Error estimates of the DtN finite element method for the exterior Helmholtz problem
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(2) 22. D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31. The DtN finite element method for the exterior Helmholtz problem has first been proposed by MacCamy–Marin [23] in 1980, who represented the DtN operator through an integral equation. Feng [4], Masmoudi [24], and Keller–Givoli [18] have derived the Fourier series representation of the DtN operator. Masmoudi [24] and Keller–Givoli [18] incorporated such a representation directly into the finite element method. The DtN finite element method for other kinds of problems concerning the Helmholtz equation has been investigated by several authors. In 1978, Fix–Marin [6], who are pioneers in the DtN finite element method, studied the under-water acoustic problem. Goldstein [11] established error estimates for the Helmholtz problem on unbounded waveguides. Bao [1] also established error estimates for the problem concerning the diffraction of a time harmonic wave incident on a periodic surface of some inhomogeneous material. In China, independently of the western world, the DtN finite element method has been suggested and developed first by Feng and Yu in 1980 and 1982, for example, see [5], where they called it the canonical boundary element method or the natural boundary element method. Yu has published many papers [31,32,30,35,17] and a monograph [33,34] (see also its review [9]) in this direction. In Ushijima’s paper [28], we can find these words: “Chinese scholars, Feng Kang, Han Houde, Yu De-hao and others should be mentioned among founders of the treatment. In western world, J.B. Keller and D. Givoli are also should be quoted.” In this paper, we establish a priori error estimates in the H 1 - and L2 -norms for the exterior Helmholtz problem. The use of the Fourier series representation of the DtN operator requires truncating the series in practical computations. So we analyze the series truncation error as well as the finite element discretization error. To the best knowledge of the author, our error estimates are new, because no error estimate treating both the truncation error and the discretization error simultaneously has been published yet (cf. [10, p. 32]). MacCamy–Marin [23] and Masmoudi [24] have derived an error estimate, but their estimate depends only on the mesh size (see also [3,17]). The counterpart of our error estimates for the Helmholtz problem on unbounded waveguides has been established by Goldstein [11] in 1982. Our error analysis roughly follows his analysis; however, we need some properties of the Hankel functions, which contain a new and important result (Lemma 4); we were inspired to prove Lemma 4 by Han–Bao [12, Lemma 3.1]. We here remark that in the error analysis of ours (and also of Goldstein), the argument of Schatz [27] plays an essential role, since the Helmholtz equation is indefinite. Analysis of the truncation error in the DtN finite element method is an important topic. For problems of the positive definite type, Yu [32] and Han–Wu [13] have first derived error estimates including the truncation error for the exterior Laplace problem in 1985. After that, such error estimates for other problems are established in many papers, for example, in Han–Wu [14] for the linear elastic problem, and in Givoli–Patlashenko–Keller [10] and Han–Bao [12] for a certain class of the linear elliptic second order boundary value problem on exterior domains and semi-infinite strips (the error estimate in [12] is more sophisticated than that in [10]). For problems of other types, we mention Lenoir–Tounsi [21] (the sea-keeping problem) and Koyama–Tanimoto–Ushijima [20] (the eigenvalue problem of the linear water wave in a water region with a reentrant corner). The remainder of this paper is organized as follows. In Section 2, we formulate the reduced problem with the DtN boundary condition. In Section 3, we introduce some properties of the Hankel functions, which are employed for establishing the error estimates in Section 4.. 2. The DtN formulation We first introduce a theorem concerning the well-posedness of problem (1). 2 (), where Theorem 1. For every compactly supported f ∈ L2 (), problem (1) has a unique solution in Hloc m Hloc () = {u|u ∈ H m (B) for all bounded open set B ⊂ }. (m ∈ N).. Here L2 () denotes the usual space of complex-valued square integrable functions on , and H m (B) denotes the usual complex Sobolev space on B (see, e.g., [22]). Proof. See [25,26].. .
