Why Repeated Doses Need a Nonlinear Profile
For linear elimination, a dose produces a concentration profile that is independent of other doses. The total profile after repeated doses can therefore be obtained by adding shifted single-dose profiles. With saturable elimination, the elimination rate depends nonlinearly on the concentration present at that time. A later dose changes that concentration and thus changes the rate at which both the new dose and the drug already in the body are eliminated. Adding isolated-dose profiles is no longer valid.
This tutorial considers repeated intravenous bolus doses in a single, well-mixed compartment. The volume \(V\) is constant; each dose \(D\) is given every \(\tau\) time units; and elimination follows a Michaelis–Menten rate law. There is no absorption delay, distribution between compartments, or additional elimination pathway. These assumptions isolate the effect of saturation on peak, trough, and accumulation. A real PBPK model may have more processes, but the same nonlinear principle applies.
The model assumes the same elimination parameters and dosing interval throughout, with no time-varying enzyme capacity. Since the elimination rate approaches \(V_{\max}\) rather than increasing without bound, elimination becomes relatively slow at high concentrations. At low concentrations, the rate is approximately proportional to concentration: \(V_{\max}C/(K_m+C)\approx (V_{\max}/K_m)C\) when \(C\) is much smaller than \(K_m\).
Concentration Decline Between Boluses
Separating variables gives an exact relation between the concentration at the start of a dose interval, \(C_{\mathrm{start}}\), the concentration \(C(t)\) later in that interval, and elapsed time. The relation is useful for calculating troughs without treating elimination as a constant-rate process.
Here, \(C_{\mathrm{start}}\) is the post-dose concentration for the interval being considered. To justify the expression, rearrange the differential equation as \(dt=-(V/V_{\max})(1+K_m/C)\,dC\). Integrating from \(C_{\mathrm{start}}\) to \(C(t)\) yields the displayed equation. The concentration at a specified time is generally obtained by solving this equation numerically; the logarithmic term means the decline is not a single exponential.
At a periodic steady dosing profile, let \(C_{\min}\) be the trough and \(C_{\max}\) the peak. The bolus raises concentration by \(D/V\), so \(C_{\max}=C_{\min}+D/V\). Applying the decline equation over one full interval gives a direct steady-state relationship.
Solving for the trough gives
A finite positive solution requires \(\tau V_{\max}/V>D/V\), equivalently \(D< V_{\max}\tau\). This condition says the dose per interval must be less than the maximum amount that could be eliminated during that interval. At equality, the limiting elimination capacity matches the dose, but the rate is strictly below \(V_{\max}\) at every finite concentration. The concentration therefore does not settle to a finite periodic profile.
Worked Example: Tracking the First Three Peaks and Troughs
Suppose a hypothetical one-compartment model has \(V=10\) L, \(V_{\max}=10\) mg/h, \(K_m=1\) mg/L, and intravenous boluses of \(D=20\) mg every \(\tau=4\) h. Each dose raises concentration by \(D/V=2\) mg/L. Assume the first dose is given when the concentration is zero.
Immediately after the first dose, the peak is \(2.000\) mg/L. The concentration after four hours solves \(4=(10/10)[2-C+1\ln(2/C)]\), with concentrations in mg/L and the resulting time in hours. Numerically, \(C\approx0.218\) mg/L: substituting gives \(2-0.218+\ln(2/0.218)\approx4.00\). This is the first trough.
The second dose raises that trough to a peak of \(0.218+2=2.218\) mg/L. Solving \(4=2.218-C+\ln(2.218/C)\) gives a second trough of approximately \(0.282\) mg/L. The third peak is therefore approximately \(2.282\) mg/L; solving the same decline equation over four hours gives a third trough of approximately \(0.302\) mg/L. The peaks and troughs are both rising toward their periodic steady values.
This calculation uses the concentration present before each dose as the starting condition for the next interval. It does not add the first-dose profile to a second isolated-dose profile: the second interval begins at a nonzero concentration, which affects the elimination rate throughout that interval.
Steady-State Profiles and the Feasibility Condition
At steady state, each interval begins and ends at the same trough, with the bolus producing the same peak on every cycle. The steady-state equation makes the effects of dose and interval explicit: the dose determines the jump between trough and peak, while the interval determines how far concentration can decline before the next dose. As \(\tau\) shortens at a fixed dose, both the steady trough and peak generally rise. As \(\tau\) lengthens, the trough falls and the peak approaches the concentration increment from a single dose.
Worked Example: Comparing Three Dosing Intervals
Use the hypothetical parameters \(V=10\) L, \(V_{\max}=10\) mg/h, \(K_m=1\) mg/L, and \(D=20\) mg. The concentration increment is \(2\) mg/L, and the feasibility limit is \(D/V_{\max}=2\) h. Thus each interval below permits a finite steady profile.
For \(\tau=4\) h, the exponent in the trough formula is \((4\times10/10-2)/1=2\). Hence \(C_{\min}=2/(\exp(2)-1)=0.313\) mg/L, and \(C_{\max}=0.313+2=2.313\) mg/L.
For \(\tau=3\) h, the exponent is \(3-2=1\), giving \(C_{\min}=2/(\exp(1)-1)=1.164\) mg/L and \(C_{\max}=3.164\) mg/L. For \(\tau=6\) h, the exponent is \(6-2=4\), giving \(C_{\min}=2/(\exp(4)-1)=0.0373\) mg/L and \(C_{\max}=2.037\) mg/L.
The shorter interval produces much greater accumulation in this example: the trough is higher, and the repeated-dose peak is higher. The comparison is consistent with the model because all intervals exceed 2 h. At \(\tau=2\) h, the mean dose rate would equal \(V_{\max}\), and there would be no finite steady profile; below 2 h, the dose rate exceeds maximum elimination capacity.
Why Linear Superposition Fails
For linear first-order elimination, \(dC/dt=-kC\), the concentration after a bolus declines exponentially. Repeated bolus profiles can then be added because the differential equation is linear: the response to the sum of doses equals the sum of responses to each dose. For a fixed interval, the steady linear peak and trough are \(C_{\max}= (D/V)/(1-\exp(-k\tau))\) and \(C_{\min}=C_{\max}\exp(-k\tau)\). These formulas rely on a concentration-independent elimination constant \(k\).
Under Michaelis–Menten elimination, the rate \(V_{\max}C/(K_m+C)\) is not a constant times \(C\). In general, the decline from a combined concentration is not the sum of the declines that would occur from each component separately. A high post-dose concentration can also be eliminated at a smaller fraction of its amount per unit time than a low concentration, extending the tail and increasing the trough from later doses. This is why the single-dose profile cannot simply be replicated and summed.
Worked Example: A Linear Approximation Misstates the Trough
For the hypothetical parameters in the previous example, the low-concentration approximation to the Michaelis–Menten rate has \(k=V_{\max}/(VK_m)=1.00\) h\(^{-1}\). If this approximate linear rate is used with \(\tau=4\) h, the linear steady-state formulas give \(C_{\max}=2/(1-\exp(-4))=2.037\) mg/L and \(C_{\min}=2.037\exp(-4)=0.0373\) mg/L.
The saturable model instead gives a steady peak of \(2.313\) mg/L and a trough of \(0.313\) mg/L. The linear approximation substantially understates the trough here. Its assumptions are not well supported over the full profile: the saturable peak is above \(K_m=1\) mg/L, not much smaller than it. Also, using a linear steady-state formula is not a valid way to add isolated profiles under the saturable model. The discrepancy illustrates why the approximation should be checked against the concentration range, rather than used automatically.
Practical Calculation, Diagnostics, and Limitations
A practical calculation can proceed by identifying the dose-induced concentration jump, applying the integrated decline relation across each interval, and checking whether successive troughs stabilize. At steady state, the explicit trough formula is available when \(D<V_{\max}\tau\). Otherwise, a finite steady periodic peak–trough profile is not possible under this model. For more complicated inputs or additional processes, numerical integration of the model equations can track dose jumps and concentration changes directly; it still must use the nonlinear rate law rather than superpose isolated-dose profiles.
Calculate \(D/V\) and add it to the concentration immediately before each bolus.
Use the integrated Michaelis–Menten relation or numerically solve the concentration differential equation over the interval.
Track successive peaks and troughs; for a proposed steady profile, verify \(D<V_{\max}\tau\) before interpreting calculated concentrations.
Diagnostics include checking that concentration decreases between boluses, jumps by exactly \(D/V\) at each dose, and has a lower trough than peak within a cycle. A computed steady trough that is negative, undefined, or based on an exponent at or below zero signals failure of the finite-steady-state condition, not a concentration prediction. If a numerical solution gives increasing troughs despite \(D<V_{\max}\tau\), check units, dose timing, and the implementation of the rate law. Model-data disagreement may also reflect assumptions that do not hold, such as a changing enzyme capacity or distribution outside the single compartment.
Constant infusion is a different input pattern: it continuously supplies drug rather than creating bolus peaks and troughs. It can inform how saturable elimination balances a sustained input, but it does not make repeated bolus profiles additive. Likewise, a linear model may be a useful approximation when concentrations remain far below \(K_m\), but it cannot represent the capacity limit at high concentrations. The present one-compartment result also does not specify how saturable metabolism is distributed across tissues; that requires a model that represents the relevant tissue processes.
Common Pitfall
A frequent error is to calculate the concentration after one dose, shift that curve by each dosing time, and add the curves to predict repeated-dose accumulation. That reasoning assumes each dose experiences the same linear elimination process regardless of the concentrations already present. With saturable elimination, the total concentration determines the rate, so the elimination of a new dose depends on residual drug. Instead, propagate the total concentration through each interval using the nonlinear equation, or solve the steady-state peak–trough relation when its assumptions and feasibility condition hold.
Check Your Understanding
Try each question, then compare your reasoning with the solutions below.
- In the bolus model, how much does concentration rise at each dose, and what determines its value?
- For \(D=15\) mg and \(V_{\max}=5\) mg/h, what interval is the feasibility boundary for a finite steady profile? What happens at that boundary?
- For fixed dose and model parameters, what happens to the steady trough when the interval is lengthened, provided a finite steady profile remains possible?
- Why can’t the linear repeated-dose superposition formula be used as an exact solution for Michaelis–Menten elimination?
Solutions. (1) The jump is \(D/V\), because an instantaneous amount \(D\) is distributed through volume \(V\). (2) The boundary is \(D/V_{\max}=15/5=3\) h. At equality, the dose per interval matches maximum elimination capacity, but elimination is below that capacity at every finite concentration, so no finite periodic steady profile exists. (3) The trough falls: a longer interval allows more time for concentration to decline before the next dose. (4) Superposition requires a linear elimination equation, in which the elimination rate is a constant times concentration. The Michaelis–Menten rate depends nonlinearly on concentration, so each dose changes the elimination of the drug already present.