Beltrami identity
| Part of a series of articles about |
| Calculus |
|---|
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
- [math]\displaystyle{ I[u]=\int_a^b L[x,u(x),u'(x)] \, dx \, , }[/math]
where [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] are constants and [math]\displaystyle{ u'(x) = \frac{du}{dx} }[/math].[1]
If [math]\displaystyle{ \frac{\partial L}{\partial x} = 0 }[/math], then the Euler–Lagrange equation reduces to the Beltrami identity,
[math]\displaystyle{ L-u'\frac{\partial L}{\partial u'}=C \, , }[/math]
where C is a constant.[2][note 1]
Derivation
By the chain rule, the derivative of L is
- [math]\displaystyle{ \frac{dL}{dx} = \frac{\partial L}{\partial x} \frac{dx}{dx} + \frac{\partial L}{\partial u} \frac{du}{dx} + \frac{\partial L}{\partial u'} \frac{du'}{dx} \, . }[/math]
Because [math]\displaystyle{ \frac{\partial L}{\partial x} = 0 }[/math], we write
- [math]\displaystyle{ \frac{dL}{dx} = \frac{\partial L}{\partial u} u' + \frac{\partial L}{\partial u'} u'' \, . }[/math]
We have an expression for [math]\displaystyle{ \frac{\partial L}{\partial u} }[/math] from the Euler–Lagrange equation,
- [math]\displaystyle{ \frac{\partial L}{\partial u} = \frac{d}{dx} \frac{\partial L}{\partial u'} \, }[/math]
that we can substitute in the above expression for [math]\displaystyle{ \frac{dL}{dx} }[/math] to obtain
- [math]\displaystyle{ \frac{dL}{dx} =u'\frac{d}{dx} \frac{\partial L}{\partial u'} + u''\frac{\partial L}{\partial u'} \, . }[/math]
By the product rule, the right side is equivalent to
- [math]\displaystyle{ \frac{dL}{dx} = \frac{d}{dx} \left( u' \frac{\partial L}{\partial u'} \right) \, . }[/math]
By integrating both sides and putting both terms on one side, we get the Beltrami identity,
- [math]\displaystyle{ L - u'\frac{\partial L}{\partial u'} = C \, . }[/math]
Applications
Solution to the brachistochrone problem
An example of an application of the Beltrami identity is the brachistochrone problem, which involves finding the curve [math]\displaystyle{ y = y(x) }[/math] that minimizes the integral
- [math]\displaystyle{ I[y] = \int_0^a \sqrt { {1+y'^{\, 2}} \over y } dx \, . }[/math]
The integrand
- [math]\displaystyle{ L(y,y') = \sqrt{ {1+y'^{\, 2}} \over y } }[/math]
does not depend explicitly on the variable of integration [math]\displaystyle{ x }[/math], so the Beltrami identity applies,
- [math]\displaystyle{ L-y'\frac{\partial L}{\partial y'}=C \, . }[/math]
Substituting for [math]\displaystyle{ L }[/math] and simplifying,
- [math]\displaystyle{ y(1+y'^{\, 2}) = 1/C^2 ~~\text {(constant)} \, , }[/math]
which can be solved with the result put in the form of parametric equations
- [math]\displaystyle{ x = A(\phi - \sin \phi) }[/math]
- [math]\displaystyle{ y = A(1 - \cos \phi) }[/math]
with [math]\displaystyle{ A }[/math] being half the above constant, [math]\displaystyle{ \frac{1}{2C^{2}} }[/math], and [math]\displaystyle{ \phi }[/math] being a variable. These are the parametric equations for a cycloid.[3]
Solution to the catenary problem
Consider a string with uniform density [math]\displaystyle{ \mu }[/math] of length [math]\displaystyle{ l }[/math] suspended from two points of equal height and at distance [math]\displaystyle{ D }[/math]. By the formula for arc length, [math]\displaystyle{ l = \int_S dS = \int_{s_1}^{s_2} \sqrt{1+y'^2}dx, }[/math] where [math]\displaystyle{ S }[/math] is the path of the string, and [math]\displaystyle{ s_1 }[/math] and [math]\displaystyle{ s_2 }[/math] are the boundary conditions.
The curve has to minimize its potential energy [math]\displaystyle{ U = \int_S g\mu y\cdot dS = \int_{s_1}^{s_2} g\mu y\sqrt{1+y'^2} dx, }[/math] and is subject to the constraint [math]\displaystyle{ \int_{s_1}^{s_2} \sqrt{1+y'^2} dx = l , }[/math] where [math]\displaystyle{ g }[/math] is the force of gravity.
Because the independent variable [math]\displaystyle{ x }[/math] does not appear in the integrand, the Beltrami identity may be used to express the path of the string as a separable first order differential equation
[math]\displaystyle{ L - \frac{\partial L}{\partial y\prime} = \mu gy\sqrt{1+y\prime ^2} + \lambda \sqrt{1+y\prime ^2} - \left[\mu gy\frac{y\prime ^2}{\sqrt{1+y\prime ^2}} + \lambda \frac{y\prime ^2}{\sqrt{1+y\prime ^2}}\right] = C, }[/math] where [math]\displaystyle{ \lambda }[/math] is the Lagrange multiplier.
It is possible to simplify the differential equation as such: [math]\displaystyle{ \frac{g\rho y - \lambda }{\sqrt{1+y'^2}} = C. }[/math]
Solving this equation gives the hyperbolic cosine, where [math]\displaystyle{ C_0 }[/math] is a second constant obtained from integration
[math]\displaystyle{ y = \frac{C}{\mu g}\cosh \left[ \frac{\mu g}{C} (x + C_0) \right] - \frac{\lambda}{\mu g}. }[/math]
The three unknowns [math]\displaystyle{ C }[/math], [math]\displaystyle{ C_0 }[/math], and [math]\displaystyle{ \lambda }[/math] can be solved for using the constraints for the string's endpoints and arc length [math]\displaystyle{ l }[/math], though a closed-form solution is often very difficult to obtain.
Notes
- ↑ Thus, the Legendre transform of the Lagrangian, the Hamiltonian, is constant along the dynamical path.
References
- ↑ Methods of Mathematical Physics. I (First English ed.). New York: Interscience Publishers, Inc.. 1953. p. 184. ISBN 978-0471504474.
- ↑ Weisstein, Eric W. "Euler-Lagrange Differential Equation." From MathWorld--A Wolfram Web Resource. See Eq. (5).
- ↑ 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.
 |  |
