Exact Solution for Inhibited Production

Why an Exact Trajectory Is Useful

An indirect response does not generally change immediately to its drug-on steady state. Production and loss continue to operate while the response moves toward a new balance. For constant fractional inhibition of production, the response follows a simple exponential trajectory. Its exact form makes it possible to predict the response at any time, estimate when an intermediate target will be reached, and recognize which targets cannot be reached under the model.

The derivation here concerns a drug effect that is held constant over time. It does not describe how a changing concentration produces a changing inhibition. The earlier material on indirect response state equations and inhibition of response production supplies the underlying rate balance; here we solve that balance explicitly.

Model and Assumptions

Definition: Let $R(t)$ be the response at time $t$, $k_{\mathrm{in}}$ the zero-order production rate, $k_{\mathrm{out}}$ the first-order loss-rate constant, and $I$ the constant fractional inhibition of production. The inhibited-production model is $$ \frac{dR}{dt}=k_{\mathrm{in}}(1-I)-k_{\mathrm{out}}R. $$ Here $k_{\mathrm{in}}>0$ has units of response per unit time, $k_{\mathrm{out}}>0$ has units of inverse time, and $0\leq I\leq 1$ is dimensionless. The initial response is $R(0)=R_{\mathrm{init}}$.

This model assumes that the drug reduces production by the same fraction at every time during the interval considered, while loss remains proportional to the current response with an unchanged $k_{\mathrm{out}}$. Both rate parameters are constant, and there is no additional delay, feedback, or response-dependent change in production or loss. The drug-on steady state is meaningful as an equilibrium under these conditions. A baseline equilibrium before inhibition is a special case, not a requirement: $R_{\mathrm{init}}$ can be any specified starting response.

With no drug effect, production and loss balance at the baseline response $R_0=k_{\mathrm{in}}/k_{\mathrm{out}}$. Under constant inhibition, production becomes $k_{\mathrm{in}}(1-I)$, so the drug-on steady state is $$ R_{\mathrm{ss}}=\frac{k_{\mathrm{in}}(1-I)}{k_{\mathrm{out}}}=(1-I)R_0. $$ This is the response at which the inhibited production rate exactly equals the loss rate. It is not necessarily the response at the instant the drug effect begins.

Deriving the Exact Solution

The steady state provides a direct way to solve the differential equation. Since $k_{\mathrm{in}}(1-I)=k_{\mathrm{out}}R_{\mathrm{ss}}$, subtracting the steady-state balance from the rate equation gives $$ \frac{dR}{dt}=-k_{\mathrm{out}}(R-R_{\mathrm{ss}}). $$ For a response different from $R_{\mathrm{ss}}$, separate variables and integrate from time 0 to $t$, and from $R_{\mathrm{init}}$ to $R(t)$:

\int_{R_{\mathrm{init}}}^{R(t)}\frac{dR}{R-R_{\mathrm{ss}}} =-k_{\mathrm{out}}\int_0^t du \quad\Longrightarrow\quad \ln\left|\frac{R(t)-R_{\mathrm{ss}}}{R_{\mathrm{init}}-R_{\mathrm{ss}}}\right| =-k_{\mathrm{out}}t.

Exponentiating and using the initial condition yields the exact solution. It also remains valid when the initial response equals the steady state:

R(t)=R_{\mathrm{ss}}+\bigl(R_{\mathrm{init}}-R_{\mathrm{ss}}\bigr)e^{-k_{\mathrm{out}}t}.

The exponential is dimensionless because $k_{\mathrm{out}}t$ is dimensionless. The difference $R(t)-R_{\mathrm{ss}}$ is the deviation from the drug-on equilibrium; it retains its initial sign and shrinks in magnitude over time. Differentiating the solution verifies the rate equation directly:

\frac{dR}{dt} =-k_{\mathrm{out}}\bigl(R_{\mathrm{init}}-R_{\mathrm{ss}}\bigr)e^{-k_{\mathrm{out}}t} =-k_{\mathrm{out}}\bigl(R(t)-R_{\mathrm{ss}}\bigr).

Worked Example: Predicting the Response After Inhibition Begins

Suppose a response has baseline equilibrium $R_0=100$ response units, $k_{\mathrm{out}}=0.200\ \mathrm{h}^{-1}$, and inhibition $I=0.600$. The response is at baseline when inhibition begins. Find the drug-on steady state and predict the response after 5 hours.

