Nonexistence of Positive Solutions for an Integral Equation Related to the Hardy-Sobolev Inequality

Acta Appl MathDOI 10.1007/s10440-014-9878-z

Nonexistence of Positive Solutions for an IntegralEquation Related to the Hardy-Sobolev Inequality

Dongyan Li · Pengcheng Niu · Ran Zhuo

Received: 15 April 2013 / Accepted: 11 March 2014© Springer Science+Business Media Dordrecht 2014

Abstract Let α and s be real numbers satisfying 0 < s < α < n. We are concerned with theintegral equation

u(x) =∫

Rn

up(y)

|x − y|n−α|y|s dy,

where n−sn−α

< p < α∗(s) − 1 with α∗(s) = 2(n−s)

n−α. We prove the nonexistence of positive

solutions for the equation and establish the equivalence between the above integral equationand the following partial differential equation

(−�)α2 u(x) = |x|−sup.

Keywords Method of moving planes in integral forms · Symmetry · Nonexistence ·Equivalence · Singular integral equations and PDEs

The third author is supported by China Scholarship Council (No. 201206060010).

This work is supported by the National Natural Science Foundation of China (Grant No. 11271299),Natural Science Foundation Research Project of Shaanxi Province (Grant No. 2012JM1014) and ChinaScholarship Council (No. 201206290110).

D. Li · P. Niu (B)Department of Applied Mathematics, Key Laboratory of Space Applied Physics and Chemistry,Ministry of Education, Northwestern Polytechnical University, Xi’an, Shaanxi, 710129, P.R. Chinae-mail: [email protected]

D. Lie-mail: [email protected]

R. ZhuoDepartment of Mathematics, Yeshiva University, New York, USAe-mail: [email protected]

mailto:[email protected]



D. Li et al.

1 Introduction

Let α and s be real numbers satisfying 0 < s < α < n, and Rn be an n-dimensional Eu-clidean space. In this paper, we consider the following integral equation related to the Hardy-Sobolev inequality

u(x) =∫

Rn

up(y)

|x − y|n−α|y|s dy. (1.1)

This integral equation is closely related to the following differential equation

(−�)α2 u(x) = |x|−sup. (1.2)

We consider solutions of PDE (1.2) in the following sense:(i) When α is any real number, and 0 < α < n, we assume

u ∈ Dα2 ,2

(Rn

) ={u

∣∣∣∫

Rn

|ξ |αu(ξ)dξ < ∞}

and satisfies∫

Rn

(−�)α4 u(−�)

α4 φdx =

∫Rn

|x|−sup(x)φ(x)dx, ∀φ ∈ C∞0

(Rn

). (1.3)

Here, ∫Rn

(−�)α4 u(−�)

α4 φdx

is defined by the Fourier transform

∫Rn

|ξ |αu(ξ)φ(ξ)dξ,

where u and φ are the Fourier transform of u and φ respectively.(ii) When α = 2m, we assume u ∈ C2m(Rn) and satisfies (1.2) in the distribution sense

∫Rn

(−�)muψ(x)dx =∫

Rn

|x|−sup(x)ψ(x)dx, ∀ψ ∈ C∞0

(Rn

).

In the case α = 2 and p = 2(n−s)

n−α− 1, (1.2) is the Euler-Lagrange equation of the ex-

tremal functions of the classical Hardy-Sobolev inequality, a special case of Caffarelli-Kohn-Nirenberg inequality [1]. The classical Hardy-Sobolev inequality asserts that for s ∈ [0,2]and 2∗(s) = 2(n−s)

n−2 , there exists C > 0 such that for u ∈ D1,2(Rn)

C

(∫Rn

|u|2∗(s)

|x|s dx

) 22∗(s) ≤

∫Rn

|∇u|2dx.

When s = 0 and p = n+αn−α

, (1.1) is reduced to the integral equation

u(x) =∫

Rn

un+αn−α (y)

|x − y|n−αdy. (1.4)

Nonexistence of Positive Solutions for an Integral Equation

Integral equation (1.4) arises as an Euler-Lagrange equation in the context of the Hardy-Littlewood-Sobolev inequality which has been explored in [6]. The regularity and extremalfunction of (1.4) were also obtained by Li in [15]. We note that (1.4) is actually equivalentto the following partial differential equation

(−�)α2 u(x) = u

n+αn−α . (1.5)

In the critical case when p = 2(n−s)

n−α− 1, the symmetry of (1.1) was studied by Lu and

Zhu [14], where they proved that

Proposition 1 Assume u(x) ∈ L2n

n−α (Rn) is a positive solution of

u(x) =∫

Rn

u2∗(s)−1(y)

|x − y|n−α|y|s dy, (1.6)

then u(x) is radially symmetric and strictly decreasing about the origin.

Now we are more concerned with the subcritical case that n−sn−α

< p < α∗(s) − 1, and ourfirst result is

Theorem 1 Suppose that u ∈ Ln(p−1)

α−s

loc (Rn) is a solution of (1.1), then u is radially symmetric

and monotonically decreasing about the origin. If u ∈ Ln(p−1)

α−s

loc (Rn), and 1 < α < n, thenu(x) ≡ 0.

To prove this result, we combine the method of moving planes and the Kelvin transform.We first prove the symmetry of positive solutions by the method of moving planes in integralforms. For other related results on integral equation and the method of moving planes inintegral forms, please see [3–5, 7–12, 16–19] and the references therein. In the applicationof method of moving planes in integral forms, one usually needs to assume some globalintegrability on the solution u. Without this condition, one is not able to carry on the methodof moving planes directly on u. Here we use a proper Kelvin transform to overcome thisdifficulty, and then based on the symmetry, we prove the nonexistence for positive solutionsto (1.1).

We also establish the equivalence between (1.1) and (1.2).

Theorem 2 (i) If α is any real number, 0 < α < n, and u ∈ D α2 ,2(Rn) is a positive weak

solution of (1.2), then a constant multiple of u(x) satisfies (1.1);(ii) If α = 2m, u ∈ C2m(Rn) is a positive distributional solution of (1.2), then a constant

multiple of u(x) satisfies (1.1).

Based on the equivalence, a series of results for integral equation (1.1) can now be appliedto (1.2).

Corollary 1 (i) If α is any real number satisfying 1 < α < n, u ∈ D α2 ,2(Rn) is a positive

weak solution of (1.2), then u(x) ≡ 0.(ii) If α = 2m, u ∈ C2m(Rn) is a positive distributional solution of (1.2), then u(x) ≡ 0.

D. Li et al.

Authors in [14] also established the equivalence between (1.6) and its correspondingPDE, however, under assumption that u ∈ D α

2 ,2(Rn). For α = 2m, by using the ideas of re-centers introduced by Chen-Li [2] in the proof of Theorem 2, we do not need to assume anyasymptotic behavior of this kind near infinity.

The paper is organized as follows. Section 2 is devoted to proving Theorem 1 by themethod of moving planes in integral forms with a proper Kelvin transform. The rest of thepaper is for the proof of Theorem 2.

For simplicity of presentation we will use C and c for a general positive constant whichis usually different in different contexts.

2 Proof of Theorem 1

As mentioned in Sect. 1, we will use the method of moving planes in integral forms toprove Theorem 1. We need the following Weighted Hardy-Littlewood-Sobolev inequality toestimate certain integral norms of the solutions.

Lemma 1 (see [14]) Let 1 < l,m < ∞, 0 < ν < n, τ + β ≥ 0, 1l+ 1

m+ ν+β+τ

n= 2 and

1 − 1m

− νn

≤ τn

< 1 − 1m

. Then the Weighted HLS inequality states

∣∣∣∣∫

Rn

∫Rn

f (x)g(y)

|x|τ |x − y|ν |y|β dxdy

∣∣∣∣ ≤ C‖f ‖Lm‖g‖Ll .

We can also write the Weighted HLS inequality in another form. Let

T g(x) =∫

Rn

g(y)

|x|τ |x − y|ν |y|β dy,

then ∥∥T g(x)∥∥

Lγ = Sup‖f ‖=1

⟨T g(x), f (x)

⟩ ≤ ‖g‖Ll ,

where 1l+ ν+β+τ

n= 1 + 1

γ, 1

m+ 1

γ= 1.

For a given real number λ, denote

Σλ = {x = (x1, x2, . . . , xn)|x1 < λ

}

and

Tλ = {x ∈ Rn|x1 = λ

}.

Let xλ = (2λ − x1, x2, . . . , xn) be the reflection of point x about the plane Tλ.Since we do not assume any integrability of u near infinity, we consider the Kelvin trans-

form of u:

u(x) = 1

|x|n−αu

(x

|x|2)

.

A direct calculation yields

u(x) = 1

|x|n−αu

(x

|x|2)


= 1

|x|n−α

∫Rn

up(y)

| x

|x|2 − y|n−α|y|s dy

=∫

Rn

up(y)

|x − y|n−α|y|s |y|n+α−(n−α)p−2sdy.

Because u ∈ Ln(p−1)

α−s

loc (Rn), for any domain Ω that is positive distance away from theorigin, we have

∫Ω

(up−1(y)

|y|β) n

α−s

dy < ∞, (2.1)

where β = n + α − (n − α)p − 2s. Then what left to prove is that u is symmetric about theorigin.

The proof of Theorem 1 The proof consists three steps. In Step 1, set uλ(x) = u(xλ), weshow that

uλ(x) ≥ u(x) a.e. on Σλ, (2.2)

for sufficiently negative λ. In Step 2, we continue to move the plane Tλ as long as (2.2) holdsup to the limiting case x1 = 0. As the direction of x1 can be chosen arbitrarily, we derivethat u(x) is symmetric and decreasing about the origin. In step 3, we will show that u ≡ 0in subcritical case.

Step 1. Let us set

ωλ = u(x) − uλ(x)

and for any ε > 0, define

Σ−λ = {

x ∈ Σλ \ Bε

(0λ

)|ωλ < 0},

where 0λ is the reflection of 0 about the plane Tλ. We will show that for λ sufficientlynegative, Σ−

λ must be measure zero.One can easily verify by the Mean Value Theorem that

u(x) − uλ(x) =∫

Σλ

[1

|x − y|n−α− 1

|xλ − y|n−α

][up(y)

|y|s |y|β − up

λ (y)

|yλ|s |yλ|β]dy

≤∫

Σ−λ

[1

|x − y|n−α− 1

|xλ − y|n−α

][up(y)

|y|s |y|β − up

λ (y)

|yλ|s |yλ|β]dy

≤∫

Σ−λ

[1

|x − y|n−α− 1

|xλ − y|n−α

][up(y) − u

p

λ (y)

|y|s |y|β]dy

≤ C

∫Σ−

λ

up−1(u(y) − uλ(y))

|x − y|n−α|y|s |y|β dy. (2.3)

Applying Lemma 1 to the above inequality, we have

‖ωλ‖Lq(Σ−λ ) ≤ C

∥∥|y|−β up−1ωλ

∥∥Lt (Σ−

λ ),

D. Li et al.

where q = n(p−1)

α−s> n−s

n−αand t = n(p−1)

(α−s)p. From the Hölder inequality, it follows


∥∥|y|−βup−1∥∥

Ln

α−s (Σ−λ )

‖ωλ‖Lq(Σ−λ ). (2.4)

By (2.1), we can choose N sufficiently large such that for λ ≤ −N , it holds

C∥∥|y|−β up−1

∥∥L

nα−s (Σ−

λ )≤ 1

2. (2.5)

Combining (2.4) and (2.5) leads to

‖ωλ‖Lq(Σ−λ ) ≤ 1

2‖ωλ‖Lq(Σ−

λ ).

This implies that Σ−λ must be measure zero. Hence (2.2) is proved.

Step 2. The inequality (2.2) provides a starting point to move the plane Tλ = {x ∈Rn|x1 = λ}. We continue to move the plane as long as (2.2) holds and show that the planewill not stop before hitting the origin. Denoting

λ0 = sup{λ|uμ(x) ≥ u(x),μ ≤ λ,∀x ∈ Σμ

},

we will prove

λ0 = 0.

In fact, suppose in the contrary that λ0 < 0, we will show that u must be symmetric aboutthe plane Tλ0 , i.e.

uλ0(x) ≡ u(x) a.e. on Σλ \ Bε

(0λ

). (2.6)

Otherwise, on Σλ0 ,

ωλ0(x) ≥ 0 but ωλ0(x) �≡ 0.

We show that in this case, the plane Tλ can be moved further to the right. More precisely,there exists an ε > 0 such that for any λ in [λ0, λ0 + ε),

uλ(x) ≥ u(x) for any x ∈ Σλ \ Bε

(0λ

).

