Mixing
Composition of solutions to ODE on a manifold
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I don't understand lemma 1.4.7 page 18 of this introduction to differential geometry (https://www.math.ens.fr/~biquard/idg2008.pdf).
$varphi_{t}left ( {x} right )$ is a solution to the ordinary differential equation $dot{c}= {X}left ( c right )$ with initial condition ${x}$, ${X}$ being a vector field on a manifold ${M}$ .
I don't understand this composition and must be confused. How would this composition work for $dot{x^{i}}={e^{t}}$
ordinary-differential-equations differential-geometry manifolds
asked Jan 28 at 22:27
ClipperClipper
214
$endgroup$
add a comment |
$begingroup$
I don't understand lemma 1.4.7 page 18 of this introduction to differential geometry (https://www.math.ens.fr/~biquard/idg2008.pdf).
$varphi_{t}left ( {x} right )$ is a solution to the ordinary differential equation $dot{c}= {X}left ( c right )$ with initial condition ${x}$, ${X}$ being a vector field on a manifold ${M}$ .
I don't understand this composition and must be confused. How would this composition work for $dot{x^{i}}={e^{t}}$
ordinary-differential-equations differential-geometry manifolds
asked Jan 28 at 22:27
ClipperClipper
214
$endgroup$
add a comment |
$begingroup$
I don't understand lemma 1.4.7 page 18 of this introduction to differential geometry (https://www.math.ens.fr/~biquard/idg2008.pdf).
$varphi_{t}left ( {x} right )$ is a solution to the ordinary differential equation $dot{c}= {X}left ( c right )$ with initial condition ${x}$, ${X}$ being a vector field on a manifold ${M}$ .
I don't understand this composition and must be confused. How would this composition work for $dot{x^{i}}={e^{t}}$
ordinary-differential-equations differential-geometry manifolds
asked Jan 28 at 22:27
ClipperClipper
214
$endgroup$
I don't understand lemma 1.4.7 page 18 of this introduction to differential geometry (https://www.math.ens.fr/~biquard/idg2008.pdf).
$varphi_{t}left ( {x} right )$ is a solution to the ordinary differential equation $dot{c}= {X}left ( c right )$ with initial condition ${x}$, ${X}$ being a vector field on a manifold ${M}$ .
I don't understand this composition and must be confused. How would this composition work for $dot{x^{i}}={e^{t}}$
ordinary-differential-equations differential-geometry manifolds
ordinary-differential-equations differential-geometry manifolds
asked Jan 28 at 22:27
ClipperClipper
214
asked Jan 28 at 22:27
ClipperClipper
214
edited Jan 29 at 10:51
Clipper
asked Jan 28 at 22:27
ClipperClipper
214
asked Jan 28 at 22:27
ClipperClipper
214
asked Jan 28 at 22:27
ClipperClipper
214
214
add a comment |
add a comment |
1 Answer
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$begingroup$
This is a easy consequence of the Picard-Lindölf theorem.
Notice that $tmapstovarphi_{t+s}(x)$ is solution of the following Cauchy problem:
$$dot{c}=X(c),c(0)=varphi_s(x),$$
but by definition $tmapstovarphi_t(varphi_s(x))$ is also a solution, whence the equality.
The claim is true only for time-independent vector-field!
Regarding the example, the equation is not the one you wrote $dot{x^i}=e^t$, but rather ${dot{x^i}}=x^i$, since the vector field is given by $X=(x_1,ldots,x_n)$ and not by $X=(e^t,ldots,e^t)$ which is not time-independent.
In this case, the flow is given by $varphicolon (x,t)mapsto e^tx$, since $tmapsto e^tx$ satisfies $dot{x^i}=x^i$ and has value $x$ at $t=0$. You can check that $varphi(x,t+s)=e^{t+s}x=e^t(e^sx)=e^tvarphi(x,s)=varphi(varphi(s,x),t)$, as claimed.
answered Jan 29 at 13:14
C. FalconC. Falcon
15.3k41951
$endgroup$
$begingroup$
Merci ! It's much clearer. I guess you meant only true for time-independent vector-field ? I'm still not sure about which definition you used to state that $tmapstovarphi_t(varphi_s(x))$ is a solution.
