Liouville Theorems for Integral Systems Related to Fractional Lane-Emden Systems in R N +

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1 Liouville Theorems for Integral Systems Related to Fractional Lane-Emden Systems in R Senping Luo & Wenming Zou Department of Mathematical Sciences, Tsinghua University, Beijing 00084, China Abstract In this paper, we consider some integral systems in the half space R obtain Liouville type theorems about the positive solutions By moving plane method in terms of the integral form, we shall see that the positive solution (u(x,, x ), v(x,, x )) of the integral systems must be independent of the first ( )-variables, ie, u = u(x ), v = v(x ) Then, combine with the order estimates about x, we reduce the problem to a sequence of algebraic systems Furthermore, we discuss the relationship between the integral system the fractional differential system related to the fractional Lane-Emden e- quations By this way, we obtain two non-existence theorems for the fractional differential system Introduction The Liouville type theorem of the positive solutions to the PDEs integral equations has been received much attention In this paper, we consider the Liouville theorem about the integral systems in the half space R : u(x) = G(x, y)u R p (y)v q (y)dy, v(x) = () G(x, y)u R p2 (y)v q2 (y)dy, where R := {x = (x,, x ) R : x > 0} is the upper Euclidean space Obviously, the system () includes the following special cases: u(x) = G(x, y)u p (y)v q (y)dy, R v(x) = G(x, y)u q (y)v p (2) (y)dy, R Zou is supported by SFC(3722, 27386) address: luosp989@63com; luosp4@mailstsinghuaeducn; wzou@mathtsinghuaeducn

2 u(x) = G(x, y)v p (y)dy, R v(x) = G(x, y)u q (y)dy R (3) Here thereafter we denote by G(x, y) := A [ s,α B,α t s ( α)/2 (s t) ( 2)/2 0 (s tb) ( 2)/2 ] db b α/2 ( b) (4) the Green function in R with the 0-Dirichlet condition, here s = x y 2 t = 4x y We note the following integral equation defined in the whole space u(x) = G(x, y)u(y) α α dy, (5) R where G(x, y) = / x y α is the Green function in the whole space R Chen, Li Ou [5] prove that every regular solution is radially symmetric monotone about some point therefore assuming the form c( 2 x x 0 ) ( α)/2 with some 2 constant c = c(, α) for some > 0, x 0 R These results give a classification of the solution to (5) Later, in Yu [20], the author obtained some Liouville theorems for the following integral equation system defined in the whole space R : u(x) = G(x, y)f(u(y))dy, x R, R { u(x) = R G(x, y)f(u(y), v(y))dy, x R, v(x) = R G(x, y)g(u(y), v(y))dy, x R It seems that few Liouville type result can be seen for the above integral system on the half space R However, the importance of establishing the Liouville theorem for the above integral systems ()-(3) on the half space R is apparent On the other h, the systems ()-(3) above are closely related to the following fractional Lane-Emden system (see Section 4 of the current paper for more details about the relationship): ( ) α 2 u = u p v q, x R, ( ) α 2 v = u p2 v q2, x R, (6) u = v = 0, x R, ( ) α 2 u = u p v q, x R, ( ) α 2 v = u q v p, x R, (7) u = v = 0, x R ; 2

3 ( ) α 2 u = v p, x R, ( ) α 2 v = u q, x R, u = v = 0, x R ; where the fractional Laplacian in R is a nonlocal operator, taking the form ( ) α u(x) u(z) 2 u(x) = C,α P V dz, x z α R (8) here 0 < α < 2 P V sts for the Cauchy principal value One can extend this definition to the more general space L α/2 := {u L loc(r u(x) ) : dx < } x α by ( ) α 2 u, φ = R R u( ) α 2 φdx, φ C 0 (R ) Given any f L loc (R ), we say that u L α/2 solves the problem ( ) α 2 u(x) = f(x), x R, if only if u( ) α 2 φdx = f(x)φ(x)dx, φ C0 (R ) R R ow we recall some backgrounds on these topics In the half space, recall the scalar equation { ( ) α 2 u = u p, x R, u = 0 x R (9) For classical solution, when α = 2, Dancer [8] proved that the equation has no bounded solution if < p < 3 About the viscosity solution, it was proved in Quaas Xia [6] that there is no positive viscosity bounded solution of (9) provided that > 2α < p < α 2α α In particular, when p 2α, if u D s,2 (R ) C(R ) is a nonnegative solution of (9), then u(x) 0, see Fall Weth [9] About the distribution solution, Chen, Fang Yang [3] show that the nonnegative locally bounded solution of (9) is u(x) 0 provided that < p α α In particular, when p = α 2 α α, if u Lloc (R ) is a nonnegative solution of (9), then it must be u(x) 0 It is natural to consider the fractional Lane-Emden system in the half space as below: ( ) α 2 u = v p, x R, ( ) α 2 v = u q, x R, (0) u = v = 0, x R 3

4 On the viscosity solution, Quaas Xia [6] has proved that, under the restriction p, q >, there is no positive viscosity bounded solution to (0) if only if p, q 2α 2α satisfy 2α p, q 2α but not both equal to 2α max{ 2α(p ) 2α(q ), pq pq } > 2α More results about the Lane-Emden system with α = 2 (ie, the usual Laplacian ), we refer the readers to Dahmani-Karami-Kerbal [7], Mitidieri [3, 4], Serin Zou [7], Quitter Souplet [5], Souplet [8] Lin [2] In the present paper, we are interested in establishing the Liouville theorems for the integral systems (), (2) (3) in the distribution sense We firstly establish the following Liouville theorem under the global integrability assumption Theorem Let r := p 2q (p )(q 2 ) α( q q 2 ), s := p 2q (p )(q 2 ) α( p 2 p ) Assume p j, q j > (j =, 2) min{r, s} > α, then the system () has no positive solution of such type u L r (R ), v L s (R ) Corollary Assume p >, q >, p q > positive solution satisfying u L pq α α (R ), v L pq α (R )} Then the system (2) has no Corollary 2 Assume p >, q >, max{ p pq, q pq } < α α, then the system (3) has no such a positive solution (u, v) belonging to L pq α(p) (R ) L pq α(q) (R ) The basic idea for proving the above results is the moving plane method of the integral form More precisely, we move the plane along the x -direction to show that the solution must be monotone increasing in x -direction then we can derive a contradiction ext, we use the Kelvin transformation to weaken the global integrability condition in Theorem We place the centers on the boundary R to ensure that the half space R is invariant under such a transformation For any point z 0 R, we consider the transform x z0 u z 0(x) := x z 0 u( α x z 0 2 z0 ) Then we have the following improved version of Theorem Theorem 2 Assume p >, q >, p q α α, then the system (2) has no positive locally bounded solution In particular, if p q = α α, then the system (2) has no positive solution in L 2 α loc (R ) 4

