6 November  2023   8:51:08.116 AM      
 
fem1d():
  FORTRAN77 version.
   A finite element solver for a 1D problem.
  
   Solve the two-point boundary value problem
  
   -d/dx ( P(x) dU(x)/dx ) + Q(x) U(x) = F(x)
  
   on the interval [XL,XR], specifying
   the value of U or U' at each end.
  
   The interval [XL,XR] is broken into NSUB =            5  subintervals
   Piecewise linear finite element functions are used.
   The number of basis functions that are nonzero
   in any element is at most NL =            2
  
   The equation is to be solved for
   X greater than XL =    0.0000000000000000       and less than XR =    1.0000000000000000     
  
   The boundary conditions are:
  
   At X=XL, U=   0.0000000000000000     
   At X=XR, U'=   1.0000000000000000     
  
   Number of quadrature points per element is            1
  
   Node      Location
  
           0   0.0000000000000000     
           1  0.20000000000000001     
           2  0.40000000000000002     
           3  0.59999999999999998     
           4  0.80000000000000004     
           5   1.0000000000000000     
  
   Subint    Length
  
           1  0.20000000000000001     
           2  0.20000000000000001     
           3  0.19999999999999996     
           4  0.20000000000000007     
           5  0.19999999999999996     
  
   Subint    Quadrature point
  
           1  0.10000000000000001     
           2  0.30000000000000004     
           3  0.50000000000000000     
           4  0.69999999999999996     
           5  0.90000000000000002     
  
   Subint  Left Node  Right Node
  
           1           0           1
           2           1           2
           3           2           3
           4           3           4
           5           4           5
  
   Number of unknowns NU =            5
  
   Node  Unknown
  
           0          -1
           1           1
           2           2
           3           3
           4           4
           5           5
 
  The tridiagonal linear system:
 
  Equation  ALEFT  ADIAG  ARITE  RHS
 
           1   0.0000000000000000        10.000000000000000       -5.0000000000000000        0.0000000000000000     
           2  -5.0000000000000000        10.000000000000000       -5.0000000000000009        0.0000000000000000     
           3  -5.0000000000000009        10.000000000000000       -4.9999999999999982        0.0000000000000000     
           4  -4.9999999999999982        10.000000000000000       -5.0000000000000009        0.0000000000000000     
           5  -5.0000000000000009        5.0000000000000009        0.0000000000000000        1.0000000000000000     
  
 Computed solution:
  
 Node    X(I)        U(X(I))
  
           0   0.0000000000000000        0.0000000000000000     
           1  0.20000000000000001       0.19999999999999979     
           2  0.40000000000000002       0.39999999999999958     
           3  0.59999999999999998       0.59999999999999931     
           4  0.80000000000000004       0.79999999999999927     
           5   1.0000000000000000       0.99999999999999911     
 
fem1d():
  Normal end of execution.
 
 6 November  2023   8:51:08.116 AM