Weak Convergence Methods for Nonlinear Partial Differential ...

1. r = 2. As Rank(ξ 1 , ξ 2 ) = 1, ξ 1 and ξ 2 are linearly dependent and so, for
some c ∈ R, B(ξ 1 )λ 2 = cB(ξ 2 )λ 2 = 0. So in fact
B(ξ 1 ) = 0 ∀λ ∈ span {λ 1 , λ 2 } ,
and, in particular, span {λ 1 , λ 2 } ⊂ Λ.
By Theorem 2.40 (also cf. Remark 2.41) the quadratic form D 2 f(y)[·, ·]
vanishes on Λ. It follows that
D 2 f(y)[λ 1 , λ 2 ] = 0.
2. Consider ϕ 1 , ϕ 2 , ϕ 3 periodic with average zero and define
u (ν) (x) = u(x) + t[λ 1 ϕ 1 (νξ 1 · x) + λ 2 ϕ 2 (νξ 2 · x) + λ 3 ϕ 3 (νξ 3 · x)],
u(x) ≡ y ∈ R m . For any t,
For |t| small we have
u (ν) ∗ ⇀ u in L ∞ and Au (ν) = Au = 0.
f(u (ν) (x)) = f(y) + tDf(y)[λ 1 ϕ 1 (νξ 1 · x) + λ 2 ϕ 2 (νξ 2 · x) + λ 3 ϕ 3 (νξ 3 · x)]
+ 1 2 t2 D 2 f(y)[λ 1 ϕ 1 (νξ 1 · x) + . . . , λ 1 ϕ 1 (νξ 1 · x) + . . .]
+ 1 6 t3 D 3 f(y)[. . ., . . .,...]
+ O(t 4 ),
where the linear term converges weakly* to zero.
3. Consider the quadratic term
D 2 f(y)[..., ...] = ∑
1≤i,j≤3
D 2 f(y)[λ i , λ j ]ϕ i (νξ i · x)ϕ j (νξ j · x).
If Rank(ξ i , ξ j ) = 1, then as in the first step we have
If Rank(ξ i , ξ j ) = 2, then
(Exercise!). Consequently,
D 2 f(y)[λ i , λ j ] = 0.
ϕ i (νξ i · x)ϕ j (νξ j · x) ∗ ⇀ 0
D 2 f(y)[..., ...]
33
∗
⇀ 0.