We study positive solutions to the integral system
{u(x)=∫RNK(y)v−p(y)|x−y|N−αdy for all x∈RN,v(x)=∫RNL(y)u−q(y)|x−y|N−βdy for all x∈RN,
where p,q>0, α,β∈(0,N) and K,L:RN→(0,∞) are continuous functions which satisfy C1(1+|x|)−γ≤K(x),L(x)≤C2(1+|x|)−γ in RN, for some γ>0 and constants C2>C1>0. We discuss the existence, nonexistence, and uniqueness of positive solutions to the above system with respect to α,β,p,q, and γ. We also classify the finite and infinite total mass solutions of the system.
Citation: Changhong Li. An integral system of Matukuma type with negative exponents[J]. Networks and Heterogeneous Media, 2025, 20(2): 566-589. doi: 10.3934/nhm.2025025
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We study positive solutions to the integral system
{u(x)=∫RNK(y)v−p(y)|x−y|N−αdy for all x∈RN,v(x)=∫RNL(y)u−q(y)|x−y|N−βdy for all x∈RN,
where p,q>0, α,β∈(0,N) and K,L:RN→(0,∞) are continuous functions which satisfy C1(1+|x|)−γ≤K(x),L(x)≤C2(1+|x|)−γ in RN, for some γ>0 and constants C2>C1>0. We discuss the existence, nonexistence, and uniqueness of positive solutions to the above system with respect to α,β,p,q, and γ. We also classify the finite and infinite total mass solutions of the system.
In this paper, we are concerned with the study of positive solutions to the following integral system:
{u(x)=∫RNK(y)v−p(y)|x−y|N−αdy for all x∈RN,v(x)=∫RNL(y)u−q(y)|x−y|N−βdy for all x∈RN, | (1.1) |
where p,q>0, α,β∈(0,N), and K,L:RN→(0,∞) are continuous functions which satisfy:
C1(1+|x|)−γ≤K(x),L(x)≤C2(1+|x|)−γ in RN, | (1.2) |
for some γ>0 and C2>C1>0.
We investigated positive solutions (u,v) of Eq (1.1), that is, functions u,v∈C(RN) such that u,v>0 and satisfying Eq (1.1) pointwise in RN. Our study is motivated by the Matukuma equation [1,2] introduced in the 1930s to describe the dynamics of globular clusters of stars. Mathematically, the Matukuma equation reads as
Δu+11+|x|2up=0 in R3 | (1.3) |
where p>1. Solutions of Eq (1.3) which tend to zero at infinity satisfy the integral equation
u(x)=∫R3u(y)p|x−y|(1+|y|2)dy for all x∈R3. |
Also, the quantity
14π∫R3up1+|x|2dx | (1.4) |
is called the total mass of the solution u of Eq (1.3) (see [3,4]). The extended integral equation
u(x)=∫RNu(y)p|x−y|N−α(1+|y|α)dy for all x∈RN,N≥3 | (1.5) |
was studied in [5], and the behavior of positive solutions was discussed in [6,7,8] (see also [9,10,11,12,13]). The case of negative exponents, that is,
u(x)=∫RNK(y)u(y)−p|x−y|N−αdy in RN(N≥1), |
was recently investigated in [14] under the conditions p>0, α∈(0,N), and C1(1+|x|)−β≤K(x)≤C2(1+|x|)−β in RN, for some β>0 and C2>C1>0. Integral systems with negative exponents were considered in the case of half-space in [15,16] and in the case of whole-space in [17,18,19] (see also [20,21,22,23,24] for further results).
The aim of the present paper is to discuss various qualitative properties of the positive solutions to Eq (1.1). We start our study of Eq (1.1) with the following nonexistence result which extends and improves Theorem 1.1 in [14].
Theorem 1.1. (Nonexistence)
Let 1≤s1,s2≤∞ be such that
max{α+N(ps2+1s1),β+N(qs1+1s2)}>γ. | (1.6) |
Then, Eq (1.1) has no positive solutions (u,v)∈Ls1(RN)×Ls2(RN).
We notice that if max{α,β}>γ, then Eq (1.6) is clearly fulfilled and we directly obtain the following corollary.
Corollary 1.2. Let 1≤s1,s2≤∞. If one of the conditions below holds,
(ⅰ) max{α,β}>γ and 1≤s1,s2≤∞;
(ⅱ) max{α,β}≥γ and 1≤s1,s2≤∞, (s1,s2)≠(∞,∞);
then Eq (1.1) has no positive solutions (u,v)∈Ls1(RN)×Ls2(RN).
Next, using some integral identities from harmonic analysis, we can construct an exact solution to Eq (1.1).
Theorem 1.3. (Exact solution)
Assume that γ>2N and let
K(x)=(1+|x|2)−γ2,L(x)=(1+|x|2)−γ2. |
Then, there exist C,D>0, and p,q>0 depending on α,β,γ, and N such that
{u(x)=C(1+|x|2)−N−α2v(x)=D(1+|x|2)−N−β2x∈RN, | (1.7) |
is an exact solution to Eq (1.1).
The condition γ>2N is needed to ensure that pq≠1, which is a restriction in our argument.
Next, we turn to the existence of a solution to Eq (1.1) in a more general framework. Given two positive functions f,g:RN→(0,∞) we use the symbol f≃g to denote that the quotient f/g is bounded between two positive constants on RN. Our result in this case is stated below.
Theorem 1.4. (Existence)
Assume max{α,β}<γ.
(ⅰ) If max{α,β}<γ<N, 0<p<γ−αγ−β, and 0<q<γ−βγ−α, then the integral system (1.1) has a solution (u,v) such that
{u(x)≃(1+|x|)−γ−α−p(γ−β)1−pq,v(x)≃(1+|x|)−γ−β−q(γ−α)1−pq. |
Moreover, (u,v)∈C0,μ(RN)×C0,ν(RN) for some μ,ν∈(0,1).
(ⅱ) If γ>N, 0<p<γ−NN−β, 0<q<γ−NN−α and pq<1, then the integral system (1.1) has a solution (u,v) such that
{u(x)≃(1+|x|)α−N,v(x)≃(1+|x|)β−N. |
Moreover, (u,v)∈C0,μ(RN)×C0,ν(RN) for some μ,ν∈(0,1).
The existence of a positive solution to Eq (1.1) is achieved using the Schauder fixed-point theorem. One major difficulty in our approach is the fact that C(RN) is not a suitable space to work on, due to its lack of compactness. Instead, we shall use an approximation argument and construct solutions to Eq (1.1) over a sequence of expanding balls in RN. Further, using some integral estimates which are universal (see Lemma 2.2 below), we are able to obtain the solution to Eq (1.1) through a limit argument. Our argument requires a restriction in the range of exponents p and q as stated in Theorem 1.4 above. Let us point out that the restrictions on p,q in Theorem 1.4(ii) are natural. They appear again in Theorem 1.5 and Theorem 1.6 below.
Next, we show that Eq (1.1) may have a unique solution in some ranges of exponents p and q. This is stated in the following result.
Theorem 1.5. (Uniqueness)
Assume 0<p<γ−NN−β and 0<q<γ−NN−α. If pq<1, then the integral system (1.1) has a unique positive solution (u,v) in C(RN)×C(RN).
Finally, and in analogy to the case of a single equation (see Eq (1.4)), we say that a solution (u,v) of system (1.1) has finite total mass if:
∫RNK(x)v(x)−pdx<∞ and ∫RNL(x)u(x)−qdx<∞. | (1.8) |
Otherwise, we say that (u,v) has infinite total mass.
Our last result in this section is stated below
Theorem 1.6. (Finite and infinite total mass solutions)
Let (u,v) be a positive solution of Eq (1.1).
(ⅰ) If 0<p<γ−NN−β and 0<q<γ−NN−α, then (u,v) has finite total mass. Moreover
lim|x|→∞|x|N−αu(x)=∫RNK(y)v(y)−pdy, | (1.9) |
lim|x|→∞|x|N−βv(x)=∫RNL(y)u(y)−qdy. | (1.10) |
(ⅱ) If either p=γ−NN−β, q>γ−NN−α or q=γ−NN−α, p>γ−NN−β, then (u,v) has infinite total mass.
The remainder of this paper is organized as follows. In Section 2, we recall some results related to the Riesz potential and its regularity. Section 3 is devoted to the proof of Theorem 1.1. Further, Theorems 1.3–1.5 are proved in Sections 4, 5, and 6. The method we develop in the article can be used to study integral systems of type (1.1) with different signs of exponents. This will be explained in Section 7. Throughout this paper, C,c,c1,c2,... denote positive constants whose values may change from line to line, unless otherwise stated. By Br, we denote the open ball in RN centered at the origin and having radius r>0.
Let f∈L1loc(RN). The Riesz potential of order γ∈(0,N) of f is given by
Iγf(x)=∫RNf(y)|x−y|N−γdy for all x∈RN. |
For α,β∈(0,N), we define
Jα,βf(x)=(∫RNf(y)|x−y|N−αdy,∫RNg(y)|x−y|N−βdy) for all x∈RN, |
and for n>0 we also set
Jα,β,nf(x)=(∫Bnf(y)|x−y|N−αdy,∫Bng(y)|x−y|N−βdy) for all x∈Bn. |
We recall the following lemma from [25] and [26].
Lemma 2.1. Assume s>1 is such that α−1<Ns<α and β−1<Ns<β. Let μ=α−Ns and ν=β−Ns. Then, the Riesz operators
Jα,β:Ls(RN)→C0,μ(RN)×C0,ν(RN) |
and
Jα,β,n:Ls(Bn)→C0,μ(¯Bn)×C0,ν(¯Bn) |
are continuous.
From [14], we recall the following integral estimates that extend to those in [27].
Lemma 2.2. Let n≥1 be an integer, α∈(0,N), and σ>α.
(ⅰ) If α<σ<N, then there exist c2>c1>0 independent of n such that for all x∈Bn, we have
c1(1+|x|)α−σ≤∫Bndy(1+|y|)σ|x−y|N−α≤c2(1+|x|)α−σ. | (2.1) |
(ⅱ) If σ>N, then there exist c2>c1>0 independent of n such that for all x∈Bn, we have
c1(1+|x|)α−N≤∫Bndy(1+|y|)σ|x−y|N−α≤c2(1+|x|)α−N. | (2.2) |
(ⅲ) There exist c2>c1>0 independent of n such that for all x∈Bn, we have
c1(1+|x|)α−Nlog(e+|x|)≤∫Bndy(1+|y|)N|x−y|N−α≤c2(1+|x|)α−Nlog(e+|x|). | (2.3) |
Assume that (u,v)∈Ls1(RN)×Ls2(RN) is a positive solution of Eq (1.1) for some 1≤s1,s2≤∞.
Case 1: s1=s2=∞. Then, u and v are bounded, and Eq (1.6) yields α>γ.
For all x∈RN∖B1, we deduce
u(x)≥C∫RNK(y)|x−y|N−αdy≥C∫|y|≥|2x|dy(1+|y|)γ|x−y|N−α. |
We notice that
|y|≥|2x|⇒|x−y|≤|x|+|y|≤2|y|. |
Together with α>γ, they yield
u(x)≥C∫|y|≥|2x|(2|y|)α−γ−Ndy≥C∞∫2|x|tα−γ−1dt=∞. | (3.1) |
This contradicts the assumption that u is bounded, implying that Eq (1.1) has no positive solution in Ls1(RN)×Ls2(RN).
