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Predator-Prey Model (Lotka-Volterra)

Recall

As long as algebra is taught in school, there will be prayer in school, Cokie Roberts

Don’t panic, The Hitchhiker’s Guide to the Galaxy

Non Linear System

Differential equations

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x0, y0), meaning that it satisfies the initial condition y(x0) = y0.

This theorem ensures that under these conditions, the solution exists and is unique near x = x0.

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

Non-linear Systems

Predator-Prey Model (Lotka-Volterra)

In ecological modeling, predator-prey interactions are crucial for understanding population dynamics within ecosystems. One classic model that describes these interactions is the Lotka-Volterra equations. This model captures the cyclical nature of predator and prey populations, such as sharks (predators) and fish (prey).

Let’s define the variables and parameters:

Predator and Prey dynamics

Predator and Prey dynamics

In the absence of prey, predators (sharks) die out due to lack of food: x’ = -ax, where a > 0 is the natural rate of predators in the absence of prey. Growth due to consuming prey: x’ -ax + bxy, where b > 0 represents the rate at which predator population increases due to consuming prey.

Without predators, prey reproduce and their population grows exponentially: y′ = cy where c > 0 is the natural growth rate of prey in the absence of predators. Prey population decreases as they are consumed by predators: y′= cy − dxy where d > 0 represents the rate at which prey population decreases due to being eaten by predators.

The system of equations modeling the interaction is: $\begin{cases} x’ = -ax + bxy \\ y’ = cy -dxy \end{cases}$

Critical points occur where the population sizes do not change over time (x′= 0 and y′= 0).

Setting the Equations to Zero: $\begin{cases} x(-a + by) = 0 ~(i)\\ y(c-dx) = 0 ~(ii)\end{cases}$

If x = 0 (no predators) ⇒[(ii)] yc = 0 ⇒[c is a positive constant] y = 0 (no fish, both populations are extinct).

Otherwise, by (i) -a + by = 0 ⇒ y = ab > 0 (a and b are positive constants). By (ii) c -dx = 0 ⇒ x = cd. It only produces two critical points: (0, 0) and (cd, ab).

Critical point: (cd, ab). Both populations are positive here, representing a non-trivial equilibrium where both predator and prey coexist.

Linearizing the System Near the Origin (0, 0)

To understand what happens near (0,0), we consider small populations of predators and prey. For very small x and y, the product terms bxy and dxy are even smaller (negligible compared to terms like −ax or cy).

Linearized System $\begin{cases} x’ = -ax \\ y’ = cy \end{cases}$

Matrix Representation $(\begin{smallmatrix}x’\\ y’\end{smallmatrix})=(\begin{smallmatrix}-a & 0\\ 0 & c\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix})$

If we have a diagonal matrix, the eigenvalues are the diagonal elements, λ1 = -a < 0, λ2 = c > 0.

For complementary sake, det(A -λI) = 0 ↭ $(\begin{smallmatrix}-a-λ & 0\\ 0 & c-λ\end{smallmatrix}) = (-a -λ)(c-λ) = 0$. Hence, λ = -a, c. Given that a > 0 and c > 0, λ1 = -a is a negative eigenvalue, and λ2 = c is a positive eigenvalue. Detailed Analysis of the Matrix with Two distinct real Eigenvalues , general solution: x(t) = x0e-at, y(t) = y0ect

Since one eigenvalue is negative and the other is positive, the critical point (0,0) is a saddle point. Saddle points are unstable because trajectories move toward the critical point along the stable direction (x-axis, predator population) and away from it along the unstable direction (y-axis, prey population). (Refer to Figure iii for a visual representation and aid in understanding it).

The Trace-Determinat Plane

Simplifying the Model

For simplicity, let’s assume that a = b = c = d = 1 (set all parameters to 1). The system becomes:

$\begin{cases} x’ = -x + xy \\ y’ = y -xy \end{cases}$

Critical points with simplified parameters: (0, 0) and (1, 1).

Linearizing at the Non-Zero Critical Point (1, 1)

In order to linearize the system at the critical point (cd, ab) = (1, 1), we are going to use the Jacobian matrix:

$J = (\begin{smallmatrix}\frac{∂x’}{∂x} & \frac{∂x’}{∂y}\\ \frac{∂y’}{∂x} & \frac{∂y’}{∂y}\end{smallmatrix}) = (\begin{smallmatrix}-1+y & x\\ -y & 1-x\end{smallmatrix})$

Evaluate at (1, 1) $(\begin{smallmatrix}-1+y & x\\ -y & 1-x\end{smallmatrix})\bigg|_{(1, 1)} = (\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix})$. The Linearized system is:

$\begin{cases} x’ = y \\ y’ = -x \end{cases}$.

The characteristic equation is obtained from det(J -λI) = $(\begin{smallmatrix}-λ & 1\\ -1 & -λ\end{smallmatrix}) = λ^2+1 = 0$. Eigenvalues are λ = ±i (purely imaginary).

