如何垂直和水平移动数学符号

Question

如何将双箭头水平移动到右侧？

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{tabular}{p{5cm}p{5cm}}
{\begin{align}
\underset{u(t), \eta(t), r}{\text{min}}&\quad\int_{0}^{t_f}y^Ty dt\nonumber\\
\text{s.t:}&\quad \int_{0}^{t_f} u^Tu\leq f_{1,k}\nonumber\\
&\quad\eta(0)=(-1,-1,0,0,0,0)\nonumber\\ 
&\quad\frac{d\eta}{dt}=A\eta+Bu\nonumber\\
&\quad t_f = 10\nonumber\\  
\raisebox{29ex}[10pt][10ex]{$\boldsymbol{\Longrightarrow}$}\nonumber
\end{align}}
&
{\begin{align}
\underset{u(t), \eta(t), r}{\text{min}}&\quad z_2(t_f)\nonumber\\
\text{s.t:}&\quad z_1(t_f)-\phi\leq0\nonumber\\
&\quad z(0)=0\nonumber\\
&\quad\frac{d\eta}{dt}=A\eta+Bu\nonumber\\
&\quad t_f = 10\nonumber
\end{align}}
\end{tabular}

\end{document}

Answer 1

您的原始显示位于顶部，而我建议的更改位于底部：

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{tabular}{p{5cm}p{5cm}}
{\begin{align}
\underset{u(t), \eta(t), r}{\text{min}}&\quad\int_{0}^{t_f}y^Ty dt\nonumber\\
\text{s.t:}&\quad \int_{0}^{t_f} u^Tu\leq f_{1,k}\nonumber\\
&\quad\eta(0)=(-1,-1,0,0,0,0)\nonumber\\ 
&\quad\frac{d\eta}{dt}=A\eta+Bu\nonumber\\
&\quad t_f = 10\nonumber\\  
\raisebox{29ex}[10pt][10ex]{$\boldsymbol{\Longrightarrow}$}\nonumber
\end{align}}
&
{\begin{align}
\underset{u(t), \eta(t), r}{\text{min}}&\quad z_2(t_f)\nonumber\\
\text{s.t:}&\quad z_1(t_f)-\phi\leq0\nonumber\\
&\quad z(0)=0\nonumber\\
&\quad\frac{d\eta}{dt}=A\eta+Bu\nonumber\\
&\quad t_f = 10\nonumber
\end{align}}
\end{tabular}


\[
  \begin{aligned}
    \min_{u(t), \eta(t), r} \quad & \int_{0}^{t_f}y^Ty \mathrm{d}t \\
    \text{s.t:} \quad & \int_{0}^{t_f} u^Tu \leq f_{1,k} \\
    & \eta(0) = (-1,-1,0,0,0,0) \\ 
    & \frac{\mathrm{d}\eta}{\mathrm{d}t} = A\eta + Bu \\
    & t_f = 10  
  \end{aligned} \qquad \Longrightarrow \qquad
  \begin{aligned}
    \min_{u(t), \eta(t), r} \quad & z_2(t_f) \\
    \text{s.t:} \quad & z_1(t_f)-\phi \leq 0 \\
    & z(0) = 0\\
    & \frac{\mathrm{d}\eta}{\mathrm{d}t} = A\eta + Bu \\
    & t_f = 10
  \end{aligned}
\]

\end{document}

注意运算符的使用\min（不需要\underset）。

aligned自动垂直居中其内容。如果你想让它们t与 vertically-centered 对齐\Longrightarrow，那么你可以使用类似

\begin{aligned}[t]
  ...
\end{aligned}
\qquad \raisebox{-3\baslelineskip}{$\Longrightarrow$} \qquad
\begin{aligned}[t]
  ...
\end{aligned}

Answer 2

这是一个使用两个minipage环境来容纳环境的解决方案。请注意右侧align*使用s来确保方程的行在两组方程中对齐。\vphantomminipage

\documentclass{article}
\usepackage{amsmath}
\newcommand\tran{{}^T\!} % well-spaced transpose op.
\begin{document}

\begin{minipage}{5cm}
\begin{align*}
\min_{u(t), \eta(t), r} \quad 
&\int_{0}^{t_f}y\tran y\, dt\\
\text{s.t: }\quad 
&\int_{0}^{t_f} u\tran u\leq f_{1,k}\\
&\eta(0)=(-1,-1,0,0,0,0)\\ 
&\frac{d\eta}{dt}=A\eta+Bu\\
& t_f = 10
\end{align*}
\end{minipage}
\qquad$\Longrightarrow$
\begin{minipage}{5cm}
\begin{align*}
\min_{u(t), \eta(t), r}\quad 
&z_2(t_f) \vphantom{\int_{0}^{t_f}}\\
\text{s.t: }\quad 
&z_1(t_f)-\phi\leq0 \vphantom{\int_{0}^{t_f}}\\
&z(0)=0\\
&\frac{d\eta}{dt}=A\eta+Bu\\
&t_f = 10
\end{align*}
\end{minipage}

\end{document}

Answer 3

也许这就是你的意思，使用更简单的代码：

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools, nccmath}

\begin{document}

\begin{align*}
  & \begin{aligned}[t]
  &\!\min_{\mathclap {u(t), \eta(t), r}}\quad\int_{0}^{t_f}y^Ty dt\\
  & \text{\footnotesize s. t. \enspace}\mathrlap{\medmath{\begin{array}[t]{|l}
  \int_{0}^{t_f} u^Tu\leq f_{1,k}\\
  \eta(0)=(-1,-1,0,0,0,0)\\
  \frac{d\eta}{dt}=A\eta+Bu\\
  t_f = 10
  \end{array}}}
  \end{aligned}
    & & \boldsymbol{\Longrightarrow}%
    & &
  \begin{aligned}[t]
  &\!\min_{\mathclap {u(t), \eta(t), r}} \quad z_2(t_f)\\
  &\text{\footnotesize s. t.\enspace}\medmath{\begin{array}[t]{|l}
  z_1(t_f)-\phi\leq0\\
  z(0)=0\\
  \frac{d\eta}{dt}=A\eta+Bu\\
  t_f = 10
  \end{array}}
  \end{aligned}
\end{align*}

\end{document} 

Answer 4

好吧，还有一个可能的解决方案：

\documentclass[border=3mm,
               preview]{standalone}%preview enable showing math
\usepackage{mathtools} % loads amsmath too

\begin{document}
\begin{align*}
\min_{u(t),\eta(t),r}
    &\quad  \int_{0}^{t_f} y^Ty dt 
        &\Longrightarrow    &&
\min_{u(t),\eta(t),r}
    &\quad   z_2(t_f)                           \\
%        
\text{s.t:}
    &\quad  \int_{0}^{t_f} u^Tu \leq f_{1,k}    
        &&&
\text{s.t:}
    &\quad   z_1(t_f)-\phi\leq                  \\
%
    &\quad  \eta(0) = (-1,-1,0,0,0,0)           
        &&&
    &\quad   z(0)=0                             \\
%
    &\quad  \frac{d\eta}{dt}=A\eta+Bu           
        &&&
    &\quad   \frac{d\eta}{dt}=A\eta+Bu          \\
%
    &\quad  t_f = 10                            
        &&&
    &\quad   t_f = 10
    \end{align*}
\end{document}

它给：

我假设，Longarrow属于第一条方程线...使用一个对齐环境，方程线垂直对齐。