1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laplacian of dyad

  1. Oct 3, 2016 #1
    1. The problem statement, all variables and given/known data

    Given the dyad formed by two arbitrary position vector fields, u and v, use indicial notation in Cartesian coordinates to prove:

    $$\nabla^2 ({\vec u \vec v}) = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v} + 2\nabla {\vec u} \cdot {(\nabla \vec v)}^T

    2. Relevant equations

    Per my professor's notes, the Laplacian of a dyad (also a tensor) is given as:
    \nabla^2 {\mathbf {S}} = \nabla \cdot {S_{ij,k} \mathbf{e_{i}e_{j}e_{k}}} = S_{ij,kk} \mathbf{e_{i}e_{j}}

    3. The attempt at a solution

    \nabla^2 {\mathbf {uv}} = (u_{i}v_{j})_{,kk} = u_{i,kk}v_{j} + u_{i}v_{j,kk} \\
    u_{i,kk}v_{j} + u_{i}v_{j,kk} = \vec v \nabla^2 {\vec u} + \vec u \nabla^2 {\vec v}

    I don't know where the following terms come from:
    2\nabla {\vec u} \cdot {(\nabla \vec v)}^T

    Does anyone have any suggestions? I feel that I am missing a step or something.
  2. jcsd
  3. Oct 3, 2016 #2


    User Avatar
    2017 Award

    Staff: Mentor

    I'm not good with coordinates, so I leave this up to you.
    But ##\nabla (\mathbf{u}\mathbf{v}) = \mathbf{u}(\nabla \mathbf{v}) + (\nabla \mathbf{u})\mathbf{v}## and the next differentiation gives you the third term (twice).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Laplacian dyad Date
Laplacian in [0,L] x [0, ∞] Jul 9, 2017
Converting Laplacian to polar coordinates Nov 18, 2016
Laplacian of vector proof Aug 30, 2016
Laplacian of the value function Jun 17, 2016