Axisymmetric solutions to the 3D Euler equations

Nonlinear Analysis 66 (2007) 1938–1948 www.elsevier.com/locate/na

Axisymmetric solutions to the 3D Euler equations✩ Shen Gang, Zhu Xiangrong ∗ Department of Mathematics, Zhejiang University, Hangzhou, China Received 28 December 2005; accepted 23 February 2006

Abstract Here we consider the 3D incompressible Euler equations with  axisymmetric velocity without swirl. First we will show that if u 0 ∈ C s ∩ L 2 , s > 1, and ω0 (x) ≤ C x12 + x22 , then there exists a unique u ∈ C([0, ∞); C s ) that solves the equation. This conclusion improves on the related results given by Majda [A.L. Bertozzi, A. Majda, Vorticity and Incompressible Flow, in: Cambridge Texts in Applied Mathematics, vol. 27, 2002; A. Majda, Vorticity and the mathematical theory of incompressible fluid flow, Comm. Pure Appl. Math. 39 (1986) S187–S220] and by Raymond [X. Saint Raymond, Remarks on axisymmetric solutions of the incompressible Euler system, Comm. Partial Differential Equations 19 (1994) 321–334]. ∈ L ∞ , then there exists a unique On the other hand, if u 0 ∈ L 2 , ω0 ∈ L ∞ and  ω0 x 12 +x 22

quasilipschitzian solution u ∈ C([0, ∞); C∗1 ). This improves on the corresponding results due to Raymond [X. Saint Raymond, Remarks on axisymmetric solutions of the incompressible Euler system, Comm. Partial Differential Equations, 19 (1994) 321–334] and to Chae and Kim [D. Chae, N. Kim, Axisymmetric weak solutions of the 3D Euler equations for incompressible fluid flows, Nonlinear Anal. 29(12) (1997) 1393–1404]. c 2006 Elsevier Ltd. All rights reserved. 

1. Introduction A 3D ideal incompressible fluid is described by the Euler equations ⎧ ∂u ⎪ + u · ∇u + ∇ p = 0 ⎨ ∂t x ∈ R 3 , t ∈ (0, ∞) ⎪ ⎩∇ · u = 0 u(x, 0) = u 0 (x) ✩ Supported by 973Project (G1999075105), NSFC (G10571156), RFDP (G20030335019), ZJNSF (RC97017). ∗ Corresponding author.

E-mail address: [email protected] (Z. Xiangrong). c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter  doi:10.1016/j.na.2006.02.034

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S. Gang, Z. Xiangrong / Nonlinear Analysis 66 (2007) 1938–1948

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where u = (u 1 (x, t), u 2 (x, t), u 3 (x, t)) and p = p(x, t) denote the velocity vector and the pressure of fluid at the point (x, t) ∈ R 3 × (0, ∞). Definition (Weak Solution). u is called a weak solution to (1) on (0, T ) with initial data u 0 if it satisfies the following conditions: ⎧ (a)  u ∈ L 2loc (R 3 × (0, T )); ⎪ ⎪ ⎪ ⎪ ⎪ u(t) · ∇φ = 0 for all φ ∈ C0∞ (R 3 ); ⎪ ⎨(b)

  T  3 ∂f ∂f k ⎪ k l ⎪ −u · dxdt = + (c) u u u 0 · f (0) ⎪ ⎪ ∂t ∂ xl ⎪ R3 R3 0 ⎪ k,l=1 ⎩ for all f ∈ C0∞ ([0, T ) × R 3 )) with ∇ · f = 0. Let er , eθ , and e3 be the standard orthogonal unit vectors defining the cylindrical coordinate system, x x x x1 1 2 2 , , 0 , eθ = , − , 0 , e3 = (0, 0, 1) er = r r r r  where r = r (x) = x 12 + x 22 . We say the velocity field u is axisymmetric if u = u r (r, x 3 , t)er + u θ (r, x 3 , t)eθ + u 3 (r, x 3 , t)e3 , i.e. if u does not depend on the θ coordinate. The velocity u θ in the eθ direction is called the swirl velocity. The corresponding vorticity 

θ ∂u θ ∂u θ u e3 + ω=− er + ωθ eθ + ∂ x3 r ∂r ∂u where ωθ = ∂u ∂ x 3 − ∂r . More details and a proof can be found in [4,9]. The general question of whether or not smooth axisymmetric solutions exist globally in time is still an open problem. Some work has been done on this problem, such as [8,11]. On the other hand, if u is an axisymmetric velocity without swirl, i.e. u θ = 0, then ω = ωθ eθ . This Euler system is similar to the 2D case. In this paper we consider the Euler system (1) with initial velocity u 0 where u 0 is an axisymmetric velocity without swirl. For the case of smooth data u 0 ∈ C s (s > 1) Majda obtained a global existence theorem in [9] and improved on it in [4]. In [4, p. 153] he proved a global existence theorem for when the initial vorticity ω0 = ω0θ eθ satisfies (A) |ω0θ | ≤ Cr (x); (B) ω0 has compact support. In [12] Raymond obtained a better global existence theorem. His result requires that u 0 ∈ C s ∩ L p (s > 1, 1 < p < ∞) and |ω0 | + |a0 (= ωr0 )| ∈ L q ∩ L ∞ (q < 3). In the first part of this paper we shall prove the following theorem. r

3

Theorem 1. Let u 0 be a divergence free, axisymmetric velocity without swirl and ω0 = curl u 0 be its initial vorticity. If they satisfy u 0 ∈ C s ∩ L 2 (s > 1),

and

|ω0 (x)| ≤ Cr (x)

where C is a constant independent of x, then there exists a unique u(x, t) ∈ C([0, ∞); C s ) ∩ Li p([0, ∞); L 2) that solves the incompressible Euler equation.

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Remark. Since ω0 (x) = 0 when r (x) = 0, i.e. x = (0, 0, x 3 ), by virtue of this theorem we know that if u 0 ∈ C 2 ∩ L 2 then there exists a solution globally in time. For when u 0 is not smooth data, Chae and Kim [7] obtained some results about the existence of axisymmetric weak solutions. Details can found in the related paper. Raymond ([12]. Theorem 3.3) obtained the global existence and uniqueness of a quasilipschitzian solution similar to the 2D case of Yudovich [16]. His result can be stated as follows. Let u 0 ∈ L 2 be a divergence free, axisymmetric velocity without swirl such that ω0 , a0 = ω0 q ∞ r ∈ L ∩ L for some q < 3. Then there exists a unique quasilipschitzian solution to the Euler system such that u ∈ C([0, ∞); C∗1 ∩ L 2 ),

∞ ω ∈ L∞ Loc ([0, ∞); L ).

In this paper we get the same results without the condition ω0 , a0 ∈ L q . Theorem 2. Let u 0 be a divergence free, axisymmetric velocity without swirl and ω0 = curl u 0 be its initial vorticity. If they satisfy ω0 ∈ L∞ (3) u 0 ∈ L 2 , ω0 ∈ L ∞ , and r (x) where r (x) is as same as above, then there exists a unique quasilipschitzian solution (axisymmetry without swirl) to the 3D Euler equation such that u ∈ C([0, ∞); C∗1 ∩ L 2 ),

∞ ω ∈ L∞ Loc ([0, ∞); L ).

