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1

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-...


0

This maybe raises more questions than it answers, but here is what I think one should do. Lets assume that at the boundary point p, the velocity u is defined as $$ u(p) = u_1 \phi_k(p) + u_2 \phi_{k+1} (p), $$ where $$\phi_k = \begin{bmatrix} \phi \\ 0 \end{bmatrix}, \quad \phi_{k+1} = \begin{bmatrix} 0 \\ \phi \end{bmatrix} $$ are corresponding nodal basis ...


2

Tensorial notations are helpful here as an alternative--and an excellent way to prove-- vector calculus identities mentionned in other answers (if you can't remember them of the top of your head). The Navier-Stokes equations for the i-th component of $v$ are: $$ \partial_t v_i + \sum_{1\le j \le 3}v_j \partial_j v_i = -\partial_i p + \sum_{1\le j \le 3}\...


3

The operation in the second equation is the divergence of the dyadic product $vv$.The dyadic product of two vectors is a square matrix. In this case, a $3\times3$ matrix where $(vv)_{ij}=v_iv_j$. The divergence of this product is given by $$\nabla\cdot(vv)=(\nabla\cdot v)v+v\cdot\nabla v$$ You can find the details of the vector identity here. The first term ...


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