rate of convergence. (Here " is the thickness of the layer and p is the degree of the approximating polynomial.) The importance of the reslts in 13 li

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1 athematical odels and ethods in Applied Sciences fc World Scientic Pblishing Company Vol. 3, o.8 (1998) THE hp FIITE ELEET ETHOD FOR SIGULARLY PERTURBED PROES I SOOTH DOAIS CHRISTOS A. XEOPHOTOS Department of athematics The Ohio State University at Lima 440 Camps Drive Lima, OH 45804, U.S.A. We consider the nmerical approximation of singlarly pertrbed elliptic problems in smooth domains. The soltion to sch problems can be decomposed into a smooth part and a bondary layer part. We present gidelines for the eective resoltion of bondary layers in the context of the hp nite element method and we constrct tensor prodct spaces that approximate these layers niformly at a near-exponential rate. Keywords Bondary layer, singlarly pertrbed problem, hp version. 1. Introdction We stdy the nite element approximation of elliptic bondary vale problems whose soltions contain bondary layers. Bondary layers are rapidly varying soltion components that have spport in a narrow neighborhood of the bondary of the domain. They arise in the stdy of plates and shells in strctral mechanics and in heat transfer problems with small thermal coecients. In 11, it was shown that bondary layers are prominent in the stdy of shells, where their eective resoltion was of otmost importance. Also, in 1, bondary layers arising in plate theory (and in particlar in the Reissner-indlin plate model) was discssed. The reslts presented here provide the grondwork for ftre directions in the approximation of bondary layers arising in plate and shell theory. In one-dimension, bondary layers have the general form " (x) = exp(?x="), x I R (1.1) where " (0; 1] is a small parameter that can approach zero. Soltions to singlarly pertrbed elliptic bondary vale problems contain fnctions sch as (1.1). The approximation of these fnctions was stdied in 13, where an hp nite element method scheme with robst convergence properties was presented. In particlar, it was shown that adding an extra element of size O(p") near the bondary gives an hp scheme that approximates bondary layers niformly in " at an exponential

2 rate of convergence. (Here " is the thickness of the layer and p is the degree of the approximating polynomial.) The importance of the reslts in 13 lies in the fact that soltions to two-dimensional problems exhibit similar eects that tend to be of one-dimensional natre. That is, the soltion contains components of the type (1.1), where x denotes the distance normal to the bondary, c.f. 1, 9. In this paper, we extend the one-dimensional ideas to two-dimensional problems over smooth domains, whose soltions contain parts of the form (; ) = C()e?=" ; (1.) where C() is smooth. In (1.), ; denote, respectively, the arc length and the normal distance to the bondary, of a point x in a neighborhood of Using the reslts of 13 we will constrct two-dimensional tensor prodct nite element spaces in the (; ) coordinates, for approximating fnctions of the type (1.). We will consider a singlarly pertrbed elliptic bondary vale problem and provide a decomposition for its soltion into a smooth part, a bondary layer part (like (1.)) and a smooth remainder. Using tensor prodct spaces, we will approximate each component of the soltion, obtaining an arbitrarily high algebraic convergence rate, niform in the energy norm. Finally, we will illstrate the validity of or reslts throgh nmerical comptations in the case when is the nit disk. (The reslts presented here are taken from 16.) Throghot this paper, H k () will denote the Sobolev space of order k 0 on a domain R, with H 0 () = L (), and kk k; ; jj k; denoting the norm and seminorm as sal. Whenever there is no confsion we omit the sbscript. The set C n () will denote continos fnctions with n continos derivatives and p () will denote the set of polynomials of degree p in each variable, over. The letters K; C (withot any dependencies on variables) will be sed to denote generic positive constants, possibly not the same in each occrrence.. The odel Problem and its Reglarity We will consider the singlarly pertrbed model bondary vale problem L " "?" " + " = f in ; (.1) " = 0 on ; where R is a smooth domain with analytic, and " (0; 1] (This model problem is the two-dimensional analog of the bondary vale problem considered in 13.) We cast (.1) into an eqivalent variational form that reads Find " H 1 0 () sch that where B " ( " ; v) = F (v) 8 v H 1 0 (); (.) B " (; v) = Z " rvr + v dxdy; (.3)

3 and Z F (v) = fvdxdy (.4) For every f L () there exists a niqe soltion " H0 1 (), to (.), for every " (0; 1] We will show that " can be decomposed into a smooth part " ; a bondary layer part and a smooth remainder. For this prpose, we dene bondary tted coordinates in a region 0 = fz??! n z z ; 0 < < 0 < min. radis of crvatre of g ; (.5) near the bondary, by the correspondence (; )! z??! n z = X()? Y 0 (); Y () + X 0 () In (.5), z = z() = (X(); Y ()) and?! n z denotes the otward nit normal at z (see Figre 1) We have the following decomposition theorem whose proof appears in the appendix (see also 8 for other decompositions).. (x,y) ρ θ Ω Fig. 1. Bondary tted coordinates. Theorem.Let " satisfy (.) and assme f H 4+ () for some Then where C 1 ([0; 1)) is a cto fnction sch that " = w " + + r " ; (.6) (r) = 1 for 0 r 0 =3 0 for r 0 =3 (.7) with (m) (r) C (0 ; m), m = 0; 1;, w " (x; y) = X " i (i) f(x; y); (.8)

