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Chapter 6: Problem 18

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. $$ \begin{aligned} &x=c_{1} e^{-2 t}+c_{2} e^{-4 t} \\ &y=c_{1} e^{-2 t}+2 c_{2} e^{-4 t} \end{aligned} $$

Short Answer

Expert verified
Answer: The equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable.

Step by step solution

01

Analyze the given expressions

We have the given expressions for the solutions x(t) and y(t): $$ \begin{aligned} &x=c_{1} e^{-2 t}+c_{2} e^{-4 t} \\\ &y=c_{1} e^{-2 t}+2 c_{2} e^{-4 t} \end{aligned} $$ Here, we can see that both x(t) and y(t) are sums of exponential functions with negative exponents. As \(t \rightarrow \infty\), exponential functions with negative exponents will approach 0.
02

Determine the behavior of x(t) and y(t) as \(t \rightarrow \infty\)

To see how the solutions behave as time goes to infinity, we will analyze the individual terms in x(t) and y(t). For x(t), we have two terms: \(c_{1} e^{-2 t}\) and \(c_{2} e^{-4 t}\). As \(t \rightarrow \infty\), both terms will approach 0, regardless of the constants \(c_1\) and \(c_2\) since the exponents are negative: $$\lim_{t\to\infty} c_{1} e^{-2 t} = 0$$ $$\lim_{t\to\infty} c_{2} e^{-4 t} = 0$$ For y(t), we also have two terms: \(c_{1} e^{-2 t}\) and \(2 c_{2} e^{-4 t}\). Similarly, as \(t \rightarrow \infty\), both terms will approach 0: $$\lim_{t\to\infty} c_{1} e^{-2 t} = 0$$ $$\lim_{t\to\infty} 2 c_{2} e^{-4 t} = 0$$
03

Classify the stability of the equilibrium point

Since both x(t) and y(t) approach 0 as \(t \rightarrow \infty\), the solution (x, y) approaches the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\). As a result, we can conclude that the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable, because the system converges towards the equilibrium point as time goes to infinity.

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Key Concepts

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

Asymptotic Stability
In the context of linear differential systems, asymptotic stability is a vital concept. It refers to how the solutions of a differential system behave as time progresses toward infinity. A system is called asymptotically stable if solutions approach an equilibrium point as time tends to infinity and stay close to it when given small disturbances.
The general solution provided in the exercise exhibits characteristics of asymptotic stability. Both solutions for \(x(t)\) and \(y(t)\) contain terms with exponential functions that have negative exponents. This means, as time \(t\) approaches infinity, these exponential terms tend to zero. Thus, the solutions \(x(t)\) and \(y(t)\) also tend towards zero.
This behavior indicates that the system gravitates naturally back to the equilibrium point when disturbed, making it asymptotically stable. The conclusion is that the equilibrium point \( \mathbf{y}_{e} = \mathbf{0} \) is asymptotically stable since all paths of \(x(t)\) and \(y(t)\) lead back to this point as time progresses.
Equilibrium Points
Equilibrium points in differential equations are the states at which the system remains constant, meaning the derivatives (rate-of-change) of the system are zero. In the exercise, it is given that the equilibrium point to consider is \( \mathbf{y}_{e} = \mathbf{0} \). This is a point where both \(x(t)\) and \(y(t)\) settle eventually, because they do not change at this point.
The inherent nature of equilibrium points in linear systems offers several insights:
  • An equilibrium point is stable if, when slightly disturbed, the system remains close to it.
  • If the system returns to the equilibrium point after a disturbance, the system is said to be asymptotically stable.
  • Unstable equilibrium points diverge away with small disturbances.
This exercise showed that \( \mathbf{y}_{e} = \mathbf{0} \) is not only stable but also asymptotically stable since both solutions for \(x(t)\) and \(y(t)\) decay towards zero over time.
Exponential Functions
The exponential functions in differential equations like those given in the exercise reveal important behavior about the system's stability and convergence.
Exponential functions in the form of \(e^{rt}\), where \(r\) is a constant, exhibit growth or decay based on the sign of \(r\).
  • If \(r\) is positive, \(e^{rt}\) grows as \(t\) increases.
  • If \(r\) is negative, \(e^{rt}\) decays as \(t\) increases, approaching zero.
In this particular exercise, the functions \(e^{-2t}\) and \(e^{-4t}\) illustrate exponential decay since both have negative exponents.
This decay towards zero signifies that any oscillations or deviations in the system's state eventually diminish, steering the system back to its equilibrium point. Such characteristics are crucial as they determine whether a linear differential system like the one in this exercise will be stable in the long run.

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Most popular questions from this chapter

In each exercise, the given system is an almost linear system at each of its equilibrium points. (a) Find the (real) equilibrium points of the given system. (b) As in Example 2, find the corresponding linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\) at each equilibrium point. (c) What, if anything, can be inferred about the stability properties of the equilibrium point(s) by using Theorem \(6.4\) ? $$ \begin{aligned} &x^{\prime}=y^{2}-x \\ &y^{\prime}=x^{2}-y \end{aligned} $$

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 2 & -3 \\ 3 & 2 \end{array}\right] \mathbf{y} $$

Use the polar equations derived in Exercise 25 to show that if $$ a_{11}=a_{22}, \quad a_{21}=-a_{12}, \quad g_{1}(\mathbf{z})=z_{1} h\left(\sqrt{z_{1}^{2}+z_{2}^{2}}\right), \quad g_{2}(\mathbf{z})=z_{2} h\left(\sqrt{z_{1}^{2}+z_{2}^{2}}\right) $$ for some function \(h\), then the polar equations uncouple into $$ \begin{aligned} r^{\prime} &=a_{11} r+r h(r) \\ \theta^{\prime} &=a_{21} . \end{aligned} $$ Note that the radial equation is a separable differential equation and the angle equation can be solved by antidifferentiation.

In each exercise, the given system is an almost linear system at each of its equilibrium points. (a) Find the (real) equilibrium points of the given system. (b) As in Example 2, find the corresponding linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\) at each equilibrium point. (c) What, if anything, can be inferred about the stability properties of the equilibrium point(s) by using Theorem \(6.4\) ? $$ \begin{aligned} &x^{\prime}=x y-1 \\ &y^{\prime}=(x+4 y)(x-1) \end{aligned} $$

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 1 & -6 \\ 2 & -6 \end{array}\right] \mathbf{y} $$
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