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8 Convergence results for the Navier-Stokes equations

This final section is entirely devoted to the particular example of the Navier-Stokes equations in the cylinder O=R×T. Our goal is to use the results of Sections 4 to 6 to obtain qualitative informations on the long-time behavior of the solutions. Without loss of generality, we fix M >0 and work in the function space XM defined by (3.2), where equations (1.1), (1.2) define an extended dissipative system satisfying assumptions (A2’), (A5) for some constants β, γ. In particular, applying Proposition 5.3 and using the explicit formula (1.9) for the energy density, we obtain the estimate which is (1.7). Thus, to complete the proof of Theorem 1.3, it remains to establish (1.8).

Let E ⊂X denote the set of equilibria of the Navier-Stokes equation (1.1) inX, namely the set of all constant velocity fields of the formu= (0, m)t, with m∈R. Given u∈X andR >0,

The following estimate will be useful :

Lemma 8.1. Fix θ ∈(0,1). There exists C5 > 0 such that, for any u ∈XM and any R ≥1, one has

dR(u,E) ≤ C5MθR1+θ2 k∇uk1L2(Bθ R) . (8.3)

Proof. We decomposeu= (0, m)t+u, whereb m=hu2i. Since∂1m=hωiand|ω| ≤M, we have On the other hand, since ub has zero mean over BR, the Sobolev embedding theorem and the Poincar´e-Wirtinger inequality imply that, if 2< p <∞,

kbukL(BR) ≤ CpR11pk∇bukLp(BR) ,

where the constant depends only on p (here we use the assumption that R ≥ 1). Moreover, interpolating between L2 and BMO and using (2.10), we obtain

k∇bukLp(BR) ≤ Cpk∇buk If we now combine (8.4) and (8.5), we obtain

dR(u,E) ≤ sup

xBR

|u(x)−(0, m(0))t| ≤ C5MθR1+θ2 k∇uk1L2(Bθ R) , whereC5 >0 depends only on θ. This is the desired estimate.

The distance (8.2) allows us to introduce the following family of neighborhoods of the set of equilibria. Given ǫ >0 and R >0, we denote

Uǫ,R = {u∈X|dR(u,E)< ǫ} .

Using estimate (8.1) and Lemma 8.1, we now show that any solution of the Navier-Stokes equation in XM spends a relatively small fraction of its lifetime outside Uǫ,R, even if ǫ > 0 is very small and R >0 very large. More precisely, we have

Proposition 8.2. Fixθ∈(0,1) andM >0. There existsC6>0such that, if u∈C0([0,∞), X) is any solution of the Navier-Stokes equations (1.1),(1.2) with initial data inXM, the following estimate holds for any ǫ >0, any R≥1, and any T >0:

ǫ,R is the characteristic function of the complement of Uǫ,R.

Proof. Using the definition of the set Uǫ,R and estimate (8.3), we easily find Z T Moreover, H¨older’s inequality and estimate (8.1) imply

Z T Combining both inequalities, we arrive at (8.6).

There are several ways to exploit the conclusion of Proposition 8.2. If we fix ǫ, R and take θ sufficiently small, we obtain the following result which already implies estimate (1.8) in Theorem 1.3.

Corollary 8.3. Any solution u∈C0([0,∞), X) of the Navier-Stokes equations (1.1), (1.2) with initial data in XM satisfies

lim sup

T→∞

1 T3+θ4

Z T

0

1Uc

ǫ,R(u(t)) dt < ∞ , for all ǫ >0, all R ≥1, and all θ >0.

It is also interesting to consider a time-dependent domain BR(T) whose size increases (suffi-ciently slowly) as T → ∞. In that case, we can still show that any solution of (1.1) converges to the set of equilibria insideBR(T), except perhaps on a sparse subset of the time axis.

Corollary 8.4. Fixa, b, c >0 such thata/2 +b < c <1/4. Ifu∈C0([0,∞), X) is any solution of the Navier-Stokes equations (1.1), (1.2) with initial data in XM, there exists C7 > 0 such that, for all T ≥1,

meas

t∈[0, T] inf

mR sup

|x1|≤Ta

sup

x2T|u(x1, x2)−(0, m)t| ≥ 1 Tb

≤ C7T34+c . (8.7) Proof. If we set R =Ta and ǫ=Tb, the quantity in the left-hand side of (8.7) is exactly the integralRT

0 1Uǫ,Rc (u(t)) dt. Using (8.6) and the fact thatR =Ta≤√

T sincea <1/2 andT ≥1, we easily obtain

Z T

0

1Uc

ǫ,R(u(t)) dt ≤ C(e(0))1−θ2 T34+a2+b+θ4(1+2a) . The conclusion now follows if we takeθ >0 small enough.

