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One-, Two- and Three-Compartment Structural Models · Tutorial 26 of 797

Deriving the Two-Compartment Eigenvalues

Derive the two positive decay rates from the two-compartment amount balances and connect them to the exponential modes of the central concentration profile.

Beginner 10 min read

What You'll Learn

  • Write the post-bolus central and peripheral amount-balance equations using the course microconstants.
  • Derive the characteristic equation and the formulas for alpha and beta.
  • Verify the rates using their sum, product, units, and positivity.
  • Explain why a central concentration profile can contain two exponential modes.
  • Recognize when the eigenvalue derivation does not match the model assumptions.

Why Two Amounts Produce Two Decay Rates

In a two-compartment model, drug can move between a central and a peripheral compartment while elimination occurs from the central compartment. Because the two amounts change together, the central concentration generally does not decline at a single rate. The rate pair \(\alpha\) and \(\beta\) comes from the coupled amount-balance equations: they are the positive decay rates associated with the system’s two independent modes.

This tutorial considers the period after an intravenous bolus, when there is no continuing input. The central and peripheral amounts are \(A_c(t)\) and \(A_p(t)\). The first-order rate constants are \(k_{10}\) for central elimination, \(k_{12}\) for transfer from central to peripheral, and \(k_{21}\) for transfer from peripheral to central. Each rate constant has units of inverse time.

Model assumptions: The amounts are governed by a linear, time-invariant two-compartment model; the three rate constants are constant and positive; elimination and intercompartmental transfer are first order; and, after the bolus, no additional dose or infusion enters the system. Central concentration is \(C_c(t)=A_c(t)/V_c\), with constant central volume \(V_c\).

Under these assumptions, the amount balances are

$$ \frac{dA_c}{dt}=-(k_{10}+k_{12})A_c+k_{21}A_p, \qquad \frac{dA_p}{dt}=k_{12}A_c-k_{21}A_p. $$

The terms in each equation describe flows into and out of that compartment. For example, central amount leaves by elimination and transfer to the peripheral compartment, and it gains amount through return from the peripheral compartment. In matrix form, with \(\mathbf A(t)=(A_c(t),A_p(t))^\mathsf{T}\),

$$ \frac{d\mathbf A}{dt}=M\mathbf A, \qquad M= \begin{pmatrix} -(k_{10}+k_{12}) & k_{21}\\ k_{12} & -k_{21} \end{pmatrix}. $$

Deriving the Decay Rates

A mode of this linear system has the form \(\mathbf A(t)=\mathbf v e^{rt}\), where \(r\) is its exponential rate and \(\mathbf v\) is a nonzero constant vector. Substitution into the matrix equation gives \(M\mathbf v=r\mathbf v\). Thus the possible values of \(r\) satisfy \(\det(rI-M)=0\).

$$ \begin{aligned} \det(rI-M) &= \begin{vmatrix} r+k_{10}+k_{12} & -k_{21}\\ -k_{12} & r+k_{21} \end{vmatrix}\\ &=(r+k_{10}+k_{12})(r+k_{21})-k_{12}k_{21}\\ &=r^2+(k_{10}+k_{12}+k_{21})r+k_{10}k_{21}. \end{aligned} $$

Both modes decay, so write their exponential rates as \(r=-\alpha\) and \(r=-\beta\), where \(\alpha\) and \(\beta\) are positive. Substituting \(r=-\gamma\) into the characteristic equation gives the equation for a positive decay rate \(\gamma\):

$$ \gamma^2-(k_{10}+k_{12}+k_{21})\gamma+k_{10}k_{21}=0. $$

Applying the quadratic formula gives the two rates. By convention, \(\alpha\) denotes the larger, faster decay rate and \(\beta\) the smaller, slower one.

Definition: For positive \(k_{10}, k_{12}, k_{21}\), the two-compartment decay rates are
$$ \alpha,\beta = \frac{(k_{10}+k_{12}+k_{21}) \mathbin{\pm} \sqrt{(k_{10}+k_{12}+k_{21})^2-4k_{10}k_{21}}}{2}, \qquad \alpha>\beta>0, $$
where the plus sign gives \(\alpha\) and the minus sign gives \(\beta\). Equivalently, \(\alpha+\beta=k_{10}+k_{12}+k_{21}\) and \(\alpha\beta=k_{10}k_{21}\).

The rates are real and distinct for positive model rate constants. In particular, the discriminant can be rewritten as

$$ (k_{10}+k_{12}+k_{21})^2-4k_{10}k_{21} =(k_{10}+k_{12}-k_{21})^2+4k_{12}k_{21}>0. $$

Their sum is positive and their product is positive. Since the two roots are real, those facts imply that both roots are positive. The strict discriminant inequality also means that the two decay rates are not equal.

