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5

From a performance view, you are always interested in preserving as much 'structure' in your grid as possible. Computations on a simplex- or a hexaedral mesh, where every cell looks like the next will be more performant, as you do not have to transform from local to global coordinates differently for each cell. Also, you do not have to save the cell ...

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The best choice for a numerical grid is the one that will most accurately approximate the solution to your problem (without being too computationally expensive). But beyond that the specific features will depend heavily on the type of problem you are trying to solve. A grid might be aesthetically pleasing because it cleverly exploits some symmetry of the ...

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Your step two is to solve the original ODE, which doesn't make sense. I'll write out the steps for applying Forward Euler to your second order ODE. Forward Euler solves the first order ODE $$M \dot{y} = f(y)$$ with the steps \begin{align} M k_1 &= f(y_n) \\ y_{n+1} &= y_n + dt \, k_1 \end{align} Let $v = \dot{u}$. You finite element ...

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You simply make sure that the initial condition satisfies the boundary conditions and that you don't add anything to the elements of $u_{n+1}$ corresponding to the boundary. In other words, you drop the rows and columns of $M$ that correspond to the boundary nodes and solve only in the interior nodes. Let me add that inhomogeneous Dirichlet boundary ...

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A general way would be to include the boundary nodes in the definition of $A$ (which will give you a matrix with more columns than rows) and derive $b$ as the contribution of the Dirichlet nodes. This way, other linear terms like convection are readily included. Assume that the matrix $A$ looks like $$A = [A_I | A_\Gamma ]$$ where $A_I$ is the (square) ...

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