On the solution of generalized time-fractional telegraphic equation

Abstract

In this article, we have introduced a nonlinear extension of the generalized time-fractional telegraph equation (TFTE). Further, an analytical solution to this equation has been obtained by using the Laplace homotopy perturbation method (LHPM). Some known results are also discussed as special cases which supports the strength and viability of the work.

Keywords:

• Diffusion equation
• Laplace homotopy perturbation method
• time-fractional telegraphic equation

Mathematics Subject Classifications:

• 34A34
• 65M06
• 26A33

1. Introduction

Fractional calculus has gained popularity in the last few decades because of the non-local behaviour from the time-fractional derivative involved in fractional partial differential equations. This non-local behaviour helps fractional differential equations to provide a more accurate model for the natural physical phenomenon, like earthquake modelling, traffic flow model and properties of viscoelastic materials then the differential equations with ordinary derivatives. We refer the reader to see the work [1–10] and references therein to know about the recent contribution of the fractional calculus in the literature. The detailed literature review of fractional calculus is provided by Podlubny [11] and Kilbas et al. [12].

Hyperbolic partial differential equations have wide applications in various fields. The telegraph equation is a particular hyperbolic partial differential equation derived from the study of signal analysis for transmission and electrical signals. It is also used for modelling of the reaction-diffusion [13] equation. The classical telegraph equation explains well the common transmission phenomenon. Still, it fails to explain abnormal diffusion like finite long transmission due to the current–voltage wave. The telegraph equation of fractional order describes such process very well. The fractional telegraph equation helps to understand the diffusion in blood flow better [14].

Due to its wide range of applications, the fractional telegraph equation was studied by many researchers (see [15–17]). Zhao and Li studied the numerical solution of the time–space fractional order telegraph equation in [18]. Kumar et al. [19,20] used analytical and numerical techniques to solve this equation. Atangana [21] worked on the stability and convergence of the time-fractional variable order telegraph equation. Huang [22] derived the analytical solution for three fundamental problems of the time-fractional telegraph equation. Wang et al. [23] used reproducing kernel for solving a class of time-fractional telegraph equations with initial value conditions. Ferreira et al. [24] obtained the first and second fundamental solutions of the time-fractional telegraph equation with Laplace or Dirac operators. Chaurasia and Dubey [25] attempted to study a more general time-fractional telegraph equation given below. (1) $a_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )=\frac{{\mathrm{\partial }}^2\xi (x,\tau )}{\mathrm{\partial }{x}^2}+h(x,\tau ),$(1) where $D_{\tau }^{\beta }$ represents the Caputo fractional derivative of order β (see [12,26]), $1/n<\beta \le 1,x$ is the space coordinate and τ is the time coordinate, $\xi (x,\tau )$ represents the cell density and $h(x,\tau )$ is the force term.

Fujita [27–29] considered the following nonlinear diffusion-like equation: $\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }\tau }=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(f(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)$where $f(\xi )$ is the diffusion term.

He studied the cases $f(\xi )=\frac{1}{1-\lambda \xi },\frac{1}{(1-\lambda \xi {)}^2},\frac{1}{1+2a\xi +b{\xi }^2}$, respectively, where $\lambda ,a,b$ are arbitrary constants. Recently, this idea has been extended in [1–3], where $f(\xi )$ is an analytic function. For example, in [3], the following generalized fractional diffusion equation has been considered: $D_{\tau }^{\beta }\xi (x,\tau )=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(f(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right),$where $f(\xi )$ is a function which can be approximated by a finite degree polynomial. Also, the classical Burger equation ${\xi }_{\tau }(x,\tau )+ϵ\xi (x,\tau ){\xi }_x(x,\tau )+\eta {\xi }_{xx}(x,\tau )=0,$has been generalized to the following fractional Burgers equation: $D_{\tau }^{\beta }\xi (\overrightarrow{x},\tau )+ϵf(\xi )\sum _{i=1}^n{D}_{x_i}^{\delta }\xi (\overrightarrow{x},\tau )=\eta \sum _{i=1}^n{D}_{x_i}^{2\delta }\xi (\overrightarrow{x},\tau )+\mathbf{G}(\overrightarrow{x},\tau ),$where $\mathbf{G}(\overrightarrow{x},\tau )$ is the force term, $\overrightarrow{\xi }=({\xi }_1,{\xi }_2,\dots ,{\xi }_n)\in {\mathbb{R}}^n$ is arbitrary, $D_{\tau }^{\beta }$ denotes the Caputo fractional derivative with respect to τ of the order $\beta \in (0,1]$ and $D_{{\xi }_i}^{\delta }$ denotes the Caputo fractional derivative with respect to ${\xi }_i$ for all $i=1,2,3,\dots$ of the order $\delta \in (\frac{1}{2},1]$.

$f(\xi )={\xi }^M,M\in \mathbb{N}$ is considered in [1] and $f(\xi )$ an analytic function function that can be approximated by a finite-degree polynomial is considered in [2].

The above ideas motivate us to consider the generalization of the TFTE on the right side of Equation (1(1) $a_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )=\frac{{\mathrm{\partial }}^2\xi (x,\tau )}{\mathrm{\partial }{x}^2}+h(x,\tau ),$(1) ) using the approach applied in [1–3] (2) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(f(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ),\end{array}$(2) where $f(\xi )$ represents a class of functions that can be approximated by a polynomial.

The nonlinear terms in a nonlinear differential equation make it difficult to solve. Analytical solutions to non-linear models are very difficult to solve by the usual way. Therefore, it is nice to have an approximate solution to these problems up to a desired accuracy. He [30] has used the homotopy perturbation method (HPM) to solve nonlinear problems. In the original perturbation technique, we need to consider a small parameter $p\in [0,1]$, but HPM does not require a small parameter. So, it provides an approximate solutions to broad range of nonlinear problems in applied science, and thus provides an approximate analytical solution. In most cases, only a few iterations are required to obtain the most accurate solutions.

A Laplace transform converts a partial differential equation into an ordinary one, making it easy to solve. Therefore, HPM combined with the Laplace transform has proven to handle many nonlinear problems more effectively (see [31,32] and others). We use the Laplace Homotopy Perturbation Method (LHPM) to solve general nonlinear time-fractional telegraph-type equations of order $n\beta$. We apply the LHPM to solve Equation (2(2) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(f(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ),\end{array}$(2) ) along with the initial condition (3) $\xi (x,0)=g_0(x),\frac{{\mathrm{\partial }}^k\xi (x,0)}{\mathrm{\partial }{t}^k}=g_k(x)\mathrm{\forall }k=1,2,\dots ,n,$(3) where $g_k(x)$ are given functions.

The organization of the paper is as follows. Section 1 is introductory, where we have given literature overview of the topic given along with the proposed problem. In Section 2, basic definitions are included which are related and are used in this paper. In Section 3, the Laplace homotopy perturbation method is applied on the considered nonlinear problem. In Section 4, we illustrate the method with some examples and special cases. The last section is the conclusion of the paper.

2. Basic definitions and preliminaries

In 1903, Mittag-Leffler generalized the exponential function to the Mittag-Leffler function $E_{\alpha }$ which is a complex function. It is defined by

Definition 2.1

$E_{\alpha }(z)=\sum _{k=0}^{\mathrm{\infty }}\frac{z^k}{\mathrm{\Gamma }(\alpha k+1)};\alpha \in \mathbb{C},\mathrm{\Re }(\alpha )>0.$

It was further generalized to the Mittag–Leffler function of two parameters and three parameters, denoted by $E_{\alpha ,\beta }$ and $E_{\alpha ,\beta }^{\gamma }$, respectively.

Definition 2.2

$E_{\alpha ,\beta }(z)=\sum _{k=0}^{\mathrm{\infty }}\frac{z^k}{\mathrm{\Gamma }(\alpha k+\beta )};\alpha ,\beta \in \mathbb{C},\mathrm{\Re }(\alpha )>0,\mathrm{\Re }(\beta )>0.$

Definition 2.3

$E_{\alpha ,\beta }^{\gamma }(z)=\sum _{k=0}^{\mathrm{\infty }}\frac{(\gamma {)}_k{z}^k}{\mathrm{\Gamma }(\alpha k+\beta )k!};\alpha ,\beta ,\gamma \in \mathbb{C},\mathrm{\Re }(\alpha )>0,\mathrm{\Re }(\beta )>0,$where $(\gamma {)}_k=\mathrm{\Gamma }(k+\gamma )/\mathrm{\Gamma }(\gamma )$.

Definition 2.4

The Laplace transform of a function $\xi (x,\tau )$ with respect to the variable t is defined by $\mathcal{L}[\xi (x,\tau )]={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-s\tau }\xi (x,\tau )\mathrm{d}t,\tau >0,x\in \mathbb{R},\mathrm{\Re }(s)>0.$

Lemma 2.1

The Laplace transform of the generalized Mittag–Leffler function for $\mathrm{\Re }(\beta )>1,\mathrm{\Re }(\alpha )>0,\mathrm{\Re }(\gamma )\ge 0$ with respect to the variable t is given by

• $\mathcal{L}[{\tau }^{\beta -1}{E}_{\alpha ,\beta }(-\lambda {\tau }^{\alpha })]=\frac{s^{\alpha -\beta }}{s^{\alpha }+\lambda }$,

• $\mathcal{L}[{\tau }^{\beta -1}{E}_{\alpha ,\beta }^{\gamma }(-\lambda {\tau }^{\alpha })]=\frac{s^{\alpha \gamma -\beta }}{(s^{\alpha }+\lambda {)}^{\gamma }}$.

Definition 2.5

For $m-1<\beta <m$, the Laplace transform of the Caputo fractional derivative is given by $\mathcal{L}\left\{{\mathcal{D}}_t^{\beta }\xi (x,\tau )\right\}(p)=\frac{1}{p^{m-\beta }}[p^m\mathcal{L}\{\xi (x,\tau )\}(p)-p^{m-1}\xi (x,0)-\cdots -{\xi }^{(m-1)}(x,0)].$

In particular, for $0<\beta \le 1$, we have $\mathcal{L}\left\{{\mathcal{D}}_t^{\beta }\xi (x,\tau )\right\}(p)=p^{\beta }\mathcal{L}\{\xi (x,\tau )\}(p)-p^{\beta -1}\xi (x,0).$

