ajcem.2016.vol5.iss5.59.pp59-
62.
[13] Ahmadi, S. A. P., Hosseinzadeh, H., & Cherati,
A. Y. (2019). A New Integral Transform for
Solving Higher Order Linear Ordinary Laguerre
and Hermite Differential Equations.
International Journal of Applied and
Computational Mathematics, 5(5), 142.
https://doi.org/10.1007/s40819-019-0712-1
[14] Patil, D. P., Gangurde, J. P., Wagh, S. N., &
Bachhav, T. P. (2022). Applications of the HY
Transform for Newtons Law of Cooling. 9(2).
[15] Patil, D. P., GulamGaus, S. A. F., Kiran, M. N.,
& Rashid, S. J. (2022). The HY Integral
Transform for Handling Exponential Growth and
Decay Problems. 9(6).
[16] Mahgoub, M. M. A. (2017). The New Integral
Transform Mohand Transform. 8.
[17] Patra, A., Baliarsingh, P., & Dutta, H. (2022).
Solution to fractional evolution equation using
Mohand transform. Mathematics and Computers
in Simulation, 200, 557570.
https://doi.org/10.1016/j.matcom.2022.04.021
[18] Mahgoub, M. M. A. (2019). The New Integral
Transform Sawi Transform. 8.
[19] Sedeeg, A. K. H. (2016). The New Integral
Transform Kamal Transform.
[20] Aggarwal, S., Sharma, N., & Chauhan, R.
(2020). Duality relations of Kamal transform
with Laplace, LaplaceCarson, Aboodh, Sumudu,
Elzaki, Mohand and Sawi transforms. SN
Applied Sciences, 2(1), 135.
https://doi.org/10.1007/s42452-019-1896-z
[21] Kumar, P., & Qureshi, S. (2020).
Laplace-Carson integral transform for exact
solutions of non-integer order initial value
problems with Caputo operator. Journal of
Applied Mathematics and Computational
Mechanics, 19(1), 5766.
https://doi.org/10.17512/jamcm.2020.1.05
[22] Makarov, A. M. (1970). Application of the
Laplace-carson integral transform method to the
theory of nonstationary flows of a viscoplastic
medium. Journal of Engineering Physics, 19(1),
870874. https://doi.org/10.1007/BF00832575
WSEAS TRANSACTIONS on SYSTEMS and CONTROL
DOI: 10.37394/23203.2024.19.9
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Volume 19, 2024
[23] Kim, H. (2017a). The Intrinsic Structure and
Properties of Laplace-Typed Integral
Transforms. Mathematical Problems in
Engineering, 2017, 18.
https://doi.org/10.1155/2017/1762729
[24] Tuan, N. M. (2023). A Study of Applied
Reduced Differential Transform Method Using
Volterra Integral Equations in Solving Partial
Differential Equations. EQUATIONS, 3, 93103.
https://doi.org/10.37394/232021.2023.3.11
[25] Abell, M. L., & Braselton, J. P. (2018).
Introductory differential equations (Fifth
edition). Elsevier/Academic Press.
[26] Elzaki, T. M., & Elzaki, S. M. (2011). On the
ELzaki Transform and Systems of Ordinary
Differential Equations.
[27] Patil, D. P., Tile, G. K., & Shinde, P. D. (2022).
Volterra Integral Equations of First Kind by
Using Anuj Transform. 4(5).
[28] Khan, F., & Khalid, M. (2023). Fareeha
transform: A new generalized Laplace
transform. Mathematical Methods in the Applied
Sciences. https://doi.org/10.1002/mma.9167.
[29] Herrmann, R. (2014). Fractional calculus: An
introduction for physicists (2nd edition). World
Scientific.
[30] Podlubny, I. (c1999.). Fractional differential
equations: An introduction to fractional
derivatives, fractional differential equations, to
methods of their solution and some of their
applications. Academic Press.
[31] Adomian, G. (1988). A review of the
decomposition method in applied mathematics.
Journal of Mathematical Analysis and
Applications, 135(2), 501544.
https://doi.org/10.1016/0022-247X(88)90170-9
[32] Wazwaz A.-M. (2009). Partial Differential
Equations and Solitary Waves Theory. Higher
Education Press; 1sted. Beijing and Berlin:
Springer; 2009.
[33] E. J. Watson. (1981). Laplace transforms and
applications, Van Nostrand Reinhold Co., New
York-London.
[34] Wazwaz, A.-M. (2005). Adomian
decomposition method for a reliable treatment of
the Bratu-type equations. Applied Mathematics
and Computation, 166(3), 652663.
https://doi.org/10.1016/j.amc.2004.06.059
[35] Kapoor, M., Shah, N. A., Saleem, S., & Weera,
W. (2022). An Analytical Approach for
Fractional Hyperbolic Telegraph Equation Using
Shehu Transform in One, Two and Three
Dimensions. Mathematics, 10(12), 1961.
https://doi.org/10.3390/math10121961
[36] Yldrm, A. (2010). Hes homotopy perturbation
method for solving the space- and
time-fractional telegraph equations. International
Journal of Computer Mathematics, 87(13),
29983006.
https://doi.org/10.1080/00207160902874653
[37] Iatkliang, T., Kaewta, S., Tuan, N. M., &
Sirisubtawee, S. (2023). Novel Exact Traveling
Wave Solutions for Nonlinear Wave Equations
with Beta-Derivatives via the sine-Gordon
Expansion Method. WSEAS TRANSACTIONS
ON MATHEMATICS, 22, 432450.
https://doi.org/10.37394/23206.2023.22.50
Contribution of Individual Authors to the
Creation of a Scientific Article (Ghostwriting
Policy)
Nguyen Minh Tuan: Conceptualization, data cura-
tion, investigation, methodology, software, visualiza-
tion, writing-original draft and writing-review and
editing.
Sanoe Koonprasert: Conceptualization, data cura-
tion, formal analysis, investigation, methodology,
project administration, resources, supervision, vali-
dation, visualization, writing-original draft, writing-
review and editing.
Phayung Meesad: Conceptualization, data curation,
formal analysis, investigation, resources, supervision,
validation, visualization, writing-original draft, and
editing.
Sources of Funding for Research Presented in a
Scientific Article or Scientific Article Itself No
funding was received for conducting this study.
Conflicts of Interest The authors
have no conflicts of interest to
declare that are relevant to the content of this
article.
Creative Commons Attribution License 4.0
(Attribution 4.0 International , CC BY 4.0)
This article is published under the terms of the
Creative Commons Attribution License 4.0
https://creativecommons.org/licenses/by/4.0/deed.en
_US
WSEAS TRANSACTIONS on SYSTEMS and CONTROL
DOI: 10.37394/23203.2024.19.9
Nguyen Minh Tuan, Sanoe Koonprasert, Phayung Meesad
E-ISSN: 2224-2856
95
Volume 19, 2024
Figure 1: 2D solution depict example 4.1.
Figure 2: 3D solution example 4.1,
α
=2.
Table 3: Solutions of example 4.1 for x=0.5.
t
α
=1.1
α
=1.2
α
=1.5
α
=1.7
α
=2
0.0 1.6487 1.6487 1.6487 1.6487 1.6487
0.1 2.1974 1.7374 1.4350 1.3896 1.3499
0.2 3.1172 2.1464 1.3112 1.1994 1.1052
0.3 4.2396 2.7909 1.2724 1.0576 0.9048
0.4 5.5136 3.6384 1.3328 0.9644 0.7408
0.5 6.9118 4.6687 1.5085 0.9274 0.6065
0.6 8.4166 5.8679 1.8152 0.9581 0.4966
0.7 10.0155 7.2254 2.2686 1.0707 0.4066
0.8 11.6991 8.7328 2.8836 1.2814 0.3329
0.9 13.4600 10.3831 3.6745 1.6077 0.2725
1.0 15.2923 12.1706 4.6550 2.0685 0.2231
Figure 3: 2D solution depict example 4.2.
