# Two Conservative Difference Schemes for Rosenau-Kawahara Equation

Authors: Jinsong Hu; Youcai Xu; Bing Hu; Xiaoping Xie
Journal: Advances in Mathematical Physics
 (2014)

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## Abstract

Two conservative finite difference schemes for the numerical solution of the initialboundary value problem of Rosenau-Kawahara equation are proposed. The difference schemes simulate two conservative quantities of the problem well. The existence and uniqueness of the difference solution are proved. It is shown that the finite difference schemes are of second-order convergence and unconditionally stable. Numerical experiments verify the theoretical results.


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## Body

## 1. Introduction

In the study of compact discrete systems, the wave-wave and wave-wall interactions cannot be described by the well-known KdV equation. To overcome this shortcoming of KdV equation, Rosenau proposed the following so-called Rosenau equation in [1, 2]:
(1)ut+uxxxxt+ux+uux=0,
which is usually used to describe the dense discrete system and simulate the long-chain transmission model through an L-C flow in radio and computer fields. Park proved the existence and uniqueness of solution to (1) in [3]. However, it is difficult to find its analytical solution. Therefore, the numerical study of (1) is very significant and attract many scholars (see, e.g., [4–10]).As the further consideration of nonlinear wave, Zuo obtained Rosenau-Kawahara equation by adding viscous term+uxxx and -uxxxxx to Rosenau equation (1) and studied the solitary solution and periodic solution of this equation in [10]. In [11], Labidi and Biswas got the integral of Rosenau-Kawahara equation by using He’s principle. Then the solitary solution and two invariance of a generalized Rosenau-Kawahara equation are investigated in [12]. To our best knowledge, there is no study on the numerical method of Rosenau-Kawahara equation. Therefore, we will study the difference approximate solution of Rosenau-Kawahara equation with the initial data and boundary conditions as follows:
(2)ut+uxxxxt+ux+αuux+uxxx-uxxxxx=0,(3)u(x,0)=u0(x),x∈[xL,xR],(4)u(xL,t)=u(xR,t)=0,ux(xL,t)=ux(xR,t)=0,uxx(xL,t)=uxx(xR,t)=0,t∈[0,T],
where α is a known constant and u0(x) is a smooth function. When α=1, the solitary wave solution of (2) is (see [10])
(5)u(x,t)=(-3512+35156205)×sech4[12-13+205(x-113205t)].
The initial boundary value problem (2)–(4) is in accordance with the Cauchy problem of (2) when -xL≫0, xR≫0. Hence, the boundary condition (4) is reasonable. It is easy to verify that (2)-(4) satisfy the following conservative quantities (see [12]):
(6)Q(t)=∫xLxR‍u(x,t)dx=∫xLxR‍u0(x)dx=Q(0),(7)E(t)=∥u∥L22+∥uxx∥L22=E(0),
where Q(0), E(0) are both constants only depending on initial data.It is well known that a reasonable difference scheme has not only high-accuracy but also can maintain some physical properties of original problem. Lots of numerical experiments show that conservative difference scheme can simulate the conservative law of initial problem well since it could avoid the nonlinear blow-up [13–24]. Moreover, it is suitable to compute for long-time. Li and Vu-Quoc pointed in [14] that in some areas, the ability to preserve some invariant properties of the original differential equation is a criterion to judge the success of a numerical simulation. Therefore, constructing the conservative difference scheme is a significant job.The rest of this paper is organized as follows. We respectively propose a two-level nonlinear Crank-Nicolson difference scheme and three-level linear difference scheme for initial boundary value problems (2)–(4) in Sections 2 and 3. We analyze its two discrete conservative laws. The existence and uniqueness of the difference solution are proved. It is shown that the finite difference schemes are of second-order convergence and unconditionally stable. In Section 4, we verify our theoretical analysis by numerical experiments.


## 2. Nonlinear Crank-Nicolson Conservative Difference Scheme

In this section, we propose a two-level nonlinear Crank-Nicolson difference scheme and give the theoretical analysis.In the rest of this paper,C denotes a general positive constant which may denote different value in different occurrence.


### 2.1. Nonlinear Difference Scheme and Its Conservative Law

Leth and τ be the uniform step size in the spatial and temporal direction, respectively. The interval [xL,xR] is divided into J equal parts, where J is a fixed positive integer. Denote h=(xR-xL)/J, xj=xL+jh(j=-2,-1,0,…,J,J+1,J+2), tn=nτ(n=0,1,2,…,N,N=[T/τ]). Let ujn be the difference approximation of u(x,t) at (xj,tn), that is, ujn≈u(xj,tn). Denote Zh0={u=(uj)|u-2=u-1=u0=uJ=uJ+1=uJ+2=0,j=-2,-1,0,1,…,J,J+1,J+2}. We define the difference operators, inner product, and norms that will be used in this paper as follows:
(8)(ujn)x=uj+1n-ujnh,(ujn)x¯=ujn-uj-1nh,(ujn)x^=uj+1n-uj-1n2h,(ujn)t=ujn+1-ujnτ,(ujn)t^=ujn+1-ujn-12τ,ujn+1/2=ujn+1+ujn2,u¯jn=ujn+1+ujn-12,〈un,vn〉=h∑j=1J-1‍ujnvjn,∥un∥2=〈un,un〉,∥un∥∞=max⁡1≤j≤J-1|ujn|.
Consider the following finite difference scheme for problems (2)–(4):
(9)(ujn)t+(ujn)xxx¯x¯t+(ujn+1/2)x^+(ujn+1/2)xx¯x^-(ujn+1/2)xxx¯x¯x^+13α{ujn+1/2(ujn+1/2)x^+[(ujn+1/2)2]x^}=0,2222(j=1,2,…,J-1,n=1,2,…,N-1),(10)uj0=u0(xj),(j=0,1,2,…,J),(11)un∈Zh0,(u0n)x^=(uJn)x^=0,(u0n)xx¯=(uJn)xx¯=0,(n=0,1,2,…,N).
The discrete boundary condition (11) is reasonable from the boundary condition (4).Lemma 1.
For any two discrete functionsu,v∈Zh0, we have
(12)〈ux,v〉=-〈u,vx¯〉,〈uxx¯,v〉=-〈ux,vx〉
from summation by parts (see [22]). Thus,
(13)〈ux,u〉=-〈u,ux¯〉,〈uxx¯,u〉=-〈ux,ux〉=-∥ux∥2.
And if (u0n)xx¯=(uJn)xx¯=0, then
(14)〈uxxx¯x¯,u〉=∥uxx∥2.The following theorem shows how the difference schemes (9)–(11) simulate the conservative law numerically.Theorem 2.
Suppose thatu0∈H02[xL,xR]; then the difference schemes (9)–(11) are conservative for discrete energy; that is,
(15)Qn=h∑j=1J-1‍ujn=Qn-1=⋯=Q0,(16)En=∥un∥2+∥uxxn∥2=En-1=⋯=E0.Proof.
Multiplyingh in the two sides of (9) and summing up for j from 1 to J-1, from boundary (11) and Lemma 1, we obtain
(17)h∑j=1J-1‍ujn+1-ujnτ=0.
From the definition of Qn, (15) is deduced from (17).
Taking the inner of (9) with 2un+1/2 (i.e., un+1+un), from boundary (11) and Lemma 1, we get
(18)∥un∥t2+∥uxxn∥t2+2〈ux^n+1/2,un+1/2〉+2〈uxx¯x^n+1/2,un+1/2〉-2〈uxxx¯x¯x^n+1/2,un+1/2〉+2α〈φ(un+1/2),un+1/2〉=0,
where φ(ujn+1/2)=(1/3){ujn+1/2(ujn+1/2)x^+[(ujn+1/2)2]x^}. On the other hand,
(19)〈ux^n+1/2,un+1/2〉=0,(20)〈uxx¯x^n+1/2,un+1/2〉=0,(21)〈uxxx¯x¯x^n+1/2,un+1/2〉=0,(22)〈φ(un+1/2),un+1/2〉=13h∑j=1J-1{ujn+1/2(ujn+1/2)x^+[(ujn+1/2)2]x^}‍ujn+1/2=13h∑j=1J-1(ujn+1/2)2‍(ujn+1/2)x^+13h∑j=1J-1[(ujn+1/2)2]x^‍ujn+1/2=13h∑j=1J-1‍(ujn+1/2)2(ujn+1/2)x^-13h∑j=1J-1(ujn+1/2)2×(ujn+1/2)x^=0.
Substituting (19)–(22) into (18), we get
(23)(∥un+1∥2-∥un∥2)+(∥uxxn+1∥2-∥uxxn∥2)=0.
From the definition of En, we get (16) by deducing (23).


