From Repeated Dosing to a Cycle Map
Under a repeated regimen, a PBPK model is driven by the same dose schedule over and over. At periodic steady state, the full concentration and amount trajectories repeat from one dosing interval to the next. Instead of checking every state at every time indefinitely, we can summarize one complete interval by a map from the state at a chosen phase in one cycle to the state at the same phase in the next.
The phase must be defined precisely. For example, let \(x_k^-\) denote the vector of PBPK states immediately before dose \(k\), with dose intervals of length \(\tau\). The state might include amounts in tissues and blood, amounts in absorption depots, and any other dynamic states needed by the model. The cycle map applies the dose, integrates the model over the interval, and returns the state immediately before the next dose.
The map is not usually a simple formula. For a PBPK model represented by ordinary differential equations, it is often an operation: apply the scheduled dose event, integrate the equations to the next cycle boundary, and return the resulting state. Bolus doses are state jumps; infusions and other scheduled inputs must be handled over their specified times. The cycle map is meaningful only when these events are represented consistently on every cycle.
The earlier material on checking steady state by repeated-dose convergence describes comparing states or exposures across successive cycles. A cycle map expresses the same repeated-dose structure as a fixed-point problem. It also makes clear what must be solved: the state at the cycle boundary, not merely one selected concentration or an average over the interval.
Why a Fixed Point Represents a Repeating Trajectory
Let \(\Phi_\tau\) denote the operation that takes the chosen starting state through one complete interval, including any dose event required by the phase convention. Then \(M(x)=\Phi_\tau(x)\). If \(M(x^*)=x^*\), the state at the boundary is unchanged after a cycle. Because the same model equations, inputs, and event schedule apply on each interval, the within-cycle solution beginning at \(x^*\) is repeated on the next interval. This gives a periodic trajectory, not a constant state: concentrations can rise and fall within each interval.
This argument assumes the model is deterministic, the dosing schedule and covariates remain unchanged across cycles, and the state vector includes every variable needed to determine future evolution. If an enzyme state, depot amount, or other history-dependent quantity is omitted, two apparently identical boundary states may not lead to identical next-cycle trajectories. The map must also use the same event timing and boundary convention each time.
A fixed point need not be unique, and its existence alone does not guarantee that repeated simulation from an arbitrary initial state will approach it. Local attraction can be assessed through the map’s derivative. For a differentiable multistate map, a fixed point is locally attracting when the eigenvalues of the Jacobian \(J_M(x^*)\) have magnitudes below one. For a scalar map, this condition is \(|M'(x^*)|<1\). This is a local condition: it describes behavior near the fixed point, not necessarily from every possible starting state.
Worked Examples
Worked Example: One-Compartment Elimination with a Bolus Dose
Consider a hypothetical one-compartment model with instantaneous bolus doses of 100 mg every 12 h and first-order elimination with \(k=0.2000\ \mathrm{h}^{-1}\). Let \(A_k^-\) be the amount in the compartment just before dose \(k\). The dose raises the amount to \(A_k^-+D\); over the following interval, elimination multiplies that amount by \(e^{-k\tau}\). Thus the cycle map is
Here \(r=e^{-k\tau}=e^{-(0.2000\ \mathrm{h}^{-1})(12\ \mathrm{h})}=e^{-2}=0.1353353\), which is dimensionless. The fixed-point equation is \(A^-=r(A^-+D)\). Rearranging gives \(A^-(1-r)=rD\), so
This is the amount immediately before a dose at periodic steady state. Immediately after the dose, the amount is \(15.6518+100=115.6518\ \mathrm{mg}\). If the hypothetical volume is 20.00 L, the corresponding pre-dose concentration is \(15.6518\ \mathrm{mg}/20.00\ \mathrm{L}=0.7826\ \mathrm{mg/L}\), and the immediate post-dose concentration is \(115.6518\ \mathrm{mg}/20.00\ \mathrm{L}=5.783\ \mathrm{mg/L}\).
The map is a contraction because its derivative is \(M'(A)=r=0.1353353\), whose magnitude is less than one. Thus deviations from the fixed-point amount shrink by this factor per cycle. This conclusion follows for this linear model and does not establish that every PBPK cycle map is contractive.
Worked Example: A Two-State Linear Cycle Map
Suppose a hypothetical two-state PBPK approximation has the cycle map \(x_{k+1}=Ax_k+b\), where the states are amounts in two compartments in mg, and
The fixed point satisfies \(x^*=Ax^*+b\), or \((I-A)x^*=b\). Writing \(x^*=(x_1^*,x_2^*)^\mathsf{T}\), the equations are \(0.4x_1^*-0.1x_2^*=10\) and \(-0.2x_1^*+0.5x_2^*=0\). The second equation gives \(x_2^*=0.4x_1^*\). Substitution into the first gives \(0.4x_1^*-0.1(0.4x_1^*)=0.36x_1^*=10\), so \(x_1^*=27.7778\ \mathrm{mg}\) and \(x_2^*=11.1111\ \mathrm{mg}\).
As a direct check, the first component of \(Ax^*+b\) is \(0.6(27.7778)+0.1(11.1111)+10\approx27.7778\ \mathrm{mg}\). The second is \(0.2(27.7778)+0.5(11.1111)\approx11.1111\ \mathrm{mg}\). Both components reproduce the proposed fixed point.
