Uniqueness theorems through the method of moving spheres

YanYan Li^{∗} and Meijun Zhu Department of Mathematics Rutgers University New Brunswick, NJ 08903

Abstract

In this paper we first classify all nonnegative solutions of −Δu = u^{\frac{n+2}{n−2}} and −Δu = 0 in ℝ_{+}^{n} (n ≥ 3) with nonlinear Neuman boundary condition \frac{∂u}{∂t} = cu^{\frac{n}{n−2}} on ∂ℝ_{+}^{n} for some c ∈ ℝ. Under a further hypothesis : u(x) = O(|x|^{2−n}) for |x| large, such classification has been given by Escobar in [12].

In ℝ^{n}, the classification was given by Obata [24], Gidas, Ni and Nirenberg [15] under such a hypothesis. Caffarelli, Gidas and Spruck [5] established the result without this hypothesis. A somewhat simpler proof was later given by Chen and Li [6]. The result of Caffarelli, Gidas and Spruck has played important roles in obtaining energy independent apriori estimates for solutions of Yamabe type equations and scalar curvature equations. See Schoen [26] and Li [21].

We believe that our results will be useful toward establishing energy independent apriori estimates for solutions of the conformal metrics with prescribed mean curvature on the boundary as studied by Escobar ([12]-[14]). Our method is very different from that of Escobar [12]. We use the method of moving spheres, a variant of the method of moving planes ([15]).

Using the same approach we also study a similar problem in ℝ_{+}^{2}. This problem in ℝ^{2} was studied by Chen and Li [6], also Chou and Wan [11].

∗Partially supported by Alfred Sloan Foundation and NSF grant DMS-9401815

1 Introduction

For n ≥ 2, R > 0, x̄ ∈ ℝ^{n}, let

ℝ_{+}^{n} = { (x_{1}, ⋯, x_{n−1}, t) | (x_{1}, ⋯, x_{n−1}) ∈ ℝ^{n−1}, t > 0 },

B_{R}(x̄) = { x ∈ ℝ^{n} | |x − x̄| < R }, B_{R} = B_{R}(0),

B_{R}^{+}(x̄) = { (x_{1}, ⋯, x_{n−1}, t) ∈ B_{R}(x̄) | t > 0 }, B_{R}^{+} = B_{R}^{+}(0).

We always use the notation x = (x′, t) ∈ ℝ_{+}^{n}.

For n ≥ 3, c ∈ ℝ, we consider

\{ \begin{matrix}−Δu & = n(n − 2)u^{\frac{n+2}{n−2}} & in ℝ_{+}^{n}, \\ \frac{∂u}{∂t} & = cu^{\frac{n}{n−2}} & on ∂ℝ_{+}^{n}. \end{matrix}  (1)

It is easy to check that for all ϵ > 0, x′_{0} ∈ ℝ^{n−1}, and t_{0} = (n − 2)^{−1}ϵc, the following functions are solutions of (1):

u(x′, t) = (\frac{ϵ}{ϵ^{2} + |(x′, t) − (x′_{0}, t_{0})|^{2}} )^{\frac{n−2}{2}} .  (2)

Theorem 1.1: Let u ∈ C^{2}(ℝ_{+}^{n}) ∩ C^{1}(_{ℝ̄_{+}^{n}}) (n ≥ 3) be any nonnegative solution of (1). Then either u ≡ 0 or u takes the form (2) for some ϵ > 0, x′_{0} ∈ ℝ^{n−1}, and t_{0} = (n − 2)^{−1}ϵc.

Almost the same proof applies to the following equation for n ≥ 3.

\{ \begin{matrix}−Δu & = 0 & in ℝ_{+}^{n}, \\ \frac{∂u}{∂t} & = cu^{\frac{n}{n−2}} & on ∂ℝ_{+}^{n}. \end{matrix}  (3)

When c < 0, for any ϵ > 0, x′_{0} ∈ ℝ^{n−1} and t_{0} = −(n−2)^{−1}ϵc, the following functions are clearly solutions of (3):

u(x′, t) = (\frac{ϵ}{(t + t_{0})^{2} + |x′ − x′_{0}|^{2}} )^{\frac{n−2}{2}} .  (4)

Theorem 1.2: Let u ∈ C^{2}(ℝ_{+}^{n}) ∩ C^{1}(_{ℝ̄_{+}^{n}}) (n ≥ 3) be any nonnegative solution of (3). When c ≥ 0, u = at + b with a, b ≥ 0, a = cb^{n/(n−2)}. When c < 0, either u ≡ 0 or u takes the form (4) for some ϵ > 0 , x′_{0} ∈ ℝ^{n−1} and t_{0} = −(n − 2)^{−1}ϵc.

We also study a two dimensional problem which is similar to (1):

\{ \begin{matrix}−Δu & = e^{u} & in ℝ_{+}^{2}, \\ \frac{∂u}{∂t} & = ce^{u/2} & on ∂ℝ_{+}^{2}. \end{matrix}  (5)

It is easy to see that for any x′_{0} ∈ ℝ, λ > 0, and t_{0} = cλ/\sqrt{2},

u(x′, t) = log \frac{8λ^{2}}{(λ^{2} + (x′ − x′_{0})^{2} + (t − t_{0})^{2})^{2}} ,  (6)

satisfies (5) and

∫_{ℝ_{+}^{2}} e^{u} < ∞, ∫_{∂ℝ_{+}^{2}} e^{u/2} < ∞.  (7)

Theorem 1.3 Let u ∈ C^{2}(ℝ_{+}^{2})∩C^{1}(_{ℝ̄_{+}^{2}}) be any solution of (5) satisfying (7). Then u takes the form (6) for some λ > 0, x′_{0} ∈ ℝ and t_{0} = cλ/\sqrt{2}.

Theorem 1.1 and Theorem 1.2 under a further hypothesis:

u(x) = O(\frac{1}{|x|^{n−2}} ), for |x| large,  (8)

were proved by Escobar in [12] in connection with his studies of conformal metrics with prescribed mean curvature on the boundary. Our method is very different from his. We prove it by using the method of moving spheres, a variant of the method of moving planes. Roughly speaking, we make reflection with respect to spheres instead of planes, and then obtain the symmetry of solutions. The method of moving spheres were used by Chou and Chu [10], Padilla [22] , Chen and Li [7]. The method of moving planes goes back to Alexandroff in his study of embedded constant mean curvature surfaces. It was then used and developed through the work of Serrin [25], Gidas, Ni and Nirenberg [15]. More recently further progress has been made, see for example [1], [2], [3], [5], [6], [19], ... and the references therein. After we essentially completed our work, we learned that Terracini [27] had given an alternative proof of Escobar′s result (i.e. Theorem 1.1 and Theorem 1.2 under a further hypothesis (8)) through the method of moving planes. In [9], Chipot, Shafrir and Fila have also presented a proof of Theorem 1.1 and Theorem 1.2. Their original proof did not address the possible singularity at infinity, i.e. they implicitly assumed a decay hypothesis as in Escobar′s result. The authors of this paper showed them the way to handle the possible singularity at infinity. After this paper was submitted, the authors were informed that Theorem 1.2 in the case c < 0, was proved independently by Ou in [28]. The case c > 0 is much more delicate in, among other things, handling the possible singularity at infinity.

In ℝ^{n}, the classification of all nonnegative solutions of −Δu = u^{\frac{n+2}{n−2}} was given by Obata [24], Gidas, Ni and Nirenberg [15] under (8). The hypothesis (8) was then removed by Caffarelli, Gidas and Spruck [5]. This latter result has played important roles in obtaining energy independent apriori estimates for solutions of Yamabe type equations and scalar curvature equations, see Schoen [26] and Li [21]. In particular, it is used in the work of Schoen [26] in obtaining the compactness of all solutions to the Yamabe problem when the manifold is not conformally equivalent to the standard sphere. This gives an alternative proof of the Yamabe problem, as well as a counting (with multiplicities) of all solutions. Yamabe type problem with prescribed mean curvature on the boundary has been studied by Escobar ([12]-[14]), and existence results are obtained by minimizing the corresponding functionals. In this case, a natural question to ask is whether or not one can obtain compactness results similar to what was done in Schoen′s work [26] for the Yamabe problem. This is the motivation of our present work. We believe that Theorem 1.1 and Theorem 1.2 will be useful in understanding the compactness problem raised above.

Some subcritical problems with nonlinear boundary conditions have been studied by Hu [18], and nonexistence of positive solutions have been proved. Such nonexistence results in ℝ^{n} was obtained by Gidas and Spruck [16].

Theorem 1.3 concerns a similar problem in ℝ_{+}^{2}. In ℝ^{2}, the problem was studied by Chen and Li [6], also Chou and Wan [11].

The paper is organized as following. In Section 2, we prove Theorem 1.1-1.2. In fact we only present the proof for Theorem 1.1 since the proof of Theorem 1.2 is very similar. This section is divided into two subsections, one treating the case c ≥ 0 and the other c < 0. The first case is more subtle than the second case, so we give more details in the first case and set the structure of the proof which will be followed in the second case, also in the proof of Theorem 1.3. In Section 3 we prove Theorem 1.3. In the proof we often apply the moving sphere method to the Kelvin transformations of the solutions. This is because we do not know a priori any asymptotic behavior of solutions at infinity. Once we have suitable asymptotic behavior of solutions at infinity, we can work on the solutions directly.

It is easy to see that our approach can be used to give somewhat different proofs of the classification results in ℝ^{n}.

2 Proof of Theorem 1.1-1.2.

Due to similarity, we only give the proof of Theorem 1.1. If u = 0 somewhere in _{ ℝ̄_{+}^{n}}, it follows from the strong maximum principle and the Hopf lemma (see [25] and [15]) that u ≡ 0. Therefore we assume throughout this section

that u > 0 in _{ ℝ̄_{+}^{n}}.

2.1 Case c ≥ 0.

Let u be a positive function satisfying the hypotheses of Theorem 1.1. We perform the Kelvin transformation on u by setting

v(x) = \frac{1}{|x|^{n−2}} u(\frac{x}{|x|^{2}} ), x ∈ ℝ_{+}^{n}.  (9)

It follows from elementary calculations that v satisfies

\{ \begin{matrix}−Δv & = n(n − 2)v^{\frac{n+2}{n−2}} & in ℝ_{+}^{n}, \\ \frac{∂v}{∂t} & = cv^{\frac{n}{n−2}} & on ∂ℝ_{+}^{n}\\{0\}, \\ v & > 0 & in \overline{ℝ}_{+}^{n} \ \{0\}. \end{matrix}  (10)

For b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1}, we define the Kelvin transformation of u centered at b by

v_{b}(x) = \frac{1}{|x|^{n−2}} u_{b}(\frac{x}{|x|^{2}} ),

where u_{b}(x) = u(x′ + b, t).

Proposition 2.1 Let u be a positive function satisfying the hypotheses of Theorem 1.1 for some c ≥ 0. Then for all b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1}, there exists λ_{b} > 0, such that

u_{b}(x) = \frac{λ_{b}^{2−n}}{|x|^{n−2}} u_{b}(\frac{λ_{b}^{−2}x}{|x|^{2}}), ∀ x ∈ ℝ_{+}^{n}.

We establish the above proposition by the moving sphere method. From the properties of the Kelvin transformation, we only need to show Proposition 2.1 for x outside B+_{\frac{1}{λ_{b}}} (b). First we need a lemma which makes it possible to

get started.

