# Linearθ-Method and Compact θ-Method for Generalised Reaction-Diffusion Equation with Delay

Authors: Fengyan Wu; Qiong Wang; Xiujun Cheng; Xiaoli Chen
Journal: International Journal of Differential Equations
 (2018)

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## Abstract

This paper is concerned with the analysis of the linearθ-method and compact θ-method for solving delay reaction-diffusion equation. Solvability, consistence, stability, and convergence of the two methods are studied. When θ∈[0,1/2), sufficient and necessary conditions are given to show that the two methods are asymptotically stable. When θ∈[1/2,1], the two methods are proven to be unconditionally asymptotically stable. Finally, several examples are carried out to confirm the theoretical results.


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## Body

## 1. Introduction

Partial functional differential equations (PFDEs) are widely used to model many natural phenomena in various scientific fields [1–9]. In order to gain a better understanding of the complicated dynamics, numerous researchers have investigated PFDEs. For instance, Garrido-Atienza and Real discussed the existence and uniqueness of solutions for delay evolution equations [10]. Mei et al. analysed the stability of travelling waves for nonlocal time-delayed reaction-diffusion equations [11]. Polyanin and Zhurov constructed exact solutions for delay reaction-diffusion equations and more complex nonlinear equations by the functional constraints method [12].However, the exact solutions are difficult to be obtained [1]. Most researchers have to seek efficient and effective numerical methods to numerically solve PFDEs. Jackiewicz and Zubik-Kowal utilised spectral collocation and waveform relaxation methods to study nonlinear delay partial differential equations [13]. Chen and Wang utilised the variational iteration method to solve a neutral functional-differential equation with proportional delays [14]. Li et al. used the discontinuous Galerkin methods to solve the delay differential equations [15–17]. Bhrawy et al. applied an accurate Chebyshev pseudospectral scheme to study the multidimensional parabolic problems with time delays [18]. Aziz and Amin employed the Haar wavelet to study the numerical solution of a class of delay differential and delay partial differential equations [19].When it comes to solving PFDEs numerically, here comes one question, that is, whether the numerical solution approximates the exact solution in a stable manner, especially for a long time. In this study, we use the following model as the test equation for analysing stability of the numerical method, which is an extension to the previous work [20–22].(1)∂∂tux,t=r1∂2∂x2ux,t+r2∂2∂x2ux,t-τ+r3ux,t+r4ux,t-τ,t>0,0<x<π,ux,t=u0x,t,-τ≤t≤0,0≤x≤π,u0,t=uπ,t=0,t≥-τ.Here and hereafter parameters r1>0 and r2>0 denote the diffusion coefficients, r3∈R, r4∈R, and τ>0 is the delay term. In particular, when r3=r4=0, the above model (1) is reduced to the original test equation in [20–22].For the case wherer3=r4=0, model (1) has been studied by many researchers [2, 20–27]. In this work, we will examine the case where r1>0, r2>0, r3∈R, and r4∈R, which is a generalisation of above mentioned work, and analyse the stability condition of the numerical method. The standard second-order central difference method and compact finite difference method are utilised to discrete the diffusion operator, respectively, and the linear θ-method is utilised to discrete the temporal direction. For convenience, we name the standard second-order central difference version as linear θ-method and the compact finite difference method version as compact θ-method. With the spectral radius condition, we consider the stability of the linear θ-method and compact θ-method, respectively.The rest of this paper is organized as follows. In Section2, we give a sufficient delay-independent condition for Problem (1) to be asymptotically stable. In Section 3, we propose the linear θ-method for solving Problem (1); solvability, stability, and convergence of the method are discussed. In Section 4, we extend the compact θ-method to solve Problem (1). In Section 5, several numerical tests are performed to validate the theoretical results.


## 2. Stability of PFDE (1)

In this section, based on Tian’s work [20], we give a sufficient condition for the trivial solution of Problem (1) to be asymptotically stable.Definition 1.
The trivial solutionux,t≡0 of PFDE (1) is called asymptotically stable if its solution u(x,t) corresponding to a sufficiently differentiable function u0(x,t) with u0(0,t)=u0(π,t)=0 satisfies(2)limt→∞⁡ux,t=0.Lemma 2 (cf. [3]).
All the roots ofpez+q-zez=0, where p and q are real, have negative real parts if and only if(1)
p<1,(2)
p<-q<a12+p2,
wherea1 is the root of g=ptan⁡g such that 0<g<π. If p=0, we take a1=π/2.Theorem 3.
Assume that the solution of Problem (1) is u(x,t)=eλteinx, where λ∈C, n∈R, x∈[0,π], and t≥0. Then the sufficient condition for the trivial solution of Problem (1) to be asymptotically stable is that(1)
τr3<1+τr1n2,(2)
τ(r3-r1n2)<τ(r2n2-r4)<a12+τ2(r3-r1n2)2,
wherea1 is the root of g=τr3-r1n2tan⁡(g) such that 0<g<π. If r3=r1n2, we take a1=π/2.Proof.
LetX=B[0,π] denote the Banach space equipped with the maximum norm, and D(A)={y∈X:y″∈X,y(0)=y(π)=0}, and Ay=y″fory∈D(A).
Letr1n2(n=1,2,⋯) be the eigenvalues of -A. According to [1, 28], if all zeros of the characteristic equations (3)fλ=λ+r1n2-r3+r2n2-r4e-λτhave negative real part, then the trivial solution is asymptotically stable. Meanwhile, if at least one zero has positive real part, then it is unstable.
Letf(λ)=0; that is, λ=r3-r1n2+(r4-r2n2)e-λτ.
Multiplying byeλτ, we have (4)λeλτ=r3-r1n2eλτ+r4-r2n2.Setting λτ=z, we get (5)zez=τr3-r1n2ez+τr4-r2n2.Denote p=τ(r3-r1n2), and q=τ(r4-r2n2), and rewrite the above equation as (6)zez=pez+q.Applying Lemma 2, if(1)
τr3<1+τr1n2,(2)
τ(r3-r1n2)<τ(r2n2-r4)<a12+τ2(r3-r1n2)2,
then the real parts of all zeros of the characteristic equations are negative. Therefore, the trivial solution of Problem (1) is asymptotically stable. Otherwise, there exists a zero λ0 whose real part is positive such that f(λ0)=0. Hence, the trivial solution is unstable. It completes the proof.


