Global nonlinear stability of steady solutions of the 3-D incompressible Euler equations with helical symmetry and with no swirl
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1 Around Vortices: from Cont. to Quantum Mech. Global nonlinear stability of steady solutions of the 3-D incompressible Euler equations with helical symmetry and with no swirl Maicon José Benvenutti (UNICAMP) Joint work with Helena Judith Nussenzveig Lopes (UFRJ) and Milton da Costa Lopes Filho (UFRJ) IMPA - Rio de Janeiro 17th March 2014
2 Introduction We extend Burton s results and also the Wan and Pulvirenti results to the 3-D helical Euler equations with no swirl: BURTON G., Global nonlinear stability for steady ideal fluid flow in bounded planar domains, Arc. Rat. M. Anal. (2005). WAN Y., PULVIRENTI M., Nonlinear stability of circular vortex patches, C., Math. Phys. (1985). Burton proved nonlinear L p -stability of steady flows of an ideal fluid in a bounded, simply connected, planar region, that are strict maximisers of kinetic energy on an isovortical surface. Wan and Pulvirenti proved nonlinear stability of circular vortex patches in circular domains.
3 Introduction We extend Burton s results and also the Wan and Pulvirenti results to the 3-D helical Euler equations with no swirl: BURTON G., Global nonlinear stability for steady ideal fluid flow in bounded planar domains, Arc. Rat. M. Anal. (2005). WAN Y., PULVIRENTI M., Nonlinear stability of circular vortex patches, C., Math. Phys. (1985). Burton proved nonlinear L p -stability of steady flows of an ideal fluid in a bounded, simply connected, planar region, that are strict maximisers of kinetic energy on an isovortical surface. Wan and Pulvirenti proved nonlinear stability of circular vortex patches in circular domains.
4 Main Ideas 1 o point Reduction by Symmetry: Following Ettinger and Titi, we use the helical symmetry and the additional geometric constraint (no swirl) to setup the reduction of the 3-D vorticity equation: ETTINGER B., TITI E., Global existence and uniqueness of weak solutions of three-dimensional Euler equations with helical symmetry in the absence of vorticity stretching, Siam J. Math. Anal. 41, (2009). 2 o point We adjust the techniques developed by Burton and also by Wan and Pulvirenti for this reduced vorticity equation.
5 Main Ideas 1 o point Reduction by Symmetry: Following Ettinger and Titi, we use the helical symmetry and the additional geometric constraint (no swirl) to setup the reduction of the 3-D vorticity equation: ETTINGER B., TITI E., Global existence and uniqueness of weak solutions of three-dimensional Euler equations with helical symmetry in the absence of vorticity stretching, Siam J. Math. Anal. 41, (2009). 2 o point We adjust the techniques developed by Burton and also by Wan and Pulvirenti for this reduced vorticity equation.
6 Helical Symmetry Notation: x R 3 we define: ρ R R ρ (x) := ρ R S ρ (x) := R ρ (x) + efiniton: A domain Ω R 3 is helical if cos ρ sen ρ 0 sen ρ cos ρ Ω = ρ R S ρ (Ω), ρ R. 0 0 ρ. x, A vector field u : Ω R 3 is helical if R ρ (u(x)) = u(s ρ (x)), x Ω, ρ R. A helical vector field is with no swirl if u(x).(x 2, x 1, 1) = 0, x Ω.
7 Helical Symmetry
8 Euler Equations Ω R 3 a helical domain: t u + [u. ]u = p Ω [0, ), div u = 0 Ω [0, ), u.η = 0 Ω [0, ). If (u, p) is a smooth solution, then (R ρ u(t, S ρ (x)), p(t, S ρ (x))) as well. DUTRIFOY, Existence globale en temps de solutions hélicoïdales des équation d Eler, C.R.A. Paris (1999). rot u = (x 2, x 1, 1)ω 3. t ω 3 + (u. )ω 3 = 0.
9 Euler Equations Ω R 3 a helical domain: t u + [u. ]u = p Ω [0, ), div u = 0 Ω [0, ), u.η = 0 Ω [0, ). If (u, p) is a smooth solution, then (R ρ u(t, S ρ (x)), p(t, S ρ (x))) as well. DUTRIFOY, Existence globale en temps de solutions hélicoïdales des équation d Eler, C.R.A. Paris (1999). rot u = (x 2, x 1, 1)ω 3. t ω 3 + (u. )ω 3 = 0.
10 Euler Equations Ω R 3 a helical domain: t u + [u. ]u = p Ω [0, ), div u = 0 Ω [0, ), u.η = 0 Ω [0, ). If (u, p) is a smooth solution, then (R ρ u(t, S ρ (x)), p(t, S ρ (x))) as well. DUTRIFOY, Existence globale en temps de solutions hélicoïdales des équation d Eler, C.R.A. Paris (1999). rot u = (x 2, x 1, 1)ω 3. t ω 3 + (u. )ω 3 = 0.
11 Euler Equations Ω R 3 a helical domain: t u + [u. ]u = p Ω [0, ), div u = 0 Ω [0, ), u.η = 0 Ω [0, ). If (u, p) is a smooth solution, then (R ρ u(t, S ρ (x)), p(t, S ρ (x))) as well. DUTRIFOY, Existence globale en temps de solutions hélicoïdales des équation d Eler, C.R.A. Paris (1999). rot u = (x 2, x 1, 1)ω 3. t ω 3 + (u. )ω 3 = 0.
12 Reduced Equations Ω o := {(x 1, x 2 ); (x 1, x 2, 0) Ω}, ξ(x 1, x 2 ) := ω 3 (x 1, x 2, 0), ( ) ( 1 + x 2 v(x 1, x 2 ) := 2 x 1 x 2 u 1 x 1 x x1 2 u 2 (x1,x 2,0) ). t ξ + (v. )ξ = 0 Ω o [0, ), div v = 0 Ω o [0, ), v.η o = 0 Ω o [0, ).
13 Reduced Equations Ω o := {(x 1, x 2 ); (x 1, x 2, 0) Ω}, ξ(x 1, x 2 ) := ω 3 (x 1, x 2, 0), ( ) ( 1 + x 2 v(x 1, x 2 ) := 2 x 1 x 2 u 1 x 1 x x1 2 u 2 (x1,x 2,0) ). t ξ + (v. )ξ = 0 Ω o [0, ), div v = 0 Ω o [0, ), v.η o = 0 Ω o [0, ).
14 Biot-Savart Law If Ω o is simply connected, we have the Biot- Savart Law: v = G[ξ]. { L[ G[ξ]] = ξ in Ωo, G[ξ] Ωo = 0. L[. ] = div [ {K(x 1, x 2 )[. ]}, x 2 K(x 1, x 2 ) = 2 x 1 x 2 1+x1 2+x 2 2 x 1 x x1 2 ].
15 Biot-Savart Law If Ω o is simply connected, we have the Biot- Savart Law: v = G[ξ]. { L[ G[ξ]] = ξ in Ωo, G[ξ] Ωo = 0. L[. ] = div [ {K(x 1, x 2 )[. ]}, x 2 K(x 1, x 2 ) = 2 x 1 x 2 1+x1 2+x 2 2 x 1 x x1 2 ].
16 Weak Solutions Definition Let p > 4 3, Ω o bounded, smooth and simply connected and ξ 0 L p (Ω o ). Then ξ L Loc ([0, ), Lp (Ω o )) is a weak solution with initial data ξ 0 if ξ(t, x)φ t (t, x)dxdt + ξ(t, x)v(t, x) φ(t, x)dxdt 0 Ω o 0 Ω o = ξ 0 (x)φ(0, x)dx, Ω o φ C0 ([0, ) Ω o).
17 Steady Solutions Theorem Let p > 4 3, Ω o = B R 2(0, R) and g L p (0, R 2 ) monotonic. Then ξ(x) = g( x 2 ) is a steady solution. Proof: Rmk: (x 2, x 1, 1)g( x 2 ) is a steady helical vorticity. Rmk 2: (x 2, x 1, 1)X (0,1) ( x ) is the helical vortex patch. ( ) ( ) L[ G[ξ]] = r G[ξ] r + 1 r G[ξ] 1+r 2 r. 1+r 2 L[ G[ξ]] = ξ and G[ξ] Ωo = 0 = div{vξ} = 0. v = G[ξ] = x (1 + x 2 ) 2 x 2 x 2 0 g(s)ds.
18 Steady Solutions Theorem Let p > 4 3, Ω o = B R 2(0, R) and g L p (0, R 2 ) monotonic. Then ξ(x) = g( x 2 ) is a steady solution. Proof: Rmk: (x 2, x 1, 1)g( x 2 ) is a steady helical vorticity. Rmk 2: (x 2, x 1, 1)X (0,1) ( x ) is the helical vortex patch. ( ) ( ) L[ G[ξ]] = r G[ξ] r + 1 r G[ξ] 1+r 2 r. 1+r 2 L[ G[ξ]] = ξ and G[ξ] Ωo = 0 = div{vξ} = 0. v = G[ξ] = x (1 + x 2 ) 2 x 2 x 2 0 g(s)ds.
19 Steady Solutions Theorem Let p > 4 3, Ω o = B R 2(0, R) and g L p (0, R 2 ) monotonic. Then ξ(x) = g( x 2 ) is a steady solution. Proof: Rmk: (x 2, x 1, 1)g( x 2 ) is a steady helical vorticity. Rmk 2: (x 2, x 1, 1)X (0,1) ( x ) is the helical vortex patch. ( ) ( ) L[ G[ξ]] = r G[ξ] r + 1 r G[ξ] 1+r 2 r. 1+r 2 L[ G[ξ]] = ξ and G[ξ] Ωo = 0 = div{vξ} = 0. v = G[ξ] = x (1 + x 2 ) 2 x 2 x 2 0 g(s)ds.
20 Steady Solutions Theorem Let p > 4 3, Ω o = B R 2(0, R) and g L p (0, R 2 ) monotonic. Then ξ(x) = g( x 2 ) is a steady solution. Proof: Rmk: (x 2, x 1, 1)g( x 2 ) is a steady helical vorticity. Rmk 2: (x 2, x 1, 1)X (0,1) ( x ) is the helical vortex patch. ( ) ( ) L[ G[ξ]] = r G[ξ] r + 1 r G[ξ] 1+r 2 r. 1+r 2 L[ G[ξ]] = ξ and G[ξ] Ωo = 0 = div{vξ} = 0. v = G[ξ] = x (1 + x 2 ) 2 x 2 x 2 0 g(s)ds.
21 Steady Solutions Theorem Let p > 4 3, Ω o = B R 2(0, R) and g L p (0, R 2 ) monotonic. Then ξ(x) = g( x 2 ) is a steady solution. Proof: Rmk: (x 2, x 1, 1)g( x 2 ) is a steady helical vorticity. Rmk 2: (x 2, x 1, 1)X (0,1) ( x ) is the helical vortex patch. ( ) ( ) L[ G[ξ]] = r G[ξ] r + 1 r G[ξ] 1+r 2 r. 1+r 2 L[ G[ξ]] = ξ and G[ξ] Ωo = 0 = div{vξ} = 0. v = G[ξ] = x (1 + x 2 ) 2 x 2 x 2 0 g(s)ds.
22 Steady Solutions Theorem Let p > 4 3, Ω o = B R 2(0, R) and g L p (0, R 2 ) monotonic. Then ξ(x) = g( x 2 ) is a steady solution. Proof: Rmk: (x 2, x 1, 1)g( x 2 ) is a steady helical vorticity. Rmk 2: (x 2, x 1, 1)X (0,1) ( x ) is the helical vortex patch. ( ) ( ) L[ G[ξ]] = r G[ξ] r + 1 r G[ξ] 1+r 2 r. 1+r 2 L[ G[ξ]] = ξ and G[ξ] Ωo = 0 = div{vξ} = 0. v = G[ξ] = x (1 + x 2 ) 2 x 2 x 2 0 g(s)ds.
23 Steady Solutions Theorem Let p > 4 3, Ω o = B R 2(0, R) and g L p (0, R 2 ) monotonic. Then ξ(x) = g( x 2 ) is a steady solution. Proof: Rmk: (x 2, x 1, 1)g( x 2 ) is a steady helical vorticity. Rmk 2: (x 2, x 1, 1)X (0,1) ( x ) is the helical vortex patch. ( ) ( ) L[ G[ξ]] = r G[ξ] r + 1 r G[ξ] 1+r 2 r. 1+r 2 L[ G[ξ]] = ξ and G[ξ] Ωo = 0 = div{vξ} = 0. v = G[ξ] = x (1 + x 2 ) 2 x 2 x 2 0 g(s)ds.
24 Steady Solutions Definition If 4 3 < p, we define E : Lp (Ω o ) R by E[ξ] := 1 ξ(x)g[ξ](x)dx. 2 Ω o Definition Let f L p (Ω o ). Then g L p (Ω o ) is a rearrangement of f if f 1 [β, ) = g 1 [β, ), β R.
25 Steady Solutions Definition If 4 3 < p, we define E : Lp (Ω o ) R by E[ξ] := 1 ξ(x)g[ξ](x)dx. 2 Ω o Definition Let f L p (Ω o ). Then g L p (Ω o ) is a rearrangement of f if f 1 [β, ) = g 1 [β, ), β R.
26 Steady Solutions Theorem Let 4 3 < p <, f Lp (Ω o ) and R the set of the rearrangements of f with the L p (Ω o ) topology. Then E : R R attains its supremum. Furthermore, if ξ R is a local maximiser, then ξ is a steady solution.