Lemma 2.1 Let v ∈ C^{2}(ℝ_{+}^{n}) ∩ C^{1}(_{ℝ̄_{+}^{n}}) \ {0} satisfy (10). Then for all

0 < ϵ < min\{1, min_{∂B_{1}^{+}∩∂B_{1}} v\}, we have v(x) ≥ \frac{ϵ}{2(c+1)} for all x ∈ \overline{B}_{1}^{+}\\{0\}.

Proof: For 0 < r < 1, we introduce an auxilary function

φ_{0}(x) = \frac{ϵ}{2(c + 1)} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{ϵt}{2}, x ∈ \overline{B}_{1}^{+} \ B_{r}^{+}.

Considering w = v − φ_{0}. Clearly w satisfies

\{ \begin{matrix}Δw & ≤ 0 & in B_{1}^{+}\\overline{B}_{r}^{+}, \\ \frac{∂w}{∂t} & = cv^{\frac{n}{n−2}} − \frac{ϵ}{2} & on ∂(\overline{B}_{1}^{+}\B_{r}^{+}) ∩ ∂ℝ_{+}^{n}. \end{matrix}  (11)

We will show that

w ≥ 0 in \overline{B}_{1}^{+} \ B_{r}^{+}.  (12)

On ∂B_{r}^{+} ∩ ∂B_{r}, w = v − (\frac{ϵ}{2(c+1)} − ϵ + \frac{ϵt}{2}) > v > 0. On ∂B_{1}^{+} ∩ ∂B_{1}, w = v − (\frac{ϵ}{2(c+1)} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{ϵt}{2}) > v − ϵ > 0. Suppose the contrary of (12), it follows from the maximum principle that there exists some x_{0} = (x′_{0}, 0) with r < |x′_{0}| < 1 such that w(x_{0}) = min_{\overline{B}_{1}^{+}\B_{r}^{+}} w < 0. It follows that \frac{∂w}{∂t}(x_{0}) ≥ 0.

Using the boundary condition of w, we have v(x_{0}) ≥ (\frac{ϵ}{2(c+1)})^{(n−2)/n} > \frac{ϵ}{2(c+1)}. It follows that

w(x_{0}) = v(x_{0}) − (\frac{ϵ}{2(c + 1)} − \frac{r^{n−2}ϵ}{|x′_{0}|^{n−2}} ) > v(x_{0}) − \frac{ϵ}{2(c + 1)} > 0,

which contradicts to w(x_{0}) < 0. Thus we have (12). For x ∈ B_{1}^{+} \ {0}, it follows from (12) that for all 0 < r < |x| we have w(x) ≥ 0. Let r → 0, Lemma 2.1 follows.

Corollary 2.1 (scaled version ) Let v ∈ C^{2}(ℝ_{+}^{n})∩C^{1}(_{ℝ̄_{+}^{n}})\{0} satisfy

(10). Then for all 0 < ϵ < min\{R^{(2−n)/2}, min_{∂B_{R}^{+}∩∂B_{R}} v\}, we have v(x) ≥ \frac{ϵ}{2(c+1)}

for all x ∈ \overline{B}_{R}^{+}\\{0\}.

Proof: We just apply Lemma 2.1 to v̄(x) = R^{\frac{n−2}{2}} v(Rx).

Proof of Proposition 2.1: Define for λ > 0, b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1},

w_{λ,b}(x) = v_{b}(x) − \frac{λ^{n−2}}{|x|^{n−2}} v_{b}(\frac{λ^{2}x}{|x|^{2}}).

In the following we always write v(x) = v_{0}(x), v_{λ}(x) = \frac{λ^{n−2}}{|x|^{n−2}} v(\frac{λ^{2}x}{|x|^{2}}), w_{λ}(x) = v(x) − v_{λ}(x). Clearly, w_{λ} satisfies:

\{ \begin{matrix}−Δw_{λ} & = c_{1}(x)w_{λ} & in B_{λ}^{+}, \\ \frac{∂w_{λ}}{∂t} & = c_{2}(x)w_{λ} & on ∂B_{λ}^{+} ∩ ∂ℝ_{+}^{n} \ \{0\}, \end{matrix}  (13)

where c_{1} = n(n + 2)ξ_{1}(x)^{\frac{4}{n−2}} , c_{2} = \frac{nc}{n−2}ξ_{2}(x)^{\frac{2}{n−2}} , ξ_{1}, ξ_{2} are two functions between v_{λ} and v.

Claim 1: For λ large enough, w_{λ}(x) ≥ 0 for all x ∈ \overline{B}_{λ}^{+} \ \{0\}.

Proof: We prove this claim by three steps.

Step 1: ∃R_{0} > 0, such that for all R_{0} ≤ |x| ≤ λ/2, we have w_{λ}(x) ≥ 0. Proof: Set A = lim_{|y|→∞} |y|^{n−2}v(y) = u(0). It is clear that for R_{0} > 0

large, we have for all R_{0} ≤ |x| ≤ λ/2 that

w_{λ}(x) = \frac{1}{|x|^{n−2}} (|x|^{n−2}v(x)) − \frac{1}{λ^{n−2}} (|\frac{λ^{2}x}{|x|^{2}}|^{n−2}v(\frac{λ^{2}x}{|x|^{2}}))
= \frac{1}{|x|^{n−2}} (A + O(\frac{1}{|x|})) − \frac{1}{λ^{n−2}} (A + O(\frac{|x|}{λ^{2}}))
≥ \frac{1}{|x|^{n−2}} \{(1 − \frac{1}{2})A − O(\frac{1}{R_{0}} )\}
> 0.

Step 1 is established.

Step 2: ∃R_{1} ≥ R_{0}, such that for R_{1} ≤ λ/2 ≤ |x| ≤ λ, we have w_{λ}(x) ≥ 0. Proof: Let g(x) = |x|^{−α} with 0 < α < n − 2, and w̄_{λ}(x) = w_{λ}(x)/g(x). It

follows from (13) that

\{ \begin{matrix}Δw̄_{λ} & +\frac{2}{g}∇g · ∇w̄_{λ} + (c_{1}(x) + \frac{Δg}{g})w̄_{λ} = 0 in B_{λ}^{+}, \\ \frac{∂w̄_{λ}}{∂t} & = c_{2}(x)w̄_{λ} on ∂B_{λ}^{+} ∩ ∂ℝ_{+}^{n} \ \{0\}. \end{matrix}  (14)

Suppose the contrary, ∃ x_{0} = (x′_{0}, t_{0}) with λ/2 ≤ |x_{0}| ≤ λ, such that w̄_{λ}(x_{0}) = min_{λ/2≤|x|≤λ}w̄_{λ}(x) < 0. Then |x_{0}| ≠ λ from the definition of w_{λ}, |x_{0}| ≠ λ/2 from

Step 1. It follows that λ/2 < |x_{0}| < λ and

v_{λ}(x_{0}) ≤ 2^{n−2}v(\frac{λ^{2}x_{0}}{|x_{0}|^{2}} ) ≤ \frac{C}{|x_{0}|^{n−2}} ,

v(x_{0}) ≤ v_{λ}(x_{0}) ≤ \frac{C}{|x_{0}|^{n−2}} .

Here and in the following, C denotes various constant independent of λ. Thus, c_{1}(x_{0}) ≤ \frac{C}{|x_{0}|^{4}} . By a direct calculation we have

\frac{Δg}{g} (x) = −\frac{α(n − 2 − α)}{|x|^{2}}.

It is clear that for R_{1} ≥ R_{0} large enough we have

c_{1}(x_{0}) + \frac{Δg}{g} (x_{0}) < 0.  (15)

For c > 0, it is clear that t_{0} > 0. When c = 0 it follows from (15) and the strong form of the Hopf lemma ([25]) that t_{0} > 0. Using (15), we reach a contradiction simply by evaluating (14) at x_{0}. Step 2 is established.

Step 3: ∃ R_{2} ≥ R_{1}, such that for λ ≥ R_{2},

w_{λ}(x) ≥ 0 for x ∈ B_{R_{0}}^{+} \ {0}.

Proof: It follows from Corollary 2.1 that v(x) > 1/C > 0 for x ∈ B_{R_{0}}^{+}\{0}. Writing

w_{λ}(x) = v(x) − \frac{1}{λ^{n−2}} (|\frac{λ^{2}x}{|x|^{2}}|^{n−2}v(\frac{λ^{2}x}{|x|^{2}})).

Step 3 follows from the fact that lim_{|y|→∞} |y|^{n−2}v(y) = u(0) and |\frac{λ^{2}x}{|x|^{2}}| ≥ λ^{2}/R_{0} ≥ R_{2}^{2}/R_{0} for all x ∈ B_{R_{0}}^{+} \ {0}.

We have verified Claim 1.

Now we define for b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1} that

λ_{b} = inf {λ > 0 | w_{μ,b}(x) ≥ 0 in B̄_{μ}^{+} \ {0} for all λ < μ < ∞}. (16)

Claim 2: There exists b̄ ∈ ℝ^{n−1}, such that λb̄ > 0.

First we prove a general lemma.

Lemma 2.2: Suppose f ∈ C^{1}(ℝ_{+}^{n}) satisfying: for all b ∈ ℝ^{n−1}, λ > 0,

f_{b}(x) − \frac{λ^{n−2}}{|x|^{n−2}} f_{b}(\frac{λ^{2}x}{|x|^{2}}) ≤ 0, ∀ x ∈ B_{λ}^{+},

where f_{b}(x) = f(x′ + b, t), ∀x = (x′, t) ∈ ℝ_{+}^{n}. Then

f(x) = f(x′, t) = f(0, t), ∀ x = (x′, t) ∈ ℝ_{+}^{n}.

Proof: For all b ∈ ℝ^{n−1}, λ > 0, set

g_{b,λ}(x) ≡ f_{b}(x) − \frac{λ^{n−2}}{|x|^{n−2}} f_{b}(\frac{λ^{2}x}{|x|^{2}}), x ∈ ℝ_{+}^{n}, |x| ≤ λ.

It is easy to see that for all b ∈ ℝ^{n−1}, x ∈ ℝ_{+}^{n}, we have

\{ \begin{matrix}g_{b,|x|}(x) & = 0, \\ g_{b,|x|}(rx) & ≤ 0, ∀ 0 < r < 1. \end{matrix}

It follows that

\frac{d}{dr}\{g_{b,|x|}(rx)\}|_{r=1} ≥ 0.

A direct computation yields

2∇f_{b}(x) · x + (n − 2)f_{b}(x) ≤ 0,

namely,

2∂_{x′}f(x′ + b, t) · x′ + 2∂_{t}f(x′ + b, t)t + (n − 2)f(x′ + b, t) ≥ 0.

Since x′, b ∈ ℝ^{n−1}, t > 0 are arbitrary, by a change of variable we have

2∂_{x′}f(x′, t) · (x′ − b) + 2∂_{t}f(x′, t)t + (n − 2)f(x′, t) ≥ 0.

Dividing the above by |b| and sending |b| to infinity, we have: for all x′ ∈ ℝ^{n−1}, t > 0, ω ∈ ℝ^{n−1}, |ω| = 1,

−∂_{x′}f(x′, t) · ω ≥ 0.

It follows that

∂_{x′}f(x′, t) = 0, ∀x′ ∈ ℝ^{n−1}, t > 0.

