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Using the Divergence Theorem to Prove Green's Theorem

  1. Feb 26, 2015 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    Prove Green's theorem
    [tex]
    \int_{\tau} (\varphi \nabla^{2} \psi -\psi\nabla^{2}\varphi)d\tau = \int_{\sigma}(\varphi\nabla\psi
    -\psi\nabla\varphi)\cdot d\vec{\sigma}[/tex]

    2. Relevant equations
    [tex] div (\vec{V})=\lim_{\Delta\tau\rightarrow 0} \frac{1}{\Delta\tau} \int_{\sigma} \vec{V} \cdot d\vec{\sigma} [/tex]
    is the divergence, and Gauss's divergence theorem says
    [tex]\int_{\tau} div(\vec{V}) d\tau=\int_{\sigma} V \cdot d\vec{\sigma} [/tex]
    [tex] div(\vec{V})=\nabla \cdot \vec{V} [/tex]
    3. The attempt at a solution
    The first thing I'll do is apply Gauss's Divergence theorem to the vectors [itex] \vec{A} = \varphi\nabla\psi [/itex] and [itex]B=\psi\nabla\varphi[/itex]
    Applied to A:
    [itex]\nabla \cdot \vec{A}=\nabla \cdot \varphi\nabla\psi =\varphi \nabla^{2} \psi \rightarrow \int_{\tau} div(\vec{A}) d\tau= \int_{\tau} \varphi \nabla^{2} \psi \, d\tau = \int_{\sigma} \varphi \nabla\psi \,\cdot d\vec{\sigma} [/itex]

    Applied to B
    [itex]\nabla \cdot \vec{B}=\nabla \cdot \psi\nabla\varphi =\psi \nabla^{2} \varphi \rightarrow \int_{\tau} div(\vec{B})\, d\tau= \int_{\tau} \psi \nabla^{2} \varphi \, d\tau = \int_{\sigma} \psi \nabla\varphi \,\cdot d\vec{\sigma}[/itex]

    And there is where I am stuck. These look like they're supposed to, but I am not sure if I'm allowed to subtract them to get Green's Theorem or if I am supposed to do something else.
     
  2. jcsd
  3. Feb 26, 2015 #2

    Ray Vickson

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    Why do you claim that ##\nabla \cdot (\varphi \nabla \psi) = \psi \nabla^2 \varphi##? Have you proved it?
     
  4. Feb 26, 2015 #3

    B3NR4Y

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    [tex]
    \begin{align*}
    \nabla \cdot \psi \nabla \varphi =\\
    \nabla \cdot \psi <\partial_1 \varphi, \partial_2 \varphi, \partial_3 \varphi> =\\
    \nabla \cdot <\psi \partial_1 \varphi,\psi \partial_2 \varphi, \psi\partial_3 \varphi> =\\ \psi\partial_{1}^{2}\varphi + \psi\partial_{2}^{2}\varphi + \psi \partial_{3}^{2}\varphi=\\
    \psi (\partial_{1}^{2}\varphi + \partial_{2}^{2}\varphi + \partial_{3}^{2}\varphi)=\\
    \psi\nabla^{2}\varphi
    \end{align*}
    [/tex]

    My logic may be wrong, but here's what I had written. I also "proved" the dot product is distributive, so I could use that for this problem.

    [tex]
    \begin{align*} (\vec{A}-\vec{B})\cdot \vec{C}=\vec{A}\cdot\vec{C}-\vec{B}\cdot\vec{C} \rightarrow \\ \vec{A}-\vec{B}=a_{i}-b_{i} \rightarrow https://www.physicsforums.com/file://\\(\vec{A}-\vec{B})\cdot [Broken] \vec{C}= (a_{i}-b_{i})c_{i} \, ; \, \, \\\vec{A}\cdot\vec{C}-\vec{B}\cdot\vec{C} = a_{i}c_{i}-b_{i}c_{i} = (a_{i}-b_{i})c_{i} = (\vec{A}-\vec{B})\cdot \vec{C}
    \end{align*}
    [/tex]

    Therefore I can use this to subtract the two integrals I got and boom, green's theorem?
     
