I am trying to non-dimensionalize the following differential equation. I am a bit stuck and unsure about how to proceed. The model is the logistic model where:
$$\frac{dP}{dT} = rP(K - P) $$
I am told that $K$ is a function of time where $K(t) = K_0 + Af(T)$ and that $A$ is a parameter where $ |A| << K_0 $, and that $f(T)$ is a dimensionless function of time. Also, we assumed that $P(T=0) = K_0$
I am told to derive the model to the form: $$\frac{dp}{dt} = p(1+\epsilon(g(t)) - p)$$
with $p(0) = 1$. I think it is pretty obvious that I am supposed to divide by $K_0$ but I am not sure how the new parameter $p$ is defined. Would I also have to divide by some parameter for time as well? I tried dividing by $r$ as well but then I couldn't come up with a definition that matches for the change of parameter in $p$ which I assume that it is also redefined.
So far, I have thought the following where $\epsilon = \frac{A}{K_0}$ and am unsure about this variable where: $t = \frac{T}{r}$. This is the only way I can some how reconcile a way for $P$ to become $p$ and $f(T)$ to become $g(t)$.
Some help would be greatly appreciated.
Thanks
Since your equation describes logistic growth, I assume that $P$ has the dimension `population density' or 'population number'. Anyway, since $P$ is substracted from $K$, the dimension of $K$ must be the same as that of $P$. Let's denote the dimension of $P$ by $[P]$. Then, we see that the right hand side of the ODE has dimension $[r] [P]^2$ -- we don't know the dimension of $r$ just yet, so we denote it by $[r]$.
The left hand side of the ODE has dimension '$[P]/[T]$'; we denote the dimension of $T$ by $[T]$ (this can be minutes, seconds, hours, whatever time unit you choose). The dimensions of both sides of the ODE must of course match, so we can deduce that \begin{equation} [r] = \frac{1}{[P][T]}. \tag{1} \end{equation}
Also, the variable $K$ (which has the same dimension as $P$), is written as a sum: $K= K_0 + A f$. Since $f$ is given to be dimensionless, we see that $K_0$ and $A$ must have the same dimension, and that dimension is therefore equal to the dimension of $P$.
You have the ODE \begin{equation} \frac{\text{d} P}{\text{d} T} = r P (K - P) = r P (K_0 + A f - P) \tag{2} \end{equation} and are told to derive the ODE \begin{equation} \frac{\text{d} p}{\text{d} t} = p(1+ \epsilon g - p).\tag{3} \end{equation} From the form of $(2)$, it seems a good idea to divide both sides of $(2)$ by $K_0$. You then obtain \begin{equation} \frac{\text{d}}{\text{d} T} \frac{P}{K_0} = r P \left(1 + \frac{A}{K_0}f - \frac{P}{K_0}\right). \end{equation} Since $K_0$ has the same dimension as $P$, the fraction $\frac{P}{K_0}$ is dimensionless. So, it seems a good idea to introduce $p = \frac{P}{K_0}$. In this new variable, you obtain the ODE \begin{equation} \frac{\text{d} p}{\text{d} T} = r K_0\, p \left(1 + \frac{A}{K_0} f - p\right). \tag{4} \end{equation} Now, we tackle the time dimension. From the dimension of $r$ as given in $(1)$, we see that \begin{equation} [r K_0] = [P] \frac{1}{[P] [T]} = \frac{1}{[T]}. \end{equation} So, by dividing both sides of $(4)$ by $r K_0$, we obtain the equation \begin{equation} \frac{1}{r K_0} \frac{\text{d} p}{\text{d} T} = p \left(1 + \frac{A}{K_0} f - p\right). \tag{5} \end{equation} Therefore, the proper way to nondimensionalise the time variable $T$ is to introduce $t = r K_0\, T$, such that $(5)$ becomes \begin{equation} \frac{\text{d} p}{\text{d} t} = p \left(1 + \frac{A}{K_0} f - p\right). \tag{6} \end{equation} Now, introducing the (nondimensional) constant $\epsilon = \frac{A}{K_0}$ and writing $f(T) = f(\frac{t}{r K_0}) := g(t)$, you obtain the required form.
Note: I more or less skipped over this, but from the above, you see that the proper definition of $p$ is $p(t) = \frac{1}{K_0} P(\frac{t}{r K_0})$.