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Chapter 6: Problem 23

In Problems, investigate the type of the critical 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 additional critical points; you can use the computational methods discussed in the application material for this section. $$ \frac{d x}{d t}=2 x-5 y+x^{3}, \quad \frac{d y}{d t}=4 x-6 y+y^{4} $$

Short Answer

Expert verified
The critical point \((0,0)\) is a stable spiral point.

Step by step solution

01

Determine the Jacobian Matrix

The given system is: \( \frac{dx}{dt} = 2x - 5y + x^3 \) and \( \frac{dy}{dt} = 4x - 6y + y^4 \). To find the Jacobian matrix at the point \((0, 0)\), we calculate the partial derivatives of the right-hand sides with respect to \(x\) and \(y\). The Jacobian is given by:\[J(x, y) = \begin{bmatrix}\frac{\partial}{\partial x}(2x - 5y + x^3) & \frac{\partial}{\partial y}(2x - 5y + x^3) \\frac{\partial}{\partial x}(4x - 6y + y^4) & \frac{\partial}{\partial y}(4x - 6y + y^4) \\end{bmatrix}\]Evaluating the partial derivatives at \((0,0)\):- \( \frac{\partial}{\partial x} (2x - 5y + x^3) = 2 \) - \( \frac{\partial}{\partial y} (2x - 5y + x^3) = -5 \)- \( \frac{\partial}{\partial x} (4x - 6y + y^4) = 4 \) - \( \frac{\partial}{\partial y} (4x - 6y + y^4) = -6 \)Hence, the Jacobian \( J(0, 0) \) is:\[\begin{bmatrix}2 & -5 \4 & -6\end{bmatrix}\]
02

Find the Eigenvalues of the Jacobian

Next, we calculate the eigenvalues of the Jacobian matrix \( \begin{bmatrix} 2 & -5 \ 4 & -6 \end{bmatrix} \).The characteristic equation is \[det(J - \lambda I) = \det\begin{bmatrix} 2 - \lambda & -5 \ 4 & -6 - \lambda \end{bmatrix} = 0\]Computing the determinant, we have:\[(2 - \lambda)(-6 - \lambda) + 20 = \lambda^2 + 4\lambda + 8 = 0\]Solving for \(\lambda\) using the quadratic formula, \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), \(c = 8\):\[lambda = \frac{-4 \pm \sqrt{16 - 32}}{2}\]\[lambda = \frac{-4 \pm \sqrt{-16}}{2} = \frac{-4 \pm 4i}{2} = -2 \pm 2i\]These are complex conjugate eigenvalues.
03

Determine the Type of Critical Point

Because the eigenvalues are complex conjugates with a negative real part (\(-2 \pm 2i\)), the critical point \((0,0)\) is a stable spiral point. This implies that trajectories will spiral into the point \((0,0)\) over time.
04

Verify with Phase Portrait

Now, construct a phase portrait using a computer algebra system or graphing calculator. This involves plotting the trajectories in the \(x-y\) plane and observing the behavior of solutions as \(t\) evolves. The phase portrait should confirm that trajectories spirals towards the origin, supporting our conclusion that it is a stable spiral point.
05

Explore Other Critical Points

Other critical points can be explored by setting \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\), beyond \((0,0)\):1. Solve \(2x - 5y + x^3 = 0\)2. Solve \(4x - 6y + y^4 = 0\)Using computational tools, these can yield other critical points, whose stability can be assessed similarly by finding Jacobian matrices and their eigenvalues at those points. The specific forms and stability types can be further illustrated by their corresponding phase trajectory shapes in the phase portrait.

