## Discontinuous Galerkin Finite Element Analysis of for the Extended Fisher-Kolmogorov Equation

Yang Xiaoxia,, Zhang Houchao

School of Mathematics and Statistics, Pingdingshan University, Henan Pingdingshan 467000

 基金资助: 国家自然科学基金.  11271340国家自然科学基金.  116713692020年度河南省高等学校重点科研项目.  20A1100302020年度河南省高等学校重点科研项目.  20B110013平顶山学院高层次人才启动基金.  PXY-BSQD-2019001

 Fund supported: the NSFC.  11271340the NSFC.  116713692020 Key Scientific Research Project of Henan Province Colleges and Universities.  20A1100302020 Key Scientific Research Project of Henan Province Colleges and Universities.  20B110013the Doctoral Starting Foundation of Pingdingshan University.  PXY-BSQD-2019001

Abstract

The discontinuous Galerkin finite element approximation schemes for the Extended Fisher-Kolmogorov (EFK) equation are studied by using the Wilson element. Without using the technique of postprocessing technique, the convergence results with order $O(h^{2})/O(h^{2}+\tau)$ for the primitive solution $u$ and intermediate variable $v=-\triangle u$ are obtained for the semi-discrete and linearized Euler fully discrete approximation schemes respectively through a new splitting technique for the nonlinear terms. The above results are just one order higher than the usual error estimates of the Wilson element. Here, $h$ and $\tau$ are parameters of the subdivision in space and time step, respectively.

Keywords： EFK equation ; Discontinuous Galerkin finite element ; Wilson element ; Semi-discrete and fully-discrete schemes ; Convergence

Yang Xiaoxia, Zhang Houchao. Discontinuous Galerkin Finite Element Analysis of for the Extended Fisher-Kolmogorov Equation. Acta Mathematica Scientia[J], 2021, 41(6): 1880-1896 doi:

## 1 引言

$$$\left\{ \begin{array}{ll} u_t+r\Delta^2 u-\Delta u+f(u) = 0 , & (X, t)\in{{\Omega} \times (0, T]}, \\ u(X, t) = \Delta u(X, t) = 0, & (X, t)\in {{\partial{\Omega}} \times (0, T]}, \\ u(X, 0) = u_{0}(X), & X\in {\Omega}, \end{array} \right.$$$

## 2 有限元的构造及性质

$\Omega$的边界$\partial{\Omega}$分别平行于$x$轴与$y$轴, $\Gamma_{h} $$\Omega 的矩形单元剖分族, 满足正则性假设. 对任何 K\in\Gamma_{h} , 设其平行于 x 轴与 y 轴的边的边长分别为 2h_1, 2h_2 , 记 h_K = \max\limits_{K\in\Gamma_{h}}\{h_1, h_2\},$$ h = \max\limits_{K\in\Gamma_{h}}\{h_K\}$.

$\varepsilon_h$表示所有单元边界所组成的集合, $E$表示单元边界, $h_E$表示$E$的长度. 规定函数$f $$E = \partial K\bigcap \partial K' 上的跳跃值和平均值分别为 [[f]] = f|_K-f|_{K'},$$ \{f\} = (f|_K+f|_{K'})/2,$其中$K $$K' 表示相邻的单元. Wilson元的形函数空间为 P_2(K) , 有限元空间为 其中 M_h|_K 由四顶点函数值以及 \frac{h_1^2}{h_1h_2}\int_K \frac{\partial^2v_h}{\partial x^2}{\rm d}X$$ \frac{h_2^2}{h_1h_2}\int_K \frac{\partial^2v_h}{\partial y^2}{\rm d}X$确定.

$$$\|v_h\|^2_h = \sum\limits_{K\in \Gamma_h}|v_h|^2_{1, K}+\sum\limits_{E\in \varepsilon _h} \bigg\{\frac{1}{h_E}\|[[v_h]]\|^2_{0, E}\bigg\} .$$$

