Radial symmetry and asymptotic behaviors of positive solutions for certain nonlinear integral equations

J. Math. Anal. Appl. 427 (2015) 307–319

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Radial symmetry and asymptotic behaviors of positive solutions for certain nonlinear integral equations ✩ Jiankai Xu a,c,∗ , Huoxiong Wu b , Zhong Tan b a b c

College of Sciences, Hunan Agriculture University, Hunan, 410128, China School of Mathematical Sciences, Xiamen University, Fujian, Xiamen 361005, China Lab of Granular computing, Minnan Normal University, Zhangzhou, Fujian 363000, China

a r t i c l e

i n f o

Article history: Received 30 October 2013 Available online 17 February 2015 Submitted by D. Wang Keywords: Integral equations Radially symmetric solutions Asymptotic behaviors Moving plane method Hardy–Littlewood–Sobolev inequality

a b s t r a c t In this paper, we investigate some properties of positive solutions for a nonlinear integral system. Using the method of moving planes in an integral form, which was recently introduced by Chen, Li, and Ou in Chen et al. (2005, 2006) [2,3], we show that all positive solutions of the integral system in certain functional spaces are radially symmetric with respect to the origin and decreasing in critical space. Furthermore, with the help of regularity lifting lemma, we obtain the optimal integrable intervals and the asymptotic estimates for such solutions around the origin and near infinity. © 2015 Elsevier Inc. All rights reserved.

1. Introduction In this paper, we consider the following non-linear integral system involving the double weighted Riesz potentials: ⎧  v p (y)uq (y) 1 ⎪ ⎪ dy, u(x) = ⎪ ⎪ |x|α |x − y|λ |y|β ⎨ n R up (y)v q (y) 1 ⎪ ⎪ dy, ⎪ v(x) = ⎪ ⎩ |x|α |x − y|λ |y|β

(1.1)

Rn

where α > 0, β > 0, 0 < λ < n with λ  λ + α + β < n, and p, q > 0 satisfying p + q = (2n − λ − 2β)/ (2α + λ). ✩ Supported by the National Natural Science Foundation of China (Nos. 11271305, 11371295, 11471041, 11126148), the Scientific Research Foundation of Hunan Provincial Education Department (No. 13C395) and the Young Foundation of Hunan Agriculture University (No. 12QN08). * Corresponding author. Fax: +86 0731 84618071. E-mail address: [email protected] (J. Xu).

http://dx.doi.org/10.1016/j.jmaa.2015.02.043 0022-247X/© 2015 Elsevier Inc. All rights reserved.

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308

In recent years, there has been tremendous interest in studying integral equation(s), which are equivalent to some partial differential equation(s) in Rn and also provide a special skill to investigate the global properties of corresponding differential equation(s). We recount some related investigations as follows. As 0 < β < n − λ, α = 0, p + q = 2(n − β)/λ − 1 and u(x) = v(x), the system (1.1) can be reduced to the following single equation: 

up+q (y) . |x − y|λ |y|β

u(x) = Rn

(1.2)

G. Lu and J. Zhu in [12] obtained some regularity results and showed that every positive solution u(x) is radially symmetric about the origin and strictly decreasing in L2n/λ (Rn ). Subsequently, under the same conditions, Y. Lei in [6] showed that for 0 < p + q ≤ (n − β)/λ, Eq. (1.2) has no positive solution and that for p + q > (n − β)/λ, the positive solution of (1.2), u ∈ L2n/λ (Rn ) is bounded and decays fast with rate |x|−λ . When α = β = 0 and λ = n − 2, the system (1.1) can be rewritten as ⎧  ⎪ ⎪ u(x) = ⎪ ⎪ ⎨ Rn ⎪ ⎪ v(x) = ⎪ ⎪ ⎩

v p (y)uq (y) dy , |x − y|n−2 (1.3)

up (y)v q (y) . |x − y|n−2

Rn

Under some integrability condition, (1.3) is equivalent to the following differential equations 

