TeX - LaTeX Asked on December 9, 2020
I just want to make "for N = 1" and "for N = 2" left justified. How can I manage this?
begin{multline}
begin{aligned}
underline{textbf{for N = 1:}}& Big{ 1 -lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{2}^k - 4lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{1}^k - lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{0}^k + 2lambda_2 left(prescript{}{l}{delta}_{1}^kright)^3Big}textcolor{red}{boldsymbol{prescript{}{l+1}{delta}_{1}^k}} + Big{ 2lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{2}^k + 4lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{1}^k + 2lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{0}^k - 2lambda_2 left(prescript{}{l}{delta}_{1}^kright)^3Big}textcolor{red}{boldsymbol{prescript{}{l+1}{delta}_{2}^k}} + Big{ -lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{2}^k - lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{0}^k Big}textcolor{red}{boldsymbol{prescript{}{l+1}{delta}_{3}^k}} = textcolor{blue}{boldsymbol{delta_{1}^{k-1}}}
underline{textbf{for N = 2:}}& Big{ 1 -lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{2}^k - 4lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{1}^k - lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{0}^k + 2lambda_2 left(prescript{}{l}{delta}_{1}^kright)^3Big}textcolor{red}{boldsymbol{prescript{}{l+1}{delta}_{1}^k}} + Big{ 2lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{2}^k + 4lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{1}^k + 2lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{0}^k - 2lambda_2 left(prescript{}{l}{delta}_{1}^kright)^3Big}textcolor{red}{boldsymbol{prescript{}{l+1}{delta}_{2}^k}} + Big{ -lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{2}^k - lambda_1 left(prescript{}{l}{delta}_{1}^kright)^3 prescript{}{l}{f}_{0}^k Big}textcolor{red}{boldsymbol{prescript{}{l+1}{delta}_{3}^k}} = textcolor{blue}{boldsymbol{delta_{1}^{k-1}}}
label{SingleDropletEqns}
end{aligned}
end{multline}
I'd like to suggest that you use aligned
environments inside equation
(not multline
) environments and use &
alignment markers. I'd also get rid of all left
and right
directives and place the For $N=1$
and For $N=2$
labels outside the math material. Finally, in order to make the delta
-terms less turgid-looking, I'd boldface only delta
but not its pre- and post- arguments.
documentclass{article}
usepackage{geometry} % set page parameters suitably
usepackage{xcolor,mathtools,bm}
begin{document}
noindent
For $N = 1$:
begin{equation}label{SingleDropletEqn1}
begin{aligned}[b]
bigl{1 - lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{2}^k
- 4lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{1}^k
- lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{0}^k
+ 2lambda_2 (prescript{}{l}{delta}_{1}^k)^3 bigr}
&textcolor{red}{prescript{}{l+1}{bm{delta}_{1}^k}}
{}+ bigl{ 2lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{2}^k
+ 4lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{1}^k
+ 2lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{0}^k
- 2lambda_2 (prescript{}{l}{delta}_{1}^k)^3 bigr}
&textcolor{red}{prescript{}{l+1}{bm{delta}_{2}^k}}
{}+ bigl{-lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{2}^k
-lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{0}^k bigr}
&textcolor{red}{prescript{}{l+1}{bm{delta}_{3}^k}}
=textcolor{blue}{bm{delta}_{1}^{k-1}}
end{aligned}
end{equation}
medskipnoindent
For $N = 2$:
begin{equation} label{SingleDropletEqn2}
begin{aligned}[b]
bigl{1 - lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{2}^k
- 4lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{1}^k
- lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{0}^k
+ 2lambda_2 (prescript{}{l}{delta}_{1}^k)^3 bigr}
&textcolor{red}{prescript{}{l+1}{bm{delta}_{1}^k}}
{}+ bigl{ 2lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{2}^k
+ 4lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{1}^k
+ 2lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{0}^k
- 2lambda_2 (prescript{}{l}{delta}_{1}^k)^3 bigr}
&textcolor{red}{prescript{}{l+1}{bm{delta}_{2}^k}}
{}+ bigl{-lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{2}^k
-lambda_1 (prescript{}{l}{delta}_{1}^k)^3 prescript{}{l}{f}_{0}^k bigr}
&textcolor{red}{prescript{}{l+1}{bm{delta}_{3}^k}}
=textcolor{blue}{bm{delta}_{1}^{k-1}}
end{aligned}
end{equation}
end{document}
Answered by Mico on December 9, 2020
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