Compact Difference Method for Time-Fractional Neutral Delay Nonlinear Fourth-Order Equation

In this paper, we present a compact finite difference method for a class of fourth-order nonlinear neutral delay sub-diffusion equations in two-dimensional space. The fourth-order problem is first transformed into a second-order system by a reduced-order method. Next by using compact operator to approximate the second order space derivatives and L2-1 σ formula to approximate the time fractional derivative, the difference scheme which is fourth order in space and second order in time is obtained. Then, the existence and uniqueness of solution, the convergence rate of and the stability of the scheme are proved. Finally, numerical results are given to verify the accuracy and validity of the scheme.

• Two-Dimensional Nonlinear Sub-Diffusion Equations
• Neutral Delay
• Compact Difference Scheme
• Convergence
• Stability

1. Introduction

It has been found that in many experiments and researches the diffusion process of many complex systems no longer satisfies Fick’s second law. Such process is called anomalous diffusion, one remarkable feature of which is that the mean square displacement of the particle and the time variable have the following power-law dependence:

〈 x 2 ( t ) 〉 ~ 2 K α Γ ( 1 + α ) t α , t → ∞ , (1.1)

where α is the anomalous diffusion exponent and K_α is the generalized diffusion coefficient. When 0 < α < 1 , we have sub-diffusion, and when 1 < α < 2 , we have super-diffusion. Various table text styles are provided. The formatter will need to create these components, incorporating the applicable criteria that follow. Anomalous diffusion process can be described by differential equations involving fractional calculus, that is, fractional order diffusion equations.

In recent years, the application of fractional diffusion equations in mechanics, physics viscoelastic mechanics [1] [2] [3] [4], porous media [5] [6], hydrology [7] and so on has been an important research topic. There are several different methods to solve the analytical solutions of some fractional diffusion equations, such as Laplace method, Fourier method, Green function method, etc. [8] [9] [10] [11]. But only certain types of fractional differential equation can be solved for exact solutions. Therefore, in most cases we need to rely on numerical methods. In the past few years, a number of different numerical methods for fractional diffusion equations have been developed [12] - [19]. In [20] [21] [22] [23], L1 formula is introduced for discretization of fractional diffusion equations, and the precision in time is O ( τ 2 − α ) . In order to improve the precision in time, Zhao and Sun [24] obtained a second order approximation of time fractional derivative by using Crank-Nicolson method. In [25], Gao and Sun proposed the L1-2 approximation formula. On the basis of L1-2 formula, Alikhanov [26] established the L2-1σ approximation formula, which has the 3 − α order uniform convergence rate.

In many applications, it is necessary to use equations which contain fourth-order derivatives in space, for example, wave propagation of light beam [27], modeling of Plane Grooves [28], the formation of ice [29] [30] and the propagation of intense laser beams through the Quer [31] [32] body. In [33], Agrawa gave the analytic solution for fourth order fractional diffusion-wave equation by means of Laplace and Fourier transform. In [34], Hu and Zhang studied finite difference method for spatial fourth-order fractional diffusion equations, and the convergence order of the scheme is O ( τ 2 − α + h 2 ) . Later, in [35], Guo and Li et al. proposed two numerical schemes for the equations in literature [34], and proved unconditional stability and the convergence order of O ( τ 2 + h 2 ) of the two schemes. In [36], Sun and Ji constructed a difference scheme of fourth order fractional diffusion equations with the first Dirichlet boundary condition, and obtained fourth order spatial accuracy. Then in [37], Zhang et al. constructed a compact finite difference scheme for fourth order fractional diffusion equations with the second Dirichlet boundary condition by using the L2-1σ formula to approximate the time fractional derivative and the compact operator to approximate spatial fourth order derivative.

Delay differential equations are widely used in many fields, such as population ecology, cell biology, control theory, economics and so on [38] [39] [40]. In [38] [39] Sarita Ndal et al. discuss finite difference scheme for one-dimensional time fractional fourth-order diffusion equation, which contains a nonlinear source function with time delay and a fourth-order space delay term. The unique solvability, stability and convergence of the scheme are proved.

In this work, we consider the following fourth-order nonlinear sub-diffusion neutral delayed equation in two-dimensional space.

D 0 c t α u ( x , y , t ) + Δ 2 u ( x , y , t − s ) = f ( x , y , t , u ( x , y , t ) , u ( x , y , t − s ) ) , ( x , y , t ) ∈ Ω × ( 0 , T ) . (1.2)

subject to initial and boundary conditions:

{ u ( x , y , t ) = φ ( x , y , t ) ,     ( x , y , t ) ∈ ∂ Ω × [ 0 , T ] , Δ u ( x , y , t ) = ψ ( x , y , t ) ,     ( x , y , t ) ∈ ∂ Ω × [ 0 , T ] , u ( x , y , t ) = ϕ ( x , y , t ) ,     ( x , y , t ) ∈ Ω × [ − s , 0 ] , (1.3)

where s > 0 is the delay, f ( x , y , t , u ( x , y , t ) , u ( x , y , t − s ) ) represents a nonlinear source term with time delay, Ω = ( 0 , L 1 ) × ( 0 , L 2 ) , α ∈ ( 0 , 1 ) , f , ϕ , ψ , φ are all given and sufficiently smooth functions.

The fractional derivative is defined in Caputo form:

D 0 c t α u ( x , y , t ) ≡ ∂ α u ( x , y , t ) ∂ t α : = 1 Γ ( 1 − α ) ∫ 0 t ( t − ξ ) − α ∂ u ( x , y , ξ ) ∂ ξ d ξ , 0 < α < 1 . (1.4)

Let m be the integer satisfying m s ≤ T ≤ ( m + 1 ) s . Define I r = ( r s , ( r + 1 ) s ) , r = − 1 , 0 , ⋯ , m − 1 , I m = ( m s , T ) , I = U q = − 1 I q .

Assume that the partial derivatives f μ ( x , y , t , μ , ν ) and f ν ( x , y , t , μ , ν ) are continuous in the ϵ 0 -neighborhood of μ and ν for a positive constant ϵ 0 . Define

c 1 = sup 0 < x < L 1 , 0 < y < L 2 , 0 < t < T | ϵ 1 | ≤ ϵ 0 , | ϵ 2 | ≤ ϵ 0 | f μ ( x , y , t , u ( x , y , t ) + ϵ 1 , u ( x , y , t − s ) + ϵ 2 ) | , (1.5)

c 2 = sup 0 < x < L 1 , 0 < y < L 2 , 0 < t < T | ϵ 1 | ≤ ϵ 0 , | ϵ 2 | ≤ ϵ 0 | f ν ( x , y , t , u ( x , y , t ) + ϵ 1 , u ( x , y , t − s ) + ϵ 2 ) | . (1.6)

In this article, we construct a compact difference scheme for the problem (1.2)-(1.3). First, the fourth-order equation is transformed into two second-order equations by introducing v = Δ u as an auxiliary variable. Next, we use the L2-1σ formula and the compact difference operator to approximate temporal Caputo derivative and spatial derivative. Doing in this way, we can obtain the compact difference scheme. Then the existence and uniqueness of the difference scheme is proved. By using mathematical induction and energy method, we prove the convergence and stability of the scheme. Numerical experiments show that the scheme can achieve convergence order O ( τ 2 + h 4 ) in discrete L 2 norm and L ∞ norm, which verify the theoretical results.

2. Notations and Preliminary

Introducing positive integers M 1 , M 2 and n, let h x = L 1 M 1 , h y = L 2 M 2 and τ = s n , be the spatial steps in x and y directions and the temporal step respectively. Define x i = i h x ( 0 ≤ i ≤ M 1 ) , y j = j h y ( 0 ≤ j ≤ M 2 ) , t k = k τ ( − n ≤ k ≤ N ) , where N = [ T t ] . Define Ω ¯ h = { ( x i , y j ) | 0 ≤ i ≤ M 1 , 0 ≤ j ≤ M 2 } , Ω h = Ω ¯ h ∩ Ω , ∂ Ω h = Ω ¯ h ∩ ∂ Ω , ω = { ( i , j ) | ( x i , y j ) ∈ Ω h } , ∂ ω = { ( i , j ) | ( x i , y j ) ∈ ∂ Ω h } , Ω t = { t k | − n ≤ k ≤ N } .