Case 2: 1≤s1,s2<∞. Without loss of generality, from Eq (1.6) we may assume
α+N(ps2+1s1)>γ. |
Let R>1. Then,
u(x)=∫RNK(y)v−p(y)|x−y|N−αdy≥C∫|y|<Rv−p(y)(1+|y|)γ|x−y|N−αdy. |
If |x|,|y|<R, then |x−y|≤2R and 1+|y|≤1+R≤2R. We estimate
u(x)≥C∫BRv−p(y)2Rγ+N−αdy=CRα−γ(1RN∫BRv−p(y)dy)≥CRα−γ(1RN∫BR(vs2(y))−ps2dy). |
By Jensen's inequality, we obtain
u(x)≥CRα−γ(1RN∫BRvs2(y)dy)−ps2=CRα−γ+Nps2(∫BRvs2(y)dy)−ps2. |
This yields
(∫BRvs2(y)dy)ps2u(x)≥CRα−γ+Nps2. |
Since v∈Ls2(RN), the above estimate implies
(∫RNvs2(y)dy)ps2u(x)≥CRα−γ+Nps2 for all x∈BR. |
This yields
u(x)≥CRα−γ+Nps2 in BR, | (3.2) |
and then
us1(x)≥CR(α−γ)s1+Nps1s2,∫BRus1(x)≥CR(α−γ)s1+Nps1s2+N. |
Since α−γ+N(ps2+1s1)>0 holds, letting R→∞ in the above estimate, we deduce
∫RNus1(x)=∞. |
This contradicts the assumption that u∈Ls1(RN), implying that Eq (1.1) has no positive solution in Ls1(RN)×Ls2(RN).
Case 3: 1≤s1<∞=s2. In this case, Eq (1.6) yields
max{α+Ns1,β+Nqs1}>γ. |
If α+Ns1>γ, as in the estimate Eq (3.1), we obtain
u(x)≥C∫|y|≥|2x||y|α−γ−Ndy=C∞∫2|x|tα−γ−1dtfor all |x|>1. |
This further yields α<γ and u(x)≥C|x|α−γ for |x|>1. Then, u(x)s1≥C|x|(α−γ)s1 for all x∈RN∖B1, which yields u∉Ls1(RN), a contradiction.
If β+Nqs1>γ, as in the estimate Eq (3.2) which we obtained in Case 2 above, we find
v(x)≥CRβ−γ+Nqs1 in BR, |
which contradicts the fact that v∈L∞(RN).
Case 4: 1≤s2<∞=s1. We use the same argument as in Case 3 above in which we replace s1 by s2 and u by v to raise a contradiction.
The proof of Corollary 1.2 follows directly from Theorem 1.1 since either of the conditions (i) and (ii) imply Eq (1.6).
We will employ the following integral identities, see [28].
∫RNdy(1+|y|2)N+α2|x−y|N−α=πN/2Γ(α2)Γ(N+α2)(1+|x|2)−N−α2 for all x∈RN. |
∫RNdy(1+|y|2)N+β2|x−y|N−β=πN/2Γ(β2)Γ(N+β2)(1+|x|2)−N−β2 for all x∈RN. |
where Γ stands for the Gamma function. Based on the α,β in Eq (1.1), we set
{A(α)=πN/2Γ(α2)Γ(N+α2),A(β)=πN/2Γ(β2)Γ(N+β2). | (3.3) |
Let us take
{K(x)=(1+|x|2)−γ2L(x)=(1+|x|2)−γ2 for all x∈RN, |
where γ>2N, and next define
{p=γ−N−αN−β>0,q=γ−N−βN−α>0. |
Since γ>max{2N,α+β}, we have pq≠1. Next, let C,D>0 be such that CDp=A(α) and DCq=A(β). We then deduce
{C=(A(β)pA(α))1/(pq−1),D=(A(α)qA(β))1/(pq−1). |
Next, we claim that
{u(x)=C(1+|x|2)−N−α2,x∈RN,v(x)=D(1+|x|2)−N−β2,x∈RN. |
is a solution of Eq (1.1). Indeed, we have
∫RNK(y)v(y)−p|x−y|N−αdy=D−p∫RNdy(1+|y|2)N+α2|x−y|N−α=D−pA(α)(1+|x|2)−N−α2=C(1+|x|2)−N−α2=u(x) for all x∈RN. |
Similarly, we have
∫RNL(y)u(y)−q|x−y|N−αdy=C−q∫RNdy(1+|y|2)N+β2|x−y|N−β=C−qA(β)(1+|x|2)−N−β2=D(1+|x|2)−N−β2=v(x) for all x∈RN. |
This concludes the proof of our result.
The existence of a positive solution to Eq (1.1) will be obtained by combining the Schauder fixed-point theorem with the properties of the Riesz potentials stated in Section 2. We split our discussion into two cases.
(ⅰ) Assume 0<p<γ−αγ−β and 0<q<γ−βγ−α. Let
κ1=γ−α−p(γ−β)1−pq>0 and κ2=γ−β−q(γ−α)1−pq>0. | (3.4) |
For n≥1, we define the closed and convex set An⊂C(¯Bn)×C(¯Bn) by
An={(u,v)∈C(¯Bn)×C(¯Bn):m1(1+|x|)−κ1≤u(x)≤M1(1+|x|)−κ1m2(1+|x|)−κ2≤v(x)≤M2(1+|x|)−κ2 in ¯Bn}, | (3.5) |
where 0<m1<M1, 0<m2<M2 are constants depending on α,β,γ,p,q, and N that will be chosen in Lemma 3.1 below.
