A NEW TYPE OF CONCENTRATION SOLUTIONS FOR A SINGULARLY PERTURBED ELLIPTIC PROBLEM

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1 TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 359, Number 4, April 2007, Pages S Article electronically published on November 22, 2006 A NEW TYPE OF CONCENTRATION SOLUTIONS FOR A SINGULARLY PERTURBED ELLIPTIC PROBLEM E. N. DANCER AND SHUSEN YAN Abstract. We prove the existence of positive solutions concentrating on some higher dimensional manifolds near the boundary of the domain for a nonlinear singularly perturbed elliptic problem. 1. Introduction The aim of this paper is to construct solutions concentrating on some higher dimensional manifolds for the following singularly perturbed elliptic problem: { 2 u + u = u,u>0, in, 1.1 u =0, on, where >0isasmallnumber. We assume that is a domain in R N, whose boundary is Lipschitz continuous, and satisfies the following condition: 1 : There is an integer m, 1<m N, such that y, if and only if y,y D, wherey =y,y, y R m, y R N m, D is a relatively open domain in R+,and R+ = { z =z 1,,z :z 1 0 }. Let us emphasize here that we do not assume that is bounded. The domain can be a bounded domain, or an exterior domain in R N, or many other unbounded domains. We assume that p satisfies p 2, 2N m +1/N m 1 if m<n 1, p 0, + ifm N 1. In view of the assumption on, we will work on the following subspace of H0 1 : H s = { u : u H0 1,uy =u y,y }. Let U be the unique solution of the following problem: v + v = v,v>0, in R, v0 = max z R vz, v H 1 R. Then Uz =U z, U < 0, and z N m/2 e z U z c>0 Received by the editors January 27, Mathematics Subect Classification. Primary 35J65. The work of the first author was partially supported by ARC. c 2006 American Mathematical Society Reverts to public domain 28 years from publication 1765 License or copyright restrictions may apply to redistribution; see

2 1766 E. N. DANCER AND SHUSEN YAN R N m x 2 x 1 D z 1 Figure 1. D is a bounded domain as z +. Moreover, Uz is nondegenerate. That is, the kernel of the operator w + w p 1U p 2 w in H 1 R is spanned by { Uz z i,i =1,, N m +1}. See [11, 15]. For any y =y,y R N, y R m, y R N m,wedenoteỹ = y,y R.Let W y =Uỹ. For any x D, let W, x y =U ỹ x. Then, W, x satisfies W, x + W, x = W, x m 1 y x 1 y ỹ x U ỹ x, in. In this paper, we assume that also satisfies the following condition: 2 : There are k different points x = x,1, x D, =1,,k, such that for =1,,k, i there is a C 2 function ψ z inr N m such that and D B δ x ={z =z 1,z :z 1 >ψ z } B δ x, D B δ x ={z =z 1,z :z 1 = ψ z } B δ x, where δ>0 is a small constant; ii x,1 = ψ x =min z B δ x ψ z > 0, and x,1 < min z B δ x ψ z. By 2, we can deduce that for each =1,,k, there is constant δ 0,δ, such that min z B δ x \B δ x ψ z > min z B δ x ψ z. We will prove that 1.1 has a solution u,whichiscloseto W, x in a small neighbourhood of y = x,1, =1,,k, and is close to zero elsewhere. Since the right-hand side of 1.2 has a singularity, we truncate W, x as follows. Let ξ C0 B δ x be a function such that ξ =1inB δ x forsomeδ δ,δ. For any x D B δ x, define W,x y =ξ y,y W,x y. License or copyright restrictions may apply to redistribution; see

3 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1767 R N m x 2 D x 1 z 1 Figure 2. D is an exterior domain Then W,x satisfies W,x + W, x = ξ W,x + f,x y in, where f,x y = ξ m 1 y x,1 y ỹ x U ỹ x 2Dξ DU ỹ x 2 U ỹ x ξ. Since z x,1 z x U z x = z 1 U z x and ξ = 0 in a neighbourhood of y = 0, it is easy to see that f,x is a smooth function in both y and x, and satisfies f,x CU ỹ x. Let P, W,x be the solution of { 2 v + v = ξ W,x f,x y, in, v =0, on. By the uniqueness, we know that P, W,x H s. Let u, v = 2 DuDv + uv, v = u, v 1/2. The main result of this paper is the following. Theorem 1.1. Assume that 1 <m<n. Suppose that satisfies 1 and 2. Then, there is an 0 > 0, such that for every 0, 0 ], 1.1 has at least one solution of the form 1.5 u = P, W,x, + ω, License or copyright restrictions may apply to redistribution; see

