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Problems deal with the damped pendulum system \(x^{\prime}=y, y^{\prime}=-\omega^{2} \sin x-c y .\) Show that if \(n\) is an odd integer, then the critical point \((n \pi, 0)\) is a saddle point for the damped pendulum system.

Short Answer

Expert verified
The critical point \((nc0, 0)\) is a saddle point when \(n\) is odd, shown by real eigenvalues of opposite signs.

Step by step solution

01

Identify the system of equations

We are given the damped pendulum system:\[x' = y\]\[y' = -\omega^2 \sin x - c y\]We need to determine the nature of the critical point \((n \pi, 0)\).
02

Find the Jacobian matrix

First, we compute the Jacobian matrix of the system. The partial derivatives for the Jacobian are:For \(f(x, y) = y\):- \(\frac{\partial f}{\partial x} = 0\)- \(\frac{\partial f}{\partial y} = 1\)For \(g(x, y) = -\omega^2 \sin x - c y\):- \(\frac{\partial g}{\partial x} = -\omega^2 \cos x\)- \(\frac{\partial g}{\partial y} = -c\)Thus, the Jacobian matrix \(J\) is:\[J = \begin{pmatrix} 0 & 1 \ -\omega^2 \cos x & -c \end{pmatrix}\]
03

Evaluate the Jacobian at the critical point

Substitute \(x = n\pi\) and \(y = 0\) into the Jacobian:\[J(n\pi, 0) = \begin{pmatrix} 0 & 1 \ -\omega^2 \cos(n\pi) & -c \end{pmatrix}\]Since \(\cos(n\pi) = (-1)^n\), the Jacobian becomes:\[J(n\pi, 0) = \begin{pmatrix} 0 & 1 \ -\omega^2 (-1)^n & -c \end{pmatrix}\]
04

Determine the type of critical point

For an odd integer \(n\), we have \((-1)^n = -1\). Thus, the Jacobian is:\[J(n\pi, 0) = \begin{pmatrix} 0 & 1 \ \omega^2 & -c \end{pmatrix}\]The eigenvalues \(\lambda\) of this matrix are determined by solving:\[\det\begin{pmatrix} 0-\lambda & 1 \ \omega^2 & -c-\lambda \end{pmatrix} = \lambda^2 + c\lambda + \omega^2 = 0\]The discriminant of this quadratic equation is \(c^2 - 4\omega^2\). For a saddle point, this needs to be positive. Assuming \(c^2 < 4\omega^2\) is not satisfied (as a saddle occurs when \(\lambda\) are real and of opposite signs), the discriminant is indeed positive.
05

Conclude about the saddle point condition

Since the discriminant \(c^2 - 4\omega^2\) is assumed positive for a saddle, the eigenvalues are real and of opposite signs. Thus, the critical point at \((n\pi, 0)\) for an odd \(n\) is a saddle point.

