Beltrami identity


Let q⁢(t) be a function ℝ→ℝ, q˙=dd⁢t⁢q, and L=L⁢(q,q˙,t). Begin with the time-relative Euler-Lagrange conditionPlanetmathPlanetmath

∂∂⁡q⁢L-dd⁢t⁢(∂∂⁡q˙⁢L)=0. (1)

If ∂∂⁡t⁢L=0, then the Euler-Lagrange condition reduces to

L-q˙⁢∂∂⁡q˙⁢L=C, (2)

which is the Beltrami identityMathworldPlanetmath. In the calculus of variationsMathworldPlanetmath, the ability to use the Beltrami identity can vastly simplify problems, and as it happens, many physical problems have ∂∂⁡t⁢L=0.

In space-relative terms, with q′:=dd⁢x⁢q, we have

∂∂⁡q⁢L-dd⁢x⁢∂∂⁡q′⁢L=0. (3)

If ∂∂⁡x⁢L=0, then the Euler-Lagrange condition reduces to

L-q′⁢∂∂⁡q′⁢L=C. (4)

To derive the Beltrami identity, note that

dd⁢t⁢(q˙⁢∂∂⁡q˙⁢L)=q¨⁢∂∂⁡q˙⁢L+q˙⁢dd⁢t⁢(∂∂⁡q˙⁢L) (5)

Multiplying (1) by q˙, we have

q˙⁢∂∂⁡q⁢L-q˙⁢dd⁢t⁢(∂∂⁡q˙⁢L)=0. (6)

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

q˙⁢∂∂⁡q⁢L+q¨⁢∂∂⁡q˙⁢L-dd⁢t⁢(q˙⁢∂∂⁡q˙⁢L)=0. (7)

Now consider the total derivative

dd⁢t⁢L⁢(q,q˙,t)=q˙⁢∂∂⁡q⁢L+q¨⁢∂∂⁡q˙⁢L+∂∂⁡t⁢L. (8)

If ∂∂⁡t⁢L=0, then we can substitute in the left-hand side of (8) for the leading portion of (7) to get

dd⁢t⁢L-dd⁢t⁢(q˙⁢∂∂⁡q˙⁢L)=0. (9)

Integrating with respect to t, we arrive at

L-q˙⁢∂∂⁡q˙⁢L=C, (10)

which is the Beltrami identity.

Title Beltrami identity
Canonical name BeltramiIdentity
Date of creation 2013-03-22 12:21:08
Last modified on 2013-03-22 12:21:08
Owner PrimeFan (13766)
Last modified by PrimeFan (13766)
Numerical id 8
Author PrimeFan (13766)
Entry type Definition
Classification msc 47A60
Related topic CalculusOfVariations
Related topic EulerLagrangeDifferentialEquation