FOR THE CR HEAT EQUATION ON HEISENBERG GROUPS
SHU-CHENG CHANG1 , JINGZHI TIE2, AND CHIN-TUNG WU3
Abstract. In this paper, we get a subgradient estimate of the CR heat equation on a closed pseudohermitian (2n + 1)-manifold. Moreover, we are able to establish Liouville-type theorems for the CR heat equation and the subgradient estimate for the CR heat kernel on (2n + 1)-dimensional Heisenberg groups.
1. Introduction
In [Ha], Hamilton established gradient estimate of the logarithm of a positive so-lution to the heat equation on closed Riemannian manifolds. Recently, Kotschwar ([K]) extended the result to complete noncompact Riemannian manifolds. In this paper, we consider the CR heat equation (1.4) with respect to the sub-Laplacian on a complete pseudohermitian (2n + 1)-manifold. By using the arguments of [LY1] and CR Bochner formula (2.1), we are able to derive the CR version of parabolic Li-Yau gradient estimate and the so-called reversed Li-Yau Harnack inequality for the pos-itive solution of the CR heat equation. Then by combining the standard parabolic Li-Yau gradient estimate, we derive a subgradient estimate of the CR heat equation on closed pseudohermitian (2n + 1)-manifold. Moreover, we are able to establish subgradient estimate of the CR heat kernel and Liouville-type theorems for the CR heat equation on (2n + 1)-dimensional Heisenberg groups. There are many important related works such as [BGG], [CY], [PP], [SC], [JSC] etc.
1991 Mathematics Subject Classi…cation. Primary 32V05, 32V20; Secondary 53C56.
Key words and phrases. Subgradient estimate, Liouville-type Theorem, Heat Kernel, Pseudo-hermitian manifold, Heisenberg Group, CR-pluriharmonic, CR-Paneitz operator, Sub-Laplacian, Li-Yau Harnack inequality.
Research supported in part by NSC of Taiwan. 1
The main key step is to derive the CR version of Bochner formula. This formula (2.1) involving a third order operator P which characterizes CR-pluriharmonic func-tions ([L1]), is hard to control. However after integrating by parts (see 1.3), we are able to relate this extra term to the CR Paneitz operator P0.
Let (M; J; ) be a pseudohermitian (2n + 1)-manifold without boundary with an oriented contact structure (see Appendix A for basic notions in pseudohermitian geometry).
De…nition 1.1. (i) A piecewise smooth curve : [0; 1]! M is said to be horizontal if 0(t)2 whenever 0(t) exists. The length of is then de…ned by
l( ) = Z 1
0
h 0(t); 0(t)i12dt:
The Carnot-Carathéodory distance between two points p; q 2 M is dc(p; q) = inffl( )j 2 Cp;qg ;
where Cp;q is the set of all horizontal curves joining p and q.
(ii) By Chow connectivity theorem [Cho], there always exists a horizontal curve joining p and q, so the distance is …nite. We say M is complete if it is complete as a metric space.
De…nition 1.2. A smooth real-valued function u in M is said to be CR-pluriharmonic function if for any point p 2 M; there is an open neighborhood U of p in M and a smooth real-valued function v on U such that @b(u + iv) = 0:
De…nition 1.3. ([L1]) Let (M2n+1; J; ) be a complete pseudohermitian manifold.
De…ne P ' = n X =1 (' + inA ' ) = (P ') ; = 1; 2; ; n which is an operator that characterizes CR-pluriharmonic functions. Here
(1.1) P ' =
n
X
=1
(' + inA ' )
and P ' = (P ') , the conjugate of P . Moreover we de…ne
which is the so-called CR Paneitz operator P0: Here b is the divergence operator that
takes (1; 0)-forms to functions by b( ) = ; and b( ) = ; . If we de…ne
@b' = ' and @b' = ' , then the formal adjoint of @b on functions (with respect
to the Levi form and the volume form d ) is @b = b.
Note that for a closed pseudohermitian (2n + 1)-manifold (M; J; ); one get (1.3) Z MhP ' + P '; d b'i d = 1 4 Z M P0' ' d :
Moreover, for a closed pseudohermitian (2n + 1)-manifold (M; J; ) of zero torsion, then ([GL])
P0' =LnLn= [( b')2+ n2T2']:
Here
Ln' = b' + inT ' = 2' :
For the details about these operators, the reader can make reference to [GL], [H] and [L1].
Remark 1.1. ([H], [GL]) (i) Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold with n 2. Then a smooth real-valued function f satis…es P0f = 0 on M
if and only if P f = 0 on M: It holds also for a closed pseudohermitian 3-manifold of zero torsion.
(ii) A smooth real-valued function f 2 L2(Hn) satis…es P
0f = 0 on Hn if and only
if P f = 0 on Hn:
(iii) Let P f = 0 as in (ii). If M is the boundary of a connected strictly pseudo-convex domain Cn+1, then f is the boundary value of a pluriharmonic function
u in : That is, @@u = 0 in : Moreover, if is simply connected, there exists a holomorphic function w in such that Re(w) = u and ujM = f:
Let (M; J; ) be a complete pseudohermitian (2n + 1)-manifold without boundary. In this paper, we consider the positive solution u(x; t) of the CR heat equation with respect to the sub-Laplacian
(1.4) @
@tu(x; t) = bu(x; t) on M [0; T ).
Proposition 1.1. Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold for n 2. If u(x; t) is the positive smooth solution of (1.4) on M [0;1). Suppose that
[2Ric (n + 2)T or](Z; Z) l0jZj2;
for all Z 2 T1;0 and l0 is a nonnegative constant. Then the function
(1.5) G = t jrb'j
2
+ 1 + 2 n 't satis…es the inequality
b @t@ G n+22n hrb';rbGi +(n+1)(n+2)2n 2tG G (n+1)(n+2)2 2n l0tjrb'j 2 8 ntu 2 P u + P u; d bu L :
Let u(x; t) be a positive solution of (1.4) on M [0;1): In section 2, it is proved that if P u = 0 at t = 0; then P u = 0 for all t on a closed pseudohermitian (2n + 1)-manifold of zero torsion. Then the extra term of CR Bochner formula (2.1) becomes
(1.6) hP u + P u; dbui = 0
on M [0;1) if P u = 0 at t = 0:
In fact, by using the arguments of [LY1], (2.1) and (1.6), we are able to derive the CR version of parabolic Li-Yau gradient estimate for the positive solution u(x; t) of (1.4) on M [0;1). Then combining the result of [CY] and Proposition 1.1, we get the following subgradient estimate of the logarithm of a positive solution to (1.4). Theorem 1.2. Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold of zero torsion and nonnegative pseudohermitian Ricci tensor. If u(x; t) is the positive solu-tion of (1.4) on M [0;1) such that
P u = 0; = 1; 2; ; n
at t = 0: Then there exist constants C1; C2 such that u satis…es the subgradient
estimate
(1.7) tjrblog uj
2
C1+ C2t
Note that the arguments of [LY1] can be extended easily to complete noncompact pseudohermitian (2n + 1)-manifold if one can have the CR version of Laplacian com-parison theorem. Indeed, this is the case for a (2n + 1)-dimensional Heisenberg group Hn (see section 5 for details). Then we have
Theorem 1.3. If u(x; t) be a positive smooth solution of (1.4)
b
@
@t u(x; t) = 0 on Hn [0; T ) with
P u = 0 at t = 0; then u satis…es the subgradient estimate
tjrblog uj
2 (n + 2)(n2+ 5n + 2)
4(n + 1) +
on Hn [0; T ) for any > 0:
Remark 1.2. For the CR Yamabe ‡ow on a closed pseudohermitian 3-manifold of zero torsion and nonnegative Tanaka-Webster curvature, we have the similar result on CR version of Li-Yau-Hamilton inequality ([CCW]).
