Startseite Technik An unconditionally stable numerical scheme for solving nonlinear Fisher equation
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An unconditionally stable numerical scheme for solving nonlinear Fisher equation

  • Vikash Vimal EMAIL logo , Rajesh Kumar Sinha und Pannikkal Liju
Veröffentlicht/Copyright: 3. September 2024
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Abstract

In this study, novel numerical methods are presented for solving nonlinear Fisher equations. These equations have a wide range of applications in various scientific and engineering fields, particularly in the biomedical sciences for determining the size of brain tumors. The challenges posed by the nonlinearity of the equations are effectively addressed through the development of numerical techniques. The nonlinearity is tackled using a combination of the method of lines and backward differentiation formulas of varied orders. This method is unconditionally stable, and its accuracy is evaluated using error norms. The methods are successfully validated against test problems with known solutions, demonstrating their superiority through comparative analyses with existing methodologies in the literature.

1 Introduction

The Fisher equation, first developed in 1937 [1], is expressed as follows in its original form:

(1) u t = μ u x x + α u δ ( 1 u ) , x ( , ) , t 0 ,

where u ( x , t ) represents the population density.

This equation is commonly referred to as the Kolmogorov–Petrovsky–Piscunov equation [2]. The first exact solution of Eq. (1) was obtained by Ablowitz and Zeppetella [3]. Various alternative formulations of Eq. (1) were subsequently explored in multiple instances [46]. The Fisher equation has found numerous applications in various scientific and engineering disciplines, as evidenced in previous literature [711]. Researchers have explored it in various forms and extended its scope, as indicated in previous studies [1216].

Various techniques have been employed to solve modified versions of the nonlinear Fisher’s reaction–diffusion equation. These approaches include the use of radial basis functions based on differential quadrature methods (RBFs with DQMs) [17], the Faedo–Galerkin method with homogeneous Dirichlet conditions [18], and the extended homogeneous balance method [19]. Furthermore, researchers have explored the application of Lie symmetries [20] and fractional extensions [21] to this equation. In recent years, there has been a concerted effort to develop efficient numerical techniques for computing solutions to the Fisher equation.

In this study, we utilize the method of lines (MOL) to semi-discretize the nonlinear reaction–diffusion equation, which effectively transforms it into a first-order ordinary differential equation (ODE). This approach, initially introduced by Schiesser in 1991 and documented in his research [22], has proven to be a powerful and effective tool for tackling time-dependent partial differential equations (PDEs). Within the MOL, we approximate spatial derivatives using finite differences. This method allows us to convert the PDEs into a set of ODEs that can be integrated over the time domain. To tackle the nonlinearity head-on, the study utilizes Newton’s method, providing implicit solutions that obviate the need for restrictive time step constraints.

We apply the finite difference method to introduce three distinct numerical approaches for addressing the generalized Fisher equation. These methods, referred to as backward differential formula (1) (BDF1), BDF2, and BDF3, involve replacing the differential equation with a discrete difference equation at each computational node. This substitution allows us to effectively employ backward differentiation techniques. After including boundary conditions in these equations, we proceed to solve the resulting system of algebraic equations. Various numerical solutions for Eq. (1) under different initial and boundary conditions have been investigated in previous studies [2325]. Within the domain of numerical error analysis, the fully discretized system, which includes BDF1, demonstrates a time accuracy of the first order and spatial accuracy of the second order. On the other hand, BDF2 achieves quadratic accuracy not only in time but also in spatial dimensions. In comparison, BDF3 attains a significant third-order accuracy in the time dimension while maintaining a second-order accuracy in spatial computations.

2 Test example

Consider the one-dimensional Fisher equation (1) for μ = 1 ,

(2) u t = 2 u x 2 + α u δ ( 1 u ) , 0 x 1 , t > 0 ,

with initial condition

u ( x , 0 ) = u 0 ( x ) , 0 x 1 ,

and the boundary conditions

u ( 0 , t ) = f 1 ( t ) , 0 t T , u ( 1 , t ) = f 2 ( t ) , 0 t T ,

where α represents the reactive factor, and T stands for the final time. Additionally, the functions u 0 ( x ) , f 1 , and f 2 denote the initial and boundary conditions.

3 Numerical scheme

In this section, we transform the PDE into an ODE with initial and boundary conditions using the MOL. The nonlinear reaction–diffusion equation, represented as Eq. (2), undergoes discretization exclusively along the spatial dimension ‘ x ’ through the utilization of the MOL [26].

3.1 Semi-discretization

The interval [ a , b ] is divided into N equal subintervals, where each subinterval maintains a uniform spacing of h = b a N . We can represent the discrete spatial points as x i = a + i h for i = 1 , 2 , , N , with the sequence: a = x 0 x 1 x N = b . When we discretize the variable ‘ x ’ using the MOLs, the second-order spatial derivative u x x is obtained.

(3) 2 u x 2 ( x i , t ) = 1 h 2 ( u i + 1 ( t ) 2 u i ( t ) + u i 1 ( t ) ) , i = 0 , 1 , 2 , N .

Substituting this into Eq. (2) yields a system of ODEs with the required initial condition.

(4) u i t = 1 h 2 ( u i + 1 ( t ) 2 u i ( t ) + u i 1 ( t ) ) + α u i ( 1 u i ) ,

(5) u ( t ) = 1 h 2 A u ( t ) , u ( 0 ) = u 0 .

The term u ( t ) = [ u 0 ( t ) , , u N ( t ) ] T stands for the vector, while u 0 = [ u 0 ( 0 ) , , u N ( 0 ) ] T represents the initial condition. Additionally, the matrix in question is a tridiagonal matrix of dimensions ( N + 1 ) * ( N + 1 ) .

A = 2 2 1 2 1 1 2 1 2 2 .

