Deriving the First-Order Absorption Profile

From Depot Input to a Concentration Profile

A first-order absorption model describes drug leaving an oral depot at a rate proportional to the amount remaining there. The resulting input to the central compartment changes over time, while elimination removes drug from the central compartment. Solving the two linked differential equations gives the concentration profile after a single dose.

The derivation is useful beyond producing a formula. It shows why the formula has two exponential terms, how bioavailability enters, and why the apparent singularity when the two rate constants are equal is removable. The equations below describe a hypothetical individual with fixed parameters; in a mixed-effects analysis, the individual parameters may vary across people, but that population layer is not needed to derive this structural profile.

Definition: Let \(A_g(t)\) be the amount of drug in the oral depot at time \(t\), and let \(A_c(t)\) be the amount in a well-mixed central compartment of volume \(V\). A single dose \(D\) is placed in the depot at time zero. The absorption rate constant is \(k_a>0\), the fraction reaching the central circulation is \(0\leq F\leq 1\), and the first-order elimination rate constant is \(k_e>0\). Concentration is \(C(t)=A_c(t)/V\). With no lag time, the model is

\frac{dA_g}{dt}=-k_a A_g,\qquad A_g(0)=D, \qquad \frac{dA_c}{dt}=F k_a A_g-k_e A_c,\qquad A_c(0)=0.

Time is measured from dosing. The depot amount has units of mass, \(k_a\) and \(k_e\) have units of inverse time, \(V\) has units of volume, and concentration has units of mass per volume. The input to the central compartment is \(F k_a A_g(t)\), in mass per time. This formulation assumes a single dose, first-order absorption, linear first-order elimination, a constant bioavailability fraction, and one well-mixed central compartment. It does not include a lag time, additional distribution compartments, or dose-dependent kinetics.

Deriving the Unequal-Rate Profile

Solving the depot equation gives \(A_g(t)=D e^{-k_a t}\). Substituting this input into the central-compartment equation yields a linear differential equation with a time-varying input. Multiplying by the integrating factor \(e^{k_e t}\) makes its left side a product derivative:

\frac{d}{dt}\left(e^{k_e t}A_c(t)\right) =F k_a D e^{(k_e-k_a)t}.

Integrate from time zero to \(t\) and use \(A_c(0)=0\). When \(k_a\ne k_e\), the integral is an exponential difference. Dividing by \(V\) gives the concentration formula.

\begin{aligned} e^{k_e t}A_c(t) &=F k_a D\int_0^t e^{(k_e-k_a)s}\,ds\\ &=F k_a D\frac{e^{(k_e-k_a)t}-1}{k_e-k_a},\\[4pt] C(t) &=\frac{F D}{V}\frac{k_a}{k_a-k_e} \left(e^{-k_e t}-e^{-k_a t}\right),\qquad k_a\ne k_e. \end{aligned}

Result: For unequal positive absorption and elimination rates, the single-dose concentration is \(C(t)=\dfrac{F D}{V}\dfrac{k_a}{k_a-k_e}(e^{-k_e t}-e^{-k_a t})\). The rate constants may occur in either order: the signed denominator and the signed exponential difference work together to give a nonnegative concentration.

The factor \(F D/V\) has concentration units, and \(k_a/(k_a-k_e)\) and each exponential are dimensionless. At \(t=0\), the two exponentials are both one, so \(C(0)=0\), as required by the initial condition that no drug is in the central compartment immediately after dosing. For positive time, the concentration is nonnegative: if \(k_a>k_e\), both \(k_a-k_e\) and \(e^{-k_e t}-e^{-k_a t}\) are positive; if \(k_a<k_e\), both are negative. In either case their ratio is positive.

Worked Example: Absorption Faster Than Elimination

Suppose a hypothetical 100 mg oral dose has \(F=1\), \(V=10\ \mathrm{L}\), \(k_a=1.2\ \mathrm{h}^{-1}\), and \(k_e=0.2\ \mathrm{h}^{-1}\). Calculate the concentration at \(t=1\ \mathrm{h}\).

