Computational Analysis for Solving the Linear Space-Fractional Telegraph Equation

Abstract

Over the last few years, there has been a significant increase in attention paid to fractional differential equations, given their wide array of applications in the fields of physics and engineering. The recent development of using fractional telegraph equations as models in some fields (e.g., the thermal diffusion in fractal media) has heightened the importance of examining the method of solutions for such equations (both approximate and analytic). The present work is designed to serve as a valuable contribution to work in this field. The key objective of this work is to propose a general framework that can be used to guide quadratic spline functions in order to create a numerical method for obtaining an approximation solution using the linear space-fractional telegraph equation. Additionally, the Von Neumann method was employed to measure the stability of the analytical scheme, which showed that the proposed method is conditionally stable. What’s more, the proposal contains a numerical example that illustrates how the proposed method can be implemented practically, whilst the error estimates and numerical stability results are discussed in depth. The findings indicate that the proposed model is highly effective, convenient and accurate for solving the relevant problems and is suitable for use with approximate solutions acquired through the two-dimensional differential transform method that has been developed for linear partial differential equations with space- and time-fractional derivatives.

Share and Cite:

Alaofi, Z. , El-Danaf, T. , Hadhoud, A. and Dragomir, S. (2022) Computational Analysis for Solving the Linear Space-Fractional Telegraph Equation. Open Journal of Modelling and Simulation, 10, 267-282. doi: 10.4236/ojmsi.2022.103014.

1. Introduction

In this context, this paper proposes a quadratic-polynomial spline-based method to obtain the numerical solution of the time-space fractional-order telegraph equation in the form:

$\frac{{\partial }^{\alpha }u}{\partial {x}^{\alpha }}=\frac{{\partial }^{2}u}{\partial {t}^{2}}+\frac{\partial u}{\partial t}+u,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}1<\alpha \le 2,$ (1)

subject to boundary conditions

$u\left(a,t\right)={\beta }_{1}\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}u\left(b,t\right)={\beta }_{2}\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}t>0$ (2)

and the initial conditions

$u\left(x,0\right)={f}_{1}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial u\left(x,0\right)}{\partial t}={f}_{2}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}a\le x\le b$ (3)

The space-fractional partial derivative of order $\alpha$ in Equation (1) is considered in the Caputo sense, defined by [5] [6],

$\frac{{\partial }^{\alpha }}{\partial {x}^{\alpha }}u\left(x,{t}_{j}\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\int }_{g}^{x}\frac{{\partial }^{n}u\left(s,{t}_{j}\right)}{\partial {x}^{n}}{\left(x-s\right)}^{n-\alpha -1}\text{d}s,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n-1<\alpha \le n.$ (4)

2. Derivation of the Method

To set up the quadratic polynomial spline method, select an integer $N>0$ and time-step size $k>0$. With $h=\frac{b-a}{N}$, then mesh points $\left({x}_{i},{t}_{j}\right)$ are ${x}_{i}=a+ih$, for each $i=0,1,\cdots ,N$, and ${t}_{j}=jk,\text{\hspace{0.17em}}k=\Delta t$ for each $j=0,1,\cdots$.

Let ${Z}_{i}^{j}$ be an approximation to $u\left({x}_{i},{t}_{j}\right)$ obtained by the segment ${P}_{i}\left(x,{t}_{j}\right)$ of the spline function passing through the points $\left({x}_{i},{Z}_{i}^{j}\right)$ and $\left({x}_{i+1},{Z}_{i+1}^{j}\right)$. Each segment has the form [30]

${P}_{i}\left(x,{t}_{j}\right)={a}_{i}\left({t}_{j}\right){\left(x-{x}_{i}\right)}^{2}+{b}_{i}\left({t}_{j}\right)\left(x-{x}_{i}\right)+{c}_{i}\left({t}_{j}\right)$ (5)

for each $i=0,1,\cdots ,N-1$. To obtain expressions for the coefficients of (5) in terms of ${Z}_{i+1/2}^{j}$, ${D}_{i}^{j}$, and ${S}_{i+1/2}^{j}$, we first define

${P}_{i}\left({x}_{i+1/2},{t}_{j}\right)={Z}_{i+1/2}^{j}$ (6)

