A generalised Ladyzhenskaya inequality and a coupled parabolic-elliptic problem

Size: px

Start display at page:

Download "A generalised Ladyzhenskaya inequality and a coupled parabolic-elliptic problem"

Stephen Farmer
6 years ago
Views:

1 A generalised Ladyzhenskaya inequality and a coupled parabolic-elliptic problem Dave McCormick joint work with James Robinson and José Rodrigo Mathematics and Statistics Centre for Doctoral Training University of Warwick February 26th, 2013

2 A coupled parabolic-elliptic MHD system We consider the following modified system of equations for magnetohydrodynamics on a bounded domain Ω R 2 : u + p = (B )B tb ε B + (u )B = (B )u, with u = B = 0 and Dirichlet boundary conditions. This is like the standard MHD system, but with the terms tu + (u )u removed. Theorem Given u 0, B 0 L 2 (Ω) with u 0 = B 0 = 0, for any T > 0 there exists a unique weak solution (u, B) with u L (0, T; L 2, ) L 2 (0, T; H 1 ) and B L (0, T; L 2 ) L 2 (0, T; H 1 ). We prove this using both a generalisation of Ladyzhenskaya s inequality, and some elliptic regularity theory for L 1 forcing.

3 Weak solutions of the Navier Stokes equations Consider the Navier Stokes equations on a domain Ω R n, n = 2 or 3: tu + (u )u u + p = 0, with u = 0, and Dirichlet boundary conditions. Theorem (Leray, 1934, and Hopf, 1951) Given u 0 L 2 (Ω) with u 0 = 0, there exists a weak solution u of the Navier Stokes equations satisfying u L (0, T; L 2 ) L 2 (0, T; H 1 ). Moreover, if n = 2, this weak solution is unique. The same is true if Ω = R n, or if Ω = [0, 1] n with periodic boundary conditions.

4 Weak solutions of NSE: existence Let u m be the mth Galerkin approximation: i.e., the solution of tu m + P m [(u m )u m ] u m + p m = 0. Taking the L 2 inner product with u m, we get Fact tu m, u m + (u m )u m, u m u m, u m + p m, u m = 0. }{{} =0 If u = 0, and u, v, w = 0 on Ω, then (u )v, w = (u )w, v. Hence 1 d 2 dt um 2 + u m 2 = 0, so integrating in time yields u m (t) 2 + t 0 u m (s) 2 ds = u m (0) 2 u 0 2. Hence u m are uniformly bounded in L (0, T; L 2 ) L 2 (0, T; H 1 ).

5 Ladyzhenskaya s inequality To get uniform bounds on tu m = u m P m [(u m )u m ], one uses: Ladyzhenskaya s inequality in 2D (1958) u L 4 c u 1/2 L 2 u 1/2 L 2. Ladyzhenskaya s inequality yields a priori bounds on the nonlinear term (u m )u m : (u m )u m φ = (u m )φ u m u m 2 L 4 φ L 2, so (u m )u m H 1 u m 2 L 4 c um L 2 u L 2, and thus (u m )u m L 2 (0, T; H 1 ), and hence tu m L 2 (0, T; H 1 ). Theorem (Aubin, 1963, and Lions, 1969) If u m L 2 (0, T; H 1 ) and tu m L 2 (0, T; H 1 ) uniformly, then a subsequence u m k u L 2 (0, T; L 2 ) (strongly).

6 Magnetic relaxation (Moffatt, 1985) Aim to construct stationary solutions of the Euler equations with non-trivial topology: (u )u + p = 0. Consider the MHD equations with zero magnetic resistivity u t u + (u )u + p = (B )B B t + (u )B = (B )u and assume that smooth solutions exist for all t 0 (open even in 2D). Energy equation 1 d ( u 2 + B 2) + u 2 = 0. 2 dt So B decreases while u = 0. Since the magnetic helicity H = A B is preserved, where B = A and A = 0, B is bounded below: ( 2 c B 4 B 2 A 2 A B) = H 2. So u(t) 0 as t (Nuñez, 2007) and B(t) B with p = (B )B.

7 Magnetic relaxation (Moffatt, 2009) The dynamics are arbitrary, so consider instead the simpler model The new energy equation is u + p = (B )B B t + (u )B = (B )u. 1 d 2 dt B 2 + u 2 = 0. So u 2 0 = u(t) 0 and B(t) B as t. Open: does u(t) 0 as t in this case? To address existence of solutions we make two simplifications: we consider 2D, and regularise the B equation: u + p = (B )B B t ε B + (u )B = (B )u.

8 A priori estimates We consider the 2D system u + p = (B )B u = 0 B t ε B + (u )B = (B )u B = 0. Toy version of 3D Navier Stokes, which in vorticity form (ω = u) is ω t ν ω + (u )ω = (ω )u. Take inner product with u in the first equation, with B in the second equation and add: We get: u 2 = (B )B, u = (B )u, B 1 d 2 dt B 2 + ε B 2 = (B )u, B 1 d 2 dt B 2 + ε B 2 + u 2 = 0. B L (0, T; L 2 ) L 2 (0, T; H 1 ), u L 2 (0, T; H 1 ).

9 A priori estimates What can we say about u? We know that B L (0, T; L 2 ). Note that [(u )v] i = u j j v i = j (u j v i ) =: (u v), since v = 0. We can write the equation for u as Elliptic regularity works for p > 1: u + p = (B )B = (B B). }{{} L 1 u + p = f, f L p = u W 2,p (e.g. Temam, 1979). If we could take p = 1 then we would expect, for RHS f with f L 1, to get u W 1,1 L 2. In fact for RHS f with f L 1 we get u L 2,, where L 2, is the weak-l 2 space.

10 L p, : weak L p spaces For f : R n R define d f (α) = µ{x : f(x) > α}. Note that f p L R p = f(x) p f(x) p α p d f (α). n {x: f(x) >α} For 1 p < set } f L p, = inf {C : d f (α) Cp = sup{γd f (γ) 1/p : γ > 0}. The space L p, (R n ) consists of all those f such that f L p, <. L p L p, x n/p L p, (R n ) but / L p (R n ). if f L p, (R n ) then d f (α) f p L p, α p. α p

11 L p, : weak L p spaces Just as with strong L p spaces, we can interpolate between weak L p spaces: Weak L p interpolation Take p < r < q. If f L p, L q, then f L r and f L r c p,r,q f p(q r)/r(q p) L p, f q(r p)/r(q p) L q,. Recall Young s inequality for convolutions: if 1 p, q, r and 1 p + 1 = 1 q + 1 r then E f L p E L q f L r. There is also a weak form, which requires stronger conditions on p, q, r: Weak form of Young s inequality for convolutions If 1 r < and 1 < p, q <, and = then p q r E f L p, E L q, f L r.

12 Elliptic regularity in L 1 Fundamental solution of Stokes operator on R 2 is E ij (x) = δ ij log x + x ix j x 2, i.e. solution of u + p = f is u = E f. Solution of u + p = f is u = E ( f) = ( E) f. Note that Thus E L 2, and so x k k E ij = δ ij x + δ ikx j + δ jk x i x ix j x k 1 2 x 2 x 4 x. f L 1 = u = E f L 2,. If we consider the problem in a bounded domain we have the same regularity. We replace the fundamental solution E by the Dirichlet Green s function G satisfying G = δ(x y) G Ω = 0. Mitrea & Mitrea (2011) showed that in this case we still have G L 2,. So on our bounded domain, u L (0, T; L 2, ).

13 Estimates on time derivatives: t B L 2 (0, T; H 1 )? Take v H 1 with v H 1 = 1. Then tb, v = ε B (u )B + (B )u, v ε B v + 2 u L 4 B L 4 v L 2. so tb H 1 ε B + 2 u L 4 B L 4. Standard 2D Ladyzhenskaya inequality gives B L 4 c B 1/2 B 1/2 ; but we only have uniform bounds on u in L 2,. If f L 4 c f 1/2 L 2, f 1/2 then which would yield tb H 1 ε B + c u 1/2 L 2, B 1/2 u 1/2 B 1/2 tb L 2 (0, T; H 1 ).