−u(x) = v p (x)uq (x), −v(x) = up (x)v q (x),

(1.4)

which are closely related to the stationary Schrödinger system with critical exponents for the Bose Einstein condensate. C. Li and L. Ma in [10] showed that for n ≥ 3, 1 ≤ p, q ≤ (n + 2)/(n − 2) and p + q = (n + 2)/ (n − 2), any positive solution pair (u, v) of system (1.5) in L2n/(n−2) (Rn ) × L2n/(n−2) (Rn ) is radially symmetric and unique. Recently, Y. Zhao and Y. Lei in [13] considered the following nonlinear system: ⎧  ⎪ ⎪ u(x) = ⎪ ⎪ ⎨ Rn ⎪ ⎪ v(x) = ⎪ ⎪ ⎩ Rn

v p (y)uq (y) dy |x − y|λ |y|β (1.5)

up (y)v q (y) dy , |x − y|λ |y|β

which is the special case of (1.1) for α = 0. The authors in [13] showed that as λ > β ≥ 0, λ + β < n and 0 < p, q satisfying p + q = (2n − λ − β)/(λ + β), any positive solution pair (u, v) of system (1.5) in L2n/(λ+β) (Rn ) × L2n/(λ+β) (Rn ) is symmetric about the origin and monotone decreasing. Additionally, if (n − λ)(λ + β) ≥ 2nβ, then the positive solution pair (u, v) of (1.5) is bounded and satisfies that  lim |x|λ u(x) =

|x|→∞

Rn

uq (y)v p (y) dy, |y|β

 lim |x|λ v(x) =

|x|→∞

Rn

up (y)v q (y) dy. |y|β

(1.6)

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Another similar integral system is ⎧  v q (y) 1 ⎪ ⎪ dy , u(x) = ⎪ α ⎪ |x| |x − y|λ |y|β ⎨ R n up (y) 1 ⎪ ⎪ dy . v(x) = ⎪ ⎪ ⎩ |x|β |x − y|λ |y|α

(1.7)

Rn

It is closely related to the best constant in the weighted Hardy–Littlewood–Sobolev inequality (see Lemma 2.1 below). The properties of positive solutions for system (1.7) have been well studied and the interested readers can consult [4,5,7–9], among numerous references, for more information. In this paper, we will prove that the pair of solutions (u, v) in (1.1) is radially symmetric about the origin and monotone decreasing in Ls (Rn ) × Ls (Rn ) with s = n(p + q − 1)/(n − λ). Moreover, we will use the regularity lifting lemma to obtain the optimal integrable intervals of the solutions. Also, by the integrability result, we will establish the asymptotic estimates of the positive solutions around the origin and near infinity. Our main results can be formulated as follows: Theorem 1.1. Assume that s = n(p + q − 1)/(n − λ), (u(x), v(x)) ∈ Ls (Rn ) × Ls (Rn ) is a pair of positive solutions of (1.1). Then (u, v) is radially symmetric and monotone decreasing about the origin. Theorem 1.2. For s = n(p + q − 1)/(n − λ), assume that u(x), v(x) ∈ Ls (Rn ) are the solutions of (1.1). Then u(x), v(x) ∈ Lr (Rn ),

∀ r ∈ (n/(α + λ), n/α),

(1.8)

where the integral interval of u(x), v(x) is optimal, i.e., ur , vr = ∞, if r ∈ / (n/(α + λ), n/α).

(1.9)

Furthermore, we have the following asymptotic estimates of u(x), v(x) at origin and infinity, respectively,  lim |x|α u(x) =

|x|→0

Rn



uq (y)v p (y) dy, |y|β+λ

lim |x|α+λ u(x) =

|x|→∞

Rn

uq (y)v p (y) dy; |y|β

(1.10)

v q (y)up (y) dy. |y|β

(1.11)

and  lim |x|α v(x) =

|x|→0

Rn



v q (y)up (y) dy, |y|β+λ

lim |x|α+λ v(x) =

|x|→∞

Rn

Remark 1.1. Comparing Theorem 1.1 with the results of (1.5) and (1.7) in [13] and [4], the pair of solutions has the same radial symmetry and is decreasing about the origin. But, the integrable intervals and asymptotic behaviors of solutions for the system (1.1) are different from those of (1.5) and (1.7). At first, the system (1.1) is distinct from the system (1.7), even if for α = β. The different structure leads to Kelvin transform used in [7–9] to obtain the decay estimate of (1.7) that is invalid for the system (1.1). Therefore, we have to look for a new way to obtain the asymptotic result. Precisely, the difference between systems (1.1) and (1.7) leads to the thoroughly different asymptotic behavior at the origin as well as at infinity. Indeed, by [7–9], we have learned that for system (1.7), when α + β ≥ 0, the pair of solutions (u, v) has the following asymptotic behaviors at the origin: u(x) 

A0 , |x|α

if

λ + (q + 1)β < n,

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and ⎧ A 1 ⎪ , ⎪ ⎪ β ⎪ |x| ⎪ ⎨ A | ln |x|| 2 , v(x)  ⎪ |x|β ⎪ ⎪ ⎪ A3 ⎪ ⎩ , |x|α(p+1)+β+λ−n

if λ + α(p + 1) < n, if λ + α(p + 1) = n,

(1.12)

if λ + α(p + 1) < n.

