ON SINGULAR PERTURBATION PROBLEMS WITH ROBIN ...

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is played by the analogous problem in a half space with Robin boundary .... unique solution to (1.16). Then, by taking v = w(y. 0. ; yN ?R), it is easy to see that lim.
ON SINGULAR PERTURBATION PROBLEMS WITH ROBIN BOUNDARY CONDITION HENRI BERESTYCKI AND JUNCHENG WEI

Abstract. We consider the following singularly perturbed elliptic problem 2 u ? u + f (u) = 0; u > 0 in ;

 @u @ + u = 0 on @ ; where f satis es some growth conditions, 0    +1, and  R N (N > 1) is a smooth and bounded domain. The cases  = 0 (Neumann problem) and  = +1 (Dirichlet problem) have been studied by many authors in recent years. We show that, there exists a generic constant  > 1 such that, as  ! 0, the least energy solution has a spike near the boundary if    , and has an interior spike near the innermost part of the domain if  >  . Central to our study is the corresponding problem on the half space. Mathematics Subject Classi cation (2000): 35B35 (primary), 35J40, 92C40(secondary). 1. Introduction In the recent years, many works have been devoted to the study of the following singularly perturbed problems: 2 u ? u + f (u) = 0; u > 0 in ; (1.1) with either Neumann boundary condition

@u = 0 on @ ; @

(1.2)

or Dirichlet boundary condition

u = 0 on @ :

(1.3) Here  R N is a smooth and bounded domain and f satis es some structure conditions. A typical f is f (u) = up; 1 < p < NN ?+22 if N  3; and 1 < p < 1 if N = 1; 2: 1

2

HENRI BERESTYCKI AND JUNCHENG WEI

In [26], [27], Ni and Takagi showed that, under some conditions on f (u), as  ! 0, the least energy solution for (1.1) with Neumann boundary condition (1.2) has a unique maximum point, say P, on @ . Moreover, H (P) ! maxP 2@ H (P ), where H (P ) is the mean curvature function on @ . On the other hand, Ni and Wei in [30] showed that, as  ! 0, the least energy solution for (1.1) with Dirichlet boundary condition (1.3) has a unique maximum point, say Q, in . Furthermore, d(Q; @ ) ! maxQ2 d(Q; @ ), where d(Q; @ ) is the distance function from Q to @ . Since then, many papers further investigated the higher energy solutions for (1.1) with either (1.2) or (1.3). These solutions are called spike layer solutions. A general principle is that the interior spike layer solutions are generated by distance functions. We refer the reader to the articles [1], [6], [8], [9], [10], [13], [14], [17], [20], [29], [31], [33], [34] and the references therein. On the other hand, the boundary peaked solutions are related to the boundary mean curvature function. This aspect is discussed in the papers [2], [5], [15], [19], [32], [35], [36], and the references therein. A good review of the subject is to be found in [25]. It is a natural question to ask what happens if we replace (1.2) or (1.3) by the following Robin boundary conditions (or boundary conditions of the third kind)

a @u @ + (1 ? a)u = 0 on @ ;

(1.4)

 @u @ + u = 0 on @ ;

(1.5)

where 0 < a < 1. Such Robin boundary conditions are particularly interesting in biological models where they often arise. We refer the reader to [7] for this aspect. The main purpose of this paper is to answer the above question. First of all, we rewrite (1.4) in the following form 2 where  = (1a?a) > 0. (The term  @u @ is a right scaling with respect to  u, as we shall see later.) We shall investigate the role of  on the properties of least-energy solutions, which we shall de ne now.

ROBIN BOUNDARY CONDITION

3

Similar to [26] and [30], we can de ne the following energy functional associated with (1.5): Z Z Z 2Z  1  2 2 J[u] := 2 jruj + 2 u ? F (u) + 2 u2; (1.6)



@

R where F (u) = 0u f (s)ds; u 2 H 1( ). Assume that f has superlinear growth. Then, for each xed , by taking a function e(x)  k for some constant k in , and choosing k large enough, we have J(e) < 0, for all  2 (0; 1). Then for xed , and for each  2 (0; 1), we can de ne the so-called mountain-pass value c; = hinf max J [h(t)] (1.7) 2? 0t1 

where ? = fh : [0; 1] ! H 1( )jh(t) is continuous; h(0) = 0; h(1) = eg. Under some further conditions on f , which we will specify later, similar to [26] or [30], c; can be characterized by c; = u60;uinf sup J[tu]; (1.8) 2H 1 ( ) t>0

which can be shown to be the least among all nonzero critical values of J. (This formulation is sometimes referred to as the Nehari manifold technique.) Moreover, c; is attained by some function u; which is then called a leastenergy solution. Here and throughout this paper, we say that a function u; achieves the maximum in (1.8) if it satis es c; = sup J(tu;): t>0

This also applies to similar variational formulations below. For xed  small, as  moves from 0 (which is (1.2)) to +1 (which is (1.3)), by the results of [26], [27] and [30], the asymptotic behavior of u; changes dramatically: a boundary spike is moved to become an interior spike. The question we shall answer is: where is the borderline of  for spikes to move inwards? Note that when N = 1, by ODE analysis, it is easy to see that the borderline is exactly at  = 1. We leave this as an exercise. So from now on, we may assume that N  2. To state our results, let us rst put some assumptions on f . Firstly, since we look for positive solution, by the maximum principle, we may extend f

4

HENRI BERESTYCKI AND JUNCHENG WEI

by assuming f (t) = 0 for t  0. We assume the same assumptions used in [26], [27] and [30]: we suppose that f : R ! R is continuous and satisfy the following structure conditions (f1) f 0 (0) = 0. (f2) For t  0, f admits the decomposition in C 1+ (R ): f (t) = f1 (t) ? f2(t) where (i) f1 (t)  0 and f2 (t)  0 with f1 (0) = f10 (0) = 0, whence it follows that f2 (0) = f20 (0) = 0 by (f1); and (ii) there is a q  1 such that f1(t)=tq is nondecreasing in t > 0, whereas f2(t)=tq is nonincreasing in t > 0, and in case q = 1 we require further that the above monotonicity condition for f1(t)=t is strict. (f3) f (t) = O(tp) as t ! +1 for some 1 < p < NN ?+22 and 1 < p < +1 if N = 2. (f4) There exists a constant  2 (0; 12 ) such that F (t)  tf (t) for t  0, R where F (t) = 0t f (s)ds. Condition (f1) is rst related to the fact that we already include the linear part (via ?u) in the equation. The assumption (f2) is technical while (f3) and (f4) are classical assumptions. (f3) is an assumption of compactness (subcritical growth) and (f4) is used for Palais-Smale condition and implies a superlinear growth for f , namely, tf1+(t) ! +1 as t ! +1 for some  > 0. Examples of f satisfying (f1)-(f4) include: N + 2 if N  3 and +1 if N = 2: f (u) = up ?auq with a  0; 1 < q < p < N ?2 To understand the location of the spikes at the boundary, an essential role is played by the analogous problem in a half space with Robin boundary condition on the boundary. Thus we rst consider 8 < u ? u + f (u) = 0; u > 0 in R N+ ; (1.9) : u 2 H 1(R N ); @u + u = 0 on @ R N + @ + 0 where R N+ = f(y ; yN )jyN > 0g. Let Z Z 1 Z 1  2 2 I[u] = N ( 2 jruj + 2 u ) ? N F (u) + 2 N u2: (1.10) R+

