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ISSN 2581-3463
REVIEW ARTICLE
A Mathematical Review to Estimate the Fate of a Drug Following Open Two
Compartment Model, When it is Administered by Intravenous Route
S. Dey
Department of Pharmacy, University of Rajshahi, Rajshahi, Bangladesh
Received: 15-03-2020; Revised: 01-05-2020; Accepted: 01-07-2020
ABSTRACT
This paper deals with the application of Laplace Transform [L(f)(s) = F(s) = e-ts
f(t)dt] to derive the
formula [(Cp
(t)= Ae-at
+ Be-bt
] and [Ct
(t)= Te-bt
-Te-at
] to estimate the fate of a drug in the mammillary body
with the prediction of the biological parameters [k, k12
, and k21
]. The numerical values of [k, k12
, and k21
]
which is estimated through the determination of the mathematical hybrid constants (A, B, a, b, T) also
predict the tissue concentration of the drug, biological half-life as well.
Key words: Compartment model, Laplace transform and inverse transform, pharmacokinetics, rate constant
INTRODUCTION
When a drug is injected through intravenous
route, initially, all the drugs present in the
plasma compartment (central compartment).
As time increases, the concentration of drug
in central compartment declines due to the
transport of the drug to peripheral compartment
(tissue compartment), elimination of the drug
from central compartment as well. Hence, in the
central compartment, there are two phases, drug
distribution phase as well as drug decline phase.
In tissue compartment (peripheral compartment),
the drug concentration increases and finally forms
an equilibrium condition with central compartment.
Hence, at any time, some fraction of drug remains
in the central compartment, in tissue compartment
as well. The drug elimination occurs from the
central compartment, that is, some fraction of
plasma drug is excreted. Moreover, some fraction
of drug is metabolized in the central compartment.
THEORY AND DISCUSSION
Methodology, when a drug is administered by
intravenous route as a bolus dose, the injected drug
distributes through central compartment as well
Address for correspondence:
Sunandan Dey
E-mail: sunandanpharm@gmail.com
as peripheral compartment. The drug elimination
occurs from the central compartment.
Laplace transform
The Laplace transform F=F(s) of a function f=f(t) is
defined by , L(f)(s) = F(s) = e-ts
f(t) dt. The integral
is evaluated with respect to t, hence, once the limits
are substituted, what is left are in terms of s.
Inverse Transform
Let f be a function and L(f)= F be its Laplace
transform. Then, by definition, f is the inverse
transform of F. This is denoted by L-1
(F)=f. As for
example, from inverse Laplace transform. We get
that L(e-kt
)= 1/(s+k), L(1) = 1/s written in the inverse
transform notation L-1
(1/(s+k))= e-kt
; L(1/s) = 1.
Transform of derivatives
If a function y=y(t). The transform of its derivative
y’ can be expressed in terms of the Laplace
transform of y: L(y’)= sL(y)–y(0)
Proof
In my proposed mathematical analysis, I
consideredboththeplasmaandtissuecompartment
elimination rate constant,
Dey: Open two compartment model
AJMS/Jul-Sep-2020/Vol 4/Issue 3 2
Let a drug be administered as intravenous bolus dose
(C0
p
), the drug concentration in plasma compartment
and tissue compartment is cp
and ct
, respectively, at
time t. The change in drug concentration in plasma
compartment and tissue compartment is dcp
and dct
,
respectively, with respect to time dt. The elimination
rate constant from central compartment is k.
According to chemical kinetics (the drug satisfies
first order kinetics),
dcp
/dt = k21
ct
–cp
k12
–kcp
(1)
dct
/dt = k12
cp
–k21
ct
(2)
By adding the Equations (1) and (2),
We get, dcp
/dt + dct
/dt = -kp
cp
(3)
Transforming the derivatives by Laplace transform
From Equation (2), sCt
(s)(t)–Ct
(0)= k12
Cp
(s)(t)–
k21
Ct
(s)(t)
Or, (s+k21
)Ct
(s)(t) = k12
Cp
(s)(t) (4); Ct
(0) = 0
(initial tissue drug concentration)
From Equation (3),
sCp
(s)(t)–Cp
(0)+sCt
(s)(t)–Ct
(0) = Kp
Cp
(s)(t)
Or, (s+k)Cp
(s)(t) + sCt
(s)(t) = Cp
0
; [Initially
Cp
(0)=C0
p
, Ct
(0)=0](5)
By solving the Equations (4) and (5),
(s+k)Cp
(s)(t)+sk12
Cp
(s)(t)/(s+k21
) = Cp
0
.
