许多物理现象的数学模型, 如结构振动、水波、声波、电磁波等的数学模型, 都可以描述为双曲型方程. 关于双曲型方程数值方法的研究, 已有许多的结果^[1-5]. 本文考虑如下非线性双曲问题

(1.1) $ \begin{eqnarray} & c(u)u_{tt}-\nabla\cdot(K\nabla u) = f, \quad (\mathit{\boldsymbol{x}}, t)\in\Omega\times J, \end{eqnarray} $

(1.2) $ \begin{eqnarray} & u(\mathit{\boldsymbol{x}}, t) = 0, \quad (\mathit{\boldsymbol{x}}, t)\in\partial\Omega\times J, \end{eqnarray} $

(1.3) $ \begin{eqnarray} &u(\mathit{\boldsymbol{x}}, 0) = u_0(\mathit{\boldsymbol{x}}), \ u_t(\mathit{\boldsymbol{x}}, 0) = u_1(\mathit{\boldsymbol{x}}), \quad \mathit{\boldsymbol{x}}\in\Omega, \end{eqnarray} $

其中$ \Omega\subset{{\Bbb R}} ^2 $为有界区域, $ \partial\Omega $为其光滑边界, $ J = (0, T] $, $ u_{tt} = \frac{\partial^2u}{\partial t^2} $, $ u_t = \frac{\partial u}{\partial t} $, $ K = K(\mathit{\boldsymbol{x}}) = (k_{ij}(\mathit{\boldsymbol{x}}))_{2\times 2} $是对称一致正定的$ 2\times2 $矩阵, $ c(u) $及其导数关于$ u $是Lipschitz连续, $ f $、$ u_0 $和$ u_1 $是充分光滑的已知函数. 该类问题在粘弹性材料中有着广泛的应用^[1].

20世纪90年代, 陈掌星^[6]提出了扩展混合有限元方法数值求解二阶椭圆型问题, 之后, 他进一步利用该方法分析了线性和拟线性椭圆型问题^[7-8]. 与此同时, Arbogast等人^[9]针对线性椭圆型问题, 提出了基于最低阶RTN元的扩展混合有限元方法. 与标准混合有限元方法相比, 扩展混合有限元方法能够同时高精度逼近未知函数、未知函数的梯度和通量. 由于该方法无需对扩散系数求逆, 因此更适应于求解扩散项较小与渗透项较低的微分方程. 在过去的几十年里, 该方法得到了广泛的应用和关注. Woodward-Dawson^[10]利用扩展混合有限元方法分析了非线性抛物型问题, 导出了半离散和全离散解的$ L^2 $误差估计. Gatica-Heuer^[11]分析了一类椭圆型方程的扩展混合有限元格式的$ L^2 $收敛性. 最近, 文献[12] 提出了一种新的扩展混合有限元方法数值求解抛物型积分微分方程. 文献[13] 使用最低阶RT元的扩展混合有限元方法研究了一类非线性和非局部抛物型问题. 受上述文献的启发, 本文研究问题(1.1)–(1.3) 扩展混合有限元半离散和全离散格式下的先验误差估计.

在本文中, $ C $表示不依赖于离散参数的正常数. $ L^p(\Omega) $表示定义在$ \Omega $的Banach空间, 用$ ||\cdot||_p $表示其范数. 对任意非负正整数$ m $, $ W^{m, p}(\Omega) = \{\mu\in L^p(\Omega), \ D^\vartheta\mu\in L^p(\Omega), |\vartheta|\leq m\} $为传统的Sobolev空间, 其范数定义为$ ||\mu||_{m, p}^p = \sum\limits_{|\vartheta|\leq m}||D^\vartheta\mu||_{L^p(\Omega)}^p $. 特别是, 当$ p = 2 $, 我们记$ H^m(\Omega) = W^{m, 2}(\Omega) $, 其范数简记为$ ||\cdot||_m = ||\cdot||_{m, 2} $, $ ||\cdot|| = ||\cdot||_{0, 2} $.

定义空间$ W = L^2(\Omega) $, $ \mathit{\boldsymbol{V}} = H({\rm div};\Omega) = \{\mathit{\boldsymbol{v}}:\mathit{\boldsymbol{v}}\in(L^2(\Omega))^2, \ \nabla\cdot \mathit{\boldsymbol{v}}\in L^2(\Omega)\}, $其模定义为

$ \|\mathit{\boldsymbol{v}}\|_{H({\rm div};\Omega)}\equiv(||\mathit{\boldsymbol{v}}||^2+||\nabla\cdot\mathit{\boldsymbol{v}}||^2)^{1/2}. $

对$ \Omega $进行拟一致三角剖分, 其单元直径不大于$ h $ ($ 0<h<1 $), 设$ W_h\times\mathit{\boldsymbol{V}}_h $为单元上的混合元空间, 其中$ W_h\times\mathit{\boldsymbol{V}}_h\subset W\times\mathit{\boldsymbol{V}} $. 本文使用Raviart-Thomas (RT$ _k $) 混合元空间^[14], 其指标$ k\geq0 $. 接下来的结论在$ RT_k $空间中成立, 即

(2.1) $ \begin{eqnarray} \nabla\cdot \mathit{\boldsymbol{v}}_h\in W_h, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{eqnarray} $

为了理论分析的需要, 引入$ L^2 $投影$ Q_{h}: W\rightarrow W_h $和$ Q_{h}^*: \mathit{\boldsymbol{V}}\rightarrow \mathit{\boldsymbol{V}}_h $, 满足

(2.2) $ \begin{equation} (\phi, w_h) = (Q_{h}\phi, w_h), \quad w_h\in W_h, \end{equation} $

(2.3) $ \begin{equation} (\varphi, \mathit{\boldsymbol{v}}_h) = (Q_{h}^*\varphi, \mathit{\boldsymbol{v}}_h), \quad \ \ \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{equation} $

考虑标准混合有限元空间的Fortin投影$ \Pi_h:(H^1(\Omega))^2\rightarrow \mathit{\boldsymbol{V}}_h $, 对任意$ \psi \in H({\rm div}, \Omega) $, 有

(2.4) $ \begin{equation} (\nabla\cdot\Pi_h \psi, w_h) = (\nabla\cdot \psi, w_h), \quad \forall w_h\in W_h. \end{equation} $

上述投影具有如下逼近性质^[14]

(2.5) $ \begin{equation} \|Q_{h}\phi\|_{0, q}\leq C\|\phi\|_{0, q}, \quad 2\leq q<\infty, \end{equation} $

(2.6) $ \begin{equation} \|\phi-Q_{h}\phi\|_{0, q}\leq C\|\phi\|_{r, q}h^r, \quad 0\leq r\leq k+1, \end{equation} $

(2.7) $ \begin{equation} \| \psi-\Pi_h \psi\|_{0, q}\leq C\| \psi\|_{r, q}h^r, \ 1/q< r\leq k+1, \end{equation} $

(2.8) $ \begin{equation} \|\nabla\cdot( \psi-\Pi_h \psi)\|_{0, q}\leq C\|\nabla\cdot \psi\|_{r, q}h^r, \ 0\leq r\leq k+1, \end{equation} $

其中$ \phi\in W^{k+1, q}(\Omega) $, $ \psi\in (W^{k+1, q}(\Omega))^2 $.

此外, 令$ c_* $, $ c^* $, $ K_* $, $ K^* $和$ K_1 $为正常数, 对函数$ c(u) $, $ K $和$ f $给出以下假设

(I) $ c_*\leq c(u)\leq c^* $;

(II) $ \forall \mathit{\boldsymbol{y}}\in{{\Bbb R}} ^2 $, 对于$ \mathit{\boldsymbol{x}}\in \bar{\Omega} $, $ K_*|\mathit{\boldsymbol{y}}|^2\leq \mathit{\boldsymbol{y}}^TK(\mathit{\boldsymbol{x}})\mathit{\boldsymbol{y}}\leq K^*|\mathit{\boldsymbol{y}}|^2 $;

(III) $ |f|, \left|\frac{\partial f}{\partial t}\right|, \left|\frac{\partial c}{\partial u}\right|, \left|\frac{\partial^2 c}{\partial u^2}\right|\leq K_1 $.

令$ \mathit{\boldsymbol{ \widetilde{p}} } = -\nabla u $, $ \mathit{\boldsymbol{p}} = K\mathit{\boldsymbol{ \widetilde{p}} } $, 则问题(1.1)–(1.3) 等价于

(2.9) $ \begin{equation} \left\{ \begin{array}{ll} c(u)u_{tt}+\nabla\cdot \mathit{\boldsymbol{ \widetilde{p}} } = f, \ (\mathit{\boldsymbol{x}}, t)\in\Omega\times J, & {\rm (a)}\\ \mathit{\boldsymbol{ \widetilde{p}} }+\nabla u = 0, \ (\mathit{\boldsymbol{x}}, t)\in\Omega\times J, & {\rm(b)}\\ \mathit{\boldsymbol{ \widetilde{p}} }-K\mathit{\boldsymbol{ \widetilde{p}} } = 0, \ (\mathit{\boldsymbol{x}}, t)\in\Omega\times J, & {\rm (c)}\\ u(\mathit{\boldsymbol{x}}, t) = 0, \ (\mathit{\boldsymbol{x}}, t)\in\partial\Omega\times J, &{\rm (d)}\\ u(\mathit{\boldsymbol{x}}, 0) = u_0(\mathit{\boldsymbol{x}}), \ u_t(\mathit{\boldsymbol{x}}, 0) = u_1(\mathit{\boldsymbol{x}}), \ \mathit{\boldsymbol{x}}\in\Omega. {\qquad}& \end{array} \right. \end{equation} $

问题(2.9) 的扩展混合变分问题为: 求$ (u, \mathit{\boldsymbol{ \widetilde{p}} }, \mathit{\boldsymbol{p}}):[0, T]\rightarrow W\times\mathit{\boldsymbol{V}} \times\mathit{\boldsymbol{V}} $, 满足

(2.10) $ \begin{eqnarray} \left\{ \begin{array}{lr} (c(u)u_{tt}, w)+(\nabla\cdot \mathit{\boldsymbol{ \widetilde{p}} }, w) = (f, w), \ \forall w\in W, {\qquad} &{\rm (a)}\\ (\mathit{\boldsymbol{ \widetilde{p}} }, \mathit{\boldsymbol{v}})-(\nabla\cdot\mathit{\boldsymbol{v}}, u) = 0, \ \forall \mathit{\boldsymbol{v}}\in \mathit{\boldsymbol{V}}, &{\rm (b)}\\ (\mathit{\boldsymbol{ \widetilde{p}} }, \mathit{\boldsymbol{v}})-(K\mathit{\boldsymbol{ \widetilde{p}} }, \mathit{\boldsymbol{v}}) = 0, \ \forall \mathit{\boldsymbol{v}}\in \mathit{\boldsymbol{V}}, &{\rm (c)}\\ u(\mathit{\boldsymbol{x}}, 0) = u_0(\mathit{\boldsymbol{x}}), \ u_t(\mathit{\boldsymbol{x}}, 0) = u_1(\mathit{\boldsymbol{x}}), \ \mathit{\boldsymbol{x}}\in\Omega. \end{array} \right. \end{eqnarray} $

定理 2.1  问题(2.9) 与(2.10) 是等价的.

证  容易得到问题(2.9) 的解是弱形式(2.10) 的解. 因此, 我们只需证明弱形式(2.10) 的解是问题(2.9) 的解. 取$ w = c(u)u_{tt}+\nabla\cdot \mathit{\boldsymbol{p}}-f $, 代入(2.10a) 式得

$ \begin{eqnarray*} \|c(u)u_{tt}+\nabla\cdot \mathit{\boldsymbol{ \widetilde{p}} }-f\|^2 = 0, \end{eqnarray*} $

从而有$ c(u)u_{tt}+\nabla\cdot \mathit{\boldsymbol{p}} = f $, 即(2.9a) 式成立. 由于$ H_0^1(\Omega)\subset \mathit{\boldsymbol{V}} $, 利用Green公式可得

(2.11) $ \begin{equation} \left\{\begin{array}{ll} (\widetilde{\mathit{\boldsymbol{ \widetilde{p}} }}, \mathit{\boldsymbol{v}})+(\nabla u, \mathit{\boldsymbol{v}}) = 0, \quad \forall \mathit{\boldsymbol{v}}\in \mathit{\boldsymbol{V}}, \\ u(\mathit{\boldsymbol{x}}, t) = 0, \quad (\mathit{\boldsymbol{x}}, t)\in\partial\Omega\times J. \end{array}\right. \end{equation} $

取$ \mathit{\boldsymbol{v}} = \mathit{\boldsymbol{ \widetilde{p}} }+\nabla u $, 代入(2.10b) 式得

(2.12) $ \begin{eqnarray} \left\{\begin{array}{ll} \widetilde{\mathit{\boldsymbol{ \widetilde{p}} }}+\nabla u = 0, \quad (\mathit{\boldsymbol{x}}, t)\in\Omega\times J, \\ u(\mathit{\boldsymbol{x}}, t) = 0, \quad (\mathit{\boldsymbol{x}}, t)\in\partial\Omega\times J. \end{array}\right. \end{eqnarray} $

即(2.9b) 式和(2.9d) 式成立. 同理, 在(2.10c) 式中, 取$ \mathit{\boldsymbol{v}} = \mathit{\boldsymbol{p}}-K\mathit{\boldsymbol{ \widetilde{p}} } $, 可得(2.9c) 式成立. 因此当(2.10) 式成立时, (2.9) 式成立. 即问题(2.9) 与(2.10) 是等价的.