The rates differ, so use the unequal-rate formula. The dose-to-volume factor is \(F D/V=100/10=10\ \mathrm{mg/L}\):

\begin{aligned} C(1) &=10\frac{1.2}{1.2-0.2} \left(e^{-0.2}-e^{-1.2}\right)\ \mathrm{mg/L}\\ &=10(1.2)(0.81873075-0.30119421)\ \mathrm{mg/L}\\ &\approx 6.210\ \mathrm{mg/L}. \end{aligned}

The signs are consistent: \(k_a-k_e=1.0\ \mathrm{h}^{-1}\) is positive, and the exponential difference is also positive. The result is a concentration, not an amount.

The Equal-Rate Case

The unequal-rate expression appears to divide by zero if \(k_a=k_e\), but the differential equations themselves remain well-defined. The correct procedure is to solve them at equal rates or take the continuous limit of the unequal-rate result. In the integral solution, if \(k_a=k_e\), the integrand \(e^{(k_e-k_a)s}\) is exactly one. Therefore:

\begin{aligned} e^{k_a t}A_c(t) &=F k_a D\int_0^t 1\,ds =F k_a D t,\\ C(t)&=\frac{F D}{V}k_a t e^{-k_a t}, \qquad k_a=k_e. \end{aligned}

The same result follows by taking a limit. Write \(k_e=k_a+\delta\), where \(\delta\) approaches zero. Then \(e^{-k_e t}-e^{-k_a t}=e^{-k_a t}(e^{-\delta t}-1)\), and \((e^{-\delta t}-1)/(-\delta)\) approaches \(t\). Also \(k_a/(k_a-k_e)=-k_a/\delta\). Their product therefore approaches \(k_a t e^{-k_a t}\), giving the equal-rate expression. This verifies that the profile changes continuously through \(k_a=k_e\); the singularity is in the unequal-rate algebraic form, not in the model.

Worked Example: Exactly Equal Rates

Consider a hypothetical 80 mg oral dose with \(F=0.8\), \(V=8\ \mathrm{L}\), and \(k_a=k_e=0.4\ \mathrm{h}^{-1}\). Find the concentration at \(t=3\ \mathrm{h}\).

Here \(F D/V=(0.8)(80)/8=8\ \mathrm{mg/L}\). Since the rate constants are equal, use the limiting formula rather than substituting them into a zero denominator:

\begin{aligned} C(3) &=8(0.4)(3)e^{-(0.4)(3)}\ \mathrm{mg/L}\\ &=9.6e^{-1.2}\ \mathrm{mg/L}\\ &=9.6(0.30119421)\ \mathrm{mg/L}\\ &\approx 2.891\ \mathrm{mg/L}. \end{aligned}

The result is positive and has concentration units. The equal-rate profile is not zero or infinite; it is the finite limit of the unequal-rate profiles.

Checking the Formula and Its Numerical Use

Several checks help catch algebraic or implementation errors before a profile is used in a model. First, at the dosing instant the central amount is zero. Second, for \(t>0\) and positive rates, the concentration is nonnegative. Third, as \(t\) becomes large, the exponential terms tend to zero; in the equal-rate case, \(t e^{-k_a t}\) also tends to zero. These checks follow from the stated single-dose model and do not require choosing which rate is larger.

The initial slope provides another useful check. From the central-compartment equation, \(dA_c/dt=F k_a D\) at \(t=0\), because \(A_c(0)=0\). Thus \(dC/dt=F k_a D/V\) at that instant. The unequal-rate formula has the same initial slope: differentiating it at zero gives \((F D/V)[k_a/(k_a-k_e)](k_a-k_e)=F k_a D/V\). The equal-rate formula also gives this slope when differentiated at zero.

Worked Example: Elimination Faster Than Absorption

For a hypothetical 140 mg dose, let \(F=1\), \(V=10\ \mathrm{L}\), \(k_a=0.25\ \mathrm{h}^{-1}\), and \(k_e=0.80\ \mathrm{h}^{-1}\). Calculate \(C(2\ \mathrm{h})\). This example checks the formula when \(k_a<k_e\).

The concentration factor is \(F D/V=140/10=14\ \mathrm{mg/L}\). Both the rate difference in the denominator and the exponential difference are negative:

\begin{aligned} C(2) &=14\frac{0.25}{0.25-0.80} \left(e^{-(0.80)(2)}-e^{-(0.25)(2)}\right)\ \mathrm{mg/L}\\ &=14\frac{0.25}{-0.55} \left(e^{-1.6}-e^{-0.5}\right)\ \mathrm{mg/L}\\ &\approx14(-0.4545455)(0.20189652-0.60653066)\ \mathrm{mg/L}\\ &\approx 2.575\ \mathrm{mg/L}. \end{aligned}

A negative concentration would signal a sign or ordering error. Here the two negative factors multiply to a positive result, as the model requires.

When \(k_a\) and \(k_e\) are close but not exactly equal, directly subtracting nearly equal exponentials and dividing by a small rate difference can lose numerical precision in finite-precision computation. For hand calculation at ordinary precision, the unequal-rate formula may still be adequate; in software, a stable exponential-difference calculation or the equal-rate limit near equality can reduce cancellation. Do not simply set the denominator to zero, and do not interpret a numerical instability as a biological discontinuity.

Assumptions, Alternatives, and Limits

This profile is the analytic solution to the stated two-compartment-in-series system: a depot supplies a single central compartment, and elimination is first order from that central compartment. Its simplicity depends on those assumptions. A measured concentration profile that systematically departs from this shape may indicate that the model structure or assumptions are inadequate; the formula alone cannot identify which alternative mechanism is responsible.

An intravenous bolus has a different input condition: the dose is in the central compartment at time zero, producing \(C(t)=D e^{-k_e t}/V\) under the same one-compartment linear elimination assumption. The oral profile instead begins with zero central amount and a time-varying absorption input, which is why it contains two exponential terms (or \(t e^{-k_a t}\) when the rates coincide). A numerical solution of the same differential equations is an alternative way to compute concentrations, and can be useful when extending the model makes a closed form inconvenient. It should agree with the analytic expressions here under these assumptions. Neither computation fixes a misspecified absorption or disposition model.

A common mistake is to assume the absorption rate must be greater than the elimination rate, then use a formula that silently changes sign or fails when \(k_a<k_e\). The derived expression works in either ordering because both the denominator and the exponential difference change sign together. Another mistake is to plug equal rates into the unequal-rate expression. The correction is to use \(C(t)=(F D/V)k_a t e^{-k_a t}\), derived from the differential equations or as their continuous limit. These checks are particularly important when the formula is coded for individual parameter values in a nonlinear mixed-effects model.

Key takeaway: The first-order oral profile follows by solving the depot equation and integrating its absorption flux into the central compartment. Use the unequal-rate expression when \(k_a\ne k_e\), and its finite limit when \(k_a=k_e\); in both cases check units, initial concentration, and nonnegativity.

Check Your Understanding

Answer each question, then compare your reasoning with the solution included.

For a single oral dose under this model, what is the central-compartment input rate at time \(t\)?
Solution: Since \(A_g(t)=D e^{-k_a t}\), the input rate is \(F k_a A_g(t)=F k_a D e^{-k_a t}\), with units of mass per time.
Why does the unequal-rate concentration remain nonnegative when \(k_a<k_e\)?
Solution: Then \(k_a-k_e<0\), and for \(t>0\), \(e^{-k_e t}-e^{-k_a t}<0\). The quotient of these two negative quantities is positive, as are \(F D/V\) and \(k_a\).
A model has \(k_a=k_e=0.6\ \mathrm{h}^{-1}\). Which concentration expression should be used?
Solution: Use \(C(t)=(F D/V)(0.6\ \mathrm{h}^{-1})t e^{-(0.6\ \mathrm{h}^{-1})t}\). The unequal-rate expression has a zero denominator at equal rates and must be replaced by its limit.
For the unequal-rate model, verify the initial concentration by setting \(t=0\) in the formula. What value results?
Solution: Both exponentials equal one at \(t=0\), so their difference is zero and \(C(0)=0\), matching \(A_c(0)=0\).