Euler–Lagrange equation

File:Tautochrone curve.gif problem. ]]

The Euler–Lagrange equation was developed in the 1750s by Euler and Lagrange in connection with their studies of the tautochrone problem. This is the problem of determining a curve on which a weighted particle will fall to a fixed point in a fixed amount of time, independent of the starting point.

Lagrange solved this problem in 1755 and sent the solution to Euler. Both further developed Lagrange's method and applied it to mechanics, which led to the formulation of Lagrangian mechanics. Their correspondence ultimately led to the calculus of variations, a term coined by Euler himself in 1766.[http://numericalmethods.eng.usf.edu/anecdotes/lagrange.pdf A short biography of Lagrange] {{webarchive|url=https://web.archive.org/web/20070714022022/http://numericalmethods.eng.usf.edu/anecdotes/lagrange.pdf |date=2007-07-14 }}

Let $(X,L)$ be a real dynamical system with $n$ degrees of freedom. Here $X$ is the configuration space and $L=L(t,\{\backslash boldsymbol\; q\}(t),\; \{\backslash boldsymbol\; v\}(t))$ the Lagrangian, i.e. a smooth real-valued function such that $\{\backslash boldsymbol\; q\}(t)\; \backslash in\; X,$ and $\{\backslash boldsymbol\; v\}(t)$ is an $n$-dimensional "vector of speed". (For those familiar with differential geometry, $X$ is a smooth manifold, and $L\; :\; \{\backslash mathbb\; R\}\_t\; \backslash times\; X\; \backslash times\; TX\; \backslash to\; \{\backslash mathbb\; R\},$ where $TX$ is the tangent bundle of $X).$

Let $\{\backslash cal\; P\}(a,b,\backslash boldsymbol\; x\_a,\backslash boldsymbol\; x\_b)$ be the set of smooth paths $\backslash boldsymbol\; q:\; [a,b]\; \backslash to\; X$ for which $\backslash boldsymbol\; q(a)\; =\; \backslash boldsymbol\; x\_a$ and $\backslash boldsymbol\; q(b)\; =\; \backslash boldsymbol\; x\_b.$

The action functional $S\; :\; \{\backslash cal\; P\}(a,b,\backslash boldsymbol\; x\_a,\backslash boldsymbol\; x\_b)\; \backslash to\; \backslash mathbb\{R\}$ is defined via

$$S[\backslash boldsymbol\; q]\; =\; \backslash int\_a^b\; L(t,\backslash boldsymbol\; q(t),\backslash dot\{\backslash boldsymbol\; q\}(t))\backslash ,\; dt.$$

A path $\backslash boldsymbol\; q\; \backslash in\; \{\backslash cal\; P\}(a,b,\backslash boldsymbol\; x\_a,\backslash boldsymbol\; x\_b)$ is a stationary point of $S$ if and only if

{{Equation box 1

|indent =:

|equation = $\backslash frac\{\backslash partial\; L\}\{\backslash partial\; q^i\}(t,\backslash boldsymbol\; q(t),\backslash dot\{\backslash boldsymbol\; q\}(t))-\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}t\}\backslash frac\{\backslash partial\; L\}\{\backslash partial\; \backslash dot\; q^i\}(t,\backslash boldsymbol\; q(t),\backslash dot\{\backslash boldsymbol\; q\}(t))\; =\; 0,$

\quad i = 1, \dots, n.

|border colour = #50C878

|background colour = #ECFCF4}}

Here, $\backslash dot\{\backslash boldsymbol\; q\}(t)$ is the time derivative of $\backslash boldsymbol\; q(t).$ When we say stationary point, we mean a stationary point of $S$ with respect to any small perturbation in $\backslash boldsymbol\; q$. See proofs below for more rigorous detail.

{{math proof|title=Derivation of the one-dimensional Euler–Lagrange equation|proof=

The derivation of the one-dimensional Euler–Lagrange equation is one of the classic proofs in mathematics. It relies on the fundamental lemma of calculus of variations.

