Mixing
Prove that this $ langle u,vrangle = int_Omega [ bigtriangledown u cdotbigtriangledown v +uv ] $ is an inner...
$begingroup$
Information needed
$$I = (a,b)$$
$$ H_0^1 (I) = { v in H^1(I) : v(a)= v(b) = 0 } $$
$$ H^1 (I) = { v: v , v' in mathbb{L}_2(I) } $$
$$ mathbb{L}_2(I) = { v:v text{ it's defined at } I text{ and } int v^2,dx < infty } $$
Prove that this is a scalar product in $H^1_0 $
Using the notation $$ langle u,vrangle = int_Omega [ bigtriangledown u cdotbigtriangledown v +uv ] $$
where $Omega$ is a bounded domain.
In the book numerical solutions of PDE by the finite element method by Claes there is this passage and I could not see clearly this statement.
Thanks to any help !
functional-analysis metric-spaces hilbert-spaces
edited Jan 30 at 14:32
Adrian Keister
5,26971933
asked Jan 30 at 12:07
SaitenSaiten
346
$endgroup$
add a comment |
$begingroup$
Information needed
$$I = (a,b)$$
$$ H_0^1 (I) = { v in H^1(I) : v(a)= v(b) = 0 } $$
$$ H^1 (I) = { v: v , v' in mathbb{L}_2(I) } $$
$$ mathbb{L}_2(I) = { v:v text{ it's defined at } I text{ and } int v^2,dx < infty } $$
Prove that this is a scalar product in $H^1_0 $
Using the notation $$ langle u,vrangle = int_Omega [ bigtriangledown u cdotbigtriangledown v +uv ] $$
where $Omega$ is a bounded domain.
In the book numerical solutions of PDE by the finite element method by Claes there is this passage and I could not see clearly this statement.
Thanks to any help !
functional-analysis metric-spaces hilbert-spaces
edited Jan 30 at 14:32
Adrian Keister
5,26971933
asked Jan 30 at 12:07
SaitenSaiten
346
$endgroup$
3
$begingroup$
Have you tried proving it yourself? An inner product has three defining properties, each of which should not be too hard to verify.
$endgroup$
– MaoWao
Jan 30 at 12:15
$begingroup$
you mean, symmetry; linearity and podsitive definiteness ?
$endgroup$
– Saiten
Jan 30 at 12:19
$begingroup$
It's bilinearity (linearity in each argument separately), not linearity. Otherwise yes.
$endgroup$
– MaoWao
Jan 30 at 12:23
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Jan 30 at 12:39
$begingroup$
Thankss for the help!
$endgroup$
– Saiten
Jan 30 at 13:24
add a comment |
$begingroup$
Information needed
$$I = (a,b)$$
$$ H_0^1 (I) = { v in H^1(I) : v(a)= v(b) = 0 } $$
$$ H^1 (I) = { v: v , v' in mathbb{L}_2(I) } $$
$$ mathbb{L}_2(I) = { v:v text{ it's defined at } I text{ and } int v^2,dx < infty } $$
Prove that this is a scalar product in $H^1_0 $
Using the notation $$ langle u,vrangle = int_Omega [ bigtriangledown u cdotbigtriangledown v +uv ] $$
where $Omega$ is a bounded domain.
In the book numerical solutions of PDE by the finite element method by Claes there is this passage and I could not see clearly this statement.
Thanks to any help !
functional-analysis metric-spaces hilbert-spaces
edited Jan 30 at 14:32
Adrian Keister
5,26971933
asked Jan 30 at 12:07
SaitenSaiten
346
$endgroup$
Information needed
$$I = (a,b)$$
$$ H_0^1 (I) = { v in H^1(I) : v(a)= v(b) = 0 } $$
$$ H^1 (I) = { v: v , v' in mathbb{L}_2(I) } $$
$$ mathbb{L}_2(I) = { v:v text{ it's defined at } I text{ and } int v^2,dx < infty } $$
Prove that this is a scalar product in $H^1_0 $
Using the notation $$ langle u,vrangle = int_Omega [ bigtriangledown u cdotbigtriangledown v +uv ] $$
where $Omega$ is a bounded domain.
In the book numerical solutions of PDE by the finite element method by Claes there is this passage and I could not see clearly this statement.
