# Solid-Wall boundary Condition

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 February 14, 2007, 10:54 Solid-Wall boundary Condition #1 CG CHEN Guest   Posts: n/a 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.

 February 14, 2007, 12:38 Re: Solid-Wall boundary Condition #2 Harish Guest   Posts: n/a 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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