Asymptotic estimates on the time derivative of entropy on a Riemannian manifold

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1 Asymptotic estimates on the time erivative of entropy on a Riemannian manifol Arian P. C. Lim a, Dejun Luo b a Nanyang Technological University, athematics an athematics Eucation, Singapore b Key Lab of Ranom Complex Structures an Data Science, Acaemy of athematics an Systems Science, Chinese Acaemy of Sciences, Beijing 9, China Abstract We consier the entropy of the solution to the heat equation on a Riemannian manifol. When the manifol is compact, we provie two estimates on the rate of change of the entropy in terms of the lower boun on the Ricci curvature an the spectral gap respectively. Our explicit computation for the three imensional hyperbolic space shows that the time erivative of the entropy is asymptotically boune by two positive constants. SC : 58J5 Keywors: Heat equation, entropy, Ricci curvature, Ornstein-Uhlenbeck operator, hyperbolic space Introuction In Perelman s solution to the Poincaré conjecture, the W -functional (i.e. Perelman s W -entropy) playe an important role. Since then, there have been many attempts to unerstan or generalize this functional to other situations, see [3, 7, 8, 9]. This work is motivate by [8, 9], where the author presente the expression for the time erivative of the entropy an stuie its properties. Our aim is to estimate the asymptotic behavior of the time erivative of the entropy. Let be a close manifol of imension n. Equip with a Riemannian metric g an efine the corresponing Laplacian g. We consier the following heat equation t u = u, u = f, (.) where f C () is a positive function. In this work we follow the probabilists convention (e.g. [, 5]) of consiering instea of the Laplacian. Let x be the Riemannian volume measure on. It is clear that u x f x, hence t u x = u x =. t Since the solution u to (.) is strictly positive, we can efine the entropy Ent(u) = u log u x. arian.lim@nie.eu.sg

2 Substituting u into the simple inequality x log x x (x ) an integrating on, we see that for all t, Ent(u) Vol() f x, (.) where Vol() is the volume of, an the equality hols if an only if f. Using the integration by parts formula, it is easy to show that t Ent(u) = u x, u where u is the graient vector fiel of u. From this, we see that if the initial value f is not constant, then Ent(u) is an increasing function of t >. Denote the Ricci curvature tensor on by Ric. The main result in this paper is Theorem.. Assume that f x = an there is a k R such that Ric k I. Then [ ] e kt t Ent(u) e kt, q nk k ; nq (n+q t), k =, (.3) where q = f f x. This result will be prove in Section, which oes not use Li-Yau s graient estimate. In [9], Ni efine Ñ(u, t) = Ent(u) Ent(Φ n), whereby Φ n is the heat kernel on R n. If the manifol has nonnegative Ricci, using Li-Yau s graient estimate, he showe that the time erivative of Ñ(u, t) is less than or equal to. In other wors, t Ent(u) t Ent(Φ n) = n t. (.4) From Theorem., we can conclue that if the Ricci curvature is boune from below by a positive constant k >, then the time erivative of the entropy has an exponential ecay as t ; however, if k <, then the time erivative of the entropy is asymptotically ominate by nk/ > as t. For a compact (without bounary) Riemannian manifol with negative Ricci curvature, the estimate given in Theorem. is not satisfactory. In Section 3 we will give an estimate on the rate of change of entropy in terms of the spectral gap, which implies that the time erivative of the entropy always has an exponential ecay, even though Ricci is boune below by a negative constant (see Proposition 3.). One might woner if a similar result hols when the manifol is non-compact. The ifficulty in this case is the justification of the existence of the entropy an the integration by parts formula. Accoring to [7, p.5], when the Ricci curvature of a non-compact manifol is nonnegative, the entropy formula has been establishe rigorously, see [3]. Equation (.4) tells us that the time erivative of the entropy will not excee n/t; furthermore, we can show that the time erivative achieves the critical value n/t if an only if is isometric to R n (see Theorem.7). However, when is non-compact an has negative Ricci curvature, the situation is ifferent. In the case of the three imensional hyperbolic space H 3, we will prove in Theorem 4. that the rate of change of entropy is asymptotically boune by two positive constants. Rate of change of the entropy uner curvature conition In this section we will provie the proof of Theorem.. In fact we can eal with more general cases. Suppose Z is a C -vector fiel on an consier the secon orer ifferential operator L = + Z.

