The main objective of the present note is to study positive solutions of the following interesting system of integral equations in Rn u(x) = Z Rn |x − y| p v(y) −q dy, v(x) = Z Rn |x − y| pu(y) −q dy, (0.1) with p, q > 0 and n > 1. Under the nonnegative Lebesgue measurability condition for solutions (u, v) of (0.1), we prove that pq = p+2n and that u and v are radially symmetric and monotone decreasing about some point. To prove this, we introduce an integral form of the method of moving spheres for systems to tackle (0.1). As far as we know, this is the first attempt to use the method of moving spheres for systems.
3 CLASSIFICATION OF SOLUTIONS FOR A SYSTEM OF INTEGRAL EQUATIONS WITH NEGATIVE EXPONENTS VIA THE METHOD OF MOVING SPHERES ´ˆ C ANH NGO ˆ QUO Dedicated to Professor Hoang Quoc Toan on the occasion of his 70th birthday A BSTRACT The main objective of the present note is to study positive solutions of the following interesting system of integral equations in Rn |x − y|p v(y)−q dy, u(x) = Rn (0.1) v(x) = |x − y|p u(y)−q dy, Rn with p, q > and n Under the nonnegative Lebesgue measurability condition for solutions (u, v) of (0.1), we prove that pq = p+2n and that u and v are radially symmetric and monotone decreasing about some point To prove this, we introduce an integral form of the method of moving spheres for systems to tackle (0.1) As far as we know, this is the first attempt to use the method of moving spheres for systems I NTRODUCTION Given p, q > and n 1, of interest in the present note is to study non-negative solutions of the following interesting system of integral equations in Rn |x − y|p v(y)−q dy, u(x) = Rn (1.1) p −q v(x) = |x − y| u(y) dy Rn 10 11 Perhaps, the motivation for studying (1.1) comes from a natural extension from one equation to a system However, its counterpart when p, q < comes from the study of the sharp constants C(n, s, p) of the following Hardy-Littlewood-Sobolev inequality 12 Rn 13 14 15 Rn f (x)g(y) dxdy |x − y|p C(n, s, p) f Lr g Ls with 1/r + 1/s = − n/p, see [Lieb83, CheLiOu05] In the special case where u = v, our interested system (1.1) becomes the following well-known integral equation |x − y|p u(y)−q dy u(x) = (1.2) Rn 16 17 18 19 20 in Rn where u > To be precise, Eq (1.2) has its root in geometry as it arises from studying quantitative properties of geometric curvatures in the conformal geometry The classification of the solutions of (1.2) and its variations thus has a long history and is important as it also provides essential ingredients in the study of the Yamabe problem and the prescribing scalar curvature problem, see [CafGidSpr89, CheLi91, WeiXu99, ZhaHao08] Date: 11th May, 2015 at 22:11 2000 Mathematics Subject Classification 35J60, 58G35, 53C21 Key words and phrases Integral equation; Symmetry; Moving sphere ˆ Q.A NGO 2 for details As can be observed, the corresponding partial differential equation to (1.2) is the following well-known semilinear equation (−∆) n+p u = u−q (1.3) n with u > in R Concerning Eq (1.2), solutions to Eq (1.2) were first studied by Xu in [Xu05] when p = and n = He provided the following classification of C entire solutions of Eq (1.2) Theorem (see [Xu05], Thm 1.1) Suppose that u is a C entire positive solution of the integral equation |x − y|u(y)−q dy u(x) = (1.4) R3 10 11 in R3 with q > 0, then q = and up to a constant multiplication, a translation and a dilation, u takes the form u(x) = + |x − x|2 12 13 14 15 16 17 18 Note that we had already dropped the factor 1/8π in (1.4) in the original statement in [Xu05, Theorem 1.1] Almost simultaneously, Li studied Eq (1.2) in its present form and obtained, among others, the following result in [Li04] Theorem (see [Li04], Thm 1.5) For n 1, p > and < q + 2n/p, let u be a nonnegative Lebesgue measurable function in Rn satisfying (1.2) Then q = + 2n/p and, for some constants a, d > and some x ∈ Rn , u(x) = a d + |x − x|2 19 p 24 Clear, Theorem already includes Theorem We take this chance to mention a recent paper [XuWuTan14] in which the authors considered a slightly perturbation of (1.2) and obtained some new phenomena Let us just go back to the very special case when n = 3, p = In this setting, Eq (1.3) is associated with some fourth order partial differential equation since it simply becomes 25 (−∆)2 u = u−q 20 21 22 23 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 (1.5) in R Although (1.4) and (1.5) are closely related, here we would like to compare the structure of solutions of (1.4) and (1.5) Clearly, we know from Theorem that it is necessary to have q = in order for (1.4) to admit solutions In addition, we further know that all solutions corresponding to the case q = are radial symmetric and asymptotically linear at infinity However, this is no longer true for (1.5) in the sense that it is not necessary to have q = for (1.5) to have radial symmetric solutions with asymptotically linearity at infinity, see [Gue12]; see also [CheFanLi13, FanChe12] for related works concerning the equivalence of integral and differential equations In the literature, to classify solutions of some partial differential equations, people usually use one of the following two methods: moving planes and moving spheres While the former method, first invented by Alexandrov in early 1950s, turns out to be powerful when dealing with partial differential equations such as (1.3) through a series of seminal works by Serrin [Ser71], Gidas, Ni, and Nirenberg [GidNiNir79], Caffarelli, Gidas, and Spruck [CafGidSpr89], Chen and Li [CheLi91], the latter method is a variant of the former one As far as we know, this method was used by Li and Zhu in [LiZhu95] A prior to the work by Li and Zhu, to make use of the method of moving planes, one needs to go through two steps: First to prove solutions are radial symmetric by using the reflection through some planes, then to classify all radial solutions by solving some transformed ordinary differential equations In the work by Li and Zhu, the authors successfully SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES 10 exploited the conformal invariance of the problem which, as a by-product, captures solutions directly rather than going the second step as in the method of moving planes, see [Li04, Section 1] It is worth noticing that all these “traditional” methods of moving planes and spheres can be used to classify solutions to, for example, (1.3) because local properties of the differential operators of the problem are well-exploited In case of lack of knowledge of local properties, it could prevent us from using some known results, see [CheLiOu05, Section 1] Now we turn our attention to problem (1.1) In the recent paper [Lei15], Lei studied positive entire solutions (u, v) of (1.1) Among other things, Lei proved the following result 12 Theorem (see [Lei15], Thm 1.2) Assume the positive entire solutions u, v ∈ C (Rn ) of (1.1) are radially symmetric about some point x ∈ Rn , then 13 u(x) ≡ v(x) ≡ a(b2 + |x − x|2 ) 11 p 14 with a, b > 15 As can be seen from the proof of [Lei15, Theorem 1.2], the assumption of radial property plays a key role because it allows us to prove that u ≡ v identically Upon having this, the form for u and v follows from known results Also in [Lei15], the author questioned the assumption of radial property of (u, v) in Theorem can be removed Motivated by the question in [Lei15], in the present note we prove Theorem without assuming the radial property of solutions (u, v) To achieve that goal, we adopt the method of moving spheres in [Li04] to