Convergence Analysis of the Energy and Helicity Preserving Scheme for Axisymmetric Flows

Transcription

1 Convegence Analysis of the Enegy Helicity Peseving Scheme fo Axisymmetic Flows Jian-Guo Liu Institute fo Physical Science Technology Depatment of Mathematics Univesity of Mayl College Pak, MD USA Wei-Cheng Wang 2 Depatment of Mathematics National Tsing Hua Univesity HsinChu, Taiwan 300 TAIWAN Running head: Analysis of Enegy Helicity Peseving Scheme Key wods: Enegy, Helicity, Jacobian, Pemutation Identity, Axisymmetic Flow, Navie Stokes, Convegence Analysis, Eo Estimate, Pole Singulaity. AMS subject classifications: 65N06, 65N35, 35J05. jliu@math.umd.edu 2 wangwc@math.nthu.edu.tw

2 Abstact We give an eo estimate fo the Enegy Helicity Peseving Scheme EHPS in second ode finite diffeence setting on axisymmetic incompessible flows with swiling velocity. With caeful detailed tuncation eo analysis nea the geometic singulaity fa field decay estimate fo the steam function, we have achieved optimal eo bound using weighted enegy estimate. A key ingedient in ou a pioi estimate is the pemutation identity associated with the Jacobians, which is also a unique featue that distinguishes EHPS fom stad finite diffeence schemes. Intoduction Axisymmetic flow is an impotant subject in fluid dynamics has become stad textbook mateials e.g. [2] as a stating point of theoetical study fo complicated flow pattens. Although the numbe of independent spatial vaiables is educed by symmety, some of the essential featue complexity of geneic 3D flows emains. Fo example, when the swiling velocity is nonzeo, thee is a voticity stetching tem pesent. This is widely believed to account fo possible singulaity fomation fo Navie-Stokes Eule flows. Fo geneal smooth initial data, it is well known that the solution emains smooth fo shot time in Eule [3] Navie-Stokes flows [4]. A fundamental egulaity esult concening the solution of the Navie-Stokes equation is given in the pioneeing wok of Caffaelli, Kohn Nienbeg [4]: The one dimensional Hausdoff measue of the singula set is zeo. As a consequence, the only possible singulaity fo axisymmetic Navie-Stokes flows would be on the axis of otation. This esult has motivated subsequent eseach activities concening the egulaity of axisymmetic solutions of the Navie-Stokes equation. Some egulaity patial egulaity esults fo axisymmetic Eule Navie-Stokes flows can be found in, fo example, [6] the efeences theein. To date, the egulaity of the Navie-Stokes Eule flows, whethe axisymmetic o not, emains a challenging open poblem. Fo a compehensive eview on the egulaity of the Navie-Stokes equation, see [5] the efeences theein. Due to the subtle egulaity issue, the numeical simulation of axisymmetic flows is also a challenging subject fo computational fluid dynamicists. The ealiest attempt of 2

3 numeical seach fo potential singulaities of axisymmetic flows dates back to the 90 s [9, 0]. In a ecent wok [7], the authos have developed a class of Enegy Helicity Peseving Schemes EHPS fo incompessible Navie-Stokes MHD equations. Thee the authos extended the voticity-steam fomulation of axisymmetic flows given in [9] poposed a genealized voticity-steam fomulation fo 3D Navie-Stokes MHD flows with coodinate symmety. In the case of axisymmetic flows, the main diffeence between EHPS the fomulation in [9] is the expession numeical discetization of the nonlinea tems. It is shown in [7] that all the nonlinea tems in the Navie-Stokes MHD equation, including convection, voticity stetching, geometic souce, Loentz foce electo-motive foce, can be witten as Jacobians. Associated with the Jacobians is a set of pemutation identities which leads natually to the consevation laws fo fist second moments. The main featue of the EHPS schemes is the numeical ealization of these consevation laws. In addition to peseving physically elevant quantities, the discete fom of consevation laws povides numeical advantages as well. In paticula, the consevation of enegy automatically enfoces nonlinea stability of EHPS. A potential difficulty associated with axisymmetic flows is the appeaance of facto which becomes infinite at the axis of otation, theefoe sensitive to inconsistent o low ode numeical teatment nea this pole singulaity. In [7], the authos poposed a second ode finite diffeence scheme hled the pole singulaity by shifting the gids half gid length away fom the oigin. Remakably, the pemutation identities theefoe the enegy helicity identities emain valid in this case. Thee ae altenative numeical teatments poposed in liteatues e.g. [0] to hle this coodinate singulaity. Howeve, igoous justifications fo vaious pole conditions ae yet to be established. The pupose of this pape is to give a igoous eo estimate of EHPS fo axisymmetic flows. To focus on the pole singulaity avoid complication caused by physical bounday conditions, we conside hee only the whole space poblem with the swiling components of velocity voticity decaying fast enough at infinity. The eo analysis of numeical methods fo NSE with nonslip physical bounday condition has been well studied. We efe the woks of Hou Wetten [], Liu Wang [24] to inteested eades. Ou poof is based on a weighted enegy estimate togethe with a caeful detailed pointwise local tuncation eo analysis. A majo ingedient in ou enegy estimate is the pemutation 3

4 identities associated with the Jacobians 4.7. These identities ae key to the enegy helicity peseving popety of EHPS fo geneal symmetic flows. Hee the same identities enable us to obtain a pioi estimate even in the pesence of the pole singulaity, see section 5 fo details. To ou knowledge, this is the fist igoous convegence poof fo finite diffeence schemes devised fo axisymmetic flows. In ou pointwise local tuncation eo estimate, a fundamental issue is the identification of smooth flows in the vicinity of the pole. Using a symmety agument, we show that if the swiling component is even in o moe pecisely, is the estiction of an even function on > 0, the vecto field is in fact singula. See Example in section 2 fo details. This is an easily ovelooked mistake that even appeaed in some eseach papes tageted at numeical seach of potential fomation of finite time singulaities. In addition to the egulaity issue at the axis of symmety, a efined decay estimate fo the steam function also plays an impotant ole in ou analysis. In geneal, the steam function only decays as Ox at infinity. Accodingly, we have selected an appopiate combination of weight functions that constitute an -homogeneous nom. As a esult, the slow decay of the steam function is compensated by the fast decay of velocity voticity. Oveall, we obtain a second ode optimal eo estimate on axisymmetic flows. The est of this pape is oganized as follows: In section 2, we intoduce a pope notion of smoothness fo the swiling component of an axisymmetic divegence fee vecto field. We show that the -deivatives of even ode must vanish on = 0 +. The same holds tue fo vecto fields in Sobolev spaces. In section 3, we show that the genealized voticity-steam fomulation is equivalent to the oiginal Navie-Stokes equation in pimitive vaiable. Thee is no exta egulaity equiement upon switching to the voticity fomulation. In section 4, we ecall the enegy helicity peseving popety fo EHPS use it to pove ou main theoem by enegy estimate in section 5. The technical details of the pointwise estimate fo the local tuncation eo is given in the Appendix. 4

5 2 Classical Sobolev Spaces fo Axisymmetic Solenoidal Vecto Fields In this section, we establish basic egulaity esults fo axisymmetic vecto fields. We will show that the swiling component of a smooth axisymmetic vecto field has vanishing even ode deivatives in the adial diection at the axis of otation. This is done in Lemma by a symmety agument. Thoughout this pape, we will be using the cylindical coodinate system whee the x-axis is the axis of otation. x = x, y = cos θ, z = sin θ. 2. A vecto field u is said to be axisymmetic if θ u x = θ u = θ u θ = 0. Hee thoughout this pape, the subscipts of u ae used to denote components athe than patial deivatives. The thee basic diffeential opeatos in cylindical coodinate system ae given by u = x ue x + ue + θue θ 2.2 u = xu x + u + θ u θ 2.3 u = e x e e θ x θ 2.4 u x u u θ Hee e x, e e θ ae the unit vectos in the x, θ diections espectively. Denote by C k s the axisymmetic divegence fee subspace of C k vecto fields: Definition : C k s = {u C k R 3, R 3, θ u x = θ u = θ u θ = 0, u = 0} 2.5 We have the following epesentation egulaity esult fo C k s : Lemma with a Fo any u C k s, thee exists a unique u, ψ such that u = ue θ + ψe θ = ψ e x x ψe + ue θ, > 0, 2.6 ux, C k R R +, 2l ux, 0 + = 0 fo 0 2l k, 2.7 5

