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Numerical solutions of generalized Atangana–Baleanu time-fractional FitzHugh–Nagumo equation using cubic B-spline functions

  • Afzaal Mubashir Hayat , Muhammad Abbas , Farah Aini Abdullah , Tahir Nazir , Hamed Ould Sidi , Homan Emadifar and Amani Alruwaili EMAIL logo
Published/Copyright: January 25, 2024

Abstract

The generalization of the classical FitzHugh–Nagumo model provides a more accurate description of the physical phenomena of neurons by incorporating both nonlinearity and fractional derivatives. In this article, we present a numerical method for solving the time-fractional FitzHugh–Nagumo equation (TFFNE) in the sense of the Atangana–Baleanu fractional derivative using B-spline functions. The proposed method employs a finite difference scheme to discretize the fractional derivative in time, while θ -weighted scheme is used to discretize the space directions. The efficiency of the scheme is demonstrated through numerical results and rate of convergence. The convergence order and error norms are studied at different values of the noninteger parameter, temporal directions, and spatial directions. Finally, the effectiveness of the proposed methodology is examined through the analysis of three applications.

1 Introduction

Splines hold a prominent position among mathematical functions frequently employed for estimation purposes. A spline is characterized as a piecewise polynomial function. In the realm of numerically interpreting ordinary differential equations, partial differential equations (PDEs), and fractional PDEs, the approach of utilizing spline functions for approximation has gained substantial popularity [1]. Fractional calculus has become a popular tool in many fields of research due to its ability to describe systems with memory, long-range dependence, and nonlocality. Traditional operators in fractional calculus have limitations. To address these issues, several researchers have proposed new operators or modifications to existing operators. The new operator developed by Caputo and Fabrizio [2] is nonsingular, which helps to alleviate this issue. However, this operator has a problem of nonlocality. The new operator developed by Atangana and Baleanu [3] helps to address both the nonlocality and the singularity issues associated with traditional operators, which makes it a valuable tool for researchers working with fractional calculus.

There are several implications for fractional derivatives in the areas of physics, mechanics, engineering, and biology [4]. In recent developments, with the use of fractional derivatives, the financial [5] and economic processes [6] have been described. Many interpretations of fractional derivatives exist, such as the geometric approach [7], informatic interpretation [8], and the economic approach [9]. Applications of fractional calculus include non-Newtonian fluid dynamics [10], rheology [11], hysteretic phenomena [12], and abnormal diffusion [13]. Studying the analytical or numerical approaches to fractional differential equations (FDEs) is extremely important since the majority of these issues may be stated as FDEs.

The FitzHugh–Nagumo equation (FNE) system has been derived by both FitzHugh [14] and Nagumo et al. [15]. Its simplicity and ability to capture key dynamics make it a valuable tool in the study of nonlinear dynamical systems and the analysis of biological phenomena related to excitation and signal propagation. The time-fractional FNE that combines diffusion and nonlinearity has attracted the interest of many scientists in the fields of neurophysiology, logistical population increase, flame spread, catalytic chemical reaction, and nuclear reactor theory, as presented in the study by Injrou et al. [16].

Nonlinear inhomogeneous time-fractional FitzHugh–Nagumo PDE is considered as in [17],

(1.1) γ p ( z , t ) t γ = ν 2 p ( z , t ) z 2 + β p ( z , t ) ( 1 p q ( z , t ) ) ( p q ( z , t ) η ) + U ( z , t ) , a z b , t [ t 0 , T ] ,

with initial condition (IC)

(1.2) p ( z , t 0 ) = ε ( z )

and boundary conditions (BCs)

(1.3) p ( a , t ) = ζ 1 ( t ) , p ( b , t ) = ζ 2 ( t ) ,

where 0 < γ < 1 , U ( z , t ) is source function, ν > 0 is kinematic viscosity, and β , η , q are parameters that β 0 , η ( 0 , 1 ) , and q > 0 . The γ t γ p ( z , t ) is taking in the sense of Atangana–Baleanu time-fractional derivative (ABTFD).

Gordon [18] used a collocation and Hopscotch finite difference scheme to approximate FNE. Dehghan et al. [19] have approximated FNE using a semi-analytic method. Keskin and Oturan [20] used reduced differential transform method to approximate PDEs. By using Adomian Decomposition Method, the exact solution has been founded by Momani and Odibat [21] of time-fractional Navier–Stokes equation. Ragab et al. [22] used homotopy analysis method to approximate time-fractional Navier–Stokes equation. A biological behavior of three-species predator–prey model involving Atangana–Baleanu fractional derivative was studied by Ghanbari et al. [23]. Approximate solution using redefined extended cubic B-spline of time-fractional telegraph equation is proposed by Amin et al. [24]. Liu et al. [25] constructed a shifted Grünwald–Letnikov scheme to discretize the Riesz derivative of the fractional FitzHugh–Nagumo model. Shih et al. [26] examined FNE with applications. Abbasbandy [27] founded the soliton solutions of FNE using the homotopy analysis scheme. The explicit solution of FNE was founded by Kakiuchi and Tchizawa [28]. Schonbek [29] examined the FNE with boundary problems. Yanagida [30] discussed the stability of traveling front solutions of FNE. A nonstandard difference method was proposed by Namjoo et al. [31]. Angadi [32] solved FNE by wavelet-based lifting methods. Olmos and Shizgal [33] developed a pseudo-spectral method to examine the FNE. Abdulazeez and Modanli [34] found analytic solution of fractional order pseudo-hyperbolic telegraph equation using modified double Laplace transform method. The effect of fractal-fractional Caputo–Fabrizio derivative on the analysis of tumor growth model is studied by [35]. Baleanu et al. [36] studied stability and dynamical analysis of generalized fractional model with a real case study. Ali et al. [37] solved FNE using Galerkin finite element approach. A polynomial differential quadrature method is used by Jiwari et al. [38] for the numerical solutions of the generalized FitzHugh–Nagumo.

The article is structured as follows: the ABTFD, Parseval’s identity (PI), and CBSFs are presented in Section 2; Section 3 shows the newly developed scheme; Sections 4 and 5 display stability and convergence of proposed approach, respectively; the effectiveness and validity of the suggested technique are examined in Section 6; and finally, Section 7 summarizes the conclusion.

2 Preliminaries

Definition 1

The ABTFD γ t γ p ( z , t ) of order γ ( 0 , 1 ) is presented by Atangana and Baleanu [3] as follows:

(2.1) γ t γ p ( z , t ) = A B ( γ ) 1 γ 0 t υ p ( z , υ ) E γ × γ 1 γ ( t υ ) γ d υ ,

where A B ( γ ) is a normalized form of function that has the property A B ( γ = 0 ) = A B ( γ = 1 ) = 1 . E γ , σ ( z ) is the Mittag-Leffler function (MLF) of two parameters with E γ , 1 ( z ) = E γ ( z ) , which is defined by Mittag-Leffler [39] as follows:

E γ , σ ( z ) = y = 0 z y Γ ( γ y + σ ) .

Some properties of MLF by fixing γ and σ are as follows:

  • E 1 , 1 ( z ) = e z ,

  • E 1 , 2 ( z ) = e z 1 z ,

  • E 2 , 1 ( z 2 ) = cosh ( z ) ,

  • E 2 , 1 ( z 2 ) = cos ( z ) ,

  • E 2 , 2 ( z 2 ) = sinh ( z ) z .

Definition 2

If Ω ˆ L 2 [ a , b ] , then PI is given as follows [40]:

(2.2) n = Ω ˆ ( n ) 2 = a b Ω ˜ ( z ) 2 d z ,

where Ω ˆ ( n ) = a b Ω ˜ ( z ) e 2 π i n z d z is Fourier transform for all integers n .

2.1 Cubic B-spline basis functions

The spatial domain [ a , b ] be divided into equal length of N subintervals with h = b a N such that { a = z 0 , z 1 , z N = b } with z r < z r + 1 , where z r = h k + z 0 and r = 0 ( 1 ) N .

Now, let P ( z , t ) be the CBSFs approach for p ( z , t ) , i.e.,

(2.3) P ( z , t ) = r = 1 N + 1 Ψ r m ( t ) B ˆ r ( z ) ,

where control points Ψ r m ( t ) to be calculated at every temporal stage and CBSFs are defined as follows:

(2.4) B ˆ r ( z ) = 1 6 h 3 ( z z r 2 ) 3 , if z [ z r 2 , z r 1 ) , 3 ( z z r 1 ) h 2 + 3 ( z z r 1 ) 2 h 3 ( z z r 1 ) 3 + h 3 , if z [ z r 1 , z r ) , h 3 + 3 h 2 ( z r + 1 z ) + 3 h ( z r + 1 z ) 2 3 ( z r + 1 z ) 3 , if z [ z r , z r + 1 ) , ( z r + 2 z ) 3 , if z [ z r + 1 , z r + 2 ) , 0 , otherwise .

Numerous geometrical features, including the geometric invariability, symmetry, the convex hull characteristic, local support, nonnegativity, and the partition of unity, are preserved in the CBSFs [41,42]. In addition, B ˆ 1 , B ˆ 0 , , B ˆ N + 1 have been constructed. Equations (2.3) and (2.4) provide the following approximations:

(2.5) ( P ) r m = 1 6 Ψ r 1 m + 4 6 Ψ r m + 1 6 Ψ r + 1 m , ( P z ) r m = 1 2 h Ψ r + 1 m + 1 2 h Ψ r 1 m , ( P z z ) r m = 1 h 2 Ψ r 1 m + 2 h 2 Ψ r m + 1 h 2 Ψ r + 1 m .

3 Illustration of the scheme

Suppose [ 0 , T ] the time domain be divided into M subintervals of equal length with Δ t = T M using { 0 = t 0 , t 1 , t M = T } with t m < t m + 1 , where t m = m Δ t and m = 0 : 1 : M . The ABTFD in Eq. (1.1) is discretized at t = t m + 1 as in [42]:

(3.1) γ t γ p ( z , t m + 1 ) = A B ( γ ) 1 γ 0 t m + 1 υ p ( z , υ ) E γ γ 1 γ ( t m + 1 υ ) γ d υ , 0 < γ < 1 , = A B ( γ ) 1 γ v = 0 m t v t v + 1 υ p ( z , υ ) E γ γ 1 γ ( t m + 1 υ ) γ d υ .

Utilizing forward difference formulation, Eq. (3.1) becomes

γ t γ p ( z , t m + 1 ) = A B ( γ ) 1 γ v = 0 m p ( z , t v + 1 ) p ( z , t v ) Δ t × t v t v + 1 E γ γ 1 γ ( t m + 1 υ ) γ d υ + λ Δ t m + 1 = A B ( γ ) 1 γ v = 0 m [ p ( z , t m v + 1 ) p ( z , t m v ) ] × ( v + 1 ) E γ , 2 γ 1 γ ( ( v + 1 ) Δ t ) γ v E γ , 2 γ 1 γ ( v Δ t ) γ + λ Δ t m + 1 = A B ( γ ) 1 γ v = 0 m [ p ( z , t m v + 1 ) p ( z , t m v ) ] ( ( v + 1 ) E v + 1 v E v ) + λ Δ t m + 1 .

Hence,

(3.2) γ t γ p ( z , t m + 1 ) = A B ( γ ) 1 γ v = 0 m w v × [ p ( z , t m v + 1 ) p ( z , t m v ) ] + λ Δ t m + 1 ,

where E v = E γ , 2 γ 1 γ ( v Δ t ) γ and w v = ( v + 1 ) E v + 1 v E v . Simple observation reveals that

  • w v > 0 and w 0 = E 1 , v = 0 : 1 : m ,

  • w 0 > w 1 > w 2 > > w v , w v 0 as v ,

  • v = 0 m ( w v w v + 1 ) + w m + 1 = ( E 1 w 1 ) + v = 1 m 1 ( w v w v + 1 ) + w m = E 1 .

In addition, the truncation error λ Δ t m + 1 is shown in [43] as follows:

λ Δ t m + 1 = A B ( γ ) 1 γ v = 0 m t v t v + 1 Δ t 2 2 p ( z , t v ) t 2 × E γ γ 1 γ ( t m + 1 υ ) γ d υ = A B ( γ ) 1 γ ( Δ t ) 2 2 v = 0 m 2 p ( z , t v ) t 2 × ( m v + 1 ) E γ , 2 γ 1 γ ( ( m v + 1 ) Δ t ) γ ( m v ) E γ , 2 γ 1 γ ( ( m v ) Δ t ) γ = A B ( γ ) 1 γ ( Δ t ) 2 2 v = 0 m 2 p ( z , t v ) t 2 × ( ( m v + 1 ) E m v + 1 ( m v ) E m v ) A B ( γ ) 1 γ ( Δ t ) 2 2 max 0 t t m 2 p ( z , t ) t 2 c 1 ,

where c 1 is constant.

(3.3) λ Δ t m + 1 ϑ ( Δ t ) 2 ,

where Ψ is constant.

Now using θ -weighted scheme and Eq. (3.2), Eq. (1.1) becomes

(3.4) A B ( γ ) 1 γ v = 0 m w v [ p ( z , t m v + 1 ) P ( z , t m v ) ] = θ [ ν p z z ( z , t m + 1 ) + β p ( z , t m + 1 ) ( 1 p q ( z , t m + 1 ) ) ( p q ( z , t m + 1 ) η ) ] + ( 1 θ ) [ ν p z z ( z , t m ) + β p ( z , t m ) ( 1 p q ( z , t m ) ) ( p q ( z , t m ) η ) ] + U ( z , t m + 1 ) .

Discretizing Eq. (3.4) along spatial direction for θ = 1 and using linearization formula as defined in [44], we obtain

(3.5) ( p q ) r m + 1 = q ( p ) r m + 1 ( p q 1 ) r m ( q 1 ) ( p q ) r m .

(3.6) ( p 3 ) r m + 1 = 3 ( p ) r m + 1 ( p 2 ) r m 2 ( p 3 ) r m , ( p 2 ) r m + 1 = 2 ( p ) r m + 1 ( p ) r m ( p 2 ) r m .

Now using q = 1 , we have

(3.7) [ E 1 2 ξ β ( 1 + η ) ( p ) r m + ξ β η + 3 ξ β ( p 2 ) r m ] ( p ) r m + 1 ξ ν ( p z z ) r m + 1 = E 1 ( p ) r m ξ β ( 1 + η ) ( p 2 ) r m + 2 ξ β ( p 3 ) r m v = 1 m w v ( ( p ) r m v + 1 ( p ) r m v ) + ξ U r m + 1 ,

where ξ = 1 γ A B ( γ ) , p k m = p ( z r , t m ) , and U r m + 1 = U ( z r , t m + 1 ) .

