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]

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 $\partial L / \partial x = 0$ , the Euler–Lagrange equation reduces to the Beltrami identity,^[2]

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

where $C$ is a constant.^[3]

Derivation

The following derivation of the Beltrami identity^[4] 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 $u'' = du' / dx = d 2 u / dx 2$ .

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 $u' \partial L /\partial u$ 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 $\partial L / \partial x = 0$ , 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.^[5]

References

↑ Courant R, Hilbert D (1953). Methods of Mathematical Physics. Vol. I (First English ed.). New York: Interscience Publishers, Inc. p. 184. ISBN 978-0471504474.
↑ Weisstein, Eric W. "Euler-Lagrange Differential Equation." From MathWorld--A Wolfram Web Resource. See Eq. (5).
↑ Thus, the Legendre transform of the Lagrangian, the Hamiltonian, is constant on the dynamical path.
↑ This derivation of the Beltrami identity corresponds to the one at — Weisstein, Eric W. "Beltrami Identity." From MathWorld--A Wolfram Web Resource.
↑ 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.

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