Boussinesq方程是常用于数值模拟波浪的理论模型。对于缓变地形的情况，基于Boussinesq浅水波方程建立的数学模型有很好的色散性和非线性性质，与波浪的实际变形情况相符合，能够准确地描述波浪的许多变形情况。

一般情况下，Boussinesq方程的解析解难以求得，需要用适当的数值方法来求解。Dougalis和Antonopoulos等人^[1-8]考虑如下Boussinesq方程：

$ \left\{ \begin{array}{l} {\eta _t} + {u_x} + {\left( {\eta u} \right)_x} + a{u_{xxx}} - b{\eta _{xxt}} = 0,\\ {u_t} + {\eta _x} + u{u_x} + c{\eta _{xxx}} - d{u_{xxt}} = 0。\end{array} \right. $

文献[1]空间离散用谱方法，时间离散用Runge-Kutta (R-K)法，模拟了小振幅长水波的双向传播。文献[2]研究了上述问题孤波解的存在唯一性。文献[3]研究了上述问题的孤立波解在小扰动和大扰动下，长时间渐近稳定性质。文献[4-7]研究了上述方程几种特殊情况的初边值问题及周期边值问题的数值解法，空间离散分别采用标准Galerkin有限元方法和样条Galerkin有限元法，证明了半离散格式的-范数最优阶误差估计，进一步用显式Euler法，改进的Euler法和显式三阶R-K法等对时间离散并证明了全离散格式的-范数最优阶误差估计。文献[8]研究了一类完全非线性Boussinesq系统的标准Galerkin方法并证明了半离散格式的-范数误差估计。

研究上述问题的数值解法时，使用有限元法的较多，计算复杂。本文采用计算上相对简单的有限差分法讨论其数值解。

本文考虑如下BBM-BBM (Benjamin-Bona-Mahony)方程的2L周期边值问题：

$ \begin{array}{l} \left\{ \begin{array}{l} {\eta _t} + {u_x} + {\left( {\eta u} \right)_x} - b{\eta _{xxt}} = 0,\;\;\;\;\;\;\;\;\;\;\;\left( {1{\rm{a}}} \right)\\ {u_t} + {\eta _x} + u{u_x} - d{u_{xxt}} = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {1{\rm{b}}} \right) \end{array} \right.\\ \eta \left( {x,0} \right) = {\eta _0}\left( x \right),u\left( {x,0} \right) = {u_0}\left( x \right)。\end{array} $

其中，η=η(x, t)表示自由水面与静止水面的距离；u=u(x, t)表示水深平均的水质点速度，x∈[-L, L]表示空间方向水质点的坐标，t≥0则表示时间变量；系数b, d的值由下述公式给出：

$ b = \frac{1}{2}\left( {{\theta ^2} - \frac{1}{3}} \right)\left( {1 - \upsilon } \right),d = \frac{1}{2}\left( {1 - {\theta ^2}} \right)\left( {1 - \mu } \right), $

υ和μ是实常数，且0≤θ≤1。

1 半离散紧致差分格式

在构造有限差分格式前，把求解区域Ω=[-L, L]×[0, T]划分成网格。记h, τ分别为空间、时间步长，h=2L/N, τ=T/M.网格线的交点称为网格节点，记为(x_j, t_n)，其中

$ {x_j} = - L + jh,j = 0,1, \cdots ,2N;N = \frac{{2L}}{h}, $

$ {t_n} = n\tau ,n = 0,1,2, \cdots ,M;M = \frac{T}{\tau }。$

记u_jⁿ=u(x_j, t_n).为叙述简便，引入如下差分算子：

$ \mathit{\boldsymbol{D}} + U_j^n = \frac{{U_{j + 1}^n - U_j^n}}{h},\mathit{\boldsymbol{D}} - U_j^n = \frac{{U_j^n - U_{j - 1}^n}}{h}, $

$ {\mathit{\boldsymbol{D}}_0}U_j^n = \frac{{U_{j + 1}^n - U_j^n}}{{2h}}, $

$ \delta _x^2U_j^n = \mathit{\boldsymbol{D}} + \mathit{\boldsymbol{D}} - U_j^n = \frac{{U_{j + 1}^n - 2U_j^n + U_{j - 1}^n}}{h}, $

