
Quero digitar uma equação matemática assim:
Mas com este código não consigo obter o resultado acima. Alguém pode me ajudar a editar meu código para obter o resultado acima?
\documentclass{book}
\usepackage{amsmath}
\begin{document}
\begin{align}
u_j^{n+1}&=u_j^{n}-a(t_n)\Delta t \dfrac{\partial u}{\partial x}
+\dfrac{(\Delta t)^2}{2}\left(a(t_n)^2\dfrac{\partial^2 u}{\partial x^2}-\dfrac{da(t_n)}{dt}\dfrac{\partial u}{\partial x}\right)\nonumber\\
&=u_j^{n}-a(t_n)\Delta t \dfrac{\partial u}{\partial x}
+a(t_n)^2\dfrac{(\Delta t)^2}{2}\dfrac{\partial^2 u}{\partial x^2}-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{2}\dfrac{\partial u}{\partial x}\nonumber\\
&=u_j^{n}-a(t_n)\Delta t \left(\dfrac{u_{j+1}^{n}-u_{j-1}^{n}}{2\Delta x}\right)+a(t_n)^2\dfrac{(\Delta t)^2}{2}\left(\dfrac{u_{j+1}^{n}-2u_{j}^{n}+u_{j-1}^{n}}{(\Delta x)^2}\right)\nonumber\\
%%%%%%%%%%%%%%%%%%%%%%
&-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{2}\left(\dfrac{u_{j+1}^{n}-u_{j-1}^{n}}{2\Delta x}\right)\nonumber\\
\begin{aligned}
&= u_j^{n}-\dfrac{1}{2}\nu_n\left(u_{j+1}^{n}-u_{j-1}^{n}\right)
+\dfrac{1}{2}\nu_n^2 \left(u_{j+1}^{n}-2u_{j}^{n}+u_{j-1}^{n}\right)
-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{4\Delta x}\left(u_{j+1}^{n}-u_{j-1}^{n}\right)
\end{aligned}
\end{align}
\end{document}
Responder1
Para o número da equação centralizado, use equation
e aligned
, em vez de align
. Para os recuos, inseri \qquad
onde necessário. Também tive que quebrar uma de suas longas filas para abrir espaço para o número eqn centralizado.
\documentclass{book}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{aligned}
u_j^{n+1}&=u_j^{n}-a(t_n)\Delta t \dfrac{\partial u}{\partial x}
+\dfrac{(\Delta t)^2}{2}\left(a(t_n)^2\dfrac{\partial^2 u}{\partial x^2}-\dfrac{da(t_n)}{dt}\dfrac{\partial u}{\partial x}\right)\\
&=u_j^{n}-a(t_n)\Delta t \dfrac{\partial u}{\partial x}
+a(t_n)^2\dfrac{(\Delta t)^2}{2}\dfrac{\partial^2 u}{\partial x^2}-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{2}\dfrac{\partial u}{\partial x}\\
&=u_j^{n}-a(t_n)\Delta t \left(\dfrac{u_{j+1}^{n}-u_{j-1}^{n}}{2\Delta x}\right)\\
&\qquad+a(t_n)^2\dfrac{(\Delta t)^2}{2}\left(\dfrac{u_{j+1}^{n}-2u_{j}^{n}+u_{j-1}^{n}}{(\Delta x)^2}\right)\\
%%%%%%%%%%%%%%%%%%%%%%
&\qquad-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{2}\left(\dfrac{u_{j+1}^{n}-u_{j-1}^{n}}{2\Delta x}\right)\\
&= u_j^{n}-\dfrac{1}{2}\nu_n\left(u_{j+1}^{n}-u_{j-1}^{n}\right)
+\dfrac{1}{2}\nu_n^2 \left(u_{j+1}^{n}-2u_{j}^{n}+u_{j-1}^{n}\right)\\
&\qquad-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{4\Delta x}\left(u_{j+1}^{n}-u_{j-1}^{n}\right)
\end{aligned}
\end{equation}
\end{document}
TERMO ADITIVO
O comentário do OP ofereceu uma interpretação diferente do que é desejado. Esperançosamente, isso corresponde ao desejo.
\documentclass{book}
\usepackage{amsmath}
\begin{document}
\begin{align}
u_j^{n+1}&=u_j^{n}-a(t_n)\Delta t \dfrac{\partial u}{\partial x}
+\dfrac{(\Delta t)^2}{2}\left(a(t_n)^2\dfrac{\partial^2 u}{\partial x^2}-\dfrac{da(t_n)}{dt}\dfrac{\partial u}{\partial x}\right)\nonumber\\
&=u_j^{n}-a(t_n)\Delta t \dfrac{\partial u}{\partial x}
+a(t_n)^2\dfrac{(\Delta t)^2}{2}\dfrac{\partial^2 u}{\partial x^2}-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{2}\dfrac{\partial u}{\partial x}\nonumber\\
&=u_j^{n}-a(t_n)\Delta t \left(\dfrac{u_{j+1}^{n}-u_{j-1}^{n}}{2\Delta x}\right)
\nonumber\\
&\qquad+a(t_n)^2\dfrac{(\Delta t)^2}{2}\left(\dfrac{u_{j+1}^{n}-2u_{j}^{n}+u_{j-1}^{n}}{(\Delta x)^2}\right)\nonumber\\
%%%%%%%%%%%%%%%%%%%%%%
&\qquad-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{2}\left(\dfrac{u_{j+1}^{n}-u_{j-1}^{n}}{2\Delta x}\right)\nonumber\\
&
\begin{aligned}
{} &= u_j^{n}-\dfrac{1}{2}\nu_n\left(u_{j+1}^{n}-u_{j-1}^{n}\right)
+\dfrac{1}{2}\nu_n^2 \left(u_{j+1}^{n}-2u_{j}^{n}+u_{j-1}^{n}\right)\\
&\qquad-\dfrac{da(t_n)}{dt}\dfrac{(\Delta t)^2}{4\Delta x}\left(u_{j+1}^{n}-u_{j-1}^{n}\right)
\end{aligned}
\end{align}
\end{document}