Abrikosov Lattices in Finite Domains

Transcription

1 Commun Math Phys (005) Dgtal Object Identfer (DOI) /s x Communcatons n Mathematcal Physcs Abrkosov Lattces n Fnte Domans Y Almog Faculty of Mathematcs, Technon - Israel Insttute of Technology, Hafa 3000, Israel eceved: 15 March 005 / Accepted: 1 May 005 Publshed onlne: 9 November 005 Sprnger-Verlag 005 Abstract: In 1957 Abrkosov publshed hs work on perodc solutons to the lnearzed Gnzburg-Landau equatons Abrkosov s analyss assumes perodc boundary condtons, whch are very dfferent from the natural boundary condtons the mnmzer of the Gnzburg-Landau energy functonal should satsfy In the present work we prove that the global mnmzer of the fully non-lnear functonal can be approxmated, n every rectangular subset of the doman, by one of the perodc soluton to the lnearzed Gnzburg-Landau equatons n the plane Furthermore, we prove that the energy of ths soluton s close to the mnmum of the energy over all Abrkosov s solutons n that rectangle 1 Introducton Consder a planar superconductng body whch s placed at a suffcently low temperature (below the crtcal one) under the acton of an appled magnetc feld Its energy s gven by the Gnzburg-Landau energy functonal whch can be represented n the followng dmensonless form [11] E = ( h h ex + ) + A dx 1 dx, (11) n whch H 1 (, C) s the superconductng order parameter, such that vares from =0 (when the materal s at a normal state) to =1 (for the purely superconductng state) The magnetc vector potental s denoted by A H 1 (, ) (the magnetc feld s, then, gven by h = A), h ex s the constant appled magnetc feld, and s the Gnzburg-Landau parameter whch s a materal property The Present address: Department of Mathematcs, Lousana State Unversty, Baton olge, LA 70803, USA

2 Y Almog superconductor les n, whch s a smooth connected doman Its Gbbs free energy s gven by E Note that E s nvarant to the gauge transformaton e ζ ; A A + ζ, (1) where ζ s any smooth functon Thus, we confne ourselves n the sequel to compettors from the space } H = (ψ, A) H 1 (, C) H 1 (, ) A satsfes (13b), (13a) (A Â) = 0 (A Â) ˆn = 0, (13b) Â = h ex x 1 î where î s a unt vector n the x drecton For suffcently large magnetc felds t s well known, both from expermental observatons [1] and from theoretcal predctons [15], that superconductvty s destroyed and the materal must be n the normal state If the appled magnetc feld s then decreased there s a crtcal feld where the materal enters the superconductng phase once agan Ths feld s called the onset feld and s denoted by H C3 At the bfurcaton from the normal state, superconductvty remans concentrated near the boundary, whch s why ths phenomenon has been termed surface superconductvty [4, 8, 9, 19, 1, 17] In the absence of boundares the crtcal feld at whch superconductvty nucleates s denoted by H C and s sgnfcantly weaker than H C3 (H C3 17 whereas H C = ) Furthermore, the bfurcatng modes are perodc lattces, named after Abrkosov [, 10, 4] that have been expermentally observed [14] It has been conjectured, therefore, by ubnsten [3] that superconductvty remans concentrated near the boundary for H C < h ex < H C3 When h ex H C (ether for large or for large domans) a bfurcaton of Abrkosov s lattces far away from the wall was conjectured [3] ecently, t has been proved both n the large lmt [, 7], and n the large doman lmt [5] that as long as H C <h ex <H C3 superconductvty remans concentrated near the boundares However, the second part of the conjecture n [3] s stll open In [5] t s shown for the global mnmzer of (11) (ψ,a ), that ψ dmnshes as h ex H C away from the boundares However, the exact structure of ψ s that lmt has never been found In [6] the bfurcaton of perodc solutons from the one-dmensonal surface superconductvty soluton ntroduced n [] was studed Nevertheless, the analyss n [6] was performed n a half-plane and under the assumpton that the solutons are perodc n the drecton parallel to the boundary In the present contrbuton we focus on the emergence of Abrkosov s lattces n fnte domans n We prove that when h ex s slghtly smaller than the global mnmzer of (11) n H can be approxmated n an approprately chosen rectangle n by one of Abrkosov s solutons Furthermore, we prove that the energy of ths soluton s close to the mnmum of the energy over all Abrkosov s solutons n that rectangle Thus, we prove the followng result Theorem 11 Let 1/5 ɛ = [1 h ex /] 1/ 1 log Let further (ψ,a ) denote the global mnmzer of (11) n H Then, denote by a rectangle n whose sde lengths are gven by

3 Abrkosov Lattces n Fnte Domans L 1 = ωn hex ; L = π ω h ex, where N(,ɛ) N and ω(,ɛ) are such that 1 ɛ 5 (L 1L ) 1/ ; r 1 < L L 1 <r, where r 1 and r are ndependent of and ɛ Fnally, let P = u H 1 loc U = u P u(x 1 + L 1,x ) = e h exl 1 x u(x 1,x ) ae u(x 1,x + L ) = u(x 1,x ) ae }, ( + u ) + φ = h } ex u φ φ Cc (), where  s gven n (13) Then, there exsts u 0 (, ɛ) U such that ψ u 0 Cɛ ψ, where (14a) J (u 0 ) nf v U J (v)[1 δ] < 0 v U, (14b) δ ɛ α/4 α<1 (14c) and Furthermore, J (u) = u 4 ɛ u (14d) A  H 1 () Cɛ 3 ; A  H () Cɛ (14e) Ths result proves that we can approxmate ψ n every rectangular subset of, and possbly even n dmnshngly small rectangles (as ), by some functon u 0 n U, whch s the space of Abrkosov s perodc solutons n Furthermore, the theorem shows that u 0 can be found by studyng the mnmzaton problem of J n U, whch s a fnte dmensonal subspace The rest of ths contrbuton s arranged as follows: n the next secton we revew some of the results obtaned for the lnear perodc problem, analyzed frst by Abrkosov [] In 3weobtan some a-pror estmates that are vald for any crtcal pont of (11) In 4 we obtan upper and lower bounds for (11) n whch enable the proof of Theorem 11 Fnally, n 5 we brefly summarze the man results of ths work and emphasze some addtonal key ponts

4 Y Almog The Perodc Problem Consder the problem where + u = h ex u n, (1) Let ω, and let L 1 = ωn hex  = h ex x 1 î () ; L = π ω h ex, where N N The perodc boundary condtons u should satsfy are gven by v(x 1 + L 1,x ) = e h exl 1 x v(x 1,x ) v(x 1,x + L ) = v(x 1,x ) We can now apply the transformaton x ( hex ) x, (3) to obtan ( +x 1 î ) v = v, v(x 1 + L 1,x ) = e ωnx v(x 1,x ), v(x 1,x + L ) = v(x 1,x ), (4a) (4b) (4c) where n the new coordnates L 1 = ωn ; L = π ω It can be easly verfed that the phase change around, where = [0,L 1 ] [0,L ] s πn Thus, the number of vortces n (ncludng multplctes) s N [13] The general soluton of (4a) and (4c) s gven n the form where g n satsfes v = n= g n (x 1 )e ωnx, g n [(x 1 nω) 1]g n = 0 The general soluton of the above ordnary dfferental equaton s g n = C n e 1 (x 1 nω) + D n G(x nω), (5) where G can be expressed n terms of Parabolc Cylnder functons [1]

5 Abrkosov Lattces n Fnte Domans From (4b) we conclude that for all x 1 we must have g n (x 1 + ωn) = g n N (x 1 ), or equvalently that C n = C n N ; D n = D n N By the lemma of emann-lebesgue we must have g n (x 1 ) 0 x 1, n and snce G(x 1 ) s unbounded n, we must have D n = 0 for all n Z Consequently, the general soluton of (4) s v = C n e 1 (x 1 nω) e ωnx, n= where C n+n = C n for all n Thus, we can wrte that where Note that and hence f n = r= v = N 1 n=0 C n f n, e (n+rn)ωx e 1 [x 1 (n+rn)ω] f n+1 (x 1,x ) = e ωx f n (x 1 ω), (6a) (6b) f n π 3/ L = 0 n N 1 (7) () ω We now defne the spaces P = u Hloc 1 u(x 1 + L 1,x ) = e ωnx } u(x 1,x ) ae u(x 1,x + L ) = u(x, (8a) 1,x ) ae ) } U = u P ( +x 1 î u +x 1 î φ = u φ φ Cc (), (8b) and state the followng result: Lemma 1 Let w P Then, w = w 0 + w, (9) where w 0 U 0 and w U0 (the orthogonal complement wth respect to the L () nner product) Furthermore, ( ) +x 1 î w 3 w (10) Proof Snce the restrcton of P to H 1 () s a closed subspace of H 1 (), and snce U 0 s a closed subspace of P (9) follows mmedately

6 Y Almog To prove (10) we recall frst that w must satsfy (4c), and hence t can be represented by the Fourer seres where by (4b) we have Consequently, w = L w = n= w n (x 1 )e ωnx, (11) w n (x 1 + Nω) = w n N (x 1 ) (1) n= L 1 0 w n dx 1 = L N 1 n=0 from whch we also obtan that w n L () for all 0 n N 1 Snce w U0 we must have w, f k = 0 0 k N 1, w n dx 1, (13) where f k s gven by (6b) Substtutng (6b) nto the above and makng use of (11) and (1) yelds w n exp 1 } (x 1 nω) dx 1 = 0 (14) We now make use of (11) to obtan ( ) +x 1 î w = L u e x / = L L 1 n= 0 N 1 n=0 w n (x 1 nω) w n dx 1 w n (x 1 nω) w n dx 1, (15) from whch we obtan that w n Hmag 1 (), where } Hmag w 1 () = w + w xw dx < It s well-known [16] that nf u xu dx = 3 u Hmag 1 () u dx Consequently, by (15) and (13) ( ) +x 1 î w 3L N 1 n=0 w n dx 1 = 3 w

7 Abrkosov Lattces n Fnte Domans 3 A pror Estmates In ths secton we obtan some a pror estmates whch should be satsfed by any soluton of the Euler-Lagrange equatons and the natural boundary condtons assocated wth (11) Thus, (ψ, A) must satsfy the equatons ( +A ψ = ψ 1 ψ ), (31a) A = ( ψ ψ ψ ψ ) + ψ A, (31b) together wth the boundary condtons +A ψ ˆn = 0 ; h = h ex (31c,d) Snce (31) are nvarant to the gauge transformaton (1) we fx the gauge (13b) for A The frst a pror estmates nclude the followng well-known results: Lemma 31 Let h ex o() Then, any soluton of (31) must satsfy Here and n the sequel, C s ndependent of and ɛ Proof See [18, 5] Let now ɛ = (1 h ex /) 1/ be postve and satsfy ρ L () < 1, (3a) h h ex C 1 ( ) C, (3b) +A ψ C (3c) L ( ) 1/5 ɛ 1 log, (33) as In ths case, accordng to the followng result, every soluton must be close to the normal state (ψ, A) = (0, Â): Lemma 3 Let ψ =ρ Any soluton of (31) must satsfy ρ 4 Cɛ 4, (34a) h h ex Cɛ 6 (34b) Furthermore, let  : be gven by (), Then, A  L p () C(p)ɛ 3, A  H () Cɛ, (34c) (34d) for any p 1

8 Y Almog Proof Let In [5] t was shown that f ρ f = J ˆ + where We note that Jˆ s contnous at ponts where ρ = 0 [5] Integratng (36) over yelds f ρ Snce by (3), we obtan f ρ f = h + 1 ρ (35) ( 1 ) ρ 4, (36) J ˆ ρ = f (37) + ρ ρ (h h ex ) ɛ ρ 1 f n 1 f n + 1 ρ (h h ex ) C, + ρ 4 C + ɛ ρ C [ ] 1/ + Cɛ ρ 4 (38) Usng (33), (34a) easly follows Furthermore, [ [ f ρ ]1/ Hence, Poncaré nequalty [0] then yelds, f ρ ] 1/ Cɛ 3 f + f (h ex ) Cɛ 3 f (h ex ) W 1,1 () Cɛ 3 Consequently, by the Gaglardo-Nrenberg nequalty we obtan that f (h ex ) Cɛ 6, from whch (34b) readly follows To prove (34c) we utlze the well-known nequalty A Â H 1 () (A Â) = h h ex (39)

9 Abrkosov Lattces n Fnte Domans and Sobolev embeddng Fnally, n order to prove (34d) we observe that by (31b), h ρ +A ψ Multplyng (31a) by ρ ψ and ntegratng by parts we obtan ρ +A ψ ρ 4 Hence, by (34a) Combnng the above wth (39) we obtan h Cɛ 4 A Â H () Cɛ We next proceed to obtan some L estmates Lemma 33 Let δ denote the doman δ =x d(x, ) δ}, and let α denote, here and n the sequel, any real number smaller than 1 Then, ψ L ( 1/ɛ ) C α ɛ α/, (310a) + A ψ C α ɛ α/, (310b) L ( 1/ɛ ) h h ex L ( 1/ɛ ) C α ɛ α (310c) Proof Let χ r denote a smooth cutoff functon satsfyng 0 x >r χ r (x) = 1 x < r χ r C χ r C r r (311) Let x 0 /(ɛ) We multply (36) by χ 1/(ɛ) (x x 0 ) and ntegrate by parts to obtan χ f χρ (h ) = χ f B 1 B 1 B 1 ρ + χρ 4, (31) B 1 ɛ ɛ ɛ ɛ where B r = B(x 0,r) For the frst ntegral on the rght-hand sde of (31) we have 1/ 1/ χ f B 1 χ ρ f B 1 B 1 ρ ɛ ɛ ɛ 1/4 1/ Cɛ 1/ 1/ f ρ ρ 4 B 1 ɛ B 1 ɛ

10 Y Almog For the second ntegral we have, utlzng (3b), χρ (h ) B 1 ρ h h ex + ρ h ex B 1 B 1 ɛ ɛ ɛ 1/ Cɛ + C 1 ɛ Let X n = B n 1 ɛ ρ 4 1/ ρ 4 B 1 ɛ ; Y n = B n 1 ɛ f ρ ρ 4 B 1 ɛ Then, by (31) we have [ ɛ X0 + Y 0 1/ ] C X1/ 1/ 1 Y 1 + ɛ X 1 (313) Snce, f ρ h ρ + ρ = +A ψ, we have by (3), X0 + Y 0 C ɛ Snce x 0 was arbtrarly chosen, we can cover B 1/ɛ by a fnte number of dscs of radus 1/(ɛ) Consequently, Substtutng n (313) yelds X 1 + Y 1 C ɛ X0 + Y 0 C ɛ 1/4 The above procedure can recursvely be appled to obtan X0 + Y 0 C nɛ α n, where α n can be determned by the recurrence relaton α n = α n 1 ; α 0 = 1 Clearly, α n 1, and hence, we can conclude that B 1 ɛ ρ 4 C α 1/ 1/ ɛ α α<1 (314)

