The Symmetry of Least-Energy Solutions

for Semilinear Elliptic Equations

Jann-Long Chern∗ and Chang-Shou Lin†

Abstract. In this paper we will apply the method of rotating planes (MRP) to investigate

the radial and axial symmetry of the least-energy solutions for semilinear elliptic equations

on the Dirichlet and Neumann problems respectively. MRP is a variant of the famous

method of moving planes (MMP). One of our main results is to consider the least-energy

solutions of the following equation

4u+K(x)up = 0, x ∈ B1,

u > 0 in B1, u|∂B1= 0,

(∗)

where 1 < p < n+2n−2 and B1 is the unit ball of R

n with n ≥ 3. Here K(x) = K(|x|) is

not assumed to be decreasing in |x|. In this paper, we prove that any least-energy solution

of (∗) is axially symmetric with respect to some direction. Furthermore, when p is close

to n+2n−2 , under some reasonable condition of K, radial symmetry is shown for least-energy

solutions. This is the example of the general phenomenon of the symmetry induced by

point-condensation. A fine estimate for least-energy solution is required for the proof of

symmetry of solutions. This estimate generalizes the result of Han [H] to the case when

K(x) is nonconstant. In constrast to previous works for this kinds of estimates, we only

assume that K(x) is continous.

1

1. Introduction

Recently in the research area of nonlinear elliptic PDEs, there have been many

works devoted to studying problems where solutions exhibit the “phenomenon of point-

condensation”. Two well-known examples are semilinear ellitptic equations involving the

Sobolev critical exponent and nonlinear elliptic equations with small diffusion coefficient.

These works show that the concentration often induces the asymptotic symmetry. For

example, spherical Harnack inequalities have been proved for blowup solutions to either

mean field equations on compact Riemann surfaces or the scalar curvature equation. These

spherical Harnack inequalities implies that blowup solutions usually are asymptotically

symmetric. Similar results were proved for spike-layer solutions of singularly elliptic Neu-

mann problem. See [CL1], [CL2], [L1], [L2] and [NT] for more precise statements. Natu-

rally, when the underlying equation is invariant under a group of transformations, we would

like to know whether solutions with point-condensation actually possess certain symmetry

which is invariant under the action of some elements of the group. In [Ln], for the mean

field equation on S2, the second author first succeeded to prove the axial symmetry for

solutions with two blowup points. In this article, we continue to study this problem.

In this paper, we first consider positive solutions of the following equation.

4u+ f(|x|, u) = 0 in B1

u|∂B1= 0,

(1.1)

where B1 is the unit ball in Rn, n ≥ 2, 4 is the Laplace operator and f(r, t) is a C1 function

of both variables r and t. The typical examples of f are K(|x|)up where 1 < p < n+2n−2

if

n ≥ 3, 1 < p if n = 2. When K(r) is decreasing in r, the famous theorem by Gidas, Ni and

Nirenberg says that any positive solution u(x) of (1.1) is radially symmetric. However, the

radial symmetry of solutions generally fails if K(r) does not decrease with respect to r for

2

all r ≤ 1. In this paper, we want to show that certain symmetry still holds for least-energy

solutions. The definition of the least-energy solutions of (1.1) is stated as follows. Consider

the variational functional

J(u) =

∫

B1

[1

2|∇u|2 − F (|x|, u+)] dx in H1

0 (B1), (1.2)

where F (r, u) =∫ u

0f(r, s) ds and u+(x) = max(0, u(x)). For the nonlinear functional J ,

we set

c∗ = infh∈Γ

max0≤t≤1

J(h(t)), (1.3)

where

Γ = h ∈ C([0, 1], H10(B1)) |h(0) = 0, h(1) = e

and e ∈ H10 (B1), e 6= 0 in B1 with J(e) = 0. To guarantee the c∗ of (1.3) to be a critical

value of J by the mountain pass lemma, the nonlinear term f is usually assumed to satisfy

the following condition.

(fa) f(r, t) = o(|t|) near t = 0 and 0 ≤ r ≤ 1;

(fb) there exist constants θ ∈ (0, 12) and U0 > 0 such that 0 < F (x, u) ≡

∫ u

0f(r, s) ds ≤

θuf(x, u) for all u ≥ U0;

(fc) |f(x, t)| ≤ Ctq for some 1 < q < n+2n−2 for large t if n ≥ 3 and 1 < q < +∞ if n = 2.

Using the above conditions (fa) through (fc) and by the well-known mountain-pass lemma

due to Ambrosetti and Rabinowitz (see[AR]), we can obtain that (1.2) possesses a positive

critical point u∗ with its critical value J(u∗) to be equal to c∗ of (1.3). Moreover, under

the additional assumption that f(r, t)/t is increasing in t, from the Lemma 3.1 in [NT], c∗

does not depend on the choice of e and is the least positive critical value of J . Therefore,

We call such u∗ to be a least-energy solution of Eq. (1.1). We remark that solutions of

3

least energy can also be obtained by minimization of

infv∈H1

0(Ω)

∫

|∇v|2

[

(∫

K(v+)p+1)+]

2p+1

,

where f(x, u) = K(x)up with maxΩ

K > 0.

Our first result is concerned with the axial symmetry of the least-energy solution of

equation (1.1).

Theorem 1.1 Suppose f satisfies the conditions (fa) through (fc) and

(fd)∂2f∂t2 (r, t) > 0 for t > 0 and for 0 ≤ r ≤ 1.

Let u be a least-energy solution of Eq. (1.1) and P0 be a maximum point of u. Then the

following conclusions hold.

(i) If P0 = O is the origin, then u is radially symmetric.

(ii) If P0 6= O, then u is axially symmetric with respect to−−→OP0 and on each sphere

Sr = x : |x| = r for 0 < r < 1, u(x) is increasing as the angle of−→Ox and

−−→OP0

decreases. In particular, u satisfies

xj∂u

∂xn(x)− xn

∂u

∂xj(x) > 0 for xj > 0, (1.4)

where P0 is assumed to locate on the positive xn−axis.

We note that, by the condition (fd), it is easy to see that ∂∂t (t

∂f(r,t)∂t − f(r, t)) =

t∂2f(r,t)∂t2

≥ 0 ∀t > 0 ∀0 ≤ r ≤ 1 and hence f(r, t)/t is increasing in t. So, by (fa), we have

f(r, t) ≥ 0 ∀t ≥ 0 ∀0 ≤ r ≤ 1. In this paper, we will use the method of rotating planes

to prove Theroem 1.1. The method of rotating planes is a variant of the famous method

of moving planes(MMP). MMP was first invented by Alexandroff and later was used by

Gidas-Ni-Nirenberg to prove the radial symmetry of positive solutions. See [BN], [CGS],

4

[GNN] and the references therein. Recently, MMP was applied to prove spherical Harnack

inequality for blowup solutions to either scalar curvature equation or mean field type

equations. See [CL1], [CL2], [L1] and [L2]. We note that MMP can not be applied to the

Neumann problem for semilinear elliptic equations. As far as the authors know, the result

concerning the of radial symmetry for the Neumann problem is very rare. Nevertheless,

our next result shows that the method of rotating planes can be employed for the Neumann

problem and the axial symmetry can be established by this method.

Our second result is about the axial symmetry of the least-energy solutions of the

Neumann problem. We consider the following equation.

d4u− u+ f(u) = 0 in B1,

u > 0 in B1

∂u

∂ν= 0 on ∂B1,

(1.5)

where d is a positive parameter and f satisfies the conditions (fa)− (fc).

