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Gradient Estimates For The CR Heat Equation On Complete Noncompact Pseudohermitian Manifolds111 This work is supported by NSFC Grants No.12171091.

Yuxin Dong, Yibin Ren and Biqiang Zhao222 E-mail addresses: [email protected]
Abstract

In this paper, we derive local and global Li-Yau type gradient estimates for the positive solutions of the CR heat equation on complete noncompact pseudo-Hermitian manifolds. As applications of the gradient estimates, we give a Harnack inequality for the positive solutions of the CR heat equation, and then obtain an upper bound estimate for the corresponding heat kernel.

1 Introduction

The method of gradient estimates is an important tool in geometric analysis, which is originated first in Yau [30] and Cheng and Yau [13] for investigating harmonic functions and further developed in Li and Yau [25] for studying solutions of the heat equation on complete Riemannian manifolds. In [25], Li and Yau established their celebrated parabolic inequality, which asserts that, if MM is an nn-dimensional complete Riemannian manifold with Ricci curvature bounded from below by K-K, for some constant K0K\geq 0 and uu is any positive solution of the heat equation

(t)u=0,(\bigtriangleup-\frac{\partial}{\partial t})u=0,

then

|u|2u2αutunα22t+nα22(α1)K\frac{|\nabla u|^{2}}{u^{2}}-\alpha\frac{u_{t}}{u}\leq\frac{n\alpha^{2}}{2t}+\frac{n\alpha^{2}}{2(\alpha-1)}K

for all t>0t>0, α>1\alpha>1. In particular, if K=0K=0, then the following more precise inequality holds

|u|2u2utun2t.\frac{|\nabla u|^{2}}{u^{2}}-\frac{u_{t}}{u}\leq\frac{n}{2t}.

Since then many improvements or generalizations of Li-Yau’s parabolic inequality have been developed on Riemannian manifolds or more general metric measure spaces, see e.g. [17], [22], [4], [7], [3], [27], [24], [2], [32], [12], [33], [31] and the references therein.

The Li-Yau’s inequality has also been generalized for non elliptic operators that include subelliptic operators on sub-Riemannian manifolds, see e.g. [8], [1], [5], [6][19], [11], [10], etc. Some of these works concern Li-Yau type inequalities on pseudo-Hermitian manifolds. In [11], Chang et al. derived a CR Li-Yau type estimate in terms of the lower bound of pseudo-Hermitian Ricci curvature essentially for closed Sasakian 3-manifolds. In [5], Baudoin and Garofalo proved, among other results, a CR Li-Yau type inequality on complete Sasakian manifolds under a curvature dimension inequality. In [9], the authors announced a CR Li-Yau gradient estimate by using a generalized curvature-dimension inequality and the maximum principle in a closed pseudo-Hermitian manifold possibly with nonvanishing torsion. Besides, they also established a Li-Yau type inequality for the sum of squares of vector fields up to higher step on a closed manifold, generalizing Cao-Yau’s result ([8]) for operators expressed as the sum of squares of vector fields of step 2. However, we don’t understand their proof for the CR part. Anyhow Cao-Yau’s inequality in [8] almost gave us in particular a Li-Yau type inequality for closed pseudo-Hermitian manifolds (See Remark 3.1).

Let’s recall briefly Cao-Yau’s work in [8] as follows. Suppose X1,.,X_{1},...., XnX_{n} are smooth vector fields on a closed manifold MM and

L=i=1nXi2X0L=\sum_{i=1}^{n}X_{i}^{2}-X_{0} (1.1)

with X0=i=1nciXiX_{0}=\sum_{i=1}^{n}c_{i}X_{i}, where cic_{i} are smooth functions on MM. Suppose X1,,XnX_{1},...,X_{n} satisfy the following conditions: for 1i,j,kn1\leq i,j,k\leq n, [Xi,[Xj,Xk]][X_{i},[X_{j},X_{k}]] can be expressed as linear combinations of X1,,XnX_{1},...,X_{n} and their brackets [X1,X2],.,[X_{1},X_{2}],....,
[Xn1,Xn][X_{n-1},X_{n}]. Cao and Yau considered a positive solution u(x,t)u(x,t) of

(Lt)u=0\left(L-\frac{\partial}{\partial t}\right)u=0 (1.2)

on M×(0,)M\times(0,\infty) and showed that there exists a constant δ0>0\delta_{0}>0 such that for any δ>δ0\delta>\delta_{0}, uu satisfies

1u2i=1n|Xiu|2δX0uuδutuC1t+C2,\frac{1}{u^{2}}\sum_{i=1}^{n}|X_{i}u|^{2}-\delta\frac{X_{0}u}{u}-\delta\frac{u_{t}}{u}\leq\frac{C_{1}}{t}+C_{2}, (1.3)

where C1C_{1} and C2C_{2} are positive constants depending on nn, δ0\delta_{0}, δ\delta and {Xi}\{X_{i}\}.

This paper is devoted to establish a Li-Yau type inequality on a complete pseudo-Hermitian manifold possibly with nonvanishing pseudo-Hermitian torsion. The pseudo-Hermitian manifolds considered here are CR manifolds of hypersurface type which admit positive definite pseudo-Hermitian structures (see §2 for the detailed definition). Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) denote a pseudo-Hermitian manifold of dimension 2m+12m+1. Here (HM,J)(HM,J) is a CR structure of type (m,1)(m,1), and θ\theta is a pseudo-Hermitian structure on MM. We find that a pseudo-Hermitian manifold carries a rich geometric structure, including an almost complex structure JJ on HMHM, the positive definite Levi form LθL_{\theta} on HMHM induced from θ\theta and JJ, the Webster metric (a Riemannian metric on MM extending LθL_{\theta}), the Reeb vector field ξ\xi on MM, the sub-Laplacian b\bigtriangleup_{b} (a subelliptic differential operator) and the horizontal gradient operator b\nabla_{b} acting on functions. Note also that the pair (HM,Lθ)(HM,L_{\theta}) is a 22-step sub-Riemannian structure, which induces a Carnot–Carathédory distance dccd_{cc} on MM. These geometric data provide us a basis to investigate Li-Yau type inequality on a pseudo-Hermitian manifold. We will consider a positive solution of the following CR heat equation

ut=bu\frac{\partial u}{\partial t}=\bigtriangleup_{b}u (1.4)

on a complete pseudo-Hermitian manifold, and establish a Li-Yau type inequality for uu. The main ingredients of Li-Yau’s method [25] or Cao-Yau’s method ([8]) involve the Bochner type formula, a parabolic differential inequality for a suitable auxiliary function and the maximum principle. For any smooth function ff on the pseudo-Hermitian manifold, one has two CR Bochner formulas for |bf|2|\nabla_{b}f|^{2} and f02f_{0}^{2} respectively, where f0=ξ(f)f_{0}=\xi(f) (see (2.8) and (2.9) in §2). Now set f=lnuf=\ln u. Following Cao-Yau’s idea, we will consider the auxiliary functions

=t(|bf|2+t2λ1(1+f02)λδft)\mathcal{F}=t\left(|\nabla_{b}f|^{2}+t^{2\lambda-1}\left(1+f_{0}^{2}\right)^{\lambda}-\delta f_{t}\right) (1.5)

or

𝒢=t(|bf|2+(1+f02)λδft)\mathcal{G}=t\left(|\nabla_{b}f|^{2}+\left(1+f_{0}^{2}\right)^{\lambda}-\delta f_{t}\right) (1.6)

according to the ranges of tt. Some parabolic differential inequalities for \mathcal{F} and 𝒢\mathcal{G} can be derived from the CR Bochner formulas. Following the technique in [25], we may multiply \mathcal{F} and 𝒢\mathcal{G} by a suitable cut-off function ϕ\phi to localize the problem. By applying the maximum principle to ϕ\phi\mathcal{F} and ϕ𝒢\phi\mathcal{G}, and using the CR sub-Laplacian comparison theorem in [15], we are able to establish the following local Li-Yau gradient estimate.

𝐓𝐡𝐞𝐨𝐫𝐞𝐦 1.1\mathbf{Theorem\ 1.1} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbk,and|A|,|bA|k1,\displaystyle Ric_{b}+2(m-2)Tor_{b}\geq-k,\ and\ |A|,|\nabla_{b}A|\leq k_{1},

and uu be a positive solution of the CR heat equation

ut=Δbu\displaystyle\frac{\partial u}{\partial t}=\Delta_{b}u

on Bp(2R)×(0,)withR1B_{p}(2R)\times(0,\infty)\ with\ R\geq 1, where Bp(r)B_{p}(r) denotes the Riemannian ball of radius rr with respect to the Webster metric gθg_{\theta}. Then for any constant 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3} and any constant δ>1+4mλ(2λ1)\delta>1+\frac{4}{m\lambda(2\lambda-1)}, there exists a constant CC depending on m,k,k1,λ,δm,k,k_{1},\lambda,\delta, such that

|bu|2u2δutuC(1+1t+1Rλ+1tRλ)\displaystyle\frac{|\nabla_{b}u|^{2}}{u^{2}}-\delta\frac{u_{t}}{u}\leq C(1+\frac{1}{t}+\frac{1}{R^{\lambda}}+\frac{1}{tR^{\lambda}}) (1.7)

on Bp(R)×(0,)B_{p}(R)\times(0,\infty).

Letting RR\rightarrow\infty in Theorem 1.1, we get immediately the global Li-Yau type gradient estimate.

𝐓𝐡𝐞𝐨𝐫𝐞𝐦 1.2\mathbf{Theorem\ 1.2} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbk,and|A|,|bA|k1,\displaystyle Ric_{b}+2(m-2)Tor_{b}\geq-k,\ and\ |A|,|\nabla_{b}A|\leq k_{1},

and uu be a positive solution of the heat equation

ut=Δbu\displaystyle\frac{\partial u}{\partial t}=\Delta_{b}u

on M×(0,)M\times(0,\infty). Then for any constant 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3} and any constant δ>1+4mλ(2λ1)\delta>1+\frac{4}{m\lambda(2\lambda-1)}, there exists a constant CC depending on m,k,k1,λ,δm,k,k_{1},\lambda,\delta, such that

|bu|2u2δutuC+Ct\displaystyle\frac{|\nabla_{b}u|^{2}}{u^{2}}-\delta\frac{u_{t}}{u}\leq C+\frac{C}{t} (1.8)

on M×(0,)M\times(0,\infty).

As applications of the above gradient estimates, we give a Harnack inequality for the positive solutions of the CR heat equation, and then obtain an upper bound estimate for the heat kernel of the CR heat equation.

𝐓𝐡𝐞𝐨𝐫𝐞𝐦 1.3\mathbf{Theorem\ 1.3} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbk,and|A|,|bA|k1,\displaystyle Ric_{b}+2(m-2)Tor_{b}\geq-k,\ and\ |A|,|\nabla_{b}A|\leq k_{1},

and uu be a positive solution of the heat equation

ut=Δbu\displaystyle\frac{\partial u}{\partial t}=\Delta_{b}u

on M×(0,)M\times(0,\infty). Then for any constant 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3} and any constant δ>1+4mλ(2λ1)\delta>1+\frac{4}{m\lambda(2\lambda-1)}, there exists a constant CC which is given by Theorem 1.2 such that for any 0<t1<t20<t_{1}<t_{2} and x,yMx,y\in M, we have

u(x,t1)u(y,t2)(t2t1)Cδexp(Cδ(t2t1)+δdcc2(x,y)4(t2t1)).\displaystyle u(x,t_{1})\leq u(y,t_{2})(\frac{t_{2}}{t_{1}})^{\frac{C}{\delta}}exp(\frac{C}{\delta}(t_{2}-t_{1})+\frac{\delta d_{cc}^{2}(x,y)}{4(t_{2}-t_{1})}). (1.9)

𝐓𝐡𝐞𝐨𝐫𝐞𝐦 1.4\mathbf{Theorem\ 1.4} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbk,and|A|,|bA|k1,\displaystyle Ric_{b}+2(m-2)Tor_{b}\geq-k,\ and\ |A|,|\nabla_{b}A|\leq k_{1},

and H(x,y,t)H(x,y,t) be the heat kernel of (1.4). Then for any constants 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3}, δ>1+4mλ(2λ1)\delta>1+\frac{4}{m\lambda(2\lambda-1)} and 0<ϵ<10<\epsilon<1, there exists constants CandC′′C^{{}^{\prime}}\ and\ C^{{}^{\prime\prime}} depending on m,k,k1,λ,δ,ϵm,k,k_{1},\lambda,\delta,\epsilon, such that H(x,y,t)H(x,y,t) satisfies

H(x,y,t)C[Vol(Bcc(x,t))]12[Vol(Bcc(y,t))]12exp(C′′ϵtdcc2(x,y)(4+ϵ)t),\displaystyle H(x,y,t)\leq C^{{}^{\prime}}[Vol(B_{cc}(x,\sqrt{t}))]^{-\frac{1}{2}}[Vol(B_{cc}(y,\sqrt{t}))]^{-\frac{1}{2}}exp(C^{{}^{\prime\prime}}\epsilon t-\frac{d^{2}_{cc}(x,y)}{(4+\epsilon)t}), (1.10)

where the Bcc(x,r)B_{cc}(x,r) is the ball with respect to Carnot-Carathe´\acute{e}odory distance. The constant CC^{{}^{\prime}}\rightarrow\infty as ϵ0\epsilon\rightarrow 0.

2 CR Bochner formulas on pseudo-Hermitian manifolds

In this section we introduce some basic notations in pseudo-Hermitian geometry (cf. [18, 29, 28] for details), and then give the CR Bochner formulas for functions on a pseudo-Hermitian manifold. Next, we will derive parabolic differential inequalities for the auxiliary functions \mathcal{F} and 𝒢\mathcal{G}.

Let M2m+1M^{2m+1} be a real 2m+12m+1 dimensional orientable CC^{\infty} manifold. A CR structure on MM is a complex subbundle H1,0MH^{1,0}M of TMTM\otimes\mathbb{C} satisfying

H1,0MH0,1M={0},[Γ(H1,0M),Γ(H1,0M)]Γ(H1,0M)H^{1,0}M\cap H^{0,1}M=\{0\},\ \ [\Gamma(H^{1,0}M),\Gamma(H^{1,0}M)]\subseteq\Gamma(H^{1,0}M) (2.1)

where H0,1M=H1,0M¯H^{0,1}M=\overline{H^{1,0}M}. Equivalently, the CR structure may also be described by the real bundle HM=Re{H1,0MH0,1M}HM=Re\{H^{1,0}M\oplus H^{0,1}M\} and an almost complex structure JJ on HMHM, where J(X+X¯)=1(XX¯)J(X+\overline{X})=\sqrt{-1}(X-\overline{X}) for any XH1,0MX\in H^{1,0}M. Then (M,HM,J)(M,HM,J) is said to be a CR manifold.

We denote by EE the conormal bundle of HMHM in TMT^{*}M, whose fiber at each point xMx\in M is given by

Ex={ωTxM|ω(HxM)=0}.\displaystyle E_{x}=\{\omega\in T_{x}^{*}M|\omega(H_{x}M)=0\}. (2.2)

It turns out that EE is a trivial line bundle. Therefore there exist globally defined nowhere vanishing sections θΓ(E)\theta\in\Gamma(E). A section θΓ(E\{0})\theta\in\Gamma(E\backslash\{0\}) is called a pseudo-Hermitian structure on MM. The Levi form LθL_{\theta} of a pseudo-Hermitian structure θ\theta is defined by

Lθ(X,Y)=dθ(X,JY)\displaystyle L_{\theta}(X,Y)=d\theta(X,JY)

for any X,YHMX,Y\in HM. The integrability condition in (2.1) implies that LθL_{\theta} is JJ-invariant, and thus symmetric. When LθL_{\theta} is positive definite on HMHM for some θ\theta, then (M,HM,J)(M,HM,J) is said to be strictly pseudoconvex. From now on, we will always assume that (M,HM,J)(M,HM,J) is a strictly pseudoconvex CR manifold endowed with θ\theta, such that LθL_{\theta} is positive definite. Then the quadruple (M,HM,J,θ)(M,HM,J,\theta) is referred to as a pseudo-Hermitian manifold.

For a pseudo-Hermitian manifold (M,HM,J,θ)(M,HM,J,\theta), due to the positivity of LθL_{\theta}, we have a sub-Riemannian structure (HM,Lθ)(HM,L_{\theta}) of step-2 on MM. We say that a Lipschitz curve γ:[0,l]M\gamma:[0,l]\rightarrow M is horizontal if γHγ(t)M\gamma^{{}^{\prime}}\in H_{\gamma(t)}M a.e. in [0,l][0,l]. For any two points p,qMp,q\in M, by the well-known theorem of Chow-Rashevsky([16, 26] ), there always exist such horizontal curves joining pp and qq. Therefore we may define the Carnot-Carathe´\acute{e}odory distance as follows:

dcc(p,q)=inf{0lLθ(γ,γ)𝑑t|γΓ(p,q)},\displaystyle d_{cc}(p,q)=inf\{\int_{0}^{l}\sqrt{L_{\theta}(\gamma^{{}^{\prime}},\gamma^{{}^{\prime}})}dt\ |\ \gamma\in\Gamma(p,q)\},

where Γ(p,q)\Gamma(p,q) denotes the set of all horizontal curves joining pp and qq. Clearly dccd_{cc} induces to a metric space structure on MM, in which its metric ball centered at xx with radius rr is given by

Bcc(x,r)={yM|dcc(y,x)<r}.\displaystyle B_{cc}(x,r)=\{y\in M\ |\ d_{cc}(y,x)<r\}.

For a pseudo-Hermitian manifold (M,HM,J,θ)(M,HM,J,\theta), it is clear that θ\theta is a contact form on MM. Consequently there exists a unique vector field ξ\xi such that

θ(ξ)=1,dθ(ξ,)=0.\displaystyle\theta(\xi)=1,\ d\theta(\xi,\cdot)=0. (2.3)

This vector field ξ\xi is called the Reeb vector field. From (2.2) and (2.3), it is easy to see that TMTM admits the following direct sum decomposition

TM=HMRξ,\displaystyle TM=HM\oplus R\xi, (2.4)

which induces a natural projection πb:TMHM\pi_{b}:TM\rightarrow HM. In terms of θ\theta and the decomposition (2.4), the Levi form LθL_{\theta} can be extended to a Riemannian metric

gθ=Lθ+θθ,\displaystyle g_{\theta}=L_{\theta}+\theta\otimes\theta,

which is called the Webster metric. We will denote by rr the corresponding Riemannian distance and by Bp(R)B_{p}(R) the Riemannian ball of radius RR centered at pp. One may extend the complex structure JJ on HMHM to an endomorphism of TMTM, still denoted by JJ, by requiring

Jξ=0.\displaystyle J\xi=0.

It is known that there exists a canonical connection \nabla on a pseudo-Hermitian manifold, called the Tanaka-Webster connection (cf. [18, 28, 29]), such that

1.\displaystyle 1. XΓ(HM)Γ(HM),foranyXΓ(TM);\displaystyle\ \nabla_{X}\Gamma(HM)\subseteq\Gamma(HM),\ for\ any\ X\in\Gamma(TM);
2.\displaystyle 2. gθ=0andJ=0;\displaystyle\ \nabla g_{\theta}=0\ and\ \nabla J=0;
3.\displaystyle 3. T(X,Y)=2dθ(X,Y)ξandT(ξ,JX)+JT(ξ,X)=0,\displaystyle\ T_{\nabla}(X,Y)=2d\theta(X,Y)\xi\ and\ T_{\nabla}(\xi,JX)+JT_{\nabla}(\xi,X)=0,
foranyX,YHM,whereTdenotesthetorsionof\displaystyle for\ any\ X,Y\in HM,\ where\ T_{\nabla}\ denotes\ the\ torsion\ of
theconnection.\displaystyle the\ connection\ \nabla.

The pseudo-Hermitian torsion of \nabla is an important pseudo-Hermitian invariant, which is an HMHM-valued 1-form defined by

τ(X)=T(ξ,X)\displaystyle\tau(X)=T_{\nabla}(\xi,X)

for any XTMX\in TM. Note that is τ\tau trace-free and self-adjoint with respect to the Webster metric gθg_{\theta} (cf. [18]). Set A(X,Y)=gθ(T(ξ,X),Y)A(X,Y)=g_{\theta}(T_{\nabla}(\xi,X),Y) for any X,YTMX,Y\in TM, then we have

A(X,Y)=A(Y,X).\displaystyle A(X,Y)=A(Y,X). (2.5)

We say that MM is Sasakian if τ=0\tau=0 (or equivalently, A=0A=0).

Let (M,HM,J,θ)(M,HM,J,\theta) be a complete pseudo-Hermitian manifold of dimension 2m+12m+1. We choose a local orthonormal frame field {eA}A=02m={ξ,e1,,em,em+1,,e2m}\{e_{A}\}_{A=0}^{2m}=\{\xi,e_{1},\cdots,e_{m},\\ e_{m+1},\cdots,e_{2m}\} with respect to the Webster metric gθg_{\theta} such that

{em+1,,e2m}={Je1,,Jem}.\displaystyle\{e_{m+1},\cdots,e_{2m}\}=\{Je_{1},\cdots,Je_{m}\}.

Set

ηα=12(eα1Jeα),ηα¯=12(eα+1Jeα),(α=1,,m).\displaystyle\eta_{\alpha}=\frac{1}{\sqrt{2}}(e_{\alpha}-\sqrt{-1}Je_{\alpha}),\quad\eta_{\bar{\alpha}}=\frac{1}{\sqrt{2}}(e_{\alpha}+\sqrt{-1}Je_{\alpha}),\ (\alpha=1,\cdots,m).

Then {ηα}α=1m\{\eta_{\alpha}\}_{\alpha=1}^{m} is a unitary frame field of H1,0MH^{1,0}M with respect to gθg_{\theta}. Let {θ1,,θm}\{\theta^{1},\cdots,\theta^{m}\} be the dual frame field of {ηα}α=1m\{\eta_{\alpha}\}_{\alpha=1}^{m}. According to the property 3 of the Tanaka-Webster connection, one may write

τ\displaystyle\tau =\displaystyle= ταηα+τα¯ηα¯\displaystyle\tau^{\alpha}\eta_{\alpha}+\tau^{\bar{\alpha}}\eta_{\bar{\alpha}}
=\displaystyle= Aβ¯αθβ¯ηα+Aβα¯θβηα¯.\displaystyle A_{\bar{\beta}}^{\alpha}\theta^{\bar{\beta}}\otimes\eta_{\alpha}+A_{\beta}^{\bar{\alpha}}\theta^{\beta}\otimes\eta_{\bar{\alpha}}.

We will also write Aαβ=Aβα¯A_{\alpha\beta}=A_{\beta}^{\bar{\alpha}} and Aα¯β¯=Aβ¯αA_{\bar{\alpha}\bar{\beta}}=A_{\bar{\beta}}^{\alpha}. Then (2.5) means that Aαβ=AβαA_{\alpha\beta}=A_{\beta\alpha} and Aα¯β¯=Aβ¯α¯A_{\bar{\alpha}\bar{\beta}}=A_{\bar{\beta}\bar{\alpha}}. From [29], we have the following structure equations of the Tanaka-Webster connection \nabla:

dθ\displaystyle d\theta =\displaystyle= 21θαθα¯,\displaystyle 2\sqrt{-1}\theta^{\alpha}\wedge\theta^{\bar{\alpha}},
dθα\displaystyle d\theta^{\alpha} =\displaystyle= θβθβα+Aα¯β¯θθβ,\displaystyle\theta^{\beta}\wedge\theta^{\alpha}_{\beta}+A_{\bar{\alpha}\bar{\beta}}\theta\wedge\theta^{\beta}, (2.6)
dθβα\displaystyle d\theta^{\alpha}_{\beta} =\displaystyle= θβγθγα+Πβα\displaystyle\theta^{\gamma}_{\beta}\wedge\theta^{\alpha}_{\gamma}+\Pi^{\alpha}_{\beta}

with

Πβα=21(θατβ¯ταθβ¯)+Rβλμ¯αθλθμ¯+Wβγ¯αθθγ¯Wβγαθθγ,\displaystyle\Pi^{\alpha}_{\beta}=2\sqrt{-1}(\theta^{\alpha}\wedge\tau^{\bar{\beta}}-\tau^{\alpha}\wedge\theta^{\bar{\beta}})+R^{\alpha}_{\beta\lambda\bar{\mu}}\theta^{\lambda}\wedge\theta^{\bar{\mu}}+W^{\alpha}_{\beta\bar{\gamma}}\theta\wedge\theta^{\bar{\gamma}}-W^{\alpha}_{\beta\gamma}\theta\wedge\theta^{{\gamma}},

where Wβγ¯α=Aγ¯,βα,Wβγα=Aβ,α¯γ¯W^{\alpha}_{\beta\bar{\gamma}}=A^{\alpha}_{\bar{\gamma},\beta},\ W^{\alpha}_{\beta\gamma}=A^{\bar{\gamma}}_{\beta,\bar{\alpha}} are the are the covariant derivatives of AA, and Rβλμ¯αR_{\beta\lambda\bar{\mu}}^{\alpha} are the components of curvature tensor of the Tanaka-Webster connection. Set

Rαβ¯=Rγαβ¯γ,\displaystyle R_{\alpha\bar{\beta}}=R_{\gamma\alpha\bar{\beta}}^{\gamma},

then Rαβ¯=Rβ¯αR_{\alpha\bar{\beta}}=R_{\bar{\beta}\alpha} (cf. [18]). For any X=aαηα+bα¯ηα¯X=a^{\alpha}\eta_{\alpha}+b^{\bar{\alpha}}\eta_{\bar{\alpha}} and Y=cβηβ+dβ¯ηβ¯HMY=c^{\beta}\eta_{\beta}+d^{\bar{\beta}}\eta_{\bar{\beta}}\in HM\otimes\mathbb{C}, we define

Ricb(X,Y)=Rαβ¯aαdβ¯+Rα¯βbα¯cβ,\displaystyle Ric_{b}(X,Y)=R_{\alpha\bar{\beta}}a^{\alpha}d^{\bar{\beta}}+R_{\bar{\alpha}\beta}b^{\bar{\alpha}}c^{\beta},

whose components are given by

Ricb(ηα,ηβ¯)\displaystyle Ric_{b}(\eta_{\alpha},\eta_{\bar{\beta}}) =\displaystyle= Rαβ¯,Ricb(ηα¯,ηβ)=Rα¯β,\displaystyle R_{\alpha\bar{\beta}},\ Ric_{b}(\eta_{\bar{\alpha}},\eta_{\beta})=R_{\bar{\alpha}\beta},
Ricb(ηα,ηβ)\displaystyle Ric_{b}(\eta_{\alpha},\eta_{{\beta}}) =\displaystyle= Ricb(ηα¯,ηβ¯)=0.\displaystyle Ric_{b}(\eta_{\bar{\alpha}},\eta_{\bar{\beta}})=0.

The 2-tensor RicbRic_{b} will be referred to as the pseudo-Hermitian Ricci tensor. For any X=Xαηα+Xα¯ηα¯X=X^{\alpha}\eta_{\alpha}+X^{\bar{\alpha}}\eta_{\bar{\alpha}} and Y=Yβηβ+Yβ¯ηβ¯HMY=Y^{\beta}\eta_{\beta}+Y^{\bar{\beta}}\eta_{\bar{\beta}}\in HM\otimes\mathbb{C}, we introduce

Torb(X,Y)\displaystyle Tor_{b}(X,Y) =\displaystyle= A(X,JY)\displaystyle A(X,JY)
=\displaystyle= 1A(Xαηα+Xα¯ηα¯,YβηβYβ¯ηβ¯)\displaystyle\sqrt{-1}A(X^{\alpha}\eta_{\alpha}+X^{\bar{\alpha}}\eta_{\bar{\alpha}},Y^{\beta}\eta_{\beta}-Y^{\bar{\beta}}\eta_{\bar{\beta}})
=\displaystyle= 1(AαβXαYβAα¯β¯Xα¯Yβ¯).\displaystyle\sqrt{-1}(A_{\alpha\beta}X^{\alpha}Y^{\beta}-A_{\bar{\alpha}\bar{\beta}}X^{\bar{\alpha}}Y^{\bar{\beta}}).

Clearly both RicbRic_{b} and TorbTor_{b} are real symmetric, fiberwise 2-tensors on HMHM.

For a C2C^{2} function f:MRf:M\rightarrow R, its differential dfdf and gradient f\nabla f can be expressed as

df=f0θ+fαθα+fα¯θα¯\displaystyle df=f_{0}\theta+f_{\alpha}\theta^{\alpha}+f_{\bar{\alpha}}\theta^{\bar{\alpha}}

and

f=f0ξ+fα¯ηα+fαηα¯,\displaystyle\nabla f=f_{0}\xi+f_{\bar{\alpha}}\eta_{\alpha}+f_{\alpha}\eta_{\bar{\alpha}},

where f0=ξ(f),fα=ηα(f),fα¯=ηα¯(f)f_{0}=\xi(f),f_{\alpha}=\eta_{\alpha}(f),f_{\bar{\alpha}}=\eta_{\bar{\alpha}}(f). Then the horizontal gradient of ff is given by

bf=fα¯ηα+fαηα¯.\displaystyle\nabla_{b}f=f_{\bar{\alpha}}\eta_{\alpha}+f_{\alpha}\eta_{\bar{\alpha}}.

Let df\nabla df be the covariant derivative of the differential dfΓ(TM)df\in\Gamma(T^{*}M) with respect to the Tanaka-Webster connection. Then df\nabla df may be expressed as

df\displaystyle\nabla df =\displaystyle= fαβθαθβ+fαβ¯θαθβ¯+fα¯βθα¯θβ+fα¯β¯θα¯θβ¯\displaystyle f_{\alpha\beta}\theta^{\alpha}\otimes\theta^{\beta}+f_{\alpha\bar{\beta}}\theta^{\alpha}\otimes\theta^{\bar{\beta}}+f_{\bar{\alpha}\beta}\theta^{\bar{\alpha}}\otimes\theta^{\beta}+f_{\bar{\alpha}\bar{\beta}}\theta^{\bar{\alpha}}\otimes\theta^{\bar{\beta}}
+f0αθθα+f0α¯θθα¯+fα0θαθ+fα¯0θα¯θ.\displaystyle+f_{0\alpha}\theta\otimes\theta^{\alpha}+f_{0\bar{\alpha}}\theta\otimes\theta^{\bar{\alpha}}+f_{\alpha 0}\theta^{\alpha}\otimes\theta+f_{\bar{\alpha}0}\theta^{\bar{\alpha}}\otimes\theta.

The following communication relations are known (see, e.g., Chapter 9 in [18], or §3 in [14]):

fαβ=fβα,fαβ¯fβ¯α=21f0δαβ,f0αfα0=fβ¯Aαβ¯.\displaystyle f_{\alpha\beta}=f_{\beta\alpha},\quad f_{\alpha\bar{\beta}}-f_{\bar{\beta}\alpha}=2\sqrt{-1}f_{0}\delta_{\alpha}^{\beta},\quad f_{0\alpha}-f_{\alpha 0}=f_{\bar{\beta}}A_{\alpha}^{\bar{\beta}}. (2.7)

The horizontal Hessian of ff is defined by

Hessb(f)\displaystyle Hess_{b}(f) =\displaystyle= (df)(πb,πb)\displaystyle(\nabla df)(\pi_{b},\pi_{b})
=\displaystyle= fαβθαθβ+fαβ¯θαθβ¯+fα¯βθα¯θβ+fα¯β¯θα¯θβ¯.\displaystyle f_{\alpha\beta}\theta^{\alpha}\otimes\theta^{\beta}+f_{\alpha\bar{\beta}}\theta^{\alpha}\otimes\theta^{\bar{\beta}}+f_{\bar{\alpha}\beta}\theta^{\bar{\alpha}}\otimes\theta^{\beta}+f_{\bar{\alpha}\bar{\beta}}\theta^{\bar{\alpha}}\otimes\theta^{\bar{\beta}}.

Consequently

|bf|2=2fαfα¯,|Hessb(f)|2=2(fαβfα¯β¯+fαβ¯fα¯β).\displaystyle|\nabla_{b}f|^{2}=2f_{\alpha}f_{\bar{\alpha}},\quad|Hess_{b}(f)|^{2}=2(f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}}+f_{\alpha\bar{\beta}}f_{\bar{\alpha}\beta}).

The sub-Laplacian of ff is defined by

Δbf\displaystyle\Delta_{b}f =\displaystyle= tr{Hessb(f)}\displaystyle tr\{Hess_{b}(f)\}
=\displaystyle= fαα¯+fα¯α.\displaystyle f_{\alpha\bar{\alpha}}+f_{\bar{\alpha}\alpha}.

From [20], [21] (see also [18] and [14]), we have the following Bochner formulas

12Δb|bf|2\displaystyle\frac{1}{2}\Delta_{b}|\nabla_{b}f|^{2} =\displaystyle= 2(fαβfα¯β¯+fαβ¯fα¯β)+fα¯(fββ¯+fβ¯β)α+fα(fββ¯+fβ¯β)α¯\displaystyle 2(f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}}+f_{\alpha\bar{\beta}}f_{\bar{\alpha}\beta})+f_{\bar{\alpha}}(f_{\beta\bar{\beta}}+f_{\bar{\beta}\beta})_{\alpha}+f_{\alpha}(f_{\beta\bar{\beta}}+f_{\bar{\beta}\beta})_{\bar{\alpha}}
+2Rαβ¯fα¯fβ+21(m2)(Aαβfα¯fβ¯Aα¯β¯fαfβ)\displaystyle+2R_{\alpha\bar{\beta}}f_{\bar{\alpha}}f_{\beta}+2\sqrt{-1}(m-2)(A_{\alpha\beta}f_{\bar{\alpha}}f_{\bar{\beta}}-A_{\bar{\alpha}\bar{\beta}}f_{\alpha}f_{\beta})
+41(fα¯f0αfαf0α¯)\displaystyle+4\sqrt{-1}(f_{\bar{\alpha}}f_{0\alpha}-f_{\alpha}f_{0\bar{\alpha}})
=\displaystyle= |Hessb(f)|2+bf,bΔbf+4Jbf,bf0\displaystyle|Hess_{b}(f)|^{2}+\langle\nabla_{b}f,\nabla_{b}\Delta_{b}f\rangle+4\langle J\nabla_{b}f,\nabla_{b}f_{0}\rangle
+(Ricb+2(m2)Torb)(bf,bf)\displaystyle+(Ric_{b}+2(m-2)Tor_{b})(\nabla_{b}f,\nabla_{b}f)

and

12Δbf02\displaystyle\frac{1}{2}\Delta_{b}f_{0}^{2} =\displaystyle= |bf0|2+f0(Δbf)0\displaystyle|\nabla_{b}f_{0}|^{2}+f_{0}(\Delta_{b}f)_{0}
+2f0(fβAβ¯α¯,α+fβ¯Aβα,α¯+fβαAβ¯α¯+fβ¯α¯Aβα)\displaystyle+2f_{0}(f_{\beta}A_{\bar{\beta}\bar{\alpha},\alpha}+f_{\bar{\beta}}A_{\beta\alpha,\bar{\alpha}}+f_{\beta\alpha}A_{\bar{\beta}\bar{\alpha}}+f_{\bar{\beta}\bar{\alpha}}A_{\beta\alpha})
=\displaystyle= |bf0|2+f0(Δbf)0+2f0ImQf,\displaystyle|\nabla_{b}f_{0}|^{2}+f_{0}(\Delta_{b}f)_{0}+2f_{0}\mathrm{Im}Qf,

where QQ is the purely holomorphic second-order operator defined by ([20])

Qf=21(Aβ¯α¯fβ)α.\displaystyle Qf=2\sqrt{-1}(A_{\bar{\beta}\bar{\alpha}}f_{\beta})_{\alpha}.

Note that the coefficient before the ’mixed term’ Jbf,bf0\langle J\nabla_{b}f,\nabla_{b}f_{0}\rangle in (2.8) is slight different from that in [21].

𝐋𝐞𝐦𝐦𝐚 2.1\mathbf{Lemma\ 2.1} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a pseudo-Hermitian manifold and uu be a positive solution of the CR heat equation (1.4). Set f=lnuf=\mathrm{ln}\ u. Then for any 0<λ10<\lambda\leq 1, we have

(Δbt)|bf|2\displaystyle(\Delta_{b}-\partial_{t})|\nabla_{b}f|^{2} \displaystyle\geq 1m(Δbf)2+4mf02+4fαβfα¯β¯\displaystyle\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}} (2.10)
2b|bf|2,bf+8bf0,Jbf\displaystyle-2\langle\nabla_{b}|\nabla_{b}f|^{2},\nabla_{b}f\rangle+8\langle\nabla_{b}f_{0},J\nabla_{b}f\rangle
+2(Ricb+2(m2)Torb)(bf,bf)\displaystyle+2(Ric_{b}+2(m-2)Tor_{b})(\nabla_{b}f,\nabla_{b}f)

and

(Δbt)(1+f02)λ\displaystyle(\Delta_{b}-\partial_{t})(1+f^{2}_{0})^{\lambda} \displaystyle\geq 2λ(2λ1)(1+f02)λ1|bf0|2\displaystyle 2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2} (2.11)
2λ(1+f02)λ1(bf02,bf2f0A(bf,bf))\displaystyle-2\lambda(1+f^{2}_{0})^{\lambda-1}(\langle\nabla_{b}f^{2}_{0},\nabla_{b}f\rangle-2f_{0}A(\nabla_{b}f,\nabla_{b}f))
+4λ(1+f02)λ1f0ImQf.\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\mathrm{Im}Qf.

𝐏𝐫𝐨𝐨𝐟\mathbf{Proof} A direct computation gives

(Δbt)f=|bf|2.\displaystyle(\Delta_{b}-\partial_{t})f=-|\nabla_{b}f|^{2}. (2.12)

Using (2.7), we may estimate the term in |Hessb(f)|2|Hess_{b}(f)|^{2}:

fαβ¯fα¯β\displaystyle f_{\alpha\bar{\beta}}f_{\bar{\alpha}\beta} \displaystyle\geq α=1mfαα¯fα¯α=14α=1m(|fαα¯fα¯α|2+|fαα¯+fα¯α|2)\displaystyle\sum\limits_{\alpha=1}^{m}f_{\alpha\bar{\alpha}}f_{\bar{\alpha}\alpha}=\frac{1}{4}\sum\limits_{\alpha=1}^{m}(|f_{\alpha\bar{\alpha}}-f_{\bar{\alpha}\alpha}|^{2}+|f_{\alpha\bar{\alpha}}+f_{\bar{\alpha}\alpha}|^{2}) (2.13)
\displaystyle\geq 14m|α=1m(fαα¯+fα¯α)|2+14α=1m|fαα¯fα¯α|2\displaystyle\frac{1}{4m}|\sum\limits_{\alpha=1}^{m}(f_{\alpha\bar{\alpha}}+f_{\bar{\alpha}\alpha})|^{2}+\frac{1}{4}\sum\limits_{\alpha=1}^{m}|f_{\alpha\bar{\alpha}}-f_{\bar{\alpha}\alpha}|^{2}
=\displaystyle= 14m(Δbf)2+mf02.\displaystyle\frac{1}{4m}(\Delta_{b}f)^{2}+mf_{0}^{2}.

Then (2.10) follows immediately from (2.8), (2.12) and (2.13). From (2.9), we get

(Δbt)f02=2|bf0|2+2f0(Δbftf)0+4f0ImQf.\displaystyle(\Delta_{b}-\partial_{t})f_{0}^{2}=2|\nabla_{b}f_{0}|^{2}+2f_{0}(\Delta_{b}f-\partial_{t}f)_{0}+4f_{0}\mathrm{Im}Qf. (2.14)

Using (2.14), we derive that

(Δbt)(1+f02)λ\displaystyle(\Delta_{b}-\partial_{t})(1+f^{2}_{0})^{\lambda}
=\displaystyle= 2λ(λ1)(1+f02)λ2(f02)α(f02)α¯+λ(1+f02)λ1(f02)αα¯+α¯α\displaystyle 2\lambda(\lambda-1)(1+f^{2}_{0})^{\lambda-2}(f^{2}_{0})_{\alpha}(f^{2}_{0})_{\bar{\alpha}}+\lambda(1+f^{2}_{0})^{\lambda-1}(f^{2}_{0})_{\alpha\bar{\alpha}+\bar{\alpha}\alpha}
t(1+f02)λ\displaystyle-\partial_{t}(1+f^{2}_{0})^{\lambda}
=\displaystyle= 4λ(λ1)(1+f02)λ2f02|bf0|2+λ(1+f02)λ1(Δbt)(f02)\displaystyle 4\lambda(\lambda-1)(1+f^{2}_{0})^{\lambda-2}f^{2}_{0}|\nabla_{b}f_{0}|^{2}+\lambda(1+f^{2}_{0})^{\lambda-1}(\Delta_{b}-\partial_{t})(f^{2}_{0})
=\displaystyle= 4λ(λ1)(1+f02)λ2f02|bf0|2+2λ(1+f02)λ1|bf0|2\displaystyle 4\lambda(\lambda-1)(1+f^{2}_{0})^{\lambda-2}f^{2}_{0}|\nabla_{b}f_{0}|^{2}+2\lambda(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}
+2λ(1+f02)λ1f0ξ(Δbfft)\displaystyle+2\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\nabla_{\xi}(\Delta_{b}f-f_{t})
+4λ(1+f02)λ1f0ImQf\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\mathrm{Im}Qf
=\displaystyle= 2λ(1+f02)λ2|bf0|2((2λ1)f02+1)\displaystyle 2\lambda(1+f^{2}_{0})^{\lambda-2}|\nabla_{b}f_{0}|^{2}((2\lambda-1)f^{2}_{0}+1)
2λ(1+f02)λ1(bf02,bf2f0A(bf,bf))\displaystyle-2\lambda(1+f^{2}_{0})^{\lambda-1}(\langle\nabla_{b}f^{2}_{0},\nabla_{b}f\rangle-2f_{0}A(\nabla_{b}f,\nabla_{b}f))
+4λ(1+f02)λ1f0ImQf\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\mathrm{Im}Qf
\displaystyle\geq 2λ(2λ1)(1+f02)λ1|bf0|2\displaystyle 2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}
2λ(1+f02)λ1(bf02,bf2f0A(bf,bf))\displaystyle-2\lambda(1+f^{2}_{0})^{\lambda-1}(\langle\nabla_{b}f^{2}_{0},\nabla_{b}f\rangle-2f_{0}A(\nabla_{b}f,\nabla_{b}f))
+4λ(1+f02)λ1f0ImQf.\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\mathrm{Im}Qf.

This completes the proof of Lemma 2.1. ∎

3 Li-Yau type gradient estimates

In this section, we derive the CR version of Li-Yau type gradient estimates. Inspired by Cao-Yau’s method in [8] for weakly elliptic operators on closed manifolds, we consider the auxiliary functions \mathcal{F} and 𝒢\mathcal{G} on complete pseudo-Hermitian manifolds, trying to establish the gradient estimates according to two cases: 0<t10<t\leq 1 and t1t\geq 1.

Let uu be a positive solution of (1.4). Recall that, if f=lnuf=\mathrm{ln}\ u, then

(Δbt)f=|bf|2.\displaystyle(\Delta_{b}-\frac{\partial}{\partial t})f=-|\nabla_{b}f|^{2}. (3.1)

As mentioned in Introduction, we consider the following auxiliary functions

=t(|bf|2+t2λ1(1+f02)λδft)=1+2\mathcal{F}=t\left(|\nabla_{b}f|^{2}+t^{2\lambda-1}\left(1+f_{0}^{2}\right)^{\lambda}-\delta f_{t}\right)=\mathcal{F}_{1}+\mathcal{F}_{2} (3.2)

and

𝒢=t(|bf|2+(1+f02)λδft),\mathcal{G}=t\left(|\nabla_{b}f|^{2}+\left(1+f_{0}^{2}\right)^{\lambda}-\delta f_{t}\right), (3.3)

where

1=t(|bf|2δft),2=t2λ(1+f02)λ,\displaystyle\mathcal{F}_{1}=t(|\nabla_{b}f|^{2}-\delta f_{t}),\ \mathcal{F}_{2}=t^{2\lambda}(1+f_{0}^{2})^{\lambda},

and δ>1\delta>1 is a constant. Using Lemma 2.1, we derive the following inequalities

(Δbt)\displaystyle(\Delta_{b}-\partial_{t})\mathcal{F} =\displaystyle= (Δbt)1+(Δbt)2\displaystyle(\Delta_{b}-\partial_{t})\mathcal{F}_{1}+(\Delta_{b}-\partial_{t})\mathcal{F}_{2} (3.4)
=\displaystyle= t(Δbt)(|bf|2δft)(|bf|2δft)\displaystyle t(\Delta_{b}-\partial_{t})(|\nabla_{b}f|^{2}-\delta f_{t})-(|\nabla_{b}f|^{2}-\delta f_{t})
+t2λ(Δbt)(1+f02)λ2λt2λ1(1+f02)λ\displaystyle+t^{2\lambda}(\Delta_{b}-\partial_{t})(1+f_{0}^{2})^{\lambda}-2\lambda t^{2\lambda-1}(1+f_{0}^{2})^{\lambda}
\displaystyle\geq 1t2λ2t+t{1m(Δbf)2+4mf02+4fαβfα¯β¯\displaystyle-\frac{\mathcal{F}_{1}}{t}-\frac{2\lambda\mathcal{F}_{2}}{t}+t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}}
+8bf0,Jbf+2(Ricb+2(m2)Torb)(bf,bf)}\displaystyle+8\langle\nabla_{b}f_{0},J\nabla_{b}f\rangle+2(Ric_{b}+2(m-2)Tor_{b})(\nabla_{b}f,\nabla_{b}f)\}
+t2λ{2λ(2λ1)(1+f02)λ1|bf0|2\displaystyle+t^{2\lambda}\{2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}
+4λ(1+f02)λ1f0A(bf,bf)\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}A(\nabla_{b}f,\nabla_{b}f)
+4λ(1+f02)λ1f0ImQf}\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\mathrm{Im}Qf\}
2tbf,b|bf|2+2δtbf,bft\displaystyle-2t\langle\nabla_{b}f,\nabla_{b}|\nabla_{b}f|^{2}\rangle+2\delta t\langle\nabla_{b}f,\nabla_{b}f_{t}\rangle
2λt2λ(1+f02)λ1bf,bf02\displaystyle-2\lambda t^{2\lambda}(1+f_{0}^{2})^{\lambda-1}\langle\nabla_{b}f,\nabla_{b}f_{0}^{2}\rangle
\displaystyle\geq 1t2λ2t+t{1m(Δbf)2+4mf02+4fαβfα¯β¯\displaystyle-\frac{\mathcal{F}_{1}}{t}-\frac{2\lambda\mathcal{F}_{2}}{t}+t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}}
+8bf0,Jbf+2(Ricb+2(m2)Torb)(bf,bf)}\displaystyle+8\langle\nabla_{b}f_{0},J\nabla_{b}f\rangle+2(Ric_{b}+2(m-2)Tor_{b})(\nabla_{b}f,\nabla_{b}f)\}
+t2λ{2λ(2λ1)(1+f02)λ1|bf0|2\displaystyle+t^{2\lambda}\{2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}
+4λ(1+f02)λ1f0A(bf,bf)\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}A(\nabla_{b}f,\nabla_{b}f)
+4λ(1+f02)λ1f0ImQf}2bf,b\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\mathrm{Im}Qf\}-2\langle\nabla_{b}f,\nabla_{b}\mathcal{F}\rangle

and

(Δbt)𝒢\displaystyle(\Delta_{b}-\partial_{t})\mathcal{G} =\displaystyle= t(Δbt)(|bf|2+(1+f02)λδft)\displaystyle t(\Delta_{b}-\partial_{t})(|\nabla_{b}f|^{2}+(1+f_{0}^{2})^{\lambda}-\delta f_{t}) (3.5)
(|bf|2+(1+f02)λδft)\displaystyle-(|\nabla_{b}f|^{2}+(1+f_{0}^{2})^{\lambda}-\delta f_{t})
\displaystyle\geq 𝒢t+t{1m(Δbf)2+4mf02+4fαβfα¯β¯\displaystyle-\frac{\mathcal{G}}{t}+t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}}
+8bf0,Jbf+2(Ricb+2(m2)Torb)(bf,bf)\displaystyle+8\langle\nabla_{b}f_{0},J\nabla_{b}f\rangle+2(Ric_{b}+2(m-2)Tor_{b})(\nabla_{b}f,\nabla_{b}f)
+2λ(2λ1)(1+f02)λ1|bf0|2\displaystyle+2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}
+4λ(1+f02)λ1f0A(bf,bf)\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}A(\nabla_{b}f,\nabla_{b}f)
+4λ(1+f02)λ1f0ImQf}\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\mathrm{Im}Qf\}
2tbf,b|bf|2+2δtbf,bft\displaystyle-2t\langle\nabla_{b}f,\nabla_{b}|\nabla_{b}f|^{2}\rangle+2\delta t\langle\nabla_{b}f,\nabla_{b}f_{t}\rangle
2λt(1+f02)λ1bf,bf02\displaystyle-2\lambda t(1+f_{0}^{2})^{\lambda-1}\langle\nabla_{b}f,\nabla_{b}f_{0}^{2}\rangle
\displaystyle\geq 𝒢t+t{1m(Δbf)2+4mf02+4fαβfα¯β¯\displaystyle-\frac{\mathcal{G}}{t}+t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}}
+8bf0,Jbf+2(Ricb+2(m2)Torb)(bf,bf)\displaystyle+8\langle\nabla_{b}f_{0},J\nabla_{b}f\rangle+2(Ric_{b}+2(m-2)Tor_{b})(\nabla_{b}f,\nabla_{b}f)
+2λ(2λ1)(1+f02)λ1|bf0|2\displaystyle+2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}
+4λ(1+f02)λ1f0A(bf,bf)\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}A(\nabla_{b}f,\nabla_{b}f)
+4λ(1+f02)λ1f0ImQf}2bf,b𝒢.\displaystyle+4\lambda(1+f^{2}_{0})^{\lambda-1}f_{0}\mathrm{Im}Qf\}-2\langle\nabla_{b}f,\nabla_{b}\mathcal{G}\rangle.

Hence we have the following Lemma 3.1.

𝐋𝐞𝐦𝐦𝐚 3.1\mathbf{Lemma\ 3.1} Let (M2m+1,θ)(M^{2m+1},\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbkand|A|,|bA|k1,\displaystyle Ric_{b}+2(m-2)Tor_{b}\geq-k\ \ and\ \ |A|,|\nabla_{b}A|\leq k_{1},

then for 12<λ<1\frac{1}{2}<\lambda<1, we have

(Δbt)\displaystyle(\Delta_{b}-\partial_{t})\mathcal{F} \displaystyle\geq 1t2λ2t2bf,b+t{1m(Δbf)2+4mf02\displaystyle-\frac{\mathcal{F}_{1}}{t}-\frac{2\lambda\mathcal{F}_{2}}{t}-2\langle\nabla_{b}f,\nabla_{b}\mathcal{F}\rangle+t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0} (3.6)
8λ(2λ1)(1+f02)1λ|bf|2t12λ2k|bf|2\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}\cdot t^{1-2\lambda}-2k|\nabla_{b}f|^{2}
2k1|bf|24k1λ(1+f02)λ1|f0||bf|2t2λ1\displaystyle-2k_{1}|\nabla_{b}f|^{2}-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}\cdot t^{2\lambda-1}
2(k1+k12)λ2(1+f02)2λ2|f02|t4λ2}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t^{4\lambda-2}\}

for 0<t10<t\leq 1 and

(Δbt)𝒢\displaystyle(\Delta_{b}-\partial_{t})\mathcal{G} \displaystyle\geq 𝒢t2bf,b𝒢+t{1m(Δbf)2+4mf02\displaystyle-\frac{\mathcal{G}}{t}-2\langle\nabla_{b}f,\nabla_{b}\mathcal{G}\rangle+t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0} (3.7)
8λ(2λ1)(1+f02)1λ|bf|22k|bf|2\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}-2k|\nabla_{b}f|^{2}
2k1|bf|24k1λ(1+f02)λ1|f0||bf|2\displaystyle-2k_{1}|\nabla_{b}f|^{2}-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}
2(k1+k12)λ2(1+f02)2λ2|f02|}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\}

for t1t\geq 1.  
𝐏𝐫𝐨𝐨𝐟\mathbf{Proof} From (3.4) and the condition, we find that

(Δbt)\displaystyle(\Delta_{b}-\partial_{t})\mathcal{F} \displaystyle\geq 1t2λ2t+t{1m(Δbf)2+4mf02+4fαβfα¯β¯\displaystyle-\frac{\mathcal{F}_{1}}{t}-\frac{2\lambda\mathcal{F}_{2}}{t}+t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}} (3.8)
+8bf0,Jbf2k|bf|2\displaystyle+8\langle\nabla_{b}f_{0},J\nabla_{b}f\rangle-2k|\nabla_{b}f|^{2}
+2λ(2λ1)(1+f02)λ1|bf0|2t2λ1\displaystyle+2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}\cdot t^{2\lambda-1}
4k1λ(1+f02)λ1|f0||bf|2t2λ1\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}\cdot t^{2\lambda-1}
8λ(1+f02)λ1|f0||fαAβ¯α¯,β|t2λ1\displaystyle-8\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|f_{\alpha}A_{\bar{\beta}\bar{\alpha},\beta}|\cdot t^{2\lambda-1}
8λ(1+f02)λ1|f0||fαβAα¯β¯|t2λ1}\displaystyle-8\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|f_{\alpha\beta}A_{\bar{\alpha}\bar{\beta}}|\cdot t^{2\lambda-1}\}
2bf,b.\displaystyle-2\langle\nabla_{b}f,\nabla_{b}\mathcal{F}\rangle.