Or, Cp
(s)(t) = Cp
0
(s+k21
)/{(s+k)(s+k21
)+sk12
}
=Cp
0
(s+k21
)/{s2
+(k12
+k21
+k)s+kk21
}
=Cp
0
(s+k21
)/{s2
+(a+b)s+ab}
Here, a+b = k12
+k21
+k
ab = kk21
By performing partial fraction,
We get, Cp
(s)(t) = Cp
0
[(a–k21
)/{(a–b)(s+a)}]+C0
p
[(k21
–b)/{(a–b)(s+b)}]
By applying inverse Laplace transform,
We get from Equation (6)
Cp
(t) = Ae-at
+Be-bt
(6)
Where, A = (a–k21
)/(b–a)
B = (k21
–b)/(a–b)
Again we get from the Equations (4) and (5)
Ct
(s)(t) = k12
Cp
0
/{(s+k)(s+k21
)+sk12
}
=K12
Cp
0
/{s2
+(k12
+k21
+k)s+kk21
}
So, Ct
(s)(t) =k12
Cp
0
/(s+a)(s+b)
Byperformingpartialfraction,theequationnumber,
We get, Ct
(s)(t) = k12
Cp
0
/(b−a)(s+a)+k12
Cp
0
/(s+b)
(a−b)
By applying the inverse Laplace transform,
We get, Ct
(t) = Te-bt
–Te-at
(7)
Where, a+b = k+k12
+k21
; ab = kk21
T = k12
Cp
0
/(a−b).
MATHEMATICAL ANALYSIS
The initial (time t = 0) plasma concentration of the
drug, Cp
0
tissue concentration Ct
0
.[1-5]
We get from
the Equation number (7) and (9) and their hybrid
constant (A, B, a, b),
Cp
0
= A+B, (8)
Ct
0
=0(9)
k21
=(Ab+Ba)/(A+B)(10)
k12
=AB(a–b)2
/(A+B)(Ab+Ba)(11)
k=ab(A+B)/(Ab+Ba)(12)
T=AB(a–b)/(Ab+Ba)(13)
By solving the Equations (10), (11), and (12);
we can get the value of the pharmacokinetic rate
constants k12
, k21
, and k. The plasma concentration
and tissue concentration of the drug at any time
can be determined from the Equation number
(6) and (7) as well.
CALCULATION
Now, let the data obtained by the blood sample at
different intervals t1
, t2
, t3
, t4
, t5
, t6
, t7
, t8
, t9
, and t10
; the
observed plasma concentration quantities are c1
, c2
,
c3
, c4
, c5
, c6
, c7
, c8
, c9
, and c10
, respectively.According
to the Equation (7), it is obviously obtained a non-
linear curve rather than a straight line. Hence, the
biexponential curve represents the distribution
phase as well as elimination phase Figure 1.
Let the intrapolated data c1
, c2
, c3
and c8
, c9
, c10
lie in the distribution phase curve and elimination
phase curve, respectively, of the biexponential
curve.
According to biexponential distribution in plasma
compartment, the initial distribution phase (Ae-at
)
is more rapid than the elimination phase (Be-bt
).[6-11]
Hence, it is considered with the increasing of time
distribution phase slows down and elimination
phase proceeds rapidly and Ae-at
approaches to
zero, Be-bt
has a finite value.
Hence, it is obtained from the Equation (9)
Or, Cp
(t)= Be-bt
.