The typical examples of f are up where 1 < p < (n + 2)/(n − 2) if n ≥ 3, and

1 < p < +∞ if n = 2. In this case, Eq. (1.5) is the steady state problem for a chemotactic

aggregation model with logarithmic sensitivity by Keller and Segel [KS]. It can be also

considered as the shadow system of some reaction-diffusion system in chemotaxis, see e.g.

[NT2].

Under the conditions (fa)− (fc), Theorem 2 and Proposition 2.2 in [LNT] guarantee

that, for each d > 0, (1.5) possesses a solution ud which is a critical point of the variational

functional

Jd(u) =1

2

∫

B1

(d|∇u|2 + u2) dx−

∫

B1

F (u+) dx ∀u ∈ H1(B1), (1.6)

where u+ = maxu(x), 0, and its critical value cd = Jd(ud) is proved to be equal to

cd = infh∈Γ

max0≤t≤1

J(h(t)) (1.7)

5

where Γ is defined as before, and cd is independent of the choice of e by the Lemma 3.1 in

[NT]. Such a critical point ud is called a least-energy solution of equation (1.5).

Our second result is in the following.

Theorem 1.2 Suppose the conditions (fa) through (fd) hold. Let u be a least-energy

solution of (1.5) and P0 be a local maximum point of u on B1. Then either u ≡ constant

in B1, or P0 6= 0 and u is axially symmetric with respect to−−→OP0 and (1.4) holds where P0

is assumed to locate on the positive xn−axis.

We give some remarks here. (I) If f satisfies the conditions (fa) through (fd), then

any nonconstant least-energy solution u must be non-radial because the origin can not be

a critical point of u. See the proof of Theorem 1.2. (II) Theorem 1.2 had been proved in

[Proposition 5.1, NT] when P0 is assumed to locate on the boundary of B1 and f(t) is an

analytic function for t > 0. After the paper is finished, a stronger version of Theorem 1.2,

that is, P0 must be on the boundary ∂B1, has been proved in [LN2].

Before stating our third result, we let Sn be the best Sobolev constant, i.e., for any

bounded domain Ω of Rn and for n ≥ 3,

Sn = infv∈H1

0 (Ω)

∫

Ω|∇v|2 dx

(∫

Ω|v|

2nn−2 dx)

n−2n

. (1.8)

It is well known that the best Sobolev constant is independent of Ω and is never achieved

by an element in H10 (Ω). For a C1 function K with max

B1

K > 0, we consider a least-energy

solution u of

4u+K(x)up = 0, x ∈ B1,

u > 0 in B1, u|∂B1= 0,

(1.9)

6

where 1 < p < n+2n−2

. Suppose ui is a least-energy solution of (1.9) with p = pi ↑n+2n−2

. It is

easy to see that ui achieves the infinimum of the variational problem,

∫

B1|∇ui|

2 dx

(∫

B1K(x)upi+1

i dx)2

pi+1

= infv∈H1

0(B1)

∫

B1|∇v|2 dx

[(∫

B1K(x)(v+)pi+1 dx)+]

2pi+1

= (1

(maxB1

K)2

pi+1

· Sn) · (1 + o(1)),(1.10)

for i sufficiently large. Let Pi be the global maximum point of ui. Since (1.8) is never

achieved by some function in H10 (B1), we have

limi→∞

ui(Pi) = +∞ and limi→∞

K(Pi) = maxB1

K(x). (1.11)

Obviously, by (1.11), the necessary condition for ui to be radially symmetric is thatK(x) =

K(|x|) and the origin is the maximum point of K. For the final result, we want to prove

what is the sufficient condition of K such that for any least-energy solution is radially

symmetric. Suppose K(x) = K(|x|) satisfies the following condition.

There exists r0 ∈ (0, 1] such that K ′(r) ≤ 0 ∀ 0 ≤ r ≤ r0

and max(0, K(r)) < K(0) ∀ r0 ≤ r ≤ 1.(Ka)

Note that (Ka) could allow K(r) ≡ K(0) for all small r > 0. In this case, it is not evident

that the maximum point Pi of ui would tend to the origin. However, we have the following.

Theorem 1.3 Suppose the condition (Ka) holds. Then there exists a small ε > 0 such

that for any least-energy solution u of (1.9) with 0 < n+2n−2

− p ≤ ε, u is radially symmetric.

Furthermore, if ( rK′(r)K(r)

)′ ≤ 0 ∀0 ≤ r ≤ r1 for some r1 ≤ r0, then (1.9) possesses a unique

least-energy solution when 0 < n+2n−2 − p ≤ ε.

The proof of Theorem 1.3 is more complicated than previous theorems. Here, the

concentration actually occurs for least-energy solutions as p tends to n+2n−2

. In order to start

7

the process of the method of rotating planes, we have to require some fine estimates for

least-energy solutions, that is, we have to show that least-energy solutions always behaves

“simply” near its blowup point. When K(x) ≡ a positive constant, this was proved by

Han [H]. However, for a nonconstant function K(x), there is additional difficulty even by

using Han’s method. In the appendix, we give a proof which is simpler in conception even

for the case when K is a constant. Since the Pohozaev identity is not employed in our

proof, we do not require any smoothness assumption on K.

We organize this paper as follow. In Section 2, we prove Theorem 1.1 by using the

method of rotating planes. By the same method, the axial symmetry of the Neumann

problem is established in Section 3. Here we emphasize that any nonconstant least-energy

solution must be non-radially symmetric. Finally we complete the proof of Theorem 1.3

in Section 4.

2. Maximum Principle via the Method of Rotating Planes

In this paper we will give the detail of the proof of Theorem 1.1.

Proof of Theorem 1.1. Let u be a least-energy solution of Eq. (1.1). We divide the

proof into the following steps.

Step 1. Let T be any hyperplane which passes the origin O. We claim that one of

the following conclusions holds :

(A) u(x) = u(x∗) ∀x ∈ B+1 ,

(B) u(x) > u(x∗) ∀x ∈ B+1 ,

(C) u(x) < u(x∗) ∀x ∈ B+1 , where B+

1 is one of half-balls of B1\T and x∗ is the reflection

point of x with respect to the hyperplane T .

8

First, we will prove that

either u(x) ≥ u(x∗) ∀x ∈ B+1

or u(x) ≤ u(x∗) ∀x ∈ B+1 .

(2.1)

Suppose the conclusions of (2.1) are not true. Then the following two sets are all nonempty.

Ω+ = x ∈ B+1 |u(x) > u(x∗),

Ω− = x ∈ B+1 |u(x) < u(x∗).

(2.2)

Let

w(x) = u(x)− u(x∗) ∀x ∈ B+1 . (2.3)

Then w satisfies

4w + fu(|x|, z(x))w = 0, x ∈ B+1

w|∂B+1

= 0,(2.4)

where z(x) is between u(x) and u(x∗). Let

Ω∗− = x∗|x ∈ Ω−. (2.5)

Set

v(x) =

w(x) if x ∈ Ω+;

dw(x∗) if x ∈ Ω∗−;

0 otherwise.

(2.6)

Choose the constant d > 0 such that

∫

B1

v(x)φ1(x) dx = 0, (2.7)

where φ1 is the first eigenfunction of the following eigenvalue problem

4φ+ fu(|x|, u)φ = −λφ, x ∈ B1,

φ|∂B1= 0.

(2.8)

9

Let λ2 be the second eigenvalue of (2.8). By the condition (fd), we easily have ∂∂t

(t∂f(r,t)∂t

−

f(r, t)) = t∂2f(r,t)∂t2 ≥ 0 ∀t > 0 ∀0 ≤ r ≤ 1 and hence f(r, t)/t is nondecreasing in t. Since u

is a least-energy solution of (1.1), by using the same method of the proof of Theorem 2.11

in [NL], we have

λ2 ≥ 0. (2.9)

By the condition (fd) and (2.2)-(2.6), it is easy to see that

4v + fu(|x|, u(x))v

≥ 0 ∀x ∈ Ω+;

≤ 0 ∀x ∈ Ω∗−;

= 0, otherwise;

6≡ 0 in B1. (2.10)

From (2.6), (2.7), (2.9),(2.10) and v 6≡ 0, we obtain

0 >

∫

B1

(−v(x)) · [4v(x) + fu(|x|, u(x))v(x)] dx

=

∫

B1

[|Dv(x)|2 − fu(|x|, u(x))v2(x)] dx ≥ 0,

(2.11)

a contradiction. This proves (2.1). By (2.1), we may assume w(x) ≥ 0 for x ∈ B+1 . Since

w(x) satisfies Eq. (2.4), by using the strong maximum principle, we have either w(x) ≡ 0

for x ∈ B+1 or w(x) > 0 for x ∈ B+

1 . Similarly, if w(x) ≤ 0, we have either w(x) ≡ 0 on

B+1 or w(x) < 0 on B+

1 This finishes step 1.

Step 2. If P0 = O, then we want to prove that u is symmetric with respect to any linear

hyperplane and then u is radially symmetric. Without loss of generality, we assume that

the hyperplane is x1 = 0, that is, we want to prove w(x) = u(x)−u(x−) ≡ 0 for x1 > 0,

where x = (x1, x′) and x− = (−x1, x

′). Suppose w(x) 6≡ 0 for x ∈ B+1 = x ∈ B1|x1 > 0.

Then, by step 1, we may assume that w(x) > 0 ∀x ∈ B+. Then, from (2.4) and applying

the Hopf lemma, we have

∂w

∂(−x1)(O) = −2

∂u

∂x1(O) < 0. (2.12)

10

However, since O is the maximal point of u in B1, we have ∂u∂x1

(O) = 0, which yields a

contradiction to (2.12). Thus, w(x) ≡ 0 and the radial symmetry of u follows readily. We

prove part (i) in Theorem 1.1.

Step 3. If P0 6= O, then we will prove that part (ii) in Theorem 1.1 hold. Without

loss of generality, we may assume−−→OP0 is the positive xn−axis. Let P0 = (0, . . . , 0, t) and

T0 be the hyperplane xn = 0. Then u(P0) ≥ u(P−0 ), and from step 1, we obtain that :

either u(x) = u(x−) ∀x ∈ B+1

or u(x) > u(x−) ∀x ∈ B+1 ,

(2.13)

where x = (x′

, xn), x− = (x′

,−xn), B+1 = x ∈ B1|xn > 0. If the former case holds,

then u(P0) = u(Q0) = maxB1

u(x), where Q0 = (0, . . . , 0,−t). Let T be any hyperplane

passing through the origin such that P0 6∈ T and B+(T ) be the half-ball of B1\T such that

P0 ∈ B+(T ). Because Q∗∗0 = Q0 6∈ B

+(T ) and u(P0) ≥ u(P ∗0 ) and u(Q∗0) ≤ u(Q∗∗0 ), where

x∗ is the reflection point of x w.r.t. T , by step 1, we must have u(x) = u(x∗) ∀x ∈ B+(T ).

Since T is any hyperplane, we conclude that u is radially symmetric in this case. If the

latter case is true, then we will prove that u is axially symmetric with respect to−−→OP0 and

the second conclusion of part (ii) in Theorem 1.1 holds. Consider any two-dimensional

plane which contains P0. For the simplicity, let us assume that the plane is spanned by

e1 = (1, 0, . . . , 0) and en = (0, . . . , 0, 1). Let lθ be the line having the angle θ with x1−

axis, and νθ, with ν0 = en, be the normal vector to the line in this plane. Set Tθ to be the

(n − 1)−dimensional linear hyperplane which passes the origin and has νθ as the normal

vector. Obviously, Tθ = (r1 cos θ, x2, . . . , xn−1, r1 sin θ)|xj ∈ R for 2 ≤ j ≤ n−1 and r1 >

0. Let Bθ be one of the half-balls of B1\Tθ which contains P0 for 0 ≤ θ < π2 . Let xθ be

the reflection point of x ∈ Bθ w.r.t. Tθ.

Set

wθ(x) = u(x)− u(xθ) ∀x ∈ Bθ. (2.14)

11

Then wθ satisfies

4wθ + cθ(x)wθ = 0 in Bθ

wθ(x) = 0 on ∂Bθ,(2.15)

where

cθ(x) =f(|x|, u(x))− f(|x|, u(xθ))

u(x)− u(x∗).

For θ = 0, we have w0(x) > 0 ∀x ∈ B0. Set

θ0 = supθ|wθ(x) ≥ 0 ∀x ∈ Bθ ∀0 ≤ θ ≤ θ ≤π

2. (2.16)

We claim that θ0 = π2. Suppose this is not true. Then, from step 1 and the definition of

θ0, we have for 0 ≤ θ < θ0,

wθ(x) > 0 for x ∈ Bθ and∂u

∂νθ(x) < 0 for x ∈ Tθ,

wθ0≡ 0 in Bθ0

.

(2.17)

Let P ∗0 be the reflection point of P0 w.r.t. Tθ0. Then P ∗0 is also a global maximum point.

Since w0(x) > 0 in B0, we have P ∗0 ∈ Tθ1for some θ1 ∈ (0, θ0) and ∇u(P ∗0 ) = 0, which

yields a contradictions to (2.17). Hence, we have

wπ2≥ 0 in Bπ

2. (2.18)

Similarly, using the above arguments, we can also obtain

w− π2≥ 0 in B−π

2. (2.19)

From (2.18) and (2.19) we deduce that

wπ2≡ 0 in Bπ

2. (2.20)

The axial symmetry follows readily from (2.20).

12

Let x = (r1 cos θ, x2, . . . , xn−1, r1 sin θ), r1 = (|x|2 − x22 − · · · − x2

n−1)1/2. Then, from

∂u

∂θ(x) =

−1

2

∂wθ

∂νθ(x) > 0 ∀ x ∈ (B1 ∩ Tθ) ∀

−π

2< θ <

π

2, (2.21)

where νθ is the outnormal of Σθ on the boundary Tθ, the monotonicity follows clearly.

From (2.20) and (2.21), we easily obtain (1.4). This proves step 3 and completes the proof

of Theorem 1.1.

3. Axial Symmetry for the Neumann Problem

In this section, we will complete the proof of Theorem 1.2.

Proof of Theorem 1.2. We divide the proof into the following steps.

Step 1. Let u be the least-energy solution of (1.5). Consider the following eigenvalue

problem

d4φ− φ+ fu(|x|, u)φ+ λφ = 0, x ∈ B1,

∂φ

∂ν= 0 on ∂B1.

(3.1)

From the conditions (fa) − (fd) and Theorem 2.11 in [NL], we obtain that the second

eigenvalue of (3.1) is nonnegative, i.e.,

λ2(u) ≥ 0. (3.2)

Let T be any hyperplane which contains the origin. Then, using the same arguments

in step 1, we also obtain that one of the following conclusion holds.

u(x) = u(x∗),

u(x) > u(x∗),

u(x) < u(x∗),

for all x ∈ B+, (3.3)

13

and

the outnormal derivative∂w0

∂ν(x) < 0 for x ∈ T \ ∂B1, (3.4)

where w0(x) = u(x)− u(x∗) and B+ is one of half-balls of B1 which is divided by T such

that the maximum point P0 ∈ B+. Step 2. We want to prove u ≡ constant if u(x) is

radially symmetric. Let x = (x1, . . . , xn) and r = |x|. If u is radial symmetry, then u

satisfies

d(u′′ +n− 1

ru′)− u+ f(u) = 0, r > 0,

u(0) = α > 0, u′(0) = 0,

(3.5)

and we have

∂u

∂x1= u′(r)

x1

rin B1 and

∂u

∂x1= 0 on ∂B1. (3.6)

Suppose u 6≡ constant, then ∂u∂x1

6≡ 0. Let w(r) be the first eigenfunction of (3.1).

