Lotka–Volterra equations

When multiplied out, the prey equation becomes

${\displaystyle {\frac {dx}{dt}}=\alpha x-\beta xy.}$

The prey are assumed to have an unlimited food supply and to reproduce exponentially, unless subject to predation; this exponential growth is represented in the equation above by the term αx. The rate of predation upon the prey is assumed to be proportional to the rate at which the predators and the prey meet, this is represented above by βxy. If either x or y is zero, then there can be no predation.

With these two terms the equation above can be interpreted as follows: the rate of change of the prey's population is given by its own growth rate minus the rate at which it is preyed upon.

The predator equation becomes

${\displaystyle {\frac {dy}{dt}}=\delta xy-\gamma y.}$

In this equation, δxy represents the growth of the predator population. (Note the similarity to the predation rate; however, a different constant is used, as the rate at which the predator population grows is not necessarily equal to the rate at which it consumes the prey). γy represents the loss rate of the predators due to either natural death or emigration, it leads to an exponential decay in the absence of prey.

Hence the equation expresses that the rate of change of the predator's population depends upon the rate at which it consumes prey, minus its intrinsic death rate.

Population dynamics for baboons and cheetahs problem mentioned aside.

Phase-space plot for the predator prey problem for various initial conditions of the predator population.

Suppose there are two species of animals, a baboon (prey) and a cheetah (predator). If the initial conditions are 10 baboons and 10 cheetahs, one can plot the progression of the two species over time; given the parameters that the growth and death rates of baboon are 1.1 and 0.4 while that of cheetahs are 0.1 and 0.4 respectively. The choice of time interval is arbitrary.

One may also plot solutions parametrically as orbits in phase space, without representing time, but with one axis representing the number of prey and the other axis representing the number of predators for all times.

This corresponds to eliminating time from the two differential equations above to produce a single differential equation

${\displaystyle {\frac {dy}{dx}}=-{\frac {y}{x}}{\frac {\delta x-\gamma }{\beta y-\alpha }}}$

relating the variables x and y. The solutions of this equation are closed curves. It is amenable to separation of variables: integrating

${\displaystyle {\frac {\beta y-\alpha }{y}}\,dy+{\frac {\delta x-\gamma }{x}}\,dx=0}$

yields the implicit relationship

${\displaystyle V=\delta x-\gamma \ln(x)+\beta y-\alpha \ln(y),}$

where V is a constant quantity depending on the initial conditions and conserved on each curve.

An aside: These graphs illustrate a serious potential problem with this as a biological model: For this specific choice of parameters, in each cycle, the baboon population is reduced to extremely low numbers, yet recovers (while the cheetah population remains sizeable at the lowest baboon density). In real-life situations, however, chance fluctuations of the discrete numbers of individuals, as well as the family structure and life-cycle of baboons, might cause the baboons to actually go extinct, and, by consequence, the cheetahs as well. This modelling problem has been called the "atto-fox problem", an atto-fox being a notional 10⁻¹⁸ of a fox.[26][27]

A less extreme example covers:

α = 2/3, β = 4/3, γ = 1 = δ. Assume x, y quantify thousands each. Circles represent prey and predator initial conditions from x = y = 0.9 to 1.8, in steps of 0.1. The fixed point is at (1, 1/2).

Population equilibrium occurs in the model when neither of the population levels is changing, i.e. when both of the derivatives are equal to 0:

${\displaystyle x(\alpha -\beta y)=0,}$

${\displaystyle -y(\gamma -\delta x)=0.}$

The above system of equations yields two solutions:

${\displaystyle \{y=0,\ \ x=0\}}$

and

${\displaystyle \left\{y={\frac {\alpha }{\beta }},\ \ x={\frac {\gamma }{\delta }}\right\}.}$

Hence, there are two equilibria.

The first solution effectively represents the extinction of both species. If both populations are at 0, then they will continue to be so indefinitely. The second solution represents a fixed point at which both populations sustain their current, non-zero numbers, and, in the simplified model, do so indefinitely. The levels of population at which this equilibrium is achieved depend on the chosen values of the parameters α, β, γ, and δ.

The stability of the fixed point at the origin can be determined by performing a linearization using partial derivatives.

The Jacobian matrix of the predator–prey model is

${\displaystyle J(x,y)={\begin{bmatrix}\alpha -\beta y&-\beta x\\\delta y&\delta x-\gamma \end{bmatrix}}.}$

and is known as the community matrix.

When evaluated at the steady state of (0, 0), the Jacobian matrix J becomes

${\displaystyle J(0,0)={\begin{bmatrix}\alpha &0\\0&-\gamma \end{bmatrix}}.}$

The eigenvalues of this matrix are

${\displaystyle \lambda _{1}=\alpha ,\quad \lambda _{2}=-\gamma .}$

In the model α and γ are always greater than zero, and as such the sign of the eigenvalues above will always differ. Hence the fixed point at the origin is a saddle point.

The stability of this fixed point is of significance. If it were stable, non-zero populations might be attracted towards it, and as such the dynamics of the system might lead towards the extinction of both species for many cases of initial population levels. However, as the fixed point at the origin is a saddle point, and hence unstable, it follows that the extinction of both species is difficult in the model. (In fact, this could only occur if the prey were artificially completely eradicated, causing the predators to die of starvation. If the predators were eradicated, the prey population would grow without bound in this simple model.) The populations of prey and predator can get infinitesimally close to zero and still recover.

Evaluating J at the second fixed point leads to

${\displaystyle J\left({\frac {\gamma }{\delta }},{\frac {\alpha }{\beta }}\right)={\begin{bmatrix}0&-{\frac {\beta \gamma }{\delta }}\\{\frac {\alpha \delta }{\beta }}&0\end{bmatrix}}.}$

The eigenvalues of this matrix are

${\displaystyle \lambda _{1}=i{\sqrt {\alpha \gamma }},\quad \lambda _{2}=-i{\sqrt {\alpha \gamma }}.}$

As the eigenvalues are both purely imaginary and conjugate to each others, this fixed point is elliptic, so the solutions are periodic, oscillating on a small ellipse around the fixed point, with a frequency ${\displaystyle \omega ={\sqrt {\lambda _{1}\lambda _{2}}}={\sqrt {\alpha \gamma }}}$ and period ${\displaystyle T=2{\pi }/({\sqrt {\lambda _{1}\lambda _{2}}})}$.

As illustrated in the circulating oscillations in the figure above, the level curves are closed orbits surrounding the fixed point: the levels of the predator and prey populations cycle and oscillate without damping around the fixed point with frequency ${\displaystyle \omega ={\sqrt {\alpha \gamma }}}$.

The value of the constant of motion V, or, equivalently, K = exp(V), ${\displaystyle K=y^{\alpha }e^{-\beta y}x^{\gamma }e^{-\delta x}}$, can be found for the closed orbits near the fixed point.

Increasing K moves a closed orbit closer to the fixed point. The largest value of the constant K is obtained by solving the optimization problem

${\displaystyle y^{\alpha }e^{-\beta y}x^{\gamma }e^{-\delta x}={\frac {y^{\alpha }x^{\gamma }}{e^{\delta x+\beta y}}}\longrightarrow \max \limits _{x,y>0}.}$

The maximal value of K is thus attained at the stationary (fixed) point ${\displaystyle \left({\frac {\gamma }{\delta }},{\frac {\alpha }{\beta }}\right)}$ and amounts to

${\displaystyle K^{*}=\left({\frac {\alpha }{\beta e}}\right)^{\alpha }\left({\frac {\gamma }{\delta e}}\right)^{\gamma },}$

where e is Euler's number.