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Not really an answer, but too short for a comment. In my experience, the best way to preserve symmetry and deal with boundary conditions is to use just internal points. "Penalization methods" (or big number approach, I know it has several names) is another possibility . If your enumeration is $x_1,\ldots,x_{n+1}$, then work just on $x_2, \ldots, x_{n}$, as ...

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It could have changed the condition number adversely, yes, which would make it harder to solve. How are you forcing this matrix to be symmetric? Should be said that if you're solving the linear problem 10 orders, you're probably doing pretty well. As whpowell pointed out in their comment matlab will use a different solver routine for symmetric vs. ...

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Using the given ODE, the boundary conditions $u(x)=0$ at $x=\pm1$, and the symmetry of the solution it is easy to write the exact analytic solution: $u(x)= a \sin(\pi x)$, for x $\in$ [1/4,1] $u(x)=- a \sin(\pi x)$, for x $\in$ [-1,-1/4] $u(x) = 1/\pi^2 + b \cos(\pi x)$, for x $\in$ [-1/4,1/4] The matching condition for the function derivative at x=1/4 ...

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It becomes a root finding problem. Find $u^{n+1}$ such that $$u^{n+1} + \Delta t \theta F(u^{n+1},t^{n+1}) = u^n + \Delta t (\theta - 1) F(u^{n},t^{n}).$$ Now, you can multiply by a test function, form your trilinear form $w(\cdot,\cdot,\cdot)$ to obtain the weak problem; Find $u^{n+1}$ in an appropriate space $V$ such that $$w(u^{n+1},u^{n+1},v) = (u^n +... 1 The boundary conditions don't depend on the choice of your basis but on the formulation you have for your problem. If you have a "standard" finite element formulation, you don't need to do anything to apply (homogeneous) Neumann boundary conditions, they are already satisfied by your system. In the most common formulation, Neumann boundary conditions are ... 1 In principle, you could try to quantify how much communication two cells K_i and K_j will have to exchange if (i) they share an edge, or (ii) share a vertex. Let's say you call this amount W_{ij}. Then the goal is to partition the mesh in such a way that (i) the partitions are of roughly equal size, and (ii) the sum of the W_{ij} over all cut edges ... 0 Inspired from this Computational Science SE post: Transform your problem from one with inhomogeneous BC to homogeneous BC. This is done by substracting any function B(x,t) with the right inhomogeneous BC from u(x,t) to create a new function h(x,t) = u(x,t) - B(x,t). For instance, take B(x,t) = \frac{2v(t)\sin(\pi x/2)}{\pi}. Your problem becomes \... 0 For stability analysis, we can assume f = 0. First we write$$ M \left( u_{k+1} - u_{k} \right) = \Delta t K u_{k+1} $$or$$ \left( M - \Delta t K \right) u_{k+1} = M u_{k} $$Take the eigendecomposition of (K, M) (there is an explicit expression in 1D), such that$$ w_i^T K w_j = \delta_{ij} \lambda_i \quad \mbox{and} \quad w_i^T M w_j = \delta_{ij}  ...

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Method of Moments is the name given to the Boundary Element Method (BEM) in the Electromagnetism community. Since you are using Green functions for BEM you get fully populated and asymmetric matrices. Some advantages that are commonly mentioned for BEM are: It reduces the dimensionality of the mesh. This might look to lead to smaller matrices but, in ...

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A lot of the papers that came out of the early stages of the MoFEM package considered these sorts of questions. You might want to look at their results.

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