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Old   November 6, 2008, 14:18
Default BC for back step
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I am having trouble deriving the streamline BC for the symmetry line for a 2-d backstep flow. The step is at the bottom, so the top boundary is a symmetry line located at y=Y

I expand phi(Y) with a taylor series and end up with:

# = delta @ = evaluate at

phi(@Y) = phi(@Y-1)+ (d(phi)/dy)#y(@Y-1) + .5(d^2(phi)/dy^2)#y^2(@Y-1)

then, v=0 -> dv/dx = 0 du/dy = 0 b/c of symmetry u=U -> du/dy = (d^2(phi)/dy^2)= 0 zeta(vorticity) = du/dy-dv/dx = 0 (d(phi)/dy)=U -> phi = U#y

So, filling back in I get for the BC

phi(@Y) = phi(@Y-1) + U#y

I had read that the streamfunction BC along a symmetry line is constant, but mine isn't. My results are almost there..I can't get the flow to wrap around the step, but all the other stream lines are matching up with data that I have.
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