Worked Example: Calculate Both Rates From Microconstants

Suppose a hypothetical model has \(k_{10}=0.30\ \mathrm{h}^{-1}\), \(k_{12}=0.40\ \mathrm{h}^{-1}\), and \(k_{21}=0.20\ \mathrm{h}^{-1}\). Find the two positive decay rates.

First calculate the sum and product terms:

$$ S=k_{10}+k_{12}+k_{21}=0.30+0.40+0.20=0.90\ \mathrm{h}^{-1}, \qquad P=k_{10}k_{21}=(0.30)(0.20)=0.0600\ \mathrm{h}^{-2}. $$

The discriminant is \(S^2-4P\):

$$ 0.90^2-4(0.0600)=0.8100-0.2400=0.5700\ \mathrm{h}^{-2}, \qquad \sqrt{0.5700}=0.754983\ \mathrm{h}^{-1}. $$

Therefore,

$$ \alpha=\frac{0.90+0.754983}{2}=0.827492\ \mathrm{h}^{-1}, \qquad \beta=\frac{0.90-0.754983}{2}=0.0725083\ \mathrm{h}^{-1}. $$

The larger rate is \(\alpha\), as required by the convention. As checks, the rates add to approximately \(0.900000\ \mathrm{h}^{-1}\) and their product is approximately \(0.0600\ \mathrm{h}^{-2}\), matching \(S\) and \(P\). Both rates have units of inverse time, just like the microconstants.

Why These Rates Govern the Central Concentration

The characteristic equation identifies the exponential modes of the whole amount system, not just the central compartment. To see the same rates directly in the central amount, let \(a=k_{10}+k_{12}\). Differentiate the central balance and use the peripheral balance to replace \(dA_p/dt\). Then use \(k_{21}A_p=dA_c/dt+aA_c\), which follows from the central balance:

$$ \begin{aligned} \frac{d^2 A_c}{dt^2} &=-a\frac{dA_c}{dt} +k_{21}\frac{dA_p}{dt}\\ &=-a\frac{dA_c}{dt}+k_{21}(k_{12}A_c-k_{21}A_p)\\ &=-(a+k_{21})\frac{dA_c}{dt}-k_{10}k_{21}A_c. \end{aligned} $$

Consequently, the central amount satisfies a second-order linear differential equation whose characteristic equation is the one already derived. Its solutions are combinations of \(e^{-\alpha t}\) and \(e^{-\beta t}\). Dividing \(A_c(t)\) by the constant \(V_c\) changes the coefficients but not the exponential rates, so central concentration has the same two possible time scales. The particular coefficients depend on the initial amounts; deriving them is a separate step.

The mode interpretation gives a useful qualification: a system can, in principle, start exactly along one eigenvector, in which case only that mode is present. For a central IV bolus, however, the initial peripheral amount is zero, and the initial amount vector is not aligned with either mode under positive rate constants. Both modes therefore contribute to the central amount profile.

Worked Example: Check the Rates and Their Meaning

For another hypothetical parameter set, let \(k_{10}=0.12\ \mathrm{h}^{-1}\), \(k_{12}=0.08\ \mathrm{h}^{-1}\), and \(k_{21}=0.30\ \mathrm{h}^{-1}\). Calculate the rates and verify their defining sum and product.

Here \(S=0.12+0.08+0.30=0.50\ \mathrm{h}^{-1}\) and \(P=(0.12)(0.30)=0.0360\ \mathrm{h}^{-2}\). The discriminant and its square root are

$$ S^2-4P=(0.50)^2-4(0.0360)=0.2500-0.1440=0.1060\ \mathrm{h}^{-2}, \qquad \sqrt{0.1060}=0.325576\ \mathrm{h}^{-1}. $$

Thus

$$ \alpha=\frac{0.50+0.325576}{2}=0.412788\ \mathrm{h}^{-1}, \qquad \beta=\frac{0.50-0.325576}{2}=0.0872120\ \mathrm{h}^{-1}. $$

Their sum is \(0.500000\ \mathrm{h}^{-1}\), and their product is approximately \(0.0360\ \mathrm{h}^{-2}\). These checks are consistent with the coefficients of the characteristic equation. Because these are rates of the coupled system, neither should automatically be identified with one individual microconstant such as \(k_{10}\) or \(k_{21}\).

Worked Example: Why a Central Bolus Excites Both Modes

Use the first hypothetical parameter set, \(k_{10}=0.30\), \(k_{12}=0.40\), and \(k_{21}=0.20\ \mathrm{h}^{-1}\). For a mode with decay rate \(\gamma\), let its amount vector be proportional to \((1,q)^\mathsf{T}\), where \(q\) is the peripheral-to-central amount ratio in that mode. The first row of \(M\mathbf v=-\gamma\mathbf v\) gives

$$ q=\frac{k_{10}+k_{12}-\gamma}{k_{21}}. $$

Using the rates calculated above gives \(q_\alpha=(0.70-0.827492)/0.20\approx-0.63746\) and \(q_\beta=(0.70-0.0725083)/0.20\approx3.13746\). In particular, neither mode has peripheral-to-central ratio zero.