5 Theorem 3 Assume < p, q α p α max{ no positive locally bounded solution to (3) pq, q pq } > 2 α 2α, then there is Remark In Theorems 2 3 above, we only require that u, v are locally bounded Furthermore, in the critical case, we may claim no solution in L 2 α loc (R ) In Section 4, we will discuss the equivalence relationship between the fractional differential system the corresponding integral system Based on those argument, we obtain the following theorems for the fractional differential systems (7) (8) Theorem 4 Assume p, q > < p q < α α If any positive solution (u, v) of the system (7) is independent of the first ( )-variables, ie, u = u(x ), v = v(x ), then the system (7) has no positive locally bounded solution Theorem 5 Assume < p, q α p α max{ pq, q pq } > 2 α 2α If any positive solution (u, v) of the system (8) is independent of the first ( )-variables, ie, u = u(x ), v = v(x ), then the system (8) has no positive locally bounded solution Remark 2 We conjecture that the assumption u = u(x ), v = v(x ) in the Theorems 4-5 can be removed, since they are indeed not needed for the integral systems 2 Proof of Theorem We will use the moving plane method in terms of the integral form To this goal, we should do some preparation Let be a positive number T := {x R x = } be the moving plane Let Σ := {x = (x,, x, x ) R 0 < x < } denote the region between the plane x = 0 the plane x = Consider the following reflection Set x = (x, x 2,, x ) x = (x,, x, 2 x ) Σ c := R \ Σ, u (x) := u(x ), w := u (x) u(x), w 2 (x) := v (x) v(x) ext, we present some properties of the Green function G(x, y) in the half space R, which is very important to our estimates to derive the relationships for fractional integral differential equation In particular, it is helpful for discussing the equivalence between the fractional differential system the corresponding integral system Lemma 2 (See [3]) Let G(x, y) be defined in (4) () For any x, y Σ, x y, we have G(x, y ) > max{g(x, y), G(x, y )} G(x, y ) G(x, y) > G(x, y) G(x, y ) 5

6 (2) For any x Σ, y Σ c, it holds G(x, y) > G(x, y) (3) Let t = 4x y be fixed s = x y 2 go to infinity, then G(x, y) t α 2 Lemma 22 (See [3]) Assume that w is a nonnegative solution of { ( ) α 2 w(x) = 0, x R, w(x) = 0, x R, then there holds either or there exists a constant c such that s 2 w(x) = 0, x R, w(x) c(x ) α 2, x R The following maximum principle for the fractional differential equation can be seen in [9] Lemma 23 (See [9]) () If u L α/2 ( ) α 2 u 0 in an open set Ω, then u is lower semi-continuous in Ω (2) Let Ω R be a bounded open set, let f be a lower-semicontinuous function in Ω such that ( ) α 2 f 0 in Ω f 0 in R \ Ω, then f 0 in R The following lemma provides an important integral estimate to the moving plane method in terms of the integral form Lemma 24 For any x Σ, () if (u, v) is a positive solution of the integral system (3), then we have the following estimate: u(x) u (x) [G(x, y ) G(x, y )][v p (y) v p (y)]dy, Σ (2) v(x) v (x) [G(x, y ) G(x, y )][u q (y) u q (y)]dy; Σ (2) if (u, v) is a positive solution of the integral system (2), then we have the following estimate: u(x) u (x) [G(x, y ) G(x, y )][u p (y)v q (y) u p (y)vq (y)]dy, Σ v(x) v (x) [G(x, y ) G(x, y )][u q (y)v p (y) u q (y)vp (y)]dy; Σ 6

7 (3) if (u, v) is a positive solution of the integral system (), then we have the following estimate: u(x) u (x) [G(x, y ) G(x, y )][u p (y)v q (y) u p (y)vq (y)]dy, Σ v(x) v (x) [G(x, y ) G(x, y )][u p2 (y)v q2 (y) u p2 (y)vq2 (y)]dy Σ Proof We only need to prove () since we can do the other two in a similar way Let Σ be the reflection of Σ about the plane T Since (u, v) is positive solution of (3), by an integral transformation, we have u(x) = G(x, y)v p (y)dy u(x ) = R = G(x, y)v p (y)dy Σ G(x, y )v p (y)dy G(x, y)v p (y)dy, Σ Σ c \ Σ R G(x, y)v p (y)dy = G(x, y)v p (y)dy Σ G(x, y )v p (y)dy G(x, y)v p (y)dy Σ Σ c \ Σ Subtract the above two equations combine with Lemma 2, we get that u(x) u(x ) = [G(x, y) G(x, y)]v p (y)dy Σ [G(x, y ) G(x, y )]v p (y)dy Σ [G(x, y) G(x, y)]v p (y)dy Σ c \ Σ [G(x, y) G(x, y)]v p (y)dy Σ [G(x, y ) G(x, y )]v p (y)dy Σ [G(x, y ) G(x, y )]v p (y)dy Σ [G(x, y ) G(x, y )]v p (y)dy Σ = [G(x, y ) G(x, y )][v p (y) v p (y)]dy Σ 7

8 The following Hardy-Littlewood-Sobolev inequality is indispensible for our estimates Lemma 25 (See [4]) Assume that 0 < α < Ω R Let g L p αp (Ω) for < p < Define α then T g(x) := Ω g(y)dy, x y α T g L p (Ω) C(, p, α) g L p αp (Ω) Proof of Theorem We divide the proof into several steps Step : We claim that for > 0 sufficiently small, there hold To prove this claim, we define w := u (x) u(x) 0 ae x Σ, w 2 := v (x) v(x) 0 ae x Σ Ω := {x Σ w < 0}; Ω 2 := {x Σ w 2 < 0} Equivalently, we will show that for > 0 sufficiently small, Ω Ω2 both have the 0 Lebesgue measure For convenience, we replace u p (y)v q (y), u p2 (y)v q2 (y) by f(u(y), v(y)), g(u(y), v(y)) respectively We start from Lemma 24, for any x Ω, by Lemma 24, we have 0 <u(x) u(x ) ( )( ) G(x, y ) G(x, y ) f(u(y), v(y)) f(u (y), v (y)) dy Σ ( ) f ( ) = G(x, y ) G(x, y ) Σ u (ϕ (y), φ (y)) u(y) u (y) dy ( ) f ( ) G(x, y ) G(x, y ) Σ v (ϕ (y), φ (y)) v(y) v (y) dy ( ) f ( ) = G(x, y ) G(x, y ) Ω u (ϕ (y), φ (y)) u(y) u (y) dy ( ) f ( ) G(x, y ) G(x, y ) Σ \Ω u (ϕ (y), φ (y)) u(y) u (y) dy [G(x, y ) G(x, y )] f Ω 2 v (ϕ (y), φ (y))[v(y) v (y)]dy ( ) f G(x, y ) G(x, y ) v (ϕ (y), φ (y))[v(y) v (y)]dy Σ \Ω 2 (22) 8