3.2 Full discretization

We divide the time interval [ 0 , T ] into M identical subintervals as follows: 0 = t 0 t 1 t M = T , where each subinterval has a length of Δ t = T M . In this scheme, we can express t n as t n = n Δ t , with n ranging from 1 to M . We numerically solve the system of first-order ODEs as described in Eq. (5) using various backward differentiation formulas (BDFs). The typical expression for a BDF of order p used to solve the subsequent initial value problem can be illustrated as follows:

u ( t ) = f ( t , u ) , u ( 0 ) = u 0

is given by

(6) u n + 1 = k = o p 1 α k u n k + Δ t β f ( t n + 1 , u n + 1 ) .

In this context, where u n + 1 = u ( x i , t n + 1 ) = [ u 0 , n + 1 , u N , n + 1 ] T , and Δ t signifies the time increment, the precise values of the coefficients α k for j = 0 , 1 , , p 1 , as well as the coefficient β , are detailed in the study by Atkinson et al. [27].

3.3 BDF of order p = 1 (BDF1)

In Eq. (6), when we set α 0 = 1 , β = 1 , and the function f ( t , u ) = 1 h 2 A u , we obtain the following expression:

(7) u n + 1 = u n + Δ t h 2 A u n + 1 , n = 0 , 1 , , M 1 ,

where u 0 represents the initial condition and is commonly referred to as the backward Euler method.

3.4 BDF of order p = 2 (BDF2)

With α 0 = 4 3 , α 1 = 1 3 , β = 2 3 , and f ( t , u ) = 1 h 2 A u , Eq. (6) becomes

(8) u n + 1 = 4 3 u n 1 3 u n 1 + 2 3 Δ t h 2 A u n + 1 , n = 1 , , M 1 .

This is the solution at the initial time level, denoted as u 1 and is computed using the BDF1 method.

3.5 BDF of order p = 3 (BDF3)

By substituting α 0 = 18 11 , α 1 = 9 11 , α 2 = 2 11 , β = 6 11 , and f ( t , u ) = 1 h 2 A u in Eq. (6), we obtain

(9) u n + 1 = 18 11 u n 9 11 u n 1 + 2 11 u n 2 + 6 11 Δ t h 2 A u n + 1 , n = 2 , , M 1 .

The solutions at the first time level, denoted as u 1 , and at the second time level, denoted as u 2 , are both obtained using the BDF1 method.

4 BDF (BDF1) scheme

(10) u t = u i 1 2 u i + u i + 1 ( Δ x ) 2 + α u i δ ( 1 u i ) , = λ u i 1 + u i ( α u i δ 1 α u i δ 2 λ ) + λ u i + 1 .

Substituting Eq. (10) in Eq. (7), we obtain

u n + 1 = u n + Δ t ( λ u i 1 + u i ( α u i δ 1 α u i δ 2 λ ) + λ u i + 1 )

λ Δ t u i 1 n + 1 + ( Δ t ( α ( u i n + 1 ) δ 1 α ( u i n + 1 ) δ 2 λ ) 1 ) u i n + 1 + λ Δ t u i + 1 n + 1 + u i n = 0 ,

where λ = 1 ( Δ x ) 2 , R i = Δ t ( α ( u i n + 1 ) δ 1 α ( u i n + 1 ) δ 2 λ ) 1 ,

I i = Δ t ( α α u i n + 1 2 λ ) 1 ,  where i = 1 , 2 , , N 1 .

The Jacobian matrix ( J ) is given by

J = I 1 λ Δ t λ Δ t I 2 λ Δ t λ Δ t I N 1 λ Δ t λ Δ t I N ,

B = R 1 λ Δ t λ Δ t R 2 λ Δ t λ Δ t R N 2 λ Δ t λ Δ t R N 1 u 1 1 u 2 1 u N 1 1 u N 1 + λ Δ t u 0 1 0 0 λ Δ t u N 1 + u 1 0 u 2 0 u N 1 0 u N 0 = 0 .

5 BDF (BDF2) scheme

Substituting Eq. (10) in Eq. (8), we obtain

u i n + 1 = 4 3 u i n 1 3 u i ( n 1 ) + 2 3 Δ t ( λ u i 1 n + 1 + u i n + 1 ( α ( u i n + 1 ) δ 1 α ( u i n + 1 ) δ 2 λ ) + λ u i + 1 n + 1 ) 2 3 Δ t λ u i 1 n + 1 + u i n + 1 × 2 3 Δ t ( α ( u i n + 1 ) δ 1 α ( u i n + 1 ) δ 2 λ ) 1 + 2 3 Δ t λ u i + 1 n + 1 + 4 3 u i n 1 3 u i n 1 = 0 ,

where A i = 2 3 Δ t ( α ( u i n + 1 ) δ 1 α ( u i n + 1 ) δ 2 λ ) 1 , H = 2 3 Δ t λ , and C i = 2 3 Δ t ( α 2 α u i n + 1 2 λ ) 1 .

J 1 is the Jacobian matrix given by

J 1 = C 1 H H C 2 H H C N 2 H H C N 1

M = A 1 H H A 2 H H A N 2 H H A N 1 u 1 2 u 2 2 u N 2 2 u N 1 2 + H u 0 1 0 0 H u N 1 1 + 4 3 u 1 1 u 2 1 u N 1 1 u N 1 1 3 u 1 0 u 2 0 u N 1 0 u N 0 = 0 .

6 BDF (BDF3) scheme

Substituting Eq. (10) in Eq. (9), we obtain

u i n + 1 = 8 11 u i n 9 11 u i ( n 1 ) + 2 11 u i ( n 2 ) + 6 11 Δ t ( λ u i 1 n + 1 + u i n + 1 ( α ( u i n + 1 ) δ 1 α ( u i n + 1 ) δ 2 λ ) + λ u i + 1 n + 1 )

6 11 Δ t λ u i 1 n + 1 + u i n + 1 × 6 11 Δ t ( α ( u i n + 1 ) δ 1 α ( u i n + 1 ) δ 2 λ ) 1 ) + 6 11 Δ t λ u i + 1 n + 1 + 2 11 u i n 2 9 11 u i n 1 + 18 11 u i n = 0 ,

where L i = 6 11 Δ t ( α ( u i n + 1 ) δ 1 α ( u i n + 1 ) δ 2 λ ) 1 , K = 6 11 Δ t λ ,

Q i = 6 11 Δ t ( α 2 α u i n + 1 2 λ ) 1 .

