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Expansion of scalar function times laplace's equation

  1. Aug 20, 2014 #1

    f \nabla^2 f = \nabla \cdot f \nabla f - \nabla f \cdot \nabla f

    where f is a scalar function.

    Can someone please show me why this is step by step.
    Feel free to use suffix notation.

    Thanks in advance.
  2. jcsd
  3. Aug 20, 2014 #2
    Are you familiar with the product rules of the del operator?
  4. Aug 20, 2014 #3
    Yes i am. I think i got the answer. I believe the appropriate rule is:

    \nabla \cdot \vec{μ} f = \nabla f \cdot \vec{μ} + f \nabla \cdot \vec{μ}

    You need to start with


    \nabla \cdot f\nabla f = \nabla f \cdot \nabla f + f(\nabla \cdot \nabla f)


    which becomes:
    \nabla \cdot f\nabla f - \nabla f \cdot \nabla f = f(\nabla \cdot \nabla f)

    Thank you for your help. :smile:
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