Thanks to any help !
functional-analysis metric-spaces hilbert-spaces
functional-analysis metric-spaces hilbert-spaces
edited Jan 30 at 14:32
Adrian Keister
5,26971933
asked Jan 30 at 12:07
SaitenSaiten
346
edited Jan 30 at 14:32
Adrian Keister
5,26971933
asked Jan 30 at 12:07
SaitenSaiten
346
edited Jan 30 at 14:32
Adrian Keister
5,26971933
edited Jan 30 at 14:32
Adrian Keister
5,26971933
edited Jan 30 at 14:32
Adrian Keister
5,26971933
5,26971933
asked Jan 30 at 12:07
SaitenSaiten
346
asked Jan 30 at 12:07
SaitenSaiten
346
asked Jan 30 at 12:07
SaitenSaiten
346
346
3
$begingroup$
Have you tried proving it yourself? An inner product has three defining properties, each of which should not be too hard to verify.
$endgroup$
– MaoWao
Jan 30 at 12:15
$begingroup$
you mean, symmetry; linearity and podsitive definiteness ?
$endgroup$
– Saiten
Jan 30 at 12:19
$begingroup$
It's bilinearity (linearity in each argument separately), not linearity. Otherwise yes.
$endgroup$
– MaoWao
Jan 30 at 12:23
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Jan 30 at 12:39
$begingroup$
Thankss for the help!
$endgroup$
– Saiten
Jan 30 at 13:24
add a comment |
3
$begingroup$
Have you tried proving it yourself? An inner product has three defining properties, each of which should not be too hard to verify.
$endgroup$
– MaoWao
Jan 30 at 12:15
$begingroup$
you mean, symmetry; linearity and podsitive definiteness ?
$endgroup$
– Saiten
Jan 30 at 12:19
$begingroup$
It's bilinearity (linearity in each argument separately), not linearity. Otherwise yes.
$endgroup$
– MaoWao
Jan 30 at 12:23
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Jan 30 at 12:39
$begingroup$
Thankss for the help!
$endgroup$
– Saiten
Jan 30 at 13:24
3
3
$begingroup$
Have you tried proving it yourself? An inner product has three defining properties, each of which should not be too hard to verify.
$endgroup$
– MaoWao
Jan 30 at 12:15
$begingroup$
Have you tried proving it yourself? An inner product has three defining properties, each of which should not be too hard to verify.
$endgroup$
– MaoWao
Jan 30 at 12:15
$begingroup$
you mean, symmetry; linearity and podsitive definiteness ?
$endgroup$
– Saiten
Jan 30 at 12:19
$begingroup$
you mean, symmetry; linearity and podsitive definiteness ?
$endgroup$
– Saiten
Jan 30 at 12:19
$begingroup$
It's bilinearity (linearity in each argument separately), not linearity. Otherwise yes.
$endgroup$
– MaoWao
Jan 30 at 12:23
$begingroup$
It's bilinearity (linearity in each argument separately), not linearity. Otherwise yes.
$endgroup$
– MaoWao
Jan 30 at 12:23
$begingroup$
Use
$langle Xrangle$ for $langle Xrangle$.$endgroup$
– Shaun
Jan 30 at 12:39
$begingroup$
Use
$langle Xrangle$ for $langle Xrangle$.$endgroup$
– Shaun
Jan 30 at 12:39
$begingroup$
Thankss for the help!
$endgroup$
– Saiten
Jan 30 at 13:24
$begingroup$
Thankss for the help!
$endgroup$
– Saiten
Jan 30 at 13:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Once you establish that the expression is well defined, you just need to verify the conditions for an inner product.
$langle u,v rangle = langle v,u rangle$.
$langle a u, vrangle = a langle u,brangle$ and $langle u+v,wrangle = langle u,wrangle+langle v,wrangle$.
$langle u,uranglege 0$ and $langle u,urangle=0 Leftrightarrow u = 0$
The third property is the trickiest. You need to show that if $u in H^1_0$ and
$$
int_I (|nabla u|^2 + u^2) = 0
$$
then $u=0$ almost everywhere in $I$.
answered Jan 30 at 12:25
PierreCarrePierreCarre
1,734212
$endgroup$
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Jan 30 at 13:17
add a comment |
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1 Answer
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$begingroup$
Once you establish that the expression is well defined, you just need to verify the conditions for an inner product.
