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Weak Convergence Methods for Nonlinear Partial Differential ...

Weak Convergence Methods for Nonlinear Partial Differential ...

Weak Convergence Methods for Nonlinear Partial Differential ...

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Although this approach does not lead to a rigorous proof of u ε ⇀ u 0 , the<br />

solution of the equation above, this multi-scale analysis is an important first<br />

step: It provides us with an explicit <strong>for</strong>mula <strong>for</strong> the homogenized coefficients.<br />

Rigorous analysis:<br />

In the second part of the section we will give a rigorous convergence result.<br />

Suppose u ε is the (unique) weak solution of (3.13). We will now also distinguish<br />

more carefully between the operator L 1 and its <strong>for</strong>mal adjoint L ∗ 1 . (Of course,<br />

L 1 = L ∗ 1 if A is symmetric.) Define χ k = χ k (y), k = 1, . . .,n, to be a weak<br />

solution of the corrector problem<br />

n∑ (<br />

∂ yi aij (y)∂ yj χ k (y) ) =<br />

i,j=1<br />

n∑<br />

∂ yi a ik (y)<br />

i=1<br />

in Q<br />

subject to periodic boundary conditions. More precisely, χ k ∈ Hper 1 (Q) is such<br />

that<br />

∫ n∑<br />

∫ n∑<br />

a ij (y) ∂ yj χ k (y) ∂ yi ψ(y) dy = ∂ yi a ik (y) ψ(y) dy (3.24)<br />

Q i,j=1<br />

Q i=1<br />

<strong>for</strong> every ψ ∈ Hper(Q). 1 Here the Hilbert space Hper(Q) 1 of Q-periodic H 1 -functions<br />

consists of the restrictions to Q of Q-periodic functions belonging to Hloc 1 (Rn ).<br />

By Fredholm’s alternative, such functions χ k do exist. Also note that χ k is determined<br />

uniquely up to an additive constant. Define the homogenized coefficients<br />

by<br />

∫ n∑<br />

ā ij := a ik (y) (δ kj + ∂ yk χ j (y)) dy.<br />

Q<br />

k=1<br />

Theorem 3.13 u ε converges to u weakly in W 1,2<br />

0 (Ω), where u is the (unique)<br />

weak solution of<br />

−<br />

n∑<br />

∂ xj (ā ij ∂ xi u(x)) = f(x) in Ω, u = 0 on ∂Ω. (3.25)<br />

i,j=1<br />

Proof. As solutions to (3.13), the sequence (u ε ) is bounded in W 1,2<br />

0 (Ω). We may<br />

there<strong>for</strong>e extract a converging subsequence u ε ⇀ u in W 1,2<br />

0 (Ω). Since the weak<br />

solution of Equation (3.25) is unique, it suffices to show that u solves (3.25).<br />

For a further subsequence (not relabeled) we have<br />

n∑<br />

i=1<br />

( ·<br />

)<br />

a ij ∂ xi u ε ⇀ ξ j (3.26)<br />

ε<br />

78

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