LIOUVILLE THEOREMS ON SOME INDEFINITE EQUATIONS

Meijun Zhu Department of Mathematics The University of British Columbia, Vancouver B. C., V6T 1Z2 Canada

Abstract In this note, we present some Liouville type theorems about the

non-negative solutions to some indefinite elliptic equations.

1 Introduction

In the study of some indefinite elliptic problems, in order to get a priori estimate one may use a standard blowup argument. This procedure leads to the following equations in ℝ^{n}:

Δu + x_{n}u^{p} = 0, u ≥ 0, in ℝ^{n}  (1.1)

where we write x = (x_{1}, x_{2}, ..., x_{n}) ∈ ℝ^{n}. Throughout this paper, we always assume n ≥ 3.

Using some test functions on S^{n}, Berestycki, Capuzzo Dolcetta and Nirenberg [1] proved the following theorem (we state a weak version in the whole space).

Theorem A. Let u(x) ∈ C^{2}(ℝ^{n}) be a solution to (1.1). If p < \frac{n+2}{n−1}, then u(x) = 0.

One observes that \frac{n+2}{n−1} is less than the critical Sobolev exponent \frac{n+2}{n−2}. In the study of prescribing scalar curvature problems, we do need to consider those equations with critical exponents. The main purpose of this paper is to study (1.1) for p = \frac{n+2}{n−2}.

We here consider a slightly general equation:

Δu + x_{n}^{k}u^{p} = 0, u ≥ 0, in ℝ^{n}  (1.2)

where and throughout this paper we assume that k is an odd positive integer. Our first result can be stated as the following.

Theorem 1.1 Let u(x) ∈ C^{2}(ℝ^{n}) be a solution to (1.2). If p = (n + 2 + 2k)/(n − 2), then u = 0.

Remark 1.1 It is obvious to see that Theorem 1.1 still holds if we change x_{n} to x_{i} for i = 1, ..., n − 1 in equation (1.2).

For the critical exponent, we have the following.

Theorem 1.2 Let u(x) ∈ C^{2}(ℝ^{n}) be a solution to (1.2). If p = (n+2)/(n−2) and the dimension n is an even number, then u = 0.

Remark 1.2 For k ≥ 3, Theorem 1.2 was included in Theorem 2.2 of [1].

We also consider a related problem in the half space with Neumann boundary condition:

\{ \begin{matrix}Δu = 0, u ≥ 0, in ℝ_{+}^{n} \\ \frac{∂u}{∂x_{n}} = x_{i}^{k}u^{q},  on ∂ℝ_{+}^{n}, \end{matrix}  (1.3)

where and through this paper we write ℝ_{+}^{n} = {(x′, x_{n}) = (x_{1}, ..., x_{n−1}, x_{n}) | x′ ∈ ℝ^{n−1}, x_{n} > 0}, i ≤ n−1. This equation can be viewed as the limit equation when we “blow up” some equations with indefinite boundary nonlinearities.

We have the following result.

Theorem 1.3 Let u ∈ C^{2}(ℝ_{+}^{n}) be a solution to (1.3). If q = (n + 2k)/(n − 2), then u(x) just depends on x_{n} and x_{i}.

Similarly, for the critical exponent, we have the following.

Theorem 1.4 Let u ∈ C^{2}(ℝ_{+}^{n}) be a solution to (1.3). If q = n/(n − 2), then u(x) just depends on x_{n} and x_{i}.

An interesting consequence of Theorem 1.4 is the following corollary.

Corollary 1.1 Let u ∈ C^{2}(ℝ_{+}^{n}) be a solution to (1.3). If q = n/(n − 2) and u(x) → 0 as |x| → ∞, then u(x) = 0.

Remark 1.3 We tend to believe that Corollary 1.1 holds without the decay assumption at infinity on u(x). We hope to clarify this point in our future study.

The proofs of Theorem 1.1 and 1.3 follow from the standard moving plane method, see for instance, [3], [4], [6] in the whole ℝ^{n} and [2], [8]-[12] in the upper half space. Here we first observe that the equations with those exponents in Theorem 1.1 and 1.3 are invariant under the Kelvin transformation, thus after we perform the Kelvin transformation on these equations, the coefficients are still monotone in any direction perpendicular to x_{n}-axis. We then apply the moving plane method as usual in these directions. The main difficulty comes from the analysis of the possible singular point and is overcome with the aides of some comparison lemmas, see Lemma 2.1 and Lemma 3.1 below. By adding dimensions, we can prove Theorem 1.2 and 1.4 as in the work of [10]. As an application, we state an existence result concerning the same equations as in [1] in the last section.

2 Proofs of Theorem 1.1 and 1.2

We first derive Theorem 1.2 from Theorem 1.1 by using the method of adding dimensions, which was introduced by us in [10].

Let u(x) ≥ 0 be a solution to (1.2). Set ũ(x_{1}, x_{2}, ..., x_{n}, x_{n+1}, ..., x_{m}) = u(x_{1}, ..., x_{n}), where m is an integer which will be chosen later. Thus ũ solves

Δũ + x_{n}^{k}ũ^{\frac{n+2}{n−2}} = 0, ũ ≥ 0, in ℝ^{m}.

