Modeling Infusions with Start and Stop Events

Why Infusion Events Need Explicit Treatment

An infusion supplies drug continuously over a specified interval. In a PBPK model, that supply is an input to the amount-balance equations, not an instantaneous addition to the state. When the infusion starts or stops, the input rate changes abruptly. The amount already in the body does not: it continues to evolve according to the model’s disposition processes.

This distinction matters in numerical integration. If a solver takes a step across an infusion start or stop without being told about the event, it may approximate the changing input as though it were smooth over that step. The resulting error can be especially visible when the step is long relative to the infusion duration or the model’s important time scales. The robust approach is to represent the rate as a time-dependent input and ensure that integration recognizes its discontinuities.

Definition: A constant-rate infusion with rate \(R_0\), start time \(t_s\), and stop time \(t_e\) has input \(R_{\mathrm{in}}(t)=R_0\) for \(t_s \leq t < t_e\), and \(R_{\mathrm{in}}(t)=0\) otherwise. The infusion duration is \(T=t_e-t_s\), and the administered amount is \(R_0T\). An infusion start or stop changes the ODE input; absent another instantaneous process, it does not cause a jump in the amount state.

One-Compartment Model and Its Assumptions

Consider a well-mixed compartment with amount \(A(t)\), volume \(V\), and concentration \(C(t)=A(t)/V\). Let \(CL\) be constant clearance and assume linear elimination, so the elimination rate is \(CL\,C(t)\). If the infusion enters this compartment directly, its amount balance is

\frac{dA}{dt}=R_{\mathrm{in}}(t)-CL\,C(t).

For constant volume, dividing by \(V\) gives the concentration equation. Define \(k=CL/V\), the first-order elimination rate constant:

\frac{dC}{dt}=\frac{R_{\mathrm{in}}(t)}{V}-kC(t), \qquad k=\frac{CL}{V}.

The closed-form calculations below require a single well-mixed compartment, constant \(V\) and \(CL\), linear elimination, and a known constant infusion rate during the interval. They also assume that there is no additional input, distribution process, or time-varying physiological change affecting this one-compartment equation. A full PBPK model can have many coupled states; in that setting the infusion is added to the appropriate amount-balance equation, while the other flows and processes remain in the model.

Deriving the Concentration Profile

During an infusion, \(R_{\mathrm{in}}(t)=R_0\), so the concentration equation is a linear ODE. Multiplication by the integrating factor \(e^{kt}\) gives

e^{kt}\frac{dC}{dt}+ke^{kt}C =\frac{R_0}{V}e^{kt} \quad\Longrightarrow\quad \frac{d}{dt}\left(e^{kt}C(t)\right) =\frac{R_0}{V}e^{kt}.

Integrate from the start time \(t_s\) to a time \(t\) during the infusion. If the concentration at the start is \(C(t_s)\), then

e^{kt}C(t)-e^{kt_s}C(t_s) =\frac{R_0}{Vk}\left(e^{kt}-e^{kt_s}\right).

Since \(Vk=CL\), rearranging yields the concentration during infusion. If the compartment is drug-free at the start, this simplifies to the familiar finite-duration infusion profile.

C(t)=C(t_s)e^{-k(t-t_s)} +\frac{R_0}{CL}\left(1-e^{-k(t-t_s)}\right), \qquad t_s\leq t\leq t_e.

At \(t_e\), the input becomes zero. The model then has \(dC/dt=-kC\), so the concentration declines exponentially from its value at the stop time:

C(t)=C(t_e)e^{-k(t-t_e)},\qquad t\geq t_e.

The infusion-only contribution approaches \(R_0/CL\) during an indefinitely continued infusion. A finite infusion reaches only the fraction \(1-e^{-kT}\) of that contribution by its stop time, if its initial concentration is zero. This follows directly by setting \(t=t_e\) in the infusion solution and using \(T=t_e-t_s\). The concentration is continuous at both events: the input changes, but the state has no instantaneous jump.

Worked Example: Concentration During and After an Infusion

Suppose a hypothetical one-compartment model has \(V=15\ \mathrm{L}\), \(CL=3\ \mathrm{L/h}\), and a \(30\ \mathrm{mg/h}\) infusion from \(t=0\) to \(t=4\ \mathrm{h}\). The initial concentration is zero. Find concentrations at 2, 4, and 6 hours.