From (2.4), it follows for all λ ∈ [λ0, λ0 + ε),


∥∥|y|−βup−1∥∥

Ln

α−s (Σ−λ )

‖ωλ‖Lq(Σ−λ ). (2.7)

By condition (2.1), we choose ε sufficiently small so that for all λ in [λ0, λ0 + ε),

C∥∥|y|−β up−1

∥∥L

nα−s (Σ−

λ )≤ 1

2. (2.8)

We postpone the proof of (2.8). Now with (2.7) and (2.8), we have

‖u − uλ‖Lq(Σ−λ ) = 0,

and therefore Σ−λ must be measure zero. Hence, for λ ∈ [λ0, λ0 + ε], we have uλ(x) ≥

u(x), x ∈ Σλ. This contradicts with the definition of λ0, and hence (2.6) must hold.


Now we show that the plane cannot stop before hitting the origin. Otherwise, if the planestops at x1 = λ0 < 0, then by (2.3) and the fact |y| > |yλ0 |, we get

u(x) − uλ0(x) =∫

Σλ0

[1

|x − y|n−α− 1

|xλ0 − y|n−α

][up(y)

|y|s |y|β − up

λ0(y)

|yλ|s |y|β]dy

<

∫Σλ0

[1

|x − y|n−α− 1

|xλ0 − y|n−α

][up(y) − u

p

λ0(y)

|y|s |y|β]dy

= 0.

This contradicts with (2.6). It follows λ0 = 0.Let us prove (2.8). For any small η > 0, choose R sufficiently large such that

C∥∥|y|−β up−1

∥∥L

nα−s (Rn\BR(0))

≤ η. (2.9)

We will prove that the measure of Σ−λ ∩BR(0) is sufficiently small if λ closes to λ0. Indeed,

we first state that in the interior of (Σλ0 \ Bε(0λ)) ∩ BR(0),

uλ0(x) > u(x). (2.10)

Otherwise, there exists some point x0 ∈ (Σλ0 \ Bε(0λ)) ∩ BR(0), such that uλ0(x0) = u(x0).From (2.3), it derives

0 = u(x0) − uλ0(x0)

=∫

Σλ0

[1

|x0 − y|n−α− 1

|xλ00 − y|n−α

][up(y)

|y|s |y|β − up

λ0(y)

|yλ0 |s |y|β]dy

< 0,

since

up(y)

|y|s+β<

up

λ0(y)

|yλ0 |s+β.

This is impossible unless u(x) ≡ 0. Hence (2.10) holds in the interior of (Σλ0 \ Bε(0λ)) ∩BR(0).

For any ι > 0, denote

Eι = {x ∈ (

Σλ0 \ Bε

(0λ

)) ∩ BR(0)|uλ0(x) − u(x) > ι},

Fι = (Σλ0 \ Bε

(0λ

) ∩ BR(0))\Eι.

One can see immediately that

limι→0

μ(Fι) = 0.

For λ > λ0, setting Dλ = (Σλ\Σλ0) ∩ BR(0), one can obtain that

(Σ−

λ ∩ BR(0)) ⊂ (

Σ−λ ∩ Eι

) ∪ Fι ∪ Dλ. (2.11)

D. Li et al.

Apparently, the measure of Dλ is small for λ close to λ0. Now we show that the measure ofΣ−

λ ∩ Eι is also small enough as λ close to λ0. In fact, for any x ∈ Σ−λ ∩ Eι, it holds

uλ(x) − u(x) = uλ(x) − uλ0(x) + uλ0(x) − u(x) < 0

and therefore

uλ0(x) − uλ(x) > uλ0(x) − u(x) > ι.

This implies

Σ−λ ∩ Eι ⊂ Hι ≡

{x ∈ (

BR(0) \ Bε

(0λ

))|uλ0(x) − uλ(x) > ι}. (2.12)

By the well-known Chebyshev inequality, we have

μ(Hι) ≤ 1

ιp+1

∫Hι

∣∣uλ0(x) − uλ(x)∣∣p+1

dx

≤ 1

ιp+1

∫BR(0)

∣∣uλ0(x) − uλ(x)∣∣p+1

dx. (2.13)

For any fixed ι, the right hand side of (2.13) can be made sufficiently small as λ ap-proaches to λ0 and the measure of Σ−

λ ∩ Eι is very small. Then by (2.12) and (2.13), themeasure of Σ−

λ ∩ BR(0) can also be made as small as we wish. With (2.9), we obtain (2.8).Step 3. We will prove that u ≡ 0. The Pohozaev type identity in integral forms is used

here (see [13]). Since u is radially symmetry, we have

u(r) = u(re) =∫

Rn

up(y)

|re − y|n−α|y|s dy

≥∫ r

0

∫∂B1

up(τ)

|re − τω|n−ατ n−1−sdωdτ

= Cup(r)

rn−α

∫ r

0τn−1−sdτ

= Cup(r)rα−s .