Choosing m = (n − 2)(2 + k)/2 + 2 (here we use the fact that n is even), we have (n + 2)/(n − 2) = (m + 2 + 2k)/(m − 2). It follows from Theorem 1.1 that ũ = 0, therefore, u = 0.

We now focus on the proof of Theorem 1.1. Later on we write x = (x′, x_{n}) and assume that u(x) solves (1.2). From the strong maximum principle, we know that either u = 0 or u > 0. We prove u = 0 by contradiction. Suppose u > 0, we aim to derive a contradiction.

Since there is no assumption on the decay rate of u at infinity, as usual, we set

v(x) = \frac{1}{|x|^{n−2}} u(\frac{x}{|x|^{2}} ).  (2.4)

Then v(x) satisfies

Δv + x_{n}^{k}v^{\frac{n+2+2k}{n−2}} = 0, v > 0, in ℝ^{n} \ \{0\}.  (2.5)

Our purpose is to obtain some symmetric properties about v(x) on the x′-hyperplane. We achieve this by using moving planes which are parallel to x_{n}-axis. Without loss of generality, we move the planes along x_{1}-direction.

Denote B_{r} as a ball of radius r with the center at the origin. Our first lemma will be used to handle the possible singular point of v(x) at the origin.

Lemma 2.1 Assume that v satisfies

\{ \begin{matrix}Δv + x_{n}^{k}v^{\frac{n+2+2k}{n−2}} = 0 in B_{1/2} \ \{0\} \\ v > 0, v ∈ C^{2}(B_{1/2} \ \{0\}). \end{matrix}

If v(x) ≥ ϵ on ∂B_{1/2} for some ϵ ≤ 1, then v(x) ≥ ϵ/2 in B_{1/2} \ {0}.

Proof: Let φ_{1}(x) = \frac{ϵ}{2} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{x_{n}^{2}ϵ}{2} in B_{1/2} \ B_{r} for some small r > 0, A(x) = v(x) − φ_{1}(x), then

ΔA(x) = −x_{n}^{k}v^{\frac{n+2+2k}{n−2}} − ϵ.

Set

\begin{matrix}S = \{x | − x_{n}^{k}v^{\frac{n+2+2k}{n−2}} − ϵ > 0, x ∈ B_{1/2} \ B_{r}\}, \\ S^{c} = \{x | − x_{n}^{k}v^{\frac{n+2+2k}{n−2}} − ϵ ≤ 0, x ∈ B_{1/2} \ B_{r}\}. \end{matrix}

In S̄, −x_{n}^{k}v^{\frac{n+2+2k}{n−2}} − ϵ ≥ 0, thus x_{n} < 0. It follows that (notice that k is odd)

v^{\frac{n+2+2k}{n−2}} ≥ \frac{ϵ}{−x_{n}^{k}} ≥ \frac{ϵ}{(1/2)^{k}} > φ_{1}.

Since \frac{n+2+2k}{n−2} > 1 and 1 > ϵ > φ_{1} in S̄, we know that v > φ_{1} for x ∈ S̄.

In S^{c}, we know ΔA ≤ 0. Also we can check that: On \{x | −x_{n}^{k}v^{\frac{n+2+2k}{n−2}} −ϵ = 0}, as in S, v > φ_{1}; On ∂B_{1/2}, v − φ_{1} ≥ ϵ − (ϵ/2 + ϵ/8) > 0; On ∂B_{r}, v − φ_{1} ≥ 0 − (ϵ/2 − ϵ + ϵ/8) > 0; That is : on ∂S^{c}, v ≥ φ_{1}. By the maximum principle, we have that v ≥ φ_{1} in S^{c}.

Therefore, we know v ≥ φ_{1} in B_{1/2} \ B_{r}. To complete the proof of the lemma it is enough to take the limit for r → 0.

Now we are ready to move the planes.

For λ < 0 we define

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

Then w_{λ}(x) satisfies

Δw_{λ} + x_{n}^{k}c(x)w_{λ} = 0 in Σ̃_{λ},  (2.6)

where c(x) = \frac{n+2+2k}{n−2}ξ^{\frac{2k+4}{n−2}} (x), ξ(x) is a positive function between v(x) an v_{λ}(x).

Proposition 2.1 There exists R > 1 such that, if λ < −R, w_{λ} ≥ 0 in Σ̃_{λ}.

Proof: As in [9], we choose an auxiliary function g(x) = |x|^{−α} with 0 < α < n − 2, and consider w̄_{λ} = w_{λ}/g.

Claim: There exists R > 1 such that, if λ < −R, w̄_{λ} ≥ 0 in Σ̃_{λ}.

Before we prove the claim, we first take care of the possible singular point of w_{λ} at the origin. Due to the fact that v(x) → 0 as |x| → ∞, we easily see that there exists a R_{1} > 0 such that if λ < −R_{1}, v_{λ}(x) ≤ \frac{1}{4} min\{min_{∂B_{1/2}(0)} v(x), 1\} in B_{1/2}\{0}. Therefore, from Lemma 2.1, we know that w_{λ}(x) ≥ 0 in B_{1/2}\{0} for λ < −R_{1}, so is w̄_{λ}.

We then prove the claim by contradiction. Assume for any λ < −R_{1}, infΣ̃_{λ} w̄_{λ}(x) < 0. From the above argument and the fact that v(x) → 0 as |x| → ∞, we know that there exists a x̄ ∈ Σ̃_{λ} such that w̄_{λ}(x̄) = infΣ̃_{λ} w̄_{λ}(x) < 0.

Direct computation shows that w̄_{λ} satisfies

Δw̄_{λ} + \frac{2}{g}∇g · ∇w̄_{λ} + (x_{n}^{k}c(x) + \frac{Δg}{g})w̄_{λ} = 0 in Σ_{λ} \ \{0\}.  (2.7)

Since w_{λ}(x̄) < 0, we know v_{λ}(x̄) > v(x̄). From (2.4) we have: there exists C > 0, such that, as |λ| is sufficiently large

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

It follows that at point x̄,

|x_{n}^{k}|c(x̄) ≤ \frac{n + 2 + 2k}{n − 2} · (\frac{C}{|x̄^{λ}|^{n−2}} )^{\frac{2k+4}{n−2}} · |x̄|^{k} ≤ C_{1}|x̄|^{−4−k}.

Noticing \frac{Δg}{g}(x̄) = −\frac{α(n−2−α)}{|x̄|^{2}}, we know that there exists R_{2} > R_{1} such that, as λ < −R_{2}, Δg(x̄)/g(x̄) + x_{n}^{k}c(x̄) < 0. Therefore, in view of the maximum principle, we know that w̄_{λ} can not attain an interior negative minimum in a neighborhood of x̄. Contradiction!

Proposition 2.1 follows from the above claim directly.

Now we define

λ_{0} = sup{λ < 0 | w_{μ}(x) ≥ 0 in Σ̃_{μ} for all − ∞ < μ < λ}.  (2.8)

Proposition 2.2 If λ_{0} < 0, then w_{λ_{0}} = 0.

Proof: We prove this proposition by contradiction. Suppose not, by the strong maximum principle we know that w_{λ_{0}}(x) > 0 in Σ̃_{λ_{0}} \ T_{λ_{0}}.

Claim: There exist some small constants: r_{0} ≤ min(|λ_{0}|/2, 1) and ϵ < 1, such that

w_{λ_{0}}(x) ≥ \frac{ϵ}{2} in B_{r_{0}}(0) \ \{0\}.

Proof of the claim: Let φ_{2}(x) = \frac{ϵ}{2} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{|x|ϵ}{2} in B_{r_{0}}(0) \ B_{r}(0) for some small r < r_{0}, where ϵ and r_{0} will be chosen. Since w_{λ_{0}} satisfies

Δw_{λ_{0}} + x_{n}^{k}c(x)w_{λ_{0}} = 0, w_{λ_{0}} > 0 in B_{r_{0}}(0) \ {0},

we know

Δ(w_{λ_{0}} − φ_{2}) = −x_{n}^{k}c(x)w_{λ_{0}} − \frac{ϵ}{2} · (n − 1)|x|^{−1} in B_{r_{0}}(0) \ B_{r}(0).

Set B_{r_{0}}(0) \ B_{r}(0)= S_{1} ∪ S_{1}^{c} such that S_{1} = {x : w_{λ_{0}} ≥ 1 > φ_{2}} ∩ B_{r_{0}}(0) \ B_{r}(0). Obviously we only need to show that the claim holds in S_{1}^{c}.

In S_{1}^{c} we know w_{λ_{0}} < 1. Since 0 < 1/C ≤ v_{λ_{0}}(x) ≤ C for x ∈ B_{r_{0}}(0) (here C is independent of r_{0} whenever we choose r_{0} ≤ |λ_{0}|/2), we have v(x) ≤ C + 1 in S_{1}^{c}. Thus, in S_{1}^{c}, there exists a constant C̄ > 0, such that

c(x) ≤ C̄ < +∞.

Now we fix r_{0} small enough such that

(n − 1)|x|^{−1} ≥ 2|x_{n}^{k}|c(x) ∀x ∈ B_{r_{0}}(0) \ B_{r}(0),  (2.9)

and choose ϵ ≤ min{min_{∂B_{r_{0}}(0)} w_{λ_{0}}, 1}. Set S_{1}^{c} = S_{1,1} ∪ S_{1,2} such that

S_{1,1} = \{x : −x_{n}^{k}c(x)w_{λ_{0}} ≥ \frac{ϵ}{2}(n − 1)|x|^{−1}, x ∈ S_{1}^{c}\}.

In S_{1,1}, from (2.9), we know that w_{λ_{0}} ≥ ϵ ≥ φ_{2}.

