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Classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point. $$ \begin{aligned} &x^{\prime}=x\left(1-x^{2}-3 y^{2}\right) \\ &y^{\prime}=y\left(3-x^{2}-3 y^{2}\right) \end{aligned} $$

Short Answer

Expert verified
(0,0) is an unstable node; (±1,0) are saddle points; (0,±√3) are stable nodes.

Step by step solution

01

Find Critical Points

To find the critical points of the system, set \(x' = 0\) and \(y' = 0\). From \(x' = x(1-x^2-3y^2) = 0\), we get either \(x = 0\) or \(1-x^2-3y^2 = 0\). From \(y' = y(3-x^2-3y^2) = 0\), we get either \(y = 0\) or \(3-x^2-3y^2 = 0\). Solving these, we find the critical points: \((0, 0)\), \((\pm 1, 0)\), and \((0, \pm \sqrt{3})\).
02

Linearization and Jacobian Matrix

To classify each critical point, we must evaluate the Jacobian matrix at these points. The Jacobian matrix is given by: \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 1-3x^2-3y^2 & -6xy \ -2xy & 3-x^2-9y^2 \end{bmatrix} \]
03

Evaluate Jacobian at (0,0)

For \((0,0)\), the Jacobian becomes: \[ J = \begin{bmatrix} 1 & 0 \ 0 & 3 \end{bmatrix} \] The eigenvalues are \(1\) and \(3\), both positive, indicating \((0,0)\) is an unstable node.
04

Evaluate Jacobian at (1,0) and (-1,0)

For \((1,0)\) and \((-1,0)\), the Jacobian is: \[ J = \begin{bmatrix} -2 & 0 \ 0 & 2 \end{bmatrix} \] The eigenvalues are \(-2\) and \(2\), one negative and one positive, thus \((1,0)\) and \((-1,0)\) are saddle points.
05

Evaluate Jacobian at (0, sqrt(3)) and (0, -sqrt(3))

For \((0, \sqrt{3})\) and \((0, -\sqrt{3})\), the Jacobian becomes: \[ J = \begin{bmatrix} -8 & 0 \ 0 & -6 \end{bmatrix} \] The eigenvalues are \(-8\) and \(-6\), both negative, indicating \((0, \sqrt{3})\) and \((0, -\sqrt{3})\) are stable nodes.

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Key Concepts

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

Plane Autonomous Systems
Plane autonomous systems are systems of differential equations where the time variable does not appear explicitly in the equations. These systems are often represented in the form:
  • \( x' = f(x, y) \)
  • \( y' = g(x, y) \)
This model captures the dynamic behavior of two interconnected variables over time.
In the exercise example, the dynamics of variables \(x\) and \(y\) are interdependent. Each variable's rate of change depends on the state of the other. This allows us to study complex behavior such as stability and oscillations without reference to explicit initial time conditions.
The focus is on understanding how the system evolves in the phase plane, which is a graphical representation of all possible states of the system.
Jacobian Matrix
The Jacobian matrix is a crucial tool when analyzing how dynamic systems behave near a critical point. It provides a linear approximation of a nonlinear system around these critical points.
In the given system, the Jacobian matrix is constructed by taking the partial derivatives of the system functions \(x'\) and \(y'\) with respect to both variables \(x\) and \(y\). Specifically, the Jacobian \( J \) for a system \( (x', y') = 0 \) is:
  • \( J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} \)
The resulting matrix captures how the rates of change of both variables affect each other.
Evaluating this matrix at critical points helps in determining the nature of these points.
Eigenvalues
Eigenvalues are fundamental in classifying the nature of critical points found via the Jacobian matrix. They are the solutions to the characteristic equation derived from a matrix, often related to understanding the qualitative behavior of the dynamical system.
Once the Jacobian matrix at a critical point is evaluated, its eigenvalues are calculated by solving the equation:
  • \( \det(J - \lambda I) = 0 \)
where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix of compatible dimension.
The sign and multiplicity of the eigenvalues determine the behavior near the critical points. Positive eigenvalues indicate instability, while negative values suggest stability. If eigenvalues are complex with a non-zero real part, the system might exhibit oscillating behavior, classifying those points as spirals.
Stability of Critical Points
Stability analysis involves studying how a system behaves when slightly disturbed from its critical points. The configuration of eigenvalues provides insight into the type of stability present at each critical point.
  • Stable Node: All eigenvalues are negative, indicating convergence towards the critical point.
  • Unstable Node: All eigenvalues are positive, indicating divergence away from the critical point.
  • Saddle Point: Mixed signs among eigenvalues, suggesting some paths lead towards and others away from the point.
  • Stable Spiral: Complex eigenvalues with negative real parts leading to a spiral inwards.
  • Unstable Spiral: Complex eigenvalues with positive real parts leading to a spiral outwards.
In the exercise provided, each critical point was classified using these definitions, showcasing how the Jacobian and eigenvalues work together to identify their nature. This process allows for a structured approach to predicting long-term behavior of systems.

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Most popular questions from this chapter

The rod of a pendulum is attached to a movable joint at a point \(P\) and rotates at an angular speed of \(\omega(\mathrm{rad} / \mathrm{s})\) in the plane perpendicular to the rod. See Figure 10.39. As a result the bob of the rotating pendulum experiences an additional centripetal force, and the new differential equation for \(\theta\) becomes $$ m l \frac{d^{2} \theta}{d t^{2}}=\omega^{2} m l \sin \theta \cos \theta-m g \sin \theta-\beta \frac{d \theta}{d t} . $$ (a) If \(\omega^{2}g / I\), show that \((0,0)\) is unstable and there are two additional stable critical points \((\pm \hat{\theta}, 0)\) in the domain \(-\pi<\theta<\pi\). Describe what occurs physically when \(\theta(0)=\theta_{0}, \theta^{\prime}(0)=0\), and \(\theta_{0}\) is small.

Classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point. $$ \begin{aligned} &x^{\prime}=-2 x+y+10 \\ &y^{\prime}=2 x-y-15 \frac{y}{y+5} \end{aligned} $$

In Problems 7-16 find all critical points of the given plane autonomous system. $$ \begin{aligned} &x^{\prime}=x^{3}-y \\ &y^{\prime}=x-y^{3} \end{aligned} $$

(a) Show that \((0,0)\) is the only critical point of the Raleigh differential equation $$ x^{*}+\epsilon\left(\frac{1}{3}\left(x^{\prime}\right)^{3}-x^{\prime}\right)+x=0 $$ (b) Show that \((0,0)\) is unstable when \(\epsilon>0\). When is \((0,0)\) an unstable spiral point? (c) Show that \((0,0)\) is stable when \(\epsilon<0\). When is \((0,0)\) a stable spiral point? (d) Show that \((0,0)\) is a center when \(\epsilon=0\).

When expressed in polar coordinates, a plane autonomous system takes the form $$ \begin{aligned} &\frac{d r}{d t}=\alpha r(5-r) \\ &\frac{d \theta}{d t}=-1 \end{aligned} $$ Show that \((0,0)\) is an asymptotically stable critical point if and only if \(\alpha<0\).

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