Integrating an inner product by parts
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I am trying to integrate the following:
$$ int langle F, -(partial_u kappa) Trangle du $$
where $F: S^1 times [0,T) to mathbb{R}^2$ is a simple closed plane curve given
by spatial parameter $u$ (mod $2pi$) and time parameter $t$. The paper I
am reading states that integrating the above by parts yields:
$$ int langle partial_u F, kappa Trangle + langle F, kappa partial_u Trangle du $$
I am not even sure how you would integrate an inner product, let alone integrate one by parts. Any help is appreciated.
Edit: Thought I would share some of the work I have done here. The closest I have come to solving this is getting the equation
$$ partial_u langle F, kappa Trangle = langle partial_u F, kappa Trangle + langle F, (partial_u kappa)Trangle + langle F,kappapartial_u Trangle $$
which gives us:
$$ langle F, -(partial_u kappa)Trangle + partial_u langle F, kappa Trangle = langle partial_u F, kappa Trangle + + langle F,kappapartial_u Trangle $$
But I have an extra term and haven't done any integration.
2nd edit: Thought I would link the paper just in case anyone wanted to get some more context. https://projecteuclid.org/download/pdf_1/euclid.jdg/1214439902. Page 7 Lemma 3.1.7
differential-geometry mean-curvature-flows
asked May 11 at 3:27
Anmol Bhullar
19611
add a comment |
up vote
1
down vote
favorite
I am trying to integrate the following:
$$ int langle F, -(partial_u kappa) Trangle du $$
where $F: S^1 times [0,T) to mathbb{R}^2$ is a simple closed plane curve given
by spatial parameter $u$ (mod $2pi$) and time parameter $t$. The paper I
am reading states that integrating the above by parts yields:
$$ int langle partial_u F, kappa Trangle + langle F, kappa partial_u Trangle du $$
I am not even sure how you would integrate an inner product, let alone integrate one by parts. Any help is appreciated.
Edit: Thought I would share some of the work I have done here. The closest I have come to solving this is getting the equation
$$ partial_u langle F, kappa Trangle = langle partial_u F, kappa Trangle + langle F, (partial_u kappa)Trangle + langle F,kappapartial_u Trangle $$
which gives us:
$$ langle F, -(partial_u kappa)Trangle + partial_u langle F, kappa Trangle = langle partial_u F, kappa Trangle + + langle F,kappapartial_u Trangle $$
But I have an extra term and haven't done any integration.
2nd edit: Thought I would link the paper just in case anyone wanted to get some more context. https://projecteuclid.org/download/pdf_1/euclid.jdg/1214439902. Page 7 Lemma 3.1.7
differential-geometry mean-curvature-flows
asked May 11 at 3:27
Anmol Bhullar
19611
1
Your work looks good. Now take integrals of everything. Do you know the general Stokes' Theorem?
– Jesse Madnick
May 11 at 8:19
If I take the integral of that equation do I not end up with an extra term? (Namely $partial_u langle F,kappa Trangle$).
– Anmol Bhullar
May 11 at 15:35
Yes, but that extra integral equals zero (by Stokes' Theorem).
– Jesse Madnick
May 11 at 18:57
I am not very familiar with differential forms sorry. What would be $omega$ in this case? (Following the notation from the wiki page you linked)
– Anmol Bhullar
May 11 at 19:22
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to integrate the following:
$$ int langle F, -(partial_u kappa) Trangle du $$
where $F: S^1 times [0,T) to mathbb{R}^2$ is a simple closed plane curve given
by spatial parameter $u$ (mod $2pi$) and time parameter $t$. The paper I
am reading states that integrating the above by parts yields:
$$ int langle partial_u F, kappa Trangle + langle F, kappa partial_u Trangle du $$
I am not even sure how you would integrate an inner product, let alone integrate one by parts. Any help is appreciated.
Edit: Thought I would share some of the work I have done here. The closest I have come to solving this is getting the equation
$$ partial_u langle F, kappa Trangle = langle partial_u F, kappa Trangle + langle F, (partial_u kappa)Trangle + langle F,kappapartial_u Trangle $$
which gives us:
$$ langle F, -(partial_u kappa)Trangle + partial_u langle F, kappa Trangle = langle partial_u F, kappa Trangle + + langle F,kappapartial_u Trangle $$
But I have an extra term and haven't done any integration.
