考虑如下一维非线性抛物问题

$ \left\{\begin{array}{ll} \frac{\partial u}{\partial t}-\frac{\partial}{\partial x}(A(u)\frac{\partial u}{\partial x})=f(u, x, t), ~~&(x, t)\in\Omega\times (0, T], \\[3mm] u(a, t)=\frac{\partial u (b, t)}{\partial x}=0, &t\in (0, T], \\[2mm] u(x, 0)=u_{0}(x), &x\in\Omega. \end{array}\right. $

(1.1)

定义 ${\Bbb R}$上的有界闭集为 $G=\{u:|u|\leq K_0\}$, 其中 $K_0$为正常数.假定系数 $A(u)$为有正的上下界的光滑函数, 并且 $A(u), A'(u), f(u, x, t)$关于变量 $u$满足Lipschitz连续条件, 即

$ \begin{equation} |g(u)-g(v)|\leq L|u-v|, ~\forall u, v\in G, \label{1.2}\end{equation} $

(1.2)

其中 $L$为正常数.

有限体积元方法^[1]作为一类求解非线性微分方程的重要工具被广泛应用到工程领域, 如流体力学、热传导问题、石油工程等.有限体积元方法最大的优点是能够保持问题的局部守恒性, 如在每个单元上的质量、动量、能量守恒, 并且形式简单, 易于编程实现，目前已有很多研究成果^[2-10].

两层网格方法是基于两个不同网格的求解非线性方程的一种离散技巧.其主要思想是:在粗网格空间上求解一个非线性问题, 得到非线性方程的粗网格解, 然后将此粗网格解作为细网格上的初始近似将细网格上的非线性问题线性化.该方法不仅具有较高的精度而且可以节省大量的计算工作量, 从而提高计算效率.两层网格方法最先由许进超教授提出^[11-12], 此后得到了大量的研究.毕春加和V. Ginting^[13]对线性和非线性椭圆问题的两层网格有限体积元方法进行了研究.陈传军等^[14-15]对带有非线性反应项的半线性抛物方程的两层网格有限体积元方法进行了研究.

在这些工作的基础上, 我们对一维非线性抛物问题的有限体积元逼近进行研究, 首先证明了解的存在性, 并得到了最优阶 $L^2$ -模和 $H^1$ -模先验误差估计, 然后构造了该问题的两层网格算法, 证明了最优阶 $H^1$ -模先验误差估计.数值算例表明所构造的两层网格算法保证了问题的精度, 并且可以节省大量的计算时间.

本文用 $C$或者带有上下标的 $C$表示与 $h$和 $\Delta t$无关的正常数, 在不同的地方表示不同的常数.

采用标准的Sobolev空间记号 $W^{s, p}(\Omega)$ ( $1\leq p\leq\infty$), 范数如下定义

$ \|u\|_{s, p}=\Big(\int_{a}^{b}\sum\limits_{i=0}^{s}|D_x^iu|^p{\rm d}x\Big)^{1\over p},   1\leq p\leq \infty. $

并记 $H^s=H^s(\Omega)=W^{k, 2}(\Omega)$, 相应的范数为 $\|\cdot\|_s=\|\cdot\|_{s, 2}=\|\cdot\|_{s, 2, \Omega}$.

首先将区间 $\Omega=[a, b]$分成 $N$个子区间: $ a=x_0<x_1<x_2<$ $\cdots<x_i<\cdots<x_N=b$, 记 $I_i=[x_{i-1}, x_i]$, $h_i=x_i-x_{i-1}$, $h=\max\limits_{1\leq i\leq N} h_{i}$ ( $i=1, 2, \cdots, N$), $T_h=\{I_i\}_{i=1}^{N}$.并引入相应的对偶剖分, 取每个小区间的中点 $x_{i-1/2}=(x_{i-1}+x_i)/2$, 记 $I_{0}^*=[x_{0}, x_{1/2}]$, $I_{i}^*=[x_{i-{1}/{2}}, x_{i+{1}/{2}}]$ ( $1\leq i \leq N-1$), $I_{N}^*=[x_{N-{1}/{2}}, x_{N}]$.则所有的 $I_{i}^*$ $(0\leq i\leq N)$构成对偶剖分 $T_h^*$, $I_{i}^*$被称为对偶体积.

记

$ H_E^1(\Omega\times [0, T])=\Big\{u(x, t)\in H^1(\Omega\times [0, T])\Big |u(a, t)={{\partial u (b, t)}\over{\partial x}}=0\Big\}.\\ {\cal U}=H^1_E(\Omega\times [0, T])\cap H^2(\Omega\times [0, T]). $

于是引入解空间 ${\cal U}_h\subset {\cal U}$, 定义在 $x_i$ $(1\leq i\leq N-1)$点的基函数为

$ \varphi_i(x)=\left\{ \begin{array}{ll} 1-h_i^{-1}(x_i-x),&x_{i-1}\leq x< x_i, \\ 1-h_{i+1}^{-1}(x-x_i),&x_{i}\leq x\leq x_{i+1}, \\ 0,&\mbox{其它.} \end{array}\right.\\ \varphi_N(x)=\left\{ \begin{array}{ll} 1-h_N^{-1}(x_N-x),&x_{N-1}\leq x\leq x_N, \\ 0,&\mbox{其它.} \end{array}\right. $

对任意的 $u_h\in {\cal U}_h$, 有 $u_h=\sum\limits_{i=1}^{N}u_i\varphi_i(x)$, 其中 $u_i=u_h(x_i)$.

