On the profiles of locally self-similar solutions for the 2D inviscid Boussinesq equations

J. Math. Anal. Appl. 484 (2020) 123671

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On the profiles of locally self-similar solutions for the 2D inviscid Boussinesq equations Luman Ju Department of Mathematics and Statistics, Jiangsu Normal University, 101 Shanghai Road, Xuzhou 221116, Jiangsu, PR China

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Article history: Received 8 April 2019 Available online 18 November 2019 Submitted by D. Wang Keywords: 2D Boussinesq equations Locally self-similar solutions Pressure profiles Energy inequality

In this paper we investigate the locally self-similar solutions of the 2D inviscid Boussinesq equations and mainly focus on the possible velocity and temperature blowup profiles. We derive the formula of the pressure profiles in terms of the velocity and temperature profiles in the case that the possible nontrivial profiles have nondecaying spatial asymptotics. Then using the local energy inequality of the profiles and iterative method, we prove some nonexistence results and obtain the energy behavior concerning these possible profiles. © 2019 Elsevier Inc. All rights reserved.

1. Introduction and main results This paper examines the initial-value problem for the following two-dimensional (2D) inviscid Boussinesq equations ⎧ ⎪ ∂t u + u · ∇u + ∇p = θe2 , ⎪ ⎪ ⎪ ⎪ ⎨ ∂t θ + u · ∇θ = 0,

(x, t) ∈ R2 × R+ ,

⎪ ∇ · u = 0, ⎪ ⎪ ⎪ ⎪ ⎩ u(0, x) = u (x), θ(0, x) = θ (x), 0 0

(1.1)

where u = (u1 , u2 ) is the velocity field, p is the scalar pressure field, θ is the temperature field and e2 = (0, 1)T is the unit vector in the vertical direction. The Boussinesq equations model the motion of lighter or denser fluid under the influence of gravitational forces and describe Raleigh-Bénard convection (see for example [28,12,8,25]). Besides their wide physical applicability, the Boussinesq equations are also of great interest in mathematics and have recently attracted enormous attention, especially the global regularity results (see for example [4,18,3,16,17,9,22,27,31,33,34,38,37]). Mathematically the 2D Boussinesq equations serve as a lower E-mail address: [email protected]. https://doi.org/10.1016/j.jmaa.2019.123671 0022-247X/© 2019 Elsevier Inc. All rights reserved.

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

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dimensional model of the three-dimensional (3D) hydrodynamics equations. Actually, the 2D Boussinesq equations retain some key features of the following classical 3D incompressible Euler equations ⎧ ⎪ ∂t u + u · ∇u + ∇p = 0, ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

(x, t) ∈ R3 × R+ ,

∇ · u = 0,

(1.2)

u(0, x) = u0 (x).

Indeed, as pointed out in [26], the inviscid 2D Boussinesq equations are analogous to the Euler equations for 3D axisymmetric swirling flows. On one hand, a recent progress concerning the 2D inviscid Boussinesq equations has been made by Elgindi and Jeong [11], where they have shown finite-time singularity formation for strong solutions when the fluid domain is a sector of angle less than π. On the other hand, assume that (u0 , θ0 ) ∈ H s (R2 ) with s > 2, then (1.1) has a unique local smooth solution (u, θ) ∈ C([0, T ); H s (R2 )) with T = T (u0 , θ0 ) > 0. Moreover, the pressure p can be expressed, up to a harmonic polynomial, by p = div div(−Δ)−1 (u ⊗ u) − ∂x2 (−Δ)−1 θ,

(1.3)

or precisely, p(x, t) = −

|u(x, t)|2 + p.v. 2



 Kij (x − y)ui (y, t)uj (y, t) dy + p.v.

R2

H(x − y)θ(y, t) dy,

(1.4)

R2

where Kij is the Calderón-Zygmund kernel given by Kij (y) =

1 2yi yj − |y|2 δij , i, j = 1, 2 2π |y|4

(1.5)

and H(y) is given by H(y) =

1 y2 . 2π |y|2

(1.6)

We remark that the Einstein convention on repeated indices is used here and thereafter. Obviously, (1.1) has an important scaling invariant property, namely, if (u, p, θ) is the solution of (1.1), so does (u(λ) , p(λ) , θ(λ) ), where u(λ) (x, t)  λα u(λ1+α t, λx), p(λ) (x, t)  λ2α p(λ1+α t, λx), θ(λ) (x, t)  λ2α+1 θ(λ1+α t, λx), for λ > 0 and α = −1. Based on this property, it is natural to consider the following locally self-similar solutions of (1.1)   x−x 1 0 , α U 1 (T − t) α+1 (T − t) α+1  x−x  1 0 p(x, t) = + d(t), 2α P 1 (T − t) α+1 (T − t) α+1  x−x  1 0 θ(x, t) = , 2α+1 Θ 1 (T − t) α+1 (T − t) α+1 u(x, t) =

(1.7) (1.8) (1.9)

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

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where (U, P, Θ) are stationary profiles, (x, t) ∈ Bρ (x0 ) × (0, T ), t < T, α > −1, x0 ∈ R2 , ρ > 0, and the solution (u, θ) remains regular outside the ball Bρ (x0 ). d(t) is a function depending only on t. We point out that the expression formula (1.8) can be derived from the formulas (1.7) and (1.9) (see Lemma 2.1 for details). If ρ = ∞, this corresponds to the “globally” self-similar solutions; while if ρ < ∞, these are the “locally” self-similar solutions. Chae [6, Theorem 3.2] studied the globally self-similar solutions of 2D Boussinesq equations. The motivation to study such a blow-up scenario comes from numerical evidence suggesting that singular solutions tend to form self-similar structures (see, e.g., [1,13,19–21,23,24,29,10]). In fact, Hou and Luo investigated the finite-time blowup of the 3D incompressible Euler equations and found a class of potentially singular solutions via some numerical studies [19–21]. Moreover, Hou-Luo [21] observed the potentially singular solutions of the 2D Boussinesq equations. If (u, p) is a distributional solution to (1.2), then the pair (U, P ) given by (1.7) and (1.8) satisfies ⎧ ⎨

α α+1 U

+

1 α+1 y

· ∇U + U · ∇U + ∇P = 0,

⎩∇ · U = 0.

(1.10)

The self-similar singular solutions of Euler equations and their properties have been intensely studied in the mathematical literature (see, e.g., [2,5–7,19–21,30,15,14,32,35]). In [15], He constructed the non-trivial solutions to (1.10) with α = 1 on the exterior domain B1c (0), and the asymptotic decay of such solutions are |U (y)|  |y|−1 and |∇U (y)|  |y|−2 . For the “locally” self-similar solutions of 3D Euler equations, Chae 1 and Shvydkoy [7] proved that if U ∈ Cloc (RN ) ∩ Lr (RN ) with r ∈ [3, ∞], then U ≡ 0 for all −1 < α < Nr N or α > 2 . In the case of α > 0, Bronzi and Shvydkoy [2] gave the expression formula of pressure for the 3 locally self-similar solutions and proved that, for some p ≥ 3, γ < p − 2, 0 < α < N/2, if U ∈ Cloc (RN ) satisfies  |U (y)|p dy  Lγ , ∀L  1, |y|≈L

then, either U ≡ 0 or 

|U (y)|2 dy ≈ LN −2α ,

∀L  1.

(1.11)

|y|≤L

Recently, for the velocity with non-decaying profiles, Xue [35] established the expression formula of pressure for the locally self-similar solutions of N-dimension Euler equations. Precisely, for α > −1, 0 ≤ δ ≤ 1, 0 < 0  1, if the velocity profile U satisfies that |y|0  |U (y)|  |y|δ ,

∀|y|  1,

then U satisfies (1.11) for all −δ ≤ α ≤ − 0 . Inspired by the previous works [2,35,36], the work presented here contributes to establish the profiles estimates of the locally self-similar solutions for the 2D inviscid Boussinesq equations. The main results can be stated as follows. Theorem 1.1. Suppose that (u, θ) ∈ C([0, T ); H  (R2 )) and θ ∈ L∞ ([0, T ); L1 (R2 )) with  > 2 are the locally self-similar solutions of (1.1) on the spacetime domain Bρ (x0 ) × (0, T ) with the form (1.7)-(1.9), and 3 2 U ∈ Cloc (R2 ), Θ ∈ Cloc (R2 ). In addition, assume that, for p ∈ (1, ∞) and for some r > p, 0 ≤ γ ≤ r − p,  |Θ(y)|r dy  Lγ , |y|≤L

∀L  1,

(1.12)

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

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and for some δ1 ∈ (0, 1), |U (y)|  |y|δ1 ,

∀|y|  1.

(1.13)

1 1 1 Then the possible scope of α to admit nontrivial temperature profiles Θ is [ 2−γ 2r − 2 , p − 2 ] and for every such an α the corresponding profiles satisfy that

 |Θ(y)|p dy ≈ L2−p(2α+1) ,

∀L  1.

(1.14)

|y|≤L

As a consequence of Theorem 1.1, we have the following corollary. Corollary 1.1. Under the assumptions of Theorem 1.1, the following statements hold true. (1) For all |y|  1, if there exists some μ > 0 such that |Θ(y)|  (2) If there exists some δ2 ∈ [0, 1) such that 1  |Θ(y)|  |y|δ2 ,

1 |y|μ ,

then Θ ≡ 0.

∀|y|  1,

then the range of index α admitting nontrivial profiles is − δ22 − sponding to each α satisfy (1.14) for every p ∈ (1, ∞).

(1.15) ≤ α ≤ − 12 , and the profiles corre-

1 2

In addition, if some more conditions are restricted on the profiles (U, Θ), then we have the estimate on the profile U . Theorem 1.2. Under the assumptions of Theorem 1.1, if δ1 ∈ [ 12 , 1) and δ2 ∈ [0, 12 ) such that 1

|y| 2  |U (y)|  |y|δ1 ,

1  |Θ(y)|  |y|δ2 ,

∀|y|  1,

(1.16)

1 2 then the only possible range of index α admitting nontrivial profiles is max{−δ1 , − 1+δ 2 } ≤ α ≤ − 2 , and for p ∈ (1, ∞), the corresponding profile Θ still satisfies (1.14) and the corresponding profile U satisfies

 |U (y)|2 dy ≈ L2−2α ,

∀L  1.

(1.17)

|y|≤L

Remark 1.1. Note that Theorem 1.2 implies that for α ∈ (− 34 , − 12 ], we can expect the corresponded “typical” nontrivial profiles as |U (y)| ≈ |y|−α , |Θ(y)| ≈ |y|−(2α+1) ,

∀|y|  1.

(1.18)

Moreover, under the suitable condition that profiles increase like (1.16) at infinity, according to the proof of Theorem 1.2, we can expect (1.18) to work for all α ∈ (−1, − 12 ]. Throughout this paper, C stands for a constant which may be different from line to line. A  B means that there is a harmless constant C such that A ≤ CB, and A ≈ B means A  B and B  A. We denote Br (x0 )  {x ∈ R2 : |x − x0 | ≤ r} and Brc (x0 )  R2 \Br (x0 ) its complement set. The rest of the paper is organized as follows. In Section 2, we establish two key lemmas. Section 3 is devoted to deriving the locally energy inequalities of the profiles. In Section 4, we carry out the proof of Theorem 1.1 and Corollary 1.1. Section 5 is devoted to the proof of Theorem 1.2.

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

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2. Two key lemmas In this section we establish two key lemmas: one concerns the justification of the expression formula of pressure p, and the other one concerns the estimates of each term involving in the pressure profiles. The two lemmas play some crucial roles in proving Theorem 1.1 and Theorem 1.2. Before proving the two key lemmas, we state the following basic energy estimates for the Boussinesq equations (1.1) u(t)L2 (R2 ) ≤ u0 L2 (R2 ) + tθ0 L2 (R2 ) ,

(2.1)

and the Lp -estimate for p ∈ [1, ∞] θ(t)Lp (R2 ) ≤ θ0 Lp (R2 )

(2.2)

for any 0 ≤ t ≤ T . Since (u, θ) ∈ C([0, T ); H  (R2 )) and θ ∈ L∞ ([0, T ); L1 (R2 )) with  > 2, the estimate (2.2) is a consequence of the incompressibility of the flow, and the estimate (2.1) can be obtained by taking the L2 -inner product of the velocity equation. 2.1. The expression formula of pressure First, let us state the main result of this subsection. Lemma 2.1. Suppose that (u, θ) ∈ C([0, T ); H  (R2 )) and θ ∈ L∞ ([0, T ); L1 (R2 )) with  > 2 are the locally 3 self-similar solutions for (1.1), and satisfy (1.7)-(1.9) in Bρ (x0 ) × (0, T ). Assume that (U, Θ) ∈ Cloc (R2 ) × 2 Cloc (R2 ) satisfies for some δ1 , δ2 ∈ [0, 1) |U (y)|  |y|δ1 ,

|Θ(y)|  |y|δ2 ,

∀|y|  M,

where M > 0 is an absolute constant. Then the corresponding pressure p has the following expression formula in Bρ (x0 ) × (0, T ):

p(x, t) =



1 (T − t)

2α α+1

P

x − x0 (T − t)

 + d(t),

1 α+1

(2.3)

2 where d(t) is a function depending only on t, and P (y) is a Cloc (R2 )-smooth scalar function (in the meaning of constant difference) expressed as

1 P (y) = − |U (y)|2 + p.v. 2  + p.v. |z|≤M



 Kij (y − z)Ui (z)Uj (z) dz +

|z|≤M



H(y − z)Θ(z) dz +

ij (y, z)Ui (z)Uj (z) dz K

|z|≥M

(2.4)

z)Θ(z) dz + A · y H(y,

|z|≥M

with A ∈ R2 some fixed constant vector (especially, A = 0 at the case {α > − 12 , δ1 < ij and H given by {α > − 12 , (U, Θ) ∈ Lp × Lq , p, q ∈ (2, ∞)}), and the kernel function K

1 2 , δ2

< 1} or

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

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ij (y, z) = K

⎧ ⎪ Kij (y − z), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪Kij (y − z) − Kij (z), ⎪ ⎨

if U ∈ Lp (R2 ), p ∈ (2, ∞), if 1  |U (z)|  |z|δ1 , δ1 ∈ [0, 12 ),

⎪ ⎪ ⎪ ⎪ ⎪ Kij (y − z) − Kij (z) − y · ∇Kij (z), ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

⎧ ⎪ H(y − z), ⎪ ⎪ ⎨ z) = H(y − z) − H(z), H(y, ⎪ ⎪ ⎪ ⎩ H(y − z) − H(z) − y · ∇H(z),

|z| ≥ M, 1

if |z| 2  |U (z)|  |z|δ1 , δ1 ∈ [ 12 , 1), |z| ≥ M,

if Θ ∈ Lq (R2 ), q ∈ (1, 2), if Θ ∈ Lq (R2 ), q ∈ [2, ∞), if 1  |Θ(z)|  |z|δ2 , δ2 ∈ [0, 1), |z| ≥ M,

where Kij (i, j = 1, 2) and H given by (1.5) and (1.6), respectively. Proof. The proof of Lemma 2.1 is largely inspired by [2, Lemma 2.1] and [35, Lemma 2.1]. Firstly, we introduce a function Π(y), which is main part of (2.4), then we show that it is a well-defined tempered distribution, and satisfies ΔΠ = −div div(U ⊗ U ) + ∂2 Θ. To this end, we denote Π(y) as   1 ij (y, z)Ui (z)Uj (z) dz Π(y)  − |U (y)|2 + p.v. Kij (y − z)Ui (z)Uj (z) dz + K 2 |z|≤M

 + p.v.



H(y − z)Θ(z) dz +

|z|≤M

|z|≥M

z)Θ(z) dz. H(y,

(2.5)

|z|≥M

Let φ ∈ Cc∞ (R) be a smooth test function with support on B1 (0), 0 ≤ φ ≤ 1, and satisfies that φ ≡ 1 on B 12 (0). For every L ≥ M , we denote φL (z) = φ( Lz ). Therefore, this allows us to deduce 1 Π(y) = − |U (y)|2 + p.v. 2 

Kij (y − z)Ui (z)Uj (z)φ4L (z) dz

R2

H(y − z)Θ(z)φ4L (z) dz

+ p.v. 



R2

+

ij (y, z) − Kij (y − z))Ui (z)Uj (z)φ4L (z) dz (K

|z|≥M



ij (y, z)Ui Uj (z)(1 − φ4L (z)) dz K

+

(2.6)

R2



+

z) − H(y − z))Θ(z)φ4L (z) dz (H(y,

|z|≥M



+

z)Θ(z)(1 − φ4L (z)) dz H(y,

R2 6

1 Πi,L (y).  − |U (y)|2 + 2 i=1 3 Since U ∈ Cloc (R2 ), by the Besov embedding, we can deduce that Π1,L (y) ∈ C β for all β < 3. According to the definition of the function H, we derive

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

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Π2,L (y) = −∂y2 (−Δ)−1 (Θ φ4L ). z 2 Due to Θ ∈ Cloc (R2 ) and φ4L (z) = φ( 4L ) ∈ Cc∞ (R), one may conclude that Π2,L (y) ∈ C β for all β < 3. For Π3,L (y) and Π5,L (y), we obtain

ij (y, z) − Kij (y − z) = K

⎧ ⎪ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −Kij (z), ⎪ ⎪ ⎨

if U ∈ Lp (R2 ), p ∈ (2, ∞), if 1  |U (z)|  |z|δ1 , δ1 ∈ [0, 12 ), |z| ≥ M,

⎪ ⎪ ⎪ ⎪ ⎪ −Kij (z) − y · ∇Kij (z), ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

z) − H(y − z) = H(y,

⎧ ⎪ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨−H(z),

1

if |z| 2  |U (z)|  |z|δ1 , δ1 ∈ [ 12 , 1), |z| ≥ M, if Θ ∈ Lq (R2 ), q ∈ (1, 2), if Θ ∈ Lq (R2 ), q ∈ [2, ∞),

⎪ ⎪ −H(z) − y · ∇H(z), ⎪ ⎪ ⎪ ⎪ ⎩

if 1  |Θ(z)|  |z|δ2 , δ2 ∈ [0, 1), |z| ≥ M.

1

We thus deduce for all y ∈ BL (0) and |z| 2  |U (z)|  |z|δ1 with δ1 ∈ [0, 1) that Π3,L (y)  L2δ1 .