Lemma 2.2 is established.

Proof of Claim 2: If λ_{b} = 0 for all b ∈ ℝ^{n−1}, we have for all b ∈ ℝ^{n−1} and λ > 0

v_{b}(x) ≥ \frac{λ^{n−2}}{|x|^{n−2}} v_{b}(\frac{λ^{2}x}{|x|^{2}}) ∀x ∈ B_{λ}^{+}.

It follows that for all b ∈ ℝ^{n−1}, and λ > 0 that

u_{b}(y) ≤ \frac{λ^{2−n}}{|y|^{n−2}} u_{b}(\frac{λ^{−2}y}{|y|^{2}} ) ∀y ∈ B_{1/λ}^{+}.

Therefore, from Lemma 2.2 we know that u(x) depends only on t. Writing u(t) = u(x), it follows that u ∈ C^{2}(0, ∞) ∩ C^{1}[0, ∞) satisfies

\{ \begin{matrix}u′′(t) + n(n − 2)u^{\frac{n+2}{n−2}} (t) & = 0, 0 < t < ∞, \\ u(t) & > 0, 0 ≤ t < ∞. \end{matrix}  (17)

An elementary phase-plane argument (writing (17) as a first order autonomous system) shows that (17) has no solution. This is a contradiction. We have verified Claim 2.

Claim 3: Suppose λ_{b} > 0 for some b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1}, then we have w_{λ_{b},b}(x) = 0, ∀ x ∈ ℝ_{+}^{n}.

Proof: Without loss of generality, we assume b = 0. Suppose the contrary of Claim 3, w_{λ_{0}} satisfies

\{ \begin{matrix}Δw_{λ_{0}} & ≤ 0 & in B_{λ_{0}}^{+}(0), \\ \frac{∂w_{λ_{0}}}{∂t} & = cv^{\frac{n}{n−2}} − cvλ_{0}^{\frac{n}{n−2}} & on ∂B_{λ_{0}}^{+} ∩ ∂ℝ_{+}^{n} \ \{0\}, \\ w_{λ_{0}} & ≥ 0, & in B̄_{λ_{0}}^{+} \ \{0\}. \end{matrix}  (18)

It follows from the strong form of the maximum principle and the Hopf lemma (using also the boundary condition of w_{λ_{0}} given in (18) ) that

\{ \begin{matrix}w_{λ_{0}}(x) > 0, & ∀ x ∈ B̄_{λ_{0}}^{+}, 0 < |x| < λ_{0}, \\ \frac{∂w_{λ_{0}}}{∂ν} (x) > 0, & ∀ x ∈ ∂B_{λ_{0}}^{+} ∩ ℝ_{+}^{n}, \end{matrix}  (19)

where ν denotes the inner unit normal of the sphere ∂B_{λ_{0}}.

Lemma 2.3: There exists some constant γ = γ(λ_{0}) > 0 such that w_{λ_{0}}(x) ≥ γ for x ∈ \overline{B}_{λ_{0}/2}^{+} \ \{0\}.

Proof: Recall that v_{λ_{0}}(x) = \frac{λ_{0}^{n−2}}{|x|^{n−2}} v(\frac{λ_{0}^{2}x}{|x|^{2}}), v_{λ_{0}}(x) → λ_{0}^{2−n}u(0) as |x| → 0. Thus ∃C_{1} > 0 such that v_{λ_{0}}(x) < C_{1} for |x| < λ_{0}. Using (19), we have min_{∂B_{λ_{0}/2}^{+}∩∂B_{λ_{0}/2}} w_{λ_{0}} ≥ ϵ, for some 0 < ϵ < 1. Without loss of generality, we

assume λ_{0} = 2. For 0 < r < 1, we introduce an auxilary function,

φ_{1}(x) = \frac{ϵμ}{2(c + 1)} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{ϵt(1 − μ)}{2}, x ∈ B_{1}^{+} \ B_{r}^{+},  (20)

where 0 < μ < 1 being chosen later. Let P (x) = w_{λ_{0}}(x) − φ_{1}(x), then P (x) satisfies

\{ \begin{matrix}ΔP (x) & ≤ 0 & in B_{1}^{+} \ B̄_{r}^{+}, \\ \frac{∂P (x)}{∂t} & = cv^{\frac{n}{n−2}} − cvλ_{0}^{\frac{n}{n−2}} − \frac{ϵ(1−μ)}{2} & on ∂(B_{1}^{+} \ B̄_{r}^{+}) ∩ ∂ℝ_{+}^{n}. \end{matrix}  (21)

We will show that

P (x) ≥ 0, ∀ x ∈ B̄_{1}^{+} \ B_{r}^{+}.  (22)

On ∂B_{1}^{+} ∩ ∂B_{1} : P (x) ≥ ϵ − (ϵ − r^{n−2}ϵ) > 0; on ∂B_{r}^{+} ∩ ∂B_{r} : P (x) > w_{λ_{0}}(x) ≥ 0. If (22) does not hold, there exists some x_{0} = (x′_{0}, t_{0}) such that, P (x_{0}) = min_{x∈\overline{B}_{1}^{+}\B_{r}^{+}} P (x) < 0. It follows from the above consideration and the

maximum principle that t_{0} = 0, r < |x′_{0}| < 1. Then we have \frac{∂P (x)}{∂t}(x_{0}) ≥ 0. From P (x_{0}) < 0, we have v(x_{0}) − v_{λ_{0}}(x_{0}) − φ_{1}(x_{0}) < 0. It follows that

v(x_{0}) < C_{2},  (23)

for some constant C_{2} = C(ϵ, C_{1}). Also

w_{λ_{0}}(x_{0}) < \frac{ϵμ}{2(c + 1)} − \frac{r^{n−2}ϵ}{|x_{0}|^{n−2}} < \frac{ϵμ}{2(c + 1)}.  (24)

From (23),(24), and the mean value theorem we have v^{\frac{n}{n−2}} (x_{0})−vλ_{0}^{\frac{n}{n−2}} (x_{0}) < C_{3}w_{λ_{0}}(x_{0}) for some constant C_{3} > 0. Combining with \frac{∂P}{∂t} (x_{0}) ≥ 0, we have

w_{λ_{0}}(x_{0}) ≥ \frac{ϵ}{2(c + 1)C_{3}} · (1 − μ).  (25)

Combining (24)and (25) we have

μ > \frac{1}{1 + C_{3}} .

If we choose μ such that 0 < μ < \frac{1}{1+C_{3}} from the beginning, We reach a contradiction. (22) is established. Let r → 0, we have proved Lemma 2.3 with γ = \frac{ϵ}{2(c+1)(C_{3}+1)} .

From the definition of λ_{0}, there exists a sequence λ_{k} → λ_{0} with λ_{k} < λ_{0}, such that

inf_{B̄_{λ_{k}}^{+} \{0}} w_{λ_{k}} < 0.

It is not difficult to see from Lemma 2.3 and the continuity of u at 0 that for k large enough, we have

w_{λ_{k}}(x) ≥ γ/2, ∀ x ∈ \overline{B}_{λ_{0}/2}^{+} \ \{0\}.

It follows that there exists x_{k} = (x′_{k}, t_{k}) ∈ B̄_{λ_{k}}^{+} \ B_{λ_{0}/2}^{+} such that

w_{λ_{k}}(x_{k}) = min_{B̄_{λ_{k}}^{+} \{0}}w_{λ_{k}} < 0.

It is clear that λ_{0}/2 < |x_{k}| < λ_{k} and, due to the boundary condition, t_{k} > 0. Hence ∇w_{λ_{k}}(x_{k}) = 0. After passing to a subsequence (still denoted as x_{k}) x_{k} → x_{0} = (x′_{0}, t_{0}). It follows that

w_{λ_{0}}(x_{0}) = 0,  ∇w_{λ_{0}}(x_{0}) = 0.  (26)

It follows from (26) and (19) that t_{0} = 0, |x′_{0}| = λ_{0}.

Lemma 2.4: If (18) and (19) hold, then \frac{∂w_{λ_{0}}}{∂ν} (x_{0}) > 0 for all x_{0} = (x′_{0}, 0), |x′_{0}| = λ_{0}.

Once we establish Lemma 2.4, we reach a contradiction due to (26), thus have verified Claim 3.

Proof of Lemma 2.4: Without loss of generality we assume λ_{0} = 1. Set Ω = {x = (x′, t) | x ∈ B_{1}^{+}\B̄_{1/2}^{+}, t < 1/4} and, for some α > max{n/2, n−3},

h(x) = ϵ(|x′|^{−α} − 1)(t + μ), φ_{2}(x) = h(x) − \frac{1}{|x|^{n−2}} h(\frac{x}{|x|^{2}} ), x ∈ Ω,

where 0 < ϵ, μ < 1 being chosen later. A direct computation yields

Δφ_{2}(x) ≥ 0, ∀ x ∈ Ω.

Consider A(x) = w_{λ_{0}}(x) − φ_{2}(x), it follows that

\{ \begin{matrix}ΔA & ≤ 0, & in Ω, \\ \frac{∂A}{∂t} & = c_{2}(x)w_{λ_{0}}(x) − \frac{∂φ_{2}}{∂t} & on ∂Ω ∩ ∂ℝ_{+}^{n}, \end{matrix}

where c_{2}(x) = \frac{nc}{n−2}ξ_{2}(x)^{\frac{2}{n−2}} , ξ_{2} is some function between v_{λ_{0}} and v. For suitably chosen ϵ and μ, we want to show

A(x) = w_{λ_{0}}(x) − φ_{2}(x) ≥ 0, ∀ x ∈ Ω.  (27)

Using (19), we can choose ϵ_{0} > 0 small enough, such that for all 0 < ϵ < ϵ_{0}, we have A(x) ≥ 0 on ∂Ω ∩ {∂B_{1/2} ∪ {t = 1/4}}. Also from the construction of φ_{2}, we know A(x) = 0 on ∂Ω ∩ ∂B_{1}. Suppose the contrary of (27), there exists some x_{1} = (x′_{1}, t_{1}) ∈ Ω̄ such that

A(x_{1}) = min_{Ω̄} A < 0.  (28)

From the above and the maximum principle, we have t_{1} = 0, 1/2 < |x′_{1}| < 1. Thus

\frac{∂A}{∂t}(x_{1}) ≥ 0.  (29)

A simple calculation yields

\frac{∂φ_{2}}{∂t} (x_{1}) = ϵ(|x′_{1}|^{−α} − 1)(|x′_{1}|^{−n+α} + 1).  (30)

Combining (29) and (30) we have

c_{2}(x_{1})w_{λ_{0}}(x_{1}) − ϵ(|x′_{1}|^{−α} − 1)(|x′_{1}|^{−n+α} + 1) ≥ 0.  (31)

It follows from (28) that

w_{λ_{0}}(x_{1}) < ϵμ(|x′_{1}|^{−α} − 1)(|x′_{1}|^{−n+α+2} + 1).  (32)

Combining (31) and (32) we have

c_{2}(x_{1})μ > 1.

So if we choose 0 < μ < min_{1/2≤|x|≤1}(c_{2}(x) + 1)^{−1} from the beginning, we reach a

contradiction. Thus (27) holds. Since we also know that A(x_{0}) = 0, we have

\frac{∂A}{∂ν}(x_{0}) ≥ 0.

