Estimates of solutions to the linear Navier-Stokes equation
aa r X i v : . [ m a t h . G M ] F e b Estimates of solutions to the linear Navier-Stokesequation
Dr. Bazarbekov Argyngazy B.
Kazakh National State University, EKSU. e-mail : [email protected]
Abstact .
The linear Navier-Stokes equations in three dimensions are given by: u it ( x, t ) − ρ △ u i ( x, t ) − p x i ( x, t ) == w i ( x, t ) , div u ( x, t ) = 0 , i = 1 , , with initial conditions: u | ( t =0) S ∂ Ω = 0 . The Green function to the Dirichlet problem u | ( t =0) S ∂ Ω = 0 of the equation u it ( x, t ) − ρ △ u i ( x, t ) = f i ( x, t ) present as: G ( x, t ; ξ, τ ) = Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) . Where Z ( x, t ; ξ, τ ) = π / ( t − τ ) / · e − ( x − ξ x − ξ x − ξ t − τ ) - is the fundamental solution to this equation [1 p.137] and V ( x, t ; ξ, τ ) -is the smooth function of variables ( x, t ; ξ, τ ) . The construction of the function G ( x, t ; ξ, τ ) is resulted inthe book [1 p.106]. By the Green function we present the Navier-Stokes equation as: u i ( x, t ) = R t R Ω (cid:18) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:19) dp ( ξ,τ ) dξ dξdτ + R t R Ω G ( x, t ; ξ, τ ) w i ( ξ, τ ) dξdτ . But div u ( x, t ) = P du i ( x,t ) dx i = 0 . Using these equations and thefollowing properties of the fundamental function: Z ( x, t ; ξ, τ ) : dZ ( x,t ; ξ,τ ) dx i = − dZ ( x,t ; ξ,τ ) dξ i , for the definition of the unknownpressure p(x,t) we shall receive the integral equation. From this integral equation we define the explicit expression of thepressure: p ( x, t ) = − ddt △ − ∗ R t R Ω P dG ( x,t ; ξ,τ ) dx i w i ( ξ, τ ) dξdτ + ρ · R t R Ω P dG ( x,t ; ξ,τ ) dx i w i ( ξ, τ ) dξdτ. By this formula thefollowing estimate: R t P (cid:13)(cid:13)(cid:13)(cid:13) ∂p ( x,τ ) ∂x i (cid:13)(cid:13)(cid:13)(cid:13) L (Ω) dτ < c · R t P k w i ( x, τ ) k L (Ω) dτ holds.2000 Mathematical Subject Classification. Primary: K ; Secondary: E ; Keywords: postulate , integral equation,Green function, Dirichlet problem, operator , smooth functions.
1. Introduction . Let Ω ⊂ R be a finite domain bounded by the Lipschitz surface ∂ Ω . Q t = Ω × [0 , t ] , x = ( x , x , x ) and u ( x, t ) = ( u i ( x, t ) i =1 , , are the vector functions.The boundary problems for linear Navier-Stokes equations were considered in the book [2p.89-96]. If ∂ Ω -is a smooth surface, then there exists a solution: u ( x, t ) : u x i x j ( x, t ) , p ( x, t ) ∈ L (Ω ′ ) × [0 , T ] where Ω ′ ⊂ Ω is a subarea of the area Ω . I do not know other works devotedto this problem.
