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 Rn 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. 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 B1is 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
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) = Z B1 [1 2|∇u| 2− F (|x|, u+)] dx in H1 0(B1), (1.2) where F (r, u) = Ru
0 f (r, s) ds and u+(x) = max(0, u(x)). For the nonlinear functional J ,
we set c∗ = inf h∈Γ0≤t≤1max J (h(t)), (1.3) where Γ = {h ∈ C([0, 1], H01(B1)) |h(0) = 0, h(1) = e} and e ∈ H1
0(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) ≡
Ru
0 f (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
least energy can also be obtained by minimization of inf v∈H1 0(Ω) R |∇v|2 h R K(v+)p+1+i 2 p+1 ,
where f (x, u) = K(x)up with max
Ω 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],
[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 Z B1 (d|∇u|2+ u2) dx − Z B1 F (u+) dx ∀u ∈ H1(B1), (1.6)
where u+ = max{u(x), 0}, and its critical value c
d = Jd(ud) is proved to be equal to
cd = inf
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= inf v∈H1 0(Ω) R Ω |∇v|2 dx (R Ω |v| 2n n−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 H1
0(Ω). For a C1 function K with maxB
1 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)
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,
R B1 |∇ui| 2 dx (R B1K(x)u pi+1 i dx) 2 pi+1 = inf v∈H1 0(B1) R B1 |∇v| 2 dx [(R B1K(x)(v +)pi+1 dx)+] 2 pi+1 = ( 1 (max ¯ B1 K)pi+12 · 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 H01(B1), we have
lim
i→∞ ui(Pi) = +∞ and i→∞lim K(Pi) = maxB¯1 K(x). (1.11)
Obviously, by (1.11), the necessary condition for uito be radially symmetric is that K(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 K0(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 Piof ui would tend to the origin. However, we have the following.
Theorem 1.3 Suppose the condition (Ka) holds. Then there exists a small > 0 such
that for any least-energy solutionu of (1.9) with 0 < n+2n−2− p ≤ , u is radially symmetric. Furthermore, if (rKK(r)0(r))0 ≤ 0 ∀0 ≤ r ≤ r1 for some r1 ≤ r0, then (1.9) possesses a unique
least-energy solution when 0 < n+2n−2 − p ≤ .
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
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 ∈ B1+, (B) u(x) > u(x∗) ∀x ∈ B+ 1,
(C) u(x) < u(x∗) ∀x ∈ B1+, where B+1 is one of half-balls of B1\T and x∗ is the reflection
First, we will prove that
either u(x) ≥ u(x∗) ∀x ∈ B1+ or u(x) ≤ u(x∗) ∀x ∈ B1+.
(2.1)
Suppose the conclusions of (2.1) are not true. Then the following two sets are all nonempty. Ω+ = {x ∈ B1+|u(x) > u(x∗)}, Ω− = {x ∈ B+1|u(x) < u(x∗)}. (2.2) Let w(x) = u(x) − u(x∗) ∀x ∈ B1+. (2.3) Then w satisfies 4w + fu(|x|, z(x))w = 0, x ∈ B1+ 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 Z
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)
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∂2∂tf (r,t)2 ≥ 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 > Z B1 (−v(x)) · [4v(x) + fu(|x|, u(x))v(x)] dx = Z 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 ∈ B1+. Since w(x) satisfies Eq. (2.4), by using the strong maximum principle, we have either w(x) ≡ 0 for x ∈ B1+ or w(x) > 0 for x ∈ B1+. Similarly, if w(x) ≤ 0, we have either w(x) ≡ 0 on B1+ or w(x) < 0 on B1+ 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, x0) and x− = (−x1, x0). Suppose w(x) 6≡ 0 for x ∈ B1+ = {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
However, since O is the maximal point of u in B1, we have ∂x∂u1(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(P0−), and from step 1, we obtain that :
either u(x) = u(x−) ∀x ∈ B1+ or u(x) > u(x−) ∀x ∈ B1+,
(2.13)
where x = (x0, xn), x− = (x
0
, −xn), B+1 = {x ∈ B1|xn > 0}. If the former case holds,
then u(P0) = u(Q0) = max ¯ B1
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(P0∗) 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 = {(r1cos θ, x2, . . . , xn−1, r1sin 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
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 = sup{θ|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 P0∗ ∈ Tθ1 for some θ1 ∈ (0, θ0) and ∇u(P
∗
0) = 0, which
yields a contradictions to (2.17). Hence, we have wπ
2 ≥ 0 in B π
2. (2.18)
Similarly, using the above arguments, we can also obtain w−π
2 ≥ 0 in B− π
2. (2.19)
From (2.18) and (2.19) we deduce that wπ
2 ≡ 0 in B π
2. (2.20)
Let x = (r1cos θ, x2, . . . , xn−1, r1sin θ), r1 = (|x|2− x22− · · · − x2n−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)
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(u00 + n − 1 r u 0) − u + f (u) = 0, r > 0, u(0) = α > 0, u0(0) = 0, (3.5) and we have ∂u ∂x1 = u0(r)x1 r in B1 and ∂u ∂x1 = 0 on ∂B1. (3.6)
Suppose u 6≡ constant, then ∂x∂u
1 6≡ 0. Let w(r) be the first eigenfunction of (3.1).