It follows from a direct computation that

\frac{∂w_{λ_{0}}}{∂ν} (x_{0}) = \frac{∂A}{∂ν}(x_{0}) + \frac{∂φ_{2}}{∂ν} (x_{0}) ≥ \frac{∂φ_{2}}{∂ν} (x_{0}) = 2αϵμ > 0

Lemma 2.4 is established, also Claim 3 as remarked earlier.

Claim 4: For all b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1}, we have λ_{b} > 0.

Proof: It follows from Claim 2 and Claim 3 that there exists some b̄ ∈ ℝ^{n−1} such that λb̄ > 0 and w_{λb̄,b̄}(x) = 0, ∀ x ∈ ℝ_{+}^{n}. It follows that

ub̄(y) = λ2−n ; b̄|y|^{2−n}ub̄(\frac{y}{λ2 ; b̄|y|^{2}} ), ∀ y ∈ ℝ_{+}^{n}.

Clearly

lim_{|y|→∞} |y|^{n−2}ub̄(y) = λ2−n ; b̄u(b̄, 0),

namely,

lim_{|y|→∞} |y|^{n−2}u(y) = λ2−n ; b̄u(b̄, 0).  (33)

Suppose the contrary of Claim 4 for some b ∈ ℝ^{n−1}, namely,

w_{λ,b}(x) = v_{b}(x) − \frac{λ^{n−2}}{|x|^{n−2}} v_{b}(\frac{λ^{2}x}{|x|^{2}}) ≥ 0, ∀ λ > 0, x ∈ B̄_{λ}^{+} \ \{0\}.

It follows that

\frac{λ^{n−2}}{|x|^{n−2}} u_{b}(\frac{x}{|x|^{2}} ) ≥ u_{b}(\frac{x}{λ^{2}} ), ∀ λ > 0, x ∈ B̄_{λ}^{+} \ \{0\}.

Fixing λ > 0 in the above and sending |x| to 0, we have (using (33))

\frac{λ^{n−2}}{λn−2 ; b̄} u(b̄, 0) ≥ u_{b}(0).

Sending λ to 0, we have

u(b, 0) = u_{b}(0) ≤ 0,

a contradiction. Claim 4 has been verified.

Proposition 2.1 follows immediately from Claim 1-4.

Lemma 2.5: Suppose f ∈ C^{1}(ℝ^{n−1}) (n ≥ 3) satisfying: ∀ b ∈ ℝ^{n−1}, there exists μ_{b} ∈ ℝ such that

f(x′ + b) = \frac{μ_{b}^{n−2}}{|x′|^{n−2}} f(\frac{μ_{b}^{2}x′}{|x′|^{2}} + b), ∀ x′ ∈ ℝ^{n−1} \ \{0\}.  (34)

Then for some a ≥ 0, d > 0, x′_{0} ∈ ℝ^{n−1},

f(x′) = (\frac{a}{|x′ − x′_{0}|^{2} + d})^{(n−2)/2}, ∀ x′ ∈ ℝ^{n−1},

or

f(x′) = −(\frac{a}{|x′ − x′_{0}|^{2} + d})^{(n−2)/2}, ∀ x′ ∈ ℝ^{n−1}.

Proof: Rewriting it as

f(x′) = \frac{μ_{b}^{n−2}}{|x′ − b|^{n−2}} f(\frac{μ_{b}^{2}(x′ − b)}{|x′ − b|^{2}} + b), ∀ x′ ∈ ℝ^{n−1} \ \{b\}.

It follows that

A := lim_{|x′|→∞} |x′|^{n−2}f(x′) = μ_{b}^{n−2}f(b), ∀ b ∈ ℝ^{n−1}.  (35)

If A = 0, it is easy to see from (34) and (35) that f ≡ 0. If A ≠ 0, both f(b) and μ_{b} can not change sign. Without loss of generality we assume that A = 1. It follows from (35) that f(b) > 0, μ_{b} > 0, ∀ b ∈ ℝ^{n−1}.

For x′ large,

f(x′) = \frac{μ_{0}^{n−2}}{|x′|^{n−2}} \{f(0) + \frac{∂f}{∂x′_{i}} (0) · \frac{μ_{0}^{2}x′_{i}}{|x′|^{2}} + ∘(\frac{1}{|x′|})\},  (36)

and

f(x′) = \frac{μ_{b}^{n−2}}{|x′ − b|^{n−2}} \{f(b) + \frac{∂f}{∂x′_{i}} (b) · \frac{μ_{b}^{2}(x′_{i} − b_{i})}{|x′ − b|^{2}} + ∘(\frac{1}{|x′|})\},  (37)

Combining (36), (37), (35) and our assumption A = 1 we have

f^{− \frac{n}{n−2}} (b) · \frac{∂f}{∂x′_{i}} (b) = f^{− \frac{n}{n−2}} (0) · \frac{∂f}{∂x′_{i}} (0) − (n − 2)b_{i},

It follows that for some x′_{0} ∈ ℝ^{n−1}, d > 0

f^{− \frac{2}{n−2}} (x′) = |x′ − x′_{0}|^{2} + d.

Lemma 2.6: For some a, d > 0, x′_{0} ∈ ℝ^{n−1},

u(x′, 0) = (\frac{a}{|x′ − x′_{0}|^{2} + d})^{(n−2)/2}, ∀ x′ ∈ ℝ^{n−1}.

Proof : It follows from Lemma 2.5 and u > 0 in ℝ^{n}.

Proof of Theorem 1.1 for c ≥ 0: Let x_{0} = (x′_{0}, −\sqrt{d}), set

φ(x) = \frac{1}{|x − x_{0}|^{n−2}} u(\frac{x − x_{0}}{|x − x_{0}|^{2}} + x_{0}),

and B = \{(x′, t) : \frac{t+\sqrt{d}}{|x′−x′_{0}|^{2}+(t+\sqrt{d})^{2}} − \sqrt{d} > 0\}. Clearly, B is a ball in ℝ^{n}. Without loss of generality, we assume a = 1 in Lemma 2.6. It follows that on ∂B

φ(x′, t) = [|x′ − x′_{0}|^{2} + (t + \sqrt{d})^{2}]^{\frac{2−n}{2}} [d + \frac{|x′−x′_{0}|^{2}}{(|x′−x′_{0}|^{2}+(t+\sqrt{d})^{2})^{2}} ]^{\frac{2−n}{2}}
= \{d[|x′ − x′_{0}|^{2} + (t + \sqrt{d})^{2}] + \frac{|x′−x′_{0}|^{2}}{|x′−x′_{0}|^{2}+(t+\sqrt{d})^{2}} \}^{\frac{2−n}{2}}
= \{\frac{d(t+\sqrt{d})}{\sqrt{d}} + \frac{\sqrt{d}|x′−x′_{0}|^{2}}{t+\sqrt{d}} \}^{\frac{2−n}{2}} = 1.

Define Q(x) = φ(x) − 1, we know that Q = 0 on ∂B and

−ΔQ = n(n − 2)(Q + 1)^{\frac{n+2}{n−2}} in B.  (38)

It follows from the maximum principle that Q > 0 in B. Applying the result of [15] we know that Q is radially symmetric about the center of B. Hence by the uniqueness of the ode solution of (38), φ(x) must take the form φ(x) = (\frac{ϵ}{ϵ^{2}+|x−x_{1}|^{2}} )^{(n−2)/2} for some ϵ > 0 and x_{1} ∈ ℝ^{n}. Then

\begin{aligned}u(y) & = \frac{1}{|y−x_{0}|^{n−2}} φ(\frac{y−x_{0}}{|y−x_{0}|^{2}} + x_{0}) \\ & = (\frac{ϵ}{(ϵ^{2}+|x_{0}−x_{1}|^{2})|y−x_{0}|^{2}+2(y−x_{0})(x_{0}−x_{1})+1})^{\frac{n−2}{2}} . \end{aligned}

Theorem 1.1 when c ≥ 0 follows immediately.

2.2 Case c < 0.

The main procedure of this case is similar to Case c ≥ 0, so we define u_{b}, v(x),v_{b}(x),w_{λ,b} and w_{λ} as before. It is clear that this case is much easier than the case C > 0 since the boundary condition has the good sign.

Proposition 2.2 Let u be a positive function satisfying the hypotheses of Theorem 1.1 for some c < 0, then for any b ∈ ℝ^{n−1}, there exists a λ_{b} > 0 such that

u_{b}(x) = \frac{λ_{b}^{2−n}}{|x|^{n−2}} u_{b}(\frac{λ_{b}^{−2}x}{|x|^{2}}) for x ∈ ℝ_{+}^{n}.

Lemma 2.7: Let v ∈ C^{2}(ℝ_{+}^{n})∩C^{1}(_{ℝ̄_{+}^{n}})\{0} satisfy (10), ϵ = min_{∂B_{R}^{+}∩∂B_{R}} v(x),

R > 0. Then we have v(x) ≥ ϵ for all x ∈ \overline{B}_{R}^{+} \ \{0\}.

Proof: Since c < 0, it follows from the maximum principle that v(x) ≥ ϵ − ϵ\frac{r^{n−2}}{|x|^{n−2}} , ∀ 0 < r < R, x ∈ \overline{B}_{R}^{+} \ B_{r}^{+}. The lemma follows by sending r to 0.

Proof of Proposition 2.2: We still prove it by four Claims. As in Case c ≥ 0, w_{λ} satisfies (13).

Claim 1: For λ large enough, w_{λ}(x) ≥ 0 for all x ∈ \overline{B}_{λ}^{+} \ \{0\}. Proof: The proof is still divided into three steps. Step 1: ∃R_{0} > 0, such that for all R_{0} ≤ |x| ≤ λ/2, we have w_{λ}(x) ≥ 0. Proof: The same as in Case c ≥ 0. Step 2: ∃R_{1} ≥ R_{0}, such that for R_{1} ≤ λ/2 ≤ |x| ≤ λ, we have w_{λ}(x) ≥ 0. Proof: Let g(x) = |z|^{−α} with 0 < α < n − 2, z = x + (0, 0, ..., \frac{λ}{4}). Also

w̄_{λ}(x) = w_{λ}(x)/g(x).

\{ \begin{matrix}Δw̄_{λ} & +\frac{2}{g}∇g · ∇w̄_{λ} + (c_{1}(x) + \frac{Δg}{g})w̄_{λ} = 0 & in B_{λ}^{+}, \\ \frac{∂w̄_{λ}}{∂t} & = (c_{2}(x) − \frac{1}{g} · \frac{∂g}{∂t})w̄_{λ} on ∂B_{λ}^{+} ∩ ∂ℝ_{+}^{n} \ \{0\}. \end{matrix}  (39)

If ∃x_{0} with λ/2 ≤ |x_{0}| ≤ λ, such that w̄_{λ}(x_{0}) = min_{λ/2≤|x_{0}|≤λ}w̄_{λ}(x) < 0. Then

x_{0}∉{|x| = λ} ∪ {|x| = λ/2}. As in Case c ≥ 0,

0 ≤ c_{1}(x_{0}) ≤ \frac{C_{1}}{|x_{0}|^{4}} , 0 ≤ −c_{2}(x_{0}) ≤ \frac{C_{2}}{|x_{0}|^{2}} .

