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Each exercise lists a linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\), where \(A\) is a real constant invertible \((2 \times 2)\) matrix. Use Theorem \(6.3\) to determine whether the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable, stable but not asymptotically stable, or unstable. $$ \begin{aligned} &x^{\prime}=-3 x+3 y \\ &y^{\prime}=x-5 y \end{aligned} $$

Short Answer

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

Step by step solution

01

Identify coefficients and form the matrix A

The given equations are: $$ \begin{aligned} &x^{\prime}=-3x+3y \\ &y^{\prime}=x-5y \end{aligned} $$ From these equations, we can get the matrix A with coefficients: $$ A = \begin{pmatrix} -3 & 3 \\ 1 & -5 \end{pmatrix} $$
02

Find the eigenvalues of the matrix A

To find the eigenvalues, we need to determine the roots of the characteristic equation, which we can obtain by taking the determinant of \((A - \lambda I)\). Here, \(I\) is the identity matrix, and \(\lambda\) represents the eigenvalues: $$ \det(A - \lambda I) = \det\begin{pmatrix} -3-\lambda & 3 \\ 1 & -5-\lambda \end{pmatrix} $$ Calculate the determinant: $$ (-3-\lambda)(-5-\lambda)-3(1) = \lambda^{2} + 8\lambda + 16 - 3 = \lambda^2 + 8\lambda + 13 $$
03

Determine the roots of the characteristic equation

Now we need to find the roots (eigenvalues) of the characteristic equation \(\lambda^2 +8\lambda + 13 = 0\). Using the quadratic formula, \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we have: $$ \lambda_{1,2} = \frac{-8 \pm \sqrt{(-8)^{2} - 4(1)(13)}}{2(1)} $$ Calculate the discriminant: $$ (-8)^{2} - 4(1)(13) = 64 - 52 = 12 $$ Thus, the eigenvalues are: $$ \lambda_{1,2} = \frac{-8 \pm \sqrt{12}}{2} $$
04

Analyze the stability using Theorem 6.3

According to Theorem 6.3, the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is: 1. Asymptotically stable if all eigenvalues of A have negative real parts. 2. Stable but not asymptotically stable if all eigenvalues of A have non-positive real parts, and at least one eigenvalue has a zero real part. 3. Unstable if at least one eigenvalue of A has a positive real part. Since both eigenvalues have a negative real part, \(\lambda_1 = \frac{-8 + \sqrt{12}}{2} < 0\) and \(\lambda_2 = \frac{-8 - \sqrt{12}}{2} < 0\), we can conclude that the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable.

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Key Concepts

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

Eigenvalues and Eigenvectors
In linear algebra, eigenvalues and eigenvectors are crucial concepts that help understand the behavior of linear transformations and systems of differential equations. An eigenvalue is a scalar that reveals how much the corresponding eigenvector is stretched or shrunk during the transformation associated with a matrix. For a given matrix \( A \), the eigenvalue \( \lambda \) and the corresponding eigenvector \( \mathbf{v} \) satisfy the equation: \[ A \mathbf{v} = \lambda \mathbf{v} \] Eigenvectors remain in the same direction after the application of the matrix \( A \), merely scaled by the eigenvalue \( \lambda \). Considering our system of linear differential equations, finding the eigenvalues of matrix \( A \) becomes the key to understanding system dynamics. Here are a few important points to note:
  • Eigenvectors often provide insight into the invariant directions of a transformation.
  • Eigenvalues help determine stability, oscillation, and other dynamic behaviors of systems.
  • The number of eigenvalues usually equals the order of the matrix.
Stability Analysis
Stability analysis is a method used to assess whether the solutions of a system of differential equations remain close to an equilibrium point over time. When dealing with linear systems, this analysis often relies on the eigenvalues of the system matrix. These eigenvalues tell us how solutions grow, decay, or oscillate. The stability of an equilibrium point can be understood as follows:
  • An **asymptotically stable** equilibrium occurs when all eigenvalues have negative real parts. Solutions converge to the equilibrium as time progresses.
  • An equilibrium is considered **stable but not asymptotically stable** if eigenvalues have non-positive real parts, with one being zero. Here, solutions neither diverge nor converge sharply.
  • An **unstable** equilibrium is identified when at least one eigenvalue has a positive real part, causing solutions to diverge from the equilibrium over time.
Understanding the stability of a system is crucial, especially in predicting long-term behavior in scientific and engineering problems.
Characteristic Equation
The characteristic equation arises from linear algebra and is foundational in finding eigenvalues of a matrix. It is a polynomial equation derived from the determinant of the matrix equation \( (A - \lambda I) \), where \( A \) is the system matrix and \( I \) is the identity matrix. Forming the characteristic equation involves:
  • Calculating the determinant of \( A - \lambda I \).
  • Setting the determinant equal to zero: \( \det(A - \lambda I) = 0 \).
  • Solving the resulting polynomial equation for \( \lambda \), which provides the eigenvalues.
For our system, we encountered a quadratic characteristic equation: \[ \lambda^2 + 8\lambda + 13 = 0 \] This quadratic equation was solved using the quadratic formula to find the eigenvalues, which are essential for determining stability and system behavior.
Equilibrium Point
An equilibrium point in a system of differential equations refers to a condition where the system does not change over time, meaning all derivatives equal zero. For our specific system described by \( \mathbf{y}^{\prime} = A \mathbf{y} \), the origin \( \mathbf{y}_e = \mathbf{0} \) is generally regarded as an equilibrium point. When analyzing equilibrium points, we look at:
  • The **system behavior** around the equilibrium point, which is heavily influenced by eigenvalues of matrix \( A \).
  • The **stability** of this point determines whether the system returns to equilibrium after a disturbance.
Finding and studying equilibrium points is important since they often represent steadied states of systems, like continuous motion or constant concentration levels in chemical reactions. In essence, understanding equilibrium points can offer significant insights into system dynamics and potential long-term outcomes.