From (3.6), we easily have∫

B1

w(|x|)∂u

∂x1(|x|) dx = 0. (3.7)

By (1.5) we obtain

d4(∂u

∂x1)−

∂u

∂x1+ f ′(u)

∂u

∂x1= 0

∂u

∂x1= 0 on ∂B1.

(3.8)

Now using (3.2), (3.7) and (3.8), we obtain that

0 ≤ λ2 = infv⊥w,v∈H1(B1)

∫

B1

[d|∇v|2 + v2 − f ′(u)v2] dx

=

∫

B1

[d|∇(∂u

∂x1)|2 + (

∂u

∂x1)2 − f ′(u)(

∂u

∂x1)2] dx

= 0

(3.9)

Since ∂u∂x1

archives the infinimum of (3.9), we obtain that

∫

∂B1

∂

∂ν(∂u

∂x1) · φ dσ = 0 ∀φ ∈ H1(B1) and φ⊥w. (3.10)

14

Set φ = x1∂

∂x1( ∂u

∂x1) + x2

∂∂x2

( ∂u∂x1

). Since φ is odd in x1, we have φ⊥w. Then we obtain

that ∂∂ν ( ∂u

∂x1) = 0 on ∂B1 and ∂u

∂x1is a solution of the Neumann problem (1.5). Hence we

have u′′(1) = 0 and, from Eq. (1.5), u(1) = f(u(1)). From Eq. (3.5) and the uniqueness

of ODE, we finally obtain that u ≡ u(1). This contradiction proves that u ≡ constant if

u(x) is radially symmetric.

Now suppose u 6≡constant. Let P0 be a maximum point of u on B1. If P0 = O, then,

from the above step 1 and using the same arguments in step 1 of the proof of Theorem 1.1,

we obtain that u is radially symmetric. By the above step 2, u ≡ constant in this case, a

contradiction. Hence P0 6= O if u is nonconstant.

Step 3. We claim that u is axially symmetric w.r.t.−−→OP0 and (1.4) holds.

Without loss of generality, we may assume−−→OP0 is the positive xn−axis and T0 is the

hyperplane xn = 0. Consider any two-dimensional plane where P0 is contained. For the

simplicity, we assume the plane is spanned by e1 = (1, 0, . . . , 0) and en = (0, . . . , 0, 1). Let

lθ be the line having the angle θ with x1− axis, and νθ be the normal vector to the line in

this plane. Set Tθ to be the (n−1)−dimensional linear hyperplane which passes the origin

and has νθ as the normal vector. Let Bθ be one of the half-balls of B1 which is divided by

Tθ and P = (0, · · · , 0, t) ∈ Bθ ∀0 ≤ θ < π2. Let Σθ denote the component of Bθ\Tθ and

xθ be the reflection point of x w.r.t. Tθ.

Set

wθ(x) = u(x)− u(xθ) ∀x ∈ Σθ. (3.11)

Clearly, wθ satisfies

d4wθ + (cθ(x)− 1)wθ = 0,

wθ(x) = 0 on Tθ and∂wθ

∂ν= 0 on ∂B1 ∪ Σθ,

(3.12)

where

cθ(x) =f(u(x))− f(u(xθ))

u(x)− u(x∗).

15

We want to prove

wθ(x) > 0 for x ∈ Σθ and 0 ≤ θ <π

2, (3.13)

and

wπ2≡ 0. (3.14)

After (3.13) and (3.14) are established, the axial symmetry and the monotonicity follow

readily. For θ = 0, from step 1 we obtain that : either w0 ≡ 0 in B+1 or w0(x) > 0 ∀x ∈ B+

1 ,

where B1 = x ∈ B1|xn > 0. If the former case holds, then using the above step 2 and

first part of step 3 in the proof of Theorem 1.1, we can conclude that u is constant. This

contradicts with u is a least-energy solution of (1.5). If the second case is true, then we set

θ0 = supθ|wθ(x) ≥ 0 ∀x ∈ Σθ ∀0 ≤ θ ≤ θ ≤π

2. (3.15)

Following the standard argument of the method of moving planes, we can prove θ0 = π2.

Since the present case is the Neumann problem and the boundary of ∂Σθ is not smooth,

we should briefly scatch the proof for the sake of completeness.

Suppose θ0 < π/2. Then, by te continuity, wθ0(x) ≥ 0 for x ∈ Σθ0

. By the above sep

1, we have wθ0(x) > 0 for x ∈ Σθ0

/∂B1. By the definition of θ0, there is a sequence of

θj > θ0 with limj→∞ θj = θ0 such that

wθj(xj) = inf

Σθj

wθj(x) < 0.

By passing to a subsequence, x0 = limj→∞ xj satisfies wθ0(x0) = 0 and ∇wθ0

(x0) = 0.

Hence we have x0 ∈ Tθ0∩ ∂B1. Since wθ0

≡ 0 on Tθ0, Dei

Dejwθ0

(x0) < 0 for any

tangent vector ei, ej on Tθ0. Since

∂wθ0

∂ν (x0) = 0, Dei

∂wθ0

∂ν (x0) = 0 for any tangent

vector ei of ∂B1 at x0. Let e1, · · · , en−1 be the base of the normal to the plane Tθ0

such that en−1 is the normal of ∂B1 at x0, and en be the normal to Tθ0. Then we have

16

Deiejwθ0

(x0) = 0 ∀1 ≤ i ≤ n, 1 ≤ j ≤ n. Thus, the Hessian of wθ0at x0 is completely

zero, which yields a contradiction to Lemma S in [GNN 2]. Therefore, θ0 = π2 , the axial

symmetry follows readily.

The monotonicity and (1.4) follow from ∂wθ

∂ν< 0 for x ∈ Tθ where ν is the outnormal

of Σθ on the boundary Tθ. This proves the results of the case P0 6= O of Theorem 1.2.

The proof of Theorem 1.2 is complete.

4. Radial Symmetry near the Critical Exponent

Proof of Theorem 1.3. Suppose that the conclusion of the first part of Theorem

1.3 does not hold. Then there exists a sequence of least-engergy solution ui of (1.9) with

p = pi ↑n+2n−2 such that ui(x) is not radially symmetric. Let Pi be a maximum point of ui.

If Pi is the origin, then we can prove that ui(x) is radially symmetric. For the detail of

the argument, see the end of the proof of Lemma 4.1 below. Under the assumption that

ui(x) is non-radial, we have Pi 6= O. We first want to prove ui is axially symmetric with

respect to−−→OPi. Note that by (1.10), ui satisfies

∫

B1|∇ui|

2dx(

∫

B1K(x)upi+1

i dx)

2pi+1

=Sn

(maxB1K)

n−2n

(1 + o(1)). (4.1)

In the following, the axial symmetry is established for solution ui satisfying (4.1). Note that

even for least-energy solutions, Theorem 1.1 can not be applied for our present situation,

because K(|x|) in not assumed to be positive in the whole ball B1.

Lemma 4.1. Suppose ui is a solution of (1.9) with p = pi and K ∈ C(B1), K(x) =

K(|x|) and maxB1K > 0. Assume (4.1) holds and Pi is a global maximum point of ui.