4 and in 0 ; X ix = " i C ki () k k=0 " k e?=" = X C i () i e?=" ; (.9) kr " k k; c" +3=?k ; 0 k + 3= (.10) with c R a constant independent of " and C ki () smooth and independent of ". 3. The Finite Element ethod The nite element approximation of (.) consists of nding S sch that B "? ; v = F (v) 8v S ; (3.1) where S H 1 0 (), is a nite dimensional sbspace of dimension. For every " (0; 1], there exists a niqe soltion S to (3.1) satisfying where the energy norm is dened as "? "; = inf vs k"? vk "; ; (3.) kk " = B "(; ) = " jj 1; + kk 0; Theorem gives a representation for the soltion " to (.1), so that the design of a nite element space to approximate " will consist of separate spaces for the approximation of w " ; ; r" given by (.8), (.9) and (.10) respectively. Ths, we will design specic nite element spaces S ; S w ; Sr for the approximation of ; w" ; r", so that by (.6) we will have an approximation for ". Then we have for any v 1 S ; v Sw ; v 3 Sr sch that v = (v 1 + v + v 3 ) S ; k "? vk "; kw "? v 1k "; +? v "; + kr "? v 3k "; ote that 0 is chosen so that f H 4+ (). If f p (), then we cold choose = p In particlar, if f is constant, we choose = 0; which gives w" = w" 0 = f(x; y) and which is precisely the fnction (1.). = 0 = C 0 () e?=" 4. The Bondary Layer Approximation First, we design the space S for the approximation of, with given by (.9), over R with analytic. To this end, let 0 be given by (.5) and set 1 = n 0 ; as seen in Figre. In practice, 0 cold be selected in the following way. Divide into sbintervals ( j ; j+1 ) ; j = 1; ; m? 1;, and draw the inward normal at j ; j = 1; ; m, of length 0 Then, connect each

5 point ( j ; j ) = ( 0 ; j ) sing the crve = 0 =constant. Frther, divide each j 0 = f(; ) 0 0; j j+1 g ; j = 1; ; m into j 0;1 ; j 0;, where j 0;1 = (; ) j j+1 ; 0 1 0p" ; j 0; = j 0 nj 0;1 In the above constrction, R is a xed constant, p is the degree of the approximating polynomial, and we assme that p" < 1=. This will dene a mesh n o m 0 = j 0;1 ; j 0; over 0. ext, let i 1 and dene the mesh = j=1 ; n i=1 be some sbdivision of 1; that is compatible with 0 ; n j 0;1 ; j 0; ; i 1 ; j = 1; m; i = 1; ; n o (4.3) Ω Ω 1 ρ 0 Ω 0 Fig.. The regions 0 and 1 ext, we consider a typical element j 0 0, and map it to the master element J = (?1; 1), as follows. Recall that [0; 0 ]; [0; L] for some 0 ; L R. Then, the linear mapping (in and ) = L? 1; = 0? 1; (4.4) will map j 0 ; from the? plane, onto J = (?1; 1) ; in the? plane. Since j 0 = j 0;1 [j 0;, we have J = J 1[J with J 1 = (?1; 1)(?1;?1 + p"), J = JnJ 1 Under the mapping (4.4), dened by (.9) becomes 0 e (; ) = e? (+1) " = e? 0 (+1) " X ix " i k=0 X ix " i k=0 C ik L ( + 1)? 0 ( + 1) k ec ki () k () " k ; " k

6 where k () is a polynomial of degree k in and ec ki () = ix Xi?j L i ikj` h ( + 1) (j) f `=0 with ikj` () smooth. In the case when f is constant, i.e. = 0, we have and in the general case of f H 4+ () ; we have e (; ) = e? 0 (+1) " I ; e 0 (; ) = e C0 () e? 0(+1) " ; (4.5) X ix k=0 ec ik () " i?k k () = e? 0 (+1) " C(; ) (4.6) The nite element space mst be chosen so that both the exponential term exp(? 0 (+ 1)=(")) and C(; ) are approximated at a sciently fast rate. For the approximation of e Cik () we will assme that ec ik () is analytic with all derivatives bonded independently of " (4.7) Then, by the reslts of 3, we have that there exist K ; K 3 R and 0 < < 1; sch that e Cik () K ; C p i (I) K 3 p ; (4.8) where I = (?1; 1) ; C p i (I) = d i d i ecki ()? er() 0;I ; i = 0; 1; and er() is the H 1 projection of Cki e () onto p (I) We begin the constrction of the space S by noting that since = 0 in n 0;? v "; =? v ";0 +? v ";n0 =? v ";0 ; (4.9) provided v = 0 in n 0. Ths, we dene the space S () = ( = (;?! p ) ; = S where m S j 0 = m j=1 j=1 v H 1 () vj j p( j 0;`); vj 0;` i = 0; 1 j 0;`; i 1 ; j = 1; ; m; i = 1; ; n; ` = 1; n j 0 ; i 1; j = 1; ; m; i = 1; ; n j 0;1 [ j 0; = 0 ; ns i=1 o ) ; (4.10) ;?! p = fp; p; ; pg ; i 1 = 1, as given by (4.3),(.5). In (4.10), we dene p as follows (; ) (; ) = r()q() with r; q polynomials p ( 0 ) = of degree p over 0 and p > We have the following reslt.

7 Lemma 4.1 Let be given by (4.6) satisfy (4.7) and let S () be the space dened by (4.10). Then for any s > 0; there exists v S sch that? v "; C" 1= p?s ; (4.11) where C = C(s) R is independent of p and "; bt depends on s. Proof. ote that by (4.9), we have for any v S k=1? v "; =? v ";0 =? v ";0; +? v ";0n 0; (4.1) Consider given by (.9), over an element `, ` 0 n 0; ; and note that nder the linear mapping (4.4), becomes e (; ) as given by (4.6); over the sqare J = (?1; 1). Let v S be sch that v (; ) = r()q (), with P q () = ip qik 1 () q ik () ; and q1 ik () ; q ik () are polynomials of degree p= in, p >. Under the mapping (4.4), v (; ) becomes ev (; ) and we have? v = e? ev ";J = C (; ) e? 0 (+1) "? er()eq() ";` = X ix 0(+1) e? " k=0 X ix k=0 e? 0 (+1) ec ik () " i?k k ()? er() X ";J ix k=1 eq 1 ik () eq ik () " Cik e () " i?k k ()? er()eq ik 1 () eq ik () ";J Set U ik () = " i?k k () e? 0 (+1) " and eq ik () = eq ik 1 ()eq ik (). Then, e? 0 (+1) " Cik e () " i?k k ()? er()eq ik 1 () eq ik () = ";J = Cik e () U ik ()? er()eq ik () ";J = Cik e () U ik ()? Cik e () eq ik () + Cik e () eq ik ()? er()eq ik () ";J Cik e () U ik ()? Cik e () eq ik () + ";J Cik e () eq ik ()? er()eq ik () ";J Cik e () ku ik ()? eq ik ()k ";J + keq ik ()k ";J Cik e ()? er() ";J ";J K Cik e () ku ik ()? eq ik ()k + keq ik ()k Cik e ()? er() where K R and I = (?1; 1). Since keq ik ()k ku ik ()? eq ik ()k + ku ik ()k ; ";J and Cik e ()? er() " Cik e ()? er() "C p 1 (I) + Cp 0 (I) 1;I + Cik e ()? er() 0;I

8 we frther obtain Cik e () U ik ()? er()eq ik () K Cik e () ";J By (4.8) we have ku ik ()? eq ik ()k + o + ku ik ()? eq ik ()k + ku ik ()k ("C p 1 (I) + Cp 0 (I)) K Cik e () + C p 0 (I) + "Cp 1 (I) ku ik ()? eq ik ()k + o +C p 0 (I) ku ik ()k + "C p 1 (I) ku ik ()k Cik e () U ik ()? er()eq ik () ";J (K + K 3 p + "K 3 p ) ku ik ()? eq ik ()k + +K 3 p ku ik ()k + "K 3 p ku ik ()k e K ku ik ()? eq ik ()k + e K3 p ku ik ()k ; where 0 < < 1 and K e ; K e R are independent of p and ". ext consider ku ik ()? eq ik ()k = " i?k k () e? 0 (+1) "? eq ik()eq 1 ik() = " i?k k () e? 0 (+1) " i?k k () e? 0 (+1) "? " i?k k () eq 1 ik() + " i?k k () eq 1 ik()? eq 1 ik()eq ik() "? eq 1 ik() + eq 1 ik() " i?k k ()? eq ik() Recall that by constrction, eq ik () is a polynomial of degree less than or eqal to p= with p >. Since " i?k k () is also a polynomial of degree k with k = 0; 1; ; i; i = 0; ;, we take eq ik () to be eqal to "i?k k () ; so that " i?k k ()? eq ik () = 0 and Combining we get ku ik ()? eq ik ()k " i?k k () e? 0 (+1) "? eq ik() 1? v ";` X ix ek3 p ku ik ()k + K e " i?k k () e? 0 (+1) k=0 Using Theorem 5.1 of 13, we have e? 0 (+1) "? eq ik() 1 K" 1= p+1= ; "? eq 1 ik() (4.13)

9 with < 1 and K R independent of p and ". (We actally get (p+1)= bt we can adjst the vale of to get p+1= as stated above). Also ku ik ()k = and (4.13) becomes? v ";` " i?k k () e? 0 (+1) " max f k ()g" i?k e? 0 (+1) " " 1= max I f k ()g ; I X ix k=0 K " 1= max I n ek3 p " 1= max I f k ()g e p ; f k ()g + K" 1= p+1= " i?k k () o (4.14) with e p = max p ; p+1= ; 0 < ; < 1 and K R independent of p and ". Then,? v ";n0; = X? v `n 0; K" 1= max I ";` f k ()g e p ; (4.15) with K R independent of p and ". ext, we consider? v ";0; Using the standard p version estimate (c.f. 4 ), we obtain? v 1;0; p?s+1 s;0; By assmption, (s) (r) C(0 ; s); C R independent of p and ". Ths, 1;0; C( 0 ; s)p?s+1 exp(? 0=")? v " s (4.16) C( 0 ; s)p?s+1 exp(? 0=") " s+t " t ; C( 0 ; s; t)" t p?s+1 ; for any t > 0. Combining (4.1), (4.14) and (4.16), we obtain? v "; =? v ";0; +? v ";0n 0; C( 0 ; s; t)" t p?s+1 + K" 1= max I C" 1= p?s ; with s > 0 arbitrary and C = C(s) R independent of p and ". f k ()g e p

10 Remark 1 ote that in the above proof, eq ik () is eqal to "i?k k () and eq ik 1 () matches exp (? 0 ( + 1) =") at =?1, so that eq ik (?1) = U ik (?1) Hence, on? v = e Cik () U ik (?1)? er()eq ik (?1) = e Cik ()? er(); where er() is the H 1 projection of e Cik (). Remark We point ot that the assmption (4.7) in Lemma 4.1 can be weakened to ec ik () C 1 () since we only need arbitrarily high, algebraic rates of convergence for the bondary layer fnctions in the proof of Lemma 4.1, rather than the exponential rate implied by the analyticity of e Cik (). 5. The Approximation of the Smooth Part and the Remainder In this section we discss the approximation of the smooth part w ", and the remainder r " ; of the weak soltion " to (.1). We will assme that the right hand P side f in (.1) is analytic over, so that w " (x; y) = " i (i) f(x; y) and r " will also be analytic over, with arbitrarily high. With f i g n i=1 some sbdivision of, we dene Sr;w as the standard nite element space of piecewise polynomials, S r;w () = v H 1 () vj i p ( i ) ; i ; (5.1) = (;?! p ) ; = f i g n i=1 ; p = fp; ; pg to be sed for the nite element approximation of w " and r" Let v 1; v be the H projections of w ", r" respectively onto S r;w \ C 1 () ; with S r;w given by (5.1) above. Then we have the standard p version estimate (c.f. 4 ), for ` = 0; 1; Also, kw " k s; = X kv 1? w " k`; Cp?s+` kw " k s; ; (5.) kv? r " k`; Cp?s+` kr " k s; (5.3) " i (i) f s; X " i (i) f s; C(); and by (.10), so that kr " k s; C"+3=?s ; kv 1? w " k "; kv 1? w " k 1; C()p?s+1 ; (5.4) kv? r " k "; kv? r " k 1; Cp?s+1 " +3=?s (5.5)

11 6. The Approximation of " We now address the approximation of the soltion " = w " + +r" ; given by (.6). Dene the space S e H 1 () as es = v = v 1 + v + v 3 v 1 S ; v ; v 3 S r;w ; (6.1) and note that v = v 1 + v + v 3 e S, satises (4.11), (5.) and (5.3). Also, kv? " k "; v 1? w " "; + v? "; + v 3? r " "; Using Theorem 4.1, (5.4) and (5.5) we obtain the following reslt. Lemma 6. Let " be the soltion to (.) as given by (.6) and let e S be the space dened by (6.1). Then for any es > 0; there exists v e S sch that v? " "; Cp?es ; (6.) where C = C(es) R is independent of p and ", bt depends on es. Proof. Theorem 4.1, (5.4) and (5.5) give v? " "; C()p?s+1 + C" 1= p?s + Cp?s+1 " +3=?s and (6.) follows with es = s? 1. ote that S e dened by (6.1) is not a sbspace of H0 1, ths the best approximation reslt (3.) has not been established yet for this space. We need to adjst the approximating polynomial v so that v j = 0 and (6.) still holds, i.e. sch that v S = S e \ H0 1 (). ote that for > 0 small, and C R independent of p and ", we have v 1 v 1? w "? w " 0; v 1 1; v 1? w " 1=+; C () p?s+1 ;? w " 3=+; C () p?s+ ; with similar estimates holding for v3? r". Also, by Remark 1 and (4.8), v? 0; = C p 0 (I) Cp ; v? 1; = C p 1 (I) Cp ; with 0 < < 1 and C R is independent of p and ". Combining, we get (since " j = 0) v = 0; v? " Cp?s+1 0; ; (6.3) v 1; = v? " 1; Cp?s+ (6.4) The following Lemma gives s the tool for adjsting the nonzero trace of v S on the bondary. (For a proof see Lemma 3.6. of 16 ).

