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A wide variety of phenomena can occur when an equilibrium point is completely degenerate, i.e., when the Jacobian is the zero matrix. We will look at just one. Consider the system $$ \begin{aligned} &x^{\prime}=x y \\ &y^{\prime}=x^{2}-y^{2} \end{aligned} $$ a) Show that the Jacobian at the origin is the zero matrix. b) Plot the solutions through the six points \((0, \pm 1)\), and \((\pm \sqrt{2}, \pm 1)\). Plot additional solutions of your choice. c) Compare what you see with the behavior of solutions near a saddle point.

Short Answer

Expert verified
a) Jacobian at origin is zero matrix. b) Plot solutions from given points. c) Compare behavior to a saddle point, noting differences due to degeneracy.

Step by step solution

01

Define the System and Equilibrium Point

The given system of differential equations is \(x' = x y\) and \(y' = x^2 - y^2\). The equilibrium point given is the origin \((0,0)\). At equilibrium points, the derivatives are zero.
02

Find the Jacobian Matrix

The Jacobian matrix \(J(x,y)\) of the system is formed by taking partial derivatives of each equation with respect to \(x\) and \(y\). For \(x' = xy\), \(\frac{\partial}{\partial x}(xy) = y\) and \(\frac{\partial}{\partial y}(xy) = x\). For \(y' = x^2 - y^2\), \(\frac{\partial}{\partial x}(x^2 - y^2) = 2x\) and \(\frac{\partial}{\partial y}(x^2 - y^2) = -2y\). Thus, the Jacobian matrix is \(\begin{pmatrix} y & x \ 2x & -2y \end{pmatrix}\).
03

Evaluate the Jacobian at the Equilibrium Point

Substitute \(x = 0\) and \(y = 0\) into the Jacobian matrix \(\begin{pmatrix} y & x \ 2x & -2y \end{pmatrix}\) to get \(\begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}\). This confirms the Jacobian at the origin is a zero matrix.
04

Plot the Solutions Through Given Points

The given points are \((0, \pm 1)\) and \((\pm \sqrt{2}, \pm 1)\). Plot the trajectories originating from these points using a direction field or by numerical simulation of the system. Observe how each trajectory behaves in proximity to these initial points.
05

Examine Additional Solution Plots

Plot additional solutions using other points near the origin or other interesting areas of the phase plane. This can help reveal more about the nature of the system and its behavior around the origin.
06

Compare with Saddle Point Behavior

A saddle point in a dynamical system is characterized by trajectories being repelled along one direction and attracted along another. Compare the behavior of solutions near the origin in this degenerate system to that of a typical saddle point. Typically, this will mean observing some hyperbolic-like behavior, but perhaps with unique characteristics due to the degeneracy.

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Key Concepts

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

Jacobian Matrix
In differential equations, the Jacobian matrix plays a crucial role, especially when examining the stability of equilibrium points. It provides a linear approximation of a dynamical system near an equilibrium. For a given system of equations, the Jacobian matrix is composed of the first derivatives of each equation with respect to each variable.
For the system given:
  • The derivative of \(x' = xy\) with respect to \(x\) results in \(y\)
  • The derivative with respect to \(y\) results in \(x\)
  • For \(y' = x^2 - y^2\), the partial with respect to \(x\) is \(2x\)
  • And with respect to \(y\) is \(-2y\)
This leads to the Jacobian matrix:\[J(x, y) = \begin{pmatrix} y & x \ 2x & -2y \end{pmatrix}\]Evaluating the Jacobian matrix at the origin \((0,0)\) yields a zero matrix, indicating the equilibrium point is completely degenerate.
This implies the standard linear analysis might not provide sufficient information on the system's behavior, so further analysis is needed.
Equilibrium Points
Equilibrium points occur where the system's derivatives equal zero, indicating no change at those points. These are crucial in understanding system dynamics because they represent steady-state solutions where the system neither evolves nor declines.
For the given system:
  • The equilibrium point is located at the origin \((0,0)\)
  • Substituting \((0,0)\) into the equations \(x' = xy\) and \(y' = x^2 - y^2\) confirms zero derivatives
Equilibrium points can be classified into various types based on their stability and how they influence nearby trajectories. Since the Jacobian at this equilibrium is the zero matrix, it tells us that standard techniques may not completely capture the dynamics in its vicinity, suggesting complex behavior at and around the origin.
Saddle Points
Saddle points are a specific type of equilibrium point characterized by particular stability properties. While some directions in the phase space are attractive, others are repulsive. This mixed stability results in a characteristic 'saddle' shape in the phase plane.
In many systems, saddle points can be identified through the eigenvalues of the Jacobian matrix at the point. A saddle point typically has both negative and positive real parts in its eigenvalues, indicating its dual nature of attracting and repelling.
In the given system:
  • Since the Jacobian at \((0,0)\) is a zero matrix, traditional eigenvalue analysis fails
  • This suggests that while the point might exhibit saddle-like behavior, it might be inherently more complex due to degeneracy
To understand the behavior around this degenerate saddle point, plotting solutions near \((0,0)\) can be insightful, showing trajectories behaving in hyperbolic or even unexpected manners.
Phase Plane Analysis
Phase plane analysis is a visual method for studying dynamical systems. It involves plotting the trajectories on a plane, providing insights into the system's behavior over time.
For our system of equations,
  • Each trajectory plotted through points like \((0, \pm 1)\) and \((\pm \sqrt{2}, \pm 1)\) reveals how solutions evolve
  • By observing these trajectories, we can infer tendencies such as attraction, repulsion, or cycling behaviors around equilibrium points
This approach is particularly useful here because of the zero Jacobian matrix at the origin. It visually demonstrates complex phenomena that are not evident from linear analysis alone. Observing the phase plane may reveal non-linear dynamics that are characteristic of degenerate systems, which can seem hyperbolic and saddle-like but with distinctive features unique to such systems. Therefore, phase plane analysis is essential for a detailed understanding of this system's intricacies.

