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Chapter 6: Problem 9

$$ \text { (Dipole fixed point) } \dot{x}=2 x y, \dot{y}=y^{2}-x^{2} $$

Short Answer

Expert verified
The given differential system is: 1. \(\dot{x} = 2xy\) 2. \(\dot{y} = y^2 - x^2\) We found the fixed point at (0,0) by setting the equations to zero and solving. The Jacobian matrix is: \[ J(x, y) = \begin{pmatrix} 2y & 2x \\ -2x & 2y \end{pmatrix}\] Evaluating at the fixed point: \[J(0, 0) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\] Calculating the eigenvalues, we get \(\lambda_1 = 0\) and \(\lambda_2 = 0\). Since both eigenvalues are zero, it is inconclusive to determine the stability using linear methods. Nonlinear analysis is required for a definitive conclusion.

Step by step solution

01

Find the fixed points

To find the fixed points, set both differential equations to zero and solve for x and y simultaneously: \(\dot{x} = 2xy = 0\) \(\dot{y} = y^2 - x^2 = 0\) From the first equation, either \(x=0\) or \(y=0\). Let's consider these cases separately: Case 1: \(x=0\) If \(x=0\), from the second equation: \(y^2 = x^2\) \(y^2 = 0\) \(\Rightarrow y = 0\) So the system has a fixed point at \((0, 0)\). Case 2: \(y=0\) If \(y=0\), we also have a fixed point at \((0, 0)\) from the first equation. This is already found in case 1. Therefore, the only fixed point of the system is at \((0, 0)\).
02

Linearize the system around the fixed points

Linearize the system by finding the Jacobian matrix. The Jacobian matrix \(J\) for this system is: \[ J(x, y) = \begin{pmatrix} \frac{\partial \dot{x}}{\partial x} & \frac{\partial \dot{x}}{\partial y}\\ \frac{\partial \dot{y}}{\partial x} & \frac{\partial \dot{y}}{\partial y} \end{pmatrix} = \begin{pmatrix} 2y & 2x \\ -2x & 2y \end{pmatrix}\] Evaluate this matrix at the fixed point \((0, 0)\): \[J(0, 0) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\]
03

Determine the stability of the fixed points

Calculate the eigenvalues of the Jacobian matrix at the fixed point by finding the determinant: \[\det{(J - \lambda I)} = \begin{vmatrix} -\lambda & 0 \\ 0 & -\lambda \end{vmatrix} = \lambda^2\] The eigenvalues are: \(\lambda_1 = 0\) \(\lambda_2 = 0\) Since both eigenvalues are equal to zero, we cannot determine the stability of the fixed point \((0, 0)\) by analyzing the eigenvalues of the Jacobian matrix. Further analysis using nonlinear methods is required to conclusively determine the stability of the fixed point.

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Key Concepts

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

Jacobian Matrix
The Jacobian matrix is a powerful tool in analyzing nonlinear dynamic systems, especially when we are interested in understanding the behavior of the system near its fixed points. A fixed point, also known as an equilibrium point, is where the system does not change as time progresses.

Mathematically, the Jacobian matrix is composed of first-order partial derivatives. It represents how each component of the system's velocity vector \( \(\dot{x}\), \(\dot{y}\) \) changes with respect to changes in the state variables \(x\) and \(y\). In our exercise, we find that the Jacobian matrix at the fixed point \( (0, 0) \) is a zero matrix. This indicates that near the fixed point, the system's behavior can't be linearly approximated simply by this matrix.

However, this matrix is central to the next steps of our analysis, linearization, and stability analysis, which help us to predict the behavior of our system under small perturbations around the fixed points.
Linearization of Differential Equations
Linearization is a method that simplifies the study of nonlinear dynamics by approximating the behavior of a system near a fixed point using linear equations. When a system is nonlinear, as in our exercise with \(\dot{x} = 2xy\) and \(\dot{y} = y^2 - x^2\), it can be complex to analyze directly. By using the Jacobian matrix at a fixed point, we can transform the nonlinear system into a linear one that is easier to work with.

