Comparing First-Order, Transit, And Zero-Order Absorption SDEs

Why Compare Absorption SDEs?

Several oral absorption mechanisms can produce similar concentrations at a few observation times while implying different behavior between those times. A first-order depot releases a declining fraction of its remaining amount; a transit chain represents delayed passage through sequential stages; and a zero-order model releases drug at a constant rate for a finite interval. Comparing their expected profiles under matched conditions separates the consequences of the assumed input shape from those of disposition.

The comparison below uses a shared central-compartment SDE. This keeps the stochastic disposition component explicit while making clear which features arise from the absorption mechanism. The mean profiles are not individual sample paths: diffusion changes path variability even when it leaves the expected drift profile unchanged.

Definition: Let $C_t$ be central concentration, $V$ the apparent volume, $k_e>0$ the elimination rate, and $I_t$ the nonnegative absorption input rate in amount per time. We use $$ dC_t=\left(\frac{I_t}{V}-k_e C_t\right)dt+\sigma_c C_t\,dW_t,\qquad C_0=0, $$ where $\sigma_c\geq 0$ is constant and $W_t$ is standard Brownian motion. The absorption mechanisms below specify $I_t$, either directly or through absorption states. We assume finite parameters and finite time intervals; the linear SDEs have unique solutions with finite first and second moments on every such interval.

The Brownian motions driving absorption states may be taken independent of one another and of $W_t$. Independence is convenient for describing the models, but is not needed for the mean-profile calculations: the stochastic integrals have expectation zero when their integrands are square-integrable and adapted. Positivity follows from the multiplicative-noise form and the nonnegative inputs in these models.

Three Absorption Mechanisms

For first-order absorption, let $A_t$ be the amount in the oral depot, with dose $D>0$ at time zero. The depot and its output are

dA_t=-k_a A_t\,dt+\sigma_a A_t\,dW_{a,t},\qquad A_0=D,\qquad I_t=k_a A_t,

where $k_a>0$, $\sigma_a\geq0$, and $W_{a,t}$ is Brownian motion. The model describes a random depot trajectory whose expected amount declines at rate $k_a$; it does not assert that every realized path follows a deterministic exponential curve.

For a three-stage transit chain, let $X_{1,t},X_{2,t},X_{3,t}$ denote amounts in successive stages. With common transit rate $k_{tr}>0$, take

\begin{aligned} dX_{1,t}&=-k_{tr}X_{1,t}\,dt+\sigma_1X_{1,t}\,dW_{1,t},\\ dX_{2,t}&=k_{tr}(X_{1,t}-X_{2,t})\,dt+\sigma_2X_{2,t}\,dW_{2,t},\\ dX_{3,t}&=k_{tr}(X_{2,t}-X_{3,t})\,dt+\sigma_3X_{3,t}\,dW_{3,t},\\ (X_{1,0},X_{2,0},X_{3,0})&=(D,0,0),\qquad I_t=k_{tr}X_{3,t}. \end{aligned}

This is a sequential-delay model: material must pass through each stage before it enters the central compartment. The zero initial values in stages 2 and 3 do not create negative amounts; each stage receives nonnegative input from the preceding stage.

For finite-duration zero-order release, the input is constant through duration $T>0$, then stops:

I_t= \begin{cases} D/T,&0\leq t<T,\\ 0,&t\geq T. \end{cases}

This formulation represents a fixed-rate release in the mean model, with stochasticity in central disposition. It is not a claim that every real formulation has a perfectly constant release rate. A model that adds random variation to the release rate must also specify how it prevents cumulative release from exceeding the available dose.

Deriving the Expected Concentration Profiles

Write $m(t)=E[C_t]$. Taking expectations in the central SDE gives $m'(t)=E[I_t]/V-k_em(t)$, provided the stochastic integral has mean zero. In the absorption models above, the linear drift equations similarly imply that the expected input is determined by the corresponding deterministic drift system. Solving the resulting linear equation gives the useful convolution form:

Result: If $E[I_t]$ is integrable on finite intervals, then $$ m(t)=\frac{1}{V}\int_0^t E[I_s]e^{-k_e(t-s)}\,ds. $$ The factor $e^{-k_e(t-s)}$ is the fraction of concentration contribution remaining at time $t$ from input at time $s$.

To justify the equation, the integrating-factor solution of the mean equation is $m(t)=e^{-k_et}m(0)+V^{-1}\int_0^t e^{-k_e(t-s)}E[I_s]ds$. Since $m(0)=0$, it is the stated formula. The expectation step is valid because each linear SDE has finite second moments on bounded time intervals; hence its Brownian integrand is square-integrable there, and its Itô integral is a mean-zero martingale. This establishes the result without exchanging an uncontrolled limit and integral. $\square$

For the first-order depot, $E[A_t]$ solves $dE[A_t]/dt=-k_aE[A_t]$, so $E[I_t]=k_aD e^{-k_at}$. If $k_a\ne k_e$, direct integration yields

m_{\mathrm{FO}}(t)=\frac{D}{V}\frac{k_a}{k_a-k_e} \left(e^{-k_et}-e^{-k_at}\right).