$langle u,v rangle = langle v,u rangle$.
$langle a u, vrangle = a langle u,brangle$ and $langle u+v,wrangle = langle u,wrangle+langle v,wrangle$.
$langle u,uranglege 0$ and $langle u,urangle=0 Leftrightarrow u = 0$
The third property is the trickiest. You need to show that if $u in H^1_0$ and
$$
int_I (|nabla u|^2 + u^2) = 0
$$
then $u=0$ almost everywhere in $I$.
answered Jan 30 at 12:25
PierreCarrePierreCarre
1,734212
$endgroup$
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Jan 30 at 13:17
add a comment |
$begingroup$
Once you establish that the expression is well defined, you just need to verify the conditions for an inner product.
$langle u,v rangle = langle v,u rangle$.
$langle a u, vrangle = a langle u,brangle$ and $langle u+v,wrangle = langle u,wrangle+langle v,wrangle$.
$langle u,uranglege 0$ and $langle u,urangle=0 Leftrightarrow u = 0$
The third property is the trickiest. You need to show that if $u in H^1_0$ and
$$
int_I (|nabla u|^2 + u^2) = 0
$$
then $u=0$ almost everywhere in $I$.
answered Jan 30 at 12:25
PierreCarrePierreCarre
1,734212
$endgroup$
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Jan 30 at 13:17
add a comment |
$begingroup$
Once you establish that the expression is well defined, you just need to verify the conditions for an inner product.
$langle u,v rangle = langle v,u rangle$.
$langle a u, vrangle = a langle u,brangle$ and $langle u+v,wrangle = langle u,wrangle+langle v,wrangle$.
$langle u,uranglege 0$ and $langle u,urangle=0 Leftrightarrow u = 0$
The third property is the trickiest. You need to show that if $u in H^1_0$ and
$$
int_I (|nabla u|^2 + u^2) = 0
$$
then $u=0$ almost everywhere in $I$.
answered Jan 30 at 12:25
PierreCarrePierreCarre
1,734212
$endgroup$
Once you establish that the expression is well defined, you just need to verify the conditions for an inner product.
$langle u,v rangle = langle v,u rangle$.
$langle a u, vrangle = a langle u,brangle$ and $langle u+v,wrangle = langle u,wrangle+langle v,wrangle$.
$langle u,uranglege 0$ and $langle u,urangle=0 Leftrightarrow u = 0$
The third property is the trickiest. You need to show that if $u in H^1_0$ and
$$
int_I (|nabla u|^2 + u^2) = 0
$$
then $u=0$ almost everywhere in $I$.
answered Jan 30 at 12:25
PierreCarrePierreCarre
1,734212
edited Jan 30 at 13:41
answered Jan 30 at 12:25
PierreCarrePierreCarre
1,734212
answered Jan 30 at 12:25
PierreCarrePierreCarre
1,734212
answered Jan 30 at 12:25
PierreCarrePierreCarre
1,734212
1,734212
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Jan 30 at 13:17
add a comment |
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Jan 30 at 13:17
$begingroup$
Use
$langle Xrangle$ for $langle Xrangle$.$endgroup$
– Shaun
Jan 30 at 13:17
$begingroup$
Use
$langle Xrangle$ for $langle Xrangle$.$endgroup$
– Shaun
Jan 30 at 13:17
add a comment |
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3
$begingroup$
Have you tried proving it yourself? An inner product has three defining properties, each of which should not be too hard to verify.
$endgroup$
– MaoWao
Jan 30 at 12:15
$begingroup$
you mean, symmetry; linearity and podsitive definiteness ?
$endgroup$
– Saiten
Jan 30 at 12:19
$begingroup$
It's bilinearity (linearity in each argument separately), not linearity. Otherwise yes.
$endgroup$
– MaoWao
Jan 30 at 12:23
$begingroup$
Use
$langle Xrangle$for $langle Xrangle$.$endgroup$
– Shaun
Jan 30 at 12:39
$begingroup$
Thankss for the help!
$endgroup$
– Saiten
Jan 30 at 13:24