Then ui is axially symmetric with respect to the axis ↔OPi

.

17

Proof. Let Pi be a maximum point of ui and assume Pi 6= O first. Since the Sobolev

constant is never achieved in H10 (B1), by (4.1), we have

K(Pi) → maxB1

K and ui(Pi) → +∞.

Without loss of generality, we may assume Pi = (0, . . . , 0, ti) for some ti > 0 and maxB1

K(x) = 1.

As in the proof of Theorem 1.1, we want to show

wi(x) = ui(x)− ui(x−) > 0 for xn > 0, (4.2)

where x− = (x1, . . . , xn−1,−xn). To prove (4.2), instead of (2.9) for Theorem 1.1, we claim

thatThere exists a constant c > 0 such that

ui(x) ≤ c Uλi(x− Pi) for x ∈ B1,

(4.3)

where Uλi(x) =

(

λi

λ2i+

|x|2

n(n−2)

)n−2

2

and λ−1i = (ui(Pi))

2n−2 .

Recall that for any λ > 0, Uλ(x) satisfies

4Uλ(x) + Un+2n−2

λ (x) = 0 in Rn, and

Uλ(0) = maxRn

Uλ(x).(4.4)

(4.3) was proved in [H] for the case K(x) ≡ a positive constant. However, it is unclear

whether the argument in [H] can be applied to the present case where K(x) is only assumed

to be continuous. For the sake of completeness, an alternative proof will be presented in

the appendix of this paper. For the moment, let us assume that (4.3) holds and we return

to the proof of (4.2). Clearly, wi(x) satisfies

4wi(x) + bi(x)wi = 0 for B+1 = x ∈ B1 | xn > 0,

wi(x) = 0 on ∂B+1 ,

(4.5)

18

where

bi(x) = K(x)upi

i (x)− upi

i (x−)

ui(x)− ui(x−).

Suppose Ωi = x ∈ B+1 | wi(x) < 0 is a non-empty set. We want to prove

|x− − Pi|n−2

2 ui(Pi) → +∞ for x ∈ Ωi (4.6)

as i→ +∞. If there is a sequence xi ∈ Ωi such that (4.6) fails. Since

|Pi|n−2

2 ui(Pi) < |x− − Pi|n−2

2 ui(Pi),

|Pi|n−2

2 ui(Pi) is bounded. By passing to a subsequence, we let ξ0 = limi→+∞

ti(ui(Pi))2

n−2

and q0 = (0, . . . , 0, ξ0). By rescaling ui, we set

Vi(y) = M−1i ui

(

M−2

n−2

i y

)

,

where Mi = ui(Pi). By elliptic estimates, Vi(y) is bounded in C2loc(R

n), thus, by passing to

a subsequence, Vi converges to U1(y− q0) in C2loc(R

n+), where U1(y) =

(

1 + |y|2

n(n−2)

)−n−22

.

Note that U1(y) is the solution of (4.4) with U1(0) = 1.

Two cases are considered separately. If ξ0 > 0, then U1(y − q0) > U1(y− − q0) for

y ∈ Rn+ = y | yn > 0. Set yi = M

−2n−2

i xi, where xi ∈ Ωi is the sequence such that

|Pi − x−i |n−2

2 ui(Pi) < +∞. Thus,

|xi|u2

n−2

i (Pi) ≤ |xi − Pi|u2

n−2

i (Pi) + |Pi|ui(Pi)2

n−2

≤ |x−i − Pi|ui(Pi)2

n−2 + |Pi|ui(Pi)2

n−2

≤ C

for some constant C. Then |yi| is bounded. Assume y0 = limi→+∞ yi. Since Vi(yi) −

Vi(y−i ) = M−1

i wi(xi) < 0, we have

U1(y0 − q0)− U1(y−0 − q0) = lim

i→+∞M−1

i wi(xi) ≤ 0,

19

which implies y0,n = 0, here y0,n is the yn-coordinate of y0. Since Vi(y)− Vi(y−) = 0 on

yn = 0, there exists ηi = (yi,1, . . . , yi,n−1, ηi,n) with ηi,n ∈ (0, yi,n) such that ∂∂yn

(Vi(ηi)−

Vi(η−i )) ≤ 0 by the mean value Theorem. By passing to the limit, it yields

0 ≥∂

∂yn(U1(y − ξ0)− U1(y

− − ξ0)) |y=y0

= 2∂U1

∂yn(y0 − ξ0) =

2

n

ξ0(

1 + |y0−ξ0|2

n(n−2)

)n > 0

a contradiction. Hence, we have ξ0 = 0.

Now we assume ξ0 = 0. To prove (4.6), as the previous step, it suffices to show that

wi(y) = N−1i wi(M

−2n−2

i y)

converges to a positive function in C2loc(R

n+), where

Ni = maxB+

1

|wi(x)| = |wi(zi)|

We first claim that

|zi|Mpi−1

2i ≤ c (4.7)

for some positive constant c > 0. Assume (4.7) does not hold. First, we assume ri → 0 as

i→ +∞. Set ri = |zi| and rescale wi by

wi(y) = N−1i wi(riy),

and wi satisfies

4wi(y) + r2i bi(riy)wi = 0, for |y| ≤1

ri,

sup|y|=1

|wi(y)| = 1.

By (4.3), for any compact set of Rn+\0, we have

|r2i bi(riy)| ≤ c1r2i |ui(riy) + ui(riy

−)|pi−1

≤ o(1)|y|−2.

20

Here, we have used the assumption limi→+∞ riMpi−1

2i = +∞ and the fact that limi→+∞M

n+2n−2·pi

i =

1. For the proof of limi→+∞Mn+2

N−2−pi

i = 1, see step 1 in the proof of A.2 in the appendix.

Hence, r2i bi(riy) converges to 0 uniformly in any compact set of Rn+\0. By elliptic

estimates and due to the assumption ri → 0, wi(y) converges to a harmonic function h(y)

in C2loc(R

n+\0), where h satisfies.

|h(y)| ≤ 1 and sup|y|=1

|h(y)| = 1. (4.8)

By the regularity theorem for bounded harmonic functions, h(y) is smooth at 0. Note

that h(y) ≡ 0 for yn = 0. By the Liouville theorem, h(y) ≡ 0 in Rn+, which yields a

contradiction to the second identity of (4.8). Hence (4.7) is established, in the first case.

Secondly, we assume |zi| ≥ δ0 > 0. Then the argument of contradiction in the above

yields that there exists r1 > 0 such that

sup|x|≤r1

|wi(x)| = o(1)Ni (4.9)

Let λ > 0 and ψ(x) > 0 be the eigenvalue and the eigenfunction of ∆ in B2 with the

Dirichlet problem and set

wi(x) =wi(x)

ψ(x)for x ∈ B1.

By a direct computation, wi(x) satosfoes

∆wi(x) + 2∇ logψ(x) · ∇wi(x) + (bi(x)− λ)wi(x) = 0

for x ∈ B1. Let xi be the maximum of |wi(x)|. By (4.9), we have |xi| ≥ r1. Clearly, (4.3)

yields |bi(xi)| = O(1)M−4

n−2

i . Applying the maximum principle at xi, we have

0 ≥ ∆wi(xi) = (λ− bi(xi))wi(xi) > 0,

21

a contradiction, where |wi(xi)| = wi(xi) is assumed. Thus, |zi| ≥ δ0 is impossible. This

proves (4.7).

Rescale wi again by

wi(y) = N−1i wi(M

−2n−2

i y).

It is easy to see that by passing to a subsequence, wi converges to w in C2loc(R

n+), where

w satisfies

4w +n+ 2

n− 2U

4n−2

1 (y)w = 0 y ∈ Rn+,

w(y) = 0 on yn = 0.