That is

u(r) ≤ Cr− α−s

p−1 .

It is easy to see∫

∂Br

r1−sup+1(x)dσ → 0, r → ∞ (2.14)

and∫

Br\B1

up+1(x)

|x|s dx ≤∫ r

1

∫∂Bτ (0)

τ− α−s

p−1 (p+1)−sdσdτ

≤∫ r

1τ

− α−sp−1 (p+1)+n−s−1

dτ

≤ C0(r

− α−sp−1 (p+1)+n−s − 1

),


where C0 = − α−sp−1 (p +1)+n− s. Since n−s

n−α< p < 2∗(s)−1, one can easily verify C0 < 0.

Thus,

∫Rn\B1

up+1(x)

|x|s dx < ∞, as r → ∞.

Moreover,

∫B1

up+1(x)

|x|s dx =∫ 1

0

∫∂Br

up+1(r)rsdσdr

≤ cup+1(0)

∫ 1

0rn−s−1dr

< ∞.

Therefore, we have

∫Rn

up+1(x)

|x|s dx < ∞.

Let

u(ρx) =∫

Rn

up(y)

|ρx − y|n−α|y|s dy. (2.15)

By elementary calculation, we have

d

dρ

(|ρx − y|α−n) = (α − n)|ρx − y|α−n−2x · (ρx − y).

Next we differentiate (2.15) respect to ρ and obtain

x · ∇u(ρx) = (α − n)

∫Rn

x · (ρx − y)up(y)

|ρx − y|n−α+2|y|s dy, x �= 0. (2.16)

Letting ρ = 1, multiplying (2.16) by up(x)

|x|s and integrating on Br \ Bε , it follows

∫Br\Bε

up(x)

|x|s(x · ∇u(x)

)dx

= 1

p + 1

∫Br\Bε

x

|x|s ∇(up+1(x)

)dx

= 1

p + 1

∫∂Br

⋃∂Bε

r1−sup+1(x)dσ − n − s

p + 1

∫Br\Bε

up+1(x)

|x|s dx. (2.17)

It is easy to see

∫∂Bε

ε1−sup+1(x)dσ ≤ cup+1(0)εn−s → 0, ε → 0. (2.18)

D. Li et al.

For each fixed r , we first let ε → 0 in (2.17), and then let r → ∞, by (2.14) and (2.18), wehave

∫Rn

up(x)

|x|s(x · ∇u(x)

)dx = − n − s

p + 1

∫Rn

up+1(x)

|x|s dx < ∞. (2.19)

On the other hand,∫

Rn

up(x)

|x|s(x · ∇u(x)

)dx = (α − n)

∫Rn

∫Rn

x · (x − y)up(x)up(y)

|x − y|n−α+2|y|s |x|s dydx

= α − n

2

∫Rn

∫Rn

x · (x − y)up(x)up(y)

|x − y|n−α+2|y|s |x|s dydx

+ α − n

2

∫Rn

∫Rn

y · (y − x)up(x)up(y)

|x − y|n−α+2|y|s |x|s dxdy

= α − n

2

∫Rn

∫Rn

up(x)up(y)

|x − y|n−α|y|s |x|s dxdy

= α − n

2

∫Rn

up+1(x)

|x|s dx, (2.20)

where we have used the fact that x · (x − y) + y · (y − x) = |x − y|2. Since p < 2 ∗ (s) − 1,by (2.19) and (2.20), we derive a contradiction if u �≡ 0. �

3 Proof of Theorem 2

In order to prove Theorem 2, we need the following lemma.

Lemma 2 Suppose that u(x) is a positive solution of

(−�)mu(x) = |x|−sup, (3.1)

where m is a positive integer. Then for any 1 ≤ j ≤ m − 1,

(−�)ju(x) > 0.

Proof of Lemma 2 Write uj = (−�)ju > 0, j = 1,2, . . . ,m − 1.Step 1. We claim um−1 > 0. Suppose in the contrary, there are two possible cases:(1) There exists x1 ∈ Rn, such that

um−1

(x1

)< 0. (3.2)

(2) um−1(x) ≥ 0 and there is a point x, such that um−1 = 0. In this case, we have−�um−1 ≤ 0, because x is a local minimum of um−1. This contradicts with

−�um−1 = |x|−sup > 0.

Hence we only need to treat the case (1).


