Why a Closed-Form Solution Is Useful
The effect site is represented as a latent equilibration state: its concentration need not equal the measured plasma concentration at a given time. The first-order effect-compartment equation describes how the effect-site concentration moves toward plasma concentration. When plasma concentration is held constant, that differential equation has an exact solution. The solution makes the role of the starting concentration explicit and lets us calculate the effect-site concentration at any later time without stepping through small time increments.
This result is useful for checking intuition as well as doing calculations. It shows the direction of movement, the rate of approach, and the long-run limit. It also provides a simple diagnostic: if the calculated effect-site concentration moves away from a constant plasma concentration, or crosses it under the model assumptions, a sign, unit, or implementation error may have occurred.
Model and Closed-Form Solution
The initial condition is arbitrary: \(C_{e,0}\) may be below, equal to, or above \(C_p^*\). The time elapsed since that condition is \(t-t_0\), not necessarily \(t\) itself. If the initial condition is specified at time zero, then \(t_0=0\); otherwise, using \(t\) in place of \(t-t_0\) would apply the wrong elapsed duration.
An equivalent form displays the weights on the initial and plasma concentrations:
For \(k_{e0}>0\) and \(t\geq t_0\), the exponential lies between zero and one. Thus the solution is a weighted average of the initial effect-site concentration and the constant plasma concentration. In particular, it remains between those two values. The formula gives \(C_e(t_0)=C_{e,0}\) at the starting time, and as elapsed time increases without bound, \(C_e(t)\) approaches \(C_p^*\).
Why the Formula Holds
The derivation follows directly from the differential equation. Subtract the constant \(C_p^*\) from both sides by defining the difference \(x(t)=C_e(t)-C_p^*\). Because \(C_p^*\) is constant, \(dx/dt=dC_e/dt\), so the model becomes a homogeneous first-order equation:
Multiplying by \(e^{k_{e0}(t-t_0)}\) gives a product whose derivative is zero:
Therefore this product is constant in time. At \(t=t_0\), its value is \(x(t_0)=C_{e,0}-C_p^*\). Solving for \(x(t)\), then adding \(C_p^*\), yields the stated solution:
This derivation depends on the plasma concentration being constant throughout the interval and on \(k_{e0}\) being constant. It also assumes the first-order effect-compartment equation applies over that interval. If plasma concentration changes, one constant value \(C_p^*\) cannot represent the input over the entire period.
Worked Example: Effect Site Initially Below Plasma
Suppose a hypothetical model has a constant plasma concentration of \(8.00\ \mathrm{mg/L}\) from \(t_0=0\), an initial effect-site concentration of \(2.00\ \mathrm{mg/L}\), and \(k_{e0}=0.120\ \mathrm{h^{-1}}\). Calculate the effect-site concentration after \(5.00\ \mathrm{h}\).
The elapsed time is \(t-t_0=5.00\ \mathrm{h}\). Substitute into the solution:
The result is above the initial \(2.00\ \mathrm{mg/L}\) and below the constant plasma concentration of \(8.00\ \mathrm{mg/L}\), as the model requires. The initial rate of change is \(k_{e0}(C_p^*-C_{e,0})=(0.120\ \mathrm{h^{-1}})(6.00\ \mathrm{mg/L})=0.720\ \mathrm{mg/L/h}\), which is positive: initially, the effect-site concentration increases.
Worked Example: Effect Site Initially Above Plasma
In a second hypothetical case, plasma concentration is constant at \(4.00\ \mathrm{mg/L}\), the effect-site concentration at \(t_0=1.00\ \mathrm{h}\) is \(10.0\ \mathrm{mg/L}\), and \(k_{e0}=0.200\ \mathrm{h^{-1}}\). Find \(C_e\) at \(t=4.00\ \mathrm{h}\).
The elapsed time is \(4.00-1.00=3.00\ \mathrm{h}\). Using the initial effect-site-minus-plasma difference gives:
The calculated value has decreased from \(10.0\ \mathrm{mg/L}\) but remains above \(4.00\ \mathrm{mg/L}\). The initial rate is \((0.200\ \mathrm{h^{-1}})(4.00-10.0)\ \mathrm{mg/L}=-1.20\ \mathrm{mg/L/h}\), consistent with movement downward toward plasma concentration.
Interpreting and Checking the Solution
The differential equation gives the instantaneous direction of change:
Because \(k_{e0}\) is positive, the derivative is positive when \(C_e<C_p^*\), negative when \(C_e>C_p^*\), and zero when \(C_e=C_p^*\). The closed-form solution is consistent with these signs: it retains the sign of \(C_{e,0}-C_p^*\), while the magnitude of that difference shrinks exponentially. The effect-site concentration approaches the constant plasma concentration without overshooting it.