J 2 is the Jacobian matrix given by

J 2 = Q 1 K K Q 2 K K Q N 2 K K Q N 1 ,

N = L 1 K K L 2 K K L N 2 K K L N 1 u 1 3 u 2 3 u N 2 3 u N 1 3 + K u 0 3 0 0 K u N 1 3 + 2 11 u 1 0 u 2 0 u N 1 0 u N 0 9 11 u 1 1 u 2 1 u N 1 1 u N 1 + 18 11 u 1 2 u 2 2 u N 1 2 u N 2 = 0 .

7 Stability analysis

In this section, we assess the stability of the proposed numerical scheme using the von Neumann method. We have examined the linear version of the Fisher equation, assuming the constant b for the nonlinear term. We obtain

f i ( u 1 , u 2 , , u N 1 , t ) = λ ( u i + 1 ( t ) 2 u i ( t ) + u i 1 ( t ) ) + b ,

where λ = 1 ( Δ x ) 2 .

Applying the time integration to Eq. (4) results in

u i ( n + 1 ) = u i n + λ Δ t 2 ( u i + 1 ( n + 1 ) 2 u i ( n + 1 ) + u i 1 ( n + 1 ) ) + λ Δ t 2 ( u i + 1 ( n ) 2 u i ( n ) + u i 1 ( n ) ) + b .

By substituting u i n = ξ n e j k α i Δ x = ξ n e j i ϕ , where j = 1 , ϕ = k α Δ x , Δ x represents the space step size, and, k α = π α where α = 1 , 2 , , N , in the given equation, we derive the amplification factor, G , as

G = ξ ( n + 1 ) ξ n = X j Y X 1 + j Y ,

where

X = 1 Δ t λ ( 1 cos ϕ ) , X 1 = 1 + Δ t λ ( 1 cos ϕ ) , Y = 0 .

As both Δ t and Δ x are positive, therefore

X 1 X = 2 Δ t λ ( 1 c o s ϕ ) = 2 Δ t ( Δ x ) 2 ( 1 cos ϕ ) 0 , and

X X 1 . So

G = G G ¯ = X 2 + Y 2 X 1 2 + Y 2 1 .

Hence the proposed scheme is unconditionally stable.

8 Error analysis

We examine the error analysis of three different numerical techniques using the backward differentiation schemes of the first, second, and third order, based on the Taylor series expansion of the completely discretized numerical method.

8.1 BDF of order one (BDF1)

This scheme, fully discretized, can be expressed in the following manner using the BDF of the order one as specified in Eq. (7),

(11) u ( x i , t n + 1 ) = u ( x i , t n ) + γ h 2 [ u ( x i 1 , t n + 1 ) 2 u ( x i , t n + 1 ) + u ( x i + 1 , t n + 1 ) ] + γ α u ( x i , t n + 1 ) ( 1 u ( x i , t n + 1 ) ) .

In this context, with γ = Δ t and h = Δ x , where i = 1 , 2 , , N 1 and n = 0 , 1 , , M 1 , the local truncation error (LTE) can be described as follows:

(12) LTE = u ( x i , t n + 1 ) u ( x i , t n ) γ h 2 × [ u ( x i 1 , t n + 1 ) 2 u ( x i , t n + 1 ) + u ( x i + 1 , t n + 1 ) ] γ α u ( x i , t n + 1 ) ( 1 u ( x i , t n + 1 ) ) .

By applying the Taylor series expansion and subsequently simplifying, we deduce the following expression:

LTE = γ u t + γ 2 2 2 u t 2 + γ 3 6 3 u t 3 γ 2 u x 2 γ h 2 12 4 u x 4 + = γ 2 2 2 u t 2 + γ 3 6 3 u t 3 γ h 2 12 4 u x 4 + = O ( γ 2 + h 2 γ )

The expression for the truncation error (TE) is as follows:

T E = γ 1 ( LTE ) = γ 2 2 u t 2 + γ 3 3 u t 3 γ h 2 12 4 u x 4 + = O ( γ + h 2 ) = O ( Δ t + ( Δ x ) 2 ) .

Numerical errors grow in proportion to both the time step size and the square of the spatial step size. As a result, the recommended completely discretized system, when combined with the first-order BDF, exhibits second-order accuracy in spatial aspects and first-order accuracy in temporal aspects.

8.2 BDF of order two (BDF2)

The LTE associated with the scheme, fully discretized, can be expressed in the following manner using the BDF of the order two as specified in Eq. (8):

LTE = u ( x i , t n + 1 ) 4 3 u ( x i , t n ) 1 3 u ( x i , t n 1 ) 2 γ 3 h 2 × [ u ( x i 1 , t n + 1 ) 2 u ( x i , t n + 1 ) + u ( x i + 1 , t n + 1 ) ] 2 3 γ α u ( x i , t n + 1 ) ( 1 u ( x i , t n + 1 ) ) .

By applying the Taylor series expansion and subsequently simplifying, we deduce the following expression:

LTE = 2 γ 3 u t + 2 γ 2 3 2 u t 2 + γ 3 9 3 u t 3 2 γ 3 2 u x 2 γ h 2 18 4 u x 4 + = 2 γ 3 u t 2 u x 2 + 2 γ 2 3 2 u t 2 + γ 3 9 3 u t 3 γ h 2 18 4 u x 4 + = O ( γ 3 + γ h 2 )

The expression for the TE is as follows:

TE = γ 1 ( L T E ) = = 2 3 u t 2 u x 2 + 2 γ 3 2 u t 2 + γ 2 9 3 u t 3 h 2 18 4 u x 4 + = O ( γ 2 + h 2 ) = O ( ( Δ t ) 2 + ( Δ x ) 2 ) .

Therefore, the errors demonstrate quadratic dependence on both the time and space step sizes.

8.3 BDF of order three (BDF3)

Similarly, we calculate the LTE for the entirely discretized system by employing the third-order BDF comparably.

LTE = u ( x i , t n + 1 ) 4 3 u ( x i , t n ) 1 3 u ( x i , t n 1 ) 2 γ 3 h 2 [ u ( x i 1 , t n + 1 ) 2 u ( x i , t n + 1 ) + u ( x i + 1 , t n + 1 ) ] 2 γ 3 u ( x i , t n + 1 ) ( 1 u ( x i , t n + 1 ) ) = O ( γ 4 + γ h 2 ) .

The expression for the TE is as follows:

TE = γ 1 ( L T E ) = O ( γ 3 + h 2 ) = O ( ( Δ t ) 3 + ( Δ x ) 2 )

The computational inaccuracies exhibit a cubic relationship with the time increment and a quadratic relationship with the spatial increment. As a result, the errors are reduced to a minimum when employing the completely discretized method with BDF3, and this finding is further substantiated by the outcomes from multiple test scenarios.

9 Numerical tests and conversations

In this section, we evaluate the accuracy of the proposed techniques [28] by computing errors in both the L 2 and L norms, as defined in Eqs (13) and (14), at designated time points.

(13) L 2 = 1 M j = 1 M ϕ j ( exact ) U j ( num ) 2 1 2 ,

(14) L = ϕ j ( exact ) U j ( num ) = max j ϕ j ( exact ) U j ( num ) .

The rate of numerical convergence for these methods has been determined using the subsequent formula:

(15) ROC log ( E ( N 2 ) E ( N 1 ) ) log ( N 1 N 2 ) ,

where the L error value obtained by employing N j number of grid points is denoted as E ( N j ) . The L 2 and L error norms are computed for Tables 14 and 16 at the final time T = 0.1 , with α = δ = 1 and a time step of Δ t = 0.00005 . The tables show that as the grid sizes increased, both L 2 and L errors decreased. The table also displays the rate of convergence (ROC) for the suggested methods across various grid sizes.

Table 14

Comparing numerical and exact solutions for Example 1 at various spatial points with Δ t = 0.00005 , T = 0.1 , and α = δ = 1

BDF1 BDF2 BDF3
N L 2 L ROC L 2 L ROC L 2 L ROC
10 1.10 × 1 0 6 9.52 × 1 0 7 4.94 × 1 0 7 3.66 × 1 0 7 3.66 × 1 0 7
15 1.03 × 1 0 6 7.81 × 1 0 7 1.65 × 1 0 0 2.20 × 1 0 7 1.63 × 1 0 7 2.00 × 1 0 00 2.44 × 1 0 7 1.66 × 1 0 7 1.74 × 1 0 00
20 1.02 × 1 0 6 7.27 × 1 0 7 1.32 × 1 0 0 1.24 × 1 0 7 9.19 × 1 0 8 1.99 × 1 0 00 3.41 × 1 0 7 2.37 × 1 0 7 1.16 × 1 0 00
Table 16

Comparison of numerical and exact solutions (BDF1, BDF2, BDF3) for Example 2 at Δ t = 0.0005 , T = 0.1 , and α = δ = 1

Computed solution Exact solution
x BDF1 BDF2 BDF3
0.1 0.2606471000 0.260738174 0.260752490 0.260738428
0.2 0.2503452690 0.250420646 0.250447850 0.250421096
0.3 0.2402488598 0.240311103 0.240348491 0.240311688
0.4 0.2303642690 0.230417726 0.230461597 0.230418385
0.5 0.2206991200 0.220747972 0.220794013 0.220748648
0.6 0.2112596750 0.211308559 0.211352270 0.211309201
0.7 0.2020521340 0.202105453 0.202142581 0.202106010
0.8 0.1930823300 0.193143852 0.193170799 0.193144276
0.9 0.1843556660 0.184428190 0.184442349 0.184428430

9.1 Test problems

We conducted various numerical experiments to compare our proposed numerical method with the exact solution. Two specific examples were analyzed, and all calculations were carried out in MATLAB.

Example 1

Consider the Fisher Eq. (2) with initial condition,

u t = u x x + u ( 1 u ) ,

subject to the initial condition

u ( x , 0 ) = 1 ( 1 + e α 6 x ) 2 , 0 t 1 ,

where the exact solution is presented in the study by Wazwaz and Gorguis [29] given by

u ( x , t ) = 1 ( 1 + e α 6 x 5 6 α t ) 2 .

We have tabulated both the exact solution and numerical results obtained at various node points for different values of α .

In this study, we compare numerical results and exact solutions derived from different BDFs. Tables 1 and 2 focus on the evaluation of three BDF methods at fixed time instances but varying spatial positions, providing insights into their spatial performance. In parallel, Tables 3 and 4 explore error comparisons under consistent time points but different spatial locations, offering valuable information about precision. Figures 1, 2, 3, 4, 5, 6 illustrate the computed solutions produced by BDF methods of first, second, and third orders, together with the exact solution, across a range of alpha values. In Figures 7 and 8, the absolute errors are depicted at the same time points but in different spatial locations. In a concurrent analysis, we expanded our comparison by using a time step of Δ t = 0.0005 , and the results are documented in Tables 5 and 6. Absolute errors, which are essential for evaluating accuracy, are presented in Tables 7 and 8. Figures 9, 10, 11, 12, 13, 14 display the numerical solutions, while Figures 15 and 16 provide a detailed view of the absolute errors. Notably, under these conditions, BDF2 demonstrates superior accuracy compared to BDF1 and BDF3.

