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Semi-analytical approximation of time-fractional telegraph equation via natural transform in Caputo derivative

  • Mamta Kapoor EMAIL logo and Samanyu Khosla
Published/Copyright: April 28, 2023
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Nonlinear Engineering
From the journal Nonlinear Engineering Volume 12 Issue 1

Abstract

In the present research study, time-fractional hyperbolic telegraph equations are solved iteratively using natural transform in one, two, and three dimensions. The fractional derivative is considered in the Caputo sense. These equations serve as a model for the wave theory process of signal processing and transmission of electric impulses. To evaluate the validity and effectiveness of the suggested strategy, a graphical comparison of approximated and exact findings is performed. Convergence analysis of the approximations utilising L ∞ has been done using tables. The suggested approach may successfully and without errors solve a wide variety of ordinary differential equations, partial differential equations (PDEs), fractional PDEs, and fractional hyperbolic telegraph equations.

Graphical abstract

Keywords: natural transform; Caputo derivative; fractional telegraph equation; convergence analysis

1 Introduction

Fractional calculus is a branch of mathematics that deals with partial order integrals, derivatives, and their equations. It was initially alluded to in a letter written by L’ Hospital to Leibnitz, essentially asking about a derivative with order ½ [1]. As fractional calculus is further developed, it found its application in many domains, which includes variational calculus [2], random walks [3], optimal control theory [4], image processing [5], etc.

Second-order hyperbolic partial differential equations (PDE) known as telegraph equation, developed by Oliver Heaviside [6], is used to elaborate voltage and current in an electric transmission line. Further generalizing such equations, we come across a particular kind of equation, known as the fractional hyperbolic telegraph equation (FHTE). Being a type of fractional PDE (FPDE), it is quite a challenge to solve such an equation without using advanced numerical and analytical techniques. FPDEs have multiple methods developed by many researchers due to their importance, e.g., Kudryashov method, Bäcklund transformation method based on the Riccati equation, and a new auxiliary ordinary differential equation have been used to obtain obtaining optical soliton solutions for the Lakshmanan–Porsezian–Daniel model with Kerr law nonlinearity [7]. Laplace transform method has been used to solve constructive equations of the fractional second grade fluid [8]. An extension of homotopy analysis method using Laplace transform, called q-homotopy analysis transform method, was used to solve time and space fractional coupled Burgers’ equations [9]. The fractional reduced differential transform method has been used to obtain approximate solutions of time fractional gas dynamics equation [10]. A hybrid method of decomposition method and Elzaki transformation has been used to solve Navier–Stokes equation [11]. Fractional order diffusion equations were solved using natural transform decomposition [12]. Shallow water forced Korteweg–DeVries equation describing free surface critical flow over a hole has been solved using q-homotopy analysis transformed technique [13]. Variable-order time-space fractional KdV–Burgers–Kuramoto equation has been solved using a pseudo-operational collocation scheme [14]. A Vieta-Fibonacci collocation method has been used to solve fractional delay integro-differential equations [15]. A method based on Laplace transform and homotopy analysis method has been used to obtain approximate solutions of time fractional nonlinear water wave equation [16]. The homotopy analysis transform method using Laplace transform has been used to obtain the approximate solution of nonlinear time-fractional Cauchy reaction–diffusion equation [17].

Multiple techniques have been used to solve FHTEs. Khan et al. [18] used a new technique of adomian decomposition method (ADM) to solve fractional order hyperbolic telegraph equations. Sweilam et al. [19] approached time-fractional telegraph equations using the Sinc-Legendre collocation method. Saadatmandi and Mohabbati [20] employed a computational tau technique based on Legendre polynomials to solve time-fractional telegraph equations. A hybrid technique using Sumudu and Fourier transforms has been used to solve time fractional telegraph equation [21]. Shehu transform has been used to obtain series solution of FHTE [22]. For solving linear time-fractional telegraph equations, two-variable Vieta-Fibonacci polynomials have been constructed and coupled with a matrix collocation scheme [23].

1.1 One-dimensional fractional telegraph equation [ 22 ]

D t β φ + ε φ ( γ , t ) + γ φ t = φ γ γ + g ( γ , t ) .

Here, β and γ are arbitrary constants and φ ( γ , t ) is the unknown function. β is the fractional order of the equation.

1.2 Two-dimensional fractional telegraph equation [ 22 ]

D t 2 β φ + 2 μ D t β φ + ε 2 φ = φ γ γ + φ ρ ρ + g ( γ , ρ , t ) ,

where φ ( γ , ρ , 0 ) = f 1 ( γ , ρ ) and φ t ( γ , ρ , 0 ) = f 2 ( γ , ρ ) .

1.3 Three-dimensional fractional telegraph equation [22]

D t 2 β φ + 2 μ D t β φ + ε 2 φ = φ γ γ + φ ρ ρ + φ σ σ + g ( γ , ρ , σ , t ) ,

where φ ( γ , ρ , σ , 0 ) = f 1 ( γ , ρ , σ ) and φ t ( γ , ρ , σ , 0 ) = f 2 ( γ , ρ , σ ) .

There exist different approaches to tackle FHTEs, and one such way is to apply an integral transform on the equation and obtain an iterative solution. Several integral transforms exist, such as Sumudu transform [24], natural transform [25], Shehu transform [26], and Elzaki transform [27]. These transforms can be applied on various techniques, such as homotopy perturbation method [28], ADM [29], reduced differential transform method [30], and generalized differential transform method [31] to obtain solutions to FHTEs. For example, Laplace Transform has been used for solving telegraph equations by implementing q-Homotopy Analysis Transform Method [32]. Natural transform was developed by Khan and Khan [25] in 2008 and was used to solve a certain kind of the fluid flow problem. Since then, it has found its application in many different cases, one such case being natural transform decomposition method (NTDM). NTDM has been used to solve fractional order partial differential equations [33] and Burgers’ equations [34]. In the present article, natural transform would be used to obtain an iterative semi analytical solution to FHTEs. Fractional order has been considered because of it being a more generalized form of hyperbolic telegraph equations, which means the integer order solutions can also be obtained using the appropriate value of fractional order. The approach is based on the one used in solving FHTEs using Shehu transform [22]. When developing the general formula, natural transform is applied on the equation. Linear and nonlinear operators are separated into sum of terms inverse natural transform is applied, which is then used to develop an iterative scheme (Tables 1 and 2).

Table 1

Natural transform of common functions [36]

F ( t ) NT [ F ( t ) ] = Q ( s , v )
1 1 1 s
2 t v s 2
3 t n − 1 Γ ( n ) v n − 1 s n
4 sin at a v s 2 + ( av ) 2
5 e at 1 s − av
6 t n − 1 e at Γ ( n ) v n − 1 ( s − av ) n
7 cos at s s 2 + ( av ) 2
Table 2

Inverse natural transform of common functions [36]

Q ( s , v ) NT − 1 [ Q ( s , v ) ] = F ( t )
1 1 s 1
2 v s 2 t
3 v n − 1 s n t n − 1 Γ ( n )
4 v s 2 + ( av ) 2 sin at a
5 1 s − av e at
6 v n − 1 ( s − av ) n t n − 1 e at Γ ( n )
7 s s 2 + ( av ) 2 cos at

1.4 Basics of natural transform

Definition 1

Natural transform of a function [35].

NT [ φ ( t ) ] = ∫ 0 ∞ e − st φ ( vt ) d t ,

where s , v ∈ ( 0 , ∞ ) .

Definition 2

Natural transform of fractional derivative in Caputo sense [35].

NT [ D t β 0 C φ ( t ) ] = s v β NT [ φ ( t ) ] − 1 s φ ( 0 ) .

Theorem 1

Linearity property of natural transform [25].

NT [ af ( t ) + bf ( t ) ] = aNT [ f ( t ) ] + bNT [ f ( t ) ] ,

where a and b are arbitrary constants and f ( t ) is a function in t .

Theorem 2

Change of scale property of natural transform [36].

NT [ f ( at ) ] = 1 a F s a , v ,

where a is an arbitrary constant, f ( t ) is a function in t , and F ( s , v ) is the natural transform of f ( t ) .

Definition 3

Fractional derivative of a function in Caputo sense [37].

D β 0 C f ( x ) = 1 Γ ( m − β ) ∫ 0 x ( x − t ) m − β − 1 f ( m ) ( t ) d t ,

where m − 1 < β ≤ m , m ∈ N , x > 0 .

2 Development of the formula

2.1 Implementation upon 1D time-fractional telegraph equation

The form of 1D time-fractional telegraph equation is as follows [22]:

D t β φ ( γ , t ) + L [ φ ( γ , t ) ] + N [ φ ( γ , t ) ] = q ( γ , t ) ,

where L is linear operator and N is nonlinear operator. D t β φ ( γ , t ) is Caputo derivative of φ ( γ , t ) . By applying natural transform on the following equation:

NT [ D t β φ ( γ , t ) ] + NT [ L [ φ ( γ , t ) ] + N [ φ ( γ , t ) ] ] = NT [ q ( γ , t ) ]

⇒ s v β NT [ φ ( γ , t ) ] − 1 s φ ( 0 )

= NT [ q ( γ , t ) ] − NT [ L [ φ ( γ , t ) ] ] − NT [ N [ φ ( γ , t ) ] ]

⇒ NT [ φ ( γ , t ) ] = 1 s φ ( 0 ) + v s β NT [ q ( γ , t ) ] − v s β ( NT [ L [ φ ( γ , t ) ] ] + NT [ N [ φ ( γ , t ) ] ] )

⇒ φ ( γ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , t ) ] − NT − 1 v s β ( NT [ L [ φ ( γ , t ) ] ] + NT [ N [ φ ( γ , t ) ] ] ) .

Now,

L [ φ ] = L ∑ r = 0 ∞ φ r ( γ , t ) = L [ φ 0 ( γ , t ) ] + ∑ r = 1 ∞ L ∑ i = 0 r φ i ( γ , t ) − L ∑ i = 0 r − 1 φ i ( γ , t )

N [ φ ] = N ∑ r = 0 ∞ φ r ( γ , t ) = N [ φ 0 ( γ , t ) ] + ∑ r = 1 ∞ N ∑ i = 0 r φ i ( γ , t ) − N ∑ i = 0 r − 1 φ i ( γ , t ) .

Therefore,

∑ k = 0 ∞ φ k ( γ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , t ) ]

− NT − 1 v s β NT L [ φ 0 ( γ , t ) ] + N [ φ 0 ( γ , t ) ] + L ∑ r = 1 ∞ φ r ( γ , t ) + N ∑ r = 1 ∞ φ r ( γ , t )

⇒ ∑ k = 0 ∞ φ k ( γ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , t ) ]

− NT − 1 v s β NT ( L [ φ 0 ( γ , t ) ] + N [ φ 0 ( γ , t ) ] )

− NT − 1 v s β NT ∑ r = 1 ∞ L [ φ r ( γ , t ) ] + ∑ r = 1 ∞ N ∑ i = 0 r φ i ( γ , t ) − N ∑ i = 0 r − 1 φ i ( γ , t )

⇒ φ 0 ( γ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , t ) ]

⇒ φ 1 ( γ , t ) = − NT − 1 v s β NT ( L [ φ 0 ( γ , t ) ] + N [ φ 0 ( γ , t ) ] )

⇒ φ r + 1 ( γ , t ) = − NT − 1 v s β NT L [ φ r ( γ , t ) ] + N ∑ i = 0 r φ i ( γ , t ) − ∑ i = 0 r − 1 φ i ( γ , t )

r = 1 , 2 , 3 , 4 , 5 , …

2.2 Implementation upon 2D time-fractional telegraph equation

The form of 2D time-fractional telegraph equation is as follows [22]:

D t β φ ( γ , ρ , t ) + L [ φ ( γ , ρ , t ) ] + N [ φ ( γ , ρ , t ) ] = q ( γ , ρ , t ) .

See Appendix A.

2.3 Implementation upon 3D time-fractional telegraph equation

The form of 3D time-fractional telegraph equation is as follows [22]:

D t β φ ( γ , ρ , σ , t ) + L [ φ ( γ , ρ , σ , t ) ] + N [ φ ( γ , ρ , σ , t ) ] = q ( γ , ρ , σ , t ) :

See Appendix B.

3 Examples and calculation

Example 1. Considered following 1D time-fractional telegraph equation [22]:

(1) φ t β = φ − 2 φ t − φ γ γ ,

where φ ( γ , 0 ) = e γ and φ t ( γ , 0 ) = − 2 e γ , 0 < β ≤ 2

φ 0 ( γ , t ) = φ ( γ , 0 ) + t φ t ( γ , 0 ) = e γ − t ( 2 e γ ) = e γ ( 1 − 2 t ) .

By applying natural transform on Eq. (1), we obtain:

NT [ φ t β ] = NT [ φ − 2 φ t − φ γ γ ]

⇒ NT [ D t β φ ( γ , t ) ] = NT [ φ − 2 φ t − φ γ γ ]

⇒ φ ( γ , t ) = NT − 1 1 s φ ( 0 ) + NT − 1 v s β ( NT [ R [ φ ] ] )

(2) ⇒ φ 0 ( γ , t ) = NT − 1 1 s φ ( 0 ) ,

⇒ φ 0 ( γ , t ) = φ ( 0 ) NT − 1 1 s

⇒ φ 0 ( γ , t ) = φ ( 0 ) = e x ( 1 − 2 t )

(3) φ r + 1 ( γ , t ) = NT − 1 v s β ( NT [ R [ φ r ] ] ) ,

r = 0 , 1 , 2 , 3 , …

R [ φ 0 ] = φ 0 − 2 ( φ 0 ) t − ( φ 0 ) γ γ = 4 e γ

By using Eq. (3), we obtain:

φ 1 ( γ , t ) = NT − 1 v s β ( NT [ 4 e γ ] )

⇒ φ 1 ( γ , t ) = 4 e γ NT − 1 v β s β + 1

⇒ φ 1 ( γ , t ) = 4 e γ t β Γ ( β + 1 )

R [ φ 1 ] = φ 1 − 2 ( φ 1 ) t − ( φ 1 ) γ γ = − 8 e γ β t β − 1 Γ ( β + 1 ) .

By using Eq. (3), we obtain:

φ 2 ( γ , t ) = NT − 1 v s β NT − 8 e γ β t β − 1 Γ ( β + 1 )

⇒ φ 2 ( γ , t ) = − 8 e γ β Γ ( β ) Γ ( β + 1 ) NT − 1 v s β NT t β − 1 Γ ( β )

⇒ φ 2 ( γ , t ) = − 8 e γ β Γ ( β ) Γ ( β + 1 ) NT − 1 v 2 β − 1 s 2 β

⇒ φ 2 ( γ , t ) = − 8 e γ β Γ ( β ) t 2 β − 1 Γ ( β + 1 ) Γ ( 2 β )

R [ φ 2 ] = φ 2 − 2 ( φ 2 ) t − ( φ 2 ) γ γ = 16 e γ β ( 2 β − 1 ) Γ ( β ) t 2 β − 2 Γ ( β + 1 ) Γ ( 2 β ) .

By using Eq. (3), we obtain:

φ 3 ( γ , t ) = NT − 1 v s β NT 16 e γ β ( 2 β − 1 ) Γ ( β ) t 2 β − 2 Γ ( β + 1 ) Γ ( 2 β )

⇒ φ 3 ( γ , t ) = 16 e γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( β + 1 ) Γ ( 2 β ) NT − 1 × v s β NT t 2 β − 2 Γ ( 2 β − 1 )

⇒ φ 3 ( γ , t ) = 16 e γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( β + 1 ) Γ ( 2 β ) NT − 1 v 3 β − 2 s 3 β − 1

⇒ φ 3 ( γ , t ) = 16 e γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 2 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 )

R [ φ 3 ] = φ 3 − 2 ( φ 3 ) t − ( φ 3 ) γ γ = − 32 e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) .

By using Eq. (3), we obtain:

φ 4 ( γ , t ) = NT − 1 v s β NT − 32 e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 )

⇒ φ 4 ( γ , t ) = − 32 e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) NT − 1 v s β NT t 3 β − 3 Γ ( 3 β − 2 )

⇒ φ 4 ( γ , t ) = − 32 e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) NT − 1 v 4 β − 3 s 4 β − 2

⇒ φ 4 ( γ , t ) = − 32 e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 )

R [ φ 4 ] = φ 4 − 2 ( φ 4 ) t − ( φ 4 ) γ γ = 64 e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) .

By using Eq. (3), we obtain:

φ 5 ( γ , t ) = NT − 1 v s β NT 64 e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 )

⇒ φ 5 ( γ , t ) = 64 e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) NT − 1 v s β NT t 4 β − 4 Γ ( 4 β − 3 )

⇒ φ 5 ( γ , t ) = 64 e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) NT − 1 v 5 β − 4 s 5 β − 3

⇒ φ 5 ( γ , t ) = 64 e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) t 5 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) Γ ( 5 β − 3 )

φ ( γ , t ) = φ 0 ( γ , t ) + φ 1 ( γ , t ) + φ 2 ( γ , t ) + φ 3 ( γ , t ) + φ 4 ( γ , t ) + φ 5 ( γ , t ) + …

⇒ φ ( γ , t ) = e γ ( 1 − 2 t ) + 4 e γ t β Γ ( β + 1 ) − 8 e γ β Γ ( β ) t 2 β − 1 Γ ( β + 1 ) Γ ( 2 β ) + 16 e γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 2 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) − 32 e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) + 64 e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) t 5 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) Γ ( 5 β − 3 ) − …

By considering α = 2 , the series reduces to:

φ ( γ , t ) = e γ 1 − 2 t 1 ! + ( 2 t ) 2 2 ! − ( 2 t ) 3 3 ! + ( 2 t ) 4 4 ! − ( 2 t ) 5 5 ! + ( 2 t ) 6 6 ! − …

⇒ φ ( x , t ) = e γ − 2 t .

Example 2. Consider the following 1D time-fractional telegraph equation [22]:

(4) φ t β = φ − φ t − φ γ γ ,

where φ ( γ , 0 ) = e γ and φ t ( γ , 0 ) = − e γ , 0 < β ≤ 2 .

φ 0 ( γ , t ) = φ ( γ , 0 ) + t φ t ( γ , 0 ) = e γ − t ( e γ ) = e γ ( 1 − t ) .

By applying natural transform on Eq. (4), we obtain:

NT [ φ t β ] = NT [ φ − 2 φ t − φ γ γ ]

⇒ NT [ D t β φ ( γ , t ) ] = NT [ φ − 2 φ t − φ γ γ ]

⇒ φ ( γ , t ) = NT − 1 1 s φ ( 0 ) + NT − 1 v s β ( NT [ R [ φ ] ] )

(5) ⇒ φ 0 ( γ , t ) = NT − 1 1 s φ ( 0 ) ,

⇒ φ 0 ( γ , t ) = φ ( 0 ) NT − 1 1 s

⇒ φ 0 ( γ , t ) = φ ( 0 ) = e x ( 1 − t )

(6) φ r + 1 ( γ , t ) = NT − 1 v s β ( NT [ R [ φ r ] ] ) ,

r = 0 , 1 , 2 , 3 , …

R [ φ 0 ] = φ 0 − ( φ 0 ) t − ( φ 0 ) γ γ = e γ .

