Weak Convergence Methods for Nonlinear Partial Differential ...

espect to powers of ε be zero! We are then led to the equations
L 1 u 0 = 0, (3.15)
L 1 u 1 + L 2 u 0 = 0, (3.16)
L 1 u 2 + L 2 u 1 + L 3 u 0 = f, (3.17)
which are to be solved for functions functions u i = u i (x, y), i = 1, 2, 3, on Ω × Q
with zero boundary conditions in x on ∂Ω and periodic boundary conditions in
y on ∂Q.
Multiplying (3.15) with u 0 itself and integrating over Q we obtain
∫
(∇ y u 0 (x, y)) T A(y)∇ y u 0 (x) = 0
Q
by partial integration. Note that there are no boundary terms due to the periodicity
of u 0 in y. But then our ellipticity assumptions shows that ∇ y u 0 = 0 and
thus u 0 = u 0 (x) is a function of x only.
Consequently,
n∑
L 2 u 0 = − ∂ yj a ij (y) ∂ xi u 0
and (3.16) implies
L 1 u 1 =
i,j=1
n∑
∂ yj a ij (y) ∂ xi u 0 .
i,j=1
In order to determine the y-dependency of u 1 , we solve the so-called corrector
problems
L 1˜χ k =
n∑
∂ yj a kj (y) in Q (3.18)
j=1
for ˜χ = ˜χ(y), k = 1, . . .,n, with periodic boundary conditions on ∂Q. Note that
this is possible by Fredholm’s alternative: As we just saw, the kernel of L 1 = L ∗ 1
with respect to periodic boundary conditions precisely consists of all constant
functions. Since ∫ n∑
∂ yj a kj (y) = 0
Q j=1
(again due to periodicity), the right hand side of (3.18) is indeed orthogonal to
all these constant functions.
This allows us to re-write the above equation as
(
L 1 u 1 − ∑ )
˜χ k ∂ xk u 0 = 0
k
76