Type of the Paper (Article


Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 4s (2025) 

 

309 https://internationalpubls.com 

Time-Fractional Hyperbolic Telegraph Equation: A Semi-Analytic Ap-

proach Using Modified Adomian Decomposition Elzaki Transform 

Method 

 

Parmeshwari Aland 1,2, Prince Singh3 

1 Research Scholar, Department of Mathematics, Lovely Professional University, Phagwara, Punjab-144411, India;  

parmeshwarialand20@gmail.com 

2 Assistant Professor, School of Engineering, Ajeenkya DY Patil University, Lohegaon, Pune-412105, India;  

facultyit415@adypu.edu.in 

3 Department of Mathematics, School of Chemical Engineering and Physical Science., Lovely Professional University, 

Phagwara, Punjab-144411, India; princesingh16092@gmail.com 

 

Article History: 

Received: 25-09-2024 

Revised: 27-11-2024 

Accepted: 07-12-2024 

Abstract:  

In this research paper, an approximate analytical solution approach known as the Modified 

Adomian Decomposition Method with the coupling of Elzaki Transform (MADETM) is de-

ployed for addressing one-dimensional, two- dimensional, and three-dimensional time-frac-

tional hyperbolic telegraph equations. The Caputo derivative operator yields the approximate 

analytical solution. The impact ness and its accuracy of the adopted method are demonstrated 

through comparison of the approximate results with the exact solutions, both presented graph-

ically by plotting its surface graph, line graph through analyzing its error. The MADETM 

proves to be a reliable and efficient tool for deriving approximate and exact solutions for a 

large class of partial differential equations (PDEs), fractional PDEs, and ordinary differential 

equations (ODEs). The considered method yields a solution in series form with low compu-

tational complexity and swiftly converges towards precise solutions. The outcomes showcase 

an effective and uncomplicated approach for examining issues across diverse scientific and 

technological domains. 

Keywords: Modified Adomian Decomposition technique, hyperbolic time fractional tele-

graph equations, Elzaki transform operator 

 

 

 



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1. Introduction 

Fractional order differential equations (FDEs) indeed have gained significance in applied mathematics, 

being applied in various systems within applied science. These equations provide a substitute approach 

to non-linear equations and have proven essential in mathematical modelling in fields such as mechan-

ics, process control, complex systems, and technology. Integral equations work as crucial role in effi-

ciently elaborate mathematical problems associated with FDEs and PDEs.  

By selecting appropriate integral transformations, one can convert FDEs and PDEs into algebraic equa-

tions, simplifying the problem-solving process. Integral transforms offer a convenient method to ad-

dress the complexity of various types of differential equations. The development and implementation 

of integral transforms, like Laplace transform, Elzaki transformation, Elzaki-Laplace transform, Shehu 

transform, and Sawi transform, Natural transforms have been instrumental in advancing research in 

this area [[1],[2],[3],[4],[5],[9]] 

Moreover, research efforts over the past few decades have extensively explored the application of 

integral transformations to solve fractional and FDEs. These studies have involved various operators 

like Caputo, Atanga Baleanu, Erdelyi-Kober, He- polynomials and Grunwald-Letnikov operation, Rie-

mann-Liouville types are leading to applications in diverse fields beyond mathematics [[11],[12],[13]]. 

Multiple integral equations, ODEs, PDEs, and fractional PDEs are solved using these transformations 

that are offered in the literature. 

A range of analytical and numeration techniques will be utilized to solve hyperbolic time. fractional 

telegraph equations. These methods encompass the Homotopy Perturbation Transform [[14]] Tech-

nique, Sinc-Collocation Technique [[15]], Adomian Decomposition Technique [[16]], q-Homotopy 

Analysis Transform Technique [[17]], Reduction Differential Transform Technique [[18]], Reproduc-

ing Kernel Method [[19]] Variational Iteration Method [[20]], and Haar Wavelet technique [[21]] 

The communication process is essential in the modern global community. Due to the extensive usage 

of radio frequency systems and microwave communication, technologies continue to get substantial 

industrial attention. Importantly, all transmission media experience the signal deficit, which must be 

measured for each medium. Telegraph equations are employed to analyses electrical signal propaga-

tion, random walks, wave propagation, and transmission line cables and similar phenomena. Heaviside 

introduced the concept of the transmission line, which will be divided into two types: guided and un-

guided. In guided media, signals are transmitted through physical systems such as copper wires, which 

carry voltage waves and higher frequency currents. Conversely, the unguided media use magnetic 

fields to transmit signals across communication channels, employing microwave communication and 

radio frequency systems, with antennas facilitating the broadcasting and reception of these electro-

magnetic waves. To increase the efficiency of telegraph transmission, cable transmission media are 



Communications on Applied Nonlinear Analysis 

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researched in controlled transmission environments. Direct information propagation between two or 

more sites is represented by a physical system using a directed transmission medium. Power and signal 

losses must be predicted or calculated because they are a necessary part of any system to enhance 

controlled communication [[22]].  

In the last few years, fractional order partial differentiation equations (PDEs) gained significant atten-

tion due to their extensive application in a variety of technical and scientific domains, from scientists 

and researchers. The fractional derivative within these models offers a high degree of flexibility, of-

fering a great tool for characterizing the inherited traits of various prototypes and their varying histo-

ries. Extensive research has been conducted to develop analytical, semi-analytical and numerical so-

lutions for solving both nonlinear and linear fractional PDEs [[23]]. 

The time-fractional telegraph equations are deployed in this article with the use of MADETM. Hyper-

bolic time fractional telegraph equations are deployed in this study through demonstration of the Mod-

ified Adomian Decomposition with coupling Elzaki Transformation Method (MADETM). Certain 

fractional-order telegraph equation models are used to determine the MADETM solutions. The method 

demonstrates increased precision and efficiency, as evidenced by graphical comparisons with the exact 

solution. The EDM solutions for fractional-order telegraph equations exhibit a high rate of conver-

gence. Consequently, this technique is promising for simplifying other fractional forms linearly and 

nonlinearly partial differentiation equations. 

• One-Dimensional fractional order telegraph equation: 

𝐷𝑡
2𝛼 [∅(𝑥 , 𝑡)] + 2𝛼𝐷𝑡

𝛼  [∅(𝑥 , 𝑡)] + 𝛽2  = ∅𝑥𝑥(𝑥 , 𝑡) + 𝑔(𝑥 , 𝑡)           0 <∝  ,              𝑥 ≤ 1 

                          ∅(𝑥 ,0) = 𝑓1(𝑥),       ∅𝑡(𝑥 ,0) = 𝑓2(𝑥) 

                               ∅(0, 𝑡) = 𝑓1(𝑡),        ∅𝑥(𝑥, 𝑡) = 𝑓2(𝑡) 

 where 𝛽 → arbitrary constants and ∅(𝑥, 𝑡) is an unknown function. 

• Two-Dimensional fractional order telegraph equation: 

𝐷𝑡
2𝛼 [∅(𝑥 , 𝑦 , 𝑡)] + 2𝛼𝐷𝑡

𝛼 [∅(𝑥 , 𝑦 , 𝑡)] + 𝛽2 [∅(𝑥 , 𝑦 , 𝑡)]

= ∅𝑥𝑥(𝑥 , 𝑦 , 𝑡) + ∅𝑦𝑦(𝑥 , 𝑦 , 𝑡) + 𝑔(𝑥 , 𝑦 , 𝑡)          

                                    0 <∝≤ 1 ,               𝑥 = 1 

With initial conditions: ∅(𝑥, 𝑦, 0) = 𝑔1(𝑥, 𝑦),        ∅𝑡(𝑥, 𝑦, 0) = 𝑔2(𝑥, 𝑦) 

 

 

 

With initial conditions:



Communications on Applied Nonlinear Analysis 

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•  

𝐷𝑡
2𝛼  [∅(𝑥. , 𝑦 , 𝑧 , 𝑡)] + 2𝛼𝐷𝑡

𝛼  [∅(𝑥 , 𝑦 , 𝑧 , 𝑡)] + 𝛽2  [∅(𝑥. , 𝑦 , 𝑧 , 𝑡)] = ∅𝑥𝑥(𝑥 , 𝑦 , 𝑧 , 𝑡) +

          ∅𝑦𝑦(𝑥 , 𝑦 , 𝑧 , 𝑡) + ∅𝑧𝑧(𝑥 , 𝑦 , 𝑧 , 𝑡) + 𝑔(𝑥 , 𝑦 , 𝑧 , 𝑡)                                              

        0 <∝≤ 1 ,                 𝑥 = 1 

With initial conditions: ∅(𝑥. , 𝑦 , 𝑧 , 𝑡) = ℎ1(𝑥 , 𝑦 , 𝑧 ),        ∅𝑡(𝑥. , 𝑦 , 𝑧 , 𝑡) = ℎ2(𝑥 , 𝑦 , 𝑧 ) 

The hyperbolic telegraph equation is broadly applied in the signal processing for transmitting wave 

theory and electric impulses. It has found various applications in biomedical sciences and aerospace. 

Researchers are particularly interested in solving problems involving fractional derivatives. Fractional-

order partial differential equations (PDEs) are essentially a type of integer-order PDEs. Fractional-

order methods yield results that converge to those of integer-order methods. 

 2. Preliminaries and Notations  

Elzaki transformations, also called Elzaki integral transformations, are algebraic equation transfor-

mations that are used to solve differential equations in ordinary form (ODEs). Elzaki Ali Elzaki, a 

mathematician from Sudan, introduced it in the 1960s. Heat conduction, fluid dynamical mechanics, 

and electrical circuits are only a few of the applied scientific and engineering domains where the Elzaki 

transformation has been effectively used. It offers an alternate strategy for resolving ODEs, especially 

in situations when existing methods are not easily able to produce analytical solutions. When analytical 

solutions are hard to come by with other approaches, applying the Elzaki transform to solve FPDEs 

can be quite helpful. This section presents a basic explanation of Elzaki transformation and offers a 

framework for transforming the problem into an algebraic form that can be solved using well-estab-

lished algebraic techniques. 

Definition 1: Fundamental Principle of Elzaki Transformation  

The exponential form of function in A series, defined by set A expressed as new transformation which 

renamed as an Elzaki Transformation [[24]] 

𝐴 = {𝑓(𝑡): ∃ 𝑀 , 𝑘1, 𝑘2 > 0, |𝑓(𝑡)| < 𝑀𝑒
|𝑡|

𝑘𝑗 , 𝑖𝑓 𝑡 ∈ (−1)𝑗 𝑋 [ 0,∞)}                              

Regarding the specified function in the set, M is defined as finite or infinite number, as  𝑘1 𝑎𝑛𝑑 𝑘2 . 

The Elzaki Transform defined as operator  𝐸(g(𝜏)) in the integral form as follows.  

 𝐸[ 𝑓(𝜏)] = 𝑣 ∫ 𝑓(𝜏)−  
𝜏

𝑣  𝑑𝜏 = 𝑇 (𝑣)    , 𝜏 ≥ 0 ,   
∞

0
𝑘1 ≤ 𝑣 ≤  𝑘2        

                                                                          

Three-dimensional fractional order telegraph equation:



Communications on Applied Nonlinear Analysis 

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The Elzaki Transformation for some functions is defined below [[24]]. 

 𝒇(𝒕) 𝑬[𝒇(𝒕)] = 𝑻(𝑣) 

1 𝟏 𝑣𝟐 

2 𝒕 𝑣𝟑 

3 𝒕𝒏 𝒏! 𝑣𝒏+𝟐 

The following conclusion was established based on the description and fundamental analyses.                         

 𝐸[𝑡𝑛] = 𝑛! 𝑣𝑛+2 

 𝐸[𝑓 ′(𝑡)] =
𝐹 (𝑣)

𝑣
− 𝑣𝑓(0)  

 𝐸[𝑓 ′′(𝑡)] =
𝐹(𝜸)

𝑣2
− 𝑓(0) − 𝑣𝑓 ′(0) 

 𝐸[𝑓(𝑛)(𝑡)] =
𝐹(𝑣)

𝑣𝑛
− ∑ 𝑣2−𝑛+𝑘𝑓(𝑘)(0) 𝑛−1

𝑘=0  

Definition 2: Caputo Fractional Elzaki Transform Operator 

The Caputo Fractional operator's Elzaki Transformation is as follows:  

𝐸 [
𝜕𝛼

𝜕𝜏𝛼
 𝑓(𝜏)] =  

𝐸 [𝑓 (𝜏)]

𝑣𝛼 
 − ∑ 𝑣𝑘−𝛼+2 𝑛−1

𝑘=0 𝑓(𝑘)(0),    𝑛 − 1 < 𝛼 ≤ 𝑛    

3. The methodology for Modified Adomian Decomposition Elzaki Transform (MADETM): 

Consider the partial differentiation equation of fractional order non-linearity. 

𝐷𝑡
𝛼  ∅(𝑥, 𝑡) + 𝑅 [∅(𝑥, 𝑡)] + 𝑁[∅(𝑥, 𝑡)] = 𝑔(𝑥, 𝑡)         𝑥, 𝑡 ≥ 0          𝑚 − 1 < 𝛼 < 𝑚        

With initial condition ∅(𝑥, 0) = 𝑓(𝑥)     

the Caputo fractional function ∅(𝑥, 𝑡) defined as: 

𝐷𝑡
𝛼  ∅(𝑥, 𝑡) =

𝜕𝛼∅(𝑥, 𝑡)

𝜕𝑡𝛼
=

{
 
 

 
 1

⌈(𝑛 − 𝛼)
∫(𝑡−𝑥)𝑛−𝛼−1

𝜕𝑛∅(𝑥, 𝑡)

𝜕𝑡𝑛
  𝑑𝑡   , 𝑛 − 1 < 𝛼 < 𝑛    

𝑡

𝑎

𝜕𝑛∅(𝑥, 𝑡)

𝜕𝑡𝑛
                                                             𝛼 = 𝑛 ∈ 𝑁

 

Where 𝐷𝑡
𝛼 ∅(𝑥, 𝑡) is Caputo fractional order derivative 𝛼 , N and are R nonlinear and linear terms 

respectively, and 𝑔 is source term.   