introduce an integral form of the method of moving spheres for solving systems of integral equations As far as we know, this is the first attempt to use the method of moving spheres for systems (with negative exponents) 16 17 18 19 20 21 22 23 24 25 26 27 28 It is worth noticing that in the same paper [Lei15], the author pointed out that J Xu has already completed a preprint where an answer for this question was addressed using the ideas in [LiZhu95] and [Xu07] Unfortunately, such a preprint has not been made available yet Nevertheless, our proof follows [Li04] which should be different from above Before discussing further, we state our main result of the present note 31 Theorem For n 1, p > and q > 0, let (u, v) be a pair of nonnegative Lebesgue measurable functions in Rn satisfying (1.1) Then q = + 2n/p and, for some constants a, b > and some x ∈ Rn , 32 u(x) = v(x) = a(b2 + |x − x|2 ) 29 30 p 33 34 35 36 37 38 39 40 41 42 43 44 45 for any x ∈ Rn Note that Theorem already includes the case q > + 2n/p which was not considered in [Li04], see Theorem above Nearly a decade ago, Chen, Li, and Ou introduced an integral form of the method of moving planes in [CheLiOu06] to classify solutions of some integral equations Since then, this new method has been used by many mathematicians for different problems under various contexts In a continued paper [CheLiOu05], Chen, Li, and Ou also introduced an integral form of the method of moving planes for system The basic idea of this new method that allows the method to go through depends heavily on some new special features possessed by integral equations Perhaps, the global form of the integral equations allows us to overcome the lack of local properties possessed by differential operators In addition, we should emphasize that the most important ingredient in this new method is to apply the Hardy-Littlewood-Sobolev inequality in a very clever way However, as pointed out ˆ Q.A NGO 4 10 11 12 13 14 15 by Lei in [Lei15], it is inappropriate to apply this new method to (1.1) since the HardyLittlewood-Sobolev inequality does not work due to the presence of the negative exponent q The primary aim of this note is to reformulate the method of moving spheres in an integral form that suits for our analysis Following the method of moving spheres by Li and Zhu, Zhang and Hao first introduced the integral form of the method of moving spheres in [ZhaHao08] to provide a new proof for [CheLiOu06, Theorem 1.1] However, the approach in [ZhaHao08] seems to be narrowed as it still requires the Hardy-LittlewoodSobolev inequality; hence it cannot be applied to our context To make our analysis doable, we analyze some conformal invariances found in [Li04] for one equation and made some necessary changes for systems For the reader’s convenience and in order to make our note self-contained, we follow the proof of [Li04, Theorem 1.5] closely with a little bit more explanation Before closing this section, we would like to mention that our integral system (1.1) is closely related to the following system of partial differential equations (−∆) 16 17 n+p u = v −q , (−∆) n+p v = u−q , (1.6) in Rn where u, v > In fact, any C solution (u, v) of (1.1) solves (1.6) 18 P RELIMINARIES AND THE METHOD OF MOVING SPHERES FOR SYSTEMS 19 In this section, we setup some preliminaries necessarily for our analysis The most important part of this section is the a prior estimates for solutions of (1.1) as stated in Lemma below Here and in what follows, by and we mean inequalities up to p, q, and dimensional constants 20 21 22 23 24 Lemma For n and p, q > 0, let (u, v) be a pair of non-negative Lebesgue measurable functions in Rn satisfying (1.1) Then there hold (1 + |y|p )u(y) 25 −q 26 27 28 31 32 33 34 dy < ∞, (2.1) Rn and