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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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One approach to convert this into an ODE is with index reduction methods. These allow you to convert high-index DAEs into low-index DAEs or ODEs. See section VII.2 of "Solving Ordinary Differential Equations II" by Hairer and Wanner. Consider a generic, Hessenberg index-2 DAE \begin{align} y' &= f(y, z) \\ 0 &= g(y) \end{align} ...

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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: $p=0$ Actually thought that we need only one value of P to pin, not really outlet P = 0; can take any point, the reason behind is that infinite number of answers are imposed by all Neuman conditions, need one point to fix the ...

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There are many variations of the idea of Taylor-Hood elements that remain stable. The Taylor-Hood element on quadrilateral and hexahedral elements are generally understood (though historically incorrect) to be the space $Q_2\times Q_1$ for velocity and pressure. But this is not optimal: We can make the pressure space larger and instead use $Q_2 \times P_{-1}... 2 Consider a FV method as an approximation of the integral conservation law. Starting from the one-dimensional, scalar conservation equation $$u_t + f(u,\nabla u)_x =0,$$ with the definitions of the step sizes$\Delta x$and$\Delta t$t_n = n \Delta t,~~~x_i = i \Delta x,~~~x_{i+1/2}=\frac{1}{2}\left(x_i + x_{i+... 2 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 ... 2 You get a divergence-free$Z_h$if you choose for example$V_h$and$Q_h$such that$\forall v_h\in V_h(div(v_h)\in Q_h)$. (An example for such a pair is the Scott-Vogelius element with$V_h=P^k,Q_h=P^{k-1}_{discont}$) Because if you take$v_h\in Z_h$it fulfills$(div(v_h),q_h)=0 \ \forall q_h\in Q_h $. Now by the special choice of$V_h$and$Q_h$above we ... 1 I don't see what you're not seeing. You've written out a perfectly good non-linear weak form of the PDE in your last two equations. If you invert$M$and apply it, then you have a non-linear equation for$\dot{u}(t)$that you can try to solve. It's a continuous in space and time PDE that you would need to discretize in space to then have only an ODE, with$M\$...

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