At the same time, (u, v) at the infinity admits the following asymptotic behaviors: u(x) 

B0 , |x|λ+α

if λq + β(q + 1) > n,

and ⎧ B 1 ⎪ , ⎪ ⎪ β+λ ⎪ |x| ⎪ ⎨ B | ln |x|| 2 , v(x)  β+λ ⎪ |x| ⎪ ⎪ ⎪ B3 ⎪ ⎩ , |x|(α+λ)(p+1)+β−n Here, we use the notation w(x)  C/|x|t to denote that

if λp + α(p + 1) > n, if p + α(p + 1) = n,

(1.13)

if p + α(p + 1) < n. lim

x→0 or ∞

|x|t w(x) = C for a function w(x), a real

number t and non-zero real number C. However, for the system (1.1), we know from (1.10) and (1.11) that the pair of solutions (u, v) has the same decay estimate at the origin as well as at infinity. This implies that the pair of solutions (u, v) in (1.1) is more closed than those of system (1.7). The other significant difference between (1.1) and (1.7) is the corresponding optimal integrable intervals of solutions. From (1.8) and (1.9), we know that the pair of solutions (u, v) in (1.1) admits the same optimal integrable intervals. However, for the pair of solutions (u, v) in system (1.7), the optimal integrable interval of u is different from one of v (see [5] for the details). Remark 1.2. We remark that the asymptotic estimates in (1.6) for the system (1.5) obtained in [13] can be regarded as the special cases of our results in (1.10) and (1.11) for α = 0 with that |x| → ∞. On the other hand, although the system (1.5) is the special case of the system (1.1) for α = 0, the solution space Ls (Rn ) × Ls (Rn ) with s = n(p + q − 1)/(n − λ) = 2n/(2α + λ) in our theorems is different from Ls (Rn ) × Ls (Rn ) with s = 2n/(λ + β) in [13] since the index p + q = (2n − λ − 2β)/(2α + λ) in this paper, even if α = 0, is different from one p + q = (2n − λ − β)/(λ + β) in [13]. Moreover, the positive solutions considered in this paper are singular, and the positive solutions considered in [13] are bounded. Therefore, our results are new and interesting, even if for α = 0. Throughout the rest of this paper, we always use the letter C, sometimes with certain parameters, to denote positive constants that may vary at each occurrence but are independent of the essential variables. 2. Preliminary In this section, we recall some standard ingredients, which will be used in the proofs of our main results. Lemma 2.1 (Weighted Hardy–Littlewood–Sobolev inequality). (Cf. [11].) Let 1 < r, s < ∞, 0 < λ < n, α + β ≥ 0, 1/r + 1/s + (λ + α + β)/n = 2 and 1 − 1/r − λ/n ≤ α/n < 1 − 1/r. Then

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311

       −α −λ −β   ≤ C(λ, α, β, n)f r gs |x| f (x) g(y)|x − y| |y| dxdy    n n 

(2.1)

R R

for all f ∈ Ls (Rn ), g ∈ Lr (Rn ). The weighted Hardy–Littlewood–Sobolev inequality can also be written in the following equivalent form: Rλ,α,β (g)p ≤ C(n, α, λ, β, p)g

np n+(n−λ)p

,

(2.2)

where p satisfies α/n < 1/p < min {λ/n, (α + λ)/n} and Rλ,α,β (g)(x) is given by Rλ,α,β (g)(x) 

1 |x|α

 Rn

g(y) dy. |x − y|λ |y|β

(2.3)

Lemma 2.2. (Cf. [1].) Let X and Y be both Banach spaces with norm  · X and  · Y , respectively. The subspace of X and Y, Z = X ∩ Y is endowed with a new norm by  · Z =

p  · pX +  · pY ,

p ∈ [1, ∞].

(2.4)

Suppose that T is a contraction map from Banach space X into itself and from Banach space Y into itself. If f ∈ X and there exists a function g ∈ Z = X ∩ Y such that f = T f + g, then f ∈ Z. 3. Proof of Theorem 1.1 In this section, we will apply the method of moving plane in integral forms recently introduced by Chen et al. in [2,3] to obtain the radial symmetry of positive solutions of system (1.1). For convenience, we set Σμ  {x = (x1 , . . . , xn ) ∈ Rn : x1 ≥ μ}, xμ  (2μ − x1 , x2 , . . . , xn ), uμ (x)  u(xμ ) and vμ (x)  v(xμ ) for a given real number μ ∈ R. For any pair of solutions (u, v) of system (1.1), it is easy to check that 

 −

uμ (x) − u(x) = Σμ

1

1 1 1 1 1 − α |xμ |α |xμ − y|λ |y μ |β |x| |x − y|λ |y μ |β

(vμp uqμ − v p uq )

1 1 − μ β v p uq + |y|β |y |  1 1 1 1 1 1 vμp uqμ + − α μ α λ μ β λ μ |x | |x − y| |y | |x| |x − y| |y |β

 1 1 1 1 1 1 p q vμ uμ dy + − α μ |xμ |α |xμ − y|λ |y μ |β |x| |x − y|λ |y μ |β 

1 1 1 − α |xμ |α |xμ − y|λ |x| |x − y|λ 1



 A1 + A2 + A3 + A4 ,

(3.1)

and 

 −

vμ (x) − v(x) = Σμ

+



1

1 1 1 1 1 − α |xμ |α |xμ − y|λ |y μ |β |x| |x − y|λ |y μ |β

1 1 1 − α μ α μ λ |x | |x − y| |x| |x − y|λ 1





1 1 − μβ β |y| |y |

(upμ vμq − up v q ) up v q

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1 1 1 1 1 1 − α up v q |xμ |α |x − y|λ |y μ |β |x| |x − y|λ |y μ |β μ μ

 1 1 1 1 1 1 p q u dy + − v μ μ |xμ |α |xμ − y|λ |y μ |β |x|α |xμ − y|λ |y μ |β +(

 B1 + B2 + B3 + B4 .

(3.2)

The proof of Theorem 1.1 will be made up of two steps. In Step 1, we compare the values of u(x) with uμ (x) and v(x) with vμ (x) on Σμ , respectively, and show that for sufficiently negative μ < 0, there holds u(x) ≥ uμ (x) and v(x) ≥ vμ (x),

∀ x ∈ Σμ − {0}.

(3.3)

In Step 2, we continuously move the plane x1 = μ along x1 direction from near negative infinity to the right as long as (3.3) holds. By moving this plane in this way, we finally show that the plane will stop at the origin. Now we turn our attention to Step 1. Step 1. Since A2 , A3 , A4 ≤ 0 and |y μ | > |y| for any y ∈ Σμ , we have   uμ (x) − u(x) ≤ Σμ

 

≤ Σu μ

1 1 1 1 − μα μ |x|α |x − y|λ |x | |x − y|λ 1 1 1 1 − μα μ |x|α |x − y|λ |x | |x − y|λ

 

+ Σv μ





1 (v p uq − v p uq )dy |y μ |β μ μ 1 (v p [uq − uq ])dy |y|β μ μ

1 1 1 1 − μα μ |x|α |x − y|λ |x | |x − y|λ



1 (uq [vμp − v p ])dy |y|β

 A11 (x) + A12 (x),

(3.4)

where Σuμ = {x ∈ Σμ | u(x) < uμ (x)} and Σvμ = {x ∈ Σμ | v(x) < vμ (x)}. Noting that p + q = (2n − λ − 2β)/(2α + λ) and s  (n(p + q − 1))/(n − λ) ∈ (n/[α + λ], n/α), by Lemma 2.1, we have A11 Ls (Σuμ ) ≤ C(n, p, q)vμp uq−1 [uμ − u] μ

ns

¯ L n+(n−λ)s (Σu μ)

≤ C(n, p, q)vμp uq−1 μ 

n

¯ L n−λ (Σu μ)

uμ − uLs (Σuμ )

s u ≤ C(n, p, q)vμ pLs (Σu ) uμ q−1 Ls (Σu ) uμ − uL (Σμ ) , μ

μ

(3.5)

and A12 Ls (Σuμ ) ≤ C(n, p, q)uq [vμp − v p ]Ls (Σvμ ) s v ≤ C(n, p, q)uqLs (Σv ) vμ p−1 Ls (Σv ) vμ − vL (Σμ ) . μ

μ

(3.6)

This combining with (3.5), implies that s u uμ − uLs (Σuμ ) ≤ C(n, p, q)vμ pLs (Σu ) uμ q−1 Ls (Σu ) uμ − uL (Σμ ) μ

+

μ

C(n, p, q)uqLs (Σv ) vμ p−1 vμ Ls (Σv μ μ)

− vLs (Σvμ ) .

