Existence of Positive Solution of a Class of Semi-linear Sub-elliptic Equation in the Entire Space IH n

Transcription

1 arxiv:math/0183v1 [math.ap] 0 Dec 00 Existence of Positive Solution of a Class of Semi-linear Sub-elliptic Equation in the Entire Space I Zhujun Zheng Institute of Mathematics, Henan University, Kaifeng, Xiufang Feng Department of Mathematics, Henan University, Kaifeng, Abstract In this paper, we study the following problem u u+u p = 0 u > 0 u(x) 0 in in ρ(x) where 1 < p < Q+, Q is the homogeneous dimension of Heisenberg group. Our main result is that this problem has at least one positive solution. Key Words and Phrases: Semilinear subelliptic equation, Heisenberg group. AMS(1991) Subject Classification: 35J60 1 Introduction Let be the Heisenberg group, where = n (Xi + Y ) is its subelliptic Laplacian operator, ρ(x) is the distance function from x to the point 0. Under the real coordinate (x 1,,x n,y 1,,y n,t), the vector field X i and Y i are defined by X i = x i +y i t Y i = y i x i t and the distance function ρ(x) is defined by i=1 i = 1,,,n. n ρ(x) = ( (x i +y i ) +t ) 1 4. i=1 i zhengzj@mail.henu.edu.cn 1

2 It is well known that {X i,y i } generate the real Lie algebra of Lie group and [X i,y i ] = 4δ ij, i,j = 1,,n. t In this Lie group, there is a group of natural dilations defined by δ λ (x,y,t) = (λx,λy,λ t), λ > 0 where x = (x 1,,x n ),y = (y 1,,y n ). With this group of dilations, the Lie group isatwostepstratifiednilpotent Liegroupofhomogeneousdimension Q = n+, and is homogeneous partial differential operator of degree. In this paper, we deal with the existence of the positive solution to the following semi-linear subelliptic equation u u+u p = 0 in u > 0 in (1) u(x) 0 ρ(x) where 1 < p Q+ Equation (1) comes from the CR-Yamabe problem(see [14]) and has been studied by several authors(see [4], [10],[1] and the references therein). In the paper [1], they studied the problem { u+u p = 0 in u > 0 in and showed that if the problem s solution is cylindrical, then it must be 0. In the works [] and [4], for 1 < p < Q+ they have gotten some results on the existence of the boundary value problem of equation (1) on the bounded domain and unbounded domain with thin condition. In these condition, the corresponding functional satisfies P.S condition, and the normal variational methods works. In the entire space,1 < p Q+, the Folland-Stein-Soblev embedding lost compactness. The corresponding functional lost P.S condition. Their methods don t work. To our knowledge, in these situation, there exists no report of progress on this problem up to now. In the Euclidean space, the similar problem was studied by many peoples(see [], [3],[11], and the references therein). In [], W-Y Ding and W-M Ni gave some beautiful results on the similar semilinear problem in Euclidean. But for our problem, as a consequence of [1], it may not have radical symmetry solution. So our problem is more subtle then them. This is one of the reasons to study the problem (1).

3 Our main result is the following theorem. Theorem 1 For 1 < p < Q+, the problem (1) has a solution u E. To proof our theorem, we first give some preliminary definition and Lemmas. For u C 0 (Hn ) the C smooth funciotn with compact support, we define norm by u = IR n H u +u () where H = ( X1,, X, Y1,, Yn ). Then we define the Folland-Stein- Sobolev space by E = C 0 ( ), the completion of C 0 (Hn ) under the norm (). This is a Hilbert space. For Ω, the completion of C 0 (Ω) in E is denoted by E(Ω), it is a Hilbert space too. These spaces have embedding theorem like the Sobolev embedding. Lemma 1.1 u E,1 < q Q, we have where C is a constant independent of u. u L q C u (3) Lemma 1. Let Ω be bounded smooth domain in, the embedding E(Ω) L p (Ω), 1 p < Q (4) is compact. We use two methods to solve the problem (1). The first method: In the Folland-Stein-Sobolev space E, we define the energy functional J(u) = 1 H u +u 1 p+1 u p+1, u E (5) Let B k be the ball B k = {x ρ(x) < k}. Denoted the completion C 0 (B k ) in E by E k, then E k E k+1 E E = E k k=1 Set J k = J Ek, choose an element u 0 E 1 E 1 E k E, such that J(u 0 ) < 0, J k (u 0 ) < 0 (6) 3

4 Let Γ,Γ k defined by Γ = {r : [0,1] E r(0) = 0, r(1) = u 0, r is continuous} r(0) Γ k = {r : [0,1] E k = 0, r(1) = u0, r is continuous} (7) Define For Γ k Γ k+1 Γ, we have c = min max I(r(t)), r Γ 0 t 1 c k = min max I k(r(t)). r Γ k 0 t 1 (8) c k c k+1 c > 0 (9) By mountain-path Lemma, we know c k is a critical value of the functional I k. Let u k be a critical point of I k corresponding the critical value, that is I k (u k ) = c k and I k (u k) = 0. By some complex estimates of u k, we shall proof c is a critical value of I, and u k u in E,u is a critical point and I(u) = c. By the maximum principle we get a positive solution of (1). the second method: Define M = {u E u p+1 = 1} E. On the manifold, we define I(u) = 1 H u +u (10) The main idea is, for I have bounded from below, we define c = inf I(u) (11) u M Then we prove that the critical c can be arrived by u M. Then by Lagrane multiplier method, we know the problem have a positive solution. For the second method of subcritical case to overcome the difficult that the functional I lost P.S condition, we give some Lions version concentration-compactness Lemmas. This is one of bones in this work. For the case 1 < p < Q+, by our proof we know, for every smooth bounded domain Ω, the Dirichlet problem ε u u+u p = 0 u > 0 u = 0 in Ω in Ω on Ω have a least energy solution u ε. Let x ε Ω,u(x ε ) = maxu(x), like J.Wei in the x Ω paper [13], we want to know what is the limdist(x ε, Ω). In one of our preparing ε 0 4 (1)

5 works ([1]), we shall proof that dist(x ε, Ω) max x Ω d(x, Ω), ε 0 and we shall publish this result elsewhere. In the following, as the Euclidean case, we shall study the effect of topology of the unbounded Ω. The Proofs of Main Theorem.1 The first method In this subsection, we shall use the mountain-path lemma and domain extension method to proof the Theorem in the subcritical exponent case 1 < p < Q+. And more, we get that the problem have a least energy solution, and proof that c = inf r Γ max 0 t 1 J(rt) can be arrived by a path r 0 Γ. This is the foundation of our paper [1]. For the Folland-Stein-Sobolev embedding E k L p+1,1 < p < Q+ q ([5]), by the standard method we have the following lemma. is compact Lemma.1 For k IN, the functional J k defined in the Hilbert E k satisfies P.S condition. For an element e E 1 E k E, e = 1, k IN, we have J k (te) = t tp+1 p+1 e p+1 dx (13) For p+1 >, we have the following Lemma.. Lemma. For any k IN, there exists an element u ( E k ) E, such k=1 that I k (u 0 ) < 0, J(u 0 ) < 0 (14) For u = 1, we have J(tu) = t tp+1 p+1 u p+1 (15) By the Lemma.1, there is a positive constant C > 0 independent of u, such that u p+1 C (16) 5

6 Combine the inequality (16) and the formula (15) we have Since p+1 >, we have the following Lemma.3. J(tu) t tp+1 p+1 C (17) Lemma.3 There is a neighborhood U k of 0 respectively in E k, and a neighborhood U of 0 in E, such that J k (u) α, J(u) α (18) for all u U k or u U respectively, where α > 0 is a positive constant..4. have From mountain path lemma and the above Lemma, we have the following Lemma Lemma.4 The value c k is a critical value of functional I k, and more we have c k c k+ c > α > 0 (19) Suppose u k is a critical point of J k corresponding the critical value c k. Then we From (0) and (1), we have J (u k )u k = u k u p+1 (0) J(u k ) = u k 1 u k p+1 = c k > α (1) p+1 c 1 p p+1 u k = c k > α () That is to say {u k } is a bounded point set in E. So there is a subset of {u k }, we still denote it by {u k }, and a point u E, such that u k u, (3) and u 0 is a weak solution of u u+u p = 0 in. By the method of Ding and Ni(see []), If we can prove u 0, then u is a critical of functional J, and J(u) = c. 6

7 Then by maximum principal, we know u is a positive solution of problem (1), and it is a positive least energy solution of it. So if we can prove u 0, our theorem is proved. Next we locus on this problem. For u k E is a solution of u u+u p = 0 in, we have u k +u k u p+1 k = 0. Then we have u k (up 1 k 1) = u k 0. Since u k 0, there must be exists ξ k, such that u k (ξ k ) = max u k 1 (4) We claim that {ξ k } is a bounded subset of. This is our next lemma. Lemma.5 The subset {ξ k } defined by (4) is a bounded subset of. Proof. For {u k } is bounded subset of E, by some standard estimates and the Folland-Stein-Sobelev embedding theorem, there is a positive constant α, such that sup u k α. So there is a large enough β > 0 such that H u k +βu k = u p k (β 1)u k 0 (5) Define function v = ce δρ(x), on where c and δ are positive number which shall be determined. For H is a order operator. So H v is a -1 order function. Then there are large positive numbers R 0, and δ > 0, such that for all ξ,ρ(ξ) > R 0, and large positive number β such that H v(ξ)+β v(ξ) 0. (6) Choose large positive number R 0, for all ξ,ρ(ξ) = R 0, we have Set β = max{β,β },, then by (5, 6, 7)we have (v u)(ξ) 0 (7) 7

8 { (v u)+β (v u) 0 in \B R0 (0), v u 0, on ( \B R0 (0) By the maximum principle, this implies that for all ξ > R 0,, for any k, The inequality implies that ξ k is bounded. u k ce δρ(ξ) (8) For the Folland-Stein-Sobolve spaces have similar embedding theorems with the Sobolev embedding and the Sub-Laplacian operator have similar characters with the Laplacian operator(see [5]), so by the method of Noussair, Ezzat S. and Swanson, Charles A(see [13]), we have the following lemma. Lemma.6. There is a subsequence of {u k } we still denote it by {u k }, such that for any bounded domain Ω, u k u in C +α (Ω), where α is a positive number. That is u k u in C +α loc (Hn ). From Lemma.5 and Lemma.6, we have the following Lemma. Lemma.7. The functional defined by (3) u 0. Proof. For {ξ k } is bounded in, so we may assume that there is a ξ 0 H k, such that ξ k ξ 0. So we have u k (ξ k ) u(ξ 0 ). By the inequality (4), we have u(ξ 0 ) 1. That is to say u 0. #. the second method In this subsection, we shall use the constraint functional method to study the problem (1). First we defined the manifold M = {u E u p+1 dx = 1} (9) On this manifold, define a functional I(u) = 1 H u +u, u M (30) It is obviously that the functional I is bounded from below. We shall study whether the functional defined by (30) arrive its minimum on the manifold M. That is we want to find a u 0 M, such that I(u 0 ) = mini(u) = α (31) u M 8

9 For the embedding E L p+1 ( ) lost compactness, so the functional I does not satisfy P.S condition. To overcome this difficult, we first transplant the Lions concentration-compactness Lemma([6,7,11]) to Heisenberg group case. Lemma..1 Let (ρ m ) m 1 be a sequence in L 1 ( ) satisfying: ρ m 0 in, ρ m = 1 (3) Then there exists a sequence (ρ mk ) k 1 satisfying one the following three possibilities: (i) (Compactness) There exists a sequence z k such that ρ nk (z) is tight, i.e ε > 0, R <, ρ nk (z)dz 1 ε; (33) z k +B R (ii) (Vanishing) lim sup ρnk (z)dz = 0,, for all R < ; k y B R (iii) (Dichotomy) There exists α (0,1) such that for all ε > 0, there k 0 1 and ρ 1 k,ρ k L1 + (Hn ) satisfying for k k 0, ρ nk (ρ 1 k +ρ k ) L 1 ε ρ 1 kdz α ε (34) and dist(suppρ 1 k,suppρ k ) +,k +, where dz = dxdydt. For the measure dxdydt on, it has translation invariant and it is a homogeneous on dilations δ λ like the measure on IR n+1. That is for u L 1 ( ),z 0, u(z)dz = u(z z0 1 )dz H n u(z λ z)dz = λ Q u(z)dz Where λ > 0. So, just like P.L.Lions[6,7], we can prove this lemma. We omit its proof here. Let {u m } M,I(u) mini(u) = α,m. By the Folland-Stein-Sobolev u M and there exists a constant c > 0, such that (35) u p c u, u E (36) So α = mini(u) > 0. u M Lemma.. For the sequence {u m }, there is a positive number {R m }, for the function ν m (z) = R 1 q m u m (δ 1 (z) (37) Rm 9

