ONE-COMPARTMENT KINETICS

Transcription

1 British Journal of Anaesthesia 1992; 69: REVIEW ARTICLE ONE-COMPARTMENT KINETICS J. NORMAN Pharmacokinetics may appear daunting, especially if the background mathematical development is presented too briefly and if the number of compartments or subdivisions of the model are too many. The aim of this review is to develop the basic model and concepts with a more rigorous mathematical treatment. The review is confined to considerations of a single compartment, but even such a simple model has its uses in conditions in which the distinction between "fast" and "slow" components is not all that significant. The model is built up to show what happens with bolus injections, with infusions, with combinations and also with a decreasing rate of input such as may be seen after an i.m. injection. The initial section gives a revision of the mathematics involved and the next is the development of the necessary equations to accompany the model. These are applied to the use of a drug such as morphine for the control of pain in the first 24 h after operation. These illustrations have been chosen to show what can happen with several common techniques. MATHEMATICAL REVISION The concept of a function In the model, several equations are developed which are examples of functions. A function is, simply, a rule by which an independent variable (usually symbolized by x or t) may be transformed uniquely into a dependent variable (the y value). Thus 8 y = 2x* is a function. The rule is that for a given value of x, y is given by squaring x and doubling that answer. Similarly, y is a somewhat more complex function but, again, for any x value a corresponding value for y is calculated by the rules given in the equation. A more general expression would be: y=kx) where x = the independent value and y the result of ~applying xhe_rule summarized by/(x). Other general expressions for a function may include:.., y=g(x) KEY WORDS Pharmacokinetics: single compartment model. Analgesics: morphine. All the symbols imply is that, for a set of values of a dependent variable, there is some rule symbolized by f(x) or g(x) which allows us to calculate the independent variable. Exponential functions The two equations given above use powers of x that is, both expect that (in part) x has to be multiplied by itself once or more often. With exponential functions, the difference from power functions is that the independent variable is the power. We shall be developing several such functions where a particular constant (e) is raised to the power of the independent variable, e is an irrational number the value of which is approximately and may be calculated with any desired degree of accuracy from the binomial expansion of: as n approaches infinity. For n = 1 the value is 2; for n = 10 the value is 2.594; for n = 100 it is and for n = it is The value of e may also be calculated from the expansion of the series: An exponential function may resemble a function such as: _ ffx or y = e«*> for there is no reason why we cannot use a function of x as the exponent; indeed we usually shall. The series giving e 1 is given by: x e = Differentiation With pharmacokinetics we are often interested in rates of change how quickly a blood concentration decreases after an i.v. injection or how quickly it increases during an i.v. infusion. Further, we can usually calculate how quickly a dependent variable (the concentration) changes and we shall want from that estimate to be able-to calculate what the actual value may be at any one instant. Rates of change are thus of vital interest. They J. NORMAN, M.B., PH.D., F.R.C.ANAES., Shackleton Department of Anaesthetics, The General Hospital, Southampton SO9 4XY. x 3

2 388 form the basis of differential calculus. There are several rules which we must remember in order to build up kinetic models. Rule 1. If the function is then the rate of change (symbolized by dy/dx) is: dy/dx = n.x"~ l The proof of this is developed easily and is found in any standard textbook (e.g. [5]). Rule 2. If the function is: where C = an arbitrary constant, the rate of change of y with x is uninfluenced by C and is given by: dy/dx = n.x"~ l Rule 3. If the function is: y = A.x" + C where A = another constant, the rate of change is given by: dy/dx = n.a.x"' 1 Rule 4. If the function is a sum of sub-functions such as y = A.x" + B.x m the rate of change (or derivative) is the sum of the two derivatives, that is: dy/dx = n.a.x"~ 1 + m.b. x m ~ l Rule 5. (The chain rule). Suppose there is a function of a function: thus if z = f{y) and y = g(x) where g(x) represents a second function, then the rate of change of z with respect to x is given by dz/dx = dz/dy*dy/dx Exponential differentiation The rules given above may be applied to exponential functions. The unique character of an exponential function is shown by differentiating: The series giving e* was given above: BRITISH JOURNAL OF ANAESTHESIA more complex exponential functions the other rules may be applied. For example, if then dx y = dx What happens when the exponent is a multiple of the independent variable? The chain rule applies. Suppose Y = A.e* This function may be looked at as being composed of the product of two parts. If we let then and z = tx.x Y = The two differential equations are: dz dz dx = a. Multiplying the two together gives: = ^ = * A.e dx dxdz This equation is of fundamental importance for kinetic understanding. If the coefficient a is positive, the process is of exponential growth and the rate of growth at any one time is proportional to the actual value at that time. If a is negative, the process is of exponential decay, where the rate of decay at any one time is again proportional to the actual value at that time. We shall meet many examples of such exponential decay in pharmacokinetics. Decay constants are often expressed in another form as half-lives. The relationship between half-life and the decay constant (commonly given the symbol ft) is given by: ~ ln(2) + or Using the rules given above: that is dv x x 2 = dx 1.2 TT273 xx 2 + dyjdx = e 1 Thus the rate of change of this exponential function is the same as the function itself. For other y? Differential equations With pharmacokinetic models it is often relatively easy to write an equation giving the rate of change of concentration of the drug; what is more difficult is to solve that equation so as to calculate what the concentration of drug is at any given time. Generally, in solving such equations, we need to have some additional information to give a known concentration at a particular time. (An easier example to think of may be that any given straight line is given by knowing its intercept on the y axis and the value for the slope. Many lines have the same slope and only when the intercept is known or can be calculated may we identify a specific line. The same is true for

3 ONE-COMPARTMENT KINETICS differential equations.) That information, together with the differential equation, leads to a solution of the problem. Most standard pharmacokinetic texts give the solutions for several input conditions such as a bolus injection or a constant infusion. Such solutions may be checked by differentiating the solution to ensure that the differential matches the theoretical prediction. An alternative approach is to develop a method of producing a more direct solution. With the differential equations in use here, the method involved is to use a Laplace transform (see Appendix, including table I). This mathematical device allows us to remove terms such as dy/dx or dc/dt, where C represents concentration, and replace them with others which can be manipulated algebraically. When an expression is finally derived, it may be converted back into concentration terms by using the inverse transform. Whilst developing the transform is difficult, there are "cook-book" recipes (e.g. [4]) which make the task easier. The use of both approaches is described here. THE ONE-COMPARTMENT MODEL The aim of this section is to develop the necessary differential equations and their solutions to allow us to calculate what the concentration of drug (morphine) is after a bolus injection, a constant infusion or a combination of the two, when we regard the body as composed of a single compartment. In addition, we shall develop the equations to describe what happens in a situation similar to that occurring Input 210 litre Cl 60 litre h' 1 Output - - Ar-«CI/V _ FIG. 1. The one-compartment model. The compartment is represented by the area of the box corresponding to the volume (K) with the lower part identified as the fraction of the volume cleared in unit time {Cl). The drug is input into the model as an instantaneous bolus, a constant rate i.v. infusion, or a combination of thete, or may leak into the compartment exponentially from another source (the " i.m." injection). The drug is cleared from the compartment at a rate (A) determined by the ratio of the clearance to the volume. 389 after an i.m. or s.c. injection, when all the drug appears eventually in the circulation. Figure 1 shows the basic model. The symbols used are those adopted by the British Journal of Anaesthesia in accordance with the proposal of Hull [2]. Assumptions In order to derive the equations, there are several assumptions that must be made to simplify the arithmetic. 1st assumption. Any drug within the compartment is mixed instantaneously throughout the compartment. Thus at any time, if we know the volume (V) of the compartment and the total amount or quantity of drug within the compartment (X) at a given time, t, then the concentration of drug at that time (C) is given by: C = f (1) 2nd assumption. The volume (V) remains constant during the period under study. 3rd assumption. The rate at which drug is removed from the compartment is proportional to the amount present. This frequently applies, especially to such processes in which the drug is removed by nitration or where it is metabolized with a large concentration of enzyme being present. Thus, in general, over a small interval of time (At): Xd on, = -At.k.Xd where Xd oul = amount being eliminated, k = a constant and Xd amount of drug present. The differential equation showing the rate of change of amount present is: dx ml ld t = -k Given assumption 2, that the volume remains the same, we can also write the equation in concentration terms (C): An,,. dc 0M /dt = - / Clearance. The clearance (C/) of a drug is best described formally as the rate at which drug is eliminated (dx ml /di) per unit concentration of the drug (C). Thus: Cl = (dx out /dt)/c In terms of units, the rate, of removal may be expressed in units of mass of substance per unit of time and the concentration in terms of mass of substance per unit of volume. The effect is diat clearances are expressed in units of volume per unit time. This last equation may be modified further from a knowledge of the relationship between amount of drug present and the volume. Thus: or, by rearrangement: c/ =. X V dt

4 390 Thus, if the volume (assumption 2) and the clearance remain constant, the drug is removed exponentially. The ratio of the clearance to the volume is the rate constant for elimination that is: k = Cl: V In general, we can control the input of drugs into the patient, but elimination depends on filtration or metabolism, or both, and is less easy to manipulate. If the rates of filtration or metabolism are proportional to the amounts (or concentrations) of drug present, we can write the following for the amount of drug removed over any short interval of time (At): AX^^-At.Cl.C or, in terms of concentrations: As we let the time interval become infinitesimally small, we can write the differential equation thus: dc C / In order to find what concentration exists at any time, we must solve this differential equation. Most textbooks give the solution without proof. The solution (what concentration exists at any given time, t (C)) is: C = (2) where C = concentration at the start of elimination (at time zero). The equation can be checked by differentiating it and using equation (2) in substitution : The alternative development is to use the Laplace transform. In essence, this transformation looks at the differential equation and, where we find the dependent variable (C) or its derivative, these are replaced by appropriate functions of a new variable (s). The relevant rules for making these changes are summarized in the Appendix. First, we rearrange equation (2) to: Any term involving only the dependent variable (or a constant multiplying the variable) is transformed to be a new function (g(s)). Any term involving a derivative of the dependent variable is transformed into an expression of the following form: C C Change dc/dt into s.g(s)- C where s = the Laplace operator, g(s) = the new function of that operator and C = the concentration present at the start of the process. The whole transformed equation becomes: BRITISH JOURNAL OF ANAESTHESIA This, in turn, can be rearranged: or s - This equation can now be transformed back to one giving the concentration at any given time (C). The table in the Appendix shows that the inverse transform of an expression of the form (l/(s a)), where a is a constant, is given by the expression e~"'. The new term a is given by Cl/ V when we transform back. The constant C remains unchanged. Thus we return to the solution given above: C=C 0.t-% t (2) Bolus injection into the single compartment This equation describes well what happens when all input into the model has ceased or, more readily, what happens after a bolus injection into the empty model. In the latter event, the initial concentration (C ) is given by dividing the dose injected (Xd) by the volume (V). Remember that the first assumption is that any drug is mixed instantaneously in the volume. Thus the solution for a bolus injection into the model which was previously empty becomes: (3) As an example, consider what happens when we give an i.v. dose of morphine (say 10 mg) to a patient. Assume the one-compartment model is valid and that for the average 70-kg adult the volume of distribution is 210 litre and the clearance about 60 litre h" 1. The equation giving the concentration at any time after the injection thus becomes (in units of ug litre" 1 ): 210 Figure 2 shows the graph of the resulting exponential decay. If the therapeutic plasma concentration of morphine for the control of postoperative pain is in the range ug litre" 1, then this injection should produce relief for about 2 h. Constant infusion into the single compartment With this variant, we need to consider in more detail the input and the output from the model. The output, as before, is given by: AX 0M = -At.Cl.C The input is given by multiplying the infusion rate (k') by the time interval (At): = At.k' and the net change is: AX = At(k'-Cl.C)

5 ONE-COMPARTMENT KINETICS Divide the top and bottom of the terms on the right hand side by V to simplify the result: 40 C CD M 30 I cg The inverse transform of the first term has already been given (see equation (2)). That for the second term is shown in table I in the Appendix. The inverted equation thus becomes: o u 20 This simplifies to: V.CV (4) 10 For a constant rate infusion when there is no drug present initially, the first term on the right hand side must be zero and the equation simplifies to: (5) Time (h) FIG. 2. Hypothetical example of the concentration of morphine in a one-compartment model when 10 mg is given as an initial bolus into a volume of 210 litre and with a clearance of 60 litre h" 1. In terms of changes in concentration, the equation changes to: and, in the limit as At becomes infinitely small, we can write: dc _k' Cl Rearrange this to get all the terms involving concentrations and rates of change on one side: dt The Laplace transforms for the terms on the left have been given already. The transform for the constant (&') on the right hand side is given by k'/s. The transformed equation is: This is solved for g(s) the the concentration: or g{s)(v.s+cl)= V.C + j v.c transformed function of. _.. k' V.s + Cl s(v.s + CI) If we look at this in more detail, at time zero before the infusion starts the expression e" = 1 (by definition) and the equation evaluates the concentration as being zero. At infinity, e~ OT evaluates as 0 and the equation gives the final concentration as being given by the ratio of the infusion rate (in mass units per unit time) divided by the clearance (volume divided by time). This makes sense, given the definition of clearance the rate of drug removal per unit concentration. Alternatively, if we know what steady state concentration we want, we can calculate the infusion rate: k' = (7 s. C/ (6) Infusions of morphine are now being used more commonly to reach a more steady state in terms of pain control. Figure 3 shows the pattern of drug concentration seen with this model with an infusion of 2.5 mg h" 1, given the same parameters as for figure 2. The concentration climbs almost to a plateau of approximately 40 ng litre" 1 by about 10 h. Remembering that the elimination half-life is given by multiplying the natural logarithm of 2 by the ratio of the clearance divided by the volume of distribution (about 2.43 h for this example) we can see that it takes four to five half-lives to reach this point. If analgesia were obtained only when the concentration reached 25 ug litre" 1, then that would take about 3 h. Combining a bolus dose and an infusion Equation" (4)-has other.uses for our model. The second term gives the concentration produced by the - infusion. The first term gives the decay in concentration that occurs for any drug already in the system. So far we have assumed the compartment is empty at the start. What happens if, at the start of the infusion, we also give a bolus dose (d) in an amount

