Beltrami identity

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The Beltrami identity is an identity in the calculus of variations. It says that a function u which is an extremal of the integral

$I(u)=\int_a^b L(x,u,u') \, dx$

satisfies the differential equation

$\frac{d}{dx}\left(L-u'\frac{\partial L}{\partial u'}\right)-\frac{\partial L}{\partial x}=0.$

In the case that L is the Lagrangian of a Mechanical system, and if L does not depend on x explicitly, which corresponds to a Lagrangian which does not explicitly depend on time, the Beltrami identity states that the Hamiltonian associated to the Lagrangian is a conserved energy.

Proof

Define the conjugate momentum p to be the partial derivative of the function L with respect to u'.

$p = {\partial L \over \partial u'}$

The Euler-Lagrange equation says

${dp\over dx} - {\partial L \over \partial u} = 0$

$p' = {\partial L \over \partial u}$

define the Hamiltonian H to be the Legendre transform of L with respect to u:

$H = p u' - L \,$

Then

$H' = {dH \over dx} = p' u' + p u'' - {\partial L \over \partial u'} u'' - {\partial L \over \partial u} u' - {\partial L \over \partial x}$

The second and third term cancel, by the definition of p, and the first and fourth terms cancel by the Euler Lagrange equation. This leaves the Beltrami identity:

$H' = - {\partial L \over \partial x}$

which is a special case of Noether's theorem.

Application

In case that L is independent of x, then the Beltrami identity states that H is constant

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

This can be used to solve the Euler-Lagrange equation, in the same way that conservation of energy can be used to solve Newton's law of motion. The condition that H is constant is an equation for the first derivative of u, while the original Euler Lagrange equation is an equation for the second derivative of u.

For example, consider the Brachistochrone problem: find the curve minimizing the integral:

$\int_0^1 {\sqrt{1+y'^2} \over \sqrt{y}} dx$

The quantity to minimize is the integral of an L which does not depend explicitly on time,

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

so the associated Hamiltonian is constant:

$H = p y' - L = {y'^2 \over \sqrt{1+y'^2} \sqrt{y}} - {\sqrt{1+y'^2}\over \sqrt{y}} = {-1 \over \sqrt{1+y'^2}\sqrt{y}}$

$\sqrt{1+y'^2}\sqrt{y} = \text{constant}$

which simplifies to the differential equation for the cycloid.

External links

http://mathworld.wolfram.com/BeltramiIdentity.html

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