The Breakdown of Superconductivity Due to Strong Fields for the Ginzburg Landau Model

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1 SIAM REVIEW Vol. 44, No., pp c 00 Socety for Industral and Appled Mathematcs The Breakdown of Superconductvty ue to Strong Felds for the Gnzburg Landau Model T. Gorg. Phllps Abstract. We study the behavor of a superconductng materal subjected to a constant appled magnetc feld, H a = he wth e =, usng the Gnzburg Landau theory.we analytcally show the exstence of a crtcal feld h, for whch, when h>h, the normal states are the only solutons to the Gnzburg Landau equatons.we estmate h.as κ 0 we derve h = O(), whle as κ we obtan h = O(κ). Key words. superconductvty, Gnzburg Landau equatons, upper crtcal felds, normal state AMS subject classfcatons. 35J60, 35J65, 35Q40 PII. S Introducton. If a superconductng body s subjected to a suffcently strong appled magnetc feld, ts ablty to act as a superconductor breaks down and only the normally conductng (resstve) state s observed. In ths paper we consder superconductvty as modeled by the Gnzburg Landau theory and establsh ths type of phenomenon. Here, superconductvty s characterzed n terms of a complex valued order parameter ψ (where ψ represents the densty of superconductng electron pars) and a vector feld A the magnetc potental. Consder a superconductng body gven by a bounded doman R n, where n = or 3 and s of class C,α for some 0 <α<. Assume the body has constant permeablty normalzed to and that the exteror conssts of a second materal wth constant permeablty µ e > 0. efne the permeablty densty as follows: µ(x) = forx, = µ e for x R n \. A magnetc feld s appled to all space n the form H a = he, where h s a postve constant and e R 3 s a fxed unt vector. The presence of produces an Publshed electroncally May, 00.Ths paper orgnally appeared n SIAM Journal on Mathematcal Analyss, Volume 30, Number, 999, pages Ths research was sponsored n part by NSF grant MS and also supported by the epartment of Energy grant E-FG0-90ER4547 through the Mdwest Superconductvty Consortum. Mathematcs epartment, Towson Unversty, 8000 York Road, Towson, M (tgorg@towson.edu). 395 epartment of Mathematcs, Purdue Unversty, West Lafayette, IN (phllps@ math.purdue.edu). 37

2 38 T. GIORGI AN. PHILLIPS nduced magnetc feld, µ curl A n R3, and a supercurrent densty j := κ (ψ ψ ψ ψ ) A ψ n. Here κ>0 s the Gnzburg Landau constant determned from the superconductng materal, and the superscrpt denotes complex conjugaton. Accordng to ths theory, the par (ψ, A) s an equlbrum state for the Gbbs free energy ( ) G(ψ, A) := ψ + Aψ κ + ( ψ ) dx (.) + µ curl A he R µ dx + γ ψ ds κ n (see [6], [6]). The constant γ 0 reflects the retardng effect of the materal n the exteror doman on the densty ψ at ; γ s taken to be zero f R n \ s a vacuum and large f the exteror s a magnetc materal. Thus, we consder pars (ψ, A) such that ψ H (; C) H (), A H loc(r n ; R n ), whch are weak solutons to ( ) κ + A ψ ψ + ψ ψ =0 n, ( ) ( ) curl µ curl A + κ (ψ ψ ψ ψ )+A ψ χ =0 nr n, (.) ( ) n κ + A ψ = γψ on, ( ) curl A he L (R n ; R 3 ). µ Here n s the outward normal to at and χ s the characterstc functon for. A prncpal feature of the energy (.) and the solutons to (.) s that they are nvarant under the gauge transformaton where (ψ, A) (ψ, A ), ψ = ψe κη, A = A + η for an arbtrary real valued functon η Hloc (Rn ). Moreover, the ntrnsc quanttes for a soluton are preserved under ths transformaton: ts densty ψ = ψ, magnetc feld µ curl A = µ curl A, current j = j, and the modulus of the dervatve ( κ + A )ψ = ( κ + A)ψ. A soluton s n the normal phase f ψ 0n. Ths s wrtten as (ψ, A) = (0,ha N ), where a N satsfes ( ) curl µ curl a N =0 nr n, (.3) ( ) µ curl a N e L (R n ; R 3 ).

