Linear Compartment Model Identifiability Workshop

Why the Two-Compartment Bolus Model Is a Useful Identifiability Test

A central concentration curve after a bolus dose often appears to offer several distinct features: an early decline, a later terminal decline, and an amplitude. The identifiability question is whether these features uniquely determine the model parameters, or whether different parameter values can generate exactly the same curve. For a linear two-compartment model, that question can be answered by deriving the curve and matching its exponential terms to the rate constants.

The analysis here concerns structural identifiability: whether ideal, noise-free observations over time uniquely determine a parameter, given the model and its assumptions. It does not imply that finite, noisy measurements will estimate the parameter precisely.

Model and assumptions: Let \(A_1(t)\) and \(A_2(t)\) be drug amounts in the central and peripheral compartments. The dose \(D\) is delivered instantaneously into the central compartment at \(t=0\), so \(A_1(0)=D\) and \(A_2(0)=0\). Drug leaves the central compartment at first-order rate \(k_{10}\), transfers from central to peripheral at rate \(k_{12}\), and returns at rate \(k_{21}\). The observed output is central concentration \(C(t)=A_1(t)/V_c\), with a constant, positive central volume \(V_c\). The model is linear, time-invariant, and has strictly positive \(k_{10}, k_{12}, k_{21}, V_c\); the dose is known and nonzero.

The governing equations are

\frac{dA_1}{dt}=-(k_{10}+k_{12})A_1+k_{21}A_2,\qquad \frac{dA_2}{dt}=k_{12}A_1-k_{21}A_2.

The model treats the dose as an instantaneous initial condition, not as a continuing input. It also assumes that the measured concentration is on an absolute scale: there is no unknown multiplicative assay factor. If those conditions change, the mapping from the observed curve to parameters can change as well.

Deriving the Output Curve

The system matrix has characteristic polynomial

\lambda^2+(k_{10}+k_{12}+k_{21})\lambda+k_{10}k_{21}.

Its two eigenvalues are negative. Write them as \(-\alpha\) and \(-\beta\), with \(\alpha>\beta>0\). Comparing coefficients gives

\alpha+\beta=k_{10}+k_{12}+k_{21},\qquad \alpha\beta=k_{10}k_{21}.

Because \(k_{12}>0\), the two decay rates are distinct, and the central output contains both modes. Thus the concentration has the form \(C(t)=A e^{-\alpha t}+B e^{-\beta t}\), where \(A\) and \(B\) are exponential coefficients. The initial condition gives \(A+B=D/V_c=C(0)\). Differentiating the output at zero gives a second relation:

-\alpha A-\beta B=C'(0) =-\frac{(k_{10}+k_{12})D}{V_c}.

Define \(f=B/(A+B)\), the fraction of the initial concentration assigned to the slower exponential. Then \(A/(A+B)=1-f\). The initial-slope equation and the eigenvalue sum yield the following inverse mapping:

k_{10}+k_{12}=(1-f)\alpha+f\beta =\alpha-f(\alpha-\beta),

k_{21}=\beta+f(\alpha-\beta),\qquad k_{10}=\frac{\alpha\beta}{k_{21}},\qquad k_{12}=\alpha-f(\alpha-\beta)-k_{10}.

The volume is recovered from the initial concentration when \(D\) is known:

V_c=\frac{D}{A+B}.

These equations are not merely a way to fit a biexponential curve. They show explicitly how the observable curve maps back to the model parameters, and therefore allow the uniqueness question to be checked.

Identifiability result: If \(D\ne 0\) is known, \(V_c\) and all three rate constants are strictly positive, and the complete noise-free central concentration curve is known for \(t\geq 0\), then \(V_c,k_{10},k_{12},k_{21}\) are uniquely determined. The curve determines its two distinct exponential rates and coefficients; those determine \(f\), and the inverse equations above determine each rate constant and \(V_c\). A zero dose would instead give \(C(t)=0\) for every parameter choice, so it cannot identify the rates or volume.

Worked Examples: Matching the Curve to the Model

Worked Example: Recovering All Parameters From the Full Curve

Suppose a hypothetical noise-free central concentration curve after a known \(100\ \mathrm{mg}\) bolus is

C(t)=7.5e^{-1.2t}+2.5e^{-0.1t}\ \mathrm{mg/L},

where time is in hours and the coefficients are in \(\mathrm{mg/L}\). The exponential rates are \(\alpha=1.2\ \mathrm{h}^{-1}\) and \(\beta=0.1\ \mathrm{h}^{-1}\), so the initial concentration is \(A+B=10\ \mathrm{mg/L}\), and \(f=B/(A+B)=2.5/10=0.25\).