${P}_{i}^{\left(1\right)}\left({x}_{i},{t}_{j}\right)={D}_{i}^{j}$ (7)

${P}_{i}^{\left(\alpha \right)}\left({x}_{i+1/2},{t}_{j}\right)=\frac{{\partial }^{\alpha }}{\partial {x}^{\alpha }}{P}_{i}\left({x}_{i+1/2},{t}_{j}\right)={S}_{i+1/2}^{j},\text{\hspace{0.17em}}\text{\hspace{0.17em}}1<\alpha \le 2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{i}<{x}_{i+1/2}\le {x}_{i+1}$ (8)

where ${a}_{i}\equiv {a}_{i}\left({t}_{j}\right),\text{\hspace{0.17em}}{b}_{i}\equiv {b}_{i}\left({t}_{j}\right),\text{\hspace{0.17em}}{c}_{i}\equiv {c}_{i}\left({t}_{j}\right),\text{\hspace{0.17em}}{d}_{i}\equiv {d}_{i}\left({t}_{j}\right)$ and $\theta =\omega h$. Equations (5), (6) and (7), give

$\frac{{h}^{2}}{4}{a}_{i}+\frac{h}{2}{b}_{i}+{c}_{i}={Z}_{i+1/2}^{j}$ (9)

${b}_{i}={D}_{i}^{j}$ (10)

Using Equations (4), (5), and (8), we obtain

$\frac{{\partial }^{\alpha }}{\partial {x}^{\alpha }}u\left({x}_{i+1/2},{t}_{j}\right)=\frac{1}{\Gamma \left(2-\alpha \right)}{\int }_{{x}_{i}}^{{x}_{i+1/2}}\frac{{\partial }^{2}{P}_{i}\left(s,{t}_{j}\right)}{\partial {x}^{2}}{\left({x}_{i+1/2}-s\right)}^{1-\alpha }\text{d}s={S}_{i+1/2}^{j}$.

This equation can be simplified as:

$\mu {a}_{i}={S}_{i+1/2}^{j}$ (11)

where $\mu =\frac{2}{\Gamma \left(3-\alpha \right)}{\left(\frac{h}{2}\right)}^{2-\alpha }$. By solving Equations (9), (10), and (11), we obtain the following expressions:

${a}_{i}=\frac{\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha -2}{S}_{i+1/2}^{j}$,

${b}_{i}={D}_{i}^{j}$,

${c}_{i}=-\frac{1}{2}\Gamma \left(3-\alpha \right){\left(\frac{h}{2}\right)}^{\alpha }{S}_{i+1/2}^{j}-\frac{h}{2}{D}_{i}^{j}+{Z}_{i+1/2}^{j}$ (12)

Spline Relations

Using the following continuity conditions at $x={x}_{i}$

${P}_{i}\left({x}_{i},{t}_{j}\right)={P}_{i-1}\left({x}_{i},{t}_{j}\right)⇒{c}_{i}={h}^{2}{a}_{i-1}+h{b}_{i-1}+{c}_{i-1}$ (13)

${P}_{i}^{\left(1\right)}\left({x}_{i},{t}_{j}\right)={P}_{i-1}^{\left(1\right)}\left({x}_{i},{t}_{j}\right)⇒{b}_{i}=2h{a}_{i-1}+{b}_{i-1}$ (14)

Using expressions in Equation (12), Equations (13) and (14) become

$\begin{array}{l}{Z}_{i+1/2}^{j}-{Z}_{i-1/2}^{j}-\frac{\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha }\left({S}_{i+1/2}^{j}-{S}_{i-1/2}^{j}\right)-\frac{h}{2}\left({D}_{i}^{j}-{D}_{i-1}^{j}\right)\\ =\frac{4\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha }{S}_{i-1/2}^{j}+h{D}_{i-1}^{j}\end{array}$ (15)

${D}_{i}^{j}-{D}_{i-1}^{j}=\frac{4\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha -1}{S}_{i-1/2}^{j}$ (16)

By solving for ${D}_{i-1}^{j}$

$h{D}_{i-1}^{j}=\left({Z}_{i+1/2}^{j}-{Z}_{i-1/2}^{j}\right)-\frac{\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha }{S}_{i+1/2}^{j}-\frac{7\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha }{S}_{i-1/2}^{j}$ (17)

