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No blow-up of the 3D incompressible Euler equations
Léo Agélas
To cite this version:
No blow-up of the 3D incompressible Euler equations
L´eo Ag´elas ∗
July 27, 2011
Abstract
One of the most challenging questions in fluid dynamics is whether the incompressible Euler equations can develop a finite-time singularity from smooth initial data. In this paper, we found that local geometry regularity of vorticity leads to a strong dynamic depletion of the nonlinear vortex stretching, thus avoiding finite-time singularity formation. Then, we prove the existence and uniqueness of global strong solutions in C([0, +∞[; Wr,q(R3))3 with q > 1, r > 1 +3
q, of the Euler equations as soon as the initial data u0 ∈ Wr,q(R3)3. This result gives a positive answer to the open problem about existence and smoothness of solutions of Euler equations.
1
Introduction
The Euler equations describe the motion of a fluid in R3. These equations have enriched many branches of mathematics, were involved in many area outside mathematical activity from weather prediction to exploding supernova (see [CONST]) and present important open physical and mathematical problems (see [CONST]). For Euler equations, non uniqueness of weak solutions has been demonstrated in [SCH], [SHN]. However, these solutions have not been obtained from a Leray solution of Navier-Stokes equations in the limit as the viscosity tends to zero (zero-viscosity limit of Navier-Stokes equations). Zero-viscosity limits of Navier-Stokes equations have only been shown to exist and to give Euler solutions in some more generalized sense known as measure-valued Euler solutions (see [DM]). Another notion of weak solutions to the Euler equations was introduced in [PLLIONS] called dissipative solutions and it was proven that they coincide with classical Euler solutions, when those exist. Recent studies indicate that the local geometric regularity of vortex lines can lead to dynamic depletion of vortex stretching [[CONST2],[CFM], [DHY],[DHY2]]. In particular, the recent results obtained in [DHY],[DHY2] show that geometric regularity of vortex lines, even in an extremely localized region containing the maximum vorticity can lead to depletion of nonlinear stretching, thus avoiding finite time singularity formation of the 3D Euler equations.
The blow-up problem for the Euler equations is a major open problem of PDE theory, of far greater physical importance than the blow-up problem for the Navier- Stokes equation, which of course is known to non-specialists because it is a Clay Millennium Problem.
It is known that if there are no singularities in the solution of the Euler equations with initial data u0 on the time interval [0, T ], then there can be no singularities in the Navier-Stokes
∗
solution with the same initial data and small enough viscosity (see [CONST3]). The regularity for large enough viscosities is also known. Unfortunately, there is a gap between the two ranges of viscosities, and it is not clear how to close it.
The blow-up of smooth solutions to the Euler equations which is one of the most challenging mathematical problems in nonlinear PDE is controlled by the time integral of the maximum magnitude of the vorticity (see [BKM], [KP]). And this criterion have been refined in [CHAE] by replacing the L∞norm of the vorticity by weaker norms close to the L∞norm, more precisely, the blow-up of smooth solutions is only controlled by the time integral of the Besov space ˙B∞,10 norm of two components of the vorticity.
Let us now introduce the Euler equations in R3 given by
( ∂u
∂t + (u · ∇)u + ∇p = 0, ∇ · u = 0,
(1)
in which u = u(x, t) = (u1(x, t), u2(x, t), u3(x, t)) ∈ R3, p = p(x, t) ∈ R denote respectively the unknown velocity field, the scalar pressure function of the fluid at the point (x, t) ∈ R3× [0, ∞[, with initial conditions,
u(x, 0) = u0(x) for a.e x ∈ R3. (2) In this paper, we prove that the maximum magnitude of the vorticity grows at most one ex-ponential in time. For this, we consider the particle-trajectory mapping X(, t) : α ∈ R3 7−→ X(α, t) ∈ R3, where X(α, t) is the location at time t of a fluid particle initially placed at the point α at time t = 0 and such that the fluid velocity u is tangent to the particle trajectory. After, we fix the time at s > 0 and we study the evolution in time from the integral of the initial fluid vorticity over an initial unit ball of fluid particles, B, to the integral of the vorticity at times s over the domain X(B, s), with the ball B such that X(B, s) contains a point where the maximum magnitude of the vorticity is reached at time s. The equation (E) expressing this evo-lution is obtained from the vorticity equation and by the fact that vorticity is stretched at time t ∈ [0, s] by ∇αX(α, t) along particle trajectories (ω(X(α, t), t) = ∇αX(α, t)ω0(α)). After some manipulations, The equation (E) makes appear at the right hand side of the equality, the main obstruction in establishing the regularity of Euler solution, the stretching term which presents itself under the form I =
Z Ω
u(y, t)ft(y)dy such that kftkL∞ ≤ kω0kL∞,
Z Ω
ft(y)dy = 0 being the consequence that ω0 is a divergence-free vector field and Ω = X(B, t). By interpreting the term I as the integral of the product between an element in BM Or(Ω), the class of functions of bounded mean oscillation only on cubes inside Ω (see Section 3), u(·, t) and an element in H1z(Ω), a hardy space on Ω (see Section 3), ft, we are then able to use all the theory about Hardy spaces on domains that has been developed in recent years (see [JSW], [MIYA], [CKS] and [CDS]). After, using a duality result between BM Or(Ω) and H1z(Ω), we obtain a fine estimate of I given by the bound J = ku(t)kBM Or(Ω)kftkH1z(Ω). Thanks to Lemmata 3.1 and 3.2, the term J is
bounded by Ckω(t)kL∞kω0kL∞, where C > 0 is an absolute constant. Furthermore, on the left
hand side of equality of (E), we have the derivative in time of the term Z
X(B,t)
equation, we infer the following inequality,
kω(s)kL∞ ≤ kω0kL∞+ C
Z s 0 kω(t)k
L∞kω0kL∞ ds.
Thanks to Gronwall Lemma, we conclude that,
kω(s)kL∞ ≤ kω0kL∞exp(Ckω0kL∞ s),
which proves that there does not exist finite time T > 0 for which Z T
0 kω(s)kL
∞ ds = +∞.
From [BKM], we know that solution of Euler equation blow up at time T if and only if the time integral of the maximum magnitude of the vorticity diverges at time T ,
Z T
0 kω(t)k
L∞dt = +∞,
then we deduce that the incompressible Euler equations cannot develop a finite-time singularity from smooth initial data.
Notations : For any function ϕ defined on R3× [0, +∞[, for all t ≥ 0, we denote by ϕ(t) the function defined on R3 by x 7−→ ϕ(x, t). For any vector x = (x1, x
2, x3) ∈ R3, we denote by |x| the norm defined by |x| =
v u u t 3 X i=1
x2i. For all vectors v = (vi)1≤i≤3 and w = (wj)1≤j≤3,
we denote by v ⊗ w the matrix with ij−elements given by (viwj)1≤i,j≤3. For any bounded subset Ω ⊂ R3, we denote by |Ω| the measure of Ω and by diam(Ω) the diameter of Ω de-fined by diam(Ω) = sup
x,y∈Ω|x − y|. For all Ω ⊂ R
3, we denote by χ
Ω, the function such that χΩ(x) = 1, ∀x ∈ Ω and χΩ(x) = 0, ∀x 6∈ Ω. For any x0 ∈ R3 and r > 0, we denote by B(x0, r), the ball of R3 of center x
0 and radius r. For the sake of simplicity, for any q ≥ 1, we denote by Lq (resp Wr,q) indifferently the Sobolev space Lq(R3)3 (resp Wr,q(R3)3) and Lq(R3)3×3 (resp Wr,q(R3)3×3). We denote by BC the class of bounded and continuous functions and by BCm the class of bounded and m times continuously derivable functions. We denote by div the
dif-ferential operator given by, div = 3 X i=1 ∂ ∂xi.
2
Preliminary
Definition 2.1 If ∆ denotes the Laplace operator, the Bessel potential space Lps , 1 < p < ∞, −∞ < α < ∞ can be defined as the space of functions (or distributions) f such that (I − ∆)α2f
belongs to the Lebesgue space Lp, normed by the corresponding Lebesgue norm. The operator (I − ∆)α2 = G−α, which for α > 0 is most easily defined by means of the Fourier transform. It
corresponds, in fact, to multiplication of the Fourier transform f of by (1 + |ξ|)α2.
For α > 0 the elements of Lpα are themselves Lp-functions, which can be represented as Bessel potentials of Lp-functions. In fact, the function (1 + |ξ|2)−α2 is then the Fourier transform of an integrable function, the Bessel kernel Gα(x). In other words, f ∈ Lpα, 1 < p < ∞, α > 0 if and only if there is a g ∈ Lp such that f (x) = R Gα(x − y)g(y)dy, where the integral is taken over all of R3 with respect to the Lebesgue measure.
In what follows, we denote Wr,q as the space Lq r.
Definition 2.2 We denote by P the matrix Leray’s projection operator with components,
Pi,j = δi,j − ∂ ∂xi ∂ ∂xj∆ −1 = δi,j − RjRk, (3)
where Rj are the Riesz transform given by Rj = ∂x∂
j(−∆)
−1 2 = 1
4π xj
|x|4⋆ (see [STE] for more details), ∆−1 is the inverse of Laplace operator given by ∆−1= − 1
4π|x|⋆ , with ⋆ the convolution operator.
The operator P, which acts on vector-valued functions, is a projection : P2 = P, annihilates gradients and maps into solenoidal (divergence-free) vectors; it is a bounded operator on (vector-valued) Lp, 1 < p < ∞ and commutes with translation. We can notice that the operator P can be written under the form,
P= I − ∇∆−1div, (4)
which yields to Helmholtz decomposition, indeed for all v ∈ Lq(R3)3, 1 < q < ∞, v = Pv + ∇q, with div Pv = 0,
q = ∆−1div v. (5)
3
Hardy space and BMO space in bounded domain in R
3We recall the definition and some of the main properties of Hardy spaces Hp(R3) introduced in [SW] ( for more details on these spaces, see [FS], see also [STE2]).
Definition 3.1 Let 0 < p < ∞ and let Ψ ∈ S(R3) the Schwartz class, satisfying Z
R3
Ψ(x) dx = 1. A tempered distribution f belongs to the Hardy space Hp(R3) if,
f∗(x) = sup
t>0|(Ψt⋆ f )(x)| ∈ L
p(R3), (6)
where Ψt(x) = t−3Ψ(t−1x).
The (quasi)-norm of Hp(R3) is defined, up to equivalence, by,
In what follows, in Definition 3.1, we choose Ψ ∈ C∞
0 (B(0, 1)) such that Z
R3
Ψ(x)dx = 1. Furthermore, an L1-function f on R3 belongs to H1(R3) if and only if its Riesz transforms Rjf all belong to L1(R3) and
kfkH1(R3) ∼=kfkL1(R3)+
3 X j=1
kRjf kL1(R3) (equivalent norms). (8)
Notice that all function f ∈ H1(R3) satisfy Z
R3
f (x) dx = 0.
In [CKS], two Hardy spaces are defined on domains Ω of R3, one which is reasonably speaking the largest, and the other which in a sense is the smallest. The largest, H1
r(Ω), arises by restricting to Ω arbitrary elements of H1(R3). The other, H1z(Ω), arises by restricting to Ω elements of H1(R3) which are zero outside Ω. Norms on these spaces are defined as follows,
kfkH1
r(Ω)= inf kF kH1(R3)
the infinimum being taken over all functions F ∈ H1(R3) such that F |Ω = f, kfkH1
z(Ω) = kF kH 1
(R3
)
where F is the zero extension of f to R3. From [CHA] (see also [AR], Theorem 1 (b2) combined with Theorem 5 (e)), the dual of H1z(Ω) is BM Or(Ω), a space of locally integrable functions with kfkBM Or(Ω)= sup Q⊂Ω 1 |Q| Z Q|f(x) − fQ| dx < ∞, (9) where fQ = 1 |Q| Z Q
f (y)dy, and the supremum is taken over all cubes Q in the domain Ω.
Lemma 3.1 Let Ω ⊂ R3a closed bounded domain and f ∈ H1(R3)∩L∞(R3) such that supp f ⊂ Ω, then, we have,
kfkH1(R3)≤ 2|Ω| kfkL∞(R3)+ kfkL1(R3).
Proof. The continuous embedding of H1(R3) ֒→ L1(R3) give us that f ∈ L1(R3). Since B(0, ǫ) → {0} as ǫ → 0, then, we can choose 0 < δ < 1 such that,
|B(0, δ) + Ω | ≤ 2|Ω|. (10) From Definition 3.1, we have,
where, I = Z R3 sup 0<t<δ|(Ψ t⋆ f )(x)| dx J = Z R3 sup t≥δ|(Ψ t⋆ f )(x)| dx.
Let us estimate the term I. For all t > 0, we have, supp Ψt⋆ f ⊂ supp f + supp Ψt. Since for all t > 0, supp Ψt ⊂ B(0, t), then, for all 0 < t < δ, we have supp Ψt ⊂ B(0, δ) and therefore supp Ψt⋆ f ⊂ Ω + B(0, δ). This implies that supp h ⊂ Ω + B(0, δ), where h is the function defined by,
h(x) = sup
0<t<δ|(Ψt⋆ f )(x)|, ∀x ∈ R
3. (12)
Moreover, for all 0 < t < δ, we have kΨt⋆ f kL∞ ≤ kΨtkL1 kfkL∞ = kfkL∞, which implies
khkL∞ ≤ kfkL∞. Then, we obtain, I = Z R3 h(x) dx = Z Ω+B(0,δ) h(x) dx ≤ khkL∞(|Ω + B(0, δ)|) ≤ 2kfkL∞|Ω|, (13)
where we have used (10).
Let us now estimate the term J. For this, we take N ≥ 1 and we bound the term J by two terms J1 and J2, J ≤ J1+ J2, (14) where, J1 = Z R3 sup δ≤t≤N|(Ψt⋆ f )(x)| dx J2 = Z R3 sup t>N|(Ψ t⋆ f )(x)| dx.
The term J1is estimate as follows, since Ψ ∈ C0∞(B(0, 1)), for all t > 0, we have Ψt∈ C0∞(B(0, t)) and then for all δ ≤ t ≤ N, Ψt ∈ C0∞(B(0, N )). Therefore, for all δ ≤ t ≤ N, we have Ψt⋆ f ∈ C∞
0 (Ω + B(0, N )).
Moreover, by recalling Ψt= t−3Ψ(t−1•), for all δ ≤ t ≤ N, for all x ∈ R3, we have, ∂Ψt(x)
∂t = −3t
−4Ψ(t−1
x) − t−5(x · ∇Ψ)(t−1x)
and since suppΨ ⊂ B(0, 1), the support of the function x 7−→ (x · ∇Ψ)(t−1x) is B(0, t), then, we deduce for all δ ≤ t ≤ N,
Then, we obtain, sup δ≤t≤N ∂Ψt ∂t L∞ ≤ 3δ−4kΨkW1,∞,
which implies for all δ ≤ t0< t1≤ N,
kΨt1 − Ψt0kL∞ ≤ sup δ≤t≤N ∂Ψt ∂t L∞ |t1− t0| ≤ 3δ−4kΨkW1,∞ |t1− t0|.
Then, we deduce that for all δ ≤ t < t′ ≤ N,
kΨt⋆ f − Ψt′⋆ f kL∞ = k(Ψt− Ψt′) ⋆ f kL∞
≤ kΨt− Ψt′kL∞kfkL1
≤ 3δ−4kΨkW1,∞kfkL1 |t − t′|.
Let ε > 0 and let us introduce a subdivision of [δ, N ] defined by a finite non-decreasing sequence of reals (τi)0≤i≤n such that τ0= δ, τn= N and for all 1 ≤ i ≤ n, |τi− τi−1| ≤ ε.
Since for all δ ≤ t ≤ N, we have Ψt⋆ f ∈ C0∞(Ω + B(0, N )), then, we get, J1 = Z Ω+B(0,N) sup δ≤t≤N|(Ψ t⋆ f )(x)| dx.
For all t ∈ [δ, N], there exists it ∈ [0, n] ∩ N such that |t − τit| ≤ ε, then we deduce for all
t ∈ [δ, N] and x ∈ Ω + B(0, N), |(Ψt⋆ f )(x)| ≤ |(Ψt⋆ f )(x) − (Ψτit ⋆ f )(x)| + |(Ψτit ⋆ f )(x)| ≤ kΨt⋆ f − Ψτit ⋆ f kL∞+ |(Ψτ it ⋆ f )(x)| ≤ 3δ−4kΨkW1,∞kfkL1|t − τi t| + |(Ψτit ⋆ f )(x)| ≤ 3δ−4kΨkW1,∞kfkL1 ε + sup 0≤i≤n|(Ψτi⋆ f )(x)|. Therefore for all x ∈ Ω + B(0, N),
sup δ≤t≤N|(Ψ t⋆ f )(x)| ≤ 3δ−4kΨkW1,∞kfkL1 ε + sup 0≤i≤n|(Ψτi⋆ f )(x)|. Then, we obtain, J1 ≤ 3|Ω + B(0, N)|δ−4kΨkW1,∞kfkL1 ε + sup 0≤i≤n Z Ω+B(0,N)|(Ψ τi⋆ f )(x)| dx
Then, we deduce,
J1≤ 3|Ω + B(0, N)|δ−4kΨkW1,∞kfkL1 ε + kfkL1, (15)
which is valid for all ε > 0, then taking the limit in (15) as ε → 0, we get,
J1 ≤ kfkL1. (16)
It remains to estimate the term J2, in fact, we have J2 → 0 as N → ∞. Indeed, for all t > N, we have, kΨt⋆ f kL∞ ≤ kΨtkL∞kfkL1 = 1 t3kΨkL∞kfkL1 ≤ N13kΨkL∞kfkL1. Then, we infer, sup t>NkΨt⋆ f kL ∞ ≤ 1 N3kΨkL∞kfkL1. (17) We introduce the sequence of functions (fN)N >1 defined for all x ∈ R3 by,
fN(x) = sup t>N|(Ψ
t⋆ f )(x)|. Then, thanks to (17), we have for all x ∈ R3,
fN(x) → 0 as N → ∞. (18)
Since, for all x ∈ R3, 0 ≤ fN(x) ≤ f∗(x) and f∗ ∈ L1(R3) due to the fact that f ∈ H1(R3), therefore, thanks to Lebesgue’s dominated convergence Theorem, we deduce,
Z R3
fN(x) dx → 0 as N → ∞, which means,
J2 → 0 as N → ∞. (19)
Then, thanks to (11), (13), (14), (16) and (19), we deduce, kfkH1
(R3
)≤ 2|Ω|kfkL∞+ kfkL1,
which conclude the proof.
Littlewood-Paley decomposition : To prove Lemma 3.2, we need to introduce the usual dyadic unity partition of Littlewood-Paley decomposition. To this end, we take an arbitrary radial function Φ in the Schwartz class S(R3) whose Fourier transform ˆΦ is such that,
supp ˆΦ ⊂ {ξ ∈ R3, |ξ| ≤ 1} and ˆΦ(ξ) ≥ 1
2 for |ξ| ≤ 5 6.
and define ϕj(x) = 23jϕ(2jx) so that ˆϕj(ξ) = ˆϕ(2−jξ) for j ∈ Z. We may assume, ∀ξ ∈ R3, ˆΦ(ξ) +X j≥0 ˆ ϕj(ξ) = 1 and X j∈Z ˆ ϕj(ξ) = 1.
For any f ∈ S′, we denote by,
∆−1f = F−1( ˆΦ ˆf ) = Φ ⋆ f,
∆jf = F−1( ˆϕjf ) = ϕˆ j ⋆ f for all j ≥ 0.
Lemma 3.2 Let u ∈ PWr,s(R3)3 with 1 < s < ∞, r > 1 + 3s, for all x ∈ R3 and y ∈ R3, there exists an absolute constant C > 0 such that,
|u(x) − u(y)| ≤ C kωkL∞|x − y|,
where ω = ∇ × u is the vorticity of u. Proof. Thanks to the Sobolev embedding,
Wm,s(R3) ֒→ BC(R3) for all m > 3 s,
we have u ∈ BC1(R3)3. By the Biot-Savart law (see Proposition 1.3.1 in [CHE]), we have for all x ∈ R3, u(x) = −4π1 Z R3 y |y|3 × ω(x + y) dy. (20) Denoting K the kernel being in the Biot-Savart law (20), we re-write (20) as u = K ⋆ ω.
We have also,
∂u
∂xj = Rj(R × ω) for j = 1, 2, 3, (21) where R = (R1, R2, R3) and Rj are the Riesz transform.
Thanks to Littlewood-Paley decomposition, we have,
u = ∞ X j=−1
∆ju. (22)
Thanks to Bernstein inequalities (see Chapter 3 [LEM], see also [MEY]), there exists a constant C > 0 such that for all j ∈ N,
k∆jukL∞ ≤ C2−jk∆j∇ukL∞. (23)
Thanks to (21) and using the same arguments as the proof of (iii) in [MIY], pages 439-440, we get that there exists a constant C2 > 0 such that for all j ∈ N,
k∆j∇ukL∞ = k∆j(R ⊗ (R × ω))kL∞
≤ C2k∆jωkL∞.
(24)
We borrow some arguments used in [VIS] to obtain our proof. Let x ∈ R3, y ∈ R3 and 3 < p < ∞,
First case : if |x − y| < 1, then, thanks to (22), (24) and (25), we get that for all N ≥ −1, |u(x) − u(y)| ≤ N X j=−1 |∆ju(x) − ∆ju(y)| + 2 ∞ X j=N +1 k∆jukL∞ ≤ N X j=−1 k∆j∇ukL∞ |x − y| + 2C1 ∞ X j=N +1 2−jk∆jωkL∞ ≤ k∆−1∇ukL∞ |x − y| + C2 N X j=0 k∆jωkL∞ |x − y| + 2C1 ∞ X j=N +1 2−jk∆jωkL∞.
Thanks to Young inequality (see Theorem IV.30 in [BRE]), there exists a real cp > 0 depending only on p, Φ such that,
k∆−1∇ukL∞ = kΦ ⋆ ∇ukL∞
≤ cpk∇ukLp. (26)
Thanks to Theorem 3.1.1 in [CHE], there exists a real dp > 0 depending only p such that,
k∇ukLp ≤ dpkωkLp. (27)
Using (26) and (27), there exists a real Cp > 0 depending only on p, Φ such that,
k∆−1∇ukL∞ ≤ CpkωkLp. (28)
Let 3 < q < ∞ and q′ > 1 such that 1
q′+1q = 1, Thanks to Young inequality (see Theorem IV.30
in [BRE]), we have for j ≥ 0,
k∆jωkL∞ = kϕj⋆ ωkL∞ ≤ kϕjkLq′kωkLq = 2 3j qkϕkL q′kωkLq (29)
Thanks to (28) and (29), we deduce,
|u(x) − u(y)| ≤ CpkωkLp |x − y| + C2kϕkLq′kωkLq |x − y| N X j=0 23jq + 2C1kϕkLq′kωkLq ∞ X j=N +1 2−j(1− 3 q) = CpkωkLp |x − y| + kϕk Lq′kωkLq(C2|x − y| 2 3 q(N +1)+ 2C 12−(1− 3 q)(N +1))
We choose N + 1 = [− log2|x − y|] and since q > 3, we deduce that there exists a constant C3> 0 such that,
|u(x) − u(y)| ≤ CpkωkLp |x − y| + C3 kϕkLq′kωkLq |x − y|1− 3
Then, taking the limit as q → ∞ in (30), we obtain,
|u(x) − u(y)| ≤ CpkωkLp |x − y| + C3 kϕkL1kωkL∞ |x − y|. (31)
Second case : if |x − y| ≥ 1, thanks to thanks to (22) and (25), we have, |u(x) − u(y)| ≤ |∆−1u(x) − ∆−1u(y)| + 2
∞ X j≥0 k∆jukL∞ ≤ k∆−1∇ukL∞ |x − y| + 2C1 ∞ X j≥0 2−jk∆jωkL∞ ≤ k∆−1∇ukL∞ |x − y| + 2C1 ∞ X j≥0 kωkL∞kϕjkL12−j ≤ CpkωkLp |x − y| + 4C1kϕkL1kωkL∞,
by using (28) and the fact that kϕjkL1 = kϕkL1 . Since |x − y| ≥ 1, we infer,
|u(x) − u(y)| ≤ CpkωkLp |x − y| + 4C1kϕkL1kωkL∞|x − y|. (32)
Then, using (31) and (32), we deduce that there exists a real Cp > 0 depending only on p, Φ and an constant C4 > 0 such that for all x ∈ R3 and y ∈ R3, we have,
|u(x) − u(y)| ≤ CpkωkLp |x − y| + C4kϕkL1kωkL∞|x − y|. (33)
Now, we use the scaling of Inequality (33), for this, we consider the function uλ = u(λ•) with λ > 0 a non-negative real. Since ∇ · u = 0, then we have also ∇ · uλ = 0 and ωλ the vorticity of uλ is given by ωλ = λω(λ•).
Let x ∈ R3 and y ∈ R3. We set x λ =
x
λ and yλ = y
λ, then using (33) with uλ instead of u, we obtain,
|uλ(xλ) − uλ(yλ)| ≤ CpkωλkLp |xλ− yλ| + C4kϕkL1kωλkL∞|xλ− yλ|. (34)
Using the expression of uλ, ωλ, xλ and yλ, from (34), we deduce,
|u(x) − u(y)| ≤ Cpλ−
3 pkωk
Lp |x − y| + C4kϕkL1kωkL∞|x − y|. (35)
Then, taking the limit as λ → ∞ in (35), we deduce,
|u(x) − u(y)| ≤ C4kϕkL1kωkL∞|x − y|. (36)
Since also ϕ is fixed, we conclude the proof.
4
Global regularity of solution of Euler equations
Using P the matrix Leray operator, Euler equations (1)-(2) can be re-written as follows, ∂u
with initial condition,
u(0) = u0. (38)
For u solution of (37)-(38), ω = ∇×u, the vorticity of u, formally satisfies the vorticity equation, ∂ω
∂t + (u · ∇)ω − (ω · ∇)u = 0, (39) with initial conditions,
ω(0) = ω0, (40)
where ω0= ∇ × u0 is the vorticity of u0. Let us give now the proof of our Theorem.
Theorem 4.1 Let u0 ∈ PWr,q(R3)3 with 1 < q < ∞, r > 2 + 3q, then there exists an unique strong solution u ∈ C([0, +∞[, PWr,q(R3))3 ∩ C1([0, +∞[, PWr−1,q(R3))3 to Euler equations (37)-(38) and moreover, there exists a constant C > 0 such that for all t ≥ 0,
kω(t)kL∞ ≤ kω0kL∞exp(Ckω0kL∞t).
Proof. Since u0 ∈ PWr,q(R3)3 and thanks to Theorem I in [KP2], Theorem 3.5 and 4.7 in [KP], Theorem 1 in [BB] or the results obtained in [BKM], we deduce that there exists a maximal time of existence strictly positive T∗ > 0 such that there exists an unique strong solution u ∈ C([0, T∗[, PWr,q(R3))3 to the Euler equations (37)-(38) and if T∗
< +∞, then we get,
Z T∗ 0 kω(t)k
L∞dt = +∞. (41)
Since u ∈ C([0, T∗[, PWr,q(R3))3, the operator P a bounded operator in Lq(R3) and thanks to Lemma X4 in [KP], from Euler equation (37), we deduce that u ∈ C1([0, T∗[, PWr−1,q(R3))3. From Helmholtz decomposition (5), we retrieve the pressure p from the velocity u with the formula,
p = −∆−1div((u · ∇)u). Let us assume that T∗
< +∞. We introduce T a non-negative real such that 0 < T < T∗. Thanks to Sobolev embedding,
Wm,q(R3) ֒→ BC(R3) for all m > 3
q, (42)
and since u ∈ C([0, T ], Wr,q(R3))3 ∩ C1([0, T ], PWr−1,q(R3))3 with r > 1 + 3
q, we deduce that u ∈ BC1([0, T ] × R3)3.
