# Gompertz growth equation

:)

Hi! I'm almost finished with a homework problem, but I cannot quite finish it. The problem is as follows:

Given the Gompertz growth equation $$\frac{dN}{dt}=K(t)N(t),\ N(0)=N_0 \\ \frac{dK}{dt}=-\alpha K(t),\ K(0)=\beta,$$ I shall first determined a closed-form expression, which will be $K(t)=\beta e^{-\alpha t}$ and $$\frac{dN}{dt}=\beta e^{-\alpha t} N(t) \Leftrightarrow \\ \int\frac{1}{N(t)}dN=\int \beta e^{-\alpha t}dt \Leftrightarrow \\ \log[N(t)]=\frac{-\beta}{\alpha} e^{-\alpha t} + C \Leftrightarrow \\ N(t)=\exp\left( \frac{-\beta}{\alpha}e^{-\alpha t}\right) \frac{N_0}{\exp\left(\frac{-\beta}{\alpha}\right)};$$ the last factor is due to the initial condition.

Then: Determine the limit $$B:=\lim_{t\rightarrow \infty}N(t),$$ it will be $B=\frac{N_0}{\exp\left(\frac{-\alpha}{\beta}\right)}$ (right?)

Then: Show that this is equivalent to $$\frac{dN}{dt}=\alpha N \log\frac{B}{N},$$ I also managed that (the LHS is $KN$, then solve the logarithm), but THEN:

For the cases $N_0>B, N_0 <B$, determine $\lim_{t\rightarrow -\infty }N(t)$. If I plug in $t=-\infty$ into $N(t)$, then I get just $0$. Or not? And with my solution for $B$, I get $\exp\left( \frac{-\beta}{\alpha}\right)<> 1$, but since $\alpha, \beta >0$, the exp cannot be smaller than one?

I either have a blonde moment, or my solutions are wrong, but magically the equivalence in the $\log$-equation works! Can someone defuse my brain? :)

-marie

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Should be $B = N_0/\exp(-\beta/\alpha) = N_0 \exp(\beta/\alpha)$. – Robert Israel May 7 '12 at 23:35
If $\beta/\alpha > 0$, there is no case $N_0 > B$. – Robert Israel May 7 '12 at 23:43
Gompertz? Gesundheit! – rschwieb May 8 '12 at 1:41
So... does that mean that the question is wrong? :/ – Marie. P. May 8 '12 at 9:22