Two Elimination Models, Two Density Calculations
A transition density describes the distribution of a concentration at a later time, conditional on its current value. For a linear one-compartment diffusion, that density can be obtained explicitly. Saturable elimination is nonlinear, so its forward equation is still available but generally does not yield a simple closed-form transition density. The distinction matters: an exact Gaussian formula is useful for the linear model, while the nonlinear model calls for careful interpretation of its forward equation and boundary flux.
Throughout, \(X_t\) denotes concentration, \(W_t\) is a standard Wiener process, and \(p(x,t)\) denotes a density with respect to concentration \(x\). The forward-equation convention used here is that positive probability flux points toward increasing \(x\).
Linear One-Compartment Transition Density
Consider a linear elimination model with constant input and additive process noise:
where \(k>0\) is the elimination rate, \(r\) is a constant input rate, \(\sigma>0\) is the diffusion coefficient, and \(x_0\) is a fixed initial concentration. This is a diffusion on the real line. Its Gaussian transition law is mathematically exact, but the model does not enforce nonnegative concentration.
To derive it, multiply the SDE by the integrating factor \(e^{kt}\). Itô’s product rule gives \(d(e^{kt}X_t)=r e^{kt}\,dt+\sigma e^{kt}\,dW_t\), since the factor \(e^{kt}\) has finite variation. Integrating from 0 to \(t\), then multiplying by \(e^{-kt}\), yields
The stochastic integral has a deterministic integrand. It is therefore normal with mean zero and variance \(\sigma^2\int_0^t e^{-2k(t-s)}\,ds=\sigma^2(1-e^{-2kt})/(2k)\). Adding its mean and variance to the deterministic terms proves the stated distribution and density. As \(t\) tends to zero, the variance tends to zero and the transition law tends to a point mass at \(x_0\), as it should.
Worked Example: Calculating a Linear Transition Density
Suppose \(k=0.5\) per day, \(r=0.2\) concentration units per day, \(\sigma=0.3\) concentration units per square-root day, and \(x_0=2\). Calculate the transition distribution after two days and its density at \(x=1\).
The mean is \(m_2=2e^{-1}+(0.2/0.5)(1-e^{-1})\). Using \(e^{-1}\approx0.367879\), this is \(0.735758+0.252848=0.988606\).
The variance is \(v_2=0.3^2(1-e^{-2})/(2(0.5))=0.09(0.864665)=0.077820\). Thus the standard deviation is approximately \(\sqrt{0.077820}=0.27896\).
At \(x=1\), the density is \((2\pi(0.077820))^{-1/2}\exp\{-(1-0.988606)^2/(2(0.077820))\}\). The denominator inside the exponential is \(0.155640\), so its exponent is approximately \(-0.000833\). The normalizing factor is approximately \(1/\sqrt{0.48896}=1.430\), giving a density of approximately \(1.429\) per concentration unit.
Saturable Elimination and Its Forward Equation
For elimination without input between doses, consider the positive-state model
where \(V_{\max}>0\), \(K_m>0\), and \(\gamma>0\). The drift represents Michaelis–Menten elimination, and the diffusion coefficient is proportional to concentration. This choice makes the stochastic term vanish at zero. The following forward equation applies in the interior \(x>0\), provided the law has a sufficiently regular density \(p(x,t)\).
Here \(b(x)=-V_{\max}x/(K_m+x)\) and \(a(x)=\gamma^2x^2\). Substitution gives
with flux \(J(x,t)=-\frac{V_{\max}x}{K_m+x}p(x,t)-\frac{\gamma^2}{2}\partial_x(x^2p(x,t))\). The derivatives act on the entire products, not only on the coefficients. In particular, state-dependent diffusion contributes both a first-derivative and a second-derivative term when the equation is expanded.
Worked Example: Evaluating Saturable-Model Flux
Take \(V_{\max}=1.2\), \(K_m=0.8\), \(\gamma=0.2\), and consider a point where \(x=1\), \(p=0.4\), and \(\partial_xp=-0.2\). Calculate the flux there. These local values illustrate the calculation; they do not by themselves define a full probability density.
At \(x=1\), \(b=-1.2/(0.8+1)=-2/3\), \(a=0.2^2(1)^2=0.04\), and \(a'=2(0.2^2)(1)=0.08\). Since \(\partial_x(ap)=a'p+ap'\), we obtain \(\partial_x(ap)=0.08(0.4)+0.04(-0.2)=0.032-0.008=0.024\).
Consequently, \(J=bp-\tfrac12\partial_x(ap)=(-2/3)(0.4)-0.5(0.024)=-0.266667-0.012=-0.278667\). The negative sign means the local net flux is toward smaller concentrations. Omitting \(a'p\) would give the wrong flux even though the diffusion coefficient itself is easy to evaluate.
Boundary Conditions and Probability Conservation
The forward equation describes local change; the flux specifies how probability crosses the ends of the state space. For a density on \((0,\infty)\), integrating the equation from \(\varepsilon\) to \(R\) gives \(\frac{d}{dt}\int_\varepsilon^R p(x,t)\,dx=J(\varepsilon,t)-J(R,t)\), whenever differentiation under the integral is justified. If \(p\) and \(\partial_t p\) are integrable and the flux has endpoint limits, taking \(\varepsilon\downarrow0\) and \(R\uparrow\infty\) shows that total probability is conserved when both endpoint flux limits are zero.