同时在对偶剖分 $T_h^*$上定义检验函数空间 ${\cal V}_h$, 其基函数为 $\varphi_i^*(x)$

$ \varphi_i^*(x)=\left\{ \begin{array}{ll} 1, \ &x_{i-1/2}\leq x\leq x_{i+1/2}, \\ 0, &\mbox{其它.} \end{array}\right.\\ \varphi_N^*(x)=\left\{ \begin{array}{ll} 1, \ &x_{N-1/2}\leq x\leq x_{N}, \\ 0, &\mbox{其它.} \end{array}\right. $

则对任意的 $v_h\in{\cal V}_h$, 有 $v_h=\sum\limits_{i=1}^{N}v_i\varphi_i^*(x)$, 其中 $v_i=v_h(x_i)$.

对任意的 $u\in {\cal U}$, 定义两个插值算子 $I_h: {\cal U}\rightarrow{\cal U}_h$和 $I_h^*: {\cal U}\rightarrow{\cal V}_h$, 有

$ I_hu=\sum\limits_{i=1}^{N}u(x_i)\varphi_i(x),    I_h^*u=\sum\limits_{i=1}^{N}u(x_i)\varphi_i^*(x). $

假定剖分为拟一致正则剖分, 如下逆不等式成立^[16]

$ \Big\|\frac{\partial \chi}{\partial x}\Big\|\leq Ch^{-1}\|\chi\|, ~~~~ \Big\|\frac{\partial \chi} {\partial x}\Big\|_{\infty}\leq Ch^{-1}\Big\|\frac{\partial \chi}{\partial x}\Big\|, ~~~~~\forall \chi\in {\cal U}_h. $

问题(1.1)相应的变分形式为:求 $u\in{\cal U}$满足

$ (u_{t}, v)+a(u;u, v)=(f(u, x, t), v), ~\forall v\in{\cal U}, $

(2.1)

其中

$ a(u;u, v)=\int_{a}^{b}A(u)\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}{\rm d}x. $

于是可以得到问题(1.1)的有限体积元逼近:求 $u_h\in{\cal U}_h$满足

$ (u_{h, t}, v_h)+a(u_h;u_h, v_h)=(f(u_h, x, t), v_h), ~\forall v_h\in{\cal V}_h, $

(2.2)

其中

$ a(u_h;u_h, v_h)=-\sum\limits_{i=1}^{N}v(x_i)A(u_h)\frac{\partial u_h}{\partial x}\Big |_{I_i^*}. $

引入插值算子 $\Pi_h:~{\cal U}_h\rightarrow{\cal V}_h$, $\forall w\in {\cal U}_h$, 有

$ \Pi_hw=\sum\limits_{i=1}^{N}w(x_i)\varphi_i^*(x). $

定义 $\varepsilon_a(\cdot;\cdot, \cdot)$和 $\varepsilon(\cdot, \cdot)$分别为

$ \varepsilon_a(\varrho;u_h, v_h)=a(\varrho;u_h, \Pi_hv_h)-a(\varrho;u_h, v_h), \forall \varrho, u_h, v_h\in{\cal U}_h, \\ \varepsilon(\vartheta, \sigma)=(\vartheta, \Pi_h\sigma)-(\vartheta, \sigma). $

定义 $\langle\cdot, \cdot\rangle$为

$ \langle\Phi, \Psi\rangle=(\Phi, \Pi_h\Psi), ~\forall~\Phi, \Psi\in{\cal U}_h. $

于是(2.2)式可以改写为

$ \langle {u_{h, t}, \chi}\rangle+a(u_h;u_h, \Pi_h\chi)=(f(u_h, x, t), \Pi_h\chi), ~\forall \chi\in{\cal U}_h. $

(2.3)

为了给出全离散有限体积元格式, 令 $\Delta t=T/J$, $J$为正整数, $t^n=n\Delta t$, $n=0, 1, \cdots, J$.于是可以得到全离散向后欧拉有限体积元格式

$ \langle\bar{\partial}u_h^n, \chi\rangle+a(u^n_h;u^n_h, \Pi_h\chi) =(f(u^n_h, x, t^n), \Pi_h\chi), \forall \chi\in{\cal U}_h. $

(2.4)

其中 $\bar{\partial}u_h^n=(u^n_h-u^{n-1}_h)/\Delta t$.

构造粗网格有限体积元空间 ${\cal U}_H$及细网格有限体积元空间 ${\cal U}_h$, 在粗网格空间求解原始的非线性有限体积元逼近, 得到粗网格有限体积元解, 将此解作为细网格空间非线性项的初值, 将非线性问题线性化, 进而在细网格空间仅求解一个线性问题, 简化计算.具体算法为

第一步  在粗网格 $T_H$上求解 $u_H^n\in{\cal U}_H$

$ \langle\bar{\partial}u_H^n, \chi\rangle+a(u^n_H;u^n_H, \Pi_H\chi)=(f(u^n_H, x, t^n), \Pi_H\chi), \forall \chi\in{\cal U}_H. $

第二步  在细网格 $T_h$上求解 $u^n_h\in{\cal U}_h$

$ \langle\bar{\partial}u_h^n, \chi\rangle+a(u^n_H;u^n_h, \Pi_h\chi)=(f(u_H^n)+f'(u_H^n)(u_h^n-u_H^n), \Pi_h\chi), ~ \forall \chi\in{\cal U}_h. $

首先给出一些有用的引理, 并讨论一维非线性抛物问题有限体积元解的存在性.

引理3.1^{[1, 6]}   $\forall u_h, v_h\in {\cal U}_h$, 存在与 $h$无关的正常数 $C$, 满足

$ (u_h, \Pi_h v_h)=(v_h, \Pi_h u_h), ~(u_h, \Pi_h v_h)\leq C\|u_h\|\|v_h\|. $

记 $|||u|||\equiv\langle u, u\rangle^{\frac{1}{2}}$, 由引理3.1可知, $|||u|||$和 $\|u\|$是等价的.

