The 3D NS equations & intermittency

The 3D NS equations & intermittency J. D. Gibbon : Imperial College London Dec 2012 J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 1 / 40

Summary of this talk Consider the 3D Navier-Stokes equations in the domain [0, L] 3 per u t + u u = ν u p + f (x) div u = div f = 0 Is intermittency the key to understanding the NSE? (a) Intermittent events are manifest as violent spiky surges away from space-time averages in both vorticity ω & strain S i,j. Spectra have non-gaussian characteristics Batchelor & Townsend 1949 [1]. (b) Are the spiky surges in vorticity smooth down to some small scale or do they cascade down to singularity? [2, 3, 4]. Could there be regular solutions that are nevertheless intermittent? Further aims of this talk (i) Weak solutions : address conditional regularity in a new way. (ii) A return to stretching & folding : compressible NS equations. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 2 / 40

Intermittency from wind tunnel data : (Hurst & Vassilicos) 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 Dissipation-range intermittency from wind tunnel turbulence where hot wire anemometry has been used to measure the longitudinal velocity derivative at a single point (D. Hurst & J. C. Vassilicos). The horizontal axis spans 8 integral time scales with Re λ 200. 0 0 100 200 300 400 500 600 J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 3 / 40

Vorticity iso-surfaces Vorticity iso-surfaces in a 512 3 sub-domain of the LANL decaying 2048 3 NS-simulation at Re λ 200. Uneven clustering results in intermittency ; Batchelor & Townsend (1949) [1]; Kuo & Corrsin (1971) [2], Sreenivasan & Meneveau (1988, 1991) [3, 4] ; Frisch (1995) [5] : courtesy of Mark Taylor & Darryl Holm. Physicists use velocity structure functions : Frisch [5] & [6, 7] u(x + r) u(x) p ens.av. r ζp J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 4 / 40

Leray s energy inequality Begin with the definitions H 0 = V u 2 dv (energy) & H 1 = V ω 2 dv (enstrophy). Leray s energy inequality shows that (ϖ 0 = νl 2 ) d 1 2 u 2 dv ν ω 2 dv + f 2 u 2 dt V V ϖ 0 u 2 dv + f 2 u 2 Now define frms 2 = L 3 f 2 2 and U2 0 = L 3 u 2 2 Gr = L 3 f rms ν 2, Re = U 0 Lν 1. lim sup t H 0 = u 2 2 L5 ϖ0 2 Gr (a pointwise bound) together with Lemma (Leray s energy inequality for weak solutions [8]) H 1 T L 3 ϖ 2 0Gr Re + O ( T 1). V T J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 5 / 40

Conditional 3D-NS regularity : a very brief history Older history : Leray (1934), Prodi (1959), Foias & Prodi (1962), Serrin ( 63) & Ladyzhenskaya (1964) [8, 11, 12, 13, 14, 15, 16] : every Leray-Hopf solution u of the 3D-NSE with u L r ((0, T ) ; L q ) is regular on (0, T ] if 2/r + 3/q = 1 q (3, ] or if u L ((0, T ) ; L p ) with p > 3. Books: Ladyzhenskaya [14], Constantin & Foias (1988) [15] ; Foias, Manley, Rosa, Temam (2001) [16]. Modern developments : i) When s = 3 : von Wahl (1983) [17] & Giga (1986) [18] proved regularity in C ( (0, T ] ; L 3) : see also Kozono & Sohr (1997) [19] & Escauriaza, Seregin & Sverák (2003) [21] & Seregin (2012) [20]. ii) Assumptions on the pressure or 1 velocity derivative : Kukavica & Ziane (2006, 2007) [22, 23], Zhou (2002) [24], Cao & Titi (2008, 2010), Cao (2010), Cao, Qin & Titi (2008) [25, 26, 27, 28], & Chen & Gala ( 11) [29] ; iii) Assumptions on the direction of vorticity : BKM (1984) [30], Constantin & Fefferman (1993) [31] ; Vasseur (2008) [32], Cheskidov & Shvydkoy (2010) [33] used Besov spaces ; J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 6 / 40

