Smooth curves and velocity

2

$begingroup$

Let $M$ be a smooth manifold.

Let $X:M to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=overline{{pin M: X_pnot=0}}$. Suppose $K$ a is compact subset of $M$.

Let $pnotin K$, and let $theta^p$ be the maximal integral curve of $X$ starting at $p$.

Why $theta^p$ is costant?

I know that $theta^p:U^pto M$ is a smooth map, with $U^p$ open interval in $mathbb{R}$, and for each $t in U^p$ I have $(theta^p)'(t)=X_{theta^p(t)}$.

The definition of $(theta^p)'(t)$ is $(theta^p)'(t):C^infty(M)to mathbb{R}$ defined by $(theta^p)'(t)(f)=frac{d}{dt}(fcirctheta^p)(t)$.

I know I have to use that $X_q=0$ for each $qnotin K$ but I can't see how.

For example, I don't know if $theta^p(t)$ is in $K$ or not, so when I have $(theta^p)'(t)=X_{theta^p(t)}$ I don't know if $X_{theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(theta^p)'(t)$ is $0$ for each $tin U^p$, then I don't know how to deduce from this that $theta^p $ is in fact costant.

differential-geometry smooth-manifolds smooth-functions

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edited Jan 6 at 17:36

Minato

asked Jan 6 at 17:30

MinatoMinato

465212

$endgroup$

add a comment | 

2

$begingroup$

Let $M$ be a smooth manifold.

Let $X:M to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=overline{{pin M: X_pnot=0}}$. Suppose $K$ a is compact subset of $M$.

Let $pnotin K$, and let $theta^p$ be the maximal integral curve of $X$ starting at $p$.

Why $theta^p$ is costant?

I know that $theta^p:U^pto M$ is a smooth map, with $U^p$ open interval in $mathbb{R}$, and for each $t in U^p$ I have $(theta^p)'(t)=X_{theta^p(t)}$.

The definition of $(theta^p)'(t)$ is $(theta^p)'(t):C^infty(M)to mathbb{R}$ defined by $(theta^p)'(t)(f)=frac{d}{dt}(fcirctheta^p)(t)$.

I know I have to use that $X_q=0$ for each $qnotin K$ but I can't see how.

For example, I don't know if $theta^p(t)$ is in $K$ or not, so when I have $(theta^p)'(t)=X_{theta^p(t)}$ I don't know if $X_{theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(theta^p)'(t)$ is $0$ for each $tin U^p$, then I don't know how to deduce from this that $theta^p $ is in fact costant.

differential-geometry smooth-manifolds smooth-functions

share|cite|improve this question

edited Jan 6 at 17:36

Minato

asked Jan 6 at 17:30

MinatoMinato

465212

$endgroup$

add a comment | 

2

2

2

1

$begingroup$

Let $M$ be a smooth manifold.

Let $X:M to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=overline{{pin M: X_pnot=0}}$. Suppose $K$ a is compact subset of $M$.

Let $pnotin K$, and let $theta^p$ be the maximal integral curve of $X$ starting at $p$.

Why $theta^p$ is costant?

I know that $theta^p:U^pto M$ is a smooth map, with $U^p$ open interval in $mathbb{R}$, and for each $t in U^p$ I have $(theta^p)'(t)=X_{theta^p(t)}$.

The definition of $(theta^p)'(t)$ is $(theta^p)'(t):C^infty(M)to mathbb{R}$ defined by $(theta^p)'(t)(f)=frac{d}{dt}(fcirctheta^p)(t)$.

I know I have to use that $X_q=0$ for each $qnotin K$ but I can't see how.

For example, I don't know if $theta^p(t)$ is in $K$ or not, so when I have $(theta^p)'(t)=X_{theta^p(t)}$ I don't know if $X_{theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(theta^p)'(t)$ is $0$ for each $tin U^p$, then I don't know how to deduce from this that $theta^p $ is in fact costant.

differential-geometry smooth-manifolds smooth-functions

share|cite|improve this question

edited Jan 6 at 17:36

Minato

asked Jan 6 at 17:30

MinatoMinato

465212

$endgroup$

Let $M$ be a smooth manifold.

Let $X:M to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=overline{{pin M: X_pnot=0}}$. Suppose $K$ a is compact subset of $M$.

Let $pnotin K$, and let $theta^p$ be the maximal integral curve of $X$ starting at $p$.

Why $theta^p$ is costant?

I know that $theta^p:U^pto M$ is a smooth map, with $U^p$ open interval in $mathbb{R}$, and for each $t in U^p$ I have $(theta^p)'(t)=X_{theta^p(t)}$.

The definition of $(theta^p)'(t)$ is $(theta^p)'(t):C^infty(M)to mathbb{R}$ defined by $(theta^p)'(t)(f)=frac{d}{dt}(fcirctheta^p)(t)$.

I know I have to use that $X_q=0$ for each $qnotin K$ but I can't see how.

