Uniqueness of Conformal Ricci Flow using Energy Methods
aa r X i v : . [ m a t h . DG ] J a n Uniqueness of Conformal Ricci Flow usingEnergy Methods
Thomas BellApril 9, 2018
Abstract
We analyze an energy functional associated to Conformal Ricci Flowalong closed manifolds with constant negative scalar curvature. Giveninitial conditions we use this functional to demonstrate the uniqueness ofboth the metric and the pressure function along Conformal Ricci Flow.
The uniqueness of Ricci Flow on closed manifolds was originally proven byHamilton [4]. Later on, Chen and Zhu proved the uniqueness on completenoncompact manifolds with bounded curvature [2]. Both proofs utilize DeTurckRicci Flow. Recently Kotschwar used energy techniques to give another proofof the uniqueness on complete manifolds [5]. Kotchwar’s proof does not relyon DeTurck Ricci Flow. A natural question is whether similar techniques canbe applied to demonstrate the uniqueness of other geometric flows. One ofthese flows we have in mind is Conformal Ricci Flow, introduced by Fischer [3].Conformal Ricci Flow is, like Ricci Flow, a weakly parabolic flow of the metricon manifolds. Unlike Ricci Flow, Conformal Ricci Flow is restricted to the classof metrics of constant scalar curvature.Let ( M n , g ) be a smooth n-dimensional Riemannian manifold with a metric g of constant scalar curvature s . Conformal Ricci Flow on M is defined asfollows: ∂g∂t = − g ( t ) + 2 s n g ( t ) − p ( t ) g ( t ) s (cid:0) g ( t ) (cid:1) = s on M × [0, T ]. (1)1ere g ( t ), t ∈ [0, T ], is a family of metrics on M with g (0) = g , s (cid:0) g ( t ) (cid:1) isthe scalar curvature of g ( t ), and p ( t ), t ∈ [0, T ], is a family of functions on M .In [3] and [6] we see that (1) is equivalent to the following system: ∂g∂t = − g ( t ) + 2 s n g ( t ) − p ( t ) g ( t ) (cid:0) ( n − g ( t ) + s (cid:1) p ( t ) = − (cid:10) Ric g ( t ) − s n g ( t ), Ric g ( t ) − s n g ( t ) (cid:11) (2)Throughout this paper we will use V to denote the following symmetric2-tensor: V ( t ) = Ric g ( t ) − s n g ( t ) + p ( t ) g ( t ) (3)In this paper we use Kotchwar’s idea to give a proof of the uniqueness ofConformal Ricci Flow for closed manifolds with metrics of constant negativescalar curvature. Such uniqueness has been observed by Lu, Qing and Zhengusing DeTurck Conformal Ricci Flow [6]. More precisely we will prove thefollowing uniqueness theorem of Conformal Ricci Flow: Theorem 1.
Let ( M n , g ) be a closed manifold with constant negative scalarcurvature s . Suppose (cid:0) g ( t ), p ( t ) (cid:1) and (cid:0) ˜ g ( t ), ˜ p ( t ) (cid:1) are two solutions of (1) on M × [0, T ] with ˜ g (0) = g (0) . Then (cid:0) ˜ g ( t ), ˜ p ( t ) (cid:1) = (cid:0) g ( t ), p ( t ) (cid:1) for ≤ t ≤ T . g ( t ) and ˜ g ( t ) Let g ( t ) and ˜ g ( t ) be as in Theorem 1. We will treat g as our background metricand ˜ g as our alternative metric. Let ∇ , ˜ ∇ be the Riemannian connections of g and ˜ g