From Three Compartments to Three Exponential Terms
A central concentration-time profile after an intravenous bolus can decline at several rates: a rapid early decline, a slower intermediate decline, and a still slower terminal decline. In a linear three-compartment model, these phases can be represented together by a tri-exponential curve. The curve is useful as a compact description of the observed profile; its coefficients are not, by themselves, a list of the model’s physiological processes.
Assume a linear mammillary model with one central compartment and two peripheral compartments, constant positive elimination and exchange rate constants, and drug administered as a bolus into the central compartment. Let \(D\) be the dose, \(V_c\) the central volume, and \(C_c(t)\) the central concentration at time \(t\). There is no absorption step, and the measured concentration is assumed to be the central concentration. The earlier material on the three-compartment model establishes its amount-balance equations and exponential eigenmodes; here we focus on the curve coefficients and what they summarize.
Each amplitude is the value that its exponential term would contribute at \(t=0\), since \(e^0=1\). It is not generally the concentration at a later time, and it is not the amount of drug in one compartment. At a time \(t\), for example, the first term contributes \(A e^{-\alpha t}\) to the total concentration. The decay rate describes how quickly that term decreases: a larger rate gives a more rapid decline. The reciprocal \(1/\alpha\) is a time scale for that term, not a compartment volume or a clearance.
These terms arise because the three-compartment amount equations form a linear system with three eigenmodes. Solving that system gives a sum of exponentials in the central concentration. The eigenvalues determine the decay rates, while the model and initial condition determine the amplitudes. Thus the tri-exponential curve summarizes the combined central response of the system. It does not assign each exponential to a separate physical compartment.
Initial-Condition Checks
The curve has two immediate checks that follow from the bolus initial condition and the amount-balance equations. Write \(X_c(t)\) for central amount and \(k_{10}\), \(k_{12}\), and \(k_{13}\) for central elimination and outward transfer rate constants. Immediately after the bolus, \(X_c(0)=D\), and the peripheral amounts are zero. Concentration is \(C_c(t)=X_c(t)/V_c\), so the curve must start at \(D/V_c\).
At time zero, there is no drug in the peripheral compartments to transfer back into the central compartment. However, drug can still leave the central compartment by elimination and by transfer out to either peripheral compartment. The central amount slope is therefore \(X_c'(0)=-(k_{10}+k_{12}+k_{13})D\), not merely \(-k_{10}D\). Dividing by \(V_c\) gives the initial concentration slope.
The second equality follows by differentiating each exponential term and setting \(t=0\). These identities provide necessary checks on a proposed curve and parameter set. They do not prove that the full curve matches a particular three-compartment model: the entire time course, not just its starting value and slope, must agree for that stronger conclusion.
Worked Example: Reading a Tri-Exponential Curve
Suppose a hypothetical central concentration curve is $$ C_c(t)=8.00e^{-1.20t}+3.00e^{-0.200t}+1.00e^{-0.0200t}\ \text{mg/L}, $$ with \(t\) measured in hours. Identify the amplitudes and decay rates, check the starting concentration, and calculate the concentration at \(t=4.00\) h.
The amplitudes are \(A=8.00\), \(B=3.00\), and \(C=1.00\) mg/L. The decay rates are \(\alpha=1.20\), \(\beta=0.200\), and \(\gamma=0.0200\) h\(^{-1}\). Their ordering is fastest to slowest, as required. At time zero, $$ C_c(0)=8.00+3.00+1.00=12.0\ \text{mg/L}. $$
At \(t=4.00\) h, calculate the three contributions separately: $$ 8.00e^{-(1.20)(4.00)}=8.00e^{-4.80}\approx0.0658\ \text{mg/L}, $$ $$ 3.00e^{-(0.200)(4.00)}=3.00e^{-0.800}\approx1.348\ \text{mg/L}, $$ $$ 1.00e^{-(0.0200)(4.00)}=1.00e^{-0.0800}\approx0.9231\ \text{mg/L}. $$ Adding them gives \(C_c(4.00)\approx0.0658+1.348+0.9231=2.337\) mg/L. The fast term has become small, while the slower terms still contribute substantially. This calculation describes the terms’ contributions; it does not show that any one term equals the concentration in a particular compartment.
