Mixing
Extension of Du-Bois-Raymond lemma to Vector Fields on a Riemannian Manifold
$begingroup$
I am trying to show the following extension of the Du Bois Raymond lemma:
Let $M$ be a smooth Riemannian Manifold and $omega: [0,1] rightarrow M$ be a $W^{1,2}$ curve on M. Consider a tangential $L^2$ vector field along $omega$ denoted by $v in L^2(omega^*TM)$.
If $$int_{0}^{1} langle v, frac{Du}{partial t} rangle text{ dt} = 0 text{for all } u in W ^{1,2}(omega^*TM) text{ with } u(0)=u(1)=0$$
Then
$$v in W^{1,2}(omega^*TM) text{ with } frac{Dv}{partial t} = 0 a.e.$$
where $frac{Du}{partial t}$ denotes the covariant derivative of $u$ along the curve $omega$.
For my purposes it would be sufficient to show this for the simplified case where $M = S^2$ and the covariant derivative becomes the projection onto the respective tangent space.
In particular (let $M=S^2$ from now on), I am baffled by my proof:
With $u$ being as above we have $$0 = int_{0}^{1} langle v, frac{Du}{partial t} rangle text{ dt} = int_0^1 langle v, u' rangle text{ dt} - int_0^1 langle v, langle u', omega rangle omega rangletext{ dt} $$
where the second expression vanishes since $v bot omega$ because $v(t)$ lies in the tangent space of $omega(t)$ to $S^2$ for all $t$ by definition. Hence
$$int_0^1 langle v, u' rangle text{ dt} = 0$$ and the regular Du Bois Raymond lemma for the ordinary derivatives as functions into $mathbb{R}^3$ becomes applicable and we have $v$ being constant (and ordinary differentiable) almost everywhere which would mean
$$v' equiv 0$$
and thus
$$frac{Dv}{partial t} = 0$$
Is this correct? I feel like I've somehow lost information on my $v$ at the end. The Du Bois Raymond yields that $v$ is constant, the extension yields that $v$ is parallel transported along the curve and this is not equivalent, isn't it?
Edit: As pointed out in the comments the proof goes wrong, when I use the classic Du-Bois-Raymond Lemma when the assumption is only tested for certain vector fields but not all as necessary. Still the extension has yet to be proven correctly.
proof-verification differential-geometry manifolds riemannian-geometry calculus-of-variations
asked Jan 30 at 21:37
NhatNhat
1,0261017
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to show the following extension of the Du Bois Raymond lemma:
Let $M$ be a smooth Riemannian Manifold and $omega: [0,1] rightarrow M$ be a $W^{1,2}$ curve on M. Consider a tangential $L^2$ vector field along $omega$ denoted by $v in L^2(omega^*TM)$.
If $$int_{0}^{1} langle v, frac{Du}{partial t} rangle text{ dt} = 0 text{for all } u in W ^{1,2}(omega^*TM) text{ with } u(0)=u(1)=0$$
Then
$$v in W^{1,2}(omega^*TM) text{ with } frac{Dv}{partial t} = 0 a.e.$$
where $frac{Du}{partial t}$ denotes the covariant derivative of $u$ along the curve $omega$.
For my purposes it would be sufficient to show this for the simplified case where $M = S^2$ and the covariant derivative becomes the projection onto the respective tangent space.
In particular (let $M=S^2$ from now on), I am baffled by my proof:
With $u$ being as above we have $$0 = int_{0}^{1} langle v, frac{Du}{partial t} rangle text{ dt} = int_0^1 langle v, u' rangle text{ dt} - int_0^1 langle v, langle u', omega rangle omega rangletext{ dt} $$
where the second expression vanishes since $v bot omega$ because $v(t)$ lies in the tangent space of $omega(t)$ to $S^2$ for all $t$ by definition. Hence
$$int_0^1 langle v, u' rangle text{ dt} = 0$$ and the regular Du Bois Raymond lemma for the ordinary derivatives as functions into $mathbb{R}^3$ becomes applicable and we have $v$ being constant (and ordinary differentiable) almost everywhere which would mean
$$v' equiv 0$$
and thus
$$frac{Dv}{partial t} = 0$$
Is this correct? I feel like I've somehow lost information on my $v$ at the end. The Du Bois Raymond yields that $v$ is constant, the extension yields that $v$ is parallel transported along the curve and this is not equivalent, isn't it?