In S_{1,2}, we have Δ(w_{λ_{0}} − φ_{2}) ≤ 0. Also one can check that: On {x : −x_{n}^{k}c(x)w_{λ_{0}} = \frac{n−1}{2}|x|^{−1}ϵ, x ∈ S^{c}\}, as in S_{1,1} we know w_{λ_{0}} ≥ φ_{2}; On ∂B_{r_{0}}(0), w_{λ_{0}} − φ_{2} ≥ ϵ − (ϵ/2 + ϵ/2) = 0; On ∂B_{r}(0), w_{λ_{0}} − φ_{2} ≥ 0 − (ϵ/2 − ϵ + ϵ/2) = 0. That is on ∂S_{1,2}, w_{λ_{0}} ≥ φ_{2}. Thus from the maximum principle, we know that w_{λ_{0}} ≥ φ_{2} in S_{1,2}.

It follows that w_{λ_{0}} ≥ φ_{2} in S_{1}^{c}. Let r → 0, we have the claim.

Now we continue the proof of Proposition 2.2. By the definition of λ_{0}, there is a sequence λ_{l} → λ_{0} with λ_{l} > λ_{0} such that infΣ̃_{λ_{l}} w_{λ_{l}} < 0. As before, we consider w̄_{λ_{l}} = w_{λ_{l}}/g with g(x) = |x|^{−α}. It follows from the above claim and w̄_{λ_{l}}(x) → 0 as |x| → ∞ that there is P_{l} such that w̄_{λ_{l}}(P_{l}) = minΣ̃_{λ_{l}} w̄_{λ_{l}}(x) < 0. Similar discussion to the proof of Proposition 2.1, we also know that P_{l} ∈ B_{R}(0) for some uniform constant R. Thus, as l → ∞, P_{l} → x̄ ∈ T_{λ_{0}}. Since |∇w̄_{λ_{l}}(P_{l})| = 0, we know ∂w̄_{λ_{0}}/∂x_{1}(x̄) = 0. On the other hand, since w̄_{λ_{0}} satisfies (2.7) and w̄_{λ_{0}} > 0 in Σ̃_{λ_{0}} \ T_{λ_{0}}, by Hopf Lemma we know \frac{∂w̄_{λ_{0}}}{∂x_{1}} (x̄) > 0. We have reached a contradiction, hence the proof of Proposition 2.2 is completed.

If λ_{0} < 0, from Proposition 2.2 we derive that lim_{x→0} v(x) exists, that is, |x|^{n−2}u(x) tends to some constant c_{0} > 0 as x → ∞. If λ_{0} = 0, we then begin to move the planes from positive x_{1}-axis to the origin and get either case 1:

lim_{|x|→∞} |x|^{n−2}u(x) = c_{0} > 0,  (2.10)

or case 2: v(x) is symmetric about origin on the x′-hyperplane (recall we can move the planes along any direction on x′-hyperplane).

The first case is ruled out by

Proposition 2.3 There exists no positive solution of (1.2) which satisfies (2.10) if p > 1.

Its proof will be given later.

In the second case, from the property of the Kelvin transformation one easily gets that u(x) is radial symmetry about the origin on the x′-hyperplane. Since we can choose the origin arbitrarily on the x′-hyperplane, we know that u(x) is independent of x′ and (1.2) becomes the following ODE:

u′′ + x_{n}^{k}u^{\frac{n+2+2k}{n−2}} = 0, u ≥ 0, in ℝ.  (2.11)

An elementary phase-plane argument shows that (2.11) has only trivial solution. Therefore the proof of Theorem 1.1 is completed once we establish Proposition 2.3.

Proof of Proposition 2.3. We again prove this proposition by contradiction. Suppose that u > 0 solves (1.2) and satisfies (2.10) for some positive constant c_{0}.

Claim: \frac{∂u}{∂x_{n}} ≥ 0 ∀x ∈ ℝ^{n}.

Clearly if the claim holds true then

lim_{x_{n}→∞} u(0, x_{n}) ≥ u(0) > 0.

This contradicts (2.10) and proves Proposition 2.3. Therefore, we only need to prove the claim under the contrary assumption (that is u > 0 solves (1.2) and satisfies (2.10)).

We use the method of moving planes again. This time we move planes along the positive x_{n}-direction.

For any λ ∈ ℝ, set

Σ_{λ} = {x | x_{n} > λ}, T_{λ} = {x | x_{n} = λ},
x^{λ} is the reflection of x about T_{λ},
u_{λ}(x) = u(x^{λ}), w_{λ} = u(x) − u_{λ}(x).

The claim can be proved through the following standard three steps. Here we outline the proof for completeness.

Step 1. There exists some constant K > 0 such that, if λ < −K, w_{λ} ≥ 0 in Σ_{λ}.

The proof of this step is similar to that of Proposition 2.1. Here there is no singular point to worry about.

Then we can define

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

Step 2. If λ_{0} ≠ +∞, then w_{λ_{0}} = 0.

This can be proved as in Proposition 2.2. Again, there is no singular point to worry about.