6 ψx, C k+ R R +, 2l ψx, 0 + = 0 fo 0 2l k b If u, ψ satisfies 2.7, 2.8 u is given by 2.6 fo > 0, then u C k s emovable singulaity at = 0. with a Poof: Pat a: Since u is axisymmetic, we can wite u = u x x, e x + u x, e + u θ x, e θ fo > 0. Note that e z x, y, z z=0 = { eθ x, y, z z=0 if y > 0 e θ x, y, z z=0 if y < Hee e z is the unit vecto in the z diection. With slight abuse of notation, we denote the components of u in Catesian cylindical coodinates by u x x, y, z = u x x, u y x, y, z = u x, cos θ u θ x, sin θ u z x, y, z = u x, sin θ + u θ x, cos θ 2.0 whee, θ is given by 2.. It follows that u z x, y, z z=0 = { uz x,, θ =y,θ=0 = u θ x, y if y > 0 u z x,, θ = y,θ=π = u θ x, y if y < 0 2. theefoe fo y > 0, u z x, y, z z=0 = u θ x, y = u θ x, y = u z x, y, z z= Since u z C k R 3 u z x, y, 0 is odd in y fom 2.2, it follows that u θ has a C k extension up to = 0 ux, := all the even -deivatives vanish at = 0 + : { uθ x, if > 0 0 if = 0 = u zx, y, 0 y=, 0 lim 0 + 2l ux, = lim y 0 + 2l y u z x, y, 0 = 2l y u z x, 0, 0 = 0, 0 2l k Hence 2.7 follows. Next we deive the epesentation 2.6. Since u is divegence fee, 2.3 gives x u x + u = 0, 6

7 we know fom stad agument that thee exists a potential φx, such that x φ = u, φ = u x 2.3 Resticting 2.0 to θ = 0, o equivalently to z = 0, y > 0, we have u x x, = u x x, y, z y=,z=0, u x, = u y x, y, z y=,z=0, u θ x, = u z x, y, z y=,z=0 2.4 Fom , it is clea that φx, C k+ R R +. Since x φx, 0 + = 0, we may, without loss of geneality, assume that φx, 0 + = 0. This also detemines ψ uniquely. Next we define ψx, = φx,, > It is easy to see that ψx, C k+ R R +, ψx, 0 + = φx, 0 + = follows fo > 0. It emains to show that lim 0 + j ψx, is finite fo j k +. To this end, we fist establish the following identity by induction: Claim: j + j ψx, 0 + = j j u x x, 0 +, j k Poof of Claim: Fom 2.3 it follows that When j =, 2.7 gives we conclude fom l Hospital s ule that thus 2.6 is veified fo j =. Fo j >, 2.7 gives j ψ j u x + j ψ = j u x, > ψ + ψ = u x 2 ψx, 0 + = u x x, 0 +, j j ψ j j 2 u x =

8 Now suppose 2.6 is validated fo j = l, l l ψx, 0 + l l 2 u x x, 0 + = 0, 2.9 fom l Hospital s ule, we can easily deive ψx, l 0 + l u x x, 0 + = l ψx, l 0 + l l u x x, 0 + thus 2.6 is validated fo j = l inductively fo all j k +. This completes the poof of the claim. We conclude fom that ψx, C k+ R R +. Moeove, following the same agument outlined in , it is easy to show that u x x, y, 0 is even in y. Thus 2.8 follows fom 2.6. This completes the poof of pat a. Pat b: Convesely, we now show the egulaity of u = ue θ + ψe θ when u, ψ satisfies Since u is axisymmetic, it suffices to check the deivatives of u on a coss section, say θ = 0, o z = 0, y 0. It is clea fom that u x x, y, 0, u y x, y, 0 u z x, y, 0 have continuous x deivatives up to ode k on y 0. It emains to estimate the y-, z- mixed deivatives. Fom we can deive the following y = cos θ sin θ θ 2.20 z = sin θ + cos θ θ 2.2 Poposition i j yf x,, θ = j F x,, θ + sin θ Gx,, θ 2.22 ii whee G consists of the deivatives of F. z 2m fx, cos θ = y z 2m+ fx, cos θ = y m l+m f a l,m z 2l l=0 m l+m+ f b l,m z 2l+ l=

9 2m z gx, sin θ = 2m z gx, sin θ = m l+m g c l,m z 2l+ l=0 m l+m g d l,m z 2l l=0 fo some constants a l,m, b l,m, c l,m d l,m Poof: Fom 2.20 the following identity cos θ sin θ θf + sin θg = cos θ F + sin θcos θ G θf sin θ θg, 2.27 it is easy to deive This poves pat i. Fo pat ii, equations esult fom substituting cos θ = y, sin θ = z followed by staight fowad calculations. We omit the details. Now we poceed to show that all the mixed deivatives of odes up to k ae also continuous on y 0. Fo simplicity of pesentation, we conside mixed deivatives pefomed in the following ode j y q z i x. We stat with j y q z i xu x analyze fo q even odd sepaately. When q = 2m +, we deive fom that j y 2m+ z i xu x x, y, 0 = y j z 2m sin θ xu i x x, θ=0,=y = j y m l=0 c l,mz 2l+ l+m i xu xx, z=0,=y 2.28 = 0 Next, when q = 2m, it follows fom 2.2, 2.22, 2.26, that j y 2m z i xu x x, y, 0 = y j z 2m sin θ xu i x x, θ=0,=y = j z 2m sin θ xu i x + sin θ G θ=0,=y = j m l=0 d l,m sin θ 2l l+m xu i xx, θ=0,=y 2.29 = d 0,m j = d 0,m j m i x u x x, =y m+ i x ψx, =y. 9

10 Fom Lemma Taylo s Theoem, we have ψx, = a x + a 3 x a 2m x 2m + R 2m+ ψ, whee a n x = n! n ψx, 0 +, R 2m+ ψ = { p R 2m+ ψx, 0 + = It follows fom diect calculation that 0 2m+ s2m ψx, s ds, 2m! m+ i xψx, = m+ i xr 2m+ ψ = m+ m+ j xψx, i = C l,m j l xr i 2m+ ψ = 2m+ l l=0 fo some constants C l,m C p,j. j Fom 2.30, 2.3 l Hospital s ule we conclude that m+ j m+ xψx, i 0 + = l=0 0, 0 p 2m p ψx, 0 +, 2m + p 2m + + j p=0 m+ l=0 m+ l=0 C l,m C p,j 2m l + j p! C l,m l i xr 2m+ ψ 2m+ l j C l,m C p,j p=0 p+l xr i 2m+ ψ 2m+ l+j p 2.3 2m++j i xψx, Since ψ C k+ R R +, it follows fom 2.29, that j y q z i xu x x, y, 0 is continuous bounded up to y = 0 + fo j + q + i k. Next we conside the mixed deivatives of u y u z. In view of 2.0, it suffices to calculate j y q z i xfx, cos θ + gx, sin θ θ=0,=y whee f g ae eithe ± x ψ o ±u. When q = 2m, it follows fom that y j z 2m xfx, i cos θ + gx, sin θ θ=0,=y = y j z 2m xfx, i cos θ + xgx, i sin θ θ=0,=y = a 0,m j m i xf =y + c 0,m j m i xg =y 0

