# 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[1]

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

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,[2]

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

where C is a constant.[3]

## Derivation

The following derivation of the Beltrami identity[4] starts with the Euler–Lagrange equation,

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

Multiplying both sides by u,

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

According to the chain rule,

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

where u′′ = du′/dx = d2u / dx2.

Rearranging this yields

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

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

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

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

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

and rearranging,

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

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

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

so that taking the antiderivative results in the Beltrami identity,

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

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

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

The integrand

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

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

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

Substituting for 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 A being half the above constant, 1/(2C ²), and φ being a variable. These are the parametric equations for a cycloid.[5]

## 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. ^ Thus, the Legendre transform of the Lagrangian, the Hamiltonian, is constant on the dynamical path.
4. ^ This derivation of the Beltrami identity corresponds to the one at — Weisstein, Eric W. "Beltrami Identity." From MathWorld--A Wolfram Web Resource.
5. ^ 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.