Taking the Elzaki Transform on both sides of Equation (2)  

E[ 𝐷𝑡
𝛼  ∅( 𝑥. , 𝑡)] + 𝐸 [𝑅 [∅(𝑥. , 𝑡)] + 𝑁[∅(𝑥 , 𝑡)] ] = 𝐸[𝑔(𝑥. , 𝑡)]    

1

𝑣𝛼
 𝐸[∅( 𝑥 , 𝑡)] − 𝑣2−𝛼  ∅(𝑥 ,0) = 𝐸[𝑔(𝑥 , 𝑡)] −  𝐸 [𝑅 [∅(𝑥 , 𝑡)] + 𝑁[∅(𝑥 , 𝑡)]] 

𝐸 [∅ (𝑥 , 𝑡)] = 𝑣2  ∅(𝑥, 0) + 𝑣𝛼  𝐸[𝑔(𝑥 , 𝑡)] − 𝑣𝛼 𝐸 [𝑅 [∅(𝑥 , 𝑡)] + 𝑁[∅(𝑥 , 𝑡)]]      

From equation (3) its Initial Conditions is: ∅(𝑥, 0) = 𝑓 (𝑥) 

𝐸  [∅ (𝑥 , 𝑡)] = 𝑣2  𝑓(𝑥) + 𝑣𝛼  𝐸[𝑔(𝑥 , 𝑡)] − 𝑣𝛼 𝐸 [𝑅 [∅(𝑥 , 𝑡)] + 𝑁[∅(𝑥 , 𝑡)]] 

(2)

(5)

(1)

(4)

(3)



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Applying inverse Elzaki Transform on Equation (5)  

𝐸−1[𝐸  [∅ (𝑥 , 𝑡)]] = 𝐸−1 [𝑣2  𝑓(𝑥) + 𝑣𝛼  𝐸[𝑔(𝑥 , 𝑡)] − 𝑣𝛼𝐸 [𝑅 [∅(𝑥 , 𝑡)] + 𝑁[∅(𝑥 , 𝑡)]]]             

∅ (𝑥 , 𝑡) = 𝐸−1 [𝑣2  𝑓(𝑥)] + 𝐸−1[𝑣𝛼 𝐸 [𝑔(𝑥 , 𝑡)]] − 𝐸−1 [𝑣𝛼𝐸 [𝑅 [∅(𝑥, 𝑡)] + 𝑁[∅(𝑥, 𝑡)]]]    

By applying MADM on right hand side of equation occurs the solution in infinite series given below. 

∅(𝑥, 𝑡) = ∑ ∅𝑛(𝑥, 𝑡)
∞
𝑛=0                    

The non-linear terms N in an Adomian polynomial represented as follows. 

𝑁 [∅(𝑥, 𝑡)] = ∑ 𝐴𝑛
∞
𝑛=0    

Where 𝐴𝑛 =
1

𝑛!
 [
𝑑𝑛

𝑑𝜆𝑛
 [𝑁 ∑ 𝜆𝑖∅𝑖

∞
𝑖=0 ]]

𝜆=0

  ; 

 𝑖 = 0,1,2, 3, …….                                                    

The nonlinear terms denoted by 𝑁  are explained with adapted modified Adomian decomposi-

tion technique for  

handling nonlinear polynomial system solution. following the utilization of the Elzaki transformation 

as specified below: 

{𝐴𝑛} = { 𝑁1( 𝑠𝑛) −  𝑁1( 𝑠𝑛−1)}      

Equation (6) is obtained by substituting Equations (7) and (8)  

∑ ∅𝑛(𝑥, 𝑡)
∞
𝑛=0 = 𝐸−1[𝑣2  𝑓(𝑥)] + 𝐸−1[𝑣𝛼 𝐸 [𝑔(𝑥, 𝑡)]] − 𝐸−1 [𝑣𝛼𝐸 [𝑅[∑ ∅𝑛(𝑥, 𝑡)

∞
𝑛=0 ] + [∑ 𝐴𝑛

∞
𝑛=0 ]]]    

Since, 𝐸−1(𝑣2  ) = 1 

∑∅𝑛(𝑥, 𝑡)

∞

𝑛=0

= 𝑓(𝑥) + 𝐸−1[𝑣𝛼 𝐸[𝑔(𝑥, 𝑡)]] − 𝐸−1 [𝑣𝛼𝐸 [𝑅 [∑∅𝑛(𝑥, 𝑡)

∞

𝑛=0

] + [∑𝐴𝑛

∞

𝑛=0

]]] 

Analysing both sides of the Equation (9) 

∅0(𝑥, 𝑡) = 𝑓(𝑥) + 𝐸
−1[𝑣𝛼 𝐸[𝑔(𝑥 , 𝑡)]] 

       ∅1(𝑥, 𝑡) = −𝐸−1[𝑣𝛼𝐸 [𝑅[∅0(𝑥, 𝑡)] + 𝐴0]] 

       ∅2(𝑥, 𝑡) = −𝐸−1[𝑣𝛼𝐸 [𝑅[∅1(𝑥, 𝑡)] + 𝐴1]] 

. 

: 

: 

 ∅𝑛+1(𝑥, 𝑡) = −𝐸−1[𝑣𝛼𝐸 [𝑅[∅n(𝑥, 𝑡)] + 𝐴n]]       

 

(6)

(8)

(10)

(9)

(7)



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The analytic solution ∅(𝑥, 𝑡) is finally approximated using truncated series: 

The following different forms of nonlinear one-dimensional, two-dimensional and three-dimensional 

fraction form telegraph equations are demonstrated with adopted technique for validation of results in 

the following applications.   

4. Application  

One-dimensional non-linear telegraph equation: 

Example 1. consider one-dimensional nonlinear telegraph equation [330]: 

𝐷𝑡
𝛼∅(𝑥, 𝑡) = ∅𝑥𝑥(𝑥 , 𝑡) + ∅𝑡(𝑥 , 𝑡) − ∅

2 + 𝑥∅∅𝑥(𝑥 , 𝑡)where   0 <∝≤ 2  (12) 

    Initial conditions: ∅(𝑥 ,0) =  𝑥,       ∅𝑡(𝑥, 0) = 𝑥          

   ∅0(𝑥 , 𝑡) = ∅(𝑥 ,0) + 𝑡 ∅𝑡(𝑥, 0) = 𝑥 + 𝑥𝑡 = 𝑥(1+ 𝑡) 

Apply the Elzaki transformation on Equation (12), 

𝐸 [∅ (𝑥 , 𝑡)] = 𝑣2  ∅(𝑥, 0) + 𝑣𝛼 [𝐸[∅𝑥𝑥 + ∅𝑡 − ∅
2 + 𝑥 ∅∅𝑥]] 

Applying inverse Elzaki Transform on above equation 

∅ (𝑥, 𝑡) = 𝐸−1[𝑣2  ∅(𝑥, 0)] + 𝐸−1 [𝑣𝛼  𝐸 [∅𝑥𝑥 + ∅𝑡 − ∅
2 + 𝑥 ∅∅𝑥]] 

∅ (𝑥, 𝑡) = ∅(𝑥, 0) + 𝐸−1[𝑣𝛼 𝐸 [𝑅[∅] + 𝑁[∅]]]       

Here, 𝐸−1(𝑣2  ) = 1 ; 𝑅[∅] = (∅𝑥𝑥 + ∅𝑡)  and 𝑁[∅] = (𝑥 ∅∅𝑥 − ∅
2) 

Appling the MADETM process on equation (14) 

              ∅0(𝑥, 𝑡) = ∅(𝑥, 0) = 𝑥(1+ 𝑡)     

Appling the recursive series as shown in equation (10), 

             ∅𝑛+1(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑅( ∅𝑛) + 𝑁( ∅𝑛)]] 