$ \mathit{\boldsymbol{D}}U_j^n = \mathit{\boldsymbol{D}} + \frac{{{h^2}}}{{12}}\delta _x^2 = \frac{{U_{j + 1}^n - 10U_j^n + {U^n}{ + _{j - 1}}}}{{12}}, $

$ \mathit{\boldsymbol{D}}U_j^n = \mathit{\boldsymbol{D}} + \frac{{{h^2}}}{6}\delta _x^2 = \frac{{U_{j + 1}^n + 4U_j^n + U_{j - 1}^n}}{6}。$

考虑如下微分方程：

$ {u_t} + f{\left( {u,\eta } \right)_x} + \beta {u_{xxt}} = 0。$

(2)

由于$ \delta _x^2u = {u_{xx}} + \frac{{{h^2}}}{{12}}{u_{xxxx}} + \frac{{{h^4}}}{{360}}{u^{\left( 6 \right)}} + O\left( {{h^6}} \right), $,

记ζ=u_xx, 则得到

$ \delta _x^2u = \zeta + \frac{{{h^2}}}{{12}}\delta _x^2\zeta - \frac{1}{{240}}{h^4}{u^{\left( 6 \right)}} + O\left( {{h^6}} \right), $

所以

$ \mathit{\boldsymbol{A}}{u_{xx}} - \delta _x^2u = \frac{{{h^4}}}{{240}}{u^{\left( 6 \right)}} + O\left( {{h^6}} \right)。$

若记ρ=u_t，又由(2)式知

$ \rho + f{\left( {u,\eta } \right)_x} + \beta {\rho _{xx}} = 0, $

所以有

$ \mathit{\boldsymbol{A}}\rho + \mathit{\boldsymbol{A}}{f_x} + \beta \delta _x^2\rho = - \frac{\beta }{{240}}{h^4}{\rho ^{\left( 6 \right)}} + O\left( {{h^6}} \right)。$

(3)

由Taylor展开可知

$ {f_x} = {\mathit{\boldsymbol{D}}_0}f - \frac{{{h^2}}}{6}{\mathit{\boldsymbol{D}}_0}\delta _x^2f + \frac{{{h^4}}}{{30}}{f^{\left( 5 \right)}} + O\left( {{h^6}} \right)。$

将上式代入(3)式可得

$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{A}}\rho + \mathit{\boldsymbol{A}}\left( {I - \frac{{{h^2}}}{6}\delta _x^2} \right){\mathit{\boldsymbol{D}}_0}f + \beta \delta _x^2\rho = }\\ { - \frac{{{h^4}}}{{30}}\mathit{\boldsymbol{A}}{f^{\left( 5 \right)}} - \frac{\beta }{{240}}{h^4}{\rho ^{\left( 6 \right)}} + O\left( {{h^6}} \right),} \end{array} $

由$ \mathit{\boldsymbol{A}} = \mathit{\boldsymbol{I}} + \frac{{{h^2}}}{{12}}\delta _x^2 $, 可得

$ \mathit{\boldsymbol{A}}{u_t} + \left( {I - \frac{{{h^2}}}{{12}}\delta _x^2} \right){\mathit{\boldsymbol{D}}_0}f + \beta \delta _x^2{u_t} = O\left( {{h^4}} \right)。$

分别取$ f = \eta + \frac{1}{2}{u^2} $和f=u+ηu，略去右端项，用U_jⁿ, H_jⁿ分别代表u_jⁿ, η_jⁿ, 的差分近似, 得到BBM-BBM方程(1)的半离散差分格式：

$ \mathit{\boldsymbol{A}}{H_t} + {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)\left( {U + HU} \right) - b\delta _x^2{H_t} = 0, $

(4a)

$ \mathit{\boldsymbol{A}}{H_t} + {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)\left( {U + \frac{1}{2}{U^2}} \right) - b\delta _x^2{H_t} = 0, $

(4b)

2 全离散格式

半离散差分方程(4)可整理为：

$ \left( {\mathit{\boldsymbol{A}} - b\delta _x^2} \right){H_t} + {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)\left( {U + HU} \right) = 0, $

(5a)

$ \left( {\mathit{\boldsymbol{A}} - d\delta _x^2} \right){U_t} + {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{{\rm{h}}^2}}}{{12}}{\rm{ \mathsf{ δ} }}_{\rm{x}}^2} \right)\left( {\mathit{\boldsymbol{H}} + \frac{1}{2}{\mathit{\boldsymbol{U}}^2}} \right) = 0, $

(5b)

记, $ {\mathit{\boldsymbol{P}}_1} = \mathit{\boldsymbol{A}} - b\delta _x^2 $, $ {\mathit{\boldsymbol{Q}}_1} = {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right) $, 方程(5a)可写为：

$ {\mathit{\boldsymbol{P}}_1}{H_t} + {\mathit{\boldsymbol{Q}}_1}\left( {HU + U} \right) = 0。$

采用C-N(Crank-Nicolson)格式进行时间离散，得到全离散格式

$ \begin{array}{*{20}{c}} {{\mathit{\boldsymbol{P}}_1}\frac{{{H^{n + 1}} - {H^n}}}{\tau } + }\\ {{\mathit{\boldsymbol{Q}}_1}\left( {\frac{1}{2}\left( {{H^n}{U^n} + {U^n} + {H^{n + 1}}{U^{n + 1}} + {U^{n + 1}}} \right)} \right) = 0。} \end{array} $

(6)

(6) 是非线性格式。

由于

$ \frac{{{u^{n + 1}} + {n^n}}}{2} = \frac{{3{u^n} - {u^{n - 1}}}}{2} + O\left( {{\tau ^2}} \right), $

(7)

可得到如下线性化格式

$ \begin{array}{*{20}{c}} {{\mathit{\boldsymbol{P}}_1}\frac{{{H^{n + 1}} - {H^n}}}{\tau } + }\\ {{\mathit{\boldsymbol{Q}}_1}\left( {\frac{3}{2}\left( {{H^n}{U^n} + {U^n}} \right) - \frac{1}{2}\left( {{H^{n - 1}}{U^{n - 1}} + {U^{n - 1}}} \right)} \right) = 0,} \end{array} $

(8)

(8) 式的截断误差为O(h⁴+τ²).整理可得：

$ \begin{array}{*{20}{c}} {{\mathit{\boldsymbol{P}}_1}{H^{n + 1}} = }\\ {{\mathit{\boldsymbol{P}}_1}{H^n} - \tau {\mathit{\boldsymbol{Q}}_1}\left( {\frac{3}{2}\left( {{{\rm{H}}^{\rm{n}}}{{\rm{U}}^{\rm{n}}} + {{\rm{U}}^{\rm{n}}}} \right) - \frac{1}{2}\left( {{{\rm{H}}^{{\rm{n}} - 1}}{{\rm{U}}^{{\rm{n}} - 1}} + {{\rm{U}}^{{\rm{n}} - 1}}} \right)} \right)。} \end{array} $

(9)

记P₂=A-dδ_x², (5b)式可以写成

$ {\mathit{\boldsymbol{P}}_2}{U_t} + {\mathit{\boldsymbol{Q}}_1}\left( {H + \frac{1}{2}{U^2}} \right) = 0。$

时间离散采用线性化C-N格式，方程(5b)的全离散格式为

$ \begin{array}{*{20}{c}} {{\mathit{\boldsymbol{P}}_2}{U^{n + 1}} = {\mathit{\boldsymbol{P}}_2}{U^n} - }\\ {\tau {\mathit{\boldsymbol{Q}}_1}\left( {\frac{3}{2}\left( {{H^n} + \frac{1}{2}{{\left( {{U^n}} \right)}^2}} \right) - \frac{1}{2}\left( {{H^{n - 1}} + \frac{1}{2}{{\left( {{U^{n - 1}}} \right)}^2}} \right)} \right)。} \end{array} $

(10)

(8) 和(10)即为BBM-BBM方程(1)的线性化全离散紧致差分格式。

3 误差分析

定义离散内积：

$ \left\langle {\eta ,u} \right\rangle = \sum\limits_{j = 0}^{N - 1} {{\eta _j}{u_j}h,} $

相应的范数为$ {\left\| \eta \right\|_0} = \sqrt {\left\langle {\eta , \eta } \right\rangle } $

