des Sciences, D´epartement de Math´ematiques, B.P 1171 Sfax 3000, Universit´e de Sfax, Tunisia.

arXiv:1701.04119v1 [math.AP] 15 Jan 2017

Abstract In this paper we study solutions, possibly unbounded and sign-changing, of the following problem −∆λ u = |x|aλ |u| p−1 u,

in Rn , n ≥ 1, p > 1, and a ≥ 0,

where ∆λ is a strongly degenerate elliptic operator, the functions λ = (λ1 , ..., λk ) : Rn → Rk satisfies some certain conditions, and |.|λ the homogeneous norm associated to the ∆λ -Laplacian. We prove various Liouville-type theorems for smooth solutions under the assumption that they are stable or stable outside a compact set of Rn . First, we establish the standard integral estimates via stability property to derive the nonexistence results for stable solutions. Next, by mean of the Pohozaev identity, we deduce the Liouville-type theorem for solutions stable outside a compact set. Liouville-type theorems, ∆λ -Laplace operator, Stable solutions, Stability outside a compact set, Pohozaev identity. PACS: Primary : 35J55, 35J65 ; Secondary : 35B65. Keywords:

1.

Introduction and main results

The Liouville type theorem is the nonexistence of solutions in the entire space or in half-space. The classical Liouville type theorem stated that a bounded harmonic (or holomorphic) function defined in entire space must be constant. This theorem, known as Liouville theorem, was first announced in 1844 by Liouville [15] for the special case of a doubly-periodic function. Later in the same year, Cauchy [3] published the first proof of the above stated theorem. This classical result has been extended to nonnegative solutions of the semilinear elliptic equation −∆u = |u| p−1 u

in Rn , p > 1,

(1.1)

in the whole space Rn by Gidas and Spruck [10, 11] see also the paper of Chen and Li [4]. They , then the above equation only has the trivial solution u ≡ 0 and this proved that if 1 < p < n+2 n−2 result is optimal. In an elegant paper, Farina [7] proved that nontrivial finite Morse index solutions n+2 (whether positive or sign changing) to (1.1) exists if and only if p ≥ pc (n) and n ≥ 11, or p = n−2 and n ≥ 3, where pc (n) is the so-called Joseph-Lundgren exponent. The study of stable solutions in the H´enon type elliptic equation : −∆u = |x|a |u| p−1 u, in Rn , p > 1 and a > −2 has been studied recently, Wang and Ye [23] gave a complete classification of stable weak solutions and those of finite Morse index solutions. Email address: [email protected] (Belgacem Rahal ) Preprint submitted to arXiv

17 janvier 2017

2

In the past years, the Liouville property has been refined considerably and emerged as one of the most powerful tools in the study of initial and boundary value problems for nonlinear PDEs. It turns out that one can obtain from Liouville-type theorems a variety of results on qualitative properties of solutions such as universal, pointwise, a priori estimates of local solutions ; universal and singularity estimates ; decay estimates ; blow-up rate of solutions of nonstationary problems, etc., see [19, 21] and references therein. Liouville-type theorems for degenerate elliptic equations have been attracted the interest of many mathematicians. The classical Liouville theorem was generalized to p-harmonic functions on the whole space Rn and on exterior domains by Serrin and Zou [22], see also [5] for related results. The Liouville theorems for some linear degenerate elliptic operators such as X-elliptic operators, Kohn-Laplacian (and more general sublaplacian on Carnot groups) and degenerate OrnsteinUhlenbeck operators were proved in [14, 13]. More recently, Yu [24] studied the equation −Lα u = f (u) in Rn1 × Rn2 , where Lα = ∆x + (1 + α)2 ∆y , α > 0 and Q = n1 + (1 + α)n2 is the homogeneous dimension of the space. Under some assumptions on the nonlinear term f , he showed that the above equation possesses no positive solutions and the main technique used is the moving plane method in the integral form. In this paper, we are concerned with the Liouville-type theorems for the following problem −∆λ u = |x|aλ |u| p−1 u, where n ≥ 1, a ≥ 0, p > 1, ∆λ = λ21 ∆x(1) + ... + λ2k ∆x(k) ,

in Rn := Rn1 × Rn2 × ... × Rnk ,

(1.2)

k 2σ1 X Y |x|λ := λ2i (x)ǫ 2j |x( j) |2 , j=1 i, j

Pk

σ = 1 + i=1 (ǫi − 1), 1 ≤ ǫ1 ≤ ... ≤ ǫk , x = (x(1) , ..., x(k) ) ∈ Rn . Here the functions λi : Rn → R 1 are continuous, strictly positive the coordinate hyperplanes, i.e. λi > 0, Q and of class C outside Q n Q i = 1, ..., k in R \ , where = {x = (x1 , ..., xn ) ∈ Rn : ni=1 xi = 0}, and ∆x(i) denotes the classical Laplacian in Rni , i = 1, ..., k. As in [12] we assume that λi satisfy the following properties : (H1 ) λ1 (x) = 1, λi (x) = λi (x(1) , ..., x(i−1) ), i = 2, ..., k. (H2 ) For every x ∈ Rn , λi (x) = λi (x∗ ), i = 1, ..., k, where x∗ = (|x(1) |, ..., |x(k) |) if x = (x(1) , ..., x(k) ). (H3 ) There exists a group of dilations {δt }t>0

δt : Rn → Rn , δt (x) = δt (x(1) , ..., x(k) ) = (tǫ1 x(1) , ..., tǫk x(k) ),

where 1 ≤ ǫ1 ≤ ǫ2 ≤ ... ≤ ǫk , such that λi is δt -homogeneous of degree ǫi − 1, i.e. λi (δt (x)) = tǫi −1 λi (x), ∀ x ∈ Rn , t > 0, i = 1, ..., k.

