DECAY ESTIMATES OF GLOBAL SOLUTIONS TO 2D INCOMPRESSIBLE INHOMOGENEOUS NAVIER-STOKES EQUATIONS WITH VARIABLE VISCOSITY

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DECAY ESTIMATES OF GLOBAL SOLUTIONS TO 2D INCOMPRESSIBLE INHOMOGENEOUS

NAVIER-STOKES EQUATIONS WITH VARIABLE VISCOSITY

Jingchi Huang, Marius Paicu

To cite this version:

Jingchi Huang, Marius Paicu. DECAY ESTIMATES OF GLOBAL SOLUTIONS TO 2D INCOM-

PRESSIBLE INHOMOGENEOUS NAVIER-STOKES EQUATIONS WITH VARIABLE VISCOS-

ITY. 2012. �hal-00765650�

DECAY ESTIMATES OF GLOBAL SOLUTIONS TO 2D INCOMPRESSIBLE INHOMOGENEOUS NAVIER-STOKES EQUATIONS WITH VARIABLE

VISCOSITY

JINGCHI HUANG AND MARIUS PAICU

Abstract. In this paper, we investigate the time decay behavior to Lions weak solution of 2D incompressible density-dependent Navier-Stokes equations with variable viscosity.

Keywords: Inhomogeneous Navier-Stokes equations, Decay estimates.

AMS Subject Classification (2000): 35Q30, 76D03

1. Introduction

The main purpose of this paper is to investigate the decay estimates for the global solutions of the following two-dimensional incompressible inhomogeneous Navier-Stokes equations with viscous coefficient depending on the density

(1.1)







∂

ρ + div(ρu) = 0, (t, x) ∈ R

× R

,

∂

(ρu) + div(ρu ⊗ u) − div(µ(ρ) M (u)) + ∇ Π = 0, div u = 0,

where ρ, u = (u

, u

) stand for the density and velocity of the fluid respectively, M (u) = ∇ u+ ∇

u, Π is a scalar pressure function, and in general, the viscosity coefficient µ(ρ) is a smooth positive function on [0, ∞ ). Such system describes a fluid which is obtained by mixing two immiscible fluids that are incompressible and that have different densities. It may also describe a fluid containing a melted substance. One may check [9] for the detailed derivation.

When µ(ρ) is independent of ρ, i.e. µ is a positive constant, and the initial density has a positive lower bound, Ladyˇzenskaja and Solonnikov [8] first addressed the question of unique solvability of (1.1). More precisely, they considered the system (1.1) in a bounded domain Ω with homoge- neous Dirichlet boundary condition for u. Under the assumption that u

∈ W

2 p,p

(Ω) (p > d) is divergence free and vanishes on ∂Ω and that ρ

∈ C

(Ω) is bounded away from zero, then they [8]

proved

• Global well-posedness in dimension d = 2;

• Local well-posedness in dimension d = 3. If in addition u

is small in W

2 p,p

(Ω), then global well-posedness holds true.

Danchin [3] proved similar well-posedness result of (1.1) in the whole space case and the initial data in the almost critical spaces. In particular, in two dimension, he proved the global well-posedness of (1.1) provided the initial data (ρ

, u

) satisfying ρ

− 1 ∈ H

(R

), ρ

≥ m > 0, and u

∈ H

(R

) for any α ∈ (0, 1) and β ∈ (0, 1].

In general, Lions [9] (see also the references therein) proved the global existence of weak solutions to (1.1) with finite energy. More precisely, given 0 ≤ ρ

∈ L

(R

), 0 < µ

≤ µ(ρ) and u

satisfying

Date: 20/Nov/2012.

1

div u

= 0, √ ρ

u

∈ L

(R

), Lions proved that (1.1) has a global weak solution so that 1

2 k p

ρ(t)u(t) k

+ µ

Z

0

k∇ u(τ ) k

dτ ≤ 1

2 k √ ρ

u

k

. Moreover, for any α and β, the Lebesgue measure

µ

x ∈ R

; α ≤ ρ(t, x) ≤ β is independent of t.

In dimension two and under the additional assumption that ρ

is close enough to a positive constant and ∇ u

∈ L

(R

), smoother weak solutions may be built. Their existence stems from a quasi- conservation law involving the norm of ∇ u ∈ L

((0, T ); L

(R

)) and of ∂

u, ∇ p ∈ L

((0, T ); L

(R

)) for any T < ∞ . For both types of weak solutions however, the problem of uniqueness and regularities of such weak solutions are big open questions even in two space dimensions.

Under the additional assumptions that

(1.2) k µ(ρ

) − 1 k

≤ ε and u

∈ H

( T

), Desjardins [4] proved the following theorem.

Theorem 1.1. Let ρ

∈ L

(T

) and div u

= 0. Then there exists ε > 0 such that under the assumption (1.2), Lions weak solutions ([9]) to (1.1) satisfy the following regularity properties hold for all T > 0 :

(1) u ∈ L

((0, T ); H

(T

)) and √ ρ∂

u ∈ L

((0, T ) × T

),

(2) ρ and µ(ρ) ∈ L

((0, T ) × T

) ∩ C([0, T ]; L

( T

)) for all p ∈ [1, ∞ ), (3) ∇ (Π − R

R

(µ M (u)

)) and ∇ ( P ⊗ Q (µ M (u)

)) ∈ L

((0, T ) × T

),

(4) Π may be renormalized in such a way that for some universal constant C

> 0, Π and

∇ u ∈ L

((0, T ); L

(T

)) for all p ∈ [4, p

), where

= 2C

k µ(ρ

) − 1 k

.

In which, we denote R as the Riesz transform: R = ∇△

. Q = ∇△

div and P = I − Q respectively denote the projection on the space of curl-free and divergence-free vector fields.

In order to investigate the global well-posedness of thus solutions, we first need to study the global-in-time type estimates. However, because of the difficulties of the continuity equation in (1.1) being of hyperbolic nature and the estimate of the diffusion term in the momentum equation, we shall first study the time decay of the solutions, which is very much motivated by [5, 10, 12].

Theorem 1.2. For 1 < p < 2, let u

∈ L

(R

) ∩ H

(R

), ρ

− 1 ∈ L

(R

) and ρ

∈ L

(R

) with a positive lower bound. We assume that (ρ, u, ∇ p) is a given Lions weak solution of (1.1) with initial data (ρ

, u

). Denote µ(1) = µ

, then under the assumption

(1.3) k µ(ρ) − µ

k

≤ ε

,

for a small constant ε

, there exists a constant C

which depends on k ρ

− 1 k

, k u

k

and k u

k

such that there hold

(1.4) k u(t) k

≤ C

(t + e)

, k∇ u(t) k

≤ C

(t + e)

, (1.5)

Z

0

k u

k

+ k div µ(ρ) M (u)

k

+ k∇ Π k

dt ≤ C

,

(1.6)

Z

(t + e)

k u

k

+ k div µ(ρ) M (u)

k

+ k∇ Π k

dt ≤ C

,

with β(p) =

(

− 1) and any ε > 0.

Remark 1.1. The first estimate of (1.4) coincides with the L

-norm decay result in [10, 12] for the weak solutions of the two-dimensional classical Navier-Stokes system, and also coincides with the result in [5] for (1.1). When µ(ρ) be a constant, we can get optimal decay of k∇ u k

with the order − 1 − 2β(p), see [6]. Notice the main ingredients of the proof in [6, 10, 12] are the usual energy estimates and the phase space analysis. In our case, due to the additional difficulties mentioned above, we not only need to apply phase space analysis, but also need more explicit energy estimates, see Proposition 2.1 below. We note also that the 3D case with constant viscosity was studied in [1].

Using energy estimates with weight in time and the Fourier splitting method of Schonbek [10] we can generalize this decay in time estimates to the 3D case with variable viscosity.

Motivated by Proposition 2.1, we have a more general result. Indeed, using interpolation ar- gument we obtain a similar decay rate of the solution, under a weaker assumption on the initial velocity.