11 Abrkosov Lattces n Fnte Domans We now apply the transformaton x (x x 0 ) to (31a) n B(x 0, 1/) to obtan ( +Ã ψ = ψ 1 ψ ) n B(0, 1), where ψ(x) = ψ((x x 0 )), Ã = A(k(x x 0 )) Standard ellptc estmates [3], together wth (1), then show that ψ H mag [B(0,1)] C ψ L [B(0,1)] = C ψ L [B(x 0,1/)] C 1/ ψ L 4 [B(x 0,1/)], wheren Snce u H mag (U) = U u + ( +Ã)u + ( +Ã) u ψ 4 L 4 [B(x 0,1/)] ρ 4 ɛ α C α B 1 α<1, (315) ɛ (whch s a rather crude estmate, but appears to be dffcult to mprove) (310a) follows mmedately from Sobolev embeddng To prove (310b) we use bootstrappng and Sobolev embeddng Fnally, to prove (310c) we notce that by (31b), h ρ + A) ψ Consequently, by (310a), (310b), (3b), and (31d) we have h h ex C α ɛ α + C ɛ We note, once agan, that (310) s not the optmal estmate Ideally, one should obtan ψ L ( 1/ɛ ) + + A ψ Cɛ, L ( 1/ɛ ) however, n vew of the crudeness of (315), (310) s the best we can obtan here Let (ɛ, ) denote a rectangle whose sdes length are gven by L 1 = ωn hex where ω, and N N are chosen such that r 1 L 1 L r ; ; L = π ω h ex, (316) 1 ɛ 5 (L 1L ) 1/,

12 Y Almog where r 1 and r are constant as and ɛ 0 accordng to (33) Denote by x 1 and x the coordnates n the respectve drectons of L 1 and L Let denote the mage of under the transformaton (3) Let P denote the restrcton of (8a) to H 1 () and P denote ts mage under the nverse of (3), e, the restrcton to H 1 () of u Hloc 1 u(x 1 + L 1,x ) = exp ωn } } h ex x u(x1,x ) ae u(x 1,x + L ) = u(x 1,x ) ae Let η denote a smooth cutoff functon satsfyng 1 x : d(x, ) 1 η = ɛ 0 x η Cɛ η C ɛ (317) \ Let (ψ,a ) denote the global mnmzer (whch depends on ɛ as well) of (11) n H (To keep the notaton consstent wth the one n the next secton, we state the rest of the results n ths secton for the global mnmzer, although they could have been stated for any soluton of (31)) Clearly, ηψ P Let U denote the restrcton of (8b) to H 1 () Let U denote ts mage under the nverse of (3), e, u P +Â u +Â φ = h ex u φ φ C c () } From the results of tfollows that U s a fnte dmensonal subspace of P Furthermore, we can now wrte ηψ = u 0 +ũ, (318) where u 0 U and ũ U The next lemma estmates the L () norm of ũ Lemma 34 Let ũ be defned by (318) Then, ũ Cɛ ψ + C α (L 1 L ) α ɛ 5 (319) Proof We frst multply (31a) by η ψ and ntegrate over to obtan +A (ηψ ) = 1 η ψ + η ψ (1 ψ ) (30) We now wrte +A (ηψ ) = +Â (ηψ ) [ ] + η (A Â) ψ ψ ψ ψ + ψ A η ψ A Â (31) For the second ntegral on the rght-hand-sde of (31) we have [ ] I = η (A Â) ψ ψ ψ ψ + ψ A [ [ η ψ A Â ]1/ +A (ηψ ) ] 1/

13 Abrkosov Lattces n Fnte Domans By (30) and Hölder nequalty we thus have, Snce by (34c), we obtan I C A Â L 4 () ψ L 4 () ψ L () C p (L 1 L ) 1 1 4p A Â L 4p () ψ L 4 () p>1 A Â L p () A Â L p () Cɛ 3, I C α (L 1 L ) α ɛ 5 α<1 (3) For the last ntegral on the rght-hand-sde of (31) we have η ψ A Â A Â L 4 () ψ L 4 () C α(l 1 L ) α ɛ 8 (33) Consequently, by (31), (3), and (33), +A (ηψ ) = +Â (ηψ ) + O((L 1 L ) α ɛ 5 ) (34) Combnng the above wth (30) and (318) we obtan +Â ũ ũ + η ψ 4 = 1 η ψ + ɛ u 0 + O((L 1 L ) α ɛ 5 ) (35) By (10), however, +Â ũ 3 h ex ũ Consequently, ũ + η ψ 4 1 η ψ + ɛ u 0 + C α (L 1 L ) α ɛ 5 (36) In vew of (317) we obtan 1 η ψ Cɛ ψ, and snce u 0 L () ψ L (), we can combne the above wth (35) to obtan (319) We note that the error term n (319) s ndeed small compared wth the frst term on the rght-hand-sde of (319) We shall demonstrate ths pont later when we obtan a lower bound for ψ L () at the end of 4 Lemma (34) bascally shows that ψ s ndeed close, n the L () sense to u 0 However, n order to obtan a mnmzaton problem for u 0 we need an estmate of (11) To ths end, we need the followng L estmate

