Categories: Mathematics, Physics.

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 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 $x$:

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**:

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.

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.

- O. Bang,
*Nonlinear mathematical physics: lecture notes*, 2020, unpublished.

© 2023 Marcus R.A. Newman,
CC BY-NC-SA 4.0.