Question

These exercises explore the question "When one of two species in a colony is desirable and the other is undesirable, is it better to use resources to nurture the growth of the desirable species or to harvest the undesirable one?" Let \(x(t)\) and \(y(t)\) represent the populations of two competing species, with \(x(t)\) the desirable species. Assume that if resources are invested in promoting the growth of the desirable species, the population dynamics are given by $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x+\mu x \\ &y^{\prime}=r(1-\alpha y-\beta x) y \end{aligned} $$ If resources are invested in harvesting the undesirable species, the dynamics are $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x \\ &y^{\prime}=r(1-\alpha y-\beta x) y-\mu y \end{aligned} $$ In (10), \(r, \alpha, \beta\), and \(\mu\) are positive constants. For simplicity, we assume the same parameter values for both species. For definiteness, assume that \(\alpha>\beta>0\). Consider system (10), which describes the strategy in which resources are invested in harvesting the undesirable species. Again assume that \(\alpha>\beta>0\). (a) Determine the four equilibrium points for the system. (b) Show that it is possible, by investing sufficient resources (that is, by making \(\mu\) large enough), to prevent equilibrium coexistence of the two species. In fact, if \(\mu>r\), show that there are only two physically relevant equilibrium points. (c) Assume \(\mu>r\). Compute the linearized system at each of the two physically relevant equilibrium points. Determine the stability characteristics of the linearized system at each of these equilibrium points. (d) System (10) can be shown to be an almost linear system at each of the equilibrium points. Use this fact and the results of part (c) to infer the stability properties of system (10) at each of the two equilibrium points of interest. (e) Sketch the direction field. Will sufficiently aggressive harvesting of species \(y\) ultimately drive undesirable species \(y\) to extinction? If so, what is the limiting population of species \(x\) ?

Short answer

Short Answer: If the harvesting rate \(\mu\) is sufficiently large (i.e., \(\mu > r\)), aggressive harvesting of the undesirable species (denoted as \(y\)) will ultimately drive them to extinction. The limiting population of species \(x\) will be \(x_{limit} = \frac{1+\frac{\mu}{r}}{\alpha}\).

Step-by-step solution

Determine the equilibrium points

To find the equilibrium points, we set both \(x'(t)\) and \(y'(t)\) to 0: $$ \begin{aligned} &r(1-\alpha x-\beta y) x + \mu x = 0 \\ &r(1-\alpha y-\beta x) y - \mu y = 0 \end{aligned} $$ Let's solve this system of equations.

Solve for x in the first equation

Factor out \(x\) from the first equation to get: $$x\left(r(1-\alpha x-\beta y) + \mu\right) = 0$$ Now, either \(x=0\), or \((1-\alpha x-\beta y) + \frac{\mu}{r} = 0\).

Solve for y in the second equation

Factor out \(y\) from the second equation to get: $$y\left(r(1-\alpha y-\beta x) - \mu\right) = 0$$ Now, either \(y=0\), or \((1-\alpha y-\beta x) - \frac{\mu}{r} = 0\).

Find all possible equilibrium points

Now that we've found the relationships between x and y when the system is at equilibrium, we can find all possible equilibrium points: 1. \((x, y) = (0, 0)\), from both \(x=0\) and \(y=0\). 2. \((x, y) = \left(0, \frac{1-\frac{\mu}{r}}{\alpha}\right)\), from \(x=0\) and \((1-\alpha y-\beta x) - \frac{\mu}{r} = 0\). 3. \((x, y) = \left(\frac{1+\frac{\mu}{r}}{\alpha}, 0\right)\), from \(y=0\) and \((1-\alpha x-\beta y) + \frac{\mu}{r} = 0\). 4. \((x, y) = \left(\frac{1+\frac{\mu}{r}}{\alpha}, \frac{1-\frac{\mu}{r}}{\alpha}\right)\), from \((1-\alpha x-\beta y) + \frac{\mu}{r} = 0\) and \((1-\alpha y-\beta x) - \frac{\mu}{r} = 0\). Notice that these points are physically relevant since they're positive.

Find the conditions for preventing equilibrium coexistence

To prevent equilibrium coexistence of the two species, we only want two physically relevant equilibrium points. This is when \(\mu > r\). For the equilibrium points we found in step 4, this is satisfied when the following conditions are true: $$ \begin{aligned} &\frac{1+\frac{\mu}{r}}{\alpha} > 0 \\ &\frac{1-\frac{\mu}{r}}{\alpha} > 0 \end{aligned} $$ These conditions are both satisfied for \(\mu > r\).

Linearize the system at the equilibrium points, assume \(\mu > r\)

Compute the Jacobian matrix \(J(x,y)\) of the system: $$ J(x, y) = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} r(1 - 2\alpha x - \beta y) + \mu & r(-\beta x) \\ r(-\beta y) & r(1 - 2\alpha y - \beta x) \end{bmatrix} $$ Next, compute the linearized system at the equilibrium points \((x_2, y_2) = \left(\frac{1+\frac{\mu}{r}}{\alpha}, 0\right)\) and \((x_3, y_3) = \left(0, \frac{1-\frac{\mu}{r}}{\alpha}\right)\).

