Euler–Lagrange equation: Difference between revisions

{{Short description|Second-order ordinarypartial differential equation describing motion of mechanical system}}

In [[Lagrangian mechanics]], according to [[Hamilton's principle]] of stationary action, the evolution of a physical system is described by the solutions to the Euler equation for the [[action (physics)#Action (functional)|action]] of the system. In this context Euler equations are usually called '''Lagrange equations'''. In [[classical mechanics]],<ref name="Goldstein"> {{cite book|author1-link=Herbert Goldstein|author2-link=Charles P. Poole|first1=H. |last1=Goldstein|first2=C.P.|last2=Poole|first3=J.|last3=Safko|title=Classical Mechanics|publisher=Addison Wesley|year=2014|edition=3rd}}</ref> it is equivalent to [[Newton's laws of motion]]; indeed, the Euler-Lagrange equations will produce the same equations as Newton's Laws. This is particularly useful when analyzing systems whose force vectors are particularly complicated. itIt has the advantage that it takes the same form in any system of [[generalized coordinate]]s, and it is better suited to generalizations. In [[classical field theory]] there is an [[classical field theory#Lagrangian dynamics|analogous equation]] to calculate the dynamics of a [[field (physics)|field]].

[[File:Tautochrone curve.gif|thumb|The Euler–Lagrange equation was in connection with their studies of the [[tautochrone]] 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.

Let <math>(X,L)</math> be a mechanical[[real dynamical system]] with <math>n</math> degrees of freedom. Here <math>X</math> is the [[configuration space (physics)|configuration space]] and <math>L=L(t,{\boldsymbol q}(t), {\boldsymbol v}(t))</math> the ''[[Lagrangian mechanics#Lagrangian|Lagrangian]]'', i.e. a smooth real-valued function such that <math>{\boldsymbol q}(t) \in X,</math> and <math>{\boldsymbol v}(t)</math> is an <math>n</math>-dimensional "vector of speed". (For those familiar with [[differential geometry]], <math>X</math> is a [[smooth manifold]], and <math>L : {\mathbb R}_t \times TX \to {\mathbb R},</math> where <math>TX</math> is the [[tangent bundle]] of <math>X).</math>

Let <math>{\cal P}(a,b,\boldsymbol x_a,\boldsymbol x_b)</math> be the set of smooth paths <math>\boldsymbol q: [a,b] \to X</math> for which <math>\boldsymbol q(a) = \boldsymbol x_a</math> and <math>\boldsymbol q(b) = \boldsymbol x_{b}x_b. </math>

The [[action (physics)|action functional]] <math>S : {\cal P}(a,b,\boldsymbol x_a,\boldsymbol x_b) \to \mathbb{R}</math> is defined via

A path <math>\boldsymbol q \in {\cal P}(a,b,\boldsymbol x_a,\boldsymbol x_b)</math> is a [[Lagrangian mechanics|stationary point]] of <math>S</math> if and only if

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

<math display="block"> J[f] = \int_a^b L(x,f(x),f'(x))\, \mathrm{d}x\ . </math>

Let <math>g_{\varepsilon} (x) = f (x) + \varepsilon \eta (x)</math> be the result of such a perturbation <math>\varepsilon \eta (x)</math> of <math>f</math>, where <math>\varepsilon</math> is small and <math>\eta (x)</math> is a differentiable function satisfying <math>\eta (a) = \eta (b) = 0</math>. Then define

<math display="block"> J_\Phi(\varepsilon) = J[f+\varepsilon\eta] = \int_a^b L(x,g_f(x)+\varepsilon\eta(x), g_f'(x)+\varepsilon\eta'(x) ) \, \mathrm{d}x = \int_a^b L_\varepsilon\, \mathrm{d}x .</math>

ItWe followsnow fromwish to calculate the [[total derivative]] of <math> \Phi</math> with respect to that''εいぷしろん''.

\frac{\mathrm&= d L_\varepsilon}{\mathrmint_a^b d\varepsilon} & =\frac{\mathrm d x}{\mathrm d\varepsilon}\frac{\partial L_L(x,f(x)+\varepsilon}{\partialeta(x), f'(x} )+ \frac{\mathrm d g_\varepsilon}{\mathrmeta'(x)) d\varepsilon}\frac{\partial, L_\varepsilon}{\partial g_\varepsilon} + \frac{\mathrm d g_\varepsilon'}{\mathrm d\varepsilon}\frac{\partial L_\varepsilon}{\partial g_\varepsilon'} \\x

& = \int_a^b \left[\eta(x) \frac{\partial L_\varepsilonL}{\partial g_{f} }(x,f(x)+\varepsilon}\eta(x),f'(x)+\varepsilon\eta'(x)) + \eta'(x) \frac{\partial L_\varepsilonL}{\partial g_f'}(x,f(x)+\varepsilon\eta(x),f'(x)+\varepsilon\eta'(x))\right] \mathrm{d}x \ . \end{align}</math>

The secondthird line follows from the fact that <math> x </math> does not depend on <math> \varepsilon </math>, i.e. <math> \frac{\mathrm {d} x}{\mathrm {d} \varepsilon} = 0</math>.

<math display="block"> \left.\frac{\mathrm{ d} J_\varepsilonPhi}{\mathrm{ d\varepsilon} \right|_{\varepsilon=0} = \int_a^b \left[ \eta(x) \frac{\partial L_\varepsilonL}{\partial g_\varepsilonf}(x,f(x),f'(x)) + \eta'(x) \frac{\partial L_\varepsilonL}{\partial g_\varepsilonf'}(x,f(x),f'(x)) \, \right]\,\mathrm{d}x = 0 \, . </math>

<math display="block"> \int_a^b \left[ \frac{\partial L}{\partial f}(x,f(x),f'(x)) - \frac{\mathrm{d}}{\mathrm{d}x} \frac{\partial L}{\partial f'}(x,f(x),f'(x)) \right] \eta(x)\,\mathrm{d}x + \left[ \eta(x) \frac{\partial L}{\partial f'}(x,f(x),f'(x)) \right]_a^b = 0 \ . </math>

<math display="block"> \int_a^b \left[ \frac{\partial L}{\partial f}(x,f(x),f'(x)) - \frac{\mathrm{d}}{\mathrm{d}x} \frac{\partial L}{\partial f'}(x,f(x),f'(x)) \right] \eta(x)\,\mathrm{d}x = 0 \, . </math>

<math display="block"> \frac{\partial L}{\partial f}(x,f(x),f'(x)) - \frac{\mathrm{d}}{\mathrm{d}x} \frac{\partial L}{\partial f'}(x,f(x),f'(x)) = 0 \, . </math>

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

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 (topology)|path]] [[arc length|length]] along the [[Curve#length of a curve|curve]] traced by ''y'' is as short as possible.

the integrand function being {{nowrap|1<math display=''"inline"> L''(''x'', ''y'', ''y''′) = \sqrt{{radic|1 + ''y''′ ²}}}^2} </math>.

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

where <math>L:\colon TM\to\mathbb{R}</math> is the Lagrangian, the statement <math>\mathrm{d} S_f=0</math> is equivalent to the statement that, for all <math>t\in [a,b]</math>, each coordinate frame [[fiber bundle|trivialization]] <math>(x^i,X^i)</math> of a neighborhood of <math>\dot{f}(t)</math> yields the following <math>\dim M</math> equations: