Derivation of Convolution PK models

Convolution PK models are used for describing a complex absorption mechanism such as modal absorption curve like a camel with double peak. I could not find an explicit derivation how come \(C(t)=f(t)*i(t)\) becomes \(\displaystyle \frac{\partial C(t)}{\partial t}=f(t) - k C(t)\). Then, I summarized the basic concept behind it here. Let's make it clear.

Formula

The formula for the first derivative of convolution integral is: \begin{align} C(t)=f(t)*g(t)=\int_0^t f(\tau)g(t-\tau)d\tau \label{eq:con} \end{align} \begin{align} \frac{\partial C(t)}{\partial t}=\int_0^t f(\tau) \frac{\partial g(t-\tau)}{\partial t} d\tau + f(t)g(0) \label{eq:der} \end{align} Let \(C(t)\) be concentration of central compartment at time \(t\). One can consider that the function \(f(t)\) is PULSE INPUT, i.e., \(\displaystyle f(t)=\frac{d r(t)}{d t}\), where \(r(t), 0 \le r(t) \le 1\) is time-varying fraction of the dose released, and the function \(g(t)\) is fraction of remaining, i.e., \(\displaystyle g(t)=e^{-kt},0 \le g(t) \le 1, \Leftrightarrow \frac{d g(t)}{d t}=-k g(t)\). The part \(f(\tau)g(t-\tau)\) describes what amount of PULSE INPUT which occurred at time \(\tau\) will be still remaining at time \(t\). Note that \(g(0)=e^{-k \cdot 0}=1\).

Simple example for better understanding - 1

Consider which one has more impact on you at the moment, 1) Deadly accident in childhood, 2) Lost \(\$\)10 yesterday. In short, the function \(f\) and \(g\) are considered to be functions for the extent of impact and the rate of unforgetting, respectively.

Simple example for better understanding - 2

For example, suppose the function \(g(t)=1\), where no elimination is assumed, then Eq.(\ref{eq:con}) becomes \(C(t)=\int_0^t f(\tau)d\tau=r(t)-r(0)=r(t)\), since the fraction of dose at time \(0\) should be \(0\).

Conclusion

Then Eq.(\ref{eq:der}) can be \begin{align} \frac{\partial C(t)}{\partial t} &=\int_0^t f(\tau) \frac{\partial e^{-k(t-\tau)}}{\partial t} d\tau + f(t)e^{-k0} \nonumber \\ &= -k\int_0^t f(\tau) e^{-k(t-\tau)} d\tau + f(t) \nonumber \\ &= -k\int_0^t f(\tau) g(t-\tau) d\tau + f(t) \nonumber \\ &= f(t) - k C(t) \label{eq:sol} \end{align} The function \(f(t)\) corresponds to the part of \(k_a C_{abs}(t)\) in a conventional model.