&x m 2 ,
{
&2. m +(a m +x 1 cos &x 2 sin ) 2. m =b m . m , for y 2 >&x m . m (& p m , y 2 )=.(q m , y 2 )=0 2 ,
2 ,( y 1 , &x m 2 )=0.
409
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
m Here p m =m+x m 1 and q m =m&x 1 . Passing to a subsequence, we may assume that
p m Ä p,
q m Ä q,
.m Ä .
in C
2 loc
a m Ä a,
&< & p
where
.
Note that .(0)=1 and . satisfies &2.+(a+x 1 cos &x 2 sin ) 2 .=b() .
for &p> y 1 >q,
y 2 # R 1. (3.16)
We consider three cases. Case 1. p=q=+. Since b()<1, (3.16) has no nontrivial bounded solution, which yields a contradiction. Case 2. p<+, q=+. Then, .( p, y 2 )=0. Make change of variables z 1 = y 2 + p cot &
a , sin
z 2 = y 1 + p,
,(z)=.( y).
We may write (3.16) as
{
&2+(z 1 sin &z 2 cos ) 2=b() =0 on R 2+ .
in R 2+ ,
(3.17)
However, since b()<1, (3.17) has no nontrivial bounded solution. In fact, if there is such a solution , choosing ' as a test function, where ' is a smooth cutoff function, we have that inf 1, 2
. # W 0 (R 2+ )
R 2+ [ {. 2 +(z 1 sin &z 2 cos ) 2 . 2 ] dx R2+ . 2 dx
b().
However, R 2+ [ {. 2 +(z 1 sin &z 2 cos ) 2 . 2 ] dx
inf
R2+ . 2dx
2 2 . # W 1, 0 (R + )
=
inf . # W 1, 2(R 2 )
R 2 [ {. 2 +(z 1 sin &z 2 cos ) 2 . 2 ] dx =1. R2 . 2dx
(3.18)
410
LU AND PAN
The last equality was obtained by changing variables y 1 =z 1 sin &z 2 cos ,
y 2 =z 1 cos +z 2 sin ,
and by using of (3.12), also see Lemma 4.3. This contradiction shows that (3.17) has no nontrivial bounded solution when b()<1. Hence Case 2 cannot happen. Case 3. p=+, q>+. In the same fashion as case 2, we show that this case can not happen neither. Summarizing the above arguments we conclude that x m  must be bounded. So v0. This completes Step 2. K By the maximum principle we see that v>0 on R 2+ _ R 2+ . So Lemma 3.4 is proved. K In Lemma 3.4 we see that (3.1) has a bounded positive solution when 0<0 and satisfies (3.1). Now we show that v # W 1, 2 (R 2+ ). Step 1. First we show that, if [v m ] is bounded in L 2 (R 2+ ), then v # W 1, 2 (R 2+ ) and is a minimizer of b(). In fact, by using the definition of b m , we see that, if [v m ] is bounded in L 2 (R 2+ ), then it is also bounded in W 1, 2 (R 2+ ). Hence, we may pass to a subsequence again to obtain vm Ä v
weakly in W 1, 2 (R 2+ ),
which implies v # W 1, 2 (R 2+ ). Integrating (3.1) we obtain that

R 2+
[{v 2 +(x 1 cos &x 2 sin ) 2 v 2 ] dx=b()
So, v is the minimizer of b() in W 1, 2 (R 2+ ).

v 2 dx. R 2+
411
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
Step 2. Denote by &v m & 2 the L 2 norm of v m . Next, we show &v m & 2 is bounded. Otherwise, set . m =v m &v m & 2 . Since max v m =1,
and
&v m & 2 Ä +,
we have . m Ä 0 in C 2loc ,
. m Ä 0 weakly in W 1, 2 (R 2+ )
and
as
Ä .
In the following, we denote M a =[x # R 2+ : x 1 cos &x 2 sin  a], L a =[x # R 2+ : x 1 cos &x 2 sin  a, 0>x 2 >2]. Since v m is the minimizer of b m , we have

R 2+
[ {. m  2 +(x 1 cos &x 2 sin ) 2 . m  2 ] dx=b m

R 2+
. m  2 dx=b m .
Therefore, for any a<0,

. m  2 dx
Ma
bm . a2
Let ' be a smooth cutoff function, '(t)=1 for t2, '(t)=0 for t1, and '$(t) 2. Set m (x)='(x 2 ) . m (x). Then,

R 2+

{ m  2dx=
R 2+
[' 2 {. m  2 + '$. m  2 +2'. m '$ 2 . m ] dx.
Since L a is bounded and . m Ä 0 in C 2loc , we have, as m Ä $ m (a)#

[ '$. m  2 +2'. m '$ 2 . m ] dx=o(1).
La
Since '$(x 2 )=0 for x 2 <2, we obtain

R 2+ "La
[ '$. m  2 +2'. m '$ 2 . m  ] dx

Ma
2
{\4+= + . 2
m
=
 2 += 2 {. m  2 dxb m
4
2
_a + (a=) += & . 2
2
2
412
LU AND PAN
Therefore,

R 2+
[ { m  2 +(x 1 cos &x 2 sin ) 2  m  2 ] dx

R 2+
[{. m  2 +(x 1 cos &x 2 sin ) 2 . m  2 ] dx
=b m
_
4 2 += 2 +$ m (a) 2+ a (a=) 2
& 4 2 _1+a + (a=) += & +$ (a).
+b m
2
2
(3.19)
m
2
On the other hand, since m # W 1,0 2(R 2+ ), as in (3.18) we have

R 2+
[ { m  2 +(x 1 cos &x 2 sin ) 2  m  2 ] dx

R 2+
=1&
 m  2 dx

R 2+
(1&' 2 ). m  2dx1&

. m  2 dx
La _ M a
1&$ m (a)&
bm , a2
(3.20)
where $ m (a) Ä 0 as m Ä +. Combining (3.19) and (3.20) gives
_
1b m 1+
5 2 + += 2 +$ m (a)+$ m (a). a 2 (a=) 2
&
Fix =<0 small enough, then let a== &32. Letting m go to infinity in the above inequality we get 1b()[1+2=+= 2 +5= 3 ].
(3.21)
However, for small enough =, the right hand in (3.21) is less than 1. This contradiction shows that &v m & is bounded. This completes Step 2, and finishes the proof of Lemma 3.5. K In the following, for 0<
x # R 2+
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
413
Lemma 3.6. When 0<0 on R 2+ , and the only bounded eigenfunctions of (3.1) associated with the first eigenvalue b() are cv . Proof. Step 1. First we have that, if 0<0. Step 3. We can show that, if v # W 1, 2 (R 2+ ) is an eigenfunction of b(), then v does not change sign. If not, denote v + =max(v, 0). Then, v + 0 and v + vanishes at some points. Multiplying (3.1) by v + and integrating, we see that v + is also a minimizer of b(). Hence, it is the weak solution of equation (3.1). By the elliptic regularity, v + is smooth. As in Step 2 we see that v + does not vanish in R 2+ , which is a contradiction. Step 4. We prove the conclusion of the lemma. Assume v is a bounded solution of (3.1) for b=b(). From Step 1, v # W 1, 2 (R 2+ ). From Step 3, we may assume v0. Then from Step 2, v>0. Using the same argument we see that, for any constant t>0, if w#v&tv 0 then w does not change sign. Set l=sup [t: v&tv 0 in R 2+ ]. Then 0l<+. Note that for any =>0, v&(l+=) v does not change sign. By the definition of l we must have v&(l+=) v 0. Hence, lv v(l+=) v . Sending = to zero we get v=lv . The proof of Lemma 3.6 is completed. K Lemma 3.7. b() is continuous in (0, ?2) and has no local minimum. Proof. Since b() is achieved for 0<
414
LU AND PAN
Make change of variables y 1 =x 1 cos ,
y 2 =x 2 sin ,
v (x)=w( y).
Then w satisfies
{
&(cos 2 21 w+sin 2 22 w)+( y 1 & y 2 ) 2 w=b() w on R 2+ . 2 w=0
in R 2+ ,
(3.22)
Obviously, b()=
inf
R 2+ [cos 2  1 , 2 +sin 2  2 , 2 +( y 1 & y 2 ) 2 , 2 ] dy R 2+ , 2 dy
, # W 1, 2(R 2+ )
.
(3.23) Suppose that there is a 0 # (0, ?2) such that b( 0 )=min 0>>?2 b(). Denote w 0 ( y)=Cv 0 , where C>0 is a constant such that &w 0 & L2 =1. Then, for all small t we have b( 0 )b( 0 +t)

R 2+
[cos2 ( 0 +t) 1 w0  2 +sin 2 ( 0 +t)  2 w0  2 +( y1 & y 2 ) 2 w0  2 ] dy

=b( 0 )+t sin(2 0)
R 2+
[ 2 w0  2 & 1 w 0  2 ] dy+O(t2 ).
So,

t sin(2 0 )
R 2+
[  2 w 0  2 &  1 w 0  2 ] dy+O(t 2 )0
for all t small, which implies that

R 2+
[  2 w 0  2 &  1 w 0  2 ] dy=0.
Thus, by (3.12) and (3.23) we obtain b( 0 )

R 2+
w 0  2dy=

+
dy 2
0

[  1 w 0  2 +( y 1 & y 2 ) 2 w 0  2 ] dy 1
&
+
0

+
dy 2

+ &
w 0  2 dy 1 =

R 2+
w 0  2 dy.
(3.24)
Since b( 0 )>1, (3.24) is impossible. This contradiction shows that b() has no local minimum in (0, ?2). We complete the proof. K
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
415
Lemma 3.8. For 0<
\
v (x)=. x 1 +
z  sin , x2 . cos
+
We have b()=
inf 1, 2
R2+ [{. 2 +[x 1 cos & sin (x 2  sin +z)] 2 . 2 ] dx R2+ . 2 dx
(R 2+ )
.#W
.
Choose z=z 0 , the unique minimum point of ;(z). Note that ;(z 0 )=; 0 . Let u(t) be the positive eigenfunction of (2.4) associated with ; 0 . Set g(x 2 )=u(x 2  sin ),
f (x 1 )=exp(& 12 x 21 cos ),
,(x)= f (x 1 ) g(x 2 ).
From Lemma 2.3(3) we see that , # W 1, 2 (R 2+ ),