Table 1

Comparison of numerical and exact solutions (BDF1, BDF2, BDF3) for Example 1 at Δ t = 0.000005 , T = 0.1 , and α = δ = 1

x BDF1 BDF2 BDF3 Exact solution
0.2 0.250420019 0.250420775 0.250421049 0.250421096
0.3 0.240310644 0.240311272 0.240311649 0.240311688
0.4 0.230417375 0.230417912 0.230418355 0.230418385
0.5 0.220747663 0.220748154 0.220748619 0.220748648
0.6 0.211308229 0.211308721 0.211309162 0.211309201
0.7 0.202105045 0.202105581 0.202105956 0.202106010
0.8 0.193143323 0.193143940 0.193144212 0.193144276
0.9 0.184427509 0.184428235 0.184428378 0.184428430
Table 2

Comparison of numerical and exact solutions (BDF1, BDF2, BDF3) for Example 1 at Δ t = 0.000005 , T = 0.1 , and α = 6 , δ = 1

x BDF1 BDF2 BDF3 Exact solution
0.1 0.3584088969 0.3584211896 0.3584221152 0.3584269144
0.2 0.3299634211 0.3299734888 0.3299752538 0.3299842051
0.3 0.3022946501 0.3023027800 0.3023052162 0.3023174246
0.4 0.2755792279 0.2755858557 0.2755887275 0.2756031473
0.5 0.2499758181 0.2499814841 0.2499845115 0.2500000000
0.6 0.2256212575 0.2256265533 0.2256294397 0.2256447723
0.7 0.2026276009 0.2026331139 0.2026355750 0.2026494300
0.8 0.1810801962 0.1810864553 0.1810882474 0.1810991715
0.9 0.1610368521 0.1610442798 0.1610452236 0.1610515941
Table 3

Error comparison (BDF1, BDF2, BDF3) for Example 1 at Δ t = 0.000005 , T = 0.1 with α = δ = 1

x BDF1 BDF2 BDF3
0.1 1.09612 × 1 0 6 1.84267 × 1 0 7 3.98471 × 1 0 8
0.2 1.07701 × 1 0 6 3.21258 × 1 0 7 4.67977 × 1 0 8
0.3 1.04363 × 1 0 6 4.16011 × 1 0 7 3.87691 × 1 0 8
0.4 1.01011 × 1 0 6 4.72709 × 1 0 7 2.99744 × 1 0 8
0.5 9.85211 × 1 0 7 4.93894 × 1 0 7 2.91792 × 1 0 8
0.6 9.71300 × 1 0 7 4.79938 × 1 0 7 3.86630 × 1 0 8
0.7 9.64053 × 1 0 7 4.28806 × 1 0 7 5.39126 × 1 0 8
0.8 9.52847 × 1 0 7 3.36153 × 1 0 7 6.40192 × 1 0 8
0.9 9.21778 × 1 0 7 1.95717 × 1 0 7 5.27188 × 1 0 8
Table 4

Error comparison (BDF1, BDF2, BDF3) for Example 1 at Δ t = 0.000005 , T = 0.1 with α = 6 , δ = 1

x BDF1 BDF2 BDF3
0.1 1.80175 × 1 0 5 5.72477 × 1 0 6 4.79917 × 1 0 6
0.2 2.07840 × 1 0 5 1.07164 × 1 0 5 8.95132 × 1 0 6
0.3 2.27745 × 1 0 5 1.46446 × 1 0 5 1.22084 × 1 0 5
0.4 2.39194 × 1 0 5 1.72915 × 1 0 5 1.44198 × 1 0 5
0.5 2.41819 × 1 0 5 1.85159 × 1 0 5 1.54885 × 1 0 5
0.6 2.35148 × 1 0 5 1.82190 × 1 0 5 1.53326 × 1 0 5
0.7 2.18291 × 1 0 5 1.63161 × 1 0 5 1.38550 × 1 0 5
0.8 1.89753 × 1 0 5 1.27162 × 1 0 5 1.09241 × 1 0 5
0.9 1.47420 × 1 0 5 7.31436 × 1 0 6 6.37055 × 1 0 6
Figure 1 
                  Error propagation of Example 1 at 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                      for 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                     .
Figure 1

Error propagation of Example 1 at T = 0.1 , Δ t = 0.000005 for α = δ = 1 .

Figure 2 
                  Error propagation of Example 1 at 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                      for 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                     .
Figure 2

Error propagation of Example 1 at T = 0.1 , Δ t = 0.000005 for α = δ = 1 .

Figure 3 
                  Error propagation of Example 1 at 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                      for 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                     .
Figure 3

Error propagation of Example 1 at T = 0.1 , Δ t = 0.000005 for α = δ = 1 .

Figure 4 
                  Error propagation of Example 1 at 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                      for 
                        
                           
                           
                              α
                              =
                              6
                           
                           \alpha =6
                        
                      and 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                     .
Figure 4

Error propagation of Example 1 at T = 0.1 , Δ t = 0.000005 for α = 6 and δ = 1 .

Figure 5 
                  Error propagation of Example 1 at 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                      for 
                        
                           
                           
                              α
                              =
                              6
                           
                           \alpha =6
                        
                      and 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                     .
Figure 5

Error propagation of Example 1 at T = 0.1 , Δ t = 0.000005 for α = 6 and δ = 1 .

Figure 6 
                  Error propagation of Example 1 at 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                      for 
                        
                           
                           
                              α
                              =
                              6
                           
                           \alpha =6
                        
                      and 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                     .
Figure 6

Error propagation of Example 1 at T = 0.1 , Δ t = 0.000005 for α = 6 and δ = 1 .

Figure 7 
                  Error comparison of Example 1 at 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                     .
Figure 7

Error comparison of Example 1 at α = δ = 1 , T = 0.1 , and Δ t = 0.000005 .

Figure 8 
                  Error comparison of Example 1 at 
                        
                           
                           
                              α
                              =
                              6
                           
                           \alpha =6
                        
                     , 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                     .
Figure 8

Error comparison of Example 1 at α = 6 , δ = 1 , T = 0.1 , and Δ t = 0.000005 .