Also \frac{Δg}{g} = −\frac{α(n−2−α)}{|z|^{2}}, −\frac{1}{g} · \frac{∂g}{∂t}|_{t=0} = \frac{αλ}{4|z|^{2}} . For λ large enough, z_{0} = x_{0} + (0, 0, ..., \frac{λ}{4}),

−\frac{α(n − 2 − α)}{|z_{0}|^{2}} + c_{1}(x_{0}) < 0,

and

\frac{αλ}{4|z_{0}|^{2}} + c_{2}(x_{0}) > 0, when x_{0} ∈ ∂ℝ_{+}^{n}.

This contradicts to the maximum principle.

Step 3: ∃ R_{2} ≥ R_{1}, such that for λ ≥ R_{2},

w_{λ}(x) ≥ 0 for x ∈ B_{R_{0}}^{+} \ {0}.

Proof: From Lemma 2.7 we have v(x) > 1/C > 0 for x ∈ B_{R_{0}}^{+} \ {0}. Step 3 follows as in Case c ≥ 0.

We have verified Claim 1.

Define λ_{b} for b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1} as in (16). Claim 2: There exists b̄ ∈ ℝ^{n−1}, such that λb̄ > 0.

Proof of Claim 2: If λ_{b} = 0 for all b ∈ ℝ^{n−1}, it follows from Lemma 2.2 that u(x) just depends on t. Writing u(t) = u(x), it follows that u ∈ C^{2}(0, ∞) ∩ C^{1}[0, ∞) satisfies (17). Claim 2 follows exactly as in Case c ≥ 0.

Claim 3: Suppose λ_{b} > 0 for some b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1}, then we have w_{λ_{b},b}(x) = 0, ∀ x ∈ ℝ_{+}^{n}.

Proof: Without loss of generality, we assume b = 0. Suppose the contrary of Claim 3, w_{λ_{0}} satisfies (18) and (19) with c < 0. It follows that

\frac{∂w_{λ_{0}}}{∂t} (x) < 0, x ∈ ∂ℝ_{+}^{n}, 0 < |x| < λ_{0}.  (40)

Lemma 2.8: There exists some constant γ = γ(λ_{0}) > 0 such that w_{λ_{0}}(x) ≥ γ for x ∈ \overline{B}_{λ_{0}/2}^{+} \ \{0\}.

Proof: Set γ = min_{∂B_{λ_{0}/2}^{+}∩∂B_{λ_{0}/2}} w_{λ_{0}}(x). It follows from (19) that γ >

0. Using (18) , (40) and the maximum principle we have w_{λ_{0}}(x) ≥ γ − γ \frac{r^{n−2}}{|x|^{n−2}} , ∀ 0 < r < λ_{0}/2, x ∈ B̄_{λ_{0}/2}^{+} \ B_{r}^{+}. Lemma 2.8 follows after sending r to 0.

From the definition of λ_{0} and arguing in a way similar to that in Case c ≥ 0, there exists some x_{0} = (x′_{0}, 0) with x′_{0} ∈ ℝ^{n−1}, |x′_{0}| = λ_{0}, such that

w_{λ_{0}}(x_{0}) = 0, \frac{∂w_{λ_{0}}}{∂ν} (x_{0}) = 0,

where ν denotes the inner unit normal of ∂B_{λ_{0}}.

Lemma 2.9: If (18) and (19) hold, then \frac{∂w_{λ_{0}}}{∂ν} (x_{0}) > 0 for all x_{0} = (x′_{0}, 0), |x′_{0}| = λ_{0}.

Proof: Without loss of generality we assume λ_{0} = 1. Set Ω = {x = (x′, t) | x ∈ B_{1}^{+} \ B̄_{1/2}^{+}, t < 1/4} and, for some α > max{n/2, n − 3},

h(x) = ϵ(|x′|^{−α} − 1)(1 + t), φ_{3}(x) = h(x) − \frac{1}{|x|^{n−2}} h(\frac{x}{|x|^{2}} ),

where ϵ > 0 will be chosen later. A direct computation yields

Δφ_{3}(x) ≥ 0, x ∈ Ω.

Considering A(x) = w_{λ_{0}}(x) − φ_{3}(x), it satisfies

\{ \begin{matrix}ΔA & ≤ 0, & in Ω, \\ \frac{∂A}{∂t} & ≤ 0 & on ∂Ω ∩ \{t = 0\}. \end{matrix}  (41)

By (19) we can choose ϵ_{0} small enough, such that for 0 < ϵ < ϵ_{0}, we have A(x) > 0 on ∂Ω ∩ {∂B_{1/2} ∪ {t = 1/4}}. Also from the construction of φ_{3} we know A(x) = 0 on ∂Ω ∩ ∂B_{1}. It follows from the maximum principle that

A ≥ 0, in Ω.  (42)

Also we know A(x_{0}) = 0, thus \frac{∂A}{∂ν}(x_{0}) ≥ 0. A direct computation yields

\frac{∂φ_{3}}{∂ν} (x_{0}) = 2αϵ.

It follows that

\frac{∂w_{λ_{0}}}{∂ν} (x_{0}) ≥ \frac{∂φ_{3}}{∂ν} (x_{0}) > 0.

Lemma 2.9 is established and Claim 3 follows immediately.

Claim 4: For all b ∈ ∂ℝ_{+}^{n} = ℝ^{n−1}, we have λ_{b} > 0.

Proof: It is exactly the same as that in Case c ≥ 0.

Proposition 2.2 follows from Claim 1-4.

The rest of the proof of Theorem 1.1 in Case c < 0 follows immediately the same way as in Case c ≥ 0.

3 Proof of Theorem 1.3

Proposition 3.1: Suppose u ∈ C^{2}(ℝ_{+}^{2}) ∩ C^{1}(_{ℝ̄_{+}^{2}}) satisfies (5) and (7).

Then sup_{ℝ_{+}^{2}} u < +∞.

Lemma 3.1: Suppose u ∈ C^{2}(B_{3}^{+})∩C^{1}(B̄_{3}^{+}) satisfies, for some c, A_{1} ∈ ℝ,

\{ \begin{matrix}−Δu & = e^{u}, & in B_{3}^{+}, \\ \frac{∂u}{∂t} & = ce^{u/2}, & on \{t = 0\} ∩ B̄_{3}^{+}, \\ u(x_{0}) & = 1, & for some x_{0} ∈ B̄_{1}^{+}, \\ u & ≤ A_{1}, & in B_{3}^{+}. \end{matrix}  (43)

Then there exists some constant C_{1} = C_{1}(c, A_{1}), such that, u(x) ≥ −C_{1} in B̄_{1}^{+}.

Proof: Set

Γ_{1} = {t = 0} ∩ B̄_{2}^{+}, Γ_{2} = {t > 0} ∩ ∂B_{2}^{+}.

Let

u_{2}(y) = \frac{c}{2π} ∫_{∂ℝ_{+}^{2}∩∂B_{3}^{+}}(log |x − y| + log |x − ȳ|)e^{u(x)/2}dx, y ∈ B̄_{2}^{+},

where ȳ is the reflection point of y about {t = 0}. A direct computation yields

\{ \begin{matrix}−Δu_{2} & = 0 & in B_{2}^{+}, \\ \frac{∂u_{2}}{∂t} & = ce^{u/2} & on Γ_{1}. \end{matrix}

Define u_{1}, u_{3} by

−Δu_{1} = e^{u}, in B_{2}^{+},
\frac{∂u_{1}}{∂t} = 0, on Γ_{1},
u_{1} = 0, on Γ_{2},
−Δu_{3} = 0,  in B_{2}^{+},
\frac{∂u_{3}}{∂t} = 0,  on Γ_{1},
u_{3} = u − u_{2}, on Γ_{2}.

Clearly u = u_{1} + u_{2} + u_{3}.

Extending u_{1} evenly, we have

\{ \begin{matrix}−Δu_{1} & = e^{u}, & in B_{2}, \\ u_{1} & = 0, & on ∂B_{2}. \end{matrix}

Since e^{u} ≤ e^{A_{1}}, it follows from the W^{2,p} estimates that

max_{B̄_{2}} |u_{1}| ≤ C(A_{1}, c).  (44)

A direct computation yields for y ∈ B_{2}^{+}

|u_{2}(y)| = \frac{|c|}{2π}| ∫_{∂ℝ_{+}^{2}∩∂B_{3}^{+}}(log |x − y| + log |x − ȳ|)e^{u(x)/2}dx|
≤ \frac{|c|}{2π}e^{A_{1}/2} ∫_{∂ℝ_{+}^{2}∩∂B_{3}^{+}} | log |x − y| + log |x − ȳ||dx
≤ C(A_{1}, c).

(45)

Reflecting u_{3} evenly, we have

\{ \begin{matrix}−Δu_{3} & = 0, & in B_{2}, \\ u_{3} & = u − u_{2}, & on ∂B_{2}. \end{matrix}

Notice that u = u_{1} + u_{2} + u_{3}, we have

u_{3}(x) = u(x) − u_{1}(x) − u_{2}(x) ≤ A_{1} + C(A_{1}, c), on B_{2}^{+},

and

u_{3}(x_{0}) = u(x_{0}) − u_{1}(x_{0}) − u_{2}(x_{0}) ≥ 1 − C(A_{1}, c).

Applying the Harnack inequality to A_{1} + C(A_{1}, c) − u_{3}, we have

min_{B_{1}^{+}} u_{3}(x) ≥ −C(c, A_{1}).  (46)

Lemma 3.1 follows from (44),(45),(46).

Lemma 3.2: For c ∈ ℝ, there exists ϵ_{0} = ϵ_{0}(c) > 0, such that for all u ∈ C^{2}(B_{2}^{+}) ∩ C^{1}(B̄_{2}^{+}) satisfying

\{ \begin{matrix}−Δu = & e^{u}, & in B_{2}^{+}, \\ \frac{∂u}{∂t} = & ce^{u/2} & on ∂B_{2} ∩ \{t = 0\}, \\ ∫_{B_{2}^{+}} e^{u} < & ϵ_{0}, \end{matrix}  (47)

we have

max_{B̄_{1/4}^{+}}u ≤ C(c).

Proof: We prove it by contradiction through a blow up argument used in the proof of Proposition 2.1 of [20]. Suppose the contrary, there exists some u_{i} satisfying (47), x_{i} ∈ B̄_{1/4}^{+}, such that u_{i}(x_{i}) → ∞.

In the following, we always denote S^{T} = S ∩ {t > T } for any set S and S^{+} = S^{0}. Considering (\frac{1}{2} − |x − x_{i}|)^{2}e^{u_{i}(x)} in B_{1/2}^{+}(x_{i}), there exist y_{i} = (s_{i}, t_{i}) ∈ \overline{B}_{1/2}^{+}(x_{i}), such that

(\frac{1}{2} − |y_{i} − x_{i}|)^{2}e^{u_{i}(y_{i})} = max_{x∈\overline{B}_{1/2}^{+}(x_{i})}(\frac{1}{2} − |x − x_{i}|)^{2}e^{u_{i}(x)}.

Let σ_{i} = \frac{1}{2}(\frac{1}{2} − |y_{i} − x_{i}|) > 0. We have

4σ_{i}^{2}e^{u_{i}(y_{i})} = max_{x∈\overline{B}_{1/2}^{+}(x_{i})}(\frac{1}{2} − |x − x_{i}|)^{2}e^{u_{i}(x)} ≥ \frac{1}{4}e^{u_{i}(x_{i})}.