This implies that the operator ∆λ is δt -homogeneous of degree two, i.e. ∆λ (u(δt (x))) = t2 (∆λ u)(δt (x)), ∀ u ∈ C ∞ (Rn ).

3

We denote by Q the homogeneous dimension of Rn with respect to the group of dilations {δt }t>0 , i.e. Q := ǫ1 n1 + ǫ2 n2 + ... + ǫk nk . The ∆λ -Laplace operator was first introduced by Franchi and Lanconelli [8], and recently reconsidered in [12] under an additional assumption that the operator is homogeneous of degree two with respect to a group dilation in Rn . It was proved in [1], that the autonomous case, i.e. a = 0, (1.2) Q has no positive classical solution if 1 < p ≤ Q−2 , with Q = ǫ1 + ǫ2 + ... + ǫn , (ni = 1, i = 1, ..., n). The ∆λ -operator contains many degenerate elliptic operators. We now give some examples of ∆λ -Laplace operators (see also [12]). We use the following notation : we split Rn as follows Rn = Rn1 × ... × Rnk and write (i) ni x = (x(1) , ..., x(k) ), x(i) = (x(i) 1 , ..., xni ) ∈ R ,

|x(i) |2 =

ni X j=1

2 |x(i) j | ,

i = 1, 2, ..., k.

We denote the classical Laplace operator in ∈ Rni by ∆x(i) =

ni X

∂2x(i) .

j=1

j

Example 1. Let α be a real positive constant and k = 2. We consider the Grushin-type operator ∆λ = ∆x + |x|2α ∆y , where λ = (λ1 , λ2) with λ1 (x) = 1,

λ2 (x) = |x(1) |α ,

x ∈ Rn1 × Rn2 .

Our group of dilations is δt (x) = δt (x(1) , x(2) ) = (tx(1) , tα+1 x(2) ), and the homogenous dimension with respect to (δt )t>0 is Q = n1 + (α + 1)n2 . Example 2. Given a multi-index α = (α1 , ..., αk−1 ), α j ≥ 1, j = 1, ..., k − 1, define ∆α := ∆x(1) + |x(1) |2α1 ∆x(2) + ... + |x(k−1) |2αk−1 ∆x(k) . Then ∆α = ∆λ with λ = (λ1 , ..., λk ) and λi = |x(i−1) |αi−1 , i = 1, ..., k. Here we agree to let |x(0) |α0 = 1. A group of dilations for which λ satisfies (H3 ) is given by δt : Rn → Rn , δt (x) = δt (x(1) , ..., x(k) ) = (tǫ1 x(1) , ..., tǫk x(k) ), with ǫ1 = 1 and ǫi = αi−1 ǫi−1 + 1, i = 2, ..., k. In particular, if α1 = ... = αk−1 = 1, the operator ∆α and the dilation δt becomes, respectively ∆α = ∆x(1) + |x(1) |2 ∆x(2) + ... + |x(k−1) |2 ∆x(k) , and δt (x) = (tx(1) , t2 x(2) , ..., tk x(k) ).

4

Example 3. Let α, β and γ be positive real constants. For the operator ∆λ = ∆x(1) + |x(1) |2α ∆x(2) + |x(1) |2β |x(2) |2γ ∆x(3) , where λ = (λ1 , λ2, λ3 ) with λ1 (x) = 1,

λ2 (x) = |x(1) |α ,

λ3 (x) = |x(1) |β |x(2) |γ ,

x ∈ Rn1 × Rn2 × Rn3 ,

we find the group of dilations δt (x) = δt (x(1) , x(2) , x(3) ) = (tx(1) , tα+1 x(2) , tβ+(α+1)γ+1 x(3) ). The aim of the present paper was to establish the Liouville-type theorems with finite Morse index for the equation (1.2). In order to state our results we need the following : Definition 1.1. We say that a solution u of (1.2) belonging to C 2 (Rn ) • is stable, if Z Z Qu (ψ) :=

