What stable memory actually promises

The Week 2 submissions turn “stable” into a sequence of testable claims about conserved pools, uncertain rates, basin geometry, pulse duration and individual molecules. CCBS · EST 5053 · Lecture 4 audience extensionEditorial synthesis of nine Week 2 student submissions · prepared with Codex · 15 September 2026 A phase portrait with two attracting points is a promising memory design. It is not yet a specification for a memory device. We still need to know which disturbances it withstands, which state a preparation reaches, how long a writing pulse must last and whether molecular fluctuations erase the result. The students' Week 2 extensions make those missing questions concrete. Some change the allowed perturbation. Others change a parameter, a hidden timescale or the object being measured. Their arguments fit together when we stop using “stable” as a single verdict and ask exactly what each calculation guarantees. We will build that specification in three stages. First define the state and the feedback that can support two outcomes. Then distinguish robustness of a family from accessibility of a particular basin. Finally ask whether a finite pulse and a finite molecular population can realize the promised memory. Names link to the accepted submissions used here. New connections and calculations are editorial synthesis, not quotations or an invented discussion. Read beside the core, the lecturer-written exposition and lineage. All numerical examples below are declared models, not fitted measurements.

1Stable against which perturbation?

A stability test must use perturbations the physical system can actually make. Kehan Lang exposes this requirement with reversible binding rather than a complicated gene circuit. For the elementary pair

E+S\xrightleftharpoons[k_-]{k_+}C

, the totals

q_E=E+C

and

q_S=S+C

are conserved. The three species therefore move on a one-dimensional slice. On that slice the exact equation is

\dot C=k_+(q_E-C)(q_S-C)-k_-C=f(C), \qquad 0\le C\le\min(q_E,q_S).

This is a coordinate reduction by conservation, not a fast-binding approximation. For positive totals and rate constants, the field points inward at both endpoints and

f'(C)=-k_+(q_E+q_S-2C)-k_-<0.

There is one equilibrium, and every point in the allowed interval approaches it. With $q_E=1$ , $q_S=3$ and $k_+=k_-=1$ in chosen units, the complex level is $(5-\sqrt{13})/2=0.6972$ and its restoring eigenvalue is $-\sqrt{13}$ .

The full three-species Jacobian also has two zero eigenvalues. They describe directions that change the conserved totals. An equilibrium for one total need not return to itself after the total is altered. That does not invalidate attraction within its original slice. Chemical reaction network theory calls these slices stoichiometric compatibility classes.¹

Kehan's example establishes the first line of a memory specification: what is held fixed during the disturbance? Adding molecules, redistributing existing molecules and changing an environment are different experiments. Their stability questions need not have the same answer.

Reduce by exact conservation before interpreting zero eigenvalues, and name which totals an admissible perturbation preserves.

2The nonlinearity need not be in production

Positive feedback permits a switch only when the complete addition–removal balance has the right shape. Xiaomeng Zhu tests this by changing the steepness of a self-activating production law:

\dot x=0.08+\frac{2.5x^n}{1+x^n}-x.

All variables and time are dimensionless. With $n=1$ , there is one positive equilibrium, 1.6291. With $n=4$ , the equilibria are 0.0801, 0.7959 and 2.5195. Their net slopes alternate negative, positive, negative. The same positive feedback sign accompanies two different repertoires.

Read next to that construction, Siheng Wang's growth-feedback example shifts the question. Must the steep nonlinearity be in transcription? Let production be constant and let protein burden slow growth:

\dot x=\alpha-r(x), \qquad r(x)=x[\delta+\lambda(x)], \qquad \lambda(x)=\frac{1}{1+x^4}.

Here $\delta$ is degradation, $\lambda(x)$ is growth dilution and $\alpha$ is production, in dimensionless units. With $\alpha=0.55$ and $\delta=0.20$ , the three equilibria are 0.4781, 1.4864 and 2.3981. The middle state is unstable.

Figure 1. A constant production rate can cross removal three times. Filled markers are stable and the open marker is unstable. The steep growth-burden relation is a modeling assumption, not a measured law for a specified strain. The feedback runs through host growth: more protein slows dilution, which allows more protein to accumulate.

The required shape has a compact test. Since $r'(x)=\delta+\lambda(x)+x\lambda'(x)$ , removal decreases locally only if

-\frac{d\log\lambda}{d\log x}>1+\frac{\delta}{\lambda}.

Slower growth alone is insufficient. Growth must fall steeply enough that the reduced removal per molecule outweighs the larger number of molecules being removed. Three crossings additionally require the production level to lie between the removal curve's local extrema.