respectively. Similarly, let R , ˜ R represent the full Riemannian curvaturetensors of g and ˜ g respectively.Let h = g − ˜ g . Let A = ∇ − ˜ ∇ . Explicitly, A ijk = Γ ijk − ˜Γ ijk whereΓ ijk and ˜Γ ijk are the Christoffel symbols of ∇ and ˜ ∇ respectively. Also let S = R − ˜ R , q = p − ˜ p .In this section we find bounds on h , A , S , q , ∇ q and ∇∇ q (see Propositions1 and 2). Throughout this paper we will use the convention X ∗ Y to denoteany finite sum of tensors of the form X · Y . We use C(X) to denote a finite sumof tensors of the form X . 2 .1 Preliminary Calculations First we calculate some useful expressions for quantities which will arise in theproofs of Propositions 1 and 2. We calculate g ij − ˜ g ij = g ik (˜ g jℓ ˜ g kℓ ) − ˜ g jℓ ( g ik g kℓ ) = − g ik ˜ g jℓ h kℓ ,i.e. g − − ˜ g − = ˜ g − ∗ h .If X is any tensor which is not a function we have (cid:0) ∇ − ˜ ∇ (cid:1) X = A ∗ X .We check this when X is a (1, 1)-tensor. Calculating in local coordinates wesee (cid:0) ∇ i − ˜ ∇ i (cid:1) X kj = ∂ i X kj − Γ ℓij X kℓ + Γ kiℓ X ℓj − ∂ i X kj + ˜Γ ℓij X kℓ − ˜Γ kiℓ X ℓj = A kiℓ X ℓj − A ℓij X kℓ = A ∗ X .If f is a function however, then we have the following: (cid:0) ∇ i − ˜ ∇ i (cid:1) f = (cid:0) g ij − ˜ g ij (cid:1) ∂ i f = − g ik ˜ g jℓ h kℓ ∂ i f = − g ik h kℓ ˜ ∇ ℓ f ,or in other words (cid:0) ∇ − ˜ ∇ (cid:1) f = h ∗ ˜ ∇ f .We now calculate ∇ ˜ g − = (cid:0) ∇ − ˜ ∇ (cid:1) ˜ g − = ˜ g − ∗ A .The following calculation will also be important. ∇ i h jk = ∇ i g jk − ∇ i ˜ g jk = − (cid:0) ∇ i − ˜ ∇ i (cid:1) ˜ g jk .Thus we have ∇ h = ˜ g ∗ A .3ow we are able to calculate the following for a function f . ∇ (cid:0) ∇ − ˜ ∇ (cid:1) f = ∇ (cid:0) h ∗ ˜ ∇ f (cid:1) = ∇ h ∗ ˜ ∇ f + h ∗ (cid:0) ∇ − ˜ ∇ (cid:1) ˜ ∇ f + h ∗ ˜ ∇ ˜ ∇ f = ˜ g ∗ A ∗ ˜ ∇ f + h ∗ A ∗ ˜ ∇ f + h ∗ ˜ ∇ ˜ ∇ f .Now let U aijkℓ = g ab ∇ b ˜ R ijkℓ − ˜ g ab ˜ ∇ b ˜ R ijkℓ (4)= g ab ( ∇ b − ˜ ∇ b ) ˜ R ijkℓ + (cid:0) g ab − ˜ g ab (cid:1) ˜ ∇ b ˜ R ijkℓ = A ∗ ˜ R + ˜ g − ∗ h ∗ ˜ ∇ ˜ R ,and we may calculate ∇ a (cid:0) g ab ∇ b R − ˜ g ab ˜ ∇ b ˜ R (cid:1) = ∇ a (cid:0) g ab ∇ b ˜ R − ˜ g ab ˜ ∇ b ˜ R (cid:1) + g ab ∇ a ∇ b (cid:0) R − ˜ R (cid:1) = div U + ∆ S .We summarize the above calculations in the following Lemma: Lemma 1.
Using the notation defined at the beginning of this section, g − − ˜ g − = ˜ g − ∗ h (5) (cid:0) ∇ − ˜ ∇ (cid:1) X = A ∗ X (6) (cid:0) ∇ − ˜ ∇ (cid:1) f = h ∗ ˜ ∇ f (7) ∇ ˜ g − = ˜ g − ∗ A (8) ∇ h = ˜ g ∗ A (9) ∇ (cid:0) ∇ − ˜ ∇ (cid:1) f = ˜ g ∗ A ∗ ˜ ∇ f + h ∗ A ∗ ˜ ∇ f + h ∗ ˜ ∇ ˜ ∇ f (10) U = A ∗ ˜ R + ˜ g − ∗ h ∗ ˜ ∇ ˜ R (11) ∇ a (cid:0) g ab ∇ b R − ˜ g ab ˜ ∇ b ˜ R (cid:1) = div U + ∆ S (12) where U is defined in (4). h , A and S In this subsection we derive bounds on the time derivatives of h , A and S . Inparticular we will prove the following proposition. Here, as well as throughout4his paper, C will denote a constant dependent only upon n while N will denotea constant with further dependencies. Proposition 1.
