In coordinates

Given bundle coordinates x^λ, yⁱ on a fiber bundle Y and the adapted coordinates x^λ, yⁱ, yⁱ_Λ, (Λ = (λ₁, ...,λ_k), |Λ| = k ≤ r) on jet manifolds J^rY, a Lagrangian L and its Euler–Lagrange operator read

${\displaystyle L={\mathcal {L}}(x^{\lambda },y^{i},y_{\Lambda }^{i})\,d^{n}x,}$

${\displaystyle \delta L=\delta _{i}{\mathcal {L}}\,dy^{i}\wedge d^{n}x,\qquad \delta _{i}{\mathcal {L}}=\partial _{i}{\mathcal {L}}+\sum _{|\Lambda |}(-1)^{|\Lambda |}\,d_{\Lambda }\,\partial _{i}^{\Lambda }{\mathcal {L}},}$

where

${\displaystyle d_{\Lambda }=d_{\lambda _{1}}\cdots d_{\lambda _{k}},\qquad d_{\lambda }=\partial _{\lambda }+y_{\lambda }^{i}\partial _{i}+\cdots ,}$

denote the total derivatives.

For instance, a first-order Lagrangian and its second-order Euler–Lagrange operator take the form

${\displaystyle L={\mathcal {L}}(x^{\lambda },y^{i},y_{\lambda }^{i})\,d^{n}x,\qquad \delta _{i}L=\partial _{i}{\mathcal {L}}-d_{\lambda }\partial _{i}^{\lambda }{\mathcal {L}}.}$

Euler–Lagrange equations

The kernel of an Euler–Lagrange operator provides the Euler–Lagrange equations δL = 0.