Can $\alpha=1$ be a bifurcation point if $\alpha\geq 1$? 
Consider the system $$x'=\alpha x-x^2-y^2, \ \ y'=-y+xy.$$

One of the equilibrium points for this system is $$(x,y)=(1,\sqrt{\alpha-1}).$$
Now for this equilibrium point to exist, $\alpha\geq 1$ (our bifurcation diagram is of the real plane). My question is, how an we determine if $\alpha=1$ is or is not a bifurcation point? Is there a method or an intuition?
 A: Here it is important to trace all the equilibrium points. Using the factorization of the second equation, the cases are 


*

*$x=1$ results in $y^2=α-1$

*$y=0$ results in $x(x-α)=0$
For $α<1$ we get two stationary points $(0,0)$ and $(α,0)$. For $α>1$, we get additionally the mentioned ones $(1,\pm\sqrt{α-1})$. The parabolic arc of the latter ones intersects with the line of the second point to form the classical fork pattern.
Let's look at the Jacobians
$$
J(x,y)=\pmatrix{α-2x&-2y\\y&x-1}\implies J(0,0)=\pmatrix{α&0\\0&-1},~~
J(α,0)=\pmatrix{-α&0\\0&α-1}
$$
so that indeed the second point changes its stability characteristic at $α=1$ from sink (without rotation) to saddle point, making it a fork-bifurcation.

For $α>1$ the stable manifold tangent to $x=α$ has the next order terms according to
$$
x=α+c_2y^2+...
\implies
2c_2y=\frac{dx}{dy}=\frac{αx-x^2-y^2}{y(x-1)}=\frac{-(α+c_2y^2)c_2y^2-y^2+...}{y(α-1+c_2y^2+...)}
\\
\implies 2c_2=\frac{-αc_2-1}{α-1},~~ 2(α-1)c_2=-αc_2-1, ~~ c_2=-\frac1{3α-2}
$$
This approximation is depicted as blue curve in the plot.
A: Here is a movie of varying $\alpha$, so you can visually see the bifurcation happening.
Click on it for a larger view to go along with the other answer.