2nd edit: Thought I would link the paper just in case anyone wanted to get some more context. https://projecteuclid.org/download/pdf_1/euclid.jdg/1214439902. Page 7 Lemma 3.1.7
differential-geometry mean-curvature-flows
asked May 11 at 3:27
Anmol Bhullar
19611
I am trying to integrate the following:
$$ int langle F, -(partial_u kappa) Trangle du $$
where $F: S^1 times [0,T) to mathbb{R}^2$ is a simple closed plane curve given
by spatial parameter $u$ (mod $2pi$) and time parameter $t$. The paper I
am reading states that integrating the above by parts yields:
$$ int langle partial_u F, kappa Trangle + langle F, kappa partial_u Trangle du $$
I am not even sure how you would integrate an inner product, let alone integrate one by parts. Any help is appreciated.
Edit: Thought I would share some of the work I have done here. The closest I have come to solving this is getting the equation
$$ partial_u langle F, kappa Trangle = langle partial_u F, kappa Trangle + langle F, (partial_u kappa)Trangle + langle F,kappapartial_u Trangle $$
which gives us:
$$ langle F, -(partial_u kappa)Trangle + partial_u langle F, kappa Trangle = langle partial_u F, kappa Trangle + + langle F,kappapartial_u Trangle $$
But I have an extra term and haven't done any integration.
2nd edit: Thought I would link the paper just in case anyone wanted to get some more context. https://projecteuclid.org/download/pdf_1/euclid.jdg/1214439902. Page 7 Lemma 3.1.7
differential-geometry mean-curvature-flows
differential-geometry mean-curvature-flows
asked May 11 at 3:27
Anmol Bhullar
19611
asked May 11 at 3:27
Anmol Bhullar
19611
edited May 11 at 7:41
user99914
asked May 11 at 3:27
Anmol Bhullar
19611
asked May 11 at 3:27
Anmol Bhullar
19611
asked May 11 at 3:27
Anmol Bhullar
19611
19611
1
Your work looks good. Now take integrals of everything. Do you know the general Stokes' Theorem?
– Jesse Madnick
May 11 at 8:19
If I take the integral of that equation do I not end up with an extra term? (Namely $partial_u langle F,kappa Trangle$).
– Anmol Bhullar
May 11 at 15:35
Yes, but that extra integral equals zero (by Stokes' Theorem).
– Jesse Madnick
May 11 at 18:57
I am not very familiar with differential forms sorry. What would be $omega$ in this case? (Following the notation from the wiki page you linked)
– Anmol Bhullar
May 11 at 19:22
add a comment |
1
Your work looks good. Now take integrals of everything. Do you know the general Stokes' Theorem?
– Jesse Madnick
May 11 at 8:19
If I take the integral of that equation do I not end up with an extra term? (Namely $partial_u langle F,kappa Trangle$).
– Anmol Bhullar
May 11 at 15:35
Yes, but that extra integral equals zero (by Stokes' Theorem).
– Jesse Madnick
May 11 at 18:57
I am not very familiar with differential forms sorry. What would be $omega$ in this case? (Following the notation from the wiki page you linked)
– Anmol Bhullar
May 11 at 19:22
1
1
Your work looks good. Now take integrals of everything. Do you know the general Stokes' Theorem?
– Jesse Madnick
May 11 at 8:19
Your work looks good. Now take integrals of everything. Do you know the general Stokes' Theorem?
– Jesse Madnick
May 11 at 8:19
If I take the integral of that equation do I not end up with an extra term? (Namely $partial_u langle F,kappa Trangle$).
– Anmol Bhullar
May 11 at 15:35
If I take the integral of that equation do I not end up with an extra term? (Namely $partial_u langle F,kappa Trangle$).
– Anmol Bhullar
May 11 at 15:35
Yes, but that extra integral equals zero (by Stokes' Theorem).
– Jesse Madnick
May 11 at 18:57
Yes, but that extra integral equals zero (by Stokes' Theorem).
– Jesse Madnick
May 11 at 18:57
I am not very familiar with differential forms sorry. What would be $omega$ in this case? (Following the notation from the wiki page you linked)
– Anmol Bhullar
May 11 at 19:22
I am not very familiar with differential forms sorry. What would be $omega$ in this case? (Following the notation from the wiki page you linked)
– Anmol Bhullar
May 11 at 19:22
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
We have a simple closed curve in the plane $F colon mathbb{S}^1 times I to mathbb{R}^2$ that is evolving in time $t in I$ according to the heat equation. We are letting $u$ denote the spatial parameter. We let $T$ and $N$ be the unit tangent and inward-pointing normal vectors, respectively.