Table 5

Comparison of numerical and exact solutions (BDF1, BDF2, BDF3) for Example 1 at Δ t = 0.0005 , T = 0.1 , and α = δ = 1

Computed solution Exact solution
x BDF1 BDF2 BDF3
0.1 0.260555346 0.260738174 0.260752490 0.260738428
0.2 0.250268689 0.250420646 0.250447850 0.250421096
0.3 0.240184531 0.240311103 0.240348491 0.240311688
0.4 0.230309036 0.230417726 0.230461597 0.230418385
0.5 0.220648412 0.220747972 0.220794013 0.220748648
0.6 0.211208986 0.211308559 0.211352270 0.211309201
0.7 0.201997201 0.202105453 0.202142581 0.202106010
0.8 0.193019543 0.193143852 0.193170799 0.193144276
0.9 0.184282411 0.184428190 0.184442349 0.184428430
Table 6

Comparison of numerical and exact solutions (BDF1, BDF2, BDF3) for Example 1 at Δ t = 0.0005 , T = 0.1 , and α = 6 , δ = 1

Computed solution Exact solution
x BDF1 BDF2 BDF3
0.1 0.357193283 0.358420734 0.358512716 0.358426914
0.2 0.328968781 0.329972662 0.330148039 0.329984205
0.3 0.301492288 0.302301712 0.302543759 0.302317425
0.4 0.274925758 0.275584698 0.275870007 0.275603147
0.5 0.249417565 0.249980376 0.250281128 0.250000000
0.6 0.225099525 0.225625602 0.225912319 0.225644772
0.7 0.202084201 0.202632381 0.202876840 0.202649430
0.8 0.180462762 0.181085967 0.181263959 0.181099172
0.9 0.160303577 0.161044038 0.161137779 0.161051594
Table 7

Error comparison (BDF1, BDF2, BDF3) for Example 1 at Δ t = 0.0005 , T = 0.1 with α = δ = 1

x BDF1 BDF2 BDF3
0.1 0.000183082 2.538 × 1 0 7 1.40619 × 1 0 5
0.2 0.000152407 4.497 × 1 0 7 2.67542 × 1 0 5
0.3 0.000127158 5.848 × 1 0 7 3.68024 × 1 0 5
0.4 0.000109349 6.594 × 1 0 7 4.32119 × 1 0 5
0.5 0.000100236 6.768 × 1 0 7 4.53644 × 1 0 5
0.6 0.000100215 6.414 × 1 0 7 4.30690 × 1 0 5
0.7 0.000108809 5.567 × 1 0 7 3.65716 × 1 0 5
0.8 0.000124734 4.238 × 1 0 7 2.65234 × 1 0 5
0.9 0.000146020 2.400 × 1 0 7 1.39185 × 1 0 5
Table 8

Error comparison (BDF1, BDF2, BDF3) for Example 1 at Δ t = 0.0005 , T = 0.1 with α = 6 , δ = 1

x BDF1 BDF2 BDF3
0.1 0.001233632 6.18020 × 1 0 6 8.58017 × 1 0 5
0.2 0.001015424 1.15434 × 1 0 5 0.000163834
0.3 0.000825137 1.57122 × 1 0 5 0.000226335
0.4 0.000677389 1.84491 × 1 0 5 0.000266860
0.5 0.000582435 1.96236 × 1 0 5 0.000281128
0.6 0.000545247 1.91708 × 1 0 5 0.000267547
0.7 0.000565229 1.70489 × 1 0 5 0.000227410
0.8 0.000636410 1.32046 × 1 0 5 0.000164788
0.9 0.000748017 7.55610 × 1 0 6 8.61852 × 1 0 5
Figure 9 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      of Example 1.
Figure 9

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = δ = 1 of Example 1.

Figure 10 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      of Example 1.
Figure 10

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = δ = 1 of Example 1.

Figure 11 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      of Example 1.
Figure 11

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = δ = 1 of Example 1.

Figure 12 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              6
                           
                           \alpha =6
                        
                     , 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                      of Example 1.
Figure 12

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = 6 , δ = 1 of Example 1.

Figure 13 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              6
                           
                           \alpha =6
                        
                     , 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                      of Example 1.
Figure 13

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = 6 , δ = 1 of Example 1.

Figure 14 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              6
                           
                           \alpha =6
                        
                     , 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                      of Example 1.
Figure 14

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = 6 , δ = 1 of Example 1.

Figure 15 
                  Error comparison of Example 1 at 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     .
Figure 15

Error comparison of Example 1 at α = δ = 1 , T = 0.1 , and Δ t = 0.0005 .

Figure 16 
                  Error comparison of Example 1 at 
                        
                           
                           
                              α
                              =
                              6
                           
                           \alpha =6
                        
                     , 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     .
Figure 16

Error comparison of Example 1 at α = 6 , δ = 1 , T = 0.1 , and Δ t = 0.0005 .

We further examined the numerical solutions of BDF1, BDF2, and BDF3 in Tables 9, 10, and 11, comparing them with findings from previous publications [30,31] under the specific conditions of Δ t = 0.00005 , T = 0.1 , and α = 6 . Our numerical results for BDF1, BDF2, and BDF3 perform better than a few existing methods. The accuracy of these BDF methods was assessed by employing the error norms L 2 and L as detailed in Tables 12 and 13, revealing that BDF3 exhibits higher accuracy values compared to BDF1 and BDF2.