As a consequence, we have the following Liouville-type theorems for CR heat equa-tion on Hn [0;1).
Corollary 1.4. Let (Hn; J; ) be the complete noncompact (2n + 1)-dimensional
Heisenberg group. If u(x; t) is a positive solution of (1.4) on Hn [0;
1) with a positive smooth CR-pluriharmonic function as an initial. Then u is a constant. Remark 1.3. It is true that there are no nontrivial positive harmonic functions on Hn: See [KS] and [CCW1] for details.
Now for any L2-function u(x; t); we may write
u(x; t) = uker(x; t) + u?(x; t)
with P0(uker(x; t)) = 0. From Lemma 2.2, we may split the CR heat equation (1.4)
into the following heat equations respectively :
(1.8) @
@tu
?= bu?
and
(1.9) @
@tuker= buker
on the complete noncompact Heisenberg group (Hn; J; )
. Observe that H(x; y; t) 2 C1(Hn Hn R+) and for any …xed y; t; H(x; y; t) 2 L2(Hn): Then for any L2 -function u(x; 0) = f (x); we have
f (x) = fker(x) + f?(x)
and
H(x; y; t) = Hker(x; y; t) + H?(x; y; t)
with P0(fker(x)) = 0 and P0(Hker(x; y; t)) = 0: Hence
u?(y; t) = Z H?(x; y; t)f?(x)dx and uker(y; t) = Z Hker(x; y; t)fker(x)dx:
As a consequence from Theorem 1.3 and Corollary 4.4, we have the following subgradient estimate of the heat kernel.
Corollary 1.5. Let H(x; y; t) be the heat kernel of (1.4) on Hn [0; T ) with
H(x; y; t) = Hker(x; y; t) + H?(x; y; t) and Hker(x; y; t) is a positive L2-function. Then
for some constant and 0 < < 1;
jrbHker(x; y; t)j C( )2t (2n+3) 2 exp d 2 c(x; y) 2(4 + )t with C( ) ! 1 as ! 0.
Remark 1.4. For simplicity, we …rst prove the Theorems of this paper for n = 1 as in section 3; 4; 5: The higher dimensional cases will be given in section 6.
Acknowledgments. The …rst named author would like to express his thanks to Prof. S.-T. Yau for constant encouragement, Prof. J.-P. Wang for valuable discussions during his visit at NCTS, Hsinchu, Taiwan. The …rst and second named authors would like to express their thanks to Prof. Jih-Hsin Cheng for his supports and Institute of Mathematics, Academia Sinica for the hospitality during the visit.
2. CR Bochner Formula and Preserving Property
In this section, we will drive the CR version of Bochner formula and the preserving property for (1.4) on a pseudohermitian (M2n+1; J; ):
We …rst derive the following CR version of Bochner formula on a complete pseudo-hermitian (M2n+1; J; ):
Lemma 2.1. Let (M2n+1; J; ) be a complete pseudohermitian manifold. For a real
smooth function u on (M; J; ), (2.1) 1 2 bjrbuj 2 = j(rH)2uj2+ (1 + n2) < rbu;rb bu >L +[2Ric (n + 2)T or]((rbu)C; (rbu)C) 4 n < P u + P u; dbu >L :
Here (rbu)C = u Z is the corresponding complex (1; 0)-vector …eld of rbu and
dbu = u + u :
Remark 2.1. In [Chi] and [CC], the CR Bochner formulae (2.1) was derived for n = 1:
Proof. First from [Gr], we have for a real function u
(2.2) 1 2 bjrbuj 2 = j(rH)2u j2+ < rbu;rb bu >L +(2Ric nT or)((rbu)C; (rbu)C) 2iPn=1(u u 0 u u 0):
We use the matrix h to raise and lower indices. In the following we always compute at one point. Then one may assume h = to lower the index. For instance, P ' = n X =1 (' + inA ' ) and u0 u 0 = n X =1 A u and iu0 = u u :
Compute iu u 0 = iu u0 iPn=1A u u = n1u Pn=1(u u ) iPn=1A u u = n1u P u + iPn=1A u u 1nPn=1(u u ) iPn=1A u u = n1u P u n1 Pn=1(u u ) and iu u 0 =conj(iu u 0): Then 2iPn=1(u u 0 u u 0) = 2n < P u + P u; dbu >L +n2 Pn; =1(u u + u u ): But <rbu;rb bu >L =Pn; =1[u (u + u ) + u (u + u ) ] =Pn; =1(u u + u u ) +Pn; =1(u u + u u ) =Pn; =1(u u + u u )+ < P u + P u; dbu >L +inPn; =1(A u u A u u ) =Pn; =1(u u + u u )+ < P u + P u; dbu >L +nT or((rbu)C; (rbu)C): It follows that (2.3) 2iPn=1(u u 0 u u 0) = 4n < P u + P u; dbu >L 2T or((rbu)C; (rbu)C) +n2 <rbu;rb bu >L :
Finally from (2.2) and (2.3), we have
1 2 bjrbuj 2 = j(rH)2u j2+ (1 + 2 n) < rbu;rb bu >L +[2Ric (n + 2)T or]((rbu)C; (rbu)C) 4 n < P u + P u; dbu >L :
Lemma 2.2. Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold of zero torsion. If u(x; t) is a solution of
b
@
@t u(x; t) = 0
on M [0;1) with P u(x; 0) = 0: Then P u(x; t) = 0 for all t 2 (0; 1):
Proof. Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold of zero torsion. From Remark 1.1, we have P0u = 0 if and only if P u = 0 and
P0u = (( b)2u + nT2u):
It follows that bP0u = P0 bu. Apply P0 to the heat equation, we obtain
b
@
@t P0u(x; t) = 0
on M [0;1) with P0u(x; 0) = 0: Hence the Lemma follows from the maximum
principle and Remark 1.1.
Lemma 2.3. Let (Hn; J; ) be the complete noncompact (2n+1)-dimensional
Heisen-berg group. If u(x; t) is a solution of
b
@
@t u(x; t) = 0
on M [0;1) with P u(x; 0) = 0: Then P u(x; t) = 0 for all t 2 [0; 1):
Remark 2.2. Since (Hn; J; ) is complete noncompact, P u is not necessarily
van-ishing even if P0u = 0. So we need to have a di¤erent proof from Lemma 2.2.