2. The formulation and proof of result.
Let Ω ⊂ R be a finite domain bounded by the Lipschitz surface ∂ Ω . Q t = Ω × [0 , t ] , x =( x , x , x ) and u ( x, t ) = ( u i ( x, t ) i =1 , , , f ( x, t ) = ( f i ( x, t ) i =1 , , - are vector functions. Here t> 0, ρ > are an arbitrary real numbers. The linear Navier-Stokes equations are given by: ∂u i ( x, t ) ∂t − ρ △ u i ( x, t ) − ∂p ( x, t ) ∂x i = w i ( x, t ) (2 , div u ( x, t ) = X i =1 ∂u i ( x, t ) ∂x i = 0 , i = 1 , , , ′ ) u ( x,
0) = 0 , u ( x, t ) | ∂ Ω × [0 ,t ] = 0 (2 , The Green function for Dirichlet problem is as follows: G ( x, t ; ξ, τ ) = Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) where Z ( x, t ; ξ, τ ) = π / ( t − τ ) / · e − ( x − ξ x − ξ x − ξ t − τ ) is the fundamental solution forthis equation [1 p.106] and V ( x, t ; ξ, τ ) is the smooth function of variables ( x, t ; ξ, τ ) . Theconstruction of the function V ( x, t ; ξ, τ ) is resulted in the book [1 p.106]. By the Greenfunction we present the Navier-Stokes equation as: u i ( x, t ) = Z t Z Ω (cid:16) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:17) dp ( ξ, τ ) dξ dξdτ + Z t Z Ω G ( x, t ; ξ, τ ) w i ( ξ, τ ) dξdτ . i = 1 , , . div u ( x, t ) = X du i ( x, t ) dx i = 0 (2 , We shall prove the following fundamental result:
Theorem 2,1
There exists the explicit expression to the pressure p(x,t), depending onthe right-hand side w i ( x, t ) : p ( x, t ) = − ddt △ − ∗ Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ + ρ · Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ where △ − - is the inverse operator to the Dirichlet problem of Laplase operator for thedomain Ω . To the solutions u i ( x, t ) , p ( x, t ) for these linear Navier-Stokes equation thefollowing estimates: k u ( x, t ) k W , ( Q t ) < c · k w ( x, t ) k L ( Q t ) Z t X (cid:13)(cid:13)(cid:13) ∂p ( x, τ ) ∂x i (cid:13)(cid:13)(cid:13) L (Ω) dτ < c · Z t X k w i ( x, τ ) k L (Ω) dτ (2 , are valid. Here and below by symbol c we denote a generic constants, independent on thesolution and right-hand side whose value is inessential to our aims, and it may change fromline to line. ◭ Postulate:
Let t > 0 be an arbitrary real number. For the proof of the estimates (2,3)we assume that: p ( x, t ) ∈ L ( W (Ω); [0 , t ]) . I.e. Z t X (cid:13)(cid:13)(cid:13) ∂p ( x, τ ) ∂x i (cid:13)(cid:13)(cid:13) L (Ω) dτ < ∞ (2 , Remark 1.
For the function p ( x, t ) ∈ L ( W (Ω); [0 , t ]) there exists p ( x, t ) | ∂ Ω ∈ L ( ∂ Ω; [0 , t ]) . ◭ Below, for the definition of the unknown pressure p ( x, t ) : p ( x, t ) ∈ L ( Q t ) we shallreceive the integral equation. And from this integral equation we shall receive the explicitexpression of the pressure p(x,t) and estimates (2,3).It is known , that G ( x, t ; ξ, τ ) = G ( ξ, t ; x, τ ) [3 p. 305] and G ( x, t ; ξ, τ ) | ξ ∈ ∂ Ω = G ( x, t ; ξ, τ ) | x ∈ ∂ Ω = 0 . Since G ( x, t ; ξ, τ ) | ξ ∈ ∂ Ω = 0 and p ( x, t ) ∈ L ( W (Ω); [0 , t ]) , integratingby part, from (2,2) we have: u i ( x, t ) = − Z t Z Ω (cid:16) dZ ( x, t ; ξ, τ ) dξ i + dV ( x, t ; ξ, τ ) dξ i (cid:17) p ( ξ, τ ) dξdτ + + Z t Z Ω G ( x, t ; ξ, τ ) w i ( ξ, τ ) dξdτ (2 , Since dZ ( x,t ; ξ,τ ) dx i = − dZ ( x,t ; ξ,τ ) dξ i and div u ( x, t ) = P du i ( x,t ) dx i = 0 , from (2,5) we obtain: △ x Z t Z Ω Z ( x, t ; ξ, τ ) p ( ξ, τ ) dξdτ − Z t Z Ω (cid:16) X d V ( x, t ; ξ, τdx i dξ i (cid:17) p ( ξ, τ ) dξdτ ++ Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ = 0 (2 , Since P ddx i dV ( x,t ; ξ,τ ) dx i = △ x V ( x, t ; ξ, τ ) , we rewrite this equation in the following form: △ x Z t Z Ω (cid:16) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:17) p ( ξ, τ ) dξdτ −− Z t Z Ω 3 X ddx i (cid:16) dV ( x, t ; ξ, τ ) dξ i + dV ( x, t ; ξ, τ ) dx i (cid:17) p ( ξ, τ ) dξdτ = (2 , − Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ Let us denote: △ ∗ u ( x, t ) = X u x i x i ( x, t ); ; △ − ∗ f ( x, t ) = Z Ω G D ( x, ξ ) f ( ξ, t ) dξT ∗ u ( x, t ) = ∂u ( x, t ) ∂t − ρ · △ u ( x, t ); ; T ∗ ∗ u ( x, t ) = − ∂u ( x, t ) ∂t − ρ △ u ( x, t ) (2 , V ∗ p ( x, t ) = Z t Z Ω 3 X ddx i (cid:16) dV ( x, t ; ξ, τ ) dξ i + dV ( x, t ; ξ, τ ) dx i (cid:17) p ( ξ, τ ) dξdτ where G D ( x ; ξ ) is the Green function of the Dirichlet problem for Laplase operator and thedomain Ω . Note that: R t R Ω (cid:16) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:17) p ( ξ, τ ) dξdτ (cid:12)(cid:12)(cid:12) x ∈ ∂ Ω = 0 . Usingthese denotes , we rewrite the basis equation (2,7): Z t Z Ω (cid:16) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:17) p ( ξ, τ ) dξdτ − △ − ∗ V ∗ p ( x, t ) == − △ − ∗ Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ (2 , Since T ∗ R t R Ω (cid:16) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:17) p ( ξ, τ ) dξdτ = p ( x, t ) , from (2,9) we have: p ( x, t ) − T ∗ △ − ∗ V ∗ p ( x, t ) = − T ∗ △ − ∗ Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ (2 , Lemma 2,1.
For any function p ( x, t ) ∈ L ( Q t ) the following equality is valid: T ( x,t ) ∗ V ∗ p ( x, t ) = d V ∗ p ( x, t ) dt − ρ · △ x V ∗ p ( x, t ) ≡ , where the operator V ∗ p ( x, t ) is defined above by the formula: V ∗ p ( x, t ) = Z t Z Ω (cid:16) X d V ( x, t ; ξ, τ ) dx i dξ i + △ x V ( x, t ; ξ, τ ) (cid:17) p ( ξ, τ ) dξdτ (2 , ◮ On the domain Ω × [0 , t ] А.Friedman has constructed in the book [1 p. 106-111] the Greenfunction of Dirichlet problem to the parabolic equation: T ∗ u ( x, t ) = u t ( x, t ) − ρ · △ x u ( x, t ) = p ( x, t ); u { t =0 } S ∂ Ω × [0 ,t ] = 0 . And present the solution to this problem as: u ( x, t ) = Z t Z Ω (cid:16) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:17) p ( ξ, τ ) dξdτ (2 , where the function V ( x, t ; ξ, τ ) by variables ( ξ, τ ) satisfies to the following equation andboundary conditions [1 p. 108]: T ∗ ( ξ,τ ) ∗ V ( x, t ; ξ, τ ) = − ddτ V ( x, t ; ξ, τ ) − ρ · △ ξ V ( x, t ; ξ, τ ) = 0 (2 , V ( x, t ; ξ, τ ) (cid:12)(cid:12)(cid:12) ( ξ,τ ) ∈ ∂ Ω × [0 ,t ] = − Z ( x, t ; ξ, τ ) (cid:12)(cid:12)(cid:12) ( ξ,τ ) ∈ ∂ Ω × [0 ,t ] (2 , ′ ) V ( x, t ; ξ, τ ) (cid:12)(cid:12)(cid:12) τ = t = V ( x, t ; ξ, t ) = 0 (2 , The formula (2,13) follows from integrating by part the following expression: Z t Z Ω (cid:16) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:17)(cid:16) u τ − ρ · △ ξ u ( ξ, τ ) (cid:17) dξdτ = u ( x, t ) , using the boundary conditions: (2,1"), (2,14’) , (2,14").The function V ( x, t ; ξ, τ ) by variables ( x, t ) satisfies to the following parabolic equation: T ( x,t ) ∗ V ( x, t ; ξ, τ ) = 0 . Then, the following equalities are valid: T ( x,t ) ∗ d V ( x, t ; ξ, τ ) dx i dξ i = ddx i T ( x,t ) ∗ dV ( x, t ; ξ, τ ) dξ i = 0 T ( x,t ) ∗ △ x V ( x, t ; ξ, τ ) = △ x ∗ T ( x,t ) ∗ V ( x, t ; ξ, τ ) = 0 Since V ( x, t ; ξ, t ) = 0 for any x ∈ Ω , ξ ∈ Ω , then d V ( x,t ; ξ,t ) dx i dξ i = 0 and △ x V ( x, t ; ξ, t ) = 0 . Therefore, from (2,12) we have T ( x,t ) ∗ V ∗ p ( x, t ) = 0 . Lemma 2,1 is proved. ◭ The Laplase and parabolic opеrаtоrs.