11 Fom , both x ψx, ux, have local expansions of the fom b x + b 3 x b 2m x 2m + R 2m+. Following the same agument above, we can show that both j y 2m z xu i y y j 2m xu i z ae continuous bounded up to y = 0 + fo j + 2m + i k. The calculations fo y j z 2m+ xu i y j y 2m+ z i xu z ae simila. This completes the poof of Pat b. In view of Lemma, we now intoduce the following function spaces: Definition 2 z C k s R R + = {fx, C k R R +, 2j fx, 0 + = 0, 0 2j k} We can ecast Lemma as Lemma C k s = {ue θ + ψe θ u C k s R R +, ψ C k+ s R R + } 2.33 In fact, we will show a countepat fo 2.33 in stad Sobolev spaces: A weak solenoidal axisymmetic vecto field admits the epesentation 2.6 with ux, ψx, in cetain weighted L 2 H k spaces. Moeove, both u ψ, togethe with cetain even ode deivatives have vanishing taces on = 0 +. We poceed with the following identity fo geneal solenoidal vecto fields: Lemma 2 If u C k R 3, R 3 H k R 3, R 3 u = 0, then u 2 H k R 3,R 3 = k l u 2 L 2 R 3,R l=0 Poof: We pove 2.34 fo l even odd sepaately. Since u = 0, it follows that u = 2 u. Thus if l is even, we can wite 2m u L 2 R 3,R 3 = 2 m u L 2 R 3,R When m = u C k R 3, we can integate by pat to get 2 u 2 = R 3 R 3 i = 3 i 2 u 2 = 3 3 i 2 u i 2 2 u = i i2 u 2 R 3 i,i 2 = R 3 i,i 2 =

12 Similaly, when m = 2, 2 2 u 2 = R i 2 u 2 = i 2 i 2 2 u i 2 3 i 2 4 u R 3 i= R 3 i,i 2,i 3,i 4 = 3 = i,i 2,i 3,i 4 = It is theefoe easy to see that 3 2 m u 2 = R 3 R 3 i i2 i3 i4 u 2. i,,i 2m = R 3 i i2m u 2 consequently fo u C k R 3, R 3, 2m k, 3 2 m u 2 L 2 R 3,R 3 = i i2m u 2 L 2 R 3,R i,,i 2m = On the othe h, if l is odd, we fist wite 2m+ u = 2 m u = m 2 m u then apply the identity v 2 L 2 R 3,R 3 = v 2 L 2 R 3,R 3 + v 2 L 2 R 3 to get 2m+ u L 2 R 3,R 3 = 2 m u L 2 R 3,R 3 = 2 m u L 2 R 3,R fom 2.36, m u 2 L 2 R 3,R 3 = 2 m i u j 2 L 2 R 3 = i i2m i u j 2 R 3 i,j= i,j= i,,i 2m = 3 = i i2m+ u 2 L 2 R 3,R 3. i,,i 2m+ =

13 It follows fom 2.35, 2.36, that u 2 H k R 3,R 3 = k 3 i il u 2 L 2 R 3,R 3 = l=0 i,,i l = k l u 2 L 2 R 3,R 3 l=0 This completes the poof of Lemma 2. In Lemma 3 Lemma 4 below, we will deive an equivalent epesentation of the Sobolev noms fo axisymmetic solenoidal vecto fields. Lemma 3 Let u C k s ψ C k+ s be epesented by u = ue θ + ψe θ with u C k s R R +. Then l u C k l s R R + 2m u = L m ue θ + L m ψe θ, if 2m k, 2m+ u = L m+ ψe θ + L m ue θ, if 2m + k, whee Moeove L := = x + 2. L m u C k 2m s R R +, if 2m k, L m+ ψ C k 2m s R R +, if 2m + k. Poof: Fo any φ C i s R R +, we have φ e θ C i s fom Lemma b. With staight fowad calculation using 2.4, it is easy to veify that fo i 2, On the othe h, it is clea that theefoe fom Lemma a, φ e θ = Lφe θ φ e θ C i 2 s, Lφ C s i 2 R R The Lemma then follows fom

14 Next we poceed to define the weighted Sobolev space fo axisymmetic solenoidal vecto fields. Fo a, b C 0 R R +, we define the weighted L 2 inne poduct nom a, b = 0 ax, bx, dxd, a 2 0 = a, a, 2.4 fo a, b C s R R +, we define the weighted H inne poduct nom [a, b] = x a, x b + a, b + a, b, a 2 = [a, a] When a C s R R + b C s R R + C 2 R R +, we also have the following identity fom integation by pat: a, Lb = [a, b]. If u = ue θ + ψe θ, with u C 0 R R + ψ C s R R +, it is easy to see that u 2 L 2 R 3,R 3 = u ψ Highe ode Sobolev noms can be defined similaly in tems of u ψ: Definition 3 Fo u = ue θ + ψe θ Cs k, u 2 H k s := u ψ 2 + Lψ u 2 + Lu Lψ 2 + = L m u L m ψ 2 + L m+ ψ L m u 2 2m k 2m+ k In view of Lemma, Lemma 2, Lemma , we have poved the following Lemma 4 If u Cs k, then u H k R 3,R 3 = u H k s Denote by C c the space of compactly suppoted functions. We can now define the Sobolev spaces fo axisymmetic solenoidal vecto fields following stad pocedue: Definition 4 L 2 sr R + := Completion of C 0 s R R + C c R R + with espect to 0 H s R R + := Completion of C s R R + C c R R + with espect to H k s := Completion of C k s C c R 3, R 3 with espect to H k s 4

15 Accodingly, we have the following chaacteization fo Hs: k Lemma 5 If u Hs, k then u admits a unique epesentation u = ue θ + ψe θ 2.44 with u L 2 sr R + ψ Hs R R +. Moeove, L m u L 2 s R R +, L m ψ H s R R +, if 2m k, 2.45 L m+ ψ L 2 s R R +, L m u H s R R +, if 2m + k, 2.46 u H k R 3,R 3 = u H k s Hee the equality in 2.44 the diffeential opeatos, L m ae ealized in the sense of distibution. The poof of Lemma 5 is based on stad density agument. We omit the details. Finally, the countepat of fo u H k s is given by 2.45, 2.46 the following tace Lemma: Lemma 6 If v H s R R +, then the tace of v on = 0 vanishes. Poof: Fo any v C R R + C c R R +, we have v vx, dx = 2 v v dx d R R R + R R + v 2 dx d v Since vx, 0 = 0 fo v C s R R +, the Lemma follows fom stad density agument. Example : Take u = ue θ with u = e x2 2 e. Note that u = O 2 nea the axis. Simila functions can be found in liteatues as initial data in numeical seach fo finite time singulaities. Although u C R R + u may appea to be a smooth vecto field, it is easy to veify that Lux, Thus fom Lemma, Lemma 5 Lemma 6, u is neithe in C 2 R 3, R 3 no in H 3 R 3, R 3. 5

16 3 Genealized Voticity-Steam Fomulation fo Axisymmetic Flows 3. Axisymmetic Fomulation of Navie-Stokes Equations In this section, we show that, unde suitable egulaity assumptions, any axisymmetic solution of the Navie-Stokes equation t u + u u + p = ν u u = 0 3. is also a solution to the following axisymmetic fomulation of Navie-Stokes solution deived fomally in [7] u t + 2 J u, ψ = ν 2 2 u, ω t + J ω, ψ = ν 2 2 ω + J u, u, 3.2 ω = 2 2 ψ, with u = ue θ + ψe θ vice vesa. The voticity fomulation fo axisymmetic flows 3.2 has appeaed in [9] with an altenative expession fo the nonlinea tems. In [7], the authos have genealized the voticity fomulation to geneal symmetic flows with the nonlinea tems ecast in Jacobians as in 3.2. Accompanied with the Jacobians is a set of pemutation identities which played a key ole in the design of enegy helicity peseving scheme in the convegence poof of the scheme. We will explain the details in section 5. The axisymmetic Navie-Stokes equation 3.2 can be fomally deived fom 3.. A smooth solution of 3. also gives ise to a smooth solution of 3.2. Moeove, fom Lemma a, the esulted solution satisfies the pole condition 2.7,2.8 automatically. In othe wods, if u C 0, T ; Cs k is a solution to 3., then the swiling components ae in the class ψt; x, C 0, T ; Cs k+ R R + ut; x, C 0, T ; Cs k R R + ωt; x, C 0, T ; Cs k R R Howeve, it is not clea whethe smooth solutions of 3.2 also give ise to smooth solutions of 3.. 6

