# Navier-stokes control volume

1. Jan 26, 2007

### verdigris

navier-stokes smoothness problem almost solved

Penny Smith has made progress with showing that smooth conditions exist for all time in a domain for the Navier Stokes equations
http://notes.dpdx.net/2006/10/06/penny-smiths-proof-on-the-navier-stokes-equations/ [Broken]

However a flaw was found in the mathematics - hopefully it can be sorted out soon.There's still the problem though of finding actual solutions to the equations!

Thanks for the info on control volume.I was just wondering if in reality there
is a real,if very small size,to the differential element.

Last edited by a moderator: Apr 22, 2017 at 3:22 PM
2. Jan 26, 2007

### FredGarvin

It's not a control volume but a differential element (infinitely small). Imagine a cube with sides that measure, in cartesian coordinates $$\delta x$$, $$\delta y$$ and $$\delta z$$.

3. Jan 26, 2007

### minger

Actually Fred, the NS equations have 4 forms:

Lagrangian (moving frame of reference)
Eulerian (stationary)
..and then...
Differential
Integral

So for example, looking simply at continuity (sorry again for my lack of latex):
Differential Lagrangian:
Dp/Dt + rho*del•V =0
Where D/Dt is the substantial derivative with respect to time

Differential Eulerian:
dp/dt + del•(rho*V) = 0
(note this form is now strongly conservative as all variables are inside of a derivative)

Integral Lagrangian:
D/Dt [Volume Integral] rho dV = 0

Integral Eulerian:
d/dt [Volume Integral] rho dV + [surface integral] rho*V dS = 0

The entire equations can be derived any of the four ways. It's easiest (at least for me) to remember one form, and then how to go from one form to another.

To answer the question, the control volume for an integral approach is simply V, there is no need to know anything else besides that.