(4.10)

Readily from (4.7), w(y) is a bounded nonzero function. Thus,

w(y) = c∂U1(y)

∂ynfor some constant c 6= 0.

From the explicit expression of U1(y), we have w(y) 6= 0 for any y ∈ Rn+ and ∂w

∂yn(y) 6= 0

for yn = 0. Let ξi = M2

n−2

i Pi. We have

∂wi(ξ−i )

∂yn= N−1

i M−2

n−2

i

(

∂wi

∂xn

)

(P−i )

= N−1i M

−2n−2

i

(

∂

∂xnui(P

−i )−

∂

∂xnui(Pi)

)

= N−1i

∂

∂ynVi(ξ

−i )−

∂

∂ynVi(ξi)

= N−1i

∂2Vi

∂y2n

(ηi)(−2ξi,n).

(4.11)

where ηi ∈ (ξ−i , ξi). Since ξi → 0 and

0 >∂2U1

∂y2n

(0) = limi→+∞

∂2Vi

∂y2n

(ηi),

(4.10) yields

∂w(0)

∂yn= lim

i→∞N−1

i

∂2Vi(ηi)

∂y2n

(−2ξi,n) ≥ 0.

22

Hence, ∂w∂yn

(y) > 0 for yn = 0 and we conclude that w(y) > 0 in Rn+. Now suppose xi ∈ Ωi.

Because, wi, the scaling of wi, converges to a positive function in Rn+, with the negative

outnormal derivative on ∂Rn+, we conclude (4.6) holds.

By (4.6), we have for x ∈ Ωi,

bi(x) ≤n+ 2

n− 2u

4n−2

i (x−) ≤ o(1)|x− − Pi|−2. (4.12)

For x ∈ Ωi, we set

wi(x) = −wi(x)|x− Pi|−α,

where 0 < α < n−22 . By a straightforward computation, wi(x) satisfies

4wi(x) + 2(∇ log |x− Pi|α · ∇wi(x)) + (bi(x)− α(n− 2− α)|x− Pi|

−2)wi(x) = 0. (4.13)

Note that wi(x) = 0 on ∂Ωi and Pi 6∈ Ωi. Let wi(x) achieves its maximum in Ωi at xi.

Then by the maximum principle and (4.11), (4.12) yields

0 = 4wi(xi) + (bi(xi)− α(n− 2− α)|x− Pi|−2)wi(xi)

≤ (bi(xi)− α(n− 2− α)|x− Pi|−2)wi(xi) < 0

when i is sufficiently large. Therefore, we have proved wi(x) > 0 in B+1 for i large.

Once (4.2) is proved, we can apply the method of rotating planes and Alexandroff

Maximum Principle in [BN] and [HL] to conclude that ui(x) is axially symmetric with

respect to xn-axis and the monotonicity

xj∂ui(x)

∂xn− xn

∂ui

∂xj(x) > 0. (4.14)

holds for xj > 0 and 1 ≤ j ≤ n−1. This ends the proof of Lemma 4.1 for the case Pi 6= O.

When Pi = O, we want to prove wi(x) ≡ 0 on B+1 . Suppose not. Then let zi be the

maximum point of |wi(x)|, and as the same proof of (4.7), we have

|zi|Mpi−1

2i ≤ c

23

for some constant c. Set wi(y) = N−1i wi(x), where Ni = wi(zi) and x = M

−2n−2

i y. It is easy

to see that wi(y) converges to a non-zero limit w(y) in C2loc(R

n+), where w(y) is a solution

of (4.9). Thus, w(y) = c ∂U1

∂ynfor some c 6= 0. However, ∇wi(0) = 0 because the origin is a

maximum point of ui. Especially,

0 =∂w

∂yn(0) = c

∂2U1

∂y2n

(0)

yields c = 0, a contradiction. Therefore, we conclude wi(x) ≡ 0, that is, ui(x) is symmetric

with respect to xn. Of course, we can prove the symmetry of ui with respect to any

hyperplane passing the origin by the same argument. Hence ui(x) is radially symmetric if

Pi = O. This completely proves Lemma 4.1.

Now we return to the proof of Theorem 1.3. now suppose Pi 6= 0. Without loss of

generality, we may assume Pi = (0, . . . , 0, ti) for some ti > 0. Set

φi(x) = x1∂ui

∂xn(x)− xn

∂ui

∂x1(x) > 0. (4.15)

Then φi(x) > 0 in B+1 = x ∈ B1 | x1 > 0, and φi satisfies

4φi + piK(|x|)upi−1i φi = 0 in B+

1 ,

φi = 0 on ∂B+1 .

(4.16)

Since upi−1i (x) uniformly converges to zero in any compact set of B+

1 \Pi, by the Harnack

inequality,

(

max|x|=r0

φi(x)

)−1

φi(x) converges to a harmonic function h in C2loc(B

+1 \Pi),

where r0 is the positive number in the condition (Ka). Since h(x) = 0 for x ∈ ∂B+1 \P0

where P0 = limi→+∞

Pi, we have h(x) has a nonremovable singularity at P0. Otherwise,

h(x) ≡ 0 on B+1 , which contradicts to the fact that max

|x|=r0

h(x) = 1. Thus, ∂h(x)∂ν

< 0 for

x ∈ ∂B+1 \x1 = 0. Hence we have

−∂φi(x)

∂ν≥ c1

(

max|x|≥r0

|φi(x)|

)

(4.17)

24

for x ∈ ∂B+1 , x1 ≥

12. and for a positive constant c1.

From (1.9), we have

4(∂ui

∂x1) + piK(|x|)upi−1

i (∂ui

∂x1) = −K ′(|x|)

x1

rupi

i in B+1 ,

∂ui

∂x1= 0 on x1 = 0.

(4.18)

By the boundary condition of ui, −∂ui

∂x1> 0 for x ∈ ∂B+

1 \x1 = 0. Since by (4.3),

Miui(x) converges to c G(x, P0) for some c > 0 where G(x, P0) is the Green function with

the singularity at P0, we have

−∂ui(x)

∂x1≥ c1M

−1i for x ∈ ∂B+

1 and x1 ≥1

2. (4.19)

By (4.16),(4.18) and (Ka), we obtain∫

∂B+1 \x1=0

∂ui

∂x1·∂φi

∂νdS = −

∫

B+1

[φi4(∂ui

∂x1)−

∂ui

∂x14φi] dx

=

∫

B+1 ∩|x|≤r0

[K ′(|x|)x1

|x|upi

i φi] dx+

∫

B+1 ∩|x|≥r0

[K ′(|x|)x1

|x|upi

i φi] dx

≤

∫

B+1 ∩|x|≥r0

[K ′(|x|)x1

|x|upi

i φi] dx

≤ C · maxB+

1 ∩|x|≥r0φi(x) ·

∫

B+1 ∩|x|≥r0

upi

i dx.

By (4.17) and (4.19), we get

C1 ·M−1i ≤ C2 ·M

−pi

i for i sufficiently large,

a contradiction. This proves Pi = O for i large.

For the uniqueness part of Theorem 1.3, we reduce (1.9) in an ODE. Here, K(r) is

continuously extended for all r ∈ [0,∞). Following conventional notations, for any fixed

p, we denote u(r;α) to be the unique radial solution of

u′′(r) +n− 1

ru′(r) +K(r)up = 0, r > 0,

u(0;α) = α > 0, u′(0;α) = 0.

