C 代码 应用有限元法 （FEM） 时使用自适应网格.rar

19 January 2020 11:30:34 AM FEM1D_ADAPTIVE C version Solve the two-point boundary value problem: -d/dx ( P(x) * dU(x)/dx ) + Q(x) * U(x) = F(x) on the interval [0,1], specifying the value of U at each endpoint. The number of basis functions per element is 2 The number of quadrature points per element is 2 Problem index = 6 "ARCTAN" problem: U(X) = ATAN((X-0.5)/A) P(X) = 1.0 Q(X) = 0.0 F(X) = 2*A*(X-0.5)/(A**2+(X-0.5)**2)**2 IBC = 3 UL = ATAN(-0.5/A) UR = ATAN( 0.5/A) A = 0.01 Arctangent problem The equation is to be solved for X greater than 0 and less than 1 The boundary conditions are: At X = XL, U = -1.5508 At X = XR, U= 1.5508 Begin new iteration with 4 nodes. Printout of tridiagonal linear system: Equation A-Left A-Diag A-Rite RHS 1 8 -4 -9.87506 2 -4 8 -4 1.38778e-16 3 -4 8 9.87506 Basic solution Node X(I) U(X(I)) Uexact Error 0 0 -1.5508 -1.5508 0 1 0.25 -1.23438 -1.53082 0.296435 2 0.5 -5.55112e-16 0 -5.55112e-16 3 0.75 1.23438 1.53082 -0.296435 4 1 1.5508 1.5508 0 ETA 0.244233 2.19633 2.19633 0.244233 Tolerance = 1.46435 Subdivide interval 2 Subdivide interval 3 Begin new iteration with 6 nodes. Basic solution Node X(I) U(X(I)) Uexact Error 0 0 -1.5508 -1.5508 0 1 0.25 -1.57269 -1.53082 -0.0418763 2 0.375 -1.55364 -1.49097 -0.0626714 3 0.5 -6.55032e-14 0 -6.55032e-14 4 0.625 1.55364 1.49097 0.0626714 5 0.75 1.57269 1.53082 0.0418763 6 1 1.5508 1.5508 0 ETA 0.00944353 0.185938 3.52685 3.52685 0.185938 0.00944353 Tolerance = 1.4889 Subdivide interval 3 Subdivide interval 4 Begin new iteration with 8 nodes. Basic solution Node X(I) U(X(I)) Uexact Error 0 0 -1.5508 -1.5508 0 1 0.25 -1.74416 -1.53082 -0.213342 2 0.375 -1.81084 -1.49097 -0.31987 3 0.4375 -1.78503 -1.41214 -0.372888 4 0.5 -2.44249e-15 0 -2.44249e-15 5 0.5625 1.78503 1.41214 0.372888 6 0.625 1.81084 1.49097 0.31987 7 0.75 1.74416 1.53082 0.213342 8 1 1.5508 1.5508 0 ETA 0.00944353 0.026573 0.292454 3.3433 3.3433 0.292454 0.026573 0.00944353 Tolerance = 1.10154 Subdivide interval 4 Subdivide interval 5 Begin new iteration with 10 nodes. Basic solution Node X(I) U(X(I)) Uexact Error 0 0 -1.5508 -1.5508 0 1 0.25 -1.56403 -1.53082 -0.0332166 2 0.375 -1.54065 -1.49097 -0.0496819 3 0.4375 -1.46981 -1.41214 -0.0576689 4 0.46875 -1.32254 -1.26109 -0.0614455 5 0.5 6.88338e-14 0 6.88338e-14 6 0.53125 1.32254 1.26109 0.0614455 7 0.5625 1.46981 1.41214 0.0576689 8 0.625 1.54065 1.49097 0.0496819 9 0.75 1.56403 1.53082 0.0332166 10 1 1.5508 1.5508 0 ETA 0.00944353 0.026573 0.0736369 0.235021 2.82238 2.82238 0.235021 0.0736369 0.026573 0.00944353 Tolerance = 0.760103 Subdivide interval 5 Subdivide interval 6 Begin new iteration with 12 nodes. Basic solution Node X(I) U(X(I)) Uexact Error 0 0 -1.5508 -1.5508 0 1 0.25 -1.51216 -1.53082 0.0186565 2 0.375 -1.46284 -1.49097 0.0281279 3 0.4375 -1.37903 -1.41214 0.0331091 4 0.46875 -1.22528 -1.26109 0.0358167 5 0.484375 -0.964992 -1.00148 0.0364909 6 0.5 4.55191e-14 0 4.55191e-14 7 0.515625 0.964992 1.00148 -0.0364909 8 0.53125 1.22528 1.26109 -0.0358167 9 0.5625 1.37903 1.41214 -0.0331091 10 0.625 1.46284 1.49097 -0.0281279 11 0.75 1.51216 1.53082 -0.0186565 12 1 1.5508 1.5508 0 ETA 0.00944353 0.026573 0.0736369 0.192415 0.41609 1.8311 1.8311 0.41609 0.192415 0.0736369 0.026573 0.00944353 Tolerance = 0.509861 Subdivide interval 6 Subdivide interval 7 Begin new iteration with 14 nodes. Basic solution Node X(I) U(X(I)) Uexact Error 0 0 -1.5508 -1.5508 0 1 0.25 -1.52873 -1.53082 0.00209243 2 0.375 -1.48768 -1.49097 0.00328175 3 0.4375 -1.40802 -1.41214 0.00412196 4 0.46875 -1.25633 -1.26109 0.00475898 5 0.484375 -0.997085 -1.00148 0.00439793 6 0.492188 -0.66027 -0.663203 0.0029329 7 0.5 2.08722e-14 0 2.08722e-14 8 0.507812 0.66027 0.663203 -0.0029329 9 0.515625 0.997085 1.00148 -0.00439793 10 0.53125 1.25633 1.26109 -0.00475898 11 0.5625 1.40802 1.41214 -0.00412196 12 0.625 1.48768 1.49097 -0.00328175 13 0.75 1.52873 1.53082 -0.00209243 14 1 1.5508 1.5508 0 ETA 0.00944353 0.026573 0.0736369 0.192415 0.410124 0.512307 0.653928 0.653928 0.512307 0.410124 0.192415 0.0736369 0.026573 0.00944353 Tolerance = 0.322026 Subdivide interval 5 Subdivide interval 6 Subdivide interval 7 Subdivide interval 8 Subdivide interval 9 Subdivide interval 10 Begin new iteration with 20 nodes. Basic solution Node X(I) U(X(I)) Uexact Error 0 0 -1.5508 -1.5508 0 1 0.25 -1.52941 -1.53082 0.00140268 2 0.375 -1.48872 -1.49097 0.00224713 3 0.4375 -1.40923 -1.41214 0.0029149 4 0.46875 -1.25763 -1.26109 0.0034657 5 0.476562 -1.1649 -1.16751 0.00261453 6 0.484375 -0.999675 -1.00148 0.00180788 7 0.488281 -0.863012 -0.86437 0.00135779 8 0.492188 -0.662316 -0.663203 0.000886981 9 0.496094 -0.372028 -0.372398 0.000370799 10 0.5 -3.44724e-14 0 -3.44724e-14 11 0.503906 0.372028 0.372398 -0.000370799 12 0.507812 0.662316 0.663203 -0.000886981 13 0.511719 0.863012 0.86437 -0.00135779 14 0.515625 0.999675 1.00148 -0.00180788 15 0.523438 1.1649 1.16751 -0.00261453 16 0.53125 1.25763 1.26109 -0.0034657 17 0.5625 1.40923 1.41214 -0.0029149 18 0.625 1.48872 1.49097 -