Deriving the Thin-Film Equation from Navier-Stokes Equations 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.
 A: For non-dimensionalization you are using $x = L\bar{x}$, $y = L\bar{y}$, and $z = H \bar{z}= \epsilon L \bar{z}$ where $\epsilon << 1$.
Using $U$ for the characteristic velocity scale in the $x-$ and $y-$directions  the characteristic scale in the $z-$ direction should be taken as $\epsilon U$ to ensure  that mass is conserved in the limit that $\epsilon \to 0$.  In other words, the dimensionless velocity components $(\bar{u}, \bar{v} , \bar{w})$ should satisfy
$$u = U\bar{u}, \quad v = U \bar{v}, \quad w = \epsilon U \bar{w},$$
so that the continuity equation
$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}= 0$$
assumes the same form in terms of dimensionless variables, viz.
$$\frac{\partial \bar{u}}{\partial \bar{x}} + \frac{\partial \bar{v}}{\partial \bar{y}} + \frac{\partial \bar{w}}{\partial \bar{z}}=0$$
Assuming the body force (gravity) term is absorbed into the pressure, the $x-$component of the Navier-Stokes equations is
$$\tag{*}\frac{\partial u}{\partial t} +  u \frac{\partial u}{\partial x}+ v \frac{\partial u}{\partial y}+  w \frac{\partial u}{\partial z}= -\frac{1}{\rho} \frac{\partial p}{\partial x} + \frac{\mu}{\rho}\left(\frac{\partial^2 u}{\partial x^2}+  \frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}\right)$$
Introducing dimensionless variables $\bar{x}, \bar{y}, \bar{z}, \bar{u} , \bar{v}, \bar{w}$ and the dimensionless time $\bar{t} = \frac{U}{L}t$ into (*), we get
$$\frac{U^2}{L}\left(\frac{\partial \bar{u}}{\partial \bar{t}} +  \bar{u} \frac{\partial \bar{u}}{\partial \bar{x}} +\bar{v} \frac{\partial \bar{u}}{\partial \bar{y}}+ \bar{w} \frac{\partial \bar{u}}{\partial \bar{z}}\right) = -\frac{1}{\rho L} \frac{\partial p}{\partial \bar{x}} + \frac{\mu U}{\rho L^2}\left(\frac{\partial^2 \bar{u}}{\partial \bar{x}^2}+  \frac{\partial^2 \bar{u}}{\partial \bar{y}^2}+\frac{1}{\epsilon^2} \frac{\partial^2 \bar{u}}{\partial \bar{z}^2}\right)$$
Multiplying both sides by $\frac{\rho L^2 \epsilon^2}{\mu U}$, we get
$$\tag{**}\epsilon ^2\frac{\rho L U}{\mu}\left(\frac{\partial \bar{u}}{\partial \bar{t}} +  \bar{u} \frac{\partial \bar{u}}{\partial \bar{x}} +\bar{v} \frac{\partial \bar{u}}{\partial \bar{y}}+ \bar{w} \frac{\partial \bar{u}}{\partial \bar{z}}\right) = -\frac{L\epsilon^2}{\mu U} \frac{\partial p}{\partial \bar{x}} + \epsilon^2\frac{\partial^2 \bar{u}}{\partial \bar{x}^2}+  \epsilon^2\frac{\partial^2 \bar{u}}{\partial \bar{y}^2}+ \frac{\partial^2 \bar{u}}{\partial \bar{z}^2} $$
For the thin-film approximation, we assume that $\epsilon << 1$ and also that the Reynolds number $\frac{\rho L U}{\mu}$ is sufficiently small so that $\epsilon^2 \frac{\rho L U}{\mu} << 1$. We choose a characteristic pressure scale $p_c = \frac{\mu U}{L \epsilon^2}$ so that with $p = p_c \bar{p}$, equation (**) reduces as $\epsilon \to 0$ to
$$ \frac{\partial \bar{p}}{\partial x} = \frac{\partial^2 \bar{u}}{\partial \bar{z}^2}$$
Non-dimensionalizing the pressure in this way ensures there is the correct balance between viscous stress and the pressure gradient driving the flow.
This should help you continue with the non-dimensionalization of the remaining Navier-Stokes equations.
