Sciencemadness Discussion Board

Differential Equation?

MagicJigPipe - 22-3-2010 at 19:37

I am trying to solve what I think is a differential equation but I can't determine how.

I am trying to "make" an equation for the amount of drug in a user's bloodstream at various times. But the catch is that the user is taking a certain dose (d) every amount of time (t). I treated it as a rate in - rate out equation where:

dQ/dt = rate in - rate out

Q = amount of drug in user

t = time elapsed

P = dosing period


HL= half-life of drug

rate in = (dose)/(P)

rate out = (??)/(P)

?? = dose + Q*(.5)^(t/HL) ??

I can't seem to get this equation into a form to solve it. It better not be separable b/c I've been looking at it for about an hour.

Is the setup correct? If so, how do I solve the diff. eq. And don't make fun of me, math isn't my strongest subject and I tend to make mistakes without seeing them readily.

Thanks in advance.



[Edited on 3-23-2010 by MagicJigPipe]

pantone159 - 22-3-2010 at 20:30

So you mean something like:
Q decreases at a rate proportional to Q, but then add a constant dose rate d (?)

That would give:
dQ/dt = -k*Q + d
where k is the decay rate, and d is the dose rate.
(k is related to the half-life by k = ln2 / half-life).

The easiest way to solve this, is to change to a new variable, which you make up, which is Q plus a constant. Say,
W = Q + b
where W is the new variable, and b is the constant (which you don't know yet.)
Then, dW/dt = dQ/dt, since b is constant, and you can rewrite as:
dW/dt = -k * (W - b) + d = -k*W + (d + k*b)
Now you pick your b: If you choose b = -d/k, then this simplifies to:
dW/dt = -k * W
(since you made up W in the first place, you are free to choose any b you like!)
This has the trivial solution:
W(t) = W(0) * exp (-k*t)
or
Q(t) - d/k = (Q(0) - d/k) * exp (-k*t)
or
Q(t) = (Q(0) - d/k) * exp (-k*t) + d/k
Notice that at t=0, you get Q = Q(0) as you should.

This is, btw, basically the same equation as for mortgage interest, except in that case, Q increases at a rate proportional to itself (interest on the debt), but with a constant decrease on top of that (constant rate of paying the note), all in all just a different sign.


MagicJigPipe - 22-3-2010 at 21:07

Thank you very much. It would've taken me hours to figure that out.

Wait, let's say a drug had a half-life of 4 hours. Then you took 10 mg every 4 hours. The limit as t approaches infinity should be, of course, 20 mg. But, at 4 hours (right after the second dose) you should have 15 mg of drug in the bloodstream. This equation, however, gives:

12.21 mg.

I'm probably writing it down wrong. Be right back.

EDIT

For example:

Using k = 0.173 (ln(2)/4)
and d = 2.5 (10mg/4hrs)

I get the lim as t->inf is 14.45 mg

with d = 4 I get 23.12 mg

I should be getting 20 mg, right?

2nd EDIT

I have found that the equation:

d*(1-exp(-kt))+Q(0)

Gives the correct results.

3rd EDIT

Nevermind. That only works when HL = Dosing period

[Edited on 3-23-2010 by MagicJigPipe]

pantone159 - 23-3-2010 at 05:14

BTW - My equation assumes that you are dosing at a constant rate, not in discrete bits, like once every hour. The continuous case is easier to analyze :)

entropy51 - 23-3-2010 at 06:13

Pharmacokinetics

First consider the simple case of a drug that is instantaneously and completely absorbed:

C = conc in plasma = total amount in body/volume of distribution

C = Q / Vd

Vd is some multiple of body mass, varies with the drug, depends strongly on lipid solubility of drug.

Q(t) = D exp (-k * t)

where D = dose ingested at time zero, say in mg

k = ln 2 / half-life = 0.693 / thalf

Now consider the case of a drug, of which a fraction f is absorbed with an absorption half life ta and ka = 0.693/ta. The drug is eliminated with a half life te and ke = 0.693/te. Then for a single dose D the plasma conc. C is:

C(t) = (f * D/Vd)[ka/(ka-ke)]*[exp(-ke * t) - exp(-ka * t)]

For a drug dose D administered at time intervals T the plasma conc. will rise to a maximum and then fall exponentially until the next dose is given. The average plasma conc. after several doses will be:

Cavg = f * D/( Vd * ke * T)

and the maximum plasma conc. will be

Cmax = f * D /[Vd * (1 - exp(- ke* T)]

There is a nice pharmacokinetics tutorial online.



[Edited on 23-3-2010 by entropy51]

DJF90 - 23-3-2010 at 07:20

Nice maths Entropy, I noticed the ln2/halflife that is present in radioactive decay, so I would think you're along the right lines in that respect.

chief - 23-3-2010 at 08:20

Integrate it, after separation of variables ...

pantone159 - 23-3-2010 at 09:16

Quote: Originally posted by chief  
Integrate it, after separation of variables ...

Separate what variables? Everything discussed here seems to be just an ODE, not a partial differential equation.

MagicJigPipe - 23-3-2010 at 10:55

I tried that with my original stuff. I don't think it was separable.

And thanks entropy, that was very helpful as well.

[Edited on 3-23-2010 by MagicJigPipe]

Magpie - 23-3-2010 at 13:18

Quote: Originally posted by MagicJigPipe  
I am trying to solve what I think is a differential equation but I can't determine how.

I am trying to "make" an equation for the amount of drug in a user's bloodstream at various times. But the catch is that the user is taking a certain dose (d) every amount of time (t). I treated it as a rate in - rate out equation where:

dQ/dt = rate in - rate out

Q = amount of drug in user

t = time elapsed

P = dosing period


Assuming the dosing, D, is continuous and constant, and the rate of depletion is proportional to Q, ie, kQ, where k is a proportionality constant.

in time interval dt:

dQ = Ddt - kQdt; then separating and integrating (easiest for me):

dQ = (D-kQ)dt

dQ/(D-kQ) = dt

Then, going to my CRC "Standard Mathematical Tables" to get the appropriate integral:

-(1/k)ln(D-kQ) (from Qi to Q) = t (from ti to t)

where Qi is the initial drug load in the user at the initial time, ti. ti is taken as = 0. ln = natural logarithm to base e.

t= -(1/k)ln(D-kQ) - (-1/k)ln(D-kQi)

t= (-1/k)ln[(D-kQ)/D-kQi]

(D-kQ)/(D-kQi) = e^(-tk)

where e = the base of natural logarithms.

Q = [D-(D-kQi)e^(-tk)]/k