What is the orthogonal complement of $H^1_0$ in $H^1$?
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Let $Omega$ be a closed domain with smooth boundary in $mathbb{R}^n$. Let $H^1_0(Omega)$ be the closure of compactly supported smooth functions under the norm $|u|_1 = int_Omega u^2 + |nabla u|^2 dx$ and let $H^1(Omega)$ be the closure of smooth, continuous functions under the same norm.
Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, |cdot|_1)$, hence has an orthogonal complement.
What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?
Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.
Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.
functional-analysis sobolev-spaces trace
asked Nov 20 at 1:34
Neal
23.3k23582
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Let $Omega$ be a closed domain with smooth boundary in $mathbb{R}^n$. Let $H^1_0(Omega)$ be the closure of compactly supported smooth functions under the norm $|u|_1 = int_Omega u^2 + |nabla u|^2 dx$ and let $H^1(Omega)$ be the closure of smooth, continuous functions under the same norm.
Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, |cdot|_1)$, hence has an orthogonal complement.
What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?
Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.
Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.
functional-analysis sobolev-spaces trace
asked Nov 20 at 1:34
Neal
23.3k23582
For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
– reuns
Nov 20 at 1:54
@reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
– Neal
Nov 20 at 2:04
That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
– reuns
Nov 20 at 2:13
@reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
– Neal
2 days ago
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $Omega$ be a closed domain with smooth boundary in $mathbb{R}^n$. Let $H^1_0(Omega)$ be the closure of compactly supported smooth functions under the norm $|u|_1 = int_Omega u^2 + |nabla u|^2 dx$ and let $H^1(Omega)$ be the closure of smooth, continuous functions under the same norm.
Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, |cdot|_1)$, hence has an orthogonal complement.
What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?
Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.
Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.
functional-analysis sobolev-spaces trace
asked Nov 20 at 1:34
Neal
23.3k23582
Let $Omega$ be a closed domain with smooth boundary in $mathbb{R}^n$. Let $H^1_0(Omega)$ be the closure of compactly supported smooth functions under the norm $|u|_1 = int_Omega u^2 + |nabla u|^2 dx$ and let $H^1(Omega)$ be the closure of smooth, continuous functions under the same norm.
Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, |cdot|_1)$, hence has an orthogonal complement.
What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?
Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.
Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.
functional-analysis sobolev-spaces trace
functional-analysis sobolev-spaces trace
asked Nov 20 at 1:34
Neal
23.3k23582
asked Nov 20 at 1:34
Neal
23.3k23582
asked Nov 20 at 1:34
Neal
23.3k23582
asked Nov 20 at 1:34
Neal
23.3k23582
asked Nov 20 at 1:34
Neal
23.3k23582
23.3k23582
For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
– reuns
Nov 20 at 1:54
@reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
– Neal
Nov 20 at 2:04
That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
– reuns
Nov 20 at 2:13
@reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
– Neal
2 days ago
add a comment |
For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
– reuns
Nov 20 at 1:54
@reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
– Neal
Nov 20 at 2:04
That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
– reuns
Nov 20 at 2:13
@reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
– Neal
2 days ago
For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
– reuns
Nov 20 at 1:54
For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
– reuns
Nov 20 at 1:54
@reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
– Neal
Nov 20 at 2:04
@reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
– Neal
Nov 20 at 2:04
That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
– reuns
Nov 20 at 2:13
That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
– reuns
Nov 20 at 2:13
@reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
– Neal
2 days ago
@reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
– Neal
2 days ago
add a comment |
1 Answer
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3
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As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$
In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$
For different inner products, you get different PDEs.
answered 2 days ago
gerw
18.7k11133
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$
In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$
For different inner products, you get different PDEs.
answered 2 days ago
gerw
18.7k11133
add a comment |
up vote
3
down vote
As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$
In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$
For different inner products, you get different PDEs.
answered 2 days ago
gerw
18.7k11133
add a comment |
up vote
3
down vote
up vote
3
down vote
As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$
In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$
For different inner products, you get different PDEs.
answered 2 days ago
gerw
18.7k11133
As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$
In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$
For different inner products, you get different PDEs.
answered 2 days ago
gerw
18.7k11133
answered 2 days ago
gerw
18.7k11133
answered 2 days ago
gerw
18.7k11133
answered 2 days ago
gerw
18.7k11133
18.7k11133
add a comment |
add a comment |
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For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
– reuns
Nov 20 at 1:54
@reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
– Neal
Nov 20 at 2:04
That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
– reuns
Nov 20 at 2:13
@reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
– Neal
2 days ago