(2.7) 1

In the case when s = 1, 2, we find that ∇sy (Π3,L (y)) = 0 in any case except that while |z| 2  |U (z)|  |z|δ1 with δ1 ∈ [ 12 , 1), 

|∇Kij (z)||U (z)|2 dz  L2δ1 −1 .

|∇1y (Π3,L (y))|  M ≤|z|≤4L

Similarly, for all y ∈ BL (0) and 1  |Θ(z)|  |z|δ2 with δ2 ∈ [0, 1), one gets Π5,L (y)  Lδ2 +1 . In the case when s = 1, 2, ∇sy (Π5,L (y)) = 0 in any case except that while 1  |Θ(z)|  |z|δ2 with δ2 ∈ [0, 1),  |∇1y (Π5,L (y))| 

|∇H(z)||Θ(z)| dz  Lδ2 . M ≤|z|≤4L

In terms of Π4,L (y), for y ∈ BL (0) and U ∈ Lp (R2 ), we obtain for p ∈ (2, ∞) that Π4,L (y) 

k=0





∞

∞

1 |U (z)|2 dz |z|2

2k L≤|z|≤2k+1 L

(2k L)−2+2(1− p ) U 2Lp

k=0

 L− p . 4

2

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For |z| ≥ M , 1  |U (z)|  |z|δ1 with δ1 ∈ [0, 12 ), then  Π4,L (y)  |z|≥2L

|y| |U (z)|2 dz  L2δ1 . |z|3

1

For all |z| ≥ M , |z| 2  |U (z)|  |z|δ1 with δ1 ∈ [ 12 , 1), then  Π4,L (y)  |z|≥2L

|y|2 |U (z)|2 dz  L2δ1 . |z|4

For s = 1, 2, y ∈ BL (0) and U ∈ Lp (R2 ) with p ∈ (2, ∞), we have 

∞

∇sy (Π4,L (y)) 

k=0

2k L≤|z|≤2k+1 L

4 1 |U (z)|2 dz  L−s− p . 2+s |z|

For 1  |U (z)|  |z|δ1 with δ1 ∈ [0, 12 ), ∀|z| ≥ M , we have   1 ∇sy (Π4,L (y))

=

∇sy 

y · ∇Kij (τ y − z)(1 − φ4L (z))Ui Uj (z) dτ dz

R2

 |z|≥2L

0

|y| |U (z)|2 dz |z|3+s

 L−s+2δ1 , 1

while |z| 2  |U (z)|  |z|δ1 with δ1 ∈ [ 12 , 1), ∀|z| ≥ M , we also obtain that   1 1 ∇sy (Π4,L (y))

=

∇sy 

 |z|≥2L

R2 0



y · ∇ Kij (τ ηy − z) · y (1 − φ4L (z))Ui Uj (z)τ dη dτ dz 2

0

|y| |U (z)|2 dz |z|4+s 2

 L−s+2δ1 . Similarly, for Θ ∈ Lq (R2 ) with q ∈ (1, 2), it yields Π6,L (y) 



∞

k=0

∞

2k L≤|z|≤2k+1 L

1 2 1 |Θ(z)| dz  (2k L)−1+2(1− q ) ΘLq  L1− q , |z| k=0

while if Θ ∈ Lq (R2 ) with q ∈ [2, ∞), then Π6,L (y) 

∞

k=0

 2k L≤|z|≤2k+1 L

∞

1 y −2+2(1− q1 ) 1− q2 q k |Θ(z)| dz  L (2 L) Θ . q  L L |z|2 k=0



L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

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For |z| ≥ M and 1  |Θ(z)|  |z|δ2 with δ2 ∈ [0, 1), one shows  Π6,L (y)  |z|≥2L

|y|2 |Θ(z)| dz  Lδ2 . |z|3

(2.8)

For s = 1, 2 and y ∈ BL (0), for Θ ∈ Lq (R2 ) with q ∈ (1, 2), we have 

∞

∇sy (Π6,L (y)) 

k=0

2 1 |Θ(z)| dz  L−s−1− q , 1+s |z|

2k L≤|z|≤2k+1 L

while Θ ∈ Lq (R2 ) with q ∈ [2, ∞), we have

  1 ∇sy (Π6,L (y))

=

∇sy

y · ∇H(τ y − z)(1 − φ4L (z))|Θ(z)| dτ dz R2

 

|z|≥2L

0

|y| |Θ(z)| dz |z|2+s

 L−s−1− q , 2

while 1  |Θ(z)|  |z|δ2 with δ2 ∈ [0, 1), ∀|z| ≥ M , we have   1 1 ∇sy (Π6,L (y)) = ∇sy   |z|≥2L

R2 0



y · ∇2 Hij (τ ηy − z) · y (1 − φ4L (z))Θ(z)τ dη dτ dz



0

|y| |Θ(z)| dz |z|3+s 2

 L−s−1+δ2 . Combining all the above estimates, Π(y) defined by (2.5) is C 2 -smooth on BL (0), and for every y ∈ BL (0), we have  1  ΔΠ = Δ − |U (y)|2 + Π1,L (y) + Π2,L (y) + Δ(Π3,L (y) + Π4,L (y) + Π5,L (y) + Π6,L (y)) 2   = −div div(U φ4L ⊗ U φ4L ) + ∂2 (φ4L Θ) = −div div(U ⊗ U ) + ∂2 Θ, c where we have used Δ(Π3,L (y) + Π4,L (y) + Π5,L (y) + Π6,L (y)) = 0 because y ∈ BL (0) and z ∈ B2L (0), Kij (y, z) and H(y, z) are harmonic for y. Our next goal is to find a pressure profile Q, which is just a distribution and satisfies the following equation

α 1 U+ y · ∇U + U · ∇U + ∇Q = Θe2 . α+1 α+1 Noting (1.7)-(1.9) and letting y

x − x0 1

(T − t) α+1

,

p(x, t)  p(y, t),

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

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we can show for |y| ≤ ρ(T − t)− α+1 (namely x ∈ Bρ (x0 )) that 1

 2α α 1 U (y) + y · ∇y U (y) + U · ∇y U (y) + ∇y (T − t) α+1 p(y, t) = Θe2 . α+1 α+1 2α

For t < T , let f (y, t) = (T − t) α+1 p(y, t), then the vector function ∇y f (y, t)  g(y) only depends on the 1 variable y on D(t)  {y : |y| ≤ ρ(T − t)− α+1 }. Thus, for every y ∈ D(t), it follows that 2α

2α

(T − t) α+1 p(y, t) − (T − t) α+1 p(0, t) = f (y, t) − f (0, t) 1 =

d f (sy, t) ds ds

0

1 y · ∇f (sy, t) ds

= 0

1 y · g(sy) ds

= 0

= : Q(y). Therefore, we arrive at  2α p(x, t) = (T − t)− α+1 Q

x − x0 1

(T − t) α+1



∀x ∈ Bρ (x0 ),

+ c(t),

(2.9)

where c(t) = p(x0 , t). We now show that Q(y) is a tempered distribution. By (1.3), we divide the original pressure into p(x, t) = p1 (x, t) + p2 (x, t), where p1 (x, t) = div div(−Δ)−1 (u ⊗ u),

p2 (x, t) = ∂2 Δ−1 θ.

According to the energy equality of the velocity field u, we infer that p1 (x, t)L1weak  u(x, t)2L2  u0 2L2 + t2 θ0 2L2  1 + T 2 . This implies that for all t < T , we have |{x : |p1 (x, t)| > λ/2}| ≤

C λ.

Notice that

p2 (x, t)L4 ≤ Cθ(t)L4/3 ≤ Cθ0 L4/3 , we have |{x : |p2 (x, t)| > λ/2}| ≤ Thus, for λ > 1, it yields |{x : |p(x, t)| > λ}| ≤

C λ.

1 C p2 (x, t)4L4 ≤ 4 . λ4 λ

ρ) > 0 such that Therefore, there exists a small η = η(C,

    π 2 2 2  x : |p(x, t)| > 1 (T − t)− (α+1)  ≤ ρ (T − t) α+1 .   η 2

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

11

1

This implies that in the ball {x : |x − x0 | ≤ ρ(T − t) α+1 }, there exists a point xt such that 2 1 (T − t)− (α+1) . η

|p(xt , t)| ≤

Considering xt and the corresponding yt =

xt −x0

1

(T −t) α+1

∈ Bρ (0) yields

|c(t)| ≤ (T − t)− α+1 |Q(yt )| + η −1 (T − t)− α+1  (T − t)− α+1 + (T − t)− α+1 , 2α

2

2α

2

2 where we used the fact |Q(yt )| ≤ C due to Q(y) ∈ Cloc (R2 ). By (2.9), we see that:

2α

1

2α

Q(y) = (T − t) α+1 p(x0 + y(T − t) α+1 , t) + (T − t) α+1 c(t),

∀|y| ≤ ρ(T − t)− α+1 . 1

Therefore, for some p˜ ∈ (4, ∞), one has 

p ˜

|Q(y)| 2 dy ρ 1 4(T −t) α+1

≤|y|≤

ρ 1 2(T −t) α+1

 (T − t)− α+1 + (T − t) 2

(α−1)p−2 ˜ α+1

+ (T − t)



pα−2 ˜ α+1

p ˜

|p(x, t)| 2 dx ρ ρ 4 ≤|x−x0 |≤ 2

2 − α+1

+ (T − t)

 (T − t)

+ (T − t)

pα−2 ˜ α+1

(α−1)p−2 ˜ α+1

+ (T − t)

|u(x, t)|p˜ dx ρ 8 ≤|x−x0 |≤ρ







pα−2 ˜ α+1

2p ˜

˜ |θ(x, t)| p+4 dx

˜  p+4 4

+ (T − t)

pα−2 ˜ α+1



p ˜

u(t)pL˜ 2 + θ(t)L2 1



(2.10)

ρ 8 ≤|x−x0 |≤ρ

 (T − t)− α+1 + (T − t)  + 2

ρ 8(T −t)1/(α+1)

≤|y|≤





(α−1)p−2 ˜ α+1

+ (T − t)

2(pα−2) ˜ α+1

|U (y)|p˜ dy

ρ (T −t)1/(α+1) 2p ˜

˜ |Θ(y)| p+4 dy

+ ρ 8(T −t)1/(α+1)

˜  p+4 4

.

