Nikolay Kirov
Research interests
Mesh generators and finite element solvers
Unstructured mesh generators and a finite element solver

### Problems solving

Example A.

We consider the following boundary value problem:

-e2Du + u = 1  in  [0,1]2,   u|G=0,
where G is the boundary of unite square. When is small, the solution has boundary layer on G.

We create Shishkin mesh for solving this problem.

h                                      1-2h                                          h n/4                                        n/2                                        n/4

h = min
ì
í
î
1
4
, 2e ln n ü
ý,
þ
n is the number of the mesh points.
The unite square is divided of 3 domains:
- D1: a central domain , [1-2h,1-2h]2, where we have to place 25% of all N mesh points on the unite square;
- D2: 4 subdomains, each is [h,1-2h]2, at the center of a square side. In this subdomain 50% nodes have to be placed;
- D3: 4 square [0,h]2 subdomains, each one has a vertex, which coincides with a square vertex, contains 25% of all nodes.
h                       1-2h                              h In the examples we fix the parameters h = 0.05 and the number of mesh points N = 1089. Then we can calculate :

h = 2e ln (N
1
2

) = lnNe
0.05
lnN
» 0.0071,   e2 = 5.1122 x 10-5.

Other mesh type is created ("special mesh'') in the following rules: at the first step a coarse triangulation is done, and at the refinement steps a size-function

sT = ì
í
î
 ST/16, if TÇG = PiPj (an edge) ST/4, if TÇG = Pi (a vertex) 2 ST else
is used.      Example 1.  Shishkin mesh 3D example

We consider the following boundary value problem:

-e2D u + u = 1     in  [0,1]3,    u|G=0,
where G is the boundary of unite 3D cube. When is small, the solution has boundary layer on G.

The unite cube is divided of 4 domains:
- D1: a central domain , [1-2h,1-2h]3, where we have to place 1/8 of all N 3D mesh points;
- D2: 6 square prism subdomains, each has a face [h,1-2h]2 at the center of a cube face and third dimension h. There we should place 1/16 nodes;
- D3: 12 square prism subdomains, each has an edge [h,1-2h], at the center of a cube edge and size h x h x 1-2h. In this subdomain 1/32 nodes have to be placed;
- D4: 8 cube [0, h]3 subdomains, each one has a vertex, which coincides with a cube vertex contains 1/64 part of all nodes.

It follows that the distribution of the nodes should be:

D1 12.5%   D2 37.5%    D3 37.5%    D4 12.5%.
The distribution of nodes (in the mesh generation step) can be controlled by size function s: R3 R:

s

2
 3 Õ i=1
sign æ
ç
ç
è
½
½
½
½
xi-
1
2
½
½
½
½
+
1
2
-h ö
÷
÷
ø
+3

=

ì
ï
í
ï
î
 1/4 in D1 1/8 in D2 1/16 in D3 1/32 in D4.

We fix the parameter h=0.1,  h=2e ln(N1/3) = 2/3ln Ne = 3/(20 lnN).

For N = 6533, e » 0.0171, e2 = 2.9157 x 10-4, we have: 30528 tetrahedra, 3963 internal nodes, 2570 boundary nodes, 49135 nonzero matrix elements, 210 min solution time.
Distribution of the nodes in subdomains is: D1 8.6%, D2 32.9%, D3 43.2%, D4 15.3%. The maximal solution value is 1.035.
For N = 46161, e » 0.0140, e2 = 1.9507 x 10-4, 46161 nodes, 244224 tetrahedra, 35887 internal nodes, 497179 nonzero matrix elements, 5 hours solution time.
Distribution of the nodes in subdomains: D1 9.5%, D2 35.8%, D3 37.4%, D4 17.3%;  the maximal solution value is 1.024.      Example 3.
- e D u + b. Ñu + cu = f
on unite square and Dirichlet boundary conditions, u|G= 0.
The exact solution is u(x, y) = xy
æ
ç
ç
è
1 - exp æ
ç
ç
è
-
1-x
e
ö
÷
÷
ø
ö
÷
÷
ø
æ
ç
ç
è
1 - exp æ
ç
ç
è
-
1-y
e
ö
÷
÷
ø
ö
÷
÷
ø

Right hand side can be calculated using the function u.

b æ
ç
ç
è
Ö2
2
æ
ç
ç
è
1 -
Ö2
2
x ö
÷
÷
ø
Ö2
2
æ
ç
ç
è
1 -
Ö2
2
y ö
÷
÷
ø
ö
÷
÷
ø
=  æ
ç
ç
è
Ö2-x
2
,
Ö2-y
2
ö
÷
÷,
ø
c = 1, e = 10-4.

There are boundary layers at x = 1 and y = 1 boundary. The thickness of the boundary layers is h = min{e lnN, 1/2}.

We define a comparison function u0(x, y) = xy
æ
ç
ç
è
1 - exp æ
ç
ç
è
-
2-x
e
ö
÷
÷
ø
ö
÷
÷
ø
æ
ç
ç
è
1 - exp æ
ç
ç
è
-
2-y
e
ö
÷
÷
ø
ö
÷
÷.
ø
Then we calculate right hand side (function f) and boundary conditions u0|G.