Thus

u_{i}(y_{i}) + 2 log σ_{i} ≥ u_{i}(x_{i}) − 2 log 4 → ∞.  (48)

Also for x ∈ \overline{B}_{σ_{i}}^{+}(y_{i}), \frac{1}{2} − |x − x_{i}| ≥ σ_{i}. Therefore

4σ_{i}^{2}e^{u_{i}(y_{i})} ≥ σ_{i}^{2} max_{\overline{B}_{σ_{i}}^{+}(y_{i})} e^{u_{i}},

namely,

u_{i}(y_{i}) ≥ max_{B̄_{σ_{i}}^{+}(y_{i})} u_{i} − log 4.

Considering

w_{i}(x) = u_{i}(μ_{i}x + y_{i}) + 2 log μ_{i},

with μ_{i} = 2e^{− \frac{u_{i}(y_{i})}{2}} . Then

−Δw_{i} = e^{w_{i}}  in B_{R_{i}}^{T_{i}},
\frac{∂w_{i}}{∂t} = ce^{w_{i}/2} on ∂B_{R_{i}}^{T_{i}} ∩ \{t = T_{i}\},
∫_{B_{R_{i}}^{T_{i}}} e^{w_{i}} < ϵ_{0},
w_{i}(x) ≤ 4 log 2 for x ∈ B̄_{R_{i}}^{T_{i}},
w_{i}(0) = 2 log 2,

where R_{i} = \frac{σ_{i}}{μ_{i}} , T_{i} = −t_{i}/μ_{i} ≤ 0. We know from (48) that R_{i} → ∞. Using Lemma 3.1 when T_{i} ≥ −1/2 and Theorem 3 of [4] when T_{i} < −1/2, we conclude that for some constant C = C(c) depending only on c , w_{i}(x) > −C(c), ∀ x ∈ \overline{B}_{1/4}^{+}. It follows that

∫_{B_{1/4}^{+}} e^{w_{i}(x)} > 1/C(c).

If we choose 0 < ϵ_{0} < 1/C(c) from the beginning, we have a contradiction. Proposition 3.1 follows from Lemma 3.2 and Theorem 3 of [4].

Next proposition is concerning the decay of u(x) at infinity.

Proposition 3.2: Suppose u ∈ C^{2}(ℝ_{+}^{2}) ∩ C^{1}(_{ℝ̄_{+}^{2}}) is a solution of (5) satisfying (7). Then

lim_{|x|→∞} \frac{u(x)}{log |x|} = d := −\frac{1}{π} ∫_{ℝ_{+}^{2}} e^{u(x)}dx + \frac{c}{π} ∫_{∂ℝ_{+}^{2}} e^{u(x)/2}dx.

Proof: Let

w(x) = \frac{1}{2π} ∫_{ℝ_{+}^{2}}(log |x − y| + log |x̄ − y| − 2 log |y|)e^{u(y)}dy
−\frac{c}{2π} ∫_{∂ℝ_{+}^{2}}(log |x − y| + log |x̄ − y| − 2 log |y|)e^{u(y)/2}dy.

It is easy to check that w satisfies

\{ \begin{matrix}Δw = & e^{u} & in ℝ_{+}^{2}, \\ \frac{∂w}{∂t} = & −ce^{u/2} & on ∂ℝ_{+}^{2}, \end{matrix}

and, using (7) and Proposition 3.1,

lim_{|x|→+∞} \frac{w(x)}{log |x|} = −d.

Considering v̄(x) = u(x) + w(x). Then

\{ \begin{matrix}Δv̄ = & 0 & in ℝ_{+}^{2}, \\ \frac{∂v̄}{∂t} = & 0 & on ∂ℝ_{+}^{2}. \end{matrix}

We extend v̄(x) to ℝ^{2} by even reflection. From Proposition 3.1, we know v̄(x) ≤ C(1 + log(|x| + 1)) for some positive constant C. Thus v̄(x) is a constant. This completes the proof.

3.1 Case c ≥ 0

Proposition 3.3 Suppose u ∈ C^{2}(ℝ_{+}^{2}) ∩ C^{1}(_{ℝ̄_{+}^{2}}) satisfies (5) and (7) with c ≥ 0. Then d = −4, and for all b ∈ ℝ, there exists some λ_{b} > 0, such that

u_{b}(x) = u_{b}(\frac{λ_{b}^{−2}x}{|x|^{2}}) − 4 log(\frac{|x|}{λ_{b}^{−1}} ), ∀ x ∈ ℝ_{+}^{2},

where u_{b}(x) = u(s + b, t), ∀ x = (s, t) ∈ ℝ_{+}^{2}.

It is crucial to show that d = −4. We will show it by obtaining contradictions when assuming d < −4 or d > −4.

First we assume d > −4.

Set

v_{b}(x) = u_{b}(\frac{x}{|x|^{2}} ) − 4 log |x|,

and

w_{λ,b}(x) = v_{b}(x) − (v_{b}(\frac{λ^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ}).

In the following we always write v(x) = v_{0}(x), v_{λ}(x) = v(\frac{λ^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ} and w_{λ}(x) = v(x) − v_{λ}(x). So w_{λ} satisfies:

\{ \begin{matrix}Δw_{λ}+ & c_{1}(x)w_{λ} = 0 & in B_{λ}^{+}, \\ \frac{∂w_{λ}}{∂t} = & c_{2}(x)w_{λ} & on ∂B_{λ}^{+} ∩ ∂ℝ_{+}^{2}, \end{matrix}  (49)

where c_{1} = e^{ξ_{1}(x)}, c_{2} = \frac{c}{2}e^{\frac{ξ_{2}}{2}} , ξ_{i} (i = 1, 2) are two functions between v_{λ} and v. We will derive a contradiction from the following four claims.

Claim 1: For λ large enough, w_{λ}(x) ≥ 0 for all x ∈ \overline{B}_{λ}^{+} \ \{0\}.

Proof: We prove this claim by three steps.

Step 1: ∃R_{0} > 0, such that for all R_{0} ≤ |x| ≤ λ/2, we have w_{λ}(x) ≥ 0. Proof: For R_{0} ≤ |x| ≤ λ/2, R_{0} large enough

w_{λ}(x) = v(x) − (v(\frac{λ^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ})
= u(\frac{x}{|x|^{2}} ) − 4 log |x| − (u(\frac{x}{λ^{2}} ) − 4 log \frac{λ^{2}|x|}{|x|^{2}} − 4 log \frac{|x|}{λ})
= O(\frac{1}{|x|}) + 4 log \frac{λ}{|x|} ≥ O(\frac{1}{|x|}) + 4 log 2 > 0.

Step 2: ∃R_{1} ≥ R_{0}, such that for R_{1} ≤ λ/2 ≤ |x| ≤ λ, we have w_{λ}(x) ≥ 0. Proof: Let g(x) = log(|x| − 1), and w̄_{λ}(x) = w_{λ}(x)/g(x). From (49) we

have

\{ \begin{matrix}Δw̄_{λ} & +\frac{2}{g}∇g · ∇w̄_{λ} + (c_{1}(x) + \frac{Δg}{g})w̄_{λ} = 0 in B_{λ}^{+}, \\ \frac{∂w̄_{λ}}{∂t} & = c_{2}(x)w̄_{λ} on \{t = 0\} ∩ \overline{B}_{λ}^{+} \ \{0\}. \end{matrix}

Suppose the contrary, ∃ x_{0} = (s_{0}, t_{0}) with λ/2 ≤ |x_{0}| ≤ λ, such that w̄_{λ}(x_{0}) = min_{λ/2≤|x|≤λ}w̄_{λ}(x) < 0. Then |x_{0}| ≠ λ from the definition of w_{λ}, |x_{0}| ≠ λ/2 from

Step 1. It follows that λ/2 < |x_{0}| < λ and

v_{λ}(x_{0}) = v(\frac{λ^{2}x_{0}}{|x_{0}|^{2}} ) − 4 log \frac{|x_{0}|}{λ}
= u(\frac{x_{0}}{λ^{2}} ) − 4 log \frac{λ^{2}}{|x_{0}|} − 4 log \frac{|x_{0}|}{λ}
≤ C − 4 log λ ≤ C_{1} − 4 log |x_{0}|,
v(x_{0}) ≤ v_{λ}(x_{0}) ≤ C_{1} − 4 log |x_{0}|.

Thus, c_{1}(x_{0}) ≤ \frac{C}{|x_{0}|^{4}} . By direct calculation we have

\frac{Δg}{g} (x) = −\frac{1}{|x_{0}|(|x_{0}| − 1)^{2} log(|x_{0}| − 1)}.

It is clear that for R_{1} ≥ R_{0} large enough we have

c_{1}(x_{0}) + \frac{Δg}{g} (x_{0}) < 0.

Arguing as in Section 2.1, we have t_{0} > 0, and reach a contradiction. Step 2 is established.

Step 3: ∃ R_{2} ≥ R_{1}, such that for λ ≥ R_{2},

w_{λ}(x) ≥ 0 for x ∈ B_{R_{0}}^{+} \ {0}.

Proof: For 0 ≤ |x| ≤ R_{0}, d > −4, λ large enough we have

w_{λ}(x) = v(x) − [v(\frac{λ^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ}]
= u(\frac{x}{|x|^{2}} ) − 4 log |x| − [u(\frac{x}{λ^{2}} ) − 4 log \frac{λ^{2}}{|x|} − 4 log \frac{|x|}{λ}]
= u(\frac{x}{|x|^{2}} ) − 4 log |x| − u(\frac{x}{λ^{2}} ) + 4 log λ.

(50)

Notice that as |x| → 0, u(\frac{x}{|x|^{2}} ) − 4 log |x| = (−d − 4 + o(1)) log |x| > 0. So we know that

u(\frac{x}{|x|^{2}} ) − 4 log |x| > −C for x ∈ B_{R_{0}}^{+} \ \{0\}.

Plugging it into (50), we have established Step 3. Claim 1 follows from Step 1-3.

Remark 3.1: We only used d > −4 in Step 3.

Now we define for b ∈ ∂ℝ_{+}^{2} = ℝ that

λ_{b} = inf{λ > 0 | w_{μ,b}(x) ≥ 0 in B̄_{μ}^{+} \ {0} for all λ < μ < ∞}.

Claim 2: ∃b̄ ∈ ℝ, such that λb̄ > 0.

Lemma 3.3: Suppose f ∈ C^{1}(ℝ_{+}^{2}) satisfying, for all b ∈ ℝ, λ > 0,

f_{b}(x) − [f_{b}(\frac{λ^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ}] ≤ 0, ∀ x ∈ B_{λ}^{+},

where f_{b}(x) = f_{b}(s, t) = f(s + b, t), ∀x = (s, t) ∈ ℝ_{+}^{2}. Then

f(x) = f(s, t) = f(0, t), ∀ x = (s, t) ∈ ℝ_{+}^{2}.

Proof: For all b ∈ ℝ, λ > 0, set

g_{b,λ}(x) ≡ f_{b}(x) − [f_{b}(\frac{λ^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ}], x ∈ ℝ_{+}^{2}, |x| ≤ λ.

It is easy to see that for all b ∈ ℝ, x ∈ ℝ_{+}^{n}, we have

\{ \begin{matrix}g_{b,|x|}(x) & = 0, \\ g_{b,|x|}(rx) & ≤ 0, ∀ 0 < r < 1. \end{matrix}

It follows that

\frac{d}{dr}\{g_{b,|x|}(rx)\}|_{r=1} ≥ 0.

A direct computation yields

2∇f_{b}(x) · x + 4 ≥ 0,

Since b is arbitry, similarly as in lemma 2.2, we can deduce

∂_{s}f(s, t) = 0, ∀s ∈ ℝ, t > 0.

We finish the proof.

Proof of Claim 2: If λ_{b} = 0 for all b ∈ ℝ, we have for all b ∈ ℝ^{n−1} and λ > 0

v_{b}(x) ≥ [v_{b}(\frac{λ^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ} ∀x ∈ B_{λ}^{+}.

It follows that for all b ∈ ℝ^{n−1}, and λ > 0 that

u_{b}(y) ≤ [u_{b}(\frac{λ^{−2}y}{|y|^{2}} ) − 4 log λ|x|] ∀y ∈ B_{1/λ}^{+}.

Therefore from Lemma 3.3 we know that u(x) depends only on t. This obviously violates (7). We have verified Claim 2.

Claim 3: Suppose λ_{b} > 0 for some b ∈ ∂ℝ_{+}^{2} = ℝ, then we have w_{λ_{b},b}(x) = 0, ∀ x ∈ ℝ_{+}^{2}.

Proof: Without loss of generality, we assume b = 0. Suppose the contrary of Claim 3, w_{λ_{0}} satisfies

\{ \begin{matrix}Δw_{λ_{0}} & ≤ 0 & in B_{λ_{0}}^{+}, \\ \frac{∂w_{λ_{0}}}{∂t} & = ce^{\frac{v}{2}} − ce^{\frac{v_{λ}}{2}} & on ∂B_{λ_{0}}^{+} ∩ ∂ℝ_{+}^{2} \ \{0\}, \\ w_{λ_{0}} & ≥ 0, & in B_{λ_{0}}^{+}. \end{matrix}  (51)

It follows from the strong maximum principle and the Hopf lemma (using also the boundary condition of w_{λ_{0}} given in (51)) that

\{ \begin{matrix}w_{λ_{0}}(x) > 0, & ∀ x ∈ B̄_{λ_{0}}^{+}, 0 < |x| < λ_{0}, \\ \frac{∂w_{λ_{0}}}{∂ν} (x) > 0, & ∀ x ∈ ∂B_{λ_{0}}^{+} ∩ ℝ_{+}^{2}, \end{matrix}  (52)

where ν denotes the inner unit normal of the sphere ∂B_{λ_{0}}.

Lemma 3.4: There exists some constant γ = γ(λ_{0}) > 0 such that w_{λ_{0}}(x) ≥ γ for x ∈ \overline{B}_{λ_{0}/2}^{+} \ \{0\}.

Proof: Notice that

\begin{aligned}v_{λ_{0}}(x) & = v(\frac{λ_{0}^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ_{0}} \\ & = u(\frac{x}{λ_{0}^{2}} ) − 4 log λ_{0} ≤ C_{1}(λ_{0}). \end{aligned}  (53)

From (52) we have min_{∂B+ ; \frac{λ_{0}}{2} ∩∂B_{λ_{0}/2}} w_{λ_{0}} ≥ ϵ, for some 0 < ϵ < 1.Without loss

of generality, we assume λ_{0} = 2. For 0 < r < 1, we introduce an auxilary function,

φ_{4}(x) = \frac{ϵμ}{2(c + 1)} − \frac{log |x|}{log r} · ϵ + \frac{ϵt(1 − μ)}{2}, x ∈ B_{1}^{+} \ B_{r}^{+},

where 0 < μ < 1 being chosen later. Let P (x) = w_{λ_{0}}(x) − φ_{4}(x), then P (x) satisfies

\{ \begin{matrix}ΔP (x) & ≤ 0 & in B_{1}^{+} \ B̄_{r}^{+}, \\ \frac{∂P (x)}{∂t} & = ce^{\frac{v}{2}} − ce^{\frac{v_{λ_{0}}}{2}} − \frac{ϵ(1−μ)}{2} & on ∂(B_{1}^{+} \ B̄_{r}^{+}) ∩ ∂ℝ_{+}^{2}. \end{matrix}

We will show that

P (x) ≥ 0, ∀ x ∈ B̄_{1}^{+} \ B_{r}^{+}.  (54)

On ∂B_{1}^{+} ∩ ∂B_{1} : P (x) ≥ ϵ − ϵ = 0; on ∂B_{r}^{+} ∩ ∂B_{r} : P (x) > w_{λ_{0}}(x) ≥ 0. If (54) does not hold, there exists some x_{0} = (s_{0}, t_{0}) such that, P (x_{0}) = min_{x∈\overline{B}_{1}^{+}\B_{r}^{+}} P (x) < 0. It follows from the above consideration and the maximum

principle that t_{0} = 0, r < |s_{0}| < 1. Then we have \frac{∂P (x)}{∂t}(x_{0}) ≥ 0.

From P (x_{0}) < 0, we have v(x_{0}) − v_{λ_{0}}(x_{0}) − φ_{4}(x_{0}) < 0. It follows from (53) that

v(x_{0}) < C_{2},  (55)

for some constant C_{2} = C_{2}(ϵ, C_{1}). Also

w_{λ_{0}}(x_{0}) < \frac{ϵμ}{2(c + 1)} − \frac{log |x|}{log r} · ϵ < \frac{ϵμ}{2(c + 1)}.  (56)

From (55),(53), and the mean value theorem we have e^{\frac{v}{2}} −e^{\frac{v_{λ_{0}}}{2}} < C_{3}w_{λ_{0}}(x_{0})for some constant C_{3} > 0. Combining with \frac{∂P}{∂t} (x_{0}) ≥ 0, we have

w_{λ_{0}}(x_{0}) ≥ \frac{ϵ}{2(c + 1)C_{3}} · (1 − μ).  (57)

Combining (56)and (57) we have

μ > \frac{1}{1 + C_{3}} .

If we choose μ such that 0 < μ < \frac{1}{1+C_{3}} from the beginning, We reach a contradiction. (54) is established. Let r → 0, we have proved Lemma 3.4 with γ = \frac{ϵ}{2(1+c)(1+C_{3})} .

From the definition of λ_{0}, there exists a sequence λ_{k} → λ_{0} with λ_{k} < λ_{0}, such that

inf_{B̄_{λ_{k}}^{+} \{0}} w_{λ_{k}} < 0.

It is not difficult to see from Lemma 3.2 and the continuity of u at 0 that for k large enough, we have

w_{λ_{k}}(x) ≥ γ/2, ∀ x ∈ \overline{B}_{λ_{0}/2}^{+} \ \{0\}.

It follows that there exists x_{k} = (s_{k}, t_{k}) ∈ B̄_{λ_{k}}^{+} \ B_{λ_{0}/2}^{+} such that

w_{λ_{k}}(x_{k}) = min_{B̄_{λ_{k}}^{+} \{0}}w_{λ_{k}} < 0.

It is clear that λ_{0}/2 < |x_{k}| < λ_{k} and, due to the boundary condition, t_{k} > 0. Hence ∇w_{λ_{k}}(x_{k}) = 0. After passing to a subsequence (still denoted as x_{k}) x_{k} → x_{0} = (s_{0}, t_{0}). It follows that

w_{λ_{0}}(x_{0}) = 0,  ∇w_{λ_{0}}(x_{0}) = 0.  (58)

It follows from (58) and (52) that t_{0} = 0, |s_{0}| = λ_{0}.

Lemma 3.5: If (51) and (52) hold. Then \frac{∂w_{λ_{0}}}{∂ν} (x_{0}) > 0 for all x_{0} = (s_{0}, 0), |s_{0}| = λ_{0}.

Once we establish Lemma 3.5, we will have reached a contradiction, thus have verified Claim 3.

Proof of Lemma 3.5: Without loss of generality we assume λ_{0} = 1, s_{0} = 1. Set Ω = {x = (s, t) | 1/2 < s^{2} + t^{2} < 1, s > 0, 0 < t < 1/4}, and

h(x) = ϵ(1 − s)(t + μ), φ_{5}(x) = h(x) − h(\frac{x}{|x|^{2}} ),

where 0 < ϵ, μ < 1 being chosen later. A direct computation yields

Δφ_{5}(x) = 0, ∀ x ∈ Ω.

Consider B(x) = w_{λ_{0}}(x) − φ_{5}(x), it follows that

\{ \begin{matrix}ΔB & ≤ 0, & in Ω, \\ \frac{∂B}{∂t} & = c_{2}(x)w_{λ_{0}}(x) − \frac{∂φ_{5}}{∂t} & on ∂Ω ∩ ∂ℝ_{+}^{2}, \end{matrix}  (59)

where c_{2} = \frac{c}{2}e^{\frac{ξ_{2}}{2}} , ξ_{2} is a function between v_{λ} and v.

For suitably chosen ϵ and μ, we want to show

B(x) = w_{λ_{0}}(x) − φ_{5}(x) ≥ 0, ∀ x ∈ Ω.  (60)

Using (52), we can choose ϵ_{0} > 0 small enough, such that for all 0 < ϵ < ϵ_{0}, we have B(x) ≥ 0 on ∂Ω ∩ {∂B_{1/2} ∪ {t = 1/4}}. Also from the construction of φ_{5}, we know B(x) = 0 on Ω ∩ ∂B_{1}. Suppose the contrary of (60), there exists some x_{1} = (s_{1}, t_{1}) ∈ Ω̄ such that

B(x_{1}) = min_{Ω̄} B < 0.  (61)

From the above and the maximum principle, we have t_{1} = 0, 1/2 < s_{1} < 1. Thus

\frac{∂B}{∂t} (x_{1}) ≥ 0.  (62)

A simple calculation yields

\frac{∂φ_{5}}{∂t} (x_{1}) = ϵ(1 − s_{1})(s_{1}^{−3} + 1).  (63)

Combining (59), (62) and (63) we have

c_{2}(x_{1})w_{λ_{0}}(x_{1}) − ϵ(1 − s_{1})(s_{1}^{−3} + 1) ≥ 0.  (64)

It follows from (61) that

w_{λ_{0}}(x_{1}) < ϵμ(1 − s_{1})(s_{1}^{−1} + 1).  (65)

Combining (64) and (65) we have

c_{2}(x_{1})μ > 1.

So if we choose 0 < μ < min_{1/2≤|x|≤1}(c_{2}(x) + 1)^{−1} from the beginning, we reach a

contradiction. Thus (60) holds. Since we also know that B(x_{0}) = 0, we have

\frac{∂B}{∂ν} (x_{0}) ≥ 0.

It follows from a direct computation that

\frac{∂w_{λ_{0}}}{∂ν} (x_{0}) = \frac{∂B}{∂ν} (x_{0}) + \frac{∂φ_{5}}{∂ν} (x_{0}) ≥ \frac{∂φ_{5}}{∂ν} (x_{0}) = 2ϵμ > 0

Lemma 3.2 is established, also Claim 3 as remarked earlier.

From Claim 2 and Claim 3, it is easy to see that d = −4. This contradicts to our assumption d > −4. We have thus proved d ≤ −4.

Lemma 3.6: For d ≤ −4, α := lim_{|x|→∞}[u(x) − d log |x|] exists and is finte. Furthermore

|u(x) − d log |x| − α| ≤ \frac{C log |x|}{|x|}, ∀ x ∈ ℝ_{+}^{2}, |x| ≥ 1.