Detailed Analysis of the Matrix with Complex Eigenvalues , The general solution to this system (λ = α±βi) is $x(t) = e^{αt}[C_1(\begin{smallmatrix}cos(βt)\\ -sin(βt)\end{smallmatrix})+ C_2(\begin{smallmatrix}sin(βt)\\ cos(βt)\end{smallmatrix})] = [λ = ±i, α = 0, β = 1] C_1(\begin{smallmatrix}cos(t)\\ -sin(t)\end{smallmatrix})+ C_2(\begin{smallmatrix}sin(t)\\ cos(t)\end{smallmatrix})$ where c1 and c2 are constant determined by initial conditions.

Both x(t) and y(t) involve sine and cosine functions. There is no exponential growth or decay factor. The solutions are periodic with a period of 2π. The amplitude of oscillations remains constant over time. Plotting x(t) versus y(t) yields closed curves (concentric circles or ellipses centered at the origin). This is characteristic of center behavior in dynamical systems.

Since there is no exponential term (no real part in eigenvalues), trajectories neither spiral inward nor outward. The system remains in perpetual oscillation around the critical point. At (1, 1), the trajectories represent oscillations of predator and prey populations around their equilibrium values. Populations return to their starting values after each cycle.

Deriving the Second-Order ODE

Starting from the linearized system:

$\begin{cases} x’ = y \\ y’ = -x \end{cases}$.

Differentiate equation (i) with respect to t: x’’ = y’. Substitute y′ from equation (ii): x’’ = -x. We get a second-Order ODE: x’’ + x = 0.

Characteristic Equation: r2 + 1 = 0 ⇒λ = ±i, purely imaginary. This implies no real part and thus no exponential growth or decay. The solution around (1,1) will involve sines and cosines, representing periodic oscillations of predator and prey populations around their equilibrium values:

General Solution: x(t) = Acos(t) + Bsin(t). Since y = x’ (i), y(t) = -Asin(t) + Bcos(t). Such a system, with purely imaginary eigenvalues, is called a center.

Trace-Determinant Analysis

Characteristic Equation: λ2-Tλ + D = 0 where T is the trace Trace (sum of diagonal elements) and D is the determinant of the Jacobian matrix.

At (1, 1): Jacobian Matrix: $(\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix})$, T = 0, D = 1, Δ = T2-4D = 02-4·1 = -4.

Interpretation

If a small parameter change makes the trace positive (T > 0), the equilibrium would become a spiral source (unstable focus), as eigenvalues gain a positive real part. If a small parameter change makes the trace negative (T < 0), the equilibrium would become a spiral sink (stable focus), as eigenvalues gain a negative real part.

$λ_{1, 2} = \frac {T}{2}±\frac{\sqrt -Δ}{2}$

Summarize:

The Trace-Determinat Plane

Centers are across the line T = 0. A small change of the parameters, I might get a sink or a spiral source. In other words, you cannot predict the behaviour when you change a little its parameters (The behavior near (1, 1) is sensitive to changes in the parameters a, b, c, d). The non-linear system at (1, 1) might exhibit center behaviour, a spiral sink, or a spiral source depending on parameter values.

$\begin{cases} x’ = -ax + bxy ~(i)\\ y’ = cy -dxy ~(ii)\end{cases}$

Another approach: Eliminating time to find trajectories

We begin with a two-dimensional nonlinear system of differential equations representing, for example, a predator-prey model:

$\begin{cases} x’ = -x + xy ~(i)\\ y’ = y -xy ~(ii)\end{cases}$

By rearranging, we can factor certain terms where it shows the system’s nonlinear nature:

$\begin{cases} x’ = x(-1 + y) ~(i)\\ y’ = y(1 -x) ~(ii)\end{cases}$

To find the trajectories (relationship between x and y), eliminate time t. Divide (ii) and (i): $\frac{dy}{dx} = \frac{y’}{x’} = \frac{y(1-x)}{x(-1+y)} = \frac{y(1-x)}{x(y-1)}$

Separate variables: $\frac{y-1}{y}dy = \frac{1-x}{x}dx ↭[\text{Simplify}] (1-\frac{1}{y})dy = (\frac{1}{x}-1)dx$

Integrating Both Sides: $\int (1-\frac{1}{y})dy = \int (\frac{1}{x}-1)dx ⇒ y -ln(y) = ln(x) -x + c_1$

Let’s exponentiate: $e^{y -ln(y)} = \frac{e^y}{e^{ln(y)}} = \frac{e^{ln(x)}}{e^x}·e^{c_1} ⇒\frac{e^y}{y} = \frac{x}{e^x}c_2 ⇒ \frac{x}{e^x}·\frac{y}{e^y} = c, c = \frac{1}{c_2}$ The integral curves (trajectories of the system) are graphs of this equation.

