Calculus of variations

Sep 2006
782
100
The raggedy edge.
Write down the Euler-Lagrange equations for critical points of the functional \(\displaystyle I(y)=\int_{x_1}^{x_2}y^2+y'y'''-(y'')^2~dx\)
I'm finding this odd since the only functions i've looked at are of the form \(\displaystyle g(x,y,y')\) not \(\displaystyle f(y',y'',y''')\).

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

Would I do it like this:

\(\displaystyle \frac{\partial f}{\partial y'''}=y'\)

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

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

Which becomes \(\displaystyle 2y''-1=0\)
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
I'm finding this odd since the only functions i've looked at are of the form \(\displaystyle g(x,y,y')\) not \(\displaystyle f(y',y'',y''')\).

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

Would I do it like this:

\(\displaystyle \frac{\partial f}{\partial y'''}=y'\)

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

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

Which becomes \(\displaystyle 2y''-1=0\)
For higher order Lagrangians, the Euler-Lagrange equation is

\(\displaystyle
\frac{\partial L}{\partial y} - \frac{d}{d x}\left(\frac{\partial L}{\partial y'}\right) + \frac{d^2}{d x^2} \left(\frac{\partial L}{\partial y''} \right) - \cdots + (-1)^n \frac{d^n}{d x^n}\left(\frac{\partial L}{\partial y^{(n)}}\right) = 0.
\)
 
  • Like
Reactions: Showcase_22