等式中的 \downarrow

Question

你能建议我一种像图中那样以向下的箭头写出方程式的方法吗？

我不明白它该如何设置。

我看到了这篇文章：

在等式下方设置箭头

如何在方程式中编码长向下箭头

---------------------------------------------更新--------------------------------------

我尝试了这个解决方案，但我并不满意，我希望方程式以相应的数字为中心。

我插入了两个解决方案，

首先（我想你会发现很多错误）：

\begin{equation*}
\begin{aligned}
\Gradu = \textit{sym}(\Gradu) + \textit{skew}(\Gradu) = \Fra{1}{2}(\Gradu + \Grad{\bar{u}})+\Fra{1}{2}(\Gradu - \Grad{\bar{u}})\\
\hspace{-10.5cm}
\Gradu = \barbII{e}+\barbII{r} \quad \text{dove :}
\begin{cases}
    \barbII{e} = \text{Velocità di Deformazione;}\\
    \barbII{r} =  \text{Velocità di Rotazione Rigida};\\
\end{cases}\\
\barbII{e} = \overbrace{\Fra{1}{3}tr(\barbII{e})\I}^{1}+\underbrace{\left(\barbII{e} -\Fra{1}{3}tr(\barbII{e})\I\right)}_{2} %\hspace{4.7cm}
\begin{cases}
1) = \text{Parte Sferica : velocità di variazione volumentrica;}\\
2) =  \text{Velocità di deformazione a volume costante};\\
\end{cases}\\
\end{aligned}
\end{equation*}

第二对应于图像的解的一部分；

\begin{gather*}
\begin{align}
\Gradu\tikzmark{enda} = \barbII{e}\tikzmark{endb} + \barbII{r}\tikzmark{endc}
\end{align}
\end{gather*}

\begin{tikzpicture}[remember picture,overlay]
\draw[->,>=latex]
([shift={(-15pt,0pt)}]pic cs:enda) |- ++(70pt,-15pt) node[right,text width=6cm] {Tensore Gradiente di Velocità Trasposto};
\draw[->,>=latex]
([shift={(-2pt,-6pt)}]pic cs:endb) |- ++(40pt,-30pt) node[right,text width=6cm] {Velocità di Deformazione};
\draw[->,>=latex]
([shift={(-2pt,-6pt)}]pic cs:endc) |- ++(25pt,-50pt) node[right,text width=6cm] {Velocità di Rotazione Rigida};
\draw[->,>=latex]
([shift={(-2pt,-6pt)}]pic cs:endb) |- ++(40pt,-80pt) node[right,text width=6cm] {
    $\barbII{e} = \Fra{1}{3}tr(\barbII{e})\I+\left(\barbII{e} -\Fra{1}{3}tr(\barbII{e})\I\right)$
};
\end{tikzpicture}    
\vspace{3cm}

\begin{equation}
\begin{aligned}
\barbII{e} = \Fra{1}{3}tr(\barbII{e})\I+\left(\barbII{e} -\Fra{1}{3}tr(\barbII{e})\I\right)
\end{aligned}
\end{equation}

但如果我添加另一个节点来解释$\barbII{e} = \Fra{1}{3}tr(\barbII{e})\I+\left(\barbII{e} -\Fra{1}{3}tr(\barbII{e})\I\right)$，它会出现错误：

\draw[->,>=latex]
    ([shift={(-2pt,-6pt)}]pic cs:endb) |- ++(40pt,-80pt) node[right,text width=6cm] {
        $\barbII{e} = \Fra{1}{3}tr(\barbII{e})\tikzmark{ende}\I+\left(\barbII{e} -\Fra{1}{3}tr(\barbII{e})\I\right)$
    };

---------------------------------------------更新 1----------------------------------

我选择了这个最终解决方案

\begin{align*}
\Gradu&=\tikzmarknode{e1}{\barbII{e}}+\barbII{r}\quad
\text{dove}~\left\{\begin{array}{rcl}
\barbII{e}&=& \text{Velocità di Deformazione;}\\
\barbII{r}&=& \text{Velocità di Rotazione Rigida};\\
\end{array}\right.\\
&\qquad\quad \tikzmarknode{e2}{\barbII{e}}=
\overbrace{ \Fra{1}{3}tr(\barbII{\tau})\I}^{\text{\ding{192}}}+
\underbrace{\left(\barbII{e} -\Fra{1}{3}tr(\barbII{\tau})\I\right)}_{\text{\ding{193}}}\tikzmark{X}\quad
\begin{array}{rcl} 
\end{array}
\end{align*}
\begin{tikzpicture}[overlay,remember picture]
\draw[thick,-latex] (e1) |- (e2);
\end{tikzpicture}
Il tensore $\barbII{e}$ rappresenta pertanto la velocità di deformazione in forma (la distorsione) ed in volume e può essere decomposto in due parti:

%%-----------------新指挥部----------------------------

\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
        \node[shape=circle,draw,inner sep=0.5pt] (char) {#1};}}

\begin{itemize}
    \item[\circled{1}]Parte Sferica : Velocità di Variazione Volumentrica;
    \item[\circled{2}]Parte Deviatoria : Velocità di Deformazione a Volume Costante;
\end{itemize}

建立一个箭距从\barbII{e}

---------------------------------------------更新2----------------------------------

另一个有趣的解决方案

\begin{equation}\label{eq063}
\begin{aligned}
    \tau_{ij}=-\textbf{p}\delta_{ij}+2\mu\left( \eij-\Fra{1}{3}e_{kk}\delta_{ij}\right) 
    \text{dove $\delta_{ij}$ = }~\left\{\begin{array}{rcl}
    1&\rightarrow& \text{se i = j;}\\
    0&\rightarrow& \text{se i $\neq$ j};\\
    \end{array}\right.\\
\end{aligned}
\end{equation}

\begin{tikzpicture}[right node/.style={rectangle,draw}]
 node{%
        $\begin{aligned}
            \tikzmarknode{ex1}e_{kk}\tikzmarknode{ex2}=tr(\barbII{e})=\tikzmarknode{ex3}\Div{\bar{u}}
        \end{aligned}$};
    \draw[->,>=latex]([shift={(-70pt,-5pt)}]pic cs:ex1-ex2) |- ++(80pt,-25pt) node[right,text width=6cm] 
    { È considerata la somma degli elementi della diagonale di $\barbII{e}$ che non coinvolge gli ij;};
    \draw[->,>=latex]([shift={(-8pt,-15pt)}]pic cs:ex3) |- ++(16pt,-52pt) node[right,text width=6cm] 
    {Velocità di Deformazione Volumetrica (Comprimibiltà);};
\end{tikzpicture}

Answer 1

由于 Sebastiano 向我发送了 ping 消息，我尝试写一个答案。我专注于箭头，不幸的是没有您的宏，所以我编造了一些东西，但您当然更喜欢使用自己的宏\barbII。此答案需要最新版本的tikzmark，它是最新标准 TeX 发行版的一部分。您可以使用它来制作公式的某些元素节点，然后您可以在覆盖中连接它们tikzpicture。

\documentclass[fleqn]{article}
\usepackage{mathtools}
\usepackage{pifont}
\usepackage{tikz}
\usetikzlibrary{tikzmark}
\newcommand{\barbII}[1]{\underset{=}{#1}}
\begin{document}
\begin{align*}
 \nabla \bar\mu^T&=\dots \\
 \nabla \bar\mu^T&=\tikzmarknode{e1}{\barbII{e}}+\barbII{\ddot r}\quad
 \text{dove}~\left\{\begin{array}{rcl}
 \barbII{e}&=& \text{\dots}\\
 \barbII{\ddot r}&=& \text{\dots}\\
 \end{array}\right.\\
 &\qquad\quad \barbII{\tikzmarknode{e2}{e}}=
 \overbrace{\dots}^{\text{\ding{192}}}+
 \underbrace{\dots}_{\text{\ding{193}}}\tikzmark{X}\quad
 \begin{array}{rcl} \tikzmarknode{i1}{\text{\ding{192}}} & & \\
 \tikzmarknode{i2}{\text{\ding{193}}} & & \\
 \end{array}
\end{align*}
\begin{tikzpicture}[overlay,remember picture]
 \draw[thick,-latex] (e1) |- (e2);
 \draw (i1) -- ([yshift=5pt]pic cs:X) -- (i2);
\end{tikzpicture}
\end{document}