(3.7)

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On the other hand, for u, v ∈ Ls (Rn ), we can choose N > 0 large enough, such that for any μ ≤ −N < 0, C(n, p, q)vμ pLs (Σu ) uμ q−1 uμ − uLs (Σuμ ) ≤ 1/4 and C(n, p, q)uqLs (Σv ) vμ p−1 ≤ 1/4, which comLs (Σu Ls (Σv μ μ) μ μ) bining with (3.7) implies that uμ − uLs (Σuμ ) ≤

1 1 uμ − uLs (Σuμ ) + vμ − vLs (Σvμ ) . 4 4

(3.8)

vμ − vLs (Σvμ ) ≤

1 1 uμ − uLs (Σuμ ) + vμ − vLs (Σvμ ) . 4 4

(3.9)

Similarly, we have

This together with (3.8) leads to that uμ − uLs (Σuμ ) = vμ − vLs (Σvμ ) = 0. Therefore Σuμ and Σvμ must be two zero-measure sets, which completes Step 1. Step 2. We continuously move x1 = μ to the right as long as (3.3) holds. Indeed, suppose that at x1 = μ0 < 0, we have, for any x ∈ Σμ0 − {0} u(x) ≥ uμ0 (x) and v(x) ≥ vμ0 (x), but u(x) ≡ uμ0 (x) or v(x) ≡ vμ0 (x). Next, we will show that the plane can be moved further to the right. Precisely, there exists an  depending on n, α, λ, β and the solution (u(x), v(x)) itself such that u(x) ≥ uμ (x) and v(x) ≥ vμ (x),

∀ x ∈ Σμ − {0},

μ ∈ [μ0 , μ0 + ).

(3.10)

Under the assumption that v(x) ≡ vμ0 (x) on Σμ0 , by (3.1) and the non-negativity of u, v, we have u(x) > uμ0 (x) in the interior of Σμ0 − {0}. Let  u0 = {x ∈ Σμ0 | u(x) ≤ uμ0 (x)} and Σ  v 0 = {x ∈ Σμ0 |v(x) ≤ vμ0 (x)}. Σ μ μ  u0 is a zero measure set in Rn . Similarly, From the analysis mentioned above, it is easy to check that the Σ μ  v 0 } = 0. This together with (3.7) and the integrability conditions u, v ∈ Ls (Rn ) ensures we also have m{Σ μ

that one can choose  small enough such that for all μ in [μ0 , μ0 + ) 1 , 4 1 ≤ , 4

1 ; 4 1 ≤ . 4

C(n, p, q)vμ pLs (Σu ) uμ q−1 Ls (Σu ) ≤

C(n, p, q)uqLs (Σv ) vμ p−1 Ls (Σv ) ≤

C(n, p, q)uμ pLs (Σv ) vμ q−1 Ls (Σv )

C(n, p, q)vqLs (Σu ) uμ p−1 Ls (Σu )

μ

μ

μ

μ

μ

μ

μ

μ

(3.11)

So (3.8) and (3.9) hold. Thus we also have uμ (x) − u(x)Ls (Σuμ ) = 0 and vμ (x) − v(x)Ls (Σvμ ) = 0,

(3.12)

 u and Σ  v must be zero. This verifies (3.10). Finally, we show that the which implies that the measures of Σ μ μ plane can’t stop before hitting the origin. On the contrary, if the plane stops at x1 = μ0 < 0, then u(x) and v(x) must be symmetric about the plane x1 = μ0 , i.e.,

J. Xu et al. / J. Math. Anal. Appl. 427 (2015) 307–319

314

u(x) = uμ0 (x) and v(x) = vμ0 (x),

∀x ∈ Σμ0 .

(3.13)

0

On the other hand, noting that |xμ | > |x| for any x ∈ Σμ0 , we have 1 u(x) − uμ0 (x) = |x|α  ≥ Rn

Rn

1 v p (y)uq (y) 1 dy − μ0 α λ β |x − y| |y| |x | 



 Rn

v p (y)uq (y) 1 dy |xμ0 − y|λ |y|β

1 v (y)uq (y) 1 1 − dy |x|α |x − y|λ |y|β |xμ0 − y|λ

1 = |x|α >



1 |x|α

  Σμ0

 

Σμ0

1 1 − μ0 λ |x − y| |x − y|λ

p



p q v p (y)uq (y) vμ0 (y)uμ0 (y) − 0 |y|β |y μ |β

 dy

 p v (y)uq (y) − vμp 0 (y)uqμ0 (y) 1 1 − dy |x − y|λ |y|β |xμ0 − y|λ

= 0,

(3.14)

which obviously contradicts with (3.13). Since the direction is arbitrary, we derive that u and v are radially symmetric about the origin and decreasing. This completes the proof of Theorem 1.1. 4. Proof of Theorem 1.2 This section is devoted to proving Theorem 1.2. We first give some prior estimates. Proposition 4.1. Assume that u, v ∈ Ls (Rn ), s = n(p + q − 1)/(n − λ), is a pair of positive solutions of weighted integral system (1.1). Then there exists a positive bounded function R(x) satisfying w(x) =

