Categories: Mathematics, Physics.

Beltrami identity

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

J[f]=x0x1L(f,f,x)dx\begin{aligned} J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x} \end{aligned}

Where LL is the Lagrangian. To find the ff that maximizes or minimizes J[f]J[f], the calculus of variations states that the Euler-Lagrange equation must be solved for ff:

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

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

dLdx=Lfdfdx+Lfdfdx+Lx\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:

dLdx=fddx(Lf)+dfdxLf+Lx=ddx(fLf)+Lx\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 dL/dx\idv{L}{x}, we arrive at an expression for the “soft” derivative L/x\ipdv{L}{x}, describing the explicit dependence of LL on xx:

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

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

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

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

Higher dimensions

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

J[f]=(x0,y0)(x1,y1)L(f,fx,fy,x,y)dxdy\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:

0=Lfddx(Lfx)ddy(Lfy)\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 xx-derivative of LL (the yy-derivative is analogous):

dLdx=Lfdfdx+Lfxdfxdx+Lfydfydx+Lx=dfdx(ddx(Lfx)+ddy(Lfy))+Lfxdfxdx+Lfydfydx+Lx=ddx(fxLfx)+ddy(fxLfy)+Lx\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 L/x\ipdv{L}{x}:

Lx=ddx(fxLfxL)+ddy(fxLfy)\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 L/x=0\ipdv{L}{x} = 0, this equation is still useful. For an off-topic demonstration of this fact, let us choose xx as the transverse coordinate, and integrate over it to get:

0=x0x1Lxdx=x0x1ddx(fxLfxL)+ddy(fxLfy)dx=[fxLfxL]x0x1+ddyx0x1(fxLfy)dx\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 yy. 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.