Proof. For simplicity, we proof it for n = 1: We …rst need the following commutation relation ([L1])
(2.4)
CI;01 CI;10 = CI;1A11 kCI;A11;1;
CI;01 CI;10 = CI;1A11 kCI;A11;1;
CI;11 CI;11 = iCI;0+ kW CI;
Here CI denotes a coe¢ cient of a tensor with multi-index I consisting of only 1 and
For @ @tu111 = ( bu)111 = (u 11+ u11)111 = (u11111 + u11111); it follows from (2.4) that
u 11111 = u11111 iu1101 = u11111 iu1110 and u 11111 = u11111 + iu0111 = (u 11111 iu1101) + iu0111 = (u 11111 iu1110) iu1101+ iu0111 = u 11111 iu1110:
Thus for L2 = b + 2iT
(2.5) @
@tu111= bu111 2iu1110 = L2u111:
This plus (2.5) imply
@
@t(P1u) = L2(P1u): Similarly for n 2; we have
@
@t(P u) = L2n(P u):
Since 2n is not an odd integer, L2n is a subelliptic operator. Then by the
unique-ness of solution to subelliptic parabolic equation, P u(x; t) = 0 for all t 2 [0; 1) if P u(x; 0) = 0:
3. Subgradient Estimate of the CR Heat Equation
In this section, we …rst establish the subgradient estimate of Theorem 1.2 for n = 1. For n 2; we refer it to section 6.
Let (M; J; ) be a closed pseudohermitian 3-manifold. By using the arguments of [LY1], we are able to derive the CR version of parabolic Li-Yau gradient estimate for the positive solution u(x; t) of (1.4) on M [0;1):
Let ' = log u: Then ' satis…es b @ @t ' = jrb'j 2 :
On the other hand, from Cao-Yau’s ([CY]) paper, one has the standard parabolic Li-Yau gradient estimate.
Proposition 3.1. ([CY, Theorem 2.1]) Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold and u(x; t) be a positive smooth solution of (1.4) on M [0;1): Then there exist constants C0; C00 and 0 > 1 such that for any 0; usatis…es the
estimate (3.1) jrbuj 2 u2 ut u C0 t + C 00 on M [0;1):
Now we derive the CR version of parabolic Li-Yau gradient estimate for the positive solution of the CR heat equation. First, we need the following Lemma.
Lemma 3.2. Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold. Let ' = ln f , for f > 0: Then (3.2) 4 P ' + P '; db' L = 4h P f +P f;dbfi L f2 2 < rb';rbjrb'j2 > +2 fbfjrb'j2
Proof. In the following, we use the Einstein convention notation. Let Q(x) = jrb'j2(x).
We compute rbQ = Q Z + Q Z = 2rb(' ' ) = 2(f 2f f +f2f f 2f f f f4 )Z + complex conjugate: It follows that
P ' = ' + inA ' = f 2f f + f3f f2f f f2f f + 2f f f f f4 + inA f f = P f f 1 2Q f f f2 = P f f 1 2Q ' f f : Thus 4 P ' + P '; db' L = 4 D (P ') + P ' ; ' + ' E L = 4 (P ') ' + P ' ' = 4 P f f 1 2Q ' f f ' +complex conjugate = 4 P f + P f; dbf L f2 2 rb';rbjrb'j 2 + 2 bf f jrb'j 2 :
This implies the Lemma.
Lemma 3.3. Let (M; J; ) be a closed pseudohermitian 3-manifold. If u(x; t) is the positive smooth solution u(x; t) of (1.4) on M [0;1). Suppose that
(2Ric 3T or)(Z; Z) k0jZj2;
for all Z 2 T1;0 and k0 is a nonnegative constant. Then the function
(3.3) F = t jrb'j
2
+ 3't
satis…es the inequality
b @ @t F 2 3hrb';rbFi + 1 9tF (F 9) + k0tjrb'j 2 8tu 2 P u + P u; dbu L :
Proof. First di¤erentiating (3.3) w.r.t. the t-variable, we have Ft = 1 tF + t jrb'j 2 + 3't t = 1 tF + t 4jrb'j 2 + 3 b' t (3.4) = 1 tF + t [8hrb';rb'ti + 3 b't] :
By using the CR version of Bochner formula (2.1) and Lemma 3.2, one obtains
bF = t bjrb'j2+ 3 b't = t[2j(rH)2'j2 + 6hrb';rb b'i +2(2Ric 3T or)((rb')C; (rb')C) 8hP ' + _ P '; db'iL + 3 b't] t[4j'11j2+ ( b')2+ 6hrb';rb b'i k0jrb'j2 (3.5) 8hP ' + _ P '; db'iL + 3 b't] = t[4j'11j2+ ( b')2+ 6hrb';rb b'i k0jrb'j 2 8u 2 P u + P u; dbu L + 4'tjrb'j2 +4 rb';rbjrb'j2 + 3 b't]:
Here we have used the inequalities
(3.6) (rH)2' 2 = 2j'11j2+1 2( b') 2+1 2' 2 0 2j'11j 2 +1 2( b') 2; and (2Ric 3T or)((rb')C; (rb')C) k0j(rb')Cj2 = k0 2 jrb'j 2 ; and 't= ut u = bu u : Applying the formula
(3.7) b' = 't jrb'j 2 = 1 3tF 4 3jrb'j 2
and combining (3.4), (3.5), we conclude b @ @t F 1 tF + t[4j'11j 2 + ( b')2+ 6hrb';rb b'i +4 rb';rbjrb'j2 8hrb';rb'ti k0jrb'j2+ 4'tjrb'j2 8u 2 P u + P u; dbu L ] = 1 tF + t[ 2 3thrb';rbFi 4 3 rb';rbjrb'j 2 + 4 j'11j 2 +( b')2 k0jrb'j 2 + 4'tjrb'j2 8u 2 P u + P u; dbu L ]:
Now it is easy to see that
rb';rbjrb'j2 = 4 Re('11'1'1) + b'jrb'j2: Thus 4 3hrb';rbjrb'j 2 i = 163 Re('11'1'1) 4 3 b'jrb'j 2 4j'11j2 169j'1j4 43 b'jrb'j2 = 4j'11j2 49jrb'j4 43 b'jrb'j2:
Here we have used the basic inequality 2 Re(zw) jzj2 + 1
jwj2 for all > 0: All
these imply b @t@ F 1tF 23hrb';rbFi + t[( b')2+83 b'jrb'j2 +329jrb'j4 k0jrb'j2 8u 2 P u + P u; dbu L ] 2 3hrb';rbFi + 1 9tF (F 9) k0tjrb'j 2 8u 2 P u + P u; dbu L :
This completes the proof of Lemma 3.3.
Theorem 3.4. Let (M; J; ) be a closed pseudohermitian 3-manifold of zero torsion and nonnegative Tanaka-Webster curvature. If u(x; t) is the positive solution of (1.4) on M [0;1) such that
P1u = 0
at t = 0: Then u satis…es the estimate
(3.8) jrbuj 2 u2 + 3 ut u 9 t on M [0;1):
Proof. Applying Lemma 3.3 to ' by setting A11= 0; k0 = 0 and P u + P u; dbu = 0: Then we have (3.9) b @ @t F 2 3hrb';rbFi + 1 9tF (F 9):
The theorem claims that F is at most 9: If not, at the maximum point (x0; t0) of F
on M [0; T ] for some T > 0;
F (x0; t0) > 9:
Clearly, t0 > 0; because F (x; 0) = 0: By the fact that (x0; t0) is a maximum point of
F on M [0; T ]; we have
bF (x0; t0) 0;
rbF (x0; t0) = 0;
and
Ft(x0; t0) 0:
Combining with (3.9), this implies
0 1
9t0
F (x0; t0)(F (x0; t0) 9);
which is a contradiction. Hence F 9 and the theorem follows.
Then by combining Proposition 3.1 and Theorem 3.4, the subgradient estimate Theorem 1.2 follows easily for n = 1.
