How do the steady states change as we change the parameter?

Question

Consider the following ODE $$ \frac{dx}{dt} = x \left(1-\frac{x}{m} - \frac{a}{1+x}\right), $$ where $a$ is a bifurcation parameter and $a\in(0,\infty)$ a positive constant. How do you find the steady states as $a$ varies? And what are the bifurcation points? I know that $\frac{dx}{dt} = 0$, however how do you move on from there?

Also what is conufusing is the $m$. It is a fixed parameter. However, how do you find the steady states if you have one parameter that varies and the other one fixed?

EDIT: This is what I got after the reading the answer below. The steady points are $x=0$ and $$ x = \frac{-(m-1)\pm \sqrt{(m-1)^2+4m(1-a)}}{2} $$ where $m>0$. Now my question is how do you determine the stability of such a complicated root? You get distinct roots when $0< a\leq1$. However as soon as $a>1$ I am not sure if I get a complex root or a distinct root because of the parameter $m$.

Moreover,how do you determine their stability? Lastly,what are their bifurcation points?

NOTE: This a dimensionless biological system.

Answer 1

I see that you have got the steady state solutions $x^* =0$ and I got other two fixed points as $x^* = \frac{(m-1) \pm \sqrt{(1-m)^2 + 4m(1-a)}}{2}$.

For $\dot{x} = f(x)$, we have here $f(x) = x(1 - \frac{x}{m} - \frac{a}{1+x})$

Differentiating $f$ wrt $x$, we have $f'(x) = 1 -\frac{2x}{m} - \frac{a}{(1+x)^2}$

The fixed point $x^*$ is stable if $f'(x^*) < 0$ and unstable if $f'(x^*) > 0$.

For the fixed point $x^* = 0$, we have $f'(0) = (1-a)$, so the fixed point is stable if $a>1$ and unstable if $a<1$.

For the next fixed point, doing this analysis seems difficult, as it turns to be tedious calculations I think.

I thought of it in a graphical way in which we can see where we get the bifurcation. Observe the discriminant in the equation of fixed points(here the discriminant is $(1-m)^2 + 4m(1-a)$), if the discriminant is greater than 0, then we have two fixed points and when the discriminant is equal to zero, we have a single fixed point and when the discriminant is less than zero, we have none of the fixed points except $x^* = 0$. so what is happening? - the two fixed points come closer a s the parameters are varied and collide and then annihilate. This situation is Saddle-node bifurcation.

So we plot the discriminant that is we plot $a $ vs. $m$, see below -

The blue region is the region where $(1-m)^2 + 4m(1-a)<0$ (No fixed points except $x^* = 1-a$), the grey region is the region where $(1-m)^2 + 4m(1-a) > 0$(two fixed points) and the boundary separating these two regions is the region/curve where we have $(1-m)^2 + 4m(1-a) = 0$ or $a = \frac{(m+1)^2}{2m}$ (one fixed point or the two fixed point collide).

So the bifurcation diagram( here known as "bifurcation set" as we are potting parameter against parameter). Just to avoid confusion - Here $x$ axis refers to the $m$ axis and the $y$ axis refers to $a$ - axis.

The bifurcation point here we can see is the locus $a = \frac{(m+1)^2}{2m}$, which is a point if $m$ is kept fixed!

Observe also that at this value of $a$, the fixed point $x^* =0$ is unstable since $a = \frac{(m+1)^2}{2m} > 1$, since $(m+1)^2 > 2m$.

Answer 2

To find the steady states you solve he equation $$ x \Bigl(1-\frac{x}{m} - \frac{a}{1+x}\Bigr)=0. $$ One solution is $x=0$. The others come from$$ 1-\frac{x}{m} - \frac{a}{1+x}=0, $$ which after some trivial algebraic manipulations leads to a quadratic equation. The solutions will depend on $m$, but since it is kept fixed, you consider only the dependence on $a$.