12 Lemma 6.3 Let S = (h 1 ; h ) (0; k) be a rectanglar element. Let z(x) be a polynomial of degree p on = (h 1 ; h ) ; and dene V (x; y) = z(x)l(y) p (S), with L(y) linear, L(0) = 1; and L(k) = 0. Then, V satises V j = zj ; V = 0 on the side opposite and kv k ";S C "k 1= jzj 1; + k 1= kzk 0; + "k?1= kzk 0; ; (6.5) where C R is independent of p and ". Using Lemma 6.3, we obtain the following reslt. Lemma 6.4 Let v S e be as in Lemma 6.. Then, there exists V S e sch that V j = v j and V "; K" 1= p?es ; (6.6) where es > 0 is arbitrary and K R is independent of p and ". Proof. Dene V (; ) = L()z(), 0 1 p" 0; 0 0 with L() linear, L(0) = 1; L( 1 p" 0) = 0 and z() = v () = v j. Let ` be any element of the mesh ; with ` \ 6= ;. Then, sing the mapping (4.4), we map ` to the rectangle J = (?1; 1) (?1;?1 + p") and V (; ) to e V (; ) = e L()ez() = el()ev () Applying Lemma 6.3, we obtain e V ";J C apping back to the? plane, we get V ";` C Using (6.3) and (6.4), we have V ";` C n" (p")?1= kezk 1;J + (p") 1= kezk 0;J o e C n " 1= p?1= ev 1;J + (p") 1= ev 0;J o n " 1= p?1= v 1;` + (p") 1= v 0;` o n " 1= p?1= p?s+ + (p") 1= p?s+1o K" 1= p?s+3= Since, V "; = X ` V ";` X ` K" 1= p?s+3= ; (6.6) follows, with es = s? 3=. We now present or main reslt. Theorem 6.1 Let " be the soltion to (.) and the soltion to (3.1). Then for any s > 0, "? "; C(s)p?s where C(s) R is independent of p and ".

13 Proof. By Lemma 6. there exists ev e S sch that ev? " "; Cp?s ; with C R independent of p and " Let e V be constrcted as in Lemma 6.4, and dene v = ev? V e Then v S = S e \ H0 1 () and "? v "; "? ev "; + e V "; Cp?s + K" 1= p?s C(s)p?s Since by (3.) is the best approximation, the assertion follows. 7. merical Reslts In this section we present the reslts of nmerical comptations for the model bondary vale problem (.1), in the case where is the nit disk. That is, we consider the problem?" + = 1 in ; (7.1) = 0 on ; where = f(r; #) 0 r 1; 0 # g, in polar coordinates, and " (0; 1], as before. The exact soltion to (7.1) is a fnction (r; #), dened in polar coordinates by (r; #) = (r) = 1? I 0(r=") I 0 (1=") ; (7.) where I 0 (z) is the modied Bessel fnction of order zero. From (7.), we see that = s + where s = 1 and (r; #) = (r) =? I0(r=") I 0(1="). Even thogh (r) is not the typical bondary layer fnction, it behaves like exp (?r="), c.f. 15. We will consider two nite element schemes for this model problem, and we will compare the relative errors in the energy norm, E = EX? F E "; k EX ; k "; verss the nmber of degrees of freedom,. Figres 3, 4 show the mesh design for the two methods considered. The rst scheme consists of 8 elements, with the inner circle taken at 1= (see Figre 3). The second scheme consists of the same nmber of elements bt with the inner circle taken at (1? p") ; i.e. an O(p") layer of elements along the edge of the bondary (see Figre 4). We performed experiments for varios vales of ", sing the software package stresscheck T, for p = 1; ; 8. De to the fact that stresscheck T is a p version software package, we are not able to perform \tre" hp version experiments.

14 Fig. 3. The mesh for Scheme 1. Fig. 4. The mesh for Scheme.

15 Scheme is designed to represent an hp version, given the software at hand. oreover the maximm allowed polynomial degree is p = 8, ths the dierence between the two methods we are considering lies only in the position of the \inner-circle" element. That is, for the rst scheme the inner circle has radis 1= and for the second scheme the inner circle has radis 1? p max " = 1? 8". This type of experiment will illstrate the necessity of an O(p") layer of elements at the edge of the bondary, in order to niformly approximate bondary layers at a sciently fast rate. Figres 5 and 6 show the relative error in the energy norm E; verss the nmber of degrees of freedom, in a log-log scale for " = 001 and " = 0001, respectively. As can be clearly seen from this comparison, the rst scheme deteriorates as "! 0, while the second scheme demonstrates the expected robstness and spectral convergence rate. In particlar, this nmerical experiment sggests that the second scheme converges at a robst exponential rate. This was not established in or analysis de to technicalities in the proofs. The observations made here establish the speriority of this hp method for problems of this type epsilon= Relative Error in the Energy orm Scheme 1 Scheme Degrees of Freedom Fig. 5. Energy norm comparison of the two schemes. We have also considered the pointwise error in the derivatives of EX and F E at the point (1; 0) on the bondary of. Strictly speaking, the previos analysis does not cover the convergence of pointwise derivatives, bt we address this isse here comptationally. Figres 7 and 8 show the comparison of the two nite element

16 10 0 epsilon= Relative Error in the Energy orm Scheme 1 Scheme Degrees of Freedom Fig. 6. Energy norm comparison for the two schemes. Derivative Convergence at (1,0), epsilon= Pointwise Error Scheme Scheme Degrees of Freedom Fig. 7. Derivative comparison for the two schemes.