Why is H 1 important? Standard contradiction strategy : assume a maximal interval of existence & uniqueness [0, T ): thus H 1 (T ) =. In the limit t T, if we can prove that H 1 < then we have a contradiction... 1 (Euler) 2Ḣ n = n u n (u u) dv c n u H n V Include viscosity & use a different route to that we used for Euler 1 2Ḣ n ν ( n u) ( n u) dv + n+1 u n 1 (u u) dv V νh n+1 + c n u H 1/2 n+1 H1/2 n ( ) νh n+1 + c n ν 1/2 u (νh n+1 ) 1/2 H 1/2 n Lemma (See [9, 10]) On [0, T ) 1 2Ḣ n νh n+1 + c n,1 u H n n 1 1 2Ḣ n 1 2νH n+1 + c n,2 ν 1 u 2 H n n 1 J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 7 / 40 V

Note : Either H n (t) c n exp t 0 u 2 dτ or H n (t) c n exp t 0 u dτ. Return to 1 2Ḣ n 1 2νH n+1 + c n,2 ν 1 u 2 H n n 1 and use a generalization of (Agmon s inequality?) for n 2 : Thus we have u c n n u 1 2(n 1) 2 u 2n 3 2(n 1) 2 H p+q n H q n ph p n+q with p = n 1, q = 1 1 1 2Ḣ n 1 2ν H1+ n 1 n H 1 n 1 1 + c n ν 1 H 1+ 1 n 2(n 1) 2n 3 2(n 1) H1 n > 1 If H 1 is bdd then so is every H n for n 2 : thus H 1 controls regularity. Problem : analytically H 1 is uncontrolled only H 1 is bounded & numerically it is highly intermittent. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 8 / 40

Why u q for q > 3? 1) For the H n+1 -term : 1 2Ḣ n 1 2νH n+1 + c n ν 1 u 2 H n n 1 n u 2 c n n+1 u a 2nq + 6 3q 2 u 1 a q a = 2(n + 1)q + 6 3q 2) For the u 2 -term : u c n n u b 6 2 u 1 b q b = 2nq + 6 3q 1 2Ḣ n 1 2ν ( ) 1/a H n + c u 2(1 a) n ν 1 Hn 1+b u 2(1 b) q q For what values of q is a 1 > 1 + b? Answer : q > 3. Problem : assumptions on u q for q 3 are unphysical. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 9 / 40

Higher moments of vorticity for the NS equations [34, 35] 1 How might we pick up intermittent behaviour? 2 Can this be related to a different form of conditional regularity? Consider higher moments of vorticity (m 1) as the frequencies ) 1/2m Ω m (t) = (L 3 ω 2m dv + ϖ 0 The basic frequency associated with the domain is ϖ 0 = νl 2. Ω 2 1 = L 3 ω 2 dv + ϖ 0 H 1 -norm V V is the enstrophy/unit volume which is related to the energy dissipation rate. The higher moments will naturally pick up events at smaller scales ϖ 0 Ω 1 (t) Ω 2 (t)... Ω m (t) Ω m+1 (t)... J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 10 / 40

Estimates in terms of Re Traditionally, NS-estimates have been found in terms of the Grashof number Gr (frms 2 = L 3 f 2 2 ) of the forcing f (x) but it would be more helpful to express these in terms of the Reynolds number Re Gr = L 3 f rms ν 2, Re = U 0 Lν 1. Doering and Foias (2002) [36] used the idea of defining U 0 as U0 2 = L 3 u 2 2 where the time average over an interval [0, T ] is defined by T g( ) T = lim sup g(0) T 1 T g(τ) dτ. T 0 Gr is fixed provided f is L 2 -bounded, while Re is the system response. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 11 / 40

Kolmogorov length from Leray s energy inequality Leray s energy inequality shows that d dt V Ω 2 1 1 2 u 2 dv ν V ω 2 dv + f 2 u 2, T ϖ2 0Gr Re + O ( T 1). Doering & Foias [36, 37] showed that NS-solutions obey Gr c Re 2 provided the spectrum of f is concentrated in a narrow-band around a single frequency or its spectrum is bounded above & below Ω 2 c 1 T ϖ2 0Re 3 + O ( T 1). ν Ω 2 1 is the time-averaged energy dissipation rate over [0, T ] and the T Kolmogorov length scale λ 1 k is estimated as k = ν Ω 2 1 T ν 3 Lλ 1 k c Re 3/4 + O ( T 1/4). λ 4 J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 12 / 40

