Beltrami identity

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


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

where $$a$$ and $$b$$ are constants and $$u'(x) = \frac{du}{dx}$$.

If $$\frac{\partial L}{\partial x} = 0$$, then the Euler–Lagrange equation reduces to the Beltrami identity,

where $C$ is a constant.

Derivation
By the chain rule, the derivative of $L$ is
 * $$ \frac{dL}{dx} = \frac{\partial L}{\partial x} \frac{dx}{dx} + \frac{\partial L}{\partial u} \frac{du}{dx} + \frac{\partial L}{\partial u'} \frac{du'}{dx} \, . $$

Because $$ \frac{\partial L}{\partial x} = 0 $$, we write
 * $$ \frac{dL}{dx} = \frac{\partial L}{\partial u} u' + \frac{\partial L}{\partial u'} u'' \, . $$

We have an expression for $$ \frac{\partial L}{\partial u}$$ from the Euler–Lagrange equation,
 * $$ \frac{\partial L}{\partial u} = \frac{d}{dx} \frac{\partial L}{\partial u'} \, $$

that we can substitute in the above expression for $$ \frac{dL}{dx}$$ to obtain
 * $$ \frac{dL}{dx} =u'\frac{d}{dx} \frac{\partial L}{\partial u'} + u''\frac{\partial L}{\partial u'} \, . $$

By the product rule, the right side is equivalent to
 * $$ \frac{dL}{dx} = \frac{d}{dx} \left( u' \frac{\partial L}{\partial u'} \right) \, . $$

By integrating both sides and putting both terms on one side, we get the Beltrami identity,
 * $$ L - u'\frac{\partial L}{\partial u'} = C \, . $$

Solution to the brachistochrone problem
An example of an application of the Beltrami identity is the brachistochrone problem, which involves finding the curve $$y = y(x)$$ that minimizes the integral


 * $$ I[y] = \int_0^a \sqrt { {1+y'^{\, 2}} \over y } dx \, . $$

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

does not depend explicitly on the variable of integration $$x$$, so the Beltrami identity applies,
 * $$L-y'\frac{\partial L}{\partial y'}=C \, .$$

Substituting for $$L$$ and simplifying,
 * $$ y(1+y'^{\, 2}) = 1/C^2 \text {(constant)} \,, $$

which can be solved with the result put in the form of parametric equations
 * $$x = A(\phi - \sin \phi) $$
 * $$y = A(1 - \cos \phi) $$

with $$A$$ being half the above constant, $$\frac{1}{2C^{2}}$$, and $$\phi$$ being a variable. These are the parametric equations for a cycloid.

Solution to the catenary problem


Consider a string with uniform density $$\mu$$ of length $$l$$ suspended from two points of equal height and at distance $$D$$. By the formula for arc length, $$l = \int_S dS = \int_{s_1}^{s_2} \sqrt{1+y'^2}dx, $$ where $$S$$ is the path of the string, and $$s_1$$ and $$s_2$$ are the boundary conditions.

The curve has to minimize its potential energy $$ U = \int_S g\mu y\cdot dS = \int_{s_1}^{s_2} g\mu y\sqrt{1+y'^2} dx, $$ and is subject to the constraint $$ \int_{s_1}^{s_2} \sqrt{1+y'^2} dx = l ,$$ where $$g$$ is the force of gravity.

Because the independent variable $$x$$ 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

$$L - y\prime \frac{\partial L}{\partial y\prime} = \mu gy\sqrt{1+y\prime ^2} + \lambda \sqrt{1+y\prime ^2} - \left[\mu gy\frac{y\prime ^2}{\sqrt{1+y\prime ^2}} + \lambda \frac{y\prime ^2}{\sqrt{1+y\prime ^2}}\right] = C,$$ where $$\lambda$$ is the Lagrange multiplier.

It is possible to simplify the differential equation as such: $$\frac{g\rho y - \lambda }{\sqrt{1+y'^2}} = C.$$

Solving this equation gives the hyperbolic cosine, where $$C_0$$ is a second constant obtained from integration

$$y = \frac{C}{\mu g}\cosh \left[ \frac{\mu g}{C} (x + C_0) \right] - \frac{\lambda}{\mu g}. $$

The three unknowns $$C$$, $$C_0$$, and $$\lambda$$ can be solved for using the constraints for the string's endpoints and arc length $$l$$, though a closed-form solution is often very difficult to obtain.