R(x) |x|α



wp+q (y) 1 dy, |x − y|λ |y|β

Rn

(4.1)

where w(x) = u(x) + v(x). Proof. Noting that there exists a constant C > 0 such that up (y)v q (y) + v p (y)uq (y) ≤ Cwp+q (y). Therefore, C w(x) ≤ |x|α

 Rn

wp+q (y) 1 dy, |x − y|λ |y|β

which, taking R(x) =

1 |x|α



w(x) wp+q (y) 1 dy Rn |x−y|λ |y|β

,

implies that (4.1), and 0 < R(x) ≤ C, Proposition 4.1 is proved. 2

∀ x ∈ Rn .

(4.2)

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Proposition 4.2. Assume that w(x) ∈ Ls (Rn ), s = n(p + q − 1)/(n − λ), and satisfies (4.1) with (4.2). Then w(x) ∈ Lr (Rn ),

∀ r ∈ (n/(α + λ), n/α).

Proof. For a given positive number A > 0, define R(x) T f (x)  |x|α

 Rn

p+q−1 wA (y) f (y) dy, |x − y|λ |y|β

where wA (x) is given by  wA (x) =

w(x), if w(x) ≥ A, or |x| ≥ A; 0, otherwise.

Write R(x) F (x)  |x|α

 Rn

(w(y) − wA (y))p+q 1 dy. |x − y|λ |y|β

Then w(x) is a solution of the following equation: f (x) = T f (x) + F (x). We first claim that T is a contraction map from Lr (Rn ) into itself, provided r satisfies α/n < 1/r < (α + λ)/n. By Lemma 2.1 and Hölder’s inequality, we conclude that p+q−1 nr T f r ≤ wA f  n+(n−λ)r ≤ wA p+q−1 n(p+q−1) f r ,

∀ r ∈ (n/(α + λ), n/α).

(4.3)

n−λ

Since w(x) ∈ Ls (Rn ), there exists a positive number A large enough satisfying wA  n(p+q−1) ≤ n−λ

1 , 2

(4.4)

which together with T being a linear operator, implies that the claim holds. Similarly, we have F r ≤ Cw − wA p+q nr(p+q) . n+(n−λ)r

Noting that s = (n(p + q − 1))/(n − λ) ∈ (n/[α + λ], n/α) and w ∈ Ls (Rn ). Therefore, taking X = Ls (Rn ), Y = Lt (Rn ) for any t ∈ (n/(α + λ), n/α) and Z = Ls (Rn ) ∩ Lt (Rn ) in Lemma 2.2, we obtain w(x) ∈ Lr (Rn ),

∀ r ∈ (n/(α + λ), n/α).

(4.5)

This completes the proof of Proposition 4.2. 2 Now we are in the position to prove Theorem 1.2. Proof of Theorem 1.2. By Proposition 4.2, it is easy to obtain (1.8). Next, we turn to (1.9) and (1.10). At first, we will show that the following improper integrals are convergent

J. Xu et al. / J. Math. Anal. Appl. 427 (2015) 307–319

316

 D0  Rn



D∞  Rn

uq (y)v p (y) dy, E0  |y|β+λ



v q (y)up (y) dy, |y|β+λ

Rn

uq (y)v p (y) dy, E∞  |y|β



Rn

(4.6)

v q (y)up (y) dy. |y|β

(4.7)

Since 1 − [(α + λ)(p + q)]/n < β/n < 1, there exists a positive number r > 1 such that 1−

1 α(p + q) 1 β (α + λ)(p + q) < <1− and < < 1. n r n r n

This together with Hölder’s inequality and Proposition 4.2, implies that  |y|>1

wp+q (y) dy ≤ |y|λ+β



wp+q (y) dy |y|β

|y|>1

⎛

⎞1/r ⎛



⎜ ≤⎝

⎟ |y|−βr dy ⎠

⎜ ⎝

|y|>1

⎞1/r



 ⎟ w(p+q)r (y)dy ⎠

< ∞,

(4.8)

|y|>1

where 1/r + 1/r = 1. Therefore, as |x| < 1 and |y| ≥ 2, we have C u(x) ≥ |x|α

 |y|>2

C(n, p, q) uq (y)v p (y) dy ≥ , |y|β+λ |x|α

(4.9)

and ⎛



⎜ uι ≥ ⎝

⎞1/ι ⎟ |x|−αι dx⎠

= ∞, ∀ ι ≥ n/α.

