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Compartmental Absorption and Transit Models · Tutorial 229 of 799

Building a Three-Segment Absorption Model

Learn to assemble and solve a three-segment absorption model, including its oral boundary input, local dissolved-drug sources, and cumulative uptake.

Intermediate 11 min read

What You'll Learn

  • Define three segment-specific luminal amount balances with transit, absorption, and dissolved-drug input.
  • Represent oral entry through a gastric state or a specified input to the first segment.
  • Solve a sequential model using convolution for time-varying inputs.
  • Calculate segmental luminal amounts and cumulative absorbed amount in a worked example.
  • Check the model's mass accounting and identify common implementation errors.
  • Compare transient ODE solutions with steady-state and cumulative-fraction calculations.

From Segment Processes to a Coupled Model

A three-segment absorption model connects processes that are often described separately: drug enters the gut, moves between adjacent segments, may dissolve into the luminal drug pool, and may be absorbed from each segment. Writing one balance for each segment makes these processes explicit and shows how an input in one location can contribute to absorption farther along the gut.

The model here tracks dissolved luminal amount in three sequential segments. It uses first-order transfer and absorption, with constant rate constants over the modeled interval. Oral entry may be represented by a gastric state or by an input rate directly into the first segment. A nonnegative source rate can also add dissolved drug to any segment, for example through dissolution represented by a separate solid-drug model.

Definition: Let \(L_i(t)\) be the dissolved luminal amount in segment \(i\), in mg, for \(i=1,2,3\). Let \(k_{t,i}\) be its first-order transit rate constant (h\(^{-1}\)), and \(k_{a,i}\) its first-order absorption rate constant (h\(^{-1}\)). Let \(r_i(t)\) be a dissolved-drug source into that segment (mg/h). Define \(\lambda_i=k_{t,i}+k_{a,i}\). Drug leaving segment 1 or 2 by transit enters the next segment; drug leaving segment 3 by transit exits the modeled gut.

The absorption rate from segment \(i\) is \(k_{a,i}L_i(t)\). Define \(A(t)\) as cumulative absorbed amount, in mg, and \(E(t)\) as cumulative amount that has transited out of segment 3, also in mg. These are cumulative amounts, not rates. The first-order assumption means that, over a short interval, each process rate is proportional to the amount currently available in its segment.

Assembling the Oral-Input and Segment Balances

For a gastric state \(G(t)\), with amount in mg, suppose gastric emptying is first order with rate constant \(k_g\) (h\(^{-1}\)). Then the rate entering segment 1 is \(u_g(t)=k_gG(t)\). In the absence of another gastric input, \(dG/dt=-k_gG\), so an initial gastric amount \(D\) gives \(G(t)=D e^{-k_gt}\). A specified input \(u(t)\) directly into segment 1 is an alternative boundary condition; it need not be generated by a gastric state.

With a gastric input and local dissolved-drug sources, the balances are:

$$ \begin{aligned} \frac{dL_1}{dt} &= u_g(t)+r_1(t)-\lambda_1L_1(t),\\ \frac{dL_2}{dt} &= k_{t,1}L_1(t)+r_2(t)-\lambda_2L_2(t),\\ \frac{dL_3}{dt} &= k_{t,2}L_2(t)+r_3(t)-\lambda_3L_3(t),\\ \frac{dA}{dt} &= k_{a,1}L_1(t)+k_{a,2}L_2(t)+k_{a,3}L_3(t),\\ \frac{dE}{dt} &= k_{t,3}L_3(t). \end{aligned} $$

For direct input to segment 1, replace \(u_g(t)\) by \(u(t)\) and omit \(G\). Initial conditions specify any amount already present at the start of the calculation. The model assumes well-mixed amounts within each segment, nonnegative rates and inputs, no return flow to an earlier segment, and no change in the rate constants during the interval. It does not by itself predict \(r_i(t)\): if a source represents dissolution, that source must be supplied by a dissolution model or data. When tracking total drug from the original dose, its donor state must also be included in the accounting.

