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arXiv:1406

.5168v1

[math.AP]19Jun2014

Qualitative properties of solutions for an integral

system related to the HardySobolev inequality

John Villavert

Department of Mathematics, University of Oklahoma

Norman, OK 73019, USA

Abstract

This article carries out a qualitative analysis on a system of inte-gral equations of the HardySobolev type. Namely, results concerningLiouville type properties and the fast and slow decay rates of positivesolutions for the system are established. For a bounded and decayingpositive solution, it is shown that it either decays with the slow rates orthe fast rates depending on its integrability. Particularly, a criteria fordistinguishing integrable solutions from other bounded and decayingsolutions in terms of their asymptotic behavior is provided. Moreover,related results on the integrability, boundedness, radial symmetry andmonotonicity of positive integrable solutions are also established. As

a result of the equivalence between the integral system and a systemof poly-harmonic equations under appropriate conditions, the resultstranslate over to the corresponding poly-harmonic system. Hence, sev-eral classical results on semilinear elliptic equations are recovered andfurther generalized.

MSC: Primary: 35B40, 35B53, 45G15, 45M05; Secondary: 35J91.

Keywords: LaneEmden equations, HardySobolev inequality; weighted HardyLittlewood

Sobolev inequality, singular integral equations; poly-harmonic equations.

1 Introduction and the main results

In this paper, we study the qualitative properties of positive solutions for anintegral system of the HardySobolev type. In particular, we consider the

email: john.villavert@gmail.com, villavert@math.ou.edu

1

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system of integral equations involving Riesz potentials and Hardy terms,

u(x) =

Rn

v(y)q

|x y|n|y|1dy,

v(x) =

Rn

u(y)p

|x y|n|y|2dy,

(1.1)

where throughout we assume thatn 3,p, q >0 withpq >1, (0, n) and1, 2 [0, ). As a result, we establish analogous properties for the closelyrelated system of semilinear differential equations with singular weights,

()/2u(x) =

v(x)q

|x|1in Rn\{0},

()/2v(x) =u(x)p

|x|2in Rn\{0},

(1.2)

since both systems are equivalent under appropriate conditions. Particu-larly, if p ,q > 1 and = 2k is an even positive integer, then a positiveclassical solution u, v C2k(Rn\{0}) C(Rn) of system (1.1), multipliedby suitable constants if necessary, satisfy the poly-harmonic system (1.2)pointwise except at the origin; and vice versa (cf. [8,39, 41]).1

Our aim is to effectively characterize the positive solutions, specificallythe ground states, in terms of their asymptotic behavior and elucidate itsconnection with Liouville type non-existence results. The motivation for

studying these properties for the HardySobolev type systems arises fromseveral related and well-known problems. For example, one problem origi-nates from the doubly weighted HardyLittlewoodSobolev (HLS) inequal-ity [38], which states that for r, s (1, ), (0, n) and 0 1+2 ,

Rn

Rn

f(x)g(y)

|x|1 |x y|n|y|2dxdy

Ci,s,,nfLr(Rn)gLs(Rn),where

n

1

r 0;

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(ii) integrable solutions ifu Lr0(Rn) andv Ls0(Rn) where

r0= n

q0and s0=

n

p0.

Definition. Let u, v be positive solutions of (1.1). Then u, v are said todecay with theslow ratesas |x| ifu(x) |x|q0 andv(x) |x|p0.Supposeq p and1 2. Thenu, v are said to decay with the fast ratesas |x| if

u(x) |x|(n)

and

v(x) |x|(n), if p(n ) +2> n;v(x) |x|(n) ln |x|, if p(n ) +2= n;

v(x) |x|(p(n)(2)), if p(n ) +2< n.

Remark 1. The conditions on the parameters in the previous definitioni.e. q p and1 2, which we will sometimes assume within our maintheorems, are not so essential. Namely, the results still remain true if we in-terchange the parameters provided thatu andv are interchanged accordinglyin the definition and the theorems.

Remark 2. In view of the equivalence with poly-harmonic systems and theregularity theory indicated by the earlier references, classical solutions of(1.1) should be understood to mean solutions belonging to C(Rn\{0}) C(Rn), where is the greatest integer function.

1.1 Main results

Theorem 1. There hold the following.

(i) Ifu, v are bounded and decaying solutions of (1.1), then there exists apositive constantC such that as|x| ,

u(x) C|x|q0 and v(x) C|x|p0.

(ii) Suppose q p and1 2 (so that q0 p0), and letu, v be positive

solutions of (1.1). Then there exists a positive constantc such that as|x| ,

u(x) c

|x|n and v(x)

c

|x|min{n,p(n)(2)}.

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We now introduce our Liouville type theorem for the HardySobolev

type system. Basically, our result states that the system has no ground stateclassical solutions, or non-negative ones besides the trivial pair u, v 0, inthe subcritical case.

