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Chapter 10: Problem 16

A Jacobian matrix and two equlibria are given. Determine if each is locally stable, unstable, or if the analysis is inconclusive. $$\begin{array}{l}{J=\left[ \begin{array}{cc}{2 x_{1}} & {-\sin x_{2}} \\\ {\cos x_{1}} & {0}\end{array}\right]} \\ {\text { (i) } \hat{x}_{1}=1, \hat{x}_{2}=-\pi} \\ {\text { (ii) } \hat{x}_{1}=1, \hat{x}_{2}=\pi}\end{array}$$

Short Answer

Expert verified
Both equilibria are unstable.

Step by step solution

01

Evaluate Jacobian at the first equilibrium point

Substitute the values \(\hat{x}_{1} = 1\) and \(\hat{x}_{2} = -\pi\) into the Jacobian matrix \(J\).\[ J_1 = \begin{bmatrix} 2(1) & -\sin(-\pi) \ \cos(1) & 0 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ \cos(1) & 0 \end{bmatrix} \]
02

Determine eigenvalues for the first equilibrium point

Calculate the eigenvalues of \(J_1\) using the determinant \(\det(J_1 - \lambda I) = 0\).\[ \det \left(\begin{bmatrix} 2 - \lambda & 0 \ \cos(1) & -\lambda \end{bmatrix}\right) = (2-\lambda)(-\lambda) = -\lambda(\lambda - 2) = 0\]The eigenvalues are \(\lambda = 0\) and \(\lambda = 2\).
03

Assess stability at the first equilibrium point

If any eigenvalue has a positive real part, the equilibrium is unstable. Since \(\lambda = 2\) is positive, the first equilibrium is unstable.
04

Evaluate Jacobian at the second equilibrium point

Substitute the values \(\hat{x}_{1} = 1\) and \(\hat{x}_{2} = \pi\) into the Jacobian matrix \(J\).\[ J_2 = \begin{bmatrix} 2(1) & -\sin(\pi) \ \cos(1) & 0 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ \cos(1) & 0 \end{bmatrix} \]
05

Determine eigenvalues for the second equilibrium point

Calculate the eigenvalues of \(J_2\) in a similar manner to \(J_1\).\[ \det \left(\begin{bmatrix} 2 - \lambda & 0 \ \cos(1) & -\lambda \end{bmatrix}\right) = -\lambda(\lambda - 2) = 0\]The eigenvalues are \(\lambda = 0\) and \(\lambda = 2\).
06

Assess stability at the second equilibrium point

Similar to the first equilibrium, if any eigenvalue has a positive real part, it indicates instability. Since \(\lambda = 2\) is positive, the second equilibrium is unstable.

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Key Concepts

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

Eigenvalues
Eigenvalues are a fundamental concept in linear algebra and have key applications in stability analysis and differential equations. When examining the system's behavior near an equilibrium point, eigenvalues of the Jacobian matrix provide crucial insights. These are essentially numbers that describe how a dynamic system behaves in the neighborhood of equilibrium.

The process of determining eigenvalues typically involves solving the characteristic equation given by \( \ ext{det}(J - \lambda I) = 0 \), where \( J \) is the Jacobian matrix and \( I \) is the identity matrix of the same size. The solutions \( \lambda \) are the eigenvalues of the matrix.
  • Real parts of eigenvalues: Indicate whether perturbations grow or diminish over time.
  • Positive real parts: Suggest that perturbations will grow, pointing towards system instability.
  • Negative real parts: Suggest that perturbations will diminish, pointing toward system stability.
Stability Analysis
Stability analysis is crucial when assessing the behavior of dynamical systems near equilibrium points. It determines whether a system will remain close to an equilibrium when subjected to small disturbances, or if it will stray away significantly. This form of analysis often relies on evaluating the eigenvalues of the Jacobian matrix.

In the context of equilibrium analysis:
  • An equilibrium point is said to be stable if all eigenvalues of the Jacobian matrix have negative real parts.
  • If even one eigenvalue has a positive real part, the equilibrium is considered unstable, indicating that disturbances could cause the system to deviate far from equilibrium.
  • A zero real part in an eigenvalue could mean the analysis is inconclusive, requiring further investigation or alternate approaches.
The analysis of eigenvalues provides a way to predict system behavior through local linear approximation, helping to foresee whether equilibrium can be maintained under small perturbations.
Equilibrium Points
Equilibrium points of a dynamical system are key positions where the system doesn't change, which means it is in a state of balance. However, just because a system is at equilibrium doesn鈥檛 mean it is stable.

Determining whether these are stable or unstable involves examining the Jacobian matrix evaluated at these points. This determines how the system evolves when started close to an equilibrium point.
  • Equilibrium can be static (all directions remain unchanged) or dynamic (system returns to equilibrium after a disturbance).
  • Stability assessment at equilibrium points needs identification of Jacobian matrix eigenvalues with their real parts.
  • Depending on eigenvalues, the equilibrium can be stable, unstable, or require further research for more complex behaviors.
By evaluating the system at equilibrium, it becomes easier to figure out its reaction to small changes and whether it will return to equilibrium or diverge away.

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Most popular questions from this chapter

Metapopulation dynamics Example 2 presents a model for a population of deer mice that is split into two patches through habitat fragmentation. The model is $$\frac{d x_{\mathrm{A}}}{d t}=-x_{\mathrm{A}}+2 x_{\mathrm{B}} \quad \frac{d x_{\mathrm{B}}}{d t}=3 x_{\mathrm{A}}-3 x_{\mathrm{B}}$$ where \(x_{\mathrm{A}}\) and \(x_{\mathrm{B}}\) are the population sizes in patches \(\mathrm{A}\) and \(\mathrm{B},\) respectively. (a) Construct the phase plane, including the nullclines. (b) Describe what happens to the population in each patch as \(t \rightarrow \infty\) if both start with nonzero sizes.

Prostate cancer During treatment, tumor cells in the prostate can become resistant through a variety of biochemical mechanisms. Some of these are reversible-the cells revert to being sensitive once treatment stops-and some are not. Using \(x_{1}, x_{2},\) and \(x_{3}\) to denote the fraction of cells that are sensitive, temporarily resistant, and permanently resistant, respectively, a simple model for their dynamics during treatment is \(\begin{aligned} d x_{1} / d t &=-a x_{1}-c x_{1}+b x_{2} \\ d x_{2} / d t &=a x_{1}-b x_{2}-d x_{2} \\ d x_{3} / d t &=c x_{1}+d x_{2} \end{aligned}\) Use the fact that \(x_{1}+x_{2}+x_{3}=1\) to reduce this to a non-homogeneous system of two linear differential equations for \(x_{1}\) and \(x_{3} .\)

7\. Cancer progression The development of many cancers, such as colorectal cancer, proceed through a series of pre- cancerous stages. Suppose there are \(n-1\) precancerous stages before developing into cancer at stage \(n .\) A simple system of differential equations modeling this is $$\begin{aligned} x_{0}^{\prime} &=-u_{0} x_{0} \\ x_{i}^{\prime} &=u_{i-1} x_{i-1}-u_{i} x_{i} \\ x_{n}^{\prime} &=u_{n-1} x_{n-1} \end{aligned}$$ where \(x_{i}\) is the fraction of the population in state \(i,\) the \(u_{i}\) 's are positive constants, and \(i=1, \ldots, n-1 .\) $$\begin{array}{l}{\text { (a) Suppose } n=2 . \text { What is the system of differential equa }} \\ {\text { tions for the three stages? }} \\ {\text { (b) Note that the variable } x_{2} \text { does not appear in the equa- }} \\\ {\text { tions for the rate of change of } x_{0} \text { or } x_{1} . \text { Consequently, }} \\ {\text { we can solve the two-dimensional system for } x_{0} \text { and }} \\ {x_{1} \text { separately. Do so, assuming that } x_{0}(0)=k \text { and }} \\ {x_{1}(0)=0 .}\end{array}$$ $$\begin{array}{l}{\text { (c) Use your solution for } x_{1}(t) \text { obtained in part (b) to write }} \\ {\text { a differential equation for } x_{2}(t) \text { . }} \\ {\text { (d) Solve the differential equation from part (c), assuming }} \\ {x_{2}(0)=0}\end{array}$$

Find all equilibria. Then find the linearization near each equilibrium. $$\begin{array}{l}{\frac{d x_{1}}{d t}=\ln x_{1}-x_{2}} \\ {\frac{d x_{2}}{d t}=x_{1}\left(1-x_{1}-x_{2}\right)}\end{array}$$

Given the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) , construct the phase plane, including the nullclines. Does the equilibrium look like a saddle, a node, or a spiral? \(A=\left[ \begin{array}{ll}{1} & {1} \\ {0} & {1}\end{array}\right]\)
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