1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
|
---
title: "Beltrami identity"
sort_title: "Beltrami identity"
date: 2022-09-17
categories:
- Physics
- Mathematics
layout: "concept"
---
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](/know/concept/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 $$\idv{L}{x}$$,
we arrive at an expression for the "soft" derivative $$\ipdv{L}{x}$$,
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. $$\ipdv{L}{x} = 0$$?
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 $$\ipdv{L}{x} = 0$$ 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 $$\ipdv{L}{x}$$:
$$\begin{aligned}
- \pdv{L}{x}
&= \dv{}{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big)
\end{aligned}$$
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 $$\ipdv{L}{x} = 0$$, 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.
## References
1. O. Bang,
*Nonlinear mathematical physics: lecture notes*, 2020,
unpublished.
|