4 1768 E. N. DANCER AND SHUSEN YAN where ω H s, x, =x,,1,x, D, andas 0, dx,, D +, x, ˆx D B δ x, with ˆx,1 = ψ ˆx = min z B δ x ψ z, and ω 2 = o. Our assumption on the boundary implies that x is the closest point to the subspace z 1 =0inD B δ x. If we interpret the assumption on the boundary in this way, we can also include the case m = N in Theorem 1.1. If m = N, then i and ii imply that D is an interval [r 1,r 2 ]inr 1 with r 1 > 0. That is, is an annulus or the exterior domain of a ball. Theorem 1.1 states that 1.1 has a solution concentrating near the inner boundary of the annulus. This is the result in [4]. There are many works in the case m = 1 since the pioneering works [16]. See for example [3, 5, 6, 7, 8, 9, 10, 12, 14, 16, 17, 18]. Except for [8], where the exterior domain problem was studied, all the other papers consider the problem in a bounded domain. To obtain the results mentioned above for the case m =1, no symmetry condition is imposed on the domain. In the case m > 1, we use the solution U of a lower dimensional problem as an approximate solution for problem 1.1. So, there is no control in some directions for the corresponding linear operator L v =: 2 p 2 v + v p 1 W, x v in H0 1. As a consequence, L v = λv, v H0 1, will have many small eigenvalues. By imposing some partial symmetry conditions on, we can get rid of the small eigenvalues if we work on the subspace H s. As far as we know, Theorem 1.1 is the first result on the existence of solution for 1.1, concentrating on an m 1-dimensional sphere. In [1, 2], Ambrosetti, Malchiodi and Ni studied the existence of solutions concentrating on spheres for some elliptic problems, assuming that the domain is either a ball or an annulus. But for 1.1, neither Theorem 1.1 nor the results in [2] gives the existence of a solution concentrating on an m 1-dimensional sphere, if the domain is an annulus and 1 <m<n. Solutions concentrating on a connected component of the boundary were constructed in [13] for the singularly perturbed Neumann problem. The techniques to prove Theorem 1.1 can also be used to study the following Neumann problem: { 2 u + u = u,u>0, in, 1.6 u n =0, on, where n is the outward unit normal of aty. We assume that is an open connected set in R N, satisfying 1 and 3 : there exists x = x 1, x D such that i there is a C 2 function ψz inr N m such that D B δ x ={z =z 1,z :z 1 <ψz } B δ x, and D B δ x ={z =z 1,z :z 1 = ψz } B δ x, where δ>0isaconstant; License or copyright restrictions may apply to redistribution; see

5 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1769 ii x 1 = ψ x =max z B δ x ψz > 0and x 1 > max z B δ x ψz. For any x D, letp,,n W, x be the solution of { 2 v + v = ξ W, x + f, x y, in, v n =0, on, where ξ C0 B δ x with 0 ξ 1; ξ =1inB δ/2 x. Then, we have Theorem 1.2. Assume that 1 <m N. Suppose that satisfies 1 and 3. Then, for any positive integer k, there is an 0 > 0, such that for every 0, 0 ], 1.6 has at least one solution of the form u = P,,N W,x, + ω, where ω H s, x, =x,,1,x, D, andas 0, dx,, D x, x,i +, x, ˆx D B δ x, with ˆx,1 = ψ ˆx +, i, = max ψ z, z B δ x and ω 2 = o. The solutions obtained in Theorem 1.2 concentrate near the boundary but not on the boundary. Our next result shows that 1.6 has a solution concentrating on several manifolds on the boundary. Theorem 1.3. Assume that 1 <m<n. Suppose that satisfies 1 and 3. Then, for any positive integer k, there is an 0 > 0, such that for every 0, 0 ], 1.6 has at least one solution of the form u = P,,N W,x, + ω, where ω H s, x, =x,,1,x, D, andas 0, x, x,i +, i, x, ˆx D B δ x, with ˆx,1 = ψ ˆx = max ψ z, z B δ x and ω 2 = o. Condition 3 implies that x has the largest distance to the subspace z 1 =0 in D B δ x. Unlike the Dirichlet problem, Theorem 1.2 shows that the Neumann problem 1.6 has solutions with several peaks clustering near the manifold y = ˆx 1,whereˆx =ˆx 1, ˆx is a maximum point of the distance function of x D to z 1 = 0. In the case that is an annulus, the results here show that the Dirichlet problem 1.1 has a solution concentrating near the inner boundary; while the Neumann problem 1.6 has a solution with several peaks clustering near the outer boundary. We will use the reduction method, together with the comparison of the energy, to prove the theorems. In this paper, we only give the proof of Theorem 1.1. For License or copyright restrictions may apply to redistribution; see

6 1770 E. N. DANCER AND SHUSEN YAN the Neumann problem, we can follow [7, 19] to make the necessary modifications. In Section 2, we will estimate the energy of the approximate solution P, W,x and thus lay the foundation for the proof of Theorem 1.1. In [16], Ni and Wei used the viscosity solution method to obtain the estimate of the energy of the approximate solutions when m = 1 and is bounded. But it seems that the viscosity solution method cannot be applied to treat the present case, due to the possible unboundedness of the domain, and/or the occurrence of a singularity of the corresponding problem in D. In this paper, we will modify the techniques developed in [7, 8] to obtain the desired estimates. The functional corresponding to 1.1 may not be well defined in H s, because the exponent p may be supercritical. Our obective is to construct solutions concentrating near the m 1-dimensional manifolds y = x,1. So we can modify the nonlinear term u in such a way that corresponding to the modified problem, the functional is well defined in H s and the modified problem has a solution concentrating near y = x,1, which is also a solution of the original problem. To this aim, we define 1.7 fy, t = 1 B t B ft, where B = {y : y, y,y D B δ x }, 1 B = 1 in B, and is zero otherwise, and { t ft +, t 1, = 1+t 1, t > 1. Now we consider the following problem: { 2 u + u = fy, u, u>0, in, 1.8 u =0, on. The functional corresponding to 1.8 is I u = 1 2 Du 2 + u 2 F y, u, 2 where F y, t = t 0 fy, τ dτ. For any y B,wehave y x,1 > 0. We see that I u is well defined in H s if p 2, 2N m +1/N m Basic estimates In this section, we give some basic estimates needed in the proof of the main result. By our assumption on, we can deduce that for each =1,,k,thereisa constant δ 0,δ such that min z B δ x \B δ x ψ z > min z B δ x ψ z. Define D = {z =z 1,z :z 1 ψ z,ψ z +γ,z B δ x }, where γ>0isa small constant. We choose γ>0so small that if x D,then 2.1 d x, D B δ x = d x, D. In this paper, we always assume that x D,ande dx, D/ 1 θ for a fixed small θ >0. License or copyright restrictions may apply to redistribution; see

7 SINGULARLY PERTURBED ELLIPTIC PROBLEM Let ϕ,x = W,x P, U,x.Thenϕ,x satisfies { 2 ϕ,x + ϕ,x =0, in, ϕ,x = W,x, on. Since ϕ,x H0 1, by the maximum principle on bounded or unbounded domains, we have Thus, we see 0 <ϕ,x Ce dx, D/. W,x Ce dx, D/ <P, W,x <W,x. k In this section, we will estimate I P,W,x. For each fixed, noting that W,x is exponentially small if y,,y / B δ x, we deduce I P, W,x = 1 ξ W,x 2 P, W,x + f,x P, W,x P, W,x p + p = 1 ξ W,x 2 W,x 1 ξ W,x 2 ϕ,x + O N m+2 1 W,x p p + W,x ϕ,x + O W,x p 2 ϕ 2,x = p 2 W,x p 2p + 1 ξ W,x 2 ϕ,x + O W,x p 2 ϕ 2,x + N m+2. So, we see that to estimate I P, W,x, we need to estimate τ,x = ξ W,x ϕ,x, and W p 2,x ϕ 2,x. First, we have the following estimate for τ,x. Lemma 2.1. Suppose that x D. Then for any small θ>0, there are C 2 > C 1 > 0, such that C 1 e 2+θd/ + O e d / τ,x C 2 e 2 θd/ + O e d /, where d = dx, D. Proof. Since B f,x yϕ,x Ce d/ U ỹ x Ce d / N m+2, License or copyright restrictions may apply to redistribution; see

8 1772 E. N. DANCER AND SHUSEN YAN we have 2.4 τ,x = ξ W,x ϕ,x B = 2 W,x + W,x f,x y ϕ,x B = 2 W,x B n ϕ,x + 2 ϕ,x n W,x + 2 ϕ,x + ϕ,x W,x + O e d/ N m+2 B = 2 W,x B n ϕ,x + 2 ϕ,x n W,x + O e d / N m+2, where n is the outward unit normal of B at y. We are now ready to get an upper bound for τ,x. We can deduce from 2.1 that 2.5 B 2 W,x n ϕ,x Ce d / U ỹ x B Ce 2 θd /. To estimate B 2 ϕ,x n W,x, we need to estimate ϕ,x n on B. Let z =x,1, 0,, 0,x, and let V y =P, W,x y + z. Then, V satisfies 2.6 V + V = η y,y B,, where η y = η y + z and η y =ξ W,x y+ f,x y. By the L p estimate for the elliptic equations, we obtain that for any θ>0smallandx 0 B,, V W 2,q B θ x 0 C η Lq B 2θ x 0 + C V Lq B 2θ x C η L B 2θ x 0 + C V L B 2θ x 0 Ce 1 θd/, since P, W,z W,z + Ce d/,wherec>0isaconstant depending on θ. So, for q>n, we obtain from 2.7 that 2.8 V C 1 B θ x 0 Ce 1 θd/, which implies 2.9 P,W,x C 1 e 1 θd/, on B. n So we obtain 2.10 ϕ,x P,W,x + W,x n n n Using 2.10, we find 2.11 C 1 e 1 θd /, on B. B 2 ϕ,x n W,x C e 21 θd /. Combining 2.4, 2.5 and 2.10, we obtain τ,x C e 21 θd/ + C O e d /. Next, we get a lower bound for τ,x. License or copyright restrictions may apply to redistribution; see

9 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1773 Since W,x =0on B \, using 2.10, we find 2.12 τ,x = 2 W,x n W,x + 2 ϕ,x n W,x + O e d /. B For x =x,1,x D, y =y,y, let ȳ = y x,1 y 1 y,y x R N.Then 2.13 B = c m 1 2 W,x n W,x = B D D y m 1 U ỹ x U ỹ x U ỹ x U ỹ x 1 n ỹ x ȳ, ỹ x ỹ x,n D, where n D is the outward unit normal of D at ỹ, andc m 1 > 0 is the area of the unit sphere in R m. Let x D be the point such that x x = d.letσ>0beasmallnumber such that, if ỹ B 1+σd x D, then As a result, c m c m 1 ỹ x ỹ x,n 1 D 2. y m 1 U ỹ x D B 1+σd x y m 1 U ỹ x D B 1+σd x c y m 1 U 2 ỹ x 2.14 D B 1+σd x c y m 1 U 2 ỹ x D B 1+θd x c e 2+θd / U θ ỹ x D B 1+θd x c 1 e 2+θd /, U ỹ x ỹ x ỹ x,n D U ỹ x where c, c and c 1 are some positive constants. On the other hand, we have y m 1 U ỹ x U ỹ x ỹ x D D\B 1+σd x ỹ x,n D Ce 1+σ2 θd / U θ ỹ x 2.15 D D\B 1+σd x Ce 1+σ2 θd/ U θ ỹ x D Ce 1+σ2 θd /. License or copyright restrictions may apply to redistribution; see