For 𝑛 = 0  

             ∅1(𝑥, 𝑡) = 𝐸
−1[𝑣𝛼𝐸 [𝑅( ∅0) + 𝑁[ ∅0]]]           

Here,    𝑅 ( ∅0) = ∅0𝑥𝑥 + ∅0𝑡 = 𝑥  

         𝑁 ( ∅0) = 𝑥 ∅0∅0𝑥 − ∅0
2 = 0 

Therefore, above equation implies,  ∅1(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑅( ∅0) + 𝑁[ ∅0]]] 

                        
 
∅1
(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑥 + 0]] = 4𝑒𝑥𝐸−1[𝑣𝛼𝐸 [𝑥]] = 𝑥𝐸−1[𝑣𝛼𝐸(1)] 

 
∅1
(𝑥, 𝑡) = 𝑥 𝐸−1(𝑣𝛼+2) = 𝑥 

   𝑡𝛼

⌈(𝛼 + 1)
 

 

(13)

(14)

(15)

(11)

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 ∅1(𝑥, 𝑡) = 𝑥 
   𝑡𝛼

⌈(𝛼+1)
   

For 𝑛 = 2 

           ∅2(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑅( ∅1) + 𝑁[ ∅1]]] 

Here, 𝑅 ( ∅1) = ∅1𝑥𝑥 + ∅1𝑡 = 𝑥 
𝛼  

⌈(𝛼+1)
  𝑡𝛼−1 

         𝑁 ( ∅1) = 𝑥 ∅0∅0𝑥 − ∅0
2 = 0 

 ∅2(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑅( ∅1) + 𝑁[ ∅1]]] = 𝐸
−1 [𝑣𝛼𝐸 [𝑥 

𝛼  

⌈(𝛼 + 1)
  𝑡𝛼−1]] 

 ∅2(𝑥, 𝑡) =  𝑥 
𝛼 ⌈𝛼 

⌈(𝛼 + 1)

   𝑡2𝛼−1

⌈(2𝛼)
 

Considering 𝑛 = 3, 4… .. 

 ∅3(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑅( ∅2) + 𝑁[ ∅2]]] =  𝑥 
𝛼 ⌈𝛼 

⌈(𝛼+1)

(2𝛼−1) ⌈(2𝛼−1) 

⌈(2𝛼)

   𝑡3𝛼−1

⌈(3𝛼−1)
    

 ∅4(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑅( ∅3) + 𝑁[ ∅3]]] = 𝑥 
𝛼 ⌈𝛼 

⌈(𝛼+1)

(2𝛼−1) ⌈(2𝛼−1) 

⌈(2𝛼)

(3𝛼−1) ⌈(3𝛼−1) 

⌈(3𝛼)

   𝑡4𝛼−1

⌈(4𝛼−1)
      

Therefore, Series representation of the solution ∅ (𝑥, 𝑡) is as follows: 

∅ (𝑥, 𝑡) =   ∅0( 𝑥, 𝑡)+∅1( 𝑥, 𝑡)+∅2( 𝑥, 𝑡) +  ∅3( 𝑥, 𝑡) +  ∅4( 𝑥, 𝑡) + ⋯…… 

∅ (𝑥 , 𝑡) = 𝑥(1+ 𝑡) + 𝑥 
   𝑡𝛼

⌈(𝛼+1)
+  𝑥 

𝛼 ⌈𝛼 

⌈(𝛼+1)

   𝑡2𝛼−1

⌈(2𝛼)
+ 𝑥 

𝛼 ⌈𝛼 

⌈(𝛼+1)

(2𝛼−1) ⌈(2𝛼−1) 

⌈(2𝛼)

   𝑡3𝛼−1

⌈(3𝛼−1)
  +

𝑥 
𝛼 ⌈𝛼 

⌈(𝛼+1)

(2𝛼−1) ⌈(2𝛼−1) 

⌈(2𝛼)

(3𝛼−1) ⌈(3𝛼−1) 

⌈(3𝛼)

   𝑡4𝛼−1

⌈(4𝛼−1)
+⋯…….    

∅ (𝑥, 𝑡) = 𝑥 [(1+ 𝑡) +
   𝑡𝛼

⌈(𝛼+1)
+

𝛼 ⌈𝛼 

⌈(𝛼+1)

   𝑡2𝛼−1

⌈(2𝛼)
+

𝛼 ⌈𝛼 

⌈(𝛼+1)

(2𝛼−1) ⌈(2𝛼−1) 

⌈(2𝛼)

   𝑡3𝛼−1

⌈(3𝛼−1)
+

𝛼 ⌈𝛼 

⌈(𝛼+1)

(2𝛼−1) ⌈(2𝛼−1) 

⌈(2𝛼)

(3𝛼−1) ⌈(3𝛼−1) 

⌈(3𝛼)

   𝑡4𝛼−1

⌈(4𝛼−1)
……… . ]   

In particular when  𝛼 = 2 , the solution is of the form: 

∅ (𝑥, 𝑡) = 𝑥 [1+
 𝑡

1!
+ 

t2

2!
+ 

t3

3!
+ 

t4

4!
+ 

t5

5!
… . . ]      

The exact solution for equation (12) is:  ∅ (𝑥, 𝑡) = 𝑥𝑒𝑡         

Example 2. Consider the following one-dimensional nonlinear telegraph equation [25] 

𝐷𝑡
𝛼∅(𝑥, 𝑡) = ∅𝑥(∅

2(𝑥, 𝑡). ∅𝑥(𝑥 , 𝑡))               (19) 

Initial condition  ∅(𝑥 ,0) =
𝑥+𝑏

2𝑐
 ; where 𝑐 > 0, 𝑎𝑛𝑑  𝑏 is arbitrary constant.    (20) 

Apply the Elzaki transformation on Equation (19) 

                   𝐸[ 𝐷𝑡
𝛼∅(𝑥, 𝑡)]  = 𝐸[2∅(𝑥, 𝑡) ∅𝑥

2 (𝑥, 𝑡) + ∅2(𝑥, 𝑡)∅𝑥𝑥(𝑥 , 𝑡)]   

(16)

(18)

(17)

(19)

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                   𝐸 [∅ (𝑥 , 𝑡)] = 𝑣2  ∅(𝑥, 0) + 𝑣𝛼 [𝐸[2∅.  ∅𝑥
2 + ∅2. ∅𝑥𝑥]]         

Applying inverse Elzaki Transform on equation (21)  

∅ (𝑥, 𝑡) = 𝐸−1[𝑣2  ∅(𝑥, 0)] + 𝐸−1 [𝑣𝛼  𝐸 [2∅. ∅𝑥
2 + ∅2. ∅𝑥𝑥]] 

∅ (𝑥, 𝑡) = ∅(𝑥, 0) + 𝐸−1[𝑣𝛼 𝐸 [𝑁1(∅) + 𝑁2(∅)]]          

Here, 𝐸−1(𝑣2  ) = 1 ; 𝑁1(∅) = (2∅. ∅𝑥
2 )  and 𝑁2(∅) = (∅2. ∅𝑥𝑥)     

Applying the MADETM process on equation (22) 

              ∅0(𝑥, 𝑡) = ∅(𝑥, 0) =
𝑥+𝑏

2𝑐
            (23) 

Applying the recursive series as shown in equation (10), 

             ∅𝑛+1(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑁1( ∅𝑛) + 𝑁2 (∅𝑛)]] 

For 𝑛 = 0  

 ∅1(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑁1( ∅0) + 𝑁2( ∅0)]]  

 
∅1
(𝑥, 𝑡) = 𝐸−1 [𝑣𝛼𝐸 [2∅0. ∅0𝑥

2 + ∅0
2. ∅0𝑥𝑥]] 

 
∅1
(𝑥, 𝑡) =

𝑥 + 𝑏

4𝑐3
 𝐸−1(𝑣𝛼+2) =

𝑥 + 𝑏

4𝑐3
 
   𝑡𝛼

⌈(𝛼 + 1)
 

 ∅1(𝑥, 𝑡) =
𝑥+𝑏

4𝑐3
 
   𝑡𝛼

⌈(𝛼+1)
    

For 𝑛 = 2 

    ∅2(𝑥, 𝑡) = 𝐸
−1[𝑣𝛼𝐸 [𝑁1( ∅1) + 𝑁2( ∅1)]] 

 ∅2(𝑥, 𝑡) = 𝐸−1 [𝑣𝛼𝐸 [2∅1. ∅1𝑥
2 + ∅1

2. ∅1𝑥𝑥]] = 𝐸−1 [𝑣𝛼𝐸 [
(𝑥 + 𝑏)

4𝑐5
 

3 𝑡𝛼   

2⌈(𝛼 + 1)
 ]] 

 ∅2(𝑥, 𝑡) =
3(𝑥 + 𝑏)

8𝑐5
 
 𝑡2𝛼   

⌈(2𝛼 + 1)
  

Considering 𝑛 = 3, 4… .. 

 ∅3(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝑁1( ∅2) + 𝑁2[ ∅2]]] =  
4 (𝑥+𝑏)

16 𝑐5
 
 𝑡3𝛼  

⌈(3𝛼+1)
          

. 

: 

Therefore, Series representation of the solution ∅ (𝑥, 𝑡) is as follows: 

∅ (𝑥, 𝑡) =   ∅0( 𝑥, 𝑡)+∅1( 𝑥, 𝑡)+∅2( 𝑥, 𝑡) +  ∅3( 𝑥, 𝑡) +  ∅4( 𝑥, 𝑡) + ⋯…… 

 

(21)

(24)

(22)

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Communications on Applied Nonlinear Analysis 

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∅ (𝑥 , 𝑡) =
𝑥+𝑏

2𝑐
+
(𝑥+𝑏)

4𝑐3
 
   𝑡𝛼

⌈(𝛼+1)
+

3(𝑥+𝑏)

8𝑐5
 
 𝑡2𝛼  

⌈(2𝛼+1)
+

4 (𝑥+𝑏)

16 𝑐5
 
 𝑡3𝛼  

⌈(3𝛼+1)
+⋯…….    

In particular when  𝛼 = 1 , the solution is of the form: 

∅ (𝑥, 𝑡) = [
𝑥+𝑏

2𝑐
+
(𝑥+𝑏) 𝑡

4𝑐3
 +

3(𝑥+𝑏) 𝑡2

16 𝑐5
 +

4 (𝑥+𝑏)  𝑡3

64 𝑐7
 + ⋯… ]         

The exact solution for equation (19) is:  ∅ (𝑥, 𝑡) =
𝑥+𝑏

2√𝑐2− 𝑡 
, 𝑡 < 𝑐2       

Example 3. Illustrate the following fractional-order one dimensional telegraph equation [23] 

       𝐷𝑡
2𝛼∅(𝑥, 𝑡) + 2 𝐷𝑡

𝛼∅(𝑥, 𝑡) + ∅(𝑥, 𝑡) = ∅𝑥𝑥(𝑥, 𝑡)   0 <∝≤ 1 , 𝑥 = 1         

With initial conditions: ∅(𝑥, 0) =  𝑒𝑥,       ∅𝑡(𝑥, 0) = −2𝑒𝑥           

Using the Elzaki transformation of Equation (27), 

                    𝐸[ 𝐷𝑡
2𝛼∅ + 2 𝐷𝑡

𝛼∅ + ∅]  = 𝐸[∅𝑥𝑥]   

Applying Elzaki Transform on above equation we get, 

           
1

𝑣𝛼
 𝐸[∅(𝑥, 𝑡)] − 𝑣2−𝛼  ∅(𝑥, 0) − 𝑣3−𝛼  ∅𝑡(𝑥, 0) = −𝐸[(∅ − ∅𝑥𝑥) − 2 𝐷𝑡

𝛼∅] 

                 𝐸[∅(𝑥, 𝑡)] = 𝑣2  ∅(𝑥, 0) + 𝑣3  ∅𝑡(𝑥, 0) − 𝑣
𝛼  𝐸[(∅ − ∅𝑥𝑥) − 2 𝐷𝑡

𝛼∅] 

Applying inverse Elzaki Transform on above equation  

              𝐸−1[𝐸 [∅(𝑥, 𝑡)]] = 𝐸−1 [𝑣2  ∅(𝑥, 0) + 𝑣3  ∅𝑡(𝑥, 0) − 𝑣
𝛼  𝐸[(∅ − ∅𝑥𝑥) − 2 𝐷𝑡

𝛼∅]] 

              ∅(𝑥, 𝑡) = 𝐸−1 [𝑣2  ∅(𝑥, 0) + 𝑣3  ∅𝑡(𝑥, 0) − 𝑣
𝛼  𝐸[𝐿(∅) − 2 𝐷𝑡

𝛼∅]] 

Appling the MADETM process on above equation 

              ∅0(𝑥, 𝑡) = ∅(0) = 𝑒
𝑥(1− 2𝑡)            (29) 

Using the recursive series as shown in equation (10), 

             ∅𝑛+1(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝐿( ∅𝑛) − 2 𝐷𝑡
𝛼  ∅𝑛]]         

For 𝑛 = 0  

             ∅1(𝑥, 𝑡) = 𝐸
−1[𝑣𝛼𝐸 [𝐿( ∅0) − 2 𝐷𝑡

𝛼  ∅0]]     

Here,    𝐿 [∅0] = ∅0𝑥𝑥 − ∅0 = 0  

Therefore, above equation implies,  ∅1(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [−2 𝐷𝑡
𝛼∅0]] 

Consider,  𝐸 [−2 𝐷𝑡
𝛼∅0] = −2 [

1

𝑣𝛼
 𝐸[∅0] − 𝑣

2−𝛼  ∅0(0)] 

       𝐸 [−2 𝐷𝑡
𝛼∅0] = −2 [[

1

𝑣𝛼
]  𝐸[𝑒𝑥(1− 2𝑡)] − 𝑣2  𝑒𝑥]       

      𝐸 [−2 𝐷𝑡
𝛼∅0] = −2𝑒𝑥 [[

1

𝑣𝛼
] [𝐸(1) − 2𝐸(𝑡)] − 𝑣2  ] = −2𝑒𝑥 [[

1

𝑣𝛼
] [𝑣2  − 2𝑣3  ] − 𝑣2  ] 

(27)

(26)

(28)

(25)

(30)

(29)

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    𝐸 [−2 𝐷𝑡
𝛼∅0] = −2𝑒𝑥 [[

1

𝑣𝛼
] [−2𝑣3  ]] = 4𝑒𝑥[ 𝑣3−𝛼  ] 

Therefore,  ∅1(𝑥, 𝑡) = 𝐸
−1 [𝑣𝛼 [4𝑒𝑥(𝑣3−𝛼  )]] = 4𝑒𝑥𝐸−1[  (𝑣3+𝛼  )] 

                   ∅1(𝑥, 𝑡) = 4𝑒𝑥
   𝑡𝛼+1

⌈(𝛼+2)
              

For 𝑛 = 1,2,3…. 