(4.10)

|x|<1

Similarly, noting that 1 − α(p + q)/n > (λ + β)/n, we can find a positive number r > 1 such that 1−

1 α(p + q) 1 λ+β (α + λ)(p + q) < <1− and > . n r n r n

(4.11)

Thus,  |y|≤1

wp+q (y) dy ≤ |y|β

 |y|≤1

⎛ ⎜ ≤⎝

wp+q (y) dy |y|λ+β



⎞1/r ⎛ ⎟ |y|−(λ+β)r dy ⎠

|y|≤1

⎜ ⎝



⎞1/r  ⎟ wr (p+q) (y)dy ⎠

< ∞.

(4.12)

|y|≤1

Hence, as |x| > 2 and |y| < 1, we conclude that C(p, q, n, α, β, λ) u(x) ≥ |x|α+λ

 |y|<1

C(p, q, n, α, β, λ) v p (y)uq (y)dy ≥ , β |y| |x|α+λ

(4.13)

J. Xu et al. / J. Math. Anal. Appl. 427 (2015) 307–319

317

which implies that uι = ∞,

∀ ι ≤ n/α + λ.

(4.14)

Combining this with (4.8), (4.10), (4.12) and (4.13), we complete the proofs of the convergence for the improper integrals in (4.6) and (4.7), and the conclusion of (1.9) for u(x). Similarly, we can get the conclusion for v(x) in (1.9). Therefore, to show Theorem 1.2, it suffices to prove (1.10) and (1.11). For fixed R > 0, we define L1 , L2 , L3 , respectively, as follows: 

v p (y)uq (y) |x|λ dy, |y|β |x − y|λ

L1 (x) 

(4.15)

BR (0)



v p (y)uq (y) |x|λ dy, |y|β |x − y|λ

L2 (x)  (Rn \B

(4.16)

R (0))\B|x|/2 (x)

and 

v p (y)uq (y) |x|λ dy. |y|β |x − y|λ

L3 (x) 

(4.17)

B |x| (x) 2

Observe that when y ∈ BR (0) and |x| > 2R, we have    v p (y)uq (y) v p (y)uq (y)  |x|λ − 1 ≤ C(λ) ∈ L1 (Rn ).  β λ |y| |x − y| |y|β By Lebesgue’s dominated convergence theorem, we obtain 

v p (y)uq (y) |y|β

lim

|x|→∞ BR (0)

which implies that lim



|x|λ − 1 dy = 0, |x − y|λ

(4.18)

lim L1 (x) = D∞ .

R→∞ |x|→∞

To obtain the second equality of (1.10), it suffices to prove lim L2 (x) = 0 and lim L3 (x) = 0. Noting |x|→∞

that y ∈ (Rn \ BR (0)) \ B|x|/2 (x) and |x| > 2R, by (4.7), we have  v p (y)uq (y) |L2 (x)| ≤ C(λ) dy → 0, as R → ∞. |y|β Rn \B

|x|→∞

(4.19)

R (0)

Next, we turn to L3 (x). By Theorem 1.1 and Proposition 4.2, we conclude that, for any s ∈ (n/(α+λ), n/α), 

 dy ≤

ws (|x|) |x|/2≤|y|≤|x|

ws (y)dy ≤ wss ≤ C < ∞, |x|/2≤|y|≤|x|

which implies that w(|x|)|x|n/s ≤ C for |x| ≥ 2R and L3 (x) ≤ C|x|

λ−β

w

p+q



(|x|/2) B|x|/2 (x)

≤ C(n)|x|n−β wp+q (|x|/2)

1 dy |x − y|λ

(4.20)

J. Xu et al. / J. Math. Anal. Appl. 427 (2015) 307–319

318

= C(n) [(|x|/2)n/s w(|x|/2)]p+q |x|n−β−n(p+q)/s ≤ C(n) |x|n−β−n(p+q)/s .

(4.21)

Noting that n−β−n

(α + β + λ − n)λ α+λ (p + q) = < 0, n 2α + λ

and taking 1/s = (α+λ)/n −ε with ε being a small positive number, we conclude that n −β −[(p +q)n]/s < 0 and lim L3 (x) = 0.