Proof: Since d ≤ −4, it’s easy to see that for |y| ≥ 1, e^{u(y)} ≤ C|y|^{−4}, e^{u(y)/2} ≤ C|y|^{−2}. From the proof of Proposition 3.2 and the above decay property of e^{u}, we have for some constant α

u(x) − d log |x| − α = −\frac{1}{2π} ∫_{ℝ_{+}^{2}}(log |x − y| + log |x̄ − y| − 2 log |x|)e^{u(y)}dy

+\frac{c}{2π} ∫_{∂ℝ_{+}^{2}}(log |x − y| + log |x̄ − y| − 2 log |x|)e^{u(y)/2}dy.

Lemma 3.6 follows from some elementary calculations.

With Lemma 3.6, we simply repeat the previous moving sphere argument for u instead of for v and we will also reach a contradiction if d < −4. One difference we need to point out is in Step 3. For this step, d < −4 for u will play the same role as d > −4 for v. The other difference is in Step 1. We leave the details to readers.

Now we have proved that d = −4. In this case, we still work with the Kelvin transformation v of u and can see that Claim 1 to Claim 3 are still valid. The reason is that, as pointed out in Remark 3.1, we only need to establish Step 3 when d = −4. But this can be done without much difficulty with the help of Lemma 3.6. We leave the details to readers.

Claim 4: For all b ∈ ℝ, we have λ_{b} > 0.

Proof: It follows from Claim 2 and Claim 3 that there exists some b̄ ∈ ℝ such that λb̄ > 0 and w_{λb̄,b̄}(x) = 0, ∀ x ∈ ℝ_{+}^{2}. That is

ub̄(x) = ub̄(\frac{x}{λ2 ; b̄|x|^{2}} ) − 4 log(λb̄|x|) ∀ x ∈ ℝ_{+}^{2}.

It follows that

lim_{|x|→∞}[ub̄(x) + 4 log |x|] = u(b̄, 0) − 4 log λb̄.

namely,

lim_{|x|→∞}[u(x) + 4 log |x|] = u(b̄, 0) − 4 log λb̄.  (66)

Suppose the contrary of Claim 4 for some b ∈ ℝ, namely,

w_{λ,b}(x) = v_{b}(x) − [v_{b}(\frac{λ^{2}x}{|x|^{2}}) − 4 log \frac{|x|}{λ}] ≥ 0, ∀ λ > 0, ∀x ∈ B̄_{λ}^{+} \ \{0\}.

It follows that

u_{b}(\frac{x}{|x|^{2}} ) ≥ u_{b}(\frac{x}{λ^{2}} ) − 4 log \frac{λ}{|x|}, ∀ λ > 0, ∀x ∈ B̄_{λ}^{+} \ \{0\}.

Fixing λ > 0 in the above and sending |x| to 0, we have (using (66))

u(b̄, 0) − 4 log λb̄ ≥ u_{b}(0) − 4 log λ.

Sending λ to 0, we get a contradiction. Claim 4 has been verified. Proposition 3.3 follows from Claim 1 to Claim 4.

Lemma 3.7: Suppose f ∈ C^{1}(ℝ) satisfies: ∀ b ∈ ℝ, there exists some μ_{b} > 0 such that

f(s + b) = f(\frac{μ_{b}^{2}}{s} + b) − 4 log(\frac{|s|}{μ_{b}}), ∀ s ∈ ℝ \ \{0\}.

Then for some s_{0} ∈ ℝ, a, d > 0,

f(s) = log(\frac{a}{(s − s_{0})^{2} + d})^{2}, ∀ s ∈ ℝ.

Proof: Considering

h(s) = e^{−f(s)/2}, s ∈ ℝ.

For all b ∈ ℝ, we have

h(s + b) = \frac{s^{2}}{μ_{b}^{2}} h(\frac{μ_{b}^{2}}{s} + b), ∀ s ∈ ℝ \ \{0\}.

For |s| large, we have

h(s) = \frac{s^{2}}{μ_{0}^{2}} \{h(0) + h′(0)\frac{μ_{0}^{2}}{s} + O(\frac{1}{s^{2}} )\},

h(s) = \frac{(s − b)^{2}}{μ_{b}^{2}} \{h(b) + h′(b)\frac{μ_{b}^{2}}{s − b} + O(\frac{1}{s^{2}} )\}.

Comparing the above two formula, we have

\frac{h(b)}{μ_{b}^{2}} = \frac{h(0)}{μ_{0}^{2}}, h′(b) − \frac{2bh(b)}{μ_{b}^{2}} = h′(0).

It follows that

h(b) = \frac{h(0)}{μ_{0}^{2}}b^{2} + h′(0)b + h(0), ∀ b ∈ ℝ.

Therefore for some s_{0}, d ∈ ℝ, a > 0, we have

h(s) = \frac{1}{a}\{(s − s_{0})^{2} + d\}.

Clearly d > 0 and Lemma 3.7 follows.

Proof of Theorem 1.3 when c ≥ 0: Using Proposition 3.3 and Lemma 3.7, we know that for some s_{0} ∈ ℝ, a, d > 0 we have

u(s, 0) = log(\frac{a}{(s − s_{0})^{2} + d})^{2}, ∀ s ∈ ℝ.

Set x_{0} = (s_{0}, −\sqrt{d}), B = \{(s, t) : \frac{t+\sqrt{d}}{|s−s_{0}|^{2}+(t+\sqrt{d})^{2}} − \sqrt{d} ≥ 0\} , and

φ(x) = u(\frac{x − x_{0}}{|x − x_{0}|^{2}} + x_{0}) − 4 log |x − x_{0}|, x ∈ B.

It is clear that B is a ball in ℝ^{2}. On the boundry of that ball ∂B = {(s, t) : \frac{t+\sqrt{d}}{|s−s_{0}|^{2}+(t+\sqrt{d})^{2}} − \sqrt{d} = 0\}, we have

φ(s, t) = 2 log a − 2 log[\frac{|s−s_{0}|^{2}}{(|s−s_{0}|^{2}+(t+\sqrt{d})^{2})^{2}} + d] − 2 log[|s − s_{0}|^{2} + (t + \sqrt{d})^{2}]
= 2 log a − 2 log\{\frac{|s−s_{0}|^{2}}{|s−s_{0}|^{2}+(t+\sqrt{d})^{2}} + d · [|s − s_{0}|^{2} + (t + \sqrt{d})^{2}]\}
= 2 log a.

Using the maximum principle and the result of Gidas, Ni and Nirenberg [15], we conclude that w − 2 log a ≥ 0 in B and w is radially symmetric about the center of B. Theorem 1.3 in this case follows basically the same way as in the proof of Theorem 1.1.

3.2 Case c < 0.

Proof of Theorem 1.3 in Case c < 0: From Section 2.2 and section 3.1, we can see that almost all steps are the same as in Section 3.1. The main difference between Case c ≥ 0 and c < 0 when n = 2 is to show Step 2 in the proof of Claim 1. Actually we can prove this step in Case c < 0 by using test fuction g(x) = log(|z| − 1) instead of g(x) = log(|x| − 1) with z = x + (0, −λ/4). The main procedure of this case is similar to Case c > 0, so we define v(x),v_{b}(x),w_{λ,b} and w_{λ} as in Section 3.1.

Proposition 3.5 If u(x) satisfies (5) and (7), then d = −4 and for all b ∈ ℝ, there exists some λ_{b} > 0, such that

u_{b}(x) = u_{b}(\frac{λ_{b}^{−2}x}{|x|^{2}}) − 4 log(\frac{|x|}{λ_{b}^{−1}} ), ∀ x ∈ ℝ_{+}^{2},

where u_{b}(x) = u(s + b, t).

Similarly we will deduce contradiction if d < −4 and d > −4.

First we assume d > −4. We will derive a contradiction as in Section 3.1. Claim 1: If λ is large enough, then w_{λ}(x) ≥ 0 for all x ∈ \overline{B}_{λ}^{+} \ \{0\}. Proof: We still prove this claim by three steps. Step 1: ∃R_{0} > 0, such that for all R_{0} ≤ |x| ≤ λ/2, we have w_{λ}(x) ≥ 0. Proof: See Section 3.1 (Step 1 in Claim 1). Step 2: ∃R_{1} ≥ R_{0}, such that for R_{1} ≤ λ/2 ≤ |x| ≤ λ, we have w_{λ}(x) ≥ 0. Proof: Let g(x) = log(|z| − 1) with z = x + (0, −\frac{λ}{4}) and w̄_{λ}(x) =

w_{λ}(x)/g(x). From (49) we have

\{ \begin{matrix}Δw̄_{λ} & +\frac{2}{g}∇g · ∇w̄_{λ} + (c_{1}(x) + \frac{Δg}{g})w̄_{λ} = 0 in B_{λ}^{+}, \\ \frac{∂w̄_{λ}}{∂t} & = (c_{2}(x) − \frac{1}{g} · \frac{∂g}{∂t})w̄_{λ} on \{t = 0\} ∩ \overline{B}_{λ}^{+} \ \{0\}. \end{matrix}  (67)

Suppose the contrary, ∃ x_{0} = (s_{0}, t_{0}) with λ/2 ≤ |x_{0}| ≤ λ, such that w̄_{λ}(x_{0}) = min_{λ/2≤|x|≤λ}w̄_{λ}(x) < 0. Then |x_{0}| ≠ λ from the definition of w_{λ},

|x_{0}| ≠ λ/2 from Step 1, and t_{0} > 0 due to the boundary condition. It follows that x_{0} ∈ B_{λ}^{+} \ B̄_{λ/2}^{+} and

v_{λ}(x_{0}) ≤ C − 4 log λ ≤ C_{1} − 4 log |x_{0}|,
v(x_{0}) ≤ v_{λ}(x_{0}) ≤ C_{1} − 4 log |x_{0}|.

Thus, when λ is large enough

|c_{1}(x_{0})| ≤ \frac{C_{1}}{|x_{0}|^{4}} , |c_{2}(x)| ≤ \frac{C_{2}}{|x_{0}|^{2}} , |x_{0}| ∼ |z_{0}| ∼ λ,

where z_{0} = x_{0} + (0, −λ/4).

By a direct calculation we have

\frac{Δg}{g} (x_{0}) = −\frac{1}{|z_{0}|(|z_{0}| − 1)^{2} log(|z_{0}| − 1)},

\frac{1}{g} · \frac{∂g}{∂t} (x_{0}) = −\frac{λ}{4|z_{0}|(|z_{0}| − 1) log(|z_{0}| − 1)}.

Then as λ large enough, we have

c_{1}(x_{0}) + \frac{Δg}{g} (x_{0}) < 0,

and

c_{2}(x) − \frac{1}{g} · \frac{∂g}{∂t} (x_{0}) > 0.

Step 2 can be established as before.

Step 3: ∃ R_{2} ≥ R_{1}, such that for λ ≥ R_{2},

w_{λ}(x) ≥ 0 for x ∈ B_{R_{0}}^{+} \ {0}.

Proof: See Section 3.1 (Step 3 in Claim 1).

The rest of the proof is very similar to that in Section 3.1, changing test functions to some standard comparision functions in the proof of Claim 3 as in Section 2.2. Then we can prove Proposition 3.5.

Theorem 1.3 in Case c < 0 follows by some simple modifications of previous argument.

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