09-Dec-2022 16:30:12

advection_pde_test()
  MATLAB/Octave version 4.2.2
  Solve the advection PDE.

  Parameter values():
    c     = 5
    t0    = 0
    tstop = 0.08
    xmin  = 0
    xmax  = 1

advection_pde_ftcs():
  Solve the constant-velocity advection PDE in 1D,
    du/dt + c du/dx = 0
  over the interval:
    0.0 <= x <= 1.0
  with periodic boundary conditions:
    u(0) = u(1)
  and advection velocity
    c = constant
  and initial condition
    u(0,x) = (10x-6)^2 (8-10x)^2 for 0.6 <= x <= 0.8
           = 0 elsewhere.
  and NX equally spaced nodes in X,
  and NT equally spaced points in T,
  using the FTCS method:
    FT: Forward Time  : du/dt = (u(t+dt,x)-u(t,x))/dt
    CS: Centered Space: du/dx = (u(t,x+dx)-u(t,x-dx))/2/dx

  Number of nodes NX = 101
  Number of time steps NT = 8001
  Graphics saved as "advection_pde_ftcs_1.png"
  Graphics saved as "advection_pde_ftcs_801.png"
  Graphics saved as "advection_pde_ftcs_1601.png"
  Graphics saved as "advection_pde_ftcs_2401.png"
  Graphics saved as "advection_pde_ftcs_3201.png"
  Graphics saved as "advection_pde_ftcs_4001.png"
  Graphics saved as "advection_pde_ftcs_4801.png"
  Graphics saved as "advection_pde_ftcs_5601.png"
  Graphics saved as "advection_pde_ftcs_6401.png"
  Graphics saved as "advection_pde_ftcs_7201.png"
  Graphics saved as "advection_pde_ftcs_8001.png"
  Graphics saved as "advection_pde_ftcs_conservation.png"

advection_pde_test()
  Normal end of execution.
09-Dec-2022 16:34:50