Notes-PhasePlane

# Equations by their stability in addition due to the

• Notes
• 28

This preview shows pages 3–7. Sign up to view the full content.

equations by their stability . In addition, due to the truly two-dimensional nature of the parametric curves, we will also classify the type of those critical points by their shapes (or, rather, by the shape formed by the trajectories about each critical point). Comment : The accurate tracing of the parametric curves of the solutions is not an easy task without electronic aids. However, we can obtain very reasonable approximation of a trajectory by using the very same idea behind the direction field, namely the tangent line approximation. At each point x = ( x 1 , x 2 ) on the ty -plane, the direction of motion of the solution curve the passes through the point is determined by the direction vector (i.e. the tangent vector ) x ′, the derivative of the solution vector x , evaluated at the given point. The tangent vector at each given point can be calculated directly from the given matrix-vector equation x ′ = Ax , using the position vector x = ( x 1 , x 2 ). Like working with a direction field, there is no need to find the solution first before performing this approximation.

This preview has intentionally blurred sections. Sign up to view the full version.

© 2008 Zachary S Tseng D -2 - 4 Given x ′ = Ax , where there is only one critical point, at (0,0): Case I. Distinct real eigenvalues The general solution is t r t r e k C e k C x 2 1 2 2 1 1 + = . 1. When r 1 and r 2 are both positive, or are both negative The phase portrait shows trajectories either moving away from the critical point to infinite-distant away (when r > 0), or moving directly toward, and converge to the critical point (when r < 0). The trajectories that are the eigenvectors move in straight lines. The rest of the trajectories move, initially when near the critical point, roughly in the same direction as the eigenvector of the eigenvalue with the smaller absolute value. Then, farther away, they would bend toward the direction of the eigenvector of the eigenvalue with the larger absolute value The trajectories either move away from the critical point to infinite-distant away (when r are both positive), or move toward from infinite-distant out and eventually converge to the critical point (when r are both negative). This type of critical point is called a node . It is asymptotically stable if r are both negative, unstable if r are both positive.
© 2008 Zachary S Tseng D -2 - 5 Two distinct real eigenvalues, both of the same sign Type: Node Stability: It is unstable if both eigenvalues are positive; asymptotically stable if they are both negative.

This preview has intentionally blurred sections. Sign up to view the full version.

© 2008 Zachary S Tseng D -2 - 6 2. When r 1 and r 2 have opposite signs (say r 1 > 0 and r 2 < 0) In this type of phase portrait, the trajectories given by the eigenvectors of the negative eigenvalue initially start at infinite-distant away, move toward and eventually converge at the critical point. The trajectories that represent the eigenvectors of the positive eigenvalue move in exactly the opposite way: start at the critical point then diverge to infinite-distant out. Every other trajectory starts at infinite-distant away, moves toward but never converges to the critical point, before changing direction and moves back to infinite-distant away.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern