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N + α > 2, γ − α ≤ −2, the above equation admits no weak solution. ... weak solution which is stable outside a compact set of RN , hence any weak solution.
COMMUNICATIONS ON PURE AND APPLIED ANALYSIS Volume 13, Number 1, January 2014

doi:10.3934/cpaa.2014.13.293 pp. 293–305

STABLE WEAK SOLUTIONS OF WEIGHTED NONLINEAR ELLIPTIC EQUATIONS

Xia Huang Department of Mathematics and Center for Partial Differential Equations East China Normal University, Shanghai 200241, China

(Communicated by Xuefeng Wang) Abstract. This paper deals with the weighted nonlinear elliptic equation  −div(|x|α ∇u) = |x|γ eu in Ω, u=0 on ∂Ω, where α, γ ∈ R satisfy N + α > 2 and γ − α > −2, and the domain Ω ⊂ RN (N ≥ 2) is bounded or not. Moreover, when α 6= 0, we prove that, for N + α > 2, γ − α ≤ −2, the above equation admits no weak solution. We also study Liouville type results for the equation in RN .

1. Introduction. In this paper, we consider the following weighted nonlinear elliptic equation  −div(|x|α ∇u) = |x|γ eu in Ω, (1.1) u=0 on ∂Ω, where α, γ ∈ R satisfy N + α > 2, γ − α > −2, and the domain Ω ⊂ RN (N ≥ 2) is bounded or not. Several physical phenomena related to equilibrium of continuous media are modeled by this elliptic problem. In order to be able to handle with media which possibly are somewhere “perfect” insulators or “perfect” conductors, we allow the coefficient |x|α to vanish somewhere or to be unbounded (see [10]). It is well-known that Farina [12] showed that Eq. (1.1) with α = γ = 0 has no stable classical solution in RN for 2 ≤ N ≤ 9. He also proved that any classical solution u of (1.1) with finite Morse index in R2 verifies eu ∈ L1 (R2 ), so it must be a solution classified by Chen and Li [4], that is   32λ2 for some λ > 0, x0 ∈ R2 . u(x) = ln (4 + λ2 |x − x0 |2 )2 When N ≥ 3, Dancer and Farina [8] proved that Eq. (1.1) with α = γ = 0 and Ω = RN , admits classical entire solutions which are stable outside a compact set of RN if and only if N ≥ 10. Later, for α = 0, γ > −2 and Ω = RN , Wang and Ye [19] have obtained the following: (i) if 2 ≤ N < 10 + 4γ, Eq. (1.1) admits no weak 2010 Mathematics Subject Classification. Primary: 35J91; Secondary: 35B35, 35B53. Key words and phrases. Stability, weak solution, weighted sobolev space, Liouville theorems, exponential nonlinearity. The author was partially supported by NSF of China (No. 10971067).

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stable solution; (ii) if 2 < N < 10 + 4γ − where γ − = min (γ, 0), Eq. (1.1) admits no weak solution which is stable outside a compact set of RN , hence any weak solution of (1.1) in RN has infinite Morse index. The motivation of our study comes also from the weighted power nonlinearity case, as: −div(|x|α ∇u) = |x|γ |u|p−1 u, which has been studied by many authors. Some Liouville type results for the cases α = 0 can be found in [7, 13, 19]. Recently, for α 6= 0, Du and Guo [9] have obtained the behavior of solutions with finite Morse index for the equation of divergence form, which comes from the nonlinear Schr¨ odinger equation with Hardy potential by using a suitable transformation, while Jeong and Lee [15] directly study the nonlinear elliptic equations with Hardy potential. In this paper, we are interested in the general non-autonomous i.e.(α 6= 0) equation (1.1). In particular, we focus on stable solutions and the finite Morse index solutions and solutions which are stable outside a compact set of RN . Recently, Cowan and Fazly [5] studied the following semi-linear elliptic equations −div(ω1 (x)∇u) = ω2 (x)f (u)

