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Advanced Nonlinear Studies

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Volume 16, Issue 1

Issues

Symmetry and Regularity of Solutions to the Weighted Hardy–Sobolev Type System

Lu Chen / Zhao Liu / Guozhen Lu
Published Online: 2015-12-08 | DOI: https://doi.org/10.1515/ans-2015-5005

Abstract

Hardy–Littlewood–Sobolev inequalities and the Hardy–Sobolev type system play an important role in analysis and PDEs. In this paper, we consider the very general weighted Hardy–Sobolev type system

u(x)=n1|x|τ|x-y|n-α|y|tf1(u(y),v(y))𝑑y,v(x)=n1|x|t|x-y|n-α|y|τf2(u(y),v(y))𝑑y,

where

f1(u(y),v(y))=λ1up1(y)+μ1vq1(y)+γ1uα1(y)vβ1(y),f2(u(y),v(y))=λ2up2(y)+μ2vq2(y)+γ2uα2(y)vβ2(y).

Only the special cases when γ1=γ2=0 and one of λi and μi is zero (for both i=1 and i=2) have been considered in the literature. We establish the integrability of the solutions to the above Hardy–Sobolev type system and the C regularity of solutions to this system away from the origin, which improves significantly the Lipschitz continuity in most works in the literature. Moreover, we also use the moving plane method of [8] in integral forms developed in [6] to prove that each pair (u,v) of positive solutions of the above integral system is radially symmetric and strictly decreasing about the origin.

Keywords: Hardy–Littlewood–Sobolev Inequality; Method of Moving Planes in Integral Forms; Radial Symmetry; Regularity

MSC 2010: 35B06; 45G15

1 Introduction

Let 0<ν<n and 1<θ,m<. The well-known classical Hardy–Littlewood–Sobolev inequality [11] states

nnf(x)g(y)|x-y|ν𝑑x𝑑yCθ,m,νfLmgLθ

for any fLm(n), gLθ(n), and for 1m+1θ+νn=2.

Hardy and Littlewood also introduced the double weighted inequality, which was later generalized by Stein and Weiss in [20]. It states

|nnf(x)g(y)|x|τ|x-y|ν|y|t𝑑y𝑑x|Ct,τ,θ,m,vfLmgLθ,(1.1)

where τ+t0,1θ+1m+ν+t+τn=2 and 1-1m-νnτn<1-1m (see also [9] for Stein–Weiss inequalities on the Heisenberg group).

One can also write the above Stein–Weiss inequality in another form. Let

T(g)(x)=nf(x)g(y)|x|τ|x-y|ν|y|t𝑑y,

then

T(g)Ls=supfLm=1T(g)(x),f(x)Ct,τ,θ,m,νgLθ,

where 1θ+ν+t+τn=1+1s, 1m+1s=1.

To obtain the best constant in the weighted inequality (1.1), one can maximize the functional

J(f,g)=nnf(x)g(y)|x|τ|x-y|ν|y|t𝑑y𝑑x

under the constraints fLm=gLθ=1. The corresponding Euler–Lagrange equations are the integral system

v1mf(x)m-1=n1|x|τ|x-y|n-α|y|tg(y)𝑑y,v2θg(x)θ-1=n1|x|t|x-y|n-α|y|τf(y)𝑑y,(1.2)

where f,g0 and v1m=v2θ=J(f,g).

Set u=c1fm-1,v=c2gθ-1,p=1m-1,q=1θ-1, then for a proper choice of constants c1 and c2, system (1.2) becomes

u(x)=n1|x|τ|x-y|n-α|y|tvq(y)𝑑y,v(x)=n1|x|t|x-y|n-α|y|τup(y)𝑑y.(1.3)

Integral system (1.3) is closely related to the following partial differential equations:

(-Δ)α2(|x|τu(x))=|x|-tvq(x),(-Δ)α2(|x|tv(x))=|x|-τup(x).(1.4)

Lei, Li and Ma [10] established the equivalence between the weighted Hardy–Sobolev type system (1.3) and the corresponding Euler–Lagrange system (1.4). Chen et al. [3] obtained the symmetry, monotonicity and regularity of solutions of (1.3). In particular, they also obtained the optimal integrability of the solutions for a class of such system. The best constant in a weighted Hardy–Littlewood–Sobolev inequality related to integral system (1.3) was studied in [5] and solutions of (1.3) for α=2,t=τ,p=q were classified. For more results related to Hardy–Sobolev type equation and system, we refer the reader to [2, 15] and the references therein.

In the special cases when τ=t=0,u=v,p=q=n+αn-α, (1.3) becomes the single equation

u(x)=n1|x-y|n-αun+αn-α(y)𝑑y.

