Beltrami identity

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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

${\displaystyle I[u]=\int _{a}^{b}L[x,u(x),u'(x)]\,dx\,,}$

where ${\displaystyle a}$ and ${\displaystyle b}$ are constants and ${\displaystyle u'(x)={\frac {du}{dx}}}$.^[1]

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

where C is a constant.^{[2][note 1]}

Derivation

By the chain rule, the derivative of L is

${\displaystyle {\frac {dL}{dx}}={\frac {\partial L}{\partial x}}{\frac {dx}{dx}}+{\frac {\partial L}{\partial u}}{\frac {du}{dx}}+{\frac {\partial L}{\partial u'}}{\frac {du'}{dx}}\,.}$

Because ${\displaystyle {\frac {\partial L}{\partial x}}=0}$, we write

${\displaystyle {\frac {dL}{dx}}={\frac {\partial L}{\partial u}}u'+{\frac {\partial L}{\partial u'}}u''\,.}$

We have an expression for ${\displaystyle {\frac {\partial L}{\partial u}}}$ from the Euler–Lagrange equation,

${\displaystyle {\frac {\partial L}{\partial u}}={\frac {d}{dx}}{\frac {\partial L}{\partial u'}}\,}$

that we can substitute in the above expression for ${\displaystyle {\frac {dL}{dx}}}$ to obtain

${\displaystyle {\frac {dL}{dx}}=u'{\frac {d}{dx}}{\frac {\partial L}{\partial u'}}+u''{\frac {\partial L}{\partial u'}}\,.}$

By the product rule, the right side is equivalent to

${\displaystyle {\frac {dL}{dx}}={\frac {d}{dx}}\left(u'{\frac {\partial L}{\partial u'}}\right)\,.}$

By integrating both sides and putting both terms on one side, we get the Beltrami identity,

${\displaystyle L-u'{\frac {\partial L}{\partial u'}}=C\,.}$

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

${\displaystyle I[y]=\int _{0}^{a}{\sqrt {{1+y'^{\,2}} \over y}}dx\,.}$

The integrand

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

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

${\displaystyle L-y'{\frac {\partial L}{\partial y'}}=C\,.}$

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

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

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

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

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

with ${\displaystyle A}$ being half the above constant, ${\displaystyle {\frac {1}{2C^{2}}}}$, and ${\displaystyle \phi }$ being a variable. These are the parametric equations for a cycloid.^[3]

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,

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

and is subject to the constraint 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

where ${\displaystyle \lambda }$ is the Lagrange multiplier.

It is possible to simplify the differential equation as such:

Solving this equation gives the hyperbolic cosine, where ${\displaystyle C_{0}}$ is a second constant obtained from integration

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.

Notes

References