We wish to find a function $f$ which satisfies the boundary conditions $f(a)\; =\; A$, $f(b)\; =\; B$, and which extremizes the functional

$$J[f]\; =\; \backslash int\_a^b\; L(x,f(x),f\text{'}(x))\backslash ,\; \backslash mathrm\{d\}x\backslash \; .$$

We assume that $L$ is twice continuously differentiable.{{harvnb|Courant|Hilbert|1953|p=184}} A weaker assumption can be used, but the proof becomes more difficult.{{Citation needed|date=September 2013}}

If $f$ extremizes the functional subject to the boundary conditions, then any slight perturbation of $f$ that preserves the boundary values must either increase $J$ (if $f$ is a minimizer) or decrease $J$ (if $f$ is a maximizer).

Let $f\; +\; \backslash varepsilon\; \backslash eta$ be the result of such a perturbation $\backslash varepsilon\; \backslash eta$ of $f$, where $\backslash varepsilon$ is small and $\backslash eta$ is a differentiable function satisfying $\backslash eta\; (a)\; =\; \backslash eta\; (b)\; =\; 0$. Then define

$$\backslash Phi(\backslash varepsilon)\; =\; J[f+\backslash varepsilon\backslash eta]\; =\; \backslash int\_a^b\; L(x,f(x)+\backslash varepsilon\backslash eta(x),\; f\text{'}(x)+\backslash varepsilon\backslash eta\text{'}(x))\backslash ,\; \backslash mathrm\{d\}x\; \backslash \; .$$

We now wish to calculate the total derivative of $\backslash Phi$ with respect to ε.

&= \int_a^b \frac{\mathrm d}{\mathrm d\varepsilon} L(x,f(x)+\varepsilon\eta(x), f'(x)+\varepsilon\eta'(x)) \, \mathrm{d}x

\\

&= \int_a^b \left[\eta(x)\frac{\partial L}{\partial {f} }(x,f(x)+\varepsilon\eta(x),f'(x)+\varepsilon\eta'(x)) + \eta'(x)\frac{\partial L}{\partial f'}(x,f(x)+\varepsilon\eta(x),f'(x)+\varepsilon\eta'(x))\right] \mathrm{d}x \ . \end{align}

The third line follows from the fact that $x$ does not depend on $\backslash varepsilon$, i.e. $\backslash frac\{\backslash mathrm\{d\}\; x\}\{\backslash mathrm\{d\}\; \backslash varepsilon\}\; =\; 0$.

When $\backslash varepsilon\; =\; 0$, $\backslash Phi$ has an extremum value, so that

$$\backslash left.\backslash frac\{\backslash mathrm\; d\; \backslash Phi\}\{\backslash mathrm\; d\backslash varepsilon\}\backslash right|\_\{\backslash varepsilon=0\}\; =\; \backslash int\_a^b\; \backslash left[\; \backslash eta(x)\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\}(x,f(x),f\text{'}(x))\; +\; \backslash eta\text{'}(x)\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\text{'}\}(x,f(x),f\text{'}(x))\; \backslash ,\backslash right]\backslash ,\backslash mathrm\{d\}x\; =\; 0\; \backslash \; .$$

The next step is to use integration by parts on the second term of the integrand, yielding

$$\backslash int\_a^b\; \backslash left[\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\}(x,f(x),f\text{'}(x))\; -\; \backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\text{'}\}(x,f(x),f\text{'}(x))\; \backslash right]\; \backslash eta(x)\backslash ,\backslash mathrm\{d\}x\; +\; \backslash left[\; \backslash eta(x)\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\text{'}\}(x,f(x),f\text{'}(x))\; \backslash right]\_a^b\; =\; 0\; \backslash \; .$$

Using the boundary conditions $\backslash eta\; (a)\; =\; \backslash eta\; (b)\; =\; 0$,

$$\backslash int\_a^b\; \backslash left[\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\}(x,f(x),f\text{'}(x))\; -\; \backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\text{'}\}(x,f(x),f\text{'}(x))\; \backslash right]\; \backslash eta(x)\backslash ,\backslash mathrm\{d\}x\; =\; 0\; \backslash ,\; .$$

Applying the fundamental lemma of calculus of variations now yields the Euler–Lagrange equation

$$\backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\}(x,f(x),f\text{'}(x))\; -\; \backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}x\}\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\text{'}\}(x,f(x),f\text{'}(x))\; =\; 0\; \backslash ,\; .$$

}}

{{math proof|title=Alternative derivation of the one-dimensional Euler–Lagrange equation|proof=

Given a functional

$$J\; =\; \backslash int^b\_a\; L(t,\; y(t),\; y\text{'}(t))\backslash ,\backslash mathrm\{d\}t$$

on $C^1([a,\; b])$ with the boundary conditions $y(a)\; =\; A$ and $y(b)\; =\; B$, we proceed by approximating the extremal curve by a polygonal line with $n$ segments and passing to the limit as the number of segments grows arbitrarily large.