R 2+
x 1 (x 2 +z 0 ) , 2 dx=0.
We compute

R2+
[ {. 2 +[x 1 cos & sin (x 2  sin +z 0 )] 2 . 2 ] dx =[cos +; 0 sin ]

. 2dx.
R2+
Hence, b()cos +; 0 sin . Since this test function . does not satisfies (3.1), we must have a strict inequality. K Proof of Theorem 3.1. We only need to show (3.3), because other conclusions follows from the above lemmas. In fact, the conclusion (1) was proved in Lemma 3.2. (2) was proved in Lemmas 3.5 and 3.6. From (3.3) and Lemma 3.7 we see that for in (0, ?2), b() is strictly decreasing and ; 0
416
LU AND PAN
From Lemma 3.8 we see that lim sup Ä ?2 b(); 0 . Denote b =lim inf b(). * Ä ?2 Then b ; 0 . We shall show that b ; 0 . * * Let j be any sequence such that j Ä ?2 and b( j ) Ä b . For simplicity * we denote j by . Recall that the eigenfunction v of (3.1) with b=b() satisfies 0
Denote z =x 1 cos &x 2 sin . As in Step 2 in the proof of Lemma 3.4, we can show that z is bounded as Ä ?2. Thus, passing to a subsequence if necessary, we may assume that z Ä z as Ä ?2. Set w ( y)=v (x + y). Then w (0)=1. By the elliptic estimates we have, after passing to a subsequence, w Ä w 1 in C 2loc as Ä ?2, and w 1 satisfies &2w 1 +( y 2 +z) 2 w 1 =bw 1
(3.25)
with b=b ; 0 . * We claim that x 2 is bounded as Ä ?2. Suppose x 2 Ä +. In this case w 1 satisfies (3.25) on R 2. But (3.25) has a nontrivial bounded solution in R 2+ only if b1. This contradiction shows that x 2 is bounded. So, we may assume x 2 Ä a0. Set w^( y)=w 1 ( y 1 , y 2 &a). Then w^ satisfies
{
&2w^ +( y 2 +z&a) 2 w^ =bw^ on R 2+ 2 w^ =0
in R 2+ ,
with b=; ; 0 . However, from Lemma 3.2, this Neumann problem has * a nontrivial bounded solution only if b;(z&a). Hence, we have ; 0 b 0 ;(z&a). Since ;(z) achieves the unique minimum ; 0 at z=z 0 , we have z&a=z 0 and b =; 0 . Hence lim Ä ?2 b()=; 0 . *
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
417
Step 2. Next we show that lim Ä 0 b()=1. Since b()<1 for all 0<
We shall show that b 0 1. Suppose not. Then b 0 <1. Let j be any sequence such that j Ä 0 and b( j ) Ä b 0 . For simplicity we denote j by . Denote z =x 1 cos &x 2 sin as above. Again as in Step 2 in the proof of Lemma 3.4, we can show that z is bounded as Ä 0. Thus, passing to a subsequence if necessary, we may assume that z Ä z as Ä 0. Set w ( y)=v (x + y). Then, w (0)=1. By the elliptic estimates we have, after passing to a subsequence, w Ä w 2 in C 2loc as Ä ?2, and w 2 satisfies &2w 2 +( y 1 +z) 2w 2 =bw 2
(3.26)
with b=b 0 and we suppose b 0 <1. Case 1. x 2 Ä + as Ä 0. Then (3.26) holds in the entire plane R 2+ . However, from the proof of Lemma 3.2 we see that (3.26) has a nontrivial bounded solution only if b1. This contradicts the assumption b 0 <1. Case 2. x 2 remains bounded as Ä 0. We may assume x 2 Ä a as Ä 0. Set w~( y)=w 2 ( y 1 , y 2 &a). Then w~ satisfies
{
&2w~ +( y 1 +z&a) 2 w~ =bw~ 2 w~ =0 on R 2+
in R 2+ ,
with b=b 0 >1. From Lemma 3.2 we see that this problem has a nontrivial bounded solution only if b1. Again we have a contradiction. Therefore we must have b 0 =1. This completes Step 2 and finishes the proof of Theorem 3.1. K
4. EIGENVALUE PROBLEMS IN R 3 AND IN R 3+ In this section, we study the eigenvalue problems involving the GinzburgLandau operator &{ 20 in R 3 and in R 3+ , where 0(x)= 12 h_X= 12 (h 2 x 3 &h 3 x 2 , h 3 x 1 &h 1 x 3 , h 1 x 2 &h 2 x 1 ).
(4.1)
Here h=(h 1 , h 2 , h 3 ) is a unit vector in R 3, X=(x 1 , x 2 , x 3 ) is the position vector. Note that the vector field given in (4.1) satisfies curl 0=h.
418
LU AND PAN
First, we consider the eigenvalue problem in R 3 &{ 20 =:
in R 3.
(4.2)
Note that (4.2) can also be written as &2+i(h_X) {+ 14 h_X 2 =:
in R 3.
Define
:(h)=
inf
# W(R 3 )

R3


{ 0  2 dx =  2 dx
inf
# W(R 3 )
R3
R3
}
i {& h_X 2

2
} dx ,
(4.3)
 2 dx
R3
2 3 2 3 where W(R 3 )=W 1, loc (R ) & L (R ). Obviously, :( \h)=  \ :(h). We shall see that :(h) does not depend on the direction of h. In fact, for any unit vector h we always have :(h)=1. We shall also see that :(h) is not achieved in W(R 3 ). Instead, there are infinitely many bounded eigenfunctions which essentially depend on 2 variables.
Theorem 4.1. For any constant unit vector h, :(h)=1 and is not achieved in W(R 3 ). There are infinitely many linearly independent bounded eigenfunctions associated with the eigenvalue 1, and they are constant along the direction of h. For every :1, (4.2) has no eigenfunction in L 2 (R 3 ). Instead, it has infinitely many linearly independent bounded eigenfunctions which, in the new variables, are given by (4.9) below. We shall also study a related eigenvalue problem on the half space R 3+ &{ 20 =*
in R 3+ ,
on R 3+ ,
({ 0 ) } &=0
(4.4)
where 0 is given in (4.1) and & is the outer normal vector to R 3+ . Similar to (4.3) we define *(h)=
inf
# W(R 3+ )
R 3+ { 0  2 dx R3+  2 dx
,
(4.5)
2 3 2 3 where W(R 3+ )=W 1, loc (R + ) & L (R + ). We shall see that ; 0 *(h)1 and *(h) depends on the direction of h. *(h)=1 if h is perpendicular to the surface R 3+ and *(h)=; 0 if h is parallel to R 3+ . We shall also see that when h is either parallel to or orthogonal to R 3+ then *(h) does not achieve in W(R 3+ ). Instead, there are bounded eigenfunctions.
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
419
Theorem 4.2. Let h be a constant unit vector, and let be the angle between h and the unit out normal of R 3. Then we have the following conclusions. (1)
Let b() be the first eigenvalue of (3.1). We have 1
if
=0 or ?,
b()
if
? 0<< , 2
;0
if
? = , 2
b(?&)
if
? <
*(h)=b()=
Therefore, *(h) is decreasing in for # [0, ?2]. (2) *(h) is achieved in L 2 (R 3+ ) if and only if h is neither perpendicular nor parallel to the surface R 3+ . In this case, (4.4) has infinitely many linearly independent L 2 eigenfunctions for *=*(h). (3) Moreover, in any case, for *=*(h), Eq. (4.4) has bounded eigenfunctions which are not in L 2 (R 3+ ). We first consider Eq. (4.2). Proof of Theorem 4.1. Let 6 be the plane determined by the equation h } X=0. Choose unit vectors e 1 and e 2 in 6 such that e 1 , e 2 and h form a righthand system. Introduce new variables y 1 , y 2 , z such that X= y 1 e 1 + y 2 e 2 +zh. Denote y=( y 1 , y 2 ) and set (x)=.( y 1 , y 2 , z). Then, Eq. (4.2) can be written as &2 y .& zz .+2i } { y .+ 14  y 2.=:.
in R 3,
(4.6)
where ( y)= 12 (&y 2 , y 1 ). :(h) can be written as :(h)=
inf
, # W(R 3 )
R3 [ {  , 2 +  z , 2 ] dx , R 3 , 2 dx
(4.7)
where {  ,={ y ,&i,, and { y =( y1 , y2 ). Note that if . only depends on y, then (4.6) is reduced to an equation in R 2 &2.+2i } {.+ 14  y 2.=:.
in R 2.
(4.8)
420
LU AND PAN
Claim 1. For any unit vector h we have :(h)=1. Proof of Claim 1. It was proved in [LP2] that the first eigenvalue of (4.8) is 1, and the set 8 1 of the bounded eigenfunctions for :=1 consists of infinitely many linearly independent functions in L 2 (R 2 ) and infinitely many linearly independent bounded functions which are not in L 2 (R 2 ). All the L 2 eigenfunctions in 8 1 can be written as
\
,( y)=exp &
 y 2 4
+ f ( y),
where f ( y) is any entire function such that exp( &  y 24) f ( y) # L 2 (R 2 ). Using this fact we easily show that :(h)1. In the new variables ( y, z), choose (x)=' exp(& y 24) as a test function, where ' is a smooth cutoff function with compact support. Then we can show that :(h)1. Hence we have :(h)=1. K For any , # 8 1 , let .( y, z)=,( y). Then . is a bounded eigenfunction of (4.6) for :=1. In the following we show that all the bounded eigenfunctions of (4.6) with :=1 must be in this form. Claim 2. When :=1, Eq. (4.6) has no nontrivial solution in L 2 (R 3 ). Therefore :(h) is not achieved in W(R 3 ). Proof of Claim 2. Let . # L 2 (R 3 ) be a nontrivial solution of Eq. (4.6) for :=1. Multiplying (4.6) by . and integrating, we have

R3
[ {  . 2 +  z . 2 ] dx=

. 2 dx.
R3
Since (see [LP2])

R2
{  .( y, z) 2 dy

.( y, z) 2 dy
for all z # R 1,
R2
we get

R3
{  . 2 dy dz

. 2 dy dz,
R3
so

R3
 z . 2 dx=0.
Note that . is a smooth function. So we have .( y, z)=.( y), and . # L 2 (R 3 ) only if .#0. K
421
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
Claim 3. All the bounded solutions of Eq. (4.6) with :=1 are in the form .( y, z)=,( y), where , is an arbitrary eigenfunction of (4.8) for :=1. Proof of Claim 3. Let . be a bounded solution of (4.6) with :=1. We shall show that z . # L 2 (R 3 ). Let ' be a smooth cutoff function with compact support. Multiplying (4.6) by ' 2. and integrating we get