## 3. Linearθ-Method

In this section, the linearθ-method is presented to solve Problem (1).DenoteΩΔt=tk∣k=-m,-m+1,⋯ as a uniform partition on the time interval [-τ,∞), where Δt=τ/m is the time step size and tk=kΔt. Denote ΩΔx=xj∣j=0,1,⋯,N as a uniform mesh on the space interval Ω=[0,π], where Δx=π/N is the space step size and xj=jΔx. Here m and N are two positive integers. Let ujk be the numerical approximation of u(xj,tk) and let V=ujk∣0≤j≤N,k≥-m be the grid function defined on ΩΔx×ΩΔt. For any grid function u∈V, we use the following notations: (7)δtujk=ujk+1-ujkΔt,δx2ujk=uj+1k-2ujk+uj-1kΔx2,ujk+1/2=12ujk+ujk+1.Now, applying the standard second-order central difference method to discrete the diffusion operator, we obtain the following linearθ-method:(8)δtujk=r11-θδx2ujk+θδx2ujk+1+r21-θδx2ujk-m+θδx2ujk-m+1+r31-θujk+θujk+1+r41-θujk-m+θujk-m+1,j=1,2,⋯,N-1,k=0,1,⋯,ujk=u0xj,tk,j=1,2,⋯,N-1,k=-m,-m+1,⋯,0,u0k=uNk=0,k=-m,-m+1,⋯.We can rewrite the linearθ-method (8) as the following matrix form:(9)ϕ0SUk+1=ϕ1SUk-ϕmSUk+1-m-ϕm+1SUk-m,where (10)Uk=u1k,u2k,⋯,uN-1kT,a=r1ΔtΔx2,b=r2ΔtΔx2,c=r3Δt,d=r4Δt,ϕ0η=1+2aθ-cθ-aθη,ϕ1η=1-2a1-θ+c1-θ+a1-θη,ϕmη=2bθ-dθ-bθη,ϕm+1η=2b1-θ-d1-θ-b1-θη,S=010…00101…00⋱⋱⋱⋱00…10100…010N-1×N-1,and the eigenvalues of matrix S are λj=2cos⁡jΔx, j=1,2,⋯,N-1.


### 3.1. Solvability of Linearθ-Method

Theorem 4.
The linearθ-method (8) is solvable and has a unique solution.Proof.
The mathematical induction is utilised to prove it. We can obtain the solution ofU1 according to the initial condition. Now, assume that the solution of Ul has been determined. Then we can derive the solution of Ul+1 with (9). It follows from (9) that the coefficient matrix of the linear system is (11)ϕ0S=1+2aθ-cθI-aθS.It is easy to verify that the matrix ϕ0(S) is symmetric positive definite. Therefore, the solution of Ul+1 is determined uniquely. By mathematical induction, the existence and uniqueness of the solution of difference system (8) are obtained immediately.


### 3.2. Asymptotic Stability of Linearθ-Method

Section2 gives the sufficient condition for the trivial solution of Problem (1) to be asymptotically stable. Next, we will analyse the numerical stability of the linear θ-method (8) under this condition.Definition 5.
A numerical method applied to Problem (1) is called asymptotically stable about the trivial solution if its approximate solution ujk corresponding to a sufficiently differentiable function u0(x,t) with u0(0,t)=u0(π,t)=0 satisfies(12)limk→∞⁡max1≤j≤N⁡ujk=0.
In order to prove that a polynomial is a Schur polynomial, the following lemma is needed.Lemma 6 (cf. [29]).
Letγm(z)=α(z)zm-β(z) be a polynomial, where α(z) and β(z) are polynomials of constant degree. Then, the polynomial γm(z) is a Schur polynomial for any m≥1 if and only if the following conditions hold: (i)
α(z)=0⇒|z|<1,(ii)
|α(z)|≥|β(z)|, for all z∈C,|z|=1,(iii)
γm(z)≠0, for all z∈C,|z|=1.Taking the analytical technique in [20–22], we know that the linear θ-method (8) is asymptotically stable about the trivial solution if and only if(13)Pm,jθz≡ϕ0λjzm+1-ϕ1λjzm+ϕmλjz+ϕm+1λjis a Schur polynomial for any m≥1.Basic calculations give(14)Pm,jθz=μjzzm-νjz,where (15)μjz=1+θa2-λj-cz-1+1-θa2-λj-c,νjz=d-b2-λjθz+1-θ.With the help of Lemma6, we obtain the following theorem when θ∈[0,1/2), which offers a sufficient and necessary condition of asymptotic stability for the linear θ-method.Theorem 7.
Suppose thata>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a(1-cos⁡Δx)-c>2b1+cos⁡Δx-d. Then the linear θ-method (8) is asymptotically stable about the trivial solution for θ∈[0,1/2) if and only if(16)1-2θa+b-c+d21+cos⁡Δx<11+cos⁡Δx,where a=r1Δt/Δx2, b=r2Δt/Δx2, c=r3Δt, and d=r4Δt.Proof.
( ⇒ ) First, let us verify item (i) of Lemma 6. According to μj(z)=0, we derive that(17)z=1-a2-λj-c1+θa2-λj-c.By the same technique used in [22], one can check that z<1.
Next, to verify the rest of items of Lemma6, that is, μj(z)>νj(z), j=1,2,⋯,N-1, for all z∈C, z=1, we define the following complex variable function:(18)w=μjzνjz=1+θa2-λj-cz-1+1-θa2-λj-cd-b2-λjθz+1-θ.
Lettingw=x+yi and z=1, after some basic calculations (see [20–22]), we find (19)minz=1,z∈C⁡w=minz=1,z∈C⁡μjzνjz=min⁡a2-λj-cb2-λj-d,2-a2-λj-c1-2θb2-λj-d1-2θ.The value of minz=1,z∈C⁡w will be discussed in the following two different cases.
Case a min z = 1 , z ∈ C ⁡ μ j z / ν j z = a 2 - λ j - c / b 2 - λ j - d . By conditions a>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a1-cos⁡Δx-c>2b1+cos⁡Δx-d, and noting that 21-cos⁡Δx≤(2-λj)≤21-cos⁡N-1Δx=21+cos⁡Δx, we have a(2-λj)-c>b(2-λj)-d>0. Therefore, for all z∈C, z=1, we find (20)μjzνjz≥minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d>1.
Case b min z = 1 , z ∈ C ⁡ μ j z / ν j z = 2 - c + a 2 - λ j 1 - 2 θ / d + b 2 - λ j 1 - 2 θ . It follows from condition (16) that (21)2-a2-λj-c1-2θ>0.
Noticing the fact that2-λj≤2(1+cos⁡(Δx)) and condition (16), we have(22)minz=1,z∈C⁡μjzνjz=2-a2-λj-c1-2θb2-λj-d1-2θ≥2-2a1+cos⁡Δx-c1-2θ2b1+cos⁡Δx-d1-2θ>1.
In brief, combiningCase a andCase b, we conclude that, for all z∈C, |z|=1, μj(z)>νj(z), j=1,2,⋯,N-1 holds, which implies that items (ii) and (iii) of Lemma 6 hold.
Now, with the help of Lemma6, we derive that the linear θ-method (8) is asymptotically stable about the trivial solution.
(⇐) Next, we prove the necessary part from two points by contradiction:(i)
Suppose that(1-2θ)[a+b-c+d/2(1+cos⁡(Δx))]=1/1+cos⁡Δx. Let m be even, j=N-1, and z=-1, and then, for z=1, we get that Pm,N-1θ-1=0, which indicates that condition (iii) of Lemma 6 does not hold. Thus, the linear θ-method (8) is not asymptotically stable.(ii)
Suppose that(1-2θ)[a+b-c+d/2(1+cos⁡(Δx))]>1/1+cos⁡Δx. Let j=N-1 and z=-1, and then, after some basic calculations, we arrive at νN-1-1>μN-1-1. This signifies that condition (ii) of Lemma 6 does not hold. Therefore, the linear θ-method (8) is not asymptotically stable.
Then, we know that (16) is a necessary condition for asymptotic stability. This completes the proof.Remark 8.
Whenr3=r4=0, the sufficient and necessary condition (16) in Theorem 7 is simplified to (23)1-2θa+b<11+cos⁡Δx,which is consistent with the previous work [20].Next, whenθ∈[1/2,1], we will prove that the linear θ-method (8) is unconditionally asymptotically stable with respect to the trivial solution.Theorem 9.
Suppose thata>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a1-cos⁡Δx-c>2b1+cos⁡Δx-d. Then the linear θ-method (8) is unconditionally asymptotically stable about the trivial solution for θ∈[1/2,1].Proof.
We will prove the theorem with Lemma6. First, it follows from μj(z)=0 that (24)z=1-a2-λj-c1+θa2-λj-c.Similar to the proof of Theorem 7, we get z<1.
Then, we check items (ii) and (iii) of Lemma6. To do that, we introduce the following complex variable function: (25)w=μjzνjz=1+θa2-λj-cz-1+1-θa2-λj-cd-b2-λjθz+1-θ.(i)
θ=1/2. Set w=x+yi and z=1. Basic calculations give (26)minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d.It follows from assumptionsa>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a1-cos⁡Δx-c>2b1+cos⁡Δx-d that a2-λj-c>b2-λj-d>0. Then, for all z∈C, z=1, we find that (27)μjzνjz≥minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d>1,which indicates that items(ii) and (iii) of Lemma 6 hold.(ii)
θ∈(1/2,1]. Similarly, we can get (28)minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d.In this case, for allz∈C, |z|=1, we also obtain that (29)μjzνjz≥minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d>1.
According to Lemma6, we conclude that the linear θ-method (8) is asymptotically stable about the trivial solution. This completes the proof of the theorem.