Divide the interval $[a,\; b]$ into $n$ equal segments with endpoints $t\_0\; =\; a,\; t\_1,\; t\_2,\; \backslash ldots,\; t\_n\; =\; b$ and let $\backslash Delta\; t\; =\; t\_k\; -\; t\_\{k\; -\; 1\}$. Rather than a smooth function $y(t)$ we consider the polygonal line with vertices $(t\_0,\; y\_0),\backslash ldots,(t\_n,\; y\_n)$, where $y\_0\; =\; A$ and $y\_n\; =\; B$. Accordingly, our functional becomes a real function of $n\; -\; 1$ variables given by

$$J(y\_1,\; \backslash ldots,\; y\_\{n\; -\; 1\})\; \backslash approx\; \backslash sum^\{n\; -\; 1\}\_\{k\; =\; 0\}L\backslash left(t\_k,\; y\_k,\; \backslash frac\{y\_\{k\; +\; 1\}\; -\; y\_k\}\{\backslash Delta\; t\}\backslash right)\backslash Delta\; t.$$

Extremals of this new functional defined on the discrete points $t\_0,\backslash ldots,t\_n$ correspond to points where

$$\backslash frac\{\backslash partial\; J(y\_1,\backslash ldots,y\_n)\}\{\backslash partial\; y\_m\}\; =\; 0.$$

Note that change of $y\_m$ affects L not only at m but also at m-1 for the derivative of the 3rd argument.

$$L(\backslash text\{3rd\; argument\})\; \backslash left(\; \backslash frac\{y\_\{m+1\}\; -\; (y\_\{m\}\; +\; \backslash Delta\; y\_\{m\})\}\{\backslash Delta\; t\}\; \backslash right)\; =\; L\; \backslash left(\backslash frac\{y\_\{m+1\}\; -\; y\_\{m\}\}\{\backslash Delta\; t\}\backslash right)\; -\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; y\text{'}\}\; \backslash frac\{\backslash Delta\; y\_m\}\{\backslash Delta\; t\}$$

$$L\; \backslash left(\; \backslash frac\{(y\_\{m\}\; +\; \backslash Delta\; y\_\{m\})\; -\; y\_\{m-1\}\}\{\backslash Delta\; t\}\; \backslash right)$$

= L \left(\frac{y_{m} - y_{m-1}}{\Delta t}\right) + \frac{\partial L}{\partial y'} \frac{\Delta y_m}{\Delta t}

Evaluating the partial derivative gives

$$\backslash frac\{\backslash partial\; J\}\{\backslash partial\; y\_m\}\; =\; L\_y\backslash left(t\_m,\; y\_m,\; \backslash frac\{y\_\{m\; +\; 1\}\; -\; y\_m\}\{\backslash Delta\; t\}\backslash right)\backslash Delta\; t\; +\; L\_\{y\text{'}\}\backslash left(t\_\{m\; -\; 1\},\; y\_\{m\; -\; 1\},\; \backslash frac\{y\_m\; -\; y\_\{m\; -\; 1\}\}\{\backslash Delta\; t\}\backslash right)\; -\; L\_\{y\text{'}\}\backslash left(t\_m,\; y\_m,\; \backslash frac\{y\_\{m\; +\; 1\}\; -\; y\_m\}\{\backslash Delta\; t\}\backslash right).$$

Dividing the above equation by $\backslash Delta\; t$ gives

$$\backslash frac\{\backslash partial\; J\}\{\backslash partial\; y\_m\; \backslash Delta\; t\}\; =\; L\_y\backslash left(t\_m,\; y\_m,\; \backslash frac\{y\_\{m\; +\; 1\}\; -\; y\_m\}\{\backslash Delta\; t\}\backslash right)\; -\; \backslash frac\{1\}\{\backslash Delta\; t\}\backslash left[L\_\{y\text{'}\}\backslash left(t\_m,\; y\_m,\; \backslash frac\{y\_\{m\; +\; 1\}\; -\; y\_m\}\{\backslash Delta\; t\}\backslash right)\; -\; L\_\{y\text{'}\}\backslash left(t\_\{m\; -\; 1\},\; y\_\{m\; -\; 1\},\; \backslash frac\{y\_m\; -\; y\_\{m\; -\; 1\}\}\{\backslash Delta\; t\}\backslash right)\backslash right],$$

and taking the limit as $\backslash Delta\; t\; \backslash to\; 0$ of the right-hand side of this expression yields

$$L\_y\; -\; \backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}t\}L\_\{y\text{'}\}\; =\; 0.$$

The left hand side of the previous equation is the functional derivative $\backslash delta\; J/\backslash delta\; y$ of the functional $J$. A necessary condition for a differentiable functional to have an extremum on some function is that its functional derivative at that function vanishes, which is granted by the last equation.

}}

A standard example{{Citation needed|reason=this example seems like a poor one as it is not even presented as a dynamical system|date=July 2023}} is finding the real-valued function y(x) on the interval [a, b], such that y(a) = c and y(b) = d, for which the path length along the curve traced by y is as short as possible.

:$\backslash text\{s\}\; =\; \backslash int\_\{a\}^\{b\}\; \backslash sqrt\{\backslash mathrm\{d\}x^2+\backslash mathrm\{d\}y^2\}\; =\; \backslash int\_\{a\}^\{b\}\; \backslash sqrt\{1+y\text{'}^2\}\backslash ,\backslash mathrm\{d\}x,$

the integrand function being $L(x,y,\; y\text{'})\; =\; \backslash sqrt\{1+y\text{'}^2\}$.

The partial derivatives of L are:

:$\backslash frac\{\backslash partial\; L(x,\; y,\; y\text{'})\}\{\backslash partial\; y\text{'}\}\; =\; \backslash frac\{y\text{'}\}\{\backslash sqrt\{1\; +\; y\text{'}^2\}\}\; \backslash quad\; \backslash text\{and\}\; \backslash quad$

\frac{\partial L(x, y, y')}{\partial y} = 0.

By substituting these into the Euler–Lagrange equation, we obtain

:$$

\begin{align}

\frac{\mathrm{d}}{\mathrm{d}x} \frac{y'(x)}{\sqrt{1 + (y'(x))^2}} &= 0 \\

\frac{y'(x)}{\sqrt{1 + (y'(x))^2}} &= C = \text{constant} \\

\Rightarrow y'(x)&= \frac{C}{\sqrt{1-C^2}} =: A \\

\Rightarrow y(x) &= Ax + B

\end{align}

that is, the function must have a constant first derivative, and thus its graph is a straight line.

The stationary values of the functional

:$$

I[f] = \int_{x_0}^{x_1} \mathcal{L}(x, f, f', f'', \dots, f^{(k)})~\mathrm{d}x ~;~~

f' := \cfrac{\mathrm{d}f}{\mathrm{d}x}, ~f'' := \cfrac{\mathrm{d}^2f}{\mathrm{d}x^2}, ~

f^{(k)} := \cfrac{\mathrm{d}^kf}{\mathrm{d}x^k}

can be obtained from the Euler–Lagrange equation{{cite book | last1=Courant | first1=R | author-link1=Richard Courant | last2=Hilbert | first2=D | author-link2=David Hilbert | title = Methods of Mathematical Physics | volume = I | edition = First English | publisher = Interscience Publishers, Inc | year = 1953 | location = New York | isbn = 978-0471504474}}

:$$

\cfrac{\partial \mathcal{L}}{\partial f} - \cfrac{\mathrm{d}}{\mathrm{d} x}\left(\cfrac{\partial \mathcal{L}}{\partial f'}\right) + \cfrac{\mathrm{d}^2}{\mathrm{d} x^2}\left(\cfrac{\partial \mathcal{L}}{\partial f''}\right) - \dots +

(-1)^k \cfrac{\mathrm{d}^k}{\mathrm{d} x^k}\left(\cfrac{\partial \mathcal{L}}{\partial f^{(k)}}\right) = 0

under fixed boundary conditions for the function itself as well as for the first $k-1$ derivatives (i.e. for all $f^\{(i)\},\; i\; \backslash in\; \backslash \{0,\; ...,\; k-1\backslash \}$). The endpoint values of the highest derivative $f^\{(k)\}$ remain flexible.