3. Derivation of the solution by LHPM

Let us assume that we can approximate the function f by a suitable nth degree polynomial P (4) $f(\xi )\approx P(\xi )=c_0+c_1\xi +c_2{\xi }^2+\cdots +c_n{\xi }^n,$(4) where $c_0,c_1,\dots ,c_n$ are constants. Substituting (4(4) $f(\xi )\approx P(\xi )=c_0+c_1\xi +c_2{\xi }^2+\cdots +c_n{\xi }^n,$(4) ) into differential equation (2(2) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(f(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ),\end{array}$(2) ), we have (5) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ).\end{array}$(5) Applying the Laplace Transform in the above equation, we obtain $\begin{array}{rl}& a_1(s^{\beta }\tilde{\xi }(x,s)-s^{\beta -1}\xi (x,0))+a_2(s^{2\beta }\tilde{\xi }(x,s)-s^{2\beta -1}\xi (x,0)-s^{2\beta -2}{\xi }^{\prime }(x,0))\\ & +a_3(s^{3\beta }\tilde{\xi }(x,s)-s^{3\beta -1}\xi (x,0)-s^{3\beta -2}{\xi }^{\prime }(x,0)-s^{3\beta -3}{\xi }^{{}^{″}}(x,0))+\cdots \\ & +a_n(s^{n\beta }\tilde{\xi }(x,s)-s^{n\beta -1}\xi (x,0)-s^{n\beta -2}{\xi }^{\prime }(x,0)-\cdots -s^{n\beta -n}{\xi }^{(n-1)}(x,0))\\ & =\mathcal{L}[\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)+h(x,\tau )].\end{array}$We rearrange the terms to get an expression for $\tilde{\xi }$ (6) $\begin{array}{rl}\tilde{\xi }(x,s)& =\frac{(a_1{s}^{\beta -1}+a_2{s}^{2\beta -1}+\cdots +a_n{s}^{n\beta -1})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\xi (x,0)\\ & +\frac{(a_2{s}^{2\beta -2}+a_3{s}^{3\beta -2}+\cdots +a_n{s}^{n\beta -2})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{\prime }(x,0)+\cdots \\ & +\frac{a_n{s}^{n\beta -n}}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{(n-1)}(x,0)\\ & +\frac{1}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\mathcal{L}[\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)+h(x,\tau )].\end{array}$(6) Let us denote $H(x,s)=\mathcal{L}\{h(x,\tau )\}$, then for $h\in [0,1]$, we construct the homotopy $\mathrm{\Xi }(x,s;h)$ for the above Equation (6(6) $\begin{array}{rl}\tilde{\xi }(x,s)& =\frac{(a_1{s}^{\beta -1}+a_2{s}^{2\beta -1}+\cdots +a_n{s}^{n\beta -1})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\xi (x,0)\\ & +\frac{(a_2{s}^{2\beta -2}+a_3{s}^{3\beta -2}+\cdots +a_n{s}^{n\beta -2})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{\prime }(x,0)+\cdots \\ & +\frac{a_n{s}^{n\beta -n}}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{(n-1)}(x,0)\\ & +\frac{1}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\mathcal{L}[\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)+h(x,\tau )].\end{array}$(6) ) as follows: (7) $\begin{array}{rl}\tilde{\mathrm{\Xi }}(x,s;h)& =\frac{1}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}H(x,s)\\ & +\frac{(a_1{s}^{\beta -1}+a_2{s}^{2\beta -1}+\cdots +a_n{s}^{n\beta -1})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\xi (x,0)\\ & +\frac{(a_2{s}^{2\beta -2}+a_3{s}^{3\beta -2}+\cdots +a_n{s}^{n\beta -2})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{\prime }(x,0)+\cdots \\ & +\frac{a_n{s}^{n\beta -n}}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{(n-1)}(x,0)\\ & +\frac{h}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\mathcal{L}\left[\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\mathrm{\Xi })\frac{\mathrm{\partial }\mathrm{\Xi }(x,\tau )}{\mathrm{\partial }x}\right)\right].\end{array}$(7) Let us assume that (8) $\mathrm{\Xi }(x,\tau ;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k(x,\tau ),\tilde{\mathrm{\Xi }}(x,s;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\tilde{\mathrm{\Xi }}}_k(x,s).$(8) Substituting (8(8) $\mathrm{\Xi }(x,\tau ;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k(x,\tau ),\tilde{\mathrm{\Xi }}(x,s;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\tilde{\mathrm{\Xi }}}_k(x,s).$(8) ) in the differential equation (7(7) $\begin{array}{rl}\tilde{\mathrm{\Xi }}(x,s;h)& =\frac{1}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}H(x,s)\\ & +\frac{(a_1{s}^{\beta -1}+a_2{s}^{2\beta -1}+\cdots +a_n{s}^{n\beta -1})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\xi (x,0)\\ & +\frac{(a_2{s}^{2\beta -2}+a_3{s}^{3\beta -2}+\cdots +a_n{s}^{n\beta -2})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{\prime }(x,0)+\cdots \\ & +\frac{a_n{s}^{n\beta -n}}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{(n-1)}(x,0)\\ & +\frac{h}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\mathcal{L}\left[\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\mathrm{\Xi })\frac{\mathrm{\partial }\mathrm{\Xi }(x,\tau )}{\mathrm{\partial }x}\right)\right].\end{array}$(7) ), (9) $\begin{array}{rl}\sum _{k=0}^{\mathrm{\infty }}{h}^k{\tilde{\mathrm{\Xi }}}_k(x,s)& =\frac{h\mathcal{L}\left[\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P\left(\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k\right)\frac{\mathrm{\partial }\left(\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k\right)}{\mathrm{\partial }x}\right)\right]}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\\ & +\frac{1}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}H(x,s)+\frac{\xi (x,0)}{s}\\ & +\frac{(a_2{s}^{2\beta -2}+a_3{s}^{3\beta -2}+\cdots +a_n{s}^{n\beta -2})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{\prime }(x,0)+\cdots \\ & +\frac{a_n{s}^{.n\beta -n}}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{(n-1)}(x,0).\end{array}$(9) We expand the right side of Equation (5(5) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ).\end{array}$(5) ) as $\begin{array}{rl}& \frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)=P^{\prime }(\xi ){\left(\frac{\mathrm{\partial }\xi }{\mathrm{\partial }x}\right)}^2+P(\xi )\frac{{\mathrm{\partial }}^2\xi }{\mathrm{\partial }{x}^2}\\ & =(c_1+2c_2\xi +\cdots +n{c}_n{\xi }^{n-1}){\xi }_x^2+(c_0+c_1\xi +c_2{\xi }^2+\cdots +c_n{\xi }^n){\xi }_{xx}\\ & =c_1\sum _{k=0}^{\mathrm{\infty }}{h}^k\sum _{k_1+k_2=k}{\xi }_{k_{1x}}{\xi }_{k_{2x}}+2c_2\sum _{k=0}^{\mathrm{\infty }}{h}^k\sum _{k_1+k_2+k_3=k}{\xi }_{k_1}{\xi }_{k_{2x}}{\xi }_{k_{3x}}+3c_3\sum _{k=0}^{\mathrm{\infty }}{h}^k\\ & \times \sum _{k_1+k_2+k_3+k_4=k}{\xi }_{k_{1x}}{\xi }_{k_{2x}}{\xi }_{k_3}{\xi }_{k_4}+\cdots +n{c}_n\sum _{k=0}^{\mathrm{\infty }}{h}^k\sum _{k_1+\cdots +k_{n+1}=k}{\xi }_{k_{1x}}{\xi }_{k_{2x}}{\xi }_{k_3}\cdots {\xi }_{k_{n+1}}\\ & +c_0\sum _{k=0}^{\mathrm{\infty }}{h}^k{\xi }_{kxx}+c_1\sum _{k=0}^{\mathrm{\infty }}{h}^k\sum _{k_1+k_2=k}{\xi }_{k_1}{\xi }_{k_{2xx}}+c_2\sum _{k=0}^{\mathrm{\infty }}{h}^k\sum _{k_1+k_2+k_3=k}{\xi }_{k_1}{\xi }_{k_2}{\xi }_{k_{3xx}}\\ & +\cdots +c_n\sum _{k=0}^{\mathrm{\infty }}{h}^k\sum _{k_1+k_2+k_3+\cdots +k_{n+1}=k}{\xi }_{k_1}{\xi }_{k_2}\cdots {\xi }_{k_n}{\xi }_{k_{n+1}xx.}\end{array}$Now, comparing like powers of h, we get the following set of differential equations: $\begin{array}{rl}{h}^0:{\tilde{\mathrm{\Xi }}}_0(x,s)& =\frac{1}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}H(x,s)\\ & +\frac{\xi (x,0)}{s}+\frac{(a_2{s}^{2\beta -2}+a_3{s}^{3\beta -2}+\cdots +a_n{s}^{n\beta -2})}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{\prime }(x,0)+\cdots \\ & +\frac{a_n{s}^{n\beta -n}}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}{\xi }^{(n-1)}(x,0),\\ h^1:{\tilde{\mathrm{\Xi }}}_1(x,s)& =\frac{1}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\mathcal{L}[c_1{\mathrm{\Xi }}_{0x}^2+2c_2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0x}^2\\ & +3c_3{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0x}^2+\cdots +n{c}_n{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{0x}^2+c_0{\mathrm{\Xi }}_{0xx}+c_1{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0xx}\\ & +c_2{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0xx}+\cdots +c_n{\mathrm{\Xi }}_0^n{\mathrm{\Xi }}_{0xx}],\\ h^2:{\tilde{\mathrm{\Xi }}}_2(x,s)& =\frac{1}{(a_1{s}^{\beta }+a_2{s}^{2\beta }+\cdots +a_n{s}^{n\beta })}\mathcal{L}[2c_1{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}\\ & +2c_2[2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}+{\mathrm{\Xi }}_{0x}^2{\mathrm{\Xi }}_1]+3c_3[2{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}+2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_1{\mathrm{\Xi }}_{0x}^2]\\ & +\cdots +n{c}_n[(n-1){\mathrm{\Xi }}_1{\mathrm{\Xi }}_0^{n-2}{\mathrm{\Xi }}_{0x}^2+2{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{1x}{\mathrm{\Xi }}_{0x}]\\ & +c_0{\mathrm{\Xi }}_{1xx}+c_1({\mathrm{\Xi }}_0{\mathrm{\Xi }}_{1xx}+{\mathrm{\Xi }}_1{\mathrm{\Xi }}_{0xx})+c_2[{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{1xx}+2{\mathrm{\Xi }}_1{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0xx}]\\ & +\cdots +c_n[{\mathrm{\Xi }}_0^n{\mathrm{\Xi }}_{1xx}+n{\mathrm{\Xi }}_1{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{0xx}]]\end{array}$and so on. Now, let $s^{\beta }={\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ be the zeroes of the polynomial $\sum _{i=1}^n{a}_i{s}^{i\beta }$.

Then, the above set of differential equations can be rewritten as $\begin{array}{rl}& h^0:{\tilde{\mathrm{\Xi }}}_0(x,s)=(\frac{Z_1}{s^{\beta }-{\lambda }_1}+\frac{Z_2}{s^{\beta }-{\lambda }_2}+\cdots +\frac{Z_n}{s^{\beta }-{\lambda }_n})H(x,s)\\ & +\frac{\xi (x,0)}{s}+s^{\beta -2}(\frac{A_1}{s^{\beta }-{\lambda }_1}+\frac{A_2}{s^{\beta }-{\lambda }_2}+\cdots +\frac{A_n}{s^{\beta }-{\lambda }_n}){\xi }^{\prime }(x,0)+\cdots \\ & +s^{\beta -n}(\frac{Y_1}{s^{\beta }-{\lambda }_1}+\frac{Y_2}{s^{\beta }-{\lambda }_2}+\cdots +\frac{Y_n}{s^{\beta }-{\lambda }_n}){\xi }^{(n-1)}(x,0),\\ & h^1:{\tilde{\mathrm{\Xi }}}_1(x,s)=(\frac{Z_1}{s^{\beta }-{\lambda }_1}+\frac{Z_2}{s^{\beta }-{\lambda }_2}+\cdots +\frac{Z_n}{s^{\beta }-{\lambda }_n})\mathcal{L}\{c_1{\mathrm{\Xi }}_{0x}^2+2c_2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0x}^2\\ & +3c_3{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0x}^2+\cdots +n{c}_n{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{0x}^2+c_0{\mathrm{\Xi }}_{0xx}+c_1{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0xx}+c_2{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0xx}+\cdots \\ & +c_n{\mathrm{\Xi }}_0^n{\mathrm{\Xi }}_{0xx}\},\\ & h^2:{\tilde{\mathrm{\Xi }}}_2(x,s)=(\frac{Z_1}{s^{\beta }-{\lambda }_1}+\frac{Z_2}{s^{\beta }-{\lambda }_2}+\cdots +\frac{Z_n}{s^{\beta }-{\lambda }_n})\mathcal{L}\{2c_1{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}\\ & +2c_2[2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}+{\mathrm{\Xi }}_{0x}^2{\mathrm{\Xi }}_1]+3c_3[2{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}+2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_1{\mathrm{\Xi }}_{0x}^2]+\cdots \\ & +n{c}_n[(n-1){\mathrm{\Xi }}_1{\mathrm{\Xi }}_0^{n-2}{\mathrm{\Xi }}_{0x}^2+2{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{1x}{\mathrm{\Xi }}_{0x}]\\ & +c_0{\mathrm{\Xi }}_{1xx}+c_1({\mathrm{\Xi }}_0{\mathrm{\Xi }}_{1xx}+{\mathrm{\Xi }}_1{\mathrm{\Xi }}_{0xx})+c_2[{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{1xx}+2{\mathrm{\Xi }}_1{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0xx}]\\ & +\cdots +c_n[{\mathrm{\Xi }}_0^n{\mathrm{\Xi }}_{1xx}+n{\mathrm{\Xi }}_1{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{0xx}]\},\end{array}$where the coefficients of partial fractions are given by $\begin{array}{rl}{A}_j& =\frac{\sum _{i=2}^n{a}_i{\lambda }_j^{i-1}}{\prod _{i,j=1}^n({\lambda }_j-{\lambda }_i)},Y_j=\frac{a_n{\lambda }_j^{n-1}}{\prod _{i,j=1}^n({\lambda }_j-{\lambda }_i)},\\ Z_j& =\frac{1}{\prod _{i,j=1}^n({\lambda }_j-{\lambda }_i)},j=1,\dots ,n,i\ne j.\end{array}$Applying the inverse Laplace transform to both sides of each equation above, we get the following result: $\begin{array}{rl}{h}^0:{\mathrm{\Xi }}_0(x,\tau )& ={\tau }^{\beta -1}[Z_1{E}_{\beta ,\beta }({\lambda }_1{\tau }^{\beta })+\cdots +Z_n{E}_{\beta ,\beta }({\lambda }_n{\tau }^{\beta })]\ast h(x,\tau )\\ & +\xi (x,0)+\tau [A_1{E}_{\beta ,2}({\lambda }_1{\tau }^{\beta })+\cdots +A_n{E}_{\beta ,2}({\lambda }_n{\tau }^{\beta })]{\xi }^{\prime }(x,0)\\ & +\cdots +{\tau }^{n-1}[Y_1{E}_{\beta ,n}({\lambda }_1{\tau }^{\beta })+\cdots +Y_n{E}_{\beta ,n}({\lambda }_n{\tau }^{\beta })]{\xi }^{(n-1)}(x,0),\\ h^1:{\mathrm{\Xi }}_1(x,\tau )& ={\tau }^{\beta -1}[Z_1{E}_{\beta ,\beta }({\lambda }_1{\tau }^{\beta })+\cdots +Z_n{E}_{\beta ,\beta }({\lambda }_n{\tau }^{\beta })]\ast [c_1{\mathrm{\Xi }}_{0x}^2\\ & +2c_2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0x}^2+3c_3{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0x}^2+\cdots +n{c}_n{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{0x}^2+c_0{\mathrm{\Xi }}_{0xx}+c_1{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0xx}\\ & +c_2{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0xx}+\cdots +c_n{\mathrm{\Xi }}_0^n{\mathrm{\Xi }}_{0xx}],\\ h^2:{\mathrm{\Xi }}_2(x,\tau )& ={\tau }^{\beta -1}[Z_1{E}_{\beta ,\beta }({\lambda }_1{\tau }^{\beta })+\cdots +Z_n{E}_{\beta ,\beta }({\lambda }_n{\tau }^{\beta })]\ast [2c_1{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}\\ & +2c_2[2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}+{\mathrm{\Xi }}_{0x}^2{\mathrm{\Xi }}_1]+3c_3[2{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}+2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_1{\mathrm{\Xi }}_{0x}^2]+\cdots \\ & +n{c}_n[(n-1){\mathrm{\Xi }}_1{\mathrm{\Xi }}_0^{n-2}{\mathrm{\Xi }}_{0x}^2+2{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{1x}{\mathrm{\Xi }}_{0x}]\\ & +c_0{\mathrm{\Xi }}_{1xx}+c_1({\mathrm{\Xi }}_0{\mathrm{\Xi }}_{1xx}+{\mathrm{\Xi }}_1{\mathrm{\Xi }}_{0xx})\\ & +c_2[{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{1xx}+2{\mathrm{\Xi }}_1{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0xx}]+\cdots +c_n[{\mathrm{\Xi }}_0^n{\mathrm{\Xi }}_{1xx}+n{\mathrm{\Xi }}_1{\mathrm{\Xi }}_0^{n-1}{\mathrm{\Xi }}_{0xx}]]\end{array}$and so on. Therefore, the approximate solution is given by (10) $\xi (x,\tau )=\underset{h\to 1}{lim}\mathrm{\Xi }(x,\tau ;h)={\mathrm{\Xi }}_0(x,\tau )+{\mathrm{\Xi }}_1(x,\tau )+{\mathrm{\Xi }}_2(x,\tau )+\cdots .$(10)

