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Structural Identifiability · Tutorial 539 of 799

Time-Scale Invariance in Dynamical Models

Learn to derive time-scale transformations in mechanistic ODEs, identify the rate combinations that observations can determine, and recognize when known timing breaks the ambiguity.

Advanced 10 min read

What You'll Learn

  • Derive the ODE transformation induced by changing the time coordinate.
  • Identify which rate constants change and which parameters remain unchanged.
  • Explain why an unknown clock can confound absolute rates with time calibration.
  • Check whether known input timing or other time references break the invariance.
  • Distinguish time-scale ambiguity from dose or exposure scaling.
  • Diagnose when a fitted rate is expressed only in arbitrary time units.

Why the Clock Matters

A mechanistic dynamical model describes how a system changes over time. Its rate constants therefore have units tied to the time coordinate: a first-order rate might be reported per hour, for example. If observations reveal the shape of a trajectory but do not establish how the recorded time relates to physical time, a faster process on a slower clock can produce the same observations as a slower process on a faster clock. The issue is not necessarily a lack of precise measurements; it can be a structural ambiguity in the model and experiment.

The central question is whether the data establish an absolute time scale, rather than only the trajectory’s shape relative to an unknown clock. The distinction matters when a rate is to be interpreted in physical units or compared across experiments.

Definition: A time-scale transformation replaces a time coordinate \(t\) by \(s=ct\), where \(c>0\). A model is invariant under this transformation when its parameters and, if necessary, its input schedule can be transformed so that it produces the same state and output values at corresponding times. If the conversion between \(s\) and physical time is unknown, the data may identify rates only relative to that conversion, not as absolute rates per physical time unit.

Deriving the Rate Transformation

Consider a state vector \(x(t)\) with dynamics

$$ \frac{dx}{dt}=f(x(t),u(t);k,\eta), \qquad x(0)=x_0. $$

Here \(u(t)\) is an external input, \(k\) denotes rate constants, and \(\eta\) denotes other parameters. Suppose the measured output is \(y(t)=h(x(t);\eta)\), with no explicit dependence on time. Define \(s=ct\) and \(z(s)=x(s/c)\). Applying the chain rule gives

$$ \frac{dz}{ds} =\frac{1}{c}\frac{dx}{dt}\bigg|_{t=s/c} =\frac{1}{c}f\bigl(z(s),u(s/c);k,\eta\bigr). $$

This equation is valid for any \(c>0\). To express it within the same model family, a parameter transformation must make the right-hand side representable by that family. In the common case where the dynamics are a sum of rate constants multiplying rate laws,

$$ \frac{dx}{dt}=\sum_{j=1}^{m} k_j g_j(x,u;\eta), $$

set \(k'_j=k_j/c\) for every \(j\), and define the transformed input \(u'(s)=u(s/c)\). Then the transformed-coordinate model is

$$ \frac{dz}{ds}=\sum_{j=1}^{m} k'_j g_j(z,u'(s);\eta). $$

The initial state is still \(z(0)=x_0\), and the output at corresponding times is unchanged: \(h(z(s);\eta)=h(x(s/c);\eta)\). If the right-hand side is locally Lipschitz in the state on the domain of interest, the initial-value problem has a unique solution there. Thus the transformed model’s solution is precisely the original trajectory expressed in the new time coordinate. Without uniqueness, the transformation still gives the same initial-value problem, but equality of unique trajectories cannot be concluded.

This rate transformation applies when the model family is closed under multiplying its entire dynamic right-hand side by \(1/c\). Parameters that do not scale the speed of the dynamics—for example, an initial concentration or a static observation-scale parameter—are not automatically divided by \(c\). A model with fixed time-dependent inputs must also transform those inputs to preserve the trajectory. That condition is important: a known input schedule can supply the time reference that observations of the states alone lack.

Key result: If the time conversion \(c\) is unknown and the model admits the transformation above, corresponding trajectories can be reproduced with \(k'_j=k_j/c\). An output trajectory by itself then does not distinguish the original rates from the transformed rates and clock. A known time reference or input schedule may remove this ambiguity.