引理3.2^{[1, 6]}   $\forall \omega_h, u_h, v_h\in {\cal U}_h$, 存在与 $h$无关的正常数 $\alpha$和 $C$, 满足

$ \alpha\|u_h\|_1^2\leq a(\omega_h;u_h, \Pi_h u_h), ~~~~a(\omega_h;u_h, \Pi_h v_h)\leq C\|u_h\|_1\|v_h\|_1. $

引理3.3^[9-10]   $\forall \vartheta\in H^1, \sigma\in{\cal U}_h$, 有

$ |\varepsilon(\vartheta, \sigma)|\leq Ch^2\Big\|\frac{\partial \vartheta} {\partial x}\Big\|\Big\|\frac{\partial \sigma}{\partial x}\Big\|, |\varepsilon(\vartheta, \sigma)|\leq C\|\vartheta\|\|\sigma\|. $

引理3.4^[9-10]   $\forall\varrho, ~u_h, ~v_h\in {\cal U}_h$, 存在与 $h$无关的正常数 $C$, 满足

$ |\varepsilon_a(\varrho;u_h, v_h)|\leq Ch^2\Big\|\frac{\partial\varrho}{\partial x}\Big\|_{\infty}\Big\|\frac{\partial u_h} {\partial x}\Big\|\Big\|\frac{\partial v_h}{\partial x}\Big\|. $

引理3.5^[9-10]   $\forall\chi, ~\varphi\in {\cal U}_h$, 存在与 $h$无关的正常数 $C$, 满足

$ |\varepsilon_a(\nu;\varphi, \chi)-\varepsilon_a(\omega;\varphi, \chi)|\leq Ch\Big\|\frac{\partial \varphi}{\partial x}\Big\|_{\infty}\Big(1+\Big\|\frac{\partial \omega}{\partial x}\Big\|_{\infty}\Big) \Big\|\frac{\partial }{\partial x}(\nu-\omega)\Big\|\Big\|\frac{\partial \chi}{\partial x}\Big\|. $

定义椭圆投影 $R_h:{\cal U}\rightarrow{\cal U}_h$, 有

$ a(u;R_hu, \chi)=a(u;u, \chi), ~~~~~\forall \chi\in{\cal U}_h. $

(3.1)

则有如下引理成立.

引理3.6^[17]  令 $w_h=R_hu, ~\rho=w_h-u$, 则存在与 $h$无关的正常数 $C$, 满足

$ \Big\|\frac{\partial w_h}{\partial x}\Big\|_{\infty}\leq C(u), ~\|\rho\|\leq C(u)h^2, ~\Big\|\frac{\partial \rho}{\partial x}\Big\|\leq C(u)h, ~\|\rho\|_{0, \infty}\leq Ch| \ln h|\|u\|_{1, \infty}. $

令 ${\cal J}_h:{\cal U}_h\rightarrow{\cal U}_h$如下定义

$ \langle{\cal J}_h\upsilon-u^{n-1}_h, \chi\rangle+\Delta ta(\upsilon;{\cal J}_h\upsilon, \Pi_h\chi) =\Delta t(f(\upsilon, x, t^n), \Pi_h\chi), ~~~~~~\forall \chi\in{\cal U}_h. $

(3.2)

如果 ${\cal J}_h$存在不动点, 则 $u^n_h=\upsilon$是方程(3.2)的解.于是可以可得如下定理.

定理3.1  在给定的条件假设下, 方程(3.2)的解是存在的.

证  利用Brouwer不动点定理进行证明

$ |||{\cal J}_h\mu-{\cal J}_h\omega|||<L|||\mu-\omega|||, ~~\forall\mu, \omega\in{\cal U}_h, ~ 0<L<1. $

对于任意的 $\mu, ~\omega, ~\chi\in{\cal U}_h$, 由(3.2)式得到

$ \begin{eqnarray}\label{3.2} &&\langle{\cal J}_h\mu-{\cal J}_h\omega, \chi\rangle+\Delta ta(\mu;{\cal J}_h\mu, \Pi_h\chi) -\Delta ta(\omega;{\cal J}_h\omega, \Pi_h\chi)\\ &=&\Delta t(f(\mu, x, t^n), \Pi_h\chi)-\Delta t(f(\omega, x, t^n), \Pi_h\chi). \end{eqnarray} $

(3.3)

令 $\chi={\cal J}_h\mu-{\cal J}_h\omega$, 于是

$ \begin{eqnarray}\label{3.3} &&|||\chi|||^2+\Delta ta(\omega;\chi, \Pi_h\chi)\\ &=&\Delta ta(\omega;{\cal J}_h\mu, \Pi_h\chi)-\Delta ta(\mu;{\cal J}_h\mu, \Pi_h\chi)+\Delta t(f(\mu, x, t)-f(\omega, x, t), \Pi_h\chi) \\ &=&\Delta t(a(\omega;{\cal J}_h\mu, \chi)-a(\mu;{\cal J}_h\mu, \chi))+\Delta t(\varepsilon_a(\mu;{\cal J}_h\mu, \chi)-\varepsilon_a(\omega;{\cal J}_h\mu, \chi)) \\ && +\Delta t(f(\mu, x, t)-f(\omega, x, t), \Pi_h\chi)\\ &\equiv &I+II+III. \end{eqnarray} $

(3.4)

对 $I$, 有

$ \begin{eqnarray} a(\omega;{\cal J}_h\mu, \chi)-a(\mu;{\cal J}_h\mu, \chi) &=&\int^b_a(A(\omega)-A(\mu))\frac{\partial {\cal J}_h\mu}{\partial x}\frac{\partial \chi}{\partial x}{\rm d}x \\ &\leq &C_1\Big\|\frac{\partial {\cal J}_h\mu}{\partial x}\Big\|_{\infty}\|\omega-\mu\|~\Big\|\frac{\partial \chi}{\partial x}\Big\|. \label{3.4}\end{eqnarray} $

(3.5)

对 $II$, 由引理3.5和逆不等式得到

$ \begin{equation} \mid II\mid\leq C_2\Delta t h\Big\|\frac{\partial (\mu-\omega)}{\partial x}\Big\|\Big\|\frac{\partial \chi}{\partial x}\Big\|\leq C_2\Delta t\|\mu-\omega\|~\Big\|\frac{\partial \chi}{\partial x}\Big\|. \label{3.5}\end{equation} $

