Question

Suppose it is known that for any \(N>0\) and \(t>0\) there exists a solution \(\varphi\) of \(\mathrm{Eq} .\) (1) such that for some \(t>0\) the inequality \(\|\varphi(t)\|>N\|\varphi(0)\|\) holds, and \(\|\varphi(0)\|<\varepsilon\). Does it follow from this that the equilibrium position \(\boldsymbol{x}=\mathbf{o}\) is unstable?

Short answer

Answer: Yes, the equilibrium position \(\boldsymbol{x} = \mathbf{o}\) is unstable because for every neighborhood around the equilibrium, there exists a trajectory that starts close to the equilibrium but leaves the neighborhood after some time.

Step-by-step solution

Write down the definition of an unstable equilibrium

An equilibrium position \(\boldsymbol{x} = \mathbf{o}\) is considered unstable if for every neighborhood of the equilibrium, there exists a trajectory that starts close to the equilibrium but leaves the neighborhood after some time. In mathematical terms, given any \(\delta > 0\), there exists an initial condition \(\|\varphi(0)\| < \delta\) such that the solution \(\varphi(t)\) moves away from the equilibrium position as \(t > 0\).

Use the given inequality to show the trajectory moves away from the equilibrium

We know that for any \(N > 0\) and \(t > 0\), there exists a solution \(\varphi\) of Eq. (1) such that for some \(t > 0\) the inequality \(\|\varphi(t)\| > N\|\varphi(0)\|\) holds, and \(\|\varphi(0)\| < \varepsilon\). Choose a neighborhood of size \(\eta\) around the equilibrium position such that \(\eta > N\delta\). This is possible because \(\delta > 0\) and \(N > 0\). Now, we have a solution \(\varphi\) with initial conditions \(\|\varphi(0)\| < \delta\) such that \(\|\varphi(t)\| > N\|\varphi(0)\|\) for some \(t > 0\). Therefore, we have: \(\|\varphi(t)\| > N\delta > \eta\) This shows that the trajectory represented by \(\varphi(t)\) moves outside the neighborhood of size \(\eta\) around the equilibrium after some time \(t > 0\).

Conclude that the equilibrium position is unstable

Since for every neighborhood around the equilibrium, there exists a trajectory that starts close to the equilibrium (with initial conditions \(\|\varphi(0)\| < \delta\)) but leaves the neighborhood after some time (as \(\|\varphi(t)\| > \eta\)), we can conclude that the equilibrium position \(\boldsymbol{x} = \mathbf{o}\) is unstable.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are fundamental in modeling the dynamics of various systems, such as population growth, heat transfer, or motion. For instance, when we're dealing with physics problems, differential equations can describe how the velocity of an object changes over time.

In our context, the stability of an equilibrium point is analyzed using differential equations. To put it simply, a differential equation will govern how a certain system changes, with the solution to this equation providing the trajectory of the system over time. In the exercise, we're examining whether a solution to such an equation indicates that the equilibrium is unstable, meaning the system will not stay close to the equilibrium after some disturbance.

Trajectory of a Solution

The 'trajectory of a solution' in the field of differential equations refers to the path that a system's state follows in a state-space as it evolves over time. This path is traced out by solutions to the differential equations representing the system. It can be visualized as a curve on a graph, where each point corresponds to the state of the system at a specific time.

In the provided exercise, the trajectory is crucial in understanding the behavior of the system near an equilibrium point. The exercise requires examining whether the trajectory stays near this equilibrium or moves away from it, which is key to determining the stability of the system. In simpler terms, just like a ball rolling on a surface, the trajectory would tell us if the ball comes to a stop (stable) or rolls away (unstable) when pushed slightly.

Stability Theory

Stability theory deals with the behavior of solutions to differential equations near an equilibrium point. An equilibrium is considered stable if a system that starts off near this point stays close to it over time. Conversely, it's unstable if any small displacement will cause the system to move away from the equilibrium. This concept is critical when we want to predict whether a system will return to a steady state after a small disturbance, such as a pendulum swinging back to its resting position after being pushed.

In our exercise, we are asked to determine the stability of an equilibrium point based on a given inequality. Stability theory guides us to conclude that, according to the given conditions, the equilibrium point is unstable. This is because there will always be a trajectory that moves away from the point, despite starting very close to it, fulfilling the condition required for instability.