Categories:
Mathematics,
Physics.
Beltrami identity
Consider a general functional J[f] of the following form,
with f(x) an unknown function:
J[f]=∫x0x1L(f,f′,x)dx
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:
0=∂f∂L−dxd(∂f′∂L)
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:
dxdL=∂f∂Ldxdf+∂f′∂Ldxdf′+∂x∂L
Substituting the Euler-Lagrange equation into the first term gives us:
dxdL=f′dxd(∂f′∂L)+dxdf′∂f′∂L+∂x∂L=dxd(f′∂f′∂L)+∂x∂L
Although we started from the “hard” derivative dL/dx,
we arrive at an expression for the “soft” derivative ∂L/∂x,
describing the explicit dependence of L on x:
−∂x∂L=dxd(f′∂f′∂L−L)
What if L does not explicitly depend on x, i.e. ∂L/∂x=0?
In that case, the equation can be integrated to give the Beltrami identity:
f′∂f′∂L−L=C
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 ∂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:
J[f]=∬(x0,y0)(x1,y1)L(f,fx,fy,x,y)dxdy
In which case the Euler-Lagrange equation takes the following form:
0=∂f∂L−dxd(∂fx∂L)−dyd(∂fy∂L)
Once again, we calculate the hard x-derivative of L (the y-derivative is analogous):
dxdL=∂f∂Ldxdf+∂fx∂Ldxdfx+∂fy∂Ldxdfy+∂x∂L=dxdf(dxd(∂fx∂L)+dyd(∂fy∂L))+∂fx∂Ldxdfx+∂fy∂Ldxdfy+∂x∂L=dxd(fx∂fx∂L)+dyd(fx∂fy∂L)+∂x∂L
This time, we arrive at the following expression for the soft derivative ∂L/∂x:
−∂x∂L=dxd(fx∂fx∂L−L)+dyd(fx∂fy∂L)
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, 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:
0=−∫x0x1∂x∂Ldx=∫x0x1dxd(fx∂fx∂L−L)+dyd(fx∂fy∂L)dx=[fx∂fx∂L−L]x0x1+dyd∫x0x1(fx∂fy∂L)dx
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.