Minimize the functional: $\int_a^b {x^2(y')^2dx}$

Question

I have to minimize the following functional:

$\int_a^b {x^2(y')^2dx}$.

I used the Euler-Lagrange equation which gives:

$-\frac{d}{dx}2x^2y'=2xy'+x^2y''=0$,

which I found out is called a Euler-Cauchy equation. Is this correct or did I do the partial derivative the wrong way? Thank you.

Answer 1

For the functional $$I=\int_a^b L(x,y,y')dx$$ the Euler-Lagrange equation is $$\frac{\partial L}{\partial y}-\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)=0$$ Note that your equation is a special case where $L$ is not a function of $y$. Therefore the equation reduces to $$\frac{\partial L}{\partial y'}=c\;\;(\text{constant})$$ So let $y'=\frac{dy}{dx}=g(x,c)$ and solve this ODE to obtain: $$y=\int_a^x g(u,c)du+c_1$$ In your case, $L=(x y')^2$ results in $$2x^2 y'=c$$ hence $g(x,c)=\frac{c}{2x^2}$ and I think you won't have any problems with the rest.