Equation (1) represents the Eulerian description of the flow. The Lagrangian formulation of Euler equations (1) describes the flow in term of a volume preserving diffeomorphism, the time dependent map X : R37−→ R3,
These maps represent marked fluid particle trajectories, α is the label of the particle, which can be seen as the location of the particle at time t = 0. The fact that the particle travel with velocity u is expressed in the system of ordinary differential equations,
∂
∂tX(α, t) = u(X(α, t), t), X(α, 0) = α. (43) The Lagrangian formulation of Euler equations and incompressibility condition given by (1) are respectively the Newton’s second law ∂t2X + ∇p(X, t) = 0 and det(∇αX) = 1 (see Proposition 1.4 in [MB]), where det(∇αX) denotes the determinant of ∇αX the jacobian matrix of X. The map X defined by Equation (43) is well a volume preserving C1-diffeomorphism from R3 on itself, indeed, since u ∈ BC1([0, T ] × R3)3, then thanks to Cauchy-Lipschitz Theorem, we deduce that the differential equation (43) admits a unique solution X ∈ C1(R3 × [0, T ])3 and for all t ∈ [0, T ], the map X(t) is a C1−diffeomorphism from R3 on itself (see Lemma 2.3.2 in [LERN] in particular the end of its proof, see also Theorem 2.10 combined with Theorem 2.17 in [TES]).
Thanks to Proposition 1.4 in [MB], we have for all t ∈ [0, T ],
the mapping X(t) : α 7−→ X(α, t) is volume preserving ,
and det(∇αX(α, t)) = 1. (44) Gathering these results, we obtain the desired result, for any t ∈ [0, T ], the mapping
X(t) is a volume preserving C1− diffeomorphism from R3 on itself. (45) Let us give an estimate about the length of any segment transformed during the time by the application X.
Let e0 = (1, 1, 1), from (43), we have for all t ∈ [0, T ] and α ∈ R3, ∇αX(α, t) = e0+
Z t
0 ∇u(X(α, τ), τ)∇
αX(α, τ ) dτ. (46)
Then, we deduce from (46), for all t ∈ [0, T ], |∇αX(α, t)| ≤√3 +
Z t
0 k∇u(τ)kL
∞ |∇αX(α, τ )| dτ.
Then, thanks to Gronwall Lemma, we obtain for all t ∈ [0, T ] and α ∈ R3, |∇αX(α, t)| ≤ √ 3 exp Z t 0 k∇u(τ)k L∞ dτ .
Therefore, we get for all t ∈ [0, T ], k∇αX(t)kL∞ ≤ √ 3 exp Z t 0 k∇u(τ)k L∞ dτ . (47)
Since, we have ω ∈ C([0, T ], Wr−1,q(R3))3 with r−1 > 1+3q, then we get ω ∈ C([0, T ]; BC1(R3))3 and also for all t ∈ [0, T ], ω(x, t) → 0 as |x| → +∞. For all t ∈ [0, T ], the application x 7−→ |ω(x, t)| is a continuous function from R3 to [0, +∞[ and since |ω(x, t)| → 0 as |x| → +∞, therefore we deduce that the function x 7−→ |ω(x, t)| reaches its maximum value for some x(t) ∈ R3.
Then, we have for all t ∈ [0, T ],
|ω(x(t), t)| = kω(t)kL∞
(R3
)3. (48)
For any s ∈]0, T ], due to (45), there exists an unique α(s) ∈ R3 such that the flow X starting from α(s) (which means X(α(s), 0) = α(s)) pass through the particle x(s) at time s, that means X(α(s), s) = x(s).
It is well known that for 3D Euler flows (see Proposition 1.8 in [MB] or Proposition page 24 in [CM]), we have for all t ∈ [0, T ] and α ∈ R3,
ω(X(α, t), t) = ∇αX(α, t)ω0(α). (49) Let us fix s ∈]0, T ] and x(s) such that (48) holds. Let R a non-negative real such that 0 < R ≤ 1 and let B(α(s), R) be the ball of center α(s) with radius R.
For any t ∈ [0, T ], we introduce the subset of R3, V
R,s(t) = X(B(α(s), R), t), thanks to (44), we have the conservation of volume,
|VR,s(t)| = |B(α(s), R)|. (50) For all 0 ≤ t ≤ s, by using the change of variables y = X(α, t) with α ∈ B(α(s), R) and since det(∇αX(α, t)) = 1, we deduce, Z VR,s(t) ω(y, t) dy = Z B(α(s),R) ω(X(α, t), t) dα. (51)
Therefore, thanks to (51), (43) and (39) we deduce that for all 0 ≤ t ≤ s, d dt Z VR,s(t) ω(y, t) dy = Z B(α(s),R) d dtω(X(α, t), t) dα = Z B(α(s),R)((ω · ∇)u)(X(α, t), t) dα. (52)
Since (ω · ∇)u = ∇u ω and thanks to (49), we deduce,
((ω · ∇)u)(X(α, t), t) = ∇u (X(α, t), t) ∇αX(α, t)ω0(α)
= ∇(u(X(α, t), t)) ω0(α). (53) Then, using (52), (53) and an integration by parts, we obtain for all 0 ≤ t ≤ s,
where n(α) is the unit normal outward to B(α(s), R) in α ∈ ∂B(α(s), R).