An IV bolus placed in the central compartment starts with amount vector \((D,0)^\mathsf{T}\), where \(D>0\) is the hypothetical dose. This vector is not aligned with either mode because each mode has a nonzero peripheral component. The initial state must therefore be expressed using both independent modes. Since each mode also has a nonzero central component, both exponential rates appear in the central amount and concentration. This argument explains the two-rate structure without calculating the profile’s specific coefficients.

Assumptions, Checks, and Alternatives

The derivation is valid only for the stated linear model. If elimination is saturable, rate constants vary with time, input continues during the period being analyzed, or a model includes additional compartments, the two-compartment post-bolus equations do not describe the system as written. In those settings, these particular formulas need not give the relevant decay rates.

Useful checks catch many algebra and implementation errors:

  • Check signs. Under positive \(k_{10},k_{12},k_{21}\), both decay rates must be positive. A negative calculated rate commonly signals a sign error in the characteristic equation or a transcription mistake.
  • Check the sum and product. The rates must satisfy \(\alpha+\beta=k_{10}+k_{12}+k_{21}\) and \(\alpha\beta=k_{10}k_{21}\), allowing for rounding.
  • Check units. Each rate must be in inverse time. The discriminant has units of inverse time squared, so its square root has units of inverse time.
  • Check the model being solved. A continuing input changes the forcing of the amount equations; it does not justify treating an infusion profile as a post-bolus, unforced decay without accounting for that input.

The eigenvalue calculation and the scalar second-order equation are two algebraically equivalent ways to obtain the rates. The matrix approach emphasizes coupled compartment modes; eliminating the peripheral amount emphasizes the central amount’s differential equation. Numerical integration is a useful alternative for generating predictions from the amount balances, especially when inputs are more complicated, but it does not remove the need to specify the correct model. An empirical biexponential curve can describe two exponential components, but by itself it does not establish that their rates arise from these particular compartmental balances.

Common Mistakes and Limitations

  • Using the eigenvalues as positive decay rates without changing signs. The eigenvalues of \(M\) are \(-\alpha\) and \(-\beta\), not \(+\alpha\) and \(+\beta\). The positive decay-rate formula results from defining the rates as the negatives of those eigenvalues.
  • Confusing microconstants with system rates. The formulas combine the three microconstants. In general, \(\alpha\) is not simply \(k_{10}\), and \(\beta\) is not simply \(k_{21}\).
  • Assuming the rates alone determine the concentration curve. The rates determine its possible exponential time scales, while initial conditions and the concentration scale determine the coefficients.
  • Ignoring assumptions because the algebra works. A real-valued pair from the quadratic formula does not prove that the two-compartment model is an appropriate description of data. Model structure and assumptions still need justification.

The formulas provide a compact mathematical result, not a guarantee that both rates can be estimated precisely from any particular dataset. Whether observed data distinguish the two components depends on the design and the information in the observations, not on the existence of two roots alone.

Key takeaway: The post-bolus two-compartment amount-balance system has characteristic polynomial \(r^2+(k_{10}+k_{12}+k_{21})r+k_{10}k_{21}\). Its two negative eigenvalues are \(-\alpha\) and \(-\beta\); the corresponding positive rates govern the two exponential modes available to the central concentration profile.

Check Your Understanding

Try each question, then compare your work with the brief solutions below.

  1. For \(k_{10}=0.20\), \(k_{12}=0.10\), and \(k_{21}=0.30\ \mathrm{h}^{-1}\), calculate \(\alpha\) and \(\beta\).
  2. Why are the eigenvalues of the amount matrix negative while \(\alpha\) and \(\beta\) are defined as positive?
  3. State the sum and product checks that apply to the two decay rates.
  4. Would the same derivation apply unchanged during a constant-rate infusion? Briefly explain.

Solutions. (1) \(S=0.60\ \mathrm{h}^{-1}\), \(P=0.0600\ \mathrm{h}^{-2}\), and \(S^2-4P=0.1200\ \mathrm{h}^{-2}\). Thus \(\alpha=(0.60+\sqrt{0.12})/2=0.473205\ \mathrm{h}^{-1}\) and \(\beta=(0.60-\sqrt{0.12})/2=0.126795\ \mathrm{h}^{-1}\). (2) The modes decay exponentially, so their exponents are \(-\alpha\) and \(-\beta\); the positive decay rates are the negatives of the eigenvalues. (3) \(\alpha+\beta=k_{10}+k_{12}+k_{21}\) and \(\alpha\beta=k_{10}k_{21}\). (4) Not unchanged as a complete concentration-profile derivation: an infusion adds an input term to the amount balances. The unforced post-bolus equations used here do not include that continuing input.