Why Assumptions Belong in an Identifiability Claim
A claim that a model’s parameters are identifiable is incomplete unless it says what is known about the experiment. An assumed known dose, an assumed zero peripheral amount at the start, or a calibrated concentration measurement can each constrain the model’s possible explanations of the observations. Change one of those assumptions, and the conclusion may change.
This tutorial uses a linear two-compartment model to make those dependencies explicit. The point is not that one set of assumptions is always correct: it is that the assumptions must match the experimental design and be stated before interpreting an identifiability result. The earlier material on global and local structural identifiability concerns uniqueness under ideal observations; here the focus is how that uniqueness depends on specified inputs, initial conditions, and observation functions.
Model and Assumptions
Let \(A_c(t)\) and \(A_p(t)\) denote amounts of drug in the central and peripheral compartments, measured in mg. Let \(k_{10}\) be the first-order central elimination rate constant, \(k_{12}\) the central-to-peripheral transfer rate constant, and \(k_{21}\) the peripheral-to-central transfer rate constant. Each rate constant is in h\(^{-1}\). The input \(u(t)\), in mg/h, enters the central compartment.
For the derivation and first examples, assume the rate constants are positive and constant, the model structure is correct, and a known bolus has already been given at time zero. Thus \(u(t)=0\) for \(t>0\), and the initial amounts \(A_c(0)=A_{c0}\) and \(A_p(0)=A_{p0}\) describe the state just after the bolus. Assume the observation is the central amount itself, \(y(t)=A_c(t)\), so its scale and units are known. We also use ideal, continuous, noise-free observations over a time interval. These are structural assumptions, not claims that real data are perfect.
A Direct Relation Between the Output and the Rates
Write \(a=k_{10}+k_{12}\). With no input after time zero, differentiating the central equation and using the peripheral equation gives a second-order equation for the observed central amount:
From the central equation, \(k_{21}A_p=A_c'+aA_c\). Substituting this expression into the differentiated equation yields
The output trajectory therefore contains information about \(S\) and \(P\), but these two quantities alone do not generally specify all three rate constants. Known initial amounts supply additional information. The initial derivative is
Since \(S=a+k_{21}\), rearranging this derivative equation gives a useful recovery formula:
This recovery requires \(A_{c0}+A_{p0}\ne 0\), known initial amounts, and enough output information to determine \(S\), \(P\), and \(r\). For nonnegative amounts, a nonzero total initial amount satisfies the denominator condition. If both initial amounts are zero and there is no later input, the output is identically zero and provides no rate information.
Worked Example: Recovering Rates After a Central Bolus
Suppose a hypothetical bolus produces \(A_{c0}=100\) mg and \(A_{p0}=0\) mg. The post-bolus central amount is observed to have initial slope \(r=-50\) mg/h, and its ideal biexponential trajectory gives \(S=0.90\) h\(^{-1}\) and \(P=0.080\) h\(^{-2}\). Assume the stated amounts are known and the model assumptions above hold.
First, the known initial state gives \(a=-r/A_{c0}=50/100=0.50\) h\(^{-1}\). Therefore,
Next, use \(P=k_{10}k_{21}\), then \(a=k_{10}+k_{12}\):
These values reproduce the supplied quantities: \(S=0.20+0.30+0.40=0.90\) h\(^{-1}\), \(P=0.20(0.40)=0.080\) h\(^{-2}\), and \(r=-(0.20+0.30)(100)=-50\) mg/h. Under the stated assumptions, the three rates are recovered.
The Zero-Central-Amount Case
Initial-state formulas need to be checked at boundary cases rather than divided through automatically. For the no-input model, taking the Laplace transform gives
When \(A_{c0}\ne 0\), the ratio of the numerator’s constant coefficient to its coefficient of \(s\) is \(k_{21}(A_{c0}+A_{p0})/A_{c0}\). That ratio is not in general \(k_{21}\); moreover, the ratio is defined only when \(A_{c0}\ne 0\). If \(A_{c0}=0\), the coefficient of \(s\) in the numerator is zero, so that ratio is undefined. The case must be handled directly, not by substituting zero into a divided expression.
Worked Example: Starting With Drug Only in the Peripheral Compartment
Suppose a hypothetical experiment has known initial amounts \(A_{c0}=0\) mg and \(A_{p0}=80\) mg. From the ideal central-amount trajectory, suppose \(S=0.90\) h\(^{-1}\), \(P=0.080\) h\(^{-2}\), and the initial slope is \(r=32\) mg/h. The slope is positive because drug moves from the initially occupied peripheral compartment into the central compartment.
Here the Laplace-transform numerator is \(k_{21}(80)\), a constant, with no \(s\)-coefficient. Its constant-to-\(s\)-coefficient ratio does not exist. The derivative-based recovery formula remains valid because the total initial amount is 80 mg:
Then \(a=S-k_{21}=0.90-0.40=0.50\) h\(^{-1}\), \(k_{10}=P/k_{21}=0.080/0.40=0.20\) h\(^{-1}\), and \(k_{12}=a-k_{10}=0.50-0.20=0.30\) h\(^{-1}\). The calculation works because the peripheral initial amount is known and nonzero, not because an undefined coefficient ratio has a value.
What Changes When Initial Amounts Are Unknown?
If the central output is measured in amount units but the initial peripheral amount is unknown, the output’s initial value and slope no longer fix the three rates by the preceding formula. Distinct parameter values can produce the same output whenever they give the same \(S\), \(P\), central initial amount, and central initial slope. The following example verifies that this is an exact ambiguity, not merely a similar-looking fit.