问题(2.10) 关于时间连续的扩展混合有限元逼近问题为: 求$ (U(t) $, $ \mathit{\boldsymbol{ \widetilde{p}} }(t), \mathit{\boldsymbol{P}}(t)): [0, T]\rightarrow W_h\times\mathit{\boldsymbol{V}}_h\times\mathit{\boldsymbol{V}}_h $, 使得

(3.1) $ \begin{eqnarray} &(U(0), w_h) = (Q_hu_0, w_h), \quad \forall w_h\in W_h, \end{eqnarray} $

(3.2) $ \begin{eqnarray} &(U_t(0), w_h) = (Q_hu_1, w_h), \quad \forall w_h\in W_h, \end{eqnarray} $

(3.3) $ \begin{eqnarray} &(\mathit{\boldsymbol{ \widetilde{P}} }(0), \mathit{\boldsymbol{v}}_h) = (u_0, \nabla\cdot\mathit{\boldsymbol{v}}_h ), \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{eqnarray} $

(3.4) $ \begin{eqnarray} &(\mathit{\boldsymbol{P}}(0), \mathit{\boldsymbol{v}}_h) = (-K\nabla u_0, \mathit{\boldsymbol{v}}_h), \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{eqnarray} $

(3.5) $ \begin{eqnarray} &(c(U(t))U_{tt}(t), w_h)+(\nabla\cdot \mathit{\boldsymbol{P}}(t), w_h) = (f(t), w_h), \quad \forall w_h\in W_h, \ \forall t>0, \end{eqnarray} $

(3.6) $ \begin{eqnarray} &(\mathit{\boldsymbol{ \widetilde{P}} }(t), \mathit{\boldsymbol{v}}_h)-(\nabla\cdot\mathit{\boldsymbol{v}}_h, U(t)) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \ \forall t>0, \end{eqnarray} $

(3.7) $ \begin{eqnarray} &(\mathit{\boldsymbol{P}}(t), \mathit{\boldsymbol{v}}_h)-(K\mathit{\boldsymbol{ \widetilde{P}} }(t), \mathit{\boldsymbol{v}}_h) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \ \forall t>0. \end{eqnarray} $

为简单起见, 时空范数$ \|\cdot\|_{L^2(0, T;L^2(\Omega))} $定义如下

$ \|u\|_{L^2(0, T;L^2(\Omega))} = \|u\|_{L^2(L^2)} = \bigg(\int_{0}^{T}||u||^2_{L^2(\Omega)}\bigg)^{\frac{1}{2}}, $

范数$ \|\cdot\|_{L^\infty(L^2)} $也有类似定义.

定理 3.1  令$ (u(t), \mathit{\boldsymbol{ \widetilde{p}} }(t), \mathit{\boldsymbol{p}}(t)) $和$ (U(t), \mathit{\boldsymbol{ \widetilde{p}} }(t), \mathit{\boldsymbol{P}}(t)) $分别是方程(2.10) 和(3.1)–(3.7) 的解. 取$ U(0) = Q_hu_0 $和$ U_t(0) = Q_hu_1 $, 则对任意给定的$ t\geq 0 $, 存在不依赖于$ h $和$ t $的常数$ C>0 $, 使得

(3.8) $ \begin{eqnarray} \|u-U\|_{L^\infty( L^2)}+\|(u-U)_t\|_{L^\infty( L^2)}+\|\mathit{\boldsymbol{ \widetilde{p}} }-\mathit{\boldsymbol{ \widetilde{P}} }\|_{L^\infty (L^2)}+\|\mathit{\boldsymbol{ \widetilde{p}} }-\mathit{\boldsymbol{P}}\|_{L^\infty (L^2)}\leq {\cal C}h^{k+1}, \end{eqnarray} $

其中$ {\cal C} = C(\|u\|_{L^2(H^{k+1})}+\|u_{tt}\|_{L^2(H^{k+1})}+\|\mathit{\boldsymbol{ \widetilde{p}} }\|_{L^2(H^{k+1})} +\|\mathit{\boldsymbol{p}}\|_{L^2(H^{k+1})}) $.

证  使用(2.1) 式和投影算子$ Q_h $, $ Q^*_h $和$ \Pi_h $, 则(2.10a)–(2.10c) 式可改写为

(3.9) $ \begin{eqnarray} &\left((c(u)u_{tt}), w_h\right)+(\nabla\cdot \Pi_h\mathit{\boldsymbol{ \widetilde{p}} }, w_h) = (f, w_h), \quad \forall w_h\in W_h, \end{eqnarray} $

(3.10) $ \begin{eqnarray} &(Q^*_h\mathit{\boldsymbol{ \widetilde{p}} }, \mathit{\boldsymbol{v}}_h)-(\nabla\cdot\mathit{\boldsymbol{v}}_h, Q_hu) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{eqnarray} $

(3.11) $ \begin{eqnarray} &(\mathit{\boldsymbol{ \widetilde{p}} }-\Pi_h\mathit{\boldsymbol{ \widetilde{p}} }, \mathit{\boldsymbol{v}}_h)+(\Pi_h\mathit{\boldsymbol{ \widetilde{p}} }, \mathit{\boldsymbol{v}}_h)-(K\mathit{\boldsymbol{ \widetilde{p}} }, \mathit{\boldsymbol{v}}_h) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{eqnarray} $

令$ \chi = U-Q_hu $, $ \mathit{\boldsymbol{ \widetilde{ξ}} } = \mathit{\boldsymbol{ \widetilde{p}} }-Q_{h}^* {\mathit{\boldsymbol{ \widetilde{p}}} }$, $ \mathit{\boldsymbol{ξ}} = \mathit{\boldsymbol{P}}-\Pi_h\mathit{\boldsymbol{p}} $, $ \eta = u-Q_hu $, $ \mathit{\boldsymbol{ \widetilde{ζ}} } = \mathit{\boldsymbol{ \widetilde{p}} }-Q_{h}^*\mathit{\boldsymbol{ \widetilde{p}} } $和$ \mathit{\boldsymbol{ζ}} = \mathit{\boldsymbol{p}}-\Pi_h\mathit{\boldsymbol{p}} $. 由(3.5)–(3.7) 式和(3.9)–(3.11) 式, 可得误差方程

(3.12) $ \begin{eqnarray} &(c(U)\chi_{tt}, w_h)+(\nabla\cdot \mathit{\boldsymbol{ξ}}, w_h) = ((c(u)-c(U))u_{tt}+c(U)\eta_{tt}, w_h), \ \forall w_h\in W_h, \end{eqnarray} $

(3.13) $ \begin{eqnarray} &(\mathit{\boldsymbol{ \widetilde{ξ}} }, \mathit{\boldsymbol{v}}_h)-(\nabla\cdot\mathit{\boldsymbol{v}}_h, \chi) = 0, \ \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{eqnarray} $

(3.14) $ \begin{eqnarray} &(\mathit{\boldsymbol{ξ}}, \mathit{\boldsymbol{v}}_h)-(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \mathit{\boldsymbol{v}}_h) = (\mathit{\boldsymbol{ ζ}}, \mathit{\boldsymbol{v}}_h)-(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \mathit{\boldsymbol{v}}_h), \ \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{eqnarray} $

将(3.13) 式两端关于时间$ t $求导, 得

(3.15) $ \begin{eqnarray} (\mathit{\boldsymbol{ \widetilde{ξ}} }_t, \mathit{\boldsymbol{v}}_h)-(\nabla\cdot\mathit{\boldsymbol{v}}_h, \chi_t) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{eqnarray} $

在(3.12), (3.14) 和(3.15) 式中分别取$ w_h = \chi_t $, $ \mathit{\boldsymbol{v}}_h = \widetilde{\mathit{\boldsymbol{ξ}}}_t $和$ \mathit{\boldsymbol{v}}_h = \mathit{\boldsymbol{ξ}} $, 并结合三式整理可得

(3.16) $ \begin{eqnarray} (c(U)\chi_{tt}, \chi_t)+(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \widetilde{\mathit{\boldsymbol{ξ}}}_t) = ((c(u)-c(U))u_{tt}+c(U)\eta_{tt}, \chi_t)-(\mathit{\boldsymbol{ ζ}}, \widetilde{\mathit{\boldsymbol{ξ}}}_t) +(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \widetilde{\mathit{\boldsymbol{ξ}}}_t). \end{eqnarray} $

对上式两端关于$ t $从0到$ t $积分, 得

(3.17) $ \begin{eqnarray} &&\frac{1}{2}\int_{0}^{t}(\frac{\rm d}{{\rm d} t}(c(U)\chi_{t}, \chi_t)+\frac{\rm d}{{\rm d} t}(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \mathit{\boldsymbol{ \widetilde{ξ}} })){\rm d}s\\ & = &\int_{0}^{t}[((c(u)-c(U))u_{tt}+c(U)\eta_{tt}, \chi_t)-(\mathit{\boldsymbol{ ζ}}, \widetilde{\mathit{\boldsymbol{ξ}}}_t) +(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \widetilde{\mathit{\boldsymbol{ξ}}}_t)]{\rm d}s+\frac{1}{2}\int_{0}^{t}(c_u(U)U_{t}\chi_{t}, \chi_t){\rm d}s. {\qquad} \end{eqnarray} $

注意$ \chi(0) = 0 $, $ \widetilde{\mathit{\boldsymbol{ξ}}}(0) $=0, 使用假设条件(I)–(III) 可得

(3.18) $ \begin{equation} \frac{1}{2}\int_{0}^{t}(\frac{\rm d}{{\rm d} t}(c(U)\chi_{t}, \chi_t) +\frac{\rm d}{{\rm d} t}(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \mathit{\boldsymbol{ \widetilde{ξ}} })){\rm d}s \geq\frac{c_*}{2}||\chi_t||^2+\frac{K_*}{2}||\mathit{\boldsymbol{ \widetilde{ξ}} }||^2, \end{equation} $

(3.19) $ \begin{eqnarray} \frac{1}{2}\int_{0}^{t}(c_u(U)U_{t}\chi_{t}, \chi_t){\rm d}s& = &\frac{1}{2}\int_{0}^{t}(c_u(U)\chi_{t}\chi_{t}, \chi_t){\rm d}s +\frac{1}{2}\int_{0}^{t}(c_u(U)Q_hu_{t}\chi_{t}, \chi_t){\rm d}s\\ &\leq& C(||\chi_t||_{L^\infty(L^\infty)}+1)\int^t_0\|\chi_t\|^2{\rm d}s. \end{eqnarray} $

由Cauchy不等式和假设条件(I)–(III), 则(3.17) 式的右端各项可估计为

$ \begin{eqnarray*} ((c(u)-c(U))u_{tt}+c(U)\eta_{tt}, \chi_t)& = &((c(u)-c(Q_hu)+c(Q_hu)-c(U))u_{tt}+c(U)\eta_{tt}, \chi_t)\nonumber\\ &\leq&(K_1(u-Q_hu+Q_hu-U)u_{tt}+c(U)\eta_{tt}, \chi_t)\nonumber\\ &\leq& C(\|\eta\|\|\chi_t\|+\|\eta_{tt}\|\|\chi_t\|+\|\chi\|\|\chi_t\|), \end{eqnarray*} $

$ (K\mathit{\boldsymbol{ \widetilde{ξ}} }, \widetilde{\mathit{\boldsymbol{ξ}}}_t)-(\mathit{\boldsymbol{ ζ}}, \widetilde{\mathit{\boldsymbol{ξ}}}_t)\leq C(\|\mathit{\boldsymbol{ ζ}}\| \|\widetilde{\mathit{\boldsymbol{ξ}}}_t\|+\|\mathit{\boldsymbol{ \widetilde{ξ}} }\| \|\widetilde{\mathit{\boldsymbol{ξ}}}_t\|), $

将上述两式进一步利用Young和$ \varepsilon $-Young不等式, 对充分小的常数$ \varepsilon>0 $, 易得

(3.20) $ \begin{eqnarray} &&((c(u)-c(U))u_{tt}+c(U)\eta_{tt}, \chi_t)+(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \widetilde{\mathit{\boldsymbol{ξ}}}_t)-(\mathit{\boldsymbol{ ζ}}, \widetilde{\mathit{\boldsymbol{ξ}}}_t)\\ &\leq& C(\|\eta\|^2+\|\eta_{tt}\|^2+\|\mathit{\boldsymbol{ ζ}}\|^2+\|\mathit{\boldsymbol{ \widetilde{ξ}} }\|^2 +\|\chi\|^2+\|\chi_t\|^2 + \varepsilon\|\widetilde{\mathit{\boldsymbol{ξ}}}_t\|^2 ). \end{eqnarray} $

结合(3.17)–(3.20) 式可得

(3.21) $ \begin{eqnarray} ||\chi_t||^2+||\mathit{\boldsymbol{ \widetilde{ξ}} }||^2\leq C\int_0^t(\|\eta\|^2+\|\eta_{tt}\|^2+\|\mathit{\boldsymbol{ ζ}}\|^2+\|\mathit{\boldsymbol{ \widetilde{ξ}} }\|^2 +\|\chi\|^2+\|\chi_t\|^2){\rm d}s. \end{eqnarray} $

在(3.14) 式中, 取$ \mathit{\boldsymbol{v}}_h = \mathit{\boldsymbol{ξ}} $, 得

(3.22) $ \begin{eqnarray} (\mathit{\boldsymbol{ξ}}, \mathit{\boldsymbol{ξ}}) = (K\mathit{\boldsymbol{ \widetilde{ξ}} }, \mathit{\boldsymbol{ξ}})+(\mathit{\boldsymbol{ ζ}}, \mathit{\boldsymbol{ξ}})-(K\mathit{\boldsymbol{ \widetilde{ξ}} }, \mathit{\boldsymbol{ξ}}). \end{eqnarray} $

由(3.22) 式可以推出

(3.23) $ \begin{eqnarray} \|\mathit{\boldsymbol{ξ}}\|\leq C(\|\mathit{\boldsymbol{ \widetilde{ξ}} }\|+\|\mathit{\boldsymbol{ ζ}}\| +\|\mathit{\boldsymbol{ \widetilde{ξ}} }\|). \end{eqnarray} $

注意到$ \chi(0) = 0 $, $ ||\chi||^2\leq T\int_{0}^{t}||\chi_t||^2{\rm d}s $, 因此在(3.21) 式的两端同时加上一项$ ||\chi||^2 $, 并使用(3.23) 式得

$ \begin{eqnarray*} ||\chi||^2+||\chi_t||^2+||\mathit{\boldsymbol{ \widetilde{ξ}} }||^2+\|\mathit{\boldsymbol{ξ}}\|^2 \leq C\int_0^t(\|\eta\|^2+\|\eta_{tt}\|^2+\|\mathit{\boldsymbol{ ζ}}\|^2+\| \mathit{\boldsymbol{ \widetilde{ξ}} }\|^2 +\|\chi\|^2+\|\chi_t\|^2){\rm d}s, \end{eqnarray*} $

上式利用Gronwall引理, 得

(3.24) $ \begin{eqnarray} \|\chi\|^2(t)+\|\chi_t\|^2(t) +\|\mathit{\boldsymbol{ \widetilde{ξ}} }\|^2(t)+\|\mathit{\boldsymbol{ξ}}\|^2(t) \leq C\int_0^t(\|\eta\|^2+\|\eta_{tt}\|^2+\|\mathit{\boldsymbol{ ζ}}\|^2+\|\mathit{\boldsymbol{ \widetilde{ξ}} }\|^2){\rm d}s. \end{eqnarray} $

对(3.24) 式中的$ t $取上确界可得

(3.25) $ \begin{eqnarray} &&\|\chi\|_{L^\infty(L^2)}^2 +\|\chi_t\|_{L^\infty(L^2)}^2 +\|\mathit{\boldsymbol{ \widetilde{ξ}} }\|_{L^\infty(L^2)}^2+\|\mathit{\boldsymbol{ξ}}\|_{L^\infty(L^2)}^2\\ &\leq& C(\|\eta\|_{L^2(L^2)}^2 +\|\eta_{tt}\|_{L^2(L^2)}^2+\|\mathit{\boldsymbol{ ζ}}\|_{L^2(L^2)}^2 +\|\mathit{\boldsymbol{ \widetilde{ξ}} }\|_{L^2(L^2)}^2 ). \end{eqnarray} $

最后, 利用三角不等式, 结合(2.6)、(2.7) 和(3.25)式, 可推出(3.8) 式成立.

对时间变量应用差分离散化, 将区间$ [0, T] $分成$ N $个相等的子区间: $ 0\leq t_0<t_1<\cdots<t_{N-1}<t_N = T $, $ t_{n+1}-t_n = \Delta t $. 为了简单起见, 令$ \varphi^n = \varphi(t_n) $, 其它的一些记号表示如下

(4.1) $ \begin{equation} \begin{array}{c} { } \varphi^{n+\frac{1}{2}} = \frac{\varphi^{n+1}+\varphi^n}{2}, \quad\varphi^{n, \frac{1}{2}} = \frac{\varphi^{n+1}+\varphi^{n-1}}{2}, {\quad} \partial_t\varphi^{n+\frac{1}{2}} = \frac{\varphi^{n+1}-\varphi^n}{\Delta t}, \\ { } \partial_t\varphi^n = \frac{\varphi^{n+1}-\varphi^{n-1}}{2\Delta t}, {\quad} \partial_{tt}\varphi^n = \frac{\varphi^{n+1}-2\varphi^n+\varphi^{n-1}}{(\Delta t)^2}. \end{array} \end{equation} $

下面的关系式容易得到

$ \partial_{tt}\varphi^n = \frac{\partial_t\varphi^{n+\frac{1}{2}}-\partial_t\varphi^{n-\frac{1}{2}}}{\Delta t}, \quad \partial_t\varphi^n = \frac{\partial_t\varphi^{n+\frac{1}{2}}+ \partial_t\varphi^{n-\frac{1}{2}}}{2}. $

问题(2.10) 的全离散逼近格式为: 求$ (u_h^{0}, \mathit{\boldsymbol{ \widetilde{p}} }_h^{0}, \mathit{\boldsymbol{p}}_h^{0})\in W_h\times\mathit{\boldsymbol{V}}_h\times\mathit{\boldsymbol{V}}_h $, 满足

(4.2) $ \begin{equation} ({u}_h^{0}, {w}_h) = (u_0, {w}_h ), \quad \forall {w}_h\in {W}_h, \end{equation} $

(4.3) $ \begin{equation} (\mathit{\boldsymbol{ \widetilde{p}} }_h^{0}, \mathit{\boldsymbol{v}}_h) = (u_0, \nabla\cdot\mathit{\boldsymbol{v}}_h ), \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{equation} $

(4.4) $ \begin{equation} (\mathit{\boldsymbol{ \widetilde{p}} }_h^{0}, \mathit{\boldsymbol{v}}_h) = (-K\nabla u_0, \mathit{\boldsymbol{v}}_h ), \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h \end{equation} $

及$ (u_h^{1}, \mathit{\boldsymbol{ \widetilde{p}} }_h^{1}, \mathit{\boldsymbol{p}}_h^{1})\in W_h\times\mathit{\boldsymbol{V}}_h\times\mathit{\boldsymbol{V}}_h $, 满足

(4.5) $ \begin{equation} \left(\frac{{u}_h^{1}-{u}_h^{-1}}{2\Delta t}, {w}_h\right) = (u_1, {w}_h ), \quad \forall {w}_h\in {W}_h, \end{equation} $

(4.6) $ \begin{equation} \left(\frac{\mathit{\boldsymbol{ \widetilde{p}} }_h^{1}-\mathit{\boldsymbol{ \widetilde{p}} }_h^{-1}}{2\Delta t}, \mathit{\boldsymbol{v}}_h\right) = (u_1, \nabla\cdot\mathit{\boldsymbol{v}}_h ), \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{equation} $

(4.7) $ \begin{equation} \left(\frac{\mathit{\boldsymbol{ \widetilde{p}} }_h^{1}-\mathit{\boldsymbol{ \widetilde{p}} }_h^{-1}}{2\Delta t}, \mathit{\boldsymbol{v}}_h\right) = (-K\nabla u_1, \mathit{\boldsymbol{v}}_h ), \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{equation} $

对$ n\geq 0 $, 求$ (u_h^{n+1}, \mathit{\boldsymbol{ \widetilde{p}} }_h^{n+1}, \mathit{\boldsymbol{p}}_h^{n+1})\in W_h\times\mathit{\boldsymbol{V}}_h\times\mathit{\boldsymbol{V}}_h $, 使得

(4.8) $ \begin{equation} (c(u_h^{n+1})\partial_{tt}u_h^n, w_h)+(\nabla\cdot \mathit{\boldsymbol{ \widetilde{p}} }^{n, \frac{1}{2}}_h, w_h) = (f^{n, \frac{1}{2}}, w_h), \quad \forall w_h\in W_h, \end{equation} $

(4.9) $ \begin{equation} (\mathit{\boldsymbol{ \widetilde{p}} }_h^{n+1}, \mathit{\boldsymbol{v}}_h)-(\nabla\cdot\mathit{\boldsymbol{v}}_h, u_h^{n+1}) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{equation} $

(4.10) $ \begin{equation} (K\mathit{\boldsymbol{ \widetilde{p}} }_h^{n+1}, \mathit{\boldsymbol{v}}_h)-(\mathit{\boldsymbol{ \widetilde{p}} }_h^{n+1}, \mathit{\boldsymbol{v}}_h) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{equation} $

定理 4.1  令$ (u^n, \mathit{\boldsymbol{ \widetilde{p}} }^n, \mathit{\boldsymbol{p}}^n) $和$ (u_h^n, \mathit{\boldsymbol{ \widetilde{p}} }_h^n, \mathit{\boldsymbol{p}}_h^n) $分别是方程(2.10) 和(4.2)–(4.10) 在$ t = t^n $时的解. 假设$ \Delta t $, $ h $充分小, 取$ u_h^0 = Q_hu^0 $, $ u_h^1 = Q_hu^1 $, 则存在常数$ C $, 使得下述估计成立

(4.11) $ \begin{eqnarray} \sup\limits_n\{||\partial_t(u-u_h)^{n-\frac{1}{2}}||+||u^n-u_h^n||+||\mathit{\boldsymbol{ \widetilde{p}} }^n-\mathit{\boldsymbol{ \widetilde{p}} }_h^n|| +||\mathit{\boldsymbol{ \widetilde{p}} }^n-\mathit{\boldsymbol{ \widetilde{p}} }_h^n||\}\leq C(h^{k+1}+\Delta t^2). \end{eqnarray} $

