Discretizing the Navier-Stokes-Equation

$$\mu \nabla ^{2}\mathbf{v}+(\mu +\lambda )\nabla (\nabla \cdot \mathbf{v})=-\mathbf{f}$$ (12)

$$\mu \left( \frac{\partial ^{2}}{\partial x^{2}}+\frac{\partial ^{... ...\frac{\partial ^{2}}{\partial z^{2}} \end{array}\right) \mathbf{v}=-\mathbf{f},$$ (13)

For the x-component of (13) we may write:

$$\mu \left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\par... ... \partial y}+\frac{\partial^2v^{(z)}}{\partial x \partial z}\right) = -f^{(x)},$$ (14)

$$(2\mu +\lambda )\frac{\partial^2v^{(x)}}{\partial x^2}+\mu \left(... ... \partial y}+\frac{\partial^2v^{(z)}}{\partial x \partial z}\right) = -f^{(x)}.$$ (15)

Discretizing this using numerical derivatives based on finite differences (8, pp. 186-189) yields

$$\begin{array}{ll} \frac{(2\mu +\lambda)}{h^2} \left( v^{(x)}_... ...{i-1,j,k-1}-v_{i+1,j,k-1}^{(z)}\right) & =-f^{(x)} \end{array}.$$ (16)

With shortcuts $a=\frac{\mu }{h^2}$, $b=\frac{\mu +\lambda }{h^2}$ follows,

$$\begin{array}{ll} v_{i,j,k}^{(x)}+\frac{(a+b)}{(6a+2b)}\left(... ...+1,j,k-1}^{(z)}\right) & =\frac{1}{(6a+2b)}f^{(x)} \end{array}.$$ (17)

y- and z- components can be obtained in a similar manner.

With $\hat{\mathbf{f}}:=\frac{1}{(6a+2b)}\mathbf{f}$, writing (12) in its discretized representation yields a linear system

$$\mathbf{Av}=\hat{\mathbf{f}}.$$ (18)