Physica D 237 (2008) 1663–1675
www.elsevier.com/locate/physd
Global well-posedness for 2D polymeric fluid models and growth estimate
Nader Masmoudi a,∗ , Ping Zhang b , Zhifei Zhang c
a Courant Institute, New York University, New York, NY 10012, United States
b Academy of Mathematics & Systems Science, CAS Beijing 100080, China
c School of Mathematical Science, Peking University, Beijing 100871, China
Available online 21 March 2008
Abstract
Motivated by [P. Constantin, N. Masmoudi, Global well-posedness for a Smoluchowski equation coupled with Navier–Stokes equations in 2D,
Comm. Math. Phys. 278 (2008) 179–191; F. Lin, Ping Zhang, Zhifei Zhang, On the global existence of smooth solution to the 2-D FENE dumbbell
model, Comm. Math. Phys. 277 (2008) 531–553], we prove the global existence of smooth solutions to a coupled microscopic–macroscopic model
for polymeric fluid in 2D under the co-rotational assumption. Furthermore, we provide an estimate on the time growth of these solutions. Our
method is general and can be applied to the co-rotational FENE model and to the rod-like model without the co-rotational assumption.
c 2008 Elsevier B.V. All rights reserved.
Keywords: Nonlinear Fokker–Planck equations; Navier–Stokes equations; Micro–macro interactions; Global well-posedness
1. Introduction
In this paper, we study the global smooth solutions for a coupled microscopic–macroscopic model for polymeric fluid in 2D.
The micro-mechanical models for polymeric liquids often consist of beads joined by springs or rods [2,8]. In the simplest case, a
molecule configuration can be described by its end-to-end vector Q. Taking into account the elastic effect together with the thermofluctuation, the distribution function ψ(t, x, Q) of molecule orientations Q satisfies a Fokker–Planck equation. The convection
velocity u satisfies the Navier–Stokes equations with an elastic stress which reflects the microscopic contribution of the polymer
molecules to the overall macroscopic flow fields. Mathematically, this system reads (one may check [12] for a formal energetic
variational derivation)

2
u t + u · ∇u + ∇ p = 4u + ∇ · τ, x ∈ R ,
2
(1.1)
∇ · u = 0, x ∈ R ,

ψt + u · ∇ψ = 4 Q ψ − ∇ Q · (S(u)Qψ − ∇ Q U ψ), (x, Q) ∈ R2 × R2 ,
where S(u) and the polymer stress τ are given by
Z
∇u − ∇u t
S(u) =
, τ=
∇ Q U ⊗ QψdQ,
2
R2
(1.2)
and U (Q) = U (|Q|2 ) is the potential.
Existence results for micro-macro models of polymeric fluids are usually limited to small time existence and uniqueness of
strong solutions [17] or global existence of weak solution [16] (see also [15] for the Oldroyd B model). In the setting when the last
∗ Corresponding author. Tel.: +1 212 529 3327; fax: +1 212 995 4121.
E-mail addresses: [email protected] (N. Masmoudi), [email protected] (P. Zhang), [email protected] (Z. Zhang).
c 2008 Elsevier B.V. All rights reserved.
0167-2789/$ - see front matter doi:10.1016/j.physd.2008.03.020
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
equation of (1.1) is formulated as a stochastic PDE, we refer to [10] (see also [9] for a polynomial force). With regards to PDE
coupled systems, we refer the reader to the earlier work [18].
In the work [12] (see also [17]), the authors proved the global existence of smooth solutions to (1.1) near equilibrium, which
is a sort of an extension of a related result for the Oldroyd model [11], and which corresponds to the Hooke dumbbell model. In
two space dimensions, Constantin, Fefferman, Titi and Zarnescu [6] proved the global existence of smooth solutions to a coupled
nonlinear Fokker–Planck and Navier–Stokes system when the convection velocity u in the Fokker–Planck equation is replaced by
a sort of time averaged one. Later this assumption was removed by Constantin and Masmoudi [7]. At the same time, Lin, Zhang
and Zhang [13] independently proved the global regularity for the 2D co-rotational FENE model. The main ideas in [7,13] are
the so-called losing derivative estimate from [1,5]. In this paper, we treat the case where Q ∈ R2 . Our observation here is that an
adaptation of this method enables us to get not only the global existence of smooth solutions but also the time growth estimate for
thus obtained solution as in (1.12). We also point out that there is a new difficulty for the model (1.1) which lies in the fact that in
order to get regularities for the solutions of (1.1), we need to do momentum estimate for the distribution function ψ(t, x, Q). One
of the main limitation of our result is the fact that we have to take the antisymmetric part of ∇u which explains the terminology
co-rotational model. A more physical model would be the case where S(u) = ∇u. However, we do not know how to get global
solutions in that case. Besides, our method of proving the time growth estimate also holds for the 2D co-rotational FENE model
and the 2D rod-like model. For the 2D rod-like model [7], there is no need for the co-rotational assumption since, we have an L ∞
bound on the extra stress tensor τ .
R
def
Let M = e−U . After renormalization, we will assume that R2 M dQ = 1. In what follows, we will seek the solution of (1.1)
with the form
√
ψ = M + M f.
(1.3)
R
R
We assume that R2 ψ(0, x, Q)dQ = 1; then formally from (1.1), there holds R2 ψ(t, x, Q)dQ = 1, which together with (1.3)
yield
Z √
M f dQ = 0.
(1.4)
R2
Plugging (1.3) into (1.1) and using the fact that S(u)Q · ∇ Q U = 0, we get the new system for (u, f ):

4u + ∇ · τ, x ∈ R2 ,

u t + u · ∇u + ∇ p =

2
∇ · u = 0, x ∈ R ,
4 U

|∇ U |2

 f t + u · ∇ f + ∇ Q · (S(u)Q f ) − 4 Q + Q − Q
f = 0,
2
4
(1.5)
(x, Q) ∈ R × R ,
with the polymer stress τ given by
Z
√
τ=
∇ Q U ⊗ Q M f dQ,
2
2
(1.6)
R2
together with the initial conditions:
u|t=0 = u 0 (x),
f |t=0 = f 0 (x, Q),
for (x, Q) ∈ R2 × R2 .
(1.7)
Before presenting the main result of this paper, we introduce some basic notations that will be used in the rest of the paper. We
shall denote R2x by R2 , and R2Q by D. Also as in [12], to avoid additional technicalities, we assume that
|Q| ≤ C(1 + |∇ Q U |),
Z
|∇ Q U |2 e−U dQ ≤ C,
D
4 Q U ≤ C + δ|∇ Q U |2 with δ < 1,
Z
|Q|4 e−U dQ ≤ C,
(1.8)
D
and
k
|∇ Q
(Q∇ Q U )|
≤ C(|Q||∇ Q U | + 1),
Z U 2
k
∇ Q Q∇ Q U e− 2 dQ ≤ C,
D
k
|∇ Q U |2 2
∇
4
QU −
Q
≤ C(1 + |∇ Q U | ),
2
(1.9)
for all integers 0 ≤ k ≤ 2. More precisely, we only need k = 0 in (1.9) for Theorem 1.1, and 0 ≤ k ≤ 2 in (1.9) is required for
Theorem 1.2. In particular, these conditions hold for the Hooke model, namely the case U = |Q|2 /2.
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
For the convenience of the readers, let us recall some basic facts about the Littlewood–Paley theory — check [3] and [4] for
def
def
more details. Let B ={ξ ∈ R2 , |ξ | ≤ 43 } and C ={ξ ∈ R2 ,
X
χ (ξ ) +
ϕ(2− j ξ ) = 1, ξ ∈ R2 .
3
4
≤ |ξ | ≤ 83 }. Let χ ∈ Cc∞ (B) and ϕ ∈ Cc∞ (C) which satisfy
j≥0
def
def
We denote h = F −1 ϕ and h̃ = F −1 χ . Then the dyadic operators ∆ j and S j can be defined as follows
Z
−j
2j
h(2 j y) f (x − y)dy, for j ≥ 0,
∆ j f = ϕ(2 D) f = 2
2
R
Z
X
−j
2j
S j f = χ (2 D) f =
∆k f = 2
h̃(2 j y) f (x − y)dy, and ∆−1 f = S0 f.
R2
−1≤k≤ j−1
(1.10)
With the introduction of ∆ j and S j , we can define the following space:
Definition 1.1. Let s ∈ R, we define
C̃ s (R2 ; L 2 (D)) = { f ∈ S 0 (R2 × D);
k f ks < ∞},
Z
1 2
2
|∆ j f | dQ k f ks = sup 2 D
j≥−1
L∞
x
with
js
Now we state our main result as follows.
Theorem 1.1. Let 1 < s < 2. Let u 0 ∈ H 1 (R2 )∩C s (R2 ), f 0 ∈ H 1 (R2 ; L 2 (D))∩ C̃ s−1 (R2 ; L 2 (D)), and |Q| f 0 ∈ L ∞ (R2 ; L 2 (D)).
Then (1.5)–(1.7) has a unique global solution (u, f ) such that for any T > 0, there hold
u ∈ C([0, +∞); H 1 (R2 ) ∩ C s (R2 )) ∩ L 2 ((0, T ); H 2 (R2 )),
f ∈ C([0, +∞); H 1 (R2 ; L 2 (D)) ∩ C̃ s−1 (R2 ; L 2 (D))),
1
∞
2
2
|Q| f ∈ L ((0, +∞) × R ; L (D)), and
∇ Q + ∇ Q U f ∈ L 2 ((0, T ); H 1 (R2 ; L 2 (D))).
2
(1.11)
Furthermore, there holds
ku(t)kC s + k f (t)ks−1 ≤ C0 (1 + ku 0 kC s + k f 0 ks−1 )exp(C0 t) ,
where C0 only depends on ku 0 k2L 2 + k f 0 k2L 2 + k(1 + |Q|) f 0 k2 ∞
∀t < ∞
L (R2 ;L 2 (D))
(1.12)
.
Remark 1.1. Compared with [7,13], the new ingredient in the above theorem is the time growth estimate (1.12), which is not
possible via a Lyapunov functional argument. Actually the method of the proof of (1.12) can be extended to the Rod-like model
(without the anti-symmetric assumption on S(u) in (1.1)) and co-rotational FENE model. But as there are already global wellposedness results in [7,13], we chose to present the estimate (1.12) for the model (1.1).
With smoother initial data together with higher momentum integrability on f 0 , we can prove further regularity of the solutions
(u, f ) obtained in Theorem 1.1. For a clear presentation, we will only give a proof of the following theorem
Theorem 1.2. Under the p
assumptions of Theorem 1.1, we assume further that u 0 ∈ H 2 (R2 ), f 0 ∈ H 2 (R2 ; L 2 (D)) and
|Q| f 0 ∈ H 1 (R2 ; L 2 (D)), 1 + |Q|2 ∇ Q f 0 ∈ L ∞ (R2 ; L 2 (D)). Then for any T > 0, we have