When $k_a=k_e$, the continuous limiting expression is $m_{\mathrm{FO}}(t)=(D/V)k_at e^{-k_at}$. This equal-rate case should be evaluated from the limit rather than by dividing by zero in the general formula.

For the three-stage chain, the expected state equations are the same linear chain without diffusion terms. Its expected output rate is $E[I_s]=Dk_{tr}^3s^2e^{-k_{tr}s}/2$. Substitution into the convolution gives, with $\delta=k_{tr}-k_e\ne0$,

m_{\mathrm{tr}}(t)=\frac{D}{V}\frac{k_{tr}^3}{2\delta^3}e^{-k_et} \left[2-e^{-\delta t}\left((\delta t)^2+2\delta t+2\right)\right].

This expression is also valid when $\delta<0$, though the integral form is often numerically safer when the rates are close. At $k_{tr}=k_e$, use the integral form or its continuous limit.

For zero-order release, integration over the active interval gives

m_{\mathrm{ZO}}(t)= \begin{cases} \dfrac{D}{Vk_e}\left(1-e^{-k_et}\right),&0\leq t<T,\\[6pt] \dfrac{D}{Vk_e}\left(1-e^{-k_eT}\right)e^{-k_e(t-T)},&t\geq T. \end{cases}

Worked Numerical Comparison

The next examples use the same dose and disposition parameters so that differences in the predicted means are attributable to absorption shape. Stochastic coefficients affect dispersion around these means, but not the mean calculations shown.

Worked Example: First-Order Depot

Take $D=100$ mg, $V=1$ L, $k_a=1$ h$^{-1}$, and $k_e=0.2$ h$^{-1}$. The mean concentration is $m_{\mathrm{FO}}(t)=125(e^{-0.2t}-e^{-t})$ mg/L.

At $t=2$ h, $e^{-0.4}=0.6703200$ and $e^{-2}=0.1353353$, so $m_{\mathrm{FO}}(2)=125(0.5349847)=66.8731$ mg/L, or $66.87$ mg/L to two decimal places.

The peak occurs where the derivative is zero: $e^{-k_at}k_a=e^{-k_et}k_e$, which gives $t_{\max}=\log(k_a/k_e)/(k_a-k_e)$. Here $t_{\max}=\log(5)/0.8=2.0118$ h. A peak close to 2 h is therefore consistent with the calculated value at 2 h.

Worked Example: Three-Stage Transit Chain

Use the same $D=100$ mg, $V=1$ L, and $k_e=0.2$ h$^{-1}$. Set $k_{tr}=3$ h$^{-1}$. The mean transit time is $3/k_{tr}=1$ h, matching the first-order model's mean absorption time $1/k_a=1$ h. Thus $\delta=2.8$ h$^{-1}$, and the leading coefficient before $e^{-0.2t}$ and the bracket is $100(27)/[2(2.8)^3]=61.4978$.

At $t=2$ h, the bracket is $2-e^{-5.6}(31.36+11.2+2)=1.835223$, and the leading coefficient after multiplication by $e^{-0.4}$ is $61.4978e^{-0.4}\approx41.2232$. Therefore $m_{\mathrm{tr}}(2)\approx41.2232(1.835223)=75.6538$ mg/L, or $75.65$ mg/L.

At $t=1$ h the formula gives approximately $53.43$ mg/L. At $t=4$ h, the bracket is $2-e^{-11.2}(125.44+22.4+2)\approx1.997951$; the leading coefficient is $61.4978e^{-0.8}\approx27.6327$. Their product is approximately $55.2089$ mg/L, or $55.21$ mg/L. These evaluations use the unrounded coefficient in the products.

Worked Example: Finite Zero-Order Release

Use $D=100$ mg, $V=1$ L, $k_e=0.2$ h$^{-1}$, and release duration $T=2$ h. The constant release rate is $D/T=50$ mg/h. At $t=1$ h, $m_{\mathrm{ZO}}(1)=250(1-e^{-0.2})=45.3173$ mg/L, or $45.32$ mg/L.

At the end of release, $m_{\mathrm{ZO}}(2)=250(1-e^{-0.4})=82.41999$ mg/L, or $82.42$ mg/L. For $t=4$ h, two hours have passed since release stopped: $m_{\mathrm{ZO}}(4)=82.41999e^{-0.4}=55.2478$ mg/L, or $55.25$ mg/L.

The concentration rises throughout release because the input is positive while the system approaches its release-phase balance $D/(TVk_e)=250$ mg/L. After $T$, the input is zero and the mean declines exponentially.

