Problem 87 - 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. \(\frac{d x}{d t}=h x-x^{2}\), where \(h\) is a constant and (a) \(h>0\), (b) \(h<0\)

Step-by-Step Solution

Answer

For \( h > 0 \), \( x = 0 \) is unstable and \( x = h \) is stable. For \( h < 0 \), \( x = 0 \) is stable and \( x = h \) is unstable.

Step 1: Finding Equilibria Points

To find the equilibrium points, set \( \frac{d x}{d t} = 0 \). This gives the equation \( h x - x^2 = 0 \). Factoring this equation gives \( x(h - x) = 0 \). The solutions are \( x = 0 \) and \( x = h \). Thus, the equilibria are at \( x = 0 \) and \( x = h \).

Step 2: Calculating the Derivative

The stability of each equilibrium point can be analyzed by taking the derivative of \( \frac{d x}{d t} = h x - x^2 \) with respect to \( x \). The derivative is \( \frac{d}{dx} (h x - x^2) = h - 2x \). This derivative will be evaluated at each equilibrium point to determine stability.

Step 3: Analyzing Stability for Equilibrium Points

Evaluate the derivative \( h - 2x \) at \( x = 0 \) and \( x = h \).- At \( x = 0 \): \( h - 2(0) = h \). - At \( x = h \): \( h - 2(h) = -h \). The sign of the derivative at each point determines stability. If it is negative, the equilibrium is stable. If it is positive, the equilibrium is unstable.

Step 4: Stability Analysis for h > 0

For \( h > 0 \):- At \( x = 0 \), since \( h > 0 \), the derivative \( h \) is positive. This means that \( x = 0 \) is an unstable equilibrium.- At \( x = h \), since \( -h < 0 \), the derivative is negative. This means that \( x = h \) is a stable equilibrium.

Step 5: Stability Analysis for h < 0

For \( h < 0 \):- At \( x = 0 \), since \( h < 0 \), the derivative \( h \) is negative. Therefore, \( x = 0 \) is a stable equilibrium.- At \( x = h \), \( -h > 0 \), so the derivative is positive. Therefore, \( x = h \) is an unstable equilibrium.

Key Concepts

Equilibrium Points

In differential equations, equilibrium points represent the values of the variable for which the system stays constant over time. These are vital in understanding how a system behaves without being influenced by changes. Finding equilibrium points is usually the first step in analyzing a differential equation. Here's a simple way to find them:

• Set the derivative of the function to zero, i.e., \( \frac{d x}{d t} = 0 \).
• Determine the values of \( x \) where this condition is met. In our case, from the equation \( h x - x^2 = 0 \), factor out \( x \), which leads to \( x(h - x) = 0 \).
• Solve the resulting factored equation to find the equilibrium points \( x = 0 \) and \( x = h \).

These equilibrium points give us foundational information about the behavior of the system, but they don't tell us about stability. That's where stability analysis comes in.

Stability Analysis

Once equilibrium points are found, stability analysis helps us discern their nature — whether they're stable or unstable. Stability essentially refers to how the system responds to small disturbances and can be analyzed using the derivative of the differential equation evaluated at these points.Here's a quick exercise on how stability works:

• Calculate the first derivative of the function that describes how \( x \) changes over time, i.e., \( \frac{d}{dx}(h x - x^2) \), which simplifies to \( h - 2x \).
• Evaluate this derivative at each of the equilibrium points obtained earlier.
• Analyze the sign of the derivative at these points:
  • If the derivative is negative, the equilibrium is stable.
  • If the derivative is positive, the equilibrium is unstable.

For \( h > 0 \):- At \( x = 0, h - 2(0) = h > 0 \), making it an unstable equilibrium since the derivative is positive.- At \( x = h, h - 2(h) = -h < 0 \), indicating stability as the derivative is negative.Similarly, for \( h < 0 \), analyze using the same derivative, ensuring solid insights into the behavior around these points.

Eigenvalue Calculation

Eigenvalue calculation is crucial in stability analysis as it provides precise information on how perturbations behave near equilibrium points. In this context, eigenvalues are derived from the linearization of the system around these points and involve evaluating the derivative at equilibrium.Here is how to perform this process:

• First, we linearize the differential equation near an equilibrium point. This gives us a system matrix from which eigenvalues are calculated.
• The matrix, in simple cases like ours, is a single value derived from \( h - 2x \) evaluated at each equilibrium point.
• If the eigenvalue is negative, perturbations disappear over time, indicating stability.
• If it's positive, perturbations grow, signaling instability.

Taking for example \( h > 0 \), at \( x = 0 \), our eigenvalue \( h > 0 \) suggests instability. At \( x = h \), \( -h < 0 \) suggests stability.This concise process helps determine not just suspected stability but quantifiable insights into how each equilibrium point reacts to small disturbances.