3 In the following, we enote by D, Hess u an HS respectively the Levi-Civita connection, the Hessian of the function u an the Hilbert-Schmit norm. Proposition.. Let f C () be a positive function an u : [, ) R + the solution to the heat equation t u = Lu, u = f >. (.) We have ( L )( ) u = u u t u u Hess u u + u( Ric( u, u) D u Z, u ). Proof. By the Weitzenböck formula, ( u ) = u, u + Hess u HS + Ric( u, u). (.) Next taking Y = u an ϕ = u in Lemma. below gives us As a result, u, Z(u) = D u Z, u + u, D Z u. Z( u ) = Z u, u = D Z u, u = Z(u) D u Z, u. (.3) Finally t ( u ) = t ( u), u = ( t u), u. Combining this with (.) an (.3), an using the equation (.), we get ( L ) ( u ) = Hess u HS + Ric( u, u) D u Z, u. (.4) t Now notice that ( ) u = u ( u ) + u (u ) + (u ), ( u ). (.5) u We have an (u ) = u u + u 3 u (u ), ( u ) = u u, ( u ) = u D u u, u = (Hess u)( u, u). u Substituting the above equalities into (.5) gives rise to ( ) u = u u ( u ) u u u + u 3 u 4 4 (Hess u)( u, u). (.6) u We also have ( ) u Z = Z, u (u ) + u ( u ) u = u u Z(u) + u Z( u ) (.7) 3

4 an Consequently, by (.6) (.8), ( L )( ) u = ( L t u u t which, by (.4), is equal to ( ) u = u t u u t u + u t ( u ). (.8) + u 4 u 3 )( u ) u u (Hess u)( u, u), u ( L ) u t ( Hess u u HS + Ric( u, u) D u Z, u ) + u 4 u 3 (Hess u)( u, u) u = ( u u ) Hess u u u + Ric( u, u) D u Z, u. The proof is complete. HS Lemma.. Let Y, Z be two C -vector fiels on an ϕ C (). We have Y, Z(ϕ) = D Y Z, ϕ + Y, D Z ϕ. Proof. By the consistency of the Levi-Civita connection D with the metric, Notice that The result follows. Y, Z(ϕ) = Y [Z(ϕ)] = Y Z, ϕ = D Y Z, ϕ + Z, D Y ϕ. Z, D Y ϕ = (Hess ϕ)(y, Z) = (Hess ϕ)(z, Y ) = Y, D Z ϕ. Now we take V C () an let Z = V. That is, we consier the heat equation t u = Lu = u + V, u, u() = f >. (.9) Define the measure µ = e V x on, where x is the volume element of. For any ϕ, ψ C (), the following integration by parts formula hols: Lϕ ψ µ = ϕ, ψ µ. (.) We have u µ f µ for all t >. Therefore t u µ = t u µ =. Define the entropy Ent(u) = u log u µ. (.) Similar to (.), we have Ent(u) µ() f µ for all t, an the equality hols if an only if f. Using the integration by parts formula (.), ((log t Ent(u) = u) t u + t ) u µ = (log u)lu µ = log u, u µ = u µ. (.) u 4

5 This tells us that if the initial value f is not a constant, then the function t Ent(u) is monotonely increasing. We want to estimate the rate of change of the entropy, i.e. tent(u), as t. To this en, we will make use of the equality prove in Proposition.. It is easy to see that u u Hess u u = u Hess(log u) HS u n (log u), (.3) HS where n is the imension of, an the equality hols if an only if Hess(log u) = ψ I with some function ψ : R + R. Therefore by Proposition., we arrive at ( L )( ) u u t u n (log u) + u( ) Ric Hess V ( u, u). (.4) Theorem.3. Assume that there is a k > such that Then for all t >, Ric Hess V k I. e kt f Ent(u) µ. t f Proof. Integrating both sies of (.4), we obtain ( L )( ) u µ u (log u) µ + k t u n By (.), we have thus the above inequality reuces to ( ) u µ t u n ( ) u L µ =, u Therefore u u µ k µ, t u u from which the esire estimate follows. u µ. u u (log u) u µ + k µ. (.5) u In the above iscussions, the compactness of the manifol is only use to guarantee that the entropy is finite an the integration by parts formula can be applie. If these are justifie, then the same argument works as well for non-compact manifols. The following is an example on R n. Example.4. Let k > an consier the general Ornstein-Uhlenbeck operator L = k x, where is the stanar Laplacian operator on R n. L is the generator of the iffusion process X t satisfying the Itô stochastic ifferential equation X t = B t k X t t, X = x, where B t is the stanar Brownian motion on R n. Solving the above equation gives us X t = e kt/ x + t 5 e k(t s)/ B s.

