Problem 48 - Calculus for Biology and Medicine

Question

Two different species of organisms may compete for the same limited resource, for example, hyenas and lions may compete for territory and prey. A large lion population will reduce the growth rate of the hyena population, and vice versa. Let the number of lions be \(x\) and the number of hyenas be \(y .\) The growth rates of the lion and hyena populations may be modeled using Lotka- Volterra equations (see Chapter 11). Suppose the growth rates are given by functions: $$ \begin{array}{l} f(x, y)=0.5 x-\frac{x^{2}}{10}-\frac{x y}{20} \\ g(x, y)=2 y-\frac{y^{2}}{20}-\frac{x y}{5} \end{array} $$ (a) Show that \(f(x, y)=0\) and \(g(x, y)=0\) when \(x=0\) and \(y=40\). The two populations are said to be in equilibrium when at this size. (b) Linearize the vector valued function $$ \mathbf{h}:(x, y) \mapsto\left[\begin{array}{l} f(x, y) \\ g(x, y) \end{array}\right] $$ when \(x\) is approximately 0, and \(y\) is approximately 40 .

Step-by-Step Solution

Answer

When \(x=0\) and \(y=40\), both functions equal zero, meaning the system is at equilibrium. Linearizing at this point yields the Jacobian \(\begin{bmatrix} -1.5 & 0 \\ -8 & -2 \end{bmatrix}\).

Step 1: Finding f(0, 40)

Substitute the values of \(x = 0\) and \(y = 40\) into the function \(f(x, y) = 0.5x - \frac{x^2}{10} - \frac{xy}{20}\). This yields \(f(0, 40) = 0.5(0) - \frac{0^2}{10} - \frac{0 \cdot 40}{20} = 0\). Thus, \(f(x, y) = 0\) at \(x = 0\) and \(y = 40\).

Step 2: Finding g(0, 40)

Substitute the values of \(x = 0\) and \(y = 40\) into the function \(g(x, y) = 2y - \frac{y^2}{20} - \frac{xy}{5}\). This yields \(g(0, 40) = 2(40) - \frac{40^2}{20} - \frac{0 \cdot 40}{5} = 80 - 80 = 0\). Thus, \(g(x, y) = 0\) at \(x = 0\) and \(y = 40\).

Step 3: Jacobian Matrix Calculation

To linearize the vector-valued function \(\mathbf{h}(x, y)\), we calculate the Jacobian matrix of partial derivatives. The partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\) are \(f_x = 0.5 - \frac{2x}{10} - \frac{y}{20}\) and \(f_y = -\frac{x}{20}\), respectively. For \(g(x, y)\), they are \(g_x = -\frac{y}{5}\) and \(g_y = 2 - \frac{2y}{20} - \frac{x}{5}\).

Step 4: Evaluating Jacobian at Point (0, 40)

Substitute \(x = 0\) and \(y = 40\) into the Jacobian matrix. For \(f_x\), substitute to get \(f_x(0, 40) = 0.5 - 0 - 2 = -1.5\). For \(f_y\), substitute to get \(f_y(0, 40) = 0\). For \(g_x\), substitute to get \(g_x(0, 40) = -8\). For \(g_y\), substitute to get \(g_y(0, 40) = 2 - 4 - 0 = -2\).

Step 5: Linearized System Representation

The linearization of \(\mathbf{h}(x, y)\) about the point \((0, 40)\) using the Jacobian matrix then results in the system described by \([\begin{matrix} -1.5 & 0 \ -8 & -2 \end{matrix}](x, y)^T\). This represents small deviations from the point of equilibrium \((0, 40)\).

Key Concepts

Equilibrium

The concept of equilibrium in the context of Lotka-Volterra equations deals with finding points where the populations of competing species do not change over time.
In simpler terms, it's when the growth rates of both species become zero as they reach a stable state.
In our given problem, the equations for the growth rates of lions and hyenas are represented by functions:

• For lions: \( f(x, y) = 0.5x - \frac{x^2}{10} - \frac{xy}{20} \)
• For hyenas: \( g(x, y) = 2y - \frac{y^2}{20} - \frac{xy}{5} \)

The equilibrium point is found when both functions are zero.After calculating with \( x = 0 \) and \( y = 40 \), we see:

• The population growth of lions, \( f(x, y) = 0 \).
• The population growth of hyenas, \( g(x, y) = 0 \).

This point represents an ideal balance between the two species where the population of each does not increase or decrease. This stable state is crucial for modeling and understanding interactions in ecosystems where such competition exists.

Jacobian matrix

The Jacobian matrix is a useful mathematical tool to study the behavior of a system near equilibrium points.
In our case, it is used to linearize the Lotka-Volterra equations at the equilibrium.To construct the Jacobian matrix, partial derivatives of the functions are taken:

• Partial derivatives of \( f(x, y) \):
  • with respect to \( x \): \( f_x = 0.5 - \frac{2x}{10} - \frac{y}{20} \)
  • with respect to \( y \): \( f_y = -\frac{x}{20} \)
• Partial derivatives of \( g(x, y) \):
  • with respect to \( x \): \( g_x = -\frac{y}{5} \)
  • with respect to \( y \): \( g_y = 2 - \frac{2y}{20} - \frac{x}{5} \)

The Jacobian matrix is then:\[J = \begin{bmatrix} f_x & f_y \ g_x & g_y \end{bmatrix}\]Evaluating the Jacobian at \( x = 0 \) and \( y = 40 \), we get:

• \( f_x(0, 40) = -1.5 \)
• \( f_y(0, 40) = 0 \)
• \( g_x(0, 40) = -8 \)
• \( g_y(0, 40) = -2 \)

Thus, the Jacobian matrix becomes:\[J(0, 40) = \begin{bmatrix} -1.5 & 0 \ -8 & -2 \end{bmatrix}\]This matrix helps to understand the nature of the equilibrium point, and it forms the basis for linearization.

Linearization

Linearization simplifies a complex system near an equilibrium point into a linear system.
It's a way to approximate the system's behavior using the Jacobian matrix derived at the equilibrium.In our context, starting from the equilibrium \((0, 40)\), we look at small deviations using the linearized system. This is useful when exact solutions are difficult to compute or when we need an approximation around a known stable state.The linearized version of the original Lotka-Volterra based system is represented as:\[\mathbf{J}(x, y)^T = \begin{bmatrix} -1.5 & 0 \ -8 & -2 \end{bmatrix}(x, y)^T\]This linear system approximates how small disturbances in the lion or hyena populations would dynamically evolve around the point \((0, 40)\).
It means if the population sizes slightly shift from the equilibrium, the Jacobian matrix predicts how these gaps evolve or stabilize.By using linearization, one can better understand local stability and predict responses to initial changes within the system.