On the Hénon-Lane-Emden conjecture

Transcription

1 On the Hénon-Lane-Emden conjecture Mostafa Fazly and Nassif Ghoussoub Department of Mathematics, University of British Columbia, Vancouver BC Canada V6T 1Z2 September 27, 2012 Abstract We consider Liouville-type theorems for the following Hénon-Lane-Emden system { u = x a v p in N, v = x b u q in N, when pq > 1, p, q, a, b 0 The main conjecture states that there is no non-trivial non-negative solution N+a whenever p, q is under the critical Sobolev hyperbola, ie + N+b > N 2 We show that this is q+1 indeed the case in dimension N = 3 provided the solution is also assumed to be bounded, extending a result established recently by Phan-Souplet in the scalar case Assuming stability of the solutions, we could then prove Liouville-type theorems in higher dimensions For the scalar cases, albeit of second order a = b and p = q or of fourth order a 0 = b and p > 1 = q, we show that for all dimensions N 3 in the first case resp, N 5 in the second case, there is no positive solution with a finite Morse index, whenever p is below the corresponding critical exponent, ie 1 < p < N+2+2a resp, 1 < p < N+4+2a Finally, we show that non-negative stable solutions of the full N 2 N 4 Hénon-Lane-Emden system are trivial provided pb a + 2 pqq + 1 N < p pqq + 1 p + 1 pqq + 1 p Mathematics Subject Classification 35J47; 35B33; 35B45; 35B08 Key words Liouville-type theorems, Non-linear elliptic systems, Finite Morse index solutions, Hénon-Lane-Emden conjecture 1 Introduction and main results We consider the following weighted system { u = x a v p in N, v = x b u q in N, 1 where pq > 1 and p, q, a, b 0 and is a subset of N, N 1 esearch partially supported by a University Graduate Fellowship at the University of British Columbia Partially supported by a grant from the Natural Sciences and Engineering esearch Council of Canada 1

2 We start by noting that in the case of the Lane-Emden scalar equation ie, when p = q and a = b = 0 on a bounded star-shaped domain N, the Pohozaev inequality shows that there is no positive solution satisfying the Dirichlet boundary condition, whenever p N+2 N 2, the critical Sobolev exponent On the other hand, a celebrated theorem by Gidas-Spruck [14] states that there is no positive solution for the Lane-Emden equation on the whole space whenever p < N+2 N 2 for N 3 This non-existence result is also optimal as shown by Gidas, Ni and Nirenberg in [13] under the assumption that u = O x 2 N, and by Caffarelli, Gidas and Spruck in [3] without the growth assumption See also Chen and Li [4] for an easier proof based on the moving planes method Also, Lin [16] using moving plane methods proved similar optimal non-existence results for p < N+4 N 4, N > 4 in the case of the fourth order Lane-Emden equation ie, when p > 1 = q and a = b = 0 In the case of the system 1, one can again use the Pohozaev identity whenever is a bounded starshaped domain in N, to establish the following non-existence result Theorem A [10, 23] Let N 3 and let N be a star-shaped bounded domain If N + a p N + b N 2, 2 q + 1 then there is no positive solution for 1 on that satisfy the Dirichlet boundary conditions By noting that the curve N+a + N+b q+1 = N 2 is the critical Sobolev hyperbola, the above theorem states that the Liouville-type result for positive solutions on bounded star-shaped domain holds when p, q is above the critical hyperbola It is therefore expected that just like the case of the scalar Lane-Emden equation p = q and a = b = 0 the non-existence of solutions on the whole space N should occur exactly when p, q is in the complementary domain, that is when it is under the critical hyperbola This is the statement of the following Hénon-Lane-Emden conjecture Conjecture 1 Suppose p, q is under the critical hyperbola, ie, Then there is no positive solution for system 1 N + a p N + b > N 2 3 q + 1 Proving such a non-existence result seems to be challenging even for the Lane-Emden conjecture ie, when a = b = 0 for systems The case of radial solutions was solved by Mitidieri [17] in any dimension, and both Mitidieri [17] and Serrin-Zou [27] constructed positive radial solutions on and above the critical 1 hyperbola, ie + 1 q+1 N 2 N, which means that the non-existence theorem is optimal for radial solutions For non-radial solutions of the Lane-Emden system, there are the results of Souto [28], Mitidieri [17] and Serrin-Zou [26] who proved the non-existence of solutions in dimensions N = 1, 2, while in dimension N = 3, Serrin-Zou [26] gave a proof for the non-existence of polynomially bounded solutions, an assumption that was removed later by Poláčik, Quittner and Souplet [22] More recently, Souplet [25] settled completely the conjecture in dimension N = 4, while providing in dimensions N 5, a more restrictive new region for the exponents p, q that insures non-existence Theorem B Souplet [25] Assume a = b = 0 i Let N = 4 and p, q > 0 If p, q satisfies then system 1 has no positive solutions 1 p q + 1 > N 2 N, 4 ii Let N 5, and p, q > 0 with pq > 1 If p, q satisfies 4, along with { p max, q + 1 } > N 3, 5 then every non-negative solution of system 1 is necessarily trivial 2

3 The Lane-Emden conjecture in dimensions N 5 is still open The Hénon-Lane-Emden conjecture is even less understood Even for the scalar case a = b and p = q ie, the Hénon equation, Gidas and Spruck in [14] solved the conjecture only for radial solutions, also showing that in this case, the non-existence result is optimal For non-radial solutions, they proved some partial results such as the non-existence of positive solutions for a 2 and p N+2 N 2 the Sobolev critical exponent for a = 0 For systems, Mitidieri [17] gave a partial solution to the conjecture for radial solutions by showing that there is no positive radial solution for 1 for all N 3 provided p, q > 1 satisfy N + min{a, b} p N + min{a, b} q + 1 > N 2, 6 ecently, Bidaut-Veron-Giacomini [2] used a Pohozaev type argument and a suitable change of variables to give a complete solution in the radial case Theorem C Bidaut-Veron-Giacomini [2] For N 3, System 1 admits a positive radial solution u, v such that u, v C 2 0, C[0, if and only if p, q is above or on the critical hyperbola, ie, when 2 holds 11 Liouville theorems for bounded non-negative solutions With the lack of progress on the full conjecture, the attention turned to showing that bounded non-negative solutions are necessarily trivial ecently, Phan and Souplet [21] showed among other things that the Hénon- Lane-Emden conjecture for the scalar case holds for bounded positive solutions in dimension N = 3 Theorem D Phan-Souplet [21] Let N = 3, a = b > 0 and p = q > 1 Assume p, q satisfies 3, then there is no positive bounded solution for the Hénon equation, ie, u = x a u p in N 7 In this note, we shall first extend the above result of Phan-Souplet [21] to the full Hénon-Lane-Emden system by showing the following 1 Theorem 1 Suppose N = 3 and p, q satisfy 3 Then, there is no positive bounded solution for 1 We also give a few partial results for the Hénon equation whether of second order or fourth order in all dimensions N 3 or N 5 We note that Miditieri and Pohozaev [19] have shown that the above result holds in higher dimension provided the following stronger condition holds: where α := b+2p+a+2 max{α, β} N 2, and β := a+2q+b+2 For that they used a rescaled test-function method as in Lemma 1 below to prove the result for p, q 1 More recently, Armstrong and Sirakov [1] proved among other things similar results for p, q > 0, by developing new maximum principle type arguments We are thankful to P Souplet for informing us of these latest developments by Armstrong and Sirakov 12 Liouville theorems for stable non-negative solutions We shall also consider in the scalar case the question of existence of solutions with finite Morse index solutions as opposed to bounded solutions For scalar equations, we get the following counterpart to the Phan-Souplet result in higher dimensions N 3 1 Upon receiving our preprint, P Souplet informed us that QH Phan has also proved the same result in dimension N = 3, as well as other interesting results in higher dimensions Our proofs are quite similar since both are essentially refinements of those of P Souplet in his groundbreaking work on the Lane-Emden conjecture for systems 3

4 Theorem 2 Let a 0, p > 1 and N 3 Then, for any Sobolev sub-critical exponent, ie, 1 < p < N a, N 2 equation 7 has no positive solution with finite Morse index We also have the following result for the fourth order equation, 2 u = x a u p in N 8 Theorem 3 Let a 0, p > 1 and N 5 Then, for any Sobolev sub-critical exponent, ie, 1 < p < N a, N 4 equation 8 has no positive solution with finite Morse index For systems, we have the following result Theorem 4 Suppose that 0 a b N 2p q Then, system 1 has no positive stable solution whenever the dimension satisfy pb a + 2 pqq + 1 N < pqq + 1 pqq p + 1 p + 1 p + 1 The case when a = b = 0 ie, the Lane-Emden system was already established by by Cowan in [5] Note that this result contains the result of Fazly in [10], who had considered the case q = 1 < p, a = b and shown the result under the condition, N < 8 + 3a a p 1, 10 which is already larger than the domain under the critical hyperbola, ie N < 4 + a + 8+4a Also, this contains the result of Wei-Ye in [29] who had considered the case q = 1 < p, a = b = 0 There are also various results for the cases where 2 < a, b < 0 and pq 1 For that we refer to [2, 10, 19, 12, 14, 15, 21] 2 Proof in the case of non-negative solutions In this section, we shall prove here Theorem 1 The main tools will be Pohozaev-type identities as well as various integral estimates The proof is heavily inspired by ideas of Souplet [25] and Serrin-Zou [26] We use Pohozaev-type identities, various integral estimates, as well as some elliptic estimates on the sphere Throughout this section, all norms refer to functions defined on the unit sphere, ie u m := u Lm S N 1 We start with the following estimate on the non-linear terms Note that for a = b = 0, this was proved by Serrin and Zou [26] via ODE techniques, and by Miditieri and Pohozaev [19] who used the following rescaled test functions approach for a, b > 2 For the sake of convenience of readers, we recall the proof Interested readers can find more details for both scalar and system cases in [24] Lemma 1 For any positive entire solution u, v of 1 and > 1, there holds x a v p C N 2 b+2p+a+2, 11 B x b u q C N 2 a+2q+b+2, 12 where the positive constant C does not depend on 4

5 Proof: Fix the following function ζ Cc 2 N with 0 ζ 1; { 1, if x < ; ζ x = 0, if x > 2; where ζ C and ζ C For fixed m 2, we have 2 { 0, if x < or x > 2; ζ m x C 2 ζ m 2, if < x < 2; For m 2, test the first equation of 1 by ζ m and integrate to get x a v p ζ m = uζ m N N = u ζ m C 2 uζ m 2 N \ Applying Hölder s inequality we get N x a v p ζ m C 2 1 x b q q q \ C N b q q 1 q 2 By a similar calculation for k 2, we obtain N x b u q ζ k C N a p p 1 p 2 x b u q ζ m 2q \ x b u q ζ m 2q \ 1/q x a v p ζ k 2p \ where 1 p + 1 p = 1 Since pq > 1, for large enough k we have 2 + k q < k 2p So, we can choose m such that 2 + k q m k 2p which means that m k 2p and k m 2q By collecting the above inequalities we get for pq > 1, N x a v p ζ m pq C [N b q q 1 q 2]pq x b u q ζ k p 1 p, 1/q C N 2 [b+2p+a+2] \ x a v p ζ m, 13 and N x b u q ζ k pq C [N a p p 1 p 2]pq x a v p ζ m C N 2 [a+2q+b+2] \ x b u q ζ k 14 q By using Hölder s inequality, we can now get the following L 1 -estimates Corollary 1 With the same assumptions as Lemma 1, we have v s C N B u t C N a+2q+b+2 s, b+2p+a+2 t, for any 0 < t < q and 0 < s < p where the positive constant C does not depend on 5

6 We now recall the following fundamental elliptic estimates Lemma 2 Sobolev inequalities on the sphere S N 1 Let N 2, integer j 1 and 1 < k < m For z W j,k S N 1, we have where and C = Cj, k, N > 0 z Lm S N 1 C D j θ z L k S N 1 + z L1 S N 1, { 1 k 1 m = j m =, N 1, if k < N 1/j, if k > N 1/j, Lemma 3 Elliptic L p -estimate on Let 1 < k < and > 0 For z W 2,k, we have Dxz 2 k C z k + 2k z k, where C = Ck, N > 0 Lemma 4 An interpolation inequality on Let > 0 For z W 2,1, we have D x z C z + 1 z, where C = CN > 0 By applying Lemma 1, Corollary 1 and Lemma 4, we obtain the following estimates on the derivatives of u and v Lemma 5 We have D x v C N 1 B D x u C N 1 where the positive constant C does not depend on a+2q+b+2, b+2p+a+2, Lemma 6 L 1 -regularity estimate on Let N > 2 and 1 k < N 2 For any z L1 we have z L k C 2+N 1 k 1 z L1 + N 1 k 1 z L1, where C = Ck, N > 0 For a = b = 0, the following Pohozaev identity has been obtained by Mitidieri [18], Serrin and Zou [26] It has also been used by Souplet in [25] Lemma 7 Pohozaev identity Suppose λ, γ satisfy λ + γ = N 2 If u, v is a positive solution of 1, then it necessarily satisfy N + a N + b p + 1 λ x a v + q + 1 γ x b u q+1 = N+a v SN 1 S p u q+1 N+b q N ur v r 2 u θ v θ N 1 S N 1 + N 1 S N 1 λu r v + γv r u N 6

7 Now, we are in the position to prove Theorem 1 Proof of Theorem 1: Since p, q satisfy 3, then we can choose λ and γ such that N+a Now, for all > 0 define N + a N + b F := p + 1 λ x a v + q + 1 γ x b u q+1 From Lemma 7, we have where and > λ and N+b q+1 > γ F C G 1 + G 2, 15 G 1 := N+a v + N+b S N 1 S N 1 u q+1, G 2 := N S N 1 Dx u + 1 u D x v + 1 v Step 1 Upper bounds for G 1 and G 2 Set m = in Lemma 2 to get for either t = p + 1 or t = q + 1 u t u C D 2 θu 1+ɛ + u 1 C 2 D 2 xu 1+ɛ + u 1, where ɛ > 0 is small enough and will be chosen later So, G 1 N+a+2 D 2 xv 1+ɛ + 2 v 1 1+p + N+b+2q+1 D 2 xu 1+ɛ + 2 u 1 1+q 16 We now look for the same type bounds for G 2 Apply Schwarz s inequality to get G 2 N Dx u + 1 u 1/2 2 Dx v + 1 v 2 S N 1 S N 1 N D x u u 1 Dx v v 1 Then, using Lemma 2 we obtain the following upper bounds It follows that D x u 2 C D θ D x u 1+ɛ + D x u 1 C D 2 xu 1+ɛ + D x u 1, D x v 2 C D θ D x v 1+ɛ + D x v 1 C D 2 xv 1+ɛ + D x v 1 G 2 N+2 D 2 xu 1+ɛ + 1 D x u u 1 D 2 x v 1+ɛ + 1 D x v v 1 17 Step 2 The following L t -estimates hold in the annulus domain \ /2 ; /2 /2 /2 /2 /2 /2 1/2 vr 1 r N 1 a+2q+b+2 N dr C, 18 ur 1 r N 1 b+2p+a+2 N dr C, 19 D x v 1 r N 1 a+2q+b+2 N 1 dr C, 20 D x u 1 r N 1 b+2p+a+2 N 1 dr C, 21 Dxv 2 1+ɛ a+2q+b+2 1+ɛ rn 1 N 2 dr C +bɛ, 22 Dxu 2 1+ɛ b+2p+a+2 1+ɛ rn 1 N 2 dr C +aɛ 23 7

8 To prove 18-21, we just apply Corollary 1 and Lemma 5 Here is for example the proof for 23 Apply Lemma 3, Corollary 1 and Lemma 1 to get /2 D 2 xu 1+ɛ 1+ɛ rn 1 dr = /2 D 2 xu 1+ɛ dx C u 1+ɛ dx + C 21+ɛ u 1+ɛ dx B 2 C aɛ x a v p1+ɛ dx + C 21+ɛ C b+2p+a+2 N 2 b+2p+a+2 N 2 C +aɛ +aɛ + C u b+2p+a+2 N 21+ɛ The proof of 22 is similar Step 3 For large enough M, define following sets; Using 23, we get C Γ 1 := {r, 2; vr 1 > M a+2q+b+2 }, Γ 2 := {r, 2; ur 1 > M b+2p+a+2 }, a+2q+b+2 1 Γ 3 := {r, 2; D x v 1 > M }, b+2p+a+2 1 Γ 4 := {r, 2; D x u 1 > M }, Γ 5 := {r, 2; Dxv 2 1+ɛ a+2q+b+2 1+ɛ > M 2 +bɛ }, Γ 6 := {r, 2; Dxu 2 1+ɛ b+2p+a+2 1+ɛ > M 2 +aɛ } b+2p+a+2 2 N+2+ aɛ b+2p+a+2 N+2+ D 2 xu 1+ɛ 1+ɛ rn 1 dr aɛ Γ 6 N 1 b+2p+a+2 2 M +aɛ = M Γ 6 1 Therefore, choosing large enough M, we get Γ 6 /7 Similarly, using 18-22, one can see Γ i /7 for 1 i 5 Hence, for each 1, we can find i=6 ˆ, 2 \ Γ i φ 24 We now have the following upper bounds on 16 and 17 for the radius ˆ given by 24; i=1 G 1 ˆ C ˆ N+a+2 ˆ a+2q+b+2 2+bɛ 1 1+ɛ + a+2q+b+2 ˆ 2 +C ˆ N+b+2q+1 ˆ b+2p+a+2 2+aɛ 1 1+ɛ + b+2p+a+2 q+1 ˆ 2, C ˆ a 1ɛ + ˆ a 1 ɛ, where [ a 1 ɛ = p a 1ɛ = q + 1 a + 2q + b bɛ 1 [ b + 2p + a aɛ 1 + ɛ 2 N + a ], p + 1 ] 1 + ɛ 2 N + b q + 1 8

9 Also, G 2 ˆ C ˆ N+2 ˆ b+2p+a+2 2+aɛ 1 1+ɛ + ˆ a+2q+b+2 2+bɛ 1 1+ɛ + b+2p+a+2 ˆ 2 a+2q+b+2 ˆ 2, where Hence, from 15 we get C ˆ a2ɛ, a 2 ɛ = N a + bɛ ɛ F C G 1 ˆ + G 2 ˆ C ηɛ, b + 2p a + 2q + 1 where η ɛ := min{a 1 ɛ, a 1ɛ, a 2 ɛ} and the positive constant C does not depend on By a straightforward calculation, we have Also, a 2 0 = N b + 2p a + 2q + 1 > 0 iff N + a p N + b q + 1 > N 2 a 1 0 > 0, iff a 10 > 0, iff a + 2q + b + 2 b + 2p + a + 2 > N + a p + 1, 25 > N + b q Now, if p and q satisfy 3, then 53 and 54 hold, and we can therefore choose η ɛ > 0 for small enough ɛ > 0 We now conclude by sending and get the contradiction 3 On solutions of the second order Hénon equation with finite Morse index We shall prove here Theorem 2 For that we recall that a critical point u C 2 of the energy functional 1 Iu := 2 u 2 1 p + 1 x a u is said to be a stable solution of 7 if for any φ Cc 1, we have I uu φ := φ 2 p x a u φ 2 0 a stable solution outside a compact set Σ if I uu φ 0 for all φ C 1 c \ Σ, also u has a Morse index equal to m 1 if m is the maximal dimension of a subspace X m of C 1 c such that I uu φ < 0 for all φ X m \ {0} a solution with Morse index m if there exist φ 1,, φ m such that X m = Span{φ 1,, φ m } C 1 c and I uu φ < 0 for all φ X m \ {0} 9

10 Note that if u is of Morse index m, then for all φ C 1 c \Σ we have I uu φ 0, where Σ = m i=1 suppφ i, and therefore u is stable outside the compact set Σ We shall need the following lemma Lemma 8 Let N and let u C 2 be a positive stable solution of 7 Set fx = x a, a > 0, then, for any 1 t < 1 + 2p + 2 pp 1 we have u 2 u t 1 + fxu t+p φ 2m C fx t+1 φ 2 t+p, 27 for all φ C 1 c with 0 φ 1 and for large enough m The constant C does not depend on and u Proof: The following proof also holds true for weak solutions The ideas are adapted from [7, 8, 9] Note first that for any stable solution of 7 and η Cc 1, we have the following: p x a u η 2 η 2, 28 x a u p η = u η 29 Test 29 on η = u t φ 2 for φ Cc 1 for an appropriate t that will be chosen later, to get x a u t+p φ 2 = u u t φ 2 = t u 2 u t 1 φ u t u φφ Apply Young s inequality 2 to u u t 1 2 φ u t+1 2 φ to obtain t ɛ u 2 u t 1 φ 2 C ɛ u t+1 φ 2 + x a u t+p φ 2 30 Now, test 28 on u t+1 2 φ to get p x a u t+p φ 2 t + 12 u 2 u t 1 φ 2 + u t+1 φ t + 1 u t u φφ t ɛ u 2 u t 1 φ 2 + C ɛ,t + C 4 ɛ,t u t+1 φ 2, where again we have used Young s inequality in the last estimate Combine now this inequality with 30 to see t ɛ t+1 2 p x a u t+p φ ɛ C ɛ + C ɛ,t + C ɛ,t u t+1 φ 2 31 t ɛ t ɛ For an appropriate choice of t, given in the assumption, we see that the coefficient in LHS is positive for ɛ small enough Therefore, replacing φ with φ m for large enough m and applying Hölder s inequality with exponents t+p t+p t+1 and we obtain x a u t+p φ 2m D ɛ,t,m x t+1 a φ 2 t+p 32 2 For any a, b, ɛ > 0, ab ɛa 2 + Cɛb 2, for some Cɛ 10

11 Note that both exponents are greater than 1 for t given in i and ii On the other hand, combining 30 and 31 gives us u 2 u t 1 φ 2 D ɛ,t u t+1 φ 2 Similarly, replace φ by φ m and apply Hölder s inequality with exponents t+p t+1 u 2 u t 1 φ 2m D ɛ,t,m x t+1 a φ 2 t+p and t+p to get This inequality and 32 finish the proof of 27 Now, we are in the position to prove the theorem Proof of Theorem 2: We proceed in the following steps Step 1: We have the following standard Pohozaev type identity on any N N + a x a u N 2 u 2 = 1 x a u x ν + x uν u 1 p p To get 33, just multiply both sides of 7 by x u, do integration by parts and collect terms Step 2: The following estimates hold: u L 2 N, x a u L 1 N u 2 x ν 33 First recall that u is stable outside a compact set Σ To prove our claim, we use 27 with the following test function ξ Cc 1 N \ Σ for > and Σ 0 ; 0, if x < 0 + 1; ξ x := 1, if < x < ; 0, if x > 2; which satisfies 0 ξ 1, ξ L \ < C and ξ L 0 +2\0 +1 < C 0 Therefore, 0+2< x < u 2 u t 1 + x a u t+p 2t+p C 0 + Ĉ N t+1 a, for all 1 t < 1 + 2p + 2 pp 1 Now, set t = 1 and send Since N < 2p+a+1, we see N u 2 < and N x a u < Step 3: The following equality holds u 2 = N x a u N 34 Multiply 7 with uζ for ζ C 1 c N which satisfies 0 ζ 1, ζ < C and ζ x := { 1, if x < ; 0, if x > 2 11

12 Then, integrate over to get x a u ζ u 2 ζ = ζ u u 35 By Hölder s inequality, we have the following upper bound for HS of 35, ζ u u 1 u x a u x a 1 = N 2 u 2 a 1 2 B2 1 u 2 1 x a u x 2a 2 B2 1 2 B2 x a u 1 Therefore, from Step 2, there exists a positive constant C independent of such that ζ u u C N 2a+ 2 Since N < 2p+a+1, we have lim ζ u u = 0 Hence 35 implies 34 Step 4: we have N + a p + 1 N 2 x a u = 0 2 N Apply Lemma 8 for t = 1 with the following test function φ Cc 1 N \ Σ for > 2 0 ; 0, if x < /2; φ x := 1, if < x < 2; 0, if x > 3; which satisfies 0 φ 1, φ L B 3 \/2 < C to get \ u 2 + x a u C N Now, define the following sets for large enough M; 2p+a+1 36 From 36, we have θ 1 := {r, 2; D x ur 2 2 > M 2p+a+1 }, θ 2 := {r, 2; ur 2p+a+1 > M a } 2p+a+1 N+ C +a 2p+a+1 N+ 2 ur rn 1 dr +a θ 2 N 1 M 2p+a+1 a = M θ 2 1 Similarly, one can show θ 1 /M By choosing M large enough we conclude θ i /3 for i = 1, 2 Therefore, for each 1, we can find i=2, 2 \ Λ i φ i=1 12

13 Now, apply Pohozaev identity, 33, with = B to see that HS converges to zero if for subcritical p, ie N < 2p+a+1 Hence, N 2 u 2 = N + a x a u 2 p + 1 N N From this and 34, we finish the proof of Step 4 emark: For the Sobolev critical case p = N+2+2a N 2, using the change of variable w := ur 1+ a 2 and applying well-known classifying-type results mentioned in the introduction for the Lane-Emden equation, one can see all radial solutions of 7 are of the following form u ɛ r := kɛɛ + r 2+a 2 N 2+a, 37 where kɛ = ɛn + an 2 N 2 22+a Then, from the classical Hardy s inequality it is straightforward to see u ɛ is stable outside a compact set 0, for an appropriate 0 Note that for 2 < a 0, by Schwarz symmetrization or rearrangement, it is shown in [12] that all radial solutions of 7 with p = N+2+2a N 2 and N > 2 are of the form 37 4 On solutions of the fourth order Hénon equation with finite Morse index We shall prove here Theorem 3 For that we recall that a critical point u of the energy functional 1 Iu := 2 u 2 1 x a u, p + 1 is said to be a stable solution of 8, if for any φ Cc 4, we have I uu φ := φ 2 p x a u φ 2 0 Similarly to the second order case, one can define the notion of stability outside a compact set, which contains the notion of solutions with finite Morse index We first prove the following estimate Lemma 9 Let N and let u C 4 be a positive stable solution of 8 Then, for large enough m, we have for all φ Cc 4 with 0 φ 1, u 2 + x a u φ 2m C x 2 a T φ, 38 where T φ := φ 2 + φ 4 + φ 2 + φ φ The constant C does not depend on and u Proof: For any stable solution of 8 and η Cc 4, we have the followings: p x a u η 2 η 2, 39 x a u p η = u η 40 Test 40 on η = uφ 2 for φ Cc 4 to get x a u φ 2 = u uφ

14 Also, test 39 on uφ and use 41 to get p 1 x a u φ 2 = uφ 2 uφ 2 x a u φ 2 u uφ 2 By a straightforward calculation, one can see that the following identity holds: uφ 2 u uφ 2 = 4 u φ 2 + u 2 φ 2 2u u φ u 2 φ φ 42 Therefore, we have p 1 x a u φ u 2 φ 2 + u 2 φ φ A simple integration by parts yields u 2 φ 2 = u u φ which then simplifies the previous inequality to become p 1 x a u φ 2 6 u u φ 2 + Therefore, u 2 φ 2 2 u u φ 2 u 2 φ 2, 43 u 2 φ φ 2 2 φ φ x a u φ 2 C u u φ 2 + u 2 Lφ, 44 where Lφ := φ φ φ φ On the other hand, from 42 and 43, one can see uφ 2 = u uφ u φ 2 + u 2 φ 2 2u u φ 2 2 u 2 div φ φ = x a u φ u u φ 2 + u 2 φ φ 2 2 φ φ By combining 44, the identity uφ = φ u+2 u φ+u φ and Young s inequality, we get the following estimate u 2 φ 2 C u u φ 2 + C u 2 Lφ Therefore, x a u + u 2 φ 2 C u u φ 2 + C u 2 Lφ Now, replacing φ with φ m for large enough m > 0 and applying Young s inequality we end up with x a u + u 2 φ 2m C u u φ 2 φ 2m 1 + C u 2 Lφ m ɛ u 2 φ 2m + C ɛ u 2 φ 4 φ 2m 2 + C u 2 Lφ m 14

15 Then, for large enough m x a u + u 2 φ 2m C u 2 φ 2m 2 T φ, 45 where T φ := φ 2 + φ 4 + φ 2 + φ φ Now, apply Hölder s inequality to get u 2 φ 2m 2 T φ = x 2a u 2 φ 2m 2 x 2a T φ x a u φ 2m x 2a T φ Choosing m large enough, say 2m 2 2 2m, from 45 we finally get the desired inequality x a u + u 2 φ 2m C x 2a T φ Proof of Theorem 3: We proceed in the following steps Step 1: We have the following standard Pohozaev type identity on any N N + a x a u N 4 u 2 1 = x a u x ν 1 u 2 x ν p p u νx u + u x u ν 46 To get 46, just multiply both sides of 8 by x u, do integration by parts and collect terms Step 2: we have u L 2 N, x a u L 1 N Since u is stable outside a compact set Σ, using 38 with the following test function ξ Cc 1 N \Σ for > and Σ 0 ; 0, if x < 0 + 1; ξ x := 1, if < x < ; 0, if x > 2; which satisfies 0 ξ 1, D i ξ L \ < C and D i ξ i L 0 +2\0 +1 < C 0 for i = 1,, 4, we get u 2 + x a u 4 C 0 + Ĉ N 2 a 0+2< x < For subcritical exponents, N < 22p+a+2, we see N u 2 < and N x a u < Step 3: The following equality holds x a u = N u 2 N 47 Multiply 8 with uζ for ζ Cc 4 which satisfies 0 ζ 1, D i ζ < C for i = 1,, 4 and i { 1, if x < ; ζ x := 0, if x > 2 15