Then we estimate certain terms in (3.8) as follows

8\displaystyle 8\langle\nabla f0b,Jbf+2λ(2λ1)(1+f02)λ1|bf0|2t2λ1{}_{b}f_{0},J\nabla_{b}f\rangle+2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}\cdot t^{2\lambda-1}
8λ(2λ1)(1+f02)1λ|bf|2t12λ,\displaystyle\geq-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}\cdot t^{1-2\lambda}, (3.9)
8λ\displaystyle-8\lambda (1+f02)λ1|f0||fαAβ¯α¯,β|t2λ1\displaystyle(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|f_{\alpha}A_{\bar{\beta}\bar{\alpha},\beta}|\cdot t^{2\lambda-1}
2k1|bf|22k1λ2(1+f02)2λ2|f02|t4λ2,\displaystyle\geq-2k_{1}|\nabla_{b}f|^{2}-2k_{1}\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t^{4\lambda-2}, (3.10)
8λ\displaystyle-8\lambda (1+f02)λ1|f0||fαβAα¯β¯|t2λ1+4fαβfα¯β¯\displaystyle(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|f_{\alpha\beta}A_{\bar{\alpha}\bar{\beta}}|\cdot t^{2\lambda-1}+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}}
2k12λ2(1+f02)2λ2|f0|2t4λ2.\displaystyle\geq-2k_{1}^{2}\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}|^{2}\cdot t^{4\lambda-2}. (3.11)

Finally we can get (3.6) from (3.8)-(3.11).

From (3.5) and the condition, we have

(Δbt)𝒢\displaystyle(\Delta_{b}-\partial_{t})\mathcal{G} \displaystyle\geq 𝒢t+t{1m(Δbf)2+4mf02+4fαβfα¯β¯\displaystyle-\frac{\mathcal{G}}{t}+t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}} (3.12)
+8bf0,Jbf2k|bf|2\displaystyle+8\langle\nabla_{b}f_{0},J\nabla_{b}f\rangle-2k|\nabla_{b}f|^{2}
+2λ(2λ1)(1+f02)λ1|bf0|2\displaystyle+2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}
4k1λ(1+f02)λ1|f0||bf|2\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}
8λ(1+f02)λ1|f0||fαAβ¯α¯,β|\displaystyle-8\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|f_{\alpha}A_{\bar{\beta}\bar{\alpha},\beta}|
8λ(1+f02)λ1|f0||fαβAα¯β¯|}\displaystyle-8\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|f_{\alpha\beta}A_{\bar{\alpha}\bar{\beta}}|\}
2bf,b𝒢.\displaystyle-2\langle\nabla_{b}f,\nabla_{b}\mathcal{G}\rangle.

Next we are going to estimate certain terms that appear in (3.12). It is easy to prove that

8\displaystyle 8\langle\nabla f0b,Jbf+2λ(2λ1)(1+f02)λ1|bf0|2{}_{b}f_{0},J\nabla_{b}f\rangle+2\lambda(2\lambda-1)(1+f^{2}_{0})^{\lambda-1}|\nabla_{b}f_{0}|^{2}
8λ(2λ1)(1+f02)1λ|bf|2,\displaystyle\geq-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}, (3.13)
8λ\displaystyle-8\lambda (1+f02)λ1|f0||fαAβ¯α¯,β|\displaystyle(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|f_{\alpha}A_{\bar{\beta}\bar{\alpha},\beta}|
2k1|bf|22k1λ2(1+f02)2λ2|f02|,\displaystyle\geq-2k_{1}|\nabla_{b}f|^{2}-2k_{1}\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|, (3.14)
8λ\displaystyle-8\lambda (1+f02)λ1|f0||fαβAα¯β¯|+4fαβfα¯β¯\displaystyle(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|f_{\alpha\beta}A_{\bar{\alpha}\bar{\beta}}|+4f_{\alpha\beta}f_{\bar{\alpha}\bar{\beta}}
2k12λ2(1+f02)2λ2|f0|2.\displaystyle\geq-2k_{1}^{2}\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}|^{2}. (3.15)

Then (3.7) can be obtained by the above estimates. ∎

Choose a cut-off function φC([0,))\varphi\in C^{\infty}([0,\infty)) such that

φ|[0,1]=1,φ|[2,)=0,C1|φ|12φ0,φ′′C1.\displaystyle\varphi|_{[0,1]}=1,\ \varphi|_{[2,\infty)}=0,\ -C_{1}^{{}^{\prime}}|\varphi|^{\frac{1}{2}}\leq\varphi^{{}^{\prime}}\leq 0,\ \varphi^{{}^{\prime\prime}}\geq-C_{1}^{{}^{\prime}}.

Set g=φ11μg=\varphi^{\frac{1}{1-\mu}} where μ<1\mu<1. Direct calculations show that

g\displaystyle g^{{}^{\prime}} =\displaystyle= 11μφμ1μφ=11μgμφ,\displaystyle\frac{1}{1-\mu}\varphi^{\frac{\mu}{1-\mu}}\varphi^{{}^{\prime}}=\frac{1}{1-\mu}g^{\mu}\varphi^{{}^{\prime}},
g′′\displaystyle g^{{}^{\prime\prime}} =\displaystyle= μ(1μ)2g2μ1φ2+11μgμφ′′.\displaystyle\frac{\mu}{(1-\mu)^{2}}g^{2\mu-1}\varphi^{{}^{\prime}2}+\frac{1}{1-\mu}g^{\mu}\varphi^{{}^{\prime\prime}}. (3.16)

Let rr be the Riemannian distance and Bp(R)B_{p}(R) denotes the Riemannian ball of radius RR centered at pp. Put

ϕ=g(rR).\phi=g(\frac{r}{R}).

Assuming R1R\geq 1 and using a comparison theorem in [15], we find that

|bϕ|2ϕ2μ\displaystyle\frac{|\nabla_{b}\phi|^{2}}{\phi^{2\mu}} =\displaystyle= |ϕ|2|br|2ϕ2μR2C2R2,\displaystyle\frac{|\phi^{{}^{\prime}}|^{2}|\nabla_{b}r|^{2}}{\phi^{2\mu}R^{2}}\leq\frac{C^{{}^{\prime}}_{2}}{R^{2}},
Δbϕϕ2μ1\displaystyle\frac{\Delta_{b}\phi}{\phi^{2\mu-1}} =\displaystyle= g′′|br|2ϕ2μ1R2+gΔbrϕ2μ1RC2R,\displaystyle\frac{g^{{}^{\prime\prime}}|\nabla_{b}r|^{2}}{\phi^{2\mu-1}R^{2}}+\frac{g^{{}^{\prime}}\Delta_{b}r}{\phi^{2\mu-1}R}\geq-\frac{C^{{}^{\prime}}_{2}}{R},

where C2C^{{}^{\prime}}_{2} is a constant depending on k,k1,μk,k_{1},\mu. Let μ=3λ1\mu=3\lambda-1, where 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3}, then

|bϕ|2ϕ6λ2C1R2,Δbϕϕ6λ3C1R,\displaystyle\frac{|\nabla_{b}\phi|^{2}}{\phi^{6\lambda-2}}\leq\frac{C_{1}}{R^{2}},\ \ \frac{\Delta_{b}\phi}{\phi^{6\lambda-3}}\geq-\frac{C_{1}}{R},

where C1C_{1} is a constant depending on k,k1,λk,k_{1},\lambda.

𝐋𝐞𝐦𝐦𝐚 3.2\mathbf{Lemma\ 3.2} Let (M2m+1,θ)(M^{2m+1},\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbkand|A|,|bA|k1.Ric_{b}+2(m-2)Tor_{b}\geq-k\ \ and\ \ |A|,|\nabla_{b}A|\leq k_{1}.

Let ϕ\phi be defined as above with R1R\geq 1. If ϕ(x)0\phi(x)\neq 0 and 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3}, then at xx, we have

(Δbt)ϕ\displaystyle(\Delta_{b}-\partial_{t})\phi\mathcal{F} \displaystyle\geq 2b(ϕ),bϕϕ13C1Rϕ6λ3ϕ1t2λϕ2t\displaystyle 2\langle\nabla_{b}(\phi\mathcal{F}),\nabla_{b}\phi\rangle\phi^{-1}-\frac{3C_{1}}{R}\phi^{6\lambda-3}\mathcal{F}-\frac{\phi\mathcal{F}_{1}}{t}-\frac{2\lambda\phi\mathcal{F}_{2}}{t}
2bf,b(ϕ)+2bf,bϕ\displaystyle-2\langle\nabla_{b}f,\nabla_{b}(\phi\mathcal{F})\rangle+2\langle\nabla_{b}f,\nabla_{b}\phi\rangle\mathcal{F}
+ϕt{1m(Δbf)2+4mf02(2k+2k1)|bf|2\displaystyle+\phi t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}-(2k+2k_{1})|\nabla_{b}f|^{2}
8λ(2λ1)(1+f02)1λ|bf|2t12λ\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}\cdot t^{1-2\lambda}
4k1λ(1+f02)λ1|f0||bf|2t2λ1\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}\cdot t^{2\lambda-1}
2(k1+k12)λ2(1+f02)2λ2|f02|t4λ2}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t^{4\lambda-2}\}

for 0t10\leq t\leq 1 and

(Δbt)ϕ𝒢\displaystyle(\Delta_{b}-\partial_{t})\phi\mathcal{G} \displaystyle\geq 2b(ϕ𝒢),bϕϕ13C1Rϕ6λ3𝒢\displaystyle 2\langle\nabla_{b}(\phi\mathcal{G}),\nabla_{b}\phi\rangle\phi^{-1}-\frac{3C_{1}}{R}\phi^{6\lambda-3}\mathcal{G}
ϕ𝒢t2bf,b(ϕ𝒢)+2bf,bϕ𝒢\displaystyle-\frac{\phi\mathcal{G}}{t}-2\langle\nabla_{b}f,\nabla_{b}(\phi\mathcal{G})\rangle+2\langle\nabla_{b}f,\nabla_{b}\phi\rangle\mathcal{G}
+ϕt{1m(Δbf)2+4mf02(2k+2k1)|bf|2\displaystyle+\phi t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}-(2k+2k_{1})|\nabla_{b}f|^{2}
8λ(2λ1)(1+f02)1λ|bf|2\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}
4k1λ(1+f02)λ1|f0||bf|2\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}
2(k1+k12)λ2(1+f02)2λ2|f02|}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\}

for t1t\geq 1.

𝐏𝐫𝐨𝐨𝐟\mathbf{Proof} From Lemma 3.1 and the properties of ϕ\phi, we have

(Δbt)ϕ\displaystyle(\Delta_{b}-\partial_{t})\phi\mathcal{F} =\displaystyle= (Δbϕ)+2bϕ,b+ϕ(Δbt)\displaystyle(\Delta_{b}\phi)\mathcal{F}+2\langle\nabla_{b}\phi,\nabla_{b}\mathcal{F}\rangle+\phi(\Delta_{b}-\partial_{t})\mathcal{F}
\displaystyle\geq (Δbϕ)+2b(ϕ),bϕϕ12|bϕ|2ϕ\displaystyle(\Delta_{b}\phi)\mathcal{F}+2\langle\nabla_{b}(\phi\mathcal{F}),\nabla_{b}\phi\rangle\phi^{-1}-\frac{2|\nabla_{b}\phi|^{2}}{\phi}\mathcal{F}
ϕ1t2λϕ2t2bf,b(ϕ)+2bf,bϕ\displaystyle-\frac{\phi\mathcal{F}_{1}}{t}-\frac{2\lambda\phi\mathcal{F}_{2}}{t}-2\langle\nabla_{b}f,\nabla_{b}(\phi\mathcal{F})\rangle+2\langle\nabla_{b}f,\nabla_{b}\phi\rangle\mathcal{F}
+ϕt{1m(Δbf)2+4mf02(2k+2k1)|bf|2\displaystyle+\phi t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}-(2k+2k_{1})|\nabla_{b}f|^{2}
8λ(2λ1)(1+f02)1λ|bf|2t12λ\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}\cdot t^{1-2\lambda}
4k1λ(1+f02)λ1|f0||bf|2t2λ1\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}\cdot t^{2\lambda-1}
2(k1+k12)λ2(1+f02)2λ2|f02|t4λ2}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t^{4\lambda-2}\}
\displaystyle\geq 2b(ϕ),bϕϕ13C1Rϕ6λ3\displaystyle 2\langle\nabla_{b}(\phi\mathcal{F}),\nabla_{b}\phi\rangle\phi^{-1}-\frac{3C_{1}}{R}\phi^{6\lambda-3}\mathcal{F}
ϕ1t2λϕ2t2bf,b(ϕ)+2bf,bϕ\displaystyle-\frac{\phi\mathcal{F}_{1}}{t}-\frac{2\lambda\phi\mathcal{F}_{2}}{t}-2\langle\nabla_{b}f,\nabla_{b}(\phi\mathcal{F})\rangle+2\langle\nabla_{b}f,\nabla_{b}\phi\rangle\mathcal{F}
+ϕt{1m(Δbf)2+4mf02(2k+2k1)|bf|2\displaystyle+\phi t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}-(2k+2k_{1})|\nabla_{b}f|^{2}
8λ(2λ1)(1+f02)1λ|bf|2t12λ\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}\cdot t^{1-2\lambda}
4k1λ(1+f02)λ1|f0||bf|2t2λ1\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}\cdot t^{2\lambda-1}
2(k1+k12)λ2(1+f02)2λ2|f02|t4λ2\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t^{4\lambda-2}\

for 0t10\leq t\leq 1 and

(Δbt)ϕ𝒢\displaystyle(\Delta_{b}-\partial_{t})\phi\mathcal{G} =\displaystyle= (Δbϕ)𝒢+2bϕ,b𝒢+ϕ(Δbt)𝒢\displaystyle(\Delta_{b}\phi)\mathcal{G}+2\langle\nabla_{b}\phi,\nabla_{b}\mathcal{G}\rangle+\phi(\Delta_{b}-\partial_{t})\mathcal{G}
\displaystyle\geq (Δbϕ)𝒢+2b(ϕ𝒢),bϕϕ12|bϕ|2ϕ𝒢\displaystyle(\Delta_{b}\phi)\mathcal{G}+2\langle\nabla_{b}(\phi\mathcal{G}),\nabla_{b}\phi\rangle\phi^{-1}-\frac{2|\nabla_{b}\phi|^{2}}{\phi}\mathcal{G}
ϕ𝒢t2bf,b(ϕ𝒢)+2bf,bϕ𝒢\displaystyle-\frac{\phi\mathcal{G}}{t}-2\langle\nabla_{b}f,\nabla_{b}(\phi\mathcal{G})\rangle+2\langle\nabla_{b}f,\nabla_{b}\phi\rangle\mathcal{G}
+ϕt{1m(Δbf)2+4mf02(2k+2k1)|bf|2\displaystyle+\phi t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}-(2k+2k_{1})|\nabla_{b}f|^{2}
8λ(2λ1)(1+f02)1λ|bf|2\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}
4k1λ(1+f02)λ1|f0||bf|2\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}
2(k1+k12)λ2(1+f02)2λ2|f02|}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\}
\displaystyle\geq 2b(ϕ𝒢),bϕϕ13C1Rϕ6λ3𝒢\displaystyle 2\langle\nabla_{b}(\phi\mathcal{G}),\nabla_{b}\phi\rangle\phi^{-1}-\frac{3C_{1}}{R}\phi^{6\lambda-3}\mathcal{G}
ϕ𝒢t2bf,b(ϕ𝒢)+2bf,bϕ𝒢\displaystyle-\frac{\phi\mathcal{G}}{t}-2\langle\nabla_{b}f,\nabla_{b}(\phi\mathcal{G})\rangle+2\langle\nabla_{b}f,\nabla_{b}\phi\rangle\mathcal{G}
+ϕt{1m(Δbf)2+4mf02(2k+2k1)|bf|2\displaystyle+\phi t\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}-(2k+2k_{1})|\nabla_{b}f|^{2}
8λ(2λ1)(1+f02)1λ|bf|2\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}
4k1λ(1+f02)λ1|f0||bf|2\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}
2(k1+k12)λ2(1+f02)2λ2|f02|}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\}

for t1t\geq 1. This completes the proof. ∎

Now we are ready to consider the first case of the gradient estimate, that is, 0<t10<t\leq 1.

