3.2.3 Simple, Distinctive Mesh Examples

The pure diffusion equation is solved with the finite element and the finite volume method using AMIGOS [128]. This allows the comparison of the solutions on identical meshes with the same linear solver. A Gaussian profile is used as the initial distribution. In two dimensions correct and identical results are obtained with both methods on Delaunay meshes. In three dimensions the finite volume method still achieves correct results on a Delaunay mesh as expected. However, the finite element method fails on the same three-dimensional Delaunay mesh. Even for such a simple test problem the finite element solution strongly violates the maximum principle. The resulting concentration reaches negative values in some areas. These areas spread out in time and the absolute value of the emerging negative concentrations is much larger than the minimal initial concentration. The relative error between the solutions of the two methods oscillates strongly and is large in regions where the concentration is negative. These negative concentrations are particularly annoying for diffusion applications in semiconductor process simulation where the concentration varies many orders of magnitude within a small area.

In the following the observed effects are investigated in terms of mesh requirements and simple mesh examples are constructed where the finite element method can be applied successfully to the diffusion problem.

Mesh Example 1:

A Delaunay mesh which is not suitable as a finite element mesh for diffusion applications.

Mesh Example 2:

A Delaunay mesh which is suitable for finite element diffusion simulation.

Mesh Example 3:

A non-Delaunay mesh with obtuse dihedral angles which is still suitable as a finite element mesh.

The three presented meshes illustrate the different scope of the Delaunay criterion and Crit. 3.3. They prove that in three dimensions the Delaunay criterion is neither sufficient nor necessary to obtain a correct finite element mesh for diffusion so that the maximum principle is fulfilled. The examples also show that a strict adherence to a sole non-obtuse angle criterion is not necessary. This important insight complements previous research [89] where one example, a Delaunay mesh insufficient for such applications, was given.

The examples were constructed by exploiting an ortho-product point distribution. A cube defined by eight points can be tetrahedralized in two qualitatively different ways.

$T_{6}$ Tessellation:

A cube is composed of six tetrahedra (Fig. 3.6).

$T_{5}$ Tessellation:

A cube is composed of five tetrahedra (Fig. 3.7).

For comparison purposes a specific tessellation $T_{6}$ is used which contains sliver elements with obtuse dihedral angles. The tessellation $T_{5}$ on the other hand does not contain such elements. Note that also $T_{6}$ tessellations exist which do not contain obtuse angles. The key idea is that all elements of both tessellations fulfill the empty circumsphere Delaunay criterion (Crit. 5.3), because all points lie on the perimeter of a single sphere. On the other hand Fig. 3.6 clearly shows that the finite element mesh requirement (Crit. 3.3) is not met by the chosen $T_{6}$ tessellation. It is only met by the $T_{5}$ tessellation, because of the total absence of obtuse dihedral angles. A simulation with AMIGOS allows to display the stiffness matrix and hence to directly check the sign of the matrix entries. The resulting stiffness matrix of a $T_{6}$ tessellation is shown in Fig. 3.8 where the entries for edge $(3,4)$ of the mesh are underlined.

Figure 3.8: Global stiffness matrix for a $T_{6}$ tessellation and local matrices of those four elements which are adjacent to edge $(3,4)$.

$\left( \begin{array}{rrrrrrrr} 5.0 & -1.\dot{6} & -1.\dot{6} & -1.\dot{6} & ... ...& & -1.\dot{6} & -1.\dot{6} & -1.\dot{6} & 4.\dot{9} \\ \end{array} \right)$

TET: 4 5 6 3

$$\left( \begin{array}{rrrr} 1.\dot{6} & -1.\dot{6} & -1.\dot... ...dot{6}} & -3.\dot{3} & -3.\dot{3} & 5 \\ \end{array} \right)$$

TET: 3 4 1 2

$$\left( \begin{array}{rrrr} 1.\dot{6} & \underline{1.\dot{6}... ...& -3.\dot{3} & 1.\dot{6} & 3.\dot{3} \\ \end{array} \right)$$

TET: 6 4 3 2

$$\left( \begin{array}{rrrr} 3.\dot{3} & 0 & -1.\dot{6} & -1... ...\dot{6} & -1.\dot{6} & 0 & 3.\dot{3} \\ \end{array} \right)$$

TET: 4 5 3 1

$$\left( \begin{array}{rrrr} 1.\dot{6} & 0 & \underline{0} & ... ....\dot{6} & -1.\dot{6} & 0 & 3.\dot{3} \\ \end{array} \right)$$

Suitable meshes for simulation are then built by stacking a large number of such tessellated cubes. The typical characteristics of each tessellation type are thereby conserved. Hence, both meshes are global Delaunay meshes and yet only one satisfies Crit. 3.3. The two fundamentally different meshes based on an identical ortho-product point cloud are depicted in Fig. 3.11 and Fig. 3.12. The finite element simulation on the $T_{6}$ type Delaunay mesh results in negative concentrations as was previously pointed out. The $T_{5}$ type Delaunay mesh which fulfills Crit. 3.3 indeed succeeds to yield the required entries in the stiffness matrix and the concentrations remain positive at any time during the transient simulation.

The most important fact however is shown by the third example. Further exploiting the ortho-product point set and its $T_{5}$ type tessellation with slightly shifted points in certain locations results in a non-Delaunay mesh which still satisfies Crit. 3.3. Figure 3.9 shows an instance of the mesh consisting of eight cubes. The point in the middle has been shifted. The Delaunay criterion is violated, because the circumspheres of several unmodified tetrahedra contain the shifted point in its interior. The dashed line in the figure marks two of the non-Delaunay triangles. The simulation using AMIGOS for the entire mesh (Fig. 3.13) shows, that the requirements for the stiffness matrix are fulfilled. For example one can consider the edge $(14,10)$ in Fig. 3.9. This edge is shared by elements which contain the shifted point and which are non-Delaunay. The matrix contributions of the six elements which are adjacent to this edge are given in Fig. 3.10. The first two matrices belong to the elements with the shifted point, and indeed possess undesirable positive off-diagonal entries. The last two matrices however belong to very well shaped elements which are able to compensate the overall sum. The resulting entry in the global stiffness matrix for the edge $(14,10)$ equals $(0.8\dot{3} + 0.8\dot{3} + 0 + 0 + (-0.8\dot{3}) + (-0.8\dot{3}) = 0 )$. Again, the concentrations do not reach negative values at all times. The shifting of a point introduces obtuse dihedral angles and positive contributions to off-diagonal elements of the stiffness matrix. However, Crit. 3.3 is satisfied and the stiffness matrix remains an M-matrix.

Figure 3.9: $T_{5}$ type tessellation with a shifted point.

Figure 3.10: Element matrices which contribute to the entry in the global stiffness matrix for the edge $(14,10)$. Due to the symmetry of the mesh the three matrices on the left and on the right side possess the same entries.

TET: 10 6 14 2

$$\left( \begin{array}{rrrr} 5 & -6.\dot{6} & \underline{0.8\... ...dot{3} & -1.\dot{6} & -0 & 0.8\dot{3} \\ \end{array} \right)$$

TET: 10 17 14 26

$$\left( \begin{array}{rrrr} 1.\dot{6} & -1.\dot{6} & \underl... ... \\ 0 & -1.\dot{6} & -0 & 1.\dot{6} \\ \end{array} \right)$$

TET: 2 10 16 14

$$\left( \begin{array}{rrrr} 2.5 & -0.8\dot{3} & -0.8\dot{3} ... ...line{-0.8\dot{3}} & -0.8\dot{3} & 2.5 \\ \end{array} \right)$$

TET: 6 10 14 18

$$\left( \begin{array}{rrrr} 10 & -6.\dot{6} & -1.\dot{6} & -... ...dot{6} & 0.8\dot{3} & -0 & 0.8\dot{3} \\ \end{array} \right)$$

TET: 10 14 17 16

$$\left( \begin{array}{rrrr} 1.\dot{6} & \underline{0} & -1.\... ...} \\ 0 & 0 & -1.\dot{6} & 1.\dot{6} \\ \end{array} \right)$$

TET: 10 18 26 14

$$\left( \begin{array}{rrrr} 2.5 & -0.8\dot{3} & -0.8\dot{3} ... ...3}} & -0.8\dot{3} & -0.8\dot{3} & 2.5 \\ \end{array} \right)$$

Figure 3.11: Delaunay mesh ($T_{6}$), 3072 tetrahedra.
Figure 3.12: Delaunay mesh ($T_{5}$), 2560 tetrahedra.
Figure 3.13: Non-Delaunay mesh, 2560 tetrahedra.