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Chapter 6: Problem 17

Determine whether the critical point \((0,0)\) is stable, asymptotically stable, or unstable. Use a computer system or graphing calculator to construct a phase portrait and direction field for the given system. Thereby ascertain the stability or instability of each critical point, and identify it visually as a node, a saddle point, a center, or a spiral point. $$ \frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x $$

Short Answer

Expert verified
The critical point \((0,0)\) is neutrally stable and is a center.

Step by step solution

01

Understanding the System

The given system of differential equations is \( \frac{d x}{d t} = y \) and \( \frac{d y}{d t} = -x \). This represents a simple harmonic oscillator, which is a common example in physics and dynamical systems.
02

Forming a Matrix Representation

We form the matrix \( A \) of the linearized system as \(A = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \). This matrix will be used to find the eigenvalues and analyze stability.
03

Finding the Eigenvalues

Calculate the eigenvalues, \( \lambda \), of matrix \( A \) by solving \(\det(A - \lambda I) = 0 \). We compute:\[ \det \left( \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \right) = \det \begin{bmatrix} -\lambda & 1 \ -1 & -\lambda \end{bmatrix} = \lambda^2 + 1 = 0 \].This gives \( \lambda = \pm i \).
04

Interpretation of Eigenvalues

The eigenvalues are \( i \) and \(-i\), which are purely imaginary numbers. This indicates that the critical point is a center.
05

Analyzing Stability

Since the eigenvalues are purely imaginary, the critical point \((0,0)\) is neutrally stable. This means trajectories neither converge toward nor diverge away from the critical point but instead form closed orbits.
06

Visual Confirmation with Phase Portrait

Using a computer system or graphing calculator, a phase portrait for this system can be generated. The direction field and phase portrait will show circular orbits around the origin. This visual confirmation aligns with our analysis, showing the origin as a center.

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Key Concepts

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

Phase Portrait
A phase portrait is a graphical representation of trajectories of a dynamical system in the phase plane. For a system described by the equations \( \frac{dx}{dt} = y \) and \( \frac{dy}{dt} = -x \), the phase portrait helps us visualize how the system behaves over time for different initial conditions.
  • Each trajectory in the phase portrait represents a solution to the differential equations.
  • The direction, shape, and density of the paths show how solutions evolve over time.
  • Phase portraits are particularly useful for identifying the nature of critical points.
In our exercise, the phase portrait shows circular orbits centered at \((0,0)\). This indicates that solutions revolve around the critical point without spiraling inward or outward. Such representations provide strong visual insights into the system's stability characteristics.
Stability Analysis
Stability analysis involves determining how a dynamical system responds to small perturbations near a critical point. In our example, the system consists of a simple harmonic oscillator with the equations \( \frac{dx}{dt} = y \) and \( \frac{dy}{dt} = -x \).
  • For stability analysis, we linearize the system around the critical point and find the eigenvalues of the resulting matrix.
  • The signs and nature (real, complex, imaginary) of these eigenvalues determine the system's stability.
  • In our exercise, the eigenvalues are purely imaginary, indicating neutral stability.
Neutral stability means that perturbations neither grow nor decay over time but rather, solutions form closed orbits. This is a defining feature of systems like our simple harmonic oscillator, where energy is conserved and oscillations continue indefinitely.
Critical Points
Critical points, or equilibrium points, are where the system remains constant if undisturbed. For our system \( \frac{dx}{dt} = y \) and \( \frac{dy}{dt} = -x \), the critical point can be identified by setting the derivatives to zero.
  • The obtained critical point in this exercise is \((0,0)\).
  • At a critical point, understanding the local behavior is crucial to assessing stability.
  • The matrix formed from the linearization around the critical point plays a key role in this assessment.
It's important to analyze the nature of every critical point using eigenvalues of the linearized system. In our case, \((0,0)\) represents a center, characterized by pure oscillations, as indicated by our phase portrait and eigenvalues.
Eigenvalues
Eigenvalues are fundamental in analyzing linear systems of differential equations, as they help determine the stability around a critical point. For a matrix \( A \), the eigenvalues are solutions to the equation \( \det(A - \lambda I) = 0 \).
  • In the solution, the matrix \( A \) is \( \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \).
  • By solving for \( \lambda \), we find the eigenvalues are \( i \) and \( -i \).
  • These purely imaginary eigenvalues tell us that the critical point is a center.
This information helps in categorizing the critical point's stability as neutral, meaning trajectories will orbit the center point in closed, circular paths without moving toward or away from the center. Understanding eigenvalues is crucial in dynamical systems analysis, giving us insights into the long-term behavior of the system.

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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 linearization of (2) at \((0,21)\) is \(u^{\prime}=-3 u\), \(v^{\prime}=-63 u-42 v\). Then show that the coefficient matrix of this linear system has negative eigenvalues \(\lambda_{1}=-3\) and \(\lambda_{2}=-42 .\) Hence \((0,21)\) is a nodal sink for the system in (2).

Problems outline an investigation of the period \(T\) of oscillation of a mass on a nonlinear spring with equation of motion $$ \frac{d^{2} x}{d t^{2}}+\phi(x)=0 $$ If \(\phi(x)=k x\) with \(k>0\), then the spring actually is linear with period \(T_{0}=2 \pi / \sqrt{k}\), Integrate once (as in Eq. (6)) to derive the energy equation $$ \frac{1}{2} y^{2}+V(x)=E $$ where \(y=d x / d t\) and $$ V(x)=\int_{0}^{x} \phi(u) d u . $$

Determine whether the critical point \((0,0)\) is stable, asymptotically stable, or unstable. Use a computer system or graphing calculator to construct a phase portrait and direction field for the given system. Thereby ascertain the stability or instability of each critical point, and identify it visually as a node, a saddle point, a center, or a spiral point. $$ \frac{d x}{d t}=2 x, \quad \frac{d y}{d t}=-2 y $$

Each of the systems in Problems 11 through 18 has a single critical point \(\left(x_{0}, y_{0}\right) .\) Apply Theorem 2 to classify this critical point as to type and stability. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait for the given system. $$ \frac{d x}{d t}=x-2 y-8, \quad \frac{d y}{d t}=x+4 y+10 $$

In Problems, show that the given system is almost linear with \((0,0)\) as a critical point, and classify this critical point as to type and stability. Use a computer system or graphing calculator to construct a phase plane portrait that illustrates your conclusion. $$ \frac{d x}{d t}=1-e^{x}+2 y, \frac{d y}{d t}=-x-4 \sin y $$
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