4. Subgradient Estimates in the Heisenberg group H1
We …rst establish Liouville-type theorems for the CR heat equation on a 3-dimensional Heisenberg group H1:Second, we derive the subgradient estimate for CR Heat Kernel
on H1:
Proposition 4.1. If u(x; t) be a positive smooth solution of (1.4) on Hn [0; T );
then u satis…es the estimate
jrbuj 2 u2 ut u n t on Hn [0; T ):
Theorem 4.2. If u(x; t) be a positive smooth solution of (1.4) ( b
@
@t)u(x; t) = 0 on H1 [0; T ) with
P1u = 0
at t = 0; then u satis…es the subgradient estimate jrbuj 2 u2 + 3 ut u 9 + t on H1 [0; T ) for any > 0:
Proof. Let B2R be a ball of radius 2R center at O 2 H1: Let ' = log u and F =
t(jrb'j2+ 3't);then sup BR jrbuj2 u2 + 3 ut u ! = sup BR F t : Let 2 C1
0 (R) be a cut-o¤ function ([DT]) such that 0 1; (t) 1 for
t 2 [0; 1]; (t) 0for t 2: We also require
(4.1) 0 0; 00 C1; and j
0j2
C2;
where C1 and C2are positive constants. Denote by dc(x)be the Carnot-Carathéodory
distance from O to x in H1. Then we de…ne (x) = dc(x)
R : It is clear that
supp B2R and jBR 1:
We want to apply the maximum principle to F: The function may not be smooth at the cut locus of O 2 H1:However, when applying the maximum principle, we can assume is di¤erentiable as in [LY1].
If F attains its maximum at (x0; t0) 2 B2R [0; T0] with 0 < T0 < T; clearly
we may assume ( F )(x0; t0) > 0 (otherwise F 0; and the theorem is true). So
x0 2 B2R; t0 > 0; and by the maximum principle, at (x0; t0)
(4.3) b( F ) 0;
and
(4.4) @
@t( F ) = Ft 0:
In the sequel, all computations will be at the point (x0; t0):By (4.2), rbF = Frb = ;
and by (4.3) 0 b( F ) = F b + bF + 2hrb ;rbFi (4.5) = F b + bF 2Fjr b j2 : By (4.1), we have (4.6) jrb j 2 = j 0j2 jrbdcj2 R2 = j 0j2 R2 C2 R2; and b = 00 jrbdcj2 R2 + 0 bdc R = 00 R2 + 0 R bdc C1 R2 p C2 R bdc:
Since in H1;we have the sub-Laplacian comparison ( ) (see the proof in next section)
( ) bdc
C dc
; for some constant C. Then
b
C3
R2:
Substituting this into (4.5) and applying Lemma 2.3 and Lemma 3.3 with A11 = 0;
k0 = 0; all these imply
0 b( F ) C3 R2F 2F jrb j2 + bF C3 R2F 2F jrb j2 + [Ft+ 2hrb';rbFi + 1 9tF (F 9)]:
Since Ft = ( F )t 0; 2 hrb';rbFi = 23Fhrb';rb i; the above inequality can be
reduced as 0 C3 R2F 2F jrb j2 +2 3Fhrb';rb i + 1 9t F (F 9);
and multiplying by ; we get 0 C3 R2 F 2F jrb j 2 +2 3F hrb';rb i + 1 9t 2F (F 9) = ( F ) C3 R2 2 jrb j2 t ! +2 3 Fhrb';rb i + 1 9t( F ) 2 ( F ) C3 R2 2 jrb j2 t ! 2 Fjrb'j jrb j + 1 9t( F ) 2
Using 0 1;and (4.6), we get
0 ( F ) C3 R2 2 C2 R2 1 t 2 3=2F p C2 R jrb'j + 1 9t( F ) 2 = ( F ) 1 t C4 R2 2 3=2 F p C2 R jrb'j + 1 9t( F ) 2 ;
where C4 = C3+ 2C2: Multiplying by t to the above inequality, then leads to
0 ( F ) 1 9 F 1 C4 R2t 2t 3=2F p C2 R jrb'j = ( F ) 1 9 F 1 C4 R2t 2pC2 R 1=2 jrb'j t : Therefore, we get F 9 + 9C4 R2 t + 18pC2 R 1=2 jrb'j t:
(i) If 't(x0; t0) < 0;then, by the Proposition 4.1, jrb'j2 jrb'j2 't 1=t and
using 0 1; we have F 9 + C4 R2t + 18pC2 R t 1=2:
Recall that all the computations are at (x0; t0) and (x0; t0) is the maximum point,
t0 T0;so we have ( F )(x; T0) ( F )(x0; t0) 9 + C4 R2T 0+18 p C2 R p T0: But 1 on BR; hence (4.7) sup x2BR (jrb'j2 + 3't)(x; T0) C4 R2 + 18pC2 R 1 p T0 + 9 T0:
Now for any …xed time t 2 (0; 1); by letting R ! 1, one obtains jrbuj 2 u2 + 3 ut u 9 t
on H1 [0; T ):
(ii) If 't(x0; t0) 0;then t1=2jrb'j F1=2:The above inequality leads to
F 18 p C2 R t 1=2( F )1=2 9 + C4 R2t 0: Hence F 9 + C4 R2t + 18pC2 R t 1=2( F )1=2: If ( F ) 1;then F 9 + C4 R2t + 18pC2 R t 1=2: Otherwise, F 9 + C4 R2t + 18pC2 R t 1=2 ( F ): For …x t, we can choose R such that 18pC2
R t
1=2 1 2, thus
F 18 + C4 R2t
and similar argument as before
(4.8) sup x2BR (jrb'j 2 + 3't)(x; T0) C4 R2 + 18 T0:
Now for any …xed time t 2 (0; 1); by letting R ! 1 such that 18pC2 R t 1=2 ! 0; one obtains jrbuj2 u2 + 3 ut u 9 + t on H1 [0; T ) for any > 0:
Then by combining the Theorem 4.2 and Proposition 4.1, Theorem 1.3 follows for n = 1 easily.
Now we will apply the subgradient estimates in Theorem 4.2 and Proposition 4.1 to obtain the following Harnack inequality for positive solutions of the CR heat equation (1.4) on H1 [0; T ).
Theorem 4.3. If u(x; t) be a positive smooth solution of (1.4)
b
@
@t u(x; t) = 0 on H1 [0; T ) with
at t = 0; then for all points x1; x2 in H1 and times 0 < t1 < t2 < T; we have the inequality t1 t2 exp d 2 c(x1; x2) 4(t2 t1) u(x2; t2) u(x1; t1) t2 t1 (3+ ) exp 3d 2 c(x1; x2) 4(t2 t1) for any > 0:
Proof. Let be a horizontal curve with (t1) = x1 and (t2) = x2: We de…ne :
[t1; t2]! M [t1; t2] by
(t) = ( (t); t):
Clearly (t1) = (x1; t1) and (t2) = (x2; t2): Let ' = log u(x; t); integrate dtd' along
;we get '(x2; t2) '(x1; t1) = Z t2 t1 d dt'dt = Z t2 t1 n h ; rb'i + 't o dt: Applying Theorem 4.2 to 't; this yields
'(x2; t2) '(x1; t1) Z t2 t1 j j jrb'j + 't dt Z t2 t1 3 4j j 2 + 3 + t dt = Z t2 t1 3 4j j 2 dt + (3 + ) log t2 t1 :
Then the right-hand side inequality in theorem 4.3 follows by taking exponentials of the above inequality. Similarly, we can also get the left-hand side inequality.
