/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Show that the equilibrium \(\lef... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that the equilibrium \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) of $$ \left[\begin{array}{c} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{ll} 2 & -4 \\ 5 & -6 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ is stable.

Short Answer

Expert verified
The equilibrium \( \begin{bmatrix} 0 \\ 0 \end{bmatrix} \) is stable because the eigenvalues of the matrix have a negative real part.

Step by step solution

01

Understand Stability of Equilibrium

The equilibrium point \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \) is stable if, when perturbed, the system returns to this point over time. Stability is determined by the eigenvalues of the matrix associated with the system.
02

Extract the System Matrix

Identify the matrix \( A \) from the system of equations: \[ A = \begin{bmatrix} 2 & -4 \ 5 & -6 \end{bmatrix} \]
03

Find the Characteristic Equation

For matrix \( A \), form the characteristic equation \( \det(A - \lambda I) = 0 \), where \( \lambda \) is an eigenvalue, and \( I \) is the identity matrix: \[ A - \lambda I = \begin{bmatrix} 2-\lambda & -4 \ 5 & -6-\lambda \end{bmatrix} \] \[ \det(A - \lambda I) = (2-\lambda)(-6-\lambda) - (5)(-4) \]
04

Simplify the Determinant

Simplify \( \det(A - \lambda I) \) to find the characteristic equation: \[ (2-\lambda)(-6-\lambda) + 20 \] \[ = \lambda^2 + 4\lambda + 8 \] This simplifies to the characteristic equation \( \lambda^2 + 4\lambda + 8 = 0 \).
05

Solve for Eigenvalues

Solve the characteristic equation for \( \lambda \): \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 4 \), \( c = 8 \). The discriminant \( \Delta = 4^2 - 4 \times 1 \times 8 = 16 - 32 = -16 \), which is negative, indicating complex eigenvalues.
06

Determine Stability from Eigenvalues

The eigenvalues are complex and can be written as \( \lambda = -2 \pm 2i \sqrt{\frac{|\Delta|}{16}} \). Since the real part of the complex eigenvalues is negative (\(-2\)), the equilibrium point is stable.

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Key Concepts

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

Eigenvalues
In the study of linear algebra and dynamic systems, eigenvalues play a crucial role. They are numbers associated with a square matrix that provide valuable insights about the behavior of linear transformations represented by matrices. An eigenvalue is determined by solving the equation \( A\vec{v} = \lambda\vec{v} \), where \( A \) is a matrix, \( \lambda \) is an eigenvalue, and \( \vec{v} \) is the corresponding eigenvector.
In the context of our system, eigenvalues help us understand the long-term behavior of the system. Specifically, they determine whether the system will converge to an equilibrium point or diverge away. This makes eigenvalues fundamental in determining the stability of dynamic systems.
When examining the matrix associated with a dynamic system, solving for eigenvalues is one of the first steps to analyze stability.
Characteristic Equation
The characteristic equation is a polynomial whose roots are the eigenvalues of a matrix. For a given square matrix \( A \), the characteristic equation is obtained by calculating the determinant of \( A - \lambda I \) and setting it to zero, resulting in \( \det(A - \lambda I) = 0 \).
For our specific example, the matrix \( A \) is:
  • \( \begin{bmatrix} 2 & -4 \ 5 & -6 \end{bmatrix} \)
Add \( \lambda I \) (where \( I \) is the identity matrix) and compute the determinant to come up with the characteristic polynomial \( \lambda^2 + 4\lambda + 8 = 0 \).
The characteristic equation is essential because it directly provides us with the eigenvalues of the matrix, which we need to determine the system's stability.
Complex Eigenvalues
When solving the characteristic equation, sometimes we encounter complex numbers as eigenvalues. These arise particularly when the discriminant of the polynomial is negative, meaning the polynomial has no real roots. Instead, the solutions are complex conjugates.
For the system matrix \( A \), we found a characteristic equation \( \lambda^2 + 4\lambda + 8 = 0 \) with a negative discriminant \(-16\). This results in complex eigenvalues \( \lambda = -2 \pm 2i \).
Complex eigenvalues often indicate oscillatory behavior in systems. However, the real part of these eigenvalues influences stability. With a negative real part in our eigenvalues, the system shows damped oscillation, suggesting stable equilibrium.
Stability of Equilibrium
In dynamic systems, stability of equilibrium refers to whether a system will return to an equilibrium state after a slight disturbance. This is crucial for determining the system's behavior over time.
For our example, we analyze the system's stability by looking at the eigenvalues of the matrix \( A \). We found that the eigenvalues are complex with a negative real part: \( \lambda = -2 \pm 2i \). This indicates that any small perturbations will diminish over time, leading the system back to equilibrium.
Stability is determined primarily by the sign of the real parts of the eigenvalues:
  • Negative real parts mean the system returns to equilibrium (stable).
  • Positive real parts suggest the system will move away from equilibrium (unstable).
  • Zero real parts require further investigation, as they could indicate neutral stability or a more complicated behavior.
In conclusion, complex eigenvalues with negative real parts assure us that our equilibrium point is stable.

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Most popular questions from this chapter

The growth rate \(r\) of a particular organism is affected by both the availability of food and the number of competitors for the food source. Denote the amount of food available at time \(t\) by \(F(t)\) and the number of competitors at time \(t\) by \(N(t)\). The growth rate \(r\) can then be thought of as a function of the two time-dependent variables \(F(t)\) and \(N(t)\). Assume that the growth rate is an increasing function of the availability of food and a decreasing function of the number of competitors. How is the growth rate \(r\) affected if the availability of food decreases over time while the number of competitors increases?

Find the Jacobi matrix for each given function. $$ \mathbf{f}(x, y)=\left[\begin{array}{c} x+y \\ x^{2}-y^{2} \end{array}\right] $$

Evaluate the Nicholson-Bailey model for the first 15 generations when \(a=0.02, c=3\), and \(b=1.5 .\) For the initial host density, choose \(N_{0}=5\), and for the initial parasitoid density, choose \(P_{0}=5\).

Refer to the negative binomial host-parasitoid model. Problems \(7,8,11\), and 12 are best done with the help of a spreadsheet, but can also be done with a calculator. The negative binomial model is a discrete-generation host- parasitoid model of the form $$ \begin{array}{l} N_{t+1}=b N_{t}\left(1+\frac{a P_{t}}{k}\right)^{-k} \\ P_{t+1}=c N_{t}\left[1-\left(1+\frac{a P_{t}}{k}\right)^{ k}\right] \end{array} $$ for \(t=0,1,2, \ldots\) Evaluate the negative binomial model for the first 10 generations when \(a=0.02, c=3, k=0.75\), and \(b=0.5 .\) For the initial host density, choose \(N_{0}=15\), and for the initial parasitoid density, choose \(P_{0}=0\)

In the negative binomial model, the fraction of hosts escaping parasitism is given by $$ f(P)=\left(1+\frac{a P}{k}\right)^{-k} $$ (a) Graph \(f(P)\) as a function of \(P\) for \(a=0.1\) and \(a=0.01\) when \(k=0.75\) (b) For \(k=0.75\) and a given value of \(P\), how are the chances of escaping parasitism affected by increasing \(a\) ?

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