in RN ,

with general positive smooth weights ω1 (x), ω2 (x). The existence or non-existence of non trivial stable sub or super solution has been established under various conditions β α on ωi , i = 1, 2. In particular, when ω1 (x) = (|x|2 + 1) 2 , ω2 (x) = (|x|2 + 1) 2 and f (u) = eu , they proved that (i) if N + α − 2 < 0, then there is no stable sub-solution for above equation; (ii) if 0 < N + α − 2 < 4(β − α + 2), then there is no stable sub-solution for above equation. The essence of this article is to obtain a useful integral estimate which is satisfied by stable weak solutions. Due to the presence of the weight function |x|α , thus, it is hard to apply the Wkp theories as was used in [8, 12, 19] to the problem (1.1) directly. To overcome this difficulty, we need to define the weighted space and employ the Caffarelli-Kohn-Nirenberg inequality instead of the Hardy inequality. Furthermore, the equation (1.1) may become to a degenerate (resp. singular) nonlinear elliptic equation with α > 0 (resp. α < 0), which also further complicates the analysis of (1.1). Note that our results hold also for the case N + α < 2, γ − α < −2 and Ω = RN . Indeed, using the transformation v(x) = u( |x|x 2 ), we can change (1.1) into an equivalent equation for v, i.e. −div(|x|α1 ∇v) = |x|γ1 ev in RN \ {0} with α1 = 4 − 2N − α, γ1 = −2N − γ. Although a solution u in RN yields only a solution v in RN \ {0}, the asymptotic behavior of v near the origin can be studied thanks to the decay of u at infinity. see [7, 19]. To describe our results more precisely, we first introduce some notations and definitions. A suitable weighted Sobolev space need to be considered because of the weight function |x|α . For α ∈ R satisfies α > 2 − N , and ϕ ∈ Cc∞ (Ω), we define Z 2 kϕkα := |x|α |∇ϕ|2 dx, Ω

and H01 (Ω, |x|α dx) := closure of Cc∞ (Ω) with respect to the k · kα − norm. 1 Remark that for N +α > 2, we have Cc1 (Ω) ⊂ H01 (Ω, |x|α dx) and u ∈ Hloc (Ω, |x|α dx) ∞ 1 α means that if for any ψ ∈ C0 (Ω \ {0}), there holds uψ ∈ H0 (Ω, |x| dx). Let us make also the meaning of weak solution more precisely.

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295

Definition 1.1. We say that u is a weak solution of −div(|x|α ∇u) = |x|γ eu in 1 Ω ⊂ RN , if u ∈ Hloc (Ω, |x|α dx) verifies |x|γ eu ∈ L1loc (Ω) and Z  |x|α ∇u∇ϕ − |x|γ eu ϕ dx = 0, holds for all ϕ ∈ Cc1 (Ω). (1.2) Ω

Definition 1.2. Let u be a weak solution of −div(|x|α ∇u) = |x|γ eu in Ω ⊂ RN . We say that u is stable if Z  Qu (ψ) := |x|α |∇ψ|2 − |x|γ eu ψ 2 dx ≥ 0, (1.3) Ω

holds for all ψ ∈ Cc1 (Ω). Remark 1. In the above definition, since |x|γ eu is nonnegative, (1.2) and (1.3) hold for any text function ϕ, ψ ∈ H01 (Ω, |x|α dx) by a density argument. Definition 1.3. We say that a weak solution u of (1.1) has Morse index which is denoted by ind(u):= l ≥ 0 if l is the maximal dimension of all subspaces Xl of Cc1 (Ω), such that Qu (ψ) < 0 for all ψ ∈ Xl \ {0}. We note that a weak solution u is stable if and only if l = 0. Furthermore, a weak solution u of (1.1) is said to be stable outside a compact set K of Ω, which means that Qu (ψ) ≥ 0 for all ψ ∈ Cc1 (Ω\K). Recall that any finite Morse index solution u of (1.1) is stable outside a compact set K ⊂ Ω. Indeed, there exists an integer l ≥ 0 and a subspace Xl := span{ϕ1 , ..., ϕl } ⊂ Cc1 (Ω) such that Qu (ϕ) < 0 for all ϕ ∈ Xl \ {0}. This implies that Qu (ψ) ≥ 0 for all ψ ∈ Cc1 (Ω\K), where K = ∪j supp(ϕj ). This paper is organized as follows. In section 2, we prove the results about nonexistence and existence of stable weak singular solution for equation (1.1). The main Liouville type results for equation (1.1) are given in section 3. 2. Nonexistence and existence of weak (stable) solutions. First for N +α > 2, we show the necessity to work with γ − α > −2, then we discuss the main results. 2.1. Nonexistence of weak solution for N + α > 2 and γ − α ≤ −2. Proposition 1. For N +α > 2, let Ω be a fixed domain containing 0, if γ −α ≤ −2, then equation (1.1) has no weak solution. Proof. Arguing by contradiction, we assume u is a weak solution of (1.1). Without loss of generality, let B(0, R) ⊂ B(0, 1) ⊂ Ω, where B(0, R) is the open ball of radials R centered at 0. Define v to be the average of u in the sense of distribution. Z 1 x v(r) = u(r, ϑ) = u(r, ϑ)dϑ, ωN = |S N −1 |, r = |x|, ϑ = . ωN S N −1 |x| By simple calculations and using the Jensen’s inequality, −div(|x|α ∇v)