Edit: As pointed out in the comments the proof goes wrong, when I use the classic Du-Bois-Raymond Lemma when the assumption is only tested for certain vector fields but not all as necessary. Still the extension has yet to be proven correctly.
proof-verification differential-geometry manifolds riemannian-geometry calculus-of-variations
asked Jan 30 at 21:37
NhatNhat
1,0261017
$endgroup$
$begingroup$
It seems to me that the regular du Bois-Reymond lemma is not applicable, as you have the equality only for vector fields $u$ which are tangent to the sphere. I guess you would need this equality to hold for every vector field $u$ in order to use the du Bois-Reymond lemma. This makes sense, as you noted yourself, as the vector field $v$ usually cannot be constant as a function to $mathbb{R}^3$.
$endgroup$
– Amitai Yuval
Jan 31 at 6:45
$begingroup$
@AmitaiYuval That is indeed correct and a good find. It also only makes sense since I am expecting $v$ to only be constant along $omega$ as well so I only have an assumptions made for the vector fields along $omega$. At least the „extension“ seems intuitive. I’m going to edit my OP with your input, thanks.
$endgroup$
– Nhat
Jan 31 at 10:18
$begingroup$
Are you familiar with a proof of the original du Bois-Reymond lemma? I'm guessing the argument you are looking for is an imitation of the one used in the original proof. The key ingredient is that the covariant derivative respects the Riemannian metric on the surface. In other words, you can use the Leibniz rule.
$endgroup$
– Amitai Yuval
Jan 31 at 11:58
$begingroup$
The original proof uses mollifications of the test function which are again test functions in the sense that they stay compactly supported. This is crucial since the function $v$ we‘re looking at is not differentiable and we can roll the mollification onto $v$ which in turn is differentiable. The same would be needed here, since we don’t know if $v$ is covariantly differentiable along $/omega$ at all: Is there a possibility for a mollification $u_/varepsilon$ such that the $u_/varepsilon$ stays a vector field along the curve $omega$.
$endgroup$
– Nhat
Jan 31 at 15:52
$begingroup$
@AmitaiYuval If there was, I think Leibniz rule and the Fundamental Lemma would work.
$endgroup$
– Nhat
Jan 31 at 15:54
|
show 1 more comment
$begingroup$
I am trying to show the following extension of the Du Bois Raymond lemma:
Let $M$ be a smooth Riemannian Manifold and $omega: [0,1] rightarrow M$ be a $W^{1,2}$ curve on M. Consider a tangential $L^2$ vector field along $omega$ denoted by $v in L^2(omega^*TM)$.
If $$int_{0}^{1} langle v, frac{Du}{partial t} rangle text{ dt} = 0 text{for all } u in W ^{1,2}(omega^*TM) text{ with } u(0)=u(1)=0$$
Then
$$v in W^{1,2}(omega^*TM) text{ with } frac{Dv}{partial t} = 0 a.e.$$
where $frac{Du}{partial t}$ denotes the covariant derivative of $u$ along the curve $omega$.
For my purposes it would be sufficient to show this for the simplified case where $M = S^2$ and the covariant derivative becomes the projection onto the respective tangent space.
In particular (let $M=S^2$ from now on), I am baffled by my proof:
With $u$ being as above we have $$0 = int_{0}^{1} langle v, frac{Du}{partial t} rangle text{ dt} = int_0^1 langle v, u' rangle text{ dt} - int_0^1 langle v, langle u', omega rangle omega rangletext{ dt} $$
where the second expression vanishes since $v bot omega$ because $v(t)$ lies in the tangent space of $omega(t)$ to $S^2$ for all $t$ by definition. Hence
$$int_0^1 langle v, u' rangle text{ dt} = 0$$ and the regular Du Bois Raymond lemma for the ordinary derivatives as functions into $mathbb{R}^3$ becomes applicable and we have $v$ being constant (and ordinary differentiable) almost everywhere which would mean
$$v' equiv 0$$
and thus
$$frac{Dv}{partial t} = 0$$
Is this correct? I feel like I've somehow lost information on my $v$ at the end. The Du Bois Raymond yields that $v$ is constant, the extension yields that $v$ is parallel transported along the curve and this is not equivalent, isn't it?
