# Is there a bifurcation node in the activation energy of chemical reactions?

I was thinking about the activation energy of chemical reactions (obviously), and I was wondering if there exists a bifurcation node somewhere in the transition state. Let me give you a link to an image: http://en.wikipedia.org/wiki/File:Activation_energy.svg

Here's what's bugging me about it. I don't have an actual equation to the curve, so I can't necessarily do a quantitative analysis. Regardless, I can at least make conjectures about it.

Personally, my instincts are telling me, provided there truly is a bifurcation node, it would be stable at the peak, which sounds odd. From the mathematical definition, stability indicating that any small perturbations from the bifurcation point will "dampen" out in time, rather than flying off into infinity. In context though, that doesn't sound correct. The reaction has to go one way or another. The peak, representing the transition state, is said to be unstable chemically.

Another issue is with the parameter of time. It has to go forward. In order to analyze it, we would more than likely have to nondimensionalize it.

These are just some of my current thoughts on it. I know the application is in regards to chemistry, but my question is regarding the mathematics of it. I'm just debating whether there is a bifurcation point (which I believe there is), and if so whether it is stable, unstable, halfstable. Basically, I'm not sure if my reasoning is correct right now. Any help would be appreciated.

Edit: Let me also add that instability does not necessarily mean that a phase point will fly off into infinity. Stability and instability can be be local. In other words, an unstable point can fly into another equilibrium point.

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What I am curious about is the stability of the transient point as a function of time. The graph does not reflect that. The instantaneous slope, as it is now, is $\frac{dV}{dq}$ where $q$ is a coordinate. The question, however, depends on $\frac{dV}{dt}$. Through the chain rule and definition of potential, one can reach the following equivalency: $\frac{dV}{dt} = -(\frac{dV}{dx})^2$. Physically, this means that $V(t)$ decreases along trajectories, and therefore always moves toward a lower potential.
In terms of the question, there is an unstable bifurcation node at the local maxima of $V(q)$, while there are stable fixed points at the minima of the same function (note these points are bounded). This intuitively makes sense both mathematically and chemically.