I'm trying to find the critical points & bifurcation point to the ODE: $$ \dot{x} = f(x;\beta)=2-\beta e^{-x^2}$$
By letting $f(x;\beta) = 0$, I found the critical points are at $x=\pm\sqrt{-ln\frac{2}{\beta}}$ (Also, I'm unsure if $x=0$ is a critical point.
Since the bifurcation point will be found when $f'(x;\beta)$ is also $0$, I'm trying to substitute x into $f'(x;\beta)$ to find $\beta$, yet I always end up with 4 = 0 which is absurd.
Can anyone help me with this? Thank you!
First of all: I don't see an ODE. Note that for $\beta < 2$ you have no (real) critical points, if $\beta = 2$ you have one and for $\beta >2$ you have two critical points. Thus, your bifurcation point might be $(x_0,\beta_0) = (0,2)$.
Now check that $\frac{\partial}{\partial x}f(0,2)=0$, $\frac{\partial}{\partial \beta}f(0,2) \neq 0$ and $\frac{\partial^2}{\partial x^2}f(0,2) \neq 0$.
Having this checked, you know that your vector field undergoes a saddle node bifurcation and the bifurcation point is $(0,2)$.
EDIT: Your critical set is a bit strange. $x=0$ can't be a critical point, you have to look for pairs $(x_0, \beta_0)$. As said before, $x=0$ is no critical point, but $(0,2)$ is. So your critical set is given by $\{(x,\beta) \in \mathbb R^2~|~ x= \pm\sqrt{-\ln\frac{2}{\beta}} \}$.