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Well-conditioned boundary integral formulations for the ... - Njit

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problems. In addition, <strong>the</strong> far field errors incurred by our implementation are of <strong>the</strong> same order as<br />

those resulting from <strong>the</strong> classical Combined Field Integral Equation counterparts. Thus, <strong>the</strong> important<br />

gains in computational complexity make <strong>the</strong> Regularized Combined Field Integral Equations<br />

discussed in this text a viable method of solution to PEC scattering problems.<br />

Acknowledgments<br />

Yassine Boubendir gratefully acknowledge support from NSF through contract DMS-1016405.<br />

Catalin Turc gratefully acknowledge support from NSF through contract DMS-1312169.<br />

6 Appendix<br />

We present in this section a proof of <strong>the</strong> technical Lemma 3.2<br />

Lemma 6.1 There exist constants C j > 0, j = 1, . . . , 4 and a number ˜k 0 > 0 such that<br />

(i) C 1 k −2 (n 2 + k 2 ) 1/2 ≤ iJ ′ n+1/2 (ik)(H(1) n+1/2 )′ (ik) ≤ C 2 k −2 (n 2 + k 2 ) 1/2<br />

(ii) 1 4 (n2 + k 2 ) −1/2 ≤ −S (1)<br />

n (ik) ≤ C 3 (n 2 + k 2 ) −1/2 + C 4 k −2 ,<br />

(iii)|J ′ n+1/2 (ik)H(1) n+1/2 (ik)| ≤ C 4k −1 , |J n+1/2 (ik)(H (1)<br />

n+1/2 )′ (ik)| ≤ C 4 k −1<br />

<strong>for</strong> all k > ˜k 0 and all n ≥ 0.<br />

Proof. (i) We begin by using <strong>the</strong> representation of <strong>the</strong> functions J n+1/2 (ik) and H (1)<br />

terms of <strong>the</strong> Bessel and Hankel functions of <strong>the</strong> third kind<br />

iJ ′ n+1/2 (ik)(H(1) n+1/2 )′ (ik) = − 2 π I′ n+1/2 (k)K′ n+1/2 (k)<br />

n+1/2<br />

(ik) in<br />

where I ′ ν(k) > 0 and K ′ ν(k) < 0 <strong>for</strong> ν ≥ 0. Using <strong>the</strong> uni<strong>for</strong>m asymptotic expansions as ν → ∞<br />

(Formulas 9.7.9 and 9.7.10 in [1])<br />

where µ = √ 1 + z 2 + ln<br />

I ν(νz)<br />

′ 1 (1 + z 2 ) 1 4<br />

∼ √ e νµ (1 + O(ν −1 ))<br />

2πν z<br />

√ π<br />

K ν(νz) ′ (1 + z 2 ) 1 4<br />

∼ −√ e −νµ (1 + O(ν −1 )) (94)<br />

2ν z<br />

z<br />

1+ √ 1+z 2 , we get that <strong>the</strong>re exists a constant N 0 > 0 such that<br />

1<br />

4 k−2 ((n + 1/2) 2 + k 2 ) 1/2 ≤ −I ′ n+1/2 (k)K′ n+1/2 (k) ≤ k−2 ((n + 1/2) 2 + k 2 ) 1/2 ,<br />

<strong>for</strong> all n > N 0 , k > 0. Given that we can choose <strong>the</strong> constant N 0 above to fur<strong>the</strong>r satify (n +<br />

1/2) 2 + k 2 < 2(n 2 + k 2 ) <strong>for</strong> all n > N 0 and all k > 0, we obtain <strong>the</strong> following estimate<br />

1<br />

4 k−2 (n 2 + k 2 ) 1/2 ≤ −I ′ n+1/2 (k)K′ n+1/2 (k) ≤ √ 2k −2 (n 2 + k 2 ) 1/2 , <strong>for</strong> all n > N 0 , k > 0. (95)<br />

40

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