$I_h:H^2(\Omega)\rightarrow M_h$为相应的插值算子.

$$$a_h(u_h, v_h)\leq \beta_0\|u_h\|_h\|v_h\|_h, \qquad \forall u_h, v_h\in M_h ,$$$

$$$a_h(v_h, v_h)\geq \alpha_0\|v_h\|_h^2, \qquad \forall v_h\in M_h.$$$

$(2.3)$式的证明知, 可取$\alpha>\frac{3}{4}C_0$以保证$\alpha_0>0$.

$$$\|\varphi_h\|_{0, 2k}\leq C(k)\|\varphi_h\|_{1, h},$$$

## 3 半离散格式的收敛性分析

$v = -\triangle u$, 问题(1.1) 等价于

$$$\left\{ \begin{array}{ll} u_t-r \triangle v+v+f(u) = 0 , & (X, t)\in{{\Omega} \times (0, T]}, \\ v+\triangle u = 0, & (X, t)\in{{\Omega} \times (0, T]}, \\ u(X, t) = v(X, t) = 0, & (X, t)\in {{\partial{\Omega}} \times (0, T]}, \\ u(X, 0) = u_{0}(X), & X\in {\Omega}. \end{array} \right.$$$

$$$\left\{ \begin{array}{ll} (u_t, w)+r(\nabla v, \nabla w)+(v, w)+(f(u), w) = 0, & \forall w\in H^1_0(\Omega), \\ (v, \varphi)-(\nabla u, \nabla \varphi) = 0, & \forall \varphi\in H^1_0(\Omega). \end{array} \right.$$$

$$$\left\{ \begin{array}{ll} (a)(u_{ht}, w_h)+r(\nabla v_h, \nabla w_h)+(v_h, w_h)+(f(u_h), w_h) = 0, & \forall w_h\in M_h, \\ (b)(v_h, \varphi_h)-(\nabla u_h, \nabla \varphi_h) = 0, & \forall \varphi_h\in M_h. \end{array} \right.$$$

$$$\left\{ \begin{array}{ll} (a)(u_{ht}, w_h)+ra_h(v_h, w_h)+(v_h, w_h)+(f(u_h), w_h) = 0, & \forall w_h\in M_h, \\ (b)(v_h, \varphi_h)-a_h(u_h, \varphi_h) = 0, & \forall \varphi_h\in M_h, \\ u_h(0) = I_hu_0(X), v_h(0) = I_h(-\Delta u_0(X)), & X\in {\Omega}, \end{array} \right.$$$

$\begin{eqnarray} \|u_h\|_h\leq [C(r)(\|v_{h}(0)\|_0^2+(S(u_h(0)), 1)]^{\frac{1}{2}}, \end{eqnarray}$

$$$\left\{ \begin{array}{ll} (a)(\theta_t, w_h)+ra_h(\zeta, w_h)+(\zeta, w_h) = -(\rho_t, w_h)-ra_h(\eta, w_h)-(\eta, w_h)\\ \quad \quad \quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad -(f(u)-f(u_h), w_h), & \forall w_h\in M_h, \\ (b)(\zeta, \varphi_h)-a_h(\theta, \varphi_h) = -(\eta, \varphi_h)+a_h(\rho, \varphi_h), & \forall \varphi_h\in M_h, \\ \end{array} \right.$$$

$\begin{eqnarray} &&ra_h(\zeta, \zeta)+(\zeta, \zeta)+a_h(\theta, \theta_t)\\ & = &-(\rho_t, \zeta)-ra_h(\eta, \zeta)-(\eta, \zeta) -(f(u)-f(u_h), \zeta)+(\eta, \theta_t)-a_h(\rho, \theta_t) \doteq \sum\limits_{i = 1}^6A_i. \end{eqnarray}$