4. Convergence analysis and error estimation

We discuss the convergence of the solution of a given problem in the following theorem.

Theorem 4.1

Consider ${\mathrm{\Xi }}_n(x,\tau )$ and $\xi (x,\tau )$ from Equation (10(10) $\xi (x,\tau )=\underset{h\to 1}{lim}\mathrm{\Xi }(x,\tau ;h)={\mathrm{\Xi }}_0(x,\tau )+{\mathrm{\Xi }}_1(x,\tau )+{\mathrm{\Xi }}_2(x,\tau )+\cdots .$(10) ) and for some constant $0<\beta <1$, if ${\mathrm{\Xi }}_n(x,\tau )\le \beta {\mathrm{\Xi }}_{n-1}(x,\tau )$, infinite series $\sum _{k=0}^{\mathrm{\infty }}{\mathrm{\Xi }}_k(x,\tau )$ converges to the solution $\xi (x,\tau )$ of the problem (2(2) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(f(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ),\end{array}$(2) )–(3(3) $\xi (x,0)=g_0(x),\frac{{\mathrm{\partial }}^k\xi (x,0)}{\mathrm{\partial }{t}^k}=g_k(x)\mathrm{\forall }k=1,2,\dots ,n,$(3) ).

The following theorem helps to truncate the approximate solution.

Theorem 4.2

The estimation of the maximum absolute truncation error of the series solution $\xi (x,\tau )=\sum _{k=0}^{\mathrm{\infty }}{\mathrm{\Xi }}_k(x,\tau )$ of the problem (2(2) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(f(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ),\end{array}$(2) )–(3(3) $\xi (x,0)=g_0(x),\frac{{\mathrm{\partial }}^k\xi (x,0)}{\mathrm{\partial }{t}^k}=g_k(x)\mathrm{\forall }k=1,2,\dots ,n,$(3) ) is given by $|\xi (x,\tau )-\sum _{i=0}^m{\mathrm{\Xi }}_i(x,\tau )|\le \frac{{\beta }^{m+1}}{(1-\beta )}\Vert {\mathrm{\Xi }}_0\Vert .$

For the detailed proofs, we refer the reader to [1].

5. Illustrative examples

Here are some examples that illustrate the strength and scope of the results obtained.

Example 5.1

For arbitrary $\lambda ,\nu$ constants, we take $a_1=\lambda ,a_2=1,f(\xi )=\nu ,h(x,\tau )=0$ in the system (2(2) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(f(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ),\end{array}$(2) )–(3(3) $\xi (x,0)=g_0(x),\frac{{\mathrm{\partial }}^k\xi (x,0)}{\mathrm{\partial }{t}^k}=g_k(x)\mathrm{\forall }k=1,2,\dots ,n,$(3) ), so we have the following form: (11) $\lambda D_{\tau }^{\beta }\xi (x,\tau )+D_{\tau }^{2\beta }\xi (x,\tau )=\nu \frac{{\mathrm{\partial }}^2\xi (x,\tau )}{\mathrm{\partial }{x}^2},$(11) along with the conditions (12) $\xi (x,0)=g_1(x),{\xi }_{\tau }(x,0)=g_2(x),\tau >0.$(12)

