Periodic Dosing and the Repeating Response Profile

Why a Repeating Profile Is Not a Single Steady State

With repeated dosing, drug effects may change from one part of a dosing cycle to another. A response can rise during one interval and fall during the next, then repeat that pattern once transients have faded. The resulting periodic steady state is a repeating trajectory, not a constant response. Its value immediately before a dose can differ substantially from its value at the end of the dosing interval.

The calculation is especially direct when the drug effect, and therefore the response production and loss rates, are constant within each interval. Each interval has an exact response update. Chaining those updates gives a recurrence from one dosing cycle to the next, whose fixed point determines the repeating profile. This uses the piecewise-constant interval representation and integrating-factor solution from the earlier material on Piecewise-Constant Exposure and Integrating-Factor Solutions.

Definition: A periodic steady-state response is a response trajectory that repeats after each dosing period $T$: $R(t+T)=R(t)$. The response itself is continuous across an interval boundary, although the rates governing its change may jump when the drug effect changes.

Exact Update Within One Constant-Effect Interval

Let $R(t)$ denote the response. In interval $j$, suppose its dynamics are

\frac{dR}{dt}=A_j-B_jR,\qquad B_j>0,

where $A_j$ is the constant response input in that interval and $B_j$ is its constant first-order loss rate. The interval equilibrium is $R_{\infty,j}=A_j/B_j$, the value approached if those rates continued indefinitely. If the interval lasts $\Delta_j$, define $q_j=\exp(-B_j\Delta_j)$. Solving the differential equation from an initial value $R_{\mathrm{start},j}$ gives

R_{\mathrm{end},j} =R_{\infty,j}+(R_{\mathrm{start},j}-R_{\infty,j})q_j.

For positive $B_j$ and duration, $0<q_j<1$. The endpoint therefore lies between the starting value and the interval equilibrium. It does not generally reach that equilibrium during a finite interval. At a boundary, the next interval begins at the preceding interval's endpoint; the response does not reset merely because the rates change.

Composing a Full Dosing Cycle

Suppose one dosing period consists of $m$ consecutive constant-effect intervals. Let $x_0$ be the response just before interval 1, and let $x_j$ be the response at the end of interval $j$. Applying the interval update successively gives

x_j=q_jx_{j-1}+(1-q_j)R_{\infty,j}.

After the final interval, the next cycle begins with $x_m$. Because the schedule repeats, the cycle map can be written $x_m=Qx_0+C$, where

Q=\prod_{j=1}^{m}q_j,\qquad C=\sum_{j=1}^{m}(1-q_j)R_{\infty,j} \prod_{k=j+1}^{m}q_k.

An empty product, as occurs for $j=m$, equals 1. The factors after interval $j$ appear because that interval's contribution is carried through all the later intervals before the cycle ends.

Theorem: If every interval has $B_j>0$ and positive duration, then $0<Q<1$ and there is a unique periodic steady-state cycle-start response: $$ x_0^*=\frac{C}{1-Q}. $$ The responses immediately after each interval are obtained by applying $x_j^*=q_jx_{j-1}^*+(1-q_j)R_{\infty,j}$. These values, together with the exact within-interval trajectories, specify the repeating response profile.

To justify the result, the cycle-start response must be unchanged by one full cycle, so $x_0^*=Qx_0^*+C$. Since $Q<1$, rearranging yields $x_0^*=C/(1-Q)$, and no other value solves this equation. For any starting value $x_0$, subtracting the fixed-point equation from the cycle map gives $x_1-x_0^*=Q(x_0-x_0^*)$. Repeating this argument yields

$$ x_n-x_0^*=Q^n(x_0-x_0^*), $$

where $x_n$ is the value at the start of cycle $n$. Thus the cycle-start response converges to the periodic value. Convergence can be slow when $Q$ is near 1, as may occur when the total loss over a cycle is small.

Worked Example: Stimulation of Production During an On Interval

Consider a hypothetical response with baseline production $k_{\mathrm{in}}=20$ response units per hour and loss rate $k_{\mathrm{out}}=0.20$ per hour. Production is stimulated to 1.5 times baseline for 4 hours, then returns to baseline for 8 hours. Find the periodic steady-state response immediately before and after the stimulated interval.

During stimulation, the input is $30$ response units per hour, so $R_{\infty,1}=30/0.20=150$ units. During the drug-free interval, $R_{\infty,2}=20/0.20=100$ units. The interval retention factors are

q_1=e^{-0.20(4)}=e^{-0.8}=0.4493,\qquad q_2=e^{-0.20(8)}=e^{-1.6}=0.2019.

Let $x_0^*$ be the pre-stimulation value. The response after stimulation is $x_1^*=150+(x_0^*-150)(0.4493)$; the response after the drug-free interval must return to $x_0^*$. The cycle map is therefore

x_0^*=q_2\{150+(x_0^*-150)q_1\}+(1-q_2)100.

Equivalently, $Q=q_1q_2=0.09072$ and $C=(1-q_1)(150)q_2+(1-q_2)(100)=16.68+79.81=96.49$ units. Hence $x_0^*=96.49/(1-0.09072)=106.1$ units. The value after stimulation is $x_1^*=150+(106.1-150)(0.4493)=130.3$ units. Thus the repeating profile rises from about 106.1 to 130.3 units during the stimulated interval, then returns to about 106.1 units during the drug-free interval. Neither value is the stimulated interval equilibrium of 150 units.

