prepare correct position for equation

Question

Hi all,

I want to write my equation which is like the attached picture. I tried to write this code below but I couldn't make "Hp" starting from the same positions.

How should I write code to have the picture like this.

\begin{eqnarray} \label{eq:5}
Se & = & \left\{
                \begin{array}{ll}
                  \frac{1}{{[1+|\alpha H_{p}|^{n}]}^{m}}     & \quad H_{p}<0\\
                  1  & \quad H_{p} \geq0
                \end{array}
              \right.   \nonumber
 \\
 \theta & = & \left\{
                \begin{array}{ll}
                  \theta_{r}+Se(\theta_{s}-\theta_{r})  & \quad H_{p}<0\\
                  \theta_{s}     & \quad H_{p} \geq0 
                \end{array}
              \right. 
\\  
C & = & \left\{
                \begin{array}{ll}
                  \frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})Se^{\frac{1}{m}}\left(1-Se^{\frac{1}{m}}\right)^{m}    & \quad H_{p}<0\\
                  0     & \quad H_{p} \geq0
                \end{array}
              \right.   \nonumber   
\\  
k_{r} & = & \left\{
                \begin{array}{ll}
                  Se^{l}\left[1-\left(1-Se^{\frac{1}{m}}\right)^{m} \right]^2   & \quad H_{p}<0\\
                  1     & \quad H_{p} \geq0
                \end{array}
              \right.   \nonumber                                            
\end{eqnarray}

Answer 1

Please do not use eqnarray! It is deprecated since years. For such things, there are cases from amsmath or the newer mathtools. Just put everything inside of an align. The trick in order to align the most right column is to use an \hphantom here. I hope, this is what you wanted to have.

% arara: pdflatex

\documentclass{article}
\usepackage{mathtools}
\newcommand{\Se}{\operatorname{\mathit{S\kern-.15em e}}}

\begin{document}
\begin{align*}
    \theta &=  
    \begin{dcases}
        \theta_{r}+\Se(\theta_{s}-\theta_{r}) & H_{p}<0 \\ 
        \mathrlap{\theta_{s}}\hphantom{\frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})\Se^{\frac{1}{m}}\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m}} & H_{p} \geq 0 
    \end{dcases}\\
    \Se &=  
    \begin{dcases}
        \frac{1}{{\bigl[1+\lvert\alpha H_{p}\rvert^{n}\bigr]}^{m}} & H_{p}<0 \\ 
        \mathrlap{1}\hphantom{\frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})\Se^{\frac{1}{m}}\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m}} & H_{p} \geq 0 
    \end{dcases} \\
    C &=
    \begin{dcases} 
        \frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})\Se^{\frac{1}{m}}\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m} & H_{p}<0 \\ 
        0 & H_{p} \geq 0 
    \end{dcases} \\
    k_{r} &= 
    \begin{dcases}
        \Se^{l}\Bigl[1-\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m} \Bigr]^2 & H_{p}<0 \\ 
        \mathrlap{1}\hphantom{\frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})\Se^{\frac{1}{m}}\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m}} & H_{p} \geq 0 
    \end{dcases}
\end{align*}
\end{document}

---

As recommended by daleif, you could also define some certain width and put your lower rows in a mathmakebox with left alignment. This would look like:

\begin{align*}
    \theta &=  
    \begin{dcases}
        \theta_{r}+\Se(\theta_{s}-\theta_{r}) & H_{p}<0 \\ 
        \mathmakebox[5cm][l]{\theta_{s}} & H_{p} \geq 0 
    \end{dcases}\\
    \Se &=  
    \begin{dcases}
        \frac{1}{{\bigl[1+\lvert\alpha H_{p}\rvert^{n}\bigr]}^{m}} & H_{p}<0 \\ 
        \mathmakebox[5cm][l]{1} & H_{p} \geq 0 
    \end{dcases} \\
    C &=
    \begin{dcases} 
        \frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})\Se^{\frac{1}{m}}\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m} & H_{p}<0 \\ 
        \mathmakebox[5cm][l]{0} & H_{p} \geq 0 
    \end{dcases} \\
    k_{r} &= 
    \begin{dcases}
        \Se^{l}\Bigl[1-\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m} \Bigr]^2 & H_{p}<0 \\ 
        \mathmakebox[5cm][l]{1} & H_{p} \geq 0 
    \end{dcases}
\end{align*}

Answer 2

Here's a solution that uses four dcases environments placed inside an aligned environment contained in an equation environment. I would actually not bother providing a common alignment of the four pairs of conditioning statements. Instead, I'd concentrate on (a) not letting some of the individual lines become too tall, e.g., by writing ...^{1/m} rather than ...^\frac{1}{m} and (b) providing a bit more vertical whitespace between the four sets of subequations.