10 1774 E. N. DANCER AND SHUSEN YAN Combining 2.13, 2.14 and 2.15, we are led to W,x n W,x c e 2+θd / B for some c > 0. Multiplying 2.2 by ϕ,x and integrating by parts, noting that W,x =0in \ B, we find ϕ,x B n W,x = 2 ϕ,x n W,x = 2 Dϕ,x 2 + ϕ 2,x > 0. Let us emphasize that we can use the integration by parts to obtain 2.17 because isc 1 at y if W,x y 0. So, from 2.12, 2.16 and 2.17, we obtain τ,x c 1 e 2+θd / + O e d /. Lemma 2.2. Let q 1,p]. Suppose that x D.Thereisaσ>0, such that 2.18 ϕ q,x W,x p q = O e q 1σd/ τ,x. Proof. Write 2.19 ϕ q,x W,x p q = ϕ q,x W,x p q + B ϕ q,x W,x p q, \ B where B = {y : y : y,y B 1 2σd x }, σ>0 is a small constant. It is easy to see that 2.20 \ B ϕ q,x W p q,x Ce q+1 2σp q θd / Ce qd / \ B W p q,x \ B W θ,x C e 2+σd /, since p>2. On the other hand, if y B,then U ỹ x c e 1+θ ỹ x / c e 1+θ1 2σd/. Thus As a result, 2.21 ϕ,x y U Ce d / ỹ x e 1+θ1 2σd / Ce σd /. ϕ q,x W,x p q = ϕ,x W ϕ,x y q 1,x B B U ỹ x ϕ,x W,x Ce σq 1d/ τ,z. Ce σq 1d / Combining 2.19, 2.20 and 2.21, we obtain We are ready to prove the main result of this section. License or copyright restrictions may apply to redistribution; see

11 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1775 Proposition 2.3. Suppose that x D, =1,,k.Then I P, W,x = A x m 1,1 + 1 τ,x + O e 2+σd/ +, 2 where A = p 2 2p c m 1 U p, c R m 1 is the area of the unit sphere in R m and σ>0 is a small constant. Proof. By 2.3 and Lemma 2.2, we have I P, W,x 2.22 = p 2 2p W,x p τ,x + O W,x p 2 ϕ 2,x + N m+2 =A x m 1, τ,x + O On the other hand, I P, W,x = = e 2+σd / + I P, W,x + P, W,xi,P, W,x i< + 1 p P, W,x p P, W,x p + + I P, W,x By Lemma A.1, we have + ξ W,x i P, W,x + f,xi yp, W,x i< + 1 p P, W,x p P, W,x p +. + P, W,xi P, W,x Ce σ/, for i. Asaresult, p P, W,x P, W,x p C i Ce σ / i for some σ > 0 small. P, W,xi P, W,x P, W,xi σ P, W,x 1 σ Ce σ /,. License or copyright restrictions may apply to redistribution; see

12 1776 E. N. DANCER AND SHUSEN YAN Similarly, we can prove W,x i P, W,x + i< f,xi P, W,x Ce σ/. So, we obtain 2.24 I P, W,x = where σ >0 is a constant. The result follows from 2.24 and I P, W,x + O e σ/, 3. Proof of the main result First we define 3.1 D = {x =x 1,,x k :x D,,,k, e 2d/ 1 θ}, where θ >0 is a fixed small constant, d = dx, D. We also define 3.2 Jx, ω =I P, W,x + ω, x D,ω H s. Let E,x,k = { ω : ω H s, ω, P,W,x x,l D, =0, =1,,k,l =1,,N m +1}, where u, v D, = z m DuDv + uv dz. D First, we will prove that for each x D,thereisanω,x E,x,k such that 3.3 Jx, ω,x ω = l=1 G l P, U,x x l, for some constants G l R, =1,,k, l =1,,N m + 1. Then, we will choose x D such that G l = 0, for all =1,,k, l =1,,N m +1. We expand Jx, ω nearω = 0 as follows: where l,x ω = Jx, ω =Jx, 0 + l,x ω+ 1 2 Q,xω+R,x ω, Q,x ω = 2 DP, W,x Dω + P, W,x ω 2 Dω 2 + ω 2 p 1 k k P, W,x + ω, P, W,x p 2 + ω2, License or copyright restrictions may apply to redistribution; see

13 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1777 and 3.6 R,x ω = F y, P, W,x + ω + F y, + k P, W,x + ω p 1 P, W,x k P, W,x p 2 + ω2. Lemma 3.1. There is a constant C>0 and σ>0 such that l,x ω C /2 k e 1+σd/ + ω. Proof. We have 3.7 l,x ω = We also have 3.8 = ξ W,x ξ W,x ω ξ W,x k W,x ω + + f,x y ω P, W,x + ω W,x k + P, W,x + k P, W,x ω + =O + O i W /2,x ϕ /2,x From Lemmas A.1 and 2.2, we obtain 3.9 W,x /2 ϕ /2,x ω = W,x /2 B ϕ /2,x ω + ω + O e δ / ω P, W,xi /2 P, W,x /2 ω. W,x /2 \B W p/2,x ϕ p/2,x /p B ω p 1/p + Ce σ / ϕ /2,x ω P, W,x + ω. P, W,x ω + W,x σ/2 \B C e 2+σd / /p C p/2 B 2 Dω 2 +ω 2 p/2 1/p ω + Ce σ / ω C /2 e 2+σ1 1/pd / ω. License or copyright restrictions may apply to redistribution; see

14 1778 E. N. DANCER AND SHUSEN YAN Similar to the proof of 2.23, we deduce that for i, P, W,xi /2 P, W,x /2 ω P, W,xi P, W,x 1/ ω Ce σ / ω. Combining 3.8, 3.9 and 3.10, we are led to W,x k ω P, W,x + ω 3.11 = /2 O e 1+σd/ + e σ/ ω. On the other hand, f,x yω f,x y 2 1/2 ω 3.12 C 1+/2 ω. Combining 3.7, 3.11 and 3.12, we obtain the result. Let Q,x be the bounded linear map E,x,k to E,x,k such that Q,x ω 1,ω 2 = 2 Dω 1 Dω 2 + ω 1 ω 2 Then we have k p 2 p 1 P, W ω,x + 1ω 2, ω 1,ω 2 E,x,k. Lemma 3.2. There are constants 0 > 0 and ρ>0 such that, for each 0, 0 ] and x D, Q,x ω ρ ω, ω E,x,k. Proof. We argue by contradiction. Suppose that there are n 0, x n D n and ω n E n,x n,k such that 3.13 Q n,x n ω n n = o1 ω n n, where o1 0asn +. From 3.13, we see k 2 p 2 n Dω n Dξ + ω n ξ p 1 P n,w n,x n, + ω nξ 3.14 =o1 ω n n ξ n, ξ E n,x n,k. In 3.14, we assume that 3.15 ω n n = n /2. For each fixed i, let ω n,i ỹ =ω n n ỹ + x n,i. License or copyright restrictions may apply to redistribution; see

15 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1779 Since x n,i,1 c > 0, from 3.15, we obtain 3.16 D ωn,i 2 + ω n,i 2 C, B R 0 for any R>0 large, where C>0isaconstant independent of R, B R 0 is the ball in R with radius R, centred at the origin. So, we may assume that there is an ω H 1 R such that, for any R>0, 3.17 ω n,i ω, weakly in H 1 B R 0, and 3.18 ω n,i ω, strongly in L 2 B R 0. Now, we prove ω =0. From 3.14, we see that ω n,i satisfies n z 1 + x n,i,1 m 1 D ω n,i Dξ + ω n,i ξ D n 3.19 p 1 n z 1 + x n,i,1 m 1 D n where and Ẽ n = { ξ : ξ ỹ x n,i n Hs, D n = {z : z R, n z + x n,i D}, V n, z =P n,w n,x n, n z + x n,i, V n, p 2 + ω n,iξ = o1 ξ n, ξ Ẽn, D n n z 1 + x n,i,1 m 1 DξD V n, x,h + ξ V n, x,h dz =0 }, for =1,,k, h =1,,N m +1. For any ξ C 0 R, we can choose a n,,h R 1, such that ξ n = ξ h=1 a n,,h V n, x,h Ẽn. For i, wehave n z 1 + x n,i,1 m 1 DξD V n, + ξ V n, dz = o1, D n x,h x,h for any ξ C0 R, since the support of V n, x,h tends to infinity as n +. On the other hand, we have n z 1 + x n,i,1 m 1 DξD V n, + ξ V n, dz = O1. D n x,h x,h So, it is easy to check that a n,,h 0asn + for i, while a n,i,h a i,h up to a subsequence. Putting ξ n into 3.19 and letting n +, notingthatx n,i,1 c > 0, we find DωDξ + ωξ p 1 U p 2 ωξ R R + h=1 a i,h R DωD U x h + ω U x h p 1 R U p 2 ω U x h =0. License or copyright restrictions may apply to redistribution; see

16 1780 E. N. DANCER AND SHUSEN YAN But DωD U + ω U p 1 U p 2 ω U =0. R x h x h R x h So, we obtain 3.20 DωDξ+ωξ U p 2 ωξ =0, ξ C0 R. R R Since U is nondegenerate, we see from 3.20 that 3.21 ω = h=1 b h U z h, for some b h R 1. On the other hand, from ω n, Ẽn, we can deduce U 3.22 DωD + ω U =0, R z h z h for h =1,,N m +1. Combining 3.21 and 3.22, we obtain ω =0. Asaresult,wehave 3.23 ωn 2 = on, i =1,,k, B i,r where B i,r = {y : y,y B n Rx n,i }. For y \ k i=1 B i,r, by Lemma A.1, we have P n,w n,x x, y = o R 1, where o R 1 0asR +. From 3.14 and 3.23, we find o n This is a contradiction. Let = ω n 2 n on o R 1n = n where α>0 is a small constant. S = { ω : ω H s, ω on o R 1n. e α ỹ x / }, Lemma 3.3. For any ω S with ω /2, we have R,x ω = O 3.24 p/2 ω p, R 3.25,xω,ξ = /2 O and 3.26 R,xωξ 1,ξ 2 =O where p is a constant with p >2. /2 ω p 2/2 ω p 2 ξ1 ξ 2, ξ License or copyright restrictions may apply to redistribution; see