Using Eq. (2.5) in Eq. (3.7), we obtain

(3.8) a 1 1 6 Ψ r 1 m + 1 + 4 6 Ψ r m + 1 + 1 6 Ψ r + 1 m + 1 ς 1 h 2 Ψ r 1 m + 1 2 h 2 Ψ r m + 1 + 1 h 2 Ψ r + 1 m + 1 = ψ 1 1 6 Ψ r 1 m + 4 6 Ψ r m + 1 6 Ψ r + 1 m v = 1 m w v 1 6 ( Ψ r 1 m v + 1 Ψ r 1 m v ) + 4 6 ( Ψ r m v + 1 Ψ r m v ) + 1 6 ( Ψ r + 1 m v + 1 Ψ r + 1 m v ) + ξ U r m + 1 ,

(3.9) 1 6 a 1 ς h 2 Ψ r 1 m + 1 + 4 6 a 1 + 2 ς h 2 Ψ r m + 1 + 1 6 a 1 ς h 2 Ψ r + 1 m + 1 = Q r m

(3.10) f r m Ψ r 1 m + 1 + g r m Ψ r m + 1 + H r m Ψ r + 1 m + 1 = Q r m , r = 0 : 1 : N , m = 0 : 1 : M ,

where a 1 = E 1 2 ξ β ( 1 + η ) ( p ) r m + ξ β η + 3 ξ β ( p 2 ) r m , ς = ξ ν , f r m = 1 6 a 1 ς h 2 , g r m = 4 6 a 1 + 2 ς h 2 , H r m = 1 6 a 1 ς h 2 , Q r m = ψ 1 1 6 Ψ r 1 m + 4 6 Ψ r m + 1 6 Ψ r + 1 m v = 1 m w v 1 6 ( Ψ r 1 m v + 1 Ψ r 1 m v ) + 4 6 ( Ψ r m v + 1 Ψ r m v ) + 1 6 ( Ψ r + 1 m v + 1 Ψ r + 1 m v ) + ξ U r m + 1 , and ψ 1 = E 1 ξ β ( 1 + η ) ( p ) r m + 2 ξ β ( p 2 ) r m . This system (Eq. (3.10)) has N + 1 linear equations with N + 3 unknowns. Two more equations can be found for unique solution from the given BCs (Eq. (1.3)).

(3.11) 1 6 Ψ 1 m + 1 + 4 6 Ψ 0 m + 1 + 1 6 Ψ 1 m + 1 = ζ 1 m + 1 1 6 Ψ N 1 m + 1 + 4 6 Ψ N m + 1 + 1 6 Ψ N + 1 m + 1 = ζ 2 m + 1

Equations (3.10)–(3.11) equations have the following form of the matrix:

(3.12) 1 6 4 6 1 6 f 0 m g 0 m H 0 m f 1 m g 1 m H 1 m f N 1 m g N 1 m H N 1 m f N m g N m H N m 1 6 4 6 1 6 Ψ 1 m + 1 Ψ 0 m + 1 Ψ 1 m + 1 Ψ N 1 m + 1 Ψ N m + 1 Ψ N + 1 m + 1 = ζ 1 m + 1 Q 0 m + 1 Q 1 m + 1 Q N 1 m + 1 Q N m + 1 ζ 2 m + 1 .

By using Eq. (1.2), the initial vector Ψ 0 = [ Ψ 1 0 , Ψ 0 0 , , Ψ N + 1 0 ] T is obtained as follows:

(3.13) ( P z ) r 0 = ε ( z r ) , r = 0 , ( P ) r 0 = ε ( z r ) , r = 0 : 1 : N , ( P z ) r 0 = ε ( z r ) , r = N .

The representation of the matrix form for the above system of equation is shown as

(3.14) G Ψ 0 = R ,

where

(3.15) 1 2 h 0 1 2 h 1 6 4 6 1 6 1 6 4 6 1 6 1 6 4 6 1 6 1 6 4 6 1 6 1 2 h 0 1 2 h Ψ 1 0 Ψ 0 0 Ψ 1 0 Ψ N 1 0 Ψ N 0 Ψ N + 1 0 = ε ( z 0 ) ε ( z 0 ) ε ( z 1 ) ε ( z N 1 ) ε ( z N ) ε ( z N ) .

Any numerical algorithm can be used to solve Eq. (3.15). Mathematica 10 is used to conduct the numerical results.

4 The stability

The numerical technique is assumed to be stable when the error does not increase during the computing process [45]. Fourier method [4648] is applied for the stability of proposed scheme. Now, let τ r m and τ ˜ r m represent the amplification component and its computation in Fourier transform, respectively κ r m is the error that can be addressed as follows:

κ r m = τ r m τ ˜ r m , m = 1 : 1 : M , r = 1 : 1 : N 1 .

Taking source function U ( z , t ) = 0 in Eq. (3.8), Eq. (3.8) becomes

(4.1) f r m Ψ r 1 m + 1 + g r m Ψ r m + 1 + H r m Ψ r + 1 m + 1 = b 1 Ψ r 1 m + b 2 Ψ r m + b 3 Ψ r + 1 m v = 1 m w v 1 6 ( Ψ r 1 m v + 1 Ψ r 1 m v ) + 4 6 ( Ψ r m v + 1 Ψ r m v ) + 1 6 ( Ψ r + 1 m v + 1 Ψ r + 1 m v ) ,

where b 1 = ψ 1 6 , b 2 = 4 ψ 1 6 , and b 3 = ψ 1 6 . From IC and BCs, we can write

(4.2) Ψ r 0 = ε ( z r ) , r = 1 : 1 : N

and

(4.3) Ψ 0 m = ζ 1 ( t m ) , Ψ N m = ζ 2 ( t m ) , m = 0 : 1 : M .

Define grid function as follows:

(4.4) Ψ m = Ψ r m , z z r h 2 , z r + h 2 , r = 1 : 1 : N 1 , 0 , z a , a + h 2 o r z b h 2 , b .

The function Ψ m ( z ) in the Fourier mode can be presented as follows:

(4.5) Ψ m ( z ) = n = F m ( n ) e 2 π t n z b a ,

where

(4.6) F m ( n ) = 1 b a a b Ψ m ( z ) e 2 π t n z b a d z , m = 0 ( 1 ) M and Ψ m = [ Ψ 1 m , Ψ 2 m , , Ψ N 1 m ] T .

Applying . 2 norm, we acquire

Ψ m 2 = h r = 1 N 1 Ψ r m 2 = a a + h 2 Ψ m 2 d z + k = 1 N 1 z k h 2 z r + h 2 Ψ m 2 d z + b h 2 b Ψ m 2 d z 1 2 = a b Ψ m 2 d z 1 2 .

From Eq. (2.2), we have [40]

a b Ψ m 2 d z = n = F m ( n ) 2 .

Hence, we acquire

(4.7) Ψ m 2 2 = n = F m ( n ) 2 .

Consider Eqs (4.1)–(4.3) present a solution in Fourier series as follows:

(4.8) Ψ r m = F m e i α r h ,

where i = 1 and α is a real. Using Eq. (4.8) in Eq. (4.1), we obtained

(4.9) f r m F m + 1 e i α h + g r m F m + 1 + H r m F m + 1 e i α h = b 1 F m e i α h + b 2 F m + b 3 F m e i α h v = 1 m w v 1 6 ( F m v + 1 e i α h F m v e i α h ) + 4 6 ( F m v + 1 F m v ) + 1 6 ( F m v + 1 e i α h F m v e i α h ) .

Utilizing the relation e i α h + e i α h = 2 cos ( α h ) and values of b 1 , b 2 , b 3 , f r m , g r m , and H r m we obtain the following equality:

(4.10) F m + 1 = F m ψ 1 6 ( 2 cos ( α h ) + 4 ) ( 2 cos ( α h ) + 4 ) a 1 6 + 2 ς ( 1 cos ( α h ) ) h 2 ( 2 cos ( α h ) + 4 ) v = 1 m w v 6 [ F m v + 1 F m v ] ( 2 cos ( α h ) + 4 ) a 1 6 + 2 ς ( 1 cos ( α h ) ) h 2

(4.11) F m + 1 = 1 a 1 ψ 1 + 24 ς sin 2 ( r h 2 ) ψ 1 h 2 ( 2 cos ( r h ) + 4 ) × F m v = 1 m w v [ F m v + 1 F m v ] ψ 1

(4.12) F m + 1 = F m σ 1 v = 1 m w v F m v + 1 F m v ψ 1 σ 1 ,

where σ 1 = a 1 ψ 1 + 24 ς sin 2 r h 2 ψ 1 h 2 ( 2 cos ( r h ) + 4 ) where ψ 1 > 0 and a 1 > 0 . Obviously σ 1 1 .

Lemma 4.1

If F m is the solution of Eq. (4.12), then F m F 0 , m = 0 ( 1 ) M .

Proof

To prove this lemma, we use mathematical induction. So, for m = 0 , Eq. (4.12) reveals

F 1 = 1 σ 1 F 0 F 0 , σ 1 1 .

