具有浓度迁移率和对数势能的修正Cahn-Hillard方程的有限元算法

Finite Element Method for the Modified Cahn-Hilliard Equation with the Concentration Mobility and the Logarithmic Potential

摘要: 为了研究具有浓度迁移率和对数势能的修正Cahn-Hilliard方程，在空间上采用混合有限元方法进行离散，时间上采用Crank-Nicolson格式进行离散，对于非线性项采用了凸分裂的方法。证明了数值方法的稳定性，并且给出了误差估计。最后，通过数值算例对理论分析进行了验证。结果表明，理论分析与数值实验相一致。

Abstract: In order to study the modified Cahn-Hilliard equation with concentration mobility and logarithmic potential energy, the mixed finite element method was used in space, and the Crank-Nicolson scheme was used in time. The convex splitting method is used for nonlinear terms. Furthermore, the stability of the numerical method is proved and the error estimate is given. Finally, a numerical example is given to verify the theoretical analysis. The results show that the theoretical analysis is consistent with the numerical experiment.

1. 引言

Cahn-Hilliard方程最早是由Cahn和Hilliard在20世纪50年代提出的 [1]，常用来描述二元合金在某种不稳定状态时相的分离和粗化现象 [2] [3] [4]。为了抑制粗化现象，Aristotelous提出修正Cahn-Hilliard方程 [5]。本文研究的具有浓度迁移率和对数势能的修正Cahn-Hilliard方程具有如下形式：

$\{\begin{array}{l}{u}_t=\nabla \cdot \left(m\left(u\right)\nabla w\right),\left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ w=-{\epsilon }^2\Delta u+\varphi \left(u\right)+\phi ,\left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ -\Delta \phi =\beta \left(u-{\bar{u}}_0\right),\left(x,t\right)\in \Omega \times \left(0,T\right),\hfill \\ {\partial }_{n}u={\partial }_{n}w=0,\left(x,t\right)\in \partial \Omega \times \left(0,T\right),\hfill \\ u\left(x,0\right)=u_0\left(x\right),x\in \Omega .\hfill \end{array}$ (1)

其中： $\Omega \in R^d,d=2$； $\epsilon$ 是给定的参数； $u_t=\frac{\partial u}{\partial t}$； $\beta >0$；n是单位外法向量。u是混合物中某种物质的浓度，w是化学势能， $\phi$ 是辅助变量， ${\bar{u}}_0=\frac{1}{\left|\Omega \right|}{\displaystyle {\int }_{\Omega }{u}_0\left(x\right)\text{d}x}$。 $m\left(u\right)$ 是浓度迁移率，定义如下 [6]：对于 $\text{0}<\sigma \le 1$，

$m\left(u\right)=\{\begin{array}{l}\frac{1}{4}\left[1-\left(1-\sigma \right){\left(2u-1\right)}^2\right],0\le u\le 1,\\ \frac{1}{8}\sigma \left[1+{\text{e}}^{\frac{-2\left(1-\sigma \right)\left(4u^2-4u\right)}{\sigma }}\right],其他.\end{array}$

容易得到，对于 $\forall u\in R$，存在 $m_1>m_2>0$，使得 $m_1\ge m\left(u\right)\ge m_2$ 成立；对于 $\forall u\in R$，存在 $M>0$，使得 $\left|m^{\prime }\left(u\right)\right|\le M$ 成立。 $\Phi \left(u\right)$ 是对数势能，定义如下 [7]：对于 $\forall \theta \in \left(0,1\right)$，

$\Phi \left(u\right)=\frac{\theta }{2}\left(\left(1+u\right)\ln \left(1+u\right)+\left(1-u\right)\ln \left(1-u\right)\right)+\frac{1}{2}\left(1-u^2\right),u\in \left(-1,1\right).$

$\varphi \left(u\right)={\Phi }^{\prime }\left(u\right)=\frac{\theta }{2}\left(\ln \left(1+u\right)-\ln \left(1-u\right)\right)-u.$

根据正则化思想，考虑下面的对数势能函数 [8]：对于 $\forall k\in \left(0,1\right)$，

$\stackrel{⌣}{\Phi }\left(u\right):=\{\begin{array}{l}\frac{\theta }{2}\left(1+u\right)\ln \left(1+u\right)+\frac{\theta }{4k}{\left(1-u\right)}^2-\frac{\theta k}{4}+\frac{\theta }{2}\left(1-u\right)\ln k+\frac{1}{2}\left(1-u^2\right),u\ge 1-k,\hfill \\ \frac{\theta }{\text{2}}\left(\left(1+u\right)\ln \left(1+u\right)-\left(1-u\right)\ln \left(1-u\right)\right)+\frac{1}{2}\left(1-u^2\right),\left|u\right|<1-k,\hfill \\ \frac{\theta }{2}\left(1-u\right)\ln \left(1-u\right)+\frac{\theta }{4k}{\left(1+u\right)}^2-\frac{\theta k}{4}+\frac{\theta }{2}\left(1+u\right)\ln k+\frac{1}{2}\left(1-u^2\right),u\le -1+k.\hfill \end{array}$