6 392 BRITISH JOURNAL OF ANAESTHESIA 50 ID g 30 CD _i C g c 8 20 o i 10 8 Time (h) FIG. 3. The one-compartment model of the morphine concentration to be achieved with a constant rate infusion of 2.5 mg h" 1. The volume and clearance arc the same as in figure 1. The half-life is 2.43 h. By 10 h, the infusion is producing a reasonably steady state that is, after about four half-lives. Xd? Obviously, the initial concentration then becomes Xd divided by the volume: C = Xd/V ' and, for any subsequent time during the infusion: _ ^ d ci. k '. <?(.. This can itself be simplified to: v a a (7) (8) This equation is worth examining in more detail. The last term on the right gives the steady state concentration achieved with the infusion. The first term is composed of two parts: the initial one giving the effect of the bolus dose and its decay, and the next the decay from the infusion. Thus we can see that we are adding the two decay processes together. Giving a bolus injection at the start of an infusion is an effective way of achieving the concentration we need rapidly and then maintaining it. To get a given steady state concentration (C 53 ) we need to know the size of the bolus dose and this must be given by multiplying the volume of distribution by the desired concentration, thus: Xd = V*O* 12 Time (h) FIG. 4. The effects of combining an infusion (2.5 mg h" 1 ) with a bolus dose (8.75 mg) given at the start of the infusion. = Concentration profile for the bolus ; = concentration profile for the infusion; = concentration profile for the combination. Stopping the infusion at 16 h leads to an exponential decay, as with the bolus alone. The infusion rate, similarly (as before), is given by the product of the clearance and the concentration: k' = CUC* Substituting C for the appropriate terms in equation (7) we derive: or C= C" Figure 4 shows how this approach would work for morphine. Suppose the target concentration were 40 ig litre" 1. The infusion rate can be calculated as 2.5 mg rr 1 and the bolus dose 8.75 mg. Figure 4 shows the two components and the sum and, also, the decay to be seen when the infusion is finished after about 16 h. There is one other modification which enlarges the range of uses of equation (7). We often want to know what happens if, after giving a bolus or infusion or combination, we give another bolus or change the rate of infusion. We must calculate the residual concentration at the times when the changes are made. This becomes the concentration (C ) at the new start rime. Changing the infusion rate gives a new value for k'. Giving a bolus dose changes the value of C by adding to it the increase in