u ∈ Cloc ([0, +∞); H 2 (R2 )) ∩ L 2 ((0, T ); H 3 (R2 )),
q
f ∈ Cloc ([0, +∞); H 2 (R2 ; L 2 (D))),
1 + |Q|2 ∇ Q f ∈ L ∞ ((0, T ) × R2 ; L 2 (D)),
1
and
∇ Q + ∇ Q U f ∈ L 2 ((0, T ); H 2 (R2 ; L 2 (D))).
2
(1.13)
As can be seen from the proof of Theorem 1.2, we can also get higher regularity, namely for any integer s, s ≥ 3, we have
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
Theorem 1.3. Under the p
assumptions of Theorem 1.1, we assume further that u 0 ∈ H s (R2 ), f 0 ∈ H s (R2 ; L 2 (D)) and
2
1
2
|Q| f 0 ∈ H (R ; L (D)), 1 + |Q|2 ∇ Q f 0 ∈ L ∞ (R2 ; L 2 (D)). Then for any T > 0, we have
u ∈ Cloc ([0, +∞); H s (R2 )) ∩ L 2 ((0, T ); H s+1 (R2 )),
q
1 + |Q|2 ∇ Q f ∈ L ∞ ((0, T ) × R2 ; L 2 (D)),
f ∈ Cloc ([0, +∞); H s (R2 ; L 2 (D))),
1
and
∇ Q + ∇ Q U f ∈ L 2 ((0, T ); H s (R2 ; L 2 (D))).
2
(1.14)
The proof of Theorem 1.3 will be left to the reader (see also Remark 2.1)
Let us end this section by some notations used in the sequel:
Notations: We will denote [a; b] the commutator between a and b, ∂ α the derivatives with respect to x variables, and
!1
Z
1
Z
2
X
p
p
α
2
kφk L p =
|φ(x, Q)| dx dQ
,
|∂ g| dx dQ
, |g|s =
R2 ×D
R2 ×D |α|≤s
and kuk H s the standard Sobolev
R norm of u when u dependsR only on the x variable. We shall use the convention ( f, g) to denote
both the inner product on R2 , R2 f g dx, and on R2 × D, R2 ×D f g dx dq. And we will denote C··· a uniform positive constant
which depends on the listed variables, but may change from line to line.
2. Proof of Theorems 1.1 and 1.2
Let us first recall the following lemma from [12], which will be constantly used in the sequel.
R √
Lemma 2.1. Let f satisfy D f MdQ = 0, and U satisfy (1.8). Then
1
,
k f k L 2 + k∇ Q U f k L 2 + k∇ Q f k L 2 ≤ C ∇
f
+
∇
U
f
Q
Q
2
2
L
1
1
2
k∇ Q U Q f k L 2 ≤ C (1 + |Q| ) 2 ∇ Q f + 2 ∇ Q U f 2 .
L
Now let us present the proof of Theorem 1.1.
Proof of Theorem 1.1. As the existence of solutions to (1.5)–(1.7) can be essentially proved from the a priori estimates, for
simplicity, we only present the a priori estimate for smooth enough solutions to (1.5)–(1.7).
Step 1. The estimate of ku(t)k H 1 and | f |1 .
Standard energy estimate applied to the first equation of (1.5) gives
X
1 d
−([∂ α ; u] · ∇u, ∂ α u) + (∂ α divτ, ∂ α u)
kuk2H 1 + k∇uk2H 1 =
2 dt
|α|≤1
≤ C(k∇uk L ∞ kuk2H 1 + | f |1 k∇uk H 1 ),
(2.1)
where we used (1.6) and (1.9) such that
1
Z 2
2
− U2 kτ k H 1 ≤
U
⊗
Qe
dQ
| f |1 ≤ C| f |1 .
∇ Q
D
Similarly, we apply ∂ α with |α| ≤ 1 to the third equation of (1.5), and take L 2 (R2 × D) inner product of the resulting equation with
∂ α f to get
"
#
X 1 1
1 d
2
2
2
α
α
4 Q U − |∇ Q U | ∂ f, ∂ f
| f | + |∇ Q f |1 −
2 dt 1
2
2
|α|≤1
X
=−
(∂ α (u · ∇ f ), ∂ α f ) + (∂ α ∇ Q · (S(u)Q f ), ∂ α f ) .
(2.2)
|α|≤1
By integration by parts, we get
1
− (4 Q U ∂ α f, ∂ α f ) = (∇ Q U ∂ α f, ∇ Q ∂ α f ).
2
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
Thus the dissipation due to the Fokker–Planck kinetic dynamics is given by
2
X 1 1
1
4 Q U − |∇ Q U |2 ∂ α f, ∂ α f = ∇ Q f + ∇ Q U f .
|∇ Q f |21 −
2
2
2
1
|α|≤1
Since div u = 0, we have
X
X
(∂ α (u · ∇ f ), ∂ α f ) =
(∂ α u · ∇ f, ∂ α f ) ≤ k∇uk L ∞ | f |21 .
|α|≤1
|α|≤1
Then, we notice that div u = 0 implies that div Q (S(u)Q) = 0. Hence, we get
X
X
(∂ α ∇ Q · (S(u)Q f ), ∂ α f ) = −
((∂ α S(u))Q f, ∂ α ∇ Q f )
|α|≤1
|α|≤1
Z
1 2
2
≤
|Q f | dQ D
k∇uk H 1 |∇ Q f |1 .
L∞
Therefore, for any δ > 0, we get by summing (2.1) and δ× (2.2)
2
1
d
2
2
2
(kuk H 1 + δ| f |1 ) + k∇uk H 1 + 2δ ∇ Q f + ∇ Q U f dt