### 3.3. Convergence of Linearθ-Method

Here and below, when we discuss the convergence of numerical methods, we will always assume that the solutionu(x,t) of Problem (1) is smooth enough and satisfies(30)∂j+k∂xj∂tkux,t≤C,0≤j≤6,0≤k≤3,C>0,where C is a constant.LetUjk=u(xj,tk), j=0,1,⋯, k=-m,-m+1,⋯. Then, we get (31)δtUjk=r11-θδx2Ujk+θδx2Ujk+1+r21-θδx2Ujk-m+θδx2Ujk-m+1+r31-θUjk+θUjk+1+r41-θUjk-m+θUjk-m+1+Rjk,where Rjk is the local truncation error. Taylor expansion yields that there exists a constant C¯ such that, for j=1,2,⋯,N-1, k=0,1,⋯,(32)Rjk≤C¯Δt2+Δx2,θ=12,C¯Δt+Δx2,0≤θ<12or12<θ≤1.Thus, the consistence of linearθ-method (8) is obtained. Now, the convergence result is presented in the following theorem.Theorem 10.
Assume that the assumptions in Theorems7 and 9 hold. Then, for k=1,2,⋯, we have the following convergent result:(33)ek≤C^Δt2+Δx2,θ=12,C^Δt+Δx2,0≤θ<12or12<θ≤1,where ek=[u1k-U1k,u2k-U2k,⋯,uN-1k-UN-1k]T and C^ is a constant that is independent of Δt,Δx.Proof.
It follows from Theorem4 that difference system (8) is solvable and has a unique solution. Moreover, the assumptions in Theorems 7 and 9 hold, signifying that the method is stable. Together with the consistence of the method, we derive that (33) holds by the Lax equivalence theorem [30, 31].


## 3.1. Solvability of Linearθ-Method

Theorem 4.
The linearθ-method (8) is solvable and has a unique solution.Proof.
The mathematical induction is utilised to prove it. We can obtain the solution ofU1 according to the initial condition. Now, assume that the solution of Ul has been determined. Then we can derive the solution of Ul+1 with (9). It follows from (9) that the coefficient matrix of the linear system is (11)ϕ0S=1+2aθ-cθI-aθS.It is easy to verify that the matrix ϕ0(S) is symmetric positive definite. Therefore, the solution of Ul+1 is determined uniquely. By mathematical induction, the existence and uniqueness of the solution of difference system (8) are obtained immediately.


