A bunch of statements on pitchfork bifurcations I want refuted or verified

Question

Here is how I understand conditions for a pitchfork bifurcation, intuitively and mathematically: I am mostly referring to the graph of x vs. a, where a is the parameter that varies and leads to bifurcations.

1. vector field function f must be odd in x. Intuitively, this makes some sense. My understanding of a pitchfork bifurcation is essentially three fixed points, two unstable and one stable, which doesn't seem like it could happen with an even function (with degree 2, two solutions with same behaviors). Is this necessary though?

2. My second intuition is that the function f should have changing partials with respect to a, i.e a point of convexity or concavity wrt. to a ($\frac{df^2}{d^2a}>0$ or $\frac{df^2}{d^2a}<0$). But on the side where the solutions are bounded (stable?), we should have changing derivative with respect to x. Of course the opposite behavior on the non concave side of the double 0 at the origin of the x vs a graph.

Answer 1

1. Not quite, but almost. A bifurcation is something that happens locally (both in the bifurcation parameter $a$, around its bifurcation value $a_*$, and in the variable $x$). Therefore, a bifurcation of a certain type (say pitchfork) has a so-called normal form. This means that, if you know that a pitchfork bifurcation takes place for some value $a_*$, you can conclude something about the local expansion of the vector field $f(x,a)$ around $(0,a_*)$. Up to some coordinate rescalings, for the pitchfork bifurcation this means that you know that \begin{equation} f(x,a) = x(a - x^2) + \mathcal{O}(x^4), \end{equation} see here. The answer to your question whether $f$ must be odd in $x$ lies in the remainder term $\mathcal{O}(x^4)$. This can be anything, and therefore $f$ is not necessarily odd in $x$ -- only locally. For example, take $f(x) = x(a - x^2) + x^4$, which has a pitchfork bifurcation at $a = 0$, but $f$ is not odd in $x$.

2. That same normal form indicates that the second derivative to the parameter $a$ does not influence the pitchfork behaviour. A vector field which is just equal to the normal form, i.e. $f(x,a) = x(a - x^2)$ is linear in $a$ and thus has vanishing second derivatives to $a$.