Open Access Published by De Gruyter Open Access August 12, 2022

# Global weak solution of 3D-NSE with exponential damping

• Jamel Benameur
From the journal Open Mathematics

## Abstract

In this paper, we prove the global existence of incompressible Navier-Stokes equations with damping α ( e β u 2 1 ) u , where we use the Friedrich method and some new tools. The delicate problem in the construction of a global solution is the passage to the limit in exponential nonlinear term. To solve this problem, we use a polynomial approximation of the damping part and a new type of interpolation between L ( R + , L 2 ( R 3 ) ) and the space of functions f such that ( e β f 2 1 ) f 2 L 1 ( R + × R 3 ) . Fourier analysis and standard techniques are used.

MSC 2010: 35-XX; 35Q30; 76D05; 76N10

## 1 Introduction

In this paper, we study the global existence of weak solution to the modified incompressible Navier-Stokes equations (NSE) in R 3 :

( S ) t u ν Δ u + u . u + α ( e β u 2 1 ) u = p in R + × R 3 , div u = 0 in R + × R 3 , u ( 0 , x ) = u 0 ( x ) in R 3 ,

where u = u ( t , x ) = ( u 1 , u 2 , u 3 ) and p = p ( t , x ) denote, respectively, the unknown velocity and the unknown pressure of the fluid at the point ( t , x ) R + × R 3 , and the viscosity of fluid ν > 0 . Also α , β > 0 denote the parameters of damping term. The fact that div u = 0 , allows to write the term ( u . u ) u 1 1 u + u 2 2 u + u 3 3 u in the following form:

div ( u u ) ( div ( u 1 u ) , div ( u 2 u ) , div ( u 3 u ) ) .

While u 0 = ( u 1 0 ( x ) , u 2 0 ( x ) , u 3 0 ( x ) ) is an initial given velocity. If u 0 is quite regular, the divergence-free condition determines the pressure p . We recall in our case it was assumed that the viscosity is unitary ( ν = 1 ) to simplify the calculations and the proofs of our results. The global existence of the weak solution of the initial value problem of classical incompressible Navier-Stokes was proved by Leray and Hopf (see [1,2, 3,4]) long before. Uniqueness remains an open problem for the dimensions d 3 . The damping is from the resistance to the motion of the flow. It describes various physical situations such as porous media flow, drag or friction effects, and some dissipative mechanisms (see [2,3,5,6] and references therein). The polynomial damping α u β 1 u is studied in [7] by Cai and Jiu, where they proved the global existence of weak solution in

L ( R + , L 2 ( R 3 ) ) L 2 ( R + , H ˙ 1 ( R 3 ) ) L β + 1 ( R + , L β + 1 ( R 3 ) ) .

The purpose of this paper is to study the well posedness of the incompressible NSEs with damping α ( e β u 2 1 ) u . We will show that the Cauchy problem ( S ) has global weak solutions for any α , β ( 0 , ) . We apply the Friedrich method to construct the approximate solutions and make more delicate estimates to proceed to compactness arguments. In particular, we obtain new more a priori estimates:

u ( t ) L 2 2 + 2 0 t u ( z ) L 2 2 d z + 2 α 0 t ( e β u ( z ) 2 1 ) u ( z ) 2 L 1 d z u 0 L 2 2 ,

comparing with the NSEs, to guarantee that solution u belongs to

L ( R + , L 2 ( R 3 ) ) L 2 ( R + , H ˙ 1 ( R 3 ) ) β ,

where

β = { f : R + × R 3 R measurable ; ( e β f 2 1 ) f 2 L 1 ( R + × R 3 ) } .

Before treating the global existence, we give the definition of solution of ( S ) .

## Definition 1.1

The function pair ( u , p ) is called a weak solution of problem ( S ) if for any T > 0 , the following conditions are satisfied:

1. u L ( [ 0 , T ] ; L σ 2 ( R 3 ) ) L 2 ( [ 0 , T ] ; H ˙ 1 ( R 3 ) ) and ( e β u 2 1 ) u 2 L 1 ( [ 0 , T ] , L 1 ( R 3 ) ) .

2. t u Δ u + u . u + α ( e β u 2 1 ) u = p in D ( [ 0 , T ] × R 3 ) : for any Φ C 0 ( [ 0 , T ] × R 3 ) such that div Φ ( t , x ) = 0 for all ( t , x ) [ 0 , T ] × R 3 and Φ ( T ) = 0 , we have

0 T ( u ; t Φ ) L 2 + 0 T ( u ; Φ ) L 2 + 0 T ( ( u . ) u ; Φ ) L 2 + α 0 T ( ( e β u 2 1 ) u ; Φ ) L 2 = ( u 0 ; Φ ( 0 ) ) L 2 .

3. div u ( x , t ) = 0 for a.e. ( t , x ) [ 0 , T ] × R 3 . ( ( ; ) L 2 is the inner product in L 2 ( R 3 ) .)

## Remark 1.2

1. If ( u , p ) is a solution of the system ( S ) , then ( e β u 2 1 ) u L loc 1 ( R + , L 1 ( R 3 ) ) . Indeed: for T > 0 , consider the subsets A T 1 = { ( t , x ) [ 0 , T ] × R 3 ; 0 < u ( t , x ) < 1 } and A T 2 = { ( t , x ) [ 0 , T ] × R 3 ; u ( t , x ) 1 } . We have

[ 0 , T ] × R 3 ( e β u 2 1 ) u = A T 1 A T 2 ( e β u 2 1 ) u A T 1 ( e β u 2 1 ) u + A T 2 ( e β u 2 1 ) u A T 1 ( e β u 2 1 ) u u 2 + A T 2 ( e β u 2 1 ) 1 u u 2 M β A T 1 u 2 + A T 2 ( e β u 2 1 ) u 2 , M β = max 0 < X 1 ( e β X 2 1 ) X M β 0 T R 3 u 2 + 0 T R 3 ( e β u 2 1 ) u 2

M β T u L ( [ 0 , T ] , L 2 ( R 3 ) ) + [ 0 , T ] × R 3 ( e β u 2 1 ) u 2 M β T u L ( [ 0 , T ] , L 2 ( R 3 ) ) + ( e β u 2 1 ) u 2 L 1 ( [ 0 , T ] , L 1 ( R 3 ) ) ,

2. which implies the desired result.

3. By this definition, the scalar pressure function p is well defined in L loc 1 ( R + , H s ( R 3 ) ) , for all s > 3 / 2 . Indeed: Formally, if p exists, then

p = ( Δ ) 1 ( div u . u + α div ( e β u 2 1 ) u ) .

So, just prove that ( Δ ) 1 ( div ( u . u ) ) , ( Δ ) 1 ( α div ( e β u 2 1 ) u ) L loc 1 ( R + , H s ( R 3 ) ) , for all s > 3 / 2 . For s > 3 / 2 , we have

( Δ ) 1 ( div ( u . u ) ) ( t ) H s ( R 3 ) = ( Δ ) 1 ( div ( div ( u u ) ) ) ( t ) H s ( R 3 ) u u ( t ) H s ( R 3 ) R 3 ( 1 + ξ 2 ) s ( u u ) ( t , ξ ) 2 d ξ 1 / 2 c s ( u u ) ( t ) L ( R 3 ) , c s = R 3 ( 1 + ξ 2 ) s d ξ 1 / 2 c s u u ( t ) L 1 ( R 3 ) c s u ( t ) L 2 ( R 3 ) 2 L loc 1 ( R + ) .

On the other hand, we have

( Δ ) 1 ( div ( e β u 2 1 ) u ) ( t ) H s ( R 3 ) R 3 ( 1 + ξ 2 ) s ξ 2 ( ( e β u 2 1 ) u ) ( t , ξ ) 2 d ξ 1 / 2 C s ( ( e β u 2 1 ) u ( t ) ) L ( R 3 ) , C s = R 3 ( 1 + ξ 2 ) s ξ 2 d ξ 1 / 2 C s ( e β u 2 1 ) u ( t ) L 1 ( R 3 ) L loc 1 ( R + ) .

