Unknown Journal Series Volume 000, Number 0, 0000

CLASSIFICATIONS OF NON-NEGATIVE SOLUTIONS TO SOME ELLIPTIC PROBLEMS

YUAN LOU AND MEIJUN ZHU

Abstract. The main purpose of this paper is to show that all non-negative solutions to Δu = 0 (or Δu = u^{p}) in the n-dimensional upper half space H = {(x′, t)|x′ ∈ ℝ^{n−1}, t > 0} with boundary condition ∂u/∂t = u^{q} on ∂H must be linear functions of t (or u ≡ 0) when n ≥ 2 and q > 1 (or n ≥ 2 and p, q > 1).

1. Introduction

Let H = {(x′, t) | x′ = (x_{1}, x_{2}, ..., x_{n−1}) ∈ ℝ^{n−1}, t > 0} be the upper half space in ℝ^{n} with n ≥ 2. We are interested in the following problem.

(1)  \{ \begin{matrix}Δu = 0, u ≥ 0 in H, \\ \frac{∂u}{∂t} = cu^{q} on ∂H. \end{matrix}

Recently there are many work concerning the symmetry property about the solutions in a half space, see, for example, [2], [4], [6], [7], [8], [9] and the references therein. In [6], Hu studied (1) when c < 0, 1 < q < n/(n − 2) and n ≥ 3. Using the moving plane method(see e.g., [5]), he proved that u ≡ 0 is the only solution. Later in [7], all solutions of (1) are classified when q = n/(n − 2) and n ≥ 3 by using the method of moving spheres regardless of the sign of c. In this note, by using some technical lemmas developed in [7] and the moving plane method we shall classify all solutions when c > 0 and q > 1. To this end, after rescaling and by the Strong Maximum Principle, it suffices to consider the following problem.

(2)  \{ \begin{matrix}Δu = 0, u > 0 in H, \\ \frac{∂u}{∂t} = u^{q} on ∂H. \end{matrix}

Our first result can be stated as follows.

Theorem 1.1. If u(x) ∈ C^{2}(H) ∩ C^{1}(H̄) solves (2) and q > 1, then u = at + b for some positive constants a, b satisfying a = b^{q}.

There are two ingredients in our proof. First, by virtue of the boundary condition we can prove Theorem 1.1 in the super-critical case, i.e., q > n/(n−2) and n ≥ 3, by the moving plane method. In the proof some technical lemmas will play important role as in [7]. Secondly, when q is sub-critical or q > 1 and n = 2, we observe that if

\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society

???-????/00 $1.00 + $.25 per page

1991 Mathematics Subject Classification. 35J65.

(2) is put into a high dimensional space setting(its meaning will be made precisely in the proof), any super-linear power becomes a super-critical power. This reduces all the cases to the super-critical one.

It turns out that these two ideas combined is also useful in treating some other problems with so-called “good sign” coefficient in front of the nonlinearity. Among them, the following problem serves as a good example.

(3)  \{ \begin{matrix}Δu = u^{p}, u ≥ 0 in H, \\ \frac{∂u}{∂t} = u^{q} on ∂H. \end{matrix}

For solutions of (3) we have the following result.

Theorem 1.2. If u(x) ∈ C^{2}(H) ∩ C^{1}(H̄) solves (3) for p, q > 1, then u ≡ 0.

Our paper is organized as follows: Theorem 1.1 will be established in section 2. In this section, we first prove Theorem 1.1 in the super-critical case (Theorem 2.1 below), and then pursue the second ingredient to complete the proof of Theorem 1.1(Theorem 2.7 below). The proof of Theorem 1.2 will be presented in section 3. In the last section we shall discuss some related things.

Acknowledgments: The research of the first author at MSRI is supported by NSF grant DMS# 9022140.

The second author would like to thank Professor H. Brezis for bringing this problem to his attention and also for his interest in this work. He also would like to thank Professor Z.C. Han for some interesting discussions. Some initial ideas stem from the joint work [7]. He would like to take this opportunity to thank his adviser Professor Y.Y. Li for his help and encouragement.

