3
Stability Conditions and the Failure of Joint Function
HYC-1979Lesson 5: Selection, Species, Will
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In an open catalytic chain under constant organization, a stable positive stationary state is not evidence of cooperative self-organization. The stability of a stationary state is governed by two conditions: a rate-constant condition and a total-concentration condition.

Stability Conditions

  • Rate-constant condition: The first member’s rate constant must dominate all downstream constants: ≫ kβ±Ό for every .
  • Total-concentration condition: The total concentration must exceed the threshold β±Όβ‚Œ2ⁿ (β±Ό)/kβ±Ό. Below this threshold, required downstream concentrations become nonpositive, so no positive stationary state exists in the concentration simplex.

Stronger dominance by the first member (larger β±Ό ratios) raises the required total concentration c0.

Example

For a two-membered chain, the conditions reduce to and . If and , then . With , the stationary state gives and . Below , no positive stationary state exists.

Why Stability Does Not Imply Integration

When both conditions are satisfied, the system may settle to a stable stationary state. However, this state remains an open, first-member-dominated chain. The open chain lacks cyclic feedback: downstream species do not return to influence the first member. Consequently, perturbations do not engage the whole chain as a joint functional unit. Therefore, the chain cannot stabilize a joint function; stability of the stationary state is not equivalent to integration into a hypercyclic collective.

In this framework, selective advantage for a molecule corresponds to a larger rate constant. A molecule with a higher k outcompetes others, but this competitive advantage does not produce a joint functionβ€”it merely shifts the stationary state toward domination by that molecule.

Stability Threshold DiagramTwo-member chain:Β  c0,min = (k1 βˆ’ k2) / k2 = k1 / k2 βˆ’ 1Two stability conditionsk1 ≫ kj and c0 > Ξ£j=2n (k1 βˆ’ kj) / kj12345612345c0,mink1 / k2Positive stable stationary state possiblec0 > c0,minNo positive stationary staterequired concentrations become nonpositivec0,min = k1/k2 βˆ’ 1Above the linePositive stable stationary statepossible: c0 > c0,minBelow the lineNo positive stationary state:required downstream concentrationsbecome nonpositive.Stability β‰  integrationThe open chain has no cyclic feedback,so perturbations never engage thewhole chain as a joint functional unit.Example: k1/k2 = 100 β†’ c0,min = 99; with c0 = 100 > 99, a stable positive stationary state exists.Stability of the stationary state is not integration: with no cyclic feedback the chain cannot stabilize a joint function.