Assume that F m F 0 for m = 1 , 2 , , M 1 , then

F m + 1 1 σ 1 F m 1 σ 1 ψ 1 v = 1 m w v ( F m v + 1 F m v ) 1 σ 1 F 0 1 σ 1 ψ 1 v = 1 m w v ( F 0 F 0 ) F 0 .

Theorem 1

The proposed technique (Eq. (3.10)) is stable unconditionally.

Proof

Using Lemma 4.1 and Eq. (4.7), we obtain

Ψ m 2 Ψ 0 2 m = 0 , 1 , , M .

This lemma shows that the developed approach is stable unconditionally.□

5 The convergence

The methodology as in [49] is used to analyze the convergence of the propagated approach. Initially, Theorem 2 is presented [50,51]:

Theorem 2

Suppose U belongs to C 2 [ a , b ] and P ( z , t ) belongs to C 4 [ a , b ] , also partition of [ a , b ] is = { a = z 0 , z 1 , z N = b } with z r < z r + 1 such that z r a = h r , where k = 0 : 1 : N . If solution curve is interpolated by unique spline P ˜ ( z , t ) at z r , then for every t 0 , there is a constant ϖ k independent of h, and for r = 0 , 1 , 2 , we have

(5.1) D r ( P ˜ ( z , t ) p ( z , t ) ) ϖ r h 4 r .

Lemma 5.1

The CBSFs { B ˆ 1 , B ˆ 0 , , B ˆ N + 1 } in Eq. (2.4) fulfills the inequality as given in [42]

(5.2) r = 1 N + 1 B ˆ r ( z ) 5 3 , 0 z 1 .

Theorem 3

The numerical approach P ( z , t ) to the analytic solution p ( z , t ) for the the time-fractional FitzHugh–Nagumo equation (TFFNE) (Eqs (1.1)–(1.3)) exists. In addition, if source function U ( z , t ) belongs to C 2 [ 0 , 1 ] as in [42], then

(5.3) p ( z , t ) P ( z , t ) ϖ ˜ h 2 , for a l l t 0 ,

for suitably small h , and ϖ ˜ > 0 is independent of h .

Proof

Let P ( z , t ) = r = 1 N + 1 V r m ( t ) B ˆ r ( z ) be the estimated spline for p ( z , t ) . Through the triangular inequality, we obtain

P ( z , t ) p ( z , t ) P ˜ ( z , t ) p ( z , t ) + P ˜ ( z , t ) p ( z , t ) .

We have from the help of Theorem 2,

(5.4) p ( z , t ) P ( z , t ) ϖ 0 h 4 + P ˜ ( z , t ) p ( z , t ) .

For the developed scheme, these L P ( z r , t ) = L p ( z r , t ) = U ( z r , t ) , where r = 0 , 1 , , N are the conditions of collocation. Suppose that

L P ˜ ( z , t ) = U ˜ ( z r , t ) , r = 0 ( 1 ) N .

Thus, L ( p ˜ ( z r , t ) p ( z r , t ) ) can be indicated for any time level as

(5.5) a 1 6 ς h 2 Φ r 1 m + 1 + 4 a 1 6 + 2 ς h 2 Φ r m + 1 + a 1 6 ς h 2 Φ r + 1 m + 1 = ψ 1 6 Φ r 1 m + 4 ψ 1 6 Φ r m ψ 1 6 ( Φ r + 1 m ) + ξ h 2 ρ r m + 1 v = 1 m w v 1 6 ( Φ r 1 m v + 1 Φ r 1 m v ) + 4 6 ( Φ r m v + 1 Φ r m v ) + 1 6 ( Φ r + 1 m v + 1 Φ r + 1 m v ) .

The description of the BCs is

1 6 Φ r 1 m + 1 + 4 6 Φ r m + 1 + 1 6 Φ r + 1 m + 1 = 0 , r = 0 , N ,

where

Φ r m = Ψ r m V r m , r = 1 : 0 : N + 1

and

ρ r m = h 2 [ U r m U ˜ r m ] , r = 0 , 1 , , N .

Now using the inequality (5.1), we obtain

ρ r m = h 2 U r m U ˜ r m ϖ h 4 .

Define ρ m = max { ρ r m ; 0 r N } , ε r m = Φ r m , and ε m = max { ε r m ; 0 r N } .

When m = 0 , Eq. (5.5) becomes

a 1 6 ς h 2 Φ r 1 1 + 4 a 1 6 + 2 ς h 2 Φ r 1 + a 1 6 ς h 2 Φ r + 1 1 = ψ 1 6 Φ r 1 0 + 4 ψ 1 6 Φ r 0 + ψ 1 6 Φ r + 1 0 + ξ h 2 ρ r 1 ,

where r = 0 , 1 , , N and, through the IC, ε 0 = 0 . Applying norms of Φ r 1 , ρ r 1 and appropriately small h , we have

ε r 1 3 ξ a 1 h 2 + 12 ς ϖ h 4 , r = 0 , 1 , , N .

We obtain ε 1 1 and ε N + 1 1 through the BCs

ε 1 1 15 ξ a 1 h 2 + 12 ς ϖ h 4 ,

ε N + 1 1 15 ξ a 1 h 2 + 12 ς ϖ h 4 ,

which implies

(5.6) ε 1 ϖ 1 h 2 ,

where ϖ 1 is independent of h .

Now we use mathematical induction on m for the proof of this theorem. Suppose that ε r i ϖ i h 2 is true for i = 1 ( 2 ) m and ϖ = max { ϖ i : i = 0 : 1 : m } , then from Eq. (5.5), we obtain the following:

a 1 6 ς h 2 Φ r 1 m + 1 + 4 a 1 6 + 2 ς h 2 Φ r m + 1 + a 1 6 ς h 2 Φ r + 1 m + 1 = ( w 0 w 1 ) 1 6 Φ r 1 m + 4 6 Φ r m + 1 6 Φ r + 1 m + ( w 1 w 2 ) 1 6 Φ r 1 m 1 + 4 6 Φ r m 1 + 1 6 Φ r + 1 m 1 + ( w m 1 w m ) 1 6 Φ r 1 1 + 4 6 Φ r 1 + 1 6 Φ r + 1 1 + w m 1 6 Φ r 1 0 + 4 6 Φ r 0 + 1 6 Φ r + 1 0 + ξ h 2 ρ r m + 1 .

Again, using norms on Φ r m and ρ r m , we obtained

ε r m + 1 3 ϖ h 4 a 1 h 2 + 12 ς ξ + v = 0 m 1 ( w v w v + 1 ) .

Similarly, we obtain the values of ε 1 m + 1 and ε N + 1 m + 1 from the BCs

ε 1 m + 1 15 ϖ h 4 a 1 h 2 + 12 ς ξ + v = 0 m 1 ( w v w v + 1 )

ε N + 1 m + 1 15 ϖ h 4 a 1 h 2 + 12 ς ξ + v = 0 m 1 ( w v w v + 1 ) .

Hence, for all m , we acquire

(5.7) ε m + 1 ϖ h 2 .

In particular,

p ˜ ( z , t ) P ( z , t ) = r = 1 N + 1 ( V r ( t ) Ψ r ( t ) ) B ˆ r ( z ) .

Therefore, from Lemma 5.1 and inequality (5.7), we obtain

(5.8) p ˜ ( z , t ) P ( z , t ) 5 3 ϖ h 2 .

Using Eq. (5.8), the inequality (5.4) gives

P ( z , t ) p ( z , t ) ϖ 0 h 4 + 5 3 ϖ h 2 = ϖ ˜ h 2 ,

where ϖ ˜ = ϖ 0 h 2 + 5 3 ϖ .□

Theorem 4

The TFFNE is convergent with BCs and IC.