### 2.2. Solvability of the Difference Scheme

In order to prove the solvability of difference scheme, we present the following Brouwer fixed point theorem [25].Lemma 3 (Brouwer fixed point theorem).
LetH be a finite dimensional inner product space; suppose that g:H→H is continuous and there exists an α>0 such that 〈g(x),x〉>0 for all x∈H with ∥x∥=α. Then there exists x*∈H such that g(x*)=0 and ∥x*∥≤α.Theorem 4.
There existsun∈Zh0(1≤n≤N) that satisfies difference schemes (9)–(11).Proof.
We use the mathematical induction to prove our theorem.
Suppose that there existu0,u1,…,un that satisfy difference scheme (9) for n≤N-1. Next we prove that there exists un+1 that satisfies difference scheme.
Letg be an operator on Zh0 defined by
(24)g(v)=2v-2un+2vxxx¯x¯-2uxxx¯x¯n+τvx^+τvxx¯x^-τvxxx¯x¯x^+ταφ(v).
Taking the inner product of (24) with v, and noticing that
(25)〈vx^,v〉=0,〈vxx¯x^,v〉=0,〈vxxx¯x¯x^,v〉=0,〈φ(v),v〉=0,
we have
(26)〈g(v),v〉=2∥v∥2-2〈un,v〉+2∥vxx∥2-2〈uxxn,vxx〉≥2∥v∥2-2∥un∥·∥v∥+2∥vxx∥2-2∥uxxn∥·∥vxx∥≥2∥v∥2-(∥un∥2+∥v∥2)+2∥vxx∥2-(∥uxxn∥2+∥vxx∥2)≥∥v∥2-(∥un∥2+∥uxxn∥2)+∥vxxn∥2≥∥v∥2-(∥un∥2+∥uxxn∥2).
Therefore, for anyv∈Zh0, if ∥v∥2=∥un∥2+∥uxxn∥2+1, then 〈g(v),v〉>0. From Lemma 3, there exists v*∈Zh0 such that g(v*)=0. Choose un+1=2v*-un. It is easy to verify that un+1 satisfies difference scheme (9).


### 2.3. Convergence, Stability, and Uniqueness of Solution

Suppose thatv(x,t) is the solution to (2)–(4). Denote vjn=v(xj,tn). Then the truncation error of difference schemes (9)–(11) is
(27)rjn=(vjn)t+(vjn)xxx¯x¯t+(vjn+1/2)x^+(vjn+1/2)xx¯x^-(vjn+1/2)xxx¯x¯x^+αφ(vjn+1/2),00(j=1,2,…,J-1,n=1,2,…,N-1),(28)vj0=u0(xj),(j=0,1,2,…,J),(29)hhhhhhvn∈Zh0,(v0n)x^=(vJn)x^=0,(v0n)xx¯=(vJn)xx¯=0,(n=0,1,2,…,N).
By Taylor expansion, we know that rjn=O(τ2+h2) as h,τ→0.Lemma 5.
Suppose thatu0∈H02[xL,xR]. Then the solution of problems (2)–(4) satisfies
(30)∥u∥L2≤C,∥ux∥L2≤C,∥uxx∥L2≤C,∥u∥L∞≤C,∥ux∥L∞≤C.Proof.
It follows from (7) that
(31)∥u∥L2≤C,∥uxx∥L2≤C.
Using Hölder inequality and Cauchy-Schwarz inequality, we can get
(32)∥u∥L22=∫xLxR‍uxuxdx=uux|xLxR-∫xLxR‍uuxxdx=-∫xLxR‍uuxxdx≤∥u∥L2·∥uxx∥L2≤12(∥u∥L22+∥uxx∥L22).
Thus,
(33)∥ux∥L2≤C,
which also yields ∥u∥L∞≤C, ∥ux∥L∞≤C from Sobolev inequality.Lemma 6 (discrete Sobolev inequality [22]).
There exist constantsC1 and C2 such that
(34)∥un∥∞≤C1∥un∥+C2∥uxn∥.Lemma 7 (discrete Gronwall inequality [22]).
Assume thatw(k), ρ(k) are nonnegative gridding functions, and ρ(k) is increasing. If C>0, and for any k(35)w(k)≤ρ(k)+Cτ∑l=0k-1w(l),
then for any k(36)w(k)≤ρ(k)eCτk.Theorem 8.
Suppose thatu0∈H02[xL,xR]. Then the solution of difference schemes (9)–(11) satisfies ∥un∥≤C, ∥uxn∥≤C, ∥uxxn∥≤C thus ∥un∥∞≤C, ∥uxn∥∞≤C(n=1,2,…,N).Proof.
From (16), we have
(37)∥un∥≤C,∥uxxn∥≤C.
From (13) and by Cauchy-Schwarz inequality
(38)∥uxn∥2≤∥un∥·∥uxxn∥≤12(∥un∥2+∥uxxn∥2)≤C.
Thus,
(39)∥uxn∥≤C,
Therefore, from Lemma 6, we obtain ∥un∥∞≤C, ∥uxn∥∞≤C.Remark 9.
Theorem8 implies that the solution of difference schemes (9)–(11) is unconditionally stable in the sense of norm ∥·∥∞.Theorem 10.
Suppose thatu0∈H02[xL,xR]. Then the solution of difference schemes (9)–(11) un converges to the solution of (2)–(4) in the sense of norm ∥·∥∞, and the convergent rate is O(τ2+h2).Proof.
Subtracting (9)–(11) from (27)–(29) and denoting ejn=vjn-ujn, we have
(40)rjn=(ejn)t+(ejn)xxx¯x¯t+(ejn+1/2)x^+(ejn+1/2)xx¯x^-(ejn+1/2)xxx¯x¯x^+αφ(vjn+1/2)-αφ(ujn+1/2),222(j=1,2,…,J-1,n=1,2,…,N-1),(41)ej0=0,(j=0,1,2,…,J),(42)en∈Zh0,(e0n)x^=(eJn)x^=0,(e0n)xx¯=(eJn)xx¯=0,(n=0,1,2,…,N).
Taking the inner product of (40) with 2en+1/2, we get
(43)〈rn,2en+1/2〉=∥en∥t2+∥exxn∥t2+〈ex^n+1/2,2en+1/2〉+〈exx¯x^n+1/2,2en+1/2〉-〈exxx¯x¯x^n+1/2,2en+1/2〉+α〈φ(vn+1/2)-φ(un+1/2),2en+1/2〉.
Similar to (19)–(21), we have
(44)〈ex^n+1/2,2en+1/2〉=0,〈exx¯x^n+1/2,2en+1/2〉=0,〈exxx¯x¯x^n+1/2,2en+1/2〉=0.
So, (43) can be rewritten as
(45)(∥en+1∥2-∥en∥2)+(∥exxn+1∥2-∥exxn∥2)=τ〈rn,2en+1/2〉-ατ〈φ(vn+1/2)-φ(un+1/2),2en+1/2〉.
As
(46)-〈φ(vn+1/2)-φ(un+1/2),2en+1/2〉=-2h3∑j=1J-1[vjn+1/2(vjn+1/2)x^-ujn+1/2(ujn+1/2)x^]‍ejn+1/2-2h3∑j=1J-1[(vjn+1/2)2-(ujn+1/2)2]x^‍ejn+1/2=-2h3∑j=1J-1[vjn+1/2(ejn+1/2)x^+ejn+1/2(ujn+1/2)x^]‍ejn+1/2+2h3∑j=1J-1[ejn+1/2(vjn+1/2+ujn+1/2)]‍(ejn+1/2)x^,
from Lemma 5 and Theorem 8, we obtain
(47)|vjn+1/2|≤C,|ujn+1/2|≤C,|(ujn+1/2)x^|≤C,(j=0,1,2,…,J).
Using Cauchy-Schwarz inequality, we have
(48)-α〈φ(vn+1/2)-φ(un+1/2),2en+1/2〉≤23C|α|h∑j=1J-1‍(|(ejn+1/2)x^|+|ejn+1/2|)|ejn+1/2|+23C|α|h∑j=1J-1‍|ejn+1/2|·|(ejn+1/2)x^|≤C(∥en+1/2∥2+∥exn+1/2∥2)≤C(∥en+1∥2+∥en∥2+∥exn+1∥2+∥exn∥2).
On the other hand,
(49)〈rn,2en+1/2〉=〈rn,en+1+en〉≤∥rn∥2+12(∥en+1∥2+∥en∥2).
Substituting (48) and (49) into (45), we get
(50)(∥en+1∥2-∥en∥2)+(∥exxn+1∥2-∥exxn∥2)≤Cτ(∥en+1∥2+∥en∥2+∥exn+1∥2+∥exn∥2)+τ∥rn∥2.
Similar to the deduction process of (38), one can get
(51)∥exn+1∥2≤12(∥en+1∥2+∥exxn+1∥2),∥exn∥2≤12(∥en∥2+∥exxn∥2).
Therefore, (50) is changed into
(52)(∥en+1∥2-∥en∥2)+(∥exxn+1∥2-∥exxn∥2)≤Cτ[∥en+1∥2+∥en∥2+∥exxn+1∥2+∥exxn∥2]+τ∥rn∥2.
Denoting Bn=∥en∥2+∥exxn∥2, (52) is equivalent to
(53)Bn+1-Bn≤Cτ(Bn+1+Bn)+τ∥rn∥2,
that is,
(54)(1-Cτ)(Bn+1-Bn)≤2CτBn+τ∥rn∥2.
Choosing sufficiently small τ such that 1-Cτ>0, we get
(55)Bn+1-Bn≤CτBn+Cτ∥rn∥2.
Summing up (55) from 0 to n-1, we get
(56)Bn≤B0+Cτ∑l=0n-1‍∥rl∥2+Cτ∑l=0n-1‍Bl.
Noticing that
(57)τ∑l=0n-1‍∥rl∥2≤nτmax⁡0≤l≤n-1∥rl∥2≤T·O(τ2+h2)2,
and B0=O(τ2+h2)2. we have
(58)Bn≤O(τ2+h2)2+Cτ∑l=0n-1Bl.
From Lemma 7, we get
(59)Bn≤O(τ2+h2)2,
that is,
(60)∥en∥≤O(τ2+h2),∥exxn∥≤O(τ2+h2).
Noticing equality (51), one can obtain
(61)∥exn∥≤O(τ2+h2).
From Lemma 6, we have
(62)∥en∥∞≤O(τ2+h2).Similar to Theorem10, we have the following theorems.Theorem 11.
The solution to difference schemes (9)–(11) is unique.


## 2.1. Nonlinear Difference Scheme and Its Conservative Law

Leth and τ be the uniform step size in the spatial and temporal direction, respectively. The interval [xL,xR] is divided into J equal parts, where J is a fixed positive integer. Denote h=(xR-xL)/J, xj=xL+jh(j=-2,-1,0,…,J,J+1,J+2), tn=nτ(n=0,1,2,…,N,N=[T/τ]). Let ujn be the difference approximation of u(x,t) at (xj,tn), that is, ujn≈u(xj,tn). Denote Zh0={u=(uj)|u-2=u-1=u0=uJ=uJ+1=uJ+2=0,j=-2,-1,0,1,…,J,J+1,J+2}. We define the difference operators, inner product, and norms that will be used in this paper as follows:
(8)(ujn)x=uj+1n-ujnh,(ujn)x¯=ujn-uj-1nh,(ujn)x^=uj+1n-uj-1n2h,(ujn)t=ujn+1-ujnτ,(ujn)t^=ujn+1-ujn-12τ,ujn+1/2=ujn+1+ujn2,u¯jn=ujn+1+ujn-12,〈un,vn〉=h∑j=1J-1‍ujnvjn,∥un∥2=〈un,un〉,∥un∥∞=max⁡1≤j≤J-1|ujn|.
Consider the following finite difference scheme for problems (2)–(4):
(9)(ujn)t+(ujn)xxx¯x¯t+(ujn+1/2)x^+(ujn+1/2)xx¯x^-(ujn+1/2)xxx¯x¯x^+13α{ujn+1/2(ujn+1/2)x^+[(ujn+1/2)2]x^}=0,2222(j=1,2,…,J-1,n=1,2,…,N-1),(10)uj0=u0(xj),(j=0,1,2,…,J),(11)un∈Zh0,(u0n)x^=(uJn)x^=0,(u0n)xx¯=(uJn)xx¯=0,(n=0,1,2,…,N).
The discrete boundary condition (11) is reasonable from the boundary condition (4).Lemma 1.
For any two discrete functionsu,v∈Zh0, we have
(12)〈ux,v〉=-〈u,vx¯〉,〈uxx¯,v〉=-〈ux,vx〉
from summation by parts (see [22]). Thus,
(13)〈ux,u〉=-〈u,ux¯〉,〈uxx¯,u〉=-〈ux,ux〉=-∥ux∥2.
And if (u0n)xx¯=(uJn)xx¯=0, then
(14)〈uxxx¯x¯,u〉=∥uxx∥2.The following theorem shows how the difference schemes (9)–(11) simulate the conservative law numerically.Theorem 2.
Suppose thatu0∈H02[xL,xR]; then the difference schemes (9)–(11) are conservative for discrete energy; that is,
(15)Qn=h∑j=1J-1‍ujn=Qn-1=⋯=Q0,(16)En=∥un∥2+∥uxxn∥2=En-1=⋯=E0.Proof.
Multiplyingh in the two sides of (9) and summing up for j from 1 to J-1, from boundary (11) and Lemma 1, we obtain
(17)h∑j=1J-1‍ujn+1-ujnτ=0.
From the definition of Qn, (15) is deduced from (17).
Taking the inner of (9) with 2un+1/2 (i.e., un+1+un), from boundary (11) and Lemma 1, we get
(18)∥un∥t2+∥uxxn∥t2+2〈ux^n+1/2,un+1/2〉+2〈uxx¯x^n+1/2,un+1/2〉-2〈uxxx¯x¯x^n+1/2,un+1/2〉+2α〈φ(un+1/2),un+1/2〉=0,
where φ(ujn+1/2)=(1/3){ujn+1/2(ujn+1/2)x^+[(ujn+1/2)2]x^}. On the other hand,
(19)〈ux^n+1/2,un+1/2〉=0,(20)〈uxx¯x^n+1/2,un+1/2〉=0,(21)〈uxxx¯x¯x^n+1/2,un+1/2〉=0,(22)〈φ(un+1/2),un+1/2〉=13h∑j=1J-1{ujn+1/2(ujn+1/2)x^+[(ujn+1/2)2]x^}‍ujn+1/2=13h∑j=1J-1(ujn+1/2)2‍(ujn+1/2)x^+13h∑j=1J-1[(ujn+1/2)2]x^‍ujn+1/2=13h∑j=1J-1‍(ujn+1/2)2(ujn+1/2)x^-13h∑j=1J-1(ujn+1/2)2×(ujn+1/2)x^=0.
Substituting (19)–(22) into (18), we get
(23)(∥un+1∥2-∥un∥2)+(∥uxxn+1∥2-∥uxxn∥2)=0.
From the definition of En, we get (16) by deducing (23).