$\begin{eqnarray} \|\theta\|_h^2 \leq Ch^4[\|u\|^2_3+\|v\|^2_2+\int^t_0(\|u\|^2_3+\|v\|^2_3+\|u_t\|_3^2+\|v_t\|_2^2){\rm d}s]. \end{eqnarray}$

$t = t_n$, 问题(3.2) 的等价形式为

$$$\left\{ \begin{array}{ll} (a)(\partial_tu^n, w)+ra_h(v^n, w)+(v^n, w)+(f(u^n), w) = (R^n_1, w), & \forall w\in H^1_0(\Omega), \\ (b)(v^n, \varphi)-a_h(u^n, \varphi) = 0, & \forall \varphi\in H^1_0(\Omega). \end{array} \right.$$$

$$$\left\{ \begin{array}{ll} (a)(\partial_tU_h^n, w_h)+ra_h(V_h^n, w_h)+(V_h^n, w_h)+(f(U_h^{n-1}), w_h) = 0, & \forall w_h\in M_h, \\ (b)(V_h^n, \varphi_h)-a_h(U_h^n, \varphi_h) = 0, & \forall \varphi_h\in M_h, \\ U^0_h = I_hu_0(X), V^0_h = I_h(-\Delta u_0(X)), & \forall X\in \Omega. \end{array} \right.$$$