The corresponding PDE is the well-known family of semi-linear equations

(-Δ)α2u(x)=un+αn-α(x).(1.5)

In particular, when n3 and α=2, equation (1.5) becomes

(-Δ)u(x)=un+2n-2(x).(1.6)

The classification of the solutions of (1.6) has provided an important ingredient in the study of the well-known Yamable problem and the prescribing scalar curvature. Caffarelli, Gidas and Spruck [1] classified the solutions of equation (1.6). Wei and Xu [21] generalized their results to the solutions of the more general equation (1.5) with α being any even number between 0 and n.

Let 0<τ+t<α<n. We consider the following weighted Hardy–Sobolev type system:

u(x)=n1|x|τ|x-y|n-α|y|tf1(u(y),v(y))𝑑y,v(x)=n1|x|t|x-y|n-α|y|τf2(u(y),v(y))𝑑y,(1.7)

where

f1(u(y),v(y))=λ1up1(y)+μ1vq1(y)+γ1uα1(y)vβ1(y),f2(u(y),v(y))=λ2up2(y)+μ2vq2(y)+γ2uα2(y)vβ2(y).

Such f1(u(y),v(y)) and f2(u(y),v(y)) are considerably more complicated when none of γi, λi and μi is zero (i=1,2). Therefore, it is more difficult to deal with these cases.

In this paper, we always assume that λi,μi,γi (i=1,2) are nonnegative constants and they are not equal to zero simultaneously. We also use the following notations:

Π1={g(x)xn,gLt11(n)Lt12(n)Lk0(n)},Π2={g(x)xn,gLt21(n)Lt22(n)Lk0(n)},

where t1i=n(pi-1)α-t-τ,t2i=n(qi-1)α-t-τ (i=1,2) and k0=n(α1+β1-1)α-t-τ=n(α2+β2-1)α-t-τ.

We first obtain the following regularity of Hardy–Sobolev type system (1.7).

For α1,β2>0, α2,β1>1, pi,qi>1 (i=1,2), if (u,v)Π1×Π2 is a pair of solutions of system (1.7), then u,vLs(n) for any nn-α+t+τ<s<.

We note that t1i,t2i and k0 cannot be compared easily, thus one cannot use interpolation to conclude what the spaces Π1 or Π2 are. Therefore, it is highly nontrivial to derive the Ls integrability of the solutions u and v in Theorem 1.1. One can also see from its proof that the above lower bound for s is the best one can get from the Hardy–Littlewood–Sobolev inequality.

Next, we will establish the C of the solutions away from the origin for the solutions u and v. This improves significantly known results in the literature where only the Lipschitz continuity of the solutions of the Hardy–Sobolev system without weight has been established, except in the special case of the single equation and when τ=0 and the critical exponent p=n+α-2tn-α (see [15]).

For α1,β2>0, α2,β1>1, nn-α+t+τ<pi,qi,αi+βi< (i=1,2), if (u,v)Π1×Π2 is a pair of solutions of system (1.7), then |x|τu(x),|x|tv(x)L(n) and u,vC(n{0}).

It is well known that the moving plane method was invented by Alexandrov in the 1950s. Then the method was further developed by Serrin [19], Gidas, Ni and Nirenberg [8], Caffarelli, Gidas and Spruck [1] and many other authors. In [6], Chen, Li and Ou developed the method of moving plane in integral forms which is quite different from the traditional moving plane method of differential equations which relies on maximum principles. The moving plane in integral forms can be easily applied to higher order equations without using maximum principles. For more results related to the moving plane in integral forms and related symmetry results and Liouville-type theorems, see, e.g., [4, 7, 12, 13, 14, 16, 17] and the references therein.

To establish the symmetry of the solutions, we use the moving plane in integral forms and the Hardy–Littlewood–Sobolev inequality to prove the following theorem.

For pi,qi,αi,βi>1 (i=1,2), assume that (u,v)Π1×Π2 is a pair of nonnegative solutions of system (1.7). Then u and v are radially symmetric and strictly decreasing about the origin.

This paper is organized as follows. In Section 2 and Section 3, we prove Theorem 1.1 and Theorem 1.2, respectively. In Section 4, we prove that each pair of solutions of (1.7) is radially symmetric and strictly decreasing about the origin, thus give the proof of Theorem 1.3.

2 Proof of Theorem 1.1

In this section, we need the following regularity lifting lemma (see [18]) and the weighted Hardy–Littlewood–Sobolev inequality to prove Theorem 1.1.

Let V be a topological vector space. Suppose there are two extended norms (i.e., the norm of an element in V might be infinity) defined on V,

X,Y:V[0,].

Let

X:={fV:fX<}andY:={fV:fY<}.

Let T be a contraction map from X into itself and from Y into itself. Assume that for any fX there exists a function gZ:=XY such that f=Tf+gX. Then fZ.

Now, we start our proof of Theorem 1.1. Denote

ua(x)={u(x)if |u(x)|>a or |x|>a,0otherwise,

and ub(x)=u(x)-ua(x).

