On the leapfrogging phenomenon in fluid mechanics Didier Smets Université Pierre et Marie Curie - Paris Based on works with Robert L. Jerrard U. of Toronto) CIRM, Luminy, June 27th 2016 1 / 22
Single vortex ring : a movie example c D. Kleckner & W. Irvine, Nature, 2013 2 / 22
Helmholtz master paper on vorticity Vortex rings : Leapfrogging : 3 / 22
Leapfrogging vortices : a movie example c T.T. Lim 4 / 22
Leapfrogging vortices : a movie example B. Balog, Q. Kriaa, S. Yehudi 5 / 22
Existence of vortex rings : a practical point of view 6 / 22
Euler equation for incompressible fluids The 3 dimensional Euler equations read { tv + v )v = p, div v = 0, where v : R 3 R R 3 is the velocity field, p : R 3 R R is the pressure field. For a number of physically meaningful flows, an important quantity is given by the vorticity which satisfies ω = curl v tω + v ω = ω v. The velocity v can always be recovered from the vorticity ω through the Biot-Savart law vt, x) = 1 y x ωt, y) dy. 4π R 3 y x 3 7 / 22
Gross-Pitaevskii equation for quantum fluids The 3 dimensional Gross-Pitaevskii equation reads where i tu u = 1 ε 2 u1 u 2 ) u : R 3 R C is the complex field function ε > 0 is a given length-scale. In the framework of thin vortex tubes, it bears some resemblance with the Euler equation. An important quantity here is given by the vorticity Ju = 1 curl ju) = curl u u). 2 In contrast with the Euler equation, the current ju) is not fully determined by the Jacobian Ju since an additional degree of freedom is present through the phase: u = ρ expiϕ) = ju) = ρ 2 ϕ, ρ 1 = Ju = 1 curl ϕ 0. 2 8 / 22
On the analogy between GP and Euler For axisymmetric solutions of GP) we have the equation for the vorticity d Ju ϕ drdz = F u, ϕ), dt H where k r F u, ϕ) := ε ij j u, k u) i ϕ + ε ij j u, k u) ik ϕ. H r H For the axisymmetric Euler equation we have d ω ϕ drdz = Fv, ϕ). dt H In contrast to Euler, for GP) we also need an equation for the compressible part of ju) : ε t u 2 1 ε = 2 r divrju)). 9 / 22
Existence of vortex rings : a mathematical point of view One looks for cylindrically symmetric traveling wave solutions of the Euler equation. These can be obtained by the variational problem maximize E := ωx)gx, x )ωx ) dνdν H H P := r 2 ωdrdz = given Cst, under constrains H ω ω0 is a transport measure of r r, where ω 0 is given. Here, H := {r, z), r 0, z R}, dν = r dr dz and G refers to the Green function of the Laplacian in cylindrical coordinates. Important contributions by Arnold 1964, Fraenkel-Berger 1974, Benjamin 1976, Friedman-Turkington 1981, Burton 2003. For the Gross-Pitaevskii, a related strategy is given by minimize E := [ 1 H 2 u 2 + 1 4ε 2 u 2 1) 2 ] rdrdz under constrain P := r 2 Ju drdz = given Cst. H Analysis by Bethuel-Orlandi-S. 2003. 10 / 22
Leapfrogging for GP : notion of reference vortex tubes Let C be a smooth oriented closed curve in R 3 and let J be the vector distribution corresponding to 2π times the circulation along C, namely J, X = 2π X τ X DR 3, R 3 ), C where τ is the tangent vector to C. To the current density J is associated the induction B, which satisfies the equations div B) = 0, curl B) = J in R 3, and is obtained from J by the Biot-Savart law. To B is then associated a vector potential A, which satisfies div A) = 0, curl A) = B in R 3, so that A = curl curl A) = J in R 3. For a H, let C a be the circle of radius ra) parallel to the xy-plane in R 3, centered at the point 0, 0, za)), and oriented so that its binormal vector points towards the positive z-axis. By cylindrical symmetry, we may write the corresponding vector potential as A a A ar, z) e θ. The expression of the vector laplacian in cylindrical coordinates yields the equation for the scalar function A a : r 2 + 1 r r 1 ) r 2 + 2 z A a = 2πδ a in H A a = 0 on H, 11 / 22
