Question: navier-stokes equations for 2D flow down a porous pipe

Hi,

I’m quite new to maple so I need a bit of help. I am trying to solve the Navier-Stokes equations (NSE) in 2D along an infinite channel  which has fluid being sucked out of it through the walls.

∂u/∂t + (u∙grad)u=-(1/ρ)grad(p)+ (nu)grad^2u

I need to write down the NSE.
u=(u,v,0) is the velocity of the fluid.
The flow is steady so ∂u/∂t=0.
I need to take the curl of the NSE to eliminate the pressure term as taking the curl of
gradp=0.

So we have:
curl[(u∙grad)u]=(nu)curl[grad^2u]

Now, because I’m using a 2D flow I need to use stream functions. So we say u=∂ψ/∂y and v=-∂ψ/∂x. where ψ=wxF(η) where w is the velocity in the y-direction at the wall and η=y/a. +a and –a are the heights of the channel walls.

Now I need to substitute all these facts into the NSE to get a forth order ODE.

I think the forth order ODE should be

F’’’’=R(F’F’’-FF’’’)                          (1)
R is the Reynolds number R=wa/(nu)

The boundary conditions are
u=0,v=-w at y=a
u=0,v=w at y=-a
therefore
F’(1)=0, F’(-1)=0, F(1)=F(-1)=1

I then need to go on to say that When R=0
We have a forth order ODE which is easy to solve
F’’’’=0 with the same boundary conditions.
I think it will give a solution of
F(η)= -(1/2)η^3+(3/2)η
I need to plot this.

I then need to go on so say that for small R we can write F as a power series.

F(η)=F_0(η)+RF_1(η)+R^2F_2(η)+…

Now I need to use this substitution in the equation (1) and equate powers of R.

For R to the power 0 we should get F_0’’’’=0
Boundary conditions are the same as before but with F_0. I need to keep looking at powers of R to get more and more accurate solutions. If anyone can even help getting me started on this it would be very greatly appreciated. I don’t expect someone to just go through all of this but I would be most grateful for any help on any part of this.

Thanks

MJ