Mixing
Uniqueness of solution for $a(u,v)=F(v)$
$begingroup$
Let $a(u,v)$ be a bilinearform on a hilbert space $mathcal{H}$ which satisfies all conditions for the Lax-Milgram Lemma.
Furthermore,
$$a(u,v)=F(v), forall vinmathcal{H}$$ for a bounded functional $F$ on $mathcal{H}$.
I don't understand why a solution $u$ has to be unique.
Using Lax-Milgram we obtain a unique linear operator $T$ such that
$$langle Tu,vrangle = F(v)$$
Then using the Riesz Theorem I can obtain another unique $winmathcal{H}$ with
$$langle Tu,vrangle = F(v)=langle v,wrangle$$
So we have that, since $w$ is unique that $Tu=w$. Why is $u$ then unique?
functional-analysis pde
asked Jan 19 at 12:40
EpsilonDeltaEpsilonDelta
6801615
$endgroup$
add a comment |
$begingroup$
Let $a(u,v)$ be a bilinearform on a hilbert space $mathcal{H}$ which satisfies all conditions for the Lax-Milgram Lemma.
Furthermore,
$$a(u,v)=F(v), forall vinmathcal{H}$$ for a bounded functional $F$ on $mathcal{H}$.
I don't understand why a solution $u$ has to be unique.
Using Lax-Milgram we obtain a unique linear operator $T$ such that
$$langle Tu,vrangle = F(v)$$
Then using the Riesz Theorem I can obtain another unique $winmathcal{H}$ with
$$langle Tu,vrangle = F(v)=langle v,wrangle$$
So we have that, since $w$ is unique that $Tu=w$. Why is $u$ then unique?
functional-analysis pde
asked Jan 19 at 12:40
EpsilonDeltaEpsilonDelta
6801615
$endgroup$
$begingroup$
Is the statement $a(u,v) = F(v)$ for all $v in H$?
$endgroup$
– Umberto P.
Jan 19 at 12:55
$begingroup$
Yes. I will make an edit.
$endgroup$
– EpsilonDelta
Jan 19 at 13:37
add a comment |
$begingroup$
Let $a(u,v)$ be a bilinearform on a hilbert space $mathcal{H}$ which satisfies all conditions for the Lax-Milgram Lemma.
Furthermore,
$$a(u,v)=F(v), forall vinmathcal{H}$$ for a bounded functional $F$ on $mathcal{H}$.
I don't understand why a solution $u$ has to be unique.
Using Lax-Milgram we obtain a unique linear operator $T$ such that
$$langle Tu,vrangle = F(v)$$
Then using the Riesz Theorem I can obtain another unique $winmathcal{H}$ with
$$langle Tu,vrangle = F(v)=langle v,wrangle$$
So we have that, since $w$ is unique that $Tu=w$. Why is $u$ then unique?
functional-analysis pde
asked Jan 19 at 12:40
EpsilonDeltaEpsilonDelta
6801615
$endgroup$
Let $a(u,v)$ be a bilinearform on a hilbert space $mathcal{H}$ which satisfies all conditions for the Lax-Milgram Lemma.
Furthermore,
$$a(u,v)=F(v), forall vinmathcal{H}$$ for a bounded functional $F$ on $mathcal{H}$.
I don't understand why a solution $u$ has to be unique.
Using Lax-Milgram we obtain a unique linear operator $T$ such that
$$langle Tu,vrangle = F(v)$$
Then using the Riesz Theorem I can obtain another unique $winmathcal{H}$ with
$$langle Tu,vrangle = F(v)=langle v,wrangle$$
So we have that, since $w$ is unique that $Tu=w$. Why is $u$ then unique?
functional-analysis pde
functional-analysis pde
asked Jan 19 at 12:40
EpsilonDeltaEpsilonDelta
6801615
asked Jan 19 at 12:40
EpsilonDeltaEpsilonDelta
6801615
edited Jan 19 at 13:38
EpsilonDelta
asked Jan 19 at 12:40
EpsilonDeltaEpsilonDelta
6801615
asked Jan 19 at 12:40
EpsilonDeltaEpsilonDelta
6801615
asked Jan 19 at 12:40
EpsilonDeltaEpsilonDelta
6801615
6801615
$begingroup$
Is the statement $a(u,v) = F(v)$ for all $v in H$?
$endgroup$
– Umberto P.
Jan 19 at 12:55
$begingroup$
Yes. I will make an edit.
$endgroup$
– EpsilonDelta
Jan 19 at 13:37
add a comment |
$begingroup$
Is the statement $a(u,v) = F(v)$ for all $v in H$?
$endgroup$
– Umberto P.
Jan 19 at 12:55
$begingroup$
Yes. I will make an edit.
$endgroup$
– EpsilonDelta
Jan 19 at 13:37
$begingroup$
Is the statement $a(u,v) = F(v)$ for all $v in H$?
$endgroup$
– Umberto P.
