Deterministic ODEs as Pharmacokinetic Reference Models

Why Use a Deterministic Reference Model?

A pharmacokinetic model often begins with a simple question: if the dose, initial condition, and model parameters were fixed, what concentration-time curve would the model produce? A deterministic ordinary differential equation (ODE) answers this by specifying how the drug amount changes over time. For the same inputs and parameters, it gives the same trajectory every time.

That repeatability makes a fixed-parameter ODE a useful reference model. It gives a clear baseline for expected elimination and accumulation. It does not, by itself, say that every observed concentration in a patient will equal the predicted value; it says what follows from the specified model and inputs.

Definition: A deterministic model is one whose future trajectory is fixed by its initial condition, inputs, and parameters. In the one-compartment model below, the amount changes continuously between doses according to a fixed elimination rate, and each instantaneous dose produces a specified jump in amount.

One-Compartment Elimination After a Dose

Let $A(t)$ be the amount of drug in a well-mixed central compartment at time $t$, and let $C(t)=A(t)/V$ be the concentration. Assume the apparent volume $V>0$ and clearance $CL>0$ are constant. With first-order elimination, the amount changes in proportion to the amount present:

\frac{dA(t)}{dt}=-\frac{CL}{V}A(t)=-kA(t), \qquad k=\frac{CL}{V}>0.

Here $k$ has units of inverse time. Suppose there are no doses during the interval considered and the initial amount is $A(0)=A_0$. The corresponding concentration is $C(0)=A_0/V$.

Result: Under these assumptions, the unique solution for $t\geq 0$ is $A(t)=A_0e^{-kt}$, or equivalently $C(t)=C(0)e^{-kt}$.

To derive the solution, multiply the ODE by the integrating factor $e^{kt}$. The product rule gives

\frac{d}{dt}\big(e^{kt}A(t)\big) =e^{kt}\left(\frac{dA(t)}{dt}+kA(t)\right)=0.

Thus $e^{kt}A(t)$ is constant on the interval. Its value at $t=0$ is $A_0$, so $A(t)=A_0e^{-kt}$. This also proves uniqueness: any differentiable solution satisfying the ODE has the same constant $e^{kt}A(t)=A_0$, and therefore must equal this expression. Dividing by the fixed $V$ gives the concentration formula. $\square$

The exponential form implies that the half-life is $t_{1/2}=\log(2)/k$: substituting this time gives $C(t_{1/2})=C(0)e^{-\log(2)}=C(0)/2$. This half-life is constant under the model because $k$ is constant.

Worked Example: Concentration After a Single Dose

A 500 mg intravenous bolus is given to a patient with $V=40$ L and $CL=4$ L/h. Assume the bolus is instantaneously and completely in the compartment, with no further input. Calculate the concentration at 6 hours and the model half-life.

The initial amount is $A_0=500$ mg, so $C(0)=500/40=12.5$ mg/L. The elimination rate constant is

k=\frac{CL}{V}=\frac{4\ \text{L/h}}{40\ \text{L}}=0.1\ \text{h}^{-1}.

At 6 hours,

C(6)=12.5e^{-0.1(6)} =12.5e^{-0.6} =12.5(0.5488116361\ldots) =6.860145451\ldots\ \text{mg/L} \approx 6.860\ \text{mg/L}.

The half-life is $\log(2)/0.1=6.9314718\ldots$ h, or approximately 6.931 h. Both calculations use the same fixed $k$; neither requires a new parameter at later times.

Representing Repeated Instantaneous Doses

Between doses, the amount follows the elimination ODE. At each instantaneous bolus, the amount jumps by the administered dose. For equal doses $D$ at times $0,\tau,2\tau,\ldots$, the jump rule is $A(n\tau^+)=A(n\tau^-)+D$, where the superscripts indicate the values immediately after and before the dose. Let $C_n^+$ denote the concentration immediately after the dose at $n\tau$. Since $C=A/V$, each dose raises concentration by $D/V$.