\documentclass{article}
\usepackage{mathtools} % for 'dcases'  environment
\newcommand\Se{\textit{Se}}
\begin{document}
\begin{equation} \label{eq:5}
\begin{aligned}
\Se  &= \begin{dcases}
        \frac{1}{[1+\lvert\alpha H_p \rvert^{n}]^{m}} & H_p<0\\
        1  &  H_p\geq0
        \end{dcases}  \\[2ex]
\theta &= \begin{dcases}
          \theta_{r}+\Se (\theta_{s}-\theta_{r})  & H_p<0\\
          \theta_{s}     & H_p\geq0
          \end{dcases}\\[2ex]
C &= \begin{dcases}
     \frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})\Se ^{1/m} \bigl(1-\Se ^{1/m}\bigr)^{m}    & H_p<0\\
     0     & H_p\geq0
     \end{dcases}\\[2ex]
k_{r} &= \begin{dcases}
         \Se^{l}\bigl[1-(1-\Se ^{1/m})^{m} \bigr]^2   & H_p<0\\
         1     & H_p\geq0
         \end{dcases}
\end{aligned}
\end{equation}
\end{document} 

---

Addendum: If you believe it's preferable to align the four conditioning pairs as well, here's a straightforward way to achieve this goal: (a) Save the longest term in the equations to a macro named, say, \longestterm and measure its width; (b) place one term each of subequations 1, 2, and 4 in a \parbox whose width is that of \longestterm.

\documentclass{article}
\usepackage{mathtools} % for 'dcases'  environment
\newcommand\Se{\textit{Se}}

\newlength\mylen
\newcommand\longestterm{\frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})\Se ^{1/m} (1-\Se ^{1/m}) ^{m}}
\settowidth\mylen{$\displaystyle\longestterm$}

\begin{document}
\begin{equation} \label{eq:5}
\begin{aligned}
\Se    &= \begin{dcases}
          \frac{1}{[1+\lvert\alpha H_p\rvert^{n}]^{m}} & H_p<0\\
          \parbox{\mylen}{$1$}  &  H_p\geq0
          \end{dcases}  \\[2ex]
\theta &= \begin{dcases}
          \theta_{r}+\Se(\theta_{s}-\theta_{r}) & H_p<0\\
          \parbox{\mylen}{$\theta_{s}$} & H_p\geq0
          \end{dcases}\\[2ex]
C      &= \begin{dcases}
          \longestterm & H_p<0\\
          0 & H_p\geq0
          \end{dcases}\\[2ex]
k_{r}  &= \begin{dcases}
          \Se^{l}\bigl[1-(1-\Se^{1/m})^{m} \bigr]^2 & H_p<0\\
          \parbox{\mylen}{$1$}  & H_p\geq0
          \end{dcases}
\end{aligned}
\end{equation}
\end{document} 

Answer 3

This is just a supplement to LaRiFaRis solution, doing the same thing with less code

\documentclass{article}
\usepackage{mathtools}
\newcommand{\Se}{\operatorname{\mathit{S\kern-.15em e}}}

\newcommand\mybox[1]{}
\begin{document}
% jsut to make this function for this patciluar case 
\renewcommand\mybox[1]{
  \mathrlap{#1}
  \hphantom{\frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})\Se^{\frac{1}{m}}\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m}}
}
\begin{align*}
    \theta &=  
    \begin{dcases}
        \theta_{r}+\Se(\theta_{s}-\theta_{r}) & H_{p}<0 \\ 
        \mybox{\theta_{s}} & H_{p} \geq 0 
    \end{dcases}\\
    \Se &=  
    \begin{dcases}
        \frac{1}{{\bigl[1+\lvert\alpha H_{p}\rvert^{n}\bigr]}^{m}} & H_{p}<0 \\ 
        \mybox{1} & H_{p} \geq 0 
    \end{dcases} \\
    C &=
    \begin{dcases} 
        \frac{\alpha
          m}{1-m}(\theta_{s}-\theta_{r})\Se^{\frac{1}{m}}\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m}
        & H_{p}<0 \\ 
        % this is the widest, so no need to do it here, but we do for
        % completness 
        \mybox{0} & H_{p} \geq 0 
    \end{dcases} \\
    k_{r} &= 
    \begin{dcases}
        \Se^{l}\Bigl[1-\bigl(1-\Se^{\frac{1}{m}}\bigr)^{m} \Bigr]^2 & H_{p}<0 \\ 
        \mybox{1} & H_{p} \geq 0 
    \end{dcases}
\end{align*}
\end{document}

Answer 4

A simple solution with eqparbox: I define an \eqmathbox command, which makes all commands with the same tag have length equal to the longest contents (two compilations required):

\documentclass{article}
\usepackage{mathtools} % for 'dcases' environment
\usepackage{eqparbox}
\newcommand\eqmathbox[2][M]{\eqmakebox[#1][l]{$\displaystyle#2$}}
\newcommand\Se{\textit{Se}}

\begin{document}

\begin{equation} \label{eq:5}
\begin{aligned}
\Se &= \begin{dcases}
          \eqmathbox{\frac{1}{[1+\lvert\alpha H_p\rvert^{n}]^{m}}} & H_p<0\\
          1 & H_p\geq0
          \end{dcases} \\[2ex]
\theta &= \begin{dcases}
          \eqmathbox{\theta_{r}+\Se(\theta_{s}-\theta_{r})}
          & H_p<0\\
            \theta_{s} & H_p\geq0
          \end{dcases}\\[2ex]
C &= \begin{dcases}
          \eqmathbox{\frac{\alpha m}{1-m}(\theta_{s}-\theta_{r})Se^{\frac{1}{m}}\left(1-Se^{\frac{1}{m}}\right)^{m}} & H_p<0\\
          0 & H_p\geq0
          \end{dcases}\\[2ex]
k_{r} &= \begin{dcases}
          \eqmathbox{\Se^{l}\bigl[1-(1-\Se^{1/m})^{m} \bigr]^2} & H_p<0\\
           1 & H_p\geq0
          \end{dcases}
\end{aligned}
\end{equation}

\end{document}