Let (cid:0) g ( t ), p ( t ) (cid:1) and (cid:0) ˜ g ( t ), ˜ p ( t ) (cid:1) be two solutions of (1) on M × [0, T ] . Using the notation defined at the beginning of this section, there existconstants N h , N A and N S such that (cid:12)(cid:12)(cid:12)(cid:12) ∂∂t h (cid:12)(cid:12)(cid:12)(cid:12) ≤ N h | h | + C (cid:0) | S | + | q | (cid:1) (13) (cid:12)(cid:12)(cid:12)(cid:12) ∂∂t A (cid:12)(cid:12)(cid:12)(cid:12) ≤ N A (cid:0) | h | + | A | (cid:1) + C (cid:0) |∇ S | + |∇ q | (cid:1) (14) (cid:12)(cid:12)(cid:12)(cid:12) ∂∂t S − ∆ S − div U (cid:12)(cid:12)(cid:12)(cid:12) ≤ N S (cid:0) | h | + | A | + | S | + | q | (cid:1) + C |∇∇ q | (15) where U is defined in (4).Proof. We start with the time derivative of h . By (1) we have ∂∂t h ij = − R ij − ˜ R ij ) + 2 s n ( g ij − ˜ g ij ) − p g ij − ˜ p ˜ g ij )= − S kkij + 2 s n h ij − (cid:2) ( p − ˜ p ) g ij + ˜ p ( g ij − ˜ g ij ) (cid:3) = − S kkij + 2 s n h ij − q g ij − p h ij .Hence ∂∂t h = C ( S ) + C ( s h ) + C ( q ) + ˜ p ∗ h and (cid:12)(cid:12)(cid:12)(cid:12) ∂∂t h (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:16)(cid:0) | s | + | ˜ p | (cid:1) | h | + | S | + | q | (cid:17) . (16)This proves (13).Recall the definition of V from (3): V ( t ) = Ric g ( t ) − s n g ( t ) + p ( t ) g ( t ). (17)We may define ˜ V similarly using our alternate metric ˜ g . Since V and ˜ V aresymmetric 2-tensors, then by [1, p. 108] we may calculate ∂∂t A kij = ˜ g kℓ (cid:0) ˜ ∇ i ˜ V jℓ + ˜ ∇ j ˜ V iℓ − ˜ ∇ ℓ ˜ V ij (cid:1) − g kℓ (cid:0) ∇ i V jℓ + ∇ j V iℓ − ∇ ℓ V ij (cid:1) . (18)5e proceed to calculate˜ g kℓ ˜ ∇ i ˜ V jℓ − g kℓ ∇ i V jℓ =˜ g kℓ ( ˜ ∇ i ˜ R jℓ ) − g kℓ ( ∇ i R jℓ ) + ˜ g kℓ ˜ ∇ i (˜ p ˜ g jℓ ) − g kℓ ∇ i ( p g jℓ )= (cid:0) ˜ g kℓ − g kℓ (cid:1) ˜ ∇ i ˜ R jℓ + g kℓ ( ˜ ∇ i − ∇ i ) ˜ R jℓ − g kℓ ∇ i ( S mmjℓ ) + δ kj ˜ ∇ i ˜ p − δ kj ∇ i p =˜ g − ∗ h ∗ ˜ ∇ ˜ R + A ∗ ˜ R + C ( ∇ S ) + h ∗ ˜ ∇ ˜ p + C ( ∇ q ), (19)where we have used (7) to get the last equality. Similarly we find˜ g kℓ ˜ ∇ j ˜ V iℓ − g kℓ ∇ j V iℓ =˜ g − ∗ h ∗ ˜ ∇ ˜ R + A ∗ ˜ R + C ( ∇ S ) + h ∗ ˜ ∇ ˜ p + C ( ∇ q ). (20)Now we consider − ˜ g kℓ ˜ ∇ ℓ ˜ V ij + g kℓ ∇ ℓ V ij =˜ g − ∗ h ∗ ˜ ∇ ˜ R + A ∗ ˜ R + C ( ∇ S ) + ˜ g kℓ ˜ g ij ˜ ∇ ℓ ˜ p − g kℓ g ij ∇ ℓ p =˜ g − ∗ h ∗ ˜ ∇ ˜ R + A ∗ ˜ R + C ( ∇ S ) + (cid:0) ˜ g kℓ − g kℓ (cid:1) ˜ g ij ˜ ∇ ℓ ˜ p + g kℓ (cid:0) ˜ g ij − g ij (cid:1) ˜ ∇ ℓ ˜ p + g kℓ g ij ( ˜ ∇ ℓ − ∇ ℓ )˜ p + g kℓ g ij ∇ ℓ (˜ p − p )=˜ g − ∗ h ∗ ˜ ∇ ˜ R + A ∗ ˜ R + C ( ∇ S ) + ˜ g − ∗ h ∗ ˜ g ∗ ˜ ∇ ˜ p + h ∗ ˜ ∇ ˜ p + C ( ∇ q ). (21)Hence by (18), (19), (20) and (21), ∂∂t A = ˜ g − ∗ h ∗ ˜ ∇ ˜ R + A ∗ ˜ R + C ( ∇ S ) + h ∗ ˜ ∇ ˜ p + C ( ∇ q ) + ˜ g − ∗ h ∗ ˜ g ∗ ˜ ∇ ˜ p and (cid:12)(cid:12)(cid:12)(cid:12) ∂∂t A (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:16)(cid:0) | ˜ g − || ˜ ∇ ˜ R | + | ˜ ∇ ˜ p | + | ˜ g − || ˜ g || ˜ ∇ ˜ p | (cid:1) | h | + | ˜ R || A | + |∇ S | + |∇ q | (cid:17) . (22)This proves (14). 6y [1, eqn. (2.67)] we have ∂∂t R ℓijk = g ℓm (cid:0) ∇ i ∇ k V jm − ∇ i ∇ m V jk − ∇ j ∇ k V im + ∇ j ∇ m V ik (cid:1) − g ℓm (cid:0) R rijk V rm + R qijm V kq (cid:1) = g ℓm (cid:0) −∇ i ∇ k R jm + ∇ i ∇ m R jk + ∇ j ∇ k R im − ∇ j ∇ m R ik (cid:1) + g ℓm (cid:0) − g jm ∇ i ∇ k p + g jk ∇ i ∇ m p + g im ∇ j ∇ k p − g ik ∇ j ∇ m p (cid:1) + g ℓm (cid:0) R rijk R rm + R rijm R kr (cid:1) − s n g ℓm (cid:0) R rijk g rm + R rijm g kr (cid:1) p + g ℓm (cid:0) R rijk g rm + R rijm g kr (cid:1) p . (23)Following the calculations in [1, p. 119-120] we have∆ R ℓijk = g ab ∇ a ∇ b R ℓijk = g ab (cid:0) −∇ a ∇ i R ℓjbk − ∇ a ∇ j R ℓbik (cid:1) = g ab (cid:0) −∇ i ∇ a R ℓjbk + R maij R ℓmbk + R maib R ℓjmk + R maik R ℓjbm − R ℓaim R mjbk − ∇ j ∇ a R ℓbik + R majb R ℓmik + R maji R ℓbmk + R majk R ℓbim − R ℓajm R mbik (cid:1) = g ℓm (cid:0) −∇ i ∇ k R jm + ∇ i ∇ m R jk + ∇ j ∇ k R im − ∇ j ∇ m R ik (cid:1) + g mr (cid:0) − R ir R ℓjmk − R jr R ℓmik )+ g ab (cid:0) R maij R ℓmbk + R maik R ℓjbm − R ℓaim R mjbk + R maji R ℓbmk + R majk R ℓbim − R ℓajm R mbik (cid:1) . (24)Combining (23) and (24) we have ∂∂t R ℓijk = ∆ R ℓijk + g mr (cid:0) R ir R ℓjmk + R jr R ℓmik (cid:1) + g ab (cid:0) − R maij R ℓmbk − R maik R ℓjbm + R ℓaim R mjbk − R maji R ℓbmk − R majk R ℓbim + R ℓajm R mbik (cid:1) + g ℓm (cid:0) − g jm ∇ i ∇ k p + g jk ∇ i ∇ m p + g im ∇ j ∇ k p − g ik ∇ j ∇ m p (cid:1) + g ℓm (cid:0) R rijk R rm + R rijm R kr (cid:1) − s n g ℓm (cid:0) R rijk g rm + R rijm g kr (cid:1) + g ℓm (cid:0) R rijk g rm + R rijm g kr (cid:1) p . (25)7ence the evolution of S is ∂∂t S ℓijk = ∆ R ℓijk − ˜∆ ˜ R ℓijk + g mr (cid:0) R ir R ℓjmk + R jr R ℓjmk (cid:1) − ˜ g mr (cid:0) ˜ R ir ˜ R ℓjmk + ˜ R jr ˜ R ℓmik (cid:1) + g ab (cid:0) − R maij R ℓmbk − R maik R ℓjbm + R ℓaim R mjbk − R maji R ℓbmk − R majk R ℓbim + R ℓajm R mbik (cid:1) − ˜ g ab (cid:0) − ˜ R maij ˜ R ℓmbk − ˜ R maik ˜ R ℓjbm + ˜ R ℓaim ˜ R mjbk − ˜ R maji ˜ R ℓbmk − ˜ R majk ˜ R ℓbim + ˜ R ℓajm ˜ R mbik (cid:1) + g ℓm (cid:0) − g jm ∇ i ∇ k p + g jk ∇ i ∇ m p + g im ∇ j ∇ k p − g ik ∇ j ∇ m p (cid:1) − ˜ g ℓm (cid:0) − ˜ g jm ˜ ∇ i ˜ ∇ k ˜ p + ˜ g jk ˜ ∇ i ˜ ∇ m ˜ p + ˜ g im ˜ ∇ j ˜ ∇ k ˜ p − ˜ g ik ˜ ∇ j ˜ ∇ m ˜ p (cid:1) + g ℓm (cid:0) R rijk R rm + R rijm R kr (cid:1) − ˜ g ℓm (cid:0) ˜ R rijk ˜ R rm + ˜ R rijm ˜ R kr (cid:1) − s n g ℓm (cid:0) R rijk g rm + R rijm g kr (cid:1) + s n ˜ g ℓm (cid:0) ˜ R rijk ˜ g rm + ˜ R rijm ˜ g kr (cid:1) + g ℓm (cid:0) R rijk g rm + R rijm g kr (cid:1) p − ˜ g ℓm (cid:0) ˜ R rijk ˜ g rm + ˜ R rijm ˜ g kr (cid:1) ˜ p . (26)Looking at the individual components, we see∆ R − ˜∆ ˜ R = g ab ∇ a ∇ b R − ˜ g ab ˜ ∇ a ˜ ∇ b ˜ R = ∇ a ( g ab ∇ b R ) − ∇ a (˜ g ab ˜ ∇ b ˜ R ) + ( ∇ a − ˜ ∇ a )(˜ g ab ˜ ∇ b ˜ R )= ∇ a (cid:0) g ab ∇ b R − ˜ g ab ˜ ∇ b ˜ R (cid:1) + ˜ g − ∗ A ∗ ˜ ∇ ˜ R , (27)while g − RR − ˜ g − ˜ R ˜ R =( g − − ˜ g − )( ˜ R ˜ R ) + g − ( RR − ˜ R ˜ R )=˜ g − ∗ h ∗ ˜ R ∗ ˜ R + g − ( R − ˜ R ) ˜ R + g − ( RR − R ˜ R )=˜ g − ∗ h ∗ ˜ R ∗ ˜ R + S ∗ ˜ R + S ∗ R , (28)8nd g − g ∇∇ p − ˜ g − ˜ g ˜ ∇ ˜ ∇ ˜ p =( g − − ˜ g − )˜ g ˜ ∇ ˜ ∇ ˜ p + g − ( g − ˜ g ) ˜ ∇ ˜ ∇ ˜ p + g − g ( ∇∇ p − ˜ ∇ ˜ ∇ ˜ p )=˜ g − ∗ h ∗ ˜ g ∗ ˜ ∇ ˜ ∇ ˜ p + h ∗ ˜ ∇ ˜ ∇ ˜ p + g − g ( ∇ − ˜ ∇ )( ˜ ∇ ˜ p ) + g − g ( ∇∇ p − ∇ ˜ ∇ ˜ p )=˜ g − ∗ h ∗ ˜ g ∗ ˜ ∇ ˜ ∇ ˜ p + h ∗ ˜ ∇ ˜ ∇ ˜ p + A ∗ ˜ ∇ ˜ p + g − g ∇ ( ∇ − ˜ ∇ )˜ p + g − g ∇∇ ( p − ˜ p )=˜ g − ∗ h ∗ ˜ g ∗ ˜ ∇ ˜ ∇ ˜ p + h ∗ ˜ ∇ ˜ ∇ ˜ p + A ∗ ˜ ∇ ˜ p + h ∗ A ∗ ˜ ∇ ˜ p + C ( ∇∇ q ), (29)where in the last equality we used (10). We also have g − gR − ˜ g − ˜ g ˜ R =( g − − ˜ g − )˜ g ˜ R + g − ( g − ˜ g ) ˜ R + g − g ( R − ˜ R )=˜ g − ∗ h ∗ ˜ g ∗ ˜ R + h ∗ ˜ R + C ( S ), (30)and lastly g − gRp − ˜ g − ˜ g ˜ R ˜ p =( g − − ˜ g − )˜ g ˜ R ˜ p + g − ( g − ˜ g ) ˜ R ˜ p + g − g ( R − ˜ R )˜ p + g − gR ( p − ˜ p )=˜ g − ∗ h ∗ ˜ g ∗ ˜ R ∗ ˜ p + h ∗ ˜ R ∗ ˜ p + S ∗ ˜ p + R ∗ q . (31)Now by (26), (27), (28), (29), (30) and (31) we see ∂∂t S = ∇ a (cid:0) g ab ∇ b R − ˜ g ab ˜ ∇ b ˜ R (cid:1) + ˜ g − ∗ A ∗ ˜ ∇ ˜ R + ˜ g − ∗ h ∗ ˜ R ∗ ˜ R + S ∗ ˜ R + S ∗ R + ˜ g − ∗ h ∗ ˜ g ∗ ˜ ∇ ˜ ∇ ˜ p + h ∗ ˜ ∇ ˜ ∇ ˜ p + A ∗ ˜ ∇ ˜ p + h ∗ A ∗ ˜ ∇ ˜ p + C ( ∇∇ q ) + ˜ g − ∗ h ∗ ˜ g ∗ ˜ R + h ∗ ˜ R + C ( S )+ ˜ g − ∗ h ∗ ˜ g ∗ ˜ R ∗ ˜ p + h ∗ ˜ R ∗ ˜ p + S ∗ ˜ p + R ∗ q .9ence by (12) we have (cid:12)(cid:12)(cid:12)(cid:12) ∂∂t S − ∆ S − div U (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18)(cid:16) | ˜ g − || ˜ R | + | ˜ g − || ˜ g || ˜ ∇ ˜ ∇ ˜ p | + | ˜ ∇ ˜ ∇ ˜ p | + | ˜ g − || ˜ g || ˜ R | + | ˜ R | + | ˜ g − || ˜ g || ˜ R || ˜ p | + | ˜ R || ˜ p | (cid:17) | h | + (cid:16) | ˜ g − || ˜ ∇ ˜ R | + | ˜ ∇ ˜ p | + | h || ˜ ∇ ˜ p | (cid:17) | A | + (cid:16) | ˜ R | + | R | + 1 + | ˜ p | (cid:17) | S | + | R || q | + |∇∇ q | (cid:19) . (32)This proves (15). Remark 1.
Upon closer observation we notice the following dependencies: N h = N h (cid:0) n , s , | ˜ p | (cid:1) , N A = N A (cid:0) n , s , | ˜ g | , | ˜ g − | , | ˜ R | , | ˜ ∇ ˜ R | , | ˜ ∇ ˜ p | (cid:1) , N S = N S (cid:0) n , s , | ˜ g | , | ˜ g − | , | h | , | R | , | ˜ R | , | ˜ ∇ ˜ R | , | ˜ p | , | ˜ ∇ ˜ p | , | ˜ ∇ ˜ ∇ ˜ p | (cid:1) . M is closed, so M × [0, T ] is compact. Thus, given two metrics g and ˜ g , all ofthese quantities will be bounded. q and its Spacial Derivatives We turn our attention now to finding bounds on the differences between ourpressure functions p and ˜ p . We have the following proposition: Proposition 2.
Let (cid:0) g ( t ), p ( t ) (cid:1) and (cid:0) ˜ g ( t ), ˜ p ( t ) (cid:1) be two solutions of (1) on M × [0, T ] . Then there exist constants N q and ˆ N q such that Z M | q | dµ ≤ N q Z M (cid:0) | h | + | A | + | S | (cid:1) dµ (33) Z M |∇ q | dµ ≤ N q Z M (cid:0) | h | + | A | + | S | (cid:1) dµ (34) Z M |∇∇ q | dµ ≤ ˆ N q Z M (cid:0) | h | + | A | + | S | (cid:1) dµ (35) Proof.
We let f represent any smooth function or tensor. In particular we willlet f be represented by the function q , the difference of the pressure functions.10ince M is compact we have Z M (cid:0) ( n − s (cid:1) ( f ) · f dµ = s Z M | f | dµ − ( n − Z M (cid:10) ∇ f , ∇ f (cid:11) dµ .Since s <
0, taking the absolute value gives (cid:12)(cid:12)(cid:12)(cid:12)Z M (cid:0) ( n − s )( f ) · f dµ (cid:12)(cid:12)(cid:12)(cid:12) = | s | Z M | f | dµ + ( n − Z M |∇ f | dµ (36)Now we deal specifically with p , ˜ p and q . By (2) we have the followingequations for the pressure functions p and ˜ p : (cid:0) ( n − s (cid:1) p = − D Ric − s n g , Ric − s n g E (37) (cid:0) ( n −
1) ˜∆ + s (cid:1) ˜ p = − D ˜Ric − s n ˜ g , ˜Ric − s n ˜ g E . (38)Now we calculate∆ p − ˜∆˜ p = g ab ∇ a ∇ b p − ˜ g ab ˜ ∇ a ˜ ∇ b ˜ p = ( g − − ˜ g − ) ˜ ∇ ˜ ∇ ˜ p + g − ( ∇ − ˜ ∇ ) ˜ ∇ ˜ p + g − ∇ ( ∇ − ˜ ∇ )˜ p + ∆( p − ˜ p )= ˜ g − ∗ h ∗ ˜ ∇ ˜ ∇ ˜ p + A ∗ ˜ ∇ ˜ p + h ∗ A ∗ ˜ ∇ ˜ p + ∆ q . (39)We also compute − D Ric − s n g , Ric − s n g E + D ˜Ric − s n ˜ g , ˜Ric − s n ˜ g E = − (cid:0) g ik g jℓ R ij R kℓ − ˜ g ik ˜ g jℓ ˜ R ij ˜ R kℓ (cid:1) + 2 s