By taking natural logarithm in both sides, ln(Cp
)
(t)=−bt+ln(B)	 (i)
This equation follows the formula y = mx+c; m
and c are slope and intercept (y-axis), respectively,
of the straight line. The numerical values of the
b1
, B1
can be easily determined from elimination
phase data (c8
, c9
, c10
) and (t8
, t9
, t9
)
According to the Equation (i), b = {ln(c8
)- ln(c9
)}/
(t9
-t8
) and B = c8
/e-bt
8
,
And the elimination half-life of the drug,
t1/2
=0.693/b.
In the distribution phase, the drug is also
eliminated concurrently. Hence, we must take into
consideration the eliminated fraction of the plasma
drug concentration to determine the distribution
Dey: Open two compartment model
AJMS/Jul-Sep-2020/Vol 4/Issue 3 3
phase hybrid constants a, A. Hence, the eliminated
fraction can be determined by, c’p
= Be-bt
,
Where, c’p
is obtained from the distribution phase
data (c1
, c2
, c3
). Hence, the residual plasma drug
concentration is cp
-c’p
(let, cp
−c’p
= c”p
) during
distribution phase and the eliminated fraction drug
concentration is c’1
, c’2
, c’3
.
Now, it is obvious that, Cp
(t) = Ae-at
+Be-bt
.
Or, Cp
(t) = Ae-at
+C’p
(t)
Or, Cp
(t)−C’p
(t) = Ae-at
Or,C”p
(t) = Ae-at
;
where, C”p
(t) = Cp
(t)−C’p
(t) is the residual plasma
drug concentration
Let the residual plasma drug concentration is c”1
,
c”2
, c”3
with respect to eliminated fraction of drug
from plasma c’1
, c’2
, c’3
By taking the natural logarithm, it is obtained,
ln(c”p
) = ln(A) − at	 (ii)
According to equation, a = {ln(c”1
)−ln(c”2
)}/(t2
−
t1
) and A = c”2
/e-at
2
RESULTS
Hence, all the pharmacokinetic factors as well as
the drug concentration in plasma as well as tissue
can be evaluated at any time.
CONCLUSION
It is obvious that we can estimate the fate of a drug
following open two compartment model at any
time and predict the biological half-life of the drug
and drug remained in peripheral compartment at
any time.
ACKNOWLEDGMENTS
The author would like to acknowledge the
Rajshahi University, Rajshahi-6205, Bangladesh,
for providing computing facilities. Thus, this
study provides intuitive information of a drug
more precisely.
REFERENCES
1.	 Gibaldi M, Perrier D. Pharmacokinetics. 2nd
ed. New
York: Marcel Dekker; 1982.
2.	 Wagner JG, Pernarowski M. Biopharmaceutics and
Relevant Pharmacokinetics. Hamilton, Illinois: Drug
Intelligence Publications; 1971.
3.	 Eichler HG, Müller M. Drug distribution. The
forgotten relative in clinical pharmacokinetics. Clin
Pharmacokinet 1998;34:95-9.
4.	 Jusko WJ, Gibaldi M. Effects of change in elimination
on various parameters of the two-compartment open
model. J Pharm Sci 1972;61:1270-3.
5.	 Loughnan PM, Sitar DS, Ogilvie RI, Neims AH. The
two-compartment open-system kinetic model: A review
of its clinical implications and applications. J Pediatr
1976;88:869-73.
6.	 Thongmoon M, Pusjuso S. The numerical solutions of
differential transform method and the Laplace transform
method for a system of differential equations. Nonlinear
Anal Hybrid Syst 2010;4:425-31.
7.	 Hermann TW. Application of Laplace transforms to a
pharmacokinetic open two-compartment body model.
Acta Pol Pharm Drug Res 2017;74:1527-31.
8.	 Hermann TW. Pharmacokinetics. Theory and Practice.
Warswa: PZWL; 2002.
9.	 Vaughan DP, Trainor A. Derivation of general equations
for linear mammillary models when the drug is
administered by different routes. J Pharm Biopharm
1975;3:203-18.
10.	Wu G. Use of a five-compartment closed model to
describe the effects of ethanol inhalation on the transport
and elimination of injected pyruvate in the rat. Alcohol
Alcohol 1997;32:555-61.
11.	 Hermann TW. Zastosowanie transformacji Laplace’a w
farmakokinetycznym modelu jednokompartmentowym.
Farm Pol 2015;71:289-94.
Central
Compartment
Tissue
compartment
k
k12
k21
Figure 1: Two compartment open model

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Mathematical Review of Drug Fate Estimation Using Open Two Compartment Model

  • 1. www.ajms.com 1 ISSN 2581-3463 REVIEW ARTICLE A Mathematical Review to Estimate the Fate of a Drug Following Open Two Compartment Model, When it is Administered by Intravenous Route S. Dey Department of Pharmacy, University of Rajshahi, Rajshahi, Bangladesh Received: 15-03-2020; Revised: 01-05-2020; Accepted: 01-07-2020 ABSTRACT This paper deals with the application of Laplace Transform [L(f)(s) = F(s) = e-ts f(t)dt] to derive the formula [(Cp (t)= Ae-at + Be-bt ] and [Ct (t)= Te-bt -Te-at ] to estimate the fate of a drug in the mammillary body with the prediction of the biological parameters [k, k12 , and k21 ]. The numerical values of [k, k12 , and k21 ] which is estimated through the determination of the mathematical hybrid constants (A, B, a, b, T) also predict the tissue concentration of the drug, biological half-life as well. Key words: Compartment model, Laplace transform and inverse transform, pharmacokinetics, rate constant INTRODUCTION When a drug is injected through intravenous route, initially, all the drugs present in the plasma compartment (central compartment). As time increases, the concentration of drug in central compartment declines due to the transport of the drug to peripheral compartment (tissue compartment), elimination of the drug from central compartment as well. Hence, in the central compartment, there are two phases, drug distribution phase as well as drug decline phase. In tissue compartment (peripheral compartment), the drug concentration increases and finally forms an equilibrium condition with central compartment. Hence, at any time, some fraction of drug remains in the central compartment, in tissue compartment as well. The drug elimination occurs from the central compartment, that is, some fraction of plasma drug is excreted. Moreover, some fraction of drug is metabolized in the central compartment. THEORY AND DISCUSSION Methodology, when a drug is administered by intravenous route as a bolus dose, the injected drug distributes through central compartment as well Address for correspondence: Sunandan Dey E-mail: sunandanpharm@gmail.com as peripheral compartment. The drug elimination occurs from the central compartment. Laplace transform The Laplace transform F=F(s) of a function f=f(t) is defined by , L(f)(s) = F(s) = e-ts f(t) dt. The integral is evaluated with respect to t, hence, once the limits are substituted, what is left are in terms of s. Inverse Transform Let f be a function and L(f)= F be its Laplace transform. Then, by definition, f is the inverse transform of F. This is denoted by L-1 (F)=f. As for example, from inverse Laplace transform. We get that L(e-kt )= 1/(s+k), L(1) = 1/s written in the inverse transform notation L-1 (1/(s+k))= e-kt ; L(1/s) = 1. Transform of derivatives If a function y=y(t). The transform of its derivative y’ can be expressed in terms of the Laplace transform of y: L(y’)= sL(y)–y(0) Proof In my proposed mathematical analysis, I consideredboththeplasmaandtissuecompartment elimination rate constant,