Rn

|∇λ ψ|2 − p

Rn

|x|aλ |u| p−1 ψ2 ≥ 0 , ∀ ψ ∈ Cc1 (Rn ),

where ∇λ = (λ1 ∇x(1) , ..., λk ∇x(k) ). • has Morse index equal to K ≥ 1 if K is the maximal dimension of a subspace XK of Cc1 (Rn ) such that Qu (ψ) < 0 for any ψ ∈ XK \{0}. • is stable outside a compact set K ⊂ Rn if Qu (ψ) ≥ 0 for any ψ ∈ Cc1 (Rn \K). Remark 1.1. a) Clearly, a solution stable if and only if its Morse index is equal to zero. b) It is well know that any finite Morse index solution u is stable outside a compact set K ⊂ Rn . Indeed, there exists m0 ≥ 1 and Xm0 := Span{φ1, ..., φm0 } ⊂ Cc1 (Rn ) such that Qu (φ) < 0 for any 0 supp(φ j ). φ ∈ Xm0 \{0}. Hence, Qu (ψ) ≥ 0 for every ψ ∈ Cc1 (Rn \K), where K := ∪mj=1 In the following, we state Liouville-type results for solutions u ∈ C 2 (Rn ) of (1.2). In what follows, we divide our study to stable solutions and solutions which are stable outside a compact set. Stable solutions set ΓM (p) = 2p − 1 + pTo state the following result we need ǫto introduceǫ some notation. We ǫk 1 2 2 p(p − 1) and denote by ΩR = B1 (0, R ) × B2(0, R ) × ... × Bk (0, R ), where Bi(0, Rǫi ) ⊂ Rni , i = 1, ..., k, the balls of center 0 and radius Rǫi . Proposition 1.1. Let u ∈ C 2 (Rn ) be a stable solution of (1.2). Then, for any γ ∈ 1, ΓM (p)), there exists a positive constant C independent of R, such that Z 2(p+γ)+(γ+1)a γ−1 |x|aλ |u| p+γ + |∇λ (|u| 2 u)|2 dx ≤ CRQ− p−1 , for all R > 0. (1.3) 1.1.

ΩR

Proposition 1.1 provides an important estimate on the integrability of u and ∇λ u. As we will see, our nonexistence results will follow by showing that the right-hand side of (1.3) vanishes under the right assumptions on p when R → +∞. More precisely, as a corollary of Proposition 1.1, we can state our first Liouville type theorem.

5

Theorem 1.1. Let u ∈ C 2 (Rn ) be a stable solution of (1.2) with, if Q ≤ 10 + 4a, +∞ √ pc (Q, a) = 2 3 (a+2) (a+2Q−2) (Q−2) −2(a+2)(a+Q)+2 if Q > 10 + 4a. (Q−2)(Q−4a−10) Then u ≡ 0.

Solutions which are stable outside a compact set In this subsection we prove some integral identities extending to the ∆λ setting the classical Pohozaev identity for semilinear Poisson equation [18]. Pohozaev identity has been extended by several authors to general elliptic equations and systems, both in Riemannian and sub-Riemannian we closely follow the context, see, e.g., [2, 9, 20] and the references therein. To prove our identities P original procedure of Pohozaev, just replacing the vector field P = ni=1 xi ∂ xi in [18], page 1410], by k X T= ǫi x(i) ∇x(i) , 1.2.

i=1

the generator of the group of dilation (δt )t≥0 in (H3 )(we say that T generates (δt )t≥0 since a function u is δt -homogeneous of degree m if and only if T u = mu). Proposition 1.2. Let u ∈ C 2 (Rn ) be a solution of (1.2) and φ ∈ Cc1 (ΩR ). If T (|x|λ ) = |x|λ , then # " # # Z " Z " |x|aλ Q−2 1 Q + a a p+1 2 2 p+1 φ= ∇λ u∇λ φT (u) + − |∇λ u| + T (φ) . |∇λ u| − |x| |u| |u| 2 p+1 λ 2 p+1 ΩR ΩR (1.4) Thanks to Proposition 1.2, we derive Theorem 1.2. Let u ∈ C 2 (Rn ) be a solution of (1.2) which is stable outside a compact set of Rn , with if Q ≤ 2, +∞ p s (Q, a) = Q+2+2a Q−2 if Q > 2. If T (|x|λ ) = |x|λ , then u ≡ 0.

2.

Example which satisfies T (|x|λ ) = |x|λ The degenerate elliptic operators we consider are of the form ∆λ = λ21 ∆x(1) + ... + λ2k ∆x(k) .

We denote by |x( j) | the euclidean norm of x( j) ∈ Rn j and assume the functions λi are of the form λi (x) =

k Y j=1

|x( j) |αi j ,

i = 1, ..., k,

(2.1)

6

such that 1) αi j ≥ 0 for i = 2, ..., k, j = 1, ..., i − 1. 2) αi j = 0 for j ≥ i. P 3) kl=1 ǫl α jl = ǫ j − 1, j = 1, ..., k with 1 = ǫ1 ≤ ǫ2 ≤ ... ≤ ǫk . Clearly, λi is δt -homogeneous of degree ǫi − 1 with respect to a group of dilations {δt }t>0 δt : Rn → Rn , δt (x) = δt (x(1) , ..., x(k) ) = (tǫ1 x(1) , ..., tǫk x(k) ). P Now, using the relation kl=1 ǫl α jl = ǫ j − 1, we get T (|x|λ ) = |x|λ is satisfied.

This paper is organized as follows. In section 3, we give the proof of Proposition 1.1 and Theorem 1.1. Section 4 is devoted to the proof of Proposition 1.2 and Theorem 1.2. 3.