The linearization is essentially a Taylor series approximation, with the Jacobian matrix providing the coefficients for the first-order terms. If the original system鈥檚 behavior around the fixed point is close to this linear model, then the analysis becomes much simpler. The result of this process provides us with a starting point for stability analysis, which often relies on understanding the eigenvalues of the linearized system.
Stability Analysis
Stability analysis is a key part of studying dynamical systems, as it helps us to understand the robustness of fixed points. It tells us whether a small perturbation will decay over time, allowing the system to return to the fixed point, or whether the perturbation will grow, moving the system away from the fixed point.

In the case of our exercise, after linearizing the system we proceeded to calculate the eigenvalues of the Jacobian matrix at the fixed point. The eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 0\) indicate that we have a case that cannot be decided using linear stability analysis alone. This is because non-negative real parts of eigenvalues suggest potential instability, but eigenvalues of zero indicate a need for further analysis with nonlinear methods.

Typically, when nonzero eigenvalues have negative real parts, the fixed point is stable, and if they have positive real parts, the fixed point is unstable. However, zero eigenvalues mean that the system may be in what's called a 'critical case,' and we must look to the system's higher-order terms or to other methods such as Lyapunov's functions or center manifold theory to make definitive conclusions about the stability situation.

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Most popular questions from this chapter

Show that each of the following systems is reversible, and sketch the phase portrait. $$ \bar{x}=y, \dot{y}=x \cos y $$

(Reversible system on a cylinder) While studying chaotic streamlines inside a drop immersed in a steady Stokes flow, Stone ct al. (I991) encountered the system $$ \dot{x}=\frac{\sqrt{2}}{4} x(x-1) \sin \phi, \quad \dot{\phi}=\frac{1}{2}\left[\beta-\frac{1}{\sqrt{2}} \cos \phi-\frac{1}{8 \sqrt{2}} x \cos \phi\right] $$ where \(0 \leq x \leq 1\) and \(-\pi \leq \phi<\pi\) Since the system is \(2 \pi\)-periodic in \(\phi\), it may be considered as a vector ficld on a cylinder. (See Section \(6.7\) for another vector field on a cylinder.) The \(x\)-axis runs along the cylinder, and the \(\phi\)-axis wraps around it. Note that the cylindrical phase space is finite, with edges given by the circles \(x=0\) and \(x=1\) a) Show that the system is reversible. b) Verify that for \(\frac{9}{8 \sqrt{2}}>\beta>\frac{1}{2}\), the system has three fixed points on the cylinder. one of which is a saddle. Show that this saddle is connected to itself by a homoclinic orbit that winds around the waist of the cylinder. Using reversibility, prove that there is a band of closed orbits sandwiched between the circle \(x=0\). and the homoclinic orbit. Sketch the phase portrait on the cylinder, and check. your results by numerical integration. c) Show that as \(\beta \rightarrow \frac{1}{\sqrt{2}}\) from above, the saddle point moves toward the circle \(x=0\), and the homoclinic orbit tightens like a noose. Show that all the closed orbits disappear when \(\beta=\frac{1}{\sqrt{2}}\). d) For \(0<\beta<\frac{1}{\sqrt{2}}\), show that there are two saddle points on the edge \(x=0\). Plot the phase portrait on the cylinder.

(Wallpaper) Consider the system \(\dot{x}=\sin y, \dot{y}=\sin x\). a) Show that the system is reversible. b) Find and classify all the fixed points. c) Show that the lines \(y=\pm x\) are invariant (any trajectory that starts on them stays on them forever). d) Sketch the phase portrait.

For cach of the following systems, find the fixed points. Then sketch the nullclines, the vector field, and a plausible phase portrait. $$ \dot{x}=x(x-y), \dot{y}=y(2 x-y) $$

(Unusual fixed points) For each of the following systems, locate the fixed points and calculate the index. (Hint: Draw a small closed curve \(C\) around the fixed point and examine the variation of the vector field on \(C\).) $$ \dot{x}=x y, \dot{y}=x+y $$
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