Edit: As pointed out in the comments the proof goes wrong, when I use the classic Du-Bois-Raymond Lemma when the assumption is only tested for certain vector fields but not all as necessary. Still the extension has yet to be proven correctly.
proof-verification differential-geometry manifolds riemannian-geometry calculus-of-variations
asked Jan 30 at 21:37
NhatNhat
1,0261017
$endgroup$
I am trying to show the following extension of the Du Bois Raymond lemma:
Let $M$ be a smooth Riemannian Manifold and $omega: [0,1] rightarrow M$ be a $W^{1,2}$ curve on M. Consider a tangential $L^2$ vector field along $omega$ denoted by $v in L^2(omega^*TM)$.
If $$int_{0}^{1} langle v, frac{Du}{partial t} rangle text{ dt} = 0 text{for all } u in W ^{1,2}(omega^*TM) text{ with } u(0)=u(1)=0$$
Then
$$v in W^{1,2}(omega^*TM) text{ with } frac{Dv}{partial t} = 0 a.e.$$
where $frac{Du}{partial t}$ denotes the covariant derivative of $u$ along the curve $omega$.
For my purposes it would be sufficient to show this for the simplified case where $M = S^2$ and the covariant derivative becomes the projection onto the respective tangent space.
In particular (let $M=S^2$ from now on), I am baffled by my proof:
With $u$ being as above we have $$0 = int_{0}^{1} langle v, frac{Du}{partial t} rangle text{ dt} = int_0^1 langle v, u' rangle text{ dt} - int_0^1 langle v, langle u', omega rangle omega rangletext{ dt} $$
where the second expression vanishes since $v bot omega$ because $v(t)$ lies in the tangent space of $omega(t)$ to $S^2$ for all $t$ by definition. Hence
$$int_0^1 langle v, u' rangle text{ dt} = 0$$ and the regular Du Bois Raymond lemma for the ordinary derivatives as functions into $mathbb{R}^3$ becomes applicable and we have $v$ being constant (and ordinary differentiable) almost everywhere which would mean
$$v' equiv 0$$
and thus
$$frac{Dv}{partial t} = 0$$
Is this correct? I feel like I've somehow lost information on my $v$ at the end. The Du Bois Raymond yields that $v$ is constant, the extension yields that $v$ is parallel transported along the curve and this is not equivalent, isn't it?
Edit: As pointed out in the comments the proof goes wrong, when I use the classic Du-Bois-Raymond Lemma when the assumption is only tested for certain vector fields but not all as necessary. Still the extension has yet to be proven correctly.
proof-verification differential-geometry manifolds riemannian-geometry calculus-of-variations
proof-verification differential-geometry manifolds riemannian-geometry calculus-of-variations
asked Jan 30 at 21:37
NhatNhat
1,0261017
asked Jan 30 at 21:37
NhatNhat
1,0261017
edited Jan 31 at 10:21
Nhat
asked Jan 30 at 21:37
NhatNhat
1,0261017
asked Jan 30 at 21:37
NhatNhat
1,0261017
asked Jan 30 at 21:37
NhatNhat
1,0261017
1,0261017
$begingroup$
It seems to me that the regular du Bois-Reymond lemma is not applicable, as you have the equality only for vector fields $u$ which are tangent to the sphere. I guess you would need this equality to hold for every vector field $u$ in order to use the du Bois-Reymond lemma. This makes sense, as you noted yourself, as the vector field $v$ usually cannot be constant as a function to $mathbb{R}^3$.
$endgroup$
– Amitai Yuval
Jan 31 at 6:45
$begingroup$
@AmitaiYuval That is indeed correct and a good find. It also only makes sense since I am expecting $v$ to only be constant along $omega$ as well so I only have an assumptions made for the vector fields along $omega$. At least the „extension“ seems intuitive. I’m going to edit my OP with your input, thanks.
