Bifurcation diagram and bifurcation value

Question

Determine the bifurcation values of $\dot{x} = x(x-r^2)$, and sketch the bifurcation diagram.

My attempt: First, we see that if $f(x_0, r_0) = Df(x_0, r_0) = 0$, then $x_0$ is a non-hyperbolic critical point and $r_0$ is a bifurcation value. We see that this only occurs when $(x_0, r_0) = (0,0)$, so this is the only bifurcation value. Now, for each $r_0\neq 0$, the solution would increase without bound for $x_0 > \sqrt{r_0}$ and $x_0 < 0$, and decrease for $0 < x_0< \sqrt{r_0}$. I don't know how to demonstrate these information on the bifurcation diagram though:( Can someone please help?

Answer 1

For the original problem, we have two critical points $x = 0 , r^2$. The following phase portraits show four different values of $r$, but note that they are identical for $\pm r$.

$r = -10$

$r = -3$

$r = -1$

$r = 0$

Answer 2

A bifurcation is usually a point where a branch of stable stationary points switches to unstable in its smooth continuation. Usually there will emerge two new stable branches of stationary points.

The only rewriting of the initial equation satisfying these criteria is $$ \dot x = x(r-x^2). $$ For $r<0$, $x_0=0$ is a stable stationary point, for $r>0$ it becomes unstable and $x_0=\pm\sqrt{r}$ are the new stationary points.

The linearization at $x_0=\sqrt{r}$ is $$ \dot u=\frac d{dt}(\sqrt r+u)=(\sqrt r+u)(-u)(2\sqrt r+u)=-2ru-3\sqrt r u^2-u^3\\=-2r·u+O(u^2) $$ so that these points remain stable for all $r>0$.