# Equation label in center (while using flalign environment)

See the output in the given image.

Question:

First, I want to give the equation label (2) in center and Second move the part of equation (2) rounded by red color in right side.

MWE

\documentclass[12pt]{article}
\usepackage[a4paper,top=0.6 in,bottom=0.6 in,left=0.6 in,right=0.6 in]{geometry}
\usepackage{amsmath}
\begin{document}
\large
\begin{flalign}
&\dfrac{\partial (u)}{\partial x} + \dfrac{\partial (v)}{\partial y}=0&& \\
&(u)\cdot\dfrac{\partial (u)}{\partial x} + (v)\cdot\dfrac{\partial (v)}{\partial y}=  u_{e}\;\dfrac{d \,u_{e}}{d\, x}+\dfrac{\mu \left(1+\frac{1}{\gamma}\right)}{\rho}\,\frac{\partial^2 u}{\partial y^2}-\dfrac{\sigma B_{0}^{2}}{\rho}+ g\;\beta_{C}\;(C-C_{\infty})&&\\[2ex]
&-\frac{\mu\,\phi}{\rho k'}\,\left(u_{e}-u\right)+ g\;\beta_{T}\;(T-T_{\infty})\nonumber \\[2ex]
&(u)\;\dfrac{\partial (T)}{\partial x} + (v)\;\dfrac{\partial (T)}{\partial y}
=\left(\dfrac{k_{2}}{\rho\,c_{p}}\right)\,\dfrac{\partial^2 T}{\partial y^2}-\dfrac{1}{\rho\,c_{p}}\;\dfrac{\partial q_{r}}{\partial y}+\dfrac{Q_{0}}{\rho\,c_{p}}\left(T-T_{\infty}\right)+\dfrac{DK_{T}}{c_{s}\,c_{p}}\,\dfrac{\partial^2 C}{\partial\,y^2}&&\\[2ex]
&(u)\;\dfrac{\partial (C)}{\partial x} + (v)\;\dfrac{\partial (C)}{\partial y} =\dfrac{D_{T}}{T_{w}}\,\dfrac{\partial^2 T}{\partial y^2}+D_{B}\;\dfrac{\partial^2 C}{\partial y^2}&&
\end{flalign}
\end{document}


• Thanks to Bernard Aug 18, 2022 at 8:36

I see no reason for using flalign. You can just use aligned for the equation needing to be split.

I removed all manual spacings, as well as all \left and \right, but not all inconsistencies (sometimes you have \partial(C) and sometimes the parentheses are missing).

I also defined \der and \pder commands that make input easier.

\documentclass[12pt]{article}
\usepackage[a4paper,top=0.6 in,bottom=0.6 in,left=0.6 in,right=0.6 in]{geometry}
\usepackage{amsmath}

\NewDocumentCommand{\der}{omm}{%
\frac{d\IfValueT{#1}{^{#1}}#2}{d#3\IfValueT{#1}{^{#1}}}%
}
\NewDocumentCommand{\pder}{omm}{%
\frac{\partial\IfValueT{#1}{^{#1}}#2}{\partial#3\IfValueT{#1}{^{#1}}}%
}

\begin{document}

\begin{align}
&\pder{(u)}{x} + \pder{(v)}{y}=0 \\
&\begin{aligned}
(u)\pder{(u)}{x} + (v)\pder{(v)}{y}=
u_{e}\der{u_{e}}{x} &+\frac{\mu \bigl(1+\frac{1}{\gamma}\bigr)}{\rho}
\pder[2]{u}{y}-\frac{\sigma B_{0}^{2}}{\rho}+ g\beta_{C}(C-C_{\infty})\\
&-\frac{\mu\phi}{\rho k'}(u_{e}-u)+ g\beta_{T}(T-T_{\infty})
\end{aligned}\\[2ex]
&(u)\pder{(T)}{x} + (v)\pder{(T)}{y}=
\left(\frac{k_{2}}{\rho c_{p}}\right)\pder[2]{T}{y}-
\frac{1}{\rho c_{p}}\pder{q_{r}}{y}+\frac{Q_{0}}{\rho c_{p}}(T-T_{\infty})+
\frac{DK_{T}}{c_{s}c_{p}}\pder[2]{C}{y}\\[2ex]
&(u)\pder{(C)}{x} + (v)\pder{(C)}{y} =\frac{D_{T}}{T_{w}}\pder[2]{T}{y}+
D_{B}\pder[2]{C}{y}
\end{align}

\end{document}


• Incredible solution. Thank you so much sir. Aug 18, 2022 at 15:13

You might make use of multlined environment to wrap a nice-looking long equation break and it will be treated as a single block so that the label will be centered.

\documentclass[12pt]{article}
\usepackage[a4paper,top=0.6 in,bottom=0.6 in,left=0.6 in,right=0.6 in]{geometry}
\usepackage{amsmath}
\usepackage{mathtools} % <- provides multlined
\begin{document}
\large
\begin{flalign}
&\dfrac{\partial (u)}{\partial x} + \dfrac{\partial (v)}{\partial y}=0&& \\
&\begin{multlined}
(u)\cdot\dfrac{\partial (u)}{\partial x} + (v)\cdot\dfrac{\partial (v)}{\partial y}=  u_{e}\;\dfrac{d \,u_{e}}{d\, x}+\dfrac{\mu \left(1+\frac{1}{\gamma}\right)}{\rho}\,\frac{\partial^2 u}{\partial y^2}-\dfrac{\sigma B_{0}^{2}}{\rho}+ g\;\beta_{C}\;(C-C_{\infty})\\[2ex]
-\frac{\mu\,\phi}{\rho k'}\,\left(u_{e}-u\right)+ g\;\beta_{T}\;(T-T_{\infty})
\end{multlined}
\\[2ex]
&(u)\;\dfrac{\partial (T)}{\partial x} + (v)\;\dfrac{\partial (T)}{\partial y}
=\left(\dfrac{k_{2}}{\rho\,c_{p}}\right)\,\dfrac{\partial^2 T}{\partial y^2}-\dfrac{1}{\rho\,c_{p}}\;\dfrac{\partial q_{r}}{\partial y}+\dfrac{Q_{0}}{\rho\,c_{p}}\left(T-T_{\infty}\right)+\dfrac{DK_{T}}{c_{s}\,c_{p}}\,\dfrac{\partial^2 C}{\partial\,y^2}&&\\[2ex]
&(u)\;\dfrac{\partial (C)}{\partial x} + (v)\;\dfrac{\partial (C)}{\partial y} =\dfrac{D_{T}}{T_{w}}\,\dfrac{\partial^2 T}{\partial y^2}+D_{B}\;\dfrac{\partial^2 C}{\partial y^2}&&
\end{flalign}
\end{document}


• Thanks for answering my question. Aug 18, 2022 at 8:36