17 Derivative Convergence at (1,0), epsilon= Pointwise Error Scheme Scheme Degrees of Freedom Fig. 8. Derivative comparison for the two schemes. schemes for E 0 = 4 dex df E dr? dr dex dr verss the nmber of degrees of freedom,, in a log-log scale. The same behavior as in the energy norm error is observed, with the rst scheme deteriorating completely as "! 0. Acknowledgment The athor wold like to thank Dr. Christoph Schwab and Dr. anil Sri for their ideas and gidance throgh the completion of this work. Appendix Here we give a proof of Theorem which we restate for convenience Let " satisfy (.) and assme f H 4+ () for some Then 3 5 r=1 " = w " + + r " ; where C 1 ([0; 1)) is a cto fnction sch that 1 for 0 r 0 =3 (r) = 0 for r 0 =3 ;

18 with (m) (r) C (0 ; m), m = 0; 1;, w " (x; y) = X " i (i) f(x; y); and in 0 ; X ix = " i C ki () k k=0 " k e?=" = X C i () i e?=" ; kr " k k; c"+3=?k ; 0 k + 3= with c R a constant independent of " and C ki () smooth and independent of ". Proof. Dene w " (x; y) = X " i (i) f(x; y) Then, sing (.1) we see that R " = "? w " satises L " R " = f? L " w " = " + (+1) f = g(x; y) (A.1) For sciently large, w " will satisfy (.1) p to the correction factor g(x; y) However, w " will not satisfy, in general, the bondary conditions. To correct this we introdce appropriate bondary layer terms as follows. Dene formally to be a fnction satisfying L " = 0; in 0 ; (A.) j =? 1X " i h (i) f i ; (formally) where, for any fnction G dened on ; [G] denotes G restricted to. We solve (A.) via series expansion, bt since is dened only in 0, we consider (A.) in bondary tted coordinates. Let () be the crvatre of and set (; ) = 1 1? () (A.3) We have, see 1, an expression for the Laplacian in bondary tted coordinates as (; ) = So, is obtained by solving? ()(; ) + 0 () 3 (; ) + (; ) + (A.4)?" + = 0; in 0 ; (A.5)

19 where is given by (A.4) and 0 is given by (.5). Using a geometric series, expand (; ), given by (A.3) as (; ) = 1X [()] j = 1X [() " b] j ; where b = =" denotes a stretched variable. Then, (A.5) becomes with?" 8 < + 1X a j j 1 + aj + a j 3 a j 1 =? [()]j+1 ; a j = (j + 1) [()]j ; a j 3 = j(j + 1) Switching to the stretched variable notation, (A.7) becomes? 1X b? " ("b) j a j 1 "?1 b + aj + a j 3 (A.6) 9 = ; + = 0; in 0 ; [()] j?1 0 () (A.7) (A.8) + = 0; in b 0 ; (A.9) where 0 b denotes the image of 0 nder the stretched variable transformation. P ow, we formally write = 1 " i Ui b (b; ), with Ui b (b; ) to be determined, and we insert it into (A.9). Eqating like powers of "; we get where Xi?1 bf i (b; ) =? b Ui b + b Ui = b Fi (b; ); i = 0; 1; ; (A.10) (b) j a j Ui?j?1 b 1 + a j Ui?j? b b + a j 3! Ui?j? b (A.11) (We set b Uj 0 for j < 0). We shall consider soltions which satisfy lim bu i = 0; b!1 (A.1) to ensre that each b Ui decays exponentially with b and is therefore negligible otside b 0. As a bondary condition for (A.10) we will consider h bui i =? (i=) f if i is even, 0 if i is odd. (A.13) Let s nd b Ui (b; ) sing (A.10) and (A.13). First, consider the case of i = 0 We have to solve b? U0 b + U0 b = 0 in 0 b ; (A.14) h i bu0 = [?f] on