(3.6)

由(1.2)式和Poincaré不等式可以得到对 $III$的估计

$ \begin{equation} |III|\leq C\Delta t\|\mu-\omega\|\|\chi\|\leq C\Delta t\|\mu-\omega\|\Big\|\frac{\partial \chi}{\partial x}\Big\|. \label{3.6}\end{equation} $

(3.7)

综合(3.5)、(3.6)和(3.7)式, 可得

$ |||\chi|||^2+\Delta t\bar{\alpha}\Big\|\frac{\partial \chi}{\partial x}\Big\|^2\leq C_3\Delta t\|\mu-\omega\|~\Big\|\frac{\partial \chi}{\partial x}\Big\|\leq C_4\Delta t\|\mu-\omega\|^2+\Delta t\bar{\alpha}\Big\|\frac{\partial \chi}{\partial x}\Big\|^2. $

于是 $|||\chi|||\leq C_5\|\mu-\omega\|$.由 $|||\cdot|||$和 $\|\cdot\|$的等价性知定理成立.

本节主要研究一维非线性抛物问题有限体积元及其两层网格有限体积元逼近的先验误差估计.首先对有限体积元逼近, 证明具有最优阶 $L^2$模和 $H^1$模误差估计.

定理4.1  令 $U^n$和 $u^n$分别为问题(2.4)和(2.1)的解, 在解 $u$满足一定的正则性假设下, 有误差估计

$ \|U^n-u^n\|\leq{\cal C}(\Delta t+h^2). $

证  令 $u^n=u(t_n)$, $w_h^n=R_hu^n$, $U^n-u^n=(U^n-w_h^n)+$ $(w_h^n-u^n)\equiv\theta^n+\rho^n$. $\forall\chi\in{\cal U}_h$, 有

$ \begin{eqnarray} (\bar{\partial}\theta^n, \chi)+a(U^n;\theta^n, \chi) &=&(\bar{\partial}U^n, \chi)+a(U^n;U^n, \chi)-(\bar{\partial}w_h^n, \chi)-a(U^n;w_h^n, \chi) \\ &=&(f(U^n), \chi)+\delta-(u^n_t, \chi)-(\bar{\partial}w_h^n-u^n_t, \chi)\\ &&-a(u^n;w_h^n, \chi)-[a(U^n;w_h^n, \chi)-a(u^n;w_h^n, \chi)], \end{eqnarray} $

(4.1)

其中

$ \delta=\varepsilon(f, \chi)-\varepsilon(\bar{\partial}U^n, \chi)-\varepsilon(U^n;U^n, \chi) =\varepsilon(f-\bar{\partial}U^n, \chi)-\varepsilon(U^n;U^n, \chi). $

由 $R_h$的定义, 得到

$ (\bar{\partial}\theta^n, \chi)+a(U^n;\theta^n, \chi)\\ =(f(U^n)-f(u^n), \chi)+\delta-(\bar{\partial}\rho^n, \chi)-(\bar{\partial}u^n-u^n_t, \chi) \\ -[a(U^n;w_h^n, \chi)-a(u^n;w_h^n, \chi)] . $

(4.2)

令 $\chi=\theta^n$, 由(1.2)式及引理3.6, 得

$ \begin{eqnarray} \frac{1}{2}\bar{\partial}\|\theta^n\|^2+\mu\Big\|\frac{\partial \theta^n}{\partial x}\Big\|^2 &\leq&C(\|U^n-u^n\|)\Big(\|\theta^n\|+\Big\|\frac{\partial \theta^n}{\partial x}\Big\|\Big) \\ && +(\|\bar{\partial}\rho^n\|+\|\bar{\partial}u^n-u^n_t\|)\|\theta^n\|+|\delta| .\end{eqnarray} $

(4.3)

由引理3.3、3.4, 可得

$ |\delta|\leq Ch^2\Big\|\frac{\partial \theta^n}{\partial x}\Big\|. $

因此

$ \begin{equation} \bar{\partial}\|\theta^n\|^2+\mu\Big\|\frac{\partial \theta^n}{\partial x}\Big\|^2\leq C(\|\theta^n\|^2+E_n), \end{equation} $

(4.4)

其中

$ E_n=\|\rho^n\|^2+\|\bar{\partial}\rho^n\|^2+\|\bar{\partial}u^n-u^n_t\|^2+h^4. $

于是

$ (1-C\Delta t)\|\theta^n\|^2\leq\|\theta^{n-1}\|^2+C\Delta t E_n. $

对于充分小的 $\Delta t$, 有

$ \|\theta^n\|^2\leq(1+C\Delta t)\|\theta^{n-1}\|^2+C\Delta t E_n. $

注意到 $\theta^0=0$, 因此

$ \begin{eqnarray} \|\theta^n\|^2 &\leq&(1+C\Delta t)^n\|\theta^{0}\|^2+C\Delta t\sum\limits_{j=1}^{n}(1+C\Delta t)^{n-j}E_j \\ &=& C\Delta t\sum\limits_{j=1}^{n}(1+C\Delta t)^{n-j}E_j. \end{eqnarray} $

(4.5)

由引理3.6可得

$ \|\bar{\partial}\rho^j\|=\Big\|\Delta t^{-1}\int_{t_{j-1}}^{t_{j}}\rho_t{\rm d}s\Big\|\leq C(u)\Delta t, \\ \|\bar{\partial}u^j-u^j_t\|=\Big\|\Delta t^{-1}\int_{t_{j-1}}^{t_{j}}(s-t_{j-1})u_{tt}(s){\rm d}s\Big\|\leq C(u)\Delta t. $

于是

$ E_j\leq C(u)(h^2+\Delta t)^2, $

因此

$ \|\theta^n\|\leq C(u)(h^2+\Delta t). $

定理得证.

