Zero-Order Drug Input

Why Model Constant-Rate Input?

A drug can enter the systemic circulation at an approximately constant rate for a limited period. An intravenous infusion is a direct example; a controlled-release product may also be represented this way if its systemic delivery is sufficiently close to constant over the interval of interest. The resulting concentration rises while drug enters, then declines after input stops. A useful structural model must describe both phases and make clear how the profile depends on input rate, duration, clearance, and volume.

The key distinction is between a constant rate of systemic input and a constant rate of drug leaving a formulation or absorption site. For an infusion, the prescribed infusion rate is the systemic input rate. For a controlled-release dose, release from the formulation is not automatically equal to systemic input: the fraction reaching the circulation and the timing of absorption also matter. This tutorial models the systemic input rate itself as constant during a defined interval.

Definition: Zero-order drug input is a model in which a fixed amount of drug enters the systemic compartment per unit time over a specified interval. Let \(R_{\mathrm{in}}\) denote that constant rate, in amount per time. In the one-compartment linear model, the central amount \(A(t)\) is eliminated at rate \(kA(t)\), where \(k=CL/V\), \(CL\) is clearance, and \(V\) is the apparent volume of distribution. Concentration is \(C(t)=A(t)/V\).

The model assumes one well-mixed compartment, constant \(CL\) and \(V\), and first-order elimination over the concentration range considered. Input begins at time zero, remains constant until time \(T\), and then stops. The basic profile below assumes no drug is present at time zero. These assumptions allow a compact derivation; if concentration data show multiple disposition phases or a changing input rate, the model may not be adequate.

Deriving the Concentration Profile

During input, the amount increases by \(R_{\mathrm{in}}\) and decreases by elimination, so its rate of change is input minus elimination. Dividing by \(V\), and using \(k=CL/V\), gives the concentration equation.

\frac{dA}{dt}=R_{\mathrm{in}}-kA,\qquad \frac{dC}{dt}=\frac{R_{\mathrm{in}}}{V}-kC,\qquad 0\le t\le T.

To solve it, multiply the concentration equation by the integrating factor \(e^{kt}\). The left side becomes the derivative of \(e^{kt}C(t)\). Integrating from zero to \(t\), with \(C(0)=0\), yields the concentration during input:

\frac{d}{dt}\left(e^{kt}C(t)\right) =\frac{R_{\mathrm{in}}}{V}e^{kt}

e^{kt}C(t)=\frac{R_{\mathrm{in}}}{Vk}\left(e^{kt}-1\right), \qquad C(t)=\frac{R_{\mathrm{in}}}{CL}\left(1-e^{-kt}\right),\quad 0\le t\le T.

Since input ceases at \(T\), the rate equation after that time is \(dC/dt=-kC\). Concentration is continuous at the stopping time: the input switches off, but the drug already in the compartment does not disappear instantaneously. Solving the elimination equation with initial value \(C(T)\) gives the post-input profile.

C(t)= \begin{cases} \displaystyle \frac{R_{\mathrm{in}}}{CL}\left(1-e^{-kt}\right), & 0\le t\le T,\\[6pt] \displaystyle \frac{R_{\mathrm{in}}}{CL}\left(1-e^{-kT}\right)e^{-k(t-T)}, & t>T. \end{cases}

The concentration at the end of input is \(C(T)=\frac{R_{\mathrm{in}}}{CL}(1-e^{-kT})\). It is the maximum concentration in this model: while input is on, the concentration approaches \(R_{\mathrm{in}}/CL\) from below; after input ends, it declines. The value \(R_{\mathrm{in}}/CL\) is the steady-state concentration that would be approached if the same rate continued indefinitely, not the concentration reached immediately or necessarily reached during a finite input.

Worked Example: An Infusion Followed by Washout

Suppose a hypothetical one-compartment drug is infused at \(30\ \mathrm{mg/h}\) for \(4\ \mathrm{h}\). Let \(CL=5\ \mathrm{L/h}\) and \(V=40\ \mathrm{L}\). Calculate the concentration at the end of infusion and \(3\ \mathrm{h}\) later. The calculation assumes constant clearance and volume, linear elimination, and an infusion rate that is constant over the stated interval.

First, \(k=CL/V=(5\ \mathrm{L/h})/(40\ \mathrm{L})=0.125\ \mathrm{h^{-1}}\). The concentration at \(T=4\ \mathrm{h}\) is

C(4)=\frac{30\ \mathrm{mg/h}}{5\ \mathrm{L/h}} \left(1-e^{-(0.125\ \mathrm{h^{-1}})(4\ \mathrm{h})}\right) =6(1-e^{-0.5})=2.361\ \mathrm{mg/L}.

Three hours after stopping, the post-input equation gives

C(7)=C(4)e^{-(0.125)(3)} =(2.3608\ \mathrm{mg/L})e^{-0.375} =1.623\ \mathrm{mg/L}.