7 ONE-COMPARTMENT KINETICS is seen. In general, equation (9) allows the solution of any combination of bolus and continuous infusion. The equivalent of an "i.m." injection So far, we have assumed that any bolus dose given is mixed instantaneously throughout the compartment, but in practice we may be giving the bolus into another site (the i.m. injection) from which it is absorbed into the main compartment. The rate at which absorption occurs is greatest when the concentration is greatest and declines with time it is another example of exponential decay. We also assume that drug does not return from the main compartment to the injection site. This is realistic, given the size of the injection site a few millilitre at the most. The concentration of drug at the injection site at a given time (C ( ) is given by: where k = the rate constant for absorption. The loss from the injection site is the same as the rate of uptake into the main compartment and we can write: or: in = At.C i.k FIG. 5. A bizarre pattern that might be expected to follow first, a bolus injection of morphine 10 mg at the start, and at 4 h, a second bolus of 10 mg, given together with an infusion of 2.5 mg h" 1 for the next 12 h. concentration produced by the bolus that is a value given by Xd/V where Xd is the dose. Thus the equation becomes: where C = residual concentration at the end of the previous section. This is a general equation which can thus be used for any bolus injection, infusion or combination. It requires us to know the clearance, the volume, the initial concentration, the sizes of any bolus and constant infusion rates. It can be simplified Xd k' ~V + a a C remains as the residual concentration present at the end of the previous phase. Figure 5 shows, with this equation, what would happen if, about 4 h after morphine 10 mg, a second bolus of 10 mg was given, with an infusion of 2.5 mg h" 1. The effect of the first -bolusis-calculated^usingequation (2) and, especially, the concentration at the end of 4 h. This value is substituted into equation (9) and the new pattern calculated. The effect of the second bolus is to increase the concentration to greater than the steady state value which is approached over the next few hours. When the infusion is stopped, the usual exponential decay (9) As always, the rate of loss from the compartment is proportional to the concentration and the loss in any small interval is given by: AX = -At.C.Cl The net change is the difference: AX = At(C l 0.c- tt.k)-c.cl To obtain concentrations, we need to divide both sides by the volume (V) and the differential equation for the instantaneous rate of change of concentration becomes: dc_c.k a d?"~p^e ~ c y As before, this can be rearranged to: dc Cl _C?k dt +c v~ v k The Laplace transforms for the terms on the left of the equation have been used before. The transform for that on the right is the one the inverse transform of which we have used to develop the formula for exponential decay. Thus the transformed equation If we assume that there is initially no drug present (that is, C is nil) the equation can be rearranged to: *(*) = Cfk 1 1 ~V~7+k Cl The inverse transform for the last two components of the right hand side is more complex than usual the terms are equivalent to the expression

8 394 BRITISH JOURNAL OF ANAESTHESIA greatest when the concentration is greatest, and that occurs early with the i.v. bolus. If the two curves were to be extended to an infinite time, the two areas under the curves would be the same. If the bioavailability after the i.m. injection was less, the area under that curve would be smaller than that under the instantaneous bolus curve. Sequential "i.m." injections. If we are looking at sequential injections, we have to keep the term C for the initial concentration. The Laplace transform then becomes: k ( \ 8 = C ' y[j^k The first term on the right transforms back as before and the second is an example we have seen, which gives the exponential decay; thus: (11) In general, this equation makes sense. At the time of the initial injection the concentration must be nil (e~ is 1) so the first term disappears. If there is already drug in the compartment, then that is the FIG. 6. Comparison of the concentration curves after either an i.v. bolus ( ) of morphine 10 mg or an i.m. injection ( ) of the same dose where the rate constant for absorption from the i.m. site is 5 h" 1 (or min" 1 ). The concentration initially is greatest after the i.v. bolus, but after about 45 min that after the i.m. injection is greater. l/((s + a)(5+ >)), where a and b are two constants which are neither zero nor equal to each other. The inverse transform is then (Appendix 1): a-y and thus the basic equation is: C O z. (10) Should we be considering a process in which not all the drug is absorbed, or some is degraded before it reaches the main compartment, then we can multiply this equation by a factor, F, which expresses the bioavailability. Figure 6 compares the effect of a bolus injection of morphine into the main compartment with that of one administered as an i.m. injection. I have assumed that the i.m. injection leads to a peak value in about 30 min. For that to happen, the rate constant (k) for the absorption has been given a value of 5 h" 1. It is obvious that the peak value is smaller than with the i.v. injection, but after about 45 min the concentration after the i.m. injection is slightly greater than that after the instantaneous bolus. What is happening is that the absolute amount of drug removed is FIG. 7. Effects of multiple i.m. injections of morphine 10 mg. After an initial bolus of 10 mg and a second one after 4 h, the dashed curve shows what might be expected with repeated injections at 4-h intervals: the concentration oscillates between 30 and 50 ug litre" 1. With an "as required" pattern, the third injection is given at 12 h and the concentrations achieved are much smaller.