16 Then, integrate over to get x a u ζ u 2 ζ = u u ζ + 2 u u ζ =: I 1 + I 2 48 By Hölder s inequality, we have the following upper bound for I 1, I 1 2 u x a u x a 2 = N 2 u 2 a 1 2 B2 2 u 2 1 x a u x 2a 2 B2 1 2 B2 x a u 1 Therefore, from Step 2, there exists a positive constant C independent of such that I 1 C N 2a+ 2 Since N < 2p+a+1, we have lim I 1 = 0 Now, we consider the second term in HS of 48 Apply Young s inequality for a given ɛ > 0 we choose it later to get I 2 ɛ u 2 + C ɛ N u 2 ζ 2, Using Green s theorem we get u 2 ζ 2 = u u ζ u 2 ζ 2 =: I 3 + I 4 2 By the same discussion as given for I 1 one can see lim I 3 = 0 For the term I 4, we apply Hölder s inequality again I 4 4 x a u 2 x a 4 x a u 2 B2 = N 2a B 4 x a u 2 x 2a 2 By Step 2 and sending to infinity we get, lim I 4 = 0 Since lim I 2 ɛ N u 2 for any ɛ > 0, we have lim I 2 = 0 Therefore, 47 follows Step 4: The following equality holds N + a p + 1 N 4 x a u = 0 2 N Apply Lemma 9 with the following test function φ Cc 1 N \ Σ for > 2 0 ; 0, if x < /2; φ x := 1, if < x < 2; 0, if x > 3; 16

17 where 0 φ 1, D i φ L B 3 \/2 < C i Then, we get \ u 2 + x a u C N On the other hand, we are interested in similar upper bounds for the following terms J 1 := u u \ and J 2 := u Dxu 2 \ For the first term, J 1, using Schwarz s inequality we have 1/2 1/2 u u < u 2 u 2 \ \ \ From standard elliptic interpolation estimates, L 2 -norm version of Lemma 4, we have u 2 \ C 2 u 2 + C 2 B 4 \/2 u 2 B 4 \/2 C 22p+2+a N = C N 22p+2+a N 2a 2 x a u N 22p+2+a N 2 22p+2+a + x a u N Since x a u < and N < 22p+2+a N, for > 1 we have u 2 C N 22p+2+a +2+2 \ Therefore, \ u u < C p N 22p+2+a Similarly for the second term, J 2, using Lemma 3, ie, Dxu 2 2 C \ u B 4 \/2 and similar type discussions one can see Now, define the following sets for large enough M; B 4 \/2 u 2 \ u D 2 xu < C p N 22p+2+a 51 Λ 1 := {r, 2; x ur 2 2 > M 22p+2+a }, Λ 2 := {r, 2; ur 22p+2+a > M a }, Λ 3 := {r, 2; x ur x ur 1 > M p N p + 22p+2+a +1 }, Λ 4 := {r, 2; x urd 2 xur 1 > M p N p + 22p+2+a }, 17

18 In the following, we shall find a bound for the measure of the above sets From 49, we have Also, from 50 22p+2+a N+ C +a 22p+2+a N+ 2 ur rn 1 dr +a Λ 2 N 1 M 22p+2+a a = M Λ 2 1 C p N+ 22p+2+a 1 2 x ur x ur 1 r N 1 dr p N+ 22p+2+a 1 Λ 3 N 1 M p N p + 22p+2+a +1 = M Λ 3 1 Similarly, from 51 and 49 we get Λ 1, Λ 4 /M By choosing M large enough we conclude Λ i /5 for i = 1,, 4 Therefore, for each 1, we can find i=4, 2 \ Λ i φ 52 Then, from the definition of and Λ i for i = 1,, 4, we have x u D xu 2 C N p 22p+2+a 1 x = x = i=1 53 x u x u C p N 22p+2+a 54 x = x = x u 2 C p N 22p+2+a 1 u C p Using 46 with = B 2 \ B, one can see u νx u < C p x = N 22p+2+a a 1 N 22p+2+a Now, applying the Pohozaev identity, 46, with = B and using 53-57, HS of 46, converges to zero if for subcritical p, ie N < 22p+2+a Hence, N 4 u 2 = N + a x a u 2 p + 1 N N From this and 47, we finish the proof of Step 4 5 On stable solutions of the Hénon-Lane-Emden system We shall prove here Theorem 4 For that we recall that a classical solution u, v of 1 is said to be pointwise stable if there exists positive smooth ζ, η such that { ζ = p x a v η in, η = q x b u q 1 58 ζ in In what follows we give the stability inequality for system 1 This inequality is the novelty here and is key tool in proving Theorem 4 The idea of geting such an inequality comes from [11] 18

19 Lemma 10 Assume that u, v is a pointwise stable solution of 1, then for any test function φ Cc 1, we have pq x a+b q 1 2 v 2 u 2 φ 2 φ 2 59 Proof: Let u, v be a pointwise stable solution of 1 in such a way that there exists positive smooth ζ, η such that 58 Multiply the first equation by φ 2 ζ 1 and the second equation by φ 2 η 1, integrate by parts and use Young s inequality to get p x a v η ζ ζ φ2 = ζ φ2 φ 2 q x b u q 1 ζ η φ2 = η η φ2 φ 2 Adding these two equations and doing simple calculations we get 2 φ 2 p x a v η ζ + q x b u q 1 ζ φ 2 2 pq x a+b q 1 2 u 2 v 2 φ 2 η The following pointwise estimate is taken from [20] As was said before, the first version of this paper was done independently of [20] and without using the following lemma of Phan This last section which uses Lemma 11 was added after his paper was posted Lemma 11 [Phan, [20]] Assume that u, v is a classical solution for 1, then for we have 0 a b N 2p q 60 x a v p + 1 q + 1 x b u q+1 61 Combining the above lemmas we conclude the following integral estimate which is a counterpart of Lemma 8 for the second order case and Lemma 9 for the fourth order case Lemma 12 For N, assume that 60 holds and that u, v is a pointwise stable solution of 1 Set θ := pqq+1 Then, for any t such that θ θ θ < t < θ + θ θ, we have for all φ Cc 2 such that 0 φ 1, x a v p u 2t 1 φ 2 C The constant C does not depend on and u, v Proof: Note first that for p q, we have θ q 2 > 1 and also θ θ < 1 < θ < θ u 2t φ 2 + φ 62 θ θ Let u, v is a pointwise stable solution of 1 Then, Lemma 10 applies and by replacing φ with u t φ in 59, where φ is a test function, we obtain pq x a+b q 1 2 v 2 u 2 u 2t φ 2 u t φ

20 ewriting the left hand side as pq x a+b 2 v 2 u q+1 u q+1 2, we get pqq + 1 p + 1 x a v p u 2t 1 φ 2 t 2 2 u 2t 1 φ 2 and using Lemma 11, ie q+1 u 2 u 2t 2 φ 2 + x a b 2 v 2 u 2t φ φ 64 To find an upper bound for the first term in the above inequality with the gradient term, we multiply both sides of the first equation in 1 to get x a v p u 2t 1 φ 2 = u u 2t 1 φ 2 = 2t 1 u 2 u 2t 2 φ 2 1 u 2t φ 2 2t Since t > 1 2, we get t 2 u 2 u 2t 2 φ 2 t2 x a v p u 2t 1 φ 2 + C t 2t 1 u 2t φ φ + φ 2 Combining this and 65 we have pqq + 1 p + 1 t2 x a v p u 2t 1 φ 2 C t 2t 1 u 2t φ φ + φ 2 65 Proof of Theorem 4: Define z := u τ for 1 < 2 θ 2 θ θ < τ < 2 θ + 2 θ θ and θ := pqq+1 Then, z C u 2 u τ 2 + x a u τ 1 v p By integrating over balls we get z C u 2 u τ 2 + C x a u τ 1 v p 66 We are now after an upper bound for the right hand side of the above inequality To control the second term, apply Lemma 8 for t := τ 2 and standard test function ζ used in the proof of Theorem 3 to get x a v p u τ 1 C 2 u τ To bound the first term, we use the first equation of the system Multiply both sides of 1 with u τ 1 ζ 2 and integrate by parts to get u 2 u τ 2 = u 2 u τ 2 ζ 2 1 x a v p u τ 1 ζ u τ ζ τ 1 ττ 1 2 B C x a v p u τ 1 + C 2 u τ Therefore, the following upper bound holds for 66, z C 2 20 u τ,

21 which means 2 z L1 C z L1 Now, applying Lemma 6 for z = u τ we get z L k C N 1 k 1 z L1, N N 2 where C = Ck, N > 0 and any 1 k < Now take 1 k i < N N 2 for 1 i n and 2 θ 2 θ θ < 2t := τk n 1! < 2 θ + 2 θ θ The notation! stands for k n 1! := n 1 i=0 k i and set k 0 = 1 By induction we have where k n = N n 1 k i i=1 k i! = N 1 k 1 n! z L kn! C k n z L1, u τkn! Let 0 < τ < q and from Corollary 1 we get and C = Ck i, N > 0 So, 1 kn! C N 1 kn! 1 u τ u τ C N pb+2+a+2 τ Therefore So, in the following dimensions u τkn! 1 kn! C τ N τkn! pb+2+a+2 67 pb a + 2 N < τk n! the right hand side of 67 converges to zero as tends to infinity Note that since τk n 1! < 2 θ+2 θ θ and k n < N N 2, we have τk n! < 2 θ + 2 θ θ N N 2 So, pb a + 2 pb a + 2 N < τk n! < θ + 2 θ θ ecall that θ := pqq+1, which completes the proof eferences [1] SN Amstrong, B Sirakov, Nonexistence of positive supersolutions of elliptic equations via the maximum principle, Comm Partial Differential Equations, 2011 to appear [2] M F Bidaut-Veron, H Giacomini; A new dynamical approach of Emden-Fowler equations and systems, Adv Differential Equations , no 11-12, [3] L A Caffarelli, B Gidas, J Spruck; Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth, Comm Pure Appl Math , no 3, [4] W X Chen, C Li; Classification of solutions of some nonlinear elliptic equations, Duke Math J , no 3,

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