(4.22)

|x|→∞

Now, we are in a position to show the first equality of (1.10). We define J1 (x) and J2 (x) as follows: 

uq (y)v p (y) dy, |x − y|λ |y|β

J1 (x)  Rn \B

(4.23)

R (0)

and  J2 (x) 

uq (y)v p (y) dy. |x − y|λ |y|β

(4.24)

BR (0)

Noting that |x| < R/2 and |y| > R, we have    1 1 1    |x − y|λ − |y|λ  ≤ C(λ) |y|λ , and  Rn \B

uq (y)v p (y) |y|β



1 1 − λ λ |x − y| |y|



 dy ≤ C(λ) Rn \B

R (0)

uq (y)v p (y) dy. |y|β+λ

(4.25)

R (0)

This together with (4.6) and Lebesgue’s dominated convergence, implies that  lim lim J1 (x) =

R→0 |x|→0

Rn

uq (y)v p (y) dy. |y|λ+β

(4.26)

Next, we turn our attention to J2 (x). We claim that lim J2 (x) = 0. By Hölder’s inequality, we have, for |x|→0

|x| < R, 

w(p+q)λ/(λ+β) (y) w(q+p)β/(λ+β) (y) dy |x − y|λ |y|β

J2 (x) ≤ BR (0)

⎛ ⎜ ≤⎝



BR (0)



w

(q+p)β/(λ+β)

|y|β

(y)

 β+λ β

⎞β/(λ+β) ⎟ dy ⎠

J. Xu et al. / J. Math. Anal. Appl. 427 (2015) 307–319

⎛ ⎜ ×⎝





w

(q+p)λ/(λ+β)

|x −

(y)

(β+λ)/λ

y|λ

319

⎞λ/(λ+β) ⎟ dy ⎠

BR (0)

 J2,1 (x) × J2,2 (x).

(4.27)

By (4.8) and (4.12), we have ⎛ ⎜ J2,1 (x) = ⎝





(q+p)

(y) w |y|β+λ



⎞β/(λ+β) ⎟ dy ⎠

→ 0, as R → 0.

(4.28)

BR (0)

On the other hand, by Hölder’s inequality, we deduce that (J2,2 (x))

λ+β λ



 =

w(q+p) (y) |x − y|β+λ

 dy

BR (0)

⎛ ⎜ ≤ ⎝



⎞1/r ⎟ |x − y|−(λ+β)r dy ⎠

× wp+q (p+q)r 

B2R (0)

≤ C(n, r, p, q)R r −(λ+β) → 0, n

as R → 0

(4.29)

since r satisfies (4.11). This implies that the claim above holds. Then, combining with (4.23)–(4.29), we obtain the first equality of (1.10). Similarly, we can get (1.11) and complete the proof of Theorem 1.2. 2 References [1] W. Chen, C. Li, Methods on Nonlinear Elliptic Equations, AIMS Ser. Differ. Equ. Dyn. Syst., vol. 4, 2010. [2] W. Chen, C. Li, B. Ou, Classification of solutions for a system of integral equation, Comm. Partial Differential Equations 30 (2005) 59–65. [3] W. Chen, C. Li, B. Ou, Classification of solutions for an integral equation, Comm. Pure Appl. Math. 54 (2006) 330–343. [4] C. Jin, C. Li, Symmetry of solutions to some systems of integral equations, Proc. Amer. Math. Soc. 134 (2006) 1661–1670. [5] C. Jin, C. Li, Qualitative analysis of some systems of equations, Calc. Var. Partial Differential Equations 26 (2006) 447–457. [6] Y. Lei, Asymptotic properties of positive solutions of the Hardy–Sobolev type equations, J. Differential Equations 254 (2013) 1774–1799. [7] Y. Lei, C. Li, C. Ma, Asymptotic radial symmetry and growth estimates of positive solutions to weighted Hardy– Littlewood–Sobolev system of integral equations, Calc. Var. Partial Differential Equations 45 (2012) 43–61. [8] Y. Lei, C. Ma, Asymptotic behavior for solutions of some integral equations, Commun. Pure Appl. Anal. 10 (2011) 193–207. [9] C. Li, J. Lim, The singularity analysis of solutions to some integral equations, Commun. Pure Appl. Anal. 6 (2007) 453–464. [10] C. Li, L. Ma, Uniqueness of positive bound states to Schrödinger systems with critical exponents, SIAM J. Math. Anal. 40 (3) (2008) 1049–1057. [11] E.H. Lieb, M. Loss, Analysis, 2nd edition, Graduate Studies in Mathematics, vol. 14, Amer. Math. Soc., Providence, RI, 2001. [12] G. Lu, J. Zhu, Symmetry and regularity of extremals of an integral equation related to the Hardy–Sobolev inequality, Calc. Var. Partial Differential Equations 42 (2011) 563–577. [13] Y. Zhao, Y. Lei, Asymptotic behavior of positive solutions of a nonlinear integral system, Nonlinear Anal. 75 (2012) 1989–1999.