Problem 3 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Equilibrium values are -1, 0, and 1. -1 and 1 are stable, 0 is unstable.

Step 1: Identify Equilibrium Values

To find the equilibrium values for the differential equation \( \frac{dy}{dx} = y^3 - y \), set the derivative equal to zero: \[ y^3 - y = 0 \]Factor the equation:\[ y(y^2 - 1) = 0 \]Further factor this equation:\[ y(y-1)(y+1) = 0 \]The equilibrium values are \( y = 0 \), \( y = 1 \), and \( y = -1 \).

Step 2: Determine Stability of Equilibrium Values

To decide stability, observe the sign of the derivative \( \frac{dy}{dx} \) in the intervals determined by the equilibrium values \(-1\), \(0\), and \(1\):- For \( y < -1 \), the expression \( y^3 - y \) is positive.- Between \( y = -1 \) and \( y = 0 \), the expression is negative.- Between \( y = 0 \) and \( y = 1 \), the expression is positive.- For \( y > 1 \), the expression is again positive.Based on this analysis, \( y = 0 \) is unstable, \( y = -1 \) is stable, and \( y = 1 \) is stable.

Step 3: Construct a Phase Line

Draw a vertical line representing \( y \) and mark the equilibrium points \( y = -1 \), \( y = 0 \), and \( y = 1 \). Indicate the sign of \( \frac{dy}{dx} \) in each interval:- From \( -\infty \) to \(-1\), mark an upward arrow (positive sign), indicating \( \frac{dy}{dx} > 0 \).- From \( -1 \) to \( 0 \), mark a downward arrow (negative sign), indicating \( \frac{dy}{dx} < 0 \).- From \( 0 \) to \( 1 \), mark an upward arrow (positive sign), indicating \( \frac{dy}{dx} > 0 \).- From \( 1 \) to \( \infty \), mark an upward arrow (positive sign), indicating \( \frac{dy}{dx} > 0 \).

Step 4: Identify Signs of \( y' \,\) and \( y'' \)

The first derivative \( \frac{dy}{dx} = y^3 - y \) is positive or negative depending on the region, as identified in the phase line.The second derivative helps determine concavity:\[ y' = y^3 - y \]\[ \frac{d^2y}{dx^2} = 3y^2 - 1 \]Check the sign of \( 3y^2 - 1 \) in each segment to understand the concavity: It is positive for \( y \) when \( y^2 > \frac{1}{3} \) and negative for \( y^2 < \frac{1}{3} \).

Step 5: Sketch Solution Curves

On a \( y \)-\( x \) graph, start with horizontal lines at equilibrium values. Use arrows to show direction (up or down) as indicated by the phase line.- At or near \( y = -1 \), solutions curve toward the equilibrium (stable node).- At or near \( y = 0 \), solutions diverge (unstable).- At or near \( y = 1 \), solutions converge potentially from either side (stable node).Change in concavity according to the sign of \( \frac{d^2y}{dx^2} \).

Key Concepts

Equilibrium Values

In the analysis of differential equations, equilibrium values are crucial as they represent constants where the system experiences no change. To identify these equilibrium states, set the derivative to zero. From the given differential equation \( \frac{dy}{dx} = y^3 - y \), setting it to zero leads to the equation \( y^3 - y = 0 \). To solve this, you factor it as \( y(y^2 - 1) = 0 \), which further factors to \( y(y-1)(y+1) = 0 \). Thus, the equilibrium values are \( y = 0 \), \( y = 1 \), and \( y = -1 \). These are points where the slope of the tangent line to the graph is zero, meaning the solution does not change.

Stability Analysis

Once equilibrium values are found, understanding their stability is essential. Stability informs whether solutions stay close to an equilibrium state after small disturbances. To determine stability, examine the sign of \( \frac{dy}{dx} = y^3 - y \) around the identified equilibrium points.

• For \( y < -1 \), \( y^3 - y \) is positive, implying the solution moves away from the equilibrium.
• Between \( y = -1 \) and \( y = 0 \), it turns negative, suggesting attraction towards equilibrium.
• Between \( y = 0 \) and \( y = 1 \), it returns to positive, causing divergence from \( y = 0 \).
• For \( y > 1 \), \( y^3 - y \) remains positive, leading to divergence from \( y = 1 \).

Consequently, \( y = -1 \) and \( y = 1 \) are stable, while \( y = 0 \) is unstable. Stability allows us to predict the system's response to small changes.

Differential Equations

Differential equations describe the relationship between a function and its derivatives, thus modeling the rate of change in systems. They are foundational in numerous fields such as physics, biology, and engineering, where changes occur continuously. In the equation \( \frac{dy}{dx} = y^3 - y \), \( y^3 - y \) is the rate of change of \( y \) with respect to \( x \). By solving these equations, you can predict future behavior or state of a system given initial conditions. Capturing these dynamics often involves identifying equilibrium points and analyzing stability, as both offer insight into long-term behavior. This exercise exemplifies how phase line analysis helps visualize these insights, facilitating a better understanding of the equilibrium points' stability and the system's overall dynamic behavior.