What does it mean for a difference equation to be made dimensionless? Given the logistic difference equation, what does it mean for it to be made "dimensionless" so that it's written in a simpler form?
$${N_{n+1}-N_n\over \Delta t}=r\cdot N_n\cdot \left(1-{N_n\over K}\right)$$
 A: Typically differential equations from physical modeling come equipped with parameters which have units. The most basic case is when $x$ has some unit (length, population, ...) and $t$ has the usual units of time. If you rescale $x$ and $t$ in an appropriate way, you can cancel out some of the parameters, in the process making the actual parameters in the new problem have no units.
For instance let's consider the logistic equation:
$$\frac{dx}{dt} = rx \left (1-\frac{x}{M} \right )$$
where $r>0$ is the growth parameter and $M>0$ is the carrying capacity. Now introduce dimensionless variables $y,s$ so that $x=x_C y$ and $t=t_C s$. (I read the "C" as "characteristic", so $t_C$ is the characteristic time scale, etc.) Now using the chain rule you get
$$\frac{dx}{dt} = \frac{dy}{ds} \frac{ds}{dt} \frac{dx}{dy} = \frac{dy}{ds} \frac{x_C}{t_C}.$$
Rearranging:
$$\frac{dy}{ds} = \frac{r t_C}{x_C} x_C y \left ( 1 - \frac{x_C y}{M} \right ) = r t_C y \left ( 1 - \frac{x_C}{M} y \right )$$
Now we want to cancel parameters by choosing $t_C,x_C$ appropriately. In this case it is easy to see what to do: take $t_C=\frac{1}{r}$ and $x_C=M$, then you get
$$\frac{dy}{ds} = y(1-y).$$
The situation is similar for difference equations, provided there is some timescale parameter $h$ or $\Delta t$ or similar.
I think the most interesting application of this procedure is in perturbative problems. For instance, consider a spring-mass system with external force $F$ and small friction:
$$m \frac{d^2 x}{dt^2} + c \frac{dx}{dt} + k x = F.$$
Doing the same procedure as before you get
$$\frac{m x_C}{t_C^2} \frac{d^2 y}{ds^2} + \frac{c x_C}{t_C} \frac{dy}{ds} + k x_C y = F.$$
The philosophy is that we keep the small parameters and rescale so that the larger parameters are $1$. So here we need $x_C = \frac{1}{k}$ and $t_C = \sqrt{\frac{m}{k}}$. The remaining parameter is $\frac{c}{\sqrt{mk}}$. Calling this $\varepsilon$ the equation becomes
$$\frac{d^2 y}{ds^2} + \varepsilon \frac{dy}{ds} + y = F.$$
So now, using only dimensional analysis, we have seen that the qualitative effect of the friction depends only on the parameter $\frac{c}{\sqrt{mk}}$. This is hinting at a fact that in this simple case is easy to see: the qualitative behavior of the system really depends on the discriminant $c^2-4mk$.
