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







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    up vote
    1
    down vote

    favorite
    2












    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







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite
      2









      up vote
      1
      down vote

      favorite
      2






      2





      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







      share|cite|improve this question















      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






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      edited 20 hours ago

























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      kaithkolesidou

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