The Lotka-Volterra model is a pair of coupled differential equations (namely, equations linked together through their dependent variables and/or their derivatives) that attempt to model populations of 2 species, namely a predator and a prey. While these equations have obvious applications in the field of biology and environmentalism, they can also be germane to any system of 2 items that regulate one another in quantity. This is the case for competing entities that are subjected to cycles (a model devised by Richard Goodwin, for instance, sought to describe the cyclical class struggle between bourgeoisie and proletariat described by Marx, and has found further application in modeling business cycles).

The equations are expressed in a very simple and elegant form:

The prey population is modeled as: \[ \frac{dx}{dt} = \alpha x - \beta xy \] The predator population is modeled as: \[ \frac{dy}{dt} = \gamma xy - \delta y \]

Let's have a closer look at the terms.

  • \(\alpha x\) refers to the birth of prey individuals. The coefficient \(\alpha\) indicates the rate of growth of the prey population.
  • \(-\beta xy\) refers to the decline of the prey population resulting from predation. The coefficient \(\beta\) refers to the rate of decline. Note the negative sign that precedes the coefficient.
  • \(\gamma xy\) refers to the predator's population growth resulting from predation.
  • \(-\delta y\) refers to the decline in predator population from natural death. Once again, note the negative sign before the \(\delta\) coefficient.

To achieve a deeper analysis of this system of differential equations, one can use a phase plane. The goal is to visualize the system's nullclines, namely the set of points where \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\).

The stationary points of the system correspond to its steady-state (where it doesn't move), found where the nullclines intersect. These intersections occur at the origin \((0,0)\) and at the point \(\left(\frac{\delta}{\gamma},\frac{\alpha}{\beta}\right)\).

(Here's the proof):

\[ \begin{aligned} \frac{dx}{dt} = 0 &\iff \alpha x - \beta xy = 0 \\ &\iff x(\alpha - \beta y) = 0 \\ &\iff x = 0 \quad \text{or} \quad y = \frac{\alpha}{\beta} \end{aligned} \] \[ \begin{aligned} \frac{dy}{dt} = 0 &\iff \gamma xy - \delta y = 0 \\ &\iff y(\gamma x - \delta) = 0 \\ &\iff y = 0 \quad \text{or} \quad x = \frac{\delta}{\gamma} \end{aligned} \]

This yields stationary points at \((0,0)\) and \(\left(\frac{\delta}{\gamma}, \frac{\alpha}{\beta}\right)\).

One must then investigate the trajectories of the system around those steady points (does the system exhibit stability, instability, etc.).

The Jacobian matrix allows us to perform that examination, and can be obtained from our system of differential equations in the following way:

\[ J = \begin{bmatrix} \frac{\partial}{\partial x}\left(\frac{dx}{dt}\right) & \frac{\partial}{\partial y}\left(\frac{dx}{dt}\right) \\ \frac{\partial}{\partial x}\left(\frac{dy}{dt}\right) & \frac{\partial}{\partial y}\left(\frac{dy}{dt}\right) \end{bmatrix} \]

Substituting for the expressions, we have:

\[ J = \begin{bmatrix} \alpha - \beta y & -\beta x \\ \gamma y & \gamma x - \delta \end{bmatrix} \]

One then evaluates the Jacobian at the stationary points.

\[ J_{(0,0)} = \begin{bmatrix} \alpha & 0 \\ 0 & -\delta \end{bmatrix} \]

We then find the matrix's eigenvalues:

\[ (\alpha - \lambda)(-\delta - \lambda) = 0 \quad \Rightarrow \quad \lambda = \alpha,\; \lambda = -\delta \]

This yields one positive and one negative eigenvalue, corresponding to a saddle point (unstable equilibrium).

Now for the second stationary point:

\[ J_{\left(\frac{\delta}{\gamma},\frac{\alpha}{\beta}\right)} = \begin{bmatrix} 0 & \frac{-\beta \delta}{\gamma} \\ \frac{\gamma \alpha}{\beta} & 0 \end{bmatrix} \]

Characteristic equation:

\[ \det(P - \lambda I) = 0 \]

(where \(\det\) denotes the determinant; solving yields eigenvalues).

Find \(\{x,y,z\} \in \mathbb{N}\) such that:

\[ x^2 + y^2 = z^2 \] \[ \begin{aligned} (a - b)^2 &= (a - b)(a - b) \\ &= a(a - b) - b(a - b) \\ &= a^2 - ab - ba + b^2 \\ &= a^2 - 2ab + b^2 \end{aligned} \]
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