= = = = = = ≥

−[∇(|x|α )∇v + |x|α ∆v] −[∇(|x|α )∇u + |x|α ∆u] −[∇(|x|α )∇u + |x|α ∆u] −div(|x|α ∇u) |x|γ eu |x|γ eu |x|γ ev .

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i.e., − div(|x|α ∇v) ≥ |x|γ ev , holds in the sense of distribution.

(2.1)

Note that Z ∂v − rN −1+α ωN v 0 (r) = − |x|α dσ ∂ν ∂B(0,r) Z Z =− div(|x|α ∇v)dx ≥ |x|γ ev dx > 0. B(0,r)

B(0,r)

0

γ u

Thus v (r) < 0 for any r ∈ (0, R). As |x| e ∈ L1loc (Ω) and 0 ∈ Ω, we must have γ > −N . Furthermore, by (2.1), for 0 < r1 < r < R, we have Z r Z r −rN −1+α v 0 (r) ≥ sN −1+γ ev ds − r1N −1+α v 0 (r1 ) ≥ sN −1+γ ev ds. r1

r1

Tending r1 to 0, we get − rN −1+α v 0 (r) ≥

Z

r

sN −1+γ ev ds

0

≥e

v(r)

Z

r N −1+γ

s

ds =

0

rN +γ ev(r) . N +γ

(2.2)

Hence −e−v v 0 (r) ≥ Cr1−α+γ , where C is a positive constant depending only on N, γ, which yields Z r −v(r) −v(r) −v(r1 ) e >e −e ≥C s1−α+γ ds. r1

Rr

1−α+γ

Letting r1 → 0, then r1 s ds → ∞ for any γ − α ≤ −2, which is a contradiction, so the proof of the Proposition 1 is finished. 2.2. The singular solution. We only consider the radial solution. Let Ω = RN , N + α > 2 and γ − α > −2. Direct calculation shows that −div(|x|α ∇us ) = |x|γ eus , so Eq. (1.1) has a singular weak solution defined by Definition 1.1 and here us (x) = (α − γ − 2)ln|x| + ln(2 − α + γ)(N − 2 + α). similar see [16]. Proposition 2. Let Ω = RN , for N ≥ 2, N + α > 2 and γ − α > −2. The function us (x) given above is a stable weak solution of (1.1) if and only if N ≥ 10 − 5α + 4γ. Instead of the Hardy inequality used in [12], we apply the Caffarelli-KohnNirenberg inequality (see [1]) to derive the stability of us (x), that is Z RN

|ϕ|q dx |x|bq

 q2

Z ≤ C(N, a, b) RN

|∇ϕ|2 dx, |x|2a

for all ϕ ∈ Cc∞ (RN )

where C(N, a, b) is a positive constant and −∞ < a
2, let HN = (N −2+α) 4 Z Z ϕ2 HN |x|α |∇ϕ|2 dx, for all ϕ ∈ H 1 (RN , |x|α dx). |x|α 2 dx ≤ |x| RN RN

(2.4)

In our case here, α a=− , 2

b=1+a=1−

α 2

and q = 2.

2 )2(b−a) of (2.3), i.e. HN is the By [3], C(N, a, b) has the optimal value ( N −2−2a best constant of (2.4). For the sake of completeness, now we give a proof of this lemma.