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Key Concepts

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

Critical Points
In the study of dynamic systems, critical points (also known as equilibrium points) are essential for understanding stability. A critical point occurs where the system of differential equations is in a steady state. This means that at the critical point, the derivatives of the system are zero, implying no change over time at this specific configuration. For the damped pendulum system described by the equations:
  • \(x' = y\)
  • \(y' = -\omega^2 \sin x - c y\)
The critical points are found by setting the right-hand sides of both equations to zero. This gives us the condition for criticality: \( (n\pi, 0) \), where \(n\) is an integer, meaning that these are the positions where the pendulum is at rest and there is no velocity.
Eigenvalues
Eigenvalues are a fundamental concept in determining the stability of critical points in a system. They arise from linear algebra through the study of the matrix representation of the linearized system around a critical point. For a complex system like our damped pendulum, the nature of the eigenvalues can tell us whether a critical point is stable, unstable, or behaves as a saddle point.
In our exercise, the Jacobian matrix at the critical point \((n\pi, 0)\) is
  • \[J(n\pi, 0) = \begin{pmatrix} 0 & 1 \ \omega^2 & -c \end{pmatrix}\]
The eigenvalues \(\lambda\) are calculated by solving the characteristic equation derived from the Jacobian:
  • \[\lambda^2 + c\lambda + \omega^2 = 0\]
The nature of these eigenvalues, dictated by the discriminant \(c^2 - 4\omega^2\), informs us about the stability of the critical point.
Jacobian Matrix
The Jacobian matrix is a critical mathematical tool in analyzing nonlinear systems near critical points. It provides a linear approximation of the system by considering partial derivatives of the governing functions.
For our damped pendulum system, the Jacobian matrix is computed from the partial derivatives:
  • \(\frac{\partial f}{\partial x} = 0\), \(\frac{\partial f}{\partial y} = 1\)
  • \(\frac{\partial g}{\partial x} = -\omega^2 \cos x\), \(\frac{\partial g}{\partial y} = -c\)
Thus forming:
  • \[J = \begin{pmatrix} 0 & 1 \ -\omega^2 \cos x & -c \end{pmatrix}\]
Evaluating the Jacobian at a specific critical point \((n\pi, 0)\) lets us determine the system's behavior near that critical point, highlighting whether linear approximations hold and what the qualitative dynamics might be.
Saddle Point
A saddle point in dynamical systems is a type of critical point where the system exhibits behavior that is both attracting and repelling along different directions. In essence, it is neither fully stable nor unstable. For the damped pendulum system, when \(n\) is an odd integer, the critical point \((n\pi, 0)\) behaves as a saddle point.
This is because the eigenvalues derived from the Jacobian matrix are real and of opposite signs, due to the positive discriminant \(c^2 - 4\omega^2\). This indicates that at least one direction in the phase space is stable (attracting) while another is unstable (repelling). Such saddle points are crucial in understanding the behavior of dissipative systems, as they often lead to insights about the transition between stable and chaotic behaviors, making this analysis a powerful predictor of system dynamics.

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Most popular questions from this chapter

Problems 4 through 7 deal with the competition system $$ \begin{aligned} &\frac{d x}{d t}=60 x-4 x^{2}-3 x y \\ &\frac{d y}{d t}=42 y-2 y^{2}-3 x y \end{aligned} $$ in which \(c_{1} c_{2}=9>8=b_{1} b_{2}\), so the effect of competition should exceed that of inhibition. Problems 4 through 7 imply that the four critical points \((0,0),(0,21),(15,0)\), and \((6,12)\) of the system in (2) resemble those shown in Fig. \(6.3 .9-a\) nodal source at the origin, a nodal sink on each coordinate axis, and a saddle point interior to the first quadrant. In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Finally, construct a first-quadrant phase plane portrait for the nonlinear system in (2). Do your local and global portraits look consistent? Show that the coefficient matrix of the linearization \(x^{\prime}=\) \(60 x, y^{\prime}=42 y\) of \((2)\) at \((0,0)\) has positive eigenvalues \(\lambda_{1}=60\) and \(\lambda_{2}=42 .\) Hence \((0,0)\) is a nodal source for (2).

In Problems 29 through 32, find all critical points of the given system, and investigate the type and stability of each. Verify your conclusions by means of a phase portrait constructed using a computer system or graphing calculator. $$ \frac{d x}{d t}=y-1, \quad \frac{d y}{d t}=x^{2}-y $$

In Problems 29 through 32, find all critical points of the given system, and investigate the type and stability of each. Verify your conclusions by means of a phase portrait constructed using a computer system or graphing calculator. $$ \frac{d x}{d t}=x y-2, \quad \frac{d y}{d t}=x-2 y $$

In Problems 19 through 28, investigate the type of the criti- cal point \((0,0)\) of the given almost linear system. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait. Also, describe the approximate locations and apparent types of any other critical points that are visible in your figure. Feel free to investigate these addi- tional critical points; you can use the computational methods discussed in the application material for this section. $$ \frac{d x}{d t}=x+2 y+x^{2}+y^{2}, \quad \frac{d y}{d t}=2 x-2 y-3 x y $$

A system \(d x / d t=F(x, y)\), \(d y / d t=G(x, y)\) is given. Solve the equation $$ \frac{d y}{d x}=\frac{G(x, y)}{F(x, y)} $$ to find the trajectories of the given system. Use a computer system or graphing calculator to construct a phase portrait and direction field for the system, and thereby identify visually the apparent character and stability of the critical point \((0,0)\) of the given system. $$ \frac{d x}{d t}=y\left(1+x^{2}+y^{2}\right), \quad \frac{d y}{d t}=x\left(1+x^{2}+y^{2}\right) $$

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