(3) D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31. 23. To seek approximate solutions of problem (1), we introduce an artificial boundary a = {x ∈ Rd ||x| = a}, where a is a positive number such that O ∪ supp f ⊂ Ba ≡ {x ∈ Rd ||x| < a}. Then problem (1) is reduced equivalently to the following problem on the bounded computational domain a = ∩ Ba : ⎧ −u − k 2 u = f in a , ⎪ ⎪ ⎨ u=0 on , (2) ⎪ ju ⎪ ⎩ = −Su on a , jr where S is the DtN operator corresponding to the outgoing radiation condition. For its definition, see [24,18]. Now we introduce the circular harmonics Yn defined by ein Yn () = √ , 2 where denotes the angular variable of an (r, ) polar coordinate system, and the spherical harmonics Ynm defined by (2n + 1) (n − |m|)! |m| Ynm (, ) = P (cos )eim , 4 (n + |m|)! n where , denote the angular variables of an (r, , ) spherical coordinate system, and Pnm are the associated Legendre functions. We define the Sobolev space H s (a ) (s > 0) by H s (a ) = { ∈ L2 (a )| s,a < ∞}, where · s,a is the norm of H s (a ) defined by ⎧ ∞ 2s 2 ⎪ if d = 2, ⎪ ⎨ a n=−∞ (1 + n )|n | 2. s,a = ∞ n ⎪ ⎪ 2 ⎩ a2 (1 + n2s )|m if d = 3, n| n=0 m=−n. m n,. respectively, are the Fourier coefficients of defined by where n and 2 n = ()Yn () d. (3). 0. and m n =. 2. d. 0. 0. (, )Ynm (, ) sin d.. (4). We here note that the DtN operator S is a bounded linear operator from H 1/2 (a ) into H −1/2 (a ) (see [24]), where H −1/2 (a ) is the set of all bounded semilinear forms on H 1/2 (a ). To formulate a weak form of problem (2), we introduce the following sesquilinear forms:. a(u, v) = (∇u · ∇v − k 2 uv) dx + s(u, v) for u, v ∈ H 1 (a ), a. s(u, v) =
(4) Su, v H −1/2 (a )×H 1/2 (a ) ⎧ ∞ (1) Hn (ka) ⎪ ⎪ un (a)vn (a) −ka (1) ⎪ ⎨ n=−∞ Hn (ka) = (1) ⎪ n ∞ ⎪ ⎪ ⎩ −ka 2 hn (ka) um (a)v m (a) n n (1) n=0 m=−n hn (ka). if d = 2, (5) if d = 3,.
(5) 24. D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31 (1). (1). where Hn and hn are the cylindrical and spherical Hankel functions of the first kind of order n, respectively, and un (a) and um n (a) denote the Fourier coefficients of u|a defined by (3) and (4), respectively. Then a weak form of (2) is written as follows: find u ∈ V such that a(u, v) = (f, v). (6). for all v ∈ V , where V = {v ∈ H 1 (a )|v = 0 on },. (u, v) = uv dx for u, v ∈ L2 (a ). a. For every f ∈ L2 (a ), problem (6) has a unique solution which is the restriction to a of the solution of problem (1) (see [24,15,16]). 3. Some properties of the Hankel functions In this section, we state three lemmas concerning properties of the Hankel functions. These lemmas will be used to establish error estimates in the next section. Lemma 2. For each x > 0, there exists a positive constant C such that. 1 H (1) (x). n. C for all n ∈ Z,. 1 + |n| Hn(1) (x). 1 h(1) (x). n. C for all n ∈ N ∪ {0},. 1 + n h(1) (x) n. (7). (8). where C depends on x, but is independent of n. Proof. For proofs of (7) and (8), see [24] and [16], respectively.. . Lemma 3. For all x > 0, we have.
