Why Three Compartments Produce Three Decay Rates
In a three-compartment model, drug can move among a central compartment and two peripheral compartments while elimination occurs from the central compartment. The amount-balance equations describe those transfers directly. The eigenvalues of that system reveal a different feature: the characteristic decay rates that combine to form the predicted concentration-time profile.
This distinction matters when interpreting a profile. The three rates are not simply the three exchange or elimination rate constants. Each is determined by the system of coupled equations as a whole. When the rates are distinct, the central concentration after an intravenous bolus can be represented as a sum of three exponential terms. The size of each term also matters: an eigenvalue supplies a rate, while the corresponding amplitude determines how much that mode contributes to the observed profile.
From Amount-Balance Equations to the Characteristic Polynomial
Use the mammillary structure with central amount \(A_1\), peripheral amounts \(A_2\) and \(A_3\), and positive first-order rate constants \(k_{10}, k_{12}, k_{21}, k_{13}, k_{31}\), all in inverse-time units. Here \(k_{10}\) is the central elimination rate constant, and each pair of exchange constants describes movement in opposite directions between the central and one peripheral compartment. The amount-balance equations are
Write these equations as \(d\mathbf{A}/dt=-B\mathbf{A}\), where \(\mathbf{A}=(A_1,A_2,A_3)^\mathsf{T}\) and
A mode proportional to \(e^{-\lambda t}\) satisfies \(B\mathbf{v}=\lambda\mathbf{v}\) for a nonzero vector \(\mathbf{v}\). Thus the decay rates are the eigenvalues of \(B\), found by solving \(\det(\lambda I-B)=0\). Expanding that determinant gives
where
The coefficients are also the sum of the three eigenvalues, the sum of their pairwise products, and their product, respectively. These identities provide useful checks on a numerical root calculation.
Why the Rates Are Positive and Real
For positive rate constants and positive central elimination, the three eigenvalues are positive real numbers. One way to justify this is to rescale the state coordinates by a diagonal similarity transformation. Choose scale factors in the ratios \(\sqrt{k_{12}/k_{21}}\) and \(\sqrt{k_{13}/k_{31}}\). Under this transformation, \(B\) becomes a symmetric matrix with diagonal entries \(k_{10}+k_{12}+k_{13}, k_{21}, k_{31}\) and off-diagonal entries \(-\sqrt{k_{12}k_{21}}\), \(-\sqrt{k_{13}k_{31}}\), and zero.
For a vector \((x_1,x_2,x_3)^\mathsf{T}\), the quadratic form of that symmetric matrix is
This is strictly positive for every nonzero vector when \(k_{10}, k_{21}, k_{31}>0\). Therefore the symmetric matrix is positive definite and has positive real eigenvalues. Similar matrices have the same eigenvalues, so \(B\) does as well. This establishes that the modes decay rather than grow. If central elimination is zero, the argument no longer guarantees strict positivity; that boundary case is outside the assumptions used here.
Worked Example: Calculating the Three Rates
Suppose a hypothetical model has \(k_{10}=0.0200\), \(k_{12}=0.300\), \(k_{21}=0.0800\), \(k_{13}=0.100\), and \(k_{31}=0.0500\ \mathrm{h}^{-1}\). All five constants are positive, and central elimination is positive, so the positivity result above applies.
First calculate the polynomial coefficients:
The characteristic equation is therefore
Numerical solution gives, in descending order, \(\lambda_1\approx0.4899\), \(\lambda_2\approx0.05723\), and \(\lambda_3\approx0.002853\ \mathrm{h}^{-1}\). Check the sum: \(0.4899+0.05723+0.002853=0.5500\ \mathrm{h}^{-1}\) to the displayed precision. The pairwise-product sum is approximately \(0.02960\ \mathrm{h}^{-2}\), and the product is approximately \(0.0000800\ \mathrm{h}^{-3}\). Agreement with all three coefficients supports the root calculation.
How the Rates Shape Concentration Over Time
For distinct eigenvalues, the solution of the linear amount-balance system is a sum of eigenmodes. Since central concentration is \(C(t)=A_1(t)/V_c\), it has the form
Here \(A_i^{*}\) denotes an exponential amplitude, not a compartment amount. The amplitudes depend on the initial condition and model parameters; the eigenvalues alone do not specify the whole curve. For a central intravenous bolus, the initial amounts are \(A_1(0)=\mathrm{Dose}\), \(A_2(0)=A_3(0)=0\), and the concentration at time zero is \(\mathrm{Dose}/V_c\). Consequently, the three amplitudes must add to that initial concentration.
When the amplitudes are positive, the fastest term loses its relative contribution first, the intermediate term may govern an ensuing portion of the curve depending on the amplitudes and rate differences, and the slowest term dominates sufficiently late in time. The terminal approximation follows by factoring out the slow term:
Because \(\lambda_1-\lambda_3>0\) and \(\lambda_2-\lambda_3>0\), the two faster contributions shrink relative to the slow term as time increases. Thus the terminal log-concentration slope approaches \(-\lambda_3\) when the slow-mode amplitude is nonzero. The rate constants are in inverse-time units, and each rate’s half-life is \(\ln(2)/\lambda_i\).
Worked Example: Relating the Rates to Half-Lives
Use the rates from the preceding hypothetical calculation. Convert each rate into the half-life of its exponential mode with \(t_{1/2}=\ln(2)/\lambda\), using \(\ln(2)\approx0.6931\).
For the fastest mode, \(t_{1/2}=0.6931/0.4899=1.415\ \mathrm{h}\). For the intermediate mode, \(t_{1/2}=0.6931/0.05723=12.11\ \mathrm{h}\). For the slowest mode, \(t_{1/2}=0.6931/0.002853=242.9\ \mathrm{h}\). The fast term decays most rapidly; the slow term persists longest. These are half-lives of the exponential modes, not three separate compartment-specific half-lives.
Worked Example: Reading a Three-Exponential Profile
Consider a hypothetical concentration curve with positive amplitudes of \(6\), \(3\), and \(1\ \mathrm{mg/L}\) and rates \(0.4899\), \(0.05723\), and \(0.002853\ \mathrm{h}^{-1}\), respectively. These amplitudes are chosen to illustrate the shape of a tri-exponential curve; they are not claimed to be the fitted amplitudes for the microconstants in the earlier example. The curve is
At \(t=0\), \(C(0)=6+3+1=10\ \mathrm{mg/L}\). At \(t=2\ \mathrm{h}\),
At \(t=24\ \mathrm{h}\), the fast contribution is nearly zero, while the other two remain:
The curve has fallen substantially, but the slow contribution remains. At late times, the two faster terms become negligible relative to the slow term, so the profile approaches a straight line on a log-concentration versus time plot, with slope near \(-0.002853\ \mathrm{h}^{-1}\). The transition times are not fixed boundaries: they depend on both rate differences and amplitudes.
Justifying the Changing Semilog Slope
For positive amplitudes, the instantaneous negative slope of the log concentration is a weighted average of the three rates. Let \(w_i(t)=A_i^{*}e^{-\lambda_i t}/C(t)\). The weights are positive and sum to one, so differentiation gives
As time passes, faster terms lose weight relative to slower terms. More formally, differentiating the weighted mean gives
The bracket is a weighted variance and cannot be negative. Thus, under the positive-amplitude condition, the magnitude of the semilog slope decreases over time: the profile bends toward a shallower terminal slope. If an amplitude is zero, that mode is absent from the observed profile even though it is an eigenvalue of the system. This is one reason the number of structural compartments need not be obvious from a concentration curve.
Checks, Alternatives, and Limitations
A practical calculation can use an eigenvalue routine on \(B\), a numerical solver for the cubic, or direct time integration of the amount-balance equations. Eigenvalue calculation exposes the three rates directly; integration produces the predicted amounts or concentrations without requiring a separate exponential decomposition. For checking a numerical result, the coefficient identities are especially useful: the computed rates should sum to \(S_1\), their pairwise products should sum to \(S_2\), and their product should equal \(S_3\), up to rounding error.
Keep the units consistent. If all microconstants are in \(\mathrm{h}^{-1}\), then \(S_1\), \(S_2\), and \(S_3\) must have units \(\mathrm{h}^{-1}\), \(\mathrm{h}^{-2}\), and \(\mathrm{h}^{-3}\), respectively, and each eigenvalue must be in \(\mathrm{h}^{-1}\). A negative computed eigenvalue, a complex root, or a large mismatch in the coefficient checks signals an input, sign, or numerical calculation problem under the positive-rate assumptions used here.
There are limits to what the rates alone reveal. A small amplitude can make a mode contribute very little to observed concentrations; an eigenvalue does not guarantee that a clearly visible phase will appear in measured data. If two eigenvalues are close, their exponentials can be difficult to distinguish over a finite time range. If eigenvalues are repeated, the usual three distinct exponential terms do not apply in their simple form: a repeated eigenvalue can produce terms involving \(t e^{-\lambda t}\), depending on the system’s eigenvectors. The standard tri-exponential interpretation therefore assumes three distinct eigenvalues and refers to modes actually present in the concentration response.
Common Mistakes
- Calling the eigenvalues the microconstants. They are system-level decay rates calculated from all the rate constants together, not one-to-one labels for elimination or exchange pathways.
- Reversing the phase order. The largest positive rate decays fastest and has the shortest half-life; the smallest positive rate decays slowest and has the longest half-life.
- Assuming rates determine the whole curve. The exponential amplitudes are also needed to calculate concentrations and determine how visible each mode is.
- Skipping coefficient checks. A set of plausible-looking roots may still be wrong. Check their sum, pairwise-product sum, and product against the cubic coefficients.
- Using a terminal slope too early. The slowest mode dominates only after faster contributions have become small relative to it; the transition depends on amplitudes as well as rates.
Check Your Understanding
Try each question before reading its solution.
- If three rates are \(0.40\), \(0.08\), and \(0.01\ \mathrm{h}^{-1}\), which one has the longest half-life, and why?
- For the polynomial \(\lambda^3-0.30\lambda^2+0.019\lambda-0.00030=0\), what should the sum and product of its three roots be?
- Why can a model eigenvalue fail to produce a clearly visible phase in measured concentration data?
- For a positive-amplitude tri-exponential curve, what happens to the magnitude of the semilog slope over time?
Solutions. (1) The rate \(0.01\ \mathrm{h}^{-1}\) has the longest half-life because \(t_{1/2}=\ln(2)/\lambda\), so a smaller positive rate gives a longer half-life. (2) The sum is \(0.30\ \mathrm{h}^{-1}\), and the product is \(0.00030\ \mathrm{h}^{-3}\), by the coefficient identities for a monic cubic. (3) Its amplitude may be small, making its contribution negligible over the observed time range; close rates can also make modes hard to distinguish. (4) Its magnitude decreases or stays constant, since its time derivative is the negative weighted variance of the rates.