记, $ {\eta _1} = \sqrt {\left\langle {D + \eta , D + \eta } \right\rangle } $，$ {\left\| \eta \right\|_1} = \sqrt {\left\| \eta \right\|_0^2 + \left| \eta \right|_1^2} $，$ {\left\| \eta \right\|_c} = \mathop {\max }\limits_{0 \le j \le N - 1} \left| {{u_j}} \right| $, $ {\left| {\left\| u \right\|} \right|^2} = \left\langle {Au, u} \right\rangle $。

由上述定义及周期性条件u_j=u_j+N，易知如下结论成立：

$ \begin{array}{*{20}{c}} {\left\langle {\delta _x^2\eta ,u} \right\rangle = - \left\langle {{\mathit{\boldsymbol{D}}_ + }\eta ,{\mathit{\boldsymbol{D}}_ + }u} \right\rangle ,\left\langle {{\mathit{\boldsymbol{D}}_0}\eta ,u} \right\rangle = - \left\langle {\eta ,{\mathit{\boldsymbol{D}}_0}u} \right\rangle ,}\\ {\left\langle {{\mathit{\boldsymbol{D}}_ + }\eta ,u} \right\rangle = - \left\langle {\eta ,\mathit{\boldsymbol{D}},u} \right\rangle \left\langle {\mathit{\boldsymbol{A}}u,v} \right\rangle = \left\langle {u,\mathit{\boldsymbol{A}}v} \right\rangle ,}\\ {\left\langle {\mathit{\boldsymbol{A}}u,u} \right\rangle = \left\| u \right\|_0^2 - \frac{{{h^2}}}{{12}}\left| u \right|_1^2,\frac{2}{3}\left\| u \right\|_0^2 \le {{\left| {\left\| u \right\|} \right|}^2} \le \left\| u \right\|_0^2。} \end{array} $

引理3.1  (逆估计)

$ \left| u \right|_1^2 \le \frac{4}{{{h^2}}}\left\| u \right\|_0^2。$

(11)

引理3.2^[9]  对任意存在常数，使得

$ {\left\| u \right\|_c} \le \varepsilon {\left\| {{\mathit{\boldsymbol{D}}_0} + u} \right\|_0} + K\left( \varepsilon \right){\left\| u \right\|_0}。$

引理3.3  (离散形式的Gronwall不等式)设α，β≥0是任意常数，序列{η_n}满足：

$ \left| {{\eta _n}} \right| \le \beta + \alpha h\sum\limits_{j = 0}^{n - 1} {{\eta _j}} $, n=k, k+1, …, nh≤T，其中h＞0是步长，则

$ \left| {{\eta _n}} \right| \le {{\rm{e}}^{\alpha T}}\left( {\beta + \alpha h{M_0}} \right),n \ge k,nh \le T $, 其中$ {M_0} = \max \left( {\left| {{\eta _0}} \right|,\left| {{\eta _1}} \right|, \cdots ,\left| {{\eta _{k - 1}}} \right|} \right) $

定理3.1  设uⁿ, ηⁿ是(1)的解，且充分光滑，Uⁿ, Hⁿ是差分方程(8) (10)的解，τ, h充分小，则有

$ {\left\| {{u^n} - {U^n}} \right\|_1} + {\left\| {{\eta ^n} - {H^n}} \right\|_1} \le C\left( {{\tau ^2} + {h^4}} \right)。$

证明  记$ {\tilde u^n} = \frac{3}{2}{u^n} - \frac{1}{2}{u^{n - 1}} $, $ {{\tilde \eta }^n} = \frac{3}{2}{\eta ^n} - \frac{1}{2}{\eta ^{n - 1}}f\left( {u, \eta } \right) = u + \eta u $, $ g\left( {u, \eta } \right) = \eta + \frac{1}{2}{u^2} $方程(1a), (1b)可表示为：

$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{A}}\frac{{{\eta ^{n + 1}} - {\eta ^n}}}{\tau } + {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)}\\ { - b\delta _x^2\frac{{{\eta ^{n + 1}} - {\eta ^n}}}{\tau } = R_1^n,} \end{array} $

(12a)

$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{A}}\frac{{{u^{n + 1}} - {u^n}}}{\tau } + {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)g\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)}\\ { - d\delta _x^2\frac{{{u^{n + 1}} - {u^n}}}{\tau } = R_2^n,} \end{array} $

