Asymptotic expansions of the error of Boundary Element Methods I: spline Galerkin methods

In this paper we analyse the existence of asymptotic expansions of the error of Galerkin methods with splines of arbitrary degree for the approximate solution of integral equations with logarithmic kernels. These expansions are obtained in terms of an interpolation operator and are useful for the application of Richardson extrapolation and for the obtention of sharper error bounds Integral representations are useful tools to solve boundary value problems in partial differential equations. Through these representations, the original problems are transformed into integral equations. Among these, integral equations on curves with logarithmic kernels have largely been object of study. These equations are closely related to potential problems in the plane. Furthermore, they can generalise to a certain type of systems which include useful representations of the solutions of several boundary value problems. Among the numericalmethods applied to solve approximately such equations, Galerkin methods have been of great importance. In LeRoux(1974), Galerkin nite element methods were analysed. Here we will deal with a family of Galerkin methods where the test-trial function are smoothest splines. The problem we are concerned with is: given a 1 periodic function u0, nd a function g such that Z 1 0 log jx( ) x(s)j2g(s)ds = u0(s); 8s: In the previous equation x is the smooth 1 periodic parametrization of the boundary of a simply connected bounded open set in the plane, which we assume to be of logarithmic capacity di erent from 1. Speci cally we prove in the following the existence of asymptotic expansions of the error of the Galerkin method like kg Qhg M X k=f(m)hkQhfkk ChM+1; where Qh is a certain operator of interpolation onto the discrete space and f(m) is an integer depending on the degree of the splines used for the method. The expansion will hold in several Sobolev and uniform norms. Several useful consequences arise from that expansion. First, it will give asymptotic expansions of the error in midpoints and nodes of the grid, allowing consequently the use of Richardson extrapolation to improve the solution or to estimate the error. Secondly, some improved error bounds are obtained in uniform norms. Finally, there exists also the possibility of applying the mentioned extrapolation method to the approximate solution of the BVP solved by our Boundary Element Method. An example of this can be found in Sayas (1994). Throughout the following, C will denote a certain constant, independent of the parameter h, possibly di erent in each occurrence. The symbolO(hk) will denote a function of the parameter h such that divided by hk remains bounded as h goes to 0. 1 Statement of the problem Throughout the following section will be the boundary of a simply connected bounded open set in R2, given by a in nitely often di erentiable 1 periodic function x : R! R2, such that jx0(s)j 6= 0; 8s 2 R and x(s) 6= x( ); 8s; ; 0 < j sj < 1: 1 Functions on will be identi ed with 1 periodic functions on R via the parametrization x. We suppose also that the logarithmic capacity of (see Sloan (1992) for instance) is not equal to 1. We consider the periodic Sobolev spaces Hr for all r 2 R, which are identi able with the Sobolev spaces on the curve , since x is smooth. For all r, Hr is a Hilbert space with the real valued inner product (v;w)r := v̂(0)ŵ(0) +X k 6=0 jkj2rv̂(k)ŵ(k); being û(k) := Z 1 0 u(s)e 2k {sds: The space H0 is equal to the space of 1 periodic functions belonging to L2(0; 1) and so are their inner products. The spaces Hr and H r form duality pairings with respect to the H0 inner product. The single-layer operator V g(s) := Z 1 0 log jx(s) x( )j2g( )d (1) is extendable as a continuous isomorphism from Hr ro Hr+1 for every r. Hence, given u0 2 H 1 2 the variational problem (P )8><>: nd g 2 H 12 such that b(g; r) = (u0; r)0; 8r 2 H 12 ; where b(g; r) := (V g; r)0; has a unique solution. Moreover, if u belongs to C1, so does g. A Galerkin method to solve approximately (P ) consists of a family of restrictions of the original problem to nite dimensional subspaces of H 12 , generally denoted Vh, that is, a sequence of problems (Ph)8><>: nd gh 2 Vh; such that; b(gh; rh) = (u0; rh)0; 8rh 2 Vh: Each of these problems can be reformulated as a system of linear equations. Let N be a positive integer and h := 1=N: We de ne a uniform partition of R with midpoints in si := ih, for all i 2 Z and therefore with nodes in si+ 12 := i+ 1 2 h: The intervals (si 12 ; si+ 12 ) will be denoted Ii:2 Let Vh be the space of 1 periodic smoothest splines of degree m with nodes on fsi 1 2 gi2Z, that is, if m = 0, Vh := fqh 2 H0 : qhjIi 2 P0; 8ig; and if m 1, Vh := fqh 2 Cm 1 : qhjIi 2 Pm; 8ig; where Ck denotes the space of real 1 periodic functions of a real variable, whose k th derivative is continuous and Pk is the set of the polynomials of degree lesser than or equal to k. The Galerkin projection of g 2 H 12 onto Vh is de ned as the unique solution of (Ph)8><>: nd gh 2 Vh such that b(gh; rh) = b(g; rh); 8rh 2 Vh; for h small enough, and denoted Ghg = gh. The operator Gh is bounded inH 12 , uniformly in the parameter h. In LeRoux (1974), Hsiao and Wendland (1977 and 1981) it is proven that provided g 2 Hm+1, then kGhg gks Chm+1 skgkm+1; (2) for all s 2 [ m 2;m]. Consequently, Gh : Hm+1 ! (Vh; k :km) is bounded uniformly in h. For all u 2 C0 we de ne its interpolate Qhu as the unique Qhu 2 Vh such that Qhu(si) = u(si); 8i; if m is even, or Qhu(si 1 2 ) = u(si 12 ); 8i; ifm is odd. This operator is well de ned by Schoenberg-Whitney's theorem. Remark that for m 1, Qh is a projection, and that for m = 0 the de nition ot Qh can be extended to C0 + Vh, where it is a projection too. We will also consider the Sobolev spaces Wm;1 := fu 2 L1 : u(j) 2 L1; 1 j mg; with the usual essential supremum norm denoted by k : km;1: From well-known properties of the interpolation operator Qh and the boundedness of Gh it is straightforward that Gh is bounded from Wm+1;1 to Wm;1, uniformly in h. 2 Asymptotic expansion of the error In what follows [ : ] denotes the Gaussian bracket, that is, [x] denotes the largest integer lesser than or equal to x. As for the function g appearing in propositions and theorems, it will be assumed, for simplicity, to belong to C1: 3 Denote f(m) := 2 m+ 2 2 = m+ 2; if m even, m+ 1; if m odd. (3) The next sections will be devoted to proving the following result. Proposition 1 For any positive integer M , there exist functions ek, as smooth as desired, such that b(Qhg g; rh) = M X k=f(m) hk(ek; rh)0 +O(hM+1)krhk 12 ; (4) for all rh 2 Vh. The expression as smooth as desiredmust be understood in the sense that for arbitrary l 0 we will be able to nd functions in C l satisfying the requirements of the sentence. Of course, for di erent values of l, the functions are a priori di erent. As a consequence of this preliminary result, we obtain the main theorem of this work. Theorem 2 For all positive integer M , there exist functions fk, as smooth as desired, such that kGhg Qhg M X k=f(m) hkQhfkkm;1 CthM+1: (5) Proof: From the de nition of the Galerkin projection, (4) holds also for all rh 2 Vh substituting g by Ghg. Taking M +m + 1 terms in Proposition 1, we have that for all rh 2 Vh, b(Qhg Ghg M+m+1 X k=f(m) hkGhg1;k; rh) = O(hM+m+2)krhk 1 2 (6) where g1;k := V 1ek. Thus, we obtain kQhg Ghg M+m+1 X k=f(m) hkGhg1;kk 1 2 ChM+m+2; (7) by the boundeness of Gh in H 1 2 . From the inverse inequalities (see Arnold and Wendland (1983) and Aubin (1972)) it follows that for all rh 2 Vh krhkm;1 Ch 1 2krhkm C 0h m 1krhk 1 2 : Then kQhg Ghg M+m+1 X k=f(m) hkGhg1;kkm;1 ChM+1; from where kQhg Ghg M X k=f(m)hkGhg1;kkm;1 ChM+1; (8) because of the uniform boundedness of Gh. 4 Since, by (8) applied to the functions g1;k, we have kGhg1;k Qhg1;kkm;1 Chf(m); then kQhg Ghg 2f(m) 1 X k=f(m) hkQhg1;kkm;1 Ch2f(m): (9) Again, (9) can be applied to the functions g1;k and that gives the existence of functions g2;k such