>> fem1d

05-Oct-2008 10:29:45

FEM1D
  MATLAB version

  Solve the two-point boundary value problem:

    -d/dx (p(x) du/dx) + q(x)*u  =  f(x)

  on an interval [xl,xr], with the values of
  u or u' specified at xl and xr.

  The interval is broken into 10 subintervals.
  The number of basis functions per element is 2

The equation is to be solved for
X greater than XL  =   and less than XR = 
The boundary conditions are:

  At X = XL, U = 0.000000
  At X = XR, U' = 1.000000

Number of quadrature points per element is 1

  Node      Location

       0      0.000000
       1      0.100000
       2      0.200000
       3      0.300000
       4      0.400000
       5      0.500000
       6      0.600000
       7      0.700000
       8      0.800000
       9      0.900000
      10      1.000000

Subint    Length

       1      0.100000
       2      0.100000
       3      0.100000
       4      0.100000
       5      0.100000
       6      0.100000
       7      0.100000
       8      0.100000
       9      0.100000
      10      0.100000

Subint    Quadrature point

       1      0.050000
       2      0.150000
       3      0.250000
       4      0.350000
       5      0.450000
       6      0.550000
       7      0.650000
       8      0.750000
       9      0.850000
      10      0.950000

Subint  Left Node  Right Node

       1       0       1
       2       1       2
       3       2       3
       4       3       4
       5       4       5
       6       5       6
       7       6       7
       8       7       8
       9       8       9
      10       9      10

  Node  Unknown

       0      -1
       1       1
       2       2
       3       3
       4       4
       5       5
       6       6
       7       7
       8       8
       9       9
      10      10

Printout of tridiagonal linear system:

Equation   ALEFT         ADIAG         ARITE         RHS

  1                   20.000000    -10.000000      0.000000
  2    -10.000000     20.000000    -10.000000      0.000000
  3    -10.000000     20.000000    -10.000000      0.000000
  4    -10.000000     20.000000    -10.000000      0.000000
  5    -10.000000     20.000000    -10.000000      0.000000
  6    -10.000000     20.000000    -10.000000      0.000000
  7    -10.000000     20.000000    -10.000000      0.000000
  8    -10.000000     20.000000    -10.000000      0.000000
  9    -10.000000     20.000000    -10.000000      0.000000
 10    -10.000000     10.000000                    1.000000

Computed solution:

    Node        X(I)          U(I)

       0      0.000000      0.000000
       1      0.100000      0.100000
       2      0.200000      0.200000
       3      0.300000      0.300000
       4      0.400000      0.400000
       5      0.500000      0.500000
       6      0.600000      0.600000
       7      0.700000      0.700000
       8      0.800000      0.800000
       9      0.900000      0.900000
      10      1.000000      1.000000

FEM1D:
  Normal end of execution.

05-Oct-2008 10:29:45
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