Nonexistence of positive solutions for the weighted higher-order elliptic system with Navier boundary condition

We establish a Liouville-type theorem for a weighted higher-order elliptic system in a wider exponent region described via a critical curve. We ﬁrst establish a Liouville-type theorem to the involved integral system and then prove the equivalence between the two systems by using superharmonic properties of the diﬀerential systems. This improves the results in (Complex Var. Elliptic Equ. 5:1436–1450, 2013) and (Abstr. Appl. Anal. 2014:593210, 2014).


Introduction
In this paper, we establish a Liouville-type theorem for the weighted 2mth-order elliptic equations coupled via the Navier boundary conditions in the half-space R n + = {x ∈ R n : x n > 0}: where m is a positive integer satisfying 0 < 2m < n, p, q ≥ 1, and α, β ≥ 0, which is closely related to the following integral system: where C n > 0, and x = (x 1 , . . ., x n-1 , -x n ) is the reflection of the point x about the ∂R n + .Similar to some integral systems or partial differential systems, the integral system (1.2) is usually divided into three cases according to the value of the exponents (p, q).We introduce the critical curve for (1.2) to determine a Liouville-type theorem.
For α, β = 0 in system (1.2),Cao and Dai [4] obtained a Liouville-type theorem in the super-and subcritical cases under some integrability conditions by the Pohozaev-type identity of integral form, and in the critical case, they showed that a pair of positive solutions to the system is rotationally symmetric about the x n -axis.Also, we mention the recent important works on the existence and asymptotic analysis of nontrivial solutions for some elliptic systems; see [24][25][26][27][28].
In the present paper, instead of (1.1), we will first establish a Liouville-type theorem for the integral system (1.2) in the supercritical case and then prove that systems (1.2) and (1.1) are equivalent by using the superharmonic properties, that is, the following two propositions.
Remark 1 Without the growth conditions Based on Propositions 1 and 2, the main result of the paper is the following theorem.
Theorem 1 Under the conditions of Proposition 1, the classical nonnegative solutions of system (1.1) must be trivial.
To prove Proposition 1, we will explore the moving plane method in integral forms by Chen, Li, and Ou [5].For the proof of Proposition 2, we first prove the superharmonic properties of systems (1.1) and then establish the equivalence between the two systems by using a technique introduced in [7] for the scalar case of higher-order equations.
Next, we will prove Propositions 1 and 2 in Sects. 2 and 3, respectively.

Proof of Proposition 1
We introduce three lemmas for the integral system (1.2) as preliminaries, and let C n = 1 there for simplicity.Denote with x reflecting x about the ∂R n + .Let x λ = (x 1 , x 2 , . . ., 2λx n ) be the reflection of the point x about the plane The following lemma on the Green function G(x, y) in λ is known.
Lemma 2.2 Let (u, v) be a nonnegative solution of (1.2).For all x ∈ λ , we have The second inequality can be obtained in the same way.
In addition, we also need the weighted Hardy-Littlewood-Sobolev inequality.

Now we can prove Proposition 1.
Proof of Proposition 1 We apply the moving-plane method in two steps.

Determine the starting position
Start from the very low end of R n + , i.e., near x n = 0. We will show that for λ sufficiently small, We will prove that B u λ and B v λ must be of zero measure, provided that λ sufficiently small.In fact, by Lemma 2.2 with the mean value theorem we have that for sufficiently small λ and Furthermore, by Lemma 2.3 with Hölder's inequality and q * 1 = q 1 q 1 -1 with the universal constant C > 0, where the supercritical inequality (1.7) with p, q ≥ 1 and α + β < 2m implies and Similarly, we have 3) It follows from (2.2) and (2.3) that and thus w λ q 1 ,B u λ = 0 by (2.4).In the same way, g λ q 2 ,B v λ = 0.This proves (2.1). 2. Move the plane to the infinity Inequalities (2.1) provide a starting point to move the plane T λ .We start from a neighborhood of λ and move the plane up as long as (2.1) holds.Define (2.5) We first prove that λ 0 = ∞.Assume for contradiction that λ 0 < ∞.We claim that Otherwise, for such λ 0 , e.g., (2.8) Denote λ := λ + with > 0 to be determined.For any small η > 0, choose R sufficiently large such that |u| q 1 (y) dy ≤ η.
It follows from Lusin's theorem and (2.8) that for any δ > 0, there exists a closed set Choosing > 0 sufficiently small, we have Let R be large and δ and small such that B u λ |u| q 1 (y) dy ≤ M |u| q 1 (y) dy ≤ 1 2 .Similarly, and, similarly, This contradicts (2.7) with (2.5).Thus (2.6) holds.This yields the contradiction that u(x) = v(x) ≡ 0 on the plane {x n = 2λ 0 }.We conclude that λ 0 = +∞, which implies that both u and v are strictly monotonically increasing with respect to x n .Moreover, we know that u ∈ L q 1 (R n + ) and v ∈ L q 2 (R n + )and for any a > 0, and hence u(x , a) = v(x , a) = 0 for all x ∈ R n-1 , a contradiction.

Proof of Proposition 2
Denote by B R (0 , the union of the flat and hemisphere parts of R .Let x * := x |x 2 | R 2 be the reflection of x about ∂B R (0), and let .
We begin with the well-known lemma.
where ν is the outward unit normal vector of R .
We follow the main idea of Chen, Fang, and Li [7] to give superharmonic properties of system (1.1).This result plays a key role in the proof of Proposition 2.
Proof We make an odd extension of u and v to the whole space.Define We will deduce a contradiction by two substeps.
(i) We first claim where ûm-i is the (mi)th average of u m-i .
Similarly to (3.9) and (3.11), we have Repeating the same argument to u m-3 , we also obtain the third average on ∂B r (x 3 ): By induction we can get the claim (3.5) for the component u.
(ii) Taking the scaling transformations we find that u μ and v μ are also nonnegative solutions of (3.3).This implies that by repeating step 2 in Part I of [7, Sect.2] a suitably large μ > 0 ensures with b 0 := p + q + 2m + n and a 0 > 0 sufficiently large.
Next, we treat the component v.
We claim that Obviously, (3.18) holds for k = 0 by choosing a 0 sufficiently large.Assume that (3.18) is true for k.We have Therefore (3.18) holds for all integer k.Now choose r = 2. Then (3.17) and (3.18) yield a contradiction that This excludes (3.4).
Here we have used the odd symmetry of u m-i+1 (x) with respect to ∂R n + and the fact that more than half of B r (x 0 ) is contained in R n + .Together with (3.19), we deduce ū m-i (r) < 0 and ūm-i (r) ≤ ūm-i (0) = u m-i (x 0 ) < 0, r ≥ 0.
Then by the second to the last equation in (3.20) we have and hence Continuing this way with mi even, we derive that This yields a contraction that < 0 as r is sufficiently large.
(ii) Assume that mi is odd.
Proof of Proposition 2 First, we show that the classical solutions of (1.2) must solve (1.1).When x ∈ ∂R n + , we have due to x = x, and thus by system (1.2) that j u(x) = 0, j = 0, 1, . . ., m -1.For x ∈ R n + , we have Next, we should prove that if (u, v) is a smooth positive solution of (1.1) with p, q ≥ 1 and α, β > 0, then a constant multiple of (u, v) satisfies (1.2).
Theorem 1], we can arrive at the same result by using the proof of Proposition 2.