## 3.2. Asymptotic Stability of Linearθ-Method

Section2 gives the sufficient condition for the trivial solution of Problem (1) to be asymptotically stable. Next, we will analyse the numerical stability of the linear θ-method (8) under this condition.Definition 5.
A numerical method applied to Problem (1) is called asymptotically stable about the trivial solution if its approximate solution ujk corresponding to a sufficiently differentiable function u0(x,t) with u0(0,t)=u0(π,t)=0 satisfies(12)limk→∞⁡max1≤j≤N⁡ujk=0.
In order to prove that a polynomial is a Schur polynomial, the following lemma is needed.Lemma 6 (cf. [29]).
Letγm(z)=α(z)zm-β(z) be a polynomial, where α(z) and β(z) are polynomials of constant degree. Then, the polynomial γm(z) is a Schur polynomial for any m≥1 if and only if the following conditions hold: (i)
α(z)=0⇒|z|<1,(ii)
|α(z)|≥|β(z)|, for all z∈C,|z|=1,(iii)
γm(z)≠0, for all z∈C,|z|=1.Taking the analytical technique in [20–22], we know that the linear θ-method (8) is asymptotically stable about the trivial solution if and only if(13)Pm,jθz≡ϕ0λjzm+1-ϕ1λjzm+ϕmλjz+ϕm+1λjis a Schur polynomial for any m≥1.Basic calculations give(14)Pm,jθz=μjzzm-νjz,where (15)μjz=1+θa2-λj-cz-1+1-θa2-λj-c,νjz=d-b2-λjθz+1-θ.With the help of Lemma6, we obtain the following theorem when θ∈[0,1/2), which offers a sufficient and necessary condition of asymptotic stability for the linear θ-method.Theorem 7.
Suppose thata>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a(1-cos⁡Δx)-c>2b1+cos⁡Δx-d. Then the linear θ-method (8) is asymptotically stable about the trivial solution for θ∈[0,1/2) if and only if(16)1-2θa+b-c+d21+cos⁡Δx<11+cos⁡Δx,where a=r1Δt/Δx2, b=r2Δt/Δx2, c=r3Δt, and d=r4Δt.Proof.
( ⇒ ) First, let us verify item (i) of Lemma 6. According to μj(z)=0, we derive that(17)z=1-a2-λj-c1+θa2-λj-c.By the same technique used in [22], one can check that z<1.
Next, to verify the rest of items of Lemma6, that is, μj(z)>νj(z), j=1,2,⋯,N-1, for all z∈C, z=1, we define the following complex variable function:(18)w=μjzνjz=1+θa2-λj-cz-1+1-θa2-λj-cd-b2-λjθz+1-θ.
Lettingw=x+yi and z=1, after some basic calculations (see [20–22]), we find (19)minz=1,z∈C⁡w=minz=1,z∈C⁡μjzνjz=min⁡a2-λj-cb2-λj-d,2-a2-λj-c1-2θb2-λj-d1-2θ.The value of minz=1,z∈C⁡w will be discussed in the following two different cases.
Case a min z = 1 , z ∈ C ⁡ μ j z / ν j z = a 2 - λ j - c / b 2 - λ j - d . By conditions a>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a1-cos⁡Δx-c>2b1+cos⁡Δx-d, and noting that 21-cos⁡Δx≤(2-λj)≤21-cos⁡N-1Δx=21+cos⁡Δx, we have a(2-λj)-c>b(2-λj)-d>0. Therefore, for all z∈C, z=1, we find (20)μjzνjz≥minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d>1.
Case b min z = 1 , z ∈ C ⁡ μ j z / ν j z = 2 - c + a 2 - λ j 1 - 2 θ / d + b 2 - λ j 1 - 2 θ . It follows from condition (16) that (21)2-a2-λj-c1-2θ>0.
Noticing the fact that2-λj≤2(1+cos⁡(Δx)) and condition (16), we have(22)minz=1,z∈C⁡μjzνjz=2-a2-λj-c1-2θb2-λj-d1-2θ≥2-2a1+cos⁡Δx-c1-2θ2b1+cos⁡Δx-d1-2θ>1.
In brief, combiningCase a andCase b, we conclude that, for all z∈C, |z|=1, μj(z)>νj(z), j=1,2,⋯,N-1 holds, which implies that items (ii) and (iii) of Lemma 6 hold.
Now, with the help of Lemma6, we derive that the linear θ-method (8) is asymptotically stable about the trivial solution.
(⇐) Next, we prove the necessary part from two points by contradiction:(i)
Suppose that(1-2θ)[a+b-c+d/2(1+cos⁡(Δx))]=1/1+cos⁡Δx. Let m be even, j=N-1, and z=-1, and then, for z=1, we get that Pm,N-1θ-1=0, which indicates that condition (iii) of Lemma 6 does not hold. Thus, the linear θ-method (8) is not asymptotically stable.(ii)
Suppose that(1-2θ)[a+b-c+d/2(1+cos⁡(Δx))]>1/1+cos⁡Δx. Let j=N-1 and z=-1, and then, after some basic calculations, we arrive at νN-1-1>μN-1-1. This signifies that condition (ii) of Lemma 6 does not hold. Therefore, the linear θ-method (8) is not asymptotically stable.
Then, we know that (16) is a necessary condition for asymptotic stability. This completes the proof.Remark 8.
Whenr3=r4=0, the sufficient and necessary condition (16) in Theorem 7 is simplified to (23)1-2θa+b<11+cos⁡Δx,which is consistent with the previous work [20].Next, whenθ∈[1/2,1], we will prove that the linear θ-method (8) is unconditionally asymptotically stable with respect to the trivial solution.Theorem 9.
Suppose thata>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a1-cos⁡Δx-c>2b1+cos⁡Δx-d. Then the linear θ-method (8) is unconditionally asymptotically stable about the trivial solution for θ∈[1/2,1].Proof.
We will prove the theorem with Lemma6. First, it follows from μj(z)=0 that (24)z=1-a2-λj-c1+θa2-λj-c.Similar to the proof of Theorem 7, we get z<1.
Then, we check items (ii) and (iii) of Lemma6. To do that, we introduce the following complex variable function: (25)w=μjzνjz=1+θa2-λj-cz-1+1-θa2-λj-cd-b2-λjθz+1-θ.(i)
θ=1/2. Set w=x+yi and z=1. Basic calculations give (26)minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d.It follows from assumptionsa>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a1-cos⁡Δx-c>2b1+cos⁡Δx-d that a2-λj-c>b2-λj-d>0. Then, for all z∈C, z=1, we find that (27)μjzνjz≥minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d>1,which indicates that items(ii) and (iii) of Lemma 6 hold.(ii)
θ∈(1/2,1]. Similarly, we can get (28)minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d.In this case, for allz∈C, |z|=1, we also obtain that (29)μjzνjz≥minz=1,z∈C⁡μjzνjz=a2-λj-cb2-λj-d>1.
According to Lemma6, we conclude that the linear θ-method (8) is asymptotically stable about the trivial solution. This completes the proof of the theorem.


## 3.3. Convergence of Linearθ-Method

Here and below, when we discuss the convergence of numerical methods, we will always assume that the solutionu(x,t) of Problem (1) is smooth enough and satisfies(30)∂j+k∂xj∂tkux,t≤C,0≤j≤6,0≤k≤3,C>0,where C is a constant.LetUjk=u(xj,tk), j=0,1,⋯, k=-m,-m+1,⋯. Then, we get (31)δtUjk=r11-θδx2Ujk+θδx2Ujk+1+r21-θδx2Ujk-m+θδx2Ujk-m+1+r31-θUjk+θUjk+1+r41-θUjk-m+θUjk-m+1+Rjk,where Rjk is the local truncation error. Taylor expansion yields that there exists a constant C¯ such that, for j=1,2,⋯,N-1, k=0,1,⋯,(32)Rjk≤C¯Δt2+Δx2,θ=12,C¯Δt+Δx2,0≤θ<12or12<θ≤1.Thus, the consistence of linearθ-method (8) is obtained. Now, the convergence result is presented in the following theorem.Theorem 10.
Assume that the assumptions in Theorems7 and 9 hold. Then, for k=1,2,⋯, we have the following convergent result:(33)ek≤C^Δt2+Δx2,θ=12,C^Δt+Δx2,0≤θ<12or12<θ≤1,where ek=[u1k-U1k,u2k-U2k,⋯,uN-1k-UN-1k]T and C^ is a constant that is independent of Δt,Δx.Proof.
It follows from Theorem4 that difference system (8) is solvable and has a unique solution. Moreover, the assumptions in Theorems 7 and 9 hold, signifying that the method is stable. Together with the consistence of the method, we derive that (33) holds by the Lax equivalence theorem [30, 31].


## 4. Extension to Compactθ-Method

In this section, we would like to use the compactθ-method with a higher convergence order in space to extend our work. We introduce the compact difference operator,(34)Ahujk=uj-1k+10ujk+uj+1k12,j=1,2,⋯,N-1,ujkj=0,N,and an important lemma below, which will be needed to construct and prove our main results.Lemma 11 (cf. [32]).
Assume thatv(x)∈C6[0,π]. Then (35)v″xj-1+10v″xj+v″xj+112-vxj-1-2vxj+vxj+1Δx2=Δx4240v6ωj,where ωj∈(xj-1,xj+1).Now, applying the compact difference operator (34) to discrete the diffusion operator, we have the compact θ-method:(36)Ahδtujk=r11-θδx2ujk+θδx2ujk+1+r21-θδx2ujk-m+θδx2ujk-m+1+r31-θujk+θujk+1+r41-θujk-m+θujk-m+1,j=1,2,⋯,N-1,k=0,1,⋯,ujk=u0xj,tk,j=1,2,⋯,N-1,k=-m,-m+1,⋯,0,u0k=uNk=0,k=-m,-m+1,⋯.The compactθ-method (36) can be rewritten in the following matrix form:(37)ψ0SUk+1=ψ1SUk-ψmSUk+1-m-ψm+1SUk-m,where (38)ψ0η=56+2aθ-cθ+112-aθη,ψ1η=56-2a1-θ+c1-θ+112+a1-θη,ψmη=2bθ-dθ-bθη,ψm+1η=2b1-θ-d1-θ-b1-θη.Similarly, the solvability, asymptotic stability, and convergence of the compactθ-method (36) can also be obtained. For conciseness, we merely list our main results and omit the details.Theorem 12.
Suppose thata>c/21-cos⁡Δx, b>d/21-cos⁡Δx, and 2a1-cos⁡Δx-c>2b1+cos⁡Δx-d. Then the compact θ-method (36) is asymptotically stable about the trivial solution for θ∈[0,1/2) if and only if(39)16+1-2θa+b-c+d21+cos⁡Δx<11+cos⁡Δx,where a=r1Δt/Δx2, b=r2Δt/Δx2, c=r3Δt, and d=r4Δt.Remark 13.
Whenr3=r4=0, the sufficient and necessary condition (39) in Theorem 12 is reduced to (40)16+1-2θa+b<11+cos⁡Δx,which is consistent with the previous work [22].Theorem 14.
Suppose thata>c/2(1-cos⁡(Δx)), b>d/2(1-cos⁡(Δx)), and 2a(1-cos⁡Δx)-c>2b(1+cos⁡(Δx))-d. Then the compact θ-method (36) is unconditionally asymptotically stable about the trivial solution for θ∈[1/2,1].Theorem 15.
Assume that the assumptions in Theorems12 and 14 hold. Then, for k=1,2,⋯, we have the following convergent result:(41)ek≤C~Δt2+Δx4,θ=12,C~Δt+Δx4,0≤θ<12or12<θ≤1,where C~ is a constant that is independent of temporal and spatial stepsizes.Remark 16.
The comparison of linearθ-method and compact θ-method applied to problem in [20] and Problem (1) is presented in Table 1.Table 1
Stability and convergence order of different methods.
θ ∈ 0 , 1 2 θ ∈ 1 2 , 1 Order Linearθ-method for problem of [20] 1 - 2 θ a + b < 1 1 + cos ⁡ Δ x Unconditionally stable 2    Compactθ-method for problem of [20] 1 6 + 1 - 2 θ a + b < 1 1 + cos ⁡ Δ x Unconditionally stable 4    Linearθ-method for problem (1) 1 - 2 θ a + b - c + d 2 1 + cos ⁡ Δ x < 1 1 + cos ⁡ Δ x Unconditionally stable 2    Compactθ-method for problem (1) 1 6 + 1 - 2 θ a + b - c + d 2 1 + cos ⁡ Δ x < 1 1 + cos ⁡ Δ x Unconditionally stable 4


## 5. Numerical Tests

In this section, several numerical experiments are carried out to illustrate the theoretical results.


### 5.1. Stability Tests of Linearθ-Method and Compact θ-Method

We use the following equation to test stability of the proposed method:(42)∂∂tux,t=∂2∂x2ux,t+0.5∂2∂x2ux,t-τ-ux,t-0.5ux,t-τ,0<t≤T,0<x<π,ux,t=sin⁡x,-τ≤t≤0,0≤x≤π,u0,t=uπ,t=0,t≥-τ.Here, let τ=1 and Δx=π/10. We set parameters r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5 such that the trivial solution of Problem (1) is asymptotically stable.


#### 5.1.1. Linearθ-Method

First, to verify the effectiveness of the sufficient and necessary condition (16) of the linear θ-method, here we choose the case where θ=0 to illustrate that. Noting that Δt=τ/m, where m>0 is an integer, and substituting parameters θ=0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5 into condition (16), we derive that the proposed method is asymptotically stable if m>30.4025. In other words, if m≥31, then the method is asymptotically stable. Meanwhile, if m≤30, then the method is not asymptotically stable. In Figure 1, we can get a pictorial understanding of that.Figure 1
Numerical solution at the different final time T for varying parameter m ( θ = 0 and τ=1). (a) m=25. (b) m=30. (c) m=31. (d) m=35.
(a) (b) (c) (d)It is easily seen from Figures1(a) and 1(b) that the numerical solution is unstable as time goes on for m=25 and 30. As shown in Figures 1(c) and 1(d), we know that the numerical solution is asymptotically stable for m=31 and 35. Furthermore, denote the left-hand side of (16) as lhs=(1-2θ)[a+b-c+d/21+cos⁡Δx], and denote the right-hand side of (16) as rhs=1/1+cos⁡Δx, and Ind=lhs-rhs. From above paragraph, we know that Ind>0 for m≤30, and Ind<0 for m≥31. The relationship between Ind and m is shown in Figures 2(a) and 2(b). All these illustrate the results in Theorem 7.Figure 2
Ind as a function of parameter m ( θ = 0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5). (a) m∈[1,30]. (b) m∈[31,1000].
(a) (b)Next, whenθ=1/2 or 1, we apply the linear θ-method and use different stepsizes to solve problem (42). Theoretically, the numerical solution is asymptotically stable by Theorem 9. Numerically, we know that the numerical solution is asymptotically stable from the plots of Figure 3, which is consistent with the theoretical result.Figure 3
Numerical solution at the different final time T for varying parameters θ and m ( τ = 1). (a) θ=0.5 and m=25. (b) θ=0.5 and m=35. (c) θ=1 and m=25. (d) θ=1 and m=35.
(a) (b) (c) (d)