$endgroup$
– Clipper
Jan 29 at 19:20
$begingroup$
Yes, indeed, for time INdependent vector field, thank you! By definition, $ccolon tmapstovarphi_t(x)$ satisfies $dot{c}=X(c)$ and $c(0)=x$, now use this for $x=varphi_s(x)$, $s$ is fixed.
$endgroup$
– C. Falcon
Jan 29 at 20:08
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is a easy consequence of the Picard-Lindölf theorem.
Notice that $tmapstovarphi_{t+s}(x)$ is solution of the following Cauchy problem:
$$dot{c}=X(c),c(0)=varphi_s(x),$$
but by definition $tmapstovarphi_t(varphi_s(x))$ is also a solution, whence the equality.
The claim is true only for time-independent vector-field!
Regarding the example, the equation is not the one you wrote $dot{x^i}=e^t$, but rather ${dot{x^i}}=x^i$, since the vector field is given by $X=(x_1,ldots,x_n)$ and not by $X=(e^t,ldots,e^t)$ which is not time-independent.
In this case, the flow is given by $varphicolon (x,t)mapsto e^tx$, since $tmapsto e^tx$ satisfies $dot{x^i}=x^i$ and has value $x$ at $t=0$. You can check that $varphi(x,t+s)=e^{t+s}x=e^t(e^sx)=e^tvarphi(x,s)=varphi(varphi(s,x),t)$, as claimed.
answered Jan 29 at 13:14
C. FalconC. Falcon
15.3k41951
$endgroup$
$begingroup$
Merci ! It's much clearer. I guess you meant only true for time-independent vector-field ? I'm still not sure about which definition you used to state that $tmapstovarphi_t(varphi_s(x))$ is a solution.
$endgroup$
– Clipper
Jan 29 at 19:20
$begingroup$
Yes, indeed, for time INdependent vector field, thank you! By definition, $ccolon tmapstovarphi_t(x)$ satisfies $dot{c}=X(c)$ and $c(0)=x$, now use this for $x=varphi_s(x)$, $s$ is fixed.
$endgroup$
– C. Falcon
Jan 29 at 20:08
add a comment |
$begingroup$
This is a easy consequence of the Picard-Lindölf theorem.
Notice that $tmapstovarphi_{t+s}(x)$ is solution of the following Cauchy problem:
$$dot{c}=X(c),c(0)=varphi_s(x),$$
but by definition $tmapstovarphi_t(varphi_s(x))$ is also a solution, whence the equality.
The claim is true only for time-independent vector-field!
Regarding the example, the equation is not the one you wrote $dot{x^i}=e^t$, but rather ${dot{x^i}}=x^i$, since the vector field is given by $X=(x_1,ldots,x_n)$ and not by $X=(e^t,ldots,e^t)$ which is not time-independent.
In this case, the flow is given by $varphicolon (x,t)mapsto e^tx$, since $tmapsto e^tx$ satisfies $dot{x^i}=x^i$ and has value $x$ at $t=0$. You can check that $varphi(x,t+s)=e^{t+s}x=e^t(e^sx)=e^tvarphi(x,s)=varphi(varphi(s,x),t)$, as claimed.
answered Jan 29 at 13:14
C. FalconC. Falcon
15.3k41951
$endgroup$
$begingroup$
Merci ! It's much clearer. I guess you meant only true for time-independent vector-field ? I'm still not sure about which definition you used to state that $tmapstovarphi_t(varphi_s(x))$ is a solution.
$endgroup$
– Clipper
Jan 29 at 19:20
$begingroup$
Yes, indeed, for time INdependent vector field, thank you! By definition, $ccolon tmapstovarphi_t(x)$ satisfies $dot{c}=X(c)$ and $c(0)=x$, now use this for $x=varphi_s(x)$, $s$ is fixed.
$endgroup$
– C. Falcon
Jan 29 at 20:08
add a comment |
$begingroup$
This is a easy consequence of the Picard-Lindölf theorem.
Notice that $tmapstovarphi_{t+s}(x)$ is solution of the following Cauchy problem:
$$dot{c}=X(c),c(0)=varphi_s(x),$$
but by definition $tmapstovarphi_t(varphi_s(x))$ is also a solution, whence the equality.
The claim is true only for time-independent vector-field!
Regarding the example, the equation is not the one you wrote $dot{x^i}=e^t$, but rather ${dot{x^i}}=x^i$, since the vector field is given by $X=(x_1,ldots,x_n)$ and not by $X=(e^t,ldots,e^t)$ which is not time-independent.
