Question

What can you say about the behavior of the solution? Are there equilibria? What stability do these equilibria have? $$y^{\prime }=1-y^2-x^2$$

Short answer

The equilibria are on the circle $y^2+x^2=1$ and are stable.

Step-by-step solution

Identify Equilibria

To find the equilibria, set the derivative $y^{\prime }=1-y^2-x^2$ to zero. This gives us the equation:$$1-y^2-x^2=0$$Which simplifies to:$$y^2+x^2=1$$This equation describes a circle centered at the origin with a radius of 1. Therefore, all points $(x,y)$ on this circle are equilibria.

Analyze Stability of Equilibria

To determine the stability, analyze if the values of $y^{\prime }$ change to positive or negative as we move away from the circle $y^2+x^2=1$. Consider outside and inside the circle separately:- If $y^2+x^2>1$, then $y^{\prime }=1-y^2-x^2<0$, indicating that outside the circle, the value decreases toward the circle.- If $y^2+x^2<1$, then $y^{\prime }=1-y^2-x^2>0$, indicating that inside the circle, the value increases toward the circle.Thus, the equilibria on the circle $y^2+x^2=1$ are stable.

Key concepts

Stability Analysis

When studying equilibria in a system, one crucial aspect is stability analysis. This analysis helps us understand whether these equilibria are stable, unstable, or somewhere in between.

To check stability, we examine what happens to the system's behavior as small disturbances occur. Specifically, do such disturbances cause the system to return to equilibrium, or do they grow and move the system away? For differential equations like the one given, we use signs of the derivative to assess changes:

• If a small change results in derivatives that "pull" back towards the equilibrium, it is considered stable.
• Conversely, if the derivative "pushes" the state away, it indicates instability.

In the given problem, equilibrium points occur along the circle defined by the equation $y^2+x^2=1$. When a point is disturbed to be outside of this circle, the derivative $y^{\prime }<0$ pulls it back in. If it moves inside the circle, $y^{\prime }>0$ also nudges it back out. This behavior confirms the circle's stability.

Differential Equations

Differential equations are a fundamental tool in mathematics for modeling how things change. They define relationships between functions and their derivatives. In other words, these equations show how a quantity depends on its rate of change.

In our exercise, the differential equation is $y^{\prime }=1-y^2-x^2$. Here, $y^{\prime }$ represents the rate of change of $y$ with respect to some parameter, such as time. This equation changes depending on the values of $x$ and $y$, and by solving it, we can learn about the system's behavior.

To solve a differential equation, we often try to find equilibrium points, where the rate of change is zero, meaning the system is not moving. Analyzing these points provides insights into the system's long-term behavior.

Circle of Equilibrium

In this particular exercise, the concept of a "circle of equilibrium" is quite literal. The equation $y^2+x^2=1$ describes a perfect circle centered at the origin with a radius of 1, forming a geometric representation of the equilibria.

This circle is important because it illustrates all combinations of $x$ and $y$ where the system is in equilibrium. Therefore, when $y^2+x^2=1$, the rate of change $y^{\prime }$ is exactly zero, meaning the system maintains its state without any tendency to move.

The circle acts as a boundary or a surface that physically represents the state of balance within the system. Having this visual cue helps in understanding how disturbances may affect the system and confirms its stability. Essentially, points on the circle contrast smoothly against points outside or inside, which helps in drawing clear inferences about their stability.