Step 3. λ_{0} = +∞.

Proof. Assume λ_{0} < ∞, from step 2 we know w_{λ_{0}} = 0. Therefore

\begin{matrix}Δu + x_{n}^{k}u^{p} = 0 in ℝ^{n} \\ Δu + (2λ_{0} − x_{n})^{k}u^{p} = 0 in ℝ^{n}. \end{matrix}

It follows that x_{n} = 2λ_{0} − x_{n} for all x_{n} ∈ ℝ. This is a contradiction and the claim is proved.

3 Proofs of Theorem 1.3 and 1.4

Theorem 1.4 can be derived from Theorem 1.3 similarly to how Theorem 1.2 was derived from Theorem 1.1, hence we only give the proof of Theorem 1.3 in this section. Without loss of generality, we only consider x_{i} = x_{n−1} in (1.3). We argue by contradiction.

Suppose u ≠ 0. Then, in view of the maximum principle and Hopf lemma, we know that u > 0 in \overline{ℝ_{+}^{n}}. Set

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

then v(x) satisfies

\{ \begin{matrix}Δv = 0, v > 0, & in ℝ_{+}^{n}, \\ \frac{∂v}{∂x_{n}} = x_{n−1}^{k}v^{\frac{n+2k}{n−2}} & on ∂ℝ_{+}^{n} \ \{0\}. \end{matrix}  (3.12)

Again, we will move the planes which are parallel to x_{n}-axis along x_{1}- direction.

In order to start to move planes, we still need the following lemma to take care of the possible singular point at the origin. We denote B_{r}^{+} := B_{r}(0) ∩ ℝ_{+}^{n}.

Lemma 3.1 Assume that v(x) satisfies

\{ \begin{matrix}Δv(x) = 0 & in B_{1/2}^{+} \\ \frac{∂v}{∂x_{n}} = x_{n−1}^{k}v^{\frac{n+2k}{n−2}} & on ∂B_{1/2}^{+} ∩ ∂ℝ_{+}^{n} \ \{0\} \\ v > 0,  v ∈ C^{2}(\overline{B_{1/2}^{+}} \ \{0\}). \end{matrix}

If v(x) ≥ ϵ on ∂B_{1/2} ∩ ℝ_{+}^{n} for some ϵ ≤ 1, then v(x) ≥ ϵ/2 in \overline{B_{1/2}^{+}} \ \{0\}.

Proof. Let φ_{3}(x) = \frac{ϵ}{2} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{x_{n}ϵ}{2} in B_{1/2}^{+} \ B_{r} for some small r. It is easy to see that v − φ_{3} satisfies

\{ \begin{matrix}Δ(v − φ_{3}) = 0 & in B_{1/2}^{+}, \\ \frac{∂(v−φ_{3})}{∂x_{n}} = x_{n−1}^{k}v^{\frac{n+2k}{n−2}} − \frac{ϵ}{2} & on ∂B_{1/2}^{+} ∩ ∂ℝ_{+}^{n} \ \{0\}, \\ v > 0,  v ∈ C^{2}(\overline{B_{1/2}^{+}} \ \{0\}). \end{matrix}

Note that v − φ_{3} > 0 on ∂B_{1/2} ∩ ℝ_{+}^{n} and ∂B_{r} ∩ ℝ_{+}^{n}. If there exists a point P_{0} such that v(P_{0}) − φ_{3}(P_{0}) = min_{\overline{B_{1/2}^{+}}\\overline{B_{r}^{+}}}(v(x) − φ_{3}(x)) < 0, by the virtue of the

maximum principle, we know that P_{0} ∈ ∂ℝ_{+}^{n}, and thus ∂(v − φ_{3})/∂x_{n}(P_{0}) ≥ 0. It follows that v^{\frac{n+2k}{n−2}} (P_{0}) ≥ ϵ, thus v(P_{0}) > φ_{3}(P_{0}), contradiction! So v−φ_{3} > 0 in \overline{B_{1/2}^{+}} \ \overline{B_{r}^{+}} for some small r. The proof is completed by taking the limit for r → 0.

Now, we can start to move the planes.

For λ < 0 we define as in Section 2 the following:

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

Then w_{λ}(x) satisfies

\{ \begin{matrix}Δw_{λ}(x) = 0 & in Σ_{λ}, \\ \frac{∂w_{λ}}{∂x_{n}} = x_{n−1}^{k}c_{1}(x)w_{λ} & on Σ̃_{λ} ∩ ∂ℝ_{+}^{n} \end{matrix}  (3.13)

where c_{1}(x) = \frac{n+2k}{n−2}ξ1^{\frac{2k+2}{n−2}} (x), ξ_{1}(x) is a positive function between v and v_{λ}.

Proposition 3.1 There exists L > 1 such that, if λ < −L, w_{λ} ≥ 0 in Σ̃_{λ}.

Proof: As in [9], we choose an auxiliary function g_{1}(x) = |z|^{−α}, where 0 < α < n − 2, z = x + (0, 0, ..., 1) and define w̄_{λ} = w_{λ}/g_{1}. We only need to show w̄_{λ} ≥ 0 in Σ̃_{λ} for λ negative enough.

Similar to the proof of Proposition 2.1, by using Lemma 3.1, we know that there exists a L_{1} > 0 such that if λ < −L_{1}, w̄_{λ} ≥ 0 in B_{1/2}^{+} \ {0}.

If for any λ < −L_{1}, infΣ̃_{λ} w̄_{λ} < 0, as in the proof of Proposition 2.1, we know that there exists a x̄ ∈ Σ̃_{λ} such that w̄_{λ}(x̄) = infΣ̃_{λ} w̄_{λ} < 0. Direct computation shows that w̄_{λ} satisfies

\{ \begin{matrix}Δw̄_{λ} + \frac{2}{g_{1}} ∇g_{1} · ∇w̄_{λ} + \frac{Δg_{1}}{g_{1}} w̄_{λ} = 0 & in Σ_{λ} \\ \frac{∂w̄_{λ}}{∂x_{n}} = (x_{n−1}^{k}c_{1}(x) − \frac{1}{g_{1}} · \frac{∂g_{1}}{∂x_{n}})w̄_{λ} & on Σ̃_{λ} ∩ ∂ℝ_{+}^{n}. \end{matrix}  (3.14)

Since Δg_{1}/g_{1} = −α(n − 2 − α)/|z|^{2} < 0, we know that x̄ ∈ ∂ℝ_{+}^{n}. Hence ∂w̄_{λ}/∂x_{n}(x̄) ≥ 0. From w_{λ}(x̄) < 0, using a similar argument to the proof of Proposition 2.1, we have |x_{n−1}^{k}|c_{1}(x̄) ≤ C|x̄|^{−2−k}. Also we know −\frac{1}{g_{1}} · \frac{∂g_{1}}{∂x_{n}} = α/|z|^{2}. It follows that if λ < −L for some large L > L_{1}, ∂w̄_{λ}/∂x_{n}(x̄) < 0. We derive a contradiction.

Proposition 3.2 If λ_{0} < 0, then w_{λ_{0}} = 0.

Proof: We prove this proposition by contradiction. Suppose not, by the virtue of the maximum principle and Hopf lemma, we know w_{λ_{0}}(x) > 0 in Σ̃_{λ_{0}} \ T_{λ_{0}}.

Claim: There exist some small constants: r_{0} ≤ min(|λ_{0}|/2, 1) and ϵ < 1, such that

w_{λ_{0}}(x) ≥ \frac{ϵ}{2} in \overline{B_{r_{0}}^{+}(0)} \ \{0\}.

Proof of the claim: Let φ_{4} = \frac{ϵ}{2} − \frac{r^{n−2}ϵ}{|x|^{n−2}} + \frac{ϵx_{n}}{2} in B_{r_{0}}^{+} \ B_{r}^{+} for some small r < r_{0}, where ϵ and r_{0} will be chosen. Since w_{λ_{0}} satisfies

\{ \begin{matrix}Δw_{λ_{0}} = 0 w_{λ_{0}} ≥ 0 & in B_{r_{0}}^{+}(0) \\ \frac{∂w_{λ_{0}}}{∂x_{n}} = x_{n−1}^{k}c_{1}(x)w_{λ_{0}} & on ∂B_{r_{0}}^{+} ∩ ∂ℝ_{+}^{n} \ \{0\}, \end{matrix}

we know that

\{ \begin{matrix}Δ(w_{λ_{0}} − φ_{4}) = 0 & in B_{r_{0}}^{+}(0) \\ \frac{∂(w_{λ_{0}}−φ_{4})}{∂x_{n}} = x_{n−1}^{k}c_{1}(x)w_{λ_{0}} − \frac{ϵ}{2} & on ∂B_{r_{0}}^{+} ∩ ∂ℝ_{+}^{n} \ \{0\}. \end{matrix}  (3.15)

We want to show that for a suitable small r_{0} and ϵ < min{min_{∂B_{r_{0}}} w_{λ_{0}}(x), 1},

w_{λ_{0}}(x) ≥ φ_{4}(x) ∀x ∈ \overline{B_{r_{0}}^{+}} \ \overline{B_{r}^{+}}.  (3.16)

If not, due to (3.15) and the fact that w_{λ_{0}} −φ_{4} ≥ 0 on (∂B_{r_{0}} ∪∂B_{r})∩ℝ_{+}^{n}, we know that there exists a P_{0} ∈ ∂ℝ_{+}^{n} such that w_{λ_{0}}(P_{0}) − φ_{4}(P_{0}) = infΣ̃_{λ_{0}}(w_{λ_{0}} − φ_{4}) < 0 and

\frac{∂(w_{λ_{0}} − φ_{4})}{∂x_{n}}(P_{0}) ≥ 0.

Note that w_{λ_{0}}(P_{0}) ≤ φ_{4}(P_{0}) < ϵ. As in the proof of Proposition 2.2, it yields that c_{1}(P_{0}) ≤ C (C is independent of r_{0} whenever we choose r_{0} ≤ |λ_{0}|/2). Now we choose r_{0} small enough, such that

r_{0}C < \frac{1}{2}.

Then, we have

\frac{∂(w_{λ_{0}} − φ_{4})}{∂x_{n}}(P_{0}) < 0.

Contradiction! Thus, we have shown that (3.16) holds for some suitable chosen r_{0} and ϵ. The proof of the claim is completed by taking limit for r → 0.

Now we continue the proof of Proposition 3.2. By the definition of λ_{0}, there exists a sequence λ_{l} → λ_{0} with λ_{l} > λ_{0} such that infΣ̃_{λ_{l}} w_{λ_{l}} < 0. As before, we consider w̄_{λ_{l}} = w_{λ_{l}}/g_{1} with g_{1}(x) defined in the proof of Proposition 3.1. It follows from the above claim, w̄_{λ_{l}}(x) → 0 as |x| → ∞ and Δg_{1}/g_{1} < 0 in (3.14) that there exists P_{l} ∈ ∂ℝ_{+}^{n} such that w̄_{λ_{l}}(P_{l}) = infΣ̃_{λ_{l}} w̄_{λ_{l}}. Therefore ∂w̄_{λ_{l}}/∂x_{1}(P_{l}) = 0. Similar discussion to the proof of Proposition 3.1 we can conclude that P_{l} ∈ B_{R}(0) for some large constant R uniformly for l. Thus as l → ∞, P_{l} → x̄ = T_{λ_{0}} ∩ ∂ℝ_{+}^{n}, and ∂w̄_{λ_{0}}/∂x_{1}(x̄) = 0.

In order to derive a contradiction, we still need the following technical lemma to take care of the corner point x̄. Without loss of generality, we assume λ_{0} = −1 and x̄ = (−1, 0, ..., 0). The proof is almost the same as that of Lemma 2.4 in [9], we include it here for completeness.

Lemma 3.2 Assume that w_{λ_{0}} satisfies (3.13), w_{λ_{0}} > 0 in Σ̃_{λ_{0}} for some λ_{0} < 0 and x̄ ∈ T_{λ_{0}} ∩ ∂ℝ_{+}^{n}, then

\frac{∂w_{λ_{0}}}{∂x_{1}} (x̄) > 0.  (3.17)

Proof. Set Ω = {x : x ∈ B_{1}^{+} \ B̄_{1/2}^{+}, x_{n} < 1/4} and for some α > max{n/2, n − 3},

h(x) = β(|x′|^{−α} − 1)(x_{n} + μ), φ_{5}(x) = h(x) − \frac{1}{|x|^{n−2}} h(\frac{x}{|x|^{2}} ), x ∈ Ω,

where 0 < β, μ < 1 will be chosen later. A direct computation yields

Δφ_{5} ≥ 0.

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

\{ \begin{matrix}ΔB ≤ 0, & in Ω, \\ \frac{∂B}{∂x_{n}} = x_{n−1}^{k}c_{1}w_{λ_{0}} − \frac{∂φ_{5}}{∂x_{n}} & on ∂Ω ∩ ∂ℝ_{+}^{n}. \end{matrix}  (3.18)

For some suitable chosen β and μ, we want to show that

B(x) ≥ 0, ∀x ∈ Ω.  (3.19)

Using w_{λ_{0}} > 0 in Σ_{λ_{0}} \ {0}, we can choose β_{0} > 0 such that for all 0 < β < β_{0}, B(x) ≥ 0 on ∂Ω ∩ {∂B_{1/2} ∪ {x_{n} = 1/4}}. Also, one can see that B(x) ≥ 0 on ∂Ω ∩ ∂B_{1}. Suppose the contrary of (3.19), there exists some x̃ = (x̃′, d) ∈ Ω̄ such that

B(x̃) = min_{Ω̄} B(x) < 0.

It follows that d = 0,

w_{λ_{0}}(x̃) < βμ((|x̃′|^{−α} − 1)(|x̃′|^{−n+α+2} + 1)  (3.20)

and

\frac{∂B}{∂x_{n}} (x̃) ≥ 0.

A simple calculation yields

\frac{∂φ_{5}}{∂x_{n}} (x̃) = β(|x̃′|^{−α} − 1)(|x̃′|^{−n+α} + 1).

Combining the above two inequalities we have

x̃_{n−1}^{k}c_{1}(x̃)w_{λ_{0}}(x̃) − β(|x̃′|^{−α} − 1)(|x̃′|^{−n+α} + 1) ≥ 0.  (3.21)

It follows from (3.20) and (3.21) that

x̃_{n−1}^{k}c_{1}(x̃)μ > 1.

If we choose 0 < μ < min_{1/2≤|x|≤1}(x̃_{n−1}^{k}c_{1}(x) + 1)^{−1} from the beginning, we reach a contradiction, thus (3.19) holds.

Notice B(x̄) = 0, we have

\frac{∂B}{∂x_{1}} (x̄) ≥ 0.

It follows that

\frac{∂w_{λ_{0}}}{∂x_{1}} (x̄) = \frac{∂B}{∂x_{1}} (x̄) + \frac{∂φ_{5}}{∂x_{1}} (x̄) ≥ \frac{∂φ_{5}}{∂x_{1}} (x̄) = 2αβμ > 0.

The proof of Lemma 3.2 is completed, so is the proof of Proposition 3.2.

Now, as in Section 2, we have two case. Case 1: λ_{0} < 0, we know that

lim_{|x|→∞} |x|^{n−2}u(x) = c_{0} > 0.  (3.22)

This case is ruled out by Proposition 3.3 There exists no positive solution of (1.3) which satisfies (3.22) if p > 1.

Its proof will be given later.

Case 2: v(x) is radial symmetric about the origin on the x′′-hyperplane (we write x = (x′′, x_{n−1}, x_{n})). Due to the property of Kelvin transformation, we know that u(x) is radial symmetry about the origin on the x′′-hyperplane. Since we can choose the origin arbitrarily on the x′′-hyperplane, we conclude that u(x) just depends on x_{n} and x_{n−1} in this case. Therefore the proof of Theorem 1.3 is completed once we establish Proposition 3.3.

Proof of Proposition 3.3. The proof is the same as that of Proposition 2.3. Assume u > 0, due to (3.22), we can apply moving planes directly along the x_{n−1}-direction, and get

\frac{∂u}{∂x_{n−1}} ≥ 0.

Then the contradiction comes from the above, u(0) > c > 0 and (3.22). We leave these details to readers.

Remark. Our method heavily depends on the invariance of the equation under the Kelvin transformation, therefore, we can only classify some equations with discrete exponents and can not prove our Theorems for all p less than or equal to the critical exponents. The natural question is: Does Theorem 1.1 still hold for any 1 < p ≤ (n + 2)/(n − 2) ?

Also it might be interesting to seek that Theorem 1.3 holds for some continuous range of q.

4 Application

The Liouville theorems we derived here are mainly applied to get some a priori estimates in the study of certain elliptic boundary value problems. As a consequence, one can obtain some existence results via blowup argument and degree theory, see for instance [7] or [1]. Let Ω ⊂ ℝ^{n}(n ≥ 3) be a bounded smooth domain, we here present an existence result concerning the same equation which was discussed in [1]:

\{ \begin{matrix}Lu + a(x)g(u) = 0, u > 0, & in Ω \\ Bu = 0 & on ∂Ω \end{matrix}  (4.23)

where, L is an uniformly elliptic linear operator:

L = a_{ij}(x)\frac{∂^{2}}{∂x_{i}∂x_{j}} + b_{i}(x)\frac{∂}{∂x_{i}} + c(x),

with a_{ij}(x) ∈ C^{2}(\overline{Ω}), b_{i}(x) ∈ C^{1}(\overline{Ω}), c(x) ∈ L^{∞} and

c_{0}|ξ|^{2} ≤ a_{ij}(x)ξ_{i}ξ_{j} ≤ C_{0}|ξ|^{2}, ∀ξ ∈ ℝ^{n}, ∀x ∈ Ω,

for some c_{0}, C_{0} > 0. The boundary operator B is one of the following

Bu := u  (4.24)
Bu := ν_{j}a_{jk}u_{x_{k}} + α(x)u,  (4.25)

where ν = (ν_{1}, ..., ν_{n}) denotes the exterior unit normal on ∂Ω, α is a given continuous nonnegative function on ∂Ω.

We assume that g is a C^{1} function on ℝ^{+} with

g(0) = g′(0) = 0 and lim_{s→∞} g(s)/s^{\frac{n+2}{n−2}} = l > 0.  (4.26)

The function a(x) which we are considering here is C^{2} and changes sign, that is, both

Ω^{+} := {x ∈ Ω : a(x) > 0} and Ω^{−} := {x ∈ Ω : a(x) < 0}

are nonempty. We assume that

Γ := \overline{Ω^{+}} ∩ \overline{Ω^{−}} ⊂ Ω and ∇a(x) ≠ 0 ∀x ∈ Γ.  (4.27)

Let λ_{1}(−L) be the principle eigenvalue of the operator −L in Ω under the boundary condition Bu = 0. Then the existence result can be stated as:

Theorem 4.1 Assume (4.26), (4.27), λ_{1}(−L) > 0 and the dimension n ≥ 3 is an even number. Then problem (4.23) has one solution.

With the aid of Theorem 1.2, we can prove the above theorem exactly in the same way as in [1]. We omit details here.

Remark 4.1 Chen and Li also obtained some existence results concerning the prescribing scalar curvature problem in [5]. However, their method can not be applied to general equation (4.23).

ACKNOWLEDGMENT

I would like to thank Y.Y.Li for bringing these problems to my attentions and for his encouragement and help through the years. I would like to thank

L. Nirenberg for his interest in this work. I also would like to thank the anonymous referee for the suggestions on my writing.

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