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 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

O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.