Implementing Blood-Flow Coupling in the Derivative Function

From Physiological Flows to Derivatives

A PBPK derivative function turns the current compartment amounts into rates of change. For a perfused organ, the blood-flow contribution is not simply “blood flow times concentration.” It is the difference between arterial delivery and venous removal. Getting both terms into the right compartments, with consistent concentration definitions, is essential: an omitted return term can create mass, while a duplicated delivery term can make the model appear to generate drug.

The formulation here assumes perfusion-limited exchange: blood leaving an organ is in instantaneous equilibrium with the organ tissue. It also assumes constant organ blood flows, no drug storage in a separate vascular subcompartment, and well-mixed organ and central compartments. These assumptions are useful for a basic derivative implementation, but they are not universal descriptions of tissue transport.

Definition: For organ \(i\), let \(A_i\) be drug amount in the organ tissue (mg), \(V_i\) its tissue volume (L), and \(C_i=A_i/V_i\) its tissue concentration (mg/L). Let \(K_{p,i}\) be the tissue-to-blood partition coefficient, defined as tissue concentration divided by equilibrated blood concentration. The venous blood concentration is therefore \(C_{v,i}=C_i/K_{p,i}\). Let \(Q_i\) be blood flow through the organ (L/h), and let \(C_a\) be the arterial blood concentration supplied to it (mg/L).

The blood-flow rate of drug entering organ \(i\) is \(Q_iC_a\), and the rate leaving in venous blood is \(Q_iC_{v,i}\). Both have units of mg/h. The net flow contribution to the organ amount derivative is their difference. To represent local irreversible elimination, use a first-order rate \(k_iA_i\), where \(k_i\) has units of h\(^{-1}\). This is a deliberately simple elimination model; it is distinct from venous outflow, which returns drug to the circulation.

\frac{dA_i}{dt} = Q_iC_a-Q_iC_{v,i}-k_iA_i = Q_i\left(C_a-\frac{A_i}{V_iK_{p,i}}\right)-k_iA_i.

The first equality makes the physical bookkeeping visible: arterial delivery enters the organ, venous removal leaves it, and elimination leaves the modeled body. The second equality substitutes the definitions of tissue and venous concentrations. A negative net flow term is possible and meaningful: it means that, at the current state, the organ returns more drug in venous blood than it receives from arterial blood.

Coupling Organs Through the Central Compartment

For a closed circulation represented by one well-mixed central amount \(A_c\), define its volume \(V_c\) and arterial concentration \(C_a=A_c/V_c\). If the central compartment receives the venous return from all modeled organs and supplies their arterial flows, its derivative is

\frac{dA_c}{dt} = \sum_{i=1}^{m}Q_iC_{v,i} - \left(\sum_{i=1}^{m}Q_i\right)C_a + R(t),

where \(R(t)\) is a drug input rate into the central compartment, in mg/h. The arterial outflow term must use the same organ flows that appear in the organ equations. In a closed circuit, total outflow and return flow are consistent with the same flows; a model with additional unrepresented flows or compartments needs corresponding terms rather than an unexplained mismatch.

Key result: With no external input, and with the same \(Q_iC_a\) and \(Q_iC_{v,i}\) terms used in the organ and central equations, blood-flow transfer cancels from the derivative of total modeled drug amount. Under the stated first-order elimination model, total amount changes only through elimination.

This cancellation is a useful implementation check. It does not mean that drug amounts in individual compartments are constant: flow can transfer substantial drug among them. It means only that internal transfer cannot change the sum of their amounts.

Why the Flow Terms Cancel

Let \(A_{\text{tot}}=A_c+\sum_i A_i\). Add the central derivative to every organ derivative. For each organ, the tissue equation contains \(+Q_iC_a-Q_iC_{v,i}\), while the central equation contains \(+Q_iC_{v,i}-Q_iC_a\). The terms cancel pairwise:

\begin{aligned} \frac{dA_{\text{tot}}}{dt} &= \left[\sum_i Q_iC_{v,i}-\left(\sum_iQ_i\right)C_a+R(t)\right] +\sum_i\left[Q_iC_a-Q_iC_{v,i}-k_iA_i\right]\\ &=R(t)-\sum_i k_iA_i. \end{aligned}

This derivation depends on matching each organ’s flow terms across the two sides of the circulation. It assumes that the stated compartments account for all relevant drug amounts and that there are no other inputs, outputs, or elimination processes. If a central elimination term is included, for example, it must also appear as a loss in the total-amount balance.

Worked Example: Calculating Two Organ Derivatives

Consider a hypothetical model with two organs and a central compartment. The current arterial concentration is \(C_a=2.0\) mg/L. Organ 1 has \(Q_1=3.0\) L/h, \(V_1=1.0\) L, \(A_1=6.0\) mg, \(K_{p,1}=2.0\), and \(k_1=0.10\) h\(^{-1}\). Organ 2 has \(Q_2=2.0\) L/h, \(V_2=2.0\) L, \(A_2=4.0\) mg, \(K_{p,2}=1.0\), and \(k_2=0.050\) h\(^{-1}\). There is no external input at this instant.

For organ 1, \(C_1=A_1/V_1=6.0\) mg/L, so \(C_{v,1}=C_1/K_{p,1}=3.0\) mg/L. Its arterial delivery is \(Q_1C_a=(3.0)(2.0)=6.0\) mg/h, its venous removal is \(Q_1C_{v,1}=(3.0)(3.0)=9.0\) mg/h, and its elimination is \(k_1A_1=(0.10)(6.0)=0.60\) mg/h. Therefore,

\frac{dA_1}{dt}=6.0-9.0-0.60=-3.60\ \text{mg/h}.

For organ 2, \(C_2=A_2/V_2=2.0\) mg/L and \(C_{v,2}=2.0/1.0=2.0\) mg/L. Delivery and venous removal are both \((2.0)(2.0)=4.0\) mg/h; elimination is \((0.050)(4.0)=0.20\) mg/h. Thus,

\frac{dA_2}{dt}=4.0-4.0-0.20=-0.20\ \text{mg/h}.

The central compartment receives \(9.0+4.0=13.0\) mg/h in venous return and supplies \((Q_1+Q_2)C_a=(3.0+2.0)(2.0)=10.0\) mg/h to the organs. Hence \(dA_c/dt=3.0\) mg/h. Adding all three derivatives gives \(3.0-3.60-0.20=-0.80\) mg/h, exactly the negative total elimination rate, \(0.60+0.20=0.80\) mg/h.

Finding Duplicated and Omitted Contributions

The most useful diagnostic is to calculate the implied total-amount derivative independently. With no input in the example above, it must equal \(-0.80\) mg/h. A discrepancy indicates a bookkeeping error, an unaccounted process, or inconsistent assumptions. Checking only whether each compartment derivative looks plausible is weaker: errors can be hidden by transfers between compartments.

Worked Example: Detecting a Duplicated Arterial Term

Suppose an implementation mistakenly adds arterial delivery to organ 1 twice, while retaining the correct venous removal and elimination. Using the preceding hypothetical values, the erroneous organ 1 derivative is

\left(\frac{dA_1}{dt}\right)_{\text{wrong}} =2Q_1C_a-Q_1C_{v,1}-k_1A_1 =2(6.0)-9.0-0.60=2.40\ \text{mg/h}.

The correct derivative was \(-3.60\) mg/h, so the error adds \(6.0\) mg/h, exactly one extra copy of \(Q_1C_a\). If the central derivative remains \(3.0\) mg/h and organ 2 remains at \(-0.20\) mg/h, the summed derivative becomes \(2.40+3.0-0.20=5.20\) mg/h. Yet the total elimination is still \(0.80\) mg/h and there is no input. The mass balance should be \(-0.80\) mg/h, so the 6.0 mg/h discrepancy exposes the duplicated delivery term.

The same check catches omitted return. If organ 1’s \(Q_1C_{v,1}=9.0\) mg/h is removed from its tissue equation but not added to the central equation, the modeled total derivative is too low by 9.0 mg/h. This is not elimination: the returning drug has simply disappeared from the equations.

Implementing and Checking the Derivative Function

At each evaluation, the function should read the current state, calculate concentrations, calculate venous concentrations, and then assemble each derivative from explicitly identified inputs and outputs. A compact implementation is easier to audit when the direction of every flow term is apparent.

Read amounts and compute concentrations.
For each organ, compute \(C_i=A_i/V_i\); compute the central arterial concentration as \(C_a=A_c/V_c\).

Compute venous concentrations and flow rates.
Use \(C_{v,i}=C_i/K_{p,i}\), then form delivery \(Q_iC_a\) and return \(Q_iC_{v,i}\), each in amount per time.

Assemble organ derivatives.
For each organ, add arterial delivery, subtract venous removal, and subtract the organ elimination rate \(k_iA_i\).

Assemble the central derivative and audit.
Add all venous returns, subtract all arterial outflows, include any external input, and check the sum of derivatives against input minus elimination.

Worked Example: Auditing Units and the Return Term

A hypothetical organ has \(Q=0.50\) L/min, \(C_a=1.2\) mg/L, \(C_v=0.80\) mg/L, and an elimination rate of \(0.10\) mg/min. Calculate its flow contribution and state the corresponding central-compartment terms.

The arterial delivery is \((0.50\ \text{L/min})(1.2\ \text{mg/L})=0.60\) mg/min. Venous removal is \((0.50\ \text{L/min})(0.80\ \text{mg/L})=0.40\) mg/min. The net flow contribution to the organ is \(0.60-0.40=0.20\) mg/min, so its amount derivative is \(0.20-0.10=0.10\) mg/min.

The central derivative must include \(+0.40\) mg/min for this organ’s venous return and \(-0.60\) mg/min for its arterial outflow. Summed across organ and central equations, the flow terms cancel: \(0.20+0.40-0.60=0\) mg/min. The remaining total change is \(-0.10\) mg/min, the stated elimination. This check also shows why a flow in L/min cannot be combined with a concentration in mg/L and an elimination rate in mg/h without first reconciling the time units.

Assumptions, Alternatives, and Failure Modes

The formulation is appropriate when organ exchange can be treated as rapid relative to changes in the modeled concentrations, blood flow is adequately represented by the specified \(Q_i\), and the chosen tissue-to-blood partition coefficient supports the venous-equilibrium approximation. It is a model assumption, not a consequence of writing an ODE. When exchange across the capillary or cell membrane is not effectively instantaneous, a permeability-limited model may represent blood and tissue concentrations separately and use an exchange flux between them. A separate vascular compartment can also be needed when vascular drug amount or transit dynamics matter.

Elimination is another modeling choice. Here, \(k_iA_i\) removes drug directly from the organ amount. Other formulations may express elimination through a clearance multiplied by a specified concentration, or may represent metabolic pathways with additional states. Those alternatives require an explicit definition of the concentration acted on and of where eliminated drug leaves the model. They should not be added on top of \(k_iA_i\) unless the model intends to include both processes.

Flow units do not match time units. Convert all rates to one time basis before evaluating derivatives; mg/min and mg/h cannot be added directly.
Concentration bases are mixed. The \(C_a\) and \(C_{v,i}\) terms must be blood concentrations on the same basis. If tissue concentration is used in place of venous blood concentration, apply the stated partition relationship.
Return is omitted or assigned to the wrong compartment. Each organ’s venous outflow must appear as a loss from that organ and a corresponding input to the receiving circulation.
A flow term is counted twice. Write each physical delivery and return once in the source and destination equations, then verify cancellation in the total amount.
Flows are inconsistent. Central outflow should agree with the flows supplying the organs in this closed-circuit formulation. If not, identify and model the missing flow path rather than silently forcing a balance.

A derivative function may produce numerical values even when its bookkeeping is wrong. Unit checks, sign checks, and total-amount balance are therefore structural diagnostics, not optional refinements. They test the equations themselves before numerical integration is considered.

Check Your Understanding

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

An organ receives \(Q=1.5\) L/h at \(C_a=0.8\) mg/L and returns blood at \(C_v=0.5\) mg/L. What is its net flow contribution to the organ amount derivative?
Why must the central equation include a venous-return term when the organ equation subtracts venous outflow?
With no input and no elimination, what should the total modeled amount derivative be in this closed-circuit model?
If \(C_i=4.0\) mg/L and \(K_{p,i}=2.0\), what venous blood concentration follows from the defined equilibrium relationship?

Solutions. 1. Delivery is \((1.5)(0.8)=1.20\) mg/h; removal is \((1.5)(0.5)=0.75\) mg/h, so net flow into the organ is \(0.45\) mg/h. 2. The returning drug has left the organ but remains in the modeled circulation; adding it centrally preserves the internal-transfer balance. 3. Zero: with no external input or elimination, all internal flow terms cancel in the sum. 4. \(C_{v,i}=C_i/K_{p,i}=4.0/2.0=2.0\) mg/L.