Finding equilibrium and bifurcation points for first order ODE's

Question

I'm trying to find the equilibrium and bifurcation points for $$\frac{d}{dx}=rx+xe^{-x}.$$ I started by noticing that $x=0$ is an equilibrium point for all $r$. Also, I then set the RHS equal to zero and solved for $x$. I found that $x=-\ln(-r)$ so this means that $r=-1$ also creates an equilibrium point. I'm not exactly sure where to take this problem further to solve for the bifurcation points. I know that I need to create a bifurcation diagram, and I think that I need to sketch a graph with $x$ in the vertical axis and $r$ as the independent variable. Is $x=-\ln(-r)$ what I want to sketch?

Answer 1

Yes. The system has trivial equilibrium point $x_0=0$ and interior equilibrium point $x^*=-\ln(-r)$ provided $r<0$. So for positive values of $r$, the interior equilibrium point does not exist.

Also you can draw the bifurcation diagram in $x-r$ plane as you mentioned, what type of bifurcation the system exhibit. An easy method for checking what type of bifurcation the system exhibit is to use MATCONT software which is based on MATLAB package. The eigenvalue of the Jacobian matrix calculated at the equilibrium point $x_0$ and $x^*$, of the system are $r-1$ and $r\ln(-r)$. So $x_0$ is locally stable for $r<-1$ and $x^*$ is locally stable for $r<0$, which is also existence condition for $x^*$. So the stability of $x_0$ changes at $r=-1$. Therefore $r=-1$ is the bifurcation point. In the above figure the point $r=-1$ is shown as a branch point.