Applying the Laplace transform with respect to the variable τ in (11(11) $\lambda D_{\tau }^{\beta }\xi (x,\tau )+D_{\tau }^{2\beta }\xi (x,\tau )=\nu \frac{{\mathrm{\partial }}^2\xi (x,\tau )}{\mathrm{\partial }{x}^2},$(11) ) $\begin{array}{rl}& \lambda [s^{\beta }\tilde{\xi }(x,s)-s^{\beta -1}\xi (x,0)]+s^{2\beta }\xi (x,s)-s^{2\beta -1}\xi (x,0)-s^{2\beta -2}{\xi }_{\tau }(x,0)\\ & =\nu {\xi }_{xx}(x,s).\end{array}$Rearranging the terms, we get $\tilde{\xi }(x,s)=\frac{\nu }{s^{\beta }(s^{\beta }+\lambda )}{\xi }_{xx}(x,s)+\frac{g_1(x)}{s}+\frac{s^{\beta -2}}{(s^{\beta }+\lambda )}{g}_2(x).$Constructing the homotopy for the above equation (13) $\tilde{\mathrm{\Xi }}(x,s;h)=[\frac{h\nu }{s^{\beta }(s^{\beta }+\lambda )}{\mathrm{\Xi }}_{xx}(x,s)]+\frac{g_1(x)}{s}+\frac{s^{\beta -2}}{(s^{\beta }+\lambda )}{g}_2(x).$(13) Let us assume that $\mathrm{\Xi }(x,\tau ;h)$ has the following series form: (14) $\mathrm{\Xi }(x,\tau ;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k(x,\tau ),\mathrm{or}\tilde{\mathrm{\Xi }}(x,s;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\tilde{\mathrm{\Xi }}}_k(x,s).$(14) Putting the series form (14(14) $\mathrm{\Xi }(x,\tau ;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k(x,\tau ),\mathrm{or}\tilde{\mathrm{\Xi }}(x,s;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\tilde{\mathrm{\Xi }}}_k(x,s).$(14) ) into Equation (13(13) $\tilde{\mathrm{\Xi }}(x,s;h)=[\frac{h\nu }{s^{\beta }(s^{\beta }+\lambda )}{\mathrm{\Xi }}_{xx}(x,s)]+\frac{g_1(x)}{s}+\frac{s^{\beta -2}}{(s^{\beta }+\lambda )}{g}_2(x).$(13) ) and then comparing the like powers of h, we get the following iterations: $\begin{array}{rl}{h}^0:{\tilde{\mathrm{\Xi }}}_0(x,s)& =\frac{g_1(x)}{s}+\frac{s^{\beta -2}}{(s^{\beta }+\lambda )}{g}_2(x),\\ h^1:{\tilde{\mathrm{\Xi }}}_1(x,s)& =\frac{\nu }{s^{\beta }(s^{\beta }+\lambda )}{\mathrm{\Xi }}_{0xx}(x,s),\\ h^2:{\tilde{\mathrm{\Xi }}}_2(x,s)& =\frac{\nu }{s^{\beta }(s^{\beta }+\lambda )}{\mathrm{\Xi }}_{1xx}(x,s),\\ & \cdots \end{array}$Taking the inverse Laplace of each of equations and using the definitions of the Mittag–Leffler function definitions 2.1–2.3, we get the following: $\begin{array}{rl}{\mathrm{\Xi }}_0(x,\tau )& =g_1(x)+\tau E_{\beta ,2}(-\lambda {\tau }^{\beta })g_2(x),\\ {\mathrm{\Xi }}_1(x,\tau )& ={\mathcal{L}}^{-1}[\frac{\nu }{s^{\beta }(s^{\beta }+\lambda )}{\mathrm{\Xi }}_{0xx}(x,s)]\\ & ={\mathcal{L}}^{-1}[\frac{\nu g_1^{″}(x)}{s^{\beta +1}(s^{\beta }+\lambda )}+\frac{\nu s^{-2}}{(s^{\beta }+\lambda {)}^2}{g}_2^{″}(x)]\\ & =\nu g_1^{″}(x){\mathcal{L}}^{-1}\left[\frac{s^{\beta -(2\beta +1)}}{(s^{\beta }+\lambda )}\right]+\nu g_2^{″}(x){\mathcal{L}}^{-1}\left[\frac{s^{2\beta -(2\beta +2)}}{(s^{\beta }+\lambda {)}^2}\right]\\ & =\nu g_1^{″}(x){\tau }^{2\beta }{E}_{\beta ,2\beta +1}(-\lambda {\tau }^{\beta })+\nu g_2(x{)}^{″}{\tau }^{2\beta +1}{E}_{\beta ,2\beta +2}^2(-\lambda {\tau }^{\beta }),\\ {\mathrm{\Xi }}_2(x,\tau )& ={\mathcal{L}}^{-1}[\frac{\nu }{s^{\beta }(s^{\beta }+\lambda )}{\mathrm{\Xi }}_{1xx}(x,s)]\\ & ={\mathcal{L}}^{-1}[\frac{{\nu }^2{g}_1^{⁗}(x)}{s^{2\beta +1}(s^{\beta }+\lambda {)}^2}+\frac{{\nu }^2{s}^{-\beta -2}}{(s^{\beta }+\lambda {)}^3}{g}_2^{⁗}(x)]\\ & ={\nu }^2{g}_1^{⁗}(x){\mathcal{L}}^{-1}\left[\frac{s^{2\beta -(4\beta +1)}}{(s^{\beta }+\lambda {)}^2}\right]+{\nu }^2{g}_2^{⁗}(x){\mathcal{L}}^{-1}\left[\frac{s^{3\beta -(4\beta +2)}}{(s^{\beta }+\lambda {)}^3}\right]\\ & ={\nu }^2{g}_1^{⁗}(x){\tau }^{4\beta }{E}_{\beta ,4\beta +1}^2(-\lambda {\tau }^{\beta })+{\nu }^2{g}_2^{⁗}(x){\tau }^{4\beta +1}{E}_{\beta ,4\beta +2}^3(-\lambda {\tau }^{\beta })\\ & \cdots \end{array}$Thus, $\begin{array}{rl}\xi (x,\tau )& ={\mathrm{\Xi }}_0+{\mathrm{\Xi }}_1+{\mathrm{\Xi }}_2+\cdots \\ & =g_1(x)+\tau E_{\beta ,2}(-\lambda {\tau }^{\beta })g_2(x)+\nu g_1^{″}(x){\tau }^{2\beta }{E}_{\beta ,2\beta +1}(-\lambda {\tau }^{\beta })\\ & +\nu g_2(x{)}^{″}{\tau }^{2\beta +1}{E}_{\beta ,2\beta +2}^2(-\lambda {\tau }^{\beta })+{\nu }^2{g}_1^{⁗}(x){\tau }^{4\beta }{E}_{\beta ,4\beta +1}^2(-\lambda {\tau }^{\beta })\\ & +{\nu }^2{g}_2^{⁗}(x){\tau }^{4\beta +1}{E}_{\beta ,4\beta +2}^3(-\lambda {\tau }^{\beta })+\cdots \end{array}$The solution can be written in the series form as given by $\begin{array}{rl}& \xi (x,\tau )=g_1(x)+g_2(x)\tau \sum _{k=0}^{\mathrm{\infty }}\frac{(-\lambda {\tau }^{\beta }{)}^k}{\mathrm{\Gamma }(\beta k+2)}+\nu g_1^{″}(x){\tau }^{2\beta }\sum _{k=0}^{\mathrm{\infty }}\frac{(-\lambda {\tau }^{\beta }{)}^k}{\mathrm{\Gamma }(\beta k+2\beta +1)}\\ & +\nu g_2(x{)}^{″}{\tau }^{2\beta +1}\sum _{k=0}^{\mathrm{\infty }}\frac{(2{)}_k(-\lambda {\tau }^{\beta }{)}^k}{\mathrm{\Gamma }(\beta k+2\beta +2)k!}+{\nu }^2{g}_1^{⁗}(x){\tau }^{4\beta }\sum _{k=0}^{\mathrm{\infty }}\frac{(2{)}_k(-\lambda {\tau }^{\beta }{)}^k}{\mathrm{\Gamma }(\beta k+4\beta +1)k!}\\ & +{\nu }^2{g}_2^{⁗}(x){\tau }^{4\beta +1}\sum _{k=0}^{\mathrm{\infty }}\frac{(3{)}_k(-\lambda {\tau }^{\beta }{)}^k}{\mathrm{\Gamma }(\beta k+4\beta +2)k!}+\cdots \end{array}$Thus for $\beta =1$, the series solution for the classical telegraphic equation is given by $\begin{array}{rl}\xi (x,\tau )& =g_1(x)+g_2(x)\tau [1-\frac{\lambda \tau }{2!}+\frac{{\lambda }^2{\tau }^2}{3!}-\frac{{\lambda }^3{\tau }^3}{4!}+\cdots ]\\ & +\nu g_1^{″}(x){\tau }^2[\frac{1}{2!}-\frac{\lambda \tau }{3!}+\frac{{\lambda }^2{\tau }^2}{4!}-\frac{{\lambda }^3{\tau }^3}{5!}+\cdots ]\\ & +\nu g_2(x{)}^{″}{\tau }^3[\frac{1}{3!}-\frac{2\lambda \tau }{4!}+\frac{3{\lambda }^2{\tau }^2}{5!}-\frac{4{\lambda }^3{\tau }^3}{6!}+\cdots ]\\ & +{\nu }^2{g}_1^{⁗}(x){\tau }^4[\frac{1}{4!}-2\frac{\lambda \tau }{5!}+3\frac{{\lambda }^2{\tau }^2}{6!}-4\frac{{\lambda }^3{\tau }^3}{7!}+\cdots ]\\ & +{\nu }^2{g}_2^{⁗}(x){\tau }^5[\frac{1}{5!}-3\frac{\lambda \tau }{6!}+6\frac{{\lambda }^2{\tau }^2}{7!}-10\frac{{\lambda }^3{\tau }^3}{8!}+\cdots ]+\cdots \end{array}$This is same as the solution obtained by Momani in [16].

