advanced38 min studyLesson 143 in Physics

When equilibria appear and exchange stability

Derive saddle-node and pitchfork normal forms, identify their nondegeneracy conditions, and recognize the role of symmetry.

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What you’ll understand

  • Track fixed points as a control parameter varies.
  • Check the local nondegeneracy conditions of a saddle-node.
  • Explain why a pitchfork usually needs a symmetry or another constraint.

A smooth change in a parameter can produce an abrupt change in the kinds of long-term motion available. An equilibrium can disappear at a threshold, or one symmetric state can give way to two alternatives. Bifurcation theory identifies that structural change. The useful calculation is not merely finding a zero eigenvalue: it is checking how the parameter and nonlinear terms resolve that degeneracy, and whether a physical symmetry protects the proposed diagram.

The saddle-node creates a pair

Consider the scalar normal form x˙=μ−x2\dot x=\mu-x^2. For positive parameter there are two fixed points: the positive root attracts locally and the negative root repels. At zero they meet; for negative parameter no equilibrium remains nearby. The exact signs depend on the chosen coordinates and direction of the parameter. What survives those conventions is pair creation or annihilation, with one stable and one unstable scalar branch on the side where the pair exists.

For a smooth scalar family f(x,μ)f(x,\mu), a generic saddle-node at the origin requires f=fx=0f=f_x=0 while fxx≠0f_{xx}\ne0 and fμ≠0f_\mu\ne0 there. The first two equalities identify the degenerate equilibrium; the inequalities ensure a quadratic state dependence and an effective parameter crossing. A zero derivative without those inequalities can describe a higher degeneracy. In higher-dimensional systems a single zero eigenvalue and suitable complementary stability permit a reduced one-dimensional description, but those additional assumptions must be checked.

x∗±=±μ,fx(x∗±)=∓2μ(μ>0)x_*^{\pm}=\pm\sqrt\mu,\qquad f_x(x_*^{\pm})=\mp2\sqrt\mu\quad(\mu>0)
For ẋ = μ − x², the positive branch is stable and the negative branch unstable. The square-root branch dependence and slowing recovery close to the threshold are local normal-form predictions.↔ Scroll wide equations horizontally.

A pitchfork is organized by symmetry

The supercritical pitchfork x˙=μx−x3\dot x=\mu x-x^3 is invariant under replacing x by its negative. Its central branch exists for every parameter. Below zero it attracts; above zero it repels and two attracting states appear at opposite signs. This makes a useful model of a symmetric device choosing either orientation. The two alternatives are related by the symmetry, not by a random term inserted into the deterministic equation; imperfect initial preparation can select which branch is reached.

That symmetry also explains fragility. Adding a small constant bias destroys the exact oddness and generally unfolds the pitchfork into a different arrangement with folds. A subcritical cubic sign instead produces unstable nearby nonzero branches on the opposite side. Stable distant states and hysteresis may require a higher-order saturating term; the destabilizing cubic alone does not prove them. Normal forms describe a neighborhood, so do not infer the entire global response curve from a truncated polynomial near threshold.

x˙=μx−x3,x∗=0 or x∗2=μ;fx=μ−3x∗2\dot x=\mu x-x^3,\qquad x_*=0\ \text{or}\ x_*^2=\mu;\qquad f_x=\mu-3x_*^2
The nonzero branches exist for μ > 0 and have derivative −2μ. A generic symmetry-preserving pitchfork also needs a nonzero parameter dependence of the linear term and a nonzero cubic coefficient.↔ Scroll wide equations horizontally.

A threshold also changes time scales

Close to an attracting branch with a very small negative derivative, disturbances decay slowly. This critical slowing can make a gradual parameter sweep appear to lag behind its equilibrium diagram. A saddle-node can also leave a bottleneck just beyond disappearance: trajectories move slowly through the place where equilibria nearly existed. An observed long pause is therefore not necessarily a surviving fixed point. Estimate rates and observe a long enough interval before labeling a state stationary.

To use these ideas in an experiment, distinguish the controlled parameter from the observed coordinate, sweep in both directions, and record the disturbance size and settling time. Bias, noise, finite sweep rate, and hidden degrees of freedom can blur a diagram. A bifurcation is a statement about the model's invariant structures as parameters change, while a measured trace includes the protocol. Matching those two requires checking symmetry and nondegeneracy, rather than assigning a familiar name to any branching picture.

Two symmetric operating states

For x˙=μx−x3\dot x=\mu x-x^3, compare the equilibria and their linear recovery rates at μ=−1\mu=-1 and μ=1/4\mu=1/4.

  1. At μ=−1\mu=-1, only x=0x=0 is real. Its derivative is −1-1, so small displacements decay initially with rate one in the chosen inverse-time units.

  2. At μ=1/4\mu=1/4, the three equilibria are 0,1/2,−1/20,1/2,-1/2. The central derivative is 1/4>01/4>0, so the symmetric state repels.

  3. At either nonzero equilibrium, fx=1/4−3/4=−1/2f_x=1/4-3/4=-1/2. Each branch attracts locally, with linear relaxation time two. Initial states on opposite sides of zero select different stable branches in this exact symmetric scalar model.

The result

The parameter crossing exchanges central stability for two stable symmetry-related states. The result is a supercritical pitchfork, and relies on maintaining the odd symmetry of the equation.

Explore the model

Interactive lab

Interactive lab · function

Inspect quadratic curvature before a fold

Change coefficient a in y=ax² and use position x to inspect the tangent slope. Predict that opposite sides of the parabola have opposite slopes for nonzero a. This diagram illustrates the quadratic curvature used in a saddle-node expansion; its controls do not add μ, move equilibria, or simulate a bifurcation.

Inspect quadratic curvature before a foldThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forCurvature supplies a necessary local ingredient of a fold, but a true bifurcation also needs parameter dependence.

Check your understanding

Which change generally destroys the exact symmetric pitchfork ẋ = μx − x³?

Try it yourself

For x˙=μ−x2\dot x=\mu-x^2 at μ=0.09\mu=0.09, find both equilibria and their linear rates. What happens to the stable relaxation time as μ approaches zero from above?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Take the square root, evaluate −2x, and invert the magnitude of the negative rate.

Compare with the solution

The roots are x=0.3x=0.3 and x=−0.3x=-0.3. Their derivatives are −0.6-0.6 and 0.60.6, so the positive root attracts and the negative root repels. The stable relaxation time is 1/(2μ)1/(2\sqrt\mu), here about 1.671.67 time units, and diverges as μ approaches zero from above. Exactly at the threshold linear recovery is inconclusive and the remaining quadratic flow is only one-sided attracting.

Take it with you

  • A saddle-node needs both degeneracy equalities and nonzero unfolding coefficients.
  • A pitchfork's symmetry matters; a generic small bias changes the diagram.
  • Equilibrium branches and finite-rate measurement traces answer different questions.

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