Problem 9 - Calculus for Biology and Medicine

Question

Find the equilibria of $$x_{t+1}=4 x_{t}^{2}+x_{t}-1, \quad t=0,1,2, \ldots$$ and use the stability criterion for an equilibrium point to determine whether they are stable or unstable.

Step-by-Step Solution

Answer

The equilibria are at \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \), both are unstable.

Step 1: Define Equilibrium Points

Equilibrium points occur where the system does not change in time, meaning \( x_{t+1} = x_t \). For the given equation \( x_{t+1} = 4x_t^2 + x_t - 1 \), setting \( x_{t+1} = x_t \) gives: \[ x_t = 4x_t^2 + x_t - 1. \] Simplifying it, we get the equation: \[ 4x_t^2 - 1 = 0. \]

Step 2: Solve for Equilibrium Points

To find \( x_t \), solve the equation from Step 1: \[ 4x_t^2 - 1 = 0. \] Factor the left-hand side: \[ (2x_t - 1)(2x_t + 1) = 0. \] This gives solutions as \( 2x_t = 1 \) or \( 2x_t = -1 \). Therefore, the equilibrium points are \( x_t = \frac{1}{2} \) and \( x_t = -\frac{1}{2} \).

Step 3: Determine Stability of Equilibrium Points

To determine stability, calculate the derivative \( f'(x) \) at each equilibrium point, where \( f(x) = 4x^2 + x - 1 \). The derivative is \( f'(x) = 8x + 1 \).

Step 4: Calculate Stability at \( x = \frac{1}{2} \)

Substitute \( x = \frac{1}{2} \) into \( f'(x) \): \[ f'\left(\frac{1}{2}\right) = 8\left(\frac{1}{2}\right) + 1 = 4 + 1 = 5. \] Since \(|f'(x)| > 1\), the equilibrium at \( x = \frac{1}{2} \) is unstable.

Step 5: Calculate Stability at \( x = -\frac{1}{2} \)

Substitute \( x = -\frac{1}{2} \) into \( f'(x) \): \[ f'\left(-\frac{1}{2}\right) = 8\left(-\frac{1}{2}\right) + 1 = -4 + 1 = -3. \] Since \(|f'(x)| > 1\), the equilibrium at \( x = -\frac{1}{2} \) is also unstable.

Key Concepts

Understanding the Stability Criterion

In mathematics, the stability criterion is a method used to determine whether an equilibrium point, when perturbed, will return to its original state or not. In simpler terms, it checks if a system can balance itself if slightly adjusted.

• Stable equilibrium points return to equilibrium after a disturbance.
• Unstable equilibrium points tend to move away from equilibrium when disturbed.

When applied to difference equations, this criterion examines the behavior of a system over discrete points in time. For a system with an equilibrium point \( x^* \), and given a function \( f(x) \), the stability can often be assessed using the derivative \( f'(x) \).

The general rule is:

• If \( |f'(x^*)| < 1 \), the system is stable at \( x^* \).
• If \( |f'(x^*)| > 1 \), the system is unstable at \( x^* \).
• If \( |f'(x^*)| = 1 \), the system's stability is inconclusive without further analysis.

Exploring Difference Equations

Difference equations are mathematical expressions that capture the relationship between a sequence of values. They are similar to differential equations but work over discrete, not continuous, points in time. The equation provided in the exercise, \( x_{t+1} = 4x_t^2 + x_t - 1 \), is an example of a first-order difference equation.

Key characteristics include:

• A recurrence relation that connects different terms in the sequence.
• The sequence \( x_t \) usually evolves step by step based on a function, in this case, \( f(x_t) = 4x_t^2 + x_t - 1 \).
• Initial conditions (such as \( x_0 \)) are often needed to derive specific future values.

Equilibrium points in these equations occur where \( x_{t+1} = x_t \), meaning the system reaches a state of balance, and future values are the same as current ones. Solving for equilibrium gives critical insight into the behavior of these systems.

The Role of Derivative Calculation in Stability Analysis

Derivative calculation plays a pivotal role in our ability to determine the stability of an equilibrium point within a system. Given a function \( f(x) \) derived from a difference equation, its derivative \( f'(x) \) reveals how sensitive the system is to changes.
For the system defined by \( f(x) = 4x^2 + x - 1 \), the derivative is \( f'(x) = 8x + 1 \). Computing this derivative at the equilibrium points \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \) helps assess each point's stability:

• Substitute \( x = \frac{1}{2} \) to find \( f'(\frac{1}{2}) = 5 \).
• Substitute \( x = -\frac{1}{2} \) to find \( f'(-\frac{1}{2}) = -3 \).

Since both have \( |f'(x)| > 1 \), the equilibrium points are unstable.

By using derivatives, we not only identify the equilibrium points but also predict their long-term behavior.