Mixing
Unique Harmonic Extension in $H^1(Omega)$
Clash Royale CLAN TAG#URR8PPP
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In lecture we had the following theorem:
Let $Omega subseteq subseteq mathbbR^n$. Then $$H^1(Omega) =
H^1_0(Omega) oplus u in H^1 : Delta u = 0$$ where $Delta u$
is understood in the distributional sense. Moreover, if $Omega$ is of class $C^1$, $uvert_partial Omega$ does admit a unique harmonic extension.
Now the proof of the above is straight forward, however, the moreover part is where I stumble: In the notes it is phrased that this is a direct consequence of the decomposition, but I do not see this. I mean, we do have $$uvert_partial Omega = u_0vert_partialOmega + u_1 vert_partial Omega = u_1vert_partial Omega$$ when $u = u_0 + u_1$ and hence $u_1$ is a harmonic extension of $uvert_partial Omega$, but why is this extension unique? The reason why I am asking this is because I need to proof that $$H^1_0(Omega) = u in H^1(Omega) : uvert_partial Omega = 0.$$
sobolev-spaces
asked 6 hours ago
TheGeekGreek
5,0093933
add a comment |Â
up vote
0
down vote
favorite
In lecture we had the following theorem:
Let $Omega subseteq subseteq mathbbR^n$. Then $$H^1(Omega) =
H^1_0(Omega) oplus u in H^1 : Delta u = 0$$ where $Delta u$
is understood in the distributional sense. Moreover, if $Omega$ is of class $C^1$, $uvert_partial Omega$ does admit a unique harmonic extension.
Now the proof of the above is straight forward, however, the moreover part is where I stumble: In the notes it is phrased that this is a direct consequence of the decomposition, but I do not see this. I mean, we do have $$uvert_partial Omega = u_0vert_partialOmega + u_1 vert_partial Omega = u_1vert_partial Omega$$ when $u = u_0 + u_1$ and hence $u_1$ is a harmonic extension of $uvert_partial Omega$, but why is this extension unique? The reason why I am asking this is because I need to proof that $$H^1_0(Omega) = u in H^1(Omega) : uvert_partial Omega = 0.$$
sobolev-spaces
asked 6 hours ago
TheGeekGreek
5,0093933
1
The last equation can be shown independently of the above using the trace operator. If this would be an option for you, too.
– Jonas Lenz
6 hours ago
@JonasLenz I am aware of this fact, but in the oral exam I should reproduce the proof given in the lecture. Sigh...
– TheGeekGreek
6 hours ago
that's unfortunate.
– Jonas Lenz
6 hours ago
I don't see why you need the trace characterization here; the uniqueness amounts to saying that if $uin H^1_0(Omega)$ satisfies $Delta u=0$ in the distributional sense, then $uequiv 0$. A proof of that amounts to justifying $int |nabla u|^2 = -int u Delta u$.
– user357151
4 hours ago
en.wikipedia.org/wiki/Dirichlet_problem - we need a Green function for the Dirichlet problem, if the existence is part of the problem...
– dan_fulea
3 hours ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In lecture we had the following theorem:
Let $Omega subseteq subseteq mathbbR^n$. Then $$H^1(Omega) =
H^1_0(Omega) oplus u in H^1 : Delta u = 0$$ where $Delta u$
is understood in the distributional sense. Moreover, if $Omega$ is of class $C^1$, $uvert_partial Omega$ does admit a unique harmonic extension.
Now the proof of the above is straight forward, however, the moreover part is where I stumble: In the notes it is phrased that this is a direct consequence of the decomposition, but I do not see this. I mean, we do have $$uvert_partial Omega = u_0vert_partialOmega + u_1 vert_partial Omega = u_1vert_partial Omega$$ when $u = u_0 + u_1$ and hence $u_1$ is a harmonic extension of $uvert_partial Omega$, but why is this extension unique? The reason why I am asking this is because I need to proof that $$H^1_0(Omega) = u in H^1(Omega) : uvert_partial Omega = 0.$$
sobolev-spaces
asked 6 hours ago
TheGeekGreek
5,0093933
In lecture we had the following theorem:
Let $Omega subseteq subseteq mathbbR^n$. Then $$H^1(Omega) =
H^1_0(Omega) oplus u in H^1 : Delta u = 0$$ where $Delta u$
is understood in the distributional sense. Moreover, if $Omega$ is of class $C^1$, $uvert_partial Omega$ does admit a unique harmonic extension.
