On 2 d incompressible Euler equations with partial damping.

Transcription

1 On 2 d incompressible Euler equations with partial damping. Wenqing Hu 1. (Joint work with Tarek Elgindi 2 and Vladimir Šverák 3.) 1. Department of Mathematics and Statistics, Missouri S&T. 2. Department of Mathematics, Princeton University. 3. School of Mathematics, University of Minnesota, Twin Cities.

2 The 2 d Navier Stokes equation on a torus. u t + u u + p ν u = f, divu = 0, u t=0 = u 0. u = u(x, t) = (u 1 (x, t), u 2 (x, t)), p = p(x, t), x T 2 = R 2 /2πZ 2.

3 The 2 d Navier Stokes equation on a torus. Solution u(x, t) can be written as a Fourier series u(x, t) = 1 2π k Z 2 û(k, t)e ikx. The Laplace operator is a damping term if we view it from the Fourier space ( u)(k) = k 2 û(k) for all k Z 2.

4 Modification of the 2 d Navier Stokes equation on a torus. Introduce the partial damping operator Y : (Ŷu)(k) = { k 2û(k), k K, 0, k K. We will assume that K is symmetric (i.e. invariant under k k) to keep the solution real valued. Ultimate goal is to study u t + u u + p νyu = f, divu = 0, u t=0 = u 0.

5 Background. Canonical (conjectural) picture due to Kraichnan 4 : the energy and enstrophy cascades which spread the excitations to other Fourier modes through the nonlinearity. Nonlinear interactions should tend to distribute energy uniformly between all degrees of freedom, and hence a system for which some of the degrees of freedom are forced while some other degrees of freedom are damped should still reach some kind of dynamical equilibrium. 5 Kraichnan s downward cascade of energy : if in the case K = {k 0, k 0 } where k 0 is one of the lowest non trivial frequencies, the solution will presumably not stay bounded generically, even if the forcing acts far away in the Fourier space. 4. Kraichnan, R.H., Inertial ranges in two dimensional turbulence, Physics of Fluids, 10(7), pp , At least if there is enough interaction between the damped and forced parts of the system.

6 Background. The simplest way to establish dynamical equilibrium is to assume that f is random, and validate the fact that the system admits a unique invariant measure. 6 Asymptotic Strong Feller property. It does not work in the case of partial damping. 6. Hairer, M., Mattingly, J.C., Ergodicity of the 2D Navier Stokes equations with degenerate stochastic forcing, Annals of Mathematics, (2) 164 (2006), no. 3., pp

7 Our problem. We study the modest case u t + u u + p νyu = 0, divu = 0, u t=0 = u 0. Case 1 : K is finite (remove damping from finitely many modes) ; Case 2 : K is co finite (i.e. Z 2 \K is finite, leave damping at finitely many modes) ; Case 3 : Both K and Z 2 \K are infinite.

8 General Result. Write Y + Z =. Regularization. (Ẑu)(k) = { k 2û(k), k K, 0, k K.

9 General Result. Theorem. Let K be any symmetric subset of Z 2. For each divergence free vector field u 0 L 2 on the torus T 2 the initial value problem (u ε ) t + u ε u ε + p ε νyu ε εzu ε = 0, divu ε = 0, u ε t=0 = u 0 has a unique solution u ε C([0, ), L 2 x) L 2 t Ḣ2 x. The solution u ε is smooth in T 2 (0, ) and satisfies the energy identity T 2 1 t 2 u ε(x, t) 2 dx + ( ν(yu ε )u ε ε(zu ε )u ε )dxdt 0 T 1 2 = T 2 2 u 0(x) 2 dx for each t 0.

10 General Result. Theorem. (continued) Moreover, if ω 0 = curlu 0 L 2, then u ε L 2 t Ḣ2 x and satisfies T 2 1 t 2 ω2 ε(x, t)dx + ( ν(yu ε )u ε ε(zu ε )u ε )dxdt 0 T 1 2 = T 2 2 ω2 0(x)dx for each t 0.

11 General Result. Weak Solution. u C([0, ), L 2 ) L t Ḣ 1 x {v, Yv L 2 t L 2 x}. Corollary. For any symmetric K Z 2 and any initial datum u 0 H 1 our problem has at least one weak solution. Leray Hopf argument.

12 Case 1. Finitely many undamped frequencies. Let κ = max k K k. v 2 (Yv, v) + κ 2 v 2 L 2 x. Uniform control of the solutions u ε in L t L 2 x L 2 t Ḣ1 x. Standard embedding argument gives existence and uniqueness of solutions to the initial value problem with u 0 L 2 in the class C([0, ), L 2 ) L 2 t Ḣ1 x.

13 Case 1. Finitely many undamped frequencies. Long time behavior : E K,E,I = v : T 2 R 2, Krasovskii LaSalle principle. v solves the 2d incompressible Euler equation and, in addition ˆv(k) = 0 for each k K, T 2 v 2 = 2E, T 2 curlv 2 = I. Question : What is the structure of the solution of 2 d Euler on T 2 that is supported on finitely many Fourier modes? Answer : Independent of time and is supported either on a line passing through the origin or on a circle centered at the origin. (Will come back to this later.) So in the case finitely many undamped modes solutions converge to steady state!.

14 Case 2. Finitely many damped frequencies. Vorticity formulation : ω t + (u )ω = νy ω. Bound Y ω L c K ω L 2 due to the finiteness of Z 2 \K. ω(t) L 2 is not increasing. Thus d dt ω(t) L c K ω 0 L 2. We arrive at the a priori estimate ω(t) L ω 0 L + c K ω 0 L 2t, t > 0. If ω 0 L, uniqueness follows from Yudovich theory.

15 The structure of Euler solution supported on finitely many Fourier modes. Question : What is the structure of the solution of 2 d Euler on T 2 that is supported on finitely many Fourier modes? Answer : Independent of time and is supported either on a line passing through the origin or on a circle centered at the origin. It must be steady state!

16 The structure of Euler solution supported on finitely many Fourier modes. 2 d Euler can be written in Fourier components as ( 1 (k 1 l 2 k 2 l 1 ) d 1 ω(m, t) = dt 4π k+l=m k 2 1 l 2 Suppose the support is a finite set of Fourier modes S. ) ω(k, t) ω(l, t).

17 The structure of Euler solution supported on finitely many Fourier modes. An (unordered) pair {k, l} of two distinct points k = (k 1, k 2 ), l = (l 1, l 2 ) Z 2 \{(0, 0)} is called degenerate if either k, l lie on the same circle centered at the origin (i.e. k1 2 + k2 2 = l l 2 2 ), or k, l lie on the same line passing through the origin (i. e. k 1 l 2 k 2 l 1 = 0). In the former case we call the pair to be c degenerate, and in the latter case we call the pair to be l degenerate. A pair which is not degenerate is called non degenerate. If the set S is not degenerate, then there exists a non zero element m Z 2 \S such that m = k + l for exactly one non degenerate (unordered) pair {k, l} S. S wakes up those modes that are not in S.

18 Figure 1: Proof that S is symmetric w.r.t. O (if not a line).

19 Figure 2: Proof that S is degenerate.

20 Thank you for your attention!