基于GIS的国外文化地理学研究热点演变与趋势

张涵; 李彦辉; 刘博

doi:10.6054/j.jscnun.2022067

基于GIS的国外文化地理学研究热点演变与趋势

广东金融学院工商管理学院，广州 510521

基金项目:

国家自然科学基金项目 41971190

国家自然科学基金项目 42001162

国家自然科学基金项目 42201264

广东省哲学社会科学规划学科共建项目 GD22XSH10

详细信息

通讯作者:
刘博，Email: 114659591@qq.com

中图分类号: K901.2
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- 文章访问数: 415
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出版历程
- 收稿日期: 2022-09-09
- 网络出版日期: 2022-12-15
- 刊出日期: 2022-10-24

The Evolution and Trend of Foreign Research Hotspots in GIS-based Cultural Geography

School of Business Administration, Guangdong University of Finance, Guangzhou 510521, China

摘要

摘要: 借助文献计量法和科学知识图谱分析法，筛选“Web of Science”核心合集数据库中基于GIS的文化地理研究文献，通过CiteSpace可视化软件绘制发文作者、学科领域和关键词等知识图谱，并进一步追溯核心文献，聚焦研究主题与热点。研究结果表明：(1)该领域研究趋势指向新型地理信息系统，国外研究的高中心性和高突现性文献集中指向批判地理信息系统、历史地理信息系统、质性地理信息系统；(2)核心文献较多采用ArcMAP制图的分层设色模式及核密度分析法来呈现文化地理信息的制图可视化；(3)基于GIS的文化地理研究具有跨学科的优势及良好的研究前景，今后的文化地理学研究在数据获取上可以借助志愿地理信息系统等开放式的地理信息技术平台，在方法上可以探索空间插值等较为精准的空间分析方法。
- 文化地理学 /
- 地理信息系统 /
- 方法论 /
- 认知论 /
- CiteSpace
Abstract: With the bibliometric method and the scientific knowledge graph analysis method, the research literature on GIS-based cultural geography in the Web of Science database is screened, the knowledge graph of authors, disciplines and keywords is drawn through the CiteSpace visualization software, and then the core literature is traced to focus on the hot research topics. The findings are as follows. First, the literature of high centrality and prominence focuses on criticizing GIS, historical GIS and qualitative GIS. Second, the core literature mainly adopts the hierarchical color setting mode of ArcMAP mapping and the kernel density analysis to present the cartographic visualization of cultural geographic information. Third, the GIS-based cultural geography research has interdisciplinary advantages and good research prospects. In the future, cultural geography research can rely on open geographic information technology platforms such as the voluntary geographic information system for data acquisition and try more accurate spatial analysis methods such as spatial interpolation.
- cultural geography /
- GIS /
- methodology /
- epistemology /
- CiteSpace

HTML全文

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 ${u_t} = \Delta u + {\varepsilon ^{-2}}(f\left( u \right) - \varepsilon \lambda (t))$ 在含无流边界条件的有界闭域上的质量守恒性，其中， $\varepsilon \lambda (t)$ 是 $f(u(\cdot, t))$ 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：

$E = {N_V}\int_V {({F_0}(} c) + k{(\nabla c)^2}){\rm{d}}V,$

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，N_V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

$E(\phi ) = \int_\varOmega {\left( {\frac{1}{2}|\nabla \phi {|^2} + F(\phi )} \right)} {\rm{d}}x$

的Cahn-Hilliard方程 ${\phi _t} = - {\Delta ^2}\phi + \Delta {F^\prime }\left( \phi \right)$ 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 ${\phi _t} = - {\Delta ^2}\phi + \Delta {F^\prime }\left( \phi \right)$ 变形为方程组

$\left\{ {\begin{array}{*{20}{l}} {{\phi _t} = \Delta \omega ,}\\ {\omega = - \Delta \phi + {F^\prime }(\phi ),} \end{array}} \right.$

(1)

其中， ${F^\prime }\left( \phi \right) = f\left( \phi \right) = {\phi ^3} - \phi , F\left( \phi \right)$ 是非线性的光滑的位势函数, Ω是 ${{\mathbb{R}}^{2}}$ 的闭集； $\phi \left( {x,t} \right)\left( {x \in \mathit{\Omega } ,t \in \left( {0,T} \right]} \right)$ 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 $\left( {f\left( x \right), g\left( x \right)} \right) = {\smallint _\mathit{\Omega } }f\left( x \right)g(x){\rm{d}}x\;(x \in \mathit{\Omega } \subseteq {\mathbb{R}^2})$ ; $f(x)$ 的的L²范数为 $\left\| f \right\| = \sqrt {\left( {f, f} \right)} ;q\left( \phi \right) = \sqrt {F\left( \phi \right) + B}$ ； $g\left( \phi \right) = 2\frac{{\rm{d}}}{{{\rm{d}}\phi }}q\left( \phi \right) = \frac{{f\left( \phi \right)}}{{\sqrt {F\left( \phi \right) + B} }}$ .

