Why a Segment Interface Needs One Shared Flux
A transit term describes drug leaving one luminal segment and entering the next. These are not two independent events: they are the donor and receiver sides of the same transfer. If a model subtracts one amount from the upstream segment but adds a different amount downstream, it creates or destroys drug at the interface. Writing both terms from one shared flux prevents that error and makes the total mass balance auditable.
Here, \(M_i(t)\) denotes the luminal drug mass in segment \(i\), measured in mg, and \(F_i(t)\) denotes the downstream mass flux from segment \(i\) to segment \(i+1\), measured in mg/h. The positive direction is downstream. These definitions concern mass, not concentration: a concentration must be multiplied by an appropriate volume flow before it represents a mass flux.
A common first-order assumption is that downstream transfer is proportional to the mass in the donor segment:
Here \(k_i\) is a transit rate constant in h\(^{-1}\), so \(F_i\) has units mg/h. This expression assumes a well-mixed donor state, a constant or specified \(k_i\), and one-way transfer proportional to donor mass. It is a modeling choice, not a requirement of mass conservation. A different flux law can be used, provided the very same interface flux is paired with opposite signs in the donor and receiver balances.
Writing the Paired Segment Balances
For two adjacent segments with no other inputs or losses, the equations are:
If first-order donor transfer is appropriate, substitute \(F_1=k_1M_1\). The signs follow from the chosen downstream direction: transit reduces the donor mass and increases the receiver mass. The equations describe a transfer between states, not removal from the modeled gut.
For a set of adjacent segments, the same bookkeeping applies at every interface. For an interior segment \(i\), the incoming flux is \(F_{i-1}\) and the outgoing flux is \(F_i\):
At a closed upstream boundary there is no incoming flux, and at a closed downstream boundary there is no outgoing flux. For that case, set \(F_0=0\) and \(F_n=0\), where \(n\) is the number of segments. These boundary conditions describe no drug crossing the system boundary. If drug enters from the stomach or exits the final gut segment, that boundary flux is not internal transit and must be represented separately.
Worked Example: First-Order Transfer Between Two Segments
Suppose a hypothetical model has \(M_1=12.0\) mg and \(M_2=3.0\) mg at a particular time. The transfer rate constant is \(k_1=0.250\) h\(^{-1}\). Assume no other inputs or losses at that instant. Calculate the flux and each segment's instantaneous rate of change.
The shared interface flux is:
The donor loses this flux, while the receiver gains it:
Their summed rate is \(-3.00+3.00=0\) mg/h. Thus, at this instant, transit redistributes the 15.0 mg across the two segments without changing the total.
Why Internal Transit Cancels
The cancellation is an algebraic consequence of pairing each interface flux. Sum the mass-balance equations over all \(n\) segments, with no external inputs or losses:
Writing out the terms makes the cancellation explicit:
Every internal interface flux appears once with a minus sign in its donor balance and once with a plus sign in its receiver balance. Under closed boundary conditions \(F_0=F_n=0\), so:
This proof does not depend on first-order kinetics. Each \(F_i\) may be a nonlinear, time-varying, or otherwise specified function. Cancellation still holds because the donor and receiver use the identical flux. The flux law affects how quickly and in what pattern mass redistributes; it does not alter the conservation result.
Worked Example: A Flux Based on Volume Flow
In a hypothetical interface calculation, luminal material moves downstream at \(Q=0.400\) L/h, and the donor concentration is \(C_1=5.00\) mg/L. Assume the stated concentration applies to the material crossing the interface and that the receiver accepts the entire transferred amount. Find the mass flux and paired terms.
A volume flow times concentration gives mass per time:
Thus the interface contributes \(-2.00\) mg/h to the donor balance and \(+2.00\) mg/h to the receiver balance. The liters cancel, leaving mg/h. The total rate due to this interface is zero. This representation and the first-order representation above are alternative ways to specify a flux; they should not be added together unless they describe distinct physical transfer processes.
External Losses Do Not Cancel as Internal Transit
Absorption, degradation, or an outflow from the final modeled segment changes the amount counted as luminal gut mass. Such a process is not canceled by a receiving gut segment unless the model explicitly includes its destination as another state. For example, let \(A_i\) be an absorption rate from segment \(i\), in mg/h, defined as nonnegative when mass leaves the lumen. Then:
Summing over segments cancels the internal interface fluxes, but leaves the external loss terms:
With closed transit boundaries, the change in total luminal mass is therefore \(-\sum_i A_i\), not zero. This does not violate conservation of total drug in a larger model if absorbed mass is tracked in another state; it means only that luminal mass alone is not conserved when absorption removes drug from the lumen.
Worked Example: Transit Plus Absorption
Consider two hypothetical gut segments. At one instant, transit from segment 1 to segment 2 is \(F_1=1.20\) mg/h. Absorption rates are \(A_1=0.30\) mg/h and \(A_2=0.20\) mg/h. There are no boundary inputs or outputs. Calculate the segment rates and the total luminal rate of change.
The balances are:
The total luminal rate is \(-1.50+1.00=-0.50\) mg/h. Equivalently, internal transit cancels and the two absorption losses sum to \(0.30+0.20=0.50\) mg/h. Segment 2 gains mass at this instant, but total luminal mass falls because absorption is an external loss from the luminal subsystem.
Assumptions, Diagnostics, and Implementation Choices
The cancellation result requires consistent accounting, rather than a particular transit mechanism. To apply it correctly, define the system boundary, state which masses are included, specify the positive direction for each interface, and use the same flux value on both sides. The simple first-order law adds assumptions about mixing and proportionality; the conservation proof does not require those assumptions.
In a numerical implementation, a useful diagnostic is to calculate the net contribution of each internal interface to the total derivative. For an interface \(i\), the donor contribution plus the receiver contribution must equal \(-F_i+F_i=0\). A nonzero residual suggests a sign error, inconsistent flux calculations, a missing receiver or donor term, or an interface that was inadvertently counted twice. Also check units: both sides of a mass-balance equation must have units of mass per time.
A second diagnostic is a total-mass check over time. In a closed system containing only internal transit, the sum \(\sum_i M_i\) should remain unchanged in the continuous model. Small discrepancies in a numerical solution can arise from solver tolerances or discretization; large or systematic discrepancies call for checking the equations and implementation. Numerical time steps can also produce negative masses if they are too large for the chosen update method. Such negative values are not physical evidence of conservation failure, but they indicate that the numerical update is not adequately resolving the transfer.
There are two common modeling choices for the interface flux. A donor-based law such as \(F_i=k_iM_i\) is compact and often appropriate for first-order transit. A flow-based law such as \(F_i=QC_i\) connects mass transfer to volume flow and concentration, when those quantities are defined consistently. These alternatives differ in their physical assumptions and inputs; neither changes the requirement for paired terms. Defining donor outflow and receiver inflow independently is not a valid alternative for a single interface unless the difference represents an explicitly modeled process such as loss or input between the states.
Common Mistakes
- Using the same sign twice. Writing \(-F_i\) in both neighboring balances removes mass from both segments. The receiver term must be \(+F_i\).
- Calculating donor and receiver terms separately. If a single transfer is intended, calculate one \(F_i\) and reuse it. Separate values can create an artificial source or sink.
- Calling absorption internal transit. Absorption leaves the luminal subsystem, so it remains in the summed luminal balance unless its destination is included as a state.
- Confusing concentration flow with mass flow. \(QC\) has units of mass per time; concentration alone does not specify the amount crossing an interface per unit time.
- Assuming conservation means each segment's mass is constant. Only the total is constant in a closed system. Individual segment masses can increase or decrease as drug moves.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A first-order interface has \(M_i=8.00\) mg and \(k_i=0.150\) h\(^{-1}\). What is the flux, and what are its contributions to donor and receiver balances?
- Explain why \(F_i=k_iM_i\) is not necessary for the internal-flux cancellation result.
- Two segments have an interface flux of \(0.80\) mg/h and absorption rates of \(0.10\) mg/h and \(0.25\) mg/h. With no boundary flows, what is the total luminal rate of change?
- A model computes an interface outflow of \(1.0\) mg/h but adds \(0.9\) mg/h to the receiving segment. What is the net change in total mass caused by this interface, and what should the model do if no loss is intended?
Solutions. (1) \(F_i=(0.150\ \mathrm{h}^{-1})(8.00\ \mathrm{mg})=1.20\) mg/h. The donor contribution is \(-1.20\) mg/h and the receiver contribution is \(+1.20\) mg/h. (2) Cancellation follows from using the identical flux with opposite signs, regardless of how that flux is calculated. First-order proportionality is an additional kinetic assumption. (3) Internal transit cancels, leaving \(-0.10-0.25=-0.35\) mg/h for total luminal mass. (4) The interface causes a net change of \(-1.0+0.9=-0.10\) mg/h, an artificial loss. If no loss is intended, use the same \(1.0\) mg/h flux as the receiving contribution, with a positive sign.