There is literature behind this distinction, but two results should not be conflated. Tan, Marguet and You observed bistability in a noncooperative self-activating circuit coupled to growth. Melendez-Alvarez and Tian later analyzed how an ultrasensitive growth response can permit bistability even with a constitutive promoter. The constant-production example here belongs to the latter modeling argument.²³

Locate the nonlinearity in the full production–removal balance, and test its slope rather than treating a feedback sign or a Hill exponent as a universal memory criterion.

3One stable matrix is not a stable family

Robustness to rate uncertainty is a stronger question than local recovery at one parameter point. Leo Sin asks which Jacobian sign patterns remain Hurwitz for every assignment of magnitudes.

A matrix is sign stable when every real matrix with exactly the same signs, including the same zero entries, has eigenvalues with strictly negative real parts. With every diagonal entry strictly negative, the characterization excludes positive two-node cycles and every directed cycle involving three or more distinct nodes. Opposite-sign two-node interactions remain allowed. The strict diagonal assumption matters.⁴

The distinction is already visible in two variables. Compare these matrices with positive parameters.

J_- = \begin{bmatrix}-a&b\\-c&-d\end{bmatrix}, \qquad J_+ = \begin{bmatrix}-a&b\\c&-d\end{bmatrix}, \qquad a,b,c,d>0.

Both traces are negative. The first determinant is $ad+bc>0$ , so every allowed magnitude is stable. The second is $ad-bc$ , so stability depends on loop strength relative to removal. A positive loop can be locally stable without being sign stable.

Independent enumeration reproduces Leo's finite counts. Among all off-diagonal sign patterns with a fixed strictly negative diagonal, the sign-stable counts are 7 of 9 in two dimensions, 223 of 729 in three, and 28,321 of 531,441 in four. These count labeled sign patterns. They do not estimate how frequently stable networks occur in biology.

Nor does adding dilution automatically make the full diagonal negative. A self-activating term can contribute a positive derivative larger than the dilution derivative. The actual Jacobian, evaluated at the actual state, must satisfy the assumption.

For memory, losing a universal guarantee can be useful. A circuit that must support two stable states cannot be understood solely through a promise that every allowed linearization restores the same state. But failure of sign stability does not itself establish multistability, oscillation or any particular nonlinear outcome.

Separate a Hurwitz test at one state from a guarantee over a sign family, and check the full diagonal before invoking a structural criterion.

4Two memories need not be equally accessible

The number of attractors does not determine how a preparation is divided between them. Changkai Li perturbs the lecture's toggle while keeping both stable expression states.

The dimensionless system is

\dot x=\frac{a_x}{1+y^2}-x, \qquad \dot y=\frac{4}{1+x^2}-y.

At $a_x=4$ , exchange symmetry makes the diagonal an invariant basin boundary. The two stable states are $(2-\sqrt3,2+\sqrt3)$ and its reflection. Increasing only $a_x$ to 4.4 preserves bistability but moves both equilibria and the separating curve.

Figure 2. A modest asymmetry changes basin accessibility before removing either memory. Colors show numerical outcomes from a 101 × 101 initial-condition grid at

a_x=4.4

. The high-first-repressor basin contains 62.02% of these seeds. The dashed diagonal is the boundary of the symmetric comparison, not the boundary of the asymmetric system. Dots mark the asymmetric equilibria, with the saddle open.

“Basin size” needs a measure. Here we sample initial concentrations uniformly on $[0,5]^2$ . A biological preparation concentrated near one corner would have a different outcome frequency. Uniform area is a geometric benchmark, not a distribution measured from cells.

The numerical classification also needs a stopping rule. We integrated to time 60 and assigned a seed only when it lay within $10^{-5}$ of a stable equilibrium. The 201 × 201 grid gives about 62.08% among resolved seeds, with one seed still unclassified near the boundary. Symmetry gives exactly half on each side in the symmetric case, excluding the diagonal seeds that approach the saddle.

The folds lie near $a_x=3.229$ and 5.796 when the other capacity remains 4. Changkai's change from 4 to 4.4 is well inside that interval. The accessible fraction changes before the count of stable states does. This is why a device can remain mathematically bistable yet become harder to initialize in its less-favored state.

Specify an initial-condition distribution and a numerical classification rule before turning basin geometry into a claim about memory accessibility.

5A switch is not a clock

The same equations can support two stable states while forbidding a periodic orbit. Zixuan Wang gives a global exclusion argument for the reduced two-protein toggle.

Write its broader form as $\dot x=p(y)-\delta_xx$ and $\dot y=q(x)-\delta_yy$ , with continuously differentiable production functions and positive constant removal rates. Its divergence throughout the positive quadrant is

\frac{\partial\dot x}{\partial x}+\frac{\partial\dot y}{\partial y} =-\delta_x-\delta_y<0.

Suppose a nonconstant periodic orbit enclosed an area. The vector field is tangent to that orbit, so its net outward flux across the boundary is zero. Green's theorem equates that flux with the area integral of the divergence, which is strictly negative. The contradiction excludes the orbit. Smoothness and a simply connected planar domain containing the enclosed region are essential to this Bendixson argument.