#### 5.1.2. Compactθ-Method

First, we choose the caseθ=0 to verify the effectiveness of the sufficient and necessary condition (39) of the compact θ-method. Under the case that θ=0, τ=1, Δx=π/10, r1=r3=1.0, and r2=r4=0.5, and noting that Δt=τ/m, by condition (39), the proposed method is asymptotically stable if and only if m>45.052. In other words, if m≥46, then the proposed method is asymptotically stable. Meanwhile, if m≤45, then the proposed method is not asymptotically stable. In order to validate it, we give a pictorial understanding of that in Figure 4.Figure 4
Numerical solution at the different final time T for varying parameter m ( θ = 0 and τ=1). (a) m=40. (b) m=45. (c) m=46. (d) m=50.
(a) (b) (c) (d)From Figures4(a) and 4(b), it is easily seen that the numerical solution is unstable as time goes on for m=40 and 45. The numerical solution is asymptotically stable for m=46 and 50 in Figures 4(c) and 4(d), respectively. Furthermore, denote the left-hand side of (39) as LHS=1/6+(1-2θ)[a+b-c+d/2(1+cos⁡Δx)], and denote the right-hand side of (39) as RHS=1/1+cos⁡Δx, and Ind=LHS-RHS. According to the analysis of above paragraph, we know that Ind>0 for m≤45, and Ind<0 for m≥46. The schematic presentation of the relationship between Ind and positive integer m is given in Figures 5(a) and 5(b). All the numerical results agree well with the findings in Theorem 12.Figure 5
Ind as a function of parameter m ( θ = 0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5). (a) m∈[1,45]. (b) m∈[46,1000].
(a) (b)Then, forθ=1/2 or 1, we apply the compact θ-method and choose different stepsizes to solve problem (42). The numerical results are shown in Figure 6. Theoretically, according to Theorem 14, the numerical solution is asymptotically stable. Numerically, from these figures, we know that the numerical solution is asymptotically stable, which confirms the theoretical result.Figure 6
Numerical solution at the different final time T for varying parameters θ and m ( τ = 1). (a) θ=0.5 and m=40. (b) θ=0.5 and m=50. (c) θ=1 and m=40. (d) θ=1 and m=50.
(a) (b) (c) (d)


### 5.2. Convergence Tests of Linearθ-Method and Compact θ-Method

We use the following equation to show our convergence results:(43)∂∂tux,t=∂2∂x2ux,t+0.5∂2∂x2ux,t-τ-ux,t-0.5ux,t-τ+fx,t,0<t≤T,0<x<π,ux,t=u0x,t,-τ≤t≤0,0≤x≤π,u0,t=uπ,t=0,t≥-τ.Here, the added term h(x,t) and the initial condition u0(x,t) are specified so that the exact solution is u(x,t)=e-tsin⁡(x).We set the parametersr1=1.0, r2=0.5, r3=-1.0, r4=-0.5, τ=0.5, and T=2 and solve problem (43) with different spatial and temporal stepsizes (Δx=π/N and Δt=τ/m).Forθ=0, we let Δt=1e-5 to guarantee that the linear θ-method (8) is asymptotically stable. When the compact θ-method (36) is applied to solve problem (43), we let Δt≈Δx4. The numerical errors and corresponding orders in different sense of norms are displayed in Table 2. Clearly, these results confirm the convergence of the two methods.Table 2
Errors and convergence orders whenθ=0, T=2, and τ=0.5.
Linearθ-method Compactθ-method N L 2-error Order L ∞-error Order N L 2-error Order L ∞-error Order 2 2.40E-02 − 1.91E-02 − 7 1.15E-03 − 8.95E-04 − 4 5.62E-03 2.09 4.48 E - 03 2.09 14 7.31 E - 05 3.98 5.83 E - 05 3.94 8 1.38 E - 03 2.02 1.10 E - 03 2.02 28 4.58 E - 06 4.00 3.65 E - 06 4.00 16 3.44 E - 04 2.01 2.75 E - 04 2.01 56 2.86 E - 07 4.00 2.28 E - 07 4.00 32 8.57 E - 05 2.01 6.84 E - 05 2.01 112 1.79E-08 4.00 1.43 E - 08 4.00Forθ=1/2, when the linear θ-method (8) is applied to solve problem (43), we let Δt≈Δx, and for the compact θ-method (36), we let Δt≈Δx2. For θ=1, when method (8) is applied to solve problem (43), we let Δt≈Δx2, and for method (36), we let Δt≈Δx4. The numerical errors and corresponding orders in different sense of norms are listed in Tables 3 and 4, respectively. It is readily found that these results confirm the convergence of the two methods. Obviously, the compact θ-method gives a better convergence result in the space.Table 3
Errors and convergence orders whenθ=0.5, T=2, and τ=0.5.
Linearθ-method Compactθ-method N L 2-error Order L ∞-error Order L 2-error Order L ∞-error Order 5 1.40E-03 − 1.06E-03 − 7.69E-04 − 5.83 E - 04 − 10 3.81E-04 1.87 3.04E-04 1.80 4.46E-05 4.11 3.56 E - 05 4.03 20 9.67E-05 1.98 7.72E-05 1.98 2.74E-06 4.03 2.19 E - 06 4.03 40 2.61E-05 1.89 2.09E-05 1.89 1.69E-07 4.01 1.35 E - 07 4.01 80 6.15E-06 2.09 4.91E-06 2.09 1.06E-08 4.00 8.44 E - 09 4.00Table 4
Errors and convergence orders whenθ=1, T=2, and τ=0.5.
Linearθ-method Compactθ-method N L 2-error Order L ∞-error Order L 2-error Order L ∞-error Order 5 2.16E-02 − 1.64 E - 02 − 5.45 E - 03 − 4.14 E - 03 − 10 4.34E-03 2.31 3.47 E - 03 2.24 2.96 E - 04 4.20 2.36 E - 04 4.13 20 9.83 E - 04 2.14 7.84 E - 04 2.14 1.82 E - 05 4.03 1.45 E - 05 4.03 40 2.38E-04 2.04 1.90 E - 04 2.04 1.13 E - 06 4.00 9.05 E - 07 4.00 80 5.92E-05 2.01 4.72 E - 05 2.01 7.09 E - 08 4.00 5.65 E - 08 4.00


## 5.1. Stability Tests of Linearθ-Method and Compact θ-Method

We use the following equation to test stability of the proposed method:(42)∂∂tux,t=∂2∂x2ux,t+0.5∂2∂x2ux,t-τ-ux,t-0.5ux,t-τ,0<t≤T,0<x<π,ux,t=sin⁡x,-τ≤t≤0,0≤x≤π,u0,t=uπ,t=0,t≥-τ.Here, let τ=1 and Δx=π/10. We set parameters r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5 such that the trivial solution of Problem (1) is asymptotically stable.