In this case, the flow is given by $varphicolon (x,t)mapsto e^tx$, since $tmapsto e^tx$ satisfies $dot{x^i}=x^i$ and has value $x$ at $t=0$. You can check that $varphi(x,t+s)=e^{t+s}x=e^t(e^sx)=e^tvarphi(x,s)=varphi(varphi(s,x),t)$, as claimed.
answered Jan 29 at 13:14
C. FalconC. Falcon
15.3k41951
$endgroup$
This is a easy consequence of the Picard-Lindölf theorem.
Notice that $tmapstovarphi_{t+s}(x)$ is solution of the following Cauchy problem:
$$dot{c}=X(c),c(0)=varphi_s(x),$$
but by definition $tmapstovarphi_t(varphi_s(x))$ is also a solution, whence the equality.
The claim is true only for time-independent vector-field!
Regarding the example, the equation is not the one you wrote $dot{x^i}=e^t$, but rather ${dot{x^i}}=x^i$, since the vector field is given by $X=(x_1,ldots,x_n)$ and not by $X=(e^t,ldots,e^t)$ which is not time-independent.
In this case, the flow is given by $varphicolon (x,t)mapsto e^tx$, since $tmapsto e^tx$ satisfies $dot{x^i}=x^i$ and has value $x$ at $t=0$. You can check that $varphi(x,t+s)=e^{t+s}x=e^t(e^sx)=e^tvarphi(x,s)=varphi(varphi(s,x),t)$, as claimed.
answered Jan 29 at 13:14
C. FalconC. Falcon
15.3k41951
edited Jan 29 at 20:07
answered Jan 29 at 13:14
C. FalconC. Falcon
15.3k41951
answered Jan 29 at 13:14
C. FalconC. Falcon
15.3k41951
answered Jan 29 at 13:14
C. FalconC. Falcon
15.3k41951
15.3k41951
$begingroup$
Merci ! It's much clearer. I guess you meant only true for time-independent vector-field ? I'm still not sure about which definition you used to state that $tmapstovarphi_t(varphi_s(x))$ is a solution.
$endgroup$
– Clipper
Jan 29 at 19:20
$begingroup$
Yes, indeed, for time INdependent vector field, thank you! By definition, $ccolon tmapstovarphi_t(x)$ satisfies $dot{c}=X(c)$ and $c(0)=x$, now use this for $x=varphi_s(x)$, $s$ is fixed.
$endgroup$
– C. Falcon
Jan 29 at 20:08
add a comment |
$begingroup$
Merci ! It's much clearer. I guess you meant only true for time-independent vector-field ? I'm still not sure about which definition you used to state that $tmapstovarphi_t(varphi_s(x))$ is a solution.
$endgroup$
– Clipper
Jan 29 at 19:20
$begingroup$
Yes, indeed, for time INdependent vector field, thank you! By definition, $ccolon tmapstovarphi_t(x)$ satisfies $dot{c}=X(c)$ and $c(0)=x$, now use this for $x=varphi_s(x)$, $s$ is fixed.
$endgroup$
– C. Falcon
Jan 29 at 20:08
$begingroup$
Merci ! It's much clearer. I guess you meant only true for time-independent vector-field ? I'm still not sure about which definition you used to state that $tmapstovarphi_t(varphi_s(x))$ is a solution.
$endgroup$
– Clipper
Jan 29 at 19:20
$begingroup$
Merci ! It's much clearer. I guess you meant only true for time-independent vector-field ? I'm still not sure about which definition you used to state that $tmapstovarphi_t(varphi_s(x))$ is a solution.
$endgroup$
– Clipper
Jan 29 at 19:20
$begingroup$
Yes, indeed, for time INdependent vector field, thank you! By definition, $ccolon tmapstovarphi_t(x)$ satisfies $dot{c}=X(c)$ and $c(0)=x$, now use this for $x=varphi_s(x)$, $s$ is fixed.
$endgroup$
– C. Falcon
Jan 29 at 20:08
$begingroup$
Yes, indeed, for time INdependent vector field, thank you! By definition, $ccolon tmapstovarphi_t(x)$ satisfies $dot{c}=X(c)$ and $c(0)=x$, now use this for $x=varphi_s(x)$, $s$ is fixed.
$endgroup$
– C. Falcon
Jan 29 at 20:08
add a comment |
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