Proof

Let p ( z , t ) be analytic solution and P ( z , t ) be the approximate solutions for the TFFNE. Therefore, the above theorem and relation (3.3) justify that there exist constants ϖ ˜ and δ such that

p ( z , t ) P ( z , t ) δ ( Δ t ) 2 + ϖ ˜ h 2 .

Hence, the proposed approach is second-order convergent in time and spatial directions.□

6 Analysis and presentation of numerical applications

Approximate results are revealed to show the perfection of the proposed methodology through L 2 ( N ) , and L ( N ) in this section that are defined as follows:

L 2 ( N ) = P ( z r , t ) p ( z r , t ) 2 = h z = 0 N p ( z r , t ) P ( z r , t ) 2 ,

L ( N ) = P ( z r , t ) p ( z r , t ) = max 0 r N p ( z r , t ) P ( z r , t ) ,

and order of convergence μ is calculated using following formula as in [42]:

μ = log L ( N ) L ( N + 1 ) log R ( N + 1 ) R ( N ) .

Every application is examined by considering A B ( γ ) = 1 .

Application 6.1

Consider the TFFNE for q = 1

γ p t γ = ν 2 p z 2 + β p ( 1 p ) ( p η ) + U ( z , t ) , a z b , 0 t T ,

with IC and BCs

(6.1) p ( z , 0 ) = 0 , p ( 0 , t ) = 0 , p ( 1 , t ) = 0 ,

and calculation of U is

U ( z , t ) = 24 1 γ ( z z 2 ) t 4 E γ , 5 γ t γ 1 γ + 2 ν t 4 β [ t 4 ( z z 2 ) ] [ 1 t 4 ( z z 2 ) ] [ t 4 ( z z 2 ) η ] .

The analytic solution is p ( z , t ) = ( z z 2 ) t 4 , where ν = β = 1 and η = 0.5 . Numerical results and absolute errors for different values of γ with N = 100 , 60, and 80, Δ t = 0.001 , and t = 1 for Application 6.1 are presented in Tables 1 and 2. For various time levels and different choices of γ , the error norms are shown in Table 3. The order of convergence μ and the error norms along spatial and time directions are represent in Tables 4 and 5. The obtained results of proposed method and analytic solution have closed commitment as in Figure 1 for different values of time t with Δ t = 0.001 . Figure 2 for N = 100 , Δ t = 0.001 , γ = 0.255 , t = 1 , and z [ 0 , 1 ] shows the 3D plot of numerical and analytic solutions. The 2D and 3D error plots are presented in Figure 3 at t = 1 .

Table 1

Errors norm at t = 1 of Application 6.1 for different values of γ with Δ t = 0.001 and N = 100

Numerical result Absolute error
z Analytic solution γ = 1 4 γ = 3 4 γ = 1 4 γ = 3 4
0.1 0.09000 0.08999 0.08999 1.67581 × 1 0 8 1.82337 × 1 0 9
0.2 0.16000 0.15999 0.15999 3.59818 × 1 0 8 4.22148 × 1 0 9
0.3 0.21000 0.20999 0.20999 5.00533 × 1 0 8 6.53577 × 1 0 9
0.4 0.24000 0.23999 0.23999 5.91083 × 1 0 8 8.11400 × 1 0 9
0.5 0.25000 0.24999 0.24999 6.22188 × 1 0 8 8.66179 × 1 0 9
0.6 0.24000 0.23999 0.23999 5.91083 × 1 0 8 8.11401 × 1 0 9
0.7 0.21000 0.20999 0.20999 5.00533 × 1 0 8 6.53578 × 1 0 9
0.8 0.16000 0.15999 0.15999 3.59818 × 1 0 8 4.22150 × 1 0 9
0.9 0.09000 0.08999 0.08999 1.85955 × 1 0 8 1.82339 × 1 0 9
Table 2

Errors norm at t = 1 for Application 6.1 with Δ t = 0.001

Numerical result Absolute error
z Analytic result γ = 2 5 , N = 60 γ = 4 5 , N = 80 γ = 2 5 , N = 60 γ = 4 5 , N = 80
0.1 0.09 0.08999 0.09000 1.64381 × 1 0 8 4.63942 × 1 0 9
0.2 0.16 0.15999 0.16000 3.18930 × 1 0 8 8.01115 × 1 0 9
0.3 0.21 0.20999 0.21000 4.44481 × 1 0 8 1.02170 × 1 0 8
0.4 0.24 0.23999 0.24000 5.25382 × 1 0 8 1.15103 × 1 0 8
0.5 0.25 0.24999 0.25000 5.53179 × 1 0 8 1.19462 × 1 0 8
0.6 0.24 0.23999 0.24000 5.25382 × 1 0 8 1.15103 × 1 0 8
0.7 0.21 0.20999 0.21000 4.44481 × 1 0 8 1.02170 × 1 0 8
0.8 0.16 0.15999 0.16000 3.18930 × 1 0 8 8.01115 × 1 0 9
0.9 0.09 0.08999 0.09000 1.64381 × 1 0 8 4.63943 × 1 0 9
Table 3

L ( N ) and L 2 ( N ) for different values of γ with Δ t = 1 2,000 , N = 100 , and 0 z 1 of Application 6.1

L ( N ) L 2 ( N )
t γ = 1 4 γ = 3 4 γ = 1 4 γ = 3 4
0.1 8.44658 × 1 0 14 3.74539 × 1 0 13 5.99959 × 1 0 14 2.66203 × 1 0 13
0.5 3.76645 × 1 0 11 4.98079 × 1 0 10 2.51322 × 1 0 11 3.55499 × 1 0 10
0.9 8.33893 × 1 0 9 2.08303 × 1 0 10 5.82464 × 1 0 9 1.30658 × 1 0 10
Table 4

Order of convergence of Application 6.1 by taking various values of Δ t = 1 m for fixed h = 1 240 and t = 9 10

Proposed method
γ m L ( N ) L 2 ( N ) μ
0.32 5 0.00079894 0.00055564
10 0.00026095 0.00018178 1.61432
20 0.00007315 0.00005101 1.83478
40 0.00001921 0.00001340 1.92919
80 4.90113 × 1 0 6 3.42021 × 1 0 6 1.97054
0.72 20 9.29997 × 1 0 6 5.78571 × 1 0 6
40 2.97431 × 1 0 6 1.89860 × 1 0 6 1.64467
80 8.19033 × 1 0 7 5.27495 × 1 0 7 1.86056
160 2.13139 × 1 0 7 1.37757 × 1 0 7 1.94213
320 5.41608 × 1 0 8 3.50547 × 1 0 8 1.97647
Table 5

Order of convergence with fixed Δ t = 1 1,000 for Application 6.1 and various choices of h = 1 n when t = 9 10

Proposed method
γ m = n L ( N ) L 2 ( N ) μ
0.32 5 0.000735108 0.000542213
10 0.000258923 0.000180519 1.50543
20 0.000073009 0.0000509183 1.82638
40 0.000019199 0.0000133951 1.92705
80 4.90060 × 1 0 6 3.41986 × 1 0 6 1.97000
0.72 20 9.27310 × 1 0 6 5.77695 × 1 0 6
20 2.97235 × 1 0 6 1.89788 × 1 0 6 1.64145
40 8.18914 × 1 0 7 5.27449 × 1 0 7 1.85982
80 2.13134 × 1 0 7 1.37755 × 1 0 7 1.94195
160 3.50549 × 1 0 8 5.41613 × 1 0 8 1.97643
Figure 1 
               Approximate and analytic solutions with 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                   of Application 6.1 at various time levels. (a) 
                     
                        
                        
                           N
                           =
                           65
                           ,
                           γ
                           =
                           0.255
                        
                        N=65,\gamma =0.255
                     
                  , (b) 
                     
                        
                        
                           N
                           =
                           85
                           ,
                           γ
                           =
                           0.255
                        
                        N=85,\gamma =0.255
                     
                  , (c) 
                     
                        
                        
                           N
                           =
                           65
                           ,
                           γ
                           =
                           0.755
                        
                        N=65,\gamma =0.755
                     
                  , and (d) 
                     
                        
                        
                           N
                           =
                           85
                           ,
                           γ
                           =
                           0.755
                        
                        N=85,\gamma =0.755
                     
                  .
Figure 1

Approximate and analytic solutions with Δ t = 0.001 of Application 6.1 at various time levels. (a) N = 65 , γ = 0.255 , (b) N = 85 , γ = 0.255 , (c) N = 65 , γ = 0.755 , and (d) N = 85 , γ = 0.755 .

Figure 2 
               3D Analytic solution and numerical solution images with 
                     
                        
                        
                           N
                           =
                           100
                        
                        N=100
                     
                  , 
                     
                        
                        
                           t
                           =
                           1
                        
                        t=1
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.225
                        
                        \gamma =0.225
                     
                  , 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                   for Application 6.1 when 
                     
                        
                        
                           0
                           ≤
                           z
                           ≤
                           1
                        
                        0\le z\le 1
                     
                  . (a) Analytic solution and (b) numerical solution.
Figure 2

3D Analytic solution and numerical solution images with N = 100 , t = 1 , γ = 0.225 , Δ t = 0.001 for Application 6.1 when 0 z 1 . (a) Analytic solution and (b) numerical solution.

Figure 3 
               2D and 3D Error images with 
                     
                        
                        
                           N
                           =
                           100
                        
                        N=100
                     
                  , 
                     
                        
                        
                           t
                           =
                           1
                        
                        t=1
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.255
                        
                        \gamma =0.255
                     
                  , 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                   for Application 6.1 when 
                     
                        
                        
                           0
                           ≤
                           z
                           ≤
                           1
                        
                        0\le z\le 1
                     
                  .
Figure 3

2D and 3D Error images with N = 100 , t = 1 , γ = 0.255 , Δ t = 0.001 for Application 6.1 when 0 z 1 .

Application 6.2

Consider the TFFNE for q = 1

γ p t γ = ν 2 p z 2 + β p ( 1 p ) ( p η ) + U ( z , t ) , a z b , 0 t T ,

with IC and BCs

(6.2) p ( z , 0 ) = 0 p ( 0 , t ) = t 5 γ , p ( 1 , t ) = exp ( 1 ) t 5 γ ,

and calculation of U is

U ( z , t ) = Γ ( 5 γ + 1 ) 1 γ exp ( z 2 ) t 5 γ E γ , 5 γ + 1 γ t γ 1 γ + 2 ν [ t 5 γ exp ( z 2 ) ] [ 1 2 z 2 ] β [ t 5 γ exp ( z 2 ) ] [ 1 t 5 γ exp ( z 2 ) ] × [ t 5 γ exp ( z 2 ) η ] .

p ( z , t ) = t 5 γ exp ( z 2 ) is the analytic solution, where ν = β = 1 and η = 0.5 . The absolute errors and the numerical outcomes for various values of z of Application 6.2 setting N = 200 , 150 , 180 , and different values of γ and Δ t at t = 1 are presented in Tables 6 and 7. The error norms for t = 0.1 , t = 0.45 , and t = 0.9 are presented in Table 8 for different values of γ with N = 175 , Δ t = 0.0005 , and z [ 0 , 1 ] . The error norms and order of convergence μ are calculated in Tables 9 and 10 in time and space directions, respectively. Figure 4 highlights the performance of analytic solution and numerical outcomes at different temporal directions. 3D plots in Figure 5 of exact and computational solution are presented. The 2D error graph and 3D error graph are shown in Figure 6. For different values of γ , the 2D graphs are shown in Figure 7 for exact and numerical solution.

Table 6

Absolute errors of Application 6.2 for different values of γ when Δ t = 1 2,500 and N = 200 at t = 1

γ = 1 4 γ = 3 4
z Analytic solution Numerical result Error Analytic solution Numerical result Error
0.1 0.99005 0.99005 4.15062 × 1 0 7 0.99005 0.99005 3.27136 × 1 0 7
0.2 0.96079 0.96079 6.03925 × 1 0 7 0.96079 0.96079 4.60952 × 1 0 7
0.3 0.91393 0.91393 6.03551 × 1 0 7 0.91393 0.91393 4.3518 × 1 0 7
0.4 0.85214 0.85214 4.65706 × 1 0 7 0.85214 0.85214 2.96126 × 1 0 7
0.5 0.77880 0.77880 2.51426 × 1 0 7 0.77880 0.77880 9.8195 × 1 0 8
0.6 0.69768 0.69768 2.43833 × 1 0 8 0.69768 0.69768 1.01663 × 1 0 7
0.7 0.61263 0.61263 1.55997 × 1 0 7 0.61263 0.61263 2.49878 × 1 0 7
0.8 0.52729 0.52729 2.40588 × 1 0 7 0.52729 0.52729 3.01590 × 1 0 7
0.9 0.44486 0.44486 1.94746 × 1 0 7 0.44486 0.44486 2.24421 × 1 0 7
Table 7

Errors with Δ t = 0.0005 of Application 6.2 at t = 1

γ = 9 20 , N = 150 γ = 13 20 , N = 180
z Analytic solution Numerical result Absolute error Analytic solution Numerical result Absolute error
0.1 0.99005 0.99005 6.69283 × 1 0 7 0.99005 0.99005 4.16680 × 1 0 7
0.2 0.96079 0.96079 1.01761 × 1 0 6 0.96079 0.96079 5.90103 × 1 0 7
0.3 0.91393 0.91393 1.00680 × 1 0 6 0.91393 0.91393 5.61936 × 1 0 7
0.4 0.85214 0.85214 7.61051 × 1 0 7 0.85214 0.85214 3.90011 × 1 0 7
0.5 0.77880 0.77880 3.86353 × 1 0 7 0.77880 0.77880 1.42579 × 1 0 7
0.6 0.69768 0.69768 6.68630 × 1 0 9 0.69768 0.69768 1.08793 × 1 0 7
0.7 0.61263 0.61263 3.14688 × 1 0 7 0.61263 0.61263 2.96796 × 1 0 7
0.8 0.52729 0.52729 4.51997 × 1 0 7 0.52729 0.52729 3.65201 × 1 0 7
0.9 0.44486 0.44486 3.57983 × 1 0 7 0.44486 0.44486 2.73714 × 1 0 7
Table 8

L ( N ) and L 2 ( N ) error norms of Application 6.2 for various values of γ with Δ t = 0.0005 , N = 160 and 0 z 1

L ( N ) L 2 ( N )
t γ = 15 20 γ = 13 20 γ = 15 20 γ = 13 20
0.1 4.58028 × 1 0 11 5.22857 × 1 0 10 2.67718 × 1 0 11 3.02405 × 1 0 10
0.45 6.76907 × 1 0 8 1.40524 × 1 0 7 3.95313 × 1 0 8 8.10530 × 1 0 8
0.9 5.41753 × 1 0 7 6.69999 × 1 0 7 3.17722 × 1 0 7 3.87817 × 1 0 7
Table 9

Order of convergence with various values of Δ t = 1 m and fixed h = 1 425 of Application 6.2 when t = 1

Proposed method
γ m L ( N ) L 2 ( N ) μ
0.32 10 0.00129431 0.000890232
20 0.000385587 0.000266753 1.74705
40 0.000104669 0.0000725964 1.88123
80 0.0000271709 0.0000188835 1.94569
160 6.84355 × 1 0 6 4.77689 × 1 0 6 1.98925
0.72 10 0.00422254 0.00289823
20 0.00158785 0.00109955 1.41103
40 0.000480743 0.000334018 1.72374
80 0.000132024 0.0000918791 1.86447
160 0.0000345204 0.0000240582 1.93527
Table 10

Order of convergence with fixed Δ t = 1 425 and different choices of h = 1 n for Application 6.2 when t = 1

