## Homework 5 collaboration area

Question from Student: Does the solution for #5 on page 151 need to be in the form of cos and sin rather than vectors and natural logs?

Question from Ryan Russon

For problems p.159: 4,7,11 are we supposed to accompany each solution with a sketch of the what is happening at the critical points or are we fine just stating what is happening at those points based on the eigenvalues of the linearized system? My confusion is stemming from the answer in the back of the book for #11 which says, "Use -cos(+- 1/2...)," you get the picture.

Answer from Steve Bell :

Those problems don't seem to ask for a sketch, so don't bother. (If you wanted to practice for the exam, testing yourself to see if you would know how to draw a sketch if you had to would be therapeutic.)

That cryptic remark about the trig identity can be used like so

$-\cos(\frac{\pi}{2}+ x)=\sin x = x -\frac{1}{3!}x ^3 +\dots \approx x$

(when x is small) to realize that the linearized system at the critical point (pi/2,0) has a 1 times y1 in that spot. I think it's easier to use the Jacobian matrix to find the first order Taylor term the way I demonstrated in class to see this.

Response from Ryan Russon

Thanks Steve! Using the Jacobian is how I approached it and it is a lot more intuitive for me to approach it that way.

Question from Christine

For problems p.159: I am looking for some suggestion for starting #7. Factoring with the two variables to get the eigen values is not working out...? Thanks!

Answer from Chris

Once you have the equations for y1' and y2', you can set them both equal to zero, and use algebra to find which combinations of y1 and y2 satisfy y1'=y2'=0 (these are your critical points). Then, linearize your equations for y1' and y2' using the Jacobian method used in class. From there, it shouldn't be hard to find the eigenvalues at each critical point.

--Dalec 21:45, 22 September 2013 (UTC)

Response from Mickey Rhoades Mrhoade

If you follow the method Dalec suggested, you will find that y1 = -y2 from the second equation and that if you sub this into the first equation, your critical points are at (0,0) and (-2,+2).  Then, linearize with the jacobian and sub in each critical point into the Jacobian matrix and find the eigenvalues.  You should obtain two eigenvectors from each critical point, and using the rules from class you should obtain the types of critical points.  - Mick

Question from Craig

Regarding #11 on p.151

When setting up my equation, I am getting the matrix {[0,1],[-2,-2]}. Working through the problem, this gives me the complex e-values of -1+i and -1-i, which corresponds to the stable spiral answer in the back of the book. However, when I go to calculate the e-vectors, I am not getting an empty bottom row for either e-value. Herein lies my confusion. This non-zero bottom row is causing for two bound variables, and a complex e-vector [-1-i,2].

Have I missed a step, or set the problem up incorrectly? Any insight would be appreciated.

Remark from Steve Bell:

Hmmm. For the eigenvalue lambda=-1+i, the system (A-lambda I)a=0 for the complex eigenvector a is

[ 1-i   1   | 0 ]
[  -2  -1-i | 0 ]


and that second row is -(1+i) times the first row. I think you do get a row of zeroes. (By the way, if you were to get TWO bound variables, the only solution to the system would be the zero solution, and that would give you a clue that you botched up finding the eigenvalues.)

--Czehrung 17:42, 23 September 2013

Response from Mickey Rhoades Mrhoade

The bottom row of eigenvectors corresponds to the deriative of the top row.  Remember that you initially set y1 = y and y2 = y'.  The y1 solution is the solution of interest.  For a second order ODE you need two initial conditions to solve at any point, so if you graph (y1,y2) what you are looking at is where the solution y will go based on the initial position and initial derivative. - Mick

Question from Ryan Leemhuis

On question 4, I get eigenvalues of 0 and 1 for the point at (0,0). When I check the point type based on the conditions in table 4.1 it technically doesn't appear to meet any conditions as q = 0. Should the node point type include 0 for q or did I do something incorrect? I did problem 5 just to have a problem to "practice" on and that seemed to go alright using the same strategy.

Answer from Eun Young :

$A = \begin{bmatrix} 4 \ \ 0 \\ 0 \ \ 1 \end{bmatrix}$ at (0,0) and so eigenvalues of A are 4 and 1. Hence, p = 5, q= 4, and $\Delta$ = 9. This shows that the critical point at (0,0) is an unstable node.

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In the review lecture yesterday, Dr. Bell only used one eigenvector in problem 4 to find a real general solution. I think I have got something mixed up, because I thought that you could only find the general solution from the rows of the eigenvectors? In the sample exam, Dr. Bell used a single eigenvector column to find the solution. I am missing the connection. Any help is appreciated.

Remark from Steve Bell:

I think you are talking about the spot where I got TWO REAL SOLUTIONS from just one complex solution. Once I get those two linearly independent real solutions, I forget I ever saw that complex solution and I'm able to write down a real valued general solution. I did another example of such a problem on Wednesday during the review.

## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy Francisco Blanco-Silva