Euler-Lagrange equations: independence of $y$ and $y'$.

Question

What is the derivative of $y$ with respect to $y'$ in general? In the context of variational calculus, we treat $y$ and $y'$ as independent variables and when taking the partial derevitives with respect to each other it gives 0. Is that always the case in general that partial derevitive of $y$ with respect to $y'$ is zero, or it's just for reasons that got to do with variational calculus that we treat $y$ and $y'$ as independent?

Answer 1

Yes, it is always zero, $y$ and $y'$ are considered as separate "variables". When you deal with a lagrangian $L$ you usually consider it with variables $(x,s,p)$ with $x \in \mathbb{R}^n$, $s \in \mathbb{R}$ and $p \in \mathbb{R}^n$ (I am talking about the scalar case here, I don't want to complicate things with vector valued functions). What you have is that $s$ represents the value of your function $u: \mathbb{R}^n \to \mathbb{R}$ while $p$ represent the partial derivatives of $u$ with respect to $x_1 , \dots, x_n$.

Notice that here we don't use $y$ and $y'$ because it would generate great confusion, it's not a wise notation at all. The best exercise to help you understand these things is to derive the Euler-Lagrange equation by yourself.