Periodic Invariant Laws for Repeated-Dose SDEs

Why Repeated Dosing Calls for a Periodic Law

A continuously infused drug in a time-homogeneous model may approach a distribution that is unchanged as time passes. Scheduled bolus dosing is different: the state evolves between doses and jumps at each dose time. Even after transients have disappeared, the distribution just after a dose need not match the distribution just before the next one. Instead, the distribution repeats across dosing cycles while changing within each cycle. This repeating pattern is a periodic invariant law.

The key is to specify the phase of observation. “The steady-state concentration distribution” is incomplete if it does not say whether concentration is measured immediately after a dose, at a fixed number of hours after dosing, or at another phase. The law can vary with that choice even when the dosing schedule and model have been unchanged for many cycles.

Definition and the Dose-Boundary Condition

Let doses occur at times \(nT\), where \(T>0\) is the dosing interval and \(n\) is an integer. Between doses, suppose the state is a well-defined Markov process with transition kernels \(P_{s,t}(x,A)\): starting from state \(x\) at time \(s\), this is the probability that the state is in the measurable set \(A\) at time \(t\). At a dose time, let \(K(x,A)\) describe the dose jump. For a deterministic bolus map \(J\), \(K(x,A)=1\) if \(J(x)\in A\), and \(0\) otherwise. We use right-continuous states, so the state at a dose time is the post-dose state.

Definition: A periodic invariant family is a family of probability distributions \(\{\mu_t\}\) that repeats with period \(T\) and is propagated by the model: for times \(s\leq t\), \(\mu_s P_{s,t}=\mu_t\) between jumps, and at each dose time \(\mu_{nT-}K=\mu_{nT+}\). Periodicity means \(\mu_{t+T}=\mu_t\), with the same convention for the pre-dose and post-dose limits at jumps. Here \(\mu P\) denotes the distribution obtained by first sampling a state with law \(\mu\), then applying the transition kernel \(P\).

This definition assumes the inter-dose process and dose kernels are well-defined on the state space, and that their transition structure repeats from one cycle to the next. It does not assume that an invariant family exists or is unique. Those properties require additional conditions on the dynamics. One useful test is to look at the process only just after each dose. The resulting cycle-to-cycle transition kernel is an ordinary Markov kernel; its invariant distribution determines the whole periodic family when that family can be propagated through the cycle.

Theorem: Suppose the model has a well-defined, repeating transition structure over each dosing interval. A periodic invariant family exists if and only if the one-cycle kernel \(Q\), taking the post-dose state at phase \(0\) to the post-dose state at phase \(T\), has an invariant probability distribution \(\mu_0\). In that case, propagating \(\mu_0\) through the interval and applying the dose jump gives the periodic family.

Proof. If a periodic invariant family exists, propagation from phase \(0\) through one full cycle returns the process to phase \(T\), which has the same post-dose law as phase \(0\) by periodicity. Thus \(\mu_0 Q=\mu_T=\mu_0\), so \(\mu_0\) is invariant for \(Q\).

Conversely, suppose \(\mu_0 Q=\mu_0\). For each phase \(0\leq s<T\), propagate \(\mu_0\) to phase \(s\) using the inter-dose dynamics, applying any dose kernels encountered along the way. This defines \(\mu_s\). At the end of the interval, the post-dose distribution is \(\mu_0 Q=\mu_0\). Repeating the same propagation in every interval therefore gives the same phase-specific laws in every cycle, including the required pre-dose to post-dose jump relation. These laws form a periodic invariant family. \(\square\)

The theorem reduces a continuous-time periodicity question to an invariant-distribution question for the cycle kernel. It does not supply a convergence rate or guarantee that every initial distribution approaches the family; those are separate questions.

Worked Examples

Worked Example: A Linear Diffusion With Repeated Bolus Doses

Consider the scalar model \(dX_t=-kX_t\,dt+\sigma\,dW_t\) between doses, with \(k>0\), and a fixed bolus \(D\) added every \(T\) time units. Write \(Y_n\) for the post-dose state in cycle \(n\). Over an interval of length \(T\), the solution has the form

Y_{n+1}=aY_n+D+\eta_n,\qquad a=e^{-kT},\qquad \eta_n\sim N\!\left(0,q\right),\qquad q=\frac{\sigma^2}{2k}(1-e^{-2kT}).

The interval noise \(\eta_n\) is independent of the state at the start of that interval, and the interval noises are independent across cycles. Since \(k,T>0\), \(0<a<1\). Taking expectations under an invariant law gives \(m=am+D\), hence

m=\frac{D}{1-a}.