From (3.6), we easily have Z B1 w(|x|)∂u ∂x1 (|x|) dx = 0. (3.7) By (1.5) we obtain d4( ∂u ∂x1 ) − ∂u ∂x1 + f0(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 = inf v⊥w,v∈H1(B1) Z B1 [d|∇v|2+ v2− f0(u)v2] dx = Z B1 [d|∇( ∂u ∂x1)| 2+ ( ∂u ∂x1) 2− f0(u)( ∂u ∂x1) 2] dx = 0 (3.9) Since ∂x∂u
1 archives the infinimum of (3.9), we obtain that
Z ∂B1 ∂ ∂ν( ∂u ∂x1 ) · φ dσ = 0 ∀φ ∈ H1(B1) and φ⊥w. (3.10)
Set φ = x1∂x∂1(∂x∂u1) + x2∂x∂2(∂x∂u1). Since φ is odd in x1, we have φ⊥w. Then we obtain
that ∂ν∂ (∂x∂u1) = 0 on ∂B1 and ∂x∂u1 is a solution of the Neumann problem (1.5). Hence we
have u00(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. T θ. 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∗) .
We want to prove wθ(x) > 0 for x ∈ Σθ and 0 ≤ θ < π 2, (3.13) and wπ 2 ≡ 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 B1+or w0(x) > 0 ∀x ∈ B1+,
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 = sup{θ|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, DeiDejwθ0(x0) < 0 for any
tangent vector ei, ej on Tθ0. Since
∂wθ0
∂ν (x0) = 0, Dˆei
∂wθ0
∂ν (x0) = 0 for any tangent
vector ˆei of ∂B1 at x0. Let {e1, · · · , en−1} be the base of the normal to the plane Tθ0
Deiejwθ0(x0) = 0 ∀1 ≤ i ≤ n, 1 ≤ j ≤ n. Thus, the Hessian of wθ0 at 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
R B1|∇ui| 2dx R B1K(x)u pi+1 i dx pi+12 = Sn (maxB1K) n−2 n (1 + o(1)). (4.1)
In the following, the axial symmetry is established for solution uisatisfying (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.
Proof. Let Pi be a maximum point of ui and assume Pi 6= O first. Since the Sobolev
constant is never achieved in H01(B1), by (4.1), we have
K(Pi) → max B1
K and ui(Pi) → +∞.
Without loss of generality, we may assume Pi = (0, . . . , 0, ti) for some ti > 0 and max B1
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− = (x
1, . . . , xn−1, −xn). To prove (4.2), instead of (2.9) for Theorem 1.1, we claim
that
There exists a constant c > 0 such that ui(x) ≤ c Uλi(x − Pi) for x ∈ B1, (4.3) where Uλi(x) = λi λ2 i+ |x|2 n(n−2) n−22 and λ−1i = (ui(Pi)) 2 n−2.
Recall that for any λ > 0, Uλ(x) satisfies
4Uλ(x) + U n+2 n−2 λ (x) = 0 in Rn, and Uλ(0) = max Rn 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 B1+ = {x ∈ B1 | xn> 0}, wi(x) = 0 on ∂B+1, (4.5)
where
bi(x) = K(x)
upi
i (x) − upii(x−)
ui(x) − ui(x−) .
Suppose Ωi = {x ∈ B1+ | 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 = lim
i→+∞ti(ui(Pi))
2 n−2
and q0 = (0, . . . , 0, ξ0). By rescaling ui, we set
Vi(y) = Mi−1ui M −2 n−2 i y ,
where Mi = ui(Pi). By elliptic estimates, Vi(y) is bounded in Cloc2 (Rn), thus, by passing to
a subsequence, Vi converges to U1(y − q0) in Cloc2 ( ¯Rn+), where U1(y) =
1 + n(n−2)|y|2 −
n−2 2
. 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
−2 n−2
i xi, where xi ∈ Ωi is the sequence such that
|Pi− x−i | n−2 2 ui(Pi) < +∞. Thus, |xi|u 2 n−2 i (Pi) ≤ |xi− Pi|u 2 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(yi−) = Mi−1wi(xi) < 0, we have
U1(y0− q0) − U1(y0−− q0) = lim i→+∞M
−1
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 ∂y∂n(Vi(ηi) −
Vi(ηi−)) ≤ 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) = Ni−1wi(M
−2 n−2
i y)
converges to a positive function in C2
loc( ¯Rn+), where
Ni = max B+1
|wi(x)| = |wi(zi)|
We first claim that
|zi|M
pi−1 2
i ≤ 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) = Ni−1wi(riy), and ˜wi satisfies
4 ˜wi(y) + ri2bi(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
|ri2bi(riy)| ≤ c1r2i|ui(riy) + ui(riy−)|pi−1
Here, we have used the assumption limi→+∞riM
pi−1 2
i = +∞ and the fact that limi→+∞M
n+2 n−2·pi
i =
1. For the proof of limi→+∞M
n+2 N−2−pi
i = 1, see step 1 in the proof of A.2 in the appendix.
Hence, r2
ibi(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 Cloc2 ( ¯Rn+\{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∇ log ψ(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
a contradiction, where | ¯wi(¯xi)| = ¯wi(¯xi) is assumed. Thus, |zi| ≥ δ0 is impossible. This proves (4.7). Rescale wi again by ˜ wi(y) = Ni−1wi(M −2 n−2 i y).
It is easy to see that by passing to a subsequence, ˜wi converges to w in Cloc2 ( ¯Rn+), where
w satisfies 4w + n + 2 n − 2U 4 n−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)
∂yn
for some constant c 6= 0.
From the explicit expression of U1(y), we have w(y) 6= 0 for any y ∈ Rn+ and ∂y∂wn(y) 6= 0
for yn = 0. Let ξi = M 2 n−2 i Pi. We have ∂ ˜wi(ξi−) ∂yn = Ni−1M −2 n−2 i ∂wi ∂xn (Pi−) = Ni−1M −2 n−2 i ∂ ∂xn ui(Pi−) − ∂ ∂xn ui(Pi) = Ni−1 ∂ ∂yn Vi(ξi−) − ∂ ∂yn Vi(ξi) = Ni−1∂ 2V i ∂y2 n (ηi)(−2ξi,n). (4.11)
where ηi ∈ (ξ−i , ξi). Since ξi → 0 and
0 > ∂ 2U 1 ∂y2 n (0) = lim i→+∞ ∂2V i ∂y2 n (ηi), (4.10) yields ∂w(0) ∂yn = lim i→∞N −1 i ∂2Vi(ηi) ∂y2 n (−2ξi,n) ≥ 0.
Hence, ∂y∂w
n(y) > 0 for yn = 0 and we conclude that w(y) > 0 in R
n
+. 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 4 n−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
4 ¯wi(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 = 4 ¯wi(¯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 B1+ 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 B1+. Suppose not. Then let zi be the
maximum point of |wi(x)|, and as the same proof of (4.7), we have
|zi|M
pi−1 2
for some constant c. Set ˜wi(y) = Ni−1wi(x), where Ni = wi(zi) and x = M
−2 n−2
i y. It is easy
to see that ˜wi(y) converges to a non-zero limit w(y) in Cloc2 ( ¯Rn+), where w(y) is a solution
of (4.9). Thus, w(y) = c ∂U1
∂yn for some c 6= 0. However, ∇ ˜wi(0) = 0 because the origin is a
maximum point of ui. Especially,
0 = ∂w ∂yn (0) = c ∂ 2U 1 ∂y2 n (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 B1+ = {x ∈ B1| x1 > 0}, and φi satisfies
4φi+ piK(|x|)upii−1φi = 0 in B1+, φi = 0 on ∂B1+. (4.16) Since upi−1
i (x) uniformly converges to zero in any compact set of ¯B1+\{Pi}, by the Harnack
inequality, max |x|=r0 φi(x) −1
φi(x) converges to a harmonic function h in Cloc2 ( ¯B1+\{Pi}),
