# Need help drawing phase portraits for coupled systems of ODEs

1. Aug 7, 2008

### Sdarcy

Okay, I know that this is probably a simple question but I've always been good at doing the complicated things and bad at doing the easy things :D

Here's what I've got:
Find the general solution for the system of coupled ODEs. Determine kind and stability of the critical point. Sketch phase portrait.

Y'1 = -3Y1 - 2Y2
Y'2 = -2Y1 - 3Y2

I constructed a system of matrices and determined the characteristics of the systems as follows:
(sorry I gave up on trying to construct the matrix with TeX, hope this still makes sense)
p = -6
q = 5
$$\Delta$$ = 16

$$\lambda$$1 = -1
$$\lambda$$2 = -5

I then applied the two values for $$\lambda$$ to the original matrix and found two vector solutions for the system as follows:
x$$^{1}$$ = (1 -1)$$^{T}$$
x$$^{2}$$ = (1 1)$$^{T}$$

So the general solutions are:
x$$^{1}$$e$$^{-t}$$
x$$^{2}$$e$$^{-5t}$$

General Solution to the system:
Y(t) = c1x1e^-t + c2x2e^-5t

(sorry for the messy equation, TeX decided it didn't like me :D)

I also determined the characteristics of the node as being a stable, attractive improper node.

Now I just have to sketch the phase plane and I have NO idea where to start with this. Its probably stupidly easy and I'm just being daft, but I don't know what I'm supposed to do next. Any help would be fantastic. Thanks.

Last edited: Aug 7, 2008
2. Aug 8, 2008

### Defennder

You have the general solution for the DE in terms of the eigenvectors, right? Start by sketching the eigenvectors at the origin of the phase plane. Then look at the dominant exponential function of t, the sign next to the t tells you the general direction of the flow lines. Also note that since you have two exp functions of t, and one decreases a lot faster than the other, that should tell you how to draw the vector flow lines at points far away from the origin.

3. Aug 8, 2008

### HallsofIvy

Staff Emeritus
Actually you've done much more than was required just to find the phase plane diagram. Finding the solution itself isn't really necessary to drawing the phase plane diagram. You see, I presume, that the only equilibrium point is (0,0). Since both eigenvalues are negative, both solutions are exponentials with negative exponents and so all solutions tend toward 0: (0,0) is a stable equilibrium and the phase plane diagram consists of lines heading toward (0,0).

If you want to be a bit more accurate than that, since (1, -1) is an eigenvector corresponding to eigenvalue -1 and (1, 1) is an eigenvector corresponding to eigenvalue -5, you know that the lines in those directions, y= -x and y= x respectively, are lines in the phase plane: further that the "flow" along y= x is 5 times as great as along y= -x so the other "lines" in the phase plane diagram will tend toward y= x as they converge on (0,0).