Bifurcations in a one-dimensional dynamical system, with two parameters

Question

I am given the following dynamical system:

$x'= -(x^3/(1+x^2))+ \lambda x^2+\mu$

I need to find all bifurcation points when $\lambda$ and $\mu$ vary on the reals, and study their type. (Note: I'm using Maple software to solve this). Please note that this is my very first attempt at solving such an exercise, I have never dealt with dynamical systems before. Any kind of help would be much appreciated.

First of all, I wanted to find all non-hyperbolic equilibrium points of the system, that is, if $f = -(x^3/(1+x^2))+ \lambda x^2+\mu $,

$\begin{cases} f(x,\lambda,\mu)=0 \\ \partial f/ \partial x \ f (x,\lambda, \mu) = 0 \end{cases}$

For some reason, Maple (a software I am not too familiar with, but I am required to use that) does not give me explicit solutions for this system. It actually just doesn't compute anything.

So, I have tried a sightly different approach, as done in this example by Hale-Kocak. By solving the second equation with respect to $\lambda$ (parameter $\mu$ vanishes after derivating), I get

$\lambda = (1/2)x(x^2+3)/(x^2+1)^2$

And by substituting in the first one, solving with respect to $\mu$,

$\mu = (1/2)x^3(x^2-1)/(x^2+1)^2$

.. and at this point I'm stuck again. I should get the relation between $\lambda$ and $\mu$ as they have done with parameters $c$ and $d$ in their example, but I can't seem to find a way to do it.

Any hint is welcome. I'm in serious need of ideas on how to start solving this.

Answer 1

I suppose that you tried by using Maple's solve command. That doesn't work very well when the number of equations is different from the number of variables being solved for. The command to use in this case is eliminate.

f:= -(x^3/(1+x^2))+lambda*x^2+mu:
eliminate({f = 0, diff(f,x) = 0}, {x});

This will return a list of two sets. The first contains an equation that expresses $x$ in terms of $\lambda$ and $\mu$. The second contains an expression which when equated to $0$ expresses the relationship between $\lambda$ and $\mu$.

Since this relationship is provided in factored form, you can immediately see that $\mu=0$ is a special case. The other factor is a polynomial fifth-degree in $\lambda$ and fourth-degree in $\mu$. I'll call it lu. I believe that solving lu for one of its variables will add great complexity to your work while not giving much benefit. Instead, we can study a bivariate polynomial in great detail without explicitly solving it. In particular, it can be plotted with plots:-implicitplot or algcurves:-plot_real_curve and some further analysis can be done with other tools in the algcurves package.

Please note that I don't know much about analyzing iterative maps. I'm providing this Answer because I know a great deal about Maple.

For plotting the bifurcations in Maple 2016, check out the new IterativeMaps package. This produces plots of far superior detail and color than the other Maple plotting commands.