A boundary element method for homogenization of periodic structures
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...On the other 1 hand, let ðχ1; χkÞ ∈ V solve (13), then the extension e χ k ∈ e V satisfying (11) is a solution to (3)....
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...Note that χk2jΓ1 1⁄4 χ1 and that e χ i ∈ H(1)ðYiÞ is the harmonic extension of χi , ie, ∫Yi∇eχki ·∇e φidy 1⁄4 0 ∀e φi ∈ H(1)0ðYiÞ: (11) Using a splitting ev 1⁄4 e Ev2 þ e φ1 þ e φ2 for some continuous extension e Ev2 of any v2∈H(Γ2) and the harmonic extensions (11), we can reduce the bilinear form in (3) to ∫Ya∇e χ k ·∇evdy 1⁄4 a1 ∫Y 1∇ e χ 1 ·∇e Ev2 dyþ a2 ∫Y 2∇e χ 2 ·∇e Ev2 dy 1⁄4 a1 ⟨S1χ1; v2jΓ1 ⟩Γ1 þ a2 ⟨S2χ k 2; v2⟩Γ2 ∀v2 ∈ H ðΓ2Þ; (12)...
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...Indeed, any pair of traces (v1,v) is extended by the harmonic extension (11) to ev ∈ H1ðYÞ such that ‖ev‖H1ðY 1Þ ≤ C 1 ‖v1‖1=2;Γ1 and ‖ev‖H1ðY Þ ≤ C ‖v‖1=2;Γ with positive constants C 1 and C ext depending only on Y1 and Y, respectively....
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...On the other hand, e χ k ∈ e V are well‐defined as harmonic extensions (11) of ðχ1; χkÞ....
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...By the Gauss theorem, we can evaluate the homogenized coefficients (2) from χk1, the Γ1‐trace of eχ k ∈ eV ¼ H1perðYÞ, ðA0Þik ¼ δik a2 þ ða1 − a2Þ∫Γ1yi n1ðyÞð ÞidsðyÞ n o − biðχk1Þ; (8) where biðvÞ: ¼ ða1 − a2Þ∫Γ1vðyÞ n1ðyÞð ÞidsðyÞ: (9) Similarly, the right‐hand side of (3) can be reduced to Γ1 using the Gauss theorem and the Γ‐periodicity of ev ∫Ya ∂ev ∂yk dy ¼ a1 ∫Γ1v1 ðn1Þk dsðyÞ þ a2 ∫Γ2v2 ðn2Þk dsðyÞ¼ ða1 − a2Þ∫Γ1v1 ðn1Þk dsðyÞ: (10) In order to derive a boundary integral version of (3), we consider restrictions and traces of χk ∈ H1perðY Þ; eχ k1 : ¼ eχ kjY1 ; eχ k2 : ¼ eχ kjY2 ; χk1 : ¼ eχ kjΓ1 ∈ H1=2ðΓ1Þ; χk2 : ¼ eχ kjΓ2 ∈ H1=2ðΓ2Þ: Note that χk2jΓ1 ¼ χk1 and that eχ ki ∈ H1ðYiÞ is the harmonic extension of χki , ie, ∫Yi∇eχki ·∇eφidy ¼ 0 ∀eφi ∈ H10ðYiÞ: (11) Using a splitting ev ¼ eEv2 þ eφ1 þ eφ2 for some continuous extension eEv2 of any v2∈H1/2(Γ2) and the harmonic extensions (11), we can reduce the bilinear form in (3) to ∫Ya∇eχ k ·∇evdy ¼ a1 ∫Y 1∇ eχ k1 ·∇eEv2 dyþ a2 ∫Y 2∇eχ k2 ·∇eEv2 dy ¼ a1 ⟨S1χk1; v2jΓ1 ⟩Γ1 þ a2 ⟨S2χ k 2; v2⟩Γ2 ∀v2 ∈ H 1=2ðΓ2Þ; (12) where the Steklov–Poincaré operators Si:H 1/2(Γi)→H −1/2(Γi) are defined as Dirichlet to Neumann maps by Green's formulae in Y1 and Y2....
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...In particular, for all v2≡(v1,v)∈H(Γ1)×H(Γ): ‖v1‖21=2;Γ1 þ ‖v‖(2)1=2;Γ ≤ v2 k k21=2;Γ2 ≤ 1þ 4maxfjΓ1j; jΓjg distðΓ1; ΓÞ ! ‖v1‖21=2;Γ1 þ ‖v‖(2)1=2;Γ ; (16)...
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...The first inequality in (16) is straightforward....
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...ðA0Þik : 1⁄4 ∫Ya y ð Þδik − ∂e χ k ∂yi y ð Þdy (2)...
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...By the Gauss theorem, we can evaluate the homogenized coefficients (2) from χ1, the Γ1‐trace of e χ k ∈ e V 1⁄4 H(1)perðYÞ, ðA0Þik 1⁄4 δik a2 þ ða1 − a2Þ∫Γ1yi n1ðyÞ ð ÞidsðyÞ n o − bðχ1Þ; (8)...
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"A boundary element method for homog..." refers background in this paper
...Owing to the Poincaré inequality, there exists some ecP : 1⁄4 ecPðY 2Þ > 0 such that ∫Y j∇evðyÞj2 dyþ ∫Γ2evðyÞdsðyÞ 2 ≥ ecP ∫Yev2ðyÞdy ∀ev ∈ H1ðY Þ; (5) and ãð·; ·Þ is elliptic on the subspace e U : 1⁄4 ev ∈ e V :∫Γ2evðyÞdsðyÞ 1⁄4 0 n o; namely, ãðev; evÞ ≥ ec ‖ev‖2H1ðYÞ ∀ev ∈ e U ; (6)...
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...Using (5) and (18), a straightforward calculation yields...
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"A boundary element method for homog..." refers background in this paper
...2 | BOUNDARY INTEGRAL FORMULATION We shall arrive at a boundary integral formulation of (3)....
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...Let e χ k ∈ e V be a solution to (3), then the pair of traces ðχk; χkÞ ∈ V solves (13)....
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...The rest of the paper is organized as follows: In Section 2, we present a direct boundary integral formulation of the auxiliary problem and prove its well‐posedness and equivalence to (3)....
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...4, auxiliary problem (3) is well‐posed on e U ....
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...Note that χk2jΓ1 1⁄4 χ1 and that e χ i ∈ H(1)ðYiÞ is the harmonic extension of χi , ie, ∫Yi∇eχki ·∇e φidy 1⁄4 0 ∀e φi ∈ H(1)0ðYiÞ: (11) Using a splitting ev 1⁄4 e Ev2 þ e φ1 þ e φ2 for some continuous extension e Ev2 of any v2∈H(Γ2) and the harmonic extensions (11), we can reduce the bilinear form in (3) to ∫Ya∇e χ k ·∇evdy 1⁄4 a1 ∫Y 1∇ e χ 1 ·∇e Ev2 dyþ a2 ∫Y 2∇e χ 2 ·∇e Ev2 dy 1⁄4 a1 ⟨S1χ1; v2jΓ1 ⟩Γ1 þ a2 ⟨S2χ k 2; v2⟩Γ2 ∀v2 ∈ H ðΓ2Þ; (12)...
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