### 5.1.1. Linearθ-Method

First, to verify the effectiveness of the sufficient and necessary condition (16) of the linear θ-method, here we choose the case where θ=0 to illustrate that. Noting that Δt=τ/m, where m>0 is an integer, and substituting parameters θ=0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5 into condition (16), we derive that the proposed method is asymptotically stable if m>30.4025. In other words, if m≥31, then the method is asymptotically stable. Meanwhile, if m≤30, then the method is not asymptotically stable. In Figure 1, we can get a pictorial understanding of that.Figure 1
Numerical solution at the different final time T for varying parameter m ( θ = 0 and τ=1). (a) m=25. (b) m=30. (c) m=31. (d) m=35.
(a) (b) (c) (d)It is easily seen from Figures1(a) and 1(b) that the numerical solution is unstable as time goes on for m=25 and 30. As shown in Figures 1(c) and 1(d), we know that the numerical solution is asymptotically stable for m=31 and 35. Furthermore, denote the left-hand side of (16) as lhs=(1-2θ)[a+b-c+d/21+cos⁡Δx], and denote the right-hand side of (16) as rhs=1/1+cos⁡Δx, and Ind=lhs-rhs. From above paragraph, we know that Ind>0 for m≤30, and Ind<0 for m≥31. The relationship between Ind and m is shown in Figures 2(a) and 2(b). All these illustrate the results in Theorem 7.Figure 2
Ind as a function of parameter m ( θ = 0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5). (a) m∈[1,30]. (b) m∈[31,1000].
(a) (b)Next, whenθ=1/2 or 1, we apply the linear θ-method and use different stepsizes to solve problem (42). Theoretically, the numerical solution is asymptotically stable by Theorem 9. Numerically, we know that the numerical solution is asymptotically stable from the plots of Figure 3, which is consistent with the theoretical result.Figure 3
Numerical solution at the different final time T for varying parameters θ and m ( τ = 1). (a) θ=0.5 and m=25. (b) θ=0.5 and m=35. (c) θ=1 and m=25. (d) θ=1 and m=35.
(a) (b) (c) (d)


### 5.1.2. Compactθ-Method

First, we choose the caseθ=0 to verify the effectiveness of the sufficient and necessary condition (39) of the compact θ-method. Under the case that θ=0, τ=1, Δx=π/10, r1=r3=1.0, and r2=r4=0.5, and noting that Δt=τ/m, by condition (39), the proposed method is asymptotically stable if and only if m>45.052. In other words, if m≥46, then the proposed method is asymptotically stable. Meanwhile, if m≤45, then the proposed method is not asymptotically stable. In order to validate it, we give a pictorial understanding of that in Figure 4.Figure 4
Numerical solution at the different final time T for varying parameter m ( θ = 0 and τ=1). (a) m=40. (b) m=45. (c) m=46. (d) m=50.
(a) (b) (c) (d)From Figures4(a) and 4(b), it is easily seen that the numerical solution is unstable as time goes on for m=40 and 45. The numerical solution is asymptotically stable for m=46 and 50 in Figures 4(c) and 4(d), respectively. Furthermore, denote the left-hand side of (39) as LHS=1/6+(1-2θ)[a+b-c+d/2(1+cos⁡Δx)], and denote the right-hand side of (39) as RHS=1/1+cos⁡Δx, and Ind=LHS-RHS. According to the analysis of above paragraph, we know that Ind>0 for m≤45, and Ind<0 for m≥46. The schematic presentation of the relationship between Ind and positive integer m is given in Figures 5(a) and 5(b). All the numerical results agree well with the findings in Theorem 12.Figure 5
Ind as a function of parameter m ( θ = 0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5). (a) m∈[1,45]. (b) m∈[46,1000].
(a) (b)Then, forθ=1/2 or 1, we apply the compact θ-method and choose different stepsizes to solve problem (42). The numerical results are shown in Figure 6. Theoretically, according to Theorem 14, the numerical solution is asymptotically stable. Numerically, from these figures, we know that the numerical solution is asymptotically stable, which confirms the theoretical result.Figure 6
Numerical solution at the different final time T for varying parameters θ and m ( τ = 1). (a) θ=0.5 and m=40. (b) θ=0.5 and m=50. (c) θ=1 and m=40. (d) θ=1 and m=50.
(a) (b) (c) (d)


## 5.1.1. Linearθ-Method

First, to verify the effectiveness of the sufficient and necessary condition (16) of the linear θ-method, here we choose the case where θ=0 to illustrate that. Noting that Δt=τ/m, where m>0 is an integer, and substituting parameters θ=0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5 into condition (16), we derive that the proposed method is asymptotically stable if m>30.4025. In other words, if m≥31, then the method is asymptotically stable. Meanwhile, if m≤30, then the method is not asymptotically stable. In Figure 1, we can get a pictorial understanding of that.Figure 1
Numerical solution at the different final time T for varying parameter m ( θ = 0 and τ=1). (a) m=25. (b) m=30. (c) m=31. (d) m=35.
(a) (b) (c) (d)It is easily seen from Figures1(a) and 1(b) that the numerical solution is unstable as time goes on for m=25 and 30. As shown in Figures 1(c) and 1(d), we know that the numerical solution is asymptotically stable for m=31 and 35. Furthermore, denote the left-hand side of (16) as lhs=(1-2θ)[a+b-c+d/21+cos⁡Δx], and denote the right-hand side of (16) as rhs=1/1+cos⁡Δx, and Ind=lhs-rhs. From above paragraph, we know that Ind>0 for m≤30, and Ind<0 for m≥31. The relationship between Ind and m is shown in Figures 2(a) and 2(b). All these illustrate the results in Theorem 7.Figure 2
Ind as a function of parameter m ( θ = 0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5). (a) m∈[1,30]. (b) m∈[31,1000].
(a) (b)Next, whenθ=1/2 or 1, we apply the linear θ-method and use different stepsizes to solve problem (42). Theoretically, the numerical solution is asymptotically stable by Theorem 9. Numerically, we know that the numerical solution is asymptotically stable from the plots of Figure 3, which is consistent with the theoretical result.Figure 3
Numerical solution at the different final time T for varying parameters θ and m ( τ = 1). (a) θ=0.5 and m=25. (b) θ=0.5 and m=35. (c) θ=1 and m=25. (d) θ=1 and m=35.
(a) (b) (c) (d)


