Assessing the Efficiency of the Homotopy Analysis Transform Method for Solving a Fractional Telegraph Equation with a Bessel Operator

Abstract

In this study, we apply the Laplace Transform Homotopy Analysis Method (LTHAM) to numerically solve a fractional-order telegraph equation with a Bessel operator. The iterative scheme developed is tested on multiple examples to evaluate its efficiency. Our observations indicate that the method generates an approximate solution in series form, which converges rapidly to the analytic solution in each instance. The convergence of these series solutions is assessed both geometrically and numerically. Our results demonstrate that LTHAM is a reliable, powerful, and straightforward approach to solving fractional telegraph equations, and it can be effectively extended to solve similar types of equations.

1. Introduction

Fractional derivatives play a fundamental role in almost all branches of applied mathematics, as they are used to replace integer-order derivatives in modeling numerous physical and engineering phenomena. Over the past decades, a variety of fractional-order mathematical models have been employed to characterize various phenomena in different scientific fields, including physics, chemistry, biology, engineering, and economics, etc. [1,2]. The mathematicians J. Liouville and M. Caputo are considered the first pioneers of the fractional calculus; J. Liouville introduced the concepts of fractional integration and differentiation in 1832 [3,4], while M. Caputo proposed the concept of the Caputo fractional derivative in 1967 [5]. In fact, their significant contributions form the cornerstone of the development of these mathematical concepts.

Despite the widespread application of fractional-order partial differential equations across different disciplines, solving these equations presents several challenges compared to traditional integer-order ones. The lack of comprehensive analytical techniques often makes it difficult to find precise solutions. Moreover, traditional numerical methods may not be directly applicable, and research in this field remains active. Various numerical methods and techniques have been proposed and developed to address these challenges. Recently, various authors have merged the well-known Laplace transform with a number of traditional numerical methods to solve plenty of fractional and non-fractional models, such as the Laplace transform Adomian decomposition method (LTADM) [6,7], the Laplace transform differential method (LTDM) [8], the Laplace transform variational iteration method (LTVIM) [9], the Laplace transform homotopy perturbation method (LTHPM) [10,11,12], the Laplace transform homotopy analysis method [13,14,15], and others. The LTHPM is developed by merging two powerful techniques, namely, the Laplace transform method [16,17,18], and the homotopy analysis method [19,20,21,22,23]. In fact, the Laplace transform is widely used to solve linear ordinary and partial differential equations, especially those with constant coefficients and defined initial conditions. For integer-order partial differential equations, it reduces the complexity of the problem, as it converts the equation into an ordinary differential equation, which is generally easier to handle analytically. Additionally, applying the Laplace transform to a partial differential equation with respect to time transforms it into an ordinary differential equation in the spatial variable.

Several versions of the fractional-order telegraph equation have been solved using different analytical and numerical techniques (see [24,25,26,27,28]). Recently, a numerical solution of a nonlinear time-fractional telegraph equation modeling neutron transport in a nuclear reactor using the nonpolynomial spline method is presented in [29]. Moreover, a study investigating the inverse problem of determining the right-hand side of a telegraph equation with the Caputo derivative in a Hilbert space is presented in [30].

In the present work, our aim is to explore the application and efficiency of the LTHAM for solving a one-dimensional fractional version of the telegraph equation. Specifically, we will utilize the LTHAM to determine homotopy analysis transform solutions for the following partial differential equation:

$${\partial }_{\tau }^{2\alpha }v(y,\tau )+2\lambda {\partial }_{\tau }^{\alpha }v(y,\tau )-{\displaystyle \frac{c^2}{y}}{\displaystyle \frac{\partial }{\partial y}}\left(y{\displaystyle \frac{\partial v}{\partial y}}\right)=f(y,\tau ),\tau \in (0,T],y\in (0,L],$$

(1)

subject to the initial conditions

$$v(y,0)=\delta \left(y\right),v_t(y,0)=\beta \left(y\right),y\in (0,L],$$

(2)

for some known functions, $\delta \left(y\right)$, $\beta \left(y\right)$.

Here, $0<\alpha \le 1$ is the fractional order, which controls the memory effects. $\lambda \ge 0$ denotes the damping coefficient. The parameter $c\ne 0$ is the stiffness constant (wave propagation speed). The function $f(y,t)$ represents an external acting force, which allows for localized or distributed inputs (e.g., sources, controls). The terms $\delta \left(y\right)$ and $\beta \left(y\right)$ are, respectively, the displacement and velocity profiles.

This fractional differential equation is a generalization of the classical telegraph equation; using Caputo fractional derivatives denoted as ${\partial }_{\tau }^{\alpha }$ describes the evolution of a wave-like quantity denoted as $v(y,t)$ over the domain $(0,L]\times [0,T],$ where the system exhibits damped oscillatory behavior (like in telegraphy or electrical circuits) but also includes memory effects due to the fractional time derivatives, meaning that the damping or wave propagation is not instantaneous. It depends on the history of the system. The spatial part, ${\displaystyle \frac{c^2}{y}}{\displaystyle \frac{\partial }{\partial y}}\left(y{\displaystyle \frac{\partial v}{\partial y}}\right)$, models how the wave spreads out in a cylindrical or radial fashion. This is the radially symmetric Laplacian in cylindrical coordinates (assuming radial symmetry only). This physically suggests that the system evolves in a cylindrical geometry, such as waves propagating radially from a point source (e.g., seismic waves or acoustic pulses in a circular membrane).

We now begin by presenting the Laplace transform of the fractional differential operator in Caputo’s sense, which is defined as follows [31,32]:

$$L\left[{\displaystyle \frac{{\partial }^{\alpha }}{\partial {\tau }^{\alpha }}}v(y,\tau )\right]=s^{\alpha }\omega (y,s)-\sum _{k=0}^{n-1}{s}^{\alpha -k-1}{v}^{\left(k\right)}(y,0^{+}),n\ge 1,$$

(3)

in which $\omega (y,s)$ represents the Laplace transform of $v(y,\tau )$.

The rest of this article is structured as follows: Section 2 introduces the basics of the LTHAM. In Section 3, we employ the LTHAM to formulate an iterative scheme to solve problems (1) and (2) numerically. In Section 4, we explore the efficiency of this scheme through a set of test examples. Finally, we present some comments and concluding remarks in Section 5.

