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MATHEMATICAL METHODS IN THE APPLIED SCIENCESMath. Meth. Appl. Sci. 2007; 30:1223–1229Published online 23 February 2007 in Wiley InterScience(www.interscience.wiley.com) DOI: 10.1002/mma.841MOS subject classification: 35Q 30; 76D 05
A remark on the decay of solutions to the 3-DNavier–Stokes equations
Yong Zhou∗,†
Department of Mathematics, East China Normal University, Shanghai 200062, China
Communicated by X. Wang
SUMMARY
In this paper we derive a decay rate of the L2-norm of the solution to the 3-D Navier–Stokes equations.Although the result which is proved by Fourier splitting method is well known, our method is new, conciseand direct. Moreover, it turns out that the new method established here has a wide application on otherequations. Copyright q 2007 John Wiley & Sons, Ltd.
KEY WORDS: Navier–Stokes equations; decay of solutions
1. INTRODUCTION
In this paper, we consider the Navier–Stokes equations in R3,⎧⎪⎪⎪⎨⎪⎪⎪⎩
�u�t
+ u · ∇u + ∇ p= �u
div u = 0
u(x, 0)= u0(x)
(1)
where u = (u1(x, t), u2(x, t), u3(x, t)) is the velocity field, p(x, t) is a scalar pressure, and u0(x)with div u0 = 0 in the sense of distribution is the initial velocity field.
The study of problem (1) has a long history. Leray [1] and Hopf [2] constructed a weak solution(so-called Leray–Hopf weak solution) to (1), u∈L∞(0, ∞; L2) ∩ L2(0, ∞; H1) for any given
∗Correspondence to: Yong Zhou, Department of Mathematics, East China Normal University, Shanghai 200062,China.
†E-mail: [email protected]
Contract/grant sponsor: IHESContract/grant sponsor: Shanghai Leading Academic Discipline, NSFC; contract/grant number: 10501012Contract/grant sponsor: Shanghai Rising-Star Program 05QMX1417
Copyright q 2007 John Wiley & Sons, Ltd. Received 14 August 2006
1224 Y. ZHOU
u0∈L2(R3) with div u0 = 0. It is proved that the weak solution is strong (and unique) locally if theinitial datum u0∈H1(R3) in addition, and the strong solution exists globally for small initial datum.
It is well known that the L2-norm of a weak solution to (1) decays to zero as time goesto infinity. A natural question is how fast does the solution decay in L2(R3). Our main theoremreads as follows.
Theorem 1.1Let u0 ∈ L2(R3) and div u0 = 0. If the solution e�t u0 to the heat equation corresponding to u0satisfies
‖e�t u0‖L2�C(1 + t)−�, t�0
for some �>0. Then there exists a weak solution u(x, t) such that
‖u(t)‖L2�C(1 + t)−�0 (2)
with �0 =min{�, 54 }.
Actually, there are many previous decay results for Leray–Hopf weak solutions, see [1, 3–6] andreferences therein. An important work was done by Wiegner [7] in 1987. He proved a decay resultfor a Leray–Hopf weak solution by the so-called Fourier splitting method, which was first appliedto parabolic conservation laws in [8]. Recently, Gallay and Wayne [9] discussed the asymptoticsfor strong solutions to (1) when initial vorticity is small with respect to a weighted norm.
The purpose of this paper is to find a new, concise and direct method to show Theorem 1.1instead of that used before. First, we present a formal proof; i.e. assume that the solution is smoothand exists globally, which can be achieved provided that the initial datum is small [4]. Then we usethe argument of Caffarelli et al. [10] to make it rigorous. So the decay rate established here is validat least for smooth solutions with small data and suitable weak solutions defined by Caffarelli,Kohn and Nirenberg. Moreover, it turns out that the new method established here has a wideapplication on other equations (see the author’s recent paper [11]).
2. PROOF
We present a formal proof first.Assume that the solution is smooth. We represent the solution by the Stokes semigroup as
u(t) = e�t u0 −∫ t
0e�(t−s)P(u · ∇u)(s) ds
= e�t u0−∫ t
0P∇e�(t−s)(u ⊗ u)(s) ds (3)
where
e�t f (x)=∫
R3Et (x − y) f (y) dy, Et (x)= 1
(4�t)−3/2e−|x |2/4t
and, for 1<q<∞, P = Pr is the bounded projection from Lq to the closed subspace Lq� of Lq
consisting of all the solenoidal vector fields.
Copyright q 2007 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2007; 30:1223–1229DOI: 10.1002/mma
DECAY OF SOLUTIONS TO THE 3-D NAVIER–STOKES EQUATIONS 1225
So directly, from (3), we have
‖u(t)‖L2�‖e�t u0‖L2 + C∫ t
0‖e�(t−s)(u · ∇u)‖L2(s) ds (4)
and
‖u(t)‖L2�‖e�t u0‖L2 + C∫ t
0‖∇e�(t−s)(u ⊗ u)‖L2(s) ds (5)
By Parseval’s equality, it is easy to prove that
‖e�t (u · ∇u)‖L2�Ct−3/4‖u‖L2‖Du‖L2 and ‖∇e�t (u ⊗ u)‖L2�Ct−5/4‖u‖2L2 (6)
However, for the completeness, we show this inequalities in the Appendix.From Equation (1), one can derive a rough estimate for ‖Du(t)‖L2 ,
‖Du(t)‖L2�C(1 + t)−1/2
which will be proved in the Appendix.Therefore, it follows from (6) that
‖u(t)‖L2�C(1 + t)−� + C∫ t
0(t − s)−3/4(1 + s)−1/2‖u(s)‖L2 ds (7)
From (7) we have
(1 + t)�‖u(t)‖L2 �C + C(1 + t)�Q(t)∫ t
0(t − s)−3/4(1 + s)−1/2−� ds
=C + C(1 + t)�Q(t)∫ t/2
0(t − s)−3/4(1 + s)−1/2−� ds
+C(1 + t)�Q(t)∫ t
t/2(t − s)−3/4(1 + s)−1/2−� ds (8)
with
Q(t) = max0�s�t
{(1 + s)�‖u(s)‖L2}
By direct computation,
∫ t/2
0(t − s)−3/4(1 + s)−1/2−� ds�Ct−3/4
⎧⎪⎪⎨⎪⎪⎩
1 if 1/2 + �>1
ln(e + t) if 1/2 + � = 1
(1 + t)1−1/2−� if 1/2 + �<1
and ∫ t
t/2(t − s)−3/4(1 + s)−1/2−� ds�C(1 + t)−�−1/2t1/4
Copyright q 2007 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2007; 30:1223–1229DOI: 10.1002/mma
1226 Y. ZHOU
Hence, if �< 34 , then
C(1 + t)�∫ t
0(t − s)−3/4(1 + s)−1/2−� ds → 0 as t →∞
So there exists a t0 sufficiently large so that
C(1 + t)�∫ t
0(t − s)−3/4(1 + s)−1/2−� ds�1
2for any t�t0
So from (8), we have
(1 + t)�‖u(t)‖L2�C + 12Q(t) for t�t0 (9)
LetQ(t) = max
t0�s�t{(1 + s)�‖u(s)‖L2}
then (9) can be reduced to
(1 + t)�‖u(t)‖L2�C + 12Q(t0) + 1
2 Q(t) for t�t0 (10)
Now taking maximum for t ∈ [t0, T ] on both sides of (10), we have
Q(T )�C + 12Q(t0) + 1
2 Q(T ) for T�t0
Therefore,
(1 + t)�‖u(t)‖L2�2C + max0�s�t0
{(1 + s)�‖u(s)‖L2}<∞
due to the energy inequality.Now, we assume �� 3
4 . Since (1 + t)−�<(1 + t)−3/5, it follows from the above step that
‖u(t)‖L2�C(1 + t)−3/5 (11)
Thanks to (6), we obtain
‖u(t)‖L2 �C(1 + t)−� + C∫ t/2
0(t − s)−5/4‖u(s)‖2L2 ds
+C∫ t
t/2(t − s)−3/4(1 + s)−1/2‖u(s)‖L2 ds (12)
Due to the rough estimate (11), it follows∫ t/2
0(t − s)−5/4‖u(s)‖2L2 ds�C
∫ t/2
0(t − s)−5/4(1 + s)−6/5 ds�Ct−5/4
So (12) reduces to
‖u(t)‖L2�C(1 + t)−�0 + C∫ t
t/2(t − s)−3/4(1 + s)−1/2‖u(s)‖L2 ds (13)
where �0 = min{�, 54 }.
Copyright q 2007 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2007; 30:1223–1229DOI: 10.1002/mma
DECAY OF SOLUTIONS TO THE 3-D NAVIER–STOKES EQUATIONS 1227
Multiplying (1 + t)�0 on the both sides of (13), we have
(1 + t)�0‖u(t)‖L2 �C + C(1 + t)�0Q(t)∫ t
t/2(t − s)−3/4(1 + s)−1/2−�0 ds
�C + CQ(t)(1 + t)−1/2t1/4
with Q(t)= max0�s�t {(1 + s)�0‖u(s)‖L2}.By the similar argument as that for �< 3
4 , we obtain that
(1 + t)�0‖u(t)‖L2�max
{2C, max
0�s�t0{(1 + s)�0‖u(s)‖L2}
}<∞
where t0 satisfies C(1 + t0)−1/2−�0 t1/40 � 12 .
This completes the formal part of the proof.To make the above arguments rigorous, apply the same proof to the ‘retarded mollification un’
defined by, which are the solutions to the following approximate equations:
�un�t
+ ��n (un)∇un + ∇ pn = �un
where ��n (un) is defined by Caffarelli et al. [10],
��n (un) =∫ ∫
R4�(y, �)un(x − �n y, t − �n�) dy d�
The function �(x, t) is smooth and has support in {x : |x |�1} × [1, 2], and ∫∫R4 � dx dt = 1.
The values of ��n (un) at time t clearly depend only on the values of un in [t−2�n, t−�n]. Asproved in [10] the un converge to a weak solution u and strongly in L2 almost everywhere in t .Since the bounds for the un are independent of n it follows that they hold for the limiting solution u.
This finishes the proof of the theorem.
APPENDIX
In this section, we assume that the solution is smooth.Now we show (6). By Parseval’s equality, we have
‖e�t (u · ∇u)‖2L2 =∫
R3e−2|�|2t | u · ∇u|2(�) d��‖ u · ∇u‖2L∞
∫R3
e−2|�|2t d�
�C‖u · ∇u‖2L1 t−3/2�C‖u‖2L2‖Du‖2L2 t
−3/2
Similarly,
‖∇e�t (u ⊗ u)‖2L2 =∫
R3e−2|�|2t |�|2|u ⊗ u|2(�) d��‖ u ⊗ u‖2L∞
∫R3
|�|2e−2|�|2t d�
�C‖u ⊗ u‖2L1 t−5/2�C‖u‖2L2‖u‖2L2 t
−5/2
where f denotes the Fourier transform of f .
Copyright q 2007 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2007; 30:1223–1229DOI: 10.1002/mma
1228 Y. ZHOU
Now, we give a rough estimate for the estimate of ‖Du‖L2 . First, we have the following energyinequality, which is valid also for weak solutions:
‖u(t)‖2L2 + 2∫ t
0‖Du(�)‖2L2 d��‖u0‖2L2 (A1)
for all t�0.Multiplying �u on both sides of (1) and doing integration by parts, we have
d
dt‖Du‖2L2�2‖u‖L∞‖Du‖L2‖�u‖L2 − 2‖�u‖2L2
Due to the Gagliardo–Nirenberg inequalities,
‖u‖L∞�C‖u‖1/4L2 ‖D2u‖3/4
L2 �C‖u‖1/4L2 ‖�u‖3/4
L2
and‖Du‖L2�C‖u‖1/2
L2 ‖D2u‖1/2L2 �C‖u‖1/2
L2 ‖�u‖1/2L2
where D2 f denotes the second order derivatives of f , and we used the following inequality, whichis easy to get by integration by parts:
‖D2u‖2L2�3‖�u‖2L2
Then, it follows from the above inequalities and (A1) that
d
dt‖Du‖2L2 �C‖u‖1/4
L2 ‖Du‖L2‖�u‖7/4L2 − 2‖�u‖2L2
=C‖u‖1/4L2 ‖Du‖1/2
L2 ‖Du‖1/2L2 ‖�u‖7/4
L2 − 2‖�u‖2L2
� 2‖�u‖2L2(C‖u‖1/2L2 ‖Du‖1/2
L2 − 1)
� 2‖�u‖2L2(C‖u0‖1/2L2 ‖Du‖1/2L2 − 1) (A2)
From the integrability of ‖Du‖2L2 , it is obvious that there exists a t0 such that
‖Du(t0)‖1/2L2 � 1
C‖u0‖L2
Then from Equation (A2) again and by the continuity of ordinary differential equations (ODEs)(it can be proved easily by contradiction argument), we have
d
dt‖Du(t)‖2L2�0 for all t�t0
Therefore,
(t − t0)‖Du(t)‖2L2�∫ t
t0‖Du(s)‖2L2 ds�‖u0‖L2
Hence‖Du(t)‖L2�C(1 + t)−1/2 for all t�0
Copyright q 2007 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2007; 30:1223–1229DOI: 10.1002/mma
DECAY OF SOLUTIONS TO THE 3-D NAVIER–STOKES EQUATIONS 1229
ACKNOWLEDGEMENTS
The author would like to thank IHES for the financial support and the warm hospitality during theauthor’s visit, when this work is finished. Thanks also to Professor Bardos and the referees for theirhelpful comments, which improve the paper greatly. This work is partially supported by Shanghai LeadingAcademic Discipline, NSFC under Grants No. 10501012 and Shanghai Rising-Star Program 05QMX1417.
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Copyright q 2007 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2007; 30:1223–1229DOI: 10.1002/mma