## 2.2. Solvability of the Difference Scheme

In order to prove the solvability of difference scheme, we present the following Brouwer fixed point theorem [25].Lemma 3 (Brouwer fixed point theorem).
LetH be a finite dimensional inner product space; suppose that g:H→H is continuous and there exists an α>0 such that 〈g(x),x〉>0 for all x∈H with ∥x∥=α. Then there exists x*∈H such that g(x*)=0 and ∥x*∥≤α.Theorem 4.
There existsun∈Zh0(1≤n≤N) that satisfies difference schemes (9)–(11).Proof.
We use the mathematical induction to prove our theorem.
Suppose that there existu0,u1,…,un that satisfy difference scheme (9) for n≤N-1. Next we prove that there exists un+1 that satisfies difference scheme.
Letg be an operator on Zh0 defined by
(24)g(v)=2v-2un+2vxxx¯x¯-2uxxx¯x¯n+τvx^+τvxx¯x^-τvxxx¯x¯x^+ταφ(v).
Taking the inner product of (24) with v, and noticing that
(25)〈vx^,v〉=0,〈vxx¯x^,v〉=0,〈vxxx¯x¯x^,v〉=0,〈φ(v),v〉=0,
we have
(26)〈g(v),v〉=2∥v∥2-2〈un,v〉+2∥vxx∥2-2〈uxxn,vxx〉≥2∥v∥2-2∥un∥·∥v∥+2∥vxx∥2-2∥uxxn∥·∥vxx∥≥2∥v∥2-(∥un∥2+∥v∥2)+2∥vxx∥2-(∥uxxn∥2+∥vxx∥2)≥∥v∥2-(∥un∥2+∥uxxn∥2)+∥vxxn∥2≥∥v∥2-(∥un∥2+∥uxxn∥2).
Therefore, for anyv∈Zh0, if ∥v∥2=∥un∥2+∥uxxn∥2+1, then 〈g(v),v〉>0. From Lemma 3, there exists v*∈Zh0 such that g(v*)=0. Choose un+1=2v*-un. It is easy to verify that un+1 satisfies difference scheme (9).


## 2.3. Convergence, Stability, and Uniqueness of Solution

Suppose thatv(x,t) is the solution to (2)–(4). Denote vjn=v(xj,tn). Then the truncation error of difference schemes (9)–(11) is
(27)rjn=(vjn)t+(vjn)xxx¯x¯t+(vjn+1/2)x^+(vjn+1/2)xx¯x^-(vjn+1/2)xxx¯x¯x^+αφ(vjn+1/2),00(j=1,2,…,J-1,n=1,2,…,N-1),(28)vj0=u0(xj),(j=0,1,2,…,J),(29)hhhhhhvn∈Zh0,(v0n)x^=(vJn)x^=0,(v0n)xx¯=(vJn)xx¯=0,(n=0,1,2,…,N).
By Taylor expansion, we know that rjn=O(τ2+h2) as h,τ→0.Lemma 5.
Suppose thatu0∈H02[xL,xR]. Then the solution of problems (2)–(4) satisfies
(30)∥u∥L2≤C,∥ux∥L2≤C,∥uxx∥L2≤C,∥u∥L∞≤C,∥ux∥L∞≤C.Proof.
It follows from (7) that
(31)∥u∥L2≤C,∥uxx∥L2≤C.
Using Hölder inequality and Cauchy-Schwarz inequality, we can get
(32)∥u∥L22=∫xLxR‍uxuxdx=uux|xLxR-∫xLxR‍uuxxdx=-∫xLxR‍uuxxdx≤∥u∥L2·∥uxx∥L2≤12(∥u∥L22+∥uxx∥L22).
Thus,
(33)∥ux∥L2≤C,
which also yields ∥u∥L∞≤C, ∥ux∥L∞≤C from Sobolev inequality.Lemma 6 (discrete Sobolev inequality [22]).
There exist constantsC1 and C2 such that
(34)∥un∥∞≤C1∥un∥+C2∥uxn∥.Lemma 7 (discrete Gronwall inequality [22]).
Assume thatw(k), ρ(k) are nonnegative gridding functions, and ρ(k) is increasing. If C>0, and for any k(35)w(k)≤ρ(k)+Cτ∑l=0k-1w(l),
then for any k(36)w(k)≤ρ(k)eCτk.Theorem 8.
Suppose thatu0∈H02[xL,xR]. Then the solution of difference schemes (9)–(11) satisfies ∥un∥≤C, ∥uxn∥≤C, ∥uxxn∥≤C thus ∥un∥∞≤C, ∥uxn∥∞≤C(n=1,2,…,N).Proof.
From (16), we have
(37)∥un∥≤C,∥uxxn∥≤C.
From (13) and by Cauchy-Schwarz inequality
(38)∥uxn∥2≤∥un∥·∥uxxn∥≤12(∥un∥2+∥uxxn∥2)≤C.
Thus,
(39)∥uxn∥≤C,
Therefore, from Lemma 6, we obtain ∥un∥∞≤C, ∥uxn∥∞≤C.Remark 9.
Theorem8 implies that the solution of difference schemes (9)–(11) is unconditionally stable in the sense of norm ∥·∥∞.Theorem 10.
Suppose thatu0∈H02[xL,xR]. Then the solution of difference schemes (9)–(11) un converges to the solution of (2)–(4) in the sense of norm ∥·∥∞, and the convergent rate is O(τ2+h2).Proof.
Subtracting (9)–(11) from (27)–(29) and denoting ejn=vjn-ujn, we have
(40)rjn=(ejn)t+(ejn)xxx¯x¯t+(ejn+1/2)x^+(ejn+1/2)xx¯x^-(ejn+1/2)xxx¯x¯x^+αφ(vjn+1/2)-αφ(ujn+1/2),222(j=1,2,…,J-1,n=1,2,…,N-1),(41)ej0=0,(j=0,1,2,…,J),(42)en∈Zh0,(e0n)x^=(eJn)x^=0,(e0n)xx¯=(eJn)xx¯=0,(n=0,1,2,…,N).
Taking the inner product of (40) with 2en+1/2, we get
(43)〈rn,2en+1/2〉=∥en∥t2+∥exxn∥t2+〈ex^n+1/2,2en+1/2〉+〈exx¯x^n+1/2,2en+1/2〉-〈exxx¯x¯x^n+1/2,2en+1/2〉+α〈φ(vn+1/2)-φ(un+1/2),2en+1/2〉.
Similar to (19)–(21), we have
(44)〈ex^n+1/2,2en+1/2〉=0,〈exx¯x^n+1/2,2en+1/2〉=0,〈exxx¯x¯x^n+1/2,2en+1/2〉=0.
So, (43) can be rewritten as
(45)(∥en+1∥2-∥en∥2)+(∥exxn+1∥2-∥exxn∥2)=τ〈rn,2en+1/2〉-ατ〈φ(vn+1/2)-φ(un+1/2),2en+1/2〉.
As
(46)-〈φ(vn+1/2)-φ(un+1/2),2en+1/2〉=-2h3∑j=1J-1[vjn+1/2(vjn+1/2)x^-ujn+1/2(ujn+1/2)x^]‍ejn+1/2-2h3∑j=1J-1[(vjn+1/2)2-(ujn+1/2)2]x^‍ejn+1/2=-2h3∑j=1J-1[vjn+1/2(ejn+1/2)x^+ejn+1/2(ujn+1/2)x^]‍ejn+1/2+2h3∑j=1J-1[ejn+1/2(vjn+1/2+ujn+1/2)]‍(ejn+1/2)x^,
from Lemma 5 and Theorem 8, we obtain
(47)|vjn+1/2|≤C,|ujn+1/2|≤C,|(ujn+1/2)x^|≤C,(j=0,1,2,…,J).
Using Cauchy-Schwarz inequality, we have
(48)-α〈φ(vn+1/2)-φ(un+1/2),2en+1/2〉≤23C|α|h∑j=1J-1‍(|(ejn+1/2)x^|+|ejn+1/2|)|ejn+1/2|+23C|α|h∑j=1J-1‍|ejn+1/2|·|(ejn+1/2)x^|≤C(∥en+1/2∥2+∥exn+1/2∥2)≤C(∥en+1∥2+∥en∥2+∥exn+1∥2+∥exn∥2).
On the other hand,
(49)〈rn,2en+1/2〉=〈rn,en+1+en〉≤∥rn∥2+12(∥en+1∥2+∥en∥2).
Substituting (48) and (49) into (45), we get
(50)(∥en+1∥2-∥en∥2)+(∥exxn+1∥2-∥exxn∥2)≤Cτ(∥en+1∥2+∥en∥2+∥exn+1∥2+∥exn∥2)+τ∥rn∥2.
Similar to the deduction process of (38), one can get
(51)∥exn+1∥2≤12(∥en+1∥2+∥exxn+1∥2),∥exn∥2≤12(∥en∥2+∥exxn∥2).
Therefore, (50) is changed into
(52)(∥en+1∥2-∥en∥2)+(∥exxn+1∥2-∥exxn∥2)≤Cτ[∥en+1∥2+∥en∥2+∥exxn+1∥2+∥exxn∥2]+τ∥rn∥2.
Denoting Bn=∥en∥2+∥exxn∥2, (52) is equivalent to
(53)Bn+1-Bn≤Cτ(Bn+1+Bn)+τ∥rn∥2,
that is,
(54)(1-Cτ)(Bn+1-Bn)≤2CτBn+τ∥rn∥2.
Choosing sufficiently small τ such that 1-Cτ>0, we get
(55)Bn+1-Bn≤CτBn+Cτ∥rn∥2.
Summing up (55) from 0 to n-1, we get
(56)Bn≤B0+Cτ∑l=0n-1‍∥rl∥2+Cτ∑l=0n-1‍Bl.
Noticing that
(57)τ∑l=0n-1‍∥rl∥2≤nτmax⁡0≤l≤n-1∥rl∥2≤T·O(τ2+h2)2,
and B0=O(τ2+h2)2. we have
(58)Bn≤O(τ2+h2)2+Cτ∑l=0n-1Bl.
From Lemma 7, we get
(59)Bn≤O(τ2+h2)2,
that is,
(60)∥en∥≤O(τ2+h2),∥exxn∥≤O(τ2+h2).
Noticing equality (51), one can obtain
(61)∥exn∥≤O(τ2+h2).
From Lemma 6, we have
(62)∥en∥∞≤O(τ2+h2).Similar to Theorem10, we have the following theorems.Theorem 11.
The solution to difference schemes (9)–(11) is unique.