9 ONE-COMPARTMENT KINETICS 395 initial value. At an infinite time after the injection, the concentration must also be nil. As with equation (8), this can be used to calculate sequentially the concentrations at any time after a series of injections provided we know the size of each injection and the time intervals. Figure 7 illustrates what might be expected after the not uncommon pattern of using i.m. injections on a "as needed" basis. Suppose the first injection is given in the postoperative recovery room and a second about 4 h later. The third is given after another 8 h. The pattern is of large fluctuations in blood concentration and the likelihood of having a less than effective concentration for much of the time. A more effective schedule is also shown, with which the drug is given at 4-h intervals. The concentration oscillates throughout the therapeutic range. CAUTION The equations developed here are for the simplest kinetic model the one-compartment system. Such a simplification may be dangerous if carried too far. In practice, it is often safer to assume at least two compartments, with a central one into which drug is 70 given leaking out into a wider space. Hull [3] has given the necessary equations for examining what happens to boluses and i.v. infusions given into such models. Figure 8 compares the plasma morphine concentrations expected from the one-compartment model dealt with here and the three-compartment model based on data given by Aitkenhead and his colleagues [1]. The initial plasma concentrations are very much greater with the three-compartment model. Nevertheless, given that it takes some time for morphine to diffuse across the blood-brain barrier, the one-compartment model does give not too unrealistic a picture. The second caveat is that, although the illustrations use doses of morphine in common use in adult patients, it is vital to remember that patients differ widely, not only in their pharmacokinetics (the way they distribute and eliminate morphine) but also in their sensitivities to the effects of the drug. Each patient must be treated with due consideration of these factors. Nevertheless, the insight gained by looking in the simplest way at what can happen with differing patterns of administration of the drug does help explain some of the variability of effect. It should also help ensure that the anaesthetist does design his dosing schedule to give the effects needed with minimal toxicity and the likelihood of a rapid recovery. To understand the model, we need to understand the mathematics behind it a 40 c o I I 30 o c o O SUMMARY The mathematical development of the equations needed to determine the plasma concentrations of drug in a one-compartment pharmacokinetic model are developed from first principles and are illustrated (with caution) by the use of morphine given by bolus or continuous i.v. injection or by a simulated i.m. injection. The equations allow the calculation of concentrations, given a knowledge of what drug is currently present, what are the sizes of the boluses by either i.m. or i.v. injection and what is the constant infusion rate. What happens when changes occur is also illustrated. The development illustrates the principles needed to apply to more complex models. APPENDIX THE LAPLACE TRANSFORM The problem in looking at a biological process in which a " rate of change" occurs is that we can often write an equation showing Time(h) FIG. 8. Comparison of the morphine concentrations predicted with" the~ one-compartment -models [---) and in the central compartment of the three-compartment model ( ), apply ingdata given by Aitkenhead and colleagues [1], using the methods of Hull [3]. There is a much greater initial peak in central compartment concentration, but the decrease is, as expected, much quicker and the concentration in the compartment after about 15 min is smaller than with the one-compartment model. (Data for the three-compartment model include a volume of distribution of 227 litre and a clearance of 64 litre h" 1.) 3.5 TABLE I. The transforms and inverse transforms used in the paper, a, b and k are all constants. C" is the value of the function at the start in our terms, the concentration at the start of the process Function Fit) and inverse F(k) where * is a constant e~" where a is constant Fit) df(f)/dt Laplace transform g(s) -1/*- k/s gi')

10 396 BRITISH JOURNAL OF ANAESTHESIA what the rate of change is. Further, we may know the starting and finishing values. Thus we have a differential equation (one involving rates of change) and some constants at the start or end, or both. Such equations can be solved by means of the transform introduced by the mathematician P. S. de Laplace around Its virtue is that it changes the function from one involving differentials into one that can be solved algebraically. When a solution is derived it can be transformed back (the inverse transformation) into the equation we need. A good guide to the theory behind the transformation has been given by Weltner and colleagues [5] and there are also several applications in Simon's book [4]. The following account is necessarily brief. Let the function to be transformed be F(t). The transformed equation is: Jo where the terms on the left and right are symbols for the transformation. What is being done is that the function is multiplied by the term e~" and then integrated from zero to infinity, J, the operator, is a number which must be positive if the integral is to converge to a definite value. REFERENCES 1. Aitkenhead AR, Vater M, Achola K, Cooper CMS, Smith G. Pharmacokinetics of single-dose i.v. morphine in normal volunteers and patients with end-stage renal failure. British Journal of Anaesthesia 1984; 56: Hull CJ. Symbols for compartmental models. British Journal of Anaesthesia 1979; 51: Hull CJ. Pharmacokinetics for Anaesthesia. Oxford: Butterworth-Heinemann Ltd, Simon W. Mathematical Techniques for Biology and Medicine. New York: Dover Publications, Inc., Weltner K, Grosjean J, Schuster P, Weber WJ. Mathematics for Engineers and Scientists. Cheltenham: Stanley Thome (Publishers) Ltd, 1986.