2. Brief Review of the Basics of LTHAM

Consider a fractional equation as follows:

$${\displaystyle \frac{{\partial }^{\alpha }}{\partial {\tau }^{\alpha }}}v(y,\tau )+\mathcal{R}v(y,\tau )+\mathcal{N}v(y,\tau )=g(y,\tau ),n-1<\alpha \le n,n=1,2,\dots ,$$

(4)

in which ${\displaystyle \frac{{\partial }^{\alpha }}{\partial {\tau }^{\alpha }}}$ indicates Caputo’s fractional operator of order $\alpha$, while $\mathcal{R}$ and $\mathcal{N}$ stand for linear and nonlinear differential operators, respectively, and g is some known function.

Now, operating with the Laplace transform on both sides of Equation (4) yields the following:

$$s^{\alpha }\mathcal{L}\left[v(y,\tau )\right]-\sum _{k=0}^{n-1}{s}^{\alpha -k-1}{v}^{\left(k\right)}(y,0)+\mathcal{L}\left[\mathcal{R}v(y,\tau )+\mathcal{N}v(y,\tau )\right]=\mathcal{L}\left[g(y,\tau )\right],$$

or

$$\mathcal{L}\left[v(y,\tau )\right]-{\displaystyle \frac{1}{s^{\alpha }}}\sum _{k=0}^{n-1}{s}^{\alpha -k-1}{v}^{\left(k\right)}(x,0)+{\displaystyle \frac{1}{s^{\alpha }}}\mathcal{L}\left[\mathcal{R}v(y,\tau )+\mathcal{N}v(y,\tau )-g(y,\tau )\right]=0.$$

Then, in view of the theory of the homotopy analysis method [19], an operator $\overline{\mathcal{N}}$ can be defined as follows:

$$\begin{array}{ccc}\overline{\mathcal{N}}\left[q(y,\tau ;\kappa )\right]\hfill & =\hfill & \mathcal{L}\left[q(y,\tau ;\kappa )\right]-{\displaystyle \frac{1}{s^{\alpha }}}\sum _{k=0}^{n-1}{s}^{\alpha -k-1}{q}^{\left(k\right)}(y,0;\kappa )+{\displaystyle \frac{1}{s^{\alpha }}}\mathcal{L}[\mathcal{R}q(y,\tau ;\kappa )\hfill \\ \hfill & +\hfill & \mathcal{N}q(y,\tau ;\kappa )-g(y,\tau )],\hfill \end{array}$$

where $\kappa \in [0,1]$, and q is some function in $y,\tau$, and $\kappa$. Thus, a zeroth-order deformation equation can be defined as follows:

$$(1-\kappa )\mathcal{L}\left[q(y,\tau ;\kappa )-v_0(y,\tau )\right]=\kappa \widehat{h}\overline{\mathcal{N}}\left[q(y,\tau ;\kappa )\right],$$

(5)

in which $\kappa \in [0,1]$ is an embedding parameter, $\mathcal{L}$ indicates the Laplace transform operator, $v_0(y,\tau )$ is an initial approximation of the analytical solution $v(y,\tau )$, $q(y,\tau ;\kappa )$ is an unknown function, and $\widehat{h}$ is a non zero parameter, it plays a fundamental role in controlling and adjusting the convergence region of the approximate series solution. Unfortunately, in the theory of the homotopy analysis method, there is no theoretical technique that can be applied easily to determine the range of the values of the parameter $\widehat{h}$, except in some limited cases when a closed form of the series solution can be obtained. But mostly, closed-form series solutions are not obtained in practice; instead, the series solutions are approximated using truncated ones. Therefore, the range of values of this parameter can be obtained graphically using the $\widehat{h}$-curve based on the graphs of truncated series solutions, namely the interval on which the $\widehat{h}$-curve is almost horizontal.

As it appears from Equation (5), one can easily deduce that, at $\kappa =0$ and $\kappa =1$, we have the following:

$$q(y,\tau ;0)=v_0(y,\tau )\mathrm{and}q(y,\tau ;1)=v(y,\tau ).$$

In other words, as $\kappa$ deforms from 0 to 1, the function $q(y,\tau ;\kappa )$ deforms from the initial guess, $v_0(y,\tau )$, towards the precise solution, $v(y,\tau )$.

On the other hand, the expansion of $q(y,\tau ;\kappa )$ in a Taylor series with respect to $\kappa$ implies the following:

$$q(y,\tau ;\kappa )=v_0(y,\tau )+\sum _{j=1}^{\infty }{v}_j(y,\tau ){\kappa }^j,$$

(6)

where

$$v_j(y,\tau )={\left.{\displaystyle \frac{1}{j!}}{\displaystyle \frac{{\partial }^{j}q(x,\tau ;\kappa )}{\partial {\kappa }^j}}\right|}_{\kappa =0}.$$

As specified in [22], if the parameter $\widehat{h}$ and the initial approximate value $v_0(y,\tau )$ are chosen properly, then the series (6) will converge at $\kappa =1$ and produce a power series solution of the problem as follows:

$$v(y,\tau )=v_0(y,\tau )+\sum _{j=1}^{\infty }{v}_j(y,\tau ).$$

Now, differentiating Equation (5) j-times with respect to $\kappa$, and then dividing the outcome by $j!$ and setting $\kappa =0$, gives the following jth order equation:

$$\mathcal{L}\left[v_j(y,\tau )-{\chi }_j{v}_{j-1}(y,\tau )\right]=\widehat{h}\mathcal{R}\left({\overrightarrow{v}}_{j-1}\right),$$

(7)

where

$$\overrightarrow{v_k}:=\overrightarrow{v_k}(y,\tau )=[v_0(y,\tau ),v_1(y,\tau ),\dots ,v_k(y,\tau )],$$

and

$$\mathcal{R}\left({\overrightarrow{v}}_{j-1}\right)={\left.{\displaystyle \frac{1}{(j-1)!}}\left\{{\displaystyle \frac{{\partial }^{j-1}}{\partial {\kappa }^{j-1}}}\overline{\mathcal{N}}\left[q(y,\tau ;\kappa )\right]\right\}\right|}_{\kappa =0}.$$

Next, operating with the inverse Laplace transform on both sides of Equation (7), we can determine the components $v_j(y,\tau )$ by implementing the following formula:

$$v_j(y,\tau )={\chi }_j{v}_{j-1}(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_{j-1}\right)\right],j=1,2,\dots ,$$

where

$${\chi }_j=\left\{\begin{array}{cc}0,\hfill & j=1,\hfill \\ 1,\hfill & j>1.\hfill \end{array}\right.$$

3. Application of LTHAM to a Fractional Telegraph Equation

In this section, the LTHAM is utilized to derive an iterative formula for solving the following fractional telegraph Equation (1):

$${\displaystyle \frac{{\partial }^{2\alpha }}{\partial {\tau }^{2\alpha }}}v(y,\tau )+2\lambda {\displaystyle \frac{{\partial }^{\alpha }}{\partial {\tau }^{\alpha }}}v(y,\tau )-{\displaystyle \frac{c^2}{y}}{\displaystyle \frac{\partial }{\partial y}}\left(y{\displaystyle \frac{\partial v}{\partial y}}\right)=g(y,\tau ),0<\alpha \le 1,y\in (0,1],\tau \in [0,1],$$

(8)

subject to the following initial conditions:

$$v(y,0)=\delta \left(y\right),v_{\tau }(y,0)=\beta \left(y\right),$$

(9)

where g, $\delta$, and $\beta$ are known functions, and $\lambda$ and c are some constants. Then, operating with the Laplace transform on both sides of Equation (8) and utilizing (3), we get the following:

$$\mathcal{L}\left[v(y,\tau )\right]-{\displaystyle \frac{1}{s}}v(y,0)-{\displaystyle \frac{1}{s^2}}{v}_{\tau }(y,0)+{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}\left[2\lambda {\displaystyle \frac{{\partial }^{\alpha }}{\partial {\tau }^{\alpha }}}v(y,\tau )-{\displaystyle \frac{c^2}{y}}{\displaystyle \frac{\partial }{\partial y}}v(y,\tau )-c^2{\displaystyle \frac{{\partial }^2}{\partial y^2}}v(y,\tau )-g(y,\tau )\right]=0.$$

Therefore, we define an operator $\widehat{\mathcal{N}}\left[q(y,\tau ;\kappa )\right]$ as follows:

$$\begin{array}{ccc}\widehat{\mathcal{N}}\left[q(y,\tau ;\kappa )\right]\hfill & =\hfill & \mathcal{L}\left[q(y,\tau ;\kappa )\right]-{\displaystyle \frac{1}{s}}v(y,0)-{\displaystyle \frac{1}{s^2}}{v}_{\tau }(y,0)+{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}[2\lambda {\displaystyle \frac{{\partial }^{\alpha }}{\partial {\tau }^{\alpha }}}q(y,\tau ;\kappa )-{\displaystyle \frac{c^2}{y}}{\displaystyle \frac{\partial }{\partial y}}q(y,\tau ;\kappa )\hfill \\ \hfill & \hfill & \\ \hfill & \hfill & -c^2{\displaystyle \frac{{\partial }^2}{\partial y^2}}q(y,\tau ;\kappa )-g(y,\tau )],\hfill \end{array}$$

which leads to the following $k^{th}$-order deformation equation:

$$\mathcal{L}\left[v_k(y,\tau )-{\chi }_k{v}_{k-1}(y,\tau )\right]=\widehat{h}\mathcal{R}\left({\overrightarrow{v}}_{k-1}\right),$$

in which

$$\begin{array}{ccc}\mathcal{R}\left({\overrightarrow{v}}_{k-1}\right)\hfill & =\hfill & \mathcal{L}\left[v_{k-1}(y,\tau )\right]-\left(1-{\chi }_k\right)\left({\displaystyle \frac{1}{s}}v(y,0)+{\displaystyle \frac{1}{s^2}}{v}_{\tau }(y,0)\right)+{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}[2\lambda {\displaystyle \frac{{\partial }^{\alpha }}{\partial {\tau }^{\alpha }}}{v}_{k-1}(y,\tau )\hfill \\ \hfill & \hfill & \\ \hfill & \hfill & -{\displaystyle \frac{c^2}{y}}{\displaystyle \frac{\partial }{\partial y}}{v}_{k-1}(y,\tau )-c^2{\displaystyle \frac{{\partial }^2}{\partial y^2}}{v}_{k-1}(y,\tau )-\left(1-{\chi }_k\right)g(y,\tau )].\hfill \end{array}$$

Hence, the terms of the analytical series solution can be obtained by applying the following iterative formula:

$$v_k(y,\tau )={\chi }_k{v}_{k-1}(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_{k-1}\right)\right],$$

(10)

and thus, the power series solution is given as follows:

$$v(y,\tau )=v_0(y,\tau )+\sum _{i=1}^{\infty }{v}_i(y,\tau ).$$

Mostly, this series solution is approximated using a truncated series of n-term partial sum as follows:

$$v^{\left[n\right]}(y,\tau )=v_0(y,\tau )+\sum _{i=1}^n{v}_i(y,\tau ).$$

4. Numerical Implementations

To test the efficiency and accuracy of the Formula (10), it will be implemented in the following set of testing examples:

Example 1.

Consider the following equation:

$${\partial }_{\tau }^{2\alpha }v(y,\tau )+2{\partial }_{\tau }^{\alpha }v(y,\tau )-{\displaystyle \frac{1}{y}}{\displaystyle \frac{\partial }{\partial y}}\left(y{\displaystyle \frac{\partial v}{\partial y}}\right)=g(y,\tau ),0<\alpha \le 1,0<y\le 1,\tau \in [0,1],$$

(11)

subject to the following conditions:

$$v(y,0)=(1-y)ln(1+y),v_{\tau }(y,0)=1,0<y\le 1,$$

(12)

where

$$g(y,\tau )=6\tau +2\left(1+3{\tau }^2\right)-{\displaystyle \frac{1}{y}}\left({\displaystyle \frac{1-y}{1+y}}+y\left(-{\displaystyle \frac{1-y}{{(1+y)}^2}}-{\displaystyle \frac{2}{1+y}}\right)-ln(1+y)\right).$$

Solution.

Suppose $v_0(y,\tau )=v(y,0)=(1-y)ln(1+y)$, and then, using (10), we get the following:

$$\begin{array}{ccc}{v}_1(y,\tau )\hfill & =\hfill & {\chi }_1{v}_0(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_0\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & \widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_0\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & \widehat{h}{\mathcal{L}}^{-1}[\mathcal{L}\left\{v_0(y,\tau )\right\}-\left({\displaystyle \frac{1}{s}}v(x,0)+{\displaystyle \frac{1}{s^2}}{v}_{\tau }(x,0)\right)\hfill \\ \hfill & \hfill & \\ \hfill & \hfill & +{\displaystyle \frac{2}{s^{\alpha }}}\left(\mathcal{L}\left\{v_0(y,\tau )\right\}-{\displaystyle \frac{1}{s}}v(y,0)\right)-{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}\{{\displaystyle \frac{1}{y}}{\left(v_0(y,\tau )\right)}_y+{\left(v_0(y,\tau )\right)}_{yy}-g(y,\tau )\}]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & \widehat{h}\left(-{\displaystyle \frac{2{\tau }^{2\alpha }}{\Gamma [1+2\alpha ]}}+\tau \left(-1-{\displaystyle \frac{6{\tau }^{2\alpha }}{\Gamma [2+2\alpha ]}}-{\displaystyle \frac{12{\tau }^{1+2\alpha }}{\Gamma [3+2\alpha ]}}\right)\right),\hfill \end{array}$$

$$\begin{array}{ccc}{v}_2(y,\tau )\hfill & =\hfill & {\chi }_2{v}_1(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_1\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & v_1(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{L}\left\{v_1(y,\tau )\right\}+{\displaystyle \frac{2}{s^{\alpha }}}\mathcal{L}\left\{v_1(y,\tau )\right\}-{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}\{{\displaystyle \frac{1}{y}}{\left(v_1(y,\tau )\right)}_y+{\left(v_1(y,\tau )\right)}_{yy}\}\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & \widehat{h}\left(-\tau -\widehat{h}\tau -{\displaystyle \frac{2\widehat{h}\lambda {\tau }^{1+\alpha }}{\Gamma [2+\alpha ]}}-{\displaystyle \frac{2(1+\widehat{h})\lambda {\tau }^{2\alpha }}{\Gamma [1+2\alpha ]}}-{\displaystyle \frac{6{\tau }^{1+2\alpha }}{\Gamma [2+2\alpha ]}}-{\displaystyle \frac{6\widehat{h}{\tau }^{1+2\alpha }}{\Gamma [2+2\alpha ]}}\right)\hfill \\ \hfill & \hfill & \\ \hfill & \hfill & -\widehat{h}\left({\displaystyle \frac{12\lambda {\tau }^{2+2\alpha }}{\Gamma [3+2\alpha ]}}-{\displaystyle \frac{12\widehat{h}\lambda {\tau }^{2+2\alpha }}{\Gamma [3+2\alpha ]}}-{\displaystyle \frac{4\widehat{h}{\lambda }^2{\tau }^{3\alpha }}{\Gamma [1+3\alpha ]}}-{\displaystyle \frac{12\widehat{h}\lambda {\tau }^{1+3\alpha }}{\Gamma [2+3\alpha ]}}-{\displaystyle \frac{24\widehat{h}{\lambda }^2{\tau }^{2+3\alpha }}{\Gamma [3+3\alpha ]}}\right).\hfill \end{array}$$

Thus, continue in this manner; the numerical series solution becomes

$$\begin{array}{ccc}v(y,\tau )\hfill & =\hfill & v_0(y,\tau )+v_1(y,\tau )+v_2(y,\tau )+\dots \hfill \\ \hfill & =\hfill & {\displaystyle -2\widehat{h}\tau -{\widehat{h}}^2\tau -\frac{2{\widehat{h}}^2\lambda {\tau }^{1+\alpha }}{\Gamma [2+\alpha ]}-\frac{2\widehat{h}(2+\widehat{h})\lambda {\tau }^{2\alpha }}{\Gamma [1+2\alpha ]}-\frac{12\widehat{h}{\tau }^{1+2\alpha }}{\Gamma [2+2\alpha ]}-\frac{6{\widehat{h}}^2{\tau }^{1+2\alpha }}{\Gamma [2+2\alpha ]}}\hfill \\ \hfill & \hfill & {\displaystyle -\frac{24\widehat{h}\lambda {\tau }^{2+2\alpha }}{\Gamma [3+2\alpha ]}-\frac{12{\widehat{h}}^2\lambda {\tau }^{2+2\alpha }}{\Gamma [3+2\alpha ]}-\frac{4{\widehat{h}}^2{\lambda }^2{\tau }^{3\alpha }}{\Gamma [1+3\alpha ]}-\frac{12{\widehat{h}}^2\lambda {\tau }^{1+3\alpha }}{\Gamma [2+3\alpha ]}}\hfill \\ \hfill & \hfill & {\displaystyle -\frac{24{\widehat{h}}^2{\lambda }^2{\tau }^{2+3\alpha }}{\Gamma [3+3\alpha ]}+(1-y)ln(1+y)+\dots .}\hfill \end{array}$$

(13)

The Figure 1 shows the $\widehat{h}$-curve determined by the approximate series solution of order 15 of Example 1. It shows that the values of the parameter $\widehat{h}$ leading to a convergent series solution are within the range $-1.7<\widehat{h}<-0.4$.

On the other hand, Figure 2 demonstrates the plots of several successive terminated approximate series solutions of Example 1 of various orders at $y=0.01$, $\widehat{h}=-0.65$, and $\alpha =0.8$. It shows that these iterative solutions converge rapidly toward a certain analytical approximate solution.

