# Autonomous non-linear systems - predator prey equation

• Apr 8th 2010, 01:15 PM
redwings6
Autonomous non-linear systems - predator prey equation
I need to state the type of each point of equilibrium of the for the following system of ODE:
y'1= y2*(y2-1)
y'2= -y1*(y1-1)

I also need to draw the phase portrait but I understand how to do that once I can figure out how to solve this. I am unsure of how to do this so any help would be appreciated
• Apr 8th 2010, 01:55 PM
Can you just re-check your equation...

Nearly all pred-prey models I've used would be

$y'_1 = y_1(y_2-1)$
$y'_2 = y_2(y_1-1)$
• Apr 8th 2010, 02:19 PM
redwings6
Yeah I know, all the ones I have solved are of that form too but this one is different. That is why I am confused
• Apr 8th 2010, 02:34 PM
Quote:

Originally Posted by redwings6
Yeah I know, all the ones I have solved are of that form too but this one is different. That is why I am confused

I would imagine just solve in the same way.

set $y'_1$ and $y'_2 = 0$ to get your 4 critical points.

Linearize the system (i.e. find the jacobian of the matrix of the system), sub in the critical points, then find eigenvalues.
• Apr 8th 2010, 02:59 PM
So....

Do you know what to do? Or would you like me to post a solution...
• Apr 8th 2010, 03:07 PM
redwings6
I understand setting it to zero to get the 4 critical points however I am unsure of how to proceed from there
• Apr 8th 2010, 03:26 PM
Quote:

Originally Posted by redwings6
I understand setting it to zero to get the 4 critical points however I am unsure of how to proceed from there

Lol all those 'other example' tire you out..?

You'll get 4 critical points.

$(0,0)$, $(0,1)$, $(1,1)$, $(1,0)$.

The Jacobian is the matrix...

$y'_1 = f(y_1, y_2)$

$y'_2 = g(y_1, y_2)$

$
\left( \begin{array}{cc}
\frac{\partial f}{\partial y_1} & \frac{\partial f}{\partial y_2} \\
\frac{\partial g}{\partial y_1} & \frac{\partial g}{\partial y_2}
\end{array} \right)$

Ah hell that keeps coming out tiny, top line is partial derivatives of f with respect to $y_1$ then $y_2$.
bottom line is partial derivatives of g with respect to $y_1$ then $y_2$.

$
\left( \begin{array}{cc}
0 & 2y_2 - 1 \\
-2y_1 + 1 & 0
\end{array} \right)$

Then sub in each critical point one at a time.

$