There are three closely related questions for the single compartment model applied to oral dosing in pharmacokinetics:

  1. The time evolution of blood concentrations, $t=0$ to $t=\infty$, given one and only one dose to the stomach at $t=0$; and,
  2. The equilibrium state time evolution of blood concentrations in between doses, $t=0$ at the instant a dose is taken and $t=\tau$ at the instant the next dose is taken (which then sets $t=0$ again), assuming regularly spaced doses, given that enough time has passed to reach that equilibrium state ($t \to \infty$); and,
  3. The fuller time evolution of blood concentrations, $t=0$ to $t=\infty$, also assuming regularly spaced doses, and also given that the first dose starts at $t=0$ (each dose occurs at $t=n\tau$, where integer $n \ge 0$ and where $\tau$ is the dose interval.)

In each of the above cases, there is an assumed single absorption rate ($k_a$) to model the transfer from the stomach (mouth/tongue, esophagus, stomach, and the rest of the gut) to the blood stream and another assumed single elimination rate ($k_e$) to model the removal from the blood stream. Let's call $A$ the amount of the dose remaining in the gut and awaiting absorption into the blood and $E$ the amount residing in the blood stream and awaiting elimination. (We'll avoid confounding issues about blood volume and concentration, effectiveness, etc.)

The set up for case 1 above is simple: \begin{equation} dA = -k_a \cdot A \, dt \\ dE = k_a\cdot A \, dt - k_e \cdot E \, dt \end{equation} and I'm able to solve this when $k_a \ne k_e$ and when $k_a = k_e$.

Just to dot the i, regarding case 1, here's the logic I've applied. (I apologize for doing the obvious here.) Assume that D is the single dose amount, which occurs at $t=0$. So the initial condition is $A_0 = D$.

For the stomach: \begin{equation} dA = -k_a \cdot A \, dt, \,\,\,\text{where $A_0=D$} \\ \frac{dA}{A} = -k_a \, dt \\ \int \frac{dA}{A} = \int -k_a \, dt \\ ln(A_t) = -k_a \cdot t + C_0 \\ A_t = D \cdot e^{-k_a \cdot t} \end{equation}

For the blood: \begin{equation} dE = k_a\cdot A \, dt - k_e \cdot E \, dt, \,\,\,\text{where $E_0=0$} \\ dE = k_a\cdot D \cdot e^{-k_a \cdot t} \, dt - k_e \cdot E \, dt \\ \frac{dE}{dt} = k_a\cdot D \cdot e^{-k_a \cdot t} - k_e \cdot E \\ \frac{dE}{dt} + k_e \cdot E = k_a\cdot D \cdot e^{-k_a \cdot t} \\ \text{setting integrating factor $\mu=e^{\int k_e \, dt}=e^{k_e \cdot t}$}, then \\ E = \frac{1}{\mu} \int_0^t \mu \cdot k_a \cdot D \cdot e^{-k_a \cdot s} \, ds \\ E = D\cdot e^{-k_e \cdot t}\cdot k_a \int_0^t e^{\left(k_e-k_a\right) \cdot s} \, ds \\ \text{which resolves one of two ways,} \\ E_t = D \cdot k \cdot e^{-k_e \cdot t} \cdot t, \,\,\,\text{where $k = k_a = k_e$} \\ E_t = D \cdot \frac{k_a}{k_a - k_e} \cdot \left( e^{-k_e \cdot t} - e^{-k_a \cdot t} \right), \,\,\,\text{where $k_a \ne k_e$} \end{equation}

I get that far without difficulty.

Per the above development, I've also found two related questions have already been asked and answered on stackexchange. These are the relatively clearly stated single compartment model, single dose, absorption rate equals elimination rate and the somewhat more confused question on the same single compartment model, single dose, absorption and elimination rates not necessarily the same. Unfortunately, neither of these address case 2 and case 3. Since I already understand case 1, those cases on stackexchange don't further my understanding at all.

I'm particularly interested in developing case 3 above for the situation where $k_a = k_e$. (I have the book answer, but not solution method, when $k_a \ne k_e$, so if I had those solution steps I could probably intercede and find the other case.) I tap out approximately where I'm able to find the equilibrium equation over time for the stomach value, $A$, for case 2, but not the equilibrium equation for $E$ (though I know what it should be, again from stock answers I've already found.)