lim |x|→∞ u(x) = |x|p −q v(y) dy, Rn lim |x|→∞ v(x) = |x|p −q u(y) dy, (2.2) Rn and u and v are bounded from below in the following sense 29 30 −q (1 + |y|p )v(y) dy < ∞, Rn u(x), v(x) + |x|p (2.3) u(x), v(x) + |x|p (2.4) and above in the following sense for all x ∈ Rn In other words, there holds + |x|p C n in R for some constant C u(x), v(x) C(1 + |x|p ) 37 Proof To prove our lemma, we first observe from (1.1) that both u and v are strictly positive everywhere in Rn and are finite within a set of positive measure Hence there exist some R > sufficiently large and some Lebesgue measurable set E ⊂ Rn such that 38 E ⊂ {y : u(y) < R, v(y) < R} ∩ B(0, R) 35 36 (2.5) SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES with meas(E) v(x) 1/R Using this, we can easily bound v from below as follows Rq |x − y|p u(y)−q dy E |x − y|p dy = E Rq |y|p dy E+x for any x ∈ Rn Choose ε > small enough and then fix it in such a way that vol(B(0, ε)) < |E|/2 Then we can estimate |y|p dy |y|p dy E+x E+x\B(0,ε) εp dy E+x\B(0,ε) = εp |E + x| − vol(B(0, ε)) From this, it is clear that v is bounded from below by some positive constant The same reason applied to u shows that there exists some constant C0 > such that u(x), v(x) > C0 10 11 14 15 16 17 18 19 (2.6) Proof of (2.3) To prove this, we first consider |x| 2R where R is defined through (2.5) Note that for every y ∈ E ⊂ B(0, R), there holds |x − y| |x| − |y| |x|/2 thanks to |x| 2R Using this we can estimate Rq v(x) 12 13 ∀x ∈ Rn vol(E) p |x| (2R)p |x − y|p dy E for any |x| 2R A similar argument also shows u(x) vol(E)(2R)−p |x|p in the region {|x| 2R} Hence, it is easy to select a large constant C > in such a way that (2.3) holds in the region {|x| 2R} Thanks to (2.6), we can further decrease C, if necessary, to obtain the estimate (2.3) in the ball {|x| 2R}; hence the proof of (2.3) follows Proof of (2.1) We only need to estimate v since u can be estimated similarly To this purpose, we first show that u−q ∈ L1 (Rn ) Clearly for some x satisfying |x| 2, there holds |x − y|p u(y) 20 −q dy = v(x) ∈ (0, +∞) Rn 21 Observer that for any y ∈ Rn \B(0, 4), |x − y| Rn \B(0,4) 23 |x − y|p u(y) dy < +∞ Rn In the small ball B(0, 4), thanks to (2.3), it is obvious to verify that u(y)−q dy 24 (1 + |y|p ))−q dy < +∞ B(0,4) 25 −q u(y)−q dy < 22 |y| − |x| > 1; hence B(0,4) −q Thus, we have just shown that u ∈ L (Rn ) In view of (2.1), it suffices to prove that |y|p u(y)−q dy < +∞ 26 (2.7) Rn 27 To see this, we again observe that |y| |y|p u(y)−q dy 28 Rn 29 |x − y|p u(y) Rn \B(0,4) −q dy < +∞ \B(0,4) In the small ball B(0, 4), it is obvious to see that |y|p u(y)−q dy 30 B(0,4) 31 2|x − y| for all y ∈ Rn \B(0, 4) Therefore, u(y)−q dy < +∞, B(0,4) thanks to u−q ∈ L1 (Rn ) From this, (2.7) follows, so does (2.1) ˆ Q.A NGO 6 Proof of (2.2) We only consider the limit |x|−p v(x) as |x| → ∞ since the limit |x|−p u(x) can be proved similarly Indeed, using (1.1), we first obtain v(x) = lim |x|→∞ |x|p |x|→∞ lim 12 13 14 15 16 17 18 19 20 21 22 (2.8) Observe that as |x| → +∞, (|x − y|/|x|) u(y)−q → u(y)−q almost everywhere y in Rn Hence we can apply the Lebesgue dominated convergence theorem to pass (2.8) to the limit to conclude (2.1) provided we can show that |x − y|p |x|−p u(y)−q is bounded by some integrable function To this end, for each |x| fixed we first split Rn = {y ∈ Rn : |x − y| 2|x|} ∪ {y ∈ Rn : |x − y| > 2|x|} = D1 ∪ D2 Then, in D1 , we immediately obtain |x − y| |x| 10 11 Rn p |x − y|p −q u(y) dy |x|p p 2p (1 + |y|p ), for any y ∈ D1 while in D2 we realize that |y| |x − y| − |x| |x − y|/2 which helps us to estimate p |x − y| |x − y|p |y|p + |y|p |x| thanks to |x| Thus, we have just proved that for any |x| 1, the following estimate |x − y| |x| p u(y)−q (1 + |y|p )u(y)−q holds Our proof now follows