For those who care, a first try at a complex problem; for those who don't, an
exercise with the delete key:

A Very Simple Model of Drug Build-up in the Bloodstream.

A note on notation:  E-mail is notoriously indifferent to formatting.
Below:  t1, t2, T1, T2, u1, u2, v1, v2, k1, k2, etc  are subscripted constants
or variables as the case may be.
        D is capital delta

The graphs mentioned at the end are available to any one able to receive them as
files.  Just ask.



        1. Some simplifying assumptions.  The amount of an ingested drug in the
bloodstream depends on the rate at which the drug is digested (passes through
the intestinal wall into the bloodstream) and the rate at which the drug
spontaneously disappears from the bloodstream either by loss through the walls
of the blood vessels or by chemical reaction.  The amount of the drug actually
removed by the brain for its own use is probably negligible.
        All drug/bloodstream solutions are very dilute.  Blood circulation is so
fast that drug transport times are negligible in comparison with the times
associated with the diffusion and loss mechanisms.  The bloodstream is
sufficiently turbulent that drug concentration is essentially constant
throughout the circulatory system.

        2.  The digestion process is taken to be mathematically equivalent to
that of diffusion through a thin wall or membrane.  The rate of diffusion is
directly proportional to the difference in concentration across the thin wall.
        Let the concentration of the drug be u(t).  Then Du(t)  =  u(t)intestine
- u(t)bloodstream is the concentration gradient across the intestinal wall.
Then the transfer rate is du/dt = k1Du(t).  Further, assume that the drug is so
dilute in the bloodstream that  u(t)bloodstream   0.  Then
        du/dt  =  k1[uo - u(t)] (1)

where uo is the initial concentration in the intestine.  This readily integrates
to
- ln(uo - u) = k1t + C.   Substituting the initial condition, u(0) = 0, gives  C
= - ln(uo).  Substituting and rearranging yields
        u(t)  =  uo[1- exp(-k1t)]       (2)

        3.  The loss process is taken to be mathematically equivalent to that of
radio-active decay.  The rate of decay is directly proportional to the amount of
material remaining.
        Let the amount of the drug in the bloodstream be v(t).  Then the decay
rate is
        dv/dt = - k2v(t).       (3)

This readily integrates to   ln(v) = - k2t + C.  The initial condition is v(0) =
vo and hence C = ln(vo).  Finally
        v(t) = vo exp(- k2t)    (4)

        4.  Combining the two processes.  In equation (4), vo is the original
amount of the drug in the bloodstream. This is not constant, however, but
depends from moment to moment on the amount supplied by digestion. This is given
by equation (2). Substituting u(t)  =  vo  gives
        v(t) =  uo[1- exp(-k1t)] [exp(- k2t)]   (5)
the desired result.
        The form of this function is correct, but care needs to be taken with
units.  In the derivation of (2), u(t) was the drug concentration, perhaps in
[mg/liter].  In (4), however,  v(t) was the amount of drug in the bloodstream
[mg].  Further, the digestion process is certainly not 100% efficient; some,
perhaps most, of the drug may pass through the intestine undigested.  All these
complications can be avoided by simply treating the time, t, and the amount of
the drug as a function of time, v(t), as being given in arbitrary units.

        5.  The constants k1 and k2 have units of inverse time.  They can be
written as
        k1  =  1/t1  and k2  =  1/t2  .

        v(t) =  uo[1- exp(-t/t1] [exp(- t/t2]   (5)

        The e-based time constants, t1 and t2, are easily converted into the more
familiar half-lives,    T(1/2)  = t ln(2).  Since T and t differ only by a
constant multiplier of ln(2):  T1/T2  = t1/t2.  In what follows only the ratio
of half-lives is important, and for that purpose the t  are equivalent to the T.

        6.  Absorption into (Capture by) the brain occurs by the same diffusion
governed process as digestion.  Therefore

        w(t)  =  wo[1- exp(-k3t)]       (6)

and as before   v(t) =  wo

so that w(t) =  uo[1- exp(-t/t1] [exp(- t/t2] [1- exp(-k3t)]    (7)

Assume that absorption by the brain is a much faster process than either
digestion or loss. Then t3 is much smaller than either t1 or t2. so that [1-
exp(-k3t)]   1, and w(t) is nearly the same as v(t) , except perhaps for a
constant factor:   w(t)  Kv(t).

        A single dose.

        Figure 1 shows the amount of the drug entering the bloodstream, and the
amount of the drug remaining over time for two different values of the loss
half-life.
        The solid line [equation (2) with t1 = 5 time units] shows the amount of
the drug in the bloodstream as a function of time for a lossless bloodstream.
Note that u(t) = 1 represents the maximum amount of the drug which will pass
through the intestine wall.  This is only a part of the quantity of drug
ingested.
        The line  - - - - - -  shows the quantity of the drug in the bloodstream
for t2 = 50 time units, and the line - - - - - - -  for t2 = 500 time units.
Note that decay loss begins immediately and thus prevents blood levels from ever
reaching uo = 1.

        Repeated doses.

        Figures 2 and 3 show the effect of regularly repeated doses for two
different situations.  In Figure 2 the decay rate half-life is ten times longer
than the digestion half-life.  In Figure 3 they are equal. Both figures also
show the effect of taking a half-dose twice as often.

********
George M. Andes
63/14 and still counting