14 Y Almog Lemma 35 Let ũ be defned n (318) Then, ũ L () C α ɛ α, (37a) +Â ũ C α ɛ α (37b) L () Proof Snce the proof s rather lengthy we dvde t nto several steps Step 1 Let χ r be gven by (311), and let x 0 1/(ɛ) (see Lemma 33 for the defnton of δ ) Clearly, χ 1/ɛ (x x 0 )ψ P Denote by u (0) 0 and ũ (0) the respectve projecton of χ 1/ɛ (x x 0 )ψ on U and U Step11 Prove that ũ (0) ɛ α C α (38) To prove (38) we repeat the same steps leadng to (36) to obtan ũ (0) + χ ψ 4 1 χ ψ + ɛ u (0) 0 + C α α ɛ5 ɛα (where χ stands for χ 1/ɛ (x x 0 )) Consequently, n vew of (311) and (314) we have for 5/6 <α<1, Step 1 Prove that ũ (0) Cɛ ψ + C α ɛ α B x 0, ɛ 1 α ɛ α ɛ5 C α ũ (0) L () C α ɛ α, (39a) L + Â ũ (0) C α ɛ α (39b) () We use standard ellptc estmates to prove (39) Snce ũ (0) = χψ u (0) 0 we have n vew of (31a), +Â ũ (0) h ex ũ(0) =ɛ χψ χ ψ ψ 1 ψ χ A Â χψ +(A Â) +A (χψ ) + χ +A ψ (330) Furthermore snce, u (0) P, we can extend t perodcally to, e, exp ω N } hex x u (0) (x 1 + L 1,x ) = u (0) (x 1,x + L ) = u (0) (x 1,x ) (331) The perodc extenson of u (0) satsfes (330) for every x f the rght-hand sde of t s extended n exactly the same manner

15 Abrkosov Lattces n Fnte Domans Applyng (3) to (330) we obtan +x 1 î ũ(0) ũ (0) = 1 1 ɛ [ɛ χ ψ χ ψ ψ ] ψ χ (1 ɛ ) 1/ Ã x 1 î χ ψ +((1 ɛ ) 1/ Ã x 1 î ) ) ( +(1 ɛ ) 1/ Ã (χ ψ ) ( ) + χ +(1 ɛ ) 1/ Ã ψ, (33) where ( ψ, Ã ) denotes (ψ,a ) n the stretched coordnates (3) To apply standard ellptc estmates we need an L estmate of the rght-hand-sde of (33) n B(x,1) for every x By (310a) and (314) we have that χ ψ 6 C α ɛ 5α α<1, B(x,1) and by (311) and (310a), ψ χ Cɛ 4 B(x,1) B(x,1) ψ C α ɛ 5α In vew of (34d) and (310) we also have (1 ɛ ) 1/ A x 1 î ( ) +(1 ɛ ) 1/ A ψ Cɛ 5α B(x,1) Furthermore, Fnally, B(x,1) B(x,1) χ ( +(1 ɛ ) 1/ A ) ψ Cɛ 3α (1 ɛ ) 1/ A x 1 î χ ψ C α ɛ 5α Combnng the above and (38), we may rely on the framework n [3] to obtan ũ (0) H (B(x,1) C α ɛ α α<1 Sobolev embeddng then yelds (39a) Bootstrappng and Sobolev embeddng prove (39b) Step Prove (37) We frst note that (37) and (39) are dfferent: whle ũ s the projecton of ηψ on U, ũ(0) s the projecton of χψ on the same space To obtan (37) we thus need to relate χ and η Let then x } M =1 denote a set of ponts n satsfyng 1 1/(ɛ) M B ( x, =1 B ( x, 1 ɛ) 1 ( 4ɛ) B 1 xj, 4ɛ) = f j

16 Y Almog Let χ } M =1 denote a set of C functons satsfyng 1 supp χ B ( x, ɛ) 1 M =1 χ = 1, x 1/(ɛ) 3 χ ɛ C, χ C ɛ Let η = M χ (333) =1 Clearly η satsfes (317) and hence, we may use t n (318) to defne u 0 and ũ Furthermore, let ũ () denote the projecton of χ ψ on U Then, Furthermore, ũ = ũ M ũ () =1 M ũ (), (334a) =1 + Â M ũ + Â ũ () =1 (334b) Snce M s a large number, we seek an estmate for ũ () when x x > /(ɛ) Step 1 Prove that ũ () (x) C α ɛ α/ N exp 1 [ 4 h ex dp (x, x ) ]} ɛ, (335a) + Â ũ () (x) CN exp 1 [ 4 dp (x, x ) ]} ɛ, (335b) where d p (x, x ) = mn x x (kl 1,jL ) j,k= 1,0,1 (335c) Snce u () 0 = ũ () for every x \ B(x, 1/(ɛ)) we prove (335) for u () 0 ecall from (6) that where f n s gven by f n = N 1 u () 0 = C n () f n(x), n=0 e (n+rn)ω h ex x e 1 [ h ex x 1 (n+rn)ω] (336) r=

17 Abrkosov Lattces n Fnte Domans Snce u () 0 s the projecton of χ ψ on U we have C n () = h exω π 3/ χ ψf n Let ˆψ k hex ω L = e ωk h ex x χ ψdx π 0 Let x = (x 1,x ) Then, snce ˆψ r (x 1) s supported n (x 1 1/(ɛ), x 1 + 1/(ɛ)) we have C () n = hex π r= By (310) and (311) we have that x1 + 1 ɛ x 1 1 ɛ ˆψ n+rn C α ɛ α/ 1 ɛl (recall that L 1/(ɛ 5 )) Thus ˆψ n+rn e 1 [ h ex x 1 (n+rn)ω] dx 1 α<1, Let C () n C αɛ α/ h ex r= 1 ɛ 1 ɛ e 1 [ h ex (x 1 x 1 ) (n+rn)ω] dx 1 Then, C () n d p1 (x 1,y)= mn x 1 y kl 1 k= 1,0,1 C αɛ α/ exp 1 ( h ex dp1 (x 1,nω/ h ex ) 1 )} ɛ (337) By (336) we have the estmate Consequently, Snce f n (x) C exp u () 0 (x) C αɛ α/ +d p1 1 ( h exdp1 x 1,nω/ h ex ) } (338) N 1 n=0 1 [ h ex dp1 ( x1,nω/ h ex ) ( x1,nω/ h ex ) 1 ]} ɛ [ dp1 (x 1,x 1 ) dp1( x1,nω/ h ex ) + dp1( x1,nω/ h ex )],