2. Proof of Theorem 1.1

2.1 Case q > n/(n − 2), n ≥ 3.

We assume n ≥ 3 and q > n/(n − 2) throughout this subsection.

Theorem 2.1. If u(x) ∈ C^{2}(H) ∩ C^{1}(H̄) solves (2) and q > n/(n − 2), then u = at + b with a = b^{q} and b > 0.

Since there is no assumption on the decay rate of u(x) at infinity, as usual we perform the Kelvin transformation on u, i.e., set

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

Then v(x) satisfies

(4)  \{ \begin{matrix}Δv = 0, v(x) > 0 in H, \\ \frac{∂v}{∂t} = |x|^{α}v^{q} on ∂H \ \{0\}, \end{matrix}

where α = (n − 2)q − n > 0. Our purpose is to obtain some symmetry properties of v(x), and we achieve this goal by using moving planes which are parallel to t-axis. Our first lemma, which is a modification of Lemma 2.1 in [7], will be used to handle the possible singular point.

CLASSIFICATIONS OF NON-NEGATIVE SOLUTIONS TO SOME ELLIPTIC PROBLEMS 3

Lemma 2.2. Let v ∈ C^{2}(H) ∩ C^{1}(H̄) \ {0} satisfy (4). Then for all 0 < ϵ <

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

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

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

Set P_{0} = v − φ_{0}. Clearly P_{0} satisfies

(5)  \{ \begin{matrix}ΔP_{0} = 0 in B_{1}^{+} \ B_{r}^{+}, \\ \frac{∂P_{0}}{∂t} = |x|^{α}v^{q} − \frac{ϵ}{2} on ∂(\overline{B_{1}^{+}} \ \overline{B_{r}^{+}}) ∩ ∂H. \end{matrix}

We will show that

(6)  P_{0} ≥ 0 in \overline{B_{1}^{+}} \ \overline{B_{r}^{+}}.

On ∂B_{r}^{+} ∩ ∂B_{r}, P_{0} = v − (\frac{ϵ}{2} − ϵ + \frac{ϵt}{2}) > v > 0; on ∂B_{1}^{+} ∩ ∂B_{1}, P_{0} = v − (\frac{ϵ}{2} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{ϵt}{2}) > v − ϵ > 0. Suppose that (6) fails, it follows from the Strong Maximum Principle that there exists some x̄ = (x̄′, 0) with r < |x̄′| < 1 such that P_{0}(x̄) = min_{\overline{B_{1}^{+}}\\overline{B_{r}^{+}}P_{0}} < 0. Therefore \frac{∂P_{0}}{∂t} (x̄) ≥ 0. By using the boundary condition of

P_{0}, we have v(x̄)^{q} ≥ \frac{ϵ}{2}. Hence

P_{0}(x̄) = v(x̄) − (\frac{ϵ}{2} − \frac{r^{n−2}ϵ}{|x̄′|^{n−2}} ) > v(x̄) − \frac{ϵ}{2} > 0,

which contradicts P_{0}(x̄) < 0. This establishes (6). For x ∈ B_{1}^{+} \{0}, it follows from (6) that for all 0 < r < |x| we have P_{0}(x) ≥ 0. Let r → 0, this finishes the proof of Lemma 2.2.

Corollary 2.3 (scaled version). Let v ∈ C^{2}(H)∩C^{1}(H̄)\{0} solve (4). Then

for all 0 < ϵ < min\{R^{(2−n)/2}, min_{∂B_{R}^{+}∩∂B_{R}} v\}, we have v(x) > \frac{ϵ}{2} for all x ∈ \overline{B_{R}^{+}} \ \{0\}.

Proof: It follows easily from applying Lemma 2.2 to v̄(x) = R^{\frac{n−2}{2}} v(Rx).

For λ < 0 we define

Σ_{λ} = {x | t > 0, x_{1} > λ}, T_{λ} = {x | t ≥ 0, x_{1} = λ},
Σ̃_{λ} = \overline{Σ}_{λ} \ \{0\}, x^{λ} is the reflection point of x about T_{λ},
v_{λ}(x) = v(x^{λ}), w_{λ} = v(x) − v_{λ}(x).

