Discretization

The physical domain is mapped onto a rectangular domain in the $(\xi^{1}, \xi^{2})$ plane, which is called the computational domain. All points of the domain are used, including the dry ones.


Using the relation (summation convection applied)

  $$\frac{\partial \varphi}{\partial x^{\beta}} = \frac{1}{\sqrt{g}} ... ...{\partial}{\partial\xi^{\gamma}}(\sqrt{g}\; a^{(\gamma)}_{\beta}\; \varphi)\;,$$ (5.7)

with $a^{(\gamma)}_{\beta}$ the components of the contravariant basevectors $\vec{a}^{(\alpha)}$ defined as

  $$\vec{a}^{(\alpha)} = \nabla \xi^{\alpha}\;,$$ (5.8)

and $\sqrt{g}$ the Jacobian of the transformation

  $$\sqrt{g} = a^{1}_{(1)} a^{2}_{(2)} - a^{2}_{(1)}a^{1}_{(2)}\;.$$ (5.9)

$\vec{a}_{(\alpha)}$ are the covariant base vectors defined by

  $$\vec{a}_{(\alpha)} = \frac{\partial \vec{x}}{\partial \xi^{\alpha}} \; .$$ (5.10)

The contravariant base vectors follow immediately from the covariant ones due to:

$$\sqrt{g}\vec{a}^{(1)} = (a^{2}_{(2)}, \; -a^{1}_{(2)})^{T}\;,$$ (5.11)

$$\sqrt{g}\vec{a}^{(2)} = (-a^{2}_{(1)}, \; a^{1}_{(1)})^{T}\;.$$ (5.12)

Application of (5.7) to equation (5.5) results in

  $$\frac{1}{\sqrt{g}} \frac{\partial}{\partial \xi^{\alpha}}(\sqrt{g} \vec{a}^{(\alpha)} \cdot (k \nabla \zeta + \vec{F})) = 0\;.$$ (5.13)

Note that $\nabla \zeta$ is a derivative in the Cartesian $(\vec{x})$ direction and not in the $\vec{\xi}$ direction.


In the remainder we shall use the local numbering as given in Figure 5.1.

Figure 5.1: Local numbering in computational domain

The points (0,0), (2,0), (0,2) and so on are the vertices of the cells. The integration cell for the finite volume method is defined by the cell $\Omega$ (-1,0), (0,-1), (1,0), (0,1).


Integrating (5.13) over this cell gives

$$\int\limits_{\Omega_{x}} \frac{1}{\sqrt{g}} \frac{\partial}{\part... ...lpha}}(\sqrt{g} \vec{a}^{(\alpha)} \cdot (k \nabla \zeta + \vec{F}))d\Omega_{x}$$

$$\int\limits_{\Omega_{\xi}} \frac{\partial}{\partial \xi^{\alpha}}(\sqrt{g}\vec{a}^{(\alpha)} \cdot (k \nabla \zeta + \vec{F}))d\Omega_{\xi}$$ (5.14)

$$\approx \sqrt{g}\vec{a}^{(1)} \cdot (k\nabla \zeta + F)\vert^{(1,... ...sqrt{g}\vec{a}^{(2)} \cdot (k \nabla \zeta + \vec{F})\vert^{(0,1)}_{(0, -1)}\;,$$

where $\Omega_{x}$ is the cell in the physical space and $\Omega_{\xi}$ the cell in the computational domain.
The four points (1,0), (0,1), (-1,0) and (0,-1) will be cell integration points. The covariant basis vectors $\vec{a}_{(\alpha)}$ are approximated by central differences

$$\vec{a}_{(2)}\vert _{(0,1)} = \vec{x}_{(0, 2)} - \vec{x}_{(0,0)}\;,$$ (5.15)

$$\vec{a}_{(1)}\vert _{(1, 0)} = \vec{x}_{(2,0)} - \vec{x}_{(0,0)}\;,$$ (5.16)

and by linear interpolation in other points. In these relations we have used that the step width in the computational domain is equal to 1.
The term $\nabla \zeta$ needs special attention. Since it concerns derivatives in the $\vec{x}$ direction, whereas all derivatives in the computational domain are in the $\vec{\xi}$ directions it is necessary to make some approximation. We approximate this term by the integration path method as outlined in Wesseling (2001).


To that end $\nabla \zeta$ is integrated in two independent directions $\xi^{1}$ and $\xi^{2}$. This yields two equations to express $\frac{\partial \zeta}{\partial x}$ and $\frac{\partial \zeta}{\partial y}$ in $\zeta$ values of neighbours.

$$(\vec{x}_{2,0} - \vec{x}_{),0})\nabla \zeta\vert _{(1, 0)} = \zeta_{2,0} - \zeta_{0,0}\;,$$ (5.17)

$$\frac{1}{2}((\vec{x}_{2,2} - \vec{x}_{2,-2}) + (\vec{x}_{0,2} - \... ...} = \frac{1}{2}((\zeta_{2,2} - \zeta_{2,-2}) + (\zeta_{0,2} - \zeta_{0,-2}))\;.$$ (5.18)

(5.17), (5.18) may be considered as two sets of equations to express $\nabla \zeta$ into $\zeta$ values. Solution of this linear system results in:

  $$\nabla \zeta \vert _{(1,0)} = \zeta \vert^{(2,0)}_{(0,0)} \vec{c}... ... (\zeta\vert^{(0,2)}_{(0, -2)} + \zeta \vert^{(2,2)}_{(2,-2)})\vec{c}^{(2)}\;,$$ (5.19)

with

$$\vec{c}^{1} = \frac{1}{C}(c^{2}_{(2)}, \; -c^{1}_{(2)})\; ; \; \vec{c}^{2} = \frac{1}{C}(-c^{2}_{(1)}, \; c^{1}_{(1)})\;,$$ (5.20)

$$C = c^{2}_{(2)}c^{1}_{(1)} - c^{2}_{(1)}c^{1}_{(2)}\;,$$ (5.21)

$$\vec{c}_{(1)} = a_{(1)}\vert _{(1,0)}\; \; \vec{c}_{(2)} = \vec{a}_{(2)}\vert _{(... ...\vert _{(0,1)} + \vec{a}_{(2)}\vert _{(2, -1)} + \vec{a}_{(2)}\vert _{(2,1)}\;.$$ (5.22)

A similar formula is applied for point (0,1). Equation (5.14) together with expression (5.19) gives one row of the discretized equation.