Open Access

Nonhomogeneous boundary value problem for ( I , J ) similar solutions of incompressible two-dimensional Euler equations

Journal of Inequalities and Applications20142014:277

https://doi.org/10.1186/1029-242X-2014-277

Received: 11 February 2014

Accepted: 27 June 2014

Published: 1 August 2014

Abstract

In this paper we introduce the ( I , J ) similar method for incompressible two-dimensional Euler equations, and obtain a series of explicit ( I , J ) similar solutions to the incompressible two-dimensional Euler equations. These solutions include all of the twin wave solutions, some new singularity solutions, and some global smooth solutions with a finite energy. We also reveal that the twin wave solution and an affine solution to the two-dimensional incompressible Euler equations are, respectively, a plane wave and constant vector. We prove that the initial boundary value problem of the incompressible two-dimensional Euler equations admits a unique solution and discuss the stability of the solution. Finally, we supply some explicit piecewise smooth solutions to the incompressible three-dimensional Euler case and an example of the incompressible three-dimensional Navier-Stokes equations which indicates that the viscosity limit of a solution to the Navier-Stokes equations does not need to be a solution to the Euler equations.

MSC:35Q30, 76D05, 76D10.

Keywords

Euler equation ( I , J ) similar methodtwin wave solutionaffine solutionexplicit smooth solutionuniquenessstability

1 Introduction

In this paper we consider the Euler equations ( σ = 0 ) or the Navier-Stokes equations below:
{ u t + ( u ) u + p = σ u , in  Ω × ( 0 , ) , div u = 0 , in  Ω × [ 0 , ) , n = 2 , 3 ,
(1.1)

where Ω R + n = { x R n | x n > 0 } ; u = u ( x , t ) = ( u 1 ( x 1 , x 2 , t ) , u 2 ( x 1 , x 2 , t ) ) and p = p ( x 1 , x 2 , t ) denote the velocity and pressure, respectively. Though there is a large amount of physics and mathematics literature on the Euler and Navier-Stokes equations, many basic questions remain open.

There are various open problems in fluid physics. The Navier-Stokes equation has been recognized as the basic equation and the very starting point of all problems in fluid physics (see [1]). One of the most significant developments related to the above problem may be the discovery of Lax pairs of the two-dimensional and three-dimensional Euler equations (see [2])
ω t + [ ψ , ω ] = 0 , ω = ψ x 1 x 1 + ψ x 2 x 2 ,
(1.2)
where the velocity u = ( u 1 , u 2 ) is determined by the stream function ψ through
u 1 = ψ x 1 , u 2 = ψ x 2 .
(1.3)

Is an exact solution of the Euler equations explicitly given via solving the vortex equations (the weak Lax pair) to the Euler equations? Since the Lax pair has still only weak meaning, one cannot get the solutions to the Euler equations from those solutions of vortex equations by the Biot-Savart law. Thus whether the integrable two-dimensional Euler equations in some stronger sense are similar to those of the three-dimensional Euler equations is still an open question. In this paper we find a so-called ( I , J ) similar method which can give some explicit smooth solutions to two-dimensional incompressible Euler equations (see Section 2). As applications of the ( I , J ) similar method, a large amount of explicit twin wave solutions are constructed in Section 3.

There are various open problems in mathematics, such as: how to establish the global existence of smooth solutions, and how to establish the blow-up solution at least when the space dimension equals three (see [3]), and so on. The study of the incompressible Navier-Stokes equations has a long history. A deeper result on the weak solution was obtained by Caffarelli et al. in [4]. On the blow-up problem of the incompressible Navier-Stokes equations, Tsai in [5] proved that the Leray self-similar solutions to (1.1) must be zero if they satisfy local energy estimates. So in Section 4 we discuss the method of determining the nonexistence of a non-constant affine solution to the two-dimensional Euler equations, which we can correctly obtain due to the ( I , J ) similar method.

The blow-up problem of the compressible Navier-Stokes equation has been established by Xin (see [6]). He proved that any smooth solution to the multidimensional Navier-Stokes equation for polytropic fluids in the absence of heat conduction will blow up in finite time if the initial density is compactly supported (see [6]).

In Section 5, we prove that incompressible two-dimensional Euler equations under a class of initial boundary values has a unique solution u ( x , t ) C ( [ 0 , ) ; L 2 ( Ω ) ) for every bounded domain Ω ¯ R + n , and we discuss the stability of solutions in Section 6.

Since it is very hard to solve the Navier-Stokes equations in a three-dimensional space, we consider the two equations in the half space case. In Section 7, we construct some explicit smooth solutions to the incompressible three-dimensional Euler and Navier-Stokes equations and an example of the three-dimensional Navier-Stokes equations which indicates that a solution to the Navier-Stokes equations does not need to tend to a solution to the Euler equations in the continuous function space on the half space.

For open problems of the Euler equations and the Navier-Stokes equations to the incompressible cases, we refer to [7, 8] for more information.

2 ( I , J ) similar method in solving the Euler equations

We first have the following definition.

Definition 2.1 A piecewise smooth solution u ( x , t ) to (1.1) is called a ( I , J ) similar solution, if
u ( x , t ) = i = 1 I α i ( t ) v i ( j = 1 J β j ( t ) M j ( x ) ) ,
(2.1)

where α i ( t ) and β j ( t ) are smooth functions on [ 0 , ) , M j ( x ) = ( M j 1 ( x ) , M j 2 ( x ) , , M j n ( x ) ) is a n-dimensional smooth vector function independent of t, and v i ( y 1 , y 2 , , y n ) is a piecewise smooth vector function from R n to R n .

Here a vector valued function f ( t ) is called piecewise smooth on [ 0 , ) , if there exist 0 < t 1 < t 2 < < t k < + such that f ( t ) is a smooth function on ( 0 , t 1 ) , ( t i , t i + 1 ) , i = 1 , 2 , , k 1 and ( t k , ) , respectively. Similarly we call a vector valued function u ( x , t ) piecewise smooth on R n if there exist 0 < r 1 < r 2 < < r k < + such that u ( x , t ) is a smooth function on { x | 0 < | x | < r 1 } , { x | r i < | x | < r i + 1 } , i = 1 , 2 , , k 1 , and { x | r k < | x | < } , respectively. We rewrite
( j = 1 J β j ( t ) M j ( x ) ) = ( j = 1 J β j ( t ) M j 1 ( x ) , j = 1 J β j ( t ) M j 2 ( x ) , , j = 1 J β j ( t ) M j n ( x ) ) = : ( y 1 , y 2 , , y n ) .
By inserting (2.1) into (1.1), we have
i = 1 I α i t v i + i = 1 I j = 1 J k = 1 n β j t M j k v i y k + i 1 , i 2 = 1 I j = 1 J k = 1 n α i 1 α i 2 β j M j k x k v i 2 k v i 1 y k + p = i = 1 I j 1 , j 2 = 1 J k , l , m , s = 1 n α i β j 1 β j 2 M j 1 m x k M j 2 s x k v i y k y l + i = 1 I j = 1 J k , l = 1 n α i β j v i y l M j l x k x k , i = 1 I j = 1 J k , l = 1 n α i β j v i y l M j l x k = 0 .
(2.2)
For the incompressible Euler equations, we take n = 2 , I = 1 , J = 2 , α 1 = c ( t ) , β 1 = β 2 = 1 , v 1 ( y 1 , y 2 ) = ( y 1 , y 2 ) , and we also set
M 1 ( x ) = ( x 2 r 2 , x 1 r 2 ) , M 2 ( x ) = ( h ( r ) x 2 , h ( r ) x 1 ) ,
(2.3)
where r = x 1 2 + x 2 2 . Then u and p must satisfy the following equation:
c ( t ) r 2 ( x 2 , x 1 ) ( c ( t ) r 2 + h ( r ) ) 2 ( x 1 , x 2 ) + p = 0 .
(2.4)
So the incompressible Euler equations (2.6) below in Theorem 2.2 have a family of ( I , J ) similar solutions,
u = ( ( c ( t ) r 2 + h ( r ) ) x 2 , ( c ( t ) r 2 + h ( r ) ) x 1 ) , p = c ( t ) arctan x 1 x 2 + F ( r , t ) , x 2 0 ,
(2.5)

where F ( r , t ) = r ( c ( t ) r 2 + h ( r ) ) 2 d r , c is an arbitrary smooth function of t, h is an arbitrary smooth function of r.

Thus we have the following result.

Theorem 2.2 Equation (2.5) is a family of ( I , J ) similar solutions to the incompressible Euler equations,
{ u t + ( u ) u + p = 0 , in R 2 × ( 0 , ) , div u = 0 , in R 2 × ( 0 , ) .
(2.6)

Remark 2.3 To the best of our knowledge, there is little known of exact solutions to vortex equations, but they are not solutions to the two-dimensional Euler equations (2.6) except for the zero solution and they did not bring about any solution to the two-dimensional Euler equations (2.6) by the Biot-Savart law as they have a singularity (see [9, 10]), as seen by using VIM (see [11]), and by a Bäcklund transformation (see [12]) method. Notice that (2.5) correctly is a family of exact solutions to the two-dimensional Euler equations (2.6).

Remark 2.4 It is interesting to get many properties by choosing c ( t ) , h ( r ) .

Example 2.5 According to [10], we see that
u = ( ( c ( t ) | x C t | 2 + h ( | x C t | ) ) x 2 + c 1 , ( c ( t ) | x C t | 2 + h ( | x C t | ) ) x 1 + c 2 ) , p = c ( t ) arctan x 1 c 1 t x 2 c 2 t + F ( | x C t | , t ) , F ( r , t ) = r ( c ( t ) r 2 + h ( r ) ) 2 d r
(2.7)

is also a solution pair for any constant vectors C R 2 . w = Q T u ( Q x , t ) , p ¯ = p ( Q x , t ) is also a solution pair for any rotation matrices Q.

Example 2.6 There are some u with finite energy only at some points, such as t = 1 . Taking c ( t ) = t , h ( r ) = 1 r 2
u = ( ( t r 2 1 r 2 ) x 2 , ( t r 2 1 r 2 ) x 1 ) , p = arctan x 1 x 2 1 2 r 2 ( t 1 ) 2 .
(2.8)

Then u ( x , 0 ) L 2 + ϵ ( R 2 B δ ) , however, for every ϵ > 0 and δ > 0 , we have u ( x , t ) L 2 + ϵ ( R 2 B δ ) but u ( x , t ) L 2 ( R 2 B δ ) , where B δ = { x R 2 | | x | < δ } .

Example 2.7 There are some explicit solutions u with singularities only at some points. Taking c ( t ) = 1 T t , h ( r ) = 1 r 2 ,
u = ( ( 1 r 2 ( T t ) 1 r 2 ) x 2 , ( 1 r 2 ( T t ) 1 r 2 ) x 1 ) , p = 1 ( T t ) 2 arctan x 1 x 2 1 2 r 2 ( 1 T t 1 ) 2 .
(2.9)

Then u is singular at r = 0 and blows up at t = T .

3 Twin wave solutions

In this section we give more explicit nonzero solutions by considering explicit twin wave solutions to the two-dimensional Euler equations. Here a twin wave solution has the form of u = u ( x 1 c 1 t , x 2 c 2 t ) . The twin wave solution is a ( I , J ) similar solution. In fact, if we take I = 1 , J = 3 , β 1 = 1 , β 2 = c 1 t , β 3 = c 2 t , M 1 = ( x 1 , x 2 ) , M 2 = ( 1 , 0 ) , M 3 = ( 0 , 1 ) , v 1 ( y 1 , y 2 ) = u ( y 1 , y 2 ) , then u = u ( x 1 c 1 t , x 2 c 2 t ) . Inserting them into (2.6), we have the following theorem.

Theorem 3.1 If the pressure is independent of x, all twin wave solutions to the two-dimensional Euler equations u = u ( x 1 c 1 t , x 2 c 2 t ) will be given by u ( x , t ) = ( v ( c 3 x 1 x 2 ( c 3 c 1 c 2 ) t ) + c 1 , c 3 v ( c 3 x 1 x 2 ( c 3 c 1 c 2 ) t ) + c 2 ) , where v is any function of c 3 x 1 x 2 ( c 3 c 1 c 2 ) t , and c 1 , c 2 , c 3 are arbitrary constants.

Proof Inserting u = u ( x 1 c 1 t , x 2 c 2 t ) and p ( x , t ) = p ( x 1 , x 2 , t ) into (2.6),
{ ( c 1 + u 1 ) u 1 x 1 + ( c 2 + u 2 ) u 1 x 2 + p x 1 = 0 , ( c 1 + u 1 ) u 2 x 1 + ( c 2 + u 2 ) u 2 x 2 + p x 2 = 0 , u 1 x 1 + u 2 x 2 = 0 .
(3.1)
Then (3.1) is rewritten as
{ ( u 1 c 1 ) ( u 1 c 1 ) x 1 + ( u 2 c 2 ) ( u 1 c 1 ) x 2 + p x 1 = 0 , ( u 1 c 1 ) ( u 2 c 2 ) x 1 + ( u 2 c 2 ) ( u 2 c 2 ) x 2 + p x 2 = 0 , ( u 1 c 1 ) x 1 + ( u 2 c 2 ) x 2 = 0 .
(3.2)
Let us take y 1 = x 1 c 1 t , y 2 = x 2 c 2 t , v 1 ( y 1 , y 2 ) = u 1 c 1 , v 2 ( y 1 , y 2 ) = u 2 c 2 . By (3.2), we have
{ v 1 v 1 y 1 + v 2 v 1 y 2 + p x 1 = 0 , v 1 v 2 y 1 + v 2 v 2 y 2 + p x 2 = 0 , v 1 y 1 + v 2 y 2 = 0 .
(3.3)

If p p ( t ) , this may be interpreted as the equations in v 1 ( y 1 , y 2 ) , v 2 ( y 1 , y 2 ) .

We now consider the following equations:
{ v 1 v 2 y 2 = v 2 v 1 y 2 , v 1 v 2 y 1 = v 2 v 1 y 1 , v 1 y 1 + v 2 y 2 = 0 .
(3.4)
So
{ v 2 = f ( y 1 ) v 1 , v 2 = g ( y 2 ) v 1 , v 1 y 1 + v 2 y 2 = 0 ,
(3.5)
where f ( y 1 ) , g ( y 2 ) are determinate functions. Hence, we have
f ( y 1 ) = g ( y 2 ) = c 3 .
Further let
{ v 1 = v ( c 3 y 1 y 2 ) , v 2 = c 3 v ( c 3 y 1 y 2 ) .
(3.6)
Therefore,
{ u 1 = v ( c 3 x 1 x 2 ( c 3 c 1 c 2 ) t ) + c 1 , u 2 = c 3 v ( c 3 x 1 x 2 ( c 3 c 1 c 2 ) t ) + c 2 ,
(3.7)

where v is any function of c 3 x 1 x 2 ( c 3 c 1 c 2 ) t , and c 1 , c 2 , c 3 are arbitrary constants. □

Example 3.2 By taking c ( t ) = 1 , h ( r ) = 1 r 2 + 1 ( 1 + r 2 ) 2 in Example 2.5, we have
u = ( 1 ( 1 + | x C t | 2 ) 2 x 2 + c 1 , 1 ( 1 + | x C t | 2 ) 2 x 1 + c 2 ) , p = 1 6 ( 1 + | x C t | 2 ) 3 .
(3.8)

These are a global smooth twin wave solutions pair for any constant vectors C R 2 . w = Q T u ( Q x , t ) , p ¯ = p ( Q x , t ) are also a twin wave solutions pair for any rotation matrices Q.

Remark 3.3 These solutions in (3.8) are symmetric only in some domains. In particular, if c 1 = c 2 , they are symmetry solutions for all t 0 , and if c 1 c 2 they are not symmetric for all t 0 and symmetric only at t = 0 . These examples show that the difference between the velocity of flow and its wave speed u C has a finite energy over R 2 , i.e. u C L 2 ( R 2 ) .

Example 3.4 There are some forms of symmetric solutions u only in some domains; for example, taking v ( ξ ) = 1 | ξ + T ( c 1 c 2 ) | 2 , c 3 = 1 ,
{ u 1 = 1 | x 1 x 2 + ( c 1 c 2 ) ( T t ) | 2 + c 1 , u 2 = 1 | x 1 x 2 + ( c 1 c 2 ) ( T t ) | 2 + c 2 , p = p ( t )
(3.9)

are some twin wave solutions to (2.6), and they form a symmetry only at t = T , or they are static.

Example 3.5 Take v ( ξ ) = 1 ( 1 + | ξ | 2 ) 2 c 1 , c 3 = 1 . Then we find that
{ u 1 = 1 ( 1 + | x 1 x 2 ( c 1 c 2 ) t | 2 ) 2 , u 2 = 1 ( 1 + | x 1 x 2 ( c 1 c 2 ) t | 2 ) 2 + c 2 c 1 , p = p ( t )
(3.10)

are global smooth twin wave solutions to the Euler equations (2.6) with finite energy in any bounded domain, but u C with infinite energy over R 2 except for the static case. If the components of the wave speed are equal, then the system is static.

Example 3.6 If we take v ( ξ ) = 1 | ξ | 2 , c 3 = 1 , then
{ u 1 = 1 | x 1 x 2 ( c 1 c 2 ) t | 2 + c 1 , u 2 = 1 | x 1 x 2 ( c 1 c 2 ) t | 2 + c 2 , p = p ( t )
(3.11)

are some twin wave solutions with singularity to the Euler equations (2.6).

Remark 3.7 These solutions in (3.11) have a singularity on the line { ( x 1 , x 2 ) | x 1 x 2 = ( c 1 c 2 ) t } for every t 0 . In particular, we have the following result:

For every given time t 0 and arbitrary line { ( x 1 , x 2 ) | A x 1 B x 2 = C , A 2 + B 2 0 } , there exist some solutions with singularity over the line { ( x 1 , x 2 ) | A x 1 B x 2 = C , A 2 + B 2 0 } .

Example 3.8 According to [10], w = u ( x C t , t ) + C , p ¯ = p ( x C t , t ) is also a solution pair for any constant vectors C R 2 . w = Q T u ( Q x , t ) , p ¯ = p ( Q x , t ) , is also a solution pair for any rotation matrices Q. w = λ τ ( x λ , t τ ) , p ¯ = λ 2 τ 2 ( x λ , t τ ) is also a solution pair.

4 Nonexistence

In this section we consider the explicit affine solution to the two-dimensional Euler equations. Here a solution u ( x , t ) is called an affine solution, if the u ( x , t ) is denoted by u ( x , t ) = ( v 1 ( x 1 c 1 t x 2 c 2 t ) , v 2 ( x 1 c 1 t x 2 c 2 t ) ) , c 2 0 . The affine solution indeed is a ( I , J ) similar solution. In fact, this is the case: β 1 = 1 , β 2 = c 1 t , β 3 = 1 , β 4 = c 2 t , M 1 = ( x 1 , 0 ) , M 2 = ( 1 , 0 ) , M 3 = ( 0 , x 2 ) , M 4 = ( 0 , 1 ) , z 1 = β 1 M 1 + β 2 M 2 , z 2 = β 3 M 3 + β 4 M 4 , w ( z 1 , z 2 ) = v ( z 1 z 2 ) . We have the following result.

Theorem 4.1 All affine solutions must be twin wave solutions. Affine solutions to the two-dimensional Euler equations are constant vectors. That is to say there does not exist a non-constant affine solution to the two-dimensional Euler equations.

Proof Since we are concerned with an affine solution here, let
u ( x , t ) = ( v 1 ( x 1 c 1 t x 2 c 2 t ) , v 2 ( x 1 c 1 t x 2 c 2 t ) ) , c 2 0 ,
by using (2.6), and letting ξ = x 1 c 1 t x 2 c 2 t , a straightforward calculation shows that
{ 1 x 2 c 2 t v 1 ξ x 1 c 1 t ( x 2 c 2 t ) 2 v 2 ξ = 0 , v ξ ( c 1 x 2 c 2 t + c 2 ( x 1 c 1 t ) ( x 2 c 2 ) 2 ) + 1 x 2 c 2 t v 1 v ξ x 1 c 1 t ( x 2 c 2 t ) 2 v 2 v ξ + p = 0 ,
(4.1)
{ v ξ ( c 1 ξ + c 2 ξ 2 + ξ v 1 ξ 2 v 2 ) + ( x 1 c 1 t ) p = 0 , v 1 ξ = ξ v 2 ξ ,
(4.2)
{ v 1 ξ ( c 1 ξ + c 2 ξ 2 + ξ v 1 ξ 2 v 2 ) + ( x 1 c 1 t ) p x 1 = 0 , v 2 ξ ( c 1 ξ + c 2 ξ 2 + ξ v 1 ξ 2 v 2 ) + ( x 1 c 1 t ) p x 2 = 0 , v 1 ξ = ξ v 2 ξ ,
(4.3)
{ v 1 ξ ( c 1 ξ + c 2 ξ 2 + ξ v 1 ξ 2 v 2 ) + ξ p ξ = 0 , v 2 ξ ( c 1 ξ + c 2 ξ 2 + ξ v 1 ξ 2 v 2 ) ξ 2 p ξ = 0 , v 1 ξ = ξ v 2 ξ ,
(4.4)
{ v 2 ξ ( c 1 ξ + c 2 ξ 2 + ξ v 1 ξ 2 v 2 ) + p ξ = 0 , v 2 ξ ( c 1 ξ + c 2 ξ 2 + ξ v 1 ξ 2 v 2 ) ξ 2 p ξ = 0 , v 1 ξ = ξ v 2 ξ .
(4.5)
Thus
p ξ = ξ 2 p ξ
and
p = p ( t ) .
(4.6)
Hence
{ v 2 ξ = 0 , v 1 ξ = ξ v 2 ξ ,
(4.7)
or
{ c 1 ξ + c 2 ξ 2 + ξ v 1 ξ 2 v 2 = 0 , v 1 ξ = ξ v 2 ξ .
(4.8)
Therefore, we get
v 1 = constant , v 2 = constant , p = p ( t ) .

 □

5 Uniqueness

In this section, we study the uniqueness of the initial boundary value problem for the Euler equations ( σ = 0 ) below:
u t + ( u ) u + p = σ u , in  R n × ( 0 , ) , div u = 0 , in  R n × [ 0 , ) , n = 2 , 3 ,
(5.1)

and we have the following result.

Theorem 5.1 Assume that Ω R + 2 is a bounded domain and inf x Ω { x 2 } = a > 0 . If c C ( R + 1 ¯ ) , h ( s ) is an arbitrary smooth function of s satisfying | h ( r ) | M 1 , | h r ( r ) | M 2 r 2 for r a > 0 , where M 1 , M 2 , and a are positive constants, and r = x 1 2 + x 2 2 . Then the following initial boundary value problem:
{ u t + ( u ) u + p = 0 , in Ω × ( 0 , ) , div u = 0 , in Ω × ( 0 , ) , u ( x , 0 ) = ( ( c 0 r 2 + h ( r ) ) x 2 , ( c 0 r 2 + h ( r ) ) x 1 ) , in Ω , p ( x , 0 ) = c 1 arctan x 1 x 2 + a r s ( c 0 s 2 + h ( s ) ) 2 d s , in Ω , u ( x , t ) = ( ( c ( t ) r 2 + h ( r ) ) x 2 , ( c ( t ) r 2 + h ( r ) ) x 1 ) , on Ω × [ 0 , ) , p ( x , t ) = c ( t ) arctan x 1 x 2 , on Ω × [ 0 , ) ,
(5.2)

has a unique smooth solution u ( C ( [ 0 , ) × Ω ) ) 2 , p C ( [ 0 , ) × Ω ) .

Proof Notice that div u ( x , 0 ) = 0 , inf x Ω { x 2 } = a > 0 , c C [ 0 , ) , and given the assumptions on h ( s ) satisfying the initial boundary value problems in (2.6), the result about existence is directly derived by Theorem 2.2. To prove the uniqueness we consider two smooth solution pairs, say u, p and v, p 1 . Let their difference be w = u v , with initial value w 0 , and let p ˜ be the difference of the corresponding pressures. Then, subtracting the equations from each other in (5.1), we have
{ w t + u w + w u w w + p ˜ = 0 , in  Ω × ( 0 , ) , div u = 0 , div w = 0 , in  Ω × [ 0 , ) , w ( x , 0 ) = ( 0 , 0 ) , in  Ω , p ˜ ( x , 0 ) = 0 , in  Ω , w | Ω = ( 0 , 0 ) , p ˜ | Ω = 0 .
(5.3)
Multiplying the first equation of (5.3) by w, integrating over Ω, and using the Gauss formula, we obtain
1 2 d d t w L 2 2 = ( w u , w ) ,
(5.4)
since (using the Gauss formula and div u = 0 , div w = 0 )
( u w , w ) = i , j = 1 2 Ω u j w i x j w i d x = Ω | w | 2 2 u n d s = 0 , ( p ˜ , w ) = i = 1 2 Ω p ˜ x i w i d x = Ω p ˜ div w d x = 0 ,

where n stands for the outward unit normal to Ω.

Similarly ( w w , w ) = Ω | w | 2 2 w n d s = 0 .

By using u = ( ( c ( t ) r 2 h ( r ) ) x 2 , ( c ( t ) r 2 h ( r ) ) x 1 ) in the term | ( w u , w ) | , we obtain
| ( w u , w ) | w L 2 2 u L ,
(5.5)
| u | = [ ( 2 c ( t ) r 4 + h r ) ] 2 r 4 + 2 ( c ( t ) r 2 + h ) 2 ( 2 + 2 ) | c ( t ) | r 2 + 2 | h | + r 2 | h r | , u L ( 2 + 2 ) | c ( t ) | a 2 + M 1 + M 2 .
(5.6)
Inserting (5.5)-(5.6) into (5.4), it follows that
d d t w L 2 2 2 ( ( 2 + 2 ) | c ( t ) | a 2 + M 1 + M 2 ) w L 2 2 .
(5.7)
Thanks to the Gronwall inequality, we have the following:
w L 2 2 exp 0 t 2 ( ( 2 + 2 ) | c ( t ) | a 2 + M 1 + M 2 ) d τ w ( x , 0 ) L 2 2 = 0 .
(5.8)

Therefore there exists a unique solution in the sense of L 2 ( 0 , T ; ( L 2 ( Ω ) ) 3 ) , T > 0 . The denseness of C ( [ 0 , T ] × Ω ) in L 2 ( 0 , T ; ( L 2 ( Ω ) ) 3 ) implies the uniqueness of the solution in the sense of C ( [ 0 , T ] × Ω ) from w = u v ( C ( [ 0 , ) × Ω ) ) 2 . We can apply the same argument on the intervals [ T , 2 T ] , [ 2 T , 3 T ] , etc., according to the uniform Gronwall lemma since c C ( R + 1 ¯ ) . We obtain the uniqueness of the solution.

