Available online at www.sciencedirect.com

J. Math. Pures Appl. 99 (2013) 409–418

www.elsevier.com/locate/matpur

A Varadhan type estimate on manifolds with time-dependent metricsand constant volume

Hiba Abdallah a,b,∗

a Institut Fourier, Université de Grenoble 1, Département de Mathématiques, B.P. 74, 38042 Saint Martin d’Hères Cedex, Franceb Laboratoire Jean Kuntzmann, Université Grenoble I, France

Received 11 January 2012

Available online 6 September 2012

Abstract

In this paper, we consider a compact n-dimensional manifold M with a time-dependent smooth Riemannian metric g(t) whosevolume is constant in t . We give a suitable form of the fundamental solution of the linear parabolic operator �g(t) − ∂

∂t, where �g(t)

stands for the time-dependent Laplacian based on g(t). We focus on the short-time behavior of the given fundamental solution,extending Varadhan’s estimate which holds in the case where the metric is fixed.© 2012 Elsevier Masson SAS. All rights reserved.

Résumé

Dans cet article, on considère une variété compacte M munie d’une famille continue de métriques riemanniennes g(t) dontle volume est constant en t . On discute la forme de la solution fondamentale de l’opérateur parabolique linéaire �g(t) − ∂

∂t, où

�g(t) est le laplacien par rapport à la métrique g(t). En étudiant le comportement asymptotique en temps petits de la solutionfondamentale, on généralise l’estimation de Varadhan qui est bien connue dans le cas où la métrique est fixe.© 2012 Elsevier Masson SAS. All rights reserved.

Keywords: Laplacian; Fundamental solution; Parametrix; Riemannian metric

1. Introduction

Let g(t), t ∈ [0, T ], be a smooth 1-parameter family of metrics on a compact n-dimensional Riemannian mani-fold M . And let �g(t) be the following differential operator acting on smooth functions on M :

�g(t) = gij (t)

[∂2

∂xi∂xj− Γ (t)kij

∂

∂xk

],

* Correspondence to: Institut Fourier, Université de Grenoble 1, Département de Mathématiques, B.P. 74, 38042 Saint Martin d’Hères Cedex,France.

E-mail address: Hiba.Abdallah@ujf-grenoble.fr.

0021-7824/$ – see front matter © 2012 Elsevier Masson SAS. All rights reserved.http://dx.doi.org/10.1016/j.matpur.2012.09.007

410 H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418

for all t ∈ [0, T ], where (gij (t))i,j is the inverse of (gij (t))i,j , and Γ (t)kij , i, j, k = 1, . . . , n, the Cristoffel symbolsof g(t). We are interested in the initial value problem⎧⎪⎨

⎪⎩(

�g(t) − ∂

∂t

)P(t, s, x, y) = 0,

limt→s

P (t, s, x, y) dvg(s)(x) = δy(x),

(1)

where dvg(s)(x) denotes the volume form of (M,g(s)).

Theorem 1.1. (See [3].) Problem (1) has a solution P which satisfies the following properties:

1. P is C∞ on M × M , and C1 on R2T , where

R2T = {

(t, s), 0 � s < t � T}.

2. P(t, s, x, y) � 0 for all x, y ∈ M and all (t, s) ∈R2T .

3. P verifies the semi-group property i.e.∫M

P(t, τ, x, y)P (τ, s, y, z) dvg(τ)(z) = P(t, s, x, y),

for all x, y ∈ M , and all (t, s) ∈ R2T .

Moreover, in [3], the author proves an asymptotic formula for the fundamental solution P as

P(t, s, x, y) = (4π(t − s)

)− n2 e

− r2s (x,y)

4(t−s)

N∑k=0

(t − s)kuk(t, x, y) + O((t − s)N− n

2), (2)

where rs is the Riemannian distance with respect to the metric g(s). Here, uk , k = 1, . . . ,N , are C∞ functions overM × M , and smooth in time by the smoothness of g(t), and verify

uk(t, x, y) = θt (x, y)r−k

r∫0

sk−1θ12t

(�g(t) − ∂

∂t

)uk−1(expy,t sξ, y) ds,

where ξ ∈ TyM is the unique tangent vector to the unit geodesic from y to x with respect to the metric g(t), x ∈B(y, i(y, t)), with i(y, t) the injectivity radius of M at the point y at time t , and

θt (x, y) =√

detg(t)

rn−1t

.

