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Chapter 11: Problem 58

We consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) Analyze the stability of the equilibrium \(\mathbf{( 0 , 0 ) , \text { and classify the }}\) equilibrium. \(A=\left[\begin{array}{ll}-2 & 2 \\ -4 & 3\end{array}\right]\)

Short Answer

Expert verified
The equilibrium is a saddle point, indicating instability.

Step by step solution

01

Define the System

The given system is described by the differential equation \(\frac{d \mathbf{x}}{dt} = A \mathbf{x}(t)\), where \( A = \begin{bmatrix} -2 & 2 \ -4 & 3 \end{bmatrix} \). The vector \( \mathbf{x}(t) \) represents the system state, here given as \( \mathbf{x} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \). We need to analyze the stability of the equilibrium at \( (0, 0) \).
02

Find the Eigenvalues of A

The stability can be determined by analyzing the eigenvalues of matrix \( A \). We find the eigenvalues \( \lambda \) by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Thus, we form the matrix \( A - \lambda I = \begin{bmatrix} -2 - \lambda & 2 \ -4 & 3 - \lambda \end{bmatrix} \), and calculate its determinant: \[ \det(A - \lambda I) = (-2 - \lambda)(3 - \lambda) - (-4)(2) \] \[ = \lambda^2 - \lambda - 2 \] Set the equation to zero: \( \lambda^2 - \lambda - 2 = 0 \).
03

Solve the Characteristic Polynomial

We solve the quadratic equation \( \lambda^2 - \lambda - 2 = 0 \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -2 \). Substituting these values, we have: \[ \lambda = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \] Thus, the eigenvalues are \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \).
04

Analyze Stability Using Eigenvalues

The stability of the equilibrium is determined by the signs of the real parts of the eigenvalues. If all eigenvalues have negative real parts, the equilibrium at \( (0,0) \) is stable. In this case, one eigenvalue is positive (\(\lambda_1 = 2\)) and one is negative (\(\lambda_2 = -1\)). This indicates that the system has a saddle point.
05

Classify the Equilibrium

Given that we have a positive and a negative eigenvalue, the equilibrium point \( (0, 0) \) is classified as a saddle point. A saddle point is stable along some directions and unstable along others, indicating that nearby trajectories will diverge away from the equilibrium along at least one direction.

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Key Concepts

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

Eigenvalues in Differential Equations
When dealing with differential equations, especially those of the form \( \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \), understanding eigenvalues becomes crucial. Eigenvalues are numbers that provide insights into the behavior of the system. They are derived by solving the characteristic equation \( \det(A - \lambda I) = 0 \).
  • Here, \( A \) is the system matrix, and \( \lambda \) represents the eigenvalues.
  • The identity matrix \( I \) is used to construct the matrix \( A - \lambda I \).
  • The determinant of \( A - \lambda I \) gives us a polynomial equation.
By solving this polynomial equation, we can determine the eigenvalues \( \lambda \). For example, in the system matrix \( A = \begin{bmatrix} -2 & 2 \ -4 & 3 \end{bmatrix} \), solving its characteristic equation yields the eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \). These values are essential for determining the system's stability.
Stability Analysis of Equilibrium
Stability analysis helps us understand how a system behaves over time, especially as it approaches equilibrium points. For linear systems, the stability of an equilibrium point can be inferred directly from the eigenvalues of the system matrix \( A \).
  • If all eigenvalues have negative real parts, the system is stable (trajectories converge to the equilibrium).
  • If at least one eigenvalue has a positive real part, the system is unstable (trajectories diverge from the equilibrium).
  • If eigenvalues have zero real parts, we need further analysis to determine stability.
In our example, the matrix \( A \) resulted in eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \). Since \( \lambda_1 \) is positive, the system cannot be stable. This differentiation enables us to predict how an initial state will evolve over time. A mix of positive and negative eigenvalues often points to a special kind of point known as a saddle point.
Understanding Saddle Points
Saddle points are fascinating aspects of stability in differential equations. They occur when an equilibrium has both stable and unstable directions. This happens when the eigenvalues are of mixed signs.
  • A mixed-sign situation indicates that trajectories move towards the equilibrium in one direction and away in another.
  • Saddle points are classified as 'unstable' because some trajectories can diverge.
  • They represent a kind of balance, being stable in some directions but not in others.
In the case of the equilibrium at \( (0,0) \) analyzed earlier, we determined it is a saddle point due to the presence of one positive eigenvalue (\( \lambda_1 = 2 \)) and one negative eigenvalue (\( \lambda_2 = -1 \)). Saddle points highlight the complex nature of dynamical systems, showing that not all aspects of a system's behavior are immediately intuitive.

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Most popular questions from this chapter

Use a graphing calculator to sketch solution curves of the given Lotka- Volterra predator-prey model in the N-P plane. That is, you should plot the level curves of the associated function \(f(N, P) .\) \(\frac{d N}{d t}=3 N-2 P N\) \(\frac{d P}{d t}=P N-P\) passing through the points: (a) \((N(0), P(0))=(1,3 / 2)\) (b) \((N(0), P(0))=(2,2)\) (c) \((N(0), P(0))=(3,1)\)

Transform the second-order differential equation $$ \frac{d^{2} x}{d t^{2}}=-\frac{1}{2} x $$ into a system of first-order differential equations.

We consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) Analyze the stability of the equilibrium \(\mathbf{( 0 , 0 ) , \text { and classify the }}\) equilibrium. \(A=\left[\begin{array}{rr}-2 & 3 \\ 1 & -4\end{array}\right]\)

We consider differential equations of the form \(\frac{d \mathbf{x}}{d t}=A \mathbf{x}(t)\) where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. \(A=\left[\begin{array}{ll}1 & -4 \\ 1 & -1\end{array}\right]\)

Assume that $$ \begin{array}{l} \frac{d N}{d t}=5 N-P N \\ \frac{d P}{d t}=P N-P \end{array} $$ (a) Show that this system has two equilibria: the trivial equilibrium \((0,0)\), and a nontrivial one in which both species have positive densities. (b) Use the eigenvalue approach to show that the trivial equilibrium is unstable. (c) Determine the eigenvalues corresponding to the nontrivial equilibrium. Does your analysis allow you to infer anything about the stability of this equilibrium? (d) Use a graphing calculator to sketch curves in the \(N-P\) plane. Also, sketch solution curves of the prey and the predator densities as functions of time.
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