In addition, denote t k − 1 + σ = ( k − 1 + σ ) τ , where σ = α 2 . Define the grid function space S h = { u | u = { u i , j ∈ R ( M 1 + 1 ) × ( M 2 + 1 ) , 0 ≤ i ≤ M 1 , 0 ≤ j ≤ M 2 } } and S h 0 = { u | u ∈ S h , u 0 , j = u M 1 , j , 0 ≤ j ≤ M 2 ; u i , 0 = u i , M 2 , 0 ≤ i ≤ M 1 } on Ω h .

Define the following difference operators for any grid functions u ∈ S h ,

δ x u i − 1 2 , j = 1 h x ( u i , j − u i − 1 , j ) ,     δ y u i , j − 1 2 = 1 h y ( u i , j − u i , j − 1 ) , δ x 2 u i , j = 1 h x 2 ( u i + 1 , j − 2 u i , j + u i − 1 , j ) ,     δ y 2 u i , j = 1 h y 2 ( u i , j + 1 − 2 u i , j + u i , j − 1 ) . (2.1)

Definition 2.1. For u ∈ S h , the compact difference operator is defined as follows

A x u i , j = { ( I + h x 2 12 δ x 2 ) u i , j , 1 ≤ i ≤ M 1 − 1 , 0 ≤ j ≤ M 2 , u i , j , i = 0     or     i = M 1 , 0 ≤ j ≤ M 2 , (2.2)

A y u i , j = { ( I + h y 2 12 δ y 2 ) u i , j , 1 ≤ j ≤ M 2 − 1 , 0 ≤ j ≤ M 1 , u i , j , j = 0   or   j = M 2 , 0 ≤ j ≤ M 1 , (2.3)

Denote

A h = A x A y ,     B h = A y δ x 2 + A x δ y 2 . (2.4)

Then define the following discrete inner products and the corresponding norms for u , v ∈ S h ,

( u , v ) = h x h y ∑ i = 1 M 1 − 1 ∑ j = 1 M 2 − 1 u i , j v i , j ,       ‖ u ‖ 2 = ( u , u ) , (2.5)

( δ x u , δ x v ) x = h x h y ∑ i = 1 M 1 − 1 ∑ j = 1 M 2 − 1 δ x u i − 1 2 , j δ x v i , j ,       ‖ δ x u ‖ x 2 = ( δ x u , δ x u ) x , (2.6)

( δ x 2 u , δ x 2 v ) x = h x h y ∑ i = 1 M 1 − 1 ∑ j = 1 M 2 − 1 δ x 2 u i , j δ x 2 v i , j ,       ‖ δ x 2 u ‖ x 2 = ( δ x 2 u , δ x 2 u ) x , (2.7)

( δ y u , δ y v ) y , ( δ y 2 u , δ y 2 v ) y , ‖ δ y u ‖ y 2 , ‖ δ y 2 u ‖ y 2 can be defined similarly. We also using the following norm

‖ u ‖ ∞ = max 1 ≤ i ≤ M 1 − 1 1 ≤ j ≤ M 2 − 1 | u i , j | . (2.8)

Some related lemmas for difference operators δ x 2 , δ y 2 , A x , A y , A h , B h are given as follows.

Lemma 2.1. [41] Denote ζ ( λ ) = ( 1 − λ ) 3 [ 5 − 3 ( 1 − λ ) 2 ] and let g ( • , y ) ∈ C 6 [ x i − h x , x i + h x ] , g ( x , · ) ∈ C 6 [ y j − h y , y j + h y ] for any function g ( x , y ) . We have

A x g x x ( x i , y j ) = δ x 2 g ( x i , y j ) + h x 4 360 ∫ 0 1 [ g x ( 6 ) ( x i − λ h , y j ) + g x ( 6 ) ( x i + λ h , y j ) ] ζ ( λ ) d λ , (2.9)

A y g y y ( x i , y j ) = δ y 2 g ( x i , y j ) + h y 4 360 ∫ 0 1 [ g y ( 6 ) ( x i , y j − λ h ) + g y ( 6 ) ( x i , y j + λ h ) ] ζ ( λ ) d λ . (2.10)

Lemma 2.2. [42] For any grid functions u , v ∈ S h 0 , we have

( A x u , v ) = ( u , A x v ) , ( A y u , v ) = ( u , A y v ) ; (2.11)

( A h u , v ) = ( u , A h v ) , ( A h u , v ) = ( u , A h v ) ; (2.12)

( δ x 2 u , v ) = ( u , δ x 2 v ) , ( δ y 2 u , v ) = ( u , δ y 2 v ) ; (2.13)

( A h u , B h v ) = ( B h u , A h v ) . (2.14)

Proof. Because the first six equalities have been proven by Q. Li in [42], we only prove the last equality. According to the definition of operator A h , B h , and applying the first and third equalities above, we get

( A h u , B h v ) = ( A x A y u , A y δ x 2 v + A x δ y 2 v ) = ( A x A y u , A y δ x 2 v ) + ( A x A y u , A x δ y 2 v ) = ( A x A y δ x 2 u , A y v ) + ( A x A y δ y 2 u , A x v ) = ( A y δ x 2 u , A x A y v ) + ( A x δ y 2 u , A x A y v ) = ( B h u , A h v ) . (2.15)

The lemma has been proved. ◻

Lemma 2.3. [41] For any grid function u ∈ S h , we have

1 3 ‖ u ‖ 2 ≤ ‖ A x u ‖ 2 ≤ ‖ u ‖ 2 ,     1 3 ‖ u ‖ 2 ≤ ‖ A y u ‖ 2 ≤ ‖ u ‖ 2 , 1 9 ‖ u ‖ 2 ≤ ‖ A h u ‖ 2 ≤ ‖ u ‖ 2 . (2.16)

Lemma 2.4. [42] For any grid function u ∈ S h , we have

2 3 ‖ u ‖ 2 ≤ ( A x u , u ) ≤ ‖ u ‖ 2 ,     2 3 ‖ u ‖ 2 ≤ ( A y u , u ) ≤ ‖ u ‖ 2 , 4 9 ‖ u ‖ 2 ≤ ( A h u , u ) ≤ ‖ u ‖ 2 . (2.17)

According to [26], the L2-1σ approximation formula is defined as

Δ τ α u i , j k − 1 + σ = τ − α Γ ( 2 − α ) [ c 0 ( k ) u i , j k − ∑ m = 1 k − 1 ( c k − m − 1 ( k ) − c k − m ( k ) ) u i , j m − c k − 1 ( k ) u i , j 0 ] . (2.18)

where,

a 0 = σ 1 − α ,   a l = ( l + σ ) 1 − α − ( l − 1 + σ ) 1 − α ,     l ≥ 1 ,

b l = 1 2 − α [ ( l + σ ) 2 − α − ( l − 1 + σ ) 2 − α ] − 1 2 [ ( l + σ ) 1 − α + ( l − 1 + σ ) 1 − α ] ,   l ≥ 1 .(2.19)

when k = 1 , c 0 ( k ) = a 0 , and when k ≥ 2 ,

c m ( k ) = { a 0 + b 1 ,     m = 0 , a m + b m + 1 − b m ,     1 ≤ m ≤ k − 2 , a m − b m ,     m = k − 1. (2.20)

Lemma 2.5. [26] For a ( t ) ∈ C 3 [ 0 , t k ] , we have the following error estimates

| D 0 c t α a ( t ) | t = t k − 1 + σ − Δ τ α a ( t ) | t = t k − 1 + σ | ≤ ( 4 σ − 1 ) σ − α 12 Γ ( 2 − α ) max 0 ≤ t ≤ t k | a ( 3 ) ( t ) | τ 3 − α . (2.21)

Lemma 2.6. [26] Let w = { w k | − n ≤ k ≤ N } be a grid function defined on Ω τ , then

( σ w k + ( 1 − σ ) w k − 1 ) D 0 c t k − 1 + σ α w ≥ 1 2 D 0 c t k − 1 + σ α ( w 2 ) . (2.22)

Lemma 2.7. [26] For α ∈ ( 0 , 1 ) , σ = 1 − α 2 and c m ( k ) ( 0 ≤ m ≤ k − 1 , k ≥ 1 ) defined by (2.4), we have the following inequalities

c m ( k ) > 1 − α 2 ( k + σ ) − α > 0 , (2.23)

c 0 ( k ) > c 1 ( k ) > c 2 ( k ) > ⋯ > c k − 2 ( k ) > c k − 1 ( k ) . (2.24)