Repeated-dose recurrence: Define $q=e^{-k\tau}$. For $k>0$ and $\tau>0$, $0<q<1$. If each dose is $D$ and there is no drug initially, then $C_0^+=D/V$, and for $n\geq 1$, $$ C_n^+=qC_{n-1}^+ + \frac{D}{V}. $$

The recurrence follows directly from the model: during the interval after dose $n-1$, concentration decays from $C_{n-1}^+$ to $qC_{n-1}^+$. The next bolus adds $D/V$. Iterating gives $C_n^+=(D/V)(1+q+\cdots+q^n)$.

Because $0<q<1$, this geometric series has a finite limit. In particular,

C_n^+=\frac{D}{V}\frac{1-q^{n+1}}{1-q} \ \longrightarrow\ C_{\mathrm{ss}}^+=\frac{D/V}{1-q} =\frac{D}{V(1-e^{-k\tau})}.

This limit is the steady-state peak, meaning the concentration immediately after a dose once the repeated-dose profile has reached its limiting cycle. Immediately before a dose, the concentration is $qC_{\mathrm{ss}}^+$, so the steady-state trough is $C_{\mathrm{ss}}^-=qC_{\mathrm{ss}}^+$. The concentration then decays exponentially between the peak and the next trough. These formulas are consequences of the same fixed ODE and jump rule; they do not require the concentration to be constant across a dosing interval.

Worked Example: Three Doses and the Steady-State Limit

A patient receives 250 mg as an instantaneous bolus every 24 hours. Let $V=50$ L and $k=0.05\ \text{h}^{-1}$. Assume no drug is present before the first dose. Find the peak after the third dose, the steady-state peak and trough, and the difference between the third-dose peak and the steady-state peak.

Each dose increases concentration by $D/V=250/50=5$ mg/L. The fraction remaining after 24 hours is

q=e^{-k\tau}=e^{-0.05(24)}=e^{-1.2} =0.3011942119\ldots

The peak after the third dose is

C_2^+=5(1+q+q^2) =5\left(1+e^{-1.2}+e^{-2.4}\right) =5(1.3919121652\ldots) =6.959560826\ldots\ \text{mg/L} \approx 6.960\ \text{mg/L}.

The limiting peak is $5/(1-q)$, and the limiting trough is $5q/(1-q)$. Using the unrounded value of $q$,

C_{\mathrm{ss}}^+=\frac{5}{1-e^{-1.2}} =7.15506\ldots\ \text{mg/L} \approx 7.155\ \text{mg/L}, \qquad C_{\mathrm{ss}}^-=\frac{5e^{-1.2}}{1-e^{-1.2}} =2.15506\ldots\ \text{mg/L} \approx 2.155\ \text{mg/L}.

The peak after three doses is below the limiting peak. Their difference is the contribution of all later doses:

C_{\mathrm{ss}}^+-C_2^+ =\frac{5q^3}{1-q} =\frac{5e^{-3.6}}{1-e^{-1.2}} =0.195503\ldots\ \text{mg/L} \approx 0.1955\ \text{mg/L}.

This agrees with direct subtraction: $7.15506\ldots-6.959560826\ldots=0.19550\ldots$ mg/L. Rounding $q$ before calculating the peaks can change the final reported digits, so the calculations retain its unrounded value.

Worked Example: A Different Dosing Interval

Suppose $D/V=2$ mg/L, $k=0.04\ \text{h}^{-1}$, and doses are given every 20 hours. Calculate the steady-state peak and trough, then check the first two peaks.

The fraction remaining after each interval is $q=e^{-0.04(20)}=e^{-0.8}=0.4493289641\ldots$, so $1-q=0.5506710359\ldots$. Therefore,

C_{\mathrm{ss}}^+=\frac{2}{1-q} =3.6319\ldots\ \text{mg/L} \approx 3.632\ \text{mg/L}, \qquad C_{\mathrm{ss}}^-=qC_{\mathrm{ss}}^+ =\frac{2q}{1-q} =1.6319\ldots\ \text{mg/L} \approx 1.632\ \text{mg/L}.