ρ ≤|y|≤ (T −t)1/(α+1)

We thus infer that there exists some m1 ∈ N such that  ρ 1 4(T −t) α+1

≤|y|≤

p ˜

|Q(y)| 2 dy  (T − t)−m1 . ρ 1 2(T −t) α+1

The deduction of (2.10) from line 2 to line 3, we have applied the following facts

(2.11)

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

12

p(x, t) |u(x, t)|2 + p.v. =− 2 |u(x, t)|2 + =− 2  +

 Kij (x − y)ui (y, t)uj (y, t) dy + p.v.

R2







p.v.

+

+

p.v.





+ ρ 8 ≤|z−x0 |≤ρ

H(x − y)θ(y, t) dy



R2

Kij (x − z)ui uj (z, t) dz



|z−x0 |≥ρ

ρ 8 ≤|z−x0 |≤ρ







+

|z−x0 |≤ ρ 8

|z−x0 |≤ ρ 8

−



H(x − z)θ(z, t) dz



|z−x0 |≥ρ

|u(x, t)|2 + p1,K (x, t) + p2,K (x, t) + p3,K (x, t) + p1,H (x, t) + p2,H (x, t) + p3,H (x, t) 2

and the corresponding estimates ρ ρ p1,K (x, t)L∞  u(x, t)2L2x , x ({ 4 ≤|x−x0 |≤ 2 })   p ˜ 2 |p2,K (x, t)| dx  |u(x, t)|p˜ dx, ρ ρ 4 ≤|x−x0 |≤ 2

ρ 8 ≤|x−x0 |≤ρ



ρ ρ p3,K (x, t)L∞  x ({ 4 ≤|x−x0 |≤ 2 })

|z−x0 |≥ρ

1 |u(z, t)|2 dz  u(x, t)2L2x , |z − x0 |2

ρ ρ p1,H (x, t)L∞  θ(x, t)L1x , x ({ 4 ≤|x−x0 |≤ 2 })

⎛



p ˜ ⎜ |p2,H (x, t)| 2 dx  ⎝

ρ ρ 4 ≤|x−x0 |≤ 2

˜ ⎞ p+4 4



2p ˜ ⎟ ˜ |θ(x, t)| p+4 dx⎠

,

ρ 8 ≤|x−x0 |≤ρ



p3,H (x, t)

ρ ρ L∞ x ({ 4 ≤|x−x0 |≤ 2 })

 |z−x0 |≥ρ

1 |θ(z, t)| dz  θ(x, t)L1x . |z − x0 |

By (2.11), we infer that Q(y) grows at most polynomially to infinity. So Q(y) is a tempered distribution. Now we show that Q and Π differ by a first order harmonic polynomial. Since they both solve the Poisson equation ΔΠ = −div div(U ⊗ U ) + ∂2 Θ = ΔQ, and are both distributions, their difference hQ−Π

(2.12)

is a harmonic polynomial. Then we show that the order of h is not greater than 1. For |y| ≤ inserting (1.7) and (1.9) into (1.4), using (2.6) and 4L = ρ(T − t)− α+1 , we have 1

2α

1

(T − t) α+1 p(x0 + y(T − t) α+1 , t)    z−x 1 1 0 2 = − |U (y)| + p.v. Kij (x0 + y(T − t) α+1 − z)(φρ (z − x0 )(Ui Uj ) dz 1 2 (T − t) α+1 2α



+(T − t) α+1

R2

1

Kij (x0 + y(T − t) α+1 − z)(1 − φρ (z − x0 ))(ui uj )(z, t) dz R2

ρ 1

4(T −t) α+1

,

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

1 − α+1

+(T − t)

 R2



2α

+(T − t) α+1

 1 H(x0 + y(T − t) α+1 − z)φρ (z − x0 )Θ

p.v.

13

z − x0 (T − t)

 dz

1 α+1

1

H(x0 + y(T − t) α+1 − z)(1 − φρ (z − x0 ))θ(z, t) dz R2

1 = − |U (y)|2 + p.v. 2



 1 Kij (x0 + y(T − t) α+1 − z)(φρ (z − x0 )(Ui Uj )

R2



+(T − t)− α+1 p.v. 1

 1 H(x0 + y(T − t) α+1 − z)φρ (z − x0 )Θ

R2



z − x0 1

(T − t) α+1

z − x0 (T − t)

dz



1 α+1

dz

+˜ p(y, t) t) + p˜(y, t), = Π(y) − Π(y, where p˜(y, t) (T − t)



2α α+1

1

Kij (x0 + y(T − t) α+1 − z)(1 − φρ (z − x0 ))(ui uj )(z, t) dz R2

+ (T − t) t)  Π(y,



2α α+1

1

H(x0 + y(T − t) α+1 − z)(1 − φρ (z − x0 ))θ(z, t) dz, R2



H(y − z)φ4L (z)Θ(z) dz

R2



ij (y, z) − Kij (y − z))φ4L (z)Ui (z)Uj (z) dz (K

+ |z|≥M



ij (y, z)(1 − φ4L (z))Ui Uj (z) dz K

+ R2



z) − H(y − z))φ4L (z)Θ(z) dz (H(y,

+ |z|≥M



+

z)(1 − φ4L (z))Θ(z) dz. H(y,

R2

Thanks to (2.9), one deduces 2α

1

(T − t) α+1 p(x0 + y(T − t) α+1 , t) = Q(y) + d(t)

(2.13)

2α

where d(t)  (T − t) α+1 c(t). Thus, it follows from (2.12)-(2.13) that t)|, |h(y) − d(t)| ≤ |˜ p(y, t)| + |Π(y,

∀|y| ≤

ρ 1

4(T − t) α+1

.

1

For p˜(y, t), since z ∈ B cρ (x0 ) and (T − t) α+1 y ∈ B ρ4 (0), we have 2

1

|Kij (x0 + y(T − t) α+1 − z)| 

1 , ρ2

1

|H(x0 + y(T − t) α+1 − z)| 

1 , ρ

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

14

which along with (2.1) and (2.2) yield 2α

2α

|˜ p(y, t)|  (T − t) α+1 (u2L2 + θL1 )  (T − t) α+1 . by the division of (2.6), we have For Π, Π(y) = Π3,L (y) + Π4,L (y) + Π5,L (y) + Π6,L (y), by (2.7)-(2.8) and L = 14 ρ(T − t)− α+1 , we get: 1

t)|  (T − t)− |Π(y,

max{2δ1 ,δ2 } α+1

,

∀y ≤

ρ 1

4(T − t) α+1

.

Since α > −1, we infer that the order of the harmonic polynomial h(y) is at most one. Otherwise, there 1 2 exists |y| ≤ 14 ρ(T − t)− α+1 such that |h(y)|  (T − t)− α+1 , which contradicts with the above conclusion. It thus implies |d(t)|  1 + (T − t) α+1 + (T − t)− 2α

max{2δ1 ,δ2 } α+1

,

which also proves (2.3)-(2.4). Therefore, we complete the proof of Lemma 2.1.  2.2. The estimates of each term involving in the pressure profiles 1 1 Lemma 2.2. Suppose U ∈ Cloc (R2 ) and Θ ∈ Cloc (R2 ) are locally regular vector field, satisfy that for 1/2 ≤ δ2 1 δ ≤ 1, 0 < δ2 < 1, 0 < b ≤ 2 + 2δ, − 2 − 2 ≤ α ≤ − 12 , 1

|y| 2  |U (y)|  |y|δ ,  |U (y)|2 dy  Lb ,

∀|y| ≥ M, ∀L ≥ M,

|y|≤L



|Θ(y)|p dy ≈ L2−p(2α+1) ,

∀L ≥ M, p ∈ (1, ∞),

|y|≤L

where M > 0 is a fixed number. Let Q be a scalar field defined from U by  Q(y) = c0 |U (y)|2 + A · y + p.v.