17 Fo example, in the case of Eule equation, an exact stationay solution to 3.2 with ν = 0 is given by ut, x, = 2 e, ω = ψ It is clea that 3.4 is in C R R +, yet the coesponding u = ue θ is only in C R 3, R 3 fo any t. In the case of Navie-Stokes equation ν > 0, 3.2 is an elliptic-paabolic system on a semi-bounded egion > 0. Fom stad PDE theoy, we need to assign bounday values fo ψ, u, ω. The zeoth ode pat of the pole condition 2.7,2.8 would suffice: ψx, 0 = ux, 0 = ωx, 0 = It is theefoe a natual question to ask if a smooth solution of 3.2, 3.5 in the class ψt; x, C 0, T ; C k+ R R + ut; x, C 0, T ; C k R R + ωt; x, C 0, T ; C k R R will give ise to a smooth solution of 3.2. In othe wods, is the pole condition 2.7,2.8 automatically satisfied if only the zeoth ode pat 3.5 is imposed? The answe to this question is affimative. We will show in Lemma 9 that ae indeed equivalent fo solutions of 3.2, 3.5. The poof is based on local Taylo expansion. We decompose the poof into seveal Lemmas. Lemma 7 If v C 2j+2 R R + 2l vx, 0 + = 0 fo 0 l j, then Poof: Since v C 2j+2 R R +, we have vx, lim 2j+ 0 + = lim 0 + 2j + 2 2j+2 vx, vx, = a x + a 3 x a 2j+ x 2j+ + R 2j+2 v 3.7 fom Taylo s Theoem. Hee R 2j+2 v = a n x = n! n vx, 0 +, 0 2j+2 s2j+ vx, s 2j +! ds 7

18 i R 2j+2 vx, 0 + = 0, 0 i 2j +, 2j+2 R 2j+2 vx, 0 + = 2j+2 vx, Fom 3.7, it follows that 2j+ vx, = 2j+ R 2j+ 2j+2v = Fom 3.8, 3.9 l Hospital s ule, we can easily deive lim 2j j+ vx, = i=0 i=0 C2j+ i i i! 2j+ i R 2j+2 v 3.9 +i C2j+ i i 2j+2 vx, 0 + = i + 2j + 2 2j+2 vx, 0 +. This completes the poof of Lemma 7. Lemma 8 If 2j k 2 ψ C k+ R R + Cs 2j R R + u C k R R + Cs 2j R R + ω C k R R + Cs 2j R R +, 3.0 then the Jacobians Ju, ψ, J ω, ψ 2 Ju, u ae in C2j s R R +. Poof: Fom 3.0, it is obvious that 2 Ju, ψ C k R R +, J ω, ψ Ck 2 R R + J u, u Ck R R +. Theefoe all thee Jacobians ae in C 2j R R +. It emains to evaluate the -deivatives of the Jacobians at x, 0 +. Fom 3.0, we have the following expansions: ψ = a x + + a 2j x 2j + R 2j+ ψ u = b x + + b 2j x 2j + R 2j+ u ω = c x + + c 2j x 2j + R 2j+ ω whee a n x, b n x, c n x = n! n ψ, u, ωx, 0 + R 2j+ v = 0 2j+ s2j vx, s ds. 2j! Befoe we continue, we fist intoduce the following notations fo bevity: 8

19 Definition 5 Fo 0 q 2j +, we define. fx, = Ôq if fx, = Cx q. 2. gx, = Õq if gx, C 2j+ R R + p gx, 0 + = 0, 0 p q, gx, q 0 + <. Using these notations, we can wite R 2j+ ψ, u, ω = Õ2j+ ψ, u, ω = Ô + Ô3 + + Ô2j + Õ2j+. 3. It is easy to veify diectly that p Õq = Õq p, p q 2j +, Õ q / p = Õq p, p q 2j +. Fom 3., we have Theefoe xψ, ψ, xu = Ô + Ô3 + + Ô2j + x Õ 2j+ u = Ô + Ô2 + + Ô2j 2 + Õ2j. Jψ, u = Ô + 2 Ô3 + + Ô2j +Õ2j Ô + Ô Ô2j + x Õ 2j+ Ô + Ô Ô2j 2 +Ô2j+3 + Ô2j Ô4j It is easy to see fom 3.2 that 2i 2 Ju, ψx, 0+ = 0, 0 i j J u, ψ C 2j 2 s R R +. The same agument applies to J ω, ψ J u, u. This completes the poof of Lemma 8. 9

20 Lemma 9 If ψ, u, ω is a solution to 3.2, 3.5 in the class 3.6 with k 3. Then ψ C k+ s R R + u C k s R R + ω C k s R R fo 0 t T. Poof: Let j be the lagest intege such that 2j k. We fist show that on 0 t T, 2l 2l 2l ψt, x, 0 + = 0 ut, x, 0 + = 0 ωt, x, 0 + = fo 0 l j. This is done by induction on l. When l = 0, 3.4 is given by the bounday condition 3.5. Suppose that 3.4 is veified fo l = j with j + j. We apply 2j 2 x,0 + on both sides of 3.2 conclude that, in view of Lemma 8, Since ν 2j 2 ν 2j 2 2j 2 ux, 0 + = 0, 2 ωx, 0 + = 0, 2 ψx, 0 + = v = 2 xv + 2 v + v, 2j 2 v = 2j+2 2 v + 2j 2 xv + 2j+ v, it follows fom Lemma 7 that 2j+2 ψx, 0 + = 2j+2 ux, 0 + = 2j+2 ωx, 0 + = 0 thus 3.4 is veified fo l = j +. We can continue the induction until 3.4 is veified fo l = j to get ψ C k+ R R + Cs 2j R R + u C k R R + Cs 2j R R + ω C k R R + Cs 2j R R To complete the poof, we poceed with k odd even sepaately. If k is odd, say k = 2m +, then j = m 3.5 can be witten as ψ C 2m+2 R R + C 2m s R R +, u C 2m+ s R R +, ω C 2m s R R

21 Since 2m 2 ψ = 2m+2 2 ψ + 2m xψ 2 + 2m+ ψ = 2m ω, 3.7 we conclude fom Lemma 7, that 2m+2 ψx, 0 = 0, theefoe ψ Cs 2m+2 R R +. Similaly, if k = 2n, then j = n we have fom 3.5 ψ C 2n+ R R + C 2n 2 s R R +, u C 2n R R + C 2n 2 s R R +, ω C 2n s R R +. Since 2n 2 = k 2, the assumption in Lemma 8 is satisfied. Theefoe we can continue the induction fo u to get 2n ux, 0 + = 0, thus u Cs 2n R R +. Finally, 2n 2 2 ψ = 2n 2 ψ + 2n 2 xψ 2 + 2n ψ = 2n 2 ω we conclude that 2n ψx, 0 + = 0 ψ C 2n+ R R + Cs 2n R R + = Cs 2n+ R R +. This completes the poof of Lemma 9. The equivalence of in tems of egulaity of classical solutions is given by Theoem I Suppose u, p be an axisymmetic solution to NSE 3. with u C 0, T ; C k s, p C 0 0, T ; C k R 3 k 3. Then thee is a solution ψ, u, ω to 3.2 in the class u = ue θ + ψe θ. ψt; x, C 0, T ; Cs k+ R R + ut; x, C 0, T ; Cs k R R + ωt; x, C 0, T ; Cs k R R + II Let ψ, u, ω be a solution to 3.2,3.5 in the class ψt; x, C 0, T ; C k+ R R + ut; x, C 0, T ; C k R R + ωt; x, C 0, T ; C k R R + with k 3. Then u = ue θ + ψe θ C 0, T ; C k s thee is a axisymmetic scala function p C 0 0, T ; C k R 3 such that u, p is a solution to NSE 3.. 2

22 Poof: Pat I: Since u C 0, T ; C k s is a solution to 3. with k 3, it follows that ω = u = ωe θ + ue θ C 0, T ; Cs k is also an axisymmetic solution to the Navie-Stokes equation in voticity fom: t ω + ω u = ν ω 3.8 Next, we expess each tem of 3.8 in the cylindical coodinate as ω = ω u = Fom , we can ewite 3.8 as t ω = t ωe θ + t ue θ, ω e 2 θ + 2 ue 2 θ, 3.20 J ω, ψ Ju, u e θ + J u, ψ e 2 θ. 3.2 ae θ + be θ = 0, 3.22 whee ω u a = ω t + J, ψ J, u ν 2 ω, 2 b = u t + J u, ψ 2 ν 2 u. 2 Fom 3.22, it follows that ax, = 0 bx, is a constant. Since bx, 0 + = 0 fom Lemma 8 Lemma 9, we conclude that bx, 0 as well. This completes the poof of pat I. Pat II: Fom Lemma 9, we know that ψ, u, ω satisfies 3.3. Theefoe Lemma applies we have u = ue θ + ψe θ C 0, T ; Cs k Next we define ω = u. Fom , we see that ω satisfies the Navie-Stokes equation in voticity fomulation 3.8. That is t u + ω u + ν ω = 0. 22