对于有限体积元方法的 $H^1$ -模误差估计, 有如下定理成立.

定理4.2  令 $U^n$和 $u^n$分别为问题(2.4)和(2.1)的解, 在解 $u$满足一定的正则性假设下, 有误差估计

$ \|U^n-u^n\|_1\leq{\cal C}(\Delta t+h). $

证  在方程(4.2)中取 $\chi=\bar{\partial}\theta^n$, 有

$ (\bar{\partial}\theta^n, \bar{\partial}\theta^n)+a(U^n;\theta^n, \bar{\partial}\theta^n) =\\ (f(U^n)-f(u^n), \bar{\partial}\theta^n)+\delta-(\bar{\partial}\rho^n, \bar{\partial}\theta^n) -(\bar{\partial}u^n-u^n_t, \bar{\partial}\theta^n)\\ -[a(U^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)]. $

(4.6)

对双线性形式 $a(\cdot;\cdot, \cdot)$, 有

$ 2\Delta ta(U^n;\theta^n, \bar{\partial}\theta^n)=a(U^n;\theta^n, \theta^n)-a(U^n;\theta^{n-1}, \theta^{n-1}) +\Delta t^2a(U^n;\bar{\partial}\theta^n, \bar{\partial}\theta^n). $

结合(4.6)式可得

$ \begin{eqnarray}\label{4.7} &&2\Delta t\|\bar{\partial}\theta^n\|^2+a(U^n;\theta^n, \theta^n)-a(U^{n};\theta^{n-1}, \theta^{n-1})+ \alpha\Delta t^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|^2 \\ &\leq &2\Delta t\Big\{ (f(U^n)-f(u^n), \bar{\partial}\theta^n)+\delta-(\bar{\partial}\rho^n, \bar{\partial}\theta^n) -(\bar{\partial}u^n-u^n_t, \bar{\partial}\theta^n)\\ &&-[a(U^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)]\Big\}, \end{eqnarray} $

(4.7)

其中

$ \begin{eqnarray} && a(U^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)= \Big(\Big[A(U^n)-A(u^n)\Big]\frac{\partial w_h^n}{\partial x}, \frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big) \\ &=&\Big(\Big[A(U^n)-A(w_h^n)\Big]\frac{\partial w_h^n}{\partial x}, \frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big)+ \Big(\Big[A(w_h^n)-A(u^n)\Big]\frac{\partial w_h^n}{\partial x}, \frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big) \\ &=&\Big(\frac{\partial }{\partial x}\Big[A(U^n)-A(w_h^n)\frac{\partial w_h^n}{\partial x}\Big], \bar{\partial}\theta^n\Big)+ \Big(\Big[A(w_h^n)-A(u^n)\Big]\frac{\partial w_h^n}{\partial x}, \frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big). \end{eqnarray} $

(4.8)

由(1.2)式及引理3.6, 可得

$ \begin{equation} \Big |\Big(\Big[A(w_h^n)-A(u^n)\Big]\frac{\partial w_h^n}{\partial x}, \frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big)\Big |\leq Ch^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|. \label{4.9} \end{equation} $

(4.9)

$ \begin{eqnarray}\label{4.10} &&\Big|\Big(\frac{\partial }{\partial x}\Big[A(U^n)-A(w_h^n)\frac{\partial w_h^n}{\partial x}\Big], \bar{\partial}\theta^n\Big)\Big| \\ &\leq& C\Big[\Big\|(A'(U^n)-A'(w_h^n))\frac{\partial w_h^n}{\partial x}\Big\|+ \Big\|A'(U^n)\frac{\partial }{\partial x}(U^n-w_h^n)\Big\|\Big]\|\bar{\partial}\theta^n\| \\ &\leq &C\Big(\|U^n-w_h^n\|+\Big\|\frac{\partial }{\partial x}(U^n-w_h^n)\Big\|\Big)\|\bar{\partial}\theta^n\|. \end{eqnarray} $

(4.10)

结合(4.9)和(4.10)式, 并利用Poincaré不等式, 有

$ |a(U^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)|\leq C \Big\|\frac{\partial }{\partial x}(U^n-w_h^n)\Big\|\|\bar{\partial}\theta^n\|+ Ch^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|. $

注意到 $\theta^n=U^n-w_h^n$, 因此

$ |a(U^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)|\leq C \Big\|\frac{\partial }{\partial x}\theta^n\Big\|\|\bar{\partial}\theta^n\|+ Ch^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|. $

于是可以得到(4.7)式右端项的估计

$ \begin{eqnarray}\label{4.11} &&2\Delta t \Big\{(f(U^n)-f(u^n), \bar{\partial}\theta^n)+\delta-(\bar{\partial}\rho^n, \bar{\partial}\theta^n) \\ &&-(\bar{\partial}u^n-u^n_t, \bar{\partial}\theta^n) -[a(U^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)]\Big\}\\ &\leq &2\Delta t\Big\{(C\|U^n-u^n\|)\|\bar{\partial}\theta^n\|+ C \Big\|\frac{\partial }{\partial x}\theta^n\Big\|\|\bar{\partial}\theta^n\|\\ &&+Ch^2\Big\|\frac{\partial }{\partial x}\bar{\partial}\theta^n\Big\|+ (\|\bar{\partial}\rho^n\|+\|\bar{\partial}u^n-u^n_t\|)\|\bar{\partial}\theta^n\|+|\delta|\Big\}, \end{eqnarray} $

(4.11)

其中

$ |\delta|\leq Ch^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|. $

由(4.11)和(4.7)式可得

$ a(U^n;\theta^n, \theta^n)-a(U^{n};\theta^{n-1}, \theta^{n-1})\leq C\Delta t\Big(\Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2+E_n\Big), $

