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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,

$${\displaystyle l=\int _{S}dS=\int _{s_{1))^{s_{2)){\sqrt {1+y'^{2))}dx,}$$

${\displaystyle S}$

${\displaystyle s_{1))$

${\displaystyle s_{2))$

The curve has to minimize its potential energy

$${\displaystyle U=\int _{S}g\mu y\cdot dS=\int _{s_{1))^{s_{2))g\mu y{\sqrt {1+y'^{2))}dx,}$$

$${\displaystyle \int _{s_{1))^{s_{2)){\sqrt {1+y'^{2))}dx=l,}$$

${\displaystyle g}$

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

$${\displaystyle L-{\frac {\partial L}{\partial y\prime ))=\mu gy{\sqrt {1+y\prime ^{2))}+\lambda {\sqrt {1+y\prime ^{2))}-\left[\mu gy{\frac {y\prime ^{2)){\sqrt {1+y\prime ^{2))))+\lambda {\frac {y\prime ^{2)){\sqrt {1+y\prime ^{2))))\right]=C,}$$

${\displaystyle \lambda }$

It is possible to simplify the differential equation as such:

$${\displaystyle {\frac {g\rho y-\lambda }{\sqrt {1+y'^{2))))=C.}$$

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.