By using Eq. (6):

φ 1 ( γ , t ) = NT − 1 v s β ( NT [ e γ ] )

⇒ φ 1 ( γ , t ) = e γ NT − 1 v β s β + 1

⇒ φ 1 ( γ , t ) = e γ t β Γ ( β + 1 )

R [ φ 1 ] = φ 1 − 2 ( φ 1 ) t − ( φ 1 ) γ γ = − e γ β t β − 1 Γ ( β + 1 ) .

By using Eq. (6):

φ 2 ( γ , t ) = NT − 1 v s β NT − e γ β t β − 1 Γ ( β + 1 )

⇒ φ 2 ( γ , t ) = − e γ β Γ ( β ) Γ ( β + 1 ) NT − 1 v s β NT t β − 1 Γ ( β )

⇒ φ 2 ( γ , t ) = − e γ β Γ ( β ) Γ ( β + 1 ) NT − 1 v 2 β − 1 s 2 β

⇒ φ 2 ( γ , t ) = − e γ β Γ ( β ) t 2 β − 1 Γ ( β + 1 ) Γ ( 2 β )

R [ φ 2 ] = φ 2 − 2 ( φ 2 ) t − ( φ 2 ) γ γ = e γ β ( 2 β − 1 ) Γ ( β ) t 2 β − 2 Γ ( β + 1 ) Γ ( 2 β ) .

By using Eq. (6), we obtain:

φ 3 ( γ , t ) = NT − 1 v s β NT e γ β ( 2 β − 1 ) Γ ( β ) t 2 β − 2 Γ ( β + 1 ) Γ ( 2 β )

⇒ φ 3 ( γ , t ) = e γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( β + 1 ) Γ ( 2 β ) NT − 1 × v s β NT t 2 β − 2 Γ ( 2 β − 1 )

⇒ φ 3 ( γ , t ) = e γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( β + 1 ) Γ ( 2 β ) NT − 1 v 3 β − 2 s 3 β − 1

⇒ φ 3 ( γ , t ) = e γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 2 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 )

R [ φ 3 ] = φ 3 − 2 ( φ 3 ) t − ( φ 3 ) γ = − e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) .

By using Eq. (6), we obtain:

φ 4 ( γ , t ) = NT − 1 v s β NT − e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 )

⇒ φ 4 ( γ , t ) = − e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) NT − 1 v s β NT t 3 β − 3 Γ ( 3 β − 2 )

⇒ φ 4 ( γ , t ) = − e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) NT − 1 v 4 β − 3 s 4 β − 2

⇒ φ 4 ( γ , t ) = − e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 )

R [ φ 4 ] = φ 4 − 2 ( φ 4 ) t − ( φ 4 ) γ γ = e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) .

By using Eq. (6), we obtain:

φ 5 ( γ , t ) = NT − 1 v s β NT e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 )

⇒ φ 5 ( γ , t ) = e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) NT − 1 v s β NT t 4 β − 4 Γ ( 4 β − 3 )

⇒ φ 5 ( γ , t ) = e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) NT − 1 v 5 β − 4 s 5 β − 3

⇒ φ 5 ( γ , t ) = e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) t 5 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) Γ ( 5 β − 3 )

φ ( γ , t ) = φ 0 ( γ , t ) + φ 1 ( γ , t ) + φ 2 ( γ , t ) + φ 3 ( γ , t ) + φ 4 ( γ , t ) + φ 5 ( γ , t ) + …

⇒ φ ( γ , t ) = e γ ( 1 − t ) + e γ t β Γ ( β + 1 ) − e γ β Γ ( β ) t 2 β − 1 Γ ( β + 1 ) Γ ( 2 β ) + e γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 2 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) − e γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) + e γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) t 5 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) Γ ( 5 β − 3 ) − … .

By substituting α = 2 , the series reduces to:

φ ( γ , t ) = e γ 1 − t 1 ! + t 2 2 ! − t 3 3 ! + t 4 4 ! − t 5 5 ! + t 6 6 ! − …

⇒ φ ( γ , t ) = e γ − t .

Example 3. Considered following nonlinear 1D time-fractional telegraph equation [22]:

(7) D t β φ = φ γ γ + φ t − φ 2 + γ φ φ γ ,

where φ ( γ , 0 ) = γ and φ t ( γ , 0 ) = γ , 0 < β ≤ 2

φ 0 ( γ , t ) = φ ( γ , 0 ) + t φ t ( γ , 0 ) = γ + t γ = γ ( 1 + t )

R [ φ ( γ , t ) ] = φ γ γ + φ t , N [ φ ( γ , t ) ] = − φ 2 + γ φ φ γ .

By applying natural transform on Eq. (7), we obtain:

NT [ D t β φ ( γ , t ) ] = NT [ φ γ γ + φ t − φ 2 + γ φ φ γ ]

⇒ φ ( γ , t ) = NT − 1 1 s φ ( 0 ) + NT − 1 v s β ( NT [ R [ φ ] + N [ φ ] ] )

(8) ⇒ φ 0 ( γ , t ) = NT − 1 1 s φ ( 0 ) ,

(9) φ 1 ( γ , t ) = NT − 1 v s β ( NT [ R [ φ 0 ] + N [ φ 0 ] ] ) ,

(10) φ r + 1 ( γ , t ) = NT − 1 v s β ( NT [ R [ φ r ] + N [ φ r ] ] ) ,

where r = 1 , 2 , 3 , … .

Consider Eq. (8):

φ 0 ( γ , t ) = φ ( 0 ) NT − 1 1 s .

⇒ φ 0 ( γ , t ) = φ ( 0 ) = γ ( 1 + t )

R [ φ 0 ] = ( φ 0 ) γ γ + ( φ 0 ) t = γ

N [ φ 0 ] = − ( φ 0 ) 2 + γ φ 0 ( φ 0 ) γ = 0 .

By using Eq. (9), we obtain:

φ 1 ( γ , t ) = NT − 1 v s β ( NT [ γ ] )

⇒ φ 1 ( γ , t ) = γ NT − 1 v s β ( NT [ 1 ] )

⇒ φ 1 ( γ , t ) = γ NT − 1 v β s β + 1

⇒ φ 1 ( γ , t ) = γ t β Γ ( β + 1 )

R [ φ 1 ] = ( φ 1 ) γ γ + ( φ 1 ) t = γ β t β − 1 Γ ( β + 1 )

N [ φ 1 ] = − ( φ 1 ) 2 + γ φ 1 ( φ 1 ) γ = 0 .

By using Eq. (10), we obtain:

φ 2 ( γ , t ) = NT − 1 v s β NT γ β t β − 1 Γ ( β + 1 )

⇒ φ 2 ( γ , t ) = γ β Γ ( β ) Γ ( β + 1 ) NT − 1 v s β NT t β − 1 Γ ( β )

By using Eq. (10), we obtain:

⇒ φ 2 ( γ , t ) = γ β Γ ( β ) Γ ( β + 1 ) NT − 1 v 2 β − 1 s 2 β

⇒ φ 2 ( γ , t ) = γ β Γ ( β ) t 2 β − 1 Γ ( β + 1 ) Γ ( 2 β )

R [ φ 2 ] = ( φ 2 ) γ γ + ( φ 2 ) t = γ β ( 2 β − 1 ) Γ ( β ) t 2 β − 2 Γ ( β + 1 ) Γ ( 2 β )

N [ φ 2 ] = − ( φ 2 ) 2 + γ φ 2 ( φ 2 ) γ = 0 .

By using Eq. (10), we obtain:

φ 3 ( γ , t ) = NT − 1 v s β NT γ β ( 2 β − 1 ) Γ ( β ) t 2 β − 2 Γ ( β + 1 ) Γ ( 2 β )

⇒ φ 3 ( γ , t ) = γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( β + 1 ) Γ ( 2 β ) NT − 1 v s β NT t 2 β − 2 Γ ( 2 β − 1 )

⇒ φ 3 ( γ , t ) = γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( β + 1 ) Γ ( 2 β ) NT − 1 v 3 β − 2 s 3 β − 1

⇒ φ 3 ( γ , t ) = γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 2 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 )

R [ φ 3 ] = ( φ 3 ) γ γ + ( φ 3 ) t = γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 )

N [ φ 3 ] = − ( φ 3 ) 2 + γ φ 3 ( φ 3 ) γ = 0 .

By using Eq. (10), we obtain:

φ 4 ( γ , t ) = NT − 1 v s β NT γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 )

⇒ φ 4 ( γ , t ) = γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) NT − 1 v s β NT t 3 β − 3 Γ ( 3 β − 2 )

⇒ φ 4 ( γ , t ) = γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) NT − 1 v 4 β − 3 s 4 β − 2

⇒ φ 4 ( γ , t ) = γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 )

R [ φ 4 ] = ( φ 4 ) γ γ + ( φ 4 ) t = γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 )

N [ φ 4 ] = − ( φ 4 ) 2 + γ φ 4 ( φ 4 ) γ = 0 .

By using Eq. (10), we obtain:

φ 5 ( γ , t ) = NT − 1 v s β NT γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 )

⇒ φ 5 ( γ , t ) = γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) NT − 1 v s β NT t 4 β − 4 Γ ( 4 β − 3 )

⇒ φ 5 ( γ , t ) = γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) NT − 1 v 5 β − 4 s 5 β − 3

⇒ φ 5 ( γ , t ) = γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) t 5 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) Γ ( 5 β − 3 )

φ ( γ , t ) = φ 0 ( γ , t ) + φ 1 ( γ , t ) + φ 2 ( γ , t ) + φ 3 ( γ , t ) + φ 4 ( γ , t ) + φ 5 ( γ , t ) + …

⇒ φ ( γ , t ) = γ ( 1 + t ) + γ t β Γ ( β + 1 ) + γ β Γ ( β ) t 2 β − 1 Γ ( β + 1 ) Γ ( 2 β ) + γ β ( 2 β − 1 ) Γ ( β ) Γ ( 2 β − 1 ) t 3 β − 2 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) + γ β ( 2 β − 1 ) ( 3 β − 2 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) t 4 β − 3 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) + γ β ( 2 β − 1 ) ( 3 β − 2 ) ( 4 β − 3 ) Γ ( β ) Γ ( 2 β − 1 ) Γ ( 3 β − 2 ) Γ ( 4 β − 3 ) t 5 β − 4 Γ ( β + 1 ) Γ ( 2 β ) Γ ( 3 β − 1 ) Γ ( 4 β − 2 ) Γ ( 5 β − 3 ) + …

By substituting α = 2 , a series solution is as follows:

φ ( γ , t ) = γ 1 + t 1 ! + t 2 2 ! + t 3 3 ! + t 4 4 ! + t 5 5 ! + t 6 6 ! + …

⇒ φ ( γ , t ) = γ e t .

Example 4. Considered following nonlinear 1D time-fractional telegraph equation [22]:

(11) D t 2 β φ + 2 D t β φ + φ = φ γ γ ,

where φ ( γ , 0 ) = e γ and φ t ( γ , 0 ) = − 2 e γ

φ 0 ( γ , t ) = φ ( γ , 0 ) + t φ t ( γ , 0 ) = e γ − t ( 2 e γ ) = e γ ( 1 − 2 t ) .

Applying natural transform on Eq. (11):

NT [ D t 2 β φ + 2 D t β φ + φ ] = NT [ φ γ γ ]

⇒ NT [ D t 2 β φ ] = NT [ ( φ γ γ − φ ) − 2 D t β φ ]

⇒ NT [ D t 2 β φ ] = NT [ R [ φ ] − 2 D t β φ ]

⇒ φ ( γ , t ) = NT − 1 1 s φ ( 0 ) + NT − 1 v s 2 β ( NT [ R [ φ ] − 2 D t β φ ] )

(12) ⇒ φ 0 ( γ , t ) = NT − 1 1 s φ ( 0 ) ,

⇒ φ 0 ( γ , t ) = φ ( 0 ) NT − 1 1 s

⇒ φ 0 ( γ , t ) = φ ( 0 ) = e x ( 1 − 2 t )

(13) φ r + 1 ( γ , t ) = NT − 1 v s 2 β ( NT [ R [ φ r ] − 2 D t β φ r ] ) ,

r = 0 , 1 , 2 , 3 , …

R [ φ 0 ] = ( φ 0 ) γ γ − φ 0 = 0 .

By using Eq. (13), we obtain:

φ 1 ( γ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 0 ] )

− 2 NT [ D t β φ 0 ] = − 2 s v β NT [ φ 0 ] − 1 s φ 0 ( 0 )

− 2 NT [ D t β φ 0 ] = − 2 s v β NT [ e γ ( 1 − 2 t ) ] − 1 s e γ

− 2 NT [ D t β φ 0 ] = − 2 e γ s v β NT [ 1 ] − 2 NT [ t ] − 1 s

− 2 NT [ D t β φ 0 ] = − 2 e γ s v β 1 s − 2 v s 2 − 1 s

− 2 NT [ D t β φ 0 ] = 4 e γ s β − 2 v β − 1

⇒ φ 1 ( γ , t ) = NT − 1 v s 2 β 4 e γ s β − 2 v β − 1

⇒ φ 1 ( γ , t ) = 4 e γ NT − 1 v β + 1 s β + 2

⇒ φ 1 ( γ , t ) = 4 e γ t β + 1 Γ ( β + 2 )

R [ φ 1 ] = ( φ 1 ) γ γ − φ 1 = 0 .

By using Eq. (13), we obtain:

φ 2 ( γ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 1 ] )

− 2 NT [ D t β φ 1 ] = − 2 s v β NT [ φ 1 ] − 1 s φ 1 ( 0 )

− 2 NT [ D t β φ 1 ] = − 2 s v β NT 4 e γ t β + 1 Γ ( β + 2 ) − 1 s ( 0 )

− 2 NT [ D t β φ 1 ] = − 8 e γ s v β NT t β + 1 Γ ( β + 2 )

− 2 NT [ D t β φ 1 ] = − 8 e γ s v β v β + 1 s β + 2

− 2 NT [ D t β φ 1 ] = − 8 e γ s − 2 v − 1

⇒ φ 2 ( γ , t ) = NT − 1 v s 2 β − 8 e γ s − 2 v − 1

⇒ φ 2 ( γ , t ) = − 8 e γ NT − 1 v 2 β + 1 s 2 β + 2

⇒ φ 2 ( γ , t ) = − 8 e γ t 2 β + 1 Γ ( 2 β + 2 )

R [ φ 2 ] = ( φ 2 ) γ γ − φ 2 = 0 .

By using Eq. (13), we obtain:

φ 3 ( γ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 2 ] )

− 2 NT [ D t β φ 2 ] = − 2 s v β NT [ φ 2 ] − 1 s φ 2 ( 0 )

− 2 NT [ D t β φ 2 ] = − 2 s v β NT − 8 e γ t 2 β + 1 Γ ( 2 β + 2 ) − 1 s ( 0 )

− 2 NT [ D t β φ 2 ] = 16 e γ s v β NT t 2 β + 1 Γ ( 2 β + 2 )

− 2 NT [ D t β φ 2 ] = 16 e γ s v β v 2 β + 1 s 2 β + 2

− 2 NT [ D t β φ 2 ] = 16 e γ s − β − 2 v − β − 1

⇒ φ 3 ( γ , t ) = NT − 1 v s 2 β 16 e γ s − β − 2 v − β − 1

⇒ φ 3 ( γ , t ) = 16 e γ NT − 1 v 3 β + 1 s 3 β + 2

⇒ φ 3 ( γ , t ) = 16 e γ t 3 β + 1 Γ ( 3 β + 2 )

R [ φ 3 ] = ( φ 3 ) γ − φ 3 = 0 .

By using Eq. (13), we obtain:

φ 4 ( γ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 3 ] )

− 2 NT [ D t β φ 3 ] = − 2 s v β NT [ φ 3 ] − 1 s φ 3 ( 0 )

− 2 NT [ D t β φ 3 ] = − 2 s v β NT 16 e γ t 3 β + 1 Γ ( 3 β + 2 ) − 1 s ( 0 )

− 2 NT [ D t β φ 3 ] = − 32 e γ s v β NT t 3 β + 1 Γ ( 3 β + 2 )

− 2 NT [ D t β φ 3 ] = − 32 e γ s v β v 3 β + 1 s 3 β + 2

− 2 NT [ D t β φ 3 ] = − 32 e γ s − 2 β − 2 v − 2 β − 1

⇒ φ 4 ( γ , t ) = NT − 1 v s 2 β − 32 e γ s − 2 β − 2 v − 2 β − 1

⇒ φ 4 ( γ , t ) = − 32 e γ NT − 1 v 4 β + 1 s 4 β + 2

⇒ φ 4 ( γ , t ) = − 32 e γ t 4 β + 1 Γ ( 4 β + 2 )

R [ φ 4 ] = ( φ 4 ) γ γ − φ 4 = 0 .

By using Eq. (13), we obtain:

φ 5 ( γ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 4 ] )

− 2 NT [ D t β φ 4 ] = − 2 s v β NT [ φ 4 ] − 1 s φ 4 ( 0 )

− 2 NT [ D t β φ 4 ] = − 2 s v β NT − 32 e γ t 4 β + 1 Γ ( 4 β + 2 ) − 1 s ( 0 )

− 2 NT [ D t β φ 4 ] = 64 e γ s v β NT t 4 β + 1 Γ ( 4 β + 2 )

− 2 NT [ D t β φ 4 ] = 64 e γ s v β v 4 β + 1 s 4 β + 2

− 2 NT [ D t β φ 4 ] = 64 e γ s − 3 β − 2 v − 3 β − 1

⇒ φ 5 ( γ , t ) = NT − 1 v s 2 β 64 e γ s − 3 β − 2 v − 3 β − 1

⇒ φ 5 ( γ , t ) = 64 e γ NT − 1 v 5 β + 1 s 5 β + 2

⇒ φ 5 ( γ , t ) = 64 e γ t 5 β + 1 Γ ( 5 β + 2 )

φ ( γ , t ) = φ 0 ( γ , t ) + φ 1 ( γ , t ) + φ 2 ( γ , t ) + φ 3 ( γ , t ) + φ 4 ( γ , t ) + φ 5 ( γ , t ) + …

⇒ φ ( γ , t ) = e γ ( 1 − 2 t ) + 4 e γ t β + 1 Γ ( β + 2 ) − 8 e γ t 2 β + 1 Γ ( 2 β + 2 ) + 16 e γ t 3 β + 1 Γ ( 3 β + 2 ) − 32 e γ t 4 β + 1 Γ ( 4 β + 2 ) + 64 e γ t 5 β + 1 Γ ( 5 β + 2 ) − …

By substituting α = 1 , a series solution is obtained:

φ ( γ , t ) = e γ 1 − 2 t 1 ! + ( 2 t ) 2 2 ! − ( 2 t ) 3 3 ! + ( 2 t ) 4 4 ! − ( 2 t ) 5 5 ! + ( 2 t ) 6 6 ! − …

⇒ φ ( γ , t ) = e γ − 2 t .