For all u,v∈An, we define
Jn(u,v)(x)=(∫BnK(y)v(y)−p|x−y|N−αdy,∫BnL(y)u(y)−q|x−y|N−βdy) for all x∈¯Bn. |
By Lemma 2.2 (ⅰ) for σ=γ−pκ2>α and σ=γ−qκ1>β, respectively, there exists c2>c1>0 independent of n≥1 such that
c1(1+|x|)−κ1≤∫BnK(y)dy(1+|y|)−pκ2|x−y|N−α≤c2(1+|x|)−κ1 for all x∈Bn, | (3.6) |
and
c1(1+|x|)−κ2≤∫BnL(y)dy(1+|y|)−qκ1|x−y|N−β≤c2(1+|x|)−κ2 for all x∈Bn. | (3.7) |
Lemma 3.1. Let
m1=(cp2c1)1pq−1,M1=(cp1c2)1pq−1,m2=(c12c1)1pq−1,M2=(cq1c2)1pq−1. | (3.8) |
where c1,c2 are defined in Eqs (3.6) and (3.7). If pq<1, then Jn(An)⊂An.
Proof. Since pq<1, it is obvious to see that m1<M1 and m2<M2. We also have
{m1=c1M−p2,M1=c2m−p2,m2=c1M−q1,M2=c2m−q1. | (3.9) |
To prove Jn(An)⊂An, let (u,v)∈An.
Since v(x)≤M2(1+|x|)−κ2 in Bn, by Eqs (3.6) and (3.9)1 we have
∫BnK(y)v(y)−p|x−y|N−αdy≥M−p2∫BnK(y)(1+|x|)−pκ2|x−y|N−αdy≥M−p2c1(1+|x|)−κ1=m1(1+|x|)−κ1 in Bn. |
Also, v(x)≥m2(1+|x|)−κ2 in Bn together with Eqs (3.6) and (3.9)2 implies
∫BnK(y)v(y)−p|x−y|N−αdy≤m−p2∫BnK(y)(1+|x|)−pκ2|x−y|N−αdy≤m−p2c2(1+|x|)−κ1=M1(1+|x|)−κ1 in Bn. |
Similarly, u(x)≤M1(1+|x|)−κ1 in Bn combined with Eqs (3.7) and (3.9)3 yields
∫BnL(y)u(y)−q|x−y|N−βdy≥M−q1∫BnL(y)(1+|x|)−qκ1|x−y|N−βdy≥M−q1c1(1+|x|)−κ2=m2(1+|x|)−κ2 in Bn. |
Finally, since u(x)≥m1(1+|x|)−κ1 in Bn, Eqs (3.7) and (3.9)4 produces
∫BnL(y)u(y)−q|x−y|N−βdy≤m−q1∫BnL(y)(1+|x|)−qκ1|x−y|N−βdy≤m−q1c2(1+|x|)−κ2=M2(1+|x|)−κ2 in Bn. |
Hence, Jn(An)⊂An.
From the definition of κ1 and κ2 in Eq (3.4), we can easily check that
α<γ−pκ2 and β<γ−qκ1. |
Thus, we can select s>1 such that
α−1<Ns<α<γ−pκ2 and β−1<Ns<β<γ−qκ1. |
Let μ=α−Ns∈(0,1) and ν=β−Ns∈(0,1). By Lemma 2.1, we obtain that
Jn=Jα,β,n:An⊂Ls(Bn)×Ls(Bn)→C0,μ(Bn)×C0,ν(Bn) is continuous |
and Jn(An)⊂An.
Hence, we can use the Schauder fixed-point theorem for Jn, which implies the existence of (un,vn)∈An such that
(un,vn)=(∫BnK(y)vn(y)−p|x−y|N−αdy,∫BnL(y)un(y)−q|x−y|N−βdy) for all x∈¯Bn. |
Next, we argue as follows.
● Since {(un,vn)} is bounded in C0,μ(¯B1)×C0,ν(¯B1). Since the embedding C0,μ(¯B1)×C0,ν(¯B1)↪C(¯B1)×C(¯B1) is compact, there exists a subsequence {(u1n,v1n)}n≥1 of {(un,vn)}n≥1 which converges in C(¯B1)×C(¯B1).
● Since {(u1n,v1n)}n≥2 is bounded in C0,μ(¯B2)×C0,ν(¯B2) and the embedding C0,μ(¯B2)×C0,ν(¯B2)↪C(¯B2)×C(¯B2) is compact, there exists a subsequence {(u2n,v2n)}n≥2 of {(u1n,v1n)}n≥2 which converges in C(¯B2)×C(¯B2).
● Since {(u2n,v2n)}n≥3 is bounded in C0,μ(¯B3)×C0,ν(¯B3) and the embedding C0,μ(¯B3)×C0,ν(¯B3)↪C(¯B3)×C(¯B3) is compact, we can find a subsequence {(u3n,v3n)}n≥3 of {(u2n,v2n)}n≥3 which converges in C(¯B3)×C(¯B3).
● Inductively, we deduce that, for all k≥1, there exists a subsequence {(ukn,vkn)}n≥k of {(uk−1n,vk−1n)}n≥k which converges in C(¯Bk)×C(¯Bk).
Let {(Un,Vn)}={(unn,vnn)}n≥1, which converges to a certain (U,V)∈C(RN)×C(RN) and fulfills
{Un(x)=∫RNK(y)V−pn(y)|x−y|N−αdy for all x∈¯Bn,Vn(x)=∫RNL(y)U−qn(y)|x−y|N−βdy for all x∈¯Bn. |
Since (Un,Vn)∈An, we can apply the Lebesgue dominated convergence theorem to obtain that (U,V) is a continuous solution of Eq (1.1). Moreover,
{U(x)=limn→∞Un(x)≃(1+|x|)−κ1 for all x∈RNV(x)=limn→∞Vn(x)≃(1+|x|)−κ2 for all x∈RN. |
Finally, from Lemma 2.1 we deduce (U,V)∈C0,μ(RN)×C0,ν(RN). This concludes the proof in (i).
(ⅱ) Assume 0<p<γ−NN−β, 0<q<γ−NN−α, and pq<1.