R3
[ {  ('.) 2 +  z ('.) 2 & '. 2 ]dx=

{' 2 . 2 dx. R3
Since

{  ('.) 2 dx

 z . 2 dx
R3

'. 2 dx,
R3
we get
R3

R3
{' 2 . 2 dx.
Choose
{
'(x)='(r)=1 if rn, 0 if rn+m,
and '$(r) 2m for n r n+m, where r= x. Then we get

 z . 2 dx
Bn

{' 2 . 2 dx
R3
4 m2

. 2 dx [nrn+m]
2n 1 4 &.& 2L[B n+m  & B n  ]=4? &.& 2L 1+ + 2 . m2 m m
_
&
First we fix n and send m to , then we send n to . So we get

R3
 z . 2 dx4? &.& 2L .
Therefore != z . # L 2 (R 3 ). Note that != z . is also a solution of (4.6). Now we use Claim 2 to conclude that != z .#0. That is, .( y, z)=,( y). Plugging .( y, z)=,( y) to (4.6) we see that ,( y) satisfies (4.8), i.e. , # 81 . K
422
LU AND PAN
Now we show that every :1 is an eigenvalue of (4.6). It has been proved in [LP2] that the eigenvalues of (4.8) are 2n&1, n=1, 3, 5, ..., and the set 8 n of bounded eigenfunctions for :=2n&1 consists of infinitely many linearly independent functions in L 2 (R 2 ) and infinitely many linearly independent bounded functions which are not in L 2 (R 2 ). For every :1 denote by [(:+1)2], the integer part of (:+1)2, and denote n=1, 2, ..., [ 12 (:+1)].
t n = :&(2n&1),
By the method of separation variables, it is easy to see that for every :1, (4.6) has solutions given by [12(:+1)]
.( y 1 , y 2 , z)=
:
(c n e itnz +c &n e &itnz ) , n ( y)
(4.9)
n=1
where c n and c &n are arbitrary complex constants, and , n # 8 n . Note that any nonzero function . given by (4.9) does not belong to L 2 (R 3 ). Claim 4. Every :1 is an eigenvalue of (4.6), and none of the associated bounded eigenfunctions belongs to L 2 (R 3 ). Proof of Claim 4. Let . # L 2 (R 3 ) be a nontrivial solution of (4.6). By the elliptic regularity we know that . is smooth. For any fixed y # R 2, let .~( y, t) be the Fourier transform of . in the variable z. Then, for fixed t # R 1, .~( y, t) is a solution in y of the following equation &2 y .~ +2i } { y .~ + 14  y 2.~ =(:&t 2 ) .~
in R 2.
(4.10)
By the facts mentioned in the above, if .~( } , t)0 then :&t 2 =2n&1, n=1, 2, } } } . Hence, :1 and spt(.~ )/R 2_[ \t n : n=1, 2, ..., [ 12 (:+1)]]. So, we can write [12(:+1)]
.~( y, t)=
:
Kn
, n ( y) : [c n, k k$(t&t n )+c &n, k k$(t+t n )] k=0
n=1
where , n is a solution of (4.8) for :=2n&1. Hence, [12(:+1)]
.( y 1 , y 2 , z)=
: n=1
Kn
, n ( y) : k=0
(&iz) k  2?
[c n, k e itn z +c &n, k e &itn z ].
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
423
Since . is bounded, we must have , n # 8 n and c n, k =c &n, k =0 for k>0. Therefore . Â L 2 (R 3 ). This contradiction shows that (4.6) has no eigenfunction in L 2 (R 3 ). K Now Theorem 4.1 is complete. K Remark 4.1. We believe that for any :1, in the ( y, z) coordinates, all the bounded eigenfunctions of (4.2) must be in the form of (4.9). Also note that, the bounded eigenfunctions of (4.2) given in (4.9) are periodic in the h direction. Now, we consider (4.4). For the sake of convenience, we choose the coordinates such that R 3+ =[(x 1 , x 2 , x 3 ): x 2 >0]. Denote /= 12 [h 2 x 1 x 3 +h 3 x 1 x 2 &h 1 x 2 x 3 ],
B=(&h 3 x 2 , 0, h 1 x 2 &h 2 x 1 ),
0=B+{/,
=e i/.,
Then, (4.4) is reduced to
{
&2.+2iB } {.+ B 2.=*. 2 .=0 on R 3+ .
in R 3+ ,
(4.11)
Note that the coefficients in (4.11) are independent of x 3 . Let h$= (h 1 , 0, h 3 ) be the projection of h in the x 3 x 1 plane. Now we rotate the coordinates in the x 3 x 1 plane such that the new x 1 axis is coincident with h$. After this type of changing variables the equation (4.11) is invariant. So in the following we may always assume h 3 =0 without loss of generality. Thus (4.4) is equivalent to
{
&2.&2i(h 2 x 1 &h 1 x 2 ) 3 .+(h 2 x 1 &h 1 x 2 ) 2 .=*. 2 .=0 on R 3+ .
in R 3+ ,
(4.12)
Also note that *(&h)=*(h), and if . is a solution for h, then its complex conjugate . is a solution for &h. Hence, we can assume h 1 0. Furthermore, by changing x 1 to &x 1 , we can also assume h 2 0. Therefore, in the following we always assume that h=(h 1 , h 2 , 0),
h 1 0,
h 2 0,
h 21 +h 22 =1.
(4.13)
Lemma 4.3. Assume that h is perpendicular to the surface R 3+ . Then the first eigenvalue *(h) of (4.4) is 1 and the bounded eigenfunctions comprise an
424
LU AND PAN
infinite dimensional subspace of L (R 3+ ). Moreover, every bounded eigenfunction is constant along the h direction. Proof.
When h 1 =0 we have h 2 =1, and (4.12) is reduced to
{
&2.&2ix 1 3 .+x 21 .=*. 2 .=0 on R 3+ .
in R 3+ ,
(4.14)
Let . be a bounded eigenfunction of (4.14) and extend it evenly in x 2 . Then . satisfies &2.&2ix 1 3 .+x 21 .=*.
in R 3.
Let ,(x)=exp(ix 1 x 3 2) .. Then , satisfies &2,+2i } { y ,+ 14  y 2 ,=*.
in R 3,
(4.15)
where y=(x 3 , x 1 ), ( y)=(&x 1 2, x 3 2). Comparing (4.15) with (4.6) we see that the first eigenvalue of (4.15) is *=1, and the eigenspace 8 1 is of infinite dimension which consists of functions of the form ,(x 3 , x 1 ). Hence when h is perpendicular to R 3+ , *(h)=1 and the bounded eigenfunctions are given by
\
.(x)=exp &
ix 1 x 3 , 1 (x 3 , x 1 ), 2
+
Now Lemma 4.3 is complete.
for any , 1 (x 3 , x 1 ) # 8 1 .
K
Lemma 4.4. Assume that h is parallel to the surface R 3+ . Then the first eigenvalue of (4.4) is ; 0 , where ; 0 is given in Lemma 2.3. There is only one linearly independent eigenfunction .. .(x) is constant along the h direction, and .(x) depends only on the distance between x and R 3+ . Proof. When h is parallel to the surface R 3+ , we may choose h= (1, 0, 0). Let !(x)=.(x 1 , x 2 , &x 3 ). Then from (4.12), ! satisfies
{
&2!&2ix 2 3 !+x 22 !=*! on R 3+ . 2 !=0
in R 3+ ,
(4.16)
When ! does not depend on x 1 , we denote !(x)=,(x 3 , x 2 ), (4.16) is reduced to a problem in the half plane R 2+ =[(x 3 , x 2 ) : x 2 <0],
{
&2,&2ix 2 3 ,+x 22 ,=*, on R 2+ . 2 ,=0
in R 2+ ,
(4.17)
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
425
From [LP2], the first eigenvalue of (4.17) is ; 0 , and the only eigenfunctions are ,=c exp(iz 0 x 3 ) u(x 2 ),
(4.18)
where z 0 >0 is the unique minimum point of ;(z), and u(t)=u z0 (t) is the positive eigenfunction of (2.4) for z=z 0 and ;=; 0 . Choose ' exp(iz 0 x 3 ) u(x 2 ) as a test function, where ' is a cutoff function with compact support, we can show that *(h)=; 0 . The functions in (4.18) are bounded eigenfunctions of (4.16) for *=; 0 . In the following we show that, all the eigenfunctions of (4.16) with *=; 0 must be in the form of (4.18). Denote y=(x 3 , x 2 ), z=x 1 . Then we write x=( y, z), where y # R 2+ and z # R 1. Write E=E( y)=(& y 2 , y 1 ), { E ,={ y ,&iE,. Then we can write (4.16) as
{
&{ 2E !& zz !=*! in R 3+ , on R 3+ . 2 !=0
(4.19)
Claim 1. When *; 0 , (4.19) has no nontrivial solution in L 2 (R 3+ ). Proof of Claim 1.
Otherwise, from (4.19) we have

[ { E ! 2 +  z ! 2 ] dx=*
R3+

! 2 dx. R3+
Since *; 0 and for every z

{ E !( y, z) 2 dy; 0
R 2+

!( y, z) 2 dy,
R 2+
we must have R3 +  z ! 2 dx=0. Hence !( y, z)=!( y). So ! Â L 2 (R 3+ ) if !0. K Now assume ! is a bounded solution of (4.19) with *=; 0 . Let '='(r) be a radial cutoff function, here r= x. Multiplying (4.19) by ' 2! and integrating we get

R3+
[{ E ('!) 2 +  z ('!) 2 &; 0 '! 2 ]dx=

R3+
'$(r) 2 ! 2 dx.
Since for every z

R 2+
{ E ('!) 2 dy; 0

'! 2 dy, R 2+
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LU AND PAN
we get

