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Chapter 11: Problem 14

Classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point. $$ \begin{aligned} &x^{\prime}=2 x-y^{2} \\ &y^{\prime}=-y+x y \end{aligned} $$

Short Answer

Expert verified
The point (0,0) is a saddle point; (1,0) is ambiguously stable.

Step by step solution

01

Identify Critical Points

To find the critical points of the system, we set the derivatives to zero: \(x' = 2x - y^2 = 0\) and \(y' = -y + xy = 0\). \ From \(y' = 0\), we get \(y(x - 1) = 0\), giving us \(y = 0\) or \(x = 1\). If \(y = 0\), substituting in \(x' = 0\) gives \(2x = 0\), so \(x = 0\). Hence, one critical point is (0,0). If \(x = 1\), \(y' = 0\) implies \(y = 0\), so the other critical point is (1,0). Thus, the critical points are (0,0) and (1,0).
02

Determine the Jacobian Matrix

The Jacobian of the system is calculated by taking the partial derivatives: \[J = \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix} = \begin{bmatrix}2 & -2y \y & x-1\end{bmatrix}\]
03

Evaluate the Jacobian at Critical Point (0,0)

Substitute \((x, y) = (0, 0)\) into the Jacobian matrix: \[J(0, 0) = \begin{bmatrix}2 & 0 \0 & -1\end{bmatrix}\]The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = -1\). Since one eigenvalue is positive and one is negative, (0, 0) is a saddle point.
04

Evaluate the Jacobian at Critical Point (1,0)

Substitute \((x, y) = (1, 0)\) into the Jacobian matrix: \[J(1, 0) = \begin{bmatrix}2 & 0 \0 & 0\end{bmatrix}\]The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 0\). Since there is one zero eigenvalue, further analysis (phase line, etc.) would be necessary to determine stability, but typically a zero eigenvalue at equilibrium suggests semi-stability.

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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 the context of autonomous systems are specific points in the plane where the system's behavior changes significantly. These points occur where the derivatives are set to zero, meaning no change occurs in either direction. For example, given the system \[ x' = 2x - y^2 \] and \[ y' = -y + xy \], we find critical points by solving these equations simultaneously.
  • Set \( x' = 0 \), resulting in \( 2x - y^2 = 0 \).
  • Set \( y' = 0 \), giving \( -y + xy = 0 \), or simplified to \( y(x - 1) = 0 \).
Thus, critical points can be identified, for this system, as (0,0) and (1,0). They are essential, as dynamics near these points often dictate the overall behavior of the system.
Jacobian Matrix
The Jacobian matrix is a mathematical tool that helps us analyze system behavior near its critical points. Essentially, it contains all possible first-order partial derivatives of a vector-valued function. For our system, the Jacobian is calculated as follows:\[J = \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix} = \begin{bmatrix}2 & -2y \ y & x-1\end{bmatrix}\]The purpose of the Jacobian is to determine how small changes at the critical points affect the system.
  • Evaluating it at different critical points provides insight into local system behavior.
  • Understanding this can help predict stability and system trajectory.
In essence, the Jacobian matrix unlocks the linear approximation of the system at any given point.
Stability Analysis
Stability analysis is a method used to determine the behavior of critical points and whether they will return to a stable state after a disturbance. The process generally involves evaluating the Jacobian matrix at critical points.For example, consider the critical point (0,0) with Jacobian: \[J(0, 0) = \begin{bmatrix}2 & 0\0 & -1\end{bmatrix}\]Analyzing the eigenvalues of this matrix, \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \), provides crucial information.
  • Positive eigenvalues indicate instability (as seen with \( \lambda_1 = 2 \)).
  • Negative eigenvalues suggest stability (as seen with \( \lambda_2 = -1 \)).
In conclusion, since one eigenvalue is positive and another is negative, the point is a saddle and inherently unstable.
Eigenvalues
Eigenvalues are a key concept when discussing the stability of critical points in a differential system. They are derived from the Jacobian matrix and indicate the rate at which each dimension of the system grows or shrinks.To find eigenvalues, we solve the characteristics equation \( \det(J - \lambda I) = 0 \). For instance, substituting the Jacobian calculated at a critical point:
  • For (0,0) with \( J(0, 0) = \begin{bmatrix}2 & 0\0 & -1\end{bmatrix}\), we derive eigenvalues of 2 and -1.
  • For (1,0) with \( J(1, 0) = \begin{bmatrix}2 & 0\0 & 0\end{bmatrix}\), the eigenvalues are 2 and 0, suggesting more complex stability analysis is needed.
Knowing the eigenvalues helps us anticipate the type of equilibrium point (node, saddle, spiral) and its stability.
Phase Plane Analysis
Phase plane analysis visually represents the trajectories of a dynamic system. By plotting the system's variables against each other, you can see how a system evolves over time.
  • The phase plane for a planar autonomous system will show paths or 'orbits' around critical points.
  • By analyzing these paths in conjunction with the Jacobian and eigenvalues, one can predict whether these points attract or repel nearby trajectories.
  • For example, in our system, a saddle point might display trajectories that diverge along one path and converge along another.
Phase plane analysis is powerful for understanding complex nonlinear system behavior, allowing visualization of stability without solving analytically complex equations.

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Most popular questions from this chapter

In Problems \(3-10\), without solving explicitly, classify the critical points of the given first-order autonomous differential equation as either asymptotically stable or unstable. All constants are assumed to be positive. $$ m \frac{d v}{d t}=m g-k v $$

The general solution of the linear system \(\mathbf{X}^{\prime}=\mathbf{A} \mathbf{X}\) is given. (a) In each case discuss the nature of the solution in a neighborhood of \((0,0)\). (b) With the aid of a graphing utility plot the solution that satisfies \(\mathbf{X}(0)=(1,1)\) $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{ll} -6 & 5 \\ -5 & 4 \end{array}\right) \\ &\mathbf{X}(t)=c_{1}\left(\begin{array}{l} 1 \\ 1 \end{array}\right) e^{-t}+c_{2}\left[\left(\begin{array}{l} 1 \\ 1 \end{array}\right) t e^{-t}+\left(\begin{array}{l} 0 \\ \frac{1}{5} \end{array}\right) e^{-t}\right] \end{aligned} $$

In Problems, find all critical points of the given plane autonomous system. $$ \begin{aligned} &x^{\prime}=x^{2} e^{y} \\ &y^{\prime}=y\left(e^{x}-1\right) \end{aligned} $$

(a) Show that \((0,0)\) is an isolated critical point of the plane autonomous system $$ \begin{aligned} &x^{\prime}=x^{4}-2 x y^{3} \\ &y^{\prime}=2 x^{3} y-y^{4} \end{aligned} $$ but that linearization gives no useful information about the nature of this critical point. (b) Use the phase-plane method to show that \(x^{3}+y^{3}=3 c x y\). This classic curve is called a folium of Descartes. Parametric equations for a folium are $$ x=\frac{3 c t}{1+t^{3}}, y=\frac{3 c t^{2}}{1+t^{3}} $$ [Hint: The differential equation in \(x\) and \(y\) is homogeneous.] (c) Use a graphing utility or a numerical solver to obtain solution curves. Based on your phase portrait, would you classify the critical point as stable or unstable? Would you classify the critical point as a node, saddle point, center, or spiral point? Explain.

Discuss the geometric nature of the solutions to the linear system \(\mathbf{X}^{\prime}=\mathbf{A X}\) given the general solution. (a) \(\mathbf{X}(t)=c_{1}\left(\begin{array}{l}1 \\ 1\end{array}\right) e^{-t}+c_{2}\left(\begin{array}{r}1 \\ -2\end{array}\right) e^{-2 t}\) (b) \(\mathbf{X}(t)=c_{1}\left(\begin{array}{r}1 \\ -1\end{array}\right) e^{-t}+c_{2}\left(\begin{array}{l}1 \\ 2\end{array}\right) e^{2 t}\)
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