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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?

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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 $u$.

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  • $\begingroup$ The total derivative notation is not appropriate as f is a function of multiple $x_i$. As far as I understand, in EL equation, we do not track $\partial x_j/\partial x_i$. $\endgroup$ – Rohit Gupta Jun 25 '18 at 4:07
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jun 25 '18 at 5:40
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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 $u_{,ij} = \frac{\partial^2 u}{\partial x_j \partial x_i}$.

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  • $\begingroup$ The problem is indicating that chain rule has to be used. Normally when partial derivative is used, we only treat the implicit dependence, not the explicit one. For example, in case of single variable Lagrangian we use total derivative to indicate that we want to use the chain rule. $\endgroup$ – Rohit Gupta Jun 23 '18 at 5:15
  • $\begingroup$ Now I reeeally see what you mean. Now I'll really think about it. $\endgroup$ – Jackozee Hakkiuz Jun 23 '18 at 5:24

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