$\stackrel{⌣}{\varphi }:=\stackrel{⌣}{\Phi }\left(u\right)=\{\begin{array}{l}\frac{\theta }{2}\ln \left(1+u\right)+\frac{\theta }{2}-\frac{\theta }{2k}\left(1-u\right)-\frac{\theta }{2}\ln k-u,u\ge 1-k,\hfill \\ \frac{\theta }{2}\left(\ln \left(1+u\right)-\ln \left(1-u\right)\right)-u,\left|u\right|<1-k,\hfill \\ -\frac{\theta }{2}\ln \left(1-u\right)-\frac{\theta }{2}+\frac{\theta }{2k}\left(1+u\right)+\frac{\theta }{2}\ln k-u,u\le -1+k.\hfill \end{array}$

在接下来的研究中，将用正则化的对数势能 $\stackrel{⌣}{\Phi }$ 及其导数 $\stackrel{⌣}{\varphi }$ 来代替 $\Phi$ 和 $\varphi$，为了简单，仍记为 $\Phi$ 和 $\varphi$。自由能函数定义为：

$E={\displaystyle {\int }_{\Omega }\left(\frac{{\epsilon }^2}{2}{\left|\nabla u\right|}^2+\Phi \left(u\right)\right)\text{d}x}+\frac{\beta }{2}{\Vert u-{\bar{u}}_0\Vert }_{H^{-1}}^2.$

2. 数值格式

$L^2\left(\Omega \right)$ 是平方可积函数空间，内积为 $\left(u,v\right)={\displaystyle {\int }_{\Omega }u\left(x\right)v\left(x\right)\text{d}x}$，相应范数为 $\Vert u\Vert ={\Vert u\Vert }_{L^2\left(\Omega \right)}=\sqrt{\left(u,u\right)}$。 $H^1\left(\Omega \right)$ 是通常的Sobolev空间，相应的半范和范数分别为

${\left|u\right|}_{H^1}={\left({\displaystyle {\int }_{\Omega }{\left|Du\right|}^2\text{d}x}\right)}^{\frac{1}{2}},{\Vert u\Vert }_{H^1}={\left({\displaystyle {\int }_{\Omega }{\left|u\right|}^2\text{d}x}+{\displaystyle {\int }_{\Omega }{\left|Du\right|}^2\text{d}x}\right)}^{\frac{1}{2}}.$

具有浓度迁移率和对数势能的修正Cahn-Hilliard方程的弱解形式为：

$\{\begin{array}{l}\left(u_t,q\right)+\left(\nabla \cdot \left(m\left(u\right)\nabla w\right),\nabla q\right)=0,\forall q\in H^1\left(\Omega \right),\hfill \\ \left(w,v\right)-{\epsilon }^2\left(\nabla u,\nabla v\right)-\left({\varphi }_1\left(u\right)-{\varphi }_2\left(u\right),v\right)-\left(\phi ,v\right)=0,\forall v\in H^1\left(\Omega \right),\hfill \\ \left(\nabla \phi ,\nabla \psi \right)-\beta \left(u-{\bar{u}}_0,\psi \right)=0,\forall \psi \in H^1\left(\Omega \right).\hfill \end{array}$ (2)

其中

${\varphi }_{\text{1}}\left(u\right)=\{\begin{array}{l}\frac{\theta }{2}\ln \left(1+u\right)+\frac{\theta }{2}-\frac{\theta }{2k}\left(1-u\right)-\frac{\theta }{2}\ln k,u\ge 1-k,\hfill \\ \frac{\theta }{2}\left(\ln \left(1+u\right)-\ln \left(1-u\right)\right),\left|u\right|<1-k,\hfill \\ -\frac{\theta }{2}\ln \left(1-u\right)-\frac{\theta }{2}+\frac{\theta }{2k}\left(1+u\right)+\frac{\theta }{2}\ln k,u\le -1+k.\hfill \end{array}$

${\varphi }_{\text{2}}\left(u\right)=u$。容易得到 $\forall u\in \left(-\infty ,\infty \right)$， $0<{{\varphi }^{\prime }}_1\le L:=\frac{\theta }{\left(2-k\right)k}$， ${{\varphi }^{\prime }}_2=1$。

2.1. 半离散格式

把时间区间 $\left[0,T\right]$ 进行剖分， $0=t_0<t_1<\cdots <t_N=T$，N是一个正整数，时间节点满足 $t_i=i\tau$， $i=0,1,\cdots ,N$， $\tau =\frac{T}{N}$ 是时间步长。接下来构造具有浓度迁移率和对数势能的修正Cahn-Hilliard方程的半离散格式：给定 $u^n$，求 $u^{n+1}$， $w^{n+1}$， ${\phi }^{n+1}$，当 $n\ge 1$ 时，满足

$\{\begin{array}{l}\left({\delta }_{\tau }{u}^{n+1},q\right)+\left(\nabla \cdot \left(m\left(u^n\right)\nabla w^{n+1}\right),\nabla q\right)=0,\forall q\in H^1\left(\Omega \right),\hfill \\ \left(w^{n+1},v\right)-{\epsilon }^2\left(\nabla u^{n+1},\nabla v\right)-\left({\phi }^{n+1},v\right)-\left({\varphi }_1\left(u^{n+1}\right)-{\varphi }_2\left(u^n\right),v\right)=0,\forall v\in H^1\left(\Omega \right),\hfill \\ \left(\nabla {\phi }^{n+1},\nabla \psi \right)-\beta \left(u^{n+1}-{\bar{u}}_0,\psi \right)=0,\forall \psi \in H^1\left(\Omega \right).\hfill \end{array}$ (3)

其中 ${\delta }_{\tau }{u}^{n+1}=\frac{u^{n+1}-u^n}{\tau }$。

2.2. 全离散格式

设 $T_h=K$ 是区域 $\Omega$ 上拟一致剖分， $h_i$ 表示网格大小， $h=\underset{0\le i\le N}{\max }{h}_i$， $S_h$ 是分片连续的有限元空间，定义 $S_h=\left\{v_h\in C\left(\Omega \right)|{v_h|}_K\in P_k\left(x,y\right),K\in T_h\right\}$， $S_h\subset H^1\left(\Omega \right)$。这里 $P_k\left(x,y\right)$ 是 $x,y$ 的次数不超过 $k\left(\in Z^{+}\right)$ 的线性多项式的集合。定义 $L_0^2:=\left\{u\in L^2\left(\Omega \right)|\left(u,1\right)=0\right\}$， ${\stackrel{\dot{}}{S}}_h:=S_h\cap L_0^2\left(\Omega \right)$。接下来构造具有浓度迁移率和对数势能的修正Cahn-Hilliard方程的全离散格式：给定 $u_h^n$，求 $u_h^{n+1}$， $w_h^{n+1}$， ${\phi }_h^{n+1}$，当 $n\ge 1$ 时，满足

$\{\begin{array}{l}\left({\delta }_{\tau }{u}_h^{n+1},q_h\right)+\left(\nabla \cdot \left(m\left(u_h^n\right)\nabla w_h^{n+1}\right),\nabla q_h\right)=0,\forall q_h\in S_h,\hfill \\ \left(w_h^{n+1},v_h\right)-{\epsilon }^2\left(\nabla u_h^{n+1},\nabla v_h\right)-\left({\phi }_h^{n+1},v_h\right)-\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_2\left(u_h^n\right),v_h\right)=0,\forall v_h\in S_h,\hfill \\ \left(\nabla {\phi }_h^{n+1},\nabla {\psi }_h\right)-\beta \left(u_h^{n+1}-{\bar{u}}_0,{\psi }_h\right)=0,\forall {\psi }_h\in S_h.\hfill \end{array}$ (4)

Ritz投影算子 $R_h:H^1\left(\Omega \right)\to S_h$ 满足 $\left(\nabla \left(u-R_{h}u\right),\nabla v\right)=0$， $\left(R_{h}u-u,1\right)=0$，其中 $u_h^0=R_h{u}_0$。

3. 稳定性分析

定义3.1： $H^{-1}$ 范数定义如下：

${\Vert v_h\Vert }_{-1,h}:=\sqrt{\left(v_h,{\left(-{\Delta }_h\right)}^{-1}{v}_h\right)}=\underset{0\ne \chi \in {\stackrel{\dot{}}{S}}_h}{\sup }\frac{\left(v,\chi \right)}{\Vert \nabla \chi \Vert },\forall v_h\in {\stackrel{\dot{}}{S}}_h.$

引理3.1 [9]：设 $\zeta ,\chi \in {\stackrel{\dot{}}{S}}_h\left(\Omega \right)$，有

${\left(\zeta ,\chi \right)}_{-1,h}:=\left(\nabla T_h\left(\zeta \right),\nabla T_h\left(\chi \right)\right)=\left(\zeta ,T_h\left(\chi \right)\right)=\left(T_h\left(\zeta \right),\chi \right).$

其中 $T_h:{\stackrel{\dot{}}{S}}_h\left(\Omega \right)\to {\stackrel{\dot{}}{S}}_h\left(\Omega \right)$ 是可逆线性算子， $\left(\nabla T_h\left(\chi \right),\nabla \nu \right)=\left(\chi ,\nu \right)$，且满足

$\left|\left(\chi ,g\right)\right|\le {\Vert \chi \Vert }_{-1,h}\Vert \nabla g\Vert ,$

其中 $g\in S_h\left(\Omega \right)$。进一步，下面估计式成立

${\Vert \zeta \Vert }_{-1,h}\le C\Vert \zeta \Vert ,\forall \zeta \in {\stackrel{\dot{}}{S}}_h.$

定理3.1：令 $\left(u_h^{n+1},w_h^{n+1}\right)$ 是(4)的解，对任意的 $\tau ,h,\epsilon >0$，下面的不等式成立

$E\left(u_h^{n+1}\right)+\tau {\Vert \sqrt{m\left(u_h^n\right)}\nabla w_h^{n+1}\Vert }^2+\frac{{\epsilon }^2}{2}{\Vert \nabla u_h^{n+1}-\nabla u_h^n\Vert }^2+\frac{\beta }{\text{2}}{\Vert u_h^{n+1}-u_h^n\Vert }_{-1,h}^2\le E\left(u_h^n\right).$ (5)

证明：在(4)的第1式中，令 $q_h=\tau w_h^{n+1}$，得

$\left(u_h^{n+1}-u_h^n,w_h^{n+1}\right)+\tau {\Vert \sqrt{m\left(u_h^n\right)}\nabla w_h^{n+1}\Vert }^2=0.$ (6)

在(4)的第2式中，令 $v_h=-\left(u_h^{n+1}-u_h^n\right)$，并运用 $2\left(a-b,a\right)=a^2-b^2+{\left(a-b\right)}^2$ 得