(12b)

其中R₁ⁿ=O(τ²+h⁴), R₂ⁿ=O(τ²+h⁴)。

数值解Uⁿ, Hⁿ满足方程：

$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{A}}\frac{{{H^{n + 1}} - {H^n}}}{\tau } + {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - }\\ {b\delta _x^2\frac{{{H^{n + 1}} - {H^n}}}{\tau } = 0,} \end{array} $

(13a)

$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{A}}\frac{{{U^{n + 1}} - {U^n}}}{\tau } + {\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - }\\ {d\delta _x^2\frac{{{U^{n + 1}} - {U^n}}}{\tau } = 0。} \end{array} $

(13b)

记ηⁿ-Hⁿ=eⁿ, uⁿ-Uⁿ=εⁿ，, 可得误差方程

$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{A}}\frac{{{{\rm{e}}^{n + 1}} - {{\rm{e}}^n}}}{\tau } - b\delta _x^2\frac{{{{\rm{e}}^{n + 1}} - {{\rm{e}}^n}}}{\tau } = }\\ {{\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)\left[ {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right] + R_1^n,} \end{array} $

(14a)

$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{A}}\frac{{{\varepsilon ^{n + 1}} - {\varepsilon ^n}}}{\tau } - d\delta _x^2\frac{{{\varepsilon ^{n + 1}} - {\varepsilon ^n}}}{\tau } = }\\ {{\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)\left[ {g\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - g\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right] + R_2^n。} \end{array} $

(14b)

方程(14a)两边与eⁿ⁺¹+eⁿ作离散内积，则有

$ \begin{array}{*{20}{c}} {\left[ {\mathit{\boldsymbol{A}}\frac{{{{\rm{e}}^{n + 1}} - {{\rm{e}}^n}}}{\tau },{{\rm{e}}^{n + 1}},{{\rm{e}}^n}} \right] - b\left[ {\delta _x^2\frac{{{{\rm{e}}^{n + 1}} - {{\rm{e}}^n}}}{\tau },{{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right] = }\\ {\left[ {{\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{n}\delta _x^2} \right)\left[ {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right],{{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right]}\\ { + \left[ {R_1^n,{{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right]。} \end{array} $

(15)

整理得到

$ \begin{array}{*{20}{c}} {\left( {\left[ {\mathit{\boldsymbol{A}}{{\rm{e}}^{n + 1}},{{\rm{e}}^{n + 1}}} \right] + b\left[ {{\mathit{\boldsymbol{D}}_ + }{{\rm{e}}^{n + 1}},{\mathit{\boldsymbol{D}}_ + }{{\rm{e}}^{n + 1}}} \right]} \right)}\\ { - \left( {\left[ {\mathit{\boldsymbol{A}}{{\rm{e}}^n},{{\rm{e}}^n}} \right] + b\left[ {{\mathit{\boldsymbol{D}}_ + }{{\rm{e}}^n},{\mathit{\boldsymbol{D}}_ + }{{\rm{e}}^n}} \right]} \right) = }\\ {\tau \left[ {{\mathit{\boldsymbol{D}}_0}\left( {I - \frac{{{h^2}}}{{12}}\delta _x^2} \right)\left[ {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right],{{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right]}\\ { + \tau \left[ {R_1^n,{{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right]。} \end{array} $

(16)

所以

$ \begin{array}{l} \left( {\left\| {{{\left| {{{\rm{e}}^{n + 1}}} \right|}^2}} \right\| + b\left| {{{\rm{e}}^{n + 1}}} \right|_1^2} \right) - \left( {\left\| {{{\left| {{{\rm{e}}^n}} \right|}^2}} \right\| + b\left| {{{\rm{e}}^n}} \right|_1^2} \right) \le \\ \tau \left| {\left[ {{\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)\left[ {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right],{{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right]} \right| + \\ \tau \left| {\left[ {R_1^n,{{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right]} \right|。\end{array} $

(17)

由逆估计式得

$ \begin{array}{l} \left| {\left[ {{\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)u,v} \right]} \right| \le {\left\| {\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)u} \right\|_0}{\left\| {{\mathit{\boldsymbol{D}}_0}v} \right\|_0} \le \\ {\left\| u \right\|_0}{\left\| {{\mathit{\boldsymbol{D}}_0}v} \right\|_0} + \frac{{{h^2}}}{{12}}{\left\| {\delta _x^2u} \right\|_0}{\left\| {{\mathit{\boldsymbol{D}}_0}v} \right\|_0} \le \\ \frac{4}{3}{\left\| u \right\|_0}{\left\| {{\mathit{\boldsymbol{D}}_0}v} \right\|_0} \le \frac{2}{3}\left\| u \right\|_0^2 + \frac{2}{3}\left\| v \right\|_1^2, \end{array} $

所以(17)右端第一项

$ \begin{array}{l} \tau \left| {\left[ {{\mathit{\boldsymbol{D}}_0}\left( {\mathit{\boldsymbol{I}} - \frac{{{h^2}}}{{12}}\delta _x^2} \right)\left[ {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right],{{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right]} \right| \le \\ \frac{2}{3}\tau \left\| {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right\|_0^2 + \\ \frac{4}{3}\tau \left( {\left\| {{{\rm{e}}^{n + 1}}} \right\|_1^2 + \left\| {{{\rm{e}}^n}} \right\|_1^2} \right), \end{array} $

(17) 右端第二项

$ \begin{array}{l} \tau \left| {\left[ {R_1^n \cdot {{\rm{e}}^{n + 1}} + {{\rm{e}}^n}} \right]} \right| \le \frac{\tau }{2}\left\| {R_1^n} \right\|_0^2 + \\ \tau \left( {\left\| {{{\rm{e}}^{n + 1}}} \right\|_0^2 + \left\| {{{\rm{e}}^n}} \right\|_0^2} \right)。\end{array} $

下面估计 $ f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right) $。由于

$ \begin{array}{*{20}{c}} {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right) = }\\ {{{\tilde U}^n} - {{\tilde u}^n} + {{\tilde H}^n}{{\tilde U}^n} - {{\tilde \eta }^n}{{\tilde u}^n} = }\\ { - {{\tilde \varepsilon }^n} + {{{\rm{\tilde e}}}^n}{{\tilde \varepsilon }^n} - {{\tilde \eta }^n}{{\tilde \varepsilon }^n} - {{{\rm{\tilde e}}}^n}{{\tilde u}^n},} \end{array} $

所以有

$ \begin{array}{*{20}{c}} {{{\left\| {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right\|}_0} \le }\\ {{{\left\| {{{\tilde \varepsilon }^n}} \right\|}_0} + {{\left\| {{{\tilde \eta }^n}{{\tilde \varepsilon }^n}} \right\|}_0} + {{\left\| {{{{\rm{\tilde e}}}^n}{{\tilde u}^n}} \right\|}_0} + {{\left\| {{{{\rm{\tilde e}}}^n}{{\tilde \varepsilon }^n}} \right\|}_0}。} \end{array} $

(18)

设存在正常数M₀, 使精确解u, η满足：

$ \mathop {\max }\limits_n {\left\| {{u^n}} \right\|_C} \le {M_0},\mathop {\max }\limits_n {\left\| {{\eta ^n}} \right\|_C} \le {M_0}。$

为了估计(18)式, 采用归纳法。

设存在h₀, τ₀，当0＜h＜h₀, 0＜τ＜τ₀时

$ {\left\| {{{\rm{e}}^m}} \right\|_c},{\left\| {{\varepsilon ^m}} \right\|_c} \le 1,0 \le m \le n。$

(19)

注意到

$ {\left\| {{{\tilde u}^n}} \right\|_c} = {\left\| {\frac{3}{2}{u^n} - \frac{1}{2}{u^{n - 1}}} \right\|_c} \le \frac{3}{2}{\left\| {{u^n}} \right\|_c} + \frac{1}{2}{\left\| {{{\tilde u}^{n - 1}}} \right\|_c}, $

