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PAGE 1

6.3 Ecological Models

Predator-prey system (Lotka-Volterra system)

\( x(t) \): prey (rabbits)

\( y(t) \): predator (wolves) predator uses prey as food source

\[ \frac{dx}{dt} = ax - pxy \]\[ \frac{dy}{dt} = -by + qxy \]
\[ a, b, p, q > 0 \]

Observations

  • Notice if there is no predator, \( \frac{dx}{dt} = ax \) exponential growth
  • If there is no prey, \( \frac{dy}{dt} = -by \) exponential decay
\[ \frac{dx}{dt} = x(a - py) \]

\( (a - py) \) acts as reduction to growth rate

\[ \frac{dy}{dt} = y(-b + qx) \]

\( (-b + qx) \) acts as boost to growth rate

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Critical Points and Jacobian

Critical points (cp): \( (0, 0), \left( \frac{b}{q}, \frac{a}{p} \right) \)

Jacobian:

\[ J(x,y) = \begin{bmatrix} a - py & -px \\ qy & -b + qx \end{bmatrix} \]

Goal: Given \( x(0), y(0) \), what happens to \( x(t), y(t) \) as \( t \to \infty \)?

Example

\[ \frac{dx}{dt} = x - 0.5xy = x(1 - 0.5y) \]\[ \frac{dy}{dt} = -0.75y + 0.25xy = y(-0.75 + 0.25x) \]

cp: (0, 0)

everybody dies

(3, 2)

stable pop.

\[ J(x,y) = \begin{bmatrix} 1 - 0.5y & -0.5x \\ 0.25y & -0.75 + 0.25x \end{bmatrix} \]
\[ J(0,0) = \begin{bmatrix} 1 & 0 \\ 0 & -0.75 \end{bmatrix} \]
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\[ \lambda = 1, -0.75 \]

Saddle

unstable

not sensitive to perturbation

\[ \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

Solutions near \( (0,0) \):

\[ \vec{x}(t) = c_1 e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{-0.75t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

If initial condition leads to \( c_1 = 0 \), then

\[ \vec{x}(t) = c_2 e^{-0.75t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} \text{ into } (0,0) \text{ along } y\text{-axis} \]

Likewise, we see solutions go away from \( (0,0) \) along \( x \)-axis.

Phase portrait showing a saddle point at (0,0) with arrows pointing in along the y-axis and out along the x-axis.

as long as \( x(0) \neq 0 \) then solutions do NOT go into \( (0,0) \)

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\[ J(3,2) = \begin{bmatrix} 0 & -1.5 \\ 0.5 & 0 \end{bmatrix} \quad \lambda = \pm \frac{\sqrt{3}}{2}i \]

center

stable

sensitive to perturbation (subject verification)

\[ \vec{v} = \begin{bmatrix} 1 \\ \pm \frac{1}{\sqrt{3}}i \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \pm i \begin{bmatrix} 0 \\ \frac{1}{\sqrt{3}} \end{bmatrix} \]
A small sketch of an elliptical orbit around a horizontal axis.
Phase portrait with concentric ellipses around a center point (3,2) on a coordinate system.

can use the flow from saddle pt to find direction of ellipse

A phase portrait showing warped, non-circular closed orbits around a fixed point in a nonlinear system.

In nonlinear picture the ellipses are generally warped.

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Predator-Prey Dynamics

\( x \) increases, then more food for \( y \) so it increases a bit later, then lots of \( y \) eating \( x \) so \( x \) declines, which means less food for \( y \) so \( y \) declines, so \( x \) has chance to increase and then the cycle repeats.

\[ \begin{aligned} x' &= ax - pxy \\ y' &= -by + gxy \end{aligned} \]
Time-series plot of prey (x) and predator (y) populations showing periodic oscillations with y lagging x.
period of variation is
\[ \frac{2\pi}{\sqrt{ab}} \]
independent of \( x(0), y(0) \)

\( y \) lags behind \( x \) by a \( \frac{1}{4} \) period

amplitude of \( x \) is
\[ \frac{kb}{g} \]
amplitude of \( y \) is
\[ \frac{k\sqrt{ab}}{p} \]
\( k \): depends on \( x(0), y(0) \)
PAGE 6

Verifying Equilibrium and Solutions

verifying \( (3, 2) \) is a center: computer graph or solve

\[ \left. \begin{aligned} x' &= ax - pxy \\ y' &= -by + gxy \end{aligned} \right\} \implies \frac{dy}{dx} = \frac{y(-b + gx)}{x(a - py)} \]
\( \vdots \)
\[ a \ln y - py + b \ln x - gx = C \]

warped ellipses.