Problem 19 - Advanced Engineering Mathematics

Question

Writing the differential equation in the form \(d y / d x=y(1-y)(1+y)\) we see that critical points are located at \(y=-1, y=0,\) and \(y=1 .\) The phase portrait is shown at the right.

Step-by-Step Solution

Answer

Critical points are at \( y = -1 \), \( y = 0 \), and \( y = 1 \). These can be evaluated for stability based on their neighboring intervals.

Step 1: Set the Differential Equation Equal to Zero

To find the critical points, set the differential equation \( \frac{dy}{dx} = y(1-y)(1+y) \) equal to zero. This means you look for values of \( y \) where the rate of change is zero.

Step 2: Solve for Critical Points

Factor the equation: \( y(1-y)(1+y) = 0 \). Solve for \( y \) by setting each factor equal to zero: \( y = 0 \), \( 1-y = 0 \) gives \( y = 1 \), and \( 1+y = 0 \) gives \( y = -1 \). The critical points are thus \( y = -1 \), \( y = 0 \), and \( y = 1 \).

Step 3: Confirm Critical Points and Analyze Stability

Verify the critical points found are correct by substituting them back into the differential equation, ensuring \( \frac{dy}{dx} = 0 \). Further, study the stability around these points by evaluating the sign of \( \frac{dy}{dx} \) near them: for example, check the intervals \( y < -1 \), \( -1 < y < 0 \), \( 0 < y < 1 \), and \( y > 1 \).

Key Concepts

Critical Points in Differential Equations

In the context of differential equations, critical points are specific values of the variable where the derivative or rate of change is equal to zero. This means that at these points, the system is in a state of equilibrium as there is no change happening. For the given equation \( \frac{dy}{dx} = y(1-y)(1+y) \), critical points are determined by setting \( \frac{dy}{dx} = 0 \).

• First, factor the expression: \( y(1-y)(1+y) = 0 \).
• To solve for \( y \), set each factor to zero. This leads to \( y = 0 \), \( 1-y = 0 \) which solves to \( y = 1 \), and \( 1+y = 0 \) which solves to \( y = -1 \).
• Thus, critical points are \( y = -1 \), \( y = 0 \), and \( y = 1 \).

These points help us understand where the function changes behavior, entering or leaving a state of equilibrium. Identifying critical points is the first step in analyzing the stability of a system.

Stability Analysis of Critical Points

Stability analysis involves examining how a system behaves in the vicinity of its critical points. For each critical point determined (\( y = -1, 0, 1 \)), we need to assess whether perturbations will cause the system to return to the equilibrium (stable), move away (unstable), or neither (neutral stability).
Here’s how you can evaluate:

• For \( y < -1 \), test a value such as \( y = -2 \). Substitute into \( \frac{dy}{dx} \) to find whether it gives a positive or negative result.
• For \( -1 < y < 0 \), test with \( y = -0.5 \) and substitute back.
• Similarly, use \( y = 0.5 \) for \( 0 < y < 1 \) and check the sign.
• Lastly, for \( y > 1 \), use a value like \( y = 2 \).

A positive \( \frac{dy}{dx} \) suggests that \( y \) increases, moving away from the critical point (unstable). If negative, \( y \) decreases, indicating stability as it drives towards the critical point. Understanding stability is crucial for predicting system behavior in scientific and engineering applications.

The Phase Portrait of a Differential Equation

A phase portrait is a graphical representation of the trajectories of a dynamic system in the phase plane. It provides a holistic view of how solutions to differential equations change over time. For the equation \( \frac{dy}{dx} = y(1-y)(1+y) \), the phase portrait captures the behavior of solutions across different initial conditions.

• Draw horizontal lines at \( y = -1 \), \( y = 0 \), and \( y = 1 \) to mark the critical points.
• Between these boundaries and beyond, sketch arrows signifying the direction of change:
  • Above \( y = 1 \), arrows point upwards indicating \( y \) grows larger.
  • Between \( y = 0 \) and \( y = 1 \), arrows point down, showing reduction in \( y \).
  • Below \( y = 0 \), arrows upwards represent increasing \( y \).
  • Below \( y = -1 \), arrows upward again indicate increasing \( y \).
• The portrait visually identifies the regions of attraction and repulsion, reflecting the stability characteristics at a glance.

By observing the phase portrait, it's easier to grasp the overall dynamics of the system across different regions, providing insight into how solutions evolve over time.