User:Salix alba/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
 * $$I[u]=\int_a^b L[x,u(x),u'(x)] \, dx \, ,$$

where $a, b$ are constants and $u&prime;(x) = du / dx$.

For the special case of $∂L / ∂x = 0$, the Euler–Lagrange equation reduces to the Beltrami identity,

where $C$ is a constant.

Derivation
The following derivation of the Beltrami identity starts with the Euler–Lagrange equation,
 * $$ \frac{\partial L}{\partial u} =\frac{d}{dx} \frac{\partial L}{\partial u'} \, . $$

Multiplying both sides by $u&prime;$,
 * $$ u'\frac{\partial L}{\partial u} =u'\frac{d}{dx} \frac{\partial L}{\partial u'} \, . $$

According to the chain rule,
 * $$ {dL \over dx} = {\partial L \over \partial u}u' + {\partial L \over \partial u'}u'' + {\partial L \over \partial x}  \, ,$$

where $u&prime;&prime; = du&prime;/dx = d^{2}u / dx^{2}$.

Rearranging this yields
 * $$ u' {\partial L \over \partial u} = {dL \over dx} - {\partial L \over \partial u'}u'' - {\partial L \over \partial x} \, . $$

Thus, substituting this expression for $u&prime; ∂L/∂u$ into the second equation of this derivation,
 * $$ {dL \over dx} - {\partial L \over \partial u'}u'' - {\partial L \over \partial x} -u'\frac{d}{dx} \frac{\partial L}{\partial u'} = 0 \, . $$

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

For the case of $∂L / ∂x = 0$, this reduces to
 * $$ {d \over dx} \left( { L - u'\frac{\partial L}{\partial u'} } \right) = 0 \,, $$

so that taking the antiderivative results in the Beltrami identity,
 * $$ L - u'\frac{\partial L}{\partial u'}   = C \,, $$

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
 * $$ I[y] = \int_0^a \sqrt { {1+y'^{\, 2}} \over y } dx \, . $$

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, 1/(2C ²), and $&phi;$ being a variable. These are the parametric equations for a cycloid.