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Chapter 6: Problem 2

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. $$ \frac{d x}{d t}=2 \sin x+\sin y, \frac{d y}{d t}=\sin x+2 \sin y $$

Short Answer

Expert verified
The critical point \((0, 0)\) is an unstable node.

Step by step solution

01

Identify the Critical Points

A critical point of a system occurs where the derivatives are simultaneously zero. Here, we need to solve the equations \( \frac{d x}{d t} = 2 \sin x + \sin y = 0 \) and \( \frac{d y}{d t} = \sin x + 2 \sin y = 0 \). Observing that \( \sin x = 0 \) and \( \sin y = 0 \) imply \( x = 0 \) and \( y = 0 \), the critical point is \((0, 0)\).
02

Linearization of the System

Next, we linearize the system near the critical point \((0, 0)\). The Jacobian matrix at \((x, y)\) for the given system is \( J(x, y) = \begin{bmatrix} 2 \cos x & \cos y \ \cos x & 2 \cos y \end{bmatrix} \). Evaluating at \((0, 0)\), we find the Jacobian matrix \( J(0, 0) = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix} \).
03

Determine the Eigenvalues

To classify the critical point, we need the eigenvalues of the Jacobian matrix. The characteristic equation is given by \( \det(J - \lambda I) = 0 \). For our matrix, the characteristic polynomial is \( \lambda^2 - 4\lambda + 3 = 0 \), which factors to \( (\lambda - 3)(\lambda - 1) = 0 \). Solving, the eigenvalues are \( \lambda_1 = 3 \) and \( \lambda_2 = 1 \).
04

Classify the Critical Point

Since both eigenvalues are positive (\(\lambda_1 = 3\) and \(\lambda_2 = 1\)), the critical point \((0, 0)\) is an unstable node. Both eigenvalues being positive indicates that trajectories are repelled away from the critical point.
05

Verify with Phase Plane Portrait

Plotting the phase plane confirms our conclusion. Using software or a graphing calculator, plot the directions specified by the system equations. Trajectories moving away from the origin confirm our classification of an unstable node.

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Key Concepts

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

Critical Points
In a system of differential equations, a critical point is where the system's derivatives are zero for the given variables. This means there are no changes at that point, making it stationary. To find these points, you set each differential equation in the system equal to zero. For example, starting with
  • \( \frac{d x}{d t} = 2 \sin x + \sin y = 0 \)
  • \( \frac{d y}{d t} = \sin x + 2 \sin y = 0 \)
We can solve to find where both conditions are satisfied. Often, trigonometric identities, such as \( \sin x = 0 \) at specific points, help simplify finding where these are true. For this particular system, the critical point happened at \((0, 0)\), as discussed in the solution.
Linearization
Linearization involves approximating a non-linear system near a critical point with a linear system. This process helps in understanding the behavior of the system near that point and is done using Taylor series expansion. To linearize, you'd create a Jacobian matrix, which contains partial derivatives of each function in the system with respect to each variable. Evaluating this at a critical point yields a matrix that represents a local linear approximation of the system.
Jacobian Matrix
The Jacobian matrix plays a crucial role in analyzing the behavior of systems of differential equations. It is composed of partial derivatives of the functions that make up the system. Each element of the Jacobian matrix, \( J(x, y) \), is found by partially differentiating the system:
  • \( \frac{\partial f}{\partial x} \text{ and } \frac{\partial f}{\partial y} \)
  • \( \frac{\partial g}{\partial x} \text{ and } \frac{\partial g}{\partial y} \)
For our particular system: \( J(x, y) = \begin{bmatrix} 2 \cos x & \cos y \ \cos x & 2 \cos y \end{bmatrix} \). At \((0,0)\), evaluating this yields \( J(0,0) = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix} \). This matrix allows us to explore the stability of the critical point.
Eigenvalues
Eigenvalues are critical in determining the stability and nature of critical points. They are derived from the Jacobian matrix by solving the characteristic equation:\[ \det(J - \lambda I) = 0 \]In our example, the Jacobian matrix at \((0,0)\) leads to the equation \( \lambda^2 - 4\lambda + 3 = 0 \), which factors into \( (\lambda - 3)(\lambda - 1) = 0 \). Solving this gives eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = 1 \). Since both are positive, it implies that nearby trajectories move away from the critical point, meaning the system is unstable at this node.
Phase Plane Portrait
A phase plane portrait visually describes the dynamics of a system of differential equations by plotting the vector field of the system. It uses arrows to show trajectory directions, reflecting how the system evolves over time.For our system, software or a graphing calculator allows visualization. By plotting the system, you can observe the nature of trajectories. In this exercise, the phase plane illustrates the prediction that trajectories move outward from \((0,0)\), confirming that it behaves as an unstable node. Visual representation makes it more intuitive to grasp the system's global behavior.

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Most popular questions from this chapter

Find and classify each of the critical points of the almost linear systems. Use a computer system or graphing calculator to construct a phase plane portrait that illustrates your findings. $$ \frac{d x}{d t}=1-e^{x-y}, \frac{d y}{d t}=2 \sin x $$

First describe the type of \(x\) - and \(y\) -populations involved \((e x\) ponential or logistic) and the nature of their interactioncompetition, cooperation, or predation. Then find and characterize the system's critical points (as to type and stability). Determine what nonzero \(x-\) and \(y\) -populations can coexist. Finally, construct a phase plane portrait that enables you to describe the long-term behavior of the two populations in terms of their initial populations \(x(0)\) and \(y(0)\). $$ \frac{d x}{d t}=2 x y-4 x, \quad \frac{d y}{d t}=x y-3 y $$

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. \(\frac{d x}{d t}=4 y\left(1+x^{2}+y^{2}\right), \quad \frac{d y}{d t}=-x\left(1+x^{2}+y^{2}\right)\)

Each of the systems in Problems 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. $$ \frac{d x}{d t}=x-2 y-8, \quad \frac{d y}{d t}=x+4 y+10 $$

Find each equilibrium solution \(x(t)=x_{0}\) of the given second-order differential equation \(x^{\prime \prime}+f\left(x, x^{\prime}\right)=0 .\) Use a computer system or graphing cal culator to construct a phase portrait and direction field for the equivalent first-order system \(x^{\prime}=y, y^{\prime}=-f(x, y) .\) Thereby ascertain whether the critical point \(\left(x_{0}, 0\right)\) looks like a center, a saddle point, or a spiral point of this system. \(x^{\prime \prime}+\left(x^{2}-1\right) x^{\prime}+x=0\)
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