于是，方程组(1)可以写成：

$\left\{ {\begin{array}{*{20}{l}} {{\phi _t} = \Delta \omega ,}\\ {\omega = - \Delta \phi + g(\phi )q(\phi ),}\\ {{q_t} = \frac{1}{2}g(\phi ){\phi _t},} \end{array}} \right.$

(2)

满足初始条件：

$\phi {|_{t = 0}} = {\phi _0},q{|_{t = 0}} = \sqrt {F({\phi _0}) + B} ,$

并满足以下其中一个边界条件：

(i) 在边界 $\partial \mathit{\Omega }$ 上， $\phi$ 和 $\omega$ 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) $F(x) \in C^{2}(-\infty, +\infty)$ ;

$\mathop {\max }\limits_{n \le k} \{ {\left\| {\phi ({t_{n + 1}})} \right\|_{{L^\infty }}},{\left\| {{\phi ^{n + 1}}} \right\|_{{L^\infty }}}\} \le {C_0}.$

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) $F(x) \in {C^3}( - \infty , + \infty )$ ;

$\mathop {\max }\limits_{n \le k} \{ {\left\| {\phi ({t_{n + 1}})} \right\|_{{L^\infty }}},{\left\| {{\phi ^{n + 1}}} \right\|_{{L^\infty }}}\} \le {C_3}.$

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t_n+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 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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| 显示表格

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.

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

下载: 全尺寸图片幻灯片

5. 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.

图 1 基于GIS的文化地理文献发文量年度分布图(1993—2022年)

Figure 1. The annual distribution of literature in the cultural geo-graphy and GIS domain from 1993 to 2022

下载: 全尺寸图片幻灯片

图 2 基于GIS的文化地理文献共被引网络图谱(1993—2022年)

Figure 2. The co-citation network map of literature in the cultural geography and GIS domain from 1993 to 2022

下载: 全尺寸图片幻灯片

图 3 基于GIS的文化地理研究关键词的时间序列图

注：图中圆圈符号越大表示该关键词被引频率越高。

Figure 3. The time series graph of keywords in GIS-based cultural geography research

下载: 全尺寸图片幻灯片

表 1 基于GIS的文化地理研究的关键文献

Table 1 The key node literature of GIS and cultural geography research

关键文献	作者	发文期刊	年份	中心性	共被引频次
《Negotiating knowledge production: the everyday inclusions, exclusions, and contradictions of participatory GIS research》	ELWOOD S	《Professional Geographer》	2006	0.02	4
《Walter Benjamin's dionysian adventures on Google Earth》	KINGSBURY P	《Geoforum》	2009	0.02	2
《Location-based services, conspicuous mo-bility, and the location-aware future》	WILSON M W	《Geoforum》	2012	0.01	3
《Feminist geographies of new spatial me-dia》	LESZCZYNSKI A	《The Canadian Geographer》	2015	0.01	2
《Affecting geospatial technologies: toward a feminist politics of emotion》	KWAN M P	《Professional Geographer》	2007	0.01	3
《New spatial media, new knowledge politics》	ELWOOD S	《Transactions of the Institute of British Geographers》	2012	0.01	2

下载: 导出CSV

表 2 基于GIS的文化地理研究的期刊发文情况

Table 2 The journal publication of GIS and cultural geography research

期刊名	发文量	影响因子(2020年)	Journal Citation Report(JCR) 分区(2020年)
《Applied Geography》	6	4.24	Q1
《Isprs International Journal of Geo Information》	4	2.90	Q2
《Journal of Geography in Higher Education》	4	2.46	Q2
《Professional Geographer》	4	2.38	Q3
《Annals of the Association of American Geographers》	3	4.68	Q1
《Journal of Archaeological Science》	3	3.22	Q1
《Transactions in GIS》	3	2.41	Q3
《Journal of Geography》	3	2.40	Q3
《Geographical Review》	3	1.58	Q4
《Progress in Human Geography》	2	10.22	Q1
《Land Use Policy》	2	5.40	Q1
《Journal of Historical Geography》	2	1.39	Q1
《Cultural Geographies》	2	3.55	Q3

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表 3 基于GIS的文化地理研究前15名突现关键词(按突现起始年排序)

Table 3 The top 15 prominent keywords in GIS-based cultural geography (in order of the beginning year of prominencne)

关键词	突现度	起始年份	终止年份
ecology	1.23	1993	2004
cultural geography	1.21	2005	2016
map	1.79	2008	2013
space	1.69	2008	2013
policy	1.61	2008	2013
knowledge	1.54	2008	2010
city	1.53	2008	2013
world	1.49	2008	2013
landscape	1.14	2008	2013
critical GI	1.32	2014	2019
geography	2.55	2017	2022
place	1.72	2017	2022
science	1.7	2017	2019
land use	1.24	2017	2022
historical geography	1.22	2020	2022

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1. 能量不变二次化法
2. 梯度流的稳定性
3. 误差估计
4. 数值例子
5. 结束语

基于GIS的国外文化地理学研究热点演变与趋势

通讯作者: 刘博，Email: 114659591@qq.com

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出版历程

The Evolution and Trend of Foreign Research Hotspots in GIS-based Cultural Geography

1. 能量不变二次化法

2. 梯度流的稳定性

3. 误差估计

4. 数值例子

5. 结束语

计量

出版历程

目录

1. 能量不变二次化法

2. 梯度流的稳定性

3. 误差估计

4. 数值例子

5. 结束语

通讯作者:
刘博，Email: 114659591@qq.com