Analysis of the embedded cell method in 2D for the numerical homogenization of metal-ceramic composite materials

Theorem 6.

Let 0 < ε < ε 0 , μ ≡ μ 0 > 0 , m ∈ N and let λ ε be given by

with λ 0 ∈ R + and λ k ∈ L ∞ ⁢ ( Ω ) for k = 0 , … , m , independent of ε. Then, for sufficiently small ε 0 there exist functions u k for k = 0 , … , m , with u 0 ∈ W l and u k ∈ W 0 for k ≥ 1 such that for the solution u ε ∈ W l of

for all v ∈ W 0 satisfies

with respect to the H 1 -norm.

Proof.

Let u 0 ∈ 𝒲 l and u k ∈ 𝒲 0 , k = 1 , … , m , respectively, be the unique weak solutions of

for all v ∈ 𝒲 0 . Then the solution u 0 is given by

and using Theorem 1, we have for u k , k = 1 , … , m ,

In particular, the λ k are independent of ε. Hence, there exist constants C k > 0 for k = 0 , … , m independently of ε such that

Now, define

which is an element of 𝒲 l . Then u approx ε satisfies

for all v ∈ 𝒲 0 , where

Therefore, R := u ε - u approx ε solves

for all v ∈ 𝒲 0 and we can use Theorem 1 again to obtain

for constants C , C ~ > 0 , independent of ε, which yields (A.2). ∎

Theorem 7.

There exists an ε 0 > 0 such that the solution u ε ∈ W l of

for all v ∈ W 0 , where 0 < ε ≤ ε 0 , λ ε = λ 0 + ε ⁢ λ pert , λ pert ∈ L ∞ ⁢ ( Ω ) , λ 0 ∈ R + and μ ≡ μ 0 ∈ R + , has the representation

which converges absolutely in H 1 for ε ∈ [ 0 , ε 0 ) , u 0 ∈ W l and u k ∈ W 0 for k ≥ 1 . Furthermore, there exists a constant C > 0 independently of k , ε such that

for all k ≥ 0 .

Proof.

Inserting the ansatz (A.9) into (A.8) yields

for k ≥ 1 and all v ∈ 𝒲 0 . Notice that u 0 ∈ 𝒲 0 is given by (A.5). By Theorem 1 there exists a C > 0 being independent of ε > 0 such that

By induction we obtain

Hence, the series (A.9) converges absolutely for

and since H 1 ⁢ ( Ω ) is a Banach space, the series also converges in H 1 . Because the left-hand-side of equation (A.8) defines a continuous bilinear form in H 1 , it follows that (A.9) is a solution of (A.8) satisfying (A.10). ∎

Lemma 6.

Let ε > 0 , λ ε ∈ L ∞ ⁢ ( Ω ) with λ ε = λ 0 + ε ⁢ λ pert , λ 0 > 0 , μ ≡ μ 0 > 0 and l > 0 . Then there exists an ε 0 > 0 such that for all ε ∈ [ 0 , ε 0 ) there exists a unique equivalent first Lamé parameter λ equiv ∈ R + , which satisfies (3.6).

Proof.

If there exists an equivalent first Lamé parameter, then, due to (2.15), (3.5) and (A.5), it has to satisfy (3.6). By analogous calculations as in the proof of Lemma 3 we obtain that under the assumptions of the lemma the tensile force F ⁢ [ λ ε , μ , l ] has the expansion

with

Recalling that ν 0 = λ 0 λ 0 + 2 ⁢ μ , we find that 2 ⁢ μ ⁢ l < F 0 < 4 ⁢ μ ⁢ l since λ 0 , μ > 0 . Therefore, for sufficiently small ε it holds that 2 ⁢ μ ⁢ l < F ⁢ [ λ ε , μ , l ] < 4 ⁢ μ ⁢ l . Hence, for sufficiently small ε 0 > 0 and ε ∈ [ 0 , ε 0 ) equation (3.6) has a unique solution λ equiv with

which proves the lemma. ∎

Lemma 7.

Let ε > 0 , λ ε ∈ L ∞ ⁢ ( Ω ) with λ ε = λ 0 + ε ⁢ λ pert , λ 0 > 0 , μ ≡ μ 0 > 0 and l > 0 . Then the equivalent first Lamé parameter λ equiv does not depend on l.

Proof.

Let L : ℝ + → H 1 ⁢ ( Ω , ℝ 2 ) be the mapping which maps a given prescribed tensile length l to the corresponding weak solution u ∈ 𝒲 l of (2.3)–(2.6). Moreover, let F u : Range ⁢ ( L ) → ℝ be defined by

Then F l : ℝ + → ℝ with F l = F u ∘ L is linear and hence, there exists an a ∈ ℝ independent of l such that F l = a ⁢ l . This yields

which is independent of l. ∎