After the first dose, the peak is $C_0^+=2$ mg/L. Twenty hours later, before the second dose, it has decayed to $2q=0.8986579282\ldots$ mg/L. The second dose adds 2 mg/L, giving $C_1^+=2q+2=2.8986579282\ldots$ mg/L. This is below the steady-state peak, as expected from accumulation over further doses.

What Makes the Profile Repeatable?

The result is repeatable because all quantities that determine the trajectory have been fixed: the initial amount, $V$, $CL$ (and hence $k$), dose size, and dose times. The ODE specifies the evolution between doses, and the jump rule specifies what happens at a dose. Once these are specified, there is no additional random input in the model.

This is a reference profile, not a claim that a real patient’s biology or every measured concentration is perfectly constant or predictable. It isolates what the chosen structural assumptions imply. That makes it useful to assess whether the basic elimination and dosing pattern is adequate before adding complexity.

Assumptions, Diagnostics, and Alternatives

The one-compartment elimination model assumes a well-mixed compartment, constant volume and clearance, linear first-order elimination, and no input between the bolus doses being considered. Repeated-dose formulas additionally assume identical doses at a fixed interval and an initially drug-free compartment for the stated finite-dose series. If the initial amount is not zero, its contribution must also be carried forward by exponential decay.

Several checks can reveal a mismatch between the model and the situation:

Check units. $CL/V$ must have units of inverse time, and $k\tau$ must be dimensionless in the exponential.
Check direction. With $k>0$, concentration must decrease between doses. A calculated increase during an interval signals a sign or implementation error.
Check the jump. A bolus of $D$ must change concentration by exactly $D/V$ under the stated assumptions.
Check limiting behavior. As $k\tau$ becomes large, $q$ approaches zero and little drug remains at the next dose. As $k\tau$ becomes small, accumulation increases and the steady-state peak formula becomes more sensitive to the parameter values.
Check the data against the assumptions. Systematic curvature inconsistent with exponential decline, changing dose intervals, or a changing clearance can indicate that the fixed-parameter one-compartment model is too simple.

A closed-form exponential is the direct alternative to numerically solving this particular ODE; both should agree when the numerical method is implemented accurately. Numerical integration becomes useful for more complicated model equations without a convenient closed form, but it does not remove the need to justify the structural assumptions. A model with additional compartments may be appropriate when distribution cannot reasonably be represented as instantaneous mixing in one compartment. Those alternatives change the model; none should be selected solely to make a curve more flexible.

A common mistake is to treat a deterministic prediction as a guarantee that observed concentrations will lie exactly on its curve. The correction is to distinguish the model’s trajectory from the observations: the ODE predicts a concentration conditional on its fixed inputs and assumptions, while disagreement may reflect inadequate assumptions or other sources of variation. The ODE alone does not identify which explanation is responsible.

Key takeaway: With fixed $V$, $CL$, initial condition, and dosing schedule, one-compartment first-order elimination produces a uniquely determined exponential profile between doses. Repeated boluses create predictable jumps and geometric accumulation to a finite steady-state peak and trough when $k>0$ and $\tau>0$.

Check Your Understanding

Work through these questions, then compare your results with the solutions.

If $CL=6$ L/h and $V=30$ L, what is $k$, and what fraction of concentration remains after 5 hours?
A 120 mg bolus is given into a one-compartment model with $V=20$ L. What concentration jump does the dose produce?
For repeated doses with $D/V=3$ mg/L and $q=0.5$, calculate the limiting peak and trough.
Why does the repeated-dose steady-state formula require $0<q<1$?

Solutions. (1) $k=6/30=0.2\ \text{h}^{-1}$. The fraction remaining after 5 hours is $e^{-0.2(5)}=e^{-1}=0.367879\ldots$, approximately 0.368. (2) The jump is $D/V=120/20=6$ mg/L. (3) $C_{\mathrm{ss}}^+=3/(1-0.5)=6$ mg/L, and $C_{\mathrm{ss}}^-=0.5(6)=3$ mg/L. (4) When $0<q<1$, the geometric series $1+q+q^2+\cdots$ converges to $1/(1-q)$. Here $q=e^{-k\tau}$, so positive $k$ and $\tau$ ensure this condition.