by observing (1 + |y|p )u(y)−q ∈ L1 (Rn ) by (2.1) Proof of (2.4) To see this, we first observe from (2.2) that there exists some large number k > 1/R such that u(x) v(y)−q dy 0, let (u, v) be a pair of non-negative Lebesgue measurable functions in Rn satisfying (1.1) Then u and v are smooth Proof Our proof is similar to that of [Li04, Lemma 5.2] Let R > be arbitrary, first we decompose u and v into the following way u(x) =u1R (x) + u2R (x) = |x − y|p v(y)−q dy, + |y| 2R |y|>2R 32 v(x) =vR (x) + vR (x) = 33 34 |x − y|p u(y)−q dy + |y| 2R |y|>2R Thanks to (2.1), we immediately see that we can continuously differentiate u2R and vR ∞ under the integral sign for any |x| < R Consequently, uR ∈ C (B(0, R)) and vR ∈ SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES 10 C ∞ (B(0, R)) In view of (2.3) and (2.4), we know that u−q ∈ L∞ (B(0, 2R)) which implies that vR is at least Hăolder continuous in B(0, R) Similar reasons tell us that u1R is also at least Hăolder continuous in B(0, R) Hence, we have just proved that u and v are at least Hăolder continuous in B(0, R), so are at least Hăolder continuous in Rn since R > is arbitrary Standard bootstrap argument shows u ∈ C ∞ (Rn ) and at the same time v ∈ C ∞ (Rn ) follows the same lines Once we have the smoothness for solutions of (1.1), we can narrow the range for q as follows Proposition For n and p, q > Then in order for (1.1) to have solutions, it is necessary to have q + 2n/p 12 Proof The proof is just a direct consequence of [HuaYu13, Theorem 1] and Lemma above, see also [Lei15, Theorem 1.1]; hence we omit its details 13 T HE METHOD OF MOVING SPHERES FOR SYSTEMS 11 14 15 As a consequence of Proposition 1, from now on, we only consider the case q 2n/p Let w be a positive function on Rn For x ∈ Rn and λ > we define 17 18 19 ∀y ∈ Rn w(ξ x,λ ), where ξ x,λ = x + λ2 dy = ξ−x λ |z − x| |y−x| λ 2n dz (3.3) |ξ x,λ − z x,λ |p v(z x,λ )−q |z−x| λ 22 |ξ x,λ − z x,λ |p = |z−x| λ λ |z − x| λ |z − x| 2n vx,λ (z)−q dz −p |ξ x,λ − y|p v(y)−q dy |y−x| λ = 24 |y−x| λ |ξ − z|p = −p λ |ξ x,λ − z x,λ | |z − x| |ξ − z| |z−x| λ λ |z − x| |ξ x,λ − y|p v(y)−q dy 2n−pq+p vx,λ (z)−q dz Similarly, we obtain λ |ξ − x| −p |ξ x,λ − y|p v(y)−q dy |y−x| λ 26 |ξ − z|p = |z−x| λ λ |z − x| dz 2n−pq Then, using the relation |z − x||ξ − x||ξ x,λ − z x,λ | = λ2 |ξ − z|, we obtain λ |ξ − x| 25 (3.2) Note that if y = z x,λ , then z = y x,λ Therefore, we have |ξ x,λ − y|p v(y)−q dy = 23 (3.1) |ξ − x| Clearly, upon the change of variable y = z x,λ , we then have 20 21 p |ξ − x| λ wx,λ (ξ) = 16 1+ 2n−pq+p vx,λ (z)−q dz ˆ Q.A NGO Lemma For any solutions (u, v) of (1.1), we have ux,λ (ξ) = λ |z − x| 2n−pq+p |ξ − z|p λ |z − x| 2n−pq+p |ξ − z|p Rn and vx,λ (ξ) = Rn for any ξ ∈ Rn Proof Using our system (1.1), we obtain u(ξ x,λ ) p |ξ x,λ − y|p v(y)−q dy Rn |ξ − z|p = ux,λ (z)−q dz p |ξ − x| λ |ξ − x| = λ ux,λ (ξ) = vx,λ (z)−q dz Rn λ |z − x| 2n−pq+p vx,λ (z)−q dz The formula for v follows the same line as above Lemma For any solutions (u, v) of (1.1), we have 10 ux,λ (ξ) − u(ξ) = λ |z − x| 2n−pq+p k(x, λ; ξ, z) v(z)−q − λ |z − x| 2n−pq+p k(x, λ; ξ, z) u(z)−q − |z−x| λ 11 12 and vx,λ (ξ) − v(ξ) = |z−x| λ 13 vx,λ (z)−q dz, ux,λ (z)−q dz, for any ξ ∈ Rn where k(x, λ; ξ, z) = 14 |ξ − x| λ p |ξ x,λ − z|p − |ξ − z|p 15 Moreover, k(x, λ; ξ, z) > for any |ξ − x| > λ > and |z − x| > λ > 16 Proof First, we observe that |ξ − z|p ux,λ (ξ) = |z−x| λ 17 18 vx,λ (z)−q dz p |ξ − x| λ + 2n−pq+p λ |z − x| |ξ x,λ − z|p v(z)−q dz |z−x| λ and |ξ x,λ − z|p v(z)−q dz u(ξ) = |z−x| λ 19 + 20 21 |ξ − x| λ p |ξ x,λ − z|p |z−x| λ 23 2n−pq+p vx,λ (z)−q dz Therefore, k(x, λ; ξ, z) v(z)−q − ux,λ (ξ) − u(ξ) = |z−x| λ 22 λ |z − x| where k(x, λ; ξ, z) = |ξ − x| λ λ |z − x| 2n−pq+p p |ξ x,λ − z|p − |ξ − z|p vx,λ (z)−q