Solving the Sequential Balances

Each segment balance is a linear first-order differential equation. For example, multiplying the first balance by the integrating factor \(e^{\lambda_1t}\), integrating from 0 to \(t\), and rearranging gives the amount in segment 1. Applying the same step to segments 2 and 3 gives a sequential solution in which each downstream input includes transit from the upstream amount.

$$ \begin{aligned} L_1(t)&=e^{-\lambda_1t}L_1(0) +\int_0^t e^{-\lambda_1(t-s)}[u_g(s)+r_1(s)]\,ds,\\ L_2(t)&=e^{-\lambda_2t}L_2(0) +\int_0^t e^{-\lambda_2(t-s)}[k_{t,1}L_1(s)+r_2(s)]\,ds,\\ L_3(t)&=e^{-\lambda_3t}L_3(0) +\int_0^t e^{-\lambda_3(t-s)}[k_{t,2}L_2(s)+r_3(s)]\,ds,\\ A(t)&=A(0)+\int_0^t\sum_{i=1}^{3}k_{a,i}L_i(s)\,ds. \end{aligned} $$

These convolution expressions are exact for the stated model. They are especially useful when an input has a simple analytic form. For more complicated inputs, the same balances can be integrated numerically as a coupled ODE system. A numerical solution should use the same input functions, initial conditions, and rate constants in every equation.

Worked Example: A Dose Entering the First Segment

Suppose a hypothetical 100 mg dose has just entered segment 1, so \(L_1(0)=100\) mg and \(L_2(0)=L_3(0)=A(0)=E(0)=0\). Let there be no further inputs or local sources. Use \(k_{t,1}=0.6\), \(k_{a,1}=0.4\), \(k_{t,2}=0.5\), \(k_{a,2}=0.5\), \(k_{t,3}=0.3\), and \(k_{a,3}=0.7\), all in h\(^{-1}\). Find luminal amounts and cumulative absorption at \(t=2\) h.

Here \(\lambda_1=\lambda_2=\lambda_3=1.0\) h\(^{-1}\). The first balance gives \(L_1(t)=100e^{-t}\) mg. Segment 2 receives \(0.6L_1(t)\), so its convolution is \(L_2(t)=60te^{-t}\) mg. Segment 3 receives \(0.5L_2(t)\), giving \(L_3(t)=15t^2e^{-t}\) mg.

At 2 h, \(e^{-2}=0.135335\). Thus \(L_1=100(0.135335)=13.53\) mg, \(L_2=60(2)(0.135335)=16.24\) mg, and \(L_3=15(2^2)(0.135335)=8.120\) mg.

For cumulative absorption, integrate each segment's absorption rate. The required integrals are \(\int_0^2 e^{-t}dt=1-e^{-2}=0.864665\), \(\int_0^2 te^{-t}dt=1-3e^{-2}=0.593994\), and \(\int_0^2 t^2e^{-t}dt=2-10e^{-2}=0.646647\). Therefore

$$ A(2)=40(0.864665)+30(0.593994)+10.5(0.646647) =59.20\text{ mg}. $$

The amount transited out by 2 h is \(E(2)=0.3\int_0^2L_3(t)dt=4.5(0.646647)=2.910\) mg. The mass check is \(13.53+16.24+8.120+59.20+2.910=100.0\) mg, up to rounding. The remaining amount is in the three luminal pools; the rest has been absorbed or has exited the modeled gut.

Adding Dissolved-Drug Sources

A local source changes the balance in the segment where it enters, then may affect later segments through transit. For example, a source in segment 2 does not contribute to segment 1, but some of it can be absorbed in segment 2 and some can transit into segment 3. This location-specific placement is important: putting every source into segment 1 changes both its exposure to segment-specific absorption and the timing of its downstream arrival.

Worked Example: Sustained Input and Local Sources

Consider a hypothetical steady operating condition with a maintained oral boundary input of \(20\) mg/h into segment 1, plus dissolved sources \(r_2=12\) mg/h and \(r_3=6\) mg/h. There is no source in segment 1. Use the rate constants from the preceding example. Assume the system has been maintained long enough to reach steady state, and find the segment amounts and absorption rate.

At steady state, each derivative of \(L_i\) is zero. In segment 1, \(0=20-1.0L_1\), so \(L_1=20\) mg. In segment 2, \(0=0.6(20)+12-1.0L_2\), so \(L_2=24\) mg. In segment 3, \(0=0.5(24)+6-1.0L_3\), so \(L_3=18\) mg.

The absorption rate is \(0.4(20)+0.5(24)+0.7(18)=8+12+12.6=32.6\) mg/h. The terminal transit rate is \(0.3(18)=5.4\) mg/h. Total input is \(20+12+6=38\) mg/h, which equals \(32.6+5.4=38\) mg/h leaving through absorption and terminal transit. If the model starts in these steady-state amounts, cumulative absorption over the next 3 h is \(32.6(3)=97.8\) mg.

A source derived from a solid pool can be written, for example, as \(r_i(t)=k_{\mathrm{dis},i}P_i(t)\), where \(P_i(t)\) is solid amount and \(k_{\mathrm{dis},i}\) is a dissolution rate constant. If \(P_i\) is not included among the modeled states, the luminal model treats \(r_i(t)\) as a supplied input rather than predicting it. A source must not be counted both as an imposed \(r_i(t)\) and as a transfer from an explicitly modeled donor state.