Assume that (ϕ,φ)Ls(n)×Ls(n) for nn-α+t+τ<s<. Define

T1(ϕ,φ)=nλ1uap1-1(y)ϕ(y)|x|τ|x-y|n-α|y|t𝑑y+n1|x|τ|x-y|n-α|y|tfa(y)φ(y)𝑑y,T2(ϕ,φ)=nμ2vaq2-1(y)φ(y)|x|t|x-y|n-α|y|τ𝑑y+n1|x|t|x-y|n-α|y|τga(y)ϕ(y)𝑑y.

where

fa(y)=μ1vaq1-1(y)+γ1uaα1(y)vaβ1-1(y)+γ1uaα1(y)vbβ1-1(y)+γ1ubα1(y)vaβ1-1(y),ga(x)=λ2uap2-1(y)+γ2vaβ2(y)uaα2-1(y)+γ2vaβ2(y)ubα2-1(y)+γ2vbβ2(y)uaα2-1(y).

Let

F(x)=nλ1ubp1-1(y)u(y)|x|τ|x-y|n-α|y|t𝑑y+nμ1vbq1-1(y)+γ1ubα1(y)vbβ1-1(y)|x|τ|x-y|n-α|y|tv(y)𝑑y,G(x)=nμ2vbq2-1(y)v(y)|x|t|x-y|n-α|y|τ𝑑y+nλ2ubp2-1(y)+γ2vbβ2(y)ubα2-1(y)|x|t|x-y|n-α|y|τu(y)𝑑y.

Denote the norm in the cross product space Ls(n)×Ls(n) by

(ϕ,φ)Ls×Ls=ϕLs+φLs,

and define the mapping T: Ls(n)×Ls(n)Ls(n)×Ls(n) by

T(ϕ,φ)=(T1(ϕ,φ),T2(ϕ,φ)).

Consider the equation

(ϕ,φ)=T(ϕ,φ)+(F,G).(2.1)

It is easy to check that (u,v) is a pair of solutions of (2.1).

In order to show (u,v)Ls(n)×Ls(n) for all s>nn-α+t+τ, we carry out the proof by two steps:

  • 1.

    T is a contracting map from Ls(n)×Ls(n) to itself for sufficiently large a.

  • 2.

    F and G belong to Ls(n).

Then from Lemma 2.1, we conclude the proof.

Step 1. For any (ϕ,φ)Ls(n)×Ls(n) with s>nn-α+t+τ, by the weighted Hardy–Littlewood–Sobolev inequality and the Minkowski inequality, we have

T1(ϕ,φ)LsC(uap1-1ϕLθ+vaq1-1φLθ+uaα1vaβ1-1φLθ+uaα1vbβ1-1φLθ+ubα1vaβ1-1φLθ)C(uaLt11p1-1ϕLs+vaLt21q1-1φLs+uaLk0α1vaLk0β1-1φLs+uaLk0α1vbLk0β1-1φLs+ubLk0α1vaLk0β1-1φLs)(2.2)

and

T2(ϕ,φ)LsC(vaq2-1φLθ+uap2-1ϕLθ+vaβ2uaα2-1ϕLθ+vaβ2ubα2-1ϕLθ+vbβ2uaα2-1ϕLθ)C(vaLt22q2-1φLs+uaLt12p2-1ϕLs+vaLk0β2uaLk0α2-1ϕLs+vaLk0β2ubLk0α2-1ϕLs+vbLk0β2uaLk0α2-1ϕLs),(2.3)

where θ=nsn+(α-t-τ)s, t1i=n(pi-1)α-t-τ,t2i=n(qi-1)α-t-τ and k0=n(αi+βi-1)(α-t-τ), i=1,2.

Since uLt1i(n)Lk0(n),vLt2i(n)Lk0(n), we may choose sufficiently large a such that

{CuaLt11p1-114,CvaLt22q2-114,C(vaLt21q1-1+uaLk0α1vaLk0β1-1+uaLk0α1vbLk0β1-1+ubLk0α1vaLk0β1-1)14,C(uaLt12p2-1+vaLk0β2uaLk0α2-1+vaLk0β2ubLk0α2-1+vbLk0β2uaLk0α2-1)14.(2.4)

Hence, combining (2), (2.3) and (2.4), we obtain

T1(ϕ,φ)Ls14(ϕLs+φLs),T2(ϕ,φ)Ls14(ϕLs+φLs).(2.5)

By (2.5), we have

T(ϕ,φ)Ls×Ls12(ϕLs+φLs),

which implies that T is a contraction map from Ls(n)×Ls(n) to itself.

Step 2. Since the estimates of F and G are similar, we only consider F:

F(x)=nλ1ubp1-1(y)u(y)|x|τ|x-y|n-α|y|tdy+nμ1vbq1-1(y)+γ1ubα1(y)vbβ1-1(y)|x|τ|x-y|n-α|y|tv(y)dy=:J1+J2.