Example 5. Considered following nonlinear 2D time-fractional telegraph equation [22]:

(14) D t 2 β φ + 3 D t β φ + 2 φ = φ γ γ + φ ρ ρ ,

where φ ( γ , ρ , 0 ) = e γ + ρ and φ t ( γ , ρ , 0 ) = − 3 e γ + ρ

φ 0 ( γ , ρ , t ) = φ ( γ , ρ , 0 ) + t φ t ( γ , ρ , 0 ) = e γ + ρ − 3 t ( e γ + ρ ) = e γ + ρ ( 1 − 3 t ) .

By applying natural transform on Eq. (14):

NT [ D t 2 β φ + 3 D t β φ + 2 φ ] = NT [ φ γ γ + φ ρ ρ ]

⇒ NT [ D t 2 β φ ] = NT [ ( φ γ γ + φ ρ ρ − 2 φ ) − 3 D t β φ ]

⇒ NT [ D t 2 β φ ] = NT [ R [ φ ] − 3 D t β φ ]

⇒ φ ( γ , ρ , t ) = NT − 1 1 s φ ( 0 ) + NT − 1 v s 2 β ( NT [ R [ φ ] − 3 D t β φ ] )

(15) ⇒ φ 0 ( γ , ρ , t ) = NT − 1 1 s φ ( 0 ) ,

⇒ φ 0 ( γ , ρ , t ) = φ ( 0 ) NT − 1 1 s

⇒ φ 0 ( γ , ρ , t ) = φ ( 0 ) = e γ + ρ ( 1 − 3 t )

(16) φ r + 1 ( γ , ρ , t ) = NT − 1 v s 2 β ( NT [ R [ φ r ] − 3 D t β φ r ] ) ,

r = 0 , 1 , 2 , 3 , …

R [ φ 0 ] = ( φ 0 ) γ γ + ( φ 0 ) ρ ρ − 2 φ 0 = 0 .

By using Eq. (16), we obtain:

φ 1 ( γ , ρ , t ) = NT − 1 v s 2 β ( − 3 NT [ D t β φ 0 ] )

− 3 NT [ D t β φ 0 ] = − 3 s v β NT [ φ 0 ] − 1 s φ 0 ( 0 )

− 3 NT [ D t β φ 0 ] = − 3 s v β NT [ e γ + ρ ( 1 − 3 t ) ] − 1 s e x + y

− 3 NT [ D t β φ 0 ] = − 3 e γ + ρ s v β NT [ 1 ] − 3 NT [ t ] − 1 s

− 3 NT [ D t β φ 0 ] = − 3 e γ + ρ s v β 1 s − 3 v s 2 − 1 s

− 3 NT [ D t β φ 0 ] = 9 e γ + ρ s β − 2 v β − 1

⇒ φ 1 ( γ , ρ , t ) = NT − 1 v s 2 β 9 e γ + ρ s β − 2 v β − 1

⇒ φ 1 ( γ , ρ , t ) = 9 e γ + ρ NT − 1 v β + 1 s β + 2

⇒ φ 1 ( γ , ρ , t ) = 9 e γ + ρ t β + 1 Γ ( β + 2 )

R [ φ 1 ] = ( φ 1 ) γ γ + ( φ 1 ) ρ ρ − 2 φ 1 = 0 .

By using Eq. (16), we obtain:

φ 2 ( γ , ρ , t ) = NT − 1 v s 2 β ( − 3 NT [ D t β φ 1 ] )

− 3 NT [ D t β φ 1 ] = − 3 s v β NT [ φ 1 ] − 1 s φ 1 ( 0 )

− 3 NT [ D t β φ 1 ] = − 3 s v β NT 9 e γ + ρ t β + 1 Γ ( β + 2 ) − 1 s ( 0 )

− 3 NT [ D t β φ 1 ] = − 27 e γ + ρ s v β NT t β + 1 Γ ( β + 2 )

− 3 NT [ D t β φ 1 ] = − 27 e γ + ρ s v β v β + 1 s β + 2

− 3 NT [ D t β φ 1 ] = − 27 e γ + ρ s − 2 v − 1

⇒ φ 2 ( γ , ρ , t ) = NT − 1 v s 2 β − 27 e γ + ρ s − 2 v − 1

⇒ φ 2 ( γ , ρ , t ) = − 27 e γ + ρ NT − 1 v 2 β + 1 s 2 β + 2

⇒ φ 2 ( γ , ρ , t ) = − 27 e γ + ρ t 2 β + 1 Γ ( 2 β + 2 )

R [ φ 2 ] = ( φ 2 ) γ γ + ( φ 2 ) ρ ρ − 2 φ 2 = 0 .

By using Eq. (16), we obtain:

φ 3 ( γ , ρ , t ) = NT − 1 v s 2 β ( − 3 NT [ D t β φ 2 ] )

− 3 NT [ D t β φ 2 ] = − 3 s v β NT [ φ 2 ] − 1 s φ 2 ( 0 )

− 3 NT [ D t β φ 2 ] = − 3 s v β NT − 27 e γ + ρ t 2 β + 1 Γ ( 2 β + 2 ) − 1 s ( 0 )

− 3 NT [ D t β φ 2 ] = 81 e γ + ρ s v β NT t 2 β + 1 Γ ( 2 β + 2 )

− 3 NT [ D t β φ 2 ] = 81 e γ + ρ s v β v 2 β + 1 s 2 β + 2

− 3 NT [ D t β φ 2 ] = 81 e γ + ρ s − β − 2 v − β − 1

⇒ φ 3 ( γ , ρ , t ) = NT − 1 v s 2 β 81 e γ + ρ s − β − 2 v − β − 1

⇒ φ 3 ( γ , ρ , t ) = 81 e γ + ρ NT − 1 v 3 β + 1 s 3 β + 2

⇒ φ 3 ( γ , ρ , t ) = 81 e γ + ρ t 3 β + 1 Γ ( 3 β + 2 )

R [ φ 3 ] = ( φ 3 ) γ γ + ( φ 3 ) ρ ρ − 2 φ 3 = 0 .

By using Eq. (16), we obtain:

φ 4 ( γ , ρ , t ) = NT − 1 v s 2 β ( − 3 NT [ D t β φ 3 ] )

− 3 NT [ D t β φ 3 ] = − 3 s v β NT [ φ 3 ] − 1 s φ 3 ( 0 )

− 3 NT [ D t β φ 3 ] = − 3 s v β NT 81 e γ + ρ t 3 β + 1 Γ ( 3 β + 2 ) − 1 s ( 0 )

− 3 NT [ D t β φ 3 ] = − 243 e γ + ρ s v β NT t 3 β + 1 Γ ( 3 β + 2 )

− 3 NT [ D t β φ 3 ] = − 243 e γ + ρ s v β v 3 β + 1 s 3 β + 2

− 3 NT [ D t β φ 3 ] = − 243 e γ + ρ s − 2 β − 2 v − 2 β − 1

⇒ φ 4 ( γ , ρ , t ) = NT − 1 v s 2 β − 243 e γ + ρ s − 2 β − 2 v − 2 β − 1

⇒ φ 4 ( γ , ρ , t ) = − 243 e γ + ρ NT − 1 v 4 β + 1 s 4 β + 2

⇒ φ 4 ( γ , ρ , t ) = − 243 e γ + ρ t 4 β + 1 Γ ( 4 β + 2 )

R [ φ 4 ] = ( φ 4 ) γ γ + ( φ 4 ) ρ ρ − 2 φ 4 = 0 .

By using Eq. (16), we obtain:

φ 5 ( γ , ρ , t ) = NT − 1 v s 2 β ( − 3 NT [ D t β φ 4 ] )

− 3 NT [ D t β φ 4 ] = − 3 s v β NT [ φ 4 ] − 1 s φ 4 ( 0 )

− 3 NT [ D t β φ 4 ] = − 3 s v β NT − 243 e γ + ρ t 4 β + 1 Γ ( 4 β + 2 ) − 1 s ( 0 )

− 3 NT [ D t β φ 4 ] = 729 e γ + ρ s v β NT t 4 β + 1 Γ ( 4 β + 2 )

− 3 NT [ D t β φ 4 ] = 729 e γ + ρ s v β v 4 β + 1 s 4 β + 2

− 3 NT [ D t β φ 4 ] = 729 e γ + ρ s − 3 β − 2 v − 3 β − 1

⇒ φ 5 ( γ , ρ , t ) = NT − 1 v s 2 β 729 e γ + ρ s − 3 β − 2 v − 3 β − 1

⇒ φ 5 ( γ , ρ , t ) = 729 e γ + ρ NT − 1 v 5 β + 1 s 5 β + 2

⇒ φ 5 ( γ , ρ , t ) = 729 e γ + ρ t 5 β + 1 Γ ( 5 β + 2 )

φ ( γ , ρ , t ) = φ 0 ( γ , ρ , t ) + φ 1 ( γ , ρ , t ) + φ 2 ( γ , ρ , t ) + φ 3 ( γ , ρ , t ) + φ 4 ( γ , ρ , t ) + φ 5 ( γ , ρ , t ) + …

⇒ φ ( γ , ρ , t ) = e γ + ρ ( 1 − 2 t ) + 9 e γ + ρ t β + 1 Γ ( β + 2 ) − 27 e γ + ρ t 2 β + 1 Γ ( 2 β + 2 ) + 81 e γ + ρ t 3 β + 1 Γ ( 3 β + 2 ) − 243 e γ + ρ t 4 β + 1 Γ ( 4 β + 2 ) + 729 e γ + ρ t 5 β + 1 Γ ( 5 β + 2 ) − …

By substituting α = 1 , a series solution is obtained:

φ ( γ , ρ , t ) = e γ + ρ 1 − 3 t 1 ! + ( 3 t ) 2 2 ! − ( 3 t ) 3 3 ! + ( 3 t ) 4 4 ! − ( 3 t ) 5 5 ! + ( 3 t ) 6 6 ! − …

⇒ φ ( x , y , t ) = e γ + ρ − 3 t .