The OP has correctly calculated that
$$frac{partial}{partial u} langle F, kappa T rangle = leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, frac{partial kappa}{partial u} T rightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$leftlangle F, -frac{partial kappa}{partial u} T rightrangle = - frac{partial}{partial u} langle F, kappa Trangle + leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$int_0^{2pi}leftlangle F, -frac{partial kappa}{partial u} T rightrangle du = - int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du + int_0^{2pi}leftlangle frac{partial F}{partial u}, kappa Trightrangle du + int_0^{2pi}leftlangle F, kappa frac{partial T}{partial u} right rangle du.$$
By the Fundamental Theorem of Calculus, the first integral on the right side is zero (because we're working with a simple closed curve):
$$int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du = left. frac{}{} langle F, kappa Trangle right|^{2pi}_0 = langle F(2pi), kappa(2pi)T(2pi) rangle - langle F(0), kappa(0)T(0) rangle = 0.$$
answered May 11 at 23:42
Jesse Madnick
19.4k562122
Thank you so much for this. I didn't realize I almost had it. Also, you are using $tau$ sometimes. Do you mean $T$?
– Anmol Bhullar
May 12 at 9:45
Yes, that was a consistent typo; it's fixed now.
– Jesse Madnick
May 12 at 22:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have a simple closed curve in the plane $F colon mathbb{S}^1 times I to mathbb{R}^2$ that is evolving in time $t in I$ according to the heat equation. We are letting $u$ denote the spatial parameter. We let $T$ and $N$ be the unit tangent and inward-pointing normal vectors, respectively.
The OP has correctly calculated that
$$frac{partial}{partial u} langle F, kappa T rangle = leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, frac{partial kappa}{partial u} T rightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$leftlangle F, -frac{partial kappa}{partial u} T rightrangle = - frac{partial}{partial u} langle F, kappa Trangle + leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$int_0^{2pi}leftlangle F, -frac{partial kappa}{partial u} T rightrangle du = - int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du + int_0^{2pi}leftlangle frac{partial F}{partial u}, kappa Trightrangle du + int_0^{2pi}leftlangle F, kappa frac{partial T}{partial u} right rangle du.$$
By the Fundamental Theorem of Calculus, the first integral on the right side is zero (because we're working with a simple closed curve):
$$int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du = left. frac{}{} langle F, kappa Trangle right|^{2pi}_0 = langle F(2pi), kappa(2pi)T(2pi) rangle - langle F(0), kappa(0)T(0) rangle = 0.$$
answered May 11 at 23:42
Jesse Madnick
19.4k562122
Thank you so much for this. I didn't realize I almost had it. Also, you are using $tau$ sometimes. Do you mean $T$?
– Anmol Bhullar
May 12 at 9:45
Yes, that was a consistent typo; it's fixed now.
– Jesse Madnick
May 12 at 22:42
add a comment |
up vote
1
down vote
accepted
We have a simple closed curve in the plane $F colon mathbb{S}^1 times I to mathbb{R}^2$ that is evolving in time $t in I$ according to the heat equation. We are letting $u$ denote the spatial parameter. We let $T$ and $N$ be the unit tangent and inward-pointing normal vectors, respectively.
The OP has correctly calculated that
$$frac{partial}{partial u} langle F, kappa T rangle = leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, frac{partial kappa}{partial u} T rightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$leftlangle F, -frac{partial kappa}{partial u} T rightrangle = - frac{partial}{partial u} langle F, kappa Trangle + leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$int_0^{2pi}leftlangle F, -frac{partial kappa}{partial u} T rightrangle du = - int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du + int_0^{2pi}leftlangle frac{partial F}{partial u}, kappa Trightrangle du + int_0^{2pi}leftlangle F, kappa frac{partial T}{partial u} right rangle du.$$
By the Fundamental Theorem of Calculus, the first integral on the right side is zero (because we're working with a simple closed curve):
$$int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du = left. frac{}{} langle F, kappa Trangle right|^{2pi}_0 = langle F(2pi), kappa(2pi)T(2pi) rangle - langle F(0), kappa(0)T(0) rangle = 0.$$
answered May 11 at 23:42
Jesse Madnick
19.4k562122
Thank you so much for this. I didn't realize I almost had it. Also, you are using $tau$ sometimes. Do you mean $T$?
– Anmol Bhullar
May 12 at 9:45
Yes, that was a consistent typo; it's fixed now.
– Jesse Madnick
May 12 at 22:42
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have a simple closed curve in the plane $F colon mathbb{S}^1 times I to mathbb{R}^2$ that is evolving in time $t in I$ according to the heat equation. We are letting $u$ denote the spatial parameter. We let $T$ and $N$ be the unit tangent and inward-pointing normal vectors, respectively.
