Question

Each exercise lists the general solution of a linear system of the form $$ \begin{aligned} &x^{\prime}=a_{11} x+a_{12} y \\ &y^{\prime}=a_{21} x+a_{22} y \end{aligned} $$ where \(a_{11} a_{22}-a_{12} a_{21} \neq 0\). Determine whether the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable, stable but not asymptotically stable, or unstable. $$ \begin{aligned} &x=c_{1} \cos 2 t+c_{2} \sin 2 t \\ &y=-c_{1} \sin 2 t+c_{2} \cos 2 t \end{aligned} $$

Short answer

Answer: The equilibrium point 𝐲𝑒=𝐲0 is stable but not asymptotically stable in the given linear system.

Step-by-step solution

Check the behavior of the system at \(\mathbf{y}_{e} = \mathbf{0}\)

Since \(\mathbf{y}_{e} = \mathbf{0}\), let's set \(x = 0\) and \(y = 0\) in the given solution. $$ \begin{aligned} 0 &= c_{1} \cos 2t + c_{2} \sin 2t \\ 0 &= -c_{1} \sin 2t + c_{2} \cos 2t \end{aligned} $$

Identify types of stability

Now we need to determine the type of stability based on the behavior of the system near the equilibrium point 1. Asymptotically Stable: A system is asymptotically stable if all initial conditions converge to the equilibrium point as time approaches infinity. 2. Stable but not Asymptotically Stable: A system is stable but not asymptotically stable if all initial conditions remain close to the equilibrium point as time approaches infinity, but they do not converge to the equilibrium point. 3. Unstable: A system is unstable if there exists at least one initial condition that diverges away from the equilibrium point as time approaches infinity. In our case, it is clear that for any non-zero constants \(c_1\) and \(c_2\), the system is oscillating and not converging to the equilibrium point, but it remains close to the equilibrium point.

Identify the type of stability for the given system

The given system has oscillatory behavior and remains close to the equilibrium point as time approaches infinity, but it does not converge to the equilibrium point. Therefore, the equilibrium point \(\mathbf{y}_{e} = \mathbf{0}\) is stable but not asymptotically stable.

Key concepts

Asymptotically Stable

Understanding the concept of asymptotic stability is fundamental in the analysis of linear systems. When discussing asymptotic stability, we are looking at how a system behaves over time in response to its initial conditions. Specifically, a system is said to be asymptotically stable when any trajectory starting from an initial condition will eventually converge to the equilibrium point as time progresses indefinitely.

Imagine dropping a marble into a bowl. If the marble eventually comes to rest at the bottom, no matter where in the bowl you initially placed it, this is akin to how an asymptotically stable system behaves—that ultimate rest point being the equilibrium. This concept is essential in fields such as control theory, where having a system settle into a desired state over time is crucial.

In the exercise, we would determine a system is asymptotically stable if, after substituting the equilibrium conditions, the solutions for the system show the variables approaching zero as time (\( t \rightarrow \)infinity). In contrast to our example problem, however, an asymptotically stable system would result in both constants (\( c_1 \)and \( c_2 \)) decaying to zero with increasing time.

Equilibrium Point

An equilibrium point in a linear system is a special solution where the rates of change of all the variables are zero. In other words, it's a point in the system where nothing changes—it's in perfect balance. This is why the equilibrium point is often referred to as a steady state or a fixed point.

Considering our standard linear system, the equilibrium point is usually denoted by \(\mathbf{y}_e\) and is a valuable tool for understanding system behavior. By analyzing the system at the equilibrium point, as done in the exercise (setting both \( x \)and \( y \)to zero), we can infer the nature of the system's stability. An equilibrium point can be stable, unstable, or asymptotically stable, influencing how a system might respond to outside disturbances or initial conditions.

The provided exercise demonstrates that the equilibrium point is at the origin \(\mathbf{y}_e = \mathbf{0}\), which is a common convention for simplifying linear system analysis and aiding in visual interpretation on phase planes.

Oscillatory Behavior

A key dynamic exhibited by some linear systems is oscillatory behavior. This behavior is characterized by variables in the system perpetually increasing and decreasing over time, similar to the motion of a pendulum or the vibrations of a plucked guitar string. Oscillatory systems are crucial in engineering and physics, underpinning the behavior of circuits, mechanical oscillators, and even the orbit of celestial bodies.

In the exercise, the provided general solution describes such oscillatory behavior through the use of sine and cosine functions, with \( c_1 \)and \( c_2 \)as constants determining the amplitude of the oscillations. Importantly, while the system always remains close to the equilibrium point, it never actually settles there. Instead, it continues in a regular, unending cycle of motion.

The significance of oscillatory behavior in the context of stability is that it indicates the system is bounded and predictable but does not reach a static state at the equilibrium. This explains why the system from the exercise is considered stable but not asymptotically stable, as the motion continues indefinitely without converging to a halt at \(\mathbf{y}_e = \mathbf{0}\).