Integration by parts Query

bugatti79

FOlks,

I am self studying a book and I have a question on

1)what the author means by the following comment "Integrating the second term in the last step to transfer differentiation from v to u"

2) Why does he perform integration by parts? I understand how but why? I can see that the last term has no derivatives of v in it.

See attached jpg.

THanks

Attached Thumbnails

 

HallsofIvy

"The second term in the last equation" is [itex](\partial F/\partial u')v'dx[/itex] and, since that is a product of a function and a derivative, a rather obvious thing to do is to integrate by parts ([itex]\int U dV= UV- \int V dU[/itex]) with [itex]U= v'[/itex] and [itex]dV= (\partial F/\partial u')dx[/itex].

bugatti79

I can see what he has done but why? Why integrate by parts?

Thanks

HallsofIvy

Because integration by parts, being the "inverse" of the product rule for derivatives, is a natural way to integrate a product. Especially here where the product is of a function and a derivative: [itex]\int u dv[/itex].

andrien

integration by part has been done to arrive at something similar to euler lagrange eqn.it just occurs many times in physics in different form may be covariant form

bugatti79

Ok, was just wondering what he meant by transfer differentiation from v to u?

Anyhow, thanks folks.

bugatti79

Actually, shouldnt it be the other way around [itex]U= (\partial F/\partial u')dx[/itex] and [itex]dV=v' [/itex]

Then [itex]V=v [/itex] since [itex]\int dV=\int v'dx [/itex] and [itex] \displaystyle \frac{dU}{dx}= d(\frac{\partial F}{\partial u'})=\frac{d}{dx} \frac{ \partial F}{\partial u'} [/itex].......?