An Error Analysis of a Numerical Scheme for the Cahn-Hilliard Equation Based on the Invariant Energy Quadratization Approach
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摘要: 基于能量不变二次化方法，构造了一个求解Cahn-Hilliard方程的线性数值格式，该线性数值格式对非线性项半显式处理，每步迭代相应的半离散化方程只需要求解一个线性方程；证明了该线性数值格式是无条件能量稳定的，而且是唯一可解的；讨论了该线性数值格式在时间方向的误差估计.数值例子表明：该线性数值格式的数值解在时间方向上基本达到二阶精度, 能够有效模拟相位变化过程.
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关键词:
- 误差分析 /
- 能量不变二次化法 /
- Cahn-Hilliard方程
Abstract: A novel linear numerical scheme for the Cahn-Hilliard equation is constructed with the invariant energy quadratization approach. All nonlinear terms in this scheme are treated semi-explicitly and the resulting semi-discrete equation forms a linear system at each time step. It is proved that the proposed scheme is energy-stable unconditionally and solvable uniquely. The error estimate of the numerical scheme for the Cahn-Hilliard equation is discussed. Numerical examples show that the numerical solution of the linear numerical scheme basically achieves the second-order accuracy in the time direction and can effectively simulate the phase change process. -
Cahn-Hilliard方程在流体力学中具有重要应用，并被许多学者广泛研究.如：应用Cahn-Hilliard理论建立非局部反应扩散模型并采用其渐进展开式与多时间尺度去分析该模型[1]；应用Cahn-Hilliard方程表示拓扑相变，并分析如何模拟3个不能混合的流在陡峭界面的运动[2]；应用Cahn-Hilliard方程描述不可压缩的流体扩散界面以及相位场[3-4]；分析Allen-Cahn方程ut=Δu+ε−2(f(u)−ελ(t))在含无流边界条件的有界闭域上的质量守恒性，其中，ελ(t)是f(u(⋅,t))的平均值，-f是双等位势函数的导子[5]；讨论容器带有密度的相位场用平均曲率流近似[6]；给出容器V的自由能量表达式：
E=NV∫V(F0(c)+k(∇c)2)dV, 其中，V表示非均匀结构或非均匀密度的各向同性的空间几何体，NV表示每单位体积的分子数，λ表示结构梯度或密度梯度，F0表示对应均匀系(齐次系统)的每分子的自由能，k是一个参数[7]；基于二阶平均向量场方法和拟谱方法，构造了具有多辛结构的复修正KdV方程新的数值格式, 证明了该格式能保方程离散的整体能量守恒特性[8]；采用能量不变二次化法来解决一些微分方程的数值近似[9-11].
YANG等[12]利用能量不变二次化方法，设计了一阶和二阶的时间离散格式, 以求解三组分Cahn-Hilliard方程，但没有考虑时间方向的误差估计.因此，本文基于能量不变二次化方法，对一类非三组分的具有能量泛函
E(ϕ)=∫Ω(12|∇ϕ|2+F(ϕ))dx 的Cahn-Hilliard方程ϕt=−Δ2ϕ+ΔF′(ϕ)构造线性数值格式，并讨论该数值格式在时间方向的误差估计.
为了便于构造线性数值格式，将Cahn-Hilliard方程ϕt=−Δ2ϕ+ΔF′(ϕ)变形为方程组
{ϕt=Δω,ω=−Δϕ+F′(ϕ), (1) 其中，F′(ϕ)=f(ϕ)=ϕ3−ϕ,F(ϕ)是非线性的光滑的位势函数, Ω是R2的闭集；ϕ(x,t)(x∈Ω,t∈(0,T])是混合物中某种物质的浓度，ω是化学势.
1. 能量不变二次化法
本节采用能量不变二次化法分析Cahn-Hillirad方程的数值近似.能量不变二次化法[9]是指通过变量代换将自由能被积函数变成新变量的二次函数，变换后的能量函数满足能耗规律(Energy Dissipation Law).本文所用的部分记号如下：f(x)与g(x)的L2内积为(f(x),g(x))=∫Ωf(x)g(x)dx(x∈Ω⊆R2); f(x)的的L2范数为‖f‖=√(f,f);q(ϕ)=√F(ϕ)+B；g(ϕ)=2ddϕq(ϕ)=f(ϕ)√F(ϕ)+B.
于是，方程组(1)可以写成：
{ϕt=Δω,ω=−Δϕ+g(ϕ)q(ϕ),qt=12g(ϕ)ϕt, (2) 满足初始条件：
ϕ|t=0=ϕ0,q|t=0=√F(ϕ0)+B, 并满足以下其中一个边界条件：
(i) 在边界∂Ω上，ϕ和ω都是周期性的；
(ii) 无流边界，即
{\partial _\mathit{\boldsymbol{n}}}\phi {|_{\partial \varOmega }} = {\partial _\mathit{\boldsymbol{n}}}\omega {|_{\partial \varOmega }} = 0, (3) 其中，n为边界上的向外法向量.
方程组(2)的3个方程分别与\omega、{\phi _t}、2q 作L2内积，整理得
\frac{{\rm{d}}}{{{\rm{d}}t}}E(\phi ,q) = - \int_\varOmega | \nabla \omega {|^2}{\rm{d}}x, 其中，E\left( {\phi , q} \right) = {\smallint _\mathit{\Omega } }\left( {\frac{1}{2}{{\left| {\nabla \phi } \right|}^2} + {q^2}} \right){\rm{d}}x ，这里 E\left( {\phi , q} \right)等价于 E\left( \phi \right).
