How many cases are there for this bifurcation?

Question

Here is the bifurcation: $$\frac{dx}{dt} = x(1-x)-h \frac{x}{1+x}$$

fixed points $$\ x = 0, -\sqrt {1-h}, + \sqrt {1-h} $$

How many cases are there?

My guess:

$$\ 0<h< 1$$ $$\ h > 1 $$ $$\ h < 0 $$

What are the cases?

Answer 1

Let $f(x,h)=x(1-x)-h\frac x{1+x}$. At $x=0$, $$ f_x(0,h)=1-h, f_{xx}(0,h)=2(h-1), f_{xxx}(0,h)=-6h.$$ If $f_x(0,h)\neq0$ or $h\neq1$, near $x=0$, the equation can be written as $$ x'=(1-h)x+O(x^2) $$ which is equivalent to the linear equation $$ x'=(1-h)x. $$ Thus if $h>1$, $0$ is stable and if $h<1$, $0$ is unstable.

If $f_x(0,h)=0$ or $h=1$, then $f_{xx}(0,1)=0,f_{xxx}(0,1)=-6\neq 0$. Near $x=0$, the equation can be written as $$ x'=-6x^3+O(x^4) $$ which is equivalent to $$x'=-6x^3.$$ Thus $0$ is stable.

Now assume $0<h<1$. At $x=\sqrt{1-h}$, $$ f_x(\sqrt{1-h},h)=\frac{2(h-1)}{1+\sqrt{1-h}}<0.$$ Thus $x=\sqrt{1-h}$ is stable. Now assume $0<h<1$. At $x=-\sqrt{1-h}$, $$ f_x(-\sqrt{1-h},h)=\frac{2(h-1)}{1-\sqrt{1-h}}<0.$$ Thus $x=-\sqrt{1-h}$ is stable.

For $x=-1$, let $u=x+1$ and then the equation becomes $$ u'=\frac{(1-u)^3}{u}=\frac{1-3u}{u}+O(u).$$ Near $u=0$, this equation is equivalent to $$u'=\frac{(1-u)^3}{u}=\frac{1-3u}{u}.$$ Thus $u=0$ is unstable or $x=-1$ is unstable.