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.

References

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 O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.

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