In the proof of uniqueness of this theorem we cannot use the standard energy method due to [16,17,6]. The idea of this proof comes from Serfati [14]. 2. Preliminaries and some lemmas Before presenting our result, we give some definitions. Now we recall the L − P decomposition in R 3 . Take χ, ϕ ∈ S such that   1 supp χˆ ⊂ {ξ : |ξ | < 1}, supp ϕˆ ⊂ ξ : < |ξ | < 2 , χˆ , ϕˆ ≥ 0; 2 3j j ϕ j (x) = 2 ϕ(2 x), χ(ξ ˆ )+ ϕˆj (ξ ) = 1. j ≥0

Definition 3 (Besov Space [3,15]). For s ∈ R, 0 < p, q ≤ ∞, the inhomogeneous Besov spaces B spq are defined as follows: ⎧ ⎫ 

1 q ∞ ⎨ ⎬ q B spq = f ∈ S  : f B spq = χ ∗ f p + 2 j s ϕ j ∗ f p <∞ ⎩ ⎭ j =0 where S  is the dual of the Schwartz space (tempered distribution). s is the H¨older space. In this note we will use C∗1 to represent the In particular, the space B∞∞ 1 . space B∞∞ The following lemma is elementary.

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Lemma 4 (Bernstein Lemma, [5, p. 16, Lemma 2.1.1]). Assume f ∈ S  and supp fˆ ⊂ {x : |x| < r }; then for p ∈ [1, ∞], it holds that ∇ f p ≤ Cr f p . Furthermore if supp fˆ ⊂ {x : r < |x| < 5r }, then for p ∈ [1, ∞], it also holds that r f p ≤ C ∇ f p . For the axisymmetric data there is a classic result (see G´erard [10, Section 0.6]). Lemma 5. Assume that u 0 ∈ C s ∩ L p (s > 1, 1 < p < ∞) is a divergence free, axisymmetric velocity without swirl. Then the incompressible Euler system has a unique solution u ∈ C([0, T ]; C s ) ∩ Li p([0, T ]; L p ) for some T > 0, and the velocity field u(t) is an axisymmetric vector without swirl for any t ∈ [0, T ]. Moreover, if a0 = ωr0 ∈ L q for some ) q q ∈ [0, ∞], then a(t) = ω(t r ∈ L with a(t) q = a0 q for all t ∈ [0, T ]. Now we establish an equality for Euler equations due to Serfati [13,14], which will be used in the proof of Theorem 2. Lemma 6. If u ∈ C([0, ∞); C∗1 ∩ L 2 ) is a weak solution to the Euler equations and the vorticity ∞ 1 ω ∈ L∞ Loc ((0, ∞); L ), then there exist some functions f i j , gi j k ∈ L (i, j, k = 1, 2, 3) such that  t i i u (x, t) = u 0 (x) + f i j ∗ (ω − ω0 )(x, t) + gi j k ∗ u j u k (x, s)ds (4) 0

j

j,k

where f i j , gi j k satisfy the following conditions: supp f i j ⊂ {x : |x| < 2} supp gi j k ⊂ {x : |x| > 1} C C | f i j (x)| ≤ |gi j k (x)| ≤ 2 |x| |x|4 C C |∇ f i j (x)| ≤ |∇gi j k (x)| ≤ . |x|3 |x|5 Proof of Lemma 6. Firstly we take a nonnegative radial function ρ ∈ C0∞ (R 3 ) such that ρ(x) = 1,

|x| < 1;

ρ(x) = 0,

|x| > 2.

Set k(x) = C |x|x 3 . By virtue of the Biot–Savart law 

u(x, t) =  =

R3 R3

k(x − y) × ω(y, t)dy  (ρk)(x − y) × ω(y, t)dy +

R3

((1 − ρ)k)(x − y) × ω(y, t)dy.

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  Obviously R 3 (ρk)(x − y) × ω(y, t)dy can be represented as ( j f i j ∗ ω(x, t))i=1,2,3 and f i j satisfies the conditions in the lemma.