## 5.1.2. Compactθ-Method

First, we choose the caseθ=0 to verify the effectiveness of the sufficient and necessary condition (39) of the compact θ-method. Under the case that θ=0, τ=1, Δx=π/10, r1=r3=1.0, and r2=r4=0.5, and noting that Δt=τ/m, by condition (39), the proposed method is asymptotically stable if and only if m>45.052. In other words, if m≥46, then the proposed method is asymptotically stable. Meanwhile, if m≤45, then the proposed method is not asymptotically stable. In order to validate it, we give a pictorial understanding of that in Figure 4.Figure 4
Numerical solution at the different final time T for varying parameter m ( θ = 0 and τ=1). (a) m=40. (b) m=45. (c) m=46. (d) m=50.
(a) (b) (c) (d)From Figures4(a) and 4(b), it is easily seen that the numerical solution is unstable as time goes on for m=40 and 45. The numerical solution is asymptotically stable for m=46 and 50 in Figures 4(c) and 4(d), respectively. Furthermore, denote the left-hand side of (39) as LHS=1/6+(1-2θ)[a+b-c+d/2(1+cos⁡Δx)], and denote the right-hand side of (39) as RHS=1/1+cos⁡Δx, and Ind=LHS-RHS. According to the analysis of above paragraph, we know that Ind>0 for m≤45, and Ind<0 for m≥46. The schematic presentation of the relationship between Ind and positive integer m is given in Figures 5(a) and 5(b). All the numerical results agree well with the findings in Theorem 12.Figure 5
Ind as a function of parameter m ( θ = 0, τ=1, Δx=π/10, r1=1.0, r2=0.5, r3=-1.0, and r4=-0.5). (a) m∈[1,45]. (b) m∈[46,1000].
(a) (b)Then, forθ=1/2 or 1, we apply the compact θ-method and choose different stepsizes to solve problem (42). The numerical results are shown in Figure 6. Theoretically, according to Theorem 14, the numerical solution is asymptotically stable. Numerically, from these figures, we know that the numerical solution is asymptotically stable, which confirms the theoretical result.Figure 6
Numerical solution at the different final time T for varying parameters θ and m ( τ = 1). (a) θ=0.5 and m=40. (b) θ=0.5 and m=50. (c) θ=1 and m=40. (d) θ=1 and m=50.
(a) (b) (c) (d)


## 5.2. Convergence Tests of Linearθ-Method and Compact θ-Method

We use the following equation to show our convergence results:(43)∂∂tux,t=∂2∂x2ux,t+0.5∂2∂x2ux,t-τ-ux,t-0.5ux,t-τ+fx,t,0<t≤T,0<x<π,ux,t=u0x,t,-τ≤t≤0,0≤x≤π,u0,t=uπ,t=0,t≥-τ.Here, the added term h(x,t) and the initial condition u0(x,t) are specified so that the exact solution is u(x,t)=e-tsin⁡(x).We set the parametersr1=1.0, r2=0.5, r3=-1.0, r4=-0.5, τ=0.5, and T=2 and solve problem (43) with different spatial and temporal stepsizes (Δx=π/N and Δt=τ/m).Forθ=0, we let Δt=1e-5 to guarantee that the linear θ-method (8) is asymptotically stable. When the compact θ-method (36) is applied to solve problem (43), we let Δt≈Δx4. The numerical errors and corresponding orders in different sense of norms are displayed in Table 2. Clearly, these results confirm the convergence of the two methods.Table 2
Errors and convergence orders whenθ=0, T=2, and τ=0.5.
Linearθ-method Compactθ-method N L 2-error Order L ∞-error Order N L 2-error Order L ∞-error Order 2 2.40E-02 − 1.91E-02 − 7 1.15E-03 − 8.95E-04 − 4 5.62E-03 2.09 4.48 E - 03 2.09 14 7.31 E - 05 3.98 5.83 E - 05 3.94 8 1.38 E - 03 2.02 1.10 E - 03 2.02 28 4.58 E - 06 4.00 3.65 E - 06 4.00 16 3.44 E - 04 2.01 2.75 E - 04 2.01 56 2.86 E - 07 4.00 2.28 E - 07 4.00 32 8.57 E - 05 2.01 6.84 E - 05 2.01 112 1.79E-08 4.00 1.43 E - 08 4.00Forθ=1/2, when the linear θ-method (8) is applied to solve problem (43), we let Δt≈Δx, and for the compact θ-method (36), we let Δt≈Δx2. For θ=1, when method (8) is applied to solve problem (43), we let Δt≈Δx2, and for method (36), we let Δt≈Δx4. The numerical errors and corresponding orders in different sense of norms are listed in Tables 3 and 4, respectively. It is readily found that these results confirm the convergence of the two methods. Obviously, the compact θ-method gives a better convergence result in the space.Table 3
Errors and convergence orders whenθ=0.5, T=2, and τ=0.5.
Linearθ-method Compactθ-method N L 2-error Order L ∞-error Order L 2-error Order L ∞-error Order 5 1.40E-03 − 1.06E-03 − 7.69E-04 − 5.83 E - 04 − 10 3.81E-04 1.87 3.04E-04 1.80 4.46E-05 4.11 3.56 E - 05 4.03 20 9.67E-05 1.98 7.72E-05 1.98 2.74E-06 4.03 2.19 E - 06 4.03 40 2.61E-05 1.89 2.09E-05 1.89 1.69E-07 4.01 1.35 E - 07 4.01 80 6.15E-06 2.09 4.91E-06 2.09 1.06E-08 4.00 8.44 E - 09 4.00Table 4
Errors and convergence orders whenθ=1, T=2, and τ=0.5.
Linearθ-method Compactθ-method N L 2-error Order L ∞-error Order L 2-error Order L ∞-error Order 5 2.16E-02 − 1.64 E - 02 − 5.45 E - 03 − 4.14 E - 03 − 10 4.34E-03 2.31 3.47 E - 03 2.24 2.96 E - 04 4.20 2.36 E - 04 4.13 20 9.83 E - 04 2.14 7.84 E - 04 2.14 1.82 E - 05 4.03 1.45 E - 05 4.03 40 2.38E-04 2.04 1.90 E - 04 2.04 1.13 E - 06 4.00 9.05 E - 07 4.00 80 5.92E-05 2.01 4.72 E - 05 2.01 7.09 E - 08 4.00 5.65 E - 08 4.00



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*Source: 6402576-2018-06-11.xml*