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Vector Calculus Proof Help Please

  1. May 23, 2004 #1
    Vector Calculus Proof Help Please :)

    Heya Ppl i have a problem i am trying to solve.

    Prove that

    (Delta) . ( (fi)F) = (fi)(Delta) . F + F . (Delta)(fi)

    were these contain GRAD DIV in my opinion but i seem to not be able to get the answer.

    F = Vector F where F = F1i + F2j + F3k is a vector field in R3 and (fi) the Greek symbol that looks similar to Theta is a fuction of x, y and z.
  2. jcsd
  3. May 23, 2004 #2
    oh and also guys the . is like the DOT PRODUCT dot not a MULTIPLICATION
  4. May 23, 2004 #3
    Do you mean:

    [tex]\nabla \cdot (\phi \mathbf{F}) = (\phi \nabla) \cdot \mathbf{F} + \mathbf{F} \cdot (\nabla \phi)[/tex]

    If you're interested, you can click on the equation image to see what code was used to make it.

    Also, the upside-down triangle representing the del operator is called "nabla," and the greek letter that looks like theta is called "phi."
    Last edited: May 23, 2004
  5. May 23, 2004 #4
    Yeh TALewis that is how it looked like except that the BRACKETS wernt around the NABLA and PHI which are in the Right Hand Side, but i think that is a okay way to group it. Now if any1 has ideas on how to solve it i would much appreciate it Thanks
  6. May 23, 2004 #5
    And um the question now after editing your Code looked like this

    [tex]\nabla \cdot (\phi \mathbf{F}) = \phi \nabla \cdot \mathbf{F} + \mathbf{F} \cdot \nabla \phi[/tex]

    But i think the way grouped in your rewrite should be the same thing, now if sum1 knows how to solve that proof :)
  7. May 24, 2004 #6
    I will try to prove it considering the x-direction only. The full result should follow easily in the other two dimensions.

    First, the left hand side:

    \nabla \cdot (\phi \mathbf{F}) &=
    \frac{\partial}{\partial x}\mathbf{i} \cdot \phi F_1 \mathbf{i}\\
    &= \frac{\partial}{\partial x}(\phi F_1)\\
    &= \phi\frac{\partial F_1}{\partial x} + F_1\frac{\partial \phi}{\partial x} \quad \mbox{(product rule)}

    Now, the first term of the right hand side:

    (\phi\nabla)\cdot\mathbf{F} &=
    \phi\frac{\partial}{\partial x}\mathbf{i} \cdot F_1\mathbf{i}\\
    &= \phi\frac{\partial F_1}{\partial x}

    The second term of the right hand side:

    \mathbf{F}\cdot(\nabla\phi) &=
    F_1\mathbf{i}\cdot\frac{\partial\phi}{\partial x}\mathbf{i}\\
    &= F_1\frac{\partial\phi}{\partial x}

    I think you should be able to see now how it all comes together.
  8. May 24, 2004 #7
    Thanks Heaps TALewis, i get the picture now! :):)
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