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