Jan 19 at 12:55
$begingroup$
Is the statement $a(u,v) = F(v)$ for all $v in H$?
$endgroup$
– Umberto P.
Jan 19 at 12:55
$begingroup$
Yes. I will make an edit.
$endgroup$
– EpsilonDelta
Jan 19 at 13:37
$begingroup$
Yes. I will make an edit.
$endgroup$
– EpsilonDelta
Jan 19 at 13:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Coercivity of $a$ implies that if $a(u,u) = 0$ then $u = 0$.
Suppose that $a(u,v) = F(v)$ and $a(w,v) = F(v)$ both hold for all $v in H$. Then
$$a(u-w,v) = 0$$ for all $v in H$ and in particular
$$a(u-w,u-w) = 0.$$ Thus $u-w = 0$.
answered Jan 19 at 14:04
Umberto P.Umberto P.
39.6k13267
$endgroup$
$begingroup$
I also know the LM-Lemma not involving coercivity, and if I am not mistaken, coercivity also implies invertibility of $T$ and then we have a unique solution.
$endgroup$
– EpsilonDelta
Jan 19 at 14:53
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
$begingroup$
Coercivity of $a$ implies that if $a(u,u) = 0$ then $u = 0$.
Suppose that $a(u,v) = F(v)$ and $a(w,v) = F(v)$ both hold for all $v in H$. Then
$$a(u-w,v) = 0$$ for all $v in H$ and in particular
$$a(u-w,u-w) = 0.$$ Thus $u-w = 0$.
answered Jan 19 at 14:04
Umberto P.Umberto P.
39.6k13267
$endgroup$
$begingroup$
I also know the LM-Lemma not involving coercivity, and if I am not mistaken, coercivity also implies invertibility of $T$ and then we have a unique solution.
$endgroup$
– EpsilonDelta
Jan 19 at 14:53
add a comment |
$begingroup$
Coercivity of $a$ implies that if $a(u,u) = 0$ then $u = 0$.
Suppose that $a(u,v) = F(v)$ and $a(w,v) = F(v)$ both hold for all $v in H$. Then
$$a(u-w,v) = 0$$ for all $v in H$ and in particular
$$a(u-w,u-w) = 0.$$ Thus $u-w = 0$.
answered Jan 19 at 14:04
Umberto P.Umberto P.
39.6k13267
$endgroup$
$begingroup$
I also know the LM-Lemma not involving coercivity, and if I am not mistaken, coercivity also implies invertibility of $T$ and then we have a unique solution.
$endgroup$
– EpsilonDelta
Jan 19 at 14:53
add a comment |
$begingroup$
Coercivity of $a$ implies that if $a(u,u) = 0$ then $u = 0$.
Suppose that $a(u,v) = F(v)$ and $a(w,v) = F(v)$ both hold for all $v in H$. Then
$$a(u-w,v) = 0$$ for all $v in H$ and in particular
$$a(u-w,u-w) = 0.$$ Thus $u-w = 0$.
answered Jan 19 at 14:04
Umberto P.Umberto P.
39.6k13267
$endgroup$
Coercivity of $a$ implies that if $a(u,u) = 0$ then $u = 0$.
Suppose that $a(u,v) = F(v)$ and $a(w,v) = F(v)$ both hold for all $v in H$. Then
$$a(u-w,v) = 0$$ for all $v in H$ and in particular
$$a(u-w,u-w) = 0.$$ Thus $u-w = 0$.
answered Jan 19 at 14:04
Umberto P.Umberto P.
39.6k13267
answered Jan 19 at 14:04
Umberto P.Umberto P.
39.6k13267
answered Jan 19 at 14:04
Umberto P.Umberto P.
39.6k13267
answered Jan 19 at 14:04
Umberto P.Umberto P.
39.6k13267
39.6k13267
$begingroup$
I also know the LM-Lemma not involving coercivity, and if I am not mistaken, coercivity also implies invertibility of $T$ and then we have a unique solution.
$endgroup$
– EpsilonDelta
Jan 19 at 14:53
add a comment |
$begingroup$
I also know the LM-Lemma not involving coercivity, and if I am not mistaken, coercivity also implies invertibility of $T$ and then we have a unique solution.
$endgroup$
– EpsilonDelta
Jan 19 at 14:53
$begingroup$
I also know the LM-Lemma not involving coercivity, and if I am not mistaken, coercivity also implies invertibility of $T$ and then we have a unique solution.
$endgroup$
– EpsilonDelta
Jan 19 at 14:53
$begingroup$
I also know the LM-Lemma not involving coercivity, and if I am not mistaken, coercivity also implies invertibility of $T$ and then we have a unique solution.
$endgroup$
– EpsilonDelta
Jan 19 at 14:53
add a comment |
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$begingroup$
Is the statement $a(u,v) = F(v)$ for all $v in H$?
$endgroup$
– Umberto P.
Jan 19 at 12:55
$begingroup$
Yes. I will make an edit.
$endgroup$
– EpsilonDelta
Jan 19 at 13:37