
1 The absorption process

1.1 Introduction

Let $$R_a(t)$$ be the rate of absorption at time $$t$$, i.e.??the rate at which the depot compartment empties and the central compartment fills up. Assuming no elimination, the system reduces to \begin{aligned} \dot{A_d}(t) & = -R_a(t) \\\ \dot{A_c}(t) & = R_a(t) \end{aligned} where $$A_d(t)$$ and $$A_c(t)$$ are, respectively, the amounts in the depot and central compartments.

If the amounts are expressed in grams and the time in hours, then, the rate of absorption is expressed in gram per hour (g/h).

Let us see some examples of absorption processes. We will consider in these examples a single dose $$D$$ administered at time $$t_D$$. The initial conditions of the system are therefore \begin{aligned} A_d(t_D) & = D \\\ A_c(t_D) & = 0 \end{aligned} Furthermore, $$A_c(t)+A_d(t)=D$$ for ant $$t\geq t_D$$ since there is no elimination here.

We will use $$D=1$$ and $$t_D=0$$ for the numerical examples.

1.2 First order absorption

The rate $$R_a(t)$$ is proportional to the amount $$A_d(t)$$: $R_a(t) = k_a \, A_d(t)$

The coefficient of proportionality $$k_a$$ is called absorption rate constant.

There exists an analytical solution for $$A_d(t)$$: $A_d(t) = D e^{-k_a \, t}$

1.3 Zero order absorption

The rate is a constant $$R_0$$: $R_a(t) = R_0 \, {\mathbf 1}_{A_d(t)>0}$ Then, the duration of absorption is $$Tk0 = 1/R_0$$.

There exists an analytical solution for $$A_d(t)$$: \begin{align} A_d(t) &= D (1 - R_0 t) & & \text{for } 0\leq t \leq Tk0 \\\ &= 0 & & \text{for } t \geq 0 \end{align}

1.4 alpha order absorption

The rate is proportional to a power of $$A_d$$: $R_a(t) = r \, (A_d(t))^\alpha \quad \quad (\alpha>0)$ First-order and zero-order absorption processes correspond, respectively, to $$\alpha=1$$ and $$\alpha=0$$.

When $$0 <\alpha <1$$, \begin{align} A_d(t) &= (D^{1/a}-b\,t)^a & & \text{for } 0\leq t \leq {D^{1/a}}/{b} \\\ &= 0 & & \text{for } t \geq {D^{1/a}}/{b} \end{align} where $$a=1/(1-\alpha)$$ and $$b=(1-\alpha)r$$.

When $$\alpha >1$$, $A_d(t) = D \left(1 - (1+b\, t)^a \right)$ where $$a=1/(1-\alpha)$$ and $$b=r(\alpha-1)D^{\alpha-1}$$.

1.5 Saturated absorption

The rate of absorption cannot exceed a given value $$V_m$$: $R_a(t) = \frac{V_m \, A_d(t)}{K_m + A_d(t)}$

2 The distribution process

2.1 Two compartments model

\begin{aligned} A_c(0) &= 1 \\\ A_p(0) &= 0 \\\ \dot{A_c}(t) &= -k_{12}A_c(t) + k_{21}A_p(t) \\\ \dot{A_p}(t) &= k_{12}A_c(t) - k_{21}A_p(t) \end{aligned}

2.2 Three compartments model

\begin{aligned} A_c(0) &= 1 \\\ A_p(0) &= 0 \\\ A_q(0) &= 0 \\\ \dot{A_c}(t) &= -k_{12}A_c(t) + k_{21}A_p(t) -k_{13}A_c(t) + k_{31}A_q(t) \\\ \dot{A_p}(t) &= k_{12}A_c(t) - k_{21}A_p(t) \\\ \dot{A_q}(t) &= k_{13}A_c(t) - k_{31}A_q(t) \end{aligned}

3 The elimination process

3.1 linear (or first order) elimination

\begin{aligned} A_c(0) &= D \\\ \dot{A_c}(t) &= -k \, A_c(t) \end{aligned}

3.2 Michaelis Menten (or saturated) elimination

\begin{aligned} A_c(0) &= D \\\ \dot{A_c}(t) &= - \frac{V_m \, A_c(t)}{V\, K_m + A_c(t)} \end{aligned}