R+

@ R+

ROBIN BOUNDARY CONDITION

As before, we de ne a mountain-pass vale for I: c = inf1 N sup I[tv]: v60;v2H (R+ ) t>0

5

(1.11)

Our rst result deals with the half space problem:

Theorem 1.1. (1) For   1, c is achieved by some function w, which is

a solution of (1.9). (2) For  large enough, c is never achieved. (3) Set  = inf fjc is achievedg: (1.12) Then  > 1 and for    , c is achieved, and for  >  , c is not achieved.

Remark 1.1. We do not know if the solution to (1.9) is unique. It would

be interesting to solve this question. Remark 1.2. It is somewhat surprising that at the borderline number  = , c is actually achieved. Remark 1.3. It is an interesting question to see how  depends on N and the nonlinearity f . Note that when N = 1,  = 1. Now consider the problem in a bounded domain. It turns out that the critical number  in Theorem (1.1) plays an essential role in the study of the asymptotic behavior of c; and u; de ned in a bounded domain. We will show here that  is the borderline between (1.2) and (1.3) in an arbitrary domain. Here is our rst result for a general domain.

Theorem 1.2. Let    and u; be a least energy solution of (1.1) with (1.5). Let x 2 be a point where u; reaches its maximum value. Then after passing to a subsequence, x ! x0 2 @ and (1) d(x; @ )= ! d0 , for some d0 > 0, (2) v;(y) = u;(x + y) ! w (y) in C 1 locally, where w attains c of (1.11) (and thus is a solution of (1.9)), (3) the associated critical value can be estimated as follows: c; = N fc ? H (x0 ) + o()g

(1.13)

6

HENRI BERESTYCKI AND JUNCHENG WEI

where c is given by (1.11), and H (x0 ) is given by the following Z 0 0 @w H (x0) = max [? y  r w  H (x0)]

(1.14) @yN where S is the set of all solutions of (1.9) attaining c, and y 0 = (y1 ; :::; yN ?1 ); r0 = ( @y@ 1 ; :::; @y @ ?1 ), (4) H (x0 ) = maxx2@ H (x). w2S

RN +

N

Remark 1.4: For Z small, it is easy to see that @w > 0; for all w 2 S ? y0  r0 w @y   RN +

N

(1.15)

and hence H (x0 ) = maxx2@ H (x). We don't know whether or not (1.15) holds for    . When  = 0, the function H (z) is called generalized mean curvature function in [4]. Remark 1.5: When   , the maximum point x is not on the boundary. Instead, it is in the order  distance away from @ . This only happens for Robin boundary problems since in the case of Neumann conditions, the maximum points lie on the boundary, while for Dirichlet condition, they are interior points which stay away from the boundary. On the other hand, when  > , a di erent asymptotic behavior appears.

Theorem 1.3. Let  >  and u; be a least energy solution of (1.1) with (1.5). Let x 2 be a point where u; reaches its maximum value. Then after passing a subsequence, we have (1) d(x; @ ) ! maxx2 d(x; @ ), (2) v; (y) := u;(x + y) ! w(y) in C 1 locally, where w is the unique solution of the following problem 8 > > w ? w + f (w) = 0 in R N ;

> < w > 0; w(0) = maxy2RN w(y); > > > : w(y) ! 0 as jyj ! +1;

(3) the associated critical value can be estimated as follows: # " 2 d ( x ; @

)  (1 + o(1))) c; = N I [w] + exp(?



(1.16)

(1.17)

ROBIN BOUNDARY CONDITION

7

where

Z Z 1 2 2 I [w] = 2 N(jrwj + w ) ? N F (w): (1.18) R R Remark 1.6: According to [16], w is radially symmetric and decreasing, i.e.,w = w(r); r = jyj; w0 (r) < 0 for r > 0. Under the conditions (f1)-(f4), it follows from the results of [18] and [3] that the solution to (1.16) is unique. The existence and location of spikes has been studied in detail in the papers [26], [27] and [30] for Dirichlet or Neumann conditions. Here we rely on the techniques developed in these works. The main new aspect that is needed here is concerned with the problem in a half space with Robin boundary condition (Theorem 1.1). Section 2 is devoted to the proof of that Theorem. Actually, the main novel aspect of this paper is to derive the existence of this separating value  given by Theorem 1.1 which governs the location of spikes. The study of spikes and their locations are carried in Section 3 for Theorem 1.2 and in Section 4 for Theorem 1.3. These two sections are somewhat more technical. We follow and use the results of the papers [26], [27] and [30]. We only carry out in detail here the new ingredients which are needed in the computations in order to deal with Robin boundary conditions. Some technical lemmas are further left to an Appendix. Throughout the paper C > 0 is a generic constant which is independent of  and  and may change from line to line.

Acknowledgments: The research of the second author is partially supported by an Earmarked Grant from RGC of Hong Kong. 2. Proof of Theorem 1.1 In this section, we analyze problem (1.9). We will make use of the concentrationcompactness method. To this end, it is important to introduce the problem at +1 which involves: Z Z 1 2 2 1 (2.1) I := I [w] = 2 N(jrwj + w ) ? N F (w); R R

8

HENRI BERESTYCKI AND JUNCHENG WEI

where w is the unique solution to (1.16). Then, by taking v = w(y0 ; yN ? R), it is easy to see that

"

#

1: sup I [ tv ] = I lim  R!+1

(2.2)

t>0

(See the proof of (2.14) below.) Hence,

c  I 1; for  > 0:

(2.3) Applying the well-known \concentration-compactness" principle of P.L. Lions [21], [22], we prove the following:

Lemma 2.1. If for some  > 0, then c is achieved.

c < I 1;

(2.4)

Proof: We simply apply Theorem V.5 (and Remark V.2) of [22]: here by (f2), we have

f (t)t?q is nondecreasing

for some q  1 and if q = 1, of [22] is satis ed.

f (t) t

is strictly increasing. The condition (500 )

Let us rst prove (1) of Theorem 1.1. To this end, we take a test function vR(y) := w(y0 ; yN ? R). By (f2), there exists a unique tR such that sup I [tvR ] = I[tR vR ]: (2.5) t>0

(See Lemma 3.1 of [26] or Lemma 5.3 of [30].) The asymptotic behavior of w at in nity is given by the following lemma

Lemma 2.2. ([12]) As r ! +1, we have ! ? 1 1 1 ( N ? 1)( N ? 3) w(r) = Ae?r r? 2 1 ? + O( ) ; N

8

r

r2

w0 (r) = ?1 ? (N ? 1) ? (N ? 1)(N ? 3) + O( 1 ); w(r) 2r 8r2 r3

where A > 0 is a generic constant.

(2.6) (2.7)

ROBIN BOUNDARY CONDITION

9

We now ready to prove (1) of Theorem 1.1.

Proof of (1) of Theorem 1.1: Set

eN = (0; :::; 0; 1)T : Let us now compute I[tR vR ] for R large. We begin with Z Z Z 2 2 2 (jrvR j + vR ) + vR = f (w(y ? ReN ))w(y ? ReN )dy R+ @ R+ R+ Z Z @w ( y ? Re ) N + w(y ? ReN ) dy +  (w(y ? ReN ))2dy @ @ R+ @ R+ ! Z Z R 0 = f (w)+O(e?(+1)R )+ w(y?ReN )w (y?ReN ) jy ? Re j +w2(y?ReN ) dy N N R @ R+ ! Z Z 0 R w = f (w)+O(e?(+1)R )+ w2(y ?ReN ) w (y ?ReN ) jy ? Re j + dy N RN @ R+ p 2 2 0 Put = jy j. So w(y ? ReN ) = w( + R ) on @ R N+ . For R >> 1, we have 2 4 p 2 2 r 2 + R = R 1 + R2 = R(1 + 2 R2 + O( R 4 )) 4 2 = R + 2 R + O( R 3 ) p which suggests that we take = Rt. We consider rst  < 1. By (2.7), we obtain that ! Z 0 w2(y ? ReN ) ww (y ? ReN ) jy ?RRe j +  dy N @ R+ ! Z +1 p N ? 1 1 R = jS N ?1j w2( 2 + R2 ) + p 2 2 (?1? p 2 2 )+O( R2 ) N ?2 d +R 2 +R 0 Z ?1 +1 2 p 2 = jS N ?1jR 2 w ( R + Rt2)( ? 1 + O( R1 ))tN ?2 dt 0 Z +1 p ?1 N ? 1 2 w2( R2 + Rt2)tN ?2 dt(1 + O( R1 )) < 0 = jS jR ( ? 1) 0 (2.8) N

N

N

N

N

N

N

N

N

N

where jS N ?1j is the area of the unit sphere S N ?1 in RN . Next we consider  = 1 + c0

R

(2.9)

10

HENRI BERESTYCKI AND JUNCHENG WEI

where c0 > 0 is to be determined. Then we have 1 Z w2(y ? Re ) w0 (y ? Re )

! R +  dy N w N jy ? Re j jS N ?1j @R+ N ! Z +1 p R 1 ( N ? 1) R c 0 = w2( 2 + R2 ) 1 + R ? p 2 2 ? 2( 2 + R2) + O( R2 ) N ?2d +R 0 Z 2 ?1 +1 2 p 2 =R 2 w ( R + Rt2 )( cR0 + 2tR ? N2?R 1 + O( R12 ))tN ?2dt 0 Z 2 ?1 2 +1 ?2R?t2 c0 1 + O( 1 ))tN ?2 dt ? =R 2 A e ( R + 2tR ? N2? R R2 0 Z +1 2 +1 ?2R 1 2 ? ?t (2c + t2 ? (N ? 1))tN ?2 dt + O( 1 )) ( =A R 2 e e 0 2 0 R Note that Z +1 Z +1 2 Z +1 2 N ? 2 ? t ? t N ? 1 (N ? 1) t e dt = e dt = 2 tN e?t2 dt N

N

N

N

0

So we obtain

0

0

! 0 w R w2(y ? ReN ) w (y ? ReN ) jy ? Re j +  dy (2.10) N @ R+ Z +1 2 Z +1 2 +1 ?2R A ? ? t N ? 2 2 = 2R e (2c0 e t ? tN e?t2 dt + O( R1 )): 0 0 Now we choose c0 > 0 so that Z +1 2 Z +1 ? t N ? 2 4c0 e t = tN e?t2 dt (2.11) Z

N

N

0

We conclude that for   1 +

Z

RN +

(jrvRj

vR

2 + 2)+

c0 R

Z

@ +

RN

0

with c0 satisfying (2.11), we have

v  2

Z

RN

f (w)w + O(e?(+1)R )

(2.12)

8 N ?1 ?1 p R < jS jR 2 ( ? 1) 0+1 w2( R2 + Rt2 )tN ?2dt(1 + O( R1 )); if  < 1; +: R jS N ?1j A2 R? 2+1 e?2R (? 1 +1 tN e?t2 dt + O( 1 )); if 1    1 + c0 : N

N

2

2 0

On the other hand, one easily gets that

Z

RN +

F (w(y ? ReN ))dy =

Z

RN

R

F (w) + O(e?(+1)R ):

R

(2.13)

ROBIN BOUNDARY CONDITION

11

Now similar to the computations done in Section 5 of [30], which we omit here, we obtain immediately that for   1 + cR0 ; and R >> 1; Z Z 1 c  sup I(tvR ) = I (tRvR ) < 2 N f (w)w ? N F (w) = I 1; t>0 R R (2.14) which, by Lemma (2.1), proves (1) of Theorem 1.1.

Remark 2.1. The inequality (2.14) does not hold for N = 1. Indeed, this

can be seen to follow from the above computations and the fact that in this case w(y)  e?jyj as jyj ! +1. However, it holds for N  2 since for N  2, w(y)  jyj? N2?1 e?jyj as jyj ! +1, and the previous computations can be carried through. Actually, we can see that the algebraic term jyj? N2?1 helps for this question. Let  be de ned by (1.12). By (2.14), for   1 + cR0 and R >> 1, c < I 1. Hence c is achieved for   1 + cR0 . Thus  > 1. There are two cases to be considered:  = +1, or 1 <  < +1. Let us rst consider  = +1. We will show that it is impossible. That is

Lemma 2.3. For  suciently large, c is not achieved. Proof: We derive this fact by contradiction. Suppose there exists a sequence of solutions w attaining c , with n ! +1. This yields that c is achieved for any  < +1 since c is an increasing function and c  I [w] (by (2.3)). Let us denote the minimizer as w. We also note that as  ! +1, c ! I [w]. Moreover, w(y) ! 0 as jyj ! +1. By the well-known moving plane method (see [16]), w(y) is symmetric in y0 , i.e., w(y0 ; yN ) = w(jy0 j; yN ). n

n

Let R > 0 be such that maxN w(y0 ; yN ) = w(0; R): y2R+

As  ! +1, the limiting problem becomes 8 < u ? u + f (u) = 0 ; u  0 in R N+ ;

: u 2 H 1(R N+ ); u = 0 on @ R N+ :

(2.15)

(2.16)

12

HENRI BERESTYCKI AND JUNCHENG WEI

By the result of Esteban and Lions [11], any solution of (2.16) must vanish identically. So R ! +1 as  ! +1. Furthermore, by concentrationcompactness and usual limiting process, we immediately have that

kw() ? w(j  ?R eN j)kH 1(R+ ) ! 0;

(2.17)

N

kw() ? w(j  ?R eN j)kL1(R+ ) ! 0;

(2.18) as  ! +1: We will show that this is impossible. This is done by expanding c to the second term (Lemma 2.5 below). To this end, we need some delicate estimates of the di erence w() ? w(j  ?R eN j). We rst project the limiting function w to R N+ with Robin boundary condition. Let PR w be the unique solution of 8 < PR w ? PR w + f (w( ? ReN )) = 0 ; PR w > 0 in R N+ ; N

: PR w 2 H 1(R N ); PR w + ?1 @P w = 0 on @ R N : + + @

(2.19)

R

Let us set

 = ?1; R = w( ? ReN ) ? PR w():

(2.20)

Then R satis es the following linear equation

8 < R ? R = 0 in RN+ ; R 2 H 1(RN+ ); : R +  @ = w( ? ReN ) +  @w(?Re ) on @ R N : R

@

@

N

(2.21)

+

The following lemma on R plays an important role. The proof of it is technical and is delayed to appendix A.

Lemma 2.4. (1). As R ! +1, ? R?1 log R(Rx) ? 0 (x) ! 0; where 0 (x) is the unique viscosity solution of the following problem 8 < jruj2 = 1 in R N+ ;

: u(x) = jx ? eN j on @ R N+ :

(2.22) (2.23)

(In fact, 0 is given explicitly by the relation 0 (x) = inf z2@RN+ (jz ? eN j + jz ? xj).)

ROBIN BOUNDARY CONDITION

13

(2). By taking a subsequence along R ! 1, the renormalized R converges in the sense that VR (y) := R(y(+ReRe)N ) ! V0 as R ! +1; (2.24) R N where V0 is a solution of the following equation  u ? u = 0; (2.25) u(0) = 1; u > 0 in R N and sup (e(1+1 )jyjVR(y))  C; for any 1 > 0: (2.26) y2B4R (0)

Let us now use Lemma (2.4) to nish the proof of Lemma (2.3). Our basic idea is to obtain some lower bound for c . In fact we will show that

Lemma 2.5. As R ! +1, we have c = I [w] + exp(?2R (1 + o(1))): (2.27) Proof: This is similar to Section 6 of [30]. Here we give a simpli ed proof.

We follow the approach in Section 3 of [9]. We rst obtain the following global estimates: w(y)  Ce?(1?)jy?R eN j (2.28) for  such that (1?)?1 < 1, where C may depend on  but is independent of R > 0. In fact, we consider the domain R N+ nB1 (ReN ). Then it follows from (2.17) that w ? (1 ? )2w  0 in R N+ nB1 (ReN ). Now we compare w with the function Ce?(1?)jy?ReN j. One then derives (2.28) from the Maximum Principle. Let v(y) := w(y + R eN ) where y 2 R N?R := f(y0 ; yN ) 2 R N jyN > ?R g: Then, by elliptic regularity theory, we have jrvj + v  Ce?4(1?)R ; for y 2 R N?R nB4R (0): So Z Z Z 1 2 2 2 c = 2 N (jrwj + w) +  N w ? N F (w) R+

@ R+

R+

14

HENRI BERESTYCKI AND JUNCHENG WEI

=

Z Z 1 = 2 N w f (w) ? N F (w) R+ R+

Z

RN ?R \B4R (0)

( 12 v f (v) ? F (v)) + O(e?3R )

(2.29)

Consider the restriction of v to the domain R N?R \ B4R (0). We decompose v = PR w + (R (R eN ))1? h(y), where 0 <  < 1 is a small but xed number and y 2 R N?R \ B4R (0). Then h satis es h ? h + f 0 (PR w)h + N + M = 0 in R N?R \ B4R (0)

and

(2.30)

@h + h = 0 on @ R N \ B (0)  4R ?R @ 



where

N = ( (R1e ))1? R  N

# " 0  f (PR w + (R (ReN ))1? h) ? f (PR w) ? f (PR w)(R (R eN ))1? h ; 



and









M = ( (R1e ))1? (f (PR w) ? f (w)): R  N 



By the mean-value property and Lemma (2.4) (see the proof of Lemma 6.1 of [30]), we see that

jNj  C (jw( ? ReN )j + jwj) jw ? w( ? R eN )j jhj and

jMj  C (R (R eN )) (w + wR) jVR (y)j  C (R (R eN )) ejyj 





(2.31) (2.32)

where  is such that 1 ?  <  < 1. Hence h satis es 8 < h ? h + f 0 (PR w)h + o(1)h + o(1)ejyj = 0 in R N?R \ B4R (0);

:

@h @

+ h = 0 on (@ R N?R ) \ B4R (0):

(2.33)

ROBIN BOUNDARY CONDITION

15

Let G (y) be the unique radial solution of u ? 2u = 0; u(0) = 1; u > 0 in R N :

(2.34)

Note that G has the following asymptotic behavior

C1jyj?

N

?1 jyj e

2

 G (y)  C2jyj?

N

?1 jyj e :

2

Set

H = G? 1h :

(2.35)

We rst claim that

kHkL1(R?

N R

\B4R (0))

 C:

(2.36)

Suppose not. Without loss of generality, we may assume that

kHkL1(R? \B4 (0)) = H(y) > 0: Observe that jyj  C . Otherwise, suppose that jyj ! +1. Then there are two possibilities:either y 2 @ (R N?R \ B4R (0)) or y 2 R N?R \ B4R (0). N R

R





Note that on @ R N?R ,





 G?1 ) + ?1 @H = 0: H(1 + ?1 @G @  @

Since on @ R N?R ; 1 + ?1 @G@ G? 1  1 ? ?1 > 0, we know by the Hopf boundary lemma that y 62 @ R N?R . If y 2 @B4R (0), then we have H  eR e?R  C , which is a contradiction. So y 2 R N?R \ B4R (0) and then H(y)  0; rH(y) = 0. Let us compute H(y). By (2.33), H satis es H + 2G? 1rG rH + (?1 + 2 + f 0 (PR w))H + o(1)H + o(1) = 0 Thus,

H(y)  o(1)H(y) + C; which yields that H (y)  C . This is a contradiction to our assumption that kHkL1(R? \B4 (0)) ! +1. Now de ne

N R

R

h  = kh k 1 h  L (R?

