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

[Jamin2112]

Homework Statement

As in title.

Homework 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.

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?

 

[voko]

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?

 

[HallsofIvy]

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

How many solutions will you have in each case?

 

[Jamin2112]

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

How many solutions will you have in each case?

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.

 

[Jamin2112]

Ok. I've figured out this question. NEXT!