$\begin{array}{l}-\left(w_h^{n+1},u_h^{n+1}-u_h^n\right)+\frac{{\epsilon }^2}{2}\left({\Vert \nabla u_h^{n+1}\Vert }^2-{\Vert \nabla u_h^n\Vert }^2+{\Vert \nabla u_h^{n+1}-\nabla u_h^n\Vert }^2\right)\\ +\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_2\left(u_h^n\right),u_h^{n+1}-u_h^n\right)+\left({\phi }_h^{n+1},u_h^{n+1}-u_h^n\right)=0.\end{array}$ (7)

在(4)的第(3)式中，令 ${\psi }_h=-T_h\left(u_h^{n+1}-u_h^n\right)$，得

$-\left(\nabla {\phi }_h^{n+1},\nabla T_h\left(u_h^{n+1}-u_h^n\right)\right)+\beta \left(u_h^{n+1}-{\bar{u}}_0,T_h\left(u_h^{n+1}-u_h^n\right)\right)=0.$ (8)

将(6)，(7)，(8)相加并利用引理3.1得

$\begin{array}{l}\tau {\Vert \sqrt{m\left(u_h^n\right)}\nabla w_h^{n+1}\Vert }^2+\frac{{\epsilon }^2}{2}\left({\Vert \nabla u_h^{n+1}\Vert }^2-{\Vert \nabla u_h^n\Vert }^2+{\Vert \nabla u_h^{n+1}-\nabla u_h^n\Vert }^2\right)\\ +\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_2\left(u_h^n\right),u_h^{n+1}-u_h^n\right)+\beta {\left(u_h^{n+1}-{\bar{u}}_0,u_h^{n+1}-u_h^n\right)}_{-1,h}=0.\end{array}$ (9)

在 $\Phi \left(u\right)$ 的定义中，用 ${\Phi }_1\left(u\right)$ 和 ${\Phi }_2\left(u\right)$ 来表示 ${\varphi }_1\left(u\right)$ 和 ${\varphi }_2\left(u\right)$ 的原函数所对应的部分，即 $\Phi \left(u\right)={\Phi }_{\text{1}}\left(u\right)-{\Phi }_{\text{2}}\left(u\right)$， ${{\Phi }^{\prime }}_1\left(u\right)={\varphi }_1\left(u\right)$， ${{\Phi }^{\prime }}_2\left(u\right)={\varphi }_2\left(u\right)$。

然后利用泰勒展开公式，得到：

${\Phi }_1\left(u_h^n\right)-{\Phi }_1\left(u_h^{n+1}\right)={\varphi }_1\left(u_h^{n+1}\right)\left(u_h^n-u_h^{n+1}\right)+\frac{1}{2}{{\varphi }^{\prime }}_1\left({\xi }_1\right){\left(u_h^n-u_h^{n+1}\right)}^2\ge {\varphi }_1\left(u_h^{n+1}\right)\left(u_h^n-u_h^{n+1}\right).$ (10)

${\Phi }_2\left(u_h^{n+1}\right)-{\Phi }_2\left(u_h^n\right)={\varphi }_2\left(u_h^n\right)\left(u_h^{n+1}-u_h^n\right)+\frac{1}{2}{{\varphi }^{\prime }}_2\left({\xi }_2\right){\left(u_h^{n+1}-u_h^n\right)}^2\ge {\varphi }_2\left(u_h^n\right)\left(u_h^{n+1}-u_h^n\right).$ (11)

结合(10)，(11) 得到：

$\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_2\left(u_h^n\right)\right)\left(u_h^{n+1}-u_h^n\right)\ge \Phi \left(u_h^{n+1}\right)-\Phi \left(u_h^n\right).$ (12)

将(12)代入(9)式，并对(9)式最后一项运用 $2\left(a-b,a\right)=a^2-b^2+{\left(a-b\right)}^2$，得

$\begin{array}{l}\tau {\Vert \sqrt{m\left(u_h^n\right)}\nabla w_h^{n+1}\Vert }^2+\frac{{\epsilon }^2}{2}\left({\Vert \nabla u_h^{n+1}\Vert }^2-{\Vert \nabla u_h^n\Vert }^2+{\Vert \nabla u_h^{n+1}-\nabla u_h^n\Vert }^2\right)\\ +\left(\Phi \left(u_h^{n+1}\right)-\Phi \left(u_h^n\right),1\right)+\frac{\beta }{2}\left({\Vert u_h^{n+1}-{\bar{u}}_0\Vert }_{-1,h}^2-{\Vert u_h^n-{\bar{u}}_0\Vert }_{-1,h}^2+{\Vert u_h^{n+1}-u_h^n\Vert }_{-1,h}^2\right)\le 0.\end{array}$ (13)

根据自由能函数的定义，则上式变为

$E\left(u_h^{n+1}\right)+\tau {\Vert \sqrt{m\left(u_h^n\right)}\nabla w_h^{n+1}\Vert }^2+\frac{{\epsilon }^2}{2}{\Vert \nabla u_h^{n+1}-\nabla u_h^n\Vert }^2+\frac{\beta }{2}{\Vert u_h^{n+1}-u_h^n\Vert }_{-1,h}^2\le E\left(u_h^n\right).$

则稳定性得证。

4. 误差估计

为了之后证明的简单，介绍下面一些符号：

${\hat{e}}_u^{n+1}:=R_h{u}^{n+1}-u_h^{n+1},{\tilde{e}}_u^{n+1}:=u^{n+1}-R_h{u}^{n+1},$

${\hat{e}}_w^{n+1}:=R_h{w}^{n+1}-w_h^{n+1},{\tilde{e}}_w^{n+1}:=w^{n+1}-R_h{w}^{n+1},$