2
1
Z
1 2
2
2
2
|Q f | dQ ≤ C(1 + k∇uk L ∞ )(kuk H 1 + | f |1 ) + δ D
k∇uk H 1 |∇ Q f |1 .
(2.3)
L∞
R
Step 2. The estimate of k D |Q f |2 dQk L ∞ .
Multiplying the f equation in (1.5) by f , we get by integrating the resulting equation over D that
2
Z
Z
Z
Z 1
1
∇ Q f + 1 ∇ Q U f dQ = − ∇ Q · (S(u)Q f ) f dQ = 0,
∂t
| f (t)|2 dQ + u · ∇
| f (t)|2 dQ +
2
2
2
D
D
D
D
which together with the standard L p estimate of the transport equation gives
Z
Z
1 1 2
2
| f (t, x, Q)|2 dQ ≤ | f 0 (x, Q)|2 dQ , for p = 2, ∞,
D
p D
p
L
L
2
Z Z
Z ∞Z Z ∇ Q f + 1 ∇ Q U f dQ dx dt ≤ 2
| f 0 |2 dQ dx.
2
2
R2 D
0
R D
and
Similar to (2.4), we have
Z
Z
1
1
2
2
∂t
|Q| | f (t)| dQ + u · ∇
|Q|2 | f (t)|2 dQ
2
2
D
D
Z Z
1
1
4 Q + 4 Q U − |∇ Q U |2 f |Q|2 f dQ = − ∇ Q · (S(u)Q f )|Q|2 f dQ.
−
2
4
D
D
Moreover, we get by integration by parts
2
Z Z
1
1
1
4 Q + 4 Q U − |∇ Q U |2 f |Q|2 f dQ =
−
|Q|2 ∇ Q f + ∇ Q U f dQ
2
4
2
D
D
Z
Z
2
− 2 | f | dQ +
∇ Q U · Q| f |2 dQ.
D
D
Using the fact that div Q (S(u)Q) = 0 and S(u) is an antisymmetric matrix, we get
Z
Z
1
∇ Q · (S(u)Q f )|Q|2 f dQ =
S(u)Q · ∇ Q | f |2 |Q|2 dQ
2 D
D
Z
=−
S(u)Q · Q| f |2 dQ = 0.
D
So we get by summing up (2.4) and η× (2.6) that
(2.4)
(2.5)
(2.6)
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
2
Z
Z
1
1
(1 + η|Q|2 )| f (t)|2 dQ + u · ∇ (1 + η|Q|2 )| f (t)|2 dQ + (1 + η|Q|2 ) ∇ Q f + ∇ Q U f dQ
2
2
D
D
D
Z
Z
= 2η | f |2 dQ − η ∇ Q U · Q| f |2 dQ.
1
∂t
2
Z
(2.7)
D
D
On the other hand, thanks to (1.8) and Lemma 2.1, we have
Z
Z
Z
(2 − ∇ Q U · Q)| f |2 dQ ≤ C (1 + |∇ Q U |2 )| f |2 dQ ≤ C
D
D
2
∇ Q f + 1 ∇ Q U f dQ
2
D
Therefore, by taking η sufficiently small, we deduce from (2.7) that
2
Z tZ
Z
1
(1 + η|Q|2 ) ∇ Q + ∇ Q U f (t, Φt 0 (x), Q) dQdt 0
(1 + η|Q|2 )| f (t, Φt (x), Q)|2 dQ +
2
0 D
D
Z
≤ (1 + η|Q|2 )| f 0 (x, Q)|2 dQ,
(2.8)
D
where Φt (x) is determined by
(d
Φt (x) = u(t, Φt (x)),
dt
Φt (x)|t=0 = x.
Thanks to Lemma 2.1 and (2.8), we get by taking δ small enough in (2.3) that
2
1
1
d
2
2
2
(kuk H 1 + δ| f |1 ) + k∇uk H 1 + δ ∇ Q f + ∇ Q U f ≤ C(1 + k∇uk L ∞ )(kuk2H 1 + | f |21 ).
dt
2
2
1
(2.9)
Step 3. Here, we estimate (u, f ) in C s × C̃ s−1 by using some estimates in the spirit of the losing derivative estimate in [5]. Given
any 0 ≤ T0 < T , let us first recall the following estimate of u from [13]: for any λ > 0, there holds
1
s−1
s
s
s
+
kuk
(2.10)
Mλ,[T
(u)
≤
Cku(T
)k
+
C
4
0 C
L [T ,T ] (L 4 ) Mλ,[T0 ,T ] (u) + C Mλ,[T0 ,T ] (τ ),
0 ,T ]
λ
0
def
s
where Mλ,[T
(u) =
0 ,T ]
0 def
Φλ (t, t ) = λ
t
Z
sup
j≥−1,t∈[T0 ,T ]
(1 + k∇u(t )k
00
t0
2 js−Φλ (t) k∆ j u(t)k L ∞ with
L∞
Z t Z
2
00 2
)dt + λ
(1 + |Q| )| f (t )| dQ 0
dt 00 ,
00
D
t
L∞
def
Φλ (t) = Φλ (t, T0 ).
On the other hand, thanks to (1.9), we have
Z
1 2
s−1
j (s−1)−Φλ (t)
2
Mλ,[T
(τ
)
≤
C
sup
2
|∆
f
(t,
x,
Q)|
dQ
j
0 ,T ]
D
j≥−1,t∈[T0 ,T ]
.
(2.11)
L∞
def
This reduces the estimate of τ to that of f . In order to do so, let us set f j = ∆ j f . Then we get by applying ∆ j to the f equation
in (1.5) that
4Q U
|∇ Q U |2
∂t f j + u · ∇ f j − 4 Q +
−
f j = − ∆ j (u · ∇ f ) − u · ∇ f j − ∆ j ∇ Q · (S(u)Q f ) .
2
4
Multiplying the above equation by f j , and integrating the resulting equation over D, we obtain
2
Z Z
Z
1
1
1
2
2
∇ Q f j + ∇ Q U f j dQ
∂t
| f j | dQ + u · ∇
| f j | dQ +
2
2
2
D
D
D
Z
Z
=−
∆ j (u · ∇ f ) − u · ∇ f j f j dQ −
∆ j ∇ Q · (S(u)Q f ) f j dQ
D
def
= I + I I.
D
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
def R
Set F j (t, x) =
D | f j (t, x,
Q)|2 dQ. Then the proof of (4.13) in [13] ensures

