Connecting Dissolution to Absorption Through Convolution

From Dissolution to a Concentration-Time Profile

A dissolution curve describes how much drug has left a dosage form under specified in vitro conditions. A plasma concentration curve describes the combined consequences of drug release, absorption, and disposition in a person. Connecting these curves requires more than matching their shapes: one must specify how in vitro time corresponds to in vivo release time and how released drug becomes a concentration over time.

The mathematical link is convolution. It combines the rate at which drug becomes available with the concentration-time response to a unit of input. This makes it a forward prediction: given a release profile, a release-time mapping, and an absorption-disposition model, calculate the resulting plasma concentration profile. It does not establish that the chosen mapping or model is biologically correct.

Definition: Let \(D_{\mathrm{vitro}}(\tau)\) be the cumulative fraction of the dose dissolved by in vitro time \(\tau\). Let \(g(\tau)\) map an in vitro release time \(\tau\) to the corresponding in vivo time \(t=g(\tau)\). If \(g\) is increasing and differentiable, the mapped cumulative in vivo release fraction is \(R(t)=D_{\mathrm{vitro}}(g^{-1}(t))\), and the in vivo release rate is \(r(t)=dR(t)/dt\).

The release rate \(r(t)\) has units of fraction of dose per unit time. It is not itself an absorption rate: released drug may remain in the gastrointestinal tract before absorption, and some may never reach the systemic circulation. Those processes require their own representation.

Mapping In Vitro Time to In Vivo Release

The time mapping is a model component, not a consequence of the dissolution test. For example, if \(g(\tau)=a\tau\), where \(a>0\), then in vivo time is \(a\) times the corresponding in vitro time. The mapped cumulative fraction and release rate are

R(t)=D_{\mathrm{vitro}}\left(\frac{t}{a}\right), \qquad r(t)=\frac{1}{a}D'_{\mathrm{vitro}}\left(\frac{t}{a}\right).

The factor \(1/a\) matters: changing the time scale changes the rate as well as the time at which a given fraction is released. More generally, differentiating the mapped cumulative profile gives

r(t)= \frac{D'_{\mathrm{vitro}}(g^{-1}(t))} {g'(g^{-1}(t))}.

This expression assumes the mapping is differentiable and strictly increasing over the time range of interest. A plateau in dissolution produces a zero release rate over the corresponding interval. If only a fraction of the dose dissolves, the mapped profile also remains below one; the model should not silently treat unreleased drug as available.

Worked Example: Applying a Linear Release-Time Mapping

Suppose a hypothetical in vitro profile is \(D_{\mathrm{vitro}}(\tau)=1-e^{-0.5\tau}\), where \(\tau\) is in hours, and the proposed mapping is \(g(\tau)=2\tau\). Calculate the mapped fraction released and release rate at in vivo time \(t=4\) hours.

The inverse mapping is \(g^{-1}(t)=t/2\). Therefore,

R(t)=1-e^{-0.5(t/2)}=1-e^{-0.25t}, \qquad r(t)=0.25e^{-0.25t}\ \text{h}^{-1}.

At \(t=4\) hours,

R(4)=1-e^{-1}=0.6321, \qquad r(4)=0.25e^{-1}=0.0920\ \text{h}^{-1}.

Under this hypothetical mapping, 63.21% of the dose has been released by 4 hours, and the instantaneous release rate then is 0.0920 dose fractions per hour. These calculations follow from the proposed mapping; they do not demonstrate that the mapping predicts an actual in vivo profile.

Deriving the Convolution

First consider a linear one-compartment disposition model with first-order elimination. A unit amount entering the systemic circulation at time zero produces concentration \(e^{-k_e u}/V\) at elapsed time \(u\), where \(V\) is the apparent volume of distribution and \(k_e\) is the elimination rate constant. Thus the disposition response per unit systemic input is

h(u)=\frac{1}{V}e^{-k_eu}, \qquad u\geq 0.

If systemic input occurs over time at rate \(q(s)\) in amount per time, an amount \(q(s)\,ds\) entering during a short interval around \(s\) contributes \(q(s)\,ds\,h(t-s)\) to concentration at time \(t\). Adding contributions from all input times gives

C(t)=\int_0^t q(s)h(t-s)\,ds.

This integral is the convolution of the input rate and the disposition response. It follows from superposition: each input increment produces its own concentration response, and the total response is their sum. Superposition requires a linear, time-invariant disposition model; a changing clearance or concentration-dependent elimination generally violates that condition.

For oral drug, release into the gastrointestinal tract need not be immediate systemic input. Consider a first-order absorption compartment. Let \(D_{\mathrm{dose}}\) be the dose, \(F_a\) the fraction of released drug eventually reaching the systemic circulation, \(k_a\) the absorption rate constant, and \(A_g(t)\) the amount in the absorption compartment. With \(r(t)\) as the fraction release rate,

\frac{dA_g(t)}{dt}=D_{\mathrm{dose}}r(t)-k_aA_g(t), \qquad \frac{dC(t)}{dt}=\frac{F_a k_a A_g(t)}{V}-k_eC(t).