2. 梯度流的稳定性
针对方程组(2)，构造如下时间离散格式：
\left\{ {\begin{array}{*{20}{l}} {\frac{{{\phi ^{n + 2}} - {\phi ^n}}}{{2\delta t}} = \Delta \frac{{{\omega ^{n + 2}} + {\omega ^n}}}{2},}\\ {\frac{{{\omega ^{n + 2}} + {\omega ^n}}}{2} = - \Delta \frac{{{\phi ^{n + 2}} + {\phi ^n}}}{2} + {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2},}\\ {\frac{{{q^{n + 2}} - {q^n}}}{{2\delta t}} = \frac{1}{2}{g^{n + 1}}\frac{{{\phi ^{n + 2}} - {\phi ^n}}}{{2\delta t}}.} \end{array}} \right. (4) 该格式有如下的能量稳定性，且该格式的解是惟一的.
定理1 (能量稳定性)方程组(4)的解是具有能量稳定的，即
\frac{{E({\phi ^{n + 1}},{q^{n + 1}}) - E({\phi ^n},{q^n})}}{{\delta t}} = - \frac{1}{4}{\left\| {\nabla {\omega ^{n + 2}} + \nabla {\omega ^n}} \right\|^2}, 其中，
\begin{array}{*{20}{l}} {E({\phi ^n},{q^n}) = }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{1}{4}({{\left\| {\nabla {\phi ^{n + 1}}} \right\|}^2} + {{\left\| {\nabla {\phi ^n}} \right\|}^2} + 2{{\left\| {\nabla {q^n}} \right\|}^2} + 2{{\left\| {\nabla {q^{n + 1}}} \right\|}^2}).} \end{array} 证明 把方程组(4)的第1个方程与2\delta t({\omega ^{n + 2}} + {\omega ^n}) 作L2内积，并使用分部积分法，整理得
({\phi ^{n + 2}} - {\phi ^n},{\omega ^{n + 1}} + {\omega ^n}) = - \delta t{\left\| {\nabla {\omega ^{n + 2}} + \nabla {\omega ^n}} \right\|^2}. (5) 方程组(4)的第2个方程与2\left( {{\phi ^{n + 2}} - {\phi ^n}} \right) 作L2内积，整理得
\begin{array}{*{20}{c}} {({\omega ^{n + 2}} + {\omega ^n},{\phi ^{n + 2}} - {\phi ^n}) = {{\left\| {\nabla {\phi ^{n + 2}}} \right\|}^2} - {{\left\| {\nabla {\phi ^n}} \right\|}^2} + }\\ {({g^{n + 1}}({q^{n + 2}} + {q^n}),{\phi ^{n + 2}} - {\phi ^n}).} \end{array} (6) 方程组(4)的第3个方程与 4\delta t{q^{n + 2}} + {q^n}作L2内积，整理得
2{\left\| {{q^{n + 2}}} \right\|^2} - 2{\left\| {{q^n}} \right\|^2} = ({g^{n + 1}}({\phi ^{n + 2}} - {\phi ^n}),{q^{n + 2}} + {q^n}). (7) 结合式(5)~(7)，有
\begin{array}{l} {\left\| {\nabla {\phi ^{n + 2}}} \right\|^2} - {\left\| {\nabla {\phi ^n}} \right\|^2} + 2{\left\| {{q^{n + 2}}} \right\|^2} - 2{\left\| {{q^n}} \right\|^2} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - \delta t{\left\| {\nabla {\omega ^{n + 2}} + \nabla {\omega ^n}} \right\|^2}, \end{array} 从而可以导出结果.证毕.
定理2 (解的唯一性)方程组(4)的解是唯一的.
证明 由方程组(4)的第3个方程可得
{q^{n + 2}} = {q^n} + \frac{1}{2}{g^{n + 1}}({\phi ^{n + 2}} - {\phi ^n}), (8) 则方程组(4)可写成
\left\{ {\begin{array}{*{20}{l}} {{\phi ^{n + 2}} - \delta t\Delta {\omega ^{n + 2}} = {Q_1},}\\ {P({\phi ^{n + 2}}) - {\omega ^{n + 2}} = {Q_2},} \end{array}} \right. (9) 其中， {Q_1} = {\phi ^n} + \delta t{\omega ^n}, {Q_2} = \Delta {\phi ^n} + {g^{n + 1}}\left( {\frac{1}{2}{g^{n + 1}}{\phi ^n} - 2{q^n}} \right) + {\omega ^n}, P\left( \phi \right) = - \Delta \phi + \frac{1}{2}{\left( {{g^{n + 1}}} \right)^2}\phi .
于是，由方程组(9)直接推导得\left( {{\phi ^{n + 2}}, {\omega ^{n + 2}}} \right)从而由式(8)推导得 {q^{n + 2}}.进一步地，当任意的φ满足边界条件(i)或条件(ii)时，有
(P(\phi ),\varphi ) = (\nabla \phi ,\nabla \varphi ) + \frac{1}{2}({({g^{n + 1}})^2}\phi ,\varphi ) = (\varphi ,P(\phi )), 则线性算子P(\phi ) 是对称的.又当{\smallint _\mathit{\Omega } }\phi {\rm{d}}x = 0 时，有
(P(\phi ),\phi ) = {\left\| {\nabla \phi } \right\|^2} + \frac{1}{2}{\left\| {{g^{n + 1}}\phi } \right\|^2} \ge 0, 则线性算子P\left( \phi \right) 是正定的.
