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Solid-Wall boundary Condition

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Old   February 14, 2007, 10:54
Default Solid-Wall boundary Condition
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CG CHEN
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Hello! I met a problem about solid wall boundary condition. In the Cartesian coordinates, the symmetry technique can be used to construct sveral rows ghost points for the boundary, then the points on the boundary are updated by the same methods as inner points. But in a general curvilinear coordinates, it seems this method cannot work because the numerical result of velocity compenent normal to the boundary will not keep the zero during the comptation.

Are there any suggestions or recommended references? Thanks in advance!

By the way, in general curvilinear coordinates, I wrote the governing equation in strong conservative form and solve it on a logical rectangle grid.
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Old   February 14, 2007, 12:38
Default Re: Solid-Wall boundary Condition
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Harish
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In the transformed domain ( assuming its a rectangular one and assuming you use finite difference), you have the velocity component U2=0.Since velocity is a contravariant vector you can relate the components in the two coordinates cartesian (u,v) and transformed ( U1,U2).

U2(x2,x1)=dx2/dx * u(x,y) +dx2/dy *v(x,y)

so your condition is dx2/dx*u(x,y)+dx2/dy*v(x,y)=0. The way to implement this would be

If (dx2/dx) is larger calculate u=-(dx2/dy)*v/(dx2/dx) and extrapolate v=2*v_2 -v_3 else v=-(dx2/dx)*u/(dx2/dy) and extrapolate u=2*u_2-u_3

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