1
2
|I | ≤ C F j (t) 2 j
X
j 0& j
Z
1 2
2
k∆ j 0 uk L ∞ | f | dQ D
and the proof of (4.14) in [13] gives
Z
1 2
|I I | ≤ C |Q|2 | f |2 dQ D
Z
|∇ Q f j |2 dQ

+ k∇uk L ∞


1
2
×
j0
X
2 k∆ j 0 uk L ∞ + 2− j
j &j
L
D
L∞

0
2 j k∆ j 0 uk L ∞ + 2− j
k∆ j 0 uk L ∞  .
F j 0 (t) 
| j 0 − j|≤5
0
22 j k∆ j 0 uk L ∞  + j 0. j
j 0& j
2
1
2
2
X
2 j0

X

X
X
j 0. j
j 0& j
Therefore, we obtain
2
Z Z
Z
1
| f j |2 dQ + 2 ∇ Q f j + ∇ Q U f j dQ
∂t
| f j |2 dQ + u · ∇
2
D
D
D

Z
1
X
2
1
≤ C F j (t) 2 2 j
k∆ j 0 uk L ∞ | f |2 dQ + k∇uk L ∞
∞
D
0
Z
2
2
|Q|
|
f
|
dQ
+ C F j 0 (t)  ,
| j 0 − j|≤5
L∞
D
L∞
1
2
X
Z
|∇ Q f j |2 dQ,
D
which together with Lemma 2.1 gives
∂t
Z
| f j |2 dQ + u · ∇
D

Z
1
2
D
| f j |2 dQ ≤ C F j (t) 2 j
X
j 0& j
Z
1 2
k∆ j 0 uk L ∞ | f |2 dQ D
Z
2
2
+ C |Q| | f | dQ D

