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Analyzing solutions of y'= r-ky using W lambert function?

  1. Jul 14, 2014 #1
    consider this IVP
    y'=r-ky , y(0)=y0
    y= (y0)e^(-kt) + (r/k)(1-e^(-kt))
    if y,y0,r,t are provided, we should be able to solve for k and that's the problem but what I'm really interested is analyzing this problem

    if we let y=0.99 (r/k) find t in terms of all other variables

    Of courses, if y0 = 0 we can see that t= -(ln 0.01)/k

    i wonder if y0 is not zero is it possible to analyze variable t using any knowledge or technology from mathematics?

    this problems is derived from real application of pharmacokinetics IV infusion where

    y= amount of drug at time t
    y0 = initial amount of drug at t = 0
    r = infusion rate
    k = elimination rate constant
    (r/k) = amount of drug at steady-state (as t --> infinity the amount of drug will approach this value and 99% of (r/k) is a good approximation of amount of drug at steady-state)

    so what I really ask is how the initial amount of drug reflects the time to reach steady-state ( for example. how much we increase y0 in order to halve the time to reach steady-state )
  2. jcsd
  3. Jul 23, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
  4. Jul 24, 2014 #3
    Let's see what I can whip up...
    [itex]y' + ky = r[/itex] is a first-order, linear, non-homogenous ODE with [itex]k \neq 0[/itex] and [itex]r \neq 0[/itex], both of which I assume to be constant.
    The general solution is (by partitioning into homogenous and particular solutions) [itex]C_0 e^{-kt} + \frac{r}{k}[/itex]. Given an initial value, [itex]y(0) = y_0[/itex], the unique solution is [itex](y_0 - \frac{r}{k})e^{-kt} + \frac{r}{k}[/itex]. (Same as yours, but rewritten.)

    You then ask what I assume to be is the time at which this solution is equal to [itex]0.99\frac{r}{k}[/itex].

    [tex](y_0 - \frac{r}{k})e^{-kt} + \frac{r}{k} = 0.99\frac{r}{k}[/tex]
    [tex](y_0 - \frac{r}{k})e^{-kt} + 0.01\frac{r}{k} = 0[/tex]
    [tex](y_0 - \frac{r}{k})e^{-kt} = -0.01\frac{r}{k}[/tex]
    [tex]e^{-kt} = -0.01\frac{\frac{r}{k}}{y_0 - \frac{r}{k}}[/tex]
    [tex]e^{-kt} = -0.01\frac{r}{k y_0 - r}[/tex]
    Immediately, we see that [itex]\frac{r}{k y_0 - r} < 0[/itex] for there to be a real-valued, finite solution.

    It's worth noting that if [itex]k = 0[/itex], we get a linear polynomial as a solution to the ODE, for which there is no steady state unless [itex]r = 0[/itex] as well, in which case it's [itex]y_0[/itex] as the solution is just a constant. If [itex]k \neq 0[/itex] and [itex]r = 0[/itex], the only steady state would be 0, for which your approximation fails to give a finite solution to the problem. If [itex]k y_0 - r = 0[/itex], the solution to the ODE would be [itex]\frac{r}{k}[/itex], which has a steady-state but the approximation fails as well.

    In any case, assuming [itex]\frac{r}{k y_0 - r} < 0[/itex], the solution is:
    [tex]e^{-kt} = -0.01\frac{r}{k y_0 - r}[/tex]
    [tex]-kt = \ln{\left(-0.01\frac{r}{k y_0 - r}\right)}[/tex]
    [tex]t = -\frac{\ln{\left(-0.01\frac{r}{k y_0 - r}\right)}}{k} = -\frac{\ln{\left(0.01\frac{r}{r - k y_0}\right)}}{k}[/tex]
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