(6) (1) H (x) Re < 0 for all ∈ R, (1) H (x)
(7). (1) hn (x) < 0 for all n ∈ N ∪ {0}. Re (1) hn (x) Proof. See [19].. (1). (1). (1). (1). Lemma 4. For any r1 > r2 > 0, |H (r1 )/H (r2 )| is a decreasing function of on [0, ∞), and further |hn (r1 )/ hn (r2 )| is a decreasing sequence of n ∈ N ∪ {0}. Proof. We first prove the former assertion. We write F (; r) = |H(1) (r)|2 . From Nicholson’s formula (see [29, p. 444]) ∞ 8 K0 (2r sinh t) cosh(2t) dt F (; r) = 2 0. ( ∈ R, r > 0),. (9).
(8) D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31. 25. where K0 is the modified Bessel function of the second kind of order zero, we see that, for each r > 0, F (· ; r) ∈ C ∞ (R) and F (· ; r) > 0. Thus, it is sufficient to show that d F (; r1 ) <0 d F (; r2 ) for all ∈ (0, ∞). This inequality holds, if and only if dF dF (; r1 )F (; r2 ) − F (; r1 ) (; r2 ) < 0. d d. (10). Hence, let us prove (10) in the following. Differentiating (9) with leads to. dF 16 ∞ (; r) = 2 K0 (2r sinh t)t sinh(2t) dt. d 0. (11). From (9) and (11), we have dF dF (; r1 )F (; r2 ) − F (; r1 ) (; r2 ) d d 128 ∞ ∞ [K0 (2r1 sinh t1 )K0 (2r2 sinh t2 )t1 sinh(2t1 ) cosh(2t2 ) = 4 0 0 − K0 (2r2 sinh t1 )K0 (2r1 sinh t2 )t1 sinh(2t1 ) cosh(2t2 )] dt1 dt2. t1 128 ∞ = 4 dt1 [K0 (2r1 sinh t1 )K0 (2r2 sinh t2 ) − K0 (2r1 sinh t2 )K0 (2r2 sinh t1 )] 0 0 × [t1 sinh(2t1 ) cosh(2t2 ) − t2 sinh(2t2 ) cosh(2t1 )] dt2 . Here, using Macdnold’s formula (see [29, p. 439]) . . Xx dt 1 ∞ t X2 + x 2 exp − − K0 K0 (X)K0 (x) = 2 0 2 2t t t. (12). (X, x > 0). and the fact that (r12 sinh2 t1 + r22 sinh2 t2 ) − (r12 sinh2 t2 + r22 sinh2 t1 ) = (r12 − r22 )(sinh2 t1 − sinh2 t2 ) > 0 for t1 > t2 > 0, we can conclude that K0 (2r1 sinh t1 )K0 (2r2 sinh t2 ) − K0 (2r1 sinh t2 )K0 (2r2 sinh t1 ) < 0. (13). for t1 > t2 > 0. Further, we have t1 sinh(2t1 ) cosh(2t2 ) − t2 sinh(2t2 ) cosh(2t1 ) = 21 {(t1 − t2 ) sinh[2(t1 + t2 )] + (t1 + t2 ) sinh[2(t1 − t2 )]} > 0. (14). for t1 > t2 > 0. From (12)–(14), we deduce (10). We next prove the latter assertion. Since (1) (1) (r), H hn (r) = 2r n+1/2 we have. h(1) (r ) r 2. n 1. (1). =. hn (r2 ). r1. (1). H. n+1/2 (r1 ). .. (1). H n+1/2 (r2 ) (1). (1). Hence, it follows from the above result that |hn (r1 )/ hn (r2 )| is a decreasing sequence of n ∈ N ∪ {0}.. .
(9) 26. D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31. 4. The DtN finite element method and its error estimates We discretize problem (6) by using the finite element method to obtain approximate solutions to problem (6) (or ¯ of finite dimensional subspaces of V , and assume that this family satisfies (1)). We introduce a family {Vh |h ∈ (0, h]} the following condition: there exist an integer p 2 and a constant C > 0 such that for all 0 < h h¯ and for every u ∈ V ∩ H p (a ) (2 p p), . inf u − vh 1,a Chp −1 u p ,a ,. (15). v h ∈Vh. where C is independent of h and u, and · p,a is the standard norm of H p (a ) defined by . v 2p, = |D v|2 dx (p ∈ N). | | p . For examples of such a family, see [2,36]. Now since the sesquilinear form s involves the infinite series, we have to truncate it in practice. So we practically solve the following problem: find uN h ∈ Vh such that a N (uN h , vh ) = (f, vh ) where, for N ∈ N,. N a (u, v) =. s (u, v) = N. a. for all vh ∈ Vh ,. (∇u · ∇v − k 2 uv) dx + s N (u, v). (16). for u, v ∈ H 1 (a ),. ⎧ (1) Hn (ka) ⎪ ⎪ ⎪ ⎪ ⎨ |n|<N −ka H (1) (ka) un (a)vn (a). if d = 2,. ⎪ (1) n N−1 ⎪ hn (ka) m ⎪ 2 ⎪ ⎩ un (a)vnm (a) −ka (1) n=0 m=−n hn (ka). if d = 3.. n. Theorem 5. Let k be an arbitrary positive number and f an arbitrary function of L2 () with compact support. Assume that O ∪ supp f ⊂ Ba0 (a0 < a). Let u be the solution of problem (1). Assume that there exists an integer l 2 such that ¯ × N satisfying h + N −1 0 , problem (16) u ∈ H l (a ). Then there exist a 0 > 0 such that for every (h, N ) ∈ (0, h] N has a unique solution uh , and moreover, if d = 2, then we have. . H (1) (ka). N m−1. u m,a + N −s+1/2 (1) (17). u − uN. R N (u; s, a0 ) , h 1,a C h. H (ka 0 ). N. . (1) N −1 m−1 −s+1/2 HN (ka) N. u − uh 0,a C(h + N ) h. u m,a + N (18). R (u; s, a0 ) ,. (1). H (ka 0 ). N where · 0,a is the usual norm of L2 (a ), m = min{p, l}, s is an arbitrary real number 21 , ⎞2 ⎛ N 2s 2 n |un (a0 )| ⎠ , R (u; s, a0 ) = ⎝a0. (19). |n| N. and positive constants 0 and C depend on k, a0 , and a , but are independent of h, N, s, f, u, and uN h . If d = 3, then (1) (1) (17) and (18) hold by replacing HN by hN , and (19) by ⎞2 ⎛ n 2⎠ n2s |um . R N (u; s, a0 ) = ⎝a02 n (a0 )| n N m=−n.
(10) D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31. 27. Before starting to prove Theorem 5, we introduce the following inequalities associated with the trace theorem:. v m−1/2,a C v m,a. for all v ∈ H m (a ) (m = 1, 2),. (20). where C is a positive constant depending on a , but independent of v, and the following sesquilinear form on H 1 (a ): ⎧ (1) Hn (ka) ⎪ ⎪ ⎪ −ka (1) if d = 2, un (a)vn (a) ⎪ ⎨ |n| N Hn (ka) N r (u, v) = (1) ⎪ n ⎪ ⎪ 2 hn (ka) um (a)v m (a) if d = 3. ⎪ −ka ⎩ n n (1) hn (ka) n N m=−n Note here that we have s(u, v) = s N (u, v) + r N (u, v). for u, v ∈ H 1 (a ).. Proof. We prove only in the case when d = 2, because the proof of the case when d = 3 is exactly the same. We first assume that problem (16) has a solution uN h . We postpone proving the well-posedness of problem (16) until completion of the derivation of (17) and (18). Set ehN = u − uN h . Then we have a N (ehN , vh ) + r N (u, vh ) = 0. (21). for all vh ∈ Vh . Note the following identical equation:. ehN 21,a = a N (ehN , ehN ) + (k 2 + 1) ehN 