We can notice that the unit normal outward to any ball of center α(s) in a point α on its boundary, can be expressed as follows,
n(α) = α − α(s)
|α − α(s)|. (55)
For α = α(s), we set,
n(α) = (1, 0, 0). (56)
In what follows, n is considered as a function from R3 to R3 defined by (55) and (56). Since ∇ · ω0 = 0, from (54), we obtain for all 0 ≤ t ≤ s,
d dt Z VR,s(t) ω(y, t) dy = Z ∂B(α(s),R) u(X(α, t), t) ω0(α) · n(α) dγ(α). (57) We integrate Equation (57) over t ∈ [0, s] and we get,
Z VR,s(s) ω(y, s) dy = Z VR,s(0) ω(y, 0) dy + Z s 0 Z ∂B(α(s),R) u(X(α, t), t) ω0(α) · n(α) dγ(α) dt. (58) We notice that VR,s(0) = B(α(s), R) and ω(y, 0) = ω0(y) for all y ∈ R3, then we obtain,
Z VR,s(s) ω(y, s) dy = Z B(α(s),R) ω0(y) dy + Z s 0 Z ∂B(α(s),R) u(X(α, t), t) ω0(α) · n(α) dγ(α) dt. (59) We integrate Inequality (59) over R ∈]0, 1] and we obtain,
Z 1 0 Z VR,s(s) ω(y, s) dy dR = Z 1 0 Z B(α(s),R) ω0(y) dy dR + Z s 0 Z 1 0 Z ∂B(α(s),R) u(X(α, t), t) ω0(α) · n(α) dγ(α) dR dt. (60)
For all t ∈ [0, s], we have, Z ∂B(α(s),R) u(X(α, t), t) ω0(α) · n(α) dγ(α) = ∂ ∂R Z B(α(s),R) u(X(α, t), t) ω0(α) · n(α) dα ! . (61) Then, we deduce, Z 1 0 Z ∂B(α(s),R) u(X(α, t), t) ω0(α) · n(α) dγ(α) dR = Z B(α(s),1) u(X(α, t), t) ω0(α) · n(α) dα. (62) Owing to (62), from (60), for all t ∈ [0, s], we obtain,
Therefore, from (63), we deduce, Z 1 0 Z VR,s(s) ω(y, s) dy dR ≤ kω0kL∞ Z 1 0 |B(α(s), R)| dR + Z s 0 Z B(α(s),1) u(X(α, t), t) ω0(α) · n(α) dγ(α) dt. (64)
On one hand, we have, Z 1 0 Z VR,s(s) ω(y, s) dy dR = Z 1 0 Z VR,s(s) ω(X(α(s), s), s) dy dR + Z 1 0 Z VR,s(s) ω(y, s) − ω(X(α(s), s), s) dy dR ≥ Z 1 0 Z VR,s(s) ω(X(α(s), s), s) dy dR − Z 1 0 Z VR,s(s) ω(y, s) − ω(X(α(s), s), s) dy dR ≥ |ω(X(α(s), s), s)| Z 1 0 |V R,s(s)| dR − Z 1 0 Z VR,s(s) |ω(y, s) − ω(X(α(s), s), s)| dy dR. (65) On the other hand, we have for all y ∈ VR,s(s),
|ω(X(α(s), s), s) − ω(y, s)| ≤ k∇ω(s)kL∞ diam(VR,s(s))
and also |ω(X(α(s), s), s)| = |ω(x(s), s)| = kω(s)kL∞, then from (65), we deduce,
Z 1 0 Z VR,s(s) ω(y, s) dy dR ≥ kω(s)kL∞ Z 1 0 |V R,s(s)| dR −k∇ω(s)kL∞ Z 1 0 diam(VR,s(s))|VR,s(s)| dR. (66)
From (64) and thanks to (66), we deduce,
Let us give now an estimate of diam(VR,s(s)). We recall VR,s(s) = X(B(α(s), R), s), then we can write diam(VR,s(s)) as follows,
diam(VR,s(s)) = sup α,β∈B(α(s),R)|X(α, s) − X(β, s)|. (69) Thanks to (47), we deduce, diam(VR,s(s)) ≤ 2√3R exp Z s 0 k∇u(τ)k L∞ dτ . (70) Therefore, we deduce, Z 1 0 diam(VR,s(s))4π 3 R 3 dR ≤ √8π 3exp Z s 0 k∇u(τ)k L∞ dτ Z 1 0 R4dR = 8π 5√3exp Z s 0 k∇u(τ)kL ∞ dτ . (71)
Thanks to (71), from (68), we obtain, π 3 kω(s)kL∞ ≤ π 3kω0kL∞+ 8π 5√3exp Z s 0 k∇u(τ)k L∞ dτ k∇ω(s)kL∞ + Z s 0 Z B(α(s),1) u(X(α, t), t) ω0(α) · n(α) dα dt. (72)
For all t ∈ [0, T ], we introduce the set Vs(t) = X(B(α(s), 1), t). Then, thanks to (44), we have, |Vs(t)| = |B(α(s), 1)| = 4π
3 . (73)
We introduce the inverse A(y, t) = X−1(y, t) called the back to label map which satisfies A(X(α, t), t) = α and X(A(y, t), t) = y.
Thanks to (44), for all t ∈ [0, s] , using the change of variable y = X(α, t) with α ∈ B(α(s), 1), we deduce,
Z Vs(t)
u(y, t) ω0(A(y, t)) · n(A(y, t)) dy = Z
B(α(s),1)
u(X(α, t), t) ω0(α) · n(α) dα. (74) Then, from (72), using (74), we deduce,
π 3 kω(s)kL∞ ≤ π 3kω0kL∞+ 8π 5√3exp Z s 0 k∇u(τ)k L∞ dτ k∇ω(s)kL∞ + Z s 0 Z Vs(t)
u(y, t) ω0(A(y, t)) · n(A(y, t)) dy dt. (75)
Let us prove now, that there exists a constant C > 0 such that for all t ∈ [0, s] Z Vs(t)
u(y, t) ω0(A(y, t)) · n(A(y, t)) dy ≤ C kω(t)kL∞kω0kL∞.
ft(y) = ω0(A(y, t)) · n(A(y, t))χVs(t)(y) for all y ∈ R3. (76) Notice that, Z R3 ft(y) dy = 0, (77) indeed, Z R3 ft(y) dy = Z Vs(t) ω0(A(y, t)) · n(A(y, t)) dy = Z B(α(s),1) ω0(α) · n(α) dα,
by using the change of variable y = X(α, t) with α ∈ B(α(s), 1). Since, we have, Z B(α(s),1) ω0(α) · n(α) dα = Z 1 0 ∂ ∂r Z B(α(s),r) ω0(α) · n(α) dα ! dr = Z 1 0 Z ∂B(α(s),r) ω0(α) · n(α) dγ(α) ! dr.