Similarly,

$h{D}_{i}^{j}=\left({Z}_{i+3/2}^{j}-{Z}_{i+1/2}^{j}\right)-\frac{\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha }{S}_{i+3/2}^{j}-\frac{7\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha }{S}_{i+1/2}^{j}$ (18)

Using expressions in Equations (17) and (18), then Equation (16) becomes

${Z}_{i+3/2}^{j}-2{Z}_{i+1/2}^{j}+{Z}_{i-1/2}^{j}=\delta \left({S}_{i+3/2}^{j}+6{S}_{i+1/2}^{j}+{S}_{i-1/2}^{j}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,\cdots ,N-2$ (19)

where $\delta =\frac{\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha }$. As $\alpha \to 2$, system (19) reduces to

${Z}_{i+3/2}^{j}-2{Z}_{i+1/2}^{j}+{Z}_{i-1/2}^{j}=\frac{{h}^{2}}{8}\left({S}_{i+3/2}^{j}+6{S}_{i+1/2}^{j}+{S}_{i-1/2}^{j}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,\cdots ,N-2.$ (20)

Remark

The truncation error for Equation (19), that is,

${T}^{*}{}_{i}^{j}=\left({u}_{i-1/2}^{j}+{u}_{i+3/2}^{j}\right)-2{u}_{i+1/2}^{j}-\delta \left({D}_{x}^{2}{u}_{i-1/2}^{j}+{D}_{x}^{2}{u}_{i+3/2}^{j}\right)-6\delta {D}_{x}^{2}{u}_{i+1/2}^{j}$

can be obtained by expanding this equation in Taylor series in terms of $u\left({x}_{i+1/2},{t}_{j}\right)$ and its derivatives as follows

${T}^{*}{}_{i}^{j}=\left({h}^{2}-8\delta \right){D}_{x}^{2}{u}_{i+1/2}^{j}+\left(\frac{{h}^{4}}{12}-\delta {h}^{2}\right){D}_{x}^{4}{u}_{i+1/2}^{j}+\left(\frac{{h}^{6}}{360}-\frac{\delta {h}^{4}}{12}\right){D}_{x}^{6}{u}_{i+1/2}^{j}+\cdots$

Since $\delta =\frac{\Gamma \left(3-\alpha \right)}{2}{\left(\frac{h}{2}\right)}^{\alpha }$ then the last expression can be simplified as

$\begin{array}{c}{T}^{*}{}_{i}^{j}={h}^{\alpha }\left({h}^{2-\alpha }-8\theta \right){D}_{x}^{2}{u}_{i+1/2}^{j}+{h}^{2+\alpha }\left(\frac{{h}^{2-\alpha }}{12}-\theta \right){D}_{x}^{4}{u}_{i+1/2}^{j}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{h}^{4+\alpha }\left(\frac{{h}^{2-\alpha }}{360}-\frac{\theta }{12}\right){D}_{x}^{6}{u}_{i+1/2}^{j}+\cdots \end{array}$

where $\theta =\frac{\Gamma \left(3-\alpha \right)}{{2}^{\alpha +1}}$. From this expression of the local truncation error, our scheme is $O\left({h}^{\alpha }\right),1<\alpha \le 2$.

${S}_{i}^{j}=\frac{{\partial }^{\alpha }{Z}_{i}^{j}}{\partial {x}^{\alpha }}=\frac{{\partial }^{2}{Z}_{i}^{j}}{\partial {t}^{2}}+\frac{\partial {Z}_{i}^{j}}{\partial t}+{Z}_{i}^{j}$ (21)

${S}_{i}^{j}=\frac{{\partial }^{\alpha }{Z}_{i}^{j}}{\partial {x}^{\alpha }}\approx \frac{{Z}_{i}^{j+1}-2{Z}_{i}^{j}+{Z}_{i}^{j-1}}{{k}^{2}}+\frac{{Z}_{i}^{j+1}-{Z}_{i}^{j-1}}{2k}+{Z}_{i}^{j}$ (22)

which can be discretised as follows:

${S}_{i-1/2}^{j}=\frac{{\partial }^{\alpha }{Z}_{i-1/2}^{j}}{\partial {x}^{\alpha }}\approx \frac{{Z}_{i-1/2}^{j+1}-2{Z}_{i-1/2}^{j}+{Z}_{i-1/2}^{j-1}}{{k}^{2}}+\frac{{Z}_{i-1/2}^{j+1}-{Z}_{i-1/2}^{j-1}}{2k}+{Z}_{i-1/2}^{j}$

${S}_{i+1/2}^{j}=\frac{{\partial }^{\alpha }{Z}_{i+1/2}^{j}}{\partial {x}^{\alpha }}\approx \frac{{Z}_{i+1/2}^{j+1}-2{Z}_{i+1/2}^{j}+{Z}_{i+1/2}^{j-1}}{{k}^{2}}+\frac{{Z}_{i+1/2}^{j+1}-{Z}_{i+1/2}^{j-1}}{2k}+{Z}_{i+1/2}^{j}$ (23)

${S}_{i+3/2}^{j}=\frac{{\partial }^{\alpha }{Z}_{i+3/2}^{j}}{\partial {x}^{\alpha }}\approx \frac{{Z}_{i+3/2}^{j+1}-2{Z}_{i+3/2}^{j}+{Z}_{i+3/2}^{j-1}}{{k}^{2}}+\frac{{Z}_{i+3/2}^{j+1}-{Z}_{i+3/2}^{j-1}}{2k}+{Z}_{i+3/2}^{j}$

Using Formulas (23) in (19) gives the following useful systems

$\begin{array}{l}A{Z}_{i-1/2}^{j+1}+B{Z}_{i+1/2}^{j+1}+A{Z}_{i+3/2}^{j+1}\\ ={A}^{*}{Z}_{i-1/2}^{j}+{B}^{*}{Z}_{i+1/2}^{j}+{A}^{*}{Z}_{i+3/2}^{j}+\stackrel{⌢}{A}{Z}_{i-1/2}^{j-1}+\stackrel{⌢}{B}{Z}_{i+1/2}^{j-1}+\stackrel{⌢}{C}{Z}_{i+3/2}^{j-1}\end{array}$ (24)

where

$A=\frac{\delta }{{k}^{2}}+\frac{\delta }{2k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{*}=1+\frac{2\delta }{{k}^{2}}-\delta \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{⌢}{A}=\frac{-\delta }{{k}^{2}}+\frac{\delta }{2k}$

$B=\frac{6\delta }{{k}^{2}}+\frac{3\delta }{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}^{*}=-2+\frac{12\delta }{{k}^{2}}-6\delta \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{⌢}{B}=\frac{-6\delta }{{k}^{2}}+\frac{3\delta }{k}$ (25)

System (24) consists of N − 2 equations in N unknowns. To get a solution to this system, we need 2-additional equations. Using the boundary conditions (2), that are ${Z}_{0}^{j}={\beta }_{1}\left(t\right),{Z}_{N+1}^{j}={\beta }_{2}\left(t\right)$, we can obtain the following equations: Suppose that ${Z}_{1/2}^{j}$ is linearly interpolated between ${Z}_{0}^{j}$ and ${Z}_{3/2}^{j}$

$-3{Z}_{1/2}^{j}+{Z}_{3/2}^{j}=-2{Z}_{0}^{j}=-2{\beta }_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}j\ge 0$ (26)

In a similar manner,

${Z}_{N-3/2}^{j}-3{Z}_{N-1/2}^{j}=-2{Z}_{N}^{j}=-2{\beta }_{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}j\ge 0$ (27)

Equation (24) implies that the (j + 1)st time step requires values from the (j)st and (j1)st time steps. This produces a minor starting problem since values for j = 0 are given by the first part in Equation (3)

${Z}_{i}^{0}=u\left({x}_{i},0\right)={f}_{1}\left({x}_{i}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,\cdots ,N$ (28)

but values for j = 0, which are needed in Equation (25) to compute ${Z}_{i}^{1}$, must be obtained from the first part in (3)

$\frac{\partial {Z}_{i}^{0}}{\partial t}={u}_{t}\left({x}_{i},0\right)={f}_{2}\left({x}_{i}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,\cdots ,N.$

One approach is to replace $\frac{\partial {Z}_{i}^{0}}{\partial t}$ by a forward-difference approximation

${f}_{2}\left({x}_{i}\right)=\frac{\partial {Z}_{i}^{0}}{\partial t}=\frac{{Z}_{i}^{1}-{Z}_{i}^{0}}{k}+o\left(k\right)$ (29)

which gives us

${Z}_{i}^{1}={Z}_{i}^{0}+k{f}_{2}\left({x}_{i}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,\cdots ,N.$ (30)

3. Stability Analysis

The Von Neumann technique will be carried out to investigate the stability of systems (23) and (24). The key part of Von Neumann analysis is to assume a solution of the form [31]

${Z}_{i}^{j}={\zeta }_{j}{\text{e}}^{\left(q\varphi h\right)}$ (31)

where $\varphi$ is the wave number, $q=\sqrt{-1}$, h is the element size, and $\zeta$ is the amplification factor of the scheme. The use of Equations (31) and (24) gives us the characteristic equation in the form

$\begin{array}{l}{\zeta }^{j+1}\left\{A{\text{e}}^{\left(\left(i-1\right)q\varphi h\right)}+B{\text{e}}^{\left(iq\varphi h\right)}+A{\text{e}}^{\left(\left(i+1\right)q\varphi h\right)}\right\}\\ ={\zeta }^{j}\left\{{A}^{*}{\text{e}}^{\left(\left(i-1\right)q\varphi h\right)}+{B}^{*}{\text{e}}^{\left(iq\varphi h\right)}+{A}^{*}{\text{e}}^{\left(\left(i+1\right)q\varphi h\right)}\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\zeta }^{j-1}\left\{\stackrel{⌢}{A}{\text{e}}^{\left(\left(i-1\right)q\varphi h\right)}+\stackrel{⌢}{B}{\text{e}}^{\left(iq\varphi h\right)}+\stackrel{⌢}{A}{\text{e}}^{\left(\left(i+1\right)q\varphi h\right)}\right\}\end{array}$ (32)

Dividing both sides of the last equation by ${\text{e}}^{\left(iq\varphi h\right)}$ and canceling the common term, which is ${\zeta }^{j-1}$, Equation (32) becomes:

$\begin{array}{l}{\zeta }^{2}\left\{A{\text{e}}^{\left(-q\varphi h\right)}+B+A{\text{e}}^{\left(q\varphi h\right)}\right\}-\zeta \left\{{A}^{*}{\text{e}}^{\left(-q\varphi h\right)}+{B}^{*}+{A}^{*}{\text{e}}^{\left(q\varphi h\right)}\right\}\\ -\left\{\stackrel{⌢}{A}{\text{e}}^{\left(-q\varphi h\right)}+\stackrel{⌢}{B}+\stackrel{⌢}{A}{\text{e}}^{\left(q\varphi h\right)}\right\}=0\end{array}$ (33)

where

$A=\frac{\delta }{{k}^{2}}+\frac{\delta }{2k}$, ${A}^{*}=1+\frac{2\delta }{{k}^{2}}-\delta$ and $\stackrel{⌢}{A}=\frac{-\delta }{{k}^{2}}+\frac{\delta }{2k}$

$B=\frac{6\delta }{{k}^{2}}+\frac{3\delta }{k}$, ${B}^{*}=-2+\frac{12\delta }{{k}^{2}}-6\delta$ and $\stackrel{⌢}{B}=\frac{-6\delta }{{k}^{2}}+\frac{3\delta }{k}$

This equation can be rewritten in the simple form

$a{\zeta }^{2}+b\zeta +c=0$ (34)

where

$a=\left(A{\text{e}}^{\left(-q\varphi h\right)}+B+A{\text{e}}^{\left(q\varphi h\right)}\right),\text{\hspace{0.17em}}b=-\left({A}^{*}{\text{e}}^{\left(-q\varphi h\right)}+{B}^{*}+{A}^{*}{\text{e}}^{\left(q\varphi h\right)}\right)$

and

$c=-\left(\stackrel{⌢}{A}{\text{e}}^{\left(-q\varphi h\right)}+\stackrel{⌢}{B}+\stackrel{⌢}{A}{\text{e}}^{\left(q\varphi h\right)}\right)$

Or

$a=B+2A\mathrm{cos}\phi ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=-{B}^{*}-2{A}^{*}\mathrm{cos}\phi ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=-\stackrel{⌢}{B}-2\stackrel{⌢}{A}\mathrm{cos}\phi ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\phi =h\varphi$

Or

$\begin{array}{l}a=\frac{\delta }{k}\left(3+\mathrm{cos}\phi \right)\left(\frac{2}{k}+1\right),\\ b=2\left(1-\mathrm{cos}\phi \right)-2\delta \left(3+\mathrm{cos}\phi \right)\left(\frac{2}{{k}^{2}}-1\right),\\ c=\frac{\delta }{k}\left(3+\mathrm{cos}\phi \right)\left(\frac{2}{k}-1\right)\end{array}$

Equation (34) is a quadratic in $\zeta$ and, hence, will have two roots, that is

${\varsigma }_{±}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

${\varsigma }_{±}=\sqrt{\frac{c}{a}}\left(-\psi ±\sqrt{{\psi }^{2}-1}\right),\psi =\frac{b}{2\sqrt{ac}}$

For the stability, we must have $|{\zeta }_{±}|\le 1$. So, we have three cases.

Case 1: The discriminant of the Quadratic equation (34) is zero, that is ${\psi }^{2}-1=0$, in that case ${\varsigma }_{±}=±\sqrt{\frac{c}{a}}=±\sqrt{\frac{2-k}{2+k}}$, $0 and the stability condition, $|{\zeta }_{±}|\le 1$, is satisfied.

Case 2: Discriminant is less than zero, that is ${\psi }^{2}-1<0$, in this case

${\varsigma }_{±}=\sqrt{\frac{c}{a}}\left(-\psi ±q\sqrt{1-{\psi }^{2}}\right)=\sqrt{\frac{2-k}{2+k}}\left(-\psi ±q\sqrt{1-{\psi }^{2}}\right)⇒$

the stability condition, $|{\zeta }_{±}|\le 1$, is satisfied.

Case 3: The discriminant is greater than zero. This means that one of ${\zeta }_{+}$ and ${\zeta }_{-}$ does not satisfy the stability condition.

Thus, for stability we must have ${\psi }^{2}-1\le 0$

$-1\le \psi \le 1$ (35)

$-1\le \frac{b}{2\sqrt{ac}}\le 1$

Since

$\sqrt{ac}>0⇒-2\sqrt{ac}\le b\le 2\sqrt{ac}$

$\begin{array}{c}-\frac{2\delta }{{k}^{2}}\left(3+\mathrm{cos}\phi \right)\sqrt{4-{k}^{2}}\le 2\left(1-\mathrm{cos}\phi \right)-2\delta \left(3+\mathrm{cos}\phi \right)\left(\frac{2}{{k}^{2}}-1\right)\\ \le \frac{2\delta }{{k}^{2}}\left(3+\mathrm{cos}\phi \right)\sqrt{4-{k}^{2}}\end{array}$

The right above inequality takes the form:

$2\left(1-\mathrm{cos}\phi \right)\le \frac{2\delta }{{k}^{2}}\left(3+\mathrm{cos}\phi \right)\left(\sqrt{4-{k}^{2}}+2-{k}^{2}\right)$

Which is satisfied for $k\ll \delta$, where h is small enough.

But the left above inequality takes the form:

$-2\left(1-\mathrm{cos}\phi \right)\le \frac{2\delta }{{k}^{2}}\left(3+\mathrm{cos}\phi \right)\left(\sqrt{4-{k}^{2}}-2+{k}^{2}\right)$

Which is satisfied for $k\ll \delta$, where h is small enough, and the method is then conditionally stable.

4. Numerical Example

In this section, a numerical example is included to illustrate the practical implementation of the proposed method.

