# PharmPK Discussion - Analytical solution for a PD model

PharmPK Discussion List Archive Index page
• On 15 Dec 2010 at 13:48:45, "Samtani, Mahesh [PRDUS]" (MSamtani.-at-.its.jnj.com) sent the message
`The following message was posted to: PharmPKHello,I am trying to find the analytical solution to this differentialequationdR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Rok1 and k2  are parameters that need to fitted, while Ro and Rmax are thebaseline and max value (which can be fitted or fixed). The response (R)increasesinitially at an exponential rate governed by the rate constants k1 andk2.Response has a S-shaped curve as a function of time and it approachesthevalue of Rmax at time approaches infinity.If there is an analytical solution to this differential equation then itmakes my life easier when trying to perform some non-linear regression.Kindly provide the integration process so I can learn how to do itmyselffor future reference. I believe that the way would beto use integration by parts (I tried hard to find the solution but keepgetting stuck).Please help,MNS`
Back to the Top

• On 16 Dec 2010 at 11:32:58, Kevin Dykstra (kevin.dykstra.at.qpharmetra.com) sent the message
`The following message was posted to: PharmPKDear Mahesh,A couple thoughts. First, that looks like a challenging differentialequation to solve, and ordinarily I would immediately resort to numericalsolution for something like this (leading to a NONMEM \$DES type of fittingproblem--very doable). Another thought would be not to mess with thedifferential equation at all, but rather to fit a closed form equation toyour response. A sigmoidal Emax model comes to mind, along the lines ofR = 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 theappropriate characteristics. The disadvantage of these is that there may notbe any particular mechanistic interpretation to these equations. On theother hand your differential equation doesn't look like it has a particularmechanism in mind either.Hope this helps.Best,KevinKevin Dykstra, PhD, FCPqPharmetra`
Back to the Top

• On 17 Dec 2010 at 07:05:55, Nick Holford (n.holford.-at-.auckland.ac.nz) sent the message
`The following message was posted to: PharmPKMahesh,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`
Back to the Top

• On 16 Dec 2010 at 14:07:23, "Samtani, Mahesh [PRDUS]" (MSamtani.aaa.its.jnj.com) sent the message
`The following message was posted to: PharmPKDear Dr. Dykstra,I really appreciate your insightful comments. This type of model isuseful for describing tumor growth. I have used:Previous model:dR/dt = k1*R*(1-(R/Rmax)^k2); R(0) = RoCurrent model:dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = RoThe first model is called the Richard's function which has an explicitfunction [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 fitsbetter. Both models have a nice property called the inflection point(for example the inflection point for the current model isk2*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 trialsimulation and wanted to use Splus (I don't know how to use ODEs inSplus). So for CTS, Splus would have to call NM for the \$DESfunctionality. Therefore, I wish I had an explicit function for thecurrent model.Thank-you,MaheshPS. I did get some feedback from other PharmPK users who have suggestedthat a solution may exist for certain values of k1 and k2. The value ofk2 in my model is 1.5 and I still don't know if a solution exists withthis particular value of k2`
Back to the Top

• On 16 Dec 2010 at 15:48:36, Leonid Gibiansky (LGibiansky.aaa.quantpharm.com) sent the message
`The following message was posted to: PharmPKEquationdR/dt = R^1.5*(1-R); R(0) = Rocan be integrated. Indeed:Int(dR/[R^1.5(1-R)])- 1/[R0^1.5(1-R0)] = tThen the problem is to take the integral1/[R^1.5*(1-R)]that can be done using the substitutionR = x^2dR / [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 of1/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 expressionF(R) = tthat will need to be solved for R in order to get R(t). I have not  checked how easy or difficult it will be.LeonidGibiansky, QuantPharm LLCweb:    www.quantpharm.com`
Back to the Top

• On 16 Dec 2010 at 16:33:20, Peter Bonate (peter.bonate.aaa.gmail.com) sent the message
`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`
Back to the Top

• On 17 Dec 2010 at 07:40:05, Kevin Dykstra (kevin.dykstra.at.qpharmetra.com) sent the message
`The following message was posted to: PharmPKMahesh,Thanks for the context -- that helps. I can certainly sympathize with thechallenges of moving between different software tools. Regarding theinflection point, I would point out that most sigmoidal relationships willalso have an inflection point, including the Emax type relationship Idescribed earlier. As you probably remember from calculus, this is the pointwhere the 2nd derivative w/r time is 0. It sounds like you are referring tothe tumor size where this inflection occurs, or R(t-inflection). Indeed,without looking closely, I suspect the inflection point in the Richard'sfunction is not the same as in your alternative model, so you are fittingsomething somewhat different anyway (might be why the 2nd model fitsbetter!). Regarding the CTS, it is certainly possible to call NONMEM fromSPLUS, and might actually be a faster alternative if the simulation islarge, since fortran handles the memory a little more efficiently than S.Good Luck!Kevin Dykstra, PhD, FCPqPharmetra`
Back to the Top

 Want to post a follow-up message on this topic? If this link does not work with your browser send a follow-up message to PharmPK@boomer.org with "Analytical solution for a PD model" as the subject Support PharmPK by using thelink to buy books etc.from Amazon.com

Copyright 1995-2011 David W. A. Bourne (david@boomer.org)