The result is below the end-of-infusion concentration, as expected during washout. The units also agree: the rate divided by clearance has units \((\mathrm{mg/h})/(\mathrm{L/h})=\mathrm{mg/L}\), and the exponential factor is dimensionless.

Using the Model for a Controlled-Release Dose

For a controlled-release dose \(D\) delivered over duration \(T\), a simple approximation is a constant release rate \(D/T\). If a constant fraction \(F\) of the dose reaches the systemic circulation during that interval, the systemic input rate is \(R_{\mathrm{in}}=FD/T\). This relationship is an assumption of the model, not a general guarantee about formulation release or absorption. Once the systemic rate is specified, the concentration equations are the same as for an infusion.

Worked Example: Converting a Controlled-Release Dose to Systemic Input

Consider a hypothetical \(300\ \mathrm{mg}\) controlled-release dose delivered over \(6\ \mathrm{h}\), with \(F=0.80\). Assume the systemic availability is constant over the interval, so systemic input is constant. Let \(CL=4\ \mathrm{L/h}\) and \(V=32\ \mathrm{L}\). Find the systemic input rate, the concentration at the end of input, the concentration \(8\ \mathrm{h}\) after input ends, and total AUC.

The amount entering systemically is \(FD=(0.80)(300\ \mathrm{mg})=240\ \mathrm{mg}\). Therefore,

R_{\mathrm{in}}=\frac{FD}{T} =\frac{240\ \mathrm{mg}}{6\ \mathrm{h}} =40\ \mathrm{mg/h},\qquad k=\frac{CL}{V}=\frac{4}{32}=0.125\ \mathrm{h^{-1}}.

At \(T=6\ \mathrm{h}\), the concentration is

C(6)=\frac{40\ \mathrm{mg/h}}{4\ \mathrm{L/h}} \left(1-e^{-(0.125)(6)}\right) =10(1-e^{-0.75}) =5.276\ \mathrm{mg/L}.

Eight hours after input ends, the washout factor is \(e^{-(0.125\ \mathrm{h^{-1}})(8\ \mathrm{h})}=e^{-1}\). Thus \(C(14)=(5.2763\ \mathrm{mg/L})e^{-1}=1.941\ \mathrm{mg/L}\). Under linear elimination with complete eventual washout, total AUC is systemic amount divided by clearance: \(240\ \mathrm{mg}/(4\ \mathrm{L/h})=60.0\ \mathrm{mg\cdot h/L}\). The input duration changes the shape of the concentration profile, but under these assumptions it does not change total AUC for a fixed systemically available amount.

Exposure, Checks, and Diagnostics

The full AUC result can be justified directly from the amount balance. Integrate \(dA/dt=R_{\mathrm{in}}(t)-CL\,C(t)\) over the complete profile, from before input to after all drug has been eliminated. The amount begins and ends at zero, so the net change in amount is zero. Consequently, total input equals \(CL\) times total AUC. For a constant systemic rate \(R_{\mathrm{in}}\) lasting \(T\), total input is \(R_{\mathrm{in}}T\), giving the result below.

0=R_{\mathrm{in}}T-CL\int_0^\infty C(t)\,dt, \qquad AUC_{0-\infty}=\frac{R_{\mathrm{in}}T}{CL}.

For a controlled-release dose under the stated assumptions, \(R_{\mathrm{in}}T=FD\), so \(AUC_{0-\infty}=FD/CL\). This is a mass-balance check on the profile, not a claim that different input patterns have identical concentrations at every time. A faster or slower input can produce a different peak and time course while preserving total exposure when the available amount and clearance are unchanged.

Worked Example: Choosing a Rate to Reach a Target Concentration

For a hypothetical drug, let \(CL=6\ \mathrm{L/h}\) and \(V=30\ \mathrm{L}\). Choose a constant input rate that produces \(4.00\ \mathrm{mg/L}\) at \(t=3\ \mathrm{h}\), then continue that rate until \(t=6\ \mathrm{h}\). Calculate the concentration at \(6\ \mathrm{h}\), the amount systemically input over six hours, and the concentration two hours after stopping.

Here \(k=6/30=0.200\ \mathrm{h^{-1}}\). Rearranging the during-input equation at \(t=3\ \mathrm{h}\) gives

R_{\mathrm{in}}= \frac{C(3)CL}{1-e^{-k(3\ \mathrm{h})}} =\frac{(4.00\ \mathrm{mg/L})(6\ \mathrm{L/h})}{1-e^{-0.600}} =\frac{24}{0.451188} =53.1929\ \mathrm{mg/h}.

The rate continues after the target time, so concentration rises further. At \(t=6\ \mathrm{h}\),

C(6)=\frac{53.1929\ \mathrm{mg/h}}{6\ \mathrm{L/h}} (1-e^{-1.200}) =6.195\ \mathrm{mg/L}.

The systemic amount input over six hours is \((53.192861\ldots\ \mathrm{mg/h})(6\ \mathrm{h})=319.2\ \mathrm{mg}\). Two hours after stopping, the concentration is

C(8)=C(6)e^{-(0.200)(2)} =(6.1953\ \mathrm{mg/L})e^{-0.400} =4.153\ \mathrm{mg/L}.