A similar result was proven by B. Chow et al. in [2, Chapter 24]. In its more practical form, P is expressed as

P(t, s, x, y) = (4π(t − s)

)− n2 e

− r2t (x,y)

4(t−s)

N∑k=0

(t − s)kuk(t, s, x, y) + O((t − s)N− n

2), (3)

where rt is the Riemannian distance with respect to the metric g(t) and uk are C∞ functions over M × M , whichdepend continuously on the two parameters s and t . In both Eqs. (2), and (3), N > n

2 +1, x, y ∈ M , and 0 � s < t � T .

In this paper, we seek a formula for the fundamental solution involving the term e− r2(x,y)

4(t−s) where r is the Riemannian

distance with respect to the metric g(0), instead of e− r2

t (x,y)

4(t−s) in formula (3). For this result to hold, we need to workwith a compact n-dimensional manifold (M,g(t)) with constant volume for every t ∈ [0, T ].

One of the main results in this paper is the following:

H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418 411

Theorem 1.2. Let g(t), t ∈ [0, T ], be a smooth family of Riemannian metrics on a compact n-dimensional Riemannianmanifold M . There exists a smooth family of Riemannian metrics h(t), t ∈ [0, T ], such that dvh(t) = dvg(0) for allt ∈ [0, T ] and such that the fundamental solution of the operator �h(t) − ∂

∂tcan be expressed in the form

P(t, s, x, y) = (4π(t − s)

)− n2 e

− r2(x,y)4(t−s)

N∑k=0

(t − s)kuk(t, s, x, y) + O((t − s)N− n

2), (4)

for N > n2 + 1, x, y ∈ M , and 0 � s < t � T , where r is the Riemannian distance of (M,g(0)), and uk , k = 0, . . . ,N ,

are functions in C∞(M × M) and at least C1 on R2T .

We then compute the first two coefficients u0 and u1. The second main result in this paper is the following:

Theorem 1.3. Let P be the fundamental solution defined by Eq. (4), then

limt→0+ t logP(t,0, x, y) = − r2(x, y)

4, uniformly in x, y ∈ M. (5)

A well known analogous result has been proven by Varadhan in [6,5], in the case where the metric is fixed. Here,to obtain formula (5), Theorem 1.3 use estimations of the heat operator applied to the parametrix and its convolutions.Whereas in [6] the author had used the Laplace transformations and in [5] a probabilistic method. Note that the methodfollowed here can be applied to the case of a fixed metric.

2. The fundamental solution

Let M be a compact n-dimensional manifold and let g(t), t ∈ [0, T ], be a smooth time-parameterized family ofRiemannian metrics on M . The aim of this section is to prove Theorem 1.2. The proof relies on a striking result,Lemma 2.1 below due to a paper of Moser [4] which considers the volume forms on a manifold.

Lemma 2.1. There exists a C∞ family of diffeomorphisms ηt such that

η∗t dvg(t) = dvg(0), for all t ∈ [0, T ].

Proof of Theorem 1.2. Let h(t) be the family of metrics given by Lemma 2.1: h(t) = η∗t dvg(t) for all t ∈ [0, T ],

h(0) = g(0) and the Hilbert space L2(M,vh(t)) is constant in t . Let G be the Euclidian heat kernel with respect to the

metric g(0), i.e. G(t, s, x, y) = (4π(t − s))− n2 e

− r2(x,y)4(t−s) , and let N be any integer greater than n

2 . In order to ensure thesmoothness on the manifold M , let η : M × M → [0,1] be the cutoff function

η(x, y) ={

1 if (x, y) ∈ Uρ/4,

0 if (x, y) ∈ M × M − Uρ/2(6)

where ρ is the injectivity radius of the manifold (M,g(0)), and

Uρ = {(x, y) ∈ M × M; r(x, y) < ρ

}.

