Assessing the Stability of Equilibrium Points in 2D using nullclines and extreme test points: the Deer-Moose model

Finding Equilibrium Points

Let's suppose we have found the equilibrium points by setting \( D' = 0 \) and \( M' = 0 \), and solved the resulting algebra to determine which points \( (D, M) \) satisfy \[ \begin{cases} \begin{aligned} D' &= 0 \\ M' &= 0 \end{aligned} \end{cases} \]

In the deer/moose model \[ \begin{cases} \begin{aligned} D' &= 3D-MD-D^2 \\ M' &= 2M-0.5MD-M^2 \end{aligned} \end{cases} \] We calculated these as \( (D, M) = (0, 0) \), \( (D, M) = (0,2) \) , \( (D, M) = (3, 0) \) and \( (D, M) = (2, 1) \). \[ \begin{cases} \begin{aligned} {\colorbox{pink}0} &= 3D-MD-D^2 \\ {\colorbox{pink}0} &= 2M-0.5MD-M^2 \end{aligned} \end{cases} \quad\Longrightarrow\quad \begin{cases} (D, M) = \textcolor{MediumSeaGreen}{(0, 0)} &\textcolor{gray}{\text{\tiny 1st Eq}} \\ (D, M) = \textcolor{MediumSeaGreen}{(0, 2)} &\textcolor{gray}{\text{\tiny 2nd Eq}} \\ (D, M) = \textcolor{MediumSeaGreen}{(3, 0)} &\textcolor{gray}{\text{\tiny 3rd Eq}} \\ (D, M) = \textcolor{MediumSeaGreen}{(2, 1)} &\textcolor{gray}{\text{\tiny 4th Eq}} \\ \end{cases} \]

Let’s graph these:

Now, to determine the stability of these four EPs, we will use the method of nullclines and extreme points.

D-nullcline

The D-nullcline is the set of points for which \( D' = 0 \). What does this look like? Well, if \( D' = 0 \), then \( D' = \textcolor{tomato}{3D-MD-D^2 = 0} \). Notice that any point at which \( D = 0 \) satisfies this, and therefore lies on the D-nullcline. \[ D'|_{D=0} = 3D-MD-D^2|_{D=0} = 3\times 0 - M\times 0 - 0^2 = 0 \] This set of points for which D = 0 is the vertical axis of this state space. We will color it blue:

Then, if \(D \neq 0\), the expression for \(D'\) could still equal zero. Assuming \( D \neq 0 \), let’s divide by it in the \( D' = 0\) expression to get \[ \begin{cases} \begin{aligned} 0 &= 3D-MD-D^2 \\ D &\neq 0 \end{aligned} \end{cases} \quad\Longrightarrow\quad \dfrac{0}{D} = \dfrac{3D-MD-D^2}{D} \quad\Longrightarrow\quad 0 = 3 - M - D \] Rearranging this into a nice form for graphing on the \( (D, M) \) axis, we have \[ M = -D +3 \] which is the equation of a straight line of slope \(-1\) and \(M\)-intercept = 3. We will also color that line blue.

The 2 blue lines make up the D nullcline.

Therefore, one of two possibilities must be the case:

• either \( D' > 0 \) (D is increasing) in region A and \( D' < 0 \) (D is decreasing) in region B,
• or the other way around.

This question is easily answered without any elaborate computations. We just need to realize that the northeast region contains all the large values of \(M\) and \(D\), so let’s just plug in huge values for \(M\) and \(D\) and see what the sign of \(D'\) is. If we look at the \(D'\) equation: \[ D' = \underset{\text{positive term}}{3D} \quad \underset{\text{negative term}}{\textcolor{red}{-MD}} \quad \underset{\text{negative term}}{\textcolor{red}{-D^2}} \] we see that the two biggest terms, \(MD\) and \(D^2\) are both negative, so if \(M\) is large and \(D\) is large, the negative terms will overwhelm the one positive term \(3D\), and so \(D'\) must be negative in region A, and positive in region B.

M-nullcline

Now let’s turn to the \(M\) nullcline \[ M' = 2M-0.5MD-M^2 \]

If \(M' = 0\) then clearly \(M = 0\) satisfies, so the \(M\) nullcline includes the whole horizontal axis, for which \(M = 0\). We color it red.

But then if \(M \neq 0\), we can divide by both sides by \(M\) to get \[ \begin{cases} \begin{aligned} 0 &= 2M - 0.5MD - M^2 \\ M &\neq 0 \end{aligned} \end{cases} \quad\Longrightarrow\quad \dfrac{0}{M} = \dfrac{2M - 0.5MD - M^2}{M} \quad\Longrightarrow\quad 0 = 2 - 0.5D - M \] which rearranges to \( M = -0.5D +2 \), which is a line of slope \(-0.5\) and M-intercept 2. Let’s color this line red, too.

Again, the red lines separate state space into two regions, C and D. Either M is increasing in C and decreasing in D or vice versa.

The question of which is easily answered again by looking at an extreme value out in the northeast extremes. If M = 100 and D = 100 then M’ is obviously negative, so M’ must be negative in C and positive in D.