9 Since u(y) u (y) < in Σ \ Ω v(y) v (y) < in Σ \ Ω 2, we have 0 <u(x) u(x ) [ ] f [ ] G(x, y ) G(x, y ) Ω u (ϕ (y), φ (y)) u(y) u (y) dy [ ] f [ ] G(x, y ) G(x, y ) Ω 2 v (ϕ (y), φ (y)) v(y) v (y) dy G(x, y ) f [ ] u (ϕ (y), φ (y)) u(y) u (y) dy Ω Ω Ω 2 G(x, y ) f v (ϕ (y), φ (y)) A,α f x y α u (ϕ (y), φ (y)) Ω 2 A,α f x y α v (ϕ (y), φ (y)) [ ] v(y) v (y) dy [ ] u(y) u (y) dy [ ] v(y) v (y) dy (23) Similarly, we have 0 <v(x) v(x ) A,α g Ω x y α u (ϕ (y), φ (y))[u(y) u (y)]dy A,α g x y α v (ϕ (y), φ (y))[v(y) v (y)]dy Ω 2 (24) Recalling that f(u(y), v(y)) = u p (y)v q (y) g(u(y), v(y)) = u p2 (y)v q2 (y), we get that 0 <u(x) u(x ) C(, α, p, q ) Ω x y α up (y)v q (y)[u(y) u (y)]dy C(, α, p, q ) x y α up (y)v q (y)[v(y) v (y)]dy Ω 2 (25) 0 <v(x) v(x ) C(, α, p 2, q 2 ) Ω x y α up2 (y)v q2 (y)[u(y) u (y)]dy C(, α, p 2, q 2 ) x y α up2 (y)v q2 (y)[v(y) v (y)]dy Ω 2 (26) 9

10 From the above estimates, we apply the Hardy-Littlewood-Sobolev inequality (see Lemma 25) the Hölder inequality to obtain that u u L r (Ω ) C u p v q (u u ) L r αr (Ω ) C u p v q (v v ) L r αr (Ω 2 ) C u p r L p (Ω s u u ) vq L q (Ω ) Lr (Ω ) C u p L r p (Ω 2 ) vq L C u p L r (Ω ) v q L s (Ω ) u u L r (Ω ) s q (Ω 2 ) v v Ls (Ω 2 ) C u p L r (Ω 2 ) v q L s (Ω 2 ) v v Ls (Ω 2 ) (27) v v Ls (Ω 2 ) C u p2 v q2 (u u ) L s αs (Ω ) C u p2 v q2 (v v ) L s αs (Ω 2 ) C u p2 r L p 2 (Ω s u u ) vq2 L q2 (Ω ) L r (Ω ) C u p L r p (Ω 2 ) vq L C u p2 L r (Ω ) v q2 L s (Ω ) u u Lr (Ω ) s q (Ω 2 ) v v L s (Ω 2 ) C u p2 L r (Ω 2 ) v q2 L s (Ω 2 ) v v L s (Ω 2 ), (28) where the integral index r, s satisfy r, s > Hence r = p 2q (p )(q 2 ) α( q q 2 ) In the special case ( ) p q = p 2 q 2 α are determined by { p r q s = r α, p 2 r q2 s = s α (29) ; s = p 2q (p )(q 2 ) (20) α( p 2 p ) ( ) p q, q p we see that r = s = pq α Hence, r, s > α p q > In another special case ( ) ( ) p q 0 p =, p 2 q 2 q 0 α 0

11 then by a similar estimate, we see that u(x) u (x) L r (Ω ) C v p (v(y) v (y)) L r αr (Ω 2 ) C v p s L p (Ω 2 ) v(y) v (y) L s (Ω 2 ) C v p L s (Ω 2 ) v(y) v (y) L s (Ω 2 ), (2) v(x) v (x) L s (Ω 2 ) C u q (u(y) u (y)) L s αs (Ω ) C u q r L q (Ω ) u(y) u (y) Lr (Ω ) C u q L r (Ω ) u(y) u (y) Lr (Ω ), (22) where r, s satisfy { p s = r α q r = s α Thus, r = (pq ) α(p), s = (pq ) α(q) Therefore, r, s > α max{ p pq, q pq } < α α Summing up the above estimates, we realize that we can deal with the systems (), (2) (3) in a unified way Therefore, in the rest of the current paper, we may not distinct these cases when we do the moving plane process From (27) (28) we get that u u L r (Ω ) v v L s (Ω 2 ( ) C u ) u p L r (Ω ) v q L s (Ω ) u p2 L r (Ω ) v q2 L s (Ω ) u L r (Ω ( ) C u ) v p L r (Ω 2 ) v q L s (Ω 2 ) u p2 L r (Ω 2 ) v q2 L s (Ω 2 ) v L s (Ω 2 ) I(u, v, ) u u Lr (Ω ) J(u, v, ) v v Ls (Ω 2 ), (23) where ( I(u, v, ) := C u ), p L r (Ω ) v q L s (Ω ) u p2 L r (Ω ) v q2 L s (Ω ) ( J(u, v, ) := C u ) p L r (Ω 2 ) v q L s (Ω 2 ) u p2 L r (Ω 2 ) v q2 L s (Ω 2 ) Since u L r (R ) v L s (R ), we can choose sufficiently small positive such that I(u, v, ) 2 J(u, v, ) 2 (24)

12 Therefore, by (23) (24), we have u u Lr (Ω ) = 0 v v Ls (Ω 2 ) = 0, hence, Ω Ω2 must be zero-measure sets Step 2: We start from such small move the plane T until w j 0(j =, 2) does not hold Define We will prove that 0 := sup{ > 0 w j ρ(x) 0, ρ, x Σ ρ, j =, 2} 0 = Assume by contradiction that 0 <, we will show that either u(x) or v(x) is symmetric about the plane T 0, ie, either w 0 0 ae x Σ 0 or w ae x Σ 0 (25) Suppose (25) does not hold, then for such a 0, we have w 0 0, w 0 0 ae on Σ 0 w 2 0 0, w ae on Σ 0 We will show that the plane can be moved further up More precisely, we will show that there exists an ε > 0 such that for all [ 0, 0 ε), w j (x) 0 ae on Σ for j =, 2 To prove this, we will resort to the inequality (23) The strategy is the following: for ε > 0 sufficiently small, for all [ 0, 0 ε), both u r (y)dy v s (y)dy Ω Ω 2 can be made so small that I(u, v, ) 2 J(u, v, ) 2 Then by (23), we have that w L r (Ω ) = 0 w 2 L r (Ω 2 ) = 0, thus, both Ω Ω2 must be zero-measure sets Hence, for those values of > 0, we have w j (x) 0 ae on Σ for j =, 2 This contradicts the definition of 0 Therefore (25) must hold By (25), we get that either u(x) = 0 or v(x) = 0 on the plane x = 2 0 (since u(x) = v(x) = 0 on R ) This contradicts our assumption u(x) > 0, v(x) > 0 in R, hence 0 =, it follows that u, v are monotone increasing with respect to x But this contradicts with u L r (R ) or v L s (R ) Hence the positive solution of system (3) dose not exist Therefore, to finish the proof of Theorem, it suffices to verify that Ω u r (y)dy 0 v s (y)dy 0 as Ω 2 0, > 0 For any small η > 0, we can choose R sufficiently large such that R \B ur (y)dy < η R R \B vs (y)dy < η Fix this R R ext we will show that the measures of Ω B R Ω 2 B R are sufficiently 2