What the Matched Profiles Show

Time (h)	First-order (mg/L)	Three-stage transit (mg/L)	Zero-order, $T=2$ h (mg/L)
1	56.36	53.43	45.32
2	66.87	75.65	82.42
4	53.88	55.21	55.25

The equal mean absorption time does not make the models interchangeable. The first-order input is greatest immediately after dosing and then decays. The transit chain delays and concentrates expected input around a later time, producing a higher mean at 2 h in this example. The zero-order model accumulates concentration throughout its fixed release period and peaks at the stopping time. Their late means are close here, but that numerical proximity does not imply identical early input or pathwise variability.

A useful second result is that all three models have the same total expected exposure when the entire dose is eventually absorbed, bioavailability is one, and $k_e>0$:

Result: For each of these models, $$ \int_0^\infty m(t)\,dt=\frac{D}{Vk_e}. $$ For the first-order depot, the expected total input is $\int_0^\infty k_aDe^{-k_at}dt=D$. For the transit chain, the expected absorption-time density integrates to one, since it is an Erlang density. For zero-order release, $\int_0^T D/T\,dt=D$. Finally, integrating the convolution and using Tonelli's theorem for its nonnegative integrand gives $$ \int_0^\infty m(t)dt =\frac{1}{V}\int_0^\infty E[I_s]\left(\int_s^\infty e^{-k_e(t-s)}dt\right)ds =\frac{1}{Vk_e}\int_0^\infty E[I_s]ds =\frac{D}{Vk_e}. $$ The nonnegativity ensures Tonelli's theorem applies, and $k_e>0$ makes the inner integral finite. $\square$

Diagnostics, Alternatives, and Limitations

Check whether the fitted model reproduces the observed timing and shape, not only total exposure. Early samples are particularly informative about immediate versus delayed input; samples around the peak distinguish a gradual rise from a near-linear rise; and sufficiently late samples help identify elimination after absorption has ended. For the zero-order model, a sharp predicted change from increasing concentration to exponential decline at $T$ is a specific structural implication to check against data.

First-order depot: A poor fit may show systematic early or peak-time residuals if absorption has a lag or a concentrated delayed input. A single exponential release is also a strong simplification of formulation behavior.
Transit chain: Too few stages may fail to represent a narrow delay; too many can make stage rate and stage count weakly distinguishable from sparse observations. Inspect whether the inferred delay is supported by data, rather than treating a flexible chain as mechanistic proof.
Zero-order release: A fixed duration and constant rate can impose an implausibly abrupt stop. If observed profiles continue rising after the fitted stop time, or decline well before it, the assumed release interval is suspect.

In all three models, compare simulated paths as well as mean profiles when evaluating an SDE. A good fit to the mean does not establish that the diffusion scale is adequate, that variability is correctly timed, or that residual dependence has been captured. Conversely, random fluctuations in individual paths do not by themselves justify changing the mean absorption mechanism. The earlier material on calibration and model comparison provides the broader framework for evaluating predictive performance.

The main alternative is to compare simpler deterministic absorption models or a different input structure, such as a lagged or parallel absorption process. The SDE formulations here add path variability around the drift-defined mechanism; they do not automatically resolve structural ambiguity. In particular, the first-order and transit diffusion terms can generate different variances even when their expected absorption times match. Zero-order release as specified here has no stochastic absorption rate, so it should not be presented as having random release trajectories. A rate-modulated zero-order extension would need an explicit positive rate process and a defensible dose constraint.

Common Pitfall

A frequent mistake is to equate “same mean absorption time” with “same concentration profile.” Here the first-order mean absorption time is $1/k_a=1$ h, and the three-stage chain's is $3/k_{tr}=1$ h. That matches one summary of the input timing, not its distribution. The first-order absorption-time density is exponential, while the transit-chain density is proportional to $s^2e^{-k_{tr}s}$; their early and late input probabilities therefore differ. The correction is to compare the full expected input functions, then propagate each through the same disposition model using the convolution, rather than infer profile equivalence from one matched parameter.

Check Your Understanding

Answer the questions, then check the solutions below.

For the numerical first-order example, calculate the expected concentration at 1 h.
Why does matching the first-order and transit mean absorption times not force their concentration means to match?
For zero-order release with $T=2$ h, what happens to the expected concentration profile after 2 h?
Under the assumptions stated, what is the total expected AUC for $D=100$ mg, $V=1$ L, and $k_e=0.2$ h$^{-1}$?
Does adding multiplicative Brownian noise to the central compartment change the mean profile? State the condition that supports your answer.

Solutions. (1) $125(e^{-0.2}-e^{-1})=125(0.8187308-0.3678794)=56.3564$ mg/L, or $56.36$ mg/L. (2) The models have different absorption-time distributions, so their input functions differ even though their means agree; convolution with elimination preserves that difference. (3) The input stops and the expected concentration declines as $m(t)=m(2)e^{-0.2(t-2)}$ for $t\geq2$. (4) $D/(Vk_e)=100/(1\cdot0.2)=500$ mg·h/L. (5) No. The adapted stochastic integrand must be square-integrable on bounded intervals so its Itô integral is a mean-zero martingale; then the expectation follows the drift equation.