Theorem 1.3. For 1 < p < 2 and 0 < α < 1, let u

∈ L

( R

) ∩ H

( R

), ρ

− 1 ∈ L

( R

) and ρ

∈ L

(R

) with a positive lower bound. We assume that (ρ, u, ∇ p) is a given Lions weak solution of (1.1) with initial data (ρ

, u

). Then under the assumption (1.3), there exists a constant C

which depends on k ρ

− 1 k

, k u

k

and k u

k

such that there hold

(1.7) k u(t) k

≤ C

(t + e)

, k∇ u(t) k

≤ C

(t + e)

, (1.8)

Z

0

k u

k

+ k div µ(ρ) M (u)

k

+ k∇ Π k

dt ≤ C

, (1.9)

Z

t

(t + e)

k u

k

+ k div µ(ρ) M (u)

k

+ k∇ Π k

dt ≤ C

, with any ε > 0 and 0 < r < α.

Remark 1.2. We note also that the 3D case with constant viscosity was studied in [1]. Using energy estimates with weight in time and the Fourier splitting method of Schonbek [10] it is possible to generalize this decay in time estimates to the 3D case with variable viscosity.

The organization of the paper. In the second section, we shall present the proof of Theorem 1.2. In Section 3, we shall prove Theorem 1.3.

2. The Proof of Theorem 1.2 First, we need some calculus inequalities which can be found in [12].

Lemma 2.1. Let m ∈ R

, 0 ≤ α < 1 and β > 0. Then (1) R

0

(s + e)

ln(s + e)

ds ≤

for m > 1, (2) there is some γ

> 0 such that R

0

(s + e)

ln(s + e)

ds ≤ γ

β

, (3) there is some γ

> 0 such that, for all t > 0,

R

0

(s + e)

ln(s + e)

ds ≤ γ

(t + e)

ln(t + e)

.

In this section, we will prove Theorem 1.2. First, we have some energy estimates.

Proposition 2.1. Let v ∈ L

( R

; L

) ∩ L

( R

; ˙ H

), div v = 0. Assume that u

∈ H

( R

) and ρ

∈ L

(R

) with positive lower bound. f (t) be a positive second-order differentiable function satisfies f

(t) ≥ 0 and f

(t) ≥ 0. (ρ, u) be the global weak solution of the linear system:

(2.1)



 



 



∂

ρ + v ∇ ρ = 0,

ρ∂

u + ρv ∇ u − div(µ(ρ) M (u)) + ∇ Π = 0, div u = 0,

(ρ, u) |

= (ρ

, u

).

Then under the assumption (1.3), we have the following estimates:

sup

0<t<∞

f (t) Z

R²

µ(ρ) |∇ u |

(t) dx + Z

0

f(t) Z

R²

C ¯ | √ ρu

|

+ | div µ(ρ) M (u)

|

+ |∇ Π |

dxdt

≤ C(f(0) k∇ u

k

+ Z

0

f

(t) Z

R²

µ(ρ) |∇ u |

dxdt) exp { C(1 + k v k

) k∇ v k

} , (2.2)

(2.3) sup

0<t<∞

f

(t) Z

R²

ρ | u |

(t) dx+

Z

f

(t) Z

R²

µ(ρ) |∇ u |

dxdt ≤ C(f

(0) k u

k

+ Z

0

f

(t) Z

R²

ρ | u |

dxdt), where C ¯ be a large enough constant.

Proof. First, we follow the line of the proof of Theorem 1.1, see [4]. By taking L

inner product of the momentum equation of (2.1) with f (t)u

and using integration by parts, we deduce that

f (t) Z

R²

| √ ρu

|

dx + f (t) Z

R²

(ρv ∇ u) · u

dx + f (t) Z

R²

µ(ρ) ∇ u : ∇ u

dx = 0.

Note that f (t)

Z

R²

µ(ρ) ∇ u : ∇ u

dx = 1 2 ∂

[f (t)

Z

R²

µ(ρ) |∇ u |

dx] − 1 2 f

(t)

Z

R²

µ(ρ) |∇ u |

dx

− 1 2 f (t)

Z

R²

∂

µ(ρ) |∇ u |

dx, and from the derivation of (29) in [4] that

− Z

R²

∂

µ(ρ) |∇ u |

dx = Z

R²

div(µ(ρ)v) |∇ u |

dx

= Z

R²

(v ∇ )u · div(µ(ρ) M (u)) dx + Z

R²

µ(ρ)tr( ∇ v ∇ u M (u)) dx

= Z

R²

(v ∇ )u · (ρu

+ ρv ∇ u + ∇ Π) dx + Z

R²

µ(ρ)tr( ∇ v ∇ u M (u)) dx, we have

d dt [f(t)

