Is the Variation of a Functional in Calculus of Variation Correctly Calculated?

[lennyleonard]

Hi everyone!

Here's my problem:
Let's suppose that we have a functional I[f,g]=\int{L(f,\dot{f},g,\dot{g},x)\,dx}.

Is it right to say that the variation of I whit respect to g (thus taking g\;\rightarrow\;g+\delta g) is \delta I=\int{[L(f,\dot{f},g+\delta g,\dot{g}+\delta \dot g,x)-L(f,\dot{f},g,\dot{g},x)]\,dx}=\int{(\frac{\partial L}{\partial g}\delta g+\frac{\partial L}{\partial \dot{g}}\delta \dot{g})\,dx}??
Thanks for your disponibility!

 

[Petr Mugver]

Yes, it is true.

 

[lennyleonard]

Thank you petr!

I had this doubt when i saw the definition of "first variation" of a functional: according to the textbook i red (and to wikipedia: http://en.wikipedia.org/wiki/First_variation" ); taken a functional J(y) it's said that I need a function h to define the variaton of y:\;y\rightarrow y+\varepsilon h where \varepsilon \in R.
Then the first variation is:\delta J=lim_{\varepsilon\rightarrow 0}\frac{J(y+\varepsilon h)-J(y)}{\varepsilon}That's very different from the form I've written before!
In fact where i took y\rightarrow y+\delta y the books take y\rightarrow y+\varepsilon h; i.e. they use an "external" function h and a parameter \varepsilon, to be sent to zero in the operation.

Is it becaouse my \delta y can actually take the place of \varepsilon h (being \delta y an infinitesimal and a function it could work as wel..doesn'it?) or it's a whole different story??
The problem is that in some exercises (see the link to wikipedia above) the result depends on h, and h alone it's not an infinitesimal, so it cannot be "replaced" by \delta y!

Is there someone who could explain that to me??

 

[henry_m]

You seem to have understood the ideas but there are some differences here in notation and level of rigour, which add up to be quite confusing when you compare the two approaches.

In your opening post, you run through the first stages of what I might call a 'physicist's derivation' of the Euler-Lagrange equations, where you expand things up to first order and write an equality when you have ignored terms of quadratic order and higher. You regard \delta y as an 'infinitesimal function' and you only bother to keep an intuitive idea of \delta I as an 'infinitesimal change' rather than defining it properly. And this is fine, up to a point. The wiki article is a little more careful.

Here, we fix a definite function h, and a scale epsilon, which we imagine to be some very small but nonzero number. Then we see how the functional changes if we change its argument by \epsilon h, similar to your \delta y as your instincts correctly suggested. Think of h as fixing the 'shape' of the perturbation and epsilon its 'size'. This change in the function is some small but finite number, and is similar to your \delta I. For epsilon very small, this change will be linear in epsilon, so by dividing through by epsilon and taking the limit, just like taking ordinary derivatives, we get a measure of the rate of change of the functional in the 'direction' of h. This is what they call \delta J, which is a number which depends on h (Just as your \delta I depends on \delta y). Notice that nowhere here have we appealed to the hand-waving ideas of 'infinitesimals'.

Hopefully that helps; if you are still confused try doing the two approaches with ordinary functions of a single real variable.

 

[lennyleonard]

You've been very kind to spend some of your time on my problem henry_m! I'm very thankful!
But please, be patient enough to see if my conclusions are right:

Basically the difference in the two method (aside for the derivations) is that the former is the "quick-straight to the result" one, which doesn't cure for the mathematical formalism and the latter is the "elegant, precise" one, which avoids (as you said) the clutter concept of differential.

Anyway basically the two things are quite the same: in "my" method I used an expansion to first order so my \delta I depends on the infinitesimal \delta g (and its derivatives):\delta I=\int{[L(g+\delta g,\dot{g}+\delta \dot{g},x)-L(g,\dot{g},x)]\;dx}=\int{(\frac{\partial L}{\partial g}\delta g+\frac{\partial L}{\partial \dot{g}}\delta \dot{g})\;dx}while in the wiki-mathematician way there's no expansion employment whatsoever and thus the result depends on the non-infinitesimal funcion h, but besides this they're equal.
In fact it is also true that, using wiki method:\delta J(h) = \int{[(\frac{\partial L}{\partial g}h+\frac{\partial L}{\partial \dot{g}}\dot{h})]\;dx}

Is this right?

 

[henry_m]

Anyway basically the two things are quite the same: in "my" method I used an expansion to first order so my \delta I depends on the infinitesimal \delta g (and its derivatives):\delta I=\int{[L(g+\delta g,\dot{g}+\delta \dot{g},x)-L(g,\dot{g},x)]\;dx}=\int{(\frac{\partial L}{\partial g}\delta g+\frac{\partial L}{\partial \dot{g}}\delta \dot{g})\;dx}while in the wiki-mathematician way there's no expansion employment whatsoever and thus the result depends on the non-infinitesimal funcion h, but besides this they're equal.
In fact it is also true that, using wiki method:\delta J(h) = \int{[(\frac{\partial L}{\partial g}h+\frac{\partial L}{\partial \dot{g}}\dot{h})]\;dx}

Is this right?

Yes, you seem to have things pretty much spot on. Just bear in mind that your \delta I, \delta g are not well-defined objects so your approach is only heuristic and can't be used in a proper proof. But this sort of argument can be a very quick and useful way to derive and motivate results, and often appears in physics literature and textbooks.

 

[lennyleonard]

Your explanations have been extremely clarifying, sir.

Thank you very very much!