Leapfrogging for GP : notion of reference vortex tubes or equivalently ) 1 div r raa) = 2πδ a in H A a = 0 on H, which can be integrated explicitly in terms of complete elliptic integrals. Up to a constant phase factor, there exists a unique unimodular map ua C H \ {a}, S 1 ) W 1,1 loc H, S1 ) such that riua, u a ) = rju a ) = ra a). In the sense of distributions in H, we have { divrju a )) = 0 curljua )) = 2πδa, and the function ua corresponds therefore to a singular vortex ring. In order to describe a reference vortex ring for the Gross-Pitaevskii equation, we shall make the notion of core more precise. In R 2, the Gross-Pitaevskii equation possesses a distinguished stationary solution called vortex : in polar coordinates, it has the special form u εr, θ) = f εr) expiθ) where the profile f ε : R + [0, 1] satisfies f ε0) = 0, f ε+ ) = 1, and rr f ε + 1 1 r fε r r 2 fε + 1 ε 2 fε1 f ε 2 ) = 0. The reference vortex ring associated to the point a H is defined to be u ε,ar, z) = f ε r, z) a ) u a r, z). 12 / 22
Leapfrogging for GP : notion of reference vortex tubes More generally, when a = {a 1,, a n} is a family of n distinct points in H, we set u a r, z) := n ua k r, z), and uε,ar, z) := k=1 n uε,a k r, z), where the products are meant in C. The field u ε,a hence corresponds to a collection of n reference vortex rings sharing the same axis and oriented in the same direction), and is the typical kind of object which we shall study the evolution of. It can be shown that Ju ε,a π Finally, classical computations lead to k=1 n Ẇ δ ai = Oε) as ε 0. i=1 1,1 H) E u ε,a) = Hεa 1,, a n) + o1), where H εa 1,, a n) := n i=1 [ ra i ) π log ra i ) ε ) + γ + π 3 log2) 2 ) + π j i ] A aj a i ), and γ is a numerical constant. 13 / 22
The ODE of leapfrogging We consider the associated hamiltonian system LF ) ȧ i t) = 1 π J a i Hε a1 t),, a nt) ), i = 1,, n, where, with a slight abuse of notation, J := ) 0 1 ra i ) 1. 0 ra i ) In addition to the hamiltonian H ε, the system LF) also conserves the momentum Pa 1,, a n) := π n r 2 a k ), which may be interpreted as the total area of the disks determined by the vortex rings. As a matter of fact, also note that k=1 Puε,a) := Juε,a r 2 drdz = π H n r 2 a k ) + o1), as ε 0, and that, at least formally, the momentum P is a conserved quantity for the Gross- Pitaevskii equation. k=1 14 / 22
Mathematical convergence results We consider axisymmetric initial data u 0 ε for GP), we write a 0 i,ε = a0 + b0 i,ε log ε and denote by ai,ε s the corresponding solution of LF )ε. We let S > 0 and fix a constant K > 0 such that bi,ε s K for all s [0, S]. We define Ju0 ε π n i=1 δ a i,ε 0 Ẇ =: ra 0 concentration scale) 1,1 [ + Euε) 0 H εa1,ε n,ε)] 0,, a0 =: Σ 0 a energy excess). Theorem Jerrard-S.) Under the above assumptions, there exist constants ε 0, σ 0, C 0 depending only on a 0, n, K) such that if ε ε 0 and if Σ 0 a + r 0 a log ε σ 0, then n ra s := Juε s π δ a s C i,ε 0 ra i=1 Ẇ 0 + 1,1 for s [0, S], where δ > 0 is arbitrary. Σ0 a log ε + ) C δ log ε 1 δ expc 0 s) 15 / 22