20,a − s N (ehN , ehN ). Taking the real part of this identity, we can get. ehN 21,a = Re{a N (ehN , ehN )} + (k 2 + 1) ehN 20,a − Re{s N (ehN , ehN )}. By virtue of Lemma 3, we have. ehN 21,a Re{a N (ehN , ehN )} + (k 2 + 1) ehN 20,a . Step 1. We show that for an arbitrary > 0, there exists a positive constant C3 ( ) such that ⎫ ⎧. ⎬. H (1) (ka) 2 ⎨. N |a N (ehN , ehN )|. ehN 21,a + C3 ( ) h2m−2 u 2m,a + N −2s+1 (1). [R N (u; s, a0 )]2 , ⎭. H (ka 0 ). ⎩ N. (22). (23). where s is an arbitrary number 21 and C3 ( ) depends on k, a0 , and a , but is independent of h, N, s, u, and uN h . By (21), we have, for all vh ∈ Vh , N N N N N a N (ehN , ehN ) = a N (ehN , u − vh ) + r N (u, uN h − vh ) = a (eh , u − vh ) + r (u, u − vh ) − r (u, eh ).. Thus, by using the trigonometric inequality, the Schwarz inequality, Lemma 2, and the trace inequality (20), we get |a N (ehN , ehN )||ehN |1,a |u − vh |1,a + k 2 ehN 0,a u − vh 0,a + C(k, a) ehN 1/2,a u − vh 1/2,a + |r N (u, u − vh )| + |r N (u, ehN )| C(k, a ) ehN 1,a u − vh 1,a + |r N (u, u − vh )| + |r N (u, ehN )|.. (24). Let us estimate the second term on the right-hand side of (24). Since O∪supp f ⊂ Ba0 , the solution u can be analytically represented as follows u(r, ) =. ∞ . (1). Hn (kr). un (a0 )Yn () (1) n=−∞ Hn (ka 0 ). on R2 \Ba0 ..
(11) 28. D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31. This implies un (a) =. (1). Hn (ka) (1). Hn (ka 0 ). un (a0 ). for all n ∈ Z. Moreover, we can see from the usual regularity argument that u|a0 ∈ H s (a0 ) for all s > 0. Thus, by the trigonometric inequality, Lemmas 2 and 4, the Schwarz inequality, and the trace inequality (20), we have, for every s 21 ,. . Hn(1) (ka). N |r (u, u − vh )|. |u (a)||(u − vh )n (a)|. ka (1). Hn (ka) n |n| N. (1). (1). −s+1/2 ka Hn (ka). Hn (ka). |n| =. (1). |n|s |un (a0 )||n|1/2 |(u − vh )n (a)|. (1). n Hn (ka). Hn (ka 0 ). |n| N. (1) −s+1/2 HN (ka) N C(k, a, a0 )N. R (u; s, a0 ) u − vh 1/2,a. (1). H (ka 0 ). N. H (1) (ka). N. N C(k, a0 , a )N −s+1/2 (1) (25). R (u; s, a0 ) u − vh 1,a ,. H (ka 0 ). N where (u − vh )n (a) are the Fourier coefficients of u − vh . In exactly the same way, we can estimate the third term on the right-hand side of (24) as follows:. (1) N N −s+1/2 HN (ka) N |r (u, eh )|C(k, a0 , a )N (26). R (u; s, a0 ) ehN 1,a .. (1). H (ka 0 ). N Combining (24)–(26) and (15) leads to |a. N. (ehN , ehN )|C(k, a0 , a ). m−1. h. ehN 1,a u m,a. +h. m−1. N. H (1) (ka). N N. R (u; s, a0 ) u m,a. (1). H (ka 0 ). −s+1/2. . H (1) (ka). N +N −s+1/2 (1) R N (u; s, a0 ) .. eN. H (ka 0 ) h 1,a. N. N. Applying the arithmetic–geometric mean inequality to each term on the right-hand side of the above inequality, we obtain (23). Step 2. We show that there exists a positive constant C4 such that. . (1) N −1 N −s+1/2 −s−1/2 HN (ka) N. eh 0,a C4 (h + N ) eh 1,a + (hN +N ) (1) (27). R (u; s, a0 ) ,. H (ka 0 ). N where s is an arbitrary number 21 and C4 depends on k, a0 , and a , but is independent of h, N, s, u, and uN h . Suppose that w ∈ V satisfies a(v, w) = (v, ehN ). (28). for all v ∈ V . Then w is the incoming solution, that is, w is the restriction to a of the solution of problem (1) where the outgoing radiation condition is replaced by the incoming radiation condition: ju + iku = 0 lim r (d−1)/2 r−→+∞ jr.
(12) D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31. 29. and f = ehN . Note here that the sesquilinear form s corresponding to the incoming radiation condition is represented by (1) (2) replacing Hn by Hn (the Hankel function of the second kind) in (5). Since Theorem 1 also holds for the incoming problem, we have w ∈ H 2 (a ) and the following a priori estimate:. w 2,a C ehN 0,a ,. (29). where C is a positive constant independent of ehN . Taking v = ehN in (28), we obtain. ehN 20,a = a(ehN , w) = a N (ehN , w) + r N (ehN , w).. (30). Subtracting (21) from (30) gives. ehN 20,a = a N (ehN , w − vh ) + r N (ehN , w) − r N (u, vh ) = a N (ehN , w − vh ) + r N (ehN , w) + r N (u, w − vh ) − r N (u, w). Employing the argument leading to (24), we can get. ehN 20,a C(k, a ) w − vh 1,a ehN 1,a + |r N (ehN , w)| + |r N (u, w − vh )| + |r N (u, w)|.. (31). Employing an argument similar to the one used in (25), we can estimate the last three terms on the right-hand side of (31) as follows: |r N (ehN , w)|C(k, a)N −1 ehN 1/2,a w 3/2,a C(k, a )N −1 ehN 1,a w 2,a ,. (1) N −s+1/2 HN (ka) N |r (u, w − vh )|C(k, a0 , a )N. R (u; s, a0 ) w − vh 1,a ,. (1). H (ka 0 ). N. H (1) (ka). N N −s−1/2 N |r (u, w)| C(k, a, a0 )N. R (u; s, a0 ) w 3/2,a. (1). H (ka 0 ). N. (1). −s−1/2 HN (ka) N C(k, a0 , a )N. (1). R (u; s, a0 ) w 2,a .. H (ka 0 ). N Collecting (31)–(34) yields. (32) (33). (34). . H (1) (ka). N. N N 2 N −s+1/2. eh 0,a C(k, a0 , a ) eh 1,a + N. R (u; s, a0 ) w − vh 1,a. (1). H (ka 0 ). N.
(13) . (1) −1 N −s−1/2 HN (ka) N + N eh 1,a + N. R (u; s, a0 ) w 2,a .. (1). H (ka 0 ). . N. Using (15) and (29), we get. . H (1) (ka). N. N N 2 N −s+1/2. eh 0,a C(k, a0 , a ) eh 1,a + N. R (u; s, a0 ) h ehN 0,a. (1). H (ka 0 ). N.
(14) . H (1) (ka). N −1 N −s−1/2 N N + N eh 1,a + N. (1). R (u; s, a0 ) eh 0,a ,. H (ka 0 ). . N. ehN 0,a ,. and further dividing by we obtain (27). Step 3. Let us collect the results above to get (17) and (18). Squaring both sides of (27) and using arithmetic-geometric mean inequality, we have ⎧ ⎫. H (1) (ka) 2 ⎨ ⎬. N. ehN 20,a 2C42 (h + N −1 )2 ehN 21,a + (hN −s+1/2 + N −s−1/2 )2 (1). [R N (u; s, a0 )]2 .. H (ka 0 ). ⎩ ⎭ N. (35).
(15) 30. D. Koyama / Journal of Computational and Applied Mathematics 200 (2007) 21 – 31. Combining (22), (23), and (35), we get ⎫ ⎧. ⎬. H (1) (ka) 2 ⎨. N. ehN 21,a . ehN 21,a + C3 ( ) h2m−2 u 2m,a + N −2s+1 (1). [R N (u; s, a0 )]2 ⎭. H (ka 0 ). ⎩ N. ⎧ ⎨. ⎫. ⎬. H (1) (ka) 2. N + C5 (h + N −1 )2 ehN 21,a + (hN −s+1/2 + N −s−1/2 )2 (1). [R N (u; s, a0 )]2 , ⎩ ⎭. H (ka 0 ). N. where C5 = 2(k 2 + 1)C42 . This implies. ⎞. H (1) (ka) 2. N {1 − − C5 (h + N −1 )2 } ehN 21,a C6 ( ) ⎝h2m−2 u 2m,a + N −2s+1 (1). [R N (u; s, a0 )]2 ⎠ ,. H (ka 0 ). ⎛. N. where C6 ( ) = C3 ( ) + (h¯ + 1)2 , and further, by taking = 21 , ⎛ ⎞. . H (1) (ka) 2 1. N − C5 (h + N −1 )2 ehN 21,a C7 ⎝h2m−2 u 2m,a + N −2s+1 (1). [R N (u; s, a0 )]2 ⎠ ,. H (ka 0 ). 2 N. ¯ × N satisfying where C7 = C6 ( 21 ). For every {h, N } ∈ (0, h] 1 2. − C5 (h + N −1 )2 41 ,. which is equivalent to h + N −1 √. 1 ≡ 0 , 4C5. we have (17). Further, from (27) and (17), we can derive (18). Step 4. We finally show the well-posedness of problem (16). For this purpose, it is sufficient to prove uniqueness of the solution of problem (16) since Vh is finite dimensional. Thus, assume now that uN h ∈ Vh is a solution of problem (16) with f = 0. Since the solution u of problem (6) with f = 0 is identically zero, it follows from (17) (or (18)) that −1 . uN 0 h = 0. Therefore we can conclude that problem (16) is well-posed when h + N Acknowledgements The author would like to thank the referees for their comments leading to improvements of this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]. G. Bao, Finite element approximation of time harmonic waves in periodic structures, SIAM J. Numer. Anal. 32 (1995) 1155–1169. P.G. Ciarlet, The Finite Element Method for Elliptic Problems, North-Holland, Amsterdam, New York, Oxford, 1978. O.G. Ernst, A finite-element capacitance matrix method for exterior Helmholtz problems, Numer. Math. 75 (2) (1996) 175–204. K. Feng, Finite element method and natural boundary reduction, in: Proceedings of the International Congress of Mathematicians, Warszawa, 1983, pp. 1439–1453. K. Feng, D.H. Yu, Canonical integral equations of elliptic boundary value problems and their numerical solutions, in: Proceedings of China–France Symposium on the Finite Element Method (1982, Beijing), Science Press, Beijing, 1983, pp. 211–252. G.J. Fix, S.P. Marin, Variational methods for underwater acoustic problems, J. Comput. Phys. 28 (2) (1978) 253–270. D. Givoli, Numerical Methods for Problems in Infinite Domains, Studies in Applied Mechanics, vol. 33, Elsevier Scientific Publishing Co., Amsterdam, 1992. D. Givoli, Exact representations on artificial interfaces and applications in mechanics, Appl. Mech. Rev. 52 (1999) 333–349. D. Givoli, Book reviews, Appl. Mech. Rev. 56 (5) (2003) B65. D. Givoli, I. Patlashenko, J.B. Keller, High-order boundary conditions and finite elements for infinite domains, Comput. Methods Appl. Mech. Eng. 143 (1997) 13–39..
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