Worked Example: A Dissolved Pulse Entering Segment 2

Suppose a hypothetical 20 mg dissolved pulse is added to segment 2 at time zero, with no other input and no initial luminal drug. For this pulse calculation, represent the instantaneous source by the post-entry condition \(L_2(0)=20\) mg; the other luminal amounts and \(A(0)\) are zero. Use the rates from the first example. Find the eventual absorbed and terminally transited amounts from this pulse.

In segment 2, the competing first-order rates are \(k_{a,2}=0.5\) h\(^{-1}\) and \(k_{t,2}=0.5\) h\(^{-1}\). The fraction absorbed there is \(0.5/(0.5+0.5)=0.5\), so \(20(0.5)=10\) mg is absorbed in segment 2 and \(10\) mg transits to segment 3. In segment 3, the absorbed fraction is \(0.7/(0.7+0.3)=0.7\), so \(10(0.7)=7.0\) mg is absorbed there and \(10(0.3)=3.0\) mg exits the modeled gut. Total absorption is \(10+7=17\) mg, and the remaining 3.0 mg exits. The calculation accounts for the entire 20 mg pulse.

Mass Accounting, Diagnostics, and Alternatives

For the three dissolved pools, cumulative absorption, and terminal transit, summing the balances cancels the transfers between segments. The resulting accounting identity is

$$ \frac{d}{dt}\left(G+\sum_{i=1}^{3}L_i+A+E\right) =\sum_{i=1}^{3}r_i(t), $$

when gastric input is generated by \(dG/dt=-k_gG\). With direct segment-1 input \(u(t)\) instead, the right-hand side is \(u(t)+\sum_i r_i(t)\), and \(G\) is omitted. This identity provides a practical check: unexplained gains or losses often indicate a missing transit term, a duplicated input, an incorrect rate unit, or an error in the cumulative absorption equation. If the \(r_i\) terms are internal dissolution transfers, the corresponding donor states must be added to the total before expecting whole-system mass conservation.

The model is appropriate when first-order rates and a well-mixed amount in each segment are reasonable approximations over the interval of interest. It does not represent spatial gradients within a segment, time-varying permeability unless rates are allowed to vary, or feedback of luminal concentration on dissolution or absorption. The source profiles and rate constants may also be uncertain; a numerically smooth solution does not establish that their values are identifiable from available observations.

ApproachUseful forMain caution
Coupled ODE integrationGeneral time-varying inputs and simultaneous luminal and cumulative outputsCheck units, initial conditions, solver accuracy, and mass accounting
Sequential convolutionAnalytic solutions for simple inputs and understanding how upstream amounts drive downstream segmentsExpressions can become cumbersome when rates or inputs vary with time
Steady-state balancesConstant inputs maintained long enough for transient changes to settleDoes not describe the time course after a single oral dose
Cumulative competing-rate fractionsEventual absorption and exit from a segment with constant first-order ratesDoes not give transient luminal amounts or absorption timing

A common mistake is to apply the segment-1 absorption fraction to the entire oral dose and call that the total absorbed amount. That fraction applies only to drug present in segment 1: the fraction that transits onward remains available for absorption in later segments. The correction is to retain the transit term in each balance and follow the resulting amount through all three segments, as in the worked calculations above.

Check Your Understanding

Try each question first, then compare with the solutions.

  1. If \(L_1=30\) mg, \(k_{t,1}=0.4\) h\(^{-1}\), and \(k_{a,1}=0.6\) h\(^{-1}\), what are the transit and absorption rates from segment 1?
  2. At steady state, segment 1 receives 10 mg/h, has \(\lambda_1=0.5\) h\(^{-1}\), and has no other input. What is \(L_1\)?
  3. A 12 mg pulse enters segment 3, where \(k_{a,3}=0.8\) h\(^{-1}\) and \(k_{t,3}=0.2\) h\(^{-1}\). How much is eventually absorbed and how much exits?
  4. Why must a dissolution source already represented by an explicit solid-drug state not also be added as an independent source to the luminal balance?

Solutions. (1) Transit is \(0.4(30)=12\) mg/h and absorption is \(0.6(30)=18\) mg/h. (2) At steady state, \(0=10-0.5L_1\), so \(L_1=20\) mg. (3) The absorbed fraction is \(0.8/(0.8+0.2)=0.8\), giving \(12(0.8)=9.6\) mg absorbed and \(12(0.2)=2.4\) mg exiting. (4) Doing both counts the same drug transfer twice, artificially increasing the modeled mass input.