For n≥1, we define the closed and convex subset Bn⊂C(¯Bn)×C(¯Bn) by
Bn={(u,v)∈C(¯Bn)×C(¯Bn):m1(1+|x|)α−N≤u(x)≤M1(1+|x|)α−Nm2(1+|x|)β−N≤v(x)≤M2(1+|x|)β−N in ¯Bn}, | (3.10) |
where 0<m1<M1, 0<m2<M2 are constants depending on α,β,γ,p,q, and N.
For all (u,v)∈Bn, we define
Jn(u,v)(x)=(∫BnK(y)v(y)−p|x−y|N−αdy,∫BnL(y)u(y)−q|x−y|N−βdy) for all x∈¯Bn. |
By Lemma 2.2 (ⅱ) applied twice for σ=γ−p(N−β)>N and for σ=γ−q(N−α)>N, respectively, there exist c2>c1>0 independent of n≥1 such that
c1(1+|x|)α−N≤∫BnK(y)dy(1+|y|)−p(N−β)|x−y|N−α≤c2(1+|x|)α−N for all x∈Bn, | (3.11) |
and
c1(1+|x|)β−N≤∫BnL(y)dy(1+|y|)−q(N−α)|x−y|N−β≤c2(1+|x|)β−N for all x∈Bn. | (3.12) |
We choose m1,m2,M1, and M2 as given by Eq (3.8) with new constants c1,c2 from Eqs (3.11) and (3.12). Then, Lemma 3.1 holds, and we deduce Jn(Bn)⊂Bn.
Next, we select s>1 such that
α−1<Ns<α<N<γ−p(N−β) and β−1<Ns<β<N<γ−q(N−α). |
Let μ=α−Ns∈(0,1) and ν=β−Ns∈(0,1). By Lemma 2.1, we obtain
Jn:Bn⊂Ls(Bn)×Ls(Bn)→C0,μ(Bn)×C0,ν(Bn) is continuous |
and Jn(Bn)⊂Bn.
Hence, we can use the Schauder fixed-point theorem for Jn, which implies the existence of (un,vn)∈Bn such that
(un,vn)=(∫BnK(y)vn(y)−p|x−y|N−αdy,∫BnL(y)un(y)−q|x−y|N−βdy) for all x∈¯Bn. |
As in part (ⅰ) above, the diagonal sequence {(Un,Vn)}={(unn,vnn)}n≥1 converges to a certain (U,V)∈C(RN)×C(RN) which fulfills
{U(x)=∫RNK(y)V−p(y)|x−y|N−αdy for all x∈¯Bn,V(x)=∫RNL(y)U−q(y)|x−y|N−βdy for all x∈¯Bn, |
Since
{U(x)=limn→∞Un(x)≃(1+|x|)α−N for all x∈RNV(x)=limn→∞Vn(x)≃(1+|x|)β−N for all x∈RN, |
we can apply Lemma 2.1, and we deduce (U,V)∈C0,μ(RN)×C0,ν(RN). This finishes the proof of our theorem.
Assume 0<p<γ−NN−β and 0<q<γ−NN−α. The existence of a positive solution to Eq (1.1) in this range of parameters was obtained in Theorem 1.4. We next discuss the uniqueness.
Let (u,v) be a positive solution of Eq (1.1). We note that if |x|>1≥|y|, then |x−y|≤|x|+|y|≤2|x|, and we combine this fact with Kv−p∈L1loc(RN) to obtain
u(x)≥∫|y|<1K(y)v(y)−p|x−y|N−αdy≥(2|x|)α−N∫|y|<1K(y)v(y)−pdy≥C|x|α−N≥C(1+|x|)α−N. |
This means that u(x)≥C(1+|x|)α−N for |x|>1.
If |x|,|y|≤1, then |x−y|≤2 and also Kv−p∈L1(B1) yield
u(x)≥∫|y|<1K(y)v(y)−p|x−y|N−αdy≥2α−N∫|y|<1K(y)v(y)−pdy≥C≥C(1+|x|)α−N. |
We have shown that
u(x)≥C(1+|x|)α−N for all x∈RN. | (3.13) |
In the same way, we obtain
v(x)≥C(1+|x|)β−N for all x∈RN. | (3.14) |
From Eqs (1.1) and (3.14), we estimate
u(x)≤∫RNK(y)(1+|y|)−p(β−N)|x−y|N−αdy≤C∫RNdy(1+|y|)γ−p(N−β)|x−y|N−αdy. |
We can now apply Eq (2.2) with σ=γ−p(N−β)>N to obtain u(x)≤C(1+|x|)α−N in RN, and similarly v(x)≤C(1+|x|)β−N in RN. Combining these two estimates with Eqs (3.13) and (3.14), we deduce
{u(x)≃(1+|x|)α−N for all x∈RN,v(x)≃(1+|x|)β−N for all x∈RN. | (3.15) |
Now let (u1,v1) and (u2,v2) be two positive solutions of Eq (1.1). From Eq (3.15), we have
u1(x)≃u2(x)≃(1+|x|)α−N, |
and so, we can find C>c>0 such that
cu1≤u2≤Cu1 in RN. |
Then, we define
M:=inf{A>1:Au1≥u2 in RN}≥1. | (3.16) |
Clearly, Mu1≥u2. Assume by contradiction that M>1. Then, u2≤Mu1 implies that for all x∈RN, we have
v2(x)=∫RNL(y)u2(y)−q|x−y|N−βdy≥M−q∫RNL(y)u1(y)−q|x−y|N−βdy=M−qv1(x). |
Hence, v2≥M−qv1 in RN, which implies
v−p2≤Mpqv−p1 in RN. |
Together with the previous estimates, we obtain
u2(x)=∫RNK(y)v2(y)−p|x−y|N−αdy≤Mpq∫RNK(y)v1(y)−p|x−y|N−αdy=Mpqu1(x) for all x∈RN. |
Thus, Mpqu1≥u2 in RN.
