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Dermal Skin Reservoirs and Finite-Dose Depletion

Learn how sequential skin compartments turn a finite dose into a time-varying systemic input, and how to calculate its peak and check mass balance.

Intermediate 11 min read

What You'll Learn

  • Define stratum-corneum and viable-skin amounts in a sequential first-order model.
  • Derive the amount in viable skin and the systemic input rate after a finite dose.
  • Find when the systemic input rate peaks, including the equal-rate limit.
  • Calculate remaining reservoir amounts and cumulative systemic delivery.
  • Diagnose model-fit and mass-balance problems and compare alternative input representations.

Why a Finite Dermal Dose Produces a Changing Input

A dermal dose does not generally enter the systemic circulation at a constant rate. Drug can first accumulate in the stratum corneum, then transfer into viable skin, and only then leave that tissue for systemic circulation. As the finite amount in the stratum-corneum reservoir is depleted, its delivery to viable skin falls. The systemic input therefore reflects both storage and sequential transfer, not just the amount initially applied.

The model below describes the post-application phase. It treats the amount that has entered the stratum corneum as the initial reservoir amount. The earlier material on dermal surface dose and skin permeation provides context for how a finite applied dose may supply that reservoir; here, the focus is what happens after loading. This distinction matters: the modeled initial skin amount is not automatically the whole applied dose, nor is it the amount that will ultimately reach systemic circulation if local loss or metabolism occurs.

Definition: Let \(A_{\mathrm{SC}}(t)\) be the amount of drug in the stratum corneum at time \(t\), and \(A_{\mathrm{V}}(t)\) the amount in viable skin. Let \(k_1\) be the first-order transfer rate constant from stratum corneum to viable skin, and \(k_2\) the first-order rate constant for exit from viable skin to systemic circulation. Both rate constants have units of inverse time. The systemic input rate is \(R_{\mathrm{sys}}(t)=k_2A_{\mathrm{V}}(t)\), in amount per unit time.

For an initial stratum-corneum amount \(M_0\), the sequential model is

$$ A_{\mathrm{SC}}(0)=M_0,\qquad A_{\mathrm{V}}(0)=0, $$
$$ \frac{dA_{\mathrm{SC}}}{dt}=-k_1A_{\mathrm{SC}},\qquad \frac{dA_{\mathrm{V}}}{dt}=k_1A_{\mathrm{SC}}-k_2A_{\mathrm{V}},\qquad R_{\mathrm{sys}}(t)=k_2A_{\mathrm{V}}(t). $$

The assumptions are that the dose loading the stratum corneum is finite and occurs before the modeled interval; each transfer is first-order with constant rate; the compartments are well mixed; and there is no additional input, local metabolism, or loss from either skin compartment. The last assumption makes the model’s eventual systemic delivery equal to \(M_0\). If some drug is lost locally, that loss must be represented separately, and eventual delivery will be less.

Solving the Sequential Reservoir Model

The stratum-corneum amount follows directly from its first-order depletion equation. Substituting that solution into the viable-skin equation gives a linear differential equation with a known input.

$$ A_{\mathrm{SC}}(t)=M_0e^{-k_1t}. $$

For \(k_1\ne k_2\), solving for viable-skin amount gives

$$ A_{\mathrm{V}}(t)= M_0\frac{k_1}{k_2-k_1} \left(e^{-k_1t}-e^{-k_2t}\right), \qquad R_{\mathrm{sys}}(t)= M_0\frac{k_1k_2}{k_2-k_1} \left(e^{-k_1t}-e^{-k_2t}\right). $$

Although the expression contains a difference whose sign depends on the ordering of the rates, \(A_{\mathrm{V}}(t)\) and \(R_{\mathrm{sys}}(t)\) are nonnegative. For example, if \(k_2>k_1\), then \(e^{-k_1t}>e^{-k_2t}\) and the denominator is positive; if \(k_2<k_1\), both the exponential difference and denominator are negative.

When the two rates are equal, the expression with denominator \(k_2-k_1\) is not used directly. Taking its limit as \(k_2\) approaches \(k_1=k\) gives \(A_{\mathrm{V}}(t)=M_0kte^{-kt}\). The systemic input then has the same time dependence, multiplied by \(k\).

Result: If \(k_1\ne k_2\), systemic input begins at zero, rises while viable-skin amount accumulates, and later declines as the finite stratum-corneum reservoir is depleted. Its peak time is \(t_{\mathrm{peak}}=\dfrac{\ln(k_2/k_1)}{k_2-k_1}\). If \(k_1=k_2=k\), the peak occurs at \(t_{\mathrm{peak}}=1/k\).

To justify the peak formula, differentiate \(A_{\mathrm{V}}(t)\). Since \(R_{\mathrm{sys}}(t)=k_2A_{\mathrm{V}}(t)\) and \(k_2\) is constant, their peaks occur at the same time. Setting the derivative to zero gives

$$ -k_1e^{-k_1t}+k_2e^{-k_2t}=0 \quad\Longrightarrow\quad e^{-k_2t}=\frac{k_1}{k_2}e^{-k_1t}. $$

Taking logarithms and solving for \(t\) yields the stated expression. It is positive for either ordering of two distinct positive rates: numerator and denominator have the same sign. In the equal-rate case, differentiating \(M_0kte^{-kt}\) shows its maximum occurs at \(1/k\).

