Categories: Mathematics, Physics.

# 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 $$\dv*{L}{x}$$, 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.

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 O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.

© Marcus R.A. Newman, CC BY-SA 4.0.
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