Then w_{λ}(x) satisfies

(7)  \{ \begin{matrix}Δw_{λ} = 0 in Σ_{λ} \\ \frac{∂w_{λ}}{∂t} ≤ c_{1}(x)w_{λ} on ∂H ∩ Σ̃_{λ}, \end{matrix}

where c_{1}(x) = q|x|^{α} · ξ_{1}^{q−1}(x), ξ_{1} is a positive function between v_{λ} and v. Now we are ready to apply the moving plane method.

Proposition 2.4. If |λ| is large enough, then w_{λ}(x) ≥ 0 for all x ∈ Σ̃_{λ}.

Proof: We argue by contradiction. Suppose that w_{λ} is negative somewhere in Σ̃_{λ}. Since v(x) → 0 as |x| → +∞, and for any fixed λ, |x^{λ}| → +∞ as |x| → +∞, we know that w_{λ}(x) → 0 as |x| → +∞. It follows from Lemma 2.2 that if |λ| is large enough, w_{λ}(x) > 0 for x near the origin. Thus there exists some point x̄ such that w_{λ}(x̄) = min_{x∈Σ̃_{λ}} w_{λ}(x) < 0. By the Strong Maximum Principle, x̄ ∈ ∂H ∩ Σ̃_{λ},

but this contradicts the boundary condition of ( 7).

Now we can define

(8)  λ_{0} = sup{λ < 0 | w_{μ}(x) ≥ 0 in Σ̃_{μ} for all − ∞ < μ < λ}.

Proposition 2.5. λ_{0} = 0.

Proof : We again argue by contradiction. Suppose that λ_{0} < 0, then we claim that

(9)  w_{λ_{0}} ≡ 0.

This will contradict (4) since α > 0. Hence from now on it suffices to prove (9) under the assumption λ_{0} < 0. Suppose that (9) fails, then by the definition of λ_{0} and the Strong Maximum Principle, w_{λ_{0}} > 0 in Σ̃ \ T_{λ_{0}}. The following result is needed to treat the singular point, and we relegate its proof to the end.

Lemma 2.6. For r_{0} < min\{\frac{1}{2}|λ_{0}|, 1\}, there exists some positive constant γ depending only on λ_{0} and r_{0} such that w_{λ_{0}}(x) > γ in B_{r_{0}}^{+}(0).

We continue the proof of Proposition 2.5. By the definition of λ_{0}, there is a sequence λ_{k} → λ_{0} with λ_{k} > λ_{0} such that inf_{Σ̃_{λ_{k}}}w_{λ_{k}} < 0. It is not difficult to see

from Lemma 2.6 and the continuity of v(x) away from the origin that for k large enough, we have w_{λ_{k}}(x) ≥ γ/2, ∀x ∈ \overline{B_{r_{0}}^{+}} \ \{0\}. We also know that w_{λ_{k}}(x) → 0 as |x| → +∞, and it follows that there exists x_{k} = (x′_{k}, t_{k}) ∈ Σ̃_{λ_{k}} \ \overline{B_{r_{0}}^{+}} such that w_{λ_{k}}(x_{k}) = min_{Σ̃_{λ_{k}}} w_{λ_{k}} < 0. It is clear that r_{0} < |x_{k}| and due to the boundary

condition, t_{k} > 0. Since w_{λ_{k}} is a harmonic function in Σ̃_{λ_{k}}, we thus reach a contradiction. Hence (9) is verified and this finishes the proof of Proposition 2.5.

Proof of Lemma 2.6. We know that w_{λ_{0}} satisfies

(10)  \{ \begin{matrix}Δw_{λ_{0}} = 0, w_{λ_{0}} > 0, in B_{r_{0}}^{+}(0) \ \{0\}, \\ \frac{∂w_{λ_{0}}}{∂t} = |x|^{α}v^{q} − |x^{λ_{0}}|^{α}v_{λ_{0}}^{q} on \overline{B_{r_{0}}^{+}}(0) ∩ ∂H \ \{0\}. \end{matrix}

Let min_{∂B_{r_{0}}^{+} (0)} w_{λ_{0}} ≥ ϵ for some 0 < ϵ < 1. Due to the continuity of v in ℝ_{+}^{n}\{0}, there

exists some positive constant C_{1} such that

(11)  v_{λ_{0}}(x) < C_{1} < +∞ for x ∈ \overline{B_{r_{0}}^{+}}(0) \ \{0\}.