where r0 is the positive number in the condition (Ka). Since h(x) = 0 for x ∈ ∂B1+\{P0}
where P0 = lim
i→+∞Pi, we have h(x) has a nonremovable singularity at P0. Otherwise,
h(x) ≡ 0 on B1+, 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)
for x ∈ ∂B1+, x1 ≥ 12. and for a positive constant c1. From (1.9), we have 4(∂ui ∂x1 ) + piK(|x|)upii−1(∂ui ∂x1 ) = −K0(|x|)x1 r u pi i in B+1, ∂ui ∂x1 = 0 on x1 = 0. (4.18)
By the boundary condition of ui, −∂x∂u1i > 0 for x ∈ ∂B1+\{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 ≥ c1Mi−1 for x ∈ ∂B1+ and x1 ≥ 1 2. (4.19) By (4.16),(4.18) and (Ka), we obtain Z ∂B1+\{x1=0} ∂ui ∂x1 · ∂φi ∂ν dS = − Z B+1 [φi4( ∂ui ∂x1 ) − ∂ui ∂x1 4φi] dx = Z B+1∩{|x|≤r0} [K0(|x|)x1 |x|u pi i φi] dx + Z B+1∩{|x|≥r0} [K0(|x|)x1 |x|u pi i φi] dx ≤ Z B+1∩{|x|≥r0} [K0(|x|)x1 |x|u pi i φi] dx ≤ C · max B+1∩{|x|≥r0} φi(x) · Z B1+∩{|x|≥r0} upi i dx. By (4.17) and (4.19), we get
C1· Mi−1 ≤ C2· Mi−pi 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
u00(r) + n − 1 r u 0(r) + K(r)up = 0, r > 0, u(0; α) = α > 0, u0(0; α) = 0. (4.20)
For p ∈ (1,n+2n−2), we set
α(p) = inf{u(0) | u(r) is a least energy solution of (4.19) in the class of radial functions of H1
0(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 u0(R(α), α)∂R(α)
∂α + φ(R(α), α) = 0. Since u0(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
φ00+ n − 1 r φ 0+ pK(r)up−1φ = 0 , 0 < r < R(α), φ(0; α) = 1 and φ0(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
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 φ0i(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) = ru0i(r) +
2 pi− 1
ui(r). (4.26)
Then, from (4.20), wi satisfies
w00i + n − 1 r w 0 i+ piK(r)upi−1w = −rK0(r)upi wi(Ri) = Riu0i(Ri) < 0. (4.27) By (4.24) and (4.27), we get Z Ri 0 rn−1(rK0(r))upi i φi dr = Z B1 [wi4φi− φi4wi] dx = Rin−1wi(Ri)φ0i(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 =
−riK0(ri)
K(ri)
. (4.29)
From the condition (rKK(r)0(r))0 ≤ 0 for 0 ≤ r ≤ r
0 and (4.29), we have CiK(r) + rK0(r) ≥ 0 if 0 ≤ r < ri ≤ r0, ≤ 0 if ri ≤ r ≤ r0. (4.30)
Case 1. If Ri ≤ r0, then, from (4.30), we obtain
0 ≤ Z Ri
0
rn−1(CiK(r) + rK0(r))upiiφi dr = Rn−1i wi(Ri)φ0i(Ri) < 0.
This proves (4.23) in this case. Case 2. If Ri > r0, then 0 < −Rin−1wi(Ri)φ0i(Ri) = − Z Ri 0 rn−1(CiK(r) + rK0(r))upiiφi dr = − Z r0 0 + − Z Ri r0 = (I) + (II). (4.31)
Since the first term (I) in (4.31) is negative, we obtain Rni|u0i(Ri)|φ0i(Ri) ≤
Z Ri
r0
rn−1(CiK(r) + rK0(r))upiiφi dr. (4.32)
By using the same arguments of (4.3), (4.17) and (4.19), we can easily obtain
|u0i(Ri)| ∼ ui(r0) ∼ α−1i , φ0i(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)upii = 0 in B2 = {|x| < 2},
such that
ui(Pi) = max ¯ B1
ui(x) → +∞, Pi → P0 for some |P0| < 1 and
Z B2 K(x)upi+1 i 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)Mi2|x − Pi|2 n−2 2 for |x| ≤ 1, (A.2) where Mi = u 2 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. lim i→+∞M σi i = 1, where σi = n+2n−2 − pi Rescaling ui by Ui(y) = Mi−1ui(Pi+ M− pi−1 2 i y). (A.3) Then Ui satisfies
4Ui(y) + Ki(y)Uipi(y) = 0 for |y| ≤ M
pi−1 2 i , where Ki(y) = K(Pi + M − 2 n−2
i y). By elliptic estimates, Ui(y) converges to U (y) in
C2
loc(Rn), where U (y) is the solution of
4U (y) + K(P0)U n+2 n−2 = 0 in Rn, U (0) = max Rn U (y) = 1. (A.4)
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 Z Rn K(P0)U 2n 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)) ≥ Z |Pi−x|≤RiM −pi−12 i K(x)upi+1 i (x)dx = M n−2 2 σi i Z |y|≤Ri Ki(y)Uipi+1(y)dy = Mn−22 σi i Sn K(P0) n−2 n !n2 (1 + o(1)) Therefore, limi→+∞M
n−2 2 σi
i ≤ 1. Step 1 follows readily.