Proof. Note that the inequality is invariant under rotation and preserves the scaling N −2+α property. Let ϕ = ϕ0 ψ, where we assume that ψ ∈ Cc∞ (RN \ {0}), ϕ0 = |x|− 2 . Then ϕ ∈ Cc∞ (RN \ {0}) and Z Z 2 α 2 |x| |∇ϕ| dx = |x|α |∇(ϕ0 ψ)| dx N N R ZR Z 2 α = |x| |ψ∇ϕ0 + ϕ0 ∇ψ| dx = |x|α ψ 2 |∇ϕ0 |2 dx N N R Z Z R α +2 |x| ϕ0 ψ∇ϕ0 ∇ψdx + |x|α ϕ20 |∇ψ|2 dx RN RN Z Z (N − 2 + α)2 ϕ2 1 = |x|α ψ 2 02 dx + |x|α ∇(ψ 2 )∇(ϕ20 )dx 4 |x| 2 RN RN Z + |x|α ϕ20 |∇ψ|2 dx RN Z Z 2 (N − 2 + α)2 1 α ϕ ≥ |x| dx − ψ 2 div(|x|α ∇(ϕ20 ))dx 4 |x|2 2 RN RN Z (N − 2 + α)2 ϕ2 ≥ |x|α 2 dx. 4 |x| RN Where in the last inequality, we have used the fact that ϕ20 satisfies the equation −div(|x|α ∇u) = 0 in RN \ {0}. Hence it is equal to zero on the support of ψ. The inequality for any test function ϕ ∈ Cc∞ (RN \{0}) has been established. Now take a cut-off function η ∈ C ∞ (R), such that η(r) ≡ 0 for r ≤ 1 and η ≡ 1 for r ≥ 2. Given ϕ ∈ Cc∞ (RN ), let ηn (x) = η(n|x|) and ϕn = ηn ϕ. Then, the following inequality holds for all ϕn ∈ Cc∞ (RN ), Z Z (N − 2 + α)2 ϕ2 |x|α n2 dx ≤ |x|α |∇ϕn |2 dx. 4 |x| RN RN Since N + α > 2, letting n → ∞, we can conclude that Z Z (N − 2 + α)2 ϕ2 |x|α 2 dx ≤ |x|α |∇ϕ|2 dx, 4 |x| RN RN

for all ϕ ∈ Cc∞ (RN )

always holds for N ≥ 2 and N + α > 2. By a density argument, i.e. Cc∞ (RN ) is dense in H 1 (RN , |x|α dx), so we are done. Now we are ready to prove Proposition 2.

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Proof. By direct calculation, Z Z Qu (us ) = |x|α |∇us |2 dx − |x|γ eus u2s dx N N ZR ZR α 2 = |x| |∇us | dx − (2 − α + γ)(N − 2 + α)|x|γ |x|α−2−γ u2s dx RN RN Z Z u2 u2 (N − 2 + α)2 |x|α s2 dx − (2 − α + γ)(N − 2 + α) |x|α s2 dx ≥ 4 |x| |x| RN RN  Z  2 (N − 2 + α)2 u = − (2 − α + γ)(N − 2 + α) |x|α s2 dx, 4 |x| RN where in the last inequality, we have used Lemma 2.1. Therefore Qu (us ) ≥ 0 if and 2 only if (N −2+α) − (2 − α + γ)(N − 2 + α) ≥ 0, i.e. N ≥ 10 − 5α + 4γ. 4 3. Liouville type results. For the convenience of the readers, in the following, we denote N 0 = N + α and τ = γ − α through out this section unless otherwise explicitly specified. We will prove the following theorems: Theorem 3.1. Let τ > −2 and Ω = RN . For 2 ≤ N 0 < 10 + 4τ , there is no stable weak solution of (1.1). Theorem 3.2. Let Ω = RN , τ > −2 and 2 < N 0 < 10+4τ − where τ − = min (τ, 0). Then (1.1) has no weak solution which is stable outside a compact set K of RN . In particular, if 2 < N 0 < 10 + 4τ − , any weak solution of (1.1) in RN has infinite Morse index. The proofs of above two theorems are based on the following useful integral estimate (3.1) which is an extension of a result in [12], [19]. Proposition 3. Let Ω be a domain (bounded or not) in RN , N 0 ≥ 2 and τ > −2. Suppose u is a stable weak solution of (1.1). Then for any integer m ≥ 5 and any β ∈ (0, 2), there exists a constant C > 0 depending only on m, α, β and γ such that  2β+1 Z Z |ψ||∇ψ| γ (2β+1)u 2m α−2βτ 2 |x| e ψ dx ≤ C |x| + |∇ψ| + |ψ||∆ψ| dx, |x| Ω Ω (3.1) holds for all functions ψ ∈ Cc∞ (Ω) verifying kψk∞ ≤ 1. Proof. As u is not assumed to be bounded, eβu ϕ is not, a priori, a suitable test function for any β > 0, even with ϕ ∈ Cc∞ (Ω). Our idea is to consider suitable truncations of eβu and proceed as in [6]. Let β > 0, k ∈ N, k ≥ β −1 and  eβt , if t ≤ k, ζk (t) = eβk  t, if t ≥ k. k We choose also another Lipschitz function ηk and a differentiable function ξk such that ηk0 = ζk02 , ξk0 = ηk a.e in R. Let  β   e2βt , if t ≤ k, ηk (t) = 22βk β  e (t − k) + e2βk , if t ≥ k, k2 2