Kij (y − z)Ui (z)Uj (z) dz

|y|≤M



+ |z|≥M

 Kij (y − z) − Kij (z) − y · ∇Kij (z) Ui (z)Uj (z) dz



 H(y − z)Θ(z) dz +

+ p.v. |z|≤M

 H(y − z) − H(z) − y · ∇H(z) Θ(z) dz,

|z|≥M

with c0 ∈ R is a constant and A ∈ R2 is some fixed constant vector. Then there holds for −1 < α < − 12 ,  |Q(y)||U (y)| dy  |y|≤L

⎧ ⎨Lb+δ + L 2b +1−2α ,

if (b, δ) = (3, 12 ),

⎩L 72 [log L], 2

if (b, δ) = (3, 12 ),

(2.14)

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

15

while for α = − 12 , it holds 

b

|Q(y)||U (y)| dy  Lb+δ + L 2 +2 [log2 L],

(2.15)

|y|≤L

where [a] denotes the largest integer less than or equal to a. Proof. We divide Q(y) into

Q(y) = c0 |U (y)| + 2

7

Qi,L (y),

i=1

where  Q1,L (y) = A · y,

Kij (y − z)Ui (z)Uj (z) dz,

Q2,L (y) = p.v. |z|≤2L



Q3,L (y) = |z|≥2L

 Kij (y − z) − Kij (z) − y · ∇Kij (z) Ui (z)Uj (z) dz,



 Kij (z) + y · ∇Kij (z) Ui (z)Uj (z) dz,

Q4,L (y) = − M ≤|z|≤2L





H(y − z)Θ(z) dz,

Q5,L (y) = p.v. |z|≤2L



Q7,L (y) =

Q6,L (y) = −

 H(z) + y · ∇H(z) Θ(z) dz,

M ≤|z|≤2L

 H(y − z) − H(z) − y · ∇H(z) Θ(z) dz.

|z|≥2L

Firstly, we directly deduce that  |U (y)|3 dy  Lb+δ , |y|≤L

 |Q1,L (y)||U (y)|dy ≤ |A|L

2

|y|≤L

 |U (y)|2 dy

 12

b

 |A|L 2 +2 .

|y|≤L

It follows from the Hölder inequality and Calderón-Zygmund theory  |Q2,L (y)||U (y)| dy ≤ |y|≤L



3

|Q2,L (y)| 2 dy

|y|≤L

|U (y)|3 dy |y|≤2L

 Lb+δ .



|y|≤L





 23

|U (y)|3 dy

 13

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

16

Using the dyadic decomposition, we find  |Q3,L (y)||U (y)| dy  L2+δ sup |Q3,L (y)| |y|≤L

|y|≤L

 L2+δ sup

|y|≤L

L

4+δ

∞

∞

k=1 k 2 L≤|z|≤2k+1 L

 |y|2 |U (z)|2 dz 4 |z|



1 (2k L)4

k=1

 L4+δ



∞ 

|U (z)|2 dz |z|≈2k L

(2k L)b−4

k=1

L

b+δ

.

Thanks to the Hölder inequality and the dyadic decomposition again, we infer that 

|Q4,L (y)||U (y)|dy  L

|y|≤L

 |U (y)|2 dy

 12

|y|≤L

|y|≤L



L

M ]

[log2 1+ 2b

L

k=−1

L

L 2k+1

≤|z|≤

L 2k

1 L  + 3 |U (z)|2 dz 2 |z| |z|

L

M ] L b−3

[log2 2+ 2b

 sup |Q4,L (y)|

2k

k=−1

⎧ 3b −1 ⎪ ⎨L 2 , 7  L 2 [log2 L], ⎪ ⎩ L 2b +2 ,

if b > 3, if b = 3, if b < 3.

For p > 2, we get  |Q5,L (y)||U (y)| dy ≤

 

|y|≤L

|Q5,L (y)| dy p

 p1  

|y|≤L

≤

 

≤

2p

 p+2   2p

2p



|Θ(z)| p+2 dz

|Θ(z)| p+2 dz

p+2 p −(2α+1) b

 L 2 +1−2α . Similarly, direct computations imply

p

|U (y)| p−1 dy

 p−1 p

|y|≤L

|z|≤2L

L

dy

 p−1 p

|y|≤L

|z|≤2L

 

|U (y)|

p p−1

p+2 2p

  |y|≤L

b

L2 L

p−2 p

|U (y)|2 dy

 12   |y|≤L

1 dy

 p−2 2p

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

 |Q6,L (y)||U (y)| dy ≤



  sup |Q6,L (y)|

|y|≤L

|y|≤L

|U (y)| dy

|y|≤L

⎛ L M

⎜

≤ sup ⎜ ⎝ |y|≤L [log2

j=−1

×

17

 



]

L 2j+1

≤|z|≤

|U (y)|2 dy

 12

L 2j

⎞  1  ⎟ |y| + 2 |Θ(z)| dz ⎟ ⎠ |z| |z|

L

|y|≤L

L



L

M ]

[log2 b 2 +1

j=−1

L 2j+1

≤|z|≤

L 2j

L

M ] 22j  

 1 |y|  + 2 |Θ(z)| dz |z| |z|

[log2

L

b 2 +1

L

j=−1

|y|≤

L

 12 L 2j

L 2j

L

M ]

[log2 b 2 +1−2α

|Θ(z)|2 dz

2j(1+2α)

j=−1



⎧ b ⎨L 2 +1−2α ,

if α ∈ (−1, − 12 ),

⎩L 2b +2 [log L], 2

if α = − 12 .

Using the dyadic decomposition, one deduces 

b

|Q7,L (y)||U (y)|dy  L 2 +1 sup |Q7,L (y)| |y|≤L

|y|≤L

L

L

b 2 +1

sup |y|≤L

b 2 +1

L

k=1 k 2 L≤|z|≤2k+1 L

∞

2−3k

∞

 |y|2 |Θ(z)|dz |z|3



L

k=1

b 2 +1



∞ 

|Θ(z)|2 dz

 12

2k+1 L

|z|≤2k+1 L

2−2k+1 L−2α

k=1 b

 L 2 +1−2α . b

b

Due to −1 < α ≤ − 12 , we have L 2 +1−2α ≥ L 2 +2 . Summing up all the above estimates, we conclude the desired estimates (2.14)-(2.15). Therefore, we finish the proof of Lemma 2.2. 

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

18

3. Locally energy inequalities of the profiles This section is devoted to establishing the following locally energy inequalities of the profiles        2α−2  −1 2α−2 −1 2 ρ 2 ρ l |U (y)| φ 4 (yl2 ) dy − l1 |U (y)| φ 4 (yl1 ) dy  2   R2 R2  |U (y)|3 + |P (y)||U (y)| ≤ C dy |y|3−2α ρ ρ 8 l1 ≤|y|≤ 4 l2

t2 +C

1 (T − t)

t1

⎛ α−1 1+α

⎝

(3.1)

⎞1/2

 |U (y)|2 φ ρ4 (y(T − t)

1 1+α

) dy ⎠

dt,

R2

and        (2α+1)p−2  (2α+1)p−2 −1 −1 p ρ p ρ l  |Θ(y)| φ (yl ) dy − l |Θ(y)| φ (yl ) dy 1 2 1 2  4 4   R2 R2  |Θ(y)|p |U (y)| dy, ≤ C |y|3−(2α+1)p

(3.2)

ρ ρ 8 l1 ≤|y|≤ 4 l2

where li = (T − ti )− α+1 , i = 1, 2. The above estimates (3.1) and (3.2) will be our starting points in much of what follows. To this end, we start with the following locally energy equalities 1



 |u(x, t2 )|2 χ(x, t2 ) dx − R2

|u(x, t1 )|2 χ(x, t1 ) dx R2

t2 

t2  |u(x, t)| ∂t χ(x, t) dx dt + 2

= t1 R2

2  |u| u + 2(p − d(t))u · ∇χ(x, t) dx dt

(3.3)

t1 R2

t2  +2

θ(x, t)u2 (x, t)χ(x, t) dx dt, t1 R2

and 

 |θ(x, t2 )| χ(x, t2 ) dx −

|θ(x, t1 )|p χ(x, t1 ) dx

p

R2

R2

t2 

t2  |θ(x, t)| ∂t χ(x, t) dx dt +

|θ|p u(x, t) · ∇χ(x, t) dx dt,

p

= t1 R2

(3.4)

t1 R2

with −∞ < t1 < t2 < T and χ ∈ D([−∞, T ] × R2 ). One may check from [6] that the above equalities can 1 be satisfied if the velocity and the temperature are regular enough, for example, u ∈ Cloc ([−∞, T ] × R2 ) ∩ ∞ 2 2 1 2 ∞ p 2 L ([−∞, T ]; L (R )) with θ ∈ Cloc ([−∞, T ] × R ) ∩ L ([−∞, T ]; L (R )). Let φ ∈ Cc∞ (R2 ) on B1 (0) be a cutoff function such that 0 ≤ φ ≤ 1 and φ ≡ 1 on B 12 (0). For R > 0, we x set φR (x) = φ( R ) and χ(x, t) = φ ρ4 (x). Then for any t1 < t2 < T , (3.3) and (3.4) can be simplified to

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671



19

 |u(x, t2 )| φ (x) dx − 2

|u(x, t1 )|2 φ ρ4 (x) dx

ρ 4

R2

R2

t2 

2  |u| u + 2(p − d(t))u · ∇φ ρ4 (x) dx dt

=

(3.5)

t1 R2

t2  +2

θ(x, t)u2 (x, t)φ ρ4 (x) dx dt, t1 R2

and 

 |θ(x, t2 )| φ ρ4 (x) dx − R2

t2  |θ(x, t1 )| φ ρ4 (x) dx =

p

|θ|p u(x, t) · ∇φ ρ4 (x) dx dt.

p

(3.6)

t1 R2

R2

Since u and θ are respectively denoted as (1.7) and (1.9) on Bρ (0) × [0, T ], for all 0 < t1 < t2 < T , we can get that 

2     x  ρ  |u(x, ti )| φ ρ4 (x) dx =  φ (x) dx U 2α 1 (T − ti ) 1+α 2  (T − ti ) 1+α  4 R    1 1 = |U (y)|2 φ ρ4 y(T − ti ) 1+α dy 2α−2 (T − ti ) 1+α 2 R  = li2α−2 |U (y)|2 φ ρ4 (yli−1 ) dy, 1

