Question

Assume that a two-dimensional autonomous system has an isolated equilibrium point at the origin and that the phase-plane solution curves consist of the family of concentric ellipses \(x^{2} / 4+y^{2}=C, C \geq 0\). (a) Apply the definition to show that the origin is a stable equilibrium point. In particular, given an \(\varepsilon>0\), determine a corresponding \(\delta>0\) so that all solutions starting within a circle of radius \(\delta\) centered at the origin stay within the circle of radius \(\varepsilon\) centered at the origin for all \(t \geq 0\). (The \(\delta\) you determine should be expressed in terms of \(\varepsilon\).) (b) Is the origin an asymptotically stable equilibrium point? Explain.

Short answer

2. Is the origin of the given system an asymptotically stable equilibrium point?

Step-by-step solution

Finding the relationship between epsilon and delta

Given that the phase-plane solution curves consist of the family of concentric ellipses $$ x^2/4 + y^2 = C ,\quad C \geq 0. $$ We need to find δ such that all solutions starting inside a circle of radius δ centered at the origin stay within a circle of radius ε centered at the origin for all \(t \geq 0\). Let's rewrite the ellipse equation in terms of x and y: $$ x^2 + 4y^2 = 4C. $$ Now, find the relation between C and δ. To do that, we can take a circle of radius δ centered at the origin. For this circle: $$ x^2 + y^2 = \delta^2. $$ As we want to find δ when the solution curve is inside the circle of radius ε, the largest value of C must satisfy: $$ 4C \leq \varepsilon^2. $$ For every ellipse with \(C \le \frac{\varepsilon^2} {4}\), we have: $$ x^2 + 4y^2 \le x^2 + y^2 + 3y^2 \le x^2 + y^2 + 3(x^2 + y^2) = \varepsilon^2. $$ Therefore, the largest ellipse that stays inside the circle of radius ε must have C satisfying $$ C \le \frac{\varepsilon^2} {4}. $$ Now we can express δ in terms of ε, we have: $$ \delta^2 \le 4C \Rightarrow \delta \le \sqrt{4C} \Rightarrow \delta=\sqrt{\varepsilon^2}. $$ So we have found a relationship between epsilon and delta: $$ \delta = \varepsilon. $$ This means that for all solutions starting within a circle of radius δ centered at the origin stay within the circle of radius ε centered at the origin for all \(t \geq 0\). Thus, by definition, the origin is a stable equilibrium point.

Determine the asymptotic stability at the origin

For the origin to be asymptotically stable, the solution should converge to the origin as time goes to infinity, meaning the ellipse solutions should get smaller and approach the origin for large t values. However, the given exercise states that the phase-plane solution curves consist of concentric ellipses with \(x^2 / 4+y^2=C\) for \(C\geq 0\). These ellipses do not approach the origin. They simply maintain a constant distance from the origin. This means the origin is not an asymptotically stable equilibrium point. In conclusion, the origin is a stable equilibrium point, but it is not an asymptotically stable equilibrium point.

Key concepts

Phase-Plane Solution Curves

When studying two-dimensional systems, particularly in the context of differential equations, an invaluable tool at our disposal is the concept of phase-plane solution curves. These curves provide a visual map of how the system evolves over time. Imagine plotting the state of a system where each point in the plane represents a solution to the differential equations at a given moment. For the exercise in question, the solution curves are defined by the family of concentric ellipses \(x^2 / 4 + y^2 = C, C \geq 0\).
Visualizing the phase plane with its solution curves offers insight into the nature of equilibrium points, which are where solution curves meet or diverge. In our exercise, the origin is an equilibrium point, as it's the center of all ellipses.
To make this concept more concrete, it's like the rings of a tree trunk, with each ring representing a possible state of the system, and the very center, in our case, the origin, represents stability. A small disturbance won’t push it away significantly, echoing the notion of stability we explored through the exercise.

Asymptotic Stability

Asymptotic stability takes the idea of equilibrium a step further by looking at how the system behaves over an infinite time horizon. For an equilibrium point to be considered asymptotically stable, any trajectory that starts near this point must not only stay near but also eventually converge to it as time tends to infinity.
In plain language, think of a marble in a bowl. If the marble is released from anywhere within the bowl, it not only stays inside but also eventually comes to rest at the bottom—the lowest point, or the point of equilibrium. In our exercise, however, the concentric ellipses imply that once our 'marble' (representing the state of the system) is set on a certain ring, it stays on that ring indefinitely, neither spiraling inward nor outward. Therefore, while the origin is a stable equilibrium point—as all points on ellipses are consistently distanced from it—it is not asymptotically stable because the ellipses represent paths that do not converge to the origin over time.

Autonomous System

An autonomous system within the context of differential equations is one where the rules governing the system's behavior are constant over time; they do not explicitly depend on time itself. Imagine programming a robot with a set of instructions that dictate its every move regardless of when you press 'start'.
For our example, the system is defined by its behavior in the phase plane without any time-driven forces changing the rules. This property is significant because it means that understanding the entire system's dynamics comes down to examining the geometry of the phase plane solution curves—those ellipses we discussed earlier. An autonomous system is predictable in this respect, and the study of these systems involves identifying and characterizing the nature of equilibrium points—like the stable, but not asymptotically stable, origin in our exercise.