Example 5.2

Let $a_1=1,a_2=a_3=\cdots =a_n=0,f(\xi )=\xi ,h(x,\tau )=0$ in Equation (5(5) $\begin{array}{rl}{a}_1{D}_{\tau }^{\beta }\xi (x,\tau )+a_2{D}_{\tau }^{2\beta }\xi (x,\tau )+\cdots +a_n{D}_{\tau }^{n\beta }\xi (x,\tau )& =\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(P(\xi )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}\right)\\ & +h(x,\tau ).\end{array}$(5) ), (15) $D_{\tau }^{\beta }\xi (x,\tau )=\frac{\mathrm{\partial }}{\mathrm{\partial }x}(\xi (x,\tau )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}),\xi (x,0)=x.$(15)

Applying the Laplace transform to Equation (15(15) $D_{\tau }^{\beta }\xi (x,\tau )=\frac{\mathrm{\partial }}{\mathrm{\partial }x}(\xi (x,\tau )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}),\xi (x,0)=x.$(15) ), we get $\begin{array}{rl}& s^{\beta }\tilde{\xi }(x,s)-s^{\beta -1}\xi (x,0)=\mathcal{L}[{\xi }_x^2(x,\tau )+\xi (x,\tau ){\xi }_{xx}(x,\tau )]\\ ⟹& \tilde{\xi }(x,s)=\frac{1}{s^{\beta }}\mathcal{L}[{\xi }_x^2(x,\tau )+\xi (x,\tau ){\xi }_{xx}(x,\tau )]+\frac{\xi (x,0)}{s}.\end{array}$The homotopy is constructed as follows: (16) $\tilde{\mathrm{\Xi }}(x,s;h)=\frac{h}{s^{\beta }}\mathcal{L}[{\xi }_x^2(x,\tau )+\xi (x,\tau ){\xi }_{xx}(x,\tau )]+\frac{\xi (x,0)}{s}.$(16) Let us write the solution in a series form as (17) $\mathrm{\Xi }(x,\tau ;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k(x,\tau ).$(17) Substituting Equation (17(17) $\mathrm{\Xi }(x,\tau ;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k(x,\tau ).$(17) ) into Equation (16(16) $\tilde{\mathrm{\Xi }}(x,s;h)=\frac{h}{s^{\beta }}\mathcal{L}[{\xi }_x^2(x,\tau )+\xi (x,\tau ){\xi }_{xx}(x,\tau )]+\frac{\xi (x,0)}{s}.$(16) ), we obtain (18) $\begin{array}{rl}\sum _{k=0}^{\mathrm{\infty }}{h}^k{\tilde{\mathrm{\Xi }}}_k(x,s)& =\frac{h}{s^{\beta }}\mathcal{L}[\sum _{k=0}^{\mathrm{\infty }}{h}^k\left(\sum _{k_1+k_2=k}{\mathrm{\Xi }}_{k_{1x}}{\mathrm{\Xi }}_{k_{2}x}\right)+\sum _{k=0}^{\mathrm{\infty }}{h}^k\left(\sum _{k_1+k_2=k}{\mathrm{\Xi }}_{k_1}{\mathrm{\Xi }}_{k_{2}xx}\right)]\\ & +\frac{\mathrm{\Xi }(x,0)}{s}.\end{array}$(18) Comparing the similar powers of h and then applying the inverse Laplace transform in each equation, we get $\begin{array}{rl}& h^0:{\tilde{\mathrm{\Xi }}}_0(x,s)=\frac{\xi (x,0)}{s}=\frac{x}{s}\\ ⟹& {\mathrm{\Xi }}_0(x,\tau )=x\\ & h^1:{\tilde{\mathrm{\Xi }}}_1(x,s)=\frac{1}{s^{\beta }}\mathcal{L}[{\mathrm{\Xi }}_{0x}^2+{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0xx}=]=\frac{1}{s^{\beta }}\\ ⟹& {\mathrm{\Xi }}_1(x,\tau )=\frac{{\tau }^{\beta }}{\mathrm{\Gamma }(\beta +1)}\\ & h^2:{\tilde{\mathrm{\Xi }}}_2(x,s)=\frac{1}{s^{\beta }}\mathcal{L}[2{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}+{\mathrm{\Xi }}_1{\mathrm{\Xi }}_{0xx}+{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{1xx}]=0\\ ⟹& {\mathrm{\Xi }}_2(x,\tau )=0\\ & \cdot \\ & \cdot \\ & \cdot \\ & h^k:{\mathrm{\Xi }}_k(x,\tau )=0\mathrm{\forall }k\ge 2.\end{array}$Hence, we get the approximate solution as $\xi (x,\tau )=x+\frac{{\tau }^{\beta }}{\mathrm{\Gamma }(\beta +1)}$, which is in the closed form.

We see that for $\beta =1$, we get the solution as $\xi (x,\tau )=x+\tau$.