R3+
 z ('!) 2 dx

'$(r) 2 ! 2 dx.
R3+
Now as in the Claim 2 in proof of Theorem 4.1 we see that z ! # L 2 (R 3+ ). Note that z ! is also a solution of (4.19) with *=; 0 . From Claim 1 we see that z !=0. Hence !(x)=,( y), where y=(x 3 , x 2 ). Then from [LP2] we see that ,( y) must be in the form of (4.18). Now we have proved that all the eigenfunctions of (4.16) with *=; 0 must be in the form of (4.18). So we see that, when h=(1, 0, 0), the first eigenvalue of (4.12) is ; 0 , and the eigenfunctions are .(x)=c exp( &iz 0 x 3 ) u(x 2 ). Now Lemma 4.4 is proved. K Lemma 4.5. Given a unit vector h, denote by the angle between h and the outer normal direction of the surface R 3+ . Let lie in (0, ?2). Then, *(h)=b(), where b() was given in Theorem 3.1. For *=*(h), (4.4) has both infinitely many linearly independent eigenfunctions in L 2 (R 3+ ) and infinitely many linearly independent bounded eigenfunctions which are not in L 2 (R 3+ ). Proof. First we consider L 2 solutions of (4.12). If . is an L 2 solution of (4.12), for fixed x 1 and x 2 we let .~(x 1 , x 2 , y 3 )=F(x 1 , x 2 , } ) be the Fourier transform of . in the variable x 3 . Then for fixed y 3 , .~ satisfies the following equation in x 1 , x 2
{
&2.~ +(h 2 x 1 &h 1 x 2 + y 3 ) 2 .~ =*.~ on R 2+ . 2 .~ =0
in R 2+ ,
(4.20)
In (4.20), let h 1 =sin and h 2 =cos , where 0<
\
.~(x 1 , x 2 , y 3 )= f ( y 3 ) v x 1 +
y3 , x2 cos
+
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
427
for some f # L 2 (R 1 ). Hence, *(h) is achieved in L 2 (R 3+ ), and all the eigenfunctions in L 2 (R 3+ ) are given by
_
\
.(x 1 , x 2 , x 3 )=F &1 f ( y 3 ) v x 1
y3 ,x cos 2
+& ,
f # L 2 (R 1 ),
(4.21)
where the inverse Fourier transform is taken in the y 3 variable. Thus we have shown that (4.12) has infinitely many eigenfunctions in L 2 (R 2+ ) when *=*(h). It is easy to verify that for any sequence [z j ] and [c j ], where z j 's are real numbers and c j are complex numbers such that 0< j=1 c j  >,
.(x 1 , x 2 , x 3 )= : c j exp(ih 2 z j x 3 ) v (x 1 +z j , x 2 )
(4.22)
j=1
is a bounded eigenfunction of (4.12) for *=*(h) which is not in L 2 (R 3+ ). The proof is completed. K Proof of Theorem 4.2. Summarizing Lemmas 4.3, 4.4, 4.5, and using Theorem 3.1, we complete the proof of Theorem 4.2. K
5. EIGENVALUE PROBLEMS IN BOUNDED DOMAINS Let 0 be a smooth bounded domain in R 3. Given a vector field A, denote by +=+(A) the first eigenvalue of the following problem
{
&{ 2A =+ in 0, on 0. ({ A ) } &+#=0
(5.1)
Then, +(A)=
inf
# W 1, 2(0)
0 { A  2 dx+# 0  2 ds . 0  2 dx
(5.2)
In this section we shall estimate the value of +(_A) for large _. By the gauge invariance of the GinzburgLandau operator we see that +(A+{/) =+(A) for any smooth function /. Making a gauge transformation if necessary, we can always assume div A=0 in 0,
A } &=0 on 0.
Denote H=curl A. For x # 0 we denote by (x) the angle between H(x) and the outer normal &(x) of 0. The main result in this section is the following
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LU AND PAN
Theorem 5.1. For any smooth bounded domain 0 in R 3 and A # C 2 (0 ) lim
_Ä
+(_A) =min min curl A(x), min curl A(x) b((x)) . _ x#0 x # 0
{
=
(5.3)
where b() was defined in Theorem 4.2. Remark 5.1. A similar result for the 2dimensional problem has been proved in [LP3], where the asymptotic estimate depends on the distribution of minimum points of curl A. While in the 3dimensional case, the estimate depends on not only the distribution of minimum points of H but also the direction of H on 0. As a consequence, we have Corollary 5.2. then
If there is a point x 0 # 0 such that H(x 0 ) } &(x 0 )=0,
lim
_Ä
+(_A) ; 0 H(x 0 ). _
Especially, if H(x)#h, a constant unit vector, then for any smooth bounded domain 0 we have lim
_Ä
+(_A) =; 0 , _
(5.4)
where ; 0 is the minimum value of ;(z) given in Lemma 2.3. Remark 5.2. When H#h, for any smooth bounded domain 0 there is a point on 0 such that h is tangent to 0 at this point. So, (5.4) immediately follows from (5.3). (5.4) indicates that in this case the leading term of +(_A) when _ is large does not depend on the direction of h. However, for bounded nonsmooth domains such as the domains with edges, or unbounded domains, this is not the case. The proof of Theorem 5.1 follows along the same lines as in [LP3]. As in [LP3], the decomposition of vector fields plays an important role. Let 0 be a smooth bounded domain in R 3 and A=(A 1 , A 2 , A 3 ) # 2 C (0 ). We shall decompose A near a point x 0 . Without loss of generality, we may assume x 0 =0. Denote by E the 3_3 matrix (= ij ), where = ij = ( i A j (0)+ j A i (0))2. Denote by X the vector (x 1 , x 2 , x 3 ) corresponding to the point x. Then, by the Taylor expansion at 0, we have
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
429
Lemma 5.3. Let 0 be a smooth domain, 0 # 0 and A # C 2 (0). Then, we have 1 A(x)= curl A(0)_X+{/+D, 2 1 1 /(x)=A(0) } X+ (EX, X)=A(0) } X+ 2 2
3
(5.5)
: = ij x i x j . i, j=1
where D(x) C(R)x 2 in the ball B R . Next, we consider the case of boundary points. We assume that 0 # 0 and express a portion of 0 around 0 by r=r( y 1 , y 2 ), r(0, 0)=0. Denote r i = i r, n=r 1 _r 2 r 1 _r 2  and g ij =r i } r j . We choose ( y 1 , y 2 ) as the normal coordinates of the portion of surface and n to be the inward normal vector of 0. Thus, g ij (0, 0)=$ ij . Denote e 1 =r 1 (0, 0), e 2 =r 2 (0, 0) and e 3 =n(0, 0). We can rotate the original coordinate system such that it is coincident with (e 1 , e 2 , e 3 ). So, X=x 1 e 1 +x 2 e 2 +x 3 e 3 . Define a map X=F( y)=r( y 1 , y 2 )+ y 3 n( y 1 , y 2 ).
(5.6)
F is a diffeomorphism in a small neighborhood of 0 and X=F( y)=Y+ O(Y 2 ) near 0, where Y= y 1 e 1 + y 2 e 2 + y 3 e 3 . In the following we shall call the map F a diffeomorphism straightening a portion of boundary around 0. For a smooth vector field A(x), we define A ( y)=A(F( y)).
(5.7)
Then, we have Lemma 5.4. Assume that 0 is a smooth bounded domain in R 3, 0 # 0, and A # C 2 (0 ). Let F( y) be the diffeomorphism straightening a portion of boundary 0 around 0 and define A ( y) by (5.7). Then, in the neighborhood of 0 we have 1 A (x)=A(F( y))= curl A(0)_Y+{/~ +D, 2 1 1 /~( y)=A(0) } Y+ (EY, Y)=A(0) } Y+ 2 2
3
(5.8)
: = ij y i y j , i, j=1
where {/~ is the gradient of /~ and D ( y)C(R) y 2 on the half ball B + R . The proof of this lemma again follows from the Taylor expansion. We omit it.
430
LU AND PAN
Proof of Theorem 5.1. As long as we have established the above decompositions of vector fields and the results in Section 4, the proof of Theorem 5.1 follows along the same lines as in [LP3]. Hence we only give very short outline here and refer interested readers to [LP3] for reference. To prove the lower bound for _ Ä , let _ be the eigenfunction such that max _  =  _ (x _ ) =1. After passing to a subsequence, we may assume x _ Ä x 0. After rescaling the limiting equation is (4.2) if x 0 # 0 and is (4.4) if x 0 # 0. In either case the vector field involved has constant curl equal to H(x 0 ). Using Theorems 4.1 and 4.2 we get the lower bound. The upper bound follows from computations by rescaling and modifying the eigenfunctions of (4.2) or (4.4). K
6. NUCLEATION Throughout this section we always assume that the condition (1.5) holds true. In the following we consider any two sequences [}] and [_] such that }, _ Ä + and _<_*(}, H 0 ),
} =a, _
lim
} Ä +
where 0a: 0 (H 0 ).
We shall show aa 0 . Then, Theorem 1 follows. For simplicity we set ==1 _}. Then, }2
a+o(1) . =2
Rewrite the functional E by E= E= (, A)=