A "special mesh'' is created in the following rules: At the first step the unite square is divided by 16 equal squares and at the next refinement steps a size-function is used. Any triangle with exact 1 vertex at the boundary x = 1 or y = 1 is split by 4 but any triangle with 2 vertexes at the boundary is split by 16. Every other triangle may split only to keep mesh consistent.

There is a significant correlation between the coefficient kd and the errors (dT = kdST). The best value is 0.05.

Errors.

Let u be the exact solution function and u be the solution obtained in the current iteration, i.e. the values ui= u(Pi) are known at every point Pi, which is a mesh node (i = 1, ... , n).

1. Absolute error (l2 error): eabs2 = åi=1n (u(Pi) - ui)2.

2. Relative error: erel2= eabs2 / (åi=1n u(Pi)2 ) = ( åi=1n (u(Pi) -  ui)2)/ (åi=1nu(Pi)2).

3. Integral error (L2 error):

eint = ó
õ
 W
(u(x, y) - u(x, y))2dxdy
 å TÎT
ó
õ
 T
(u(x, y)- u(x, y))2dxdy »
 å TijkÎT
1
3
æ
ç
ç
è
(u(Pij)-
1
2
( ui + uj))2+ (u(Pjk)-
1
2
( uj + uk))2 + (u(Pki) -
1
2
( uk + ui))2 ö
÷
÷
ø
ST,
where Pij = 1/2(Pi + Pj), ST  is the area of T.

4. Maximal error (l¥ error): emax= maxi=1,... , n |u(Pi) - ui|.

Tables contain: node numbers    eabs   erel     eint    emax.

Regular mesh

```     9  1.737e+01  6.947e+01  5.174e+01  1.737e+01
145  3.588e+01  1.043e+01  2.549e+00  1.657e+01
1089  2.110e+01  2.074e+00  1.503e-01  5.105e+00
4225  1.033e+01  4.957e-01  1.150e-02  1.995e+00
16641  4.183e+00  9.921e-02  1.048e-03  6.665e-01```
Shishkin mesh h = 3.5 x 10-4
`  1089  7.934e+01  5.391e+00  8.320e-01  9.077e+00`
Special mesh
```    25  2.720e+01  3.108e+01  2.019e+01  1.771e+01
63  3.308e+01  1.171e+01  2.825e+00  1.869e+01
171  2.110e+01  3.098e+00  4.002e-01  8.660e+00
470  8.668e+00  7.314e-01  1.430e-02  1.975e+00
1088  3.907e+00  2.009e-01  1.610e-03  6.257e-01
2573  3.786e+00  1.230e-01  3.950e-04  3.603e-01
6616  3.113e+00  6.590e-02  1.697e-04  2.579e-01```
The next tables present the comparison function errors.

Regular mesh

```     9  4.579e-03  3.664e-03  1.306e-03  4.579e-03
145  3.911e-03  9.427e-04  6.723e-07  8.088e-04
1089  2.839e-03  2.541e-04  2.216e-08  3.567e-04
4225  1.252e-03  5.732e-05  1.341e-09  8.267e-05
16641  4.626e-04  1.072e-05  8.074e-11  1.623e-05```
Shishkin mesh h = 3.5 x 10-4
`  1089  2.105e-02  9.642e-04  8.272e-07  3.762e-03`
Special mesh
```    25  2.026e-03  1.080e-03  8.149e-05  1.483e-03
63  5.544e-02  1.437e-02  7.475e-05  2.655e-02
171  7.160e-02  8.925e-03  6.852e-05  2.203e-02
470  9.520e-02  6.993e-03  6.455e-05  2.126e-02
1088  1.108e-01  5.034e-03  6.481e-05  2.131e-02
2573  1.173e-01  3.339e-03  5.819e-05  1.827e-02
6616  1.570e-01  2.893e-03  5.819e-05  1.827e-02```      Example 4. Complicated domain problems in 3D.
- div (µ  grad  f) = 0   on  R3,
with Dirichlet boundary conditions
f| {z = z0} = z0,
i.e. the boundary condition is a linear function in the direction z and a constant in the directions x and y.

The domain W consists of two parts W0 Ì W, W0 << W and W \W0.

µ =  ì
í
î
 2.5 in W0 1 in W\W0

Case 1. W = [0, 1]3, W0 = (0.5, 0.5) + [0, 0.05]3
Case 2. W = [-60, 60]2 x[0, 120], W0 = ([-31, 31]2\[-29, 29]2) x [57.5, 62.5].
Case 3. Let K be the unite 16-gone. Then W=120*Kx [-60, 60],
W0 = (62 x K)\(58*K) x [-2.5, 2.5].        In order to present intersection of the solution u and a plane x=const, we calculate and visualize the "difference''
u - ul = u(x, y, z) - z.          This document was translated from LATEX by HEVEA.