Английская Википедия:Beltrami identity
Шаблон:Short description Шаблон: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>I[u]=\int_a^b L[x,u(x),u'(x)] \, dx \, ,</math>
where <math>a</math> and <math>b</math> are constants and <math>u'(x) = \frac{du}{dx}</math>.[1]
If <math>\frac{\partial L}{\partial x} = 0</math>, then the Euler–Lagrange equation reduces to the Beltrami identity,
Шаблон:Equation box 1 where Шаблон:Math is a constant.[2][note 1]
Derivation
By the chain rule, the derivative of Шаблон:Math is
- <math> \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> \frac{\partial L}{\partial x} = 0 </math>, we write
- <math> \frac{dL}{dx} = \frac{\partial L}{\partial u} u' + \frac{\partial L}{\partial u'} u \, . </math>
We have an expression for <math> \frac{\partial L}{\partial u}</math> from the Euler–Lagrange equation,
- <math> \frac{\partial L}{\partial u} = \frac{d}{dx} \frac{\partial L}{\partial u'} \, </math>
that we can substitute in the above expression for <math> \frac{dL}{dx}</math> to obtain
- <math> \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> \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> 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>y = y(x)</math> that minimizes the integral
- <math> I[y] = \int_0^a \sqrt { {1+y'^{\, 2}} \over y } dx \, . </math>
The integrand
- <math> L(y,y') = \sqrt{ {1+y'^{\, 2}} \over y } </math>
does not depend explicitly on the variable of integration <math>x</math>, so the Beltrami identity applies,
- <math>L-y'\frac{\partial L}{\partial y'}=C \, .</math>
Substituting for <math>L</math> and simplifying,
- <math> 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>x = A(\phi - \sin \phi) </math>
- <math>y = A(1 - \cos \phi) </math>
with <math>A</math> being half the above constant, <math>\frac{1}{2C^{2}}</math>, and <math>\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>\mu</math> of length <math>l</math> suspended from two points of equal height and at distance <math>D</math>. By the formula for arc length, <math display=block>l = \int_S dS = \int_{s_1}^{s_2} \sqrt{1+y'^2}dx, </math> where <math>S</math> is the path of the string, and <math>s_1</math> and <math>s_2</math> are the boundary conditions.
The curve has to minimize its potential energy <math display=block> 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 display=block> \int_{s_1}^{s_2} \sqrt{1+y'^2} dx = l ,</math> where <math>g</math> is the force of gravity.
Because the independent variable <math>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 display=block>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>\lambda</math> is the Lagrange multiplier.
It is possible to simplify the differential equation as such: <math display=block>\frac{g\rho y - \lambda }{\sqrt{1+y'^2}} = C.</math>
Solving this equation gives the hyperbolic cosine, where <math>C_0</math> is a second constant obtained from integration
<math display=block>y = \frac{C}{\mu g}\cosh \left[ \frac{\mu g}{C} (x + C_0) \right] - \frac{\lambda}{\mu g}. </math>
The three unknowns <math>C</math>, <math>C_0</math>, and <math>\lambda</math> can be solved for using the constraints for the string's endpoints and arc length <math>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
- ↑ Шаблон:Cite book
- ↑ 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 — Шаблон:Cite book