Moreover, in

Table 1. Iterative solutions of Example 1 produced by a terminated series solution with the first n terms of the resulting approximate series solution, $v^{\left[n\right]}$, at $y=0.01$ and $\tau =0.1$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.6	1	$0.148526$	$0.113400$	$0.0935031$	$0.0824746$
	5	$0.181403$	$0.147728$	$0.127451$	$0.116170$
	10	$0.181310$	$0.147725$	$0.127582$	$0.116441$
	15	$0.181310$	$0.147724$	$0.127581$	$0.116440$
	20	$0.181310$	$0.147724$	$0.127581$	$0.116440$
	30	$0.181310$	$0.147724$	$0.127581$	$0.116440$
−0.7	1	$0.171639$	$0.130658$	$0.107445$	$0.0945786$
	5	$0.181320$	$0.147753$	$0.127603$	$0.116436$
	10	$0.181310$	$0.147724$	$0.127581$	$0.116440$
	15	$0.181310$	$0.147724$	$0.127581$	$0.116440$
	20	$0.181310$	$0.147724$	$0.127581$	$0.116440$
	30	$0.181310$	$0.147724$	$0.127581$	$0.116440$
−1.0	1	$0.240977$	$0.182433$	$0.149271$	$0.130890$
	5	$0.181531$	$0.147749$	$0.127584$	$0.116441$
	10	$0.181310$	$0.147724$	$0.127581$	$0.116440$
	15	$0.181310$	$0.147724$	$0.127581$	$0.116440$
	20	$0.181310$	$0.147724$	$0.127581$	$0.116440$
	30	$0.181310$	$0.147724$	$0.127581$	$0.116440$

,

Table 2. Iterative solutions of Example 1 produced via a terminated series solution with the first n terms of the resulting approximate series solution, $v^{\left[n\right]}$, at $y=0.5$, $\tau =0.5$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.6	1	$1.401040$	$1.164880$	$0.983745$	$0.847383$
	5	$1.276900$	$1.132670$	$1.010230$	$0.909232$
	10	$1.277140$	$1.133100$	$1.010270$	$0.908675$
	15	$1.277140$	$1.133100$	$1.010270$	$0.908680$
	20	$1.277140$	$1.133100$	$1.010270$	$0.908680$
	30	$1.277140$	$1.133100$	$1.010270$	$0.908680$
−0.7	1	$1.600760$	$1.325240$	$1.113910$	$0.954825$
	5	$1.277230$	$1.133190$	$1.010100$	$0.908462$
	10	$1.277140$	$1.133100$	$1.010270$	$0.908680$
	15	$1.277140$	$1.133100$	$1.010270$	$0.908680$
	20	$1.277140$	$1.133100$	$1.010270$	$0.908680$
	30	$1.277140$	$1.133100$	$1.010270$	$0.908680$
−0.9	1	$2.000190$	$1.645950$	$1.374250$	$1.169710$
	5	$1.304150$	$1.137230$	$1.010640$	$0.908656$
	10	$1.277120$	$1.133100$	$1.010270$	$0.908680$
	15	$1.277140$	$1.133100$	$1.010270$	$0.908680$
	20	$1.277140$	$1.133100$	$1.010270$	$0.908680$
	30	$1.277140$	$1.133100$	$1.010270$	$0.908680$

and

Table 3. Iterative solutions of Example 1 produced via a terminated series solution with the first n terms of the approximate series solution, $v^{\left[n\right]}$, at $y=0.9$, $\tau =1$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.6	1	$4.16671$	$3.54813$	$3.01015$	$2.55051$
	5	$3.17977$	$2.87886$	$2.58876$	$2.31516$
	10	$3.17874$	$2.8778$	$2.58972$	$2.31752$
	15	$3.17874$	$2.8778$	$2.58972$	$2.31751$
	20	$3.17874$	$2.8778$	$2.58972$	$2.31751$
	30	$3.17874$	$2.8778$	$2.58972$	$2.31751$
−0.7	1	$4.85046$	$4.12879$	$3.50114$	$2.9649$
	5	$3.17945$	$2.87646$	$2.59062$	$2.31842$
	10	$3.17874$	$2.8778$	$2.58972$	$2.31751$
	15	$3.17874$	$2.8778$	$2.58972$	$2.31751$
	20	$3.17874$	$2.8778$	$2.58972$	$2.31751$
	30	$3.17874$	$2.8778$	$2.58972$	$2.31751$
−0.9	1	$6.21797$	$5.29011$	$4.48313$	$3.79367$
	5	$3.93687$	$3.10888$	$2.65127$	$2.33097$
	10	$3.16343$	$2.87722$	$2.58971$	$2.31751$
	15	$3.17879$	$2.8778$	$2.58972$	$2.31751$
	20	$3.17874$	$2.8778$	$2.58972$	$2.31751$
	30	$3.17874$	$2.8778$	$2.58972$	$2.31751$

, we present the values of approximate solutions of Example 1 obtained via terminated series solutions of distinct orders at different values of y, $\tau$, and various values of $\widehat{h}$, $\alpha$, and n. The tables show that these values converge rapidly towards an analytical solution of the equation after just a few iterations.

Example 2.

Consider the following equation:

$${\partial }_{\tau }^{2\alpha }v(y,\tau )+2{\partial }_{\tau }^{\alpha }v(y,\tau )-{\displaystyle \frac{1}{y}}{\displaystyle \frac{\partial }{\partial y}}\left(y{\displaystyle \frac{\partial v}{\partial y}}\right)=g(y,\tau ),0<\alpha \le 1,0<y\le 1,\tau \in [0,1],$$

(14)

subject to the conditions:

$$v(y,0)=sin\left(y\right),v_{\tau }(y,0)=0,0<y\le 1,$$

(15)

where

$$g(y,\tau )=6\tau +6{\tau }^2-{\displaystyle \frac{cos\left(x\right)}{x}}+sin\left(x\right).$$

Solution.