(4.20)

25

For p ∈ (1, n+2n−2

), we set

α(p) = infu(0) | u(r) is a least energy solution of (4.19)

in the class of radial functions of H10 (B1)

We claim that there exists a small ε > 0 such that for the solution u(r;α) with

α ≥ α(p) and 0 < n+2n−2 − p ≤ ε, there is a R(α) ∈ (0,∞) satisfying

u(r;α) > 0 for r ∈ [0, R(α)) and

u(R(α), α) = 0.(4.21)

Furthermore, R(α) decreases with respect to α whenever α ≥ α(p). Since R(α(p)) = 1,

the uniqueness follows readily from the claim.

To prove the claim, we let

φ(r, α) :=∂u

∂α(r;α). (4.22)

We claim

φ(R(α), α) < 0 for α ≥ α(p) and 0 <n+ 2

n− 2− p ≤ ε. (4.23)

Suppose (4.23) holds. By differentiating (4.20) with respect to α, we have

u′(R(α), α)∂R(α)

∂α+ φ(R(α), α) = 0.

Since u′(R(α), α) < 0, (4.23) yields ∂R(α)∂α < 0. Obviously (4.21) and the decrease of R(α)

follows. Thus, it suffices for us to show (4.23).

Recall that φ satisfies the linearized equation

φ′′ +n− 1

rφ′ + pK(r)up−1φ = 0 , 0 < r < R(α),

φ(0;α) = 1 and φ′(0, α) = 0.

(4.24)

By the choice of α(p), we see that φ(r;α(p)) changes sign only once and φ(R(α(p)), α(p)) ≤

0. Now suppose (4.23) fails for any small ε > 0. Then there is a sequence of αi → +∞ as

26

i → +∞ such that φi(r) := φ(r;αi) of (4.24) with pi ↑n+2n−2

and φi(r) changes sign only

once and φi(Ri) = 0, where Ri = R(αi). Let ri be the first zero of φi. Then φi(r) > 0 for

r ∈ (0, ri) and φi(r) < 0 for r ∈ (ri, Ri). Clearly,

Ri ≤ 1 and φ′i(Ri) > 0. (4.25)

For the simplicity of notations, we let ui(r) ≡ u(r;αi) denote the solution of (4.20) with

p = pi.

To yield a contradiction, we set

wi(r) = ru′i(r) +2

pi − 1ui(r). (4.26)

Then, from (4.20), wi satisfies

w′′i +n− 1

rw′i + piK(r)upi−1w = −rK ′(r)upi

wi(Ri) = Riu′i(Ri) < 0.

(4.27)

By (4.24) and (4.27), we get

∫ Ri

0

rn−1(rK ′(r))upi

i φi dr =

∫

B1

[wi4φi − φi4wi] dx

= Rn−1i wi(Ri)φ

′i(Ri) < 0.

(4.28)

Here (4.25) is used. Recall that ri is the first zero of φi. By scaling in (4.24), we easily

have ri → 0 as i→ +∞. Let

Ci =−riK

′(ri)

K(ri). (4.29)

From the condition ( rK′(r)K(r) )′ ≤ 0 for 0 ≤ r ≤ r0 and (4.29), we have

CiK(r) + rK ′(r)

≥ 0 if 0 ≤ r < ri ≤ r0,

≤ 0 if ri ≤ r ≤ r0.(4.30)

Two cases are discussed separately.

27

Case 1. If Ri ≤ r0, then, from (4.30), we obtain

0 ≤

∫ Ri

0

rn−1(CiK(r) + rK ′(r))upi

i φi dr = Rn−1i wi(Ri)φ

′i(Ri) < 0.

This proves (4.23) in this case.

Case 2. If Ri > r0, then

0 < −Rn−1i wi(Ri)φ

′i(Ri) = −

∫ Ri

0

rn−1(CiK(r) + rK ′(r))upi

i φi dr

=

(

−

∫ r0

0

)

+

(

−

∫ Ri

r0

)

= (I) + (II).

(4.31)

Since the first term (I) in (4.31) is negative, we obtain

Rni |u

′i(Ri)|φ

′i(Ri) ≤

∫ Ri

r0

rn−1(CiK(r) + rK ′(r))upi

i φi dr. (4.32)

By using the same arguments of (4.3), (4.17) and (4.19), we can easily obtain

|u′i(Ri)| ∼ ui(r0) ∼ α−1i , φ′i(Ri) ∼ φi(r0)and Ci is small. (4.33)

Hence, (4.32) yields

α−1i ≤ c α−pi

i ,

a contradiction. This ends the proof of the claim (4.23), and the uniqueness follows. Hence

we have finished the proof of Theorem 1.3.

Appendix

In this appendix, we consider a sequence of solutions ui of

4ui +K(x)upi

i = 0 in B2 = |x| < 2,

ui = 0 on ∂B2,

28

such that

ui(Pi) = maxB1

ui(x) → +∞, Pi → P0 for some |P0| < 1 and

∫

B2

K(x)upi+1i dx =

(

Sn

K(P0)n−2

n

)n2

(1 + o(1)), (A.1)

where K(x) ∈ C(B2) and K(P0) > 0 and pi ↑n+2n−2 . We want to prove that there exists a

constant c > 0 such that

ui(x) ≤ c

Mi

1 + K(P0)n(n−2)M

2i |x− Pi|2

n−22

for |x| ≤ 1, (A.2)

where Mi = u2

n−2

i (Pi).

We note that when K(x) ≡ a positive constant, (A.2) was proved by Han [H]. Here

we will present a proof of (A.2), which is simpler than [H] even for the case of constant K.

This proof does not employ the Pohozaev identity. Thus, the smooth assumption of K is

not required.

Proof of A.2. We divide the proof into several steps.

Step 1. limi→+∞

Mσi

i = 1, where σi = n+2n−2

− pi

Rescaling ui by

Ui(y) = M−1i ui(Pi +M

−pi−1

2i y). (A.3)

Then Ui satisfies

4Ui(y) +Ki(y)Upi

i (y) = 0 for |y| ≤Mpi−1

2i ,

where Ki(y) = K(Pi + M− 2

n−2

i y). By elliptic estimates, Ui(y) converges to U(y) in

C2loc(R

n), where U(y) is the solution of

4U(y) +K(P0)Un+2n−2 = 0 in R

n,

U(0) = maxRn

U(y) = 1.(A.4)

29

Then by a theorem of Caffarelli-Gidas-Spruck [CGS], we have U(y) = (1+ K(P0)n(n−2) |y|

2)−n−2

2

and∫

Rn

K(P0)U2n

n−2 (y)dy =

(

Sn

K(P0)n−2

n

)n2

Choose Ri → +∞ as i → +∞ such that Ui(y) converges to U(y) uniformly for |y| ≤ Ri.

Then(

Sn

K(P0)n−2

n

)n2

(1 + o(1)) ≥

∫

|Pi−x|≤RiM−

pi−12

i

K(x)upi+1i (x)dx

= Mn−2

2 σi

i

∫

|y|≤Ri

Ki(y)Upi+1i (y)dy

= Mn−2

2 σi

i

(

Sn

K(P0)n−2

n

)n2

(1 + o(1))

Therefore, limi→+∞Mn−2

2 σi

i ≤ 1. Step 1 follows readily.

Set

mi = inf|x|≤1

ui(x).

To Prove (A.2), we have to compare mi and M−1i . First, we claim

Step 2. there exists a constant c such that

M−1i ≤ c mi.

Consider G(x) = M−1i (|Pi − x|2−n − 1) for |Pi − x| ≤ 1. Note that by rescaling (A.3)

and step 1, we have

ui(x) ≥ c Mi for |x− Pi| = M− 2

n−2

i .

Since ui(x) is superharmonic, by the maximum principle,

c G(x) ≤ ui(x)

In particular,

ui(x) ≥ c M−1i for |x| =

1

2,

30

where step 2 follows immediately.

