Why of this bound $||u_{m0}|| leq ||u_0||$, where $u_{m0}$ is a projection of $u_0$?
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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
asked Nov 19 at 0:56
João Paulo Andrade
315
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
asked Nov 19 at 0:56
João Paulo Andrade
315
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
asked Nov 19 at 0:56
João Paulo Andrade
315
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
asked Nov 19 at 0:56
João Paulo Andrade
315
asked Nov 19 at 0:56
João Paulo Andrade
315
edited Nov 19 at 18:48
asked Nov 19 at 0:56
João Paulo Andrade
315
asked Nov 19 at 0:56
João Paulo Andrade
315
asked Nov 19 at 0:56
João Paulo Andrade
315
315
add a comment |
add a comment |
1 Answer
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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
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
answered Nov 21 at 3:19
Pedro
10.2k23066
add a comment |
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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
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
answered Nov 21 at 3:19
Pedro
10.2k23066
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
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
answered Nov 21 at 3:19
Pedro
10.2k23066
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
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
answered Nov 21 at 3:19
Pedro
10.2k23066
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
$$begin{aligned}
|u_{m0}|^2&=(u_{m0},u_{m0})\
&=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
&=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
&=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
&=sum_{i=1}^m|(u_0,w_i)|^2\
&leq sum_{i=1}^infty|(u_0,w_i)|^2\
&=|u_0|^2
end{aligned}$$
answered Nov 21 at 3:19
Pedro
10.2k23066
edited Nov 21 at 3:42
answered Nov 21 at 3:19
Pedro
10.2k23066
answered Nov 21 at 3:19
Pedro
10.2k23066
answered Nov 21 at 3:19
Pedro
10.2k23066
10.2k23066
add a comment |
add a comment |
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