Exact Solution for Stimulated Production

Why the Exact Trajectory Matters

When a drug persistently stimulates response production, the response does not usually jump immediately to its new steady state. Production changes at once in the model, but the response accumulates over time while ongoing loss continues. An exact expression for that trajectory makes it possible to predict the response at a chosen time, determine its limiting level, and check whether the shape of observed data is consistent with the assumed mechanism.

The earlier material on indirect response state equations and baseline equilibrium established the production-loss structure. Here, the focus is its exact solution when stimulation is constant. The key result is that the response approaches a higher equilibrium exponentially, with the rate of approach governed by the first-order loss-rate constant.

Model and Assumptions

Definition: Let $R(t)$ be the response at time $t$, $k_{\mathrm{in}}>0$ the baseline production rate, and $k_{\mathrm{out}}>0$ the first-order loss-rate constant. Under a constant production-stimulation factor $S\ge 0$, the model is $$ \frac{dR(t)}{dt}=k_{\mathrm{in}}(1+S)-k_{\mathrm{out}}R(t), \qquad R(0)=R_{\mathrm{init}}. $$ Here $S$ is dimensionless, $k_{\mathrm{in}}$ has units of response per time, and $k_{\mathrm{out}}$ has units of inverse time. The model assumes that stimulation is constant from time zero onward, production is multiplied by $1+S$, loss remains first-order with an unchanged $k_{\mathrm{out}}$, and there is one response state with no instantaneous jump at the time stimulation begins.

The initial response $R_{\mathrm{init}}$ may be any finite value in the model; in typical applications it is nonnegative. If the response was at baseline equilibrium before stimulation, then $R_{\mathrm{init}}=R_0$, where $R_0=k_{\mathrm{in}}/k_{\mathrm{out}}$. The derivation below does not require that initial condition. The constant-effect assumption means that the same $S$ applies throughout the time interval being considered; the result is not an exact solution for an effect that changes over time.

Deriving the New Equilibrium and Trajectory

At equilibrium, the response is constant, so its time derivative is zero. Setting the right-hand side of the model to zero gives

0=k_{\mathrm{in}}(1+S)-k_{\mathrm{out}}R_{\mathrm{ss}}, \qquad R_{\mathrm{ss}}=\frac{k_{\mathrm{in}}(1+S)}{k_{\mathrm{out}}}.

Because $k_{\mathrm{out}}>0$, this equilibrium is finite and unique for the stated parameters. To derive the trajectory, define the deviation from the new equilibrium as $D(t)=R(t)-R_{\mathrm{ss}}$. Since $R_{\mathrm{ss}}$ is constant, the model implies

\frac{dD(t)}{dt} =\frac{dR(t)}{dt} =k_{\mathrm{in}}(1+S)-k_{\mathrm{out}}\bigl(D(t)+R_{\mathrm{ss}}\bigr) =-k_{\mathrm{out}}D(t),

where the equilibrium identity $k_{\mathrm{in}}(1+S)=k_{\mathrm{out}}R_{\mathrm{ss}}$ cancels the constant terms. The initial deviation is $D(0)=R_{\mathrm{init}}-R_{\mathrm{ss}}$. One way to solve this equation without dividing by the initial deviation is to multiply it by $e^{k_{\mathrm{out}}t}$. The product rule then gives

\frac{d}{dt}\left(e^{k_{\mathrm{out}}t}D(t)\right) =e^{k_{\mathrm{out}}t}\left(\frac{dD(t)}{dt}+k_{\mathrm{out}}D(t)\right)=0.

Therefore $e^{k_{\mathrm{out}}t}D(t)$ is constant and equals $D(0)$. This argument also covers $D(0)=0$: the product remains zero and $D(t)=0$ for all $t$. Equivalently, in that special case the differential equation has the constant solution $D(t)=0$. Thus no division by $D(0)$ is needed. For $D(0)\ne 0$, separation and integration would give the same exponential solution, but that division-based argument must be restricted to the nonzero case.

D(t)=D(0)e^{-k_{\mathrm{out}}t}, \qquad R(t)=R_{\mathrm{ss}}+ \left(R_{\mathrm{init}}-R_{\mathrm{ss}}\right)e^{-k_{\mathrm{out}}t}.

This formula holds for $t\ge 0$. Since $k_{\mathrm{out}}>0$, $e^{-k_{\mathrm{out}}t}$ decreases toward zero, so $R(t)$ approaches $R_{\mathrm{ss}}$. If $R_{\mathrm{init}}<R_{\mathrm{ss}}$, the response rises monotonically toward equilibrium; if $R_{\mathrm{init}}>R_{\mathrm{ss}}$, it falls monotonically toward it. If $R_{\mathrm{init}}=R_{\mathrm{ss}}$, it remains there. The response reaches the equilibrium exactly at a finite time only in this last case; otherwise the exponential deviation approaches zero asymptotically.

Worked Example: Stimulation Beginning at Baseline

Suppose a hypothetical response has $k_{\mathrm{in}}=12$ response units per hour and $k_{\mathrm{out}}=0.300$ per hour. Initially the response is at baseline, and a constant stimulation factor $S=1.50$ begins at time zero. Find the baseline and stimulated equilibria, then calculate the response after 2.00 hours.

The baseline equilibrium, before stimulation, is $R_0=k_{\mathrm{in}}/k_{\mathrm{out}}$. Substitution gives

R_0=\frac{12\ \text{response units/hour}}{0.300\ \text{hour}^{-1}} =40.0\ \text{response units}.

The stimulated production rate is $12(1+1.50)=30.0$ response units per hour. The new equilibrium is therefore

R_{\mathrm{ss}}=\frac{12(1+1.50)}{0.300} =100\ \text{response units}.

Since $R_{\mathrm{init}}=R_0=40.0$ response units, the trajectory at $t=2.00$ hours is

\begin{aligned} R(2.00) &=100+(40.0-100)e^{-0.300(2.00)}\\ &=100-60.0e^{-0.600}\\ &\approx 100-60.0(0.5488)\\ &=67.07\ \text{response units}. \end{aligned}

The prediction is between the initial response of 40.0 and the stimulated equilibrium of 100 response units, as expected for an increasing trajectory. Its initial slope is $30.0-0.300(40.0)=18.0$ response units per hour, also equal to $k_{\mathrm{out}}(R_{\mathrm{ss}}-R_{\mathrm{init}})=0.300(60.0)=18.0$ response units per hour.

Worked Example: An Initial Response Above or Below Baseline

Use the same hypothetical parameters, $k_{\mathrm{in}}=12$ response units per hour, $k_{\mathrm{out}}=0.300$ per hour, and $S=1.50$, but suppose the response at time zero is $20.0$ response units. Calculate the response after 1.50 hours and determine its direction of change.

The stimulated equilibrium remains $100$ response units. Substituting the new initial condition into the general solution gives

\begin{aligned} R(1.50) &=100+(20.0-100)e^{-0.300(1.50)}\\ &=100-80.0e^{-0.450}\\ &\approx100-80.0(0.6376)\\ &=48.99\ \text{response units}. \end{aligned}

Because $20.0<100$, the response rises toward the stimulated equilibrium. As a check, the initial slope is $30.0-0.300(20.0)=24.0$ response units per hour. The derivative from the trajectory at time zero is $0.300(100-20.0)=24.0$ response units per hour, so the two calculations agree.

Worked Example: Time to Reach a Specified Response

For the baseline-starting hypothetical case in the first example, determine when the response first reaches 80.0 response units. The trajectory is $R(t)=100-60.0e^{-0.300t}$, with time in hours.

Set $R(t)=80.0$ and solve for the exponential term:

80.0=100-60.0e^{-0.300t}, \qquad e^{-0.300t}=\frac{20.0}{60.0}=\frac{1}{3}.

Taking natural logarithms gives $-0.300t=\ln(1/3)=-\ln(3)$, so

t=\frac{\ln(3)}{0.300\ \text{hour}^{-1}} =3.662\ \text{hours}.

This target is reached at a finite time because it lies strictly between the initial response, 40.0 response units, and the new equilibrium, 100 response units. The equilibrium itself is approached asymptotically rather than reached at a finite time.

Checks, Comparisons, and Limitations

Several simple checks help detect algebraic mistakes or a poor match between the model and the data. The predicted equilibrium must satisfy $k_{\mathrm{in}}(1+S)=k_{\mathrm{out}}R_{\mathrm{ss}}$. At time zero, the trajectory must give $R(0)=R_{\mathrm{init}}$, because $e^0=1$. Its initial slope must equal the model’s production-minus-loss rate, $k_{\mathrm{in}}(1+S)-k_{\mathrm{out}}R_{\mathrm{init}}$. Finally, for positive $k_{\mathrm{out}}$, the difference $R(t)-R_{\mathrm{ss}}$ must keep its initial sign while shrinking in magnitude.