Negative divergence means area contracts. It does not mean every pair of trajectories gets closer. At the symmetric toggle's saddle, one eigenvalue is approximately +0.311 and the other −2.311. One direction expands even though their sum is −2.

This exclusion is about the declared two-dimensional model. Promoter dynamics, mRNA, growth or explicit delay can enlarge the state and remove the planar proof. The appropriate response to a simulated oscillation is therefore to inspect the equations and integration, then ask whether a different state model is actually being used.

Use the divergence and domain hypotheses to exclude planar cycles, without confusing area contraction with pointwise attraction or extending the proof to a larger model.

6A pulse must last long enough

Destroying the ON equilibrium during a pulse does not guarantee that the pulse erases the memory. Peng Su adds the missing clock to the lecture's fold diagram.

Let $p(x)=0.15+3x^4/(1+x^4)$ and $\dot x=p(x)-d(t)x$ , in dimensionless units. With $d=1$ , the OFF state, threshold and ON state are 0.1516, 0.6810 and 3.1186. A pulse raises removal to $d_p$ before restoring the original value.

The ON branch disappears at the upper fold $d_p=1.82488$ . For a stronger pulse starting at the original ON state, the trajectory moves downward. The writing question is whether it crosses the restored system's threshold before the pulse ends.

Figure 3. Removing an attractor briefly need not remove its memory. Both pulses double removal at time 5, beyond the upper fold. A pulse lasting 3 time units ends above the restored threshold and returns to ON. A pulse lasting 7 ends below it and stays OFF. The dashed line is the unstable threshold after restoration.

Separating variables computes the threshold duration without guessing a time step. For $d_p>1.82488$ , define

T_c(d_p)=\int_{x_{\mathrm{threshold}}}^{x_{\mathrm{ON}}} \frac{dx}{d_p x-p(x)}.

The denominator is the downward speed during the pulse. The integral is the travel time from the initial ON state to the restored basin boundary. Deterministic erasure requires $T>T_c$ . At exact equality the ideal solution lands on the unstable boundary, so numerical error or a small disturbance decides the subsequent side.

Figure 4. Memory erasure has an amplitude–duration boundary. Independent quadrature gives critical durations 11.873, 4.003, 1.598 and 0.611 at pulse removal rates 1.85, 2, 2.5 and 4. The units are dimensionless, not minutes measured in a cell. Close to the fold, the small downward speed creates a bottleneck.

A static bifurcation diagram answers which equilibria exist while a parameter is held fixed. Peng's question concerns a finite protocol that leaves one vector field and returns to it. Its answer requires a trajectory between the two switching times. The original genetic-toggle experiments likewise used transient induction and then tested whether a state persisted after the inducer was removed.⁵

Turn a static fold into a finite-pulse prediction by integrating the travel time to the basin boundary of the restored system.

7The same curve can hide another clock

Steady occupancy does not determine how quickly a promoter reaches that occupancy. Álvaro Bátrez asks what the toggle's Hill curves leave out when binding is no longer fast. That question directly challenges the duration calculation: a hidden slow state can change how far a finite pulse moves the system.

To isolate the point without assuming negligible binding depletion is exact chemistry, use a deliberately simpler occupancy model. Let $p$ be the fraction of promoters active under an externally clamped regulator. Its new steady active fraction is $p_\infty$ , and its relaxation rate is $r>0$ . Normalize protein production and removal to obtain

\dot p=r(p_\infty-p), \qquad \dot x=p-x.

This is an effective two-state occupancy description with an undepleted regulator reservoir. It is not a full mass-action model of a finite transcription-factor pool. Both promoter transition rates may be scaled together to change $r$ while keeping their ratio, and hence $p_\infty$ , fixed.

Suppose the system starts at $p=x=1$ , and the regulator changes the steady active fraction to 0.2. For $r\ne1$ , direct integration gives

x(t)=0.2+0.8e^{-t} +\frac{0.8}{1-r}\left(e^{-rt}-e^{-t}\right).

Figure 5. Identical endpoints do not imply identical responses. The promoter rates are 0.2 and 20 per protein lifetime. Both models end at protein 0.2. Reaching 90% of the final change takes 12.63 versus 2.35 protein lifetimes. Instantaneous occupancy gives 2.30. These are new editorial calculations illustrating Álvaro's question, not a reconstruction of his full toggle simulation.

A steady-state titration cannot distinguish these two clocks. A step response can. Once a memory is written by a finite pulse, the distinction is operational rather than cosmetic: a reduced model can have the correct fixed points and still give the wrong writing time.

Test a hidden-state reduction against the transient observable you need, rather than accepting matching steady curves as sufficient validation.