0
+
{ 
{ (1=2 ) A  2 +
1 }2 curl A&H 0  2 + (  2 &1) 2 dx 4 = 2
=
#  2 ds,
0
and set E(=)=inf (, A) # W E= (, A). Denote the minimizers by ( =, A = ). Then, ( =, A = ) satisfy
{
&{ 2(1=2 ) A =} 2 (1&  2 ) , curl 2 (A&F)== 2I({ (1=2 ) A )
in 0
(6.1)
and ({ (1=2 ) A ) } &+#=0,
curl (A&F)_&=0
on 0.
(6.2)
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
431
We call ( =, A = ) the minimal solution of (6.1) and (6.2). Due to the gauge invariance of the GinzburgLandau system, we may always assume that div A = =0 in 0,
A = } &=0
on 0.
As in [LP4] Section 3 we have Proposition 6.1 (Global Estimates). Let ( =, A = ) be the minimal solutions of system (6.1) and (6.2). Then, we have & { (1=2 ) A= =  & W 1, 2(0)
C . =2
For any 1
&A =& C1, :(0) C(:),
(6.3)
where the constants C, C( p), and C(:) are independent of =. Next, we consider nonhomogeneous field H(x)=_H 0 (x) and prove that the order parameter concentrates at the minimum points of H 0 (x). Let : 0 =: 0 (H 0 ) be defined by (1.6). More precise information can be obtained in the following two cases (I)
: 0 =min H 0 (x) > min H 0 (x) b((x)), x#0
(B)
x # 0
: 0 = min H 0 (x) b((x))>min H 0 (x). x # 0
x#0
We shall prove that if (I) holds then the interior nucleation occurs; if (B) holds then the boundary nucleation occurs and the superconducting layer is located in (0) m with peaks lying near the boundary within a distance o(=). Let x = # 0 be the maximum point of  =  and denote * = =& =& L(0) =  = (x = ). After passing to a subsequence, we may assume lim = Ä 0 x = =x 0. Denote \= H 0 (x 0 ) and h=H 0 (x 0 )\. Define i . = ( y)=exp & A = (x = ) } y = (x = +=y). =
\
+
432
LU AND PAN
Using Proposition 6.1 and the rescaling arguments in [LP4] we obtain Theorem 6.2 (Concentration of Order Parameter). holds. Then as = Ä 0, curl A = Ä H 0 & =& L(0) Ä 0.
Assume that (1.5)
in C : (0),
On the set 0"(0 m _ (0) m ) it holds that = (x) Ä 0. & =& L(0)
(6.4)
Moreover, If (I) holds, then x 0 # 0 m , H 0 (x 0 ) =min 0 H 0 (x), and (6.4) holds on 0"0 m . After passing to a subsequence, .= Ä ,0 & & L(0) =
: in C 2, loc ,
(6.5)
where , 0 ( y \) is an eigenfunction of (4.2). If (B) holds, then x 0 # (0) m , dist(x =, 0)=o(=), H 0 (x 0 ) =min 0 H 0 (x), and (6.4) holds on 0"(0) m . After straightening a portion of boundary around x 0 , . = & =& L(0) converges to , 0 , here , 0 ( y \) is an eigenfunction of (4.4) for *=b((x 0 )). Remark 6.1. If min x # 0 H 0 (x) =min x # 0 H 0 (x) b((x)), then both interior and boundary nucleations may happen. Proof of Theorem 1 and Theorem 2. It follows from Theorem 6.2 and the details are omitted here. We refer interested readers to [LP4] for reference.
APPENDIX: ESTIMATES OF H C3 In this section we give estimates for H C3 . Assume the superconducting material occupies a bounded smooth domain 0 in R 3. Let the applied magnetic field be H=_h, where h is a unit constant vector. Assume h is tangential to 0 at P 0 . Denote e 1 =(1, 0, 0), e 2 =(0, 1, 0), e 3 =(0, 0, 1). After rotating the coordinate system we may assume P 0 is the origin and h=e 1 . A portion U of the surface 0 around the origin can be represented as r=r(y), here and after we always denote y=( y 1 , y 2 ). Denote r j = j r( y), r ij = ij r( y), etc. Let n=r 1 _r 2 r 1 _r 2 . We may choose ( y 1 , y 2 ) such that
433
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
n is the inward normal of 0, and n(0, 0)=e 3 . Denote g ij ( y)=r i ( y) } r j ( y). Denote the elements of the inverse of the matrix (g ij ) by g ij and denote g=det( g ij ). We may choose the isothermal coordinates ( y 1 , y 2 ) such that r 1 (0, 0)=h=e 1 , r 2 (0, 0)=e 2 , and g 12 #0,
g ij (0, 0)=$ ij ,
g ij, k (0, 0)# k g ij (0, 0)=0
for all i, j, k. (A.1)
Here $ ij is the Kronecker symbol. Then g ii =1g ii and g= g 11 g 22 . Denote by 0 ij the coefficients of the second fundamental form of 0. Denote by } 1 and } 2 the principal curvatures. In the isothermal coordinates, if y 1  and y 2 curves are the lines of curvature, then 0 11 =} 1 , 0 22 =} 2 , and 0 12 =0. To state our estimates for H C3 (}, h) we need to introduce several notations. First we introduce a function Q(P) on 0 Q(P)=
1 2 54 3
{
_
0 211, 1 + 0 12, 2 +
0 11, 1 +0 22, 1 4
2
& _
+ 0 12, 1 +
0 11, 2 4
2
&=
14
.
(A.2) If y 1  and y 2 curves are lines of curvature, then Q=
1 2 54 3
{
1  1 } 1  2 +  1 H r  2 +  2 } 1  2 4
=
14
,
here H r is the mean curvature, and Q(P 0 )=0 if and only if {} 1  +  1 } 2  =0 at P 0 . For a cylinder, if y 1 curve is in the direction of axis of the cylinder, then Q(P)#0 on the surface. So in some sense Q(P 0 ) measures locally the noncylindrical property of the surface near P 0 . Let u be the positive eigenfunction of (2.4) for z=z 0 and ;=; 0 . Denote &u& 22 =

+
u 2 (z) dz
0
and C1=
1 &u& 22
{
u 2 (0) & 2
C2=
u 2 (0) , 2 &u& 22
C3=
1 2 &u& 22

+
=
(z+z 0 ) 3 u 2 dz ,
0
(A.3)

+
(z+z 0 ) 2 u 2 dz.
0
Note that C 1 >0, see Lemma 2.3(4).
434
LU AND PAN
Denote E1 =
inf ` # C 02 [0, 1)
E2 =
inf 2
` # C 0 [0, 1)
10 [r`$(r) 2 +r 5` 2 (r)] dr , 10 r` 2 (r) dr
(A.4)
10 [r`$(r) 2 +r 7` 2 (r)] dr , 10 r` 2 (r) dr
where C 20[0, 1)=[` # C 2[0, 2]: `(1)=0]. Let ` j be the minimizer of E j and denote q k ( j)=2?

