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Differentiation of an integral

  1. May 25, 2012 #1
    differentiation of a functional

    Where [tex]\phi = \phi(x)[/tex] and the functional [tex]F=F(\phi(x)) = \int d^d x [\frac{1}{2}K^2(\bigtriangledown\phi)^2+ V (\phi)][/tex]

    , the author says the derivative with respect to phi gives

    [tex]\frac {\partial F} {\partial \phi(x)} = -K^2\bigtriangledown^2\phi + V'(\phi)[/tex]

    I'm not seeing this. Could anyone give me some tips for writing down the differentiation more explicitly? I'm trying to work it out by writing F(phi+ dphi) and subtracting F(phi) as detailed in http://julian.tau.ac.il/bqs/functionals/node1.html, but I can't seem to make it come out correctly. This is the chemical potential in the Cahn-Hilliard equation, if it's important.

    Thanks in advance!
     
    Last edited: May 25, 2012
  2. jcsd
  3. May 26, 2012 #2
    ooh, that's a tough one, i feel like i'm close, do you have draft, maybe i could give a hint
     
  4. May 28, 2012 #3
    First, of all, I don't understand the d^d x notation in the original expression. Does the superscript d mean something or is it a typo?

    Lets ignore that superscript and see what we can find. Let [itex]\phi_{\epsilon}=\phi + \epsilon \psi[/itex] be a variation of the function, and take the limit as epsilon goes to 0. I am going to assume that there is some kind of Dirichlet condition on the boundary of the domain of this variational problem. We are going to need it to do integration by parts.

    Let D be the name of the domain and [itex] \Gamma [/itex] its boundary.

    [itex] \int_D (1/2)K^2 |\nabla \phi_{\epsilon}|^2+V(\phi_{\epsilon}) \, dx[/itex]

    Taking the derivative with respect to epsilon and setting epsilon=0, we get

    [itex] \int_D K^2 \nabla \phi \cdot \nabla \psi + V'(\phi)\psi \, dx[/itex]

    For the first term use integration by parts, and you end up with

    [itex] \int_D \left(-K^2 \nabla^2\phi+V'(\phi)\right)\psi\, dx [/itex]

    If you are not familiar with that version of integration by parts it is equivalent to the divergence theorem:

    [itex] \int_D \text{div}\, (\psi \nabla \phi)\, dx = \int_{\Gamma} (\psi \nabla \phi)\cdot \vec{n}\, d\sigma[/itex]

    The right side is zero because of the Dirichlet condition on phi which forces psi to be zero on the boundary. Use the product rule to expand the left side and you have your integration by parts formula.
     
  5. May 28, 2012 #4
    Going back to your original question, I would suggest doing variations like this:

    [itex] \phi+\epsilon\psi[/itex]

    epsilon is a real variable, so you can use the regular rules of differentiation, Taylor expansion, w/e. In particular, you can calculate the derivative w.r.t. epsilon and set epsilon =0. That gives you the variation in the direction of psi.
     
    Last edited: May 28, 2012
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