Weak Convergence Methods for Nonlinear Partial Differential ...

Using that ∂ xi u ε ζ = ∂ xi (u ε ζ) − u ε ∂ xi ζ and that ϕ ε k solves (3.28) and thus
we arrive at
∫ ∫
f ζ ϕ ε k =
Ω
∫
Ω i,j=1
n∑ ( x
a ij
ε
Ω i,j=1
n∑ ( x
a ij
ε
)
∂ xi (u ε ζ) ∂ xj ϕ ε k dx = 0,
) ∫
∂ xi u ε ∂ xj ζ −
Ω
n∑ ( x
)
a ij u ε ∂ xi ζ ∂ xj ϕ ε k
ε
. (3.29)
i,j=1
Now by the definition of ϕ ε k and by uε ⇀ u in W 1.2 we have
strongly in L 2 as ε → 0. Also,
n∑ ( x
)
a ij ∂ xj ϕ ε
ε
k(x) =
j=1
ϕ ε k → x k and u ε → u
n∑ ( x
)( ( x
))
a ij δ kj + ∂ j χ k ⇀ ā ik
ε
ε
j=1
as the L 2 -weak limit of a highly oscillating sequence. Together with (3.26) we
find by first sending ε to 0 in (3.29) and then using (3.27) with ϕ = x k ζ
∫
Ω j=1
n∑
∫
ξ j x k ∂ xj ζ −
Ω i=1
n∑
∫
ā ik u ∂ xi ζ =
∫
=
Ω
∫
f x k ζ =
Ω j=1
Ω
ξ · ∇(x k ζ)
n∑
∫
ξ j x k ∂ xj ζ +
Ω j=1
n∑
ξ j ζ δ jk
and so
∫
Ω
n∑
i=1
∫
ā ik ∂ xi u ζ = −
Ω
n∑
∫
ā ik u ∂ xi ζ
i=1
Ω
ξ k ζ.
Since ζ was arbitrary, this yields
ξ k =
n∑
ā ik ∂ xi u.
i=1
But then again from (3.27) we obtain that for every ϕ ∈ W 1,2
0 (Ω)
∫ n∑
∫ n∑
∫
ā ik ∂ xi u ∂ xk ϕ = ξ k ∂ xk ϕ = f ϕ.
Ω
i,k=1
Ω
k=1
Ω
This proves that u is indeed a weak solution of (3.25).
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