Since M>1 and 0<pq<1, we have M>Mpq>1, and this is a contradiction of the fact that M is the infimum value defined in Eq (3.16). Hence, M=1 and u1≥u2. Swapping u1 with u2 in the above argument, we deduce u2≥u1, and thus u1≡u2. From Eq (1.1), we also have v1≡v2. This concludes the proof of the uniqueness of a positive solution to Eq (1.1).
First, we start with the proof of the finite total mass solution.
(ⅰ) Recall that in the proof of Theorem 1.5, we established that (u,v) satisfies:
u(x)≥C(1+|x|)α−N and v(x)≥C(1+|x|)β−N | (3.17) |
in RN, for some c>0. Using Eq (3.17) and 0<p<γ−NN−β, we have
∫RNK(x)v(x)−pdx≤C∫RN(1+|x|)−γ+p(N−β)dx≤C∞∫0(1+t)N−γ+p(N−β)−1dt<∞. |
Similarly,
∫RNL(x)u(x)−qdx≤C∞∫0(1+t)N−γ+q(N−α)−1dt<∞. |
Hence, (u,v) has finite total mass.
Next, in order to establish Eq (1.9), we note that
|x|N−αu(x)=∫RNK(y)v(y)−p(|x||x−y|)N−αdy for all x∈RN. |
Using Fatou's lemma, we infer that
lim|x|→∞inf|x|N−αu(x)≥∫RNK(y)v(y)−plim|x|→∞inf(|x||x−y|)N−αdy=∫RNK(y)v(y)−pdy. | (3.18) |
For the converse inequality, take ε∈(0,1) small and write
∫RNK(y)v(y)−p(|x||x−y|)N−αdy=S1+S2+S3. | (3.19) |
where
S1=∫|y|≤ε|x|K(y)v(y)−p(|x||x−y|)N−αdy,S2=∫ε|x|<|y|≤2|x|K(y)v(y)−p(|x||x−y|)N−αdy,S3=∫|y|>2|x|K(y)v(y)−p(|x||x−y|)N−αdy. |
We note that |y|≤ε|x| implies |x−y|≥|x|−|y|≥(1−ε)|x|, so |x||x−y|≤11−ε, and thus
S1≤1(1−ε)N−α∫RNK(y)v(y)−pdy. | (3.20) |
Next, using Eq (3.17) and the estimate on K(y), we have
S2≤C∫ε|x|<|y|≤2|x|(1+|y|)−γ+p(N−β)(|x||x−y|)N−αdy,≤C(1+ε|x|)−γ+p(N−β)|x|N−α∫ε|x|<|y|≤2|x|dy|x−y|N−α,≤C(1+ε|x|)−γ+p(N−β)|x|N−α∫|x−y|≤3|x|dy|x−y|N−α,=C(1+ε|x|)−γ+p(N−β)|x|N. | (3.21) |
Since p<γ−NN−β, we have S2→0 as |x|→∞.
In order to estimate S3, we note that |y|>2|x| yields |x−y|≥|y|−|x|≥|x|, so |x||x−y|≤1, and then
S3≤∫|y|>2|x|K(y)v(y)−pdy→0as|x|→∞. | (3.22) |
Now, using Eqs (3.19)–(3.22), we deduce
lim|x|→∞sup|x|N−αu(x)≤1(1−ε)N−α∫RNK(y)v(y)−pdy. |
Since ε>0 was arbitrarily chosen, this yields
lim|x|→∞sup|x|N−αu(x)≤∫RNK(y)v(y)−pdy. | (3.23) |
From Eqs (3.18) and (3.23), we complete the proof of Eq (1.9). The proof of Eq (1.10) follows similarly.
(ⅱ) Assume, without loss of the generality, that p=γ−NN−β, q>γ−NN−α, and let ε>0 be such that
q>γ−N+εN−α. | (3.24) |
Using Eq (3.17) and Lemma 2.2 (ⅲ), we find:
u(x)≤∫RNK(y)v(y)−p|x−y|N−αdy,≤C∫RNdy(1+|y|)γ−p(N−β)|x−y|N−αdy,≤C(1+|x|)α−Nlog(e+|x|)inRn. |
This last inequality and Eq (3.24) yields,
∫RNL(x)u(x)−qdy≥C∫RN(1+|x|)−γ+q(N−α)log−q(e+|x|)dx,=C∞∫0tN−1(1+t)−γ+q(N−α)log−q(e+t)dt,≥C∞∫1tN−γ+q(N−α)−1log−q(e+t)dt,≥C∞∫1tε−1log−q(e+t)dt,=∞. |
Hence, (u,v) has infinite total mass. This concludes our proof.
In this section we explain how our approach can be used to investigate the existence of solutions to the system
{u(x)=∫RNK(y)vp(y)|x−y|N−αdy for all x∈RN,v(x)=∫RNL(y)u−q(y)|x−y|N−βdy for all x∈RN, | (4.1) |
where p,q>0, α,β∈(0,N), and K,L:RN→(0,∞) are continuous functions which satisfy Eq (1.2). Our main result in this section is stated below.
Theorem 4.1. (Existence)
Assume max{α,β}<γ.
(ⅰ) If γ>N, 0<q<γ−NN−α, and pq<1, then the integral system (4.1) has a solution (u,v) such that
{u(x)≃(1+|x|)α−N,v(x)≃(1+|x|)β−N. |
Moreover, (u,v)∈C0,μ(RN)×C0,ν(RN) for some μ,ν∈(0,1).
(ⅱ) If max{α,β}<γ<N, 0<p<N−γγ−β, and 0<q<γ−βγ−α, then the integral system (4.1) has a solution (u,v) such that
{u(x)≃(1+|x|)−γ−α+p(γ−β)1+pq,v(x)≃(1+|x|)−γ−β−q(γ−α)1+pq. |
Moreover, (u,v)∈C0,μ(RN)×C0,ν(RN) for some μ,ν∈(0,1).
Proof. (ⅰ) Assume γ>N, 0<q<γ−NN−α, and pq<1.
For n≥1, we define the closed and convex subset An⊂C(¯Bn)×C(¯Bn) by
An={(u,v)∈C(¯Bn)×C(¯Bn):m1(1+|x|)α−N≤u(x)≤M1(1+|x|)α−Nm2(1+|x|)β−N≤v(x)≤M2(1+|x|)β−N in ¯Bn}, | (4.2) |
where 0<m1<M1, 0<m2<M2 are constants depending on α,β,γ,p,q, and N.
For all (u,v)∈Bn we define
Jn(u,v)(x)=(∫BnK(y)v(y)p|x−y|N−αdy,∫BnL(y)u(y)−q|x−y|N−βdy) for all x∈¯Bn. |
By Lemma 2.2 (ⅱ) applied twice for σ=γ+p(N−β)>N and for σ=γ−q(N−α)>N, respectively, there exist c2>c1>0 independent of n≥1 such that
c1(1+|x|)α−N≤∫BnK(y)dy(1+|y|)p(N−β)|x−y|N−α≤c2(1+|x|)α−N for all x∈Bn, | (4.3) |
and
c1(1+|x|)β−N≤∫BnL(y)dy(1+|y|)−q(N−α)|x−y|N−β≤c2(1+|x|)β−N for all x∈Bn. | (4.4) |
Lemma 4.2. Let
m1=(cp2q+pq2cp+11)1p2q2−1,M1=(cp2q+pq1cp+12)1p2q2−1,m2=(cpq+q2cpq2+11)1p2q2−1,M2=(cpq+q1cpq2+12)1p2q2−1. | (4.5) |
where c1,c2 are defined in Eqs (4.3) and (4.4). If pq<1, then Jn(An)⊂An.
Proof. Since pq<1, it is obvious to see that m1<M1 and m2<M2. We also have
{m1=c1mp2,M1=c2Mp2,m2=c1M−q1,M2=c2m−q1. | (4.6) |
To prove Jn(An)⊂An, let (u,v)∈An.
Since v(x)≤M2(1+|x|)β−N in Bn, by Eqs (4.3) and (4.6)2 we have
∫BnK(y)v(y)p|x−y|N−αdy≤Mp2∫BnK(y)(1+|x|)p(N−β)|x−y|N−αdy≤Mp2c2(1+|x|)α−N=M1(1+|x|)α−N in Bn. |
Also, v(x)≥m2(1+|x|)β−N in Bn, and by Eqs (4.3) and (4.6)1 we have
∫BnK(y)v(y)p|x−y|N−αdy≥mp2∫BnK(y)(1+|x|)p(N−β)|x−y|N−αdy≥mp2c1(1+|x|)α−N=m1(1+|x|)α−N in Bn. |
Similarly, u(x)≤M1(1+|x|)α−N in Bn combined with Eqs (4.4) and (4.6)3 yields
∫BnL(y)u(y)−q|x−y|N−βdy≥M−q1∫BnL(y)(1+|x|)−q(N−α)|x−y|N−βdy≥M−q1c1(1+|x|)α−N=m2(1+|x|)α−N in Bn. |
Finally, since u(x)≥m1(1+|x|)α−N in Bn, Eqs (4.4) and (4.6)4 produces
∫BnL(y)u(y)−q|x−y|N−βdy≤m−q1∫BnL(y)(1+|x|)−q(N−α)|x−y|N−βdy≤m−q1c2(1+|x|)α−N=M2(1+|x|)α−N in Bn. |
Hence, Jn(An)⊂An.
Let us recall that
γ+p(N−β)>N and γ−q(N−α)>N. |
Next, we select s>1 such that
α−1<Ns<α<N<γ+p(N−β) and β−1<Ns<β<N<γ−q(N−α). |
Let μ=α−Ns∈(0,1) and ν=β−Ns∈(0,1). By Lemma 2.1, we obtain that
Jn=Jα,β,n:An⊂Ls(Bn)×Ls(Bn)→C0,μ(Bn)×C0,ν(Bn) is continuous |
and Jn(An)⊂An.
Hence, we can use the Schauder fixed-point theorem for Jn, which implies the existence of (un,vn)∈An such that
(un,vn)=(∫BnK(y)vn(y)p|x−y|N−αdy,∫BnL(y)un(y)−q|x−y|N−βdy) for all x∈¯Bn. |
We next argue as follows.
● Since {(un,vn)} is bounded in C0,μ(¯B1)×C0,ν(¯B1). Since the embedding C0,μ(¯B1)×C0,ν(¯B1)↪C(¯B1)×C(¯B1) is compact, there exists a subsequence {(u1n,v1n)}n≥1 of {(un,vn)}n≥1 which converges in C(¯B1)×C(¯B1).
● Since {(u1n,v1n)}n≥2 is bounded in C0,μ(¯B2)×C0,ν(¯B2) and the embedding C0,μ(¯B2)×C0,ν(¯B2)↪C(¯B2)×C(¯B2) is compact, there exists a subsequence {(u2n,v2n)}n≥2 of {(u1n,v1n)}n≥2 which converges in C(¯B2)×C(¯B2).
● Since {(u2n,v2n)}n≥3 is bounded in C0,μ(¯B3)×C0,ν(¯B3) and the embedding C0,μ(¯B3)×C0,ν(¯B3)↪C(¯B3)×C(¯B3) is compact, we can find a subsequence {(u3n,v3n)}n≥3 of {(u2n,v2n)}n≥3 which converges in C(¯B3)×C(¯B3).
● Inductively, we deduce that, for all k≥1, there exists a subsequence {(ukn,vkn)}n≥k of {(uk−1n,vk−1n)}n≥k which converges in C(¯Bk)×C(¯Bk).
