Beltrami identity

The Beltrami identity, named after Eugenio Beltrami, is a simplified and less general version of the Euler–Lagrange equation in the calculus of variations.

The Euler–Lagrange equation serves to extremize action functionals of the form

$I[u]=\int_a^b L[x,u(x),u'(x)] \, dx \, ,$

where a, b are constants and u′(x) = du / dx.

For the special case of ∂L / ∂x = 0, the Euler–Lagrange equation reduces to the Beltrami identity,

$L-u'\frac{\part L}{\part u'}=C \, ,$

where C is a constant.

Derivation

The following derivation of the Beltrami identity starts with the Euler–Lagrange equation,

$\frac{\part L}{\part u} =\frac{d}{dx} \frac{\part L}{\part u'} \, .$

Multiplying both sides by u′,

$u'\frac{\part L}{\part u} =u'\frac{d}{dx} \frac{\part L}{\part u'} \, .$

According to the chain rule,

${dL \over dx} = {\part L \over \part u}u' + {\part L \over \part u'}u'' + {\part L \over \part x} \, ,$

where u′′ = du′/dx = d²u / dx².

Rearranging this yields

$u' {\part L \over \part u} = {dL \over dx} - {\part L \over \part u'}u'' - {\part L \over \part x} \, .$

Thus, substituting this expression for u′ ∂L/∂u into the second equation of this derivation,

${dL \over dx} - {\part L \over \part u'}u'' - {\part L \over \part x} -u'\frac{d}{dx} \frac{\part L}{\part u'} = 0 \, .$

By the product rule, the last term is re-expressed as

$u'\frac{d}{dx}\frac{\partial L}{\partial u'}=\frac{d}{dx}\left( \frac{\partial L}{\partial u'}u' \right)-\frac{\partial L}{\partial u'}u'' \, ,$

and rearranging,

${d \over dx} \left( { L - u'\frac{\part L}{\part u'} } \right) = {\part L \over \part x} \, .$

For the case of ∂L / ∂x = 0, this reduces to

${d \over dx} \left( { L - u'\frac{\part L}{\part u'} } \right) = 0 \, ,$

so that taking the antiderivative results in the Beltrami identity,

$L - u'\frac{\part L}{\part u'} = C \, ,$

where C is a constant.

Application

An example of an application of the Beltrami identity is the Brachistochrone problem, which involves finding the curve y = y(x) that minimizes the integral

$I[y] = \int_0^a \sqrt { {1+y'^{\, 2}} \over y } dx \, .$

The integrand

$L(y,y') = \sqrt{ {1+y'^{\, 2}} \over y }$

does not depend explicitly on the variable of integration x, so the Beltrami identity applies,

$L-y'\frac{\part L}{\part y'}=C \, .$

Substituting for L and simplifying,

$y(1+y'^{\, 2}) = 1/C^2 ~~\text {(constant)} \, ,$

which can be solved with the result put in the form of parametric equations

$x = A(\phi - \sin \phi)$

$y = A(1 - \cos \phi)$

with A being half the above constant, 1/(2C ²), and φ being a variable. These are the parametric equations for a cycloid.