Problem 25 - Advanced Engineering Mathematics

Question

Find the critical points and phase portrait of the given autonomous first- order differential equation. Classify each critical point as asymptotically stable, unstable, or semi-stable. By hand, sketch typical solution curves in the regions in the \(x y\) -plane determined by the graphs of the equilibrium solutions. $$ \frac{d y}{d x}=y^{2}\left(4-y^{2}\right) $$

Step-by-Step Solution

Answer

Critical points are at \(y = -2, 0, 2\). \(y = -2\) is unstable, \(y = 0\) is semi-stable, and \(y = 2\) is asymptotically stable.

Step 1: Determine Equilibrium Solutions

To find the critical points, set the equation \(\frac{dy}{dx} = 0\). The given differential equation is \(\frac{dy}{dx} = y^2(4-y^2)\). Set this equal to zero:\[y^2(4 - y^2) = 0.\]This implies \(y^2 = 0\) or \(4 - y^2 = 0\). Solving these equations, we find that \(y = 0\), \(y = 2\), and \(y = -2\) are the equilibrium solutions.

Step 2: Determine Sign of \(\frac{d y}{d x}\) in Intervals

The critical points divide the y-axis into four intervals: \((-\infty, -2)\), \((-2, 0)\), \((0, 2)\), and \((2, \infty)\). Test a point in each interval to determine the sign of \(\frac{d y}{d x}\):- For \((-\infty, -2)\), test \(y = -3\): \((-3)^2(4 - (-3)^2) = 9(-5) < 0\).- For \((-2, 0)\), test \(y = -1\): \((-1)^2(4 - (-1)^2) = 1(3) > 0\).- For \((0, 2)\), test \(y = 1\): \(1^2(4 - 1^2) = 1(3) > 0\).- For \((2, \infty)\), test \(y = 3\): \(3^2(4 - 3^2) = 9(-5) < 0\).

Step 3: Classify Critical Points

Based on the behaviors around the critical points, classify them:- For \(y = -2\), \(\frac{d y}{d x}\) changes from negative to positive, indicating an unstable equilibrium.- For \(y = 0\), \(\frac{d y}{d x}\) changes from positive to positive, indicating a semi-stable equilibrium.- For \(y = 2\), \(\frac{d y}{d x}\) changes from positive to negative, indicating an asymptotically stable equilibrium.

Step 4: Sketch Phase Portrait

To sketch the phase portrait, plot the y-axis, and indicate the equilibrium points at \(y = -2\), \(y = 0\), and \(y = 2\).- Draw solution curves indicating movement upward in the intervals \((-2, 0)\) and \((0, 2)\), and downward in the intervals \((-\infty, -2)\) and \((2, \infty)\).- Mark arrows on these solution curves to indicate the direction of flow as determined by the sign of \(\frac{d y}{d x}\). The equilibrium points and arrows illustrate the nature of stability for each interval.

Key Concepts

Phase Portrait

A phase portrait is an invaluable visual tool used in the analysis of differential equations. It provides students with a graphical representation of the behavior of solutions and their interrelationships. Here, the phase portrait helps illustrate how different solution curves flow toward or away from equilibrium points on the y-axis.
A typical phase portrait plots several trajectories, each representing a solution of the differential equation in the phase plane (the plane formed by the dependent variable and its derivative). In this example, \(\frac{dy}{dx} = y^2(4 - y^2)\), the phase portrait reveals the pathways of solution curves by highlighting intervals where \(\frac{dy}{dx}\) is positive or negative.
Arrows are used to depict the direction of solution flow—upward where the derivative is positive, and downward where it's negative. Critical points, marked clearly on this portrait, anchor the stability analysis by acting as nodes where the direction of solution changes.

Equilibrium Solutions

Equilibrium solutions, or critical points, are specific values where the solution of a differential equation does not change over time. These are found by setting the derivative of your function to zero and solving for the dependent variable.
In our case, by setting \(\frac{dy}{dx} = y^2(4 - y^2) = 0\), we identified the equilibrium solutions: \(y = 0\), \(y = 2\), and \(y = -2\). These critical points correspond to the values where the system is in a state of constant solution, meaning no change occurs, thus providing the base for further stability classification.
Understanding these solutions is crucial because they indicate potential stable and unstable regions in the phase portrait. They signify where trajectories converge (stable) or diverge (unstable), identifying them as focal points in dynamic system behavior.

Stability Analysis

Stability analysis involves examining how solution curves behave around equilibrium points. This step is essential to understand the long-term behavior of a dynamic system described by a differential equation.
In this analysis, critical points are classified as:

• Asymptotically stable: The solutions converge to the equilibrium point over time. This occurs at \(y = 2\).
• Unstable: Solutions diverge from the equilibrium point. This is seen at \(y = -2\).
• Semi-stable: The system is stable on one side and unstable on the other. In our equation, \(y = 0\) showcases this behavior.

To determine these, we look at the sign changes in \(\frac{dy}{dx}\) around the critical points. Such analyses are instrumental in predicting the behavior of real-world systems described by similar equations.

First-Order Differential Equations

First-order differential equations involve the first derivatives of functions, encapsulating the rate of change of some quantity. These equations are fundamental in modeling various dynamic processes that change over time.
They are essential tools in fields such as physics, biology, and economics, among others, where understanding rates of change and behavior patterns are vital. Our example equation \(\frac{dy}{dx} = y^2(4 - y^2)\) is autonomous, meaning it depends only on the variable and its derivative, not on the independent variable directly.
This distinction simplifies the process of finding equilibrium solutions and performing stability analysis. Handling these equations efficiently provides a profound understanding and management of both theoretical and practical systems.