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Writing a Pharmacokinetic SDE · Tutorial 32 of 100

Nonlinear Clearance With Process Noise

Learn to combine saturable elimination with a chosen state-dependent process-noise term and interpret the resulting pharmacokinetic SDE.

Intermediate 12 min read

What You'll Learn

  • Express Michaelis–Menten elimination as a drift in central amount.
  • Specify a saturating state-dependent diffusion coefficient and check its units.
  • Derive the model’s short-time conditional mean and variance.
  • Distinguish continuous process noise from deterministic input and constant diffusion.
  • Identify boundary behavior, modeling assumptions, and simulation diagnostics.
  • Work through numerical constructions and check their local moment implications.

From Saturable Elimination to a Stochastic Amount Model

Michaelis–Menten elimination makes the drug-removal rate depend nonlinearly on concentration: it rises at low concentration and approaches a maximum at high concentration. A stochastic model adds a separate question: how should random variation in the central amount change with the amount already present? The drift and diffusion answer different questions, so each must be specified explicitly.

Let \(A_t\) be the drug amount in a central compartment, \(V\) its fixed volume, and \(C_t=A_t/V\) its concentration. Let \(V_{\max}>0\) be the maximum elimination rate, in amount per unit time, and \(K_m>0\) the Michaelis–Menten constant, in amount per unit volume. With no input, the deterministic elimination rate is \(V_{\max}C_t/(K_m+C_t)\).

Definition: The central-amount model considered here is the Itô SDE $$ dA_t=b(A_t)\,dt+g(A_t)\,dW_t, $$ where \(W_t\) is a standard Brownian motion and $$ b(a)=-\frac{V_{\max}a}{K_m V+a}, \qquad g(a)=\frac{\sigma_{\max}a}{K_N+a}, \qquad a\geq 0. $$ Here \(K_N>0\) is a chosen amount scale for process noise and \(\sigma_{\max}>0\) has units amount per square root of time. The negative drift represents elimination; the positive diffusion coefficient sets the magnitude of continuous random fluctuations.

The diffusion form is a modeling choice, not a consequence of Michaelis–Menten kinetics. It is selected here because it is state-dependent, equals zero at zero amount, increases with amount, and approaches a finite value at high amount. The noise scale \(K_N\) need not equal the elimination scale \(K_mV\). The model is written in the Itô convention.

Constructing and Checking the SDE

The elimination term follows by converting concentration to amount. Substituting \(C_t=A_t/V\) gives $$ \frac{V_{\max}C_t}{K_m+C_t} = \frac{V_{\max}(A_t/V)}{K_m+A_t/V} = \frac{V_{\max}A_t}{K_mV+A_t}. $$ Elimination removes amount, so this rate enters the drift with a minus sign. In particular, at low amount the rate is approximately \(V_{\max}A_t/(K_mV)\); at high amount it approaches \(V_{\max}\).

For the chosen diffusion, \(a/(K_N+a)\) is dimensionless. Thus \(g(a)\) has units amount per square root of time, as required because \(dW_t\) has units square root of time. The drift has units amount per time, matching \(dt\). A dimensional mismatch in either coefficient makes the SDE incoherent even if its algebra looks plausible.

1
Choose the state.
Here the state is central amount \(A_t\), not concentration. Keep its units consistent throughout.
2
Write the mass-removal rate.
Convert concentration in the Michaelis–Menten formula using \(C_t=A_t/V\), then make the elimination contribution negative.
3
Specify process noise.
Choose \(g(a)\), its state dependence, and its parameter units. Do not infer it from the drift.
4
State the driver and convention.
Use a standard Brownian motion \(W_t\) and identify the equation as Itô.

Worked Example: Writing the Model at a Given Amount

Take \(V_{\max}=120\) mg/h, \(K_m=2\) mg/L, \(V=20\) L, \(K_N=40\) mg, and \(\sigma_{\max}=6\) mg/\(\sqrt{\text{h}}\). At \(A_t=40\) mg, calculate concentration, drift, and diffusion.

The concentration is \(C_t=40/20=2\) mg/L. The elimination rate is \(120(2)/(2+2)=60\) mg/h, so the drift is \(-60\) mg/h. Equivalently, \(K_mV=40\) mg and \(120(40)/(40+40)=60\) mg/h.

The diffusion coefficient is \(6(40)/(40+40)=3\) mg/\(\sqrt{\text{h}}\). Thus, at this state, the SDE has drift \(-60\) mg/h and diffusion \(3\) mg/\(\sqrt{\text{h}}\). For a short interval \(h=0.01\) h, its first-order conditional mean change is approximately \(-60(0.01)=-0.6\) mg and its first-order conditional variance is approximately \(3^2(0.01)=0.09\) mg\(^2\). These are local approximations, not exact finite-interval moments.

Local Meaning of Drift and Diffusion

For this model, the local mean rate is the drift, while the local variance rate is the square of the diffusion coefficient. The following result states that precisely. It applies to the bounded, globally Lipschitz coefficients used here, and to a solution adapted to the filtration of \(W_t\). To ensure standard global existence and uniqueness on the physical state space, define \(a^+=\max(a,0)\) and extend the coefficients to real \(a\) by replacing every \(a\) in their formulas with \(a^+\). These extensions are bounded and globally Lipschitz.

Result: For the extended model and any fixed time \(t\), as \(h\downarrow0\), $$ \frac{1}{h}E[A_{t+h}-A_t\mid\mathcal F_t]\longrightarrow b(A_t), \qquad \frac{1}{h}E[(A_{t+h}-A_t)^2\mid\mathcal F_t]\longrightarrow g(A_t)^2 $$ in \(L^1\). Here \(\mathcal F_t\) is the information available by time \(t\).

Justification. Write \(\Delta A=D+M\), where \(D=\int_t^{t+h}b(A_s)\,ds\) and \(M=\int_t^{t+h}g(A_s)\,dW_s\). Let \(|b|\leq B\), \(|g|\leq G\), and let \(L\) be a Lipschitz constant for \(b\). The Itô integral has conditional mean zero, so $$ \frac{1}{h}E[\Delta A\mid\mathcal F_t] = \frac{1}{h}E\left[\int_t^{t+h}b(A_s)\,ds\middle|\mathcal F_t\right]. $$ The expected absolute difference between this quantity and \(b(A_t)\) is at most $$ \frac{L}{h}\int_t^{t+h}E|A_s-A_t|\,ds \leq \frac{L}{h}\int_t^{t+h}\left(B(s-t)+G\sqrt{s-t}\right)ds, $$ which tends to zero. The bound follows from the integral equation and the Itô isometry.

For the second limit, the conditional Itô isometry gives \(E[M^2\mid\mathcal F_t]=E[\int_t^{t+h}g(A_s)^2\,ds\mid\mathcal F_t]\). The function \(g^2\) is Lipschitz because \(g\) is bounded and Lipschitz. The same continuity bound therefore shows that this term divided by \(h\) converges in \(L^1\) to \(g(A_t)^2\). The remaining terms vanish after division by \(h\): \(E[D^2]\leq B^2h^2\), \(E[M^2]\leq G^2h\), and Cauchy–Schwarz gives \(E|2DM|\leq2BGh^{3/2}\). This proves both limits. \(\square\)

This calculation explains why the diffusion coefficient is not itself a variance rate: the local variance rate is \(g(a)^2\). It also clarifies the interpretation of the selected noise. The noise amplitude depends on the current amount, but it is not a random elimination rate inserted into the drift.

Worked Example: Comparing Amount Levels

Use the parameter values from the previous example, so \(K_mV=K_N=40\) mg. Evaluate the elimination rate and diffusion coefficient at 10, 40, and 160 mg.

At \(A=10\) mg, elimination is \(120(10)/(40+10)=24\) mg/h and diffusion is \(6(10)/(40+10)=1.2\) mg/\(\sqrt{\text{h}}\). The local variance rate is \(1.2^2=1.44\) mg\(^2\)/h.

At \(A=40\) mg, elimination is \(120(40)/(40+40)=60\) mg/h and diffusion is \(6(40)/(40+40)=3\) mg/\(\sqrt{\text{h}}\). The local variance rate is \(9\) mg\(^2\)/h.

At \(A=160\) mg, elimination is \(120(160)/(40+160)=96\) mg/h and diffusion is \(6(160)/(40+160)=4.8\) mg/\(\sqrt{\text{h}}\). The local variance rate is \(23.04\) mg\(^2\)/h. Both functions increase over these values, but their limits differ: elimination approaches 120 mg/h, and diffusion approaches 6 mg/\(\sqrt{\text{h}}\). The matching scale values in this example are a convenient choice, not a required relationship.

Boundary Behavior and Modeling Choices

The chosen extension also has a useful boundary property. Starting with \(A_0\geq0\), its solution remains nonnegative. Indeed, the extended coefficients are globally Lipschitz, so the SDE has a pathwise unique strong solution. If the continuous process reaches zero, the solution that stays identically zero thereafter satisfies the SDE from that time, because both coefficients vanish at zero. Pathwise uniqueness forces the original solution to agree with it thereafter. A continuous path cannot become negative without first reaching zero; hence it cannot cross into negative amounts. Zero is absorbing in this specification. This fact is specific to the chosen zero-input model and should be considered when deciding whether the boundary behavior is suitable.