Then, p is well defined in L loc 1 ( R + , H s ( R 3 ) ) , s > 3 / 2 .

4. If ( u ( t , x ) , p ( t , x ) ) is a solution of ( S ) , then ( u ( t , x ) , p ( t , x ) + h ( t ) ) is also the solution of ( S ) . Then, if we look for p in the space L loc 1 ( R + , H 2 ( R 3 ) ) , we obtain h = 0 and we can express p in terms of u .

In our case of exponential damping, we are trying to find more regularity of Leray solution in p L p ( R + , L p ( R 3 ) ) . In particular, we give a new energy estimate. Our main result is the following.

## Theorem 1.3

Let u 0 L 2 ( R 3 ) be a divergence-free vector fields, then there is a global solution of ( S ) u L ( R + , L 2 ( R 3 ) ) C ( R + , H 2 ( R 3 ) ) L 2 ( R + , H ˙ 1 ( R 3 ) ) β . Moreover, for all t 0 ,

(1.1) u ( t ) L 2 2 + 2 0 t u ( z ) L 2 2 d z + 2 α 0 t ( e β u ( z ) 2 1 ) u ( z ) 2 L 1 d z u 0 L 2 2 .

## Remark 1.4

1. If ( u , p ) is a solution of ( S ) given by Theorem 1.3, then by (1.1) and Proposition 2.1(3), we obtain

lim t 0 u ( t ) u 0 L 2 = 0 .

2. The fact ( e β u 2 1 ) u 2 L 1 ( R + , L 1 ( R 3 ) ) implies that u 4 p < L p ( R + , L p ( R 3 ) ) . Indeed, we have

0 ( e β u ( t ) 2 1 ) u ( t ) 2 L 1 d t = k = 1 β k k ! 0 u ( t ) L 2 k + 2 2 k + 2 d t .

3. The fact u L ( R + , L 2 ( R 3 ) ) C ( R + , H 2 ( R 3 ) ) implies that u C b ( R + , H r ( R 3 ) ) , for all r > 0 .

The remainder of our paper is organized as follows. In Section 2, we provide some notations, definitions, and preliminary results. Section 3 is devoted to prove Theorem 1.3, and this proof is done in two steps. In the first, we give a general result of equicontinuity, and in the second, we apply the Friedritch method to construct a global solution of ( S ) , and the only delicate point is to show the convergence of the nonlinear part ( e β u φ ( n ) 2 1 ) u φ ( n ) to ( e β u 2 1 ) u . To do this, we approximate ( e β u φ ( n ) 2 1 ) u φ ( n ) by k = 1 m β k k ! u φ ( n ) 2 k u φ ( n ) , and we use the convergence in L loc p ( R + , L loc p ( R 3 ) ) given by technical lemmas.

## 2 Notations and preliminary results

### 2.1 Notations

• For R > 0 : B R = { x R 3 ; x < R } .

• For m N : P m ( X ) = k = 1 m X k k ! . Clearly, for each R > 0 , we have P m ( β R 2 ) ( e β R 2 1 ) , as m + .

• For R > 0 , the Friedritch operator J R is defined by

J R ( D ) f = 1 ( 1 B R ( ξ ) f ^ ) .

• The Leray projector P : ( L 2 ( R 3 ) ) 3 ( L 2 ( R 3 ) ) 3 is defined by

( P f ) = f ^ ( ξ ) f ^ ( ξ ) ξ ξ ξ ξ = M ( ξ ) f ^ ( ξ ) ; M ( ξ ) = δ k , l ξ k ξ l ξ 2 1 k , l 3 .

• For R > 0 , we define the operator A R ( D ) by

A R ( D ) f = P J R ( D ) f = 1 ( M ( ξ ) 1 B R ( ξ ) f ^ ) .

• If p , q ( 1 , ) and Ω 1 R n , Ω 2 R m are two open subsets, and we define L p ( Ω 1 , L q ( Ω 2 ) ) by: f L p ( Ω 1 , L q ( Ω 2 ) ) if f : Ω 1 × Ω 2 C , ( x 1 , x 2 ) f ( x 1 , x 2 ) is measurable and

f L p ( Ω 1 , L q ( Ω 2 ) ) = f L q ( Ω 2 ) L p ( Ω 1 )

is well defined and finite. Particularly, if p = q , then L p ( Ω 1 , L q ( Ω 2 ) ) = L p ( Ω 1 × Ω 2 ) and

f L p ( Ω 1 , L p ( Ω 2 ) ) = f L p ( Ω 1 × Ω 2 ) .

### 2.2 Preliminary results

In this section, we collect some classical results, and we provide some lemmas, which are well suited to the study of the system ( S ) .

### Proposition 2.1

[8] Let H be Hilbert space.

1. If ( x n ) is a bounded sequence of elements in H , then there is a subsequence ( x φ ( n ) ) such that

( x φ ( n ) y ) ( x y ) , y H .

2. If x H and ( x n ) is a bounded sequence of elements in H such that

( x n y ) ( x y ) , y H ,

then x liminf n x n .

3. If x H and ( x n ) is a bounded sequence of elements in H such that

( x n y ) ( x y ) , y H

and

limsup n x n x ,

then lim n x n x = 0 .

### Lemma 2.2

[9] Let s 1 and s 2 be two real numbers and d N .

1. If s 1 < d / 2 and s 1 + s 2 > 0 , there exists a constant C 1 = C 1 ( d , s 1 , s 2 ) , such that: if f , g H ˙ s 1 ( R d ) H ˙ s 2 ( R d ) , then f g H ˙ s 1 + s 2 d 2 ( R d ) and

f g H ˙ s 1 + s 2 d 2 C 1 ( f H ˙ s 1 g H ˙ s 2 + f H ˙ s 2 g H ˙ s 1 ) .

2. If s 1 , s 2 < d / 2 and s 1 + s 2 > 0 , there exists a constant C 2 = C 2 ( d , s 1 , s 2 ) such that: if f H ˙ s 1 ( R d ) and g H ˙ s 2 ( R d ) , then f g H ˙ s 1 + s 2 d 2 ( R d ) and

f g H ˙ s 1 + s 2 d 2 C 2 f H ˙ s 1 g H ˙ s 2 .

### Lemma 2.3

For m N and β > 0 , there is c m , β > 0 such that

( P m ( β z 2 ) ) 2 c m , β ( e β z 2 1 ) z 2 , z R .

### Proof

It suffices to prove that ( P m ( β z 2 ) ) 2 ( e β z 2 1 ) 1 z 2 is bounded on ( 0 , ) .□

### Lemma 2.4

For m N and β > 0 , there is C m , β > 0 such that

P m ( β x 2 ) x P m ( β y 2 ) y C m , β ( P m ( β x 2 ) + P m ( β y 2 ) ) x y , x , y R 3 .

### Proof

Suppose that y x . We have

P m ( β x 2 ) x P m ( β y 2 ) y = P m ( β x 2 ) ( x y ) + k = 1 m β k k ! ( x 2 k y 2 k ) y .

For k { 1 , , m } , we have

x 2 k y 2 k y 2 k x y ( x 2 k 1 + y 2 k 1 ) y 2 m x y ( x 2 k 1 y + y 2 k ) 2 m x y ( x 2 k + y 2 k ) ,

which implies the desired result.□

### Lemma 2.5

For all s > d / 2 : L 1 ( R d ) H s ( R d ) . Moreover, we have

f H s ( R d ) σ s , d f L 1 ( R d ) , f H s ( R d ) ,

where σ s , d = R d ( 1 + ξ 2 ) s d ξ 1 / 2 .

### Proof

Using the fact s > d / 2 , we obtain

R d ( 1 + ξ 2 ) s f ^ ( ξ ) 2 d ξ R d ( 1 + ξ 2 ) s d ξ f ^ L ( R d ) 2 R d ( 1 + ξ 2 ) s d ξ f L 1 ( R d ) 2 ,

which ends the proof.□

### Lemma 2.6

Let ε ( 0 , 1 ) and d N such that d 2 , then there is a constant C > 0 such that If ( f , g ) H ε 2 ( R d ) × H 1 ε 2 ( R d ) , then f g H 1 ε d 2 ( R d ) and

f g H 1 ε d 2 ( R d ) C f H ε 2 ( R d ) g H 1 ε 2 ( R d ) .

### Proof

We have

f g H 1 ε d 2 2 f g H ˙ 1 ε d 2 2 ξ ξ 2 1 ε d 2 η f ^ ( η ) g ^ ( ξ η ) 2 ξ ξ 2 1 ε d 2 η f ^ ( ξ η ) g ^ ( η ) 2 ξ ξ 2 1 ε d 2 η ( 1 + ξ η 2 ) ε / 4 f ^ ( ξ η ) ( 1 + ξ η 2 ) ε / 4 g ^ ( η ) 2 .

By using the elementary inequality

( 1 + ξ η 2 ) ε 4 ( 1 + 2 ξ 2 + 2 η 2 ) ε 4 6 ε 4 ( 1 + ξ ε 2 + η ε 2 ) ,

we obtain

f g H 1 ε d 2 2 C ξ ξ 2 1 ε d 2 η ( 1 + ξ η 2 ) ε 4 f ^ ( ξ η ) ( 1 + ξ ε 2 + η ε / 2 ) g ^ ( η ) 2 C ( I 1 + I 2 + I 3 ) ,

with

I 1 = ξ ξ 2 1 ε d 2 η ( 1 + ξ η 2 ) ε 4 f ^ ( ξ η ) g ^ ( η ) 2 = u 1 v 1 H ˙ 1 ε d 2 2 , u 1 = 1 ( ( 1 + ξ 2 ) ε 4 f ^ ( ξ ) ) , v 1 = 1 ( g ^ ( ξ ) ) , I 2 = ξ ξ 2 1 ε 2 d 2 η ( 1 + ξ η 2 ) ε 4 f ^ ( ξ η ) g ^ ( η ) d η 2 d ξ = u 2 v 2 H ˙ 1 ε 2 d 2 2 , u 2 = 1 ( ( 1 + ξ 2 ) ε 4 f ^ ( ξ ) ) , v 2 = 1 ( g ^ ( ξ ) ) ,

and

I 3 = ξ ξ 2 1 ε d 2 η ( 1 + ξ η 2 ) ε 4 f ^ ( ξ η ) η ε 2 g ^ ( η ) 2 = u 3 v 3 H ˙ 1 ε d 2 2 , u 3 = 1 ( 1 + ξ 2 ) ε 4 f ^ ( ξ ) , v 3 = 1 ξ ε 2 g ^ ( ξ ) .

By applying Lemma 2.2 with the following choices:

( s 1 , t 1 ) = ( 0 , 1 ε ) , ( s 2 , t 2 ) = 0 , 1 ε 2 , ( s 3 , t 3 ) = ( 0 , 1 ε ) ,

we obtain

u i v i H ˙ s i + t i d 2 ( R d ) C u i H ˙ s i ( R d ) v i H ˙ t i ( R d ) , i { 1 , 2 , 3 } .

Then

I 1 C u 1 H ˙ 0 2 v 1 H ˙ 1 ε 2 C f H ε 2 2 g H ˙ 1 ε 2 C f H ε 2 2 g H 1 ε 2 C f H ε 2 2 g H 1 ε 2 2 , I 2 C u 2 H ˙ 0 2 v 2 H ˙ 1 ε 2 2 C f H ε 2 2 g H 1 ε 2 2 , I 3 C u 3 H ˙ 0 2 v 3 H ˙ 1 ε 2 C f H ε 2 2 g H ˙ 1 ε 2 2 C f H ε 2 2 g H 1 ε 2 2 ,

which ends the proof of the Lemma.□

### Lemma 2.7

Let p ( 1 , + ) and Ω be an open subset of R 4 . If ( f n ) is a bounded sequence in L 2 ( Ω ) L p ( Ω ) , then there is a subsequence ( f h ( n ) ) and f L 2 ( Ω ) L p ( Ω ) such that

lim n ( f h ( n ) g ) L 2 = ( f g ) L 2 , g L 2 ( Ω ) , f L p liminf n f h ( n ) L p , f L 2 liminf n f h ( n ) L 2 .