$$$\|u^j-U_h^j\|_h+\|v^j-V_h^j\|_h = O(h^2+\tau).$$$

记$u^j-U_h^j = (u^j-I_hu^j)+(I_hu^j-U_h^j)\doteq\rho^j+\theta^j, $$v^j-V_h^j = (v^j-I_hv^j)+(I_hv^j-V_h^j)\doteq\eta^j+\zeta^j. 由(4.1) 和(4.2) 式得误差方程为 $$\left\{ \begin{array}{ll} (a)(\partial_t\theta^j, w_h)+ra_h(\zeta^j, w_h)+(\zeta^j, w_h) = -(\partial_t\rho^j, w_h)\\ \quad -ra_h(\eta^j, w_h)-(\eta^j, w_h)-(f(u^j)-f(U_h^{j-1}), w_h)+(R_1^j, w_h), & \forall w_h\in M_h, \\ (b)(\zeta^j, \varphi_h)-a_h(\theta^j, \varphi_h) = -(\eta^j, \varphi_h)+a_h(\rho^j, \varphi_h), & \forall \varphi_h\in M_h.\\ \end{array} \right.$$ 根据泰勒公式, 易得 $$\|R_1^j\|_0^2\leq C\tau^2\|u_{tt}\|^2_{L^\infty(L^2(\Omega))}.$$ 由于 U^0_h = I_hu_0(X), $$\|u_0-U^0_h\|_{0, p} = \|u_0-I_hu_0\|_{0, p}\leq Ch^2\|u_0\|_{2, p}, \quad 1\leq p<\infty.$$ 首先, 我们用数学归纳法证明存在 h'>0, \tau'>0 , 使得当 h<h', \tau<\tau' 时, 对于任何 0\leq j\leq N $$\|\theta^j\|_h\leq C'(h^2+\tau),$$ 这里, C' 是与 j, h, \tau 无关的常数. j = 0 时, 由于 \theta^0 = 0 , (4.7) 式成立. 设对任何 j\leq n-1 , (4.7) 式成立, 那么应用(2.4) 式有 于是, 当 \tau< \tau_1\leq1/(2CC'+1), h< h_1\leq\sqrt{1/(2CC'+1)} 时, 有 \begin{eqnarray} \|\theta^j\|_{0, 2k}< 1, k = 1, 2, 3, 4. \end{eqnarray} j = n 时, 在(4.4) 式取 w_h = \zeta^n, \varphi_h = \partial_t\theta^n , \begin{eqnarray} &&\tau ra_h(\zeta^n, \zeta^n)+\tau \|\zeta^n\|_0^2+\tau a_h(\theta^n, \partial_t\theta^n)\\ & = &-\tau(\partial_t\rho^n, \zeta^n) -r\tau a_h(\eta^n, \zeta^n)-\tau(\eta^n, \zeta^n)-\tau(f(u^n)-f(U_h^{n-1}), \zeta^n)\\ &&+\tau(R_1^n, \zeta^n)+\tau(\eta^n, \partial_t\theta^n)-\tau a_h(\rho^n, \partial_t\theta^n)\doteq\sum^7_{i = 1}F_i. \end{eqnarray} 下面对 F_i(i = 1, \cdot\cdot\cdot, 7) 进行估计. 首先注意到 \varphi, \varphi_t\in L^2(\Omega) 时, 有 \begin{eqnarray} \|\partial_t\varphi^n\|^2_0 \leq \frac{C}{\tau}\int^{t_n}_{t_{n-1}}\|\varphi_t\|^2_0{\rm d}s \end{eqnarray} 成立 (n = 1, \cdots , N) , 则有 利用(2.2) 式及(4.5) 式, 可得 注意到 $$(\varphi^n, \partial_tr^n) = \partial_t(\varphi^n, r^n)-(\partial_t\varphi^n, r^{n-1})$$ \begin{eqnarray} \|\partial_{t}\varphi^n\|_h^2 \leq\frac{C}{\tau}\int^{t_{n}}_{t_{n-1}}\|\varphi_t\|^2_h{\rm d}s (n = 1, \cdots , N) \end{eqnarray} 成立, 则有 由于 而由Sobolve嵌入定理, (2.4) 式及(4.8) 式有 于是 将上述关于 F_i(i = 1, \cdot\cdot\cdot, 7) 的估计代入(4.9) 式, 注意到 \theta(0) = 0 , (2.3) 式及 \begin{eqnarray} a_h(\varphi^n, \partial_t\varphi^n) = \frac{1}{2}\partial_ta_h(\varphi^n, \varphi^n)+\frac{1}{2\tau}a_h(\varphi^n-\varphi^{n-1}, \varphi^n-\varphi^{n-1}) \end{eqnarray} 成立, 再将 n 替换为 i , 并关于 i 从1到 n 求和可得 对上式应用离散的Gronwall引理, 存在正常数 C_1, \tau_2 , 使得当 \tau< \tau_2 时有 C'\geq C_1, \tau'\leq \min \tau_i(1\leq i\leq 2), h'\leq h_1, 则对于任意的 0\leq j\leq N , (4.7) 式成立. 其次, 我们证明存在 h''>0, \tau''>0 , 使得当 h<h'', \tau<\tau'' 时, 对于任何 0\leq j\leq N $$\|\zeta^j\|_h\leq C''(h^2+\tau),$$ 这里, C'' 是与 j, h, \tau 无关的常数. j = 0 时, 由于 \zeta^0 = 0 , (4.14) 式成立. 设对任何 j\leq n-1 , (4.14) 式成立. j = n 时, 首先由问题 (4.4) 可得 $$\left\{ \begin{array}{ll} {\rm (a)}{\quad} (\partial_t\theta^n, w_h)+ra_h(\zeta^n, w_h)+(\zeta^n, w_h) = -(\partial_t\rho^n, w_h)\\ \quad -ra_h(\eta^n, w_h)-(\eta^n, w_h)-(f(u^n)-f(U_h^{n-1}), w_h)+(R_1^n, w_h), & \forall w_h\in M_h, \\ {\rm (b)}{\quad} (\partial_t\zeta^n, \varphi_h)-a_h(\partial_t\theta^n, \varphi_h) = -(\partial_t\eta^n, \varphi_h)+a_h(\partial_t\rho^n, \varphi_h), & \forall \varphi_h\in M_h.\\ \end{array} \right.$$ 在(4.15(a)) 式中取 w_h = \partial_t\zeta^n, (4.15(b)) 式中取 \varphi_h = \partial_t\theta^n , 两式相减后两边同乘 2\tau , 再将 n 替换为 i 并关于 i 从1到 n 求和则有 \begin{eqnarray} &&2\tau r\sum^n_{i = 1}a_h(\zeta^i, \partial_t\zeta^i)+2\tau \sum^n_{i = 1}(\zeta^i, \partial_t\zeta^i)+2\tau \sum^n_{i = 1}a_h(\partial_t\theta^i, \partial_t\theta^i) {}\\ & = &-2\tau \sum^n_{i = 1}(\partial_t\rho^i, \partial_t\zeta^i) -2\tau r\sum^n_{i = 1}a_h(\eta^i, \partial_t\zeta^i)-2\tau \sum^n_{i = 1}(\eta^i, \partial_t\zeta^i) {}\\ &&-2\tau \sum^n_{i = 1}(f(u^i)-f(U_h^{i-1}), \partial_t\zeta^i) +2\tau \sum^n_{i = 1}(R_1^i, \partial_t\zeta^i)+2\tau \sum^n_{i = 1}(\partial_t\eta^i, \partial_t\theta^i){}\\ &&-2\tau \sum^n_{i = 1}a_h(\partial_t\rho^i, \partial_t\theta^i)\doteq\sum^7_{i = 1}Q_i. \end{eqnarray} 下面对 Q_i(i = 1, \cdot\cdot\cdot, 7) 进行估计. 利用(2.2) 式, 并注意到 $$(\partial_t\varphi^i, \partial_tr^i) = \frac{(\partial_t\varphi^i, r^i)-(\partial_t\varphi^{i-1}, r^{i-1})}{\tau}-(\frac{\varphi^i-2\varphi^{i-1}+\varphi^{i-2}}{\tau^2}, r^{i-1})$$ $$\bigg\|\frac{\varphi^i-2\varphi^{i-1}+\varphi^{i-2}}{\tau^2} \bigg\|_0^2\leq\frac{C}{\tau}\int^{t_i}_{t_{i-2}}\|\varphi_{tt}\|_0^2{\rm d}s,$$ 利用(4.5) 式, 并注意到 则有 再次利用(4.10) 式、(4.12) 式和(2.2) 式, 同时注意到引理2.1成立, 可得 \begin{eqnarray} Q_6+Q_7 &\leq&C\tau \sum^n_{i = 