Problem 27

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. $$ \frac{d y}{d x}=y \ln (y+2) $$

Step-by-Step Solution

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Answer

Critical points are $ y = 0 $ (unstable) and $ y = -1 $ (asymptotically stable).

1Step 1: Identify Critical Points

To find the critical points of the differential equation $ \frac{dy}{dx} = y \ln(y+2) $, we set $ \frac{dy}{dx} = 0 $. This gives us the equation $ y \ln(y+2) = 0 $. For this product to be zero, either $ y = 0 $ or $ \ln(y+2) = 0 $ must be true.

2Step 2: Solve for Critical Points

First, consider $ y = 0 $ as a critical point. Next, solve $ \ln(y+2) = 0 $, which implies $ y+2 = 1 $, giving us $ y = -1 $. Thus, the critical points are $ y = 0 $ and $ y = -1 $.

3Step 3: Analyze Stability of Critical Points

To analyze stability, consider the sign of $ \frac{dy}{dx} = y \ln(y+2) $ around each critical point. For $ y = 0 $, if $ y > 0 $, then $ y \ln(y+2) > 0 $, hence solution curves move away from the critical point, indicating that $ y = 0 $ is unstable. For $ y < 0 $ and near $ y = 0 $, $ \ln(y+2) $ becomes negative and $ y $ is negative, making $ y \ln(y+2) > 0 $, also unstable. For $ y = -1 $, if $ y > -1 $, $ y $ will be negative but small, and $ \ln(y+2) $ negative, causing $ y \ln(y+2) < 0 $, suggesting solutions move toward $ y = -1 $. If $ y < -1 $, both $ y $ and $ \ln(y+2) $ are negative, causing $ y \ln(y+2) > 0 $, suggesting solutions move toward $ y = -1 $. Thus, $ y = -1 $ is asymptotically stable.

4Step 4: Draw Phase Portrait

In the phase portrait, draw horizontal lines at $ y = 0 $ and $ y = -1 $ representing the critical points. Use arrows to show that solutions diverge from $ y = 0 $ (unstable) and converge to $ y = -1 $ (asymptotically stable). Sample solution curves can be sketched showing the trajectory behavior: moving away from $ y = 0 $ and moving toward and stabilizing at $ y = -1 $.

Key Concepts

Critical PointsPhase PortraitStability AnalysisEquilibrium Solutions

Critical Points

In the study of differential equations, finding critical points is a crucial step. Critical points occur where the derivative is zero, indicating that the function doesn’t change at that particular point. In our example, we solve the equation $ \frac{dy}{dx} = y \ln(y+2) = 0 $. The product is zero if either $ y = 0 $ or $ \ln(y+2) = 0 $. Thus, for the equation $ \ln(y+2) = 0 $, we solve for $ y $ and find that $ y = -1 $. We conclude that the critical points are $ y = 0 $ and $ y = -1 $. These points are where the solution does not change, crucial for further stability analysis.

Phase Portrait

A phase portrait provides a visual depiction of solution behaviors over time. Even without computing exact trajectories, we can map out the nature of solutions around critical points. In the depicted phase portrait for our differential equation:

We highlight the equilibrium solutions with horizontal lines at $ y = 0 $ and $ y = -1 $.
With arrows, movements are shown where solutions diverge from $ y = 0 $ and converge at $ y = -1 $.
A typical sketch shows trajectory behavior—how solutions move away from or toward critical points.

This visualization helps in understanding the behavior of solutions and identifying potential long-term behaviors.

Stability Analysis

Stability analysis involves determining how small changes in the system can affect the evolution of the system. For stability, we analyze what happens to solutions as they approach or move away from the critical points.

Unstable Point: For $ y = 0 $, if $ y > 0 $ or $ y < 0 $, $ \ln(y+2) $ ensures $ y \ln(y+2) > 0 $, indicating solutions are diverging, classifying $ y = 0 $ as unstable.
Asymptotically Stable Point: At $ y = -1 $, solutions behave differently: when $ y > -1 $ or $ y < -1 $, calculations show that solutions converge, classifying $ y = -1 $ as asymptotically stable.

Stability analysis helps predict how systems behave over time and which points may potentially attract solutions.

Equilibrium Solutions

Equilibrium solutions are the constant solutions of a differential equation, providing insight into the behavior of systems over time. For $ \frac{dy}{dx} = y \ln(y+2) $, we previously found our critical points to be $ y = 0 $ and $ y = -1 $. These are not only critical points but also equilibrium solutions where the rate of change of $ y $ is zero. At these points, the system remains in its current state:

At $ y = 0 $: Although it’s an equilibrium point, it's unstable. Any minor change causes the solutions to deviate.
At $ y = -1 $: It is a stable equilibrium where solutions tend to settle down, signifying reliable equilibrium states.

Understanding equilibrium solutions is vital to comprehend how dynamical systems maintain or change their states under varying conditions.

Problem 27

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