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Question on the Laplace Beltrami operator expressed by mean curvature vector: Missing term
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Disclaimer: The only course that I have seen in differential geometry is an introduction to differential geometry of manifolds and so I've never dealed with Laplace-Beltrami operator in the past.
I currently started to study a paper and at a point I came across with the following:
Let $Gamma$ be a $C^2-$regular boundary of an open, bounded and
connected subset of $mathbb R^3$. Define the function: $widetilde u:
B(x_0,r) cap Gamma to mathbb R$ as: $widetilde u=frac{(alpha
+1)theta}{16}{vert x-x_0vert}^2$
Then the Laplace-Beltrami operator of $widetilde u$ is given by:
$-Delta widetilde u= frac{(alpha +1)theta}{16}(-4- H(x) cdot
(x-x_0))(*)$ where $H$ denotes the mean curvature vector of $Gamma$.
After some research I found the following formula for calculating the Laplace-Beltrami operator:
$Delta_{Gamma} f=Delta f-{nabla}^2 f(eta,eta)+H_{Gamma}nabla
f$ where $Delta$ is the usual Euclidean Laplacian and ${nabla}^2 $
denotes the hessian.
I would really appreciate if somebody could explain to me why the term ${nabla}^2 widetilde u (eta,eta)$ is missing from $(*)$.
Thanks in advance!
multivariable-calculus differential-geometry laplacian
asked yesterday
kaithkolesidou
946411
add a comment |
up vote
1
down vote
favorite
Disclaimer: The only course that I have seen in differential geometry is an introduction to differential geometry of manifolds and so I've never dealed with Laplace-Beltrami operator in the past.
I currently started to study a paper and at a point I came across with the following:
Let $Gamma$ be a $C^2-$regular boundary of an open, bounded and
connected subset of $mathbb R^3$. Define the function: $widetilde u:
B(x_0,r) cap Gamma to mathbb R$ as: $widetilde u=frac{(alpha
+1)theta}{16}{vert x-x_0vert}^2$
Then the Laplace-Beltrami operator of $widetilde u$ is given by:
$-Delta widetilde u= frac{(alpha +1)theta}{16}(-4- H(x) cdot
(x-x_0))(*)$ where $H$ denotes the mean curvature vector of $Gamma$.
After some research I found the following formula for calculating the Laplace-Beltrami operator:
$Delta_{Gamma} f=Delta f-{nabla}^2 f(eta,eta)+H_{Gamma}nabla
f$ where $Delta$ is the usual Euclidean Laplacian and ${nabla}^2 $
denotes the hessian.
I would really appreciate if somebody could explain to me why the term ${nabla}^2 widetilde u (eta,eta)$ is missing from $(*)$.
Thanks in advance!
multivariable-calculus differential-geometry laplacian
asked yesterday
kaithkolesidou
946411
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Disclaimer: The only course that I have seen in differential geometry is an introduction to differential geometry of manifolds and so I've never dealed with Laplace-Beltrami operator in the past.
I currently started to study a paper and at a point I came across with the following:
Let $Gamma$ be a $C^2-$regular boundary of an open, bounded and
connected subset of $mathbb R^3$. Define the function: $widetilde u:
B(x_0,r) cap Gamma to mathbb R$ as: $widetilde u=frac{(alpha
+1)theta}{16}{vert x-x_0vert}^2$
Then the Laplace-Beltrami operator of $widetilde u$ is given by:
$-Delta widetilde u= frac{(alpha +1)theta}{16}(-4- H(x) cdot
(x-x_0))(*)$ where $H$ denotes the mean curvature vector of $Gamma$.
After some research I found the following formula for calculating the Laplace-Beltrami operator:
$Delta_{Gamma} f=Delta f-{nabla}^2 f(eta,eta)+H_{Gamma}nabla
f$ where $Delta$ is the usual Euclidean Laplacian and ${nabla}^2 $
denotes the hessian.
I would really appreciate if somebody could explain to me why the term ${nabla}^2 widetilde u (eta,eta)$ is missing from $(*)$.
Thanks in advance!
multivariable-calculus differential-geometry laplacian
asked yesterday
kaithkolesidou
946411
Disclaimer: The only course that I have seen in differential geometry is an introduction to differential geometry of manifolds and so I've never dealed with Laplace-Beltrami operator in the past.
I currently started to study a paper and at a point I came across with the following:
Let $Gamma$ be a $C^2-$regular boundary of an open, bounded and
connected subset of $mathbb R^3$. Define the function: $widetilde u:
B(x_0,r) cap Gamma to mathbb R$ as: $widetilde u=frac{(alpha
+1)theta}{16}{vert x-x_0vert}^2$
Then the Laplace-Beltrami operator of $widetilde u$ is given by:
$-Delta widetilde u= frac{(alpha +1)theta}{16}(-4- H(x) cdot
(x-x_0))(*)$ where $H$ denotes the mean curvature vector of $Gamma$.
After some research I found the following formula for calculating the Laplace-Beltrami operator:
$Delta_{Gamma} f=Delta f-{nabla}^2 f(eta,eta)+H_{Gamma}nabla
f$ where $Delta$ is the usual Euclidean Laplacian and ${nabla}^2 $
denotes the hessian.
I would really appreciate if somebody could explain to me why the term ${nabla}^2 widetilde u (eta,eta)$ is missing from $(*)$.
Thanks in advance!
multivariable-calculus differential-geometry laplacian
multivariable-calculus differential-geometry laplacian
asked yesterday
kaithkolesidou
946411
asked yesterday
kaithkolesidou
946411
edited 20 hours ago
asked yesterday
kaithkolesidou
946411
asked yesterday
kaithkolesidou
946411
asked yesterday
kaithkolesidou
946411
946411
add a comment |
add a comment |
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