Problem 22

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. $$ \frac{d y}{d x}=y^{2}-y^{3} $$

Step-by-Step Solution

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Answer

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

1Step 1: Identify Equilibrium Solutions

To find the critical points, set the derivative equal to zero: $ \frac{dy}{dx} = y^2 - y^3 = 0 $. This simplifies to $ y^2(1 - y) = 0 $. The solutions to this are $ y = 0 $ and $ y = 1 $. These are the equilibrium (critical) points.

2Step 2: Determine Stability of Each Critical Point

To classify the stability of each critical point, analyze the sign of $ \frac{dy}{dx} = y^2(1 - y) $ near the critical points.- For $ y < 0 $: $ \frac{dy}{dx} > 0 $ (as $ y^2(1 - y) > 0 $)- Near $ y = 0 $ from below: $ \frac{dy}{dx} < 0 $ for $ 0 < y < 1 $- For $ y > 1 $: $ \frac{dy}{dx} < 0 $This indicates $ y = 0 $ is unstable (solution flows away from $ y = 0 $) and $ y = 1 $ is asymptotically stable (solution moves towards $ y = 1 $).

3Step 3: Sketch Phase Portrait

The phase portrait can be sketched by understanding the flow of the solution.- Near $ y = 0 $, the arrows should point away from the line since the point is unstable.- Near $ y = 1 $, arrows should point toward $ y = 1 $ since it's stable.- Above $ y = 1 $, the arrows should point downwards back to $ y = 1 $ since solutions decrease in this region.

4Step 4: Classify Critical Points

Thus, we classify the critical points as: $ y = 0 $ (unstable), $ y = 1 $ (asymptotically stable). This indicates the behavior around each equilibrium solution in the $ y $-axis on the phase line.

Key Concepts

Critical PointsPhase PortraitStability AnalysisEquilibrium Solutions

Critical Points

In differential equations, critical points are where the rate of change of the system is zero. They are found by setting the derivative equal to zero. In our exercise, the differential equation is $ \frac{d y}{d x}=y^{2}-y^{3} $ If we set this equation to zero to find critical points, we get: $ y^{2}(1-y) = 0 $.

The solution to this is $ y = 0 $ and $ y = 1 $, which are our critical points.
These are the points where the system doesn't change, i.e., equilibrium points.

Understanding critical points is crucial because they help us know where the potential changes in the behavior of the system emerge. Finding these points is usually the first step in stability analysis and drawing phase portraits.

Phase Portrait

A phase portrait is a visual representation of the possible trajectories of a dynamical system. It provides a sketch of how solutions to a differential equation evolve over time.For the differential equation $ \frac{d y}{d x}=y^{2}-y^{3} $, the phase portrait helps us see how solutions behave near the critical points found earlier.

If you're near $ y = 0 $, the arrows in the phase portrait point away because this is an unstable point.
At $ y = 1 $, arrows direct towards this point, showing it's stable.
Above $ y = 1 $, trajectories fall back towards $ y = 1 $.

This graphical representation makes it easier for learners to visualize solutions without solving the differential equation explicitly each time. Phase portraits are valuable tools in understanding the behavior of complex systems.

Stability Analysis

Stability analysis is about determining how critical points behave under small disturbances. A stable point means that small changes will return to this point over time. In our exercise:

For $ y = 0 $, small deviations grow away from 0. Thus, $ y = 0 $ is classified as unstable.
For $ y = 1 $, any slight disturbance will decay back to the exact value, making $ y = 1 $ asymptotically stable.

To conduct a simple analysis, you examine the derivative's sign around critical points.

If the derivative is positive (the tangent is pointing up), the critical point is unstable.
If the derivative is negative (tangent points down), it's stable.

This method helps efficiently predict long-term behavior of dynamical systems.

Equilibrium Solutions

An equilibrium solution in a differential equation is a constant solution where the derivative is zero. In simple terms, it's a flat line in time, indicating no change.For our system $ \frac{d y}{d x}=y^{2}-y^{3} $, we determined equilibrium solutions as $ y = 0 $ and $ y = 1 $. These points help us understand the nature of our system and act as reference benchmarks.

Equilibrium $ y = 0 $ displays an unstable behavior, as nearby solutions diverge.
Equilibrium $ y = 1 $ means solutions gravitate towards it, signifying this point is stable.

Grasping equilibrium solutions is fundamental for students, making complex mathematical ideas more accessible by pinpointing behavior descriptions at specific points in the phase space. This understanding leads to insights about system dynamics in practical applications.

Problem 22

Other exercises in this chapter

Problem 22

Solve the given initial-value problem. $$ \left(e^{x}+y\right) d x+\left(2+x+y e^{y}\right) d y=0, \quad y(0)=1 $$

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Problem 22

$\left(e^{x}+e^{-x}\right) \frac{d y}{d x}=y^{2}$

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Problem 22

Use Euler's method with step size $h=0.1$ to approximate $y(1.2)$ where $y(x)$ is a solution of the initial-value problem \(y^{\prime}=1+x \sqrt{y}, y(1)=

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Problem 22

In Problems 1-22, solve the given differential equation by separation of variables. $$ \left(e^{x}+e^{-x}\right) \frac{d y}{d x}=y^{2} $$

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