Theorem 2. System(1.1)does not admit any bounded and decaying positiveclassical solutions whenever the following subcritical condition holds

n 11 +q

+n 2

1 +p > n . (1.6)

Remark 3. In[39], the author proved the following non-existence theorembut without any boundedness or decaying assumption on positive solutions.

Theorem A. System (1.1) has no positive solution if eitherpq (0, 1] orwhenpq >1 andmax{p0, q0} n .

Interestingly, Theorem A and Theorem 2 are reminiscent of the non-existence results of Serrin and Zou [36] for the LaneEmden system. Weshould also mention the earlier work in [13], which also obtained similarLiouville theorems among other interesting results.

Theorem 2 also applies to integrable solutions as well. In particular,we later show that integrable solutions are indeed radially symmetric anddecreasing about the origin. Therefore, the proof of Theorem 2 can beadopted in this situation to get the following.

Corollary 1. System (1.1) does not admit any positive integrable solutionswhenever the subcritical condition (1.6) holds.

The next theorem concerns the properties of integrable solutions. Moreprecisely, it states that we can distinguish integrable solutions from theother ground states with slower decay rates in that integrable solutionsare equivalently characterized as the bounded positive solutions which decaywith the fast rates. In dealing with integrable solutions, Corollary1indicatesthat we can restrict our attention to the non-subcritical case:

q0+p0 n , (1.7)

or equivalently n 11 +q

+n 2

1 +p n . (1.8)

Then, the theorem further asserts that there are also no positive integrablesolutions in the supercritical case.

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Theorem 3. Suppose q p, 1 2 (so that q0 p0), and let u, v be

positive solutions of (1.1) satisfying the non-subcritical conditon (1.8).(i) Then u, v are integrable solutions if and only if u, v are bounded anddecay with thefast ratesas |x| :

u(x) |x|(n);v(x) |x|(n), if p(n ) +2> n;

v(x) |x|(n) ln |x|, if p(n ) +2= n;

v(x) |x|(p(n)(2)), if p(n ) +2< n.

(ii) Ifu, v are integrable solutions of (1.1), then the critical case,

n 11 +q

+n 2

1 +p =n , (1.9)

necessarily holds.

Remark 4. (i) Notice that if1 =2 = 0, the HardySobolev type system(1.1) coincides with the HLS system (1.4). Indeed, our main results on theintegrable solutions, including the subsequent results below on the integrabil-ity, boundedness, radial symmetry and monotonicity of solutions, coincidewith and thus extend past results of[7,20](also see [4,21, 22,42] for closelyrelated results).

(ii) To further illustrate their connection, we remark that the integrablesolutions in the unweighted case turn out to be the finite-energy solutions ofthe HLS system (1.4) i.e. the critical points (u, v)Lp+1(Rn) Lq+1(Rn)

for the HLS functional.(iii) In the weighted case, however, system(1.1)cannot be recovered from

(1.4)due to their different singular weights. In fact, the asymptotic behaviorof positive solutions between the two are not the same (cf. [23,25]).

Our last main result asserts that if u, v are bounded and positive butare not integrable solutions, then they decay with almost the slow ratesand we conjecture that they actually do converge with the slow rates. Ofcourse, if this conjecture were true, then Theorem2would hold without anyadditional decaying assumption on solutions. However, if, in addition, u, vare decaying solutions, then they do indeed decay with the slow rates.

Theorem 4. Let u, v be bounded positive solutions of (1.1). Then therehold the following.

(i) There does not exist a positive constant c such that

either u(x) c(1 + |x|)1 or v(x) c(1 + |x|)2 ,

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where1 < q0, 2< p0.

(ii) Ifu, v are not integrable solutions, thenu, vdecay with rates not largerthan the slow rates. Namely, there does not exist a positive constantCsuch that

either u(x) C(1 + |x|)3 or v(x) C(1 + |x|)4 ,

where3 > q0, 4> p0.

(iii) Ifu, v are not integrable solutions but are decaying solutions, thenu, vmust necessarily have the slow rates as |x| .

Remark 5. For the sake of conciseness, rather than formally stating the

corresponding results for the poly-harmonic system (1.2) as corollaries, weonly point out that our main results for the integral system do translate overto (1.2) provided the equivalence conditions described earlier are satisfied.

The remaining parts of this paper are organized in the following way.In section 2, some preliminary results are established in which Theorem1 is an immediate consequence of. Then, we apply an integral form of aPohozaev type identity to prove Theorem 2. Section 3 establishes severalkey properties of integrable solutions: the integrability, boundedness andconvergence properties of integrable solutions, and we show the integrablesolutions are radially symmetric and decreasing about the origin. Using

these properties, we prove Theorem3in section4. The paper then concludeswith section5,which contains the proof of Theorem4.

2 Slow decay rates and the non-existence theorem

Proposition 1. Letu, vbe bounded and decaying positive solutions of (1.1).

(i) There exists a positive constantC such that as |x|

u(x) C|x|q0 and v(x) C|x|p0.

(ii) Moreover, the improper integrals,Rn

u(x)p+1

|x|2dx and

Rn

v(x)q+1

|x|1dx,

are finite provided that the subcritical condition (1.6) holds.

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Proof. For |x| >2R with R >0 suitably large,

u(x)

B|x|(0)\BR(0)

v(y)q

|x y|n|y|1dy

Cv(x)q|x|1nB|x|(0)\BR(0)

dt Cv(x)q|x|1 . (2.1)

Similarly, we can show

v(x) Cu(x)p|x|2, (2.2)

and combining (2.1) with (2.2) gives us

u(x) Cv(x)q|x|1 Cu(x)pq|x|q(2)+(1),v(x) Cu(x)p|x|2 Cv(x)pq |x|p(1)+(2).

Indeed, these estimates imply

u(x) C|x|q0 and v(x) C|x|p0 as |x| .

In addition,Rn

u(x)p+1

|x|2dx

BR(0)

u(x)p+1

|x|2dx+

BR(0)C

u(x)p+1

|x|2dx

C1+C2 R

tq0(p+1)+n2dt

t < ,

since (1.6) implies q0(p+ 1) +n 2 =n (q0+ p0)< 0. Likewise,Rn

v(x)q+1

|x|1 dx < using similar calculations, and this completes the proof.

Proposition 2. Letu, v be bounded positive solutions of (1.1). Then thereexists a positive constantC >0 such that as|x| ,

u(x) C

(1 + |x|)n and v(x)

C

(1 + |x|)min{n, p(n)(2)}.

Proof. For y B1(0), we can find a C >0 such that

C

B1(0)

v(y)q

|y|1dy,

B1(0)

u(y)p

|y|2dy < .

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Thus for x B1(0)C, we have

u(x)

B1(0)

v(y)q

|x y|n|y|1dy

C

(1 + |x|)n

B1(0)

v(y)q

|y|1dy

C

(1 + |x|)n. (2.3)

Similarly, we can show

v(x) C

(1 + |x|)n.

Then, with the help of estimate (2.3), we get

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

u(y)p

|x y|n|y|2 dy

C

(1 + |x|)p(n)+2

|x|/20

tdt

t =

C

(1 + |x|)p(n)(2).

Proof of Theorem1. This is a direct consequence of Proposition 1(i) andProposition 2.

Proof of Theorem2. We proceed by contradiction. That is, assume u, vare b ounded and decaying positive classical solutions. First, notice that

integration by parts impliesBR(0)

v(x)q

|x|1(x v(x)) +

u(x)p

|x|2(x u(x)) dx

= 1

1 +q

BR(0)

x

|x|1 (v(x)q+1) dx+

1

1 +p

BR(0)

x

|x|2 (u(x)p+1) dx

= n 1

1 +q

BR(0)

v(x)q+1

|x|1dx

n 21 +p

BR(0)

u(x)p+1

|x|2dx

+ R

1 +q

BR(0)

v(x)q+1

|x|1ds+

R

1 +p

BR(0)

u(x)p+1

|x|2ds

By virtue of Proposition 1(ii), we can find a sequence {Rj} such that asRj ,

Rj

BRj (0)

v(x)q+1

|x|1ds, Rj

BRj (0)

u(x)p+1

|x|2ds 0.

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Thus, we obtain the identityRn

v(x)q

|x|1(x v(x)) + u(x)

p

|x|2(x u(x)) dx

=

n 11 +q

+n 2

1 +p

Rn

v(x)q+1

|x|1dx, (2.4)

where we used the fact thatRn

u(x)p+1

|x|2dx=

Rn

Rn

u(x)pv(z)q

|x z|n|x|2 |z|1dzdx=

Rn

v(x)q+1

|x|1dx.

From the first equation, we write

u(x) =Rn

v(y)q

|x y|n|y|1 dy= 1

Rn

v(z)q

|x z|n|z|1 dz.

Differentiating this rescaled equation with respect to then taking = 1gives us

x u(x) = ( 1)

Rn

v(z)q

|x z|n|z|1dz+

Rn

qv(z)q1(z v)

|x z|n|z|1dz

= ( 1)u(x) +

Rn

z v(z)q

|x z|n|z|1dz. (2.5)

Note that an integration by parts yields

BR(0)

z v(z)q

|x z|n

|z|1

dz

=R

BR(0)

v(z)q

|x z|n|z|1ds (n 1)

BR(0)

v(z)q

|x z|n|z|1dz

(n )

BR(0)

(z (x z))v(z)q

|x z|n+2|z|1dz,

By virtue ofRn

v(y)q

|xz|n|z|1dz < ,we can find a sequence {Rj} such that

Rj

BRj (0)

v(z)q

|x z|n|z|1dS 0 as Rj

and thus obtainRn

z v(z)q

|x z|n|z|1dz= (n 1)

Rn

v(z)q

|x z|n|z|1dz

(n )

Rn

(z (x z))v(z)q

|x z|n+2|z|1dz.