As a consequence of Theorem 4.3 and [CY], we have
Corollary 4.4. Let H(x; y; t) be a L2-heat kernel of (1.4) on H1 [0; T ). Then for
some constant and 0 < < 1; we have the inequality H(x; y; t) C( ) V (Bx( p t))exp d2 c(x; y) (4 + )t C( ) t2 exp d2 c(x; y) (4 + )t with C( ) ! 1 as ! 0.
Remark 4.1. Here we use the volume V (Bx(R)) CR(2n+2) in an (2n+1)-dimensional
Heisenberg group Hn ([DT]). One should compare this result with [BGG].
Then Corollary 1.5 follows easily from Theorem 1.3 and Corollary 4.4.
5. Sub-Laplacian of Carnot-Caratheodory Distance on Heisenberg groups Hn
In this section, we prove the Sub-Laplacian comparison ( ) as in previous section. We consider the following two vector …elds de…ned on R3 with coordinates (x; t) =
(x1; x2; t): X1 = @ @x1 + 2ax2 @ @t and X2 = @ @x2 2ax1 @ @t with a > 0. It is easy to check that
[X1; X2] = 4a
@ @t: Now we consider the following operator:
H = 1 2(X 2 1 + X 2 2)
The vector …elds X1, X2and T = @t@ and the operator H are left-invariant with
re-spect to the “Heisenberg translation": for (x; t) = (x1; x2; t) and (y; s) = (y1; y2; s)2
R3,
(x; t) (y; s) = (x1+ y1; x2+ y2; t + s + 2a[x2y1 x1y2]):
Actually, the above multiplicative law de…nes a group structure on R3 which we call the 1-dimensional Heisenberg group with (x; t) 1 = ( x; t).
Remark 5.1. By comparing the previous notations, we …rst put some conventions as followings: for a = 12 Z1 = 1 2(X1 iX2) and Z1 = 1 2(X1+ iX2) and J (X1) = X2 and J (X2) = X1 and b = H:
The symbol of H is H(x; ; ) = 1 2( 1+ 2ax2 ) 2+ 1 2( 2 2ax1 ) 2 = 1 2( 2 1+ 2 2);
where 1 = 1+ 2ax2 and 2 = 2 2ax1 .
In this notation, Hamilton-Jacobi equations for the bicharacteristic curve (x1(s); x2(s); t(s); 1(s); 2(s); (s)) take the form:
(5.1)
_x1(s) = @H@
1 = 1+ 2ax2 = 1(s);
_x2(s) = @H@
2 = 2 2ax1 = 2(s);
_t(s) = @H@ = ( 1+ 2ax2 )(2ax2) ( 2 2ax1 )(2ax1) = 2a( 1x1 2x2);
_
1(s) = @x@H1 = (2a )( 2 2ax1 ) = (2a ) 2;
_
2(s) = @H
@x2 = (2a )( 1+ 2ax2 ) = (2a ) 1;
_ (s) = @H @t = 0;
where the dot denotes dsd. We let s run along the ray from 0 to a point 2 C. Because of group invariance we need to consider paths relative to the origin and a point (x; t) = (x1; x2; t)only, and assume boundary conditions
(5.2) x1(0) = 0; x2(0) = 0; x1( ) = x1; x2( ) = x2; t( ) = t:
Then it is easy to see that the Hamiltonian, 1 2_x 2 1(s) + 1 2_x 2 2(s) = H(x; ; ) = H0 1 2( 1(0) 1(0) + 2(0) 2(0)):
is constant along a given bicharacteristic. The projection of the bicharacteristic curve onto the base is a subRiemannian geodesic connecting the point (x; t) to the origin.
>From (5.1), we know that (s) = (0) = and we may take it to be the free parameter. Equations (5.1) imply that
_ 1 = _1+ 2a _x2 = 2a 2+ 2a 2 = 4a 2; _ 2 = _2 2a _x1 = 2a 1 2a 1 = 4a 1: Hence, 1(s) = cos(4a s) 1(0) + sin(4a s) 2(0); 2(s) = sin(4a s) 1(0) + cos(4a s) 2(0):
Therefore, we may solve for x(s) as a function of x, and , and then solve for t(s) as a function of x, t, and . Here are the calculations.
x1(s) = Z s 0 1( )d = 1 4a f 2(s) 2(0)g = 1 4a f sin(4a s) 1(0) + [cos(4a s) 1] 2(0)g = sin(2a s) 2a fcos(2a s) 1(0) + sin(2a s)g and x2(s) = 1 4a f 1(s) 1(0)g = 1 4a f[cos(4a s) 1] 1(0) + sin(4a s) 2(0)g = sin(2a s) 2a f cos(2a s) 1(0) + sin(2a s)g Therefore, 2 4 1(0) 2(0) 3 5 = 2a sin(2a ) 2 4 cos(2a ) sin(2a ) sin(2a ) cos(2a ) 3 5 2 4 x1 x2 3 5 It follows that H0 = 1 2( 1(0) 1(0) + 2(0) 2(0)) = (2a )2 2 sin2(2a )(x 2 1+ x 2 2) = (2a )2 2 sin2(2a )kxk 2:
When = 0, we have (s) = (0), x(s) = (0)s and t(s) = t(0). Substituting these calculations into (5.1), we have
t t(s) = 2aRs [ 1( )x2( ) 2( )x1( )] d = 21 Rs [1 cos(4a )]d 21(0) + 22(0) = ( s)sin2a2(2a2 )kxk2 a 2 sin(4a ) sin(4a s) sin2(2a ) kxk2:
Theorem 5.1. The solution of equations (5.1) with boundary conditions (5.2) is
(5.3)
x1(s) = sin(2asin(2a s))fcos[2a (s )]x1 + sin[2a (s )]x2g ;
= sin(2asin(2a s))f sin[2a (s )]x1 + cos[2a (s )]x2g ;
= ha2sin(4asin2) sin(4a s)(2a ) ( s) 2a 2 sin2(2a )
i (x2
1+ x22) t:
The value of the Hamiltonian H on this path is H0 =
2a2 2 sin2(2a )(x
2
Next (5.3) yields
t t(0) = a (2a )kxk2; where we set
(z) = z
sin2z cot z: The action integral associated to the Hamiltonian curve is
S(x; t; ; ) = Z 0 ( 2 X j=1 j(s) _xj(s) + _t(s) H(x(s); (s); ) ) ds:
H is homogeneous of degree 2 with respect to ( 1; 2; ), so
(5.4) S = Z 0 ( 2 X j=1 j @H @ j + @H @ H ) ds = Z 0 (2H H)ds = H0:
>From formulas (5.3), we have the following theorem: Theorem 5.2. The action integral S(x; t; ; ) is given by
S(x; t; ; ) = (2a ) 2 2 sin2(2a )kxk 2; = [t t(0)] + a cot(2a )(x21+ x22); 2 h 0; a :
It is convenient to …x , = 1. Then the Hamiltonian paths are determined entirely by the parameter . We may take the end points to be (0; 0) and (x; t). Then must satisfy
t = a (2a )(x21+ x22) = a (2a )kxk2:
It can be shown that is a monotone increasing di¤eomorphism of the interval ( ; ) onto R. On each interval (m ; (m + 1) ), m = 1; 2; : : : , has a unique critical point zm. On this interval decreases strictly from +1 to (zm) and then
increases strictly from (zm) to +1. Now the complete picture of the geodesics in
given in the following two theorems.
