分数阶微分方程(Fractional Differential Equation, FDEs) 在现实生活中有着广泛的应用，如在化学、生物学、物理学、控制论、力学、图像处理、聚合物流变学、空气动力学等领域备受关注^[1-2]。文献[3]研究了分数阶微积分在物理学中的应用，包括高分子物理、流变学、生物物理学、热力学等。文献[4]研究了一类广义分数阶偏微分方程的解，时间上是希尔弗时间分数阶导数，空间上是广义拉普拉斯算子。常见的分数阶微积分，如Riemann-Liouville 微积分、Caputo微积分等，都备受学者们的关注。Caputo-Hadamard型分数阶导数，常用于描述材料疲劳断裂的力学规律，应用十分广泛。文献[5]通过修正的拉普拉斯变换和傅里叶正弦变换得到线性Caputo-Hadamard分数阶导数偏微分方程的解析解，并且证明了其差分格式在时间方向上的精度为$O({N^{ - \min \{ r\alpha ,2 - \alpha \} }}) $，其中$ \alpha \in (0,1) $为分数阶导数的阶，N表示时间方向上的网格数。文献[6]研究了Hadamard型分数阶微分方程解的对数衰减和稳定性。文献[7]采用修正的拉普拉斯变换和傅里叶正弦变换，得到了具有初始奇异性的Caputo-Hadamard分数阶扩散波方程的解析解，并研究了其正则性。此外，文献[8]提出了Caputo-Hadamard分数阶慢扩散方程的误差卷积结构分析方法，证明了差分格式在时间方向上的精度为$O({N^{ - {\text{min}}\{ r\alpha ,2\} }}) $。

本文将考虑下述二维Caputo-Hadamard时间分数阶慢扩散方程的高阶有限差分格式。

$ \begin{split} &{}_{{\rm{CH}}}D_{a,t}^\alpha\left( {x,y,t} \right) - \Delta u\left( {x,y,t} \right) = f\left( {x,y,t} \right),\\ &\quad \left( {x,y} \right) \in \varOmega ,0 \lt \alpha \lt 1,a \lt t \leqslant T \end{split}$

(1)

$ \begin{array}{*{20}{r}} {\begin{array}{*{20}{r}} {u(x,y,a) = \phi (x,y) ,\quad (x,y) \in \overline \varOmega = \varOmega \cup \partial \varOmega } \end{array}} \end{array}$

(2)

$ \begin{array}{*{20}{r}} {u(x,y,t) = \varphi (x,y,t) ,\quad (x,y) \in \partial \varOmega } \end{array} $

(3)

式中：$ \Delta $是二维拉普拉斯算子，$\varOmega = (0,{L_1}) \times (0,{L_2}) $，$ \partial \varOmega $是$ \varOmega $的边界，$f(x,y,t) $，$\phi (x,y) $，$\varphi (x,y,t) $是已知函数，${}_{{\text{CH}}}D_{a,t}^\alpha $是Caputo-Hadamard时间分数阶导数：

$ \begin{array}{*{20}{c}} \begin{aligned} {}_{{\text{CH}}}D_{a,t}^\alpha u(x,y,t) & = \frac{1}{{\Gamma (1 - \alpha ) }} \times \int_a^t {{{\left( {\ln \frac{t}{s}} \right) }^{ - \alpha }}\delta u(x,y,s) \frac{{{\text{d}}s}}{s}} ，\\ \end{aligned} \end{array} $

$ 0 \lt a \lt t,\quad {\delta ^k} = {(s\frac{{\text{d}}}{{{\text{d}}s}}) ^k},k \in {\mathbb{Z}^ + } $

式中，$ \Gamma $是伽马函数。

由于高维问题可以通过ADI方法化为一系列独立的一维问题，并且能够在一定程度上提高计算效率^[9]，因此采用紧致的ADI有限差分法处理该模型。对于ADI格式，不少学者都做过相应的研究，见文献[10-13]。但关于Caputo-Hadamard分数阶导数的ADI有限差分法的研究相对较少，因此对模型(1) ~(3) 建立ADI紧致差分格式是有意义的。

本文框架如下：在第1节中首先介绍了一些符号和引理，并且对模型(1)~(3)构造一个ADI紧致差分格式。在第2节中，利用数学归纳法给出求解格式的稳定性和收敛性证明。在第3节中，给出算例，对具体模型进行数值求解，验证第2节的理论结果。

1 离散与格式建立

在建立ADI紧致差分格式之前，首先引入一些记号。在空间方向上，对于给定的正整数$ {M_1} $，$ {M_2} $，记${h_1} = {{{L_1}} \mathord{\left/ {\vphantom {{{L_1}} {{M_1}}}} \right. } {{M_1}}}$，${x_i} = i{h_1}$，$ 0 \leqslant i \leqslant {M_1} $，${h_2} = {{{L_2}} \mathord{\left/ {\vphantom {{{L_2}} {{M_2}}}} \right. } {{M_2}}}$，${y_j} = j{h_2}$，$ 0 \leqslant j \leqslant {M_2} $。

