Example 3.4 (ODEs for competitive populations) Imagine a colony of rabbits (\(r\) and sheep \(s\) with \(r,s\ge 0\) denoting the population size in normalized coordinates so that one unit represents many animals and we are justified in approximating the population size as a continuous variable. A model of the birth/death rates is \begin{align*} \dot{r} & = r (3-r-s), \qquad \text{ rabbits} \\ \dot{s} & = s(4 -2r-s), \qquad \text{ sheep} \end{align*} where \(s,r \ge 0\).

We can sketch the phase portrait in three stages: first find the stationary points, then determine their types and the local phase portrait assuming the linear approximation is valid, and then put this information together to create a consistent global phase portrait.

Stationary Points: \(\dot r=0\) if \(r=0\) or \(r+s=3\) whilst \(\dot s =0\) if \(s=0\) or \(2r+s=4\). Hence the stationary points are \begin{align*} r & = 0, \; \; s=0, \qquad \qquad (0,0) \\ r & = 0, \; \; s=4, \qquad \qquad (0,4) \\ s & = 0, \; \; r=3, \qquad \qquad (3,0) \end{align*} together with the solution of the simultaneous equations \[ r+s = 3, \qquad 2r+ s = 4 \] if they exist. Solving gives a fourth stationary point, \(r=1\) and \(s=2\), i.e. \((1,2)\). Note that the \(r\)-axis and the \(s\)-axis are invariant. Linearisation: \[ {D} f(\underline{x}) = \begin{pmatrix} 3 - 2r - s & -r \\ -2s & 4-2r-2s \\ \end{pmatrix} \] At \((0,0):\) \[ {D} f(0,0)= \begin{pmatrix} 3& 0 \\ 0&4 \\ \end{pmatrix}. \] The eigenvalues are \(3\) and \(4\) with eigenvectors \(\begin{pmatrix} 1 \cr 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \cr 1 \end{pmatrix}\) respectively, so it is an unstable node, with almost all solutions tangential to the \(r\)-axis and the local solution is as sketched in Figure 3.11).

At \((0,4):\) \({D} f(0,4)= \begin{pmatrix} -1& 0 \\ -8& -4 \\ \end{pmatrix} \) The eigenvalues are \(-1,-4\) so it is a stable node. The eigenvectors are \( e_{-1}=\begin{pmatrix} 3 \cr {-8} \end{pmatrix}\) and \(\begin{pmatrix} 0 \cr 1\end{pmatrix}\) respectively, so almost all solutions are tangential to \( e_{-1}\) at the stationary point. See Figure 3.11.

At \((3,0):\) \({D} f(3,0)= \begin{pmatrix} -3& -3 \\ 0& -2 \\ \end{pmatrix} \) so the eigenvalues are \(-3\) with eigenvector \(\begin{pmatrix} 1 \cr 0 \end{pmatrix}\) and \(-2\) with eigenvector \(e_2\begin{pmatrix} {-3}\cr 1 \end{pmatrix}\). So it is a stable node and almost all solutions are tangential to \(e_{2}\) at the stationary point.

At \((1,2):\) \({D} f(1,2)= \begin{pmatrix} -1& -1 \\ -4&-2 \\ \end{pmatrix} \) so the characteristic equation is \((s+1)(s+2)-4=0\) or \(s^2 + 3s - 2 = 0\), i.e. \(s_\pm = \frac{-3 \pm \sqrt{17}}{2}\). Since \(s_{+}>0\) and \(s_{-}<0\) is a saddle and the eigenvectors are \(e_\pm=\begin{pmatrix} {-1} \cr {s_\pm+1} \end{pmatrix}\), so \(e_+\) slopes downwards and \(e_-\) slopes upwards.

Putting the information together suggests the global phase portrait of Figure 3.12. The important features are the separatrices which separates solutions tending to \((0,4)\) from those approaching \((3,0)\).and the tangential approach to approximate stationary points.

Example 3.5 (ODEs for mutualistic interactions) Imagine a colony of bees (\(b\)) and flower (\(f\)) with \(b,f\ge 0\) denoting the population size in normalized coordinates so that one unit represents many animals and we are justified in approximating the population size as a continuous variable. Bees fly from flower to flower gathering nectar for food, and the flowers also benefit from the bees for pollination. This is a typical example of mutualistic interaction. A model for their population is \begin{equation}\label{eq:bf} \dot{b} = (3-3b+f)b,\qquad \dot{f} = (1+b-f)f. \end{equation} All the stationary points are \((0,0),(0,1),(1,0),(2,3)\). By evaluating the Jacobian \[ Df(b,f) = \begin{pmatrix} 3-6b+f & b \cr f & 1+b-2f \end{pmatrix} \] we have the following classification:

• (0,0): unstable node
• (1,0): saddle node
• (0,1): saddle node
• (2,3): stable node