Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We are asked to analyze the one-dimensional system $\frac{dx}{dt}=f(x)=x-rx(1-x)=rx^2+(1-r)x$.

The system has a fixed point for all values of $r$ at $x^{\star}=0$, and the algebraic form shows the system undergoes a transcritical bifurcation at $r_{c}=1$ (alternatively, we could apply the tangency condition for standard bifurcations (saddle-node, transcritical, and pitchfork) to get the $r_{c}$ value).

A quick graphical analysis (or, perhaps a linear stability analysis) of the different cases let's us quickly sketch the associated dynamics near $x_{c}=0$ as $r$ varies (e.g. bifurcation diagram, vector flows, etc.). It is easy to see after doing this that the system indeed undergoes a transcritical bifurcation at $(x_{c},r_{c})=(0,1)$ (stabilities interchange). That's fine, I understand this well enough.

My question is this: there is clearly another bifurcation at $(x_{c},r_{c})=(0,0)$. For $r<0$ there is an unstable fixed point at the origin, and a positive stable fixed point which tends to $x=1$ as $r\to-\infty$ and to $+\infty$ as $r\to0$. Once $r=0$, there is only an unstable fixed point at the $x=0$. Once $r>0$, the stable fixed point that was at $+\infty$ becomes a stable point at $-\infty$ and tends to $0$ as $r\to1$, and this is of course the transcritical bifurcation.

So what is going on at $(x_{c},r_{c})=(0,0)$? The text we're using seems to completely ignore this bifurcation point, and simply plots it and makes no further comment about it. I looked at solutions to this very problem from other sources, and they also make no detailed analysis of this point. (See here at the very last problem, for example: http://www.personal.psu.edu/axm62/math449%20sheets/math449HW3soln.pdf).

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

Solutions with positive initial conditions are not always finite:

  • A solution with $x(0) > 0$ and $ r < 0$ has $\displaystyle \lim_{t \to \infty}x(t) = \frac{1 }{r }-1$ for all $t > 0$.
  • A solution with $x(0) > 0$ and $ r > 0$ has $\displaystyle \lim_{t \to \infty}x(t) = \infty$.

Solutions with negative initial conditions behave similarly:

  • A solution with $x(0) < 0$ and $ r < 0$ has $\displaystyle \lim_{t \to -\infty}x(t) = -\infty$.
  • A solution with $x(0) < 0$ and $ r > 0$ has $\displaystyle \lim_{t \to -\infty}x(t) = \frac{1 }{r }-1$ for all $t < 0$.

The equilibrium points for your system \[ \dot{x} = f(x,r) =rx^2 + (1-r)x \] are $x = 0, \frac{1 }{r }-1$. The stability depends on the eigenvalues of sign of
\[ \frac{df}{dx} = 2rx + (1-r)\]

The signs are $\frac{df}{dx}(0) = 1-r$ and $\frac{df}{dx}\left( \frac{1 }{r }-1 \right) = 3(1-r) $ both of which jumps at $r = 1$.

At $r = 0$, no equilibrium points are created or destroyed and their stability does not change. Instead the point $\frac{1 }{r }-1$ "whips around infinity" from the positive to the negative side.


We can solve this system explicitly \[ \int (1-r) dt = \int \frac{dx}{x(rx + (1-r))} =\int \frac{dx}{x} - \int \frac{r\, dx}{rx + (1-r)} \] If we integrate both sides can solve for $x$ to get: \[ (1-r)t + C= \ln |x| - \ln |rx + (1-r)| = -\ln \left|r + \frac{1-r}{x} \right|\]

share|improve this answer
    
So would $r=0$ not be considered a bifurcation point? I've heard both definitions used; one is where it's only when new fixed points are created, destroyed, or stabilities changed; whereas another definition was where any qualitative behavior change occurs (in the case $r=0$, only the ladder applies as you showed). And if it is considered a bifurcation, does it have a formal name like some of the others? –  Taylor Martin Jul 9 '12 at 10:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.