For example, I don't know if $theta^p(t)$ is in $K$ or not, so when I have $(theta^p)'(t)=X_{theta^p(t)}$ I don't know if $X_{theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(theta^p)'(t)$ is $0$ for each $tin U^p$, then I don't know how to deduce from this that $theta^p $ is in fact costant.

differential-geometry smooth-manifolds smooth-functions

differential-geometry smooth-manifolds smooth-functions

share|cite|improve this question

edited Jan 6 at 17:36

Minato

asked Jan 6 at 17:30

MinatoMinato

465212

share|cite|improve this question

edited Jan 6 at 17:36

Minato

asked Jan 6 at 17:30

MinatoMinato

465212

share|cite|improve this question

share|cite|improve this question

edited Jan 6 at 17:36

Minato

edited Jan 6 at 17:36

Minato

edited Jan 6 at 17:36

Minato

asked Jan 6 at 17:30

MinatoMinato

465212

asked Jan 6 at 17:30

MinatoMinato

465212

asked Jan 6 at 17:30

MinatoMinato

465212

465212

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

1

$begingroup$

Since

$p notin K, tag 1$

$X(p) = 0; tag 2$

let

$theta^p(t) tag 3$

be the unique solution of

${theta^p}'(t) = X(theta^p(t)) tag 4$

such that

$theta^p(0) = p; tag 5$

we note that the constant curve

$phi^p(t) = p, ; forall t in Bbb R, tag 6$

satisfies

${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$

we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have

$theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$

share|cite|improve this answer

answered Jan 6 at 19:09

Robert LewisRobert Lewis

45.1k23065

$endgroup$

add a comment | 

2

$begingroup$

$theta^p$ is constant because the constant function is a solution of the differential equation.

share|cite|improve this answer

answered Jan 6 at 17:44

MindlackMindlack

3,28217

$endgroup$

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Since

$p notin K, tag 1$

$X(p) = 0; tag 2$

let

$theta^p(t) tag 3$

be the unique solution of

${theta^p}'(t) = X(theta^p(t)) tag 4$

such that

$theta^p(0) = p; tag 5$

we note that the constant curve

$phi^p(t) = p, ; forall t in Bbb R, tag 6$

satisfies

${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$

we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have

$theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$

share|cite|improve this answer

answered Jan 6 at 19:09

Robert LewisRobert Lewis

45.1k23065

$endgroup$

add a comment | 

1

$begingroup$

Since

$p notin K, tag 1$

$X(p) = 0; tag 2$

let

$theta^p(t) tag 3$

be the unique solution of

${theta^p}'(t) = X(theta^p(t)) tag 4$

such that

$theta^p(0) = p; tag 5$

we note that the constant curve

$phi^p(t) = p, ; forall t in Bbb R, tag 6$

satisfies

${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$

we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have

$theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$

share|cite|improve this answer

answered Jan 6 at 19:09

Robert LewisRobert Lewis

45.1k23065

$endgroup$

add a comment | 

1

1

1

$begingroup$

Since

$p notin K, tag 1$

$X(p) = 0; tag 2$

let

$theta^p(t) tag 3$

be the unique solution of

${theta^p}'(t) = X(theta^p(t)) tag 4$

such that

$theta^p(0) = p; tag 5$

we note that the constant curve

$phi^p(t) = p, ; forall t in Bbb R, tag 6$

satisfies

${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$

we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have

$theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$

share|cite|improve this answer

answered Jan 6 at 19:09

Robert LewisRobert Lewis

45.1k23065

$endgroup$

Since

$p notin K, tag 1$

$X(p) = 0; tag 2$

let

$theta^p(t) tag 3$

be the unique solution of

${theta^p}'(t) = X(theta^p(t)) tag 4$

such that

$theta^p(0) = p; tag 5$

we note that the constant curve

$phi^p(t) = p, ; forall t in Bbb R, tag 6$

satisfies

${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$

we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have

$theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$

share|cite|improve this answer

answered Jan 6 at 19:09

Robert LewisRobert Lewis

45.1k23065

share|cite|improve this answer

share|cite|improve this answer

answered Jan 6 at 19:09

Robert LewisRobert Lewis

45.1k23065

answered Jan 6 at 19:09

Robert LewisRobert Lewis

45.1k23065

answered Jan 6 at 19:09

Robert LewisRobert Lewis

45.1k23065

45.1k23065

add a comment | 

add a comment | 

2

$begingroup$

$theta^p$ is constant because the constant function is a solution of the differential equation.

share|cite|improve this answer

answered Jan 6 at 17:44

MindlackMindlack

3,28217

$endgroup$

add a comment | 

2

$begingroup$

$theta^p$ is constant because the constant function is a solution of the differential equation.

share|cite|improve this answer

answered Jan 6 at 17:44

MindlackMindlack

3,28217

$endgroup$

add a comment | 

2

2

2

$begingroup$

$theta^p$ is constant because the constant function is a solution of the differential equation.

share|cite|improve this answer

answered Jan 6 at 17:44

MindlackMindlack

3,28217

$endgroup$

$theta^p$ is constant because the constant function is a solution of the differential equation.

share|cite|improve this answer

answered Jan 6 at 17:44

MindlackMindlack

3,28217

share|cite|improve this answer

share|cite|improve this answer

answered Jan 6 at 17:44

MindlackMindlack

3,28217

answered Jan 6 at 17:44

MindlackMindlack

3,28217

answered Jan 6 at 17:44

MindlackMindlack

3,28217

3,28217

add a comment | 

add a comment | 

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