13 small for close to 0 (ie, ε 0) Firstly, u(x ) u(x) [G(x, y ) G(x, y )][u p (y)vq (y) up (y)v q (y)]dy Σ [G(x, y) G(x, y)]u p (y)v q (y)dy Σ c \ Σ [G(x, y) G(x, y)]u p (y)v q (y)dy Σ c \ Σ Similarly, we have v(x ) v(x) [G(x, y) G(x, y)]u p2 (y)v q2 (y)dy Σ c \ Σ It follows that in the interior of Σ For any δ > 0, let It is obvious that w j 0 (x) > 0, j =, 2 A δ = {x Σ 0 B R w 0 > δ}, B δ = (Σ 0 B R ) \ A δ lim m(b δ) = 0 δ 0 For > 0, let D = (Σ \ Σ 0 ) B R, then (Ω B R ) (Ω A δ ) B δ D δ, (26) it is easy to see that lim meas(d ) = 0 We show that the measure of Ω 0 A δ can be sufficiently small as is close to 0 In fact, for any x Ω A δ, w (x) = u (x) u(x) = u (x) u 0 (x) u 0 u(x) < 0, hence it follows that u 0 (x) u (x) > w 0 > δ, (Ω A δ ) E δ := {x B R u 0 (x) u (x) > δ}, thus meas(e δ ) δ r E δ u 0)(x) u (x) r dx, δ r B R u 0 (x) u (x) r dx Therefore, for each fixed δ, meas(e δ ) 0 as 0 Combine with (26), the measure of Ω B R can be made arbitrarily small, that is, Ω u r (y)dy 0 as 0 The proof of v s (y)dy 0 as Ω 2 0 is similar This completes the proof of Theorem 3

14 3 The proofs of Theorems 2-3: Liouville theorem under locally bounded assumption In this section, we will use Kelvin type transformation to derive Liouville-type theorems of (3) under much weak conditions, ie, the solution (u, v) is only locally bounded in the subcritical case locally integrable in the critical case Instead of applying the moving plane method to (u, v) directly, we carry on the Kelvin type transformation of (u, v) Let us start from the Kelvin transform on the half space R For z 0 R, let x z0 u z 0(x) := x z 0 u( α x z 0 2 z0 ) v z 0(x) := x z0 x z 0 v( α x z 0 2 z0 ) be the Kelvin type transformations of (u, v) centered at z 0 By a straight forward calculation, we have u z 0 = x z α G( x z0 R x z 0 2 z0, y)u p (y)v q (y)dy (3) = G(x, y) up z (y)v q 0 z (y) 0 y z 0 dy β R v z 0 = x z α = R R G( x z0 x z 0 2 z0, y)u p2 (y)v q2 (y)dy G(x, y) up2 z 0 (y)v q2 z 0 (y) y z 0 β2 dy, (32) for x R \ B ε (z 0 ), ε > 0, where β j := ( α)(τ p j q j ), p j q j τ := α α for j =, 2 We consider the critical subcritical cases separately Case : The critical case p j q j = α α, j =, 2 If (u(x), v(x)) is a solution of the system u(x) = G(x, y)u R p (y)v q (y)dy, v(x) = (33) G(x, y)u R p2 (y)v q2 (y)dy, by (3) (32), we know that the Kelvin transformations (u z0, v z0 ) is also a solution of the above integral system Since u, v L 2 α loc (R ), for any bounded domain Ω which has a positive distance away from z 0, we must have u 2 α z (y)dy <, v 2 α 0 z (y)dy < 0 Ω Ω 4

15 We distinct two subcases Subcase : If there is a z 0 = (z 0,, z 0, 0) R such that u z 0(x), v z 0(x) is bounded near z 0, then by the Kelvin transform, we have Therefore, u(y) = R v(y) = y z 0 α u y z0 z0( y z 0 2 z0 ), y z 0 α v y z0 z0( y z 0 2 z0 ) u(y) = O( ), as y, y α v(y) = O( ), as y, y α Since u, v L 2 α loc (R ), together with the above estimates, we have that u 2 α (y)dy <, v 2 α (y)dy < Hence, we still can move the plane of (u, v) as the arguments in the proof of Theorem derive the non-existence of positive solution of (3) Subcase 2: For all z 0 = (z, 0, z 0, 0) R, if either u z 0 or v z 0 is unbounded near z 0, inspired by [3], we will carry on the moving plane on (u z 0, v z 0) in R to prove that it is rotationally symmetric about the line passing through z 0 parallel to the x -axis From this, we will conclude that u = u(x ), ie, u is independent of the first ( ) variables x,, x, as we will see in the last step, this will contradict the convergence of the integrals G(x, y)u p (y)v q (y)dy G(x, y)u p2 (y)v q2 (y)dy R For the simplicity, we denote (u z 0, v z 0) by (u, v) For a real number, define the reflection as follows R R Σ = {x = (x,, x ) R x < } x = (2 x, x 2,, x ) For x, y Σ, x y, by the representation formula of the Green function (see (4)), it is easy to check that G(x, y) = G(x, y ), G(x, y) = G(x, y ), G(x, y ) > G(x, y ) (34) Since we have the follwing integral identity, u(x) = G(x, y)u p (y)v q (y)dy Σ G(x, y )u p (y)vq (y)dy, Σ 5

16 u(x ) = G(x, y)u p (y)v q (y)dy Σ G(x, y )u p (y)vq (y)dy Σ Combine with (34), it deduces that ( ) u(x) u(x ) = G(x, y Σ ) G(x, y ) u p (y)vq (y)dy ( ) G(x, y) G(x Σ, y) u p (y)v q (y) ( )( = G(x, y) G(x Σ, y) u p (y)v q (y) u p (y)vq )dy (y) Similarly, we have the counterpart for v, that is, v(x) v(x ) = (35) ( )( G(x, y) G(x Σ, y) u p2 (y)v q2 (y) u p2 (y)vq2 )dy (y) (36) Step : We will show that for sufficiently negative, the following hold: w (x) := u (x) u(x) 0 ae x Σ ; w 2 (x) := v (x) v(x) 0 ae x Σ (37) Define Ω j := {x Σ \ B ε ((z 0 ) ) w j (x) < 0}, j =, 2, where (z 0 ) is the reflection of z 0 about the plane T = {x R x = } Equivalently, we will show that for sufficiently negative, Ω Ω 2 must be zero- 6

17 measure sets In fact, for x Ω, we have 0 <u(x) u (x) = [G(x, y) G(x, y )][u p (y)v q (y) u p (y)vq (y)]dy Σ = [G(x, y) G(x, y )][u p (y)v q (y) u p (y)v q (y) Σ u p (y)v q (y) up (y)vq (y)]dy = [G(x, y) G(x, y )]u p (y)[v q (y) v q (y)]dy Ω 2 [G(x, y) G(x, y )]u p (y)[v q (y) v q (y)]dy Σ \ Ω 2 [G(x, y) G(x, y )]v q (y) u (y)[up p (y)]dy Ω [G(x, y) G(x, y )]v q (y) u (y)[up p (y)]dy Σ \ Ω [G(x, y) G(x, y )]u p (y)[v q (y) v q (y)]dy Ω 2 [G(x, y) G(x, y )]v q (y) u (y)[up p (y)]dy Ω Here we have used that v(y) v (y) in Σ \ Ω 2 u(y) u (y) in Σ \ Ω, we continue the above inequality 0 <u(x) u (x) [G(x, y) G(x, y )]u p (y)[v q (y) v q (y)]dy Ω 2 [G(x, y) G(x, y )]v q (y) u (y)[up p (y)]dy Ω q [G(x, y) G(x, y )]u p (y)v q (y)[v(y) v (y)]dy Ω 2 p Ω [G(x, y) G(x, y )]u p (y)v q (y)[u(y) u (y)]dy C(, α, q ) Ω 2 x y α up (y)v q (y)[v(y) v (y)]dy C(, α, p ) x y α up (y)v q (y)[u(y) u (y)]dy Ω 7

18 Similarly, for x Ω 2, there holds 0 <v(x) v (x) C(, α, q 2 ) Ω 2 x y α up2 (y)v q2 (y)[v(y) v (y)]dy C(, α, p 2 ) x y α up2 (y)v q2 (y)[u(y) u (y)]dy Ω Applying the Hardy-Littlewood-Sobolev inequality the Hölder inequality to obtain that w L r ( Ω ) C up v q w r L αr ( Ω ) Similarly, we can derive that here r, s satisfy u p v q w 2 L r αr ( Ω 2 ) C u p L r ( Ω ) v q L s ( Ω ) w Lr ( Ω ) C u p L r ( Ω 2 ) v q L s ( Ω 2 ) w2 L r ( Ω 2 ) w 2 L r ( Ω 2 ) C u p2 L r ( Ω ) v q2 L s ( Ω ) w Lr ( Ω ) p C u p2 L r ( Ω 2 ) v q2 L s ( Ω 2 ) w2 Lr ( Ω 2 ), r q s = r α, p 2 r q 2 s = s α (38) Thus, we have the following estimate w Lr ( Ω ) w2 Lr ( Ω ) I(u, v, ) w Lr ( Ω ) J(u, v, ) w2 Ls ( Ω 2 ), (39) where I(u, v, ) := C u p L r ( Ω ) v q L s ( Ω ) C u p2 L r ( Ω ) v q2 L s ( Ω ), J(u, v, ) := C u p L r ( Ω 2 ) v q L s ( Ω 2 ) C u p2 L r ( Ω 2 ) v q2 L s ( Ω 2 ) Since p j q j = α α, j =, 2, (38) holds, by a direct calculation, we see that r = p 2q (p )(q 2 ) = 2 α( q q 2 ) α, s = p 2q (p )(q 2 ) = 2 α( p 2 p ) α On the other h, since u, v L 2 α loc (R ), we can choose a sufficiently large such that for, we can make sure that I(u, v, ) 2 J(u, v, ) 2 ow (39) implies that w L r ( Ω ) = 0 w2 L s ( Ω 2 ) = 0 8

19 therefore Ω Ω 2 must be zero-measure sets Step 2: Define 0 := sup{ z 0 w j ρ(x) 0, j =, 2, ρ, x Σ ρ } We will prove that 0 z 0 ε for any ε > 0 small On the contrary, if 0 < z 0 ε, we claim that there hold w 0 (x) 0 ae x Σ 0 \ B ε ((z 0 ) 0 ), or w 2 0 (x) 0 ae x Σ 0 \ B ε ((z 0 ) 0 ), (30) where (z 0 ) 0 is the reflection of z 0 about the plane T 0 Suppose (30) is not true, then for such 0 < z 0 ε, we have w 0 (x) 0 but w 0 (x) 0 ae on Σ 0 \ B ε ((z 0 ) 0 ) w 2 0 (x) 0 but w 2 0 (x) 0 ae on Σ 0 \ B ε ((z 0 ) 0 ) From the estimate (39), we can choose δ small enough such that for all [ 0, 0 δ), we make sure u 2 Ω α (y) u 2 Ω α (y) so small that I(u, v, ) 2 J(u, v, ) 2 In fact, for any small η > 0, ε > 0, we can choose R sufficiently large such that (R \Bε(z0 ))\B R u 2 α (y)dy < η, (R \Bε(z0 ))\B R v 2 α (y)dy < η Fix this R then show that the measures of Ω B R Ω 2 B R are sufficiently small when is closing to 0 By (35) (36), we have that w j 0 (x) > 0 for j =, 2 in the interior of Σ 0 \ B ε ((z 0 ) 0 ), the remaining part of the proof for I(u, v, ) 2 J(u, v, ) 2 is similar as that in Theorem ow from (39), we have w j L 2 2 = 0, j =, 2 ( Ω j ) Therefore, Ω j must be zero-measure sets for j =, 2 Hence for these values [ 0, 0 δ), there holds w j (x) 0 ae x Σ \ B ε ((z 0 ) ), ε > 0 This contradicts the definition of 0, hence our claim (30) holds Say for example, u(x) = u 0 (x) ae x Σ 0 \ B ε ((z 0 ) 0 ) On the other h, since u is singular at z 0, u must be singular at (z 0 ), this is a contradiction since z 0 is the only singular point of u Hence, 0 z 0 ε further we have w j z 0 (x) 0 j =, 2 ae x Σ z 0 Since the integral system (3) is invariant under the transform (u (x), v (x)) = (u( x), v( x)), 9

20 we can move the plane from x = to the left derive that w j z 0 (x) 0 j =, 2 ae x Σ z 0 Therefore, w j z 0 (x) = 0 j =, 2 ae x Σ z 0 Case 2: The subcritical case: p j q j α α for j =, 2, (p q, p 2 q 2 ) ( α α, α α ) By (3) (32), we have u(x) = G(x, y) up (y)v q (y) Σ y z 0 dy G(x, y Σ ) up (y)vq (y) β y z 0 dy, β u(x ) = G(x (y)v, y) up q (y) Σ y z 0 dy G(x Σ, y ) up (y)vq (y) β y z 0 dy β Combine with (34), we obtain that u(x) u (x) = [G(x, y) G(x (y)v, y)] up q (y) Σ y z 0 dy β [G(x, y Σ ) G(x, y )] up (y)vq (y) y z 0 dy β = [G(x, y) G(x (y)v, y)][ up q (y) Σ v(x) v (x) = [G(x, y) G(x (y)v, y)] up2 q2 (y) Σ y z 0 dy β Step : For any ε > 0, define y z 0 up (y)vq (y) β y z 0 ]dy, β [G(x, y Σ ) G(x, y )] up2 (y)vq2 (y) y z 0 dy, β = [G(x, y) G(x (y)v, y)][ up2 q2 (y) Σ y z 0 up2 (y)vq2 (y) β y z 0 ]dy β Ω j := {x Σ \ B ε ((z 0 ) ) w j (x) = u (x) u(x) < 0}, j =, 2 ext we show that for sufficiently negative, Ω j are zero-measure sets for j =, 2 20

