# Multivariable IBP in the variation of a functional

1. Aug 24, 2015

### avikarto

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.

2. Aug 28, 2015

### ShayanJ

For any function g(x) , we have dg=g'(x) dx. Now $\frac{\partial A}{\partial f'}$ is itself a function of x so we have $d(\frac{\partial A}{\partial f'})=\frac{d}{dx}\frac{\partial A}{\partial f'} dx$.

3. Aug 31, 2015

### avikarto

A silly oversight. Thanks.