# Differential Equation - Plotting Phase Plane

1. May 26, 2009

### cse63146

1. The problem statement, all variables and given/known data

Plot the phase plane of: y(t) = e-6t(2 1) + e-3t(1 -1)

2. Relevant equations

3. The attempt at a solution

I know it's a sink because $$\lambda_2 < \lambda_1 < 0$$.

http://img35.imageshack.us/img35/8108/sink.th.jpg [Broken]

but the phase plane in the answer is slanted to the right. Does anyone know why?

Last edited by a moderator: May 4, 2017
2. May 26, 2009

### Staff: Mentor

I'm going by memory from a lot of years ago, so I might be wrong. I seem to recall that the axes in the phase plane were the eigenvectors, which appear to be [2 1] and [1 -1] in your problem statement.

3. May 26, 2009

### cse63146

So here's how the answer in the textbook appears (roughly):

http://img16.imageshack.us/img16/5559/sinki.th.jpg [Broken]

so the top left corner of the axis corresponds to the (2,1) eigenvector and the bottom right corner corresponds to the (1, -1) eigenvector?

Last edited by a moderator: May 4, 2017
4. May 26, 2009

### Staff: Mentor

No, the axis running from the lower left to upper right corresponds to [2 1], and the other axis corresponds to [1 -1]. If I'm anywhere close to being right on this, the axis with the negative slope has a slope of -1 (like in your drawing), but the one with positive slope has a slope of 1/2 (which is less than what you show).

5. May 26, 2009

### cse63146

I understand now. Thanks for your help.

6. May 26, 2009

### cse63146

Sorry, got another question. If the eigenvalues were 0 and 5, would it be a source?

7. May 26, 2009

### Staff: Mentor

I think so, but again, I studied this a long time ago, and don't have the text I used in front of me (Dynamical Systems, Steven Smale). As I recall, besides sinks and sources, you could also have a situation where the phase diagram showed circular or elliptic orbits. I don't recall offhand what the conditions were on the eigenvalues.

8. May 26, 2009

### cse63146

that's a center and it's when you have a root bi.

9. May 27, 2009

### cse63146

Sorry to bother you again.

for y'' + 4y' + 13y = 0

The general solution would be y(t) = k1e-2tcost3t + k2e-2tsin3t + (27/145)cos2t + (24/145)sin2t.

It's a spiral sink, so would the phase plane (with inital conditions y(0) = 1, and v(0) = -2)look like this:

http://img13.imageshack.us/img13/8230/spiralsink.th.jpg [Broken]

not sure what to do since the eigenvectors have complex numbers.

and would the y(t) v(t) graphis look like this:

http://img13.imageshack.us/img13/4392/yvgraph.th.jpg [Broken]

Last edited by a moderator: May 4, 2017
10. May 27, 2009

### Staff: Mentor

I won't be able to respond to your question about the phase plane question until I can review the text I mentioned.

For the DE above, the solution would be just y(t) = k1e-2tcos3t + k2e-2tsin3t. Are you showing me just the homogeneous equation? Is the nonhomogeneous equation something like this -- y'' + 4y' + 13y = cos2t ?

11. May 27, 2009

### cse63146

Oops. You're right. it was just the homogenous part. I was looking at a different question (which the nonhomogenous part).

12. May 27, 2009

### Staff: Mentor

Since you're given y(0) and v(0), you should be able to solve for k1 and k2 in the two exponential terms.

I'll take a look at my text tonight and brush up on the phase plane business. It's been 30 years since I looked at it last, so I'm pretty rusty on these ideas.

13. May 28, 2009

### Staff: Mentor

According to my source, Differential Equations, Dynamical Systems, and Linear Algebra - Hirsch and Smale -- when you have two complex eigenvalues with negative real parts (your situation), you will have clockwise orbits spiraling into the origin.

Your phase plane (the first figure in post 9) should, I believe, show one trajectory starting at 1 on the y axis, and spiraling clockwise into the origin. Another trajectory should start at -2 on the v axis, I believe, and do the same. So your phase plane has more-or-less the right appearance, but is a little off in some details.

Your graph of y(t) and v(t) is probably OK, as both functions tend to zero as t increases, but y(t) starts at 1 (when t is 0) and v(t) starts at -2 (when t is 0).

Mark

Last edited by a moderator: May 4, 2017
14. May 28, 2009

### cse63146

So there should be 2 spirals, one starting at y(0) = 1, and another spiral starting at v(0) = -2, and both going towards the origin?

15. May 28, 2009

### Staff: Mentor

Right.

16. May 28, 2009

### cse63146

would the phase portrait look the same if (for example) the original equation was y'' + 4y' + 13y = 3cost2t ?

17. May 28, 2009

### Staff: Mentor

Let me get back to you. There is a fair amount of mental translation that I need to do, to convert your differential equations (the homogeneous one and the related nonhomogeneous one) into systems of first-order DEs, which is how things are presented in the text I cited.

Your 2nd order homogeneous DE is equivalent to a system of two 1st order DEs, which can be solved by matrix techniques. Your 2nd order nonhomogeneous DE is equivalent to a 4th order homogeneous DE, which is equivalent to a system of four 1st order DEs, which also can be solved by the use of matrices.

Your phase planes have axes marked as y and v (= dy/dx), while the ones I'm looking have axes of x1 and x2.

Going from a 2nd order DE to two 1st order DEs involves introducing a new variable v = y'. Going from a 2nd order nonhomogeneous DE to a 4th order homogeneous DE uses the concept of "annihilators" that I mentioned a while back in another thread.

Lurking in the background of all this are a lot of concepts from Linear Algebra, an area of mathematics that is inextricably intertwined with differential equations.

18. May 28, 2009

### cse63146

If it helps, the general solution would be y(t) = k1e-2tcost3t + k2e-2tsin3t + (27/145)cos2t + (24/145)sin2t.

19. May 28, 2009

### Staff: Mentor

Yes, I remember. You showed this several posts back.

20. May 29, 2009

### Staff: Mentor

This is the question I said I'd come back to. The short answer is I don't think so. This nonhomogeneous 2nd order equation is equivalent to the homogeneous 4th order equation
(D2 + 4D + 13)(D2 + 4)y = 0

The characteristic equation is (r2 + 4r + 13)(r2 + 4) = 0, and the roots of this equation are r = -2 +/-3i, +/-2i. These are also the eigenvalues, and the corresponding eigenfunctions are e(-2 +/-3i)t and e(+/-2i)t. With the usual sleight of hand of taking suitable linear combinations to eliminate the imaginary parts, the eigenfunctions can be written as e-2tsin3t, e-2tcos3t, sin2t, and cos 2t.

Unlike the homogeneous 2nd order problem (which is equivalent to a system of two first-order DEs, and for which the solution space is two-dimensional), for the nonhomogeneous 2nd order system above, the solution space is four-dimensional. The eigenfunctions y1,2 = e(-2 +/-3i)t of the homogeneous DE span a two-dimensional space, and you can plot a graph of y1 vs. y2, which is nearly what you did in your phase portrait of y vs v (= y').

With the nonhomogeneous 2nd order problem (or the related homogeneous 4th problem), the eigenfunctions span a four-dimensional space, so right away there are problems in visualizing it. The highest dimension phase portrait shown in the text I've been looking at is three.

So anyway, based on a relatively quick perusal of my book, I don't believe that you can get a phase portrait for the nonhomogeneous equation you gave. It's possible that the authors are doing something to look at a projection of a higher dimension portrait down to a couple of dimensions, but I didn't catch that in my quick perusal.