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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 that illustrates how the solutions to a system of differential equations evolve over time. In math terms, it shows the direction and stability of trajectories in the phase plane (commonly the \(x-y\) plane).
Each curve, or trajectory, in the phase portrait represents a solution to the system, where the direction of time is usually shown by arrows. Some important features of phase portraits include:
  • Trajectories: These are the paths that the solutions take in the phase space.
  • Equilibrium Points: Fixed points where solutions do not change over time, also known as critical points.
  • Stability: Indicates how trajectories behave over time, whether they move towards or away from equilibrium points.
Constructing a phase portrait for the given system involves plotting these trajectories to see how they interact with the critical point \((0,0)\). Since this critical point is a stable spiral, the phase portrait will show trajectories spiraling into the origin, meaning all paths move inwards towards this point as time progresses.
Critical Points
Critical points, in the context of differential equations, are the points where the rate of change of the system is zero. This means if the system were to start at this point, it would remain there indefinitely without any time progression.
For our system, defined by the equations \( \frac{dx}{dt} = 2x - 5y + x^3 \) and \( \frac{dy}{dt} = 4x - 6y + y^4 \), the critical point \((0,0)\) was found when both equations are set to zero. Understanding critical points helps us to:
  • Identify where equilibrium occurs in the system.
  • Determine the local behavior of the system at those points, such as whether they attract or repel trajectories.
To confirm stability, we compute the eigenvalues of the Jacobian matrix at each critical point. If the real parts of all eigenvalues are negative, the critical point is stable. At \((0,0)\), these eigenvalues are complex conjugates with negative real parts, characterizing a stable spiral point.
Jacobian Matrix
The Jacobian matrix in differential equations is essential for analyzing the local behavior of a system near its critical points. It is a matrix composed of first-order partial derivatives of the system's functions with respect to its variables.
For the given system, the Jacobian matrix at a point \((x, y)\) is formed by:
\[ J(x, y) = \begin{bmatrix} \frac{\partial}{\partial x}(2x - 5y + x^3) & \frac{\partial}{\partial y}(2x - 5y + x^3) \ \frac{\partial}{\partial x}(4x - 6y + y^4) & \frac{\partial}{\partial y}(4x - 6y + y^4) \end{bmatrix} \]
Evaluating this matrix at \((0,0)\) gives us important insights into the dynamics around this critical point, where the matrix was calculated as:
\[ \begin{bmatrix} 2 & -5 \ 4 & -6 \end{bmatrix} \]
An essential step involves finding the eigenvalues of this matrix, as they describe the nature and stability of critical points. Complex eigenvalues with a negative real part imply oscillations that decay over time, indicating a stable spiral behavior around the origin, as witnessed in the phase portrait observations.

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Most popular questions from this chapter

In the case of a two-dimensional system that is \(n o t\) almost linear, the trajectories near an isolated critical point can exhibit a considerably more complicated structure than those near the nodes, centers, saddle points, and spiral points discussed in this section. For example, consider the system $$ \begin{aligned} &\frac{d x}{d t}=x\left(x^{3}-2 y^{3}\right) \\ &\frac{d y}{d t}=y\left(2 x^{3}-y^{3}\right) \end{aligned} $$ having \((0,0)\) as an isolated critical point. This system is not almost linear because \((0,0)\) is not an isolated critical point of the trivial associated linear system \(x^{\prime}=0, y^{\prime}=0\). Solve the homogeneous first-order equation $$ \frac{d y}{d x}=\frac{y\left(2 x^{3}-y^{3}\right)}{x\left(x^{3}-2 y^{3}\right)} $$ to show that the trajectories of the system in (16) are folia of Descartes of the form $$ x^{3}+y^{3}=3 c x y $$ where \(c\) is an arbitrary constant (Fig. 6.2.14).

$$ \begin{aligned} &\frac{d x}{d t}=60 x-4 x^{2}-3 x y \\ &\frac{d y}{d t}=42 y-2 y^{2}-3 x y \end{aligned} $$ 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 (3) at \((12,6)\) is \(u^{\prime}=-36 u-\) \(48 v, v^{\prime}=-12 u-18 v\). Then show that the coefficient matrix of this linear system has eigenvalues \(\lambda_{1}=-27+\) \(3 \sqrt{73}\) and \(\lambda_{2}=-27-3 \sqrt{73}\), both of which are negative. Hence \((12,6)\) is a nodal sink for the system in (3).

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. \(x^{\prime \prime}+\left(x^{2}-1\right) x^{\prime}+x=0\)

In Problems, 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. $$ \frac{d x}{d t}=x-y, \quad \frac{d y}{d t}=x^{2}-y $$

Solve each of the linear systems to 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. \(\frac{d x}{d t}=x, \quad \frac{d y}{d t}=3 y\)
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