n (cid:0) g ij R ij − ˜ g ij ˜ R ij (cid:1) = − ( g − − ˜ g − )˜ g − ˜ R ˜ R − g − ( g − − ˜ g − ) ˜ R ˜ R − g − g − ( R − ˜ R ) ˜ R − g − g − R ( R − ˜ R ) + 2 s n ( g − − ˜ g − ) ˜ R + 2 s n g − ( R − ˜ R )=˜ g − ∗ ˜ g − ∗ h ∗ ˜ R ∗ ˜ R + ˜ g − ∗ h ∗ ˜ R ∗ ˜ R + S ∗ ˜ R + S ∗ R + ˜ g − ∗ h ∗ ˜ R + C ( S ). (40)Combining (37), (38), (39) and (40), we see that q satisfies the following11lliptic equation at each time t ∈ [0, T ]: Lq = (cid:0) ( n − s (cid:1) ( q )= ˜ g − ∗ h ∗ ˜ ∇ ˜ ∇ ˜ p + A ∗ ˜ ∇ ˜ p + h ∗ A ∗ ˜ ∇ ˜ p + ˜ g − ∗ ˜ g − ∗ h ∗ ˜ R ∗ ˜ R + ˜ g − ∗ h ∗ ˜ R ∗ ˜ R + S ∗ ˜ R + S ∗ R + ˜ g − ∗ h ∗ ˜ R + C ( S ) (41)Hence | Lq | = (cid:12)(cid:12)(cid:0) ( n − s (cid:1) ( q ) (cid:12)(cid:12) ≤ N (cid:0) | h | + | A | + | S | (cid:1) . (42)To find estimates for q and ∇ q , we combine (36) and (42): | s | Z M | q | dµ + ( n − Z M |∇ q | dµ = (cid:12)(cid:12)(cid:12)(cid:12)Z M (cid:0) ( n − s (cid:1) ( q ) · q dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z M N (cid:0) | h | + | A | + | S | (cid:1) | q | dµ ≤ | s | Z M | q | dµ + N Z M (cid:0) | h | + | A | + | S | (cid:1) dµ .Thus | s | Z M | q | dµ + ( n − Z M (cid:12)(cid:12) ∇ q | dµ ≤ N Z M (cid:0) | h | + | A | + | S | (cid:1) dµ ,and we proved (33) and (34).To find an appropriate bound for |∇∇ q | we must turn to Interior RegularityTheory for Elliptic PDE. From (41) we see that Lq = f is an Elliptic Equation.We then have the following estimate from [7, p. 229]. | q | H ( W ) ≤ K (cid:0) | Lq | L ( M ) + | q | H ( M ) (cid:1) ,where W is any compactly supported open subset of M and K depends onlyupon the coefficients of the operator L , the subset W and the manifold M .Since M is a closed manifold we may in fact choose W = M . Thus we have | q | H ( M ) ≤ K (cid:0) | Lq | L ( M ) + | q | H ( M ) (cid:1) . (43)12pon squaring both sides we observe Z M |∇∇ q | dµ ≤ | q | H ( M ) ≤ K (cid:18)Z M | Lq | dµ + | q | H ( M ) (cid:19) . (44)Now (33) and (34) imply that | q | H ( M ) ≤ N Z M (cid:0) | h | + | A | + | S | (cid:1) dµ . (45)Combining (42), (44) and (45) we have Z M |∇∇ q | dµ ≤ N Z M (cid:0) | h | + | A | + | S | (cid:1) dµ ,and we proved (35). Remark 2.
