Boundary Value Problems
Volume 2010 (2010), Article ID 543503, 18 pages
doi:10.1155/2010/543503
Research Article

Two Conservative Difference Schemes for the Generalized Rosenau Equation

1School of Mathematics and Computer Engineering, Xihua University, Chengdu, Sichuan 610039, China
2School of Science, Southwest University of Science and Technology, Mianyang, Sichuan 621010, China

Received 31 October 2009; Accepted 26 January 2010

Academic Editor: Sandro Salsa

Copyright © 2010 Jinsong Hu and Kelong Zheng. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Numerical solutions for generalized Rosenau equation are considered and two energy conservative finite difference schemes are proposed. Existence of the solutions for the difference scheme has been shown. Stability, convergence, and priori error estimate of the scheme are proved using energy method. Numerical results demonstrate that two schemes are efficient and reliable.

1. Introduction

Consider the following initial-boundary value problem for generalized Rosenau equation:

𝑢 𝑡 + 𝑢 𝑥 𝑥 𝑥 𝑥 𝑡 + 𝑢 𝑥 + ( 𝑢 𝑝 ) 𝑥 [ ] [ ] , = 0 , 𝑥 0 , 𝐿 , 𝑡 0 , 𝑇 ( 1 . 1 ) with an initial condition

𝑢 ( 𝑥 , 0 ) = 𝑢 0 [ ] , ( 𝑥 ) , 𝑥 0 , 𝐿 ( 1 . 2 ) and boundary conditions

𝑢 ( 0 , 𝑡 ) = 𝑢 ( 𝐿 , 𝑡 ) = 0 , 𝑢 𝑥 𝑥 ( 0 , 𝑡 ) = 𝑢 𝑥 𝑥 [ ] , ( 𝐿 , 𝑡 ) = 0 , 𝑡 0 , 𝑇 ( 1 . 3 ) where 𝑝 2 is a integer.

When 𝑝 = 2 , (1.1) is called as usual Rosenau equation proposed by Rosenau [1] for treating the dynamics of dense discrete systems. Since then, the Cauchy problem of the Rosenau equation was investigated by Park [2]. Many numerical schemes have been proposed, such as 𝐶 1 -conforming finite element method by Chung and Pani [3], discontinuous Galerkin method by Choo et al. [4], orthogonal cubic spline collocation method by Manickam [5], and finite difference method by Chung [6] and Omrani et al. [7]. As for the generalized case, however, there are few studies on theoretical analysis and numerical methods.

It can be proved easily that the problem (1.1)–(1.3) has the following conservative law:

𝐸 ( 𝑡 ) = 𝑢 2 𝐿 2 + 𝑢 𝑥 𝑥 2 𝐿 2 = 𝐸 ( 0 ) . ( 1 . 4 ) Hence, we propose two conservative difference schemes which simulate conservative law (1.4). The outline of the paper is as follows. In Section 2, a nonlinear difference scheme is proposed and corresponding convergence and stability of the scheme are proved. In Section 3, a linearized difference scheme is proposed and theoretical results are obtained. In Section 4, some numerical experiments are shown.

2. Nonlinear Finite Difference Scheme

Let and 𝜏 be the uniform step size in the spatial and temporal direction, respectively. Denote 𝑥 𝑗 = 𝑗 ( 0 𝑗 𝐽 ) , 𝑡 𝑛 = 𝑛 𝜏 ( 0 𝑛 𝑁 ) , 𝑢 𝑛 𝑗 𝑢 ( 𝑥 𝑗 , 𝑡 𝑛 ) , and 𝑍 0 = { 𝑢 = ( 𝑢 𝑗 ) 𝑢 0 = 𝑢 𝑗 = 0 , 𝑗 = 0 , 1 , 2 , , 𝐽 } . Define