Weak solution result JDG [34] The NSE have the invariance property under the transformations x = ɛx, t = ɛ 2 t, u = ɛu and p = ɛ 2 p. The Ω m scale as Ω αm m = ɛω α m m α m = 2m 4m 3. Definition (D m = ( ϖ 1 0 Ω ) αm m where αm = 2m ) 4m 3 Theorem (JDG [34]) Weak solutions of the 3D-Navier-Stokes equations satisfy D m T c Re 3 + O ( T 1), 1 m, where ϖ 0 = νl 2 and c is a uniform constant. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 13 / 40

Proof of Theorem Lemma (Foias, Guillopé & Temam (1981) [38]) Let A n = ( ν 2 H n ) 1 2n 1, then weak solutions of 3D-NS satisfy A n+1 T c 1,n A n T + c 2,n A 1 T Thus all the weak solution properties enjoyed by the H 1 -norm also hold for A n+1 T. Moreover, A n T c n L 1 Re 3 + O ( T 1) u T c N Lϖ 0 Re 3 + O ( T 1) Proof : Use the inequality : u c n n u 1 2(n 1) 2 u 2n 3 2(n 1) 2 & insert this into 1 2Ḣ n 1 2νH n+1 + c n ν 2 u 2 H n J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 14 / 40

FGT proof contd... Divide the result by H 2n 2n 1 n H n+1 H 2n 2n 1 n Now we write T c n H & time average to get 1 (2n 1)(2n 2) n ν 2 2n+1 An+1 T = The ν-terms cancel and we find H 2n 3 2n 2 1 c n ν 2 T ( H n+1 H 2n 2n 1 n H n+1 H 2n 2n 1 n (2n 1) 2 ) 1 2n+1 1 2n+1 (2n 2)(2n+1) A n+1 T c n A n T A 1 T 2n 2 T 2n 1 An 1 H 2n (2n 1)(2n+1) n T ( ) ν 2 2n 2n+1 2n 1 An T 2n 3 (2n 2)(2n+1) T A 1 2n 3 2n 2 T (2n 1) 2 + 2n 3 = (2n 2)(2n + 1) & thus (*) becomes linear. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 15 / 40 ( )

Theorem proof contd... H n = V n u 2 dv = V n 1 ω 2 dv ω 2m c N,m n 1 ω A 2 ω 1 A 3(m 1) 2, A = 2m(n 1), n 3 Raise to the power α m and time-average ( ) 1 ω α m 2m c αm T n,m H 1 2n 1 n c αm n,m H 1 2n 1 n An explicit u.b. in terms of Re is available only if 2 Aαm(2n 1) H 1 2 (1 A)αm 1 T 1 2 Aαm(2n 1) (1 A)αm 1 1 2 Aαm(2n 1) 2 aαm(2n 1) H1 T T (1 A)α m 2 aα m (2n 1) = 1 α m = 2m 4m 3, uniformly in n c N,m can be minimized by choosing n = 3 : finite when m =. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 16 / 40

A hierarchy of length scales Based on the definition of the inverse Kolmogorov length λ 1 k, a generalization of this to a hierarchy of inverse lengths λ 1 m is : Theorem ( ) Lλ 1 2αm m := Dm T, α m = 2m 4m 3. For 1 m, the size of the Lλ 1 m Lλ 1 m are estimated as ) c Re 3/2αm + O (T 1/2αm. Averages need no point-wise estimates that the soln of the regularity problem would require. Weak solutions exist but are not unique. m 1 9/8 3/2 2 3... 3/2α m 3/4 1 3/2 15/8 9/4... 3 Computationally hard to get beyond m = 1 : m = 9/8 (Re 1 ) is close to modern resolutions why is resolution so difficult? [39] J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 17 / 40

A summary 1 Clearly, the object H 1 2n 1 n c n L 1 ν 2 2n 1 Re 3 + O ( T 1) is special in the weak solution class. 2 Following from this the exponent α m in the definition ( ) D m = ϖ 1 0 Ω αm m is also special. Scaling properties are the reason. 3 Could this be shown up in numerical computations? J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 18 / 40

