The implementation of dry points

Dry points complicate the implementation considerably.


For the dry points itself there is no problem. In fact we make the corresponding row of the matrix, as well as the right hand side element completely equal to zero.


Dry points in the neighbourhood of wet points, however, also influence the matrix for the wet point. Consider for example the integration point (1,0) in Figure 5.3.

Figure 5.3: Dry point (2, 0) and wet point (0, 0)

If (0,0) is a wet point and (2,0) a dry point then we assume that at point (1,0) we have a Neumann boundary condition. This means in fact that the contribution of the integration point (1,0) to the matrix and right hand side is equal to zero. With respect to the evaluation of the gradient of $\zeta$ with the integration path method one sided differences are applied for those formulas involving $\zeta_{(2,0)}$. This process is applied for all transitions from wet to dry points. As a consequence, in the case of a situation like in Figure 5.4 we make $\nabla \zeta$ for point 2 zero.

Figure 5.4: Wet points $\bullet$ enclosed by a row of dry points $\times$

The reason is that in point 2 it is only possible to evaluate $\frac{\partial \zeta}{\partial \xi^{1}}$ and not $\frac{\partial \zeta}{\partial \xi^{2}}$, and hence we have too few information to express $\nabla \zeta$ in neighbour values.