Problems solving

We consider the following boundary value problem:

We create Shishkin mesh for solving this problem.

 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 :

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

We consider the following boundary value problem:

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:

We fix the parameter h=0.1,  h=2e ln(N^1/3) = 2/3ln N,  e = 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.

Right hand side can be calculated using the function u.

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

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 k_d and the errors (d_T = k_dS_T). 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 u_i= u(P_i) are known at every point P_i, which is a mesh node (i = 1, ... , n).

1. Absolute error (l₂ error): e_abs² = å_i=1ⁿ (u(P_i) - u_i)².

2. Relative error: e_rel²= e_abs² / (å_i=1ⁿ u(P_i)² ) = ( å_i=1ⁿ (u(P_i) -  u_i)²)/ (å_i=1ⁿu(P_i)²).

3. Integral error (L₂ error):

4. Maximal error (l_¥ error): e_max= max_{i=1,... , n} |u(P_i) - u_i|.

Tables contain: node numbers    e_abs   e_rel     e_int    e_max.

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

  1089  7.934e+01  5.391e+00  8.320e-01  9.077e+00

    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

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

  1089  2.105e-02  9.642e-04  8.272e-07  3.762e-03

    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

The domain W consists of two parts W₀ Ì W, W₀ << W and 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 - u_l = u(x, y, z) - z.

This document was translated from L^AT_EX by H^EV^EA.

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