2 July 2014 8:28:20.671 AM FEM2D_BVP_SERENE_TEST FORTRAN77 version Test the FEM2D_BVP_SERENE library. TEST01 Solve - del ( A del U ) + C U = F on the unit square with zero boundary conditions. A1(X,Y) = 1.0 C1(X,Y) = 0.0 F1(X,Y) = 2*X*(1-X)+2*Y*(1-Y). U1(X,Y) = X * ( 1 - X ) * Y * ( 1 - Y ) The grid uses 5 by 5 nodes. The number of nodes is 21 I J X Y U Uexact Error 1 1 0.0000 0.0000 0.0000 0.0000 0.00E+00 2 1 0.2500 0.0000 0.0000 0.0000 0.00E+00 3 1 0.5000 0.0000 0.0000 0.0000 0.00E+00 4 1 0.7500 0.0000 0.0000 0.0000 0.00E+00 5 1 1.0000 0.0000 0.0000 0.0000 0.00E+00 1 2 0.0000 0.2500 0.0000 0.0000 0.00E+00 3 2 0.5000 0.2500 0.0486 0.0469 0.18E-02 5 2 1.0000 0.2500 0.0000 0.0000 0.00E+00 1 3 0.0000 0.5000 0.0000 0.0000 0.00E+00 2 3 0.2500 0.5000 0.0486 0.0469 0.18E-02 3 3 0.5000 0.5000 0.0590 0.0625 0.35E-02 4 3 0.7500 0.5000 0.0486 0.0469 0.18E-02 5 3 1.0000 0.5000 0.0000 0.0000 0.00E+00 1 4 0.0000 0.7500 0.0000 0.0000 0.00E+00 3 4 0.5000 0.7500 0.0486 0.0469 0.18E-02 5 4 1.0000 0.7500 0.0000 0.0000 0.00E+00 1 5 0.0000 1.0000 0.0000 0.0000 0.00E+00 2 5 0.2500 1.0000 0.0000 0.0000 0.00E+00 3 5 0.5000 1.0000 0.0000 0.0000 0.00E+00 4 5 0.7500 1.0000 0.0000 0.0000 0.00E+00 5 5 1.0000 1.0000 0.0000 0.0000 0.00E+00 l1 error = 0.501605E-03 L2 error = 0.805291E-03 H1S error = 0.123351E-01 TEST02 Basis function checks. The matrix Aij = V(i)(X(j),Y(j)) should be the identity. V(i)(X(j),Y(j)) Col 1 2 3 4 5 Row 1: 1.00000 0.00000 -0.00000 -0.00000 -0.00000 2: -0.00000 1.00000 0.00000 0.00000 0.00000 3: 0.00000 0.00000 1.00000 0.00000 -0.00000 4: 0.00000 -0.00000 -0.00000 1.00000 0.00000 5: -0.00000 0.00000 0.00000 -0.00000 1.00000 6: 0.00000 -0.00000 -0.00000 0.00000 0.00000 7: 0.00000 0.00000 0.00000 -0.00000 -0.00000 8: -0.00000 -0.00000 -0.00000 0.00000 0.00000 Col 6 7 8 Row 1: -0.00000 -0.00000 0.00000 2: 0.00000 -0.00000 -0.00000 3: -0.00000 0.00000 0.00000 4: 0.00000 -0.00000 -0.00000 5: 0.00000 0.00000 0.00000 6: 1.00000 -0.00000 -0.00000 7: 0.00000 1.00000 0.00000 8: 0.00000 0.00000 1.00000 The vectors dVdX(1:8)(X,Y) and dVdY(1:8)(X,Y) should both sum to zero for any (X,Y). Random evaluation point is (0.4368 , 4.913 ) dVdX dVdY 1 -.1022 0.1378 2 1.077 0.3414 3 -.9749 0.9334 4 -.8355E-01 -1.427 5 -.4667E-02 0.4932 6 0.4920E-01 -.3414 7 -.4453E-01 0.2608 8 0.8355E-01 -.3987 Sum: -.9714E-16 0.1110E-15 TEST03 Solve - del ( A del U ) + C U = F on the unit square with zero boundary conditions. A1(X,Y) = 0.0 C1(X,Y) = 1.0 F1(X,Y) = X * ( 1 - X ) * Y * ( 1 - Y ). U1(X,Y) = X * ( 1 - X ) * Y * ( 1 - Y ) This example is contrived so that the system matrix is the WATHEN matrix. The grid uses 5 by 5 nodes. The number of nodes is 21 Wathen elementary mass matrix: Col 1 2 3 4 5 Row 1: 6.00000 -6.00000 2.00000 -8.00000 3.00000 2: -6.00000 32.0000 -6.00000 20.0000 -8.00000 3: 2.00000 -6.00000 6.00000 -6.00000 2.00000 4: -8.00000 20.0000 -6.00000 32.0000 -6.00000 5: 3.00000 -8.00000 2.00000 -6.00000 6.00000 6: -8.00000 16.0000 -8.00000 20.0000 -6.00000 7: 2.00000 -8.00000 3.00000 -8.00000 2.00000 8: -6.00000 20.0000 -8.00000 16.0000 -8.00000 Col 6 7 8 Row 1: -8.00000 2.00000 -6.00000 2: 16.0000 -8.00000 20.0000 3: -8.00000 3.00000 -8.00000 4: 20.0000 -8.00000 16.0000 5: -6.00000 2.00000 -8.00000 6: 32.0000 -6.00000 20.0000 7: -6.00000 6.00000 -6.00000 8: 20.0000 -6.00000 32.0000 I J X Y U Uexact Error 1 1 0.0000 0.0000 0.0000 0.0000 0.00E+00 2 1 0.2500 0.0000 0.0000 0.0000 0.00E+00 3 1 0.5000 0.0000 0.0000 0.0000 0.00E+00 4 1 0.7500 0.0000 0.0000 0.0000 0.00E+00 5 1 1.0000 0.0000 0.0000 0.0000 0.00E+00 1 2 0.0000 0.2500 0.0000 0.0000 0.00E+00 3 2 0.5000 0.2500 0.0488 0.0469 0.20E-02 5 2 1.0000 0.2500 0.0000 0.0000 0.00E+00 1 3 0.0000 0.5000 0.0000 0.0000 0.00E+00 2 3 0.2500 0.5000 0.0488 0.0469 0.20E-02 3 3 0.5000 0.5000 0.0586 0.0625 0.39E-02 4 3 0.7500 0.5000 0.0488 0.0469 0.20E-02 5 3 1.0000 0.5000 0.0000 0.0000 0.00E+00 1 4 0.0000 0.7500 0.0000 0.0000 0.00E+00 3 4 0.5000 0.7500 0.0488 0.0469 0.20E-02 5 4 1.0000 0.7500 0.0000 0.0000 0.00E+00 1 5 0.0000 1.0000 0.0000 0.0000 0.00E+00 2 5 0.2500 1.0000 0.0000 0.0000 0.00E+00 3 5 0.5000 1.0000 0.0000 0.0000 0.00E+00 4 5 0.7500 1.0000 0.0000 0.0000 0.00E+00 5 5 1.0000 1.0000 0.0000 0.0000 0.00E+00 l1 error = 0.558036E-03 L2 error = 0.781250E-03 H1S error = 0.124347E-01 WATHEN matrix (permuted) Col 1 2 3 4 5 Row 1: 6.00000 -6.00000 2.00000 -6.00000 -8.00000 2: -6.00000 32.0000 -6.00000 20.0000 20.0000 3: 2.00000 -6.00000 6.00000 -8.00000 -6.00000 4: -6.00000 20.0000 -8.00000 32.0000 16.0000 5: -8.00000 20.0000 -6.00000 16.0000 32.0000 6: 2.00000 -8.00000 3.00000 -6.00000 -8.00000 7: -8.00000 16.0000 -8.00000 20.0000 20.0000 8: 3.00000 -8.00000 2.00000 -8.00000 -6.00000 Col 6 7 8 Row 1: 2.00000 -8.00000 3.00000 2: -8.00000 16.0000 -8.00000 3: 3.00000 -8.00000 2.00000 4: -6.00000 20.0000 -8.00000 5: -8.00000 20.0000 -6.00000 6: 6.00000 -6.00000 2.00000 7: -6.00000 32.0000 -6.00000 8: 2.00000 -6.00000 6.00000 FEM2D_BVP_SERENE_TEST Normal end of execution. 2 July 2014 8:28:20.683 AM