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The smooth solution turned out to have BC's applied in the following way: Walls and inlet: $\frac{\partial p}{\partial n}=0$ Outlet: $p=0$


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Disclaimer: I wrote this answer in a rush, may not be up to the standards. Also, I don't have enough information about your problem to give more detailed advice. However, the information presented here should be enough to get you going. I will suggest you to use MINRES with an appropriate preconditioner. For a moment, let's assume that this is not a twofold ...


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For both schemes: I am going to lump $B$ and $C$ into a single matrix $B$ for convenience of communicating the idea. Likewise, $y$ and $z$ are now $y$, and $F_y$ and $F_z$ are $F_y$. $Ax+B^Ty=F_x$ or $x = A^{-1}(F_x-B^Ty)$ $Bx=F_y$ or $BA^{-1}(F_x-B^Ty)=F_y$ $BA^{-1}B^Ty=BA^{-1}F_x-F_y$ Note that $BA^{-1}B^T$ is symmetric and invertible, and can be ...


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