23 Thus thee exists a function p : 0, T C k R 3 such that t u + ω u + ν ω = p 3.23 In othe wods, u, p satisfies the NSE 3.. Since u C 0, T ; C k s, it follows fom 3.23 that p C 0 0, T ; C k 2 s. In addition, we can futhe assign pt on a efeence point x 0, 0 so that p C 0 0, T ; C k R 3. By constuction, the left h side of 3.23 is axisymmetic theefoe so is p. In paticula Theefoe θ p e θ = θ θp = 0. p = ax, θ + bx, Since p is continuous single-valued, we conclude that a = 0. axisymmetic. This completes the poof of theoem. In othe wods, p is 3.2 Regulaity Assumption on Solutions of NSE The focus of this pape is the convegence ate of EHPS in the pesence of the pole singulaity. To sepaate difficulties avoid complications intoduced by physical boundaies, we only conside the whole space poblem with exact solution decaying fast at infinity. To be moe specific, we conside the initial data ux, 0 ωx, 0 to be smooth with compact suppots. Since 3.2 is a tanspot diffusion equation fo u ω with initially finite speed of popagation, we expect u ω to be essentially compactly suppoted, at least fo shot time. Fo linea tanspot diffusion equations with initial data smooth compactly suppoted, the solution togethe with its deivatives will decay faste than polynomials at infinity fo t > 0. Some igoous esults concening the spatial decay ate fo the solutions of axisymmetic flows can be found in [6] the efeence theein. In paticula, it is shown in [6] that both u ω decay algebaically at infinity as long as this is the case initially. Hee we make a stonge, yet plausible assumption along this diection. The pecise fom of ou assumption is fomulated in tems of weighted noms is less stingent than the analogy we daw fom linea tanspot diffusion equations, see Assumption below. 23

24 To quantify ou assumption, we fist intoduce a family of -homogeneous composite noms coesponding function spaces which tun out to be natual fo ou pointwise enegy estimate: Definition 6 a l,α,β = C k,α,β s l +l 2 =l + α + x β l x l 2 u L R R + a k,α,β = = {ax, C k s 0 l k a k l,α l,β R R +, a k,α,β < } In section 5, we will show that EHPS is second ode accuate povided the solution satisfies { t ψ, ψ, ω C 0 0, T ; C 4,α+ 7 2,β s u C 0 0, T ; C 4,2α+2,2β s Cs 4,2α+2,2β, α > 2, β > Cs 4,2α+2,2β Cs,2,0 In view of 3.24, we fomulate ou egulaity assumption as Assumption t ψ, ψ, ω C 0 0, T ; C 4,γ,δ s, u C 0 0, T ; Cs 4,5,δ, γ > 4, δ > Although we expect u, ω thei deivatives to decay faste than any polynomial at infinity, the same expectation is not pactical fo ψ. As will be shown below, geneically ψ only decays like Ox at infinity. This is elated to the decay ate of the fundamental solution of Poisson equation in 3D the vanishing zeoth moments of y, z components of voticity. To see this, we stat with the integal expession fo ψ. Fom the voticity-steam elation the identification ψ = ω ψx, = ψ y x, y, 0 y=, ωx, = ω y x, y, 0 y=, we can deive the following integal fomula fo ψ [20]: ψx, = 0 ωx, Kx x,, dx d

25 whee Kx x,, = 2π 4π 0 = 2 2 π π 2 0 cos θ x x dθ 2 + cos θ 2 + cos θ 2 cos 2 θ dθ 3.27 ρ + ρ ρ + +ρ ρ 2 ± = x x 2 + ± cos θ 2 + cos θ 2 As a consequence, we have the following fa field estimate fo K: Lemma 0 l x m Kx x,, C l,m x, x l m as x Poof: We will deive a fa field estimate fo the integ in We fist conside a typical tem with whee x 0, 0 c 0 ae some constants. With the change of vaiables we can ewite the x deivatives by lim x l x m ρ ρ 2 = x x c = σ cos λ x x 0 = σ sin λ ρ = σ2 + c 2 0 = σ σ σ2 + c λ λ σ2 + c 2 0 = cos λ σ ρ x ρ = x σ2 + c 2 0 = x σ σ σ2 + c x λ λ σ2 + c 2 0 = sin λ σ ρ Theefoe by induction, l x m ρ = P l,m cos λ, sin λq l,m σ, ρ whee P l,m cos λ, sin λ is a polynomial of degee l + m in its aguments Q l,m σ, ρ a ational function of σ ρ of degee l m. By degee of a ational function we mean the degee of the numeato subtacting the degee of the denominato. Since σ = O x ρ = O x 2 + 2, we conclude that l x m ρ = O x l m. 25

26 We can now apply the agument above Leibniz s ule to get l x m J l,m ρ + ρ ρ + + ρ = j P l,m l,m j cos λ +, sin λ +, cos λ, sin λ Q j σ +, ρ +, σ, ρ, whee J l,m is a finite intege σ ± ρ ± ae defined by l,m P j, Q l,m j ± cos θ = σ ± cos λ ± x x 0 = σ ± sin λ ± ae polynomials ational functions of degees l + m, 2 l m in thei aguments espectively. The Lemma follows by integating θ ove 0, π in We close this section by noting that ψ suffes fom slow decay ate at infinity as a consequence of 3.26 Lemma 0. Moe pecisely, ψx, Ox in geneal. This may seem to aise the question whethe Assumption is ealizable at all. Indeed, a moe efined calculation using Lemma 0 shows that the ange of γ δ in 3.25 is not void, povided ω decays fast enough at infinity: Poposition 2 If γ + δ < k + 2 t ω, ω C k,γ,δ s t ψ, ψ C k,γ,δ s. fo sufficiently lage γ δ, then 4 Enegy Helicity Peseving Scheme In this section, we outline the deivation of the discete enegy helicity identities fo EHPS. A key ingedient in the deivation is the efomulation of nonlinea tems into Jacobians. The details can be found in [7]. We intoduce the stad notation: x x φx +, φx 2 2 D x φx, =,, D φx, = x D x φx, = φx + x, φx x,, D φx, = 2 x h = D x, D, h = D, D x. φx, + φx, 2 The finite diffeence appoximation of 2 the Jacobians ae given by 2 hψ = D x D x ψ + D D ψ 2 φx, + φx,. 2, 26

27 J h f, g = { h f 3 h g + h f h g + h g } h f Altogethe, the finite diffeence vesion of EHPS is: t u h + 2 J h u h, ψ h = ν 2 h 2 u h t ω h + J h ωh, ψ h = ν 2 h 2 ω h + J h uh, u h ω h = 2 h + 2 ψ h To deive the discete enegy helicity identity, we fist intoduce the following discete analogue of weighted inne poducts [a, b] h = i= j= the coesponding noms whee a, b h = i= j= D x ad x b i 2,j + ab i,j x 4.3 i= j= D ad b i,j 2 x + a, b h 4.4 a 2 0,h = a, a h, a 2,h = [a, a] h = h a 2 0,h + a 2 0,h 4.5 = + 2 j= j= j=2 4.6 the gids have been shifted [8] to avoid placing the gid point on the axis of otation: x i = i x, i = 0, ±, ±2,, j = j, j =, 2, The evaluation of D 2 h tems in 4.2 at j = involves the dependent vaiables u h, ψ h, ω h the stetching facto h 3 = θ = at the ghost points j = 0. In view of Lemma, we impose the following eflection bounday condition acoss the axis of otation: u h i, 0 = u h i,, ψ h i, 0 = ψ h i,, ω h i, 0 = ω h i,. 4.8 Futhemoe, we take even extension fo the coodinate stetching facto h 3 = θ = which appeas in the evaluation of the Jacobians at j = : h 3 i, 0 = h 3 i,