Z

R²

µ(ρ) |∇ u |

dx] + f (t) Z

R²

| √ ρu

|

dx . f

(t)

Z

R²

µ(ρ) |∇ u |

dx + f(t) Z

R²

| √ ρv ∇ u |

dx +f(t)

Z

R²

µ(ρ) |∇ v ||∇ u |

dx + f (t) Z

R²

Π∂

v

∂

u

dx . Recall that

− µ

△ u = div (µ(ρ) − µ

) M (u)

− div µ(ρ) M (u) , so that we have

µ

∂

u

= R

P

R (µ(ρ) − µ

) M (u)

− R

P

R µ(ρ) M (u) .

Estimating it in the L

( R

) and using the Gagliardo-Nirenberg inequality, we can write k∇ u k

. k P ⊗ Q (µ(ρ) − µ

) M (u)

k

+ k P ⊗ Q µ(ρ) M (u) k

. k µ(ρ) − µ

k

k∇ u k

+ k P ⊗ Q µ(ρ) M (u) k

1 2

L²

k∇

P ⊗ Q µ(ρ) M (u) k

1 2

L²

Finally, using (1.3) and the conservation of the momentum, we obtain that k∇ u k

. k∇ u k

1 2

L²

k P (ρu

+ ρv ∇ u) k

1 2

L²

,

Now letting ( −△ )

R operate on the equation of momentum, we get that Π = R

R

µ(ρ)(∂

u

+ ∂

u

)

+ ( −△ )

R (ρu

+ ρv ∇ u).

It follows that

k Π − R

R

(µ(ρ) M (u)) k

. k∇ (Π − R

R

(µ(ρ) M (u))) k

. k ρu

+ ρv ∇ u k

. We obtain that

Z

R²

Π∂

v

∂

u

dx

≤ k∇ v k

k∇ u k

+ k Π − R

R

(µ(ρ) M (u)) k

k ∂

v

∂

u

k

≤ k∇ v k

k∇ u k

k ρu

+ ρv ∇ u k

, so that

f (t) Z

R²

Π∂

u

∂

u

dx

≤ C

f(t) k∇ v k

k∇ u k

+ εf(t)( k √ ρu

k

+ k v ∇ u k

).

f (t) Z

R²

µ(ρ) |∇ v ||∇ u |

dx ≤ Cf(t) k∇ v k

k∇ u k

≤ C

f (t) k∇ v k

k∇ u k

+ εf(t)( k √ ρu

k

+ k v ∇ u k

).

Also

k v ∇ u k

≤ k v k

k∇ u k

≤ k v k

k∇ v k

k∇ u k

k ρu

+ ρv ∇ u k

, f(t)

Z

R²

| √ ρv ∇ u |

dx ≤ C

f(t) k v k

k∇ v k

k∇ u k

+ εf(t) k √ ρu

k

. Consequently,

d dt [f (t)

Z

R²

µ(ρ) |∇ u |

dx] + f (t) Z

R²

| √ ρu

|

dx . f

(t)

Z

R²

µ(ρ) |∇ u |

dx + f (t) k∇ u k

k∇ v k

(1 + k v k

).

(2.4)

Second, we multiply −

f (t) div µ(ρ) M (u)

with the momentum equation of (2.1) and integrate in R

to get that

f (t) Z

R²

µ(ρ) ∇ u : ∇ u

dx + f (t) Z

R²

| div µ(ρ) M (u)

|

dx

= f (t) Z

R²

v ∇ u div µ(ρ) M (u)

dx + f(t) Z

R²

1

ρ ∇ Π div µ(ρ) M (u) dx.

The first term of both side have been dealt in previous. For the last term, we have

Z

R²

1

ρ ∇ Π div µ(ρ) M (u) dx

.

Z

R²

|∇ Π − div µ(ρ) M (u)

|

dx .

Z

R²

| ρu

+ ρv ∇ u |

dx.

