# Calculus of variations

#### Showcase_22

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
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

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

Showcase_22