Let's call our functional $$F[f]=\int dx\:A\left(x,f,f',f''...\right)$$ We know that the variation of F can be written as $$\delta F=\int dx\:\left[\frac{\partial A}{\partial f}\delta f+\frac{\partial A}{\partial f'}\delta f'+...\right]$$ If i wanted to get everything in terms of delta f in order to use the definition of the functional derivative, I would have to use integration by parts on the second and further terms. Looking at the second term as an example, $$\int dx \frac{\partial A}{\partial f'}\delta f'$$ we could take $$dv=dx\delta f'\:\:,\:\: u=\frac{\partial A}{\partial f'}$$ The IBP then transforms the term to $$-\int \delta f\:\:d\left(\frac{\partial A}{\partial f'}\right)$$ My questions is: how do we recover a dx from this internal derivative? I know it should be able to transform to $$\frac{d}{dx} (partials) dx$$ but I don't see why. Thanks.(adsbygoogle = window.adsbygoogle || []).push({});

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Multivariable IBP in the variation of a functional

Loading...

Similar Threads - Multivariable variation functional | Date |
---|---|

A Maximization Problem | Jan 31, 2018 |

A Time differentiation of fluid line integrals | Apr 7, 2017 |

I Multi-dimensional Integral by Change of Variables | Feb 12, 2017 |

I Help needed; problematic integral | Feb 6, 2017 |

I Find potential integrating on segments parallel to axes | Nov 17, 2016 |

**Physics Forums - The Fusion of Science and Community**