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How to properly align my continuous equations?

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}

This is the output of this code

One Answer

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.

enter image description here

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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