证  利用(2.1) 式和投影算子$ Q_h $, $ Q^*_h $和$ \Pi_h $, 则(2.10a)–(2.10c) 式可改写为

(4.12) $ \begin{equation} \left((c(u)u_{tt})^{n, \frac{1}{2}}, w_h\right)+(\nabla\cdot \Pi_h\mathit{\boldsymbol{ \widetilde{p}} }^{n, \frac{1}{2}}, w_h) = (f^{n, \frac{1}{2}}, w_h), \quad \forall w_h\in W_h, \end{equation} $

(4.13) $ \begin{equation} (Q^*_h\mathit{\boldsymbol{ \widetilde{p}} }^{n+1}, \mathit{\boldsymbol{v}}_h)-(\nabla\cdot\mathit{\boldsymbol{v}}_h, Q_hu^{n+1}) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{equation} $

(4.14) $ \begin{equation} (K\mathit{\boldsymbol{ \widetilde{p}} }^{n+1}, \mathit{\boldsymbol{v}}_h)-(\mathit{\boldsymbol{ \widetilde{p}} }^{n+1}-\Pi_h\mathit{\boldsymbol{ \widetilde{p}} }^{n+1}, \mathit{\boldsymbol{v}}_h) = (\Pi_h\mathit{\boldsymbol{ \widetilde{p}} }^{n+1}, \mathit{\boldsymbol{v}}_h), \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{equation} $

记$ \mu^n = Q_hu^n-u_h^n $, $ \varpi^n = Q_h^*\mathit{\boldsymbol{ \widetilde{p}} }^n-\mathit{\boldsymbol{ \widetilde{p}} }_h^n $, $ \omega^n = \Pi_h\mathit{\boldsymbol{p}}^n-\mathit{\boldsymbol{p}}_h^n $, $ \varsigma^n = u^n-Q_hu^n $, $ \tau^n = \mathit{\boldsymbol{ \widetilde{p}} }^{n}-Q_h^*\mathit{\boldsymbol{ \widetilde{p}} }^{n} $和$ \varrho^n = \mathit{\boldsymbol{p}}^{n}-\Pi_h\mathit{\boldsymbol{p}}^{n} $. 将(4.12)–(4.14) 式减去(4.8)–(4.10)式, 整理可得

(4.15) $ \begin{eqnarray} &&(c(u_h^{n+1})\partial_{tt}\mu^n, w_h)+(\nabla\cdot \omega^{n, \frac{1}{2}}, w_h){}\\ & = &(c(u_h^{n+1})\partial_{tt}u^n-(c(u)u_{tt})^{n, \frac{1}{2}} -c(u_h^{n+1})\partial_{tt}\varsigma^n, w_h), \ \forall w_h\in W_h, \end{eqnarray} $

(4.16) $ \begin{equation} (\varpi^{n+1}, \mathit{\boldsymbol{v}}_h)-(\mu^{n+1}, \nabla\cdot \mathit{\boldsymbol{v}}_h) = 0, \ \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{equation} $

(4.17) $ \begin{equation} (K\varpi^{n+1}, \mathit{\boldsymbol{v}}_h)-(\omega^{n+1}, \mathit{\boldsymbol{v}}_h) = (\varrho^{n+1}, \mathit{\boldsymbol{v}}_h)-(K\tau^{n+1}, \mathit{\boldsymbol{v}}_h), \ \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{equation} $

由(4.1)式, 方程(4.16) 和(4.17) 可写成

(4.18) $ \begin{equation} (\partial_t\varpi^{n}, \mathit{\boldsymbol{v}}_h)-(\partial_t\mu^{n}, \nabla\cdot \mathit{\boldsymbol{v}}_h) = 0, \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h, \end{equation} $

(4.19) $ \begin{equation} (K\varpi^{n, \frac{1}{2}}, \mathit{\boldsymbol{v}}_h)-(\omega^{n, \frac{1}{2}}, \mathit{\boldsymbol{v}}_h) = (\varrho^{n, \frac{1}{2}}, \mathit{\boldsymbol{v}}_h)-(K\tau^{n, \frac{1}{2}}, \mathit{\boldsymbol{v}}_h), \quad \forall \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{equation} $

取$ w_h = \partial_t\mu^n $, $ \mathit{\boldsymbol{v}}_h = \omega^{n, \frac{1}{2}} $和$ \mathit{\boldsymbol{v}}_h = \partial_t\varpi^{n} $分别代入(4.15), (4.18) 和(4.19) 式中, 然后整理可得

(4.20) $ \begin{eqnarray} &&(c(u_h^{n+1})\partial_{tt}\mu^n, \partial_t\mu^n)+(K\varpi^{n, \frac{1}{2}}, \partial_t\varpi^{n})\\ & = &(c(u_h^{n+1})\partial_{tt}u^n-(c(u)u_{tt})^{n, \frac{1}{2}}-c(u_h^{n+1})\partial_{tt}\varsigma^n, \partial_t\mu^n) +(\varrho^{n, \frac{1}{2}}-K\tau^{n, \frac{1}{2}}, \partial_t\varpi^{n}). \end{eqnarray} $

此外, 在(4.19) 式中, 取$ \mathit{\boldsymbol{v}}_h = \partial_t\omega^{n} $, 得

(4.21) $ \begin{eqnarray} (\omega^{n, \frac{1}{2}}, \partial_t\omega^{n}) = (K\varpi^{n, \frac{1}{2}}, \partial_t\omega^n)+(K\tau^{n, \frac{1}{2}}, \partial_t\omega^{n})-(\varrho^{n, \frac{1}{2}}, \partial_t\omega^{n}). \end{eqnarray} $

结合(4.20) 和(4.21)式, 可得

(4.22) $ \begin{eqnarray} &&(c(u_h^{n+1})\partial_{tt}\mu^n, \partial_t\mu^n)+(K\varpi^{n, \frac{1}{2}}, \partial_t\varpi^{n})+(\omega^{n, \frac{1}{2}}, \partial_t\omega^{n})\\ & = &(c(u_h^{n+1})\partial_{tt}u^n-(c(u)u_{tt})^{n, \frac{1}{2}}-c(u_h^{n+1})\partial_{tt}\varsigma^n, \partial_t\mu^n) +(\varrho^{n, \frac{1}{2}}, \partial_t\varpi^{n})\\ &&-(K\tau^{n, \frac{1}{2}}, \partial_t\varpi^{n}) +(K\varpi^{n, \frac{1}{2}}, \partial_t\omega^n)+(K\tau^{n, \frac{1}{2}}, \partial_t\omega^{n})-(\varrho^{n, \frac{1}{2}}, \partial_t\omega^{n}). \end{eqnarray} $