6 Therefore X t is a Gaussian ranom vector with mean e kt/ x an covariance matrix e kt k I, an the transition ensity of the process is [ ] k n/ ( p(t, x, y) = ( e kt exp k y e kt/ x ) ) ( e kt. ) Let V (x) = k 4 x ; then V (x) = kx/ an the invariant measure µ(x) = e k x / x. We have Ent(p(t,, y)) = n log ( e kt ) k I (t) + k ( e kt ) I (t), (.6) where I (t) = R n p(t, x, y) µ(x) an I (t) = R n p(t, x, y) y e kt/ x µ(x). Since µ is invariant uner the transition semigroup of X t, we have I (t) = e k y /. Next, irect computations lea to I (t) = e k y / ( e kt ) [( e kt ) y + nk ] e kt. Substituting the above two equalities into (.6), we obtain Ent(p(t,, y)) = [ ( ) ] / e k y n log k ( e kt ) + k( e kt ) y + ne kt. Consequently, t Ent(p(t,, y)) = [ ] / nke kt e k y e kt + k y e kt. From this expression, we see that if y, the rate of ecay is exactly the one that we get in Theorem.3, since Ric HessV = k I. Therefore the estimate in Theorem.3 is sharp. If y =, the rate of ecay of the entropy is nk Ent(p(t,, )) = t ( e kt ) e kt. oreover, we have lim k t Ent(p(t,, y)) = n t, which is the time erivative of the entropy of the heat kernel on R n. Now we prove Theorem., the case when V =. Proof of Theorem.. Remark that now µ = x, i.e. the volume measure. We still have (.5), or u x u (log u) u x + k x. t u n u For simplicity of notation, we enote by q t = x. Then the above inequality can be written as q t t u (log u) x + kq t. (.7) n u u 6

7 By the Cauchy inequality an since u x = f x =, ( u log u x) u log u x u x = u log u x. (.8) Using the integration by parts formula, we have u log u x = u, log u x = q t. Therefore (.7) becomes q t t q t n kq t. (.9) Now we solve the above ifferential inequality. If k =, then q t t q t n, from which we easily obtain that q t nq n+q t. In the case k, we ivie both sies of (.9) by qt an get q t t n k. q t This is equivalent to q t q t t n + kq t. Therefore ( e kt q ) e kt t t n. Integrating this inequality from to t leas to The proof is now complete. e kt q t We have the following simple observations. q nk ( e kt ). Corollary.5. Assume the conitions of Theorem.. (i) If k >, then the time erivative of the entropy has an exponential ecay as t ; (ii) if k <, then the time erivative of the entropy is asymptotically ominate by nk/ as t. Proof. The assertions follow irectly from Theorem.. We can also obtain an estimate on the rate of change of entropy from Hamilton s graient estimate for heat equations (see [4] or [, Corollary 3.3]). Proposition.6. Let u be a solution to the stanar heat equation (.). k R such that Ric k I. Then for all t >, ( )[ t Ent(u) t k Vol() + ( log(sup f) ) ] f x. Assume there is oreover, if Ric, then we can take k = in the above estimate. 7