14 Generalised Ladyzhenskaya inequality In 2D, f 2 L 4 c f L 2 f L 2. Proof: (i) write f 2 = 2 f j f dx j and integrate (f 2 ) 2. (ii) use the Sobolev embedding Ḣ 1/2 L 4 and interpolation in Ḣ s : f L 4 c f Ḣ1/2 c f 1/2 L 2 f 1/2 Ḣ 1. In fact, using the theory of interpolation spaces: Since Ḣ 1 BMO in 2D, this yields f L 4 c f 1/2 L 2, f 1/2 BMO. f L 4 c f 1/2 L 2, f 1/2 Ḣ 1. Besides the proof using interpolation spaces, we can also prove this directly using Fourier transforms.

15 Bounded mean oscillation Let f Q := 1 f(x) dx denote the average of f over a cube Q Q Q Rn. Define f BMO := sup Q 1 Q Q f(x) f Q dx, where the supremum is taken over all cubes Q R n. Let BMO(R n ) denote the set of functions f : R n R for which f BMO <. f BMO = 0 = f is constant (almost everywhere). L BMO and f BMO 2 f ; log x BMO but is unbounded. Ḣ n/2 BMO and f BMO c f Ḣn/2 in R n, even though Ḣ n/2 L. W 1,n BMO, by Poincaré s inequality: let B be a ball of radius r, then 1 f(x) f B dx cr ( 1/n Df dx c Df dx) n c f B B r n W 1,n, B B so f BMO c f W 1,n.

16 Interpolation spaces For 0 θ 1 one can define an interpolation space X θ := [X 0, X 1 ] θ in such a way that f Xθ c f 1 θ X 0 f θ X 1. (Note that f X1 c f X 1.) Theorem (Bennett & Sharpley, 1988) L p, = [L 1, BMO] 1 (1/p) for 1 < p < ; so L 2, = [L 1, BMO] 1/2. Write B = [L 1, BMO] 1 and note that f B c f BMO. Reiteration Theorem If A 0 = [X 0, X 1 ] θ0 and A 1 = [X 0, X 1 ] θ1 then for 0 < θ < 1 [A 0, A 1 ] θ = [X 0, X 1 ] (1 θ)θ0 +θθ 1. So L 3, = [L 2,, B] 1/3 and L 6, = [L 2,, B] 2/3, and hence f L 4 c f 1/2 L 3, f 1/2 L 6, c[c f 2/3 L 2, f 1/3 B ]1/2 [c f 1/3 L 2, f 2/3 B ]1/2 = c f 1/2 f 1/2 L 2, B c f 1/2 f 1/2 L 2, BMO.

17 Generalised Ladyzhenskaya inequality: direct method There exists a Schwartz function φ such that if ˆf is supported in B(0, R), f = φ 1/R f, where φ 1/R (x) = R n φ(rx). (Take φ with ˆφ = 1 on B(0, 1); then F [φ 1/R f] = F [φ 1/R ]ˆf = ˆf.) Lemma (Weak-strong Bernstein inequality) Suppose that supp(ˆf) B(0, R). Then for 1 p < q < f L q cr n(1/p 1/q) f L p,. Using the weak form of Young s inequality, for = 1 + 1, q r p f L q, = φ 1/R f L q, c φ 1/R L r f L p,, and noting that φ 1/R L r = cr n(1 1/r), it follows that f L 1, cr n(1/p 1) f L p, and f L 2q, cr n(1/p 1/2q) f L p,. Finally interpolate L q between L 1, and L 2q,.

18 Generalised Ladyzhenskaya inequality: direct method To show write f(x) = k R f L 4 c f 1/2 L 2, f 1/2 Ḣ 1 ˆf(k)e 2πik x dk } {{ } f + k R Now using the weak-strong Bernstein inequality and using the embedding Ḣ 1/2 L 4, Thus f L 4 cr 1/2 f L 2,, f + 2 L 4 c f+ 2 Ḣ 1/2 = c c R ˆf(k)e 2πik x dk. } {{ } f + k R k R c R f 2 Ḣ 1. k ˆf(k) 2 dk k 2 ˆf(k) 2 dk f L 4 cr 1/2 f L 2, + cr 1/2 f Ḣ1, and choosing R = f Ḣ1/ f L 2, yields the inequality.

19 Generalised Gagliardo Nirenberg inequalities It is not hard to generalise this direct proof to prove the following: Theorem Take 1 q < p < and s > n if f L q, Ḣ s then f L p and ( ) 1 1. There exists a constant c 2 p p,q,s such that f L p c p,q,s f θ L q, f 1 θ H s for every f L q, Ḣ s, where 1 p = θ q + (1 θ) ( 1 2 s n ). With a little work, in the case s = n/2 we can generalise this: Theorem Take 1 q < p <. There exists a constant c p,q such that if f L q, BMO then f L p and f L p c p,q f q/p L q, f 1 q/p BMO for every f L q, BMO.

20 Global existence of weak solutions Take B m (0) = P m B(0) and consider the Galerkin approximations: u m + p = P m (B m )B m tb m ε B m + P m (u m )B m = P m (B m )u m. The B m equation is a Lipschitz ODE on a finite-dimensional space, so it has a unique solution. By repeating the a priori estimates on these smooth equations (now rigorous) we can obtain estimates uniform in n: and B m L (0, T; L 2 ) L 2 (0, T; H 1 ), tb m L 2 (0, T; H 1 ) u m L (0, T; L 2, ) L 2 (0, T; H 1 ). By Aubin Lions, a subsequence of the Galerkin approximations B m B strongly in L 2 (0, T; L 2 ). Hence, by elliptic regularity, u m u L 2, B m B m B B L 1 B m B L 2( B m L 2 + B L 2), hence u m u strongly in L 2 (0, T; L 2, ). This is enough to show the nonlinear terms converge weak- in L 2 (0, T; H 1 ), and hence that (u, B) is a weak solution (i.e. a solution with equality in L 2 (0, T; H 1 )).

21 Conclusion Similar arguments to the a priori estimates show uniqueness of weak solutions, and so: Theorem Given u 0, B 0 L 2 (Ω) with u 0 = B 0 = 0, for any T > 0 there exists a unique weak solution (u, B) with u L (0, T; L 2, ) L 2 (0, T; H 1 ) and B L (0, T; L 2 ) L 2 (0, T; H 1 ). What about ε = 0? Try looking at more regular solutions and taking the limit ε 0 to get local existence Assume regularity and show that u(t) 0 as t (Moffatt)?

22 Selected references D. S. McCormick, J. C. Robinson and J. L. Rodrigo, Generalised Gagliardo Nirenberg inequalities using weak Lebesgue spaces and BMO, submitted. D. S. McCormick, J. C. Robinson and J. L. Rodrigo, Existence and uniqueness for a coupled parabolic-elliptic model with applications to magnetic relaxation, in preparation. H. K. Moffatt (1985), Magnetostatic equilibria and analogous Euler flows of arbitrarily complex topology. I. Fundamentals, J. Fluid Mech. 159, H. K. Moffatt (2009), Relaxation routes to steady Euler flows of complex topology, slides of talk given during MIRaW Day, Warwick, June M. Núñez (2007), The limit states of magnetic relaxation, J. Fluid Mech. 580, L. Grafakos (2008), Classical Fourier Analysis and Modern Fourier Analysis. Springer, New York, GTM 249 and 250. C. Bennett and R. Sharpley (1988), Interpolation of Operators. Academic Press, New York.

Decay in Time of Incompressible Flows

J. math. fluid mech. 5 (23) 231 244 1422-6928/3/3231-14 c 23 Birkhäuser Verlag, Basel DOI 1.17/s21-3-79-1 Journal of Mathematical Fluid Mechanics Decay in Time of Incompressible Flows Heinz-Otto Kreiss,