The variance \(v\) must satisfy \(v=a^2v+q\), so

v=\frac{q}{1-a^2}=\frac{\sigma^2}{2k}.

At phase \(s\) after the dose, where \(0\leq s<T\), the state is the decayed post-dose state plus the diffusion accumulated during \(s\). Therefore its law is Gaussian with mean \(e^{-ks}m\) and variance \(e^{-2ks}v+\frac{\sigma^2}{2k}(1-e^{-2ks})=v\). The variance is constant across phase in this model, but the mean is not.

Worked Example: Calculating Two Different Phases

Take \(k=1\), \(T=2.4\), \(D=10\), and \(\sigma=1\). Then \(a=e^{-2.4}\approx0.0907179533\), and the invariant post-dose mean is

m=\frac{10}{1-e^{-2.4}} =\frac{10}{0.9092820467} \approx 10.997688 \approx 10.998.

At phase \(s=0.8\), the mean is

e^{-0.8}m =e^{-0.8}\frac{10}{1-e^{-2.4}} \approx 0.449328964\times10.997688 \approx 4.941580 \approx 4.942.

The invariant variance is \(\sigma^2/(2k)=0.5\) at both phases. Thus the post-dose law is \(N(10.998,0.5)\) to three decimal places, while the law \(0.8\) time units after the dose is \(N(4.942,0.5)\). They are different distributions, although both belong to the same periodic invariant family.

Worked Example: Checking the Pre-Dose and Post-Dose Laws

Use the same linear model, but take \(k=1\), \(T=1\), \(D=2\), and \(\sigma=1\). The post-dose invariant mean is

m=\frac{2}{1-e^{-1}}\approx\frac{2}{0.632121}\approx3.164.

Immediately before the next dose, the mean is \(e^{-1}m\approx0.368\times3.164\approx1.164\). The deterministic dose adds \(2\), so the post-dose mean becomes \(1.164+2=3.164\), as required for periodicity. For the variance, \(v=0.5\) throughout the interval. The pre-dose variance is also \(0.5\), and a deterministic dose changes the mean but not the variance. This verifies the boundary condition for both moments; the full Gaussian distribution is likewise translated by the dose map.

Why the Gaussian Cycle Law Is Invariant

The moment equations above identify a candidate invariant Gaussian law. The following derivation verifies the distribution itself and shows uniqueness for this affine Gaussian cycle model. Suppose \(Y_{n+1}=aY_n+b+\eta_n\), where \(|a|<1\), \(b\) is fixed, and the independent errors satisfy \(\eta_n\sim N(0,q)\), with \(q\geq0\). If \(\phi(t)=E[e^{itY}]\) is the characteristic function of an invariant law, invariance implies

\phi(t)=\exp\!\left(itb-\frac{qt^2}{2}\right)\phi(at).

Iterating this identity \(n\) times gives

\phi(t)= \exp\!\left( itb\sum_{j=0}^{n-1}a^j -\frac{qt^2}{2}\sum_{j=0}^{n-1}a^{2j} \right)\phi(a^nt).

Because \(|a|<1\), \(a^nt\to0\). Every characteristic function is continuous at zero and has value \(1\) there, so \(\phi(a^nt)\to1\). The two finite sums converge to \(1/(1-a)\) and \(1/(1-a^2)\), respectively. Consequently, any invariant characteristic function must be

\phi(t)=\exp\!\left( \frac{itb}{1-a}-\frac{qt^2}{2(1-a^2)} \right).

This is the characteristic function of \(N\!\left(b/(1-a),q/(1-a^2)\right)\). Substitution into the one-step equation verifies that this Gaussian law is invariant. Since any invariant law must have this characteristic function, it is unique. Taking \(b=D\) gives the repeated-dose result above. \(\square\)

The argument requires neither a density for the initial state nor an interchange of a limit and an integral: it uses the functional identity for characteristic functions and their continuity at zero. The condition \(|a|<1\) is essential to this argument and to the contraction of the cycle dynamics.

When an Invariant Law Does Not Exist

Worked Example: Diffusion and Repeated Additive Increments Without Elimination

Consider \(dX_t=\sigma\,dW_t\) between doses, with \(\sigma>0\), and add a fixed amount \(D\) at each interval of length \(T\). The cycle equation is \(Y_{n+1}=Y_n+D+\eta_n\), where \(\eta_n\sim N(0,\sigma^2T)\). Suppose an invariant probability law had characteristic function \(\phi\). Its invariance would imply

\phi(t)=e^{itD-\sigma^2Tt^2/2}\phi(t).

