I'm trying to practice my ability to non-dimensionalize equations, and I am interested whether we can always non-dimensionalize certain differential equations.
Take the system of equations:
\begin{equation} \frac{dR}{dt} = \phi R(\frac{\alpha}{\beta} - 1) - \delta R\\ \frac{dS}{dt} = \pi \rho R - \psi S \end{equation}
where $R$ is some resource (e.g. measured in kilograms), $S$ is some stock of product (measured in product units), $\phi$ is the rate of resource growth (with dimensions 1/time), $\alpha$ is the product price (price per kg), $\beta$ is the cost of resource production (price per kg), $\delta$ is a resource loss rate (dimensions 1/time), $\pi$ is a rate of stock production (dimensions 1/time), $\rho$ is a conversion factor from resource into product units (dimensions = product units/kg) and $\psi$ is a product loss rate (e.g. consumption, 1/time).
Introducing the new variables:
\begin{equation} R = R^* \hat{R}, S = S^* \hat{S}, t = t^* \hat{t} \end{equation}
where terms with an asterisks are the scaling parameters and those with a hat are new dimensionless variables. I re-write the system as:
\begin{equation} \frac{dR^* \hat{R}}{dt^* \hat{t}} = \phi R^* \hat{R}(\frac{\alpha}{\beta} - 1) - \delta R^* \hat{R}\\ \frac{dS^* \hat{S}}{dt^* \hat{t}} = \pi \rho R^* \hat{R} - \psi S^* \hat{S} \end{equation}
After simplifying we have:
\begin{equation} \frac{d\hat{R}}{d\hat{t}} = \phi t^* \hat{R}(\frac{\alpha}{\beta} - 1) - \delta t^* \hat{R}\\ \frac{d\hat{S}}{d\hat{t}} = \pi \rho R^* t^* \hat{R} \frac{1}{S^*} - \psi t^* \hat{S} \end{equation}
We could define the dimensionless time scale as $\hat{t} = \phi t$ but how could we make $\hat{R}$ and $\hat{S}$ dimensionless?
For instance, defining $S^* = \rho$ makes the units of $\hat{S} =$ product units/(product units/kg) = kg. I also can't see how to define $R^*$.
Does anyone have any ideas?
That depends completely on your personal choice of how the normal form should look like. With your first decision $ϕt^*=1$ the first equation is completely normalized, there remain no free parameters. $$ \frac{d\hat R}{d\hat t}=(\tfracαβ-1)\hat R - \tfracδϕ\hat R $$ In the second equation one could now likewise demand $S^*=πρR^∗t^∗$ to get the first coefficient to $1$, then there still remains one degree of freedom between $R^*$ and $S^*$. The coefficient of the second term does not contain any of them, so that this remains a free scale factor. You could set $R^*=R_0$ with the initial value, so that $S^*=\frac{πρ}ϕR_0$. $$ \frac{d\hat S}{d\hat t}=\hat R-\tfracψϕ\hat S $$