Set
mi = inf
|x|≤1ui(x).
To Prove (A.2), we have to compare mi and Mi−1. First, we claim
Step 2. there exists a constant c such that
Mi−1 ≤ c mi.
Consider G(x) = Mi−1(|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 Mi−1 for |x| =
1 2,
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→+∞supB¯1(ui(x)|x − Pi|
n
pi−1) = +∞, then there is a local maximum point
Qi of ui(x) such that the rescaling of ui with the center Qi,
˜
Ui(y) = ˜Mi−1ui(Qi+ ˜M−
pi−1 2
i y) with M˜i = 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|−1u¯i(|x − Pi|) (A.6)
where ¯ui(r) is the average of ui over the sphere |x − Pi| = r. Set
vi(t) = ¯ui(r)r
n−2
2 with r = et
By a straightforward computation, vi(t) satisfies
vi00(t) − n − 2 2 2 vi(t) + ˆKi(t)v n+2 n−2 i = 0, (A.7) where
ˆ Ki(t) = Z − |x−Pi|=t K(x)u−σi i u n+2 n−2 i (x)dσ ! (¯un+2n−2(r))−1, and (R
− 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 ≤ ˆKi(t) ≤ c2
for t ≤ 0. By the rescaling (A.3), we see that vi(t) has a first local maximum at t = ti =
−n−22 log Mi+ 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 ˆsi ∈ (si, 0]. By (A.7), vi(ˆsi) ≥ c > 0
for some constant c > 0. By the spherical Harnack inequality, |vi0(t)| ≤ c1. Thus, there
exists δ0 > 0 such that
vi(t) ≥ c 2 if |t − ˆsi| ≤ δ0. Therefore, Z Ti u 2n n−2 i (x)dx ≥ c1 > 0, (A.8) where Ti = {x | eˆsi−δ0 ≤ |x − Pi| ≤ eˆsi+δ0}. However, Z |Pi−x|≤esi u 2n 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 Mi−1|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 0 to be a fixed small positive number. By
ti ≤ t ≤ ˆti and vi(ˆti) = 0. If 0 is small enough, then by (A.7), vi(t) has no critical point
for t ∈ (ˆti, 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 ˆti ≤ 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−2 2 t ≥ c2 e− n−2 2 siv i(si) = c2u¯i(esi). (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) ≤ c4u¯i(ri) ri
|x| n−2
≤ c4Mi−1|x|2−n, (A.12)
for eti = r
i ≤ |x − Pi| ≤ esi where ˆti = ti+ c(ε0), ti = −n−22 log Mi, and ¯ui(ri) ∼ Mi are
used. The second inequality of (1) in Lemma A implies ¯
for ti ≤ t ≤ si. Thus, together with (A.12) and (A.11), we have
mi ∼ Mi−1(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 Cloc2 (B2\{P0}), by passing to a subsequence,
ui(x)/mi converges to a positive harmonic function h(x) in Cloc2 (B2\{P0}). For any δ > 0,
− Z |x−P0|=δ ∂h ∂ν(x)dσ = − limi→+∞ Z |x−P0|≤δ 4(ui(x)/mi)dx = 1 mi Z |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 Z |x−Pi|≤M −n−22 i R K(x)upi i dx ≤ c miMi Z |y|≤R U n+2 n−2 i (y)dy → 0
by (A.14). By using (A.12), we have 1 mi Z M −2 n−2 i R≤|x−Pi|≤esi K(x)upi i dx ≤ c mi M− n+2 n−2 i Z |x−Pi|≥M −n−22 i R |x − Pi|−(n+2)dx ≤ c mi M− n+2 n−2 i (M − 2 n−2 i R)−2 = c1 miMi R−2
→ 0 by (A.14) again. By (A.11), the last term can be estimated by 1 mi Z |x−Pi|≥esi Kupi i dx ≤ c1m − 4 n−2 i → 0. Thus, Z |x−P0|=δ ∂h ∂νdσ = 0 for any δ > 0,
which implies h is smooth at 0. Since h(x) vanishes on the boundary of B2, h(x) ≡ 0 on
B2, which contradicts to inf ¯ B1
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.
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Jann-Long Chern∗ and Chang-Shou Lin†
∗Department of Mathematics, National Central University, Chung-Li 32054, Taiwan
†Department of Mathematics, National Chung-Cheng University, Minghsiung, Chia-Yi,
Taiwan
