Math 6410 • ODE

\textbf{Two Species Competition Model.}
Consider the simplified competition model for two populations \(x(t)\ge0\) and \(y(t)\ge0\),
\[
\begin{cases}
  \dot{x}=x(1-x-3y),\\
  \dot{y}=3y(1-y-2x).
\end{cases}
\]
Its equilibrium points are \((0,0)\), \((0,1)\), \((1,0)\) and \((.4,.2)\). The equilibria
\((0,1)\) and \((1,0)\) are stable, but \((0,0)\) is unstable. This system has a seperatrix, a
curve which divides the first quadrant of the plane into two basins of attraction
corresponding to the two stable equilibria. Thus one species or the other wins out,
depending which has the starting advantage. As a first step in showing that \((.4,.2)\)
lies on the seperatrix, prove that in an open neighborhood \(O\) of \((.4,.2)\) there is a
\(C^1\) curve \(\sigma\) through \((.4,.2)\) with the property that trajectories starting on one
side of the \(\sigma\) in \(O\), stay on that side of \(\sigma\) while in \(O\).