其中

$ E_n=\|U^n-u^n\|^2+\|\bar{\partial}\rho^n\|^2+\|\bar{\partial}u^n-u^n_t\|^2+h^4/\Delta t. $

将 $a(U^{n};\theta^{n-1}, \theta^{n-1})$移到方程的右端, 结合(1.2)式可得

$ \alpha\Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2\leq a(U^n;\theta^n, \theta^n)\leq a(U^{n};\theta^{n-1}, \theta^{n-1})+C\Delta t\Big(\Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2+E_n\Big).\\ (1-C\Delta t)\Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2\leq Ca(U^{n};\theta^{n-1}, \theta^{n-1})+C\Delta tE_n \leq C\Big\|\frac{\partial }{\partial x}\theta^{n-1}\Big\|^2+C\Delta tE_n. $

当 $\Delta t$充分小时有

$ \Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2\leq(1+C\Delta t)\|\frac{\partial }{\partial x}\theta^{n-1}\|^2 +C\Delta t E_n. $

注意到 $\theta^0=0$, 因此

$ \begin{eqnarray} \Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2 &\leq&(1+C\Delta t)^n \Big\|\frac{\partial }{\partial x}\theta^{0}\Big\|^2+C\Delta t\sum\limits_{j=1}^{n}(1+C\Delta t)^{n-j}E_j \\ &= &C\Delta t\sum\limits_{j=1}^{n}(1+C\Delta t)^{n-j}E_j. \end{eqnarray} $

(4.12)

由定理4.1及对 $E_j$的估计, 有

$ \Big\|\frac{\partial }{\partial x}\theta^n\Big\|\leq{\cal C}(\Delta t+h+\frac{h^2}{\Delta t^{1/2}}). $

取 $h^2/\Delta t$为常数, 有

$ \Big\|\frac{\partial }{\partial x}\theta^n\Big\| \leq{\cal C}(\Delta t+h). $

结合引理3.6及三角不等式, 有

$ \|U^n-u^n\|_1\leq{\cal C}(\Delta t+h). $

证毕.

下面将给出两层网格解和其相应的粗网格解之间的误差, 在定理4.3的证明中将起到关键作用.

引理4.1  令 $U^n$和 $U_H^n$分别为两层网格算法解和其相应的粗网格解, 在解 $u$满足一定的正则性假设下, 有误差估计

$ \begin{equation}\|U^n-U_H^n\|\leq{\cal C}(\Delta t+H^2).\label{lemma4.1} \end{equation} $

(4.13)

证  令 $U^n-u^n=U^n-w_h^n+w_h^n-$ $u^n\equiv\theta^n+\rho^n.$对两层网格有限体积元格式, 有误差估计式

$ (\bar{\partial}\theta^n, \chi)+a(U_H^n;\theta^n, \chi) =(\bar{\partial}U^n, \chi)+a(U_H^n;U^n, \chi)-(\bar{\partial}w_h^n, \chi)-\\ a(U_H^n;w_h^n, \chi)\\ =(f(U_H^n)+f'(U_H^n)(U^n-U_H^n), \chi)+\delta-(u^n_t, \chi)-(\bar{\partial}w_h^n-u^n_t, \chi)\\ -a(u^n;w_h^n, \chi)-[a(U_H^n;w_h^n, \chi)-a(u^n;w_h^n, \chi)], $

(4.14)

其中 $\delta=\varepsilon(f(U_H^n)+f'(U_H^n)(U^n-U_H^n), $ $\chi)-\varepsilon(\bar{\partial}U^n, \chi)- \varepsilon(U_H^n;U^n, \chi)$.

由 $R_h$的定义, 可得

$ (\bar{\partial}\theta^n, \chi)+a(U_H^n;\theta^n, \chi) =(f(U_H^n)+\\ f'(U_H^n)(U^n-U_H^n)-f(u^n), \chi)+\delta-(\bar{\partial}\rho^n, \chi)\\ -(\bar{\partial}u^n-u^n_t, \chi)-[a(U_H^n;w_h^n, \chi)-a(u^n;w_h^n, \chi)]. $

(4.15)

在上式中取 $\chi=\theta^n$.对上式右端第一项, 利用Taylor展示, 有

$ f(u^n)=f(U_H^n)+f'(U_H^n)(u^n-U_H^n)+\frac{1}{2}f''(\tilde{u}^n)(u^n-U_H^n)^2, $

其中 $\tilde{u}^n$为介于 $u^n$和 $U_H^n$之间的值.于是

$ f(U_H^n)+f'(U_H^n)(U^n-U_H^n)-f(u^n)=f'(U_H^n)(U^n-u^n)-\\ \frac{1}{2}f''(\tilde{u}^n)(u^n-U_H^n)^2.\\ \|(u^n-U_H^n)^2\|^2 \leq \|u^n-U_H^n\|^2_{0, \infty}\|u^n-U_H^n\|^2\\ \leq (\|u^n-R_Hu^n\|_{0, \infty}+\|R_Hu^n-U_H^n\|_{0, \infty})^2\|u^n-U_H^n\|^2. $

(4.16)

由定理4.1、引理3.6及逆不等式的性质, 可得

$ \begin{eqnarray} \|(u^n-U_H^n)^2\|^2 &\leq& C\Big[H|\ln H|+H^{-1}(\Delta t+H^2)\Big]^2(\Delta t+H^2)^2\\ & \leq &C\Big[H|\ln H|\Delta t+H^3|\ln H|+H^{-1}(\Delta t)^2+2H\Delta t+H^3\Big]^2. \end{eqnarray} $

(4.17)

取 $H$和 $\Delta t$满足 $H^{-1}\Delta t<\tilde{C}$, 有

$ \|(u^n-U_H^n)^2\|^2 \leq C(\Delta t+H^3|\ln H|)^2. $

对于(4.15)式右端最后一项, 有

$ \begin{eqnarray} && a(U_H^n;w_h^n, \theta^n)-a(u^n;w_h^n, \theta^n)\\ &=& a(U_H^n;w_h^n, \theta^n)-a(U^n;w_h^n, \theta^n)+ a(U^n;w_h^n, \theta^n)-a(u^n;w_h^n, \theta^n). \end{eqnarray} $