For any s>nn-α+t+τ, we apply the weighted Hardy–Littlewood–Sobolev inequality and the Minkowski inequality to obtain

J1LsCubp1-1uLθCubLk1p1-1uLk2,J2LsCvbq1-1vLθ+Cubα1vbβ1-1vLθCvbLk3q1-1vLk4+CubLk5α1vbLk6β1-1vLk7,

where

p1-1k1+1k2=q1-1k3+1k4=α1k5+β1-1k6+1k7=n+(α-t-τ)sns=1s+α-t-τn.

Since ubLp(n) and vbLp(n) for 0<p, the constants k1,k3,k5,k6 can be chosen arbitrarily. In view of u,vLk0(n), we may choose k2=k4=k7=k0 such that

1s=p1-1k1+n-(α-t-τ)k0nk0=q1-1k3+n-(α-t-τ)k0nk0=α1k5+β1-1k6+n-(α-t-τ)k0nk0.(2.6)

We carry out the process by two cases.

Case 1. If k0nα-t-τ, for any s>nn-α+t+τ, there exist k1,k3,k5,k6>0 such that (2.6) is valid. Then we obtain

FLs=J1Ls+J2Ls<for any s>nn-α+t+τ.

Similarly, we have

GLs<for any s>nn-α+t+τ.

By Lemma 2.1, we conclude that (u,v)Ls(n)×Ls(n) for any s>nn-α+t+τ.

Case 2. If k0<nα-t-τ, for any nn-α+t+τ<snk0n-(α-t-τ)k0, we have

FLs<,GLs<.

By Lemma 2.1 again, we derive that (u,v)Ls(n)×Ls(n) for any nn-α+t+τ<snk0n-(α-t-τ)k0. Repeating the above process with k0 replaced by k01=nk0n-(α-t-τ)k0, we can obtain that (u,v)Ls(n)×Ls(n) for any nn-α+t+τ<snk01n-(α-t-τ)k01. After a few steps, there exists j0 such that k0j0nα-t-τ. This returns to Case 1.

Therefore, combining Case 1 and Case 2, we conclude that (u,v)Ls(n)×Ls(n) for any s>nn-α+t+τ. This completes the proof of Theorem 1.1.

3 Proof of Theorem 1.2

In this section, we adopt ideas from Lu and Zhu [15] to prove Theorem 1.2. The proof is carried out by two parts.

Part I. We show that |x|τu(x),|x|tv(x)L(n). Since the estimates of |x|τu(x) and |x|tv(x) are similar, we only discuss |x|τu(x). The function |x|τu(x) can be written as

|x|τu(x)=nλ1up1(y)|x-y|n-α|y|t𝑑y+nμ1vq1(y)|x-y|n-α|y|t𝑑y+nγ1uα1(y)vβ1(y)|x-y|n-α|y|t𝑑y=:λ1A(x)+μ1B(x)+γ1C(x).

We claim that A(x)L<. Write

|A(x)|Ba(0)|u(y)|p1|x-y|n-α|y|tdy+nBa(0)|u(y)|p1|x-y|n-α|y|tdy=:A1(x)+A2(x).(3.1)

We first estimate A1(x). If xnB2a(0), by the Hölder inequality and uLs for any s>nn-α+t+τ, we have

A1(x)Ba(0)|u(y)|p1|x-y|n-α|y|t𝑑yBa(0)|u(y)|p1|y|n-α+t𝑑y<.(3.2)

If xB2a(0), we also have

A1(x)Ba(0)|u(y)|p1|y|n-α+t𝑑y+B3a(x)|u(y)|p1|x-y|n-α+t𝑑y<.(3.3)

Combining (3.2) and (3.3), for any xn, we derive that

A1(x)=Ba(0)|u(y)|p1|x-y|n-α|y|t𝑑y<.(3.4)

Next, it suffices to estimate A2(x). For xn, we have

nBa(0)|u(y)|p1|x-y|n-α|y|t𝑑y1atBa(x)|u(y)|p1|x-y|n-α𝑑y+(nBa(0))(nBa(x))|u(y)|p1|x-y|n-α|y|t𝑑y.(3.5)

In virtue of the Hölder inequality, it is easy to see that

1atBa(x)|u(y)|p1|x-y|n-α𝑑y<.(3.6)

We also have

(nBa(0))(nBa(x))|u(y)|p1|x-y|n-α|y|t𝑑y<(3.7)

since

(nBa(0))(nBa(x))|u(y)|p1|x-y|n-α|y|t𝑑ynBa(0)|u(y)|p1|y|n-α+t𝑑y+nBa(x)|u(y)|p1|x-y|n-α+t𝑑y.

Combining (3.5), (3.6) and (3.7), we deduce that

A2(x)=nBa(0)|u(y)|p1|x-y|n-α|y|t𝑑y<.(3.8)

Therefore, from (3.1), (3.4) and (3.8), we can derive that A(x)L(n). Similarly, we can also show that

B(x)L(n)andC(x)L(n).