For this particular SDE, zero is not reached in finite time from a positive initial value. Before any possible hit of zero, Itô’s formula gives \(d\log X_t=[-V_{\max}/(K_m+X_t)-\gamma^2/2]dt+\gamma\,dW_t\). The drift in brackets is bounded because \(0\leq V_{\max}/(K_m+X_t)\leq V_{\max}/K_m\). On any finite time interval, its time integral and the Brownian path are finite. Thus \(\log X_t\) cannot tend to \(-\infty\) at a finite first hitting time, as a hit of zero would require. This establishes the claimed nonattainment for positive initial conditions. In a regular density setting, zero limiting flux is the appropriate conservation statement; a Dirichlet condition \(p(0,t)=0\) is not a general substitute for it.
The conservation calculation also gives a useful check on a derivation or numerical solution. If the mass \(\int_0^\infty p(x,t)\,dx\) changes despite zero endpoint flux, inspect the sign convention for \(J\), the factor \(1/2\), and differentiation of \(a(x)p(x,t)\). The calculation presumes enough regularity and integrability to interchange differentiation and integration; it does not justify those conditions for every proposed density.
What the Linear Approximation Captures—and Misses
Worked Example: Comparing Low and High Concentrations
Let \(V_{\max}=1.2\) and \(K_m=0.8\). The saturable elimination drift is \(b(x)=-1.2x/(0.8+x)\). At low concentrations, its first-order approximation is \(-\left(V_{\max}/K_m\right)x=-1.5x\); at high concentrations, the elimination rate approaches the constant \(V_{\max}=1.2\).
At \(x=0.08\), the exact drift is \(-1.2(0.08)/(0.88)=-0.10909\), compared with the low-concentration approximation \(-1.5(0.08)=-0.12\). The relative difference in magnitudes is \((0.12-0.10909)/0.10909=0.10\), or 10%.
At \(x=0.8\), the exact drift is \(-0.96/1.6=-0.6\), while the low-concentration approximation is \(-1.2\); the approximation is no longer close. At \(x=8\), the exact drift is \(-9.6/8.8=-1.09091\), close to the high-concentration limiting drift \(-1.2\). The transition density of the saturable model still does not follow from these drift comparisons: a nonlinear drift and multiplicative noise require solving or approximating their own forward equation.
The linear Gaussian formula is the main explicit alternative when linear drift and additive noise are scientifically reasonable. Its strengths are an exact transition density and transparent dependence on time and parameters. Saturable elimination instead represents a concentration-dependent elimination rate; its forward equation retains that nonlinearity and the state dependence of diffusion, but typically needs numerical or other specialized methods to obtain a density.
A common mistake is to replace \(b(x)\) by a linear approximation and then use the Gaussian density as though it were exact for saturable elimination. The approximation is local: the low-concentration slope is \(-V_{\max}/K_m\), whereas the drift approaches a constant at high concentrations. Another pitfall is to assume that a Gaussian density is automatically physically suitable because it integrates to one. The linear additive-noise model has support on the whole real line, so it can assign positive probability to negative concentrations. Its suitability depends on whether that limitation is acceptable for the intended use; the formula itself does not impose a positivity constraint.
For either model, useful diagnostics include checking that the density is nonnegative and integrates to one, verifying the initial distribution and time-zero limit, and checking endpoint flux and total mass over time. For the linear model, the calculated mean and variance should match the explicit formulas. For the saturable model, also verify that the flux differentiates the full product \(a(x)p(x,t)\), and that numerical boundary treatment does not spuriously create or remove probability.
Check Your Understanding
Work each problem, then compare your reasoning with the solutions.
- For the linear model with \(r=0\), \(k=1\), \(\sigma=0.2\), \(x_0=1\), and \(t=\log 2\), find the transition mean and variance.
- For the saturable model, state \(b(x)\), \(a(x)\), and the interior forward equation.
- Explain why zero flux and an absorbing boundary are not interchangeable.
- For \(b=-0.5\), \(a=0.09\), \(a'=0.06\), \(p=2\), and \(p'=-1\) at a point, calculate the flux.
Solutions. 1. Since \(e^{-kt}=1/2\), the mean is \(1/2\). The variance is \(0.2^2(1-e^{-2t})/2=0.04(1-1/4)/2=0.015\). 2. \(b(x)=-V_{\max}x/(K_m+x)\), \(a(x)=\gamma^2x^2\), and \(\partial_t p=-\partial_x(bp)+\tfrac12\partial_{xx}(ap)\) for \(x>0\). 3. Zero flux retains probability in the state space; an absorbing boundary removes probability upon hitting it. 4. \(\partial_x(ap)=a'p+ap'=0.06(2)+0.09(-1)=0.03\), so \(J=bp-\tfrac12\partial_x(ap)=(-0.5)(2)-0.015=-1.015\).