3. Construction of the Compact Difference Scheme

In this section, we will construct the compact difference scheme of the problem (1.2)-(1.3). Let v ( x , y , t ) = Δ u ( x , y , t ) . Then the problem (1.2)-(1.3) is equivalent to

D 0 c t α u ( x , y , t ) + Δ v ( x , y , t ) + Δ v ( x , y , t − s ) = f ( x , y , t , u ( x , y , t ) , u ( x , y , t − s ) ) ,     ( x , y , t ) ∈ Ω × [ 0 , T ] , (3.1)

v ( x , y , t ) = Δ u ( x , y , t ) ,     ( x , y , t ) ∈ Ω × [ 0 , T ] , (3.2)

v ( x , y , t − s ) = Δ u ( x , y , t − s ) ,     ( x , y , t ) ∈ Ω × [ 0 , s ] , (3.3)

u ( x , y , t ) = φ ( x , y , t ) ,     ( x , y , t ) ∈ ∂ Ω × [ 0 , T ] , (3.4)

v ( x , y , t ) = ψ ( x , y , t ) ,     ( x , y , t ) ∈ ∂ Ω × [ 0 , T ] , (3.5)

u ( x , y , t ) = ϕ ( x , y , t ) ,     ( x , y , t ) ∈ Ω × [ − s , 0 ] . (3.6)

In addition, we define the following grid functions for the exact solutions u and v

U i , j k = u ( x i , y j , t k ) , V i , j k = v ( x i , y j , t k ) , 0 ≤ i ≤ M 1 , 0 ≤ j ≤ M 2 , − n ≤ k ≤ N . (3.7)

Considering the Equations (3.1)-(3.3) at the grid point ( x i , y j , t k − 1 + σ ) and applying operators A h , we have

A h ∂ α u ( x i , y j , t k − 1 + σ ) ∂ t α + A h ∂ 2 v ( x i , y j , t k − 1 + σ ) ∂ x 2 + A h ∂ 2 v ( x i , y j , t k − 1 + σ ) ∂ y 2   + A h ∂ 2 v ( x i , y j , t k − 1 + σ − n ) ∂ x 2 + A h ∂ 2 v ( x i , y j , t k − 1 + σ − n ) ∂ y 2 = A h f ( x i , y j , t k − 1 + σ , u ( x i , y j , t k − 1 + σ ) , u ( x i , y j , t k − 1 + σ − n ) ) , (3.8)

A h v ( x i , y j , t k − 1 + σ ) = A h ∂ 2 u ( x i , y j , t k − 1 + σ ) ∂ x 2 + A h ∂ 2 u ( x i , y j , t k − 1 + σ ) ∂ y 2 , (3.9)

A h v ( x i , y j , t k − 1 + σ − n ) = A h ∂ 2 u ( x i , y j , t k − 1 + σ − n ) ∂ x 2 + A h ∂ 2 u ( x i , y j , t k − 1 + σ − n ) ∂ y 2 . (3.10)

From Taylor’s series expansion, we can get

U i , j k − 1 + σ = u ( x i , y j , t k − 1 + σ ) = ( σ + 1 ) U i , j k − 1 − σ ( σ + 1 ) U i , j k − 2 + O ( τ 2 ) , (3.11)

U i , j k − 1 + σ − n = u ( x i , y j , t k − 1 + σ − n ) = σ U i , j k − n + ( 1 − σ ) U i , j k − 1 − n + O ( τ 2 ) . (3.12)

Then the nonlinear source term f ( x , y , t , u ( x , y , t ) , u ( x , y , t − s ) ) can be approximated by the following formula

f ( x i , y j , t k − 1 + σ , u ( x i , y j , t k − 1 + σ ) , u ( x i , y j , t k − 1 + σ − n ) ) = f ( x i , y j , t k − 1 + σ , ( σ + 1 ) U i , j k − 1 − σ U i , j k − 2 , σ U i , j k − n + ( 1 − σ ) U i , j k − n − 1 ) + O ( τ 2 ) . (3.13)

By using Lemma 2.1, we have

A h ∂ 2 u ( x i , y j , t k − 1 + σ ) ∂ x 2 = σ A h ∂ 2 u ( x i , y j , t k ) ∂ x 2 + ( 1 − σ ) A h ∂ 2 u ( x i , y j , t k − 1 ) ∂ x 2 + O ( τ 2 + h x 4 ) = A y δ x 2 U i , j k − 1 + σ + O ( τ 2 + h x 4 ) , (3.14)

A h ∂ 2 u ( x i , y j , t k − 1 + σ ) ∂ y 2 = A x δ y 2 U i , j k − 1 + σ + O ( τ 2 + h y 4 ) , (3.15)

A h ∂ 2 v ( x i , y j , t k − 1 + σ ) ∂ x 2 + A h ∂ 2 v ( x i , y j , t k − 1 + σ ) ∂ y 2 = A y δ x 2 V i , j k − 1 + σ + A x δ y 2 V i , j k − 1 + σ + O ( τ 2 + h x 4 + h y 4 ) = B h V i , j k − 1 + σ + O ( τ 2 + h x 4 + h y 4 ) , (3.16)

A h ∂ 2 v ( x i , y j , t k − 1 + σ − n ) ∂ x 2 + A h ∂ 2 v ( x i , y j , t k − 1 + σ − n ) ∂ y 2 = A y δ x 2 V i , j k − 1 + σ − n + A x δ y 2 V i , j k − 1 + σ − n + O ( τ 2 + h x 4 + h y 4 ) = B h V i , j k − 1 + σ − n + O ( τ 2 + h x 4 + h y 4 ) . (3.17)

Substituting Equations (3.13)-(3.17) into (3.8)-(3.10), and approximating the time fractional derivative by L2-1σ formula (2.18), we can get

τ − α Γ ( 2 − α ) [ c 0 ( k ) A h U i , j k − ∑ m = 1 k − 1 ( c k − m − 1 ( k ) − c k − m ( k ) ) A h U i , j m − c k − 1 ( k ) A h U i , j 0 ]   + B h V i , j k − 1 + σ + B h V i , j k − 1 + σ − n = A h f ( x i , y j , t k − 1 + σ , ( σ + 1 ) U i , j k − 1 − σ U i , j k − 2 , σ U i , j k − n + ( 1 − σ ) U i , j k − n − 1 ) + ( R 1 ) i , j k , (3.18)

A h V i , j k − 1 + σ = B h U i , j k − 1 + σ + ( R 2 ) i , j k , (3.19)

A h V i , j k − 1 + σ − n = B h U i , j k − 1 + σ − n + ( R 3 ) i , j k , (3.20)

where

| ( R 1 ) i , j k | + | ( R 2 ) i , j k | + | ( R 3 ) i , j k | ≤ c ^ ( τ 2 + h x 4 + h y 4 ) . (3.21)

where c ^ is a positive constant that does not depend on τ andh.

Omitting R 1 k , R 2 k and R 3 k and in Equations (3.18)-(3.20), and using numerical solution u i , j k , v i , j k to replace the exact solution U i , j k , V i , j k , the following difference scheme for problem (1.2)-(1.3) is obtained

τ − α Γ ( 2 − α ) [ c 0 ( k ) A h u i , j k − ∑ m = 1 k − 1 ( c k − m − 1 ( k ) − c k − m ( k ) ) A h u i , j m − c k − 1 ( k ) A h u i , j 0 ]   + B h v i , j k − 1 + σ + B h v i , j k − 1 + σ − n = A h f ( x i , y j , t k − 1 + σ , ( σ + 1 ) u i , j k − 1 − σ u i , j k − 2 , σ u i , j k − n + ( 1 − σ ) u i , j k − n − 1 ) ,       1 ≤ i ≤ M 1 − 1 ,   1 ≤ j ≤ M 2 − 1 ,   1 ≤ k ≤ N − 1 , (3.22)

1 ≤ i ≤ M 1 − 1 , 1 ≤ j ≤ M 2 − 1 , 1 ≤ k ≤ N − 1 ,

A h v i , j k − 1 + σ = B h u i , j k − 1 + σ , (3.23)

A h v i , j k − 1 + σ − n = B h u i , j k − 1 + σ − n , (3.24)

with the following discrete initial and boundary conditions

u i , j k = φ ( x i , y j , t k ) ,     ( i , j ) ∈ ω ,   − n ≤ k ≤ 0 , (3.25)

v i , j k = ψ ( x i , y j , t k ) ,     ( i , j ) ∈ ∂ ω ,   1 ≤ k ≤ N , (3.26)

u i , j k = ϕ ( x i , y j , t k ) ,     ( i , j ) ∈ ∂ ω ,   1 ≤ k ≤ N . (3.27)

4. Analysis of the Compact Difference Scheme

In this section, we analyze the unique solvability, convergence and stability of the difference scheme (3.22)-(3.27).

4.1. Solvability

Theorem 4.1. (Solvability) The compact difference scheme (3.22)-(3.27) is uniquely solvable.

Proof. Noting that the difference scheme (3.19)-(3.24) is linear, we only need to prove that the homogeneous system has only zero solutions. Therefore we can suppose u l = 0 , − n < l < k − 1 , f = φ = ψ = ϕ = 0 , and consider the following homogeneous problem

A h Δ τ α u i , j k − 1 + σ + B h v i , j k − 1 + σ + B h v i , j k − 1 + σ − n = 0 , (4.1)

A h v i , j k − 1 + σ = B h u i , j k − 1 + σ , (4.2)

v i , j k − 1 + σ − n = B h u i , j k − 1 + σ − n . (4.3)

Taking the inner product ( ⋅ , ⋅ ) with u k − 1 + σ and v k − 1 + σ on both sides of (4.1) and (4.2) respectively, we have

( Δ τ α B h v k − 1 + σ , u k − 1 + σ ) + ( B h v k − 1 + σ , u k − 1 + σ ) + ( B h v k − 1 + σ − n , u k − 1 + σ ) = 0 , (4.4)

( A h v k − 1 + σ , v k − 1 + σ ) = ( B h u k − 1 + σ , v k − 1 + σ ) . (4.5)

And then by Lemma 2.2 and (4.5), we have

( B h v k − 1 + σ , u k − 1 + σ ) = ( B h u k − 1 + σ , v k − 1 + σ ) = ( A h v k − 1 + σ , v k − 1 + σ ) . (4.6)

Substituting (4.6) into (4.4), we have

( Δ τ α A h u k − 1 + σ , u k − 1 + σ ) + ( A h v k − 1 + σ , v k − 1 + σ ) + ( B h v k − 1 + σ − n , u k − 1 + σ ) = 0. (4.7)

Using (2.18) and noticing u 0 = u 1 = ⋯ = u k − 1 = 0 , we have

( Δ τ α A h u k − 1 + σ , u k − 1 + σ ) = σ τ − α Γ ( 2 − α ) c 0 ( k ) ( A h u k , u k ) , (4.8)

( A h v k − 1 + σ , v k − 1 + σ ) = σ ( A h v k , v k ) , (4.9)

( B h v k − 1 + σ − n , u k − 1 + σ ) = 0. (4.10)

Substituting (4.8)-(4.10) into (4.7) and using Lemma 2.4, we have

4 σ τ − α 9 Γ ( 2 − α ) c 0 ( k ) ‖ u k ‖ 2 + 4 σ 9 ‖ v k ‖ 2 ≤ σ τ − α Γ ( 2 − α ) c 0 ( k ) ( A h u k , u k ) + σ ( A h v k , v k ) = 0. (4.11)

From Lemma 2.4 and noticing c 0 ( k ) > 0 , we get ‖ u k ‖ = ‖ v k ‖ = 0 , which immediately gives u k = 0 , v k = 0 .

The proof is completed. ◻

4.2. Convergence

Lemma 4.2. [43] Let { z k } , { g k } be two nonnegative sequences. If

z k ≤ K + ∑ i = 0 k g i z i ,   k ≥ 0 , (4.12)

it holds that

z k ≤ K ⋅ exp ( ∑ i = 0 k g i ) , k ≥ 0. (4.13)

where K is a nonnegative constant.

Theorem 4.2. (Convergence) Let { ( U i , j k , V i , j k ) | 0 ≤ i ≤ M 1 , 0 ≤ j ≤ M 2 , − n ≤ k ≤ N } be the solution of problem (3.1)-(3.6) and { ( u i , j k , v i , j k ) | 0 ≤ i ≤ M 1 , 0 ≤ j ≤ M 2 , − n ≤ k ≤ N } be the solution of the difference scheme (3.22)-(3.27) have

max 0 ≤ k τ ≤ T { ‖ e k ‖ 2 + ‖ e ^ k − 1 + σ ‖ 2 } ≤ C ( τ 2 + h x 4 + h y 4 ) 2 . (4.14)

where e i , j k = u i , j k − U i , j k , e ^ i , j k = v i , j k − V i , j k .

Proof. Subtracting Equations (3.18)-(3.20) from (3.22)-(3.27), we can get the following error equations

A h Δ τ α e i , j k − 1 + σ + B h e ^ i , j k − 1 + σ + B h e ^ i , j k − 1 + σ − n = A h F ˜ i , j k − 1 + σ + ( R 1 ) i , j k ,   1 ≤ i ≤ M 1 − 1 ,   1 ≤ j ≤ M 2 − 1 ,   1 ≤ k ≤ N − 1 , (4.15)

A h e ^ i , j k − 1 + σ = B h e i , j k − 1 + σ + ( R 2 ) i , j k , (4.16)

A h e ^ i , j k − 1 + σ − n = B h e i , j k − 1 + σ − n + ( R 3 ) i , j k , (4.17)

e i , j k = 0 ,   0 ≤ i ≤ M 1 ,   0 ≤ j ≤ M 2 ,   − n ≤ k ≤ 0 , (4.18)

e i , j k = 0 ,   e ^ i , j k = 0 ,   i = 0   or   M 1 ,   j = 0   or   M 2 , 1 ≤ k ≤ N , (4.19)

F ˜ i , j k − 1 + σ = f ( x i , y j , t k − 1 + σ , ( σ + 1 ) u i , j k − 1 − σ u i , j k − 2 , σ u i , j k − n + ( 1 − σ ) u i , j k − n − 1 )     − f ( x i , y j , t k − 1 + σ , ( σ + 1 ) U i , j k − 1 − σ U i , j k − 2 , σ U i , j k − n + ( 1 − σ ) U i , j k − n − 1 ) . (4.20)

Taking the inner product ( ⋅ , ⋅ ) with A h e k − 1 + σ on both sides of (4.15), and with A h e ^ k − 1 + σ , A h e ^ k − 1 + σ − n on both sides of (4.16) and (4.17) respectively, we have

( A h Δ τ α e k − 1 + σ , A h e k − 1 + σ ) + ( B h e ^ k − 1 + σ , A h e k − 1 + σ ) + ( B h e ^ k − 1 + σ − n , A h e k − 1 + σ ) = ( A h F ˜ k − 1 + σ , A h e k − 1 + σ ) + ( R 1 k , A h e k − 1 + σ ) , (4.21)

( A h e ^ k − 1 + σ , A h e ^ k − 1 + σ ) = ( B h e k − 1 + σ , A h e ^ k − 1 + σ ) + ( R 2 k , A h e ^ k − 1 + σ ) , (4.22)

( A h e ^ k − 1 + σ − n , A h e ^ k − 1 + σ − n ) = ( B h e k − 1 + σ − n , A h e ^ k − 1 + σ − n ) + ( R 3 k , A h e ^ k − 1 + σ − n ) . (4.23)

Using Lemma 2.4 we get

( B h e ^ k − 1 + σ , A h e k − 1 + σ ) = ( A h e ^ k − 1 + σ , A h e ^ k − 1 + σ ) − ( R 2 k , A h e ^ k − 1 + σ ) (4.24)

and

( B h e ^ k − 1 + σ − n , A h e k − 1 + σ ) = ( A h e ^ k − 1 + σ − n , A h e ^ k − 1 + σ − n ) − ( R 3 k , A h e ^ k − 1 + σ − n ) . (4.25)

Substituting (4.28)-(4.29) into (4.21), we have

( Δ τ α A h e k − 1 + σ , A h e k − 1 + σ ) + ( A h e ^ k − 1 + σ , A h e ^ k − 1 + σ ) + ( A h e ^ k − 1 + σ − n , A h e ^ k − 1 + σ − n ) = ( A h F ˜ k − 1 + σ , A h e k − 1 + σ ) + ( R 1 k , A h e k − 1 + σ ) + ( R 2 k , A h e ^ k − 1 + σ ) + ( R 3 k , A h e ^ k − 1 + σ − n ) . (4.26)

Using Lemma 2.6, we get

( Δ τ α A h e k − 1 + σ , A h e k − 1 + σ ) ≥ 1 2 Δ τ α ‖ A h e k − 1 + σ ‖ 2 . (4.27)

Then substituting the above inequality into (4.26), we can get the following inequality

1 2 Δ τ α ‖ A h e k − 1 + σ ‖ 2 + ‖ A h e ^ k − 1 + σ ‖ 2 + ‖ Δ A h e ^ k − 1 + σ − n ‖ 2 ≤ ( A h F ˜ k − 1 + σ , A h e k − 1 + σ ) + ( R 1 k , A h e k − 1 + σ ) + ( R 2 k , A h e ^ k − 1 + σ ) + ( R 3 k , A h e ^ k − 1 + σ − n ) . (4.28)

Applying Cauchy inequality to terms on the right-hand side of (4.28), we can get

( A h F ˜ k − 1 + σ , A h e k − 1 + σ ) ≤ 1 2 ‖ A h F ˜ k − 1 + σ ‖ 2 + 1 2 ‖ A h e k − 1 + σ ‖ 2 , (4.29)

( R 1 , A h e k − 1 + σ ) ≤ 1 2 ‖ R 1 k ‖ 2 + 1 2 ‖ A h e k − 1 + σ ‖ 2 , (4.30)

( R 2 , A h e ^ k − 1 + σ ) ≤ 1 2 ‖ R 2 k ‖ 2 + 1 2 ‖ A h e ^ k − 1 + σ ‖ 2 , (4.31)

( R 3 , A h e ^ k − 1 + σ − n ) ≤ 1 2 ‖ R 3 k ‖ 2 + 1 2 ‖ A h e ^ k − 1 + σ − n ‖ 2 . (4.32)

Substituting (4.29)-(4.32) into (4.28), we have

1 2 Δ τ α ‖ A h e k − 1 + σ ‖ 2 + 1 2 ‖ A h e ^ k − 1 + σ ‖ 2 ≤ 1 2 ‖ A h F ˜ k − 1 + σ ‖ 2 + ‖ A h e k − 1 + σ ‖ 2 + 1 2 ‖ R 1 k ‖ 2 + 1 2 ‖ R 2 k ‖ 2 + 1 2 ‖ R 3 k ‖ 2 . (4.33)

Noticing that ‖ e k ‖ ∞ = 0 , − n ≤ k ≤ 0 , we have ‖ e k ‖ ∞ ≤ ϵ 0 2 , − n ≤ k ≤ 0 . For 0 ≤ k ≤ l , 0 ≤ l ≤ N − 1 , let

‖ e k ‖ ∞ ≤ ϵ 0 2 . (4.34)

Then we will prove that (4.34) is also true for k = l + 1 .

According to (4.34), we have

| F ˜ i , j k − 1 + σ | = | f ( x i , y j , t k − 1 + σ , ( σ + 1 ) u i , j k − 1 − σ u i , j k − 2 , σ u i , j k − n + ( 1 − σ ) u i , j k − n − 1 )     − f ( x i , y j , t k − 1 + σ , ( σ + 1 ) U i , j k − 1 − σ U i , j k − 2 , σ U i , j k − n + ( 1 − σ ) U i , j k − n − 1 ) | ≤ c 1 | ( σ + 1 ) e i , j k − 1 − σ e i , j k − 2 | + c 2 | σ e i , j k − n + ( 1 − σ ) e i , j k − n − 1 | ≤ c 1 ( σ + 1 ) | e i , j k − 1 | + c 1 σ | e i , j k − 2 | + c 2 σ | e i , j k − n | + c 2 ( 1 − σ ) | e i , j k − n − 1 | . (4.35)

Using Lemma 2.3 and inequality (4.35), we can obtain that

‖ A h F ˜ k − 1 + σ ‖ 2 ≤ 2 c 1 2 ( σ + 1 ) 2 ‖ e k − 1 ‖ 2 + 2 c 1 2 σ 2 ‖ e k − 2 ‖ 2     + 2 c 2 2 σ 2 ‖ e k − n ‖ 2 + 2 c 2 2 ( 1 − σ ) 2 ‖ e k − n − 1 ‖ 2 , (4.36)

and

‖ A h e k − 1 + σ ‖ 2 ≤ ‖ e k − 1 + σ ‖ 2 ≤ 2 ( 1 + σ ) 2 ‖ e k − 1 ‖ 2 + 2 σ 2 ‖ e k − 2 ‖ 2 . (4.37)

Using (2.18) and substituting (4.36)-(4.37) into (4.33), we get

c 0 ( k ) ‖ A h e k ‖ 2 + μ ‖ A h e ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ A h e m ‖ 2 + c k − 1 ( k ) ‖ A h e 0 ‖ 2 + 2 μ ( c 1 2 + 2 ) ( σ + 1 ) 2 ‖ e k − 1 ‖ 2     + 2 μ ( c 1 2 + 2 ) σ 2 ‖ e k − 2 ‖ 2 + 2 μ c 2 2 σ 2 ‖ e k − n ‖ 2 + 2 μ c 2 2 ( 1 − σ ) 2 ‖ e k − n − 1 ‖ 2     + μ ‖ R 1 k ‖ 2 + μ ‖ R 2 k ‖ 2 + μ ‖ R 3 k ‖ 2 , (4.38)

where

μ = τ α Γ ( 2 − α ) = T α Γ ( 1 − α ) ( 1 − α ) N − α < T α Γ ( 1 − α ) ( 1 − α ) ( k − α 2 ) − α < 2 c k − 1 k T α Γ ( 1 − α ) . (4.39)

Substituting (4.39) into (4.38), we get

c 0 ( k ) ‖ A h e k ‖ 2 + μ ‖ A h e ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ A h e m ‖ 2 + c k − 1 ( k ) [ ‖ A h e 0 ‖ 2

      + 2 T α Γ ( 1 − α ) ( 2 ( c 1 2 + 2 ) ( σ + 1 ) 2 ‖ e k − 1 ‖ 2 + 2 ( c 1 2 + 2 ) σ 2 ‖ e k − 2 ‖ 2       + 2 c 2 2 σ 2 ‖ e k − n ‖ 2 + 2 c 2 2 ( 1 − σ ) 2 ‖ e k − n − 1 ‖ 2 + ‖ R 1 k ‖ 2 + ‖ R 2 k ‖ 2 + ‖ R 3 k ‖ 2 ) ] . (4.40)

Using Lemma 2.3 and Lemma 2.7, we can get the following inequality

1 9 c 0 ( k ) ‖ e k ‖ 2 + μ 9 ‖ e ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ e m ‖ 2 + c 0 ( k ) [ 2 T α Γ ( 1 − α ) ( 2 ( c 1 2 + 2 ) ( σ + 1 ) 2 ‖ e k − 1 ‖ 2     + 2 ( c 1 2 + 2 ) σ 2 ‖ e k − 2 ‖ 2 + 2 c 2 2 σ 2 ‖ e k − n ‖ 2 + 2 c 2 2 ( 1 − σ ) 2 ‖ e k − n − 1 ‖ 2     + ‖ R 1 k ‖ 2 + ‖ R 2 k ‖ 2 + ‖ R 3 k ‖ 2 ] (4.41)

Dividing (4.41) by 1 9 c 0 ( k ) on both sides, we have

‖ e k ‖ 2 + μ c 0 ( k ) ‖ e ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 9 c 0 ( k ) ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ e m ‖ 2 + 9 [ 2 T α Γ ( 1 − α ) ( 2 ( c 1 2 + 2 ) ( σ + 1 ) 2 ‖ e k − 1 ‖ 2     + 2 ( c 1 2 + 2 ) σ 2 ‖ e k − 2 ‖ 2 + 2 c 2 2 σ 2 ‖ e k − n ‖ 2 + 2 c 2 2 ( 1 − σ ) 2 ‖ e k − n − 1 ‖ 2     + ‖ R 1 k ‖ 2 + ‖ R 2 k ‖ 2 + ‖ R 3 k ‖ 2 ] . (4.42)

Then, letting

C 1 = 36 T α Γ ( 1 − α ) max { ( c 1 2 + 2 ) ( σ + 1 ) 2 , ( c 1 2 + 2 ) σ 2 , c 2 2 σ 2 , c 2 2 ( 1 − σ ) 2 , c ^ 2 / 2 } ,(4.43)

we can get the following inequality

‖ e k ‖ 2 + μ c 0 ( k ) ‖ e ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 9 c 0 ( k ) ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ e m ‖ 2 + C 1 [ ‖ e k − 1 ‖ 2 + ‖ e k − 2 ‖ 2       + ‖ e k − n ‖ 2 + ‖ e k − n − 1 ‖ 2 + ( τ 2 + h x 4 + h y 4 ) 2 ] . (4.44)

From (2.18), we have c 0 ( k ) ≤ a 0 + b 1 . So inequality (4.44) can be arranged as follows

‖ e k ‖ 2 + μ a 0 + b 1 ‖ e ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 9 c 0 ( k ) ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ e m ‖ 2 + C 1 [ ‖ e k − 1 ‖ 2 + ‖ e k − 2 ‖ 2       + ‖ e k − n ‖ 2 + ‖ e k − n − 1 ‖ 2 + ( τ 2 + h x 4 + h y 4 ) 2 ] . (4.45)

Using Lemma 2.7, we know that 9 c 0 ( k ) ( c k − m − 1 ( k ) − c k − m ( k ) ) is a nonnegative sequence. Then applying Lemma 4.2, we get

‖ e k ‖ 2 + μ a 0 + b 1 ‖ e ^ k − 1 + σ ‖ 2 ≤ C 1 ( τ 2 + h x 4 + h y 4 ) 2 exp [ 4 C 1 + ∑ m = 1 k − 1 9 c 0 ( k ) ( c k − m − 1 ( k ) − c k − m ( k ) ) ] = C 1 exp [ 4 C 1 + 9 ( 1 − c k − 1 ( k ) c 0 ( k ) ) ] ⋅ ( τ 2 + h x 4 + h y 4 ) 2 ≤ C 2 2 ( τ 2 + h x 4 + h y 4 ) 2 , (4.46)

where C 2 = C 1 exp ( 4 C 1 + 9 ) . Letting C 3 = min { 1 , μ a 0 + b 1 } and C = C 2 2 C 3 , we get

‖ e k ‖ 2 + ‖ e ^ k − 1 + σ ‖ 2 ≤ C ( τ 2 + h x 4 + h y 4 ) 2 , 1 ≤ k ≤ l + 1. (4.47)

Let h = max { h x , h y } and assume that τ = O ( h 2 ) . Then for sufficiently smell h, we obtain

‖ e l + 1 ‖ ∞ ≤ c h − 1 ‖ e l + 1 ‖ ≤ c h − 1 ( τ 2 + h 4 ) = O ( h 3 ) ≤ ϵ 0 2 , (4.48)

where c = C 1 2 . Thus we know that (4.34) holds for k = l + 1 . According to mathematical induction, then (4.34) holds for all 1 ≤ k ≤ N and the theorem is proved. ◻

4.3. Stability

Let { p i , j k | 0 ≤ i ≤ M 1 , 0 ≤ j ≤ M 2 , − n ≤ k ≤ N } and { q i , j k | 0 ≤ i ≤ M 1 , 0 ≤ j ≤ M 2 , − n ≤ k ≤ N } are the solution of the following system

A h Δ τ α p i , j k − 1 + σ + B h q i , j k − 1 + σ + B h q i , j k − 1 + σ − n = A h f ( x i , y j , t k − 1 + σ , ( σ + 1 ) p i , j k − 1 − σ p i , j k − 2 , σ p i , j k − n + ( 1 − σ ) p i , j k − n − 1 ) , (4.49)

A h q i , j k − 1 + σ = B h p i , j k − 1 + σ , (4.50)

A h q i , j k − 1 + σ − n = B h p i , j k − 1 + σ − n ,   1 ≤ i ≤ M 1 − 1 ,   1 ≤ j ≤ M 2 − 1 ,   1 ≤ k ≤ N − 1 , (4.51)

p i , j k = φ ( x i , y j , t k ) + ρ i , j k ,   ( i , j ) ∈ ω ,   − n ≤ k ≤ 0 , (4.52)

q i , j k = ψ ( x i , y j , t k ) ,   ( i , j ) ∈ ∂ ω ,   1 ≤ k ≤ N , (4.53)

p i , j k = ϕ ( x i , y j , t k ) ,   ( i , j ) ∈ ∂ ω ,   1 ≤ k ≤ N , (4.54)

where ρ i , j k is the perturbation of φ ( x i , y j , t k ) .

Let θ i , j k = u i , j k − p i , j k , θ ^ i , j k = v i , j k − q i , j k , 0 ≤ i ≤ M 1 ,   0 ≤ j ≤ M 2 ,   − n ≤ k ≤ N .

Theorem 4.3. (Stability) There exists a positive integer c 5 , such that

max 0 ≤ k τ ≤ T { ‖ θ k ‖ 2 + ‖ θ ^ k − 1 + σ ‖ 2 } ≤ c 5 max − n ≤ k ≤ 0 ‖ ρ k ‖ 2 . (4.55)

Proof. We subtract (3.22)-(3.27) from (4.49)-(4.54), we have

A h Δ τ α θ i , j k − 1 + σ + B h θ ^ i , j k − 1 + σ + B h θ ^ i , j k − 1 + σ − n = A h F ^ i , j k − 1 + σ , (4.56)

A h θ ^ i , j k − 1 + σ = B h θ i , j k − 1 + σ , (4.57)

A h θ ^ i , j k − 1 + σ − n = B h θ i , j k − 1 + σ − n , (4.58)

θ i , j k = ρ i , j k ,   0 ≤ i ≤ M 1 ,   0 ≤ j ≤ M 2 ,   − n ≤ k ≤ 0 . (4.59)

where

F ^ i , j k − 1 + σ = f ( x i , y j , t k − 1 + σ , ( σ + 1 ) u i , j k − 1 − σ u i , j k − 2 , σ u i , j k − n + ( 1 − σ ) u i , j k − n − 1 )     − f ( x i , y j , t k − 1 + σ , ( σ + 1 ) p i , j k − 1 − σ p i , j k − 2 , σ p i , j k − n + ( 1 − σ ) p i , j k − n − 1 ) . (4.60)

Taking the inner product ( ⋅ , ⋅ ) with A h θ k − 1 + σ on both side of (4.56), and with A h θ ^ k − 1 + σ on both side of (4.57)-(4.58) respectively, we get

( A h Δ τ α θ k − 1 + σ , A h θ k − 1 + σ ) + ( B h θ ^ k − 1 + σ , A h θ k − 1 + σ ) + ( A h θ ^ k − 1 + σ − n , A h θ k − 1 + σ ) = ( A h F ^ k − 1 + σ , A h θ k − 1 + σ ) , (4.61)

( A h θ ^ k − 1 + σ , A h θ ^ k − 1 + σ ) = ( B h θ k − 1 + σ , A h θ ^ k − 1 + σ ) , (4.62)

( A h θ ^ k − 1 + σ − n , A h θ ^ k − 1 + σ ) = ( B h θ k − 1 + σ − n , A h θ ^ k − 1 + σ ) . (4.63)

Similar to (4.24)-(4.39) for (4.61)-(4.63), we obtain:

c 0 ( k ) ‖ A h θ k ‖ 2 + μ ‖ A h θ ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ A h θ m ‖ 2     + c k − 1 ( k ) [ ‖ A h θ 0 ‖ 2 + 2 T α Γ ( 1 − α ) ( c 1 2 + 2 ) ( σ + 1 ) 2 ‖ θ k − 1 ‖ 2     + ( c 1 2 + 2 ) σ 2 ‖ θ k − 2 ‖ 2 + c 2 2 σ 2 ‖ θ k − n ‖ 2 + c 2 2 ( 1 − σ ) 2 ‖ θ k − n − 1 ‖ 2 ] . (4.64)

By equation (4.59), we can get

‖ θ 0 ‖ 2 ≤ max − n ≤ k ≤ 0 ‖ ρ k ‖ 2 . (4.65)

Applying Lemma 2.3 and Lemma 2.7, then dividing both sides by 1 9 c 0 ( k ) , we get the following inequality

‖ θ k ‖ 2 + μ c 0 ( k ) ‖ θ ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 9 c 0 ( k ) ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ θ m ‖ 2 + 9 [ max − n ≤ k ≤ 0 ‖ ρ k ‖ 2       + 2 T α Γ ( 1 − α ) ( ( c 1 2 + 2 ) ( σ + 1 ) 2 ‖ θ k − 1 ‖ 2 + ( c 1 2 + 2 ) σ 2 ‖ θ k − 2 ‖ 2       + c 2 2 σ 2 ‖ θ k − n ‖ 2 + c 2 2 ( 1 − σ ) 2 ‖ θ k − n − 1 ‖ 2 ) ] . (4.66)

According to the inequality c 0 ( k ) ≤ a 0 + b 1 . Therefore, inequality (4.66) can be tidied up

‖ θ k ‖ 2 + μ a 0 + b 1 ‖ θ ^ k − 1 + σ ‖ 2 ≤ ∑ m = 1 k − 1 9 c 0 ( k ) ( c k − m − 1 ( k ) − c k − m ( k ) ) ‖ θ m ‖ 2         + c 3 ( max − n ≤ k ≤ 0 ‖ ρ k ‖ 2 + ‖ θ k − 1 ‖ 2 + ‖ θ k − 2 ‖ 2 + ‖ θ k − n ‖ 2 + ‖ θ k − n − 1 ‖ 2 ) . (4.67)

Letting

c 3 = 18 max { T α Γ ( 1 − α ) ( c 1 2 + 2 ) ( σ + 1 ) 2 , T α Γ ( 1 − α ) ( c 1 2 + 2 ) σ 2 ,                               T α Γ ( 1 − α ) c 2 2 σ 2 , T α Γ ( 1 − α ) c 2 2 ( 1 − σ ) 2 , 1 2 } . (4.68)

Applying the Lemma 4.1, we can get

‖ θ k ‖ 2 + μ a 0 + b 1 ‖ θ ^ k − 1 + σ ‖ 2 ≤ c 3 max − n ≤ k ≤ 0 ‖ ρ k ‖ 2 exp [ 4 c 3 + ∑ m = 1 k − 1 9 c 0 ( k ) ( c k − m − 1 ( k ) − c k − m ( k ) ) ] = c 3 exp [ 4 c 3 + 9 ( 1 − c k − 1 ( k ) c 0 ( k ) ) ] ⋅ max − n ≤ k ≤ 0 ‖ ρ k ‖ 2 ≤ c 4 2 max − n ≤ k ≤ 0 ‖ ρ k ‖ 2 , (4.69)

where c 4 = c 3 exp ( 4 c 3 + 9 ) , according to C 3 = min { 1 , μ a 0 + b 1 } and let c 5 = c 4 2 C 3 , so we have

‖ θ k ‖ 2 + ‖ θ ^ k − 1 + σ ‖ 2 ≤ c 5 max − n ≤ k ≤ 0 ‖ ρ k ‖ 2 . (4.70)

◻

5. Numerical Experiments

In this section, we verify the validity and accuracy of the compact difference scheme by two numerical examples. We choose Ω = ( 0 , 1 ) 2 and T = 1 for all examples. The error between exact solution and numerical solution in discrete L ∞ norm and discrete L 2 norm are given as follows

E ∞ ( h , τ ) = ‖ u N − U N ‖ ∞ ,   E L 2 ( h , τ ) = ‖ u N − U N ‖ . (5.1)

The convergence orders are given

O r d e r ( h ) = log 2 ( E L 2 ( 2 h , τ ) E L 2 ( h , τ ) ) ,   O r d e r ( τ ) = log 2 ( E L 2 ( h , 2 τ ) E L 2 ( h , τ ) ) . (5.2)

Example 1. In this example, we choose

f ( x , y , t , u ( x , y , t ) , u ( x , y , t − s ) ) = u 2 ( x , y , t ) + u ( x , y , t − s ) + G ( x , y , t ) , (5.3)

where

G ( x , y , t ) = 1 Γ ( 5 ) Γ ( α + 5 ) t 4 sin ( π x ) sin ( π y ) + 4 π 4 t 4 + α sin ( π x ) sin ( π y )     + 4 π 4 ( t − 0.2 ) 4 + α sin ( π x ) sin ( π y ) − ( t 4 + α sin ( π x ) sin ( π y ) ) 2     − ( t − 0.2 ) 4 + α sin ( π x ) sin ( π y ) , (5.4)

ϕ ( x , y , t ) = t 4 + α sin ( π x ) sin ( π y ) ,   0 ≤ x ≤ 1 ,   0 ≤ y ≤ 1 ,   t ∈ [ − 0.2 , 0 ] , (5.5)

φ ( x , y , t ) = ψ ( x , y , t ) = 0 ,     ( x , y , t ) ∈ ∂ Ω × [ 0 , 1 ] . (5.6)

The exact solution of the problem is u ( x , y , t ) = t 4 + α sin ( π x ) sin ( π y ) .

We use the compact difference scheme (3.22)-(3.24) to solve the above problem. The errors and convergence order in discrete L 2 norm and L ∞ norm are given in Table 1 and Table 2. Obviously, the spatial accuracy of order O ( h 4 ) is consistent with our theoretical results. In Table 3 and Table 4, the numerical

Table 1

The computational error and convergence order in spatial dimension for U at T = 1 for Example 1 using present scheme

α	h	‖ u − U ‖ ∞	Order ≈	‖ u − U ‖	Order ≈
0.3 0.6 0.9	1/20 1/30 1/40 1/50 1/20 1/30 1/40 1/50 1/20 1/30 1/40 1/50	7.7704e−07 1.5497e−07 4.9198e−08 2.0183e−08 5.9755e−06 1.1819e−06 3.7415e−07 1.5328e−07 9.8734e−06 1.9511e−06 6.1742e−07 2.5291e−07	— 3.9763 3.9884 3.9930 — 3.9968 3.9982 3.9992 — 3.9990 3.9996 3.9997	3.8685e−07 7.7155e−08 2.4494e−08 1.0048e−08 2.9861e−06 5.9065e−07 1.8697e−07 7.6599e−08 4.9352e−06 9.7526e−07 3.0862e−07 1.2641e−07	— 3.9762 3.9884 3.9932 — 3.9966 3.9984 3.9991 — 3.9990 3.9995 4.0000

Table 2

The computational error and convergence order in spatial dimension for V at T = 1 for Example 1 using present scheme

α	h	‖ v − V ‖ ∞	Order ≈	‖ v − V ‖	Order ≈
0.3 0.6 0.9	1/20 1/30 1/40 1/50 1/20 1/30 1/40 1/50 1/20 1/30 1/40 1/50	6.5721e−05 1.3006e−05 4.1178e−06 1.6872e−06 1.6833e−04 3.3277e−05 1.0532e−05 4.3143e−06 2.4525e−04 4.8456e−05 1.5333e−05 6.2803e−06	— 3.9954 3.9978 3.9985 — 3.9980 3.9990 3.9996 — 3.9994 3.9997 4.0001	3.8685e−07 7.7155e−08 2.4494e−08 1.0048e−08 2.9861e−06 5.9065e−07 1.8697e−07 7.6599e−08 4.9352e−06 9.7526e−07 3.0862e−07 1.2641e−07	— 3.9954 3.9977 3.9988 — 3.9981 3.9990 3.9995 — 3.9994 3.9999 4.0000

Table 3

The computational error and convergence order in spatial dimension for U at T = 1 for Example 1 using central difference scheme

α	h	‖ u − U ‖ ∞	Order ≈	‖ u − U ‖	Order ≈
0.3 0.6 0.9	1/20 1/30 1/40 1/50 1/20 1/30 1/40 1/50 1/20 1/30 1/40 1/50	4.1206e−03 1.8305e−03 1.0295e−03 6.5883e−04 4.1082e−03 1.8263e−03 1.0274e−03 6.5755e−04 4.0921e−03 1.8201e−03 1.0241e−03 6.5550e−04	— 2.0012 2.0005 2.0003 — 1.9994 1.9996 1.9999 — 1.9981 1.9990 1.9995	2.0602e−03 9.1521e−04 5.1427e−04 3.2940e−04 2.0540e−03 9.1310e−04 5.1366e−04 3.2876e−04 2.0459e−03 9.1001e−04 5.1202e−04 3.2773e−04	— 2.0012 2.0036 1.9964 — 1.9994 1.9997 1.9998 — 1.9980 1.9991 1.9995