N R

\B4R (0))

:

16

HENRI BERESTYCKI AND JUNCHENG WEI

Note that since jyj  C , one sees that kHkL1(RN?R \B4R (0))  khkL1(RN?R \B4R (0)) and hence khkL1(RN?R \B4R (0)) ! +1. It is easy to see that by (2.33), 1 (R N ), where h satis es h  ! h0 in Cloc 0 h0 ? h0 + f 0 (w)h0 = 0; jh0j  Cejyj : (2.37) P By Lemma 6.5 of [30], h0 = Nj=1 aj @y@wj for some constants aj . However, by de nition ry h (0) = ( (R1e ))1? (rw(ReN ) ? rPR w(ReN )) R

 N

= ( (R1e ))1? (R (ReN )rVR (0)) ! 0: R  N Hence ry h0 (0) = 0, which implies that aj = 0; j = 1; :::; N; h0 = 0. This 1 and h   (y)  C; jyj  C . contradicts the fact that h  ! h0 in Cloc Hence (2.36) holds. Now we can carry the computation as in Section 6 of [30]. By (2.29) Z ( 21 PR wf (PR w) ? F (PR w)) c = N R?R \B4R (0) Z   1 ( 2 PR wf 0 (PR w)? 12 f (PR w))h+o(R (ReN )) +(R (R eN ))1? N R?R \B4R (0) Note that Z 0 ( P R  wf (PR w ) ? f (PR w ))h N

=

=

Z

RN ?R \B4R (0)

RN ?R \B4R (0)

=

R?R \B4R (0)

Z

[(?h +h?N ?M )PR w?(f (PR w)?f (w))h?(?PR w+PR w)h)]

Z

? \B4R (0)

RNR

Hence by Lemma (2.4)

c =

[(f 0 (PR w)h)PR w ? f (PR w)h]

Z

RN ?R \B4R (0)

PR w(?M) + O((R (R eN ))1? ) 



( 21 PR wf (w) ? F (PR w)) + o(R (R eN ))]

Z 1 = ( 2 wf (w) ? F (w)) + c1R (ReN ) + o(R (R eN ))] N R





(2.38)

ROBIN BOUNDARY CONDITION

where

17

Z 1 c1 = N( 2 f (w)V0) > 0 R

by Lemma 4.7 of [30]. By Lemma (2.4), R (ReN ) = e?R R (eN )(1+o(1)) = e?2R(1+o(1)) . Lemma (2.5) is thus proved. This lemma shows that for  large enough

c > I [w]

(2.39)

which is impossible since we have shown that c  I [w]. Lemma (2.3) is therefore proved. Next we consider the case 1 <  < +1. We claim that

Lemma 2.6. For the value  =  , the in mum c in (1.11) is achieved. Proof: Like Lemma (2.3), we prove it by contradiction. Suppose there exists a sequence of solutions w attaining c , with n % ; n <  . n

n

Suppose that for  =  > 1, c is not achieved. This implies that c = I 1 and by concentration-compactness, there exists R ! +1 such that as

 %  ;  <  

kw ? w( ? ReN )kH 1(R+ ) ! 0; kw ? w( ? ReN )kL1(R+ ) ! 0: N

N

(2.40)

The remaining of the proof is exactly the same as that of Lemma (2.3). Note that in the proof of Lemma (2.3), we only used the property that  = 1 < 1. (2) and (3) of Theorem 1.1 now follow from Lemma (2.3) and Lemma (2.6).

18

HENRI BERESTYCKI AND JUNCHENG WEI

3. Proof of Theorem 1.2 Let    and let u; be a least-energy solution de ned in Section 1. We now study the asymptotic behavior of u; for    as  ! 0. We will prove Theorem 1.2 in this section. We rst derive an upper bound for c;.

Lemma 3.1. Let   . Then for  suciently small, we have c;  N fc ? H (P ) + o()g; (3.1) for any P 2 @ , where c is given in (1.11) and H (P ) is the generalized mean curvature de ned by (1.14).

Proof: Fix any P 2 @ .

By Theorem 1.1 (3), c is achieved by some function w 2 H 1(R N+ ), which solves (1.9). Let

S = fwjw solves (1.9) and achieves cg: (3.2) Since c  I 1, we see that for any w 2 S , Z Z Z 2 2 2 (jrwj + w) +  w ? 2 F (w)  2c = 2I 1:

(3.3) On the other hand, since w solves (1.9), we derive the following identity @ RN+

RN +

Z

RN +

(jrwj

2+

w

2) +

RN +



Z

@ RN+

w = 2

Z

RN +

f (w)w:

(3.4)

Using the assumption (f4), from (3.3) and (3.4), we see that there exists some constant C independent of  such that

Z

RN +

(jrwj2 + w2 )  C; 8w 2 S; 8 > 0:

This shows that the set S is a compact set. So for each xed P 2 @ , there exists a w 2 S such that

" Z # @w @w 0 0   ? y r w @y H (P ) : H (P ) = ? y r w @y H (P ) = wmax 2S N N R+ R+ (3.5) Z

0 0

N





N

ROBIN BOUNDARY CONDITION

19

Now we choose the w which achieves the maximum at (3.5). Multiplying (1.9) by jy0 j2 @y@wN and integrating by parts, we obtain that Z 0 0 @w 1 Z @w ) ? N y r w @y  = 2 N jy0 j2( 12 jrwj2 + 21 w2 ? F (w) + w @y N N @ R+ R+ So we have another expression for H (P ) Z 1 @w )H (P )  H (P ) = 2 N jy0 j2( 12 jrwj2 + 21 w2 ? F (w) + w @y N @ R+ (3.6) Let (3.7) v(x) = w( x ? P ) be our test function. We now compute J [v]. Without loss of generality, we may assume that P = 0 and that there is a smooth transformation  : \ Br0 (0) ! R N+ such that (0) = 0; r(0) = 0, where r0 > 0 is a small number. Moreover, @ \ Br0 (0) = f(x0 ; xN )jxN > (x0 )g. Now de ne y0 = x0 ; yN = xN ? (x0 ) (3.8) Then we have Z (2 jrvj2 + v2)

# NX ?1 1 @ 0 0 2 2 ? = N ( (jrwj2 + w2 ) + @yN (jrwj + w) 2 i;j=1 ij (0)yi yj ) + O(e ) R+ Z Z  N 2 2 =  ( (jrwj + w) ? 2 H (P ) (jrwj2 + w2 )jy0 j2 + o()) Z "



c 

N

@ RN+

RN +

Similarly we have

Z



F (v

) = N (

On the other hand

(3.9)

Z 1 02 H ( P ) F ( w ) j y j + o()) F ( w ) ?   2 @ RN+ RN + (3.10)

Z

Z

Z

0  ( y  v =  v y ;  ) )dx0 @

@

Z 0 0 (y ) N 2 =  v (y ;  )dy0 @ R+ 2

N

2( 0

20

HENRI BERESTYCKI AND JUNCHENG WEI

= N ( So

J [v

] = N (J [w

Z

@ RN+

w ? H (P ) 2

Z

@ RN+

 jy 0 j2 + o()): w @w @y N

(3.11)

Z 1 02 1 2 + 1 w 2 ?F (w )+w @w )+o()) j y j ( jr w j  ]?H (P )    2 @RN+ 2 2  @yN

By (3.6), this nishes the proof of (3.1). Next, we shall obtain a lower bound for c;. Since we do not know if the solution to (1.9) is unique, we use an idea of [4], where a simpli ed proof of the results of [27] and [30] is obtained. Let u; be a least energy solution of (1.1) with (1.5) and u;(x) = maxx2  u;(x). Then, by the Maximum Principle, we know that x 2 . Moreover, if d(x;@ ) were not bounded, then, we would have u; ! w( x?x ) and lim inf!0 ?N c;  I [w], by the same argument as in [26]. Thus d(x;@ ) is bounded. Let

v;(y) = u;(x + y); y 2  = fyjy + x 2 g:

(3.12)

Then v; converges in the H 1-sense to w, a least energy solution of (1.9). Moreover, as in [26], for certain positive constants a and b, we have

v;(y)  ae?bjyj:

(3.13)

Let x~ be the closest point to x on @ . After a rotation and a translation, we may also assume that x~ = 0 and that a xed neighborhood \ Br0 (0) of the set can be represented as the set f(x0 ; xN )jxN > (x0 )g with (0) = 0; r(0) = 0. For an open set , we de ne Z Z Z  1 2 2 2 J;[v] = 2 (jrvj + v ) + 2 v ? F (v); v 2 H 1()  @  (3.14) From the variational characterization of c; = J[u;], we infer that

?N J[u;]  ?N J[tu;] = J; [tv;]: 

for all t > 0.

(3.15)

ROBIN BOUNDARY CONDITION

21

Let V = B r0 (0). We now de ne v~ on R N+ \ V as follows

8 > v;(y0 ; yN ); if (y0 ) > 0; < v~(y0 ; yN ) = > : v;(y0 ; (y 0 ) ) + @v@y (yN ? (y 0 ) ); if (y0 ) < 0: (3.16) Then v~(y0 ; yN ) is a function de ned on R N+ \ V . By (3.13), we may extend v~(y0 ; yN ) to the whole R N+ . We still denote such extension as v~(y0 ; yN ). ; N

Then

J; [tv;]  JR+ \V [tv~] + J \V nR+ [tv;] ? JR+ \V n [tv~] = I1 + I2 + I3 N



N



N



(3.17)

where Ii ; i = 1; 2; 3 are de ned at the last equality. Let us choose t = t so that J;RN+ \V [tv~] maximizes in t. Then by the de nition of c and the exponential decay of v;, we get that

J;R+ \V [tv~]  c + O(e? ) c 

N

Moreover

t = 1 + o(1)

Next, =

Z B r0 (0)

dy

0

I3 =

I2 = J \V nR+ [tv;] 

Z

0

N

1 jrt v j2 + 1 (t v )2 ? F (t v ))(y0 ; y )dy + O(e? c )  ;  ; N N 2  ;

? 0) (2

 (y 

Z

and

(3.18)

Z 0  ( y )  2 (x0 ; 0)d + t v;(y ;  ) ? t2 v; 0 0 @ \f (y )0g 2 2



Note that



0



p



1 + jr(y0 )j2 = 1 + O(2jy0 j2 ) So d = (1 + O(2jy0 j2))dx0 :

22

HENRI BERESTYCKI AND JUNCHENG WEI

Since v ! w in C 1 locally with uniform exponential decay, by the Lebesgue's convergence theorem, I2 + I3 ! ? 1 H (~x ) Z jy0 j2( 1 jrw j2 + 1 w2 ? F (w ) + w @w )dy0    2  @RN+ 2  2  @yN (3.19) Thus we conclude that

?N c;  c ? H (x) + o():

(3.20)

Now comparing (3.1) and (3.20), we have proved Theorem 1.2. 4. Proof of Theorem 1.3 Let  >  and u; be a least energy solution. Let u;(x) = maxx2 u;(x). We rst observe that d(x; @ ) ! +1: (4.1)



d(x ;@ ) 

In fact, suppose not. Then  C and after taking a subsequence, d (x ;@ ) x ! x0 2 @ ;  ! d0  0. Set v;(y) = u;(x + y); y 2  = fyjy + x 2 g. Let v;(y) ! w in H 1-sense, as  ! 0, where w is also a solution of (1.9). By a simple test function w( x? P ) with P 2 , we see that lim ?N c;  I [w] !0 and whence I [w]  I [w] (4.2) On the other hand, Theorem 1.1 (3) yields that for  >  ; c = I [w]. Hence w attains c, which is impossible by Theorem 1.1 (3). Thus (4.1) holds. By the same argument as in [26], [27] or [30], we have that ku;(x + y) ? w(y)kL1(  ) ! 0: (4.3) As in Section 3, we rst obtain an upper bound for c;.

Lemma 4.1. For  suciently" small, we have c;  N I [w] + e

? 2d(P;@ ) (1+o(1)) 

#

(4.4)

ROBIN BOUNDARY CONDITION

23

for any P 2 .

Proof: This follows immediately from Proposition 5.1 of [30] if we take the test function P w (de ned in [30]) and note that Z (P w)2 = 0: ;P

@

;P

Next we shall obtain a lower bound for c;:

Lemma 4.2. For  suciently small, we have " c; = N I [w] + e

? 2d(x ;@ ) (1+o(1)) 

#

:

(4.5)

Now Theorem 1.3 follows from Lemma (4.1) and Lemma (4.2). The remaining of the paper is devoted to the proof of Lemma (4.2). Let x ! x0 2  . There are two cases to be considered: case I, x0 2 @ and case II, x0 2 . Let us st consider case II. That is, x0 2 . In the end, we will show that case I can be transformed to case II. Like in [30], x any P 2 and de ne w;P to be the unique solution of the following problem with Robin boundary conditions: 8 2 <  w;P ? w;P + f (w( x? P )) = 0 in ; (4.6) : w;P +  @w;P = 0 on @ : @ Put (4.7) ';P (x) = w( x ? P ) ? w;P : Then ';P satis es 8 2 ' ? ' = 0 in

;P ;P < (4.8) : ' +  @';P = w( x?P ) +  @w( x? P ) on @ : ;P @  @ On @ , we have @w( x? P ) x ? P w(  ) +  @ = w( x ? P ) + w0 ( x ? P ) < xjx??P;Pj >

24

HENRI BERESTYCKI AND JUNCHENG WEI

= w( x ? P )( ? < xjx??P;Pj > + O( d(P;@ ) ))  ( ? 1 ? )w( x ? P ) where  ?  ? 1 > 0. Therefore, there exist two positive constants C1 and C2 such that C1';P;1  ';P  C2 ';P;1 (4.9) where ';P;1 satis es 8 2 <  ';P;1 ? ';P;1 = 0 in ; (4.10) : ';P;1 + ?1 @';P;1 = w( x?P ) on @ : @  The study of (4.10) depends on the following lemma.