We shall now construct a function

SN : Uρ/2 ×R2T → R,

such that the function defined by

PN(t, s, x, y) = η(x, y)SN(t, s, x, y), 0 � s < t � T

is a parametrix for the heat equation (see for more explanations [1, Chapter III]). For this, let SN be of the form

SN(t, s, x, y) = G(t, s, x, y)

N∑(t − s)kuk(t, s, x, y), 0 � s < t � T , (7)

k=1

412 H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418

where the functions uk , k = 1, . . . ,N , are at least C1 in the first two space variables, C∞ in the last two time variables,and satisfy {

�t SN(t, s, x, y) = O((t − s)N

)G,

limt→s

SN(t, s, x, y) dv0(x) = δy(x),(8)

where �t denotes the “heat operator” �h(t) − ∂∂t

.Using Duhamel’s formula, we seek a fundamental solution of the form

P(t, s, x, y) = PN(t, s, x, y) +t∫

s

dτ

∫M

PN(t, s, x, z)Φ(t, s, z, y) dv0(z).

Now, following a convergent iterative procedure (described in details in [1, Chapter III]), we look for Φ as the sum∑∞m=1 Φm, with the functions Φm defined recursively by

Φ1(t, s, x, y) =�tPN(t, s, x, y),

and

Φm+1(t, s, x, y) =t∫

s

dτ

∫M

�tPN(t, τ, x, z)Φm(τ, s, z, y) dv0(z).

First, from the definition of the parametrix, we can easily obtain the estimate∣∣�tPN(t, s, x, y)∣∣ � C1(t − s)N− n

2 , (9)

where the constant C1 depends only on the metric g(0). Inductively,

∣∣Φm(t, s, x, y)∣∣ � (C1)

mVol(M)m−1

(m − 1)!(N − n2 + 1)m−1

(t − s)m(N− n2 +1)−1, (10)

where Vol(M) is the volume of (M,g(0)). Therefore,∑∞

m=1 Φm converges absolutely and uniformly.Furthermore, with this choice of Φ we have �tP = 0. And since PN satisfies the δ-property and the remainder

term defined by

ΘN(t, s, x, y) =t∫

s

dτ

∫M

PN(t, s, x, z)Φ(t, s, z, y) dv0(z),

converges to zero when t goes to s, we have

limt→s

P (t, s, x, y) dvg(s)(x) = δy(x).

Setting u = ηu, the proof of Theorem 1.2 is complete. �3. The first two coefficients of the fundamental solution

The aim of this section is to compute the first two coefficients u0, u1 of the fundamental solution P(t, s, x, y) ats = 0 and (x, y) ∈ Uρ/4. We denote u0(t,0, x, y) and u1(t,0, x, y) by u0(t, x, y) and u1(t, x, y) respectively.

Corollary 3.1. For all t ∈ [0, T ] and all (x, y) ∈ Uρ/4, we have

u0(t, x, y) = θ− 1

20 (x, y)ef (x,y), (11)

where

f (x, y) =n∑ (

1

8w

ij

1 (y)xi(x)xj (x) + 1

12

(∇kwij

1

)(y)xi(x)xj (x)xk(x)

)+ Λ(x,y). (12)

i,j=1

H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418 413

In Eq. (12), {xi}ni=1 are the geodesic normal coordinates centered in y, w1 is a 2-tensor defined by

wij

1 = −hik(0)

(∂h

∂t(0)

)kl

hlj (0),

and Λ is a smooth function defined on Uρ such that Λ(x,y) = O(r(x, y)), ∇iΛ(x, y) = O(r(x, y)) and∇i∇jΛ(x, y) = O(r(x, y)), i, j = 1, . . . , n. In particular,

u0(t, x, x) = 1, for all x ∈ M and t ∈ [0, T ].

Proof. By definition

�h(t) = hij (t)

[∂2

∂xi∂xj− Γ (t)kij

∂

∂xk

].

