# How to vertically centered number equation in "align", and breaking line equation such that indent after equal sign?

I want typing math equation like this: But with this code I can't get the result like above. Anyone can help me to edit my code to get the above result?

\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}


For the equation number centered, use equation and aligned, rather than align. For the indents, I inserted \qquad where needed. I also had to break up one of your long lines to make room for the centered eqn number.

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


The OP's comment offered a different interpretation of what is desired. Hopefully, this matches the desire.

\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\\