20 Assme U0 b = C 0 () G 0 (b). Then, (A.14) gives?c 0 () G 00 0 (b) + C 0 () G 0 (b) = 0, ths G 0 (b) = K e?b The bondary condition gives K = [?f] C 0(), so that bu 0 = [?f] e?b = [?f] e?=" (A.15) ote that U0 b (b; ) is a fnction of the type (1.), i.e. a prodct of a smooth fnction in ; and the exponential e?=". We will show that Ui b is of this type for all i = 0; 1;. In particlar, for i ; bu i (b; ) = e?b ix ix Xi?j k=0 `=0 ikj`()b k ` ` h i (j) f ; (A.16) where ikj`() are smooth fnctions of, depending only on. We will say that a fnction is of type (m; n) if it is a sm of terms of the form () b k ` ` h (j) f i e?b ; with k; j; ` ; satisfying k m; j + ` n; and () smooth depending only on We wish to show that b Ui (b; ) is of type (i; i) for any i. The proof is by indction on i. The reslt is known for b U0 from (A.15). We assme the reslt holds for 0; 1; ; ; i? 1 and establish it for b Ui ote that b Fi dened by (A.11) is of type (i? 1; i). To obtain b Ui (b; ), we solve the ordinary dierential eqation dened by (A.10) - (A.13), sing standard techniqes sch as the method of ndetermined coecients. Since the right hand side of (A.10), b Fi, is of type (i? 1; i) and the dierentiation is with respect to b, the soltion b Ui will be of the form i P n=0 a n b n, with a n () the ndetermined coecients. It is then easy to see that Ui b (b; ) is of type (i; i) as desired. This establishes (A.16). ext, dene By (A.16), we have = e?=" X Letting C ki () = i P i?j P `=0 " i ix X = " i Ui b (A.17) ix Xi?j k=0 `=0? ikj`() ` ` (j) f k ` h i ikj`() " (j) f ` ; we get X ix = " i C ki () k " k e?=" ; k=0 and this establishes (.9). We will now obtain bonds on To this end dene the region 0; = fz??! n z z ; 0 =3 < < 0 g ; (A.18)

21 and note that (r) dened by (.7), satises 1 on 0 n 0;, where 0 was dened by (.5).Also let so that U ijk` = ikj`() ` bu i (b; ) = ` ix k=0 `=0 h i (j) f e?b ; ix Xi?j b k U ijk` (A.19) Lemma 4.6 of 1 states that there exists a constant K depending on sch that for any s = 0; 1; ; sx K" s;0 1=?s " n ku ijk`k n+b; ; (A.0) and for any t > 0 b k a+b U ijk` b a b n=0 b k a+b U ijk` K t " s;0; 1=+t?s b a b sx n=0 " n ku ijk`k n+b; (A.1) Using (A.0) with a = b = 0 and U ijk` dened by (A.19), we obtain Ths, b Ui s;0 b k U ijk` s;0 sx K" 1=?s ix k=0 `=0 n=0 K" 1=?s sx n=0 ix Xi?j b k U ijk` s;0 K" 1=?s sx n=0 K" 1=?s sx K ix n=0 `=0 " n ikj`() ` i " n h (j) f ix Xi?j " n h i (j) f ix " n h i (j) f " h i 1=?s (j) f Using (A.1) with a = b = 0; we get for any t > 0 b k U ijk` s;0; sx K" 1=+t?s n=0 K" 1=+t?s sx n=0 ` h (j) f n+`; n+`; n+i?j; + " h 1= (j) f i?j; " n ikj`() ` i " n h (j) f ` i n; i (A.) s+1?j; h i (j) f n; n+`;

22 This gives b Ui s;0; K" 1=+t?s sx Hence n=0 K" 1=+t?s sx s;0; K K ix X n=0 ix Xi?j " n `=0 h i (j) f ix " n h i (j) f " h i 1=+t?s (j) f " h i i+t+1=?s (i) f n+`; n+i?j; + " 1=+t i?j; h (j) f i i + " 1=+t+ 0; h (i) f (A.3) s+i?j; s+; (A.4) for any t > 0. It remains to establish the desired bonds on the remainder, which is dened by ; r " = "? w "? (A.5) ote that r " satises the bondary vale problem? L " r " = g? L " in ; r " = 0 on (A.6) This gives?? kr " k "; = B " (r " ; r" ) = (g; r ) "? L " ; r " where (; ) denotes the sal L () inner prodct. Since = 1 on 0 n 0; 0 on n 0 we have kr " k "; kgk 0; kr" k "; + L" +? L" 0;0; kr" k "; 0;0n 0; kr " k "; + which gives kr " k "; kgk 0; + L" 0;0n 0; +? L" 0;0; kgk 0; + L" 0;0 + "? 0;0; + kgk 0; + L" 0;0 + C " ;0; + 0;0; 0;0; (A.7)

23 Let s estimate L" s;0 ; s = 0; 1; We have L " =?" + (A.8) Write X ( =?" " i Ui b b "?? "?1 () (b; ) Ui b b + (b; ) Ui b () 3 (b; ) b Ui () (; ) = (; ) = 3 (; ) = kx kx kx ) + X " i b Ui a j 1 j + k+1 () k+1 (; ); k+1 a j j + k+1 () (; ); k+1 a j 3 j + k+1 0 () 3 (; ) ; where a j i ; i = 1; ; 3 are given by (A.8), and for ` = 1; ; 3 8 k+1 < ` (; ) = Pt k =? i and sbstitte in (A.8) to get L " =?? + X X X " i b Ui b 0 " i?i X! 1R 0? k+1 k+1 `(s; ) (1?s)k k! ds for k 0 X `(; ) for k =?1 0 " i?i X 1 ("b) j a j Ui? b ("b) j a j 1 A? X " i b Ui? " +1 () (b; ) b U b?" +1 0 () 3 (b; ) b U?1?" + 0 () 3 (b; ) U b?? X +"b 0 () b Ui?1 b 0 " i?i X 1 A? ("b) j a j 3 b Ui? 1 A +? "+1 (b; ) b U?1?? " + (b; ) b U? " i ("b) (()?i+1?i+1 " () Ui?1 b b 3?i+1 b Ui? )?i+1 Ui? b + +

