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

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

Therefore,

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

i.e.,

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

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 $C \ne 0$ ,

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

Therefore,

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

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