Table 9

Numerical and exact solutions for Example 1 are compared at different points of space at Δ t = 0.00005 , T = 0.1 for α = 6 , δ = 1

T x DQM [31] BDF1 Exact solution
0.5 0.25 0.81847 0.818313 0.818393
0.75 0.72592 0.725715 0.725824
1.0 0.25 0.98293 0.982909 0.982919
0.75 0.97208 0.972055 0.972071
Table 10

Numerical and exact solutions for Example 1 are compared at different points of space at Δ t = 0.00005 , T = 0.1 for α = 6 , δ = 1

T x DQM [31] BDF2 Exact solution
0.5 0.25 0.81847 0.818394 0.818393
0.75 0.72592 0.725826 0.725824
1.0 0.25 0.98293 0.982918 0.982919
0.75 0.97208 0.972070 0.972071
Table 11

Numerical and exact solutions for Example 1 are compared at different points of space at Δ t = 0.00005 , T = 0.1 for α = 6 , δ = 1

T x DQM [31] BDF3 Exact solution
0.5 0.25 0.81847 0.818395 0.818393
0.75 0.72592 0.725826 0.725824
1.0 0.25 0.98293 0.982918 0.982919
0.75 0.97208 0.972070 0.972071
Table 12

Assessing differences using the L 2 and L error norm for varying values of T = 0.1 and Δ t = 0.000005 , as illustrated in Example 1

α = δ = 1 α = 6 and δ = 1
T L 2 L L 2 L
BDF1 0.002677279 0.008466298 2.02786 × 1 0 5 2.41819 × 1 0 5
BDF2 0.002677279 0.008466298 1.34806 × 1 0 5 1.85159 × 1 0 5
BDF3 0.002677279 0.008466298 1.13446 × 1 0 5 1.54885 × 1 0 5
Table 13

Assessing differences using the L 2 and L error norm for varying values of T = 0.1 and Δ t = 0.0005 , as illustrated in Example 1

α = δ = 1 α = 6 and δ = 1
T L 2 L L 2 L
BDF1 0.000131399 0.000183082 0.000796513 0.001233631
BDF2 4.96013 × 1 0 7 6.76 × 1 0 7 1.42598 × 1 0 5 1.9624 × 1 0 5
BDF3 3.20668 × 1 0 5 4.5365 × 1 0 5 0.000198383 0.000281128

Example 2

Consider the following Fisher’s equation-2 in domain [0, 1] for α = δ = 1 :

(16) u t = u x x + u ( 1 u )

with initial condition

(17) u ( x , 0 ) = 1 2 tanh α 2 2 α + 4 x + 1 2 2 α .

The exact solution is presented in [12,30] by

(18) u ( x , t ) = 1 2 tanh α 2 2 α + 4 x α + 4 2 α + 4 t + 1 2 2 α .

In Tables 15 and 16, we calculate the numerical solutions for Example 2, considering different time steps, specifically Δ t = 0.000005 and Δ t = 0.0005 , while maintaining constants T = 0.1 and α = 1 . Furthermore, Tables 17 and 18 provide the absolute errors for this example. Additionally, Figures 17, 18, 19, 20, 21, 22 include both 2D and 3D plots that illustrate the numerical solution in comparison to the exact solution, while Figures 23 and 24 depict 2D and 3D plots of the absolute errors.

Table 15

Comparison of numerical and exact solutions (BDF1, BDF2, BDF3) for Example 2 at Δ t = 0.000005 , T = 0.1 , and α = δ = 1

Computed solution Exact solution
x BDF1 BDF2 BDF3
0.1 0.260736413 0.2607382437 0.260738388 0.260738428
0.2 0.250419251 0.2504207746 0.250421049 0.250421096
0.3 0.240310001 0.2403112721 0.240311649 0.240311688
0.4 0.230416819 0.2304179122 0.230418355 0.230418385
0.5 0.220747153 0.2207481544 0.220748619 0.220748648
0.6 0.211307719 0.2113087208 0.211309162 0.211309201
0.7 0.202104493 0.2021055807 0.202105956 0.202106010
0.8 0.193142693 0.1931439398 0.193144212 0.193144276
0.9 0.184426775 0.1844282347 0.184428378 0.184428430
Table 17

Error Comparison (BDF1, BDF2, BDF3) for Example 2 at Δ t = 0.000005 , T = 0.1 with α = δ = 1

x BDF1 BDF2 BDF3
0.1 2.01499 × 1 0 6 1.84222 × 1 0 7 3.98220 × 1 0 8
0.2 1.84504 × 1 0 6 3.21283 × 1 0 7 4.67833 × 1 0 8
0.3 1.68720 × 1 0 6 4.15963 × 1 0 7 3.87630 × 1 0 8
0.4 1.56573 × 1 0 6 4.72690 × 1 0 7 2.99897 × 1 0 8
0.5 1.49567 × 1 0 6 4.93901 × 1 0 7 2.92009 × 1 0 8
0.6 1.48139 × 1 0 6 4.79973 × 1 0 7 3.86734 × 1 0 8
0.7 1.51614 × 1 0 6 4.28834 × 1 0 7 5.39340 × 1 0 8
0.8 1.58283 × 1 0 6 3.36198 × 1 0 7 6.39982 × 1 0 8
0.9 1.65563 × 1 0 6 1.95727 × 1 0 7 5.27266 × 1 0 8
Table 18

Error Comparison (BDF1, BDF2, BDF3) for Example 2 at Δ t = 0.0005 , T = 0.1 with α = δ = 1

x BDF1 BDF2 BDF3
0.1 9.132700 × 1 0 5 2.538 × 1 0 7 1.40619 × 1 0 5
0.2 7.582690 × 1 0 5 4.497 × 1 0 7 2.67542 × 1 0 5
0.3 6.308980 × 1 0 5 5.848 × 1 0 7 3.68024 × 1 0 5
0.4 5.411580 × 1 0 5 6.594 × 1 0 7 4.32119 × 1 0 5
0.5 4.952830 × 1 0 5 6.768 × 1 0 7 4.53644 × 1 0 5
0.6 4.952580 × 1 0 5 6.414 × 1 0 7 4.30690 × 1 0 5
0.7 5.387586 × 1 0 5 5.567 × 1 0 7 3.65716 × 1 0 5
0.8 6.194570 × 1 0 5 4.238 × 1 0 7 2.65234 × 1 0 5
0.9 7.276420 × 1 0 5 2.400 × 1 0 7 1.39185 × 1 0 5
Figure 17 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      for Example 2.
Figure 17

Error propagation at Δ t = 0.000005 , T = 0.1 , and α = δ = 1 for Example 2.

Figure 18 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      for Example 2.
Figure 18

Error propagation at Δ t = 0.000005 , T = 0.1 , and α = δ = 1 for Example 2.

Figure 19 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      for Example 2.
Figure 19

Error propagation at Δ t = 0.000005 , T = 0.1 , and α = δ = 1 for Example 2.

Figure 20 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      for Example 2.
Figure 20

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = δ = 1 for Example 2.

Figure 21 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      for Example 2.
Figure 21

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = δ = 1 for Example 2.

Figure 22 
                  Error propagation at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      for Example 2.
Figure 22

Error propagation at Δ t = 0.0005 , T = 0.1 , and α = δ = 1 for Example 2.

Figure 23 
                  Error comparison at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.000005
                           
                           \Delta t=0.000005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      for Example 2.
Figure 23

Error comparison at Δ t = 0.000005 , T = 0.1 , and α = δ = 1 for Example 2.

Figure 24 
                  Error comparison at 
                        
                           
                           
                              Δ
                              t
                              =
                              0.0005
                           
                           \Delta t=0.0005
                        
                     , 
                        
                           
                           
                              T
                              =
                              0.1
                           
                           T=0.1
                        
                     , and 
                        
                           
                           
                              α
                              =
                              δ
                              =
                              1
                           
                           \alpha =\delta =1
                        
                      for Example 2.
Figure 24

Error comparison at Δ t = 0.0005 , T = 0.1 , and α = δ = 1 for Example 2.

In Tables 19, 20, and 21, we evaluate the numerical solutions of BDF1, BDF2, and BDF3 compared to a few existing methods [30,31]. This assessment is performed under specific conditions, including Δ t = 0.00005 , T = 0.1 , and α = 1 . BDF1, BDF2, and BDF3 exhibit more accuracy and precision compared to a few existing methods (Table 22).

Table 19

Comparing numerical results with exact solutions for Example 2 at various spatial positions, with a time step of Δ t = 0.00005 and a total time of T = 0.1 , while considering α = δ = 1 .

T x DQM [31] BDF1 Exact solution
0.5 0.25 0.33412 0.334071 0.334094
0.75 0.27838 0.278332 0.278353
1.0 0.25 0.45576 0.455714 0.455739
0.75 0.39544 0.395387 0.395411
Table 20

Comparing numerical results with exact solutions for Example 2 at various spatial positions, with a time step of Δ t = 0.00005 and a total time of T = 0.1 , while considering α = δ = 1

T x DQM [31] CFD6 [30] BDF2 Exact solution
0.5 0.25 0.33412 0.334094 0.334094 0.334094
0.75 0.27838 0.278353 0.278353 0.278353
1.0 0.25 0.45576 0.455739 0.455739 0.455739
0.75 0.39544 0.395411 0.395411 0.395411
Table 21

Comparing numerical results with exact solutions for Example 2 at various spatial positions, with a time step of Δ t = 0.00005 and a total time of T = 0.1 , while considering α = δ = 1

T x DQM [31] CFD6 [30] BDF3 Exact solution
0.5 0.25 0.33412 0.334094 0.334094 0.334094
0.75 0.27838 0.278353 0.278353 0.278353
1.0 0.25 0.45576 0.455739 0.455739 0.455739
0.75 0.39544 0.395411 0.395411 0.395411
Table 22

Comparing numerical and exact solutions for Example 2 at various spatial points with Δ t = 0.00005 , T = 0.1 , and α = δ = 1

BDF1 BDF2 BDF3
N L 2 L ROC L 2 L ROC L 2 L ROC
10 1.10 × 1 0 6 9.52 × 1 0 7 4.94 × 1 0 7 3.66 × 1 0 7 3.66 × 1 0 7
15 1.03 × 1 0 6 7.81 × 1 0 7 1.65 × 1 0 0 2.20 × 1 0 7 1.63 × 1 0 7 2.00 × 1 0 00 2.44 × 1 0 7 1.66 × 1 0 7 1.74 × 1 0 00
20 1.02 × 1 0 6 7.27 × 1 0 7 1.32 × 1 0 0 1.24 × 1 0 7 9.19 × 1 0 8 1.99 × 1 0 00 3.41 × 1 0 7 2.37 × 1 0 7 1.16 × 1 0 00

10 Conclusion

To transform the nonlinear Fisher equation into a set of first-order ODEs, we employ a semi-discretization method along the variable “ x ” using the MOLs, and then, by utilizing first order, second order, and third-order BDFs, we can efficiently resolve this system of first-order ODEs. The numerical results generated by BDF2 and BDF3 exhibit higher accuracy compared to those produced by BDF1. In our evaluation of errors, we observe that BDF-3 achieves third-order accuracy with O ( ( Δ t ) 3 + ( Δ x ) 2 ) , BDF2 achieves second-order precision with O ( ( Δ t ) 2 + ( Δ x ) 2 ) , and BDF1 provides first-order accuracy with O ( Δ t + ( Δ x ) 2 ) . Remarkably, for smaller values of Δ t , the error diminishes, demonstrating that the computed results remain reasonably close to the exact solution. This approach effectively addresses the nonlinear, time-dependent Fisher equation.

Acknowledgements

The authors are very thankful to the reviewers for their valuable comments and suggestions, which improved the quality of this study.

  1. Funding information: The authors state that there is no funding involved.

  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 declare that there is no conflict of interest.

  4. Data availability statement: All data, generated or used during the study, are available within the article.

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Received: 2023-10-27
Revised: 2024-03-22
Accepted: 2024-04-02
Published Online: 2024-09-03

© 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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