Why to write the Navier-Stokes equation with dimensionless quantities?

Question

The Navier-Stokes equation is

$$\rho \dfrac{D\mathbf{u}}{Dt} = -\nabla p+(\lambda+\mu)\nabla(\nabla\cdot\mathbf{u})+\mu\nabla^2\mathbf{u}$$

Then if the flow is incomprresible, and the fluid is homogenous ($\rho$ is constant in space) then $\rho$ will be constant in time and $\rho = \rho_0$ so that we can write this as

$$\rho_0 \dfrac{D\mathbf{u}}{Dt} = - \nabla p + \mu\nabla^2\mathbf{u}$$

From this we can introduce numbers $U$ and $L$ in order to write $\bar{x}_i = x_i/L$ and $\bar{\mathbf{u}} = \mathbf{u}/U$ so that $\bar{t} = t/T$ where $T = L/U$. Then we obtain the equation

$$\dfrac{\partial \bar{\mathbf{u}}}{\partial \bar{t}} + (\bar{\mathbf{u}}\cdot \nabla)\bar{\mathbf{u}} = -\nabla \bar{p} + \dfrac{1}{R}\nabla^2\bar{\mathbf{u}}$$

Where $R = LU/\nu$ is the Reynolds number. That's fine, but what's the motivation to write the equation like that? What do we gain from just changing the coordinates by rescaling them? Many books says that the importance of the Reynolds number comes from writing the equation with dimensionless quantities, but why is that?

Answer 1

Every time, when you deal with differential equations, the first step is to put it into dimensionless form. There are more reasons for that. First, "small" and "large" has no meaning in dimensional forms, since you can always change the system of units. Second, nature knows no units.

Now, when there is no exact solution (this is often the case), you can neglect terms, if they are small, and you get an approximation to your problem. This leads us to perturbation theory. For instance, you get Stokes flow when the Reynolds number is much smaller than unity, and the linear viscous terms are much larger than the non-linear terms, therefore you can find asymptotic solutions for complex problems.

In $\Pi$-theorem you can also derive relationships based on units of quantities.

In many cases of fluid dynamics, relationships have only dimensional reasons.