Abstract

CHAPTER 4. THEORY 100
denote this dependence by Tu(x, k − k ′ ). We have for k>0that
G2(un)(x, k) − G2(um)(x, k) = 4Cτ
hk
− 4Cτ
hk
x Kmax
0
−Kmax
x Kmax
0
−Kmax
This is the same as looking at three separate terms, where
G2(un)(x, k) − G2(um)(x, k) = 4Cτ
x
hk 0
+ 4Cτ
x
e
hk 0
C(z−x)
Kmax
k
−Kmax
+ 4Cτ
x
e
hk
C(z−x)
k
0
e C(z−x)
k
Kmax
e C(z−x)
k um(z,k ′ )Tum(z,k − k ′ )dk ′ dz
e C(z−x)
k un(z,k ′ )Tun(z,k − k ′ )dk ′ dz.
−Kmax
Tū(z,k − k ′ )[um(z,k ′ ) − un(x, k)] dk ′ dz
[Tum(z,k − k ′ ) − Tū(z,k − k ′ )] [um(z,k ′ ) − un(x, k)] dk ′ dz
Kmax
−Kmax
un(x, k)[Tum(z,k − k ′ ) − Tun(z,k − k ′ )] dk ′ dz
= G 1 (x, k)+G 2 (x, k)+G 3 (x, k).
Let δ>0. For k>δ, (as in the case for the G1 term) we have the first term, G 1 (x, k),
as an integral operator whose input is an L 2 ([0,L] × [−Kmax,Kmax]) function with a
kernel function given by
4Cτ
hk
e C(z−x)
k Tū(z,k − k ′ ).
For k > δ, the kernel function is continuous. Therefore, the integral operator is
a compact map from L 2 ([0,L] × [−Kmax,Kmax]) to C([0,L] × [δ, Kmax]). Since
un(x, k) ⇀ ū(x, k) weaklyinL 2 ([0,L] × [−Kmax,Kmax]), then the first term con-
verges uniformly to zero. So we can make the first term as small as we want for
δ