Problem 4 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Define Equilibrium

An equilibrium point occurs when a system remains constant over time. For the given equation \(x_{t+1}=\frac{2}{3}-\frac{2}{3} x_{t}^{2}\), an equilibrium point \(x^*\) satisfies \(x^* = x_{t+1} = x_t\).

Step 2: Set Up Equilibrium Equation

Substitute \(x^*\) into the equation: \(x^* = \frac{2}{3} - \frac{2}{3} (x^*)^2\).

Step 3: Simplify and Rearrange

Rearrange the equation: \( (x^*)^2 = \frac{2}{3} - x^* \). This leads to the quadratic equation \((x^*)^2 + x^* - \frac{2}{3} = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 1\), \(c = -\frac{2}{3}\) to find the roots. Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-\frac{2}{3}) = \frac{16}{3}\). The roots are \(x^* = \frac{-1 \pm \sqrt{\frac{16}{3}}}{2}\).

Step 5: Evaluate the Roots

Compute the two roots. \(x^*_1 = \frac{-1 + 2\sqrt{12/3}}{2} = \frac{-1 + \sqrt{4}}{2} = \frac{-1 + 2}{2} = \frac{1}{2}\). Then, \(x^*_2 = \frac{-1 - 2}{2} = \frac{-3}{2}\). The equilibria points are \(x^* = \frac{1}{2}\) and \(x^* = -\frac{3}{2}\).

Step 6: Determine Stability Criterion

The equilibrium is stable if \(|f'(x^*)| < 1\), where \(f(x) = \frac{2}{3} - \frac{2}{3} x^2\). Find the derivative: \(f'(x) = -\frac{4}{3}x\).

Step 7: Evaluate Stability of \(x^* = \frac{1}{2}\)

Substitute \(x^* = \frac{1}{2}\) into the derivative: \(f'(\frac{1}{2}) = -\frac{4}{3} \cdot \frac{1}{2} = -\frac{2}{3}\). Since \(|-\frac{2}{3}| = \frac{2}{3} < 1\), \(x^* = \frac{1}{2}\) is stable.

Step 8: Evaluate Stability of \(x^* = -\frac{3}{2}\)

Substitute \(x^* = -\frac{3}{2}\) into the derivative: \(f'(-\frac{3}{2}) = -\frac{4}{3} \cdot (-\frac{3}{2}) = 2\). Since \(|2| = 2 > 1\), \(x^* = -\frac{3}{2}\) is unstable.

Key Concepts

Stability Criterion

In mathematical dynamics, determining whether an equilibrium point is stable or unstable is crucial for understanding the system's behavior over time. Stability ensures that small deviations from the equilibrium point do not lead to drastic changes in the system's state. For the equation given in the exercise, an equilibrium point is said to be stable if:

• Any small perturbation or deviation around this point results in the system returning to equilibrium.
• The mathematical criterion to check for stability around an equilibrium point \(x^*\) involves evaluating the magnitude of the derivative of the function at that point.

For the function \(f(x) = \frac{2}{3} - \frac{2}{3} x^2\), the derivative is \(f'(x) = -\frac{4}{3}x\). An equilibrium \(x^*\) is stable if \(|f'(x^*)| < 1\) and unstable if \(|f'(x^*)| \geq 1\). This is derived from the linear approximation of the function near the equilibrium point.

Quadratic Equation

A quadratic equation is any equation of the form \(ax^2 + bx + c = 0\), where \(a eq 0\). In step 3, we rearranged the equilibrium condition \((x^*)^2 + x^* - \frac{2}{3} = 0\) into the standard quadratic form. Solving this equation gives the potential equilibrium points, which are solutions to the system where change ceases.

• The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is used to find these solutions.
• The discriminant, \(b^2 - 4ac\), helps determine the nature of the roots: real and distinct, real and repeated, or complex.

In our exercise, the discriminant \(\frac{16}{3} > 0\) indicates two distinct real roots. This tells us there are two potential equilibrium points: \(x^* = \frac{1}{2}\) and \(x^* = -\frac{3}{2}\). Understanding these roots provides insight into possible states where the system can naturally settle.

Derivative

The concept of a derivative is fundamental in calculus. It measures how a function changes as its input changes, providing the rate of change of the function with respect to one of its variables. In the context of stability, the derivative helps determine whether deviations from equilibrium grow, shrink, or remain the same.

• For the function \(f(x) = \frac{2}{3} - \frac{2}{3} x^2\), the derivative \(f'(x) = -\frac{4}{3}x\) tells us how sensitive the next state \(x_{t+1}\) is to changes in \(x_t\).
• The sign of \(f'(x)\) and its magnitude affect the stability: a derivative close to zero means minor changes, while a larger derivative can indicate a more rapid shift away from equilibrium.

For each equilibrium point, substituting them into the derivative provides a key criterion for assessing stability. \(x^* = \frac{1}{2} \) yields \(-\frac{2}{3}\) (stable), while \(x^* = -\frac{3}{2} \) results in \(2\) (unstable). These results demonstrate the power of derivatives in understanding dynamic systems.