3 ON THE BREAKOWN OF SUPERCONUCTIVITY 39 Such a soluton s called a normal state. It s unquely determned by µ and up to a gauge transformaton; that s, (.3) unquely determnes curl a N. Let κ be fxed. We denote h as the upper crtcal feld for the body, where h := nf{h : normal states are the only solutons to (.) for all h>h }. For the case n whch the body s a bounded doman R 3, we prove the followng statement: Let R 3. Gven κ, µ e, and γ we have h = h(κ, µ e,γ,) < (see Theorem 3.). We show that the normal nducton s contnuous on. In the case that t does not vansh on, we can estmate h. If curl a N 0 n R 3, then there are constants m, φ 0, dependng on µ e and so that ( m ) (.4) h(κ, µ e,γ,) max κ,φκ (see Theorem 3.9). In the classc case where µ e =, t follows that curl a N e, and as such (.4) apples. If µ, then there are constants m and φ such that h max( m κ,φκ) (see Corollary 3.0). We also consder the case of a cylndrcal doman of the form R, where the cross secton s a bounded doman n R wth a C,α boundary and the appled feld H a = he = he 3 s perpendcular to the cross secton. From symmetry the problem reduces to one n two dmensons. We consder ψ(x, y) for (x, y) and A =(A (x, y),a (x, y)) for (x, y) R. The functonal (.) then represents the Gbbs free energy per unt length for the cylnder. We prove the followng theorem. Let R be a cylndrcal body n a parallel appled feld he 3. Gven κ, µ e, and γ there s a fnte upper crtcal feld h, so that f h>h, then the only soluton to (.) wth n =s normal. Moreover, there s a constant φ(µ e, ) so that h(κ, µ e,γ,) max( κ,φκ) (see Theorem.9). Fnally, we consder the case of small κ. We prove the followng result. Let n =wth µ e > 0 or n =3wth µ e =. Then h = O() as κ 0 (see Theorem 4.). It s of nterest to compare these results wth conjectures made by physcsts. For κ fxed, de Gennes and Sant-James have studed the local problem of determnng the smallest value of h for whch all normal states are stable for h h. The nfmum, denoted as h c3, s the value for whch t s possble to have a famly of superconductng solutons bfurcate away from the normal state. In [5] Sant-James and de Gennes dscussed the case of an nfnte slab d <x<d, <y,z< n R 3. The symmetry of the doman reduced the lnear analyss to a one-dmensonal problem. They gave an ansatz for determnng h c3 and predcted lm κ h c3 /κ = c 0 for some constant <c 0 <. Ths can be compared wth our estmates for h from Theorems.9 and 3.9. We have h c3 h = O(κ) asκ. For small κ, physcsts have predcted that h = O() as κ 0 for a slab of fnte thckness d <x<d, <y,z< and that h = O(κ )asκ 0 for the nfntely thck slab <x<0, <y,z< (see [5], [8], and [4]). Our estmate from Theorem 4. gves the result h = O() as κ 0 for our doman. We next comment on past analytc work. In [3] and [4], Bolley and Bolley and Helffer made the ansatz for the slab rgorous and proved asymptotc estmates for h c3.

4 40 T. GIORGI AN. PHILLIPS In [5] they obtaned partal results for estmatng an upper crtcal feld for the slab. They consdered a partcular famly of one-dmensonal functons. For each fxed κ, they showed there s a fnte upper crtcal feld when consderng only solutons n ths famly. In [] Bauman, Phllps, and Tang estmated h c3 for the case of a crcular cylnder, B r R. Ths estmate s relevant here as t plays a central role n our analyss of h for general domans. The central estmates from ths paper were appled by Bauman et al. n [] to nvestgate phase transtons n lqud crystals. A strong analogy exsts between the normal-superconductng phase transton characterzed by (.) and the nematcsmectc phase transton n lqud crystals. The phases for the latter are descrbed n terms of a complex-valued wave functon Ψ and the molecular drector feld n. These are the analogues of the order parameter ψ and the vector potental A, respectvely. Smectc structure s observed through Ψ, and one has Ψ 0 n the nematc phase. The free energy characterzng the nematc-smectc phase transton s the Landau de Gennes energy ntroduced by de Gennes n [9]. The de Gennes model was motvated by the formal analoges between the phase transtons for lqud crystals and superconductvty. There are, however, sgnfcant dfferences between the two theores. For example, the Gnzburg Landau energy (.) s gauge nvarant, whereas the Landau de Gennes energy s not. Furthermore, the drector feld n s requred to satsfy the constrant n =, whereas the correspondng term from superconductvty, A, s not. In secton we consder cylndrcal domans and establsh Theorem.9. In secton 3 we extend these deas to treat bounded domans n R 3. In secton 4 we estmate h for small κ.. Superconductvty wthn an Infntely Long Cylnder n a Parallel Feld. Let (ψ, A) be a weak soluton to (.) wth n = and R. Recall that H a = he 3 s perpendcular to the cross secton. We frst examne the magnetc nducton, curl A, n c. Lemma.. Let (ψ, A) satsfy (.). Then curl A s constant n each component of c. Moreover, curl A = µ e he 3 n the unbounded component. Proof. From (.) we see that curl(curl A) =0 n each component of c. Snce curl(curl A) =( y ( x A y A ), x ( x A y A ), 0), we have that curl A =( x A y A )e 3 s constant n each of these components. The last asserton follows from the fourth equaton n (.). We now determne curl a N. Lemma.. A normal state exsts. Moreover, any normal state (0,ha N ) satsfes curl a N = µe 3. Proof. Consder w =Γ (µ µ e ), where Γ (x) = π ln( x ), x =(x, y). The functon µ µ e has bounded support. As a result, w s well defned wth w Hloc (R ) and w =(µ µ e )nr. Set b N =( w y,w x )+ µe ( y, x). Then curl b N = µe 3 and we see that b N s a weak soluton to (.3), that s, R µ curl b N curl ϕdx = 0 for all ϕ H (R ; R ) such that ϕ has bounded support. Thus, a normal state exsts. Suppose a N s another weak soluton. Takng the dfference of the equatons for b N and a N,weget