${\hat{e}}_{\phi }^{n+1}:=R_h{\phi }^{n+1}-{\phi }_h^{n+1},{\tilde{e}}_{\phi }^{n+1}:={\phi }^{n+1}-R_h{\phi }^{n+1},$

$\sigma \left(u^{n+1}\right)={\delta }_{\tau }{R}_h{u}^{n+1}-u_t^{n+1}.$

对于 $\left(u,w,\phi \right)$，做如下正则性假设：

$u\in L^{\infty }\left(0,T;W^{1,5}\left(\Omega \right)\right)\cap L^{\infty }\left(0,T;H^{r+1}\left(\Omega \right)\right),$

$w\in L^{\infty }\left(0,T;H^{r+1}\left(\Omega \right)\right),$

$\phi \in L^{\infty }\left(0,T;H^{r+1}\left(\Omega \right)\right).$

引理4.2 [10]：Ritz投影算子 $R_h$ 满足下面的估计：

${\Vert R_{h}u\Vert }_1\le C{\Vert u\Vert }_1,\forall u\in H^1\left(\Omega \right),$

$\Vert u-R_{h}u\Vert +h{\Vert u-R_{h}u\Vert }_1\le C{h}^{r+1}{\Vert u\Vert }_{r+1},\forall u\in H^{r+1}\left(\Omega \right).$

引理4.3 [11]：假设 $\left(u,w\right)$ 是(2)的解，则有下面估计：

${\Vert \sigma \left(u^{n+1}\right)\Vert }^2\le C{h}^{2r+2}+C{\tau }^2.$

定理4.2：设初始问题(2)，全离散格式(4)的解分别是 $\left(u,w\right)$ 和 $\left(u_h^{n+1},w_h^{n+\text{1}}\right)$，存在常数 $C,\tau ,h$，则有估计式

${\displaystyle \underset{i=1}{\overset{n}{\sum }}C\tau {\Vert \nabla {\hat{e}}_w^{i+1}\Vert }^2+{\epsilon }^2{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+\beta }{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\le C{\tau }^2+C{h}^{2r}.$ (14)

证明：对于(3)式，令 $q=q_h$， $v=v_h$， $\psi ={\psi }_h$，利用 $R_h$ 的性质，在 $t=t_{n+1}$，有

$\{\begin{array}{l}\left({\delta }_{\tau }{R}_h{u}^{n+1},q_h\right)+\left(m\left(u^{n+1}\right)\nabla R_h{w}^{n+1},\nabla q_h\right)=\left(\sigma \left(u^{n+1}\right),q_h\right),\\ {\epsilon }^2\left(\nabla R_h{u}^{n+1},\nabla v_h\right)-\left(R_h{w}^{n+\text{1}},v_h\right)+\left(R_h{\phi }^{n+1},v_h\right)\\ =\left({\tilde{e}}_w^{n+\text{1}},v_h\right)-\left({\tilde{e}}_{\phi }^{n+1},v_h\right)-\left({\varphi }_1\left(u^{n+1}\right)-{\varphi }_2\left(u^{n+1}\right),v_h\right),\\ \left(\nabla R_h{\phi }^{n+1},\nabla {\psi }_h\right)-\beta \left(R_h{u}^{n+\text{1}}-{\bar{u}}_0,{\psi }_h\right)=\beta \left({\tilde{e}}_u^{n+1},{\psi }_h\right).\end{array}$ (15)

(15)式减去(4)式得

$\{\begin{array}{l}\left({\delta }_{\tau }{\hat{e}}_u^{n+1},q_h\right)+\left(m\left(u^{n+1}\right)\nabla R_h{w}^{n+1},\nabla q_h\right)-\left(m\left(u_h^n\right)\nabla w_h^{n+1},\nabla q_h\right)=\left(\sigma \left(u^{n+1}\right),q_h\right),\\ {\epsilon }^2\left(\nabla {\hat{e}}_u^{n+1},\nabla v_h\right)-\left({\hat{e}}_w^{n+1},v_h\right)+\left({\hat{e}}_{\phi }^{n+1},v_h\right)\\ =\left({\tilde{e}}_w^{n+1},v_h\right)-\left({\tilde{e}}_{\phi }^{n+1},v_h\right)-\left({\varphi }_1\left(u^{n+1}\right)-{\varphi }_2\left(u^{n+1}\right),v_h\right)+\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_2\left(u_h^n\right),v_h\right),\\ \left(\nabla {\hat{e}}_{\phi }^{n+1},\nabla {\psi }_h\right)-\beta \left({\hat{e}}_u^{n+1},{\psi }_h\right)=\beta \left({\tilde{e}}_u^{n+1},{\psi }_h\right).\end{array}$ (16)

在(16)式中，令 $q_h={\hat{e}}_w^{n+1}$， $v_h={\delta }_{\tau }{\hat{e}}_u^{n+1}$， ${\psi }_h=-T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)$，然后三式相加得

$\begin{array}{l}\left(m\left(u^{n+1}\right)\nabla R_h{w}^{n+1},\nabla {\hat{e}}_w^{n+1}\right)-\left(m\left(u_h^n\right)\nabla w_h^{n+1},\nabla {\hat{e}}_w^{n+1}\right)\\ +\frac{{\epsilon }^2}{2\tau }\left({\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2-{\Vert \nabla {\hat{e}}_u^n\Vert }^2+{\Vert \nabla {\hat{e}}_u^{n+1}-\nabla {\hat{e}}_u^n\Vert }^2\right)+\frac{\beta }{2\tau }\left({\Vert {\hat{e}}_u^{n+1}\Vert }_{-1,h}^2-{\Vert {\hat{e}}_u^n\Vert }_{-1,h}^2+{\Vert {\hat{e}}_u^{n+1}-{\hat{e}}_u^n\Vert }_{-1,h}^2\right)\\ =\left(\sigma \left(u^{n+1}\right),{\hat{e}}_w^{n+1}\right)+\left({\tilde{e}}_w^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)-\left({\tilde{e}}_{\phi }^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)-\beta \left({\tilde{e}}_u^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\\ -\left({\varphi }_1\left(u^{n+1}\right)-{\varphi }_2\left(u^{n+1}\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)+\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_2\left(u_h^n\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right).\end{array}$ (17)

其中

$\begin{array}{l}\left(m\left(u^{n+1}\right)\nabla R_h{w}^{n+1},\nabla {\hat{e}}_w^{n+1}\right)-\left(m\left(u_h^n\right)\nabla w_h^{n+1},\nabla {\hat{e}}_w^{n+1}\right)\\ =\left(\left(m\left(u_h^n\right)+m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\right)\nabla R_h{w}^{n+1},\nabla {\hat{e}}_w^{n+1}\right)-\left(m\left(u_h^n\right)\nabla w_h^{n+1},\nabla {\hat{e}}_w^{n+1}\right)\\ ={\Vert \sqrt{m\left(u_h^n\right)}\nabla {\hat{e}}_w^{n+1}\Vert }^2+\left(m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\nabla R_h{w}^{n+1},\nabla {\hat{e}}_w^{n+1}\right).\end{array}$ (18)

则(17)式变为

$\begin{array}{l}{\Vert \sqrt{m\left(u_h^n\right)}\nabla {\hat{e}}_w^{n+1}\Vert }^2+\frac{{\epsilon }^2}{2\tau }\left({\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2-{\Vert \nabla {\hat{e}}_u^n\Vert }^2+{\Vert \nabla {\hat{e}}_u^{n+1}-\nabla {\hat{e}}_u^n\Vert }^2\right)\\ +\frac{\beta }{2\tau }\left({\Vert {\hat{e}}_u^{n+1}\Vert }_{-1,h}^2-{\Vert {\hat{e}}_u^n\Vert }_{-1,h}^2+{\Vert {\hat{e}}_u^{n+1}-{\hat{e}}_u^n\Vert }_{-1,h}^2\right)\\ =\left(\sigma \left(u^{n+1}\right),{\hat{e}}_w^{n+1}\right)+\left({\tilde{e}}_w^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)-\left({\tilde{e}}_{\phi }^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)-\beta \left({\tilde{e}}_u^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\\ -\left({\varphi }_1\left(u^{n+1}\right)-{\varphi }_2\left(u^{n+1}\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)+\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_2\left(u_h^n\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\\ -\left(m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\nabla R_h{w}^{n+1},\nabla {\hat{e}}_w^{n+1}\right)\\ ={\displaystyle \underset{i=1}{\overset{6}{\sum }}{G}_i}.\end{array}$ (19)

接下来，根据 Cauchy-Schwarz不等式，庞加莱不等式和Young不等式估计 $G_i$：

$\begin{array}{c}{G}_1=\left(\sigma \left(u^{n+1}\right),{\hat{e}}_w^{n+1}\right)\\ \le \left|\left(\sigma \left(u^{n+1}\right),{\hat{e}}_w^{n+1}\right)\right|\\ \le C\Vert \sigma \left(u^{n+1}\right)\Vert \Vert \nabla {\hat{e}}_w^{n+1}\Vert \\ \le \frac{C^2}{m_1}{\Vert \sigma \left(u^{n+1}\right)\Vert }^2+\frac{m_1}{4}{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2\\ \le C{\tau }^2+C{h}^{2r+2}+\frac{m_1}{4}{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2.\end{array}$ (20)

$\begin{array}{c}{G}_{\text{2}}=\left({\tilde{e}}_w^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\\ \le \left|\left({\tilde{e}}_w^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right|\\ \le C\Vert \nabla {\tilde{e}}_w^{n+1}\Vert {\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}\\ \le \frac{4C^2}{\alpha }{\Vert \nabla {\tilde{e}}_w^{n+1}\Vert }^2+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\\ \le C{h}^{2r}+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2.\end{array}$ (21)

$\begin{array}{c}{G}_{\text{3}}=-\left({\tilde{e}}_{\phi }^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\\ \le \left|\left({\tilde{e}}_{\phi }^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right|\\ \le C\Vert \nabla {\tilde{e}}_{\phi }^{n+1}\Vert {\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}\\ \le \frac{4C^2}{\alpha }{\Vert \nabla {\tilde{e}}_{\phi }^{n+1}\Vert }^2+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\\ \le C{h}^{2r}+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2.\end{array}$ (22)

$\begin{array}{c}{G}_{\text{4}}=-\beta \left({\tilde{e}}_u^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\\ \le \beta \left|\left({\tilde{e}}_u^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\right|\\ \le C\beta \Vert {\tilde{e}}_u^{n+1}\Vert \Vert T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\Vert \\ \le C\beta \Vert \nabla {\tilde{e}}_u^{n+1}\Vert {\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}\\ \le \frac{4C^2{\beta }^2}{\alpha }{\Vert \nabla {\tilde{e}}_u^{n+1}\Vert }^2+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\\ \le C{h}^{2r}+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2.\end{array}$ (23)

运用泰勒展开，有

$\begin{array}{c}\Vert \nabla \tau {\delta }_{\tau }u\left(t\right)\Vert ={\Vert {\displaystyle {\int }_{t-\tau }^t\nabla u_s\left(s\right)\text{d}s}\Vert }^2\\ \le {\left({\displaystyle {\int }_{t-\tau }^t\Vert \nabla u_s\left(s\right)\Vert \text{d}s}\right)}^2\\ \le {\left({\displaystyle {\int }_{t-\tau }^t{\Vert \nabla u_s\left(s\right)\Vert }^3\text{d}s}\right)}^{\frac{2}{3}}{\left({\displaystyle {\int }_{t-\tau }^t{1}^{\frac{3}{2}}\text{d}s}\right)}^{\frac{4}{3}}\\ \le {\tau }^{\frac{4}{3}}{\left({\displaystyle {\int }_{t-\tau }^t{\Vert \nabla u_s\left(s\right)\Vert }^3\text{d}s}\right)}^{\frac{2}{3}}\\ \le C{\tau }^2.\end{array}$ (24)

因此

$\begin{array}{c}{G}_{\text{5}}=-\left({\varphi }_1\left(u^{n+1}\right)-{\varphi }_2\left(u^{n+1}\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)+\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_2\left(u_h^n\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\\ =\left({\varphi }_1\left(u_h^{n+1}\right)-{\varphi }_1\left(u^{n+1}\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)+\left({\varphi }_2\left(u^{n+1}\right)-{\varphi }_2\left(u_h^n\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\\ =-\left({{\varphi }^{\prime }}_1\left({\xi }_3\right)\left(u^{n+1}-u_h^{n+1}\right),{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)+\left(u^{n+1}-u_h^n,{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\\ \le L\left|\left(u^{n+1}-u_h^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right|+\tau \left({\delta }_{\tau }{u}^{n+1},{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)+\left(u^n-u_h^n,{\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\end{array}$

$\begin{array}{c}\le \frac{4L^2}{\alpha }{\Vert \nabla \left(u^{n+1}-u_h^{n+1}\right)\Vert }^2+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\\ +\Vert \nabla \tau {\delta }_{\tau }{u}^{n+1}\Vert {\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}+\Vert \nabla \left(u^n-u_h^n\right)\Vert {\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}\\ \le C{h}^{2r}+C{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2+\frac{4}{\alpha }{\Vert \nabla \tau {\delta }_{\tau }{u}^{n+1}\Vert }^2+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\\ +C{h}^{2r}+C{\Vert \nabla {\hat{e}}_u^n\Vert }^2+\frac{\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\\ \le C{\tau }^2+C{h}^{2r}+C{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+C{\Vert \nabla {\hat{e}}_u^n\Vert }^2+\frac{3\alpha }{16}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2.\end{array}$ (25)

$\begin{array}{c}{G}_{\text{6}}=-\left(m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\nabla R_h{w}^{n+1},\nabla {\hat{e}}_w^{n+1}\right)\\ \le \Vert m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\nabla R_h{w}^{n+1}\Vert \Vert \nabla {\hat{e}}_w^{n+1}\Vert \\ \le \frac{1}{m_1}{\Vert m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\nabla R_h{w}^{n+1}\Vert }^2+\frac{m_1}{4}{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2\\ \le \frac{C{M}^2}{m_1}{\Vert \nabla \left(u^{n+1}-u_h^n\right)\Vert }^2{\Vert \nabla w^{n+1}\Vert }_{-1,h}^2+\frac{m_1}{4}{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2\\ \le C{\tau }^2+C{h}^{2r}+C{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+\frac{m_1}{4}{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2.\end{array}$ (26)

则(19)式变为

$\begin{array}{l}{\Vert \sqrt{m\left(u_h^n\right)}\nabla {\hat{e}}_w^{n+1}\Vert }^2+\frac{{\epsilon }^2}{2\tau }\left({\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2-{\Vert \nabla {\hat{e}}_u^n\Vert }^2+{\Vert \nabla {\hat{e}}_u^{n+1}-\nabla {\hat{e}}_u^n\Vert }^2\right)\\ +\frac{\beta }{2\tau }\left({\Vert {\hat{e}}_u^{n+1}\Vert }_{-1,h}^2-{\Vert {\hat{e}}_u^n\Vert }_{-1,h}^2+{\Vert {\hat{e}}_u^{n+1}-{\hat{e}}_u^n\Vert }_{-1,h}^2\right)\\ \le C{\tau }^2+C{h}^{2r}+C{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+C{\Vert \nabla {\hat{e}}_u^n\Vert }^2+\frac{m_1}{\text{2}}{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2+\frac{3\alpha }{\text{8}}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2.\end{array}$ (27)

在(16)式的第一式中令 $q_h=T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)$，有

$\begin{array}{c}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2=\left(\sigma \left(u^{n+1}\right),T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)-\left(m\left(u^{n+1}\right)\nabla R_h{w}^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\\ +\left(m\left(u_h^n\right)\nabla w_h^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\\ =\left(\sigma \left(u^{n+1}\right),T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)-\left(\left(m\left(u_h^n\right)+m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\right)\nabla R_h{w}^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\\ +\left(m\left(u_h^n\right)\nabla w_h^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\\ =\left(\sigma \left(u^{n+1}\right),T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)-\left(m\left(u_h^n\right)\nabla {\hat{e}}_w^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\end{array}$