So we get

d dt [f(t)

Z

R²

µ(ρ) |∇ u |

dx] + f (t) Z

R²

| div µ(ρ) M (u)

|

dx . f

(t)

Z

R²

µ(ρ) |∇ u |

dx + f(t) k √ ρu

k

+ f (t) k∇ u k

k∇ v k

(1 + k v k

),

along with (2.4), for a large constant ¯ C d

dt [f (t) Z

R²

µ(ρ) |∇ u |

dx] + f (t) Z

R²

C ¯ | √ ρu

|

+ | div µ(ρ) M (u)

|

dx

≤ C(f

(t) Z

R²

µ(ρ) |∇ u |

dx + f(t) k∇ u k

k∇ v k

(1 + k v k

)).

Recall that

∇ Π = div µ(ρ) M (u)

− ρu

− ρv ∇ u, and note that v ∈ L

(L

) ∩ L

( ˙ H

), so that

Z

0

(1 + k v k

) k∇ v k

dt ≤ (1 + k v k

) k∇ v k

, and (2.2) holds.

The same strategy can be held for f

(t)u, we have 1

2 d dt [f

(t)

Z

R²

| √ ρu |

dx] + f

(t) Z

R²

µ(ρ) |∇ u |

dx = 1 2 f

(t)

Z

R²

| √ ρu |

dx, so that

sup

0<t<∞

f

(t) Z

R²

ρ | u |

(t) dx+

Z

0

f

(t) Z

R²

µ(ρ) |∇ u |

dxdt ≤ C(f

(0) k u

k

+ Z

0

f

(t) Z

R²

ρ | u |

dxdt).

According these two energy estimates, letting v = u, we can prove Theorem 1.2.

Proof of Theorem 1.2: We get the standard energy estimate to (1.1) that d

dt k √ ρu(t) k

+ k∇ u(t) k

≤ 0.

Using Schonbek’s strategy, we obtain

(2.5) d

dt k √ ρu(t) k

+ g

(t) k √ ρu(t) k

≤ M g

(t) Z

S(t)

| u(t, ξ) ˆ |

dξ, where S(t)

= { ξ : | ξ | ≤

q

2

g(t) } and g(t) satisfying g(t) . (1 + t)

. We rewrite the momentum equation of (1.1) as

u(t) = e

u

+ Z

0

e

P

div (µ(ρ) − µ

) M (u)

+ (1 − ρ)u

− ρu ∇ u (s) ds.

Taking Fourier transform with respect to x variables leads to

| u(t, ξ) ˆ | . e

| u ˆ

(ξ) | + Z

0

e

| ξ ||F

(µ(ρ) − µ

) M (u)

| + |F

(1 − ρ)u

− ρu ∇ u

| ds, which implies that

Z

S(t)

| u(t, ξ) ˆ |

dξ . Z

S(t)

e

| u ˆ

(ξ) |

dξ + g

(t)(

Z

0

kF

(µ(ρ) − µ

) M (u)

k

ds)

+g

(t)(

Z

0

kF

(1 − ρ)u

− ρu ∇ u

k

ds)

. (2.6)

Note that u

∈ L

for 1 < p < 2, one has (2.7)

Z

S(t)

e

| u ˆ

(ξ) |

dξ . ( Z

S(t)

e

dξ)

k u ˆ

(ξ) k

. k u

k

(1 + t)

,

where

=

− 1,

+

= 1.

Note that u ∈ L

(L

) ∩ L

( ˙ H

) and u

∈ L

(L

), we have (

Z

0

kF

(µ(ρ) − µ

) M (u)

k

ds)

≤ ( Z

0

k (µ(ρ) − µ

) M (u) k

ds)

≤ k µ(ρ) − µ

k

( Z

0

k∇ u k

ds)

≤ C k ρ

− 1 k

k u

k

(1 + t),

( Z

0

kF

(1 − ρ)u

k

ds)

≤ k 1 − ρ k

( Z

0

k u

k

ds)

≤ C k ρ

− 1 k

k∇ u

k

(1 + t), (

Z

0

kF

ρu ∇ u

k

ds)

≤ k ρu k

( Z

0

k∇ u k

ds)

≤ C k u

k

(1 + t).

