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\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation*}
\partial_t u(t,x)=\left\{
\begin{array}{l l}
\displaystyle{\left(\partial_t \xi^t(x)R'(\xi^t(x))-\partial_x G(t,x)R(\xi^t(x)\right)\exp\left(-\int_{\xi^t(x)}^t\partial_x G(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right)},&(t,x)\in \Omega_1\\
\displaystyle{\left(\partial_t\chi^t(x)u'_0(\chi^t(x))-u_0(\chi^t(x)\partial_x G(t,x)\right)\exp\left(-\int_{t_0}^t\partial_xG(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right)},&(t,x)\in \Omega_2\\
0,&(t,x)\in\Omega_3
\end{array}
\right.
\end{equation*}
\begin{equation*}
\partial_x(Gu)(t,x)=\left\{
\begin{array}{l l}
\displaystyle{\left(\partial_xG(t,x)R(\xi^t(x))+ G(t,x)\partial_x\xi^t(x)R'(\xi^t(x))\right)\exp\left(-\int_{\xi^t(x)}^t\partial_x G(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right)},&(t,x)\in \Omega_1\\
\displaystyle{\left(\partial_xG(t,x)u_0(\chi^t(x))+G(t,x)\partial_x\chi^t(x)u'_0(\chi^t(x))\right)\exp\left(-\int_{t_0}^t\partial_xG(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right)},&(t,x)\in \Omega_2\\
0,&(t,x)\in\Omega_3
\end{array}
\right.
\end{equation*}
\end{document}
答案1
由于每行都有一个很长的常用表达式,我建议您给它起一个名字:
以下是代码(使用dcases
):
\documentclass{article}
\usepackage{mathtools} % for \shortintertext and \dcases
\begin{document}
For clarity, let us define
\[
E(a)=\exp\biggl(-\int_a^t\partial_x G(s,\Phi_{(t,x)}(s))\,\mathrm{d}s\biggr).
\]
Then we have
\begin{align*}
\partial_t &u(t,x)=\\
&\begin{dcases}
\bigl(\partial_t \xi^t(x)R'(\xi^t(x))-\partial_x G(t,x)R(\xi^t(x)\bigr)E\bigl(\xi^t(x)\bigr),&(t,x)\in \Omega_1\\
\bigl(\partial_t\chi^t(x)u'_0(\chi^t(x))-u_0(\chi^t(x)\partial_x G(t,x)\bigr)E(t_0),&(t,x)\in \Omega_2\\
0,&(t,x)\in\Omega_3
\end{dcases}\\
\shortintertext{and}
\partial_x&(Gu)(t,x)=\\
&\begin{dcases}
\bigl(\partial_xG(t,x)R(\xi^t(x))+ G(t,x)\partial_x\xi^t(x)R'(\xi^t(x))\bigr)E\bigl(\xi^t(x)\bigr),&(t,x)\in \Omega_1\\
\bigl(\partial_xG(t,x)u_0(\chi^t(x))+G(t,x)\partial_x\chi^t(x)u'_0(\chi^t(x))\bigr)E(t_0),&(t,x)\in \Omega_2\\
0,&(t,x)\in\Omega_3
\end{dcases}
\end{align*}
\end{document}
答案2
基本上没有机会将这些公式放入标准文本宽度中。
你可以拆分大方程式,我展示了两种不同的对齐方法。
\documentclass{article}
\usepackage{amsmath,mathtools}
\newcommand{\diff}{\mathop{}\!\mathrm{d}}
\begin{document}
\begin{equation*}
\partial_t u(t,x)=
\begin{dcases}
\begin{aligned}[b]
&\bigl(\partial_t \xi^t(x)R'(\xi^t(x))-\partial_x G(t,x)R(\xi^t(x)\bigr) \\
&\qquad\cdot
\exp\biggl(-\int_{\xi^t(x)}^t\partial_x G(s,\Phi_{(t,x)}(s))\diff s\biggr),
\end{aligned}
&(t,x)\in \Omega_1\\[1.5ex]
\begin{aligned}[b]
&\bigl(\partial_t\chi^t(x)u'_0(\chi^t(x))-u_0(\chi^t(x)\partial_x G(t,x)\bigr) \\
&\qquad\cdot
\exp\biggl(-\int_{t_0}^t\partial_xG(s,\Phi_{(t,x)}(s))\diff s\biggr),
\end{aligned}
&(t,x)\in \Omega_2\\[1.5ex]
0,&(t,x)\in\Omega_3
\end{dcases}
\end{equation*}
\begin{equation*}
\partial_t u(t,x)=
\begin{dcases}
\begin{aligned}[b]
\bigl(\partial_t \xi^t(x)R'(\xi^t(x))-\partial_x G(t,x)R(\xi^t(x)\bigr) \\
\cdot
\exp\biggl(-\int_{\xi^t(x)}^t\partial_x G(s,\Phi_{(t,x)}(s))\diff s\biggr)&,
\end{aligned}
&(t,x)\in \Omega_1\\[1.5ex]
\begin{aligned}[b]
