# 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,

$L-u'{\frac {\partial L}{\partial u'}}=C\,,$ where C is a constant.

## Derivation

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

${\frac {\partial L}{\partial u}}={\frac {d}{dx}}{\frac {\partial L}{\partial u'}}\,.$ Multiplying both sides by u,

$u'{\frac {\partial L}{\partial u}}=u'{\frac {d}{dx}}{\frac {\partial L}{\partial u'}}\,.$ According to the chain rule,

${dL \over dx}={\partial L \over \partial u}u'+{\partial L \over \partial u'}u''+{\partial L \over \partial x}\,,$ where u′′ = du′/dx = d2u / dx2.

Rearranging this yields

$u'{\partial L \over \partial u}={dL \over dx}-{\partial L \over \partial u'}u''-{\partial L \over \partial x}\,.$ Thus, substituting this expression for u′ ∂L/∂u into the second equation of this derivation,

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

${d \over dx}\left({L-u'{\frac {\partial L}{\partial u'}}}\right)={\partial L \over \partial x}\,.$ For the case of L / ∂x = 0, this reduces to

${d \over dx}\left({L-u'{\frac {\partial L}{\partial u'}}}\right)=0\,,$ so that taking the antiderivative results in the Beltrami identity,

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

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