Worked Examples

Worked Example: A First-Order Decay on an Uncalibrated Clock

Suppose a hypothetical system follows \(dX/dt=-kX\), with \(X(0)=80\) concentration units and \(k=0.300\ \mathrm{h}^{-1}\). The physical-time solution is \(X(t)=80e^{-0.300t}\). Its values at \(t=0,1,2\) hours are \(80\), \(80e^{-0.300}\approx59.27\), and \(80e^{-0.600}\approx43.90\) concentration units.

Now let the instrument’s time coordinate be \(s=2t\), where the conversion to hours has not been independently established. The transformed rate is

$$ k'=\frac{k}{2}=\frac{0.300}{2}=0.150\ \text{per instrument time unit}. $$

The model in \(s\) is \(dZ/ds=-0.150Z\). At instrument times \(s=0,2,4\), it predicts \(80\), \(80e^{-0.150(2)}=80e^{-0.300}\approx59.27\), and \(80e^{-0.150(4)}=80e^{-0.600}\approx43.90\) concentration units. The two descriptions agree at corresponding times. If the instrument times are uncalibrated, the curve alone does not establish whether the physical rate is \(0.300\ \mathrm{h}^{-1}\) or some other rate paired with a different time conversion. If the observations are instead known to be at exactly 0, 1, and 2 hours, that clock calibration supplies the scale and this particular ambiguity is removed.

Worked Example: Two Rate Constants Share the Same Time Scale

Consider the hypothetical sequence \(A\mathrel{\mathop{\longrightarrow}^{k_1}}B\mathrel{\mathop{\longrightarrow}^{k_2}}\text{loss}\), with \(A(0)=1\), \(B(0)=0\), \(k_1=0.800\ \mathrm{h}^{-1}\), and \(k_2=0.400\ \mathrm{h}^{-1}\). The equations are \(dA/dt=-k_1A\) and \(dB/dt=k_1A-k_2B\). Their solutions include

$$ A(t)=e^{-k_1t}, \qquad B(t)=\frac{k_1}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right). $$

At \(t=1\) hour, \(A(1)=e^{-0.800}\approx0.4493\). For \(B\), keeping the exponentials unrounded until the final calculation,

$$ B(1)=\frac{0.800}{0.400-0.800} \left(e^{-0.800}-e^{-0.400}\right) =-2\left(e^{-0.800}-e^{-0.400}\right) \approx0.4420. $$

Both rates must change together under a time rescaling. For \(s=4t\), \(k'_1=0.800/4=0.200\) and \(k'_2=0.400/4=0.100\), in inverse instrument time units. At \(s=4\), the exponents are \(-0.200(4)=-0.800\) and \(-0.100(4)=-0.400\), so the predictions are again \(A\approx0.4493\) and \(B\approx0.4420\). The ratio \(k_1/k_2=2\) is unchanged, but neither absolute rate is fixed by the corresponding uncalibrated trajectory.

Worked Example: A Known Input Time Can Anchor the Clock

Suppose a hypothetical state \(X\), measured in concentration units, satisfies \(dX/dt=-0.500X+1.00u(t)\), where time is in hours and \(X(0)=0\). The input \(u(t)\) is one from \(t=0\) to \(t=1\) hour and zero afterward. During the input, \(X(1)=2(1-e^{-0.500})\approx0.7869\) concentration units; at \(t=2\), it has declined to \(X(2)=0.7869e^{-0.500}\approx0.4773\).

Under \(s=2t\), the transformed equation has coefficients \(0.250\) per instrument time unit and \(0.500\) concentration units per instrument time unit. To preserve the trajectory, its transformed input must remain on from \(s=0\) to \(s=2\), because \(u'(s)=u(s/2)\). But if the input is independently known to end at \(s=1\), the transformed schedule is not the known schedule. For example, at \(s=1.5\), the transformed schedule \(u(s/2)\) is still on, whereas a schedule known to end at \(s=1\) is off. The proposed \(c=2\) transformation therefore fails to preserve the specified experiment. A known input transition time can anchor the time coordinate, provided the input schedule is genuinely known in the same calibrated time units as the observations.