\bigl(\partial_t\chi^t(x)u'_0(\chi^t(x))-u_0(\chi^t(x)\partial_x G(t,x)\bigr) \\
\cdot
\exp\biggl(-\int_{t_0}^t\partial_xG(s,\Phi_{(t,x)}(s))\diff s\biggr)&,
\end{aligned}
&(t,x)\in \Omega_2\\[1.5ex]
0,&(t,x)\in\Omega_3
\end{dcases}
\end{equation*}
\end{document}
请注意dcases
避免使用array
和进行笨拙的构造\displaystyle
。
也不\mbox{d}
是获取(直立)差分符号的最佳方法。使用\mathrm{d}
,但我还建议使用命令的更好方法\diff
。
答案3
一种可能性……
\documentclass{article}
\usepackage{amsmath}
\usepackage{tabstackengine}
\TABstackMath
\TABstackMathstyle{\displaystyle}
\renewcommand\stacktype{L}
\renewcommand\stackalignment{r}
\setstackgap{L}{24pt}
\begin{document}
\begin{equation*}
\partial_t u(t,x)=\left\{
\begin{array}{l l}
\tabbedstackanchor{\bigl(\partial_t \xi^t(x)R'(\xi^t(x))-\partial_x
G(t,x)R(\xi^t(x)\bigr)\times}{\exp\left(-\int_{\xi^t(x)}^t\partial_x
G(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right)\quad},&(t,x)\in \Omega_1\\[30pt]
\tabbedstackanchor{\bigl(\partial_t\chi^t(x)u'_0(\chi^t(x))-u_0(\chi^t(x)
\partial_x G(t,x)\bigr)\times}{\exp\left(-\int_{t_0}^t\partial_xG(s,
\Phi_{(t,x)}(s))\,\mbox{d}s\right)\quad },&(t,x)\in \Omega_2\\[24pt]
0,&(t,x)\in\Omega_3
\end{array}
\right.
\end{equation*}
\begin{equation*}
\partial_x(Gu)(t,x)=\left\{
\begin{array}{l l}
\tabbedstackanchor{\bigl(\partial_xG(t,x)R(\xi^t(x))+G(t,x)\partial_x
\xi^t(x)R'(\xi^t(x))\bigr)\times}{\exp\left(-\int_{\xi^t(x)}^t\partial_x
G(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right)\quad },&(t,x)\in \Omega_1\\[30pt]
\tabbedstackanchor{\bigl(\partial_xG(t,x)u_0(\chi^t(x))+G(t,x)\partial_x
\chi^t(x)u'_0(\chi^t(x))\bigr)\times}{\exp\left(-\int_{t_0}^t\partial_xG
(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right)\quad },&(t,x)\in \Omega_2\\[24pt]
0,&(t,x)\in\Omega_3
\end{array}
\right.
\end{equation*}
\end{document}
或者,正如 daleif 所建议的,
\documentclass{article}
\usepackage{amsmath}
\usepackage{tabstackengine}
\TABstackMath
\TABstackMathstyle{\displaystyle}
\renewcommand\stacktype{L}
\renewcommand\stackalignment{l}
\setstackgap{L}{24pt}
\begin{document}
\begin{equation*}
\partial_t u(t,x)=\left\{
\begin{array}{l l}
\tabbedstackanchor{\bigl(\partial_t \xi^t(x)R'(\xi^t(x))-\partial_x
G(t,x)R(\xi^t(x)\bigr)}{\quad\times\exp\left(-\int_{\xi^t(x)}^t\partial_x
G(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right)},&(t,x)\in \Omega_1\\[30pt]
\tabbedstackanchor{\bigl(\partial_t\chi^t(x)u'_0(\chi^t(x))-u_0(\chi^t(x)
\partial_x G(t,x)\bigr)}{\quad\times\exp\left(-\int_{t_0}^t\partial_xG(s,
\Phi_{(t,x)}(s))\,\mbox{d}s\right) },&(t,x)\in \Omega_2\\[24pt]
0,&(t,x)\in\Omega_3
\end{array}
\right.
\end{equation*}
\begin{equation*}
\partial_x(Gu)(t,x)=\left\{
\begin{array}{l l}
\tabbedstackanchor{\bigl(\partial_xG(t,x)R(\xi^t(x))+G(t,x)\partial_x
\xi^t(x)R'(\xi^t(x))\bigr)}{\qquad\times\exp\left(-\int_{\xi^t(x)}^t\partial_x
G(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right) },&(t,x)\in \Omega_1\\[30pt]
\tabbedstackanchor{\bigl(\partial_xG(t,x)u_0(\chi^t(x))+G(t,x)\partial_x
\chi^t(x)u'_0(\chi^t(x))\bigr)}{\qquad\times\exp\left(-\int_{t_0}^t\partial_xG
(s,\Phi_{(t,x)}(s))\,\mbox{d}s\right) },&(t,x)\in \Omega_2\\[24pt]
0,&(t,x)\in\Omega_3
\end{array}
\right.
\end{equation*}
\end{document}