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Critical points

  1. Sep 22, 2008 #1
    x' = y(1 + x - y2)
    y' = x(1 + y - x2).

    C.P. => (x,y) s.t. f(x) = 0

    [tex]f(x) = x' = 0 => y = 0[/tex]

    [tex]f(x) = y' = 0 => x = 0[/tex]

    => (0,0) is c.p.

    plug x = 0 into x':

    [tex] x' = y(1 - y^{2}) => y = 0, y =\ ^{+}_{-}1 [/tex]

    => (0,1), (0,-1) are critical points.

    plug y = 0 into y':

    [tex] y' = x(1 - x^{2}) => x = 0, x =\ ^{+}_{-}1 [/tex]

    => (1,0), (-1,0) are critical points.

    Can someone help me find the 2 critical points that I am missing?

    Thanks.
     
  2. jcsd
  3. Sep 22, 2008 #2
    You have

    x' = y(1 + x - y2)
    y' = x(1 + y - x2).

    Think of these as

    x' = a*b
    y' = c*d

    Therefore for x' = 0 you can have a = 0 OR b = 0 and for y' = 0 you can have c = 0 OR d = 0 i.e. your possible pairs are

    1) a = 0, c = 0
    2) a = 0, d = 0
    3) b = 0, c = 0
    4) b = 0, d = 0

    this translates to

    1) y = 0, x = 0
    2) y = 0, 1 + y - x^2 = 0 i.e. 1 - x^2 = 0, i.e. x = +/-1
    3) 1 + x - y^2 = 0, x = 0 i.e. y = +/-1

    Now the ones you are missing i.e.

    4) 1 + x - y^2 = 0 and 1 + y - x^2 = 0

    Using the first equation you get x = y^2 - 1 so plug that into the 2nd equation and you get 1 + y - (y^2 - 1)^2 = 0 which you can solve.

    Does that help?
     
  4. Sep 22, 2008 #3
    Yeah, it really does. Thanks a lot. For some reason I had trouble with this same thing back in intermediate DE too. Thanks for giving me a better perspective.
     
  5. Sep 22, 2008 #4
    What if I have

    x' = 16x2 + 9y2 - 25
    y' = 16x2 + 162

    Would i find x(y) when x' = 0 and plug into y'?
     
  6. Sep 22, 2008 #5
    I assume your 2nd equation is y' = 16x^2 + 16y^2?

    You would set both equal to 0 like normally

    x' = 0 gives you 16x^2 = 25 - 9y^2

    y' = 0 gives you 16x^2 + 16y^2 = 0, subst. that in and you get

    25 - 9y^2 + 16y^2 = 0 i.e. 25 + 7y^2 = 0 which doesn't seem to have any solutions over R but are you allowed to work with C?

    Remember you are essentially finding where these 2 functions(f(x,y) = x' and g(x,y) = y') so you can use whatever method you want, subst., putting them in a matrix if they are linear, etc.
     
  7. Sep 23, 2008 #6
    The back of the book claims (1,1), (-1,-1), (-1,1), (1,-1) are the critical points.

    Aaaaaaaaaaaand, typo :*(

    y' = 16x2 - 16y2.

    Sorry about that; I was really tired and frustrated when I posted earlier.
     
  8. Sep 23, 2008 #7
    Ok so now you have

    x' = 16x^22 + 9y^2 - 25
    y' = 16x^2 - 16y^2,

    As I said, set both equal to 0, the first one gives you 16x^2 = 25 - 9y^2, plug that into the 2nd one and you get 25 - 9y^2 - 16y^2 = 0, can you solve that?
     
  9. Sep 23, 2008 #8
    Yeah; I got it. Thanks for your help.
     
    Last edited: Sep 23, 2008
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