Let

u(r) = 1

|∂Br(x1)|∫

∂Br (x1)

u(x)ds

be the average of u.Then for r > 0, we have ⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

−�u = u1,

−�u1 = u2,

· · · · · ·−�um−2 = um−1,

−�um−1 > 0.

(3.3)

By the Holder inequality, we have

−�um−1 ≥ (r + ∣∣x1

∣∣)−sup > 0. (3.4)

Integrating both sides of (3.4) from 0 to r , we get

u′m−1 < 0.

In view of (3.2),

um−1(r) ≤ um−1(0) = um−1

(x1

)< 0, for any r ≥ 0. (3.5)

Then from the last second equation of (3.3), we have

−r1−n(rn−1u′

m−2(r))′ = um−1(r) < um−1(0) := −c,

where c is a positive constant. Integrating yields

u′m−2 > cr and um−2 ≥ um−2(0) + cr2, for any r > 0.

Hence there exists an r1 > 0, such that um−2(r1) > 0. Take a point x2 with |x2| = r1 as thenew center and make average, then

¯u(r) = 1

|∂Br(x2)|∫

∂Br (x2)

u(x)ds.

Moreover,

¯um−2(0) > 0. (3.6)

Now, ¯u still satisfies (3.3) and

−� ¯um−1(r) ≥ (r + ∣∣x2 − x1

∣∣ + ∣∣x1∣∣)−s ¯up > 0.

By the same argument, we also have

¯um−2(r) ≥ ¯um−2(0) + cr2, for any r ≥ 0.

(3.6) and the above result yields

¯um−2(r) > 0, for any r ≥ 0.

D. Li et al.

Then we obtain

¯um−1(r) < 0, ¯um−2(r) > 0, for any r ≥ 0.

Continuing this way, after a few steps of re-centers (denote the result by u), we get

(−1)j um−j (r) > 0, for any r ≥ 0 (3.7)

and

−�um−1(x) ≥ (r + ∣∣xm−1 − xm−2

∣∣ + · · · + ∣∣x1∣∣)−s

up > 0. (3.8)

Now we show that m must be even. In fact, if m is odd, then (3.7) means u2(r) <

0, for any r ≥ 0 and hence �u1 > 0. Integrating twice yields

u1(r) > u1(0) := c > 0.

By the first equation in (3.3), we have

�u < −c

which implies

u(r) < u(0) − cr2.

Hence, there exists r0 > 0, such that u(r0) < 0. This contradicts with the positiveness of u.Hence in the following, we only need to consider the case when m is even.

Let

uμ(x) = μ2m−sp−1 u(μx)

be the re-scaling of u. Equation (3.1) is invariant under this re-scaling, and for any μ, uμ isstill a positive solution of (3.1).

From (3.7), it follows

u(r) > u(0) = c0 > 0.

Then we choose a sufficiently large μ such that for any a0 > 0,

u(r) ≥ a0 ≥ a0rσ0 , for any r ∈ [0,1]. (3.9)

Here σ0 > 1 and σ0p ≥ 2m + n. Combining with (3.8), it gets

−r1−n(rn−1u′

m−1

)′ ≥ ap

0 rσ0p(r + ∣∣xm−1 − xm−2

∣∣ + · · · + ∣∣x1∣∣)−s

≥ ap

0 rσ0p(1 + ∣∣xm−1 − xm−2

∣∣ + · · · + ∣∣x1∣∣)−s

≥ c0ap

0 rσ0p,

where c0 is a positive constant less than 1. Integrating both sides twice yields

um−1(r) ≤ − c0ap

0

(σ0p + n)(σ0p + 2)rσ0p+2.


Substituting this into (3.3) and integrating by the same way, we arrive at

u(r) ≥ c0ap

0

(σ0p + n + 2m)2mrσ0p+2m ≥ c0a

p

0

(σ0p + n + 2m)2m+nrσ0p+2m+n.

Let us denote

σk = 2σk−1p ≥ σk−1p + n + 2m and ak = c0ap

k−1

(2σk−1p)n+2m.

Obviously,

u(r) ≥ a1rσ1 .