9 Appendix

9.1 Proof of Lemma 2.1

We start from (2.2) and recall that Pjk = δjk+RjRk, where R1, R2 are the Riesz transforms.

It follows that ∇ ·et∆P(u⊗v) =∇ ·et∆(u⊗v) +W(t, u, v), where Wj(t, u, v) =

X2

k,ℓ=1

et∆RjRkuvk = X2

k,ℓ=1

Z

t

jkes∆uvkds , j= 1,2 .

In the last equality, we used the fact that RjRk∆ = −∂jk for j, k = 1,2. Now, for any g∈L(O) and anyt >0, we know thatet∆g∈C(O) and k∂αet∆gkL ≤Cαt−|α|/2kgkL for all α= (α1, α2)∈N2 with|α|=|α1|+|α2|>0. Ifu, v ∈BUC(O), we thus have

k∇ ·et∆P(u⊗v)kL ≤ C

√tkukLkvkL+ Z

t

C

s3/2kukLkvkLds ≤ C

√tkukLkvkL , which proves (2.3). The same argument shows that∇ ·et∆P(u⊗v)∈BUC(O).

9.2 Proof of Proposition 2.2

Given u0 ∈BUC(O) with divu0 = 0, we take R > 0 and T > 0 such that 2ku0kL ≤ R and 4C0RT1/2 < 1, where C0 > 0 is the constant in Lemma 2.1. We introduce the Banach space X=C0([0, T],BUC(O)) equipped with the norm

kukX = sup

0tTku(t)kL(O) ,

and we set BR= {u ∈X| kukX ≤R}. For all u ∈X and all t∈[0, T], we denote by (F u)(t) the expression in the right-hand side of (2.1).

If u∈BR, then by Lemma 2.1 we have for allt∈[0, T] : k(F u)(t)kL ≤ ku0kL+

Z t

0

C0

√t−sku(s)k2Lds ≤ ku0kL+ 2C0T1/2R2 ≤ R . Thus F maps BR into itself, and a similar calculation shows that kF u−F vkX ≤ κku−vkX

for all u, v ∈BR, where κ = 4C0RT1/2 <1. Thus Eq. (2.1) has a unique solution in BR, and applying Gronwall’s lemma it is easy to verify thatu is also the unique solution of (2.1) in the whole spaceX. Finally, proceeding as in [8], one can prove that t1/2∇u∈Cb0((0, T],BUC(O)).

9.3 Proof of Lemma 2.3

We first observe that, sinceωb has zero average in the vertical variable, the Biot-Savart formula (2.6) can be written in the equivalent formub=∇K∗ω, whereb

K(x1, x2) = K(x1, x2)−|x1|

2 , (x1, x2)∈O.

Now it is easy to verify that K ∈L1(O) and∂jK ∈L1(O) for j = 1,2, see [1]. Using Young’s inequality, we deduce

kbu1kL ≤ k∂2KkL1kbωkL , kbu2kL ≤ k∂1KkL1kbωkL , (9.1) and the first inequality in (2.9) follows since kbωkL ≤2kωkL.

The next step is to establish the formula (2.8) for the pressure. The easiest way is to use the identity

div((u· ∇)u) = ∆(u21) + 2∂2(ωu1) , (9.2) which holds for any divergence-free vector field u = (u1, u2)t with vorticity ω = ∂1u2−∂2u1. If we define p = −u21 −2∂2K ∗(ωu1), where K is the fundamental solution of the Laplace operator, it follows immediately from (9.2) that −∆p = div((u · ∇)u), which is the desired result. Alternatively, it is possible to derive (2.8) directly from the formal expression (1.2).

Now, using (2.8), (9.1), and the fact that u1 =ub1, we find

kpkL ≤ kbu1k2L+ 2k∂2KkL1kωkLkbu1kL ≤ Ck∂2Kk2L1kωk2L . This proves the second inequality in (2.9).

Finally, to estimate ∇u, we observe that

1u1 = −∂2u2 = R1R2ω , ∂1u2 =−R12ω , ∂2u1=R22ω ,

and we use the well-known fact that the Riesz operators are bounded fromL(O) to BMO(O),

see [14, Chapter IV]. This concludes the proof of Lemma 2.3.

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