Comparing Drug-On Steady States Across Models

Why Compare Drug-On Steady States?

A drug can change an indirect response by changing its production rate or its loss rate. The response may eventually settle at a level different from baseline, but the direction and size of that change depend on which rate is multiplied. Comparing the four canonical models at steady state makes this distinction explicit, while keeping the turnover system itself simple.

Here, “drug-on steady state” means the long-run equilibrium predicted if the drug effect is held constant. It is a model-based endpoint, not a claim that every dosing regimen produces a sustained plateau. The baseline equilibrium is \(R_0=k_{\mathrm{in}}/k_{\mathrm{out}}\), where \(k_{\mathrm{in}}>0\) is the zero-order production rate and \(k_{\mathrm{out}}>0\) is the first-order loss rate. This tutorial compares how a fixed multiplier changes that equilibrium.

Fixed Multipliers and Equilibrium

For the usual inhibited and stimulated models, an inhibition multiplier is \(q=1-I\), where \(I\) is the inhibited fraction and \(0\leq I<1\). A stimulation multiplier is \(q=1+S\), where \(S\geq0\) is the stimulation increment. Thus inhibition gives \(0<q\leq1\), and stimulation gives \(q\geq1\). Keeping \(q\) explicit makes it easier to compare the same numerical multiplier across different affected rates.

The equilibrium equations follow by setting the derivative to zero. For a production effect, \(0=qk_{\mathrm{in}}-k_{\mathrm{out}}R_{\mathrm{ss}}\), so \(R_{\mathrm{ss}}=qk_{\mathrm{in}}/k_{\mathrm{out}}=qR_0\). For a loss effect, \(0=k_{\mathrm{in}}-qk_{\mathrm{out}}R_{\mathrm{ss}}\), so \(R_{\mathrm{ss}}=k_{\mathrm{in}}/(qk_{\mathrm{out}})=R_0/q\), provided \(q>0\).

These equilibria are stable under the stated conditions. With production multiplied, subtract the equilibrium equation from the differential equation to obtain \(d(R-R_{\mathrm{ss}})/dt=-k_{\mathrm{out}}(R-R_{\mathrm{ss}})\). With loss multiplied, the corresponding equation is \(d(R-R_{\mathrm{ss}})/dt=-qk_{\mathrm{out}}(R-R_{\mathrm{ss}})\). Both deviations decay exponentially when their loss coefficient is positive. This verifies that the calculated equilibrium is approached from any finite initial response, rather than merely being a value at which the derivative happens to vanish.

The Four Canonical Predictions

The four models are obtained by applying an inhibition or stimulation multiplier to one of the two rates. For inhibited production, \(q=1-I\) multiplies \(k_{\mathrm{in}}\), so the steady state is \((1-I)R_0\), below baseline when \(I>0\). For inhibited loss, the same form of multiplier acts on \(k_{\mathrm{out}}\), giving \(R_0/(1-I)\), above baseline when \(0<I<1\). For stimulated production, \(q=1+S\) multiplies \(k_{\mathrm{in}}\), giving \((1+S)R_0\), above baseline when \(S>0\). For stimulated loss, it multiplies \(k_{\mathrm{out}}\), giving \(R_0/(1+S)\), below baseline when \(S>0\).

The table shows a key contrast: the same multiplier has a direct effect on the steady state if it changes production, but a reciprocal effect if it changes loss. A multiplier below one reduces production but also reduces the rate at which response is removed. Consequently, inhibition of production lowers the equilibrium, whereas inhibition of loss raises it. A multiplier above one gives the reverse pair of directions.

Comparing Magnitudes Carefully

The direction alone does not describe the full prediction. Under a multiplier \(q=0.50\), production gives \(0.50R_0\), while loss gives \(2R_0\). For the hypothetical baseline of 40 response units, these are 20 and 80 response units, respectively. The production model is 20 units below baseline; the loss model is 40 units above baseline. The responses are not symmetric around baseline, even though both use the same multiplier.

For multipliers equally far from one in additive terms, reciprocal loss effects are generally not equal in magnitude to direct production effects. For example, \(q=1.50\) gives \(1.50R_0\) under stimulated production but \(R_0/1.50=0.667R_0\) under stimulated loss. In contrast, for a multiplier near one, the two effects have approximately equal-sized deviations in opposite directions. To see why, write \(q=1+\delta\), with \(\delta\) small. Then \(1/q=1/(1+\delta)\approx1-\delta\), so \(qR_0\approx(1+\delta)R_0\) while \(R_0/q\approx(1-\delta)R_0\). This is a local approximation, not an exact equality for larger effects.

Assumptions, Diagnostics, and Limits

These steady-state expressions assume a single response compartment with constant \(k_{\mathrm{in}}\) and \(k_{\mathrm{out}}\), a fixed effect multiplier, and no additional feedback, saturation, or time-varying process that changes the rate equations. They also assume \(R_0\) is defined by a positive production rate and a positive loss rate. A proposed multiplier should be consistent with the effect being modeled: the standard inhibited-production or inhibited-loss fraction cannot exceed one, and a loss multiplier must be strictly positive to yield a finite equilibrium.

The boundary \(q=0\) deserves special care. For production multiplied by zero, the equation becomes \(dR/dt=-k_{\mathrm{out}}R\), and the response approaches zero. For loss multiplied by zero, \(dR/dt=k_{\mathrm{in}}\): production continues while no response is lost, so \(R\) increases linearly and there is no finite steady state. The formula \(R_0/q\) is undefined there, correctly signaling that the equilibrium calculation does not apply.

In an analysis, check both the algebra and the model conditions. A calculated negative response, a nonpositive loss multiplier, or a claimed finite equilibrium when loss has been set to zero indicates an invalid parameterization or a formula applied outside its conditions. Also distinguish a measured late-time value from a true equilibrium: a finite observation period may end before the response has settled, even when the model has a stable equilibrium. The fixed-multiplier calculation is not a substitute for evaluating whether the drug effect was effectively constant for long enough.

A steady-state endpoint has an important limitation: different mechanisms can predict the same endpoint. For example, an observed rise may be compatible with inhibited loss or stimulated production, depending on the multipliers. The endpoint alone does not establish which rate changed. A transient analysis uses the time course and the model’s response dynamics, whereas a steady-state comparison isolates the long-run consequence of a specified constant effect. Neither approach removes the need to assess whether its assumptions fit the data.