1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bifurcation point of x' = r + x/2 - x/(x+1)

  1. Jan 25, 2013 #1
    1. The problem statement, all variables and given/known data

    As in title.

    2. Relevant equations

    My book has a very shaky definition of what a bifurcation point. Basically, I need to play around with r and see how the system changes.

    3. The attempt at a solution

    x' = 0 when x = 1/2 - r ± √((r-1/2)2 - 2r)

    d/dx (x') = 1/2 - 1/(x+1)2, so
    d/dx (x') > 0 when -1 - √2 < x < -1 + √2, and d/dx (x') < 0 when x < - 1 - √2 or x > -1 + √2.

    I'm trying to combine these to find the ranges of r that I need to look at. Any ideas?
     
  2. jcsd
  3. Jan 25, 2013 #2
    You have found that the equation has stationary solutions that depend on the parameter r. How many solutions are there at a given parameter value? Are there any special values?
     
  4. Jan 25, 2013 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since we are talking about real values of x, you will have a problem where [itex](r- 1/2)^2- 2r< 0[/itex]. For what values of x is that negative? positive? 0?

    How many solutions will you have in each case?
     
  5. Jan 25, 2013 #4
    0 (real) solutions when it's negative, 1 solution when it's 0, 2 solutions when it's positive.
    Since (r - 1/2)2 - 2r = r2 - 3r + 1/4 = (r -3/2 + √2)(r - 3/2 - √2), we have that (r - 1/2)2 - 2r > 0 when r > 3/2 + √2 or r <3/2 - √2; r = 0 when x = 3/2 - √2 or x = 3/2 + √2; r < 0 when 3/2 - √2 < x < 3/2 + √2.
     
  6. Jan 25, 2013 #5
    Ok. I've figured out this question. NEXT!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook