fluid dynamics - Is there any way to get around the rule that drag $propto v^2$?

I was looking at a question on aviation.stackexchange and an interesting answer I found included that even today, we have no physical way to get around the fact that drag of a body is roughly the square of velocity. Theoretically is there any way to get around this?

Answer

It comes from the Navier-Stokes equation.

More specifically, it's the ram pressure term which goes as $-\nabla (\frac{1}{2}\rho u^2)$. This gives a force per unit mass, since the LHS o the N-S equation is usually $\rho \frac{\partial \mathbf{u}}{\partial t}$.

The ram pressure is also referred to inertial terms.

Physically, in order to move at speed $u$, you need to "move away" the air in front of you, i.e. you need to provide it with kinetic energy $\frac{1}{2}m u^2$.

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Maths

Let's consider the total derivative of the fluid velocity $\mathbf{v}$ in the presence of a pressure gradient $\nabla P$ and an external field $\phi$, which could be gravity:

Let's use tensor notation:

$$\frac{\mathrm{d} (\rho v_i)}{\mathrm{d} t} = \rho \frac{\mathrm{d} v_i}{\mathrm{d} t} + v_i \frac{\mathrm{d}\rho}{\mathrm{d} t}.$$ The equation of motion is $$\frac{\mathrm{d}(\rho\mathbf{v})}{\mathrm{d} t} = -\nabla P - \rho \nabla \phi,$$ and the total derivative $\mathrm{d}_t = \partial_t + v_i \partial_i$.

Hence we get:

$$\frac{\partial \rho v_i}{\partial t} = -\rho v_j \partial_j v_i - \partial_i P -\rho \partial_i \phi - v_i \partial(\rho v_j) \\ = -\partial_j(\rho v_j v_i + P\delta_{ij}) -\rho \delta_i \phi$$ so that $$\frac{\partial \rho v_i}{\partial t} = -\partial_j T_{ij} - \rho \partial_i \phi.$$

$T_{ij}$ is the stress energy tensor, the first term of which is the ram pressure due to the bulk motion of a fluid, and the second term is the pressure introduced by thermal effect (e.g. denser regions).