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Integration by parts with a dxdy

  1. Jul 23, 2012 #1
    1. The problem statement, all variables and given/known data

    The functional ##I(u,v)=\int_\Omega F(x,y,u,v,u_x,u_y, v_x,v_y)dxdy##

    The partial variation of a functional is given as

    ## \displaystyle \delta I =\int_\Omega (\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_x} \delta u_x+\frac{\partial F}{\partial u_y} \delta u_y+\frac{\partial F}{\partial v} \delta v+\frac{\partial F}{\partial v_x} \delta v_x+\frac{\partial F}{\partial v_y} \delta v_y)dxdy##

    2. Relevant equations
    3. The attempt at a solution

    The next step in developing this equation is to integrate by parts the 2nd, 3rd, 5th and 6th terms.

    1) Why do we not integrate the 1st term since F and ##\delta u## are both functions of x and y, ie a product..? Similarly for the 4th term?

    2) For the second term I let ##U= \frac{\partial F}{\partial u_x}## and ## \displaystyle dV=\frac{\partial \delta u}{\partial x}## Therefore

    ## \displaystyle \int_\Omega (\frac{\partial F}{\partial u_x} \frac{\partial \delta u_x}{\partial x} )dxdy=\frac{ \partial F}{\partial u_x} \delta u - \int_\Omega [\frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy=\int_\Omega [\frac{\partial}{\partial x}(\frac{ \partial F}{\partial u_x} \delta u) - \frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy##

    We seem to be integrating wrt x, how do we justify not doin anything with dy...? Becasue F and u are also functions of y..? Thanks
     
  2. jcsd
  3. Jul 23, 2012 #2
    You might not have *fully* understood the derivation of Euler-Lagrange equation for the classic example of
    [tex]S(t)=\int_{a}^{b}L(t,x,\dot{x})dt[/tex]. I'm using the classical mechanics convention here where x is function of t, and [itex]\dot{x}=dx/dt[/itex], L is Lagrangian.

    The integration by part of the selected terms does the following two good things:
    1) Make use of the nice boundary conditions of the perturbation function (vanished at both ends)
    2) Leave the rest parts to be handled by the Fundamental Lemma of Calculus Of Variations -- Which gives you the Oiler-Lagrange Equation.


     
    Last edited: Jul 23, 2012
  4. Jul 23, 2012 #3
    I see, if I had of read on further I would have spotted this thanks.

    Just on a technical note regarding integrating parts. How do we treat this dxdy arrangement since we do not have a double integral such that we can integrate the inner and then the outer...? We know u is a function of both x and y according to post 1 but we seem to have done nothing with dy...? thanks
     
  5. Jul 24, 2012 #4
    You are welcome! This happens a lot in self-studying.

    We don't actually integrate the functional because u,v, and their partial derivatives are unknown. Or did I miss your question?

    We are dealing with the terms related to first variation of [itex]\frac{\partial u}{\partial x}[/itex]. y is treated as constant.
     
    Last edited: Jul 24, 2012
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