# Navier-stokes simplifications

1. Nov 30, 2013

### joshmccraney

hey pf!

so i have a question concerning navier-stokes equations in a boundary layer, which, as a refresher, is $$\frac {D \vec{V}}{Dt} = - \nabla P + \nu \nabla^2 \vec{V}$$ where we know the x-component of $\nabla^2 \vec{V}$ may be re-wrote as $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$ (i dismiss the $z$ component due to two dimensions). define the vertical displacement as $\delta$ and the horizontal length scale as $L$

okay, now for the question: we know $\frac{\partial^2 u}{\partial x^2}$ disappears, as it is not of much importance via order of magnitude. my question is, in boundary layer analysis, is this true from the fact that $\delta < L \implies \delta^2 << L^2$ and thus the $\frac{\partial^2 u}{\partial x^2}$ component can be thought of as insignificant (from the large denominator) compared to the $\frac{\partial^2 u}{\partial y^2}$ component?

if so, when we leave the boundary layer are we going to assume that the double partial over x is still insignificant, or are we allowed to assume this (assuming same flow and geometry, just outside the BL)

thanks!!

2. Aug 21, 2015

### jlefevre76

Uhhh, it would be nice to have a figure of what your system is or some kind of information describing what's going on in your system. You can't always say that (d^2u/dx^2) is 0. If du/dx is constant or weakly dependent on x, then you may make that assumption.

Can I assume y is the direction to and from a plate, and x is the direction along the plate? So, that would seem the most logical to me. I'm not sure if the u velocity varies linearly inside the boundary layer or not, I honestly don't remember, it's been a while. However, outside of the boundary layer you can definitely assume the second derivative in the x direction is 0. That is, assuming flow on a plate as I described.