Analysis of the leapfrogging system for two vortex rings When n = 2, the system LF ) may be analyzed in great details. Since P is conserved and since H ε is invariant by a joint translation of both rings in the z direction, it is classical to introduce the variables η, ξ) by { r 2 a 1 ) = P 2 η r 2 a 2 ) = P 2 + η, ξ = za 1) za 2 ), and to draw the level curves of the function H ε in those two real variables, the momentum P being considered as a parameter. Size of order O1) P = 2 log ε = 10 0.5 Pass through η 0 Attract then repel Leapfrogging log log ε O ) log ε 0.5 Equilibria corresponding to a pair of traveling vortices 1 4 2 0 2 4 ξ 16 / 22
Collision of vortex rings c T.T. Lim 17 / 22
Strategy of the proof We rely mostly on the already mentioned evolution equation for the Jacobian d F u, ϕ) Ju ϕ drdz =, ds H log ε where k r F u, ϕ) := ε ij j u, k u) i ϕ + ε ij j u, k u) ik ϕ. H r H We wish to prove that Ju remains well concentrated around a sum of Dirac masses F u, ϕ) is a good approximation of the ode LF) ε. It is clear that both are not independent and we shall use a common Gronwall argument on r s a. We will decompose j u, k u ) = j u k u + ju) j ju) k u u ) ju) = j u k u + j u j ) ju) + j u j + j ) j ), j k where j is a suitable approximation of ju). ju) u j ) j ) ) ju) + j k u k k j ) j 18 / 22
Some elements in the proof: excess energy and concentration Proposition There exist constants ε 1, σ 1, C 1 > 0, depending only on n, r 0 and K, with the following properties. If ε ε 1 and Σ r a := Σ a + r a log ε σ 1 log ε, then there exist ξ 1,, ξ n in H such that and H\ i Bξ i,ε 2 3 ) where we have written Moreover, Note the pointwise identity r ξ := Ju π n δ ξi Ẇ 1,1 C 1 ε log ε C 1 e C 1Σ r a, i=1 [ r e ε u ) + ju) u ju ξ ) 2 ] C 1 Σξ + ε 1 3 log ε C 1 e C 1Σ r a ), Σ ξ := [Eu) H εξ 1,, ξ n)] +. Σ ξ Σ a + C 1 r a log ε + C 1 ε log ε C 1 e C 1Σ r a. e εu) = 1 2 ju ξ ) 2 + juξ ju) ) u juξ )) + e ε u ) + 1 ju) juξ 2 u ) 2. 19 / 22
Some elements in the proof: approximation including the core Idea : smoothen u ξ not at scale ε but at the larger scale r ξ i.e. at the best known localization scale). It actually suffices to regularize ju ξ ), we call it j u ξ ). Proposition In addition to the statement of the previous Proposition, [ r e ε u ) + ju) ] j uξ H u ) 2 C 1 Σ r a + log log ε ). e εu) = 1 2 j uξ ) 2 + j uξ ju) ) u j uξ )) + e ε u ) + 1 ju) j uξ 2 u ) 2. 20 / 22
Some elements in the proof: get rid of the main translation Recall that and write Then where H εa 1,, a n) := n i=1 [ ra i ) π log ra i ) ε ) + γ + π 3 log2) 2 ) + π a i := r 0 + rb i ), z 0 + zb i ) ), i = 1,, n. log ε log ε j i ] A aj a i ), H εa 1,, a n) = Γ εr 0, n) + W ε,r0 b 1,, b n) + o1) as ε 0, 1) W ε,r0 b 1,, b n) = π i=1 i j Also, expansion of the squares leads directly to n Pa 1,, a n) = πnr0 2 + 2πr rb i ) 0 + π i=1 log ε and therefore H ε where log ε P 2r 0 n rb i ) log ε πr 0 log b i b j. 2) n i=1 rb i ) 2 log ε, ) a 1,, a n) = π 2 nr 0 log ε + Γ εr 0, n) + πr 0 W b 1,, b n) + o1), as ε 0, W b 1,, b n) := i j log b i b j 1 2r 2 0 n rb i ) 2. i=1 21 / 22
Some elements in the proof: get rid of the main translation Proposition For σ sufficiently small, if Σ a + H εa 1,, a n) H εξ,, ξ n) σ log ε, then [ ] 1 Σ ξ 2Σ a + C r a log ε + log ε + log ε Pu) Pa 1,, a n) For a quantity f write δf := f a 1,, a n) f ξ 1,, ξ n). By the triangle inequality we have ) δh ε δ H ε + and also log ε P 2r 0 log ε δp, 2r 0 δp Pu) Pξ 1,, ξ n) + Pu) Pa 1,, a n). In view of the discussion in the previous slide δ H ε log ε P 2r 0 ) C ξ 1 a 1,, ξ n a n) log ε. We also have a good control on P since it involves only the Jacobian : log ε 2r 0 Pu) Pξ 1,, ξ n) C 1 + Σ ξ ) 2 C 1 + Σ a + δh ε) 2 2r 0 log ε 2r 0 log ε and we may then absorb the last term involving δh ε in the left-hand side above. 22 / 22