Is there a bounded domain on which Poincaré's inequality does not hold?
Suppose that $U$ is a bounded domain in $mathbb{R}^n$. Poincaré's inequality states that (for $U$ sufficiently "nice") there exists a constant $C>0$ such that if $uin H^1(U)$ satisfies $int_U u = 0$ then
$$
int_U lvert urvert^2 leq Cint_U lvert nabla u rvert^2
$$
I was wondering if it was possible to find an example of a bounded domain on which the inequality does not hold. The proof for this inequality in Evan's PDE requires $U$ to be an extension domain. So I though that maybe the inequality would be false on a domain as simple as $U = {(x,y) : 0<x<1, 0<y<x^2}$.
I tried to construct a sequence of functions $u_kin H^1(U)$ satisfying $int_U u_k = 0$ and such that
$$
frac{int_U lvert u_krvert^2}{int_U lvert nabla u_k rvert^2} toinfty
$$
but unforunately I was unable to obtain such a sequence.
Is the inequality always true on bounded domains? Or is it indeed possible to find a counter-example
functional-analysis pde sobolev-spaces examples-counterexamples
asked Nov 27 at 0:41
Quoka
1,208212
add a comment |
Suppose that $U$ is a bounded domain in $mathbb{R}^n$. Poincaré's inequality states that (for $U$ sufficiently "nice") there exists a constant $C>0$ such that if $uin H^1(U)$ satisfies $int_U u = 0$ then
$$
int_U lvert urvert^2 leq Cint_U lvert nabla u rvert^2
$$
I was wondering if it was possible to find an example of a bounded domain on which the inequality does not hold. The proof for this inequality in Evan's PDE requires $U$ to be an extension domain. So I though that maybe the inequality would be false on a domain as simple as $U = {(x,y) : 0<x<1, 0<y<x^2}$.
I tried to construct a sequence of functions $u_kin H^1(U)$ satisfying $int_U u_k = 0$ and such that
$$
frac{int_U lvert u_krvert^2}{int_U lvert nabla u_k rvert^2} toinfty
$$
but unforunately I was unable to obtain such a sequence.
Is the inequality always true on bounded domains? Or is it indeed possible to find a counter-example
functional-analysis pde sobolev-spaces examples-counterexamples
asked Nov 27 at 0:41
Quoka
1,208212
What is your definition of "domain"?
– gerw
Nov 27 at 7:05
I think Evans takes domain to be “open and connected”. @gerw
– DaveNine
Nov 27 at 8:04
add a comment |
Suppose that $U$ is a bounded domain in $mathbb{R}^n$. Poincaré's inequality states that (for $U$ sufficiently "nice") there exists a constant $C>0$ such that if $uin H^1(U)$ satisfies $int_U u = 0$ then
$$
int_U lvert urvert^2 leq Cint_U lvert nabla u rvert^2
$$
I was wondering if it was possible to find an example of a bounded domain on which the inequality does not hold. The proof for this inequality in Evan's PDE requires $U$ to be an extension domain. So I though that maybe the inequality would be false on a domain as simple as $U = {(x,y) : 0<x<1, 0<y<x^2}$.
I tried to construct a sequence of functions $u_kin H^1(U)$ satisfying $int_U u_k = 0$ and such that
$$
frac{int_U lvert u_krvert^2}{int_U lvert nabla u_k rvert^2} toinfty
$$
but unforunately I was unable to obtain such a sequence.
Is the inequality always true on bounded domains? Or is it indeed possible to find a counter-example
functional-analysis pde sobolev-spaces examples-counterexamples
asked Nov 27 at 0:41
Quoka
1,208212
Suppose that $U$ is a bounded domain in $mathbb{R}^n$. Poincaré's inequality states that (for $U$ sufficiently "nice") there exists a constant $C>0$ such that if $uin H^1(U)$ satisfies $int_U u = 0$ then
$$
int_U lvert urvert^2 leq Cint_U lvert nabla u rvert^2
$$
I was wondering if it was possible to find an example of a bounded domain on which the inequality does not hold. The proof for this inequality in Evan's PDE requires $U$ to be an extension domain. So I though that maybe the inequality would be false on a domain as simple as $U = {(x,y) : 0<x<1, 0<y<x^2}$.
