Steady-State Consistency in Closed and Eliminating Systems

Why Steady-State Checks Matter

A model can conserve drug overall and still have compartments that are changing. Conversely, a system can have constant amounts in every compartment while drug continues to enter, move through the body, and leave by elimination. These distinctions matter when debugging PBPK equations: a flat total amount alone does not show that the distribution has reached steady state, and a constant concentration during infusion does not mean that no drug is moving.

The central check is to write the amount balance for each compartment. At steady state, each compartment has zero net accumulation. Summing those balances then gives a whole-system check: a closed distribution system conserves total amount, while a system with elimination at steady state must have total input equal to total elimination. The balance is a necessary accounting condition, not proof that every biological assumption in a model is correct.

Deriving the Mass-Balance Checks

Let \(A_i\) denote the drug amount in compartment \(i\), with units such as mg. Each compartment balance adds the rates entering it and subtracts the rates leaving it. A transfer from compartment \(i\) to compartment \(j\) appears once as an outflow and once as an inflow. When all compartment balances are summed, those internal transfer rates cancel.

Here \(A_{\mathrm{tot}}=\sum_i A_i\), and \(R_{\mathrm{in}}\) and \(R_{\mathrm{out}}\) are the external rates entering and leaving the modeled system. The rates must use the same amount-per-time units. In a closed distribution system both are zero, so the total amount is conserved. In a system with elimination, \(R_{\mathrm{out}}\) includes the elimination rate. If the system has reached steady state, every \(dA_i/dt=0\), so their sum is zero as well and total input must equal total output.

The sum is a useful check, but it is weaker than checking every compartment. In a closed system, internal transfers can change individual amounts while leaving the total unchanged. At steady state, the individual balances must also be zero. A nonzero transfer can persist at steady state if it is balanced by another rate in the same compartment; steady state does not require every flux to be zero.

Closed Distribution: Transfer Without Loss

Consider two compartments with amounts \(A_1\) and \(A_2\). Suppose drug transfers from compartment 1 to compartment 2 at rate \(k_{12}A_1\), and returns at rate \(k_{21}A_2\). The rate constants \(k_{12}\) and \(k_{21}\) have units of inverse time. There is no external input or elimination.

Adding the equations gives \(d(A_1+A_2)/dt=0\), confirming conservation of total amount. At steady state, each equation must equal zero. Both yield \(k_{12}A_1=k_{21}A_2\), so the forward and reverse transfer rates match. This relationship determines the steady distribution, while the initial total amount determines the total available to distribute. This particular model assumes linear, reversible transfer with fixed rate constants and no other processes.

Eliminating Systems: Input Must Balance Output

Now suppose a system receives drug at a constant rate and eliminates it. For a one-compartment model with amount \(A\), volume \(V\), clearance \(CL\), and constant input rate \(R_{\mathrm{in}}\), concentration is \(C=A/V\) and the elimination rate is \(CL\cdot C\). Here \(V\) is in L, \(C\) in mg/L, \(CL\) in L/h, and the rates in mg/h. With fixed volume and linear clearance, the amount balance is:

At steady state, \(dA/dt=0\), so \(R_{\mathrm{in}}=CL\cdot C_{\mathrm{ss}}\), where \(C_{\mathrm{ss}}\) is the steady-state concentration. This equality follows from the model’s amount balance; it does not require that drug cease to enter or be eliminated. It assumes constant input and fixed clearance and volume over the interval considered.

Checking Each Compartment When Drug Flows Through

In a multicompartment system, summing balances checks the whole-system accounting, but the individual equations show how local fluxes balance. A compartment may have a nonzero net transfer that is exactly matched by its input or elimination. This is a steady state with ongoing throughput, not an equilibrium in which all fluxes have stopped.

Practical Checks, Alternatives, and Failure Modes

A direct balance check is usually the clearest way to assess a proposed steady state: calculate each inflow and outflow in consistent amount-per-time units, form the net rate for each compartment, and then check the system-wide sum. A numerical simulation offers a complementary check by showing whether amounts or concentrations have stopped changing over the interval of interest. A flat-looking curve alone can be misleading if the time step is too large, the plotted scale hides change, or the calculation has not run long enough. The balance equations identify the accounting condition; the simulation shows the behavior produced by the implemented model.

Steady state and equilibrium are not interchangeable. In the closed two-compartment example, the net transfer between the pair is zero at steady state. In the infusion example, input and elimination continue at equal rates. In the two-compartment throughput example, the central-to-peripheral transfer also continues. A steady state refers to unchanging compartment amounts; equilibrium generally implies no net driving flux for the processes being considered. Which description applies depends on the modeled processes.

• Only checking the total: A closed model can have a constant total while amounts redistribute. Check every compartment derivative, not only their sum.
• Requiring all fluxes to vanish: At steady state, net accumulation is zero; a compartment’s inflows can balance its outflows even when both are nonzero.
• Comparing rates with incompatible units: Convert terms to the same amount-per-time units before adding or comparing them. For example, \(CL\cdot C\) has units of amount per time, not amount.
• Applying the constant-input balance to changing conditions: If input or model parameters vary with time, a stationary state may not exist. A momentary equality of total input and output alone does not establish that every compartment is steady.
• Ignoring model scope: The total-output check must include all modeled routes that remove drug. An omitted sink can make the equation appear inconsistent even when the implemented compartment equations are internally balanced.

The general balance \(dA_{\mathrm{tot}}/dt=R_{\mathrm{in}}-R_{\mathrm{out}}\) does not depend on linear elimination. If elimination is nonlinear, total input must still equal total output at steady state, but the concentration satisfying that equality must be found using the specified elimination function. The linear clearance examples above rely specifically on elimination rate \(CL\cdot C\).