Can $\alpha=1$ be a bifurcation point if $\alpha\geq 1$?

Question

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?

Answer 1

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.

Answer 2

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.