# How to properly align my continuous equations?

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}^k\right)^3 \prescript{}{l}{f}_{2}^k - 4\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{1}^k - \lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{0}^k + 2\lambda_2 \left(\prescript{}{l}{\delta}_{1}^k\right)^3\Big\}\textcolor{red}{\boldsymbol{\prescript{}{l+1}{\delta}_{1}^k}} + \\ \Big\{ 2\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{2}^k + 4\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{1}^k + 2\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{0}^k - 2\lambda_2 \left(\prescript{}{l}{\delta}_{1}^k\right)^3\Big\}\textcolor{red}{\boldsymbol{\prescript{}{l+1}{\delta}_{2}^k}} + \\ \Big\{ -\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{2}^k - \lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^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}^k\right)^3 \prescript{}{l}{f}_{2}^k - 4\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{1}^k - \lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{0}^k + 2\lambda_2 \left(\prescript{}{l}{\delta}_{1}^k\right)^3\Big\}\textcolor{red}{\boldsymbol{\prescript{}{l+1}{\delta}_{1}^k}} + \\ \Big\{ 2\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{2}^k + 4\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{1}^k + 2\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{0}^k - 2\lambda_2 \left(\prescript{}{l}{\delta}_{1}^k\right)^3\Big\}\textcolor{red}{\boldsymbol{\prescript{}{l+1}{\delta}_{2}^k}} + \\ \Big\{ -\lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^3 \prescript{}{l}{f}_{2}^k - \lambda_1 \left(\prescript{}{l}{\delta}_{1}^k\right)^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}


• Welcome to TeX.SE.
– Mico
Nov 21, 2020 at 17:17

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 $N = 1$:
\label{SingleDropletEqn1} \begin{aligned}[b] \bigl\{1 - \lambda_1 (\prescript{}{l}{\delta}_{1}^k)^3 \prescript{}{l}{f}_{2}^k - 4\lambda_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 + 2\lambda_2 (\prescript{}{l}{\delta}_{1}^k)^3 \bigr\} &\textcolor{red}{\prescript{}{l+1}{\bm{\delta}_{1}^k}} \\ {}+ \bigl\{ 2\lambda_1 (\prescript{}{l}{\delta}_{1}^k)^3 \prescript{}{l}{f}_{2}^k + 4\lambda_1 (\prescript{}{l}{\delta}_{1}^k)^3 \prescript{}{l}{f}_{1}^k + 2\lambda_1 (\prescript{}{l}{\delta}_{1}^k)^3 \prescript{}{l}{f}_{0}^k - 2\lambda_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}

\medskip\noindent
For $N = 2$:
\label{SingleDropletEqn2} \begin{aligned}[b] \bigl\{1 - \lambda_1 (\prescript{}{l}{\delta}_{1}^k)^3 \prescript{}{l}{f}_{2}^k - 4\lambda_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 + 2\lambda_2 (\prescript{}{l}{\delta}_{1}^k)^3 \bigr\} &\textcolor{red}{\prescript{}{l+1}{\bm{\delta}_{1}^k}} \\ {}+ \bigl\{ 2\lambda_1 (\prescript{}{l}{\delta}_{1}^k)^3 \prescript{}{l}{f}_{2}^k + 4\lambda_1 (\prescript{}{l}{\delta}_{1}^k)^3 \prescript{}{l}{f}_{1}^k + 2\lambda_1 (\prescript{}{l}{\delta}_{1}^k)^3 \prescript{}{l}{f}_{0}^k - 2\lambda_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{document}

• Thank you very much. Though, my interface is giving me couple of errors like "Undefined Control Sequence". I avoided using "\usepackage{geometry} % set page parameters suitably" because I am using a sample template my uni provided. Nov 21, 2020 at 17:31
• @Peclet - An "Undefined Control Sequence" message doesn't come along all alone. Do please tell which control sequences aren't defined. Is it maybe \bm? If so, did you load the bm package?
– Mico
Nov 21, 2020 at 17:34
• I am sorry, I have no solid knowledge on TeX, I had to search for \bm before replying you. Yes, problem was \bm so I loaded that one. Now it seems like it is working properly. I sincerely thank you for your effort and time. Nov 21, 2020 at 17:42