19.2 Sensitivity to parameters with the trace-determinant

Figure 19.2 is even more useful if your system consists of parameters that aren’t specified. Consider the following lynx - hare system:

\[\begin{equation} \begin{split} \frac{dH}{dt} &= r H - b HL \\ \frac{dL}{dt} &=ebHL -dL \end{split} \end{equation}\]

We have studied this system several times. We know that the steady states of this system are \((H,L)=(0,0)\) and \(\displaystyle \left( \frac{d}{eb}, \frac{r}{b} \right)\) in this model. Let’s compute the Jacobian at each of these systems:

\[\begin{align} J_{(0,0)} &=\begin{pmatrix} r & 0 \\ 0 & -d \end{pmatrix} \\ J_{(\frac{d}{eb}, \frac{r}{b})} &=\begin{pmatrix} r-\frac{d}{e} & -\frac{d}{e} \\ er & 0 \end{pmatrix} \end{align}\]

Let’s analyze the Jacobian at each of the the equilibrium solutions.

• \((0,0)\) equilibrium solution: We have tr\((J)=r-d\) and \(\det(J)=-rd\). Since \(r\) and \(d\) are both positive parameters the determinant will always be negative, so no matter what the origin will be a saddle.
• \(\displaystyle (\frac{d}{eb}, \frac{r}{b})\) equilibrium solution: We have tr\(\displaystyle (J)=r-\frac{d}{e}\) and \(\det(J)=d\). The determinant is positive, however we can see that the stability of this equilibrium solution will be dependent on the trace. If the trace is positive, or \(\displaystyle r > \frac{d}{e}\) than the equilibrium solution will be unstable. If \(\displaystyle r < \frac{d}{e}\) it will be stable. You can derive even stronger boundaries between a spiral source or spiral sink as well.

Notice how these relationships give you a sense for what you could expect in terms of a solution even before computing eigenvalues!