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Euler–Lagrange equation: Difference between revisions - Wikipedia

Euler–Lagrange equation: Difference between revisions

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==History==
[[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.
 
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.<ref>[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 }}</ref>
 
==Statement==
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
<math display="block"> S[\boldsymbol q] = \int_a^b L(t,\boldsymbol q(t),\dot{\boldsymbol q}(t))\, dt.</math>
 
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We wish to find a function <math>f</math> which satisfies the boundary conditions <math>f(a) = A</math>, <math>f(b) = B</math>, and which extremizes the functional
<math display="block"> J[f] = \int_a^b L(x,f(x),f'(x))\, \mathrm{d}x\ . </math>
 
We assume that <math>L</math> is twice continuously differentiable.<ref name='CourantP184'>{{harvnb|Courant|Hilbert|1953|p=184}}</ref> A weaker assumption can be used, but the proof becomes more difficult.{{Citation needed|date=September 2013}}
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If <math>f</math> extremizes the functional subject to the boundary conditions, then any slight perturbation of <math>f</math> that preserves the boundary values must either increase <math>J</math> (if <math>f</math> is a minimizer) or decrease <math>J</math> (if <math>f</math> is a maximizer).
 
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>
where <math> L_\varepsilon = L(x, \, g_\varepsilon (x), \, g_\varepsilon' (x) ) </math> .
 
We now wish to calculate the [[total derivative]] of <math> J_\varepsilonPhi</math> with respect to ''εいぷしろん''.
<math display="block"> \begin{align}\frac{\mathrm{d} J_\varepsilonPhi}{\mathrm{d} \varepsilon} &= \frac{\mathrm d}{\mathrm d\varepsilon}\int_a^b L_L(x,f(x)+\varepsilon\eta(x), \mathrm{d}f'(x = \int_a^b \frac{\partial L_)+\varepsilon}{\partial \varepsilon}eta'(x)) \, \mathrm{d}x. </math>\\
&= \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}</math>
 
The third 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>.
It follows from the total derivative that
<math display="block"> \begin{align}
\frac{\partial L_\varepsilon}{\partial \varepsilon} & =\frac{\partial x}{\partial \varepsilon}\frac{\partial L_\varepsilon}{\partial x} + \frac{\partial g_\varepsilon}{\partial \varepsilon}\frac{\partial L_\varepsilon}{\partial g_\varepsilon} + \frac{\partial g_\varepsilon'}{\partial\varepsilon}\frac{\partial L_\varepsilon}{\partial g_\varepsilon'} \\
& = \frac{\partial g_\varepsilon}{\partial \varepsilon}\frac{\partial L_\varepsilon}{\partial g_\varepsilon}+\frac{\partial g'_\varepsilon}{\partial \varepsilon}\frac{\partial L_\varepsilon}{\partial g'_\varepsilon} \\
& = \eta(x) \frac{\partial L_\varepsilon}{\partial g_\varepsilon} + \eta'(x) \frac{\partial L_\varepsilon}{\partial g_\varepsilon'} \ .
\end{align}</math>
 
The second line follows from the fact thatWhen <math>\varepsilon x= 0</math> does not depend on, <math> \varepsilon Phi</math>, i.e.has <math>an \frac{\partial[[extremum]] x}{\partialvalue, \varepsilon}so = 0</math>.that
<math display="block"> \left.\frac{\mathrm d J_\varepsilonPhi}{\mathrm d\varepsilon}\right|_{\varepsilon=0} = \int_a^b \left[ \eta(x) \frac{\partial L}{\partial f}(x,f(x),f'(x)) + \eta'(x) \frac{\partial L}{\partial f'}(x,f(x),f'(x)) \,\right]\,\mathrm{d}x = 0 \ .</math>
 
So
<math display="block"> \frac{\mathrm{d} J_\varepsilon}{\mathrm{d} \varepsilon} = \int_a^b \left[\eta(x) \frac{\partial L_\varepsilon}{\partial g_\varepsilon} + \eta'(x) \frac{\partial L_\varepsilon}{\partial g_\varepsilon'} \, \right]\,\mathrm{d}x \, . </math>
When <math>\varepsilon = 0</math> we have <math>g_{\varepsilon} = f</math>, <math>L_{\varepsilon} = L(x,f(x),f'(x))</math> and <math>J_{\varepsilon}</math> has an [[extremum]] value, so that
<math display="block"> \left.\frac{\mathrm d J_\varepsilon}{\mathrm d\varepsilon}\right|_{\varepsilon=0} = \int_a^b \left[ \eta(x) \frac{\partial L}{\partial f} + \eta'(x) \frac{\partial L}{\partial f'} \,\right]\,\mathrm{d}x = 0 \ .</math>
 
The next step is to use [[integration by parts]] on the second term of the integrand, yielding
<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>
 
Using the boundary conditions <math>\eta (a) = \eta (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>
 
Applying the [[fundamental lemma of calculus of variations]] now yields the Euler–Lagrange equation
<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=Alternative derivation of the one-dimensional Euler–Lagrange equation|proof=
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==Example==
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.
:<math> \text{s} = \int_{a}^{b} \sqrt{\mathrm{d}x^2+\mathrm{d}y^2} = \int_{a}^{b} \sqrt{1+y'^2}\,\mathrm{d}x,</math>
the integrand function being <math display="inline"> L(x,y, y') = \sqrt{1+y'^2} </math>.
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\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)}.
</math>
Euler-Lagrange equations can also be written in a coordinate-free form as <ref>{{Cite web |last=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 |website=Cambridge University Press |language=en |isbn=9780521636360}}</ref>
 
:<math>
\mathcal{L}_\Delta \theta_L=dL
</math>
where <math>\theta_L</math> is the canonical momenta [[One-form (differential geometry)|1-form]] corresponding to the Lagrangian <math>L</math>. The vector field generating time translations is denoted by <math>\Delta</math> and the [[Lie derivative]] is denoted by <math>\mathcal{L}</math>. One can use local charts <math>(q^\alpha,\dot{q}^\alpha)</math> in which <math>\theta_L=\frac{\partial L}{\partial \dot{q}^\alpha}dq^\alpha</math> and <math>\Delta:=\frac{d}{dt}=\dot{q}^\alpha\frac{\partial}{\partial q^\alpha}+\ddot{q}^\alpha\frac{\partial}{\partial \dot{q}^\alpha}</math> 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.
 
==See also==
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{{DEFAULTSORT:Euler-Lagrange Equation}}
[[Category:Eponymous equations of mathematics]]
[[Category:Eponymous equations of physics]]
[[Category:Ordinary differential equations]]
[[Category:Partial differential equations]]