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Chapter 6: Problem 6

Find and classify each of the critical points of the almost lin. ear systems in Problems Use a computer system or graphing calculator to construct a phase plane portrait that illustrates your findings. $$ \frac{d x}{d t}=y, \frac{d y}{d t}=\sin \pi x-y $$

Short Answer

Expert verified
Critical points are at \((n, 0)\); classify them via eigenvalues: saddle (even \(n\)), spiral (odd \(n\)).

Step by step solution

01

Identify the System

The given system of differential equations is: \( \frac{dx}{dt} = y \) and \( \frac{dy}{dt} = \sin(\pi x) - y \). These equations describe the rate of change of \(x\) and \(y\) over time \(t\).
02

Find Critical Points

To find critical points, set \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \). This leads to the system: \( y = 0 \) and \( \sin(\pi x) - y = 0 \). Substituting \( y = 0 \) into the second equation gives \( \sin(\pi x) = 0 \). The solutions are \( x = n \), where \( n \) is any integer. Thus, the critical points are \( (x, y) = (n, 0) \).
03

Linearize the System

Linearize the system around the critical points \( (n, 0) \) using the Jacobian matrix. Calculate the partial derivatives: \(J = \begin{bmatrix}0 & 1 \\pi \cos(\pi x) & -1\end{bmatrix}.\) At each critical point \( (n, 0) \), where \( \cos(\pi n) = (-1)^n \), the Jacobian becomes \( J_n = \begin{bmatrix} 0 & 1 \ (-1)^n \pi & -1 \end{bmatrix}. \)
04

Classify Critical Points using Eigenvalues

Find the eigenvalues of the Jacobian \( J_n \) at each critical point. The characteristic equation is \( \lambda^2 + \lambda + (-1)^n \pi = 0 \). Solving it using the quadratic formula, \( \lambda = \frac{-1 \pm \sqrt{1 - 4(-1)^n \pi}}{2} \). This yields: - If \( n \) is even, \( \lambda = \frac{-1 \pm \sqrt{1 + 4\pi}}{2} \).- If \( n \) is odd, \( \lambda = \frac{-1 \pm \sqrt{1 - 4\pi}}{2} \).

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Key Concepts

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

Critical Points
Critical points in a system of differential equations are the points where the rates of change are zero. In simpler terms, these are the points where nothing is changing over time.
To find critical points, you set the derivatives to zero, solving for coordinates where the system stabilizes.
  • In our example, the system is given by the equations \( \frac{dx}{dt} = y \) and \( \frac{dy}{dt} = \sin(\pi x) - y \).
  • Setting \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) both equal to zero provides \( y = 0 \) and \( \sin(\pi x) = 0 \).
When the sine function is zero, it happens at integer multiples of \( \pi \), thus \( x = n \), where \( n \) is an integer.
Hence, the critical points are \( (x, y) = (n, 0) \). These points serve as the baseline for phase plane analysis.
Jacobian Matrix
The Jacobian matrix is a useful tool in analyzing the stability of critical points in nonlinear systems. It involves finding the partial derivatives of the functions and constructing a matrix that represents those derivatives.
In our system, the Jacobian matrix, \( J \), is calculated as follows:
  • The partial derivative of \( \frac{dx}{dt} = y \) with respect to \( x \) and \( y \) yields elements \( 0 \) and \( 1 \)
  • The partial derivative of \( \frac{dy}{dt} = \sin(\pi x) - y \) gives \( \pi \cos(\pi x) \) with respect to \( x \) and \( -1 \) with respect to \( y \)
The Jacobian matrix thus is:
\[ J = \begin{bmatrix} 0 & 1 \ \pi \cos(\pi x) & -1 \end{bmatrix} \]
At the critical points \( (n, 0) \), where \( \cos(\pi n) = (-1)^n \), the Jacobian simplifies to \( J_n = \begin{bmatrix} 0 & 1 \ (-1)^n \pi & -1 \end{bmatrix} \). This matrix helps determine how small changes affect the system around those critical points.
Phase Plane Portrait
A phase plane portrait graphically represents the trajectories of a dynamic system in the plane defined by state variables, like \( x \) and \( y \). These visual aids help identify the behavior near critical points, showing how the system evolves over time.
In our exercise, the phase plane portrait displays:
  • The directions the system takes as \( t \) progresses
  • The nature of trajectories around each critical point \( (n, 0) \)
With a computational tool or a graphing calculator, a phase plane portrait can be created to illustrate behaviors like spirals, nodes, or saddles, giving insights into the system's dynamics. This visualization complements the analytical work done with the Jacobian matrix and the eigenvalues, reinforcing the understanding of system stability and motion.
Eigenvalues
Eigenvalues derived from the Jacobian matrix provide insight into the local dynamics around critical points. These values help classify the nature and stability of these points.
In our step-by-step solution, calculating eigenvalues involves solving the characteristic equation of the Jacobian matrix \( J_n \):
  • For even \( n \): \( \lambda = \frac{-1 \pm \sqrt{1 + 4\pi}}{2} \)
  • For odd \( n \): \( \lambda = \frac{-1 \pm \sqrt{1 - 4\pi}}{2} \)
The sign and magnitude of the eigenvalues determine the type of stability:
  • A real positive eigenvalue indicates an unstable direction.
  • A real negative eigenvalue suggests stability.
  • Complex eigenvalues with positive real parts correspond to spirals or unstable foci.
  • Complex eigenvalues with negative real parts indicate stable foci.
Understanding these eigenvalues helps in predicting the behavior of a system around its critical points.

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Most popular questions from this chapter

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. $$ \frac{d x}{d t}=4 x-5 y+3, \quad \frac{d y}{d t}=5 x-4 y+6 $$

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 $$ V(x)=\int_{0}^{x} \phi(u) d u . $$

Consider the linear system $$ \frac{d x}{d t}=\epsilon x-y, \quad \frac{d y}{d t}=x+\epsilon y . $$ Show that the critical point \((0,0)\) is (a) a stable spiral point if \(\epsilon<0 ;\) (b) a center if \(\epsilon=0 ;\) (c) an unstable spiral point if \(\epsilon>0\). Thus small perturbations of the system \(x^{\prime}=-y, y^{\prime}=x\) can change both the type and stability of the critical point. Figures \(6.2 .18(\mathrm{a})-(\mathrm{e})\) illustrate the loss of stability that occurs at \(\epsilon=0\) as the parameter increases from \(\epsilon<0\) to \(\epsilon>0\)

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. $$ \frac{d x}{d t}=x-2 y-8, \quad \frac{d y}{d t}=x+4 y+10 $$
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