2

R2

R2

and 

p     x   ρ |θ(x, ti )| φ ρ4 (x) dx = Θ  φ 4 (x) dx  1 (2α+1)p   1+α 1+α (T − t ) (T − ti ) i 2 R    1 1 p ρ 1+α y(T − t dy = |Θ(y)| φ ) i (2α+1)p−2 4 (T − ti ) 1+α 2 R  (2α+1)p−2 = li |Θ(y)|p φ ρ4 (yli−1 ) dy, 1

p

R2

R2

where li = (T − ti )− 1+α , i = 1, 2. Plugging (1.7), (1.9) and (2.3) into (3.5) and (3.6) yields 1

t   2    

2

   ρ (x) dx dt |u| u + 2 p − d(t) u (x, t) · ∇φ   4   2 t1 R

t2 ≤

1 3α

t1

(T − t) 1+α t2

+2

    U  R2

1 3α

t1

(T − t) 1+α

x (T − t)

    P  R2

3   |∇φ ρ (x)| dx dt 1  4 1+α x

(T − t)

   U 1  1+α

x (T − t)

  |∇φ ρ (x)| dx dt 1  4 1+α

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

20

t2 ≤ t1



1 (T − t)

3α−2 1+α

 1 |U (y)|3 + 2|P (y)||U (y)| |∇φ ρ4 (y(T − t) 1+α )| dy dt,

R2

and  t2      ρ (x) dx dt θ(x, t)u (x, t)φ 2   4 t1 R2

t2 ≤

⎛ θ(t)L2 ⎝

t1

⎞ 12



|u(x, t)|2 φ ρ4 (x) dx⎠ dt

R2

t2 ≤ θ0 L2 t1

1 α (T − t) 1+α ⎛

t2 ≤ C

⎛

1 α−1

(T − t) 1+α

t1

⎝

    ⎝ U  R2

⎞ 12 2  x  φ ρ (x) dx⎠ dt 1  4 (T − t) 1+α ⎞ 12

 |U (y)|2 φ ρ4 (y(T − t)

1 1+α

) dy ⎠ dt,

R2

and t   2     p  ρ |θ| u(x, t) · ∇φ 4 (x) dx dt    2 t1 R

t2 ≤ t1

1 (T − t)

t2 ≤ t1

(2α+1)p+α 1+α

p          x x     Θ  U  |∇φ ρ4 (x)| dx dt 1 1     1+α 1+α (T − t) (T − t) 2

R



1 (T − t)

p

1

|Θ(y)| |U (y) ||∇φ ρ4 (y(T − t) 1+α )| dy dt.

(2α+1)p+α−2 1+α

R2

Making use of the property of ∇φ ρ4 , integrating on the t-variable and denoting   1 ρ 1 ρ 1 1+α ≤ (T − t) , B(t)  t : ≤ 8 |y| 4 |y| we deduce for any 0 < t1 < t2 < T that  t   2   

2

   ρ (x) dx dt |u| u + 2 p − d(t) u (x, t) · ∇φ   4   2 t1 R

t2



≤ C t1

ρ ρ 8 l1 ≤|y|≤ 4 l2



≤ C ρ ρ 8 l1 ≤|y|≤ 4 l2

and

|U (y)|3 + |P (y)||U (y)| 1B(t) dy dt |y|2−3α

|U (y)|3 + |P (y)||U (y)| dy, |y|3−2α

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

t   2   t2   p   ρ |θ| u(x, t) · ∇φ 4 (x) dx dt ≤ C    ρ 2 t1 R

t1



ρ 8 l1 ≤|y|≤ 4 l2



21

|Θ(y)|p |U (y)| 1B(t) dy dt |y|2−(2α+1)p−α

|Θ(y)|p |U (y)| dy, |y|3−(2α+1)p

≤C ρ ρ 8 l1 ≤|y|≤ 4 l2

where 1B(t) denotes the classical index function on B(t). According to (1.7) and (1.9), one may obtain the desired estimates (3.1) and (3.2). 4. The proof of Theorem 1.1 and Corollary 1.1 4.1. The proof of Theorem 1.1 Firstly, thanks to (2.1) and (2.2), using change of variables yields 

1 (T − t)

p(2α+1) 1+α

|x|≤ρ

  p   x 1   Θ  dx = 1 p(2α+1)−2   1+α (T − t) (T − t) 1+α

 |Θ(y)|p dy |y|≤ρ(T −t)

−

1 1+α

 1. Denoting L = ρ(T − t)− 1+α with t ∈ [0, T ), we infer for α > −1 that 1

 |Θ(y)|p dy  L2−p(2α+1) ,

∀L  1.

(4.1)

|y|≤L

Noticing α >

1 p

−

1 2

and letting L → ∞, we deduce from (4.1)   lim

L→∞



|y|≤L

which implies that Θ ≡ 0 for all α >

1 p



|Θ(y)| dy = p

|Θ(y)|p dy = 0, R2

− 12 . Then, we show that

Θ ≡ 0, for all − 1 < α <

1 2−γ − . 2r 2

(4.2)

By (1.12) and the Hölder inequality, we deduce (2α+1)p−2



l2



|Θ(y)|p φ ρ4 (yl2−1 ) dy ≤ l2

(2α+1)p−2

|Θ(y)|p dy

|y|≤ ρ 4 l2

R2

⎛ (2α+1)p−2

 l2

⎜ ⎝



|y|≤ ρ 4 l2 (2α+1− 2−γ r )p

 l2

→0

⎞ pr ⎟ 2(1− p ) |Θ(y)|r dy ⎠ l2 r

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

22

as l2 → ∞. Letting l2 → ∞ and ρ8 l1 = L  1, plugging them into (3.2) and using the value of φ ρ4 , one gets 

1 L2−(2α+1)p

 |Θ(y)|p dy 

|y|≤L

|y|≥L

|Θ(y)|p |U (y)| dy. |y|3−p(2α+1)

It follows from (1.13) and (4.1) that  |y|≥L

∞

|Θ(y)|p |U (y)| 1 dy  k L)3−p(2α+1) |y|3−p(2α+1) (2 k=0





∞

 |Θ(y)|p |U (y)| dy 2k L≤|y|≤2k+1 L



1

(2k L)3−δ1 −p(2α+1) k=0 ∞

k=0

|Θ(y)|p dy |y|∼2k+1 L

1 (2k L)1−δ1

 L−(1−δ1 ) . Then we obtain the following refined estimate than (4.1)  |Θ(y)|p dy  L2−p(2α+1)−(1−δ1 ) ,

∀L  1.

|y|≤L

Using (4.3) and repeating the process above imply  |y|≥L

∞

|Θ(y)|p |U (y)| 1 dy  3−p(2α+1) k 3−p(2α+1) |y| (2 L) k=0





∞

 |Θ(y)|p |U (y)| dy 2k L≤|y|≤2k+1 L



1

(2k L)3−δ1 −p(2α+1) k=0

|Θ(y)|p dy |y|∼2k+1 L

∞

1 k L)2(1−δ1 ) (2 k=0

 L−2(1−δ1 ) . This further yields  |Θ(y)|p dy  L2−p(2α+1)−2(1−δ1 ) ,

∀L  1.

|y|≤L

Repeating the above process for n times, one may deduce that  |y|≥L

which implies

|Θ(y)|p |U (y)| dy  L−n(1−δ1 ) , |y|3−p(2α+1)

(4.3)

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

23

 |Θ(y)|p dy  L2−p(2α+1)−n(1−δ1 ) → 0,

as n → ∞,

∀L  1.

(4.4)

|y|≤L

 By this we can infer that R2 |Θ(y)|p dy ≡ 0, and then we get (4.2). 1 Finally, let us show the desired estimate (1.14) for 2−γ 2r − 2 ≤ α < that Θ ≡ 0, we claim that for L  1,

1 p

− 12 . By (4.1), under the assumption



1 L2−p(2α+1)

|Θ(y)|p dy  1.

(4.5)

|y|≤L

In fact, (4.5) will be established by contradiction. Suppose that (4.5) is not true, then there exists a series Li  1 (i = 1, 2, · · · ) such that 

1

|Θ(y)|p dy → 0,

2−p(2α+1)

Li

as Li → ∞.

|y|≤Li

Denoting l2 = Li → ∞ and ρ8 l1 = L  1, and plugging them into (3.2), we find that 

 |Θ(y)|p dy  L2−p(2α+1)

|y|≤L

|y|≥L

|Θ(y)|p U (y) dy. |y|3−p(2α+1)

Making use of the above argument yields (4.4), which further gives Θ ≡ 0. This contradicts with Θ ≡ 0. Combining (4.5) and (4.1) yields the desired (1.14). This ends the proof of Theorem 1.1. 4.2. The proof of Corollary 1.1 2 For the first part, since Θ ∈ Cloc (R2 ) and

|Θ(y)| 

1 , |y|μ

∀|y| ≥ L0 ,

(4.6)

we find that for r > max{ μ2 , 2}, 

 |Θ(y)|r dy  |y|≤L

 |Θ(y)|r dy +

|y|≤L0

|y|≥L0

1 dy  1, |y|μr

∀L  1,

which corresponds to (1.12) as γ = 0. Denote p1  max{ μ2 , 2}, and let p = p1 , r = p1 + 1, γ = 0 in Theorem 1.1, which obviously contains the situation of (4.6), then the range of α admitting nontrivial profiles is that p12+1 ≤ α ≤ p21 . On the other hand, let p = p1 + 2, r = p1 + 3, γ = 0 in Theorem 1.1, which also contains (4.6), then the range of α admitting nontrivial profiles is p12+3 ≤ α ≤ p12+2 . Since [ p12+1 , p21 ] ∩ [ p12+3 , p12+2 ] = ∅, we see that Θ ≡ 0. Next we show the second part. Let L0 > 0 be a constant such that (1.15) establishes for all |y| ≥ L0 . 2 Then under the assumption of (1.12) and Θ ∈ Cloc (R2 ), we get that for every p ∈ (1, ∞) and some r > p 

 |Θ(y)| dy 

|y|≤L

 |Θ(y)| dy +

r

r

|y|≤L0

l0 ≤|y|≤L

|y|rδ2 dy  Lrδ2 +2 ,

∀L  1.

