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Gauss - Green theorem for Sobolev $H^1$ space

1

$begingroup$

I know the Gauss-Green theorem:

Let $U subset mathbbR^n$ be an open, bounded set with $∂U$ being $C^1$. Suppose $u ∈ C^1(bar U)$, then
$$∫_U u_x_i dx = int_∂U u nu^i dS,$$

where $nu=(nu^1,…nu^n)$ denotes the outward-pointing unit normal vector field to the region $U$.

My question is:
How to prove that this theorem is true with the weaker assumption that $u in H^1(U)$?

pde sobolev-spaces

share|cite|improve this question

asked Mar 26 at 8:20

WawMathWawMath

61

$endgroup$

add a comment | 

1

$begingroup$

I know the Gauss-Green theorem:

Let $U subset mathbbR^n$ be an open, bounded set with $∂U$ being $C^1$. Suppose $u ∈ C^1(bar U)$, then
$$∫_U u_x_i dx = int_∂U u nu^i dS,$$

where $nu=(nu^1,…nu^n)$ denotes the outward-pointing unit normal vector field to the region $U$.

My question is:
How to prove that this theorem is true with the weaker assumption that $u in H^1(U)$?

pde sobolev-spaces

share|cite|improve this question

asked Mar 26 at 8:20

WawMathWawMath

61

$endgroup$

add a comment | 

$begingroup$

I know the Gauss-Green theorem:

Let $U subset mathbbR^n$ be an open, bounded set with $∂U$ being $C^1$. Suppose $u ∈ C^1(bar U)$, then
$$∫_U u_x_i dx = int_∂U u nu^i dS,$$

where $nu=(nu^1,…nu^n)$ denotes the outward-pointing unit normal vector field to the region $U$.

My question is:
How to prove that this theorem is true with the weaker assumption that $u in H^1(U)$?

pde sobolev-spaces

share|cite|improve this question

asked Mar 26 at 8:20

WawMathWawMath

61

$endgroup$

share|cite|improve this question

asked Mar 26 at 8:20

WawMathWawMath

61

share|cite|improve this question

asked Mar 26 at 8:20

WawMathWawMath

61

asked Mar 26 at 8:20

WawMathWawMath

61

asked Mar 26 at 8:20

WawMathWawMath

61

1 Answer
1

active

oldest

votes

0

$begingroup$

Hint

$$mathcal C^1(bar U)text is dense in H^1(U).$$

Edit

$mathcal C^1(bar U)$ dense in $H^1(U)$ mean that if $uin H^1(U)$, there exist a sequence $(u_n)subset mathcal C^1(bar U)$ s.t. $$|u_n-u|_H^1(U)undersetnto infty longrightarrow 0.$$
Therefore $$int_U|u_n-u|^2+sum_i=1^nint_U|partial _i u_n-partial _i u|^2=0.$$

In paticular, since $U$ is bounded, $$left|int_Upartial _i u_n-int_U partial _iuright|^2leq Cint _U |partial _iu_n-partial _iu|^2,$$ by Jensen's inequality. Then you can get one limit. For $$lim_nto infty int_partial Uu_nnu _i=int_partial Uunu ^i,$$ it's a consequence of the continuity of the trace operator.

share|cite|improve this answer

edited Mar 26 at 9:58

answered Mar 26 at 8:44

user657324user657324

60110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, so I've got that there exists a sequence $(u^(m)) in C^1(bar U)$ such that $u^(m) to u$ in $H^1(U)$. Since $(u^(m)) in C^1(bar U)$, I've got $$int_U u_x_i^(m) dx = int_partial U u^(m) nu^i dS.$$ Now I need to go to the limit under the integral sign but I have no idea what to use.
  $endgroup$
  – WawMath
  Mar 26 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WawMath: I edited my answer.
  $endgroup$
  – user657324
  Mar 26 at 9:41
  

  
  
  
  

add a comment | 

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0

$begingroup$

Hint

$$mathcal C^1(bar U)text is dense in H^1(U).$$

Edit

$mathcal C^1(bar U)$ dense in $H^1(U)$ mean that if $uin H^1(U)$, there exist a sequence $(u_n)subset mathcal C^1(bar U)$ s.t. $$|u_n-u|_H^1(U)undersetnto infty longrightarrow 0.$$
Therefore $$int_U|u_n-u|^2+sum_i=1^nint_U|partial _i u_n-partial _i u|^2=0.$$

In paticular, since $U$ is bounded, $$left|int_Upartial _i u_n-int_U partial _iuright|^2leq Cint _U |partial _iu_n-partial _iu|^2,$$ by Jensen's inequality. Then you can get one limit. For $$lim_nto infty int_partial Uu_nnu _i=int_partial Uunu ^i,$$ it's a consequence of the continuity of the trace operator.

share|cite|improve this answer

edited Mar 26 at 9:58

answered Mar 26 at 8:44

user657324user657324

60110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, so I've got that there exists a sequence $(u^(m)) in C^1(bar U)$ such that $u^(m) to u$ in $H^1(U)$. Since $(u^(m)) in C^1(bar U)$, I've got $$int_U u_x_i^(m) dx = int_partial U u^(m) nu^i dS.$$ Now I need to go to the limit under the integral sign but I have no idea what to use.
  $endgroup$
  – WawMath
  Mar 26 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WawMath: I edited my answer.
  $endgroup$
  – user657324
  Mar 26 at 9:41
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint

$$mathcal C^1(bar U)text is dense in H^1(U).$$

Edit

$mathcal C^1(bar U)$ dense in $H^1(U)$ mean that if $uin H^1(U)$, there exist a sequence $(u_n)subset mathcal C^1(bar U)$ s.t. $$|u_n-u|_H^1(U)undersetnto infty longrightarrow 0.$$
Therefore $$int_U|u_n-u|^2+sum_i=1^nint_U|partial _i u_n-partial _i u|^2=0.$$

In paticular, since $U$ is bounded, $$left|int_Upartial _i u_n-int_U partial _iuright|^2leq Cint _U |partial _iu_n-partial _iu|^2,$$ by Jensen's inequality. Then you can get one limit. For $$lim_nto infty int_partial Uu_nnu _i=int_partial Uunu ^i,$$ it's a consequence of the continuity of the trace operator.

share|cite|improve this answer

edited Mar 26 at 9:58

answered Mar 26 at 8:44

user657324user657324

60110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, so I've got that there exists a sequence $(u^(m)) in C^1(bar U)$ such that $u^(m) to u$ in $H^1(U)$. Since $(u^(m)) in C^1(bar U)$, I've got $$int_U u_x_i^(m) dx = int_partial U u^(m) nu^i dS.$$ Now I need to go to the limit under the integral sign but I have no idea what to use.
  $endgroup$
  – WawMath
  Mar 26 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WawMath: I edited my answer.
  $endgroup$
  – user657324
  Mar 26 at 9:41
  

  
  
  
  

add a comment | 

$begingroup$

Hint

$$mathcal C^1(bar U)text is dense in H^1(U).$$

Edit

$mathcal C^1(bar U)$ dense in $H^1(U)$ mean that if $uin H^1(U)$, there exist a sequence $(u_n)subset mathcal C^1(bar U)$ s.t. $$|u_n-u|_H^1(U)undersetnto infty longrightarrow 0.$$
Therefore $$int_U|u_n-u|^2+sum_i=1^nint_U|partial _i u_n-partial _i u|^2=0.$$

In paticular, since $U$ is bounded, $$left|int_Upartial _i u_n-int_U partial _iuright|^2leq Cint _U |partial _iu_n-partial _iu|^2,$$ by Jensen's inequality. Then you can get one limit. For $$lim_nto infty int_partial Uu_nnu _i=int_partial Uunu ^i,$$ it's a consequence of the continuity of the trace operator.

share|cite|improve this answer

edited Mar 26 at 9:58

answered Mar 26 at 8:44

user657324user657324

60110

$endgroup$

share|cite|improve this answer

edited Mar 26 at 9:58

answered Mar 26 at 8:44

user657324user657324

60110

answered Mar 26 at 8:44

user657324user657324

60110

answered Mar 26 at 8:44

user657324user657324

60110