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I am wondering how to use the Gompertz model in modelling drug diffusion:
$$x(t) = x_\mathrm{max}\exp(-a\exp(b\log t)).$$
$x_\mathrm{max}$ is to be taken as $100,$ $a$ is the undissolved proportion at time $t = 1.$ But what is $b?$ I've seen it variously described as half of the max asymptote, the $m$ of the linear portion of the graph, or the $m$ divided by $\mathrm e.$
I can see at Gompertz function they use the formula below. See the link for Graphs of Gompertz curves, showing the effect of varying one of a,b,c while keeping the others constant
$$f(t)=a\mathrm{e}^{-b\mathrm{e}^{-ct}}$$ where
- ''a'' is an asymptote, since $ \lim_{t \to \infty} a\mathrm{e}^{-b\mathrm{e}^{-ct }}=a\mathrm{e}^0=a$
- ''b'' sets the displacement along the ''x''-axis (translates the graph to the left or right). When ''b'' = log(2), f(0) = a/2, also called the halfway point.
- ''c'' sets the growth rate (''y'' scaling)
- ''e'' is Euler's Number (''e'' = 2.71828...)
The halfway point is found by solving $f(t) = a/2$ for $t$. $$t_{hwp} = -\frac{\ln(\ln(2)/b)}{c}$$
Regardless of if the equation $$x(t) = x_\mathrm{max}\exp(-a\exp(b\log t)).$$ in the question is the true one, analyzing the equation as is, being aware ift has different shape than the above.:
It can be rewritten as $$x(t)=x_\mathrm{max} \cdot \exp{\left(-a \cdot t^b\right)}$$
If $\ce{b=1}$, the equation reduces to the classical equation of the kinetic of the first order and the half-time $t_{1/2}=\frac{\ln 2}{a}$, where an asymptote at any point in time crosses x axis after $t_{1/2}$.
$$x(t)=x_\mathrm{max} \cdot \exp{\left(-a \cdot t \right)}$$
The parameter $\mathrm{b}$ affects the time evolution of the diffusion kinetic model..
$b = 1$ keeps the relative decrease with time constant.
$b \gt 1$ accelerates the relative decrease with time.
$b \lt 1$ decelerates the relative decrease with time.
Illustrative table about the $b$ value influence.
$\begin{array}{|c|c|c|c|} \hline & b & b & b \\ \hline \hline t & 0.8 & 1 &1.2 \\ 0 & 100 & 100 & 100 \\ 1 & 50.0 & 50.0 & 50.0 \\ 2 & 29.9 & 25.0 & 20.3 \\ 3 & 18.8 & 12.5 & 7.5 \\ \hline \end{array}$
$t$ was padded by small value to avoid log 0. Results were rounded.
$a$ was set to $\ln{2}$ to have $t_{1/2}=1$ for $b=1$.
The Gomperts rate eqn for a species population $n$ can be written as $\displaystyle \frac{dn}{dt}=-kn\ln\left(\frac{n}{a}\right)$ with initial values at time zero, $n=n_0$ and $a>0,\; k>0$. The solution is
$$ n=a e^{\ln(n_0/a)e^{-kt} }$$
At long times $t\to\infty$ the inner exponential tends to zero and so $a$ represents the limiting or maximum population $n\to a$. The rate constant $k$ limits how quickly the final population is reached.
The equation can also be written as $\displaystyle n=a(n_0/a)^{e^{-kt}}$ or $\displaystyle \ln(n/a)=\ln(n_0/a)e^{-kt}$ which may be easier to understand.
