A Collocation Procedure for Treating the Time-Fractional FitzHugh–Nagumo Differential Equation Using Shifted Lucas Polynomials

Abstract

This work employs newly shifted Lucas polynomials to approximate solutions to the time-fractional Fitzhugh–Nagumo differential equation (TFFNDE) relevant to neuroscience. Novel essential formulae for the shifted Lucas polynomials are crucial for developing our suggested numerical approach. The analytic and inversion formulas are introduced, and after that, new formulas that express these polynomials’ integer and fractional derivatives are derived to facilitate the construction of integer and fractional operational matrices for the derivatives. Employing these operational matrices with the typical collocation method converts the TFFNDE into a system of algebraic equations that can be addressed with standard numerical solvers. The convergence analysis of the shifted Lucas expansion is carefully investigated. Certain inequalities involving the golden ratio are established in this context. The suggested numerical method is evaluated using several numerical examples to verify its applicability and efficiency.

1. Introduction

Special functions are essential in various fields, including mathematics, engineering, and physics. These functions possess certain characteristics that enhance their significance in numerous applications. Several books illustrate the outcomes and uses of specific functions; one can consult [1,2,3,4]. In particular, the Fibonacci and Lucas polynomials are important in different fields. These polynomials and their associated numbers are crucial due to their diverse applications in biology, physics, statistics, and computer science. Many authors studied several extensions and variations of these polynomials; see [5,6]. In [7], a numerical approach was followed for treating the advection-diffusion equations based on mixed Fibonacci–Lucas polynomials. The generalized Lucas polynomials were utilized in [8] to solve telegraph equations using the Galerkin method. Vieta–Lucas polynomials were employed to treat some variable-order fractional differential equations (FDEs) in [9]. Vieta–Fibonacci polynomials were utilized in [10], along with a matrix approach, to treat a variable-order reaction advection–diffusion equation in two dimensions. The same polynomials were used in [11], along with the wavelet collocation method, to treat a certain fractional-order system. The authors of [12] used modified Lucas polynomials to treat some FDEs. A finite difference method was applied in [13] to numerically solve the time-fractional Burgers’ equation using Lucas polynomials. In [14], an algorithm utilizing Lucas polynomials was applied to treat the generalized Benjamin–Bona–Mahony–Burgers equation. In [15], the authors used Lucas polynomials for the Cauchy integral equations.

Fractional calculus studies integration and derivatives for orders that are not whole numbers. FDEs help explain various significant phenomena in the applied sciences. Some applications of the different FDEs can be found in [16,17]. Many models in the applied sciences can be described using different FDEs. For example, fractional models capture memory- and route-dependency in financial data, modeling stock prices; see [18]. Some other models appear in applications related to chemical reactions; see, for example, [19]. Some other models appear in quantum mechanics; see, for example, [20]. In addition, these equations are vital in epidemiology, particularly in modeling disease spread in populations; see, for example, [21].

The analytical solutions for such equations are often unavailable, necessitating numerical methods. Various methods are used to treat multiple FDEs. The operational matrices technique was utilized in several works, such as those referenced in [22,23]; a collocation method was used in [24,25,26]; a Laplace domian decomposition method was applied in [27]; and wavelet methods were used in [28,29]. A predictor–corrector method was applied in [30] for treating some FDEs. In [31], the authors applied the homotopy method for non-linear FDEs. In [32], some implicit methods were applied to treat multi-dimensional distributed-order fractional–integro DEs. Another implicit difference approach was followed in [33] to treat the time-fractional Kuramoto–Sivashinsky equation. One-dimensional time-fractional Burgers equations were handled in [34]. A finite difference method was applied in [35] for some FDEs arising in engineering. The cubic B-spline collocation method was applied in [36] to solve certain non-linear FDEs.

Because of its central position in mathematical physics, the FitzHugh–Nagumo (F-N) equation has recently attracted much interest from mathematicians and physicists. Numerous domains use this equation, including nuclear reactor theory, neurophysiology, branching Brownian motion, flame propagation, and logistic population growth [37]. The time-fractional FitzHugh–Nagumo differential equation (TFFNDE) is an extension of the F-N equation. The following fractional differential equation models it [38]:

$$D_t^{\zeta }\chi ={\chi }_{\xi \xi }+\chi (\chi -\theta )(1-\chi ),0<\zeta <1,$$

(1)

constrained with

$$\begin{array}{c}\hfill \chi (\xi ,0)={\chi }_0\left(\xi \right),0<\xi \le 1,\end{array}$$

(2)

$$\begin{array}{c}\hfill \chi (0,t)={\chi }_1\left(t\right),\chi (1,t)={\chi }_2\left(t\right),0<t\le 1,\end{array}$$

(3)

where $\chi$ is a function of both $\xi$ and t, i.e., $\chi =\chi (\xi ,t)$ and $\theta$ is a real number. Several authors treated these equations. For example, in [39], semi-analytical solutions were presented for the TFFNDE. In [40], the non-linear TFFNDE was treated. A numerical algorithm was proposed in [41]. A finite difference method was applied in [42] to solve the TFFNDE in irregular domains.

Spectral methods can significantly influence numerical analysis; see [1,2]. A key concept behind these methods is that approximations comprise selected special functions or polynomials. Differential and integral equations can be numerically solved using three spectral approaches, each with variations. It is possible to select different trial and test functions for each variant. The Galerkin approach implies selecting the trial functions that match their underlying conditions; see, for example, [43,44,45]. The tau approach differs from the approach of the Galerkin method, in that one may choose any polynomials as basis functions; see, for example, [46,47]. The collocation method is advantageous due to its applicability for all differential equations; see, for example, [48,49,50,51].

When solving ordinary DEs and FDEs, deriving explicit derivative expressions as combinations of their original ones is pivotal in many numerical approaches. These formulas yield operational matrices of integer derivatives and FDs to solve different DEs. These operational matrices transform the equations under investigation into matrix systems that can be treated using suitable solvers. For example, in [52], the authors established two types of operational matrices of derivatives for some modified Chebyshev polynomials to treat certain singular Emden–Fowler equations. In [53], the authors used a Laguerre operational matrix to treat FDEs with a non-singular kernel. In [54], the authors used Vieta–Fibonacci operational matrices for some variable-order fractional–integro DEs.

This paper is confined to introducing and utilizing the shifted Lucas polynomials to treat the TFFNDE. Our suggested collocation approach relies on some novel formulas of these polynomials. To our knowledge, this is the first utilization of these polynomials in the scope of numerical solutions of DEs. The following is a summary of the article’s main points:

• Introducing the shifted Lucas polynomials.
• Deriving the essential formulas of these polynomials, particularly their series representation and inversion formula.
• Deriving the integer derivatives and FDs of these polynomials.
• Studying the convergence of our approximate solution using the shifted Lucas expansion.
• Designing the collocation procedure for treating the TFFNDE.
• Testing our numerical algorithm by presenting some examples.

To the best of our knowledge, the main contributions and the novelty of this paper can be outlined in the following points:

• A new theoretical background to the shifted Lucas polynomials. More precisely, new forms of their power form representation, inversion formula, and integer- and fractional-derivative formulas will be developed.
• These new formulas provide new insights into using these polynomials in numerical analysis.
• The study of the convergence of the double-shifted Lucas expansion is new.
• In addition, we expect that the introduced polynomials will open new horizons for using other non-orthogonal polynomials in numerical analysis.

The advantages of the presented approach can be listed as follows:

• By choosing the shifted Lucas polynomials as the basis functions, a few terms of the retained modes make it possible to produce approximations with excellent precision.
• Less calculation is required to obtain the desired approximate solution.

We also note that the presented collocation algorithm for treating the TFFNDE is new, which motivates us to analyze it.

The following is the structure of the paper. Section 2 presents some fundamentals and essential formulas. Section 3 introduces new shifted Lucas polynomials and develops some new formulas of these polynomials. A collocation approach is followed in Section 4 to solve the TFFNDE. A study of the convergence of the shifted Lucas expansion is given in Section 5. Three test problems are given in Section 6. Finally, some conclusions are reported in Section 7.

