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Prove the Laplacian of Function g is equal to g

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-2-17_16-33-47.png

    2. Relevant equations
    the gradient of g is (d/dx,d/dy,d/dz)
    the divergence of g is d/dx+d/dy+d/dz

    3. The attempt at a solution
    When I run through even using only a few terms to see if I can get the final result of it equaling g I end up with u^2 terms as coefficients and this seems to me as if it is no longer equal.
     
  2. jcsd
  3. Feb 17, 2016 #2

    Samy_A

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    1) Can you show what you get as result?
    2) What does it mean that u is a unit vector?
    (3) In relevant equations, that should actually be partial derivatives.)
     
    Last edited by a moderator: Feb 17, 2016
  4. Feb 17, 2016 #3
    1) u1u2...un*e^(u1x1+u2x2+...unxn)
    2) u is a unit vector as in it can be any vector with a magnitude of 1
     
  5. Feb 17, 2016 #4

    Samy_A

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    How did you get that?
    Let's start with the beginning: what is ## \displaystyle \frac {\partial g}{\partial x_1}##?
     
  6. Feb 17, 2016 #5
    A partial derivative of g with respect to x. So we take the gradient and it becomes grad(g)=(u1e^(u1x1)+u2e^u2x2....) and so on. The thing I gave previously is after taking the divergence of the gradient.
     
  7. Feb 17, 2016 #6
    upload_2016-2-17_16-48-2.png
    Its fine now. In case anyone else needs this for future reference.
     
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