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A few years ago I developed a solution to the Navier-Stokes equations and as of yet have not been able to locate a similar version of the solution. I would like to know if anyone has seen a solution like this or can spot any significant errors.

The version of the equations I worked with are as follows, where I set $\nu = 1$:

$\partial_tu+u\partial_xu+v\partial_yu+w\partial_zu = -\partial_xp + \nu(\partial_{xx}u+\partial_{yy}u+\partial_{zz}u)$

$\partial_tv+u\partial_xv+v\partial_yv+w\partial_zv = -\partial_yp + \nu(\partial_{xx}v+\partial_{yy}v+\partial_{zz}v)$

$\partial_tw+u\partial_xw+v\partial_yw+w\partial_zw = -\partial_zp + \nu(\partial_{xx}w+\partial_{yy}w+\partial_{zz}w)$

$\partial_xu+\partial_yv+\partial_zw=0$


The original solution I came up with is (recognizing that $\sin^2(x)+\cos^2(x)=1$, which is important to keep expanded in order to easily see the cancellations):

$p =-(\sin^2(x) + \cos^2(x))e^{-\frac{1}{2}(|y|+|z|+|t|)}$

$u =\sin(x)e^{-\frac{1}{2}(|y|+|z|+|t|)}$; $v =\cos(x)e^{-\frac{1}{2}(|y|+|z|+|t|)}$; $w =\cos(x)e^{-\frac{1}{2}(|y|+|z|+|t|)}$

I have checked to see if this works several times, but have always wondered if I made a mistake, the relevant derivations are below, where $cos(x)=c_x$ and $sin(x) =s_x$.



$\partial_x{u} =c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_x{v} =-s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_x{w} =-s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$

$\partial_y{u} =\frac{-1}{2}s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_y{v} =\frac{-1}{2}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_y{w} =\frac{-1}{2}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$

$\partial_z{u} =\frac{-1}{2}s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_z{v} =\frac{-1}{2}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_z{w} =\frac{-1}{2}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$


$\partial_t{u} =\frac{-1}{2}s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_t{v} =\frac{-1}{2}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_t{w} =\frac{-1}{2}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$



$\partial_{xx}{u} =-s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_{xx}{v} =-c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_{xx}{w} =-c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$

$\partial_{yy}{u} =\frac{1}{4}s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_{yy}{v} =\frac{1}{4}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_{yy}{w} =\frac{1}{4}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$

$\partial_{zz}{u} =\frac{1}{4}s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_{zz}{v} =\frac{1}{4}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\partial_{zz}{w} =\frac{1}{4}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$


$\sum{} =\frac{-1}{2}s_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\sum{} =\frac{-1}{2}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$; $\sum{} =\frac{-1}{2}c_xe^{-\frac{1}{2}(|y|+|z|+|t|)}$



$u\partial_x{u} =s_xc_xe^{-(|y|+|z|+|t|)}$; $u\partial_x{v} =-s^2_xe^{-(|y|+|z|+|t|)}$; $u\partial_x{w} =-s^2_xe^{-(|y|+|z|+|t|)}$

$v\partial_y{u} =\frac{-1}{2}s_xc_xe^{-(|y|+|z|+|t|)}$; $v\partial_y{v} =\frac{-1}{2}c^2_xe^{-(|y|+|z|+|t|)}$; $v\partial_y{w} =\frac{-1}{2}c^2_xe^{-(|y|+|z|+|t|)}$

$w\partial_z{u}=\frac{-1}{2}s_xc_xe^{-(|y|+|z|+|t|)}$; $w\partial_z{v}=\frac{-1}{2}c^2_xe^{-(|y|+|z|+|t|)}$; $w\partial_z{w}=\frac{-1}{2}c^2_xe^{-(|y|+|z|+|t|)}$


$\sum{}=(s_xc_x-s_xc_x)e^{-(|y|+|z|+|t|)}$;$\sum{}=-(s^2_x + c^2_x)e^{-(|y|+|z|+|t|)}$;$\sum{}=-(s^2_x + c^2_x)e^{-(|y|+|z|+|t|)}$



I have played a little bit with the thought that there are actually three versions of the solution, one oriented for each axis as illustrated below:

Velocity matrix

Where the pressure component can be added as follows:

velocity with pressure

Note: Just an addendum, as pointed out in the comments, my earliest version of this was sans the absolute value symbols, which allows for the solution to approach infinite values in negative coordinates. My initial thought was that the absolute values were a sufficient constraint, but as pointed out in the comments this results in a discontinuity in some of the derivatives when values are set to zero. I have not yet explored if this is a real singularity or a coordinate one, since the solution appears singularity free in the positive domain.