First calculate \(k=CL/V=(3\ \mathrm{L/h})/(15\ \mathrm{L})=0.2\ \mathrm{h^{-1}}\), and \(R_0/CL=(30\ \mathrm{mg/h})/(3\ \mathrm{L/h})=10\ \mathrm{mg/L}\). At 2 hours, the infusion has run for 2 hours:

C(2)=10(1-e^{-(0.2)(2)}) =10(1-e^{-0.4}) =3.2968\ \mathrm{mg/L}.

At the stop time, the infusion has run for 4 hours:

C(4)=10(1-e^{-(0.2)(4)}) =10(1-e^{-0.8}) =5.5067\ \mathrm{mg/L}.

Two hours after the infusion stops, the concentration has declined from that stop-time value:

C(6)=C(4)e^{-(0.2)(6-4)} =5.506710\ldots\,e^{-0.4} =3.6913\ \mathrm{mg/L}.

The infusion contributes drug only through hour 4. Continuing the input after that time would give a different and incorrect profile.

Implementing Start and Stop Events

In a numerical PBPK implementation, the right-hand side should evaluate the infusion rate at the current integration time. For this single infusion, a piecewise definition is sufficient. If the system supports known event times, include \(t_s\) and \(t_e\) as integration boundaries or otherwise force the solver to recognize them. Integrate up to the event, change the input rule, and continue from the same state value. The derivative can differ on the two sides of the event; the state remains continuous.

Before the start.
Use \(R_{\mathrm{in}}(t)=0\) for \(t<t_s\), while the model continues to evolve under its existing disposition processes.

At the start.
Set the input to \(R_0\) for the infusion interval. Do not add \(R_0\) directly to the amount state; its units are amount per time.

At the stop.
Set the input to zero at \(t_e\), retain the current amount, and continue integrating the disposition equations.

The inequality convention at the exact stop time is useful for coding, but a single time point has zero duration and does not change the continuous solution. What matters numerically is that the solver not approximate a substantial interval using the wrong input. In a larger PBPK system, add \(R_{\mathrm{in}}(t)\) to the amount derivative for the compartment receiving the infusion, with any required bioavailability or unit conversion stated explicitly.

Worked Example: Infusion Starting After Time Zero

For a hypothetical model, let \(V=12\ \mathrm{L}\), \(CL=2.4\ \mathrm{L/h}\), and \(R_0=24\ \mathrm{mg/h}\). The subject has zero concentration before an infusion starts at \(t_s=1\ \mathrm{h}\); it stops at \(t_e=3\ \mathrm{h}\). Calculate concentration at \(t=2\), \(3\), and \(4\ \mathrm{h}\).

Here \(k=2.4/12=0.2\ \mathrm{h^{-1}}\) and \(R_0/CL=24/2.4=10\ \mathrm{mg/L}\). At \(t=2\), the infusion duration so far is \(2-1=1\ \mathrm{h}\):

C(2)=10(1-e^{-0.2(1)}) =10(1-0.8187308) =1.8127\ \mathrm{mg/L}.

At the stop time, the infusion duration is 2 hours:

C(3)=10(1-e^{-0.2(2)}) =10(1-0.6703200) =3.2968\ \mathrm{mg/L}.

At \(t=4\), input has been zero for one hour, so

C(4)=C(3)e^{-0.2(1)} =3.296799\ldots(0.8187308) =2.6992\ \mathrm{mg/L}.

The one-hour period before infusion start does not appear as an infusion duration in either calculation. If there were a nonzero concentration at \(t_s\), its residual contribution would be \(C(t_s)e^{-k(t-t_s)}\), as in the general solution.

Checking Amount Balance and Common Failure Modes

A useful independent check is conservation of amount. During the infusion, the amount administered over duration \(T\) is \(R_0T\). At the stop time, this administered amount must equal the increase in amount stored in the compartment plus the amount eliminated during the interval, assuming zero initial amount and no other processes. This check can expose a missing input, a unit error, or an infusion that was accidentally continued past its stop time.

Worked Example: Verifying an Infusion by Mass Balance

Consider a hypothetical one-compartment model with \(V=20\ \mathrm{L}\), \(CL=4\ \mathrm{L/h}\), and an infusion rate of \(18\ \mathrm{mg/h}\) for 3 hours from an initial amount of zero. Find the stop-time concentration and check the amount balance.