𝐏𝐫𝐨𝐩𝐨𝐬𝐢𝐭𝐢𝐨𝐧 3.3\mathbf{Proposition\ 3.3} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbkand|A|,|bA|k1.\displaystyle Ric_{b}+2(m-2)Tor_{b}\geq-k\ \ and\ \ |A|,|\nabla_{b}A|\leq k_{1}.

and uu be a positive solution of the CR heat equation

ut=Δbu\displaystyle\frac{\partial u}{\partial t}=\Delta_{b}u

on Bp(2R)×(0,1]B_{p}(2R)\times(0,1] with R1R\geq 1. Then for any constant 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3} and any constant δ>1+4mλ(2λ1)\delta>1+\frac{4}{m\lambda(2\lambda-1)}, there exists a constant C3C_{3}^{{}^{\prime}} depends on m,k,k1,λ,δm,k,k_{1},\lambda,\delta, such that

|bu|2u2+t2λ1(1+u02u2)λδutuC3t(1+1Rλ)\displaystyle\frac{|\nabla_{b}u|^{2}}{u^{2}}+t^{2\lambda-1}(1+\frac{u^{2}_{0}}{u^{2}})^{\lambda}-\delta\frac{u_{t}}{u}\leq\frac{C_{3}^{{}^{\prime}}}{t}(1+\frac{1}{R^{\lambda}}) (3.17)

on Bp(R)×(0,1]B_{p}(R)\times(0,1].

𝐏𝐫𝐨𝐨𝐟\mathbf{Proof} Let (x1,t1)(x_{1},t_{1}) be the maximum point of ϕ\phi\mathcal{F} on Bp(2R)×[0,1]B_{p}(2R)\times[0,1]. Without loss of generality, we may assume that (ϕ)(x1,t1)>0(\phi\mathcal{F})(x_{1},t_{1})>0, otherwise the conclusion follows trivially. At (x1,t1)(x_{1},t_{1}), we have (ϕ)=0,t(ϕ)0\nabla(\phi\mathcal{F})=0,\ \partial_{t}(\phi\mathcal{F})\geq 0 and Δb(ϕ)0.\Delta_{b}(\phi\mathcal{F})\leq 0. Using Lemma 3.2 and evaluating the inequality at (x1,t1)(x_{1},t_{1}), we obtain

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ3ϕ1t12λϕ2t1+2bf,bϕ\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-3}\mathcal{F}-\frac{\phi\mathcal{F}_{1}}{t_{1}}-\frac{2\lambda\phi\mathcal{F}_{2}}{t_{1}}+2\langle\nabla_{b}f,\nabla_{b}\phi\rangle\mathcal{F}
+ϕt1{1m(Δbf)2+4mf02\displaystyle+\phi t_{1}\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}
8λ(2λ1)(1+f02)1λ|bf|2t112λ2k|bf|2\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}\cdot t_{1}^{1-2\lambda}-2k|\nabla_{b}f|^{2}
2k1|bf|24k1λ(1+f02)λ1|f0||bf|2t12λ1\displaystyle-2k_{1}|\nabla_{b}f|^{2}-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}\cdot t_{1}^{2\lambda-1}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λ2}.\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda-2}\}.

Multiplying ϕt1\phi t_{1} yields

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ222C1R|bf|ϕ3λt1\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}-\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F} (3.19)
+ϕ2t12{1m(Δbf)2+4mf02\displaystyle+\phi^{2}t^{2}_{1}\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}
8λ(2λ1)(1+f02)1λ|bf|2t112λ2k|bf|2\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}\cdot t_{1}^{1-2\lambda}-2k|\nabla_{b}f|^{2}
2k1|bf|24k1λ(1+f02)λ1|f0||bf|2t12λ1\displaystyle-2k_{1}|\nabla_{b}f|^{2}-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}\cdot t_{1}^{2\lambda-1}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λ2}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda-2}\}
\displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ222C1R|bf|ϕ3λt1\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}-\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F}
+ϕ2t12{1m(Δbf)2+4mf022ϵ|bf|4\displaystyle+\phi^{2}t^{2}_{1}\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}-2\epsilon|\nabla_{b}f|^{4}
16ϵλ2(2λ1)2(1+f02)22λt124λ(2k+2k1)|bf|2\displaystyle-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{2-4\lambda}-(2k+2k_{1})|\nabla_{b}f|^{2}
4ϵ1k12λ2(1+f02)2λ2|f0|2t14λ2\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}\cdot t_{1}^{4\lambda-2}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λ2},\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda-2}\},

since

8λ(2λ1)(1+f02)1λ|bf|2t112λ\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}\cdot t_{1}^{1-2\lambda} (3.20)
\displaystyle\geq ϵ|bf|416ϵλ2(2λ1)2(1+f02)22λt124λ\displaystyle-\epsilon|\nabla_{b}f|^{4}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}\cdot t_{1}^{2-4\lambda}
4k1λ(1+f02)λ1|f0||bf|2t12λ1\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}\cdot t_{1}^{2\lambda-1} (3.21)
\displaystyle\geq ϵ|bf|44ϵ1λ2k12(1+f02)2λ2f02t14λ2,\displaystyle-\epsilon|\nabla_{b}f|^{4}-4\epsilon^{-1}\lambda^{2}k_{1}^{2}(1+f_{0}^{2})^{2\lambda-2}f_{0}^{2}\cdot t_{1}^{4\lambda-2},

where ϵ\epsilon is a constant to be determined. Hence we have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ222C1R|bf|ϕ3λt1\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}-\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F} (3.22)
+ϕ2t12{1m(|bf|2ft)2+4mf022ϵ|bf|4\displaystyle+\phi^{2}t^{2}_{1}\{\frac{1}{m}(|\nabla_{b}f|^{2}-f_{t})^{2}+4mf^{2}_{0}-2\epsilon|\nabla_{b}f|^{4}
16ϵλ2(2λ1)2(1+f02)22λt124λ(2k+2k1)|bf|2\displaystyle-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{2-4\lambda}-(2k+2k_{1})|\nabla_{b}f|^{2}
4ϵ1k12λ2(1+f02)2λ2|f0|2t14λ2\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}\cdot t_{1}^{4\lambda-2}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λ2}.\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda-2}\}.

Note that it is difficult to estimate \mathcal{F} directly. Let us recall the method of Cao-Yau [8] for weakly elliptic operators on closed manifolds. Translating their idea to pseudo-Hermitian case, they actually tried to control \mathcal{F} by either δ0|bf|2δft\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t} (δ>δ0>1\delta>\delta_{0}>1) or t12λ1(1+f02)λt_{1}^{2\lambda-1}(1+f^{2}_{0})^{\lambda}. However, Cao-Yau’s estimates cannot be applied directly to give the required inequalities in the complete noncompact case. We have to treat some extra terms appearing in (3.22), e.g., 2C1R|bf|ϕ3λt1\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F}. Our discussion will be divided into two cases according to the sign of ftf_{t} as follows.

𝐂𝐚𝐬𝐞𝐀\mathbf{Case\ A} ft<0f_{t}<0 at the maximum point (x1,t1)(x_{1},t_{1}).

In this case, we have

(|bf|2ft)2|bf|4+ft2.\displaystyle(|\nabla_{b}f|^{2}-f_{t})^{2}\geq|\nabla_{b}f|^{4}+f^{2}_{t}.

Then (3.22) becomes

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+2mϕ2t12f02+ϕ2t122m(|bf|2ft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+2m\phi^{2}t_{1}^{2}f_{0}^{2}+\frac{\phi^{2}t_{1}^{2}}{2m}(|\nabla_{b}f|^{2}-f_{t})^{2} (3.23)
+ϕ2t12{(12m2ϵ)|bf|42k|bf|22k1|bf|2}\displaystyle+\phi^{2}t^{2}_{1}\{(\frac{1}{2m}-2\epsilon)|\nabla_{b}f|^{4}-2k|\nabla_{b}f|^{2}-2k_{1}|\nabla_{b}f|^{2}\}
+ϕ2t12{2mf0216ϵλ2(2λ1)2(1+f02)22λt124λ\displaystyle+\phi^{2}t^{2}_{1}\{2mf_{0}^{2}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{2-4\lambda}
4ϵ1k12λ2(1+f02)2λ2|f0|2t14λ2\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}\cdot t_{1}^{4\lambda-2}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λ2}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda-2}\}
+ϕ2t1212mft22C1R|bf|ϕ3λt1.\displaystyle+\phi^{2}t^{2}_{1}\frac{1}{2m}f_{t}^{2}-\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F}.

Following the idea in [8], we want to control \mathcal{F} by either δ0|bf|2δft\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t} (δ>δ0>1\delta>\delta_{0}>1) or t12λ1(1+f02)λt_{1}^{2\lambda-1}(1+f^{2}_{0})^{\lambda}.

(𝐀𝟏)\mathbf{(A1)} Suppose δ0|bf|2δftt12λ1(1+f02)λ\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\geq t_{1}^{2\lambda-1}(1+f^{2}_{0})^{\lambda}. Clearly

t1(|bf|2δft+δ0|bf|2δft)=t1((δ0+1)|bf|22δft).\displaystyle\mathcal{F}\leq t_{1}(|\nabla_{b}f|^{2}-\delta f_{t}+\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})=t_{1}((\delta_{0}+1)|\nabla_{b}f|^{2}-2\delta f_{t}).

Consequently the last term of (3.23) can be estimate by

2C1R|bf|ϕ3λt1\displaystyle\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F} (3.24)
\displaystyle\leq 2C1R(δ0+1)t12(ϕ|bf|2)32+4C1Rt12δ(ϕ|bf|2)12|ϕft|\displaystyle\frac{2C_{1}}{R}(\delta_{0}+1)t_{1}^{2}(\phi|\nabla_{b}f|^{2})^{\frac{3}{2}}+\frac{4C_{1}}{R}t_{1}^{2}\delta(\phi|\nabla_{b}f|^{2})^{\frac{1}{2}}\cdot|\phi f_{t}|
\displaystyle\leq 2C1R(δ0+1)t12(ϕ|bf|2)32+ϕ2t1212mft2+8mC12R2δ2t12ϕ|bf|2.\displaystyle\frac{2C_{1}}{R}(\delta_{0}+1)t_{1}^{2}(\phi|\nabla_{b}f|^{2})^{\frac{3}{2}}+\phi^{2}t^{2}_{1}\frac{1}{2m}f_{t}^{2}+\frac{8mC_{1}^{2}}{R^{2}}\delta^{2}t_{1}^{2}\phi|\nabla_{b}f|^{2}.

Noting that 0<t110<t_{1}\leq 1, (3.23) and (3.24) yield that

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+2mϕ2t12f02+ϕ2t122m(|bf|2ft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+2m\phi^{2}t_{1}^{2}f_{0}^{2}+\frac{\phi^{2}t_{1}^{2}}{2m}(|\nabla_{b}f|^{2}-f_{t})^{2} (3.25)
+t12{(12m2ϵ)(ϕ|bf|2)22(k+k1+4mC12R2δ2)ϕ|bf|2\displaystyle+t^{2}_{1}\{(\frac{1}{2m}-2\epsilon)(\phi|\nabla_{b}f|^{2})^{2}-2(k+k_{1}+\frac{4mC_{1}^{2}}{R^{2}}\delta^{2})\phi|\nabla_{b}f|^{2}
2C1R(δ0+1)(ϕ|bf|2)32}\displaystyle-\frac{2C_{1}}{R}(\delta_{0}+1)(\phi|\nabla_{b}f|^{2})^{\frac{3}{2}}\}
+ϕ2{2m(t1f0)216ϵλ2(2λ1)2(1+f02)22λt144λ\displaystyle+\phi^{2}\{2m(t_{1}f_{0})^{2}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{4-4\lambda}
4ϵ1k12λ2(1+f02)2λ2|f0|2t14λ\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}\cdot t_{1}^{4\lambda}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λ}.\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda}\}.

Let ϵ<14m\epsilon<\frac{1}{4m}. By (3.25), we have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+ϕ2t122m(|bf|2ft)2C2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+\frac{\phi^{2}t_{1}^{2}}{2m}(|\nabla_{b}f|^{2}-f_{t})^{2}-C_{2} (3.26)
t12{C2+C2R2+C2R4},\displaystyle-t^{2}_{1}\{C_{2}+\frac{C_{2}}{R^{2}}+\frac{C_{2}}{R^{4}}\},

where C2C_{2} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Let x=ϕ(δ0|bf|2δft)(x1,t1)x=\phi(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})(x_{1},t_{1}), then

ϕ1t1x,ϕ2t1x,ϕ(|bf|2ft)1δx.\displaystyle\phi\mathcal{F}_{1}\leq t_{1}x,\ \phi\mathcal{F}_{2}\leq t_{1}x,\ \phi(|\nabla_{b}f|^{2}-f_{t})\geq\frac{1}{\delta}x.

Therefore we find that

0t122mδ2x2(2λ+1)t1x6C1Rt12xt12(C2+C2R2+C2R4)C2,\displaystyle 0\geq\frac{t_{1}^{2}}{2m\delta^{2}}x^{2}-(2\lambda+1)t_{1}x-\frac{6C_{1}}{R}t_{1}^{2}x-t_{1}^{2}(C_{2}+\frac{C_{2}}{R^{2}}+\frac{C_{2}}{R^{4}})-C_{2}, (3.27)

which implies that

t1xC3(1+1R),\displaystyle t_{1}x\leq C_{3}(1+\frac{1}{R}),

and

ϕ2t1x2C3(1+1R),\displaystyle\phi\mathcal{F}\leq 2t_{1}x\leq 2C_{3}(1+\frac{1}{R}), (3.28)

where C3C_{3} is a constant depending on m,k,k1,λ,δ0,δm,k,k_{1},\lambda,\delta_{0},\delta.

(𝐀𝟐)\mathbf{(A2)} Suppose δ0|bf|2δftt12λ1(1+f02)λ\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\leq t_{1}^{2\lambda-1}(1+f^{2}_{0})^{\lambda}. Then

ϕ\displaystyle\phi\mathcal{F} =\displaystyle= ϕt1(δ0|bf|2+t12λ1(1+f02)λδft+(1δ0)|bf|2)\displaystyle\phi t_{1}(\delta_{0}|\nabla_{b}f|^{2}+t_{1}^{2\lambda-1}(1+f_{0}^{2})^{\lambda}-\delta f_{t}+(1-\delta_{0})|\nabla_{b}f|^{2})
\displaystyle\leq 2ϕt12λ(1+f02)λ.\displaystyle 2\phi t_{1}^{2\lambda}(1+f_{0}^{2})^{\lambda}.

Since ft<0,t11f_{t}<0,t_{1}\leq 1 and λ>12\lambda>\frac{1}{2}, the assumption (A2) implies

(1+f02)λδ0|bf|2.\displaystyle(1+f_{0}^{2})^{\lambda}\geq\delta_{0}|\nabla_{b}f|^{2}.

The last term of (3.23) is bounded by

2C1R|bf|ϕ3λt1\displaystyle\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F} \displaystyle\leq 4C1Rϕ3λt12|bf|(1+f02)λ\displaystyle\frac{4C_{1}}{R}\phi^{3\lambda}t^{2}_{1}|\nabla_{b}f|\cdot(1+f_{0}^{2})^{\lambda}
\displaystyle\leq 4C1Rδ0ϕ3λt12(1+f02)32λ\displaystyle\frac{4C_{1}}{R\sqrt{\delta_{0}}}\phi^{3\lambda}t^{2}_{1}\cdot(1+f_{0}^{2})^{\frac{3}{2}\lambda}
=\displaystyle= 4C1Rδ0t12[ϕ2(1+f02)]32λ.\displaystyle\frac{4C_{1}}{R\sqrt{\delta_{0}}}t^{2}_{1}\cdot[\phi^{2}(1+f_{0}^{2})]^{\frac{3}{2}\lambda}.

Hence (3.23) becomes

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+2mϕ2t12f02\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+2m\phi^{2}t_{1}^{2}f_{0}^{2} (3.29)
+t12{(12m2ϵ)(ϕ|bf|2)22(k+k1)ϕ|bf|2}\displaystyle+t^{2}_{1}\{(\frac{1}{2m}-2\epsilon)(\phi|\nabla_{b}f|^{2})^{2}-2(k+k_{1})\phi|\nabla_{b}f|^{2}\}
+{2m(ϕt1f0)216ϵλ2(2λ1)2(1+f02)22λt144λϕ2\displaystyle+\{2m(\phi t_{1}f_{0})^{2}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{4-4\lambda}\phi^{2}
4ϵ1k12λ2(1+f02)2λ2|f0|2t14λϕ2\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}\cdot t_{1}^{4\lambda}\phi^{2}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λϕ2\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda}\phi^{2}
4C1Rδ0t12[ϕ2(1+f02)]32λ}.\displaystyle-\frac{4C_{1}}{R\sqrt{\delta_{0}}}t^{2}_{1}\cdot[\phi^{2}(1+f_{0}^{2})]^{\frac{3}{2}\lambda}\}.\

Letting ϵ<14m\epsilon<\frac{1}{4m} and noting that 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3}, we have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+2mϕ2t12f02C4(1+1R),\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+2m\phi^{2}t_{1}^{2}f_{0}^{2}-C_{4}(1+\frac{1}{R}),

where C4C_{4} is a constant depending on m,k,k1,δ0,λm,k,k_{1},\delta_{0},\lambda. Set y=ϕ|f0|y=\phi|f_{0}|, then

02mt12y2(2λ+1+6C1R)(t1y)2λC5(1+1R+1R4),\displaystyle 0\geq 2mt_{1}^{2}y^{2}-(2\lambda+1+\frac{6C_{1}}{R})(t_{1}y)^{2\lambda}-C_{5}(1+\frac{1}{R}+\frac{1}{R^{4}}), (3.30)

which yields that

t1y\displaystyle t_{1}y\leq C6(1+1R),\displaystyle C_{6}(1+\frac{1}{\sqrt{R}}),

hence

ϕ2ϕt12λ(1+f02)λC7(1+1Rλ),\displaystyle\phi\mathcal{F}\leq 2\phi t_{1}^{2\lambda}(1+f^{2}_{0})^{\lambda}\leq C_{7}(1+\frac{1}{R^{\lambda}}), (3.31)

where C5,C6,C7C_{5},C_{6},C_{7} are constants depending on m,k,k1,δ0,λm,k,k_{1},\delta_{0},\lambda.