Thus we accomplish the estimate of Part I.

Part II. We show that u(x),v(x)C(n{0}). One only need to prove that |x|τu(x)C(n{0}). For any xn{0}, we choose a ball B3r(x) with radius 3r such that {0}B¯3r(x)=, then

|x|τu(x)=nλ1up1(y)|x-y|n-α|y|t𝑑y+nμ1vq1(y)|x-y|n-α|y|t𝑑y+nγ1uα1(y)vβ1(y)|x-y|n-α|y|t𝑑y=:λ1A(x)+μ1B(x)+γ1C(x).

We only show that A(x), C(x)C(n{0}). We split A(x) into two parts, i.e. A(x) can be written as

A(x)=nB3r(x)λ1up1(y)|x-y|n-α|y|t𝑑y+B3r(x)λ1up1(y)|x-y|n-α|y|t𝑑y.(3.9)

We show that

P(x)=nB3r(x)up1(y)|x-y|n-α|y|t𝑑yC(n{0}).

Let

Q(x,y)=up1(y)|x-y|n-α|y|tχnB3r(x).

For fixed xn, if h is small enough, we have

|Q(x+hei,y)-Q(x,y)h|=|up1(y)|y|t(χnB3r(x+hei)|x+hei-y|n-α-χnB3r(x)|x-y|n-α)h|C|u(y)|p1χnB3r(x+θ¯hei)|y|t|x+θ¯hei-y|n-α+1C|u(y)|p1|y|trn-α+1χBϵ(0)+C|u(y)|p1|y|t|x-y|n-α+1χn(Br(x)Bϵ(0)),(3.10)

where ei={0,,1,,0} is the i-th unit vector, 0<θ¯<1 and ϵ is sufficiently small such that B3r(x)Bϵ(0)=. For such fixed r and ϵ, it is easy to verify

Bϵ(0)|u(y)|p1|y|trn-α+1<(3.11)

since uLs(n) for any s>nn-α+t+τ. Consider that

n(Br(x)Bϵ(0))|u(y)|p1|y|t|x-y|n-α+1𝑑ynBr(x)|u(y)|p1|x-y|n-α+1+t𝑑y+nBϵ(0)|u(y)|p1|y|n-α+1+t𝑑y,

and uLs(n) for any s>nn-α+t+τ. By the Hölder inequality, we have

nBr(x)|u(y)|p1|x-y|n-α+1+t𝑑y<andnBϵ(0)|u(y)|p1|y|n-α+1+t𝑑y<.

Thus, we conclude that

n(Br(x)Bϵ(0))|u(y)|p1|y|t|x-y|n-α+1𝑑y<.(3.12)

Combining (3.10), (3.11), (3.12) and Lebesgue’s dominated convergence theorem, we obtain that P(x)C1(n{0}). Continuing this process, we can derive that

P(x)C(n{0}).(3.13)

By the standard singular integral estimate (see [11]), for any σ<α, we have

B2r(x)Bϵ(0)|u(y)|p1|y|t|x-y|n-α𝑑yCσ(n{0}).(3.14)

Combining (3.9), (3.13) and (3.14), we obtain that A(x)Cσ(n{0}). By the bootstrap technique, we can prove that A(x)C(n{0}).

For the estimate of C(x), similarly, we also can split C(x) into two parts, i.e. C(x) can be written as

C(x)=nB3r(x)μ1uα1(y)vβ1(y)|x-y|n-α|y|t𝑑y+B3r(x)μ1uα1(y)vβ1(y)|x-y|n-α|y|t𝑑y.(3.15)

We show that

P*(x)=nB3r(x)uα1(y)vβ1(y)|x-y|n-α|y|t𝑑yC(n{0}).

Let

Q*(x,y)=uα1(y)vβ1(y)|x-y|n-α|y|tχnB3r(x).

For fixed xn, if h is small enough, we use the same estimate as (3.10) to obtain

|Q*(x+hei,y)-Q*(x,y)h|<C{|u(y)|α1|v(y)|β1|y|trn-α+1χBϵ(0)+|u(y)|α1|v(y)|β1|y|t|x-y|n-α+1χn(Br(x)Bϵ(0))}.

For such fixed r and ϵ, it is easy to verify that

Bϵ(0)|u(y)|α1|v(y)|β1|y|trn-α+1<.

Note that

n(Br(x)Bϵ(0))|u(y)|α1|v(y)|β1|y|t|x-y|n-α+1𝑑ynBr(x)|u(y)|α1|v(y)|β1|x-y|n-α+1+t𝑑y+nBϵ(0)|u(y)|α1|v(y)|β1|y|n-α+1+t𝑑y(3.16)

and u,vLs(n) for any s>nn-α+t+τ. We have

nBr(x)|u(y)|α1|v(y)|β1|x-y|n-α+t+1𝑑y(nBr(x)(|u(y)|α1|v(y)|β1)m1𝑑y)1/m1(nBr(x)1|x-y|m2(n-α+t+1)𝑑y)1/m2uLm1(α1+β1)α1vLm1(α1+β1)β1(nBr(x)1|x-y|m2(n-α+t+1)𝑑y)1/m2<,(3.17)

where 1m1+1m2=1.

