PlanetMath: Beltrami identity

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Beltrami identity

(Definition)

Let be a function $\mathbb{R}\to \mathbb{R}$ , $\dot{q} = \frac{d}{d{t}}{q}$ , and $L = L(q, \dot{q}, t)$ . Begin with the time-relative Euler-Lagrange condition

$\displaystyle \frac{\partial}{\partial {q}} L - \frac{d}{d{t}}\left(\frac{\partial}{\partial {\dot{q}}} L\right) = 0.$

(1)

If $\frac{\partial}{\partial {t}}L = 0$ , then the Euler-Lagrange condition reduces to

$\displaystyle L - \dot{q}{\frac{\partial}{\partial {\dot{q}}} L} = C,$

(2)

which is the Beltrami identity. In the calculus of variations, the ability to use the Beltrami identity can vastly simplify problems, and as it happens, many physical problems have $\frac{\partial}{\partial {t}}L = 0$ .

In space-relative terms, with $q' :=\frac{d}{d{x}}q$ , we have

$\displaystyle \frac{\partial}{\partial {q}} L - \frac{d}{d{x}}{\frac{\partial}{\partial {q'}} L} = 0.$

(3)

If $\frac{\partial}{\partial {x}}L = 0$ , then the Euler-Lagrange condition reduces to

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

(4)

To derive the Beltrami identity, note that

$\displaystyle \frac{d}{d{t}}\left(\dot{q}\frac{\partial}{\partial {\dot{q}}}L\r... ...{q}}}L + \dot{q}\frac{d}{d{t}}\left(\frac{\partial}{\partial {\dot{q}}}L\right)$

(5)

Multiplying (1) by $\dot{q}$ , we have

$\displaystyle \dot{q}\frac{\partial}{\partial {q}} L - \dot{q}\frac{d}{d{t}}\left(\frac{\partial}{\partial {\dot{q}}} L\right) = 0.$

(6)

Now, rearranging (5) and substituting in for the rightmost term of (6), we obtain

$\displaystyle \dot{q}\frac{\partial}{\partial {q}} L + \ddot{q}\frac{\partial}{... ...L - \frac{d}{d{t}}\left(\dot{q}\frac{\partial}{\partial {\dot{q}}}L\right) = 0.$

(7)

Now consider the total derivative

$\displaystyle \frac{d}{d{t}}L(q, \dot{q}, t) = \dot{q}\frac{\partial}{\partial ... ... \ddot{q}\frac{\partial}{\partial {\dot{q}}}L + \frac{\partial}{\partial {t}}L.$

(8)

If $\frac{\partial}{\partial {t}}L = 0$ , then we can substitute in the left-hand side of (8) for the leading portion of (7) to get

$\displaystyle \frac{d}{d{t}}L - \frac{d}{d{t}}\left(\dot{q}\frac{\partial}{\partial {\dot{q}}}L\right) = 0.$

(9)

Integrating with respect to

, we arrive at

$\displaystyle L - \dot{q}{\frac{\partial}{\partial {\dot{q}}} L} = C,$

(10)

which is the Beltrami identity.

"Beltrami identity" is owned by Mravinci. [ full author list (2) | owner history (2) ]

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Cross-references: side, derivative, terms, calculus of variations, Euler-Lagrange condition, function
There are 3 references to this entry.

This is version 5 of Beltrami identity, born on 2002-02-16, modified 2006-08-10.
Object id is 2013, canonical name is BeltramiIdentity.
Accessed 6961 times total.

Classification:

AMS MSC:

47A60 (Operator theory :: General theory of linear operators :: Functional calculus)

Pending Errata and Addenda

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