2019
DOI: 10.3934/cpaa.2019064
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Some Remarks on regularity criteria of Axially symmetric Navier-Stokes equations

Abstract: Two main results will be presented in our paper. First, we will prove the regularity of solutions to axially symmetric Navier-Stokes equations under a log supercritical assumption on the horizontally radial component u r and vertical component u z , accompanied by a log subcritical assumption on the horizontally angular component u θ of the velocity. Second, the precise Green function for the operator −(∆ − 1 r 2 ) under the axially symmetric situation, where r is the distance to the symmetric axis, and some w… Show more

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Cited by 9 publications
(9 citation statements)
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References 26 publications
(41 reference statements)
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“…Obviously, there exists a number 0 < R * 5 (M 0 , c * , α) ≤ min{1/6, R * 2 } such that 2c(M 0 , c * ) ln , being valid for all natural numbers m, and references to [24], [9], and [11].…”
Section: Introductionmentioning
confidence: 99%
“…Obviously, there exists a number 0 < R * 5 (M 0 , c * , α) ≤ min{1/6, R * 2 } such that 2c(M 0 , c * ) ln , being valid for all natural numbers m, and references to [24], [9], and [11].…”
Section: Introductionmentioning
confidence: 99%
“…We cannot pretend to cite all good works in this direction. For example, let us mention papers: [3][4][5]9,[11][12][13][19][20][21][26][27][28][29][30][31], and [15].…”
Section: (): V-volmentioning
confidence: 99%
“…It follows Lemma 2.9 and from (3.5) that ) ∈ H 2 . Therefore, one can use the results of [12,30], and [15], see also [17], [4], on the Cauchy problem for the Navier-Stokes system (1.1) in R 3 ×]T − R 2 * , T [ and conclude that v is a strong solution in the interval ]0, T [.…”
Section: Proof Of Proposition 14mentioning
confidence: 99%
“…It is well-konwn that this potential is a borderline one where the regularity theory differs from the standard one. For the regularity and mean value inequality of this equation in R d , we can refer to Z.Li&Q.Zhang [19], B.Wong&Q.Zhang [27] and Z.Li&X.Pan [17]. Actually, the inverse-square potential term A |x| 2 helps with it.…”
Section: Remark 16mentioning
confidence: 99%