Example 6. Considered following nonlinear 3D time-fractional telegraph equation [22]:

(17) D t 2 β φ + 2 D t β φ + 3 φ = φ γ γ + φ ρ ρ + φ σ σ ,

where φ ( γ , ρ , σ , 0 ) = sinh γ sinh ρ sinh σ and φ t ( γ , ρ , σ , 0 ) = − sinh γ sinh ρ sinh σ , 0 < β ≤ 1 .

φ 0 ( γ , ρ , σ , t ) = φ ( γ , ρ , σ , 0 ) + t φ t ( γ , ρ , σ , 0 ) = sinh γ sinh ρ sinh σ − t ( sinh γ sinh ρ sinh σ ) = sinh γ sinh ρ sinh σ ( 1 − t ) .

By applying natural transform on Eq. (17), we obtain:

NT [ D t 2 β φ + 2 D t β φ + 3 φ ] = NT [ φ γ γ + φ ρ ρ + φ σ σ ]

⇒ NT [ D t 2 β φ ] = NT [ ( φ γ γ + φ ρ ρ + φ σ σ − 3 φ ) − 2 D t β φ ]

⇒ NT [ D t 2 β φ ] = NT [ R [ φ ] − 2 D t β φ ]

⇒ φ ( γ , ρ , σ , t ) = NT − 1 1 s φ ( 0 ) + NT − 1 v s 2 β ( NT [ R [ φ ] − 2 D t β φ ] )

(18) ⇒ φ 0 ( γ , ρ , σ , t ) = NT − 1 1 s φ ( 0 ) ,

⇒ φ 0 ( γ , ρ , σ , t ) = φ ( 0 ) NT − 1 1 s

⇒ φ 0 ( γ , ρ , σ , t ) = φ ( 0 ) = sinh γ sinh ρ sinh σ ( 1 − t )

(19) φ r + 1 ( γ , ρ , σ , t ) = NT − 1 v s 2 β ( NT [ R [ φ r ] − 2 D t β φ r ] ) ,

r = 0 , 1 , 2 , 3 , …

R [ φ 0 ] = ( φ 0 ) γ γ + ( φ 0 ) ρ ρ + ( φ 0 ) σ σ − 3 φ 0 = 0 .

By using Eq. (19), we obtain:

φ 1 ( γ , ρ , σ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 0 ] )

− 2 NT [ D t β φ 0 ] = − 2 s v β NT [ φ 0 ] − 1 s φ 0 ( 0 )

− 2 NT [ D t β φ 0 ] = − 2 s v β NT [ sinh γ sinh ρ sinh σ ( 1 − t ) ] − 1 s sinh γ sinh ρ sinh σ

− 2 NT [ D t β φ 0 ] = − 2 sinh γ sinh ρ sinh σ s v β NT [ 1 ] − NT [ t ] − 1 s

− 2 NT [ D t β φ 0 ] = − 2 sinh γ sinh ρ sinh σ s v β 1 s − 1 v s 2 − 1 s

− 2 NT [ D t β φ 0 ] = 2 sinh γ sinh ρ sinh σ s β − 2 v β − 1

⇒ φ 1 ( γ , ρ , σ , t ) = NT − 1 v s 2 β 2 sinh γ sinh ρ sinh σ s β − 2 v β − 1

⇒ φ 1 ( γ , ρ , σ , t ) = 2 sinh γ sinh ρ sinh σ NT − 1 v β + 1 s β + 2

⇒ φ 1 ( γ , ρ , σ , t ) = 2 sinh γ sinh ρ sinh σ t β + 1 Γ ( β + 2 )

R [ φ 1 ] = ( φ 1 ) γ γ + ( φ 1 ) ρ ρ + ( φ 1 ) σ σ − 3 φ 1 = 0 .

By using Eq. (19), we obtain:

φ 2 ( γ , ρ , σ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 1 ] )

− 2 NT [ D t β φ 1 ] = − 2 s v β NT [ φ 1 ] − 1 s φ 1 ( 0 )

− 2 NT [ D t β φ 1 ] = − 2 s v β NT 2 sinh γ sinh ρ sinh σ t β + 1 Γ ( β + 2 ) − 1 s ( 0 )

− 2 NT [ D t β φ 1 ] = − 4 sinh γ sinh ρ sinh σ s v β NT t β + 1 Γ ( β + 2 )

− 2 NT [ D t β φ 1 ] = − 4 sinh γ sinh ρ sinh σ s v β v β + 1 s β + 2

− 2 NT [ D t β φ 1 ] = − 4 sinh γ sinh ρ sinh σ s − 2 v − 1

⇒ φ 2 ( γ , ρ , σ , t ) = NT − 1 v s 2 β − 4 sinh γ sinh ρ sinh σ s − 2 v − 1

⇒ φ 2 ( γ , ρ , σ , t ) = − 4 sinh γ sinh ρ sinh σ NT − 1 v 2 β + 1 s 2 β + 2

⇒ φ 2 ( γ , ρ , σ , t ) = − 4 sinh γ sinh ρ sinh σ t 2 β + 1 Γ ( 2 β + 2 )

R [ φ 2 ] = ( φ 2 ) γ γ + ( φ 2 ) ρ ρ + ( φ 2 ) σ σ − 3 φ 2 = 0 .

By using Eq. (19), we obtain:

φ 3 ( γ , ρ , σ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 2 ] )

− 2 NT [ D t β φ 2 ] = − 2 s v β NT [ φ 2 ] − 1 s φ 2 ( 0 )

− 2 NT [ D t β φ 2 ] = − 2 s v β NT − 4 sinh γ sinh ρ sinh σ t 2 β + 1 Γ ( 2 β + 2 ) − 1 s ( 0 )

− 2 NT [ D t β φ 2 ] = 8 sinh γ sinh ρ sinh σ s v β NT t 2 β + 1 Γ ( 2 β + 2 )

− 2 NT [ D t β φ 2 ] = 8 sinh γ sinh ρ sinh σ s v β v 2 β + 1 s 2 β + 2

− 2 NT [ D t β φ 2 ] = 8 sinh γ sinh ρ sinh σ s − β − 2 v − β − 1

⇒ φ 3 ( γ , ρ , σ , t ) = NT − 1 v s 2 β 8 sinh γ sinh ρ sinh σ s − β − 2 v − β − 1

⇒ φ 3 ( γ , ρ , σ , t ) = 8 sinh γ sinh ρ sinh σ NT − 1 v 3 β + 1 s 3 β + 2

⇒ φ 3 ( γ , ρ , σ , t ) = 8 sinh γ sinh ρ sinh σ t 3 β + 1 Γ ( 3 β + 2 )

R [ φ 3 ] = ( φ 3 ) γ γ + ( φ 3 ) ρ ρ + ( φ 3 ) σ σ − 3 φ 3 = 0 .

By using Eq. (19), we obtain:

φ 4 ( γ , ρ , σ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 3 ] )

− 2 NT [ D t β φ 3 ] = − 2 s v β NT [ φ 3 ] − 1 s φ 3 ( 0 )

− 2 NT [ D t β φ 3 ] = − 2 s v β NT 8 sinh γ sinh ρ sinh σ t 3 β + 1 Γ ( 3 β + 2 ) − 1 s ( 0 )

− 2 NT [ D t β φ 3 ] = − 16 sinh γ sinh ρ sinh σ s v β NT t 3 β + 1 Γ ( 3 β + 2 )

− 2 NT [ D t β φ 3 ] = − 16 sinh γ sinh ρ sinh σ s v β v 3 β + 1 s 3 β + 2

− 2 NT [ D t β φ 3 ] = − 16 sinh γ sinh ρ sinh σ s − 2 β − 2 v − 2 β − 1

⇒ φ 4 ( γ , ρ , σ , t ) = NT − 1 v s 2 β − 16 sinh γ sinh ρ sinh σ s − 2 β − 2 v − 2 β − 1

⇒ φ 4 ( γ , ρ , σ , t ) = − 16 sinh γ sinh ρ sinh σ NT − 1 v 4 β + 1 s 4 β + 2

⇒ φ 4 ( γ , ρ , σ , t ) = − 16 sinh γ sinh ρ sinh σ t 4 β + 1 Γ ( 4 β + 2 )

R [ φ 4 ] = ( φ 4 ) γ γ + ( φ 4 ) ρ ρ + ( φ 4 ) σ σ − 3 φ 4 = 0 .