On one hand, for all r > 0, we have, Z ∂B(α(s),r) ω0(α) · n(α) dγ(α) = Z B(α(s),r)∇ · ω0 (α)dα.
On the other, we have ∇ · ω0 = 0, then, we deduce, Z
B(α(s),1)
ω0(α) · n(α) dα = 0, and therefore, we get
Z R3
ft(y) dy = 0. From (76), we get kftkL∞ ≤ kω0kL∞, moreover since
supp ft ⊂ Vs(t) and thanks to (73), we get for all q0 > 1, kftkLq0 ≤ kω0kL∞(4π
3 )
1
q0. Therefore
for all q0 > 1, ft∈ Lq0(R3) and Z
R3
ft(y) dy = 0, then, we deduce that ft∈ H1(R3) (see [STE2], chapter III, §1, 1.2.4).
For any t ∈ [0, s], since the space BMOr(Vs(t)) is the dual of H1z(Vs(t)) (see Section 3), then we obtain, Z Vs(t)
u(y, t) ω0(A(y, t)) · n(A(y, t)) dy = Z Vs(t) u(y, t)ft(y) dy ≤ ku(t)kBM Or(Vs(t)) kftkH1z(Vs(t)) = ku(t)kBM Or(Vs(t)) kftkH1(R3). (78)
Due to the equivalence of norms in Hardy space H1(R3) and thanks to Lemma 3.1, there exists an absolute constant ̺ > 0 such that,
kftkH1(R3) ≤ ̺ (kftkL1 + 2|Vs(t)| kftkL∞)
From (76), we obtain, kftkH1(R3) ≤ 3̺|Vs(t)| kω0kL∞. Thanks to (73), we deduce, kftkH1(R3)≤ 4π̺ kω0kL∞. (80) From (9), we have, ku(t)kBM Or(Vs(t)) = sup Q⊂Vs(t) 1 |Q| Z Q|u(x, t) − uQ(t)| dx , (81) where uQ(t) = 1 |Q| Z Q
u(y, t)dy, and the supremum is taken over all cubes Q in the domain Vs(t) whose sides are parallel to the coordinate axes. For all x ∈ Q ⊂ Vs(t), using Lemma 3.2, we get,
|u(x, t) − uQ(t)| ≤ 1 |Q| Z Q|u(x, t) − u(y, t)| dy ≤ kω(t)kL∞diamQ.
Then, from (81), we deduce,
ku(t)kBM Or(Vs(t)) ≤ kω(t)kL∞ sup
Q⊂Vs(t)
diamQ. (82)
Since, for all Q cube of R3, the measure of Q is given by,
|Q| = diam(Q)√ 3
3 .
For all Q ⊂ Vs(t), cube of R3, we have, |Q| ≤ |Vs(t)|, and we deduce, diamQ ≤√3|Vs(t)|
1
3. (83)
Then, using (82), (83) and (73), we deduce,
ku(t)kBM Or(Vs(t)) ≤ √ 3 4π 3 1 3 kω(t)kL∞. (84)
Then, using (80) and (84), from (78), we deduce that there exists an absolute constant C > 0 such that, Z Vs(t)
u(y, t) ω0(A(y, t)) · n(A(y, t)) dy ≤ C kω(t)kL∞kω0kL∞. (85)
Using (85), from (75), we deduce that there exists an absolute constant C1 > 0 such that,
which is valid for all s ∈]0, T ] and therefore valid for all s ∈]0, T∗[.
Now, we use the scaling property of Euler equations (37) and the vorticity equations (39), in other words, for all λ > 0 and β > 0, the couple of vector-valued functions (uλ,β, ωλ,β) defined for all (z, σ) on R3× 0,T ∗ λβ by, uλ,β(z, σ) = λu(βz, λβσ) ωλ,β(z, σ) = λβω(βz, λβσ) (87) are respectively solution of Euler equations (37) and vorticity equations (39) for the initial data λu0(β ·) and λβω(β ·).
Then, Inequality (86) holds for the couple (uλ,β, ωλ,β) and we get for all σ ∈ 0,T ∗ λβ , kωλ,β(σ)kL∞ ≤ kωλ,β 0 kL∞ + C1 k∇ωλ,β(σ)kL∞ exp Z σ 0 k∇u λ,β (θ)kL∞ dθ +C1kωλ,β0 kL∞ Z σ 0 kω λ,β (θ)kL∞ dθ. (88)
For all s ∈ [0, T∗[, setting σ = s
λβ in (88) and using (87), we deduce,
kω(s)kL∞ ≤ kω0kL∞+ C 1 β k∇ω(s)kL∞ exp Z s 0 k∇u(τ)k L∞ dτ +C1kω0kL∞ Z s 0 kω(t)k L∞ dt, (89)
then taking the limit as β → 0 in (89), we obtain for all s ∈ [0, T∗[, kω(s)kL∞ ≤ kω0kL∞+ C1kω0kL∞
Z s 0 kω(t)k
L∞ dt.
Then, thanks to Gronwall Lemma, we deduce for all s ∈ [0, T∗[,
kω(s)kL∞ ≤ kω0kL∞exp(C1kω0kL∞s). (90) Then, we have Z T∗ 0 kω(s)k L∞ ds ≤ 1 C1
exp(C1kω0kL∞T∗) < +∞, which leads to a contradiction
with (41), therefore T∗
= +∞ and Inequality (90) holds for all s ≥ 0, which conclude the proof.
Remark 4.1 Theorem 4.1 have been established for all 1 < q < ∞ and r > 2 +3q, in fact, the result holds also for any r > 1 + 3q, indeed since C∞
0 (R3) is dense in Wr,q(R3) and thanks to Theorem 3.6 in [KP], then Theorem 4.1 follows for any r > 1 +3q.
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