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A Div(gradΦ) question?

  1. Apr 3, 2016 #1
    Φ=3y12+2y22-3y1y32

    Find: div(gradΦ).

    I broke it down to find the partial derivative of all 3 coordinates y1,y2 and y3, so using the continuity equation:

    y1=6y1-3y32
    y2=4y2
    y3=6y3y1

    However, I'm not sure if this correct, would the first part give the gradient? Could someone confirm or solve this?

    Regards
     
  2. jcsd
  3. Apr 3, 2016 #2
    what is the gradient of a function ? its the slope of the curve representing the function/

    divergence also has a meaning in calculus.
     
  4. Apr 3, 2016 #3

    SteamKing

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    The divergence of the gradient of a function is also known as the Laplacian of that function:

    https://en.wikipedia.org/wiki/Laplace_operator

    What you have calculated so far is:

    $$\frac{∂Φ}{∂y_1} = 6y_1-3y_3^2$$
    $$\frac{∂Φ}{∂y_2}=4y_2$$
    $$\frac{∂Φ}{∂y_3}=-6y_3y_1$$

    or

    ##∇Φ = (\frac{∂Φ}{∂y_1},\frac{∂Φ}{∂y_2},\frac{∂Φ}{∂y_3}) = (6y_1-3y_3^2,4y_2,-6y_1y_3)##

    The final step is

    ##div(grad Φ) = ∇ ⋅ ∇Φ = \frac{∂^2Φ}{∂y_1^2}+\frac{∂^2Φ}{∂y_2^2}+\frac{∂^2Φ}{∂y_3^2}##

    (Note: edited to fix sign error in third term.)
     
    Last edited: Apr 3, 2016
  5. Apr 3, 2016 #4
    Just to add, I believe you should get 10-6y1.

    I think you forgot a negative sign there.
     
  6. Apr 5, 2016 #5
    You wrote
    ##y_1=6y_1-3y_3^2##
    when you should have written
    ##\frac{\partial\Psi}{\partial y_1}=6y_1-3y_3^2##.
    I don't know if you made a mistake in understanding or used a lazy abuse of notation, but you really need to avoid misuse of the equality expression. If you want to write it as a semantic pairing of the left and right, but not an equality, you are better off using a colon or comma, or making a table.
     
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