Why an Injection Can Produce Delayed Input
An intramuscular dose may leave the syringe as a solution yet form a precipitate after dilution in tissue fluid. A suspension already contains particles that must dissolve. In either case, drug must become locally dissolved before it can cross into systemic circulation. If that dissolution is slow relative to absorption from the local dissolved pool, the injection site acts as a release-limiting depot: systemic input can be much slower and more delayed than a model that absorbs the whole dose directly.
The key modelling distinction is between release from the solid depot and absorption from the dissolved depot. These are separate processes unless absorption is sufficiently rapid that dissolved drug does not accumulate locally. The equations below describe mass at the injection site; the systemic input rate is the amount entering the circulation per unit time.
A Sequential Dissolution–Absorption Model
First consider an injection that leaves a solid depot of mass \(D_s\), with no dissolved drug initially and no local loss. Assume the solid dissolves at a rate proportional to its remaining mass, and dissolved drug is absorbed at a rate proportional to its local mass:
The first equation says that dissolution transfers mass out of the solid pool at rate \(k_dM_s\). The second says that this transfer enters the local dissolved pool, from which drug leaves for the systemic circulation at rate \(k_aM_l\). Thus \(k_dM_s\) is not generally the systemic input rate: it is the rate entering the dissolved pool. The two rates are equal only at times when the dissolved pool is not changing.
Solving the first equation gives \(M_s(t)=D_se^{-k_dt}\). Substituting this into the second equation and solving with \(M_l(0)=0\) gives, for \(k_a\ne k_d\):
When \(k_a=k_d=k\), the corresponding solution is \(M_l(t)=D_skte^{-kt}\), so \(R_{\mathrm{sys}}(t)=D_sk^2te^{-kt}\). These expressions are nonnegative for positive rate constants; when the denominator \(k_a-k_d\) is negative, the difference of exponentials is negative as well.
Peak Timing and Mass Balance
For unequal rates, the peak systemic input occurs where the derivative of \(R_{\mathrm{sys}}(t)\) is zero. Differentiating the expression above and setting the result to zero gives the condition \(k_ae^{-k_at}=k_de^{-k_dt}\). Rearranging yields:
This time is positive for either ordering of \(k_a\) and \(k_d\). At the peak, \(dM_l/dt=0\), so the dissolved pool is momentarily constant and \(k_dM_s=k_aM_l=R_{\mathrm{sys}}\). This equality at the peak does not mean that dissolution flux and systemic input are identical throughout the profile.
With no local loss, all the initial solid mass eventually dissolves and is absorbed. More formally, integrating the local mass balance gives \(M_s(t)+M_l(t)+\int_0^t R_{\mathrm{sys}}(u)\,du=D_s\). As \(t\) tends to infinity, both local pools tend to zero, so cumulative systemic input tends to \(D_s\). If an overall systemic availability fraction \(F\) is applied after the local absorption step, the systemic input and eventual systemic amount are multiplied by \(F\).
Worked Example: A Dose That Precipitates at the Injection Site
Suppose a hypothetical 100 mg injection rapidly partitions into 70 mg of precipitated drug and 30 mg that remains dissolved. Let \(k_d=0.20\ \mathrm{h}^{-1}\), \(k_a=1.00\ \mathrm{h}^{-1}\), and assume no local loss or additional precipitation. Calculate systemic input at 1 hour and check the eventual amount absorbed.
The 30 mg initially dissolved component is absorbed directly from its local pool, contributing \(30e^{-k_at}\) mg to \(M_l(t)\), and hence \(k_a(30)e^{-k_at}\) to systemic input. The 70 mg solid component contributes the sequential input from the derived model. At \(t=1\ \mathrm{h}\):
The first term is \(30(0.367879)=11.036\ \mathrm{mg/h}\). The second is \(17.5(0.818731-0.367879)=7.890\ \mathrm{mg/h}\). Therefore \(R_{\mathrm{sys}}(1)=18.926\ \mathrm{mg/h}\), or about \(18.93\ \mathrm{mg/h}\). With no local loss, both initial pools are eventually absorbed, so total systemic input integrated over time is 30 mg plus 70 mg, or 100 mg.
Worked Example: Comparing a Precipitated Depot With Direct First-Order Absorption
Consider a hypothetical 80 mg solid depot with \(k_d=0.10\ \mathrm{h}^{-1}\) and \(k_a=0.80\ \mathrm{h}^{-1}\). Assume no local loss and compare the time course with direct first-order absorption of 80 mg at \(k_a=0.80\ \mathrm{h}^{-1}\). For a fair comparison of systemic amounts, apply \(F=0.80\) to both models.
For the sequential model, the peak time is:
At that time, the local dissolution flux is \(k_dM_s(t_{\max})\). Substitution gives \(0.10(80)e^{-0.10(2.971)}\approx5.944\ \mathrm{mg/h}\). At the peak, this equals the local absorption flux \(k_aM_l\), by the peak condition derived above. Applying \(F=0.80\), the systemic input peak is \(0.80(5.944)=4.755\ \mathrm{mg/h}\).
For direct first-order absorption, systemic input is \(R_{\mathrm{sys}}(t)=F k_aD e^{-k_at}\). At 4 hours:
If the direct model is instead written with the bioavailability factor incorporated into the available dose, its rate constant and dose expression are \(0.80(80)e^{-0.80(4)}=64e^{-3.2}\approx2.609\ \mathrm{mg/h}\). This latter expression treats the coefficient \(0.80\) as the absorption rate constant and applies no additional \(F\). Do not multiply by \(F\) twice. Under that convention the direct model's initial input is \(0.80(80)=64\ \mathrm{mg/h}\); under the matched-\(F\) convention above it is \(51.2\ \mathrm{mg/h}\). Either direct model begins with its largest input at time zero, unlike a pure solid depot, whose systemic input begins at zero and rises to a later peak. The conventions must be stated before comparing numerical rates.
