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 \( {\displaystyle u'(x)={\frac {du}{dx}}}\) .[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
The following derivation of the Beltrami identity 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 \( {\displaystyle u''={\frac {du'}{dx}}={\frac {d^{2}u}{dx^{2}}}}.\)
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 \( {\displaystyle u'{\frac {\partial L}{\partial u}}} \) 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
\( u'\frac{d}{dx}\frac{\partial L}{\partial u'}=\frac{d}{dx}\left( \frac{\partial L}{\partial u'}u' \right)-\frac{\partial L}{\partial u'}u'' \, , \)
and rearranging,
\( {\displaystyle {d \over dx}\left({L-u'{\frac {\partial L}{\partial u'}}}\right)={\partial L \over \partial x}\,.}\)
For the case of \( {\displaystyle {\frac {\partial L}{\partial x}}=0},\) 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.[3]
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
\( I[y] = \int_0^a \sqrt { {1+y'^{\, 2}} \over y } dx \, . \)
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,
\( {\displaystyle L-y'{\frac {\partial L}{\partial y'}}=C\,.}\)
Substituting for L {\displaystyle L} 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, \( {\displaystyle {\frac {1}{2C^{2}}}\) , and \( \phi \) being a variable. These are the parametric equations for a cycloid.[4]
Notes
Thus, the Legendre transform of the Lagrangian, the Hamiltonian, is constant along the dynamical path.
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).
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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