Proposed method
γ n L ( N ) L 2 ( N ) μ
0.32 10 0.000246743 0.000146946
20 0.0000630247 0.0000364128 1.96902
40 0.0000150676 8.73748 × 1 0 6 2.06447
80 3.08653 × 1 0 6 1.86245 × 1 0 6 2.28739
160 8.13505 × 1 0 7 4.79792 × 1 0 7 1.92376
0.72 5 0.000949882 0.000562242
10 0.000235108 0.00014029 2.01443
20 0.0000571683 0.0000333155 2.04004
40 0.0000108836 6.87791 × 1 0 6 2.39306
80 3.92854 × 1 0 6 2.48103 × 1 0 6 1.47009
Table 11

Absolute errors of Application 6.3 for different values of γ when Δ t = 1 5,000 and N = 500 at t = 1

γ = 1 4 γ = 3 4
z Analytic solution Numerical result Error Analytic solution Numerical result Error
0.1 0.25882 0.25882 4.18042 × 1 0 7 0.258819 0.258819 3.81068 × 1 0 7
0.2 0.50000 0.49999 7.99684 × 1 0 7 0.50000 0.49999 7.28667 × 1 0 7
0.3 0.70711 0.70711 1.11019 × 1 0 6 0.70711 0.70711 1.01104 × 1 0 6
0.4 0.86602 0.86602 1.32205 × 1 0 6 0.86603 0.86602 1.20349 × 1 0 6
0.5 0.96593 0.96592 1.41713 × 1 0 6 0.96593 0.96593 1.29023 × 1 0 6
0.6 1.0000 0.99999 1.38544 × 1 0 6 1.00000 0.99999 1.26263 × 1 0 6
0.7 0.96593 0.96592 1.22299 × 1 0 6 0.96593 0.96593 1.11692 × 1 0 6
0.8 0.86603 0.86602 9.31030 × 1 0 7 0.86603 0.86603 8.53002 × 1 0 7
0.9 0.70711 0.70711 5.17630 × 1 0 7 0.70711 0.70711 4.76236 × 1 0 7
Table 12

Errors with Δ t = 0.0004 of Application 6.3 at t = 1

γ = 1 4 , N = 250 γ = 3 4 , N = 200
z Analytic solution Numerical result Absolute error Analytic solution Numerical result Absolute error
0.1 0.25882 0.25882 1.67219 × 1 0 6 0.25882 0.25882 2.40548 × 1 0 6
0.2 0.50000 0.49999 3.19879 × 1 0 6 0.50000 0.49999 4.60242 × 1 0 6
0.3 0.70711 0.70710 4.44086 × 1 0 6 0.70711 0.70710 6.39185 × 1 0 6
0.4 0.86603 0.86602 5.28832 × 1 0 6 0.86603 0.86602 7.61635 × 1 0 6
0.5 0.96593 0.96592 5.66862 × 1 0 6 0.96593 0.96592 8.17223 × 1 0 6
0.6 1.00000 0.99999 5.54189 × 1 0 6 1.00000 0.99999 8.00148 × 1 0 6
0.7 0.96593 0.96592 4.89208 × 1 0 6 0.96592 0.96592 7.07799 × 1 0 6
0.8 0.86603 0.86602 3.72419 × 1 0 6 0.86603 0.86602 5.40301 × 1 0 6
0.9 0.70711 0.70711 2.07056 × 1 0 6 0.70711 0.70710 3.01425 × 1 0 6
Table 13

For Application 6.3 the error norms for different choices of γ when Δ t = 1 2,000 , N = 175 , and 0 z 1

L ( N ) L 2 ( N )
t γ = 7 20 γ = 17 20 γ = 7 20 γ = 17 20
0.15 6.75807 × 1 0 8 7.61423 × 1 0 9 4.82336 × 1 0 8 5.44521 × 1 0 9
0.45 1.3548 × 1 0 6 4.92105 × 1 0 7 9.66342 × 1 0 7 3.5158 × 1 0 7
0.75 5.61998 × 1 0 6 3.57981 × 1 0 6 4.00719 × 1 0 6 2.55469 × 1 0 6
Figure 4 
               At different time levels, the approximate and analytic solutions for Application 6.2. (a) 
                     
                        
                        
                           N
                           =
                           65
                           ,
                           γ
                           =
                           0.255
                        
                        N=65,\gamma =0.255
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                  , (b) 
                     
                        
                        
                           N
                           =
                           85
                           ,
                           γ
                           =
                           0.255
                        
                        N=85,\gamma =0.255
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                  , (c) 
                     
                        
                        
                           N
                           =
                           65
                           ,
                           γ
                           =
                           0.755
                        
                        N=65,\gamma =0.755
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                  , and (d) 
                     
                        
                        
                           N
                           =
                           85
                           ,
                           γ
                           =
                           0.755
                        
                        N=85,\gamma =0.755
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                  .
Figure 4

At different time levels, the approximate and analytic solutions for Application 6.2. (a) N = 65 , γ = 0.255 , and Δ t = 0.001 , (b) N = 85 , γ = 0.255 , and Δ t = 0.001 , (c) N = 65 , γ = 0.755 , and Δ t = 0.001 , and (d) N = 85 , γ = 0.755 , and Δ t = 0.001 .

Figure 5 
               For Application 6.2, the 3D analytic and numerical solutions images when 
                     
                        
                        
                           N
                           =
                           200
                        
                        N=200
                     
                  , 
                     
                        
                        
                           t
                           =
                           1
                        
                        t=1
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.75
                        
                        \gamma =0.75
                     
                  , 
                     
                        
                        
                           Δ
                           t
                           =
                           
                              
                                 1
                              
                              
                                 
                                    
                                    2,500
                                    
                                 
                              
                           
                        
                        \Delta t=\frac{1}{\hspace{0.1em}\text{2,500}\hspace{0.1em}}
                     
                  , and 
                     
                        
                        
                           0
                           ≤
                           z
                           ≤
                           1
                        
                        0\le z\le 1
                     
                  . (a) Analytic solution and (b) numerical solution.
Figure 5

For Application 6.2, the 3D analytic and numerical solutions images when N = 200 , t = 1 , γ = 0.75 , Δ t = 1 2,500 , and 0 z 1 . (a) Analytic solution and (b) numerical solution.

Figure 6 
               2D and 3D error graphs with 
                     
                        
                        
                           N
                           =
                           200
                        
                        N=200
                     
                  , 
                     
                        
                        
                           t
                           =
                           1
                        
                        t=1
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.75
                        
                        \gamma =0.75
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           
                              
                                 1
                              
                              
                                 
                                    
                                    2,500
                                    
                                 
                              
                           
                        
                        \Delta t=\frac{1}{\hspace{0.1em}\text{2,500}\hspace{0.1em}}
                     
                   for Application 6.2 when 
                     
                        
                        
                           0
                           ≤
                           z
                           ≤
                           1
                        
                        0\le z\le 1
                     
                  .
Figure 6

2D and 3D error graphs with N = 200 , t = 1 , γ = 0.75 , and Δ t = 1 2,500 for Application 6.2 when 0 z 1 .

Figure 7 
               2D Graph with different values of 
                     
                        
                        
                           γ
                        
                        \gamma 
                     
                   and 
                     
                        
                        
                           N
                           =
                           100
                        
                        N=100
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           
                              
                                 1
                              
                              
                                 1000
                              
                           
                        
                        \Delta t=\frac{1}{1000}
                     
                   for Application 6.2 when 
                     
                        
                        
                           0
                           ≤
                           z
                           ≤
                           1
                        
                        0\le z\le 1
                     
                  . (a) 
                     
                        
                        
                           t
                           =
                           0.5
                        
                        t=0.5
                     
                   and (b) 
                     
                        
                        
                           t
                           =
                           0.9
                        
                        t=0.9
                     
                  .
Figure 7

2D Graph with different values of γ and N = 100 , and Δ t = 1 1000 for Application 6.2 when 0 z 1 . (a) t = 0.5 and (b) t = 0.9 .

