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6.3 Ecological models

predator-prey system (Lotka-Volterra eqs)

Let:

  • x(t): population of prey (rabbits)
  • y(t): population of predator (wolves)

For constants \( a, b, p, q > 0 \):

\[ \begin{aligned} \frac{dx}{dt} &= ax - pxy \\ \frac{dy}{dt} &= -by + qxy \end{aligned} \]

What do they say?

  • If there is no \( y \), then \( \frac{dx}{dt} = ax \) — exponentially growing
  • If there is no \( x \), then \( \frac{dy}{dt} = -by \) — declines exponentially
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\[ \begin{aligned} \frac{dx}{dt} &= x(a - py) \quad \text{(wolves reduce rabbits' growth rate)} \\ \frac{dy}{dt} &= y(-b + qx) \quad \text{(rabbits boosting wolves growth rate)} \end{aligned} \]

Critical pts:

\( (0, 0) \) (both die out), and \( (\frac{b}{q}, \frac{a}{p}) \) (both at stable pop.)

Jacobian:

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

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

Example

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

Critical points (cp): \( (0, 0), (3, 2) \)

As \( t \to \infty \), what happens to \( x \)? to \( y \)?

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Stability Analysis of Linearized Systems

The Jacobian matrix for the system is given by:

\[ J = \begin{bmatrix} 1 - 0.5y & -0.5x \\ 0.25y & -0.75 + 0.25x \end{bmatrix} \]

Analysis at Equilibrium Point (0,0)

\[ J(0,0) = \begin{bmatrix} 1 & 0 \\ 0 & -0.75 \end{bmatrix} \]

Eigenvalues:

\( \lambda = 1, -0.75 \)

  • Saddle
  • Unstable
  • Not sensitive to perturbation

Eigenvectors:

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

Solution Curves near (0,0)

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

Suppose initial condition is such that \( c_1 = 0 \) then solution goes to \( (0,0) \) along y-axis.

If \( c_2 = 0 \), solutions leave \( (0,0) \) along x-axis.

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If initial \( x \) is not zero, then as \( t \to \infty \), we do NOT see both populations go to zero.

Phase portrait near origin showing saddle point behavior with trajectories approaching along y and leaving along x.

Analysis at Equilibrium Point (3,2)

\[ J(3,2) = \begin{bmatrix} 0 & -1.5 \\ 0.5 & 0 \end{bmatrix} \]

Eigenvalues:

\( \lambda = \pm \frac{\sqrt{3}}{2} i \)

  • Pure imaginary
  • Center
  • Stable
  • Sensitive to perturbation (subject to verification)

Eigenvectors:

\[ \vec{v} = \begin{bmatrix} 1 \\ \pm \frac{1}{\sqrt{3}} i \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \pm i \begin{bmatrix} 0 \\ \frac{1}{\sqrt{3}} \end{bmatrix} \]

Ellipses Centered at (3,2)

Direction consistent with saddle point.

Phase portrait showing concentric elliptical orbits centered at (3,2) in the first quadrant.
Diagram of an ellipse on a coordinate system with horizontal and vertical axes.
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More Complete Picture

The phase portrait below illustrates the relationship between prey and predator populations over time, showing closed orbits around a central equilibrium point.

Phase portrait showing concentric closed orbits of predator y versus prey x populations.

As \(x\) increases, more food is available to \(y\), so \(y\) increases. But as \(y\) increases, it eats more \(x\), so \(x\) eventually declines, which means less food for \(y\), so \(y\) declines, which then allows \(x\) to grow and the cycle repeats.

The time-series plot below shows the oscillating populations of prey and predator, highlighting that the predator population lags behind the prey population.

Time-series plot of prey and predator populations showing oscillating waves with predator lagging.

Predator lags behind prey.

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Population Dynamics Properties

  • Same period for both: \(\frac{2\pi}{\sqrt{ab}}\) independent of initial condition.
  • Phase shift (lag) is \(\frac{1}{4}\) of that.

Amplitude of \(x\) (prey) is \(\frac{Kb}{g}\)

\(K\): some # depending on initial condition

Amplitude of \(y\) (predator) is \(\frac{K\sqrt{ab}}{p}\)

Finally, to verify the CP (critical point), we plot nonlinear phase diagram or solve:

\[x' = x(a - py)\]\[y' = y(-b + gx)\]
\(\vdots\)
\[\frac{dy}{dx} = \frac{y(-b + gx)}{x(a - py)}\]
\(\vdots\)
\[a \ln y - py + b \ln x - gx = C\]

Warped ellipses