STABLE WEAK SOLUTIONS OF WEIGHTED NONLINEAR ELLIPTIC EQUATIONS

299

and  2βt  e , if t ≤ k, 4 ξk (t) = 2βk 2βk β e  e (t − k)2 + e2βk (t − k) + , if t ≥ k. 2k 2 2 4 1 1 Since u ∈ Hloc (Ω, |x|α dx), by definition, there hold ζk (u), ηk (u) ∈ Hloc (Ω, |x|α dx) ∞ for all k ∈ N, thus the function ψ = ζk (u)ϕ for ϕ ∈ Cc (Ω) can be used as a test function in the stability inequality Qu (ψ) ≥ 0. we get that

Z

|x|γ eu ζk2 (u)ϕ2 dx ≤



Z

2

|x|α |∇ (ζk (u)ϕ)| dx



Z =

2

α

2

Z

|x| |∇ (ζk (u))| ϕ dx + |x|α ζk2 (u)|∇ϕ|2 dx Ω Z  1 ζk2 (u)div |x|α ∇(ϕ2 ) dx. − 2 Ω

(3.2)



Therefore |x|γ eu ζk2 (u) ∈ L1loc (Ω). On the other hand, by using the equation (1.2), we have Z |x|α |∇(ζk (u))|2 ϕ2 dx ZΩ Z = |x|α ζk02 (u)|∇u|2 ϕ2 dx = |x|α ∇ηk (u)∇uϕ2 dx Ω Ω Z Z (3.3) α 2 α 2 = |x| ∇u∇(ηk (u)ϕ )dx − |x| ∇uηk (u)∇(ϕ )dx Ω ZΩ Z  = |x|γ eu ηk (u)ϕ2 dx + ξk (u)div |x|α ∇(ϕ2 ) dx. Ω



Combining (3.2) and (3.3), there hold Z

|x|γ eu ζk2 (u)ϕ2 dx ≤



Z

Z |x|γ eu ηk (u)ϕ2 dx + |x|α ζk2 (u)|∇ϕ|2 dx Ω Ω Z   ζk2 (u) + ξk (u) − div |x|α ∇(ϕ2 ) dx. 2 Ω

Moreover, direct calculation shows that  ηk (t) ≤

β 1 + 2 4k



ζk2 (t)

in R.

Therefore, we get  Z β 1 1− − |x|γ eu ζk2 (u)ϕ2 dx 2 4k Ω (3.4)  Z Z  ζk2 (u) α 2 2 α 2 div(|x| ∇(ϕ ))dx := I + II ≤ |x| ζk (u)|∇ϕ| dx + ξk (u) − 2 Ω Ω

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X. HUANG

Now for any ψ ∈ Cc∞ (Ω) with kψk∞ ≤ 1 in Ω, we set ϕ = ψ m , m ∈ N∗ . Using ζk (u) ≤ eβu and H¨ older’s inequality, there holds Z Z I= |x|α ζk2 (u)|∇ϕ|2 dx = m2 |x|α ζk2 (u)ψ 2(m−1) |∇ψ|2 dx Ω Ω Z   2β ≤C |x|α eu ζk2 (u) 2β+1 ψ 2(m−1) |∇ψ|2 dx Ω

Z

|x|γ eu ζk2 (u)|ψ|

≤C

(m−1)(2β+1) β

2β Z  2β+1

α−2βτ

|x|

dx

|∇ψ|

4β+2

1  2β+1

dx

.





Take m ≥ 5, so that (m − 1)(2β + 1) ≥ 2mβ for any β ∈ (0, 2). As kψk∞ ≤ 1, we obtain Z I= |x|α ζk2 (u)|∇ϕ|2 dx Ω

Z ≤C

|x|γ eu ζk2 (u)|ψ|2m dx



2β  2β+1 Z

α−2βτ

|x|

|∇ψ|

4β+2

(3.5)

1  2β+1

dx

.



Meanwhile, for any β ∈ (0, 2) there exists C > 0 depending only on β such that   2β 2β ∀s ≥ k, e 2β+1 s ≥ Ce 2β+1 k 1 + (s − k)2 2

because et ≥ 1 + t + t2 for t ≥ 0. We deduce that for any β ∈ (0, 2), there exists C > 0 (independent of k ∈ N∗ ) satisfying 2   2β ξk (u) − ζk (u) ≤ C eu ζk2 (u) 2β+1 in R. 2 As  div |x|α ∇(ϕ2 ) =2mα|x|α−2 ψ 2m−1 x∇ψ + 2m(2m − 1)|x|α ψ 2m−2 |∇ψ|2 + 2m|x|α ψ 2m−1 ∆ψ, proceeding as above (see also [19]), applying once again H¨older’s inequality, using kψk∞ ≤ 1, we get Z   ζk2 (u) α 2 div(|x| ∇(ϕ ))dx ξ (u) − k 2 Ω   Z 2β  u 2  2β+1 |ψ||∇ψ| ≤C e ζk (u) |x|α ψ 2m−2 + |∇ψ|2 + |ψ||∆ψ| dx |x| Ω 2β Z  2β+1 (3.6) ≤C |x|γ eu ζk2 (u)|ψ|2m dx × Ω

Z Ω

|x|α−2βτ



1 2β+1 ! 2β+1 |ψ||∇ψ| + |∇ψ|2 + |ψ||∆ψ| dx . |x|

Combining (3.4)–(3.6), there hold  2β+1 Z β 1 1− − |x|γ eu ζk2 (u)ψ 2m dx 2 4k Ω  2β+1 Z |ψ||∇ψ| α−2βτ 2 ≤C |x| + |∇ψ| + |ψ||∆ψ| dx. |x| Ω

STABLE WEAK SOLUTIONS OF WEIGHTED NONLINEAR ELLIPTIC EQUATIONS

301

1 Fixed β ∈ (0, 2), letting k → ∞, we can get 1− β2 − 4k ≥ δ > 0 for some positive constant δ. The proof of (3.1) is completed by the monotone convergence theorem.

Now we are ready to prove Theorem 3.1. Proof of Theorem 3.1. Suppose that (1.1) admits a stable weak solution u in RN and 2 ≤ N 0 < 10 + 4τ. Choose β ∈ (0, 2) such that N 0 − 2βτ − 2(2β + 1) < 0. For every R > 0, consider the function φR (x) = φ(R−1 |x|), where φ ∈ Cc∞ (R) satisfies 0 ≤ φ ≤ 1 in R, φ(t) = 1 if |t| ≤ 1 and φ(t) = 0 if |t| ≥ 2. Applying Proposition 3 with ψ = φR , we have Z |x|γ e(2β+1)u dx B(0,R)

Z

|x|γ e(2β+1)u φ2m R dx

≤ B(0,2R)

Z ≤C

|x|

α−2βτ



|φR ||∇φR | +|∇φR |2 + |φR ||∆φR | |x|

2β+1 dx

B(0,2R)

≤ CRN

0

−2βτ −2(2β+1)

,

∀R > 0.

Taking R → ∞, we have Z

|x|γ e(2β+1)u dx = 0,

RN

which is impossible, so we are done. For the proof of Theorem 3.2, we need the following lemma. Lemma 3.3. Suppose that (1.1) admits a weak solution u which is stable outside a compact set K ⊂ B(0, R0 ) in RN , then we have: (1) For any β ∈ (0, 2) and any r > 3R0 , it holds Z 0 |x|γ e(2β+1)u dx ≤ A + BrN −2βτ −2(2β+1) , (3.7) 2R0 0 is depending only on N 0 , β, τ, R0 , but independent of y, R. Proof. For r > 3R0 > 0, let φ and φr be as in the previous proof and define ψr = φr − φR0 . Notice that ψr ∈ Cc∞ (RN \K) and 0 ≤ ψr ≤ 1 in RN , so we can be defined text function ψr by ηR0 ∈ Cc∞ (RN ) such that 0 ≤ ηR0 ≤ 1 in B(0, 3R0 )

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X. HUANG

and ηR0 (|x|) = 0, if |x| ≤ R0 and ηR0 (|x|) = 1, if |x| ≥ 2R0 . Hence Proposition 3 (applied with m = 5 and test function ψr ) implies that for all r > 3R0 , Z Z |x|γ e(2β+1)u dx ≤ |x|γ e(2β+1)u ψr2 dx RN \K

2R0 , if α ≥ 0, 2 − ε0 2

and 2β2 + 1 ≥ 2β2 + 1 + β2 τ − ≥ θ1 :=

N0 N0 , if α < 0. > 2 − ε0 2

N Let |y| > 4R0 and R = |y| older’s inequality 4 , so B(y, 2R) ⊂ R \B(0, R0 ). By the H¨ and (3.8). If α ≥ 0, we have

Z

(|x|γ eu )θ dx

B(y,2R)

 2β2β1 +1−θ +1

 2β θ+1 



1

Z  ≤

γ (2β1 +1)u

|x| e

1

Z  dx

B(y,2R)

 0  ≤C RN −2β1 τ −2(2β1 +1)

|x|

 

2β1 γθ 2β1 +1−θ

 dx

B(y,2R) θ 2β1 +1



2β1 γθ

RN + 2β1 +1−θ

 2β2β1 +1−θ +1 1

=CRN +αθ−2θ . Meanwhile, if α < 0, it holds Z 0 (|x|γ eu )θ1 dx ≤ C1 RN +αθ1 −2θ1 ≤ C1 RN −2θ1 . B(y,2R)

STABLE WEAK SOLUTIONS OF WEIGHTED NONLINEAR ELLIPTIC EQUATIONS

That is, ∀ |y| > 4R0 and R = |y| 4 , Z (|x|γ eu )θ dx ≤ CRN +αθ−2θ , if α ≥ 0,

303

(3.9)

B(y,2R)

and

Z

(|x|γ eu )θ1 dx ≤ C1 RN

0

−2θ1

, if α < 0.

(3.10)

B(y,2R)

Now we claim lim |x|2+τ eu = 0.

(3.11)

|x|→∞

To prove that, we need the following result of Pucci and Serrin in [17], i.e. a Harnack type inequality (see also Theorem 8.17 or P. 209 in [14]): 0

N N Lemma 3.4. Let θ = 2−ε , θ1 = 2−ε for some ε0 ∈ (0, 2). Assume that nonnega0 0 N α 1 tive function η ∈ Hloc (R , |x| dx) satisfies

−div(|x|α ∇η) ≤ a(x)η

in B(y, 2R) ⊂ RN ,

where a(x) ∈ Lθ (B(y, 2R)) (resp. Lθ1 (B(y, 2R))) with α ≥ 0 (resp. α < 0), and there exist positive constants m, M such that 0 < m ≤ |x|α ≤ M in B(y, 2R). Then, for all q > 1, we have N

kηkL∞ (B(y,R)) ≤ CST R− q kηkLq (B(y,2R)) ,

(3.12)

where CST is a positive constant depending on N , q, α, M m , θ and ε0 ka(x)k Rε0 −α ka(x)kLθ (B(y,2R)) (resp. N , q, α, M , θ and R 1 Lθ1 (B(y,2R)) ). m We set ω = eλu , β1 :=

N −2+α N0 − 2 2β1 + 1 N +γ = and λ := = . 2(2 − α + γ) 2(2 + τ ) 2 2(2 + τ )

Since 2 < N 0 < 10 + 4τ , then β1 ∈ (0, 2) and N 0 − 2β1 τ − 2(2β1 + 1) = 0. In (3.7) taking β = β1 and tending R to ∞, we obtain Z |x|γ ω 2 dx < C. (3.13) |x|≥2R0

We also have −div(|x|α ∇ω) − λ|x|γ eu ω = −λ2 |x|α |∇u|2 ω ≤ 0

in D0 (RN ).

In order to apply the above results, we consider points y ∈ RN , such that |y| > N 8R0 and set R = |y| 8 . Using the estimate (3.9) and (3.10), as θ = 2−ε0 provided α ≥ 0, it holds 1

Rε0 −α kλ|x|γ eu kLθ (B(y,2R)) ≤ CλRε0 −α (RN +αθ−2θ ) θ e ε0 −α+ Nθ +α−2 = C. e = CR

(3.14)

Similarly, when α < 0, we have Rε0 kλ|x|γ eu kLθ1 (B(y,2R)) ≤ C1 λRε0 (RN

0

−2θ1

1

N0

f1 Rε0 + θ1 −2 = C f1 . ) θ1 = C

Note that for any x ∈ B(y, 2R), by the triangle inequality, we have |y| − |x − y| ≤ |x| ≤ |y| + |x − y|

(3.15)

304

X. HUANG

and thus 6R ≤ |x| ≤ 10R. This implies that 6α Rα ≤ |x|α ≤ 10α Rα ,

α ≥ 0,

if

and 10α Rα ≤ |x|α ≤ 6α Rα , if α < 0. This proves that the constant CST is independent of both y and R. Now, applying the Pucci and Serrin’s result, i.e. Lemma 3.4 with η = ω, a(x) = λ|x|γ eu and q = 2. Or we can apply Harnack inequality (see [14] P. 209) on each ball B(y, R) with α α α α |y| > 8R0 and R = |y| 8 and the λ in [14] is 6 R or 10 R . By (3.12) and (3.13), we get ω(y) ≤ CST R

−N 2

kωkL2 (B(y,2R)) ≤ CST R = o(R−

−N 2

N +γ 2

R

−γ 2

γ

k|x| 2 ωkL2 (B(y,2R))

as |y| → ∞.

)

N +γ

1

Consequently, eu(y) = ω(y) λ = o(R− 2λ ) = o(R−(2+τ ) ) as |y| → ∞. Hence the claim (3.11) holds true. To finish the proof, we consider v = v(|x|), defined as the mean value of the 2 solution u over unit spheres. Fix M1 > 0 large enough such that 2+τ − (N 0 −2)M >0 1 (recall that τ > −2). By (3.11), there exists RM1 > 0 such that −div(|x|α ∇v) = −[∇(|x|α )∇v + |x|α ∆v] = −[∇(|x|α )∇u + |x|α ∆u] 1 , ∀r ≥ RM1 . = −div(|x|α ∇u) = rγ eu ≤ M1 r2−α i.e. −

1 rN −1

(rN

0

−1 0

v (r))0 ≤

1 . M1 r2−α

Integrating from RM1 to r, we deduce that 0

rN

0

−1 0

v (r) ≥ −C −

rN −2 , M1 (N 0 − 2)

i.e.

C 1 − , ∀r ≥ RM1 . rN 0 −1 M1 (N 0 − 2)r As N 0 > 2, there exists R0 > RM1 such that 2 , ∀r ≥ R0 . v 0 (r) ≥ − M1 (N 0 − 2)r v 0 (r) ≥ −

Integrating in [R0 , r], we get r2+τ ev(r) ≥ Cr

2 2+τ − (N 0 −2)M

1

,

∀r ≥ R0 .

By Jensen’s inequality, we have sup (|x|2+τ eu(x) ) = r2+τ sup (eu(x) ) ≥ r2+τ ev(r) ≥ Cr |x|=r

2 2+τ − (N 0 −2)M

1

→ ∞, as r → ∞,

|x|=r

which clearly contradicts (3.11), so the proof is completed. Acknowledgments. I would like to thank Professors Dong Ye and Feng Zhou for carefully guiding me into this problem. I am grateful to the anonymous referees for a careful reading and valuable comments and many helpful remarks concerning the exposition of the results presented here.

STABLE WEAK SOLUTIONS OF WEIGHTED NONLINEAR ELLIPTIC EQUATIONS

305

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Received December 2012; revised April 2013. E-mail address: [email protected]

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