1}(\|\partial_t\eta^i\|_0^2+\|\partial_t\rho^i\|_h^2)+\frac{\alpha_0}{2}\tau\sum^n_{i = 1}\|\partial_t\theta^i\|_h^2\\ &\leq&Ch^4 (\|v_t\|_{L^{\infty}(H^2(\Omega))}^2+\|u_t\|_{L^{\infty}(H^3(\Omega))}^2)+\frac{\alpha_0}{2}\tau\sum^n_{i = 1}\|\partial_t\theta^i\|_h^2. \end{eqnarray} 而利用(4.11) 式有 \begin{eqnarray} Q_4 & = &-2\tau\sum^n_{i = 1}(f(u^i)-f(u^{i-1}), \partial_t\zeta^i)-2(f(u^{n-1})-f(U_h^{n-1}), \zeta^n)\\ &&+2\tau\sum^n_{i = 2}(\frac{(f(u^{i-1})-f(U_h^{i-1}))-(f(u^{i-2})-f(U_h^{i-2}))}{\tau}, \zeta^{i-1})\\ &\doteq& Q_{41}+Q_{42}+Q_{43} \end{eqnarray} \begin{eqnarray} Q_{41} & = &-2(f(u^{n})-f(u^{n-1}), \zeta^n)+2\sum^n_{i = 2}(f(u^{i})-2f(u^{i-1})+f(u^{i-2}), \zeta^{i-1})\\ &\doteq& Q_{411}+Q_{412}. \end{eqnarray} 类似于 F_{41} 的估计有 而利用泰勒公式可得 将上面两式的估计代入(4.21) 式, 可得 利用 F_{42} 的估计技巧及(4.7) 式可得 \begin{eqnarray} Q_{43} & = &2\tau\sum^n_{i = 2}(f'(\lambda^{i-1}_1)\partial_tu^{i-1}-f'(\lambda^{i-1}_2)\partial_tU_h^{i-1}, \zeta^{i-1})\\ & = &2\tau\sum^n_{i = 2}((f'(\lambda^{i-1}_1)-f'(\lambda^{i-1}_2))\partial_tu^{i-1}, \zeta^{i-1}) +2\tau\sum^n_{i = 2}(f'(\lambda^{i-1}_2)(\partial_tu^{i-1}-\partial_tU_h^{i-1}), \zeta^{i-1})\\ &\doteq& Q_{431}+Q_{432}, \end{eqnarray} 其中, \lambda^{i-1}_1 = u^{i-2}+\sigma_1(u^{i-1}-u^{i-2}) , \lambda^{i-1}_2 = U_h^{i-2}+\sigma_2(U_h^{i-1}-U_h^{i-2}) , 0<\sigma_1, \sigma_2<1 . 由于对任意的正整数 1\leq q<+\infty 有Sobolve嵌入定理, H^1(\Omega)\hookrightarrow L^q(\Omega) , H^3(\Omega)\hookrightarrow W^{2, q}(\Omega) 成立, 从而当 q = 2, 4, 6, 8 时有 因此当 h<h_2\leq\frac{1}{\sqrt{2(C_2+1)}}, \tau<\tau_3\leq\frac{1}{2(C_2+1)} 时, 对于 q = 2, 4, 6, 8 利用 b^2-a^2 = (b-a)^2+2(b-a)a , 可得当 h<h_2, \tau<\tau_3 时, 有 \begin{eqnarray} Q_{431} & = &-6\tau\sum^n_{i = 2}(\partial_tu^{i-1}((\lambda^{i-1}_2-\lambda^{i-1}_1)^2+2(\lambda^{i-1}_2-\lambda^{i-1}_1)\lambda^{i-1}_1), \zeta^{i-1})\\ &\leq&C\tau\sum^n_{i = 2}(\|u_t\|^2_{0, \infty}\|\lambda^{i-1}_2-\lambda^{i-1}_1\|^4_{0, 4}+\|u\|_{0, \infty}^2\|\lambda^{i-1}_2-\lambda^{i-1}_1\|_{0}^2) +C\tau\sum^n_{i = 2}\|\zeta^{i-1}\|^2_h\\ &\leq& C(h^4+\tau^2)+C\tau\sum^n_{i = 2}\|\zeta^{i-1}\|^2_h. \end{eqnarray} 再利用 可得 \begin{eqnarray} Q_{432} & = &12\tau\sum^n_{i = 2}(\lambda^{i-1}_1(\lambda^{i-1}_2-\lambda^{i-1}_1)(\partial_tu^{i-1}-\partial_tU_h^{i-1}), \zeta^{i-1})\\ &&+6\tau\sum^n_{i = 2}((\lambda^{i-1}_1-\lambda^{i-1}_2)^2(\partial_tu^{i-1}-\partial_tU_h^{i-1}), \zeta^{i-1})\\ &&+2\tau\sum^n_{i = 2}(f'(\lambda^{i-1}_1)(\partial_tu^{i-1}-\partial_tU_h^{i-1}), \zeta^{i-1})\\ &\leq&C\tau \bigg[\sum^n_{i = 2}\|\partial_tu^{i-1}-\partial_tU_h^{i-1}\|_{0, 4}\|\zeta^{i-1}\|_0+\sum^n_{i = 2}\|\partial_tu^{i-1}-\partial_tU_h^{i-1}\|_{0}\|\zeta^{i-1}\|_0 \bigg]\\ &\leq&C\tau\sum^n_{i = 2}\frac{1}{\tau^{\frac{1}{2}}}\bigg(\int^{t_{i-1}}_{t_{i-2}}\|\rho_t\|^4_{0, 4}{\rm d}s\bigg)^{\frac{1}{2}} +C\tau\sum^n_{i = 2}\frac{1}{\tau}\bigg(\int^{t_{i-1}}_{t_{i-2}}\|\rho_t\|^2_{0}{\rm d}s\bigg)\\ &&+\frac{\alpha_0}{2}\tau\sum^n_{i = 2}\|\partial_t\theta^{i-1}\|^2_{h} +C\tau\sum^n_{i = 2}\|\zeta^{i-1}\|_0^2\\ &\leq&Ch^4\|u_t\|^2_{L^{\infty}(H^3(\Omega))}+\frac{\alpha_0}{2}\tau\sum^n_{i = 2}\|\partial_t\theta^{i-1}\|^2_{h} +C\tau\sum^{n-1}_{i = 1}\|\zeta^{i}\|_h^2. \end{eqnarray} 将(4.23) 式和(4.24) 式的估计代入(4.22) 式可得, 当 h<h_2, \tau<\tau_3 时, 有 Q_{4i}(i = 1, 2, 3) 的估计代入(4.20) 式, 可得 Q_{i}(i = 1, 2, \cdot\cdot\cdot, 7) 的估计代入(4.16) 式, 同时利用(4.13) 式的思想及(2.3) 式可得 对上式应用离散的Gronwall引理, 存在正常数 C_3, \tau_4 , 使得当 \tau< \tau_4 , h< h_2 时有 C''\geq \max\limits_{2\leq i\leq3} C_i, \tau''\leq \min\limits_{1\leq i\leq4}{\tau_i}, h''\leq \min\limits_{1\leq i\leq2}{h_i}, 则对于任意的 0\leq j\leq N , (4.14) 式成立. 综上, 再利用引理2.1和三角不等式, 易知(4.3) 式成立. 注4.1 对比发现, 本文得到了 u, v 的在半离散格式及向后欧拉全离散格式下的误差估计, 与文献[14]和[15]同阶, 但去掉了文献[14]和[15]中 \tau = O(h^2) 这一条件. 另外, 本文对 u, v 的光滑度的要求 (u, v\in H^3(\Omega)) 也比文献[14]及[15] (u, v\in H^5(\Omega)) 要低. 注4.2 在定理4.1的证明中, 对于非线性项 Q_4 的估计是至关重要的, 这里采用了一个新的分裂技巧. 这也是本文的创新点之一. ## 5 数值实验 在这一节, 我们将给出一个数值算例来验证理论分析的正确性和算法的有效性. 考虑如下EFK方程 $$\left\{ \begin{array}{ll} u_t+5\Delta^2 u-\Delta u+f(u) = g(X, t) , & (X, t)\in{{\Omega} \times (0, T]}, \\ u(X, t) = \Delta u(X, t) = 0, & (X, t)\in {{\partial{\Omega}} \times (0, T]}, \\ u(X, 0) = u_{0}(X), & X\in {\Omega}, \end{array} \right.$$ 其中 \Omega = \left[ {0, 1} \right]^{2} , f(u) = u^3-u,$$ g(X, t)$是由方程(5.1)和真解

 $m^{2}$ $\left\| {u^{n}-U_h^{n} } \right\|_h$ 收敛阶 $\left\| {v^{n}-V_h^{n} } \right\|_h$ 收敛阶 $4^{2}$ 0.000046035 — 0.005970921 — $8^{2}$ 0.000012875 1.8381 0.001790764 1.7374 $16^{2}$ 0.000003264 1.9799 0.000472460 1.9223 $32^{2}$ 0.000000793 2.0417 0.000117908 2.0025 $64^{2}$ 0.000000195 2.0240 0.000029285 2.0094

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