1
t k+1` j (t) 2 dt.
(A.5)
0
In the following propositions 0 ij etc. take the values at P 0 . Proposition A.1. Assume h is a unit vector which is tangential to 0 at P 0 . Assume 0 211 +0 212 {0 at P 0 . Then for } large H C3 (}, h)
} E1 & (30 211 +40 212 ) 13 } 13 ; 0 2; 53 0
+
1 (C 2 0 11 +C 1 0 22 &#)+O(} &13 ). ; 32 0
Proposition A.2. Assume h is a unit vector which is tangential to the line of curvature of 0 at P 0 . Let } 1 be the principal curvature of 0 in the direction of h. (1)
Assume } 1 {0 at P 0 , then
H C3 (}, h) (2)
} E1 & ;0 2
3} 21 ; 50
\ +
13
} 13 +
1 (C 2 } 1 +C 1 } 2 &#)+O(} &13 ). ; 32 0
Assume at P 0 we have } 1 =0, but {} 1  +  1 } 2  {0. Then
H C3 (}, h)
} 1 + ; 0 ; 32 0
_
C1 }2+
C 3 q 2 (2) } 22 &2Q(P 0 ) E 2 &# +O(} &12 ). Q(P 0 ) q 0 (2)
&
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
435
(3) Assume at P 0 we have } 1 =0, {} 1 =0 and 1 } 2 =0. Then when } 2 {0, for any m>0 fixed and } large we have H C3 (}, h)}; 0 +
C 3 q 2 (1) } 22 C 1 } 2 &# (log }) 2m + +O((log }) &2m ); 32 ; 0 q 0 (1) ; 32 0
and if } 2 =0 H C3 (}, h)
} # & +O(} &19 ). ; 0 ; 32 0
Remark A.1. Note that in the estimates, the leading term does not depend on the direction of the applied field, but the other terms do. The estimates depend on the direction of h and the geometry of the surface 0 at the tangential point P 0 Assume 0 is a ball of radius R. From Proposition A.2(1)
Example 1. we have H C3 (}, h)
} E1 & ;0 2
3 R ; 50
\ +
13
2
} 13 +
1 ; 32 0
\
C 1 +C 2 &# +O(} &13 ). R
+
Example 2. Assume that a portion of 0 is a cylindrical surface S which is generated by a line L (the axis of the cylinder, which is parallel to the x 3 axis) along a closed curve 1 in x 1 x 2 plane. Let } r be the relative curvature of 1. When h is parallel to the axis L of the cylinder, we have } 1 =0, Q(P)=0. In this case we can choose P 0 to be the maximum point of the curvature } r . Denote by } r* the maximum value of } r . Then from Proposition A.2(3) we have H C3 (}, h)
2 } C 3 q 2 (1) } * C 1 } *&# r r (log }) 2m + + 32 +O((log }) &2m ) ;0 ; 0 q 0 (1) ; 32 0
for any m>0. On the other hand, if h is orthogonal to the axis of the cylinder, we can choose P 0 the minimum point of } r . Denote by } r the * minimum value of } r . Then from Proposition A.2(1) we have H C3 (}, h)
} E1 & ;0 2
3} 2r * ; 50
\ +
13
} 13 +
1 ; 32 0
(C 2 } r &#)+O(} &13 ). *
This example shows that the direction of the applied field has an important influence on the value of H C3 . Now we begin to prove the propositions.
436
LU AND PAN
Choose a vector field A such that curl A=h. By the gauge invariant property of the GinzburgLandau functional, we may assume A=(0, &x 3 , 0). We shall select a test function which is supported near the origin, and calculate the energy. For this purpose we shall choose a new coordinates ( y 1 , y 2 , z) and carry out computations in the new coordinates. In the following ( y 1 , y 2 ) are the isothermal coordinates on 0. Define a map X=F( y, z)=r( y 1 , y 2 )+zn( y 1 , y 2 ). Then F is a diffeomorphism in a small neighborhood of 0. In the following the indices i, j, k, l, m etc. run from 1 to 2, and we also take the summation convention, that is, when the summation is taken over repeated indices, the summation symbol is omitted. We use r j and n j to denote j r and j n. For scalar functions, we use f , j to indicate the partial derivative in y j . Since r 1 , r 2 are orthogonal everywhere, we have r ij =1 sij r s +0 ij n, n i =&g ss0 is r s ,
r ijk =: sijk r s +; ijk n, n ij = p kij r k +q ij n,
r ijkl =# m ijkl r m +_ ijkl n,
n ijk q=t lijk r l +{ ijk n,
where 1 kij are the Christoffel symbols, and : lijk =1 lij, k +1 sij 1 lsk & g ll0 ij 0 kl , ; ijk =0 ij, k +1 sij 0 sk , m s m mm , #m ijkl =: ijk, l +: ijk 1 sl &; ijk 0 lm g
_ ijkl =: sijk 0 ls +; ijk, l , p kij =&g kk, j0 ik & g kk0 ik, j & g ss0 is 1 ksj , q ij =&g ss0 is 0 sj , t lijk = p lij, k + p sij 1 lsk &q ij 0 kl g ll, { ijk = p sij 0 sk +q ij, k . By Tailor expansion we get the following formulas 3
r( y 1 , y 2 )= : a k e k , k=1
3
r i ( y 1 , y 2 )= : b ki e k , k=1
3
n( y 1 , y 2 )= : c k e k , k=1
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
437
where 1 1 a k = y k + 1 kij(0, 0) y i y j + : kijl(0, 0) y i y j y l 2 6 +
1 k # (0, 0) y i y j y l y m +O(  y 5 ), 4! ijlm
1 1 1 a 3 = 0 ij (0, 0) y i y j + ; ijl (0, 0) y i y j y l + _ ijlm (0, 0) y i y j y l y m +O(  y 5 ); 2 6 4! 1 1 b ki =$ ik +1 kij(0, 0) y j + : kijl(0, 0) y j y l + # kijlm(0, 0) y j y l y m +O( y 4 ), 2 6 1 1 b 3i =0 ij (0, 0) y j + ; ijl (0, 0) y j y l + _ ijlm (0, 0) y j y l y m +O(  y 4 ); 2 6 1 1 c k =&g kk (0, 0) 0 kj (0, 0) y j + p kij(0, 0) y i y j + t kijl(0, 0) y i y j y l +O(  y 4 ), 2 6 1 1 c 3 =1+ q ij (0, 0) y i y j + { ijl (0, 0) y i y j y l +O( y 4 ). 2 6 Here  y = y 21 + y 22 . Write X= 3j=1 x j e j . Recall that r j (0, 0)=e j and n(0, 0)=e 3 . We have x k =a k +c k z,
x 3 =a 3 +c 3 z.
Next we calculate the first fundamental form of F. 1 F=[1& g 110 11 z]r 1 & g 220 12 zr 2 , 2 F=&g 110 12 zr 1 +(1& g 220 22 z)r 2 , z F=n, Hence G 11 = 1 F } 1 F= g 11[1& g 110 11 z] 2 + g 22[0 12 z] 2, G 12 = 1 F } 2 F=&20 12[1&H r z] z, G 22 = 2 F } 2 F= g 11[0 12 z] 2 + g 22[1& g 220 22 z] 2, G 23 = 2 F } z F=0, G 33 = z F } z F=1,
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LU AND PAN
and G=det(G ij )= g[1&2H r z+K r z 2 ] 2. Here H r is the mean curvature of 0 and K r is the Gauss curvature. Now we see that, for the orthogonal coordinates ( y 1 , y 2 ) on the surface 0, ( y 1 , y 2 , z) are orthogonal coordinates in a neighborhood of 0 if and only if 0 12 #0, that is, if and only if the y 1  and y 2 curves are the lines of curvature of 0. Denote by G ij the elements of the inverse of the matrix (G ij ). Then G 11 =
G 22 , G
G 12 =&
G 12 , G
G 22 =
G 11 , G
G 23 =0,
G 33 =1.
For the vector field A=(0, &x 3 , 0)=&x 3 e 2 , we denote 3
A ( y, z)=A(F( y, z))= : G ija i j F, i, j=1
here 3 = z , and a j =A } j F=&(a 3 +c 3 z) d 2j ,
j=1, 2, 3.
In the following we shall carry out computations for small =. For simplicity in the following we denote g ij (0, 0), 0 ij (0, 0), 1 kij(0, 0), H r (0, 0), K r (0, 0), etc. by g ij , 0 ij , 1 kij , H r and K r etc. Note that in the isothermal coordinates we have, when ( y 1 , y 2 )=(0, 0), 1 kij =0, k ij
: lijk =1 lij, k &0 ij 0 kl ,
p =&0 ik, j ,
q ij =&0 is 0 sj ,
; ijk =0 ij, k , { ijk =&0 ks 0 si, j &0 js 0 si, k &0 is 0 sj, i .
When =>0 small, we have the following expansions: G(=y, =z)=1&4=H r z+ f 0 = 2 += 3O(  y 3 + z 3 ), G(=y, =z) 12 =1&2=H r z+ f 1 = 2 += 3O(  y 3 + z 3 ), G(=y, =z) &12 =1+2=H r z+ f &1 = 2 += 3O( y 3 + z 3 ), G 11 (=y, =z)=1&2=0 11 z+ f 11 = 2 += 3O( y 3 + z 3 ), G 12 (=y, =z)= &2=0 12 z+ f 12 = 2 += 3O(  y 3 + z 3 ), G 22 (=y, =z)=1&2=0 22 z+ f 22 = 2 += 3O( y 3 + z 3 ),  G(=y, =z) G 11 (=y, =z)=1+=[0 11 &0 22 ] z+ p 1 = 2 += 3O(  y 3 + z 3 ),  G(=y, =z) G 12 (=y, =z)=2=0 12 z+ p 2 = 2 += 3O( y 3 + z 3 ),  G(=y, =z) G 22 (=y, =z)=1+=[0 22 &0 11 ] z+ p 3 = 2 += 3O(  y 3 + z 3 ),
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
439
here f 0 = 12 [ g 11, ij + g 22, ij ] y i y j &4H r, j y j z+[4H 2r +2K r ] z 2, f 1 = 14 [ g 11, ij + g 22, ij ] y i y j &2H r, j y j z+K r z 2, f &1 =& 14 [ g 11, ij + g 22, ij ] y i y j +2H r, j y j z+[4H 2r &K r ] z 2, f 11 = 12 g 11, ij y i y j &20 11, j y j z+[0 211 +0 212 ] z 2, f 12 =&20 12, j y j z+20 12 H r z 2, f 22 = 12 g 22, ij y i y j &20 22, j y j z+[0 212 +0 222 ] z 2, p 1 = 14 [ g 22, ij & g 11, ij ] y i y j +[0 11, j &0 22, j ] y j z +[0 211 &0 11 0 22 +20 212 ] z 2, p 2 =20 12, j y j z+2H r 0 12 z 2, p 3 = 14 [ g 11, ij & g 22, ij ] y i y j +[0 22, j &0 11, j ] y j z +[0 222 &0 11 0 22 +20 212 ] z 2. Write i F=d ki e k +d 3i e 3 ,
z F=d k3 e k +d 33 e 3 .
Recall that we have chosen ( y 1 , y 2 ) to be the isothermal coordinates on 0. We have d k1 =b k1 &[ g 110 11 b k1 + g 220 12 b k2 ] z =$ 1k &[$ 1k 0 11 +$ 2k 0 12 +[$ 1k 0 11, i +$ 2k 0 12, i ] y i ]z + 12 [1 k1i, j &0 1i 0 jk ] y i y j +  y 2 O(  y + z ), d k2 =b k2 &[ g 110 12 b k1 + g 220 22 b k2 ] z =$ 2k &[$ 1k 0 12 +$ 2k 0 22 +[$ 1k 0 12, i +$ 2k 0 22, i ] y i ]z + 12 [1 k2i, j &0 2i 0 jk ] y i y j +  y 2 O(  y + z ), d 31 =b 31 &[ g 110 11 b 31 + g 220 12 b 32 ] z =0 1i y i +0 1i, j y i y j &[0 11 0 1i +0 12 0 2i ] y i z+  y 2 O( y + z ), d 32 =b 32 &[ g 110 12 b 31 + g 220 22 b 32 ] z =0 2i y i +0 2i, j y j y k &[0 12 0 1i +0 22 0 2i ] y i z+  y 2 O(  y + z ), d k3 =c k , d 33 =c 3 .
440
LU AND PAN
For the vector field A=(0, &x 3 , 0)=&x 3 e 2 , we can compute a j now. a 1 =0 12 z 2 +0 12, j y j z+ 12 [0 1i 0 2j +0 12 0 ij &1 21i, j ] y i y j z +  y 2 O( y 2 +z 2 ), a 2 =&z+0 22 z 2 & 12 0 ij y i y j +! 3 +! 4 + y 2 O( y 2 + z 2 ), a 3 =0 2j y j z+ 12 0 2k 0 ij y i y j y k +0 2i, j y i y j z+  y 3 O( y + z). Here ! 3 and ! 4 are homogeneous polynomials, ! 3 =& 16 0 ij, k y i y j y k + 12 , ij y i y j z+0 22, j y j z 2, ! 4 =& 14 [0 ij : 22lm + 16 _ ijlm ] y i y j y l y m + 16 ijk y i y j y k z+ 12 + ij y i y j z 2, and the coefficients are given by , ij =0 22 0 ij &q ij &: 22ij =0 22 0 ij +0 2i 0 2j +0 is 0 sj &1 22i, j , ijk =30 22, k 0 ij +0 22 ; ijk &{ ijk &# 22ijk =30 22, k 0 ij +0 22 0 ij, k +0 ks 0 si, j +0 js 0 si, k +0 is 0 sj, k &# 22ijk , + ij =0 12 : 21ij +0 22 : 22ij +0 22, ij + g 22 , ij 0 22 +q ij =0 12[1 21i, j &0 1i 0 2j ]+0 22[1 22i, j &0 2i 0 2j ] +0 22, ij + g 22 , ij 0 22 &0 is 0 sj . Set 1 /= 12 0 2j y j z 2 & 16 0 22 y 32 + 12 0 2i, j y i y j z 2 + 12 0 2k 0 ij y i y j y k z& 24 0 22, 2 y 42 .
(A.6)
Then we introduce a vector field A = 3i, j=1 G ija^ i j F, where a i = i /+a^ i ,
i=1, 2, 3.
Then a^ 1 = 12 0 12 z 2 +0 12, j y j z& 12 [20 12, 1 y 1 +(0 12, 2 +0 22, 1 ) y 2 ] z 2 & 12 [0 1i 0 2j &1 21i, j ] y i y j z+  y 2 O(  y 2 + z 2 ), a^ 2 =&z+ 12 0 22 z 2 & 12 0 11 y 21 &0 12 y 1 y 2 +! 3 +! 4 +  y 2 O(  y 2 + z 2 ), a^ 3 =  y 3 O(  y + z ),
441
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
where ! 3 and ! 4 are homogeneous polynomials, ! 3 =& 16
0 ij, k y i y j y k + 12 [0 1i 0 1j &1 22i, j ] y i y j z
: (i, j, k){(2, 2, 2)
+ 12 [(0 22, 1 &0 12, 2 ) y 1 +0 22, 2 y 2 ] z 2, ! 4 =& 14 [0 ij : 22lm + 16 _ ijlm ] y i y j y l y m + 16 ijk y i y j y k z+ 12 + ij y i y j z 2. (A.7) Set b i ( y, z)=
a^ i (=y, =z) , =
i=1, 2, 3.
Then b 1 ==
_
&
=2 1 0 12 z 2 +0 12, j y j z & [20 12, 1 y 1 +(0 12, 2 +0 22, 1 ) y 2 ] z 2 2 2
&
=2 [0 1i 0 2j &1 21i, j ] y i y j z+O(= 3 ) y 2 (  y 2 + z 2 ), 2
= b 2 =&z+ [0 22 z 2 &0 11 y 21 &20 12 y 1 y 2 ]+= 2! 3 += 2! 4 2 +O(= 4 ) y 2 ( y 2 + z 2 ), b 3 =O(= 3 )  y 3 O(  y + z ), Now we choose a test function (x) such that in the ( y 1 , y 2 , z) coordinates =e i/( y, z),
y z
\ =, =+ ,
,( y, z)='( y) ' 3 (z) exp(iz 0 y 2 ) u(z),
(A.8)
where u is the positive eigenfunction of (2.4) with z=z 0 and ;=; 0 . ' 3 is a smooth cutoff function such that ' 3 (z)=1 for zR 3 = and ' 3 (z)=0 for z>2R 3 = for some R 3 >0. '( y)='(r), r=  y. '(r) is a smooth function with compact support. Both ' and ' 3 depend on =. Denote Uk =

