J. Math. Study doi: 10.4208/jms.v50n1.17.02 Vol. 50, No. 1, pp. 17-27 March 2017 Characterization by Symmetry of Solutions of a Nonlinear Subelliptic Equation on the Heisenberg Group Yazhou Han 1, Xiaodong Wang 2 and Meijun Zhu 3, 1 Department of Mathematics, College of Science, China Jiliang University, Hangzhou, 310018, P.R. China. 2 Department of Mathematics, Michigan State University, East Lansing, MI 48824, USA. 3 Department of Mathematics, The University of Oklahoma, Norman, OK 73019, USA. Received March 25, 2016; Accepted February 20, 2017 Abstract. In this paper we prove that two calculus lemmas, which are used in the method of moving sphere for classifying certain constant curvature equation, also hold on Heisenberg group H n. AMS subject classifications: 35H20, 22E30 Key words: Classification, moving sphere method, Heisenberg group. 1 Introduction The Heisenberg group H n consists of the set { } C n R (z,t) : z(z 1,,z n ) C n, t R with the multiplication law (z,t)(z,t )(z+z,t+t +2Im(z z )), where z z n j1 z jz j. As usual, we write z j x j + 1y j. The Lie algebra is spanned by the left invariant vector fields T t, X j +2y j x j t, Y j 2x j, j1,,n. y j t Corresponding author. Email addresses: yazhou.han@gmail.com (Y. Han), xwang@math.msu.edu (X. Wang), mzhu@ou.edu (M. Zhu) http://www.global-sci.org/jms 17 c 2017 Global-Science Press
18 Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 The sub-laplacian is defined by H n j1 (X 2 j +Y2 j ). In this paper, we shall study the following semilinear subelliptic equation H uu p, u 0, in H n, (1.1) where and throughout the paper we always assume that p > 1. This equation is closely related to the study of the CR Yamabe problem. The CR Yamabe problem on a strictly pseudonconvex CR manifold is analogous to the original Yamabe problem on compact manifolds. As the first step, Jerison and Lee [8] proved a sharp Sobolev inequality on the CR sphere S 2n+1. Moreover they classified all extremal functions of the sharp Sobolev inequality. Since S 2n+1 is CR equivalent to the Heisenberg group via the Cayley transform, their classification can be stated as following. Theorem 1.1. Let u C 2 (H n ) be a solution to equation (1.1) with p Q+2 Q 2 (where Q2n+2 is the homogeneous dimension), and suppose u L 1+p. Then where µ C n, λ C, and Imλ> µ 2 /4. u(z,t) t+ 1 z 2 +µ z+λ n, (1.2) The proof hinges on a complicated and remarkable identity they discovered with the help of a computer program. It is motivated by Obata s classic work [16] in the Riemannian case. When restricted on R n, Obata s theomem has the following version. Theorem 1.2. Let u C 2 (R n ) be a positive solution for the following equation and suppose u(x)o( x 2 n ) for large x. Then for some λ>0 and x 0 R n. uu n+2 n 2, (1.3) λ (n 2)/2 u(x) x x 0 2 +λ 2, We recall that there is another completely different approach to such classification problems. This is the method of moving plane initiated by Gidas, Ni and Nirenberg [6]. In particular they reproved the above theorem using this method. Later, Caffarelli, Gidas and Spruck [4] classified all positive solutions to equation (1.3) by sharpening the method of moving plane. Their result plays a crucial role in establishing the compactness results for the solution set to Yamabe problem, see, e.g., Schoen [17], Li and Zhu [14],
Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 19 Li and Zhang [11 13], Kuri, et al. [9], Brendle [3]. More recently Li and Zhu developed moving sphere method and gave a simpler proof of this classification result. Moving sphere method is essentially equivalent to moving plane method, but is more convenient. In particular, a simpler alternative proof for the result of Caffarelli, Gidas and Spruck via the method of moving sphere was given in Li s lecture note [10], and first appeared in Zhu s thesis [19]. In this proof via the method of moving sphere, one fully explores the conformal invariance of the equation and obtains certain conformally invariant identity with respect to any point. Solutions to equation (1.3) can then be derived directly from the identity via two key calculus lemmas (refered as Li-Zhu lemmas [15]). It is natural question if one can adapt moving plane method or moving sphere method on the Heisenberg group. Such a method might enable us to classify all positive solutions of (1.1) without any integrability condition or growth condition at infinity. More specifically, we have the following conjecture. Conjecture 1.1. Let u C 2 (H n ) be a solution to Eq. (1.1). If p Q+2 Q 2 (where Q2n+2 is the homogeneous dimension), then u(z,t) t+ 1 z 2 +µ z+λ n, (1.4) where µ C n, λ C, and Imλ> µ 2 /4; If p (1, Q+2 Q 2 ), then u(z,t)0. There have been attepmpts by several mathematicians in the direction. The method of moving plane was adapted by Birindelli and Prajapat [2] to study equation (1.1). Under the assumption of cylindrical symmetry (i.e. u only depends on z and t), they proved that the only nonnegative solution to equation (1.1) for subcritical power 1< p< Q+2 Q 2 is the trivial one. The method of moving plane does not work for the general case since it only yields partial invariant properties of solutions. Other partial results for subcritical case were obtained by Birindelli et al. [1], Garofalo and Lanconelli [5] and Xu [18]. Here we shall use the approach based on the method of moving sphere for the conjecture. For any ξ(z,t) H n, the norm function ρ(ξ) is defined as ρ(ξ)( z 4 +t 2 ) 1 4. It is well known that d(η,ξ) : ρ(η 1 ξ) defines a distance on H n. On H n we have the contact 1-form θ dt+ n 1 zj dz j z j dz j. j1 The study of the equation (1.1) when p(q+2)/(q 2) must use its invarance under contact transformations on H n : if Φ is a diffeomorphism of H n s.t. Φ θ J 1/(n+1) Φ θ, where J Φ > 0 is the Jacobian determinant of Φ, and u is a solution of (1.1), then so is uφj n 2(n+1) Φ. Modulo U(n), all the contact transformations of H n are generated by
20 Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 left translations L η (ξ)ηξ, with L ηθ θ, dialations δ λ (z,t)(λz,λ 2 t),λ>0, with δλ θ λ2 θ, z the inversion T(z,t) z 2, t, with T θ 1 1t z 4 +t 2 z 4 +t 2 θ. We first exam the solution to (1.1) for p(q+2)/(q 2) with center at the origin: u(ξ) 1+ z 2 + 1t n. Using the left translation operator L ξ0, we can obtain the following more general solution v(ξ) : ul ξ0 (ξ) z 2 + 1t+2z 0 z+1+ z 0 2 + n 1t 0. Using the inversion T and dialation, we know that ( λ 4 ) n/2 v λ (ξ) : z 4 +t 2 v(δ λ 2T(ξ)) is also a solution. We compute λ 2 z v λ (ξ)v z 2 1t, λ4 t λ 2 n z 4 +t 2 z 2 + 1t λ ( z 4 2 ) 1t z 2 1t 2 + 2λ2 z 0 z z 2 + 1t +1+ z 0 2 + n 1t 0 λ 2 z 2 + 1t ( λ2 +2z 0 z+ 1+ z 0 2 + ) z 2 + 1t n 1t 0 λ 2 z 2 + λ 4 1t+ (1+ z 0 2 + 2λ )+ 2 n z 0 z 1t 0 (1+ z 0 2 + ) 1t 0 λ 2 ( 1+ z 0 2 + ). 1t 0 We choose λλ 1+ z 0 2 + 1/2 1t 0. Then n v λ (ξ) z 2 + 2λ 2 z 0 z 1t+ (1+ z 0 2 + )+1+ z 0 2 1t 0 1t 0 ul ξ 0 (ξ), n n
Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 21 where In summary, we have proved or, equivalently (e iθ 0 z 0, t 0 ), e iθ 0 1+ z 0 2 1t 0 1+ z 0 2. (1.5) 1t 0 λ(ξ0 ) 2n ul ρ(ξ) δ ξ0 λ(ξ0) 2T(ξ)uL (ξ), λ(ξ0 ) 2n 0,ξ) ul δ ξ0 λ(ξ0) 2TL 1 (ξ)u(ξ). For a general solution vul 1 η, this implies that λ(ξ0 ) 2n d(ηξ 0,ξ) vl δ ηξ0 λ(ξ0) 2TL 1 η(ξ)v(ξ). Set η 0 η,η 0 ηξ 0. We obtain for λλ() λ 2n d(η 0,ξ) vl δ η0 λ 2TL 1 η 0 (ξ)v(ξ). In this paper, we will show the converse. Theorem 1.3. For any (z 0,t 0 ) H n, let ξ 0 be given by (1.5). If u C1 (H n ) satisfies: for all λ>0, λ ν u(ξ) 0,ξ) ul δ ξ0 λ 2TL 1 (ξ), d(,ξ) λ>0, (1.6) holds for a fixed parameter ν, then u const. Theorem 1.4. Assume that u C 1 (H n ) attaines its maximal value at the origin, and satisfies: for any (z 0,t 0 ) H n and ξ 0 be given by (1.5), there exist a positive constant λλ() such that λ ν u(ξ) 0,ξ) ul δ ξ0 λ(ξ0) 2TL 1 (ξ), ξ H n, (1.7) holds for a fixed parameter ν, then u(ξ) is given by the following form where C is a constant. ( ( z u(ξ)c 2 +λ(0) 2) ) 2 ν +t 2 4 C z 2 +λ(0) 2 + 1t Remark 1.5. It is clear that the assumption is invariant under left translation: for any η H n the function vul 1 η satisfies the same assumption, though with a different correspondence ξ 0. ν 2 (1.8)
