Questions: Transcritical and Pitchfork Bifurcations

Question 1

In the transcritical bifurcation ẋ = rx - x², x = 0 is always a fixed point. What happens at r = 0?

• A. The fixed point at x = 0 disappears
• B. A new fixed point is created at x = 0
• C. The fixed points x = 0 and x = r exchange stability — for r < 0, x = 0 is stable and x = r is unstable; for r > 0, x = 0 becomes unstable and x = r becomes stable
• D. Both fixed points become stable simultaneously

Answer: (C)

The two fixed points of ẋ = rx - x² = x(r - x) are x = 0 and x = r. For r < 0, the eigenvalue at x = 0 is r < 0 (stable) and at x = r it's -r > 0 (unstable). For r > 0, the eigenvalue at x = 0 is r > 0 (unstable) and at x = r it's -r < 0 (stable). At r = 0, the fixed points collide and exchange stability. Neither is created or destroyed — they pass through each other on the bifurcation diagram.

Question 2

The supercritical pitchfork bifurcation ẋ = rx - x³ models a system that spontaneously breaks symmetry. Which physical scenario best illustrates this?

• A. A ball rolling in a single-well potential that gradually deepens
• B. An Euler column under increasing compressive load — below the critical load, the straight configuration is stable; above it, the column buckles to one side or the other
• C. A pendulum with increasing friction that gradually stops oscillating
• D. A chemical reaction that reaches equilibrium faster as temperature increases

Answer: (B)

The Euler buckling problem has perfect x → -x symmetry (buckling left or right are equally likely). Below the critical load, the symmetric (straight) state is stable. Above it, the straight state becomes unstable and two stable buckled states appear symmetrically. This is exactly the supercritical pitchfork: one symmetric equilibrium loses stability and bifurcates into two asymmetric stable equilibria plus the now-unstable original. The symmetry of the problem dictates the pitchfork structure.

Question 3

A subcritical pitchfork bifurcation is more dangerous than a supercritical one because the emerging branches are unstable, leaving no nearby stable state after the bifurcation.

• T. True
• F. False

Answer: True

In a supercritical pitchfork, the new branches born at the bifurcation are stable — the system transitions smoothly to a nearby asymmetric state. In a subcritical pitchfork, the new branches are unstable. When the symmetric state loses stability, there is no nearby stable equilibrium to catch the system. The state must jump to a distant attractor, often resulting in a sudden, large-amplitude transition. This makes subcritical bifurcations associated with catastrophic events, similar to the saddle-node.

Question 4

Why can't a pitchfork bifurcation occur in a generic system without symmetry?

Model answer: The pitchfork requires f(x) = -f(-x) symmetry so that whenever x* is a fixed point, -x* is also one. Without this symmetry, small perturbations break the pitchfork into a saddle-node plus an isolated branch — the two branches no longer meet at the same point. Generically, the cubic term x³ that creates the pitchfork requires the quadratic term x² to vanish, which only happens when the system has the x → -x symmetry. Breaking the symmetry 'unfolds' the pitchfork into structurally stable components.

This is the concept of structural stability and unfolding. The pitchfork is structurally unstable — adding a small asymmetric perturbation (like εx²) destroys the simultaneous creation of two branches. In applications, perfect pitchforks are rare unless enforced by physical symmetry (like left-right symmetry in buckling, or up-down symmetry in convection). Imperfect pitchforks, where slight asymmetry disconnects the branches, are more common in practice.