# Beltrami identity

The Beltrami identity, named after Eugenio Beltrami, is a special case of the Euler–Lagrange equation in the calculus of variations.

Eugenio Beltrami

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

${\displaystyle I[u]=\int _{a}^{b}L[x,u(x),u'(x)]\,dx\,,}$

where ${\displaystyle a}$ and ${\displaystyle b}$ are constants and ${\displaystyle u'(x)={\frac {du}{dx}}}$.[1]

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

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

where C is a constant.[2][note 1]

## Derivation

By the chain rule, the derivative of L is

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

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

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

Then by combining this with the Euler–Lagrange equation,

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

We get the following expression,

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

By the product rule, the right side is equivalent to

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

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

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

## Applications

### Solution to the brachistochrone problem

The solution to the brachistochrone problem is the cycloid.

An example of an application of the Beltrami identity is the brachistochrone problem, which involves finding the curve ${\displaystyle y=y(x)}$  that minimizes the integral

${\displaystyle I[y]=\int _{0}^{a}{\sqrt {{1+y'^{\,2}} \over y}}dx\,.}$

The integrand

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

does not depend explicitly on the variable of integration ${\displaystyle x}$ , so the Beltrami identity applies,

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

Substituting for ${\displaystyle L}$  and simplifying,

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

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

${\displaystyle x=A(\phi -\sin \phi )}$
${\displaystyle y=A(1-\cos \phi )}$

with ${\displaystyle A}$  being half the above constant, ${\displaystyle {\frac {1}{2C^{2}}}}$ , and ${\displaystyle \phi }$  being a variable. These are the parametric equations for a cycloid.[3]

## Notes

1. ^ Thus, the Legendre transform of the Lagrangian, the Hamiltonian, is constant along the dynamical path.

## References

1. ^ Courant R, Hilbert D (1953). Methods of Mathematical Physics. Vol. I (First English ed.). New York: Interscience Publishers, Inc. p. 184. ISBN 978-0471504474.
2. ^ Weisstein, Eric W. "Euler-Lagrange Differential Equation." From MathWorld--A Wolfram Web Resource. See Eq. (5).
3. ^ This solution of the Brachistochrone problem corresponds to the one in — Mathews, Jon; Walker, RL (1965). Mathematical Methods of Physics. New York: W. A. Benjamin, Inc. pp. 307–9.