𝐂𝐚𝐬𝐞𝐁\mathbf{Case\ B} ft0f_{t}\geq 0 at the maximal point (x1,t1)(x_{1},t_{1}).

(𝐁𝟏)\mathbf{(B1)} Suppose δ0|bf|2δft0\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\geq 0. We observe that

(|bf|2ft)2\displaystyle(|\nabla_{b}f|^{2}-f_{t})^{2} =\displaystyle= {1δ(δ0|bf|2δft)+(1δ0δ)|bf|2}2\displaystyle\{\frac{1}{\delta}(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})+(1-\frac{\delta_{0}}{\delta})|\nabla_{b}f|^{2}\}^{2} (3.32)
\displaystyle\geq 1δ2(δ0|bf|2δft)2+(1δ0δ)2|bf|4.\displaystyle\frac{1}{\delta^{2}}(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})^{2}+(1-\frac{\delta_{0}}{\delta})^{2}|\nabla_{b}f|^{4}.

Thus (3.22) becomes

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+ϕ2t12mδ2(δ0|bf|2δft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+\frac{\phi^{2}t_{1}^{2}}{m\delta^{2}}(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})^{2} (3.33)
+ϕ2t12{((δδ0)2mδ22ϵ)|bf|42(k+k1)|bf|2}\displaystyle+\phi^{2}t^{2}_{1}\{(\frac{(\delta-\delta_{0})^{2}}{m\delta^{2}}-2\epsilon)|\nabla_{b}f|^{4}-2(k+k_{1})|\nabla_{b}f|^{2}\}
+ϕ2t12{4mf0216ϵλ2(2λ1)2(1+f02)22λt124λ\displaystyle+\phi^{2}t_{1}^{2}\{4mf^{2}_{0}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{2-4\lambda}
4ϵ1k12λ2(1+f02)2λ2|f0|2t14λ2\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}\cdot t_{1}^{4\lambda-2}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λ2}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda-2}\}
2C1R|bf|ϕ3λt1.\displaystyle-\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F}.

In this case, the proof is almost the same as that for the case of ft<0f_{t}<0.

(𝐁𝟏\mathbf{(B1}-𝟏)\mathbf{1)} Suppose δ0|bf|2δftt12λ1(1+f02)λ\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\geq t_{1}^{2\lambda-1}(1+f^{2}_{0})^{\lambda}. Using the assumption that ft0f_{t}\geq 0, we have

t1(|bf|2δft+δ0|bf|2δft)=t1((δ0+1)|bf|2).\displaystyle\mathcal{F}\leq t_{1}(|\nabla_{b}f|^{2}-\delta f_{t}+\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})=t_{1}((\delta_{0}+1)|\nabla_{b}f|^{2}).

Hence the last term of (3.33) can be estimated by

2C1R|bf|ϕ3λt12C1R(δ0+1)t12(ϕ|bf|2)32.\displaystyle\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F}\leq\frac{2C_{1}}{R}(\delta_{0}+1)t_{1}^{2}(\phi|\nabla_{b}f|^{2})^{\frac{3}{2}}. (3.34)

Therefore (3.33) and (3.34) yield that

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+ϕ2t12mδ2(δ0|bf|2δft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+\frac{\phi^{2}t_{1}^{2}}{m\delta^{2}}(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})^{2} (3.35)
+t12{((δδ0)2mδ22ϵ)ϕ2|bf|42(k+k1)ϕ2|bf|2\displaystyle+t^{2}_{1}\{(\frac{(\delta-\delta_{0})^{2}}{m\delta^{2}}-2\epsilon)\phi^{2}|\nabla_{b}f|^{4}-2(k+k_{1})\phi^{2}|\nabla_{b}f|^{2}
2C1R(δ0+1)(ϕ|bf|2)32}\displaystyle-\frac{2C_{1}}{R}(\delta_{0}+1)(\phi|\nabla_{b}f|^{2})^{\frac{3}{2}}\}
+ϕ2{4m(t1|f0|)216ϵλ2(2λ1)2(1+f02)22λt144λ\displaystyle+\phi^{2}\{4m(t_{1}|f_{0}|)^{2}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{4-4\lambda}
4ϵ1k12λ2(1+f02)2λ2|f0|2t14λ\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}\cdot t_{1}^{4\lambda}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λ}.\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda}\}.

Let ϵ<(δδ0)22mδ2\epsilon<\frac{(\delta-\delta_{0})^{2}}{2m\delta^{2}}. By (3.35), we have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+ϕ2t12mδ2(δ0|bf|2δft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+\frac{\phi^{2}t_{1}^{2}}{m\delta^{2}}(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})^{2} (3.36)
t12{C8+C8R4}C8,\displaystyle-t^{2}_{1}\{C_{8}+\frac{C_{8}}{R^{4}}\}-C_{8},

where C8C_{8} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Let x=ϕ(δ0|bf|2δft)(x1,t1)x=\phi(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})(x_{1},t_{1}), then

ϕ1t1x,ϕ2t1x,ϕ(|bf|2ft)1δx.\displaystyle\phi\mathcal{F}_{1}\leq t_{1}x,\ \phi\mathcal{F}_{2}\leq t_{1}x,\ \phi(|\nabla_{b}f|^{2}-f_{t})\geq\frac{1}{\delta}x.

Therefore we find that

0t12mδ2x2(2λ+1)t1x6C1Rt12xt12(C8+C8R4)C8,\displaystyle 0\geq\frac{t_{1}^{2}}{m\delta^{2}}x^{2}-(2\lambda+1)t_{1}x-\frac{6C_{1}}{R}t_{1}^{2}x-t_{1}^{2}(C_{8}+\frac{C_{8}}{R^{4}})-C_{8}, (3.37)

which implies that

t1xC9(1+1R),\displaystyle t_{1}x\leq C_{9}(1+\frac{1}{R}),

and

ϕ2t1x2C9(1+1R),\displaystyle\phi\mathcal{F}\leq 2t_{1}x\leq 2C_{9}(1+\frac{1}{R}), (3.38)

where C9C_{9} is a constant depending on m,k,k1,λ,δ0,δm,k,k_{1},\lambda,\delta_{0},\delta.

(𝐁𝟏\mathbf{(B1}-𝟐)\mathbf{2)} Suppose δ0|bf|2δftt12λ1(1+f02)λ\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\leq t_{1}^{2\lambda-1}(1+f^{2}_{0})^{\lambda}. Then

ϕ\displaystyle\phi\mathcal{F} =\displaystyle= ϕt1(δ0|bf|2+t12λ1(1+f02)λδft+(1δ0)|bf|2)\displaystyle\phi t_{1}(\delta_{0}|\nabla_{b}f|^{2}+t_{1}^{2\lambda-1}(1+f_{0}^{2})^{\lambda}-\delta f_{t}+(1-\delta_{0})|\nabla_{b}f|^{2})
\displaystyle\leq 2ϕt12λ(1+f02)λ.\displaystyle 2\phi t_{1}^{2\lambda}(1+f_{0}^{2})^{\lambda}.

The last term of (3.33) can be estimated by

2C1R|bf|ϕ3λt1\displaystyle\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{F} \displaystyle\leq 4C1Rϕ3λt12|bf|(1+f02)λ\displaystyle\frac{4C_{1}}{R}\phi^{3\lambda}t^{2}_{1}|\nabla_{b}f|\cdot(1+f_{0}^{2})^{\lambda}
\displaystyle\leq C10Rϕ3λt12[|bf|3+(1+f02)32λ]\displaystyle\frac{C_{10}}{R}\phi^{3\lambda}t^{2}_{1}\cdot[|\nabla_{b}f|^{3}+(1+f_{0}^{2})^{\frac{3}{2}\lambda}]
\displaystyle\leq C10Rt12[(ϕ|bf|2)32+|ϕ2(1+f02)|32λ],\displaystyle\frac{C_{10}}{R}t^{2}_{1}\cdot[(\phi|\nabla_{b}f|^{2})^{\frac{3}{2}}+|\phi^{2}(1+f_{0}^{2})|^{\frac{3}{2}\lambda}],

where we use the Young’s inequality ab13a3+23b32(a,b0)ab\leq\frac{1}{3}a^{3}+\frac{2}{3}b^{\frac{3}{2}}(a,b\geq 0) in the second inequality. Hence (3.33) becomes

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+2mϕ2t12f02\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+2m\phi^{2}t_{1}^{2}f_{0}^{2} (3.39)
+t12{((δδ0)2mδ22ϵ)(ϕ|bf|2)22(k+k1)ϕ|bf|2\displaystyle+t^{2}_{1}\{(\frac{(\delta-\delta_{0})^{2}}{m\delta^{2}}-2\epsilon)(\phi|\nabla_{b}f|^{2})^{2}-2(k+k_{1})\phi|\nabla_{b}f|^{2}
C10R(ϕ|bf|2)32}\displaystyle-\frac{C_{10}}{R}(\phi|\nabla_{b}f|^{2})^{\frac{3}{2}}\}
+{2m(ϕt1|f0|)216ϵλ2(2λ1)2(1+f02)22λt144λϕ2\displaystyle+\{2m(\phi t_{1}|f_{0}|)^{2}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{4-4\lambda}\phi^{2}
4ϵ1k12λ2(1+f02)2λ2|f0|2t14λϕ2\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}\cdot t_{1}^{4\lambda}\phi^{2}
2(k1+k12)λ2(1+f02)2λ2|f02|t14λϕ2\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\cdot t_{1}^{4\lambda}\phi^{2}
C10Rt12[ϕ2(1+f02)]32λ}.\displaystyle-\frac{C_{10}}{R}t^{2}_{1}\cdot[\phi^{2}(1+f_{0}^{2})]^{\frac{3}{2}\lambda}\}.\

Choosing ϵ<(δδ0)22mδ2\epsilon<\frac{(\delta-\delta_{0})^{2}}{2m\delta^{2}} and noting that λ<23\lambda<\frac{2}{3} and t11t_{1}\leq 1, we have

03C1Rϕ6λ2t1ϕ212λϕ22+2mϕ2t12f02C11(1+1R+1R4),\displaystyle 0\geq-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+2m\phi^{2}t_{1}^{2}f_{0}^{2}-C_{11}(1+\frac{1}{R}+\frac{1}{R^{4}}), (3.40)

where C11C_{11} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Set y=ϕ|f0|y=\phi|f_{0}|. Then

02mt12y2(2λ+1+6C1R)(t1y)2λC12(1+1R+1R4),\displaystyle 0\geq 2mt_{1}^{2}y^{2}-(2\lambda+1+\frac{6C_{1}}{R})(t_{1}y)^{2\lambda}-C_{12}(1+\frac{1}{R}+\frac{1}{R^{4}}),

which yields that

t1yC13(1+1R),\displaystyle t_{1}y\leq C_{13}(1+\frac{1}{\sqrt{R}}),

hence

ϕ2ϕt12λ(1+f02)λC14(1+1Rλ),\displaystyle\phi\mathcal{F}\leq 2\phi t_{1}^{2\lambda}(1+f^{2}_{0})^{\lambda}\leq C_{14}(1+\frac{1}{R^{\lambda}}), (3.41)

where C12,C13,C14C_{12},C_{13},C_{14} are constants depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda.

(𝐁𝟐)\mathbf{(B2)} Suppose δ0|bf|2δft0\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\leq 0. In this case, we can assume that

(δ01)|bf|2t12λ1(1+f02)λ.\displaystyle(\delta_{0}-1)|\nabla_{b}f|^{2}\leq t_{1}^{2\lambda-1}(1+f_{0}^{2})^{\lambda}. (3.42)

Otherwise

\displaystyle\mathcal{F} =\displaystyle= t1(|bf|2+t12λ1(1+f02)λδft)\displaystyle t_{1}(|\nabla_{b}f|^{2}+t_{1}^{2\lambda-1}(1+f_{0}^{2})^{\lambda}-\delta f_{t})
\displaystyle\leq t1(δ0|bf|2δft)0,\displaystyle t_{1}(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})\leq 0,

and thus the conclusion of Proposition 3.3 follows trivially. From (3.18) and (3.42), we have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1ϕ212λϕ22+2mϕ2t12f02\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+2m\phi^{2}t^{2}_{1}f^{2}_{0} (3.43)
+{(2m8λ(2λ1)(δ01))ϕ2t12f028λ(2λ1)(δ01)\displaystyle+\{(2m-\frac{8}{\lambda(2\lambda-1)(\delta_{0}-1)})\phi^{2}t_{1}^{2}f^{2}_{0}-\frac{8}{\lambda(2\lambda-1)(\delta_{0}-1)}
(2k+2k1)t12λ+1δ01ϕ2(1+f02)λ\displaystyle-(2k+2k_{1})\frac{t_{1}^{2\lambda+1}}{\delta_{0}-1}\phi^{2}(1+f_{0}^{2})^{\lambda}
4k1λt14λδ01ϕ2(1+f02)2λ1|f0|\displaystyle-\frac{4k_{1}\lambda t_{1}^{4\lambda}}{\delta_{0}-1}\phi^{2}(1+f_{0}^{2})^{2\lambda-1}|f_{0}|
(2k1+2k12)λ2ϕ2(1+f02)2λ2f02t14λ\displaystyle-(2k_{1}+2k_{1}^{2})\lambda^{2}\phi^{2}(1+f^{2}_{0})^{2\lambda-2}f^{2}_{0}\cdot t_{1}^{4\lambda}
2C1t13λ+12δ01R[ϕ2(1+f02)]32λ}.\displaystyle-\frac{2C_{1}t_{1}^{3\lambda+\frac{1}{2}}}{\sqrt{\delta_{0}-1}R}[\phi^{2}(1+f_{0}^{2})]^{\frac{3}{2}\lambda}\}.

Choosing δ0>1+4mλ(2λ1)\delta_{0}>1+\frac{4}{m\lambda(2\lambda-1)} and noting that λ<23\lambda<\frac{2}{3}, we obtain

03C1Rϕ6λ2t1ϕ212λϕ22+2mϕ2t12f02C15(1+1R),\displaystyle 0\geq-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{F}-\phi^{2}\mathcal{F}_{1}-2\lambda\phi^{2}\mathcal{F}_{2}+2m\phi^{2}t_{1}^{2}f_{0}^{2}-C_{15}(1+\frac{1}{R}), (3.44)

where C15C_{15} is a constant depending on m,k,k1,δ0,λm,k,k_{1},\delta_{0},\lambda. Set y=ϕ|f0|y=\phi|f_{0}|, we get

02mt12y2(2λ+3C1R)(t1y)2λC16(1+1R),\displaystyle 0\geq 2mt_{1}^{2}y^{2}-(2\lambda+\frac{3C_{1}}{R})(t_{1}y)^{2\lambda}-C_{16}(1+\frac{1}{R}),

which yields that

t1yC17(1+1R),\displaystyle t_{1}y\leq C_{17}(1+\frac{1}{\sqrt{R}}),

and

ϕϕt12λ(1+f02)λC18(1+1Rλ),\displaystyle\phi\mathcal{F}\leq\phi t_{1}^{2\lambda}(1+f^{2}_{0})^{\lambda}\leq C_{18}(1+\frac{1}{R^{\lambda}}),

where C16,C17,C18C_{16},C_{17},C_{18} are constants depending on m,k,k1,δ0,λm,k,k_{1},\delta_{0},\lambda.

From the above discussion, we conclude that

(ϕ)(x,t)C19(1+1Rλ)\displaystyle(\phi\mathcal{F})(x,t)\leq C_{19}(1+\frac{1}{R^{\lambda}}) (3.45)

on Bp(2R)×[0,1]B_{p}(2R)\times[0,1], where C19C_{19} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. It follows from (3.45) that

(x,t)C19(1+1Rλ)\displaystyle\mathcal{F}(x,t)\leq C_{19}(1+\frac{1}{R^{\lambda}}) (3.46)

on Bp(R)×[0,1]B_{p}(R)\times[0,1]. ∎  

The remaining part of this section is devoted to the case of t1t\geq 1, in which we will consider the auxiliary function

𝒢=t(|bf|2+(1+f02)λδft).\displaystyle\mathcal{G}=t(|\nabla_{b}f|^{2}+(1+f_{0}^{2})^{\lambda}-\delta f_{t}).

The argument for this case is almost the same as that for 0<t10<t\leq 1. Note that at t=1t=1, we have

(,1)=𝒢(,1).\displaystyle\mathcal{F}(\cdot,1)=\mathcal{G}(\cdot,1).