5 (.) ON THE BREAKOWN OF SUPERCONUCTIVITY 4 R µ curl(b N a N ) curl ϕdx =0. Let E be the unbounded component of c. From Lemma. we have curl b N = curl a N outsde the bounded set E c. As a result we can take ϕ such that ϕ = b N a N n a neghborhood of E c n (.). Whence, curl b N curl a N. We can now show that a weak soluton has a gauge-equvalent representatve that satsfes a Sobolev estmate. Lemma.3. Let (ζ,b) and (0,ha N ) be weak solutons to (.). Then there s a weak soluton (ψ, A) that s gauge equvalent to (ζ,b) such that (.) A ha N dx C 0 curl(a ha N ) R dx, where C 0 depends only on. Proof. Set curl(b ha N )=fe 3. From Lemmas. and., we have that the support of f s contaned n the bounded set E c and f L (R ). Set w =Γ f; then standard estmates on the Newtonan potental gve w Hloc (R ), w = Γ f, w L (E c ) C 0 () f L (E c ), and w = f (see []). Thus, settng à =( w y,w x ) we have à H loc (R ; R ) and curl à = we 3 = curl(b ha N ). Let A = à + ha N. Then curl(b A) =0. Hence, A = B + η for some η Hloc (R ) and A ha N dx w dx C 0 f dx = C 0 curl(a ha N ) dx. E c E c R We need the followng property for weak solutons. Proposton.4. (see [0]). Let (ψ, A) be a weak soluton to (.); then ψ almost everywhere n. Next we wrte the weak formulaton of (.): (.3) ( ) ( ψ + Aψ ϕ + Aϕ κ κ = γ ψϕ ds κ curl A curl B dx + R µ ) dx + for any ϕ H (), ( ψ )ψϕ dx [ ] κ (ψ ψ ψ ψ )+A ψ B dx =0 for any B H (R ; R ) wth bounded support. Consderng κ ψ + Aψ for (ψ, A) H () Hloc (R ; R ), we have [( ) ] Re ψ + Aψ ψ = κ κ (ψ ψ ψ ψ )+A ψ, [( ) ] Im ψ + Aψ ψ κ = κ ψ. Thus, (.4) {[ ] [ ]} ψ ψ + Aψ = κ κ (ψ ψ ψ ψ )+A ψ ψ + κ ψ ψ

6 4 T. GIORGI AN. PHILLIPS for almost every x such that ψ 0. Moreover, snce ψ = 0 almost everywhere on the set {ψ =0}, t s consstent to defne the term n braces equal to zero on ths set. We conclude that (.4) holds almost everywhere. Lemma.5. Let (ψ, A) and (0,ha N ) be weak solutons satsfyng (.). Then there s a constant C = C (,µ e ) such that (.5) ( + κha N )ψ dx C κ ψ dx. Proof. Let ϕ = ψ n the frst equaton of (.3). Usng (.4), Proposton.4, and γ 0, we obtan ( [ ] κ ψ ) + κ (ψ ψ ψ ψ )+A ψ ψ dx (.6) ( ) = κ + A ψ dx ( ψ ) ψ dx ψ dx. Ths nequalty s also vald for n = 3. Consder the second equaton n (.) for the solutons (ψ, A) and (0,ha N ). Takng the dfference of ther respectve weak equatons we have [ ] R µ curl(a ha N) curl B dx = κ (ψ ψ ψ ψ )+A ψ ψ ψ Bdx. Usng (.6) and the Cauchy Schwarz nequalty, we see R µ curl(a ha N) curl B dx ε ψ dx + ε ψ B dx for any ε>0. Let B be such that B = A ha N n E c, where E s the unbounded component of c. Then snce curl(a ha N )=0 n E and ψ we derve R µ curl(a ha N) dx ε ψ dx + ε A ha N dx. Combnng ths nequalty wth (.), we see that we can take ε suffcently small so that (.7) A ha N dx M ψ dx for some constant M = M(dam,µ e ). Next we wrte ( ) ( ) (.8) κ + A ψ = κ + ha N ψ +(A ha N )ψ. We wll use the elementary nequalty (.9) c b c + b for c, b C.

7 ON THE BREAKOWN OF SUPERCONUCTIVITY 43 Let ( κ + A)ψ = b and ( κ + ha N)ψ = c. Then usng (.6) and (.8) we derve ( ) κ + ha N ψ dx ψ dx + A ha N ψ dx. Snce ψ, we can apply (.7) to obtan ( + κha N )ψ dx ( + M)κ ψ dx. We set C = ( + M) and the lemma s proved. We see that f a superconductng state (.e., a soluton wth ψ 0) exsts, then (.5) mples that the prncpal egenvalue for ( +κha N ) on s bounded by C κ. We wll show that there exsts a constant φ such that f h>max( κ,φκ), then the prncpal egenvalue s greater than C κ. It then follows for such κ and h that there are only normal solutons to (.). The correspondng egenfunctons are expected to take the form of a boundary layer. The followng lemma gves a way of measurng to what extent functons can concentrate near the boundary. For a set O we defne the τ-neghborhood n O of O by O τ = {x O : dst(x, O) <τ}. Lemma.6. Let O be a bounded doman n R n wth a C boundary. Gven λ 0 > 0, there s a constant d(λ 0, O) > 0 such that whenever (.0) f dx λ f dx for some f H (O) wth λ λ 0, then (.) f dx O O O O\O dλ f dx. Proof. Let N k=0f k be an open cover for O such that F 0 O and such that for each k, k N, wehave F k O = {(x,x n ):g k (x ) <x n <g k (x )+δ, x <δ }, where δ and δ are postve constants, (x,x n ) are sutably rotated and translated coordnates, and g k (x )=x n characterzes O F k. We can further assume wthout loss of generalty that g k ( ) s defned for x δ, g k <, and (.) {(x,x n ):g k (x ) <x n <g k (x )+4δ, x < δ } O, {(x,x n ):g k (x ) 4δ <x n <g k (x ), x < δ } R n \ O. Let f H (O), 0 t, v δ, and fx k. We have f ds f ds {x n g k (x )=v} F k {x n g k (x )=t} F k f (x,g k (x )+v) f (x,g k (x )+t) + g k dx { x δ } O F k (f ) x n dx.