The Dirichlet problem for Laplase operatorhas infinite set of the own functions u n ( x ) and own numbers λ n > △ u n ( x ) = − λ n · u n ( x ); | u n ( x ) | L (Ω) = 1 , ( u n ( x ) , u m ( x )) L (Ω) = 0 . The each solution u(x,t) tothe homogeneous parabolic equation T ∗ u ( x, t ) = 0 present as: u ( x, t ) = ∞ X c n · e − λ n · ρ · t · u n ( x ) (2 , It follows from the decomposition of the function u(x,t) on the own functions { u n ( x ) } . Lemma 2,2.
For any function p ( x, t ) ∈ L ( Q t ) is valid the following equality: T ( x,t ) ∗ △ − ∗ V ∗ p ( x, t ) ≡ , ◮ Since △ ∗△ − ∗ f ( x, t ) = △ x ∗ R Ω G D ( x, ξ ) f ( ξ, t ) dξ = f ( x, t ) , it follows by the definitionof the operators T, △ − that: T ∗ △ − ∗ V ∗ p ( x, t ) = ddt △ − ∗ V ∗ p ( x, t ) − ρ · V ∗ p ( x, t ) (2 , As it is proved in Lemma 2.1: T ∗ V ∗ p ( x, t ) = 0 . Decomposing the function V ∗ p ( x, t ) onthe own functions to Laplase operator, we shall receive his representation as: V ∗ p ( x, t ) = ∞ X c n · e − λ n · ρ · t · u n ( x ) (2 , ′ ) Since △ − u n ( x ) = − λ n · u n ( x ) , then △ − ∗ V ∗ p ( x, t ) = − P ∞ c n λ n · e − ρ · λ n · t · u n ( x ) . Substitutingthis function to the formula (2,17), we shall receive that: T ( x,t ) ∗△ − ∗ V ∗ p ( x, t ) ≡ . Lemma2.2 is proved. ◭ From the equation (2,10) , Lemma 2.2 and formula (2,17) we shall receive the followingexplicit expression to pressure function p(x,t), depending on the right-hand side w (x,t): p ( x, t ) = − T ∗ △ − ∗ Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ = (2 , − ddt △ − ∗ Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ + ρ · Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ Let us notice that by this formula follows the following formula: △ x p ( x, t ) = P w i ( x ) dx i . Thesame formula follows from the Navier-Stokes equation. It is obvious , that: Z t Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ ∈ W , ( Q t ) It follows by the second formula (2,18) the following estimate: Z t (cid:16)(cid:13)(cid:13)(cid:13) T ∗ △ − ∗ Z τ Z Ω 3 X dG ( x, t ; ξ, τ ) dx i w i ( ξ, τ ) dξdτ (cid:13)(cid:13)(cid:13) W (Ω) (cid:17) dτ < c | w ( x, τ ) | L ( Q t ) Differentiating the equation (2,18) by x i , from this estimate we obtain an estimate (2,3). Remark 2.
By the formula (2,18) we find dp ( x,t ) dx i , i = 1 , , , depending on the functions w i ( x, t ) , and substitute these functions to the Navier-Stokes equation (2,2): u i ( x, t ) = R t R Ω (cid:16) Z ( x, t ; ξ, τ ) + V ( x, t ; ξ, τ ) (cid:17) dp ( ξ,τ ) dξ dξdτ + R t R Ω G ( x, t ; ξ, τ ) w i ( ξ, τ ) dξdτ . We find theexplicit expression to the solutions u i ( x, t ) by the right-hand side w i ( x, t ) . Theorem (2,1)is proved. ◭◭