I also understand there isn't a single approach, but actually several areas in mathematics which can be applied depending upon where my comfort level is at. For these purposes, assume ONE year of undergrad mathematics: basically MTH 251, 252, and 253. This means NO serious use of Laplace transforms (though a glancing familiarity with the idea), no familiarity with variations of parameters methods, and also very little experience with the application of matrix methods here (though I understand them as applied to solving combinations of finite linear equations.)

I need help both with the setup and the solution process for case 3 but case 2 might be used to prepare for it or else as a check to see if it falls out of the solution for case 3 where $t \to \infty$. (Case 1, of course, should match up with case 3's equation where $0 \le t < \tau$.)

I apologize in advance for any perceived lack of effort or clarity in posing this question. Please accept my assurance that I've spent dozens of hours testing my own skills and attempting to find an appropriate resource that I could learn from before posting here. I'm now at the point of hoping there is someone interested enough to help educate me about solving this case-3 problem.

For those needing/wanting to dig into my motivation:

This pharmacokinetics question actually develops because my daughter suffers from grand mal seizures and takes drugs to help manage them. This is my personal interest here and has nothing whatever to do with homework! (It's been several decades since my first year calculus coursework.) I've taken the time to read what I could over many days' worth of searches (and hand to paper work as well, of course) and still find that I'm out of my depth in being able to develop the solutions myself.

That's frustrating because, while I can find some answers stated in tables for certain questions, I can't create my own solutions to related questions as I simply lack the insight either in properly forming the problems or else find I'm lacking the methods in solving them.

It may help that I have found elsewhere (without knowing how it is achieved) that a solution for case 3, where $k_a \ne k_e$, is: \begin{equation} E_t = D \cdot \frac{k_a}{k_a - k_e} \cdot \left[ \left( \frac{1 - e^{-n \cdot k_e \cdot \tau}}{1 - e^{-k_e \cdot \tau}} \right) \cdot e^{-k_e \cdot t} - \left( \frac{1 - e^{-n \cdot k_a \cdot \tau}}{1 - e^{-k_a \cdot \tau}} \right) \cdot e^{-k_a \cdot t} \right] \end{equation} where $n$ is the number of doses.

In the case where $n=1$, you can see how this nicely resolves into: \begin{equation} E_t = D \cdot \frac{k_a}{k_a - k_e} \cdot \left( e^{-k_e \cdot t} - e^{-k_a \cdot t} \right) \end{equation} which is the solution for case 1 where $k_a \ne k_e$ (but not where $k_a = k_e$, where it is instead $E_t = D \cdot t \cdot e^{-k \cdot t}$, where $k = k_a = k_e$)

In the case where $n \to \infty$, you can see how this nicely resolves into: \begin{equation} E_t = D \cdot \frac{k_a}{k_a - k_e} \cdot \left[ \frac{e^{-k_e \cdot t}}{1 - e^{-k_e \cdot \tau}} - \frac{e^{-k_a \cdot t}}{1 - e^{-k_a \cdot \tau}} \right] \end{equation} which is the solution for case 2 where $k_a \ne k_e$ (but not where $k_a = k_e$.)

Note that this case 3 situation (where I can't develop my own work, sadly) only works where $k_a \ne k_e$. I'd like to find the answer where $k_a = k_e$. But I'd also, of course, like to know how to solve it either way. Not just the answers, but the approach.

  • 2
    $\begingroup$ That's a lot to read... could you perhaps just write down the differential equation you're trying to solve? $\endgroup$ Mar 23 '16 at 21:56
  • $\begingroup$ If I could correctly phrase the differential equations, I'd be halfway there I think. I'm worried that if I write one down, it will be wrong and lead you and others in the wrong direction. But I'll try and give it a shot. This involves instantaneous changes, as each dose is added. So I know that while the result may be integrable, it won't be differentiable at each point of dosage. $\endgroup$
    – jonk
    Mar 23 '16 at 22:00
  • $\begingroup$ Oh, so you want help writing down a differential equation to describe a particular situation? $\endgroup$ Mar 23 '16 at 22:00
  • $\begingroup$ Yeah, that would be helpful. Keep in mind that it involves regular doses -- these are instantaneous changes to $A$. $\endgroup$
    – jonk
    Mar 23 '16 at 22:01
  • $\begingroup$ Just focus on case 3. It's not too complicated to understand. There is a regular, periodic input to $A$, which is dose $D$, separated by a period $\tau$. $A(t=0) = D$. (Goes from 0 to D, instantly.) Also, assume $E(t=0)=0$. From $A$, there is an instantaneous flow outward towards $E$, at the rate of $k_a \cdot A \, dt$. This reduces $A$ but increases $E$. Also, there is another flow outward from $E$, at the rate of $k_e \cdot E \, dt$. That's it. I have a problem setting up this Dirac-like impulse at the input as well as solving this system in the case both for $k_a = k_e$ and $k_a \ne k_e$. $\endgroup$
    – jonk
    Mar 23 '16 at 22:08

The (slightly modified) expression that you already have, represents the concentration of the drug in plasma after the $n$-th dose is given (that is, parameter $t$ starts from zero after $(n-1)$ intervals $\tau$ passed),

\begin{align} C_n(t) &= \frac{\zeta_0}{k_a - k_e} \cdot \left[ \left( \frac{1 - \exp(-k_e n\tau)} {1 - \exp(-k_e \tau)} \right) \cdot \exp(-k_e t)\right. \\ &-\left.C_n(i+1,\tau,k_a,\zeta_0)( \left( \frac{1-\exp(-k_a n\tau)} {1 - \exp(-k_a \tau)} \right) \cdot \exp(-k_a t) \right] \tag{1}\label{1} . \end{align}

Here $\zeta_0$ is the universal model constant, the initial slope of the single-dose concentration curve.

\begin{align} \zeta_0&=C_0\cdot k_a=\frac{D}{V}\cdot k_a , \end{align}

It encapsulates the tricky parameters like the apparent volume of distribution $V$, the apparent initial concentration $C_0$ as well as provides a symmetry to the expression and invariance to possible flip-flop condition (when it happens that $k_a<k_e$).

So, if we define a function $f$ as \begin{align} f(x)&= \zeta_0 \cdot \left( \frac{1 - \exp(-x n\tau)} {1 - \exp(-x \tau)} \right) \cdot \exp(-x t) , \end{align}

then expression \eqref{1} in terms of $f$ is \begin{align} \frac{f(k_e)-f(k_a)}{k_a-k_e} &= -\frac{f(k_a)-f(k_e)}{k_a-k_e} . \end{align}

In order to handle the case when $k_a=k_e$, we just need to find a limit

\begin{align} C_n(t)|_{k_e=k_a}&= -\lim_{k_e\to k_a}\frac{f(k_a)-f(k_e)}{k_a-k_e} \tag{2}\label{2} . \end{align} Cn(i+1,tau,ka,zeta0)( But \eqref{2} is just a definition of the derivative of $f$, hence

\begin{align} C_n(t)|_{k_e=k_a} &=-f'(x)|_{x=k_a} , \end{align}

which we can find to be

\begin{align} C_n(t)|_{k_e=k_a}&= \frac{\zeta_0\exp(-k_a t)}{1-\exp(-k_a\tau)}\cdot \left[ \frac{(1-\exp(-k_a n\tau))\tau\exp(-k_a\tau)}{1-\exp(-k_a\tau)} + t-(n\tau+t)\exp(-k_a n\tau ) \right] \tag{3}\label{3} . \end{align}

It's easy to check that for $n=1$ expression \eqref{3} gives \begin{align} C_1(t)|_{k_e=k_a}&= \zeta_0t\exp(-k_at) , \end{align} as expected.

An illustration for $\tau=3$, $k_a=k_e=0.9$, $\zeta_0=1.3$, $n=1,\dots,6$:

enter image description here

As for the origin of \eqref{1}, it's just a sum of of geometric progression:

\begin{align} \left( \frac{1 - \exp(-k_e n\tau)} {1 - \exp(-k_e \tau)} \right) \cdot \exp(-k_e t) &=\sum_{m=0}^{n-1} \exp(-k_e m\tau)\exp(-k_e t) \\ &=\sum_{m=0}^{n-1} \exp(-k_e (m\tau+t)) , \end{align}

that is, it's a sum of all $n$ individual single-dose curves at a point of $(n-1)\tau+t$.


An alternative way to get the same expression is to start from single-dose expression

\begin{align} C_1(t)|_{k_a=k_e} &= \zeta_0\,t\exp(-k_a t) \end{align}

and just calculate the cumulative effect of multiple doses as

\begin{align} C_n(t)|_{k_a=k_e} &= \sum_{m=0}^{n-1} \zeta_0(m\tau+t)\exp(-k_a(m\tau+t)) . \end{align}

  • $\begingroup$ Your answer is clear and addresses what I asked for (and more.) Thanks! It is too bad it will still be some days before I can give some practice time to deepen it into me. But my intuition says you've hit on exactly what I was asking for. I sincerely thank you for what you offered me! $\endgroup$
    – jonk
    Jul 2 '18 at 1:09
  • $\begingroup$ @jonk: I hope you will find the answers you are looking for. $\endgroup$
    – g.kov
    Jul 2 '18 at 8:17
  • $\begingroup$ In the above, I already got what I needed. I was able to convince our Neurology team about dosing times and it has served us well for 2 years. I wanted to know, because I love math on its own and I felt incomplete here. But our daughter is now gone from 35-40 seizures a year to about 4. So some success already. My neurologists are not very good at understanding dosing. $\endgroup$
    – jonk
    Jul 2 '18 at 9:00

Here is an analysis for the third case.

Let $\mathbb{A} =\left\lbrack \begin{array}{cc}-k_a&0\\ k_a& -k_e \end{array} \right\rbrack$,the state vector $$X(t) = \left\lbrack \begin{array}{c} A(t)\\ E(t) \end{array}\right\rbrack,$$ the input matrix $\mathbb{B} = \left\lbrack \begin{array}{c}1 \\0 \end{array} \right\rbrack$, and the input $$u(t) =D \sum_{i=0}^\infty \delta(t-i \tau)$$ Then the system (I think) is described by the simple linear time-invariant system of ODEs as follows.

$$ \frac{\mathrm{d}}{\mathrm{d}t} X(t) = \mathbb{A} X(t) + \mathbb{B} u(t) $$

For a system this small, it actually is possible to write a closed form solution but it would involves using Eigenvalues and Eigenvectors. I will skip all this and write the solution, assuming zero initial conditions.

$$A(t) = D \sum_{i=0}^\infty e^{k_a(t-i\tau)} \Theta(t-i\tau)$$ and $$E(t) = D \frac{k_a}{k_a-k_e} \sum_{i=0}^\infty \left( e^{k_a(t-i\tau)}-e^{k_e(t-i\tau)} \right) \Theta(t-i\tau),$$

Where $\Theta$ is the Heaviside function (unit step). I know you are more interested in knowing how to get the solution, and I can attempt to show you, but it does involve the Laplace Transform.


I include the work, so others can check it.

Taking the Laplace Transform of the evolution equation we get

$$(sI-\mathbb{A})X(s) = \mathbb{B} U(s)$$

So we get \begin{align*} X(s) &= (sI-\mathbb{A})^{-1} \mathbb{B} U(s)\\ &= D \left\lbrack\begin{array}{c} (s+k_a)^{-1}\\ k_a(s+k_a)^{-1}(s+k_e)^{-1}\end{array} \right\rbrack \sum_{i=0}^\infty e^{i s \tau} \end{align*}

And upon taking the inverse Laplace Transform of this, we get our answer. Note that we resist the temptation to write $\sum_{i=0}^\infty e^{i s\tau}$ as $$\frac{e^s\tau}{e^{s\tau}-1}$$

In the case where $k_a=k_e = k$ we get

$$A(t) = D\sum_{i=0}^\infty e^{-k(t-i\tau))} \Theta(t-i\tau)$$

and $$E(t) = D \sum_{i=0}^\infty e^{-k(t-i\tau)}k(t-i\tau) \Theta(t-i\tau)$$

  • $\begingroup$ I'm not sure I have a solution equation for the case where $k_a = k_e$. It doesn't look that way to me. $\endgroup$
    – jonk
    Mar 24 '16 at 6:00
  • $\begingroup$ I also don't understand the approach(es.) Though, I think I may take a shot at studying Laplace more. Perhaps it is enough for me to know that it can be applied if I spend some time there. $\endgroup$
    – jonk
    Mar 24 '16 at 6:04
  • $\begingroup$ In the case where $k_a= k_e$ the form of the solution changes. Notice that attempting to use my submitted solution in that case will result in division by zero. $\endgroup$
    – fred
    Mar 24 '16 at 10:37
  • $\begingroup$ I am reasonably confident of the correctness of this approach. It reproduces the first case. $\endgroup$
    – fred
    Mar 24 '16 at 10:47
  • $\begingroup$ I'm not disagreeing with you. I'm not skilled enough to do so, for now. But if you read the title of my question you will see I'm interested in the case where $k_a = k_e$. $\endgroup$
    – jonk
    Mar 24 '16 at 20:21

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