# Critical Points and Linearization

• Sep 26th 2009, 05:34 PM
dsprice
Critical Points and Linearization
Problem:

Determine the location and type of all critical points by linearization.

y1' = -y1 + y2 - (y2)**2
y2' = -y1 - y2

From the above, I set both equal to zero and determined the critical points to be (0,0) and (-2, 2).

This is where I get a little bit lost as to what to do next. It seems to me that I need to drop non-linear terms and find the eigenvalues.

y1' = -y1 + y2
y2' = -y1 - y2

In matrix form:

| -1-lambda 1 |
| -1 -1-lambda |

I compute lambda to be (-1 +/- i)

This is where I get completely lost and could use some help. Thanks in advance.
• Sep 27th 2009, 04:18 AM
shawsend
You have to linearize the system at each equilibrium point so it's not just dropping the non-linear terms. The linearized system is the Jacobian at each equilibrium point. You can calculate the Jacobian right? It's just the partials of the right side:

$\displaystyle \left(\begin{array}{cc}\displaystyle\frac{\partial f}{\partial x}(x_0,y_0) & \displaystyle\frac{\partial f}{\partial y}(x_0,y_0) \\ \displaystyle\frac{\partial g}{\partial x}(x_0,y_0) & \displaystyle\frac{\partial f}{\partial y}(x_0,y_0)\end{array}\right)$

At the point (0,0), the eigenvalues are $\displaystyle (-1\pm i)$. That's a spirial sink according to the standard naming convention.

Find "Differential Equations" by Blanchard, Devaney, and Hall. It's the best intro to the subject.

Now, use Mathematica to plot the phase portrait:

Code:

StreamPlot[{-x + y - y^2, -x - y}, {x, -2, 2}, {y, -5, 5}]
The origin looks like a spirial sink to me.
• Sep 29th 2009, 05:13 PM
dsprice
Re: Critical Points and Linearization
Got it! Thanks! Took me a while to learn more about how and when to determine the Jacobian, but now it looks fairly obvious to me what the process steps are in this type of problem. Wish my text book was as clear! (Happy)