由归纳假设(19)式可得到

$ \begin{array}{l} {\left\| {{{\tilde \eta }^n}{{\tilde \varepsilon }^n}} \right\|_0} \le {\left\| {{{\tilde \eta }^n}} \right\|_c}{\left\| {{{\tilde \varepsilon }^n}} \right\|_0} \le 2{M_0}{\left\| {{{\tilde \varepsilon }^n}} \right\|_0},\\ {\left\| {{{{\rm{\tilde e}}}^n}{{\tilde u}^n}} \right\|_0} \le {\left\| {{{\tilde u}^n}} \right\|_c}{\left\| {{{{\rm{\tilde e}}}^n}} \right\|_0} \le 2{M_0}{\left\| {{{{\rm{\tilde e}}}^n}} \right\|_0},\\ {\left\| {{{{\rm{\tilde e}}}^n}{{\tilde \varepsilon }^n}} \right\|_0} \le {\left\| {{{{\rm{\tilde e}}}^n}} \right\|_c}{\left\| {{{\tilde \varepsilon }^n}} \right\|_0} \le 2{M_0}{\left\| {{{\tilde \varepsilon }^n}} \right\|_0}, \end{array} $

所以

$ \begin{array}{l} \left\| {f\left( {{{\tilde U}^n},{{\tilde H}^n}} \right) - f\left( {{{\tilde u}^n},{{\tilde \eta }^n}} \right)} \right\|_0^2 \le \\ C\left( {\left\| {{\varepsilon ^n}} \right\|_0^2 + \left\| {{\varepsilon ^{n - 1}}} \right\|_0^2 + \left\| {{{\rm{e}}^n}} \right\|_0^2{{\rm{e}}^{n - 1}}_0^2} \right). \end{array} $

在后面的证明中，用表示广义正常数，不同式子中可取不同值。

由(17)式及上述估计式可得

$ \begin{array}{l} \left( {{{\left| {\left\| {{{\rm{e}}^{n + 1}}} \right\|} \right|}^2} + b\left| {{{\rm{e}}^{n + 1}}} \right|_1^2} \right) - \left( {{{\left| {\left\| {{{\rm{e}}^n}} \right\|} \right|}^2} + b\left| {{{\rm{e}}^n}} \right|_1^2} \right) \le \\ C\tau \left\| {R_1^n} \right\|_0^2 + C\tau \left( {\left\| {{{\rm{e}}^{n + 1}}} \right\|_1^2 + \left\| {{{\rm{e}}^n}} \right\|_1^2 + \left\| {{{\rm{e}}^n} - 1} \right\|_1^2 + \left\| {{\varepsilon ^n}} \right\|_0^2 + } \right.\\ \left. {\left\| {{\varepsilon ^{n - 1}}} \right\|_0^2} \right)。\end{array} $

对上式中的n累加可得

$ \begin{array}{l} \min \left\{ {\frac{2}{3},b} \right\}\left\| {{{\rm{e}}^{n + 1}}} \right\|_1^2 \le {\left| {\left\| {{{\rm{e}}^0}} \right\|} \right|^2} + b\left| {{{\rm{e}}^0}} \right|_1^2 + \\ C\tau \sum\limits_{m = 0}^n {\left\| {{\varepsilon ^m}} \right\|_0^2} + C\tau \sum\limits_{m = 0}^n {\left\| {R_1^m} \right\|_0^2} , \end{array} $

整理得

$ \begin{array}{l} \left\| {{{\rm{e}}^{n + 1}}} \right\|_1^2 \le \\ C\left\| {{{\rm{e}}^0}} \right\|_1^2 + C\tau \sum\limits_{m = 0}^{n + 1} {\left\| {{{\rm{e}}^m}} \right\|_1^2} + C\tau \sum\limits_{m = 0}^n {\left\| {{\varepsilon ^m}} \right\|_0^2} + C\tau \sum\limits_{m = 0}^n {\left\| {R_1^m} \right\|_0^2} 。\end{array} $

(20a)

类似地，成立：

$ \begin{array}{l} {\left\| {{\varepsilon ^{n + 1}}} \right\|^2} \le \\ C\left\| {{\varepsilon ^0}} \right\|_1^2 + C\tau \sum\limits_{m = 0}^{n + 1} {\left\| {{\varepsilon ^m}} \right\|_1^2} + C\tau \sum\limits_{m = 0}^n {\left\| {{{\rm{e}}^m}} \right\|_0^2} + C\tau \sum\limits_{m = 0}^n {\left\| {R_2^m} \right\|_0^2} 。\end{array} $