Let $v_0(y,\tau )=v(y,0)=sin\left(y\right)$, and then, using (10), we get the following:

$$\begin{array}{ccc}{v}_1(y,\tau )\hfill & =\hfill & {\chi }_1{v}_0(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_0\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & \widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_0\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & \widehat{h}{\mathcal{L}}^{-1}[\mathcal{L}\left\{v_0(y,\tau )\right\}-\left({\displaystyle \frac{1}{s}}v(x,0)+{\displaystyle \frac{1}{s^2}}{v}_{\tau }(x,0)\right)\hfill \\ \hfill & \hfill & \\ \hfill & \hfill & +{\displaystyle \frac{2}{s^{\alpha }}}\left(\mathcal{L}\left\{v_0(y,\tau )\right\}-{\displaystyle \frac{1}{s}}v(y,0)\right)-{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}\{{\displaystyle \frac{1}{y}}{\left(v_0(y,\tau )\right)}_y+{\left(v_0(y,\tau )\right)}_{yy}-g(y,\tau )\}]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & -6\widehat{h}{\tau }^{1+2\alpha }\left({\displaystyle \frac{1}{\Gamma [2+2\alpha ]}}+{\displaystyle \frac{2\lambda \tau }{\Gamma [3+2\alpha ]}}\right),\hfill \end{array}$$

$$\begin{array}{ccc}{v}_2(y,\tau )\hfill & =\hfill & {\chi }_2{v}_1(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_1\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & v_1(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{L}\left\{v_1(y,\tau )\right\}+{\displaystyle \frac{2}{s^{\alpha }}}\mathcal{L}\left\{v_1(y,\tau )\right\}-{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}\{{\displaystyle \frac{1}{y}}{\left(v_1(y,\tau )\right)}_y+{\left(v_1(y,\tau )\right)}_{yy}\}\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & 6\widehat{h}{\tau }^{1+2\alpha }\left(-{\displaystyle \frac{1+\widehat{h}}{\Gamma \left[2\right(1+\alpha \left)\right]}}+2\lambda \left(-{\displaystyle \frac{(1+\widehat{h})\tau }{\Gamma [3+2\alpha ]}}+\widehat{h}{\tau }^{\alpha }\left(-{\displaystyle \frac{1}{\Gamma [2+3\alpha ]}}-{\displaystyle \frac{2\lambda \tau }{\Gamma [3+3\alpha ]}}\right)\right)\right).\hfill \end{array}$$

Therefore, the numerical series solution is given as follows:

$$\begin{array}{ccc}v(y,\tau )\hfill & =\hfill & v_0(y,\tau )+v_1(y,\tau )+v_2(y,\tau )+\dots \hfill \\ \hfill & =\hfill & {\displaystyle sin\left(y\right)-\frac{6\widehat{h}(2+\widehat{h}){\tau }^{1+2\alpha }}{\Gamma \left[2\right(1+\alpha \left)\right]}-\frac{12\widehat{h}(2+\widehat{h})\lambda {\tau }^{2+2\alpha }}{\Gamma [3+2\alpha ]}-\frac{12{\widehat{h}}^2\lambda {\tau }^{1+3\alpha }}{\Gamma [2+3\alpha ]}-\frac{24{\widehat{h}}^2{\lambda }^2{\tau }^{2+3\alpha }}{\Gamma [3+3\alpha ]}+\dots .}\hfill \end{array}$$

(16)

The Figure 3 shows the $\widehat{h}$-curve resulting from the approximate series solution of order 11 of Example 2. The figure shows that the values of the parameter $\widehat{h}$ leading to a convergent series solution are within the range of $-1.4<\widehat{h}<-0.4$.

On the other hand, Figure 4 demonstrates plots of several successive approximate series solutions of Example 2 of various orders at $y=0.5$, $\widehat{h}=-0.7$, and $\alpha =0.7$. It follows from this figure that these iterative solutions converge rapidly towards an analytical solution.

In

Table 4. Iterative solutions of Example 2 produced via a terminated series solution with the first n terms of the approximate series solution, $v^{\left[n\right]}$ at $y=0.01$, and $\tau =0.1$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.6	1	$0.0199563$	$0.015090$	$0.0125678$	$0.0112793$
	5	$0.0231361$	$0.0171987$	$0.0138177$	$0.0119695$
	10	$0.0231409$	$0.0172115$	$0.0138314$	$0.0119800$
	15	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	20	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	30	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
−0.7	1	$0.0216157$	$0.0159384$	$0.0129958$	$0.0114925$
	5	$0.0231418$	$0.0172113$	$0.0138301$	$0.0119785$
	10	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	15	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	20	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	30	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
−0.9	1	$0.0249345$	$0.0176351$	$0.0138518$	$0.011919$
	5	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	10	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	15	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	20	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$
	30	$0.0231409$	$0.0172115$	$0.0138315$	$0.0119801$

,

Table 5. Iterative solutions of Example 2 produced via a terminated series solution with the first n terms of the approximate series solution, $v^{\left[n\right]}$ at $y=0.5$, $\tau =0.5$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.5	1	$0.832957$	$0.726161$	$0.649525$	$0.595339$
	5	$0.899558$	$0.799786$	$0.718445$	$0.653751$
	10	$0.899256$	$0.799430$	$0.718375$	$0.654101$
	15	$0.899258$	$0.799431$	$0.718371$	$0.654094$
	20	$0.899257$	$0.799431$	$0.718371$	$0.654094$
	30	$0.899257$	$0.799431$	$0.718371$	$0.654094$
−0.7	1	$0.974370$	$0.824855$	$0.717565$	$0.641705$
	5	$0.899281$	$0.799415$	$0.718356$	$0.654110$
	10	$0.899257$	$0.799431$	$0.718371$	$0.654094$
	15	$0.899257$	$0.799431$	$0.718371$	$0.654094$
	20	$0.899257$	$0.799431$	$0.718371$	$0.654094$
	30	$0.899257$	$0.799431$	$0.718371$	$0.654094$
−0.9	1	$1.115780$	$0.923549$	$0.785605$	$0.688071$
	5	$0.902940$	$0.799764$	$0.718380$	$0.654093$
	10	$0.899256$	$0.799431$	$0.718371$	$0.654094$
	15	$0.899257$	$0.799431$	$0.718371$	$0.654094$
	20	$0.899257$	$0.799431$	$0.718371$	$0.654094$
	30	$0.899257$	$0.799431$	$0.718371$	$0.654094$

and

Table 6. Iterative solutions of Example 2 produced via a terminated series solution with the first n terms of the approximate series solution, $v^{\left[n\right]}$ at $y=0.9$, $\tau =1$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.6	1	$3.196730$	$2.701220$	$2.289910$	$1.953870$
	5	$2.774130$	$2.492430$	$2.232430$	$1.996060$
	10	$2.773730$	$2.492660$	$2.233000$	$1.996180$
	15	$2.773730$	$2.492660$	$2.233000$	$1.996170$
	20	$2.773730$	$2.492660$	$2.233000$	$1.996170$
	13	$2.773730$	$2.492660$	$2.233000$	$1.996170$
	30	$2.773730$	$2.492660$	$2.233000$	$1.996170$
−0.7	1	$3.598960$	$3.020870$	$2.541010$	$2.148960$
	5	$2.773280$	$2.492690$	$2.233200$	$1.996060$
	10	$2.773730$	$2.492660$	$2.233000$	$1.996170$
	15	$2.773730$	$2.492660$	$2.233000$	$1.996170$
	20	$2.773730$	$2.492660$	$2.233000$	$1.996170$
	13	$2.773730$	$2.492660$	$2.233000$	$1.996170$
	30	$2.773730$	$2.492660$	$2.233000$	$1.996170$
−0.9	1	$4.403430$	$3.660170$	$3.043200$	$2.539140$
	5	$2.998080$	$2.544000$	$2.242180$	$1.997250$
	10	$2.771110$	$2.492600$	$2.233000$	$1.996170$
	15	$2.773740$	$2.492660$	$2.233000$	$1.996170$
	20	$2.773730$	$2.492660$	$2.233000$	$1.996170$
	30	$2.773730$	$2.492660$	$2.233000$	$1.996170$

, we present the values of approximate solutions of Example 2 obtained via terminated series solutions of distinct orders at different values of y and $\tau$ and various values of $\widehat{h}$, $\alpha$, and n. The tables show that these values converge rapidly towards an analytical solution of the equation after just a few iterations.

Example 3.

Consider the following equation:

$${\partial }_{\tau }^{2\alpha }v(y,\tau )+2{\partial }_{\tau }^{\alpha }v(y,\tau )-{\displaystyle \frac{1}{y}}{\displaystyle \frac{\partial }{\partial y}}\left(y{\displaystyle \frac{\partial v}{\partial y}}\right)=10+20\tau -{\displaystyle \frac{e^y}{y}}(1+y),0<\alpha \le 1,0<y\le 1,\tau \in [0,1],$$

(17)

subject to the conditions

$$v(y,0)=e^y+3,v_{\tau }(y,0)=0,0<y\le 1.$$

(18)

Solution.

Assume $v_0(y,\tau )=v(y,0)=e^y+3$, and then, using (10), we get the following:

$$\begin{array}{ccc}{v}_1(y,\tau )\hfill & =\hfill & {\chi }_1{v}_0(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_0\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & \widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_0\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & \widehat{h}{\mathcal{L}}^{-1}[\mathcal{L}\left\{v_0(y,\tau )\right\}-\left({\displaystyle \frac{1}{s}}v(x,0)+{\displaystyle \frac{1}{s^2}}{v}_{\tau }(x,0)\right)\hfill \\ \hfill & \hfill & \\ \hfill & \hfill & +{\displaystyle \frac{2}{s^{\alpha }}}\left(\mathcal{L}\left\{v_0(y,\tau )\right\}-{\displaystyle \frac{1}{s}}v(y,0)\right)-{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}\{{\displaystyle \frac{1}{y}}{\left(v_0(y,\tau )\right)}_y+{\left(v_0(y,\tau )\right)}_{yy}-10-20\tau +{\displaystyle \frac{e^y}{y}}+e^y\}]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & -10\widehat{h}{\tau }^{2\alpha }\left({\displaystyle \frac{1}{\Gamma [1+2\alpha ]}}+{\displaystyle \frac{2\lambda \tau }{\Gamma [2+2\alpha ]}}\right),\hfill \end{array}$$

$$\begin{array}{ccc}{v}_2(y,\tau )\hfill & =\hfill & {\chi }_2{v}_1(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{R}\left({\overrightarrow{v}}_1\right)\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & v_1(y,\tau )+\widehat{h}{\mathcal{L}}^{-1}\left[\mathcal{L}\left\{v_1(y,\tau )\right\}+{\displaystyle \frac{2}{s^{\alpha }}}\mathcal{L}\left\{v_1(y,\tau )\right\}-{\displaystyle \frac{1}{s^{2\alpha }}}\mathcal{L}\{{\displaystyle \frac{1}{y}}{\left(v_1(y,\tau )\right)}_y+{\left(v_1(y,\tau )\right)}_{yy}\}\right]\hfill \\ \hfill & \hfill & \\ \hfill & =\hfill & 10\widehat{h}{\tau }^{2\alpha }\left(-{\displaystyle \frac{2\widehat{h}\tau }{\Gamma \left[2\right(1+\alpha \left)\right]}}-{\displaystyle \frac{1+\widehat{h}}{\Gamma [1+2\alpha ]}}+2\left(-{\displaystyle \frac{\tau }{\Gamma [2+2\alpha ]}}+\widehat{h}{\tau }^{\alpha }\left(-{\displaystyle \frac{1}{\Gamma [1+3\alpha ]}}-{\displaystyle \frac{2\tau }{\Gamma [2+3\alpha ]}}\right)\right)\right).\hfill \end{array}$$

Hence, the approximate power series solution is given as follows:

$$\begin{array}{ccc}v(y,\tau )\hfill & =\hfill & v_0(y,\tau )+v_1(y,\tau )+v_2(y,\tau )+\dots \hfill \\ \hfill & =\hfill & {\displaystyle 3+e^y-\frac{20{\widehat{h}}^2\lambda {\tau }^{1+2\alpha }}{\Gamma \left[2\right(1+\alpha \left)\right]}-\frac{10\widehat{h}(2+\widehat{h}){\tau }^{2\alpha }}{\Gamma [1+2\alpha ]}-\frac{40\widehat{h}\lambda {\tau }^{1+2\alpha }}{\Gamma [2+2\alpha ]}-\frac{20{\widehat{h}}^2\lambda {\tau }^{3\alpha }}{\Gamma [1+3\alpha ]}-\frac{40{\widehat{h}}^2{\lambda }^2{\tau }^{1+3\alpha }}{\Gamma [2+3\alpha ]}+\dots .}\hfill \end{array}$$

(19)

The Figure 5 shows the $\widehat{h}$-curve resulting from the approximate series solution of Example 3 of order 9. The figure shows that the values of the parameter $\widehat{h}$ leading to a convergent series solution are within the range of $-1.3<\widehat{h}<-0.35$.

