Problem 79 - Calculus for Biology and Medicine

Question

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d y}{d t}=y(2-y)(y-3) $$

Step-by-Step Solution

Answer

Equilibria are \( y = 0, 2, 3 \); \( y = 2 \) is stable, \( y = 0 \) and \( y = 3 \) are unstable.

Step 1: Set the differential equation to zero

To find the equilibria, we need to set the differential equation to zero. Thus, we solve \( y(2-y)(y-3) = 0 \).

Step 2: Solve the equilibrium equation

The equation \( y(2-y)(y-3) = 0 \) can be solved by setting each factor equal to zero: \( y = 0 \), \( 2-y = 0 \), and \( y-3 = 0 \). This gives the equilibria as \( y = 0 \), \( y = 2 \), and \( y = 3 \).

Step 3: Find the Jacobian matrix

The differential equation is equivalent to the system \( \frac{d y}{d t} = f(y) = y(2-y)(y-3) \). The derivative \( f'(y) \) is the Jacobian for this one-dimensional system.

Step 4: Calculate the derivative

First, expand \( y(2-y)(y-3) \) to \( y(y^2 - 5y + 6) \) which simplifies to \( y^3 - 5y^2 + 6y \). The derivative \( f'(y) = 3y^2 - 10y + 6 \) is calculated.

Step 5: Evaluate the derivative at equilibria

Now, substitute each equilibrium point into \( f'(y) \) to find whether each is stable (eigenvalue negative) or unstable (eigenvalue positive):- \( f'(0) = 6 \)- \( f'(2) = -2 \)- \( f'(3) = 3 \)

Step 6: Determine stability of equilibria

Evaluate the sign of \( f'(y) \) at each equilibrium:- At \( y = 0 \), \( f'(0) = 6 \), which is positive, suggesting an unstable equilibrium.- At \( y = 2 \), \( f'(2) = -2 \), which is negative, suggesting a stable equilibrium.- At \( y = 3 \), \( f'(3) = 3 \), which is positive, suggesting an unstable equilibrium.

Key Concepts

Equilibrium Points

In the context of differential equations, finding equilibrium points is key to understanding the behavior of the system. Equilibrium points are values of the variable where the system remains constant over time, meaning here, when the derivative \( \frac{dy}{dt} \) equals zero.

To identify these points, set the differential equation to zero and solve for the unknown. For the given equation \( y(2-y)(y-3) = 0 \), the solutions are found by setting each factor of the equation to zero:

• \( y = 0 \) when the whole term vanishes.
• \( 2-y = 0 \) implies \( y = 2 \).
• \( y-3 = 0 \) leads to \( y = 3 \).

These solutions tell us that we have three equilibrium points at \( y = 0, 2, \) and \( 3 \).

This step is foundational, as only when we know where the equilibrium lies can we further evaluate their stability.

Jacobian Matrix

The Jacobian matrix is crucial in determining how small changes in a system state affect its dynamics. While often used in multi-variable systems, for a one-dimensional case like ours, it simplifies into just computing the derivative of the function on the right side of the differential equation with respect to \( y \).

For the differential equation \( \frac{dy}{dt} = y(2-y)(y-3) \), we have the function \( f(y) = y^3 - 5y^2 + 6y \). The derivative, \( f'(y) \), acts as the Jacobian matrix in this scenario:

• First, expand the expression \( y(2-y)(y-3) \).
• Efficiently compute \( f'(y) = 3y^2 - 10y + 6 \).

This derivative helps us evaluate changes around equilibrium points. In higher dimensions, the Jacobian matrix behaves similarly but involves partial derivatives for each variable and function.

Stability Analysis

Stability analysis involves evaluating the equilibrium points to ascertain whether small perturbations will return to the equilibrium, making it stable, or diverge, indicating instability.

Evaluate the derivative \( f'(y) \) at each equilibrium point:

• At \( y = 0 \), \( f'(0) = 6 \), suggests it's an unstable point since the derivative is positive.
• At \( y = 2 \), \( f'(2) = -2 \) indicates a stable equilibrium because the derivative is negative.
• At \( y = 3 \), \( f'(3) = 3 \), which is positive, also suggesting instability.

In simple terms, if \( f'(y) \) is positive at an equilibrium point, a tiny increase in \( y \) will cause \( \frac{dy}{dt} \) to rise, driving \( y \) further away. Conversely, a negative \( f'(y) \) means a small perturbation will push the system back to the equilibrium, ensuring stability.