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Hence, inserting this into (2.5) yields

x u(x) = (n )u(x) (n )

Rn

(z (x z))v(z)q

|x z|n+2|z|1dz. (2.6)

Similar calculations on the second integral equation will lead to

x v(x) = (n )v(x) (n )

Rn

(z (x z))u(z)p

|x z|n+2|z|2dz. (2.7)

Now multiply (2.6) and (2.7) by |x|2u(x)p and |x|1v(x)q, respectively,sum the resulting equations together and integrate over Rn to get

Rn

v(x)q

|x|1 (x v(x)) dx+Rn

u(x)p

|x|2 (x v(x)) dx

= (n )

Rn

v(x)q+1

|x|1+

u(x)p+1

|x|2dx

(n )

Rn

Rn

(z (x z) +x (z x))u(z)pv(x)q

|x z|n+2|z|2 |x|1dzdx.

By noticing thatz (x z) + x (z x) = |x z|2, we obtain the Pohozaevtype identity

Rnv(x)q

|x|1(xv(x))+

u(x)p

|x|2(xv(x)) dx= (n)Rn

v(x)q+1

|x|1dx. (2.8)

Inserting (2.4) into (2.8) yieldsn 11 +q

+n 2

1 +p (n )

Rn

v(x)q+1

|x|1dx= 0,

but this contradicts with (1.6). This completes the proof of the theorem.

3 Properties of integrable solutions

3.1 An equivalent form of the weighted HLS inequality

The following estimate is a consequence of the doubly weighted HLS in-equality by duality, and it is the version of the weighted HLS inequality weapply in this paper.

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Lemma 1. Letp, q (1, ), (0, n) and0 1+2 , and define

If(x) .=

Rn

f(y)

|x|1 |x y|n|y|2dy.

ThenIfLq(Rn) Ci,p,,nfLp(Rn),

where 1p 1q =

(1+2)n and

1q

nn A or |x| > A, uA = 0 ifu A and|x| A; we givevA the analogous definition. Consider the integral operatorT = (T1, T2) where

T1g(x) =

Rn

vA(y)q1g(y)

|x y|n|y|1dy and T2f(x) =

Rn

uA(y)p1f(y)

|x y|n|y|2dy

for f Lr(Rn) and g Ls(Rn). By virtue of (3.1), applying the weightedHardyLittlewoodSobolev inequality followed by Holders inequality gives

us

T1gLr Cvq1A g nrn+r(1)

CvAq1s0 gs,

T2fLs Cup1A f nsn+s(2)

CuAp1r0 fr.

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We may choose A sufficiently large so that

CvAq1s0 , CuA

p1r0 1/2

and the operator T(f, g) = (T1g, T2f), equipped with the norm

(f1, f2)Lr(Rn)Ls(Rn) = f1r+ f2s,

is a contraction map from Lr(Rn) Ls(Rn) to itself

for all (1

r,1

s) I

.= (a b,

n

n ) (0,

n

n a+b).

Thus, T is also a contraction map from Lr0(Rn) Ls0(Rn) to itself since

( 1r0 , 1s0 ) I. Now define

F =

Rn

(v vA)q(y)

|x y|n|y|1dy and G=

Rn

(u uA)p(y)

|x y|n|y|2dy.

Then the weighted HLS inequality implies that (F, G) Lr(Rn) Ls(Rn)and since (u, v) satisfies

(f, g) =T(f, g) + (F, G),

applying Lemma 2.1 from[18] yields

(u, v) Lr(Rn) Ls(Rn) for all (1

r ,1

s ) I. (3.2)

Step 2: Extend the interval of integrability I.First, we claim that

qn

n a+b

1n

> a b. (3.3)

If this holds true, we can then apply the weighted HLS inequality to obtain

ur Cvq nr

n+r(1) Cvq nrq

n+r(1).

Then, sincev Ls

(Rn

) for all 1

s (0,n

n a+b), we obtain thatu Lr

(Rn

)for 1r (0, q{nn a+ b}

1n ), where we are using (3.3) for the last

interval to make sense. Combining this with (3.2) yields

u Lr(Rn) for all 1

r

0,n

n

. (3.4)

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Likewise, the weighted HLS inequality implies

vs Cup nspn+s(2)

.

Thus, combining this with (3.4) yields

v Ls(Rn) for all 1

s

0, minn

n ,

p(n ) ( 2)

n

.

It remains to verify the claim (3.3). To do so, notice that (1.7) implies that

qn

n a+b

1n

> qb 1

n =

q 2q+ 1n(pq 1)

=(qp) +p 2q+ 1

n(pq 1) >

(qp) +1p 2q+2 1n(pq 1)

=(qp) 1(1 p) 2(q 1)

n(pq 1) =a b.

This completes the proof.

