# Bifurications

• February 10th 2010, 10:34 AM
Len
Bifurications
Is it possible to find a function g(y) (either continuous or discontinuous) such that the one-parameter family of differential equations

dy/dt=g(y)+a

satisfies both the following statements?

for all a<= -1, the differential equation has exactly one equilibrium point which is a sink.

for all a>= 1, the differential equation has exactly three equalibria , two sources and one sink.

If so give a rough sketch, if not, why not?

I've sketched a ton of graphs now and can't get one to satisfy, also I have no idea how to explain then if it don't exist, why that is.

Other than guessing is there a way for me to determine possible functions?
• February 10th 2010, 10:59 AM
Thomas154321
$y'=y(y^2-a+1)$ satisfies your criteria but not the form g(y) +a. I'm not sure one can exist.
• February 10th 2010, 11:06 AM
Len
Quote:

Originally Posted by Thomas154321
$y'=y(y^2-a+1)$ satisfies your criteria but not the form g(y) +a. I'm not sure one can exist.

Thanks for the reply. I need one of the form g(y)+a tho or I need to prove it doesn't exist, not really sure how to prove that it doesn't exist which I'm now leaning towards. :P
• February 10th 2010, 11:14 AM
Jester
Quote:

Originally Posted by Len
Is it possible to find a function g(y) (either continuous or discontinuous) such that the one-parameter family of differential equations

dy/dt=g(y)+a

satisfies both the following statements?

for all a<= -1, the differential equation has exactly one equilibrium point which is a sink.

for all >= 1, the differential equation has exactly three equalibria , two sources and one sink.

If so give a rough sketch, if not, why not?

I've sketched a ton of graphs now and can't get one to satisfy, also I have no idea how to explain then if it don't exist, why that is.

Other than guessing is there a way for me to determine possible functions?

Did you really want $a \le -1$ and $a \ge 1$?
• February 10th 2010, 11:42 AM
Thomas154321
Oh I didn't even notice that. My answer was for both equalities -1. Basically I've only ever seen a single eq. splitting to 3 by having a factor of y through the whole thing as then you get y=0 for all a. If you solve the eqn g(y)=-a you need a function that has one solution for a<-1 and 3 for a>1. this isn't possible if you want g to be continuously differentiable (or even continuous only I think).
• February 10th 2010, 11:51 AM
Len
Quote:

Originally Posted by Danny
Did you really want $a \le -1$ and $a \ge 1$?

Fixed in original post, thanks. Still need help with the problem.
• February 10th 2010, 12:21 PM
Jester
Quote:

Originally Posted by Len
Fixed in original post, thanks. Still need help with the problem.

Sorry but I don't see a difference.
• February 10th 2010, 12:29 PM
Len
Quote:

Originally Posted by Danny
Sorry but I don't see a difference.

I don't understand what you mean then? I thought you were referring to how I omitted the a in "a" >=1 in my original post.

I have a test Friday and this was one of the emphasized review questions and I have no idea how to solve it. I scan of the question:

http://img713.yfrog.com/img713/48/17question14.jpg