Separate Solid and Dissolved Drug States

Why Keep Solid and Dissolved Drug Separate?

A segmental absorption model may need to represent drug that has entered the gut lumen but is not yet available for absorption. A solid particle can move through a segment without dissolving, while dissolved drug can be absorbed or carried onward in the luminal contents. If both forms are represented by one state, these pathways cannot be assigned distinct rates without adding extra bookkeeping.

The central accounting rule is simple: dissolution removes mass from the solid state and adds the same mass to the dissolved state. It does not itself remove drug from the gut or count as absorption. Transit moves mass between segments, and absorption removes dissolved mass from the lumen. Keeping these roles distinct makes it possible to check whether a model conserves mass.

For the balances below, each segment is treated as well mixed for each state. Transit and absorption are represented by first-order rates, and dissolution is represented by a first-order rate acting on solid drug. Rate constants have units of inverse time, such as h\(^{-1}\), so a rate constant multiplied by an amount has units of mass per time. The model assumes no degradation, precipitation, or other loss unless an additional process is explicitly added. These are modeling assumptions, not universal properties of a formulation.

Writing the Two Balances

Let \(k_{\mathrm{dis},i}\) be the dissolution rate constant in segment \(i\), \(k_{S,i}\) the solid transit rate constant, \(k_{D,i}\) the dissolved transit rate constant, and \(k_{a,i}\) the absorption rate constant. The corresponding segment-level rates are

Here \(T_{S,i}\) and \(T_{D,i}\) are the rates leaving segment \(i\) by transit. They may enter the next segment, or, for a terminal segment, leave the modeled gut. A solid transit rate and dissolved transit rate need not be equal: whether they should be equal is a model choice supported by the system being represented.

Write \(I_{S,i}\) and \(I_{D,i}\) for external inputs directly into the solid and dissolved states of segment \(i\). Let \(T_{S,i-1}\) and \(T_{D,i-1}\) denote the incoming transit rates from the preceding segment. For the first segment, these incoming terms are zero unless an upstream input is explicitly represented. Then the balances are

The dissolution term has opposite signs in the two equations. Absorption appears only as a loss from \(D_i\), because this model makes dissolved drug—not solid drug—the state available for uptake. If a systemic model receives absorbed drug, the luminal absorption rate can serve as its input; it must not also be counted as a second luminal loss.

Worked Example: Rates in One Segment

Initial Dose and Ongoing Input Are Different

A dose placed in a state at time zero is an initial condition, not a continuing input. For an immediate solid dose of \(Q\) mg in the first segment, set \(S_1(0)=Q\) and do not also add \(Q\) mg as an input rate. An ongoing input, in contrast, has units of mass per time and appears in \(I_{S,i}\) or \(I_{D,i}\). For example, a specified solid input rate of 5 mg/h contributes 5 mg/h to the solid balance for as long as that input is applied.

Why the Total-Mass Check Works

The balance structure can be justified directly. Add the solid and dissolved equations for one segment. The two dissolution terms cancel because they represent the same mass moving between states:

Now sum over all modeled segments. Each transit rate between two modeled segments appears once as an output from one segment and once as an input to the next, so it cancels in the total. If \(M=\sum_i(S_i+D_i)\) is the total luminal amount, the result is

where \(N\) is the last modeled segment and its outgoing transit rates leave the modeled gut. Thus, the gut inventory changes only through external input, terminal exit, and absorption. This check does not require dissolution to be first order; any dissolution law cancels in the same way if it is entered once as a loss from solid and once as an equal gain to dissolved.

Diagnostics, Alternatives, and Limitations

A useful diagnostic is to calculate every named flux before calculating derivatives. For each segment, check that the derivative units are mass per time, that \(R_{\mathrm{dis},i}\) is nonnegative when \(S_i\) is nonnegative, and that \(R_{a,i}\) is nonnegative when \(D_i\) is nonnegative. Then add the derivatives across both states and compare the result with the external inputs and true losses. A mismatch often indicates that a transfer was counted as a system loss, a dissolution term was entered with the wrong sign, or absorption was subtracted twice.

A common alternative is a single luminal state with one effective removal or absorption rate. This can be simpler when the distinction between solid and dissolved drug is not important to the question or when dissolution is very fast relative to the other modeled processes. But it cannot independently represent solid transit and dissolved absorption. Assuming instantaneous dissolution is another simplification: it effectively removes the solid-state delay and is inappropriate when dissolution may materially limit the amount available for absorption.

The separate-state model is still an approximation. First-order dissolution assumes a rate proportional to the solid amount, and first-order transit and absorption assume rates proportional to the corresponding state amounts. If dissolution depends on changing solubility, surface area, or other conditions, a different dissolution law may be needed. Such changes alter the flux expression, but not the accounting rule: dissolution must be equal and opposite in the two state balances. Numerical simulations should also be checked for implausible negative amounts, which may signal an implementation or solver problem.