3.3 Integrable solutions are ground states

Theorem 6. Ifu, v are positive integrable solutions of (1.1), thenu, v arebounded and converge to zero as |x| .

Proof. We prove this in two steps. The first step shows the boundedness ofintegrable solutions and the second step verifies the decay property.

Part 1: uand v are in L

(Rn

).By exchanging the order of integration and choosing a suitably smallc >0, we can write

u(x) C c

0

Bt(x)

|y|1v(y)q dy

tndt

t +

c

Bt(x)

|y|1v(y)q dy

tndt

t

.

=I1+I2.

(i) We estimate I1 first and assume |x| >1, since the case where |x| 1 canbe treated similarly. Then Holders inequality yields

Bt(x)v(y)q

|y|1dy Ct1|Bt(x)|

11/vq

for >1. Then choose suitably large so that q= 1/ is sufficiently smalland sovq L(Rn) as a result of Theorem 5. Thus,

I1 Cvq

c0

|Bt(x)|1q

tn+1dt

t C

c0

t1nqdt

t < .

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(ii) Ifz B(x), thenBt(x) Bt+(z). From this, observe that for (0, 1)

and z B(x),

I2 C

c

Bt(x)

|y|1v(y)q dy

tndt

t

C

c

Bt+(z)

|y|1v(y)q dy

(t+)n

t+t

n+1 d(t+)t+

C(1 +)n+1 c+

Bt(z)

|y|1v(y)q dy

tndt

t Cu(z).

These estimates for I1 and I2 yield

u(x) C1+C2u(z) for z B(x).

Integrating this estimate on B(x) then applying Holders inequality givesus

u(x) C1+C2

B(x)

u(z) dz C1+C2|B(x)|1/r0ur0 < .

Hence,u is bounded in Rn. Using similar calculations, we can also show vis bounded in Rn.Part 2: u(x), v(x) 0 as |x| .

Choose x Rn. Then for each > 0, there exists a sufficiently small

>0 such that 0

Bt(x)

|y|1v(y)q dy

tndt

t Cvq

0

tdt

t < .

Likewise, for|x z| < , we have

Bt(x)

|y|1v(y)q dy

tndt

t

Bt+(z)

|y|1v(y)q dy

(t+)n

t+t

n+1 d(t+)t+

C 0

Bt(z)

|y|1v(y)q dy

tn dtt =C u(z).

Therefore, the last two estimates imply

u(x) +Cu(z) for z B(x),

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and since u Lr0(Rn), we obtain

u(x)r0 = 1

|B(x)|

B(x)

u(x)r0 dz C1r0 +C2

1

|B(x)|

B(x)

u(z)r0 dz 0

as |x| and 0. Hence, lim|x|u(x) = 0, and similar calcula-tions will show lim|x|v(x) = 0. This completes the proof.

3.4 Radial symmetry and monotonicity

Theorem 7. Letu, v be positive integrable solutions of (1.1). Thenu andv are radially symmetric and monotone decreasing about the origin.

We employ the integral form of the method of moving planes (cf. [9, 11])to prove this result, but first, we introduce some preliminary tools. For R, define

.

= {x= (x1, . . . , xn) | x1 },

let x = (2 x1, x2, . . . , xn) be the reflection ofx across the plane .

={x1=} and let u(x) =u(x

) and v(x) =v(x).

Lemma 2. There holds

u(x) u(x) =

1|x y|n

1

|x y|n

1|y|1

(v(y)q v(y)q) dy

1

|x y|n

1

|x y|n 1

|y|1

1

|y|1v(y)q dy;

v(x) v(x) =

1|x y|n

1

|x y|n

1|y|2

(u(y)p u(y)p) dy

1|x y|n

1

|x y|n

1|y|2

1

|y|2

u(y)p dy.

Proof. We only prove the first identity since the second identity followssimilarly. By noticing |x y| = |x y|, we obtain

u(x) =

v(y)q

|x y|n|y|1dy+

v(y)q

|x y|n|y|1dy,

u(x) =

v(y)q|x y|n|y|1

dy+

v(y)q|x y|n|y|1

dy.

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Taking their difference yields

u(x) u(x) =

v(y)q|x y|n|y|1

v(y)q

|x y|n|y|1

dy

v(y)q|x y|n|y|1

v(y)q

|x y|n|y|1

dy

=

1|x y|n

1

|x y|n

v(y)q|y|1

dy

1|x y|n

1

|x y|n

v(y)q|y|1

dy,

and the result follows accordingly.

Proof of Theorem7. Step 1: We claim that there exists N >0 such that if N, there hold

u(x) u(x) and v(x) v(x). (3.5)

Define u.

= {x | u(x) > u(x)} and v

.= {x | v(x) > v(x)}.

From Lemma2, the mean-value theorem and since 1|x y|n

1

|x y|n

1|y|1

1

|y|1

0 for y ,

u(x) u(x)

1|x y|n

1

|x y|n 1

|y|1 (v(y)q

v(y)q

) dy

v

v(y)q1

|x y|n|y|1(v(y) v(y)) dy.

