I have a solution to a discretized differential equation given by
f(i)
where i is a spatial index. How can I find the difference between the solution at each adjacent time step? To be more clear:
The solution is defined by an array
real,dimension(0:10) :: f
I discretize the differential equation and solve it by stepping forward in time. If the time index is k, a portion of my code is
do k=1,25
do i = 1,10
f(i) = f(i+1)+f(i-1)+f(i)
end do
end do
I can print the solution, f(i) at each time step k by the following code
print*, "print f(i) for k=, k
print "(//(5(5x,e22.14)))", f
How can I find the difference between the solution at each adjacent time step? That is, time steps k+1,k. I will store this value in a new array g, which has a dimension given by
real,dimension(0:10) :: g
So I am trying to find
!g(i)=abs(f(i;k+1)-f(i;k))...Not correct code.
How can I do this? What is the way to implement this code? I am not sure how to do this using if /then statements or whatever code would need be needed to do this. Thanks
Typically, in explicit time integration methods or iterative methods, you have to save the last time-step last solution, the current time-step solution and possibly even some more.
So you have
real,dimension(0:10) :: f0, f
where f0 is the previous value
You iterate your Jacobi or Gauss-Seidel discretization:
f = f0
do k=1,25
do i = 1,9
f(i) = f(i+1)+f(i-1)+f(i)
end do
max_diff = maxval(abs(f-f0))
if (diff small enough) exit
f0 = f
end do
If you have a time-evolving problem like a heat equation:
f = f0
do k=1,25
do i = 1,9
f(i) = f0(i) + dt * viscosity * (f0(i+1)+f0(i-1)+f0(i))
end do
max_diff = maxval(abs(f-f0))
f0 = f
end do