18 Y Almog we have that u () 0 (x) C αɛ α/ N 1 [ ]} 4 h ex dp1 x1,x 1 ɛ (339) To prove that u () 0 decays n the x drecton we use the fact that w () 0 = e h exx 1 x u () 0 to satsfy the problem () h ex x î 1 w w () 0 (x 1 + L 1,x ) = w () 0 (x 1,x ) 0 = h ex w() 0 w () 0 (x 1,x + L ) = w () 0 (x 1,x )e h exl x 1 Consequently, w () 0 must decay n the x drecton accordng to (339) We thus obtan u () 0 (x) C αɛ α/ N 1 [ ]} 4 h ex dp1 x,x ɛ (340) Combnng (340) and (339) yelds (335a), from whch one can easly prove (335b) usng standard ellptc estmates Step Prove (37) Substtutng (39) and (335) nto (334) we obtan [ ] ũ(x) C α ɛ α + Ne 1 ɛ Snce L 1 L we have N Cɛ Consequently, n vew of (33), (37) s proved 4 Upper and Lower Bounds All the results of the prevous secton could have been formulated for any soluton of the Euler-Lagrange equatons In ths secton we concentrate, however, on the energy functonal ( E (ψ, A) = ψ + ψ 4 + h h ex + ) ψ + Aψ (41) We obtan estmates for t n terms of the reduced functonal J whch appears n Theorem 11 We start by provng the followng upper bound

19 Abrkosov Lattces n Fnte Domans Lemma 41 Let η be defned by (333), and let u 0 and ũ be defned by (318) Then, for all v U E (ηψ,a ) J (v) + C α ɛ α 1 (L 1L ) 1/, (4a) where J (v) = ɛ v + 1 v 4 (4b) Furthermore, [ ũ 4 C α ɛ α u C α ɛ α (L 1L ) 1/ ] + (L 1 L ) α ɛ 7α (43) Proof Followng [5], let η denote a smooth cutoff functon satsfyng 1 x η = 0 x A η Cɛ η C ɛ, (44) 1 x \ d(x, ) ɛ 3 where A k = x \ k 1 d(x, ) k } ɛ ɛ Let further ηv ψ p d(x, ) 3 = ɛ or x ηψ d(x, ) ɛ 3, and x \ (45a) where v p denotes a perodc extenson, accordng to (331), of some v U, and à satsfes  x A1 à = A d(x, ) ɛ (45b) and x \ and à h ex L () Cɛ, (45c) (cf [5]) In vew of the above, snce (ψ,a ) s the mnmzer of E n H we have Consequently, E( ψ,ã) E(ψ,A ) 0 E( ψ,ã) E(ψ,A ) = E (v, Â) E (ψ,a ) + A +à ψ +A ψ + à h ex A h ex A + 1 [ ψ 4 ψ 4] [ ψ ψ ], (46) A A

20 Y Almog where A = A 1 A A 3 By (45) and (310) we have à h ex C α ɛ 4α (L 1L ) 1/ (47) A ɛ Moreover, by (44) and (310) we have A 3 +à ψ +A ψ C α ɛ α 1 (L 1L ) 1/ (48) To estmate A 1 +à ψ +A ψ, we choose frst v U such that whch yelds In ths case Let A 1 v 4 ɛ 4 L 1 L, (49) v ɛ L 1 L + ηv p = A 1 + ηv p + v p = η v p + 1 A 1 η v p A 1 (1 + Cɛ ) v p A 1 v p l = v p (x 1 + l 1,x + l ), where 0 l 1 L 1 and 0 l L Clearly, v p l dl 1 dl = A v A Consequently, nf v p (l 1,l ) l A v Cɛ L 1L A

21 Abrkosov Lattces n Fnte Domans Denote by (l 1m,l m ) the values of l 1 and l whch mnmze the L (A) norm of v p l let and v m = v p (x 1 + l 1m,x + l m ) If we choose v p = v m n (45a) we obtan A 1 +Â ηv m Cɛ L 1L Combnng the above wth (48) we obtan A +Ã ψ C α ɛ α 1 (L 1L ) 1/ (410) Combnng (46), (47), (49), and (410) we obtan Snce E (ψ,a ) E (v m, Â) + C α ɛ α 1 (L 1L ) 1/ E (v m, Â) = J (v m ) = J (v), (4) s proved as long as (49) holds To prove (4) for any v U, we wrte J (γ v) = ɛ γ v + 1 γ 4 v 4 It s easy to show that for a gven v U, J (γ v) s mnmal for γ0 = ɛ v v 4 Let w = γ 0 v Then, w 4 = ɛ w, (411) and hence Consequently, w ɛ L 1 L E (ψ,a ) J (w) + C α ɛ α 1 (L 1L ) 1/ J (v) + C α ɛ α 1 (L 1L ) 1/, whch proves (4) Let η be gven by (333) It s easy to show that E (ηψ,a ) E (ψ,a ) + C α ɛ α 1 (L 1L ) 1/ (41)

