(0,y)g_1(y) dy\\
& = \int_0^1 \left( \frac{s_0 - r_0 y}{|r_0|} + \frac{s_1 - r_1 y}{|r_1|} \right) \frac{G(y)}{y(1-y)}dy
\end{align*}
Now, since $r_0$ and $r_1$ have opposite signs, we have for all $y \in (0,1)$ 
\[
\frac{s_0 - r_0 y}{|r_0|} + \frac{s_1 - r_1 y}{|r_1|} = \frac{s_0}{|r_0|} + \frac{s_1}{|r_1|},
\]
which implies that
\[
\lambda_1(\nu_1) = \left( \frac{s_0}{|r_0|} + \frac{s_1}{|r_1|}\right) \int_0^1 \frac{G(y)}{y(1-y)} dy. 
\]
and concludes the proof.
\end{proof}
The invasion rates for species 1, as well as for species 2 and 3 are summed up in Table \ref{tab:invasion}. As for $C_1$, $C_2$ and $C_3$ are positive constant that may be computed. However, since it is sufficient to know the signs of the invasion rates, it does not really matter to have the exact expression of the $C_i$.
 

\begin{table}
\begin{center}
\begin{tabular}{c|c|c|c||c|c|c}
       & $\mu_1$ & $\mu_2$ & $\mu_3$ & $\nu_1$ & $\nu_2$ & $\nu_3$ \\ \hline
      1 & 0 & $\Lambda_1^3 $ & $\Lambda_0^2$ & $C_1 ( \frac{s_1}{|r_1|} + \frac{s_2}{|r_2|})$ & 0 & 0\\ \hline
      2 & $\Lambda_0^3$ & 0 & $\Lambda_0^1$ & 0 & $C_2 ( \frac{r_1}{|s_1|} + \frac{r_2}{|s_2|})$ & 0\\ \hline
      3 & $\Lambda_1^2$ & $\Lambda_1^1$ & 0 & 0 & 0 &  $C^3 ( -\frac{r_1}{|s_1-r_1|} - \frac{r_2}{|s_2-r_2|})$
    \end{tabular}
    
    \end{center}
    \caption{Invasion rates}
    \label{tab:invasion}
  \end{table}  
\subsection{Environment favours and disadvantages alternately two species, the other one stay neutral over time.}







With no loss of generality, we assume the  environment promotes and disadvantages alternately the first and the second species. So without regard to order, only one configuration is possible:
\begin{equation}
         s_1>0>r_1, \qquad
       r_2>0>s_2.
\end{equation}

In this configuration, many different situations are possible for the long time behavior. In the following we highlight some of them. Mainly, we discuss about the situation where the neutral species (the third in the last configuration) invade the community. Then we prove that persistence of the three species is not possible with only two environments.


\subsubsection{The neutral species invades the community}
In this section, we use results of Section \ref{sec:suffinvade} to provide sufficient conditions for the invasion of the neutral species. We also give an example of parameters satisfying theses conditions.

\begin{theorem}
\label{thm:neutralinvade}
Assume that $\Lambda_0^1<0$ and $\Lambda_0^2<0$. Set 
\[ \Lambda = \max(\Lambda_0^1, \Lambda_0^2) < 0
\]
Then, for all $(x,y) \in \mathscr{E} \setminus \mathscr{E}_0^3$, and all $\mathbf{s} \in E$,
\[
\mathbb{P}_{(x,y,\mathbf{s})} \left( \limsup_{t \to \infty} \frac{1}{t} \log \|(X_t,Y_t)\| \leq \Lambda \right) =1.
\]
\end{theorem}
\begin{proof}
With the notations of Theorem \ref{thm:suffinvade},  $\Lambda = \Lambda_3$, and thus, by this theorem,
for all $\alpha \in (\Lambda, 0)$, there exists $\eta > 0$ and a neighbourhood $\mathscr{U}$ of $0$ such that, for all $(x,y) \in \mathscr{U}$ and all $\mathbf{s} \in E$,
\[
\mathbb{P}_{(x,y,\mathbf{s})} \left( \limsup_{t \to \infty} \frac{1}{t} \log \|(X_t,Y_t)\| \leq \alpha \right) \geq \eta.
\]
Now, we prove that the point $0$ is accessible from $\mathscr{E}_+^3$. By Lemma \ref{lem:accessconjug}, $0$ is accessible if and only if it is accessible for the vector fields given by $g_s(x,y) =( x(s^1 - (s^1 x + r^1y)), y(r^1 - (x^1x + r^1y)))$. Consider the convex combination $g_p = p_1 g_{s_1} + p_2 g_{s_2}$. Then $g_p = g_{\mathbf{s}_p}$, where $\mathbf{s}_p = p_1 \mathbf{s}_1 + p_2 \mathbf{s}_2$. Now, since $\Lambda_0^1$ and $ \Lambda_0^2$ are negative,  $\mathbf{s}_p = (s_p, r_p, 0)$. Hence, by Theorem \ref{th:deterministe} and Lemma \ref{lem:accessconjug}, the flow generated by $g_p$ converges to $0$ for all initial condition in $\mathscr{E}^3_+$. In particular, $0$ is accessible from $\mathscr{E}^3_+$ with the vector fields $g_s$, hence it is accessible for $\big((X_t, Y_t)\big)_{t>0}$. 

Finally, as for the proof of Theorem \ref{th:alwaysthebest}, we have to show that the face $\mathscr{E}_0^3$ is repulsive. We use Theorem \ref{thm:persistence}. We know that $\mathscr{P}_{erg}(\mathscr{E}_0^3)$ contain $\mu_1$, $\mu_2$ and possibly $\nu_3$. Furthermore, by Lemma \ref{lem:invasionrates}, Proposition \ref{prop:oneD} and the assumption that $\Lambda_0^1<0$ and $\Lambda_0^2<0$, we have that
\[
\lambda_3(\mu_1) = \Lambda_1^2 > 0 \quad \mbox{and} \quad \lambda_3(\mu_2) = \Lambda_1^1 > 0.
\]
Moreover, since we are in the situation 
\[
 \left\{
    \begin{array}{ll}
       s_1>0>r_1 \\
       r_2>0>s_2
    \end{array},
\right.
\]
the assumption $ \Lambda_0^1 = r_1+r_2<0 $ 
  and $    \Lambda_0^2 = s_1+s_2<0$
is equivalent to   $-r_1>r_2>0$ and $ -s_2>s_1>0$, which implies $r_1s_2>r_2s_1$. Hence, by Proposition \ref{prop : impossiblepersis} below, $\lambda_3(\nu_3) > 0$ which concludes the proof.
 
\end{proof}






\begin{ex}
Take $s_1 = 1/3$, $r_1 = - 1/3$, $s_2=-3/8$ and $r_2 = 1/4$. Choose $q_1 = q_2$ so that $p_1 = p_2 = \frac{1}{2}$. Then
\[
\Lambda_0^1 = - \frac{1}{24}; \quad \Lambda_0^2 = - \frac{1}{48}; \quad \Lambda_0^3 = - \frac{1}{5},
\]
thus conclusion of Theorem \ref{thm:neutralinvade} holds. Illustrations are given in Figure \ref{fig:neutraltriangle}. 





\begin{figure}[