# Beltrami identity

The Beltrami identity, named after Eugenio Beltrami, is a special case 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 $a$ and $b$ are constants and $u'(x)={\frac {du}{dx}}$ .

If ${\frac {\partial L}{\partial x}}=0$ , then the Euler–Lagrange equation reduces to the Beltrami identity,

$L-u'{\frac {\partial L}{\partial u'}}=C\,,$ where C is a constant.[note 1]

## Derivation

By the chain rule, the derivative of L is

${\frac {dL}{dx}}={\frac {\partial L}{\partial x}}{\frac {dx}{dx}}+{\frac {\partial L}{\partial u}}{\frac {du}{dx}}+{\frac {\partial L}{\partial u'}}{\frac {du'}{dx}}\,.$

Because ${\frac {\partial L}{\partial x}}=0$ , we write

${\frac {dL}{dx}}={\frac {\partial L}{\partial u}}u'+{\frac {\partial L}{\partial u'}}u''\,.$

Then by combining this with the Euler–Lagrange equation,

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

We get the following expression,

${\frac {dL}{dx}}=u'{\frac {d}{dx}}{\frac {\partial L}{\partial u'}}+u''{\frac {\partial L}{\partial u'}}\,.$

By the product rule, the right side is equivalent to

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

By integrating both sides and putting both terms on one side, we get the Beltrami identity,

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

## Applications

### Solution to the brachistochrone problem

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, ${\frac {1}{2C^{2}}}$ , and $\phi$  being a variable. These are the parametric equations for a cycloid.