For a unit amount released at time zero, the fraction remaining in the absorption compartment after elapsed time \(v\) is \(e^{-k_av}\), so the systemic absorption rate is proportional to \(k_ae^{-k_av}\). Combining that absorption process with the one-compartment disposition response gives the concentration response per unit amount released:

h_{\mathrm{oral}}(u) = \frac{F_a}{V} \frac{k_a}{k_a-k_e} \left(e^{-k_eu}-e^{-k_au}\right), \qquad u\geq 0,\quad k_a\ne k_e.

A release increment at time \(s\) therefore contributes according to its age \(u=t-s\). Integrating those contributions gives the oral convolution:

C(t)=D_{\mathrm{dose}}\int_0^t r(s)h_{\mathrm{oral}}(t-s)\,ds.

If \(k_a=k_e\), the expression for \(h_{\mathrm{oral}}\) is evaluated by its limit, \(h_{\mathrm{oral}}(u)=(F_a/V)k_au e^{-k_au}\). If absorption is effectively instantaneous, the model instead reduces to direct systemic input at the release rate, with the appropriate systemic availability factor. These are different assumptions about absorption, not different definitions of convolution.

Key result: A convolution-based concentration prediction requires both a mapped in vivo release rate and a specified response to each release increment. In the first-order absorption and one-compartment example, that response is \(h_{\mathrm{oral}}\); in another model, the response function must match that model’s absorption and disposition assumptions.

Worked Concentration Predictions

Worked Example: Immediate Uptake With First-Order Elimination

For a hypothetical formulation, suppose \(D_{\mathrm{dose}}=100\) mg, \(F_a=0.80\), \(V=20\) L, and \(k_e=0.20\ \text{h}^{-1}\). Assume mapped release is constant at \(r(t)=0.10\ \text{h}^{-1}\) from 0 to 5 hours and zero thereafter. For this example only, assume released drug enters the systemic circulation immediately. Calculate \(C(3)\) and \(C(8)\).

The assumptions are a constant fraction released per hour over the stated interval, complete specification of release after 5 hours, and linear one-compartment disposition with constant \(V\) and \(k_e\). The released fraction by 5 hours is \(0.10\times5=0.50\), so the profile is internally consistent with a total released fraction of one-half.

Immediate uptake gives \(q(s)=D_{\mathrm{dose}}F_a r(s)\). Substituting into the disposition convolution,

C(t)=\frac{D_{\mathrm{dose}}F_a}{V} \int_0^{\min(t,5)}0.10e^{-0.20(t-s)}\,ds = 4(0.10)\frac{1-e^{-0.20\min(t,5)}}{0.20} \quad (t\leq5),

For \(t=3\) hours, this gives

C(3)=4(0.10)\frac{1-e^{-0.20(3)}}{0.20} =2(1-e^{-0.6}) =0.9024\ \text{mg/L}.

At 8 hours, release ended at 5 hours, so the integral must include only input from 0 to 5 hours:

C(8)=4(0.10)\int_0^5e^{-0.20(8-s)}\,ds =2\left(e^{-0.6}-e^{-1.6}\right) =0.6938\ \text{mg/L}.

The later concentration is lower because elimination continues after release ends. The calculation does not include a gastrointestinal absorption delay; adding one would change the response function and the predicted values.

Worked Example: Including First-Order Absorption

Use a hypothetical dose of 100 mg with \(F_a=0.80\), \(V=20\) L, \(k_a=1.0\ \text{h}^{-1}\), and \(k_e=0.20\ \text{h}^{-1}\). Suppose mapped release is \(r(t)=0.25\ \text{h}^{-1}\) from 0 to 2 hours, then zero. Calculate the concentration at 2 hours using the first-order absorption kernel.

The rate integrates to \(0.25\times2=0.50\) of the dose, so the profile releases one-half of the dose. The rate constants are positive and unequal, as required for the displayed kernel. The parameter values and constant release interval define the hypothetical model completely.

At \(t=2\), substitute the kernel into the convolution. With \(u=2-s\), the integration limits remain 0 to 2:

C(2) =100(0.25)\frac{0.80}{20} \int_0^2 \frac{1.0}{1.0-0.20} \left(e^{-0.20u}-e^{-1.0u}\right)du.

Evaluating the integral,

\int_0^2\frac{1}{0.8} \left(e^{-0.20u}-e^{-1.0u}\right)du = \frac{1}{0.8} \left[ \frac{1-e^{-0.4}}{0.20}-(1-e^{-2}) \right] =0.9797\ \text{h}.

Therefore,

C(2)=100(0.25)\frac{0.80}{20}(0.9797) =0.9797\ \text{mg/L}.