In a neighborhood of t = 0, we can write �h(t) as

�h(t) = �h(0) +N∑

m=1

tm(w

ijm∇i∇j + vk

m∇k

) + O(tN+1)( ∂2

∂xi∂xj+ ∂

∂xk

),

where vm = vkm

∂∂xk is a vector field such that

vkm = −h(0)ij (Am)kij −

∑p+q=m

wijp (Aq)kij ,

and

(Ap)kij = 1

2h(0)kl

[∂

∂xi

(1

p!∂ph

∂tp(0)

)j l

+ ∂

∂xj

(1

p!∂ph

∂tp(0)

)il

− ∂

∂xl

(1

p!∂ph

∂tp(0)

)ij

]

+ 1

2

∑m+q=p

wklm

[∂

∂xi

(1

q!∂qh

∂tq(0)

)j l

+ ∂

∂xj

(1

q!∂qh

∂tq(0)

)il

− ∂

∂xl

(1

q!∂qh

∂tq(0)

)ij

].

With a careful calculation, we have

�t SN = G(t − s)N�uN + G

N∑m=1

tm(t − s)Nwijm∇i∇j uN + G

N∑m=1

tm(t − s)Nvlm∇luN .

From Eq. (8) we obtain

1

2r∂ log θ0

∂r(. , y)u0(t, . , y) + r

∂u0

∂r(t, . , y) − 1

16w

ij

1 ∇i r2∇j r

2u0(t, . , y) = 0,

and

∂u0

∂s

(γ (s), y

)u−1

0

(t, γ (s), y

) = −1

2

∂ log θ

∂s

(γ (s), y

) + 1

4rw1

(γ (s), γ (s)

)(13)

where γ : [0, r(x, y)] → M is a geodesic connecting y to x, verifying γ (0) = y and |γ (s)| = 1 for all s ∈ [0, r(x, y)].Finally, we integrate along γ and we use the fact that ∇r(γ (s), y) = γ (s) to get

u0(t, x, y) = θ− 1

20 (x, y) exp

( r(x,y)∫0

1

4sw1

(γ (s), γ (s)

)ds

). �

Remark 3.2. If we denote by ak(s) the sum∫

uk(s, x, x) dv0(x), then a0(0) = Vol(M,g(0)).

M

414 H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418

Corollary 3.3. For all t ∈ [0, T ] and all (x, y) ∈ Uρ/4, we have

u1(t, x, y) = r−1u0(t, x, y)

r∫0

�θ− 1

20 .θ

12

0

(γ (s), y

)ds

+ r−1u0(t, x, y)

r∫0

�f(γ (s), y

) + ⟨∇f(γ (s), y

),∇f

(γ (s), y

)⟩ + 〈∇ log θ0,∇f 〉(γ (s), y)ds

− r−1u0(t, x, y)

r∫0

[1

2w1

(∇r2(γ (s), y),∇(logu0)

(γ (s), y

))

− 1

4

(w

ij

1 ∇i∇j r2(γ (s), y

) + vl1∇lr

2(γ (s), y))]

ds, (14)

where γ : [0, r(x, y)] → M is the geodesic that joins y to x such that γ (0) = y and |γ (s)| = 1, for all s ∈ [0, r(x, y)].In particular,

u1(t, y, y) = τ0

6+ 1

2

n∑i=1

wii1 (y) + 1

4

n∑i,j=1

h(0)ijwij

1 (y) for all y ∈ M.

Proof. By Eq. (8) we have

r∂u1

∂r+

(1 + r

2

∂(log θ0)

∂r− 1

16w

ij

1 ∇i r2∇j r

2)

u1 = �u0 − 1

2w

ij

1 ∇i r2∇j u0 − 1

4

(w

ij

1 ∇i∇j r2 + vl

1∇lr2)u0. (15)

With the method of variation of parameters, we get

r∂u1

∂r+

(1 + r

2

∂(log θ0)

∂r− 1

16w

ij

1 ∇i r2∇j r

2)

u1 = 0, (16)

u1(t, x, y) = c(x, y)r−1(x, y)u0(t, x, y), (17)

where the function c ∈ C∞(M × M) verifies

∂c

∂r= u−1

0 �u0 − 1

2u−1

0 wij

1 ∇i r2∇j u0 − 1

4

(w

ij

1 ∇i∇j r2 + vl

1∇lr2).