A practical diagnostic is to compare measured responses with the predicted direction, curvature, and limiting level, while accounting for measurement error. A sustained response that overshoots the predicted equilibrium, reverses direction, or approaches a different plateau may indicate that the constant-stimulation assumptions are inadequate. Possible explanations include changing stimulation, a changing loss rate, additional response processes, or measurement issues. Such patterns are prompts to examine the model; by themselves they do not identify which assumption failed.

Approach or mechanism	What it represents	Implication for this solution
Exact stimulated-production solution	Constant multiplier on production with first-order loss unchanged	Equilibrium changes; the exponential rate of approach is $k_{\mathrm{out}}$
Numerical integration of the same model	Approximation to the differential equation using time steps	Can check the exact calculation, but introduces step-size and numerical-error considerations
Stimulated loss	A drug effect on response removal rather than on production	A different mechanism and equation; it should not be interpreted using the stimulated-production equilibrium formula

The exact solution is useful because it gives the response at any time without a numerical solver, and it exposes how the initial condition and parameters affect the trajectory. Its simplicity depends on its assumptions. If stimulation is not constant, if loss is not first-order, or if another state or delay is needed to describe the response, this formula may not apply. It also does not establish that production stimulation is the biological mechanism merely because the curve fits observed responses: competing models can sometimes describe similar patterns.

Common Mistakes

Using the baseline equilibrium as the stimulated equilibrium. With $S>0$, production is $k_{\mathrm{in}}(1+S)$, so the equilibrium is $k_{\mathrm{in}}(1+S)/k_{\mathrm{out}}$, not $k_{\mathrm{in}}/k_{\mathrm{out}}$.
Changing the decay rate in the exponential. Under this model the exponential rate is $k_{\mathrm{out}}$, because stimulation changes production while the loss term remains $k_{\mathrm{out}}R$.
Dividing by a zero initial deviation. If $R_{\mathrm{init}}=R_{\mathrm{ss}}$, then $D(0)=0$; the solution is simply $D(t)=0$. The exponential solution follows directly from the differential equation or integrating-factor argument, without dividing by $D(0)$.
Interpreting the plateau as an observed finite-time value. Except when the response starts at equilibrium, the trajectory approaches $R_{\mathrm{ss}}$ asymptotically. A finite observation may be close to the plateau without being exactly at it.

Key takeaway: Under constant stimulation of production and unchanged first-order loss, the new equilibrium is $k_{\mathrm{in}}(1+S)/k_{\mathrm{out}}$, and the response’s deviation from that equilibrium decays as $e^{-k_{\mathrm{out}}t}$. The formula applies to any initial response, including one equal to the new equilibrium.

Check Your Understanding

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

If $k_{\mathrm{in}}=8.00$ response units per day, $k_{\mathrm{out}}=0.200$ per day, and $S=0.500$, what is the stimulated equilibrium?
For the parameters in question 1, with $R_{\mathrm{init}}=40.0$ response units, write the trajectory and calculate $R(2.00)$ days.
What happens if $R_{\mathrm{init}}=R_{\mathrm{ss}}$? Explain why a separation argument dividing by $D(0)$ is not needed.
For a response starting below $R_{\mathrm{ss}}$, state the sign of its initial slope and whether it can cross the equilibrium at a finite time under this model.

Solutions. (1) $R_{\mathrm{ss}}=8.00(1+0.500)/0.200=60.0$ response units. (2) The trajectory is $R(t)=60.0+(40.0-60.0)e^{-0.200t}=60.0-20.0e^{-0.200t}$ response units, with $t$ in days. Thus $R(2.00)=60.0-20.0e^{-0.400}\approx60.0-20.0(0.6703)=46.59$ response units. (3) The deviation $D(0)$ is zero, and $D'=-k_{\mathrm{out}}D$ has the constant solution $D(t)=0$, so the response remains at equilibrium. The integrating-factor derivation also gives this result directly and never divides by $D(0)$. (4) Since $R_{\mathrm{init}}<R_{\mathrm{ss}}$, the initial slope is $k_{\mathrm{out}}(R_{\mathrm{ss}}-R_{\mathrm{init}})>0$. The deviation is $(R_{\mathrm{init}}-R_{\mathrm{ss}})e^{-k_{\mathrm{out}}t}$, which remains negative for every finite $t$; therefore the response approaches but does not cross the equilibrium.