𝑢 𝑛 𝑗 𝑥 = 𝑢 𝑛 𝑗 + 1 𝑢 𝑛 𝑗 , 𝑢 𝑛 𝑗 𝑥 = 𝑢 𝑛 𝑗 𝑢 𝑛 𝑗 1 , 𝑢 𝑛 𝑗 = 𝑢 ̂ 𝑥 𝑛 𝑗 + 1 𝑢 𝑛 𝑗 1 , 𝑢 2 𝑛 𝑗 𝑡 = 𝑢 𝑗 𝑛 + 1 𝑢 𝑛 𝑗 𝜏 , 𝑢 𝑛 𝑗 ̂ 𝑡 = 𝑢 𝑗 𝑛 + 1 𝑢 𝑗 𝑛 1 , 𝑢 2 𝜏 𝑛 𝑗 𝑥 𝑥 = 𝑢 𝑛 𝑗 + 1 2 𝑢 𝑛 𝑗 + 𝑢 𝑛 𝑗 1 2 , 𝑢 𝑛 𝑗 = 𝑢 𝑗 𝑛 + 1 + 𝑢 𝑗 𝑛 1 2 , 𝑢 𝑗 𝑛 + 1 / 2 = 𝑢 𝑗 𝑛 + 1 + 𝑢 𝑛 𝑗 2 , ( 𝑢 𝑛 , 𝑣 𝑛 ) = 𝐽 1 𝑗 = 0 𝑢 𝑛 𝑗 𝑣 𝑛 𝑗 , 𝑢 𝑛 2 = ( 𝑢 𝑛 , 𝑢 𝑛 ) , 𝑢 𝑛 = m a x 0 𝑗 𝐽 1 | | 𝑢 𝑛 𝑗 | | , ( 2 . 1 ) and in the paper, 𝐶 denotes a general positive constant which may have different values in different occurrences.

Since ( 𝑢 𝑝 ) 𝑥 = ( 𝑝 / ( 𝑝 + 1 ) ) [ 𝑢 𝑝 1 𝑢 𝑥 + ( 𝑢 𝑝 ) 𝑥 ] , then the following finite difference scheme is considered:

𝑢 𝑛 𝑗 𝑡 + 𝑢 𝑛 𝑗 𝑥 𝑥 𝑥 𝑥 𝑡 + 𝑢 𝑗 𝑛 + 1 / 2 + 𝑝 ̂ 𝑥 𝑢 𝑝 + 1 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 + 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 𝑢 ̂ 𝑥 = 0 , ( 2 . 2 ) 0 𝑗 = 𝑢 0 𝑥 𝑗 𝑢 , 0 𝑗 𝐽 1 , ( 2 . 3 ) 𝑛 0 = 𝑢 𝑛 𝐽 𝑢 = 0 , 𝑛 0 𝑥 𝑥 = 𝑢 𝑛 𝐽 𝑥 𝑥 = 0 . ( 2 . 4 )

Lemma 2.1 (see [8]). For any two mesh functions, 𝑢 , 𝑣 𝑍 0 , one has 𝑢 𝑗 𝑥 , 𝑣 𝑗 𝑢 = 𝑗 , 𝑣 𝑗 𝑥 , 𝑣 𝑗 , 𝑢 𝑗 𝑥 𝑥 𝑣 = 𝑗 𝑥 , 𝑢 𝑗 𝑥 , 𝑢 𝑗 , 𝑢 𝑗 𝑥 𝑥 𝑢 = 𝑗 𝑥 , 𝑢 𝑗 𝑥 𝑢 = 𝑥 2 . ( 2 . 5 ) Furthermore, if ( 𝑢 𝑛 0 ) 𝑥 𝑥 = ( 𝑢 𝑛 𝐽 ) 𝑥 𝑥 = 0 , then 𝑢 𝑗 , 𝑢 𝑗 𝑥 𝑥 𝑥 𝑥 = 𝑢 𝑥 𝑥 2 . ( 2 . 6 )

Theorem 2.2. Suppose 𝑢 0 𝐻 2 0 [ 0 , 𝐿 ] , then the scheme (2.2)–(2.4) is conservative for discrete energy, that is, 𝐸 𝑛 = 𝑢 𝑛 2 + 𝑢 𝑛 𝑥 𝑥 2 = 𝐸 𝑛 1 = = 𝐸 0 . ( 2 . 7 )

Proof. Computing the inner product of (2.2) with 2 𝑢 𝑛 + 1 / 2 , according to boundary condition (2.4) and Lemma 2.1, we have 1 𝜏 𝑢 𝑛 + 1 2 𝑢 𝑛 2 + 1 𝜏 𝑢 𝑛 + 1 𝑥 𝑥 2 𝑢 𝑛 𝑥 𝑥 2 + 𝑢 𝑗 𝑛 + 1 / 2 ̂ 𝑥 , 2 𝑢 𝑗 𝑛 + 1 / 2 + 𝑃 1 , 2 𝑢 𝑗 𝑛 + 1 / 2 = 0 , ( 2 . 8 ) where 𝑃 1 = 𝑝 𝑢 𝑝 + 1 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 + 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 ̂ 𝑥 . ( 2 . 9 )
According to
𝑢 𝑗 𝑛 + 1 / 2 ̂ 𝑥 , 2 𝑢 𝑗 𝑛 + 1 / 2 𝑃 = 0 , ( 2 . 1 0 ) 1 , 2 𝑢 𝑗 𝑛 + 1 / 2 = 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑢 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 + 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 = 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑢 𝑗 𝑛 + 1 / 2 𝑝 𝑢 𝑛 + 1 / 2 𝑗 + 1 𝑢 𝑛 + 1 / 2 𝑗 1 + 𝑢 𝑛 + 1 / 2 𝑗 + 1 𝑝 𝑢 𝑛 + 1 / 2 𝑗 1 𝑝 𝑢 𝑗 𝑛 + 1 / 2 = 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑢 𝑛 + 1 / 2 𝑗 + 1 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 + 𝑢 𝑗 𝑛 + 1 / 2 𝑝 𝑢 𝑛 + 1 / 2 𝑗 + 1 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑢 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑛 + 1 / 2 𝑗 1 + 𝑢 𝑛 + 1 / 2 𝑗 1 𝑝 𝑢 𝑗 𝑛 + 1 / 2 = 0 , ( 2 . 1 1 ) we obtain 𝑢 𝑛 + 1 2 𝑢 𝑛 2 + 𝑢 𝑛 + 1 𝑥 𝑥 2 𝑢 𝑛 𝑥 𝑥 2 = 0 . ( 2 . 1 2 ) By the definition of 𝐸 𝑛 , (2.7) holds.

To prove the existence of solution for scheme (2.2)–(2.4), the following Browder fixed point Theorem should be introduced. For the proof, see [9].

Lemma 2.3 (Browder fixed point Theorem). Let 𝐻 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 2.4. There exists 𝑢 𝑛 𝑍 0 satisfying the difference scheme (2.2)–(2.4).

