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Use a finite volume method. Define $$\delta_x \phi(x,y) = \frac{\phi(x+\Delta x/2,y) - \phi(x-\Delta x/2,y)}{\Delta x}$$ $$a_x \phi(x,y) = \frac{\phi(x+\Delta x/2,y) + \phi(x-\Delta x/2,y)}{2}$$ etc. For example, consider $\tau_{xx}$ which is required at $(i+1/2,j)$. $$(\tau_{xx})_{i+1/2,j} = \mu_{i+1/2,j} \left[ \frac{4}{3} \delta_x u_{i+1/2,j} - \... 1 Just open up the parentheses, e.g., \partial_{x} (\alpha \partial_{x} v) = (\partial_{x} \alpha) (\partial_x v) + \alpha \partial^2_x v, where \alpha=\mu or \alpha=\mu u etc., and apply your central differences: (\partial_{x} \alpha) (\partial_x v) = (\alpha_{i+1}-\alpha_{i-1})(v_{i+1}-v_{i-1})/(4h^2); \alpha \partial^2_x v = \alpha_i (v_{i+1}+v_{i-... 2 Solving the Riemann problem at the interface gives you the values of the solution at this interface just after t_n. Then, the basic assumption is that this interface state remains constant during the whole time step, which is true for piece-wise constant reconstruction (i.e. the solution is assumed uniform in each cell), and as long as the CFL condition is ... 2 If (x_i,y_i) is the centroid of the triangle, then$$ \frac{1}{\Delta_i}\int_{\Delta_i}{p(x,y)}\,dx\,dy = p(x_i,y_i) = \bar{\psi}_i $$This is mid-point quadrature, which is exact for an affine function. The centroid of a triangle is the arithmetic average of its three vertices. Note that$$ x_i = \frac{1}{\Delta_i} \int_{\Delta_i} x dx dy = \frac{1}{3}\...

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Have you plotted the solution using a visualization package or just a cut through the middle of your domain? Sometimes just looking at the solution you're generating will give you an insight into where it's arising from. Based on your description, I'd say there's a mistake in your implementation of your boundary condition. ETA: The last sentence is probably ...

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Upwinding is not necessary. The best known and most-used limited central scheme is Nessyahu-Tadmor Nessyahu, H. and Tadmor, E., 1990. Non-oscillatory central differencing for hyperbolic conservation laws. Journal of computational physics, 87(2), pp.408-463. However a price is paid in that the contact discontinuity is badly smeared. Google Scholar will turn ...

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