Let us again consider Example 2.6. We have
( w u , w ) = ( t 1 ) Ω [ w 1 2 2 x 1 x 2 r 4 + w 1 w 2 2 ( x 1 2 x 2 2 ) r 4 + 2 x 1 x 2 r 4 w 2 2 ] d x = ( t 1 ) Ω ( w 2 2 w 1 2 ) 2 x 1 x 2 r 4 + w 1 w 2 2 ( x 1 2 x 2 2 ) r 4 d x = ( t 1 ) Ω 1 r 2 [ ( w 2 2 w 1 2 ) sin ( 2 θ ) + 2 w 1 w 2 cos ( 2 θ ) ] d x | t 1 | a 2 w L 2 2 .
(5.9)
Using (5.9) for the right-hand side of (5.4), we obtain
d d t w L 2 2 2 | t 1 | a 2 w L 2 2 .
(5.10)
Thanks to the Gronwall inequality again, we have the following result:
w L 2 2 exp 0 t 2 | t 1 | a 2 d τ w ( x , 0 ) L 2 2 = 0 .
(5.11)

Thus we prove the uniqueness of the solution.

On taking a more in-depth look, such as considering Ω = { ( x 1 , x 2 ) | ( 2 1 ) 2 x 2 2 x 1 2 x 2 2 } , we have
( w u , w ) ( t 1 ) Ω ( w 1 2 x 1 2 x 2 2 2 x 1 x 2 r 4 + w 2 2 x 1 2 x 2 2 + 2 x 2 x 2 r 4 ) d x = 2 2 ( t 1 ) Ω ( w 1 2 sin ( π 4 2 θ ) r 2 + w 2 2 sin ( π 4 + 2 θ ) r 2 ) d x 0 , t 1 .
(5.12)
Thus we obtain
d d t w L 2 2 0 .
(5.13)
We get
w L 2 2 w ( x , 0 ) L 2 2 = 0 , uniformly for  t 1 .
(5.14)

It clarifies that the uniqueness of the solution is possible even as t if the right scope is chosen in Ω. □

6 Analysis of stability between the equations and its vortex equation

In this section we discuss the stability of the solution, respectively, in C ( Ω ) and L 2 ( Ω ) for the problem (5.1).

Let v, p 1 , and v 0 denote the solution pair of a slight disturbance and the initial boundary value of v. Let w = u v be the difference of u and v, with initial value w 0 ( x ) 0 in the sense of L 2 ( Ω ) and boundary value u v in the sense of L 2 ( Ω ) L ( Ω ) , t ( 0 , T ] , and let p ˜ be the difference of the corresponding pressure. Then, subtracting one equation from the other, if
{ w t + u w + w u w w + p ˜ = 0 , div u = 0 , div w = 0 , w ( x , 0 ) = w 0 ( x ) , w | Ω = ψ ( x , t )
(6.1)
for ( x , t ) Ω × ( 0 , T ) , and Ω = { r a | a > 0 } { ( x 1 , x 2 ) | x 1 2 x 2 2 } . Multiplying this by w, integrating over Ω, and using the Gauss formula, we obtain
d d t w L 2 2 2 | ( u w , w ) | + 2 | ( w u , w ) | + 2 | ( w w , w ) | .
(6.2)
Using the Gauss formula and div u = 0 , div w = 0 ,
( p ˜ , w ) = i = 1 2 Ω p ˜ x i w i d x = Ω p ˜ div w d x = 0 , | ( u w , w ) | = | i , j = 1 2 Ω u j w i x j w i d x | = | Ω | w | 2 2 u n d s | 1 2 u L ( Ω ) w L 2 ( Ω ) 2 ,

where n stands for the outward unit normal to Ω.

Similarly,
| ( w u , w ) | ( ( 2 + 2 ) | c ( t ) | a 2 + M 1 + M 2 ) w L 2 2 , | ( w w , w ) | = | Ω | w | 2 2 w n d s | 1 2 w L ( Ω ) w L 2 ( Ω ) 2 .
Combining the formulas above, we have
d d t w L 2 2 2 ( ( 2 + 2 ) | c ( t ) | a 2 + M 1 + M 2 ) w L 2 2 + 2 ( u L ( Ω ) + w L ( Ω ) ) w L 2 ( Ω ) 2 .
(6.3)
Since u v in the sense of L 2 ( Ω ) L ( Ω ) , we can make, for every given ε > 0 ,
( u L ( Ω ) + w L ( Ω ) ) w L 2 ( Ω ) 2 ε .
Using the Gronwall inequality in (6.3), for every t [ 0 , T ] ,
w L 2 2 exp 0 t 2 ( ( 2 + 2 ) | c ( τ ) | a 2 + M 1 + M 2 ) d τ w 0 ( x ) L 2 2 + 2 ε 0 t exp τ t 2 ( ( 2 + 2 ) | c ( s ) | a 2 + M 1 + M 2 ) d s d τ 0 ,
(6.4)