17 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1781 Proof. Let p 2, 2N/N 2 be a constant with p <min3,p. For any ω S, we have ωy 1 2 if y \ k B. Since k P,W,x 1 2 for any y \ k B, we deduce that F y, P, W,x + ω F y, P, W,x k P, W ω 1 k,x + 2 p 1 p 2 P, W,x + ω2 = 1 p P, W,x + ω p 1 p P, W,x + p + k P, W,x + ω 1 k 2 p 1 p 2 P, W,x + ω2. By the definition of R,x ω, we see that for any ω S, 3.27 R,x ω C C ω min3,p + e θα/ k B =C ω min3,p C ω min3,p + C k B ω p k B ω min3,p + Ce θ/ p ω p, \ k B where θ >0 is a small constant. For any, let ωy =ωy + x. Then ω min3,p = c 0 z 1 m 1 ω min3,p dz B D B δ x C ω min3,p dz = C 3.28 D B δ x B, ω min3,p C D ω 2 + ω 2 minp,3/2 B, =C 2 Dω 2 + ω 2 min3,p/2 D B δ x C min3,p/2 ω min3,p, ω min3,p where B, = {z : z + x D B δ x }. In the last relation, we use y c>0 for y B. Combining 3.27 and 3.28, we obtain Now, we prove We have R,xω,ξ C ω min,2 ξ 3.29 =C ω min,2 ξ + C k B \ ω min,2 ξ, k B License or copyright restrictions may apply to redistribution; see

18 1782 E. N. DANCER AND SHUSEN YAN from which we can prove 3.25 by using the same techniques as in the proof of We can prove 3.26 in a similar way. Let us point out that 3.24, 3.25 and 3.26 are not true in the whole space H s. So, we need to carry out the reduction procedure in a closed subset of E,x,k. Proposition 3.4. There is an 0 > 0 such that for each 0, 0 ],thereexists a C 1 -map ω,x : D H s such that ω,x E,x,k, 3.3 holds for some constants G l. Moreover, we have 3.30 ω,x C σ+n m+2/2, where σ>0 is a constant. Proof. By Lemma 3.1, we know that there is a l,x E,x,k, such that l,x,ω = l,xω, ω E,x,k. Thus, solving 3.30 is equivalent to solving 3.31 l,x + Q,x ω + R,xω =0, in E,x,k. By Lemma 3.2, Q,x is invertible. So we can write 3.31 as 3.32 ω = G,x ω =: Q 1,xl,x Q 1,xR,xω. Let S = { ω : ω H s, ω α e α ỹ x /, ω N m+2/2}, where α>0 is a small constant. Now, we prove that G,x is a contraction map from S s to S. By 3.26, we see that for any ω 1,ω 2 S, 3.33 G,x ω 1 G,x ω 2 C R,xω 1 R,xω 2 C p 2/2 ω 1 ω 2. Thus, G,x is a contraction map. For any ω S,wehave G,x ω C l,x + C R,xω 3.34 C l,x + C /2 /2. For any x D, by Lemma 3.1, we have l,x C /2 k e 1+σd/ + C /2 1 θ1+σ +, which, together with 3.34, gives 3.35 since p >2. G,x ω C /2 1 θ1+σ + + /2 C σ1+n m+2/2 N m+2/2, License or copyright restrictions may apply to redistribution; see

19 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1783 To finish the proof of G,x ω S, we need to prove G,x ω α e α ỹ x /. Let ω 1 = G,x ω. Then, we have Q,x ω 1 = l,x R ω, in E,x,k, which is equivalent to Q,xω 1,ξ 3.36 = h=1 + l,x,ξ + R ω,ξ G h P, W,x x,h,ξ, for some G h R 1. We claim that there is a σ>0 such that 3.37 G h C σ+3/2, =1,,k, h=1,,n m +1. In fact, letting ξ = P,W,xi x in 3.36, using Lemma A.3, we can solve the i, h linear system to obtain G h C 1 /2 ω 1 + l,x + R ω C 1 / σ+/2 C σ+3/2. Rewrite 3.36 as 3.38 = 2 ω 1 + ω 1 p 1 k P, W,x p 2 + ω 1 ξ W,x + f,x y + k P, W,x + + f y, + =:G,x y h=1 P, W,x + ω f y, G h g,x y x,h P, W,x f y, P, W,x ω where fy, t is the function defined in 1.7, and g,x y =W,x + f,x y. Since ω S,wesee ω 1 2 in \ k B.Thus 3.39 f y, P, W,x + ω f y, P, W,x f y, P, W,x C ω, where p >2 is a constant. License or copyright restrictions may apply to redistribution; see

20 1784 E. N. DANCER AND SHUSEN YAN On the other hand, we have 3.40 = ξ W,x W,x + k P, W,x + P, W,x + O + e δ / + k P, W,x + P, W,x + =O e d/ U p 2 ỹ x. U p 2 ỹ x Direct calculations show that 3.41 g,x y C 1 U ỹ x ỹ x + CU. x,h Combining 3.39, 3.40 and 3.41, we find that 3.42 G,x y C C e d / U p 2 ỹ x α e α ỹ x /, + ω + σ+1/2 U ỹ x if α>0 is small enough. Since y c>0ify B,wecanprovethat 3.43 ω 1 C α, in k B. In fact, let g,x y, t = t + k p 2t. P, W,x Then g,x y, t C t and 2 ω 1 = G,x y+ g,x y, ω 1, in B. Since ω 1 H s, we can rewrite the above equation as 2 where B = D B δ x. z m 1 1 z ω 1 = z m 1 1 G,x z+ g,x z, ω 1, in z B, License or copyright restrictions may apply to redistribution; see