8 Proof. By [, Corollary 3.3], we have u ( ) u t k log A u, where A = sup [,t] u = sup f. Therefore u u ( ) t k u log sup f u. Integrating both sies on, we get u ( ) x u t k u log sup f u x ( )[ ] = t k log(sup f) u x + Ent(u) ( )[ t k Vol() + ( log(sup f) ) ] f x, (.) where the last inequality follows from inequality (.). The proof is now complete. It is interesting to compare the two estimates given in Theorem. an Proposition.6. We assume Vol() = an f x = for simplicity of notations. We istinguish three cases: (a) k <. By Theorem., tent(u) is asymptotically ominate by nk/, epening only on the imension an curvature; while by Proposition.6, the asymptotic constant is k log(sup f) which epens on the initial conition. (b) k =. By Theorem., t Ent(u) nq (n+q t) n t. The estimate given by Proposition.6 is t Ent(u) t log(sup f). (c) k >. Theorem. gives us an exponential ecay: t Ent(u) q e kt ; while Proposition.6 only leas to a polynomial ecay: t Ent(u) t log(sup f). To complete this section, we briefly iscuss the entropy of the heat kernel on a non-compact Riemannian manifol with nonnegative Ricci curvature. As mentione in the Introuction (see also [8, p.9]), the integration by parts formula can be justifie rigorously, thanks to Li-Yau s graient estimate [6]. Therefore if u is the heat kernel of, in the same way one can show that t Ent(u) n t. The next theorem is an analogue of [8, Theorem.4], which says that if the Ricci curvature is nonnegative, then the ecay rate of n t is achieve if an only if = Rn. Theorem.7. Let be a non-compact Riemannian manifol with nonnegative Ricci curvature, an u its heat kernel. Then if an only if is isometric to R n. t Ent(u) = n t (.) Proof. We only have to prove the necessity part. Suppose (.) hols. Since has nonnegative Ricci curvature, it is stochastically complete, that is, u x = for all t > (see e.g. [5, Theorem 4..4]). By the proof of Theorem. (remark that now k = ), we must have equalities in (.3) an (.8). The equality in (.3) will imply that (log u) = nψ for some function 8

9 ψ : R + R. Next, the equality in (.8) means that ψ(t) is a function inepenent of x. Using the integration by parts formula, u nψ(t) = u (log u) x = x = u t Ent(u) = n t, therefore ψ(t) = t an (log u) = n t. The rest of the proof is similar to that of [8, Theorem.4]. By Varahan s large eviation formula (see [5, Theorem 5..]), lim t t log u(t, x, y) = (x, y), where : R + is the Riemannian istance function. Thus, From (.) we euce that (x, y) = lim t t log u(t, x, y) = n. (.) A x (r) V x (r) = n, where A x (r) an V x (r) enote respectively the area of B x (r) an the volume of B x (r) = {y : (x, y) r}. This implies that V x (r) is the same as the volume function of Eucliean balls. The equality case of the volume comparison theorem gives us = R n. 3 Rate of change of the entropy in terms of the spectral gap In this section we will obtain an estimate on the time erivative of the entropy in terms of the spectral gap. For simplicity, set V =. Then we have the Sturm-Liouville ecomposition of the funamental solution, Φ(t, x, y) = e λjt/ ϕ j (x)ϕ j (y) (3.) j= whereby ϕ j are the eigenfunctions of with eigenvalue λ j, = λ < λ <, λ j. In particular, each eigenvalue has finite multiplicity an each ϕ j is smooth. Furthermore, {ϕ j } j= form an orthonormal basis for the L () an ϕ = if Φ(t, x, y) y, see [, p.39]. The series in (3.) converges absolutely an uniformly. Given any solution u to the heat equation with initial ata f, we can write where An u(t, x) = f(y)φ(t, x, y) y = c j = t u(t, x) = u(t, x) = e λjt/ c j ϕ j (x), (3.) j= f(y)ϕ j (y) y. e λjt/ λ j c j ϕ j (x). (3.3) The series in (3.) an (3.3) converge absolutely an uniformly for t >. Write f, g = fg x an f = f, f. j= 9

