Consider a general functional of the following form,
with an unknown function:
Where is the Lagrangian.
To find the that maximizes or minimizes ,
the calculus of variations
states that the Euler-Lagrange equation must be solved for :
We now want to know exactly how depends on the free variable ,
since it is a function of , and .
Using the chain rule:
Substituting the Euler-Lagrange equation into the first term gives us:
Although we started from the “hard” derivative ,
we arrive at an expression for the “soft” derivative ,
describing the explicit dependence of on :
What if does not explicitly depend on , i.e. ?
In that case, the equation can be integrated to give the Beltrami identity:
Where is a constant.
This says that the left-hand side is a conserved quantity in ,
which could be useful to know.
If we insert a concrete expression for ,
the Beltrami identity might be easier to solve for than the full Euler-Lagrange equation.
The assumption is justified;
for example, if is time, it means that the potential is time-independent.
Above, a 1D problem was considered, i.e. depended only on a single variable .
Consider now a 2D problem, such that is given by:
In which case the Euler-Lagrange equation takes the following form:
Once again, we calculate the hard -derivative of (the -derivative is analogous):
This time, we arrive at the following expression for the soft derivative :
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 , this equation is still useful.
For an off-topic demonstration of this fact,
let us choose as the transverse coordinate, and integrate over it to get:
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 .
While not as elegant as the 1D Beltrami identity,
the above 2D counterpart still fulfills the same role.
- O. Bang,
Nonlinear mathematical physics: lecture notes, 2020,