10 such that sup z B 1 (z) ν m q (w)dw = 1 = B 1 (0) Proof. For u m {u m },r > 0,z r m Hn, we define ν m q dw (38) From (35), we have u r m = rn/q u m (δ1(z zm r )) (39) r u r m q = r n u m (δ1(zzm r )) q = u m q = 1 (40) r So there exists a R m, for every z m, B 1 (z m) u p+1 m q dz = B Rm (0) u m q dz = 1 (41) Define ν m (z) = Rm /nu m(δ 1 z). From the formula (41), we have Rm sup z B 1 (z) ν m q dx = B 1 (0) ν m q dx = 1 Let ρ m = ν m q, then ρ m L 1 ( ), and ρ m = 1. From Lemma.., we know case (ii) in Lemma..1 can t occurs. We declare that the case (iii) can t also. That is our following lemma. Lemma..3 For the function ρ m L 1 ( ) defined above, there is z m, such that ρ m (z zm 1 ) is tight, i.e. there exists a number R > 0 large enough, such that z m B R (0) #. ρ m (z)dz 1 ε (4) Proof. By th Lemma..1 and Lemma.., we only need prove the case(iii) in Lemma..1 does t occur. On contrary, there is a number β (0,λ) such that for all ε > 0, there exist m 0 1 and ρ 1 m,ρ m L1 ( ) satisfies for m > m 0, ρ m (ρ 1 m +ρ m ) m ε ρ 1 mdz β ε ρ m dz (1 β) ε (43) and dist(suppρ 1 m,suppρ m ) +. 10

11 Choose r m > 0, such that suppρ 1 m B r m (0), suppρ m Hn \B rm (0), and r m + as m +. Set ϕ C 0 (B (0)) such that ϕ 1 in B 1 (0),1 ϕ 1 and let ϕ m ( x r m ). Decompose Then H ν m + ν m = ν m = ϕ m ν m +(1 ϕ m )ν m H (ϕ m ν m ) + (ϕ m ν m ) + H (1 ϕ m )ν m ) + (1 ϕ m )ν m ) + (ϕ m ν m ) (1 ϕ m )ν m ) + ϕ m ν m (1 ϕ m )ν m (44) Next we estimate the last two terms in formula (44) respectively. H (ϕ m ν m ) H ((1 ϕ m )ν) H (ϕ m ν m ) H (1 ϕ m )ν m ) = 1 H (ϕ m ν m ) H ((1 ϕ m )ν m )) B rm (0)\B rm (0) [ B rm (0)\B rm (0) = 1 [ B rm (0)\B rm (0) H (ϕ m ν m ) H ((1 ϕ m )ν m ) H (ϕ m ν m ) + B rm (0)\B rm (0) H ((1 ϕ m )ν m ) ] { H (ϕ m ν m + H ϕ m H ν m ϕ m ν m +ϕ m m ν m ) + h ϕ m ν m Hϕ m H ν m ϕ m ν m +(1 ϕ m ) m ν m } c c B rm (0)\B rm (0) B rm (0)\B rm (0) ν m + H ν m ν m p+1 Then we have H (ϕ m ν m ) ((1 ϕ m )ν m )+ ϕ m ν m (1 ϕ m )ν m c B rm (0)\B rm (0) ν m p+1 (45) From the proof of Lemma..1, we have ε > 0, m 0, such that for m > m 0, ν m p+1 β B rm (0) ν m p+1 1 β +ε \B rm (0) (46) 11

12 So by the inequalities of (43) and (46), we have B rm (0)\B rm (0) ν m p+1 c[ ν m p+1 (ρ 1 m +ρ+m )]+ε (47) The inequality (47) means that B rm (0)\B rm (0) ν m p+1 = o(1) (48) where o(1) 0,m +. Combine the formula (48),(44) and the inequality (45) we have H ν m + ν m = ϕ m ν m + (1 ϕ m )ν m +o(1) (49) By the Folland-Stein-Sobolev embedding and formula (49), we have ν m = ϕ m ν m + (1 ϕ m )ν m +o(1) S( ϕ m ν m p+1 L p+1 + (1 ϕ m )ν m p+1 )+o(1) S( ρ 1 m ) 1 p+1 +( ρ m ) p+ )+o(1) (50) S(β p+1 +(1 β) p+1 )+o(1) By the define of α and the independence of domain of the best Folland-Stein- Sobolev constant we know S = α. And by the define ν m we have ν m α(m ). So by the inequality (50)we have α α(β p+1 +(1 β) p+1 ) (51) For p+1 < 1,0 < β < 1, we get α > α, that is a contradiction. So the case (iii) of Lemma..1 can t occur. Theorem..1 There is a subsequence of {ν m }, we still denote it by {ν m }, there exists a point u 0 M, such that ν m ν 0 in E, and I(ν 0 ) = α Proof: For{ν m }isbounded ine, we have asubsequence ofit, andwe still denote it by {ν m }, and there exists a ε,ν 0 E such that ν m ν 0. For ε < 1, and Lemma.1., we get (z m B R (0)) B 1 (0), so the points sequence {z m } is a bounded 1