           ∅2(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [−2 𝐷𝑡
𝛼∅1]] = −8𝑒𝑥

   𝑡2𝛼+1

⌈(2𝛼+2)
 

           ∅3(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [−2 𝐷𝑡
𝛼∅2]] = 16𝑒𝑥

   𝑡3𝛼+1

⌈(3𝛼+2)
   

          ∅4(𝑥, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [−2 𝐷𝑡
𝛼∅3]] = −32𝑒𝑥

   𝑡4𝛼+1

⌈(4𝛼+2)
        

. 

: 

: 

Therefore, Series representation of the solution ∅ (𝑥, 𝑡) is as follows: 

  ∅ (𝑥, 𝑡) =   ∅0(𝑥, 𝑡)+∅1(𝑥, 𝑡)+∅2(𝑥, 𝑡) +  ∅3(𝑥, 𝑡) +  ∅4(𝑥, 𝑡) + ⋯…… 

∅ (𝑥, 𝑡) = 𝑒𝑥(1− 2𝑡) + 4𝑒𝑥
   𝑡𝛼+1

⌈(𝛼+2)
− 8𝑒𝑥

   𝑡2𝛼+1

⌈(2𝛼+2)
+ 16𝑒𝑥

   𝑡3𝛼+1

⌈(3𝛼+2)
− 32𝑒𝑥

   𝑡4𝛼+1

⌈(4𝛼+2)
……….  

∅ (𝑥, 𝑡) = 𝑒𝑥 [(1− 2𝑡) + 4
   𝑡𝛼+1

⌈(𝛼+2)
− 8

   𝑡2𝛼+1

⌈(2𝛼+2)
+ 16

   𝑡3𝛼+1

⌈(3𝛼+2)
− 32

   𝑡4𝛼+1

⌈(4𝛼+2)
……… . ]  

In particular when  𝛼 = 1 , the solution is in the form: 

∅ (𝑥, 𝑡) = 𝑒𝑥 [1−
2𝑡

1!
+
(2𝑡)2

2!
−
(2𝑡)3

3!
+
(2𝑡)4

4!
−
(2𝑡)5

5!
… . . ]                                                               

The exact solution for equation (27) is:  ∅ (𝑥, 𝑡) = 𝑒𝑥−2𝑡                                                            

 

Two-dimensional fractional telegraph equation: 

Example 4. Considering the two-dimensional fractional telegraph equation as Follows [23]: 

         𝐷𝑡
2𝛼∅ + 3 𝐷𝑡

𝛼∅ + 2∅ = ∅𝑥𝑥 + ∅𝑦𝑦          0 <∝ ≤ 1     

Initial conditions: ∅(𝑥 , 𝑦, 0) =  𝑒  𝑥+𝑦,       ∅𝑡(𝑥, 𝑦, 0) = −3𝑒𝑥+𝑦           (35)  

Apply the Elzaki transformation of Equation (34), 

            𝐸[ 𝐷𝑡
2𝛼∅ + 3 𝐷𝑡

𝛼∅ + 2∅]  = 𝐸[∅ 𝑥𝑥 + ∅ 𝑦𝑦]    

     𝐸 [ 𝐷𝑡
2𝛼∅]  = −𝐸[∅𝑥𝑥 + ∅𝑦𝑦 − 3 𝐷𝑡

𝛼∅ − 2∅]       

(31)

(34)

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(33)

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Communications on Applied Nonlinear Analysis 

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1

𝑣𝛼
 𝐸[∅(𝑥, y, 𝑡)] − 𝑣2−𝛼  ∅(𝑥 , 𝑦 ,0) − 𝑣3−𝛼  ∅𝑡(𝑥, 𝑦, 0) = −𝐸[(∅𝑥𝑥 + ∅𝑦𝑦 − 2∅) − 3 𝐷𝑡

𝛼∅] 

𝐸[∅(𝑥, y, 𝑡)] = 𝑣2  ∅(𝑥, 𝑦, 0) + 𝑣3  ∅𝑡(𝑥, 𝑦, 0) − 𝑣
𝛼 𝐸[(∅𝑥𝑥 + ∅𝑦𝑦 − 2∅) − 3 𝐷𝑡

𝛼∅] 

Applying inverse Elzaki Transform.  

𝐸−1 [𝐸 [∅(𝑥 , y, 𝑡)]] = 𝐸−1 [𝑣2  ∅(𝑥, 𝑦, 0) + 𝑣3  ∅𝑡(𝑥, 𝑦, 0) − 𝑣
𝛼 𝐸[(∅𝑥𝑥 + ∅𝑦𝑦 − 2∅) − 3 𝐷𝑡

𝛼∅]] 

∅(𝑥, y, 𝑡) = 𝐸−1 [𝑣2  ∅(𝑥, 𝑦 ,0) + 𝑣3  ∅𝑡(𝑥 , 𝑦, 0) − 𝑣
𝛼  𝐸[(∅𝑥𝑥 + ∅𝑦𝑦 − 2∅) − 3 𝐷𝑡

𝛼∅]] 

 ∅(𝑥, y, 𝑡) = 𝐸−1[𝑣2  ∅(𝑥, 𝑦, 0) + 𝑣3  ∅𝑡(𝑥, 𝑦, 0) − 𝑣
𝛼  𝐸[𝐿(∅) − 3 𝐷𝑡

𝛼∅]]          

Appling the MADETM process on equation (36) 

                               ∅0(𝑥, y, 𝑡) = 𝐸−1[𝑣2  ∅(𝑥 , 𝑦, 0 ) + 𝑣3  ∅𝑡(𝑥, 𝑦, 0 )] 

                             ∅0 (𝑥, y, 𝑡) = 𝐸−1[𝑣2  𝑒𝑥+𝑦 + 𝑣3  (−3𝑒𝑥+𝑦)] 

                             ∅0(𝑥, y, 𝑡) = 𝐸−1[𝑣2  𝑒𝑥+𝑦 − 𝑣3  (3𝑒𝑥+𝑦)] 

∅0(𝑥, y, 𝑡) = ∅(0) = 𝑒
𝑥+𝑦(1− 3𝑡)    

∅ 𝑛+1(𝑥, 𝑦, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝐿( ∅𝑛) − 3 𝐷𝑡
𝛼  ∅𝑛]]               

For 𝑛 = 0 

 ∅1 (𝑥, 𝑦, 𝑡) = 𝐸
−1 [𝑣𝛼𝐸 [𝐿( ∅0) − 3 𝐷𝑡

𝛼  ∅0]] 

Here,  𝐿 [∅0] = (∅0𝑥𝑥 + ∅0𝑦𝑦 − 2∅0) = 0 

                        Therefore, above equation implies,   ∅1(𝑥, 𝑦, 𝑡) = 𝐸−1 [𝑣𝛼 𝐸 [−3 𝐷𝑡
𝛼 ∅0]] 

                              ∅1(𝑥, 𝑦, 𝑡) = −𝐸
−1[𝑣𝛼𝐸 [−3 𝐷𝑡

𝛼∅0]] = 9𝑒𝑥+𝑦
   𝑡𝛼+1

⌈(𝛼+2)
    

For 𝑛 = 1,2,3…. 

