Beltrami’s Identity

The Beltrami’s identity

F - y' \frac{\partial F}{\partial y'} = C

where C is a constant, is a reduced form of Euler-Lagrange equation for the special case when F does not dependent explicitly on x.

For F = F(y, y'),

\frac{dF}{dx} = \frac{\partial F}{\partial y} y' + \frac{\partial F}{\partial y'} y''.\quad\quad\quad(1)

From Euler-Lagrange equation

\frac{\partial F}{\partial y} - \frac{d}{dx}(\frac{\partial F}{\partial y'}) = 0,

we have

\frac{\partial F}{\partial y} = \frac{d}{dx}(\frac{\partial F}{\partial y'}).\quad\quad\quad(2)

Substituting (2) into (1) gives

\frac{dF}{dx} = \frac{d}{dx}(\frac{\partial F}{\partial y'}) y' + \frac{\partial F}{\partial y'} y''.\quad\quad\quad(3)

Consequently, when (3) is expressed as

\frac{dF}{dx} = \frac{d}{dx}(\frac{\partial F}{\partial y'}) y' + \frac{\partial F}{\partial y'} \frac{dy'}{dx},

it becomes clear that

\frac{dF}{dx} = \frac{d}{dx}(\frac{\partial F}{\partial y'}y').


\frac{d}{dx}(F-\frac{\partial F}{\partial y'}y') = 0.


F -  y' \frac{\partial F}{\partial y'} = C.

In “A Relentless Pursuit”, we derived differential equation

\frac{dy}{dx} = \frac{\sqrt{y^2-C_1^2}}{\pm C_1}\quad\quad\quad(4)

from Euler-Lagrange equation without taking advantage of the fact that F=y\sqrt{1+(y')^2} has no explicit dependency on x. The derivation was mostly done by a CAS.

Let’s derive (4) from Beltrami’s Identity. This time, we will not use CAS.

For F=y\sqrt{1+(y')^2}, the Beltrami’s Identity is

y\sqrt{1+(y')^2} - y'\cdot\left(y\cdot\frac{1}{2}\frac{2y'}{\sqrt{1+(y')^2}}\right) = C.

It simplifies to

\frac{y}{\sqrt{1+(y')^2}} = C.

Further simplification yields

C^2(y')^2 = y^2-C^2.

For C \ne 0,

(y')^2 = \frac{y^2-C^2}{C^2}.


\frac{dy}{dx} = \pm \sqrt{\frac{y^2-C^2}{C^2}}=\frac{\sqrt{y^2-C^2}}{\pm C}.

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