The OP has correctly calculated that
$$frac{partial}{partial u} langle F, kappa T rangle = leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, frac{partial kappa}{partial u} T rightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$leftlangle F, -frac{partial kappa}{partial u} T rightrangle = - frac{partial}{partial u} langle F, kappa Trangle + leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$int_0^{2pi}leftlangle F, -frac{partial kappa}{partial u} T rightrangle du = - int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du + int_0^{2pi}leftlangle frac{partial F}{partial u}, kappa Trightrangle du + int_0^{2pi}leftlangle F, kappa frac{partial T}{partial u} right rangle du.$$
By the Fundamental Theorem of Calculus, the first integral on the right side is zero (because we're working with a simple closed curve):
$$int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du = left. frac{}{} langle F, kappa Trangle right|^{2pi}_0 = langle F(2pi), kappa(2pi)T(2pi) rangle - langle F(0), kappa(0)T(0) rangle = 0.$$
answered May 11 at 23:42
Jesse Madnick
19.4k562122
We have a simple closed curve in the plane $F colon mathbb{S}^1 times I to mathbb{R}^2$ that is evolving in time $t in I$ according to the heat equation. We are letting $u$ denote the spatial parameter. We let $T$ and $N$ be the unit tangent and inward-pointing normal vectors, respectively.
The OP has correctly calculated that
$$frac{partial}{partial u} langle F, kappa T rangle = leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, frac{partial kappa}{partial u} T rightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$leftlangle F, -frac{partial kappa}{partial u} T rightrangle = - frac{partial}{partial u} langle F, kappa Trangle + leftlangle frac{partial F}{partial u}, kappa Trightrangle + leftlangle F, kappa frac{partial T}{partial u} right rangle!,$$
whence
$$int_0^{2pi}leftlangle F, -frac{partial kappa}{partial u} T rightrangle du = - int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du + int_0^{2pi}leftlangle frac{partial F}{partial u}, kappa Trightrangle du + int_0^{2pi}leftlangle F, kappa frac{partial T}{partial u} right rangle du.$$
By the Fundamental Theorem of Calculus, the first integral on the right side is zero (because we're working with a simple closed curve):
$$int_0^{2pi}frac{partial}{partial u} langle F, kappa Trangle, du = left. frac{}{} langle F, kappa Trangle right|^{2pi}_0 = langle F(2pi), kappa(2pi)T(2pi) rangle - langle F(0), kappa(0)T(0) rangle = 0.$$
answered May 11 at 23:42
Jesse Madnick
19.4k562122
edited Nov 20 at 18:23
answered May 11 at 23:42
Jesse Madnick
19.4k562122
answered May 11 at 23:42
Jesse Madnick
19.4k562122
answered May 11 at 23:42
Jesse Madnick
19.4k562122
19.4k562122
Thank you so much for this. I didn't realize I almost had it. Also, you are using $tau$ sometimes. Do you mean $T$?
– Anmol Bhullar
May 12 at 9:45
Yes, that was a consistent typo; it's fixed now.
– Jesse Madnick
May 12 at 22:42
add a comment |
Thank you so much for this. I didn't realize I almost had it. Also, you are using $tau$ sometimes. Do you mean $T$?
– Anmol Bhullar
May 12 at 9:45
Yes, that was a consistent typo; it's fixed now.
– Jesse Madnick
May 12 at 22:42
Thank you so much for this. I didn't realize I almost had it. Also, you are using $tau$ sometimes. Do you mean $T$?
– Anmol Bhullar
May 12 at 9:45
Thank you so much for this. I didn't realize I almost had it. Also, you are using $tau$ sometimes. Do you mean $T$?
– Anmol Bhullar
May 12 at 9:45
Yes, that was a consistent typo; it's fixed now.
– Jesse Madnick
May 12 at 22:42
Yes, that was a consistent typo; it's fixed now.
– Jesse Madnick
May 12 at 22:42
add a comment |
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1
Your work looks good. Now take integrals of everything. Do you know the general Stokes' Theorem?
– Jesse Madnick
May 11 at 8:19
If I take the integral of that equation do I not end up with an extra term? (Namely $partial_u langle F,kappa Trangle$).
– Anmol Bhullar
May 11 at 15:35
Yes, but that extra integral equals zero (by Stokes' Theorem).
– Jesse Madnick
May 11 at 18:57
I am not very familiar with differential forms sorry. What would be $omega$ in this case? (Following the notation from the wiki page you linked)
– Anmol Bhullar
May 11 at 19:22