Consider the following linear space-fractional telegraph equation [4]

$\frac{{\partial }^{1.5}u}{\partial {x}^{1.5}}=\frac{{\partial }^{2}u}{\partial {t}^{2}}+\frac{\partial u}{\partial t}+u,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x>0$ (36)

Subject to the initial condition

$u\left(x,0\right)=0$ (37)

and boundary conditions

$\begin{array}{c}u\left(0.0125,t\right)\approx \mathrm{exp}\left(-t\right)\left(1+0.0125\right)+\frac{{0.0125}^{1.5}}{\Gamma \left(5/2\right)}+\frac{{0.0125}^{2.5}}{\Gamma \left(7/2\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{0.0125}^{3}}{\Gamma \left(4\right)}+\frac{{0.0125}^{4}}{\Gamma \left(5\right)}+\cdots \end{array}$ (38)

and

$\begin{array}{c}u\left(1.0125,t\right)\approx \mathrm{exp}\left(-t\right)\left(1+1.0125\right)+\frac{{1.0125}^{1.5}}{\Gamma \left(5/2\right)}+\frac{{1.0125}^{2.5}}{\Gamma \left(7/2\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{1.0125}^{3}}{\Gamma \left(4\right)}+\frac{{1.0125}^{4}}{\Gamma \left(5\right)}+\cdots \end{array}$ (39)

Then the exact solution is

$u\left(x,t\right)\approx \mathrm{exp}\left(-t\right)\left(1+x+\frac{{x}^{1.5}}{\Gamma \left(5/2\right)}+\frac{{x}^{2.5}}{\Gamma \left(7/2\right)}+\frac{{x}^{3}}{\Gamma \left(4\right)}+\frac{{x}^{4}}{\Gamma \left(5\right)}+\cdots \right).$ (40)

Tables 1-3 illustrate the comparison between our method, developed in Section 3 and other existing methods [4] and [23] with $k=0.00005$, $h=0.025$, $t=0.05,0.1,0.15$ and $\alpha =1.5$.

Table 4 and Table 5 illustrate the comparison between our method, developed in Section 3 and other existing methods [4] and [23] with $k=0.00005$, $h=0.025$, $t=0.05,0.1$ and $\alpha =1.75$.

Using Tables 1-5 and Figures 1-6, it can be seen that the obtained approximate numerical solutions are in good agreement with the approximate solutions obtained using methods [4] and [23] for all values of x and t.

Figure 1. The comparison between our method and method [4] and [23] when t = 0.05, k = 0.000005, and h = 0.025 and α = 1.5.

Figure 2. The comparison between our method and method [4] and [23] whent = 0.1, k = 0.000005, and h = 0.025 and α = 1.5.

Figure 3. The comparison between our method and methods [4] and [23] when t = 0.15, k = 0.000005, and h = 0.025 and α = 1.5.

Figure 4. Comparison between our numerical method and methods [4] and [23] when t = 0.05, k = 0.00005, and h = 0.025, α = 1.75.

Table 1. Comparison between our numerical method and methods [4] and [23] whent = 0.05, k = 0.000005, and h = 0.025 and α = 1.5.

Table 2. Comparison between our numerical method and methods [4] and [23] when t = 0.1, k = 0.000005, and h = 0.025 and α = 1.5.

Figure 5. The 3-D behavior of the numerical solutions from t = 0.0005 to t = 0.05, k = 0.0005, and h = 0.025, α = 1.5.

Table 3. Comparison between our numerical method and methods [4] and [23] whent = 0.15, k = 0.000005, and h= 0.025 and α = 1.5.

Table 4. Comparison between our numerical method and methods [4] and [23] when t = 0.05, k = 0.00005, and h = 0.025, α = 1.75.

Figure 6. The 3-D behavior of the numerical solutions fromt = 0.0005 to t = 0.05, k = 0.0005, and h = 0.025, α = 1.75.

Table 5. Comparison between our numerical method and methods [4] and [23] when t = 0.1, k = 0.00005, and h = 0.025, α = 1.75.

5. Conclusion

In the present work, a numerical approach to solving the linear space-fractional telegraph equation has been proposed based on the quadratic polynomial spline. Von-Neumann stability analysis was performed, with the findings revealing that the model has high conditional stability. The numerical example is effective and supports the theoretical analysis that the numerical approach is accurate and effective in solving the time-space fractional-order telegraph equation. For the values of x and t, the approximate numerical solutions identified in this work are in line with approximate solutions acquired using the [4] and [23] methods. Additionally, it is important to note that the local truncation error of our proposed model is $O\left({h}^{\alpha }\right),1<\alpha \le 2$. It is reasonable to conclude that the proposed method is efficient and effective in identifying approximate solutions for many different linear partial differential equations of fractional order.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.