Proof. By the mathematical induction, for 𝑛 𝑁 1 , assume that 𝑢 0 , 𝑢 1 , , 𝑢 𝑛 satisfy (2.2)–(2.4). Next we prove that there exists 𝑢 𝑛 + 1 satisfying (2.2)–(2.4).
Define a operator 𝑔 on 𝑍 0 as follows:
𝑔 ( 𝑣 ) = 2 𝑣 2 𝑢 𝑛 + 2 𝑣 𝑥 𝑥 𝑥 𝑥 2 𝑢 𝑛 𝑥 𝑥 𝑥 𝑥 + + 𝜏 𝑣 ̂ 𝑥 𝜏 𝑝 𝑣 𝑝 + 1 𝑗 𝑝 1 𝑣 𝑗 + 𝑣 ̂ 𝑥 𝑗 𝑝 . ̂ 𝑥 ( 2 . 1 3 ) Taking the inner product of (2.13) with 𝑣 , we get 𝑣 𝑣 ̂ 𝑥 , 𝑣 = 0 , 𝑗 𝑝 1 𝑣 𝑗 𝑣 ̂ 𝑥 + [ 𝑗 ) 𝑝 ̂ 𝑥 , 𝑣 = 0 , ( 𝑔 ( 𝑣 ) , 𝑣 ) = 2 𝑣 2 2 ( 𝑢 𝑛 𝑣 , 𝑣 ) + 2 𝑥 𝑥 2 𝑢 2 𝑛 𝑥 𝑥 , 𝑣 𝑥 𝑥 2 𝑣 2 2 𝑢 𝑛 𝑣 𝑣 + 2 𝑥 𝑥 2 𝑢 2 𝑛 𝑥 𝑥 𝑣 𝑥 𝑥 2 𝑣 2 𝑢 2 + 𝑣 2 𝑣 + 2 𝑥 𝑥 2 𝑢 𝑥 𝑥 2 + 𝑣 𝑥 𝑥 2 𝑣 2 𝑢 𝑛 2 + 𝑢 𝑥 𝑥 2 + 𝑣 𝑥 𝑥 2 𝑣 2 𝑢 𝑛 2 + 𝑢 𝑛 𝑥 𝑥 2 . ( 2 . 1 4 ) Obviously, for all 𝑣 𝑍 0 , ( 𝑔 ( 𝑣 ) , 𝑣 ) 0 with 𝑣 2 = 𝑢 𝑛 2 + 𝑢 𝑛 𝑥 𝑥 2 + 1 . It follows from Lemma 2.3 that there exists 𝑣 𝑍 0 which satisfies 𝑔 ( 𝑣 ) = 0 . Let 𝑢 𝑛 + 1 = 2 𝑣 𝑢 𝑛 , it can be proved that 𝑢 𝑛 + 1 is the solution of the scheme (2.2)–(2.4).

Next, we discuss the convergence and stability of the scheme (2.2)–(2.4). Let 𝑣 ( 𝑥 , 𝑡 ) be the solution of problem (1.1)–(1.3), 𝑣 𝑛 𝑗 = 𝑣 ( 𝑥 𝑗 , 𝑡 𝑛 ) , then the truncation of the scheme (2.2)–(2.4) is

𝑟 𝑛 𝑗 = 𝑣 𝑛 𝑗 𝑡 + 𝑣 𝑛 𝑗 𝑥 𝑥 𝑥 𝑥 𝑡 + 𝑣 𝑗 𝑛 + 1 / 2 + 𝑝 ̂ 𝑥 𝑣 𝑝 + 1 𝑗 𝑛 + 1 / 2 𝑝 1 𝑣 𝑗 𝑛 + 1 / 2 + 𝑣 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 ̂ 𝑥 . ( 2 . 1 5 ) Using Taylor expansion, we know that 𝑟 𝑛 𝑗 = 𝑂 ( 𝜏 2 + 2 ) holds if 𝜏 , 0 .

Lemma 2.5. Suppose that 𝑢 𝐻 2 0 [ 0 , 𝐿 ] , then the solution of the initial-boundary value problem (1.1)–(1.3) satisfies 𝑢 𝐿 2 𝑢 𝐶 , 𝑥 𝐿 2 𝐶 , 𝑢 𝐶 . ( 2 . 1 6 )

Proof. It follows from (1.4) that 𝑢 𝐿 2 𝑢 𝐶 , 𝑥 𝑥 𝐿 2 𝐶 . ( 2 . 1 7 ) Using Hölder inequality and Schwartz inequality, we get 𝑢 𝑥 2 𝐿 2 = 𝐿 0 𝑢 𝑥 𝑢 𝑥 𝑑 𝑥 = 𝑢 𝑢 𝑥 | | 𝐿 0 𝐿 0 𝑢 𝑢 𝑥 𝑥 𝑑 𝑥 = 𝐿 0 𝑢 𝑢 𝑥 𝑥 𝑑 𝑥 𝑢 𝐿 2 𝑢 𝑥 𝑥 𝐿 2 1 2 𝑢 2 𝐿 2 + 𝑢 𝑥 𝑥 2 𝐿 2 . ( 2 . 1 8 ) Hence, 𝑢 𝑥 𝐿 2 𝐶 . According to Sobolev inequality, we have 𝑢 𝐶 .

