# Necessary Conditions for Existence Results of Some Integral System

Authors: Yongxia Hua; Xiaohui Yu
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

In this paper, we give some necessary conditions for the existence of positive solutions for integral systems.


---

## Body

## 1. Introduction

In this paper, we study the necessary condition for the existence of positive solutions for the following integral system:(1)u(x)=∫ℝN‍|x-y|λv(y)pdyinℝN,v(x)=∫ℝN‍|x-y|λu(y)qdyinℝN,
where λ, p,  and  q are real parameters.As for one single equation(2)u(x)=∫ℝN‍|x-y|λu(y)pdyinℝN,
there are a lot of results of this problem. If λ=α-N with 0<α<N, then problem (2) is equivalent to the following differential equation:
(3)(-Δ)α/2u(x)=u(x)pinℝN.
This problem has been widely studied in the past few years. For example, in order to answer a question raised by Lieb in [1], the authors studied the symmetric property and the uniqueness of solutions for problem (2) in [2]. Later, they studied the integral system (1) in [3]. Also, after the work of [2], Li studied the general form of (2) in [4]. For the case λ<0, he obtained similar results to [2] but with less regularity requirement. For the case λ>0, he shows that if problem (2) has a nonnegative solution in ℝN and λ(p+1)+2N≥0, then q=-1-2N/λ. The main ingredients in these papers are the moving plane method and moving sphere method based on the maximum principle of integral forms. This method has been widely used in other works. For example, inspired by these works, the author studied the Liouville-type theorems for problems (1) and (2) with general nonlinearities in [5, 6]. For further results of this type of integral equations, see [7–18], and so forth. We note that all these results concern the cases λ<0 and p>0. A natural question is whether similar results hold for λ>0 or p<0. We note that the case λ<0 and p>0 is quite different from the case λ>0 or p<0. Generally speaking, the moving plane method or the moving sphere method does not work in the latter case, so we have to look for other methods. In a recent paper [19], the author give a sufficient and necessary condition for the existence of positive solutions for problem (2) with λ>0. Based on some integral estimates, the author proved that problem (2) possesses a positive solution if and only if λp=-(λ+2N). Inspired by the work of [19], we first study the integral system (1) with λ>0. Our main result is the following theorem.Theorem 1.
Suppose thatλ>0 and problem (1) possesses a C1 positive solution; then
(4)11+p+11+q=-λN.As forλ<0, we have the following nonexistence result.Theorem 2.
If-N<λ<0, then problem (1) possesses no C1 positive solution provided that p<0 or q<0.This paper is organized as follows. We prove Theorem1 in Section 2. The proof of Theorem 2 is completed in Section 3.


## 2. Proof of Theorem1

We first claim thatu(x)∈Lq+1(ℝN) and v(x)∈Lp+1(ℝN).In fact, we infer from(5)v(0)=∫ℝN‍|y|λu(y)qdy≥∫ℝN∖B1(0)‍|y|λu(y)qdy≥∫ℝN∖B1(0)‍u(y)qdy
that u(x)∈Lq(ℝN). Also, it follows from (1) that
(6)|x|λu(x|x|2)=∫ℝN‍|x|λ|x|x|2-y|λv(y)pdy=∫ℝN‍|y|λ|x-y|y|2|λvp(y)dy.
Now taking limit in (6) by letting x→0, we obtain
(7)lim|x|→0[|x|λu(x|x|2)]=∫ℝN‍vp(y)dy<∞.
We point out that we can take the limit under the integral sign because of the dominated convergence theorem. In fact, we note that when λ>0 and |x|≤1, we have
(8)|x-y|y|2|λ≤(|x|+1|y|)λ≤(1+1|y|)λ.
It is easy to check that |y|λ(1+1/|y|)λvp(y)∈L1(ℝN).It follows from (7) that there exist R>0 and C>0 such that
(9)C-1|x|λ≤u(x)≤C|x|λ
for |x|≥R. Finally, we have
(10)∫ℝN∖BR(0)‍uq+1(x)dx=∫ℝN∖BR(0)‍uqudx≤C∫ℝN‍|x|λuqdx=Cv(0),
which further implies that u(x)∈Lq+1(ℝN). Similarly, we have v(x)∈Lp+1(ℝN).Next, we can prove as in [19] that
(11)∇u(x)=λ∫ℝN‍|x-y|λ-2(x-y)vp(y)dy,∇v(x)=λ∫ℝN‍|x-y|λ-2(x-y)uq(y)dy
in the sense of distribution. Hence, we infer from (11) that
(12)∇u1+q(x)=(1+q)uq(x)λ·∫ℝN‍|x-y|λ-2(x-y)vp(y)dy,∇v1+p(x)=(1+p)vp(x)λ·∫ℝN‍|x-y|λ-2(x-y)uq(y)dy.Now we choose a cut-off functionη∈C∞([0,+∞)) satisfying 0≤η≤1, 0≤|η′|≤2,  η(t)=1 for t≤1 and η(t)=0 for t≥2. For any R>0, if we multiply (12) by η(|x|/R)x and integrate over ℝN, then we get
(13)∫ℝN‍η(|x|R)vp(x)〈x,∇v(x)〉dx=∫ℝN‍η(|x|R)vp∫ℝN‍λ|x-y|λ-2〈x,x-y〉u(y)qdydx,(14)∫ℝN‍η(|x|R)uq(x)〈x,∇u(x)〉dx=∫ℝN‍η(|x|R)uq(x)∫ℝN‍λ|x-y|λ-2〈x,x-y〉v(y)pdydx.
While the left-hand side of (13) equals
(15)∫ℝN‍η(|x|R)vp(x)〈x,∇v(x)〉dx=11+p∫ℝN‍η(|x|R)〈x,∇v1+p(x)〉dx=-N1+p∫ℝN‍η(|x|R)v1+p(x)dx-11+p∫ℝN‍〈∇(η(|x|R)),x〉v1+pdx,
it follows from
(16)〈∇(η(|x|R)),x〉≤2|x|R
that
(17)∫ℝN‍〈∇(η(|x|R)),x〉v1+pdx≤4∫R≤|x|≤2R‍v1+pdx⟶0
as R→∞ by v∈L1+p(ℝN). Thus we conclude that
(18)limR→∞∫ℝN‍η(|x|R)vp(x)〈x,∇v(x)〉dx=-N1+p∫ℝN‍v1+p(x)dx.
While the right-hand side of (13) equals
(19)λ∫ℝN‍η(|x|R)vp(x)·∫ℝN‍|x-y|λ-2〈x,x-y〉u(y)qdydx=λ2∫ℝN‍η(|x|R)vp(x)·∫ℝN‍|x-y|λ-2〈x-y,x-y〉u(y)qdydx+λ2∫ℝN‍η(|x|R)vp(x)·∫ℝN‍|x-y|λ-2〈x+y,x-y〉u(y)qdydx=λ2∫ℝN‍η(|x|R)v1+p(x)dx+λ2∫ℝN‍η(|x|R)vp(x)·∫ℝN‍|x-y|λ-2〈x+y,x-y〉u(y)qdydx,
it can be checked as in [19] that
(20)∫ℝN‍∫ℝN‍|x-y|λ-2〈x+y,x-y〉u(y)qv(x)pdydx<∞.
Hence, by letting R→∞ in (19) we get
(21)limR→∞λ∫ℝN‍η(|x|R)vp(x)·∫ℝN‍|x-y|λ-2〈x,x-y〉u(y)qdydx=λ2∫ℝN‍v1+p(x)dx+λ2·∫ℝN‍∫ℝN‍|x-y|λ-2·〈x+y,x-y〉uq(y)vp(x)dydx.
We infer from (13), (18), and (21) that
(22)-N1+p∫ℝN‍v1+p(x)dx=λ2∫ℝN‍v1+p(x)dx+λ2∫ℝN‍∫ℝN‍|x-y|λ-2·〈x+y,x-y〉uq(y)vp(x)dydx.
Similarly, we can prove that
(23)-N1+p∫ℝN‍u1+q(x)dx=λ2∫ℝN‍u1+q(x)dx+λ2∫ℝN‍∫ℝN‍|x-y|λ-2·〈x+y,x-y〉uq(x)vp(y)dydx=λ2∫ℝN‍u1+q(x)dx-λ2∫ℝN‍∫ℝN‍|x-y|λ-2·〈x+y,x-y〉uq(y)vp(x)dxdy.
The above two equations imply that
(24)-N1+p∫ℝN‍u1+q(x)dx-N1+p∫ℝN‍v1+p(x)dx=λ2[∫ℝN‍u1+q(x)dx+∫ℝN‍v1+p(x)dx].
On the other hand, since
(25)v(x)1+p=v(x)p∫ℝN‍|x-y|λuq(y)dy,
we have
(26)∫ℝN‍v1+p(x)dx=∫ℝN‍∫ℝN‍|x-y|λuq(y)v(x)pdydx
by taking into account that v(x)∈L1+p(ℝN). Similarly, we have
(27)∫ℝN‍u1+q(x)dx=∫ℝN‍∫ℝN‍|x-y|λuq(y)v(x)pdydx.
Then it follows from (26) and (27) that
(28)∫ℝN‍u1+q(x)dx=∫ℝN‍v1+p(x)dx.
Finally, we infer from (24) and (28) that
(29)11+p+11+q=-λN.
This completes the proof of Theorem 1.


## 3. Proof of Theorem2

We assume thatq<0 without loss of generality. First, we note that by Lemma  3.11.3 in [20], we have, for all r>0,
(30)1ωNrN∫Br(0)‍u(x)dx=∫ℝN‍{1ωNrN∫Br(0)‍|x-y|λdx}vp(y)dy≤C∫ℝN‍|y|λvp(y)dy=Cu(0).
Similarly, we have
(31)1ωNrN∫Br(0)‍v(x)dx≤Cv(0)
for any r>0.If we chooseα=q/(q-1), β=1-q, and  δ=(q-1)/q, then we can infer from the Holder inequality that
(32)1=1ωNrN∫Br(0)‍u(x)-αu(x)αdx≤(1ωNrN∫Br(0)‍u(x)-αβdx)1/β·(1ωNrN∫Br(0)‍u(x)αδdx)1/δ=(1ωNrN∫Br(0)‍u(x)qdx)1/(1-q)·(Cu(0))q/(q-1).
That is,
(33)(Cu(0))q≤1ωNrN∫Br(0)‍uq(x)dx.
Since λ<0, so if |x|<r, then we have rλ<|x|λ. Multiplying both sides of (33) by ωNrN+λ, we get
(34)CqωNrN+λu(0)q≤ωNrN+λ1ωNrN∫Br(0)‍uq(x)dx=rλ∫Br(0)‍uq(x)dx≤∫Br(0)‍|x|λuq(x)dx=v(0).
Since -N<λ<0, we have N+λ>0. Hence the left-hand side of (34) goes to infinity as r→∞, which is a contradiction. This completes the proof.



---

*Source: 504282-2013-08-18.xml*