8 44 T. GIORGI AN. PHILLIPS Integratng n t from 0 to δ and then dvdng by δ gves f ds ( ) f dx +δ f f dx. {x n g k (x )=v} F k δ O O Next ntegrate v from 0 to d λ, where 0 <d<λ 0δ / s to be determned. We derve (.3) {0<x n g k (x ) d λ } F k f dx d ( ) f dx +δ f f dx. λδ O O From the assumptons on g k, for k N we have { (.4) F k O d F k 0 <x n g k (x ) d } λ λ f d λ s small enough. Indeed, f ths s false for some k we can fnd x =(x,x n ) such that x δ, x n >g k (x )+ d λ, and y =(y,y n ) O such that x y < d λ. If d λ < mn(δ,δ ), we have x y < d λ <δ, mplyng that y < δ and y n x n < d λ <δ. We clam that y n = g k (y ). In fact, y n g k (y ) y n x n + x n g k (x ) + g k (x ) g k (y ), and each term on the rght s bounded by δ. We have shown ths for the frst one. Ths s true for the second snce x F k O. For the last term we use g k < and x y < d λ <δ. Thus, y n g k (y ) < 3δ. From (.) we see that the only possblty for such a y O s y n = g k (y ). As a result, g k (x ) g k (y ) g k (x ) x n g k (y ) x n > d λ d λ = d λ. On the other hand, snce g k < wehave g k (x ) g k (y ) < x y < d λ, and ths s a contradcton. Usng (.3) and (.4) and summng on k for k N, we obtan ( ) f d dx M f dx + λ f dx + λ f dx, λ O O O O dλ where M = M (δ,n). Usng (.0) we have O dλ f dx M d f dx, O where M = M (δ,n,λ 0 ). Settng d = mn( M, λ0δ f dx f dx. O dλ O, λ0δ The asserton (.) follows from ths nequalty. We wll use the followng result from [] for B r (0) R. ), we conclude that

9 ON THE BREAKOWN OF SUPERCONUCTIVITY 45 Proposton.7. There s a contnuous functon σ( ) :t [0, ) R wth σ(t) > 0 for t>0 for whch lm σ(t) exsts wth 0 < lm σ(t) <, and such that t t (.5) B r(0) ) ( + ω ( y, x) ζ for all ζ H (B r (0)) and ω 0. Indeed, n [, secton ], t s shown that nf ζ L = B r ζ W, (B r;c) ( + ω dx ω σ(ωr) ζ dx B r(0) ) ( y, x) ζ dx ω σ, where σ = σ(ωr). Furthermore, σ s characterzed by σ(t) = nf n z σ(t, n), where for each n, σ(t, n) s analytc and postve on 0 <t<. Moreover, lm t 0 σ(t, 0)=0. In [, secton 6], t s also shown that σ(t) = mn σ(t, n) for 0 t n 0. 0 n n 0 As a result, t follows that σ(t) s postve and contnuous. The lm t σ(t) s analyzed n [, secton 6], as well. Remark. Ifb s another vector feld such that b H (B r (0); R ) wth curl b = e 3, then (.5) s also vald wth ( y, x) replaced by b. Indeed, we can defne a functon q H (B r (0)) such that q = b ( y, x). Wth ths we can defne a local gauge transformaton ζ = ζe ωq, b = ( y, x)+ q, for whch ζ H (B r (0)) provded ζ H (B r (0)). Moreover, ) ( + ω b)ζ = ( + ω ( y, x) ζ and ζ = ζ so that (.6) B r(0) ( + ω b)ζ dx ω σ(ωr) ζ dx B r(0) for all ζ H (B r (0)), b H (B r (0); R ) such that curl b = e 3. We now derve an estmate smlar to (.6) for provded ω s bounded away from zero. Lemma.8. Gven m>0, there s a constant C = C (m, ), 0 <C, such that f ω m, then (.7) C ω ζ dx ( + ω b)ζ dx for all ζ H () and b H (; R ) for whch curl b = e 3. Proof. Let ζ H () such that ζ dx > 0 and ( + ω b)ζ dx ω ζ dx