Suppose u(r) ≥ ak−1rσk−1 and go through the entire process as above, we get

u(r) ≥ akrσk

where we calculate

ak = c(pk−1+pk−2+···+p+1)(n+2m)

0 apk

0

(2p)(pk−1+2pk−2+···+(k−1)p+k)(n+2m)σ(pk−1+pk−2+···+p+1)(n+2m)

0

= cpk−1p−1

0 apk

0

(2p)(pk+1−p

(p−1)2− k

p−1 )(n+2m)σ

pk−1p−1 (n+2m)

0

≥ cpk−1p−1

0 apk

0

(2p)(

pk+1

(p−1)2)(n+2m)

σpk

p−1 (n+2m)

0

= c−2p−10

(c

1p−10 a0

(2p)(

p

(p−1)2)(n+2m)

σn+2mp−1

0

)pk

.

Taking

a0 = 2(2p)

(p

(p−1)2)(n+2m)

σn+2mp−1

0

c1

p−10

,

it follows

u(1) ≥ ak ≥ c−2p−10 2pk → ∞ as k → ∞.

This is obviously impossible. Therefore, we must have um−1(x) > 0.Step 2. Now we will show um−j > 0 for j = 2,3, . . . ,m− 1. Suppose that there exists x∗

such that um−j (x∗) < 0. Through the same arguments as in Step 1, we find that the signs of

uj are alternating. Therefore, it yields −�u < 0. It follows

u ≥ c > 0.

D. Li et al.

By (3.8), we have

−�um−1 ≥ (r + ∣∣xm−1 − xm−2

∣∣ + · · · + ∣∣x1∣∣)−s

up

≥ c > 0.

A direct integration leads to

um−1(R) ≤ um−1(0) − cR2, for R > 0.

Then there must exists R0 such that um−1(R0) < 0. This is impossible. Therefore we com-plete the proof of Lemma 2. �

Proof of Theorem 2Case 1. For α = 2m, fix x0 ∈ Rn. Let φ(x) be the solution of the following equation

{(−�)mφ(x) = δ(x), x ∈ Br(x

0),

φ = �φ = · · · = �m−1φ = 0, on ∂Br(x0).

(3.10)

By the maximum principle, it is easy to see

∂

∂ν

[(−�)jφ

] ≤ 0, j = 0,1, . . . ,m − 1. (3.11)

Multiplying both sides of (1.2) by φ and integrating on Br(x0), we have

∫Br (x0)

|x|−sup(x)φ(x)dx =∫

Br (x0)

(−�)mu(x)φ(x)dx

= u(x0

) +m−1∑j=0

∫∂Br (x0)

(−�)ju∂

∂ν

[(−�)m−1−jφ

]ds

≤ u(x0

), (3.12)

here we have used Lemma 2 and (3.11). By the results in [2], we see

φ(x) → c

|x − x0|n−2m, r → ∞ and

∣∣∣∣ ∂

∂ν

[(−�)jφ

]∣∣∣∣ ≤ cr2m−n−1−2j . (3.13)

Then let r → ∞, it yields∫

Rn

up(x)

|x − x0|n−2m|x|s dx ≤ u(x0

)< ∞.

Thus, there exists a sequence rk , such that

∫∂Brk

(x0)

|x|−sup(x)

rn−2m−1k

ds → 0. (3.14)

Using the Holder inequality, it implies

∫∂Brk

(x0)

up(x)

rn−1k

ds ≤ rs−2m

p

k

(∫∂Brk

(x0)

|x|−sup(x)

rn−2m−1k

) 1p

.


Noting s ≤ 2m and (3.14), we deduce that as rk → ∞,∫

∂Brk(x0)

up(x)

rn−1k

ds → 0.

Set

(−�)ju = vj , j = 1,2, . . . , k − 1.

Using the similar argument as in [2] we have, as r → ∞,

∫Rn

(−�)ju

|x − x0|n−2jdx ≤ u

(x0

)< ∞.

Summing j from 1 to m − 1 yields

∫Rn

m−1∑j=1

(−�)ju

|x − x0|n−2jdx < ∞.

Hence, there exists a subsequence of rk (still denoted by itself), such that as rk → ∞,

∫∂Br (x0)

m−1∑j=1

(−�)ju

|x − x0|n−2j−1dx → 0. (3.15)

Combining (3.12), (3.13), (3.14), (3.15) and letting rk → ∞, we have

u(x0

) = c

∫Rn

up(y)

|x0 − y|n−α|y|s dy.

Case 2. α is a positive real value. We define the positive solutions in the distributionsense, that is u ∈ D α

2 ,2(Rn). The proof in this case is quite similar to that in [14]. For thecompleteness and the convenience of the reader, we include it here.