Similarly, we can also obtain

nBϵ(0)|u(y)|α1|v(y)|β1|y|n-α+1+t𝑑y<.(3.18)

Combining (3.16), (3.17) and (3.18), we can derive that

n(Br(x)Bϵ(0))|u(y)|α1|v(y)|β1|y|t|x-y|n-α+1𝑑y<.

By Lebesgue’s dominated convergence theorem, we conclude that

P*(x)C1(n{0}).(3.19)

Combining (3.15), (3.19) and the standard singular integral estimate (see [11]), for any σ<α, we have

B2r(x)Bϵ(0)|u(y)|α1|v(y)|β1|y|t|x-y|n-α𝑑yCσ(n{0}).

Thus we derive that C(x)Cσ(n{0}). By the bootstrap technique, we can prove that C(x)C(n{0}). Then the proof of Part II is accomplished.

4 Proof of Theorem 1.3

In this section, we will use the method of moving plane in integral forms introduced in [6] to prove that each pair (u,v) of solutions of (1.7) is radially symmetric and strictly decreasing about the origin. In order to prove our theorem, we first introduce some notations. For any real number λ, let the moving plane be Tλ={xx1=λ} and denote Σλ={x=(x1,x2,,xn)x1λ}. Let xλ=(2λ-x1,x2,,xn) be the reflection of the point x about the moving plane Tλ, and define uλ(x)=u(xλ) and vλ(x)=v(xλ).

If (u,v) is a pair of nonnegative solutions of (1.7), for any xΣλ and xy, we have

u(x)-u(xλ)=Σλ(1|xλ|τ|xλ-y|n-α-1|x|τ|x-y|n-α)(1|yλ|t-1|y|t)f1(u(y),v(y))𝑑y+Σλ(1|x|τ-1|xλ|τ)(1|x-y|n-α|yλ|t+1|xλ-y|n-α|yλ|t)f1(uλ(y),vλ(y))𝑑y+Σλ(1|xλ|τ|xλ-y|n-α|yλ|t-1|x|τ|x-y|n-α|yλ|t)(f1(uλ(y),vλ(y))-f1(u(y),v(y))dy

and

v(x)-v(xλ)=Σλ(1|xλ|t|xλ-y|n-α-1|x|t|x-y|n-α)(1|yλ|τ-1|y|τ)f2(u(y),v(y))𝑑y+Σλ(1|x|t-1|xλ|t)(1|x-y|n-α|yλ|τ+1|xλ-y|n-α|yλ|τ)f2(uλ(y),vλ(y))𝑑y+Σλ(1|xλ|t|xλ-y|n-α|yλ|τ-1|x|t|x-y|n-α|yλ|τ)(f2(uλ(y),vλ(y))-f2(u(y),v(y))dy.

Proof.

By (1.7), we have

u(x)=Σλ1|x|τ|x-y|n-α|y|tf1(u(y),v(y))𝑑y+Σλ1|x|τ|x-yλ|n-α|yλ|tf1(uλ(y),vλ(y))𝑑y

and

u(xλ)=Σλ1|xλ|τ|xλ-y|n-α|y|tf1(u(y),v(y))𝑑y+Σλ1|xλ|τ|xλ-yλ|n-α|yλ|tf1(uλ(y),vλ(y))𝑑y.

Since |xλ-y|=|x-yλ| and |xλ-yλ|=|x-y|, we have

u(x)-u(xλ)=Σλ(1|x|τ|x-y|n-α|y|t-1|xλ|τ|xλ-y|n-α|y|t)f1(u(y),v(y))𝑑y+Σλ(1|x|τ|x-yλ|n-α|yλ|t-1|xλ|τ|xλ-yλ|n-α|yλ|t)f1(uλ(y),vλ(y))𝑑y=Σλ(1|xλ|τ|xλ-y|n-α-1|x|τ|x-y|n-α)(1|yλ|t-1|y|t)f1(u(y),v(y))𝑑y+Σλ(1|x|τ-1|xλ|τ)(1|x-y|n-α|yλ|t+1|xλ-y|n-α|yλ|t)f1(uλ(y),vλ(y))𝑑y+Σλ(1|xλ|τ|xλ-y|n-α|yλ|t-1|x|τ|x-y|n-α|yλ|t)(f1(uλ(y),vλ(y))-f1(u(y),v(y))dy.

Similarly, we can also obtain the second formula. This completes the proof of Lemma 4.1. ∎

Now we start to prove Theorem 1.3, the proof will be carried out by two steps.

Step 1. We compare the values of uλ(x) and u(x), vλ(x) and v(x). For λ sufficiently negative, we are going to show that

uλ(x)-u(x)0,vλ(x)-v(x)0for all xΣλ.(4.1)

Define

Σλu={xΣλuλ(x)-u(x)>0},Σλv={xΣλvλ(x)-v(x)>0}.

We will prove that for sufficiently negative λ, both Σλu and Σλv must be empty, and hence (4.1) holds.

According to the mean value theorem and Lemma 4.1, for any xΣλu, we have

u(xλ)-u(x)Σλ(1|x|τ|x-y|n-α|yλ|t-1|xλ|τ|xλ-y|n-α|yλ|t)(f1(uλ(y),vλ(y))-f1(u(y),v(y)))𝑑yλ1p1Σλuuλp1-1(y)(uλ(y)-u(y))|x|τ|x-y|n-α|y|t𝑑y+μ1q1Σλvvλq1-1(y)(vλ(y)-v(y))|x|τ|x-y|n-α|y|t𝑑y+γ1α1Σλuuλα1-1(y)vλβ1(y)(uλ(y)-u(y))|x|τ|x-y|n-α|y|t𝑑y+γ1β1Σλvuα1(y)vλβ1-1(y)(vλ(y)-v(y))|x|τ|x-y|n-α|y|t𝑑y=:A1(x)+A2(x)+A3(x)+A4(x).

For any s>nn-α+t+τ, we apply the weighted Hardy–Littlewood–Sobolev inequality and the Hölder inequality to the above inequality and obtain

A1Ls(Σλu)Cuλp1-1(uλ-u)Lnsn+(α-t-τ)s(Σλu)CuλLt11(Σλu)p1-1uλ-uLs(Σλu),(4.2)A2Ls(Σλv)Cvλq1-1(vλ-v)Lnsn+(α-t-τ)s(Σλv)CvλLt21(Σλv)q1-1vλ-vLs(Σλv),(4.3)A3Ls(Σλu)Cuλα1-1vλβ1(uλ-u)Lnsn+(α-t-τ)s(Σλu)Cuλα1-1vλβ1Lnα-t-τ(Σλu)uλ-uLs(Σλu)CuλLk0(Σλu)α1-1vλLk0(Σλu)β1uλ-uLs(Σλu)(4.4)

and

A4Ls(Σλu)Cuα1vλβ1-1(vλ-v)Lnsn+(α-t-τ)s(Σλv)Cuα1vλβ1-1Lnα-t-τ(Σλv)vλ-vLs(Σλv)CuLk0(Σλv)α1vλLk0(Σλv)β1-1vλ-vLs(Σλv),(4.5)

where t11=nα-t-τ(p1-1), t21=nα-t-τ(q1-1) and k0=nα-t-τ(α1+β1-1).

Combining (4.2), (4.3), (4.4) and (4.5), we obtain

uλ-uLs(Σλu)CuλLt11(Σλu)p1-1uλ-uLs(Σλu)+CvλLt21(Σλv)q1-1vλ-vLs(Σλv)+CuλLk0(Σλu)α1-1vλLk0(Σλu)β1uλ-uLs(Σλu)+CuLk0(Σλv)α1vλLk0(Σλv)β1-1vλ-vLs(Σλv)C(uλLt11(Σλu)p1-1+uλLk0(Σλu)α1-1vλLk0(Σλu)β1)uλ-uLs(Σλu)+C(vλLt21(Σλv)q1-1+uLk0(Σλv)α1vλLk0(Σλv)β1-1)vλ-vLs(Σλv).(4.6)

Similarly, we can also deduce that

vλ-vLs(Σλv)C(uλLt12(Σλu)p2-1+uλLk0(Σλu)α2-1vλLk0(Σλu)β2)uλ-uLs(Σλu)+C(vλLt22(Σλv)q2-1+uLk0(Σλv)α2vλLk0(Σλv)β2-1)vλ-vLs(Σλv),(4.7)

where t12=nα-t-τ(p2-1), t22=nα-t-τ(q2-1) and k0=nα-t-τ(α2+β2-1).

In virtue of the condition (u,v)Π1×Π2, we can choose sufficiently negative λ such that

{CuλLt11(Σλu)p1-1+CuλLk0(Σλu)α1-1vλLk0(Σλu)β1<14,CvλLt21(Σλv)q1-1+CuLk0(Σλv)α1vλLk0(Σλv)β1-1<14,CuλLt12(Σλu)p2-1+CuλLk0(Σλu)α2-1vλLk0(Σλu)β2<14,CvλLt22(Σλv)q2-1+CuLk0(Σλv)α2vλLk0(Σλv)β2-1<14.(4.8)

By (4.6), (4.7) and (4.8), we have

uλ-uLs(Σλu)=0,vλ-vLs(Σλv)=0,

which implies that Σλu and Σλv must be of measure zero. Therefore, both Σλu and Σλv must be empty sets.

Step 2. Inequality (4.1) provides a starting point to move the plane Tλ={xnx1=λ}. Now we start from the negative infinity of the x1-axis and move the plane to the right as long as (4.1) holds. Define

λ0=sup{λuμ(x)u(x),vμ(x)v(x),μλ for all xΣμ\{0}}.

We will prove

λ0=0.

Suppose on the contrary that λ0<0. We will then show that u and v must be symmetric about the plane Tλ0, that is,

uλ0(x)u(x),vλ0(x)v(x)for all xΣλ0.(4.9)

Otherwise, we may assume on Σλ0,

u(x)uλ0(x),v(x)vλ0(x),butuλ0(x)u(x)orvλ0(x)v(x).

In the case of uλ0(x)u(x) on Σλ0, by Lemma 4.1, we can obtain vλ0(x)v(x). Furthermore, in virtue of Lemma 4.1, we can derive u(x)>uλ0(x) and v(x)>vλ0(x) in the interior of Σλ0.

Next, we will show that the plane can be moved further to the right. More precisely, there exists an ε>0 such that for any λ[λ0,λ0+ε),

u(x)uλ(x),v(x)vλ(x)for all xΣλ\{0}.

Let

Σλ0u¯={xΣλ0u(x)uλ0(x)},Σλ0v¯={xΣλ0v(x)vλ0(x)}.

Obviously, both Σλ0u¯ and Σλ0v¯ have measure zero, and

limλλ0ΣλuΣλ0u¯,limλλ0ΣλvΣλ0v¯.

This together with the integrability condition (u,v)Π1×Π2 ensures that one can choose ε small enough such that, for all λ[λ0,λ0+ε),

CuλLt11(Σλu)p1-1+CuλLk0(Σλu)α1-1vλLk0(Σλu)β1<14,CvλLt21(Σλv)q1-1+CuLk0(Σλv)α1vλLk0(Σλv)β1-1<14,CuλLt12(Σλu)p2-1+CuλLk0(Σλu)α2-1vλLk0(Σλu)β2<14,CvλLt22(Σλv)q2-1+CuLk0(Σλv)α2vλLk0(Σλv)β2-1<14.

Using the similar estimates as (4.6) and (4.7), we can derive that

u-uλLs(Σλu)=|v-vλLs(Σλv)=0,

thus Σλu and Σλv must be of measure zero, which verifies (4.9). Finally, we show that the plane cannot stop before hitting the origin. On the contrary, if the plane stops at x1=λ0<0, then u(x) and v(x) must be symmetric about the plane x1=λ0, that is

u(x)=uλ0(x),v(x)=vλ0(x)for all xΣλ0.(4.10)

In virtue of |xλ0-y|>|x-y|, |x-yλ0|>|x-y|, |xλ0|>|x| and |yλ0|>|y|, we have

u(x)-u(xλ0)=Σλ0(1|x|τ|x-y|n-α|y|t-1|xλ0|τ|xλ0-y|n-α|y|t)f1(u(y),v(y))𝑑y+Σλ0(1|x|τ|x-yλ0|n-α|yλ0|β-1|xλ0|τ|x-y|n-α|yλ0|t)f1(uλ0(y),vλ0(y))𝑑y>Σλ01|x|τ(1|x-y|n-α|y|t-1|xλ0-y|n-α|y|t)f1(u(y),v(y))𝑑y+Σλ01|x|τ(1|x-yλ0|n-α|yλ0|t-1|x-y|n-α|yλ0|t)f1(uλ0(y),vλ0(y))𝑑y=Σλ01|x|τ(1|x-y|n-α-1|xλ0-y|n-α)(1|y|t-1|yλ0|t)f1(u(y),v(y))𝑑y>0,

which contradicts (4.10). Since the x1 direction can be chosen arbitrary, we conclude that both u(x) and v(x) are radially symmetric and strictly decreasing about the origin. This completes the proof of Theorem 1.3.

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About the article

Received: 2015-06-16

Accepted: 2015-08-16

Published Online: 2015-12-08

Published in Print: 2016-02-01


Funding Source: National Natural Science Foundation of China

Award identifier / Grant number: 11371056

Funding Source: National Science Foundation

Award identifier / Grant number: DMS-1301595

Funding Source: Simons Foundation

Award identifier / Grant number: Simons Fellowship

The research of the first and second authors was partly supported by the NNSF of China (grant no. 11371056), the research of the third author was partly supported by US NSF grant DMS-1301595 and a Simons Fellowship from the Simons Foundation.


Citation Information: Advanced Nonlinear Studies, Volume 16, Issue 1, Pages 1–13, ISSN (Online) 2169-0375, ISSN (Print) 1536-1365, DOI: https://doi.org/10.1515/ans-2015-5005.

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