28 We will show in the emaining sections that the extensions indeed give ise to a discete vesion of enegy helicity identity optimal local tuncation eo. As a consequence, fully second ode accuacy of EHPS is justified fo axisymmetic flows. Remak At fist glance, the extension 4.9 may seem to contadict 4.7 on the ghost points j = 0. A less ambiguous estatement of 4.9 is to incopoate it into 4.2 as t u h + J 2 h u h, ψ h = ν 2 h u 2 h ω t ω h + J h h, ψ h = ν 2 h u ω 2 h + J h h, u h on x i, j, j 4.0 ω h = 2 h + 2 ψ h The following identities ae essential to the discete enegy helicity identity the eo estimate: Lemma Suppose a, b, c satisfies the eflection bounday condition define T h a, b, c := 3 ai, 0 = ai,, bi, 0 = bi,, ci, 0 = ci, i= j= c h a h b + a h b h c + b h c h a. 4. i,j Then c i,j J h a, b i,j = T h a, b, c, 4.2 i= j= a, 2 h + 2 b h = [a, b] h. 4.3 Poof: We fist deive 4.2. In view of 4. 4., it suffices to show that c h a h b = i,j j i c h b h a = i j i,j a h c h b 4.4 b h c h a

29 o, since thee is no bounday tems in the x diection, simply i= j= with f = c g = b D x a a D x b. It is staight fowad to veify that i= j= f D g i,j = f D g i,j = i= j= i= j= g D f i,j g D f i,j 4.6 f i,0 g,0 + g i,0 f i,. i= In the deivation of the discete enegy helicity identities see below, a typical tiplet a, b, c is given by, say, a = ψ h, b = u h c = u h. Fom the eflection bounday condition , we have f i,0 = f i,, g i,0 = g i,. This gives 4.6, theefoe 4.4, Next we deive 4.3. Fom the identity j= f j g j+ 2 = 0, it is easy show that i= j= g j = 2 j= a i,j D D b i,j = f j f j g j 2 2 f + f 0 g 2 i= j= D a i,j 2 j D b 2 i,j. 2 Theefoe 4.3 follows. Fom 4., we can easily deive the pemutation identities T h a, b, c = T h b, c, a = T h c, a, b, T h a, b, c = T h b, a, c. 4.7 Moeove, fom 4.2, 4.3, we can easily deive υ, t u h h + T h u h, ψ h, υ = ν υ, 2 h 2 u h h [ϕ, t ψ h ] h + T h ω h, ψ h, ϕ = ν ϕ, 2 h 2 ω h h + T h u h, u h, ϕ 4.8 ξ, ω h h = [ξ, ψ h ] h 29

30 fo all υ, ϕ ξ satisfying υi, 0 = υi,, ϕi, 0 = ϕi,, ξi, 0 = ξi,. As a diect consequence of the pemutation identity 4.7, we take υ, ϕ = u h, ψ h in 4.8 ecove the discete enegy identity d dt 2 u h, u h h + [ψ h, ψ h ] h + ν[u h, u h ] h + ω h, ω h h = Similaly, the discete helicity identity d dt u h, ω h h + ν[u h, ω h ] h ω h, 2 h 2 u h h = follows by taking υ, ϕ = ω h, u h in 4.8. Remak 2 In the pesence of physical boundaies, the no-slip bounday condition gives u n = τ ψ = 0, u τ = n ψ = 0, u e 3 = u = whee τ = n e 3 e 3 is the unit vecto in θ diection. When the coss section Ω is simply connected, 4.2 eads: u = 0, ψ = 0, n ψ = 0 on Ω It can be shown that the enegy helicity identities 4.9, 4.20 emains valid the pesence of physical bounday conditions [7]. The numeical ealization of the no-slip condition 4.22 intoduced in [7] is second ode accuate seems to be new even fo usual 2D flows. The convegence poof fo this new bounday condition will be epoted elsewhee. 5 Enegy Estimate the Main Theoem In this section, we poceed with the main Theoem of eo estimate. We denote by ψ h, u h, ω h the numeical solution satisfying t u h + 2 J h u h, ψ h = ν 2 h 2 u h t ω h + J h ωh, ψ h = ν 2 h 2 ω h + J h uh, u h 5. ω h = 2 h + 2 ψ h 30

31 ψ, u, ω the exact solution to 3.2, t u + 2 J h u, ψ = ν 2 h 2 u + E t ω + J h ω, ψ = ν 2 h 2 ω + J h u, u + E ω = 2 h + 2 ψ + E 3 whee the local tuncation eos E j can be deived by subtacting 3.2 fom 5.2: E = 2 J h Ju, ψ ν 2 h 2 u E 2 = J h J ω, ψ ν 2 h 2 ω J h J u, u 5.3 E 3 = 2 h 2 ψ Fom , we see that t u u h + 2 J hu, ψ J h u h, ψ h = ν 2 h 2 u u h + E 5.4 t ω ω h + J h ω, ψ J h ωh, ψ h = ν 2 h 2 ω ω h + J h u, u J h uh, u h + E2 5.5 ω ω h = 2 h + 2 ψ ψ h + E Fo nonlinea poblems, it is quite unusual that such an equality in consevative fom can be deived fo finite diffeence schemes. In ou case, the eflecting bounday condition play an impotant ole in the deivation of the following equality: Poposition 3 2 t u u h 2 0,h + ψ ψ h 2,h + ν u u h 2,h + ω ω h 2 0,h = u u h, E h + ψ ψ h, E 2 t E 3 h T u uh h, u u h, ψ T h ψ ψ h, ω ω h, ψ + T h ψ ψh, u, u u h 5.7 Poof: We take the weighted inne poduct of u u h with 5.4 to get 2 t u u h 2 0,h + u u h, J 2 h u, ψ J h u h, ψ h h = ν u u h, 2 h 2 u u h h + u u h, E h

32 The second tem on the left h side of 5.8 can be ewitten as u u h, 2 J h u, ψ J h u h, ψ h h = T u uh h, u, ψ T u uh h, u h, ψ h = T h u uh In addition, fom 4.3 Thus, u u h, ψ ψ h + T u uh h, u u h, ψ + T u uh h, u, ψ ψ h. 5.9 ν u u h, 2 h 2 u u h h = ν[u u h, u u h ] h = ν u u h 2,h. 2 t u u h 2 0,h T u uh h, u u h, ψ ψ h + ν u u h 2,h = u u h, E h T u uh h, u u h, ψ T u uh h, u, ψ ψ h. 5.0 Similaly, we take the weighted inne poduct of ψ ψ h with 5.5 poceed as to get 2 t ψ ψ h 2,h + T h = T h ψ ψ h, u u h ψ ψ h, ω ω h, ψ + ν ω ω h 2 0,h + T h ψ ψh, u, u u h, u u h +T h ψ ψ h, u u h, u + ψ ψ h, E 2 t E 3 h. Next, we apply 4.3 twice to get 5. ν ψ ψ h, 2 h ω ω h 2 h = ν 2 h ψ ψ h, ω ω 2 h h = ν ω ω h 2 0,h 5.7 follows. This completes the poof of the Poposition. We poceed to the estimate of the ight h side of 5.7. We stat with the following elementay identities: Poposition 4 Define Ãxf i,j = 2 f i+,j + f i,j, Ãf i,j = 2 f i,j+ + f i,j. The following estimates hold fo j : D a C Ãa + C D a

33 D a Ãa C + C D a Ãa Cà a 5.4 D a à a, x D x a Ãx a 5.5 Remak 3 As in Remak, the stetching facto in the aguments of left h side of satisfy the even extension 4.9. A moe pecise statement fo, say, 5.2 is given by D a i,j C Ãa i,j + C j D a i,j, j. Fo simplicity of pesentation, we will adopt the expession as in though est of the pape. Poof of Poposition 4: It is easy to veify that D fg = Ãf D g + Ãg D f, Dx fg = Ãxf D x g + Ãxg D x f A staight fowad calculation shows that à j C j, D j C à j C j, D j C 2 j fo j. The estimates then follows. The poof fo 5.5 is also staight fowad. We now poceed to estimate the tilinea foms on the ight h side of 5.7 Lemma 2 Fo a, b c C s R R +, we have T h a, b, c C a,h b,h c,2,0 5.6 T h a, b, c C a 0,h b,h c 2,2,

