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The following message was posted to: PharmPK
Hello,
I am trying to find the analytical solution to this differential
equation
dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Ro
k1 and k2 are parameters that need to fitted, while Ro and Rmax are the
baseline and max value (which can be fitted or fixed). The response (R)
increases
initially at an exponential rate governed by the rate constants k1 and
k2.
Response has a S-shaped curve as a function of time and it approaches
the
value of Rmax at time approaches infinity.
If there is an analytical solution to this differential equation then it
makes my life easier when trying to perform some non-linear regression.
Kindly provide the integration process so I can learn how to do it
myself
for future reference. I believe that the way would be
to use integration by parts (I tried hard to find the solution but keep
getting stuck).
Please help,
MNS
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The following message was posted to: PharmPK
Dear Mahesh,
A couple thoughts. First, that looks like a challenging differential
equation to solve, and ordinarily I would immediately resort to numerical
solution for something like this (leading to a NONMEM $DES type of fitting
problem--very doable). Another thought would be not to mess with the
differential equation at all, but rather to fit a closed form equation to
your response. A sigmoidal Emax model comes to mind, along the lines of
R = Rmax*(t^hill) / (t50^hill + t^hill)
This has three parameters, but I wouldn't see a particular problem there.
Alternatively, I can easily imagine a combination of exponentials giving the
appropriate characteristics. The disadvantage of these is that there may not
be any particular mechanistic interpretation to these equations. On the
other hand your differential equation doesn't look like it has a particular
mechanism in mind either.
Hope this helps.
Best,
Kevin
Kevin Dykstra, PhD, FCP
qPharmetra
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The following message was posted to: PharmPK
Mahesh,
There is no closed form solution to this kind of ODE. However, there are some ways of solving it that are faster than using conventional ODE solvers. However such methods are only readily available in specialized software such as MatLab. Steve Duffull has worked with these methods so you may want to contact him for details.
Nick
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The following message was posted to: PharmPK
Dear Dr. Dykstra,
I really appreciate your insightful comments. This type of model is
useful for describing tumor growth. I have used:
Previous model:
dR/dt = k1*R*(1-(R/Rmax)^k2); R(0) = Ro
Current model:
dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Ro
The first model is called the Richard's function which has an explicit
function [R = Rmax *(M/(M-1))^(1/K2); where M ((R0^K2)/((R0^K2)-(Rmax^K2)))*exp(K1*K2*time)]
Unfortunately based on objective function value the current model fits
better. Both models have a nice property called the inflection point
(for example the inflection point for the current model is
k2*Rmax/(1+k2)). The inflection point has a nice mechanistic meaning.
Once the tumor exceeds a certain critical size (i.e. inflection point)
the growth slows down due to lack of space, shortage of nutrients etc.
I have the model working in NONMEM. I want to take it to clinical trial
simulation and wanted to use Splus (I don't know how to use ODEs in
Splus). So for CTS, Splus would have to call NM for the $DES
functionality. Therefore, I wish I had an explicit function for the
current model.
Thank-you,
Mahesh
PS. I did get some feedback from other PharmPK users who have suggested
that a solution may exist for certain values of k1 and k2. The value of
k2 in my model is 1.5 and I still don't know if a solution exists with
this particular value of k2
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The following message was posted to: PharmPK
Equation
dR/dt = R^1.5*(1-R); R(0) = Ro
can be integrated. Indeed:
Int(dR/[R^1.5(1-R)])- 1/[R0^1.5(1-R0)] = t
Then the problem is to take the integral
1/[R^1.5*(1-R)]
that can be done using the substitution
R = x^2
dR / [R^1.5(1-R)] = 2x dx/ (x^3*(1-x^2)] 2 dx/ (x^2*(1-x^2)]
The rest is algebra that should result in combination of
1/x and ln(1+x) and ln(1-x)
k1 and Rmax can be added to the process easily (although more tediously).
You will end up with the expression
F(R) = t
that will need to be solved for R in order to get R(t). I have not checked how easy or difficult it will be.
Leonid
Gibiansky, QuantPharm LLC
web: www.quantpharm.com
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When I have a problem like this I look at Maxima (http://maxima.sourceforge.net/), a freeware computer algebra program that can do many functions, not just integration, to see if it can find a solution. It's pretty handy. And free.
Peter Bonate
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The following message was posted to: PharmPK
Mahesh,
Thanks for the context -- that helps. I can certainly sympathize with the
challenges of moving between different software tools. Regarding the
inflection point, I would point out that most sigmoidal relationships will
also have an inflection point, including the Emax type relationship I
described earlier. As you probably remember from calculus, this is the point
where the 2nd derivative w/r time is 0. It sounds like you are referring to
the tumor size where this inflection occurs, or R(t-inflection). Indeed,
without looking closely, I suspect the inflection point in the Richard's
function is not the same as in your alternative model, so you are fitting
something somewhat different anyway (might be why the 2nd model fits
better!). Regarding the CTS, it is certainly possible to call NONMEM from
SPLUS, and might actually be a faster alternative if the simulation is
large, since fortran handles the memory a little more efficiently than S.
Good Luck!
Kevin Dykstra, PhD, FCP
qPharmetra
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