21 Indeed, we consider x Ω, then we have Hence, 0 <u(x) u (x) = [G(x, y) G(x, y)][ up (y)v q (y) Σ y z 0 up (y)vq (y) β y z 0 ]dy β [G(x, y) G(x Σ, y)] up (y)v q (y) u p (y)v q (y) y z 0 dy β Ω 2 Σ [G(x, y) G(x, y)] up (y)v q (y) up (y)vq (y) y z 0 dy β [G(x, y) G(x, y)] up (y)(vq(y) v q (y)) y z 0 β Σ \ Ω 2 Ω [G(x, y) G(x, y)] up (y)(v q (y) v q (y)) y z 0 β dy [G(x, y) G(x, y)] vq (y)(up (y) u p (y)) y z 0 β dy Σ \ Ω Ω [G(x, y) G(x, y)] vq (y)(up (y) u p (y)) y z 0 β dy 0 <u(x) u (x) (y)v q (y)(v(y) v G(x, y) up (y)) Ω 2 y z 0 β G(x, y) up (y)v q (y)(u(y) u (y)) y z 0 β C Ω 2 Similarly for v, we have Ω u p (y)v q (y)(v(y) v (y)) x y α y z 0 β u p (y)v q (y)(u(y) u (y)) x y α y z 0 β 0 <v(x) v (x) u p2 (y)v q2 (y)(v(y) v (y)) C Ω 2 x y α y z 0 β2 u p2 (y)v q2 (y)(u(y) u (y)) x y α y z 0 β2 Ω 2

22 Therefore, Hence, we have u(x) u (x) L r ( Ω ) C up v q (u u ) y z 0 β L r α r ( Ω ) C up v q (v v ) y z 0 β L r α r ( Ω 2 ), C up v q y z 0 β L α ( Ω ) u u L r ( Ω ) C up v q y z 0 β L α ( Ω 2 ) v v L r ( Ω 2 ), v(x) v (x) L r ( Ω 2 ) C up2 v q2 (u u ) y z 0 β2 L r α r ( Ω ) C up2 v q2 (v v ) y z 0 β2 L r α r ( Ω 2 ) C up2 v q2 y z 0 β2 L α ( Ω ) u u L r ( Ω ) C up2 v q2 y z 0 β2 L α ( Ω 2 ) v v L r ( Ω 2 ) where u(x) u (x) L r ( Ω ) v(x) v (x) L r ( Ω 2 ) I(u, v, p, q) u u L r ( Ω ) J(u, v, p, q) v v L r ( Ω 2 ), (3) I(u, v, p, q) = C up v q y z 0 β L α ( Ω ) C up2 v q2 y z 0 β2 L α ( Ω ), v q J(u, v, p, q) = C up y z 0 β L C v q2 up2 α ( Ω 2 ) y z 0 β2 L α ( Ω 2 ) On the other h, when x z 0, u(x) = x z0 x z 0 u( α x z 0 2 z0 ) x z 0 α Hence, v(x) = x z0 x z 0 v( α x z 0 2 z0 ) x z 0 α ( u(y) p v q (y) ) α y z 0 β y z 0 = ( α)(pq )β) y z

23 ( u(y)p v q (y) y z 0 β ) α y z 0 ( α)(pq )β) = y z 0 2, as y z 0 Therefore, I(u, v, p, q) is convergent to zero as By a similar calculation we see that J(u, v, p, q) is also convergent to zero as Therefore, we may choose sufficiently large, such that for, by (3), we obtain that u u L r ( Ω ) v v L r ( Ω 2 ) = 0 Thus, Ω j must be zero-measure sets for j =, 2 It follows that Step 2: Define w j (x) 0 ae x Σ, j =, 2 0 = sup{ z 0 w j ρ(x) 0, ρ, x Σ ρ, j =, 2} The rest is similar to that of the critical case We omit the details here to avoid repetition Thus we have w j 0 (x) = 0, j =, 2, ae x Σ 0, 0 = z 0 That is, u, v are symmetric about the plane T 0 Since we can take the arbitrary orthogonal transform y = (y, y 2,, y ) = A(x, x 2,, x ), where A is an orthogonal matrix, we can derive that u, v are symmetric with respect to 0 in any direction by the same procedure as that is Step Step 2 of case It follows that u, v are radially symmetric about x = (x,, x ) with respect to the point z 0 Furthermore, since z 0 is arbitrary on the upper half plane parallel to x direction, we conclude that u, v are independent of x = (x,, x ) That is, u = u(x ) Finally, we show that this yields a contradiction with the convergence of the following integrals (this idea is borrowed from [3]): G(x, y)u p (y)v q (y)dy G(x, y)u p2 (y)v q2 (y)dy R R 23

24 For fixed x R, letting R be large enough, we have > u(x ) = u p (y )v q (y ) G(x, y)dy dy 0 R u p (y )v q (y ) G(x, y)dy dy C C C R R R R Similarly, we have u p (y )v q (y )y α/2 u p (y )v q (y )y α/2 R \B R (0) R \B R (0) x y u p (y )v q (y )y α/2 dy x y dy dy R/ x y τ 2 (τ 2 ) /2 dτdy (32) > v(x ) C R u p2 (y )v q2 (y )y α/2 dy (33) These imply that there exists a sequence {y k } as k, such that u p (y k )v q (y k )(y k ) α/2 0, u p2 (y k )v q2 (y k )(y k ) α/2 0 (34) For any x = (0,, 0, x ) R, as in (32) (33), we derive that > u(x ) c 0 > v(x ) c u p (y )v q (y )y α/2 u p2 (y )v q2 (y )y α/2 x y dy x α/2 (35) x y dy x α/2 (36) Let x = 2R be sufficiently large, by (35) (36), we deduce that > u(x ) c 0 0 u p (y )v q (y )y α/2 C 0 (2R) α/2 C (2R) α/2 = C x α/2, 0 dy x α/2 x y u p (y )v q (y )y α/2 dy (37) 24

25 similarly > v(x ) c 0 ow, we suppose that 0 u p2 (y )v q2 (y )y α/2 C 0 (2R) α/2 C (2R) α/2 = C x α/2 0 dy x α/2 x y u p2 (y )v q2 (y )y α/2 dy (38) u(x ) Cx a k v(x ) Cx b k, where a k b k are to be determined later From (37) (38), we see that a 0 = α 2 b 0 = α 2 Then by (35), (37), (36) (38), whenever x = 2R is sufficiently large, we obtain that R u(x ) C y pa kq b k y α 2 x y dy x α 2 (39) similarly R 2 Cx pa kq b k α v(x ) Cx p2a kq 2b k α (320) Thus we have { a k = p a k q b k α, a 0 = α 2 ; b k = p 2 a k q 2 b k α, b 0 = α 2 (32) We will show that by an appropriate selection of k, (39) (320) will contradicts with (34) under some condition about p j q j for j =, 2 By the transformation α(q c k = a k q 2) α(p (p )(q 2 ) p 2q, d k = b k 2 p ) (p )(q 2 ) p 2q, (32) becomes { ck = p c k q d k, c 0 = α 2 α(q q 2) (p )(q 2 ) p 2q ; d k = p 2 c k q 2 d k, d 0 = α 2 α(p 2 p ) (p )(q 2 ) p 2q By the matrix iteration, we observe that ( ck ) ( ) k ( ) p q = c0 d k q 2 d 0 p 2 On the other h, (34) is equivalent to that there exists a sequence {y k } as k satisfying (y k ) a k 0 (y k ) b k 0 (322) ow we consider two special cases corresponding to two important models 25

