I want to start with the general Navier-Stokes equation in 3-D: $$ \frac{\partial \vec{u}}{\partial t} + (\vec{u}\cdot\nabla)\vec{u}=-\frac{1}{\rho}\nabla p+\frac{\mu}{\rho}\nabla^2\vec{u}+g\sin{\alpha} \vec{i} -g\cos{\alpha} \vec{k} $$ $$ \nabla\cdot\vec{u}=0 $$ and derive the thin-film equation: $$ \frac{\partial h}{\partial t} =\frac{1}{3\mu}\nabla\cdot[h^3(\rho g\cos{\alpha}\nabla h-\frac{\mu}{\rho}\nabla\kappa-\rho g\sin{\alpha} \vec{i}] $$

I understand that this process is nothing more than using non-dimensionalization and magnitude approximations of the relevant terms to simplify, however, I am getting lost in the details.

The first thing I did was create a characteristic scale for $x,y,z$ where $L$ and $H$ serve as characteristic scales for the in-plane coordinates and normal coordinate respectively. I also rewrote the velocity vector as $\vec{u}=<\vec{v},w>$. Now, we have a thin film so we can let $\epsilon=\frac{H}{L}<<1$ since $H <<L$ (by definition of a thin-film). So it follows that: $$\frac{\partial^2\vec{v}}{\partial x^2}\approx\frac{\partial^2\vec{v}}{\partial y^2}=\frac{1}{L^2}\frac{\partial^2\vec{v}}{\partial\bar{y}^2}<<\frac{1}{H^2}\frac{\partial^2\vec{v}}{\partial\bar{z}^2}=\frac{\partial^2\vec{v}}{\partial\bar{z}^2} $$

This then tells us that we can neglect the in-plane (x and y) partial derivatives since they are much smaller in magnitude. Thus we have our first approximation: $$ \frac{\mu}{\rho}\nabla^2\vec{u} \approx \frac{\mu}{\rho}\frac{\partial^2\vec{u}}{\partial\bar{z}^2} $$ The book I am trying to follow, Elementary Fluid Dynamics by D.J. Acheson, then goes on to describe how $$ (\vec{u}\cdot\nabla)\vec{u} \sim \frac{U^2}{L} (1,1,\frac{H}{L}) $$ is of the given magnitude. He then goes on to just conclude that this term can be neglected completely. I am having a little bit of trouble grasping these steps and their implications. Help would be much appreciated!

It is also right around this step, that the assumption is made that pressure is only a function of $x$ and $y$, not $z$. This seems like a pretty crucial step in reducing the number of equations, but it doesn't really seem too intuitive to me.

  • $\begingroup$ Please let me know if this question feels too vague as currently written. Perhaps I can add some extra details if neeed. $\endgroup$
    – joseph
    Jun 16 at 13:56

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