我这里有一个可行的示例。我想一次显示一个 tikzpicture 和一个方程式。我不知道如何在这里只使用覆盖。
\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{shapes, shadows, arrows}
\tikzset{
basic/.style = {draw, text width=2cm, drop shadow, font=\sffamily, rectangle},
root/.style = {basic, rounded corners=2pt, thin, align=center,
fill=green!30},
level 2/.style = {basic, rounded corners=6pt, thin,align=center, fill=green!60,
text width=8em},
level 3/.style = {basic, thin, align=left, fill=pink!60, text width=6.5em}
}
\mode<presentation>
\begin{document}
\begin{frame}
\begin{tikzpicture}
level 1/.style={sibling distance=40mm},
edge from parent/.style={->,draw},
>=latex]
% root of the the initial tree, level 1
\node[root](start) {Mass Balance};
\draw[->,very thick](0,0) (1.2,0)--(1.9,0);
\draw[->,very thick](-1.8,0)--(-1.1,0);
\draw[->,very thick](0,0.6)--(0,1.4);
\draw[->,very thick](0,-1.4)--(0,-0.6);
\node[right of =start,xshift=2.5cm](spatialpos){$q_c(t,x+\Delta x,\tau)$};
\node[left of =start,xshift=-1.7cm](spatialneg){$q_c(t,x,\tau)$};
\node[above of =start,yshift=0.8cm](agepos){$c(t,x,\tau+\Delta \tau)$};
\node[below of =start,yshift=-0.8cm](ageneg){$c(t,x,\tau)$};
\draw[->,very thick](-4,-3)--(-4,0);
\draw[->,very thick](-4,-3)--(-0.5,-3);
\node[below of = ageneg,yshift=0.2cm](spatial){Spatial dim};
\node[left of = spatialneg,yshift=0.4cm](age){Age dim.};
\end{tikzpicture}
\begin{align*}
\rho \Delta x \Delta y \Delta z \Delta \tau \partial_t c_i(t,x,\tau) \\
&= \rho \Delta x \Delta y \Delta z \Delta \tau (p_i-d_i) \\
&- \rho \Delta y, \Delta z \Delta \tau [q_{i,x}(t,x+\Delta x/2, y, z, \tau)\\
&\qquad - q_{i,x}(t,x - \Delta x/2, y, z, \tau)]\\
&- \rho \Delta x, \Delta z \Delta \tau [q_{i,y}(t,x,y+\Delta y/2, y, z, \tau)\\
&\qquad - q_{i,y}(t,x,y - \Delta y/2, z, z, \tau)]\\
&- \rho \Delta x \Delta y \Delta \tau[q_{i,z}(t,x,y,z+\Delta z/2, \tau) \\
&\qquad - q_{i,z}(t,x,y,z-\Delta z/2, \tau)]\\
&- \rho \Delta x \Delta y \Delta \tau[q_{i,z}(t,x,y,z+\Delta z/2, \tau) \\
&\qquad - q_{i,z}(t,x,y,z-\Delta z/2, \tau)] \\
\end{align*}
\end{frame}
\end{document}
答案1
语法\only是
\only<overlay specification>{text}
或者
\only{text}<overlay specification>
tikzpicture并且您几乎可以在 os 位置使用任何东西（特别是或显示的方程式） text。
\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{shapes, shadows, arrows}
\tikzset{
basic/.style = {draw, text width=2cm, drop shadow, font=\sffamily, rectangle},
root/.style = {basic, rounded corners=2pt, thin, align=center,
fill=green!30},
level 2/.style = {basic, rounded corners=6pt, thin,align=center, fill=green!60,
text width=8em},
level 3/.style = {basic, thin, align=left, fill=pink!60, text width=6.5em}
}
\mode<presentation>
\begin{document}
\begin{frame}
\only<1>{\begin{tikzpicture}
level 1/.style={sibling distance=40mm},
edge from parent/.style={->,draw},
>=latex]
% root of the the initial tree, level 1
\node[root](start) {Mass Balance};
\draw[->,very thick](0,0) (1.2,0)--(1.9,0);
\draw[->,very thick](-1.8,0)--(-1.1,0);
\draw[->,very thick](0,0.6)--(0,1.4);
\draw[->,very thick](0,-1.4)--(0,-0.6);
\node[right of =start,xshift=2.5cm](spatialpos){$q_c(t,x+\Delta x,\tau)$};
\node[left of =start,xshift=-1.7cm](spatialneg){$q_c(t,x,\tau)$};
\node[above of =start,yshift=0.8cm](agepos){$c(t,x,\tau+\Delta \tau)$};
\node[below of =start,yshift=-0.8cm](ageneg){$c(t,x,\tau)$};
\draw[->,very thick](-4,-3)--(-4,0);
\draw[->,very thick](-4,-3)--(-0.5,-3);
\node[below of = ageneg,yshift=0.2cm](spatial){Spatial dim};
\node[left of = spatialneg,yshift=0.4cm](age){Age dim.};
\end{tikzpicture}}
\only<2>{\begin{align*}
\rho \Delta x \Delta y \Delta z \Delta \tau \partial_t c_i(t,x,\tau) \\
&= \rho \Delta x \Delta y \Delta z \Delta \tau (p_i-d_i) \\
&- \rho \Delta y, \Delta z \Delta \tau [q_{i,x}(t,x+\Delta x/2, y, z, \tau)\\
&\qquad - q_{i,x}(t,x - \Delta x/2, y, z, \tau)]\\
&- \rho \Delta x, \Delta z \Delta \tau [q_{i,y}(t,x,y+\Delta y/2, y, z, \tau)\\
&\qquad - q_{i,y}(t,x,y - \Delta y/2, z, z, \tau)]\\
&- \rho \Delta x \Delta y \Delta \tau[q_{i,z}(t,x,y,z+\Delta z/2, \tau) \\
&\qquad - q_{i,z}(t,x,y,z-\Delta z/2, \tau)]\\
&- \rho \Delta x \Delta y \Delta \tau[q_{i,z}(t,x,y,z+\Delta z/2, \tau) \\
&\qquad - q_{i,z}(t,x,y,z-\Delta z/2, \tau)] \\
\end{align*}}
\end{frame}
\end{document}