By using Eq. (19), we obtain:

φ 5 ( γ , ρ , σ , t ) = NT − 1 v s 2 β ( − 2 NT [ D t β φ 4 ] )

− 2 NT [ D t β φ 4 ] = − 2 s v β NT [ φ 4 ] − 1 s φ 4 ( 0 )

− 2 NT [ D t β φ 4 ] = − 2 s v β NT − 16 sinh γ sinh ρ sinh σ t 4 β + 1 Γ ( 4 β + 2 ) − 1 s ( 0 )

− 2 NT [ D t β φ 4 ] = 32 sinh γ sinh ρ sinh σ s v β NT t 4 β + 1 Γ ( 4 β + 2 )

− 2 NT [ D t β φ 4 ] = 32 sinh γ sinh ρ sinh σ s v β v 4 β + 1 s 4 β + 2

− 2 NT [ D t β φ 4 ] = 32 sinh γ sinh ρ sinh σ s − 3 β − 2 v − 3 β − 1

⇒ φ 5 ( γ , ρ , σ , t ) = NT − 1 v s 2 β 32 sinh γ sinh ρ sinh σ s − 3 β − 2 v − 3 β − 1

⇒ φ 5 ( γ , ρ , σ , t ) = 32 sinh γ sinh ρ sinh σ NT − 1 v 5 β + 1 s 5 β + 2

⇒ φ 5 ( γ , ρ , σ , t ) = 32 sinh γ sinh ρ sinh σ t 5 β + 1 Γ ( 5 β + 2 )

φ ( γ , ρ , σ , t ) = φ 0 ( γ , ρ , σ , t ) + φ 1 ( γ , ρ , σ , t ) + φ 2 ( γ , ρ , σ , t ) + φ 3 ( γ , ρ , σ , t ) + φ 4 ( γ , ρ , σ , t ) + φ 5 ( γ , ρ , σ , t ) + …

⇒ φ ( γ , ρ , σ , t ) = sinh γ sinh ρ sinh σ ( 1 − t ) + 2 sinh γ sinh ρ sinh σ t β + 1 Γ ( β + 2 ) − 4 sinh γ sinh ρ sinh σ t 2 β + 1 Γ ( 2 β + 2 ) + 8 sinh γ sinh ρ sinh σ t 3 β + 1 Γ ( 3 β + 2 ) − 16 sinh γ sinh ρ sinh σ t 4 β + 1 Γ ( 4 β + 2 ) + 32 sinh γ sinh ρ sinh σ t 5 β + 1 Γ ( 5 β + 2 ) − … .

By substituting α = 1 , a series solution is obtained:

φ ( γ , ρ , σ , t ) = sinh γ sinh ρ sinh σ 1 − t 1 ! + 2 t 2 2 ! − 4 t 3 3 ! + 8 t 4 4 ! − 16 t 5 5 ! + 32 t 6 6 ! − …

= sinh γ sinh ρ sinh σ 2 2 − 2 t 1 ! + 4 t 2 2 ! − 8 t 3 3 ! + 16 t 4 4 ! − 32 t 5 5 ! + 64 t 6 6 ! − …

= sinh γ sinh ρ sinh σ 2 1 + 1 − 2 t 1 ! + ( 2 t ) 2 2 ! − ( 2 t ) 3 3 ! + ( 2 t ) 4 4 ! − ( 2 t ) 5 5 ! + ( 2 t ) 6 6 ! − …

⇒ φ ( γ , ρ , σ , t ) = sinh γ sinh ρ sinh σ 2 ( 1 + e − 2 t ) .

4 Graphical and tabular discussion

Figure 1 compares approximate and exact findings for Example 1 at t = 1, 2, and 3. At t = 1, 2, and 3, it is discovered that the approximation and exact graphs are very compatible with one another. In Figure 2, approximate and exact findings for Example 2 are matched for t = 1, 2, and 3. At t = 1, 2, and 3, it is discovered that the approximation and exact graphs are coherent with one another. Figure 3 compares approximate and exact findings for Example 3 at t = 1, 2, and 3. At t = 1, 2, and 3, it is discovered that the approximation and exact graphs are identical with one another. Regarding Example 4, approximate and exact results are matched at t = 1, 2, and 3 in Figure 4. At t = 1, 2, and 3, it is observed that the approximation and exact graphs can coexist peacefully. It is to be noted that Figures 1 and 4 are the same due to same exact and approximate solutions in Examples 1 and 4. Regarding Example 5, the contour graphs of approximate and exact findings at t = 1 are shown in Figure 5. At t = 1, it is determined that the approximated and exact graphs are identical with one another. Figure 6 shows contour graphs of approximate and exact results for Example 5 at t = 2. At t = 2, it is established that the approximate and exact graphs coexist. Regarding Example 5, the contour graphs of approximate and exact findings at t = 3 are shown in Figure 7. At t = 3, it is discovered that the approximation and exact graphs are compatible with one another.

Figure 1 Examining the differences between approximate and exact findings for Example 1 at t t = 1, 2, and 3.
Figure 1

Examining the differences between approximate and exact findings for Example 1 at t = 1, 2, and 3.

Figure 2 Comparison of approximate and exact findings for Example 2 at t t = 1, 2, and 3.
Figure 2

Comparison of approximate and exact findings for Example 2 at t = 1, 2, and 3.

Figure 3 Comparison of approximate and exact findings for Example 3 at t t = 1, 2, and 3.
Figure 3

Comparison of approximate and exact findings for Example 3 at t = 1, 2, and 3.

Figure 4 Comparison of approximate and exact findings for Example 4 at t t = 1, 2, and 3.
Figure 4

Comparison of approximate and exact findings for Example 4 at t = 1, 2, and 3.

Figure 5 Comparison of findings for Example 5 at t t = 1 between approximate and exact outcomes.
Figure 5

Comparison of findings for Example 5 at t = 1 between approximate and exact outcomes.

Figure 6 A comparison of roughly and precise outcomes for Example 5 at t t = 2.
Figure 6

A comparison of roughly and precise outcomes for Example 5 at t = 2.

Figure 7 Comparison of findings for Example 5 at t t = 3 between approximate and exact outcomes.
Figure 7

Comparison of findings for Example 5 at t = 3 between approximate and exact outcomes.

We define N as the number of terms in the series solution, thus making it a parameter for convergence. Each graph considers N = 35 , and each table considers the same range for the variables as considered in the graphs of the respective examples. In Table 3, the absolute value of L ∞ error at t = 1.1 , 1.2 , and 1.3 regarding Examples 1 and 4 has been provided. As N is increased, the error reduces up to 10 − 15 . In Table 4, the absolute value of L ∞ error at t = 1.1 , 1.2 , and 1.3 regarding Example 2 has been provided. As N is increased to 30, the error is reduced up to 10 − 16 . In Table 5, the absolute value of L ∞ error at t = 1.1 , 1.2 , and 1.3 regarding Example 3 has been provided. As N is increased to 30, the error is reduced up to 10 − 16 . In Table 6, the absolute value of L ∞ error at t = 0.1 , 0.2 , and 0.3 regarding Example 2 has been provided. As N is increased to 20, the error is reduced up to 10 − 11 . It is confirmed that the suggested approach produces convergent solutions with a respectable order of convergence over a wide variety of time levels.

Table 3

Analysis of the errors in Examples 1 and 4

N L ∞ e rror
t = 1.1 t = 1.2 t = 1.3
10 3 . 35618 × 10 − 4 8 . 6161 × 10 − 4 2 . 049098 × 10 − 3
20 7 . 48457 × 10 − 13 4 . 62771 × 10 − 12 2 . 46437 × 10 − 11
30 1 . 97065 × 10 − 15 3 . 02536 × 10 − 15 2 . 7478 × 10 − 15
↓ ↓ ↓
Up to 10 − 15 Up to 10 − 15 Up to 10 − 15
Table 4

Analysis of the errors in Example 2

N L ∞ e rror
t = 1.1 t = 1.2 t = 1.3
10 1 . 7788 × 10 − 7 4 . 59681 × 10 − 7 1 . 10032 × 10 − 6
20 1 . 11022 × 10 − 15 6 . 66134 × 10 − 16 8 . 88178 × 10 − 16
30 1 . 11022 × 10 − 15 6 . 66134 × 10 − 16 8 . 88178 × 10 − 16
↓ ↓ ↓
Up to 10 − 15 Up to 10 − 16 Up to 10 − 16
Table 5

Analysis of the errors in Example 3

N L ∞ e rror
t = 1.1 t = 1.2 t = 1.3
10 7 . 863 × 10 − 8 2 . 06631 × 10 − 7 5 . 02971 × 10 − 7
20 4 . 44089 × 10 − 16 8 . 88178 × 10 − 16 8 . 88178 × 10 − 16
30 4 . 44089 × 10 − 16 8 . 88178 × 10 − 16 8 . 88178 × 10 − 16
↓ ↓ ↓
Up to 10 − 16 Up to 10 − 16 Up to 10 − 16
Table 6

Analysis of the errors in Example 5

N L ∞ e rror
t = 0.1 t = 0.2 t = 0.3
10 7 . 13044 × 10 − 9 1 . 40856 × 10 − 5 1 . 189793 × 10 − 3
15 7 . 27596 × 10 − 11 5 . 82077 × 10 − 11 1 . 32422 × 10 − 9
20 7 . 27596 × 10 − 11 5 . 82077 × 10 − 11 4 . 36557 × 10 − 11
↓ ↓ ↓
Up to 10 − 11 Up to 10 − 11 Up to 10 − 11

5 Conclusion

In the present article, the approximated and exact solutions of FHTEs in 1D, 2D, and 3D were obtained using an iterative approach using natural transform. Via graphical representation, it was observed that the approximate and exact graphs of Examples 1, 2 3, and 4 each taken at t = 1 , 2 , 3 were found to have good compatibility. Approximate and exact graphs of Example 5 taken at t = 0.1 , 0.2 , 0.3 were found to have good compatibility. Thus, it was claimed that approximate and exact solutions have good compatibility across a wide range of time levels. The absolute values of L ∞ errors were provided in tabular discussion, and convergence to the exact solution was observed. Examples 1 and 4 showed convergence up to 10 − 15 as N was increased. Example 2 showed convergence up to 10 − 16 as N was increased. Example 3 showed convergence up to 10 − 16 as N was increased. Example 5 showed convergence up to 10 − 11 as N was increased. It is thus claimed that the proposed approach is a useful method to tackle FHTEs and fractional differential equations in general.

  1. Funding information: The authors state no funding involved.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors have no conflict of interest.