For any φ ∈ C∞0 (Rn), let ψ(x) = ∫

Rnφ(y)

|x−y|n−α dy and (−�)α2 ψ = φ. Then ψ ∈ D α

2 ,2(Rn).Thus, (1.3) holds for ψ :

∫Rn

(−�)α4 u(−�)

α4 ψdx =

∫Rn

|x|−sup(x)ψ(x)dx.

Integration by parts for the left hand side and exchange the order of integration for the righthand side yields

∫Rn

u(x)φ(x)dx =∫

Rn

(∫Rn

|y|−sup(y)

|x − y|n−αdy

)φ(x)dx.

This implies

u(x) =∫

Rn

|y|−sup(y)

|x − y|n−αdy.

On the other hand, we assume that u ∈ Hα2 (Rn) is a solution of the integral equation.

Making a Fourier transform on both sides, for some constant c > 0, we have that,

u(ξ) = c|ξ |−α|y|−sup(y)(ξ).

D. Li et al.

It follows∫

Rn

(−�)α4 u(−�)

α4 φdx = c

∫Rn

|ξ |αu(ξ)φ(ξ)dξ

= c

∫Rn

|y|−sup(y)(ξ)φdξ

= c

∫Rn

|y|−sup(y)(y)φdy.

Hence, u(x) is a solution of

(−�)α2 u = c|x|−sup(x)

in the sense of distributions. �

References

1. Caffarelli, L., Kohn, R., Nirenberg, L.: First order interpolation inequalities with weights. Compos. Math.53(3), 259–275 (1984)

2. Chen, W., Li, C.: Super polyharmonic property of solutions for PDE systems and its applications. Com-mun. Pure Appl. Anal. 12, 2497–2514 (2013)

3. Chen, W., Li, C.: Regularity of solutions for a system of integral equations. Commun. Pure Appl. Anal.4, 1–8 (2005)

4. Chen, W., Li, C.: An integral system and the Lane-Emden conjecture. Discrete Contin. Dyn. Syst. 4,1167–1184 (2009)

5. Chen, W., Li, C., Lim, J.: Weighted Hardy-Littlewood-Sobolev inequalities. Discrete Contin. Dyn. Syst.12, 347–354 (2005)

6. Chen, W., Li, C., Ou, B.: Classification of solutions for an integral equation. Commun. Pure Appl. Math.59, 330–343 (2006)

7. Chen, W., Li, C., Ou, B.: Classification of solutions for a system of integral equations. Commun. PartialDiffer. Equ. 30, 59–65 (2005)

8. Fang, Y., Chen, W.: A Liouville type theorem for poly-harmonic Dirichlet problems in a half space. Adv.Math. 229, 2835–2867 (2012)

9. Fang, Y., Zhang, J.: Nonexistence of positive solution for an integral equation on Rn+ . Commun. PureAppl. Anal. 12, 663–678 (2013)

10. Jin, C., Li, C.: Symmetry of solution to some systems of integral equations. Proc. Am. Math. Soc. 134,1661–1670 (2006)

11. Jin, C., Li, C.: Quantitative analysis of some system of integral equations. Calc. Var. Partial Differ. Equ.26, 447–457 (2006)

12. Li, C., Ma, L.: Uniqueness of positive bound states to Shrodinger systems with critical exponents. SIAMJ. Math. Anal. 40, 1049–1057 (2008)

13. Li, C., Lei, Y.T.: Sharp criteria of Liouville type for some nonlinear systems. Preprint14. Lu, G.Z., Zhu, J.Y.: Symmetry and regularity of extremals of an integral equation related to the Hardy-

Sobolev inequality. Calc. Var. Partial Differ. Equ. 42(3–4), 563–577 (2011)15. Li, Y.Y.: Remark on some conformally invariant integral equations: the method of moving spheres. J.

Eur. Math. Soc. 6, 153–180 (2004)16. Lin, C.S.: A classification of solutions of a conformally invariant fourth order equation in Rn. Comment.

Math. Helv. 73, 206–231 (1998)17. Ma, L., Chen, D.Z.: A Liouville type theorem for an integral system. Commun. Pure Appl. Anal. 5,

855–859 (2006)18. Ma, L., Chen, D.Z.: Radial symmetry and monotonicity for an integral equation. J. Math. Anal. Appl.

342, 943–949 (2008)19. Zhou, R., Li, D.: A system of integral equations on half space. J. Math. Anal. Appl. 381, 392–401 (2011)

Documents

Nonexistence of Positive Solutions for an Integral Equation Related to the Hardy-Sobolev Inequality