SLIDE 18 Symmetric parareal: some basic remarks - 1
At the initial iteration (k = 0), we use the symmetric version of G: xk=0
n+1/2 = G−1 −∆T/2(xk=0 n
), xk=0
n+1 = G∆T/2(xk=0 n+1/2).
we next iterate over k: xk+1
n+1/2
= G−1
−∆T/2
n
− F−∆T/2(xk
n+1/2) + G−∆T/2(xk n+1/2)
xk+1
n+1
= G∆T/2(xk+1
n+1/2) + F∆T/2(xk n+1/2) − G∆T/2(xk n+1/2)
The flow is symmetric in the following sense: if, for any k and any n, (x0
n, . . . , xk n, xk+1 n
) − → (x0
n+1, . . . , xk n+1, xk+1 n+1)
by the previous integrator, then (x0
n+1, . . . , xk n+1, xk+1 n+1) −
→ (x0
n, . . . , xk n, xk+1 n
) by the exact same algorithm reversing the time.
The University of Chicago – p. 15