Theorem 5.3. There are …nitely many geodesics that join the origin to (x; t) if and only if x 6= 0. These geodesics are parametrized by the solutions of
and their lengths increase strictly with . There is exactly one such geodesic if and only if
jtj < a (z1)kxk2;
and the number of geodesics increases without bound as akxkjtj2 ! 1.
The square of the length of the geodesic associated to a solution of (5.5) is
(5.6)
2S(x;jtj; 1; ) = sin(2a )2(2a )2 (x21+ x22)
= sin(2a )2(2a )2
(x21+x22) (x2 1+x22)+jtj=a h jtj a + (x 2 1+ x22) i
= sin(2a )2(2a )2 1 1+ (2a ) h jtj a + (x 2 1+ x22) i = (2a ) jtja +kxk2 ;
where (0) = 2 and otherwise (z) = z 2 sin2z 1 1 + (z) = z2
z + sin2z sin z cos z:
Consequently, if 2a 2 (m ; (m + 1) ) the length d of the geodesic satis…es m2 2 (m + 1) + 2 jtj a +kxk 2 < (d )2 < (m + 1) 2 2 m jtj a +kxk 2 :
When x = 0, we need to …nd the Hamiltonian paths connecting the origin to (0; t), i.e., x1(1) = 0, x2(1) = 0, t(1) = t. This implies that 1(1) = 1(0)and 2(1) = 2(0).
It follows that
1(1) = cos(4a ) 1(0) + sin(4a ) 2(0) = 1(0); 2(1) = sin(4a ) 1(0) + cos(4a ) 2(0) = 2(0)
This implies that
sin(4a ) = 0; and cos(4a ) = 1 i.e., 2a = m ; with m = 1; 2; 3; : : : : In this case, t = 1 2 ( 2 1(0) + 2 2(0));
therefore, 6= 0 and m 6= 0 in (5.4). We also know that d2m = m jtj
a : Summarizing, we have the following theorem.
Theorem 5.4. The geodesics that join the origin to a point (0; 0; t) have lengths d1,
d2, d3,..., where
d2m = m jtj a :
Since x1(1) = x2(1) = 0, we may use ( 1(0); 2(0)) to obtain the geodesics as
follows: x(m)1 (s) = 1 2m f sin(2m s) 1(0) + [cos(2m s) 1] 2(0)g = t 4am 1 2 sin(2m s) 1(0) k (0)k + [1 cos(2m s)] 2(0) k (0)k ; where k (0)k = q 2 1(0) + 2 2(0). Similarly, we have x(m)2 (s) = 1 2m f[cos(2m s) 1] 1(0) + sin(2m s) 2(0)g = t 4am 1 2 [cos(2m s) 1] 1(0) k (0)k + sin(2m s) 2(0) k (0)k ; and t(m)(s) = [2m s sin(2m s)] t 2m :
This shows that for each …xed m, m = 1; 2; : : : , the geodesics (x(m)1 (s); x(m)2 (s); t(m)(s))
can be parametrized by a unit vector (0)=k (0)k on the unit circle. These curves lie in a cylinder around the t-axis whose radius is O(1=pm).
A special case of (5.6) is the square of the Carnot-Caratheodory distance [dc(x; t)]2:
[dc(x; t)]2 = 2S(x;jtj; 1; c) = 2a c sin(2a c) 2 kxk2 = (2a c) jtj a +kxk 2 ;
where c is the unique solution of a (2a )kxk2 =jtj in the interval [0; =2a).
Intro-duce a new parameter = 2a c. Then the Carnot-Caratheodory distance between
the origin and point (x1; x2; t) can be expressed as
dc(x; t) =
sin kxk with a ( )kxk
2 =
jtj and 2 [0; ): We will compute Hdc(x; t). In polar coordinates,
H = 1 2( @2 @r2 + 1 r @2 @r@ + 1 r2 @2 @ 2) + 2a @2 @t@ + 2a 2r2 @ 2 @t2:
Since dc(x; t)depends only on r = kxk = p x2 1+ x22, we have Hdc(x; t) = ( 1 2 @2 @r2 + 2a 2r2 @2 @t2)dc(r; t):
Introduce a new variable u = jtj=ar2, then dc(r; t) := fc(r; u) =
sin r where u satis…es u = ( ) =
sin cos sin2 : Hence Hdc(r; t) = ( 1 2 @2 @r2 + 2 r2 @2 @u2)fc(r; u) = 2 r @2 @u2(sin );
where u is given by u = ( ). Let g( ) = sin . Then dg du = dg d d du and d2g du2 = d2g d 2 ( d du) 2+ dg d d2 du2: We next compute ddg, dd2g2, d du and d2 du2. dg d = sin cos sin2 and d2g d 2 =
(1 + cos2 ) 2 sin cos
sin3 : Next u = ( ) implies 1 = 0( )d du; d du = 1 0( ) and d2 du2 = 00( ) ( 03 :
We now compute 0( ) and 00( ) from ( ) = csc2 cot .
02 2 csc2 cot + csc2 = 2 csc2 (1 cot ):
and
002 cot (1 cot ) + 2 csc2 ( csc2 cot )
= 2 csc2 [ (3 cot2 + 1) 3 cot ] We …nally compute hfc(r; u) = 2r d 2 du2g( ). Hfc(r; u) = 2 r d2g d 2 ( d du) 2 + dg d d2 du2 = 2 r d2g d 2 1 ( 02 dg d 00( ) ( 03 = 2 r( 02 d2g d 2 dg d 00( ) 0( )
We shall compute the term in [: : : ] in term of …rst. d2g d 2 dg d 00( ) 0( ) = (1 + cos 2 ) 2 sin cos sin3 sin cos sin2 2 csc2 [ (3 cot2 + 1) 3 cot ] 2 csc2 (1 cot ) = (1 + cos 2 ) 2 sin cos sin3 (3 cot2 + 1) 3 cot sin = (1 + cos
2 ) 2 sin cos (3 cos2 + sin2 ) + 3 cos sin
sin3 =sin cos cos
2 sin3 Hence we have Hfc(r; u) = 2 r( 02 d2g d 2 dg d 00( ) 0( )
= sin cos cos
2
sin3
1
2r csc4 (1 cot )2
= (1 cot ) sin cos 2r csc (1 cot )2 = sin 2 cos 2r(1 cot ) Since dc = sin r, (5.7) Hdc = 1 2dc sin2 cos sin cos :
We next study the function F ( ) = 2(sinsin2 coscos ) where is given by ar2 ( ) = t with ( ) = sin cos
sin2 :
The function F ( ) is smooth on the interval [0; ], decreasing from [0; m] and
in-creasing from [ m; ]. m is the unique critical point of F ( ) inside the interval (0; ). F (0) = 3, F ( =2) = F ( ) = 0.
As r ! 0 with t > 0 …xed, ! and the equation ar2 ( ) = t implies
ar2 t
sin cos
sin2 = 1: This shows that
! and sin (a
t )
1=2r as r
This implies (5.7) makes sense when r = 0. This corresponds to = . All these imply
bdc = Hdc
3 dc
and then the Sub-Laplacian comparison ( ) follows.
We now turn to the study of the (2n + 1)-dimensional Heisenberg group Hn. The
manifold is R2n R and the group law is given by (x; t) (y; s) = (x + y; t + s + 2
n
X
j=1
aj[x2jy2j 1 x2j 1y2j])
for a1; a2; ; an are positive constants and numbered so that
0 < a1 a2 an:
The vector …elds
X2j 1 = @ @x2j 1 + 2ajx2j @ @t X2j = @ @x2j 2ajx2j 1 @ @t T = @ @t
are left-invariant and generate the Lie algebra. The associated Heisenberg sub-Laplacian is H = 1 2 2n X j=1 Xj2: The symbol of H is H(x; ; ) = 1 2 n X j=1 [( 2j 1+ 2ajx2j )2+ ( 2j 2ajx2j 1 )2 = 1 2( 2 1+ 2 2):
We can …nd the bicharacteristic curve connecting the point (x; t) to the origin by solving the associated Hamilton’s equations which take essentially the same form as (5.1). We will just list the formulae that we need and refer to [BGG] for details. The value of the Hamiltonian H on the bicharacteristic curve is the constant:
H0 = n X j=1 2a2j 2 sin2(2aj ) rj2
with r2
j = x22j 1 + x22j. The analogue of (5.5) is follows:
t =
n
X
j=1
aj (2aj )r2j:
The action integral S(x; t; ; ) takes a similar form: S(x; t; ; ) = n X j=1 4 a2j 2 sin2(2aj ) r2j = t + n X j=1 aj cot(2aj )r2j:
When we study the classical action and Carnot-Caratheodory distance, we set = 1. In the case of x 6= 0, there are …nitely many geodesics from the origin to (x; t). The geodesics are indexed by the solutions of
(5.8) jtj =
n
X
j=1
aj (2aj )r2j
and their lengths increase with . The Carnot-Caratheodory distance from the origin to (x; t) is
d2(x; t) = 2S(x;jtj; 1; c)
where c is the unique solution of (5.8) in the interval [0; =2an).
In the isotropic case a1 = a2 = = an, the results of the previous computations
for n = 1 carry over with no change.
6. Subgradient Estimate on Higher Dimensional Pseudohermitian Manifolds
Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold for n 2. In this sec-tion, we derive the CR version of parabolic Li-Yau gradient estimate for the positive solution u(x; t) of (1.4) on M [0;1) for n 2.
First, we derive the following inequalities which we need in the proof of Proposition 1.1.
Lemma 6.1. Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold. Let f be a smooth real-valued function on M: Then
j(rH)2fj2 2 n X ; =1 jf j2+ 2 n X ; =1 6= jf j2+ 1 2 n X =1 jf + f j2:
Proof. Since j(rH)2f j2 = 2Pn ; =1(f f + f f ) = 2Pn; =1(jf j2+ jf j2) = 2(Pn; =1jf j2+Pn ; =1 6= jf j 2+Pn =1jf j2) and Pn =1jf j 2 = 1 4 Pn =1(jf + f j 2+ f2 0) = 14Pn=1jf + f j2+ n 4f 2 0: It follows that j(rH)2fj2 = 2(Pn; =1jf j2+Pn; =1 6= jf j 2) + 1 2 Pn =1jf + f j 2+ n 2f 2 0 2(Pn; =1jf j2+Pn; =1 6= jf j 2) + 1 2 Pn =1jf + f j 2:
Lemma 6.2. Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold for n 2. Let f be a smooth real-valued function on M: Then
rbf;rbjrbfj2 (n + 2) n X ; =1 jf j2 + (n + 2) n X ; =1 6= jf j2 + bf +jrbfj 2 jrbfj 2 +(n + 2)(n 1) 4(n + 1) n X =1 jf + f j2: Proof. We …rst derive (6.1) hrbf;rbjrbfj2i = 4Pn; =1Re(f f f + f f f ) = 4 Re(Pn; =1f f f +Pn; =1 6= f f f ) + 2Pn=1(f + f )jf j2 (n + 2)(Pn; =1jf j2+Pn ; =1 6= jf j 2) + 4 n+2 Pn ; =1jf j2jf j2 +n+24 Pn; =1 6= jf j 2 jf j2 + 2Pn =1(f + f )jf j2 = (n + 2)(Pn; =1jf j2+Pn ; =1 6= jf j 2) + 1 n+2jrbfj 4 +n+24 Pn; =1 6= jf j 2 jf j2 + 2Pn =1(f + f )jf j2:
Now we compute the last term in the above inequality. Pn =1(f + f )jf j 2 = [Pn=1(f + f )](Pn=1jf j2) Pn =1(f + f )( Pn =1 6= jf j 2) 1 2 bfjrbfj 2 +Pn=1jf + f j(Pn=1 6= jf j 2) 1 2 bfjrbfj 2 + (n 1)(n+2)8(n+1) Pn=1jf + f j2 +(n 1)(n+2)2(n+1) Pn=1(Pn=1 6= jf j 2)2:
Substituting the above inequality into (6.1), one obtains
hrbf;rbjrbfj2i (n + 2)(Pn; =1jf j2 +Pn; =1 6= jf j 2) + bfjrbfj2 +(n 1)(n+2)4(n+1) Pn=1jf + f j2 +n+21 jrbfj4 +(n+2)(n 1)4(n+1) Pn=1(Pn=1 6= jf j 2)2+ 4 n+2 Pn ; =1 6= jf j 2 jf j2 (n + 2)(Pn; =1jf j2 +Pn ; =1 6= jf j 2) + bfjrbfj 2 +(n+2)(n 1)4(n+1) Pn=1jf + f j2 + 1 n+2jrbfj 4 +(n+2)(n 1)4(n+1) Pn=1(Pn=1 6= jf j 2)2+Pn ; =1 6= jf j 2 jf j2 = (n + 2)(Pn; =1jf j2 +Pn ; =1 6= jf j 2) + bfjrbfj2 +(n+2)(n 1)4(n+1) Pn=1jf + f j2 + jrbfj4:
Here we have used the identity
Pn =1 Pn =1 6= jf j 2 2 +Pn; =1 6= jf j 2 jf j2 = (n 1) (Pn =1jf j 2)2 = n 1 4 jrbfj 4 :
This completes the proof of the Lemma.
Now we can derive the following Proposition 1.1 which is exact form of Lemma 3.3 for n 2.