2. Some Fundamentals and Important Formulas

This section presents some properties of the fractional calculus involving Caputo’s fractional derivative (FD). In addition, we will present some fundamentals and important formulas related to Lucas polynomials and their shifted ones, which will be essential in what follows.

2.1. An Account on Caputo’s FD

Definition 1.

Caputo’s FD is defined as [55]:

$${\displaystyle D_z^{\zeta }Y\left(z\right)=\frac{1}{\Gamma (r-\zeta )}{\int }_0^z{(z-t)}^{r-\zeta -1}{Y}^{\left(r\right)}\left(t\right)dt,\zeta >0,z>0,}$$

(4)

$$r-1⩽\zeta <r,r\in \mathbb{N}.$$

In addition, we have

$$\begin{array}{cc}\hfill D_z^{\zeta }C& =0,\left(C\mathrm{is}\mathrm{a}\mathrm{constant}\right),\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill D_z^{\zeta }{z}^k& =\left\{\begin{array}{cc}0,\hfill & \mathrm{if}k\in {\mathbb{N}}_{0}andk<\lceil \zeta \rceil ,\hfill \\ {\displaystyle \frac{\Gamma (k+1)}{\Gamma (k+1-\zeta )}}{z}^{k-\zeta },\hfill & \mathrm{if}k\in {\mathbb{N}}_{0}andk\ge \lceil \zeta \rceil ,\hfill \end{array}\right.\hfill \end{array}$$

(6)

where $\mathbb{N}=\{1,2,\dots \},{\mathbb{N}}_0=\{0,1,2,\dots \}$, and $\lceil \zeta \rceil$ denotes the ceiling function.

2.2. A Brief Account of Lucas Polynomials

Lucas polynomials may be constructed using the recurrence relation [56]

$$L_m\left(t\right)=t{L}_{m-1}\left(t\right)+L_{m-2}\left(t\right),L_0\left(t\right)=2,L_1\left(t\right)=t,m\ge 2,$$

(7)

alternatively, using the series

$$L_m\left(t\right)=m\sum _{\mathsf{\ell }=0}^{\lfloor {\displaystyle \frac{m}{2}}\rfloor }{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{m-\mathsf{\ell }}{\mathsf{\ell }}\right)}{m-\mathsf{\ell }}}{t}^{m-2\mathsf{\ell }},m\ge 1.$$

(8)

They also may be expressed using the following Binet’s form:

$$L_m\left(t\right)={\displaystyle \frac{{\left(t+\sqrt{t^2+4}\right)}^m+{\left(t-\sqrt{t^2+4}\right)}^m}{2^m}},m\ge 0.$$

(9)

The following theorem explicitly expresses the derivatives of Lucas polynomials as a combination of Lucas polynomials.

Theorem 1

([57]). Let $p,r\in \mathbb{Z}\ge 0$ with $r\ge p$. We have the following expression:

$$D^p{L}_r\left(\xi \right)=r\sum _{\mathsf{\ell }=0}^{⌊{\displaystyle \frac{r-p}{2}}⌋}{c}_{r-2\mathsf{\ell }-p}{(-1)}^{\mathsf{\ell }}\left({\displaystyle \genfrac{}{}{0pt}{}{\mathsf{\ell }+p-1}{\mathsf{\ell }}}\right){(r-\mathsf{\ell }-p+1)}_{p-1}{L}_{r-2\mathsf{\ell }-p}\left(\xi \right),$$

(10)

where

$$c_s=\left\{\begin{array}{cc}{\displaystyle \frac{1}{2}},\hfill & s=0,\hfill \\ 1,\hfill & s\ge 1.\hfill \end{array}\right.$$

(11)

3. Introducing Shifted Lucas Polynomials

In this section, we introduce the shifted Lucas polynomials that will be utilized as basis functions to propose our numerical algorithm. These polynomials are defined as

$$L_k^{\ast }\left(\xi \right)=L_k(2\xi -1).$$

(12)

Finding the counterpart of the standard Lucas formulas for the shifted Lucas polynomials is useful. The following two theorems exhibit the series form of $L_k^{\ast }\left(\xi \right)$ and its inverse formula.

Theorem 2.

Let $i\in {\mathbb{Z}}^{+}$. The series form of $L_i^{\ast }\left(\xi \right)$ has the following form:

$$L_i^{\ast }\left(\xi \right)=\sum _{r=0}^i{\zeta }_{r,i}{\xi }^r,$$

(13)

where

$${\zeta }_{r,i}={(-1)}^{i+r}{2}^r\left({\displaystyle \genfrac{}{}{0pt}{}{i}{r}}\right) {}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{1}{2}}(r-i),{\displaystyle \frac{1}{2}}(1-i+r)\\ 1-i\end{array}|-4\right).$$

(14)

Proof. 

From the series in (8), we can write the series of $L_i^{\ast }\left(\xi \right)$ as

$$L_i^{\ast }\left(\xi \right)=i\sum _{k=0}^{⌊{\displaystyle \frac{i}{2}}⌋}{\displaystyle \frac{{(1+i-2k)}_{k-1}}{k!}}{(2\xi -1)}^{i-2k},$$

(15)

which can be written using the binomial theorem in the form

$$L_i^{\ast }\left(\xi \right)=i{(-1)}^i\sum _{k=0}^{⌊{\displaystyle \frac{i}{2}}⌋}{\displaystyle \frac{{(1+i-2k)}_{k-1}}{k!}}\sum _{r=0}^{i-2k}{(-2)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{i-2k}{r}}\right){\xi }^r,$$

(16)

which is written alternatively as

$$L_i^{\ast }\left(\xi \right)={(-1)}^{i}i\sum _{r=0}^i{(-2)}^r\sum _{\mathsf{\ell }=0}^{⌊{\displaystyle \frac{i}{2}}⌋}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{i-2\mathsf{\ell }}{r}\right){(1+i-2\mathsf{\ell })}_{\mathsf{\ell }-1}}{\mathsf{\ell }!}}{\xi }^r.$$

(17)

The last formula has the form

$$L_i^{\ast }\left(\xi \right)={(-1)}^i\sum _{r=0}^i{(-2)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{i}{r}}\right){}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{1}{2}}(r-i),{\displaystyle \frac{1}{2}}(1-i+r)\\ 1-i\end{array}|-4\right){\xi }^r,$$

(18)

and thus, Theorem 2 is now proven. □

Theorem 3.

The inverse formula of (13) is given

$${\xi }^k=\sum _{s=0}^k{\lambda }_{s,k}{L}_s^{\ast }\left(\xi \right),$$

(19)

where

$${\lambda }_{s,k}=\left\{\begin{array}{cc}{\displaystyle \frac{{(-1)}^{\frac{k-s}{2}}k!c_s}{2^k\frac{k-s}{2}!\frac{k+s}{2}!}{}_{2}F_1\left(\begin{array}{c}\frac{1}{2}(-k-s),\frac{1}{2}(-k+s)\\ \frac{1}{2}\end{array}|-\frac{1}{4}\right),}\hfill & (s+k)even,\hfill \\ {\displaystyle \frac{{(-1)}^{\frac{1}{2}(k-s-1)}k!c_s}{2^k\left(\frac{1}{2}(k-s-1)\right)!\left(\frac{1}{2}(k+s-1)\right)!}{}_{2}F_1\left(\begin{array}{c}\frac{1}{2}(1-k-s),\frac{1}{2}(1-k+s)\\ \frac{3}{2}\end{array}|-\frac{1}{4}\right),}\hfill & (s+k)odd.\hfill \end{array}\right.$$

(20)

Proof. 

The proof of Formula (19) is lengthy but straightforward, obtained by induction on k. □

In the following theorem, we present the integer derivatives of the shifted Lucas polynomials as combinations of their original ones.