$endgroup$
– Nhat
Jan 31 at 10:18
$begingroup$
Are you familiar with a proof of the original du Bois-Reymond lemma? I'm guessing the argument you are looking for is an imitation of the one used in the original proof. The key ingredient is that the covariant derivative respects the Riemannian metric on the surface. In other words, you can use the Leibniz rule.
$endgroup$
– Amitai Yuval
Jan 31 at 11:58
$begingroup$
The original proof uses mollifications of the test function which are again test functions in the sense that they stay compactly supported. This is crucial since the function $v$ we‘re looking at is not differentiable and we can roll the mollification onto $v$ which in turn is differentiable. The same would be needed here, since we don’t know if $v$ is covariantly differentiable along $/omega$ at all: Is there a possibility for a mollification $u_/varepsilon$ such that the $u_/varepsilon$ stays a vector field along the curve $omega$.
$endgroup$
– Nhat
Jan 31 at 15:52
$begingroup$
@AmitaiYuval If there was, I think Leibniz rule and the Fundamental Lemma would work.
$endgroup$
– Nhat
Jan 31 at 15:54
|
show 1 more comment
$begingroup$
It seems to me that the regular du Bois-Reymond lemma is not applicable, as you have the equality only for vector fields $u$ which are tangent to the sphere. I guess you would need this equality to hold for every vector field $u$ in order to use the du Bois-Reymond lemma. This makes sense, as you noted yourself, as the vector field $v$ usually cannot be constant as a function to $mathbb{R}^3$.
$endgroup$
– Amitai Yuval
Jan 31 at 6:45
$begingroup$
@AmitaiYuval That is indeed correct and a good find. It also only makes sense since I am expecting $v$ to only be constant along $omega$ as well so I only have an assumptions made for the vector fields along $omega$. At least the „extension“ seems intuitive. I’m going to edit my OP with your input, thanks.
$endgroup$
– Nhat
Jan 31 at 10:18
$begingroup$
Are you familiar with a proof of the original du Bois-Reymond lemma? I'm guessing the argument you are looking for is an imitation of the one used in the original proof. The key ingredient is that the covariant derivative respects the Riemannian metric on the surface. In other words, you can use the Leibniz rule.
$endgroup$
– Amitai Yuval
Jan 31 at 11:58
$begingroup$
The original proof uses mollifications of the test function which are again test functions in the sense that they stay compactly supported. This is crucial since the function $v$ we‘re looking at is not differentiable and we can roll the mollification onto $v$ which in turn is differentiable. The same would be needed here, since we don’t know if $v$ is covariantly differentiable along $/omega$ at all: Is there a possibility for a mollification $u_/varepsilon$ such that the $u_/varepsilon$ stays a vector field along the curve $omega$.
$endgroup$
– Nhat
Jan 31 at 15:52
$begingroup$
@AmitaiYuval If there was, I think Leibniz rule and the Fundamental Lemma would work.
$endgroup$
– Nhat
Jan 31 at 15:54
$begingroup$
It seems to me that the regular du Bois-Reymond lemma is not applicable, as you have the equality only for vector fields $u$ which are tangent to the sphere. I guess you would need this equality to hold for every vector field $u$ in order to use the du Bois-Reymond lemma. This makes sense, as you noted yourself, as the vector field $v$ usually cannot be constant as a function to $mathbb{R}^3$.
$endgroup$
– Amitai Yuval
Jan 31 at 6:45
$begingroup$
It seems to me that the regular du Bois-Reymond lemma is not applicable, as you have the equality only for vector fields $u$ which are tangent to the sphere. I guess you would need this equality to hold for every vector field $u$ in order to use the du Bois-Reymond lemma. This makes sense, as you noted yourself, as the vector field $v$ usually cannot be constant as a function to $mathbb{R}^3$.
$endgroup$
– Amitai Yuval
Jan 31 at 6:45
$begingroup$
@AmitaiYuval That is indeed correct and a good find. It also only makes sense since I am expecting $v$ to only be constant along $omega$ as well so I only have an assumptions made for the vector fields along $omega$. At least the „extension“ seems intuitive. I’m going to edit my OP with your input, thanks.