通过方程组(4)的第1个方程与1作L2内积，可推导出
\int_\varOmega {{\phi ^{n + 2}}} {\rm{d}}x = \int_\varOmega {{\phi ^n}} {\rm{d}}x = \cdots = \int_\varOmega {{\phi ^0}} {\rm{d}}x. 令 {v_\phi } = \frac{1}{{|\mathit{\Omega } |}}\int_\mathit{\Omega } {{\phi ^0}} {\rm{d}}x, {v_\omega } = \frac{1}{{|\mathit{\Omega } |}}\int_\mathit{\Omega } {{\omega ^{n + 2}}} {\rm{d}}x并记
{\hat \phi ^{n + 2}} = {\phi ^{n + 2}} - {v_\phi },{\hat \omega ^{n + 2}} = {\omega ^{n + 2}} - {v_\omega }. 结合方程组(9)，可知({\widehat \phi ^{n + 2}}{\rm{, }}{\widehat \omega ^{n + 2}}) 是以下方程组的解:
\left\{ {\begin{array}{*{20}{l}} {\phi - \delta t\Delta \omega = f{,_1}}\\ {P(\phi ) - \omega - {v_\omega } = {f_2},} \end{array}} \right. (10) 其中, {\smallint _\mathit{\Omega } }{\rm{d}}x = 0, {\smallint _\mathit{\Omega } }{\rm{d}}x = 0.
记逆算子 u = {\Delta ^{-1}}\rho 如下:
\left\{ {\begin{array}{*{20}{l}} {\Delta u = \rho ,}\\ {\int_\varOmega u {\rm{d}}x = 0,} \end{array}} \right. 其中, u满足边界条件(i)或条件(ii).将 - {\Delta ^{ - 1}} 作用到方程组(10)的第1个方程，整理得到
- {\Delta ^{ - 1}}\phi + \delta tP(\phi ) - \delta t{v_\omega } = - {\Delta ^{ - 1}}{f_1} + \delta t{f_2}. (11) 记式(11)为Γ\phi =Ζ，则对任意满足{\smallint _\mathit{\Omega } }\phi {\rm{d}}x = 0、{\smallint _\mathit{\Omega } }\varphi {\rm{d}}x = 0 的 \phi 、φ，有
\begin{array}{l} (\varGamma \phi ,\varphi ) = ( - {\Delta ^{ - 1}}\phi + \delta tP(\phi ) - \delta t{v_\omega },\varphi ) = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (\phi , - {\Delta ^{ - 1}}\varphi + \delta tP(\varphi ) - \delta t{v_\omega }) = (\phi ,\varGamma \varphi ). \end{array} 而
\begin{array}{*{20}{c}} {(\varGamma \phi ,\phi ) = ( - {\Delta ^{ - 1}}\phi + \delta tP(\phi ) - \delta t{v_\omega },\phi ) = }\\ {\left\| \phi \right\|_{{H^{ - 1}}}^2 + \delta t(P(\phi ),\phi ) \ge 0,} \end{array} 则Γ是对称的，也是正定的，从而由Lax-Milgram定理[13]可知方程组(9)存在唯一解.证毕.
3. 误差估计
定义误差e_\phi ^n = \phi \left( {{t_n}} \right) - {\phi ^n}, e_\omega ^n = \omega \left( {{t_n}} \right) - {\omega ^n}, e_q^n = q\left( {{t_n}} \right) - {q^n}.记误差函数
{r_1^{n + 1} \buildrel \Delta \over = \frac{{\phi ({t_{n + 2}}) - \phi ({t_n})}}{{2\delta t}} - {\phi _t}({t_{n + 1}}),} {r_2^{n + 1} \buildrel \Delta \over = \frac{{q({t_{n + 2}}) - q({t_n})}}{{2\delta t}} - {q_t}({t_{n + 1}}),} {r_3^{n + 1} \buildrel \Delta \over = q({t_{n + 1}}) - \frac{{q({t_{n + 2}}) + q({t_n})}}{2},} {r_4^{n + 1} \buildrel \Delta \over = \omega ({t_{n + 1}}) - \frac{{\omega ({t_{n + 2}}) + \omega ({t_n})}}{2},} {r_5^{n + 1} \buildrel \Delta \over = \phi ({t_{n + 1}}) - \frac{{\phi ({t_{n + 2}}) + \phi ({t_n})}}{2}.} 应用泰勒展式，有
\begin{array}{*{20}{l}} {\left\| {r_1^{n + 1}} \right\| \le C\delta {t^2},\left\| {r_2^{n + 1}} \right\| \le C\delta {t^2},\left\| {r_3^{n + 1}} \right\| \le C\delta {t^2},}\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left\| {r_4^{n + 1}} \right\| \le C\delta {t^2},\left\| {r_5^{n + 1}} \right\| \le C\delta {t^2}.} \end{array} 引理1 设函数F\left( x \right) 及 \phi 满足以下条件：
(a) F(x) > - A, A < B, \forall x \in ( - \infty , + \infty );
(b) F(x) \in C^{2}(-\infty, +\infty);
(c) 存在一个正的常数C0，使得
\mathop {\max }\limits_{n \le k} \{ {\left\| {\phi ({t_{n + 1}})} \right\|_{{L^\infty }}},{\left\| {{\phi ^{n + 1}}} \right\|_{{L^\infty }}}\} \le {C_0}. 则有
\begin{array}{*{20}{l}} {\mathop {\max }\limits_{n \le k} \{ {{\left\| {F({\chi ^{n + 1}})} \right\|}_{{L^\infty }}},{{\left\| {f({\chi ^{n + 1}})} \right\|}_{{L^\infty }}},{{\left\| {{f^\prime }({\chi ^{n + 1}})} \right\|}_{{L^\infty }}},}\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {{\left\| {\sqrt {F({\chi ^{n + 1}}) + B} } \right\|}_{{L^\infty }}}\} \le {C_1},} \end{array} (12) 其中，\chi^{n+1}=\varepsilon \phi\left(t_{n+1}\right)+(1-\varepsilon) \phi^{n+1}, \varepsilon \in[0, 1] .进而有
\left\| {g(\phi ({t_{n + 1}})) - {g^{n + 1}}} \right\| \le {C_2}\left\| {\phi ({t_{n + 1}}) - {\phi ^{n + 1}}} \right\|, 其中，C2是依赖于C0、C1、A、B的常数.
证明 由条件(c)可知 {\chi ^{n + 1}}是一致有界的，再结合条件(b)，可以得到式(12).应用拉格朗日中值定理，有
\begin{array}{l} \left\| {g(\phi ({t_{n + 1}})) - {g^{n + 1}}} \right\| = \left\| {\frac{{f(\phi ({t_{n + 1}}))}}{{\sqrt {F(\phi ({t_{n + 1}})) + B} }} - \frac{{f({\phi ^{n + 1}})}}{{\sqrt {F({\phi ^{n + 1}}) + B} }}} \right\| = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left\| {\frac{{f(\phi ({t_{n + 1}}))\sqrt {F({\phi ^{n + 1}}) + B} - f({\phi ^{n + 1}})\sqrt {F(\phi ({t_{n + 1}})) + B} }}{{\sqrt {F(\phi ({t_{n + 1}})) + B} \sqrt {F({\phi ^{n + 1}}) + B} }}} \right\| \le \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left\| {\frac{{f(\phi ({t_{n + 1}}))(\sqrt {F({\phi ^{n + 1}}) + B} - \sqrt {F(\phi ({t_{n + 1}})) + B} )}}{{\sqrt {F(\phi ({t_{n + 1}})) + B} \sqrt {F({\phi ^{n + 1}}) + B} }}} \right\| + \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left\| {\frac{{\sqrt {F(\phi ({t_{n + 1}})) + B} (f(\phi ({t_{n + 1}})) - f({\phi ^{n + 1}}))}}{{\sqrt {F(\phi ({t_{n + 1}})) + B} \sqrt {F({\phi ^{n + 1}}) + B} }}} \right\| \le \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{l}} {\left( {\frac{{{C_1}}}{{2\sqrt {{{(B - A)}^3}} }}\left\| {f(\chi _1^{n + 1})} \right\|\left\| {\phi ({t_{n + 1}}) - {\phi ^{n + 1}}} \right\| + } \right.}\\ {\left. {\frac{{{C_1}}}{{B - A}}\left\| {{f^\prime }(\chi _2^{n + 1})} \right\|\left\| {\phi ({t_{n + 1}}) - {\phi ^{n + 1}}} \right\|} \right) \le }\\ {{C_2}\left\| {\phi ({t_{n + 1}}) - {\phi ^{n + 1}}} \right\|.} \end{array} \end{array} 引理2 设函数 F\left( x \right)及 \phi 满足以下条件：
(a) F(x) > - A, A < B, \forall x \in ( - \infty , + \infty );
(b) F(x) \in {C^3}( - \infty , + \infty );
(c) 存在一个正的常数C3，使得
\mathop {\max }\limits_{n \le k} \{ {\left\| {\phi ({t_{n + 1}})} \right\|_{{L^\infty }}},{\left\| {{\phi ^{n + 1}}} \right\|_{{L^\infty }}}\} \le {C_3}. 则有
{\mathop {\max }\limits_{n \le k} \{ {{\left\| {F({\chi ^{n + 1}})} \right\|}_{{L^\infty }}},{{\left\| {f({\chi ^{n + 1}})} \right\|}_{{L^\infty }}},{{\left\| {{f^\prime }({\chi ^{n + 1}})} \right\|}_{{L^\infty }}},} {{{\left\| {{f^{\prime \prime }}({\chi ^{n + 1}})} \right\|}_{{L^\infty }}},{{\left\| {\sqrt {F({\chi ^{n + 1}}) + B} } \right\|}_{{L^\infty }}}\} \le {C_4},} 其中， \chi^{n+1}=\varepsilon \phi\left(t_{n+1}\right)+(1-\varepsilon) \phi^{n+1}, \varepsilon \in[0, 1]进而有
\left\| {\nabla g(\phi ({t_{n + 1}})) - \nabla {g^{n + 1}}} \right\| \le {C_5}\left\| {\phi ({t_{n + 1}}) - {\phi ^{n + 1}}} \right\|, 这里的C5是依赖于C3、C4、A、B的常数.
引理2的证明过程与引理1的类似，在此略.
引理3 设 \left\{ {{u^n}} \right\}_{n = 0}^{N - 2}是Ω上的函数列，则有
\left\| {{u^{n + 2}}} \right\| \le \sum\limits_{m = 0}^n {\left\| {{u^{m + 2}} + {u^m}} \right\|} + \left\| {{u^0}} \right\|. 证明 使用数学归纳法证明.当n=0时，易证得
\left\| {{u^2}} \right\| \le \left\| {{u^2} + {u^0}} \right\| + \left\| {{u^0}} \right\|. 假设n=k－1时，有
\left\| {{u^{k + 1}}} \right\| \le \sum\limits_{m = 0}^{k - 1} {\left\| {{u^{m + 2}} + {u^m}} \right\|} + \left\| {{u^0}} \right\|. 则n=k时，有
\begin{array}{*{20}{l}} {\left\| {{u^{k + 2}}} \right\| - \left\| {{u^0}} \right\| = \left\| {{u^{k + 2}}} \right\| - \left\| {{u^{k + 1}}} \right\| + \left\| {{u^{k + 1}}} \right\| - \left\| {{u^0}} \right\| \le }\\ \begin{array}{l} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \sum\limits_{m = 0}^{k - 1} {\left\| {{u^{m + 2}} + {u^m}} \right\| + \left\| {{u^{k + 2}}} \right\| - \left\| {{u^{k + 1}}} \right\|} \le \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \sum\limits_{m = 0}^{k - 1} {\left\| {{u^{m + 2}} + {u^m}} \right\|} + \left\| {{u^{k + 2}} + {u^{k + 1}}} \right\| = \sum\limits_{m = 0}^k {\left\| {{u^{m + 2}} + {u^m}} \right\|} . \end{array} \end{array} 则引理3得证.
为了更好地证明方程组(2)的数值近似的误差估计，定义v为:
v = \mathop {\max }\limits_{0 \le t \le T} {\left\| {\phi (t)} \right\|_{{L^\infty }}} + 1. 引理4 设 F(x) > - A, A < B, \forall x \in ( - \infty , + \infty )，F(x) \in {C^3}( - \infty , + \infty ) ，方程组(2)的精确解满足正则性假设条件 \phi \in L^{\infty}\left(0, T ; H^{2}(\mathit{\Omega })\right) \cap L^{\infty}\left(0, T ; W^{1, \infty}(\mathit{\Omega })\right), \phi_{t} \in L^{2}\left(0, T ; H^{1}(\mathit{\Omega })\right) \cap L^{\infty}\left(0, T ; L^{\infty}(\mathit{\Omega })\right), q \in L^{\infty}(0, T ; \left.W^{1, \infty}(\mathit{\Omega })\right), \omega \in L^{\infty}\left(0, T ; H^{1}(\mathit{\Omega })\right), q_{t t}, \phi_{tt} \in L^{2}\left(0, T ; L^{2}(\mathit{\Omega })\right) ，则对于正数 {\tau _0}，当\delta t < {\tau _0} 时，有
{\left\| {{\phi ^n}} \right\|_{{L^\infty }}} \le v\quad (n = 0,1,2, \cdots ;K = T/(\delta t)). 证明 使用数学归纳法证明.当n=0时，易知 {\left\| {{\phi ^0}} \right\|_{L\infty }} \le v.现假设{\left\| {{\phi ^{k + 1}}} \right\|_{L\infty }} \le v 成立，下面将要证明 {\left\| {{\phi ^{k + 2}}} \right\|_{L\infty }} \le v.先不考虑方程组(4)，而把方程组(2)在tn+1处重新表述，将方程组(2)的3个方程分别与φ、θ、ψ作L2内积，整理得
\left( {\frac{{e_\phi ^{n + 2} - e_\phi ^n}}{{2\delta t}},\varphi } \right) = - \left( {\nabla \frac{{e_\omega ^{n + 2} + e_\omega ^n}}{2},\nabla \varphi } \right) + (r_1^{n + 1} + \Delta r_4^{n + 1},\varphi ), (13) \begin{array}{*{20}{c}} {\left( {\frac{{e_\omega ^{n + 2} + e_\omega ^n}}{2},\theta } \right) = - (r_4^{n + 1} + \Delta r_5^{n + 1},\theta ) - \left( {\Delta \frac{{e_\phi ^{n + 2} + e_\phi ^n}}{2},\theta } \right) + }\\ {\left( {g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - {g^{n + 1}}\frac{{{q^{n + 1}} + {q^n}}}{2},} \right)\theta ,} \end{array} (14) \begin{array}{l} \left( {\frac{{e_q^{n + 2} - e_q^n}}{{2\delta t}},\psi } \right) = (r_2^{n + 1},\psi ) + \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{1}{2}\left( {g(\phi ({t_{n + 1}})){\phi _t}({t_{n + 1}}) - {g^{n + 1}}\frac{{{\phi ^{n + 2}} - {\phi ^n}}}{{2\delta t}},\varphi } \right). \end{array} (15) 在式(13)中，令 \varphi = 2\delta t\left( {e_\omega ^{n + 2} + e_\omega ^n} \right), 得
\begin{array}{*{20}{c}} {(e_\phi ^{n + 2} - e_\phi ^n,e_\omega ^{n + 2} + e_\omega ^n) + \delta t{{\left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\|}^2} = }\\ {2\delta t(r_1^{n + 1} + \Delta r_4^{n + 1},e_\omega ^{n + 2} + e_\omega ^n).} \end{array} (16) 在式(13)中，令\varphi = 2\delta t\left( {e_\phi ^{n + 2} + e_\phi ^n} \right) , 得
\begin{array}{l} {\left\| {e_\phi ^{n + 2}} \right\|^2} - {\left\| {e_\phi ^n} \right\|^2} = - \delta t(\nabla e_\omega ^{n + 2} + \nabla e_\omega ^n,\nabla e_\phi ^{n + 2} + \nabla e_\phi ^n) + \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 2\delta t(r_1^{n + 1} + \Delta r_4^{n + 1},e_\phi ^{n + 2} + e_\phi ^n). \end{array} (17) 在式(14)中，令 \theta = 2\left( {e_\phi ^{n + 2} + e_\phi ^n} \right), 得
\begin{array}{*{20}{l}} {(e_\omega ^{n + 2} + e_\omega ^n,e_\phi ^{n + 2} - e_\phi ^n) = }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - 2(r_4^{n + 1} + \Delta r_5^{n + 1},e_\phi ^{n + 2} - e_\phi ^n) + {{\left\| {\nabla e_\phi ^{n + 2}} \right\|}^2} - {{\left\| {\Delta e_\phi ^n} \right\|}^2} + }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 2\left( {g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2},e_\phi ^{n + 2} - e_\phi ^n} \right).} \end{array} (18) 在式(14)中，令 \theta = 2\delta t\left( {e_\omega ^{n + 2} + e_\omega ^n} \right), 得
\begin{array}{l} \delta t{\left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\|^2} = - 2\delta t(r_4^{n + 1} + \Delta r_5^{n + 1},e_\omega ^{n + 2} + e_\omega ^n) + \delta t(\nabla (e_\phi ^{n + 2} + e_\phi ^n),\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \nabla (e_\omega ^{n + 2} + e_\omega ^n)) + 2\delta t\left( {g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - } \right.\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left. {{\kern 1pt} {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2},e_\omega ^{n + 2} + e_\omega ^n} \right). \end{array} (19) 在式(15)中，令 \psi = 4\delta t\left( {e_q^{n + 2} + e_q^n} \right), 得
\begin{array}{*{20}{l}} {2{{\left\| {e_q^{n + 2}} \right\|}^2} - 2{{\left\| {e_q^n} \right\|}^2} = 4\delta t(r_2^{n + 1},e_q^{n + 2} + e_q^n) + }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 2\delta t\left( {g(\phi ({t_{n + 1}})){\phi _t}({t_{n + 1}}) - {g^{n + 1}}\frac{{{\phi ^{n + 2}} - {\phi ^n}}}{{2\delta t}},e_q^{n + 2} + e_q^n} \right).} \end{array} (20) 组合式(16)~(20)，并整理得
\begin{array}{l} \left\| {e_\phi ^{n + 2}} \right\|_1^2 - \left\| {e_\phi ^n} \right\|_1^2 + 2{\left\| {e_q^{n + 2}} \right\|^2} - 2{\left\| {e_q^n} \right\|^2} + \delta t\left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\|_1^2 = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{l}} {2\delta t(r_1^{n + 1} + \Delta r_4^{n + 1} - r_4^{n + 1} - \Delta r_5^{n + 1},e_\omega ^{n + 2} + e_\omega ^n) + }\\ {2\delta t(r_1^{n + 1} + \Delta r_4^{n + 1},e_\phi ^{n + 2} + e_\phi ^n) + }\\ {2(r_4^{n + 1} + \Delta r_5^{n + 1},e_\phi ^{n + 2} - e_\phi ^n) + 4\delta t(r_2^{n + 1},e_q^{n + 2} + e_q^n) - } \end{array}\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{l}} {2\left( {g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2},e_\phi ^{n + 2} - e_\phi ^n} \right) + }\\ {2\delta t\left( {g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2},e_\omega ^{n + 2} + e_\omega ^n} \right) + }\\ {2\delta t\left( {g(\phi ({t_{n + 1}})){\phi _t}({t_{n + 1}}) - {g^{n + 1}}\frac{{{\phi ^{n + 2}} - {\phi ^n}}}{{2\delta t}},e_q^{n + 2} + e_q^n} \right).} \end{array} \end{array} (21) 注意到
\begin{array}{l} g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{l}} {q(\phi ({t_{n + 1}}))(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) + }\\ {{g^{n + 1}}\left( {q(\phi ({t_{n + 1}})) - \frac{{{q^{n + 2}} + {q^n}}}{2}} \right) = }\\ {q(\phi ({t_{n + 1}}))(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) + }\\ {{g^{n + 1}}\left( {q(\phi ({t_{n + 1}})) - \frac{{q(\phi ({t_{n + 2}})) + q(\phi ({t_n}))}}{2}} \right) + }\\ {{g^{n + 1}}\left( {\frac{{q(\phi ({t_{n + 2}})) + q(\phi ({t_n}))}}{2} - \frac{{{q^{n + 2}} + {q^n}}}{2}} \right) = }\\ {q(\phi ({t_{n + 1}}))(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) + {g^{n + 1}}r_3^{n + 1} + {g^{n + 1}}\frac{{e_q^{n + 2} + e_q^n}}{2},} \end{array} \end{array} 与
\begin{array}{l} g(\phi ({t_{n + 1}})){\phi _t}({t_{n + 1}}) - {g^{n + 1}}\frac{{{\phi ^{n + 2}} - {\phi ^n}}}{{2\delta t}} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{l}} {{\phi _t}({t_{n + 1}})(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) + {g^{n + 1}}\left( {{\phi _t}({t_{n + 1}}) - \frac{{{\phi ^{n + 2}} - {\phi ^n}}}{{2\delta t}}} \right) = }\\ {{\phi _t}({t_{n + 1}})(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) - {g^{n + 1}}r_1^{n + 1} + {g^{n + 1}}\frac{{e_\phi ^{n + 2} - e_\phi ^n}}{{2\delta t}}.} \end{array} \end{array} 应用Gronwall和Young不等式，式(21)等号的右端可以放缩如下:
\begin{array}{l} 2\delta t(r_1^{n + 1} + \Delta r_4^{n + 1} - r_4^{n + 1} - \Delta r_5^{n + 1},e_\omega ^{n + 2} + e_\omega ^n) \le \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} C\delta {t^5} + \frac{{\delta t}}{4}{\left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\|^2}, \end{array} (22) {2\delta t(r_1^{n + 1} + \Delta r_4^{n + 1},e_\phi ^{n + 2} + e_\phi ^n) \le C\delta {t^5} + \delta t{{\left\| {e_\phi ^{n + 2} + e_\phi ^n} \right\|}^2},} (23) {4\delta t(r_2^{n + 1} + e_\phi ^{n + 2} + e_\phi ^n) \le C\delta {t^5} + \delta t{{\left\| {e_\phi ^{n + 2} + e_\phi ^n} \right\|}^2},} (24) \begin{array}{*{20}{l}} {2(r_4^{n + 1} + \Delta r_5^{n + 1},e_\phi ^{n + 2} - e_\phi ^n) = }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 4\delta t\left( {r_4^{n + 1} + \Delta r_5^{n + 1},\Delta \frac{{e_\omega ^{n + 2} + e_\omega ^n}}{2} + r_1^{n + 1} + \Delta r_4^{n + 1}} \right) \le }\\ {\quad C\delta {t^5} + \frac{{\delta t}}{4}{{\left\| {\nabla e_\omega ^{n + 2} + \nabla e_\omega ^n} \right\|}^2}.} \end{array} (25) 应用引理1, 式(21)的右端非线性项估计如下:
\begin{array}{l} - 2\left( {g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2},e_\phi ^{n + 2} - e_\phi ^n} \right) + \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{l}} {2\delta t\left( {g(\phi ({t_{n + 1}})){\phi _t}({t_{n + 1}}) - {g^{n + 1}}\frac{{{\phi ^{n + 2}} - {\phi ^n}}}{{2\delta t}},e_q^{n + 2} + e_q^n} \right) = }\\ { - 2(q(\phi ({t_{n + 1}}))(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) + {g^{n + 1}}r_3^{n + 1} + }\\ {\left. {{g^{n + 1}}\frac{{e_q^{n + 2} + e_q^n}}{2},e_\phi ^{n + 2} - e_\phi ^n} \right) + 2\delta t({\phi _t}({t_{n + 1}})(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) - }\\ {\left. {{g^{n + 1}}r_1^{n + 1} + {g^{n + 1}}\frac{{e_\phi ^{n + 2} - e_\phi ^n}}{{2\delta t}},e_q^{n + 2} + e_q^n} \right) = } \end{array}\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{l}} { - 2(q(\phi ({t_{n + 1}}))(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) + {g^{n + 1}}r_3^{n + 1},e_\phi ^{n + 2} - e_\phi ^n) + }\\ {2\delta t({\phi _t}({t_{n + 1}})(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) - {g^{n + 1}}r_1^{n + 1},e_q^{n + 2} + e_q^n) \le }\\ {C\delta t(\delta {t^4} + {{\left\| {e_\phi ^{n + 2} + e_\phi ^n} \right\|}^2} + \left\| {e_\phi ^{n + 1}} \right\|_1^2) + \frac{{\delta t}}{4}{{\left\| {\nabla e_\omega ^{n + 2} + \nabla e_\omega ^n} \right\|}^2},} \end{array} \end{array} (26) \begin{array}{l} 2\delta t\left( {g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2},e_\omega ^{n + 2} + e_\omega ^n} \right) = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 2\delta t\left( {q(\phi ({t_{n + 1}}))(g(\phi ({t_{n + 1}})) - {g^{n + 1}}) + {g^{n + 1}}r_3^{n + 1} + } \right.\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left. {{g^{n + 1}}\frac{{e_q^{n + 2} + e_q^n}}{2},e_\omega ^{n + 2} + e_\omega ^n} \right) \le C\delta t(\delta {t^4} + {\left\| {e_\phi ^{n + 1}} \right\|^2} + {\left\| {e_q^{n + 2} + e_q^n} \right\|^2}) + \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{{\delta t}}{4}{\left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\|^2}. \end{array} (27) 将式(22)~(27)代入式(21)，整理得
\begin{array}{*{20}{c}} {\left\| {e_\phi ^{n + 2}} \right\|_1^2 - \left\| {e_\phi ^n} \right\|_1^2 + 2{{\left\| {e_q^{n + 2}} \right\|}^2} - 2{{\left\| {e_q^n} \right\|}^2} + \frac{{\delta t}}{2}\left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\|_1^2 \le }\\ {C\delta {t^5} + C\delta t(\left\| {e_\phi ^{n + 1}} \right\|_1^2 + {{\left\| {e_q^{n + 2} + e_q^n} \right\|}^2} + {{\left\| {e_\phi ^{n + 2} + e_\phi ^n} \right\|}^2}).} \end{array} (28) 在式(28)中，令n=0, 1, 2…, k，并求和，整理得
\begin{array}{*{20}{l}} {\left\| {e_\phi ^{k + 2}} \right\|_1^2 + \left\| {e_\phi ^{k + 1}} \right\|_1^2 + 2{{\left\| {e_q^{k + 2}} \right\|}^2} + 2{{\left\| {e_q^{k + 1}} \right\|}^2} + }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{{\delta t}}{2}\sum\limits_{n = 0}^k {\left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\|_1^2} \le \left\| {e_\phi ^1} \right\|_1^2 + 2{{\left\| {e_q^1} \right\|}^2} + C\delta {t^4} + }\\ {\quad C\delta t\sum\limits_{n = 0}^k {(\left\| {e_\phi ^{n + 1}} \right\|_1^2 + {{\left\| {e_\phi ^{n + 2}} \right\|}^2} + {{\left\| {e_\phi ^n} \right\|}^2} + {{\left\| {e_q^{n + 2}} \right\|}^2} + {{\left\| {e_q^n} \right\|}^2})} .} \end{array} 容易证得
\left\|e_{\phi}^{1}\right\|_{1}^{2}+2\left\|e_{q}^{1}\right\|^{2} \leqslant C \delta t^{4} 应用Gronwall不等式，存在一个正数 {\tau _1}，当 \delta t < {\tau _1}时，有
\begin{array}{*{20}{c}} {\left\| {e_\phi ^{k + 2}} \right\|_1^2 + \left\| {e_\phi ^{k + 1}} \right\|_1^2 + {{\left\| {e_q^{k + 2}} \right\|}^2} + {{\left\| {e_q^{k + 1}} \right\|}^2} + }\\ {\delta t\sum\limits_{n = 0}^k {\left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\|_1^2} \le {C_6}\delta {t^4}.} \end{array} (29) 由方程组(2)、(4)知
\begin{array}{*{20}{l}} {\left\| {\Delta e_\phi ^{n + 2} + \Delta e_\phi ^n} \right\| \le 2{{\left\| {r_1^{n + 1}} \right\|}^2} + 2{{\left\| {r_5^{n + 1}} \right\|}^2} + \left\| {e_\omega ^{n + 2} + e_\omega ^n} \right\| + }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left\| {g(\phi ({t_{n + 1}}))q(\phi ({t_{n + 1}})) - {g^{n + 1}}\frac{{{q^{n + 2}} + {q^n}}}{2}} \right\| \le {C_7}\delta {t^2}.} \end{array} 由引理3可知
\left\| {\Delta e_\phi ^{k + 2}} \right\| \le \sum\limits_{n = 0}^k {\left\| {\Delta e_\phi ^{n + 2} + \Delta e_\phi ^n} \right\| + \left\| {\Delta e_\phi ^0} \right\|} \le {C_8}\delta t. 进而有
\begin{array}{l} {\left\| {{\phi ^{k + 2}}} \right\|_{{L^\infty }}} \le {\left\| {e_\phi ^{k + 2}} \right\|_{{L^\infty }}} + {\left\| {\phi ({t_{k + 2}})} \right\|_{{L^\infty }}} \le \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {C_\varOmega }\left\| {e_\phi ^{k + 2}} \right\|_1^{\frac{1}{2}}\left\| {e_\phi ^{k + 2}} \right\|_2^{\frac{1}{2}} + {\left\| {\phi ({t_{k + 2}})} \right\|_{{L^\infty }}} \le \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {C_9}\delta {t^{\frac{3}{2}}} + {\left\| {\phi ({t_{k + 2}})} \right\|_{{L^\infty }}}. \end{array} 若 {C_9}\delta {t^{\frac{3}{2}}} \le 1，则 \left\|\phi^{k+2}\right\|_{L^{\infty}} \leqslant 1+\left\|\phi\left(t_{k+2}\right)\right\|_{L^{\infty}} \leqslant v.证毕.