在时间方向上，对于给定的正整数N，对区间$ [a,T] $进行分割，记${t_k} = a{\left( {{T \mathord{\left/ {\vphantom {T a}} \right. } a}} \right) ^{k{\text{/}}N}}$，其中$ 0 \leqslant k \leqslant N $。相应地，对区间$[{\text{ln}}a,\ln T]$进行分割，记$\ln {t_k} = \ln a + \left( {K/N} \right) \left( {\ln \left( {T/a} \right)} \right)$。

定义$\tau = {\text{ln}}{t_k} - \ln {t_{k - 1}} = {{(\ln T - \ln a) } / N}$，$1 \leqslant k \leqslant N$，$ {\ln {t_{n - \theta }} = \theta \ln {t_{n - 1}} + (1 - \theta ) \ln {t_n}} $，$\theta = \alpha /2$。

对于任意的网格函数$u = \{ {u_{ij}}\mid 0 \leqslant i \leqslant {M_1},0 \leqslant j \leqslant {M_2}\} $，定义以下记号

$ \begin{gathered} {\delta _x}{u_{i - {1 / 2}}} = {{({u_{ij}} - {u_{i - 1,j}}) } / {{h_1}}} \\ \delta _x^2{u_{ij}} = {{({\delta _x}{u_{i + {1 / 2},j}} - {\delta _x}{u_{i - {1 / 2},j}}) } / {{h_1}}} \\ \end{gathered} $

${\mathcal{H}_x}{u_{ij}} = \left\{ \begin{array}{l} (1 + \dfrac{{h_1^2}}{{12}}\delta _x^2){u_{ij}},\;\;1 \le i \le {M_1} - 1,0 \le j \le {M_2}\\ {u_{ij}},\;\;i = 0或{M_1},0 \le j \le {M_2} \end{array} \right.$

相应地，对于空间上$y$方向的记号也有类似的定义。对于紧算子${\mathcal{H}_x}$、${\mathcal{H}_y}$的平方根分别用${\mathcal{A}_x}$、${\mathcal{A}_y}$来表示。对于任意的网格函数$ u $、$ v $进一步定义为

$\begin{array}{l} \mathcal{H}{u_{ij}} = {\mathcal{H}_x}{\mathcal{H}_y}{u_{ij}},\quad {\rm{\Lambda }}{u_{ij}} = ({\mathcal{H}_y}\delta _x^2 + {\mathcal{H}_x}\delta _y^2){u_{ij}}\\ \left\langle {u,v} \right\rangle = {h_1}{h_2}\displaystyle \sum \limits_{i = 0}^{{M_1} - 1} \displaystyle \sum \limits_{j = 0}^{{M_2} - 1} {u_{ij}}{v_{ij}},\quad \parallel u{\parallel ^2} = \left\langle {u,u} \right\rangle \end{array}$

$\begin{array}{c} \left\langle {{\delta _x}u,{\delta _x}v} \right\rangle = {h_1}{h_2}\displaystyle\sum\limits_{i = 0}^{{M_1}} {\displaystyle\sum\limits_{j = 0}^{{M_{2 - 1}}} {{\delta _{{x}}}{u_{i - 1/2,}}} } {}_j{\delta _x}{v_{i - 1/2,j}}\\ {\left\| {{\delta _x}u} \right\|^2} = \left\langle {{\delta _x}u,{\delta _x}u} \right\rangle \end{array}$

接下来将推导模型(1)~(3)的ADI紧致差分格式。

参考文献[8]，对Caputo-Hadamard时间分数阶导数$_{{\text{CH}}}D_{a,t}^\alpha u(t) $在点$ {t_{n - \theta }} $处用$ {L_{\ln ,2 - {1_\sigma }}} $进行离散可得

$ \begin{aligned} _{{\text{CH}}}D_\tau ^\alpha u({t_{n - \theta }}) =& A_0^{(n) }u({t_n}) - \displaystyle \sum \limits_{k = 1}^{n - 1} (A_{n - k - 1}^{(n) } - A_{n - k}^{(n) }) u({t_k}) - \\ &A_{n - 1}^{(n) }u({t_0}) + {({R_t}) ^{n - \theta }} \\ \end{aligned} $

式中，离散卷积核$A_{n - k}^{(n) }$的定义如下：当$ n = 1 $时，有$A_0^{(1) } = a_0^{(1) }$；当 $ n \geqslant 2 $时，有

${A_{n - k}^{(n) } = \left\{ {\begin{array}{*{20}{l}} {a_0^{(n) } + b_1^{(n) },}{k = n} \\ {a_{n - k}^{(n) } - b_{n - k + 1}^{(n) } - b_{n - k}^{(n) },}{2 \leqslant k \leqslant n - 1} \\ {a_{n - 1}^{(n) } - b_{n - 1}^{(n) },}{k = 1} \end{array}} \right.} $