10 Proposition 3.. Assume is a close Riemannian manifol. Let f C (), f >, an u be the solution to the heat equation with u() = f. Then t Ent(u) e λ t/ f Vol() ( log inf f + log sup f ). Here, Vol() is the volume of the manifol. Proof. By our assumption, f is continuous. Thus, f = f, ϕ j ϕ j = λ j f, ϕ j ϕ j = λ j c j ϕ j, j= j= an f = f, f = j= λ j c j <. By (3.3), u = j= j= e λjt λ jc j e λ t λ jc j = e λt f. Next by (3.), it is clear that log u x Vol() ( log inf f + log sup f ). Now the estimate follows from Cauchy s inequality: t Ent(u) = log u u ν log u u j= e λ t/ f Vol() ( log inf f + log sup f ). 4 Entropy of the heat kernel on the three imensional hyperbolic space H 3 In this section we consier the three imensional hyperbolic space H 3 of constant sectional curvature k <. Let (, ) be the Riemannian istance function on H 3. The heat kernel on H 3 has the following explicit formula (see [, p.5]): h(t, x, y) = e (x,y) /t (t) 3/ k (x, y) sinh k (x, y) ekt/, x, y H 3. (There is a mistake in the formula for h(t, x, y) given in [5, Example 5..3]: there the last factor is e t (k = ), rather than e t/.) Let V H 3 be the volume measure on H 3 an efine the entropy Ent(h(t, x, )) = h(t, x, y) log h(t, x, y) V H 3(y), H 3 (t, x) (, ) H 3. The main result in this section is Theorem 4.. k( log ) lim t Ent(h(t, x, )) lim t t t Ent(h(t, x, )) k( + log ).

11 To prove this theorem, we nee some preparations. We enote by κ = k for the simplification of notations. Using polar coorinates the metric on H 3 is expresse as s = r + κ (sinh κr) θ, where θ is the stanar volume measure on the sphere S. Hence for any integrable function f : H 3 R, the following equality hols: ( ) (sinh κr) f(x) V H 3(x) = f(rθ) θ H 3 S κ r. (4.) We have where Ent(h(t, x, )) = 3 log(t) + κ t + I (t) + I (t), (4.) I (t) = h(t, x, y)(x, y) V t H 3(y), (4.3) H 3 sinh κ(x, y) I (t) = h(t, x, y) log V H 3 κ(x, y) H 3(y). (4.4) We first compute I (t). Using the expression of h(t, x, y) an by the formula (4.), we have ( I (t) = t(t) 3/ e κ t/ = κt 5/ e I κ t/ (t), S e r /t ) κr (sinh κr) sinh κr r θ κ r where I (t) = e r /t r 3 sinh κr r. Here we collect some results for later use. Lemma 4.. Let α(t) = κt / e r / r. Then e r /t sinh κr r = t / e κt/ α(t), e r /t cosh κr r = t/ e κt/, e r /t r sinh κr r = κt3/ e κt/, e r /t r cosh κr r = t + κt 3/ e κ t/ α(t), e r /t r sinh κr r = κt + t 3/ (κ t + )e κt/ α(t), e r /t r cosh κr r = t3/ (κ t + )e κt/, e r /t r 3 sinh κr r = κt5/ (κ t + 3)e κt/, e r /t r 3 cosh κr r = t (κ t + ) + κt 5/ (κ t + 3)e κ t/ α(t), e r /t r 4 sinh κr r = κt 3 (κ t + 5) + t 5/ (κ 4 t + 6κ t + 3)e κ t/ α(t).

12 Proof. We only prove the first two equalities. The others can be prove using the integration by parts formula. We have e r /t e κr r = e κ t/ e r /t r = t / e κ t/ e r / r κt κt / an Hence e r /t e κr r = e κ t/ e r /t sinh κr r = κt / t/ e κ t/ Next since the function r cosh κr is even, κt e r /t cosh κr r = = e r /t r = t / e κ t/ e r / r. κt / κt / e r / r = t / e κ t/ α(t). e r /t cosh κr r e r /t e κr r = t/ e κ t/. By Lemma 4., we obtain I (t) = κt 5/ e κ t/ Now we consier I (t). Again by (4.), I (t) = κ t 3/ e κ t/ sinh κr κt5/ (κ t + 3)e κ t/ = (κ t + 3). (4.5) e r /t r sinh κr log sinh κr κr r. (4.6) Due to the presence of the term log κr, we are unable to compute I (t) explicitly. In the sequel we inten to fin some estimates on it. Define ξ(t) = κ t 3/ e κ t/ an η(t) = e r /t sinh κr r sinh κr log r. (4.7) κr We have ξ (t) = κ t + 3 κ t 5/ e κ t/ Therefore, to estimate I (t), it is enough to estimate η(t) an η (t). The following lemma gives the key ingreient. Lemma 4.3. For any r >, + r < e r r < + r. <. (4.8) Proof. () Let ϕ(s) = e s se s. Then ϕ() = an ϕ (s) = se s >. Hence ϕ(s) > for all s >. This implies + s e s se s > s, which leas to the first inequality by taking s = r.