Thus, applying the weighted HLS inequality followed by Holders inequalitygives us

u uLr0(u) Cvq1 (v v)

Lnr0

n+r0(1) (v)

Cvq1Ls0 (v)

v vLs0 (v) Cvq1

Ls0(C)v vLs0 (v). (3.6)

Similarly, there holds

v vLs0 (v) Cup1 (u u)

Lns0

n+s0(2) (u)

Cup1Lr0 (u)

u uLr0 (u) Cup1

Lr0(C)u uLr0(u). (3.7)

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By virtue of (u, v) Lr0(Rn) Ls0(Rn), we can choose N suitably large

such that for N, there holds

C2up1Lr0 (C

)

vq1Ls0 (C

)

1/2.

Therefore, combining (3.6) with (3.7) gives us u uLr0 (u)

12u uLr0 (u),

v vLs0 (v) 12v vLs0 (v),

which further implies that u uLr0(u) = 0 and v vLs0 (v)

= 0.Thus, both u and

v have measure zero and are therefore empty. This

concludes the proof of the first claim.Step 2: We can move the plane to the right provided that (3.5) holds.Let

0.

= sup{ | u(x) u(x), v(x) v(x)}.

We claim that u, v are symmetric about the plane 0 i.e.

u0(x) =u(x) and v0(x) =v(x) for all x 0.

On the contrary, assume 0 0 and for all x 0 ,

u0(x) u(x) and v0(x) v(x), but u0(x) =u(x) or v0(x) =v(x).

But we will show this implies the plane can be moved further to the right,thereby contradicting the definition of0. Namely, there is a small > 0such that

u(x) u(x) and v(x) v(x), x , for all [0, 0+). (3.8)

In the case, say v0(x) =v(x) on 0 , Lemma2indicates u0(x)< u(x) inthe interior of 0. Define

u0.

= {x 0 | u0(x) u(x)} and v0

.= {x 0 | v0(x) v(x)}.

Then u0 and v0

have measure zero and

lim0

u u0 and lim0

v v0.

Let denote the reflection of the set about the plane . Accordingto the integrability ofu and v, we can choose a suitably small such that

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for all [0, 0+ ), uLr0 ((u)) andvLs0 ((v)

) are sufficiently small.

Therefore, (3.6) and (3.7) imply

u uLr0(u) Cvq1Ls0((v)

)v vLs0 (v) 12v vLs0(v),

v vLs0 (v) Cup1Lr0((u)

)u uLr0(u) 12u uLr0(u),

and we deduce that u and v are both empty. This proves (3.8). Hence,

we conclude that u and v are symmetric and decreasing about the plane0.

Step 3: We assert that u and v are radially symmetric and decreasingabout the origin.

First, notice that 0 is indeed equal to zero. Otherwise, if0 0 such that

u(x) =A0+ o(1)

|x|n for x Rn\BR(0), (4.1)

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and we shall often invoke this property in establishing the decay rates for

v(x).

Proof. (i) Without loss of generality, we assume 1 > 0 since the prooffor the unweighted case is similar but far simpler. For each R > 0, sincev L(Rn) and 1< n, we have

BR(0)

v(y)q

|y|1dy < .

So it remains to showBR(0)C

v(y)q

|y|1 dy < . There are two cases to consider.

(1.) Assumen p(n ) ( 2). It is clear thatq p,1 2 and(1.8) imply q n+21n . Choose > 0 with ( 21, 1). Then

set = n+n+21 and = n++21 so that

1+ 1 = 1,lq > nn , and > n1 .

Therefore, Holders inequality and Theorem5implyBR(0)C

v(y)q

|y|1dy

BR(0)C

1

|y|1 1

dy 1

BR(0)Cv(y)q dy

1/

C

Rtn1

dt

t

1

BR(0)Cv(y)q dy

1/ Cv

qq < .

(2.) Assume n > p(n ) ( 2). For small >0, take = nn1+

and

=

n

1 so that

1

+

1

= 1. From the non-subcritical condition (1.8)and since pq >1, we get

q(n 2) + (n 1)

q(1 +p) =

n 1q(1 +p)

+n 2

1 +p 0.

Then, combining this with the decreasing property ofv yields

v(|x|/2)s|x|n C

B|x|/2(0)\B|x|/4(0)

v(y)s dy < . (4.6)

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We claim that

|x|n

J3= o(1) as |x| . (4.7)

To do so, we consider two cases.Case 1. Letn p(n ) ( 2). Then (4.6) implies that

v(|x|/2)q |x|q(n) C,

which when combined with (4.5), yields

|x|q(n)(1)J3 Cv(|x|/2)q |x|q(n) C.