Determine stability characteristics of each equilibrium point

The stability of equilibrium points depends on the eigenvalues of the Jacobian matrix evaluated at the equilibrium points. If both eigenvalues have negative real parts, the equilibrium is stable; if at least one eigenvalue has a positive real part, the equilibrium is unstable. Evaluate the Jacobian at the two equilibrium points when \(\mu > r\) and find the eigenvalues by calculating the determinant: $$ \begin{aligned} &\text{For }(x_2, y_2): J_2 = \begin{bmatrix} - r + \mu & -\beta \left(\frac{1+\frac{\mu}{r}}{\alpha}\right) \\ 0 & r(1 + \beta \left(\frac{1+\frac{\mu}{r}}{\alpha}\right)) \end{bmatrix}, \quad \text{eigenvalues: }\lambda_{2,1} = -r + \mu, \lambda_{2,2} = r(1 + \beta \left(\frac{1+\frac{\mu}{r}}{\alpha}\right)) \\ &\\ &\text{For }(x_3, y_3): J_3 = \begin{bmatrix} r(1 - \alpha \left(\frac{1-\frac{\mu}{r}}{\alpha}\right)) + \mu & 0 \\ -\beta \left(\frac{1-\frac{\mu}{r}}{\alpha}\right) & 0 \end{bmatrix}, \quad \text{eigenvalues:}\lambda_{3,1} = r(1 - \alpha\left(\frac{1-\frac{\mu}{r}}{\alpha}\right)) + \mu, \lambda_{3,2} = 0 \\ \end{aligned} $$ As \(\mu > r\), \(\lambda_{2,1} > 0\). Thus \((x_2, y_2)\) has an unstable equilibrium. For \((x_3, y_3)\), since \(\lambda_{3,2} = 0\), the stability property cannot be determined by the linearized system.

Sketch the direction field

Using the information from previous steps, we can sketch the direction field. By the analysis, we can conclude that if the harvesting rate \(\mu\) is sufficiently large (i.e., \(\mu > r\)), aggressive harvesting of the undesirable species, \(y\), will ultimately drive them to extinction. The limiting population of species x will be \(x_{limit} = \frac{1+\frac{\mu}{r}}{\alpha}\).

Key concepts

Equilibrium Points

In the study of predator-prey dynamics, equilibrium points represent the stable populations of species where their numbers do not change over time. For our problem, we set the derivatives of the populations of the two species to zero. This gives us conditions under which neither population will increase or decrease. The system’s four equilibrium points are as follows:

• (0, 0) - Both species are extinct.
• (0, \(\frac{1-\frac{\mu}{r}}{\alpha}\)) - The desirable species is extinct, but the undesirable one is at a steady state.
• (\(\frac{1+\frac{\mu}{r}}{\alpha}\), 0) - The undesirable species is extinct, but the desirable one maintains a stable population.
• (\(\frac{1+\frac{\mu}{r}}{\alpha}\), \(\frac{1-\frac{\mu}{r}}{\alpha}\)) - Both species coexist in equilibrium.

For these points to be physically meaningful, they should be positive. If we invest sufficiently in resources (making \(\mu\) large), some equilibrium points may not be feasible, reducing potential coexistence scenarios.

Stability Analysis

Stability analysis allows us to determine whether populations near an equilibrium point will return to equilibrium or move away. In simpler terms, stability indicates whether the equilibrium is attractive or repulsive. By examining the Jacobian matrix at different equilibrium points, we can find eigenvalues, which help us assess stability.

If an equilibrium has all negative eigenvalues, it is stable, meaning that a slight disturbance will only result in temporary divergence before returning to the equilibrium. Conversely, if any eigenvalue has a positive real part, the equilibrium is unstable, meaning that disturbances will lead to larger deviations.

In our case, when investing more (\(\mu > r\)), the origin (0,0) remains unstable because small deviations can lead to population growth. For (\(\frac{1+\frac{\mu}{r}}{\alpha}\), 0), stability analysis shows mixed results, often reflecting that the desirable species prevails while the undesirable one could potentially vanish given aggressive interventions.

Linearization

Linearization is a technique to simplify nonlinear systems by approximating them around equilibrium points using linear equations. This is achieved by using the Jacobian matrix, which contains the partial derivatives of the system’s equations.

For our predator-prey model, the Jacobian matrix is calculated to assess how small perturbations affect the populations near each equilibrium. The matrix elements depend on parameters \(r\), \(\alpha\), and \(\beta\), which describe growth and interspecies competition.

Once the Jacobian matrix is evaluated at an equilibrium point, its determinant and trace help find eigenvalues. These eigenvalues are crucial in determining stability. A linearized analysis is practical because it reduces complex system behaviors near equilibrium into a manageable form, often using simpler algebraic methods.

Population Modeling

Population modeling in predator-prey dynamics involves using mathematical equations to describe changes in populations over time. The core focus is to predict behaviors based on intrinsic growth rates and interaction coefficients, such as competition and predation.

In our model, the growth rate \(r\), competition coefficients \(\alpha\) and \(\beta\), and harvesting coefficient \(\mu\) dictate how populations evolve. By manipulating these parameters, we simulate various population outcomes. For instance, increasing \(\mu\) represents aggressive management strategies that control the undesirable species, potentially driving it to extinction.

Such models assist in making informed resource allocation decisions. They suggest when nurturing a desirable species or controlling an undesirable one can maintain ecological balance. Additionally, this modeling helps predict long-term population trajectories, offering insights into stable, growing, or declining population dynamics.