定义

$ \begin{array}{*{20}{c}} {a_0^{(n) } = \dfrac{1}{\tau }\displaystyle \int \nolimits_{{t_{n - 1}}}^{{t_{n - \theta }}} \delta {\varpi _n}(s) \frac{{{\text{d}}s}}{s}} \\ a_{n - k}^{(n) } = \dfrac{1}{\tau }\displaystyle \int \nolimits_{{t_{k - 1}}}^{{t_k}} \delta {\varpi _n}(s) \frac{{{\text{d}}s}}{s},\quad 1 \leqslant k \leqslant n - 1 \\ b_{n - k}^{(n) } = \dfrac{1}{{{\tau ^2}}}\displaystyle \int \nolimits_{{t_{k - 1}}}^{{t_k}} \delta {\varpi _n}(s) (\ln s - \ln {t_{k - \frac{1}{2}}}) \frac{{{\text{d}}s}}{s},\quad {1 \leqslant k \leqslant n - 1} \end{array} $

式中：$\delta {\varpi _n}(s) : = {{{{(\ln {t_{n - \theta }} - \ln s) }^{ - \alpha }}} / {\varGamma (1 - \alpha ) }}$，$\ln {t_{k - {1 / 2}}}: = {{(\ln {t_{k - 1}} + \ln {t_{k + 1}}) } / 2}$。

在给出截断误差之前，先介绍以下引理。

引理1^[14]　当$ 0 < \alpha < 1 $, $u(t) \in {C^3}[{t_0},{t_n}]$时，有

$ \begin{split} _{{\rm{CH}}}D_\tau ^\alpha u({t_{n - \theta }}) =& A_0^{(n) }u({t_n}) - \displaystyle \sum \limits_{k = 1}^{n - 1} (A_{n - k - 1}^{(n) } - A_{n - k}^{(n) }) u({t_k}) - \\ &A_{n - 1}^{(n) }u({t_0}) + {({R_t}) ^{n - \theta }},\quad n \ge 1 \end{split} $

(4)

式中：${({R_t}) ^{n - \theta }} = O({\tau ^{3 - \alpha }}) $。

引理2^[15]　若满足$\begin{array}{*{20}{r}} {\xi (s) = 5{{(1 - s) }^3} - 3{{(1 - s) }^5}} \end{array}$，$f(x) \in {C^6}[{x_{i - 1}},{x_{i + 1}}]$，有

$ \begin{gathered} \frac{{f''({x_{i - 1}}) + 10f''({x_i}) + f''({x_{i + 1}}) }}{{12}} = \\ \quad \frac{{f({x_{i - 1}}) - 2f({x_i}) + f({x_{i + 1}}) }}{{h_1^2}} + \\ \end{gathered} $

$ \frac{{h_1^4}}{{360}}\displaystyle \int \nolimits_0^1 [{f^{(6) }}({x_i} - s{h_1}) + {f^{(6) }}({x_i} + s{h_1}) ]\xi (s) {\text{d}}s $

通过引理2可得到

$ \begin{split} &\mathcal{H}({u_{xx}}({x_i},{y_j}) + {u_{yy}}({x_i},{y_j}) ) = {\mathcal{H}_y}\delta _x^2u({x_i},{y_j}) + \\ &\qquad\quad{\mathcal{H}_x}\delta _y^2u({x_i},{y_j}) + {({R_{xy}}) _{ij}} \\ \end{split} $

(5)

式中：$\begin{array}{*{20}{r}} {\mid {{({R_{xy}}) }_{ij}}\mid \leqslant C(h_1^4 + h_2^4) } \end{array}$。

当$({x_i},{y_j}) \in \varOmega ,\; 1 \leqslant n \leqslant N$时，不妨记函数

$ \begin{array}{l}u_{ij}^n = u({x_i},{y_j},{t_n}) ,\quad ({u_{xx}}) _{ij}^n = {u_{xx}}({x_i},{y_j},{t_n}) \\ ({u_{yy}}) _{ij}^n = {u_{yy}}({x_i},{y_j},{t_n}) ,\quad f_{ij}^n = f({x_i},{y_j},{t_n}) \end{array} $

考虑式(1)在点$({x_i},{y_j},{t_{n - \theta }}) $处的情况，可以得到

$ {}_{\rm{CH}}D_{{{a}},t}^{{\alpha}}u_{ij}^{n - \theta } - \Delta u_{ij}^{n - \theta } = f_{ij}^{n - \theta } $

(6)

结合式(4)~式(6)可近似地表示为

${\mathcal{H}_{{\rm{CH}}}}D_\tau ^\alpha u_{ij}^{n - \theta } - {\rm{\Lambda }}u_{ij}^{n - \theta } = \mathcal{H}f_{ij}^{n - \theta } + R_{ij}^{n - \theta }$

(7)

其中有

$\begin{array}{l} \begin{array}{*{20}{r}} {{\rm{\Lambda }}u_{ij}^{n - \theta } = \theta {\rm{\Lambda }}u_{ij}^{n - 1} + (1 - \theta ){\rm{\Lambda }}u_{ij}^n} \end{array}\\ \begin{array}{*{20}{r}} {R_{ij}^{n - \theta } = ({R_{xy}})_{ij}^{n - \theta } - \mathcal{H}({R_t})_{ij}^{n - \theta }} \end{array} \end{array} $