34 Poof: We begin with the poof of 5.6. Fist we exp the left h side as T h a, b, c = c, 3 2 h a h b h + a, h b h c h + b, h c h a h = 3 I + I 2 + I 3 estimate the I j s tem by tem. We have theefoe I C c, Ãa I = c 2, h a h b h = c, D a D x b D x a D b h, + D a D x b + Ãb + D b D x a h C a,h b,h c 0,,0 follows fom 5.2, Hölde inequality the estimate c = c c 0,,0. Next, I 2 C a, Ãb + D b D x c h + C a, D c D x b h + C a, A c D x b h = C a, Ãb + D b D x c h + C a, D c D x b h + C a, A c D x b h C a,h b,h c 0,,0 + c,2,0 C a,h b,h c,2,0 The estimate fo I 3 is simila. Next we poceed with 5.7. Since T h a, b, c = a, 2 J hb, c h a 0,h 2 J hb, c 0,h, we fist give a pointwise estimate of the integ J h b, c: 3J h b, c = D b D x c D x b D c + D b D x c D x b D c + D x c D b D c D x b = D bi + Ã D x c D x bi + Ãx D c + Ã Ãxb D Dx c +Ãx Ãc D x D b + D x cãx D b D cã D x b = D bi + Ã D x c D x bi + Ãx D c + Ã Ãxb D Dx c + 2 x2 D Dx bdxc D Dx bdc 2 + D x cãx D b D cã D x b

35 Hee I is the identity opeato we have used the identities à x = 2 x2 D 2 x + I, à = 2 2 D 2 + I in the second equality of 5.8. Fom 5.2, we have D bi + à D x c C 2 D b + Ãb x c L 5.9 D x bi + Ãx D c C 2 D x b c + c L C2 D x b c 0,0,0 + c,, Fom , we can similaly deive the emaining tems in 5.8: à Ãxb D Dx c C 2 à + Ãx b x c L C 2 à + Ãx b c,,0 + c 2,2,0, x2 D Dx bdxc 2 C x2 à D x bdxc 2 C 2 à D x b c 2,2, D Dx bdc 2 C à D x b 2 c L C 2 à D x b c 2,2, Ãx D b D x c C 2 Ãx D b x c L C 2 à x D b c,, à D x b D c C 2 à D x b c L C 2 à D x b c,, Fom , we can estimate the weighted L 2 nom of 2 J h b, c by 5.7 follows. 2 J hb, c 0,h C D x b + D b + b 0,h c 2,2,0 C b,h c 2,2,0 Fom Poposition 3, we can deive 2 t u u h 2 0,h + ψ ψ h 2,h + ν u u h 2,h + ω ω h 2 0,h u u h, E h + ψ ψ h, E 2 t E 3 h + C u u h 0,h u u h,h ψ 2,2, C ω ω h 0,h ψ ψ h,h ψ 2,2,0 + C ψ ψ h,h u u h,h u,2,0. Since a 0,h a,h, 35

36 we can futhe estimate the fist two tems on the ight h side of 5.26 by u u h, E h = u u h, E h ν 4 u u h 2,h + ν E 2 0,h ψ ψ h, E 2 t E 3 h ψ ψ h 2,h + E 2 t E 3 2 0,h We now conclude fom Hölde s inequality to get 2 t u u h 2 0,h + ψ ψ h 2,h + ν u u 4 h 2,h + ω ω h 2 0,h ψ ψ h 2,h + C ν E 2 0,h + E 2 2 0,h + te 3 2 0,h + C ν u u h 2 0,h ψ 2 2,2, C ν ψ ψ h 2,h ψ 2 2,2,0 + C ν ψ ψ h 2,h u 2,2,0 With 5.27, it emains to estimate the local tuncation eos E 0,h, E 2 0,h t E 3 0,h. We summaize the esults in Lemma 3 below. The poof will be given in the Appendix. Lemma 3 Let ψ, u, ω be a solution of the axisymmetic Navie Stokes equation 3.2 with t ψ, ψ, u, ω C 0 0, T ; C 4 s E, E 2, E 3 defined by 5.2. Then we have the following pointwise local tuncation eo estimate fo α, β R: x E C + 2α + x 2β x E 2 C + 2α + x 2β Finally, we have the eo estimate: ψ 4,α+ 7 2,β u 4,α+ 7 2,β + u 4,2α+2,2β ψ 4,α+ 7 2,β ω 4,α+ 7 2,β + u 2 4,α+ 7 2,β + ω 4,2α+2,2β x t E 3 C + 2α + x 2β t ψ 4,2α+2,2β 5.30 Theoem 2 Let ψ, u, ω be a solution of the axisymmetic Navie Stokes equation 3.2 satisfying t ψ, ψ, ω C 0 0, T ; C 4,γ,δ s, u C 0 0, T ; Cs 4,5,δ, γ > 4, δ >

37 Then sup [0,T ] T u uh 2 0,h + ψ ψ h,h 2 + u u h 2,h + ω ω h 2 0,hdt C x whee C = Cψ, u, ν, T. Poof: Fom Lemma 3, we have E 2 0,h + E 2 2 0,h + te 3 2 0,h C x Since i= j= i= j= 0 j x + j 4α + x i 4β ψ, u, ω 4 + u, ω, 4,α+ 7 2,β t ψ 2 4,2α+2,2β. j x + j 4α + x i is convegent if α >, β >, it follows that 4β 2 4 E 2 0,h + E 2 2 0,h + t E 3 2 0,h C x ψ, u, ω 4 4,γ,δ + u, ω, t ψ 2 4,γ,δ 5.33 povided γ > 4, δ > 2. We conclude that, unde assumption 5.3, we have ψ Cs 2,2,0, u Cs,2,0 2 t u u h 2 0,h + ψ ψ h 2,h + ν u u 4 h 2,h + ω ω h 2 0,h C u u h 2 0,h + C ψ ψ h 2,h + C x4 + 4 follows fom inequality. The eo estimate 5.32 then follows fom Gonwall s 6 Appendix: Local Tuncation Eo Analysis Poof of Lemma 3 In this section, we poceed with the local tuncation eo estimate. All the assetions in Lemmas 4-7 ae pointwise estimates on the gid points x i, j, j. Fo bevity, we omit the indices i, j wheneve it is obvious. We stat with the estimates of the diffusion tems in 5.3. Lemma 4 If a C 4 s R R + α, β R, we have 2 h 2 a x α + x β a 4,α+2,β 37

38 Poof: Since a Cs 4 R R +, the odd extension of a given by { ax,, if 0 ãx, = ax,, if < 0 is in C 4 R 2. It follows that 2 ha = D 2 x + D 2 + D a = 2 a + 2 x2 xa 4 ξ, a x,η a x,η2 is valid fo all j with ξ x x, x + x η, η 2, +. Thus 2 h 2 a C x x a ξ, + 4 a x,η + 3 a x,η 2 C x a 4,α+2,β + a 3,α+,β + a 4,α+2,β+ a 3,α+,β + α+ + ξ β +η + a 3,α+,β+ a 2,α,β α+ + x β +η 2 α + x β C x α + x β a 4,α+2,β Next we poceed with the estimates fo the Jacobians, stating with thei typical factos: Lemma 5 Fo a C 4 s R R +, α, β R, we have Poof: D x a = x a + O x2 + α + x β a 3,α,β 6. D x a = x a + O 2 x 2 + α + x β a 3,α,β 6.2 D a = a + O α + x β a 3,α+3,β 6.3 D a = a + O 2 We begin with Since + α + x β a 3,α+3,β 6.4 D x x f = x2 6 3 xf ξ,, ξ x x, x + x, 38

39 it follows that D x x a x 2 = 3 6 x a C x 2 ξ, + α + x a β 3,α,β D x x a = x2 6 x 3 2 a ξ, C 2 x 2 + α + x a β 3,α,β Fo , the estimate is a little moe complicated due to ou eflection bounday condition We estimate fo j > j = sepaately. When j >, we have D f = f x,η, η, +. Theefoe we have D a 2 = 3 a 6 C 2 x,η 3 + α + x a β 3,α+3,β D a C a x,η When j =, we have a j= In addition, since = 2 D a j= = 6.3 follows. C 2 a + α + x β 3,α+3,β + a 2,α+2,β + a,α+,β = C 2 3 a C 2 j= 3 a a can be poved similaly,, we apply 4.9 to get = C 2 a2 + a C α + x β a,α+,β 3 + α + x β a 0,α,β D a j= = a C 2 a 2 C 2 a a a a 2 = 3 4 a a, C 2 + α + x a 0,α,β, β C 2 + α + x a 0,α,β. β 39

