# Asymptotic Analysis and Error Estimate for Rosenau-Burgers Equation

Authors: Jun Zhang; Zixin Liu; Fubiao Lin; Jianjun Jiao
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

In this work, the asymptotic stability result for Rosenau-Burgers equation is established, under appropriate assumptions on steady state eigenvalue problem and the forcing function. In addition, we propose and analyze a linearized numerical method for solving this nonlinear Rosenau-Burgers equation. We prove that the numerical scheme is unconditionally stable, and the error estimate shows that the numerical method is in the order ofO(Δt2+N2-m), where Δt, N, and m are, respectively, step of time, polynomial degree, and regularity of u. Numerical examples are illustrated to verify the theoretical results.


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## Body

## 1. Introduction

In the research of dynamic dense discrete system, it was shown that the KdV equation can not completely describe the interaction between waves and waves, and in order to overcome the shortcoming of this KdV equation, Rosenau [1, 2] proposed the following Rosenau equation: (1)ut+uxxxxt+ux+uux=0.For further consideration of dissipation in dynamic system, the term -αuxx is included in the above equatoin. The resulting new equation is called the Rosenau-Burgers equation: (2)ut+uxxxxt-αuxx+ux+uux=0.Park [3] discussed existence and uniqueness properties for the Rosenau-Burgers equation. Mei & Liu [4–7] studied the large-time behavior of the solution for the Rosenau-Burgers equation, and they obtained some estimates for the asymptotic solution. Kinami [8] & Mei [9] considered the asymptotic behavior of solution for Benjamin-Bona-Mahony-Burgers equation. They proved that the solution asymptotically converges to 0.Besides the theoretical analysis, some recent contributions focus on using numerical methods to approximate the solutions of Rosenau-Burgers equation. Chung [10] & Sank [11] introduced finite element Galerkin method for solving a Rosenau equation. They obtained the existence and uniqueness of solutions and the error estimates of the solutions are also discussed. Chung [12], Omrani [13], and Feng [14] presented a conservative finite difference scheme for solving Rosenau equation. Convergence, stability, and error estimate of full discrete scheme are also proved. Manickam et al. [15] applied orthogonal cubic spline collocation method to approximate Rosenau equation, and the error estimates are obtained in both L2 norm and L∞ norm.In this paper, we consider the following Rosenau-Burgers equation:(3)ut+uxxxxt-αuxx+ux+uux=fx,t,t∈0,∞,(4)ux,0=u0x,(5)ux,t=ux+2L,t,where α>0.In this article, the convergence of the Rosenau-Burgers equation to its steady-state problem is discussed, under the assumption that the corresponding linearized steady-state eigenvalue problem has a positive minimum eigenvalue and appropriate condition on the forcing function. Moreover, we present numerical method to solve Rosenau-Burgers equation, and the proposed scheme is performed by combining Crank-Nicolson approach in time and Fourier-spectral in space. Our rigorous analysis result shows that the scheme is unconditionally stable, and the numerical method leads to second order in time and spectral accuracy in space.The rest of the article is organized in the following way. Section2 will study asymptotic behaviour of Rosenau-Burgers equation. In Section 3, we will discuss stability and error estimate for the full discrete scheme. In Section 4, we present some numerical experiments to illustrate the validity of the numerical method. The conclusions of this paper are given in Section 5.


