Beltrami identity

Applications

Solution to the brachistochrone problem

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

${\displaystyle I[y]={\int }_0^a\sqrt{\frac{1+y^{\prime 2}}{y}}dx.}$

The integrand

${\displaystyle L(y,y^{\prime })=\sqrt{\frac{1+y^{\prime 2}}{y}}}$

does not depend explicitly on the variable of integration ${\displaystyle x}$, so the Beltrami identity applies,

${\displaystyle L-y^{\prime }\frac{\mathrm{\partial }L}{\mathrm{\partial }{y}^{\prime }}=C.}$

Substituting for ${\displaystyle L}$ and simplifying,

${\displaystyle y(1+y^{\prime 2})=1/C^2\text{(constant)},}$

which can be solved with the result put in the form of parametric equations

${\displaystyle x=A(\varphi -\sin \varphi )}$

${\displaystyle y=A(1-\cos \varphi )}$

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

Solution to the catenary problem

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

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

Because the independent variable ${\displaystyle x}$ 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

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

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

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

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

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