We observe the following dependencies: N q = N q (cid:0) n , s , | ˜ g − | , | h | , | R | , | ˜ R | , | ˜ ∇ ˜ p | , | ˜ ∇ ˜ ∇ ˜ p | (cid:1) ˆ N q = ˆ N q (cid:0) n , s , | ˜ g − | , | h | , | R | , | ˜ R | , | ˜ ∇ ˜ p | , | ˜ ∇ ˜ ∇ ˜ p | , K (cid:1) where K is from (43). Now we shall approximate the energy E ( t ) = Z M (cid:0) | h | + | A | + | S | (cid:1) dµ . (46)We also define the following: H ( t ) = Z M | h | dµ (47) A ( t ) = Z M | A | dµ (48) S ( t ) = Z M | S | dµ (49) D ( t ) = Z M |∇ S | dµ (50)13ote that E ( t ) = H ( t ) + A ( t ) + S ( t ). We now estimate the evolution ofthe energy functional under Conformal Ricci Flow, E ′ ( t ), by first estimating theevolutions of H , A and S . H ( t ) In [6], Lu, Qing and Zheng give the evolution of the volume element underConformal Ricci Flow: ∂∂t dµ g ( t ) = − np ( t ) dµ g ( t ) (51)Hence by (13) and (47) we have H ′ ( t ) ≤ N Z M | h | dµ + Z M (cid:28) ∂h∂t , h (cid:29) dµ ≤ N H ( t ) + Z M | h | (cid:12)(cid:12)(cid:12)(cid:12) ∂h∂t (cid:12)(cid:12)(cid:12)(cid:12) dµ ≤ N H ( t ) + N Z M (cid:0) | S || h | + | h | + | q || h | (cid:1) dµ .Now we know that N (cid:0) | S || h | + | q || h | (cid:1) ≤ N (cid:0) | h | + | S | + | q | (cid:1) . Hence H ′ ( t ) ≤ N H ( t ) + N Z M (cid:0) | S | + | q | (cid:1) dµ ≤ N H ( t ) + N Z M (cid:0) | S | + | h | + | A | (cid:1) dµ ≤ N H ( t ) + N S ( t ) + N A ( t ) = N E ( t ). (52) A ( t ) By (14), (48) and (51) we have A ′ ( t ) ≤ N A ( t ) + Z M | A | (cid:12)(cid:12)(cid:12)(cid:12) ∂A∂t (cid:12)(cid:12)(cid:12)(cid:12) dµ ≤ N A ( t ) + Z M (cid:16) N | h || A | + N | A | + C |∇ S || A | + C |∇ q || A | (cid:17) dµ .Now N | h || A | + C |∇ S || A | + C |∇ q || A | ≤ N | h | + N | A | + |∇ S | + |∇ q | .14ence we have that A ′ ( t ) ≤ N A ( t ) + Z M (cid:16) N | h | + N | A | + |∇ S | + |∇ q | (cid:17) dµ ≤ N A ( t ) + N H ( t ) + D ( t ) + N Z M (cid:0) | h | + | A | + | S | (cid:1) dµ ≤ N A ( t ) + N H ( t ) + N S ( t ) + D ( t ) = N E ( t ) + D ( t ). (53) S ( t ) By (15), (49) and (51) we have S ′ ( t ) ≤ N Z M | S | dµ + Z M (cid:28) ∂S∂t , S (cid:29) dµ ≤ N S ( t ) + Z M (cid:16) (cid:10) ∆ S + div V , S (cid:11) + N (cid:0) | h | + | A | + | S | + | q | (cid:1) | S | + C |∇∇ q || S | (cid:17) dµ ≤ N S ( t ) + Z M (cid:16) (cid:10) ∆ S + div V , S (cid:11) + N (cid:0) | h | + | A | + | S | + | q | + |∇∇ q | (cid:1)(cid:17) dµ .Now by (33) and (35) we have S ′ ( t ) ≤ N S ( t ) + N H ( t ) + N A ( t )+ Z M (cid:18) (cid:10) ∆ S + div V , S (cid:11) + N (cid:0) | A | + | S | + | h | (cid:1)(cid:19) dµ ≤ N S ( t ) + N H ( t ) + N A ( t ) + Z M (cid:10) ∆ S + div V , S (cid:11) dµ .Upon integrating by parts we get S ′ ( t ) ≤ N E ( t ) − Z M (cid:10) ∇ S + V , ∇ S (cid:11) dµ ≤ N E ( t ) − Z M |∇ S | dµ + Z M | V ||∇ S | dµ .Now we know that2 | V ||∇ S | ≤ |∇ S | + | V | ≤ |∇ S | + N (cid:0) | h | + | A | (cid:1) ,15ence S ′ ( t ) ≤ N E ( t ) + N Z M (cid:0) | h | + | A | (cid:1) dµ − Z M |∇ S | dµ ≤ N E ( t ) − D ( t ). (54) Now we are ready to prove Theorem 1:
Proof.
By (54), (52) and (53) we know that H ′ ( t ) ≤ N E ( t ), A ′ ( t ) ≤ N E ( t ) + D ( t ), S ′ ( t ) ≤ N E ( t ) − D ( t ),so E ′ ( t ) ≤ N E ( t ).Our initial condition ˜ g (0) = g (0) tells us that at t = 0 we have | h | = | A | = | S | = 0. Therefore by the smoothness and integrability of our solutions we knowlim t → + E ( t ) = 0,so by Gronwall’s Inequality we know that E ≡ T ]. Thus for t ∈ [0, T ] wehave that h ≡ g ( t ) ≡ ˜ g ( t ). Also, E ≡ A ≡ S ≡
0, so (33)forces q ≡
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