The rate constant is \(k=4/20=0.2\ \mathrm{h^{-1}}\), and \(R_0/CL=18/4=4.5\ \mathrm{mg/L}\). Thus,

C(3)=4.5(1-e^{-0.2(3)}) =4.5(1-e^{-0.6}) =4.5(0.4511884) =2.0303\ \mathrm{mg/L}.

The amount present at the stop time is \(A(3)=VC(3)=(20\ \mathrm{L})(2.0303476\ \mathrm{mg/L})=40.60695\ \mathrm{mg}\). The administered amount is \((18\ \mathrm{mg/h})(3\ \mathrm{h})=54\ \mathrm{mg}\). Therefore the amount eliminated is \(54-40.60695=13.39305\ \mathrm{mg}\), which is nonnegative and completes the balance.

Common implementation failures include using a rate in milligrams per hour as though it were an amount in milligrams, mixing minutes with hours, applying the input to the wrong compartment, and failing to stop the infusion. Another failure is forcing the state to zero or adding a jump at an infusion event. Neither is appropriate for a continuous infusion under this model. At a stop event, the concentration at the stop is the initial condition for the post-infusion decline.

A practical diagnostic is to inspect both the input and state near each event. The input should switch at the specified times; the amount and concentration should not jump merely because an infusion starts or stops. Check that concentration rises while the infusion is active when the input exceeds elimination, and that it declines after stopping under first-order elimination. These checks are conditional on the stated model and initial concentration; a pre-existing concentration can affect the direction of change during infusion.

How Infusion Modeling Differs from Alternatives

Approach	What it represents	Important limitation
Finite-duration infusion input	A constant rate applied only between specified start and stop times	Requires correct event handling and units
Indefinite constant input	A rate that remains active throughout the modeled interval	Does not represent a finite stop unless the input is switched off
Bolus state jump	An amount added instantaneously to the state	Does not represent drug delivered gradually over time
Closed-form one-compartment solution	An exact profile under the assumptions stated above	Does not replace numerical integration for a general coupled PBPK system

A bolus can sometimes approximate an infusion when the infusion duration is negligible relative to the time scales relevant to the question, but it changes the early concentration profile and is not the same input. Conversely, the one-compartment formula is an excellent check for a numerical implementation under matching assumptions, but a PBPK model with multiple coupled compartments generally requires integration of its full ODE system.

Key takeaway: Specify the infusion rate as a time-dependent input, make its start and stop times visible to the solver, and keep the amount state continuous at those events. Verify units and compare a simple one-compartment implementation with the closed-form profile whenever its assumptions apply.

Check Your Understanding

Answer each question, then compare your work with the solutions below.

A drug-free one-compartment model has \(V=10\ \mathrm{L}\), \(CL=2\ \mathrm{L/h}\), and receives \(8\ \mathrm{mg/h}\) for 1 hour. What is the concentration at the stop time?
For the same model, suppose the infusion stops after 2 hours. What concentration remains 1 hour later?
What changes at an infusion stop: the input, the amount state, or both?
In a one-compartment model with \(k=0.25\ \mathrm{h^{-1}}\), how long does an infusion need to run for its infusion-only concentration contribution to reach 90% of \(R_0/CL\)?
Explain why using a bolus jump of \(R_0T\) is not generally equivalent to infusing at rate \(R_0\) over duration \(T\).

Solutions. (1) Here \(k=2/10=0.2\ \mathrm{h^{-1}}\) and \(R_0/CL=8/2=4\ \mathrm{mg/L}\). Thus \(C(1)=4(1-e^{-0.2})=0.7251\ \mathrm{mg/L}\). (2) At 2 hours, \(C(2)=4(1-e^{-0.4})=1.3187\ \mathrm{mg/L}\). After one hour without input, \(C(3)=1.3187e^{-0.2}=1.0797\ \mathrm{mg/L}\). (3) The input changes from \(R_0\) to zero; the amount state remains continuous and then continues to decline under disposition. (4) Solve \(1-e^{-kT}=0.90\), giving \(T=-\ln(0.10)/0.25=9.210\ \mathrm{h}\). (5) A bolus adds the full amount instantly and creates an immediate state jump. An infusion supplies the same total amount gradually, so elimination occurs while it is being delivered and the concentration-time profiles generally differ.