## 2. Asymptotic Behaviour of Rosenau-Burgers Equation

In this section, we will investigate the asymptotic behavior of the solution ast→∞. Let limt→∞⁡u(x,t)=u∞(x), where u∞ is the steady state solution of (3)-(5), satisfying(6)-αuxx∞+ux∞+u∞ux∞=f∞,t∈0,∞,(7)u∞x=u∞x+2L,where uxx∞=limt→∞⁡uxx(x,t),ux∞=limt→∞⁡ux(x,t),f∞(x)=limt→∞⁡f(x,t).(A1) Assume the eigenvalue problem,(8)-αϕxx+ϕx+ux∞ϕ=λϕ,(9)ϕx=ϕx+2L,has a minimum eigenvalue λ0>0.Note that, for anyϕ∈H2∩Hper1, it holds that (10)αϕx2+∫02Lux∞ϕ2dx=λϕ2≥λ0ϕ2.Then, we have (11)-αϕxx+ϕx+u∞ϕx+ux∞ϕ,ϕ=αϕx2+12∫02Lux∞ϕ2dx,≥λ02ϕ2+α2ϕx2≥α2ϕx2.Poincaré inequality [16]: For any φ∈Hper2, there holds ∂xkφ≤1/λ1∂xk+1φ,k=0,1, where λ1=π2/L2 is the minimum eigenvalue of the eigenvalue problem: -φxx=λφ, with φ(x)=φ(x+2L).Letz=u-u∞, from (3)-(5) and (6)-(7), we have(12)zt-zxxxxt-αzxx+zx+zzx+u∞zx+zux∞=F,t∈0,∞,(13)zx,0=u0-u∞=z0,(14)zx,t=zx+2L,t,where F=f-f∞.The weak formulation of (12)-(14) is to find z(t)∈Hper2(R), such that (15)zt,v+zxxt,vxx+αzx,vx+zx,v+zzx,v+u∞zx+zux∞,v=F,v,∀v∈Hper2.(A2)u0∈Hper2, and ∫0t‍F(s)2ds≤M, M>0.Theorem 1.
Under assumptions (A1) and (A2), there holds(16)zHper22≤2αL2π2+L4π4∫0tFs2ds+π2+L2π2z0Hper22;moreover, for given ϵ>0, there exists T>0, such that (17)zt2+π2L2zxt2≤ϵ,∀t>T.Proof.
Setv=z in (15), and we obtain (18)12ddtz2+zxx2+αzx2+u∞zx+ux∞z,z=F,z.Apply inequality (11), and Young’s inequality is to arrive at (19)ddtz2+zxx2+αzx2≤2λ1Fzx≤2αλ1F2+α2zx2.Now, integrating with respect to time from 0 to t, we obtain (20)z2+zxx2+α2∫0tzxs2ds≤2αλ1∫0tFs2ds+z02+z0xx2.Applying the Poincaré inequalities, we obtain (21)z2+zxx2=z2+L2π2+L2zxx2+π2π2+L2zxx2≥z2+L2π2+L2π2L2zx2+π2π2+L2zxx2≥π2π2+L2zHper22.Substiuting (21) into (20), we have (22)zHper22≤2αL2π2+L4π4∫0tFs2ds+π2+L2π2z0Hper22.On the other hand, using Poincaré inequality again, we have (23)α2∫0tzxs2ds=α2∫0tπ4π4+L4zxs2+L4π4+L4zxs2ds≥α2∫0tπ4π4+L4zxs2+L2π2π4+L4zs2ds=γ∫0tzs2+π2L2zxs2ds,where γ=απ2L2/2(π4+L4). Substituting (23) into (20) yields (24)zt2+π2L2zxt2+γ∫0tzs2+π2L2zxs2ds≤2αλ1∫0tFs2ds+z0Hper22.That is, (25)limt→∞zt2+π2L2zxt2=0.Hence, we finish the proof of (17).