21 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1785 Let ω 1 z =ωz + x. Then z1 + x,1 m 1 ω 1 z z =z 1 + x,1 m 1 G,x z + x,1 + g,x z + x,1, ω 1, in B,, where B, = {z : z + x B }. For any x 0 B,, by 3.35, we have ω 1 2 = B ω 1 2 C ω 1 2 C 1+2σ. B 1 x 0 So, by 3.42, we have ω 1 L B 1 x 0 C ω 1 L2 B 2 x 0 + C G,x L B 2 x 0 C α. Thus, 3.43 follows. Let p 2η, a y = P, W,x where η is a C 1 function, such that η =0ify k B. It is easy to see that a y 0 uniformly in as 0. From 3.43, we have p ω 1 +1 p 1a ω 1 = G,x y+o α W,x Similar to the proof of Lemma A.1, we can prove that v = α satisfies 2 v +1 p 1a v = 1 p 2a y α 2 α 1 2 α e α ỹ x / e α ỹ x / p 2. e α ỹ x / G,x y + O α W,x Since ω 1 H s, by the maximum principle, we obtain ω 1 v = α e α ỹ x /. Thus, ω 1 S. We have proved that G,x is a contraction map from S into itself. By the contraction mapping theorem, we know that there is a ω,x S,x, such that ω,x = G,x ω,x. Moreover, by 3.35, ω,x C σ+n m+2/2. License or copyright restrictions may apply to redistribution; see

22 1786 E. N. DANCER AND SHUSEN YAN We will choose x D, such that the corresponding G l in 3.3 are all zero. Proof of Theorem 1.1. Let Kx =Jx, ω,x, x D. Consider the following problem: 3.45 min Kx. x D Let x D be a minimum point of We will prove that x is an interior point of D.Thus,x is a critical point of Kx. It follows from Propositions 3.4 and 2.3 that for any x D, Kx =Jx, 0 + O N m+2+σ 3.46 =A x m 1,1 + 1 τ,x + O k e 2+σd/ +, 2 Let x, = x,,1,x,, x,,1 = x,1 + L ln ψ x, x, = x, x = x,1,, x,k, where L>0 is a large constant. Then x D. By Lemma 2.1, we see that τ, x = O N m+3, if L>0 is large. So, from 3.46, we obtain 3.47 K x =A x m 1,1 + O ln. Suppose that x D.Ife 2dx,, D/ = 1 θ for some, then, by 3.46, Kx A x m 1, θ1 θ + O N m+2 >K x, 2 since 1 θ1 + θ < 1ifθ>0issmallenough.Thisisacontradiction. Suppose that x D \{x : e 2dx, D/ = 1 θ, for some }. Then x,,1 x,1 + β for some, whereβ>0 is a small constant. So, by 3.46, Kx x m 1,1 + c + O >K x, where c > 0 is a small constant. This is a contradiction. So x is an interior point of D.Asaresult, DKx =0. We claim that for this x, the corresponding G h = 0, = 1,,k, h = 1,...,. So k P,W,x, +ω,x is a solution of 1.8. Since fy, t =0 if t 0, we see that k P,W,x, + ω,x is positive. But ω,x S. Thus ω,x 1 2 in \ k B,whichgives f k y, P, W,x, + ω,x = P, W,x, + ω.,x + License or copyright restrictions may apply to redistribution; see

23 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1787 As a result, k P,W,x, + ω,x is a solution of 1.1. Now, we prove G h =0. Since 3.48 Kx x ih = I = = l=1 l=1 l=1 P, W,x, + ω,x, P,W,x,i + ω,x x ih x ih G l P, W,x, x l G l P, W,x, x l, P,W,x,i x ih G il 2 P, W,x,i x il x ih,ω,x, P,W,x,i x ih + ω,x x ih we obtain 3.49 l=1 l=1 G l P, W,x, x l, P,W,x,i x ih G il 2 P, W,x,i x il x ih,ω,x =0. We claim P, W,x,i x il x ih = O 2+/2. Assume this at the moment. Then 2 P, W,x,i x il x ih,ω,x = O 2+/2 ω,x = o N m 1, which, together with Lemma A.3 and 3.49, implies G l =0, =1,,k, = 1,,N m +1. Now we prove We have P,W,xi + P,W,xi x il x ih x il x ih = 2 ξ W,x x il x i + f,x y, in, ih and P, W,xi x il x ih =0 on. License or copyright restrictions may apply to redistribution; see

24 1788 E. N. DANCER AND SHUSEN YAN Multiplying 3.51 by P,W,xi x il x ih and integrating by parts, we obtain 2 P, W,x,i 2 C 2 ξ W,x x il x i + W,x 2 P, W,x,i ih x il x ih C 2 ξ W,x i + W,x 2 1/2 2 P, W,x,i x il x ih C 2+/2 2 P, W,x,i. x il x ih Thus, 3.50 follows. Appendix A. From the choice of x,theterme dx, D/ is only algebraically small. To obtain a solution concentrating near several manifolds, we need to prove that P, W,x is exponentially small outside a small neighbourhood of the set {y : y = x,1,y = x }. Lemma A.1. Let θ>0 be any small constant. There is a constant C>0such that P, W,x Ce 1 θ ỹ x /. Proof. Let v = Ce 1 θ ỹ x /,wherec>0 is a large constant. Then A.1 2 v + v 1 2 Cθe 1 θ ỹ x /. On the other hand, we have A.2 2 P, W,x + P, W,x =ξ W,x + f,x y =ξ W,x + O ξ e ỹ x /. But if C>0islarge enough, then 1 A.3 2 Cθe 1 θ ỹ x / ξ W,x + O ξ e ỹ x /. By the maximum principle, we obtain from A.1, A.2 and A.3 that P, W,x v = Ce 1 θ ỹ x /. Lemma A.2. Let θ>0 be any small constant. There is a constant C>0such that P,W,x C 1 e 1 θ ỹ x /. x h Proof. We know that P,W,x x h A.4 and 2 P,W,x x h satisfies + P,W,x = ξ W,x x h x + f,x y h, in, A.5 P, W,x x h =0, on. License or copyright restrictions may apply to redistribution; see