as u v in the sense of L 2 ( Ω ) L ( Ω ) and w 0 L 2 ( Ω ) 2 0 . So we reach the stability of the solution in finite time.

Remark 6.1 Adopting the same method as we use in the proof of Theorem 5.1, we obtain
d d t w L 2 2 2 ( u L ( Ω ) + w L ( Ω ) ) w L 2 ( Ω ) 2 , w L 2 2 2 t ε + 2 w ( x , 0 ) L 2 2 0 , as  u v  in the sense of  L 2 ( Ω ) L ( Ω ) ,
(6.5)

for t [ 0 , T ] .

7 Explicit solution to ( 3 + 1 ) -dimensional Navier-Stokes equation

We now give an improvement of the example in reference [13].

Example 7.1 Let 0 < T , and consider the initial problem for the Navier-Stokes equation,
{ u t + ( u ) u + p = σ u , in  R 3 × ( 0 , T ) , div u = 0 , in  R 3 × ( 0 , T ) , u ( x , 0 ) = u 0 = ( u 10 , u 20 , u 30 ) , p ( x , 0 ) = p 0 , u i 0 , p 0 C ( [ 0 , T 1 ] × B M ) for any  0 < T 1 < T , 0 < M < + ,
(7.1)
where B M = { x R 3 | | x | < M } . Suppose
u 10 = 1 T ( 1 + c 1 exp ( 1 12 σ T s 2 1 σ T s + c ) ) , u 20 = 1 T ( 1 + c 2 exp ( 1 12 σ T s 2 1 σ T s + c ) ) , u 30 = 1 T ( 1 + ( c 1 + c 2 ) exp ( 1 12 σ T s 2 1 σ T s + c ) ) , p 0 = 1 T ( 1 2 T s + c 3 ) , s = i = 1 3 ( x i x i 0 ) .
(7.2)
Then for arbitrary constants c 1 , c 2 , c 3 , and c, (7.1) has a class of smooth blow-up solutions at finite time T,
u = ( u 1 , u 2 , u 3 ) , p ,
(7.3)
u i , p C ( [ 0 , T 1 ] × B M ) for any 0 < T 1 < T , 0 < M < . We have
{ u 1 = 1 T t ( 1 + c 1 exp ( 1 12 σ ( T t ) s 2 1 σ T t s + c ) ) , u 2 = 1 T t ( 1 + c 2 exp ( 1 12 σ ( T t ) s 2 1 σ T t s + c ) ) , u 3 = 1 T t ( 1 + ( c 1 + c 2 ) exp ( 1 12 σ ( T t ) s 2 1 σ T t s + c ) ) , p = 1 T t ( 1 2 T t s + c 3 ) .
(7.4)
Moreover, the initial function satisfies the second equation,
div u 0 = 0 , in  R 3 .
(7.5)
Here a solution of (7.1) is called a smooth blow-up solution at finite time T, if u i , p C ( [ 0 , T 1 ] × B M ) W m 1 , q 1 ( 0 , T 1 ; W m 2 , q 2 ( B M ) ) for any T 1 ( 0 , T ) , any nonnegative integer numbers m 1 , m 2 , any positive real numbers q 1 , q 2 , but
lim t T u W m , q ( B M ) = + , lim t T p W m , q ( B M ) = + ,
(7.6)

for some nonnegative integer numbers m and positive real numbers q, M.

Remark 7.2 This example indicates that the C solution of the Navier-Stokes equation does not always tend to a solution of the Euler equation.

Declarations

Acknowledgements

The third author is grateful to Professor Boling Guo and Professor Zhouping Xin for their support, and he also thanks Dr Jihui Wu for help in writing this paper. The work is supported by National Natural Science Foundation of China (No. 11161057).

Authors’ Affiliations

(1)
Department of Mathematics, Yunnan Nationalities University
(2)
School of Mathematics & Statistics, Zhengzhou University
(3)
Institute of Mathematics, Yunnan Normal University
(4)
Department of Mathematics, Tongji University

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© Song et al.; licensee Springer. 2014

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