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Most popular questions from this chapter

Duffing 's equation is $$ m x^{\prime \prime}+c x^{\prime}+k x+l x^{3}=F(t) $$ When \(k>0\) this equation models a vibrating spring, which could be soft \((l<0)\) or hard \((l>0)\) (See Student Project #1 in Chapter 8). When \(k<0\) the equation arises as a model of the motion of a long thin elastic steel beam that has its top end embedded in a pulsating frame (the \(F(t)\) term), and its lower end hanging just above two magnets which are slightly displaced from what would be the equilibrium position. We will be looking at the unforced case (i.e. \(F(t)=0\) ), with \(m=1\). The system corresponding to Duffing's equation is available in the Gallery menu. a) This is the case of a hard spring with \(k=16\), and \(l=4\). Use pplane 6 to plot the phase planes of some solutions with the damping constant \(c=0,1\), and 4. In particular, find all equilibrium points. b) Do the same for the soft spring with \(k=16\) and \(l=-4\). Now there will be a pair of saddle points. Find them and plot the stable/unstable orbits. c) Now consider the case when \(k=-1\), and \(l=1\). For each of the cases \(c=0, c=0.2\), and \(c=1\), use pplane6 to analyze the system. In particular, find all equilibrium points and determine their types. Plot stable/unstable orbits where appropriate, and plot typical orbits. d) With \(c=0.2\) in part c), there are two sinks. Determine the basins of attraction of each of these sinks. Indicate these regions on a print out of the phase plane.

In Exercises \(21-30\), perform steps i) and ii) for the linear system $$ \mathbf{x}^{\prime}=\mathbf{A x}, $$ where \(\mathbf{A}\) is the given matrix. i) Find the type of the equilibrium point at the origin. Do this without the aid of any technology or pplane6. You may, of course, check your answer using pplane. ii) Use pplane6 to plot several solution curves - enough to fully illustrate the behavior of the system. You will find it convenient to use the "linear system" choice from the Gallery menu. If the eigenvalues are real, use the "Keyboard input" option to include straight line solutions starting at \(\pm 10\) times the eigenvectors. If the equilibrium point is a saddle point, compute and plot the separatrices. $$ \left[\begin{array}{rr} -1 & 0 \\ 0 & -1 \end{array}\right] $$

If the eigenvalues and eigenvectors of a planar, autonomous, linear system are complex, then there are no straight line solutions. Use \([\mathrm{v}, \mathrm{e}]=\mathrm{eig}(\mathrm{A})\) to demonstrate that the eigenvalues and eigenvectors of the system \(x^{\prime}=2.9 x+2.0 y, y^{\prime}=-5.0 x-3.1 y\) are complex, then use pplane6 to show that solution trajectories spiral in the phase plane.

This problem involves the default system in pplane6, i.e., the planar system that appears when pplane6 is first started. a) Find and plot all interesting features, i.e., equilibrium points, separatrices, limit cycles, and any other features that appeal to you. Make a list of such to turn in with the plot. b) Find the linearization of the system at \((0,0)\). Enter that into pplane 6 and plot a few orbits. c) Go back to the default system. According to theory, the linearization should approximate the original system in a small enough neighborhood of \((0,0)\). Use the Zoom in square option to find a small enough rectangle containing \((0,0)\), where the behavior of the system is closely approximated by the linear system you analyzed in part b). d) Redo parts b) and c) for the other equilibrium points of the default system.

Consider the nonlinear system $$ \begin{aligned} &x^{\prime}=x(1-x)-x y, \\ &y^{\prime}=y(2-y)+2 x y . \end{aligned} $$ Show that \((-1 / 3,4 / 3)\) is an equilibrium point of the system. a) Without the use of technology, calculate the Jacobian of the system at the equilibrium point \((-1 / 3\), \(4 / 3)\). What is the equation of the linearization at this equilibrium point? Use [v,e]=eig( \(\mathrm{J})\) to find the eigenvalues and eigenvectors of this Jacobian. b) Enter the system in pplane6. Find the equilibrium point at \((-1 / 3,4 / 3)\). Does the data in the Equilibrium point data window agree with your findings in part (a)? Note: The eigenvalues of the Jacobian predict classification of the equilibrium point. In this case, the point \((-1 / 3,4 / 3)\) is a saddle because the eigenvalues are real and opposite in sign. c) Display the linearization. Does the equation of the linearization agree with your findings in part (a)?

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