+ k∇uk L ∞
1
2
F j 0 (t) 
| j 0 − j|≤5
L∞
2

X

L∞
X
j0
2 k∆ j 0 uk L ∞ + 2− j
j 0& j
X
2
2 j0
k∆ j 0 uk L ∞  .
(2.12)
j 0. j
Set
(s−1)
Mλ,[T0 ,T ] ( f )
=
sup
j≥−1,t∈[T0 ,T ]
Z
2
|
f
|
dQ
j
!1
2
2 j (s−1)−2Φλ (t) 2
D
.
L∞
Using that 1 < s < 2 and using the standard L p estimate for the transport equations, we get after multiplying (2.12) with
22 j (s−1)−2Φλ (t) that
Z
Z t
(s−1)
s
2
−2Φλ (t,t 0 ) 2
0 2
dt 0
)|
f
(t
)|
dQ
[Mλ,[T0 ,T ] ( f )]2 ≤ sup 22 j (s−1) kF j (T0 )k L ∞ + C[Mλ,[T
(u)]
sup
2
(1
+
|Q|
∞
0 ,T ]
D
t∈[T0 ,T ] T0
j≥−1
L
Z t
0
(s−1)
+ C[Mλ,[T0 ,T ] ( f )]2 sup
2−2Φλ (t,t ) (1 + k∇u(t 0 )k L ∞ )dt 0
t∈[T0 ,T ] T0
C
(s−1)
s
≤ k f (T0 )ks−1 + ([Mλ,[T0 ,T ] ( f )]2 + [Mλ,[T
(u)]2 ),
0 ,T ]
λ
from which, we deduce for λ sufficiently large that
(s−1)
[Mλ,[T0 ,T ] ( f )]2 ≤ Ck f (T0 )k2s−1 +
C
[M s
(u)]2 .
λ λ,[T0 ,T ]
(2.13)
Plugging (2.13) in (2.11) and using (2.10), we get
s−1
Mλ,[T
(τ ) ≤ Ck f (T0 )ks−1 +
0 ,T ]
s
(u) ≤ C(ku(T0 )kC s
Mλ,[T
0 ,T ]
C
s
Mλ,[T
(u), and
1
0 ,T ]
λ2
+ k f (T0 )ks−1 ) + Ckuk L 4
[T0 ,T ] (L
4)
s
(u).
Mλ,[T
0 ,T ]
Step 4. The closed estimates of ku(t)k H 1 , | f (t)|1 , ku(t)kC s and k f (t)ks−1 .
(2.14)
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
First, thanks to (2.9), we get by applying Gronwall inequality that
2
Z t Z
1
1 t
0
2
0 2
2
0 ∇ Q + ∇ Q U f (t ) dt 0
k∇u(t )k H 1 dt + δ
ku(t)k H 1 + δ| f (t)|1 +
2 0
2
0
1
Z t
(1 + k∇u(t 0 )k L ∞ )dt 0 .
≤ (ku 0 k2H 1 + δ| f 0 |21 ) exp C
(2.15)
0
On the other hand, taking T0 = 0 in (2.14), we find for λ 1
Mλs (u)(T ) ≤ C B0 + Ckuk L 4
[0,T ] (L
4)
Mλs (u)(T ),
(2.16)
def
s
s
where Mλs (u)(T ) = Mλ,[0,T
] (u), and B0 = ku 0 kC + k f 0 ks−1 . Next using (1.6), (1.9) and (2.5) and applying Lemma 2.1, we get
that
Z ∞
kτ k2L 2 dt ≤ k f 0 k2L 2 .
0
This bound, together with the proof of Theorem 3.1 of [14] imply that u ∈ C([0, T ]; L 2 (R2 ))∩ L 2 ([0, T ]; H 1 (R2 )) for any T < ∞.
Moreover, there holds
Z t
Z t
(2.17)
kτ (t 0 )k2L 2 dt 0 ≤ C(ku 0 k2L 2 + k f 0 k2L 2 ),
k∇u(t 0 )k2L 2 dt 0 ≤ ku 0 k2L 2 +
ku(t)k2L 2 +
0
0
from which and the standard interpolation inequality we get
1
kuk L 4
4
[0,T ] (L )
1
≤ Ckuk L2 ∞
[0,T ]
(L 2 )
k∇uk 2 2
L [0,T ] (L 2 )
.
Hence, we deduce that there exists a small number δ which only depends on (ku 0 k2L 2 + k f 0 k2L 2 ) such that if T < δ, then
kuk L 4 (L 4 ) ≤ 12 . Hence, using (2.16) we get that
[0,T ]
Mλs (u)(T ) ≤ 2C B0 .
(2.18)
With (2.18), we get by repeating step 4 of [13] and using (2.13) that
sup ku(t)kC s ≤ C(T )
and
t∈[0,T ]
sup k f (t)ks−1 ≤ C(T )
t∈[0,T ]
for some constant C(T ) which depends on B0 and k(1 + |Q|) f 0 k L ∞ (R2 ;L 2 (D)) . Moreover, (2.15) ensures that (1.5)–(1.7) has a
unique solution (u, f ) on [0, T ] with
2
Z t
Z t ∇ Q + 1 ∇ Q U f (t 0 ) dt 0
ku(t)k2H 1 ∩C s + | f (t)|21 + k f (t)k2s−1 +
k∇u(t 0 )k2H 1 dt 0 +
2
0
0
1
q
(2.19)
≤ C T, ku 0 k H 1 ∩C s , | f 0 |1 , k f 0 ks−1 , k 1 + |Q|2 f 0 k L ∞ (L 2 ) , and
x
Q
q
q
k 1 + |Q|2 f (t)k L ∞ (L 2 ) ≤ Ck 1 + |Q|2 f 0 k L ∞ (L 2 ) .
x
x
Q
Q
Now let us denote T ∗ the maximal time of existence of (u, f ) such that (2.19) holds. Then if T ∗ < ∞, we have
lim ku(t)kC s = ∞.
(2.20)
t→T ∗
Otherwise, supt∈[0,T ∗ ] ku(t)kC s < ∞. Then thanks to (2.15), we have
2 #
Z T∗ "
1
2
2
2
sup [ku(t)k H 1 + | f (t)|1 ] +
k∇u(t)k H 1 + ∇ Q + ∇ Q U f (t) dt < ∞,
2
0
t∈[0,T ∗ ]
1
and (2.13) gives
(s−1)
Mλ,[0,T ∗ ] ( f ) < ∞.
Notice that from (2.8), we deduce that
"
Φλ (T ) ≤ CλT
∗
∗
1 + sup ku(t)kC s
t∈[0,T ∗ ]
Z
2
2
+ (1 + |Q| )| f 0 (x, Q)| dQ D
#
,
L∞
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
therefore, we obtain
sup k f (t)ks−1 < ∞,
t∈[0,T ∗ ]
which contradicts the assumption that T ∗ is the maximal time of existence.
In what follows, we shall prove that
q
∗
2
sup ku(t)kC s ≤ C T , ku 0 k H 1 ∩C s , k f 0 ks−1 , k 1 + |Q| f 0 k L ∞ (L 2 ) ,
x
t∈[0,T ∗ )
(2.21)
Q
which contradicts (2.20), and therefore T ∗ = ∞. The main idea of the proof of (2.21) follows the argument of step 4 of [13] (see
also [7]). The new observation here is that not only can we prove (2.21), but we also get a time growth estimate for ku(t)kC s and
k f (t)ks−1 for t large enough. For completeness, we are going to present the details here.
Let T < T ∗ with T ∗ − T 1. Then thanks to (2.17) and the proof of Theorem 3.3 of [5] (namely, Lemma 3.5 of [5]),
we conclude that for any ε > 0, there exists δ0 > 0 which only depends on C(ku 0 k2L 2 + k f 0 k2L 2 ) such that for T0 < T and
0 < T − T0 ≤ δ0 , there holds
def
k∇uke
L1
[T0 ,T ] (C
0)
= sup 2 j k∆ j uk L 1 ([T0 ,T ];L ∞ (R2 )) < ε.
(2.22)
j≥−1
On the other hand, we can always take δ0 ≤ δ and hence, thanks to (2.13) and (2.14), a proof similar to the proof of (2.18)
ensures that
(s−1)
s
Mλ,[T
(u) + Mλ,[T0 ,T ] ( f ) ≤ C B(T0 ),
0 ,T ]
def
where B(t) = ku(t)kC s + k f (t)ks−1 .
Therefore, we get by using (2.8) that
Z t
0
0
B(t) ≤ C0 B(T0 ) exp C0
(1 + k∇u(t )k L ∞ ) dt ,
for t ∈ [T0 , T ],
T0
where C0 depends on ku 0 k2L 2 + k f 0 k2L 2 , k(1 + |Q|) f 0 k2 ∞
L (R2 ;L 2 (D))
log(e + B(t)) ≤ C0 + log(e + B(T0 )) + C0
Z
t
and may change from one line to an other. It follows that
(1 + k∇u(t 0 )k L ∞ )dt 0 ,
for t ∈ [T0 , T ],
(2.23)
T0
Next, we will assume that δ0 /2 ≤ T − T0 ≤ δ0 which allows us to replace the first C0 of the right hand side by C0 (T − T0 ).
Using Littlewood–Paley theory, for any integer N , which will be determined later on, we can decompose u into three parts: low
frequency part, middle frequency part, and high frequency part:
u = ∆−1 u +
N
X
∆ju +
X
∆ j u,
j>N
j=0
which implies
N
X
k∇uk L ∞ ≤ Ckuk L 2 +
k∆ j ∇uk L ∞ + C2−N (s−1) kukC s ,
for s > 1,
j=0
from which, we deduce that
Z t
k∇u(t 0 )k L ∞ dt 0 ≤ Ckuk L 1
[T0 ,t] (L
T0
2)
+ (N + 1)k∇uke
L1
[T0 ,t] (C
s ,
+ C(T − T0 )2−N (s−1) kuk L ∞
[T ,t] (C )
0
0)
for T0 ≤ t ≤ T.
(2.24)
s ≤ 2, i.e.
Now we choose N such that 1 ≤ 2−N (s−1) kuk L ∞
[T ,t] (C )
0
log(e + kuk
N∼
s
L∞
[T0 ,t] (C )
(s − 1) log 2
)
.
Then, we get by substituting (2.24) into (2.23) that
log(e + max B(t)) ≤ log(e + B(T0 )) + C0 kuk L 1
t∈[T0 ,T ]
[T0 ,T ] (L
2)
+ C0 (T − T0 ) + C0 k∇uke
L1
[T0 ,T ] (C
0)
log(e + max B(t)).
t∈[T0 ,T ]
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
Thus, if we choose ε small enough, we get by using (2.17) and (2.22) that
log(e + max B(t)) ≤ 2 log(e + B(T0 )) + C0 (1 + ku 0 k L 2 + k f 0 k L 2 )(T − T0 )
t∈[T0 ,T ]
≤ 2 log(e + B(T0 )) + C0 (T − T0 )
(2.25)
for δ0 /2 < T − T0 < δ0 . This implies that
log(e + B(t)) ≤ 22t/δ0 (log(e + B(0)) + C0 ),
for any t < ∞. This proves (1.12) and the proof of Theorem 1.1 is ended.
Now let us turn to the proof of Theorem 1.2.
Proof of Theorem 1.2. As the regularity of the solution (u, f ) constructed in Theorem 1.1 can essentially be obtained by estimates,
here we only present the a priori estimates for smooth enough solutions of (1.5)–(1.7).
p
Step 1. The estimate of | 1 + |Q|2 f (t)|1 .
We first get by a similar proof of (2.2) that
X Z 1
1
1 d
|Q f |21 −
−∆ Q + 4 Q U − |∇ Q U |2 ∂ α f, |Q|2 ∂ α f dQ
2 dt
2
4
|α|≤1 D
i
Xh
(∂ α (u · ∇ f ), |Q|2 ∂ α f ) + (∂ α ∇ Q · (S(u)Q f ), |Q|2 ∂ α f ) .
=−
|α|≤1
It is easy to observe that
X
(∂ α (u · ∇ f ), |Q|2 ∂ α f )| ≤ k∇uk L ∞ |Q f |21 ,
|α|≤1
and as S(u)Q · Q = 0, we get by integration by parts and using (1.8) that
X
X
|(∂ α ∇ Q · (S(u)Q f ), |Q|2 ∂ α f )| =
|((∂ α S(u))Q · ∇ Q f, |Q|2 ∂ α f )|
|α|≤1
|α|=1
Z
1 2
2
≤
|Q∇ Q f | dQ D
k∇uk H 1 ||Q|2 f |1
L∞
≤ CkQ∇ Q f k L ∞ (L 2 ) k∇uk H 1 |(1 + |Q∇ Q U |) f |1 .
x
Q
Therefore, thanks to Lemma 2.1, (2.4) and (2.7), we obtain
q
2
q
d
1
2
2
2
| 1 + |Q| f |1 + 1 + |Q| ∇ Q f + ∇ Q U f dt
2
1
q
≤ C(k∇uk L ∞ | 1 + |Q|2 f |21 + kQ∇ Q f k2L ∞ (L 2 ) k∇uk2H 1 ).
x
(2.26)
Q
p
p
This reduces the estimate of | 1 + |Q|2 f |1 to that of k 1 + |Q|2 ∇ Q f k L ∞ (L 2 ) . In order to do so, we first apply ∇ Q to the third
x
Q
equation of (1.5) and integrate the resulting equation over D to get
Z
Z
Z
1
1
2
2
∂t
|∇ Q f | dQ + u · ∇
|∇ Q f | dQ +
∇ Q (S(u)Q · ∇ Q f ) · ∇ Q f dQ
2
2
D
D
ZD 1
1
−
∇ Q 4 Q + 4 Q U − |∇ Q U |2 f · ∇ Q f dQ = 0.
2
4
D
Similar to (2.6), we can get by integration by parts
Z
∇ Q (S(u)Q · ∇ Q f ) · ∇ Q f dQ = 0,
D
and
−
Z 1
1
∇ Q 4 Q + 4 Q U − |∇ Q U |2 f · ∇ Q f dQ
2
4
D
2
Z Z 1
1
1
2
4 Q 4 Q U − |∇ Q U |
=
| f |2 dQ.
∇ Q + 2 ∇ Q U ∇ Q f dQ + 4
2
D
D
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
Therefore, we obtain
Z Z
Z
1
2
2
|∇ Q f | dQ + 2 ∇ Q + ∇ Q U ∇ Q
∂t
|∇ Q f | dQ + u · ∇
2
D
D
D
Z 1
1
4 Q 4 Q U − |∇ Q U |2 | f |2 dQ
=−
2 D
2
Z
≤ C (1 + |∇ Q U |2 )| f |2 dQ,
2
f dQ
(2.27)
D
where in the last step we used the assumption (1.9) on U .
On the other hand, we get by a similar proof of (2.27) that
Z
Z
Z
1
2
2
2
2
2
|Q| |∇ Q f | dQ + 2 |Q| ∇ Q + ∇ Q U ∇ Q
∂t
|Q| |∇ Q f | dQ + u · ∇
2
D
D
D
Z
Z
≤ C (1 + |∇ Q U Q|)|∇ Q f |2 dQ + C (1 + |∇ Q U |2 |Q|2 )| f |2 dQ.
D
2
f dQ
(2.28)
D
So, for any η > 0, we get by summing up (2.4), η× (2.27) and η2 × (2.28) that
Z
Z
2
2
2
∂t [| f | + η(1 + η|Q| )|∇ Q f | ]dQ + u · ∇ [| f |2 + η(1 + η|Q|2 )|∇ Q f |2 ]dQ
D
D
2 #
2
Z " 1
1
2
∇ Q + ∇ Q U f + η(1 + η|Q| ) ∇ Q + ∇ Q U ∇ Q f dQ
+2
2
2
D
Z
Z
Z
2
2
2
2
2
2
2
≤ C η (1 + |∇ Q U | )| f | dQ + η
(1 + |∇ Q U · Q|)|∇ Q f | dQ + C (1 + |∇ Q U | |Q| )| f | dQ .
D
D
(2.29)
D
Thanks to Lemma 2.1, we get by taking η small enough in (2.29) that
Z
Z
∂t [| f |2 + η(1 + η|Q|2 )|∇ Q f |2 ]dQ + u · ∇ [| f |2 + η(1 + η|Q|2 )|∇ Q f |2 ]dQ
D
D
2 #
2
Z "
1
1
∇ Q + ∇ Q U f + η(1 + η|Q|2 ) ∇ Q + ∇ Q U ∇ Q f dQ
+
2
2
D
1
≤ C (1 + |Q| ) ∇ Q + ∇ Q U
2
D
Z
2
2
f dQ,
from which and (2.8), we deduce
Z
Z
2
2
2
[| f |2 + η(1 + η|Q|2 )|∇ Q f |2 ]dQ ≤ [|
f
|
+
η(1
+
η|Q|
)|∇
f
|
]dQ
Q 0
0
∞
D
D
L
Z tZ
+C
(1 + |Q|2 )|∇ Q f (t 0 , Φt 0 (x), Q)|2 dQdt 0
0
Z
D
[(1 + |Q|2 )| f 0 |2 + η(1 + η|Q|2 )|∇ Q f 0 |2 ]dQk L ∞ ,
≤ Ck
(2.30)
D
which together Theorem 1.1 and (2.26) implies that
2
Z t q
q
1 + |Q|2 ∇ Q f + 1 ∇ Q U f (t 0 ) dt 0 ≤ C(t),
| 1 + |Q|2 f (t)|21 +
2
0
1
for ∀t < ∞,
(2.31)
where C(t) < ∞ for any fixed t < ∞.
Step 2. The estimate of | f (t)|2 .
We first get by a similar proof of (2.1) and (2.2) that
1 d
kuk2H 2 + k∇uk2H 2 ≤ C(k∇uk L ∞ kuk2H 2 + | f |2 k∇uk H 2 ),
2 dt
(2.32)
2
h
i
1
1 d 2 2
2
2 k∂ f k L 2 + ∇ Q ∂ f + ∇ Q U ∂ f = − (∂ 2 (u · ∇ f ), ∂ 2 f ) + (∂ 2 ∇ Q · (S(u)Q f ), ∂ 2 f ) .
2 dt
2
L2
(2.33)
and
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N. Masmoudi et al. / Physica D 237 (2008) 1663–1675
As div u = 0, we get by applying Lemma 3.1 of [13] that
|(∂ 2 (u · ∇ f ), ∂ 2 f )| = |([∂ 2 ; u]∇ f, ∂ 2 f )|
≤ C(k∇uk L ∞ | f |2 + k f k L ∞ (L 2 ) k∇uk H 2 )| f |2 .
x
Q
Note that ∇ Q · (S(u)Q) = 0, we have
(∂ 2 (S(u)Q∇ Q f ), ∂ 2 f ) = (∂ 2 S(u)Q f, ∇ Q ∂ 2 f ) + (∂ S(u)Q∂ f, ∇ Q ∂ 2 f ),
Observe that Ẇ 1,1 (R2 ) ,→ L 2 (R2 ), we obtain
Z Z
1
1
1
2
2
|Q∂ f Q∂ f | dQ dx
kQ∂ f k L 4 (L 2 ) ≤ C
≤ C|Q f |12 kQ∂ 2 f k L2 2 ,
x
Q
R2
R2
from which, we deduce that
1
1
k∂ S(u)Q∂ f k L 2 ≤ k∂ S(u)k L 4 kQ∂ f k L 4 (L 2 ) ≤ C(k∇uk L ∞ |Q f |1 ) 2 (k∇uk H 2 kQ∂ 2 f k L 2 ) 2 ,
x
Q
and
1
|(∂ 2 (S(u)Q∇ Q f ), ∂ 2 f )| ≤ C(t)(kQ f k L ∞ (L 2 ) k∇uk H 2 + (k∇uk H 2 kQ∂ 2 f k L 2 ) 2 )k∇ Q ∂ 2 f k L 2 ,
x
Q
where we used Theorem 1.1 and (2.31). Therefore, we obtain
2
1
d 2 2
2
≤ Cε (t)(| f |2 + k∇uk2 2 ) + ε(kQ∂ 2 f k2 2 + k∇ Q ∂ 2 f k2 2 ).
∇
+
k∂ f k L 2 + 2 ∇
U
∂
f
(t)
Q
Q
2
2
L
H
L
dt
2
L
Lemma 2.1 applied gives
2
3
1
d 2 2
2
≤ Cε (t)(| f |2 + k∇uk2 2 )
∇
+
k∂ f k L 2 + ∇
U
∂
f
(t)
Q
Q
2
2
H
dt
2
2
L
(2.34)
by taking ε small enough.
Now let us fix any T < ∞ and δ > 0 such that δ × Cε (T ) ≤ 21 . Then we get by summing up (2.32) and δ× (2.34) that
2
1
d
2
2
2
2
2 (kuk H 2 + δk∂ f k L 2 ) + k∇uk H 2 + δ ∇ Q + ∇ Q U ∂ f ≤ Cε (t)(kuk2H 2 + δk f k22 ),
dt
2
L2
for t ≤ T. Gronwall inequality applied gives
2 #
Z T"
1
2
k∇u(t)k2H 2 + δ sup (ku(t)k2H 2 + δk∂ 2 f (t)k2L 2 ) +
∇ Q + 2 ∇ Q U ∂ f (t) 2 dt ≤ C(T ).
0
t∈[0,T ]
L
Due to the arbitrariness of T , this completes the proof of Theorem 1.2.
Remark 2.1. In this remark, we explain how we can use the same argument to prove Theorem 1.3 when s ≥ 3. The only difficulty
is to control (∂ s (S(u)Q∇ Q f ), ∂ s f ). We have to use Leibniz formula to compute the first terms. If the s derivatives hit on f ,
then the term cancel as above. The term (∂ s−1 S(u)Q∂ f, ∇ Q ∂ s f ) can be treated exactly as above. Let us only explain how to treat
4
s−1 f in L 2 L 4 L 2 and putting ∇ ∂ s f in
(∂ S(u)Q∂ s−1 f, ∇ Q ∂ s f ). We can control this term by putting ∂ S(u) in L ∞
Q
t L x , putting Q∂
t x Q
2
2
2
L t L x L Q . We leave the details to the reader.
Acknowledgments
Nader Masmoudi is partially supported by and NSF grant DMS 0703145. This work was carried out when Ping Zhang visited
Courant Institute of New York University. He would like to thank the hospitality of the Institute. Ping Zhang is partially supported
by NSF of China under Grant 10525101 and 10421101, National 973 project and the innovation grant from Chinese Academy of
Sciences; and Zhifei Zhang is supported by NSF of China under Grant 10601002.
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