Why Model Clearance as Periodic?
Clearance can vary with clock time when physiological processes that affect drug elimination follow a daily rhythm. A periodic model represents a pattern that repeats, rather than a one-way increase or decrease over study time. This distinction matters: a transient trend may resemble part of a cycle if observations cover only a narrow time window, but it does not establish that the pattern repeats.
The model below describes a clearance pattern with a specified period, often 24 hours for a circadian hypothesis. It separates three features: the average clearance over a full cycle, the size of the oscillation, and the clock time at which clearance peaks. It can be used in a pharmacokinetic model by replacing a constant clearance in the elimination term with clearance evaluated at the current time.
The cosine equals 1 at the peak time, so the maximum clearance is \(CL_{\mathrm{mean}}(1+A)\). It equals −1 half a period later, so the minimum is \(CL_{\mathrm{mean}}(1-A)\). The absolute amplitude—the distance from the mean to either extreme—is \(A\,CL_{\mathrm{mean}}\), in clearance units. Thus, \(A=0.30\) means that the peak is 30% above the mean and the trough is 30% below it. It does not mean that the peak-to-trough difference is 30%; that difference is \(2A\,CL_{\mathrm{mean}}\).
The average over a complete cycle is exactly \(CL_{\mathrm{mean}}\): the cosine has integral zero over one period. This is a cycle average, not a claim that clearance is constant or that a short observation interval will have that average. The phase is encoded by \(t_{\mathrm{peak}}\); changing the time origin changes its numerical value, so the origin and time zone used for clock time should be stated.
From Clearance to Concentration
For a one-compartment model with constant volume \(V\), no drug input over an interval, and linear elimination, concentration satisfies \(dC/dt=-CL(t)C/V\). The solution depends on the clearance accumulated over the interval, not just on clearance at its beginning or end. This is the key link between a periodic clearance function and observed concentrations.
Substituting the periodic function gives a closed-form integral. Define \(\omega=2\pi/\tau\). Then
To justify the result, integrate the constant term to obtain \(CL_{\mathrm{mean}}(t-t_0)\). For the oscillating term, the derivative of \(\sin(\omega(u-t_{\mathrm{peak}}))/\omega\) is \(\cos(\omega(u-t_{\mathrm{peak}}))\). Evaluating this antiderivative at the two interval endpoints yields the sine difference above. The accumulated clearance therefore has both a duration component and a phase-dependent component.
Worked Example: Interpreting Mean, Amplitude, and Peak Time
Suppose a hypothetical model has \(CL_{\mathrm{mean}}=5.00\) L/h, \(A=0.30\), \(\tau=24\) h, and \(t_{\mathrm{peak}}=6\) h. Find the clearance at 6 h and 18 h, and interpret the parameters.
At 6 h, the cosine argument is zero: \(CL(6)=5.00[1+0.30\cos(0)]=5.00(1.30)=6.50\) L/h. At 18 h, the time is half a period after the peak, so the cosine argument is \(\pi\): \(CL(18)=5.00[1+0.30\cos(\pi)]=5.00(0.70)=3.50\) L/h.
The cycle-average clearance is 5.00 L/h. The absolute amplitude is \(0.30(5.00)=1.50\) L/h, giving a peak 1.50 L/h above the mean and a trough 1.50 L/h below it. The peak-to-trough difference is 3.00 L/h. These values describe the hypothetical model, not a real study.
Worked Example: Effect of a Clearance Rhythm on Concentration
Consider a hypothetical one-compartment model with \(V=40.0\) L and \(C(0)=2.00\) mg/L. Let \(CL_{\mathrm{mean}}=4.00\) L/h, \(A=0.25\), \(\tau=24\) h, and \(t_{\mathrm{peak}}=4\) h. There is no input from 0 to 12 h. Calculate \(C(12)\), checking the assumptions for the interval.
The stated constant volume, linear elimination, and absence of input allow the integrated solution to be used. Here \(\omega=2\pi/24=\pi/12\) radians per hour. The accumulated clearance is
The sine terms are \(\sin(2\pi/3)=0.8660\) and \(\sin(-\pi/3)=-0.8660\). Also, \(0.25/(\pi/12)=0.9549\) h. Thus, the accumulated clearance is \(4.00[12+0.9549(1.732)]=54.62\) L. Dividing by \(V\) gives \(54.62/40.0=1.365\), a dimensionless exponent. Therefore \(C(12)=2.00\exp(-1.365)=0.510\) mg/L.
For comparison, replacing the periodic clearance by its mean for the whole interval would give \(2.00\exp[-(4.00)(12)/40.0]=0.602\) mg/L. The difference arises because this particular 12-hour interval includes more time at relatively high clearance than at low clearance. Over a full 24-hour interval, the cosine contribution to accumulated clearance is zero, but that does not make every shorter interval equivalent to constant mean clearance.
Sampling Times and Identification of the Rhythm
A model can be written down without the data being informative enough to estimate its parameters. For a known period, one useful equivalent form is
This follows by expanding the shifted cosine. The coefficients satisfy \(\beta_0=CL_{\mathrm{mean}}\), \(\beta_c=CL_{\mathrm{mean}}A\cos(\omega t_{\mathrm{peak}})\), and \(\beta_s=CL_{\mathrm{mean}}A\sin(\omega t_{\mathrm{peak}})\). Consequently, the absolute amplitude is \(\sqrt{\beta_c^2+\beta_s^2}\), the relative amplitude is that quantity divided by \(\beta_0\), and the peak phase is obtained from the pair of coefficients with the correct quadrant. This form makes clear why observations at distinct phases are needed: the data must distinguish the constant, cosine, and sine components.
Worked Example: Comparing Phase Coverage in Two Sampling Designs
Suppose clearance could be observed directly at four times in a 24-hour cycle. Compare sampling at 0, 6, 12, and 18 h with sampling four times at 6 h. Use the coefficient representation with columns \(1\), \(\cos(2\pi t/24)\), and \(\sin(2\pi t/24)\).
For the four distinct times, the design rows are \((1,1,0)\), \((1,0,1)\), \((1,-1,0)\), and \((1,0,-1)\). These span three independent directions, so the constant level and both rhythmic components are distinguishable in this idealized direct-clearance setting. For example, at 0, 6, 12, and 18 h, the model \(CL(t)=5.00+1.50\sin(2\pi t/24)\) gives 5.00, 6.50, 5.00, and 3.50 L/h, respectively.
With four observations all at 6 h, every row is \((1,0,1)\). The data then provide only the combination \(\beta_0+\beta_s\); they cannot separately determine the mean and the sine coefficient, and they contain no information about the cosine coefficient at that phase. Repeated measurements may improve precision at that one phase, but they do not replace coverage of the cycle.
This rank argument is a design illustration, not a claim that observed concentrations are direct clearance measurements. In a pharmacokinetic model, concentration depends on the clearance history and dosing history, so information also depends on the structural model, dose timing, and observation schedule. Nevertheless, phase coverage remains essential: repeated concentrations at one clock time are generally weak evidence for a rhythm.
When possible, plan observations across multiple phases of the proposed cycle and across repeated cycles, while recording clock time consistently. Sampling at phases that are too similar makes the periodic components difficult to separate. If the period is also estimated rather than fixed, sparse or short-duration data can make different periods and phases produce similar predictions. A proposed 24-hour period should therefore be scientifically justified, and uncertainty about it should not be hidden by treating a weakly identified estimate as precise.
Diagnostics, Alternatives, and Limitations
After fitting, inspect predictions and residuals against clock time or phase within the proposed cycle, as well as against study time. A credible periodic model should capture a repeating pattern without leaving systematic phase-related residual structure. Also check whether fitted clearance remains positive and whether the estimated amplitude and phase are supported by observations from both higher- and lower-clearance portions of the cycle. Very large uncertainty, a boundary estimate near zero amplitude, or substantial changes under modest changes in initial values are warnings that the rhythm is weakly identified.
A periodic model is not interchangeable with every empirical time function. A monotonic or transient function may be more appropriate when the scientific hypothesis is induction, de-induction, or another nonrepeating process. A periodic function is attractive when repetition is plausible and the data cover the cycle; a nonperiodic function avoids imposing repetition but cannot represent a stable daily oscillation as directly. The earlier material on empirical time functions for clearance provides the broader context for time-dependent alternatives. Model choice should reflect mechanism and observation duration, not merely which curve fits a limited set of points best.
The simple cosine also imposes a smooth, symmetric rise and fall, one peak and one trough per period, and a fixed period. Real rhythms may have sharper peaks, changing amplitude, or phase shifts. A more flexible periodic shape can sometimes be considered, but it requires enough phase coverage and can be difficult to distinguish from noise or other time trends. Adding flexibility is not a substitute for informative sampling. The model here also does not by itself represent between-subject variation in rhythm parameters; that is a separate extension.
State the period, time origin, units, and whether time refers to clock time or elapsed time.
Report the cycle mean, relative or absolute amplitude, and peak phase; verify that the implied clearance remains positive.
Assess phase coverage, residual patterns, uncertainty, and sensitivity to the period and starting values.
Check Your Understanding
Try each question, then compare your reasoning with the solutions below.
- For \(CL_{\mathrm{mean}}=8\) L/h and \(A=0.20\), find the peak, trough, and peak-to-trough difference.
- If the period is 24 h and clearance peaks at 3 h, how many hours later is the trough?
- Why can repeated observations at the same phase fail to identify the mean and amplitude separately?
- In a no-input interval covering exactly one full period, what is the accumulated clearance under this model?
Solutions. (1) Peak \(=8(1.20)=9.60\) L/h; trough \(=8(0.80)=6.40\) L/h; difference \(=3.20\) L/h. (2) The trough occurs half a period after the peak, \(24/2=12\) h later, at 15 h. (3) At a single phase, clearance depends on a fixed combination of the mean and the phase-specific oscillating component; repeated values at that phase do not provide the distinct phase information needed to separate them. (4) The cosine integrates to zero over a complete period, so accumulated clearance is \(CL_{\mathrm{mean}}\tau\), with units of clearance multiplied by time.