Let {(Un,Vn)}={(unn,vnn)}n≥1 which converges to a certain (U,V)∈C(RN)×C(RN) and fulfills
{Un(x)=∫RNK(y)Vpn(y)|x−y|N−αdy for all x∈¯Bn,Vn(x)=∫RNL(y)U−qn(y)|x−y|N−βdy for all x∈¯Bn. |
Since (Un,Vn)∈An, we can apply the Lebesgue dominated convergence theorem to obtain that (U,V) is a continuous solution of Eq (4.1). Moreover,
{U(x)=limn→∞Un(x)≃(1+|x|)α−N for all x∈RNV(x)=limn→∞Vn(x)≃(1+|x|)β−N for all x∈RN. |
Finally, from Lemma 2.1 we deduce (U,V)∈C0,μ(RN)×C0,ν(RN). This concludes the proof in (ⅰ).
(ⅱ) Assume max{α,β}<γ<N, 0<p<N−γγ−β, and 0<q<γ−βγ−α. Let
κ1=γ−α+p(γ−β)1+pq>0 and κ2=γ−β−q(γ−α)1+pq>0. | (4.7) |
For n≥1, we define the closed and convex set Bn⊂C(¯Bn)×C(¯Bn) by
Bn={(u,v)∈C(¯Bn)×C(¯Bn):m1(1+|x|)−κ1≤u(x)≤M1(1+|x|)−κ1m2(1+|x|)−κ2≤v(x)≤M2(1+|x|)−κ2 in ¯Bn}, | (4.8) |
where 0<m1<M1, 0<m2<M2 are constants depending on α,β,γ,p,q, and N that will be chosen in Lemma 4.2 below.
For all u,v∈Bn, we define
Jn(u,v)(x)=(∫BnK(y)v(y)p|x−y|N−αdy,∫BnL(y)u(y)−q|x−y|N−βdy) for all x∈¯Bn. |
Since 0<q<γ−βγ−α, we can check
γ−κ1q=γ−(γ−α)q+pq(γ−β)1+pq>γ−(γ−β)+pq(γ−β)1+pq>β. |
Also, by N>γ, we deduce γ−κ1q<N.
Next, since 0<p<N−γγ−β, we obtain
γ+κ2p=γ+(γ−β)p−pq(γ−α)1+pq,<γ+(N−γ)+pq(α−γ)1+pq,<γ+(N−γ)+pq(N−γ)1+pq,<N. |
Also, by max{α,β}<γ<N, it follows that γ+κ2p>α.
Now we have β<γ−κ1q<N and α<γ+κ2p<N. Hence, by Eq (3.7), there exists c2>c1>0 independent of n≥1 such that
c1(1+|x|)−κ1≤∫BnK(y)dy(1+|y|)pκ2|x−y|N−α≤c2(1+|x|)−κ1 for all x∈Bn, | (4.9) |
and
c1(1+|x|)−κ2≤∫BnL(y)dy(1+|y|)−qκ1|x−y|N−β≤c2(1+|x|)−κ2 for all x∈Bn. | (4.10) |
We choose m1,m2,M1, and M2 as given by Eq (4.5) with new constants c1,c2 from Eqs (4.9) and (4.10). Then, Lemma 4.2 holds, and we deduce Jn(Bn)⊂Bn.
Thus, we can select s>1 such that
α−1<Ns<α<γ+pκ2 and β−1<Ns<β<γ−qκ1. |
Let μ=α−Ns∈(0,1) and ν=β−Ns∈(0,1). By Lemma 2.1, we obtain that
Jn:Bn⊂Ls(Bn)×Ls(Bn)→C0,μ(Bn)×C0,ν(Bn)is continuous |
and Jn(Bn)⊂Bn.
Hence, we can use the Schauder fixed-point theorem for Jn, which implies the existence of (un,vn)∈Bn such that
(un,vn)=(∫BnK(y)vn(y)p|x−y|N−αdy,∫BnL(y)un(y)−q|x−y|N−βdy) for all x∈¯Bn. |
As in part (ⅰ) above, the diagonal sequence {(Un,Vn)}={(unn,vnn)}n≥1 converges to a certain (U,V)∈C(RN)×C(RN), which fulfills
{U(x)=∫RNK(y)Vp(y)|x−y|N−αdy for all x∈¯Bn,V(x)=∫RNL(y)U−q(y)|x−y|N−βdy for all x∈¯Bn, |
Since
{U(x)=limn→∞Un(x)≃(1+|x|)−κ1 for all x∈RNV(x)=limn→∞Vn(x)≃(1+|x|)−κ2 for all x∈RN, |
we can apply Lemma 2.1 to deduce (U,V)∈C0,μ(RN)×C0,ν(RN). This finishes the proof of our theorem.
With a similar approach, we can also obtain the existence of solutions to the system
{u(x)=∫RNK(y)vp(y)|x−y|N−αdy for all x∈RN,v(x)=∫RNL(y)uq(y)|x−y|N−βdy for all x∈RN, | (4.11) |
where p,q>0, α,β∈(0,N), and K,L:RN→(0,∞) are continuous functions which satisfy Eq (1.2).
Our main result regarding the system (4.11) reads as follows.
Theorem 4.3. (Existence)
Assume max{α,β}<γ.
(ⅰ) If max{α,β}<γ<N, p>N−γN−β, q>N−γN−α, and pq<1, then the integral system (4.11) has a solution (u,v) such that
{u(x)≃(1+|x|)α−N,v(x)≃(1+|x|)β−N. |
Moreover, (u,v)∈C0,μ(RN)×C0,ν(RN) for some μ,ν∈(0,1).
(ⅱ) If γ>N and pq<1, then the integral system (4.11) has a solution (u,v) such that
{u(x)≃(1+|x|)α−N,v(x)≃(1+|x|)β−N. |
Moreover, (u,v)∈C0,μ(RN)×C0,ν(RN) for some μ,ν∈(0,1).
The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.s
The author would like to thank the anonymous a for the careful reading of the manuscript which led to the current form.
This publication has emanated from research conducted with the financial support of Taighde Éireann-Research Ireland under Grant number 18/CRT/6049.
The author declares there is no conflict of interest.
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