在式 (7)两边同时加上小量项$\lambda _n^2{( 1 - \theta ) ^2}\delta _x^2\delta _y^2{}_{{\text{CH}}}D_\tau ^\alpha u_{ij}^{n - \theta } = O({\tau ^{2\alpha }}) $，${\lambda _n} = \dfrac{1}{{A_0^{(n) }}}$，可以得到

$ \begin{split} &{\mathcal{H}_{{\rm{CH}}}}D_\tau ^\alpha u_{ij}^{n - \theta } - {\rm{\Lambda }}u_{ij}^{n - \theta } + \lambda _n^2{(1 - \theta ) ^2}\delta _x^2\delta _y^2{}_{{\rm{CH}}}D_\tau ^\alpha u_{ij}^{n - \theta } =\\ &\mathcal{H}f_{ij}^{n - \theta } + {(R_{ij}^{n - \theta }) ^ * }\end{split} $

(8)

式中：${(R_{ij}^{n - \theta }) ^ * } = R_{ij}^{n - \theta } + \lambda _n^2{(1 - \theta ) ^2}\delta _x^2\delta _y^2{}_{{\text{CH}}}D_\tau ^\alpha u_{ij}^{n - \theta }$。

在式(8)中忽略截断误差同时用近似解$U_{ij}^{n - \theta }$去代替精确解$u_{ij}^{n - \theta }$，可以得到如下差分格式：

$\begin{array}{l} \begin{array}{*{20}{r}} \begin{array}{ccccc} & [{\mathcal{H}_x} - {\lambda _n}(1 - \theta )\delta _x^2][{\mathcal{H}_y} - {\lambda _n}(1 - \theta )\delta _y^2]U_{ij}^n = \\ & {\lambda _n}[\mathcal{H} + \lambda _n^2{(1 - \theta )^2}\delta _x^2\delta _y^2]\Bigg[ {\displaystyle\sum\limits_{k = 1}^{n - 1} {(A_{n - k - 1}^{(n)} - A_{n - k}^{(n)})U_{ij}^k} - } \Bigg. \end{array} \end{array}\\ \Bigg. {A_{n - 1}^{(n)}U_{ij}^0} \Bigg] + \theta {\rm{\Lambda }}U_{ij}^{n - 1} + \mathcal{H}f_{ij}^{n - \theta },({x_i},{y_j}) \in \varOmega {\rm{,}}1 \le n \le N\\[-12pt] \end{array}$

(9)

$ U_{ij}^0 = \phi ({x_i},{y_j}) {\text{,}} \quad ({x_i},{y_j}) \in \overline \varOmega $

(10)

$ U_{ij}^n = \varphi ({x_i},{y_j},{t_n}) ,\quad ({x_i},{y_j}) \in \partial \varOmega ,1 \leqslant n \leqslant N $

(11)

求解$U_{ij}^n$是由2个独立的一维度问题所决定的。当$1 \leqslant i \leqslant {M_1} - 1, 1 \leqslant j \leqslant {M_2} - 1$时，定义中间变量为

$ U_{ij}^ * = [{\mathcal{H}_y} - {\lambda _n}(1 - \theta ) \delta _y^2]U_{ij}^n $

对于固定的$j \in \{ 1,2,\cdots,{M_2} - 1\} $，$U_{ij}^ * $可由下述格式求解出来

$\begin{split} &\qquad\qquad\qquad[{\mathcal{H}_x} - {\lambda _n}(1 - \theta )\delta _x^2]U_{ij}^ * = \\ &\quad{\lambda _n}[\mathcal{H} + \lambda _n^2{(1 - \theta )^2}\delta _x^2\delta _y^2]\left[ {\displaystyle\sum\limits_{k = 1}^{n - 1} {(A_{n - k - 1}^{(n)} - A_{n - k}^{(n)})U_{ij}^k} - } \right. \\ &\left. {A_{n - 1}^{(n)}U_{ij}^0} \right] + \theta {\rm{\Lambda }}U_{ij}^{n - 1} + \mathcal{H}f_{ij}^{n - \theta },({x_i},{y_j}) \in \varOmega {\rm{,}}1 \le n \le N \end{split}$

(12)

$ U_{0j}^ * = \left[ {{\mathcal{H}_y} - {\lambda _n}(1 - \theta ) \delta _y^2} \right]U_{0j}^n $

(13)

$ U_{{M_1}j}^ * = \left[ {{\mathcal{H}_y} - {\lambda _n}(1 - \theta ) \delta _y^2} \right]U_{{M_1}j}^n $

(14)

当$U_{ij}^ * $求解出来，对于固定的$i \in \{ 1,2,\cdots,{M_1} - 1\} {\mkern 1mu} $，可以从以下格式求解出$U_{ij}^n$。

$ \begin{array}{*{20}{r}} {[{\mathcal{H}_y} - {\lambda _n}(1 - \theta ) \delta _y^2]U_{ij}^n = U_{ij}^ * ,\;\; 1 \leqslant j \leqslant {M_2} - 1} \end{array} $

(15)

$ U_{i0}^n = \varphi ({x_i},{y_0},{t_n}) ,\;\; U_{i{M_2}}^n = \varphi ({x_i},{y_{{M_2}}},{t_n}) $

(16)

不难看出，在每个时间层上，差分格式(12) ~(14) 和(15) ~(16) 所对应的系数矩阵是严格对角占优的，所以该差分格式唯一可解。

2 ADI格式的稳定性与收敛性

在给出稳定性和收敛性的证明之前，先介绍一些需要用到的引理。

引理3^[10]　对于任意的网格函数$ u $，下列不等式成立

$ \begin{gathered} \left\langle {{{\mathcal{H}}_{{\text{CH}}}}D_\tau ^\alpha {u^{n - \theta }},{u^{n - \theta }}} \right\rangle \geqslant \begin{array}{*{20}{r}} {\dfrac{1}{2}\displaystyle \sum \limits_{k = 1}^n A_{n - k}^{(n) }{\nabla _\tau }(\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2}) } \end{array} \\ \left\langle {\delta _x^2\delta _{y{\text{CH}}}^2D_\tau ^\alpha {u^{n - \theta }},{u^{n - \theta }}} \right\rangle \geqslant \begin{array}{*{20}{r}} {\dfrac{1}{2}\displaystyle \sum \limits_{k = 1}^n A_{n - k}^{(n) }{\nabla _\tau }(\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}) } \end{array} \\ \end{gathered} $

式中：$\begin{array}{*{20}{r}} {{\nabla _\tau }{u^k} = {u^k} - {u^{k - 1}}} \end{array}$。

引理4^[12] 　对于任意的网格函数$ u $，不等式$\left\langle - {\rm{\Lambda }}{u^{n - \theta }}, {u^{n - \theta }} \right\rangle \ge 0$成立。

引理5　令$0 \lt \alpha \lt 1$，参考文献[10]，可以得到：

(1) 离散卷积核$A_{n - k}^{(n) }$是单调的：

$ \begin{array}{*{20}{r}} {A_0^{(n) } \gt A_1^{(n) } \gt \cdots \gt A_k^{(n) } \gt \cdots \gt A_n^{(n) }} \end{array}。$

(2) 离散卷积核$A_{n - k}^{(n) }$是有界的：

$ \begin{gathered} \begin{array}{*{20}{r}} {A_0^{(n) } \leqslant \dfrac{{24}}{{11\tau }}\displaystyle \int \nolimits_{{t_{n - 1}}}^{{t_n}} {\omega _{1 - \alpha }}(\ln {t_n} - \ln s) \frac{{{\text{d}}s}}{s}, 1 \leqslant k \leqslant n} \end{array} \\ A_{n - k}^{(n) } \leqslant \dfrac{4}{{11\tau }}\displaystyle \int \nolimits_{{t_{k - 1}}}^{{t_k}} {\omega _{1 - \alpha }}(\ln {t_n} - \ln s) \frac{{{\text{d}}s}}{s}, 1 \leqslant k \leqslant n \\ \end{gathered} $

式中：${\omega _\beta }(t) : = {{{t^\beta }} \mathord{\left/ {\vphantom {{{t^\beta }} {\Gamma (\beta ) }}} \right. } {\Gamma (\beta ) }}$。

(3) 首项$A_0^{(n) }$大于第二项：

$ \begin{array}{*{20}{r}} {\dfrac{{1 - 2\theta }}{{1 - \theta }}A_0^{(n) } - A_1^{(n) } \geqslant 0, n \geqslant 2} \end{array} $

定理1(稳定性估计)　假设$u_{ij}^n$是如下差分格式的解

$\begin{split} &{\mathcal{H}_{{\rm{CH}}}}D_\tau ^au_{ij}^{n - \theta } - {\rm{\Lambda }}u_{ij}^{n - \theta } + \lambda _n^2{(1 - \theta )^2}\delta _x^2\delta _y^2{}_{{\rm{CH}}}D_\tau ^\alpha u_{ij}^{n - \theta } =\\ &\qquad\quad{{\cal Hf}_{ij}^{n - \theta }} ,({x_i},{y_j}) \in \varOmega {\rm{,}}\quad 1 \le n \le N \end{split}$

(17)

$ u_{ij}^0 = \phi ({x_i},{y_j}) {\text{,}} ({x_i},{y_j}) \in \overline \varOmega $

(18)

$ u_{ij}^n = 0, ({x_i},{y_j}) \in \partial \varOmega , 1 \leqslant n \leqslant N $

(19)

那么当$1 \leqslant n \leqslant N$时，有

$ \begin{array}{c} \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} \leqslant \\ {\mathop {\max }\limits_{1 \leqslant i \leqslant n} } \left\{ {\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _i^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} + } \right. \\ \end{array} $

$ \left. {{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{i - \theta }}\parallel } \mathord{\left/ {\vphantom {{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{i - \theta }}\parallel } {\mathcal{A}_{i - 1}^{(i) }}}} \right. } {{A}_{i - 1}^{(i) }}}} \right\} $