If the problem involves finding several functions ($f\_1,\; f\_2,\; \backslash dots,\; f\_m$) of a single independent variable ($x$) that define an extremum of the functional

:$$

I[f_1,f_2, \dots, f_m] = \int_{x_0}^{x_1} \mathcal{L}(x, f_1, f_2, \dots, f_m, f_1', f_2', \dots, f_m')~\mathrm{d}x

~;~~ f_i' := \cfrac{\mathrm{d}f_i}{\mathrm{d}x}

then the corresponding Euler–Lagrange equations are{{cite book |last=Weinstock |first=R. |year=1952 |title=Calculus of Variations with Applications to Physics and Engineering |url=https://archive.org/details/calculusofvariat00wein |url-access=registration |publisher=McGraw-Hill |location=New York }}

:$$

\begin{align}

\frac{\partial \mathcal{L}}{\partial f_i} - \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial \mathcal{L}}{\partial f_i'}\right) = 0 ; \quad i = 1, 2, ..., m

\end{align}

A multi-dimensional generalization comes from considering a function on n variables. If $\backslash Omega$ is some surface, then

: $$

I[f] = \int_{\Omega} \mathcal{L}(x_1, \dots , x_n, f, f_{1}, \dots , f_{n})\, \mathrm{d}\mathbf{x}\,\! ~;~~

f_{j} := \cfrac{\partial f}{\partial x_j}

is extremized only if f satisfies the partial differential equation

: $\backslash frac\{\backslash partial\; \backslash mathcal\{L\}\}\{\backslash partial\; f\}\; -\; \backslash sum\_\{j=1\}^\{n\}\; \backslash frac\{\backslash partial\}\{\backslash partial\; x\_j\}\backslash left(\backslash frac\{\backslash partial\; \backslash mathcal\{L\}\}\{\backslash partial\; f\_\{j\}\}\backslash right)\; =\; 0.$

When n = 2 and functional $\backslash mathcal\; I$ is the energy functional, this leads to the soap-film minimal surface problem.

If there are several unknown functions to be determined and several variables such that

: $$

I[f_1,f_2,\dots,f_m] = \int_{\Omega} \mathcal{L}(x_1, \dots , x_n, f_1, \dots, f_m, f_{1,1}, \dots , f_{1,n}, \dots, f_{m,1}, \dots, f_{m,n}) \, \mathrm{d}\mathbf{x}\,\! ~;~~

f_{i,j} := \cfrac{\partial f_i}{\partial x_j}

the system of Euler–Lagrange equations is

: $$

\begin{align}

\frac{\partial \mathcal{L}}{\partial f_1} - \sum_{j=1}^{n} \frac{\partial}{\partial x_j}\left(\frac{\partial \mathcal{L}}{\partial f_{1,j}}\right) &= 0_1 \\

\frac{\partial \mathcal{L}}{\partial f_2} - \sum_{j=1}^{n} \frac{\partial}{\partial x_j}\left(\frac{\partial \mathcal{L}}{\partial f_{2,j}}\right) &= 0_2 \\

\vdots \qquad \vdots \qquad &\quad \vdots \\

\frac{\partial \mathcal{L}}{\partial f_m} - \sum_{j=1}^{n} \frac{\partial}{\partial x_j}\left(\frac{\partial \mathcal{L}}{\partial f_{m,j}}\right) &= 0_m.

\end{align}

If there is a single unknown function f to be determined that is dependent on two variables x₁ and x₂ and if the functional depends on higher derivatives of f up to n-th order such that

: $$

\begin{align}

I[f] & = \int_{\Omega} \mathcal{L}(x_1, x_2, f, f_{1}, f_{2}, f_{11}, f_{12}, f_{22},

\dots, f_{22\dots 2})\, \mathrm{d}\mathbf{x} \\

& \qquad \quad

f_{i} := \cfrac{\partial f}{\partial x_i} \; , \quad

f_{ij} := \cfrac{\partial^2 f}{\partial x_i\partial x_j} \; , \;\; \dots

\end{align}

then the Euler–Lagrange equation is

:$$

\begin{align}

\frac{\partial \mathcal{L}}{\partial f}

& - \frac{\partial}{\partial x_1}\left(\frac{\partial \mathcal{L}}{\partial f_{1}}\right)

- \frac{\partial}{\partial x_2}\left(\frac{\partial \mathcal{L}}{\partial f_{2}}\right)

+ \frac{\partial^2}{\partial x_1^2}\left(\frac{\partial \mathcal{L}}{\partial f_{11}}\right)

+ \frac{\partial^2}{\partial x_1\partial x_2}\left(\frac{\partial \mathcal{L}}{\partial f_{12}}\right)

+ \frac{\partial^2}{\partial x_2^2}\left(\frac{\partial \mathcal{L}}{\partial f_{22}}\right) \\

& - \dots

+ (-1)^n \frac{\partial^n}{\partial x_2^n}\left(\frac{\partial \mathcal{L}}{\partial f_{22\dots 2}}\right) = 0

\end{align}

which can be represented shortly as:

:$$

\frac{\partial \mathcal{L}}{\partial f} +\sum_{j=1}^n \sum_{\mu_1 \leq \ldots \leq \mu_j} (-1)^j \frac{\partial^j}{\partial x_{\mu_{1}}\dots \partial x_{\mu_{j}}} \left( \frac{\partial \mathcal{L} }{\partial f_{\mu_1\dots\mu_j}}\right)=0

wherein $\backslash mu\_1\; \backslash dots\; \backslash mu\_j$ are indices that span the number of variables, that is, here they go from 1 to 2. Here summation over the $\backslash mu\_1\; \backslash dots\; \backslash mu\_j$ indices is only over $\backslash mu\_1\; \backslash leq\; \backslash mu\_2\; \backslash leq\; \backslash ldots\; \backslash leq\; \backslash mu\_j$ in order to avoid counting the same partial derivative multiple times, for example $f\_\{12\}\; =\; f\_\{21\}$ appears only once in the previous equation.

If there are p unknown functions f_i to be determined that are dependent on m variables x₁ ... x_m and if the functional depends on higher derivatives of the f_i up to n-th order such that

:$$

\begin{align}

I[f_1,\ldots,f_p] & = \int_{\Omega} \mathcal{L}(x_1, \ldots, x_m; f_1,\ldots,f_p; f_{1,1},\ldots,

f_{p,m}; f_{1,11},\ldots, f_{p,mm};\ldots; f_{p,1\ldots 1}, \ldots, f_{p,m\ldots m})\, \mathrm{d}\mathbf{x} \\

& \qquad \quad

f_{i,\mu} := \cfrac{\partial f_i}{\partial x_\mu} \; , \quad

f_{i,\mu_1\mu_2} := \cfrac{\partial^2 f_i}{\partial x_{\mu_1}\partial x_{\mu_2}} \; , \;\; \dots

\end{align}

where $\backslash mu\_1\; \backslash dots\; \backslash mu\_j$ are indices that span the number of variables, that is they go from 1 to m. Then the Euler–Lagrange equation is

:$$

\frac{\partial \mathcal{L}}{\partial f_i} +\sum_{j=1}^n \sum_{\mu_1 \leq \ldots \leq \mu_j} (-1)^j \frac{\partial^j}{\partial x_{\mu_{1}}\dots \partial x_{\mu_{j}}} \left( \frac{\partial \mathcal{L} }{\partial f_{i,\mu_1\dots\mu_j}}\right)=0

where the summation over the $\backslash mu\_1\; \backslash dots\; \backslash mu\_j$ is avoiding counting the same derivative $f\_\{i,\backslash mu\_1\backslash mu\_2\}\; =\; f\_\{i,\backslash mu\_2\backslash mu\_1\}$ several times, just as in the previous subsection. This can be expressed more compactly as

:$$

\sum_{j=0}^n \sum_{\mu_1 \leq \ldots \leq \mu_j} (-1)^j \partial_{ \mu_{1}\ldots \mu_{j} }^j \left( \frac{\partial \mathcal{L} }{\partial f_{i,\mu_1\dots\mu_j}}\right)=0

{{Main|Lagrangian (field theory)}}

Let $M$ be a smooth manifold, and let $C^\backslash infty([a,b])$ denote the space of smooth functions $f\backslash colon\; [a,b]\backslash to\; M$. Then, for functionals $S\backslash colon\; C^\backslash infty\; ([a,b])\backslash to\; \backslash mathbb\{R\}$ of the form

:$$

S[f]=\int_a^b (L\circ\dot{f})(t)\,\mathrm{d} t

where $L\backslash colon\; TM\backslash to\backslash mathbb\{R\}$ is the Lagrangian, the statement $\backslash mathrm\{d\}\; S\_f=0$ is equivalent to the statement that, for all $t\backslash in\; [a,b]$, each coordinate frame trivialization $(x^i,X^i)$ of a neighborhood of $\backslash dot\{f\}(t)$ yields the following $\backslash dim\; M$ equations:

:$$

\forall i:\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial X^i}\bigg|_{\dot{f}(t)}=\frac{\partial L}{\partial x^i}\bigg|_{\dot{f}(t)}.

Euler-Lagrange equations can also be written in a coordinate-free form as {{Cite book |last1=José |last2=Saletan |year=1998 |title=Classical Dynamics: A contemporary approach |url=https://www.cambridge.org/in/academic/subjects/physics/general-and-classical-physics/classical-dynamics-contemporary-approach,%20https://www.cambridge.org/in/academic/subjects/physics/general-and-classical-physics |access-date=2023-09-12 |publisher=Cambridge University Press |language=en |isbn=9780521636360}}

:$$

\mathcal{L}_\Delta \theta_L=dL

where $\backslash theta\_L$ is the canonical momenta 1-form corresponding to the Lagrangian $L$. The vector field generating time translations is denoted by $\backslash Delta$ and the Lie derivative is denoted by $\backslash mathcal\{L\}$. One can use local charts $(q^\backslash alpha,\backslash dot\{q\}^\backslash alpha)$ in which $\backslash theta\_L=\backslash frac\{\backslash partial\; L\}\{\backslash partial\; \backslash dot\{q\}^\backslash alpha\}dq^\backslash alpha$ and $\backslash Delta:=\backslash frac\{d\}\{dt\}=\backslash dot\{q\}^\backslash alpha\backslash frac\{\backslash partial\}\{\backslash partial\; q^\backslash alpha\}+\backslash ddot\{q\}^\backslash alpha\backslash frac\{\backslash partial\}\{\backslash partial\; \backslash dot\{q\}^\backslash alpha\}$ and use coordinate expressions for the Lie derivative to see equivalence with coordinate expressions of the Euler Lagrange equation. The coordinate free form is particularly suitable for geometrical interpretation of the Euler Lagrange equations.

{{Wiktionary|Euler–Lagrange equation}}

• Lagrangian mechanics
• Hamiltonian mechanics
• Analytical mechanics
• Beltrami identity
• Functional derivative

{{Reflist}}

• {{springer|title=Lagrange equations (in mechanics)|id=p/l057150}}
• {{MathWorld|urlname=Euler-LagrangeDifferentialEquation|title=Euler-Lagrange Differential Equation}}
• {{planetmath| urlname=CalculusOfVariations| title=Calculus of Variations}}
• {{cite book |last=Gelfand |first=Izrail Moiseevich |author-link=Israel Gelfand |title=Calculus of Variations |publisher=Dover |year=1963 |isbn=0-486-41448-5}}
• Roubicek, T.: ''[https://web.archive.org/web/20150510023928/http://www.wiley-vch.de/books/sample/3527411887_c17.pdf Calculus of variations]. Chap.17 in: [https://web.archive.org/web/20150510021514/http://www.wiley-vch.de/publish/en/books/forthcomingTitles/MA00/3-527-41188-7/?sID=nrgsqk516u2v9ffab8u7io1dq4 Mathematical Tools for Physicists]. (Ed. M. Grinfeld) J. Wiley, Weinheim, 2014, {{ISBN|978-3-527-41188-7}}, pp. 551–588.

{{Leonhard Euler}}

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{{DEFAULTSORT:Euler-Lagrange Equation}}

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