22 Y Almog By (34), (318), and (10) we obtan E (ηψ,a ) ɛ u η 4 ψ 4 + ũ C α (L 1 L ) α ɛ 5 (413) Combnng the above and (4) wth v = u 0 we obtan that ũ 1 [ u C α ɛ α 1 (L 1L ) 1/ ] + (L 1 L ) α ɛ 5 By (37) we then have ũ 4 C α ɛ α u C α [ ɛ α (L 1L ) 1/ + (L 1 L ) α ɛ 7α ] Usng (43) we can obtan a lower bound for E (ψ,a ) n terms of the reduced functonal J Lemma 4 Let η be gven by (333) and u 0 be gven by (318) Then, [ E (ψ,a ) J (u 0 ) C α ɛ α 1 (L 1L ) 1/ ] + (L 1 L ) α ɛ 5 (414) for all α<1 Proof By (41) and (413) we have that E (ψ,a ) ɛ u η 4 ψ 4 C α (L 1 L ) α ɛ 5 C α ɛ α 1 (L 1L ) 1/ (415) for all α<1 Snce η 4 ψ 4 u u 0 3 ũ, we have by (43) and Hölder nequalty that η 4 ψ 4 u 0 4 [1 C α ɛ α/ ] By (43) we also have that C α [ (L 1 L ) α ɛ 5 + ɛ α 1 (L 1L ) 1/ ] 1/4 [ ] 3/4 u 0 4 (416) u 0 L 4 () ηψ L 4 () + ũ L 4 () ηψ L 4 () [ +C α ɛ α/ u 0 4 L 4 () + (L 1L ) α ɛ 5 + ɛ α 1 (L 1L ) 1/ ] 1/4 (417)

23 Abrkosov Lattces n Fnte Domans To estmate u 0 L 4 (), we thus need an estmate for ηψ L 4 () By (36), (333), and (10) we have that [ η 4 ψ 4 ɛ η ψ + C α (L 1 L ) α ɛ 5 + ɛ α 1 (L 1L ) 1/ ], (recall that u 0 L () ηψ L ()) Consequently, [ η 4 ψ 4 ɛ 4 L 1 L + C α (L 1 L ) α ɛ 5 + ɛ α 1 (L 1L ) 1/ ] We note that by (316) the frst term on the rght-hand-sde of the above nequalty s much greater than the second one f α s suffcently close to 1, such that (L 1 L ) 1 α ɛ Consequently and hence, by (417) we obtan ηψ 4 ɛ 4 L 1 L, u 0 4 ɛ 4 L 1 L Substtutng n (416) and then n (415) we obtan (414) Proof of Theorem 11 Combnng (4) and (414) we obtan [ J (u 0 ) J (v) + C α ɛ α 1 (L 1L ) 1/ ] + (L 1 L ) α ɛ 5 (418) To prove (14a) we thus need an estmate for nf v U J (v) Let β = nf B(v) := L 1 L nf v 4 v U ( v ) (419) v U Snce U s fnte-dmensonal, t s easy to show that there exsts w U satsfyng w 4 = βl 1 L w (40) Furthermore, snce β s nvarant to the transformaton w γw for every γ, we can choose w such that (411) s satsfed Combnng (411) and (40) we obtan w = ɛ L 1L β Thus, n vew of (411) J (w) = 1 ɛ w = ɛ 4 L 1L β

24 Y Almog In [, 10, 4] B(v) was calculated n varous cases In partcular, t was found that when v s the well-known square lattce, e, when where f n s gven by (6b), that N 1 v = C f n, n=0 B(v) 118 ndependently of N and the scale of It follows therefore, n vew of (419) that nf v U J (v) Cɛ 4 L 1 L, (41) where C>0 Substtutng (41) nto (418) we obtan (14b) To prove (14a) we wrte [ ] ψ u 0 η 1 ψ + ũ The rght-hand-sde of the above nequalty can be bounded utlzng (319) and (317) to obtan ψ u 0 Cɛ u 0 + C α (L 1 L ) α ɛ 5 + C α ɛ α 1 (L 1L ) 1/ (4) Snce u 0 L () ψ L () we need a lower bound for u 0 L () to complete the proof of (14a) To ths end we use (14b) and (41) to obtan from whch we obtan β L 1 L u 0 4 L () ɛ u 0 L () J (u 0 ) Cɛ 4 L 1 L, u 0 L () Cɛ L 1 L Substtutng n (4) proves (14a) 5 Concluson Let (, ɛ) = [0,L 1 ] [0,L ], where L 1 and L are gven by (316) In the prevous sectons the followng man results were proved: 1 We proved that the L () dstance of ψ from the space of Abrkosov solutons n, U s much smaller than the L () norm of ψ We proved that the energy, whch s gven by (14d), of the projecton of ηψ on U, where η s gven by (333), s approxmately the mnmum over all U of (14d)

