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
DE
If those are your differential equation and boundary conditions, the solution is simply F(y) = 1. But I think you want F(-1) = -1. Here's how I might proceed.
Your differential equation:
Make it into a series in powers of R
Extract individual differential equations for each power up to R^5.
Define a function to make boundary conditions for f[k].
Solve the boundary value problems for each power.
> for k from 1 to 6 do f[k-1]:= unapply(rhs(dsolve({des[k],bc(k-1)})),s) end do;Here is a plot of the solutions for R=0 (in red) and R=20 (in blue).
thanks very much. that was a
thanks very much. that was a big help. I understand how this works but there is one part i don't understand. would you mind explaining it to me please?
for k from 1 to 6 do f[k-1]:= unapply(rhs(dsolve({des[k],bc(k-1)})),s) end do;in this for loop what does the unapply do? this is the only section i was confused on.
thanks
mj
unapply
The unapply command used here in conjunction with the rhs(right hand side) command takes the output from dsolve and turns it into a "functional" operator. The help pages have examples that clarify this command. ?unapply to open up the help page. Without this, f[0](s) would return an emplty plot.
Regards,
Georgios Kokovidis
Dräger Medical
thank you very much
thank you very much