The spherical Harnack inequality is very important in the study of the blowup behavior

of ui. usually, this is a difficult step to prove. However, by the energy assumption (A.1),

we can prove

Step 3. There exists a constant c > 0 such that

ui(x)|x− Pi|2

pi−1 ≤ c for |x| ≤ 1. (A.5)

Because, if limi→+∞ supB1(ui(x)|x−Pi|

npi−1 ) = +∞, then there is a local maximum point

Qi of ui(x) such that the rescaling of ui with the center Qi,

Ui(y) = M−1i ui(Qi + M

−pi−1

2i y) with Mi = ui(Qi)

converges to U(y) of (A.4), where |Qi − Pi|n−2

2 Mi → +∞ as i→ +∞. Thus, ui possesses

at least two bubbles, a contradiction to (A.1). The existence of Qi can be proved by

employing the method of localizing blowup points by R. Schoen. Since the method is

well-known now, we refer the proof to [CL1] and [CL2].

By (A.5), we have the spherical Harnack inequality,

ui(x) ≤ ui(x− Pi) and

|∇u(x)| ≤ c |x− Pi|−1ui(|x− Pi|)

(A.6)

where ui(r) is the average of ui over the sphere |x− Pi| = r. Set

vi(t) = ui(r)rn−2

2 with r = et

By a straightforward computation, vi(t) satisfies

v′′i (t)−

(

n− 2

2

)2

vi(t) + Ki(t)vn+2n−2

i = 0, (A.7)

where

31

Ki(t) =

(

∫

−|x−Pi|=t

K(x)u−σi

i un+2n−2

i (x)dσ

)

(un+2n−2 (r))−1,

and (∫

− denotes the average of integration over the sphere |x−Pi| = r. By step 1 and step

2, we have uσi

i (x) uniformly converges to 1 for |x| ≤ 1. Therefore, 0 < c1 ≤ Ki(t) ≤ c2

for t ≤ 0. By the rescaling (A.3), we see that vi(t) has a first local maximum at t = ti =

− 2n−2 logMi + c0 for some constant c0. Let si > ti be the first local minimum point unless

vi(t) is decreasing for ti ≤ t ≤ 0. In the latter case, we set si = 0.

Step 4. If si < 0, then vi(t) is increasing for si < t ≤ 0.

If not, then vi(t) has a local maximum at some point si ∈ (si, 0]. By (A.7), vi(si) ≥ c > 0

for some constant c > 0. By the spherical Harnack inequality, |v′i(t)| ≤ c1. Thus, there

exists δ0 > 0 such that

vi(t) ≥c

2if |t− si| ≤ δ0.

Therefore,∫

Ti

u2n

n−2

i (x)dx ≥ c1 > 0, (A.8)

where Ti = x | esi−δ0 ≤ |x− Pi| ≤ esi+δ0. However,

∫

|Pi−x|≤esi

u2n

n−2

i (x)dx =

(

Sn

K(P0)n−2

n

)n2

(1 + o(1)).

Together with (A.8), it yields a contradition to (A.1).

Step 5. There exists T0 ≤ 0 such that si ≥ T0. Furthermore,

ui(x) ≤ c M−1i |x− Pi|

2−n for M− 2

n−2

i ≤ |x− Pi| ≤ eT0 . (A.9)

To prove step 5, we recall an ODE result from [CL2] and [CL3]. See Lemma 5.1

in [CL2] or Lemma 3.2 in [CL3]. Assume ε0 to be a fixed small positive number. By

rescaling as in (A.3), there is a unique ti = ti + c(ε0) > ti such that vi(t) is decreasing for

32

ti ≤ t ≤ ti and vi(ti) = ε0. If ε0 is small enough, then by (A.7), vi(t) has no critical point

for t ∈ (ti, si), where we recall that si is the first minimum point after ti.

Lemma A. There exists a constant c such that the following statements hold.

(1) For ti ≤ t0 ≤ t1 ≤ si, vi satisfies

t1 − t0 ≤2

n− 2log

vi(t0)

vi(t1)+ c1 and

si − t0 ≥2

n− 2log

vi(t0)

vi(si).

(2) For si ≤ t ≤ 0,

(t− si)− c1 ≤2

n− 2log

vi(t)

vi(si)≤ (t− si).

From (2), we have for t ≥ si,

ui(et) = vi(t)e

−n−22 t ≥ c2 e

−n−22 sivi(si) = c2ui(e

si). (A.10)

Since ui(r) is decreasing in r, by (A.10) together with the spherical Harnack inequality,

we have for some positive constant c3,

mi ∼ ui(x) ∼ min|x−Pi|=esi

ui(x) for |x− Pi| ≥ si. (A.11)

From the first inequality of (1) of Lemma A, we have

ui(x) ≤ c4ui(ri)

(

ri|x|

)n−2

≤ c4M−1i |x|2−n, (A.12)

for eti = ri ≤ |x− Pi| ≤ esi where ti = ti + c(ε0), ti = − 2n−2 logMi, and ui(ri) ∼ Mi are

used. The second inequality of (1) in Lemma A implies

ui(et) ≥ c5M

−1i (esi)2−n

33

for ti ≤ t ≤ si. Thus, together with (A.12) and (A.11), we have

mi ∼M−1i (esi)2−n. (A.13)

Now suppose si → −∞. Then by (A.13),

miMi → +∞ as i→ +∞. (A.14)

Since ui(x)/mi is uniformly bounded in C2loc(B2\P0), by passing to a subsequence,

ui(x)/mi converges to a positive harmonic function h(x) in C2loc(B2\P0). For any δ > 0,

−

∫

|x−P0|=δ

∂h

∂ν(x)dσ = − lim

i→+∞

∫

|x−P0|≤δ

4(ui(x)/mi)dx

=1

mi

∫

|x−P0|≤δ

K(x)upi

i dx

To estimate the right hand side, we decompose the domain into three parts: For any large

R > 0,

1

mi

∫

|x−Pi|≤M− 2

n−2i

R

K(x)upi

i dx ≤c

miMi

∫

|y|≤R

Un+2n−2

i (y)dy → 0

by (A.14). By using (A.12), we have

1

mi

∫

M−2

n−2i

R≤|x−Pi|≤esi

K(x)upi

i dx ≤c

miM

−n+2n−2

i

∫

|x−Pi|≥M− 2

n−2i

R

|x− Pi|−(n+2)dx

≤c

miM

−n+2n−2

i (M− 2

n−2

i R)−2

=c1

miMiR−2

→ 0 by (A.14) again. By (A.11), the last term can be estimated by

1

mi

∫

|x−Pi|≥esi

Kupi

i dx ≤ c1m− 4

n−2

i → 0.

Thus,∫

|x−P0|=δ

∂h

∂νdσ = 0 for any δ > 0,

34

which implies h is smooth at 0. Since h(x) vanishes on the boundary of B2, h(x) ≡ 0 on

B2, which contradicts to infB1

h(x) = 1. Hence step 5 is proved. Clearly, (A.2) is equivalent

to (A.9). Therefore, (A.2) is proved completely.

Acknowledgement. This work was done when the first author visited the National

Center for Theoretical Sciences of NSC, in Taiwan. He would like to thank the center for

its kind hospitality.

35

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Jann-Long Chern∗ and Chang-Shou Lin†

∗Department of Mathematics, National Central University, Chung-Li 32054, Taiwan

chern@math.ncu.edu.tw

37

†Department of Mathematics, National Chung-Cheng University, Minghsiung, Chia-Yi,

Taiwan

cslin@math.ccu.edu.tw

38