Notation of Euler Lagrange Equations for multiple independent variables

Question

Let $x_{1}, ..., x_{n}$ be several independent variables. Let $u = u(x_{1}, ..., x_{n})$ be a function. Let $L(x_{1}, ..., x_{n},u,\partial_{1}u,...,\partial_{n}u)$ be the Lagrangian whose action we want to extremize.

Euler Lagrange equation is stated as $$\frac{\partial L}{\partial u} ~=~ \sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}\frac{\partial L}{\partial (\partial_{i}u)} .$$

The problem is that the partial derivative with respect to $x_i$ is not really a partial derivative, rather it includes both implicit and explicit dependence on $x_i$. The notation is really unclear on this issue. Is there a way of indicating this dependency in the notation? Or is there another way of writing Euler Lagrange Equation in which this issue is bypassed altogether?

Answer 1

Yes, there exist a notation: Use total derivatives $$\frac{\mathrm{d}}{\mathrm{d} x^i} ~=~ \frac{\partial}{\partial x^i} ~+~ \partial_iu \frac{\partial }{\partial u} ~+~ \sum_j\partial_i\partial_ju \frac{\partial }{\partial (\partial_ju)} ~+~ \sum_{j\leq k}\partial_i\partial_j\partial_ku \frac{\partial }{\partial (\partial_j\partial_ku)} ~+~\ldots$$ so that the EL eqs. read $$0~=~\frac{\partial L}{\partial u} ~-~ \sum_i\frac{\mathrm{d}}{\mathrm{d} x^i}\frac{\partial L}{\partial (\partial_iu)} ~+~ \sum_{i\leq j}\frac{\mathrm{d}}{\mathrm{d} x^i}\frac{\mathrm{d}}{\mathrm{d} x^j}\frac{\partial L}{\partial (\partial_i\partial_ju)} ~-~\ldots. $$ Here the ellipsis "$\ldots$" denote possibly higher-order derivative terms in case the Lagrangian $L$ depends on higher-order derivatives of $u$.

Answer 2

I think almost everybody would use the chain rule as a first thought, since one cannot change $x_i$ without inducing a change in $u$ or any of the $u_{,i}$s. $$\frac{\partial}{\partial x_j}\frac{\partial L}{\partial u_{,i}} = \frac{\partial^2 L}{\partial x_j \partial u_{,i}} + \frac{\partial^2 L}{\partial u\partial u_{,i}}u_{,j} + \sum_k\frac{\partial L}{\partial u_{,k}\partial u_{,i}}u_{,kj}$$ Where $u_{,i} = \frac{\partial u}{\partial x_i}$ and $u_{,ij} = \frac{\partial^2 u}{\partial x_j \partial x_i}$.