The implied production rate is $k_{\mathrm{in}}=k_{\mathrm{out}}R_0=(0.200\ \mathrm{h}^{-1})(100\ \text{response units})=20.0$ response units/h. The drug-on steady state is $$ R_{\mathrm{ss}}=(1-I)R_0=(1-0.600)(100)=40.0\ \text{response units}. $$

Using $R_{\mathrm{init}}=100$ response units and $t=5.00$ h, $$ R(5)=40.0+(100-40.0)e^{-(0.200)(5.00)} =40.0+60.0e^{-1} =62.1\ \text{response units}. $$

The response has fallen from 100 toward 40 response units but remains above the drug-on steady state at 5 hours. Its initial slope is $-k_{\mathrm{out}}(R_{\mathrm{init}}-R_{\mathrm{ss}})=-(0.200)(60.0)=-12.0$ response units/h. The response is not predicted to fall at this initial rate indefinitely; the rate of decline becomes smaller as the response approaches 40.

Target Attainment and Reachability

A target time can be calculated by setting the exact trajectory equal to a specified $R_{\mathrm{target}}$. If the target lies strictly between $R_{\mathrm{init}}$ and $R_{\mathrm{ss}}$, rearranging gives

t_{\mathrm{target}}= \frac{1}{k_{\mathrm{out}}} \ln\left(\frac{R_{\mathrm{init}}-R_{\mathrm{ss}}} {R_{\mathrm{target}}-R_{\mathrm{ss}}}\right).

For a target strictly between the initial response and the steady state, the ratio inside the logarithm is positive and greater than 1, so the calculated time is positive and finite. The trajectory approaches $R_{\mathrm{ss}}$ without reaching it in finite time unless it starts there. A target beyond $R_{\mathrm{ss}}$ in the direction of travel is not reached: the response moves toward the steady state but does not cross it. Targets outside the interval between $R_{\mathrm{init}}$ and $R_{\mathrm{ss}}$ are therefore not attained under this constant-effect model.

Worked Example: Time to Reach an Intermediate Target

Use the conditions in the previous example: $R_{\mathrm{init}}=100$ response units, $R_{\mathrm{ss}}=40.0$ response units, and $k_{\mathrm{out}}=0.200\ \mathrm{h}^{-1}$. Calculate the time to reach 70.0 response units, and assess whether the response reaches 40.0 or 30.0 response units.

Because $70.0$ lies strictly between 100 and 40.0, the target-time formula applies: $$ t_{70}=\frac{1}{0.200\ \mathrm{h}^{-1}} \ln\left(\frac{100-40.0}{70.0-40.0}\right) =5.00\ \mathrm{h}\times\ln(2) =3.47\ \mathrm{h}. $$

Thus the trajectory reaches 70.0 response units after approximately 3.47 hours. It approaches 40.0 response units asymptotically and does not reach that steady-state value at a finite time from this different initial value. It also does not reach 30.0 response units: that target lies beyond the steady state in the direction of decline, and the trajectory cannot cross the steady state.

Starting Away From Baseline

The solution depends on the actual response at the start of the constant-inhibition interval, not on an assumption that the response is at baseline. This distinction matters when the drug effect begins after a prior perturbation or when the observed starting value differs from the baseline equilibrium. The formula remains the same; only $R_{\mathrm{init}}$ changes.

Worked Example: A Nonbaseline Starting Response

Suppose $k_{\mathrm{in}}=12.0$ response units/h, $k_{\mathrm{out}}=0.300\ \mathrm{h}^{-1}$, and $I=0.250$. At the time inhibition begins, the response is $R_{\mathrm{init}}=50.0$ response units. Calculate the response after 4.00 hours and the time to reach 35.0 response units.

The uninhibited baseline equilibrium is $R_0=12.0/0.300=40.0$ response units. Under inhibition, production is $12.0(1-0.250)=9.00$ response units/h, giving $$ R_{\mathrm{ss}}=\frac{9.00\ \text{response units/h}}{0.300\ \mathrm{h}^{-1}} =30.0\ \text{response units}. $$

The predicted response after 4.00 hours is $$ R(4)=30.0+(50.0-30.0)e^{-(0.300)(4.00)} =30.0+20.0e^{-1.20} =36.0\ \text{response units}. $$

The target 35.0 response units lies between the starting value and the steady state, so it is reachable at a finite time: $$ t_{35}=\frac{1}{0.300\ \mathrm{h}^{-1}} \ln\left(\frac{50.0-30.0}{35.0-30.0}\right) =\frac{\ln(4)}{0.300\ \mathrm{h}^{-1}} =4.62\ \mathrm{h}. $$ This calculation uses the observed starting response of 50.0, not the baseline equilibrium of 40.0.

