1 Dimensional System Bifurcation - Classification Question 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).
 A: 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:
\[ (1-r)t + C= \ln |x|  - \ln |rx + (1-r)|  = -\ln \left|r + \frac{1-r}{x} \right|\] 