证明　用${u^{n - \theta }}$ 同时对(17) 式两边作内积，可以得到

$\begin{split} &\qquad\langle {{\mathcal{H}_{{\rm{CH}}}}D_\tau ^\alpha {u^{n - \theta }},{u^{n - \theta }}} \rangle - \langle {{\rm{\Lambda }}{u^{n - \theta }},{u^{n - \theta }}} \rangle + \\ &\langle {\lambda _n^2{{(1 - \theta )}^2}\delta _x^2\delta _y^2{}_{{\rm{CH}}}D_\tau ^\alpha {u^{n - \theta }},{u^{n - \theta }}} \rangle = \langle {\mathcal{H}{f^{n - \theta }},{u^{n - \theta }}} \rangle \end{split}$

(20)

根据引理3和引理4可以得到

$ \begin{gathered} \begin{array}{*{20}{r}} {\langle {{{\mathcal{H}}_{{\text{CH}}}}D_\tau ^\alpha {u^{n - \theta }},{u^{n - \theta }}} \rangle \geqslant \dfrac{1}{2}\displaystyle \sum \limits_{k = 1}^n A_{n - k}^{(n) }{\nabla _\tau }(\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2}) } \end{array} \\ \langle {\delta _x^2\delta _{y{\text{CH}}}^2D_\tau ^\alpha {u^{n - \theta }},{u^{n - \theta }}} \rangle \geqslant \dfrac{1}{2}\displaystyle \sum \limits_{k = 1}^n A_{n - k}^{(n) }{\nabla _\tau }(\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}) \\ \end{gathered} $

$\left\langle { - {\rm{\Lambda }}{u^{n - \theta }},{u^{n - \theta }}} \right\rangle \ge 0$

式(20)最后一项可放缩为

$ \begin{aligned} &\langle {\mathcal{H}{f^{n - \theta }},{u^{n - \theta }}} \rangle = \langle {{\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }},{\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}} \rangle \leqslant \\ &\qquad \parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel \parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}\parallel \\ \end{aligned} $

因此, 可以得到

$ \begin{array}{*{20}{l}} \begin{gathered} \qquad \frac{1}{2}\displaystyle \sum \limits_{k = 1}^n A_{n - k}^{(n) }{\nabla _\tau }(\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2}) + \frac{1}{2}\lambda _n^2{(1 - \theta ) ^2} \times \\ \displaystyle \sum \limits_{k = 1}^n A_{n - k}^{(n) }{\nabla _\tau }(\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}) \leqslant \parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel \parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}\parallel \\ \end{gathered} \end{array} $

将上式进行展开和移项后可得

$ \begin{array}{l} \dfrac{1}{2}A_0^{(n) }\left[ {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} - \parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - 1}}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2} \times } \right. \\ \left. {\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2} - \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^{n - 1}}{\parallel ^2}} \right] \leqslant \\ \parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel \parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}\parallel - \\ \dfrac{1}{2} \displaystyle\sum\limits_{k = 1}^{n - 1} {A_{n - k}^{(n) }{\nabla _\tau }(\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2}) } - \dfrac{1}{2} \lambda _n^2{(1 - \theta ) ^2} \times \\ \displaystyle\sum\limits_{k = 1}^{n - 1} {A_{n - k}^{(n) }{\nabla _\tau }(\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}) } \\ \end{array} $

从而可以得到

$\begin{array}{l} \dfrac{1}{2}A_0^{(n) }\left[ {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} \right] \leqslant \\ \dfrac{1}{2}A_0^{(n) }\left[ {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - 1}}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^{n - 1}}{\parallel ^2}} \right] + \\ \parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel \parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}\parallel - \dfrac{1}{2} \displaystyle\sum\limits_{k = 1}^{n - 1} {A_{n - k}^{(n) }{\nabla _\tau }\left[ {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2} + } \right.} \\ \left. {\lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}} \right] \\ \end{array}$

对上式两边同时除以${{A_0^{(n) }} \mathord{\left/ {\vphantom {{A_0^{(n) }} 2}} \right. } 2}$可以得到

$ \begin{split} &\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{(1 - \theta ) ^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2} \leqslant \\ & \displaystyle \sum \limits_{k = 1}^{n - 1} \frac{{A_{n - k - 1}^{(n) } - A_{n - k}^{(n) }}}{{A_0^{(n) }}}\left[ {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\left. {\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}} \right]} \right. + \\ &\dfrac{{A_{n - 1}^{(n) }}}{{A_0^{(n) }}}\left[ {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} \right] + \frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel }}{{{A}_0^{(n) }}} \times \\ & \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^{n - \theta }}{\parallel ^2}}\\[-15pt] \end{split} $

(21)

定义

$ \begin{split} &\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^m}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^m}{\parallel ^2}} =\\ &\mathop {\max }\limits_{1 \leqslant i \leqslant n - 1} \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^i}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^i}{\parallel ^2}} \end{split}$

如果不等式

$ \begin{split} &\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} \leqslant \\ &\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^m}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^m}{\parallel ^2}} \end{split}$

成立，那么结论显然成立。

相反，若

$ \begin{split} &\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} \geqslant \\ &\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^m}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^m}{\parallel ^2}} \end{split}$