10 46 T. GIORGI AN. PHILLIPS for some ω, ω m. If no such ζ exsts, then (.7) s vald wth C = and we are fnshed. From (.4) we see ζ ( + ω b)ζ. Thus, ζ dx ω ζ dx. As a result, we can apply Lemma.6 to conclude that (.8) ζ dx ζ dx. Next we choose a cover for \ d ω \ dω consstng of a fnte collecton of dsks {B d (x k ), ω k =,...,N(ω)}, each contaned n n such a way that N(ω) k= χ B dω (x k ) K, where K s ndependent of ω. We see ( + ω b)ζ dx K N k= B d ω (x k ) N k= Usng Proposton.7, the last term bounds K ω σ(d) N k= B dω (x k ) ζ dx K (d)ω ( + ω b)ζ dx B dω (x k ) ( + ω b)ζ dx. \ dω ζ dx K ω ζ dx, where the fnal nequalty follows from (.8). Set C = K /. Ths chan of nequaltes establshes the lemma. We now establsh the prncpal result n ths secton. Here we prove the exstence of an upper crtcal feld h and obtan a bound for t as κ and κ 0. Theorem.9. There s a constant φ = φ(µ e, ) so that f h>max( κ,φκ), then any weak soluton for (.) wth n =s normal. Proof. Let (0, a N ) be a normal state for (.3) and (ψ, A) be a weak soluton for (.). A state s normal f and only f ts entre gauge equvalence class s normal. Therefore, we can assume wthout loss of generalty that (ψ, A) and (0,ha N ) satsfy (.). Set ω = hκ. Then ω by hypothess. We apply (.7) wth m =to derve (.9) C hκ ψ dx ( + hκa N )ψ dx, and by Lemma.5, the rght-hand sde of (.9) s bounded by C κ ψ dx. Let φ = C /C. We have h ψ dx φκ ψ dx. By assumpton, h>φκ. Hence t must hold that ψ dx =0.

11 ON THE BREAKOWN OF SUPERCONUCTIVITY Three-mensonal Bodes. In ths secton, we consder a superconductng body gven by a bounded doman R 3 subjected to a unform appled feld H a = he. We wll assume wthout loss of generalty that e = e 3 throughout ths secton. enote by Ȟ (R 3 ) the completon of C0 (R 3 ; R 3 ) wth respect to the norm ) B Ȟ (R 3 ) (R = B dx. 3 One can show that elements B Ȟ (R 3 ) satsfy the followng relatonshps: (3.) B L 6 (R 3 ;R 3 ) θ B Ȟ (R 3 ), where θ s ndependent of B, and (3.) B Ȟ (R 3 ) = R 3 ( dv B + curl B )dx (see []). In order to represent magnetc felds we need the followng lemma. Lemma 3.. Let g L (R 3 ; R 3 ) such that dv g =0n (R 3 ). Then there s a unque u Ȟ (R 3 ) such that curl u = g and dv u =0. Proof. Consder k Γ g, where Γ(x) =Γ 3 (x) = 3ω 3 x s the Newtonan potental for R 3 and k 3. We clam that k Γ g Ȟ (R 3 ). To see ths, let us frst assume that g has bounded support. Then k Γ g exsts as a weakly sngular ntegral and and k Γ g = O( x ) ( k Γ g) = O( x 3 )as x. Let ϕ R (x) be a standard C cutoff functon such that ϕ R = for x R and ϕ R = 0 for x R +. It follows drectly that {ϕ R(n) ( k Γ g)} s a Cauchy sequence n Ȟ (R 3 ) that converges to k Γ g pontwse for any sequence R(n). Thus, k Γ g Ȟ (R 3 ), assumng g has bounded support. Fnally, by standard L -sngular ntegral theory, (3.3) k Γ g Ȟ (R 3 ) g L (R 3 ;R 3 ), and as a consequence, k Γ g Ȟ (R 3 ) for all g L (R 3 ; R 3 ). efne u : g L (R 3 ; R 3 ) Ȟ (R 3 )by u(g) = ( Γ g 3 3 Γ g, 3 Γ g Γ g 3, Γ g Γ g ). Let g ε be a mollfcaton of g; then g ε g n L as ε 0, and dv g ε = 0 for each ε>0. We defne g ε,r = ϕ R g ε. Note that dv(g ε,r )= ϕ R g ε. If we choose sequences ε(n) 0 and R(n) as n, then g n = ϕ R(n) g ε(n) C 0 (R 3 ; R 3 ), g n g n L, and dv g n 0nL as n. Set w n =Γ g n. These are well defned snce the g n have bounded support. Usng (3.3), we see that curl w n = u(g n ) u(g) nȟ as n.

12 48 T. GIORGI AN. PHILLIPS Consder (3.4) curl u(g n )= curl curl w n = w n (dv w n )=g n (dv w n ). We know that (dv w n )= Γ (dv g n ) 0 n Ȟ (R 3 )asn snce dv g n 0n L. Thus, usng (3.) we conclude that (dv w n ) 0 n L 6 as n. Furthermore, we have curl u(g n ) curl u(g) and g n g n L as n. As a consequence, curl u(g) =g n R 3. Snce u(g n )= curl w n we have dv u(g n ) = 0, whch mples dv u(g) =0. Fnally, usng (3.) we see that u s unque n Ȟ (R 3 ). We can apply the precedng lemma to characterze weak solutons. Lemma 3.. Let (ζ,b) be a weak soluton to (.). Then there s a gaugeequvalent soluton (ψ, A) such that dv A =0and (A µeh ( y, x, 0)) Ȟ (R 3 ). Moreover, f ( ψ, Ã) s another such soluton, then ψ = aψ for some a C, a =, and à = A. Proof. Set g = (curl B µ e he 3 ) L (R 3 ; R 3 ). From the prevous lemma there s a unque element u Ȟ (R 3 ) such that curl u = g and dv u = 0. Therefore, we fnd A = u + µeh ( y, x, 0). We now characterze the normal state n three dmensons. Lemma 3.3. There s a unque normal state satsfyng (.3) such that (a N µ e ( y, x, 0)) Ȟ (R 3 ) and dv a N =0. Proof. Consder the strctly convex functonal E(b) =G(0, b)+ (dv b) dx = R 3 dx + R 3 ( µ µ curl b e 3 + (dv b) ) dx for the class S = {b :(b µe ( y, x, 0)) Ȟ (R 3 )}. A unque equlbrum exsts that also mnmzes E( ). Let b be ths equlbrum. If dv b 0, then by Lemma 3. we can fnd another vector feld b S such that curl b = curl b and dv b = 0. Ths would mply E( b) <E( b), whch s mpossble. Thus, dv b = 0. It follows that b satsfes (.3), and as a result a normal state (0, a N ) exsts. Conversely, a normal state satsfyng the hypothess s an equlbrum for E( ), and so a N s unque. Recall that the nducton curl a N, not a N, s the physcally relevant quantty. Below we show that t s unquely determned. Lemma 3.4. Let (0, a N ) be a normal state. Then curl a N s unquely determned, curl a N s harmonc n R 3 \, and Moreover, f dv a N = 0, then curl a N C,α () C,α ( c ). a N C,α () C,α ( c ). Proof. Usng Lemma 3. we see that any normal state s gauge equvalent to the normal state descrbed n Lemma 3.3. Snce a gauge transformaton leaves the curl of a vector feld nvarant, we conclude that curl a N s unquely determned for solutons to (.3). We can use the frst equaton n (.3) to prove that there exsts a functon p H loc (R3 ) such that curl a N = µ p. Snce (3.5) dv(µ p) =0nR 3