  4. Data availability statement: All the data are included within the manuscript.

Appendix A

A.1 Implementation upon 2D time-fractional telegraph equation

The form of 2D time-fractional telegraph equation is as follows [22]:

D t β φ ( γ , ρ , t ) + L [ φ ( γ , ρ , t ) ] + N [ φ ( γ , ρ , t ) ] = q ( γ , ρ , t ) .

where L is linear operator and N is nonlinear operator. D t β φ ( γ , ρ , t ) is Caputo derivative of φ ( γ , ρ , t ) . By applying the natural transform on equation, we obtain:

NT [ D t β φ ( γ , ρ , t ) ] + NT [ L [ φ ( γ , ρ , t ) ] + N [ φ ( γ , ρ , t ) ] ] = NT [ q ( γ , ρ , t ) ]

⇒ s v β NT [ φ ( γ , ρ , t ) ] − 1 s φ ( 0 ) = NT [ q ( γ , ρ , t ) ] − NT [ L [ φ ( γ , ρ , t ) ] ] − NT [ N [ φ ( γ , ρ , t ) ] ]

⇒ NT [ φ ( γ , ρ , t ) ] = 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , t ) ] − v s β ( NT [ L [ φ ( γ , ρ , t ) ] ] + NT [ N [ φ ( γ , ρ , t ) ] ] )

⇒ φ ( γ , ρ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , t ) ] − NT − 1 v s β ( NT [ L [ φ ( γ , ρ , t ) ] ] + NT [ N [ φ ( γ , ρ , t ) ] ] ) .

Now,

L [ φ ] = L ∑ r = 0 ∞ φ r ( γ , ρ , t ) = L [ φ 0 ( γ , ρ , t ) ] + ∑ r = 1 ∞ L ∑ i = 0 r φ i ( γ , ρ , t ) − L ∑ i = 0 r − 1 φ i ( γ , ρ , t )

N [ φ ] = N ∑ r = 0 ∞ φ r ( γ , ρ , t ) = N [ φ 0 ( γ , ρ , t ) ] + ∑ r = 1 ∞ N ∑ i = 0 r φ i ( γ , ρ , t ) − N ∑ i = 0 r − 1 φ i ( γ , ρ , t ) .

Therefore,

∑ k = 0 ∞ φ k ( γ , ρ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , t ) ]

− NT − 1 v s β L [ φ 0 ( γ , ρ , t ) ] + N [ φ 0 ( γ , ρ , t ) ] + L ∑ r = 1 ∞ φ r ( γ , ρ , t ) + N ∑ r = 1 ∞ φ r ( γ , ρ , t )

⇒ ∑ k = 0 ∞ φ k ( γ , ρ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , t ) ]

− NT − 1 v s β ( L [ φ 0 ( γ , ρ , t ) ] + N [ φ 0 ( γ , ρ , t ) ] )

− NT − 1 v s β ∑ r = 1 ∞ L [ φ r ( γ , ρ , t ) ] + ∑ r = 1 ∞ N ∑ i = 0 r φ i ( γ , ρ , t ) − N ∑ i = 0 r − 1 φ i ( γ , ρ , t )

⇒ φ 0 ( γ , ρ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , t ) ]

⇒ φ 1 ( γ , ρ , t ) = − NT − 1 v s β ( L [ φ 0 ( γ , ρ , t ) ] + N [ φ 0 ( γ , ρ , t ) ] )

⇒ φ r + 1 ( γ , ρ , t ) = − NT − 1 v s β L [ φ r ( γ , ρ , t ) ] + N ∑ i = 0 r φ i ( γ , ρ , t ) − ∑ i = 0 r − 1 φ i ( γ , ρ , t )

r = 1 , 2 , 3 , 4 , 5 , …

Appendix B

A.2 Implementation upon 3D time-fractional telegraph equation

The form of 3D time-fractional telegraph equation is as follows [22]:

D t β φ ( γ , ρ , σ , t ) + L [ φ ( γ , ρ , σ , t ) ] + N [ φ ( γ , ρ , σ , t ) ] = q ( γ , ρ , σ , t ) .

where L is a linear operator and N is a nonlinear operator. D t β φ ( γ , ρ , σ , t ) is Caputo derivative of φ ( x , y , z , t ) . By applying natural transform on the equation:

NT [ D t β φ ( γ , ρ , σ , t ) ] + NT [ L [ φ ( γ , ρ , σ , t ) ] + N [ φ ( γ , ρ , σ , t ) ] ] = NT [ q ( γ , ρ , σ , t ) ]

⇒ s v β NT [ φ ( γ , ρ , σ , t ) ] − 1 s φ ( 0 ) = NT [ q ( γ , ρ , σ , t ) ] − NT [ L [ φ ( γ , ρ , σ , t ) ] ] − NT [ N [ φ ( γ , ρ , σ , t ) ] ]

⇒ NT [ φ ( γ , ρ , σ , t ) ] = 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , σ , t ) ] − v s β ( NT [ L [ φ ( γ , ρ , σ , t ) ] ] + NT [ N [ φ ( γ , ρ , σ , t ) ] ] )

⇒ φ ( γ , ρ , σ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , σ , t ) ] − NT − 1 v s β ( NT [ L [ φ ( γ , ρ , σ , t ) ] ] + NT [ N [ φ ( γ , ρ , σ , t ) ] ] ) .

Now,

L [ φ ] = L ∑ r = 0 ∞ φ r ( γ , ρ , σ , t ) = L [ φ 0 ( γ , ρ , σ , t ) ] + ∑ r = 1 ∞ L ∑ i = 0 r φ i ( γ , ρ , σ , t ) − L ∑ i = 0 r − 1 φ i ( γ , ρ , σ , t )

N [ φ ] = N ∑ r = 0 ∞ φ r ( γ , ρ , σ , t ) = N [ φ 0 ( γ , ρ , σ , t ) ] + ∑ r = 1 ∞ N ∑ i = 0 r φ i ( γ , ρ , σ , t ) − N ∑ i = 0 r − 1 φ i ( γ , ρ , σ , t ) .

Therefore,

∑ k = 0 ∞ φ k ( γ , ρ , σ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , σ , t ) ]

− NT − 1 v s β L [ φ 0 ( γ , ρ , σ , t ) ] + N [ φ 0 ( γ , ρ , σ , t ) ] + L ∑ r = 1 ∞ φ r ( γ , ρ , σ , t ) + N ∑ r = 1 ∞ φ r ( γ , ρ , σ , t )

⇒ ∑ k = 0 ∞ φ k ( γ , ρ , σ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , σ , t ) ]

− NT − 1 v s β ( L [ φ 0 ( γ , ρ , σ , t ) ] + N [ φ 0 ( γ , ρ , σ , t ) ] )

− NT − 1 v s β ∑ r = 1 ∞ L [ φ r ( γ , ρ , σ , t ) ] + ∑ r = 1 ∞ N ∑ i = 0 r φ i ( γ , ρ , σ , t ) − N ∑ i = 0 r − 1 φ i ( γ , ρ , σ , t )

⇒ φ 0 ( γ , ρ , σ , t ) = NT − 1 1 s φ ( 0 ) + v s β NT [ q ( γ , ρ , σ , t ) ]

⇒ φ 1 ( γ , ρ , σ , t ) = − NT − 1 v s β ( L [ φ 0 ( γ , ρ , σ , t ) ] + N [ φ 0 ( γ , ρ , σ , t ) ] )

⇒ φ r + 1 ( γ , ρ , σ , t ) = − NT − 1 v s β L [ φ r ( γ , ρ , σ , t ) ] + N ∑ i = 0 r φ i ( γ , ρ , σ , t ) − ∑ i = 0 r − 1 φ i ( γ , ρ , σ , t )

r = 1 , 2 , 3 , 4 , 5 , …

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Received: 2023-02-13
Revised: 2023-03-08
Accepted: 2023-03-12
Published Online: 2023-04-28

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/nleng-2022-0289
Keywords for this article
natural transform; Caputo derivative; fractional telegraph equation; convergence analysis
Creative Commons
BY 4.0

Articles in the same Issue

  1. Research Articles
  2. The regularization of spectral methods for hyperbolic Volterra integrodifferential equations with fractional power elliptic operator
  3. Analytical and numerical study for the generalized q-deformed sinh-Gordon equation
  4. Dynamics and attitude control of space-based synthetic aperture radar
  5. A new optimal multistep optimal homotopy asymptotic method to solve nonlinear system of two biological species
  6. Dynamical aspects of transient electro-osmotic flow of Burgers' fluid with zeta potential in cylindrical tube
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  8. Overlapping grid SQLM for third-grade modified nanofluid flow deformed by porous stretchable/shrinkable Riga plate
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  13. Bond performance between recycled concrete and steel bar after high temperature
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  21. Cilia and electroosmosis induced double diffusive transport of hybrid nanofluids through microchannel and entropy analysis
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  25. Pricing weather derivatives in an uncertain environment
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  27. Novel soliton structures of truncated M-fractional (4+1)-dim Fokas wave model
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  29. Derivation of septic B-spline function in n-dimensional to solve n-dimensional partial differential equations
  30. Development of a gray box system identification model to estimate the parameters affecting traffic accidents
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  34. NASA DART mission: A preliminary mathematical dynamical model and its nonlinear circuit emulation
  35. Nonlinear dynamic responses of ballasted railway tracks using concrete sleepers incorporated with reinforced fibres and pre-treated crumb rubber
  36. Two-component excitation governance of giant wave clusters with the partially nonlocal nonlinearity
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  38. Engineering fault intelligent monitoring system based on Internet of Things and GIS
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  53. Research on optimization of crane fault predictive control system based on data mining
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  69. Study of time-fractional delayed differential equations via new integral transform-based variation iteration technique
  70. Exhaustive study on post effect processing of 3D image based on nonlinear digital watermarking algorithm
  71. A versatile dynamic noise control framework based on computer simulation and modeling
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