Figure 4: 3D solution example 4.2,
α
=2.
Table 4: Solutions of example 4.2 for x=0.5.
t
α
=1.1
α
=1.2
α
=1.5
α
=1.7
α
=2
0.0 0.5000 0.5000 0.5000 0.5000 0.5000
0.1 0.5808 0.5788 0.5779 0.5751 0.5526
0.2 0.6685 0.6616 0.6567 0.6521 0.6107
0.3 0.7597 0.7470 0.7350 0.7285 0.6749
0.4 0.8535 0.8348 0.8125 0.8038 0.7459
0.5 0.9494 0.9246 0.8896 0.8778 0.8244
0.6 1.0472 1.0163 0.9664 0.9504 0.9111
0.7 1.1465 1.1099 1.0432 1.0217 1.0069
0.8 1.2472 1.2052 1.1204 1.0918 1.1128
0.9 1.3493 1.3023 1.1981 1.1610 1.2298
1.0 1.4525 1.4010 1.2766 1.2294 1.3591
APPENDIX
WSEAS TRANSACTIONS on SYSTEMS and CONTROL
DOI: 10.37394/23203.2024.19.9
Nguyen Minh Tuan, Sanoe Koonprasert, Phayung Meesad
E-ISSN: 2224-2856
96
Volume 19, 2024
Figure 5: 2D solution depict example 4.3.
Figure 6: 3D solution example 4.3,
α
=2.
Table 5: Solutions of example 4.3 for x=0.5,y=0.5.
t
α
=1.1
α
=1.2
α
=1.5
α
=1.7
α
=2
0.0 0.2500 0.2500 0.2500 0.2500 0.2500
0.1 0.1826 0.1805 0.1771 0.1760 0.2763
0.2 0.1303 0.1235 0.1107 0.1063 0.3054
0.3 0.0912 0.0774 0.0514 0.0416 0.3375
0.4 0.0634 0.0399 -0.0033 -0.0199 0.3730
0.5 0.0461 0.0081 -0.0582 -0.0829 0.4122
0.6 0.0396 -0.0201 -0.1198 -0.1551 0.4555
0.7 0.0459 -0.0465 -0.1965 -0.2476 0.5034
0.8 0.0684 -0.0715 -0.2983 -0.3747 0.5564
0.9 0.1125 -0.0944 -0.4363 -0.5542 0.6149
1.0 0.1854 -0.1130 -0.6228 -0.8074 0.6796
Figure 7: 2D solution depict example 4.4.
Figure 8: 3D solution example 4.4,
α
=1.
Table 6: Solutions of example 4.4 for t=0.5.
x
α
=1.1
α
=1.2
α
=1.5
α
=1.7
α
=2
0.0 0.6065 0.6065 0.6065 0.6065 0.6065
0.1 0.7098 0.6986 0.6796 0.6737 0.6703
0.2 0.8129 0.7932 0.7573 0.7449 0.7408
0.3 0.9125 0.8845 0.8327 0.8144 0.8187
0.4 1.0087 0.9715 0.9026 0.8785 0.9048
0.5 1.1022 1.0540 0.9649 0.9344 1.0000
0.6 1.1938 1.1323 1.0181 0.9795 1.1052
0.7 1.2848 1.2072 1.0611 1.0118 1.2214
0.8 1.3765 1.2795 1.0934 1.0297 1.3499
0.9 1.4703 1.3505 1.1147 1.0317 1.4918
1.0 1.5679 1.4214 1.1250 1.0166 1.6487
WSEAS TRANSACTIONS on SYSTEMS and CONTROL
DOI: 10.37394/23203.2024.19.9
Nguyen Minh Tuan, Sanoe Koonprasert, Phayung Meesad
E-ISSN: 2224-2856
97
Volume 19, 2024
