Mixing
anti Lie derivative
$begingroup$
If $f: mathbb{R} to mathbb{R}$ is a continuous function, there is an antiderivative $g(x)=int_a^x f(t)dt$ and
$$frac{d}{dx}g(x)=f(x).$$
I want to know if a higher dimensional generalization of this holds.
If $f: mathbb{R}^n to mathbb{R}$ is a continuous function and $X$ is a smooth vector field on $mathbb{R}^n$, are there function $g : mathbb{R}^n to mathbb{R}$ such that
$$Xg=f$$
?
calculus pde indefinite-integrals lie-derivative
asked Jan 28 at 17:59
satoukibisatoukibi
16718
$endgroup$
add a comment |
$begingroup$
If $f: mathbb{R} to mathbb{R}$ is a continuous function, there is an antiderivative $g(x)=int_a^x f(t)dt$ and
$$frac{d}{dx}g(x)=f(x).$$
I want to know if a higher dimensional generalization of this holds.
If $f: mathbb{R}^n to mathbb{R}$ is a continuous function and $X$ is a smooth vector field on $mathbb{R}^n$, are there function $g : mathbb{R}^n to mathbb{R}$ such that
$$Xg=f$$
?
calculus pde indefinite-integrals lie-derivative
asked Jan 28 at 17:59
satoukibisatoukibi
16718
$endgroup$
2
$begingroup$
no: $n=2$, $X=xpartial_y-ypartial_x$, $f=1$ (the orbits are circles)
$endgroup$
– user8268
Jan 28 at 18:03
$begingroup$
Your generalization already fails for $Bbb R$, as $X$ could vanish at some point.
$endgroup$
– user98602
Jan 28 at 21:56
add a comment |
$begingroup$
If $f: mathbb{R} to mathbb{R}$ is a continuous function, there is an antiderivative $g(x)=int_a^x f(t)dt$ and
$$frac{d}{dx}g(x)=f(x).$$
I want to know if a higher dimensional generalization of this holds.
If $f: mathbb{R}^n to mathbb{R}$ is a continuous function and $X$ is a smooth vector field on $mathbb{R}^n$, are there function $g : mathbb{R}^n to mathbb{R}$ such that
$$Xg=f$$
?
calculus pde indefinite-integrals lie-derivative
asked Jan 28 at 17:59
satoukibisatoukibi
16718
$endgroup$
If $f: mathbb{R} to mathbb{R}$ is a continuous function, there is an antiderivative $g(x)=int_a^x f(t)dt$ and
$$frac{d}{dx}g(x)=f(x).$$
I want to know if a higher dimensional generalization of this holds.
If $f: mathbb{R}^n to mathbb{R}$ is a continuous function and $X$ is a smooth vector field on $mathbb{R}^n$, are there function $g : mathbb{R}^n to mathbb{R}$ such that
$$Xg=f$$
?
calculus pde indefinite-integrals lie-derivative
calculus pde indefinite-integrals lie-derivative
asked Jan 28 at 17:59
satoukibisatoukibi
16718
asked Jan 28 at 17:59
satoukibisatoukibi
16718
asked Jan 28 at 17:59
satoukibisatoukibi
16718
asked Jan 28 at 17:59
satoukibisatoukibi
16718
asked Jan 28 at 17:59
satoukibisatoukibi
16718
16718
2
$begingroup$
no: $n=2$, $X=xpartial_y-ypartial_x$, $f=1$ (the orbits are circles)
$endgroup$
– user8268
Jan 28 at 18:03
$begingroup$
Your generalization already fails for $Bbb R$, as $X$ could vanish at some point.
$endgroup$
– user98602
Jan 28 at 21:56
add a comment |
2
$begingroup$
no: $n=2$, $X=xpartial_y-ypartial_x$, $f=1$ (the orbits are circles)
$endgroup$
– user8268
Jan 28 at 18:03
$begingroup$
Your generalization already fails for $Bbb R$, as $X$ could vanish at some point.
$endgroup$
– user98602
Jan 28 at 21:56
2
2
$begingroup$
no: $n=2$, $X=xpartial_y-ypartial_x$, $f=1$ (the orbits are circles)
$endgroup$
– user8268
Jan 28 at 18:03
$begingroup$
no: $n=2$, $X=xpartial_y-ypartial_x$, $f=1$ (the orbits are circles)
$endgroup$
– user8268
Jan 28 at 18:03
$begingroup$
Your generalization already fails for $Bbb R$, as $X$ could vanish at some point.
$endgroup$
– user98602
Jan 28 at 21:56
$begingroup$
Your generalization already fails for $Bbb R$, as $X$ could vanish at some point.
$endgroup$
– user98602
Jan 28 at 21:56
add a comment |
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$begingroup$
no: $n=2$, $X=xpartial_y-ypartial_x$, $f=1$ (the orbits are circles)
$endgroup$
– user8268
Jan 28 at 18:03
$begingroup$
Your generalization already fails for $Bbb R$, as $X$ could vanish at some point.
$endgroup$
– user98602
Jan 28 at 21:56