Kerr s initial condition [40] Figure: The figure shows the NS evolution of very long anti-parallel vortices with a localized perturbation in a turbulent flow at three times. The first at t = 16, which is the time of the first reconnection, is shown on the left ; t = 96 lies in the middle and is the time of the second set of reconnections. The formation of the first vortex rings at t = 256 is on the right and shows the cumulation of this process after two sets of secondary reconnections have released two sets of vortex rings. The full domain in the vortical y direction reaches from y = 16π 50 to y = 16π 50. The partial domains shown are y 10 for t = 16, 0 y 25 for t = 96 and 0 y 16π 50 for t = 256. The inset of the upper left vorticity near y = 0 at t = 16 demonstrates that on the y = 0 symmetry plane the usual head-tail configuration forms. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 19 / 40

Recent results of Kerr & DGKP [40, 41] Figure: Plot of D m for 3D-NS with anti-parallel i.d. with resolution 1024 512 4096. Note the ordering of the D m [40, 41]. Note : D m+1 < D m a surprise with no cross-overs!! J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 20 / 40

A differential inequality for D m Proposition (JDG (2010) [35]) Ω m ϖ 0 Ω m { ϖ 0 Ω m { 1 c 1,m 1 c 3,m ( Ωm+1 Ω m ( Ωm+1 Ω m ) βm + c 2,m ( Ωm+1 Ω m ) m+1 Ω m } ) } βm ( ) + c 4,m ϖ 1 0 Ω 2αm m ( ) where β m = 4 m(m + 1) 3 ( 1 1 ) β m = 2. α m+1 α m Proof : For the Laplacian term (Ω m+1 /Ω m ) βm with A m = ω m, one shows 1 d ω 2m dv c m A m 2 dv 2m dt V J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 21 / 40 V

Proposition proof contd... A simple Sobolev inequality gives A m 2(m+1)/m c m A m 3 2(m+1) 2 A m 2m 1 2(m+1) 2 which means that *** V (ω m ) 2 dv 1 Ω 4m(m+1)/3 m+1 c m From the Euler lectures we proved Ω 2m(2m 1)/3 m = 1 c m ( Ωm+1 Ω m ) 4m(m+1)/3 Ω 2m m so we end up with ( ) m+1 Ωm+1 Ω m c 1,m Ω 2 m, 1 m <. Ω m Ω m ϖ 0 Ω m { 1 c 1,m ( Ωm+1 Ω m ) βm + c 2,m ( Ωm+1 Ω m ) m+1 Ω m } J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 22 / 40

Another look at conditional regularity : unforced NS Now convert into the D m -notation : Lemma (JDG [42]) For 1 m < on [0, T ), the satisfy Ḋ m D 3 m D m (t) = ( ) ϖ 1 0 Ω αm m { ( ) ρm Dm+1 ϖ 1,m + ϖ 2,m}, D m where ρ m = 2 3 m(4m + 1) and c n,m (n = 1, 2, 3) ϖ 1,m = ϖ 0 α m c 1 1,m ϖ 2,m = ϖ 0 α m c 2,m ϖ 3,m = ϖ 0 α m c 3,m. are algebraically increasing with m. For forced-ns add ϖ 3,m Gr D m. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 23 / 40

First integration From the previous Theorem, solutions come under control pointwise provided D m+1 (t) c ρm D m (t), where c ρm = [c 1,m c 2,m ] 1/ρm. Theorem For any value of 1 m < and 0 < ε < 2 if the integral condition is satisfied t 0 t Dm+1dτ ε c ε,ρm Dmdτ ε then D m (t) D m (0) on [0, t], where c ε,ρm = [c 1,m c 2,m ] ε/ρm. Remark The ε close to zero case suggests the conjecture a logarithmic form might be the sharp result but the general ε case appears next. 0 J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 24 / 40

Proof of Theorem Proof. Divide the differential inequality by Dm 3 ε t [D m (t)] ε 2 [D m (0)] ε 2 ϖ (ε) 1,m 0 and integrate to obtain ( Dm+1 D m ) ρm t Dmdτ ε ϖ (ε) 2,m Dmdτ ε. 0 A Hölder inequality then easily shows that (ϖ (ε) n,m = (2 ε)ϖ n,m ) t ( Dm+1 0 D m D m (t) [D m(0)] ε 2 + ϖ (ε) ) ρm D ε mdτ 1,m ( ) t ρm/ε 0 Dε m+1 dτ ( ). t (ρm ε)/ε 0 Dε mdτ ( ) t ρm/ε 0 Dε m+1 dτ ( t 0 Dε mdτ ) (ρm ε)/ε ϖ(ε) 2,m t 0 Dmdτ ε 1 2 ε J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 25 / 40