Assumptions, Diagnostics, and Alternatives

The invariance argument is conditional, not a claim that every dynamical model has an unknown time scale. Check the model and experimental design. The state equations must permit the required rate transformation; the observation function must not introduce an independent calibrated clock that changes under the transformation; and any input must be transformed along with the time coordinate. For a unique trajectory, the initial-value problem also needs an appropriate uniqueness condition, such as local Lipschitz continuity of the right-hand side in the state over the relevant domain.

Useful diagnostic questions include:

  • Are observation times known in physical units, or are they only labels from an uncalibrated clock?
  • Does multiplying every dynamic rate by the same positive factor and inversely rescaling time leave the model’s output unchanged?
  • Are input transition times, sampling intervals, or other external time references independently known?
  • Do rate estimates change inversely when the time unit changes, as dimensional consistency requires?

A numerical fitting algorithm may return a particular rate even when the absolute rate is structurally ambiguous. A selected starting value, constraint, or optimizer convention can choose one point among equivalent parameter-and-clock descriptions; it does not make that rate identifiable from the observations. In contrast, a poorly estimated rate with calibrated times may be a practical estimation problem rather than this exact invariance. The distinction is whether the model admits an exact transformation preserving ideal observations, not simply whether a fitted confidence interval is wide.

Time-scale invariance differs from dose or exposure scaling: the latter concerns rescaling an input’s magnitude, while time-scale invariance changes the speed of the dynamics and the units attached to rates. Multiple outputs can help distinguish state mechanisms, but they do not by themselves calibrate a shared unknown clock if every output is observed on that same clock. A known event time, an independently measured rate, or a calibrated observation schedule can supply the missing reference. If no such reference exists, it may be more informative to report rates in instrument-time units or estimate a rate relative to a specified reference rate than to present an unsupported absolute rate.

Common Pitfall

A common error is to say, “The data are dense and the trajectory is smooth, so the absolute rate must be identifiable.” More observations can define the curve more precisely without telling us how its time axis maps to hours or seconds. The correction is to test whether a change of clock paired with the corresponding change of rates preserves every observed output. If it does, denser observations on that same uncalibrated clock do not alone resolve the absolute scale. Conversely, do not claim ambiguity when the experiment includes a known time reference: a fixed, independently calibrated input schedule may make the transformed model inconsistent with the design.

Key takeaway: Under a rescaling \(s=ct\), rate constants that multiply the dynamic rate laws transform as \(k'_j=k_j/c\). This creates an absolute-rate ambiguity only when the clock conversion and relevant input timing are not independently anchored and the model otherwise admits the transformation.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. If \(s=3t\) and a model has rate constants \(k_1=0.60\ \mathrm{h}^{-1}\) and \(k_2=0.15\ \mathrm{h}^{-1}\), what are the transformed rates? What happens to their ratio?
  2. For \(dX/dt=-kX\), explain why an unknown time conversion can confound \(k\) with the clock even if many exact observations are available.
  3. A known input pulse ends at 5 hours. Under \(s=2t\), when must the transformed pulse end to preserve the same experiment? Why can a separately known end time of 5 in \(s\)-units break this transformation?
  4. Why is a unique-trajectory conclusion justified for a locally Lipschitz ODE but not automatically for every possible state-dependent right-hand side?

Solutions. (1) \(k'_1=0.60/3=0.200\) and \(k'_2=0.15/3=0.0500\), in inverse \(s\)-units; their ratio remains \(0.200/0.0500=4\), equal to \(0.60/0.15\). (2) The same trajectory \(X(t)=X(0)e^{-kt}\) is written in \(s\) as \(X(0)e^{-(k/ c)s}\); without knowing \(c\), the observations determine the rate relative to that clock, not necessarily \(k\) per physical time unit. (3) The transformed pulse must end at \(s=2(5)=10\). If its end is independently known to be \(s=5\), the transformed schedule does not match the known experiment, so this \(c=2\) transformation does not preserve it. (4) Local Lipschitz continuity in the state ensures uniqueness of the initial-value solution on the relevant domain. Without a uniqueness condition, the same initial-value problem can have multiple solutions, so equality of the problem statements alone does not establish equality of a unique trajectory.

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