I tried to construct a sequence of functions $u_kin H^1(U)$ satisfying $int_U u_k = 0$ and such that
$$
frac{int_U lvert u_krvert^2}{int_U lvert nabla u_k rvert^2} toinfty
$$
but unforunately I was unable to obtain such a sequence.
Is the inequality always true on bounded domains? Or is it indeed possible to find a counter-example
functional-analysis pde sobolev-spaces examples-counterexamples
functional-analysis pde sobolev-spaces examples-counterexamples
asked Nov 27 at 0:41
Quoka
1,208212
asked Nov 27 at 0:41
Quoka
1,208212
asked Nov 27 at 0:41
Quoka
1,208212
asked Nov 27 at 0:41
Quoka
1,208212
asked Nov 27 at 0:41
Quoka
1,208212
1,208212
What is your definition of "domain"?
– gerw
Nov 27 at 7:05
I think Evans takes domain to be “open and connected”. @gerw
– DaveNine
Nov 27 at 8:04
add a comment |
What is your definition of "domain"?
– gerw
Nov 27 at 7:05
I think Evans takes domain to be “open and connected”. @gerw
– DaveNine
Nov 27 at 8:04
What is your definition of "domain"?
– gerw
Nov 27 at 7:05
What is your definition of "domain"?
– gerw
Nov 27 at 7:05
I think Evans takes domain to be “open and connected”. @gerw
– DaveNine
Nov 27 at 8:04
I think Evans takes domain to be “open and connected”. @gerw
– DaveNine
Nov 27 at 8:04
add a comment |
1 Answer
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The classical counter-example is called "Rooms and passages" or "Rooms and corridors". You take a sequence of squares $R_n$ of side-length $1/n^p$ and in each of them, you assign $u$ to be a large constant, say $n^q$. Then you connect each square with a narrow corridor and in this corridor, you take $u$ to be affine. If the corridors are narrow enough the gradient will be in $L^2$ but the function will not be in $L^2$.
The details are here
Room and Passages
answered Dec 8 at 23:20
Gio67
12.4k1626
add a comment |
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1 Answer
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The classical counter-example is called "Rooms and passages" or "Rooms and corridors". You take a sequence of squares $R_n$ of side-length $1/n^p$ and in each of them, you assign $u$ to be a large constant, say $n^q$. Then you connect each square with a narrow corridor and in this corridor, you take $u$ to be affine. If the corridors are narrow enough the gradient will be in $L^2$ but the function will not be in $L^2$.
The details are here
Room and Passages
answered Dec 8 at 23:20
Gio67
12.4k1626
add a comment |
The classical counter-example is called "Rooms and passages" or "Rooms and corridors". You take a sequence of squares $R_n$ of side-length $1/n^p$ and in each of them, you assign $u$ to be a large constant, say $n^q$. Then you connect each square with a narrow corridor and in this corridor, you take $u$ to be affine. If the corridors are narrow enough the gradient will be in $L^2$ but the function will not be in $L^2$.
The details are here
Room and Passages
answered Dec 8 at 23:20
Gio67
12.4k1626
add a comment |
The classical counter-example is called "Rooms and passages" or "Rooms and corridors". You take a sequence of squares $R_n$ of side-length $1/n^p$ and in each of them, you assign $u$ to be a large constant, say $n^q$. Then you connect each square with a narrow corridor and in this corridor, you take $u$ to be affine. If the corridors are narrow enough the gradient will be in $L^2$ but the function will not be in $L^2$.
The details are here
Room and Passages
answered Dec 8 at 23:20
Gio67
12.4k1626
The classical counter-example is called "Rooms and passages" or "Rooms and corridors". You take a sequence of squares $R_n$ of side-length $1/n^p$ and in each of them, you assign $u$ to be a large constant, say $n^q$. Then you connect each square with a narrow corridor and in this corridor, you take $u$ to be affine. If the corridors are narrow enough the gradient will be in $L^2$ but the function will not be in $L^2$.
The details are here
Room and Passages
answered Dec 8 at 23:20
Gio67
12.4k1626
answered Dec 8 at 23:20
Gio67
12.4k1626
answered Dec 8 at 23:20
Gio67
12.4k1626
answered Dec 8 at 23:20
Gio67
12.4k1626
12.4k1626
add a comment |
add a comment |
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What is your definition of "domain"?
– gerw
Nov 27 at 7:05
I think Evans takes domain to be “open and connected”. @gerw
– DaveNine
Nov 27 at 8:04