A log-theorem Theorem For any value of 1 m <, if the integral condition is satisfied t 0 ( ) 1 + Zm ln dτ 0, Z m = D m+1 /D m (1) c 4,m with c 4,m = [ 2 ρm 1 (1 + c 1,m c 2,m ) ] ρ 1 m, then D m (t) D m (0) on [0, t]. Divide by D 3 m & integrate to obtain (Z m = D m+1 /D m ) 1 ( [Dm (t)] 2 [D m (0)] 2) t ϖ 1,m 2 ϖ 1,m 2 ρm 1 where C m = c 1,m c 2,m & (1 + Z m ) ρm 0 t 0 { ( [1 + Zm ρm ] 1 + ϖ 2,m ϖ 1,m )} dτ { [1 + Zm ] ρm 2 ρm 1 (1 + C m ) } dτ, 2 ρm 1 (1 + Z ρm m ). J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 26 / 40

Proof of log-theorem : Jensen s inequality gives 1 t t 0 exp F (τ) dτ exp ( ) 1 t F(τ) dτ t 0 F = ρ m ln(1 + Z m ) the RHS can be written as 1 t = 1 t t 0 t exp { [1 + Zm ] ρm 2 ρm 1 (1 + C m ) } dτ 0 { exp [ρm ln(1 + Z m )] 2 ρm 1 (1 + C m ) } dτ [ ρ m t t 0 ln(1 + Z m ) dτ ] exp [ρ m ln c 4,m ] which gives the advertized result. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 27 / 40

Lower bound assumption in the forced case : JDG [42, 43] Theorem On [0, t] if there exists a value of m lying in the range 1 < m <, with initial data [D m (0)] 2 < C m Gr m (1 m 4), for which the integral lies on or above the critical value ( c m t Gr 2δ m ) t + η 2 D m dτ where η 2 η 1 Gr 2(δm 1) and where γ m = α m+1 2(m 2 1). 0 m = 2 [δ m (2 + ρ m γ m ) ρ m γ m ] with 1 + 2ρ m γ m 2(2 + ρ m γ m ) < δ m < 1, then D m (t) decays exponentially on [0, t]. Remark δ m 1 2 for large m. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 28 / 40

Proof Inclusion of the forcing modifies the differential inequality for D m to { Ḋ m Dm 3 1 ( ) ρm Dm+1 + ϖ 2,m} + ϖ 3,m Gr D m, ϖ 1,m D m To proceed, divide by D 3 m (the ε = 0 case of the previous Theorem) 1 2 d dt ( ) D 2 m ( ) X m D 2 m ϖ2,m X m = ϖ 1,m ( Dm+1 D m ) ρm D 2 m ϖ 3,m Gr. The final result is { exp 2 } t [D m (t)] 2 0 X m dτ [D m (0)] 2 t 2ϖ 2,m 0 exp { 2 τ 0 X m dτ } dτ. ( ) t t X m dτ ϖ 1,m t 1 0 D ρm m+1 dτ ( 0 ) ϖ t ρm 2 3,mt Gr. 0 D m dτ J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 29 / 40

Proof cont... The integral is estimated as t 0 ( τ ) exp 2 X m dτ dτ 0 ( 1 exp [ 2ϖ1,m c m t Gr m ]) 2 c m ϖ 1,m Gr m and so the denominator of the estimate for D m satisfies Denominator [D m (0)] 2 c 2,mc 1,m ( 1 exp [ 2ϖ1,m c m t Gr m ]) This can never go negative if (2 c m ) Gr m [D m (0)] 2 > c 1,m c 2,m (2 c m ) 1 Gr m which means D m (0) < C m Gr 1 2 m, Because 1 m 4 this is large initial data. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 30 / 40

The four phases of intermittency. D m (t).. Gr 2 upper bound of D m... t Gr 2δ m critical lower-bound of D m... t potential singularity... t 0 t 1 t 2 t J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 31 / 40

Conclusions I 1 The comparative time-integral log-criterion for regularity in the unforced case shown in Theorem 3 needs enough intervals where D m+1 > D m. 2 Why do numerical experiments display the ordering of the D m in the reverse order than that expected? Is the regime physcially D m+1 > D m reachable, e.g. at higher Re? 3 The conditional control that comes from the critical lower bound on D m t in Theorem 4 in the forced case. 4 In the latter, even if the solution remain regular, intermittent behaviour is inevitable. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 32 / 40

