Modeling a wave on a string using finite differences

Based on problem 3.7 chapter 3 of Introduction to Seismology (Shearer)

(COMPUTER) In the case of plane-wave propagation in the x direction within a uniform medium, the homogeneous momentum equation (3.9) for shear waves can be expressed as

screenshot-from-2016-12-14-15-42-21

,where u is the displacement. Write a computer program that uses finite differences to solve this equation for a bar 100 km in length, assuming β = 4 km/s. Use dx = 1 km for the length spacing and dt = 0.1 s for the time spacing. Assume a source-time function  at u (50 km) of the form:

screenshot-from-2016-12-14-15-42-16

Apply a stress-free boundary condition at u(0 km) and a fixed boundary condition at u (100 km).  Approximate the second derivatives using the finite difference scheme:

screenshot-from-2016-12-14-15-42-25

Plot u(x) at 4 s intervals from 1 to 33 s. Verify that the pulses travel at velocities of 4 km/s. What happens to the reflected pulse at each endpoint? What happens when the pulses cross?

Solutions can be downloaded here.

!**********************************************************************
!* *
!* 1-D Plane wave propagation with fixed and free boundaries * 
!* *
!**********************************************************************

!----initialzie t, dx, dt, tlen, beta, and u1, u2, u3 arrays----c

implicit none
 integer:: i, counter, nx, nx1
 integer:: it, ntmax
 real(8):: t, dt, dt2, dx, dx2, beta, beta2, tlen, tmax, rhs
 real*8, allocatable:: u1(:), u2(:), u3(:)
 character(10):: data
 character(4):: nf
!----parameters ----c
! dt must be less than (dx/beta) for numerical stability

nx = 100 ! number of grid
 dt = 0.10d0 ! time interval
 dx = 1.0d0 ! grid interval
 beta = 4.0d0 ! wave velocity
 tlen = 5.0d0 ! time length of wave source
 ntmax = 1000 ! maximum time step (finish at it = ntmax, or)
 tmax = 33.0d0 ! maximum calculation time (finish at t< tmax)

!---- allocate dimension variables ----c

nx1 = nx + 1
 allocate (u1(nx1), u2(nx1), u3(nx1))

!---- initialize t, counter, u1, u2, u3 ----c

t = 0.0d0
 counter = 1
 
 do i = 1, nx1
 u1(i) = 0.0d0
 u2(i) = 0.0d0
 u3(i) = 0.0d0
 end do

!---- calculate c^2, dt^2, dx^2 ----c
 beta2 = beta**2
 dt2 = dt**2
 dx2 = dx**2

! ============= Time marching loop ===========
 
 do it = 1, ntmax

t = t + dt

!---- calculate u3(i) ----c

do i= 2, nx
 rhs=beta2*2*(u2(i+1)-2.0d0*u2(i)+u2(i-1))/dx2
 u3(i)=dt2*rhs+2.0d0*u2(i)-u1(i)
 end do

!---- free boundary (du/dx=0) ----c
 u3(1)=u3(2)
!---- fixed boundary (u=0) ----c
 u3(nx1)=0.0d0
!---- source time function c----c

if (t.le.tlen) then
 u3(51) = sin(3.1415927*t/tlen)**2
 end if

!---- change new and old variables ----c

do i=1,nx1
 u1(i) = u2(i)
 u2(i) = u3(i)
 end do
!---- make data file ----c
 write(nf,'(i3.3)') counter
 data = "data"//nf

open(7,file=data,status='replace')
 do i=1, nx1
 write(7,*) i, u2(i)
 end do
!---- output u2 at desired intervals, stop when t is big enough

write(6,'(a5,1x,i3,1x,f6.3)') 'loop', counter, t
 
 if (t.gt.tmax) exit

counter = counter + 1

end do

stop
 end

The following is used to plot the results:

#!/usr/bin/bash
gnuplot << eof
set term gif animate
set output "animate.gif"
n=9 #n frames
set xrange [0:102]
set yrange [-2:2]
i=0
list = system('ls data*')
do for [file in list] {
 plot file w lines lt 1 lw 1.5 t sprintf("t=%i sec",i/10)
 i = i + 1
}
set output
eof

animate

Nguyen Cong Nghia – IESAS

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