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On the other hand,  ((1 − ρ)k)(x − y) × (ω(y, t) − ω0 (y))dy R3  t = ((1 − ρ)k)(x − y) × ωt (y, s)dyds. R3

0

Note that div u = 0; if {i, j, k} = {1, 2, 3} we have j

ωtk = ∂i u t − ∂ j u it = ∂i (u l ∂l u j ) − ∂ j (u l ∂l u i ) l

l

(∂il2 (u l u j ) − ∂ 2j l (u l u i )). = l

Now it is obvious that gi j k can be taken as the linear combinations of D 2 ((1 − ρ)k). It is easy to check that gi j k satisfies the conditions in this lemma.  3. Proof of Theorem 1 By virtue of Lemma 5 and the result of Beale, Kato and Majda [2] as adapted to H¨older classes by Bahouri and Dehman [1], it suffices to prove that  t ω(s) ∞ < ∞. (6) 0

 ) θ Set r (x) = x 12 + x 22 , x ∈ R 3 , a(x, t) = ω(x,t r(x) . It is easy to check that ω = ω eθ for axisymmetric velocity without swirl. Due to Majda’s work [9] we know that ∂t a + u · ∇a = 0.

(7)

Suppose X (x, t) solves the initial value problem ∂ X (x, t) = u(X (x, t), t), X (x, 0) = x, x ∈ R 3 . (8) ∂t Then a(X (x, t), t) = a0 (x); as div u = 0, we can show that a(t) q = a0 q for any q ∈ [1, ∞]. Now we can get a key equality: ω(X (x, t), t) = r (X (x, t))a(X (x, t), t) = r (X (x, t))a0 (x) + ω(X (x, t0 ), t0 ) − r (X (x, t0 ))a(X (x, t0 ), t0 ) = ω(X (x, t0 ), t0 ) + (r (X (x, t)) − r (X (x, t0 )))a0 (x).

(9)

From (9) it is easy to check that ω(t) ∞ ≤ ω(t0 ) ∞ + |r (X (x, t)) − r (X (x, t0 ))| a0 ∞ ≤ ω(t0 ) ∞ + |X (x, t) − X (x, t0 )| a0 ∞  t ≤ ω(t0 ) ∞ + |u(X (x, s), s)|ds a0 ∞ t0

≤ ω(t0 ) ∞ + a0 ∞



t t0

u(s) ∞ ds.

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On the other hand, for the smooth solution u ∈ C s ∩ L 2 to the incompressible Euler equation the following energy equality holds: u(t) 2 = u 0 2 .

(11)

Using the Biot–Savart law we have ϕ j ∗ u(t) ∞ ≤ C2− j ϕ j ∗ ω(t) ∞ ≤ C2− j ω(t) ∞ . Now we can show that ∞ u(t) ∞ ≤ χ ∗ u(t) ∞ + ϕ j ∗ u(t) ∞ j =0

≤ C u(t) 2 +

∞

2− j ϕ j ∗ ω(t) ∞

j =0

≤ C( u 0 2 + ω(t) ∞ ).

(12)

From (10)–(12), taking t0 = 0 we get

  t ω(s) ∞ ds. ω(t) ∞ ≤ ω0 ∞ + a0 ∞ t u 0 2 + 0

By virtue of the Gr¨onwall inequality it is derived that ω(t) ∞ ≤ ( ω0 ∞ + u 0 2 ) exp(Ct a0 ∞ ) − u 0 2 . So it can be shown that ω(t) ∞ ≤ CeCt ;

u(t) ∞ ≤ CeCt

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where the constant C does not depend on t. Thus we complete the proof of Theorem 1. 4. Proof of Theorem 2 We will show the global existence of the weak solution to (1) by establishing an a priori estimate for an approximate solution sequence generated by regularizing initial data. First we establish an a priori estimate for a regular solution. Next we use strong convergence to obtain the solution to (1).  Take a radial function ρ ∈ C0∞ (B(0, 1)) with ρ ≥ 0, R 3 ρ(x)dx = 1. Set ρn (x) = n 3 ρ(nx), u 0,n = ρn ∗ u 0 . For any n ∈ N, u 0,n ∈ C ∞ and u 0,n 2 ≤ u 0 2 . Now we check that a0,n ∞ ≤ C a0 ∞ . Set Px as the orthogonal projection of x on the axis of symmetry; then |x − Px| = r (x) and by symmetry we have ρn ∗ ω0 (Px) = 0 for any x ∈ R 3 . When r (x) > n2 , if ρn (x − y) > 0 then r (y) < 2r (x), so we have    r (y) ω0 (y)  ρn (x − y) dy  ≤ 2 a0 ∞ . |a0,n (x)| =  r (x) r (y) When r (x) ≤ n2 , if ρn (x − y) > 0 then r (y) < n4 ; we can get |ρn ∗ ω0 (x)| |ρn ∗ ω0 (x) − ρn ∗ ω0 (Px)| = r (x) r (x)    1  = (ρn (x − y) − ρn (Px − y))ω0 (y)dy   r (x)

