Gauss - Green theorem for Sobolev $H^1$ space Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proof of the Gauss-Green TheoremGauss–Ostrogradsky formula for Distributionsclarification in definition of outward normal derivativeIntegration by parts in Sobolev spaceThe Gauss-Green theorem for unbounded domainApplication of Gauss divergence theorem or Grren's formula in n dimensionAbout two subspaces of (1,2)-Sobolev spaceShow that $int rcdot n ds$ equals three time the volume of $omega$.Strong derivative+GreenEvans' PDE Chapter 5 Problem 7 (trace inequality through Gauss-Green)
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Gauss - Green theorem for Sobolev $H^1$ space
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proof of the Gauss-Green TheoremGauss–Ostrogradsky formula for Distributionsclarification in definition of outward normal derivativeIntegration by parts in Sobolev spaceThe Gauss-Green theorem for unbounded domainApplication of Gauss divergence theorem or Grren's formula in n dimensionAbout two subspaces of (1,2)-Sobolev spaceShow that $int rcdot n ds$ equals three time the volume of $omega$.Strong derivative+GreenEvans' PDE Chapter 5 Problem 7 (trace inequality through Gauss-Green)
$begingroup$
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
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
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
asked Mar 26 at 8:20
WawMathWawMath
61
$endgroup$
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
pde sobolev-spaces
asked Mar 26 at 8:20
WawMathWawMath
61
asked Mar 26 at 8:20
WawMathWawMath
61
asked Mar 26 at 8:20
WawMathWawMath
61
asked Mar 26 at 8:20
WawMathWawMath
61
asked Mar 26 at 8:20
WawMathWawMath
61
61
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
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.
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.
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.
answered Mar 26 at 8:44
user657324user657324
60110
$endgroup$
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.
answered Mar 26 at 8:44
user657324user657324
60110
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
60110
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@WawMath: I edited my answer.
$endgroup$
– user657324
Mar 26 at 9:41
add a comment |
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