Lemma 2.6 (Discrete Sobolev's inequality [10]). There exist two constant 𝐶 1 and 𝐶 2 such that 𝑢 𝑛 𝐶 1 𝑢 𝑛 + 𝐶 2 𝑢 𝑛 𝑥 . ( 2 . 1 9 )

Lemma 2.7 (Discrete Gronwall inequality [10]). Suppose 𝑤 ( 𝑘 ) , 𝜌 ( 𝑘 ) are nonnegative mesh functions and 𝜌 ( 𝑘 ) is nondecreasing. If 𝐶 > 0 and 𝑤 ( 𝑘 ) 𝜌 ( 𝑘 ) + 𝐶 𝜏 𝑘 1 𝑙 = 0 𝑤 ( 𝑙 ) , 𝑘 , ( 2 . 2 0 ) then 𝑤 ( 𝑘 ) 𝜌 ( 𝑘 ) 𝑒 𝐶 𝜏 𝑘 , 𝑘 . ( 2 . 2 1 )

Theorem 2.8. Suppose 𝑢 0 𝐻 2 0 [ 0 , 𝐿 ] , then the solution 𝑢 𝑛 of (2.2) satisfies 𝑢 𝑛 𝐶 , 𝑢 𝑛 𝑥 𝐶 , which yield 𝑢 𝑛 𝐶 ( 𝑛 = 1 , 2 , , 𝑁 ) .

Proof. It follows from (2.7) that 𝑢 𝑛 𝑢 𝐶 , 𝑛 𝑥 𝑥 𝐶 . ( 2 . 2 2 ) Using Lemma 2.1 and Schwartz inequality, we get 𝑢 𝑛 𝑥 2 𝑢 𝑛 𝑢 𝑛 𝑥 𝑥 1 2 𝑢 𝑛 2 + 𝑢 𝑛 𝑥 𝑥 2 𝐶 . ( 2 . 2 3 ) According to Lemma 2.6, we have 𝑢 𝑛 𝐶 .

Theorem 2.9. Suppose 𝑢 0 𝐻 2 0 [ 0 , 𝐿 ] , then the solution 𝑢 𝑛 of the scheme (2.2)–(2.4) converges to the solution of problem (1.1)–(1.3) and the rate of convergence is 𝑂 ( 𝜏 2 + 2 ) .