25 Abrkosov Lattces n Fnte Domans We note that the above results do not show that ψ s nearly perodc, nasmuch as every functon n L () can be approxmated by a perodc functon Nevertheless, snce the energy of the above projecton s close to the mnmum of J over U,we can obtan an approxmaton of ψ by studyng a much smpler mnmzaton problem than the mnmzaton of (11) n H It s wdely beleved that the mnmzer of J n U (whch s a fnte dmensonal space), s the well-known trangular lattce [, 10, 4], as long as N, n (316), s even If n addton to that, any u U whose energy s close to the mnmum must be close, n some sense to the trangular lattce, then ψ s ndeed nearly perodc It seems worthwhle to note here that the drecton of the lattce cannot be determned by the energy consderatons appled n the prevous secton Thus f [ u 0 (x) = u 0(Qx) ; Â 0 = h ex Q, x1] where Q sa rotaton matrx, then, snce the cells affected by the rotaton are only those near the boundary, we have that E (u 0, Â ) E (u 0, Â) C (L 1L ) 1/ Clearly, the above error s ndstngushable by the lower and upper bounds, (414) and (4), obtaned n 4 Fnally, we note that the lmtatons (33) could have been replaced by the weaker assumptons 1 ɛ4 ; (L 1 L ) 1/ 1 ɛ 4, f only we could overcome the crudeness of the estmate (315) However, to extend the analyss to the case ɛ 4 O(1/), a completely dfferent approach s necessary, snce n that case the surface energy, whch s of O(1/) s at least equally mportant to E whch s of O(ɛ 4 L 1 L ) eferences 1 Abramowtz, M, Stegun, IA: Handbook of Mathematcal Functons New York: Dover, 197 Abrkosov, AA: On the magnetc propertes of superconductors of the second group Sovet Phys JETP, 5, (1957) 3 Agmon, S, Dougls, A, Nrenberg, L: Estmates near the boundary for solutons of partal dfferental Commum Pure Appl Math 17, 35 9 (1964) 4 Almog, Y: On the bfurcaton and stablty of perodc solutons of the Gnzburg-Landau equatons n the plane Sam J Appl Math 61, (000) 5 Almog, Y: Non-lnear surface superconductvty for type II superconductors n the large doman lmt Arch at Mech Anal 165, (00) 6 Almog, Y: The loss of stablty of surface superconductvty J Math Phys 45, (004) 7 Almog, Y: Non-lnear surface superconductvty n the large lmt ev Math Phys 16, (004) 8 Bauman, P, Phlps, D, Tang, Q: Stable nucleaton for the Gnzburg-Landau system wth an appled magnetc feld Arch at Mech Anal 14, 1 43 (1998) 9 Bernoff, A, Sternberg, P: Onset of superconductvty n decreasng felds for general domans J Math Phys 39, (1998) 10 Chapman, SJ: Nucleaton of superconductvty n decreasng felds I European J Appl Math 5, (1994)

26 Y Almog 11 Chapman, SJ: Asymptotc analyss of the Gnzburg-Landau model of superconductvty: reducton to a free boundary model Quart Appl Math 53, (1995) 1 del Pno, M, Felmer, PL, Sternberg, P: Boundary concentraton for egenvalue problems related to the onset of the superconductvty Commun Math Phys 10, (000) 13 Elot, CM, Matano, H, Q, T: Zeros of complex Gnzburg-Landau order parameters wth applcatons to superconductvty European J Appl Math 5, (1994) 14 Essmann, U, Träuble, H: The drect observaton of ndvdual flux lnes n type II superconductors Phys Lett A4, (1967) 15 Gorg, T, Phlps, D: The breakdown of superconductvty due to strong felds for the Gnzburg- Landau model SIAM J Math Anal 30, (1999) 16 Helffer, B: Sem-Classcal Analyss for the Schrödnger Operator and Applcatons, No 1336 n Lecture Notes n Mathematcs, Berln-Hedelberg New York: Sprnger-Verlag, Helffer, B, Morame, A: Magnetc bottles n connecton wth superconductvty J Funct Anal 185, (001) 18 Helffer, B, Pan, XB: Upper crtcal feld and locaton of nucleaton of surface superconductvty Ann Inst H Poncaré Anal Non Lnéare 0, (003) 19 Lu, K, Pan, XB: Gauge nvarant egenvalue problems n and + Trans Am Math Soc 35, (000) 0 Maz ja, VG: Sobolev spaces Sprnger Seres n Sovet Mathematcs, Berln: Sprnger-Verlag, 1985 Translated from the ussan by T O Shaposhnkova 1 Messner, W, Ochsenfeld, : Naturwssenschaffen 1, 787 (1933) Pan, XB: Surface superconductvty n appled magnetc felds above h C Commum Math Phys 8, (00) 3 ubnsten, J: Sx lectures on superconductvty In: Boundares nterfaces and transtons, M Delfour, (ed), Vol 13 of CM proceedngs and lecture notes, Provdence, I: Am Math Soc, 1998, pp Sant-James, D, de Gennes, P: Onset of superconductvty n decreasng felds Phys Let 7, (1963) 5 Sander, E, Serfaty, S: The decrease of bulk-superconductvty near the second crtcal feld n the Gnzburg-Landau model SIAM J Math Anal 34, (003) Communcated by P Constantn