Why of this bound $||u_{m0}|| leq ||u_0||$, where $u_{m0}$ is a projection of $u_0$?
up vote
1
down vote
favorite
I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.
Consider $u_m=sum_{i=1}^m g_{im}(t)w_i$ an approximation of $u$ via galerkin method, where $w_i$ are the basis and $g_{im}(t)$ the coeficients, $u_m(0)=u_{m0}$ and $u(0)=u_0$.
The space $V$ is basecally the closure, in $H^1_0(Omega)$, of the set of all function $C^infty$ with support on a open set $Omega subset mathbb{R}^2$, with it's divergent being null.
Let ${w_j}$ be a base of $V$ which we can take $u_{m0}$ as a projection in $V$ of $u_0$ spanned by $w_1,w_2, cdots,w_m $.
So, why this bound is valid,
$||u_{m0}|| leq ||u_0||$?
$||cdot|| $ is the norm in $H^1_0$.
functional-analysis analysis pde galerkin-methods
add a comment |
up vote
1
down vote
favorite
I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.
Consider $u_m=sum_{i=1}^m g_{im}(t)w_i$ an approximation of $u$ via galerkin method, where $w_i$ are the basis and $g_{im}(t)$ the coeficients, $u_m(0)=u_{m0}$ and $u(0)=u_0$.
The space $V$ is basecally the closure, in $H^1_0(Omega)$, of the set of all function $C^infty$ with support on a open set $Omega subset mathbb{R}^2$, with it's divergent being null.
Let ${w_j}$ be a base of $V$ which we can take $u_{m0}$ as a projection in $V$ of $u_0$ spanned by $w_1,w_2, cdots,w_m $.
So, why this bound is valid,
$||u_{m0}|| leq ||u_0||$?
$||cdot|| $ is the norm in $H^1_0$.
functional-analysis analysis pde galerkin-methods
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.
Consider $u_m=sum_{i=1}^m g_{im}(t)w_i$ an approximation of $u$ via galerkin method, where $w_i$ are the basis and $g_{im}(t)$ the coeficients, $u_m(0)=u_{m0}$ and $u(0)=u_0$.
The space $V$ is basecally the closure, in $H^1_0(Omega)$, of the set of all function $C^infty$ with support on a open set $Omega subset mathbb{R}^2$, with it's divergent being null.
Let ${w_j}$ be a base of $V$ which we can take $u_{m0}$ as a projection in $V$ of $u_0$ spanned by $w_1,w_2, cdots,w_m $.
So, why this bound is valid,
$||u_{m0}|| leq ||u_0||$?
$||cdot|| $ is the norm in $H^1_0$.
functional-analysis analysis pde galerkin-methods
I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.
Consider $u_m=sum_{i=1}^m g_{im}(t)w_i$ an approximation of $u$ via galerkin method, where $w_i$ are the basis and $g_{im}(t)$ the coeficients, $u_m(0)=u_{m0}$ and $u(0)=u_0$.
The space $V$ is basecally the closure, in $H^1_0(Omega)$, of the set of all function $C^infty$ with support on a open set $Omega subset mathbb{R}^2$, with it's divergent being null.
Let ${w_j}$ be a base of $V$ which we can take $u_{m0}$ as a projection in $V$ of $u_0$ spanned by $w_1,w_2, cdots,w_m $.
So, why this bound is valid,
$||u_{m0}|| leq ||u_0||$?
$||cdot|| $ is the norm in $H^1_0$.
functional-analysis analysis pde galerkin-methods
functional-analysis analysis pde galerkin-methods
edited Nov 19 at 18:48
asked Nov 19 at 0:56
João Paulo Andrade
315
315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:
Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:
${e_j}_{jin J}$ is maximal.
$ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.
$ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.
$overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.
Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.
From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".
Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:
Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:
${e_j}_{jin J}$ is maximal.
$ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.
$ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.
$overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.
Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.
From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".
Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
add a comment |
up vote
1
down vote
accepted
Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:
Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:
${e_j}_{jin J}$ is maximal.
$ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.
$ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.
$overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.
Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.
From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".
Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:
Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:
${e_j}_{jin J}$ is maximal.
$ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.
$ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.
$overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.
Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.
From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".
Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:
Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:
${e_j}_{jin J}$ is maximal.
$ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.
$ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.
$overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.
Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.
From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".
Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
edited Nov 21 at 3:42
answered Nov 21 at 3:19
Pedro
10.2k23066
10.2k23066
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004367%2fwhy-of-this-bound-u-m0-leq-u-0-where-u-m0-is-a-projection-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown