Problem 5 - Calculus for Biology and Medicine

Question

Find the equilibria of $$x_{t+1}=\frac{3}{5} x_{t}^{2}-\frac{2}{5}, \quad t=0,1,2, \ldots$$ and use the stability criterion for an equilibrium point to determine whether they are stable or unstable.

Step-by-Step Solution

Answer

Equilibrium points are \( x^* = 2 \) (unstable) and \( x^* = -\frac{1}{3} \) (stable).

Step 1: Define Equilibrium

An equilibrium point of the equation is where the value of the function does not change over time. In mathematical terms, this means finding the value of \( x \) such that \( x_{t+1} = x_t = x^* \) (where \( x^* \) is an equilibrium point). For the given equation, set \( x_{t+1} = x_t \) to find equilibrium points.

Step 2: Set the Equation for Equilibrium

Set the equation for equilibrium by equating \( x_{t+1} \) to \( x_t \):\[ x^* = \frac{3}{5} (x^*)^2 - \frac{2}{5} \] Simplify the equation by multiplying all terms by 5 to eliminate the denominator:\[ 5x^* = 3(x^*)^2 - 2 \] Rearrange it to form a standard quadratic equation:\[ 3(x^*)^2 - 5x^* - 2 = 0 \]

Step 3: Solve the Quadratic Equation

Use the quadratic formula to solve for \( x^* \) in the equation:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 3 \), \( b = -5 \), and \( c = -2 \). Substitute these values into the formula:\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-2)}}{6} \] This simplifies to:\[ x = \frac{5 \pm \sqrt{25 + 24}}{6} \]\[ x = \frac{5 \pm \sqrt{49}}{6} \]\[ x = \frac{5 \pm 7}{6} \] This gives the solutions \( x^* = 2 \) and \( x^* = -\frac{1}{3} \), which are the equilibrium points.

Step 4: Determine Stability of Equilibrium Points

To determine the stability, compute the derivative of \( f(x) = \frac{3}{5}x^2 - \frac{2}{5} \) as \( f'(x) \). This is given by:\[ f'(x) = \frac{d}{dx}\left(\frac{3}{5}x^2 \right) = \frac{6}{5}x \]Evaluate \( f'(x) \) at each equilibrium point. For stability, \(|f'(x^*)| < 1\).

Step 5: Evaluate Stability at Equilibrium Points

For \( x^* = 2 \):\[ f'(2) = \frac{6}{5} \cdot 2 = \frac{12}{5} = 2.4 \]Since \(|2.4| > 1\), this equilibrium is unstable.For \( x^* = -\frac{1}{3} \):\[ f'(-\frac{1}{3}) = \frac{6}{5} \cdot (-\frac{1}{3}) = -\frac{6}{15} = -\frac{2}{5} \]Since \(|-\frac{2}{5}| < 1\), this equilibrium is stable.

Key Concepts

Understanding Quadratic Equations in Equilibrium

At the heart of this exercise is the quadratic equation. This form of equation is central to many mathematical and real-world applications. A quadratic equation is generally expressed in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In our specific problem, this is formed during the equilibrium determination stage.
We derived the equation \( 3(x^*)^2 - 5x^* - 2 = 0 \) by setting the equilibrium condition \( x_{t+1} = x_t \), where \( x_t \) is a point that doesn't change over time. This standard quadratic format makes it easier to use the quadratic formula to find solutions.
This formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is a powerful tool for solving equations of the second degree. Used correctly, it provides all potential solutions for \( x \), as seen with roots \( x^* = 2 \) and \( x^* = -\frac{1}{3} \).

The Stability Criterion Explained

Once equilibrium points are found, the next step involves determining their stability. It is essential to understand if these points are stable or unstable because it tells us about the behavior of a system over time. The stability criterion defines stability by examining the absolute value of the derivative \(|f'(x^*)|\) at the equilibrium point.
The rule is simple yet vital:

• If \(|f'(x^*)| < 1\), the equilibrium point is stable. This means any small disturbance will cause the system to return to equilibrium.
• If \(|f'(x^*)| > 1\), the point is unstable, indicating that the system will diverge away with disturbances.

In our scenario, calculating the derivative and evaluating it at equilibrium points helps in applying this criterion.

The Role of Derivatives in Stability Analysis

Derivatives are a fundamental part of calculus, describing how a function changes. In the context of stability, derivatives allow us to measure the sensitivity of the function \( f(x) \) at equilibrium points. For a function \( f(x) = \frac{3}{5}x^2 - \frac{2}{5} \), its derivative is
\[ f'(x) = \frac{d}{dx}\left(\frac{3}{5}x^2 \right) = \frac{6}{5}x \]
This derivative provides the rate of change of the function with respect to \( x \).
By evaluating this derivative at each equilibrium point, as done with \( x^* = 2 \) and \( x^* = -\frac{1}{3} \), we can check the stability. If the absolute value of \( f'(x^*) \) is less than 1, the point is stable; otherwise, it is unstable.

Determining Equilibrium Points

To understand where a system remains unchanged over iterations, we find equilibrium points. These points mean that the value of the system at one time is equal to that at the next time: \( x_{t+1} = x_t = x^* \). It's like finding the root of a fixed point function.
For our function \( x_{t+1} = \frac{3}{5}x_t^2 - \frac{2}{5} \), equating it and setting \( x_{t+1} = x_t \), we derived
\[ x^* = \frac{3}{5}(x^*)^2 - \frac{2}{5} \]
This simplifies further into the quadratic form \( 3(x^*)^2 - 5x^* - 2 = 0 \). Solving this gives us possible equilibrium values like \( x^* = 2 \) and \( x^* = -\frac{1}{3} \).
These are points where the system can either settle (stable) or diverge (unstable) as determined by further analysis using stability criteria.