Editing Euler–Lagrange equation (section)

==Statement==
Let <math>(X,L)</math> be a [[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 X \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. </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>

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

{{Equation box 1
|indent =:
|equation = <math>\frac{\partial L}{\partial q^i}(t,\boldsymbol q(t),\dot{\boldsymbol q}(t))-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q^i}(t,\boldsymbol q(t),\dot{\boldsymbol q}(t)) = 0,
\quad i = 1, \dots, n.</math>
|border colour = #50C878
|background colour = #ECFCF4}} 
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 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 <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}}

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>f + \varepsilon \eta</math> be the result of such a perturbation <math>\varepsilon \eta</math> of <math>f</math>, where <math>\varepsilon</math> is small and <math>\eta</math> is a differentiable function satisfying <math>\eta (a) = \eta (b) = 0</math>. Then define
<math display="block"> \Phi(\varepsilon) =  J[f+\varepsilon\eta] = \int_a^b L(x,f(x)+\varepsilon\eta(x), f'(x)+\varepsilon\eta'(x))\, \mathrm{d}x \ .</math>

We now wish to calculate the [[total derivative]] of <math> \Phi</math> with respect to ''ε''.
<math display="block">\begin{align}\frac{\mathrm{d} \Phi}{\mathrm{d} \varepsilon} &= \frac{\mathrm d}{\mathrm d\varepsilon}\int_a^b L(x,f(x)+\varepsilon\eta(x), f'(x)+\varepsilon\eta'(x)) \, \mathrm{d}x \\
&= \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>.

When <math>\varepsilon = 0</math>, <math>\Phi</math> has an [[extremum]] value, so that
<math display="block"> \left.\frac{\mathrm d \Phi}{\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>

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>
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{{math proof|title=Alternative derivation of the one-dimensional Euler–Lagrange equation|proof=
Given a functional
<math display="block">J = \int^b_a L(t, y(t), y'(t))\,\mathrm{d}t</math>
on <math>C^1([a, b])</math> with the boundary conditions <math>y(a) = A</math> and <math>y(b) = B</math>, we proceed by approximating the extremal curve by a polygonal line with <math>n</math> segments and passing to the limit as the number of segments grows arbitrarily large.

Divide the interval <math>[a, b]</math> into <math>n</math> equal segments with endpoints <math>t_0 = a, t_1, t_2, \ldots, t_n = b</math> and let <math>\Delta t = t_k - t_{k - 1}</math>. Rather than a smooth function <math>y(t)</math> we consider the polygonal line with vertices <math>(t_0, y_0),\ldots,(t_n, y_n)</math>, where <math>y_0 = A</math> and <math>y_n = B</math>. Accordingly, our functional becomes a real function of <math>n - 1</math> variables given by
<math display="block">J(y_1, \ldots, y_{n - 1}) \approx \sum^{n - 1}_{k = 0}L\left(t_k, y_k, \frac{y_{k + 1} - y_k}{\Delta t}\right)\Delta t.</math>

Extremals of this new functional defined on the discrete points <math>t_0,\ldots,t_n</math> correspond to points where
<math display="block">\frac{\partial J(y_1,\ldots,y_n)}{\partial y_m} = 0.</math>

Note that change of <math> y_m </math> affects L not only at m but also at m-1 for the derivative of the 3rd argument. 
<math display="block"> L(\text{3rd argument}) \left( \frac{y_{m+1} - (y_{m} + \Delta y_{m})}{\Delta t} \right) = L \left(\frac{y_{m+1} - y_{m}}{\Delta t}\right) - \frac{\partial L}{\partial y'} \frac{\Delta y_m}{\Delta t} 
</math><math display="block">L \left( \frac{(y_{m} + \Delta y_{m}) - y_{m-1}}{\Delta t} \right) 
= L \left(\frac{y_{m} - y_{m-1}}{\Delta t}\right) + \frac{\partial L}{\partial y'} \frac{\Delta y_m}{\Delta t}</math>

Evaluating the partial derivative gives
<math display="block">\frac{\partial J}{\partial y_m} = L_y\left(t_m, y_m, \frac{y_{m + 1} - y_m}{\Delta t}\right)\Delta t + L_{y'}\left(t_{m - 1}, y_{m - 1}, \frac{y_m - y_{m - 1}}{\Delta t}\right) - L_{y'}\left(t_m, y_m, \frac{y_{m + 1} - y_m}{\Delta t}\right).</math>

Dividing the above equation by <math>\Delta t</math> gives
<math display="block">\frac{\partial J}{\partial y_m \Delta t} = L_y\left(t_m, y_m, \frac{y_{m + 1} - y_m}{\Delta t}\right) - \frac{1}{\Delta t}\left[L_{y'}\left(t_m, y_m, \frac{y_{m + 1} - y_m}{\Delta t}\right) - L_{y'}\left(t_{m - 1}, y_{m - 1}, \frac{y_m - y_{m - 1}}{\Delta t}\right)\right],</math>
and taking the limit as <math>\Delta t \to 0</math> of the right-hand side of this expression yields
<math display="block">L_y - \frac{\mathrm{d}}{\mathrm{d}t}L_{y'} = 0.</math>

The left hand side of the previous equation is the [[functional derivative]] <math>\delta J/\delta y</math> of the functional <math>J</math>. 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.
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