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8 Lorentz Invariance and Special Relativity
where v = ctanh? From the identity cosh?2 = 1 ? tanh2 we see that ? ? cosh? = 1/ p 1?v2/c2 We see from the ?rst equation that the origin of the primed coordinate system x? = 0 corresponds to x= vtwhich means that the origin of the primed system is moving on the x-axis at the speed v We say that the primed system is boosted by
Eigenvalue Comparison on Bakry-Emery Manifolds
To apply Theorem 1 2 as in [9] we consider ¯ Ds the eigenfunction on ? 2 D 2 with the corresponding eigenvalue ¯ aD Let xt =Ce? ¯ aD t ¯D s Let wx be the ?rst non-constant eigenfunction of f and let vxt =e? 1twx Since ¯D s is an odd function (by adding an eigenfunction s with ?s one can always obtain one) we do have 0t = 0
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2 Principal Normal nˆ Proved earlier that if a(t) = const then a ·da/dt = 0 So ˆt =ˆt(s) ˆt = const ?ˆt·dˆt/ds = 0 Hence the principal normal nˆ is de?ned from ?nˆ = dˆt/ds where ? ? 0 is the curve’s curvature n t d ds t s increasing 3 The Binormal ˆb The third member of a local r-h set is the binormal ˆb =ˆt×n
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Communications in Partial Differential Equations
Publication details, including instructions for authors and subscription information: http://www.tandfonline.com/loi/lpde20Eigenvalue Comparison on Bakry-Emery Manifolds
Ben Andrews a & Lei Ni b
a Mathematical Sciences Institute, Australia National University, Canberra, AustraliaMathematical Sciences Center, Tsinghua University, Beijing, China and Morningside Center for Mathematics, Chinese Academy of Sciences, Beijing, China b Department of Mathematics, University of California at San Diego, La Jolla, California, USA Accepted author version posted online: 03 Apr 2012.Version of record first published: 18 Oct 2012.To cite this article: Ben Andrews & Lei Ni (2012): Eigenvalue Comparison on Bakry-Emery Manifolds, Communications in
Partial Differential Equations, 37:11, 2081-2092
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ISSN 0360-5302 print/1532-4133 online
DOI: 10.1080/03605302.2012.668602
Eigenvalue Comparison on Bakry-Emery Manifolds
BEN ANDREWS
1AND LEI NI
2 1 Mathematical Sciences Institute, Australia National University, Canberra, Australia; Mathematical Sciences Center, Tsinghua University, Beijing, China and Morningside Center for Mathematics, Chinese Academy ofSciences, Beijing, China
2 Department of Mathematics, University of California at San Diego,La Jolla, California, USA
We prove a comparison theorem on the modulus of continuity of the solution of a heat equation with a drifting term on Bakry-Emery manifolds. A direct consequence of the result is an alternate proof of an eigenvalue comparison result of Bakry-Qian. Examples are given to show that the estimate is sharp. Discussions on an explicit lower estimate for the corresponding ODE and an application to the diameter lower bound for gradient shrinking solitons are also included. KeywordsBakry-Emery manifolds; Eigenvalue estimates; Harmonic oscillator; Heat equation; Modulus of continuity; Ricci solitons; Weber"s equation. Mathematics Subject ClassificationPrimary 35P15, 35J10; Secondary 35K05,58J35.
1. A Lower Bound for the First Eigenvalue of the Drift Laplacian
Recall that?M?g?f?, a triple consisting of a manifoldM, a Riemannian metricg and a smooth functionf, is called a gradient Ricci soliton if the Ricci curvature and the Hessian offsatisfy: Rc ij +f ij =ag ij ?(1.1) It is called shrinking, steady, or expanding soliton ifa>0,a=0ora<0 respectively. More generally?M?g?f?is called a Bakry-Emery manifold if the so-called Bakry-Emery Ricci tensor Rc ij +f ij ≥ag ij for somea??. In this paper we apply the modulus of continuity estimates developed in [1-3] to give a different proof of an eigenvalue comparison estimate on Bakry-Emery manifolds for the operator? f ??-???·???f?on strictly convex??Mwith diameterDand smooth boundary. This result was first proved in [5, Theorem 14], which serves as a Received November 20, 2011; Accepted February 16, 2012 Address correspondence to Lei Ni, Department of Mathematics, University of California at San Diego, La Jolla, CA 92093, USA; E-mail: lni@math.ucsd.edu2081Downloaded by [University of California, San Diego] at 10:28 02 April 2013
2082 Andrews and Ni
generalization to the earlier works of Payne-Weinberger[10], Li-Yau[7] and Zhong- Yang [12]. An more recent result of this kind was obtained in [6]. Here in fact in Theorem 1.2 we extend a comparison theorem of [3] on the modulus of continuity to manifolds with lower bound on the Bakry-Emery Ricci tensor. This new result gives sharp modulus of continuity comparison between the solution to t f and the corresponding sup-solution to a heat equation on a certain interval. This implies the eigenvalue comparison result of Bakry- Qian, Theorem 1.1, since first eigenvalue determines the rate of convergence to equilibrium. As in [6] applying to the soliton setting, this implies a lower diameter estimate for nontrivial gradient shrinking solitons (which improves [6]). We remark here that the eigenvalue estimate we obtain is sharp for?M?g?f?satisfying theBakry-Emery-Ricci lower bound Rc
ij +f ij ≥ag ij , but presumably is not so for Ricci solitons where the Bakry-Emery-Ricci tensor is constant, and so we expect that our diameter bound is also not sharp. We discuss the sharpness of the eigenvalue inequality in Section 2. Before we state the result, we define a corresponding 1-dimensional eigenvalue problem. On?- D 2 D 2 ?we consider the functionals D 2 D 2 e a 2 s 2 2 ds?and????=?????e a 2 s 2 2 ds? namely the Dirichlet energy with weight e a 2 s 2 and its Rayleigh quotient. The associated elliptic operator is? a d 2 ds 2 -as d ds . Let¯? a?D be the first non-zeroNeumann eigenvalue of?