    Last edited by a moderator: May 7, 2017
  5. Feb 26, 2015 #4

    Dick

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    Are you assuming ##\psi## is a constant? If not don't you have to differentiate it too? Use the product rule.
     
  6. Feb 26, 2015 #5

    Ray Vickson

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    Essentially, part of what you claim is that
    [tex] \frac{d}{dx} \left( f(x) \frac{dg(x)}{dx} \right) = f(x) \frac{d^2 g(x)}{dx^2} [/tex]
    Do you honestly believe that?
     
    Last edited by a moderator: May 7, 2017
  7. Feb 26, 2015 #6

    B3NR4Y

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    Oh, I'm dumb.
    [tex]
    \begin{align*}
    \nabla \cdot \vec{A} &= \nabla \cdot \psi \nabla\varphi\\
    &=\nabla \cdot \psi <\partial_{1}\varphi \, , \, \partial_{2} \varphi \, , \, \partial_3\varphi>\\
    &=\nabla \cdot <\psi\partial_{1}\varphi \, , \, \psi\partial_{2} \varphi \, , \, \psi\partial_3\varphi>\\
    &=(\partial_{1}\psi \, \partial_{1}\varphi + \psi \partial_{1}^2 \varphi)+(\partial_{2}\psi \, \partial_{2}\varphi + \psi \partial_{2}^2 \varphi)+(\partial_{3}\psi \, \partial_{3}\varphi + \psi \partial_{3}^2 \varphi)\\
    &=\partial_{i}\psi\partial_{i}\varphi+\psi\partial_{i}^2 \varphi \\
    &=\nabla\psi \cdot \nabla \psi +\psi\nabla^2\varphi

    \end{align*}
    [/tex]

    And for B:
    [tex]
    \begin{align*}
    \nabla \cdot \vec{B} &= \nabla \cdot \varphi \nabla\psi\\
    &=\nabla \cdot \varphi <\partial_{1}\psi \, , \, \partial_{2} \psi \, , \, \partial_3\psi>\\
    &=\nabla \cdot <\varphi\partial_{1}\psi \, , \, \varphi\partial_{2} \psi \, , \, \varphi\partial_3\psi>\\
    &=(\partial_{1}\varphi \, \partial_{1}\psi + \varphi \partial_{1}^2 \psi)+(\partial_{2}\psi \, \partial_{2}\psi + \varphi \partial_{2}^2 \psi)+(\partial_{3}\varphi \, \partial_{3}\psi + \varphi \partial_{3}^2 \psi)\\
    &=\partial_{i}\varphi\partial_{i}\psi+\varphi\partial_{i}^2 \psi \\
    &=\nabla\varphi \cdot \nabla \psi +\varphi\nabla^2\psi

    \end{align*}
    [/tex]

    And therefore the integrals I had before from Gauss's Divergence Theorem are
    [tex]
    \begin{align*}
    A\rightarrow &\int_\tau (\nabla\psi \cdot \nabla \varphi +\psi\nabla^2\varphi) \, d\tau = \int_{\sigma} \psi\nabla\varphi \cdot d \vec{\sigma} \\
    B\rightarrow &\int_\tau (\nabla\varphi \cdot \nabla \psi +\varphi\nabla^2\psi) \, d\tau = \int_{\sigma} \varphi\nabla\psi \cdot d \vec{\sigma}

    \end{align*}
    [/tex]

    and THEN I subtract, the dot product part at the beginning cancels and bam I have green's theorem?
     
  8. Feb 26, 2015 #7

    Dick

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    Sounds much better.
     
  9. Feb 26, 2015 #8

    B3NR4Y

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    Thank you both for your help, if it weren't for you guys I would have just subtracted my two original equations and gotten what looks right, but is a wrong answer. And would have been befuddled when I got a bad grade (most of the time we don\t get our assignments back despite the class only having about 13 students.)
     
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