Example 5.3

Slow Diffusion

For $f(\xi )={\xi }^2$ and $\xi (x,0)=\frac{x+b}{2c}$, where $c>0,b$ are arbitrary constants, (19) $D_{\tau }^{\beta }\xi (x,\tau )=\frac{\mathrm{\partial }}{\mathrm{\partial }x}({\xi }^2(x,\tau )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}),\xi (x,0)=\frac{x+b}{2c}.$(19)

Applying the Laplace transform to Equation (19(19) $D_{\tau }^{\beta }\xi (x,\tau )=\frac{\mathrm{\partial }}{\mathrm{\partial }x}({\xi }^2(x,\tau )\frac{\mathrm{\partial }\xi (x,\tau )}{\mathrm{\partial }x}),\xi (x,0)=\frac{x+b}{2c}.$(19) ), we get $\begin{array}{rl}& s^{\beta }\tilde{\xi }(x,s)-s^{\beta -1}\xi (x,0)=\mathcal{L}[2\xi (x,\tau ){\xi }_x^2(x,\tau )+{\xi }^2(x,\tau ){\xi }_{xx}(x,\tau )]\\ ⟹& \xi (x,s)=\frac{1}{s^{\beta }}\mathcal{L}[2\xi (x,\tau ){\xi }_x^2(x,\tau )+{\xi }^2(x,\tau ){\xi }_{xx}(x,\tau )]+\frac{\xi (x,0)}{s}\end{array}$The homotopy is constructed as follows: (20) $\tilde{\mathrm{\Xi }}(x,s;h)=\frac{h}{s^{\beta }}\mathcal{L}[2\mathrm{\Xi }(x,\tau ){\mathrm{\Xi }}_x^2(x,\tau )+{\mathrm{\Xi }}^2(x,\tau ){\mathrm{\Xi }}_{xx}(x,\tau )]+\frac{\xi (x,0)}{s}.$(20) Let us write the solution in series form as (21) $\mathrm{\Xi }(x,\tau ;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k(x,\tau ).$(21) Substituting Equation (21(21) $\mathrm{\Xi }(x,\tau ;h)=\sum _{k=0}^{\mathrm{\infty }}{h}^k{\mathrm{\Xi }}_k(x,\tau ).$(21) ) into Equation (20(20) $\tilde{\mathrm{\Xi }}(x,s;h)=\frac{h}{s^{\beta }}\mathcal{L}[2\mathrm{\Xi }(x,\tau ){\mathrm{\Xi }}_x^2(x,\tau )+{\mathrm{\Xi }}^2(x,\tau ){\mathrm{\Xi }}_{xx}(x,\tau )]+\frac{\xi (x,0)}{s}.$(20) ), we obtain $\begin{array}{rl}\sum _{k=0}^{\mathrm{\infty }}{h}^k{\tilde{\mathrm{\Xi }}}_k(x,\tau )& =\frac{h}{s^{\beta }}\mathcal{L}[2\sum _{k=0}^{\mathrm{\infty }}{h}^k\left(\sum _{k_1+k_2+k_3=k}{\mathrm{\Xi }}_{k_1}{\mathrm{\Xi }}_{k_{2x}}{\mathrm{\Xi }}_{k_{3}x}\right)\\ & +\sum _{k=0}^{\mathrm{\infty }}{h}^k\left(\sum _{k_1+k_2+k_3=k}{\mathrm{\Xi }}_{k_1}{\mathrm{\Xi }}_{k_2}{\mathrm{\Xi }}_{k_{3}xx}\right)]+\frac{\xi (x,0)}{s}.\end{array}$Comparing the similar powers of h and applying the inverse Laplace of each equation, we get $\begin{array}{rl}& h^0:{\tilde{\mathrm{\Xi }}}_0(x,s)=\frac{\xi (x,0)}{s}\\ ⟹& {\mathrm{\Xi }}_0(x,\tau )=\frac{x+b}{2c}\\ & h^1:{\tilde{\mathrm{\Xi }}}_1(x,s)=\frac{1}{s^{\beta }}\mathcal{L}[2{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0x}^2+{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{0xx}]=\frac{x+b}{4c^3{s}^{\beta +1}}\\ ⟹& {\mathrm{\Xi }}_1(x,\tau )=\frac{{\tau }^{\beta }}{\mathrm{\Gamma }(\beta +1)}\frac{(x+b)}{4c^3}\\ & h^2:{\tilde{\mathrm{\Xi }}}_2(x,s)=\frac{1}{s^{\beta }}\mathcal{L}[4{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0x}{\mathrm{\Xi }}_{1x}+2{\mathrm{\Xi }}_1{\mathrm{\Xi }}_{0x}^2+2{\mathrm{\Xi }}_1{\mathrm{\Xi }}_0{\mathrm{\Xi }}_{0xx}+{\mathrm{\Xi }}_0^2{\mathrm{\Xi }}_{1xx}]\\ & =\frac{1}{s^{\beta }}\frac{(x+b)}{4c^5}\mathcal{L}\left[\frac{3{\tau }^{\beta }}{2\mathrm{\Gamma }(\beta +1)}\right]\\ ⟹& {\mathrm{\Xi }}_2(x,\tau )=\frac{{\tau }^{2\beta }}{\mathrm{\Gamma }(2\beta +1)}\frac{3(x+b)}{8c^5}.\end{array}$and so on. Thus, the approximate solution is given by $\xi (x,\tau )=\frac{x+b}{2c}+\frac{{\tau }^{\beta }}{\mathrm{\Gamma }(\beta +1)}\frac{(x+b)}{4c^3}+\frac{{\tau }^{2\beta }}{\mathrm{\Gamma }(2\beta +1)}\frac{3(x+b)}{8c^5}+\cdots$For $\beta =1$, we have $\xi (x,\tau )=\frac{x+b}{2c}+\frac{(x+b)\tau }{4c^3}+\frac{3(x+b){\tau }^2}{16c^5}+\cdots$ which is Maclaurin's formula of the exact solution (see [33]), $\xi (x,\tau )=\frac{x+b}{2\sqrt{c^2-\tau }},\tau <c^2$.

6. Conclusion

We have considered the non-linear extension of the time-fractional telegraphic equation by introducing a non-linear term $f(\xi )$. The well-known Homotopy perturbation method along with the Laplace transform have been applied to obtain an analytic approximate solution of the considered generalized nonlinear TFTE along with the given initial conditions. Some well-known examples are presented as special cases to illustrate the strength of the method applied and the liability of the concept.

Acknowledgments

The authors are thankful to the reviewers for their constructive comments to improve this paper. All authors contributed equally in the preparation of the paper.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

The authors extend their appreciation to the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University (IMSIU) for funding and supporting this work through Research Partnership Program no. RP-21-09-06.