27 Proof: G : L p L p is compact strictly positive symmetric linear operator. E[ξ] := 1 2 Ω o ξ(x)g[ξ](x)dx attains its supremum on the rearrangements of f and if ξ is a maximiser then ξ = φ(g[ξ]) for some monotonic function φ : R R. BURTON G., Rearrangements of functions, maximization of convex funcionals, and vortex rings, Math Anal (1987). div{vξ} = div{v.φ(g[ξ])} = (v. )φ(g[ξ]) = φ (G[ξ])(v. )G[ξ] = φ (G[ξ])( G[ξ]. )G[ξ] = 0.
28 Proof: G : L p L p is compact strictly positive symmetric linear operator. E[ξ] := 1 2 Ω o ξ(x)g[ξ](x)dx attains its supremum on the rearrangements of f and if ξ is a maximiser then ξ = φ(g[ξ]) for some monotonic function φ : R R. BURTON G., Rearrangements of functions, maximization of convex funcionals, and vortex rings, Math Anal (1987). div{vξ} = div{v.φ(g[ξ])} = (v. )φ(g[ξ]) = φ (G[ξ])(v. )G[ξ] = φ (G[ξ])( G[ξ]. )G[ξ] = 0.
29 Existence of Weak Solutions Theorem Let be p > 4 3, Ω o R 2 bounded, smooth and simply connected and ξ 0 L p (Ω o ). Then there exists ξ L ([0, ), L p (Ω o )) weak solution with initial data ξ 0.
30 Stability for Steady Solutions Theorem Let 2 p <, Ω o R 2 bounded, smooth and simply connected and f L p (Ω o ). Let ξ 0 the strict maximiser of E : R R. Then ɛ > 0, δ > 0, such that if ξ(0) L p (Ω o ), ξ(0) ξ 0 L p (Ω 0 ) < δ and ξ L ([0, ), L p (Ω o )) is a solution with initial data ξ(0) given by the previous theorem, then ξ(t) ξ 0 L p (Ω 0 ) < ɛ, t 0.
31 Proof: Let ɛ > 0, 0 < δ < ɛ 2 and ξ L ([0, ), L p (Ω o )) a weak solution such that ξ(0) ξ 0 L p (Ω o) < δ. Let τ such that τ + div (vτ) t = 0, τ(0) = ξ 0. (τ ξ) + div (v(τ ξ)) t = 0, τ(0) ξ(0) = ξ 0 ξ(0). ξ(t) ξ 0 L p (Ω o) ξ(t) τ(t) L p (Ω o) + τ(t) ξ 0 L p (Ω o) ɛ 2 + τ(t) ξ 0 L p (Ω o).
32 Proof: Let ɛ > 0, 0 < δ < ɛ 2 and ξ L ([0, ), L p (Ω o )) a weak solution such that ξ(0) ξ 0 L p (Ω o) < δ. Let τ such that τ + div (vτ) t = 0, τ(0) = ξ 0. (τ ξ) + div (v(τ ξ)) t = 0, τ(0) ξ(0) = ξ 0 ξ(0). ξ(t) ξ 0 L p (Ω o) ξ(t) τ(t) L p (Ω o) + τ(t) ξ 0 L p (Ω o) ɛ 2 + τ(t) ξ 0 L p (Ω o).
33 Proof: Let ɛ > 0, 0 < δ < ɛ 2 and ξ L ([0, ), L p (Ω o )) a weak solution such that ξ(0) ξ 0 L p (Ω o) < δ. Let τ such that τ + div (vτ) t = 0, τ(0) = ξ 0. (τ ξ) + div (v(τ ξ)) t = 0, τ(0) ξ(0) = ξ 0 ξ(0). ξ(t) ξ 0 L p (Ω o) ξ(t) τ(t) L p (Ω o) + τ(t) ξ 0 L p (Ω o) ɛ 2 + τ(t) ξ 0 L p (Ω o).
