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Variational calculus Euler lagrange Equation

  1. Nov 13, 2013 #1
    I am trying to understand an example from my textbook "applied finite element analysis" and in the variational calculus, Euler lagrange equation example I cant seem to understand the following derivation in one of its examples
    ∫((dT/dx)(d(δT)/dx))dx= ∫((dT/dx)δ(dT/dx))dx= ∫((1/2)δ(dT/dx)^2)dx

    limits from 0 to 5.

    My question here is how in the last part of the derivation 1/2 appears out of the blue where the integratal remains intact..
    If anyone know the answer.. kindly refer me some examples also
  2. jcsd
  3. Nov 13, 2013 #2
    essentially it is a restatement of: xdx = 1/2 d(x^2)
  4. Nov 14, 2013 #3
    So this is a reverse step?
  5. Nov 14, 2013 #4
    Hi All,
    I found it easier to master the functional derivative concept by performing the calculations out, leaving aside for a moment the symbol $$\delta$$ to denote variations, a useful notation that might though hide the mechanics of what is going on.
    Then your example becomes transparent: the functional derivative of the functional $$\int_{\Omega} y'^{2} \mathrm{d}\Omega$$ in the direction of the test function $$\mu$$ (in other terms, introducing a variation $$\hat{y} = y(x) + \epsilon \mu (x)$$ according to the definition equals
    $$lim_{\epsilon \to 0}\frac {\int_{\Omega} [\hat{y}'^2 - y'^ 2] \mathrm{d}\Omega}{\epsilon}$$
    $$lim_{\epsilon \to 0}\frac {\int_{\Omega} [{y'^{2} + 2\epsilon \mu' y'+\epsilon^2 \mu^{2}} - y' ^2] \mathrm{d}\Omega}{\epsilon}$$
    some terms cancel, and you are left with the result you wnated to confirm: it turns out the differentiation of functions and functional share common operational properties, such as xdx = 1/2 d(x^2).
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