+
z ku 2 (z) dz,
0
D k =2?

+
r k+1 '$(r) 2 dr,
0
Q k =2?

+
0
r k+1 '(r) 2 dr.
442
LU AND PAN
Note that D k and Q k depend on = through '. In the following ' is chosen such that Qk<
Dk <
Recall that H C3 _ , and +( =12 A)=} 2 for ==1 }_ . Hence to get a * * lower bound of H C3 we need to derive an upper bound for +( =12 A). Also 1 note that +( =2 A)(I+#P)6, where I=

{ (1=2 ) A  2 dx 0 3
== 6= P=

:  G(=y, =z) G jk (=y, =z)[ j ,&ib j ,][ k ,&ib k , ] dy dz,
R 3+ j, k=1

0

0
 2dx== 3

, 2  G(=y, =z) dy dz,
R 3+
 2ds== 2

R2
, 2 dy.
Here we denote dy=dy 1 dy 2 . In the following computations we use frequently the fact that u exponentially decays at . We have I 11 #

R 3+
 G(=y, =z) G 11 (=y, =z)  1 ,&ib 1 , 2 dy dz
=&u& 22 D 0 +=(0 11 &0 22 ) U 1 D 0 += 2C 11, 2 Q 2 7
\
+
+O = 2D 2 += 2Q 0 += 3D 3 += 3Q 3 + : = jQ j+1 , j=4
(A.9)
where C 11, 2 = 12 {0 12  2U 2 . I 22 #

(A.10)
 G(=y, =z)G 22 (=y, =z)  2 ,&ib 2 , 2 dy dz
R 3+
=&u& 22 D 0 +Q 0

+
(z+z 0 ) 2 u 2 dz
0
4
+ : =j j=1

R 3+
u 2A 22, j dy dz+O
\
11
+
: = jQ j+2 , j=5
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
443
here we use the notations T 1 =(z+z 0 )[0 11 y 21 +20 12 y 1 y 2 &0 22 z 2 ], T 2 = 14 (0 11 y 21 +20 12 y 1 y 2 &0 22 z 2 ) 2 &2(z+z 0 ) ! 3 , T 3 =&[0 11 y 21 +20 12 y 1 y 2 &0 22 z 2 ] ! 3 &2(z+z 0 ) ! 4 , T 4 =! 23 &[0 11 y 21 +20 12 y 1 y 2 &0 22 z 2 ] ! 4 , A 22, 1 =' 2[T 1 +(0 22 &0 11 ) z(z+z 0 ) 2 ]+(0 22 &0 11 ) z'$ 2, A 22, 2 =' 2[T 2 +(0 22 &0 11 ) zT 1 +(z+z 0 ) 2p 3 ]+ '$ 2p 3 , A 22, 3 =' 2[T 3 +(0 22 &0 11 ) zT 2 + p 3 T 1 ] +O( y 3 + z 3 )[' 2 (z+z 0 ) 2 + '$ 2 ], A 22, 4 =' 2[T 4 +(0 22 &0 11 ) zT 3 + p 3 T 2 +O( y 3 + z 3 )[T 1 &2(z+z 0 )]]. In the following computations for I 22 we use frequently the fact

+
(z+z 0 ) u 2 (z) dz=0.
(A.11)
0
Then we have
   
R 3+
R 3+
R 3+
R 3+
u 2A 22, 1 dy dz=&C 22, 1 Q 0 +(0 22 &0 11 ) U 1 D 0 ,
u 2A 22, 2 dy dz=C 22, " 2 Q 4 +C $22, 2 D 2 +C 22, 2 Q 2 +O(Q 0 +D 0 ), (A.12) u 2A 22, 3 dy dz=C 22, 3 U 1 Q 4 +O(Q 3 +D 3 ),
u 2A 22, 4 dy dz=C 22, 4 Q 6 +O(Q 5 ),
444
LU AND PAN
where C 22, 1 =0 11 C22, " 2=

+
z(z+z 0 ) 2 u 2 dz+0 22 z 0 
0

+
z(z+z 0 ) u 2 dz,
0
1 [30 211 +40 212 ] &u& 22 , 32
1 C $22, 2 = [2g 11 &2g 22 ] &u& 22 , 8 1 C 22, 2 = [2g 11 +2g 22 +4(0 11 0 22 &20 211 &0 222 )] 8

+
(z+z 0 ) 2 u 2 dz
0
1 & 0 11 0 22 U 2 , 4 C 22, 3 =
1 [20 11[31 212, 1 +1 222, 2 ]+40 12[1 212, 2 &1 222, 1 ]&90 311 32
(A.13)
+30 211 0 22 &140 11 0 212 &40 212 0 22 ], C 22, 4 =
&u& 22 Q6

1
R2
{36 _
0 ij, k y i y j y k ] 2 ' 2
: (i, j, k){(2, 2, 2)
1 1 + (0 11 y 21 +20 12 y 1 y 2 ) 0 ij : 22lm + _ ijlm y i y j y l y m 4 6
_
+
&
1 [0 11 y 21 +20 12 y 1 y 2 ] 2 [ g 11, ij & g 22, ij ] y i y j dy. 16
=
Here, when computing C 22, 2 , we used (A.11), and noted that, in the isothermal coordinates, 1 221, 1 = g 22, 11 2, 1 222, 2 = g 22, 22 2. From (A.12) we have I 22 =&u& 22 D 0 +Q 0

+
(z+z 0 ) 2 u 2 dz+=[&C 22, 1 Q 0 +(0 22 &011 ) U 1 D 0 ]
0
+= 2[C"22, 2 Q 4 +C$22, 2 D 2 +C 22, 2 Q 2 ]+= 3C 22, 3 U 1 Q 4 += 4C 22, 4 Q 6 11
+O(= 2Q 0 += 2D 0 += 3Q 3 += 3D 3 += 4Q 5 + : = jQ j+2 ), j=5
(A.14)
445
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
Next I 12 =R

R 3+
 G(=y, =z) G 12 (=y, =z)[ 1 ,&ib 1 ,][ 2 ,&ib 2 , ] dy dz
=&C 12, 1 Q 1 = 2 +O(= 3 )(Q 3 +D 3 ), (A.15) I 33 =

R 3+
 G(=y, =z)  3 ,&ib 3 , 2 dy dz
=&u$& 2 Q 0 &=C 33, 1 Q 0 += 2C 33, 2 Q 2 +O(= 2Q 0 += 3Q 3 ), where 0 212 2

C 33, 1 =2H r

C 12, 1 =
+
z 3 (z+z 0 ) u 2 (z) dz,
0 +
zu$ 2 dz,
(A.16)
0
1 C 33, 2 = [2g 11 +2g 22 ] &u$& 2. 8 Summarizing the above we get I 1 # = =

{ (1=2 ) A  2 dx
0
=I 11 +I 22 +2I 12 +I 33 =&u$& 2 Q 0 +Q 0

+ 0
(z+z 0 ) 2 u 2 dz+2 &u& 22 D 0 &=[C 22, 1 +C 33, 1 ] Q 0
+= 2[C" 22, 2 Q 4 +[C 11, 2 +C 22, 2 +C 33, 2 ] Q 2 &2C 12, 1 Q 1 +C$22, 2 D 2 ] += 3C 22, 3 U 1 Q 4 += 4C 22, 4 Q 6 +O(= 2[Q 0 +D 2 +D 0 ]+= 3[Q 3 +D 3 ] 7
11
+ : = jQ j+1 + : = jQ j+2 ). j=4
j=5
Choose ' such that D k <
\
7
11
+
+O = 2[Q 0 +D 2 ]+= 3[Q 3 +D 3 ]+ : = jQ j+1 + : = jQ j+2 . j=4
j=5
(A.17)
446
LU AND PAN
Next we compute 6 and P. 6 = =3

, 2  G(=y, =z) dy dz R 3+
=&u& 22 Q 0 &=C 4, 1 Q 0 += 2C 4, 2 Q 2 +O(= 2Q 0 += 3Q 3 ), P== 2

, 2 dy== 2 &u& 22 Q 0 +O(= 4Q 0 ).
R2
(A.18)
where

C 4, 1 =2H r
+
zu 2dz,
0
C 4, 2 = 18 [2g 11 +2g 22 ]&u& 22 .
(A.19)
Now we can to estimate the ratio (I+#P)6. Recall that ' has compact support. So we assume '(r)=0 for r<\, where \ may depend on =. Denote '(r)=`
r
\ \+ .
Then `(t)=0 for t<1, and D k =d k \ k,
Q k =q k \ k+2,
t k+1 `$(t) 2 dt,
q k =2?
where d k =2?