Then we deduce from (2.5) to (2.7) that d

dt k √ ρu(t) k

+ g

(t) k √ ρu(t) k

. g

(t)(1 + t)

+ g

(t)(1 + t) + g

(t)(1 + t) . g

(t)(1 + t)

+ g

(t)(1 + t).

Taking g

(t) =

, then e

= ln

(t + e) and ln

(t + e) k u(t) k

. k u

k

+ Z

0

[ ln(s + e)

(s + e)

+ (s + e)

] ds . ln(t + e),

which gives

(2.8) k u(t) k

. ln

(t + e).

Now we improve the estimate (2.8).

We choose f (t) = t + e in (2.2), then we have sup

0<t<∞

(t + e) k∇ u(t) k

+ Z

0

(t + e) k u

k

dt ≤ C k u

k

exp { C k u

k

} , so that

( Z

0

kF

(1 − ρ)u

k

ds)

≤ k 1 − ρ k

( Z

0

k u

k

ds)

≤ C k ρ

− 1 k

Z

(s + e) k u

k

ds Z

0

(s + e)

ds . ln(t + e).

( Z

0

kF

ρu ∇ u

k

ds)

≤ ( Z

0

k u(s) k

k∇ u(s) k

ds)

. (t + e) ln

(t + e).

We plug these estimate into (2.6) and take g

(t) =

, then e

= ln

(t + e) and ln

(t + e) k u(t) k

.

Z

[ ln

(s + e)

(s + e)

+ ln

(s + e)

(s + e)

+ (s + e)

] ds

. ln(t + e),

which implies

(2.9) k u(t) k

. ln

(t + e).

So that

Z

(t + e)

k u k

dt ≤ C

. We choose f

(t) = ln(t + e) in (2.3), then get

sup

0<t<∞

ln(t + e) k u(t) k

+ Z

0

ln(t + e) k∇ u k

dt

≤ C k u

k

+ Z

0

(t + e)

k u k

dt

≤ C

.

Consequently, we take f (t) = (t + e) ln(t + e) in (2.2), obtain that sup

0<t<∞

(t + e) ln(t + e) k∇ u(t) k

+ Z

0

(t + e) ln(t + e) k u

k

dt

≤ C( k∇ u

k

+ Z

0

(ln(t + e) + 1) k∇ u k

dt) exp { C k u

k

}

≤ C

, which implies

(2.10) k∇ u(t) k

. (t + e)

ln

(t + e).

Combining (2.9) and (2.10), we get the revised estimates, (

Z

0

k u k

k∇ u k

ds)

. ( Z

0

(s + e)

ln

(s + e) ds)

. (t + e) ln

(t + e),

( Z

0

k 1 − ρ k

k u

k

ds)

. ( Z

0

(s + e) ln(s + e) k u

k

ds)(

Z

(s + e)

ln

(s + e) ds) . ln ln(t + e)

.

Substituting these two estimates in (2.6), and taking g

(t) =

, then e

= ln

(t+e) and

ln

(t + e) k u(t) k

. k u

k

+ Z

0

[ ln

(s + e)

(s + e)

+ ln

(s + e) ln ln(t + e) (s + e)

+ 1

s + e ] ds . ln(t + e),

from which, we obtain

(2.11) k u(t) k

. ln

(t + e).

We choose f

(t) = ln

(t + e) in (2.3), then sup

0<t<∞

ln

(t + e) k u(t) k

+ Z

0

ln

(t + e) k∇ u k

dt

≤ C( k u

k

+ Z

0

(t + e)

ln(t + e) k u(t) k

dt)

≤ C( k u

k

+ Z

0

(t + e)

ln

(t + e) dt)

≤ C

.