  • 2. Dey: Open two compartment model AJMS/Jul-Sep-2020/Vol 4/Issue 3 2 Let a drug be administered as intravenous bolus dose (C0 p ), the drug concentration in plasma compartment and tissue compartment is cp and ct , respectively, at time t. The change in drug concentration in plasma compartment and tissue compartment is dcp and dct , respectively, with respect to time dt. The elimination rate constant from central compartment is k. According to chemical kinetics (the drug satisfies first order kinetics), dcp /dt = k21 ct –cp k12 –kcp (1) dct /dt = k12 cp –k21 ct (2) By adding the Equations (1) and (2), We get, dcp /dt + dct /dt = -kp cp (3) Transforming the derivatives by Laplace transform From Equation (2), sCt (s)(t)–Ct (0)= k12 Cp (s)(t)– k21 Ct (s)(t) Or, (s+k21 )Ct (s)(t) = k12 Cp (s)(t) (4); Ct (0) = 0 (initial tissue drug concentration) From Equation (3), sCp (s)(t)–Cp (0)+sCt (s)(t)–Ct (0) = Kp Cp (s)(t) Or, (s+k)Cp (s)(t) + sCt (s)(t) = Cp 0 ; [Initially Cp (0)=C0 p , Ct (0)=0](5) By solving the Equations (4) and (5), (s+k)Cp (s)(t)+sk12 Cp (s)(t)/(s+k21 ) = Cp 0 . Or, Cp (s)(t) = Cp 0 (s+k21 )/{(s+k)(s+k21 )+sk12 } =Cp 0 (s+k21 )/{s2 +(k12 +k21 +k)s+kk21 } =Cp 0 (s+k21 )/{s2 +(a+b)s+ab} Here, a+b = k12 +k21 +k ab = kk21 By performing partial fraction, We get, Cp (s)(t) = Cp 0 [(a–k21 )/{(a–b)(s+a)}]+C0 p [(k21 –b)/{(a–b)(s+b)}] By applying inverse Laplace transform, We get from Equation (6) Cp (t) = Ae-at +Be-bt (6) Where, A = (a–k21 )/(b–a) B = (k21 –b)/(a–b) Again we get from the Equations (4) and (5) Ct (s)(t) = k12 Cp 0 /{(s+k)(s+k21 )+sk12 } =K12 Cp 0 /{s2 +(k12 +k21 +k)s+kk21 } So, Ct (s)(t) =k12 Cp 0 /(s+a)(s+b) Byperformingpartialfraction,theequationnumber, We get, Ct (s)(t) = k12 Cp 0 /(b−a)(s+a)+k12 Cp 0 /(s+b) (a−b) By applying the inverse Laplace transform, We get, Ct (t) = Te-bt –Te-at (7) Where, a+b = k+k12 +k21 ; ab = kk21 T = k12 Cp 0 /(a−b). MATHEMATICAL ANALYSIS The initial (time t = 0) plasma concentration of the drug, Cp 0 tissue concentration Ct 0 .[1-5] We get from the Equation number (7) and (9) and their hybrid constant (A, B, a, b), Cp 0 = A+B, (8) Ct 0 =0(9) k21 =(Ab+Ba)/(A+B)(10) k12 =AB(a–b)2 /(A+B)(Ab+Ba)(11) k=ab(A+B)/(Ab+Ba)(12) T=AB(a–b)/(Ab+Ba)(13) By solving the Equations (10), (11), and (12); we can get the value of the pharmacokinetic rate constants k12 , k21 , and k. The plasma concentration and tissue concentration of the drug at any time can be determined from the Equation number (6) and (7) as well. CALCULATION Now, let the data obtained by the blood sample at different intervals t1 , t2 , t3 , t4 , t5 , t6 , t7 , t8 , t9 , and t10 ; the observed plasma concentration quantities are c1 , c2 , c3 , c4 , c5 , c6 , c7 , c8 , c9 , and c10 , respectively.According to the Equation (7), it is obviously obtained a non- linear curve rather than a straight line. Hence, the biexponential curve represents the distribution phase as well as elimination phase Figure 1. Let the intrapolated data c1 , c2 , c3 and c8 , c9 , c10 lie in the distribution phase curve and elimination phase curve, respectively, of the biexponential curve. According to biexponential distribution in plasma compartment, the initial distribution phase (Ae-at ) is more rapid than the elimination phase (Be-bt ).[6-11] Hence, it is considered with the increasing of time distribution phase slows down and elimination phase proceeds rapidly and Ae-at approaches to zero, Be-bt has a finite value. Hence, it is obtained from the Equation (9) Or, Cp (t)= Be-bt . By taking natural logarithm in both sides, ln(Cp ) (t)=−bt+ln(B) (i) This equation follows the formula y = mx+c; m and c are slope and intercept (y-axis), respectively, of the straight line. The numerical values of the b1 , B1 can be easily determined from elimination phase data (c8 , c9 , c10 ) and (t8 , t9 , t9 ) According to the Equation (i), b = {ln(c8 )- ln(c9 )}/ (t9 -t8 ) and B = c8 /e-bt 8 , And the elimination half-life of the drug, t1/2 =0.693/b. In the distribution phase, the drug is also eliminated concurrently. Hence, we must take into consideration the eliminated fraction of the plasma drug concentration to determine the distribution