1
0

1
t k+1` 2 (t) dt.
0
Then I =; 0 &u& 22 &=[C 22, 1 +C 33, 1 ] Q0 = +
1 6 4 [2 &u& 22 d 0 += 2[C" 22, 2 q 4 \ +(C 11, 2 +C 22, 2 +C 33, 2 ) q 2 \ ] q0 \ 2
+C 22, 3 U 1 q 4 = 3\ 6 +C 22, 4 q 6 = 4\ 8 ]
\
+O = 2[q 0 +d 2 ]+= 3[q 3 \ 3 +d 3 \] 7
11
+
+ : d j+1 = j\ j+1 + : d j+2 = j\ j+2 . j=4
j=5
(A.20)
447
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
Case 1. 0 211 +0 212 <0. Let ` achieve E 1 , where E 1 was defined in (A.4). Choose \=
_
2 &u& 22 C 22, 2 = 2
&
16
=
2= &13 . [30 211 +40 212 ] 16
We get 1 I =; 0 &u& 22 + [30 211 +40 212 ] 13 E 1 &u& 22 = 23 Q0 = 2 &=[C 22, 1 +C 33, 1 ]+O(= 43 ) 6 =&u& 22 &C 4, 1 =+O(= 43 ), Q0 =3 Hence the first eigenvalue +
1
\= A+ 2
=
I+#P 6 1 1 ; 0 + [30 211 +40 212 ] 13 E 1 = 23 &(M 1 &#) =+O(= 43 ) , =2 2
{
=
where M1= = +
1 [C 22, 1 +C 33, 1 &; 0 C 4, 1 ] &u& 22 0 11 &u& 22

0 22 &u& 22

+
z[(z+z 0 ) 2 u 2 + u$ 2 &; 0 u 2 ] dz
0 +
z[ z 0 (z+z 0 ) u 2 + u$ 2 &; 0 u 2 ] dz.
0
Using the equation (2.4) and (A.11) we see that M 1 =C 2 0 11 +C 1 0 22 , where C 2 and C 1 were given in (A.3). So +
1
1
2
2
1
\= A+ = { ; +2 (30 0
2 11
+40 212 ) 13 E 1 = 23
=
&(C 2 0 11 +C 1 0 22 &#) =+O(= 43 ) .
(A.21)
448
LU AND PAN
Case 2. 0 11 =0, 0 12 =0. Then C22, " 2 =C 22, 3 =0 and C 22, 4 =
&u& 22 36Q 6

R2
_
2
0 ij, k y i y j y k
: (i, j, k){(2, 2, 2)
, & ' dy=2[Q(P )] &u&(A.22) 2
4
0
2 2
where Q(P) was defined in (A.2). Subcase 2.1.
0 11 =0, 0 12 =0, Q(P 0 )<0.
Choose \=
1  Q(P 0 ) =
.
Let ` achieve E 2 , where E 2 was defined in (A.4). From (A.20) we get I =; 0 &u& 22 += &(C 22, 1 +C 33, 1 ) Q0 =
_
+
q 2 (C 11, 2 +C 22, 2 +C 33, 2 ) +2E 2 Q(P 0 ) &u& 22 +O(= 32 ), q 0 Q(P 0 )
&
C q 6 =&u& 22 += &C 4, 1 + 4, 2 2 +O(= 32 ). Q0 = 3 Q(P 0 ) q 0
_
&
Hence +
1 1 A 2 2 = =
\ + {
_
; 0 &= C 1 0 22 +
M2 &2Q(P 0 ) E 2 &#]+O(= 32 ) , Q(P 0 ) (A.23)
=
where M2 =
[ ; 0 C 4, 2 &C 11, 2 &C 22, 2 &C 33, 2 ] q 2 q 2 = [C 3 0 222 &U 2 2 &u& 22 {0 12  ] q 0 &u& 22 q0
since 0 11 =0 12 =0. Here C 3 was given in (A.3). Subcase 2.2. 0 11 =0, 0 12 =0, Q(P 0 )=0. For simplicity we also assume 0 12, 1 =0 12, 2 =0. So M 2 =C 3 q 2 0 222 q 0 .
449
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
Subcase 2.2.1.
If 0 22 {0, we choose \=
1 =
m
1 =
_ \ +& , log
m>0.
Then 2m
1 q2 I =; 0 &u& 22 + (C 11, 2 +C 22, 2 +C 33, 2 ) = log Q0 = q0 = &(C 22, 1 +C 33, 1 ) =+
6 C 4, 2 q 2 =&u& 22 + 3 Q0 = q0
2 &u& 22 d 0 = 1 2m q 0 log =
_ \ +& 1 = log _ \ = +& &C
_ \ +& 1 +O = \ _log \ = +& + , 7m
32
2m
4, 1
1 =
3m
\ _ \ +& + .
=+O = 32 log
Hence +
1 1 1 C 3 q 2 0 222 A ; & = log 0 2 2 = = q0 =
\ + { +
q0 Subcase 2.2.2.
2m
_ \ +& &[C 0 &#] = 1 +O = \ _log \ = +& += . (A.24) 1
22
7m
2d 0 = 1 log =
32
_ \ +&
2m
If 0 22 =0, we choose \== &a,
1 2
Then +
1 2d 1 A 2 ; 0 +#=+ 0 = 2a +O(= 5&7a ) . 2 = = q0
\ + {
=
Especially if we choose a=59, \== &59, then +
1
1
2
2
\= A+ =
[; 0 +#=+O(= 109 )].
(A.25)
Proof of Proposition A.1. For fixed } we choose =<0 such that 1= 2 =_ }, where _ was defined in (2.3). Then } 2 =+( =12 A). * *
450
LU AND PAN
From (A.21) we get = 2} 2 ; 0 + 12 [30 211 +40 212 ] 13E 1 = 23 &[C 2 0 11 +C 1 0 22 &#] =+O(= 43 ). So _ 1 1 1 * & 2 [30 211 +40 212 ] 13E 1 = 23 + 2 [C 2 0 11 +C 1 0 22 &#] =+O(= 43 ). } ; 0 2; 0 ;0 and since H C3 _ we get * H C3
} } 13 1 1 & 53 (30 211 +40 212) 13 E 1 + 32 (C 2 0 11 +C 1 0 22&#)+O 13 . ; 0 2; 0 ;0 } (A.26)
\ +
This proves Proposition A.1.
K
Proof of Proposition A.2. Assume h is tangential to the line of curvature of 0 at P 0 . Then the y 1  and y 2 curves are the lines of curvature. Hence 0 12 #0, 0 11 =} 1 and 0 22 =} 2 are the principal curvatures of 0 at P 0 . When } 1 {0, from (A.26) we get H C3
} E1 & ;0 2
3} 21 ; 50
\ +
13
} 13 +
1 ;
32 0
(C 2 } 1 +C 1 } 2 &#)+O
1
\} + . 13
(A.27)
So the conclusion (1) in the proposition follows. Again assume h is tangential to a line of curvature at P 0 , and assume the principal } 1 at this direction is zero, i.e. 0 11 =0. Assume Q(P 0 )<0. Since 0 12 #0 and 0 22 =} 2 , from (A.23) we get +
1 1 C 3 q 2 } 22 A ; &= C } + &2Q(P 0 ) E 2 &# +O(= 32 ) , 0 1 2 =2 =2 Q(P 0 ) q 0
\ + {
_
&
=
where C 3 and C 3 were given in (A.3). Again using H C3 _ we obtain * H C3
1 } 1 C 3 q 2 } 22 + 32 C 1 } 2 + &2Q(P 0 ) E 2 &# +O 12 . ;0 ; 0 Q(P 0 ) q 0 }
_
So the conclusion (2) follows. Assume {} 1 =0 and 1 } 2 =0. So Q(P 0 )=0.
& \ +
(A.28)
SURFACE NUCLEATION OF SUPERCONDUCTIVITY
451
When } 2 {0, from (A.24) we have H C3
} C 3 q 2 } 22 C 1 } 2 &# 2d 0 = + 32 (log }) 2m + & 32 32 ;0 ;0 q0 ;0 ; 0 q 0 (log }) 2m +O
\
(log }) 7m }
+.
(A.29)
When } 2 =0, from (A.25) we get H C3
1 } # & +O 19 ; 0 ; 32 } 0
\ +
(A.30)
So the conclusion (3) follows. Now Proposition A.2 is proved. K
ACKNOWLEDGMENTS This work was partially done when the second author visited the Department of Mathematics at Brigham Young University in the winter semester in 1997. He thanks the Department for hospitality. This work was partially supported by National Science Foundation Grant DMS9622853 (to Lu) and by National Natural Science Foundation of China, Science Foundation of the Ministry of Education of China, and Zhejiang Provincial Natural Science Foundation of China (to Pan).
REFERENCES [BBH] F. Bethuel, H. Brezis, and E. Helein, ``GinzburgLandau Vortices,'' Birkhauser, Boston BaselBerlin, 1994. [BPT] P. Bauman, D. Phillips, and Q. Tang, Stable nucleation for the GinzburgLandau system with an applied magnetic field, Arch. Rational Mech. Anal. 142 (1998), 143. [BR] F. Bethuel and T. Riviere, Vortices for a variational problem related to superconductivity, Ann. Inst. H. Poincare, Anal. Non Lineaire 12 (1995), 243303. [BS] A. Bernoff and P. Sternberg, Onset of superconductivity in decreasing fields for general domains, J. Math. Phys. 39 (1998), 12721284. [C] S. J. Chapman, Nucleation of superconductivity in decreasing fields, European J. Appl. Math. 5 (1994), part 1, 449468; part 2, 468494. [CHO] S. J. Chapman, S. D. Howison, and J. R. Ockendon, Macroscopic models for superconductivity, SIAM Rev. 34 (1992), 529560. [CK] S. Chanillo and M. Kiessling, Symmetry of solutions of GinzburgLandau equations, C. R. Acad. Sci. Paris Ser. I 321 (1995), 10231026. [DGP] Q. Du, M. Gunzburger, and J. Peterson, Analysis and approximation of the GinzburgLandau model of superconductivity, SIAM Rev. 34 (1992), 481. [dG] P. G. De Gennes, ``Superconductivity of Metals and Alloys,'' Benjamin, New York, 1966. [E] W. E, Dynamics of vortices in GinzburgLandau theories and applications to superconductivity, Physica D 77 (1994), 383404.
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LU AND PAN
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