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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

Expert verified
Answer: The equilibrium point 饾惒饾憭=饾惒0 is stable but not asymptotically stable in the given linear system.

Step by step solution

01

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. $$ \begin{aligned} 0 &= c_{1} \cos 2t + c_{2} \sin 2t \\ 0 &= -c_{1} \sin 2t + c_{2} \cos 2t \end{aligned} $$
02

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.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攖hat 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鈥攊t'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 \(\mathbf{y}_e = \mathbf{0}\).

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Most popular questions from this chapter

Consider the initial value problem $$ \frac{d}{d t}\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right]=\left[\begin{array}{c} \frac{5}{4} y_{1}^{1 / 5}+y_{2}^{2} \\ 3 y_{1} y_{2} \end{array}\right], \quad\left[\begin{array}{l} y_{1}(0) \\ y_{2}(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] $$ For the given autonomous system, the two functions \(f_{1}\left(y_{1}, y_{2}\right)=\frac{5}{4} y_{1}^{1 / 5}+y_{2}^{2}\) and \(f_{2}\left(y_{1}, y_{2}\right)=3 y_{1} y_{2}\) are continuous functions for all \(\left(y_{1}, y_{2}\right)\). (a) Show by direct substitution that $$ y_{1}(t)=\left\\{\begin{array}{lr} 0, & -\infty

Consider the system $$ \begin{aligned} &x^{\prime}=x+y^{2}-x y^{n} \\ &y^{\prime}=-x+y^{-1} . \end{aligned} $$ The slope of the phase-plane trajectory passing through the point \((x, y)=(1,2)\) is \(\frac{1}{6}\). Determine the exponent \(n\).

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. $$ \begin{aligned} &x=c_{1} e^{-2 t} \cos 2 t+c_{2} e^{-2 t} \sin 2 t \\ &y=-c_{1} e^{-2 t} \sin 2 t+c_{2} e^{-2 t} \cos 2 t \end{aligned} $$

Assume the given autonomous system models the population dynamics of two species, \(x\) and \(y\), within a colony. (a) For each of the two species, answer the following questions. (i) In the absence of the other species, does the remaining population continuously grow, decline toward extinction, or approach a nonzero equilibrium value as time evolves? (ii) Is the presence of the other species in the colony beneficial, harmful, or a matter of indifference? (b) Determine all equilibrium points lying in the first quadrant of the phase plane (including any lying on the coordinate axes). (c) The given system is an almost linear system at the equilibrium point \((x, y)=(0,0)\). Determine the stability properties of the system at \((0,0)\). $$ \begin{aligned} &x^{\prime}=-x-x^{2} \\ &y^{\prime}=-y+x y \end{aligned} $$

Consider the nonlinear scalar differential equation \(x^{\prime \prime}=1-(1+x)^{3 / 2}\). An equation having this structure arises in modeling the bobbing motion of a floating parabolic trough. (a) Let \(y=x^{\prime}\) and rewrite the given scalar equation as an equivalent first order system. (b) Show that the system has a single equilibrium point at \(\left(x_{e}, y_{e}\right)=(0,0)\). (c) Determine the linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\), and analyze its stability properties. (d) Assume that the system is an almost linear system at equilibrium point \((0,0)\). Does Theorem \(6.4\) provide any information about the stability properties of the nonlinear system obtained in part (a)? Explain.

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