26 Model : ( ) p q = p 2 q 2 ( ) p q q p This case is corresponding to the model of (2) Since the matrix ( ) ( ) ( ) ( p q p q 0 = q p 0 p q d k ) can be diagonalize, we have ( ) ( ) ( ) ( ) ck (p q) k ( ) 0 c0 = 0 (p q) k Combine with c 0 = d 0 = α 2 then α pq, by a straightforward computation, we have c k = d k = 2 [(p q)k (p q) k ]( α 2 a k = b k = 2 [(p q)k (p q) k ]( α 2 d 0 α p q ), α p q ) α p q When < p q < 2α 2 α, it is easy to see that lim k a k = lim k b k = When k is large enough, this contradicts with (322) we get the final contradiction for this model Therefore, for such model, there is no positive solution for system (2) This completes the proof of Theorem 2 Model 2: ( ) p q = p 2 q 2 ( ) 0 p, q 0 which is corresponding to system (3) Since ( ) 2 ( ) ( ) 0 p pq 0 0 = = pq, q 0 0 pq 0 hence, Therefore, ( ) 2k ( ) ( ) 2k ( ) 0 p = (pq) k 0 p 0 p, = (pq) k 0 q 0 q 0 q 0 0 c 2k = p k q k d 0, c 2k = p k q k c 0, d 2k = p k q k c 0, d 2k = p k q k d 0, a 2k = p k q k α( p) d 0 pq, a 2k = p k q k α( p) c 0 pq, b 2k = p k q k c 0 α( q) pq, b 2k = p k q k d 0 α( q) pq, where c 0 = α 2 α(p) pq, d 0 = α 2 α(q) pq ow under the assumption pq >, if α(p) pq > 2 α α(q) 2α or pq > 2 α 2α, we have a contradiction with (322) Thus, for such a model, we complete the proof of Theorem 3 26

27 4 Equivalence between the integral system () the corresponding fractional system (6) Clarifying the relationship between the integral system () the corresponding fractional system (6) is much difficult We need additional delicate estimates to exclude the extra possibilities which lead to un-equivalence between the two systems We starts by dealing with the fractional Lane-Emden system in the half space as below: ( ) α 2 u = v p, x R, ( ) α 2 v = u q, x R, (4) u = v = 0, x R, which is associated with the corresponding integral system u(x) = G(x, y)v p (y)dy, R v(x) = G(x, y)u q (y)dy R Then we consider the next fractional system in the half space ( ) α 2 u = u p v q, x R, ( ) α 2 v = u q v p, x R, u = v = 0, x R which is associated with corresponding integral system u(x) = G(x, y)u p (y)v q (y)dy, R v(x) = G(x, y)u q (y)v p (y)dy R (42) (43) (44) The above two cases are contained in the following more general fractional system ( ) α 2 u = u p v q, x R, ( ) α 2 v = u p2 v q2, x R, (45) u = v = 0, x R, corresponding to u(x) = G(x, y)u R p (y)v q (y)dy, v(x) = G(x, y)u R p2 (y)v q2 (y)dy (46) ext, we consider the equivalence between (45)-(46) We assume (u, v) is the locally bounded solution of (45) Set w j R (x) := G R (x, y)u pj (y)v qj (y)dy, j =, 2, B R (p R ) 27

28 where G R (x, y) is the Green function on B R (P R ) P R = (0,, 0, R) Hence w j R (x) are well-defined when u, v are locally bounded Furthermore, we have { ( ) α 2 w j R (x) = upj (x)v qj (x), x B R (P R ), w j R (x) = 0, x B R(P R ) Let Q R (x) := u(x) w R (x) Q2 R (x) := v(x) w2 R (x), then we have the following equation { ( ) α 2 Q j R (x) = 0, x B R(P R ), Q j R (x) 0, x B R(P R ) Letting R in the above equations, we get that u(x) G(x, y)u R p (y)v q (y)dy v(x) G(x, y)u R p2 (y)v q2 (y)dy Denote that for j =, 2, then w j (x) = R G(x, y)u pj (y)v qj (y)dy, (47) { ( ) α 2 w j (x) = u pj (x)v qj (x), x R, w j (x) = 0, x B R (P R ), x R Let Q (x) = u(x) w (x) Q 2 (x) = v(x) w 2 (x), we arrive at that ( ) α 2 Q j (x) = 0, x R, Q j (x) 0, x R, Q j (x) = 0, x R, for j =, 2 Deduced from the maximum principle for fractional differential equation, we have either u(x) = w (x), x R or u(x) w (x) c(x ) α 2 for some constant c The same way applies to v, we have either v(x) = w 2 (x), x R or v(x) w 2 (x)c(x ) α 2 for some constant c Thus, we may see that there are four possibilities Possibility : u(x) = w (x), v(x) = w 2 (x), for x R Recall the definitions in (47), the above equations are u(x) = G(x, y)u R p (y)v q (y)dy, v(x) = G(x, y)u R p2 (y)v q2 (y)dy It is exactly what we want, that is, the solutions of the fractional differential system (45) are also the solutions of the integral system (46) Hence, the Liouville type theorems we established in Sections 2 3 can be applied to the corresponding fractional differential system (45) 28

29 Possibility 2: u(x) = w (x) = G(x, y)u R p (y)v q (y)dy, v(x) w 2 (x) C(x ) α 2 = G(x, y)u R p2 (y)v q2 (y)dy C(x ) α 2 So far, we have no direct method to exclude this possibility However, if the solution u = u(x,, x ), v = v(x,, x )) of the differential system (45) are independent of the first ( )-variables, ie, u = u(x ), v = v(x ), then we can reduce the system to a sequence of systems, just as the proof of Theorems 2 3, then exclude this possibility under some proper assumptions on p j, q j, j =, 2 In fact, in the spirit of the proofs of Theorems 2 3, we have u(x ) Cx α 2, v(x ) Cx α 2 By a similar estimate, we have the following recursion formulas { u(x ) Cx pa kq b k α, v(x ) Cx p2a kq 2b k α If we assume that u(x ) Cx a k, v(x ) Cx b k, thus we have the recursion formulas { a k = p a k q b k α, a 0 = α 2, b k = p 2 a k q 2 b k α, b 0 = α 2 (48) By the transform { ck = a k d k = b k α(q q 2) (p )(q 2 ) p 2q, α(p 2 p ) (p )(q 2 ) p 2q, we deduce that { ck = p c k q d k, c 0 = a 0 d k = p 2 c k q 2 d k, d 0 = b 0 We rewrite it in the matrix form: hence we derive that ( ck d k ( ck ) ( ) ( ) p q = ck, p 2 q 2 α(q q 2) (p )(q 2 ) p 2q, α(p 2 p ) (p )(q 2 ) p 2q d k ) ( ) k ( ) p q = c0 d k q 2 d 0 p 2 Model : ( ) p q = p 2 q 2 ( ) p q q p 29