Worked Example: Two Rate Sets With the Same Central Output
Take the central trajectory from the first example: \(A_{c0}=100\) mg, \(r=-50\) mg/h, \(S=0.90\) h\(^{-1}\), and \(P=0.080\) h\(^{-2}\). One explanation is \(k_{10}=0.20\), \(k_{12}=0.30\), \(k_{21}=0.40\) h\(^{-1}\), with \(A_{p0}=0\) mg. Now consider a second explanation:
Choose its unknown initial peripheral amount to match the observed slope. With \(a=k_{10}+k_{12}=0.70\) h\(^{-1}\), the required amount is
For this second explanation, \(S=0.40+0.30+0.20=0.90\) h\(^{-1}\), \(P=0.40(0.20)=0.080\) h\(^{-2}\), and \(A_c'(0)=-0.70(100)+0.20(100)=-50\) mg/h. Both explanations have the same central initial amount and derivative and satisfy the same second-order equation. A second-order linear equation with the same coefficients and the same initial value and derivative has the same solution, so they generate the same central output for all \(t\ge 0\). With the peripheral initial amount unreported and unknown, these observations do not uniquely distinguish the two rate sets.
Observation Functions and Experimental Inputs
The observation function also belongs in the assumptions. If the measured quantity is central concentration \(y=A_c/V_c\), with known central volume \(V_c\), the amount can be reconstructed as \(A_c=V_cy\), and the preceding amount-based calculation applies. If \(V_c\) is unknown, the output is on an unknown scale. That does not automatically make the rate constants unidentifiable: multiplying the whole central trajectory by a constant leaves its homogeneous differential equation unchanged. But the amount scale or volume cannot be inferred from that shape alone when the administered amount is also unknown.
Worked Example: Unknown Concentration Scale, Known Rate Information
Suppose the observation is \(y=qA_c\), where the positive scale \(q\) is unknown and has units of L\(^{-1}\). The initial state has \(A_{p0}=0\), and both compartments follow the same no-input model. If the measured trajectory has \(y(0)=5.0\) mg/L and \(y'(0)=-2.5\) mg/L/h, then its normalized initial slope is
The scale cancels. If the trajectory also determines \(S=0.90\) h\(^{-1}\) and \(P=0.080\) h\(^{-2}\), the calculations give \(k_{21}=S-0.50=0.40\) h\(^{-1}\), \(k_{10}=P/k_{21}=0.20\) h\(^{-1}\), and \(k_{12}=0.30\) h\(^{-1}\). The rates can be recovered under these assumptions even though the observation scale is unknown. The scale itself is not determined by this rate calculation; an independently known dose or calibrated observation may supply the additional information needed to determine it.
A known, time-varying input can also be part of an identifiability analysis, but it must be represented in the model. The homogeneous equation derived above applies after the input has stopped. An unrecorded input is different: it is an additional unknown function that may explain changes in the output otherwise attributed to the rates. Likewise, a single output that fails to excite a model mode may not reveal all the information available under a different initial state or input profile.
Checks, Alternatives, and Limitations
Before using a rate-recovery calculation, check that the stated observation matches the modeled quantity, the input history is known, and the initial-state assumptions are justified. Then check that the observed trajectory contains enough information to estimate its dynamic coefficients. In this model, a single visible exponential mode cannot generally provide both \(S\) and \(P\); the argument above relies on a trajectory that reveals the second-order dynamics. Even when parameters are structurally identifiable, finite, noisy, sparsely sampled data can make estimates unstable. That is a practical estimation problem distinct from the ideal uniqueness question.
The direct time-domain derivation here is one approach. A mechanistic fit to the full state equations can incorporate a known input and observation function directly, while an input-output equation emphasizes what the measured trajectory constrains. Neither approach makes an unknown dose, an unmeasured initial amount, or an uncalibrated observation scale known by calculation. More measurements or a better justified experimental design may be needed. Structural conclusions are also specific to this model: adding peripheral elimination, nonlinear exchange, or time-varying rates changes the equations and requires a new analysis.
Check Your Understanding
Answer each question using the assumptions stated in the question; brief solutions follow.
- For \(A_{c0}=60\) mg, \(A_{p0}=0\) mg, \(r=-18\) mg/h, \(S=0.50\) h\(^{-1}\), and \(P=0.040\) h\(^{-2}\), recover the three rate constants.
- If \(A_{c0}=0\) and \(A_{p0}=40\) mg, why is the constant-to-\(s\)-coefficient ratio of the Laplace numerator unavailable? What quantity can be used instead to recover \(k_{21}\)?
- Why can two different rate sets produce the same central output if the initial peripheral amount is unknown?
- Does an unknown constant multiplier in the observation \(y=qA_c\) necessarily prevent recovery of the rate constants? State a condition under which it cancels.
Solutions. (1) \(a=-r/A_{c0}=18/60=0.30\) h\(^{-1}\), so \(k_{21}=S-a=0.20\) h\(^{-1}\), \(k_{10}=P/k_{21}=0.040/0.20=0.20\) h\(^{-1}\), and \(k_{12}=a-k_{10}=0.10\) h\(^{-1}\). (2) The numerator is \(k_{21}A_{p0}\), with zero coefficient of \(s\), so the ratio is undefined. Use \(r=k_{21}A_{p0}\), giving \(k_{21}=r/40\) when the slope is known. (3) An unknown initial peripheral amount can adjust the initial slope for a different choice of rates; if \(S\), \(P\), central initial amount, and slope remain the same, the central output solves the same second-order equation with the same initial conditions. (4) No. For a no-input trajectory with \(A_{p0}=0\) and \(A_{c0}>0\), \(y'(0)/y(0)=-(k_{10}+k_{12})\), so the unknown constant multiplier cancels; this does not by itself determine that multiplier or an absolute amount scale.