## 3. A Linear Conservative Difference Scheme

In this section, we propose a three-level linear conservative difference scheme for (2)–(4) and give the theoretical analysis.


### 3.1. Linear Difference Scheme and Its Conservative Law

Consider the following finite difference scheme for problems (2)–(4):
(63)(ujn)t^+(ujn)xxx¯x¯t^+(u¯jn)x^+(u¯jn)xx¯x^-(u¯jn)xxx¯x¯x^+13α[ujn(u¯jn)x^+(ujnu¯jn)x^]=0,22222(j=1,2,…,J-1,n=1,2,…,N-1),(64)uj0=u0(xj),(j=0,1,2,…,J),(65)un∈Zh0,(u0n)x^=(uJn)x^=0,(u0n)xx¯=(uJn)xx¯=0,(n=0,1,2,…,N).The discrete boundary condition (65) is reasonable from the boundary condition (4).The following theorem shows how the difference schemes (63)–(65) simulate the conservative law as follows.Theorem 12.
Suppose thatu0∈H02[xL,xR]; then the difference schemes (63)–(65) are conservative for discrete energy; that is,
(66)Qn=h2∑j=1J-1‍(ujn+1+ujn)+h6τ∑j=1J-1‍ujn(ujn+1)x^=Qn-1=⋯=Q0,(67)En=12(∥un+1∥2+∥un∥2)+12(∥uxxn+1∥2+∥uxxn∥2)=En-1=⋯=E0.Proof.
Multiplyingh in both sides of (63) and summing up for j, from boundary (65) and Lemma 1, we obtain
(68)h∑j=1J-1‍ujn+1-ujn-12τ+13h∑j=1J-1‍ujn(u¯jn)x^=0.
Notice that
(69)h∑j=1J-1‍ujn(u¯jn)x^=h2∑j=1J-1‍ujn(ujn+1)x^+h2∑j=1J-1‍ujn(ujn-1)x^=h2∑j=1J-1‍ujn(ujn+1)x^-h2∑j=1J-1‍ujn-1(ujn)x^.
From the definition ofQn, (66) is deduced from (68) and (69).
Taking the inner product of (63) with 2u¯n, from boundary (65) and Lemma 1, we obtain
(70)∥un∥t^2+∥uxxn∥t^2+2〈u¯x^n,u¯n〉+2〈u¯xx¯x^n,u¯n〉-2〈u¯xxx¯x¯x^n,u¯n〉+2α〈ψ(un,u¯n),un〉=0,
where ψ(ujn,u¯jn)=(1/3)[ujn(u¯jn)x^+(ujnu¯jn)x^]. On the other hand,
(71)〈u¯x^n,u¯n〉=0,(72)〈u¯xx¯x^n,u¯n〉=0,(73)〈u¯xxx¯x¯x^n,u¯n〉=0,(74)〈ψ(un,u¯n),u¯n〉=13h∑j=1J-1[ujn(u¯jn)x^+(ujnu¯jn)x^]‍u¯jn=13h∑j=1J-1ujnu¯jn(u¯jn)x^‍+13h∑j=1J-1(ujnu¯jn)x^‍u¯jn=13h∑j=1J-1‍ujnu¯jn(u¯jn)x^-13h∑j=1J-1‍ujnu¯jn(u¯jn)x^=0.
Substituting (71)–(74) into (70), we obtain
(75)(∥un+1∥2-∥un-1∥2)+(∥uxxn+1∥2-∥uxxn-1∥2)=0.
From the definition of En, we get (67) by deducing (75).


### 3.2. Solvability of the Difference Scheme

Theorem 13.
There exists unique solution for difference schemes (63)–(65).Proof.
Use the mathematical induction to prove it. It is obvious thatu0 is uniquely determined by the initial condition (64). We also can get u1 by (9)–(11). Now suppose u0,u1,…,un(n≤N-1) is solved uniquely. Consider the equation of (63) for un+1(76)12τujn+1+12τ(ujn+1)xxx¯x¯+12(ujn+1)x^+12(ujn+1)xx¯x^-12(ujn+1)xxx¯x¯x^+12αψ(ujn,ujn+1)=0.
Taking the inner product of (76) with un+1, from boundary condition (65) and Lemma 1, we get
(77)12τ∥un+1∥2+∥uxxn+1∥2+12〈ux^n+1,un+1〉+12〈uxx¯x^n+1,un+1〉-12〈uxxx¯x¯x^n+1,un+1〉+12α〈ψ(un,un+1),un+1〉=0.
Similar to (71)–(73), we have
(78)〈ux^n+1,un+1〉=0,(79)〈uxx¯x^n+1,un+1〉=0,(80)〈uxxx¯x¯x^n+1,un+1〉=0.
Notice that
(81)12〈ψ(un,un+1),un+1〉=16h∑j=1J-1[ujn(ujn+1)x^+(ujnujn+1)x^]ujn+1‍=16h∑j=1J-1ujnujn+1(ujn+1)x^‍+16h∑j=1J-1(ujnujn+1)x^‍ujn+1=16h∑j=1J-1‍ujnujn+1(ujn+1)x^-16h∑j=1J-1‍ujnujn+1(ujn+1)x^=0.
Substituting (78)–(81) into (77), we get
(82)∥un+1∥2+∥uxxn+1∥2=0;
that is, (76) only admits zero solution. Therefore, there exists a unique ujn+1 that satisfies (63).


