Problem 47 - Calculus for Biology and Medicine

Question

The Lotka-Volterra equations are often used to model the links between a particular of prey organisms (e.g., sardines) and a population of predatory organisms (e.g., sharks), (see Chapter 11.) In a particular ecosystem we will use \(u\) to represent the number of sharks and \(v\) to represent the number of sardines. Suppose the growth rate of the shark population is $$ f(u, v)=-0.5 u+\frac{u v}{100} $$ and of the sardine population is $$ g(u, v)=3 v-10 u v $$ (a) Show that if \(u=0.3\) and \(v=50\), then \(f(u, v)=0\), and \(g(u, v)=0\). (The populations are said to be in equilibrium.) (b) Find the linear approximation of the vector valued function $$ \mathbf{h}:(u, v) \mapsto\left[\begin{array}{l} f(u, v) \\ g(u, v) \end{array}\right] $$ if \(u\) is close to \(0.3\) and \(v\) is close to 50 .

Step-by-Step Solution

Answer

At \((0.3, 50)\), both populations are in equilibrium since \(f(0.3, 50) = 0\) and \(g(0.3, 50) = 0\). The linear approximation near this point is determined by the Jacobian matrix.

Step 1: Calculate f(u, v) at (0.3, 50)

To find \( f(u, v) \) at \( u = 0.3 \) and \( v = 50 \), substitute these values into the function: \[ f(0.3, 50) = -0.5 \times 0.3 + \frac{0.3 \times 50}{100} \]. Calculate: \[ -0.5 \times 0.3 = -0.15 \], and \[ \frac{0.3 \times 50}{100} = 0.15 \]. Therefore, \[ f(0.3, 50) = -0.15 + 0.15 = 0 \].

Step 2: Calculate g(u, v) at (0.3, 50)

Next, calculate \( g(u, v) \) at \( u = 0.3 \) and \( v = 50 \) by substituting in the given values: \[ g(0.3, 50) = 3 \times 50 - 10 \times 0.3 \times 50 \]. Calculate: \[ 3 \times 50 = 150 \] and \[ 10 \times 0.3 \times 50 = 150 \]. Thus, \[ g(0.3, 50) = 150 - 150 = 0 \].

Step 3: Function is in equilibrium

Since \( f(0.3, 50) = 0 \) and \( g(0.3, 50) = 0 \), the populations are in equilibrium at these values.

Step 4: Find the Jacobian matrix

The Jacobian matrix for the system of equations \( \mathbf{h}(u, v) = [f(u, v), g(u, v)]^T \) at \( (0.3, 50) \) involves partial derivatives. Compute \[ \frac{\partial f}{\partial u} = -0.5 + \frac{v}{100}, \qquad \frac{\partial f}{\partial v} = \frac{u}{100} \]. Compute \[ \frac{\partial g}{\partial u} = -10v, \qquad \frac{\partial g}{\partial v} = 3 - 10u \].

Step 5: Evaluate the Jacobian matrix at (0.3, 50)

Substitute \( u = 0.3 \) and \( v = 50 \) into the partial derivatives: \[ \frac{\partial f}{\partial u} = -0.5 + \frac{50}{100} = 0 \], \[ \frac{\partial f}{\partial v} = \frac{0.3}{100} = 0.003 \]. For \( g(u, v) \): \[ \frac{\partial g}{\partial u} = -10 \times 50 = -500 \], \[ \frac{\partial g}{\partial v} = 3 - 10 \times 0.3 = 0 \]. The Jacobian is: \[ \begin{bmatrix} 0 & 0.003 \ -500 & 0 \end{bmatrix} \].

Step 6: Linear approximation function

The linear approximation of \( \mathbf{h}(u, v) \) near \( (0.3, 50) \) is given by: \[ \mathbf{h}(u, v) \approx \mathbf{h}(0.3, 50) + J(0.3, 50) \cdot \begin{bmatrix} u - 0.3 \ v - 50 \end{bmatrix} \], where \( J(0.3, 50) \) is the Jacobian matrix. Therefore, the linear approximation is: \[ \begin{bmatrix} 0 \ 0 \end{bmatrix} + \begin{bmatrix} 0 & 0.003 \ -500 & 0 \end{bmatrix} \cdot \begin{bmatrix} u - 0.3 \ v - 50 \end{bmatrix} \].

Key Concepts

Equilibrium Points

In the context of the Lotka-Volterra model, equilibrium points are where the population sizes remain steady over time. This means there is no net growth or decline in the populations of predators and prey. In mathematical terms, this occurs when both functions representing the growth rates of sharks and sardines equal zero. For the given equations:

• The sharks' growth rate function is denoted by \( f(u, v) = -0.5u + \frac{uv}{100} \).
• The sardines' growth rate function is \( g(u, v) = 3v - 10uv \).

By substituting \( u = 0.3 \) and \( v = 50 \) into these functions, we obtain \( f(0.3, 50) = 0 \) and \( g(0.3, 50) = 0 \), confirming that the populations are in equilibrium at this point. This specific situation shows no change in the populations, meaning the system has reached a balanced state where the number of sharks and sardines stabilizes over time. This concept is fundamental in understanding dynamic systems and ecological models.

Jacobian Matrix

The Jacobian matrix is a crucial component in analyzing the behavior of a system near its equilibrium point. It provides a linear representation of how small changes in the variables affect the system's functions. For a system defined by two functions, the Jacobian matrix is a 2x2 matrix of partial derivatives.

• Calculate the partial derivatives of each function with respect to both \( u \) and \( v \).
• The matrix entries for \( f(u, v) \) are \( \frac{\partial f}{\partial u} = -0.5 + \frac{v}{100} \) and \( \frac{\partial f}{\partial v} = \frac{u}{100} \).
• For \( g(u, v) \), they are \( \frac{\partial g}{\partial u} = -10v \) and \( \frac{\partial g}{\partial v} = 3 - 10u \).

At the equilibrium point \( (0.3, 50) \), substituting these values results in the Jacobian matrix: \[\begin{bmatrix}0 & 0.003 \-500 & 0\end{bmatrix}\]This matrix helps in determining the local dynamics around the equilibrium, allowing insights into stability and potential oscillations in population sizes.

Linear Approximation

Linear approximation is a method used to estimate the behavior of a function near a given point using its tangent line. In the case of two-variable functions, as in our predator-prey model, we employ the linear approximation to understand how small deviations from equilibrium affect the system. By using the Jacobian matrix evaluated at the equilibrium point \( (0.3, 50) \), we construct a first-order approximation of the system. The formula for linear approximation is: \[\mathbf{h}(u, v) \approx \mathbf{h}(0.3, 50) + J(0.3, 50) \cdot \begin{bmatrix} u - 0.3 \ v - 50 \end{bmatrix}\]This means that the function \( \mathbf{h}(u, v) \) can be approximated as a linear transformation of deviations \( (u-0.3, v-50) \) from the equilibrium point. Such an approximation provides a simplified view of the system's response to small changes, which is useful for understanding local behavior and assessing stability.