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14-Sep-2024 17:20:08

fisher_pde_ftcs():
  Solve the KPP Fisher PDE using ftcs()
    du/dt = uxx + u * ( 1 - u )
  over the spatial interval:
    0 <= x <= 1
  and time interval
     0 <= t <= 0.05
  with boundary conditions:
    u(0) = ( cos(100*t(j)) + 1.5 ) / 2.5 (Dirichlet)
    u'(1) = 0.0 (Neumann)
  and initial condition
    u(0,x) = cos(10*x).
  Space steps = 51
  Space interval is [0,1]
  Space increment is 0.02

  Time steps = 801
  Time interval is [0,0.05]
  Time increment is 6.25e-05
  Graphics saved as "fisher_pde_ftcs_initial.png"
  Graphics saved as "fisher_pde_ftcs_final.png"
  Movie saved as "fisher_pde_ftcs.avi"

fisher_pde_ftcs():
  Normal end of execution.
14-Sep-2024 17:20:36