Deriving the Checks and Interpreting the Phases
The initial-condition identities require only the curve form and the central amount balance. They can be verified directly, without solving for the eigenvalues. At \(t=0\), each exponential equals one, giving \(A+B+C\). For the slope, differentiation gives $$ \frac{dC_c(t)}{dt}=-\alpha A e^{-\alpha t}-\beta B e^{-\beta t}-\gamma C e^{-\gamma t}. $$ Setting \(t=0\) yields \(-(\alpha A+\beta B+\gamma C)\). Independently, the central amount-balance equation at \(t=0\) has zero inward-transfer terms because both peripheral amounts start at zero. Its outward-transfer and elimination terms remain, giving $$ X_c'(0)=-(k_{10}+k_{12}+k_{13})D. $$ Dividing by \(V_c\) and equating the two expressions for \(C_c'(0)\) proves the slope check.
The derivative identity is a weighted sum: the fast rate is multiplied by its amplitude, the intermediate rate by its amplitude, and the slow rate by its amplitude. It is not correct to treat the initial slope as elimination alone. Distribution out of the central compartment contributes immediately, even though no drug has yet returned from the peripheral compartments.
The phases are often useful for describing what happens over different portions of the profile. Early after a bolus, a large, rapidly declining term may contribute strongly. Later, that term can be negligible while a slower term dominates. The labels “fast,” “intermediate,” and “terminal” describe the relative decline rates of terms in the central concentration curve. They do not, on their own, specify a distinct anatomical site or a single physiological mechanism for each term.
Worked Example: Checking a Curve Against the Initial Slope
Consider a hypothetical dose \(D=120\) mg and central volume \(V_c=10.0\) L. Suppose the central rate constants are \(k_{10}=0.100\) h\(^{-1}\), \(k_{12}=0.500\) h\(^{-1}\), and \(k_{13}=0.250\) h\(^{-1}\). A proposed tri-exponential curve has amplitudes \(8.00\), \(3.00\), and \(1.00\) mg/L, with rates \(1.1975\), \(0.200\), and \(0.0200\) h\(^{-1}\).
First, its initial concentration is \(8.00+3.00+1.00=12.0\) mg/L. The bolus and volume imply \(D/V_c=120/10.0=12.0\) mg/L, so the initial-concentration check passes.
Next, the curve’s weighted rate sum is $$ (1.1975)(8.00)+(0.200)(3.00)+(0.0200)(1.00) =9.580+0.600+0.0200=10.20\ \text{mg/L/h}. $$ Thus the curve’s initial concentration slope is \(-10.20\) mg/L/h. The amount-balance rates predict $$ (k_{10}+k_{12}+k_{13})\frac{D}{V_c} =(0.100+0.500+0.250)(12.0) =(0.850)(12.0)=10.20\ \text{mg/L/h}. $$ The predicted slope is also \(-10.20\) mg/L/h, so the initial-slope check passes. Passing these two checks establishes consistency with these initial conditions, but does not establish that the curve matches the full model at every time.
Worked Example: Comparing Contributions Later in Time
Use the curve from the first worked example to find each term’s contribution at \(t=10.0\) h and the corresponding total. Also calculate the half-life of each individual exponential term.
The contributions at \(10.0\) h are $$ 8.00e^{-(1.20)(10.0)}\approx0.0000491\ \text{mg/L}, $$ $$ 3.00e^{-(0.200)(10.0)}=3.00e^{-2.00}\approx0.406\ \text{mg/L}, $$ $$ 1.00e^{-(0.0200)(10.0)}=e^{-0.200}\approx0.819\ \text{mg/L}. $$ Their sum is approximately \(1.225\) mg/L. At this time the fastest term contributes very little; the two slower terms account for nearly all of the total.