Proof. Subtracting (2.15) from (2.2) and letting 𝑒 𝑛 𝑗 = 𝑣 𝑛 𝑗 𝑢 𝑛 𝑗 , we have 𝑟 𝑛 𝑗 = 𝑒 𝑛 𝑗 𝑡 + 𝑒 𝑛 𝑗 𝑥 𝑥 𝑥 𝑥 𝑡 + 𝑒 𝑗 𝑛 + 1 / 2 + 𝑝 ̂ 𝑥 𝑣 𝑝 + 1 𝑗 𝑛 + 1 / 2 𝑝 1 𝑣 𝑗 𝑛 + 1 / 2 + 𝑣 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 𝑝 ̂ 𝑥 𝑢 𝑝 + 1 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 + 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 . ̂ 𝑥 ( 2 . 2 4 ) Computing the inner product of (2.24) with 2 𝑒 𝑛 + 1 / 2 , and using ( ( 𝑒 𝑗 𝑛 + 1 / 2 ) ̂ 𝑥 , 2 𝑒 𝑗 𝑛 + 1 / 2 ) = 0 , we get 𝑟 𝑛 𝑗 , 2 𝑒 𝑛 + 1 / 2 = 1 𝜏 𝑒 𝑛 + 1 2 𝑒 𝑛 2 + 1 𝜏 𝑒 𝑛 + 1 𝑥 𝑥 2 𝑒 𝑛 𝑥 𝑥 2 + 𝑄 1 + 𝑄 2 , 2 𝑒 𝑛 + 1 / 2 , ( 2 . 2 5 ) where 𝑄 1 = 𝑝 𝑣 𝑝 + 1 𝑗 𝑛 + 1 / 2 𝑝 1 𝑣 𝑗 𝑛 + 1 / 2 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 , 𝑄 ̂ 𝑥 2 = 𝑝 𝑣 𝑝 + 1 𝑗 𝑛 + 1 / 2 𝑝 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 . ̂ 𝑥 ( 2 . 2 6 ) According to Lemma 2.5, Theorem 2.8, and Schwartz inequality, we have 𝑄 1 , 2 𝑒 𝑛 + 1 / 2 = 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑣 𝑗 𝑛 + 1 / 2 𝑝 1 𝑣 𝑗 𝑛 + 1 / 2 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 𝑒 ̂ 𝑥 𝑗 𝑛 + 1 / 2 = 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑣 𝑗 𝑛 + 1 / 2 𝑝 1 𝑒 𝑗 𝑛 + 1 / 2 𝑒 ̂ 𝑥 𝑗 𝑛 + 1 / 2 + 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑣 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 𝑝 1 𝑢 𝑗 𝑛 + 1 / 2 𝑒 ̂ 𝑥 𝑗 𝑛 + 1 / 2 = 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑣 𝑗 𝑛 + 1 / 2 𝑝 1 𝑒 𝑗 𝑛 + 1 / 2 𝑒 ̂ 𝑥 𝑗 𝑛 + 1 / 2 + 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑒 𝑗 𝑛 + 1 / 2 𝑝 2 𝑘 = 0 𝑣 𝑗 𝑛 + 1 / 2 𝑝 2 𝑘 𝑢 𝑗 𝑛 + 1 / 2 𝑘 𝑢 𝑗 𝑛 + 1 / 2 𝑒 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝐶 𝐽 1 𝑗 = 0 | | | 𝑒 𝑗 𝑛 + 1 / 2 | | | | | 𝑒 ̂ 𝑥 𝑗 𝑛 + 1 / 2 | | + 𝐶 𝐽 1 𝑗 = 0 | | | | 𝑒 𝑗 𝑛 + 1 / 2 2 | | | | 𝑒 𝐶 𝑥 𝑛 + 1 / 2 2 + 𝑒 𝑛 + 1 / 2 2 𝑒 𝐶 𝑥 𝑛 + 1 2 + 𝑒 𝑛 𝑥 2 + 𝑒 𝑛 + 1 2 + 𝑒 𝑛 2 , ( 2 . 2 7 ) 𝑄 2 , 2 𝑒 𝑛 + 1 / 2 = 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑣 𝑗 𝑛 + 1 / 2 𝑝 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 / 2 𝑝 𝑒 ̂ 𝑥 𝑗 𝑛 + 1 / 2 = 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑣 𝑗 𝑛 + 1 / 2 𝑝 𝑢 𝑗 𝑛 + 1 / 2 𝑝 𝑒 𝑗 𝑛 + 1 / 2 ̂ 𝑥 = 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑒 𝑗 𝑛 + 1 / 2 𝑝 1 𝑘 = 0 𝑣 𝑗 𝑛 + 1 / 2 𝑝 1 𝑘 𝑢 𝑗 𝑛 + 1 / 2 𝑘 𝑒 𝑗 𝑛 + 1 / 2 𝑒 ̂ 𝑥 𝐶 𝑥 𝑛 + 1 / 2 2 + 𝑒 𝑛 + 1 / 2 2 𝑒 𝐶 𝑥 𝑛 + 1 2 + 𝑒 𝑛 𝑥 2 + 𝑒 𝑛 + 1 2 + 𝑒 𝑛 2 . ( 2 . 2 8 ) Furthermore, 𝑟 𝑛 𝑗 , 2 𝑒 𝑛 + 1 / 2 = 𝑟 𝑛 𝑗 , 𝑒 𝑛 + 1 + 𝑒 𝑛 𝑟 𝑛 2 + 1 2 𝑒 𝑛 + 1 2 + 𝑒 𝑛 2 . ( 2 . 2 9 ) Substituting (2.27)–(2.29) into (2.25), we get 𝑒 𝑛 + 1 2 𝑒 𝑛 2 + 𝑒 𝑛 + 1 𝑥 𝑥 2 𝑒 𝑛 𝑥 𝑥 2 𝑒 𝐶 𝜏 𝑛 + 1 2 + 𝑒 𝑛 2 + 𝑒 𝑥 𝑛 + 1 2 + 𝑒 𝑛 𝑥 2 + 𝜏 𝑟 𝑛 2 . ( 2 . 3 0 ) Similarly to the proof of (2.23), we have 𝑒 𝑥 𝑛 + 1 2 1 2 𝑒 𝑛 + 1 2 + 𝑒 𝑛 + 1 𝑥 𝑥 2 , 𝑒 𝑛 𝑥 2 1 2 𝑒 𝑛 2 + 𝑒 𝑛 𝑥 𝑥 2 , ( 2 . 3 1 ) and (2.30) can be rewritten as 𝑒 𝑛 + 1 2 𝑒 𝑛 2 + 𝑒 𝑛 + 1 𝑥 𝑥 2 𝑒 𝑛 𝑥 𝑥 2 𝑒 𝐶 𝜏 𝑛 + 1 2 + 𝑒 𝑛 2 + 𝑒 𝑛 + 1 𝑥 𝑥 2 + 𝑒 𝑛 𝑥 𝑥 2 + 𝜏 𝑟 𝑛 2 . ( 2 . 3 2 ) Let 𝐵 𝑛 = 𝑒 𝑛 2 + 𝑒 𝑛 𝑥 𝑥 2 , then (2.32) is written as follows: 𝐵 ( 1 𝐶 𝜏 ) 𝑛 + 1 𝐵 𝑛 2 𝐶 𝜏 𝐵 𝑛 + 𝜏 𝑟 𝑛 2 . ( 2 . 3 3 ) If 𝜏 is sufficiently small which satisfies 1 𝐶 𝜏 > 0 , then 𝐵 𝑛 + 1 𝐵 𝑛 𝐶 𝜏 𝐵 𝑛 + 𝐶 𝜏 𝑟 𝑛 2 . ( 2 . 3 4 ) Summing up (2.34) from 0 to 𝑛 1 , we have 𝐵 𝑛 𝐵 0 + 𝐶 𝜏 𝑛 1 𝑙 = 0 𝑟 𝑙 2 + 𝐶 𝜏 𝑛 1 𝑙 = 0 𝐵 𝑙 . ( 2 . 3 5 ) Noticing 𝜏 𝑛 1 𝑙 = 0 𝑟 𝑙 2 𝑛 𝜏 m a x 0 𝑙 𝑛 1 𝑟 𝑙 2 𝜏 𝑇 𝑂 2 + 2 2 , ( 2 . 3 6 ) and 𝑒 0 = 0 , we have 𝐵 0 = 𝑂 ( 𝜏 2 + 2 ) 2 . Hence 𝐵 𝑛 𝜏 𝑂 2 + 2 2 + 𝐶 𝜏 𝑛 1 𝑙 = 0 𝐵 𝑙 . ( 2 . 3 7 ) According to Lemma 2.7, we get 𝐵 𝑛 𝑂 ( 𝜏 2 + 2 ) 2 , that is, 𝑒 𝑛 𝜏 𝑂 2 + 2 , 𝑒 𝑛 𝑥 𝑥 𝜏 𝑂 2 + 2 . ( 2 . 3 8 ) It follows from (2.31) that 𝑒 𝑛 𝑥 𝜏 𝑂 2 + 2 . ( 2 . 3 9 ) By using Lemma 2.6, we have 𝑒 𝑛 𝜏 𝑂 2 + 2 . ( 2 . 4 0 ) This completes the proof of Theorem 2.9.

Similarly, the following theorem can be proved.

Theorem 2.10. Under the conditions of Theorem 2.9, the solution of the scheme (2.2)–(2.4) is stable by .

3. Linearized Finite Difference Scheme

In this section, we propose a linear-implicit finite difference scheme as follows:

𝑢 𝑛 𝑗 ̂ 𝑡 + 𝑢 𝑛 𝑗 𝑥 𝑥 𝑥 𝑥 ̂ 𝑡 + 𝑢 𝑛 𝑗 + 𝑝 ̂ 𝑥 𝑢 𝑝 + 1 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 + 𝑢 ̂ 𝑥 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 ̂ 𝑥 = 0 . ( 3 . 1 )

Theorem 3.1. Suppose 𝑢 0 𝐻 2 0 [ 0 , 𝐿 ] , then the scheme (3.1), (2.3), and (2.4) are conservative for discrete energy, that is, 𝐸 𝑛 = 1 2 𝑢 𝑛 + 1 2 + 𝑢 𝑛 2 + 1 2 𝑢 𝑛 + 1 𝑥 𝑥 2 + 𝑢 𝑛 𝑥 𝑥 2 + 𝜏 𝐽 1 𝑗 = 0 𝑢 𝑛 𝑗 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 = 𝐸 𝑛 1 𝐸 = = 0 . ( 3 . 2 )

Proof. Computing the inner product of (3.1) with 2 𝑢 𝑛 , we have 1 𝑢 2 𝜏 𝑛 + 1 2 𝑢 𝑛 1 2 + 1 𝑢 2 𝜏 𝑛 + 1 𝑥 𝑥 2 𝑢 𝑛 1 𝑥 𝑥 2 + 𝑢 𝑛 𝑗 ̂ 𝑥 , 2 𝑢 𝑛 𝑗 + 𝑃 2 , 2 𝑢 𝑛 𝑗 = 0 , ( 3 . 3 ) where 𝑃 2 = 𝑝 𝑢 𝑝 + 1 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 + 𝑢 ̂ 𝑥 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 . ̂ 𝑥 ( 3 . 4 ) According to Lemma 2.1, we get 𝑢 𝑛 𝑗 ̂ 𝑥 , 2 𝑢 𝑛 𝑗 = 𝑢 𝑛 𝑗 ̂ 𝑥 , 𝑢 𝑗 𝑛 + 1 + 𝑢 𝑛 𝑗 ̂ 𝑥 , 𝑢 𝑗 𝑛 1 = 𝑢 𝑛 𝑗 ̂ 𝑥 , 𝑢 𝑗 𝑛 + 1 𝑢 𝑛 𝑗 , 𝑢 𝑗 𝑛 1 ̂ 𝑥 = 𝐽 1 𝑗 = 0 𝑢 𝑛 𝑗 𝑢 ̂ 𝑥 𝑗 𝑛 + 1 𝐽 1 𝑗 = 0 𝑢 𝑗 𝑛 1 𝑢 ̂ 𝑥 𝑛 𝑗 , 𝑃 ( 3 . 5 ) 2 , 2 𝑢 𝑛 𝑗 = 2 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑢 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 + 𝑢 ̂ 𝑥 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 ̂ 𝑥 𝑢 𝑛 𝑗 = 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑢 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 + 1 𝑢 𝑛 𝑗 1 + 𝑢 𝑛 𝑗 + 1 𝑝 1 𝑢 𝑛 𝑗 + 1 𝑢 𝑛 𝑗 1 𝑝 1 𝑢 𝑛 𝑗 1 𝑢 𝑛 𝑗 = 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑢 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 + 1 𝑢 𝑛 𝑗 𝑢 𝑛 𝑗 + 1 𝑝 1 𝑢 𝑛 𝑗 + 1 𝑢 𝑛 𝑗 𝑝 𝑝 + 1 𝐽 1 𝑗 = 0 𝑢 𝑛 𝑗 1 𝑝 1 𝑢 𝑛 𝑗 𝑢 𝑛 𝑗 1 𝑢 𝑛 𝑗 𝑝 1 𝑢 𝑛 𝑗 𝑢 𝑛 𝑗 1 = 0 . ( 3 . 6 ) Adding (3.3) and (3.5) to (3.6), we obtain 1 𝑢 2 𝜏 𝑛 + 1 2 𝑢 𝑛 2