24 aking se of the identity X?i X F (i; j) = X ix F (i? j; j); we get L " = X a j 3 " i 4? b Ui b? b Ui?j? ix ) # + Ui b (b; ) b U?1 (b) (a j j Ui?j?1 b 1 + a j Ui?j? b b +? " +1 () (b; ) U b b () 3 (b; ) b U?1 b?" + (b; ) U + 0 () 3 (b; ) U b X? " +1 (b)?i+1 (()?i+1 " () b Ui?1 b + +??i+1 b Ui?!?! + "b 0 ()??i+1 3 b Ui?? ) ote that by (A.10), (A.11) the term in brackets vanishes. Ths, L " =?" +1 ( () (b; ) b U + 0 () 3 (b; ) + X +"b 0 () b U?1 b + (b; ) + " b U! + (b) "()?i+1?i+1 " () Ui?1 b b 3?i+1 b Ui? Using the bonds (A.), (A.3) on #) b Ui s;0 b U?1?i+1 are smooth, we obtain L" s;0 K(f; ; s)" +3=?s Then, by (A.7), (A.9) and (A.4) we have 8 t > 0 kr " k "; kgk 0; + L" 0;0 + C! b + " U +?i+1 Ui? b + + and the fact that a j i ; i = 1; ; 3 and (A.9) " ;0; + C 0;0;

25 K(; ; f) n" + + " +3= + " to K(; ; f)" +3= with K(; ; f) independent of ". This gives kr " k 0; K(; ; f)"+3= ; kr " k 1; K(; ; f)"+1= (A.30) (A.31) We wold like now to obtain bonds on r " in any Sobolev norm kk s ; s = ; 3; ; similar to (A.30), (A.31). From (A.6) we have?" r " + r " = G in ; (A.3) r " = 0 on ; where? G = g + "? Let bx = x="; by = y=" denote stretched variables, so that br " (A.3) becomes (A.33) = " r ". Then?br " + br " = b G in b ; (A.34) br " = 0 on b ; where b is the image of nder the stretched variables transformation. ote that (A.34) is an elliptic bondary vale problem, ths we have the \shift" estimate kbr " k k; b K G b k?; ; k = ; 3; b with K R We need to investigate the relationship between so that (A.35) can be sed in obtaining bonds for kr " k k; We have, Z Z bg (bx; by)dbx dby = 1 " which gives b G 1;b = b G 0;b = b b G 0; b = 1 " kgk 0; Similarly, G b bx ( = 1 " Z 0;b + G b by " G x 0;b = Z b + " G y G (x; y)dxdy; 8 <! G b bx (bx; by) + ) dxdy; (A.35) b G k; b and kgk k;,! G b 9 = by (bx; by) dbx dby ;

26 which gives G b 1; = jgj b 1; In general, we get for k = 0; 1; G b k; = " k?1 jgj b k; (A.36) ote that br " also satises (A.36). ow look at b G k?;b = k? X K G b n X k? = i;b " (i?1) jgj i; "? kgk 0; + "(k?3) kgk k?; o ; ths, b G k?;b K n"?1 kgk 0; + " (k?3) kgk k?; o (A.37) Using (A.37) together with the shift estimate (A.35), we obtain o kbr " k k; b K n"?1 kgk 0; + " (k?3) kgk k?; Since br " satises (A.36), we frther have o kr " k k; "(1?k) K n"?1 kgk 0; + " (k?3) kgk k?; which yields kr " k k; K n "?k kgk 0; + "? kgk k?; o (A.38) Recall that by assmption, (s) (r) C(0 ; s); with C R independent of p and ". Then, by (A.33) and the above observation, we have kgk s;? g? L" s; kgk s; +? L" s; kgk s; +? L" s;0; + L" s;0n 0; kgk s; + "? s;0; + C(s; ) s;0; + L" s;0 kgk s; + " C(s; ) s+;0; + C(s; ) s;0; + L" Using (A.1) and (A.4) we frther obtain 8 t > 0 kgk s; K(; ; f; s)( n K(; ; f; s)" +3=?s s;0 " + + " t?s + " t?s + " +3=?so ; (A.39) with K R, independent of ". Using (A.39) and (A.38) we have kr " k k; K (; ; f; s) n"?k " +3= + "? " +3=?k+o (A.40) K (; ; f; s) " +3=?k

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