How do you calculate the time until the steady-state of a drug?

Question

I encountered a mathematically intriguing conundrum, in that it's related to medicine but is centered around mathematics.

Suppose drug A has a half-life in the body of 30 hours. The patient takes 40mg once per day. How long is it then until the patient then reaches steady-state dosage in the body (within some tolerance interval)?

The answer should be about one week, but I'm trying to understand the math behind "why".

Answer 1

Here's a short simulation that show the process. You can find more about how to simulate a half decay process here

You can notice that the first dose of 40 mg should decay to 20 mg in 30 hours, but the process is cut by the new dose.

The daily oscillation reaches a steady state around the seventh day (you can find small changes in the data, but not appreciable at the naked eye)

import matplotlib.pylab as plt
import numpy as np

ndays = 12
ttot = 24*ndays 
dt = 0.25

time = np.arange(0, ttot, dt)
Concentration = np.zeros(len(time))

conc = 0
tau = 30./np.log(2)  ## exponential decay time constant from half-life value
dosis = 40

for k, t in enumerate(time):
    if t%24 == 0:   ## If the time is multiple of 24 hours, the dosis is given
        conc += dosis 
    Concentration[k] = conc
    ### update concentration decay
    conc = conc*(1 - dt/tau)

ax = plt.subplot(111)
ax.plot(time, Concentration)
ax.set_xticks(np.arange(0, ndays * 24, 24 ))
ax.set_xticklabels(['%d'%d for d in np.arange(ndays)])
ax.set_xlabel('Days')
ax.set_ylabel('Dosis remaining (AU)')
plt.savefig('Steady_state.png', dpi = 300)

Answer 2

heracho's answer does a good job showing the realistic timing. However you can do a pretty good approximation with a simple differential equation, where you approximate the daily dose as a continuous input. If $C(t)$ is the concentration of the drug at time $t$, then

$$ \frac{dC}{dt} = D - \frac{\ln{2}}{t_{1/s}} C$$ where $D$ is the dose rate, 40 mg/day. The steady-state condition is reached when $\frac{dC}{dt} \to 0$, which tells you that the (average) steady state concentration is $$C_{ss} = D\frac{t_{1/2}}{\ln{2}} \approx 72.13\mathrm{mg}$$ which agrees pretty well with heracho's plot.

The full solution to the differential equation, assuming $C(0)=0$ is $$ C(t) = C_{ss}\cdot(1-\mathrm{e}^{-t\ln{2}/t_{1/2}})$$

From here you can see clearly the answer to the "when do we reach steady state?" question is driven by the exponential term, which is negligible by ~5 half-lives.

Having stolen @heracho's plotting script and added the following lines:

conc_ss = dosis*(tau/24)
concavg = conc_ss*(1-np.exp(-time/tau))
...
ax.plot(time, concavg, label='continuous dose approximation')

you can see that this approximation follows the mean of the true concentration pretty well after just a few half-lives.

Answer 3

Let's take a moment to think about what it means for a drug dosed multiple times to reach "steady-state": "steady-state" would mean that it doesn't matter much if you've been taking the drug for N or N-1 days, if you've reached "steady-state" the concentration of drug is steady from day to day (of course, with a 30-hour half-life, it's not going to be particularly stable within a day, assuming relatively rapid uptake).

It's also important to recognize that simple exponential decay is a scenario that occurs when the fates of particles are independent of each other. Therefore, one way we can think about comparing day N to day N-1 would be to consider how much drug is still around on day N from the first dose, because that's the only difference in the history of doses.

For a drug with a 30-hour half-life, after 7 days you could say 7 days * 24 hours/day * 1 half-life/30 hours = 5.6 half-lives have passed.

After 5.6 half-lives, you could say 0.5^5.6 = about 0.02 or 2% of a drug is remaining relative to the original concentration.

Another way to say that is that the difference at the end of the 7th day between having had the drug for 7 days vs. having had it for 6 days would be that 2% remaining from day 1. That seems roughly reasonable to consider a "steady-state".

This is all assuming simple exponential kinetics, other assumptions will change the nature of the problem. It's also important to recognize that for a drug with a 30-hour half-life given once daily, the concentration will never really be "steady" but will still have a clear oscillation after each dosage, depending on the kinetics of release/uptake.

Answer 4

I like @thegreatmu's answer, but I think we can get closer to the original problem. During the day, the concentration decays according to $$ \frac{\partial C}{\partial t} = - \frac{\log(2)}{t_{1/2}} C $$ pretty much like @thegreatmu's equation, but without the term $D$, because during the day, no medicine is added.

At a fixed point during every day, say at breakfast, the medicine is ingested, and the concentration instantly increases by the dose $D$. At such moments, whe have both the low concentration $C_-$ and the high concentration $C_+ = C_- + D$.

Let us measure time in days (the half-time is 1.25 days), and let us take breakfast-time to be the moments $t=0$, $t=1$, $t=2$, etcetera.

The concentration at any time $n-1\leq t < n$ in the $n$-th day is given by $$ C(t) = \exp\left(- \frac{\log(2) }{t_{1/2}} (t-n+1)\right) C_+(n), $$ so just before taking the medicine again, we have the low concentration $$ C_-(n+1) = \exp\left(- \frac{\log(2) }{t_{1/2}} \right) C_+(n). $$ and the high concentration is given by $$ C_+(n+1) = D + \exp\left(- \frac{\log(2) }{t_{1/2}} \right) C_+(n). $$ From here it is pssible to calculate the high concentration for any day $n$: \begin{eqnarray} C_+(0) = D &,& C_+(1) = D\left(1 + \exp\left(- \frac{\log(2) }{t_{1/2}} \right)\right), \cdots \nonumber \end{eqnarray} leading to \begin{eqnarray} C_+(n) &=& D\sum_{j=0}^{n} \exp\left(- \frac{\log(2) }{t_{1/2}} \right)^j = D\frac{1-\exp\left(- \frac{\log(2) }{t_{1/2}} \right)^{n+1}} {1-\exp\left(- \frac{\log(2) }{t_{1/2}} \right)} \nonumber \end{eqnarray} and at $n-1\leq t < n$, any moment in the $n$-th day, the concentration is given by \begin{eqnarray} C(t) &=& D\frac{1-\exp\left(- \frac{\log(2) }{t_{1/2}} \right)^{n}} {1-\exp\left(- \frac{\log(2) }{t_{1/2}} \right)} \exp\left(- \frac{\log(2) }{t_{1/2}} (t-n+1)\right) \nonumber \end{eqnarray} we can compare the current concentration $C(t)$ to the concentration $C(t+m)$, the same time of day, but $m$ days later: \begin{eqnarray} C(t) = \frac{1-\exp\left(- \frac{\log(2) }{t_{1/2}} \right)^{n}} {1-\exp\left(- \frac{\log(2) }{t_{1/2}} \right)^{n+m}} C(t+m) = \left(1-\exp\left(- \frac{\log(2) }{t_{1/2}} \right)^{n}\right)C(t+\infty), \nonumber \end{eqnarray} where $C(t+\infty)$ means 'at this time of day, in steady state'.

Therefore, we can say that the concentration is in steady state when \begin{eqnarray} \exp\left(- \frac{\log(2) }{t_{1/2}} \right)^{n} < \epsilon, \nonumber \end{eqnarray} where $\epsilon$ is small, for instance $\epsilon = 0.01$ when you consider 'in steady state' to mean 'less that one percent away from steady state'. This will be after $n$ days, where $n$ is given by \begin{eqnarray} n > {}^2\log(1/\epsilon)~t_{1/2}. \nonumber \end{eqnarray} In the case where the half-time is $1.25$ days, and $\epsilon=0.01$, we get $n > 8.3$: just over a week. After 7 days, steady stage is approached up to 2%.