25 SINGULARLY PERTURBED ELLIPTIC PROBLEM 1789 On the other hand, it is easy to check that A.6 W,x x + f,x y C 1 W,x + f,x y. h Thus we can prove this lemma in a similar way as in Lemma A.1. Lemma A.3. We have P, W,x, x l where c >0 is a constant. Proof. We have A.7 P, W,x,, P,W,x,i x l x ih p 2 W,x, =p 1 ξ W,x, P {,W,x,i x = c N m 1, i =, l = h, ih o N m 1, otherwise, x l, P,W,x,i x ih P, W,x,i + f,x y x l x ih =p 1 p 2 W,x, P, W ξ W,x,i,x + O x l x ih U ỹ x P, W,x,i. x ih It follows from Lemma A.2 that U ỹ x P, W,x,i x ih A.8 C 1 U ỹ x e 1 θ ỹ x / C N m. Using Lemma A.2, A.7 and A.8, we conclude that if i, then P, W,x,, P,W,x,i x l x = on m 1. ih It remains to study the case i =. Using 2.2, we deduce ϕ,x i C 1 e di/. x ih Thus, A.9 = = p 2 ξ i W W,x,i P, W,x,i,x i x il x ih W,x p 2 W,x,i W,x,i i + O N m 1 e d i/ x il x ih { c N m 1 + O N m 1 e d i/, l = h, O N m 1 e d i/, l h. License or copyright restrictions may apply to redistribution; see

26 1790 E. N. DANCER AND SHUSEN YAN References [1] A. Ambrosetti, A. Malchiodi and W. M. Ni, Singularly perturbed elliptic equations with symmetry: Existence of solutions concentrating on sphere. I, Comm. Math. Phys., , MR c:35014 [2] A. Ambrosetti, A. Malchiodi and W. M. Ni, Singularly perturbed elliptic equations with symmetry: Existence of solutions concentrating on sphere. II, Indiana University Math. J., , MR c:35015 [3] D. Cao, E. N. Dancer, E. Noussair and S. Yan, On the existence and profile of multi-peaked solutions to singularly perturbed semilinear Dirichlet problems, Discrete and Continuous Dynamical Systems, 21996, MR m:35095 [4] E. N. Dancer, Some singularly perturbed problems on annuli and a counterexample to a problem of Gidas, Ni and Nirenberg, Bull. London Math. Soc., , MR m:35016 [5] E. N. Dancer and J. Wei, On the effect of domain topology in a singular perturbation problem. Topological Methods in Nonlinear Anal., , MR a:35012 [6]E.N.DancerandS.Yan,A singularly perturbed elliptic problem in bounded domains with nontrivial topology, Adv.Diff.Equations,41999, MR d:35009 [7] E.N.DancerandS.Yan,Interior and boundary peak solutions for a mixed boundary value problem, Indiana Univ. Math. J., , MR f:35146 [8] E.N.DancerandS.Yan,Singularly perturbed elliptic problem in exterior domains, J.Diff. Int. Equations, , MR c:35022 [9] E. N. Dancer and S. Yan, Effect of the domain geometry on the existence of multipeak solutions for an elliptic problem, Top. Meth. Nonlinear Anal., , MR b:35106 [10] M. Del Pino and P. Felmer, Spike-layered solutions of singularly perturbed elliptic problems in a degenerate setting, Indiana Univ. Math. J., , MR b:35027 [11] M. K. Kwong, Uniqueness of positive solutions of u + u = u p in R n, Arch. Rat. Mech. Anal , MR d:35015 [12] Y. Y. Li and L. Nirenberg, The Dirichlet problem for singularly perturbed elliptic equations, Comm. Pure Appl. Math , MR g:35014 [13] A. Malchiodi and M. Montenegro, Boundary concentration phenomena for a singularly perturbed elliptic problem, Comm. Pure Appl. Math., , MR g:35005 [14] E. S. Noussair and S. Yan, The effect of the domain geometry in singular perturbation problems, Proc. London Math. Soc., , MR m:35013 [15] W. M. Ni and I. Takagi, Locating the peaks of least energy solutions to a semilinear Neumann problem, Duke Math. J , MR h:35072 [16] W. M. Ni and J. Wei, On the location and profile of spike-layer solutions to singularly perturbed semilinear Dirichlet problems, Comm. Pure Appl. Math , MR g:35077 [17] J. Wei, On the construction of single-peaked solutions to a singularly perturbed semilinear Dirichlet problem, J. Diff. Eqns., , MR f:35015 [18] J. Wei, On the effect of domain geometry in singular perturbation problems, Differential Integral Equations, , MR c:35032 [19] S. Yan, On the number of interior multipeak solutions for singularly perturbed Neumann problems, Topological Methods in Nonlinear Anal., , MR c:35024 School of Mathematics and Statistics, University of Sydney, NSW 2006, Australia address: normd@maths.usyd.edu.au School of Mathematics, Statistics and Computer Science, The University of New England, Armidale, NSW 2351, Australia address: syan@turing.une.edu.au License or copyright restrictions may apply to redistribution; see