Now the proof of the above is straight forward, however, the moreover part is where I stumble: In the notes it is phrased that this is a direct consequence of the decomposition, but I do not see this. I mean, we do have $$uvert_partial Omega = u_0vert_partialOmega + u_1 vert_partial Omega = u_1vert_partial Omega$$ when $u = u_0 + u_1$ and hence $u_1$ is a harmonic extension of $uvert_partial Omega$, but why is this extension unique? The reason why I am asking this is because I need to proof that $$H^1_0(Omega) = u in H^1(Omega) : uvert_partial Omega = 0.$$
sobolev-spaces
asked 6 hours ago
TheGeekGreek
5,0093933
asked 6 hours ago
TheGeekGreek
5,0093933
asked 6 hours ago
TheGeekGreek
5,0093933
asked 6 hours ago
TheGeekGreek
5,0093933
5,0093933
1
The last equation can be shown independently of the above using the trace operator. If this would be an option for you, too.
– Jonas Lenz
6 hours ago
@JonasLenz I am aware of this fact, but in the oral exam I should reproduce the proof given in the lecture. Sigh...
– TheGeekGreek
6 hours ago
that's unfortunate.
– Jonas Lenz
6 hours ago
I don't see why you need the trace characterization here; the uniqueness amounts to saying that if $uin H^1_0(Omega)$ satisfies $Delta u=0$ in the distributional sense, then $uequiv 0$. A proof of that amounts to justifying $int |nabla u|^2 = -int u Delta u$.
– user357151
4 hours ago
en.wikipedia.org/wiki/Dirichlet_problem - we need a Green function for the Dirichlet problem, if the existence is part of the problem...
– dan_fulea
3 hours ago
add a comment |Â
1
The last equation can be shown independently of the above using the trace operator. If this would be an option for you, too.
– Jonas Lenz
6 hours ago
@JonasLenz I am aware of this fact, but in the oral exam I should reproduce the proof given in the lecture. Sigh...
– TheGeekGreek
6 hours ago
that's unfortunate.
– Jonas Lenz
6 hours ago
I don't see why you need the trace characterization here; the uniqueness amounts to saying that if $uin H^1_0(Omega)$ satisfies $Delta u=0$ in the distributional sense, then $uequiv 0$. A proof of that amounts to justifying $int |nabla u|^2 = -int u Delta u$.
– user357151
4 hours ago
en.wikipedia.org/wiki/Dirichlet_problem - we need a Green function for the Dirichlet problem, if the existence is part of the problem...
– dan_fulea
3 hours ago
1
1
The last equation can be shown independently of the above using the trace operator. If this would be an option for you, too.
– Jonas Lenz
6 hours ago
The last equation can be shown independently of the above using the trace operator. If this would be an option for you, too.
– Jonas Lenz
6 hours ago
@JonasLenz I am aware of this fact, but in the oral exam I should reproduce the proof given in the lecture. Sigh...
– TheGeekGreek
6 hours ago
@JonasLenz I am aware of this fact, but in the oral exam I should reproduce the proof given in the lecture. Sigh...
– TheGeekGreek
6 hours ago
that's unfortunate.
– Jonas Lenz
6 hours ago
that's unfortunate.
– Jonas Lenz
6 hours ago
I don't see why you need the trace characterization here; the uniqueness amounts to saying that if $uin H^1_0(Omega)$ satisfies $Delta u=0$ in the distributional sense, then $uequiv 0$. A proof of that amounts to justifying $int |nabla u|^2 = -int u Delta u$.
– user357151
4 hours ago
I don't see why you need the trace characterization here; the uniqueness amounts to saying that if $uin H^1_0(Omega)$ satisfies $Delta u=0$ in the distributional sense, then $uequiv 0$. A proof of that amounts to justifying $int |nabla u|^2 = -int u Delta u$.
– user357151
4 hours ago
en.wikipedia.org/wiki/Dirichlet_problem - we need a Green function for the Dirichlet problem, if the existence is part of the problem...
– dan_fulea
3 hours ago
en.wikipedia.org/wiki/Dirichlet_problem - we need a Green function for the Dirichlet problem, if the existence is part of the problem...
– dan_fulea
3 hours ago
add a comment |Â
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1
The last equation can be shown independently of the above using the trace operator. If this would be an option for you, too.
– Jonas Lenz
6 hours ago
@JonasLenz I am aware of this fact, but in the oral exam I should reproduce the proof given in the lecture. Sigh...
– TheGeekGreek
6 hours ago
that's unfortunate.
– Jonas Lenz
6 hours ago
I don't see why you need the trace characterization here; the uniqueness amounts to saying that if $uin H^1_0(Omega)$ satisfies $Delta u=0$ in the distributional sense, then $uequiv 0$. A proof of that amounts to justifying $int |nabla u|^2 = -int u Delta u$.
– user357151
4 hours ago
en.wikipedia.org/wiki/Dirichlet_problem - we need a Green function for the Dirichlet problem, if the existence is part of the problem...
– dan_fulea
3 hours ago