(20b)

由(20a)、(20b)可得

$ \begin{array}{l} \left\| {{\varepsilon ^{n + 1}}} \right\|_1^2 + \left\| {{{\rm{e}}^{n + 1}}} \right\|_1^2 \le \\ C\left( {\left\| {{{\rm{e}}^0}} \right\|_1^2 + \left\| {{\varepsilon ^0}} \right\|_1^2} \right) + C\tau \sum\limits_{m = 0}^{n + 1} {\left( {\left\| {{{\rm{e}}^m}} \right\|_1^2 + \left\| {{\varepsilon ^m}} \right\|_1^2} \right)} + \\ C\tau \sum\limits_{m = 0}^n {\left( {\left\| {R_1^m} \right\|_0^2 + \left\| {R_2^m} \right\|_0^2} \right)} , \end{array} $

取τ充分小，利用Gronwall不等式及初值误差为零得到

$ \begin{array}{l} {\left\| {{\varepsilon ^{n + 1}}} \right\|_1} + {\left\| {{{\rm{e}}^{n + 1}}} \right\|_1} \le \\ CT\mathop {\max }\limits_m \left( {{{\left\| {R_1^m} \right\|}_0} + {{\left\| {R_2^m} \right\|}_0}} \right) \le C\left( T \right)\left( {{\tau ^2} + {h^4}} \right)。\end{array} $

为了完成定理的证明，需验证‖eⁿ⁺¹‖_c≤1，‖εⁿ⁺¹‖_c≤1成立。

由引理2可知

$ \begin{array}{l} {\left\| {{\varepsilon ^{n + 1}}} \right\|_c} \le C{\left\| {{\varepsilon ^{n + 1}}} \right\|_1} \le C\left( {{\tau ^2} + {h^4}} \right),\\ {\left\| {{{\rm{e}}^{n + 1}}} \right\|_c} \le C{\left\| {{{\rm{e}}^{n + 1}}} \right\|_1} \le C\left( {{\tau ^2} + {h^4}} \right), \end{array} $

取τ，h充分小，可以使下式成立：

$ {\left\| {{{\rm{e}}^{n + 1}}} \right\|_c} \le 1,{\left\| {{\varepsilon ^{n + 1}}} \right\|_c} \le 1。$

证毕。

4 数值算例

考虑方程^[10]

$ \left\{ \begin{array}{l} {\eta _t} + {u_x} + {\left( {\eta u} \right)_x} - \frac{1}{6}{\eta _{xxt}} = 0,\\ {u_t} + {\eta _x} + u{u_x} - \frac{1}{6}{u_{xxt}} = 0。\end{array} \right. $

其精确解为：

$ \begin{array}{*{20}{c}} {\eta \left( {x,t} \right) = \frac{{15}}{4}\left( { - 2 + \cosh \left( {3\sqrt {\frac{2}{5}} \left( {x - kt - {x_0}} \right)} \right)} \right) \cdot }\\ {{\rm{sec}}{{\rm{h}}^4}\left( {\frac{{3\left( {x - kt - {x_{10}}} \right)}}{{\sqrt {10} }}} \right),}\\ {u\left( {x,t} \right) = 3k{\rm{sec}}{{\rm{h}}^2}\frac{3}{{\sqrt {10} }}\left( {x - kt - {x_0}} \right),k = \frac{5}{2}。} \end{array} $

由于|x|→∞时，η→0，u→0，所以我们可以取-100≤x≤100，0≤t≤5, 近似以200为一个周期。

表 1, 2分别给出了取不同h时，η和u的数值解的L^∞-范数、L²-范数误差及其收敛阶，计算终止时间为T=5，时间步长取充分小，τ=10^-6，数值结果表明，空间误差达到了四阶。

图 1展示了不同时刻的η的波形。容易看出，波形大致不变往x轴正方向移动。为验证差分格式关于时间步长的收敛阶，取空间步长h=10^-4。表 3, 4分别给出了取不同τ时，η和u的数值解的L^∞-范数、L²-范数误差及其收敛阶。数值结果表明，时间误差达到了二阶。图 2, 3分别给出T=5时的数值解及误差曲线。