For a single exponential \(A e^{-\alpha t}\), its term has halved when \(e^{-\alpha t}=1/2\), so its term-specific half-life is \(t_{1/2}=\ln(2)/\alpha\). The three values are \(\ln(2)/1.20=0.578\) h, \(\ln(2)/0.200=3.47\) h, and \(\ln(2)/0.0200=34.7\) h. These are half-lives of the separate exponential terms. The sum of the terms is not generally a single exponential, so these values are not, individually, the half-life of the total concentration curve.
Choosing a Representation and Recognizing Its Limits
A tri-exponential curve and a compartmental differential-equation model are two ways to represent the same type of concentration-time behavior, but they emphasize different quantities. The exponential form makes phase contributions and decay rates explicit. The compartmental form specifies volumes, clearances or rate constants, and the directions of transfer between amounts. Neither representation should be interpreted beyond the information it contains.
| Representation | What it states directly | Useful check or caution |
|---|---|---|
| Tri-exponential curve | Three amplitudes and three decay rates for central concentration over time | Check \(A+B+C=D/V_c\) and the initial weighted rate sum against the amount balance |
| Compartment amount equations | How elimination and transfer terms change each compartment amount | Use the model equations to account for all outward and inward flows at the initial time |
The curve representation is compact, but its coefficients depend on the model and the observation being described. The amplitude is not an amount, and a decay rate is not automatically a directly measured physiological rate constant. Also, the starting value and slope are only two constraints; they cannot verify all six curve coefficients or all model parameters. A curve should therefore be checked against its stated assumptions and, where a compartment model is claimed, against the model’s full predicted profile.
Practical warning signs include a proposed set of rates that is not ordered as claimed, inconsistent units (for example, rates in h\(^{-1}\) used with time in days), or amplitudes whose sum disagrees materially with the known bolus concentration \(D/V_c\). A slope mismatch can reveal an omitted outward transfer term or an inconsistent parameter set. Conversely, matching the initial checks does not ensure that later predictions are correct. Those checks are useful diagnostics, not a substitute for evaluating the full concentration-time curve.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A hypothetical curve is \(C_c(t)=5.00e^{-0.800t}+2.00e^{-0.100t}+1.00e^{-0.0100t}\) mg/L, with \(t\) in hours. State the amplitudes and decay rates, and find \(C_c(0)\).
- For the curve in question 1, calculate the magnitude of its initial slope in mg/L/h.
- A hypothetical bolus dose is 90.0 mg and \(V_c=10.0\) L. A proposed curve’s amplitudes sum to 8.00 mg/L. Does its initial value pass the bolus check? Explain.
- Why does the central concentration’s initial decline include \(k_{12}\) and \(k_{13}\), as well as \(k_{10}\), when the peripheral amounts are initially zero?
Solutions. (1) The amplitudes are 5.00, 2.00, and 1.00 mg/L; the rates are 0.800, 0.100, and 0.0100 h\(^{-1}\). The initial concentration is \(5.00+2.00+1.00=8.00\) mg/L. (2) The weighted sum is \((5.00)(0.800)+(2.00)(0.100)+(1.00)(0.0100)=4.210\) mg/L/h, so the initial slope is \(-4.210\) mg/L/h. (3) The dose and volume imply \(D/V_c=90.0/10.0=9.00\) mg/L. Since the amplitudes sum to 8.00 mg/L, the proposed curve fails the initial-value check by 1.00 mg/L. (4) Zero peripheral amounts make the inward transfer terms zero at time zero, but drug is already in the central compartment and can flow outward through both pathways. Thus \(X_c'(0)=-(k_{10}+k_{12}+k_{13})D\), and the concentration slope is that amount slope divided by \(V_c\).