|a0,n (x)| =

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r (y) |ω0 (y)| |ρn (x − y) − ρn (Px − y)| dy r (x) r (y)  4 a0 ∞ 3 dw n |ρ(z − w) − ρ(Pz − w)| 3 ≤ nr ( nz ) n  C a0 ∞ r (z) ∇ρ ∞ dw ≤ r (z) |w−z|<2 ≤ C a0 ∞ . ≤

From Theorem 1, for any n there exists a solution u n . In the proof of Theorem 1, we have the following uniform estimates: u n (t) C∗1 ≤ C u n (t) ∞ + sup ϕ j ∗ ωn (t) ∞ ≤ C u n (t) ∞ + 2 ωn (t) ∞ ≤ CeCt (14) j

 ωn (t) − ωn (s) ∞ ≤ a0 ∞

t s

u(τ ) ∞ dτ ≤ CeCt (t − s).

(15)

(14) and (15) are the direct consequence of (13). Thus by Lemma 6 we can get   t j k |gi j k ∗ u n u n (x, τ )|dτ u n (t) − u n (s) ∞ ≤ C | f i j ∗ (ωn (t) − ωn (s))(x)| + s

i, j,k

  t 2 fi j 1 ωn (t) − ωn (s) ∞ + gi j k 1 u n ∞ dτ ≤C s

i, j,k

≤ CeCt (t − s). As u

1

2 B∞∞

(16)

≤ u C∗1 for any u ∈ C∗1 , from (14) and (16) we know that when |x − y| < 1, 0 <

s < t < s + 1, it holds that 1

|u n (x, t) − u n (y, s)| < C((t − s) + |x − y| 2 )eCt , i.e. {u n } is equicontinuous in R 3 × [0, T ] for any T > 0. By the Ascoli theorem there exists a subsequence of {u n } that converges to a function u in CLoc (R 3 × [0, ∞)). It is easy to check that u is a weak solution and satisfies u(t) C∗1 ≤ CeCt .

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Now we turn to the proof of uniqueness. The idea in our proof came from [14]. First, we point out a basic fact: |u(x) − u(y)| ≤ C u C∗1 |x − y|(1 − ln |x − y|), |x − y| < 1.

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The proof can be found in [5] (p. 31, Proposition 2.3.7). Consider the initial value problem ∂ X (x, t) = u(X (x, t), t), X (x, 0) = x, x ∈ R 3 . (19) ∂t From (17), (18) and Theorem 5.2.1 in [5, p. 90] (or [14]), we can show that there exists a unique −Ct solution X (x, t) to (19) with X (t) − Id ∈ C e . Suppose u 1 , u 2 are two weak solutions to (1) with the same initial data u 0 and u 1 , u 2 ∈ C([0, ∞); C∗1 ).

S. Gang, Z. Xiangrong / Nonlinear Analysis 66 (2007) 1938–1948

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Let X j be the solution of (19) for u j . Set l(t) = sup0 0}. As l(t) is continuous, l(t0 ) = 0. Without loss of generality, we can set t0 = 0 if t0 < ∞. In the remainder of this paper we prove that when t is small enough then l(t) = 0. This fact shows that inf{t : l(t) > 0} > 0, which means t0 = ∞, i.e. l(t) ≡ 0. Obviously h(0) = 0 and  t  t |u 1 (X 1 (x, s), s) − u 2 (X 2 (x, s), s)|ds ≤ h(s)ds. l(t) ≤ sup x

0

0

t

Take t small enough that 0 h(s)ds + t < 1e and u 1 (s) C∗1 + u 2 (s) C∗1 < C for any s ∈ [0, t]. Now we give the relation between h(t) and k(t). First we see that k(t) = sup |u 1 (X 1 (y, t), t) − u 2 (X 1 (y, t), t)| y

≤ sup(|u 1 (X 1 (y, t), t) − u 2 (X 2 (y, t), t)| + |u 2 (X 1 (y, t), t) − u 2 (X 2 (y, t), t)|) y

≤ h(t) + Cl(t) ln

1 . l(t)

(20)

1 . l(t)

(21)

Similarly we have h(t) ≤ k(t) + Cl(t) ln

We begin to compute k(t) with Lemma 6. For simplicity of exposition, we reformulation (4) as



t

u(x, t) = u 0 (x) + F ∗ (ω − ω0 )(x, t) +

G ∗ (u ⊗ u)(x, s)ds

(22)

0

where F, G, u ⊗ u denote matrices or vectors; |u 1 (x, t) − u 2 (x, t)|



t

≤ |F ∗ (ω1 − ω2 )(x, t)| + = A + B.

|G ∗ (u 1 ⊗ u 1 − u 2 ⊗ u 2 )(x, s)|ds

0

We estimate A as A = |F ∗ ω1 (x, t) − F ∗ ω2 (x, t)|      =  (F(x − X 1 (y, t))ω1 (X 1 (y, t), t) − F(x − X 2 (y, t))ω2 (X 2 (y, t), t))dy    (F(x − X 1 (y, t))ω1 (X 1 (y, t), t) ≤ 2  |x−X 1 (y,t )|≤|x−X 2(y,t )|   − F(x − X 2 (y, t))ω2 (X 2 (y, t), t))dy    (F(x − X 1 (y, t))ω1 (X 1 (y, t), t) ≤ 2  |x−X 2 (y,t )|<3l(t )

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  − F(x − X 2 (y, t))ω2 (X 2 (y, t), t))dy    + 2  (F(x − X 1 (y, t))ω1 (X 1 (y, t), t) |x−X 1 (y,t )|≥2l(t )   − F(x − X 2 (y, t))ω2 (X 2 (y, t), t))dy  ≤ A1 + A2 .

(24)

In the above inequality we use the fact |X 1 (y, t) − X 2 (y, t)| ≤ l(t). It is easy to see that  A1 ≤ 2 (|F(x − X 1 (y, t))| ω1 (t) ∞ + |F(x − X 2 (y, t))| ω2 (t) ∞ )dy |x−X 2 (y,t )|<3l(t )  |F(x − z)|( ω1 (t) ∞ + ω2 (t) ∞ )dy ≤2 |x−z|<3l(t )  1 dz ≤ Cl(t). (25) ≤C 2 |x−z|<3l(t ) |x − z| We recall the equality (9) (let t0 = 0) ω j (X j (x, t), t) = ω0 (x) + (r (X j (x, t)) − r (x))a0 (x). Now A2 can be computed as   A2 = 2 

(F(x − X 1 (y, t))ω1 (X 1 (y, t), t)   − F(x − X 2 (y, t))ω2 (X 2 (y, t), t))dy   ≤C (|F(x − X 1 (y, t)) − F(x − X 2 (y, t))||ω1 (X 1 (y, t), t)| |x−X 1 (y,t )|≥2l(t )

|x−X 1 (y,t )|≥2l(t )

+ |F(x − X 2 (y, t))||ω1 (X 1 (y, t), t) − ω2 (X 2 (y, t), t)|)dy

 l(t) ≤C ω1 (t) ∞ dz |x − z|3 |x−z|≥2l(t )   + |F(x − z)||r (X 1 (y, t)) − r (X 2 (y, t))|dz |x−z|≥2l(t )  1 1 ≤ Cl(t) ln +C l(t)dz 2 l(t) 2>|x−z|≥2l(t ) |x − z| 1 1 + l(t)) ≤ Cl(t) ln . (26) ≤ C(l(t) ln l(t) l(t) t Now it is only left to estimate B. Set L(t) = 0 h(s). As x ln 1x is increasing on (0, 1e ), from (20) we have  t G ∗ (u 1 ⊗ u 1 − u 2 ⊗ u 2 ) ∞ ds B ≤ 0  t G 1 ( u 1 ∞ + u 2 ∞ ) u 1 − u 2 ∞ ds ≤ 0

S. Gang, Z. Xiangrong / Nonlinear Analysis 66 (2007) 1938–1948

  t 1 ≤C h(s) + l(s) ln ds l(s) 0 

1 ≤ C L(t) + t L(t) ln L(t) 1 ≤ C L(t) ln . L(t)

1947

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1 Taking everything together we have k(t) ≤ C L(t) ln L(t ) , so it can be obtained that  t h(s)ds L(t) = 0   t 1 ≤ k(s) + Cl(s) ln ds l(s) 0  t 1 ds. (28) L(s) ln ≤C L(s) 0 t Set φ(x) = x ln 1x (x > 0), f (t) = C 0 φ(L(s))ds(< 1e ). Then f (0) = 0. As L(0) = 0 and φ(x) is increasing on [0, 1e ] we can get that

d f (t) = φ(L(t)) ≤ φ( f (t)). dt t  f (t ) dx So we have t ≥ 0 φ(d ff (s) (s)) = 0 φ(x) . On the other hand for any y > 0 we have  y dx dx = ∞ φ(x) 0 which implies f (t) ≡ 0 and so also L(t) ≡ 0 which yields l(t) ≡ 0. Now we can conclude that u1 = u2. References [1] H. Bahouri, B. Dehman, Remarques sur l’apparition de singularit e´ s dans les e´ coulements eul´eriens incompressibles a` donn´ee initiale h¨old´erienne, J. Math. Pures Appl. (9) 73 (4) (1994) 335–346. [2] J.T. Beale, T. Kato, A. Majda, Remarks on the breakdown of smooth solutions for the 3D Euler equations, Comm. Math. Phys. 94 (1) (1984) 61–66. [3] J. Bergh, J. L¨ofstr¨om, Interpolation Spaces, Springer-Verlag, Berlin, New York, Heidelberg, 1976. [4] A.L. Bertozzi, A. Majda, Vorticity and Incompressible Flow, in: Cambridge Texts in Applied Mathematics, vol. 27, 2002. [5] J.-Y. Chemin, Perfect Incompressible Fluids, Clarendon Press, Oxford, 1998. [6] J.-Y. Chemin, Sur le mouvement des particules d’un fluide parfait incompressible bidimensionnel, Invent. Math. 103 (3) (1991) 599–629. [7] D. Chae, N. Kim, Axisymmetric weak solutions of the 3D Euler equations for incompressible fluid flows, Nonlinear Anal. 29 (12) (1997) 1393–1404. [8] D. Chae, N. Kim, On the breakdown of axisymmetric smooth solutions for the 3-D Euler equations, Comm. Math. Phys. 178 (2) (1996) 391–398. [9] A. Majda, Vorticity and the mathematical theory of incompressible fluid flow, Comm. Pure Appl. Math. 39 (1986) S187–S220. [10] P. G´erard, R´esultats r´ecents sur les fluides parfaits incompressibles bidimensionnels (d’apr`es J.-Y. Chemin et J.-M. Delort), S´eminaire Bourbaki, vol. 1991/92. [11] R. Grauer, T.C. Sideris, Numerical computation of 3D incompressible ideal fluid with swirl, Phys. Rev. Lett. 67 (1991) 3511–3514.

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