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Chapter 10: Problem 29

(a) Show that the plane autonomous system $$ \begin{aligned} &x^{\prime}=-x+y-x^{3} \\ &y^{\prime}=-x-y+y^{2} \end{aligned} $$ has two critical points by sketching the graphs of \(-x+y-\) \(x^{3}=0\) and \(-x-y+y^{2}=0\). Classify the critical point at \((0,0)\). (b) Show that the second critical point \(\mathbf{X}_{1}=(0.88054,1.56327)\) is a saddle point.

Short Answer

Expert verified
The system has critical points at \((0,0)\) and \((0.88054, 1.56327)\); (0,0) is an unstable center, and (0.88054, 1.56327) is a saddle point.

Step by step solution

01

Identify Critical Points

Critical points of a plane autonomous system are found where the derivatives \(x'\) and \(y'\) are both zero. Therefore, we solve the equations:\[-x + y - x^3 = 0\] \[-x - y + y^2 = 0\] simultaneously to find the critical points.
02

Solve the First Equation

Rearranging the first equation \(-x + y - x^3 = 0\), we find:\[y = x + x^3\]
03

Substitute into the Second Equation

Substitute \(y = x + x^3\) into the second equation:\[-x - (x + x^3) + (x + x^3)^2 = 0\]Simplifying gives:\[-2x - x^3 + x^2 + 2x^4 + x^6 = 0\]
04

Simplify and Solve for x

To solve \(-2x - x^3 + x^2 + 2x^4 + x^6 = 0\), rearrange and simplify:\[x^6 + 2x^4 + x^2 - x^3 - 2x = 0\]Factor out \(x\):\[x(x^5 + 2x^3 + x - x^2 - 2) = 0\]This gives a solution for \(x = 0\), and then solve the quintic equation for another root.
05

Verify Critical Points

Substituting \( x = 0 \) into \( y = x + x^3 \) gives \( y = 0 \), yielding the critical point \((0,0)\).Find roots of the quintic to get numeric solutions such as another critical point \( (0.88054, 1.56327) \).
06

Classify Critical Point at (0,0)

To classify the critical point \((0,0)\), calculate the Jacobian matrix \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} -1 - 3x^2 & 1 \ -1 & -1 + 2y \end{bmatrix} \]At \((0,0)\), the Jacobian becomes \[ J = \begin{bmatrix} -1 & 1 \ -1 & -1 \end{bmatrix} \]Calculate the eigenvalues to determine stability: solving the characteristic equation results in purely imaginary eigenvalues \((i, -i)\) indicating an unstable center.
07

Analyze Saddle Point at Second Critical Point

Substitute \( (0.88054, 1.56327) \) into the Jacobian:\[ J = \begin{bmatrix} -1 - 3(0.88054)^2 & 1 \ -1 & -1 + 2(1.56327) \end{bmatrix} \]Calculate the eigenvalues. If they have opposite signs, the critical point \((0.88054, 1.56327)\) is a saddle point.

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Key Concepts

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

Autonomous Systems
In mathematics, an autonomous system is a set of ordinary differential equations that do not explicitly depend on the independent variable, often time. In simpler terms, the equations describe how the system evolves based purely on its current state without external time-dependent inputs. This makes autonomous systems particularly interesting because their behavior is governed entirely by internal dynamics.
  • An example of this would be the system of equations: \( x' = f(x, y) \) and \( y' = g(x, y) \). Here, \( x' \) and \( y' \) are functions of \( x \) and \( y \) alone.
  • The system does not rely on time \( t \) or any other external parameter.
As such systems are quite common in modeling natural phenomena, understanding their behavior is crucial for predicting the stability and trajectory of the solutions over time.
Critical Points
Critical points in differential equations are where the system reaches equilibrium. This occurs when both derivatives, often denoted as \( x' \) and \( y' \), are zero, meaning there is no change in the system state. Finding these points helps in analyzing the behavior of the system.
  • In our example, to find critical points, we need to solve \(-x + y - x^3 = 0\) and \(-x - y + y^2 = 0\) simultaneously.
  • The critical points represent states where the system could potentially stabilize or change its nature.
These points are fundamental in determining how a dynamical system behaves, such as whether an object remains at rest, moves periodically, or changes phase.
Jacobian Matrix
The Jacobian matrix is a key concept in determining the behavior around critical points of a dynamical system. It consists of first-order partial derivatives of the system functions and provides an approximation of how the system behaves near these points.
The matrix is expressed as: \[J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix}\]
  • For our example, at a critical point \((0,0)\), the Jacobian matrix becomes \( J = \begin{bmatrix} -1 & 1 \ -1 & -1 \end{bmatrix} \).
  • This helps determine the linear approximation of the system near this critical point.
By analyzing the Jacobian, we can predict the system's response to small perturbations around the critical points.
Eigenvalues
Eigenvalues are essential for understanding the stability of critical points in a system of differential equations. They are derived from the Jacobian matrix and are used to determine how solutions to the system behave over time.
  • Eigenvalues are solutions to the characteristic equation obtained from the determinant \(|J - \lambda I| = 0\), where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix.
  • The sign and nature (real vs. imaginary) of the eigenvalues often determine the stability of the critical point.
In the example exercise, calculating the eigenvalues for the Jacobian matrix at different critical points indicates whether the system behaves as a stable node, saddle point, or center in those regions.
Stability Analysis
Stability analysis revolves around understanding how small disturbances affect a system. By using the Jacobian matrix and its eigenvalues at critical points, we determine if a system will return to equilibrium, or diverge away.
There are different types of stability to consider:
  • If all eigenvalues have negative real parts, the point is a stable node (attracts trajectories).
  • If any eigenvalue has a positive real part, the point is unstable.
  • If one positive and one negative eigenvalue exists, the point is a saddle point, indicating instability in one direction.
For instance, our exercise identifies \((0,0)\) as an unstable center due to imaginary eigenvalues, while the point \((0.88054, 1.56327)\) is classified as a saddle point, offering insights into how the system behaves around these points.

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Most popular questions from this chapter

A solution of the nonlinear second-order differential equation \(x^{\prime \prime}+x-x^{3}=0\) satisfies \(x(0)=0\) and \(x^{\prime}(0)=v_{0}\). Use the phaseplane method to determine when the resulting solution is periodic.

Use the phase-plane method to show that the solution to the nonlinear second- order differential equation \(x^{\prime \prime}+2 x-x^{2}=0\) that satisfies \(x(0)=1\) and \(x^{\prime}(0)=0\) is periodic.

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. $$ \begin{aligned} &x^{\prime}=x\left(1-x^{2}-3 y^{2}\right) \\ &y^{\prime}=y\left(3-x^{2}-3 y^{2}\right) \end{aligned} $$

In Problems 7-16 find all critical points of the given plane autonomous system. $$ \begin{aligned} &x^{\prime}=y^{2}-x \\ &y^{\prime}=x^{2}-y \end{aligned} $$

The nonlinear differential equation \(x^{\prime \prime}+x=1+e x^{2}\) arises in the analysis of planetary motion using relativity theory. Classify (if possible) all critical points of the corresponding plane autonomous system.
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