$endgroup$
– Nhat
Jan 31 at 10:18
$begingroup$
@AmitaiYuval That is indeed correct and a good find. It also only makes sense since I am expecting $v$ to only be constant along $omega$ as well so I only have an assumptions made for the vector fields along $omega$. At least the „extension“ seems intuitive. I’m going to edit my OP with your input, thanks.
$endgroup$
– Nhat
Jan 31 at 10:18
$begingroup$
Are you familiar with a proof of the original du Bois-Reymond lemma? I'm guessing the argument you are looking for is an imitation of the one used in the original proof. The key ingredient is that the covariant derivative respects the Riemannian metric on the surface. In other words, you can use the Leibniz rule.
$endgroup$
– Amitai Yuval
Jan 31 at 11:58
$begingroup$
Are you familiar with a proof of the original du Bois-Reymond lemma? I'm guessing the argument you are looking for is an imitation of the one used in the original proof. The key ingredient is that the covariant derivative respects the Riemannian metric on the surface. In other words, you can use the Leibniz rule.
$endgroup$
– Amitai Yuval
Jan 31 at 11:58
$begingroup$
The original proof uses mollifications of the test function which are again test functions in the sense that they stay compactly supported. This is crucial since the function $v$ we‘re looking at is not differentiable and we can roll the mollification onto $v$ which in turn is differentiable. The same would be needed here, since we don’t know if $v$ is covariantly differentiable along $/omega$ at all: Is there a possibility for a mollification $u_/varepsilon$ such that the $u_/varepsilon$ stays a vector field along the curve $omega$.
$endgroup$
– Nhat
Jan 31 at 15:52
$begingroup$
The original proof uses mollifications of the test function which are again test functions in the sense that they stay compactly supported. This is crucial since the function $v$ we‘re looking at is not differentiable and we can roll the mollification onto $v$ which in turn is differentiable. The same would be needed here, since we don’t know if $v$ is covariantly differentiable along $/omega$ at all: Is there a possibility for a mollification $u_/varepsilon$ such that the $u_/varepsilon$ stays a vector field along the curve $omega$.
$endgroup$
– Nhat
Jan 31 at 15:52
$begingroup$
@AmitaiYuval If there was, I think Leibniz rule and the Fundamental Lemma would work.
$endgroup$
– Nhat
Jan 31 at 15:54
$begingroup$
@AmitaiYuval If there was, I think Leibniz rule and the Fundamental Lemma would work.
$endgroup$
– Nhat
Jan 31 at 15:54
|
show 1 more comment
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$begingroup$
It seems to me that the regular du Bois-Reymond lemma is not applicable, as you have the equality only for vector fields $u$ which are tangent to the sphere. I guess you would need this equality to hold for every vector field $u$ in order to use the du Bois-Reymond lemma. This makes sense, as you noted yourself, as the vector field $v$ usually cannot be constant as a function to $mathbb{R}^3$.
$endgroup$
– Amitai Yuval
Jan 31 at 6:45
$begingroup$
@AmitaiYuval That is indeed correct and a good find. It also only makes sense since I am expecting $v$ to only be constant along $omega$ as well so I only have an assumptions made for the vector fields along $omega$. At least the „extension“ seems intuitive. I’m going to edit my OP with your input, thanks.
$endgroup$
– Nhat
Jan 31 at 10:18
$begingroup$
Are you familiar with a proof of the original du Bois-Reymond lemma? I'm guessing the argument you are looking for is an imitation of the one used in the original proof. The key ingredient is that the covariant derivative respects the Riemannian metric on the surface. In other words, you can use the Leibniz rule.
$endgroup$
– Amitai Yuval
Jan 31 at 11:58
$begingroup$
The original proof uses mollifications of the test function which are again test functions in the sense that they stay compactly supported. This is crucial since the function $v$ we‘re looking at is not differentiable and we can roll the mollification onto $v$ which in turn is differentiable. The same would be needed here, since we don’t know if $v$ is covariantly differentiable along $/omega$ at all: Is there a possibility for a mollification $u_/varepsilon$ such that the $u_/varepsilon$ stays a vector field along the curve $omega$.
$endgroup$
– Nhat
Jan 31 at 15:52
$begingroup$
@AmitaiYuval If there was, I think Leibniz rule and the Fundamental Lemma would work.
$endgroup$
– Nhat
Jan 31 at 15:54