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

24

p+2 Let γ = δ2 r + 2, namely, δ2 = γ−2 r , and we need to make γ = δ2 r + 2 ≤ r − p, thus we take that r = 1−δ2 . Thus by using the result of Theorem 1.1, it is not difficult to show that the range of α admitting nontrivial 1 1 1 profiles is − δ22 − 12 = 2−γ 2r − 2 ≤ α ≤ p − 2 . Moreover, the profiles corresponding to such α satisfy that

 |Θ(y)|p dy ≈ L2−p(2α+1) ,

∀L  1.

(4.7)

|y|≤L

On the other hand, by the lower bound of (1.12), we have  |Θ(y)|p dy  L2 ,

∀L  1.

|y|≤L

Compared with (4.7), this implies the range − 12 < α ≤ p1 − 12 is not available. Therefore, the range of α admitting nontrivial profiles is − δ22 − 12 ≤ α ≤ − 12 , and the profiles corresponding to each α satisfy (4.7). We thus complete the proof of Corollary 1.1. 5. The proof of Theorem 1.2 2 We first consider the case: −1 < α < max{−δ1 , − 1+δ 2 }. By Corollary 1.1 we know that as −1 < α < there exists no nontrivial density profile Θ. As a result, α belonging to this scope can be excluded. 2 2 We may as well suppose that −δ1 > − 1+δ (Otherwise, it’s already proved), that is δ1 < 1+δ 2 2 . Consider 1+δ2 the case − 2 ≤ α < −δ1 . By (1.16), we have 2 − 1+δ 2 ,

 |U (y)|2 dy  l22δ1 +2α → 0,

l22α−2

as l2 → ∞.

|y|≤ ρ 4 l2

Thus, denoting l1 = ρ8 L, l2 → ∞ (namely, t2 → T ) in (3.1) yields 

 |U (y)| dy  L 2

2−2α

|y|≤L

|y|≥L

|U (y)|3 + |P (y)||U (y)| dy |y|3−2α

T +L





1

2−2α

|U (y)|2 dy

α−1

t1

(T − t) α+1

|y|≤ ρ 4 (T −t)

−

 12

dt,

1 1+α

where P (y) is given by (2.4). We make use of the estimates of Lemma 2.2 to control the above quantities. Actually, we have  |U (y)|2 dy  L2+2δ1 ,

∀L  1.

(5.1)

|y|≤L

Denoting L = ρ4 (T − t1 )− α+1 , we get that 1

T 2−2α

L

t1



1 (T − t)

 |U (y)|2 dy

α−1 α+1

|y|≤ ρ 4 (T −t)

−

1 1+α

 12

dt

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

T  L

1

2−2α t1

T  L

2−2α

(T − t)

25

1+δ1

α−1 α+1

(T − t)−

(T − t)− 1+α dt

α+δ1 1+α

dt

t1

 L1+δ1 −2α .

(5.2)

Using the dyadic decomposition gives  L2−2α |y|≥L

= L2−2α

|U (y)|3 + |P (y)||U (y)| dy |y|3−2α 

∞

k=0 k 2 L≤|y|≤2k+1 L −1

 L

∞

|U (y)|3 + |P (y)||U (y)| dy |y|3−2α

 2

k(2α−3)

k=0

(5.3)

 |U (y)|3 + |P (y)||U (y)| dy.

2k L≤|y|≤2k+1 L

Based on (1.16) and (5.1), we find that  |U (y)|3 dy  (2k L)2+3δ1 . |y|≤2k+1 L 2 By means of the estimates of Lemma 2.2, we have for − 1+δ ≤ α ≤ −δ1 with 2

 |U (y)||P (y)| dy  |y|≤2k+1 L

1 2

≤ δ1 < 1,

⎧ ⎨(2k L)2+3δ1 + (2k L)2+δ1 −2α ,

if δ1 > 12 ,

⎩(2k L) 72 [log (2k L)] + (2k L) 52 −2α , 2

if δ1 = 12 ,

 (2k L)2+δ1 −2α . Plugging the above estimates into (5.3) yields  2−2α

L

|y|≥L

∞

|U (y)|3 + |P (y)||U (y)| 1+δ1 −2α dy  L 2−k(1−δ1 )  L1+δ1 −2α . |y|3−2α k=0

2 Thus for − 1+δ ≤ α < −δ1 , we first get a rough estimate 2



|U (y)|2 dy  L1+δ1 −2α = L2+2δ1 −(1+δ1 +2α) .

(5.4)

|y|≤L

Since δ2 < 12 ≤ δ1 , we obviously have 1 + δ1 + 2α ≥ 1 + δ1 − (1 + δ2 ) > 0, which yields that (5.4) is better than (5.1). Next we apply (5.4) to find a refined estimate. Similar to (5.2), we have T 2−2α

L

t1



1 (T − t)

 |U (y)|2 dy

α−1 α+1

|y|≤ ρ 4 (T −t)

−

1 1+α

 12

dt

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

26

T

1

 L

2−2α

(T − t)

t1

T  L

(T − t)−

α−1 α+1

1+δ1 −α 2 1+α

dt

1−δ1

(T − t) 2(1+α) dt

2−2α t1

1−δ1 2

 L1−3α−

.

Notice that when b = 1 + δ1 − 2α, we have b + δ1 < Lemma 2.2 yields

b 2

+ 1 − 2α due to −1 < α < −δ1 . This fact along with

 |U (y)||P (y)| dy  (2k L)

3+δ1 2

−3α

.

|y|≤2k+1 L

It also follows from (1.16) and (5.4) that 

|U (y)|3 dy  (2k L)1+2δ1 −2α  (2k L)

3+δ1 2

−3α

.

|y|≤2k+1 L

Plugging the above estimates into (5.3) leads to  L2−2α |y|≥L

∞

1+δ1 3−δ1 1+δ1 |U (y)|3 + |P (y)||U (y)| dy  L 2 −3α 2−k( 2 +α)  L 2 −3α . 3−2α |y| k=0

2 Thus, for − 1+δ ≤ α < −δ1 , we have the following refined estimate 2

 |U (y)|2 dy  L

3+δ1 2

−3α

= L2+2δ1 −(1+ 2 )(1+δ1 +2α) . 1

|y|≤L

Repeating above process for n times, and using (5.4), we check that 

|U (y)|2 dy  L2+2δ1 −(1+ 2 +···+ 2n−1 )(1+δ1 +2α) = L−4α+ 1

1

1+δ1 +2α 2n

.

|y|≤L

Thanks to δ2 <

1 2

2 and − 1+δ ≤ α < −δ1 , we deduce by taking n large enough 2

 |U (y)|2 dy  L3−0 ,

∀L  1,

(5.5)

|y|≤L

where 0 > 0 is a constant satisfying 0 < 0 <

1−2δ2 2 .



 |U (y)| dy 

which obviously contradicts with (5.5).

1

(|y| 2 )2 dy  L3 ,

2

|y|≤L

Then by (1.16), we get that

M ≤|y|≤L

∀L  1,

(5.6)

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

27

2 Next we consider the case α ≥ max{−δ1 , − 1+δ 2 }. By Corollary 1.1, we can narrow the scope of α into 1+δ2 1 max{−δ1 , − 2 } ≤ α ≤ − 2 . Then we show as α in the scope above that

 |U (y)|2 dy  L2−2α ,

∀L  1.

(5.7)

|y|≤L

By (2.1) and (1.7), we derive 

1 2α

(T − t) 1+α

|x−x0 |≤ρ

 2  x−x  1  0 U  dx = 1 2α−2   (T − t) 1+α (T − t) 1+α

 |U (y)|2 dy |y|≤ρ(T −t)

−

1 1+α

 1. Denoting L = ρ(T − t)− 1+α yields 1

 |U (y|2 dy  L2−2α ,

∀L  1,

|y|≤L

which implies (5.7). 1 2 To show (1.17), it suffices to show that for max{−δ1 , − 1+δ 2 } ≤ α ≤ − 2 , there holds  |U (y)|2 dy  L2−2α ,

∀L  1.

(5.8)

|y|≤L

We prove (5.8) by contradiction. Assume that (5.8) is not true, then there exists a series Lk  1 such that  |U (y)|2 dy → 0,

L2α−2 k

as Lk → ∞.

|y|≤Lk

Letting ρ4 l2 = Lk → ∞, ρ4 l1 = 2L > 0 and inserting them into (3.1), one concludes 

 |U (y)| dy  L 2

2−2α

|y|≤L

|y|≥L

|U (y)|3 + |P (y)||U (y)| dy |y|3−2α ⎛

T + L2−2α t1

1 (T − t)

α−1 1+α

⎝



⎞ 12

(5.9)

|U (y)|2 φ ρ4 (y(T − t) 1+α ) dy ⎠ dt. 1

R2

Next, we divide the remainder proof into two cases, namely, α < − 12 and α = − 12 . We begin with the first 1 2 case: max{−δ1 , − 1+δ 2 } ≤ α < − 2 . The first term at the right hand side of (5.9) can be established by using the dyadic decomposition of (5.3) and Lemma 2.2 with b = 2 − 2α. The second term at the right hand side of (5.9) can be established by using the similar argument used in deriving (5.2). More precisely, we derive  2−2α

L

|y|≥L −1

 L

∞

k=0

|U (y)|3 + |P (y)||U (y)| dy |y|3−2α  2

k(2α−3) |y|≈2k L

 |U (y)|3 + |P (y)||U (y)| dy

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

28

 L1+δ1 −2α

∞

2−k(1−δ1 )

k=0

 L1+δ1 −2α , T L





1

2−2α

|U (y)|2 dy

α−1

t1

(T − t) α+1 T

 L

1

2−2α t1

|y|≤ ρ 4 (T −t)

(T − t)

−

 12

dt

1 1+α

(T − t)− 1+α dt 1−α

α−1 α+1

 L2−2α (T − t1 )  L1−3α . Due to δ1 ≥ −α, it leads to 

|U (y)|2 dy  L1+δ1 −2α .