Observe that

u−10 �u0 = θ

12

0 �θ− 1

20 + e−f �ef + 2θ

12

0

⟨∇θ− 1

20 ,∇ef

⟩e−f

= θ12

0 �θ− 1

20 + �f + 〈∇f,∇f 〉 + 〈∇ log θ0,∇f 〉,

thus

c(x, y) =r∫

0

θ12 �θ− 1

2(γ (s), y

) + �f(γ (s), y

) + ⟨∇f(γ (s), y

),∇f

(γ (s), y

)⟩ + 〈∇ log θ,∇f 〉(γ (s), y)ds

−r∫

0

[1

2w1

(∇r2(γ (s), y),∇(logu0)

(γ (s), y

)) − 1

4

(w

ij

1 ∇i∇j r2(γ (s), y

) + vl1∇lr

2(γ (s), y))]

ds + α.

And since by Eq. (17) we have c(x, y) = u1(t,x,y)ru0(t,x,y)

, then

α = limr→0

u1(t, x, y)r

u0(t, x, y)= 0,

therefore implying Eq. (14).

H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418 415

Now by Berger, Gauduchon and Mazet [1, pp. 221–222], we have

limr(x,y)→0

r−1θ− 1

20 (x, y)

r(x,y)∫0

�θ− 1

20 (γ (s), y).θ

12

0 (γ (s), y) ds = τ0

6,

where τ0 is the scalar curvature of (M,h(0)). Finally, for every y ∈ M , we obtain

u1(t, y, y) = τ0

6+ 1

2

n∑i=1

wii1 (y) + 1

4

n∑i,j=1

h(0)ijwij

1 (y),

since

limr→0

r−1u0(t, x, y)

[ r∫0

�f(γ (s), y

) + ⟨∇f(γ (s), y

),∇f

(γ (s), y

)⟩ + 〈∇ log θ0,∇f 〉(γ (s), y)ds

−r∫

0

[1

2w1

(∇r2(γ (s), y),∇(logu0)

(t, γ (s), y

)) − 1

4

(w

ij

1 ∇i∇j r2(x, y) + vl

1∇lr2(x, y)

)]ds

]

= �f (y, y) + ⟨∇f (y, y),∇f (y, y)⟩ + ⟨∇ log θ0(y, y),∇f (y, y)

⟩− 1

2w1

(∇r2(y, y),∇(logu0)(y, y)) + 1

4

(w

ij

1 ∇i∇j r2(y, y) + vl

1∇lr2(y, y)

)= 1

2

n∑i=1

wii1 (y) + 1

4

n∑i,j=1

h(0)ijwij

1 (y). �

4. Correlation between the logarithm of the fundamental solution and the Riemannian distance

This section is devoted to the proof of Theorem 1.3. The goal is to understand the behavior of the fundamentalsolution P(t,0, x, y) of a family of compact Riemannian manifolds (M,g(t)) whose volume is constant in time ast → 0+. For this purpose, we need a sharper estimate on the parametrix than (9). The largest part of the proof consistsof the two following lemmas which hold more generally for 0 � s < t � T :

Lemma 4.1. There exists a constant C2 > 0 such that for all x, y ∈ M , (t, s) ∈R2T , 0 < ε < 1, one has

�tPN(t, s, x, y) � C2ε−(N+1)e

− r2(x,y)(4+ε)(t−s) (t − s)N− n

2 . (18)

Proof. Let 0 < ε < 1 and (t, s) ∈R2T .

a) For (x, y) ∈ (M × M − Uρ2) we have,

∣∣�tPN(t, s, x, y)∣∣ = 0 � C2ε

−(N+1)e− r2(x,y)