The Liouville theorem for stable solutions : proof of Theorem 1.1

In this section we prove all the results concerning the classification of stable solutions, i.e., Proposition 1.1 and Theorem 1.1. First, to prove Proposition 1.1, we need the following technical Lemma. Let R > 0, Ω2R = B1 (0, 2Rǫ1 ) × B2 (0, 2Rǫ2 ) × ... × Bk (0, 2Rǫk ), where Bi (0, 2Rǫi ) ⊂ Rni , i = 1, ..., k, and consider k functions ψ1,R ,..., ψk,R such that ! ! r(1) r(k) (1) (k) ψ1,R (r ) = ψ1 ǫ1 , ..., ψk,R (r ) = ψk ǫk , R R with ψ1,R , ... ψk,R ∈ Cc∞ ([0, +∞)), 0 ≤ ψ1,R , ... ψk,R ≤ 1, 1 in [0, 1], ψi (t) = 0 in [2, +∞),

and for some constant C > 0 and ψ1,R , ... ψk,R satisfy ∇x(1) ψ1,R ≤ CR−ǫ1 , ..., ∇x(k) ψk,R ≤ CR−ǫk ,

∆x(1) ψ1,R ≤ CR−2ǫ1 , ..., ∆x(k) ψk,R ≤ CR−2ǫk ,

where r(i) = |x(i) |, i = 1, ..., k.

Lemma 3.1. (1) There exists a constant C > 0 independent of R such that a) |λi (x)| ≤ CRǫi −1 , ∀ x ∈ Ω2R , i = 1, ..., k. Q b) |∇λ ψR |2 + |∆λ ψR | ≤ CR−2 , where ψR = ki=1 ψi,R .

(2) The homogeneous norm, |.|λ , is δt -homogeneous of degree one, i.e. |δt (x)|λ = t|x|λ , ∀ x ∈ Rn , t > 0. (3) There exists a constant C > 0 independent of R such that |x|λ ≤ CR, ∀ x ∈ Ω2R .

7

Proof. Proof of (1) a). For any x = (x(1) , ..., x(k) ) ∈ Ω2R , we have x(i) ∈ Bi (0, 2Rǫi ), i = 1, ..., k, this implies |x(i) | ≤ 2, i = 1, ..., k. Therefore, if we write Rǫi ! x(k) x(1) ǫk (1) (k) ǫ1 x = (x , ..., x ) = R × ǫ1 , ..., R × ǫk , R R (1) (k) and let y = (y(1) , ..., y(k) ) = Rx ǫ1 , ..., Rx ǫk , then y ∈ Ω2 . Hence by assumption (H3 ) made on functions λi , we get λi (x) = λi (Rǫ1 y(1) , ..., Rǫk y(k) ) = Rǫi −1 λi (y(1) , ..., y(k) ) = Rǫi −1 λi (y).

(3.1)

Moreover, since λi , i = 1, ..., k are continuous, then |λi (y)| ≤ C, ∀ y ∈ Ω2 .

(3.2)

Therefore, from (3.1) and (3.2), we obtain |λi (x)| ≤ CRǫi −1 , ∀ x ∈ Ω2R , i = 1, ..., k. Proof of (1) b). Using assumption (H2 ) made on functions λi , i = 1, ..., k, with r = (r(1) , ..., r(k) ) = (|x(1) |, ..., |x(k) |), we have λ1 (r) = 1, λi (r) = λi (r(1) , ..., r(i−1) ), ∀ i = 2, ..., k. If we denote by ψR =

Qk

i=1

ψi,R , we get

∇λ ψR = (λ1 (r)∇x(1) ψR , ..., λk (r)∇x(k) ψR ) k k−1 Y Y = λ1 (r)∇x(1) ψ1,R ψi,R , ..., λk (r)∇x(k) ψk,R ψi,R , i=2

and

i=1

∆λ ψR = λ21 (r)∆x(1) ψR + ... + λ2k (r)∆x(k) ψR k k−1 Y Y 2 2 = λ1 (r)∆x(1) ψ1,R ψi,R + ... + λk (r)∆x(k) ψk,R ψi,R . i=2

i=1

Since |λi (r)| = |λi (x)| ≤ CRǫi −1 , ∀ x ∈ Ω2R , i = 1, ..., k, then there exists a constant C > 0 independent of R such that |∇λ ψR |2 ≤ CR−2 and |∆λ ψR | ≤ CR−2 .

8

Proof of (2). Let x ∈ Rn . The homogeneity of the functions λi implies that Pk 1 k 2(1+ i=1 (ǫi −1)) X Y 2 2 ǫ j ( j) 2 (λi (δt (x))) ǫ j |t x | |δt (x)|λ : = j=1 i, j

k Pk 1 X Y 2(1+ i=1 (ǫi −1)) = t2ǫ j t2(ǫi −1) (λi (x))2 ǫ 2j |x( j) |2 j=1 i, j

Pk 1 k Y X 2(1+ i=1 (ǫi −1)) Pk = t2(1+ i=1 (ǫi −1)) (λi (x))2 ǫ 2j |x( j) |2 j=1 i, j

= t|x|λ

(3.3)