For every \(t\ne0\), \(\left|e^{itD-\sigma^2Tt^2/2}\right|=e^{-\sigma^2Tt^2/2}<1\). The displayed equality then forces \(\phi(t)=0\) for every \(t\ne0\), which contradicts continuity at zero because \(\phi(0)=1\). Thus this cycle kernel has no invariant probability distribution, and there is no periodic invariant family. The example shows why periodic dosing alone is not enough: the inter-dose dynamics must also control the cycle-to-cycle state sufficiently to permit an invariant law.

Diagnostics, Alternatives, and Limitations

In applications, the most useful diagnostic is to compare distributions at a fixed dose phase across cycles, rather than pooling observations taken at different phases. For a candidate periodic family, check that the post-dose law propagated through the interval gives the claimed pre-dose law, and that applying the dose kernel returns the post-dose law. In a model with a known cycle kernel, also check invariance directly: applying \(Q\) to the proposed post-dose distribution should reproduce it. A mismatch may indicate a wrong dose-jump convention, an error in the transition distribution, or that the claimed law is not invariant.

Three objects are often confused:

Object	What it describes	Does it remain unchanged in continuous time?
Time-invariant stationary distribution	A single law preserved at every time under the process dynamics	Yes, when the process and law meet the stationarity conditions
Cycle-sampled invariant distribution	The law at one chosen phase, such as immediately post-dose	Only across successive cycles at that same phase
Periodic invariant family	A law for each phase, repeated from cycle to cycle	It repeats every interval but generally changes within the interval

A phase-averaged distribution can summarize measurements collected uniformly across the dosing interval, but it is a mixture of phase-specific laws, not the law at a particular phase. It should not be substituted for a trough or post-dose distribution. Conversely, if observations are always taken at the same phase, the cycle-sampled invariant law is directly relevant. A time-invariant stationary distribution is the natural alternative only when the model dynamics and the law are genuinely time-invariant; periodic jumps generally break that condition.

Existence and uniqueness depend on the cycle kernel. The affine Gaussian example has a unique invariant law because its cycle multiplier contracts. In nonlinear PK systems, a candidate periodic law may lack a closed form, and checking moments alone does not establish that the full distribution is invariant. State constraints, boundary behavior, and possible non-uniqueness also require model-specific treatment. Finally, an invariant family describes a distribution that is preserved by propagation; it does not by itself say how quickly a particular initial distribution approaches that family.

Common Mistakes

Calling a phase-specific law time-invariant. A post-dose law can repeat from cycle to cycle while differing from the law later in the interval. State the phase whenever reporting the distribution.
Checking only the mean. Matching a periodic mean does not prove invariance of the distribution. Verify the cycle kernel or, when possible, the full transition law.
Applying the dose at the wrong boundary. With a post-dose state convention, the cycle map includes evolution over the interval followed by the next dose. Keep the pre-dose and post-dose laws distinct.
Assuming periodic forcing guarantees a periodic invariant law. The no-elimination example has a periodic schedule but no invariant cycle distribution. Existence is a property of the dynamics, not of the calendar alone.

Key takeaway: A periodic invariant law is a phase-indexed family of distributions that propagates through the SDE and dose jumps and repeats after each dosing interval. Its distribution at one chosen phase is an invariant law for the corresponding cycle-sampled kernel; other phases generally have different distributions.

Check Your Understanding

Answer each question, then compare your reasoning with the brief solutions.

What is the difference between a periodic invariant family and a time-invariant stationary distribution?
For the cycle model \(Y_{n+1}=0.5Y_n+3+\eta_n\), with independent \(\eta_n\sim N(0,4)\), find the invariant mean and variance.
In a deterministic model with cycle map \(Y_{n+1}=Y_n+2\), can a probability distribution be invariant? Explain.
Why is a mixture of laws from different dosing phases not generally the post-dose invariant law?

Solutions. (1) A periodic family repeats after each dosing interval but can change within the interval; a time-invariant stationary law is unchanged at all times. (2) The mean is \(3/(1-0.5)=6\). The variance satisfies \(v=0.5^2v+4\), so \(0.75v=4\) and \(v=16/3\). (3) No. Every cycle shifts the state by \(2\); an invariant probability distribution would have to be unchanged under a nonzero translation, which no probability distribution on the real line can satisfy. (4) The mixture includes values from phases other than immediately post-dose, so it does not describe the law at that specific phase; it is an average over phase rather than a phase-specific law.