---
title: "Beltrami identity"
sort_title: "Beltrami identity"
date: 2022-09-17
categories:
- Physics
- Mathematics
layout: "concept"
---

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 $$\idv{L}{x}$$,
we arrive at an expression for the "soft" derivative $$\ipdv{L}{x}$$,
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. $$\ipdv{L}{x} = 0$$?
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 $$\ipdv{L}{x} = 0$$ 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 $$\ipdv{L}{x}$$:

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

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 $$\ipdv{L}{x} = 0$$, 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.