Theorem 4

([57]). Let $i,r$ be non-negative integers with $i\ge r$. The following derivatives of $L_i^{\ast }\left(\xi \right)$ are given by

$$D^r{L}_i^{\ast }\left(\xi \right)=\sum _{k=0}^{i-r}{G}_{k,i,r}{L}_k^{\ast }\left(\xi \right),$$

(21)

where

$$G_{k,i,r}={\displaystyle \frac{{\eta }_{k,i,r}{c}_k{(-1)}^{\frac{1}{2}(i-k-r)}i{2}^r{\left(\frac{1}{2}(2+i+k-r)\right)}_{-1+r}{\left(r\right)}_{\frac{1}{2}(i-k-r)}}{\left(\frac{1}{2}(i-k-r)\right)!}},$$

with

$${\eta }_{k,i,r}=\left\{\begin{array}{cc}1,\hfill & (i-k-r)even,\hfill \\ 0,\hfill & otherwise,\hfill \end{array}\right.$$

and $c_k$ is as defined in (11).

Proof. 

It is a direct consequence of (10), only replacing $\xi$ with $(2\xi -1)$. □

Collary 1.

The second derivative of $L_i^{\ast }\left(\xi \right)$ has the following expression:

$${\displaystyle \frac{d^2{L}_i^{\ast }\left(\xi \right)}{d{\xi }^2}}=\sum _{k=0}^{i-2}{(-1)}^{{\displaystyle \frac{i-k+2}{2}}}{c}_{k}i(i-k)(i+k){\eta }_{k,i,2}{L}_k^{\ast }\left(\xi \right),i\ge 2.$$

(22)

Proof. 

It is a special case of Theorem 4 for $r=2$. □

Remark 1.

Consider the vector ${\mathcal{L}}^{\ast }\left(\xi \right)={[L_0^{\ast }\left(\xi \right),L_1^{\ast }\left(\xi \right),\dots ,L_{\mathcal{M}}^{\ast }\left(\xi \right)]}^T$. The second derivative of ${\mathcal{L}}^{\ast }\left(\xi \right)$ may be expressed in matrix form as

$${\displaystyle \frac{d^2{\mathcal{L}}^{\ast }\left(\xi \right)}{d{\xi }^2}}=\mathit{H}{\mathcal{L}}^{\ast }\left(\xi \right),$$

(23)

where $\mathit{H}=\left(h_{i,k}\right)$ is the operational matrix of derivatives of the order ${(\mathcal{M}+1)}^2$, and the entries of this matrix may be expressed as

$$h_{i,k}=\left\{\begin{array}{cc}{(-1)}^{{\displaystyle \frac{i-k+2}{2}}}{c}_{k}i(i-k)(i+k),\hfill & ifi\ge k+2,(i+k)even,\hfill \\ 0,\hfill & otherwise,\hfill \end{array}\right.$$

(24)

and $c_k$ is as defined in (11).

The following theorem explicitly expresses the fractional derivatives of the shifted Lucas polynomials. This formula and the formula of the integer derivatives will be the fundamental keys to deriving our proposed algorithm in this paper.

Theorem 5.

The following formula holds for $\zeta \in (0,1)$

$$D_t^{\zeta }{L}_j^{\ast }\left(t\right)=t^{-\zeta }\left(\sum _{n=0}^j{\mathcal{U}}_{n,j}^{\zeta }{L}_n^{\ast }\left(t\right)-{\rho }_j\right),$$

(25)

where

$${\mathcal{U}}_{n,j}^{\zeta }=\sum _{s=n}^j{\displaystyle \frac{s!{\zeta }_{s,j}{\lambda }_{n,s}}{\Gamma (L-\zeta +1)}},$$

(26)

${\zeta }_{r,i}$  is defined in (14) and

$${\rho }_j={\displaystyle \frac{{\zeta }_{0,j}}{\Gamma (1-\zeta )}}.$$

(27)

Proof. 

Formula (13) allows one to write $D_t^{\zeta }{L}_j^{\ast }\left(t\right)$ as

$$D_t^{\zeta }{L}_j^{\ast }\left(t\right)=\sum _{n=1}^j{\displaystyle \frac{n!{\zeta }_{n,j}}{\Gamma (n-\zeta +1)}}{t}^{n-\zeta }.$$

(28)

Now, Equation (28) can be rewritten using Equation (19) as

$$D_t^{\zeta }{L}_j^{\ast }\left(t\right)=t^{-\zeta }\sum _{n=1}^j\sum _{s=0}^n{\displaystyle \frac{n!{\zeta }_{n,j}{\lambda }_{s,n}}{\Gamma (n-\zeta +1)}}{L}_s^{\ast }\left(t\right),$$

(29)

which has the alternative form

$$D_t^{\zeta }{L}_j^{\ast }\left(t\right)=t^{-\zeta }\left(\sum _{n=0}^j{\mathcal{U}}_{n,j}^{\zeta }{L}_j^{\ast }\left(t\right)-{\rho }_j\right),$$

(30)

where

$${\mathcal{U}}_{n,j}^{\zeta }=\sum _{s=n}^j{\displaystyle \frac{s!{\zeta }_{s,j}{\lambda }_{n,s}}{\Gamma (s-\zeta +1)}},$$

(31)

and

$${\rho }_j={\displaystyle \frac{{\zeta }_{0,j}}{\Gamma (1-\zeta )}}.$$

(32)

This completes the proof. □

Remark 2.

If ${\mathcal{L}}^{\ast }\left(t\right)={[L_0^{\ast }\left(t\right),L_1^{\ast }\left(t\right),\dots ,L_{\mathcal{M}}^{\ast }\left(t\right)]}^T$, then the FD of ${\mathcal{L}}^{\ast }\left(t\right)$ may be expressed in matrix form as

$$D_t^{\zeta }{\mathcal{L}}^{\ast }\left(t\right)=t^{-\zeta }(\mathcal{U}{\mathcal{L}}^{\ast }\left(t\right)-\mathit{\rho }),$$

(33)

where $\mathit{\rho }={[{\rho }_0,{\rho }_1,\dots ,{\rho }_{\mathcal{M}}]}^T$. Also, $\mathcal{U}=\left({\mathcal{U}}_{n,j}^{\zeta }\right)$ is the operational matrix of the FD of the order ${(\mathcal{M}+1)}^2$, and the entries of this matrix can be represented by (26).

4. Collocation Procedure for the TFFNDE

This section is confined to analyzing a numerical algorithm to solve (1) governed by the initial and boundary conditions (2)–(3).

The Algorithm of the Method

Define

$${\mathcal{A}}_N({\rm Y})=\mathrm{span}\{L_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right):0\le i,j\le \mathcal{M}\},$$

${\rm Y}=(0,1)\times (0,1)$.

Consequently, any $\overline{\chi }(\xi ,t)\in {\mathcal{A}}_N({\rm Y})$ can be represented as

$$\overline{\chi }=\overline{\chi }(\xi ,t)=\sum _{i=0}^{\mathcal{M}}\sum _{j=0}^{\mathcal{M}}{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)={\left({\mathcal{L}}^{\ast }\left(\xi \right)\right)}^T\mathit{C}{\mathcal{L}}^{\ast }\left(t\right),$$

(34)

where ${\mathcal{L}}^{\ast }\left(\xi \right)={[L_0^{\ast }\left(\xi \right),L_1^{\ast }\left(\xi \right),\dots ,L_{\mathcal{M}}^{\ast }\left(\xi \right)]}^T,{\mathcal{L}}^{\ast }\left(t\right)={[L_0^{\ast }\left(t\right),L_1^{\ast }\left(t\right),\dots ,L_{\mathcal{M}}^{\ast }\left(t\right)]}^T,$ and $\mathit{C}={\left(c_{ij}\right)}_{0\le i,j\le \mathcal{M}}$ is the unknown matrix whose order is ${(\mathcal{M}+1)}^2$.

