C 代码 应用有限元法（FEM），与偶然性元素， 矩形中的二维边界值问题 （BVP）.rar

19 January 2020 11:30:35 AM FEM2D_BVP_SERENE_TEST C 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 ) Number of X grid values NX = 5 Number of Y grid values NY = 5 I J X Y U Uexact Error 0 0 0.000000 0.000000 0.000000 0.000000 0.000000e+00 1 0 0.250000 0.000000 0.000000 0.000000 0.000000e+00 2 0 0.500000 0.000000 0.000000 0.000000 0.000000e+00 3 0 0.750000 0.000000 0.000000 0.000000 0.000000e+00 4 0 1.000000 0.000000 0.000000 0.000000 0.000000e+00 0 1 0.000000 0.250000 0.000000 0.000000 0.000000e+00 2 1 0.500000 0.250000 0.048631 0.046875 1.755618e-03 4 1 1.000000 0.250000 0.000000 0.000000 0.000000e+00 0 2 0.000000 0.500000 0.000000 0.000000 0.000000e+00 1 2 0.250000 0.500000 0.048631 0.046875 1.755618e-03 2 2 0.500000 0.500000 0.058989 0.062500 3.511236e-03 3 2 0.750000 0.500000 0.048631 0.046875 1.755618e-03 4 2 1.000000 0.500000 0.000000 0.000000 0.000000e+00 0 3 0.000000 0.750000 0.000000 0.000000 0.000000e+00 2 3 0.500000 0.750000 0.048631 0.046875 1.755618e-03 4 3 1.000000 0.750000 0.000000 0.000000 0.000000e+00 0 4 0.000000 1.000000 0.000000 0.000000 0.000000e+00 1 4 0.250000 1.000000 0.000000 0.000000 0.000000e+00 2 4 0.500000 1.000000 0.000000 0.000000 0.000000e+00 3 4 0.750000 1.000000 0.000000 0.000000 0.000000e+00 4 4 1.000000 1.000000 0.000000 0.000000 0.000000e+00 l1 norm of error = 0.000501605 L2 norm of error = 0.000805291 Seminorm of error = 0.0123351 TEST02 Basis function checks. The matrix Aij = V(j)(X(i),Y(i)) should be the identity. 1.0000 -0.0000 0.0000 0.0000 -0.0000 0.0000 0.0000 -0.0000 0.0000 1.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 -0.0000 0.0000 1.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 -0.0000 0.0000 0.0000 1.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 1.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 -0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 1.0000 0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 1.0000 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 (0.436837,4.91264) dVdX dVdY 0 -0.1022 0.1378 1 1.077 0.3414 2 -0.9749 0.9334 3 -0.08355 -1.427 4 -0.004667 0.4932 5 0.0492 -0.3414 6 -0.04453 0.2608 7 0.08355 -0.3987 Sum: -9.714e-17 1.11e-16 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. Number of X grid values NX = 5 Number of Y grid values NY = 5 Wathen elementary mass matrix: Col: 0 1 2 3 4 Row 0: 6 -6 2 -8 3 1: -6 32 -6 20 -8 2: 2 -6 6 -6 2 3: -8 20 -6 32 -6 4: 3 -8 2 -6 6 5: -8 16 -8 20 -6 6: 2 -8 3 -8 2 7: -6 20 -8 16 -8 Col: 5 6 7 Row 0: -8 2 -6 1: 16 -8 20 2: -8 3 -8 3: 20 -8 16 4: -6 2 -8 5: 32 -6 20 6: -6 6 -6 7: 20 -6 32 I J X Y U Uexact Error 0 0 0.000000 0.000000 0.000000 0.000000 0.000000e+00 1 0 0.250000 0.000000 0.000000 0.000000 0.000000e+00 2 0 0.500000 0.000000 0.000000 0.000000 0.000000e+00 3 0 0.750000 0.000000 0.000000 0.000000 0.000000e+00 4 0 1.000000 0.000000 0.000000 0.000000 0.000000e+00 0 1 0.000000 0.250000 0.000000 0.000000 0.000000e+00 2 1 0.500000 0.250000 0.048828 0.046875 1.953125e-03 4 1 1.000000 0.250000 0.000000 0.000000 0.000000e+00 0 2 0.000000 0.500000 0.000000 0.000000 0.000000e+00 1 2 0.250000 0.500000 0.048828 0.046875 1.953125e-03 2 2 0.500000 0.500000 0.058594 0.062500 3.906250e-03 3 2 0.750000 0.500000 0.048828 0.046875 1.953125e-03 4 2 1.000000 0.500000 0.000000 0.000000 0.000000e+00 0 3 0.000000 0.750000 0.000000 0.000000 0.000000e+00 2 3 0.500000 0.750000 0.048828 0.046875 1.953125e-03 4 3 1.000000 0.750000 0.000000 0.000000 0.000000e+00 0 4 0.000000 1.000000 0.000000 0.000000 0.000000e+00 1 4 0.250000 1.000000 0.000000 0.000000 0.000000e+00 2 4 0.500000 1.000000 0.000000 0.000000 0.000000e+00 3 4 0.750000 1.000000 0.000000 0.000000 0.000000e+00 4 4 1.000000 1.000000 0.000000 0.000000 0.000000e+00 l1 norm of error = 0.000558036 L2 norm of error = 0.00078125 Seminorm of error = 0.0124347 Wathen matrix: Col: 0 1 2 3 4 Row 0: 6 -6 2 -6 -8 1: -6 32 -6 20 20 2: 2 -6 6 -8 -6 3: -6 20 -8 32 16 4: -8 20 -6 16 32 5: 2 -8 3 -6 -8 6: -8 16 -8 20 20 7: 3 -8 2 -8 -6 Col: 5 6 7 Row 0: 2 -8 3 1: -8 16 -8 2: 3 -8 2 3: -6