Weak Convergence Methods for Nonlinear Partial Differential ...

4. We finally investigate the cubic term
D 3 f(y)[. . ., . . .,...] = ∑ i,j,k
D 3 f(y)[λ i , λ j , λ k ]ϕ i (νξ i · x)ϕ j (νξ j · x)ϕ k (νξ k · x).
Fix i, j, k ∈ {1, 2, 3}. If two of the vectors ξ 1 , ξ 2 , ξ 3 are parallel, say ξ i ‖ ξ j
(w.l.o.g.), then as in the first step
D 2 f(y)[λ i , λ j ] = 0
∀y.
This remains true if y is replaced by y + tλ k . So
D 3 f(y)[λ i , λ j , λ k ] = d dt∣ D 2 f(y + tλ k )[λ i , λ j ] = 0.
t=0
If, on the other hand, ξ i is not parallel to ξ j for i ≠ j, then one can choose
ϕ 1 , ϕ 2 , ϕ 3 in such a way as to have
ϕ 1 (νξ 1 · x)ϕ 2 (νξ 2 · x)ϕ 3 (νξ 3 · x) ∗ ⇀ c ≠ 0
(Exercise!). As by assumption l = lim f(u (ν) ) = f(y), we obtain
0 = lim ∑ i,j,k
D 3 f(y)[λ i , λ j , λ k ]ϕ i ϕ j ϕ k
= 3 limD 3 f(y)[λ 1 , λ 2 , λ 3 ]ϕ 1 ϕ 2 ϕ 3
= 3cD 3 f(y)[λ 1 , λ 2 , λ 3 ].
Corollary 2.43 Let dim span Λ = d and choose coordinates so that
span Λ = {y ∈ R m : y d+1 = . . . = y d = 0} .
□
If l = f(u) for every sequence satisfying (H), then f is a polynomial in y 1 , ..., y d
of degree at most min{n, d} whose coefficients depend on y d+1 , ..., y m .
Proof. Suppose (e 1 , ..., e d ) ∈ Λ is a basis of span Λ and (e 1 , ..., e n ) is a basis of
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