(4+ε)(t−s) (t − s)N− n2 .

b) For (x, y) ∈ Uρ4

, Eq. (8) gives

∣∣�tPN(t, s, x, y)∣∣ = ∣∣�t SN(t, s, x, y)

∣∣ � C3e− r2(x,y)

4(t−s) (t − s)N− n2

� C3ε−(N+1)e

− r2(x,y)(4+ε)(t−s) (t − s)N− n

2 ,

where C3 > 0 is a constant independent of (t, s) ∈R2T and x, y ∈ M .

c) For (x, y) ∈ Uρ − Uρ , there exist constants C4, C5 and C6, all independent of t, s, x and y, such that

2 4

416 H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418

∣∣�tPN(t, s, x, y)∣∣ � η(x, y)�t SN(t, s, x, y) + (

�h(t)η(x, y))SN(t, s, x, y) + 2

⟨dη(x, y), dSN(t, s, x, y)

⟩h(t)

� C4(|�t SN | + |SN | + 2|∇SN |)(t, s, x, y)

� C5G(t, s, x, y)((t − s)N + 1 + (t − s)−1)

� C6e− r2(x,y)

(4+ε)(t−s) e− εr2(x,y)

4(4+ε)(t−s) (t − s)N− n2(1 + (t − s)−N + (t − s)−N−1).

And since r(x, y) � ρ4 , inequalities above imply

∣∣�tPN(t, s, x, y)∣∣ � C6e

− r2(x,y)(4+ε)(t−s) e

− ερ2

64(4+ε)(t−s) (t − s)N− n2(1 + (t − s)−N + (t − s)−N−1).

Remarking that

e− ερ2

64(4+ε)(t−s) (t − s)−k =(

ερ2

64(4 + ε)

)−k(ερ2

64(4 + ε)(t − s)

)k

e− ερ2

64(4+ε)(t−s)

�(

ερ2

64(4 + ε)

)−k

kke−k

� ε−k

(320

ρ2

)k

kke−k,

then ∣∣�tPN(t, s, x, y)∣∣ � C6e

− r2(x,y)(4+ε)(t−s) (t − s)N− n

2 ε−(N+1).

Letting C2 = max{C3,C6} we complete the proof of Lemma 4.1. �An immediate consequence of Lemma 4.1 is the following

Lemma 4.2. There exists a constant CΘ > 0 such that for all (t, s) ∈R2T , x, y ∈ M , and 0 < ε < 1, we have

∣∣ΘN(t, s, x, y)∣∣ � CΘe

− r2(x,y)(4+ε)(t−s) (t − s)N− n

2 + 12 ε−(N+ n

2 + 12 )eCΘε−(N+1)(t−s)

N− n2 +1

. (19)

Proof. First, let us prove by induction that for all m ∈N∗, x, y ∈ M , (t, s) ∈ R

2T , and 0 < ε < 1, we have

∣∣Φm(t, s, x, y)∣∣ � Cm

2 ε−m(N+1)Vol(M)m−1

(m − 1)!(N − n2 + 1)m−1

(t − s)m(N− n2 +1)−1e

− r2(x,y)(4+ε)(t−s) , (20)

where C2 is the constant given in Lemma 4.1.It is clearly true in the case m = 1 (see estimation (18)). Assume it holds for m ∈ N

∗, and let us prove it to m + 1.We have∣∣Φm+1(t, s, x, y)

∣∣�

Cm+12 ε−(m+1)(N+1)Vol(M)m−1

(m − 1)!(N − n2 + 1)m−1

(t − s)N− n2

t∫s

dτ

∫M

e− r2(x,z)

(4+ε)(t−τ ) e− r2(z,y)

(4+ε)(τ−s) (τ − s)m(N− n2 +1)−1 dv0(z).