Interpretation, Diagnostics, and Alternatives

Under the model, the direction of change follows from the starting response relative to $R_{\mathrm{ss}}$. If $R_{\mathrm{init}}>R_{\mathrm{ss}}$, the response declines monotonically; if $R_{\mathrm{init}}<R_{\mathrm{ss}}$, it increases monotonically; if they are equal, it remains constant. These predictions follow from $$ \frac{dR}{dt}=-k_{\mathrm{out}}(R-R_{\mathrm{ss}}), $$ so the slope always points toward the steady state. A trajectory that crosses the predicted steady state or reverses direction while the inhibition and model parameters are assumed constant is inconsistent with this model.

A useful diagnostic is to compare the observed trajectory with the predicted exponential, including its initial value, direction, curvature, and long-run level. Systematic residual patterns, a changing apparent rate of approach, or an estimated steady state inconsistent with the rate balance can indicate that an assumption is inadequate. Possible causes include inhibition that changes over time, a changing loss rate, delayed drug action, response feedback, measurement error, or an incorrectly specified starting response. Such patterns do not by themselves identify which cause is responsible.

The exact solution is preferable to numerical integration when the stated constant-rate model applies: it gives the trajectory without discretization error and provides a transparent target-time calculation. A numerical solver can be useful when a more complex model has been specified, but it does not make an incorrect constant-inhibition assumption valid. An empirical exponential fit can summarize a trajectory, but without the rate-balance assumptions it does not establish that production inhibition is the mechanism. In particular, this inhibited-production solution should not be substituted for a different mechanism that changes response loss.

The model’s simplicity is also its limitation. It predicts a single exponential approach to a fixed steady state and cannot represent multiple response compartments or a time-varying inhibition unless those features are added explicitly. A fitted trajectory may appear exponential over a limited observation window even when the underlying mechanism is more complicated. Mechanistic interpretation therefore depends on the design, measured exposure, and adequacy of the full model, not on the curve shape alone.

Common mistake: “The response approaches its steady state, so it cannot reach any value on the way there.” This is incorrect. It reaches every target strictly between its initial value and the steady state in finite time. Only the steady state itself is approached asymptotically from a different initial value; a target beyond that steady state in the direction of travel is not reached.

Worked Example: Checking the Halfway Deviation

Let $R_0=60.0$ response units, $k_{\mathrm{out}}=0.250\ \mathrm{h}^{-1}$, $I=0.400$, and $R_{\mathrm{init}}=R_0$. Find the time at which the difference between the response and the drug-on steady state has fallen to half its initial value, and calculate the response then.

The drug-on steady state is $R_{\mathrm{ss}}=(1-0.400)(60.0)=36.0$ response units. The deviation is $R(t)-R_{\mathrm{ss}}=(R_{\mathrm{init}}-R_{\mathrm{ss}})e^{-k_{\mathrm{out}}t}$. It is half its initial value when $e^{-k_{\mathrm{out}}t}=0.5$, so $$ t=\frac{\ln(2)}{0.250\ \mathrm{h}^{-1}}=2.77\ \mathrm{h}. $$ At that time, $R(t)=36.0+(60.0-36.0)(0.5)=48.0$ response units. This is halfway between the initial response and the steady state, not halfway between zero and the initial response.

Check Your Understanding

Answer each question using the constant fractional inhibition model. Solutions follow the questions.

If $R_0=80.0$ response units and $I=0.300$, what is the drug-on steady state?
For $R_{\mathrm{init}}=70.0$, $R_{\mathrm{ss}}=40.0$, and $k_{\mathrm{out}}=0.200\ \mathrm{h}^{-1}$, is a target of 55.0 response units reached in finite time? Calculate the time.
Under the same conditions as question 2, is a target of 35.0 response units reached? Explain.
What does the solution predict if $R_{\mathrm{init}}=R_{\mathrm{ss}}$?

Solutions. (1) $R_{\mathrm{ss}}=(1-0.300)(80.0)=56.0$ response units. (2) The target lies between the initial response and the steady state, so it is reached: $t=(1/0.200)\ln[(70.0-40.0)/(55.0-40.0)]=5.00\ln(2)=3.47$ hours. (3) No. The target is below the steady state in the direction of decline, and the trajectory approaches 40.0 response units without crossing it. (4) The deviation from steady state is initially zero, so $R(t)=R_{\mathrm{ss}}$ at all times under the constant-effect model.