This example also illustrates why the duration must be specified: the rate was selected to reach the target at three hours, but continuing input for another three hours raises the concentration above that target.

Several checks help identify a calculation or model problem. At \(t=0\), the predicted concentration is zero for the assumed initial condition. During input, it must rise toward \(R_{\mathrm{in}}/CL\), never exceed that asymptote under the model, and at \(T\) it must join continuously to the post-input curve. After stopping, the log concentration has slope \(-k\) against time if elimination remains first order and the one-compartment assumption is adequate. Units should also be consistent: \(k\) has units of inverse time, and \(R_{\mathrm{in}}/CL\) has concentration units.

In data, systematic curvature in the post-input log concentration profile, an apparent concentration jump at the stop time, or a rising concentration after input has stopped can indicate a mismatch between the assumed profile and observations, incorrect dose or time records, or a more complex disposition or input process. Such patterns are diagnostic clues, not proof of a particular alternative model. Sampling times must capture the input period and washout well enough to distinguish a changing input from elimination.

Alternatives and Limits of the Approximation

The main alternatives differ in the shape of input or disposition they represent. An instantaneous bolus places the dose in the compartment at one time and predicts an immediate concentration change; it is unsuitable when delivery is spread over a meaningful duration. First-order absorption describes input that declines in proportion to the amount remaining at an absorption site, rather than a fixed rate. The earlier material on constant-rate IV infusion gives the infusion case; the equations here apply the same constant systemic input logic to a finite interval and to controlled-release input when its assumptions are reasonable.

A zero-order input model is most defensible when the rate delivered to the systemic circulation is approximately constant over the modeled interval. A formulation may release drug at a constant rate while absorption is delayed or variable, so the systemic input need not be constant. Likewise, if concentration data show multiple disposition phases, the one-compartment elimination assumption can be inadequate even when the input rate is well characterized. More elaborate structural models should be considered only when data and design can support the added parameters.

Confusing release with systemic entry. A constant formulation release rate does not by itself establish a constant systemic input rate. The model requires the latter, or an explicit assumption connecting the two.
Treating \(R_{\mathrm{in}}/CL\) as the finite-duration peak. This is the asymptote for indefinitely continued input. For finite \(T\), the end concentration includes the factor \(1-e^{-kT}\).
Allowing concentration to jump when input stops. Input rate changes at \(T\), but concentration remains continuous; it then declines according to elimination.
Using total dose without accounting for availability. For a controlled-release dose, systemic input is \(FD\) under the constant-fraction assumption, not necessarily the full administered dose.

Key takeaway: With constant systemic input \(R_{\mathrm{in}}\) for duration \(T\), linear one-compartment disposition gives \(C(t)=\frac{R_{\mathrm{in}}}{CL}(1-e^{-kt})\) during input and \(C(t)=C(T)e^{-k(t-T)}\) afterward. The model’s usefulness depends on whether its assumptions describe systemic delivery and observed disposition adequately.

Check Your Understanding

Work through each item, then compare with the solutions below.

A hypothetical infusion runs at \(18\ \mathrm{mg/h}\), with \(CL=3\ \mathrm{L/h}\), \(V=20\ \mathrm{L}\), and duration \(2\ \mathrm{h}\). What is the end-of-input concentration?
A concentration is \(5.00\ \mathrm{mg/L}\) at the instant input stops. If \(k=0.100\ \mathrm{h^{-1}}\), what is it \(3\ \mathrm{h}\) later?
A \(180\ \mathrm{mg}\) controlled-release dose is delivered over \(3\ \mathrm{h}\), with \(F=0.75\). Under the constant systemic availability assumption, what is \(R_{\mathrm{in}}\)?
Why is \(R_{\mathrm{in}}/CL\) not generally the concentration at the end of a finite input?
For a fixed systemically available amount and the same linear clearance, does changing input duration change total \(AUC_{0-\infty}\)? Explain.

Solutions. (1) \(k=3/20=0.150\ \mathrm{h^{-1}}\). Thus \(C(2)=(18/3)(1-e^{-(0.150)(2)})=6(1-e^{-0.300})=1.555\ \mathrm{mg/L}\). (2) \(C=5.00e^{-(0.100)(3)}=3.704\ \mathrm{mg/L}\). (3) The systemically available amount is \(FD=(0.75)(180\ \mathrm{mg})=135\ \mathrm{mg}\), so \(R_{\mathrm{in}}=135/3=45.0\ \mathrm{mg/h}\). (4) \(R_{\mathrm{in}}/CL\) is the indefinite-input asymptote. At finite duration \(T\), the end concentration is \((R_{\mathrm{in}}/CL)(1-e^{-kT})\), which is below that asymptote for positive finite \(T\). (5) No. Under the assumptions, mass balance gives \(AUC_{0-\infty}=\text{systemically available amount}/CL\); duration changes the time course, not total AUC.

What You'll Learn