a , which is the minimum of?amongW 1?2 -functions with zero average. Theorem 1.1(Bakry-Qian).Let?be a compact manifoldM, or a bounded strictly convex domain inside a complete manifoldM, satisfying thatRc ij +f ij ≥ag ij . Assume thatDis the diameter of?. Then the first non-zero Neumann eigenvalue?? 1 of the operator? f is at least¯? a?D Proof. The key is to adapt Theorem 2.1 of [3] to this setting. Recall that?is a 2?? d M ?x?y? 2Theorem 1.2.Letv?x?t?be a solution to
?v t=?v-2?X??v?(1.2) with2X=?f. Assume alsov?x?t?satisfies the Neumann boundary condition. Suppose thatv?x?0?has a modulus of continuity? 0 ?s???0? D 2 ?→?with? 0 ?0?=0and ?0 >0on?0? D 2 ?. Assume further that there exists a function??s?t? ? ?0? D 2 such that (i)??s?0?=? 0 ?s?on?0?D/2?; (ii) t -as? on?0?D/2?×? (iii)? ?s?t? >0on?0? D 2(iv)??0?t?≥0for eacht≥0.Downloaded by [University of California, San Diego] at 10:28 02 April 2013
Eigenvalue Comparison 2083
Here? s st 2 ?s 2 ??s?t?. Then??s?t?is a modulus of the continuity ofv?x?t? for allt>0. Proof. The proof of Theorem 1.2 is a modification of the argument of Theorem 2.1 in [3]. Precisely, consider ?x?y?t??v?y?t?-v?x?t?-2??r?x?y? 2?t? -?e tIt suffices to show that?
ad absurdum. Assume that there exists?x 0 ?y 0 ?t 0 ?such that? ?·?·?t?=0 for the first time. Namely? ?x?y?t?achieves its maximum over?×?×?0?t 0 ?at?x 0 ?y 0 ?t 0 The strictly convexity, the Neumann boundary condition satisfied byv?x?t?, and the positivity of? rule out the possibility that the maximum can be attained at?x 0 ?y 0 ?????×??. For the interior pair?x 0 ?y 0 ?where the maximum of?is attained, pick a frame?e i ?atx 0 and parallel translate it along a minimizing geodesic ??s? ? ?0?d?→Mjoiningx 0 withy 0 . Still denote it by?e i ?. We may also arrange e n ?s?. Let?E i ?be the frame at?x 0 ?y 0 ?(inT ?x 0 ?y 0 ?×?) defined asE i =e i ?e i n =e n ??-e n ?. Direct calculations show that at?x 0 ?y 0 ?t 0 t- n j=1 2 E j E j ?x?y?t?=-???f?y??? ?-??f?x??? ?n-1 i=1 2 E i E i r?x?y?-2? t +2? -?e t Here we have used the first variation???·?·?t 0 ?=0at?x 0 ?y 0 ?which implies the identities ??v??y?t 0 ?d? ??v??x?t 0 ?0??Now choose the variational vector fieldV
i ?s?=e i ?s?, the parallel transport ofe i along??s?, the second variation computation gives that n-1 i=1 2 E i E i d 0 Rc?? ?ds?(1.3)Hence at?x
0 ?y 0 ?t 0 ?we have that t- n j=1 2 E j E j d 0 2 f+Rc??? ?ds-2? t +2? -?e t ar?x?y?-2? t +2? -?e t <0?Downloaded by [University of California, San Diego] at 10:28 02 April 20132084 Andrews and Ni
Here we have usedd=r?x?y?ands=
r?x?y? 2 . This contradicts with that at?x 0 ?y 0 ?t 0 since it is the first time??x?y?t?=0, ?t? ?x 0 ?y 0 ?t 0 ≥0? 2 E i E i ?x 0 ?y 0 ?t 0The above argument works well as long as?x
0 ?y 0 ?is not conjugate to each other along the minimizing geodesic??s?since we need this condition in establishing (1.3).quotesdbs_dbs15.pdfusesText_21[PDF] english movies dialogues pdf
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