40 Theefoe In addition, a j= 2 a a follows. D a C 2 j= + α + x a β 0,α,β j= C 2 + α + x β a,α+,β + a 0,α,β, We now continue with the pointwise estimates fo the Jacobi tems J ha, b Ja, b J a h, b J a, b. Since 3 J 2 h a, b = D x a D b D a D x b + D a D x b b D a + D 2 2 b D x a 2 a D x b, 3J a h, b = D x a D b D a D x b + D a D x b b D a + D b D x a a D x b, it suffices to estimate the tems in individually. We summaize them as the following Lemma 6 If a, b C 4 s R R + α, α 2, β, β 2 R, then Poof: D a D x b a xb + D b D x a b x a C x α +α 2 + x β +β 2 a 3,α + 5 2,β b 3,α ,β 2 D x a D b x a b + D x a D b xa b C x α +α 2 + x β +β 2 a 3,α + 5 2,β b 4,α ,β 2 D a D x b a x b + D 2 a D x b 2 a x b C x α +α 2 + x β +β 2 a 3,α + 5 2,β b 4,α ,β 2 Since ae valid fo any α, β R, we have D x a = x a + O x2 + α +λ + x β a 3,α +λ,β 6.0 D x a = x a + O 2 x 2 + α +λ + x β a 3,α +λ,β

41 D a = a + O α +λ + x β a 3,α +λ+3,β 6.2 D a = a + O 2 + α +λ + x β a 3,α +λ+3,β 6.3 D x b = x b + O x2 + α 2+µ + x β 2 b 3,α2 +µ,β D x b = x b + O 2 x 2 + α 2+µ + x β 2 b 3,α2 +µ,β D b = b + O α 2+µ + x β 2 b 3,α2 +µ+3,β D b = b + O 2 fo any λ, µ R. We apply 6.2, 6.5 with λ = 2, µ = 5 2 D a D x b a xb = D a D x b a D x b + a D x b a xb = O D x b α 2 + x β Moeove, since it follows that + α 2+µ + x β 2 b 3,α2 +µ+3,β to get a 3,α + 5 2,β + O a 3 x 2 + α x β 2 3 a a + α x β,α + 2,β D x b = x bξ, 2 b + α x β,α ,β 2 D a D x b a xb b 3,α ,β 2 C x α +α 2 + x β +β 2 a 3,α + 5 2,β b,α 2 + 2,β 2 + a,α + 2,β b 3,α ,β 2 C x α +α 2 + x β +β 2 a 3,α + 5 2,β b 3,α ,β 2 4

42 Similaly, fom , we have D x a D b x a b = D x a D b D x a b + D x a b x a b = O D x a 2 + α x β 2 b 3,α ,β 2 + O b x 2 + α x β a 3,α + 5 2,β C 2 + α +α 2 + x β +β 2 a,α + 2,β b 3,α ,β 2 + C x2 + α +α 2 + x β +β 2 a 3,α + 5 2,β b,α2 + 2,β 2 C x α +α 2 + x β +β 2 a 3,α + 5 2,β b 3,α ,β 2 This gives 6.7. Fo 6.8, we have D x f D g x f g = D x f D g + D x x f g = x f D g ξ, + 6 x2 3 xf g x,η 6.8 = x f D g ξ, + f D x g ξ, + 6 x2 3 xf g x,η. We poceed with individual tems in 6.8 with f = a µ =, we have 2 a x D b C x a 2 + α x β 2 g = b. Fom 6.5 with b 3,α ,β 2 C 2 + α +α 2 + x β +β 2 a,α + 2,β b 3,α ,β 2 Similaly, fom 6.7 a D x b C 2 a + α x β 2 x b 3,α ,β 2 C 2 + α +α 2 + x β +β 2 a 0,α 2,β b 4,α ,β 2 42

43 Theefoe, x 2 3 x a b x,η C x 2 x 3 a b + a 3 x b C x 2 + α +α 2 + x β +β 2 C x 2 + α +α 2 + x β +β 2 a 3,α + 5 2,β b 4,α ,β 2 a 3,α + 5 2,β b,α 2 + 2,β 2 + a 0,α 2,β b 3,α ,β 2 D x a D b x a b C x α +α 2 + x β +β 2 a 3,α + 5 2,β b 4,α ,β 2 Similaly, we have D x a D b xa b C x α +α 2 + x β +β 2 a 3,α + 5 2,β b 4,α ,β 2 This gives 6.8. We continue with 6.9. Fo the fist tem, we wite D a D x b a x b = D a D x x b + D a x b Since a, b C 4 s R R +, by extending a, b to an odd function acoss = 0, we see that the extended a D x b is in C 4 R 2, theefoe D a D x x b = a D x x b x,η = a D x x b + a D x x b x,η = x2 6 a 3 xb ξ,η + a 3 x b ξ2,η D a D x x b C x 2 + α +α 2 + x β +β 2 a,α + 2,β b 4,α ,β 2. Similaly, the extended a x b is in C 3 R 2, we have D a x b = a x b x,η C 2 + α +α 2 + x β +β 2 a 3,α + 5 2,β b 4,α ,β 2 43

44 The second tem in 6.9 can be teated similaly, D 2 a D x b 2 a x b = D 2 a D x x b + D 2 a x b 6.9 Again, the extensions of 2 a D x x b 2 a x b ae in C 3 R 2 C 4 R 2 espectively, we can diectly estimate these two tems by D 2 a D x x b = 2 a D x x b x,η = 2 a D x x b + 2 a D x x b x,η x,η = x2 6 a + 2a xb 3 ξ,η + a 3 x b ξ2,η 6.20 D 2 a x b = a x b x,η = a x b x,η Fom 6.9, , we easily have D 2 a D x x b C x 2 a + α +α 2 + x β +β 2,α + 2,β b 3,α ,β + a 2 0,α 2,β b 4,α ,β C x 2 + α +α 2 + x β +β 2 a,α + 2,β b 4,α ,β 2 D 2 a x b C 2 + α +α 2 + x β +β 2 a 3,α + 5 2,β b 4,α ,β Fom , we conclude that D 2 a D x b 2 a x b C x α +α 2 + x β +β 2 a 3,α + 5 2,β b 4,α ,β 2 This gives 6.9 completes the poof of Lemma 6. As a diect consequence of Lemma 6, we have the pointwise estimate fo the Jacobians 44

45 Lemma 7 If a, b Cs 4 R R +, then J ha, b Ja, b C x α +α 2 + x β +β 2 a 4,α + 7 2,β b 4,α ,β 2 a a J h, b J, b C x α +α 2 + x β +β 2 a 4,α + 7 2,β b 4,α ,β 2 fo any α, α 2, β, β 2 R. Fom 5.3, Lemma 4 Lemma 7, we can easily deive This completes the poof of Lemma 3. Acknowledgments J.G. Liu is sponsoed in pat by NSF gant DMS W.C. Wang is sponsoed in pat by NSC of Taiwan unde gant numbe M Refeences [] A. Aakawa, Computational design fo long-tem numeical integation of the equations of fluid motion: two dimensional incompessible flow. Pat I, J. Compute. Phys [2] G. K. Batchelo, An intoduction to fluid dynamics, Cambidge Univesity Pess, Cambidge, 999. [3] J.T. Beale, T. Kato A. Majda, Remaks on the beakdown of smooth solutions fo the 3-D Eule equations, Comm. Math. Phys , [4] L. Caffaelli, R. Kohn L. Nienbeg, Patial egulaity of suitable weak solutions of the Navie-Stokes equations, Comm. Pue Appl. Math [5] D. Chae, Remaks on the Helicity of the 3-D Incompessible Eule Equations Comm. Math. Phys [6] D. Chae J. Lee, On the egulaity of axisymmetic solutions of the Navie-Stokes equations, Math. Z