8A stable mean is not a quiet cell

Deterministic attraction concerns a concentration trajectory, not the absence of molecular fluctuations. Lipei Fang makes the distinction using the simplest production–removal process.

Let molecules enter at a constant propensity $\alpha$ and leave independently with total propensity $\gamma n$ when the count is $n$ . These are effective birth and death events. The mean count $m(t)$ obeys the exact equation $\dot m=\alpha-\gamma m$ because both propensities are affine in count.

At stationarity, balance between adjacent count states gives $\alpha P_n=\gamma(n+1)P_{n+1}$ . Iterating and normalizing yields

P_n=e^{-\bar n}\frac{\bar n^n}{n!}, \qquad \bar n=\frac{\alpha}{\gamma}, \qquad \operatorname{Var}(n)=\bar n.

For Lipei's $\alpha=2$ molecules per minute and $\gamma=0.2\,\mathrm{min}^{-1}$ , the mean is 10 and the relaxation time is 5 minutes. The stationary standard deviation is $\sqrt{10}=3.16$ molecules, giving a coefficient of variation of 31.6%.

Figure 6. The fixed point predicts the center of a distribution, not a frozen molecule count. Bars show the exact stationary Poisson probabilities of the stated linear birth–death model. This is an analytic distribution, not a selected stochastic trajectory. Bursts and nonlinear feedback can change both its shape and the relation between mean and variance.

Gillespie's stochastic formulation makes individual reaction events explicit under its well-mixed assumptions.⁶ It does not make every deterministic equation exact for the mean. With nonlinear propensities, averaging the rate is generally different from evaluating the rate at the average state.

For a bistable device, the next question is not merely the width of the fluctuations. It is the probability of leaving the intended basin during the storage interval. The one-state Poisson example cannot answer that escape problem, but it establishes why a deterministic basin diagram alone cannot answer it either.

Distinguish a stable mean from a stationary count distribution, and formulate memory reliability as an escape probability over a specified time.

9Specify the memory you need

A usable memory claim must name the state, the protocol and the tolerated failures. The submissions provide different pieces of that specification.

First state what is fixed: the conserved totals, environment and model resolution. Then state what supports two outcomes: the complete feedback and addition–removal shapes. Rate uncertainty, initial-state uncertainty and molecular fluctuations are separate disturbances, so each needs its own test.

Next specify the writing protocol. A parameter beyond a fold is not enough without duration. A duration predicted by a reduced model is not enough unless the hidden kinetics are fast for that protocol. Successful writing is not enough without a storage time and an acceptable error probability.

This changes how we read the original phase portrait. Its two attracting points are one verified component of a larger claim. Conservation tells us which directions to test. Structure tells us which rate changes a theorem covers. Basins tell us which preparations succeed. Dynamics tells us how to write, and stochastic kinetics tells us how long the written state can be trusted.

Write a memory specification that separates admissible perturbations, parameter robustness, initialization, writing time and stochastic retention.

Models and provenance

The nine named contributions are accepted files in the pinned Week 2 corpus. Attribution identifies each student's question or construction. It does not imply that the students coauthored or approved this editorial synthesis.

Independent code, the input checksums, the contribution ledger and figure data are retained in the instructor's course project under lectures/research/week02-extension-synthesis/. Root finding, sign-pattern enumeration, pulse quadrature and piecewise time integration are independently implemented. Basin percentages are numerical estimates with the grid and stopping rule stated above. No student script was executed.

References

F. Horn and R. Jackson, “General mass action kinetics,” Archive for Rational Mechanics and Analysis 47, 81–116 (1972). DOI. Stoichiometric geometry and stability within compatibility classes.
C. Tan, P. Marguet and L. You, “Emergent bistability by a growth-modulating positive feedback circuit,” Nature Chemical Biology 5, 842–848 (2009). DOI. Full-text results distinguish noncooperative self-activation from growth-mediated feedback.
J. R. Melendez-Alvarez and X.-J. Tian, “Emergence of qualitative states in synthetic circuits driven by ultrasensitive growth feedback,” PLOS Computational Biology 18, e1010518 (2022). DOI. The constitutive-promoter model and its nonmonotone removal curve.
C. Jeffries, V. Klee and P. van den Driessche, “When is a matrix sign stable?” Canadian Journal of Mathematics 29, 315–326 (1977). DOI. Theorems 1–2 and the strictly negative diagonal special case.
T. S. Gardner, C. R. Cantor and J. J. Collins, “Construction of a genetic toggle switch in Escherichia coli,” Nature 403, 339–342 (2000). DOI. Model, pulse protocols and post-induction memory tests.
D. T. Gillespie, “Exact stochastic simulation of coupled chemical reactions,” Journal of Physical Chemistry 81, 2340–2361 (1977). DOI. Event-level stochastic kinetics and its relation to deterministic descriptions.