Problem 3 - Thomas Calculus

Question

In Exercises \(1-8\) a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of \(y^{\prime}\) and \(y^{\prime \prime}\) c. Sketch several solution curves. $$ \frac{d y}{d x}=y^{3}-y $$

Step-by-Step Solution

Answer

Equilibrium points: \( y = 0 \) (stable), \( y = \pm 1 \) (unstable).

Step 1: Find Equilibrium Points

To identify equilibrium values, we set \( \frac{dy}{dx} = 0 \). This gives us the equation \( y^3 - y = 0 \). Factoring, we get \( y(y^2 - 1) = 0 \), which simplifies to \( y(y-1)(y+1) = 0 \). Thus, the equilibrium points are \( y = 0, 1, -1 \).

Step 2: Determine Stability of Equilibria

We analyze the stability using the derivative \( \frac{dy}{dx} = y^3 - y \). The derivative at the equilibrium points is given by \( f'(y) = 3y^2 - 1 \). Calculating for each point:- At \( y = 0 \), \( f'(0) = -1 \) (stable).- At \( y = 1 \), \( f'(1) = 2 \) (unstable).- At \( y = -1 \), \( f'(-1) = 2 \) (unstable).

Step 3: Construct the Phase Line

The phase line is constructed by indicating the direction of \( y' \) at each interval:- For \( y < -1 \), \( y^3 - y > 0 \) implying increasing \( y \).- For \( -1 < y < 0 \), \( y^3 - y < 0 \) implying decreasing \( y \).- For \( 0 < y < 1 \), \( y^3 - y < 0 \) implying decreasing \( y \).- For \( y > 1 \), \( y^3 - y > 0 \) implying increasing \( y \).Arrows: \( ... \rightarrow (-1) \leftarrow (0) \rightarrow (1) \rightarrow ... \).

Step 4: Analyze Signs of First and Second Derivatives

For the first derivative, use the phase line from the previous step. For the second derivative, take the derivative of \( y'(y^3 - y) \):\( y'' = 3y^2 - 1 \).- At \( y < -1 \), \( y'' > 0 \) (concave up).- At \( -1 < y < 0 \), \( y'' < 0 \) (concave down).- At \( 0 < y < 1 \), \( y'' < 0 \) (concave down).- At \( y > 1 \), \( y'' > 0 \) (concave up).

Step 5: Sketch Solution Curves

Using the information from the phase line and the sign analysis of the derivatives, sketch the solution curves. Plot equilibrium points at \( y = -1, 0, 1 \). Show the direction of the solution curves:- Solutions approach \( y = 0 \) from \( -1 \) and \( +1 \) indicating stability.- Solutions diverge from \( y = 1 \) and \( y = -1 \) indicating instability.Draw curves with appropriate concavity as determined by \( y'' \):- Curve is concave up outside \( y = -1 \) and \( y = 1 \), and concave down between \(-1, 0, \) and \(0, 1\).

Key Concepts

Equilibrium Points

Equilibrium points, also known as fixed points, occur where the rate of change is zero, i.e., \( \frac{dy}{dx} = 0 \). For the given differential equation \( \frac{dy}{dx} = y^3 - y \), we can find these points by setting the expression equal to zero: \[ y^3 - y = 0. \] By factoring, we get \( y(y^2 - 1) = 0 \), which simplifying further reveals the equilibrium points as \( y = 0, 1, -1 \).
These are critical points since, at these values, the system does not change or "move" over time. Identifying equilibrium points help us understand where solutions might hover indefinitely.

Stability Analysis

Once equilibrium points are identified, determining their stability is crucial. We assess stability by evaluating the first derivative of the function, \( f'(y) \), at these points. For \( \frac{dy}{dx} = y^3 - y \), the derivative is \[ f'(y) = 3y^2 - 1. \] Stability depends on whether \( f'(y) \) is positive or negative:

• At \( y = 0 \), \( f'(0) = -1 \), indicating a negative slope which implies stability (solutions return to this point).
• At \( y = 1 \), \( f'(1) = 2 \), a positive value, indicating instability (solutions tend to diverge).
• At \( y = -1 \), \( f'(-1) = 2 \), also positive, suggesting instability.

A stable equilibrium point corresponds to values where small disturbances will cause the system to return to equilibrium, while an unstable point means disturbances cause the solutions to drift away.

Phase Line

A phase line is a simple graphical tool to summarize the behavior of solutions in a one-dimensional autonomous differential equation. It provides a visual picture of increasing and decreasing intervals of the function.
For \( \frac{dy}{dx} = y^3 - y \), evaluate the intervals around each equilibrium:

• For \( y < -1 \), the function \( y^3 - y > 0 \) indicating an increase.
• For \( -1 < y < 0 \), \( y^3 - y < 0 \) represents a decline.
• For \( 0 < y < 1 \), the decrease continues as \( y^3 - y < 0 \).
• For \( y > 1 \), \( y^3 - y > 0 \) signifies growth.

This can be illustrated with arrows on the phase line: From \(-1\, \rightarrow\, (-1) \leftarrow\, (0) \rightarrow\, (1) \rightarrow\).
The arrows help in understanding the tendencies of the solution curves over those segments.

Solution Curves

Sketching solution curves involves integrating the insights from the phase line and stability analysis. Start by marking the equilibrium points \( y = 0, 1, -1 \). These points act as anchors for our solutions.
Directions and nature of the curves are influenced by the intervals of growth and decay identified in the phase line analysis:

• For \( y < -1 \) and \( y > 1 \), the curves are concave upward, indicating solutions move away from \( y = -1 \) and \( y = 1 \), showing these areas include directional changes in growth.
• Solutions move towards \( y = 0 \) from both sides, confirming its stability.
• Concave down patterns are present between the intervals of (-1, 0) and (0, 1), in line with the phase line findings.

Plotting these, you can see how solutions approach, stabilize, or move away, giving a holistic sense of the system's dynamics. These visualizations help connect the mathematics with intuitive understanding.