Adding Local Loss or Ongoing Precipitation
Dissolved drug may be lost locally, for example through degradation or a route of removal not represented as systemic absorption. With first-order local loss \(k_lM_l\), the balances become:
For a dose initially all in the solid pool and \(k_a+k_l\ne k_d\), the dissolved mass has the same sequential form with \(k_a\) replaced by \(k_a+k_l\). The fraction of dissolved drug that ultimately enters the systemic circulation is \(k_a/(k_a+k_l)\), because absorption and local loss are competing first-order exits from that pool. This fraction is not an additional bioavailability term to apply if the model already represents those pathways explicitly.
Worked Example: Quantifying the Effect of Local Loss
Suppose a hypothetical 50 mg solid depot has \(k_d=0.25\ \mathrm{h}^{-1}\), \(k_a=0.50\ \mathrm{h}^{-1}\), and \(k_l=0.25\ \mathrm{h}^{-1}\). Calculate the systemic input at 2 hours and the eventual amount absorbed systemically.
Here the total exit rate from the dissolved pool is \(k_a+k_l=0.75\ \mathrm{h}^{-1}\). Thus:
Systemic input is \(k_aM_l(2)=0.50(9.585)=4.793\ \mathrm{mg/h}\). Eventually, the proportion of the 50 mg that enters systemic circulation is \(0.50/(0.50+0.25)=2/3\). The eventual systemic amount is therefore \(50(2/3)=33.33\ \mathrm{mg}\); the remaining 16.67 mg is lost locally. The total is consistent with mass balance.
If an initially dissolved injection precipitates gradually after entering tissue, a simple extension allows reversible transfer between the local dissolved and solid pools. Under constant pseudo-first-order rates, precipitation at rate \(k_{\mathrm{prec}}M_l\) and dissolution at rate \(k_dM_s\) give:
Precipitation is a transfer between local pools, not a loss of total drug. These equations assume the precipitation rate can be represented by a constant proportional rate; a solubility threshold or concentration-dependent nucleation may require a different formulation. The parameters \(k_{\mathrm{prec}}\) and \(k_d\) should not be interpreted as measured physical constants unless the model and data support that interpretation.
Assumptions, Diagnostics, and Alternatives
The sequential model is useful when its assumptions are plausible: each rate is approximately first order over the simulated range; the depot is adequately represented by one well-mixed solid pool; rate constants do not change with time; and there is no unmodelled input, loss, or return from systemic circulation to the injection site. The no-loss version also assumes that all dissolved drug can ultimately be absorbed. A heterogeneous suspension, changing particle surface area, limited solubility, local inflammation, or changing tissue perfusion may violate these simplifications.
Useful checks are based on the predicted shape and mass balance. A pure solid depot predicts zero systemic input at time zero, a delayed rise, and a later decline. A fitted profile that requires a large instantaneous systemic input may indicate an initial dissolved fraction or a separate rapid component. If measured exposure is lower than the administered amount can support, include a justified loss or availability term rather than allowing a dissolution rate to absorb the discrepancy. If the model predicts negative local mass, systemic input, or cumulative amount, there is an implementation or parameterization error. Also verify that rate constants use the same time unit as the time variable and that mass and mass-per-time quantities are not interchanged.
A direct first-order absorption model is simpler and may be sufficient when dissolution is rapid relative to local absorption, or when the available data do not distinguish these processes. It represents input as a single exponential decline from the available dose and does not separately predict solid mass. The sequential model is more mechanistic about the injection site and can represent a delayed peak, but it has more parameters and requires information capable of distinguishing dissolution from absorption. A lag-time model can shift an otherwise specified input profile, but does not itself represent a solid pool or ongoing dissolution. These alternatives answer different questions; added mechanistic detail is valuable only when it is defensible and useful for the prediction task.
Check Your Understanding
Use the mass-balance equations above. Each question includes a concise solution for checking your reasoning.
- For an initially solid dose with \(k_d=0.10\ \mathrm{h}^{-1}\) and \(k_a=0.80\ \mathrm{h}^{-1}\), is systemic input zero or positive at time zero? Solution: It is zero because \(M_l(0)=0\), so \(R_{\mathrm{sys}}(0)=k_aM_l(0)=0\).
- For that same model, what determines the peak time? Solution: Set the derivative of the systemic input to zero, giving \(k_ae^{-k_at}=k_de^{-k_dt}\), and hence \(t_{\max}=\ln(k_a/k_d)/(k_a-k_d)\), about 2.971 h.
- If \(k_a=k_l\), what fraction of the dissolved pool's eventual exit is systemic absorption? Solution: It is \(k_a/(k_a+k_l)=1/2\), so half of the drug that reaches the dissolved pool is eventually absorbed systemically under the stated constant-rate model.
- Why is \(k_dM_s\) not generally the systemic input? Solution: It is the rate entering the local dissolved pool. Some of that mass can remain in the pool or be lost locally; systemic input is \(k_aM_l\).
- In the reversible precipitation model, does precipitation remove drug from the total local mass? Solution: No. It transfers mass from \(M_l\) to \(M_s\); only absorption and any explicitly modelled loss reduce total local mass.