The units are concentration: dose times fraction released per hour times availability divided by volume, multiplied by an integrated response in hours. A model assuming immediate systemic uptake would predict \(1.648\ \text{mg/L}\) at 2 hours for these release and disposition parameters. The difference illustrates why the assumed absorption process must be stated rather than hidden inside the dissolution-to-release mapping.

Assumptions, Diagnostics, and Alternatives

A convolution can be computed exactly and still give a poor prediction if its inputs are not suitable. The principal assumptions for using it predictively are:

Meaningful dissolution input. The in vitro test must measure a release process relevant to the formulation and conditions of use. A dissolution profile is not automatically an in vivo release profile.
Valid time mapping. The mapping must be monotone over the modeled range and appropriate to the formulation, dose, and population to which predictions will be applied. It must not create an impossible release rate or cumulative fraction.
Specified absorption and availability. The absorption model and \(F_a\) must be appropriate; the model must account for any lag, incomplete absorption, or other process that materially changes systemic input.
Linear, stable disposition. The response to each input increment must be additive and must not change with time or input amount. The simple kernel also assumes one-compartment disposition and first-order elimination.
Compatible conditions. Dose, food conditions, and other factors affecting release or absorption should be sufficiently comparable for the proposed prediction. A mapping estimated under one condition cannot be presumed valid under another.

Diagnostics begin with the inputs. Check that \(D_{\mathrm{vitro}}(\tau)\) is nondecreasing and bounded between zero and one, and that the mapped \(R(t)\) preserves those properties. Check that the release rate is nonnegative and integrates to the mapped released fraction. Inspect whether the predicted concentration curve has plausible timing and units, and whether residual discrepancies against observed concentrations show systematic features such as a consistent peak-time shift or a late-time mismatch. Such patterns can indicate a poor time mapping, an inadequate absorption model, or disposition that is not represented by the chosen kernel. They do not identify a unique cause on their own.

Convolution is a forward calculation, whereas direct empirical comparisons summarize observed concentration-time data without requiring this mechanistic chain. A compartmental model can represent absorption and disposition explicitly, but still needs a defensible input function. Convolution offers a transparent way to combine a proposed input with a specified response; it cannot rescue an implausible input, prove a causal mechanism, or establish predictive performance merely because the arithmetic is correct.

Approach	What it provides	Main limitation for this question
Convolution prediction	Concentration-time profile implied by a release rate and response function	Depends on the release mapping and the assumed absorption-disposition response
Direct empirical comparison	Comparison of observed concentration summaries or profiles	Does not by itself translate an in vitro release profile into a new concentration prediction
Compartmental input model	Explicit representation of absorption and disposition processes	Requires a specified or estimated input and suitable structural assumptions

Common Pitfall and Practical Interpretation

Wrong reasoning: “The in vitro curve reaches 50% dissolution at 4 hours, so the plasma concentration at 4 hours must reflect 50% absorption.” The first quantity is a cumulative fraction dissolved at an in vitro time. It is not necessarily the cumulative in vivo release fraction at that clock time, and neither quantity directly equals the fraction absorbed or the plasma concentration.

Correction: Specify the in vitro-to-in vivo time mapping, differentiate the mapped cumulative profile to obtain a release rate, and then apply an absorption-disposition response through convolution. Keep the quantities distinct: release is not absorption, and concentration depends on the history of systemic input and elimination. A plausible prediction also requires evidence that the mapping and response assumptions are suitable for the intended use.

The convolution framework is valuable because it makes each link explicit. It also exposes where uncertainty or model failure can enter: the dissolution measurement, time mapping, absorption assumptions, and disposition response. Prediction should therefore be judged against relevant concentration data, not by the mathematical form alone.

Check Your Understanding

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

If \(D_{\mathrm{vitro}}(\tau)=1-e^{-0.4\tau}\) and \(g(\tau)=2\tau\), what is the mapped release rate at \(t=2\) hours?
Why does the convolution integrate over earlier release times rather than use only the release rate at the observation time?
A formulation releases at \(0.20\ \text{h}^{-1}\) for 3 hours and then stops. What fraction of the dose has been released by 3 hours, and what is the release rate after 3 hours?
Name two assumptions that would be violated if elimination became concentration-dependent.

Solutions. (1) The mapped cumulative fraction is \(R(t)=1-e^{-0.2t}\), so \(r(t)=0.2e^{-0.2t}\ \text{h}^{-1}\). At 2 hours, \(r(2)=0.2e^{-0.4}=0.1341\ \text{h}^{-1}\). (2) Drug released at each earlier time contributes to the current concentration according to the response since its release; summing those contributions gives the integral. The instantaneous rate alone omits the input history. (3) The released fraction is \(0.20\times3=0.60\). After 3 hours, the rate is zero under the stated profile. (4) Concentration-dependent elimination violates linearity, so superposition fails; it also means the response to an input increment is not a fixed time-invariant kernel independent of the rest of the concentration profile.