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( )∂ ∂wUsing (A.3) and Lemma A.2, we deduce in a similar way that∂x iA ij ∂x j∈ H s−1 (R d +) and that( )∥ ∂ ∂w ∥∥∥H∥ A ij A max B(w), for all 1 ≤ i ≤ d and 1 ≤ j ≤ d − 1. (A.5)∂x i ∂x j s−1 (R d + )We can now use the PDE (3.1) to get(a similar)bound for (i, j) = (d, d). Since F ∈ H s−1 (R d +),∂ ∂wit follows from (A.4) and (A.5) that∂x dA dd ∂x d∈ H s−1 (R d +) and that( )∥ ∂ ∂w ∥∥∥H∥ A dd A max B(w) + ‖F ‖∂x d ∂x H s−1 d s−1 (R d + ) (R d + ) A max B(w) .Analogously to (A.4) we can prove that( )∥ ∂ ∂w ∥∥∥H∥ A dd∂x i ∂x d s−1 (R d + ) A max B(w), for all 1 ≤ i ≤ d − 1.∂wHence, we can finally apply Lemma A.1 to get that A dd ∂x d∥ ∥ A ∂w ∥∥∥2d∑( )∥ dd∂ ∂w ∥∥∥2∂x d∥ A dd∂x i ∂x dH s (R d + )i=1H s−1 (R d + ) +∈ H s (R d +) and that∥ ∥ A ∂w ∥∥∥2dd A 2 max B(w) 2 .∂x d L 2 (R d + )∂wBy applying Lemma A.2 again, this time with b := 1/A dd and v := A dd ∂x d, we deduce that∂w∂x d∈ H s (R d +) and that∥ ∣ ∥ ∥ ∂w ∥∥∥H ∥1 ∣∣∣C ∥∥∥ ∂w ∥∥∥L∂x d∣A dd + 1 ∥ ∥ ∥∥∥ ∂w ∥∥∥HA dd s (R d + ) A dd t (R d + ) ∂x d 2 (R d + ) A min ∂x d s (R d + )|A dd | C t (R d + )A 2 minA max |w| H 1 (R d + ) + 1A min∥ ∥∥∥A dd∂w∂x d∥ ∥∥∥H s (R d + ) A maxA minB(w) .To finish the proof we use this bound together with (A.3) and apply once more Lemma A.1 toshow that w ∈ H 1+s (R d +) and‖w‖ H 1+s (R d + ) A )max(|A|A 2 C t (R d +min,S d(R)) |w| H 1 (R d + ) + ‖F ‖ H s−1 (R d + ) + A max‖w‖A H 1 (R d ).min+A.3 Step 3 – The Case D BoundedWe can now prove Proposition 3.1 using Lemmas 3.2 and 3.3 in two successive steps. We recallthat D was assumed to be C 2 . Let (D i ) 0≤i≤m be a covering of D such that the (D i ) 0≤i≤m are openand bounded, D ⊂ ∪ m i=0 D i, ∪ m i=1 (D i ∩ ∂D) = ∂D, D 0 ⊂ D.Let (χ i ) 0≤i≤m be a partition of unity subordinate to this cover, i.e. we have χ i ∈ C ∞ (R d , R + )with compact support supp(χ i ) ⊂ D i , such that ∑ pi=0 χ i = 1 on D. We denote by u the solutionof (2.3) and split it into u = ∑ pi=0 u i, with u i = uχ i . We treat now separately u 0 and then u i ,1 ≤ i ≤ p, using Section A.1 and A.2, respectively.Lemma A.4. u 0 belongs to H 1+s (D) and‖u 0 ‖ H 1+s (D) ‖a‖ C t (D)a 2 ‖f‖ H s−1 (D) .min25
Proof. Since supp(u 0 ) ⊂ D 0 , we have that u 0 ∈ H0 1 (D) and it is the weak solution of the newequation −div(a∇u 0 ) = F on D, whereF := fχ 0 + a∇u · ∇χ 0 + div(au∇χ 0 ) on D.To apply Lemma 3.2 we will now extend all terms to R d , but continue to denote them the same.The terms u 0 and fχ 0 + a∇u · ∇χ 0 can both be extended by 0. Their extensions will belongto H 1 (R d ) and H s−1 (R d ), respectively. It follows from Lemma A.2 that au ∈ H s (D) and so ifwe continue au∇χ 0 by 0 on R d , the extension belongs to H s (R d ), since supp(χ 0 ) is