"정렬"에서 수직으로 가운데에 숫자 방정식을 배치하고 등호 뒤에 들여쓰기가 되도록 줄 방정식을 끊는 방법은 무엇입니까?

Question

다음과 같이 수학 방정식을 입력하고 싶습니다. 하지만 이 코드로는 위와 같은 결과를 얻을 수 없습니다. 위의 결과를 얻기 위해 내 코드를 편집하는 데 도움을 줄 수 있는 사람이 있나요?

\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}

Answer 1

방정식 번호가 중앙에 있는 경우에는 대신 equation및 를 사용하십시오 . 들여쓰기를 위해 필요한 곳에 삽입했습니다 . 또한 중앙에 있는 eqn 번호를 위한 공간을 만들기 위해 긴 줄 중 하나를 끊어야 했습니다.alignedalign\qquad

\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}

---

부록

OP의 의견은 원하는 것에 대해 다른 해석을 제시했습니다. 희망적으로 이것이 욕구와 일치하기를 바랍니다.

\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}