13 set. That is implies that tere is a subsequence of {z m }, we still denote it by {z m } and a point z 0, such that z m z 0,(m ). Then we have ν m p+1 > 1 ε (5) z 0 B 1+R (0) From the Folland-Stein-Sobolev emedding, we know there is a subsequence {ν m } we still denote it by {ν m }, such that ν m ν 0,m in H 1, (B 1+R (0)) and ν 0 p+1 > 1 ε (53) From the Fatou Lemma we know z 0 B 1+R (0) m ν 0 p+1 lim Combine the inequalities of (53) and (54), we have This implies that ν m ν 0,m in E. So we have ν m p+1 = 1 (54) ν 0 p+1 = 1 (55) I(ν 0 ) = lim m I(ν m) = α By the Lagrange multiplier, there is a positive number, such that Set u 0 = λ p 1 u, then we have ν 0 ν 0 +λν p 0 = 0 u 0 u 0 +u p 0 = 0 By the maximum principle, we get our our main theorem 1.1. Acknowledgments I would like thanks Prof. Zhouping Xin for his inviting me to visit IMS and useful discussions. I would like to thanks Prof. Juncheng Wei for he let me notice this problem and helpful discussions. I would like thanks Prof. Changfeng Gui, Prof. Yongsheng Li, Prof. Quanshen Jiu and Prof. Jiabao Su for their useful discussions. This work is finished during my visiting IMS of The Chinese University of Hong Kong. And is partially supported by the Zheng Ge Ru Foundation, Mathematical Tianyuan Fund with grand No and the Natural Science Foundation of Educational Committee of Henan Province with grand Number

14 References [1] Zhujun Zheng, J. Wei, Spike-Lagered solutions of singularly pefurbed subelliptic problem in Heisenberg in a degenerate setting. In preparing. [] Wei-Yue & Wei-Ming Ni, On the existence of positive entire solutions of a semilinear elliptic equation, Archive for Rational Mechanics and Analysis, V.91,1986, [3] Berestycki,H,& P-L. Lions, Nonlinear scala field equations I,II, Arch Rational Meth. Anal. 8(1983), , [4] I. Birindelli, I.Capuzzo Dolcetta, Indefinite semi-linear equations on the Heisenberg group: A priori bounds and existence, Commun. in Partial Differential Equations, 3(7&8)(1998), [5] G. Folland and E.Stein, Estimates for the b complex and analysis on the Heisenberg group, Commun. Pure Appl. Math. 7(1974), [6] Lions. P-L. The concentration - compactness principle in the calculus of variations. The locally compact case. Part I. Ann. Inst. H. Poincare 1(1984) [7] Lions. P-L. The concentration - compactness principle in the calculus of variations. The locally compact case. Part II. Ann. Inst. H. Poincare 1(1984) [8] Lions. P-L. The concentration - compactness principle in the calculus of variations. The locally compact case. Part I. Rev. Mat. Iberoamericano 1.1 (1985), [9] Lions. P-L. The concentration - compactness principle in the calculus of variations. The locally compact case. Part II. Rev. Mat. Iberoamericano 1. (1985), [10] Zhang Jihui, Ph.D Thesis, Lanzhou University, [11] Micheal Struwe. Variational Methods, Springer-Verlag Berlin Heidelberg, 000. [1] I. Birindelli, J. Prajapat, Nonlinear liouville theorems in the Heisenberg group via the moving plane. Commun. in PDE, 4(9&10), (1999). 14

15 [13] Ni, Wei-Ming and Wei, Juncheng, On the location and profile of spike-layer solutions to singularly perturbed semilinear Dirichlet problems. Comm. Pure Appl. Math. 48 (1995), no. 7, [14] D.S. Jerison, J. M. Lee, The Yamabe problem on CR manifolds, J. Differential Geometry 5(1987), [15] Jan Chabrowski, Weak Convergence Methods for Semilinear Elliptic Equations, World Scienticfic,