Bifurcation Points and Stability Analysis for x' = x(r - ex)

[Jamin2112]

Homework Statement

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

Homework 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."

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' = -xe^x, 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' = -xe^x = 0 iff x = 0, since r - e^x = 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.

 

[pasmith]

Looks to be correct apart from this:

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 x &lt; 0, then e^x &lt; 1 so x&#039; = x(1 - e^x) &lt; 0. Thus the origin is stable from above (x > 0), but unstable from below (x < 0).

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.

Yes.

Has your textbook mentioned that a fixed point of x&#039; = f(x) is stable if
\frac{\mathrm{d}f}{\mathrm{d}x} &lt; 0
and unstable if
\frac{\mathrm{d}f}{\mathrm{d}x} &gt; 0?

 

[Jamin2112]

Has your textbook mentioned that a fixed point of x&#039; = f(x) is stable if
\frac{\mathrm{d}f}{\mathrm{d}x} &lt; 0
and unstable if
\frac{\mathrm{d}f}{\mathrm{d}x} &gt; 0?

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