A useful check is to substitute the solution back into the model. Differentiating gives \(dC_e/dt=-k_{e0}(C_{e,0}-C_p^*)e^{-k_{e0}(t-t_0)}\). Meanwhile, \(C_p^*-C_e(t)=-(C_{e,0}-C_p^*)e^{-k_{e0}(t-t_0)}\). Multiplying this difference by \(k_{e0}\) reproduces the derivative, and at \(t=t_0\) the formula reproduces \(C_{e,0}\). Both the equation and the initial condition are therefore satisfied.
Worked Example: Using a Nonzero Time Origin
For a hypothetical constant plasma concentration of \(9.00\ \mathrm{mg/L}\), suppose \(C_e(2.00\ \mathrm{h})=3.00\ \mathrm{mg/L}\) and \(k_{e0}=0.0800\ \mathrm{h^{-1}}\). Calculate the effect-site concentration at \(7.00\ \mathrm{h}\).
The interval is \(7.00-2.00=5.00\ \mathrm{h}\), not \(7.00\ \mathrm{h}\). Therefore:
This value lies between the starting effect-site concentration \(3.00\ \mathrm{mg/L}\) and plasma concentration \(9.00\ \mathrm{mg/L}\). Using \(7.00\ \mathrm{h}\) as the elapsed time would incorrectly treat the stated initial condition as if it had been measured at time zero.
Assumptions, Diagnostics, and Alternatives
Before applying this expression, verify that the input is genuinely constant over the interval being calculated, that the initial effect-site concentration refers to the interval's starting time, and that the time units match those used for \(k_{e0}\). For example, a rate in \(\mathrm{h^{-1}}\) must be multiplied by elapsed time in hours to produce a dimensionless exponent. Concentrations in the plasma and effect site must also be expressed in compatible units.
The exact solution is a direct calculation for this constant-input case; it avoids numerical approximation and makes the effect of the initial condition transparent. A numerical differential-equation solver is an alternative when the input or model is more complicated, but for this simple constant-input equation it should agree with the closed form up to numerical error. A disagreement here is a reason to check the time origin, sign, units, and initial condition before attributing the difference to pharmacology.
The formula is not a general solution for a plasma concentration that varies over time. In particular, inserting a single plasma value into this expression across an interval with changing plasma concentration can give a misleading result. The model's assumptions also limit interpretation: the solution describes the effect-site state under first-order equilibration, not by itself the resulting pharmacological response.
A practical diagnostic is to check three things after calculating: the value at zero elapsed time must equal the supplied initial value; the value at a later time must remain between the initial and constant plasma concentrations; and the initial direction of movement must agree with \(k_{e0}(C_p^*-C_{e,0})\). Failure of any check usually signals an arithmetic or setup error, or that the constant-input model does not describe the interval.
Check Your Understanding
Work each problem using the constant-plasma solution. Complete solutions are included below the questions.
- For \(C_p^*=6.00\ \mathrm{mg/L}\), \(C_{e,0}=2.00\ \mathrm{mg/L}\) at \(t_0=0\), and \(k_{e0}=0.100\ \mathrm{h^{-1}}\), calculate \(C_e\) at \(t=4.00\ \mathrm{h}\).
- For \(C_p^*=3.00\ \mathrm{mg/L}\), \(C_e(2.00\ \mathrm{h})=7.00\ \mathrm{mg/L}\), and \(k_{e0}=0.250\ \mathrm{h^{-1}}\), calculate \(C_e(6.00\ \mathrm{h})\) and state whether it is increasing or decreasing initially.
- A calculation gives \(C_e(t)>C_p^*\) at a later time even though \(C_{e,0}<C_p^*\), with \(k_{e0}>0\) and constant plasma concentration. Is that compatible with this model? Explain.
- If \(C_{e,0}=C_p^*\), what does the model predict for all \(t\geq t_0\)?
Solutions. (1) \(C_e(4.00)=6.00+(2.00-6.00)e^{-(0.100)(4.00)}=6.00-4.00(0.6703)=3.319\ \mathrm{mg/L}\). This lies between 2.00 and 6.00 mg/L. (2) The elapsed time is \(6.00-2.00=4.00\ \mathrm{h}\), so \(C_e(6.00)=3.00+(7.00-3.00)e^{-(0.250)(4.00)}=3.00+4.00(0.3679)=4.472\ \mathrm{mg/L}\). Initially the derivative is \((0.250)(3.00-7.00)=-1.00\ \mathrm{mg/L/h}\), so it is decreasing. (3) No. The solution is a weighted average of \(C_{e,0}\) and \(C_p^*\), with nonnegative weights summing to one, so it cannot exceed \(C_p^*\) when it starts below it. Check the sign, time units, and inputs. (4) The initial difference is zero, so \(C_e(t)=C_p^*\) throughout the interval.