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Why do we need projection in the definition of the Stokes operator?

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1

1

$begingroup$

$DeclareMathOperator{div}{div}$
$defbu{mathbf{u}}$
Let $D$ be the square $[0,1]^2$ and consider the following space:
$$
V:={bu: buin H^2(D)^2, div bu=0, u|_{partial D}=0 }.
$$

Introduce also
$$
H:={bu: buin L^2(D)^2, div bu=0}.
$$

Then the Stokes operator is defined as an operator $Vto H$ as
$$
Abu:=-P_LDeltabu,
$$
where $P_L$ is the projection on $H$.

I don't understand, why the Stokes operator is not equal to minus Laplacian? Indeed, if $buin V$, then $div bu=0$ and thus
$$
div Deltabu=u^1_{xxx}+u^1_{yyx}+u^2_{xxy}+u^2_{yyy}=(u^1_x+u^2_y)_{xx}+(u^1_x+u^2_y)_{yy}=Deltadivbu=0.
$$
In this case, $Delta buin H$ and thus the projection is not needed.

Question: what is wrong in my reasoning? Why the Stokes operator is not equal to Laplacian?
Related question: Can one write explicitly the eigenfunctions of Stokes operator in this simple situation? At least can maybe one bound $lambda_1$, the smallest eigenvalue, from below?

functional-analysis operator-theory orthonormal laplacian

share|cite|improve this question

edited Mar 15 at 14:42

Oleg

asked Mar 14 at 21:37

OlegOleg

301212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the extra exponents $2$ ($H^2(D)^2$, $L^2(D)^2$)??
  $endgroup$
  – Ted Shifrin
  Mar 14 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin $mathbf{u}=(u_1,u_2)$ is a vector. Each of its 2 coordinates $u_1$ and $u_2$ belongs to $H^2(D)$. Thus, $mathbf{u}in H^2(D)^2$.
  $endgroup$
  – Oleg
  Mar 14 at 21:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I see ... So, offhand, my answer to your question is that you need $C^3$ to interchange partial derivatives, and you don't have $C^3$ functions.
  $endgroup$
  – Ted Shifrin
  Mar 14 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin Then why in the periodic case (periodic boundary conditions) the Stokes operator is equal to Laplacian?
  $endgroup$
  – Oleg
  Mar 14 at 22:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: In the definition of $H$, in what sense is the divergence equal to zero?
  $endgroup$
  – maxmilgram
  Mar 15 at 17:42
  

  
  
  
  

 | 
show 6 more comments

1

1

$begingroup$

$DeclareMathOperator{div}{div}$
$defbu{mathbf{u}}$
Let $D$ be the square $[0,1]^2$ and consider the following space:
$$
V:={bu: buin H^2(D)^2, div bu=0, u|_{partial D}=0 }.
$$

Introduce also
$$
H:={bu: buin L^2(D)^2, div bu=0}.
$$

Then the Stokes operator is defined as an operator $Vto H$ as
$$
Abu:=-P_LDeltabu,
$$
where $P_L$ is the projection on $H$.

I don't understand, why the Stokes operator is not equal to minus Laplacian? Indeed, if $buin V$, then $div bu=0$ and thus
$$
div Deltabu=u^1_{xxx}+u^1_{yyx}+u^2_{xxy}+u^2_{yyy}=(u^1_x+u^2_y)_{xx}+(u^1_x+u^2_y)_{yy}=Deltadivbu=0.
$$
In this case, $Delta buin H$ and thus the projection is not needed.

Question: what is wrong in my reasoning? Why the Stokes operator is not equal to Laplacian?
Related question: Can one write explicitly the eigenfunctions of Stokes operator in this simple situation? At least can maybe one bound $lambda_1$, the smallest eigenvalue, from below?

functional-analysis operator-theory orthonormal laplacian

share|cite|improve this question

edited Mar 15 at 14:42

Oleg

asked Mar 14 at 21:37

OlegOleg

301212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the extra exponents $2$ ($H^2(D)^2$, $L^2(D)^2$)??
  $endgroup$
  – Ted Shifrin
  Mar 14 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin $mathbf{u}=(u_1,u_2)$ is a vector. Each of its 2 coordinates $u_1$ and $u_2$ belongs to $H^2(D)$. Thus, $mathbf{u}in H^2(D)^2$.
  $endgroup$
  – Oleg
  Mar 14 at 21:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I see ... So, offhand, my answer to your question is that you need $C^3$ to interchange partial derivatives, and you don't have $C^3$ functions.
  $endgroup$
  – Ted Shifrin
  Mar 14 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin Then why in the periodic case (periodic boundary conditions) the Stokes operator is equal to Laplacian?
  $endgroup$
  – Oleg
  Mar 14 at 22:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: In the definition of $H$, in what sense is the divergence equal to zero?
  $endgroup$
  – maxmilgram
  Mar 15 at 17:42
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

$DeclareMathOperator{div}{div}$
$defbu{mathbf{u}}$
Let $D$ be the square $[0,1]^2$ and consider the following space:
$$
V:={bu: buin H^2(D)^2, div bu=0, u|_{partial D}=0 }.
$$

Introduce also
$$
H:={bu: buin L^2(D)^2, div bu=0}.
$$

Then the Stokes operator is defined as an operator $Vto H$ as
$$
Abu:=-P_LDeltabu,
$$
where $P_L$ is the projection on $H$.

I don't understand, why the Stokes operator is not equal to minus Laplacian? Indeed, if $buin V$, then $div bu=0$ and thus
$$
div Deltabu=u^1_{xxx}+u^1_{yyx}+u^2_{xxy}+u^2_{yyy}=(u^1_x+u^2_y)_{xx}+(u^1_x+u^2_y)_{yy}=Deltadivbu=0.
$$
In this case, $Delta buin H$ and thus the projection is not needed.

Question: what is wrong in my reasoning? Why the Stokes operator is not equal to Laplacian?
Related question: Can one write explicitly the eigenfunctions of Stokes operator in this simple situation? At least can maybe one bound $lambda_1$, the smallest eigenvalue, from below?

functional-analysis operator-theory orthonormal laplacian

share|cite|improve this question

edited Mar 15 at 14:42

Oleg

asked Mar 14 at 21:37

OlegOleg

301212

$endgroup$

share|cite|improve this question

edited Mar 15 at 14:42

Oleg

asked Mar 14 at 21:37

OlegOleg

301212

share|cite|improve this question

edited Mar 15 at 14:42

Oleg

asked Mar 14 at 21:37

OlegOleg

301212

asked Mar 14 at 21:37

OlegOleg

301212

asked Mar 14 at 21:37

OlegOleg

301212