How to get BC for Psi in the computing domain?

Pierre Forges

July 26, 2000 Dear colleagues,

Notations ========= (x,y) Cartesian coordinates. (r,s) Curvilinear coordinates. P Stream Function

P is defined as follow: u = dP/dy = dP/dr dr/dy + dP/ds ds/dy [1] v =-dP/dx =-dP/dr dr/dx - dP/ds ds/dx [2]

Inlet Boundary condition ======================== For 0<y<ymax u(0,y)=1

v(0,y)=0

Now if I go to the computing domain, I have:

1=dP/dr dr/dy + dP/ds ds/dy 0=dP/dr dr/dx + dP/ds ds/dx

Two equations with two unknowns: dP/dr and dP/ds

Solving for dP/ds and integrating from s=0 to s>0 yields:

P(r=0,s)= Int_{0}^{y(s)}

[dr/dx] /[dr/dx ds/dy - dr/dy ds/dx] dy

I have used my 2 BC, namely u=1 and v=0, to get a Dirichlet BC for the stream function P. I think it's the way to do it. Let's see now another boundary condition.

Outlet Boundary condition ========================= For x=xmax and 0<y<ymax du/dx=0

dv/dx=0 In the computing domain I have:

du/dr dr/dx + du/ds ds/dx=0 dv/dr dr/dx + dv/ds ds/dx=0

Using eqs [1] and [2] yields:

[d^2P/dr^2 dr/dy + d^2P/drds ds/dy] dr/dx + [d^2P/drds dr/dy + d^2P/ds^2 ds/dy] ds/dx=0 [3]

[d^2P/dr^2 dr/dx + d^2P/drds ds/dx] dr/dx + [d^2P/drds dr/dx + d^2P/ds^2 ds/dx] ds/dx=0 [4]

From these two equations, how can I get my boundary condition for the sream function?

I cannot isolate dP/ds from eqs [3] and [4]!

Same remark applies when I have the following BC: du/dy=0 and v=0. It's a symmetry BC.

Thank you so much in advance for replying to my question.

My very best regards,

Dr. Pierre Forges

UAE University Mech. Eng. Dept. pforges@uaeu.ac.ae