Worked Example: Peak Input When the Reservoir Depletes Quickly

Suppose a hypothetical \(100\) mg has entered the stratum corneum. Let \(k_1=2.00\ \mathrm{h}^{-1}\) and \(k_2=0.500\ \mathrm{h}^{-1}\). Calculate the time and magnitude of peak systemic input.

First, the peak time is

$$ t_{\mathrm{peak}}= \frac{\ln(0.500/2.00)}{0.500-2.00} =\frac{\ln(0.250)}{-1.50} =0.9242\ \mathrm{h}. $$

At this time, the peak condition gives \(e^{-k_2t}=(k_1/k_2)e^{-k_1t}\), with \(k_1/k_2=4\), not its square root. Substituting into the viable-skin amount formula, or evaluating that formula directly, gives

$$ A_{\mathrm{V}}(0.9242)= 100\frac{2.00}{0.500-2.00} \left(e^{-2.00(0.9242)}-e^{-0.500(0.9242)}\right) \approx 63.00\ \mathrm{mg}. $$

Therefore the systemic input rate at the peak is \(R_{\mathrm{sys}}=0.500(63.00)\approx31.50\ \mathrm{mg/h}\). The amount in viable skin is not itself a rate; multiplying it by \(k_2\) converts that stored amount into systemic input per hour.

Finite Dose, Depletion, and Mass Balance

A finite dose determines the starting amount \(M_0\), and the compartment equations determine how that amount is distributed over time. If the applied amount is \(D_{\mathrm{app}}\) and a fraction \(f_{\mathrm{SC}}\) is represented as entering the stratum corneum, then \(M_0=f_{\mathrm{SC}}D_{\mathrm{app}}\). The fraction is a model input or estimate, not an automatic consequence of the two-compartment equations. It should not be confused with systemic bioavailability: in this no-loss model, all of \(M_0\) eventually exits viable skin, but the applied dose may be larger than \(M_0\).

Cumulative systemic delivery by time \(t\) can be calculated by integrating the input rate. A useful alternative is mass balance: the cumulative amount already delivered is the initial skin amount minus what remains in both skin compartments.

$$ D_{\mathrm{sys}}(t)=\int_0^t R_{\mathrm{sys}}(u)\,du =M_0-A_{\mathrm{SC}}(t)-A_{\mathrm{V}}(t). $$

This identity follows by adding the two compartment equations: \(d(A_{\mathrm{SC}}+A_{\mathrm{V}})/dt=-k_2A_{\mathrm{V}}=-R_{\mathrm{sys}}\). Integrating from zero to \(t\), and using \(A_{\mathrm{SC}}(0)+A_{\mathrm{V}}(0)=M_0\), gives the result. As \(t\) tends to infinity, both skin amounts tend to zero for positive rates, so \(D_{\mathrm{sys}}(t)\) tends to \(M_0\).

Worked Example: From Applied Dose to Skin Reservoir

Suppose a hypothetical \(120\) mg finite application is represented by \(f_{\mathrm{SC}}=0.60\) entering the stratum corneum. Let \(k_1=0.500\ \mathrm{h}^{-1}\) and \(k_2=0.100\ \mathrm{h}^{-1}\). Find the initial modeled amount, peak time, and peak systemic input.

The modeled reservoir amount is \(M_0=0.60(120)=72.0\) mg. The peak time is

$$ t_{\mathrm{peak}}= \frac{\ln(0.100/0.500)}{0.100-0.500} =\frac{\ln(0.200)}{-0.400} =4.024\ \mathrm{h}. $$

At that time, \(e^{-k_1t}\approx e^{-2.012}=0.1338\) and \(e^{-k_2t}\approx e^{-0.4024}=0.6687\). Thus

$$ A_{\mathrm{V}}(t_{\mathrm{peak}}) =72.0\frac{0.500}{0.100-0.500}(0.1338-0.6687) \approx48.15\ \mathrm{mg}, \qquad R_{\mathrm{sys}}(t_{\mathrm{peak}}) =0.100(48.15) \approx4.815\ \mathrm{mg/h}. $$

The peak is delayed because drug must transfer through the stratum corneum and accumulate in viable skin. The \(48.15\) mg is the amount in viable skin at the peak, not the cumulative amount that has already reached systemic circulation.

Worked Example: Checking Skin Amounts and Cumulative Delivery

For a separate hypothetical dose, let \(M_0=40.0\) mg, \(k_1=0.400\ \mathrm{h}^{-1}\), and \(k_2=0.200\ \mathrm{h}^{-1}\). Calculate the compartment amounts and cumulative systemic delivery at \(t=4.00\) h.

