Problem 20 - Advanced Engineering Mathematics

Question

The Lotka-Volterra predator-prey model assumes that, in the absence of predators, the number of prey grows exponentially. If we make the alternative assumption that the prey population grows logistically, the new system is $$ \begin{aligned} &x^{\prime}=-a x+b x y \\ &y^{\prime}=-c x y+\frac{r}{K} y(K-y), \end{aligned} $$ where \(a, b, c, r\), and \(K\) are positive and \(K>a / b\). (a) Show that the system has critical points at \((0,0),(0, K)\), and \((\hat{x}, \hat{y})\), where \(\hat{y}=a l b\) and \(c \hat{x}=\frac{r}{K}(K-\hat{y})\). (b) Show that the critical points at \((0,0)\) and \((0, K)\) are saddle points, whereas the critical point at \((\hat{x}, \hat{y})\) is either a stable node or a stable spiral point. (c) Show that \((\hat{x}, \hat{y})\) is a stable spiral point if \(\hat{y}<\frac{4 b K^{2}}{r+4 b K}\). Explain why this case will occur when the carrying capacity \(K\) of the prey is large.

Step-by-Step Solution

Answer

The system has critical points \((0,0), (0, K), (\hat{x}, \hat{y})\). Each is a saddle except for \((\hat{x}, \hat{y})\), which is stable if certain conditions are met.

Step 1: Identify the Critical Points

To find the critical points of the system, we set each of the derivatives to zero:For the first equation, \(x' = -ax + bxy = 0\), we factor out \(x\) which gives \(-a + by = 0\). Hence, \(y = \frac{a}{b}\) if \(x eq 0\), and \(x = 0\) otherwise.For the second equation, \(y' = -cxy + \frac{r}{K}y(K-y) = 0\), we factor out \(y\) which gives \(-cx + \frac{r}{K}(K - y) = 0\). Hence, \(x = \frac{r}{cK}(K - y)\) if \(y eq 0\), and \(y = 0\) otherwise.Setting these equations, we solve for the critical points as follows:- \((0, 0)\) when both \(x = 0\) and \(y = 0\).- \((0, K)\) when \(x = 0\) and \(y = K\).- \((\hat{x}, \hat{y})\) where \(\hat{y} = \frac{a}{b}\) and \(c\hat{x} = \frac{r}{K}(K - \hat{y})\).

Step 2: Determine Stability Using Jacobian Matrix

To determine the nature of each critical point, we compute the Jacobian matrix:\[J = \begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y}\end{pmatrix} = \begin{pmatrix} -a + by & bx \ -cy & -cx - \frac{r}{K}(2y - K)\end{pmatrix}\]- At \((0,0)\), substituting \(x=0, y=0\) into \(J\), we find the eigenvalues to determine saddle property.- At \((0,K)\), substituting \(x=0, y=K\) into \(J\), we find the eigenvalues; since one is positive and one is negative, it's a saddle point.

Step 3: Analyze the Critical Point \((\hat{x}, \hat{y})\)

For the critical point \((\hat{x}, \hat{y})\), we substitute \(\hat{x}\) and \(\hat{y}\) into the Jacobian:The trace and determinant of the Jacobian determine the type of stability. The trace condition (\(tr(J) < 0\)) and determinant condition (\(det(J) > 0\)) indicate stability. Solving these with conditions on \(\hat{y}\) gives the type of node or spiral.

Step 4: Determine Spiral Point Condition

For \((\hat{x}, \hat{y})\) to be a stable spiral, we need specific conditions on the trace and determinant of the Jacobian. Calculate the specific condition:The critical point is a stable spiral if \(\hat{y} < \frac{4bK^2}{r + 4bK}\). Explain that a large carrying capacity \(K\) results in this condition being more likely fulfilled, often occurring when prey resources are abundant, causing \(c\hat{x}\) and \(\hat{y}\) values favoring spiral dynamics.

Key Concepts

Predator-Prey Dynamics

The Lotka-Volterra model is famous for describing the interactions between predators and prey through mathematical equations. This model is a dynamic system that observes how the populations of predators and prey affect each other over time. The idea behind this model is that predator populations depend on prey populations for food and energy. In the simplest terms, when prey is abundant, the predator population can grow since there is more food available.
However, if predator numbers grow too high, the prey population may decrease because they are being consumed at a greater rate. This, in turn, can result in a decrease in predators as their food source diminishes, creating a cyclical dynamic between these two population groups.
In the variations of the model discussed in the exercise, logistic growth is applied to the prey population rather than exponential. This assumes that environmental factors impose a carrying capacity, known as the maximum population size that the environment can sustain indefinitely given its resources.

Critical Points

Critical points in a differential equations system are values where the change in population is zero. In mathematical terms, these are found by setting derivatives of the system's equations to zero. These points are important as they reveal the system's equilibrium states where neither the predator nor prey populations grow or shrink.

• At the origin, (0,0), both predator and prey population levels are zero.
• At (0, K), the prey population is at its carrying capacity with no predators.
• At another point (\(\hat{x}, \hat{y}\)), populations are balanced at certain non-zero levels.

Critical points help to understand how a system can behave under different scenarios. In this exercise, identifying these points involves solving equations by setting predator and prey growth rates to zero and determining the values of populations where this occurs.

Stability Analysis

When analyzing critical points, understanding their stability is essential. Stability analysis tells us whether a small change in population around these points will return to the equilibrium or move away. If the system returns to equilibrium, the point is stable; if not, it is unstable.
In this exercise, through the Jacobian matrix, the stability of critical points is determined by examining the sign of the eigenvalues.

• Stable points have negative eigenvalues that ensure small disturbances diminish over time.
• Unstable points have at least one positive eigenvalue, causing disturbances to increase.

Points (0,0) and (0,K) are identified as saddle points in this exercise, which are typically unstable because they have a mix of positive and negative eigenvalues. The stability analysis segments provide insights into how populations would react to small environmental changes or other disturbances near these critical points.

Jacobian Matrix

The Jacobian matrix is a mathematical tool used to linearize a non-linear system near critical points. In the context of the Lotka-Volterra model, this matrix helps to decipher the type and stability of each critical point.
Constructed from the system's partial derivatives, the Jacobian collects information about how each state variable (e.g., prey or predator population) affects the rates of change of the others. This matrix helps calculate eigenvalues, which are critical in determining stability. - The matrix: \[ J = \begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{pmatrix} \]By evaluating the trace and determinant of the Jacobian at each critical point, we can infer whether the point is stable, unstable, a node, or a spiral. This transforms an otherwise complex non-linear system into a manageable linear approximation, offering clarity into the system's behavior dynamics.