则有

$ \begin{gathered} \begin{array}{*{20}{l}} { \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^{n - \theta }}{\parallel ^2}} } \end{array} \leqslant \\ \max \left\{ {\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - 1}}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^{n - 1}}{\parallel ^2}} ,} \right. \\ \left. {\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} } \right\} \leqslant \\ \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} \end{gathered} $

式(21) 两边同时除以

$ \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}}$

可以得到

$ \begin{gathered} \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} \leqslant \\ \displaystyle\sum\limits_{k = 1}^{n - 1} {\frac{{A_{n - k - 1}^{(n) } - A_{n - k}^{(n) }}}{{A_0^{(n) }}}\frac{{\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}}}{{\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} }}} + \\ \dfrac{{A_{n - 1}^{(n) }}}{{A_0^{(n) }}}\frac{{\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}}}{{\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} }} + \\ \end{gathered} $

$ \begin{gathered} \frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel }}{{{A}_0^{(n) }}}\sqrt {\frac{{\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^{n - \theta }}{\parallel ^2}}}{{\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}}}} \leqslant \\ \sum\limits_{k = 1}^{n - 1} {\frac{{A_{n - k - 1}^{(n) } - A_{n - k}^{(n) }}}{{A_0^{(n) }}}\frac{{\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}}}{{\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}} }}} + \\ \end{gathered} $

$ \begin{split} & \frac{{A_{n - 1}^{(n) }}}{{A_0^{(n) }}}\frac{{\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}}}{{\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} }} + \\ & \frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel }}{{{A}_0^{(n) }}}\sqrt {\frac{{\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^{n - \theta }}{\parallel ^2}}}{{\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^{n - \theta }}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^{n - \theta }}{\parallel ^2}}}} \end{split} $

因此可以得到

$ \begin{gathered} \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} \leqslant \\ \displaystyle \sum \limits_{k = 1}^{n - 1} \frac{{A_{n - k - 1}^{(n) } - A_{n - k}^{(n) }}}{{A_0^{(n) }}}\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}} + \\ \dfrac{{A_{n - 1}^{(n) }}}{{A_0^{(n) }}}\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} + \frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel }}{{A_0^{(n) }}} \\ \end{gathered} $

(22)

现在, 猜想下列不等式成立

$ {\begin{split} &\qquad\qquad \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}} \leqslant \\ &\mathop {\max }\limits_{1 \leqslant i \leqslant k} \left\{ {\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _i^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} + \frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{i - \theta }}\parallel }}{{A_{i - 1}^{(i) }}}} \right\} \end{split}}$

并且用数学归纳法去证明它。

当$k = 1$时，通过式(22) 可以得到

$ \begin{gathered} \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^1}{\parallel ^2} + \lambda _1^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^1}{\parallel ^2}} \leqslant \\ \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _1^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} + \frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{1 - \theta }}\parallel }}{{A_0^{(1) }}} \\ \end{gathered} $

结论显然成立。

假设当$k = 1,2,\cdots,n - 1$时猜想均成立。那么当$k = n$时，由式(22) 可以得到

$ \begin{aligned} & \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^n}{\parallel ^2}} \leqslant \\ & {\displaystyle \sum \limits_{k = 1}^{n - 1} \frac{{A_{n - k - 1}^{(n) } - A_{n - k}^{(n) }}}{{A_0^{(n) }}}\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^k}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^k}{\parallel ^2}} } + \\ & \frac{{A_{n - 1}^{(n) }}}{{A_0^{(n) }}}\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} + \frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{n - \theta }}\parallel }}{{A_n^{(n) }}} \leqslant \\ & \frac{{A_0^{(n) } - A_{n - 1}^{(n) }}}{{A_0^{(n) }}}\mathop {{\text{max}}}\limits_{1 \leqslant i \leqslant n - 1} \left\{ \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _i^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} +\right.\\ & \left.\frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{i - \theta }}\parallel }}{{A_{i - 1}^{(i) }}} \right\} + \frac{{A_{n - 1}^{(n) }}}{{A_0^{(n) }}} \mathop {{\text{max}}}\limits_{1 \leqslant i \leqslant n - 1} \\ & \left\{ \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _i^2{{(1 - \theta ) }^2} \parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} +\frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{i - \theta }}\parallel }}{{A_{i - 1}^{(i) }}} \right\} \leqslant \\ & \mathop {{\text{max}}}\limits_{1 \leqslant i \leqslant n - 1} \left\{ \sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{u^0}{\parallel ^2} + \lambda _i^2{{( 1 - \theta ) }^2} \parallel {\delta _x}{\delta _y}{u^0}{\parallel ^2}} +\frac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{f^{i - \theta }}\parallel }}{{A_{i - 1}^{(i) }}} \right\} \end{aligned} $

证毕。

定理 2(收敛性估计)　假设$ u(x,y,t) $是模型(1)~(3)的解，对于差分格式(9)~(11)的解是$\{ u_{ij}^{n - \theta }\mid 0 \leqslant i \leqslant {M_1} - 1,0 \leqslant j \leqslant {M_2} - 1,0 \leqslant n \leqslant N\}$。