### 3.3. Convergence and Stability of the Difference Scheme

Suppose thatv(x,t) is the solution to (2)–(4). Denote vjn=v(xj,tn). Then the truncation error of difference schemes (63)–(65) is as follows:
(83)rjn=(vjn)t^+(vjn)xxx¯x¯t^+(v¯jn)x^+(v¯jn)xx¯x^-(v¯jn)xxx¯x¯x^+αψ(vjn,v¯jn),hhh(j=1,2,…,J-1,n=1,2,…,N-1),(84)vj0=u0(xj),(j=0,1,2,…,J),(85)vn∈Zh0,(v0n)x^=(vJn)x^=0,(v0n)xx¯=(vJn)xx¯=0,(n=0,1,2,…,N).
By Taylor expansion, we know that rjn=O(τ2+h2) as h,τ→0.Theorem 14.
Suppose thatu0∈H02[xL,xR]. Then the solution of difference schemes (63)–(65) satisfies ∥un∥≤C, ∥uxn∥≤C, ∥uxxn∥≤C; thus ∥un∥∞≤C, ∥uxn∥∞≤C(n=1,2,…,N).Proof.
From (67), we know that
(86)∥un∥≤C,∥uxxn∥≤C.
By (38), one can get
(87)∥uxn∥≤C.
Then, from Lemma 6, we get ∥un∥∞≤C, ∥uxn∥∞≤C.Remark 15.
Theorem14 implies that the solution of difference schemes (63)–(65) is unconditionally stable in the sense of norm ∥·∥∞.Theorem 16.
Suppose thatu0∈H02[xL,xR]. Then the solution of difference schemes (63)–(65) un converges to the solution of (2)–(4) in the sense of norm ∥·∥∞, and the convergent order is O(τ2+h2).Proof.
Subtracting (63)–(65) from (83)–(85) and letting ejn=vjn-ujn, we have
(88)rjn=(ejn)t^+(ejn)xxx¯x¯t^+(e¯jn)x^+(e¯jn)xx¯x^-(e¯jn)xxx¯x¯x^+αψ(vjn,v¯jn)-αψ(ujn,u¯jn),(j=1,2,…,J-1,n=1,2,…,N-1),(89)ej0=u0(xj),(j=0,1,2,…,J),(90)en∈Zh0,(e0n)x^=(eJn)x^=0,(e0n)xx¯=(eJn)xx¯=0,(n=0,1,2,…,N).
Taking the inner product of the two sides of (88) with 2e¯n, and using boundary condition (90) and Lemma 1, we obtain
(91)〈rn,2e¯n〉=∥en∥t^2+∥exxn∥t^2+〈e¯x^n,2e¯n〉+〈e¯xx¯x^n,2e¯n〉-〈e¯xxx¯x¯x^n,2e¯n〉+α〈ψ(vn,v¯n)-ψ(un,u¯n),2e¯n〉.
Similar to (71)–(73), we have
(92)〈e¯x^n,2e¯n〉=0,〈e¯xx¯x^n,2e¯n〉=0,〈e¯xxx¯x¯x^n,2e¯n〉=0.
Then (91) is changed into
(93)(∥en+1∥2-∥en-1∥2)+(∥exxn+1∥2-∥exxn-1∥2)=2τ〈rn,2e¯n〉-2ατ〈ψ(vn,v¯n)-ψ(un,u¯n),2e¯n〉.
By Lemma 5 and Theorem 14, it is shown that
(94)|v¯jn|≤C,|ujn|≤C,|(ujn)x^|≤C,(j=0,1,2,…,J).
Using Cauchy-Schwarz inequality, we have
(95)-α〈ψ(vn,v¯n)-ψ(un,u¯n),2e¯n〉=-2h3α∑j=1J-1[vjn(v¯jn)x^-ujn(u¯jn)x^]‍e¯jn-2h3α∑j=1J-1[(vjnv¯jn)x^-(ujnu¯jn)x^]‍e¯jn=-2h3α∑j=1J-1[vjn(v¯jn)x^-ujn(u¯jn)x^]‍e¯jn+2h3α∑j=1J-1‍(vjnv¯jn-ujnu¯jn)(e¯jn)x^=-2h3α∑j=1J-1[ejn(v¯jn)x^+ujn(e¯jn)x^]‍e¯jn+2h3α∑j=1J-1(ejnv¯jn+ujne¯jn)(ejn)x^‍≤C(∥en∥2+∥e¯n∥2+∥e¯xn∥2)≤C(∥en+1∥2+∥en∥2+∥en-1∥2+∥exn+1∥2+∥exn-1∥2).
On the other hand,
(96)〈rn,2e¯n〉=〈rn,en+1+en-1〉≤∥rn∥2+12(∥en+1∥2+∥en-1∥2).
Substituting (95)-(96) into (93), we could obtain
(97)(∥en+1∥2-∥en-1∥2)+(∥exxn+1∥2-∥exxn-1∥2)≤Cτ(∥en+1∥2+∥en∥2+∥en-1∥2+∥exn+1∥20000000+∥exn∥2+∥exn-1∥2)+2τ∥rn∥2.
Using the deduce process which is similar to (38), we have
(98)∥exn+1∥2≤12(∥en+1∥2+∥exxn+1∥2),∥exn∥2≤12(∥en∥2+∥exxn∥2),∥exn-1∥2≤12(∥en-1∥2+∥exxn-1∥2).
Then (97) is changed into
(99)(∥en+1∥2-∥en-1∥2)+(∥exxn+1∥2-∥exxn-1∥2)≤Cτ[∥en+1∥2+∥en∥2+∥en-1∥2+∥exxn+1∥2000000+∥exxn∥2+∥exxn-1∥2]+2τ∥rn∥2.
Let Bn=∥en+1∥2+∥exxn+1∥2+∥en∥2+∥exxn∥2, (99) can be rewritten as follows:
(100)Bn-Bn-1≤Cτ(Bn+Bn-1)+2τ∥rn∥2,
that is,
(101)(1-Cτ)(Bn-Bn-1)≤2CτBn-1+2τ∥rn∥2.
By choosing τ that is small enough such that 1-Cτ>0, then
(102)Bn-Bn-1≤CτBn-1+Cτ∥rn∥2.
Summing up (102) from 1 to n, we obtain
(103)Bn≤B0+Cτ∑l=1n‍∥rl∥2+Cτ∑l=0n-1‍Bl.
Notice that
(104)τ∑l=0n-1‍∥rl∥2≤nτmax⁡0≤l≤n-1∥rl∥2≤T·O(τ2+h2)2.
Firstly by selecting other two-level scheme with second-order convergence (such as the scheme (9)–(11)) to get u1, we have
(105)B0=O(τ2+h2)2.
Therefore
(106)Bn≤O(τ2+h2)2+Cτ∑l=0n-1‍Bl.
Using Lemma 7,we obtain
(107)Bn≤O(τ2+h2)2,
that is,
(108)∥en∥≤O(τ2+h2),∥exxn∥≤O(τ2+h2).
Combining with (98), we get
(109)∥exn∥≤O(τ2+h2).
Then by Lemma 6, it is shown that
(110)∥en∥∞≤O(τ2+h2).


## 3.1. Linear Difference Scheme and Its Conservative Law

Consider the following finite difference scheme for problems (2)–(4):
(63)(ujn)t^+(ujn)xxx¯x¯t^+(u¯jn)x^+(u¯jn)xx¯x^-(u¯jn)xxx¯x¯x^+13α[ujn(u¯jn)x^+(ujnu¯jn)x^]=0,22222(j=1,2,…,J-1,n=1,2,…,N-1),(64)uj0=u0(xj),(j=0,1,2,…,J),(65)un∈Zh0,(u0n)x^=(uJn)x^=0,(u0n)xx¯=(uJn)xx¯=0,(n=0,1,2,…,N).The discrete boundary condition (65) is reasonable from the boundary condition (4).The following theorem shows how the difference schemes (63)–(65) simulate the conservative law as follows.Theorem 12.
Suppose thatu0∈H02[xL,xR]; then the difference schemes (63)–(65) are conservative for discrete energy; that is,
(66)Qn=h2∑j=1J-1‍(ujn+1+ujn)+h6τ∑j=1J-1‍ujn(ujn+1)x^=Qn-1=⋯=Q0,(67)En=12(∥un+1∥2+∥un∥2)+12(∥uxxn+1∥2+∥uxxn∥2)=En-1=⋯=E0.Proof.
Multiplyingh in both sides of (63) and summing up for j, from boundary (65) and Lemma 1, we obtain
(68)h∑j=1J-1‍ujn+1-ujn-12τ+13h∑j=1J-1‍ujn(u¯jn)x^=0.
Notice that
(69)h∑j=1J-1‍ujn(u¯jn)x^=h2∑j=1J-1‍ujn(ujn+1)x^+h2∑j=1J-1‍ujn(ujn-1)x^=h2∑j=1J-1‍ujn(ujn+1)x^-h2∑j=1J-1‍ujn-1(ujn)x^.
From the definition ofQn, (66) is deduced from (68) and (69).
Taking the inner product of (63) with 2u¯n, from boundary (65) and Lemma 1, we obtain
(70)∥un∥t^2+∥uxxn∥t^2+2〈u¯x^n,u¯n〉+2〈u¯xx¯x^n,u¯n〉-2〈u¯xxx¯x¯x^n,u¯n〉+2α〈ψ(un,u¯n),un〉=0,
where ψ(ujn,u¯jn)=(1/3)[ujn(u¯jn)x^+(ujnu¯jn)x^]. On the other hand,
(71)〈u¯x^n,u¯n〉=0,(72)〈u¯xx¯x^n,u¯n〉=0,(73)〈u¯xxx¯x¯x^n,u¯n〉=0,(74)〈ψ(un,u¯n),u¯n〉=13h∑j=1J-1[ujn(u¯jn)x^+(ujnu¯jn)x^]‍u¯jn=13h∑j=1J-1ujnu¯jn(u¯jn)x^‍+13h∑j=1J-1(ujnu¯jn)x^‍u¯jn=13h∑j=1J-1‍ujnu¯jn(u¯jn)x^-13h∑j=1J-1‍ujnu¯jn(u¯jn)x^=0.
Substituting (71)–(74) into (70), we obtain
(75)(∥un+1∥2-∥un-1∥2)+(∥uxxn+1∥2-∥uxxn-1∥2)=0.
From the definition of En, we get (67) by deducing (75).