22 Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 2 Proof We first note, by the triangle inequality, that as ρ(ξ). We also have as ρ(ξ). Proof of Theorem 1.4. Using (1.7), we define 0,ξ)4 ρ(ξ) 4 1+O(ρ(ξ) 1 ), (2.1) L ξ0 δ λ(ξ0 ) 2TL 1 (ξ), D : lim ρ(ξ) ν u(ξ)λ ν u. (2.2) ρ(ξ) + If D0, then u 0, and we are done. Without loss of generality, we assume D1. Further, if ν 0, then u D, and we are done. On the other hand, the case ν <0 can easily be reduced to the case of ν>0 by using a reversed CR inversion. We thus assume that ν>0 in the following proof. Taking 0 yields λ(0) ν u(ξ) uδ ρ(ξ) λ(0) 2T(ξ). As u attains its maximum at the origin, we have for ξ large λ(0) ν u(ξ) [u(0)+o( f(ξ))] ρ(ξ) where Moreover, λ(ξ0 ) ν u(ξ) u( 0 (,ξ) λ(ξ0 ) 0,ξ) ρ(ξ) ν +ρ(ξ) ν O( f(ξ)), (2.3) f(ξ) δ λ(0) 2T(ξ) 2 λ(0)4 z 2 z 4 +t 2 + λ(0)8 t 2 ( z 4 +t 2 ) 2. δ λ(ξ0 ) 2TL 1 ξ 0 ) ν [ u+ u ) (ξ) z ()λ 2 L 1 z z 2 1t + u z ()λ 2 L 1 z z 2 1t + u t ()λ 4 L 1 t z 4 +t 2 + u ] t () 2λ 2 Im(z 0 L 1 z z 2 1t )+O(g(ξ)), (2.4)
Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 23 where g(ξ) λ() 2 L 1 z z 2 2 1t + λ() 4 L 1 t z 4 +t 2+2λ() 2 Im(z 0 L 1 z z 2 1t )) 2. Subtracting (2.4) from (2.3), and using (2.1) and (2.2), we have [ d(,ξ) ν ν+2 u λ z ()L 1 z z 2 1t ν+2 u +λ z ()L 1 z z 2 1t +λ(ξ ν+4 u 0) t ()L 1 t z 4 +t 2 ] ν+2 u +2λ t ()Im(z 0 L 1 z z 2 1t ) + [ d(,ξ) ν O(g(ξ)) ρ(ξ) ν O( f(ξ)) ] d(,ξ) ν +ρ(ξ) ν [ ρ(ξ) ν ν z 2 Re(z 0 e iθ 0 z) ρ(ξ) 4 3ν 2 z 0 2 z 2 ρ(ξ) 4 ν 2 +ν tim(z 0e iθ0 z) ρ(ξ) 4 ]. (2.5) t 0 t ρ(ξ) 4+o(ρ(ξ) 2 ) Taking ξ(z 0 e iθ 0,t) in (2.5), we have ( d(,ξ) ν ν+4 u λ t () t+t ) 0 t+t 0 2 + [ d(,ξ) ν O(g(ξ)) ρ(ξ) ν O( f(ξ)) ] [ ρ(ξ) ν ν z 0 4 ν ] t 0 t 2 z 0 4 +t 2 2 z 0 4 +t 2+o(ρ(ξ) 2 ). Multiplying both sides by ρ(ξ) ν+2 and letting t +, we obtain which yields, by (2.2), ν+4 u λ t () ν 2 t 0, t (u 4 ν )(ξ0 )2t 0. It follows that there exists a function φ(z, z) such that u(ξ)(φ(z, z)+t 2 ) ν 4. (2.6)
24 Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 Consequently, we know that u t (z, z,0)0. On the other hand, let (z 0,z 0,0)(x 0,y 0,0) and ξ(z, z,0)(x,y,0) in (2.5), we have [ u d(,ξ) ν λ ν+2 z (ξ z z 0 0) z z 0 2 2 1(xy 0 x 0 y) + u ] z (ξ z z 0 0) z z 0 2 +2 1(xy 0 x 0 y) + [ d(,ξ) ν O(g(ξ)) ρ(ξ) ν O( f(ξ)) ] [ ρ(ξ) ν ν z 2 Re(z 0 z) ρ(ξ) 4 3ν z 0 2 z 2 ] 2 ρ(ξ) 4 +o(ρ(ξ) 2 ). (2.7) Choose y0 in (2.7). Multiplying both sides by ρ(ξ) ν+1, and sending x +, we get Similarly, choosing x0 in (2.7), we obtain Combining (2.8) with (2.9), we have [ u λ ν+2 z ()+ u ] z () νx 0. (2.8) [ u 1λ ν+2 z () u ] z () νy 0. (2.9) ν+2 u λ z () ν 2 z 0, ν+2 u λ z () ν 2 z 0. This yields, that φ(z, z) in (2.6) satisfies φ(z, z) z z, φ(z, z) z z. Therefore, we conclude that there exists a constant C such that In fact, from (2.2), we know C λ(0) 2. u(ξ)(( z 2 +C ) 2 +t 2 ) ν 4. Proof of Theorem 1.3. Step I: We first show that u(ξ)u(z, z), that is, function u(ξ) is independent of variable t. For any given t,τ R (t τ) and any b>1, let ξ(z 0,z 0,t), ζ(z 0,z 0,τ) H n, t 0 t 0 (b) τ+b(t+τ) and λλ(b) 0,ζ)d(,ζ),
Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 25 where (0,0,t 0 ) and ξ 0 (0,0, t 0). A direct computation leads to d(,ξ)( z 0 4 +(b 1) 2 t+τ 2 ) 1 4, 0,ζ)( z 0 4 +b 2 t+τ 2 ) 1 4, λ 2 ( z 0 4 +(b 1) 2 t+τ 2 ) 1 4 ( z0 4 +b 2 t+τ 2 ) 1 4 < 0,ζ) 2. Thus, we know, via (1.6), that Noting that ( δ λ 2TL 1 (ζ) lim b + lim b + λ 0 1,,ζ) λ 2 z 0 z 0 2 ib(t+τ) λ ν u(ζ) u( 0,ζ) δ λ 2TL 1 ξ 0 ) (ζ). (2.10) λ 2 z 0 z 0 2 ib(t+τ), λ 2 z 0 z 0 2 +ib(t+τ),t 0 λ4 b(t+τ) z 0 4 +b 2 t+τ 2 { 1z0, if t+τ> 0, 1z 0, if t+τ< 0, t 0 λ4 b(t+τ) z 0 4 +b 2 t+τ 2 τ+ b(t+τ) ( z 0 4 +b 2 (t+τ) 2 ) 1 2 ( ) ( z 0 4 +b 2 (t+τ) 2 ) 1 2 ( z0 4 +(b 1) 2 (t+τ) 2 ) 1 2 b(t+τ) s(t+τ) 2 τ+ ( z 0 4 +b 2 (t+τ) 2 ) 2 1 ( z 0 4 +s 2 (t+τ) 2 ) 1 2 τ+(t+τ)t (as b + ), (b 1< s<b) we obtain, by letting b + in (2.10), that { u(z 0,z 0,τ) u( 1z 0, 1z 0,t), if t+τ> 0, u(z 0,z 0,τ) u( 1z 0, 1z 0,t) if t+τ< 0. For any t 1,t 2 R, we choose t 3 R such that t 1 +t 3 >0 and t 2 +t 3 >0. Repeating the above process, we have u(z 0,z 0,t 1 ) u( 1z 0, 1z 0,t 3 ) u( z 0, z 0,t 2 ) u( 1z 0, 1z 0,t 3 ) u(z 0,z 0,t 2 ). Thus, u is independent on the variable t because of the arbitrariness of t 1 and t 2. Step II: We show u(ξ)u( z e iθ, z e iθ,t)u(e iθ,e iθ ) : ϕ(θ). ),
26 Y. Han, X. Wang and M. Zhu / J. Math. Study, 50 (2017), pp. 17-27 For given two different positive numbers b, c (with b c) and any s>1, choose a a(s) c+s(b c). Let (az 0,az 0,0), ξ (bz 0,bz 0,0) and ζ (cz 0,cz 0,0). Then (az 0,az 0,0). let λλ(s) d(,ξ) 0,ζ). A direct computation yields Thus (using the fact s > 1) d(,ξ) (1 s)(b c)z 0, d(,ζ) s(b c)z 0, λ s(1 s) (b c)z 0. δ λ 2TL 1 (ζ)ξ, λ lim s + 0 1.,ζ) d(,ξ)<λ< 0,ζ), So, by (1.6), we have λ ν u(cz 0,cz 0,0) 0,ζ) u(bz 0,bz 0,0). Sending s +, we have u(cz 0,cz 0,0) u(bz 0,bz 0,0), that is, u(z,z,0) is independent of z. Step III: Finally, we show ϕ(θ) const. By (1.6), we know that, for any two positive constants r, t satisfying r 4 +t 2 λ 4 > 0, ϕ(θ)u(re iθ,re iθ,t) λ 4 ν 4 r 4 +t 2 u δ λ 2T(re iθ,re iθ,t) λ 4 ν ( 4 λ 2 re iθ r 4 +t 2 u r 2 it, λ2 re iθ ) r 2 +it, λ4 t r 4 +t 2 λ 4 ν 4 r 4 +t 2 ϕ(θ+θ ), (2.11) where θ arctan t r 2. In particular, choosing λ 4 r 4 +t 2, we have ϕ(θ) ϕ(θ+θ ). Since r, t are arbitrary, we conclude that ϕ(θ) const.
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