Proof. First di¤erentiating (1.5) w.r.t. the t-variable, we have Gt = 1 tG + t[jrb'j 2 + (1 + 2 n)'t]t = 1 tG + t[2(1 + 1 n)jrb'j 2 + (1 + 2 n) b']t (6.2) = 1 tG + t[4(1 + 1 n)hrb';rb'ti + (1 + 2 n) b't]:
By using the CR version of Bochner formula (2.1) and Lemma 3.2, one obtains
bG = t bjrb'j 2 + (1 + 2 n) b't = t[2j(rH)2'j2+ 2(1 + 2 n)hrb';rb b'i +2[2Ric (n + 2)T or]((rb')C; (rb')C) 8 nhP ' + _ P '; db'iL + (1 + 2 n) b't] t[2j(rH)2'j2+ 2(1 + 2 n)hrb';rb b'i l0jrb'j 2 (6.3) 8 nhP ' + _ P '; db'iL + (1 + 2 n) b't] = t[2j(rH)2'j2+ 2(1 + 2 n)hrb';rb b'i l0jrb'j 2 8 nu 2 P u + P u; d bu L + 4 n'tjrb'j 2 +4 n rb';rbjrb'j 2 + (1 + 2 n) b't]: Here we have used the inequalities
[2Ric (n + 2)T or]((rb')C; (rb')C) l0j(rb')Cj2 = l0 2 jrb'j 2 and 't= ut u = bu u : Applying the formula
(6.4) b' = 't jrb'j2 = n (n + 2)tG 2(n + 1) n + 2 jrb'j 2
and combining (6.2), (6.3), we conclude b @ @t G 1 tG + t[2j(r H )2'j2+ 2(1 + 2 n)hrb';rb b'i + 4 n rb';rbjrb'j 2 4(1 + 1 n)hrb';rb'ti l0jrb'j 2 + 4 n'tjrb'j 2 8 nu 2 P u + P u; d bu L ] = 1 tG 2n (n + 2)hrb';rbGi + t[2j(r H)2' j2 4 n + 2 rb';rbjrb'j 2 l0jrb'j2+ 4 n'tjrb'j 2 8 nu 2 P u + P u; d bu L ]:
Now, by Lemma 6.1, Lemma 6.2, Cauchy-Schwarz inequality and applying the for-mula (6.4), we …nal have
b @ @t G 2n (n + 2)hrb';rbGi + t[ 2 n + 1 n X =1 j' + ' j2+ 8 n(n + 2)'tjrb'j 2 l0jrb'j 2 8 nu 2 P u + P u; d bu L ] 1 tG 2n (n + 2)hrb';rbGi + t[ 2 n(n + 1)( b') 2+ 8 n(n + 2)'tjrb'j 2 l0jrb'j2 8 nu 2 P u + P u; d bu L ] 1 tG = 2n (n + 2)hrb';rbGi + t[ 2n (n + 1)(n + 2)2t2G 2+ 8 n(n + 2)2jrb'j 4 l0jrb'j2 8 nu 2 P u + P u; d bu L ] 1 tG This completes the proof of Proposition 1.1.
Following the same proof as in Theorem 3.4. We have the following result.
Theorem 6.3. Let (M; J; ) be a closed pseudohermitian (2n + 1)-manifold of zero torsion and nonnegative Tanaka-Webster curvature for n 2. If u(x; t) is the positive solution of (1.4) on M [0;1) such that
at t = 0: Then u satis…es the estimate jrbuj 2 u2 + n + 2 n ut u (n + 1)(n + 2)2 2n 1 t on M [0;1):
Following the same proof as in Theorem 4.2. We have the following result. Theorem 6.4. If u(x; t) be a positive smooth solution of (1.4)
b
@
@t u(x; t) = 0 on Hn [0; T ) with
P u = 0 at t = 0; then u satis…es the subgradient estimate
jrbuj 2 u2 + n + 2 n ut u [ (n + 1)(n + 2)2 2n + ] 1 t on Hn [0; T ) for any > 0:
Then by combining the Theorem 6.4 and Proposition 4.1, Theorem 1.3 follows easily for all n.
Appendix A.
We will give a brief introduction to pseudohermitian geometry (see [L1] for more details). Let (M; J; ) be a complete pseudohermitian 3-manifold, where is a contact form and J is a CR structure compatible with the contact bundle = ker . The CR structure J decomposes C into the direct sum of T1;0 and T0;1 which are
eigenspaces of J with respect to i and i, respectively. The Levi form h ; i is the Hermitian form on T1;0 de…ned by hZ; W i = i d ; Z^ W . We can extend h ; i
to T0;1 by de…ning Z; W = hZ; W i for all Z; W 2 T1;0. The Levi form induces
naturally a Hermitian form on the dual bundle of T1;0, and hence on all the induced
tensor bundles. Integrating the hermitian form (when acting on sections) over M with respect to the volume form d = ^ d , we get an inner product on the space of sections of each tensor bundle.
Let fT; Z1; Z1g be a frame of T M C, where Z1 is any local frame of T1;0; Z1 =
Z1 2 T0;1and T is the Reeb vector …eld of : That is, the unique vector …eld such that
(T ) = 1; d (T; ) = 0. Then n ; 1; 1o, the coframe dual to fT; Z1; Z1g, satis…es
(A.1) d = ih11 1^ 1
for some positive function h11. Actually we can always choose Z1 such that h11= 1;
hence, throughout this paper, we assume h11= 1
The pseudohermitian connection of (J; ) is the connection r on T M C (and extended to tensors) given in terms of a local frame Z1 2 T1;0 by
rZ1 = 11 Z1; rZ1 = 11 Z1; rT = 0;
where 11 is the 1-form uniquely determined by the following equations:
d 1 = 1^ 11+ ^ 1 1
0 mod 1 0 = 11+ 11;
(A.2)
where 1 is the pseudohermitian torsion. Put 1 = A1
1 1. The structure equation
for the pseudohermitian connection is
(A.3) d 11 = W 1^ 1+ 2iIm(A11;1 1^ );
where W is the Tanaka-Webster curvature.
We will denote components of covariant derivatives with indices preceded by comma; thus write A1
1;1 1 ^ . The indices f0; 1; 1g indicate derivatives with respect to
fT; Z1; Z1g. For derivatives of a scalar function, we will often omit the comma,
for instance, '1 = Z1'; '11= Z1Z1' 11(Z1)Z1'; '0 = T 'for a (smooth) function.
For a real function ', the subgradient rb is de…ned by rb' 2 and hZ; rb'i =
d'(Z) for all vector …elds Z tangent to contact plane. Locally rb' = '1Z1+ '1Z1.
We can use the connection to de…ne the subhessian as the complex linear map
and
(rH)2'(Z) =rZrb'
and the sublaplacian b de…ned as the trace of the subhessian b' = T r (rH)2' = ('11+ '11):
Finally de…ne, for all Z = x1Z1 2 T1;0,
Ric(Z; Z) = W x1x1
T or(Z; Z) = 2Re iA11x1x1:
Let (M; J; ) be a complete pseudohermitian (2n + 1)-manifold with n 2:Then for ; ; n=1, the coframe dual to fT; Z ; Z gn=1, satis…es
d = ih ^
for some hermitian matrix of functions h :Then (A.2) becomes
d = ^ + ^
0 mod
dh = + ;
where is the pseudohermitian torsion and = A . For the curvature forms
= d ^ ;
satisfy the structure equation
(A.4) = R ^ + 2iIm(W ^ ) + i ^ i ^ :
Here the pseudohermitian Ricci tensor is
R = R
and the pseudohermitian scalar (Tanaka-Webster curvature) is W = R :
= d = R ^ + 2iIm(W ^ ): Note that
W = A ; :
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1Department of Mathematics, National Tsing Hua University, Hsinchu, Taiwan
30013, R.O.C.
E-mail address: [email protected]
2Department of Mathematics, University of Georgia, Athens, GA 30602-7403, U.S.A.
E-mail address: [email protected]
3Department of Applied Mathematics, National PingTung University of
Educa-tion, PingTung, Taiwan 90003, R.O.C. E-mail address: [email protected]