## 3.2. Solvability of the Difference Scheme

Theorem 13.
There exists unique solution for difference schemes (63)–(65).Proof.
Use the mathematical induction to prove it. It is obvious thatu0 is uniquely determined by the initial condition (64). We also can get u1 by (9)–(11). Now suppose u0,u1,…,un(n≤N-1) is solved uniquely. Consider the equation of (63) for un+1(76)12τujn+1+12τ(ujn+1)xxx¯x¯+12(ujn+1)x^+12(ujn+1)xx¯x^-12(ujn+1)xxx¯x¯x^+12αψ(ujn,ujn+1)=0.
Taking the inner product of (76) with un+1, from boundary condition (65) and Lemma 1, we get
(77)12τ∥un+1∥2+∥uxxn+1∥2+12〈ux^n+1,un+1〉+12〈uxx¯x^n+1,un+1〉-12〈uxxx¯x¯x^n+1,un+1〉+12α〈ψ(un,un+1),un+1〉=0.
Similar to (71)–(73), we have
(78)〈ux^n+1,un+1〉=0,(79)〈uxx¯x^n+1,un+1〉=0,(80)〈uxxx¯x¯x^n+1,un+1〉=0.
Notice that
(81)12〈ψ(un,un+1),un+1〉=16h∑j=1J-1[ujn(ujn+1)x^+(ujnujn+1)x^]ujn+1‍=16h∑j=1J-1ujnujn+1(ujn+1)x^‍+16h∑j=1J-1(ujnujn+1)x^‍ujn+1=16h∑j=1J-1‍ujnujn+1(ujn+1)x^-16h∑j=1J-1‍ujnujn+1(ujn+1)x^=0.
Substituting (78)–(81) into (77), we get
(82)∥un+1∥2+∥uxxn+1∥2=0;
that is, (76) only admits zero solution. Therefore, there exists a unique ujn+1 that satisfies (63).


## 3.3. Convergence and Stability of the Difference Scheme

Suppose thatv(x,t) is the solution to (2)–(4). Denote vjn=v(xj,tn). Then the truncation error of difference schemes (63)–(65) is as follows:
(83)rjn=(vjn)t^+(vjn)xxx¯x¯t^+(v¯jn)x^+(v¯jn)xx¯x^-(v¯jn)xxx¯x¯x^+αψ(vjn,v¯jn),hhh(j=1,2,…,J-1,n=1,2,…,N-1),(84)vj0=u0(xj),(j=0,1,2,…,J),(85)vn∈Zh0,(v0n)x^=(vJn)x^=0,(v0n)xx¯=(vJn)xx¯=0,(n=0,1,2,…,N).
By Taylor expansion, we know that rjn=O(τ2+h2) as h,τ→0.Theorem 14.
Suppose thatu0∈H02[xL,xR]. Then the solution of difference schemes (63)–(65) satisfies ∥un∥≤C, ∥uxn∥≤C, ∥uxxn∥≤C; thus ∥un∥∞≤C, ∥uxn∥∞≤C(n=1,2,…,N).Proof.
From (67), we know that
(86)∥un∥≤C,∥uxxn∥≤C.
By (38), one can get
(87)∥uxn∥≤C.
Then, from Lemma 6, we get ∥un∥∞≤C, ∥uxn∥∞≤C.Remark 15.
Theorem14 implies that the solution of difference schemes (63)–(65) is unconditionally stable in the sense of norm ∥·∥∞.Theorem 16.
Suppose thatu0∈H02[xL,xR]. Then the solution of difference schemes (63)–(65) un converges to the solution of (2)–(4) in the sense of norm ∥·∥∞, and the convergent order is O(τ2+h2).Proof.
Subtracting (63)–(65) from (83)–(85) and letting ejn=vjn-ujn, we have
(88)rjn=(ejn)t^+(ejn)xxx¯x¯t^+(e¯jn)x^+(e¯jn)xx¯x^-(e¯jn)xxx¯x¯x^+αψ(vjn,v¯jn)-αψ(ujn,u¯jn),(j=1,2,…,J-1,n=1,2,…,N-1),(89)ej0=u0(xj),(j=0,1,2,…,J),(90)en∈Zh0,(e0n)x^=(eJn)x^=0,(e0n)xx¯=(eJn)xx¯=0,(n=0,1,2,…,N).
Taking the inner product of the two sides of (88) with 2e¯n, and using boundary condition (90) and Lemma 1, we obtain
(91)〈rn,2e¯n〉=∥en∥t^2+∥exxn∥t^2+〈e¯x^n,2e¯n〉+〈e¯xx¯x^n,2e¯n〉-〈e¯xxx¯x¯x^n,2e¯n〉+α〈ψ(vn,v¯n)-ψ(un,u¯n),2e¯n〉.
Similar to (71)–(73), we have
(92)〈e¯x^n,2e¯n〉=0,〈e¯xx¯x^n,2e¯n〉=0,〈e¯xxx¯x¯x^n,2e¯n〉=0.
Then (91) is changed into
(93)(∥en+1∥2-∥en-1∥2)+(∥exxn+1∥2-∥exxn-1∥2)=2τ〈rn,2e¯n〉-2ατ〈ψ(vn,v¯n)-ψ(un,u¯n),2e¯n〉.
By Lemma 5 and Theorem 14, it is shown that
(94)|v¯jn|≤C,|ujn|≤C,|(ujn)x^|≤C,(j=0,1,2,…,J).
Using Cauchy-Schwarz inequality, we have
(95)-α〈ψ(vn,v¯n)-ψ(un,u¯n),2e¯n〉=-2h3α∑j=1J-1[vjn(v¯jn)x^-ujn(u¯jn)x^]‍e¯jn-2h3α∑j=1J-1[(vjnv¯jn)x^-(ujnu¯jn)x^]‍e¯jn=-2h3α∑j=1J-1[vjn(v¯jn)x^-ujn(u¯jn)x^]‍e¯jn+2h3α∑j=1J-1‍(vjnv¯jn-ujnu¯jn)(e¯jn)x^=-2h3α∑j=1J-1[ejn(v¯jn)x^+ujn(e¯jn)x^]‍e¯jn+2h3α∑j=1J-1(ejnv¯jn+ujne¯jn)(ejn)x^‍≤C(∥en∥2+∥e¯n∥2+∥e¯xn∥2)≤C(∥en+1∥2+∥en∥2+∥en-1∥2+∥exn+1∥2+∥exn-1∥2).
On the other hand,
(96)〈rn,2e¯n〉=〈rn,en+1+en-1〉≤∥rn∥2+12(∥en+1∥2+∥en-1∥2).
Substituting (95)-(96) into (93), we could obtain
(97)(∥en+1∥2-∥en-1∥2)+(∥exxn+1∥2-∥exxn-1∥2)≤Cτ(∥en+1∥2+∥en∥2+∥en-1∥2+∥exn+1∥20000000+∥exn∥2+∥exn-1∥2)+2τ∥rn∥2.
Using the deduce process which is similar to (38), we have
(98)∥exn+1∥2≤12(∥en+1∥2+∥exxn+1∥2),∥exn∥2≤12(∥en∥2+∥exxn∥2),∥exn-1∥2≤12(∥en-1∥2+∥exxn-1∥2).
Then (97) is changed into
(99)(∥en+1∥2-∥en-1∥2)+(∥exxn+1∥2-∥exxn-1∥2)≤Cτ[∥en+1∥2+∥en∥2+∥en-1∥2+∥exxn+1∥2000000+∥exxn∥2+∥exxn-1∥2]+2τ∥rn∥2.
Let Bn=∥en+1∥2+∥exxn+1∥2+∥en∥2+∥exxn∥2, (99) can be rewritten as follows:
(100)Bn-Bn-1≤Cτ(Bn+Bn-1)+2τ∥rn∥2,
that is,
(101)(1-Cτ)(Bn-Bn-1)≤2CτBn-1+2τ∥rn∥2.
By choosing τ that is small enough such that 1-Cτ>0, then
(102)Bn-Bn-1≤CτBn-1+Cτ∥rn∥2.
Summing up (102) from 1 to n, we obtain
(103)Bn≤B0+Cτ∑l=1n‍∥rl∥2+Cτ∑l=0n-1‍Bl.
Notice that
(104)τ∑l=0n-1‍∥rl∥2≤nτmax⁡0≤l≤n-1∥rl∥2≤T·O(τ2+h2)2.
Firstly by selecting other two-level scheme with second-order convergence (such as the scheme (9)–(11)) to get u1, we have
(105)B0=O(τ2+h2)2.
Therefore
(106)Bn≤O(τ2+h2)2+Cτ∑l=0n-1‍Bl.
Using Lemma 7,we obtain
(107)Bn≤O(τ2+h2)2,
that is,
(108)∥en∥≤O(τ2+h2),∥exxn∥≤O(τ2+h2).
Combining with (98), we get
(109)∥exn∥≤O(τ2+h2).
Then by Lemma 6, it is shown that
(110)∥en∥∞≤O(τ2+h2).