On the other hand, Figure 6 demonstrates plots of several successive approximate series solutions of Example 3 of various orders at $y=0.3$, $\widehat{h}=-0.7$, and $\alpha =0.75$. It follows from this figure that these iterative solutions converge rapidly towards an analytical solution.

In

Table 7. Iterative solutions of Example 3 produced via a terminated series solution with the first n terms of the approximate series solution, $v^{\left[n\right]}$, at $y=0.01$, $\tau =0.1$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.7	1	$4.44735$	$4.25309$	$4.14250$	$4.08095$
	5	$4.47737$	$4.29257$	$4.17407$	$4.10195$
	10	$4.47734$	$4.29253$	$4.17408$	$4.10199$
	15	$4.47734$	$4.29253$	$4.17408$	$4.10199$
	30	$4.47734$	$4.29253$	$4.17408$	$4.10199$
−1	1	$4.63477$	$4.35725$	$4.19927$	$4.11134$
	5	$4.47788$	$4.29256$	$4.17408$	$4.10199$
	10	$4.47734$	$4.29253$	$4.17408$	$4.10199$
	15	$4.47734$	$4.29253$	$4.17408$	$4.10199$
	30	$4.47734$	$4.29253$	$4.17408$	$4.10199$
−1.1	1	$4.69724$	$4.39197$	$4.21819$	$4.12147$
	5	$4.48227$	$4.29325$	$4.17420$	$4.10201$
	10	$4.47733$	$4.29253$	$4.17408$	$4.10199$
	15	$4.47734$	$4.29253$	$4.17408$	$4.10199$
	30	$4.47734$	$4.29253$	$4.17408$	$4.10199$

,

Table 8. Iterative solutions of Example 3 produced via a terminated series solution with the first n terms of the approximate series solution, $v^{\left[n\right]}$ at $y=0.5$, $\tau =0.5$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.5	1	$7.52187$	$6.80952$	$6.24618$	$5.81108$
	6	$7.72845$	$7.19403$	$6.70699$	$6.27365$
	12	$7.72905$	$7.19307$	$6.70462$	$6.27141$
	18	$7.72904$	$7.19308$	$6.70462$	$6.27141$
	30	$7.72904$	$7.19308$	$6.70463$	$6.27141$
−0.7	1	$8.67112$	$7.67384$	$6.88516$	$6.27602$
	6	$7.72898$	$7.19316$	$6.70458$	$6.27133$
	8	$7.72904$	$7.19308$	$6.70463$	$6.27140$
	12	$7.72904$	$7.19308$	$6.70463$	$6.27141$
	18	$7.72904$	$7.19308$	$6.70463$	$6.27141$
	30	$7.72904$	$7.19308$	$6.70463$	$6.27141$
−1	1	$10.3950$	$8.97032$	$7.84364$	$6.97344$
	6	$7.51528$	$7.16007$	$6.70011$	$6.27085$
	12	$7.72801$	$7.19306$	$6.70463$	$6.27141$
	18	$7.72904$	$7.19308$	$6.70463$	$6.27141$
	30	$7.72904$	$7.19308$	$6.70463$	$6.27141$

and

Table 9. Iterative solutions of Example 3 produced via a terminated series solution with the first n terms of the approximate series solution, $v^{\left[n\right]}$, at $y=0.9$, $\tau =1$, and different values of $\widehat{h}$, n, and $\alpha$.

			$\mathit{\alpha }$
		0.6	0.7	0.8	0.9
$\widehat{\mathit{h}}$	$\mathit{n}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$	${\mathit{v}}^{\left[\mathit{n}\right]}$
−0.5	1	$14.1231$	$12.8392$	$11.6473$	$10.5723$
	5	$13.0699$	$12.4454$	$11.8010$	$11.1441$
	10	$13.0752$	$12.4549$	$11.8038$	$11.1342$
	15	$13.0752$	$12.4548$	$11.8038$	$11.1344$
	20	$13.0752$	$12.4548$	$11.8038$	$11.1344$
	30	$13.0752$	$12.4548$	$11.8038$	$11.1344$
−0.7	1	$17.5885$	$15.7910$	$14.1224$	$12.6174$
	5	$13.0768$	$12.4518$	$11.8064$	$11.1352$
	10	$13.0752$	$12.4548$	$11.8038$	$11.1344$
	15	$13.0752$	$12.4548$	$11.8038$	$11.1344$
	20	$13.0752$	$12.4548$	$11.8038$	$11.1344$
	30	$13.0752$	$12.4548$	$11.8038$	$11.1344$
−0.9	1	$21.0539$	$18.7428$	$16.5975$	$14.6625$
	5	$15.2320$	$13.0382$	$11.9296$	$11.1535$
	10	$13.0321$	$12.4535$	$11.8038$	$11.1344$
	15	$13.0753$	$12.4548$	$11.8038$	$11.1344$
	20	$13.0752$	$12.4548$	$11.8038$	$11.1344$
	30	$13.0752$	$12.4548$	$11.8038$	$11.1344$

we present the values of approximate solutions of Example 3 obtained via terminated series solutions of distinct orders at different values of y, $\tau$, and various values of $\widehat{h}$, $\alpha$, and n. The tables show that these values converge rapidly towards an analytical solution of the equation after just a few iterations.

5. Conclusions

In this article, numerical solutions of a fractional-order telegraph equation have been determined using the LTHAM. The derived iterative scheme was applied to three test examples to investigate its efficiency and accuracy. The implementation of this scheme provided approximate series solutions that converge rapidly for each example. The convergence of these approximate series solutions was examined both geometrically and numerically. Figure 2, Figure 4 and Figure 6 demonstrate the rapid convergence of these solutions as the order of the truncated series is increased. Additionally, numerical results in Table 1, Table 2, Table 3, Table 4, Table 5, Table 6, Table 7, Table 8 and Table 9 support this conclusion and show the rapid convergence of the obtained approximate solutions for various values of the independent variables and the parameter $\widehat{h}$. Our analysis of these results confirms the accuracy and efficiency of the derived numerical scheme, establishing the LTHAM as an effective and reliable analytical technique for solving the fractional telegraph equation and similar equations.