                             ∅2(𝑥, 𝑦, 𝑡) = −𝐸
−1[𝑣𝛼𝐸 [−3 𝐷𝑡

𝛼∅1]] = −27𝑒𝑥+𝑦
   𝑡2𝛼+1

⌈(2𝛼+2)
 

                             ∅3(𝑥, 𝑦, 𝑡) = −𝐸
−1[𝑣𝛼𝐸 [−3 𝐷𝑡

𝛼∅2]] = 81 𝑒𝑥+𝑦
   𝑡3𝛼+1

⌈(3𝛼+2)
    

                                                    ∅4(𝑥, 𝑦, 𝑡) = −𝐸−1[𝑣𝛼𝐸 [−3 𝐷𝑡
𝛼∅3]] = −243 𝑒𝑥+𝑦

   𝑡4𝛼+1

⌈(4𝛼+2)
      

. 

: 

: 

Therefore, Series form ∅ (𝑥 , 𝑦 , 𝑡) will be: 

∅ (𝑥, 𝑦 , 𝑡) =   ∅0(𝑥 , 𝑦, 𝑡)+∅1(𝑥, 𝑦 , 𝑡)+∅2(𝑥, 𝑦, 𝑡) +  ∅3(𝑥, 𝑦, 𝑡) +  ∅4(𝑥, 𝑦, 𝑡) + ⋯…… 

(39)

(36)

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∅ (𝑥, 𝑦, 𝑡) = 𝑒𝑥+𝑦(1− 3𝑡) + 9𝑒𝑥+𝑦
   𝑡𝛼+1

⌈(𝛼+2)
− 27𝑒𝑥+𝑦

   𝑡2𝛼+1

⌈(2𝛼+2)
+ 81 𝑒𝑥+𝑦

   𝑡3𝛼+1

⌈(3𝛼+2)
−

243 𝑒𝑥+𝑦
   𝑡4𝛼+1

⌈(4𝛼+2)
……….   

∅ (𝑥, 𝑦, 𝑡) = 𝑒𝑥+𝑦 [(1− 3𝑡) + 9
   𝑡𝛼+1

⌈(𝛼+2)
− 27

   𝑡2𝛼+1

⌈(2𝛼+2)
+ 81

   𝑡3𝛼+1

⌈(3𝛼+2)
− 243 

   𝑡4𝛼+1

⌈(4𝛼+2)
……… . ]  

When  𝛼 = 1 ,  the approximate solution will be in the form: 

∅  (𝑥, 𝑦, 𝑡) = 𝑒𝑥+𝑦 [1−
3𝑡

1!
+
( 3𝑡 )2

2!
−
( 3𝑡 )3

3!
+
( 3𝑡)4

4!
−
( 3𝑡 )5

5!
+⋯ . . ]       

The exact solution for equation (34) implies ∅ (𝑥, 𝑦, 𝑡) = 𝑒𝑥+𝑦−3𝑡        (41) 

Three-dimensional fractional telegraph equation: 

Example 5. The 3D telegraph equation of fractional order is to be considered [23] 

𝐷𝑡
2𝛼∅ + 2 𝐷𝑡

𝛼∅ + 3∅ = ∅𝑥𝑥 + ∅𝑦𝑦 + ∅𝑧𝑧           0 <∝ ≤ 1    

 With initial conditions: ∅(𝑥, 𝑦, 𝑧, 0) = sinh 𝑥 sinh𝑦 sinh 𝑧 

                                           ∅𝑡(𝑥, 𝑦, 𝑧, 0) = − sinh 𝑥  sinh 𝑦 sinh 𝑧      

Apply the Elzaki transformation of Equation (42), 

            𝐸[ 𝐷𝑡
2𝛼∅ + 2 𝐷𝑡

𝛼∅ + 3∅]  = 𝐸[∅ 𝑥𝑥 + ∅ 𝑦𝑦 + ∅ 𝑧𝑧]   

      𝐸[ 𝐷𝑡
2𝛼∅]  = −𝐸[∅𝑥𝑥 + ∅𝑦𝑦 + ∅𝑧𝑧 − 2 𝐷𝑡

𝛼∅ − 3∅]    

1

𝑣𝛼
 𝐸[∅(𝑥, y, z, 𝑡)] − 𝑣2−𝛼  ∅(𝑥 , 𝑦, 𝑧, 0) − 𝑣3−𝛼  ∅𝑡(𝑥, 𝑦 , 𝑧, 0)

= −𝐸[(∅𝑥𝑥 + ∅𝑦𝑦 + ∅𝑧𝑧 − 3∅) − 2 𝐷𝑡
𝛼∅] 

     𝐸[∅(𝑥, y, z, 𝑡)] = 𝑣2  ∅(𝑥, 𝑦 , 𝑧 ,0) + 𝑣3  ∅𝑡(𝑥, 𝑦, 𝑧, 0) − 𝑣
𝛼  𝐸[(∅𝑥𝑥 + ∅𝑦𝑦 + ∅𝑧𝑧 −  3∅) −

2 𝐷𝑡
𝛼∅] 

Applying inverse Elzaki Transform  

𝐸−1 [𝐸 [∅(𝑥, y, z, 𝑡)]]

= 𝐸−1[𝑣2  ∅(𝑥, 𝑦, 𝑧, 0) + 𝑣3  ∅𝑡(𝑥, 𝑦, 𝑧, 0) − 𝑣
𝛼  𝐸(∅𝑥𝑥 + ∅𝑦𝑦 + ∅𝑧𝑧 − 3∅) − 2 𝐷𝑡

𝛼∅] 

∅(𝑥, y, z, 𝑡) = 𝐸−1[𝑣2  ∅(𝑥, 𝑦, 𝑧, 0) + 𝑣3  ∅𝑡(𝑥, 𝑦, 𝑧, 0) − 𝑣
𝛼  𝐸(∅𝑥𝑥 + ∅𝑦𝑦 + ∅𝑧𝑧 − 3∅) − 2 𝐷𝑡

𝛼∅] 

∅(𝑥, y, z, 𝑡) = 𝐸−1 [𝑣2  ∅(𝑥, 𝑦, 𝑧, 0) + 𝑣3  ∅𝑡(𝑥, 𝑦, 𝑧, 0) − 𝑣
𝛼  𝐸[𝐿(∅) − 2  𝐷𝑡

𝛼∅]]           

Appling the MADETM process on equation (44)  

   ∅0(𝑥, y, z, 𝑡) = 𝐸−1[𝑣2  ∅(𝑥, 𝑦, 𝑧, 0) + 𝑣3  ∅𝑡(𝑥, 𝑦, 𝑧, 0)] 

  ∅0(𝑥, 𝑦, 𝑧, 𝑡) = 𝐸
−1[𝑣2  sinh 𝑥 sinh𝑦 sinh 𝑧 + 𝑣3  (− sinh 𝑥  sinh𝑦 sinh 𝑧)] 

  ∅0(𝑥, 𝑦, 𝑧, 𝑡) = 𝐸
−1[𝑣2  sinh 𝑥 sinh𝑦 sinh 𝑧 − 𝑣3  (sinh 𝑥  sinh𝑦 sinh 𝑧)] 

  ∅0(𝑥, 𝑦, 𝑧, 𝑡) = ∅(0) = sinh 𝑥 sinh𝑦 sinh 𝑧 (1− 𝑡)       

(44)

(40)

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  ∅𝑛+1(𝑥, 𝑦, 𝑧, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝐿( ∅𝑛) − 2𝐷𝑡
𝛼 ∅𝑛]]           

For 𝑛 = 0  

         ∅1(𝑥, y, z, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [𝐿( ∅0) − 2 𝐷𝑡
𝛼 ∅0]]    

         Here,  𝐿 [∅0] = (∅0𝑥𝑥 + ∅0𝑦𝑦 + ∅0𝑧𝑧 − 3∅0) = 0 

Therefore, above equation implies,   ∅1(𝑥, 𝑦, 𝑧, 𝑡) = 𝐸−1[𝑣𝛼𝐸 [−2 𝐷𝑡
𝛼∅0]] 

           ∅1(𝑥, 𝑦, 𝑧, 𝑡) = −𝐸−1[𝑣𝛼𝐸 [−2 𝐷𝑡
𝛼∅0]] = 2 sinh 𝑥  sinh𝑦 sinh 𝑧

   𝑡𝛼+1

⌈(𝛼+2)
     

For 𝑛 = 1,2,3…. 

          ∅2(𝑥, 𝑦, 𝑧, 𝑡) = −𝐸−1[𝑣𝛼𝐸 [−2 𝐷𝑡
𝛼∅1]] = −4 sinh 𝑥  sinh𝑦 sinh 𝑧

   𝑡2𝛼+1

⌈(2𝛼+2)
 

         ∅3(𝑥, 𝑦, 𝑧, 𝑡) = −𝐸−1[𝑣𝛼𝐸 [−2 𝐷𝑡
𝛼∅2]] = 8 sinh 𝑥  sinh𝑦 sinh 𝑧

   𝑡3𝛼+1

⌈(3𝛼+2)
    

         ∅4(𝑥, 𝑦, 𝑧, 𝑡) = −𝐸−1[𝑣𝛼𝐸 [−3 𝐷𝑡
𝛼∅3]] = −16 sinh 𝑥 sinh𝑦 sinh 𝑧

   𝑡4𝛼+1

⌈(4𝛼+2)
    

. 

: 

: 

Therefore, Series form ∅ (𝑥, 𝑦, 𝑧 , 𝑡) will be: 

∅ (𝑥, 𝑦, 𝑧, 𝑡) =   ∅0 ( 𝑥, 𝑦, 𝑧, 𝑡)+∅1 (𝑥, 𝑦, 𝑧, 𝑡)+∅2 (𝑥, 𝑦, 𝑧, 𝑡) +  ∅3(𝑥, 𝑦, 𝑧, 𝑡) + ⋯…… 

∅ (𝑥, 𝑦, 𝑧, 𝑡) = sinh 𝑥 sinh𝑦 sinh 𝑧 (1− 𝑡) + 2 sinh 𝑥 sinh𝑦 sinh 𝑧
   𝑡𝛼+1

⌈(𝛼 + 2)

− 4 sinh 𝑥  sinh𝑦 sinh 𝑧
   𝑡2𝛼+1

⌈(2𝛼 + 2)
+ 8 sinh 𝑥  sinh𝑦 sinh 𝑧

   𝑡3𝛼+1

⌈(3𝛼 + 2)

− 16 sinh 𝑥  sinh𝑦 sinh 𝑧
   𝑡4𝛼+1

⌈(4𝛼 + 2)
………. 

∅ (𝑥, 𝑦, 𝑧, 𝑡) = sinh 𝑥 sinh𝑦 sinh 𝑧 [(1− 𝑡) + 2
   𝑡𝛼+1

⌈(𝛼+2)
− 4

   𝑡2𝛼+1

⌈(2𝛼+2)
+ 8

   𝑡3𝛼+1

⌈(3𝛼+2)
−−16 

   𝑡4𝛼+1

⌈(4𝛼+2)
……… . ]   

When  𝛼 = 1 , the following approximate solution will represent as: 

∅ (𝑥, 𝑦, 𝑧, 𝑡) = sinh 𝑥 sinh𝑦 sinh 𝑧 [1− 𝑡 +
2t2

2!
−

4 𝑡3

3!
+

8𝑡4

4!
−

16t5

5!
… . . ]                          (48) 

 

∅ (𝑥, 𝑦, 𝑧, 𝑡) =
sinh 𝑥 sinh𝑦 sinh 𝑧

2
[2−

2t

1!
+

4t
2

2!
−

8 𝑡3

3!
+

16𝑡4

4!
−

32 𝑡5

5!
… . . ] 

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Communications on Applied Nonlinear Analysis 

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∅ (𝑥, 𝑦, 𝑧, 𝑡) =
sinh 𝑥 sinh𝑦 sinh 𝑧

2
[2 −

2t

1!
+
( 2𝑡)2

2!
−
(2𝑡 )3

3!
+
( 2𝑡)4

4!
−
(2𝑡 )5

5!
… . . ] 

∅ (𝑥, 𝑦, 𝑧, 𝑡) =
sinh𝑥 sinh𝑦 sinh 𝑧

2
(1+ 𝑒1−2𝑡)      

The precise answer to the equation (42) is:  

 ∅ (𝑥, 𝑦, 𝑡) =
sinh𝑥 sinh𝑦 sinh 𝑧

2
(1+ 𝑒1−2𝑡)          

5. Graphical Discussion 

In This discussion, the graphical simulation is shown to validate the results between the approximate 
solution calculated by adopted technique and exact solution exist are expressed for said applications. 
Example 1, the approximation solution and exact solution outcomes are compared at 𝑡 = 1, 2 𝑎𝑛𝑑 3 at 

𝛼 = 2 shown in Figure 1.

 

Figure 2. . shows the surface graph of the approximate and exact solu-
tions for Example 1 at 𝛼 = 2.  The error surface graph for Example 1 is 
shown in Figure 2.. Additionally, Figure 2. a illustrates a line graph for 
Example 1, displaying the approximate solution, exact solution, and the 
absolute error considering 𝑡 = 1. 

 

Figure 3.  displays the surface graph for Example 2, showcasing both 
the approximate and exact solutions at 𝛼 = 1  Figure 3.  represents the 
corresponding error surface graph for Example 2. Furthermore, Figure 3.  
features a line graph for Example 2, which highlights the approximate and 
exact solutions, as well as the absolute error, evaluated at 𝑡 = 1. In Figure 
4. a : Comparison of Approximate solutions and Exact solutions at 𝛼 = 1  
for Example 3 

Figure 2. a : Line Plot for Exact, Approximate & Absolute Error Example 1 

Figure 3. c : Line Plot for Exact, Approximate & Absolute Error Example 2 

(49)



324

c



325

2



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c



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Figure 5. a : Comparison of Approximate solution and Exact solution profiles at 𝛼 = 1  for Exam-

ple 4 

 

Figure 6. a : Comparing of Approximate solution and Exact solutions at 𝛼 = 1 for Example 5 

 

 

Figure 5. c : Line Plot for Exact, Approximate & Absolute Error Example 4 

Figure 5. c 

 : Line Plot for Exact, Approximate & Absolute Error Example 4 

 

 

Figure 6. b : Error Plot between Exact- Appro. Solution Example 5 

327

4



328



329



330



331