(4.18)

对上式右端第一项有

$ \begin{eqnarray}\label{4.18} a(U_H^n;w_h^n, \theta^n)-a(U^n;w_h^n, \theta^n) & =&\Big(\Big[A(U_H^n)-A(U^n)\Big]\frac{\partial w_h^n}{\partial x}, \frac{\partial \theta^n}{\partial x}\Big)\\ & \leq& C\| U_H^n-U^n\|\Big\|\frac{\partial \theta^n}{\partial x}\Big\|\\ & \leq &C(\| U_H^n-u^n\|+\|u^n-U^n\|)\Big\|\frac{\partial \theta^n}{\partial x}\Big\|. \end{eqnarray} $

(4.19)

由定理4.1可得

$ \| U_H^n-u^n\|\leq C(\Delta t+H^2). $

于是有

$ \begin{eqnarray} &&\frac{1}{2}\bar{\partial}\|\theta^n\|^2+\mu\Big\|\frac{\partial \theta^n}{\partial x}\Big\|^2\\ &\leq & C_1(\|U^n-u^n\|)(\|\theta^n\|+\Big\|\frac{\partial \theta^n}{\partial x}\Big\|)+ C_2(\Delta t+H^2)\Big\|\frac{\partial \theta^n}{\partial x}\Big\|\\ & &+C_3(\Delta t+H^3|\ln H|)^2\|\theta^n\|+C_4(\|\bar{\partial}\rho^n\|+\|\bar{\partial}u^n-u^n_t\|)\|\theta^n\|+\delta. \end{eqnarray} $

(4.20)

由引理3.3、3.4可得

$ |\delta|\leq Ch^2\Big\|\frac{\partial \theta^n}{\partial x}\Big\|. $

于是有

$ \begin{equation} \bar{\partial}\|\theta^n\|^2+\mu\Big\|\frac{\partial \theta^n}{\partial x}\Big\|^2\leq C(\|\theta^n\|^2+E_n), \end{equation} $

(4.21)

其中

$ E_n=\|\rho^n\|^2+\|\bar{\partial}\rho^n\|^2+\|\bar{\partial}u^n-u^n_t\|^2+h^4+(\Delta t+H^2+H^6|\ln H|^2)^2. $

当 $\Delta t$充分小时, 有下式成立

$ \|\theta^n\|^2\leq(1+C\Delta t)\|\theta^{n-1}\|^2+C\Delta t E_n. $

注意到 $\theta^0=0$, 因此

$ \begin{equation} \|\theta^n\|^2 \leq C\Delta t\sum\limits_{j=1}^{n}(1+C\Delta t)^{n-j}E_j. \end{equation} $

(4.22)

由引理3.6, 可得

$ \|\bar{\partial}\rho^j\|=\Big\|\Delta t^{-1}\int_{t_{j-1}}^{t_{j}}\rho_t{\rm d}s\Big\|\leq C(u)\Delta t, \\ \|\bar{\partial}u^j-u^j_t\|=\Big\|\Delta t^{-1}\int_{t_{j-1}}^{t_{j}}(s-t_{j-1})u_{tt}(s){\rm d}s\Big\|\leq C(u)\Delta t. $

于是有

$ \|\theta^n\|\leq C(u)(h^2+H^2+H^3|\ln H|+\Delta t)\leq C(u)(h^2+H^2+\Delta t). $

由引理3.6, 定理4.1及三角不等式, 可得结论成立

$ \begin{equation}\|U^n-U_H^n\|=\|\theta^n\|+\|\rho^n\|+\| u^n-U_H^n\|\leq{\cal C}(\Delta t+h^2+H^2)\\ \leq{\cal C}(\Delta t+H^2).\label{mainlemma} \end{equation} $

(4.23)

证毕.

下面我们将给出一维非线性抛物方程两层网格有限体积元逼近的 $H^1$ -模误差估计.

定理4.3  令 $U^n$和 $u^n$分别为两层网格有限体积元格式和方程(2.1)的解, 在解 $u$满足一定的正则性假设下, 有误差估计

$ \|U^n-u^n\|_1\leq{\cal C}(\Delta t+h+H^2). $

证  在误差方程(4.15)中取 $\chi=\bar{\partial}\theta^n$, 可得

$ \begin{eqnarray}\label{4.15} &&(\bar{\partial}\theta^n, \bar{\partial}\theta^n)+a(U^n_H;\theta^n, \bar{\partial}\theta^n) \\ &=&(f(U_H^n)+f'(U_H^n)(U^n-U_H^n)-f(u^n), \bar{\partial}\theta^n)+\delta-(\bar{\partial}\rho^n, \bar{\partial}\theta^n)\\ &&-(\bar{\partial}u^n-u^n_t, \bar{\partial}\theta^n)-[a(U_H^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)] .\end{eqnarray} $

(4.24)

对双线性形式的估计式有

$ \begin{equation}2\Delta ta(U_H^n;\theta^n, \bar{\partial}\theta^n)=a(U_H^n;\theta^n, \theta^n)-a(U_H^n;\theta^{n-1}, \theta^{n-1}) +\\ \Delta t^2a(U_H^n;\bar{\partial}\theta^n, \bar{\partial}\theta^n).\label{right}\end{equation} $

(4.25)