$\begin{array}{l}-\left(m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\nabla R_h{w}^{n+1},T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\right)\\ \le \Vert \sigma \left(u^{n+1}\right)\Vert \Vert T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\Vert +{\Vert m\left(u_h^n\right)\Vert }_{L^{\infty }}\Vert \nabla {\hat{e}}_w^{n+1}\Vert \Vert \nabla T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\Vert \\ +\Vert m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\nabla R_h{w}^{n+1}\Vert \Vert \nabla T_h\left({\delta }_{\tau }{\hat{e}}_u^{n+1}\right)\Vert \\ \le \frac{3}{2}{\Vert \sigma \left(u^{n+1}\right)\Vert }^2+\frac{1}{2}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2+\frac{3m_1^2}{2}{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2+\frac{3}{2}{\Vert m^{\prime }\left({\xi }_3\right)\left(u^{n+1}-u_h^n\right)\nabla R_h{w}^{n+1}\Vert }^2\\ \le C{\tau }^2+C{h}^{2r}+C{h}^{2r+2}+\frac{3m_1^2}{2}{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2+C{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+\frac{1}{2}{\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2.\end{array}$ (28)

则

${\Vert {\delta }_{\tau }{\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\le C{\tau }^2+C{h}^{2r}+3m_1^2{\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2+C{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2.$ (29)

将(29)式代入(27)式得

$\begin{array}{l}{\Vert \sqrt{m\left(u_h^n\right)}\nabla {\hat{e}}_w^{n+1}\Vert }^2+\frac{{\epsilon }^2}{2\tau }\left({\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2-{\Vert \nabla {\hat{e}}_u^n\Vert }^2\right)+\frac{\beta }{2\tau }\left({\Vert {\hat{e}}_u^{n+1}\Vert }_{-1,h}^2-{\Vert {\hat{e}}_u^n\Vert }_{-1,h}^2\right)\\ \le C{\tau }^2+C{h}^{2r}+C{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+C{\Vert \nabla {\hat{e}}_u^n\Vert }^2+\left(\frac{m_1}{\text{2}}+\frac{9m_1^2\alpha }{8}\right){\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2.\end{array}$ (30)

在(30)式中取适当的 $\alpha \left(0<\alpha <\frac{4\left(2m_2-m_1\right)}{9m_1^2}\right)$，并乘以 $\text{2}\tau$ 得：

$\begin{array}{l}C\tau {\Vert \nabla {\hat{e}}_w^{n+1}\Vert }^2+{\epsilon }^2\left({\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2-{\Vert \nabla {\hat{e}}_u^n\Vert }^2\right)+\beta \left({\Vert {\hat{e}}_u^{n+1}\Vert }_{-1,h}^2-{\Vert {\hat{e}}_u^n\Vert }_{-1,h}^2\right)\\ \le C\tau {\tau }^2+C\tau h^{2r}+C{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+C{\Vert \nabla {\hat{e}}_u^n\Vert }^2.\end{array}$ (31)

上式从1到n求和，根据离散的Gronwall不等式，有：

${\displaystyle \underset{i=1}{\overset{n}{\sum }}C\tau {\Vert \nabla {\hat{e}}_w^{i+1}\Vert }^2+{\epsilon }^2{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }^2+\beta }{\Vert \nabla {\hat{e}}_u^{n+1}\Vert }_{-1,h}^2\le C{\tau }^2+C{h}^{2r}.$ (32)

5. 数值实验

在数值实验部分，采用一些数值算例验证理论分析的正确性和有效性。选择初始条件为

$u_0=0.5+0.17\cos \left(\pi x\right)\cos \left(2\pi y\right)+0.2\cos \left(3\pi x\right)\cos (\; \pi \; y\; )$

计算区域为 $\left[-1,1\right]\times \left[-1,1\right]$。在表1~4中，选择固定的参数 $\tau =\frac{1}{1000}$，变化的网格步长为 $h=\frac{1}{16}$， $h=\frac{1}{32}$， $h=\frac{1}{64}$， $h=\frac{1}{128}$， $h=\frac{1}{256}$， $k=0.01$。相对误差 $\Vert {\hat{e}}_u\Vert$ 和 ${\Vert {\hat{e}}_u\Vert }_{H^1}$ 的空间收敛阶接近于2和1，与理论部分得到的收敛阶保持一致，且不同的 $\theta$ 和 $\beta$ 对收敛阶影响不大。

Table 1. ε = 0.09 , θ = 0.1 , β = 0

表1. $\epsilon =0.09,\theta =0.1,\beta =0$

Table 2. ε = 0.09 , θ = 0.1 , β = 1

表2. $\epsilon =0.09,\theta =0.1,\beta =1$

Table 3. ε = 0.09 , θ = 0.7 , β = 0

表3. $\epsilon =0.09,\theta =0.7,\beta =0$

Table 4. ε = 0.09 , θ = 0.7 , β = 1

表4. $\epsilon =0.09,\theta =0.7,\beta =1$

6. 结论

本文研究了具有浓度迁移率和对数势能的修正Cahn-Hilliard方程，在理论分析中证明了它的稳定性和误差估计，并给出了数值算例验证了它的结论。

基金项目

山西省自然科学基金(No:201901D111123)。