将(4.22) 式两边同乘以$ 2\Delta t $, 再让$ n = 1, 2, \cdots, l-1 $ ($ 1<l<N $) 相加, 可得

(4.23) $ \begin{eqnarray} &&2\Delta t\sum\limits_{n = 1}^{l-1}\{(c(u_h^{n+1})\partial_{tt}\mu^n, \partial_t\mu^n)+(K\varpi^{n, \frac{1}{2}}, \partial_t\varpi^{n}) +(\omega^{n, \frac{1}{2}}, \partial_t\omega^{n})\}\\ & = &2\Delta t\sum\limits_{n = 1}^{l-1}(c(u_h^{n+1})\partial_{tt}u^n-(c(u)u_{tt})^{n, \frac{1}{2}}-c(u_h^{n+1})\partial_{tt}\varsigma^n, \partial_t\mu^n) {}\\ && +2\Delta t\sum\limits_{n = 1}^{l-1}\{(\varrho^{n, \frac{1}{2}}, \partial_t\varpi^{n}) -(K\tau^{n, \frac{1}{2}}, \partial_t\varpi^{n})\}\\ && +2\Delta t\sum\limits_{n = 1}^{l-1}\{(K\varpi^{n, \frac{1}{2}}, \partial_t\omega^n)+(K\tau^{n, \frac{1}{2}}, \partial_t\omega^{n}) -(\varrho^{n, \frac{1}{2}}, \partial_t\omega^{n})\}\\ & = &F_1+F_2+F_3, \end{eqnarray} $

其中

$ F_1 = 2\Delta t\sum\limits_{n = 1}^{l-1}(c(u_h^{n+1})\partial_{tt}u^n-(c(u)u_{tt})^{n, \frac{1}{2}}-c(u_h^{n+1})\partial_{tt}\varsigma^n, \partial_t\mu^n), $

$ F_2 = 2\Delta t\sum\limits_{n = 1}^{l-1}\{(\varrho^{n, \frac{1}{2}}, \partial_t\varpi^{n}) -(K\tau^{n, \frac{1}{2}}, \partial_t\varpi^{n})\}, $

$ F_3 = 2\Delta t\sum\limits_{n = 1}^{l-1}\{(K\varpi^{n, \frac{1}{2}}, \partial_t\omega^n)+(K\tau^{n, \frac{1}{2}}, \partial_t\omega^{n}) -(\varrho^{n, \frac{1}{2}}, \partial_t\omega^{n})\}. $

注意到$ \partial_t\mu^{\frac{1}{2}} = \frac{\mu^1-\mu^0}{\Delta t} $. 取初始函数$ u_h^0 = Q_hu^0 $, $ u_h^1 = Q_hu^1 $, 则有$ \partial_t\mu^{\frac{1}{2}} = 0 $. 在(4.16) 式中, 取$ n = 0 $, $ \mathit{\boldsymbol{v}}_h = \varpi^{1} $, 可得$ \varpi^{1} $=0. 此外, 由(4.2)–(4.7)式, 我们推出$ \varpi^0 = \omega^0 = \omega^{1} $=0. 因此, 使用假设条件(I) 和(II), (4.23) 式中的左端项可以估计为

(4.24) $ \begin{eqnarray} &&2\Delta t\sum\limits_{n = 1}^{l-1}\{(c(u_h^{n+1})\partial_{tt}\mu^n, \partial_t\mu^n) +(K\varpi^{n, \frac{1}{2}}, \partial_t\varpi^{n})+(\omega^{n, \frac{1}{2}}, \partial_t\omega^{n})\}\\ & = &\frac{1}{2}\sum\limits_{n = 1}^{l-1}\{(K(\varpi^{n+1}-\varpi^{n-1}), \varpi^{n+1}+\varpi^{n-1})+(\omega^{n+1}-\omega^{n-1}, \omega^{n+1}+\omega^{n-1})\} \\ &&+\sum\limits_{n = 1}^{l-1}(c(u_h^{n+1})(\partial_t\mu^{n+\frac{1}{2}}-\partial_t\mu^{n-\frac{1}{2}}), \partial_t\mu^{n+\frac{1}{2}} -\partial_t\mu^{n-\frac{1}{2}}) \\ &\geq&c_*\|\partial_t\mu^{l-\frac{1}{2}}\|^2+\frac{K_*}{2}(\|\varpi^{l}\|^2+\|\varpi^{l-1}\|^2)+\frac{1}{2}(\|\omega^{l}\|^2+\|\omega^{l-1}\|^2). \end{eqnarray} $

我们也注意到$ \partial_{tt}u^n = \frac{u^{n+1}-2u^n+u^{n-1}}{(\Delta t)^2} $, 使用泰勒级数对$ u^{n+1} $和$ u^{n-1} $在$ u(t^n) $处展开, 得

$ \begin{eqnarray*} u^{n+1}& = &u(t^n)+\Delta t\frac{\partial u}{\partial t}(t^n)+\frac{\Delta t^2}{2}\frac{\partial^2 u}{\partial t^2}(t^n)+\frac{\Delta t^3}{6}\frac{\partial^3 u}{\partial t^3}(t^n)\\ &&+\frac{1}{6}\int_{t^n}^{t^{n+1}}(t^{n+1}-t)^3\frac{\partial^4 u}{\partial t^4}(t^n){\rm d}t, \\ u^{n-1}& = &u(t^n)-\Delta t\frac{\partial u}{\partial t}(t^n)+\frac{\Delta t^2}{2}\frac{\partial^2 u}{\partial t^2}(t^n)-\frac{\Delta t^3}{6}\frac{\partial^3 u}{\partial t^3}(t^n)\\ &&+\frac{1}{6}\int_{t^{n-1}}^{t^{n}}(t^{n-1}-t)^3\frac{\partial^4 u}{\partial t^4}(t^n){\rm d}t, \end{eqnarray*} $

上述两式相加有

(4.25) $ \begin{eqnarray} u^{n+1}+u^{n-1} = 2u^n+\Delta t^2\frac{\partial^2 u}{\partial t^2}(t^n)-\frac{1}{6}\int_{-\Delta t}^{\Delta t}(|t|-\Delta t)^3\frac{\partial^4 u}{\partial t^4}(t^n+t){\rm d}t. \end{eqnarray} $

利用(4.25) 式可得

$ \begin{eqnarray*} \partial_{tt}u^n-u_{tt}^n = -\frac{1}{6\Delta t^2}\int_{-\Delta t}^{\Delta t}(|t|-\Delta t)^3\frac{\partial^4 u}{\partial t^4}(t^n+t){\rm d}t. \end{eqnarray*} $

从而有

(4.26) $ \begin{eqnarray} \|\partial_{tt}u^n-u_{tt}^n\|^2&\leq& C\Delta t^2\int_\Omega\left[\int_{t^n-\Delta t}^{t^n+\Delta t}\frac{\partial^4u}{\partial t^4}(t)\right]^2{\rm d}t \leq C\Delta t^3\int_{t^n-\Delta t}^{t^n+\Delta t}\left\|\frac{\partial^4u}{\partial t^4}(t)\right\|^2{\rm d}t\\ &\leq &C\Delta t^4\left\|\frac{\partial^4u}{\partial t^4}(t)\right\|^2_{L^\infty(L^2)}. \end{eqnarray} $

接下来, 我们对(4.23) 式中的右端各项分别进行估计, 首先, 关于第一项$ F_1 $, 考虑到

$ \|\partial_t\mu^{n}\|^2\leq\|\partial_t\mu^{n+\frac{1}{2}}\|^2+\|\partial_t\mu^{n-\frac{1}{2}}\|^2, $

