Mixing
Product from function in $W^{1,2}_0(Omega)$ and function in $W^{1,2}(Omega)$
$begingroup$
I was wandering what i say about product
$$
ueta
$$
where $uin W^{1,2}(Omega)$ and $etain W^{1,2}_0(Omega)$.
In particular, when i can say that
$$
uetain W^{1,2}_0(Omega).
$$
Is it necessary to make more assumptions about u?
sobolev-spaces
asked Feb 2 at 12:28
RevzoraRevzora
1135
$endgroup$
add a comment |
$begingroup$
I was wandering what i say about product
$$
ueta
$$
where $uin W^{1,2}(Omega)$ and $etain W^{1,2}_0(Omega)$.
In particular, when i can say that
$$
uetain W^{1,2}_0(Omega).
$$
Is it necessary to make more assumptions about u?
sobolev-spaces
asked Feb 2 at 12:28
RevzoraRevzora
1135
$endgroup$
$begingroup$
Look out for Hölder's inequality. As it is you will probably only get $u eta in W_0^{1,1}$. For the result you want, you'd need $u in W^{1,infty}$.
$endgroup$
– Fritz
Feb 3 at 13:20
$begingroup$
@Marvin: If you first use Sobolev's embedding theorem, one can show a little bit more than $W^{1,1}_0(Omega)$. And you do not need $u in W^{1,infty}$ since one could also assume more regularity of $eta$.
$endgroup$
– gerw
Feb 4 at 7:36
$begingroup$
@gerw True, you are correct that one can show more. And yes, I thought it is only allowed to change the regularity of $u$ in the way the question was proposed. Good answer, +1.
$endgroup$
– Fritz
Feb 4 at 10:19
add a comment |
$begingroup$
I was wandering what i say about product
$$
ueta
$$
where $uin W^{1,2}(Omega)$ and $etain W^{1,2}_0(Omega)$.
In particular, when i can say that
$$
uetain W^{1,2}_0(Omega).
$$
Is it necessary to make more assumptions about u?
sobolev-spaces
asked Feb 2 at 12:28
RevzoraRevzora
1135
$endgroup$
I was wandering what i say about product
$$
ueta
$$
where $uin W^{1,2}(Omega)$ and $etain W^{1,2}_0(Omega)$.
In particular, when i can say that
$$
uetain W^{1,2}_0(Omega).
$$
Is it necessary to make more assumptions about u?
sobolev-spaces
sobolev-spaces
asked Feb 2 at 12:28
RevzoraRevzora
1135
asked Feb 2 at 12:28
RevzoraRevzora
1135
asked Feb 2 at 12:28
RevzoraRevzora
1135
asked Feb 2 at 12:28
RevzoraRevzora
1135
asked Feb 2 at 12:28
RevzoraRevzora
1135
1135
$begingroup$
Look out for Hölder's inequality. As it is you will probably only get $u eta in W_0^{1,1}$. For the result you want, you'd need $u in W^{1,infty}$.
$endgroup$
– Fritz
Feb 3 at 13:20
$begingroup$
@Marvin: If you first use Sobolev's embedding theorem, one can show a little bit more than $W^{1,1}_0(Omega)$. And you do not need $u in W^{1,infty}$ since one could also assume more regularity of $eta$.
$endgroup$
– gerw
Feb 4 at 7:36
$begingroup$
@gerw True, you are correct that one can show more. And yes, I thought it is only allowed to change the regularity of $u$ in the way the question was proposed. Good answer, +1.
$endgroup$
– Fritz
Feb 4 at 10:19
add a comment |
$begingroup$
Look out for Hölder's inequality. As it is you will probably only get $u eta in W_0^{1,1}$. For the result you want, you'd need $u in W^{1,infty}$.
$endgroup$
– Fritz
Feb 3 at 13:20
$begingroup$
@Marvin: If you first use Sobolev's embedding theorem, one can show a little bit more than $W^{1,1}_0(Omega)$. And you do not need $u in W^{1,infty}$ since one could also assume more regularity of $eta$.
$endgroup$
– gerw
Feb 4 at 7:36
$begingroup$
@gerw True, you are correct that one can show more. And yes, I thought it is only allowed to change the regularity of $u$ in the way the question was proposed. Good answer, +1.
$endgroup$
– Fritz
Feb 4 at 10:19
$begingroup$
Look out for Hölder's inequality. As it is you will probably only get $u eta in W_0^{1,1}$. For the result you want, you'd need $u in W^{1,infty}$.
$endgroup$
– Fritz
Feb 3 at 13:20
$begingroup$
Look out for Hölder's inequality. As it is you will probably only get $u eta in W_0^{1,1}$. For the result you want, you'd need $u in W^{1,infty}$.
