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Bifurcation question

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Just making sure I understand this stuff. One question on my homework asks about x' = x(r - ex).

    2. Relevant equations

    Definition. From my textbook: "The qualitative structure of the flow can change as parameters are varied. In particular, fixed points can be created or destroyed, or their stability can change. These qualitative changes in the dynamics are called bifurcations, and the parameter values at which they occur are called bifurcation points."


    3. The attempt at a solution

    So I want to think about how this system changes when r takes on different values.

    If r = 0, then we have x' = -xex, which means x' = 0 iff x = 0. As x increases beyond 0, x' decreases, and as x decreases beyond 0, x' increases. So x* = 0 is the only fixed point and is stable.

    If r < 0, then, still, x' = -xex = 0 iff x = 0, since r - ex = 0 does not have a solution. As x increases beyond 0, x' decreases, and as x decreases beyond 0, x' increases. So, again, x* = 0 is the only fixed point and is stable.

    If r = 1, then then x' = 0 iff x = 0. If x < 0, then x' > 0, and if x > 0, then x' < 0. As x increases beyond 0, x' decreases, and as x decreases beyond 0, x' increases. So x* = 0 is the only fixed point and is stable.

    If 0 < r < 1 or 1 < r, then x' = 0 iff x = 0 or ln(r).

    ^ That's the part that's giving me trouble. Do I have to break it into cases of 0 < r < 1 and 1 < r????? I'm afraid I'll have to. This question is very annoying.
     
  2. jcsd
  3. Jan 23, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    Looks to be correct apart from this:

    If [itex]x < 0[/itex], then [itex]e^x < 1[/itex] so [itex]x' = x(1 - e^x) < 0[/itex]. Thus the origin is stable from above (x > 0), but unstable from below (x < 0).

    Yes.

    Has your textbook mentioned that a fixed point of [itex]x' = f(x)[/itex] is stable if
    [tex]\frac{\mathrm{d}f}{\mathrm{d}x} < 0[/tex]
    and unstable if
    [tex]\frac{\mathrm{d}f}{\mathrm{d}x} > 0?[/tex]
     
  4. Jan 23, 2013 #3

    Yeah, but I still need to draw the d*** diagrams, so I gotta know whether x' > 0 or x' < 0 in these ranges.
     
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