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I'm looking for rigorous discussions on the derivation of the Euler-Lagrange equation for field as it is usually discussed in classical field theory books. More precisely, if the action is given by:

$$S(\phi) = \int \mathcal{L}(\phi, \partial_{x_{i}}\phi) d^{4}\vec{x}$$ where $\vec{x} = (x_{1},x_{2},x_{3},x_{4}=t)\in \mathbb{R}^{4}$ and $\partial_{x_{i}}$ denotes, generically, any of its partial derivatives, then I'm looking for a rigorous derivation of the Euler-Lagrange equation: $$\sum_{i=1}^{4}\frac{\partial}{\partial x_{i}}\bigg{(}\frac{\partial \mathcal{L}}{\partial (\partial_{x_{i}}\phi)}\bigg{)}-\frac{\partial \mathcal{L}}{\partial \phi} = 0 $$

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    $\begingroup$ What is unsatisfactory with the usual derivation that makes them non-rigorous? (Specifically, are you worried about the use of the compactly support perturbations? Are you worried about the fact that $S(\phi)$ is, for most field theories, necessarily infinite?) $\endgroup$ Jan 29, 2021 at 2:09
  • $\begingroup$ @WillieWong this is one reason. But also, all derivations I know come from physics books and thus there are some calculuations which seem unclear or poorly justified to me. $\endgroup$
    – MathMath
    Jan 29, 2021 at 13:00
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    $\begingroup$ Most of my books are in my office; hopefully Igor Khavkine sees your question since I know for sure he has an answer to it. The correct formulation is also described in his paper arxiv.org/abs/1210.0802 but I don't know how much you would like the Jet language. I am also pretty sure that this is explained in Christodoulou's Action Principle and PDEs (Princeton Univ. Press). But I don't have my copy with me to give you a precise page reference. $\endgroup$ Jan 29, 2021 at 21:33
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    $\begingroup$ @Buzz the derivation can be made rigorous (see e.g. mathoverflow.net/a/349234/11211, mathoverflow.net/q/273254/11211 and physics.stackexchange.com/a/256496/16767). The problem is just the terminology employed in (some of) the physical literature - one is in fact looking for critical points of a family of action functionals over each compact region of space-time. Once properly defined, these are bona fide smooth functionals on the space of smooth field configurations, which on its turn can be endowed with a proper infinite-dimensional smooth manifold structure. $\endgroup$ Feb 2, 2021 at 2:40
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    $\begingroup$ A rigorous discussion starts with an accurate statement rather than a proof ("derivation"). One should specify which exactly space of functions $\phi$ is considered and so on. E.g. in the reference recommended by Carlo Beenakker only C^2 smooth fucntions and C^2 smooth variations are considered. But minimizers even of smooth actions often cannot be found among smooth functions. $\endgroup$ Feb 5, 2021 at 22:17

2 Answers 2

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See page 16 and following of Coordinate-free derivation of the Euler–Lagrange equations and identification of global solutions via local behavior by Elsa Hansen (2005).

Results concerning $C^2$-minimizing curves on manifolds are presented. A coordinate- free derivation of the Euler–Lagrange equation is presented. Using a variational approach, two vector fields are defined along the minimizing curve; the tangent to the curve $\dot{γ}$, and the infinitesimal variation $\delta\sigma$. The derivation presented involves complete lifts of arbitrary extensions of these vector fields and it is shown that the derivation is independent of the particular choice of extensions. Special care is also taken to ensure that the derivation does not require any additional differentiability constraints, other than $\gamma$ being of class $C^2$.

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    $\begingroup$ The reference concerns minimizing curves rather than submanifolds, and thus concerns mechanics rather than field theory. But the method of the proof is similar in both cases. $\endgroup$ Feb 5, 2021 at 22:11
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Let $\pi:N\rightarrow M$ be a smooth fibered manifold. Let $m=\dim M$ and $m+n=\dim N$. Assume $M$ to be orientable and oriented. The interpretation is that $M$ is the "spacetime manifold" and the fields in question are smooth sections of $\pi$. Consider the jet prolongations $J^r(\pi)$ of the fibered manifold with projections $\pi^r:J^r(\pi)\rightarrow M$ and $\pi^r_s:J^r(\pi)\rightarrow J^s(\pi)$ for $r\ge s$.

A Lagrangian of order $r$ on $\pi$ can be identified with an $m$-form $L\in\Omega^m_{r,\mathrm{hor}}(\pi)$ on $J^r(\pi)$, which is horizontal with respect to $\pi^r$.

Let's say that an $m+k$-form $\omega\in\Omega^{m+k}_r(\pi)$ on $J^r(\pi)$ is maximally horizontal if there exists a volume form $\mu\in\Omega^m(M)$ (on $M$) and a $k$-form $\omega^\prime\in\Omega^k_r(\pi)$ such that$$ \omega=\omega^\prime\wedge(\pi^r)^\ast\mu. $$

A source form on $J^r(\pi)$ is a maximally horizontal $m+1$-form that is also $\pi^r_0$-horizontal. Write $\mathrm{Src}_r(\pi)$ for the set of all source forms on $J^r(\pi)$. One can establish that $$ \mathrm{Src}_r(\pi)\cong\Gamma((\pi^r_0)^\ast( V(\pi)\otimes_N \Lambda^m(M))), $$

where $V(\pi)$ is the vertical tangent bundle of $\pi:N\rightarrow M$.


This is a good place to consider coordinates. Let $(U,x^i,u^\sigma)$ be a fibered chart on $N$. If $\phi\in\Gamma_{\mathrm{loc}}(\pi)$ is a local section whose range is in $U$, its components in the fibered chart are the functions $\phi^\sigma:=u^\sigma\circ\phi$.

Let $U_r:=(\pi^r_0)^{-1}(U)\subseteq J^r(\pi)$, and for each $0\le|I|\le r$ define $$ u^\sigma_I(j^r_p\phi)=\partial_I(u^\sigma\circ\phi)(p). $$ Here $I=(i_1,...,i_k)$ (if $|I|=k$) is a multiindex. Then $(U_r,x^i,u^\sigma,u^\sigma_i,\dots,u^\sigma_{i_1,...,i_r})$ is a chart on $J^r(\pi)$ (in a slightly generalized sense because the lower indices in $u^\sigma_{i_1,...,i_k}$ are symmetric, so these are not all independent).

Then if $L\in\Omega^m_{r,\mathrm{hor}}(\pi)$ is a Lagrangian, in any one fibered chart it looks like $$ L=\mathcal L(x,u_{(0)},\dots,u_{(r)})\mathrm dx, $$

where $\mathrm dx=\mathrm dx^1\wedge\dots\wedge\mathrm dx^m$ and $u_{(k)}=(u^\sigma_{i_1,...,i_k})$. So clearly this is indeed like a classical Lagrangian.

Then $\Delta\in\mathrm{Src}_r(\pi)$ if and only if in any one fibered chart it looks like $$ \Delta=\Delta_\sigma(x,u_{(0)},\dots,u_{(r)})\mathrm du^\sigma\wedge\mathrm dx. $$

Given a source form $\Delta\in\mathrm{Src}_r(\pi)$, the corresponding source equation is $$ \Delta\circ j^r\phi=0,\quad\phi\in\Gamma_{\mathrm{loc}}(\pi), $$ which in coordinates takes the form $$ \Delta_\sigma(x,\phi(x),\phi_{(1)}(x),\dots,\phi_{(r)}(x))=0,\quad\phi_{(k)}=(\partial_I\phi^\sigma)_{|I|=k}. $$


A form $\omega\in\Omega^k_r(\pi)$ on $J^r(\pi)$ is contact if $(j^r\phi)^\ast\omega=0$ for any local section $\phi$. The set $\mathcal C\Omega_r(\pi)=\bigoplus_{q\in\mathbb Z}\mathcal C\omega^q_r(\pi)$ of all contact forms on $J^r(\pi)$ is a homogeneous and differential ideal in the exterior algebra on $J^r(\pi)$. A form $\omega$ is $k$-contact if $\omega\in\mathcal C^k\Omega_r(\pi)$, i.e. it belongs to the $k$th power of the contact ideal.

One can then introduce maps $p_k:\Omega^q_r(\pi)\rightarrow\Omega^q_{r+1}(\pi)$ which raise the order of forms by one, satisfy $$ (\pi^{r+1}_r)^\ast=\sum_{k=0}^\infty p_k, $$ and they have the property that $p_0\omega$ is horizontal, $p_k\omega$ is $k$-contact, and $p_kp_l=\delta_{k,l}(\pi^{r+2}_r)^\ast$.

An $m$-form $\Lambda\in\Omega^m_r(\pi)$ on $J^r(\pi)$ is a Lepage form if $p_1\mathrm d\Lambda$ is a source form. If $L\in\Omega^m_{r,\mathrm{hor}}(\pi)$ is a Lagrangian of order $r$ and $\Lambda\in\Omega^m_s(\pi)$ is a Lepage form of order $s$, then we say that $\Lambda$ is a Lepage equivalent or Lepage extension of $L$ if $L\cong p_0\Lambda$, where "$\cong$" means that equivalence is meant up to pullback along an appropriate jet projection.

The following can be shown:

  • If $\Lambda$ and $\Lambda^\prime$ are both Lepage extensions of $L$ (even if of different order), then $p_1\mathrm d\Lambda=p_1\mathrm d\Lambda^\prime$.
  • Every Lagrangian $L$ of order $r$ admits a global Lepage extension of order $2r-1$. It is non-unique however.

Then the Euler-Lagrange operator $$ E:\Omega^{m}_{r,\mathrm{hor}}(\pi)\rightarrow\mathrm{Src}_{2r}(\pi) $$ is defined by $$ E(L)=p_1\mathrm d\Lambda, $$ where $\Lambda$ is any Lepage extension of $L$.

This does implement the calculus of variations for the following reason. If $\Omega\subseteq M$ is a compact domain with boundary and $\phi\in\Gamma_\Omega(\pi)$ is a smooth section on it, then the action functional determined by $L$ is $$ S(\Omega,L,\phi)=\int_\Omega (j^r\phi)^\ast L=\int_\Omega (j^{2r-1}\phi)^\ast\Lambda, $$ where $\Lambda$ is any Lepage extension of order $2r-1$. Since $L-\Lambda$ is a contact form, we could replace $L$ by any Lepage extension. It may be shown that every variation of the section $\phi$ can be described in terms of a vertical vector field $X\in\mathcal D_{\mathrm{vert}}(\pi)$, and then the variation if the action is given by the Lie derivative along the prolongation: $$ \delta_X S(\Omega,L,\phi)=\int_\Omega(j^{2r-1}\phi)^\ast\partial_{j^{2r-1}X}\Lambda=\int_\Omega(j^{2r-1}\phi)^\ast\left[j^{2r-1}X\rfloor\mathrm d\Lambda+\mathrm d j^{2r-1}X\rfloor\Lambda\right]. $$

If we assume the variation has support within $\Omega$, the second term vanishes by the Stokes theorem and the first term can be rewritten as $$ \delta_X S(\Omega,L,\phi)=\int_\Omega(j^{2r}\phi)^\ast\left( j^{2r}X\rfloor E(L)+j^{2r}X\rfloor\mathrm{K}\right), $$ where $\mathrm{K}$ is $2$-contact, so its contraction with the vector field is contact and hence its pullback along the section vanishes, and as $E(L)$ is a source form, $j^{2r}X\rfloor E(L)$ actually only depends on $X$ at each point, so there are no further "integrations by parts" to be done.

In coordinates $$ j^{2r}X\rfloor E(L)=X^\sigma E_\sigma(L)\mathrm du^\sigma\wedge\mathrm dx,\qquad E_\sigma(L)=\sum_{|I|=0}^r(-d)_I\frac{\partial\mathcal L}{\partial u^\sigma_I}. $$


The point of view used in this answer is due to D. Krupka (see e.g. Krupka - Introduction to Global Variational Geometry). There are other similar approaches using the infinite jet prolongation of a fibered manifold (see e.g. Anderon - The Variational Bicomplex), in which the Euler-Lagrange form is obtained by applying a suitable splitting operator to the abstract variation of the Lagrangian. Moving on to the Lepage extension essentially performs this splitting implicitly.

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