Conclusions II In the super-critical forced case singularities are not possible when t 0 D m dτ c m ( t Gr 2δ m + η 2 )? In the sub-critical forced case singularities are still possible when t 0 D m dτ < c m ( t Gr 2δ m + η 2 )? Potential singularities that raise the value of the integral past critical can be ruled out but not those that contribute little or nothing to the integral. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 33 / 40

A return to the B = q θ field & compressible NS 1 In our lectures on stratified incompressible Euler we found that with the two passive scalars θ and q = ω θ Dθ Dt = 0 and Dq Dt = 0 the evolution of the B = q θ vector is governed by t B = curl (u B). 2 Does the stretching & folding idea survive dissipation? 3 How about the compressible NS equations? Consider the incompressible NS equations coupled to the θ-field Du Dt + θ k = Re 1 u p, Dθ Dt = ( σre ) 1 θ. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 34 / 40

Stretching & folding : the B-field for incompressible NS The PV defined by q = ω θ evolves according to Dq Dt ( Dω = Dt + ω ( (σre) 1 θ ) ) ω u θ + ω ( ) Dθ Dt = div ( Re 1 u θ + (σre) 1 ω θ ), The trick of Haynes & McIntyre [46, 47] Define a pseudo-velocity U : q ( U u ) = Re 1( u θ + σ 1 ω θ ) t q + div (q U) = 0. Not that there are problems when q = 0. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 35 / 40

Remarks on the pseudo-velocity U Note that from and t q + div (q U) = 0 q ( U u ) = Re 1( u θ + σ 1 ω θ ) 1 q is the PV density ; 2 div U 0 but nevertheless div U = Re 1 div [ q 1 (... ) ] ; 3 Strictly speaking U is not a physical velocity (Danielsen 1990 [48]), but U can still be considered as a pseudo-velocity. 4 What of θ? Does t θ + U θ = 0? t θ + U θ = t θ + u θ q 1 Re θ { u θ + σ 1 ω θ } = t θ + u θ ( σre ) 1 θ = 0. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 36 / 40

Formal result for the B-field Theorem (See JDG & Holm [44, 45]) t q + div ( q U ) = 0, t θ + U θ = 0, and B = q θ satisfies the stretching relation t B curl (U B) = D. where the divergence-less vector D and U are defined by D = (q div U) θ. q(u u) = Re 1 { u θ + σ 1 ω θ }, q 0. Moreover, for any surface S(U) moving with the flow U, one finds d B ds = D ds. dt S(U) S(U) J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 37 / 40

Compressible NS equations I (JDG & Holm [49]) The 3D compressible Navier-Stokes equations are : [50] ρ Du = µ u ϖ, ϖ = p (µ/3 + µ v )div u, Dt Dρ Dt + ρ div u = 0, D Dt = t + u, Dθ c v = p Dt ρ div u + Q, p = R ρ θ. Here, u denotes the spatial fluid velocity, µ is the shear viscosity and µ v is the volume viscosity, both of which are taken as constitutive constants of the fluid. For definiteness, we have chosen an ideal gas equation of state to relate pressure p, temperature θ, and mass density ρ. In addition, R is the gas constant, c v the specific heat is constant and Q is the heating rate, which we may assume is known. Note : The pressure depends on the two thermodynamic variables ρ and θ. J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 38 / 40

Compressible NS equations II Assume solutions exist : t ω curl (u ω) = µρ 1 ω + ρ 1 This invites a projection against ρ q = ω ρ [ )] µ u (ϖ u2 2 which satisfies a geometric relation reminiscent of Ertel s Theorem ( ) ( ) Dq Dω Dρ = ω u ρ + ω Dt Dt Dt t q + div J = 0 J = qu + ωρ div u µ u (ln ρ) + φ f (ρ). J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 39 / 40

Compressible NS equations III In terms of the pseudo-velocity U = J/q U = u + q 1 {ωρ div u µ u (ln ρ) + φ f (ρ)}. t q + div (q U) = 0 t ρ + U ρ = 0. Note that div U 0. There could also be a change of topology when q = 0. With B = q ρ a direct computation shows that B satisfies where D = (q div U) ρ. t B curl (U B) = D J. D. Gibbon : Imperial College London () The 3D NS equations & intermittency Dec 2012 40 / 40

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