need a symbolic calculation

[Pavel Solin]

Hi,
  I would like very much to do a calculation but do
not have much time for it myself. If you have some
time and like the chain rule of differentiation, please
consider helping me. The model consists of standard
incompressible Navier-Stokes equations. The domain
is a rectangular channel with a symmetrically placed
circular obstacle (see attachment). The inlet velocity
(left edge), all other boundary conditions, and the initial
conditions are symmetric wrt. the horizontal channel axis.
I would like to show that IF a (nonstationary) velocity
and pressure are a solution, then also this velocity and
pressure flipped about the channel axis are a solution. 
From a physicist's point of view this is clear since
everything is symmetric, but a mathematician needs to
do a calculation. This should be done by defining the
"flipping" precisely as a geometric transformation,
inserting this transformation into the original solution,
substituting the composed functions into the original
equations, and using chain rule. After performing the
chain rule, it should be possible to use the fact that
the original velocity and pressure are a solution.
I have other three or four high-priority topics on
my list, and would be grateful for any help.

Best,

Pavel

--
Pavel Solin
University of Nevada, Reno
http://hpfem.math.unr.edu/people/pavel/
Hermes project: http://hpfem.org/
FemHub project: http://femhub.org/

[Ondrej Certik]

I am l looking into it. So far I only managed to proof that the
flipped velocity+pressure is not a solution, so I guess I made a
mistake somewhere. :)

Ondrej

[Pavel Solin]

Ondrej,
  please write this up in Latex when you have time.
I know that besides myself, some important CFD
people will be extremely interested in it.

Pavel

[Pavel Solin]

I did the calculation today on the bus to Prague. It was

straightforward. I showed that if u(x,y,t), v(x,y,t), p(x,y,t)

is a solution to the N-S equations (with no gravity force),

then also u(x,-y,t), -v(x,-y,t), p(x,-y,t) is a solution. This

the case when x-axis is the axis of symmetry. Of course

the domain and all boundary conditions have to be symmetric

about the x-axis. Thus if the solution to the N-S equations

is unique (which is still an open problem worth $1M), then

in a symmetric case like the channel with the hole, the

solution must be symmetric.

 

Pavel

[Ondrej Certik]

On Sat, Jun 6, 2009 at 2:05 PM, Pavel Solin<so...@unr.edu> wrote:
> I did the calculation today on the bus to Prague. It was

From Pilsen? I did lots of good stuff done on that bus! :)

> straightforward. I showed that if u(x,y,t), v(x,y,t), p(x,y,t)
> is a solution to the N-S equations (with no gravity force),
> then also u(x,-y,t), -v(x,-y,t), p(x,-y,t) is a solution. This

Right, I am getting the same thing now. I was showing that if
u(x,y,t), v(x,y,t), p(x,y,t)
is a solution, then what about u(x,-y,t), v(x,-y,t), p(x,-y,t) and it
turns out it isn't.

> the case when x-axis is the axis of symmetry. Of course
> the domain and all boundary conditions have to be symmetric
> about the x-axis. Thus if the solution to the N-S equations
> is unique (which is still an open problem worth $1M), then
> in a symmetric case like the channel with the hole, the
> solution must be symmetric.

So from this it follows that the asymmetric solutions that we are
getting (with asymmetric mesh) are all wrong. It would be interesting
to run full hp-adaptivity with the asymmetric mesh --- it should
hopefully converge to a symmetric solution.

Also it would be very interesting to study how to get the asymmetric
solution when varying the boundary conditions and/or the position or
the shape of the obstacle.

Ondrej

[Pavel Solin]

Hi Ondrej,

On Mon, Jun 8, 2009 at 7:51 AM, Ondrej Certik <ond...@certik.cz> wrote:

On Sat, Jun 6, 2009 at 2:05 PM, Pavel Solin<so...@unr.edu> wrote:
> I did the calculation today on the bus to Prague. It was

From Pilsen? I did lots of good stuff done on that bus! :)

This one was from Podluzi to Prague, on my way to ECCOMAS.

> straightforward. I showed that if u(x,y,t), v(x,y,t), p(x,y,t)
> is a solution to the N-S equations (with no gravity force),
> then also u(x,-y,t), -v(x,-y,t), p(x,-y,t) is a solution. This

Right, I am getting the same thing now. I was showing that if

u(x,y,t), v(x,y,t), p(x,y,t)

is a solution, then what about u(x,-y,t), v(x,-y,t), p(x,-y,t)  and it
turns out it isn't.

> the case when x-axis is the axis of symmetry. Of course
> the domain and all boundary conditions have to be symmetric
> about the x-axis. Thus if the solution to the N-S equations
> is unique (which is still an open problem worth $1M), then
> in a symmetric case like the channel with the hole, the
> solution must be symmetric.

So from this it follows that the asymmetric solutions that we are
getting (with asymmetric mesh) are all wrong. It would be interesting
to run full hp-adaptivity with the asymmetric mesh --- it should
hopefully converge to a symmetric solution.

I did this in my lecture as advertisement for adaptivity on
dynamical meshes. It does eliminate mesh dependence
to a really great extent, but not completely (which I did not show).
Concretely, the symmetric exact solution to this N-S example is
so unstable that sooner or later the approximation will diverge
away from it even with adaptivity on dynamical meshes.
This is a tough example. With our current adaptivity, all refinements
take place close to the obstacle (and in boundary layer). The mesh
remains coarse at the outlet edge. However, it seems to me that
small errors caused there propagate upstream and cause
a major error. Thus we should implement the new adaptivity
which is free of the global reference solution, and see how it
performs.

Pavel

Also it would be very interesting to study how to get the asymmetric
solution when varying the boundary conditions and/or the position or
the shape of the obstacle.

Ondrej