### Proof

Let q the conjugate exponent of Hölder inequality

1 p + 1 q = 1 .

The sequence ( f n ) is bounded in the Hilbert space L 2 ( Ω ) , and then there is a subsequence ( f h ( n ) ) and f L 2 ( Ω ) such that

lim n ( f h ( n ) g ) L 2 = ( f g ) L 2 , g L 2 ( Ω ) .

By Hölder inequality, we obtain

( f h ( n ) g ) L 2 f h ( n ) L p g L q , n N ,

which implies

( f g ) L 2 ( liminf n f h ( n ) L p ) g L q .

Particularly, for g L 1 ( Ω ) L ( Ω ) , we have

(2.1) ( f g ) L 2 ( liminf n f h ( n ) L p ) g L q .

For k N , put the following functions:

g ( x ) = f ( x ) ¯ f ( x ) f ( x ) q 1 , if f ( x ) 0 , 0 , if F ( x ) = 0

and

g k ( x ) = 1 A k ( x ) g ( x ) ,

where

A k = { x Ω ; g ( x ) k } B k .

Clearly, g k L 1 ( Ω ) L ( Ω ) and 1 A k 1 Ω almost every where. Applying inequality (2.1) to g k , we obtain

Ω 1 A k ( x ) f ( x ) p d x ( liminf n f h ( n ) L p ) Ω 1 A k ( x ) f ( x ) p d x 1 / q

and

Ω 1 A k ( x ) f ( x ) p d x 1 / p liminf n f h ( n ) L p .

As A k A k + 1 and

1 A k ( x ) f ( x ) p f ( x ) p , a.e x Ω ,

then Monotone Convergence Theorem implies that f L p ( Ω ) and

f L p liminf n f h ( n ) L p ,

which ends the proof of the first inequality.

The last result is given by Proposition 2.1.□

### Remark 2.8

Let p ( 1 , + ) and Ω be an open subset of R 4 . If ( f n ) is a bounded sequence in L 2 ( Ω ) L p ( Ω ) and f L 2 ( Ω ) such that

lim n ( f n g ) L 2 = ( f g ) L 2 , g L 2 ( Ω ) .

Then f L p ( Ω ) and

f L 2 liminf n f n L 2 , f L p liminf n f n L p .

### Lemma 2.9

Let T > 0 and ( f n ) be a bounded sequence in L T 2 ( H 1 ( R 3 ) ) such that

f n f strongly in C ( [ 0 , T ] , H loc 4 ( R 3 ) ) .

Then

f n f strongly in L 2 ( [ 0 , T ] , L loc 2 ( R 3 ) ) .

### Proof

Combining the inclusion C ( [ 0 , T ] , H loc 4 ( R 3 ) ) L 2 ( [ 0 , T ] , H loc 4 ( R 3 ) ) and the interpolation

L 2 ( [ 0 , T ] , H loc 4 ( R 3 ) H 1 ( R 3 ) ) L 2 ( [ 0 , T ] , L loc 2 ( R 3 ) ) ,

we obtain the desired result.□

### Lemma 2.10

Let T > 0 and p 0 ( 2 , + ) . If ( f n ) is a bounded sequence in L T 2 ( H 1 ( R 3 ) ) L T p 0 ( L p 0 ( R 3 ) ) such that

f n f strongly in C ( [ 0 , T ] , H loc 4 ( R 3 ) ) ,

then f L T p 0 ( L p 0 ( R 3 ) ) and

f n f strongly in L p ( [ 0 , T ] , L loc p ( R 3 ) ) , 2 p < p 0 .

### Proof

By Lemma 2.9, we obtain

f n f strongly in L 2 ( [ 0 , T ] , L loc 2 ( R 3 ) ) .

By interpolation, between L p 0 ( [ 0 , T ] , L p 0 ( R 3 ) ) and L 2 ( [ 0 , T ] , L loc 2 (