Recall thatq n+21n , which implies thatq(n ) ( 1) n 1>n . Thus, by choosing suitably small and sending |x| in the

previous estimate, we obtain (4.7).Case 2. Letn > p(n ) ( 2). Then (4.6) implies

v(|x|/2)q |x|q(p(n)(2)) C,

which, when combined with (4.5), gives us

|x|q[p(n)(2)](1)J3 C.

It is easy to check that (4.2) implies that

q[p(n ) ( 2)] ( 1)> n .

Assertion (4.7) follows by sending |x| in the last estimate.Notice that (4.7) implies that

lim|x|

J3= 0. (4.8)

Hence, (4.3),(4.4) and (4.8) imply

lim|x|

|x|nu(x) =A0,

and this completes the proof of the proposition.

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4.2 Fast decay rates for v(x)

Proposition 4. Ifp(n ) +2> n, then

(i) A1.

=

Rn

u(y)p

|x|2dy < ;

(ii) lim|x|

|x|nv(x) =A1.

Proof. Since p > n2n , (i) follows from Theorem5and Holders inequalitysimilar to the proof of Proposition 3(i).

To prove (ii), write

|x|n

v(x) =BR(0)

|x|nu(y)p

|x y|n|y|2 dy

+

Rn\BR(0)

|x|n[A0+ o(1)]p

|x y|n|y|p(n)+2dy

.=J4+J5.

For a large R > 0, if y BR(0), then lim|x||x|n

|xy|n = 1, and the

Lebesgue dominated convergence theorem implies that

limR

lim|x|

J4 = limR

lim|x|

BR(0)

|x|nu(y)p

|x y|n|y|2dy= A1. (4.9)

Likewise, since p(n ) +2> n,

J5 C

R

tnp(n)2dt

t = o(1) as R for suitably large |x|.

Particularly,Rn\BR(0)

|x|n

|x y|n|y|p(n)+2dy= o(1) as |x| , R .

Hence, assertion (ii) follows from this and (4.9).

Proposition 5. Ifp(n ) +2= n, then

lim|x|

|x|n

ln |x| v(x) =Ap0|S

n1|,

where |Sn1| is the surface area of the(n 1)-dimensional unit sphere.

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Proof. From (4.1), we have for large R >0,

|x|n

ln |x| v(x) =

1

ln |x|

BR(0)

|x|nu(y)p

|x y|n|y|2dy

+(A0+ o(1))

p

ln |x|

Rn\BR(0)

|x|n

|x y|n|y|ndy.

Indeed, (4.9) implies that

1

ln |x|

BR(0)

|x|nu(y)p

|x y|n|y|2dy= o(1) as |x| .

Thus, it only remains to show that

1

ln |x|

Rn\BR(0)

|x|n

|x y|n|y|ndy |Sn1| as |x| . (4.10)

Indeed, for large R >0 and c (0, 1/2), polar coordinates and a change ofvariables give us

1

ln |x|

Rn\BR(0)

|x|n

|x y|n|y|ndy =

1

ln |x|

cR|x|

Sn1

1

r|e rw|ndSwdr

+ 1

ln |x|

Rn\Bc(0)

1

|z|n|e z|ndz ,

where e is a unit vector in Rn. Clearly, the integral in the second term canbounded above by a positive constant depending only on c, sincen < nfor z neare and n +n > n near infinity. Thus, we deduce that

1

ln |x|

Rn\Bc(0)

1

|z|n|e z|ndz = o(1) as |x| .

For r (0, c), it is also clear that 1 c |e rw| 1 + c. Therefore, thereexists (1, 1) such that |e rw| = 1 +c, which leads to

1

ln |x| c

R

|x|Sn1

1

r|e rw|

ndwdr=

|Sn1|

(1 +c)

n

ln |x|

(ln c ln R+ ln |x|)

|Sn1|

(1 +c)n as |x| .

Hence, by sendingc 0, we obtain (4.10) and this concludes the proof.

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Proposition 6. Ifp(n ) +2< n, then

A2.

=Ap0

Rn

dz

|z|p(n)+2 |e z|n < .

Moreover,lim

|x||x|p(n)(2)v(x) =A2.

Proof. According to the non-subcritical condition, we have that n21+p nnear infinity;n < nneare; andp(n ) + 2 < nnear the origin. Thus, we conclude that A2< .

For large R >0, we use (4.1) to write

|x|p(n)(2)v(x) = |x|p(n)+2nBR(0)

|x|nu(y)p|x y|n|y|2

dy

+ (A0+ o(1))p

Rn\BR(0)

|x|(p+1)(n)+2n

|x y|n|y|p(n)+2dy. (4.11)

Indeed, ify BR(0), then

|x|p(n)+2nBR(0)

|x|nu(y)p

|x y|n|y|2dy C|x|p(n)+2n 0

as |x| . Likewise, as |x|

Rn\BR(0)

|x|(p+1)(n)+2n

|x y|n|y|p(n)+2dy=

Rn\BR/|x|(0)

dz|z|p(n)+2 |e z|n

A2Ap0

.

Inserting these calculations into (4.11) leads to the desired result.