## 3. Time-Discrete and Error Analysis

In this section, we present the semidiscrete scheme for the solution of (3)-(5). First, we introduce a Crank-Nicholson method to discrete time. Let K be a positive integer, Δt=T/K is the time step, and tn=nΔt,n=0,1,…,K-1 are the mesh points.Consider the following time-discrete scheme:(26)un+1-unΔt+∂xun+1/2-1Δt∂x4un+1-∂x4un+162∂xun+1/23un-un-1+un+1/2∂x3un-un-1-α∂x2un+1/2=0,n≥1.We have the following stability result.Theorem 2.
The semidiscrete scheme (26) is unconditionally stable, such that (27)Eun+1≤Eu0,n=0,1,…,K-1,where (28)Eun=un2+∂x2un2.Proof.
Taking the inner product of (26) with 2Δtun+1/2, we get (29)un+102-un2+∂x2un+12-∂x2un2+2Δtα∂xun+1/22=0.That is, (30)un+12+∂x2un+12≤un2+∂x2un2.This concludes the proof.We will consider Fourier-Galerkin spectral method for the discretization equations (27). We will present some error estimate for full-discretization schemes. First, let us define (31)SN=u∣ux=∑k≤N/2u^kek,ek=e-ikx.Denote πN:L2→SN to be the L2-projection operator which satisfies (32)πNϕ-ϕ,ψ=0,∀ψ∈SN.We also define the H2-projection operator πN1:H2→SN by (33)∂x2πN1ϕ-ϕ,∂x2ψ=0,∀ψ∈SN.We have the following estimate [17]: (34)ϕ-πNϕs≤cNs-mϕm,∀ϕ∈Hm,m≥0.Consider the full-discretization Fourier-Galerkin spectral method to (26) as follows: find uNn+1∈SN, such that (35)1ΔtuNn+1-uNn,ψN+∂xuNn+1/2,ψN-1Δt∂x4uNn+1-∂x4uNn,ψN+162∂xuNn+1/23uNn-uNn-1+uNn+1/2∂x3uNn-uNn-1,ψN-α∂x2uNn+1/2,ψN=0,n≥1,ψN∈SN.Theorem 3.
Let{uNn}n=1M-1 be the solutions of (26), and we derive that (36)uNn+12+∂x2uNn+12≤uN02+∂x2uN02.We denote the truncation errorrn+1/2(x)=r1n+1/2(x)+r2n+1/2(x); here, (37)r1n+1/2x≔ux,tn+1-u·,tnΔt-∂tux,tn+1/2,r2n+1/2x≔∂x4ux,tn+1-∂x4u·,tnΔt-∂x2∂tux,tn+1/2.From Taylor expansion, we have (38)rn+1/202≤cΔt4.We also define the following error functions: (39)e~n=πNutn-uNn,e^n=utn-πNutn,en=e^n+e~n.We show the error estimate of the full-discretization problem (35) in the following theorem.Theorem 4.
Supposeu(x,t) is the exact solution of (3)-(5) and {uNn}n=1M-1 are the solutions of (38); then, we have(40)Eu·,tn-uNn≤cΔt2+N2-m.Proof.
Subtracting (35) from a reformulation of (3) at tn+1/2, we obtain (41)e~Nn+1-e~NnΔt,ψN+∂xe~Nn+1/2,ψN+1Δt∂x2e~Nn+1-∂x2e~Nn,∂xψN+α∂xe~Nn+1/2,∂xψN+utn+1/2∂xutn+1/2-162∂xuNn+1/23uNn-uNn-1+uNn+1/2∂x3uNn-uNn-1,ψN=rn+1/2,ψN+πN1-I∂xu·,tn+1/2,ψN+1ΔtπN1-Iu·,tn+1-u·,tn,ψN+1ΔtπN1-I∂x2u·,tn+1-∂x2u·,tn,ψN,ψN∈SN.Let ψN=2Δte~Nn+1/2, and(42)Ee~n+1-Ee~n≤2Δtrn+1/2e~Nn+1/2+2ΔtπN1-I∂xu·,tn+1e~Nn+1/2+2πN1-Iu·,tn+1-u·,tne~Nn+1/2+2πN1-I∂xu·,tn+1-∂xu·,tn∂xe~Nn+1/2+2Δtu·,tn+1/2∂xu·,tn+1/2-162∂xuNn+13uNn-uNn-1+uNn+1∂x3uNn-uNn-1e~Nn+1≕Q1+Q2+Q3+Q4.Using Cauchy-Schwarz and Young’s inequality, we have (43)Q1≤Δtrn+1/22+ΔtπN1-I∂xu·,tn+1/22+2Δte~Nn+1/22,Q2≤∫tntn+1πN1-I∂tu·,t2dt+Δte~Nn+1/22,Q3≤∫tntn+1πN1-I∂x2∂tu·,t2+Δt∂x2e~Nn+1/22,Q4≤ΔtD12+D22+e~Nn+1/22,where (44)D1=23u·,tn+1/2∂xu·,tn+1/2-133uNn-uNn-1∂xuNn+1/2=23u·,tn+1/2∂xu·,tn+1/2-133u·,tn-u·,tn-1∂xu·,tn+1/2+133u·,tn-u·,tn-1∂xu·,tn+1/2-133uNn-uNn-1∂xu·,tn+1/2+133uNn-uNn-1∂xu·,tn+1/2-133uNn-uNn-1∂xuNn+1/2,D2=13utn+1/2∂xutn+1-13uNn+1/2∂x3uNn-uNn-1=13utn+1/2∂xu·,tn+1-13u·,tn+1/2∂x3u·,tn-u·,tn-1+13u·,tn+1/2∂x3u·,tn-u·,tn-1-13uNn+1/2∂x3u·,tn-u·,tn-1+13uNn+1/2∂x3u·,tn-u·,tn-1-13uNn+1/2∂x3uNn-uNn-1.Applying Taylor expansion and Young’s inequality, (45)13D102≤Δt3∂xu·,tn+1/2∞2∫tntn+1∂t2u·,t02dt+∂xu·,tn+1/2∞23eNn-eNn-12+3uNn-uNn-1∞2∂xeNn+1/22,≤cΔt4+3eNn-eNn-12+∂xeNn+1/22.13D202≤Δt3u·,tn+1/2∞2∫tntn+1∂x∂t2u·,t2dt+∂x3u·,tn-u·,tn-1∞2eNn+1/22+uNn+1/2∞2∂x3eNn-eNn-102≤cΔt4+∂x3eNn-eNn-12+eNn+1/22.Then, we obtain (46)Ee~n+1-Ee~n≤Δtrn+1/22+ΔtπN1-I∂xu·,tn+12+2Δte~Nn+12+∫tntn+1πN1-I∂tu·,t2dt+Δte~Nn+12,+∫tntn+1πN1-I∂x∂tu·,t2dt+Δt∂xe~Nn+12,+cΔt∂x3eNn-eNn-12+3eNn-eNn-12+eNn+1/22+∂xeNn+1/22.Adding up for n=1,…,k and using the following inequality, (47)∂xuNn2≤cuNn2+∂x2uNn2.we have (48)Ee~n+1≤Ee~1+cΔt4+N4-2m+cΔt∑n=1kEe~n,For the first step, it is to verify that (49)Ee~1≤cΔt4+N4-2m.By using Gronwall inequality, we have(50)Ee~n+1≤cΔt4+N4-2m.Therefore, estimate (42) is proved.


