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Math Help - Bifurications

  1. #1
    Len
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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?
    Last edited by Len; February 10th 2010 at 11:51 AM.
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    y'=y(y^2-a+1) satisfies your criteria but not the form g(y) +a. I'm not sure one can exist.
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  3. #3
    Len
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    Quote Originally Posted by Thomas154321 View Post
    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
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    Quote Originally Posted by Len View Post
    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.

    Thanks in advance.

    Other than guessing is there a way for me to determine possible functions?
    Did you really want a \le -1 and a \ge 1?
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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).
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  6. #6
    Len
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    Quote Originally Posted by Danny View Post
    Did you really want a \le -1 and a \ge 1?
    Fixed in original post, thanks. Still need help with the problem.
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    Quote Originally Posted by Len View Post
    Fixed in original post, thanks. Still need help with the problem.
    Sorry but I don't see a difference.
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  8. #8
    Len
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    Quote Originally Posted by Danny View Post
    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:

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