4.3 Characterization of integrable solutions

Proposition 7. Let u, v be positive solutions of (1.1) satisfying (1.7). Ifu, v are bounded and decay with the fast rates as |x| , then u, v areintegrable solutions.

Proof. Supposeu, v are bounded and decay with the fast rates. From (1.7),it is clear that (n )r0 > n. Thus,

Rnu(x)r0 dx C+

Rn\BR(0)

u(x)r0 dx C1+C2

R

tn(n)r0dt

t < .

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Similarly, Rnv(x)s0 dx < ifv decays with rate |x|(n). Ifv decays with

the rate |x|n

ln |x|, then we can find a suitably large R >0 and small >0for which (ln |x|)s0 |x| for |x| > R. Then, we also getn (n)s0+

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Proposition 8. Let 1 < q0 and 2 < p0. Then there does not exist a

positive constantc such that eitheru(x) c(1 + |x|)1 or v(x) c(1 + |x|)2.

Proof. Assume that there exists such ac >0 in which,

u(x) c(1 + |x|)1 where 1< q0.

Then there holds for large x,

v(x)

B|x|/2(x)

u(y)p

|x y|n|y|2dy c(1 + |x|)a1 ,

where b0 = 1 and a1 = pb0 + 2. Thus, inserting this into the firstintegral equation yields

u(x)

B|x|/2(x)

v(y)q

|x y|n|y|1dy c(1 + |x|)b1 ,

whereb1= qa1 + 1. By inductively repeating this argument, we arriveat

v(x) c(1 + |x|)aj and u(x) c(1 + |x|)bj ,

where

aj+1= pbj +2 and bj =qaj +1 for j= 1, 2, 3, . . . .

A simple calculation yields

bk = qak +1 = q(pbk1 +2) +1=pqbk1 ((1 +q) (1+2q))

= (pq)2bk2 ((1 +q) (1+2q))(1 +pq)

...

= (pq)kb0 ((1 +q) (1+2q))(1 +pq+ (pq)2 +. . .+ (pq)k1)

= (pq)kb0 ((1 +q) (1+2q))(pq)k 1

pq 1 = (pq)k(b0 q0) +q0.

Since pq >1 and b0 =1 < q0, we can find a sufficiently large k0 such thatbk0 0,

v(x) c

Rn\BR(0)

u(y)p

|x y|n|y|2dy c

Rn\BR(0)

|y|pbk0

|x y|n|y|2dy

c

R

t2pbk0dt

t = .

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Hencev(x) = , which is impossible. Similarly, if there exists a c >0 such

thatv(x) c(1 + |x|)2 where 2< p0,

then we can apply the same iteration argument to conclude u(x) = forlarge x and this completes the proof.

Proposition 9. There hold the following.

(i) Let3> q0 and4> p0. Ifu, v are not integrable solutions, then theredoes not exist a positive constantCsuch that either,

u(x) C(1 + |x|)3 or v(x) C(1 + |x|)4 .

(ii) Ifu, v are not integrable solutions but are decaying solutions i.e.

u(x) |x|1 and v(x) |x|2

for some1, 2 > 0, then they necessarily have the slow rates1 = q0and2= p0.

Proof. (i) On the contrary, assume there exists a C > 0 such that u(x) C(1 + |x|)3. Then n r03< 0 and we calculate that

Rn

u(x)r0 dx=

BR(0)

u(x)r0 dx+

Rn\BR(0)

u(x)r0 dx

C1+C2 R

tnr03 dtt

< ,

which contradicts the assumption that u, v are integrable solutions. Like-wise, if there exists a C > 0 such that v(x) C(1 +|x|)4 , a similarargument shows that v Ls0(Rn), which is a contradiction.

(ii) Now suppose that u, v are not integrable solutions but are decayingsolutions. In view of Theorem1(i), it only remains to show u(x) c|x|q0

and v(x)c|x|p0 as |x| . Thus, assume for large x, u(x) c|x|1

and v(x) c|x|2 for some 1, 2 > 0. Then we can deduce that1 = q0,since if1

.=b0> q0, we can then use the iteration argument in the proof of

Proposition 8to get for large x, u(x) c|x|bk with bk as k .

Thus, by choosing bk0 sufficiently large and if|x| >2Rfor a suitable choiceofR >0, then

v(x) c

B|x|(0)\BR(0)

u(y)p

|x y|n|y|2dy c

|x|R

t2bk0pdt

t c.

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Therefore,v(x) c >0 as |x| , which is impossible. Likewise, we can

apply the same argument to arrive at 2 =p0. This completes the proof ofthe proposition.

Proof of Theorem4. Part (i) follows immediately from Proposition 8 andparts (ii) and (iii) follow from Proposition 9.

Acknowledgment:This work was completed during a visiting appointment at theUniversity of Oklahoma, and the author would like to express his sincere appreci-ation to the university and the Department of Mathematics for their hospitality.

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