13 () Let ψ(r) = r e r re r. Then ψ() = an ψ (r) = +e r +re r. We have ψ () = an ψ (r) = 4re r < for any r >. Therefore ψ (r) <. As a result, ψ(r) < for all r >. This implies that + r e r re r < r, which is equivalent to the secon inequality. Remark 4.4. Fix any β (, ). In the same way we can show that for r (, β ), it hols e r r > +βr e r. oreover, it is clear that +βr when r is sufficiently large. Therefore the two inequalities in Lemma 4.3 are sharp in this sense. Notice that thus by Lemma 4.3, κr + log + κr sinh κr κr < log sinh κr κr r < = eκr ( e κr ), κr < κr + log Now we can obtain the upper an lower boun on η(t)., r >. (4.9) + κr Lemma 4.5. η(t) > κ t + κt 3/ (κ t + )e κt/ α(t) κt3/ e κt/ log(κ t + 4), ( ) η(t) < κ t + κt 3/ (κ t + )e κt/ α(t) κt3/ e κt/ log + κ tα(t). Proof. By the efinition (4.7) an the inequality (4.9), we have where by Lemma 4., η (t) = κ η(t) > η (t) η (t), (4.) e r /t r sinh κr r = κ t + κt 3/ (κ t + )e κ t/ α(t) (4.) an η (t) = e r /t r log( + κr) sinh κr r. Define the measure µ(r) = κ t 3/ e κt/ e r /t r sinh κr r on [, ). By Lemma 4., µ is a probability. Notice that the function r log( + κr) is concave, thus by Jensen s inequality, ( η (t) = κt3/ e κ t/ ) log( + κr) µ(r) κt3/ e κt/ log + κ r µ(r). Again by Lemma 4., r µ(r) = κ t 3/ e κ t/ e r /t r sinh κr r = t/ e κt/ + κ (κ t + )α(t). β 3

14 It is easy to show that t/ e κt/ κ e κ an α(t). Hence Consequently r µ(r) κ + κt + κ = κt + 3 κ. η (t) κt3/ e κ t/ log(κ t + 4). Combining this with (4.) an (4.), we obtain the first inequality. Now we prove the secon inequality. By (4.9), we have where η (t) is efine in (4.) an Define the measure η (t) = η(t) < η (t) η (t), (4.) e r /t r log( + κr) sinh κr r. ν(r) = t / e κ t/ α(t) e r /t sinh κr r on [, ). Then by Lemma 4., ν is also a probability. Notice that the function r r log(+κr) is convex on [, ), thus by Jensen s inequality, By Lemma 4., we have Therefore η (t) = t / e κt/ α(t) r log( + κr) ν(r) ( ) ( t / e κt/ α(t) r ν(r) log + κ r ν(r) = t / e κt/ α(t) e r /t r sinh κr r = η (t) ( ) κt3/ e κt/ log + κ tα(t). Combining this with (4.) an (4.) gives the secon inequality. Note that η (t) = t e r /t r 3 sinh κr log In the same way we can prove the bouns on η (t). Lemma 4.6. sinh κr κr η (t) > κ t(κ t + 5) + κt/ (κ 4 t + 6κ t + 3)e κ t/ α(t) ) r ν(r). κtα(t). r. (4.3) κt/ (κ t + 3)e κ t/ ( t / (κ t + 5) log + κ κ e κt/ + t + 3 κ 4 t + 6κ ) t + 3 κ α(t), t + 3 η (t) < κ t(κ t + 5) + κt/ (κ 4 t + 6κ t + 3)e κt/ α(t) ( κt/ (κ t + 3)e κt/ log + κ t(κ t + 3)e κ t/ κt / + (κ t + )e κ t/ α(t) ). 4