13 ON THE BREAKOWN OF SUPERCONUCTIVITY 49 and µ s constant on the components of R 3 \, the functon p (and thus curl a N ) s harmonc n each component. We apply the results from [3, Chap. 5, secton 4] to the soluton p for (3.5), to derve that p C,α () C,α ( c ), and as a consequence, curl a N C,α () C,α ( c ). Assume dv a N = 0. Let U be an open neghborhood of and consder w H (U; R 3 ) such that w = curl a N n U. From [3, Chap. 5], we have w C,α () C,α (U\). The dentty curl(curl w)+ w = (dv w) nu yelds from whch we obtan curl(curl w + a N )= (dv w), curl curl(curl w + a N )=0 n (U). By hypothess, dv(curl w + a N ) = 0. Whence, from the dentty above, (curl w + a N ) = curl curl(curl w + a N )=0 n (U). Ths mples (curl w + a N ) C (U), and we conclude that a N C,α () C,α ( c ). Consder the case µ. Gven e, we can fnd a lnear functon a(x) such that curl a e. Clearly a satsfes (.3). It follows from the prevous lemma, then, that curl a N e when µ. We now derve a Sobolev estmate analogous to Lemma.3. Lemma 3.5. Let (ζ,b) be a weak soluton to (.). Let (ψ, A) be the gaugeequvalent soluton found n Lemma 3. and (0,ha N ) be the normal state found n Lemma 3.3. Then there s a constant C 0 dependng only on such that A ha N dx C 0 curl(a ha N ) R dx. 3 Proof. Usng (3.) and (3.) we see A ha N L 6 (R 3 ;R 3 ) θ (A ha N ) L (R 3 ;R 3 ) = θ curl(a ha N ) L (R 3 ;R 3 ). Snce s bounded we have A ha N L (;R 3 ) M() A ha N L 6 (;R 3 ) and the lemma follows. We proceed to derve the three-dmensonal counterpart to Lemma.5. Lemma 3.6. Let (ψ, A) and (0,ha N ) be as n Lemma 3.5. Then there s a constant C = C (,µ e ) so that ( + hκa N )ψ dx C κ ψ dx. Proof. We proceed just as n Lemma.5 to obtan R µ [curl(a ha N) curl B]dx ε ψ dx + ε 3 ψ B dx

14 50 T. GIORGI AN. PHILLIPS for any ε>0 and B H (R 3 ; R 3 ) wth bounded support. However, snce A ha N Ȟ (R 3 ), we can take B = B j A ha N n Ȟ (R 3 )asj. As a result, we have R µ curl(a ha N) dx ε ψ dx + ε ψ A ha N dx. 3 The remander of the proof s just as before. We next gve a three-dmensonal analogue for the egenvalue estmate from []. Let v R 3 \{0} such that v = and x 0 R 3. Let T (x 0,r,v) be a cylnder wth central axs parallel to v, heght r, and whose mddle cross secton s the dsk of radus r wth center x 0. Lemma 3.7. Let b H (T (x 0,r,v); R 3 ) such that curl b = v. Then (3.6) ( + ω b)ζ dx ω σ(ωr) ζ dx T for all ζ H (T ), where σ( ) s as n Proposton.7. Proof. We frst transfer the problem to T (0,r,e 3 )=B r (0) ( r, r). Let Q SO(3) such that e 3 = Qv, and set y(x) =Q(x x 0 ). Then y : x T (x 0,r,v) T (0,r,e 3 ). Gven ζ H (T (x 0,r,v)) we defne ξ(y) =ζ(x(y)). Then ( y + ω bq t )ξ(y) =( x + ω b)ζ(x)q t. By changng varables we see that (3.6) s equvalent to showng the followng nequalty: ( + ω bq t )ξ dy ω σ(ωr) ξ dy. T (0,r,e 3) T (0,r,e 3) For any w R 3,wehave w t Q t w t Q t w t Qw curl y (bq t ) = det y = det x Q t = det x = w curl x b. bq t bq t b Therefore, curl y (bq t )=Q(curl x b)=qv = e 3. By changng the gauge f necessary, we can assume bq t = ( y, x, 0). Then ( + ω bq t )ξ dy (( x, y, 0) + ω bq t )ξ dy T (0,r,e 3) T (0,r,e 3) r ) = (( x, y, 0) + ω ( y, x, 0) ξ(x, y, z) r B r dx dy dz r ω σ(ωr) ξ(x, y, z) dx dy dz r B r = ω σ(ωr) ξ dy, T (0,r,e 3) where we have appled Proposton.7 for each r z r. T