Finally, we take f (t) = (t + e) ln

(t + e) in (2.2) to get that sup

0<t<∞

(t + e) ln

(t + e) k∇ u(t) k

+ Z

0

(t + e) ln

(t + e) k u

k

dt

≤ C

k∇ u

k

+ Z

0

ln(t + e) + ln

(t + e)

k∇ u k

dt

exp { C k u

k

}

≤ C

. Consequently, we obtain

( Z

0

k u

k

dt)

≤ ( Z

0

(t + e) ln

(t + e) k u

k

dt)(

Z

(t + e)

ln

(t + e) dt) ≤ C

. Which is the same for ∇ Π, div µ(ρ) M (u)

∈ L

(R

; L

), and gives (1.5). Recall that (

Z

0

k u k

k∇ u k

ds)

≤ ( Z

0

k u k

ds)(

Z

0

k∇ u k

ds) . Z

0

k u k

ds,

substituting these estimates into (2.6), noting that 2β(p) ∈ (0, 1), and taking g

(t) =

with α large enough, then we get

(t + e)

k u(t) k

. k u

k

+ Z

0

(s + e)

Z

0

k u(τ ) k

dτ ds +

Z

(s + e)

ds + Z

0

(s + e)

ds . (t + e)

+

Z

(s + e)

Z

0

k u(τ ) k

dτ ds.

For t ≥ 1, we define y(t)

=

Z

(s + e)

k u(s) k

ds, Y (t)

= max { y(s) : 1 ≤ s ≤ t } , I(t)

=

Z

0

k u(s) k

ds.

Then one has

I(t) =

Z

0

k u(s) k

ds + Z

t−[t]

k u(s) k

ds

≤ C +

[t]−1

X

j=0

Z

t−j−1

k u(s) k

(s + e)

(s + e)

ds

≤ C +

[t]−1

X

j=0

Y (t)(t − j)

. C + Y (t)(t + e)

, (2.12)

from which, we infer that

y(t) . (t + e)

+ Z

0

(s + e)

Y (s) ds.

Then, applying Gronwall’s inequality, we have (2.13) Y (t) . (t + e)

+

Z

(s + e)

ds . (t + e)

.

Plunging (2.13) into (2.12) gives rise to I(t) . (t + e)

, we obtain (t + e)

k u(t) k

. (t + e)

+

Z

(s + e)

ds . (t + e)

,

which gives the first inequality of (1.4).

Go back to (2.3), we choose f

(t) such that R

0

f

(t) k u(t) k

dt is finite. For example, we let f

(t) = (t + e)

for any ε > 0, (or f

(t) = (t + e)

ln

(t + e) for any α > 1,) then f (t) = (t + e)

. Finally, we get (1.6) and the second inequality of (1.4).

3. The Proof of Theorem 1.3

The proof of Theorem 1.3 is very similar to Theorem 1.2. We should estimate every term in terms of k u

k

instead of k u

k

. First, we choose f (t) = t + e and t in (2.2), get that

sup

0<t<∞

(t + e) k∇ u(t) k

+ Z

0

(t + e) k u

k

dt ≤ C k u

k

exp { C(1 + k v k

) k∇ v k

} , and

sup

0<t<∞

t k∇ u(t) k

+ Z

0

t k u

k

dt ≤ C k u

k

exp { C(1 + k v k

) k∇ v k

} . By interpolation, and let v = u, we get that

(3.1) sup

0<t<∞

(t + e)

t

k∇ u(t) k

+ Z

0

(t + e)

t

k u

k

dt ≤ C k u

k

exp { C k u

k

} . So that

( Z

0

kF

(1 − ρ)u

k

ds)

≤ k 1 − ρ k

( Z

0

k u

k

ds)

≤ C k ρ

− 1 k

Z

s

(s + e)

k u

k

ds Z

0

s

(s + e)

ds

≤ C

t

, (

Z

0

kF

(µ(ρ) − µ

) M (u)

k

ds)

≤ ( Z

0

k (µ(ρ) − µ

) M (u) k

ds)

≤ k µ(ρ) − µ

k

( Z

0

k∇ u k

ds)

≤ C k ρ

− 1 k

k u

k

(1 + t), (

Z

0

kF

ρu ∇ u

k

ds)

≤ k ρu k

( Z

0

k∇ u k

ds)

≤ C k u

k

(1 + t).

From which, we can deduce that d

dt k √ ρu(t) k

+ g

(t) k √ ρu(t) k

≤ C

g

(t)(1 + t)

+ g

(t)(1 + t) + g

(t)(1 + t) + g

(t)t

≤ C

g

(t)(1 + t)

+ g

(t)(1 + t) + g

(t)t

.