34 Proof: For M = R Lp w, β > 0 such that { sup E[ζ] ; ζ M [B L p (Ω o)(ξ 0, ɛ ] } c 2 ) < E[ξ 0 ] β. ξ(t) is E-conservative and δ small enough = E[τ(t)] E[ξ 0 ] E[τ(t)] E[ξ(t)] + E[ξ(0)] E[ξ 0 ] β 2 + β 2 = β, t 0. E[τ(t)] E[ξ 0 ] β, t 0. = τ(t) / [ B L p (Ω o)(ξ 0, ɛ 2 )] c, t 0. ξ(t) ξ 0 L p (Ω o) ɛ 2 + τ(t) ξ 0 L p (Ω o) ɛ 2 + ɛ = ɛ, t 0. 2
35 Proof: For M = R Lp w, β > 0 such that { sup E[ζ] ; ζ M [B L p (Ω o)(ξ 0, ɛ ] } c 2 ) < E[ξ 0 ] β. ξ(t) is E-conservative and δ small enough = E[τ(t)] E[ξ 0 ] E[τ(t)] E[ξ(t)] + E[ξ(0)] E[ξ 0 ] β 2 + β 2 = β, t 0. E[τ(t)] E[ξ 0 ] β, t 0. = τ(t) / [ B L p (Ω o)(ξ 0, ɛ 2 )] c, t 0. ξ(t) ξ 0 L p (Ω o) ɛ 2 + τ(t) ξ 0 L p (Ω o) ɛ 2 + ɛ = ɛ, t 0. 2
36 Proof: For M = R Lp w, β > 0 such that { sup E[ζ] ; ζ M [B L p (Ω o)(ξ 0, ɛ ] } c 2 ) < E[ξ 0 ] β. ξ(t) is E-conservative and δ small enough = E[τ(t)] E[ξ 0 ] E[τ(t)] E[ξ(t)] + E[ξ(0)] E[ξ 0 ] β 2 + β 2 = β, t 0. E[τ(t)] E[ξ 0 ] β, t 0. = τ(t) / [ B L p (Ω o)(ξ 0, ɛ 2 )] c, t 0. ξ(t) ξ 0 L p (Ω o) ɛ 2 + τ(t) ξ 0 L p (Ω o) ɛ 2 + ɛ = ɛ, t 0. 2
37 Proof: For M = R Lp w, β > 0 such that { sup E[ζ] ; ζ M [B L p (Ω o)(ξ 0, ɛ ] } c 2 ) < E[ξ 0 ] β. ξ(t) is E-conservative and δ small enough = E[τ(t)] E[ξ 0 ] E[τ(t)] E[ξ(t)] + E[ξ(0)] E[ξ 0 ] β 2 + β 2 = β, t 0. E[τ(t)] E[ξ 0 ] β, t 0. = τ(t) / [ B L p (Ω o)(ξ 0, ɛ 2 )] c, t 0. ξ(t) ξ 0 L p (Ω o) ɛ 2 + τ(t) ξ 0 L p (Ω o) ɛ 2 + ɛ = ɛ, t 0. 2
38 Proof: For M = R Lp w, β > 0 such that { sup E[ζ] ; ζ M [B L p (Ω o)(ξ 0, ɛ ] } c 2 ) < E[ξ 0 ] β. ξ(t) is E-conservative and δ small enough = E[τ(t)] E[ξ 0 ] E[τ(t)] E[ξ(t)] + E[ξ(0)] E[ξ 0 ] β 2 + β 2 = β, t 0. E[τ(t)] E[ξ 0 ] β, t 0. = τ(t) / [ B L p (Ω o)(ξ 0, ɛ 2 )] c, t 0. ξ(t) ξ 0 L p (Ω o) ɛ 2 + τ(t) ξ 0 L p (Ω o) ɛ 2 + ɛ = ɛ, t 0. 2
39 Other Results Theorem Let 2 p <, Ω o = B R 2(0, R), g L p (0, R 2 ) monotonic and ξ 0 (x) = g( x 2 ). Then ɛ > 0, δ > 0, such that if ξ(0) L p (Ω o ), ξ 0 ξ(0) L p (Ω o) δ and ξ is a solution with initial data ξ(0), then ξ 0 ξ(t) L p (Ω o) ɛ, t 0. T [ξ] = x 2 ξ(x)dx. Ω o
40 Other Results Theorem Let 2 p <, Ω o = B R 2(0, R), g L p (0, R 2 ) monotonic and ξ 0 (x) = g( x 2 ). Then ɛ > 0, δ > 0, such that if ξ(0) L p (Ω o ), ξ 0 ξ(0) L p (Ω o) δ and ξ is a solution with initial data ξ(0), then ξ 0 ξ(t) L p (Ω o) ɛ, t 0. T [ξ] = x 2 ξ(x)dx. Ω o
41 Other Results Theorem Let Ω o = B R 2(0, R). Then ɛ > 0, δ > 0 such that if 0 Ω o, λ R, ξ(0) = λx 0, ξ(0) X BR 2 (0,1) L 1 (Ω o) < δ and ξ L ([0, ), L (Ω o )) the solution with initial data ξ(0), then ξ(t) X BR 2(0,1) L 1 (Ω o) < ɛ, t 0.
42 Thank You Very Much!
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