15 Proof. The proofs are similar to Lemma 4.5, hence we only give a sketch here. By (4.3) an (4.9), we have η (t) > η 3 (t) η 4 (t), (4.4) where, by Lemma 4., an η 3 (t) = κ t e r /t r 4 sinh κr r = κ t(κ t + 5) + κt/ (κ 4 t + 6κ t + 3)e κ t/ α(t) (4.5) η 4 (t) = t e r /t r 3 log( + κr) sinh κr r. As in Lemma 4.5, using the concavity of r log( + κr), we can get η 4 (t) κt/ (κ t + 3)e κ t/ ( t / (κ t + 5) log + κ κ e κt/ + t + 3 κ 4 t + 6κ ) t + 3 κ α(t). t + 3 Together with (4.4) an (4.5) gives us the first inequality. Next, by (4.9), η (t) < η 3 (t) η 4 (t), (4.6) where η 3 (t) is efine in (4.5) an η 4 (t) = t e r /t r 3 log( + κr) sinh κr r. Using the convexity of the function r r log( + κr), we can show that η 4 (t) ( κt/ (κ t + 3)e κt/ κ t(κ ) t + 3)e κ t/ log + κt / + (κ t + )e κt/. α(t) Now the secon inequality follows from the above estimate an (4.6), (4.5). Finally we are reay to prove Theorem 4.. Proof of Theorem 4.. First we consier the upper limit of the time erivative of the entropy. By (4.8) an Lemma 4.5, we have ξ (t)η(t) < κ κ t + 3 t / e κ t/ α(t) κ 4 t + 4κ t + 3 t + κ t + 3 t log(κ t + 4). Recall that α(t) is efine in Lemma 4. an lim t α(t) =. An by Lemma 4.6, we have ξ(t)η (t) < κ κ t + 5 t / e κ t/ + α(t) κ 4 t + 6κ t + 3 t κ t + 3 t log ( + κ t(κ t + 3)e κ t/ κt / + (κ t + )e κ t/ α(t) ). 5

16 Summing up the above two estimates, we obtain t( ξ(t)η(t) ) < κ t / e κ t/ + From this it is easy to see that κ α(t) + κ t + 3 log t + κ t + 4 κ t(κ t+3)e κ t/ κt / +(κ t+)e κ t/ α(t). Similarly we have lim t ( ) ξ(t)η(t) κ + κ t log = κ ( + log ). lim t By (4.) an (4.5) (4.7), ( ) ξ(t)η(t) κ κ t log = κ ( log ). t Ent(h(t, x, )) = 3 t + κ + ( ) ξ(t)η(t), t Noting that κ = k, we finally complete the proof. References [] arc Arnauon an Anton Thalmaier, Li-Yau type graient estimates an harnack inequalities by stochastic analysis. Probabilistic approach to geometry, 9 48, Av. Stu. Pure ath., 57, ath. Soc. Japan, Tokyo,. [] Issac Chavel. Eigenvalues in Riemannian geometry, Pure an Applie athematics, vol. 5, Acaemic Press Inc., Orlano, FL, 984. [3] B. Chow, S.-C. Chu, D. Glickenstein, C. Guenther, J. Isenberg, T. Ivey, D. Knopf, P. Lu, F. Luo an L. Ni, The Ricci Flow: Techniques an Applications, Part II, athematical Surveys an onographs, AS, Provience, RI, 7. [4] R. S. Hamilton, A matrix Harnack estimate for the heat equation. Comm. Anal. Geom. (993), no., 3 6. [5] Elton P. Hsu, Stochastic analysis on manifols, Grauate Stuies in athematics, vol. 38, American athematical Society, Provience, RI,. [6] P. Li an S.-T. Yau, On the parabolic kernel of the Schröinger operator. Acta. ath. 56 (986), [7] Brett Kotschwar an Lei Ni, Local graient estimates of p-harmonic functions, /H-flow, an an entropy formula. Ann. Sci. c. Norm. Supr. (4) 4 (9), no., 36. [8] Lei Ni, The entropy formula for linear heat equation. J. Geom. Anal. 4 (4), no., 87. [9] Lei Ni, Aena to The entropy formula for linear heat equation. J. Geom. Anal. 4 (4), no.,