Intervals Can Change the Loss Rate Instead

The same recurrence applies when the drug changes loss rather than production. For example, inhibition of loss can be represented by $dR/dt=k_{\mathrm{in}}-B_jR$, with a smaller $B_j$ during the drug-effect interval. That interval has a higher equilibrium $k_{\mathrm{in}}/B_j$ and, for a given duration, a larger retention factor. The response then moves more slowly toward its interval equilibrium.

Worked Example: Periodic Inhibition of Loss

In a hypothetical system, $k_{\mathrm{in}}=10$ response units per hour and baseline loss is $0.10$ per hour, giving a baseline response of $10/0.10=100$ units. Loss is inhibited by 50% for 6 hours in each 24-hour cycle, followed by 18 hours at baseline loss. Calculate the periodic steady-state response before and after the inhibited-loss interval.

During inhibition, $B_1=0.05$ per hour and $R_{\infty,1}=10/0.05=200$ units. During the drug-free interval, $B_2=0.10$ per hour and $R_{\infty,2}=100$ units. Thus

q_1=e^{-0.05(6)}=0.7408,\qquad q_2=e^{-0.10(18)}=0.1653.

The cycle map has $Q=(0.7408)(0.1653)=0.1225$ and $C=(1-0.7408)(200)(0.1653)+(1-0.1653)(100)=8.57+83.47=92.04$ units. The pre-inhibition value is $x_0^*=92.04/(1-0.1225)=104.9$ units. At the end of the 6-hour interval, $x_1^*=200+(104.9-200)(0.7408)=129.5$ units. The response therefore rises during the inhibited-loss interval, but it does not reach 200 units; during the subsequent 18 hours it falls back to the cycle-start value.

Checking the Calculation and Choosing an Approach

A useful check is to reconstruct the entire cycle: each interval's endpoint should follow its update equation, and the final endpoint in a periodic profile must equal the initial endpoint. In the examples, stimulated production raises the interval equilibrium, while inhibited loss raises it by reducing $B_j$; in both cases the stated within-interval direction agrees with the calculated endpoints.

The exact recurrence is preferable to treating the response as if it instantly reached each interval equilibrium. It also preserves the timing of the distinct effects. A single constant average effect is a simpler approximation, but generally changes both the equilibrium and the response trajectory; averaging rates does not in general reproduce the sequence of exponential interval updates. If the effect changes continuously within intervals, the piecewise-constant calculation is an approximation whose adequacy depends on how well the interval values represent that changing effect. The integrating-factor solution for time-varying effects is an alternative when rates vary continuously.

The recurrence assumes a fixed repeating schedule, constant $A_j$ and $B_j$ within each interval, no instantaneous jump in $R$ at a boundary, and $B_j>0$. It does not cover response-dependent nonlinearities within a phase, cycle-to-cycle changes in the schedule, or random variation in the drug effect. If computed endpoints violate the update equation, the cycle does not close, or measurements show a systematic within-interval pattern inconsistent with the assumed rates, revisit the interval definitions and assumptions rather than interpreting the calculated profile as a reliable prediction.

Common pitfall: It is incorrect to set the response at the start of each interval equal to that interval's equilibrium. That assumes the response reaches equilibrium instantaneously. Instead, carry the preceding endpoint forward using $R_{\mathrm{end},j}=R_{\infty,j}+(R_{\mathrm{start},j}-R_{\infty,j})q_j$, and impose equality only between the end and start of the complete cycle at periodic steady state.

Worked Example: Checking Convergence Across Cycles

Use the stimulated-production schedule from the first worked example, but suppose the response at the start of the first cycle is the baseline value, 100 units. The cycle map is $x_{n+1}=0.09072x_n+96.49$. After one cycle,

x_1=(0.09072)(100)+96.49=105.6\text{ units}.

After two cycles, $x_2=(0.09072)(105.6)+96.49=106.1$ units (approximately 106.07 before rounding). The periodic value is 106.1 units. This is consistent with the convergence result: the difference from the periodic value is multiplied by about 0.09072 per cycle. The within-cycle profile in the first cycle is not yet the repeating profile, even though the response is already close to its cycle-start steady-state value.

Check Your Understanding

Answer each question, then compare with the solutions below.

For one interval with $R_{\infty}=80$ units, $q=0.5$, and starting response 40 units, calculate the endpoint.
For two intervals with $q_1=0.4$, $R_{\infty,1}=120$ units, $q_2=0.25$, and $R_{\infty,2}=80$ units, calculate the periodic cycle-start response and the value after interval 1.
Why is the periodic response unique when every interval has positive duration and $B_j>0$?
If $Q=0.8$, what fraction of the initial cycle-start error from periodic steady state remains after three cycles?

Solutions. (1) $R_{\mathrm{end}}=80+(40-80)(0.5)=60$ units. (2) The cycle map has $Q=(0.4)(0.25)=0.10$ and $C=(1-0.4)(120)(0.25)+(1-0.25)(80)=18+60=78$ units. Thus $x_0^*=78/(1-0.10)=86.67$ units, and the value after interval 1 is $120+(86.67-120)(0.4)=106.7$ units. (3) Each $q_j=e^{-B_j\Delta_j}$ lies strictly between 0 and 1, so their product $Q$ does too. The cycle map has exactly one fixed point $C/(1-Q)$. (4) The remaining fraction is $Q^3=(0.8)^3=0.512$, or 51.2% of the initial error.