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Most popular questions from this chapter

Each exercise lists a nonlinear system \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\), where \(A\) is a constant ( \(2 \times 2\) ) invertible matrix and \(\mathbf{g}(\mathbf{z})\) is a \((2 \times 1)\) vector function. In each of the exercises, \(\mathbf{z}=\mathbf{0}\) is an equilibrium point of the nonlinear system. (a) Identify \(A\) and \(\mathbf{g}(\mathbf{z})\). (b) Calculate \(\|\mathbf{g}(\mathbf{z})\|\). (c) Is \(\lim _{\mid \mathbf{z} \| \rightarrow 0}\|\mathbf{g}(\mathbf{z})\| /\|\mathbf{z}\|=0\) ? Is \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\) an almost linear system at \(\mathbf{z}=\mathbf{0}\) ? (d) If the system is almost linear, use Theorem \(6.4\) to choose one of the three statements: (i) \(\mathbf{z}=\mathbf{0}\) is an asymptotically stable equilibrium point. (ii) \(\mathbf{z}=\mathbf{0}\) is an unstable equilibrium point. (iii) No conclusion can be drawn by using Theorem \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=9 z_{1}+5 z_{2}+z_{1} z_{2} \\ &z_{2}^{\prime}=-7 z_{1}-3 z_{2}+z_{1}^{2} \end{aligned} $$

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

Introduce polar coordinates and transform the given initial value problem into an equivalent initial value problem for the polar variables. Solve the polar initial value problem, and use the polar solution to obtain the solution of the original initial value problem. If the solution exists at time \(t=1\), evaluate it. If not, explain why. $$ \begin{aligned} &x^{\prime}=y-x \ln \left[x^{2}+y^{2}\right], \quad x(0)=e / \sqrt{2} \\ &y^{\prime}=-x-y \ln \left[x^{2}+y^{2}\right], \quad y(0)=e / \sqrt{2} \end{aligned} $$

These exercises explore the question "When one of two species in a colony is desirable and the other is undesirable, is it better to use resources to nurture the growth of the desirable species or to harvest the undesirable one?" Let \(x(t)\) and \(y(t)\) represent the populations of two competing species, with \(x(t)\) the desirable species. Assume that if resources are invested in promoting the growth of the desirable species, the population dynamics are given by $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x+\mu x \\ &y^{\prime}=r(1-\alpha y-\beta x) y \end{aligned} $$ If resources are invested in harvesting the undesirable species, the dynamics are $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x \\ &y^{\prime}=r(1-\alpha y-\beta x) y-\mu y \end{aligned} $$ In (10), \(r, \alpha, \beta\), and \(\mu\) are positive constants. For simplicity, we assume the same parameter values for both species. For definiteness, assume that \(\alpha>\beta>0\). Consider system (10), which describes the strategy in which resources are invested in harvesting the undesirable species. Again assume that \(\alpha>\beta>0\). (a) Determine the four equilibrium points for the system. (b) Show that it is possible, by investing sufficient resources (that is, by making \(\mu\) large enough), to prevent equilibrium coexistence of the two species. In fact, if \(\mu>r\), show that there are only two physically relevant equilibrium points. (c) Assume \(\mu>r\). Compute the linearized system at each of the two physically relevant equilibrium points. Determine the stability characteristics of the linearized system at each of these equilibrium points. (d) System (10) can be shown to be an almost linear system at each of the equilibrium points. Use this fact and the results of part (c) to infer the stability properties of system (10) at each of the two equilibrium points of interest. (e) Sketch the direction field. Will sufficiently aggressive harvesting of species \(y\) ultimately drive undesirable species \(y\) to extinction? If so, what is the limiting population of species \(x\) ?

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 \(6.4\) ? $$ \begin{aligned} &x^{\prime}=(x-y)(y+1) \\ &y^{\prime}=(x+2)(y-4) \end{aligned} $$

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