compact inD. Using the fact that div is linear and continuous from H s (R d ) to H s−1 (R d ) (cf. [18, Remark6.3.14(b)] and the proof of (A.4) above), we can deduce that the divergence of the extension ofau∇χ 0 is in H s−1 (R d ), leading to an extension of F on R d , which belongs to H s−1 (R d ).Let ψ ∈ C ∞ (R d , [0, 1]) such that ψ = 0 on D 0 and ψ = 1 on ˜D c , where ˜D is an open set suchthat D 0 ⊂ ˜D and ˜D ⊂ D. We use the following extension of a from D 0 to all of R d :{ a(x)(1 − ψ(x)) + amin ψ(x), if x ∈ D,a(x) :=a min ψ(x), otherwise.This implies that a ∈ C t (R d ), and for any x ∈ R d , a min ≤ a(x) ≤ a(x) and |a| C t (R d ) ‖a‖ C t (D) .Using these extensions, we have that −div(a∇u 0 ) = F in D ′ (R d ). Indeed, for any v ∈ D(R d ),∫∫a(x)∇u 0 (x)∇v(x) = a(x)∇u 0 (x)∇v(x) for any v ∈ D(R d ),R d Dsince supp(u 0 ) is included in the open bounded set D 0 , which implies that ∇u 0 = 0 on D c 0 and a = aon D 0 . Since u ∈ H 1 0 (D), we have by Poincaré’s inequality that ‖u‖ L 2 (D) |u| H 1 (D). Therefore itfollows from Lemma 2.1 that|u 0 | H 1 (R d ) ≤ |u| H 1 (D)‖χ 0 ‖ ∞ + ‖u‖ L 2 (D)‖∇χ 0 ‖ ∞ ‖f‖ H s−1 (D)a min.Since χ 0 ∈ C ∞ (R d ), using Lemma A.2 and the linearity and continuity of div from H s (R d ) toH 1−s (R d ) we further get‖F ‖ H s−1 (R d ) ≤ ‖fχ 0 ‖ H s−1 (R d ) + ‖a∇u · ∇χ 0 ‖ H s−1 (R d ) + ‖div(au∇χ 0 )‖ H s−1 (R d )‖f‖ H s−1 (D) + a max |u| H 1 (D) + |a| C t (D)‖u‖ L 2 (D) + a max ‖u‖ H s (D)‖a‖ C t (D)‖f‖a H s−1 (D).minWe can now apply Lemma 3.2 with A = aI d and w = u 0 to show that u 0 ∈ H 1+s (R d ) and‖u 0 ‖ H 1+s (R d ) 1 ()|a|a C t (D) |u 0| H 1 (R d ) + ‖F ‖ H s−1 (R d ) + ‖u 0 ‖ H 1 (R d ) ‖a‖ C t (D)min a 2 ‖f‖ H s−1 (D) .minThe hidden constant depends on the choices of χ 0 and ψ and on the constant in Poincaré’s inequality,which depends on the shape and size of D, but not on a.Let us now treat the case of u i , 1 ≤ i ≤ m.Lemma A.5. For 1 ≤ i ≤ m, u i ∈ H 1+s (D) and‖u i ‖ H 1+s (D) a max‖a‖ C t (D)a 3 ‖f‖ H s−1 (D).min26
Page 1 and 2: BICSBath Institute for Complex Syst
Page 3 and 4: only Hölder continuous with expone
Page 5 and 6: for almost all ω ∈ Ω. The diffe
Page 7 and 8: 2.4 Truncated Karhunen-Loève Expan
Page 9 and 10: 3.1 Regularity of the SolutionPropo
Page 11 and 12: Theorem 3.4. Let Assumptions A1-A3
Page 13 and 14: Corollary 3.10. Let Assumptions A1-
Page 15 and 16: 3.4 Truncated FieldsCombining the r
Page 17 and 18: Since all the expectations E[Y l ]
Page 19 and 20: Figure 1: Let d = 1 and σ 2 = 1 in
Page 21 and 22: Figure 4: Left: Number of samples N
Page 23 and 24: therefore we getA min ‖(R i h )
Page 25: ∥∂which means that2 ˜w∂x i
Page 29 and 30: We now extend A i to R d +. Since w

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