Proof of (3). For any x = (x(1) , ..., x(k) ) ∈ Ω2R , we have x(i) ∈ Bi (0, 2Rǫi ), i = 1, ..., k, this implies |x(i) | ≤ 2, i = 1, ..., k. Therefore, if we write Rǫi ! x(k) x(1) ǫk (1) (k) ǫ1 x = (x , ..., x ) = R × ǫ1 , ..., R × ǫk , R R (1) (k) and let y = (y(1) , ..., y(k) ) = Rx ǫ1 , ..., Rx ǫk , then y ∈ Ω2 (0). Using (3.3), we get |x|λ = |(Rǫ1 y(1) , ..., Rǫk y(k) )|λ = R|(y(1) , ..., y(k) )|λ = R|y|λ . Since |λi (y)| ≤ C, ∀ y ∈ Ω2 , i = 1, ..., k, then there exists a constant C > 0 independent of R such that |x|λ ≤ CR, ∀ x ∈ Ω2R . This completes the proof of Lemma 3.1. Proof of Proposition 1.1. The proof follows the main lines of the demonstration of proposition 4 in [7], with more modifications. We split the proof into four steps : Step 1. For any φ ∈ Cc2 (Rn ) we have Z Z 2 Z γ−1 γ + 1 (γ + 1) a p+γ 2 2 2 |x|λ |u| φ dx + |u|γ+1 ∆λ (φ2 )dx. (3.4) |∇λ (|u| 2 u)| φ dx = 4γ 4γ n n n R R R Multiply equation (1.2) by |u|γ−1 uφ2 and integrate by parts to find Z Z Z 2 γ−1 2 2 γ−1 γ |∇λ u| |u| φ dx + ∇λ u∇λ (φ )|u| u dx = |x|aλ |u| p+γ φ2 dx, Rn

therefore Z

Rn

|x|aλ |u| p+γ φ2 dx

Rn

Rn

Z Z γ−1 1 4γ 2 2 ∇λ (|u|γ+1 )∇λ (φ2 )dx |∇λ (|u| 2 u)| φ dx + = (γ + 1)2 Rn γ + 1 Rn Z Z γ−1 1 4γ 2 2 2 |u|γ+1 ∆λ (φ2 )dx. |∇λ (|u| u)| φ dx − = (γ + 1)2 Rn γ + 1 Rn

9

Identity (3.4) then follows by multiplying the latter identity by the factor

(γ+1)2 . 4γ

Step 2. For any φ ∈ Cc2 (Rn ) we have !Z ! # " Z (γ + 1)2 γ+1 1 2 a p+γ 2 2 γ+1 p− |x|λ |u| φ dx ≤ − ∆λ (φ ) dx. |∇λ φ| + |u| 4γ 4γ 2 Rn Rn

(3.5)

γ−1

The function |u| 2 uφ belongs to Cc1 (Rn ), and thus it can be used as a test function in the quadratic form Qu . Hence, the stability assumption on u gives Z Z γ−1 a p+γ 2 (3.6) p |x|λ |u| φ dx ≤ |∇λ (|u| 2 uφ)|2 dx. Rn

Rn

A direct calculation shows that the right hand side of (3.6) equals to # Z " γ−1 1 2 2 2 γ+1 γ+1 2 2 |u| |∇λ φ| + |∇λ (|u| u)| φ + ∇λ φ ∇λ (|u| ) dx 2 Rn # " Z Z γ−1 1 2 2 γ+1 |∇λ (|u| 2 u)|2 φ2 dx. (3.7) = |∇λ φ| − ∆λ (φ ) dx + |u| 2 Rn Rn From (3.6) and (3.7), we obtain that # " Z Z Z γ−1 1 2 a p+γ 2 2 γ+1 |∇λ (|u| 2 u)|2 φ2 dx. p |x|λ |u| φ dx ≤ |∇λ φ| − ∆λ (φ ) dx + |u| 2 Rn Rn Rn

(3.8)

Putting this back into (3.4) gives !Z ! # " Z (γ + 1)2 γ+1 1 2 a p+γ 2 2 γ+1 p− − ∆λ (φ ) dx. |x|λ |u| φ dx ≤ |∇λ φ| + |u| 4γ 4γ 2 Rn Rn , 2} there exists a constant C(p, m, γ) > Step 3. For any γ ∈ 1, ΓM (p)) and any integer m ≥ max{ p+γ p−1 0 depending only on p, m and γ Z Z −(γ+1)a p+γ a p+γ 2m |x|λ |u| ψR dx ≤ C(p, m, γ) |x|λ p−1 |∇λ ψR |2 + |ψR ||∆λ ψR | p−1 dx, (3.9) Rn

Z

Rn

|∇λ (|u|

Rn

γ−1 2

u)|2 ψ2m R dx

≤ C(p, m, γ)

Z

−(γ+1)a

Rn

|x|λ p−1

|∇λ ψR |2 + |ψR ||∆λ ψR |

p+γ p−1

Q where ψR = ki=1 ψi,R . Moreover, the constant C(p, m, γ) can be explicitly computed. From (3.5), we obtain that Z Z Z a p+γ 2 γ+1 2 α |x|λ |u| φ dx ≤ |u| |∇λ φ| + β |u|γ+1 ∆λ φdx. Rn

where we have set α = p − γ ∈ 1, ΓM (p)).

Rn

(γ+1)2 4γ

and β =

1−γ . 4γ

dx,

(3.10)

(3.11)

Rn

Notice that α > 0 and β < 0, since p > 1 and

10

Now, we set φ = ψmR . The function φ belongs to Cc2 (Rn ), since m ≥ 2 and m is an integer, hence it can be used in (3.11). A direct computation gives Z Z a p+γ 2m α |x|λ |u| ψR dx ≤ |u|γ+1 ψR2m−2 m2 |∇λ ψR |2 + βm(m − 1)|∇λ ψR |2 + βmψR ∆λ ψR dx, Rn

Rn

(3.12)

hence Z

Rn

|x|aλ |u| p+γ ψ2m R dx

≤ C1

Z

Rn

|u|γ+1 ψR2m−2 |∇λ ψR |2 + |ψR ||∆λ ψR | dx,

(3.13)

2

> − βm ≥ 0. with C1 = m +βm(m−1) α α An application of Young’s inequality yields Z Z a p+γ 2m |x|λ |u| ψR ≤ C1 |u|γ+1 ψR2m−2 |∇λ ψR |2 + |ψR ||∆λ ψR | dx n Rn ZR (γ+1)a −(γ+1)a = C1 |x|λ p+γ |u|γ+1 ψR2m−2 |x|λ p+γ |∇λ ψR |2 + |ψR ||∆λ ψR | dx RnZ Z p+γ −(γ+1)a p+γ (p − 1) C1 γ+1 a p+γ (2m−2) γ+1 + ≤ |x|λ |u| ψR |x|λ p−1 |∇λ ψR |2 + |ψR ||∆λ ψR | p−1 . p + γ Rn p+γ Rn (3.14) (2m−2) p+γ

γ+1 , 2} implies (2m − 2) p+γ ≥ 2m and thus ψR ≤ ψ2m At this point we notice that m ≥ max{ p+γ R p−1 p−1 n n in R , since 0 ≤ ψR ≤ 1 everywhere in R . Therefore, we obtain Z Z Z −(γ+1)a p+γ γ+1 (p − 1) C1 p−1 a p+γ 2m a p+γ 2m |x|λ |u| ψR ≤ |x|λ |u| ψR + |x|λ p−1 |∇λ ψR |2 + |ψR ||∆λ ψR | . n p + γ p + γ n n R R R

The latter immediately implies Z Z −(γ+1)a p+γ a p+γ 2m |x|λ |u| ψR dx ≤ C1 |x|λ p−1 |∇λ ψR |2 + |ψR ||∆λ ψR | p−1 dx, Rn

(3.15)

Rn

which proves inequality (3.9) with C(p, m, γ) = C1 . To prove (3.10), we combine (3.4) and (3.5). This leads to Z Z Z γ−1 2 2 γ+1 2 |u| |∇λ φ| dx + B |∇λ (|u| 2 u)| φ dx ≤ A |u|γ+1 φ∆λ φdx, Rn

Rn

2

2

Rn

where A = (γ+1) + (γ+1) > 0 and B = β(γ+1) + (γ+1) ∈ R. 4γα 2γ 4γα 2γ Now, we insert the test function φ = ψmR in the latter inequality to find, Z Z γ−1 2 2m 2 |∇λ (|u| u)| ψR dx ≤ |u|γ+1 ψR2m−2 Am2 |∇λ ψR |2 + Bm(m − 1)|∇λ ψR |2 + BmψR ∆λ ψR dx, Rn

Rn

11

and hence Z

Rn

|∇λ (|u|

γ−1 2

u)|2 ψ2m R dx

≤ C2

Z

Rn

|u|γ+1 ψR2m−2 |∇λ ψR |2 + |ψR ||∆λ ψR | dx,

(3.16)

with C2 = max{Am2 + Bm(m − 1), |B|m} > 0. Using H¨older’s inequality in (3.16) yields p−1 ! γ+1 Z ! p+γ Z Z −(γ+1)a p+γ p+γ p+γ γ−1 (2m−2) γ+1 p−1 p−1 2 2 2m a p+γ |∇λ ψR | + |ψR ||∆λ ψR | |∇λ (|u| 2 u)| ψR ≤ C2 |x|λ |x|λ |u| ψR Rn

Rn

Rn

≤ C2

Z

Rn

|x|aλ |u| p+γ ψ2m R

! γ+1 Z p+γ

p−1

−(γ+1)a p−1

Rn

|x|λ

! p+γ p+γ p−1 2 . |∇λ ψR | + |ψR ||∆λ ψR |

Finally, inserting (3.15) into the latter we obtain Z Z −(γ+1)a 1+γ p+γ γ−1 p−1 2 2m 2 |x|λ p−1 |∇λ ψR |2 + |ψR ||∆λ ψR | p−1 dx, |∇λ (|u| u)| ψR dx ≤ C2C1 Rn

Rn

which gives the desired inequality (3.10). Step 4. For any γ ∈ 1, ΓM (p)), there exists a constant C > 0 independent of R such that Z 2(p+γ)+(γ+1)a γ−1 |x|aλ |u| p+γ + |∇λ (|u| 2 u)|2 dx ≤ CRQ− p−1 , ∀ R > 0.

(3.17)

ΩR

The proof of (3.17) follows immediately by adding inequality (3.9) to inequality (3.10) and using Lemma 3.1. Proof of Theorem 1.1. By Proposition 1.1, there exists a positive constant C independent of R such that Z 2(p+γ)+a(γ+1) (3.18) |x|aλ |u| p+γ ≤ CRQ− p−1 . ΩR

Then it suffices to show that we can always choose a γ ∈ 1, ΓM (p)), such that Q − 2(p+γ)+a(γ+1) < 0. p−1 Therefore, by letting R → +∞ in (3.18), we deduce that Z |x|aλ |u| p+γ = 0, Rn

which implies that u ≡ 0 in Rn . Next, we claim that, under the assumptions on the exponent p assumed in Theorem 1.1, we can always choose γ ∈ [1, ΓM (p)) such that Q−

2(p + γ) + a(γ + 1) < 0. p−1

(3.19)

As in [7], we consider separately the case Q ≤ 10 + 4a and the case Q > 10 + 4a. Case 1. Q ≤ 10 + 4a and p > 1. In this case we have 2(p + ΓM (p)) + a(ΓM (p) + 1) > 2(3p − 1 + 2(p − 1)) + a(2p + 2(p − 1) > (10 + 4a)(p − 1)

12

and therefore Q−

2(p + ΓM (p)) + a(ΓM (p) + 1) < Q − (10 + 4a) ≤ 0. p−1

The latter inequality and the continuity of the function x 7→ Q − the existence of γ ∈ [1, ΓM (p)) satisfying (3.19).

2(p+x)+a(x+1) p−1

(3.20) immediately imply

Case 2. Q > 10 + 4a and 1 < p < pc (Q, a). In this case we consider the real-valued function x 7→ g(x) := 2(x+ΓM (x))+a(Γ(x)+1) on (1, +∞). Since g is strictly decreasing function satisfying x−1 limx→1+ g(x) = +∞ and limx→+∞ g(x) = 10 + 4a, there exists a unique p0 > 1 such that Q = g(p0 ). We claim that p0 = pc (Q, a). Indeed, p Q = g(p) ⇔ (Q − 2)(p − 1) − (4 + 2a)p = (4 + 2a) p(p − 1) ⇔ (Q − 10 − 4a)(Q − 2)p2 + (−2(Q − 2)2 + 4(a + 2)(Q + a))p + (Q − 2)2 = 0, which implies that (Q − 10 − 4a)(Q − 2)p20 + (−2(Q − 2)2 + 4(a + 2)(Q + a))p0 + (Q − 2)2 = 0,

(3.21)

(Q − 2)(p0 − 1) − (4 + 2a)p0 > (4 + 2a)(p0 − 1).

(3.22)

p (Q − 2)2 − 2(a + 2)(a + Q) + 2 (a + 2)3 (a + 2Q − 2) p1 = = pc (Q, a), (Q − 2)(Q − 4a − 10)

(3.23)

p (Q − 2)2 − 2(a + 2)(a + Q) − 2 (a + 2)3 (a + 2Q − 2) p2 = < p0 , (Q − 2)(Q − 4a − 10)

(3.24)

and

The roots of (3.21)

while (3.22) easily implies p0 >

Q−6−2a Q−4a−10

> p2 . This proves that p0 = p1 . Hence p (Q − 2)2 − 2(a + 2)(a + Q) + 2 (a + 2)3 (a + 2Q − 2) pc (Q, a) = (Q − 2)(Q − 4a − 10)

as claimed. Since we have just proven that g(pc (Q, a)) = Q and g is a strictly decreasing function, it follows that ∀ 1 < p < pc (Q, a), Q < g(p). Now we can conclude as in the first case, i.e, the continuity of x 7→ Q − implies the existence of γ ∈ [1, ΓM (p)) satisfying (3.19).

(3.25) 2(p+x)+a(x+1) p−1

immediately

13 4.

The Liouville theorem for solutions which are stable outside a compact set of R n : proof of Theorem 1.2 In this section, we prove Proposition 1.2 and Theorem 1.2.

Proof of Proposition 1.2. Let u ∈ C 2 (Rn ) be a solution of (1.2) and φ ∈ Cc1 (ΩR ). Multiplying equation (1.2) by T (u)φ and integrating by parts in ΩR , we obtain Z Z − ∆λ uT (u)φdx = − ∆λ u ǫ j x( j) ∇x( j) u φdx ΩR Z ΩR = λ2i ∇x(i) u ∇x(i) ǫ j x( j) ∇x( j) uφ dx ZΩR Z 2 = λi ∇x(i) u ǫ j δi j ∇x( j) u φdx + λ2i ∇x(i) u ǫ j x( j) ∇x(i) (∇x( j) u) φdx ΩR ΩR Z + λ2i ∇x(i) u ǫ j x( j) ∇x( j) u ∇x(i) φdx ΩR

:= I1 + I2 + I3 ,

(4.1)

Here and in the sequel, we use the Einstein summation convention : an index occurring twice in a product is to be summed from 1 up to the space dimension. Obviously Z λ2i ∇x(i) u ǫ j δi j ∇x( j) u φdx I1 : = ZΩR = λ2i |∇x(i) u|2 ǫi φdx. (4.2) ΩR

Moreover, an integration by parts in I2 gives Z I2 : = λ2i ∇x(i) uǫ j x( j) ∇x(i) (∇x( j) u) φdx ΩR Z Z Z 2 2 ( j) 2 2 = − ∇x( j) (λi ) |∇x(i) u| ǫ j x φdx − I2 − λi |∇x(i) u| ǫ j n j φdx − λ2i |∇x(i) u|2 ǫ j x( j) ∇x( j) φdx ΩR Z ΩR Z ZΩR |∇λ u|2 φdx − |∇λ u|2 T (φ)dx. = −2 λi |∇x(i) u|2 T (λi )φdx − I2 − Q ΩR

ΩR

ΩR

Since λi is δt -homogeneous of degree ǫi − 1, then T (λi ) = (ǫi − 1)λi . Hence Z Z Z 2 2 2 |∇λ u| φdx − |∇λ u|2 T (φ)dx I2 = −2 (ǫi − 1)λi |∇x(i) u| φdx − I2 − Q ΩR ΩR ΩR Z Z 2 = (2 − Q) |∇λ u| φdx − 2I1 − I2 − |∇λ u|2 T (φ)dx. ΩR

ΩR

Then 2−Q I2 = 2

Z

1 |∇λ u| φdx − I1 − 2 ΩR 2

Z

ΩR

|∇λ u|2 T (φ)dx.

(4.3)

14

It is easily seen that I3 : = =

Z

ZΩR

ΩR

λ2i ∇x(i) uǫ j x( j) ∇x( j) u∇x(i) φdx ∇λ u∇λ φT (u)dx.

(4.4)

Hence, by (4.1), Z Z Z Z 1 2−Q 2 2 − ∆λ uT (u)φdx = |∇λ u| φdx − |∇λ u| T (φ)dx + ∇λ u∇λ φT (u)dx. 2 2 ΩR ΩR ΩR ΩR (4.5) On the other hand, an integration by parts gives Z Z 1 a p−1 |x|aλ ∇x( j) (|u| p+1 )ǫ j x( j) φdx |x|λ |u| uT (u)φdx = p + 1 ΩR ΩR Z Z Z Q a 1 a p+1 a−1 p+1 =− |x| |u| φ − |x| |u| T (|x|λ )φ − |x|a |u| p+1 T (φ)dx. p + 1 ΩR λ p + 1 ΩR λ p + 1 ΩR λ If T (|x|λ ) = |x|λ , then Z |x|aλ |u| p−1 uT (u)φdx =

Z 1 |x|a ∇x( j) (|u| p+1 )ǫ j x( j) φdx p + 1 ΩR λ Z Z 1 Q+a a p+1 |x| |u| φ − |x|a |u| p+1 T (φ)dx. = − p + 1 ΩR λ p + 1 ΩR λ

ΩR

Clearly (1.4) follows directly from (4.5) and (4.6).

(4.6)

Proof of Theorem 1.2. Let u be a solution of (1.2) which is stable outside a compact set. We begin defining some smooth compactly supported functions which will be used several times in the sequel. More precisely, for R∗ > 0, we choose a function ζi,R ∈ Cc2 (Rni ), i = 1, ..., k, 0 ≤ ζi,R ≤ 1, everywhere on Rni and ζi,R (x(i) ) = 0 if |x(i) | < R∗ + 1 or |x(i) | > 2Rǫi , ζi,R (x(i) ) = 1 if R∗ + 2 < |x(i) | < Rǫi , |∇ (i) ζ |2 + |∆ (i) ζ | ≤ CR−2ǫi for {Rǫi < |x(i) | < 2Rǫi }. x i,R x i,R

The rest of the proof splits into several steps. Step 1. Let p > 1. There exists R∗ > 0 such that for every γ ∈ 1, ΓM (p)) and every Rǫi > R∗ + 2, we have Z 2(p+γ)+(γ+1)a γ−1 a p+γ 2 2 |x|λ |u| + |∇λ (|u| u)| dx ≤ CR∗ + CRQ− p−1 , (4.7) Σ0 (R)

where Σ0 (R) = ΩR \B1 (0, R∗ + 2) × ... × Bk (0, R∗ + 2), CR∗ and C are positive constants depending on p, γ, R∗ but not on R.

15

Since u is stable outside a compact set of Rn , there exists R∗ > 0 such that, similar to that of Proposition 1.1 we derive Z Z −a(γ+1) p+γ γ−1 a p+γ 2 2 |x|λ |u| + |∇λ (|u| u)| dx ≤ C(p, m, γ) |x|λ p−1 |∇λ ζR |2 + |ζR ||∆λ ζR | p−1 dx Rn

Σ0 (R)

where ζR =

Qn

i=1 ζi,R .

≤ CR∗ + CR

Q− 2(p+γ)+(γ+1)a p−1

,

Hence, the desired integral estimate (4.7) follows.

Step 2. If Q = 2 and 1 < p < +∞ or Q ≥ 3 and 1 < p