The volume is \(V_c=D/C(0)=100\ \mathrm{mg}/(10\ \mathrm{mg/L})=10\ \mathrm{L}\). Next,

k_{21}=0.1+0.25(1.2-0.1)=0.375\ \mathrm{h}^{-1},

k_{10}=\frac{(1.2)(0.1)}{0.375} =\frac{0.12}{0.375} =0.320\ \mathrm{h}^{-1},

and

k_{12}=1.2-0.25(1.1)-0.320 =0.925-0.320 =0.605\ \mathrm{h}^{-1}.

All inferred rates are positive, as required. As a check, their sum is \(0.320+0.605+0.375=1.300\ \mathrm{h}^{-1}\), equal to \(\alpha+\beta=1.2+0.1=1.3\ \mathrm{h}^{-1}\). Their product \(k_{10}k_{21}=(0.320)(0.375)=0.120\ \mathrm{h}^{-2}\) equals \(\alpha\beta=(1.2)(0.1)=0.12\ \mathrm{h}^{-2}\). Under the stated assumptions, the full curve identifies these parameters.

Worked Example: What a Normalized Curve Does and Does Not Identify

Suppose the only available curve is normalized by its initial value:

\frac{C(t)}{C(0)}=0.75e^{-1.2t}+0.25e^{-0.1t}.

This still supplies \(\alpha=1.2\ \mathrm{h}^{-1}\), \(\beta=0.1\ \mathrm{h}^{-1}\), and \(f=0.25\). The rate calculations are therefore the same as in the first example: \(k_{21}=0.375\ \mathrm{h}^{-1}\), \(k_{10}=0.320\ \mathrm{h}^{-1}\), and \(k_{12}=0.605\ \mathrm{h}^{-1}\). But normalization removes the absolute value of \(C(0)\). Without that value, \(V_c=D/C(0)\) cannot be calculated, even if the dose is known.

If both dose and volume are unknown, the absolute curve identifies only their ratio \(D/V_c\), not the two quantities separately. For example, a hypothetical dose of \(100\ \mathrm{mg}\) with \(V_c=10\ \mathrm{L}\), and a dose of \(200\ \mathrm{mg}\) with \(V_c=20\ \mathrm{L}\), both give \(C(0)=10\ \mathrm{mg/L}\). With the same rate constants, they produce the same concentration curve. The exponential shape identifies the rates, but the scale alone cannot separate dose from volume.

Worked Example: Terminal-Phase Data Do Not Determine the Fast Rate

Suppose hypothetical late-time observations establish a terminal component \(2.5e^{-0.1t}\ \mathrm{mg/L}\), and the initial concentration is known to be \(10\ \mathrm{mg/L}\). These observations determine \(\beta=0.1\ \mathrm{h}^{-1}\), \(B=2.5\ \mathrm{mg/L}\), and \(f=0.25\), but they do not determine the fast rate \(\alpha\).

One possible completion is \(\alpha=1.2\ \mathrm{h}^{-1}\), which gives the first example’s rates \(k_{21}=0.375\), \(k_{10}=0.320\), and \(k_{12}=0.605\), all in \(\mathrm{h}^{-1}\). A different completion, \(\alpha=1.0\ \mathrm{h}^{-1}\), gives

k_{21}=0.1+0.25(1.0-0.1)=0.325\ \mathrm{h}^{-1},

k_{10}=\frac{(1.0)(0.1)}{0.325} =0.307692\ \mathrm{h}^{-1},

k_{12}=1.0-0.25(0.9)-0.307692 =0.467308\ \mathrm{h}^{-1}.

Both parameter sets have positive rates and the same terminal component and initial concentration, but their fast exponential terms differ: one is \(7.5e^{-1.2t}\), the other \(7.5e^{-1.0t}\), in \(\mathrm{mg/L}\). If only the terminal phase is observed, the fast component is absent from the available information. The full curve is needed to distinguish these possibilities.

Why the Inverse Mapping Is Unique

The key step is that a sum of two distinct exponentials with nonzero coefficients has a unique representation when the rates are ordered. One way to see this is to examine the late-time limit. For \(C(t)=Ae^{-\alpha t}+Be^{-\beta t}\), with \(\alpha>\beta\) and \(B\ne0\),

\lim_{t\to\infty}-\frac{d}{dt}\log|C(t)|=\beta, \qquad \lim_{t\to\infty}e^{\beta t}C(t)=B.

Subtracting \(Be^{-\beta t}\) from the known curve leaves \(Ae^{-\alpha t}\), from which \(\alpha\) and \(A\) are determined when \(A\ne0\). Thus the full curve fixes \(\alpha,\beta,A,B\), and therefore \(C(0)=A+B\) and \(f=B/(A+B)\). The inverse formulas then fix each rate constant. Finally, the known nonzero dose and nonzero \(C(0)\) fix \(V_c\). This argument is for the ideal complete curve; estimating the late-time limit from a short or noisy record is a different problem.