that kQhg Ghg 3f(m) 1 X k=f(m) hkQhg2;kkm;1 Ch3f(m) and thus the statement of the theorem follows by induction. Corollary 3 Let h be the set of the midpoints of the grid. Then for all 0 l m, there exists a sequence of constants fFk;l(z) : z 2 hg, independent of h, such that sup z2 h j(Ghg)(l)(z) g(l)(z) M X k=2[m l+2 2 ]hkFk;l(z)j = O(hM+1) for all positive integer M . Let h be the set of the nodes of the grid. Then for all 0 l m 1, there exists a sequence of constants fF k;l(z) : z 2 hg, independent of h, such that sup z2 h j(Ghg)(l)(z) g(l)(z) M X k=2[m l+2 2 ]hkF k;l(z)j = O(hM+1) for all positive integer M . Proof: The rst result is obvious for m even and l = 0, since the operator Qh interpolates in the midpoints The other cases follow from the existence a set of real numbers fCl;ng such that for all u smooth enough sup z2 h j(Qhu)(l)(z) u(l)(z) M X n=m+1 l hnCl;nu(l+n)(z)j = O(hM+1): (10) where Cl;n = 0 if n is odd. A proof of these expansions can be found in Sayas (1994) and is a straightforward consequence of the properties of interpolation with periodic splines. The trivial case of the second result is when m is odd and l = 0, since we are interpolating in the nodes of the grid. Expansions similar to (10) hold also in the nodes for l m 1 and consequently the statement of the corollary is a simple consequence of (5). Both sequences of coe cients fFk;l(z)gz2 h and fF k;l(y)gy2 h are unique since the dependence of h has been done away with once we restrict ourselves to nodes or midpoints. Moreover, they are pointwise values of functions which can be taken as smooth as desired, since they are linear combinations of derivatives of g and of the functions fk. Remark that these functions are not unique, but only their values in the nodes or midpoints and that expansions are di erent in h and h.5 Corollary 4 For m even we have the following superconvergence results kGhg Qhgk0;1 Chm+2; (11) and for m odd, kGhg gk0;1 Chm+1 (12) Remark that (11) when m = 0 improves a similar superconvergence result obtained by Shu (1992 and 1994), which was given in L2 norm. For m odd the order of approximation in the uniform norm between Ghg and Qhg is the same as that of the interpolation and therefore (12) holds. That supposes that we get the same order of convergence in the norm of L1 as what we had before only in the weaker L2 norm. Since for m even the order of approximation between the Galerkin projection and the interpolation is higher than that between the interpolate and the function, we do not obtain order m+2 for the Galerkin method in L1 norm. However, there is a simple way of de ning h such that k h gk0;1 Chm+2; (13) when m is even. In order to do that, consider the periodic spline h of degree m+1 with nodes in the points si (that is in the midpoints of the original grid) such that h(si) = Ghg(si); 8i: Then sup 1 i N j h(si) g(si)j = O(hm+2); and by classical results of interpolation with splines, (13) follows. An important aspect of the Galerkin method is left undealt in this work. This one is the need of numerical integration for the coe cients and right-hand side of the linear system solved to obtain the Galerkin solution. In Crouzeix and Sayas (1995) we propose a family of fully discrete numerical methods based on the previously studied method and which fall into the frame of the Galerkin collocation methods, de ned in Hsiao, Kopp and Wendland (1980). Some numerical tests are given there. 3 The piecewise constant case The aim of this section is to prove Proposition 1 in the simplest case of the piecewise constant functions, that is, when m = 0. The proofs for the other cases will include a great part of what is done here. In the present case we have b(Qhg g; rh) = Z 1 0 rh( ) N Xi=1 ZIi(g(si) g(s)) log jx( ) x(s)j2ds!d : By Taylor's theorem and the boundedness properties of the operator V , it can be easily seen that for every positive integer M and for all rh 2 Vh, b(Qhg g; rh) = M X k=1 ( 1)k k! Z 1 0 rh( ) k( ;h)d +O(hM+1)krhk 12 ; (14) 6 where k( ;h) := N Xi=1 ZIi bk( ; s)(s si)kds and bk( ; s) := log jx( ) x(s)j2g(k)(s): In the following ' 2 D(R) will be an even cut-o function supported in [ 1 3; 13] and such that ' 1 in [ 1 6; 1 6 ]. For all positive integer p we de ne ak;p( ; s) := log jx( ) x(s)j2 '( s) log( s)2 g(k)(s)+ (15) +'( s) log( s)2 g(k)(s) p Xl=0( 1)l g(k+l)( ) l! ( s)l! : Then ak;p 2 Cp(D), where D := f( ; s) : j sj < 1g. If we de ne k;l( ;h) := ( 1)l g(k+l)( ) l! 1 X i= 1 ZIi(s si)k( s)l'( s) log( s)2ds it follows that for all 2 Ij and for every j k( ;h) = p Xl=0 k;l( ;h) + X i2I(N;j) ZIi ak;p( ; s)(s si)kds; (16) where I(N; j) := 8><>: fj q; j q + 1; : : : ; j + qg if N = 2q + 1, fj q; j q + 1; : : : ; j + q 1g if N = 2q. Note that this choice of indices implies that in (16), for h 1=6; j sj 1 2 + h 2 3 : Proposition 5 For all k 2 there are functions k;n 2 Cp n; n = 0; : : : ; p 1, independent of h, such that uniformly for all 2 Ij and for all j X i2I(N;j) ZIi ak;p( ; s)(s si)kds = p 1 X n=0 hk+n k;n( ) +O(hk+p): In addition, k;n 0 if k + n is odd. Proof: We de ne by induction a biparametric family of polynomials as follows Pk;0(u) := uk; Pk;n+1(u) := Z u 1 2 (Ck;n Pk;n(t))dt; 7 for n 0, where Ck;n := Z 1 2 12 Pk;n(t)dt: It is easy to see that Ck;n = 0 when k + n is odd, that Pk;n s si h = Ck;n h d ds Pk;n+1 s si h and that Pk;n+1 1 2 = Pk;n+1 1 2 = 0 for all n 0. Then by induction we get ZIi ak;p( ; s)(s si)kds = hk ZIi ak;p( ; s)Pk;0 s si h ds = = p 1 X n=0Ck;nhk+n ZIi @nak;n @sn ( ; s)ds+ hk+p ZIi @pak;n @sp ( ; s)Pk;p s si h ds: We set k;n( ) := Ck;n X i2I(N;j) ZIi @nak;p @sn ( ; s)ds; if 2 Ij; for n p 1. Since @p @spak;p is uniformly bounded for j sj 2 3 , it follows X i2I(N;j) ZIi ak;p( ; s)(s si)kds p 1 X n=0hk+n k;n( ) = O(hk+p); for all 2 Ij and for all j. We remark that for s 2 [ 1 2 h; 1 2 + h], from (15) and '( s) = 0, ak;p( ; s) = bk( ; s) = bk( ; s+ 1) = ak;p( ; s+ 1); so that k;n( ) = Ck;n Z + 12 12 @nak;p @sn ( ; s)ds from where it is obvious that k;n 2 Cp n. The only restrictions on the smoothness of the functions ek in the statement of Proposition 1 come from this term. Obviously if we want all functions to belong to C l it is enough to take p = l k +M + 2 (see also Proposition 8). We consider now the other terms in (16). By the Taylor expansion of g(k+l) and the following change of variables in Ii Ij, u = s si h ; v = sj h ; 8 we can see that ZIj k;l( ;h)d = n X r=0 ( 1)l r!l! g(k+l+r)(sj)hk+r+1 k;r;l(h) +Wn+1;k;l;j(h) (17) where k1 ;k2;k3(h) := h 1 X m= 1Z 12 12 Z 12 1 2 uk1vk2 ((u v+m)h)k3 '((u v+m)h) log j(u v+m)2h2jdudv (18) and Wn+1;k;l;j(h) := ( 1)l k!l!(n+ 1)! ZIj g(k+l+n+1)( ) 0@ 1 X i= 1 ZIi log( s)2( s)l'( s)(s si)k( sj)n+1ds1Ad ; being a certain point in Ij for each 2 Ij. Remarking that in the previous sum indices vary from N to N , it is simple to prove that jWn+1;k;l;j(h)j Chn+k+1: (19) It is easy to observe that if k1 + k2 + k3 is odd, then k1;k2;k3(h) = 0: The other cases are studied in the following lemma, whose proof (due to its length and purely technical character) will be given in the next section. Lemma 6 If k1 + k2 + k3 is even, then there is a set of constants (independent of h, but depending on k1; k2 and k3) such that, for all n 0 k1;k2;k3(h) = C0 + C1h+ C2h2 + : : :+ Cnhn +O(hn+1): Then repeated applications of Lemma 6 prove the existence of a sequence of functions fck;rgr 0 C1, such that p Xl=0 ZIj k;l( ;h)d = n 1 X r=0 hk+r+1ck;r(sj) +O(hk+n+1) (20) for all n. Proposition 7 There exists a sequence fdk;rgr 0 C1 such that p Xl=0 Z 1 0 k;l( ;h)rh( )d = n 1 X r=0 hk+r(dk;r; rh)0 +O(hk+n)krhk 12 for all rh 2 Vh and for all n. 9 Proof: This is a simple consequence of (20), together with the following argument. Let us rstly remark that for all rh 2 Vh, h N Xj=1 jrh(sj)j = krhkL1(0;1) Ch 1 2 krhk 1 2 by the inverse inequalities. There exist constants D1; : : : ;DM such that p(0) = Z 12 1 2 p(x)dx+ M Xl=1Dl Z 12 12 p(2l)(x)dx; for all p 2 P2M+1: Therefore, for any 2 C1, we have h (sj) = ZIj (s)ds+ M Xl=1 h2lDl ZIj (2l)(s)ds+O(h2M+2); uniformly for all j. Taking one more term we get that for all rh 2 Vh, [ ; rh]h = ( ; rh)0 + M Xl=1 h2lDl( (2l); rh)0 +O(h2M+2)krhk 12 ; (21) where [ ; rh]h := h N Xj=1 (sj)rh(sj): Obviously this applies to all the ck;r, which proves the result. Finally, from (16), Proposition 5 and Proposition 7, the next expansion follows readily. Proposition 8 For all p, there exists a set of functions fek;rgp 1 r=0 such that ek;r 2 Cp r and that Z 1 0 k( ;h)rh( )d = p 1 X r=0 hk+r(ek;r; rh)0 +O(hk+p)krhk 12 for all rh 2 Vh. If k is odd, by Proposition 5 ZIj k( ;h)d = p Xl=0 ZIj k;l( ;h)d +O(hk+2); In addition, taking n = 1 in (17) and (19) ZIj k;l( ;h)d = ( 1)l l! g(k+l)(sj)hk+1 k;0;l(h) +O(hk+2) and since k;0;l(h) = O(h) for all l when k is odd ( nal remark in the next section), then for all rh 2 Vh, Z 1 0 k( ;h)rh( )d = O(hk+1)krhk 1 2 ; that is, ek;0( ) 0. Now Proposition 1 follows readily from this last Proposition and (14). Since only the term containing 1 would appear in the rst order term, and e1;0 0, the expansion begins with the second power of h. 10 4 Asymptotic study of some coe cients In the previous section Lemma 6, that is the asymptotic expansion in powers of h of k1;k2;k3(h) (see (18) for its de nition) when k1 + k2 + k3 is even, was left unproven. We will prove this in a more general case, which will include also similar expansions appearing in the study of the error of the Galerkin Boundary ElementMethod with splines of higher degree. Throughout this section Pk1 will denote a certain polynomial of degree k1 such that Pk1( t) = ( 1)k1Pk1(t); and for j > k1, Pj(t) := tk1 jPk1 (t): By n we will denote the n+1 fold convolution of the characteristic function of ( 1 2 ; 12). Consequently, n is a positive even function whose support is [ n+1 2 ; n+1 2 ], lesser than or equal to 1 everywhere in R and with unit integral. We now de ne k1;k2;k3(h) := hX m2ZZ 1 2 12 Z n+1 2 n+1 2 Pk1(u)vk2 n(v) ((u v+m)h)k3 '((u v+m)h) log j(u v+m)hj2dudv where it is understood that the domain of integration of the rst variable (here u) appears in the rst integral sign (here from 12 to 1 2). In the particular case n = 0, Pk1(u) = uk1, this expression coincides with (18). As in that particular situation, this coe cient is zero when k1 + k2 + k3 is odd, which follows easily by a simple change of variable, so we will restrict our study to k1 + k2 + k3 even. In the lemmas of this section the sequences of real constants fCkg, independent of h, are di erent in each occurence. Henceforth we consider xed (but arbitrary) values of k1; k2 and k3 such that k1 + k2 + k3 is even. Lemma 9 If we de ne ~ k1;k2;k3(h) := hX m2Z'(mh)Z 12 1 2 Z n+1 2 n+1 2 Pk1(u)vk2 n(v) ((u v+m)h)k3 log j(u v+m)hj2dudv; then there exists a set of constants, independent of h, such that k1;k2;k3(h) = ~ k1;k2;k3(h) + C1h+ : : :+ Crhr +O(hr+1); for all r 1. Proof: Let us de ne a function (x; y) := log jx+ yj2('(x+ y) '(x)); 11 which belongs to D(R [ 1 12; 1 12 ]), because ' 1 on [ 16; 1 6]. Then if h is assumed to be lesser than 1 3n+6 , k1;k2;k3(h) ~ k1;k2;k3(h) = h Z 12 12 Z n+1 2 n+1 2 Pk1(u)vk2 n(v) 0@X m2Z((u v+m)h)k3 (mh; (u v)h)1Adudv: By the Euler-Maclaurin formula, that di erence equals Z 1 2 1 2 Z n+1 2 n+1 2 Pk1 (u)vk2 n(v) (u; v;h)dudv+O(hr+1); for all r, where (u; v;h) := Z 1 1 (x; (u v)h) (x+ (u v)h)k3 dx is a C1 function of the variable h, for h h0 (so that h(u v) is inside the domain where is smooth, for all (u; v) in the domain of integration). Noticing that (x; 0) 0; the Taylor expansion of as a function of h around zero gives the statement of the lemma. Expanding the polynomials appearing in the de nition of ~ we obtain ~ k1 ;k2;k3(h) = X i1+i2+i3=k3 k3! i1!i2!i3!( 1)i2hi1+i2 k1+i1;k2+i2;i3(h); (22) being j1;j2;j3(h) := hX m2Zxj3 m'(xm)Z 12 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v) log j(u v+m)2h2jdudv; and xm := mh for all m 2 Z. That is, ~ k1;k2;k3(h) is a linear combination of coe cients of the form j1;j2;j3(h). When j1 + j2 + j3 is odd j1+j2+j3(h) = 0: However, only the other case happens in (22), when k1 + k2 + k3 is even. We will prepare the way to the asymptotic study of these terms with two merely technical lemmas. Lemma 10 There are two constants C0 and C1, independent of h, such that log h2 + 2 1 X m=1'(xm) log x2m = C0 h + C1 +O(hr); for all r 1. 12 Proof: We denote (h) := 1 X m=1'(xm) log x2m = 1 X m=1'(mh) log(mh)2; and recall that '(h) = 1 if h < 1 6 . Then for those values of h and denoting xm+ 1 2 = (m+ 12)h, (h) 1 2 log h2 = 1 2'(h) log h2 + 1 X m=2'(xm) log x2m = = 1 h 1 X m=1 Z xm+1 xm '(x) log x2dx+ 1 X m=1 Z xm+1 xm (x xm+ 1 2 )('(x) log x2)0dx! = = 1 h Z 1 0 '(x) log x2dx log h2 + 2 + 1 h 1 X m=1 Z xm+1 xm (x xm+ 1 2 )('(x) log x2)0dx: Let a(x) be a smooth function de ned on [0;1) and set A(x) := Z x 0 a(s)ds: Then, applying integration by parts, Z xm+1 xm (x xm+ 12 )a(x)dx = h2 (A(xm+1) +A(xm)) Z xm+1 xm A(x)dx; from where, by the Euler-Maclaurin formula for A in the interval [h;Mh], M 1 X m=1 Z xm+1 xm (x xm+ 1 2 )a(x)dx = = l Xj=1 2jh2j a(2j 2)(h) a(2j 2)(Mh) + h2l+2R2l+2; (23) where jR2l+2j C Z Mh h ja(2l+1)(x)jdx: We use (23) with a(x) = '0(x) log x2, compactly supported in (0;+1) and vanishing in [0; h] for h 1 6, by which we obtain that for all r 1 X m=1 Z xm+1 xm (x xm+ 1 2 )'0(x) log x2dx = O(hr): Finally, we study 1 h 1 X m=1 Z xm+1 xm (x xm+ 1 2 )'(x) 1 x dx+ 1 h 1 X m=1 Z xm+1 xm (x xm+ 1 2 ) x dx: 13 The second term of that expression can be shown, through simple calculations, to be equal to the convergent sum := 1 X m=1 1 (m+ 12) log m+ 1 m and for the rst one, we simply have to take a(x) = (1 '(x))=x in (23) and M !1 to see that it is an O(hr) for all r. To get the statement of the lemma, we only need to remark that log h2 + 2 1 X m=1'(xm) log x2m = 2 h Z 1 0 '(x) log x2dx+ 4 + 2 +O(hr) for all r, which gives the result. Lemma 11 For all k even and positive, there exists Ck such that h 1 X m=1 xkm'(xm) log x2m = Z 1 0 xk'(x) log x2dx+ Ckhk+1 +O(hr) for all r. Proof: Let us denote for h > 0 k(h) := h 1 X m=1 xkm'(xm) log x2m = h 1 X m=1(mh)k'(mh) log(mh)2; and for h = 0, k(0) := Z 1 0 xk'(x) log x2dx: Then, since xk'(x) log x2 is Riemann integrable in (0;1), it follows that k is rightcontinuous in h = 0. Besides, k 2 C1(0;1) since for all h only a nite number of terms are not zero. Thus, deriving k we get that 0 k(h) = k + 1 h k(h) + 1 X m=1(mh)k+1'0(mh) log(mh)2 + 2 1 X m=1(mh)k'(mh): Given that xk+1'0(x) log x2 2 C1(0;1) and has compact support in (0;1), by the EulerMaclaurin formula, h 1 X m=1 xk+1 m '0(xm) log x2m = Z 1 0 xk+1'0(x) log x2dx+O(hr) for all r. Likewise, h 1 X m=1xkm'(xm)=Z 1 0 xk'(x)dx M Xl=1 2lh2l xk'(x) (2l 1) x=0+O(h2M+2); and since k is even and ' 1 in a neighbourhood of x = 0, all the terms in the previous sum are zero. 14 Hence 0 k(h) = k+1 h k(h) + 1 h Z 1 0 xk+1'0(x) log x2dx+ 2 h Z 1 0 xk'(x)dx+O(hr); for all r and integrating by parts we can see that k satis es the following di erential equation 0 k(h) = k + 1 h ( k(h) k(0)) +O(hr): Let then k(h) := k(h) k(0); which is smooth on (0;1) and continuous in h = 0. Obviously, k(0) = 0 and 0k(h) = k + 1 h k(h) +O(hr) for all r. Therefore, if we de ne Ck(h) := k(h) hk+1 ; which is also smooth on (0;1) it follows that C 0 k(h)hk+1 = O(hr) for all r, which implies that C 0 k can be extended by continuity to C 0 k(0) = 0. Thus, there exists also a continuous extension of Ck(h) to h = 0, say Ck(0) = Ck and consequently, by the Mean Value Theorem and the de nition of Ck(h), k(h) = Ckhk+1 +O(hr) for all r, which proves the result. After these lemmas, we begin the study of j1;j2;j3(h), rst when j3 = 0 and afterwards when j3 is positive. As we have already indicated, we will restrict ourselves to the case j1 + j2 + j3 even, being the coe cient equal to zero otherwise. Lemma 12 For all j1; j2, there is a set of constants, independent of h but depending on j1 and j2, such that j1;j2;0(h) = C0 + C1h+ : : :+ Crhr +O(hr+1); for all r 1. Proof: We have j1;j2;0(h) =hX m2Z Z 1 2 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v)'(mh) log j(u v+m)hj2dudv; 15 so taking apart the term corresponding to m = 0 and using the symmetry between those of m and m, j1;j2;0(h) = h Z 12 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v) log(u v)2dudv+ +h log h2 + 2 1 X m=1'(mh) log(mh)2!Z 12 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v)dudv + +2h 1 X m=1'(mh) Z 1 2 1 2 Z n+1 2 n+1 2 Pj1 (u)vj2 n(v) log 1+u v m 2 dudv; and applying there Lemma 10, j1;j2;0(h) = C 0 0 + hC 0 1+ +4h 1 X m=1'(mh) Z 12 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v) log 1 + u v m dudv +O(hr) for all positive integer r. We consider the remainder in the Taylor expansion of the logarithm, Rr+1(x) := log j1 + xj+ r X j=1 ( 1)j j xj; which is a locally integrable function. If m > n+2 2 , Rr+1 u v m C(r) ju vjr+1 mr+1 C(r) n+ 2 2m r+1 (24) for all (u; v) 2 [ 1 2; 1 2] [ n+1 2 ; n+1 2 ]. We write '(y) log j1 + xj = x'(y) +R2(x) + (1 '(y))0@ r X j=2 ( 1)j j xj Rr+1(x)1A ; and get h 1 X m=1'(mh) Z 12 1 2 Z n+1 2 n+1 2 Pj1 (u)vj2 n(v) log 1 + u v m dudv = = h 1 X m=1 '(mh) m D1 + h 1 X m=1 Z 1 2 1 2 Z n+1 2 n+1 2 Pj1(u)vj2 n(v)R2 u v m dudv + + r X j=2 ( 1)j j h 1 X m=1 (1 '(mh)) mj !Dj h 1 X m=1(1 '(mh)) Z 1 2 12 Z n+1 2 n+1 2 Pj1 (u)vj2 n(v)Rr+1 u v m dudv; where Dj := Z 12 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v)(u v)jdudv; (25) 16 for any j 0. Since j1 + j2 is even, D1 = 0. The series in the second term of the previous expression converges, because of (24). For the third one, we apply again the Euler-Maclaurin formula (recall we denote xm = mh) h 1 X m=1 1 '(mh) mj = hj+1 1 X m=1 1 '(xm) xjm = hj Z 1 0 1 '(x) xj dx+O(hr+1); for all r, since all the derivatives of (1 '(x))=xj are integrable on (0;1) if j 2 and their integrals are zero. The last term can be bounded by Chr+1 h 1 X m=1 1 '(xm) xr+1 m ! = O(hr+1); which completes the proof. Lemma 13 For all j1; j2; j3, there is a set of constants, independent of h but depending on j1; j2; j3, such that j1;j2;j3(h) = C0 + C1h+ : : :+ Crhr +O(hr+1); for all r 1. Proof: We recall that j1;j2;j3(h) := hX m2Zxj3 m'(xm)Z 12 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v) log j(u v+m)2h2jdudv: As we did in the case j3 = 0 we separate terms j1;j2;j3(h) = 0@h X m2Zxj3 m'(xm) log x2m1AD0 + 4~ j1;j2;j3(h); with D0 given by (25) with j = 0 and ~ j1;j2;j3(h) :=h 1 X m=1xj3 m'(xm)Z 1 2 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v) log 1+u v m dudv; since j1 + j2+ j3 is assumed to be even. Lemma 11 gives the expansion in powers of h of the rst term in case that j3 is even. If j3 is odd, that quantity is null. For the second term we write '(y) log j1 + xj = '(y)0@j3+1 X j=1 ( 1)j j xj1A+Rj3+2(x) (1 '(y))Rj3+2(x) = = '(y)0@j3+1 X j=1 ( 1)j j xj1A+Rj3+2(x) + (1 '(y))0 @ r X j=j3+2 ( 1)j j xj Rr+1(x)1A 17 as in Lemma 12, and thus ~ j1;j2;j3(h) = j3 1 X j=1 ( 1)j j hj h 1 X m=1xj3 j m '(xm)!Dj ( 1)j3 j3 hj3 h 1 X m=1'(xm)!Dj3+ + ( 1)j3 j3+1 hj3+1 h 1 X m=1 '(xm) xm !Dj3+1 + + hj3+1Z 12 12 Z n+1 2 n+1 2 Pj1 (u)vj2 n(v) 1 X m=1mj3Rj3+2 u v m !dudv+ + r X j=j3+2 hj ( 1)j j h 1 X m=1xj3 j m (1 '(xm))!Dj h 1 X m=1xj3 m(1 '(xm))Z 12 12 Z n+1 2 n+1 2 Pj1(u)vj2 n(v)Rr+1 u v m dudv; where the coe cients Dj (depending only on j; j1 and j2) are given by (25). By the Euler-Maclaurin formula, the rst, second and fth terms have the following asymptotic behaviour h 1 X m=1 xj3 j m '(xm) = Z 1 0 xj3 j'(x)dx+O(hl); j < j3; h 1 X m=1'(xm) = Z 1 0 '(x)dx h2 +O(hl); h 1 X m=1(xm)j3 j(1 '(xm)) = Z 1 0 xj3 j(1 '(x))dx+O(hl); j > j3 + 1; for all l. Being j1 + j2 + j3 + 1 odd, Dj3+1 = 0, and hence, the third term is zero. The series in the fourth term converges, since for all m > n+2 2 , (24) holds, and then mj3Rj3+2 u v m C 0 1 m2 for all (u; v) in the domain of integration. Similar reasons show that the last one is also an O(hr+1): Finally, the gathering of Lemma 9, (22), Lemmas 12 and 13 proves easily the expansion of in powers of h, a particular case of which is Lemma 6. Lemma 14 For all k1; k2; k3 such that k1 + k2 + k3 is even, there is a set of constants, independent of h but depending on (ki)3i=1, such that k1;k2;k3(h) = C0 + C1h+ : : :+ Crhr +O(hr+1) for all r 1, for h su ciently small. 