These are contour curves of $xe^{-x}·ye^{-y} = h(x, y)$ (Refer to Figure i for a visual representation and aid in understanding it)

Relation Between Non-linear Systems and First-order ODE’s

Consider the function $f(u) = ue^{-u}$. This function reaches its maximum at 1 where f(1) = 1e. Similarly, $ve^{-v}$ has a maximum at 1, too. h(x, y) is the product of both functions, h(x, y) = f(x)·f(y), hence h(x, y) has a maximum at (1, 1). At (1, 1), the value is $h(1, 1) = (\frac{1}{e})^2$, that is, as a figure of speech, the peak of the mountain.

As we move away from (1, 1), either by letting x or y go to 0 or approach infinity, h(x,y) tends to zero. Thus, the level set (contour curves) h(x, y) = $xe^{-x}·ye^{-y} = C$ formed closed curves around (1, 1). These are concentric closed curves that move clockwise around (1, 1). These represent the closed orbits in the phase plane, confirming the oscillatory behavior of the system (populations oscillate around the equilibrium point (1, 1)).

Population dynamic.

  1. High Predator, Low Prey: When predator numbers (x) are high and prey (y) are low, predators quickly consume the remaining prey. Prey population declines further due to intense predation. Eventually, with not enough prey to sustain them, predators also decline.
  2. Low Predator, High Prey: If predator numbers are low, predation pressure is reduced. The prey population can grow more easily because fewer are eaten. This increase in prey, in turn, supports a recovery in the predator population since there is now more food available.
  3. Repeatedly cycling through these scenarios leads to an endless cycle of population rises and falls. The system’s dynamics reflects natural population oscillations observed in predator-prey relationships. The system’s dynamics show that neither population settles to a constant nonzero value; instead, they oscillate around the equilibrium.

Introducing External Factors: Fishing with a Constant Rate k

Now consider modifying the system to include an external harvesting (fishing with nets) at a constant rate k:

$\begin{cases} x’ = -ax + bxy -kx ~(i)\\ y’ = cy -dxy -ky ~(ii)\end{cases}$ where the terms −kx and -ky represents additional deaths due to fishing.

Combining terms:

$\begin{cases} x’ = -(a+k)x + bxy ~(i)\\ y’ = (c-k)y -dxy ~(ii)\end{cases}$

New Critical Point:

  1. From x′ = 0: -(a+k)x + bxy = 0 ⇒x[−(a+k)+by]=0 ⇒ x = 0 or y = $\frac{a+k}{b}$
  2. From y’ = 0: cy -dxy -ky = 0 ⇒ y[(c−k)−dx]=0 ⇒ y = 0 or x = $\frac{c-k}{d}$

The old critical point was $(\frac{c}{d}, \frac{a}{b})$. Critical point shifts to: $(\frac{c-k}{d}, \frac{a+k}{b})$ (Refer to Figure ii for a visual representation and aid in understanding it).

Relation Between Non-linear Systems and First-order ODE’s

Effect on Populations:

The end result of fishing with nets is that it reduces the predator (shark) population and increases the prey (fish) population.

Fishing removes prey and predators directly (-ky, -kx). Reduced predator numbers lead to fewer prey being eaten. The indirect positive effect outweighs the direct negative effect, resulting in a higher prey and lower predator population at equilibrium.

Statement of Volterra’s Principle: If both predator and prey populations are subjected to a proportional external mortality (e.g., fishing, pollution) that affects them equally, the prey population will increase on average, while the predator population will decrease on average.

The Lotka-Volterra equations provide valuable insights into predator-prey dynamics, demonstrating how interactions between species can lead to oscillatory behaviors. By analyzing the system’s critical points, linearizing near these points, and understanding the effects of external factors like fishing, we gain a deeper appreciation for the complex balance within ecosystems.

Analogy: Mosquito Plague and Pesticides

Suppose a fish pond experiences a mosquito infestation. What should we do? Let’s assume that the decision is to spray pesticides (e.g., DDT) to kill those pesky mosquitoes.

Mosquito Plague and Pesticides

Unintended Consequences. The result is quite the same as before because:

  1. Pesticides harm both mosquitoes and fish. When pesticides are applied, they do not discriminate between target and non-target species. Both mosquitoes and fish in the pond are affected, leading to unintended mortality in fish populations.
  2. Fish are natural predators of mosquitoes. Fish help control mosquito populations by preying on their larvae. Reducing the fish population diminishes this natural control mechanism.
  3. Reducing the fish population can lead to a mosquito population increase due to reduced predation. With fewer fish to consume mosquito larvae, mosquito numbers can rebound or even increase, exacerbating the original infestation problem.

In our analysis, the equilibrium (critical point) moves in the same direction, so the population of mosquitoes (prey) increases as the population of fish (predator) decreases, hence spraying pesticides may worsen the mosquito problem. This analysis highlights the complexity of ecological interventions.

Interventions like fishing or pesticide use can have profound and sometimes unintended effects. Careful analysis and consideration of ecological interactions are essential for sustainable management.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
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  8. MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
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