# Calculus of variations

• May 16th 2010, 05:25 AM
Showcase_22
Calculus of variations
Quote:

Write down the Euler-Lagrange equations for critical points of the functional $I(y)=\int_{x_1}^{x_2}y^2+y'y'''-(y'')^2~dx$
I'm finding this odd since the only functions i've looked at are of the form $g(x,y,y')$ not $f(y',y'',y''')$.

I only thought the Euler-Lagrange equations were defined the the first type of function, not the second.

Would I do it like this:

$\frac{\partial f}{\partial y'''}=y'$

$\frac{\partial f}{\partial y''}=2(y'')$

So the E-L equation is $2(y'')-\frac{d}{dy'}(y')=0$

Which becomes $2y''-1=0$
• May 16th 2010, 06:55 AM
Jester
Quote:

Originally Posted by Showcase_22
I'm finding this odd since the only functions i've looked at are of the form $g(x,y,y')$ not $f(y',y'',y''')$.

I only thought the Euler-Lagrange equations were defined the the first type of function, not the second.

Would I do it like this:

$\frac{\partial f}{\partial y'''}=y'$

$\frac{\partial f}{\partial y''}=2(y'')$

So the E-L equation is $2(y'')-\frac{d}{dy'}(y')=0$

Which becomes $2y''-1=0$

For higher order Lagrangians, the Euler-Lagrange equation is

$
\frac{\partial L}{\partial y} - \frac{d}{d x}\left(\frac{\partial L}{\partial y'}\right) + \frac{d^2}{d x^2} \left(\frac{\partial L}{\partial y''} \right) - \cdots + (-1)^n \frac{d^n}{d x^n}\left(\frac{\partial L}{\partial y^{(n)}}\right) = 0.
$