15 ON THE BREAKOWN OF SUPERCONUCTIVITY 5 We go on to prove the three-dmensonal counterpart to the egenvalue estmate n Lemma.8. Lemma 3.8. Let (0, a N ) be the normal state from Lemma 3.3. Assume curl a N 0 n. Then there exst constants m and 0 <C so that f ω m, t holds that (3.7) C ω ζ dx ( + ω a N )ζ dx for all ζ H (). Proof. We argue as n Lemma.8. There exsts a constant d>0 so that, gven ξ H () and ω, ether (3.7) s true wth C = and ξ = ζ or (3.8) ξ dx ξ dx. \ dω Assume the latter. In ths case we cover \d by a famly of cylnders {T k : k = ω,...,n(ω)} such that T k = T (x k, d ω, curl a N(x k )/ curl a N (x k ) ) wth T k for each k and N k= χ T k K, where K s ndependent of ω for ω. As a consequence, (3.9) ( + ω a N )ξ dx K N k= T k ( + ω a N )ξ dx. In each T k we wrte a N (x) =l k (x)+q k (x), where l k (x) =a N (x k )+ a N (x k ) (x x k ). Note that curl l k (x) = curl a N (x k ). Usng (.9), for each k we obtan ( + ω a N )ξ dx ( + ω l k )ξ dx ω 4 q k ξ dx. T k T k T k From Lemma 3.7 we have T k ( + ω l k )ξ dx ω curl a N (x k ) σ ω M 0 T k ξ dx, ( ) curl a N (x k ) d ξ dx T k where M 0 > 0 depends on nf curl a N > 0 and the structure of σ( ) (see Proposton.7). Snce a N C,α () wehave q k M (dam T k ) +α M ω α, where M s ndependent of k. As a result, we see for each k that (3.0) ( ) ( + ω a N )ξ M0 dx T k ω M ω α ξ dx M 0 T k 4 ω ξ dx, T k provded ω m = m(,µ e ) suffcently large.

16 5 T. GIORGI AN. PHILLIPS From (3.9) and (3.0), then, ( + ω a N )ξ dx M 3 ω N k= T k ξ dx for some M 3 > 0 ndependent of the cover. Usng \d N ω k= T k and (3.8), we derve ξ dx ξ dx ξ dx. \ dω N k= T k Settng C = M3, we have our lemma. The followng theorem s proved n the same manner as Theorem.9. We establsh the exstence of h and derve an upper bound for t provded curl a N does not vansh on. Theorem 3.9. Assume that curl a N 0 n. There are constants m and φ, dependng on and µ e, so that f h>max( m κ,φκ), then any weak soluton to (.) wth n =3s normal. For the case µ, we have curl a N, and we can recover the followng result. Corollary 3.0. If µ e =, then there exst constants m and φ dependng on so that f h>max( m κ,φκ), then any weak soluton to (.) wth n =3s normal. In general, one does not know whether or not curl a N vanshes somewhere n. Nevertheless, snce curl a N s harmonc n t can vansh only on a small set. Lemma 3.. Let (0, a N ) be a normal state. Then L 3 ({x : curl a N (x) = 0}) =0. Proof. Snce curl a N s harmonc n, ether L 3 ({x : curl a N (x) =0}) =0 or curl a N 0 n. From Lemma 3.4, we know that there s a functon p C,α () C,α ( c ) C(R 3 ) such that Hence, p satsfes curl a N = µ p n R 3. (3.) { dv(µ p) =0 nr 3, ( p e 3 ) L (R 3 ). Assume that curl a N 0 n. Then p = constant = p 0 n. It follows from the frst equaton n (3.) that p solves p =0 n c, p = p 0, p n =0 on c, where n s the exteror normal to. The Cauchy problem has the unque soluton p = p 0. Ths would contradct the second equaton n (3.). To conclude, we show that there s a fnte upper crtcal feld for each κ. Theorem 3.. Let κ, µ e, and γ be fxed. There s a constant h = h(κ, µ e,γ,) so that f h>h, then any weak soluton to (.) wth n =3s normal. Proof. Let (0, a N ) be as n Lemma 3.3. Assume that there exsts a sequence {(ψ j, A j )}, where for each j the par s as n Lemma 3.5, solvng (.) wth h = h j for whch lm j h j = and ψ j dx > 0.

17 ON THE BREAKOWN OF SUPERCONUCTIVITY 53 Set ϕ j (x) = ψ j (x) / ψ j L (). From (.6) we have ϕ j dx κ, and we can fnd a subsequence ϕ j ϕ 0 n L () asj wth ϕ 0 L () =. From Lemma 3. the set Q = {x : curl a N (x) =0} s a closed set of measure zero. It follows that there exsts a ball B r \Q such that B r ϕ 0 dx δ >0 and nf Br curl a N > 0. Note that for j suffcently large we have ψ j dx δ ψ j dx. B r For each j we cover B r by a fnte famly of cylnders, {T (x k,ω j, curl a N (x k )/ curl a N (x k ) )} = {T kj, k N(j)}, such that ωj = h jκ, x k B r, and such that the famly has overlap of at most K unformly n x ndependent of j. Forj suffcently large, each T kj B r and, as n Lemma 3.8, we derve ( + ωj a N )ψ j dx ωj M 0 ψ j dx, T kj T kj where M 0 > 0 depends on nf Br curl a N and a N C,α () but not on j. Whence we can wrte N(j) ( + h j κa N )ψ j dx K ( + h j κa N )ψ j dx T kj k= M h j κ ψ j dx M h j κ N(j) k= T kj ψ j dx M h j κ B r ψ j dx, where M and M are postve and depend on nf Br curl a N, a N, and δ. Lemma 3.6 we know ( + h j κa N )ψ j dx C κ ψ j dx, From whch leads to M h j κ ψ j dx C κ ψ j dx for all j suffcently large. Snce h j as j, we must have ψ j dx = 0 for j large, and ths s a contradcton. 4. Estmates for Small κ. In ths secton we consder h for small κ n cases where curl a N e n. Theorem 4.. Let n =wth µ e > 0 or n =3wth µ e =. Then h = O() as κ 0. Proof. We consder the case n = 3. The argument for n = s dentcal.

18 54 T. GIORGI AN. PHILLIPS Let κ and assume that (ψ, A) solves (.) wth ψ 0. From (.6) we have ψ dx κ ψ dx ψ dx. If we apply Lemma.6 wth λ = λ 0 =, we fnd that there s a constant d>0so that ψ dx ψ dx. \ d We take r, 0<r<d, to be determned and cover \ d by a famly of cylnders {T (x k,r,e)} such that the cylnders have fnte overlap ndependent of r and each s contaned n. Then, argung just as n Lemma.8, there exsts a constant C > 0 for whch C hκ σ((hκ) r) ψ dx ( + hκa N )ψ dx. Applyng Lemma 3.6 to the rght-hand sde and recallng that ψ dx > 0, the followng nequalty holds: (4.) C hκ σ((hκ) r) C κ. From Corollary 3.0 we have hκ M for some M < for all κ. We now choose r so that M r. The reason for ths choce s that f 0 <t, then σ(t) s the prncpal egenvalue for the Sturm Louvlle problem g (s)+ g (s) s g(s) = σ(t)g(s) for 0 <s<t, s 4 g (t) =0andg s bounded (see [, secton 6]). As such, from [7, Prop. 3.4], we have σ(t) = t 8 + o(t )ast 0. It follows that there s a constant M > 0 so that M t σ(t) for 0 t. Combnng ths estmate wth (4.), we derve h C C M r for 0 <κ. Snce ths holds for all superconductng solutons, we conclude that h C C M r for 0 <κ. The length scale for varatons n superconductng solutons s κ. Because of ths, t s of nterest to consder domans wth dmensons comparable to κ. To ths end, let R n and defne the dlated doman (κ) ={x R n : κx }.

19 ON THE BREAKOWN OF SUPERCONUCTIVITY 55 Then dam((κ)) = κ dam(). Let µ so that curl a N e n (κ). Consder (ψ, A) satsfyng (.) on (κ). Let and ψ(x) =ψ(κ x) for x Ã(x) =A(κ x) for x R 3. Then ( ψ, Ã) satsfes ( + Ã) ψ ψ + ψ ψ =0 n, ( curl à = κ (4.) ( ψ ψ ψ ψ ) )+à ψ χ n R 3, n ( + Ã) ψ = γ ψ on, (curl à hκ e) L (R 3 ; R 3 ). Assume that 0 <κ. Argung as n Lemma.5, takng nto account the multple κ of the rght-hand sde of the second equaton n (4.), we obtan the analogue of (.7), à hκ a N dx κ 4 M ψ dx, where M = M(dam ). Usng κ we then obtan the analogue of (.5), ( + hκ a N ) ψ dx C κ 4 ψ dx. It follows as n Theorem.9, then, that there s a constant φ so that f hκ φκ 4 then ψ 0. Returnng to (ψ, A), we conclude that h(κ,,γ,(κ)) = O(κ 3 ) as κ 0. As we noted n the ntroducton, there are superconductng solutons for the case of the slab <x<0, <y,z< for whch h dverges as κ 0 (see [5]). However, ths occurs at the slower rate h = O(κ ). REFERENCES [] P. Bauman, M.C. Calderer, C. Lu, and. Phllps, The Phase Transton between Chral Nematc and Smectc A Lqud Crystals, preprnt. [] P. Bauman,. Phllps, and Q. Tang, Stable nucleaton for the Gnzburg Landau system wth an appled magnetc feld, Arch.Ratonal Mech.Anal., 4 (998), pp. 43. [3] C. Bolley, Modélzaton de champ de retard à la condensaton d un supraconducteur par un problème de bfurcaton, Math.Modél.Anal.Numér., 6 (99), pp [4] C. Bolley and B. Helffer, An applcaton of sem-classcal analyss to the asymptotc study of the supercoolngfeld of a superconductngmateral, Ann.Inst.H.Poncaré, 58 (993), pp [5] C. Bolley and B. Helffer, On the asymptotcs of the crtcal felds for the Gnzburg Landau equaton, n Progress n Partal fferental Equatons: The Metz Surveys, Ptman Res. Notes Math.Ser.34, Longman Scentfc and Techncal, Harlow, UK, 994, pp.8 3.

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