dz, SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES The representation of vx,λ (ξ) − v(ξ) can be found similarly Finally, the positivity of the kernel k for any |ξ − x| > λ and |z − x| > λ is clear due to the following magic formula |ξ − x| |ξ x,λ − z|2 − |ξ − z|2 = λ2 − |z − x|2 λ2 − |ξ − x|2 λ λ Thus the proof follows For future usage, we note that |ξ −x||ξ x,λ −z| = |z −x||z x,λ −ξ|; hence we can rewrite the kernel k as follows p |z − x| k(x, λ; ξ, z) = |ξ − z x,λ |p − |ξ − z|p λ Therefore, each component of ∇ξ k(x, λ; ξ, z) can be easily calculated as the following ∂ξi k(x, λ; ξ, z) =p |z − x| λ p |ξ − z x,λ |p−2 ξi − (z x,λ )i − p|ξ − z|p−2 (ξi − zi ) |z − x| =p λ p |ξ − x| x,λ |ξ − z| |z − x| (3.4) p−2 ξi − (z x,λ )i − p|ξ − z|p−2 (ξi − zi ) 10 In particular, we would have |z − x|2 |ξ − x|p−2 x,λ |ξ − z|p−2 |ξ|2 − z x,λ · ξ λp − p|ξ − z|p−2 (|ξ|2 − z · ξ) ∇k(x, λ; ξ, z) · ξ =p 11 (3.5) 13 In the following lemma, we prove that the method of moving spheres can get started starting from a very small radius 14 Lemma For each x ∈ Rn , there exists λ0 (x) > such that 15 ux,λ (y) 12 16 u(y), for any point y and any λ such that |y − x| vx,λ (y) v(y) λ with < λ < λ0 (x) 19 Proof Since u is a positive C function and p > 0, there exists some r0 > small enough such that p ∇y |y − x|− u(y) · (y − x) < 20 for all < |y − x| < r0 Consequently, by definition 17 18 ux,λ (y) = 21 |y − x| λ p p u(y x,λ ) p =|y − x| |y x,λ − x|− u(y x,λ ) >u(y) 22 23 24 25 26 27 28 29 for all < λ < |y − x| < r0 Note that in the previous estimates, we made use of the fact that if |y − x| > λ then |y x,λ − x| < λ For small λ0 ∈ (0, r0 ) and for < λ < λ0 , we have p |y − x| ux,λ (y) inf u u(y) λ B(x,r0 ) for all |y − x| r0 Hence, we have just shown that ux,λ (y) u(y) for all point y and any λ such that |y − x| λ with < λ < λ0 A similar argument shows that vx,λ (y) v(y) for all point y and any λ such that |y − x| λ with < λ < λ1 for some λ1 ∈ (0, r1 ) Simply setting λ0 (x) = min{λ0 , λ1 } we obtain the desired result ˆ Q.A NGO 10 For each x ∈ Rn we define λ(x) = sup {µ > : ux,λ (y) u(y), vx,λ (y) v(y), ∀0 < λ < µ, |y − x| λ} In view of Lemma above, we get < λ(x) +∞ In the next few lemmas, we show that whenever λ(x) is finite for some point x, we can write down precisely the form of solutions (u, v) Lemma If λ(x0 ) < ∞ for some point x0 ∈ Rn then ux0 ,λ(x0 ) ≡ u, vx0 ,λ(x0 ) ≡ v in Rn In addition, we obtain q = + 2n/p Proof By the definition of λ(x0 ), we know that ux0 ,λ(x0 ) (y) 10 11 for any |y − x0 | u(y), vx0 ,λ(x0 ) (y) v(y) (3.6) λ(x) In view of Lemma 4, we obtain ux0 ,λ(x0 ) (y) − u(y) k(x0 , λ(x0 ); y, z) v(z) = 12 −q − |z−x0 | λ(x0 ) λ(x0 ) |z − x0 | 2n−pq+p vx0 ,λ(x0 ) (z)−q dz, (3.7) 13 and vx0 ,λ(x0 ) (y) − v(y) 14 k(x0 , λ(x0 ); y, z) u(z)−q − = |z−x0 | λ(x0 ) λ(x0 ) |z − x0 | 2n−pq+p ux0 ,λ(x0 ) (z)−q dz, (3.8) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 for any y ∈ Rn Keep in mind that 2n − pq + p 0, there are two possible cases: Case Either ux0 ,λ(x0 ) (y) = u(y) or vx0 ,λ(x0 ) (y) = v(y) for any |y − x0 | λ(x0 ) Without loss of generality, we assume that the formal case occurs Using (3.7) and the positivity of the kernel k, we get that 2n − pq + p = and that vx0 ,λ(x0 ) (y) = v(y) for any |y − x0 | λ(x0 ) Hence again by (3.7) we conclude that ux0 ,λ(x0 ) (y) = u(y) in the whole Rn A similar argument also shows that vx0 ,λ(x0 ) (y) = v(y) in Rn and we are done Case Or ux0 ,λ(x0 ) (y) > u(y) and vx0 ,λ(x0 ) (y) > v(y) for any |y − x0 | λ(x0 ) In this case, we derive a contradiction by showing that in can slightly move spheres a little bit over λ(x0 ) which then violates the definition of λ(x0 ) To this purpose, still using (3.7), in the region |z − x0 | (3.6) that v(z)−q − λ(x0 ) |z − x0 | λ(x0 ), first we know from 2n−pq+p vx0 ,λ(x0 ) (z)−q v(z)−q − vx0 ,λ(x0 ) (z)−q Hence, by the positivity of the kernel k, we can estimate ux0 ,λ(x0 ) (y) − u(y) |z−x0 | λ(x0 ) k(x0 , λ(x0 ); y, z) v(z)−q − vx0 ,λ(x0 ) (z)−q dz (3.9) SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES 11 Estimate of ux0 ,λ − u outside B(x0 , λ(x0 ) + 1) Using the Fatou lemma, from (3.9) we obtain lim inf |y|−p (ux0 ,λ(x0 ) − u)(y) |y|→∞ lim inf |y|→∞ |z−x0 | λ(x0 ) |y|−p k(x0 , λ(x0 ); y, z) v(z)−q − vx0 ,λ(x0 ) (z)−q dz |z| λ(x0 ) |z−x0 | λ(x0 ) p −1 v(z)−q − 2n−pq+p λ(x0 ) |z − x0 | vx0 ,λ(x0 ) (z)−q dz >0, where we have used the following formula k(x0 , λ(x0 ); y, z) = |y|p lim inf |y|→∞ 10 11 14 15 16 17 18 19 20 21 22 |y x0 ,λ − z|p − λ(x0 ) |y − z| |y| p |z| λ(x0 ) k(x0 , λ(x0 ); y, z) |y|p p − |y|p while As a consequence, outside a large ball, we would have (ux0 ,λ(x0 ) − u)(y) in that ball and outside of B(x0 , λ(x0 ) + 1) we would also have (ux0 ,λ(x0 ) − u)(y) |y|p thanks to the smoothness of ux0 ,λ(x0 ) − u and our assumption ux0 ,λ(x0 ) (y) > u(y) Therefore, there exists some ε1 > such that (ux0 ,λ(x0 ) − u)(y) 12 13 p to obtain the following estimate |y − x0 | |y| ε1 |y|p for all |y − x0 | λ(x0 ) + Recall that ux0 ,λ (y) = (|x0 − y|/λ)p u(y x0 ,λ ); hence there exists some ε2 ∈ (0, ε1 ) such that (ux0 ,λ − u)(y) =(ux0 ,λ(x0 ) − u)(y) + (ux0 ,λ − ux0 ,λ(x0 ) )(y) ε1 p ε1 |y|p + (ux0 ,λ − ux0 ,λ(x0 ) )(y) |y| (3.10) λ(x0 ) + and all λ ∈ (λ(x0 ), λ(x0 ) + ε2 ) Repeating the above for all |y − x0 | arguments shows that (3.10) is also valid for vx0 ,λ − v, that is ε1 p (vx0 ,λ − v)(y) |y| (3.11) for a possibly new constant ε1 > Estimate of ux0 ,λ − u inside B(x0 , λ(x0 ) + 1) Now for ε ∈ (0, ε2 ) to be determined later and for λ ∈ (λ(x0 ), λ(x0 ) + ε) ⊂ (λ(x0 ), λ(x0 ) + ε2 ) and for λ |y − x0 | λ(x0 ) + 1, from (3.9), we estimate k(x0 , λ; y, z)[v(z)−q − vx0 ,λ (z)−q ]dz (ux0 ,λ − u)(y) |z−x0 | λ(x0 ) k(x0 , λ; y, z)[v(z)−q − vx0 ,λ (z)−q ]dz λ(x0 )+1 |z−x0 | λ k(x0 , λ; y, z)[v(z)−q − vx0 ,λ (z)−q ]dz + λ(x0 )+3 |z−x0 | λ(x0 )+2 23 λ(x0 )+1 |z−x0 | λ k(x0 , λ; y, z)[vx0 ,λ(x0 ) (z)−q − vx0 ,λ (z)−q ]dz k(x0 , λ; y, z)[v(z)−q − vx0 ,λ (z)−q ]dz + λ(x0 )+3 |z−x0 | λ(x0 )+2 =I + II ˆ Q.A NGO 12 As we shall see later, I + II by term provided ε > is small We now estimate I and II term (z) δ1 for Estimate of II Thanks to (3.11), there exists δ1 > such that v −q − vx−q ,λ any λ(x0 ) + |z − x0 | λ(x0 ) + Note by the definition of k given in Lemma that k(x0 , λ; y, z) = k(0, λ; y − x0 , z − x0 ) and from (3.5) there holds ∇y k(0, λ; y, z) · y |z| for all λ(x0 ) + independent of ε such that 12 13 16 δ2 (|y| − λ) k(0, λ; y, z) for all λ(x0 ) λ |y| λ(x0 )+1 and all λ(x0 )+2 |z| λ(x0 )+3 Simply replacing y by y − x0 and z by z − z0 and making use of the rule k(x0 , λ; y, z) = k(0, λ; y − x0 , z − x0 ), we obtain with the same constant δ2 > as above the following estimate k(x0 , λ; y, z) 14 15 = p|y − z|p−2 |z|2 − |y|2 > λ(x0 ) + Hence, there exists some constant δ2 > is 10 11 |y|=λ for all λ(x0 ) λ |y − x0 | Thus, we have just proved that |z − x0 | λ(x0 ) + and all λ(x0 ) + δ1 δ2 (|y − x0 | − λ) II 17 δ2 (|y − x0 | − λ) λ(x0 ) + dz (3.12) λ(x0 )+3 |z−x0 | λ(x0 )+2 18 Estimate of I To estimate I, we first observe that |vx0 ,λ(x0 ) −q − v −q |(z) 19 20 for all λ(x0 ) |z − x0 | λ λ − λ(x0 ) λ(x0 ) + and all λ(x0 ) λ(x0 ) + ε and that k(0, λ; y − x0 , z)dz k(x0 , λ; y, z)dz = λ |z−x0 | λ+1 λ ε λ |z| λ+1 λ |z| λ+1 21 |y − x0 | λ p − |(y − x0 )0,λ − z|p dz (|(y − x0 )0,λ − z|p − |(y − x0 ) − z|p )dz + λ |z| λ+1 C(|y − x0 | − λ) + C|(y − x0 )0,λ − (y − x0 )| C(|y − x0 | − λ) 22 where C > is constant independent of ε Thus, we obtain I 23 −Cε k(x0 , λ; y, z)dz (3.13) λ(x0 )+1 |z−x0 | λ 24 25 Thus, by combining (3.13) and (3.12), it follows that for some small ε > we have (ux0 ,λ − u)(y) dz − Cε (|y − x0 | − λ) δ1 δ2 λ(x0 )+3 |z−x0 | λ(x0 )+2 26 27 28 29 for λ(x0 ) λ λ(x0 ) + ε and λ |y − x0 | λ(x0 ) + Estimates of ux0 ,λ −u and vx0 ,λ −v when |y−x0 | λ(x0 )+1 Combining the preceding estimate for ux0 ,λ − u in the ball B(x0 , λ(x0 ) + 1) and (3.10) above gives (ux0 ,λ − u)(y) SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES 13 for λ(x0 ) λ λ(x0 ) + ε and λ |y − x0 | Again by repeating the whole procedure above for the difference vx0 ,λ − v, we can conclude that (vx0 ,λ − v)(y) for λ(x0 ) λ λ(x0 ) + ε and λ |y − x0 | where ε could be smaller if necessary; thus giving us a contradiction to the definition of λ(x0 ) In the last lemma of the current section, we prove that whenever λ(x0 ) < ∞ for some point x0 ∈ Rn , there must hold λ(x) < ∞ for any point x ∈ Rn Lemma If λ(x0 ) < ∞ for some point x0 ∈ Rn then λ(x) < ∞ for any point x ∈ Rn ; hence ux,λ(x) ≡ u, vx,λ(x) ≡ v for all x ∈ Rn in Rn Proof Suppose that there exists some x0 ∈ Rn such that λ(x0 ) < ∞, then by Lemma and for |y| sufficiently large, we have −p |y| u(y) = |y|−p ux0 ,λ(x0 ) (y) −p = |y| −p lim |y| |y|→∞ lim |y| ux,λ (y) |y − x0 | (3.14) v(y) = λ(x0 )−p v(x0 ) u(y), vx,λ (y) for all < λ < λ(x) and all x, y such that |y − x| thanks to (3.14), one can easily see that −p lim inf |y| |y|→∞ 17 y − x0 (3.15) Let x ∈ Rn be arbitrary By the definition of λ(x) we get 14 16 u x0 + λ(x0 ) u(y) = λ(x0 )−p u(x0 ) −p |y|→∞ 15 p Repeating the above argument then gives 12 13 |y − x0 | which implies 10 11 y − x0 u x0 + λ(x0 ) |y − x0 | |y| = λ(x0 )−p −p λ(x0 ) |y − x0 | u(y) −p lim inf |y| |y|→∞ −p = lim inf |y| |y|→∞ v(y) λ Then by a direct computation and ux,λ (y) λ |y − x| −p u x + λ2 y−x (3.16) |y − x| = λ−p u(x) 19 for all < λ < λ(x) Combining (3.14) and (3.16) gives λ(x0 )−p u(x0 ) λ−p u(x) for all < λ < λ(x) From this, we obtain λ(x) < ∞ for all x ∈ Rn as claimed 20 P ROOF OF T HEOREM 21 To conclude Theorem 4, we first recall the following two calculus lemmas from [Li04] These two lemmas have been used repeatedly in many works 18 22 ˆ Q.A NGO 14 Lemma For ν ∈ R and f a function defined on Rn and valued in [−∞, +∞] satisfying λ |y − x| 10 11 12 ν x + λ2 f y−x |y − x| f (y) for all x, y satisfying |x − y| > λ > Then f is constant or is identical to infinity Lemma For ν ∈ R and f a continuous function in Rn Suppose that for every x ∈ Rn , there exists λ(x) > such that λ(x) |y − x| Then for some a ν f x + λ(x) y−x ∀y ∈ Rn \ {x} = f (y), |y − x| 0, d > and x ∈ Rn f (x) = ±a ν d + |x − x| Now, to prove Theorem 4, we consider the two possible cases: Case If λ(x) = ∞ for any x ∈ Rn then ux,λ (y) u(y) for all λ > and for any x, y satisfying |y − x| λ By Lemma 8, u must be constant Similarly, v is also constant However, this is not the case since constant solutions not solve (1.1) 14 Case If there exists some x0 ∈ Rn such that λ(x0 ) < ∞, then by Lemma 7, we deduce that λ(x) < ∞ for any point x ∈ Rn Then Lemma tells us that u is of the form 15 u(x) = a1 (b21 + |x − x1 |2 ) 13 p 16 (4.1) n for some a1 , d1 > and some point x1 ∈ R Similarly v is of the following form p 17 18 19 20 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U, Classification of solutions for a system of integral equations, Comm Partial Differential Equations 30 (20 05), pp 59–65 1, 26 30 31 32 33 34 35 36 SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD. .. setting, Eq (1 .3) is associated with some fourth order partial differential equation since it simply becomes 25 (−∆ )2 u = u−q 20 21 22 23 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 (1.5)... solving systems of integral equations As far as we know, this is the first attempt to use the method of moving spheres for systems (with negative exponents) 16 17 18 19 20 21 22 23 24 25 26 27 28