及使用(2.6)式, (4.26) 式和Young不等式, 可得

(4.27) $ \begin{eqnarray} &&2\Delta t\sum\limits_{n = 1}^{l-1}(c(u_h^{n+1})\partial_{tt}u^n-(c(u)u_{tt})^{n, \frac{1}{2}}-c(u_h^{n+1})\partial_{tt}\varsigma^n, \partial_t\mu^n)\\ & = &2\Delta t\sum\limits_{n = 1}^{l-1}((c(u_h^{n+1})-c(Q_hu^{n+1})+c(Q_hu^{n+1})-c(u^{n+1}))\partial_{tt}u^n \\ &&+c(u^{n+1}) (\partial_{tt}u^n-u_{tt}^n)+c(u^{n+1})u_{tt}^n-(c(u)u_{tt})^{n, \frac{1}{2}}-c(u_h^{n+1})\partial_{tt}\varsigma^n, \partial_t\mu^n)\\ &\leq& C\{h^{2k+2}+\Delta t^4+\Delta t\sum\limits_{n = 1}^{l}\|\partial_t\mu^{n-\frac{1}{2}}\|^2+\Delta t\sum\limits_{n = 1}^{l-1}\|\mu^{n}\|^2\}. \end{eqnarray} $

关于第二项$ F_2 $, 利用(2.7) 式和$ \varepsilon $-Young不等式, 对充分小$ \varepsilon>0 $, 有

(4.28) $ \begin{eqnarray} 2\Delta t\sum\limits_{n = 1}^{l-1}\{(\varrho^{n, \frac{1}{2}}, \partial_t\varpi^{n}) -(K\tau^{n, \frac{1}{2}}, \partial_t\varpi^{n})\} \leq C \{h^{2k+2}+\varepsilon\Delta t\sum\limits_{n = 1}^{l-1}\|\partial_t\varpi^{n}\|^2\}. \end{eqnarray} $

同样, 关于第三项$ F_3 $可以推出

(4.29) $ \begin{eqnarray} &&2\Delta t\sum\limits_{n = 1}^{l-1}\{(K\varpi^{n, \frac{1}{2}}, \partial_t\omega^n)+(K\tau^{n, \frac{1}{2}}, \partial_t\omega^{n}) -(\varrho^{n, \frac{1}{2}}, \partial_t\omega^{n})\}\\ &\leq &C \{h^{2k+2}+\Delta t\sum\limits_{n = 1}^{l-1}(\|\varpi^{n}\|^2+\|\varpi^{n-1}\|^2)+\varepsilon\Delta t\sum\limits_{n = 1}^{l-1}\|\partial_t\omega^n\|^2\}. \end{eqnarray} $

(4.16) 式可写成

(4.30) $ \begin{equation} (\varpi^{n+1}, \mathit{\boldsymbol{v}}_h)+(\nabla\mu^{n+1}, \mathit{\boldsymbol{v}}_h) = 0, \ \mathit{\boldsymbol{v}}_h\in \mathit{\boldsymbol{V}}_h. \end{equation} $

在(4.30) 式中, 取$ \mathit{\boldsymbol{v}}_h = \nabla u_h^{n+1} $, 得

$ |(\nabla \mu^{n+1}, \nabla \mu^{n+1})| = \|\nabla \mu^{n+1}\|^2 = |(\varpi^{n+1}, \nabla \mu^{n+1})| \leq\|\varpi^{n+1}\|\|\nabla \mu^{n+1}\|. $

使用离散的Sobolev嵌入不等式^[15], 有

(4.31) $ \begin{equation} \|\mu^{n+1}\|\leq C_1\|\varpi^{n+1}\|, \end{equation} $

其中$ C_1 $与区域$ \Omega $和$ p $ ($ 1\leq p<\infty $) 有关. 将(4.24), (4.27)–(4.29) 和(4.31) 式结合可得

(4.32) $ \begin{eqnarray} &&\|\partial_t\mu^{l-\frac{1}{2}}\|^2+\|\mu^{l}\|^2+\|\varpi^{l}\|^2+\|\varpi^{l-1}\|^2+\|\omega^{l}\|^2+\|\omega^{l-1}\|^2\\ &\leq & C\{\Delta t^4+h^{2k+2}+\Delta t\sum\limits_{n = 1}^{l-1}(\|\varpi^{n}\|^2+\|\varpi^{n-1}\|^2+\|\mu^{n}\|^2+\|\partial_t\mu^{n-\frac{1}{2}}\|^2)\}. \end{eqnarray} $

应用Gronwall不等式, 得

(4.33) $ \begin{eqnarray} \|\partial_t\mu^{l-\frac{1}{2}}\|^2+\|\mu^l\|^2+\|\omega^l\|^2+\|\varpi^{l}\|^2 \leq C\{\Delta t^4+h^{2k+2}\}. \end{eqnarray} $

最后, 使用三角不等式, 结合(2.6), (2.7) 和(4.33)式, 推出(4.11) 式成立.

本节我们给出一个数值例子来验证理论分析结果. 在数值计算中, 使用简单的迭代法来处理(4.2)–(4.10) 式中的非线性项, 即用前一次迭代得到的解$ u^{k} $的值来替换本次迭代的非线性项$ c(u^{k+1}) $中$ u^{k+1} $的值, 其中$ k $为迭代指数. 这里, 非线性项$ c(u^{k+1})u^{k+1} $用$ c(u^{k})u^{k+1} $来逼近. 此外, 我们选择$ \epsilon = 10^{-8} $作为迭代算法的控制精度.

考虑以下二阶非线性双曲型方程

$ \begin{eqnarray*} \left\{\begin{array}{ll} (1+u^2)u_{tt}-\nabla\cdot(K\nabla u) = f, \quad &(\mathit{\boldsymbol{x}}, t)\in\Omega\times J, \\ u(\mathit{\boldsymbol{x}}, 0) = u_t(\mathit{\boldsymbol{x}}, 0) = 0, \quad \quad \quad \quad& \mathit{\boldsymbol{x}}\in\Omega, \\ u(\mathit{\boldsymbol{x}}, t) = 0, \quad \quad \quad \quad\ \ \quad \quad \quad &(\mathit{\boldsymbol{x}}, t)\in\partial\Omega\times J, \end{array}\right. \end{eqnarray*} $

其中$ \mathit{\boldsymbol{x}} = (x_1, x_2)^T $, $ \Omega = [0, 1]^2 $, $ J = (0, 1] $, $ K = \frac{1}{10^5}\left( \begin{array}{cc} 1+ x_1^2&0 \\ 0& 1+ x_2^2\\ \end{array} \right) $, $ f $是由精确解$ u $的取值来决定的, 该问题的精确解为$ u(\mathit{\boldsymbol{x}}, t) = t^2\cos(\pi t)\sin^2(\pi x_1)\sin^2(\pi x_2) $.

考虑$ k = 1 $的Raviart-Thomas混合元空间(RT$ _1 $). 空间区域$ \Omega $采用具有拟一致网格步长的三角形剖分, 分别取$ h = \{\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}\} $进行计算, 即每次空间网格$ h $的选取为前一次的一半. 此外, 采用一个固定的时间步长$ \Delta t = 1.0e-3 $. 数值计算结果如表 1和图 1所示. 在表 1中, 我们观察到实验误差收敛阶接近$ 2 $, 这与理论分析结果一致. 在图 1中, 数值解$ u_h $与精确解$ u $的图像几乎一致.