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The question would be easier to follow if you specified a time range and a domain on which you want to solve the equations, along with initial and boundary conditions (granted, given your solution, one can always reverse engineer these conditions, but it raises the question of their physical significance). –  Christoph B. Apr 12 '13 at 9:32

2 Answers 2

up vote 4 down vote accepted

Solutions of the form

$$ cos(x_i)e^{-x_j}$$

are common specific solutions of the Navier-Stokes equations in simplified (not simple) problems. These are however problems where inertia is ignored, which you include. (Please note that I am using index notation, with $i,j\in\{1,2,3\}$). $x_j$ is then the wall normal direction. This is actually quite well related to the remarks of Jaime.

I did not check your solutions, and I am not very familiar with the literature on this topic, but it is interesting that the same trick also works in three dimensions. The only thing is, that you found three separate solutions, and a linear combination of the three is most likely no longer a solution due to the non-linearity of the equations.

You will probably also find that a solution in the form

$$ u_n(\vec{x})=cos(k_nx_i)e^{-l_nx_j}$$

also satisfies the equations. In the cases without inertia, you can write the solution as a linear combination and you get:

$$ u=\sum_n u_n(\vec{x})$$

where $k_n$ and $l_n$ take control of the different sizes of flow features.

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Thanks, this is definitely something I will be exploring. –  Hal Swyers Jan 31 '13 at 10:53

There is something fishy about your result: all your quantities have a $e^{-\frac{1}{2}|t|}$ factor in them. So the movement those equations describe starts off at $t=-\infty$ completely stationary and with no pressure gradient, starts moving without an external force, and eventually dies out when $t=\infty$. That doesn't seem very consistent with what one would expect from conservation of energy, especially in a dissipative setting.

There is the obvious thing that I don't see you taking into account that

$$\frac{d}{dt}e^{-\frac{1}{2}|t|} = \left\{ \begin{array}{l l} -\frac{1}{2} e^{-\frac{1}{2}|t|} & \quad \text{if $t >0$}\\ \frac{1}{2} e^{-\frac{1}{2}|t|} & \quad \text{if $t <0$} \end{array} \right.$$

an issue that predates $y$ and $z$ as well. I haven't checked it in detail, but I think this basically invalidates your results outside the $y, z, t >0$ region.

It seems to me that objection could be overcome by removing all the absolute values, although then your solution could not be extended to $-\infty$ on $t$, $y$ and $z$ without having $\infty$ values. It´s still a pretty impressive result, if no one else can find any other flaws in it.

EDIT If you limit your extension to $y, z >0$ then there is the added issue that you have a very unnatural condition at $y=z=0$, since fluid has to be supplied or removed along the $x$ axis at a precise rate to support your solution.

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Thanks, this is a good catch. My original version was sans absolute values, and was definitely only valid in the positive regimes, since it rapidly approached infinity in negative regimes, the velocity vectors themselves appeared pretty smooth when I added the absolute value constraint www.wolframalpha.com/input/?i=sin(x)*(e^((-1%2F2)(|t|))) however I missed that shift in the derivative www.wolframalpha.com/input/?i=dsin(x)*(e^((-1%2F2)(|t|)))%2Fdt –  Hal Swyers Jan 30 '13 at 10:42
    
What are your thoughts on this being a coordinate singularity? If I have a chance over the weekend I want to check to see if this will go away in other coordinate systems or if it will hold. My thinking is that one might be able to at least isolate the singularity to a single point and then find some way to remove it. –  Hal Swyers Jan 31 '13 at 10:56

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