(5.10)

|y|≤L

Using (5.10), repeating above process and noting that ⎧   ⎨b + δ 1 , b max b + δ1 , + 1 − 2α = ⎩ b + 1 − 2α, 2 2

if b ≥ 2 − 4α − 2δ1 , if b < 2 − 4α − 2δ1 ,

we infer T L

t1

(T − t)





1

2−2α

|U (y)|2 dy

α−1 α+1

|y|≤ ρ 4 (T −t)

−

 12

dt  L

1+δ1 2

−3α

1 1+α

and  L2−2α |y|≥L



|U (y)|3 + |P (y)||U (y)| dy |y|3−2α

⎧ ⎨L2(δ1 −α) , 1 ⎩L 1+δ 2 −3α ,

if

1−3δ1 2

≤ α < − 12 , δ1 ∈ ( 23 , 1),

1−3δ1 1 2 if max{−δ1 , − 1+δ 2 } ≤ α < min{ 2 , − 2 }.

It thus follows that  |U (y)|2 dy  |y|≤L

⎧ ⎨L1+δ1 −2α−(1−δ1 ) ,

if

⎩L1+δ1 −2α− 1+δ12+2α ,

1−3δ1 1 2 if max{−δ1 , − 1+δ 2 } ≤ α < min{ 2 , − 2 }.

1−3δ1 2

2 2 ≤ α < − 12 , α ≥ − 1+δ 2 , δ1 ∈ ( 3 , 1),

2 Starting with (5.11) and repeating the process, we further obtain for α ≥ − 1+δ that 2

(5.11)

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

 |y|≤L

⎧ 1+δ −2α−2(1−δ1 ) ⎪ , ⎪L 1 ⎪ ⎨ 1+δ +2α 1−δ1 1 |U (y)|2 dy  L1+δ1 −2α− 2 − 2 , ⎪ ⎪ ⎪ ⎩ 1+δ1 −2α−( 12 + 14 )(1+δ1 +2α) , L

29

if 1 − 2δ1 ≤ α < − 12 , δ1 ∈ ( 34 , 1), if

1−3δ1 2

≤ α ≤ 1 − 2δ1 , δ1 ∈ ( 23 , 1),

1 1 if − δ1 ≤ α < min{ 1−3δ 2 , − 2 }.

2 Repeating above process for n + 1 times, we show for α ≥ − 1+δ that 2

 |U (y)|2 dy |y|≤L

⎧ n+3 1 n+2 ⎪ L1+δ1 −2α−(n+1)(1−δ1 ) , if n+1 ⎪ 2 − 2 δ1 ≤ α < − 2 , δ1 ∈ ( n+3 , 1), ⎪ ⎪ ⎪ 1+δ1 +2α ⎪ n (n+3)δ1 ⎪ 1 ⎪ L1+δ1 −2α− 2 − 2 (1−δ1 ) , if n2 − (n+2)δ ≤ α ≤ n+1 , δ1 ∈ ( n+1 ⎪ 2 2 − 2 n+2 , 1), ⎪ ⎨  ··· ··· ⎪ ⎪ ⎪ 1 ⎪ ⎪L1+δ1 −2α−(1− 21n )(1+δ1 +2α)− 1−δ 2n , if 12 − 32 δ1 ≤ α ≤ 1 − 2δ1 , δ1 ∈ ( 23 , 1), ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎩L1+δ1 −2α−(1− 2n+1 )(1+δ1 +2α) 1 1 , if − δ1 ≤ α < min{ 1−3δ 2 , − 2 }.

1+δ2 1 1 1 For δ1 ∈ [ 12 , 23 ], we have [−δ1 , min{ 1−3δ 2 , − 2 }) = [−δ1 , − 2 ). Therefore, according to max{−δ1 , − 2 } ≤ 1 1 α < − 2 with δ2 < 2 , we get by taking n large enough



|U (y)|2 dy  L−4α+

1+δ1 +2α 2n+1

 L3−0 ,

(5.12)

|y|≤L

where 0 > 0 is a small constant. Thus this contradicts with (5.6). n+2 n+1 n+3 1 1 n n+2 n+1 + For δ1 ∈ ( n+1 n+2 , n+3 ], n ∈ N , we have 2 − 2 δ1 ≥ − 2 , and α ∈ [−δ1 , − 2 ) ⊂ [ 2 − 2 δ1 , 2 − 2−4δ1 1−3δ1 n+3 1 3 2 δ1 ] ∪ · · · ∪ [ 2 − 2 δ1 , 2 ] ∪ [−δ1 , 2 ). Therefore, repeating above process for m + n + 1 times (we 1 2 may assume m > n), we show that for max{−δ1 , − 1+δ 2 } ≤ α < −2  |U (y)|2 dy |y|≤L



⎧ 1+δ −2α−(1− 1 )(1+δ +2α) 1 2m+1 ⎪ , L 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨· · · · · · 1 ⎪ ⎪ L1+δ1 −2α−(1− 2m+n )(1+δ1 +2α) , ⎪ ⎪ ⎪ ⎪ 1 ⎩ 1+δ1 −2α−(1− m+n+1 )(1+δ1 +2α) 2 , L

(n+2)δ1 2

n 2

−

if

1 2

− 32 δ1 ≤ α ≤ 1 − 2δ1 , δ1 ∈ ( 23 , 34 ],

if − δ1 ≤ α <

≤α≤

n+1 2

1−3δ1 2 , δ1

−

(n+3)δ1 , δ1 2

if

n+2 ∈ ( n+1 n+2 , n+3 ],

∈ [ 12 , 23 ].

Taking m large enough, we also have (5.12), which contradicts with (5.6). Notice that  δ1 ∈

       1 2 2 3 n+1 n+2 1 , ∪ , ··· ∪ , ∪···= ,1 , 2 3 3 4 n+2 n+3 2

1 1 1 2 the desired estimate (5.8) holds true for max{−δ1 , − 1+δ 2 } ≤ α < − 2 for δ1 ∈ [ 2 , 1) and δ2 ∈ [0, 2 ). 1 Finally, we consider the case α = − 2 . According to (5.2), we find that

T 3

L

t1

 1 (T − t)−3



−2 |y|≤ ρ 4 (T −t)

|U (y)|2 dy

 12

5

dt  L 2 .

L. Ju / J. Math. Anal. Appl. 484 (2020) 123671

30

Notice that  max b + δ1 ,

b +2 2

 =

⎧ ⎨b + δ 1 ,

if b ≥ 4 − 2δ1 ,

⎩ b + 2, 2

if b < 4 − 2δ1 ,

we deduce by Lemma 2.2 that  2−2α

L

|y|≥L

 L−1

∞

|U (y)|3 + |P (y)||U (y)| dy |y|3−2α  2k(2α−3)

k=0

 L1+δ1 −2α



|U (y)|3 + |P (y)||U (y)| dy

|y|≈2k L ∞

2−k(1−δ1 )

k=0

 L

2+δ1

.

Thus for δ1 ∈ [ 12 , 1), one has  |U (y)|2 dy  L2+δ1 , |y|≤L

which gives  |U (y)|2 dy  L3−0 ,

(5.13)

|y|≤L

where 0 = 1 − δ1 > 0. Therefore, (5.13) contradicts with (5.6) for the case α = − 12 . Consequently, we complete the proof of Theorem 1.2. Acknowledgments The author would like to express her sincere gratitude to the reviewer for his (her) kind comments and suggestions, which substantially helped improve the quality of the paper. This work was supported by the Postgraduate Research & Practice Innovation Program of Jiangsu Province. References [1] O. Boratav, R. Pelz, Direct numerical simulation of transition to turbulence from a high-symmentry initial condition, Phys. Fluids 6 (1994) 2757–2784. [2] A. Bronzi, R. Shvydkoy, On the energy behavior of locally self-similar blow-up for the Euler equation, Indiana Univ. Math. J. 64 (2015) 1291–1302. [3] C. Cao, J. Wu, Global regularity for the 2D anisotropic Boussinesq equations with vertical dissipation, Arch. Ration. Mech. Anal. 208 (2013) 985–1004. [4] D. Chae, Global regularity for the 2D Boussinesq equations with partial viscosity terms, Adv. Math. 203 (2006) 497–513. [5] D. Chae, Nonexistence of self-similar singularities for the 3D incompressible Euler equations, Comm. Math. Phys. 273 (2007) 203–215. [6] D. Chae, On the self-similar solutions of the 3D Euler and the related equations, Comm. Math. Phys. 305 (2011) 333–349. [7] D. Chae, R. Shvydkoy, On formation of a locally self-similar collapse in the incompressible Euler equations, Arch. Ration. Mech. Anal. 209 (2013) 999–1017. [8] P. Constantin, C.R. Doering, Infinite Prandtl number convection, J. Stat. Phys. 94 (1999) 159–172.

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