The positivity assumptions support the interpretation of the modes and the inverse mapping. In the model, \(\alpha>\beta>0\), and with positive transfer in both directions and positive elimination, both exponential coefficients are nonzero. The recovered rates must also be positive. If data imply a zero or negative rate through the matching equations, then the assumed model and observations are not mutually consistent; numerical fitting should not be used to disguise that conflict.

Diagnostics, Alternatives, and Limitations

A useful structural check is to reconstruct the curve from the inferred parameters and verify both coefficient identities and the rate identities \(\alpha+\beta=k_{10}+k_{12}+k_{21}\) and \(\alpha\beta=k_{10}k_{21}\). Also check that \(A+B=C(0)\), that the initial slope agrees with \(-(k_{10}+k_{12})C(0)\), and that all rates and \(V_c\) satisfy the stated positivity conditions. These checks detect algebraic or unit errors, but they do not establish that real measurements are noise-free or that the model is appropriate.

The main alternatives depend on what is observed. With the full absolute concentration curve and known dose, direct coefficient matching identifies all four parameters under these assumptions. With a normalized curve, the rate constants remain identifiable but the volume does not. With unknown dose and volume, the ratio \(D/V_c\) is available from the amplitude, but the two scale parameters are confounded. With terminal-phase measurements alone, only the slow rate and its coefficient can be determined; the fast rate and the corresponding rates need not be unique. A single concentration measurement, or a small set of measurements that does not resolve both phases, generally supplies less information still.

Structural identifiability should not be confused with practical estimability. Even though the full ideal curve uniquely determines \(\alpha\) and \(\beta\), finite observations may not resolve a fast phase that occurs before the first sample, or two phases that are close relative to measurement precision. Such data limitations can make parameter estimates unstable without changing the structural result. Likewise, an unknown assay scale, non-instantaneous dosing, additional distribution compartments, or time-varying clearance violates assumptions used in this derivation and requires a different model and identifiability analysis.

A common mistake is to conclude that observing a terminal slope identifies the model’s rate constants because the terminal rate is clearly visible. The terminal slope determines \(\beta\), not all three rate constants: the second worked example shows two positive parameter sets with the same terminal component and initial concentration. The correction is to match the entire resolved curve, including its fast component, and to distinguish rate information from amplitude-scale information.

Key takeaway: For a known, nonzero bolus and a complete absolute central concentration curve, the ordered exponential rates and coefficients determine \(V_c,k_{10},k_{12},k_{21}\) uniquely in the positive-rate two-compartment model. Removing the absolute scale or the fast phase removes specific information and can leave parameters confounded.

Check Your Understanding

Use the stated model assumptions. Solutions follow each question so that the identifiability reasoning can be checked directly.

If \(D=0\), what concentration curve is observed, and can it identify any of the rates?
Solution: The initial amounts are both zero and the linear system remains at zero, so \(C(t)=0\) for all \(t\). This curve is identical for every positive choice of rates and volume; it identifies none of them.
A full normalized curve gives \(\alpha=0.8\ \mathrm{h}^{-1}\), \(\beta=0.2\ \mathrm{h}^{-1}\), and \(f=0.3\). Calculate \(k_{21}\).
Solution: \(k_{21}=\beta+f(\alpha-\beta)=0.2+0.3(0.8-0.2)=0.2+0.18=0.38\ \mathrm{h}^{-1}\).
If the full absolute curve is known but \(D\) is unknown, which quantity is obtained from the initial concentration?
Solution: Since \(C(0)=D/V_c\), the curve supplies the ratio \(D/V_c\). Without an independent value for dose or volume, it cannot separate them. The exponential shape can still identify the rates under the model assumptions.
Why is a known terminal slope alone insufficient to identify \(k_{10},k_{12},k_{21}\)?
Solution: The terminal slope supplies \(\beta\), but the inverse formulas also require the fast rate \(\alpha\) and the relative coefficient \(f\). Different positive choices of \(\alpha\) can yield different rate constants while preserving the same \(\beta\); terminal-phase information alone does not fix the full curve.
For a complete curve with \(A+B=8\ \mathrm{mg/L}\) and known dose \(D=80\ \mathrm{mg}\), what is \(V_c\)?
Solution: \(V_c=D/(A+B)=80\ \mathrm{mg}/(8\ \mathrm{mg/L})=10\ \mathrm{L}\). This volume calculation uses the absolute initial concentration and the known nonzero dose.