Now, the residual $\mathbf{Re}(\xi ,t)$ of Equation (1) has the form

$$\mathbf{Re}(\xi ,t)=D_t^{\zeta }\overline{\chi }-{\overline{\chi }}_{\xi \xi }-\overline{\chi }(\overline{\chi }-\theta )(1-\overline{\chi }).$$

(35)

Using Remarks 1 and 2 and the expansion (34), we can write $\mathbf{Re}(\xi ,t)$ given by (35) as in the following matrix form:

$$\begin{array}{cc}\hfill \mathbf{Re}(\xi ,t)& ={\mathcal{L}}^{\ast }{\left(\xi \right)}^T\mathit{C}{t}^{-\zeta }(\mathcal{U}{\mathcal{L}}^{\ast }\left(t\right)-\mathit{\rho })-{\left[\mathit{H}{\mathcal{L}}^{\ast }\left(\xi \right)\right]}^T\mathit{C}{\mathcal{L}}^{\ast }\left(t\right)\hfill \\ & -\left({\mathcal{L}}^{\ast }{\left(\xi \right)}^T\mathit{C}{\mathcal{L}}^{\ast }\left(t\right)\right)({\mathcal{L}}^{\ast }{\left(\xi \right)}^T\mathit{C}{\mathcal{L}}^{\ast }\left(t\right)-\theta )(1-{\mathcal{L}}^{\ast }{\left(\xi \right)}^T\mathit{C}{\mathcal{L}}^{\ast }\left(t\right)).\hfill \end{array}$$

(36)

Applying the spectral collocation method at some collocation points $({\xi }_i,t_j)=\left({\displaystyle \frac{i}{\mathcal{M}+1}},{\displaystyle \frac{j}{\mathcal{M}+1}}\right)$, yields

$$\mathbf{Re}\left({\displaystyle \frac{i}{\mathcal{M}+1}},{\displaystyle \frac{j}{\mathcal{M}+1}}\right)=0,i=1,2,\dots ,\mathcal{M}-1,j=1,2,\dots ,\mathcal{M}.$$

(37)

Furthermore, the constraints (2) and (3) lead to

$$\begin{array}{cc}& {\mathcal{L}}^{\ast }{\left({\displaystyle \frac{i}{\mathcal{M}+1}}\right)}^T\mathcal{C}{\mathcal{L}}^{\ast }\left(0\right)={\chi }_0\left({\displaystyle \frac{i}{\mathcal{M}+1}}\right),i=1,2,\dots ,\mathcal{M}+1,\hfill \\ & {\mathcal{L}}^{\ast }{\left(0\right)}^T\mathcal{C}{\mathcal{L}}^{\ast }\left({\displaystyle \frac{j}{\mathcal{M}+1}}\right)={\chi }_1\left({\displaystyle \frac{j}{\mathcal{M}+1}}\right),j=1,2,\dots ,\mathcal{M},\hfill \\ & {\mathcal{L}}^{\ast }{\left(1\right)}^T\mathcal{C}{\mathcal{L}}^{\ast }\left({\displaystyle \frac{j}{\mathcal{M}+1}}\right)={\chi }_2\left({\displaystyle \frac{j}{\mathcal{M}+1}}\right),j=1,2,\dots ,\mathcal{M}.\hfill \end{array}$$

(38)

We may use Newton’s iterative approach to numerically treat the non-linear system of Equations (37) and (38) with the dimension ${(\mathcal{M}+1)}^2$ in the unknowns $c_{ij}$.

5. The Convergence and Error Analysis

Lemma 1.

Let $t\in [0,1]$. This inequality holds:

$$L_i^{\ast }\left(t\right)\le 2{\varphi }^i,\forall i\ge 0,$$

where $\varphi ={\displaystyle \frac{1+\sqrt{5}}{2}}$ that represents the golden ratio.

Proof. 

Since $L_i^{\ast }\left(t\right)\le L_i^{\ast }\left(1\right)=L_i$, and ${\varphi }^i={\displaystyle \frac{L_i+\sqrt{5}{F}_i}{2}}$, where $L_i$ and $F_i$ are the Lucas and Fibonacci numbers, respectively. Then, the following estimation can be obtained:

$$L_i^{\ast }\left(t\right)\le 2{\varphi }^i,\forall i\ge 0,$$

and this proves the lemma. □

Lemma 2.

Consider the infinitely differentiable function $g\left(t\right)$ at the origin. We have

$$g\left(t\right)=\sum _{n=0}^{\infty }\sum _{m=n}^{\infty }{\displaystyle \frac{g^{\left(m\right)}\left(0\right){\lambda }_{n,m}}{m!}}{L}_n^{\ast }\left(t\right),$$

(39)

where ${\lambda }_{n,m}$ is as defined in (20).

Proof. 

Consider the following expansion of $g\left(t\right)$:

$$g\left(t\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{g^{\left(k\right)}\left(0\right)}{k!}}{t}^k.$$

(40)

Using Formula (19), the previous equation can be expressed as

$$g\left(t\right)=\sum _{k=0}^{\infty }\sum _{r=0}^k{\displaystyle \frac{g^{\left(k\right)}\left(0\right){\lambda }_{r,k}}{k!}}{L}_r^{\ast }\left(t\right),$$

(41)

that can also be expressed as

$$g\left(t\right)=\sum _{n=0}^{\infty }\sum _{m=n}^{\infty }{\displaystyle \frac{g^{\left(m\right)}\left(0\right){\lambda }_{n,m}}{m!}}{L}_n^{\ast }\left(t\right),$$

(42)

that finalizes the proof. □

Theorem 6.

If $g\left(t\right)$ is defined on $[0,1]$ and $|g^{\left(n\right)}\left(0\right)|\le {\mu }^n,n>0,$ where $\mu >0$, and ${\displaystyle g\left(t\right)=\sum _{n=0}^{\infty }{a}_n{L}_n^{\ast }\left(t\right)}$; then, we obtain

$$|a_n|\lesssim {\displaystyle \frac{{\mu }^n}{n!}},$$

(43)

where $y\lesssim z$, meaning that there exists a generic constant c such that $y\le cz$. Moreover, the series converges absolutely.

Proof. 

Using Lemma 2, we obtain

$$a_n=\sum _{m=n}^{\infty }{\displaystyle \frac{g^{\left(m\right)}\left(0\right){\lambda }_{n,m}}{m!}},$$

(44)

$$\begin{array}{cc}\hfill |a_n|& \le \sum _{m=n}^{\infty }{\displaystyle \frac{{\mu }^m\left|{\lambda }_{n,m}\right|}{m!}}\hfill \\ \hfill & \le \sum _{m=n}^{\infty }\left\{\begin{array}{cc}{\displaystyle \frac{2^{-m}{\mu }^m}{\left(\frac{m-n}{2}\right)!\left(\frac{m+n}{2}\right)!}}{}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{1}{2}}(-m-n),{\displaystyle \frac{n-m}{2}}\\ {\displaystyle \frac{1}{2}}\end{array}|-{\displaystyle \frac{1}{4}}\right),\hfill & (n+m)\mathrm{even},\hfill \\ {\displaystyle \frac{2^{-m}{\mu }^m}{\Gamma \left(\frac{1}{2}(m-n+1)\right)\Gamma \left(\frac{1}{2}(m+n+1)\right)}}{}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{1}{2}}(-m-n+1),{\displaystyle \frac{1}{2}}(-m+n+1)\\ {\displaystyle \frac{3}{2}}\end{array}|-{\displaystyle \frac{1}{4}}\right),\hfill & (n+m)\mathrm{odd}.\hfill \end{array}\right.\hfill \end{array}$$

(45)

Now, for all $0\le n<\infty$ and $n\le m<\infty$, the following estimations can be deduced:

$$\begin{array}{cc}& {\displaystyle \frac{2^{-m}{\mu }^m}{\left(\frac{m-n}{2}\right)!\left(\frac{m+n}{2}\right)!}}{}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{1}{2}}(-m-n),{\displaystyle \frac{n-m}{2}}\\ {\displaystyle \frac{1}{2}}\end{array}|-{\displaystyle \frac{1}{4}}\right)\lesssim {\displaystyle \frac{{\mu }^n}{n!}},\hfill \\ & {\displaystyle \frac{2^{-m}{\mu }^m}{\Gamma \left(\frac{1}{2}(m-n+1)\right)\Gamma \left(\frac{1}{2}(m+n+1)\right)}}{}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{1}{2}}(-m-n+1),{\displaystyle \frac{1}{2}}(-m+n+1)\\ {\displaystyle \frac{3}{2}}\end{array}|-{\displaystyle \frac{1}{4}}\right)\lesssim {\displaystyle \frac{{\mu }^n}{n!}}.\hfill \end{array}$$

(46)

Consequently, we obtain the following estimation:

$$|a_n|\lesssim {\displaystyle \frac{{\mu }^n}{n!}}.$$

(47)

To demonstrate the theorem’s second part, since

$$\sum _{n=0}^{\infty }|a_n{L}_n^{\ast }\left(t\right)|=\sum _{n=0}^{\infty }|a_n||L_n^{\ast }\left(t\right)|\lesssim \sum _{n=0}^{\infty }{\displaystyle \frac{{\mu }^n{\varphi }^n}{n!}}=e^{\mu \varphi },$$

(48)

therefore, the series converges absolutely. □

Theorem 7.

