MATH
ExercisesSolutions

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

• Notes
• 32
• 100% (1) 1 out of 1 people found this document helpful

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

11

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

–2 –1 0 1 2 y –1 –0.5 0.5 1 x Figure 11: Phase plot of system for Problem 1 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

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

(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 .
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