---
title: "Beltrami identity"
firstLetter: "B"
publishDate: 2022-09-17
categories:
- Physics
- Mathematics

date: 2022-09-10T13:39:06+02:00
draft: false
markup: pandoc
---

# Beltrami identity

Consider a general functional $J[f]$ of the following form,
with $f(x)$ an unknown function:

$$\begin{aligned}
    J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x}
\end{aligned}$$

Where $L$ is the Lagrangian.
To find the $f$ that maximizes or minimizes $J[f]$,
the [calculus of variations](/know/concept/calculus-of-variations/)
states that the Euler-Lagrange equation must be solved for $f$:

$$\begin{aligned}
    0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f'} \Big)
\end{aligned}$$

We now want to know exactly how $L$ depends on the free variable $x$,
since it is a function of $x$, $f(x)$ and $f'(x)$.
Using the chain rule:

$$\begin{aligned}
    \dv{L}{x}
    = \pdv{L}{f} \dv{f}{x} + \pdv{L}{f'} \dv{f'}{x} + \pdv{L}{x}
\end{aligned}$$

Substituting the Euler-Lagrange equation into the first term gives us:

$$\begin{aligned}
    \dv{L}{x}
    &= f' \dv{x} \Big( \pdv{L}{f'} \Big) + \dv{f'}{x} \pdv{L}{f'} + \pdv{L}{x}
    \\
    &= \dv{x} \bigg( f' \pdv{L}{f'} \bigg) + \pdv{L}{x}
\end{aligned}$$

Although we started from the "hard" derivative $\dv*{L}{x}$,
we arrive at an expression for the "soft" derivative $\pdv*{L}{x}$,
describing the *explicit* dependence of $L$ on $x$:

$$\begin{aligned}
    - \pdv{L}{x}
    = \dv{x} \bigg( f' \pdv{L}{f'} - L \bigg)
\end{aligned}$$

What if $L$ does not explicitly depend on $x$, i.e. $\pdv*{L}{x} = 0$?
In that case, the equation can be integrated to give the **Beltrami identity**:

$$\begin{aligned}
    \boxed{
        f' \pdv{L}{f'} - L
        = C
    }
\end{aligned}$$

Where $C$ is a constant.
This says that the left-hand side is a conserved quantity in $x$,
which could be useful to know.
If we insert a concrete expression for $L$,
the Beltrami identity might be easier to solve for $f$ than the full Euler-Lagrange equation.
The assumption $\pdv*{L}{x} = 0$ is justified;
for example, if $x$ is time, it means that the potential is time-independent.


## Higher dimensions

Above, a 1D problem was considered, i.e. $f$ depended only on a single variable $x$.
Consider now a 2D problem, such that $J[f]$ is given by:

$$\begin{aligned}
    J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
\end{aligned}$$

In which case the Euler-Lagrange equation takes the following form:

$$\begin{aligned}
    0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f_x} \Big) - \dv{y} \Big( \pdv{L}{f_y} \Big)
\end{aligned}$$

Once again, we calculate the hard $x$-derivative of $L$ (the $y$-derivative is analogous):

$$\begin{aligned}
    \dv{L}{x}
    &= \pdv{L}{f} \dv{f}{x} + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
    \\
    &= \dv{f}{x} \bigg( \dv{x} \Big( \pdv{L}{f_x} \Big) + \dv{y} \Big( \pdv{L}{f_y} \Big) \bigg)
    + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
    \\
    &= \dv{x} \Big( f_x \pdv{L}{f_x} \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big) + \pdv{L}{x}
\end{aligned}$$

This time, we arrive at the following expression for the soft derivative $\pdv*{L}{x}$:

$$\begin{aligned}
    - \pdv{L}{x}
    &= \dv{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big)
\end{aligned}$$

Due to the derivatives, this cannot be cleanly turned into an analogue of the 1D Beltrami identity,
and therefore we use that name only in the 1D case.

However, if $\pdv*{L}{x} = 0$, this equation is still useful.
For an off-topic demonstration of this fact,
let us choose $x$ as the transverse coordinate, and integrate over it to get:

$$\begin{aligned}
    0
    &= - \int_{x_0}^{x_1} \pdv{L}{x} \dd{x}
    \\
    &= \int_{x_0}^{x_1} \dv{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
    \\
    &= \Big[ f_x \pdv{L}{f_x} - L \Big]_{x_0}^{x_1} + \dv{y} \int_{x_0}^{x_1} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
\end{aligned}$$

If our boundary conditions cause the boundary term to vanish (as is often the case),
then the integral on the right is a conserved quantity with respect to $y$.
While not as elegant as the 1D Beltrami identity,
the above 2D counterpart still fulfills the same role.



## References
1.  O. Bang,
    *Nonlinear mathematical physics: lecture notes*, 2020,
    unpublished.