Why Piecewise Solutions Are Useful
Plasma concentration changes over time, while the effect-site concentration responds with a delay. In calculations, plasma data are often represented by consecutive intervals, with plasma concentration held constant within each interval. The useful feature of this representation is that the first-order effect-compartment equation has an exact solution on every interval. Applying that solution successively gives the effect-site concentration at later times without numerically stepping through the differential equation.
The key is to carry the effect-site concentration at the end of one interval forward as the initial condition for the next. A change in plasma concentration changes the target toward which the effect site moves; it does not, by itself, cause an instantaneous jump in effect-site concentration.
Deriving the Interval Update
Within one interval, write \(C_p(t)=C_{p,j}\) and measure elapsed time from the interval start, so \(u=t-t_j\). The model becomes a linear differential equation with constant input: \[ \frac{dC_e}{du}+k_{e0}C_e=k_{e0}C_{p,j}. \] Subtracting the constant target \(C_{p,j}\) gives \[ \frac{d}{du}\{C_e-C_{p,j}\}=-k_{e0}\{C_e-C_{p,j}\}. \] Its solution is an exponential decay of the difference between effect-site and plasma concentration: \[ C_e(t_j+u)-C_{p,j}=\{C_e(t_j)-C_{p,j}\}e^{-k_{e0}u}. \] Setting \(u=\Delta_j\) yields the interval update in the definition. This derivation also explains why the same formula applies whether the effect site is moving upward or downward: the signed difference \(C_e-C_{p,j}\) shrinks toward zero.
For a sequence of intervals, the recursion is applied in order. The value calculated at \(t_{j+1}\) becomes the initial value in the next interval. At a boundary where plasma concentration changes, \(C_e\) remains continuous under this model: the differential equation has a finite change in its forcing term, not an instantaneous impulse to the effect compartment.
Record its start and end times, its constant \(C_{p,j}\), and its duration \(\Delta_j\).
Take \(C_e(t_j)\) from the initial condition or the preceding interval’s unrounded result.
Calculate \(C_{p,j}+\{C_e(t_j)-C_{p,j}\}e^{-k_{e0}\Delta_j}\).
Confirm that the result lies between the interval’s starting effect-site concentration and \(C_{p,j}\), then use it in the next interval.
Worked Recursive Calculations
Worked Example: Two Successive Plasma Levels
Hypothetical data. Suppose \(C_e(0)=0\) mg/L and \(k_{e0}=0.200\) h\(^{-1}\). Plasma concentration is 10 mg/L from 0 to 2 h, then 4 mg/L from 2 to 5 h. Calculate \(C_e\) at 2 h and 5 h.
For the first interval, \(\Delta_1=2\) h and \(C_{p,1}=10\) mg/L. The update is \[ C_e(2)=10+(0-10)e^{-(0.200)(2)} =10\{1-e^{-0.4}\}. \] Using \(e^{-0.4}=0.670320046\), \[ C_e(2)=10(1-0.670320046)=3.29679954\ \text{mg/L}, \] or 3.297 mg/L to four significant figures.
For the second interval, the initial value is the unrounded \(C_e(2)\), the new target is 4 mg/L, and \(\Delta_2=3\) h: \[ C_e(5)=4+(3.29679954-4)e^{-(0.200)(3)}. \] Since \(e^{-0.6}=0.548811636\), \[ C_e(5)=4+(-0.70320046)(0.548811636) =3.6140754\ \text{mg/L}. \] Thus \(C_e(5)=3.614\) mg/L. The effect-site concentration is still below 4 mg/L at 5 h because it is approaching the second interval’s target from below.
Worked Example: A Declining Input After a High Level
Hypothetical data. Let \(C_e(0)=2.00\) mg/L and \(k_{e0}=0.500\) h\(^{-1}\). Suppose plasma concentration is 8.00 mg/L for 1 h, then falls to 0 mg/L for the next 2 h. Find \(C_e\) at 1 h and 3 h.
On the first interval, \[ C_e(1)=8+(2-8)e^{-0.5} =8-6(0.606530660) =4.36081604\ \text{mg/L}. \] The value is between the initial concentration of 2 mg/L and the plasma target of 8 mg/L, as expected.
On the second interval, \(C_{p,2}=0\) mg/L and \(\Delta_2=2\) h. The preceding interval’s result is the initial condition: \[ C_e(3)=0+(4.36081604-0)e^{-(0.500)(2)} =4.36081604(0.367879441) =1.60425457\ \text{mg/L}. \] Therefore, \(C_e(3)=1.604\) mg/L. It remains positive and has declined toward zero without reaching it in finite time.
Worked Example: Several Intervals With Convenient Exponential Factors
Hypothetical data. Let \(C_e(0)=0\) mg/L and \(k_{e0}=\ln(2)\) h\(^{-1}\). Plasma concentration is 8 mg/L for 1 h, 2 mg/L for the next 2 h, and 6 mg/L for the following 1 h. Calculate the effect-site concentration at each interval boundary.
Because \(e^{-\ln(2)\cdot 1}=1/2\), the first interval gives \[ C_e(1)=8+(0-8)\left(\frac{1}{2}\right)=4\ \text{mg/L}. \] For the second interval, the two-hour decay factor is \(e^{-2\ln(2)}=1/4\): \[ C_e(3)=2+(4-2)\left(\frac{1}{4}\right) =2.5\ \text{mg/L}. \] For the third interval, with one hour at a target of 6 mg/L, \[ C_e(4)=6+(2.5-6)\left(\frac{1}{2}\right) =4.25\ \text{mg/L}. \] The boundary values are therefore 4.00 mg/L at 1 h, 2.50 mg/L at 3 h, and 4.25 mg/L at 4 h. Each update begins with the previous boundary value; the plasma changes do not reset \(C_e\).
Assumptions, Checks, and Failure Modes
The exactness of the update is conditional on the model and the interval representation. It assumes \(k_{e0}\) is constant over the intervals, the first-order equation describes equilibration, and plasma concentration is constant within each stated interval. The initial \(C_e\) must also be specified. If plasma observations are used to define the intervals, treating them as constant between observations is a modeling approximation, not a fact established by the recursion.
Several checks follow directly from the solution. For \(k_{e0}>0\) and \(\Delta_j\geq 0\), the exponential factor \(r_j=e^{-k_{e0}\Delta_j}\) satisfies \(0<r_j\leq1\). The endpoint can be written as \[ C_e(t_{j+1})=r_jC_e(t_j)+(1-r_j)C_{p,j}. \] It is therefore a weighted average of the starting effect-site concentration and the interval’s plasma concentration. It cannot overshoot either value during that interval. If \(C_e(t_j)<C_{p,j}\), the endpoint must be at least \(C_e(t_j)\) and no greater than \(C_{p,j}\); if \(C_e(t_j)>C_{p,j}\), it must lie between \(C_{p,j}\) and \(C_e(t_j)\). These checks are useful for finding sign errors, wrong interval lengths, and mistaken exponential factors.
Units provide another check. The product \(k_{e0}\Delta_j\) must be dimensionless, so a rate in h\(^{-1}\) requires time in hours. Concentrations on both sides of the update must use the same units. Keep full precision during the recursion and round only reported results. Rounding an intermediate boundary value can accumulate error when many intervals are chained.
- Using the new plasma value as the initial effect-site value. A plasma jump changes the target, not the effect-site state. Use the prior interval’s endpoint as \(C_e(t_j)\).
- Reversing the difference. The correct term is \(C_e(t_j)-C_{p,j}\). With a positive input and \(C_e\) initially below it, this term is negative, so the update moves upward toward the target.
- Mixing time units. A duration of 30 minutes is 0.5 h, not 30 h, when \(k_{e0}\) is specified per hour.
- Assuming the formula makes the input representation exact. The interval solution is exact for constant \(C_p\) within each interval; it does not remove errors introduced by approximating a varying plasma profile as flat.
How This Approach Relates to Alternatives
For piecewise-constant plasma concentrations, recursive exact updates are preferable to a numerical time-stepping approximation when only interval endpoints or within-interval predictions are required. A numerical integrator approximates the differential equation using smaller steps; its result can depend on step size and solver tolerances. The closed-form interval update avoids that integration error under the stated constant-input assumption.
The same first-order model can also be written as a convolution of past plasma concentrations with an exponential weighting function. That representation emphasizes the contribution of the entire prior input history. Recursion is the convenient computational form when the input is supplied as successive constant intervals: it summarizes the relevant history in the current \(C_e\), rather than requiring the full integral to be recalculated at each boundary.
Replacing changing plasma concentration by a single average over a long interval is a different approximation. In general, it need not produce the same endpoint as applying the exact updates to the separate intervals, because earlier input levels have more time to influence the effect site than later levels. For example, each interval’s contribution is weighted through its duration and the subsequent exponential decay. Preserve the actual interval sequence when those changes matter for the intended prediction.
The method is limited by the first-order effect-compartment model and by how well the piecewise input represents plasma concentration. It does not establish that the model is biologically adequate, determine an appropriate \(k_{e0}\), or describe the concentration-response relationship. If plasma concentration varies substantially within an interval, use a finer representation or an appropriate continuous input and solve the model accordingly. The recursion’s arithmetic can be exact while its input approximation remains poor.
Check Your Understanding
Answer each question, then compare with the worked solutions below.
- With \(C_e(0)=1\) mg/L, \(C_p=6\) mg/L for 2 h, and \(k_{e0}=0.300\) h\(^{-1}\), calculate \(C_e(2)\).
- If \(C_e=4\) mg/L and plasma concentration is 0 mg/L for 1.5 h, with \(k_{e0}=\ln(2)\) h\(^{-1}\), calculate the endpoint.
- Starting at \(C_e(0)=3\) mg/L, plasma concentration is 7 mg/L for 1 h and then 1 mg/L for 1 h. Let \(k_{e0}=\ln(2)\) h\(^{-1}\). Calculate \(C_e\) at both boundaries.
- Why is it generally not valid to replace several distinct plasma intervals by their unweighted average before applying the update?
Solutions. (1) Apply the constant-input solution: \[ C_e(2)=6+(1-6)e^{-(0.300)(2)} =6-5e^{-0.6} =6-5(0.548811636) =3.25594182\ \text{mg/L}. \] Thus \(C_e(2)=3.256\) mg/L, which lies between 1 and 6 mg/L. (2) The decay factor is \(e^{-(\ln 2)(1.5)}=2^{-1.5}=0.353553391\), so \(C_e=0+4(0.353553391)=1.41421356\) mg/L, or 1.414 mg/L. (3) The first interval gives \(C_e(1)=7+(3-7)(1/2)=5\) mg/L. The second gives \(C_e(2)=1+(5-1)(1/2)=3\) mg/L. (4) The plasma levels occur at different times, and the effect-site response weights earlier and later inputs differently through exponential equilibration. Applying the interval updates preserves that order and timing; a single average generally does not.