Problem 4 - Calculus for Biology and Medicine

Question

Suppose that $$\frac{d y}{d x}=y(2-y)(y-3)$$ (a) Find the equilibria of this differential equation. (b) Graph \(d y / d x\) as a function of \(y\), and use your graph to discuss the stability of the equilibria. (c) Compute the eigenvalues associated with each equilibrium, and discuss the stability of the equilibria.

Step-by-Step Solution

Answer

Equilibria are at \( y=0, 2, 3 \). Stable: \( y=3 \), Unstable: \( y=2 \). \( y=0 \) is weakly stable.

Step 1: Set the Derivative Equal to Zero

To find the equilibria, we need to set \( \frac{d y}{d x} = 0 \). This is because equilibria occur where the rate of change of \(y\) with respect to \(x\) is zero. For our function, this means solving the equation \( y(2-y)(y-3) = 0 \).

Step 2: Solve the Equation for y

The equation \( y(2-y)(y-3) = 0 \) suggests multiple solutions (roots). Solve for \(y\):1. \( y = 0 \) 2. \( 2 - y = 0 \rightarrow y = 2 \) 3. \( y - 3 = 0 \rightarrow y = 3 \). Thus, the equilibria are at \( y = 0, 2, 3 \).

Step 3: Graph \( \frac{d y}{d x} \) as a Function of y

Sketch the graph of \( \frac{d y}{d x} = y(2-y)(y-3) \) against \(y\). Note where the curve crosses the \(y\)-axis; these crossings represent the equilibria points (\(y = 0, 2, 3\)). The sign of the function on either side of these points will help determine stability.

Step 4: Analyze Stability

From the graph, note the behavior of \( \frac{d y}{d x} \) around each equilibrium:- Near \( y = 0 \), the function crosses from negative to positive, indicating stability.- Near \( y = 2 \), the function crosses from positive to negative, indicating instability.- Near \( y = 3 \), the function crosses from negative to positive, indicating stability.

Step 5: Compute the Jacobian Matrix and Eigenvalues at Each Equilibrium

For each equilibrium, we compute the derivative (Jacobian) of \( \frac{d y}{d x} = y(2-y)(y-3) \) with respect to \(y\), evaluated at the equilibrium.1. At \( y = 0 \): \( f'(y) = 6y - 5y^2 \). Calculate \( f'(0) = 6*0 = 0 \).2. At \( y = 2 \): Calculate \( f'(2) = 6*2 - 5\times 2^2 = 12 - 20 = -8 \).3. At \( y = 3 \): Calculate \( f'(3) = 6*3 - 5*3^2 = 18 - 45 = -27 \).

Step 6: Determine Stability from Eigenvalues

The equilibrium point at \( y = 0 \) with a derivative \( f'(0) = 0 \) requires further analysis (e.g., higher derivatives), often signaling a center or a saddle, weakly stable.- For \( y = 2 \), since \( f'(2) = -8 \), this eigenvalue is negative, indicating instability.- For \( y = 3 \), since \( f'(3) = -27 \), the negative eigenvalue suggests stability.

Key Concepts

Equilibrium Points

In a differential equation, an **equilibrium point** represents a value where the system is at rest, meaning the rate of change is zero. For the equation \( \frac{d y}{d x} = y(2-y)(y-3) \), we determine equilibrium by setting \( \frac{d y}{d x} = 0 \). This involves solving \( y(2-y)(y-3) = 0 \), which breaks down into simpler equations:

• \( y = 0 \)
• \( 2-y = 0 \rightarrow y = 2 \)
• \( y-3 = 0 \rightarrow y = 3 \)

These solutions \( y = 0, 2, 3 \) are the equilibrium points, where the function is not changing, thus providing stability analysis indicators later on.

Stability Analysis

**Stability analysis** checks if small changes in the system revert back or diverge further from equilibrium points. Once the equilibria for \( \frac{d y}{d x} = y(2-y)(y-3) \) are known as \( y = 0, 2, 3 \), we explore how the function behaves around these points.Near \( y = 0 \), the function switches from negative \( - \) to positive \( + \), suggesting stability. If disturbances occur, the system tends to return to equilibrium. For \( y = 2 \), the function transitions from positive \( + \) to negative \( - \), indicating instability, as disturbances drive the system away. Lastly, at \( y = 3 \), it again shifts from negative \( - \) to positive \( + \), reinforcing stability.

Graphical Representation

A **graphical representation** illuminates the nature of equilibrium points by visually depicting \( \frac{d y}{d x} = y(2-y)(y-3) \) against \( y \). You'll observe the graph crossing the \( y \)-axis at the equilibria: \( y = 0, 2, 3 \).- At \( y = 0 \), the graph's curve slides from below to above the axis.- At \( y = 2 \), it dips from above to below.- At \( y = 3 \), it resurfaces from below to above.
These graphical crossovers convey stability: equilibrium is stable if the curve reverts, and unstable if it departs.

Eigenvalues

**Eigenvalues** in differential equations provide more nuanced stability information for equilibrium points. Using the derivative of our function, known as the Jacobian, evaluated at each equilibrium yields these values.- For \( y = 0 \), \( f'(y) = 6y - 5y^2 \) evaluates to zero. A zero eigenvalue suggests additional checks, like higher-order derivatives, are necessary to determine stability.- At \( y = 2 \), \( f'(2) = -8 \) is negative, suggesting strong instability.- At \( y = 3 \), \( f'(3) = -27 \) is also negative, indicating stability.
By analyzing eigenvalues, we confirm our graphical findings and predict how small disturbances will affect the system's balance around these critical points.