结合(4.24)和(4.25)式可得

$ \begin{eqnarray}\label{4.23} &&2\Delta t\|\bar{\partial}\theta^n\|^2+a(U^n_H;\theta^n, \theta^n)-a(U^{n}_H;\theta^{n-1}, \theta^{n-1})+ \alpha\Delta t^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|^2 \\ &\leq& 2\Delta t \Big\{(f(U_H^n)+f'(U_H^n)(U^n-U_H^n)-f(u^n), \bar{\partial}\theta^n)+\delta- (\bar{\partial}\rho^n, \bar{\partial}\theta^n)\\ &&-(\bar{\partial}u^n-u^n_t, \bar{\partial}\theta^n)-[a(U_H^n;w_h^n, \bar{\partial}\theta^n) -a(u^n;w_h^n, \bar{\partial}\theta^n)]\Big\}, \end{eqnarray} $

(4.26)

其中

$ \begin{eqnarray} && a(U_H^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)\\ &=& a(U_H^n;w_h^n, \bar{\partial}\theta^n)-a(U^n;w_h^n, \bar{\partial}\theta^n) +a(U^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n). \end{eqnarray} $

(4.27)

由定理4.2可得

$ \Big|a(U^n;w_h^n, \bar{\partial}\theta^n)-a(u^n;w_h^n, \bar{\partial}\theta^n)\Big|\leq C \Big\|\frac{\partial }{\partial x}\theta^n\Big\|\|\bar{\partial}\theta^n\|+ Ch^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|. $

由(4.19)式及引理4.1, 有

$ \Big|a(U_H^n;w_h^n, \bar{\partial}\theta^n)-a(U^n;w_h^n, \bar{\partial}\theta^n)\Big| \leq C(\Delta t+H^2)\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|. $

于是可得对(4.26)式右端项的估计式

$ \begin{eqnarray}\label{4.25} &&2\Delta t \Big\{(f(U_H^n)+f'(U_H^n)(U^n-U_H^n)-f(u^n), \bar{\partial}\theta^n)+\delta- (\bar{\partial}\rho^n, \bar{\partial}\theta^n)\\ &&-(\bar{\partial}u^n-u^n_t, \bar{\partial}\theta^n)-[a(U_H^n;w_h^n, \bar{\partial}\theta^n) -a(u^n;w_h^n, \bar{\partial}\theta^n)]\Big\}\\ &\leq& 2\Delta t\Big\{(C\|U^n-u^n\|)\|\bar{\partial}\theta^n\|+C(\Delta t+H^3|\ln H|)^2\|\bar{\partial}\theta^n\| +C \Big\|\frac{\partial }{\partial x}\theta^n\Big\|\|\bar{\partial}\theta^n\|\\ &&+Ch^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\| +C(\Delta t+H^2)\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|\\ &&+(\|\bar{\partial}\rho^n\|+\|\bar{\partial}u^n-u^n_t\|)\|\bar{\partial}\theta^n\|+Ch^2\Big\|\frac{\partial }{\partial x}(\bar{\partial}\theta^n)\Big\|\Big\}, \end{eqnarray} $

(4.28)

由(4.26)和(4.28)式, 可得

$ a(U^n_H;\theta^n, \theta^n)-a(U^{n}_H;\theta^{n-1}, \theta^{n-1})\leq C\Delta t\Big(\Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2+E_n\Big), $

其中

$ E_n=\|U^n-u^n\|^2+\|\bar{\partial}\rho^n\|^2+\|\bar{\partial}u^n-u^n_t\|^2+h^2+(\Delta t+H^2)^2. $

将 $a(U^{n}_H;\theta^{n-1}, \theta^{n-1})$移到右端, 整理可得

$ \alpha\Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2\leq a(U^n_H;\theta^n, \theta^n)\leq a(U^{n}_H;\theta^{n-1}, \theta^{n-1})+C\Delta t\Big(\Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2+E_n\Big).\\ (1-C\Delta t)\Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2\leq Ca(U^{n}_H;\theta^{n-1}, \theta^{n-1})+C\Delta tE_n \leq C\Big\|\frac{\partial }{\partial x}\theta^{n-1}\Big\|^2+C\Delta tE_n. $

于是

$ \Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2\leq(1+C\Delta t)\Big\|\frac{\partial }{\partial x}\theta^{n-1}\Big\|^2 +C\Delta t E_n. $

注意到 $\theta^0=0$, 因此

$ \begin{equation} \Big\|\frac{\partial }{\partial x}\theta^n\Big\|^2 \leq C\Delta t\sum\limits_{j=1}^{n}(1+C\Delta t)^{n-j}E_j. \end{equation} $

(4.29)

于是有

$ \Big\|\frac{\partial }{\partial x}\theta^n\Big\|\leq{\cal C}(\Delta t+h+H^2). $

结合引理3.6及三角不等式, 可得到本定理的结论

$ \|U^n-u^n\|_1\leq{\cal C}(\Delta t+h+H^2). $

证毕.

本节将给出数值算例来验证理论结果, 问题(1.1)的空间区域选取为 $[0, 1]$, 时间区间也选为 $[0, 1]$, 令 $A(u)=1/(1+u^2)$, $u(x, t)=x{\rm e}^{-x+t}$, $f(u)=u^2+g(x, t)$, $g(x, t)$可以由真解求出.对问题(1.1)的向后欧拉全离散有限体积元格式(2.4)采用迭代法来求解, 取不同的空间网格步长及相应的时间步长, 从 $T=0$计算到 $T=1$, 记NI为迭代次数, 结果如下表所示.

表 1表明有限体积元逼近的 $L^2$ -模和 $H^1$ -模分别具有二阶和一阶的收敛阶, 验证了定理4.1和定理4.2的结论.令 $H=1/N\ (N=4, 8, 16, 32, 64)$, $h=H^2$, $\Delta t=h$, 我们可以得到两层网格有限体积元逼近的误差估计.

从表 2和表 3可以看出, 两层网格有限体积元逼近关于 $H$具有二阶的收敛阶, 验证了定理4.3的结论.当满足关系 $h=O(H^2)$时, 我们可以得到最优阶 $H^1$ -模误差估计, 但是可以节省大量CPU计算时间.在实际计算过程中, 我们可以按照 $h<H$来选取空间步长以满足误差的要求.