定理3 设F(x) > - A, A < B, \forall x \in ( - \infty , + \infty ) ， F(x) \in {C^3}( - \infty , + \infty )，Cahn-Hilliard方程组的精确解满足正则性假设条件\phi \in {L^\infty }\left( {0, T;{H^2}(\mathit{\Omega } )} \right) \cap {L^\infty }(0, \left. {T;{W^{1, \infty }}(\mathit{\Omega } )} \right), {\phi _t} \in {L^2}\left( {0, T;{H^1}(\mathit{\Omega } )} \right) \cap {L^\infty }\left( {0, T;{L^\infty }(\mathit{\Omega } )} \right) , q \in {L^\infty }\left( {0, T;{W^{1, \infty }}(\mathit{\Omega } )} \right), \omega \in {L^\infty }\left( {0, T;{H^1}(\mathit{\Omega } )} \right), {q_{tt}}, {\phi _{tt}} \in {L^2}\left( {0, T;{L^2}(\mathit{\Omega } )} \right) ，则有
\begin{array}{l} {\left\| {\phi ({t_{k + 2}}) - {\phi ^{k + 2}}} \right\|_1} + \left\| {q(\phi ({t_{k + 2}})) - q({\phi ^{k + 2}})} \right\| + \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \delta t\sum\limits_{n = 0}^k {{{\left\| {\omega ({t_{n + 2}}) + \omega ({t_n}) - ({\omega ^{n + 2}} + {\omega ^n})} \right\|}_1}} \le C\delta {t^2}. \end{array} 证明 因为{\left\| {{\phi ^n}} \right\|_{L\infty }} \le v, 0 \le n \le T/\delta t , 采用相同的方法，可以证得式(29), 从而定理3得证.
4. 数值例子
下面采用数值例子验证理论分析的准确性.实验中，为了测试Cahn-Hilliard方程数值解的时间精度，选取如下初始值：
{\phi (x,y,0) = \frac{1}{2}\left( {1 + \tanh \left( {\frac{{R - 0.15}}{\varepsilon }} \right)} \right),\varepsilon = 0.01,} {R = \sqrt {{{(x - 0.5)}^2} + {{(y - 0.5)}^2}} .} 由基于能量不变二次化法的Cahn-Hilliard方程的数值解在L2范数下的误差和时间精度(表 1)可知：Cahn-Hilliard方程数值解在时间方向上基本达到二阶精度，从而验证了定理3的准确性.
表 1 能量不变二次化法的Cahn-Hilliard方程的数值结果Table 1. The numerical results of invariant energy quadratization approach of Cahn-Hilliard equationδt L2范数下的误差 阶 0.01 0.008 235 — 0.005 0.005 224 1.651 101 0.002 5 0.002 406 1.752 124 0.001 25 0.001 126 1.861 113 0.000 625 0.000 722 1.924 021 0.000 312 5 0.000 465 1.945 233 为了清楚看到相位变化过程，取T=1, 计算区域为Ω=[0, 1]×[0, 1], 初始条件为 \phi \left( {x, t = 0} \right) = {10^{ - 3}}{\rm{rand}}( - 1, 1).由基于能量不变二次化法的Cahn-Hilliard方程的数值解在t=1与t=2时刻的相位图(图 1)可知：所构造的数值格式能够有效地模拟Cahn-Hilliard方程的相位变化过程.
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图 1 数值解在t=1与t=2时刻的相位图Figure 1. The phase diagram of the numerical results on t=1 and t=25. 结束语
基于能量不变二次化法，本文对Cahn-Hilliard方程构造有效的时间离散数值格式.对该数值格式的时间误差进行的分析结果表明：该数值格式在时间方向上是二阶精度的.数值例子也验证了该分析结果的准确性.本文所构造的时间离散数值格式能够使得非线性项离散化且在时间水平上保持能量稳定，能够解决Cahn-Hilliard方程数值近似中遇到的主要难题，为进一步考虑Cahn-Hilliard方程在时间和空间方向上同时离散数值逼近奠定一定的基础.
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图 1 数值解在t=1与t=2时刻的相位图
Figure 1. The phase diagram of the numerical results on t=1 and t=2
表 1 能量不变二次化法的Cahn-Hilliard方程的数值结果
Table 1 The numerical results of invariant energy quadratization approach of Cahn-Hilliard equation
δt L2范数下的误差 阶 0.01 0.008 235 — 0.005 0.005 224 1.651 101 0.002 5 0.002 406 1.752 124 0.001 25 0.001 126 1.861 113 0.000 625 0.000 722 1.924 021 0.000 312 5 0.000 465 1.945 233 
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