17.5 Predator prey with logistic growth

Let’s take a look at another model developed from the lynx-hare system. We assumed that the hare grow exponentially (notice the term $rH$ in their equation.) However we can modify their growth rate to be a logistic growth function with carrying capacity $K$:

$$\begin{equation} \begin{split} \frac{dH}{dt} &= r H \left( 1- \frac{H}{K} \right) - b HL \\ \frac{dL}{dt} &=ebHL -dL \end{split} \tag{17.4} \end{equation}$$

Through a rescaling of Equation (17.4) with the variables $\displaystyle x=\frac{H}{K}$, $\displaystyle y=\frac{L}{r/b}$ and $T = r t$ we can rewrite Equation (17.4) as:

$$\begin{equation} \begin{split} \frac{dx}{d T} &= x(1-x) - xy \\ \frac{dy}{d T} &=\frac{ebK}{r}xy -\frac{d}{r}y \end{split} \tag{17.5} \end{equation}$$

In order to analyze the Jacobian matrix for Equation (17.5) we will need to compute several partial derivatives:

$$\begin{equation} \begin{split} \frac{\partial}{\partial x} \left( f(x,y) \right) &= \frac{\partial}{\partial x} \left( x(1-x) - xy \right) = 1-2x-y \\ \frac{\partial}{\partial y} \left( f(x,y) \right) &= \frac{\partial}{\partial y} \left( x(1-x) - xy \right) = -x \\ \frac{\partial}{\partial x} \left( g(x,y) \right) &= \frac{\partial}{\partial x} \left( \frac{ebK}{r}xy -\frac{d}{r}y \right) = \frac{ebK}{r}y \\ \frac{\partial}{\partial y} \left( g(x,y) \right) &= \frac{\partial}{\partial y} \left( \frac{ebK}{r}xy -\frac{d}{r}y \right) = \frac{ebK}{r}x -\frac{d}{r} \end{split} \end{equation}$$

So now we can construct the Jacobian matrix:

$$\begin{equation} J_{(x,y)} = \begin{pmatrix} 1-2x-y & -x \\ \frac{ebK}{r}y & \frac{ebK}{r}x -\frac{d}{r} \end{pmatrix} \end{equation}$$

The notation $J_{(x,y)}$ signifies the Jacobian matrix evaluated at the equilibrium solution $(x,y)$. Sometimes computing the Jacobian matrix is a good first step so then you are ready to compute the equilibrium solutions. In the exercises you will determine equilibrium solutions and visualize the Jacobian matrix.