I am running numerical simulations of a 2 dimensional ODE system such that:
$$ \frac{\partial M}{\partial t}= (1-e^{-M\phi})\alpha U M - \beta D M\\ \frac{\partial D}{\partial t}=\beta D M - \gamma D $$ where the variables $\{M,U,D\}\in[0,1]$ and $M+U+D=1$
All parameters ($\alpha,\phi,\beta,\gamma$) are independent of $t$.
When $\phi$ approaches zero, the above system behaves exactly as the system below: $$ \frac{\partial M}{\partial t}= \alpha U M^2 - \beta D M\\ \frac{\partial D}{\partial t}=\beta D M - \gamma D $$
My college maths date far back, but if I am not missing anything $\lim_{\phi\to0} 1-e^{-M\phi}=0$. Then could the behaviour of the system when $\phi$ approaches zero be related to the bounded nature of M,U, and D?
Although this result matches my expectations, I am clueless as to how this occurs...
Thanks in advance if anyone has a clue.
Expanding out as a Taylor series, $$e^{-M \phi} = 1 -M \phi + \frac {(M \phi)^2}{2!}-... $$and so $$1-e^{M \phi} = M \phi - \frac {(M \phi)^2}{2!}+... $$ As $\phi$ approaches $0$, $1-e^{M \phi}$ approaches $M \phi$, as the higher order terms are negligibly small in comparison. This means that: $\frac{1-e^{-M \phi}}{\phi} $ approximates $M$ as $\phi $ goes to $0$, so no, the statement in your title isn't generally true.