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%\title{Leguerinel model}
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\noindent
\begin{large}
\textbf{Leguerinel model} equation is:
\end{large}
\noindent
 \begin{eqnarray*}
\begin{array}{l}
 \displaystyle \log_{10}(N_t) = K - A^p
\end{array}
\end{eqnarray*}
\noindent
where
 \begin{eqnarray*}
\begin{array}{l}
\displaystyle A=\frac{t}{ \delta \times B} \\[3mm]
\end{array}
\end{eqnarray*}
\noindent
with
 \begin{eqnarray*}
\begin{array}{l}
      \displaystyle B = 10^{C^2+D^2} \\[3mm]
     \displaystyle C= \frac{pH - pHopt}{ZpH}\\[3mm]
      \displaystyle D = \frac{Aw - Awopt}{ ZAw}\\[3mm]
\end{array}
\end{eqnarray*}

 $t$ is time, $log_{10}$ is base 10 logarithm. The parameters  to estimate are $\delta$  and $p$. Five parameters are fixed:
$K$, $pHopt$, $Awopt$, $ZpH$, $ZAw$. Two environmental variables are fixed: $pH$ and $Aw$.
\newline
The noisy output is defined as:
\begin{displaymath}
\log_{10}(N_t)=\mathcal{N}(\log_{10}(N_t), \%noise)
\end{displaymath}
i.e random number from the normal distribution with mean parameter $\log_{10}(N_t)$ and standard deviation parameter $\%noise$.

\paragraph{}
\begin{flushleft}
\textbf{Example of Leguerinel curve}\\
Time unit is mn. Maximal time is 15mn.
 $\delta=1.8$, $p=5$,
 $K=2$, $pHopt=6.96$, $Awopt=0.985$, $ZpH=2.18$, $ZAw=0.092$,
$pH=6.96$ and $Aw=0.9$.
\end{flushleft}

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%    \caption{General form of the curve.}
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