I am really stuck in this problem, please help me out.
From my model i get following differential equation,
$\frac{df(t)}{dt} = 4 f(t) (1-X^{-1}-f(t))$
It is a sigmoidal equation and gives following solution,
$ f(t) = \frac{A}{1-e^{-k_{app} (t - t_{max})}} \\ $
Where,
$A = 1-X^{-1} \rightarrow$ equilibrium value i.e. saturation value.
$k_{app} = 4 (1-X^{-1}) \rightarrow$ Effective growth rate.
$t_{max} = \frac{1}{4 (1-X^{-1})} \ln\left( \frac{1-X^{-1}-f(0)}{f(0)}\right) \rightarrow$ time at which maximum growth rate occurs.
From this i calculation lag time i.e. time after which growth starts.
$t_{lag} = \frac{1}{4(1-X^{-1})} \ln \left( \frac{1-X^{-1}-f(0)}{f(0)}\right) - \frac{1}{2 (1-X^{-1})} $
But the property of lag time for my system is that for $X \rightarrow \infty$ my $\t_{lag}=0$ and hence i have to subtract $\frac{1}{4} \ln \left( \frac{1-f(0)}{f(0)} \right) -\frac{1}{2}$ from my lag time and then my model perfectly fits with the experimental data.
But now i have a problem that in order to accomodate my actual lag time in sigmoidal solution i have to renormalize my $t_{max}$ and $k_{app}$ as follows,
$t_{max} = \frac{1}{4 (1-X^{-1})} \ln\left( \frac{1-X^{-1}-f(0)}{f(0)}\right) - \frac{1}{4} \ln \left( \frac{1-f(0)}{f(0)} \right)$
$k_{app} = 4 (X-1)$
After these adjustments to the final solution to logistic equation my entire model fits experimental data just fine.
But then I cant seem to figure out that what kind of differential equation will give me the solution that fits my model after renormalizing $t_{max}$ and $k_{app}$.
Kindly help me out with this. Its almost killing my more than a years work.
Thanks,
Nitin