30 Similar to the proof of Theorems 2 3, ( ) ck = ( ) (c0 d 0 )(p q) k (c 0 d 0 )(p q) k d k 2 (c 0 d 0 )(p q) k (d 0 c 0 )(p q) k Here c 0 d 0 = a 0 b 0 α(2 q q 2 p 2 p ) 2α = α (p )(q 2 ) p 2 q p q On the other h, also by the spirit of the proof of Theorems 2 3, we can show that there exists a sequence {y k } as k, such that (y k ) a k 0 (y k ) b k 0 (49) If c 0 d 0 > 0, for sufficiently large k, there will be a contradiction To guarantee c 0 d 0 > 0, we have the following cases: when α < 2, then c 0 d 0 > 0 if p q > ; when 0 < α <, then c 0 d 0 > 0 if < p q < α α Hence we can exclude this possibility when α < 2 p q > or when 0 < α < < p q < α α Model 2: ( ) p q = p 2 q 2 ( ) 0 p q 0 In fact, instead of the above method, we can exclude this possibility by the following method (see the proof of Theorem 4 in [3]) under the assumption q > Since q >, by the property of G(x, y), we have u(x) = G(x, y)v q (y)dy R C R G(x, y)y α 2 q = (40) Possibility 3: u(x) w (x) C(x ) α 2 = G(x, y)u R p (y)v q (y)dy C(x ) α 2, v(x) = w 2 (x) = G(x, y)u R p2 (y)v q2 (y)dy Dealing with this possibility is similar to that of possibility 2, we just state the conclusions here, Model : ( ) ( ) p q p q = p 2 q 2 q p we can exclude this possibility when α < 2 p q >, or when 0 < α < < p q < α α Model 2: ( p q p 2 q 2 ) = ( ) 0 p q 0 30

31 we can exclude this possibility if p > Possibility 4: u(x) w (x) C(x ) α 2 = G(x, y)u R p (y)v q (y)dy C(x ) α 2, v(x) w 2 (x) C(x ) α 2 = G(x, y)u R p2 (y)v q2 (y)dy C(x ) α 2 By the same assumption, that is, u = u(x ), v = v(x ), then just as the same arguments in possibility 2 3, we can exclude this possibility More precisely, Model : ( ) ( ) p q p q = p 2 q 2 q p In this case, we get a 0 = α 2, b 0 = α 2, c 0 d 0 = α 2α p q Thus, c 0 d 0 > 0 when p q > Under this hypothesis, we may exclude this possibility Model 2: ( ) ( ) p q 0 p = p 2 q 2 q 0 In this model, c 0 = α α(p ) pq, d α(q ) 0 = pq, then c 0 > 0 d 0 > 0 if pq > Under this assumption, we may get rid of this possibility In a summary, we have the following results: Lemma 4 Assume that 0 < α < < p q < α α or that α < 2 p q > If any positive solution (u, v) of the system (43) is independent of the first ( )-variables, ie, u = u(x ), v = v(x ), then any positive locally bounded solution of (43) is also the the solution of the corresponding integral system (44) Similarly we have Lemma 42 Suppose p >, q > If any positive solution (u, v) of the system (4) is independent of the first ( )-variables, ie, u = u(x ), v = v(x ), then any positive locally bounded solution of (4) is also the the solution of the corresponding integral system (42) ow, Theorems 4-5 follow from Theorems 2-3 Lemmas 42-4 Acknowledgement: We thanks the referees for the careful reading valuable comments 3

32 References [] L A Caffarelli, B Gidas, J Spruck, Asymptotic symmetry local behavior of semilinear elliptic equations with critical Sobolev growth Comm Pure Appl Math, 42(3):27 297, 989 [2] L A Caffarelli, T Jin, S Yannick J Xiong, Local analysis of solutions of fractional semi-linear elliptic equations with isolated singularities Arch Ration Mech Anal, 23(): , 204 [3] W Chen, Y Fang, R Yang, Liouville theorems involving the fractional Laplacian on a half space Adv Math, 274:67 98, 205 [4] W Chen C Li, Methods on nonlinear elliptic equations, volume 4 of AIMS Series on Differential Equations & Dynamical Systems American Institute of Mathematical Sciences (AIMS), Springfield, MO, 200 [5] W Chen, C Li, B Ou, Classification of solutions for an integral equation Comm Pure Appl Math, 59(3): , 2006 [6] W X Chen C Li, Classification of solutions of some nonlinear elliptic equations Duke Math J, 63(3):65 622, 99 [7] Z Dahmani, F Karami, S Kerbal, onexistence of positive solutions to nonlinear nonlocal elliptic systems J Math Anal Appl, 346():22 29, 2008 [8] E Dancer, Some notes on the method of moving planes Bull Austral Math Soc, 46(3): , 992 [9] M M Fall T Weth, onexistence results for a class of fractional elliptic boundary value problems J Funct Anal, 263(8): , 202 [0] B Gidas, W M i, L irenberg, Symmetry of positive solutions of nonlinear elliptic equations in R n In Mathematical analysis applications, Part A, volume 7 of Adv in Math Suppl Stud, pages Academic Press, ew York-London, 98 [] B Gidas J Spruck, Global local behavior of positive solutions of nonlinear elliptic equations Comm Pure Appl Math, 34(4): , 98 [2] C S Lin, A classification of solutions of a conformally invariant fourth order equation in R n Comment Math Helv, 73(2):206 23, 998 [3] E Mitidieri, A Rellich type identity applications Comm Partial Differential Equations, 8(-2):25 5, 993 [4] E Mitidieri, onexistence of positive solutions of semilinear elliptic systems in R Differential Integral Equations, 9(3): , 996 [5] P Poláčik, P Quittner, P Souplet, Singularity decay estimates in superlinear problems via Liouville-type theorems I Elliptic equations systems Duke Math J, 39(3): , 2007 [6] A Quaas A Xia, Liouville type theorems for nonlinear elliptic equations systems involving fractional Laplacian in the half space Calc Var Partial Differential Equations, 52(3-4):64 659,

33 [7] J Serrin H Zou, on-existence of positive solutions of Lane-Emden systems Differential Integral Equations, 9(4): , 996 [8] P Souplet, The proof of the Lane-Emden conjecture in four space dimensions Adv Math, 22(5): , 2009 [9] L Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator Comm Pure Appl Math, 60():67 2, 2007 [20] X Yu, Liouville type theorems for integral equations integral systems Calc Var Partial Differential Equations, 46(-2):75 95,

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