Differential equation on manifold
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Suppose that $Msubsetmathbb{R}^n$ is a smooth Riemannian manifold. Let $f:Mrightarrow T_xM$ be a smooth function from the manifold to its tangent space. Suppose that the solution to $dot{x}=f(x)$ with $x(0)in M$ exists for all $t$ in some set $S$.
I would like a reference for the fact that $x(t)in M$ for all $tin S$.
ordinary-differential-equations reference-request riemannian-geometry
asked Jan 20 at 19:17
AsdfAsdf
417213
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add a comment |
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Suppose that $Msubsetmathbb{R}^n$ is a smooth Riemannian manifold. Let $f:Mrightarrow T_xM$ be a smooth function from the manifold to its tangent space. Suppose that the solution to $dot{x}=f(x)$ with $x(0)in M$ exists for all $t$ in some set $S$.
I would like a reference for the fact that $x(t)in M$ for all $tin S$.
ordinary-differential-equations reference-request riemannian-geometry
asked Jan 20 at 19:17
AsdfAsdf
417213
$endgroup$
$begingroup$
Are you sure this is correct? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 19:23
2
$begingroup$
By continuity the set $S_0 = {t: x(t) in M}$ is closed. Using charts and local existence and uniqueness for ODE's, it is also easy to prove that $S_0$ is open. Therefore $S_0$ equals the connected component of $S$ that contains $t=0$.
$endgroup$
– Hans Engler
Jan 20 at 19:36
1
$begingroup$
Bony–Brezis theorem.
$endgroup$
– Evgeny
Jan 21 at 21:24
$begingroup$
@Evgeny: Thank you, this is what I was looking for.
$endgroup$
– Asdf
Jan 22 at 22:27
add a comment |
$begingroup$
Suppose that $Msubsetmathbb{R}^n$ is a smooth Riemannian manifold. Let $f:Mrightarrow T_xM$ be a smooth function from the manifold to its tangent space. Suppose that the solution to $dot{x}=f(x)$ with $x(0)in M$ exists for all $t$ in some set $S$.
I would like a reference for the fact that $x(t)in M$ for all $tin S$.
ordinary-differential-equations reference-request riemannian-geometry
asked Jan 20 at 19:17
AsdfAsdf
417213
$endgroup$
Suppose that $Msubsetmathbb{R}^n$ is a smooth Riemannian manifold. Let $f:Mrightarrow T_xM$ be a smooth function from the manifold to its tangent space. Suppose that the solution to $dot{x}=f(x)$ with $x(0)in M$ exists for all $t$ in some set $S$.
I would like a reference for the fact that $x(t)in M$ for all $tin S$.
ordinary-differential-equations reference-request riemannian-geometry
ordinary-differential-equations reference-request riemannian-geometry
asked Jan 20 at 19:17
AsdfAsdf
417213
asked Jan 20 at 19:17
AsdfAsdf
417213
asked Jan 20 at 19:17
AsdfAsdf
417213
asked Jan 20 at 19:17
AsdfAsdf
417213
asked Jan 20 at 19:17
AsdfAsdf
417213
417213
$begingroup$
Are you sure this is correct? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 19:23
2
$begingroup$
By continuity the set $S_0 = {t: x(t) in M}$ is closed. Using charts and local existence and uniqueness for ODE's, it is also easy to prove that $S_0$ is open. Therefore $S_0$ equals the connected component of $S$ that contains $t=0$.
$endgroup$
– Hans Engler
Jan 20 at 19:36
1
$begingroup$
Bony–Brezis theorem.
$endgroup$
– Evgeny
Jan 21 at 21:24
$begingroup$
@Evgeny: Thank you, this is what I was looking for.
$endgroup$
– Asdf
Jan 22 at 22:27
add a comment |
$begingroup$
Are you sure this is correct? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 19:23
2
$begingroup$
By continuity the set $S_0 = {t: x(t) in M}$ is closed. Using charts and local existence and uniqueness for ODE's, it is also easy to prove that $S_0$ is open. Therefore $S_0$ equals the connected component of $S$ that contains $t=0$.
$endgroup$
– Hans Engler
Jan 20 at 19:36
1
$begingroup$
Bony–Brezis theorem.
$endgroup$
– Evgeny
Jan 21 at 21:24
$begingroup$
@Evgeny: Thank you, this is what I was looking for.
$endgroup$
– Asdf
Jan 22 at 22:27
$begingroup$
Are you sure this is correct? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 19:23
$begingroup$
Are you sure this is correct? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 19:23
2
2
$begingroup$
By continuity the set $S_0 = {t: x(t) in M}$ is closed. Using charts and local existence and uniqueness for ODE's, it is also easy to prove that $S_0$ is open. Therefore $S_0$ equals the connected component of $S$ that contains $t=0$.
$endgroup$
– Hans Engler
Jan 20 at 19:36
$begingroup$
By continuity the set $S_0 = {t: x(t) in M}$ is closed. Using charts and local existence and uniqueness for ODE's, it is also easy to prove that $S_0$ is open. Therefore $S_0$ equals the connected component of $S$ that contains $t=0$.
$endgroup$
– Hans Engler
Jan 20 at 19:36
1
1
$begingroup$
Bony–Brezis theorem.
$endgroup$
– Evgeny
Jan 21 at 21:24
$begingroup$
Bony–Brezis theorem.
$endgroup$
– Evgeny
Jan 21 at 21:24
$begingroup$
@Evgeny: Thank you, this is what I was looking for.
$endgroup$
– Asdf
Jan 22 at 22:27
$begingroup$
@Evgeny: Thank you, this is what I was looking for.
$endgroup$
– Asdf
Jan 22 at 22:27
add a comment |
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$begingroup$
Are you sure this is correct? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 19:23
2
$begingroup$
By continuity the set $S_0 = {t: x(t) in M}$ is closed. Using charts and local existence and uniqueness for ODE's, it is also easy to prove that $S_0$ is open. Therefore $S_0$ equals the connected component of $S$ that contains $t=0$.
$endgroup$
– Hans Engler
Jan 20 at 19:36
1
$begingroup$
Bony–Brezis theorem.
$endgroup$
– Evgeny
Jan 21 at 21:24
$begingroup$
@Evgeny: Thank you, this is what I was looking for.
$endgroup$
– Asdf
Jan 22 at 22:27