𝐏𝐫𝐨𝐩𝐨𝐬𝐢𝐭𝐢𝐨𝐧 3.4\mathbf{Proposition\ 3.4} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbkand|A|,|bA|k1.\displaystyle Ric_{b}+2(m-2)Tor_{b}\geq-k\ \ and\ \ |A|,|\nabla_{b}A|\leq k_{1}.

and uu be a positive solution of the CR heat equation

ut=Δbu\displaystyle\frac{\partial u}{\partial t}=\Delta_{b}u

on Bp(2R)×[1,T]B_{p}(2R)\times[1,T] with R1R\geq 1 and T>1T>1. Then for any constant 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3} and any constant δ>1+4mλ(2λ1)\delta>1+\frac{4}{m\lambda(2\lambda-1)}, there exists a constant C4C_{4}^{{}^{\prime}} depending on m,k,k1,λ,δm,k,k_{1},\lambda,\delta, such that

|bu|2u2+(1+u02u2)λδutuC4(1+1t+1Rλ)\displaystyle\frac{|\nabla_{b}u|^{2}}{u^{2}}+(1+\frac{u^{2}_{0}}{u^{2}})^{\lambda}-\delta\frac{u_{t}}{u}\leq C_{4}^{{}^{\prime}}(1+\frac{1}{t}+\frac{1}{R^{\lambda}}) (3.47)

on Bp(R)×[1,T]B_{p}(R)\times[1,T].

𝐏𝐫𝐨𝐨𝐟\mathbf{Proof} Let (x1,t1)(x_{1},t_{1}) be the maximum point of ϕ𝒢\phi\mathcal{G} on M×[1,T]M\times[1,T]. We may assume that ϕ𝒢\phi\mathcal{G} is positive at (x1,t1)(x_{1},t_{1}) and t1>1t_{1}>1, otherwise the result follows trivially. Evaluating the inequality in Lemma 3.2 for ϕ𝒢\phi\mathcal{G} at (x1,t1)(x_{1},t_{1}) gives the following

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ3𝒢ϕ𝒢t1+2bf,bϕ𝒢\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-3}\mathcal{G}-\frac{\phi\mathcal{G}}{t_{1}}+2\langle\nabla_{b}f,\nabla_{b}\phi\rangle\mathcal{G} (3.48)
+ϕt1{1m(Δbf)2+4mf02\displaystyle+\phi t_{1}\{\frac{1}{m}(\Delta_{b}f)^{2}+4mf^{2}_{0}
8λ(2λ1)(1+f02)1λ|bf|22(k+k1)|bf|2\displaystyle-\frac{8}{\lambda(2\lambda-1)}(1+f_{0}^{2})^{1-\lambda}|\nabla_{b}f|^{2}-2(k+k_{1})|\nabla_{b}f|^{2}
4k1λ(1+f02)λ1|f0||bf|2\displaystyle-4k_{1}\lambda(1+f^{2}_{0})^{\lambda-1}|f_{0}|\cdot|\nabla_{b}f|^{2}
2(k1+k12)λ2(1+f02)2λ2|f02|}.\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\}.

Multiplying (3.48) by ϕt1\phi t_{1} and using Cauchy-Schwarz inequality, we have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢2C1R|bf|ϕ3λt1𝒢\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}-\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{G} (3.49)
+ϕ2t12{1m(|bf|2ft)2+4mf022ϵ|bf|42(k+k1)|bf|2\displaystyle+\phi^{2}t^{2}_{1}\{\frac{1}{m}(|\nabla_{b}f|^{2}-f_{t})^{2}+4mf^{2}_{0}-2\epsilon|\nabla_{b}f|^{4}-2(k+k_{1})|\nabla_{b}f|^{2}
16ϵλ2(2λ1)2(1+f02)22λ\displaystyle-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}
(2k1+2k12+4ϵ1k12)λ2(1+f02)2λ2f02}.\displaystyle-(2k_{1}+2k_{1}^{2}+4\epsilon^{-1}k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}f_{0}^{2}\}.

The proof of Proposition 3.4 is almost same as that for Proposition 3.3. In following, we only show some necessary modifications when we try to control 𝒢\mathcal{G}. The discussion is similarly divided into the following cases.

𝐂𝐚𝐬𝐞𝐀~\mathbf{Case\ \tilde{A}} ft<0f_{t}<0 at the maximum point (x1,t1)(x_{1},t_{1}) of ϕ𝒢\phi\mathcal{G}. Corresponding to (3.23), we get the following inequality:

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢+2mϕ2t12f02+ϕ2t122m(|bf|2ft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}+2m\phi^{2}t_{1}^{2}f_{0}^{2}+\frac{\phi^{2}t_{1}^{2}}{2m}(|\nabla_{b}f|^{2}-f_{t})^{2} (3.50)
+ϕ2t12{(12m2ϵ)|bf|42k|bf|22k1|bf|2}\displaystyle+\phi^{2}t^{2}_{1}\{(\frac{1}{2m}-2\epsilon)|\nabla_{b}f|^{4}-2k|\nabla_{b}f|^{2}-2k_{1}|\nabla_{b}f|^{2}\}
+ϕ2t12{2mf0216ϵλ2(2λ1)2(1+f02)22λ\displaystyle+\phi^{2}t^{2}_{1}\{2mf_{0}^{2}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}
(2k1+2k12+4ϵ1k12)λ2(1+f02)2λ2f02}\displaystyle-(2k_{1}+2k_{1}^{2}+4\epsilon^{-1}k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}f_{0}^{2}\}
+ϕ2t1212mft22C1R|bf|ϕ3λt1𝒢.\displaystyle+\phi^{2}t^{2}_{1}\frac{1}{2m}f_{t}^{2}-\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{G}.

(𝐀~𝟏)\mathbf{(\tilde{A}1)} Suppose δ0|bf|2δft(1+f02)λ\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\geq(1+f^{2}_{0})^{\lambda}. Let ϵ<14m\epsilon<\frac{1}{4m}. Similar to getting (3.26) from (3.25), we obtain from (3.50) that

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢+ϕ2t122m(|bf|2ft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}+\frac{\phi^{2}t_{1}^{2}}{2m}(|\nabla_{b}f|^{2}-f_{t})^{2} (3.51)
C20t12{1+1R2+1R4},\displaystyle-C_{20}t^{2}_{1}\{1+\frac{1}{R^{2}}+\frac{1}{R^{4}}\},

where C20C_{20} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Let x=ϕ(δ0|bf|2δft)(x1,t1)x=\phi(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})(x_{1},t_{1}), then

ϕ𝒢2t1x,ϕ(|bf|2ft)1δx.\displaystyle\phi\mathcal{G}\leq 2t_{1}x,\ \phi(|\nabla_{b}f|^{2}-f_{t})\geq\frac{1}{\delta}x.

Consequently we have

0t122mδ2x22t1x6C1Rt12xC20t12(1+1R2+1R4).\displaystyle 0\geq\frac{t_{1}^{2}}{2m\delta^{2}}x^{2}-2t_{1}x-\frac{6C_{1}}{R}t_{1}^{2}x-C_{20}t_{1}^{2}(1+\frac{1}{R^{2}}+\frac{1}{R^{4}}). (3.52)

This implies that

t1xC21(1+t1+t1R),\displaystyle t_{1}x\leq C_{21}(1+t_{1}+\frac{t_{1}}{R}),

and

ϕ𝒢2t1x2C21(1+t1+t1R),\displaystyle\phi\mathcal{G}\leq 2t_{1}x\leq 2C_{21}(1+t_{1}+\frac{t_{1}}{R}), (3.53)

where C21C_{21} is a constant depending on m,k,k1,λ,δ0,δm,k,k_{1},\lambda,\delta_{0},\delta.

(𝐀~𝟐)\mathbf{(\tilde{A}2)} Suppose δ0|bf|2δft(1+f02)λ\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\leq(1+f^{2}_{0})^{\lambda}. Then

ϕ𝒢2ϕt1(1+f02)λ.\displaystyle\phi\mathcal{G}\leq 2\phi t_{1}(1+f_{0}^{2})^{\lambda}.

Similar to getting (3.29) from (3.23), we have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢+2mϕ2t12f02\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}+2m\phi^{2}t_{1}^{2}f_{0}^{2} (3.54)
+t12{(12m2ϵ)(ϕ|bf|2)22(k+k1)ϕ|bf|2}\displaystyle+t^{2}_{1}\{(\frac{1}{2m}-2\epsilon)(\phi|\nabla_{b}f|^{2})^{2}-2(k+k_{1})\phi|\nabla_{b}f|^{2}\}
+t12{2m(ϕ|f0|)216ϵλ2(2λ1)216ϵλ2(2λ1)2(ϕ|f0|)44λ\displaystyle+t^{2}_{1}\{2m(\phi|f_{0}|)^{2}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(\phi|f_{0}|)^{4-4\lambda}
(2k1+2k12+4ϵ1k12)λ2(ϕ|f0|)4λ2}\displaystyle-(2k_{1}+2k_{1}^{2}+4\epsilon^{-1}k_{1}^{2})\lambda^{2}(\phi|f_{0}|)^{4\lambda-2}\}
4C1Rδ0[ϕ2(1+f02)]32λ}.\displaystyle-\frac{4C_{1}}{R\sqrt{\delta_{0}}}[\phi^{2}(1+f_{0}^{2})]^{\frac{3}{2}\lambda}\}.\

Choosing ϵ<14m\epsilon<\frac{1}{4m} and noting that 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3}, (3.54) implies that

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢+2mϕ2t12f02C22t12(1+1R),\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}+2m\phi^{2}t_{1}^{2}f_{0}^{2}-C_{22}t_{1}^{2}(1+\frac{1}{R}),

where C22C_{22} is a constant depending on m,k,k1,δ0,λm,k,k_{1},\delta_{0},\lambda. Set y=ϕ|f0|y=\phi|f_{0}|. Then

02mt12y2(2t1+6C1t12R)y2λC23t12(1+1R),\displaystyle 0\geq 2mt_{1}^{2}y^{2}-(2t_{1}+\frac{6C_{1}t_{1}^{2}}{R})y^{2\lambda}-C_{23}t_{1}^{2}(1+\frac{1}{R}),

where C23C_{23} is a constant depending on m,k,k1,δ0,λm,k,k_{1},\delta_{0},\lambda. Since t1>1t_{1}>1, we have the following inequality

02my2(2+6C1R)y2λC23(1+1R+1R4),\displaystyle 0\geq 2my^{2}-(2+\frac{6C_{1}}{R})y^{2\lambda}-C_{23}(1+\frac{1}{R}+\frac{1}{R^{4}}),

which yields that

y\displaystyle y \displaystyle\leq C24(1+1R+1R2+R12(λ1))\displaystyle C_{24}(1+\frac{1}{\sqrt{R}}+\frac{1}{R^{2}}+R^{\frac{1}{2(\lambda-1)}})
\displaystyle\leq C25(1+1R),\displaystyle C_{25}(1+\frac{1}{\sqrt{R}}),

and thus

ϕ𝒢2ϕt1(1+f02)λC25(1+1Rλ)t1,\displaystyle\phi\mathcal{G}\leq 2\phi t_{1}(1+f^{2}_{0})^{\lambda}\leq C_{25}(1+\frac{1}{R^{\lambda}})t_{1}, (3.55)

where C24,C25C_{24},C_{25} are constants depending on m,k,k1,δ0,λm,k,k_{1},\delta_{0},\lambda.

𝐂𝐚𝐬𝐞𝐁~\mathbf{Case\ \tilde{B}} ft0f_{t}\geq 0 at maximum point (x1,t1)(x_{1},t_{1}) of ϕ𝒢\phi\mathcal{G}.

(𝐁~𝟏)\mathbf{(\tilde{B}1)} Suppose δ0|bf|2δft0\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\geq 0. We have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢+ϕ2t12mδ2(δ0|bf|2δft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}+\frac{\phi^{2}t_{1}^{2}}{m\delta^{2}}(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})^{2} (3.56)
+ϕ2t12{((δδ0)2mδ22ϵ)|bf|42(k+k1)|bf|2}\displaystyle+\phi^{2}t^{2}_{1}\{(\frac{(\delta-\delta_{0})^{2}}{m\delta^{2}}-2\epsilon)|\nabla_{b}f|^{4}-2(k+k_{1})|\nabla_{b}f|^{2}\}
+ϕ2t12{4mf0216ϵλ2(2λ1)2(1+f02)22λt124λ\displaystyle+\phi^{2}t_{1}^{2}\{4mf^{2}_{0}-\frac{16}{\epsilon\lambda^{2}(2\lambda-1)^{2}}(1+f_{0}^{2})^{2-2\lambda}t_{1}^{2-4\lambda}
4ϵ1k12λ2(1+f02)2λ2|f0|2\displaystyle-4\epsilon^{-1}k^{2}_{1}\lambda^{2}(1+f^{2}_{0})^{2\lambda-2}|f_{0}|^{2}
2(k1+k12)λ2(1+f02)2λ2|f02|}\displaystyle-2(k_{1}+k_{1}^{2})\lambda^{2}(1+f_{0}^{2})^{2\lambda-2}|f_{0}^{2}|\}
2C1R|bf|ϕ3λt1𝒢.\displaystyle-\frac{2C_{1}}{R}|\nabla_{b}f|\cdot\phi^{3\lambda}t_{1}\mathcal{G}.

(𝐁~𝟏\mathbf{(\tilde{B}1}-𝟏)\mathbf{1)} Suppose δ0|bf|2δft(1+f02)λ\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\geq(1+f^{2}_{0})^{\lambda}. Similar to (3.36), we have

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢+ϕ2t12mδ2(δ0|bf|2δft)2\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}+\frac{\phi^{2}t_{1}^{2}}{m\delta^{2}}(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})^{2} (3.57)
t12{C26+C26R},\displaystyle-t^{2}_{1}\{C_{26}+\frac{C_{26}}{R}\},

where C26C_{26} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Let x=ϕ(δ0|bf|2δft)(x1,t1)x=\phi(\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t})(x_{1},t_{1}). Then we may get

t1xC27(1+t1+t1R),\displaystyle t_{1}x\leq C_{27}(1+t_{1}+\frac{t_{1}}{R}),

that is,

ϕ𝒢2t1x2C27(1+t1+t1R),\displaystyle\phi\mathcal{G}\leq 2t_{1}x\leq 2C_{27}(1+t_{1}+\frac{t_{1}}{R}), (3.58)

where C27C_{27} is a constant depending on m,k,k1,λ,δ0,δm,k,k_{1},\lambda,\delta_{0},\delta.

(𝐁~𝟏\mathbf{(\tilde{B}1}-𝟐)\mathbf{2)} Suppose δ0|bf|2δft(1+f02)λ\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\leq(1+f^{2}_{0})^{\lambda}. Corresponding to (3.40), we get the following

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢+2mϕ2t12f02C28(1+1R),\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}+2m\phi^{2}t_{1}^{2}f_{0}^{2}-C_{28}(1+\frac{1}{R}),

where C28C_{28} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Set y=ϕ|f0|y=\phi|f_{0}|. Then

02mt12y2(2t1+6C1t12R)y2λC29t12(1+1R),\displaystyle 0\geq 2mt_{1}^{2}y^{2}-(2t_{1}+\frac{6C_{1}t_{1}^{2}}{R})y^{2\lambda}-C_{29}t_{1}^{2}(1+\frac{1}{R}), (3.59)

where C29C_{29} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Since t1>1t_{1}>1, we find that

02my2(2+6C1R)y2λC30(1+1R),\displaystyle 0\geq 2my^{2}-(2+\frac{6C_{1}}{R})y^{2\lambda}-C_{30}(1+\frac{1}{R}),

which yields that

y\displaystyle y\leq C31(1+1R),\displaystyle C_{31}(1+\frac{1}{\sqrt{R}}),

hence

ϕ𝒢2ϕt1(1+f02)λC32(1+1Rλ)t1,\displaystyle\phi\mathcal{G}\leq 2\phi t_{1}(1+f^{2}_{0})^{\lambda}\leq C_{32}(1+\frac{1}{R^{\lambda}})t_{1}, (3.60)

where C30,C31,C32C_{30},C_{31},C_{32} are constants depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda.

(𝐁~𝟐)\mathbf{(\tilde{B}2)} Suppose δ0|bf|2δft0\delta_{0}|\nabla_{b}f|^{2}-\delta f_{t}\leq 0. Assume that δ0>1+4mλ(2λ1)\delta_{0}>1+\frac{4}{m\lambda(2\lambda-1)}. Similar to getting (3.44) from (3.42) and (3.43), we obtain from (3.48) that

0\displaystyle 0 \displaystyle\geq 3C1Rϕ6λ2t1𝒢ϕ2𝒢+2mϕ2t12f02C33(1+1R),\displaystyle-\frac{3C_{1}}{R}\phi^{6\lambda-2}t_{1}\mathcal{G}-\phi^{2}\mathcal{G}+2m\phi^{2}t_{1}^{2}f_{0}^{2}-C_{33}(1+\frac{1}{R}),

where C33C_{33} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Set y=ϕ|f0|y=\phi|f_{0}|, we get

02mt12y2(2t1+3C1t12R)y2λC34(1+1R)\displaystyle 0\geq 2mt_{1}^{2}y^{2}-(2t_{1}+\frac{3C_{1}t_{1}^{2}}{R})y^{2\lambda}-C_{34}(1+\frac{1}{R})

which yields that

yC35(1+1R),\displaystyle y\leq C_{35}(1+\frac{1}{\sqrt{R}}),

and thus

ϕ𝒢ϕt1(1+f02)λC36(1+1Rλ)t1,\displaystyle\phi\mathcal{G}\leq\phi t_{1}(1+f^{2}_{0})^{\lambda}\leq C_{36}(1+\frac{1}{R^{\lambda}})t_{1}, (3.61)

where C34,C35,C36C_{34},C_{35},C_{36} are constants depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda.

From the above discussion, we conclude that

ϕ(x)𝒢(x,t)C37(1+t1+t1Rλ),\displaystyle\phi(x)\mathcal{G}(x,t)\leq C_{37}(1+t_{1}+\frac{t_{1}}{R^{\lambda}}), (3.62)

on Bp(2R)×[1,T]B_{p}(2R)\times[1,T], where C37C_{37} is a constant depending on m,k,k1,δ,δ0,λm,k,k_{1},\delta,\delta_{0},\lambda. Consequently

𝒢(x,t)C37(1+t1+t1Rλ)\displaystyle\mathcal{G}(x,t)\leq C_{37}(1+t_{1}+\frac{t_{1}}{R^{\lambda}}) (3.63)

on Bp(R)×[1,T]B_{p}(R)\times[1,T]. In particular, we have

𝒢(x,T)C37(1+t1+t1Rλ)C37(1+T+TRλ)\displaystyle\mathcal{G}(x,T)\leq C_{37}(1+t_{1}+\frac{t_{1}}{R^{\lambda}})\leq C_{37}(1+T+\frac{T}{R^{\lambda}}) (3.64)

on Bp(R)B_{p}(R). Since T(>1)T(>1) is arbitrary, this gives (3.47). ∎

Combining Propositions 3.3 and 3.4, we may obtain Theorem 1.1. Clearly Theorem 1.2 follows from Theorem 1.1 by letting RR\rightarrow\infty. We would like to end this section by the following remark.

𝐑𝐞𝐦𝐚𝐫𝐤 3.1.\mathbf{Remark\ 3.1.} If MM is a closed pseudo-Hermitian manifold with the same properties as in Theorem 1.1, we may carry out the above argument, without using the cut-off function, to deduce the following result: Let uu be a positive solution of the CR heat equation on M2m+1M^{2m+1}. Then for any constant 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3} and any constant δ>1+4mλ(2λ1)\delta>1+\frac{4}{m\lambda(2\lambda-1)}, there exists a constant CC depending on m,k,k1,λ,δm,k,k_{1},\lambda,\delta, such that

|bu|2u2δutuC+Ct\displaystyle\frac{|\nabla_{b}u|^{2}}{u^{2}}-\delta\frac{u_{t}}{u}\leq C+\frac{C}{t} (3.65)

on M×(0,)M\times(0,\infty). We should point out that the sub-Laplacian Δb\Delta_{b} can only be expressed as (1.1) locally, that is,

Δb=A=12meA2A=12meAeA,\displaystyle\Delta_{b}=\sum\limits_{A=1}^{2m}e_{A}^{2}-\sum\limits_{A=1}^{2m}\nabla_{e_{A}}e_{A}, (3.66)

where {eA}A=12m\{e_{A}\}_{A=1}^{2m} is the local frame field given in §2\S 2, and \nabla is the Tanaka-Webster connection. In general, one cannot express Δb\Delta_{b} as (3.66) by global vector fields. Hence, although the method for the closed case follows essentially from Cao and Yau [8], their result cannot be applied directly to get the estimate (3.65).

𝐑𝐞𝐦𝐚𝐫𝐤 3.2.\mathbf{Remark\ 3.2.} Note that 1+4mλ(2λ1)>1+18m1+\frac{4}{m\lambda(2\lambda-1)}>1+\frac{18}{m} for 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3}. Let M2m+1M^{2m+1} be either a closed pseudo-Hermitian manifold or a complete noncompact pseudo-Hermitian manifold with the same properties as in Theorem 1.1. Then we have the following Li-Yau type estimate: Let uu be a positive solution of the CR heat equation on MM. Then for any δ>1+18m\delta>1+\frac{18}{m}, there exists a constant CC depending on m,k,k1,δm,k,k_{1},\delta, such that

|bu|2u2δutuC+Ct\displaystyle\frac{|\nabla_{b}u|^{2}}{u^{2}}-\delta\frac{u_{t}}{u}\leq C+\frac{C}{t} (3.67)

on M×(0,)M\times(0,\infty).

4 Harnack inequality and heat kernel estimates

In this section, we derive the CR version of Harnack’s inequality for the positive solutions of the CR heat equation and deduce an upper bound for the heat kernel.

𝐏𝐫𝐨𝐨𝐟𝐨𝐟𝐓𝐡𝐞𝐨𝐫𝐞𝐦 1.3\mathbf{Proof\ of\ Theorem\ 1.3} Let γ:[t1,t2]M\gamma:[t_{1},t_{2}]\rightarrow M be a horizontal curve joining xx and yy, i.e. γ(t1)=x,γ(t2)=y\gamma(t_{1})=x,\gamma(t_{2})=y. Define a map η:[t1,t2]M×[t1,t2]\eta:[t_{1},t_{2}]\rightarrow M\times[t_{1},t_{2}] by

η(t)=(γ(t),t).\displaystyle\eta(t)=(\gamma(t),t).

Let f=lnuf=\mathrm{ln}\ u with uu being a positive solution of the CR heat equation. Integrating ddtf\frac{d}{dt}f along η\eta, we get

f(y,t2)f(x,t1)=t1t2ddtf𝑑t=t1t2(γ˙,bf+ft)𝑑t.\displaystyle f(y,t_{2})-f(x,t_{1})=\int_{t_{1}}^{t_{2}}\frac{d}{dt}fdt=\int_{t_{1}}^{t_{2}}(\langle\dot{\gamma},\nabla_{b}f\rangle+f_{t})dt.

Applying Theorem 1.2 gives

f(y,t2)f(x,t1)\displaystyle f(y,t_{2})-f(x,t_{1}) \displaystyle\geq t1t2(γ˙,bf+1δ|bf|2CδCδt)𝑑t,\displaystyle\int_{t_{1}}^{t_{2}}(\langle\dot{\gamma},\nabla_{b}f\rangle+\frac{1}{\delta}|\nabla_{b}f|^{2}-\frac{C}{\delta}-\frac{C}{\delta t})dt,
\displaystyle\geq t1t2(δ|γ˙|24+Cδ+Cδt)𝑑t.\displaystyle-\int_{t_{1}}^{t_{2}}(\frac{\delta|\dot{\gamma}|^{2}}{4}+\frac{C}{\delta}+\frac{C}{\delta t})dt.

Choosing a curve γ\gamma with |γ˙|=dcc(x,y)t2t1|\dot{\gamma}|=\frac{d_{cc}(x,y)}{t_{2}-t_{1}}, we obtain

lnu(y,t2)u(x.t1)Cδ(t2t1)Cδlnt2t1δdcc2(x,y)4(t2t1).\displaystyle\mathrm{ln}\frac{u(y,t_{2})}{u(x.t_{1})}\geq-\frac{C}{\delta}(t_{2}-t_{1})-\frac{C}{\delta}\mathrm{ln}\frac{t_{2}}{t_{1}}-\frac{\delta d_{cc}^{2}(x,y)}{4(t_{2}-t_{1})}.

Taking exponentials of the above inequality, we can complete the proof. ∎

A mean value type inequality follows immediately from Theorem 1.3.

𝐂𝐨𝐫𝐨𝐥𝐥𝐚𝐫𝐲 4.1\mathbf{Corollary\ 4.1} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a complete noncompact pseudo-Hermitian manifold with

Ricb+2(m2)Torbk,and|A|,|bA|k1,\displaystyle Ric_{b}+2(m-2)Tor_{b}\geq-k,\ and\ |A|,|\nabla_{b}A|\leq k_{1},

and uu be a positive solution of the heat equation

ut=Δbu\displaystyle\frac{\partial u}{\partial t}=\Delta_{b}u

on M×(0,)M\times(0,\infty). Then for any constant 12<λ<23\frac{1}{2}<\lambda<\frac{2}{3} and any constant δ>1+4mλ(2λ1)\delta>1+\frac{4}{m\lambda(2\lambda-1)}, there exists a constant CC which is given by Theorem 1.2 such that for any 0<t1<t20<t_{1}<t_{2} and xMx\in M, we have

u(x,t1)\displaystyle u(x,t_{1}) \displaystyle\leq [Vol(Bcc(x,r))]12(Bcc(x,r)u2(y,t2)𝑑y)12(t2t1)Cδ\displaystyle[Vol(B_{cc}(x,r))]^{-\frac{1}{2}}\left(\int_{B_{cc}(x,r)}u^{2}(y,t_{2})dy\right)^{\frac{1}{2}}(\frac{t_{2}}{t_{1}})^{\frac{C}{\delta}} (4.1)
exp(Cδ(t2t1)+δr24(t2t1)).\displaystyle\cdot exp(\frac{C}{\delta}(t_{2}-t_{1})+\frac{\delta r^{2}}{4(t_{2}-t_{1})}).

Our next goal of this section is to derive an upper estimate for the heat kernel. For any x,yMx,y\in M and t>0t>0, let us set ρ(x,y,t)=12tdcc2(x,y)\rho(x,y,t)=\frac{1}{2t}d^{2}_{cc}(x,y). It is known that

|bdcc|21\displaystyle|\nabla_{b}d_{cc}|^{2}\leq 1

in the weak sense, where bdcc\nabla_{b}d_{cc} denotes the horizontal gradient of dccd_{cc} with respect to either xx or yy ([23]). Define g(x,y,t)=ρ(x,y,(1+2α)Tt)g(x,y,t)=-\rho(x,y,(1+2\alpha)T-t), where α,T\alpha,T are constants to be determined later. A direct computation shows that

12|bg|2+gt0.\displaystyle\frac{1}{2}|\nabla_{b}g|^{2}+g_{t}\leq 0. (4.2)

𝐋𝐞𝐦𝐦𝐚 4.2\mathbf{Lemma\ 4.2} Let (M2m+1,HM,J,θ)(M^{2m+1},HM,J,\theta) be a complete pseudo-Hermitian manifold. Suppose H(x,y,t)H(x,y,t) is the heat kernel of (1.4). Let

x(y,t)=S1H(y,z,t)H(x,z,T)𝑑z\displaystyle\mathcal{F}_{x}(y,t)=\int_{S_{1}}H(y,z,t)H(x,z,T)dz

for any fixed xMx\in M and any nonempty subset S1MS_{1}\subset M. Then for any 0ts<(1+2α)T0\leq t\leq s<(1+2\alpha)T and any nonempty subset S2MS_{2}\subset M, we have

S2x2(z,s)𝑑z\displaystyle\int_{S_{2}}\mathcal{F}_{x}^{2}(z,s)dz \displaystyle\leq S1H2(x,z,T)𝑑zsupzS1exp(2ρ(x,z,(1+2α)T))\displaystyle\int_{S_{1}}H^{2}(x,z,T)dz\sup\limits_{z\in S_{1}}exp(-2\rho(x,z,(1+2\alpha)T)) (4.3)
supzS2exp(2ρ(x,z,(1+2α)Ts)).\displaystyle\cdot\sup\limits_{z\in S_{2}}exp(2\rho(x,z,(1+2\alpha)T-s)).

𝐏𝐫𝐨𝐨𝐟\mathbf{Proof} The proof is almost the same as in [25] by using (4.2). ∎

𝐏𝐫𝐨𝐨𝐟𝐨𝐟𝐓𝐡𝐞𝐨𝐫𝐞𝐦 1.4\mathbf{Proof\ of\ Theorem\ 1.4} Let x\mathcal{F}_{x} be defined as in Lemma 4.2, and let S1=Bcc(y,t),S2=Bcc(x,t)S_{1}=B_{cc}(y,\sqrt{t}),S_{2}=B_{cc}(x,\sqrt{t}). Applying Corollary 4.1 to x\mathcal{F}_{x} with t1=Tt_{1}=T and t2=(1+αT)t_{2}=(1+\alpha T), we have

(Bcc(y,t)H2(x,z,T)𝑑z)2=x2(x,T)\displaystyle(\int_{B_{cc}(y,\sqrt{t})}H^{2}(x,z,T)dz)^{2}=\mathcal{F}_{x}^{2}(x,T) (4.4)
\displaystyle\leq [Vol(Bcc(x,t))]1Bcc(x,t)x2(z,(1+α)T)𝑑z(1+α)2Cδ\displaystyle[Vol(B_{cc}(x,\sqrt{t}))]^{-1}\int_{B_{cc}(x,\sqrt{t})}\mathcal{F}_{x}^{2}(z,(1+\alpha)T)dz\cdot(1+\alpha)^{\frac{2C}{\delta}}
exp(2CδαT+δt2αT).\displaystyle\cdot exp(\frac{2C}{\delta}\alpha T+\frac{\delta t}{2\alpha T}).

By Lemma 4.2, we deduce from (4.4) that

x2(x,T)\displaystyle\mathcal{F}_{x}^{2}(x,T) \displaystyle\leq [Vol(Bcc(x,t))]1Bcc(y,t)H2(x,z,T)𝑑z(1+α)2Cδ\displaystyle[Vol(B_{cc}(x,\sqrt{t}))]^{-1}\int_{B_{cc}(y,\sqrt{t})}H^{2}(x,z,T)dz\cdot(1+\alpha)^{\frac{2C}{\delta}}
exp(2CδαT+(δ+1)t2αTinfzBcc(y,t)ρ(x,z,(1+2α)T)).\displaystyle\cdot exp(\frac{2C}{\delta}\alpha T+\frac{(\delta+1)t}{2\alpha T}-\inf\limits_{z\in B_{cc}(y,\sqrt{t})}\rho(x,z,(1+2\alpha)T)).

Hence

Bcc(y,t)H2(x,z,T)𝑑z\displaystyle\int_{B_{cc}(y,\sqrt{t})}H^{2}(x,z,T)dz\leq [Vol(Bcc(x,t))]1(1+α)2Cδexp(2CδαT+(δ+1)t2αT)\displaystyle[Vol(B_{cc}(x,\sqrt{t}))]^{-1}(1+\alpha)^{\frac{2C}{\delta}}exp(\frac{2C}{\delta}\alpha T+\frac{(\delta+1)t}{2\alpha T})
exp(infzBcc(y,t)ρ(x,z,(1+2α)T)).\displaystyle\cdot exp(-\inf\limits_{z\in B_{cc}(y,\sqrt{t})}\rho(x,z,(1+2\alpha)T)).

Applying Corollary 4.1 once again and letting T=(1+α)tT=(1+\alpha)t, we have

H2(x,y,t)\displaystyle H^{2}(x,y,t)
\displaystyle\leq [Vol(Bcc(y,t))]1Bcc(y,t)H2(x,y,T)𝑑z(1+α)2Cδexp(2Cδαt+δ2α)\displaystyle[Vol(B_{cc}(y,\sqrt{t}))]^{-1}\int_{B_{cc}(y,\sqrt{t})}H^{2}(x,y,T)dz(1+\alpha)^{\frac{2C}{\delta}}\ exp(\frac{2C}{\delta}\alpha t+\frac{\delta}{2\alpha})
\displaystyle\leq [Vol(Bcc(y,t))]1[Vol(Bcc(x,t))]1(1+α)4Cδ\displaystyle[Vol(B_{cc}(y,\sqrt{t}))]^{-1}[Vol(B_{cc}(x,\sqrt{t}))]^{-1}(1+\alpha)^{\frac{4C}{\delta}}
exp(4Cδ(α+2)αt+(α+2)δ+12α(1+α)infzBcc(y,t)ρ(x,z,(1+2α)T)).\displaystyle\cdot exp\left(\frac{4C}{\delta}(\alpha+2)\alpha t+\frac{(\alpha+2)\delta+1}{2\alpha(1+\alpha)}-\inf\limits_{z\in B_{cc}(y,\sqrt{t})}\rho(x,z,(1+2\alpha)T)\right).

Similar to [8], we set 4(1+α)2(1+2α)=4+ϵ4(1+\alpha)^{2}(1+2\alpha)=4+\epsilon, then

infzBcc(y,t)ρ(x,z,(1+2α)T)2dcc2(x,y)(4+ϵ)t2(1+α)(4+ϵ)α.\displaystyle\inf\limits_{z\in B_{cc}(y,\sqrt{t})}\rho(x,z,(1+2\alpha)T)\geq\frac{2d_{cc}^{2}(x,y)}{(4+\epsilon)t}-\frac{2(1+\alpha)}{(4+\epsilon)\alpha}. (4.5)

This completes the proof. ∎

𝐑𝐞𝐦𝐚𝐫𝐤 4.1.\mathbf{Remark\ 4.1.} A similar result for subelliptic operators on closed manifolds was given in [8].

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Yuxin Dong
SchoolofMathematicalScienceSchool\ of\ Mathematical\ Science
and
LaboratoryofMathematicsforNonlinearScienceLaboratory\ of\ Mathematics\ for\ Nonlinear\ Science
FudanUniversityFudan\ University
Shanghai 200433,Shanghai\ 200433, P.R.ChinaP.R.\ China
yxdong@fudan.edu.cn[email protected]

Yibin Ren
CollegeofMathematicsCollege\ ofMathematics
PhysicsandInformationEngineeringPhysics\ and\ Information\ Engineering
ZhejiangNormalUniversityZhejiang\ Normal\ University
Jinhua 321004,Jinhua\ 321004, P.R.ChinaP.R.\ China
allenry@outlook.com[email protected]

Biqiang Zhao
ShanghaiCenterForShanghai\ Center\ For
MathematicalScienceMathematical\ Science
FudanUniversityFudan\ University
Shanghai 200433,Shanghai\ 200433, P.R.ChinaP.R.\ China
17110840003@fudan.edu.cn[email protected]