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Chapter 9: Problem 11

show that the given system has no periodic solutions other than constant solutions. $$ d x / d t=x+y+x^{3}-y^{2}, \quad d y / d t=-x+2 y+x^{2} y+y^{3} / 3 $$

Short Answer

Expert verified
Question: Show that the given system has no periodic solutions other than constant solutions: $$ \begin{cases} \frac{dx}{dt} = x + y + x^3 - y^2 \\ \frac{dy}{dt} = -x + 2y + x^2 y + \frac{y^3}{3} \end{cases} $$ Answer: The given system has no periodic solutions other than constant solutions because the only equilibrium point \((0, 0)\) is an unstable node, as shown by analyzing the eigenvalues of the Jacobian matrix at this point. Since there are no stable equilibrium points, there cannot be any periodic solutions around them.

Step by step solution

01

Find equilibrium points

The equilibrium points (or constant solutions) of the system are the points \((x, y)\) where the derivatives are zero: $$\begin{cases} \frac{dx}{dt} = x + y + x^3 - y^2 = 0 \\ \frac{dy}{dt} = -x + 2y + x^2 y + \frac{y^3}{3} = 0 \end{cases}$$ It's clear that \((0,0)\) is an equilibrium point.
02

Jacobian matrix

Compute the Jacobian matrix of the system, which contains the partial derivatives of the functions with respect to x and y: $$ J = \begin{bmatrix} \frac{\partial(dx/dt)}{\partial x} & \frac{\partial(dx/dt)}{\partial y} \\ \frac{\partial(dy/dt)}{\partial x} & \frac{\partial(dy/dt)}{\partial y} \end{bmatrix}. $$ $$ J(x, y) = \begin{bmatrix} 1 + 3x^2 & 1 - 2y \\ -1 + 2xy & 2 + x^2 + y^2 \end{bmatrix}. $$
03

Find the Jacobian matrix at the equilibrium point (0,0)

Evaluate the Jacobian matrix at the equilibrium point \((0, 0)\): $$ J(0, 0) = \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix}. $$
04

Compute eigenvalues

To analyze the stability of the equilibrium point, compute the eigenvalues of the Jacobian matrix at the point \((0, 0)\). We have to solve the characteristic equation $$\text{det}(J - \lambda I) = 0$$ where \(I\) is the identity matrix. Using the Jacobian matrix at \((0,0)\): $$\text{det}(J - \lambda I) = \text{det}\left(\begin{bmatrix} 1 - \lambda & 1 \\ -1 & 2 - \lambda \end{bmatrix}\right) = (1-\lambda)(2-\lambda) - (-1)(1) = \lambda^2 - 3\lambda + 1 = 0$$.
05

Analyze stability

Solve the quadratic equation \(\lambda^2 - 3\lambda + 1 = 0\). By applying the quadratic formula, we get: $$\lambda_{1,2} = \frac{3 \pm \sqrt{3^2 - 4 \times 1 \times 1}}{2 \times 1} = \frac{3 \pm \sqrt{5}}{2}.$$ The eigenvalues are real and positive, which means that the equilibrium point \((0, 0)\) is an unstable node. Since the only equilibrium point is unstable, the conclusion is that the given system doesn't have any periodic solutions other than constant solutions.

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Key Concepts

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

Equilibrium Points
Equilibrium points in differential equations are like the 'resting positions' in a dynamic system where all changes cease. To find these points, we set the derivatives to zero because these are the spots where the system doesn't change over time.

In our case, the system of equations given by \(\frac{dx}{dt} = x + y + x^3 - y^2\) and \(\frac{dy}{dt} = -x + 2y + x^2y + \frac{y^3}{3}\) provides the equilibrium point \( (0,0) \), by solving for when both derivatives equate to zero. Recognizing these points is an essential first step in determining the behavior of a system over time and plays a crucial role in analyzing the system's stability.
Jacobian Matrix
The Jacobian matrix is a powerful tool representing how a dynamical system changes around an equilibrium point. It's constructed from the partial derivatives of the system's equations.

For our differential equation system, this involves calculating \(\frac{\partial(dx/dt)}{\partial x}\), \(\frac{\partial(dx/dt)}{\partial y}\), \(\frac{\partial(dy/dt)}{\partial x}\), and \(\frac{\partial(dy/dt)}{\partial y}\) to assemble the Jacobian matrix. Simply put, it captures the essence of the system's dynamics in a matrix form, which we can analyze further to understand the behavior of the system near the equilibrium points.
Eigenvalues
Eigenvalues are a set of scalars associated with a system of linear equations or a matrix. They play a central role in understanding the behavior of differential equations. Once we compute the Jacobian, we can find its eigenvalues by solving the characteristic equation \(\text{det}(J - \lambda I) = 0\), where \(I\) denotes the identity matrix.

