Problem 10 - Calculus for Biology and Medicine

Question

Use the stability criterion to characterize the stability of the equilibria of $$ x_{t+1}=\frac{3}{5} x_{t}^{2}-\frac{2}{5}, \quad t=0,1,2, \ldots $$

Step-by-Step Solution

Answer

Both equilibria \(x = \frac{4}{3}\) and \(x = -1\) are unstable.

Step 1: Identify the equilibrium points

To find the equilibrium points of the function, solve the equation \(x = \frac{3}{5}x^2 - \frac{2}{5}\). Set \(x_{t+1} = x_t = x\) for equilibrium. Rearranging gives the quadratic equation \(\frac{3}{5}x^2 - x - \frac{2}{5} = 0\). Solve for \(x\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = \frac{3}{5}\), \(b = -1\), and \(c = -\frac{2}{5}\).

Step 2: Solve the quadratic equation

Substitute \(a\), \(b\), and \(c\) into the quadratic formula: \[x = \frac{1 \pm \sqrt{1^2 - 4\left(\frac{3}{5}\right)\left(-\frac{2}{5}\right)}}{2\left(\frac{3}{5}\right)}\]. Simplify under the square root: \(1 + \frac{24}{25} = \frac{49}{25}\). Take the square root: \(\sqrt{\frac{49}{25}} = \frac{7}{5}\). Compute: \[x = \frac{1 \pm \frac{7}{5}}{\frac{6}{5}}\]. This results in two equilibrium points: \(x = \frac{8}{6} = \frac{4}{3}\) and \(x = \frac{-6}{6} = -1\).

Step 3: Determine stability using derivatives

The stability of an equilibrium point of a discrete dynamical system \(x_{t+1} = f(x_t)\) is determined by the derivative of \(f(x)\). Compute \(f'(x)\) by differentiating \(f(x) = \frac{3}{5}x^2 - \frac{2}{5}\): \(f'(x) = \frac{6}{5}x\).

Step 4: Evaluate the derivative at the equilibrium points

At \(x = \frac{4}{3}\), compute \(f'(\frac{4}{3}) = \frac{6}{5} \times \frac{4}{3} = \frac{24}{15} = \frac{8}{5}\). Since \(\left|f'(\frac{4}{3})\right| = \frac{8}{5} > 1\), this equilibrium is unstable. At \(x = -1\), compute \(f'(-1) = \frac{6}{5} \times (-1) = -\frac{6}{5}\). Since \(\left|f'(-1)\right| = \frac{6}{5} > 1\), this equilibrium is also unstable.

Key Concepts

Quadratic Equation

A quadratic equation is an essential algebraic expression which takes the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants where \( a eq 0 \). The quadratic equation might sound complex, but it follows a simple form and is notably simple to solve using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our scenario, we worked on solving the equation \( \frac{3}{5}x^2 - x - \frac{2}{5} = 0 \), where the coefficients are identified as:

• \( a = \frac{3}{5} \)
• \( b = -1 \)
• \( c = -\frac{2}{5} \)

By substituting these into the quadratic formula, you can find the solutions \( x = \frac{4}{3} \) and \( x = -1 \). These solutions are critical, as they represent values where the function equals zero and help us find equilibrium points.

Equilibrium Points

Equilibrium points are key in analyzing dynamical systems, representing states where a system remains constant over time. To locate equilibrium points in a function like \( x_{t+1} = f(x_t) \), we set \( x = f(x) \). Here, we solved the equation \( x = \frac{3}{5}x^2 - \frac{2}{5} \) for \( x \), which is the condition for equilibrium.

Solving this involves:

• Balancing the formula such that \( x_{t+1} = x_t \), essentially seeking where the input equals the output.
• Turning the condition into a quadratic equation, which we solve to find potential equilibrium points.

In our exercise, solving this gave us two equilibrium points: \( x = \frac{4}{3} \) and \( x = -1 \). These values are the potential resting points of the system, and determining their stability is the next step.

Discrete Dynamical System

A discrete dynamical system is a framework where changes occur at distinct points in time—think of steps or hops from one state to another. In mathematics, these systems are represented by equations like \( x_{t+1} = f(x_t) \).

Such systems are particularly useful in modeling real-world processes where events unfold in stages or periods. For instance, population models and economic forecasts apply this concept. In our example, the function \( x_{t+1} = \frac{3}{5} x_{t}^{2} - \frac{2}{5} \) describes how \( x \), the state, evolves over time.

Understanding stability within these systems involves using the derivative of \( f(x) \) to assess the behavior at equilibrium points:

• At \( x = \frac{4}{3} \) and \( x = -1 \), we calculated the derivative \( f'(x) = \frac{6}{5}x \).
• We evaluated the derivative at these points to determine stability. A magnitude greater than one, \(|f'(x)| > 1\), suggests instability.

This approach showed both equilibrium points in our problem are unstable, indicating that small deviations would lead to larger shifts away from equilibrium.