The incompressible Navier-Stokes equations in vacuum
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1 The incompressible, Université Paris-Est Créteil Piotr Bogus law Mucha, Warsaw University Journées Jeunes EDPistes 218, Institut Elie Cartan, Université de Lorraine March 23th, 218
2 Incompressible Navier-Stokes equations in an open subset Ω R d : { ut + div(u u) µ u + P = in R + Ω (NS) : divu = in R + Ω. 1 t Energy balance : 2 u(t) µ u 2 2 dτ = 1 2 u 2 2. Leray solutions : any divergence free u L 2 (Ω) generates at least one global weak solution satisfying the energy inequality. d = 2 : Those solutions are unique. d = 3 : Those solutions are unique in a suitable class of smoother solutions.
3 The inhomogeneous incompressible Navier-Stokes equations read: (ρu) t + div(ρu u) µ u + P = in R + Ω (INS) : divu = in R + Ω ρ t + div(ρu) = in R + Ω. 1 t Energy balance : 2 ρ(t) u(t) µ u 2 2 dτ = 1 2 ρ u 2 2. Conservation of L p norms of the density and of inf ρ(t). Global weak solutions with finite energy for any pair (ρ, u ) such that ρ L (Ω) with ρ, and ρ u L 2 (Ω) with divu =. Even if d = 2, uniqueness in the class of finite energy solutions is a widely open question.
4 Is (INS) a good model for mixture of non-reacting fluids? 1 Can we solve uniquely (INS) if ρ is discontinuous across an interface? 2 Is the solution unique for such a ρ? 3 Can we allow for vacuum regions? 4 Is the regularity of interfaces preserved during the evolution? Having (at least) u L 1 (, T ; L (Ω)) is fundamental both for preserving the density patch structure and for proving the uniqueness. Even for d = 2 and for the heat equation, having just u L 2 (Ω) does not ensure u L 1 (, T ; L (Ω)).
5 The main statement Theorem (Global existence and uniqueness in the 2D torus) Consider any data (ρ, u ) in L (T 2 ) H 1 (T 2 ) with ρ and divu =. Then System (INS) supplemented with data (ρ, u ) admits a unique global solution (ρ, u, P ) that satisfies the energy equality, the conservation of total mass and momentum, ρ L (R + ; L ), u L (R + ; H 1 ), ρut, 2 u, P L 2 (R + ; L 2 ) and also, for all 1 r < 2, 1 m < and T, ( tp ), 2 ( tu) L (, T ; L r ) L 2 (, T ; L m ). Furthermore, we have ρu C(R + ; L 2 ) and ρ C(R + ; L p ) for all p <. Similar statement in 2D bounded domains, if taking u H 1 (Ω). Similar statement in the 3D case, under some smallness condition on u.
6 Definition (of a weak solution) The pair (ρ, u) is a weak solution of (INS) if for all t [, T ) and function φ in Cc ([, T ) Ω; R), we have t t ρ(t), φ(t) ρ, φ ρ, tφ dτ ρu, φ dτ =, (1) t u, φ dτ =, (2) and for all divergence-free function ϕ C c ([, T ) Ω; R d ), t ρ(t)u(t), ϕ(t) ρ u, ϕ ρu, tϕ dτ t t ρu u, ϕ dτ + µ u, ϕ dτ = (3) where, designates the distribution bracket in Ω.
7 Main steps of the proof: 1 Global-in-time estimates for Sobolev regularity of u. 2 Sobolev regularity of u t and time weights. 3 Shift of regularity and integrability : from time to space variable. 4 The existence scheme. 5 Lagrangian coordinates and uniqueness. Assume with no loss of generality that T 2 ρ dx = µ = 1 and T 2 ρ u dx =.
8 Step 1: global-in-time Sobolev estimates Take the L 2 scalar product of ρ(u t + u u) u + P = with u t : 1 d u 2 dx + ρ u t 2 dx 1 ρ u t 2 dx + 1 ρ u u 2 dx. 2 dt T 2 T 2 2 T 2 2 T 2 From u + P = (ρu t + ρu u) and div u =, we have ) 2 u 2 L 2 + P 2 L 2 = ρ( tu+u u) 2 L 2 2ρ (T ρ u t 2 dx+ ρ u u 2 dx 2 T 2 Hence d u 2 dx + 1 ρ u t 2 dx + 1 ( 2 dt T 2 2 T 2 4ρ u 2 L 2 + P 2 ) 3 L 2 ρ u u 2 dx. 2 T 2 Apply Hölder and Gagliardo-Nirenberg inequality, and use ρ ρ := ρ L : T 2 ρ u u 2 dx ρ u 2 L 4 u 2 L 4 Cρ u L 2 u 2 L 2 2 u L ρ 2 u 2 L 2 + C(ρ ) 3 u 2 L 2 u 2 L 2 u 2 L 2. If ρ ρ >, then u 2 L 2 ρ 1 ρu 2 L 2. Combine the basic energy inequality with Gronwall lemma.
9 Step 1: global-in-time Sobolev estimates (continued) Lemma (B. Desjardins, 1997) If T 2 ρ dx = 1 and T 2 ρz dx = then ( T 2 ρz 4 dx ) 1 ( 2 C ρz L 2 z L 2 log 2 1 e + ρ 1 2 L 2 + ρ z 2 L 2 ρz 2 L 2 ) (4) Write ρ u u 2 dx ( ) 1 ρ ρ u 4 2 dx u 2 T 2 T 2 L 4 and use (4) with z = u, energy balance and ab a 2 /2 + b 2 /2 : ρ u u 2 dx 1 T 2 12ρ 2 u 2 L 2 ( +C(ρ ) 2 ρ u 2 L 2 u 2 L 2 u 2 L 2 log e+ ρ 1 2 L 2 + ρ u 2 L 2 ρ u 2 L 2 )
10 Step 1: global-in-time Sobolev estimates (continued) We eventually get d X fx log(e + X), dt with f(t) := C u(t) 2 L 2 for some suitable C = C(ρ, u ) and X(t) := u(t) 2 dx + 1 ( ρ u t ( 2 T 2 2 T 2 4ρ u 2 + P 2)) dx. Hence (e + X(t)) (e + X()) exp( t f(τ) dτ) (e + X()) exp (C ρ u 2 L 2 ) However X < does not imply u L 1 loc (R +; L (T 2 )).
11 Step 2: Sobolev regularity of u t Take the L 2 scalar product of ρ(u t + u u) u + P = with u tt : 1 d ρ u t 2 dx + u t 2 ( dx = ρtut ρ tu u ρu t u ) u t dx. 2 dt T 2 T 2 T 2 For ( ρu t) t= to be defined, we need the compatibility condition u = ρ g + P with g L 2. (5) Condition (5) is not needed if we compensate the singularity at time by some power of t: take the scalar product of ρ(u t + u u) u + P = with tu tt : t ρt u t L 2 + τ u t 2 L 2 dτ h(t), where h is a nondecreasing nonnegative function with h() = (use Step 1 and energy identity).
12 Step 3: Shift of regularity from time to space variable Step 1 gives u L (R + ; L 2 ), u L 2 (R + ; H 1 ), P, ρu t L 2 (R + T 2 ). Step 2 gives ρt u t L loc (R +; L 2 ) and t u t L 2 loc (R +; L 2 ). Step 3: Use Stokes equation: { t u + t P = t ρut t ρu u, div t u =. Steps 1,2 + embedding imply that for all T >, the right-hand side is almost in L 2 (, T ; L ). Hence so do 2 t u and t P. This implies that u L 1 loc (, T ; L ).
13 Step 4: The existence scheme Smooth out ρ and make it positive : ρ n C 1 (T 2 ; ], [) with ρ n ρ in L weak. Solve (INS) with data (ρ n, u ). From prior works (e.g. Ladyzhenskaya & Solonnikov (78) and others) we get a global smooth solution (ρ n, u n ). From the previous steps, we get uniform estimates. By interpolation, we have (u n ) n N is bounded in H 1 8 ([, T ] T 2 ) for all T >. Use compact embeddings to get strong convergence for (u n ) n N and pass to the limit. ρ n ρ in C([, T ]; L 2 ) (similar as for weak solutions).
14 Step 4: Lagrangian coordinates and uniqueness Lagrangian coordinates: ρ(t, y) := ρ(t, x), ū(t, y) := u(t, x) and P (t, y) := P (t, x) with x := X(t, y) where X is the flow of u defined by t Hence DX(t, y) = Id + Dū(τ, y) dτ. t X(t, y) = y + u(τ, X(τ, y)) dτ. The INS equations in Lagrangian coordinates: ρ is time independent. (ū, P ) satisfies { ρ ū t div(a T A ū) + T A P =, div(aū) = T A : ū =, with + ( t k A = (D yx) 1 = ( 1) k Dū(τ, ) dτ). k=
15 Step 4: Lagrangian coordinates and uniqueness (continued) Consider two solutions (ρ 1, u 1, P 1 ) and (ρ 2, u 2, P 2 ) for the same data (ρ, u ). Lagrangian coordinates: (ρ i, u i, P i ) (ρ, ū i, P i ), i = 1, 2. (δu, δp ) := (u 2 u 1, P 2 P 1 ) fulfills { ρ δu t div(a 1 T A 1 δu) + T A 1 δp = div ( (A 2 T A 2 A 1 T A 1 ) ū2 ) + T δa P 2, div(a 1 δu) = div(δa ū 2 ) = T δa : ū 2. Decompose δu into δu = w + z, where is a suitable solution to: and z fulfills z() = and div(a 1 w) = div(δa ū 2 ) = T δa : ū 2, w t= =, ρ z t div(a 1 T A 1 z) + T A 1 δp = div ( (A 2 T A 2 A 1 T A 1 ) ū2 ) + T δa P 2 ρ w t + div(a 1 T A 1 z), div(a 1 z) =.
16 Step 4: Lagrangian coordinates and uniqueness (continued) We have div(a 1 w) = g where A 1 Id, det A 1 = 1 and g = divr. This may be recast as Φ(v) = v with After some computation: T Φ(v) := 1 div((id A 1 )v + R). T w 2 L 2 dt c(t ) δu 2 L 2 dt with lim c(t ) =. T (6) From ρ z t div(a T 1 A 1 z) + T A 1 δp = and div(a 1 z) =, we get 1 d ρ z 2 dx + T A 1 z 2 dx =. 2 dt T 2 T 2 After some computation: sup T T ρ z(t) 2 L 2 + z 2 L 2 dt c(t ) δu 2 L 2 dt. (7) t [,T ] Putting (6) and (7) together yields δu whence w, ρ z and z. Finally δu.
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