$endgroup$
– Fritz
Feb 3 at 13:20
$begingroup$
@Marvin: If you first use Sobolev's embedding theorem, one can show a little bit more than $W^{1,1}_0(Omega)$. And you do not need $u in W^{1,infty}$ since one could also assume more regularity of $eta$.
$endgroup$
– gerw
Feb 4 at 7:36
$begingroup$
@Marvin: If you first use Sobolev's embedding theorem, one can show a little bit more than $W^{1,1}_0(Omega)$. And you do not need $u in W^{1,infty}$ since one could also assume more regularity of $eta$.
$endgroup$
– gerw
Feb 4 at 7:36
$begingroup$
@gerw True, you are correct that one can show more. And yes, I thought it is only allowed to change the regularity of $u$ in the way the question was proposed. Good answer, +1.
$endgroup$
– Fritz
Feb 4 at 10:19
$begingroup$
@gerw True, you are correct that one can show more. And yes, I thought it is only allowed to change the regularity of $u$ in the way the question was proposed. Good answer, +1.
$endgroup$
– Fritz
Feb 4 at 10:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $u , eta in W_0^{1,2}(Omega)$ you have to prove three things:
$u , eta in L^2(Omega)$,
$nabla( u , eta ) in L^2(Omega)$, and- the trace of $u , eta$ is zero.
The first is easy to get and the last one follows basically from the fact that the trace of $eta$ is zero. The second one is hardest and needs additional assumptions. From the product rule, it is sufficient to check that $u , nabla eta$ and $eta , nabla u$ are in $L^2$. These can be achieved by combining Sobolev's embedding theorem with Hölder's inequality. For example, you it is sufficient to assume one of the following (this list is not exhaustive):
$u in W^{1,infty}(Omega)$,
$eta in W^{1,infty}(Omega)$,
$u, eta in L^infty(Omega)$,
$u, eta in W^{1,p}(Omega)$ with $p$ large enough (depending on the dimension).
answered Feb 4 at 7:35
gerwgerw
20.1k11334
$endgroup$
$begingroup$
Thank you for your answer. I know to prove first and second point (with assumptions) but now it is not clear to me how to prove the last point (that is the trace is zero).
$endgroup$
– Revzora
Feb 4 at 14:36
1
$begingroup$
You can approximate $eta$ by compactly supported functions. The product of these functions with $u$ will have zero trace.
$endgroup$
– gerw
Feb 4 at 18:58
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
For $u , eta in W_0^{1,2}(Omega)$ you have to prove three things:
$u , eta in L^2(Omega)$,
$nabla( u , eta ) in L^2(Omega)$, and- the trace of $u , eta$ is zero.
The first is easy to get and the last one follows basically from the fact that the trace of $eta$ is zero. The second one is hardest and needs additional assumptions. From the product rule, it is sufficient to check that $u , nabla eta$ and $eta , nabla u$ are in $L^2$. These can be achieved by combining Sobolev's embedding theorem with Hölder's inequality. For example, you it is sufficient to assume one of the following (this list is not exhaustive):
$u in W^{1,infty}(Omega)$,
$eta in W^{1,infty}(Omega)$,
$u, eta in L^infty(Omega)$,
$u, eta in W^{1,p}(Omega)$ with $p$ large enough (depending on the dimension).
answered Feb 4 at 7:35
gerwgerw
20.1k11334
$endgroup$
$begingroup$
Thank you for your answer. I know to prove first and second point (with assumptions) but now it is not clear to me how to prove the last point (that is the trace is zero).
$endgroup$
– Revzora
Feb 4 at 14:36
1
$begingroup$
You can approximate $eta$ by compactly supported functions. The product of these functions with $u$ will have zero trace.
$endgroup$
– gerw
Feb 4 at 18:58
add a comment |
$begingroup$
For $u , eta in W_0^{1,2}(Omega)$ you have to prove three things:
$u , eta in L^2(Omega)$,
$nabla( u , eta ) in L^2(Omega)$, and- the trace of $u , eta$ is zero.
The first is easy to get and the last one follows basically from the fact that the trace of $eta$ is zero. The second one is hardest and needs additional assumptions. From the product rule, it is sufficient to check that $u , nabla eta$ and $eta , nabla u$ are in $L^2$. These can be achieved by combining Sobolev's embedding theorem with Hölder's inequality. For example, you it is sufficient to assume one of the following (this list is not exhaustive):
$u in W^{1,infty}(Omega)$,
$eta in W^{1,infty}(Omega)$,
$u, eta in L^infty(Omega)$,
$u, eta in W^{1,p}(Omega)$ with $p$ large enough (depending on the dimension).
answered Feb 4 at 7:35
gerwgerw
20.1k11334
$endgroup$
$begingroup$
Thank you for your answer. I know to prove first and second point (with assumptions) but now it is not clear to me how to prove the last point (that is the trace is zero).