If a function ${\displaystyle \chi =\chi (\xi ,t)={\chi }_1\left(\xi \right){\chi }_2\left(t\right)=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)}$, with $|{\chi }_1^{\left(i\right)}\left(0\right)|\le {\mu }_1^i$ and $|{\chi }_2^{\left(i\right)}\left(0\right)|\le {\mu }_2^i$, where ${\mu }_1,{\mu }_2$ are positive constants. We obtain

$$|c_{ij}|\lesssim {\displaystyle \frac{{{\mu }_1}^i{{\mu }_2}^j}{i!j!}}.$$

(49)

Moreover, the series converges absolutely.

Proof. 

Using Lemma 2 along with the assumptions of the theorem $\chi ={\chi }_1\left(\xi \right){\chi }_2\left(t\right)$, we can write

$$\chi =\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }\sum _{s=i}^{\infty }\sum _{r=j}^{\infty }{\displaystyle \frac{{\chi }_1^{\left(s\right)}\left(0\right){\chi }_2^{\left(r\right)}\left(0\right){\lambda }_{j,r}{\lambda }_{i,s}}{s!r!}}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right).$$

(50)

Now, the expansion coefficients $c_{ij}$ can be written as

$$c_{ij}=\sum _{s=i}^{\infty }\sum _{r=j}^{\infty }{\displaystyle \frac{{\chi }_1^{\left(s\right)}\left(0\right){\chi }_2^{\left(r\right)}\left(0\right){\lambda }_{j,r}{\lambda }_{i,s}}{s!r!}}.$$

(51)

Using the assumption $|{\chi }_1^{\left(i\right)}\left(0\right)|\le {\mu }_1^i$ and $|{\chi }_2^{\left(i\right)}\left(0\right)|\le {\mu }_2^i$, we obtain

$$|c_{ij}|\le \sum _{s=i}^{\infty }{\displaystyle \frac{{\mu }_1^s|{\lambda }_{i,s}|}{s!}}\times \sum _{r=j}^{\infty }{\displaystyle \frac{{\mu }_2^r|{\lambda }_{j,r}|}{r!}}.$$

(52)

To obtain the desired outcome, we follow the same procedure outlined in the proof of Theorem 6. □

Theorem 8.

The following upper estimate on the truncation error holds if χ meets the hypothesis of Theorem 7:

$$|\chi -\overline{\chi }|\lesssim {\displaystyle \frac{e^{\varphi ({\mu }_1+{\mu }_2)}{\varphi }^{\mathcal{M}}({\mu }_1^{\mathcal{M}}+{\mu }_2^{\mathcal{M}})}{\mathcal{M}!}}.$$

(53)

Proof. 

The definitions of $\chi$ and $\overline{\chi },$ allows us to write

$$\begin{array}{cc}\hfill |\chi -\overline{\chi }|& =\left|\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)-\sum _{i=0}^{\mathcal{M}}\sum _{j=0}^{\mathcal{M}}{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)\right|\hfill \\ & \le \left|\sum _{i=0}^{\mathcal{M}}\sum _{j=\mathcal{M}+1}^{\infty }{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)\right|+\left|\sum _{i=\mathcal{M}+1}^{\infty }\sum _{j=0}^{\infty }{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)\right|.\hfill \end{array}$$

(54)

We use Theorem 7, Lemma 1, and the following inequalities

$$\begin{array}{cc}& \sum _{i=0}^{\mathcal{M}}{\displaystyle \frac{{\mu }_1^i{\varphi }^i}{i!}}={\displaystyle \frac{e^{{\mu }_1\varphi }\Gamma (\mathcal{M}+1,\lambda \varphi )}{\Gamma (\mathcal{M}+1)}}<e^{{\mu }_1\varphi },\hfill \\ & \sum _{i=\mathcal{M}+1}^{\infty }{\displaystyle \frac{{\mu }_1^i{\varphi }^i}{i!}}={\displaystyle \frac{e^{{\mu }_1\varphi }(\Gamma (\mathcal{M}+1)-\Gamma (\mathcal{M}+1,{\mu }_1\varphi ))}{\Gamma (\mathcal{M}+1)}}<{\displaystyle \frac{e^{{\mu }_1\varphi }{\left({\mu }_1\varphi \right)}^{\mathcal{M}}}{\mathcal{M}!}},\hfill \\ & \sum _{i=0}^{\infty }{\displaystyle \frac{{\mu }_1^i{\varphi }^i}{i!}}=e^{{\mu }_1\varphi },\hfill \end{array}$$

(55)

where $\Gamma (.)$, $\Gamma (.,.)$ denote, respectively, gamma and upper incomplete gamma functions [58], to obtain the following desired estimation:

$$|\chi -\overline{\chi }|\lesssim {\displaystyle \frac{e^{\varphi ({\mu }_1+{\mu }_2)}{\varphi }^{\mathcal{M}}({\mu }_1^{\mathcal{M}}+{\mu }_2^{\mathcal{M}})}{\mathcal{M}!}},$$

which finalizes the proof. □

Remark 3.

We can say that the truncation error estimate has the factorial convergence rate. This can be seen from the inequality (53) in Theorem 8, noting that the factorial term grows very quickly—about the same as $C^M$ for a positive constant C.

Theorem 9.

(Stability) Under the assumptions of Theorem 7, and considering the following two approximate solutions for (1) governed by (2)–(3):

$$\overline{\chi }=\sum _{i=0}^{\mathcal{M}}\sum _{j=0}^{\mathcal{M}}{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right),\overline{\overline{\chi }}=\sum _{i=0}^{\mathcal{M}+1}\sum _{j=0}^{\mathcal{M}+1}{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right),$$

we have

$$|\overline{\overline{\chi }}-\overline{\chi }|\lesssim {\displaystyle \frac{{\varphi }^{\mathcal{M}+1}{e}^{\varphi }\left(e^{{\mu }_1}{\mu }_2^{\mathcal{M}+1}+e^{{\mu }_2}{\mu }_1^{\mathcal{M}+1}\right)}{(\mathcal{M}+1)!}}.$$

(56)

Proof. 

We have

$$\begin{array}{cc}\hfill |\overline{\overline{\chi }}-\overline{\chi }|& =\left|\sum _{i=0}^{\mathcal{M}+1}\sum _{j=0}^{\mathcal{M}+1}{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)-\sum _{i=0}^{\mathcal{M}}\sum _{j=0}^{\mathcal{M}}{c}_{ij}{L}_i^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)\right|\hfill \\ & =\left|{\displaystyle \sum _{i=0}^{\mathcal{M}}{c}_{i,\mathcal{M}+1}{L}_i^{\ast }\left(\xi \right)L_{\mathcal{M}+1}^{\ast }\left(t\right)+\sum _{j=0}^{\mathcal{M}+1}{c}_{\mathcal{M}+1,j}{L}_{\mathcal{M}+1}^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)}\right|\hfill \\ & \le \left|{\displaystyle \sum _{i=0}^{\mathcal{M}}{c}_{i,\mathcal{M}+1}{L}_i^{\ast }\left(\xi \right)L_{\mathcal{M}+1}^{\ast }\left(t\right)}\right|+\left|{\displaystyle \sum _{j=0}^{\mathcal{M}+1}{c}_{\mathcal{M}+1,j}{L}_{\mathcal{M}+1}^{\ast }\left(\xi \right)L_j^{\ast }\left(t\right)}\right|.\hfill \end{array}$$

(57)

Using Theorem 7 along with Lemma 1, we obtain

$$|\overline{\overline{\chi }}-\overline{\chi }|\lesssim {\displaystyle \frac{{\left({\mu }_2\varphi \right)}^{\mathcal{M}+1}}{(\mathcal{M}+1)!}}\sum _{i=0}^{\mathcal{M}}{\displaystyle \frac{{\left({\mu }_1\varphi \right)}^i}{i!}}+{\displaystyle \frac{{\left({\mu }_1\varphi \right)}^{\mathcal{M}+1}}{(\mathcal{M}+1)!}}\sum _{j=0}^{\mathcal{M}+1}{\displaystyle \frac{{\left({\mu }_2\varphi \right)}^j}{j!}}.$$

(58)

In virtue of the inequality in (55), we obtain

$$\sum _{i=0}^{\mathcal{M}}{\displaystyle \frac{{\left({\mu }_1\varphi \right)}^i}{i!}}<e^{{\mu }_1\varphi },\sum _{j=0}^{\mathcal{M}+1}{\displaystyle \frac{{\left({\mu }_2\varphi \right)}^j}{j!}}<e^{{\mu }_2\varphi }.$$

(59)

Based on the two inequalities in (59) and (58), we obtain

$$|\overline{\overline{\chi }}-\overline{\chi }|\lesssim {\displaystyle \frac{{\varphi }^{\mathcal{M}+1}{e}^{\varphi }\left(e^{{\mu }_1}{\mu }_2^{\mathcal{M}+1}+e^{{\mu }_2}{\mu }_1^{\mathcal{M}+1}\right)}{(\mathcal{M}+1)!}},$$

(60)

and accordingly, (56) is proved. □

6. Illustrative Examples

This section provides numerical examples to illustrate that the suggested approach to solving the TFFNDE is accurate and efficient. To ensure the proposed technique’s accuracy and efficacy, the results obtained by utilizing it are contrasted with precise solutions or with other well-established numerical approaches.

Example 1

([38,59]). Consider the time-fractional FitzHugh–Nagumo equation:

$$D_t^{\zeta }\chi ={\chi }_{\xi \xi }+\chi (\chi -\theta )(1-\chi ),0<\zeta <1,$$

(61)

constrained with:

$$\begin{array}{cc}& \chi (\xi ,0)={\displaystyle \frac{1}{2}}tanh\left({\displaystyle \frac{\xi }{2\sqrt{2}}}\right)+{\displaystyle \frac{1}{2}},0<\xi \le 1,\hfill \\ & \chi (0,t)={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}tanh\left({\displaystyle \frac{t}{4}}\right),\chi (1,t)={\displaystyle \frac{1}{2}}tanh\left({\displaystyle \frac{1}{4}}\left(\sqrt{2}-t\right)\right)+{\displaystyle \frac{1}{2}},t>0,\hfill \end{array}$$

(62)

where $\chi ={\displaystyle \frac{1}{2}}tanh\left({\displaystyle \frac{1}{4}}\left(\sqrt{2}\xi -t\right)\right)+{\displaystyle \frac{1}{2}}$ is the exact solution to (61) at $\theta =1$.

Table 1 presents a comparison between the solution obtained using our approach at $\mathcal{M}=11$ and the techniques given in [38,59] for various values of ζ. In addition,

Table 2. $L^{\infty }$ errors of Problem 1.

t	Gaussians [38]				Our Method at $\mathcal{M}=11$
	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{75}$	$\mathit{\zeta }=\mathbf{1}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{75}$	$\mathit{\zeta }=\mathbf{1}$
$0.2$	$6.166\times {10}^{-4}$	$3.146\times {10}^{-4}$	$1.287\times {10}^{-4}$	$3.641\times {10}^{-5}$	$2.57\times {10}^{-10}$	$3.44\times {10}^{-10}$	$4.31\times {10}^{-10}$	$5.38\times {10}^{-10}$
$0.4$	$2.992\times {10}^{-4}$	$2.302\times {10}^{-4}$	$1.413\times {10}^{-4}$	$1.571\times {10}^{-4}$	$2.34\times {10}^{-10}$	$2.39\times {10}^{-10}$	$2.05\times {10}^{-10}$	$1.43\times {10}^{-10}$
$0.6$	$4.323\times {10}^{-5}$	$7.298\times {10}^{-5}$	$5.034\times {10}^{-5}$	$1.599\times {10}^{-4}$	$2.08\times {10}^{-10}$	$1.81\times {10}^{-10}$	$1.22\times {10}^{-10}$	$2.76\times {10}^{-11}$
$0.8$	$9.455\times {10}^{-5}$	$1.617\times {10}^{-5}$	$4.743\times {10}^{-5}$	$6.206\times {10}^{-5}$	$1.91\times {10}^{-10}$	$1.52\times {10}^{-10}$	$9.01\times {10}^{-11}$	$4.707\times {10}^{-12}$
1	$7.906\times {10}^{-6}$	$8.048\times {10}^{-6}$	$2.664\times {10}^{-6}$	$6.820\times {10}^{-6}$	$2.15\times {10}^{-10}$	$2.49\times {10}^{-10}$	$3.64\times {10}^{-10}$	$5.46\times {10}^{-10}$

presents a comparison of the $L^{\infty }$ error at various ζ values between our technique with $\mathcal{M}=11$ and the method described in [38]. These tables demonstrate that the results of this method are extremely close to the exact solution. Figure 1 displays the absolute errors (AEs) at various values of $\mathcal{M}$ with ζ set to 0.85.

Table 3. The MAE of Problem 1 at $\zeta =0.4$.

$\mathcal{M}$	3	5	7	9	11
MAE	$7.73971\times {10}^{-5}$	$8.4387\times {10}^{-7}$	$7.85026\times {10}^{-9}$	$1.68627\times {10}^{-10}$	$1.70361\times {10}^{-10}$

displays the maximum absolute error (MAE) when $\zeta =0.4$. The results demonstrate that our approach is able to achieve a high accuracy with small values of $\mathcal{M}$.

Table 1. Different approximate solutions of Problem 1.

		Method in [59]		Gaussians [38]		Our Method at $\mathcal{M}=11$
$(\mathit{\xi },\mathit{t})$	Exact	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$
(0.1,0.2)	0.492678192949	0.427418	0.454935	0.492466	0.492564	0.492678192917	0.492678192905
(0.1,0.4)	0.467722618671	0.411555	0.429688	0.467632	0.467645	0.467722618641	0.467722618639
(0.1,0.6)	0.442927492940	0.401291	0.410894	0.442952	0.442910	0.442927492914	0.442927492915
(0.1,0.6)	0.418413552133	0.393583	0.395550	0.418448	0.418426	0.418413552108	0.418413552111
(0.3,0.2)	0.528003672504	0.461640	0.489905	0.527493	0.527735	0.528003672408	0.528003672375
(0.3,0.4)	0.503032971388	0.445267	0.464133	0.502797	0.502842	0.503032971299	0.503032971292
(0.3,0.6)	0.478047131210	0.434619	0.444777	0.478089	0.477992	0.478047131130	0.478047131136
(0.3,0.8)	0.453170661978	0.426595	0.428860	0.453249	0.453186	0.453170661904	0.453170661915
(0.5,0.2)	0.563050917309	0.496159	0.524966	0.562434	0.562736	0.563050917145	0.563050917089
(0.5,0.4)	0.538313096311	0.479367	0.498899	0.538014	0.538083	0.538313096160	0.538313096152
(0.5,0.6)	0.513385148789	0.468375	0.479131	0.513427	0.513312	0.513385148654	0.513385148666
(0.5,0.8)	0.488390434671	0.460051	0.462744	0.488485	0.488400	0.488390434546	0.488390434567
(0.7,0.2)	0.597479692513	0.530655	0.559780	0.596963	0.597225	0.597479692278	0.597479692199
(0.7,0.4)	0.573213594762	0.513551	0.533657	0.572956	0.573026	0.573213594547	0.573213594540
(0.7,0.6)	0.548589854921	0.502269	0.513645	0.548627	0.548530	0.548589854730	0.548589854751
(0.7,0.8)	0.523725855053	0.493674	0.496910	0.523812	0.523734	0.523725854878	0.523725854911
(0.9,0.2)	0.630973661925	0.564813	0.594013	0.630756	0.630870	0.630973661699	0.630973661623
(0.9,0.4)	0.607399960009	0.547522	0.568079	0.607291	0.607325	0.607399959803	0.607399959800
(0.9,0.6)	0.583314828533	0.536019	0.548003	0.583335	0.583293	0.583314828350	0.583314828375
(0.9,0.8)	0.558825335316	0.527198	0.531062	0.558867	0.558831	0.558825335148	0.558825335184

Table 1. Different approximate solutions of Problem 1.

		Method in [59]		Gaussians [38]		Our Method at $\mathcal{M}=11$
$(\mathit{\xi },\mathit{t})$	Exact	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$
(0.1,0.2)	0.492678192949	0.427418	0.454935	0.492466	0.492564	0.492678192917	0.492678192905
(0.1,0.4)	0.467722618671	0.411555	0.429688	0.467632	0.467645	0.467722618641	0.467722618639
(0.1,0.6)	0.442927492940	0.401291	0.410894	0.442952	0.442910	0.442927492914	0.442927492915
(0.1,0.6)	0.418413552133	0.393583	0.395550	0.418448	0.418426	0.418413552108	0.418413552111
(0.3,0.2)	0.528003672504	0.461640	0.489905	0.527493	0.527735	0.528003672408	0.528003672375
(0.3,0.4)	0.503032971388	0.445267	0.464133	0.502797	0.502842	0.503032971299	0.503032971292
(0.3,0.6)	0.478047131210	0.434619	0.444777	0.478089	0.477992	0.478047131130	0.478047131136
(0.3,0.8)	0.453170661978	0.426595	0.428860	0.453249	0.453186	0.453170661904	0.453170661915
(0.5,0.2)	0.563050917309	0.496159	0.524966	0.562434	0.562736	0.563050917145	0.563050917089
(0.5,0.4)	0.538313096311	0.479367	0.498899	0.538014	0.538083	0.538313096160	0.538313096152
(0.5,0.6)	0.513385148789	0.468375	0.479131	0.513427	0.513312	0.513385148654	0.513385148666
(0.5,0.8)	0.488390434671	0.460051	0.462744	0.488485	0.488400	0.488390434546	0.488390434567
(0.7,0.2)	0.597479692513	0.530655	0.559780	0.596963	0.597225	0.597479692278	0.597479692199
(0.7,0.4)	0.573213594762	0.513551	0.533657	0.572956	0.573026	0.573213594547	0.573213594540
(0.7,0.6)	0.548589854921	0.502269	0.513645	0.548627	0.548530	0.548589854730	0.548589854751
(0.7,0.8)	0.523725855053	0.493674	0.496910	0.523812	0.523734	0.523725854878	0.523725854911
(0.9,0.2)	0.630973661925	0.564813	0.594013	0.630756	0.630870	0.630973661699	0.630973661623
(0.9,0.4)	0.607399960009	0.547522	0.568079	0.607291	0.607325	0.607399959803	0.607399959800
(0.9,0.6)	0.583314828533	0.536019	0.548003	0.583335	0.583293	0.583314828350	0.583314828375
(0.9,0.8)	0.558825335316	0.527198	0.531062	0.558867	0.558831	0.558825335148	0.558825335184

Figure 1. The AE of Problem 1 at various $\mathcal{M}$ when $\zeta =0.85$.

Figure 1. The AE of Problem 1 at various $\mathcal{M}$ when $\zeta =0.85$.

Example 2

([38,59]). Consider the time-fractional FitzHugh–Nagumo equation:

$$D_t^{\zeta }\chi ={\chi }_{\xi \xi }+\chi (\chi -\theta )(1-\chi ),0<\zeta <1,$$

(63)

constrained with:

$$\begin{array}{cc}& \chi (\xi ,0)={\displaystyle \frac{1}{e^{-\frac{\xi }{\sqrt{2}}}+1}},0<\xi \le 1,\hfill \\ & \chi (0,t)={\displaystyle \frac{1}{e^{\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)t}+1}},\chi (1,t)={\displaystyle \frac{1}{e^{\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)t-\frac{1}{\sqrt{2}}}+1}},t>0,\hfill \end{array}$$

(64)

where $\chi ={\displaystyle \frac{1}{1+e^{\left(\frac{1}{\sqrt{2}}-\sqrt{2}\theta \right)t-\frac{\xi }{\sqrt{2}}}}}$ is the exact solution of (63).

Table 4. Comparison of AE of Problem 2.

	Method in [59]		Gaussians [38]		Our Method at $\mathcal{M}=11$
$(\mathit{\xi },\mathit{t})$	$\mathit{\zeta }=\mathbf{1},\mathit{\theta }=-\mathbf{1}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5},\mathit{\theta }=\mathbf{0}.\mathbf{45}$	$\mathit{\zeta }=\mathbf{1},\mathit{\theta }=-\mathbf{1}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5},\mathit{\theta }=\mathbf{0}.\mathbf{45}$	$\mathit{\zeta }=\mathbf{1},\mathit{\theta }=-\mathbf{1}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5},\mathit{\theta }=\mathbf{0}.\mathbf{45}$
(0.001, 0.001)	$1.5\times {10}^{-3}$	$2.8\times {10}^{-2}$	$1.019\times {10}^{-9}$	$8.717\times {10}^{-10}$	$4.19\times {10}^{-10}$	$5.71\times {10}^{-13}$
(0.002, 0.002)	$3.0\times {10}^{-3}$	$4.1\times {10}^{-2}$	$8.782\times {10}^{-11}$	$4.335\times {10}^{-10}$	$5.21\times {10}^{-10}$	$7.68\times {10}^{-13}$
(0.003, 0.003)	$4.5\times {10}^{-3}$	$5.3\times {10}^{-2}$	$1.306\times {10}^{-9}$	$1.436\times {10}^{-8}$	$5.51\times {10}^{-10}$	$1.21\times {10}^{-12}$
(0.004, 0.004)	$6.0\times {10}^{-3}$	$6.2\times {10}^{-2}$	$3.324\times {10}^{-9}$	$6.656\times {10}^{-8}$	$5.35\times {10}^{-10}$	$1.86\times {10}^{-12}$
(0.005, 0.005)	$7.5\times {10}^{-3}$	$6.9\times {10}^{-2}$	$4.515\times {10}^{-9}$	$1.806\times {10}^{-9}$	$4.94\times {10}^{-10}$	$2.67\times {10}^{-12}$
(0.006, 0.006)	$9.1\times {10}^{-3}$	$8.0\times {10}^{-2}$	$5.898\times {10}^{-9}$	$4.174\times {10}^{-9}$	$4.42\times {10}^{-10}$	$3.61\times {10}^{-12}$
(0.007, 0.007)	$1.0\times {10}^{-2}$	$8.7\times {10}^{-2}$	$1.112\times {10}^{-8}$	$2.086\times {10}^{-9}$	$3.90\times {10}^{-10}$	$4.67\times {10}^{-12}$
(0.008, 0.008)	$1.2\times {10}^{-2}$	$9.4\times {10}^{-2}$	$1.287\times {10}^{-8}$	$3.540\times {10}^{-10}$	$3.44\times {10}^{-10}$	$5.84\times {10}^{-12}$
(0.009, 0.009)	$1.3\times {10}^{-2}$	$1.0\times {10}^{-2}$	$6.502\times {10}^{-9}$	$7.412\times {10}^{-10}$	$3.06\times {10}^{-10}$	$7.10\times {10}^{-12}$
(0.01, 0.01)	$1.5\times {10}^{-2}$	$1.1\times {10}^{-2}$	$1.438\times {10}^{-9}$	$1.064\times {10}^{-9}$	$2.77\times {10}^{-10}$	$8.44\times {10}^{-12}$

compares the AE between our method at $\mathcal{M}=11$ and those in [38,59] at various ζ. Also,

Table 5. $L^{\infty }$ errors of Problem 2.

t	Hardy’s Multiquadric [38]		Our Method at $\mathcal{M}=11$
	$\mathit{\zeta }=\mathbf{1},\mathit{\theta }=-\mathbf{1}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5},\mathit{\theta }=\mathbf{0}.\mathbf{45}$	$\mathit{\zeta }=\mathbf{1},\mathit{\theta }=-\mathbf{1}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5},\mathit{\theta }=\mathbf{0}.\mathbf{45}$
$0.02$	$2.005\times {10}^{-8}$	$6.588\times {10}^{-10}$	$1.01458\times {10}^{-9}$	$2.64812\times {10}^{-11}$
$0.04$	$1.239\times {10}^{-8}$	$4.185\times {10}^{-10}$	$1.06183\times {10}^{-9}$	$4.54189\times {10}^{-11}$
$0.06$	$9.868\times {10}^{-8}$	$4.331\times {10}^{-10}$	$8.74395\times {10}^{-10}$	$5.95113\times {10}^{-11}$
$0.08$	$2.402\times {10}^{-9}$	$5.097\times {10}^{-10}$	$6.6398\times {10}^{-10}$	$7.04576\times {10}^{-11}$
$0.11$	$1.269\times {10}^{-9}$	$2.450\times {10}^{-10}$	$5.44992\times {10}^{-10}$	$7.93403\times {10}^{-11}$

compares $L^{\infty }$ error at different ζ between our method at $\mathcal{M}=11,$ and that in [38]. These tables demonstrate that the results of this method are extremely close to the exact solution. Figure 2 displays the AE at different ζ when $\theta =0.5$. The findings show that our strategy is advantageous for good accuracy outcomes with small $\mathcal{M}$ values.

Example 3

([38,59]). Consider the time-fractional FitzHugh–Nagumo equation

$$D_t^{\zeta }\chi ={\chi }_{\xi \xi }+\chi (\chi -\theta )(1-\chi ),0<\zeta <1,$$

(65)

constrained with:

$$\begin{array}{cc}& \chi (\xi ,0)={\displaystyle \frac{1}{2}}tanh\left({\displaystyle \frac{\xi }{2\sqrt{2}}}\right)+{\displaystyle \frac{1}{2}},0<\xi \le 1,\hfill \\ & \chi (0,t)={\displaystyle \frac{1}{2}}tanh\left({\displaystyle \frac{t}{4}}\right)+{\displaystyle \frac{1}{2}},\chi (1,t)={\displaystyle \frac{1}{2}}tanh\left({\displaystyle \frac{1}{4}}\left(t+\sqrt{2}\right)\right)+{\displaystyle \frac{1}{2}},t>0,\hfill \end{array}$$

(66)

where $\chi ={\displaystyle \frac{1}{2}}tanh\left({\displaystyle \frac{1}{4}}\left(t+\sqrt{2}\xi \right)\right)+{\displaystyle \frac{1}{2}}$ is the exact solution of this problem at $\theta =0$.

Table 6. $L^{\infty }$ errors of Problem 3.

t	Inverse Quadric [38]				Our Method at $\mathcal{M}=8$
	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{75}$	$\mathit{\zeta }=\mathbf{1}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{25}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{5}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{75}$	$\mathit{\zeta }=\mathbf{1}$
$0.2$	$9.508\times {10}^{-4}$	$6.348\times {10}^{-4}$	$2.569\times {10}^{-4}$	$8.712\times {10}^{-6}$	$1.651\times {10}^{-9}$	$2.692\times {10}^{-9}$	$4.263\times {10}^{-9}$	$6.516\times {10}^{-9}$
$0.4$	$7.424\times {10}^{-4}$	$7.799\times {10}^{-4}$	$4.515\times {10}^{-4}$	$3.395\times {10}^{-5}$	$7.081\times {10}^{-9}$	$1.118\times {10}^{-8}$	$1.710\times {10}^{-8}$	$2.526\times {10}^{-8}$
$0.6$	$5.776\times {10}^{-4}$	$5.470\times {10}^{-4}$	$4.562\times {10}^{-4}$	$6.511\times {10}^{-5}$	$2.611\times {10}^{-8}$	$3.975\times {10}^{-8}$	$5.896\times {10}^{-8}$	$8.502\times {10}^{-8}$
$0.8$	$3.268\times {10}^{-4}$	$6.663\times {10}^{-5}$	$1.641\times {10}^{-4}$	$2.612\times {10}^{-4}$	$8.312\times {10}^{-8}$	$1.230\times {10}^{-7}$	$1.782\times {10}^{-8}$	$2.520\times {10}^{-7}$
1	$1.173\times {10}^{-6}$	$3.610\times {10}^{-6}$	$4.250\times {10}^{-6}$	$3.064\times {10}^{-6}$	$2.377\times {10}^{-7}$	$3.420\times {10}^{-7}$	$4.835\times {10}^{-7}$	$6.722\times {10}^{-7}$

gives $L^{\infty }$ error for our method at $\mathcal{M}=11$ and that given in [38] at different ζ. Figure 3 displays the AE (left) and approximate solution (right) at $\zeta =0.6,\mathcal{M}=8$. Also,

Table 7. The MAE of Problem 3 for $0<\xi <1$.

t	Our Method at $\mathcal{M}=8$
	$\mathit{\zeta }=\mathbf{0}.\mathbf{1}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{4}$	$\mathit{\zeta }=\mathbf{0}.\mathbf{8}$
$0.1$	$6.2346\times {10}^{-10}$	$1.03016\times {10}^{-9}$	$2.12072\times {10}^{-9}$
$0.2$	$1.22167\times {10}^{-9}$	$2.22069\times {10}^{-9}$	$4.65531\times {10}^{-9}$
$0.3$	$2.55917\times {10}^{-9}$	$4.63569\times {10}^{-9}$	$9.46916\times {10}^{-9}$
$0.4$	$5.29984\times {10}^{-8}$	$9.3525\times {10}^{-9}$	$1.85421\times {10}^{-8}$
$0.5$	$1.05325\times {10}^{-8}$	$1.81054\times {10}^{-8}$	$3.49558\times {10}^{-8}$
$0.6$	$2.00591\times {10}^{-8}$	$3.37066\times {10}^{-8}$	$6.35952\times {10}^{-8}$
$0.7$	$3.67467\times {10}^{-8}$	$6.05659\times {10}^{-8}$	$1.11982\times {10}^{-7}$
$0.8$	$6.50699\times {10}^{-8}$	$1.05455\times {10}^{-8}$	$1.91455\times {10}^{-7}$
$0.9$	$1.12333\times {10}^{-8}$	$1.78502\times {10}^{-7}$	$3.18677\times {10}^{-7}$

presents the MAE for $0<\xi <1$. Figure 4 shows the AE (left) and approximate solution (right) at $\zeta =0.9,\mathcal{M}=8$. This figure shows that the proposed method is accurate for small values of $\mathcal{M}$.

7. Concluding Remarks

This work presented an effective approach for approximating solutions to the TFFNDEs arising in neurology using the shifted Lucas polynomials. We established and utilized the operational matrices of these polynomials’ integer derivatives and FDs to design the proposed numerical method. We used numerical experiments to show that the presented numerical technique is applicable and effective. As far as we know, the theoretical results of the shifted Lucas polynomials and their application to the solution of TFFNDE are innovative. We anticipate that the theoretical background in this paper will be the key to dealing with other types of DEs. We think employing such polynomials and related non-orthogonal polynomials will open up new horizons in the context of numerical solutions of different DEs. In future work, we plan to employ the theoretical conclusions provided in this research and appropriate spectral approaches to address various forms of FDEs that emerge in the applied sciences.