I am working on the construction of a bifurcation diagram of the following differential equation:
$$\dot{x} = x^3 - x + \lambda$$
I have solved for the fixed points which can be solved by setting the differential equation equal to $0$, which is:
$$ 0 = x^3 - x + \lambda$$ $$ x = \frac{1\pm\sqrt{(-1)^2 -4(1)(\lambda)}}{2}$$
I have graph this in mathematica. I was wondering if this is right?
You find the fixed points of $\dot{x} = f(x)$ by setting $\dot{x} = 0$, not $f'(x) = 0$. So we want to analyze the zeros of $f(x) = x^3 - x + \lambda$.
It's easiest to think about the graph of $g(x) = x^3 - x$. I recommend drawing some sketches to accompany the following argument. Adding $\lambda$ just shifts this graph up and down. By sketching this graph, it's clear that $g$ has only one zero when $|\lambda|$ is very large. If we think about "starting" $\lambda$ at negative infinity and moving it up, eventually the local max of $f$ at $-1/\sqrt{3}$ will cross the $x$-axis. This happens exactly at $\lambda = -g(-1/\sqrt{3})$. For $\lambda$ slightly larger than this value, the graph of $f$ crosses the x-axis three times, so there are three fixed points. Once $\lambda > g(1/\sqrt{3})$, the local minimum of $f$ is above the $x$-axis, so there is again only one fixed point.
So, to summarize, we have one fixed point for $\lambda \in (-\infty, -g(-1/\sqrt{3}))$ branching into 3 fixed points for $\lambda \in (-g(-1/\sqrt{3}), g(1/\sqrt{3}))$ and then collapsing back into 1 fixed point for $\lambda \in (g(1/\sqrt{3}), \infty)$. We can determine the stability of these fixed points by looking at the sign of $f$ on either side of a fixed point. The right-most fixed point is always unstable, since $f$ is positive to the right of this point and negative to the left of it, so the flow pushes away from this point. Similarly, the left-most fixed point is also always unstable, but the middle fixed point (while it exists) is stable.