$endgroup$
– Revzora
Feb 4 at 14:36
1
$begingroup$
You can approximate $eta$ by compactly supported functions. The product of these functions with $u$ will have zero trace.
$endgroup$
– gerw
Feb 4 at 18:58
add a comment |
$begingroup$
For $u , eta in W_0^{1,2}(Omega)$ you have to prove three things:
$u , eta in L^2(Omega)$,
$nabla( u , eta ) in L^2(Omega)$, and- the trace of $u , eta$ is zero.
The first is easy to get and the last one follows basically from the fact that the trace of $eta$ is zero. The second one is hardest and needs additional assumptions. From the product rule, it is sufficient to check that $u , nabla eta$ and $eta , nabla u$ are in $L^2$. These can be achieved by combining Sobolev's embedding theorem with Hölder's inequality. For example, you it is sufficient to assume one of the following (this list is not exhaustive):
$u in W^{1,infty}(Omega)$,
$eta in W^{1,infty}(Omega)$,
$u, eta in L^infty(Omega)$,
$u, eta in W^{1,p}(Omega)$ with $p$ large enough (depending on the dimension).
answered Feb 4 at 7:35
gerwgerw
20.1k11334
$endgroup$
For $u , eta in W_0^{1,2}(Omega)$ you have to prove three things:
$u , eta in L^2(Omega)$,
$nabla( u , eta ) in L^2(Omega)$, and- the trace of $u , eta$ is zero.
The first is easy to get and the last one follows basically from the fact that the trace of $eta$ is zero. The second one is hardest and needs additional assumptions. From the product rule, it is sufficient to check that $u , nabla eta$ and $eta , nabla u$ are in $L^2$. These can be achieved by combining Sobolev's embedding theorem with Hölder's inequality. For example, you it is sufficient to assume one of the following (this list is not exhaustive):
$u in W^{1,infty}(Omega)$,
$eta in W^{1,infty}(Omega)$,
$u, eta in L^infty(Omega)$,
$u, eta in W^{1,p}(Omega)$ with $p$ large enough (depending on the dimension).
answered Feb 4 at 7:35
gerwgerw
20.1k11334
answered Feb 4 at 7:35
gerwgerw
20.1k11334
answered Feb 4 at 7:35
gerwgerw
20.1k11334
answered Feb 4 at 7:35
gerwgerw
20.1k11334
20.1k11334
$begingroup$
Thank you for your answer. I know to prove first and second point (with assumptions) but now it is not clear to me how to prove the last point (that is the trace is zero).
$endgroup$
– Revzora
Feb 4 at 14:36
1
$begingroup$
You can approximate $eta$ by compactly supported functions. The product of these functions with $u$ will have zero trace.
$endgroup$
– gerw
Feb 4 at 18:58
add a comment |
$begingroup$
Thank you for your answer. I know to prove first and second point (with assumptions) but now it is not clear to me how to prove the last point (that is the trace is zero).
$endgroup$
– Revzora
Feb 4 at 14:36
1
$begingroup$
You can approximate $eta$ by compactly supported functions. The product of these functions with $u$ will have zero trace.
$endgroup$
– gerw
Feb 4 at 18:58
$begingroup$
Thank you for your answer. I know to prove first and second point (with assumptions) but now it is not clear to me how to prove the last point (that is the trace is zero).
$endgroup$
– Revzora
Feb 4 at 14:36
$begingroup$
Thank you for your answer. I know to prove first and second point (with assumptions) but now it is not clear to me how to prove the last point (that is the trace is zero).
$endgroup$
– Revzora
Feb 4 at 14:36
1
1
$begingroup$
You can approximate $eta$ by compactly supported functions. The product of these functions with $u$ will have zero trace.
$endgroup$
– gerw
Feb 4 at 18:58
$begingroup$
You can approximate $eta$ by compactly supported functions. The product of these functions with $u$ will have zero trace.
$endgroup$
– gerw
Feb 4 at 18:58
add a comment |
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$begingroup$
Look out for Hölder's inequality. As it is you will probably only get $u eta in W_0^{1,1}$. For the result you want, you'd need $u in W^{1,infty}$.
$endgroup$
– Fritz
Feb 3 at 13:20
$begingroup$
@Marvin: If you first use Sobolev's embedding theorem, one can show a little bit more than $W^{1,1}_0(Omega)$. And you do not need $u in W^{1,infty}$ since one could also assume more regularity of $eta$.
$endgroup$
– gerw
Feb 4 at 7:36
$begingroup$
@gerw True, you are correct that one can show more. And yes, I thought it is only allowed to change the regularity of $u$ in the way the question was proposed. Good answer, +1.
$endgroup$
– Fritz
Feb 4 at 10:19