In our exercise, we derived the eigenvalues \(\lambda_{1,2} = \frac{3 \pm \sqrt{5}}{2} \). These values give us critical information about the system's tendencies near the equilibrium, such as whether solutions will grow, decay, or oscillate.
Stability Analysis
Stability analysis in the context of differential equations is concerned with understanding how a system behaves after small perturbations around an equilibrium point. The nature and sign of the eigenvalues of the Jacobian at an equilibrium point determine stability.

For the equilibrium point \( (0,0) \) in our exercise, the eigenvalues are real and positive, which indicates that any small perturbation will compound over time, making \( (0,0) \) an unstable node. Stability analysis helps us conclude that our system won't settle into a periodic orbit 鈥� instead, it will depart from the equilibrium, which confirms the absence of non-constant periodic solutions.

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Most popular questions from this chapter

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r^{2}\left(1-r^{2}\right), \quad d \theta / d t=1 $$

The equation of motion of an undamped pendulum is \(d^{2} \theta / d t^{2}+\omega^{2} \sin \theta=0,\) where \(\omega^{2}=g / L .\) Let \(x=\theta, y=d \theta / d t\) to obtain the system of equations $$ d x / d t=y, \quad d y / d t=-\omega^{2} \sin x $$ (a) Show that the critical points are \((\pm n \pi, 0), n=0,1,2, \ldots,\) and that the system is almost lincar in the neighborhood of cach critical point. (b) Show that the critical point \((0,0)\) is a (stable) center of the corresponding linear system. Using Theorem 9.3.2 what can be said about the nonlinear system? The situation is similar at the critical points \((\pm 2 n \pi, 0), n=1,2,3, \ldots\) What is the physical interpretation of these critical points? (c) Show that the critical point \((\pi, 0)\) is an (unstable) saddle point of the corresponding linear system. What conclusion can you draw about the nonlinear system? The situation is similar at the critical points \([\pm(2 n-1) \pi, 0], n=1,2,3, \ldots\) What is the physical interpretation of these critical points? (d) Choose a value for \(\omega^{2}\) and plot a few trajectories of the nonlinear system in the neighborhood of the origin. Can you now draw any further conclusion about the nature of the critical point at \((0,0)\) for the nonlinear system? (e) Using the value of \(\omega^{2}\) from part (d) draw a phase portrait for the pendulum. Compare your plot with Figure 9.3 .5 for the damped pendulum.

In this problem we indicate how to show that the trajectories are ellipses when the eigen- values are pure imaginary. Consider the system $$ \left(\begin{array}{l}{x} \\\ {y}\end{array}\right)^{\prime}=\left(\begin{array}{ll}{a_{11}} & {a_{12}} \\\ {a_{21}} & {a_{22}}\end{array}\right)\left(\begin{array}{l}{x} \\\ {y}\end{array}\right) $$ (a) Show that the eigenvalues of the coefficient matrix are pure imaginary if and only if $$ a_{11}+a_{22}=0, \quad a_{11} a_{22}-a_{12} a_{21}>0 $$ (b) The trajectories of the system (i) can be found by converting Eqs. (i) into the single equation $$ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a_{21} x+a_{22} y}{a_{11} x+a_{12} y} $$ Use the first of Eqs. (ii) to show that Eq. (iii) is exact. (c) By integrating Eq. (iii) show that $$ a_{21} x^{2}+2 a_{22} x y-a_{12} y^{2}=k $$ where \(k\) is a constant. Use Eqs. (ii) to conclude that the graph of Eq. (iv) is always an ellipse. Hint: What is the discriminant of the quadratic form in Eq. (iv)?

Consider the autonomous system $$ d x / d t=x, \quad d y / d t=-2 y+x^{3} $$ (a) Show that the critical point \((0,0)\) is a saddle point. (b) Sketch the trajectories for the corresponding linear system and show that the trajectory for which \(x \rightarrow 0, y \rightarrow 0\) as \(t \rightarrow \infty\) is given by \(x=0\). (c) Determine the trajectories for the nonlinear system for \(x \neq 0\) by integrating the equation for \(d y / d x\). Show that the trajectory corresponding to \(x=0\) for the linear system is unaltered, but that the one corresponding to \(y=0\) is \(y=x^{3} / 5 .\) Sketch several of the trajectories for the nonlinear system.

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=(1+x) \sin y, \quad d y / d t=1-x-\cos y $$
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