Application 6.3

Consider the TFFNE for q = 1

γ p t γ = ν 2 p z 2 + β p ( 1 p ) ( p η ) + U ( z , t ) , a z b , 0 t T ,

with IC and BCs

(6.3) p ( z , 0 ) = 0 p ( 0 , t ) = 0 , p ( 1 , t ) = t 2 γ + 2 2 ,

and calculation of U is

U ( z , t ) = Γ ( 2 γ + 3 ) 1 γ sin 5 z π 6 t 2 γ + 2 E γ , 2 γ + 3 γ t γ 1 γ + ν 5 π 6 2 t 2 γ + 2 sin 5 z π 6 β t 2 γ + 2 sin 5 z π 6 1 t 2 γ + 2 sin 5 z π 6 × t 2 γ + 2 sin 5 z π 6 η .

p ( z , t ) = t 2 γ + 2 sin [ 5 z π 6 ] is analytic solution, where ν = β = 1 and η = 0.5 . The computational results and absolute errors at different values of z establishing Δ t = 0.0002 , Δ t = 0.0004 , N = 500 , 250, 200, γ = 0.25 , 0.75 , and t = 1 for Application 6.3 are displayed in Tables 11 and 12. The error norms at t = 0.15 , t = 0.45 , and t = 0.75 are shown in Table 13 for different values of γ relying on N = 175 , Δ t = 0.0005 , and z [ 0 , 1 ] . The order of convergence μ and error norms are displayed in Tables 14 and 15 in time direction and space direction. Figure 8 provides a description of exact values and computational outcomes at different time levels. 3D precision of the existing approach is demonstrated by graphs of numerical results and analytical solutions in Figure 9. Figure 10 demonstrates the 2D and 3D error description, proving the method’s efficiency. Figure 11 shows the graphs of analytic and numerical solutions with different values of γ .

Table 14

Order of convergence with fixed h = 1 425 and taking different values of Δ t = 1 m for Application 6.3 when t = 1

Proposed method
γ m L ( N ) L 2 ( N ) μ
0.32 10 0.00398681 0.00268049
20 0.00134032 0.000906445 1.57266
40 0.00038427 0.000260435 1.80238
80 0.00010198 0.0000691498 1.91383
160 0.0000254313 0.0000172158 2.00361
0.72 10 0.00588917 0.00396847
20 0.00210514 0.00142527 1.48415
40 0.000623368 0.000422822 1.75576
80 0.000168799 0.000114559 1.88478
160 0.0000431439 0.0000292582 1.96808
Table 15

Order of convergence with different values of h = 1 n and Δ t = 1 625 for Application 6.3 when t = 1

Proposed method
γ n L ( N ) L 2 ( N ) μ
0.32 10 0.00358166 0.00256983
20 0.000896036 0.00064259 1.99900
40 0.000223318 0.000159799 2.00446
80 0.0000544964 0.0000390408 2.03487
160 0.0000122878 8.84781 × 1 0 6 2.14893
0.72 10 0.00336907 0.00241909
20 0.000842198 0.000604066 2.00012
40 0.000208842 0.000149542 2.01175
80 0.0000499815 0.0000358637 2.06295
160 0.0000102688 7.44260 × 1 0 6 2.28312
Figure 8 
               For Application 6.3, the analytic and numerical solutions at various temporal direction. (a) 
                     
                        
                        
                           N
                           =
                           65
                           ,
                           γ
                           =
                           0.255
                        
                        N=65,\gamma =0.255
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                  , (b) 
                     
                        
                        
                           N
                           =
                           85
                           ,
                           γ
                           =
                           0.255
                        
                        N=85,\gamma =0.255
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                  , (c) 
                     
                        
                        
                           N
                           =
                           65
                           ,
                           γ
                           =
                           0.755
                        
                        N=65,\gamma =0.755
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                  , and (d) 
                     
                        
                        
                           N
                           =
                           85
                           ,
                           γ
                           =
                           0.755
                        
                        N=85,\gamma =0.755
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.001
                        
                        \Delta t=0.001
                     
                  .
Figure 8

For Application 6.3, the analytic and numerical solutions at various temporal direction. (a) N = 65 , γ = 0.255 , and Δ t = 0.001 , (b) N = 85 , γ = 0.255 , and Δ t = 0.001 , (c) N = 65 , γ = 0.755 , and Δ t = 0.001 , and (d) N = 85 , γ = 0.755 , and Δ t = 0.001 .

Figure 9 
               3D Analytic solution and numerical solution images with 
                     
                        
                        
                           N
                           =
                           500
                        
                        N=500
                     
                  , 
                     
                        
                        
                           t
                           =
                           1
                        
                        t=1
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.25
                        
                        \gamma =0.25
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           0.0002
                        
                        \Delta t=0.0002
                     
                   for Application 6.3 when 
                     
                        
                        
                           0
                           ≤
                           z
                           ≤
                           1
                        
                        0\le z\le 1
                     
                  . (a) Analytic solution and (b) numerical solution.
Figure 9

3D Analytic solution and numerical solution images with N = 500 , t = 1 , γ = 0.25 , and Δ t = 0.0002 for Application 6.3 when 0 z 1 . (a) Analytic solution and (b) numerical solution.

Figure 10 
               For Application 6.3, the 2D and 3D error plots when 
                     
                        
                        
                           N
                           =
                           500
                        
                        N=500
                     
                  , 
                     
                        
                        
                           t
                           =
                           1
                        
                        t=1
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.25
                        
                        \gamma =0.25
                     
                  , 
                     
                        
                        
                           Δ
                           t
                           =
                           0.0002
                        
                        \Delta t=0.0002
                     
                  , and 
                     
                        
                        
                           0
                           ≤
                           z
                           ≤
                           1
                        
                        0\le z\le 1
                     
                  .
Figure 10

For Application 6.3, the 2D and 3D error plots when N = 500 , t = 1 , γ = 0.25 , Δ t = 0.0002 , and 0 z 1 .

Figure 11 
               2D Graph with different values of 
                     
                        
                        
                           γ
                        
                        \gamma 
                     
                   and 
                     
                        
                        
                           N
                           =
                           100
                        
                        N=100
                     
                  , and 
                     
                        
                        
                           Δ
                           t
                           =
                           
                              
                                 1
                              
                              
                                 
                                    
                                    1,000
                                    
                                 
                              
                           
                        
                        \Delta t=\frac{1}{\hspace{0.1em}\text{1,000}\hspace{0.1em}}
                     
                   for Application 6.3 when 
                     
                        
                        
                           0
                           ≤
                           z
                           ≤
                           1
                        
                        0\le z\le 1
                     
                  . (a) 
                     
                        
                        
                           t
                           =
                           0.5
                        
                        t=0.5
                     
                   and (b) 0.9.
Figure 11

2D Graph with different values of γ and N = 100 , and Δ t = 1 1,000 for Application 6.3 when 0 z 1 . (a) t = 0.5 and (b) 0.9.

7 Conclusion

In the current article, through a numerical strategy based on CBSFs, an effective solution to the TFFNE involving the ABTFD has been achieved. The ABTFD has been approximated using the usual finite difference formulation, and CBSFs are used for interpolating the solution curve in the spatial direction. The approach suggested in the current study is novel and offers an acceptable level of accuracy. The proposed scheme is stable unconditionally having second-order temporal and spatial convergence. The current algorithm is shown to be more effective, simple, and allowable when implemented to numerical application. Future work should focus on expanding the scope, analyzing the algorithm’s properties in more detail, and exploring its applicability in real-world scenarios.

Acknowledgments

The authors extend their appreciation to the Deanship of Scientific Research at Northern Border University, Arar, KSA for funding this research work through the project number NBU-FFR-2024-885-01.

  1. Funding information: The authors extend their appreciationto the Deanship of Scientific Research at Northern Border University, Arar, KSA for funding this research work through the project number NBU-FFR-2024-885-01.

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

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

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Received: 2023-07-31
Revised: 2023-09-05
Accepted: 2024-01-10
Published Online: 2024-01-25

© 2024 the author(s), published by De Gruyter

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

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