Using the fact that

r2(x, z)

(t − τ)+ r2(z, y)

(τ − s)� (r(x, z) + r(z, y))2

(t − s)� r2(x, y)

(t − s), (21)

we get

e− r2(x,z)

(4+ε)(t−τ ) e− r2(z,y)

(4+ε)(τ−s) � e− r2(x,y)

(4+ε)(t−s) , (22)

which in turn yields the estimate

H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418 417

∣∣Φm+1(t, s, x, y)∣∣ � Cm+1

2 ε−(m+1)(N+1)Vol(M)m

(m − 1)!(N − n2 + 1)m−1

e− r2(x,y)

(4+ε)(t−s) (t − s)N− n2

t∫s

(τ − s)m(N− n2 +1)−1 dτ

= Cm+12 ε−(m+1)(N+1)Vol(M)m

m!(N − n2 + 1)m

e− r2(x,y)

(4+ε)(t−s) (t − s)(m+1)(N− n2 +1)−1.

This proves inequality (20). Now, summing all over n ∈N∗, we obtain

∞∑m=1

∣∣Φm(t, s, x, y)∣∣ � C2e

− r2(x,y)(4+ε)(t−s) (t − s)N− n

2 ε−(N+1) exp

(C2ε

−(N+1)Vol(M)(t − s)N− n2 +1

N − n2 + 1

).

Finally, there is a constant C7 depending only on N,T and the metric g(0) such that∣∣PN(t, τ, x, z)∣∣ � C7(t − τ)−

n2 e

− r2(x,z)4(t−τ ) .

This implies the existence of a constant C8 independent of t, s, x, y and ε such that

∣∣PN(t, τ, x, z)∣∣ � C8(t − τ)−αε− n

2 +αr−n+2α(x, z) exp

(− r2(x, z)

(4 + ε)(t − τ)

),

which in turn leads to the following estimate∣∣ΘN(t, s, x, y)∣∣

� C8e− r2(x,y)

(4+ε)(t−s) (t − s)N− n2 ε−(N+1+ n

2 −α)e

Cε−(N+1)Vol(M)(t−s)N− n

2 +1

N− n2 +1

t∫s

dτ

∫B(x,

ρ2 )

(t − τ)−αr−n+2α(x, z) dv0(z).

Using polar coordinates, we have

∫B(x,

ρ2 )

r−n+2α(x, z) dv0(z) � c

ρ2∫

0

r2α−1 dr,

where the constant c depends on the metric g(0) = h(0). We accomplish our proof by choosing α = 12 . �

Proof of Theorem 1.3. Let UN be the function defined by

UN(t, s, x, y) =N∑

k=1

(t − s)kuk(t, s, x, y), 0 � s < t � T . (23)

Combining inequalities (23) and (19), we have

(4πt)−n2∣∣UN(t,0, x, y)

∣∣ + er2(x,y)

4t

∣∣ΘN(t,0, x, y)∣∣ � C9t

− n2 e

εr2(x,y)4(4+ε)t ε−(N+ n

2 + 12 )eCΘε−(N+1)t

N− n2 +1

,

where C9 = CN,T (4π)− n2 + CΘT N+ 1

2 and

CN,T = max{UN(t,0, x, y), (t, x, y) ∈ [0, T ] × M × M

}.

With ε = t1

N+1 we obtain

t log((4πt)−

n2∣∣UN(t,0, x, y)

∣∣ + er2(x,y)

4t

∣∣ΘN(t,0, x, y)∣∣) � t log

(C9t

− n2 −1+ −n−1

2(N+1)) + D2

M

16t

1N+1 + CΘtN− n

2 +1.

Using the equality

P(t,0, x, y) = e−r2(x,y)

4t[UN(t,0, x, y) + e

r2(x,y)4t ΘN(t,0, x, y)

]and letting t → 0, we conclude Theorem 1.3. �

418 H. Abdallah / J. Math. Pures Appl. 99 (2013) 409–418

Acknowledgements

This paper is dedicated to Professors Mustafa Jazar and Nazih Mokaddem for their precious support, and to theLebanese National Council for Scientific Research (CNRS-Lebanon) for financing my doctoral studies at Universitéde Grenoble. I would also like to thank professor Bennet Chow and Gérard Besson for their inestimable help and forthe many useful discussions.

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