Beltrami Identity

The **Beltrami identity**, named after the Italian mathematician Eugenio Beltrami (1835 - 1900), is a simplified and less general version of the Euler–Lagrange equation in the **calculus of variations**.

## **Problem**

Show that the Euler-Lagrange equation

d

d
t

(

∂
L

∂

y
˙

)

−

∂
L

∂
y

=
0

{\displaystyle \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{y}} \right)-\frac{\partial L}{\partial y} = 0 }

can be written as

d

d
t

[

L
−

y
˙

∂
L

∂

y
˙

]

−

∂
L

∂
t

=
0

{\displaystyle \frac{d}{dt} \left[L - \dot{y} \frac{\partial L}{\partial \dot{y}}\right] - \frac{\partial L}{\partial t} = 0 }

Then show that if

L

{\displaystyle L}

 does not explicitly depend on

t

{\displaystyle t}

, then

L
−

y
˙

∂
L

∂

y
˙

=
C

{\displaystyle L - \dot{y} \frac{\partial L}{\partial \dot{y}} = C }

 where

C

{\displaystyle C}

 is constant.

## **Hint**

Use the Newtonian time derivative shorthand:

(
1
)

d
y

d
t

=

y
˙

(
2
)

d
2

y

d

t
2

=

d

y
˙

d
t

=

y
¨

{\displaystyle \begin{align}
&(1) \; \frac{dy}{dt} = \dot{y} \\[5 pt]
&(2) \; \frac{d^2 y}{dt^2} = \frac{d\dot{y}}{dt} = \ddot{y}
\end{align} }

## **Solution**

Note that the total derivative

d
L

d
t

=

∂
L

∂
t

∂
L

∂
y

d
y

d
t

∂
L

∂

y
˙

d

y
˙

d
t

{\displaystyle \frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial y} \frac{dy}{dt} + \frac{\partial L}{\partial \dot{y}} \frac{d \dot{y}}{dt} }

can be expressed as

d
L

d
t

=

∂
L

∂
t

∂
L

∂
y

y
˙

∂
L

∂

y
˙

y
¨

{\displaystyle \frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial y} \dot{y} + \frac{\partial L}{\partial \dot{y}} \ddot{y} }

and because

d

d
t

[

y
˙

∂
L

∂

y
˙

]

=

y
˙

d

d
t

(

∂
L

∂

y
˙

)

∂
L

∂

y
˙

y
¨

{\displaystyle \frac{d}{dt} \left[\dot{y} \frac{\partial L}{\partial \dot{y}}\right] = \dot{y} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{y}} \right) + \frac{\partial L}{\partial \dot{y}} \ddot{y} }

we discover that

d

d
t

[

L
−

y
˙

∂
L

∂

y
˙

]

−

∂
L

∂
t

=
0

{\displaystyle \frac{d}{dt} \left[L - \dot{y} \frac{\partial L}{\partial \dot{y}}\right] - \frac{\partial L}{\partial t} = 0 }

is equivalent to

∂
L

∂
t

∂
L

∂
y

y
˙

∂
L

∂

y
˙

y
¨

−

[

y
˙

d

d
t

(

∂
L

∂

y
˙

)

∂
L

∂

y
˙

y
¨

]

−

∂
L

∂
t

=

{\displaystyle \frac{\partial L}{\partial t} + \frac{\partial L}{\partial y} \dot{y} + \frac{\partial L}{\partial \dot{y}} \ddot{y} - \left[\dot{y} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{y}} \right) + \frac{\partial L}{\partial \dot{y}} \ddot{y}\right] - \frac{\partial L}{\partial t} = 0. }

Because

∂
L

∂
y

y
˙

−

y
˙

d

d
t

(

∂
L

∂

y
˙

)

=
0

{\displaystyle \frac{\partial L}{\partial y} \dot{y} - \dot{y} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{y}} \right) = 0 }

−

y
˙

[

d

d
t

(

∂
L

∂

y
˙

)

−

∂
L

∂
y

]

=
0

{\displaystyle - \dot{y} \left[\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{y}} \right)-\frac{\partial L}{\partial y} \right] = 0 }

d

d
t

(

∂
L

∂

y
˙

)

−

∂
L

∂
y

=
0

{\displaystyle \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{y}} \right)-\frac{\partial L}{\partial y} = 0 }

we discover that

d

d
t

[

L
−

y
˙

∂
L

∂

y
˙

]

−

∂
L

∂
t

=
0

{\displaystyle \frac{d}{dt} \left[L - \dot{y} \frac{\partial L}{\partial \dot{y}}\right] - \frac{\partial L}{\partial t} = 0 }

 is equivalent to the Euler-Lagrange equation

d

d
t

(

∂
L

∂

y
˙

)

−

∂
L

∂
y

=
0

{\displaystyle \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{y}} \right)-\frac{\partial L}{\partial y} = 0 }

.

If

L

{\displaystyle L}

 does not explicitly depend on

t

{\displaystyle t}

, then

∂
L

∂
t

=
0

{\displaystyle \frac{\partial L}{\partial t} = 0 }

 and

d

d
t

[

L
−

y
˙

∂
L

∂

y
˙

]

−

∂
L

∂
t

=
0

{\displaystyle \frac{d}{dt} \left[L - \dot{y} \frac{\partial L}{\partial \dot{y}}\right] - \frac{\partial L}{\partial t} = 0 }

 becomes

d

d
t

[

L
−

y
˙

∂
L

∂

y
˙

]

=
0

{\displaystyle \frac{d}{dt} \left[L - \dot{y} \frac{\partial L}{\partial \dot{y}}\right] = 0 }

.

So by integration, we get

L
−

y
˙

∂
L

∂

y
˙

=
C

{\displaystyle L - \dot{y} \frac{\partial L}{\partial \dot{y}} = C }

where

C

{\displaystyle C}

 is constant.