Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider $y=y(x)$ that extremises the functional

$I[y]=\int_{a}^{b} f(x, y, y', y'')dx$

so that $y$ satisfies the E-L equation:

$\dfrac{\partial f}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial f}{\partial y'})+\dfrac{d^2}{dx^2}(\dfrac{\partial f}{\partial y''})=0$

And as an exercise we were meant to show:

$\dfrac{d}{dx}[f-y'(\dfrac{\partial f}{\partial y'}-\dfrac{d}{dx}(\dfrac{\partial f}{\partial y'}))-y'\dfrac{\partial f}{\partial y'}]=\dfrac{\partial f}{\partial x} (1)$

I've tried for a while to get the above but can't somehow...I might have copied incorrectly, can someone first tell me whether (1) is correct in the first place (dashes in the right places etc.)? Thanks!

EDIT: Okay, seems like I've copied incorrectly in lectures, (1) should be:

$\dfrac{d}{dx}[f-y'(\dfrac{\partial f}{\partial y'}-\dfrac{d}{dx}(\dfrac{\partial f}{\partial y''}))-y''\dfrac{\partial f}{\partial y''}]=\dfrac{\partial f}{\partial x}$

then it's just straightforward computation,as Avitus says. Solved.

share|improve this question
    
What did you try? It looks to me a quite long but straightforward computation with repeated chain rules...is this what you did? For example, can you explicitly write $\frac{df}{dx}$ using the chain rule? –  Avitus Jul 21 '13 at 18:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.