## 4. Numerical Results

This section presents several numerical examples to confirm the accuracy and applicability of schemes (35) for solving Rosenau-Burgers equation. Fisrt, we let(51)uNn=∑k=-N/2N/2-1u^kn+1exp⁡-iπkxLAnd we obtain the following linear system: (52)1Δtu^kn+1-u^kn1+πkL4+iπkL+απkL2u^kn+1/2+1623uNn-uNn-1∂xuNn+1/2+∂x3uNn-uNn-1uNn+1/2k=0,where u^k or {u}k represents the kth mode Fourier coefficient of the function u.


### 4.1. Verification of the Temporal Convergence Order

We use the following quantity to compute the convergence rate in time direction [18]: (53)p=log2uNn,2Δt-uN2n,Δt0uN2n,Δt-uN4n,Δt/20.The full discrete problem (52) is solved in Ω=(0,2π) with T=1 and N=60. Tables 1–3 display the temporal convergence orders for different values of α and initial condition u0. As shown in Tables 1–3, our numerical scheme (52) is of the 2nd order accuracy in time, which is confirmed with the theoretical result in Theorem 4.Table 1
Temporal convergence orders foru0=cos⁡x.
α∖Δt Δ t=1.00E-1 Δ t=5.00E-2 Δ t=1.00E-2 Δ t=5.00E-3 α = 0 1.9933 2.0010 2.0009 2.0005 α = 1 1.9428 1.9712 1.9942 1.9972 α = 3 1.8833 1.9262 1.9825 1.9910 α = 5 1.7717 1.8503 1.9629 1.9810Table 2
Temporal convergence orders foru0=sin⁡x+cos⁡x.
α∖Δt Δ t=1.00E-1 Δ t=5.00E-2 Δ t=1.00E-2 Δ t=5.00E-3 α = 0 1.9870 1.9992 2.0007 2.0005 α = 1 1.9329 1.9667 1.9934 1.9968 α = 3 1.8715 1.9187 1.9806 1.9901 α = 5 1.7528 1.8371 1.9593 1.9791Table 3
Temporal convergence orders foru0=sin⁡3xcos⁡2x.
α∖Δt Δ t=1.00E-1 Δ t=5.00E-2 Δ t=1.00E-2 Δ t=5.00E-3 α = 0 1.9959 2.0008 2.0006 2.0004 α = 1 1.9542 1.9767 1.9953 1.9978 α = 3 1.8928 1.9325 1.9840 1.9918 α = 5 1.7906 1.8635 1.9664 1.9828


### 4.2. Asymptotic Properties of Solutions

In order to verify the asymptotic behavior ofz(x,t), the effect of α and initial condition u0 will be investigated. Let α=1 and f=exp⁡-tcos⁡x-exp⁡-tsin⁡x+exp⁡-2tcos⁡xsin⁡x; then, the exact solution is u(x,t)=exp⁡-tsin⁡x, and u∞=f∞=0. The numerical result in Figure 1 shows the convergence of the solution u(x,t) to its steady-state solution u∞. When α=1, and f=exp⁡-t+1cos⁡x-2exp⁡-tcos⁡x-sin⁡xexp⁡-t+1-cos⁡xsin⁡x(exp⁡(-t)+1)2 in (3), the exaction solution is u=exp⁡-t+1cos⁡x, the steady-state solution is u∞=cos⁡x, and f∞=cos⁡x-sin⁡x-sin⁡xcos⁡x. It is easy to verify that both u∞ and F satisfy the assumptions of (A1) and (A2). Figure 2 indicates that the numerical solution of u(x,t) converges to u∞ as t→∞. In Figure 3, we plot the numerical solution at α=0, we can observe that u(x,t) oscillates around 0. In Figures 4-5, we observed that the numerical solutions decay almost to zero when α becomes larger. This implies that the impact of α is significant in the long time behavior. The figures are for z(x,t) indicated in the result discussions.Figure 1
Approximation solutions ofz(x,t) at α=1 and u0=sin⁡x.Figure 2
Approximation solutions ofz(x,t) at α=1 and u0=cos⁡x.Figure 3
The graph of approximation solutions atα=0,Δt=0.01,N=60,u0=sin⁡3xcos⁡2x.Figure 4
The graph of approximation solutions atα=1,Δt=0.01,N=60,u0=sin⁡3xcos⁡2x.Figure 5
The graph of approximation solution atα=5,Δt=0.01,N=60,u0=sin⁡3xcos⁡2x.The reference solutions were obtained by using the numerical scheme (35) with N=128 and δt=5×10-5. We present L2 and H1 errors for u. The results are summarized in Figure 6, in which we conclude that the full-discrete scheme (35) reaches the second-order convergence rate in time direction. In order to test the spatial spectral accuracy, error functions for N are given in Figure 7, which show that our numerical method has good convergence behavior. We observe it reaches spectral convergence in space, if the error in spatial direction is negligible.Figure 6
Errors in time direction withu0=sin⁡3xcos⁡2x.Figure 7
SpatialL2 and H2 errors in space direction with u0=sin⁡3xcos⁡20x.


## 4.1. Verification of the Temporal Convergence Order

We use the following quantity to compute the convergence rate in time direction [18]: (53)p=log2uNn,2Δt-uN2n,Δt0uN2n,Δt-uN4n,Δt/20.The full discrete problem (52) is solved in Ω=(0,2π) with T=1 and N=60. Tables 1–3 display the temporal convergence orders for different values of α and initial condition u0. As shown in Tables 1–3, our numerical scheme (52) is of the 2nd order accuracy in time, which is confirmed with the theoretical result in Theorem 4.Table 1
Temporal convergence orders foru0=cos⁡x.
α∖Δt Δ t=1.00E-1 Δ t=5.00E-2 Δ t=1.00E-2 Δ t=5.00E-3 α = 0 1.9933 2.0010 2.0009 2.0005 α = 1 1.9428 1.9712 1.9942 1.9972 α = 3 1.8833 1.9262 1.9825 1.9910 α = 5 1.7717 1.8503 1.9629 1.9810Table 2
Temporal convergence orders foru0=sin⁡x+cos⁡x.
α∖Δt Δ t=1.00E-1 Δ t=5.00E-2 Δ t=1.00E-2 Δ t=5.00E-3 α = 0 1.9870 1.9992 2.0007 2.0005 α = 1 1.9329 1.9667 1.9934 1.9968 α = 3 1.8715 1.9187 1.9806 1.9901 α = 5 1.7528 1.8371 1.9593 1.9791Table 3
Temporal convergence orders foru0=sin⁡3xcos⁡2x.
α∖Δt Δ t=1.00E-1 Δ t=5.00E-2 Δ t=1.00E-2 Δ t=5.00E-3 α = 0 1.9959 2.0008 2.0006 2.0004 α = 1 1.9542 1.9767 1.9953 1.9978 α = 3 1.8928 1.9325 1.9840 1.9918 α = 5 1.7906 1.8635 1.9664 1.9828


## 4.2. Asymptotic Properties of Solutions

In order to verify the asymptotic behavior ofz(x,t), the effect of α and initial condition u0 will be investigated. Let α=1 and f=exp⁡-tcos⁡x-exp⁡-tsin⁡x+exp⁡-2tcos⁡xsin⁡x; then, the exact solution is u(x,t)=exp⁡-tsin⁡x, and u∞=f∞=0. The numerical result in Figure 1 shows the convergence of the solution u(x,t) to its steady-state solution u∞. When α=1, and f=exp⁡-t+1cos⁡x-2exp⁡-tcos⁡x-sin⁡xexp⁡-t+1-cos⁡xsin⁡x(exp⁡(-t)+1)2 in (3), the exaction solution is u=exp⁡-t+1cos⁡x, the steady-state solution is u∞=cos⁡x, and f∞=cos⁡x-sin⁡x-sin⁡xcos⁡x. It is easy to verify that both u∞ and F satisfy the assumptions of (A1) and (A2). Figure 2 indicates that the numerical solution of u(x,t) converges to u∞ as t→∞. In Figure 3, we plot the numerical solution at α=0, we can observe that u(x,t) oscillates around 0. In Figures 4-5, we observed that the numerical solutions decay almost to zero when α becomes larger. This implies that the impact of α is significant in the long time behavior. The figures are for z(x,t) indicated in the result discussions.Figure 1
Approximation solutions ofz(x,t) at α=1 and u0=sin⁡x.Figure 2
Approximation solutions ofz(x,t) at α=1 and u0=cos⁡x.Figure 3
The graph of approximation solutions atα=0,Δt=0.01,N=60,u0=sin⁡3xcos⁡2x.Figure 4
The graph of approximation solutions atα=1,Δt=0.01,N=60,u0=sin⁡3xcos⁡2x.Figure 5
The graph of approximation solution atα=5,Δt=0.01,N=60,u0=sin⁡3xcos⁡2x.The reference solutions were obtained by using the numerical scheme (35) with N=128 and δt=5×10-5. We present L2 and H1 errors for u. The results are summarized in Figure 6, in which we conclude that the full-discrete scheme (35) reaches the second-order convergence rate in time direction. In order to test the spatial spectral accuracy, error functions for N are given in Figure 7, which show that our numerical method has good convergence behavior. We observe it reaches spectral convergence in space, if the error in spatial direction is negligible.Figure 6
Errors in time direction withu0=sin⁡3xcos⁡2x.Figure 7
SpatialL2 and H2 errors in space direction with u0=sin⁡3xcos⁡20x.


## 5. Conclusion

This article studies the asymptotic stability of Rosenau-Burgers equation. Under suitable assumptions on an eigenvalue problem and the forcing function, the convergence order of the Rosenau-Burgers equation to its steady-state problem is derived. Then, we propose a linearized numerical method for solving this nonlinear Rosenau-Burgers equation. The numerical method is combined with a finite difference scheme in time and Fourier-spectral method in space. We have derived a discrete stability inequality and error estimate for the numerical scheme which leads to2nd-order accuracy in time and spectral accuracy in space. Numerical examples are illustrated to verify the theoretical results.



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