定义$ {e_{ij}^{n - \theta } = u({x_i},{y_j},{t_{n - \theta }}) - u_{ij}^{n - \theta }} $，$0 \leqslant i \leqslant {M_1} - 1$，$0 \leqslant j \leqslant {M_2} - 1$，$0 \leqslant n \leqslant N$。那么存在正常数$ C $，使得$ \parallel {e^n}\parallel \leqslant C({\tau ^{2\alpha }} + h_1^4 + h_2^4) $，$1 \leqslant n \leqslant N$成立。

证明　易得下述误差方程

$\begin{array}{l} {\mathcal{H}_{{\rm{CH}}}}D_2^2e_{ij}^{n - \theta } - {\rm{\Lambda }}e_{ij}^{n - \theta } + \lambda _n^2{(1 - \theta )^2}\delta _x^2\delta _y^2{}_{{\rm{CH}}}D_\tau ^ne_{ij}^{n - \theta } = \\ \left\{ \begin{array}{l}{(R_t^{n - \theta })^ * },\quad ({x_i},{y_j}) \in \varOmega ,\quad 1 \le n \le N \\ {e_{ij}^0 = 0,\quad ({x_i},{y_j}) \in \overline \varOmega }\\ {e_{ij}^n = 0,\quad ({x_i},{y_j}) \in \partial \varOmega ,\quad 1 \le n \le N} \end{array} \right.\end{array}$

通过类似于得到(16) 式的证明，可以得到

$ {\begin{split} &\qquad\qquad\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{e^n}{\parallel ^2} + \lambda _n^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{e^n}{\parallel ^2}} \leqslant \\ &{\mathop {\max }\limits_{1 \leqslant i \leqslant n} \left\{ {\sqrt {\parallel {\mathcal{A}_x}{\mathcal{A}_y}{e^0}{\parallel ^2} + \lambda _i^2{{(1 - \theta ) }^2}\parallel {\delta _x}{\delta _y}{e^0}{\parallel ^2}} + \dfrac{{2\parallel {\mathcal{A}_x}{\mathcal{A}_y}{{(R_t^{i - \theta }) }^ * }\parallel }}{{A_{i - 1}^{(i) }}}} \right\}} \end{split}}$

由引理3和引理4可以得到$ {\parallel {e^n}\parallel \leqslant C({\tau ^{2\alpha }} + h_1^4 + h_2^4) } $，$1 \leqslant n \leqslant N$，其中$ C $是一个正常数。

3 数值实验

本节将给出算例验证上述结论，记$E(M,N) = \mathop {{\text{max}}}\limits_{0 \leqslant k \leqslant N} \parallel {e^k}\parallel $，时间方向和空间方向的收敛阶分别为

$ \begin{gathered} \begin{array}{*{20}{r}} {{\text{Rate}}1 = {{\log }_2}\left( {\dfrac{{E(M,N{\text{/}}2) }}{{E(M,N) }}} \right) } \end{array}\quad \\ {\text{Rate}}2 = {\log _2}\left( {\dfrac{{E(M{\text{/}}2,N) }}{{E(M,N) }}} \right) \\ \end{gathered} $

考虑以下模型

$\begin{array}{l} {}_{{\rm{CH}}}D_{a,t}^\alpha u\left( {x,y,t} \right) - \Delta u\left( {x,y,t} \right) = f\left( {x,y,t} \right),\\ (x,y) \in \varOmega = (0,{\text{π}} ) \times (0,{\text{π}} ),1 < t \le 2\\ u(x,y,1) = 0,(x,y) \in \overline \varOmega \\ u\left( {x,y,t} \right) = 0,\left( {x,y} \right) \in \partial \varOmega ,1 < t \le 2 \end{array}$

式中：$\begin{array}{*{20}{r}} {f(x,y,t) = 2\sin (x) \sin (y) {{(\ln t) }^2} + } \end{array}$$ \dfrac{2}{{\Gamma (3 - \alpha ) }}\sin (x) \sin (y) \times {(\ln t) ^{2 - \alpha }}$，模型的精确解$u(x,y,t) = \sin (x) \sin (y) {(\ln t) ^2}$。

表1中列出了当空间网格数分别为${M_1} = 100$，${M_2} = 100$时，不同时间步长$\alpha = 0.4,0.6,0.8$下的误差以及收敛阶。不难看出数值收敛阶验证了紧致ADI格式在时间方向的收敛阶为$ 2\alpha $。表2列出了固定时间网格数N = 1000，当α = 0.9时不同空间步长下的误差和收敛结果，该算例验证了空间收敛精度为4阶。

4 结论

本文研究了二维 Caputo-Hadamard时间分数阶微分方程的有限差分方法。在时间方向上用$ {L_{\ln ,2 - {1_\sigma }}} $公式进行离散达到$2\alpha $阶精度；空间方向上用二阶中心差商公式进行离散，再采用紧算子使得空间达到4阶精度。此外，为了建立紧致ADI差分格式，在原模型的两边加入满足条件的小量项。根据离散系数的性质，利用数学归纳法和能量法得到差分格式的稳定性和收敛性。最后，进行数值实验来验证理论结果。