Calculus of Variations – Euler Lagrange Equation.

This is an approach that enables one to optimize an integral peformance measure of a path whose endpoints are given.

A path $f$ is this instance is a sufficiently smooth real-valued function defined on an bounded interval $[a,b]$ such that $f(a)=f_a,f(b)=f_b$ . The performance measure is described as an integral

$J(f) =\int_a^b L(x,f(x),f'(x))dx$

where prime denoted total derivative with respect to $x$ . Any pair of admissible paths $f,g$ are related by $f - g = \eta$ where $\eta$ is a smooth path whose endpoints are such that $\eta(a)=\eta(b)=0$ . Suppose $f$ is an optimal path. Given a zero-endpoint path $\eta$ , the behavior of the performance measure for small pertubations is given by

$\Phi(\epsilon)=J(f+\epsilon\eta) = \int_a^b L(x,f(x)+\epsilon \eta(x),f'(x)+\epsilon \eta'(x))dx$

Since $f$ is optimal, $\Phi'(0) = 0$ . This gives, on using integration by parts,

$\Phi'(0) = \int_a^b L_f \eta + L_{f'} \eta' dx = \int_a^b L_f \eta - (L_{f'})' \eta dx + L_{f'}(b)\eta(b)-L_{f'}(a)\eta(a)$

Due to the boundary constraints on the path $\eta$ ,

$\Phi'(0) = \int_a^b \left( L_f - (L_{f'})' \right) \eta dx$

Since this condition is satisfied for all such $\eta$ , it follows that the integrand is zero, that is

$L_f - (L_{f'})' = 0$

This is the celebrated Euler-Lagrange equation. If the Lagrangian $L$ does not explicitly depend on $x$ , then using the chain rule of differentiation of products,

$f' \left( L_f - (L_{f'})' \right) = L' - f''L_{f'} - f'(L_{f'})' = L' - (f'L_{f'})' = \left(L - f' L_{f'}\right)' = 0$

Hence $L-f'L_{f'} = c$ for some constant $c$ . This condition may be simpler to use in some instances.

Shortest Distance Between Points

For one of the simplest application of this result, consider the minimization of length of a curve between two points in the plane, $(a,f_a),(b,f_b)$ . The Lagrangian is the arc length, which is given by $L(x,f(x),f'(x)) = \sqrt{1 + f'(x)^2}$ and does not explicitly depend on $x$ . Hence using the EL equation specialized to this case,

$(1 + f'^2)^{1/2} - (f')^2 (1+f'^2)^{-1/2} = c$

$(1+f'^2)^{-1/2}\left( 1 + f'^2 - (f')^2 \right)= c$

Hence $f'$ is a constant and so the shortest distance between two points is a straight line.

Shortest Travel Time Between Points

This is the celebrated Brachistochrone problem. Consider a fixed-mass particle forced to travel in a gravity field without friction along a smooth curve connecting two points. The speed $v$ of the particle at any depth is given by $mgy = \frac{1}{2} mv^2$ . The speed of the particle is aligned with the tangent of the curve. Therefore in time $dt$ the particle travels $vdt = \left(1+{y'(x)}^2\right)^{1/2} dx$ . Hence starting at position $(a,f_a)$ at time 0, the particle will reach position $(b,f_b)$ at time $T$ where

$T = \int_0^T dt = \int_a^b v^{-1} \left(1+{y'(x)}^2\right)^{1/2} dx = \int_a^b \left(\frac{1+y'^2}{2gy}\right)^{1/2} dx \propto \int_a^b \left(\frac{1+y'^2}{y}\right)^{1/2} dx$

The EL equation yields the requirement $yy'^2 = C^2 - y$ which can be solved standard ode techniques.

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