{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ExercisesSolutions

11 2 1 1 2 y 1 05 05 1 x figure 11 phase plot of

Info icon This preview shows pages 11–15. Sign up to view the full content.

View Full Document Right Arrow Icon
11
Image of page 11

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

View Full Document Right Arrow Icon
–2 –1 0 1 2 y –1 –0.5 0.5 1 x Figure 11: Phase plot of system for Problem 1 12
Image of page 12
2. Purpose: To study fixed points and linearizations. From [Str94], 6.3 . Excercise: For each of the following systems, find the fixed points, classify them, sketch the neigh- boring trajectories, and try to fill in the rest of the phase portrait. (a) ˙ x = x - y, ˙ y = x 2 - 4 (b) ˙ x = sin y, ˙ y = x - x 3 (c) ˙ x = 1 + y - e x , ˙ y = x 3 - y (d) ˙ x = y + x - x 3 , ˙ y = - y (e) ˙ x = sin y, ˙ y = cos x (f) ˙ x = xy - 1 , ˙ y = x - y 3 Solution: For the following systems, the fixed points are found by solving ˙ x = 0, ˙ y = 0. The linearized system about the fixed point(s) is obtained by evaluating the Jacobian at those point(s). The eigenvalues of the resulting matrix are used to classify the fixed point(s). The phase portrait for each system in shown in the solution to Exercise 3. (a) ˙ x = x - y, ˙ y = x 2 - 4 The fixed points are (2 , 2) and ( - 2 , - 2). The Jacobian of the system is bracketleftbigg 1 - 1 2 x 0 bracketrightbigg . Classifying the fixed points using the eigenvalues, (2 , 2) : bracketleftbigg 1 - 1 4 0 bracketrightbigg . Eigenvalues are 1 2 ± 1 2 I 15, and thus (2 , 2) is an unstable spiral . ( - 2 , - 2) : bracketleftbigg 1 - 1 - 4 0 bracketrightbigg . Eigenvalues are 2 . 5616 and - 1 . 5616, and thus ( - 2 , - 2) is a saddle point . (b) ˙ x = sin y, ˙ y = x - x 3 The fixed points are (0 , nπ ), (0 , (2 n + 1) π ), (1 , nπ ), (1 , (2 n + 1) π ), ( - 1 , nπ ), and ( - 1 , (2 n + 1) π ). The Jacobian of the system is bracketleftbigg 0 cos y 1 - 3 x 2 0 bracketrightbigg . Classifying the fixed points using the eigenvalues, (0 , nπ ) : bracketleftbigg 0 1 1 0 bracketrightbigg . Eigenvalues are 1 and - 1, and thus (0 , nπ ) is a saddle point . (1 , nπ ) , ( - 1 , nπ ) : bracketleftbigg 0 1 - 2 0 bracketrightbigg . Eigenvalues are I 2 and - I 2, and thus a fixed point of this form is a center . (0 , (2 n + 1) π ) : bracketleftbigg 0 - 1 1 0 bracketrightbigg . Eigenvalues are I and - I , and thus (0 , (2 n + 1) π ) is a center . (1 , (2 n + 1) π ) , ( - 1 , (2 n + 1) π ) : bracketleftbigg 0 - 1 - 2 0 bracketrightbigg . Eigenvalues are 2 and - 2, and thus a fixed point of this form is a center . (c) ˙ x = 1 + y - e x , ˙ y = x 3 - y The only fixed point is (0 , 0). The Jacobian of the system is bracketleftbigg e x 1 3 x 2 - 1 bracketrightbigg . Classifying the fixed points using the eigenvalues, (0 , 0) : bracketleftbigg 1 1 0 - 1 bracketrightbigg . Eigenvalues are 1 and - 1, and thus (0 , 0) is a saddle point . (d) ˙ x = y + x - x 3 , ˙ y = - y The fixed points are (0 , 0), (1 , 0), and ( - 1 , 0). The Jacobian of the system is bracketleftbigg 1 - 3 x 2 1 0 - 1 bracketrightbigg . Classifying the fixed points using the eigenvalues, (0 , 0) : bracketleftbigg 1 1 0 - 1 bracketrightbigg . Eigenvalues are 1 and - 1, and thus (0 , 0) is a saddle point . 13
Image of page 13

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

View Full Document Right Arrow Icon
(1 , 0) , ( - 1 , 0) : bracketleftbigg - 2 1 0 - 1 bracketrightbigg . Eigenvalues are - 2 and - 1, and thus (1 , 0) and ( - 1 , 0) are stable nodes . (e) ˙ x = sin y, ˙ y = cos x The fixed points are ((2 n + 1) π/ 2 , 2 ). The Jacobian of the system is bracketleftbigg 0 cos y - sin x 0 bracketrightbigg .
Image of page 14
Image of page 15
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern