satisfies your criteria but not the form g(y) +a. I'm not sure one can exist.

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- February 10th 2010, 10:34 AM #1

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## 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.

Thanks in advance.

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

- February 10th 2010, 10:59 AM #2

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- February 10th 2010, 11:06 AM #3

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- February 10th 2010, 11:14 AM #4

- February 10th 2010, 11:42 AM #5

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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 #6

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- February 10th 2010, 12:21 PM #7

- February 10th 2010, 12:29 PM #8

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