Lemma 4.3. (Lemma 3.8 of [34]) Suppose that d(P; @ ) > d0 for some

d0 > 0. Let 'D;P be the unique solution of 8 2 D <  ';P ? 'D;P = 0 in ; (4.11) : 'D = w( x?P ) on @ : ;P  Then for any arbitrarily small  > 0, the following holds for  suciently

small

@'D;P j @ j  (1 + )'D;P : (4.12) From Lemma (4.3), we infer that on @ , @'D 'D;P + ?1 @;P  'D;P (1 + ?1(1 + ))  (1 + ?1(1 + ))w( x ? P )

and

@'D 'D;P + ?1 @;P  'D;P (1 ? ?1(1 + ))  (1 ? ?1 (1 + ))w( x ? P )

By comparison principle, it is straightforward to derive the following lemma.

Lemma 4.4. There exist two positive constants C1 and C2 such that C1'D;P  ';P  C2 'D;P (4.13)

where 'D;P satis es (4.11).

The study of (4.11) is contained in Section 4 of [30]. By Lemma 4.6 of [30] and Lemma (4.4) , we derive the following convergence results.

ROBIN BOUNDARY CONDITION

25

Lemma 4.5. (i) ';x (x + y)=';x (x) ! V0 (y) locally, where V0(y) is a 



solution of (2.25). Moreover, for any  > 0, (ii) As  ! 0,

sup e?(1+)jyj jV(y) ? V0(y)j ! 0

(4.14)

?  log(';x (x)) ! 2d(x0 ; @ ):

(4.15)

y2 



From Lemma (4.5), we can now prove Theorem 1.3. This is similar to the proof of Lemma (2.5). We rst obtain the following global estimates: (1?)jx?x j

u;  Ce? (4.16) for  such that (1 ? )?1 < 1, where C may depend on  but is independent of  > 0. In fact, we consider the domain 1 := nBR(x) where R is large. Then we have 2 u; ? (1 ? )2u;  0 in 1 . Now we compare u; with (1? )j ? j the function Ce? . The estimate (4.16) follows from the Maximum 

 x x 

Principle. De ne v;(y) := u;(x + y) = w;x + (';x (x))1? (y). Substituting the expression into the equation for u; yields that  satis es  ?  + f 0 (w;x ) + N + M = 0 in 

(4.17)

and

@ +  = 0 on @

  @ where  is the outerward normal on @  , " 1 N = (' (x ))1? f (w;x + (';x (x))1? ) ? f (w;x ) 





;x 

#

0

(4.18)

?f (w;x )(';x (x))1?  ; 

and



M = (' (1x ))1? (f (w;x ) ? f (w)): ;x  



(4.19)

26

HENRI BERESTYCKI AND JUNCHENG WEI

By the mean-value theorem and Lemma (4.5), we have that (4.20) jNj  C (jw;x j + jv;j) jv; ? w;x j jj and jMj  C (';x (x )) (jwj + jw;x j) jV(y)j  C (';x (x)) ejyj (4.21) for any 1 ?  <  < 1. Hence  satis es 8 <  ?  + f 0 (w;x ) + o(1) + o(1)ejyj = 0 in ;

:

@ @

(4.22) +  = 0 on @  : As in the proof of Lemma (2.5) of Section 2, we set  = G? 1; (4.23) where G satis es (2.34). Similar to the proof of (2.36), we obtain the following bound kkL1(  )  C: (4.24) Now we can compute in the same manner as in Section 6 of [30] " Z # Z Z 1 2 ) +  2 ? ?N c; = ?N 2 (2 jrv;j2 + v; v; F (v;)

@

Z Z 1 = 2 v;f (v;) ? F (v;)

 Z 1 = ( 2 w;x f (w;x ) ? F (w;x )) Z 1 +(';x (x))1? ( 2 w;x f 0 (w;x ) ? 12 f (w;x )) + o(';x (x)):

 Note that Z (w;x f 0 (w;x ) ? f (w;x ))

Z



=

=

Z





[(f 0 (w;x ))w;x ? f (w;x )]

[(? +  ? N ? M)w;x ? (f (w;x ) ? f (w)) ? (?w;x + w;x ))] =

Z



w;x (?M ) + O((';x (x))1? ): 



ROBIN BOUNDARY CONDITION

Hence

c;

Z 1 ( 2 w;x f (w) ? F (w;x )) + o(';x (x))]

= N [









Z 1 ( wf (w) ? F (w)) + c1';x (x) + o(';x (x))] N2

= N [ where

27



R

Z 1 c1 = N( 2 f (w)V0) > 0



(4.25)

R

by Lemma 4.7 of [30]. This proves (4.5) in Case II. Finally, let us consider the rst case: x0 2 @ . That is, we assume that d := d(x; @ ) ! 0; x ! x0 2 @ . We now show that this case can be transformed to case II by a suitable change of variables. Let R > 3 be a large but xed number. Let J; be de ned at (3.14) and c;; = v60;vinf max J [tv]: 2H 1 () t>0 ; Let

 = d :

(4.26)

By (4.16) and elliptic regularity theory, it is easy to see that jv;j  Ce?R(1?) ; x 2 n R (4.27) where R = \ BR(x). Now let u be the function de ned as the unique solution of the following problem 8 2 <  u ? u + f (u) = 0 in R ; (4.28) :  @u + u = 0 on @ R : @ From (4.27), it follows that ju ? uj  Ce?R Hence u satis es the equation on  R : 8 2 <  u ? u + f (u) + O(e?R ) = 0 in R ; (4.29) :  @u + u = 0 on @ R : @

28

HENRI BERESTYCKI AND JUNCHENG WEI

Now we rescale R as follows: x = x + dx; x 2 R = ( R ? x)=d Then we have c;; R = dN c;;R where  = d . Note that by (4.1)  ! 0 and

(4.30) (4.31)

c;; = J; (u) = J; (u) + O(N e? 2 ) In the new domain R, u attain its global maximum at x where d(x; @ R ) ! 1. Moreover, u satis es the following equation 8 2 <  u ? u + f (u) + O(e?  ) = 0 in R; (4.32) : @u + u = 0 on @  : R

R

R 

R

R 



@



R

We are now in the case II with the new domain R and the new function u. Now following the same proof as in case II, we obtain that (4.33) c;;R = N (I [w] + e? 2 (1+o(1)) ) Substituting (4.33) into (4.31), we obtain the lower bound (4.5) in Case I.

Now comparing (4.5) and (4.4) allows us to conclude and derive the proof of Theorem 1.3. 5. Appendix A: Proof of Lemma (2.4) We devote this appendix to the proof of Lemma (2.4). Our main idea is to use vanishing viscosity method as in Section 4 of [30] and Section 3 of [34]. (For vanishing viscosity method, we refer to [23].) Since our domain is R N+ which is unbounded, we have to work with a suciently large domain and then take a limit. We begin with the following observation: for  < 1 and y 2 @ R N+ , ? ReN )  w(y ? Re ) C1w(y ? ReN )  w(y ? ReN ) +  @w(y @ N (5.1)

ROBIN BOUNDARY CONDITION

29

for some constants C1; C2 > 0. Comparison principle yields that

C11R  R  C21R where 1R satis es

 1 ? 1 = 0 in R N ; 1 2 H 1(R N ) + R + R R1 1R +  @@ = w( ? ReN ) on @ R N+ : R

(5.2)

This implies that in order to prove Lemma (2.4), it is enough to consider 1R . To study 1R , we introduce another problem: x a large number M > 4, let 2R be the unique solution of 8 2 ? 2 = 0 in R N \ B (Re ); MR N < R R + (5.3) : 2 +  @2R = w( ? Re ) on @ (R N \ B (Re )): N MR N R + @

Since 1R  w(y ? ReN ), comparison principle (see similar arguments leading to (2.28)) gives

j1R ? 2R j  Ce?MR :

(5.4)

We have reduced our problem to consider 2R only. To study 2R, we compare 2R with the following function: let 3R be the unique solution of the following problem 8 3 3 < R ? R = 0 in R N+ \ BMR (ReN ); (5.5) : 3 = w( ? ReN ) on @ (R N \ BMR (ReN )): R + Put

y = Rx; R?1 = ; (x) = ? log 3R(y): Then (x) satis es 8 <  ? jr j2 + 1 = 0 in R N+ \ BM (eN );

: = ? log w( ?e ) on @ (R N \ BM (eN )): + N

We shall prove

Lemma A:

(5.6)

(5.7)

30

HENRI BERESTYCKI AND JUNCHENG WEI

M (1) As R ! +1, (x) ! M 0 (x), where 0 (x) is the unique viscosity solution of the following problem 8 < jruj2 = 1 in R N+ \ BM (eN ); (5.8) : u = jx ? eN j on @ (R N+ \ BM (eN )) In fact, M 0 can be written explicitly

M0 (x) =

inf

z2@ (RN+ \BM (eN ))

(jz ? eN j + jz ? xj):

(2) There exists a positive constant C > 0 such that

kr kL1(R+ \B

 C: (3) On @ (R N+ \ BM (eN )), we have as R ! +1, N

e

M ( N ))

@ ! @ M0 : @ @

(5.9) Lemma (2.4) now follows from Lemma A: in fact, by (5.9), we have that on @ (R N+ \ BMR(ReN )), 3 R j  (1 +  )3 j @ R @ for any  small, which implies that R 3 C13R  3R +  @ @  C2R 3

Therefore by comparison principle, we derive

C23R  2R  C23R; y 2 R N+ \ BMR (ReN ): Observing that for M large enough, we have M0 (x) = 0 (x) = inf (jz ? eN j + jz ? xj); x 2 B4 (eN ):

(5.10)

z2@ RN+

(5.11) The rest of the proof of Lemma (2.4) is similar to that of Lemma 4.6 of [30]. It remains to prove Lemma A. To prove Lemma A, it is enough to prove (3) of Lemma A: in fact, suppose (3) is true, then we have kr kL1(@(RN+ \BM (eN )))  C and simple computations shows that (jr j2) ? 2 r  r(jr j2 )  0 in R N+ \ BM (eN ): (5.12)

ROBIN BOUNDARY CONDITION

31

So by the Maximum Principle, (2) of Lemma A holds. From (2) and by taking a limiting process as in Appendix A of [30], we obtain (1) of Lemma A. To prove (3) of Lemma A, we follow the proof of Lemma 3.7 of [34]. The key fact is that for any M > 0, there exists a constant 0 < lM < 1 such that

j ? log w( z1 ? eN ) + log w( z2 ? eN )j < lM jz1 ? z2 j

(5.13) for all z1 ; z2 2 @ R N+ ; jz1j; jz2j  M , where lM is independent of . (This corresponds to Lemma 3.5 of [34].) Then we follow the proof of Lemma 3.7 of [34]. (See similar arguments in Section 8.3 of [23].) We omit the details. References [1] P. Bates and G. Fusco, Equilibria with many nuclei for the Cahn-Hilliard equation, J. Di . Eqns 160 (2000), 283-356. [2] P. Bates, E.N. Dancer and J. Shi, Multi-spike stationary solutions of the CahnHilliard equation in higher-dimension and instability, Adv. Di . Eqns 4 (1999), 1-69. [3] C.C. Chen and C.S. Lin, Uniqueness of the ground state solution of  + ( ) = 0 in RN  3, Comm. PDE. 16(1991), 1549-1572. [4] M. Del Pino and P. Felmer, Spike-layered solutions of singularly perturbed elliptic problems in a degenerate setting, Indiana Univ. Math. J. 48 (1999), no. 3, 883{ 898. [5] M. Del Pino, P. Felmer and J. Wei, On the role of mean curvature in some singularly perturbed Neumann problems, SIAM J. Math. Anal. 31 (1999), 63-79. [6] M. Del Pino, P. Felmer and J. Wei, On the role of distance function in some singularly perturbed problems, Comm. PDE 25(2000), 155-177. [7] A. Dillon, P.K. Maini and H.G. Othmer, Pattern formation in generalized Turing systems, I. Steady-state patterns in systems with mixed boundary conditions, J. Math. Biol. 32(1994), 345-393. [8] M. del Pino, P. Felmer and J. Wei, Mutiple peak solutions for some singular perturbation problems, Cal. Var. PDE 10(2000, 119-134. [9] E.N. Dancer and J. Wei, On the location of spikes of solutions with two sharp layers for a singularly perturbed semilinear Dirichlet problem, J. Di . Eqns 157(1999), 82-101. [10] E.N. Dancer and S. Yan, Multipeak solutions for a singular perturbed Neumann problem, Paci c J. Math. 189(1999), 241-262. [11] M.J. Esteban and P. L. Lions, Existence and non-existence results for semilinear elliptic problems in unbounded domains, Proc. Roy. Soc. Edinburgh Sect. A 93 (1982), 1-14. [12] R. Gardner and L.A. Peletier, The set of positive solutions of semilinear equations in large balls, Proc. Roy. Soc. Edinburgh Sect. A 104 (1986), 53-72. u

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Department of Mathematics, The Chinese University of Hong Kong, Hong Kong

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