How to properly align my continuous equations?

Question

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}

Answer 1

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 
         - 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{equation}

\medskip\noindent
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 
         - 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{equation}
\end{document}