  • 3. Dey: Open two compartment model AJMS/Jul-Sep-2020/Vol 4/Issue 3 3 phase hybrid constants a, A. Hence, the eliminated fraction can be determined by, c’p = Be-bt , Where, c’p is obtained from the distribution phase data (c1 , c2 , c3 ). Hence, the residual plasma drug concentration is cp -c’p (let, cp −c’p = c”p ) during distribution phase and the eliminated fraction drug concentration is c’1 , c’2 , c’3 . Now, it is obvious that, Cp (t) = Ae-at +Be-bt . Or, Cp (t) = Ae-at +C’p (t) Or, Cp (t)−C’p (t) = Ae-at Or,C”p (t) = Ae-at ; where, C”p (t) = Cp (t)−C’p (t) is the residual plasma drug concentration Let the residual plasma drug concentration is c”1 , c”2 , c”3 with respect to eliminated fraction of drug from plasma c’1 , c’2 , c’3 By taking the natural logarithm, it is obtained, ln(c”p ) = ln(A) − at (ii) According to equation, a = {ln(c”1 )−ln(c”2 )}/(t2 − t1 ) and A = c”2 /e-at 2 RESULTS Hence, all the pharmacokinetic factors as well as the drug concentration in plasma as well as tissue can be evaluated at any time. CONCLUSION It is obvious that we can estimate the fate of a drug following open two compartment model at any time and predict the biological half-life of the drug and drug remained in peripheral compartment at any time. ACKNOWLEDGMENTS The author would like to acknowledge the Rajshahi University, Rajshahi-6205, Bangladesh, for providing computing facilities. Thus, this study provides intuitive information of a drug more precisely. REFERENCES 1. Gibaldi M, Perrier D. Pharmacokinetics. 2nd ed. New York: Marcel Dekker; 1982. 2. Wagner JG, Pernarowski M. Biopharmaceutics and Relevant Pharmacokinetics. Hamilton, Illinois: Drug Intelligence Publications; 1971. 3. Eichler HG, Müller M. Drug distribution. The forgotten relative in clinical pharmacokinetics. Clin Pharmacokinet 1998;34:95-9. 4. Jusko WJ, Gibaldi M. Effects of change in elimination on various parameters of the two-compartment open model. J Pharm Sci 1972;61:1270-3. 5. Loughnan PM, Sitar DS, Ogilvie RI, Neims AH. The two-compartment open-system kinetic model: A review of its clinical implications and applications. J Pediatr 1976;88:869-73. 6. Thongmoon M, Pusjuso S. The numerical solutions of differential transform method and the Laplace transform method for a system of differential equations. Nonlinear Anal Hybrid Syst 2010;4:425-31. 7. Hermann TW. Application of Laplace transforms to a pharmacokinetic open two-compartment body model. Acta Pol Pharm Drug Res 2017;74:1527-31. 8. Hermann TW. Pharmacokinetics. Theory and Practice. Warswa: PZWL; 2002. 9. Vaughan DP, Trainor A. Derivation of general equations for linear mammillary models when the drug is administered by different routes. J Pharm Biopharm 1975;3:203-18. 10. Wu G. Use of a five-compartment closed model to describe the effects of ethanol inhalation on the transport and elimination of injected pyruvate in the rat. Alcohol Alcohol 1997;32:555-61. 11. Hermann TW. Zastosowanie transformacji Laplace’a w farmakokinetycznym modelu jednokompartmentowym. Farm Pol 2015;71:289-94. Central Compartment Tissue compartment k k12 k21 Figure 1: Two compartment open model