The stratum-corneum amount is

$$ A_{\mathrm{SC}}(4)=40.0e^{-0.400(4.00)} =40.0e^{-1.60} \approx8.076\ \mathrm{mg}. $$

The viable-skin amount is

$$ A_{\mathrm{V}}(4)=40.0\frac{0.400}{0.200-0.400} \left(e^{-1.60}-e^{-0.800}\right) =-80.0(0.2019-0.4493) \approx19.79\ \mathrm{mg}. $$

Mass balance gives cumulative delivery of \(40.0-8.076-19.79\approx12.13\) mg by \(4.00\) h. The instantaneous systemic input rate at that time is \(0.200(19.79)\approx3.959\ \mathrm{mg/h}\). These quantities have different meanings: cumulative delivery is an amount, while the instantaneous rate is amount per time.

Diagnostics, Alternatives, and Limitations

A fitted model should satisfy basic physical checks. Predicted amounts and rates must remain nonnegative, and the sum of the remaining skin amounts and cumulative systemic delivery must equal \(M_0\) under the stated no-loss assumptions. With \(A_{\mathrm{V}}(0)=0\), systemic input must begin at zero; a fitted profile that starts with an immediate nonzero input may indicate that this initial condition is unsuitable. The predicted input should rise and then decline for a finite reservoir. Persistent late input, an unexplained early spike, or negative fitted amounts can signal model mismatch, poor parameter constraints, or data that do not resolve the sequential process.

The two-compartment model is not the only representation. A single first-order absorption model can be simpler when the data support a delayed, declining input but cannot distinguish storage in two skin layers. A constant-rate input may be reasonable over a restricted interval when input is approximately flat, but it does not represent depletion of a finite reservoir over the full time course. More spatially detailed diffusion models can represent gradients within skin, at the cost of additional parameters and data requirements. The sequential compartments are a useful intermediate representation: they preserve storage and transfer while remaining analytically transparent.

The model has important limitations. First-order rates assume transfer is proportional to the amount present, which may fail if skin binding, saturation, changing permeability, or formulation changes matter. A single well-mixed viable-skin compartment ignores spatial gradients and variation across application sites. The initial loading \(M_0\) and transfer rates can also be difficult to distinguish from systemic concentration data alone: a smaller reservoir with faster transfer may resemble a larger reservoir with slower transfer over a limited observation window. Independent skin measurements or informative prior knowledge may be needed to constrain them. Effects of occlusion or wash-off require additional processes and are not represented by this post-loading model.

A Practical Calculation Sequence

1
Set the modeled initial amount.
Define \(M_0\) as the finite amount entering the stratum-corneum reservoir, and state how it relates to the applied dose.
2
Specify transfer rates and units.
Use positive \(k_1\) and \(k_2\), with the same time unit for both rates and for all reported times.
3
Calculate amounts and input.
Evaluate \(A_{\mathrm{SC}}(t)\), \(A_{\mathrm{V}}(t)\), and \(R_{\mathrm{sys}}(t)=k_2A_{\mathrm{V}}(t)\) separately.
4
Check the profile and mass balance.
Confirm nonnegative amounts, the expected delayed peak, and \(D_{\mathrm{sys}}(t)+A_{\mathrm{SC}}(t)+A_{\mathrm{V}}(t)=M_0\).
Common pitfall: The peak condition is \(k_2e^{-k_2t}=k_1e^{-k_1t}\), so \(e^{-k_2t}=(k_1/k_2)e^{-k_1t}\). Replacing the rate ratio by its square root gives an incorrect peak calculation. Another frequent error is to report \(A_{\mathrm{V}}(t)\) as systemic input; the correct rate is \(k_2A_{\mathrm{V}}(t)\), with units of amount per time.

Check Your Understanding

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

  1. If \(M_0=50.0\) mg and \(k_1=0.250\ \mathrm{h}^{-1}\), what is \(A_{\mathrm{SC}}(4.00\ \mathrm{h})\)?
  2. For \(k_1=0.300\ \mathrm{h}^{-1}\) and \(k_2=0.600\ \mathrm{h}^{-1}\), calculate the peak time.
  3. If \(A_{\mathrm{V}}=12.0\) mg and \(k_2=0.150\ \mathrm{h}^{-1}\), what is the systemic input rate?
  4. Under the no-loss assumptions, \(M_0=30.0\) mg, \(A_{\mathrm{SC}}=5.0\) mg, and \(A_{\mathrm{V}}=9.0\) mg at a particular time. How much has been delivered systemically?

Solutions. (1) \(A_{\mathrm{SC}}=50.0e^{-0.250(4.00)}=50.0e^{-1}=18.39\) mg. (2) \(t_{\mathrm{peak}}=\ln(0.600/0.300)/(0.600-0.300)=\ln(2)/0.300=2.310\) h. (3) \(R_{\mathrm{sys}}=0.150(12.0)=1.80\) mg/h. (4) Mass balance gives \(D_{\mathrm{sys}}=30.0-5.0-9.0=16.0\) mg.