## 4. Numerical Simulations

In this section, letα=1, xL=-40, xR=100, T=40, and
(111)u0(x)=(-3512+35156205)sech4(112-13+205x).
We denote the nonlinear Crank-Nicolson conservative difference schemes (9)–(11) as Scheme I and the linear three-level conservative difference schemes (63)–(65) as Scheme II. For some different values of τ and h, we list errors of Scheme I and Scheme II at several different time in Tables 1 and 2, respectively. We verify the two-order accuracy of the difference scheme in Tables 3 and 4 by using the method of [26, 27]. The numerical simulation of two conservative quantities (6) and (7) is listed in Tables 5 and 6.Table 1
The errors of Scheme I at various time steps with varioush and τ.
τ
=
h
=
0.1
τ
=
h
=
0.05
τ
=
h
=
0.025
∥
e
n
∥
∥
e
n
∥
∞
∥
e
n
∥
∥
e
n
∥
∞
∥
e
n
∥
∥
e
n
∥
∞
t
=
10
2.159730
e
-
4
7.520810
e
-
5
5.401591
e
-
5
1.880684
e
-
5
1.354230
e
-
5
4.702168
e
-
6
t
=
20
4.160331
e
-
4
1.421127
e
-
4
1.040460
e
-
4
3.554032
e
-
5
2.603381
e
-
5
8.886828
e
-
6
t
=
30
5.977491
e
-
4
1.996835
e
-
4
1.494989
e
-
4
4.994370
e
-
5
3.739276
e
-
5
1.248752
e
-
5
t
=
40
7.633356
e
-
4
2.502663
e
-
4
1.909241
e
-
4
6.260150
e
-
5
4.774838
e
-
5
1.565301
e
-
5Table 2
The errors of Scheme II at various time steps with varioush and τ.
τ
=
h
=
0.1
τ
=
h
=
0.05
τ
=
h
=
0.025
∥
e
n
∥
∥
e
n
∥
∞
∥
e
n
∥
∥
e
n
∥
∞
∥
e
n
∥
∥
e
n
∥
∞
t
=
10
4.302763
e
-
4
1.477136
e
-
4
1.076811
e
-
4
3.697042
e
-
5
2.693880
e
-
5
9.259288
e
-
6
t
=
20
8.320777
e
-
4
2.820308
e
-
4
2.082744
e
-
4
7.060528
e
-
5
5.207086
e
-
5
1.767754
e
-
5
t
=
30
1.199482
e
-
3
3.985869
e
-
4
3.002761
e
-
4
9.979291
e
-
5
7.504636
e
-
5
2.497799
e
-
5
t
=
40
1.535826
e
-
3
5.014481
e
-
4
3.845124
e
-
4
1.255601
e
-
4
9.606941
e
-
5
3.141830
e
-
5Table 3
The verification of the second convergence of Scheme I.
∥
e
n
(
h
,
τ
)
∥
/
∥
e
2
n
(
h
/
2
,
τ
/
2
)
∥
∥
e
n
(
h
,
τ
)
∥
∞
/
∥
e
2
n
(
h
/
2
,
τ
/
2
)
∥
∞
τ
=
h
=
0.1
τ
=
h
=
0.05
τ
=
h
=
0.025
τ
=
h
=
0.1
τ
=
h
=
0.05
τ
=
h
=
0.025
t
=
10
—
3.998318
3.988313
—
3.998969
3.999234
t
=
20
—
3.998548
3.996548
—
3.998630
3.999538
t
=
30
—
3.998359
3.998072
—
3.998171
3.999489
t
=
40
—
3.998109
3.998457
—
3.997769
3.999324Table 4
The verification of the second convergence of Scheme II.
∥
e
n
(
h
,
τ
)
∥
/
∥
e
2
n
(
h
/
2
,
τ
/
2
)
∥
∥
e
n
(
h
,
τ
)
∥
∞
/
∥
e
2
n
(
h
/
2
,
τ
/
2
)
∥
∞
τ
=
h
=
0.1
τ
=
h
=
0.05
τ
=
h
=
0.025
τ
=
h
=
0.1
τ
=
h
=
0.05
τ
=
h
=
0.025
t
=
10
—
3.995835
3.997182
—
3.995455
3.993009
t
=
20
—
3.995103
3.999826
—
3.994473
3.994248
t
=
30
—
3.994599
4.001208
—
3.994140
3.995232
t
=
40
—
3.994217
4.002444
—
3.993690
3.996400Table 5
Whenα=1, numerical simulations on the conservation invariant Qn.
τ
=
h
=
0.1
τ
=
h
=
0.05
τ
=
h
=
0.025
Scheme I
Scheme II
Scheme I
Scheme II
Scheme I
Scheme II
t
=
0
4.120893241330
4.120918187462
4.120893214999
4.120899452450
4.120893201770
4.120894761190
t
=
10
4.120892312503
4.120918835632
4.120892200166
4.120898816174
4.120892129735
4.120894227434
t
=
20
4.120893031520
4.120921893266
4.120892362862
4.120899447408
4.120892167548
4.120894662524
t
=
30
4.120893107802
4.120925120669
4.120892370454
4.120900136940
4.120892149140
4.120895106600
t
=
40
4.120892997828
4.120924996792
4.120892342248
4.120900073781
4.120892155018
4.120895399389Table 6
Whenα=1, numerical simulations on the conservation invariant En.
τ
=
h
=
0.1
τ
=
h
=
0.05
τ
=
h
=
0.025
Scheme I
Scheme II
Scheme I
Scheme II
Scheme I
Scheme II
t
=
0
0.836201094485
0.836201094485
0.836201181724
0.836201181724
0.836201203535
0.836201203535
t
=
10
0.836201095943
0.836201096547
0.836201194492
0.836201202559
0.836201310312
0.836201469358
t
=
20
0.836201095943
0.836201096662
0.836201194483
0.836201206568
0.836201308626
0.836201594118
t
=
30
0.836201095952
0.836201096780
0.836201194484
0.836201210387
0.836201309421
0.836201718674
t
=
40
0.836201095955
0.836201096906
0.836201194369
0.836201214174
0.836201309604
0.836201843455Finally, a numerical simulation figure comparison ofu(x,t) at various time steps with τ=h=0.05 is as follows. (see Figures 1 and 2).Figure 1
Whenα=1 and τ=h=0.05, a numerical simulation figure on u(x,t) of Scheme I at various times.Figure 2
Whenα=1 and τ=h=0.05, a numerical simulation figure on u(x,t) of Scheme II at various times.Numerical simulations show that the finite difference Schemes I and II in this paper are efficient. The calculation results of Scheme I are slightly better than Scheme II. But iterative numerical calculation is not required, Scheme II can save computing time.



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*Source: 217393-2014-03-18.xml*