Worked Example: Adding a Deterministic Input

Suppose a constant infusion supplies \(R=30\) mg/h while the amount is \(40\) mg, using the parameters above. The input contributes \(+R\) to the drift, while elimination remains \(-60\) mg/h. Therefore $$ dA_t=\left(30-\frac{120A_t}{40+A_t}\right)dt +\frac{6A_t}{40+A_t}\,dW_t. $$

At \(A_t=40\) mg, the drift is \(30-60=-30\) mg/h and the diffusion is \(3\) mg/\(\sqrt{\text{h}}\). The local conditional mean change over \(0.01\) h is approximately \(-0.30\) mg, and the local conditional variance is approximately \(9(0.01)=0.09\) mg\(^2\). The infusion is a deterministic input in the drift; it does not replace or rescale the Brownian term.

A constant additive diffusion \(g(a)=\sigma\) is a different model: it allows the same absolute fluctuation size at low and high amounts, including at zero. A proportional diffusion \(g(a)=\lambda a\) instead makes the noise grow without bound as amount grows. The bounded saturating choice here is between those patterns: it vanishes at zero and approaches a finite absolute noise scale. Each choice encodes a distinct assumption; none follows automatically from the nonlinear clearance law.

The construction also has limitations. The scalar Brownian driver represents continuous process variation in this one-compartment state, not uncertainty in \(V_{\max}\), \(K_m\), or the administered dose. A single diffusion coefficient cannot represent every possible source of variability. Furthermore, high-amount diffusion saturates by design, whether or not that behavior is supported by data. Parameters for drift and noise may be difficult to identify separately from sparse concentration measurements, so model adequacy must be judged against the sampling design and observed variability.

Diagnostics and Common Pitfalls

  • Dropping the volume conversion. The elimination denominator in amount units is \(K_mV+A_t\), not \(K_m+A_t\). The latter adds quantities with incompatible units.
  • Putting a positive sign on elimination. The rate is positive as a removal magnitude, but its contribution to the amount drift is negative.
  • Using a variance rate as the diffusion coefficient. If a local variance rate is \(q(a)\), the corresponding diffusion magnitude is \(\sqrt{q(a)}\), not \(q(a)\).
  • Assuming the noise law is implied by clearance. Michaelis–Menten elimination specifies the drift only. The state dependence and scale of \(g\) require a separate modeling decision.
  • Ignoring units or boundary behavior. Check that \(b\) is amount/time and \(g\) is amount/\(\sqrt{\text{time}}\); also check whether the coefficient choices behave plausibly near zero and at high amount.

In numerical work, unusually large negative simulated amounts, sensitivity to the time step, or a poor match between fitted and observed variability are warning signs. A coarse Euler step can produce values inconsistent with the continuous-time boundary behavior even when the specified coefficients vanish at zero. Such a result calls for checking the discretization and step size, not silently changing the elimination drift. Residual variability should also be assessed across amount levels: a model with increasing \(g(A_t)\) predicts a changing local variance rate \(g(A_t)^2\).

Key takeaway: Specify saturable clearance as a negative amount drift, then choose and justify the process diffusion independently. For the example here, clearance saturates at \(V_{\max}\), while the state-dependent diffusion saturates at \(\sigma_{\max}\); their local mean and variance implications are determined by \(b(A_t)\) and \(g(A_t)^2\), respectively.

Check Your Understanding

Try each question, then compare your work with the solutions below.

  1. For \(K_m=1.5\) mg/L, \(V=10\) L, \(V_{\max}=50\) mg/h, and \(A=15\) mg, calculate the elimination drift.
  2. If \(g(a)=4a/(20+a)\) in units of mg/\(\sqrt{\text{h}}\), calculate the diffusion and local variance rate at \(a=20\) mg.
  3. Explain why a constant diffusion coefficient does not have the same state dependence as the model in this tutorial.
  4. Suppose deterministic input is \(12\) mg/h and elimination at the current state is \(20\) mg/h. What is the drift at that state, before accounting for the Brownian increment?

Solutions. (1) \(K_mV=1.5(10)=15\) mg, so the elimination magnitude is \(50(15)/(15+15)=25\) mg/h and the drift is \(-25\) mg/h. (2) \(g(20)=4(20)/(20+20)=2\) mg/\(\sqrt{\text{h}}\); the local variance rate is \(2^2=4\) mg\(^2\)/h. (3) A constant coefficient assigns the same absolute noise magnitude at every state, whereas \(4a/(20+a)\) changes with amount and tends to zero as amount tends to zero. (4) The drift is input minus elimination, \(12-20=-8\) mg/h; the diffusion term remains separate.

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