18 Remark: In the proof of Lemma 12 it can be easily seen that if k3 is odd, then k1;k2;k3(h) = 4~ k1 ;k2;k3(h) = O(h) which means that in that situation k1;k2;k3(h) = O(h); since from (22), ~ k1;k2;k3(h) = k1;k2;k3(h) +O(h): 5 Smoothest splines of arbitrary degree The object of this section is the proof of Proposition 1 for other spline spaces than the step functions, studied in Section 3. Most of the proofs will be basically those of the piecewise constant functions. For the de nitions of the spaces and of the interpolation operator Qh, see the introductory section. An important point is the existence of a local expansion of Qhg g in each interval Ii, uniform in s and i, given by Qhg(s) g(s) = M X n=m+1 g(n)(s)hnPn s si h +O(hM+1) where fPn(t)gn 2 is a sequence of polynomials of degree n each, depending only on the degree m of the spline space, such that Pn( t) = ( 1)nPn(t): A proof of that can be found in Sayas (1994). That allows us to obtain an expansion similar to (14), namely, for every positive integer M and for all rh 2 Vh, b(Qhg g; rh) = M X k=m+1 Z 1 0 rh( ) k( ;h)d +O(hM+1)krhk 1 2 ; (26) where k( ;h) := hk N Xi=1 ZIi bk( ; s)Pk s si h ds and bk( ; s) := log jx( ) x(s)j2g(k)(s): Following the notations of Section 3, let ' 2 D(R) be an even cut-o function, such that ' 1 in [ 1 6; 1 6 ] and whose support is in [ 13; 1 3]. Let then k;l( ;h) := ( 1)l g(k+l)( ) l! hk 1 X i= 1 ZIiPk s si h ( s)l'( s) log( s)2ds: Following the proof of Proposition 5 (note that bk( ; s) remains unchanged), the next result can be easily proven. 19 Proposition 15 For all k m + 1 and for all positive integer p, there are functionsfk;ngp 1n=0, independent of h, such that k;n 2 Cp n and thatk( ;h) = pXl=0 k;l( ;h) + p 1Xn=0 hk+n k;n( ) +O(hk+p);(27)uniformly for all . Moreover, if k + n is odd, k;n 0.As in Section 3, p is xed depending on the desired smoothness of the functions el inProposition 1. Let us denote zj :=8><>: sj; if m even,sj 12 ; if m odd,that is fzjg is the set of the interpolation nodes de ning the operator Qh. Let m be asin Section 4, letj( ) := m s zjhand letJj := zj m+ 12 h; zj + m+ 12 hbe its support.Proposition 16 There is a sequence of functions fck;rgr 0 C1, such thatpXl=0 ZJj k;l( ;h) j( )d = n 1Xr=0 hk+r+1ck;r(zj) +O(hk+n+1)(28)for all n and for all j.Proof: By Taylor expansions of g(k+l), the change of variablesu = s sih ; v = s zjh ;in Ii Jj, following the same steps as in the piecewise constant case and using Lemma14 (a generalization of Lemma 6, proven in Section 4), it follows thatZJj k;l( ;h) j( )d = n 1Xr=0 ( 1)lr!l! g(k+l+r)(zj)hk+r+1 k;r;l(h) +O(hk+n+1):for all j; l and n. From the expansions of the coe cients k;r;l(h) in powers of h given inLemma 14 the statement of the proposition follows.Proposition 17 There exists a sequence fdk;rgr 0 C1 such thatpXl=0 Z 10 k;l( ;h)rh( )d = n 1Xr=0 hk+r(dk;r; rh)0 +O(hk+n)krhk 12for all rh 2 Vh and for all n.20 Proof: This is a simple consequence of the previous result. If rh 2 Vh, thenrh( ) = NXj=1 rh;j j( );where the basic functions j are the 1-periodic extensions of j outside Jj . Because of awell-known property of the B-spline basis (see for instance de Boor (1978)), and by theinverse inequalitiesh NXj=1 jrh;jj Ch NXj=1 jrh(zj)j C0krhk0 C 00h12krhk 12 :From the existence of constants D1; : : : ;DM such thatp(0) =ZR p(x) m(x)dx+ MXl=1Dl ZR p(2l)(x) m(x)dxfor all p 2 P2M+1, it follows, like in Proposition 7, that for all 2 C1 and for all M[ ; rh]h = ( ; rh)0 + MXl=1 h2lDl( (2l); rh)0 +O(h2M+2)krhk 12 ;(29)where[ ; rh]h := h NXj=1 (sj)rh;j:Since from Proposition 16 it follows thatpXl=0 Z 10 k;l( ;h)rh( )d = n 1Xr=0 hk+r[ck;r; rh]h +O(hk+n+1)krhk 12the result is a straightforward consequence of (29).Obviously, from (26), Propositions 15 and 17, Proposition 1 follows readily.When m is even, for the same reasons as in Section 3, we have that for all rh 2 Vh,Z 10 m+1( ;h)rh( )d =O(hm+2)krhk 12 ;since m + 1 is odd. 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