Starting from \(x_0=(0,0)^\mathsf{T}\ \mathrm{mg}\), direct cycle iteration gives \(x_1=(10,0)^\mathsf{T}\ \mathrm{mg}\), \(x_2=(16,2)^\mathsf{T}\ \mathrm{mg}\), and \(x_3=(19.8,4.2)^\mathsf{T}\ \mathrm{mg}\). These values move toward the fixed point. The eigenvalues of \(A\) are \(0.7\) and \(0.4\), both with magnitude below one, so perturbations in this linear map decay with repeated cycles.
Worked Example: Iteration and a Nonlinear Fixed Point
Consider the hypothetical scalar map \(z_{k+1}=M(z_k)=0.5z_k+3/(1+z_k)\), with \(z_0=0\). It is a teaching example of a nonlinear cycle map, not a parameterization of a particular drug. The fixed point solves \(z=0.5z+3/(1+z)\). Multiplication by \(1+z\) gives \(0.5z(1+z)=3\), or \(z^2+z-6=0\). The roots are \(z=2\) and \(z=-3\); restricting to nonnegative states leaves \(z^*=2\).
Iteration from zero gives \(z_1=3\), then \(z_2=0.5(3)+3/(1+3)=2.25\). Next,
Applying the map again, without rounding the value of \(z_3\), gives
The derivative is \(M'(z)=0.5-3/(1+z)^2\). At \(z^*=2\), \(M'(2)=0.5-3/9=1/6\), so the fixed point is locally attracting. The iterates shown are consistent with this local behavior. The exact solution is \(z^*=2\); iteration is a way to approach it, not a substitute for checking that the map and its fixed point are correctly defined.
Computing and Checking a PBPK Periodic State
There are two common computational routes. Direct repeated-dose simulation starts from an initial state and applies the cycle map repeatedly. A fixed-point method instead seeks a root of \(F(x)=M(x)-x\), so that \(F(x^*)=0\). The first route is simple and closely follows dosing; the second can be useful when convergence by repeated iteration is slow. Either route depends on evaluating the cycle map correctly, including all dose events and state variables.
Choose a boundary such as immediately before each dose, and include all dynamic states needed to predict the next cycle.
Apply doses and other scheduled inputs at their specified times, integrate through the interval, and return the state at the same phase.
Iterate \(x_{k+1}=M(x_k)\), or solve \(M(x)-x=0\). Retain the full state vector when assessing whether successive cycles agree.
Apply one more cycle to the proposed state and check that the returned state agrees within a prespecified numerical and practical tolerance. Inspect the within-cycle trajectory as well as the boundary state.
A successful fixed-point calculation should be consistent with direct repeated-dose simulation when both use the same model, dose schedule, phase convention, and numerical settings. Agreement is a useful cross-check, not proof that the model is physiologically correct. For example, a solver could return a small residual for a cycle map that omits a depot state or applies a dose at the wrong boundary.
Diagnostics, Alternatives, and Limitations
Common warning signs include a state that oscillates across cycles, drifts rather than settling, becomes negative when it represents an amount, or appears converged only because a large absolute tolerance hides changes in small states. Periodic dosing can produce genuine cycle-to-cycle oscillations during the approach to steady state; compare the same phase in each cycle before interpreting them. Also compare clinically relevant within-cycle outputs, such as peak and trough concentrations, since matching a boundary state alone does not verify that the intended dose events were applied correctly.
If iteration does not settle, do not assume that no periodic fixed point exists. The map may be weakly attracting, may have more than one fixed point, or may not be locally attracting from the chosen initial state. A fixed-point root solver may locate a solution even when direct iteration is slow, but it can also converge to an unintended root or fail if the map is poorly conditioned. Inspect the residual \(M(x)-x\), the plausibility of the state, and the resulting trajectory. A different phase convention can also change the numerical vector representing the fixed point, even though it describes the same periodic trajectory when handled consistently.
Direct repeated-dose simulation is intuitive and naturally reveals transient behavior, but may require many cycles. Solving \(M(x)-x=0\) can reduce the number of cycles, but requires a root-finding strategy and careful handling of state bounds and multiple solutions. For a linear map \(M(x)=Ax+b\), solving \((I-A)x=b\) is direct when \(I-A\) is nonsingular. These approaches are alternatives for computing the cycle-boundary state; neither replaces checks of event timing, state completeness, or trajectory plausibility.
The cycle-map formulation does not guarantee that a unique, stable periodic state exists for every model and regimen. It also does not make numerical integration error disappear: the map is evaluated by computation, so a small fixed-point residual may partly reflect the accuracy of that evaluation. The next step in numerical work is to assess solver convergence and numerical error separately from the fixed-point definition.
Check Your Understanding
Answer each question, then compare your reasoning with the brief solutions.
- For a cycle map \(M\), what equation defines a periodic steady state?
- A simulation compares only the pre-dose concentration between consecutive cycles. Why is that insufficient to establish periodic steady state in a multistate PBPK model?
- For the map \(A_{k+1}^-=r(A_k^-+D)\), what is the fixed-point amount in terms of \(r\) and \(D\), provided \(r\ne1\)?
- For a scalar differentiable map, what does \(|M'(x^*)|<1\) imply, and what does it not imply?
- Name one reason that a fixed-point root solver and direct repeated-dose iteration might produce apparently different results.
Solutions. (1) The condition is \(M(x^*)=x^*\). (2) Other states, such as tissue or depot amounts, may still differ and can affect the next cycle; the full boundary state must be checked. (3) Rearranging \(A^*=r(A^*+D)\) gives \(A^*(1-r)=rD\), hence \(A^*=rD/(1-r)\). (4) It implies local attraction near the fixed point; it does not guarantee convergence from every starting state or establish global uniqueness. (5) The methods may use different event timing, state phase, or numerical settings, or a root solver may have found a different fixed point.