An elegant way to write long equations in LaTex?

Question

As mentioned in the title: I would like to know if there is an elegant way to plot long mathematical expressions in a latex document in a very "coherent" way.

I have no specifically idea what is the best solution for that: so I am just asking for people who are familiar with long equations in a latex document: How do you manage to write it?

For instance : I have these equations :

But you can notice that the expressions are not structured in an homogeneous way: what could you advise?

\documentclass{article}
\usepackage{amsmath}
\usepackage{enumitem}

\begin{document}

\begin{enumerate}
\newpage
\item \textbf{En $i$} 
\begin{align}
h_{i}^{n} = - \color{red}h_{i-1}^{n+1} \color{black} \left(\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i}^{n} + h_{i-1}^{n}}{2} \right)^3 \right) &+ \color{red} h_{i}^{n+1} \color{black} \left (1+  \frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i}^{n} + h_{i-1}^{n}}{2} \right)^3 + \frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i+1}^{n} + h_{i}^{n}}{2} \right)^3 \right) \notag \\
- \color{red}h_{i}^{n+1} \color{black}\left(\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i}^{n} + h_{i+1}^{n}}{2} \right)^3  \right) \notag
\end{align}

\item \textbf{En $i = (N-1)$} 
\begin{align}
h_{N-1}^{n} = - \color{red} h_{N-2}^{n+1} \color{black} \left(\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{N-1}^{n} + h_{N-2}^{n}}{2} \right)^3 \right) + \color{red} h_{N-1}^{n+1} \color{black} \left (1+  \frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} \right)^3 + \frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{N}^{n} + h_{N-1}^{n}}{2} \right)^3 \right) \notag \\
- \color{red} h_{N}^{n+1} \color{black} \left(\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} \right)^3  \right) \notag
\end{align}

\end{enumerate}
\end{document}

Answer 1

With align* and equations in three lines:

\documentclass{article}
\usepackage{xcolor}
\usepackage{amsmath}
\usepackage{enumitem}

\begin{document}

\begin{enumerate}
\newpage
\item \textbf{En $i$}
\begin{align*}
h_{i}^{n} 
    = &{} - {\color{red}h_{i-1}^{n+1}} \Biggl(\frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{i}^{n} + h_{i-1}^{n}}{2} \biggr)^3 \Biggr)   \\
      &{} + {\color{red} h_{i}^{n+1}} \Biggl(1+\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i}^{n} + h_{i-1}^{n}}{2}\right)^3 +
          \frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{i+1}^{n} + h_{i}^{n}}{2}\biggr)^3 \Biggr)   \\
      &{} + {\color{red}h_{i}^{n+1}} \Biggl(\frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{i}^{n} + h_{i+1}^{n}}{2} \biggr)^3  \Biggr)
\end{align*}

\item \textbf{En $i = (N-1)$}
\begin{align*}
h_{N-1}^{n} 
    = &{} - {\color{red} h_{N-2}^{n+1}} \biggl(\frac{\Delta t}{(\Delta x)^2}
        \Bigl(\frac{h_{N-1}^{n} + h_{N-2}^{n}}{2} \Bigr)^3\biggr)     \\
      &{} - {\color{red} h_{N-1}^{n+1}} \left (1+  \frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} \right)^3 + \frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{N}^{n} + h_{N-1}^{n}}{2} \right)^3 \right)       \\
      &{} - {\color{red} h_{N}^{n+1}} 
        \biggl(\frac{\Delta t}{(\Delta x)^2} \Bigl(\frac{h_{N-1}^{n}+ h_{N}^{n}}{2} \Bigr)^3  \Biggr)
\end{align*}

\end{enumerate}
\end{document}

Answer 2

Here's an answer that's similar to the one by @Zarko in that it uses 3 rather than 2 lines per equation. It differs in the use of \textcolor instead of \color, the replacement of the needlessly large outer round parentheses with smaller, but adequately sized square brackets, and a "snugging up" of the power-3 terms to the respective closing parentheses.

\documentclass{article}
\usepackage{amsmath,xcolor}
\begin{document}

\begin{enumerate}

\item \textbf{En} $i$
\begin{align*}
h_{i}^{n} 
= -\textcolor{red}{h_{i-1}^{n+1}} 
 &\biggl[\frac{\Delta t}{(\Delta x)^2} 
  \biggl(\frac{h_{i}^{n} + h_{i-1}^{n}}{2} 
  \biggr)^{\!\!3}\, \biggr] \\
 {}+\textcolor{red}{h_{i}^{n+1}} 
 &\biggl[1+  \frac{\Delta t}{(\Delta x)^2} 
  \biggl(\frac{h_{i}^{n} + h_{i-1}^{n}}{2} \biggr)^{\!\!3} 
  + \frac{\Delta t}{(\Delta x)^2} 
  \biggl(\frac{h_{i+1}^{n} + h_{i}^{n}}{2} 
  \biggr)^{\!\!3}\, \biggr]  \\
 {}-\textcolor{red}{h_{i}^{n+1}} 
 &\biggl[\frac{\Delta t}{(\Delta x)^2} 
  \biggl(\frac{h_{i}^{n} + h_{i+1}^{n}}{2} 
  \biggr)^{\!\!3} \, \biggr] 
\end{align*}

\item \textbf{En} $i = (N-1)$
\begin{align*}
h_{N-1}^{n} 
= -\textcolor{red}{h_{N-2}^{n+1}} 
 &\biggl[\frac{\Delta t}{(\Delta x)^2} 
  \biggl(\frac{h_{N-1}^{n} + h_{N-2}^{n}}{2} 
  \biggr)^{\!\!3}\, \biggr] \\
 {}+\textcolor{red}{h_{N-1}^{n+1}} 
 &\biggl[1+  \frac{\Delta t}{(\Delta x)^2} 
  \biggl(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} \biggr)^{\!\!3} 
  + \frac{\Delta t}{(\Delta x)^2} 
  \biggl(\frac{h_{N}^{n} + h_{N-1}^{n}}{2} 
  \biggr)^{\!\!3}\, \biggr]  \\
 {}-\textcolor{red}{h_{N}^{n+1}} 
 &\biggl[\frac{\Delta t}{(\Delta x)^2} 
  \biggl(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} 
  \biggr)^{\!\!3} \, \biggr] 
\end{align*}

\end{enumerate}
\end{document}

Answer 3

Here is a possible solution, loading geometry to have more decent margins, and using the [wide=0pt] key from enumitem. For the second item, I also propose a solution with multline*.

    \documentclass{article}
    \usepackage[showframe]{geometry}
    \usepackage{xcolor}
    \usepackage{amsmath}
    \usepackage{enumitem}

    \begin{document}

    \begin{enumerate}[wide=0pt]
    \newpage
    \item \textbf{En $i$}
    \begin{align*}
    h_{i}^{n} = - \color{red}h_{i-1}^{n+1} \color{black} \left(\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i}^{n} + h_{i-1}^{n}}{2} \right)^{\!\!3} \right) & + \color{red} h_{i}^{n+1} \color{black} \biggl (1+ \frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i}^{n} + h_{i-1}^{n}}{2} \right)^{\!\!3} + \frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i+1}^{n} + h_{i}^{n}}{2} \right)^{\!\!3} \biggr) \\
    %{} & \phantom{ = }
     & - \color{red}h_{i}^{n+1} \color{black}\left(\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{i}^{n} + h_{i+1}^{n}}{2} \right)^{\!\!3} \right) \notag
    \end{align*}

    \item \textbf{En $i = (N-1)$}
     \begin{align*}
    h_{N-1}^{n} = & - \color{red} h_{N-2}^{n+1} \color{black} \Biggl(\frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{N-1}^{n} + h_{N-2}^{n}}{2} \biggr)^{\!\!3} \Biggr) \\
    & + \color{red} h_{N-1}^{n+1} \color{black} \Biggl(1+ \frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} \biggr)^{\!\!3} %\\
     + \frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{N}^{n} + h_{N-1}^{n}}{2} \biggr)^{\!\!3}\Biggr) \\
     & - \color{red} h_{N}^{n+1} \color{black} \left(\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} \right)^{\!\!3} \right)
    \end{align*}
     \begin{multline*}
    h_{N-1}^{n} = - \color{red} h_{N-2}^{n+1} \color{black} \Biggl(\frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{N-1}^{n} + h_{N-2}^{n}}{2} \biggr)^{\!\!3} \Biggr)% \\
     + \color{red} h_{N-1}^{n+1} \color{black} \Biggl(1+ \frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} \biggr)^{\!\!3} \\
     + \frac{\Delta t}{(\Delta x)^2} \biggl(\frac{h_{N}^{n} + h_{N-1}^{n}}{2} \biggr)^{\!\!3}\Biggr) %\\
     - \color{red} h_{N}^{n+1} \color{black} \left(\frac{\Delta t}{(\Delta x)^2} \left(\frac{h_{N-1}^{n} + h_{N}^{n}}{2} \right)^{\!\!3} \right)
    \end{multline*}
    \end{enumerate}

    \end{document} 

Answer 4

Structure with \newcommmand

Based on Zarko's answer, I introduced some \newcommand definitions to reduce the amount repeating markup to highlight the differences between the different terms:

\documentclass{article}
\usepackage{xcolor}
\usepackage{amsmath}
\usepackage{enumitem}

\newcommand{\dtdxtwo}{\frac{\Delta t}{(\Delta x)^2}}
\newcommand{\redh}[1]{{\color{red}h_{#1}^{n+1}}}
\newcommand{\hcube}[2]{\biggl(\frac{h_{#1}^{n} + h_{#2}^{n}}{2} \biggr)^3}
\newcommand{\aterm}[2]{\dtdxtwo \hcube{#1}{#2}}

\begin{document}
\begin{enumerate}
  \item \textbf{En} $i$
    \begin{align*}
      h_{i}^{n} 
      = &{} - \redh{i-1} \Biggl(\aterm{i}{i-1}\Biggr) \\
        &{} + \redh{i} \Biggl(1+\aterm{i}{i-1}+\aterm{i+1}{i}\Biggr) \\
        &{} + \redh{i} \Biggl(\aterm{i}{i+1}\Biggr)
    \end{align*}
  \item \textbf{En} $i = N - 1$
    \begin{align*}
      h_{N-1}^{n}
      = &{} - \redh{N-2} \Biggl(\aterm{N-1}{N-2}\Biggr) \\
        &{} - \redh{N-1} \Biggl(1+\aterm{N-1}{N}+\aterm{N}{N-1}\Biggr) \\
        &{} - \redh{N} \Biggl(\aterm{N-1}{N}\Biggr)
   \end{align*}
\end{enumerate}
\end{document}

My feeling is that it is much easier to spot inconsistencies, sign or index errors in this markup. Also, differences in repeating terms can be expressed more clearly.

Finally, this opportunity for naming makes it possible to express equations with meaningful terms.