Table 4

The computational error and convergence order in spatial dimension for V at T = 1 for Example 1 using central difference scheme

α	h	‖ u − U ‖ ∞	Order ≈	‖ u − U ‖	Order ≈
0.3 0.6 0.9	1/20 1/30 1/40 1/50 1/20 1/30 1/40 1/50 1/20 1/30 1/40 1/50	4.0633e−02 1.8075e−02 1.0170e−02 6.5100e−03 4.0389e−02 1.7992e−02 1.0129e−02 6.4848e−03 4.0071e−02 1.7870e−02 1.0064e−02 6.4445e−03	— 1.9978 1.9990 1.9992 — 1.9943 1.9971 1.9985 — 1.9916 1.9958 1.9975	2.0306e−02 9.0329e−03 5.0826e−03 3.2533e−03 2.0184e−02 8.9914e−03 5.0617e−03 3.2407e−03 2.0025e−02 8.9304e−03 5.0293e−03 3.2205e−03	— 1.9978 1.9989 1.9994 — 1.9943 1.9972 1.9983 — 1.9916 1.9959 1.9976

results of the central difference scheme are compared with those of the theoretical scheme. From the following Tables 1-4, it is easy to find that the compact difference scheme can achieve higher accuracy than the central difference scheme. (Figure 1)

Example 2. In this example, we choose

f ( x , y , t , u ( x , y , t ) , u ( x , y , t − s ) ) = 4 u ( x , y , t ) + u 2 ( x , y , t ) + u ( x , y , t − s ) + G ( x , y , t ) . (5.7)

Where

G ( x , y , t ) = ( Γ ( α + 4 ) Γ ( 4 ) − t 3 + 2 α exp ( x + y ) ) t 3 exp ( x + y )     + 3 ( t − 0.2 ) 3 + α exp ( x + y ) , (5.8)

u ( x , y , t ) = t 3 + α exp ( x + y ) ,   0 ≤ x ≤ 1 ,   0 ≤ y ≤ 1 ,   t ∈ [ − 0.2 , 0 ] , (5.9)

u ( 0 , y , t ) = t 3 + α exp ( y ) ,     u ( 1 , y , t ) = t 3 + α exp ( 1 + y ) , (5.10)

u ( x , 0 , t ) = t 3 + α exp ( x ) ,     u ( x , 1 , t ) = t 3 + α exp ( 1 + x ) ,   0 ≤ t ≤ 1. (5.11)

The exact solution of the problem is u ( x , y , t ) = t 3 + α exp ( x + y ) .

In this example, we use the formula (3.13) to calculate the nonlinear source term. The error of the L 2 and L ∞ norms for α = 0.3 , 0.6 , 0.9 and convergence orders in the spatial directions are listed in Table 5 and Table 6, from which can easy to see that the convergence order in space of the compact difference scheme (3.22)-(3.24) we proposed reaches the fourth-order accuracy, which is in agreement with our theoretical results. In Table 7 and Table 8, we compare the numerical results of the central difference scheme with those of the theoretical scheme in the spatial direction. From these tables, we can see that the accuracy of the central difference scheme in the spatial direction is O ( h 2 ) which is less accurate than the compact difference scheme. (Figure 2)

Table 5

The computational error and convergence order in spatial dimension for U at T = 1 for Example 2 using present scheme

α	h	‖ u − U ‖ ∞	Order ≈	‖ u − U ‖	Order ≈
0.3 0.6 0.9	1/5 1/10 1/20 1/40 1/5 1/10 1/20 1/40 1/5 1/10 1/20 1/40	2.5267e−04 1.7291e−05 1.0927e−06 6.8846e−08 2.5690e−04 1.7444e−05 1.0993e−06 6.9182e−08 2.5159e−04 1.6961e−05 1.0660e−06 6.6948e−08	— 3.8692 3.9841 3.9884 — 3.8804 3.9881 3.9900 — 3.8908 3.9919 3.9930	1.3605e−04 8.8958e−06 5.6198e−07 3.5215e−08 1.3832e−04 8.9745e−06 5.6541e−07 3.5387e−08 1.3546e−04 8.7266e−06 5.4826e−07 3.4246e−08	— 3.9349 3.9845 3.9963 — 3.9460 3.9885 3.9980 — 3.9563 3.9925 4.0009

Table 6

The computational error and convergence order in spatial dimension for V at T = 1 for Example 2 using present scheme

α	h	‖ u − U ‖ ∞	Order ≈	‖ u − U ‖	Order ≈
0.3 0.6 0.9	1/5 1/10 1/20 1/40 1/5 1/10 1/20 1/40 1/5 1/10 1/20 1/40	4.8790e−03 3.1855e−04 2.0141e−05 1.2671e−06 4.9616e−03 3.2140e−04 2.0265e−05 1.2733e−06 4.8584e−03 3.1242e−04 1.9642e−05 1.2318e−06	— 3.9370 3.9833 3.9905 — 3.9484 3.9873 3.9923 — 3.9589 3.9915 3.9951	2.6858e−03 1.7607e−04 1.1126e−05 6.9719e−07 2.7313e−03 1.7764e−04 1.1195e−05 7.0064e−07 2.6742e−03 1.7269e−04 1.0852e−05 6.7782e−07	— 3.9311 3.9841 3.9962 — 3.9426 3.9880 3.9980 — 3.9529 3.9922 4.0009

Table 7

The computational error and convergence order in spatial dimension for U at T = 1 for Example 2 using central difference scheme

α	h	‖ u − U ‖ ∞	Order ≈	‖ u − U ‖	Order ≈
0.3 0.6 0.9	1/5 1/10 1/20 1/40 1/5 1/10 1/20 1/40 1/5 1/10 1/20 1/40	9.9835e−04 3.0475e−04 7.9857e−05 2.0236e−05 9.9241e−04 3.0417e−04 7.9744e−05 2.0208e−05 9.9449e−04 3.0395e−04 7.9607e−05 2.0166e−05	— 1.7119 1.9321 1.9805 — 1.7061 1.9314 1.9804 — 1.7101 1.9329 1.9810	5.5526e−04 1.6940e−04 4.4341e−05 1.1211e−05 5.5209e−04 1.6909e−04 4.4281e−05 1.1197e−05 5.5319e−04 1.6897e−04 4.4207e−05 1.1175e−05	— 1.7127 1.9337 1.9837 — 1.7071 1.9330 1.9836 — 1.7110 1.9344 1.9840

Table 8

The computational error and convergence order in spatial dimension for V at T = 1 for Example 2 using central difference scheme

α	h	‖ u − U ‖ ∞	Order ≈	‖ u − U ‖	Order ≈
0.3 0.6 0.9	1/5 1/10 1/20 1/40 1/5 1/10 1/20 1/40 1/5 1/10 1/20 1/40	7.2392e−03 9.6939e−04 1.8474e−04 4.2611e−05 7.3520e−03 9.8069e−04 1.8699e−04 4.3163e−05 7.3015e−03 9.8548e−04 1.8974e−04 4.3988e−05	— 2.9007 2.3916 2.1162 — 2.9063 2.3908 2.1151 — 2.8893 2.3768 2.1089	3.9869e−03 5.3685e−04 1.0242e−04 2.3571e−05 4.0485e−03 5.4294e−04 1.0363e−04 2.3858e−05 4.0198e−03 5.4535e−04 1.0508e−04 2.4285e−05	— 2.8927 2.3900 2.1194 — 2.8985 2.3894 2.1189 — 2.8819 2.3757 2.1134

6. Conclusion

In this work, we constructed a linearized compact difference scheme for a non-linear sub-diffusion time-delay equation in two-dimensional space. The global convergence order of the scheme is O ( τ 2 + h x 4 + h y 4 ) . We linearize the nonlinear term and prove the uniqueness of the scheme by proving that the corresponding homogeneous problem has only zero solutions. The convergence of the scheme under the discrete L 2 norm is obtained by the discrete energy method, and the stability of the scheme is also proved. Numerical experiments have been conducted to show the robustness and accuracy of the proposed numerical scheme. In a word, our scheme is very effective to solve a class of fourth-order nonlinear sub-diffusion neutral delayed equation. In further research, we will apply the proposed method to nonlinear neutral time-delay sub-diffusion equations with variable fractional order derivative.

Acknowledgements

The authors would like to thank editor and referees for their valuable advices for the improvement of this article.

Conflicts of Interest

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

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