Let

φ_{1}(x) = \frac{ϵμ}{2} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{ϵt(1 − μ)}{2} in B_{r_{0}}^{+}(0) \ B_{r}^{+}(0),

where μ < 1 will be chosen later. Set P_{1}(x) = w_{λ_{0}}(x) − φ_{1}(x). P_{1}(x) satisfies

(12) \{ \begin{matrix}ΔP_{1}(x) = 0 in B_{r_{0}}^{+}(0) \ B_{r}^{+}(0) \\ \frac{∂P_{1}(x)}{∂t} = |x|^{α}v^{q} − |x^{λ_{0}}|^{α}v_{λ_{0}}^{q} − \frac{ϵ(1 − μ)}{2} on \overline{B_{r_{0}}^{+}}(0) \ \overline{B_{r}^{+}}(0) ∩ ∂H. \end{matrix}

On ∂B_{r_{0}}^{+}(0) ∩ ∂B_{r_{0}}(0) : P_{1}(x) ≥ ϵ − (ϵ − \frac{r^{n−2}ϵ}{|x|^{n−2}} ) > 0; on ∂B_{r}^{+}(0) ∩ ∂B_{r}(0) : P_{1}(x) > w_{λ_{0}}(x) ≥ 0. If there exists some minimum point x̄ of P_{1}(x) such that P_{1}(x̄) < 0, we must have x̄ ∈ \overline{B_{r_{0}}^{+}}(0) \ \overline{B_{r}^{+}}(0) ∩ ∂H and \frac{∂P_{1}(x)}{∂t}(x̄) ≥ 0. From P_{1}(x̄) < 0 we have v(x̄) − v_{λ_{0}}(x̄) − φ_{1}(x̄) < 0, and then

(13)  v(x̄) < C_{2} < +∞

for some constant C_{2} depending only on C_{1}. Again by P_{1}(x̄) < 0,

(14)  w_{λ_{0}}(x̄) < \frac{ϵμ}{2} − \frac{r^{n−2}ϵ}{|x̄|^{n−2}} < \frac{ϵμ}{2}.

By (11), (13) and the Mean Value Theorem we have |x̄|^{α}v^{q}(x̄) − |x̄^{λ}|^{α}v_{λ_{0}}^{q}(x̄) ≤ C_{3}w_{λ_{0}}(x̄) for some positive constant C_{3} depending only on C_{1}, C_{2} and λ_{0}. Thus from \frac{∂P_{1}}{∂t} (x̄) ≥ 0 we have

(15)  w_{λ_{0}}(x̄) ≥ \frac{ϵ}{2C_{3}} · (1 − μ).

Combining (14)and (15) we have

\frac{ϵμ}{2} > \frac{ϵ}{2C_{3}} · (1 − μ),

i.e., μ > \frac{1}{1+C_{3}} . If μ is chosen in such a way that μ < \frac{1}{1+C_{3}} from the beginning, we

can then deduce that P_{1}(x) ≥ 0 in \overline{B_{r_{0}}^{+}}(0) \ \overline{B_{r}^{+}}(0). Let r → 0, we thus establish Lemma 2.6 by choosing γ = \frac{ϵμ}{2} for some μ < 1/(1 + C_{3}).

Proof of Theorem 2.1.

By Proposition 2.5 and also by moving planes from the positive direction of x_{1}, we see that v(x′, t) is symmetric with respect to x_{1}. Clearly the above argument also applies to any direction perpendicular to t-axis, therefore we conclude that v(x′, t) is symmetric with respect to x′. It follows that u(x′, t) is also symmetric with respect to x′ due to the inverse Kelvin transformation. Since we can choose the origin arbitrarily on the hyperplane t = 0, it is easy to see that u(x′, t) is independent of x′. This reduces (2) to solving an ordinary differential equation, and the proof of Theorem 2.1 is now complete.

2.2 Case 1 < q ≤ n/(n − 2), n ≥ 3 or q > 1, n = 2.

Theorem 2.7. If u(x) ∈ C^{2}(H) ∩ C^{1}(H̄) solves (2), 1 < q ≤ n/(n − 2) and n ≥ 3, or q > 1 and n = 2, then u = at + b for some positive constants a, b with a = b^{q}.

Proof: It suffices to show that u(x) is independent of x′. For fixed q > 1 and n ≥ 2, we choose a positive integer m so large that q > \frac{n+m}{n+m−2}. Set

w(x_{1}, x_{2}, ..., x_{n−1}, x_{n+1}, ..., x_{n+m}, t) = u(x_{1}, ..., x_{n−1}, t),

where x_{i} ∈ ℝ for i = 1, ..., n − 1, n + 1, ..., n + m, t ≥ 0. Then w satisfies

\{ \begin{matrix}Δw = 0, w ≥ 0, in ℝ_{+}^{n+m}, \\ \frac{∂w}{∂t} = w^{q} on ∂ℝ_{+}^{n+m}. \end{matrix}

By the choice of m we see that q is a super-critical power in this new setting. Then Theorem 2.1 applies and we know that w is independent of x_{i} for i = 1, ..., n − 1, n + 1, ..., n + m. Thus u is independent of x_{i} for i = 1, ..., n − 1.

3. Proof of Theorem 1.2

In this section we focus on establishing the following result.

Theorem 3.1. Suppose that n ≥ 3. If u(x) ∈ C^{2}(H) ∩ C^{1}(H̄) solves (3) for q > n/(n − 2) and p > (n + 2)/(n − 2), then u ≡ 0.

Remark 3.2. Theorem 1.2 follows from Theorem 3.1 and some argument similar to the proof of Theorem 2.7.

Remark 3.3. When p = (n + 2)/(n − 2) and q = n/(n − 2), Theorem 1.2 has already been established in [4].

As usual we set

v(x) = \frac{1}{|x|^{n−2}} u(\frac{x}{|x|^{2}} ),

then v(x) satisfies

(16)  \{ \begin{matrix}Δv = |x|^{τ} v^{p}, v(x) > 0 in H, \\ \frac{∂v}{∂t} = |x|^{β}v^{q} on ∂H \ \{0\}, \end{matrix}

where τ = p(n − 2) − (n + 2) > 0, β = q(n − 2) − n > 0. We need the following lemma to take care of the possible singular point.

Lemma 3.4. Let v ∈ C^{2}(H) ∩ C^{1}(H̄) \ {0} satisfy (16). Then for all 0 < ϵ <

min\{1, min_{∂B_{1/2}^{+}∩∂B_{1/2}} v\}, we have v(x) > \frac{ϵ}{2} for all x ∈ \overline{B_{1/2}^{+}} \ \{0\}.

Proof: Set

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

and P_{2}(x) = v(x) − φ_{2}(x). Then we have

(17)  \{ \begin{matrix}ΔP_{2} = |x|^{τ} v^{p} − ϵ in B_{1/2}^{+} \ B_{r}^{+}, \\ \frac{∂P_{2}}{∂t} = |x|^{β}v^{q} − \frac{ϵ}{2} on ∂(\overline{B_{1/2}^{+}} \ \overline{B_{r}^{+}}) ∩ ∂ℝ_{+}^{n}. \end{matrix}

We want to show that

(18)  P_{2} ≥ 0 in \overline{B_{1/2}^{+}} \ \overline{B_{r}^{+}}.

Let

\begin{matrix}S = \{x : |x|^{τ} v^{p} − ϵ > 0\} ∩ (B_{1/2}(0) \ B_{r}(0)), \\ S^{c} = \{x : |x|^{τ} v^{p} − ϵ ≤ 0\} ∩ (B_{1/2}(0) \ B_{r}(0)). \end{matrix}

In S, since 1 ≥ |x| > 0, we have v^{p}(x) > ϵ/|x|^{τ} > ϵ. It follows that v(x) > ϵ ≥ φ_{2} for x ∈ S, i.e., P_{2} ≥ 0 in S; in S^{c}, ΔP_{2} ≤ 0. Therefore arguing similarly as in the proof of Lemma 2.2, we can show that P_{2} ≥ 0 in S^{c}. This thus establishes (18). Let r → 0, we obtain Lemma 3.4.

For λ < 0, we recall that Σ_{λ}, T_{λ}, x^{λ}, v_{λ} and w_{λ} are defined as in section 2. It is easy to check that w_{λ}(x) satisfies

(19)  \{ \begin{matrix}Δw_{λ} ≤ c_{2}(x)w_{λ} in Σ_{λ}, \\ \frac{∂w_{λ}}{∂t} ≤ c_{3}(x)w_{λ} on ∂H ∩ Σ̃_{λ}, \end{matrix}

where c_{2}(x) = p|x|^{τ} · ξ_{2}^{p−1}(x), c_{3}(x) = q|x|^{β} · ξ_{3}^{q−1}(x), ξ_{2} and ξ_{3} are two functions between v_{λ} and v. In the following we again apply the moving plane method.

Proposition 3.5. If |λ| is large enough, then w_{λ}(x) ≥ 0 for all x ∈ Σ̃_{λ}.

Proof: Suppose that w_{λ}(x) is negative somewhere in Σ̃_{λ}. Note that w_{λ}(x) → 0 as x → ∞, thus by Lemma 3.4 we know for some |λ| large enough, there exists x̄_{λ} such that w_{λ}(x̄_{λ}) = min_{x∈Σ̃_{λ}} w_{λ}(x) < 0. Since c_{2}(x) is non-negative, by the Strong

Maximum Principle we know that x̄_{λ} ∈ ∂H ∩ Σ̃_{λ}, but this contradicts to the boundary condition in (19). This finishes the proof.

Then we can define

λ_{0} = sup{λ < 0 | w_{μ}(x) ≥ 0 in Σ̃_{μ} for all − ∞ < μ < λ}.

Proposition 3.6. λ_{0} = 0.

Proof : The proof is similar to that of Proposition 2.5. It suffices to show that if λ_{0} < 0, then

(20)  w_{λ_{0}} ≡ 0.

Suppose that (20) fails, then by the Strong Maximum Principle we have w_{λ_{0}} > 0 in Σ̃ \ T_{λ_{0}}. The following lemma is needed to treat the possible singular point, and we postpone its proof till the end.

Lemma 3.7. For r_{0} < min\{\frac{1}{2}|λ_{0}|, 1/2\} , there exists some positive constant γ depending only on λ_{0} and r_{0} such that w_{λ_{0}}(x) > γ in B_{r_{0}}^{+}(0).

We now continue the proof of Proposition 3.6. By the definition of λ_{0}, there exists a sequence λ_{k} → λ_{0} with λ_{k} > λ_{0} such that inf_{Σ̃_{λ_{k}}}w_{λ_{k}} < 0. From Lemma 3.7

and the continuity of v(x) away from the origin it follows that for k large enough, there exists x_{k} = (x′_{k}, t_{k}) ∈ Σ̃_{λ_{k}} \ \overline{B_{r_{0}}^{+}} such that

(21)  w_{λ_{k}}(x_{k}) = min_{Σ̃_{λ_{k}}} w_{λ_{k}} < 0.

Clearly |x_{k}| > r_{0} and t_{k} > 0 as from Lemma 3.7 and the boundary condition. But (21) contradicts to the Strong Maximum principle since c_{2}(x) ≥ 0 in (19). This finishes the proof of Proposition 3.6.

Proof of Lemma 3.7: Set

φ_{3}(x) = \frac{ϵμ}{2} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{ϵt(1 − μ)}{2} + \frac{ϵt^{2}(1 − μ)}{2} in B_{r_{0}}^{+}(0) \ B_{r}^{+}(0),

where ϵ is some small positive constant such that min_{∂B_{r_{0}}^{+} (0)} w_{λ_{0}} ≥ ϵ. μ < 1 will be

chosen later in the proof. Let P_{3}(x) = w_{λ_{0}}(x) − φ_{3}(x), then P_{3}(x) satisfies the following equation.

(22) \{ \begin{matrix}ΔP_{3}(x) = |x|^{τ} v^{p} − |x^{λ_{0}}|^{τ} v_{λ_{0}}^{p} − ϵ(1 − μ) in B_{r_{0}}^{+}(0) \ B_{r}^{+}(0) \\ \frac{∂P_{3}(x)}{∂t} = |x|^{τ} v^{q} − |x^{λ_{0}}|^{τ} v_{λ_{0}}^{q} − \frac{ϵ(1 − μ)}{2} on \overline{B_{r_{0}}^{+}}(0) \ \overline{B_{r}^{+}}(0) ∩ ∂H. \end{matrix}

On ∂B_{r_{0}}^{+}(0) ∩ ∂B_{r_{0}}(0) and ∂B_{r}^{+}(0) ∩ ∂B_{r}(0), we know P_{3}(x) ≥ 0. Let x̄ be one of the minimum points of P_{3}(x). If P_{3}(x̄) < 0, arguing similarly as in the proof of Lemma 2.6 we know that there exists some μ_{1} = μ_{1}(λ_{0}) > 0 such that x̄ can not be on ∂H for all μ ≤ μ_{1}. Hence for μ ≤ μ_{1}, if P_{3}(x̄) < 0, then ΔP_{3}(x̄) ≥ 0. Using the same method as in the proof of Lemma 2.6, we can derive a contradiction if μ ≤ μ_{2} for some μ_{2} = μ_{2}(λ_{0}) > 0. Set μ = min{μ_{1}, μ_{2}}, then we know that P_{3}(x) ≥ 0 in \overline{B_{r_{0}}^{+}}(0) \ \overline{B_{r}^{+}}(0). Let r → 0, we thus obtain Lemma 3.7 with γ = ϵμ/2.

Proof of Theorem 3.1: Arguing similarly as in the proof of Theorem 2.1, we can show that u(x) depends only on t, and thus (3) becomes

(23)  \{ \begin{matrix}u′′ = u^{p}, u > 0 for t > 0 \\ u′(0) = u^{q}(0). \end{matrix}

Since p > 1, it is easy to check that (23) has no global solution (see [4] for details).

4. Miscellaneous Results

4.1 Sign-changing solutions.

Brezis [1] proved that u = 0 is the only solution to the following entire space problem.

\{ \begin{matrix}− Δu + |u|^{p−1}u = 0 in ℝ^{n}, \\ u ∈ L_{loc}^{p}(ℝ^{n}), p > 1. \end{matrix}

Thus one may wonder whether for q > 1, all solutions to the following equation

(24)  \{ \begin{matrix}Δu = 0 in H, \\ \frac{∂u}{∂t} = |u|^{q−1}u on ∂H \end{matrix}

depend only on t. It turns out that this is not always the case as the following example shows.

Example 4.1. u(x) = x_{1}^{3}t − x_{1}t^{3} + x_{1} solves (24) with q = 3.

In view of this example, it seems that the set of sign-changing solutions to (24) is more complicated than that of non-negative solutions.

4.2 Some other results.

Consider the following equation:

(25)  \{ \begin{matrix}Δu − u^{p} = 0, u ≥ 0 in ℝ^{n}, \\ u ∈ C^{2}(ℝ^{n}), 1 < p < +∞. \end{matrix}

The result of Brezis in [1] implies the following theorem.

Theorem 4.2. If u(x) solves (25), then u ≡ 0.

We would like to present a different proof of Theorem 4.2 here. This does not mean that our method is simpler or more flexible (actually it seems that the moving plane method does not apply to sign-changing solutions). The point which we want to make is that the sign of the coefficient for the nonlinear term in (25) is a “good” sign, which allows us to apply the moving plane method for the super-critical case.

The proof of Theorem 4.2 is similar to that of Theorem 1.2. It suffices to prove it for the case when p > (n + 2)/(n − 2), n ≥ 3, since all the other cases can be treated similarly as in the proof of Theorem 2.7. As before we let

v(x) = \frac{1}{|x|^{n−2}} u(\frac{x}{|x|^{2}} ),

then v(x) satisfies

(26)  Δv − |x|^{δ}v^{p} = 0, v(x) > 0 in ℝ^{n} \ {0}

where δ = p(n−2)−(n+2) > 0. We set to show that v(x) is symmetric with respect to the origin. The following result will be needed to take care of the singularity.

Lemma 4.3. Assume that v(x) satisfies (26). If for some 0 < ϵ < 1, v(x) ≥ ϵ on ∂B_{1}(0), then v(x) ≥ ϵ/2 in B_{1}(0) \ {0}.

Proof: For small r, Set

φ_{4}(x) = \frac{ϵ}{2} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{t^{2}ϵ}{2}, x ∈ B_{1}(0) \ B_{r}(0).

Then Δϕ = ϵ, and Δ(u − ϕ) = |x|^{δ}u^{p} − ϵ. Let

\begin{matrix}S = \{x : |x|^{δ}u^{p} − ϵ > 0\} ∩ (B_{1}(0) \ B_{r}(0)), \\ S^{c} = \{x : |x|^{δ}u^{p} − ϵ ≤ 0\} ∩ (B_{1}(0) \ B_{r}(0)). \end{matrix}

Since |x| ≤ 1 in S, thus we have u^{p} > ϵ/|x|^{δ} > ϵ. It follows that u(x) > ϵ ≥ φ_{4} for x ∈ S; in S^{c}, Δ(u − φ_{4}) ≤ 0. It is easy to check the following facts: On {x : |x|^{δ}u^{p} − ϵ = 0} and ∂B_{1}, u ≥ φ_{4}; On ∂B_{r}, u > φ_{4}. By the Maximum Principle, we know that u ≥ φ_{4} in S^{c}. Thus we always have u ≥ φ_{4} in B_{1} \ B_{r}. Let r → 0, we obtain u ≥ \frac{ϵ}{2}. This proves Lemma 4.3.

Once we establish Lemma 4.3, we can apply the moving plane method to prove Theorem 4.2. We refer readers to [3], [5] for details about such kind of process.

In our proofs of Theorems 1.1 and 1.2, the assumption p, q > 1 plays an essential role in the adding dimension argument. In the following we present a simple example to show that Theorem 4.2 breaks down for p = 1. The same example also serves as a counterexample to Theorem 1.2 for p = q = 1.

Example 4.4. u(x) = e^{t} solves (25) with p = 1.

Moreover, there exist positive solutions to (25) which depend on more than one variable.

Example 4.5. u(x) = e^{(x_{1}+x_{2})/\sqrt{2}} solves (25) with p = 1.

Thus it seems interesting to ask whether Theorem 1.1 still holds for q = 1.

References

1. H. Brezis, Semilinear equations in ℝ^{N} without condition at infinity., Appl. Math. Optim 12 (1984), 271-282.

2. H. Berestycki, L. Caffarelli and L. Nirenberg, Symmetry for elliptic equations in a half space, Boundary Value Problems for Partial Differential Equations, ed by J.L.Lions et al. Masson, Paris (1993), 27-42.

3. W. Chen and C. Li, Classification of solutions of some nonlinear elliptic equations, Duke Math. J. 63 (1991), 615-623.

4. M. Chipot, I. Shafrir and M. Fila, On the solutions to some elliptic equations with nonlinear Neumann boundary conditions, Advances in Diff. Equs. 1 (1996), 91-110.

5. B. Gidas, W.M. Ni and L. Nirenberg, “Symmetry of positive solutions of nonlinear elliptic equations in R^{n}” in Mathematical Analysis and Applications Part A, ed. L. Nachbin, Adv. Math. Suppl. Stud. 7, Academic Press, New York, 1981,, 369-402.

6. B. Hu, Nonexistence of a positive solution of the Laplace equation with a nonlinear boundary condition, Differential and Integral equations 7 (1994), 301-313.

7. Y.Y. Li and M. Zhu, Uniqueness theorems through the method of moving spheres, Duke math J. 80 (1995), 383-417.

8. B. Ou, Positive harmonic functions on the upper half space satisfying a nonlinear boundary condition, Differential and Integral equations 9 (1996), 1157-1164.

9. S. Terracini, Symmetry properties of positive solutions to some elliptic equations with nonlinear boundary conditions, Differential and Integral equations 8 (1995), 1911-1922.

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